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Physics (042) · Class XII · Chapter-wise Board Questions

CBSE Board papers 2022 to 2026, plus CBSE Sample Question Papers. Chapters as per CBSE Curriculum 2026-27. 1158 questions. Click Reveal answer under any question to see the official marking scheme.

Marks
Unit I · Electrostatics

1. Electric Charges and Fields

Q1.2026 Main 55/3/1 · Q11 mark

A particle of mass m and charge q starts from rest and moves in an electric field E = E0 i. After travelling a distance x in the field along x-axis, the kinetic energy of the particle will be :

  1. q E0 x2
  2. q E0 x
  3. q2 E0 x
  4. q2 E02 x
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Official CBSE Marking Scheme · CBSE Board 2026 (Main), Set 55/3/1

(B) q E0 x

Q2.2026 Main 55/5/1 · Q11 mark

Two small identical metallic balls having charges q and – 2q are kept far at a separation r. They are brought in contact and then separated at distance r2. Compared to the initial force F, they will now :

  1. attract with a force F2
  2. repel with a force F2
  3. repel with a force F
  4. attract with a force F
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Official CBSE Marking Scheme · CBSE Board 2026 (Main), Set 55/5/1

(B) repel with a force F/2

Q3.2026 Supp 55/7/1 · Q21 mark

A conducting sphere of radius R has a charge Q on it. Which of the following graphs correctly represents the value of electric field E in the range 0 ≤ x < ∞, where x is the distance of the point from the centre of the sphere ?

  1. Graph A: E constant up to R then decreasing
  2. Graph B: E rising to a peak at R then decreasing
  3. Graph C: E zero up to R then decreasing
  4. Graph D: E rising linearly to R then decreasing
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Official CBSE Marking Scheme · CBSE Board 2026 (Supplementary), Set 55/7/1

(C)

Graph C: E zero inside, decreasing outside
Q4.2025 Main 55/1/1 · Q11 mark

Figure shows variation of Coulomb force (F) acting between two point charges with 1r2, r being the separation between the two charges (q1, q2) and (q2, q3). If q2 is positive and least in magnitude, then the magnitudes of q1, q2 and q3 are such that

Graph of F versus 1/r² for the pairs (q1,q2) and (q2,q3)
  1. q2 < q3 < q1
  2. q3 < q1 < q2
  3. q1 < q2 < q3
  4. q2 < q1 < q3
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Q5.2025 Main 55/2/1 · Q11 mark

Two charges –q each are placed at the vertices A and B of an equilateral triangle ABC. If M is the mid-point of AB, the net electric field at C will point along

  1. CA
  2. CB
  3. MC
  4. CM
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Q6.2025 Main 55/2/1 · Q81 mark

Which one of the following statements is correct ?
Electric field due to static charges is

  1. conservative and field lines do not form closed loops.
  2. conservative and field lines form closed loops.
  3. non-conservative and field lines do not form closed loops.
  4. non-conservative and field lines form closed loops.
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Q7.2025 Main 55/4/1 · Q11 mark

A body acquires charge 8·0 × 10−12 C. The mass of the body :

  1. increases by 4·5 × 10−7 kg
  2. decreases by 1·0 × 10−6 kg
  3. decreases by 4·55 × 10−23 kg
  4. increases by 9·1 × 10−23 kg
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The official marking scheme for this paper is not available in the source folder.

Q8.2025 Supp 55/S/1 · Q11 mark

A small metallic sphere S of charge +q is placed exactly at a point midway between two-point charges A (+Q) and B (+Q) where Q >> q. If the sphere is slightly displaced towards charge A and released, then :

  1. it will move further towards A
  2. it will move towards B
  3. it will oscillate about its original position
  4. it will not move at all
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Official CBSE Marking Scheme · CBSE Board 2025 (Supplementary), Set 55/S/1

(C) It will oscillate about its original position

Q9.2025 Supp 55/S/1 · Q91 mark

A charge Q is enclosed by a spherical Gaussian surface of radius R. If the radius is doubled, then the total electric flux through the surface :

  1. becomes four times
  2. remains the same
  3. becomes half
  4. becomes twice
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Official CBSE Marking Scheme · CBSE Board 2025 (Supplementary), Set 55/S/1

(B) Remains the same

Q10.2024 Main 55/1/1 · Q11 mark

A thin plastic rod is bent into a circular ring of radius R. It is uniformly charged with charge density λ. The magnitude of the electric field at its centre is :

  1. λ2ε0R
  2. Zero
  3. λ4πε0R
  4. λ4ε0R
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Official CBSE Marking Scheme · CBSE Board 2024 (Main), Set 55/1/1

(B) Zero

Q11.2024 Main 55/2/1 · Q11 mark

Two charged particles P and Q, having the same charge but different masses mP and mQ, start from rest and travel equal distances in a uniform electric field E in time tP and tQ respectively. Neglecting the effect of gravity, the ratio tPtQ is :

  1. mPmQ
  2. mQmP
  3. mPmQ
  4. mQmP
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Official CBSE Marking Scheme · CBSE Board 2024 (Main), Set 55/2/1

(C) mPmQ

Q12.2024 Main 55/4/1 · Q21 mark

Two identical small conducting balls B1 and B2 are given –7 pC and + 4 pC charges respectively. They are brought in contact with a third identical ball B3 and then separated. If the final charge on each ball is –2 pC, the initial charge on B3 was

  1. –2 pC
  2. –3 pC
  3. –5 pC
  4. –15 pC
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Official CBSE Marking Scheme · CBSE Board 2024 (Main), Set 55/4/1

(B) –3 pC

Q13.2024 Main 55/5/1 · Q131 mark

Assertion (A) : Equal amount of positive and negative charges are distributed uniformly on two halves of a thin circular ring as shown in figure. The resultant electric field at the centre O of the ring is along OC.

Reason (R) : It is so because the net potential at O is not zero.

Thin circular ring with positive charge on the upper half (A side) and negative charge on the lower half (C side), centre O, points A, B, C, D marked
  1. If both Assertion (A) and Reason (R) are true and Reason (R) is correct explanation of Assertion (A).
  2. If both Assertion (A) and Reason (R) are true and Reason (R) is not the correct explanation of Assertion (A).
  3. If Assertion (A) is true but Reason (R) is false.
  4. If both Assertion (A) and Reason (R) are false.
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Official CBSE Marking Scheme · CBSE Board 2024 (Main), Set 55/5/1

(C) Assertion (A) is true but Reason (R) is false.

Q14.2023 Main 55/1/1 · Q11 mark

A point charge situated at a distance ‘r’ from a short electric dipole on its axis, experiences a force F. If the distance of the charge is ‘2r’, the force on the charge will be :

  1. F/16
  2. F/8
  3. F/4
  4. F/2
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Official CBSE Marking Scheme · CBSE Board 2023 (Main), Set 55/1/1

(B) F/8

Q15.2023 Main 55/2/1 · Q11 mark

The magnitude of the electric field due to a point charge object at a distance of 4.0 m is 9 N/C. From the same charged object the electric field of magnitude, 16 N/C will be at a distance of

  1. 1 m
  2. 2 m
  3. 3 m
  4. 6 m
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Official CBSE Marking Scheme · CBSE Board 2023 (Main), Set 55/2/1

(C) 3 m

Q16.2023 Main 55/3/1 · Q11 mark

An electron experiences a force (1·6 × 10-16 N) i in an electric field E. The electric field E is :

  1. (1·0 × 103 NC) i
  2. – (1·0 × 103 NC) i
  3. (1·0 × 10-3 NC) i
  4. – (1·0 × 10-3 NC) i
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Official CBSE Marking Scheme · CBSE Board 2023 (Main), Set 55/3/1

(B) – (1·0 × 103 NC) i

Q17.2023 Main 55/3/1 · Q21 mark

Which one of the following is not a scalar quantity ?

  1. Electric field
  2. Voltage
  3. Resistivity
  4. Power
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Official CBSE Marking Scheme · CBSE Board 2023 (Main), Set 55/3/1

(A) Electric Field

Q18.2023 Main 55/4/1 · Q11 mark

Two charges q1 and q2 are placed at the centres of two spherical conducting shells of radius r1 and r2 respectively. The shells are arranged such that their centres are d [ > (r1 + r2)] distance apart. The force on q2 due to q1 is :

  1. 14π ε0 q1q2d2
  2. 14π ε0 q1q2(d − r1)2
  3. Zero
  4. 14π ε0 q1q2[d − (r1 + r2)]2
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Official CBSE Marking Scheme · CBSE Board 2023 (Main), Set 55/4/1

(C) Zero

Q19.2023 Main 55/5/1 · Q11 mark

An electric dipole of length 2 cm is placed at an angle of 30° with an electric field 2 × 105 N/C. If the dipole experiences a torque of 8 × 10-3 Nm, the magnitude of either charge of the dipole, is

  1. 4 μC
  2. 7 μC
  3. 8 mC
  4. 2 mC
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Official CBSE Marking Scheme · CBSE Board 2023 (Main), Set 55/5/1

(A) 4 μC

Q20.SQP 2025-26 · Q161 mark

Assertion (A): For three point charges to be in equilibrium, they must be collinear.

Reason (R): One of the three charges must have different polarity than rest of the two.

  1. Both Assertion and Reason are true and Reason is the correct explanation of Assertion.
  2. Both Assertion and Reason are true but Reason is not the correct explanation of Assertion.
  3. Assertion is true but Reason is false.
  4. Both Assertion and Reason are false.
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Official CBSE Marking Scheme · CBSE Sample Question Paper 2025-26

(B) Both Assertion and Reason are true but Reason is not the correct explanation of Assertion.

If three point charges are in equilibrium then forces acting on each charges should be linearly opposite.

Q21.SQP 2026-27 · Q11 mark

A point charge +Q is placed at the centre of a spherical Gaussian surface of radius R. If the radius of the sphere is doubled, the electric flux through the surface will

  1. become half
  2. become double
  3. remain the same
  4. become four times
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Official CBSE Marking Scheme · CBSE Sample Question Paper 2026-27

(C) remain the same

According to Gauss’s Law, electric flux, Φ = Qε0

Flux depends only on the enclosed charge, not on the shape or size of the Gaussian surface. Therefore, doubling the radius does not change the flux.

Q22.2026 Main 55/4/1 · Q172 marks

An electric dipole consists of two point charges +1 μC and −1 μC, held 10 cm apart. It is subjected to a uniform electric field of 100 N/C. Calculate the amount of work done in turning the dipole from its position of stable equilibrium to the position of unstable equilibrium, in the field.

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Official CBSE Marking Scheme · CBSE Board 2026 (Main), Set 55/4/1

Work done in turning the dipole in an external field = − pE(cos θ2 − cos θ1) [½]

Given : θ1 = 0° (stable equilibrium) and θ2 = 180° (unstable equilibrium) [½]

Net work done in turning the dipole from its position of stable equilibrium to the position of unstable equilibrium = 2pE = 2(2qa)E

= 2 × 1 × 10−6 × 10 × 10−2 × 100 [½]

Net work done = 2 × 10−5 J [½]

Q23.2024 Main 55/5/1 · Q17 (option 1)2 marks

Four point charges of 1 μC, –2 μC, 1 μC and –2 μC are placed at the corners A, B, C and D respectively, of a square of side 30 cm. Find the net force acting on a charge of 4 μC placed at the centre of the square.

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Official CBSE Marking Scheme · CBSE Board 2024 (Main), Set 55/5/1

Diagram showing direction of forces [1]

Square ABCD with charges +1 μC at A, –2 μC at B, +1 μC at C, –2 μC at D and 4 μC at centre O, with force vectors F_OA, F_OB, F_OC, F_OD

OA = OB = OC = OD = r

Net force on charge 4μC

F = FOA + FOB + FOC + FOD [½]

FOA = – FOC ⇒ FOA + FOC = 0

FOB = – FOD ⇒ FOB + FOD = 0

F = 0 [½]

Alternatively

FOA = FOC = 9×109 × 4×10–6 × 1×10–6(152×10–2)2 = 0.8 N

FOB = FOD = 1.6 N [½]

F1 = FOA – FOC = 0

F2 = FOB – FOD = 0

Net Force F = 0 [½]

Q24.2024 Main 55/5/1 · Q17 (OR option)2 marks

Three point charges, 1 pC each, are kept at the vertices of an equilateral triangle of side 10 cm. Find the net electric field at the centroid of triangle.

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Official CBSE Marking Scheme · CBSE Board 2024 (Main), Set 55/5/1

Diagram [1]

Equilateral triangle ABC with 1 pC at each vertex, centroid O and field vectors E_OA, E_OB, E_OC, E_BC

qA = qB = qC = 1pC

AO = BO = CO = r

|EOA| = |EOB| = |EOC|

EBC = EOB + EOC

EBC = EOB2 + EOC2 + 2EOBEOC cos120° [½]

EBC = EOB , EOA = – EBC

Net electric field EO = EOA + EBC

EO = 0 [½]

Alternatively

EOA = EOB = EOC = 2.7 NC-1 [½]

EBC = EOB2 + EOC2 + 2EOBEOC cos120°

= EOB

As EBC = – EOA

EBC + EOA = 0 [½]

Net electric field is zero.

Alternatively

|EOA| = |EOB| = |EOC|

Electric field vectors are making an angle of 120° with each other. They make a closed polygon. So vector sum of all electric field vectors will be zero.

E = 0

Q25.2024 Supp 55/S/1 · Q21 (OR option)2 marks

Two point charges of 3 µC and 4 µC are kept in air at (0·3 m, 0) and (0, 0.3 m) in x-y plane. Find the magnitude and direction of the net electric field produced at the origin (0, 0).

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Official CBSE Marking Scheme · CBSE Board 2024 (Supplementary), Set 55/S/1
Fields E1 and E2 at the origin due to the two charges and their resultant E

E1 = kq1r12(–i) = 9×109 × 3×10–6(0.3)2(–i) = 3×105(–i) NC–1 [½]

E2 = kq2r22(–j) = 9×109 × 4×10–6(0.3)2(–j) = 4×105(–j) NC–1 [½]

E = E1 + E2

E = E12 + E22 = 5×105 NC–1 [½]

tan θ = 43

θ = tan–1(43) inclination with respect to the x-axis (in III quadrant). [½]

Q26.2023 Main 55/4/1 · Q192 marks

Two identical dipoles are arranged in x-y plane as shown in the figure. Find the magnitude and the direction of net electric field at the origin O.

Two identical dipoles along x-axis (-q at left, +q at right) and y-axis (+q at top, -q at bottom), each charge at distance a from origin O
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Official CBSE Marking Scheme · CBSE Board 2023 (Main), Set 55/4/1

Finding the magnitude of electric field 1½; finding direction of net electric field ½

Dipoles BA and DC with fields E1 along OB, E2 along OD and net field E_net at 225°

Dipole moment due to dipole BA is p1 & dipole moment due to dipole DC is p2.

Electric field E1 due to p1 is along OB.

Electric field E2 due to p2 is along OD. [½]

Magnitude of resultant electric field |Enet| = 22 E

Since |E1| = |E2| = E = 14π ε0 · qa2 [½]

Enet = 14π ε0 · 22 qa2 [½]

Direction of Enet is 225° to x-axis. [½]

Alternatively :

Considering CB and AD as dipoles; E1 and E2 at O pointing toward lower left

Considering CB as dipole, electric field at O

E1 = 2kq × (a / 2)[(a/2)2 + (a/2)2]3/2 = 2 kq × aa3 (½ + ½)3/2 = 2 kqa2 [½]

Similarly considering AD as another dipole, electric field at O

E2 = 2kq × (a / 2)[(a/2)2 + (a/2)2]3/2 = 2 kqa2 [½]

Enet = E1 + E2 = 2 kqa2 + 2 kqa2 = 22 kqa2 [½]

Direction of Enet is 225° to x-axis. [½]

Q27.2023 Main 55/5/1 · Q222 marks

Depict the orientation of an electric dipole in (a) stable and (b) unstable equilibrium in an external uniform electric field.

Write the potential energy of the dipole in each case.

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Official CBSE Marking Scheme · CBSE Board 2023 (Main), Set 55/5/1

(a) Stable equilibrium: dipole moment p along E (θ = 0°) [½]

Dipole p parallel to field E, theta = 0

U = – pE [½]

(b) Unstable equilibrium: dipole moment p opposite to E (θ = 180°) [½]

Dipole p antiparallel to field E, theta = 180

U = pE [½]

Note: Award ½ mark, if a student mentions the value of θ as 0° and 180° without drawing the diagram.

Q28.2023 Supp 55/C/1 · Q222 marks

The inward and the outward electric flux through a Gaussian surface are 2ϕ and ϕ respectively.

(a) What is the net charge enclosed by the surface ?

(b) If the net outward flux through the surface were zero, can it be concluded that there were no charges inside the surface ? Justify your answer.

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Official CBSE Marking Scheme · CBSE Board 2023 (Supplementary), Set 55/C/1

(a) Net outward flux = −2Φ + Φ [½]

= −Φ [½]

Charge enclosed = −Φ εo [½]

(b) No [½]

There may be charges (positive and negative) and net charge is zero. [½]

Q29.SQP 2025-26 · Q192 marks

A charge q is placed inside a sphere of radius ‘a’ filled with water and another charge 2q is placed inside cube of side ‘2a’ which is vacuumed inside. Find the ratio of the flux linked with the sphere to that linked with the cube. (Take relative permittivity of water as 80)

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Official CBSE Marking Scheme · CBSE Sample Question Paper 2025-26

From Gauss’s theorem

Φ = qεrεo [Where εr is relative permittivity of medium inside Gaussian surface] [½]

For sphere,

Φsphere = qεwaterεo ……………..(i) [½]

For cube

Φcube = 2qεo …………………(ii)

Dividing (i) by (ii)

ΦsphereΦcube = 12εwater = 1160 [1]

Q30.2026 Main 55/1/1 · Q22 (option 1)3 marks

(a) Using Gauss’s law, deduce an expression for electric field at a point due to a uniformly charged infinite plane thin sheet.

(b) Two large thin plane sheets, each having surface charge density σ are held close and parallel to each other in air. What is the net electric field at a point (i) inside and (ii) outside, the sheets ?

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Official CBSE Marking Scheme · CBSE Board 2026 (Main), Set 55/1/1

(a)

Gaussian pillbox through an infinite plane sheet of surface charge density σ

[½] (figure)

As seen from the figure, only the two faces 1 and 2 will contribute to the flux. Therefore, flux E.ΔS through both the surfaces are equal and add up. [½]

Therefore, net flux through the Gaussian surface is 2EA. The charge enclosed by the closed surface is σA.

2EA = σAε0 [½]

E = σ2ε0 n [½]

(b)

Two parallel sheets each of charge density σ with regions marked inside and outside

(i) Ein = 0 [½]

(ii) Eout = σε0 [½]

Q31.2026 Main 55/1/1 · Q243 marks

Figure shows a narrow beam of electrons entering with a velocity of 3 × 107 m/s, symmetrically through the space between two parallel horizontal plates P1P1′ and P2P2′ kept 2 cm apart.

Electron beam entering midway between parallel plates P1P1′ and P2P2′ and deflected towards end P2′

If each plate is 3 cm long, calculate the potential difference V applied between the plates so that the beam just strikes the end P2′.

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Official CBSE Marking Scheme · CBSE Board 2026 (Main), Set 55/1/1

Since a = qEm = qVmL [½]

t = xux [½]

y = 12(qVmL)(xux)2 [½]

V = 2ymLux2ex2 [½]

V = 2 × 1×10–2 × 9.1×10–31 × 2×10–2 × (3×107)21.6×10–19 × (3×10–2)2 [½]

V = 2275 Volt [½]

Q32.2026 Supp 55/7/1 · Q28 (option 1)3 marks

Two infinitely long thin straight wires AB and CD with charge densities – 12 μCm and + 12 μCm are lying parallel to y-axis at points (–1 m, 0, 0) and (1 m, 0, 0) respectively, as shown in the figure. Find the net electric field E at points P1 (– 2m, 0, 0), O (0, 0, 0) and P2 (2m, 0, 0).

Two parallel line charges AB and CD with points P1, O and P2 on the x-axis
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Official CBSE Marking Scheme · CBSE Board 2026 (Supplementary), Set 55/7/1

EP1 = EAB + ECD = λAB2πε0r1i + λCD2πε0r2(–i) [½]

= 2 × 9 × 109 × 12 × 10-6 (1 – 13) i

EP1 = 1.44 × 105 i N/C [½]

EO = EAB + ECD = λAB2πε0r1(–i) + λCD2πε0r2(–i) [½]

= 2 × 9 × 109 × 12 × 10-6 (1 + 1)(–i)

EO = – 4.32 × 105 i N/C [½]

EP2 = EAB + ECD = λAB2πε0r1(–i) + λCD2πε0r2(i) [½]

= 2 × 9 × 109 × 12 × 10-6 (– 13 + 1) i

= 1.44 × 105 i N/C [½]

Q33.2025 Main 55/5/1 · Q28 (OR option)3 marks

Two infinitely long straight wires ‘1’ and ‘2’ are placed d distance apart, parallel to each other, as shown in the figure. They are uniformly charged having charge densities λ and – λ2 respectively. Locate the position of the point from wire ‘1’ at which the net electric field is zero and identify the region in which it lies.

Two parallel line charges 1 (λ) and 2 (−λ/2) separated by d, with regions A, B, C
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Q34.2025 Supp 55/S/1 · Q223 marks

Define electric flux. Is it a scalar or a vector quantity ?

A point charge q is kept at a distance of d2 directly above the centre of a square of side d. Use Gauss’ law to obtain the expression for the electric flux through the square.

If the point charge is now moved to a point at a distance ‘d’ from the centre of the square and the side of the square is doubled, explain how the electric flux through the square will be affected.

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Official CBSE Marking Scheme · CBSE Board 2025 (Supplementary), Set 55/S/1

Electric flux : The electric flux may be defined as the number of electric lines of force crossing through a surface normal to the surface. [½]

φ = ∮ E · ds

Electric flux is a scalar quantity. [½]

Draw a cube of side d such that it completely encloses the charge q.

Cube of side d with the point charge q at distance d/2 above the centre of the top face

Total flux through the cube, φE = qenε0 [½]

Flux through one face (square) = qen6ε0 [½]

If a charge is now moved to the point of a distance d from the center of square and side of the square is doubled, then electric flux through square remains unchanged because electric flux through a closed surface depends only on the amount of charge contained inside the closed surface and is independent of the size of the Gaussian surface. [1]

Q35.2024 Main 55/1/1 · Q223 marks

A cube of side 0·1 m is placed, as shown in the figure, in a region where electric field E = 500 x i exists. Here x is in meters and E in NC−1. Calculate :

Cube of side 0.1 m placed on the x axis with its left face at x = 0.1 m, axes x, y, z shown

(a) the flux passing through the cube, and
(b) the charge within the cube.

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Official CBSE Marking Scheme · CBSE Board 2024 (Main), Set 55/1/1

(a) φL = EL · A = − [500 × 0.1] × [(0.1)2] = − 0.5 N m2 C−1 [½]

φR = ER · A = [500 × 0.2] × [(0.1)2] = 1 N m2 C−1 [½]

Net flux = φL + φR = 0.5 N m2 C−1 [1]

(b) flux, φ = qεo [½]

charge, q = φ × εo = 0.5 εo = 4.4 × 10−12 C [½]

Q36.2024 Main 55/5/1 · Q233 marks
  1. Define the term ‘electric flux’ and write its dimensions.
  2. A plane surface, in shape of a square of side 1 cm is placed in an electric field E = (100 NC) i such that the unit vector normal to the surface is given by n = 0.8i + 0.6k. Find the electric flux through the surface.
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Official CBSE Marking Scheme · CBSE Board 2024 (Main), Set 55/5/1

(a) It is the measure of the total number of electric field lines passing through a surface normally. [1]

Alternatively

Surface integral of electric field over a surface.

Alternatively

φE = E · A

[ML3T–3A–1] [½]

(b) φE = E · A [½]

= (100i) · (10–4 n)

= (100i) · (0.8i + 0.6k)10–4 [½]

= 8×10–3 Nm2C-1 [½]

Q37.2024 Supp 55/S/1 · Q24 (OR option)3 marks

Two large plane sheets P1 and P2 having charge densities + σ and – 3 σ respectively are arranged parallel to each other as shown in the figure. Find the net electric field (E) at points A, B and C.

Parallel sheets P1 (+σ) and P2 (–3σ) with points A, B, C marked along a line
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Official CBSE Marking Scheme · CBSE Board 2024 (Supplementary), Set 55/S/1

Electric field at A (EA)

EA = E1 + E2 [½]

= σ2ε0(–i) + 3σ2ε0(i)

EA = σε0(i) [½]

Electric field at B (EB)

EB = E1 + E2 [½]

= σ2ε0i + 3σ2ε0i

= 2σε0i [½]

Electric field at C (EC)

EC = E1 + E2 [½]

= σ2ε0i + 3σ2ε0(–i)

= σε0(–i) [½]

Q38.2024 Supp 55/S/1 · Q294 marks

Dipoles, whether electric or magnetic, are characterised by their dipole moments, which are vector quantities. Two equal and opposite charges separated by a small distance constitute an electric dipole, while a current carrying loop behaves as a magnetic dipole. Electric dipoles create electric fields around them. Electric dipoles experience a torque when placed in an external electric field.

(i) Two identical electric dipoles, each consisting of charges – q and + q separated by distance d, are arranged in x-y plane such that their negative charges lie at the origin O and positive charges lie at points (d, 0) and (0, d) respectively. The net dipole moment of the system is : [1]

  1. – q d (i + j)
  2. q d (i + j)
  3. q d (i – j)
  4. q d (j – i)

(ii) E1 and E2 are magnitudes of electric field due to a dipole, consisting of charges –q and +q separated by distance 2a, at points r (>> a) (1) on its axis, and (2) on equatorial plane, respectively. Then (E1E2) is : [1]

  1. 14
  2. 12
  3. 2
  4. 4

(iii) An electric dipole of dipole moment 5.0 × 10–8 Cm is placed in a region where an electric field of magnitude 1.0 × 103 N/C acts at a given instant. At that instant the electric field E is inclined at an angle of 30° to dipole moment P. The magnitude of torque acting on the dipole, at that instant is : [1]

  1. 2.5 × 10–5 Nm
  2. 5.0 × 10–5 Nm
  3. 1.0 × 10–4 Nm
  4. 2.0 × 10–6 Nm

(iv) Answer either (a) or (b) :

(a) An electron is revolving with speed v around the proton in a hydrogen atom, in a circular orbit of radius r. The magnitude of magnetic dipole moment of the electron is : [1]

  1. 4 evr
  2. 2 evr
  3. 12 evr
  4. 14 evr

(b) A square loop of side 5.0 cm carries a current of 2.0 A. The magnitude of magnetic dipole moment associated with the loop is : [1]

  1. 1.0 × 10–3 Am2
  2. 5.0 × 10–3 Am2
  3. 1.0 × 10–2 Am2
  4. 5.0 × 10–2 Am2
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Official CBSE Marking Scheme · CBSE Board 2024 (Supplementary), Set 55/S/1

(i) (B) qd(i + j) [1]

(ii) (C) 2 [1]

(iii) (A) 2.5 × 10–5 Nm [1]

(iv) (a) (C) 12evr   [1]

OR

(iv) (b) (B) 5.0 × 10–3 Am2 [1]

Q39.2023 Main 55/4/1 · Q34 (option 1)4 marks

Electrostatics deals with the study of forces, fields and potentials arising from static charges. Force and electric field, due to a point charge is basically determined by Coulomb’s law. For symmetric charge configurations, Gauss’s law, which is also based on Coulomb’s law, helps us to find the electric field. A charge/a system of charges like a dipole experience a force/torque in an electric field. Work is required to be done to provide a specific orientation to a dipole with respect to an electric field.

Answer the following questions based on the above :

(a) Consider a uniformly charged thin conducting shell of radius R. Plot a graph showing the variation of |E| with distance r from the centre, for points 0 ≤ r ≤ 3R.

(b) The figure shows the variation of potential V with 1r for two point charges Q1 and Q2, where V is the potential at a distance r due to a point charge. Find Q1Q2.

Graph of V versus 1/r: line Q1 at 60° and line Q2 at 30° to the 1/r axis

(c) An electric dipole of dipole moment of 6 × 10−7 C-m is kept in a uniform electric field of 104 N/C such that the dipole moment and the electric field are parallel. Calculate the potential energy of the dipole.

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Official CBSE Marking Scheme · CBSE Board 2023 (Main), Set 55/4/1

(a)

Graph of |E| versus r: zero for r < R, jumps to maximum at r = R and falls as 1/r^2 up to r = 3R

[1]

(b) ∵ V = kQ/r

Slope of graph is proportional to Q [½]

Q1Q2 = tan 60°tan 30° = 3 [½]

(c) U = − pE cos θ [½]

θ = 0° [½]

U = − (6 × 10−7) × (104)

U = − 6 × 10−3 J [1]

Q40.2023 Main 55/4/1 · Q34 (OR option)4 marks

Electrostatics deals with the study of forces, fields and potentials arising from static charges. Force and electric field, due to a point charge is basically determined by Coulomb’s law. For symmetric charge configurations, Gauss’s law, which is also based on Coulomb’s law, helps us to find the electric field. A charge/a system of charges like a dipole experience a force/torque in an electric field. Work is required to be done to provide a specific orientation to a dipole with respect to an electric field.

Answer the following questions based on the above :

(a) Consider a uniformly charged thin conducting shell of radius R. Plot a graph showing the variation of |E| with distance r from the centre, for points 0 ≤ r ≤ 3R.

(b) The figure shows the variation of potential V with 1r for two point charges Q1 and Q2, where V is the potential at a distance r due to a point charge. Find Q1Q2.

Graph of V versus 1/r: line Q1 at 60° and line Q2 at 30° to the 1/r axis

(c) An electric dipole of dipole moment p is initially kept in a uniform electric field E such that p is perpendicular to E. Find the amount of work done in rotating the dipole to a position at which p becomes antiparallel to E.

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Official CBSE Marking Scheme · CBSE Board 2023 (Main), Set 55/4/1

(a)

Graph of |E| versus r: zero for r < R, jumps to maximum at r = R and falls as 1/r^2 up to r = 3R

[1]

(b) ∵ V = kQ/r

Slope of graph is proportional to Q [½]

Q1Q2 = tan 60°tan 30° = 3 [½]

(c) ∵ Work done W = − pE (cos θ2 − cos θ1) [½]

where θ2 = 180°, θ1 = 90° [½]

⟹ W = − pE (cos 180° − cos 90°) [½]

W = + pE [½]

Q41.2026 Main 55/1/1 · Q31 (option 1)5 marks

(a) An electric dipole consists of two point charges q and –q separated by a distance 2a. Derive an expression for the electric field E due to this dipole at a point distant r from the centre of the dipole on the equatorial plane. Write the expression for the electric field at a far off point, i.e. r >> a.

(b) A dipole is placed in x-y plane such that charges q and –q are located at x = a and x = b respectively. There exists an electric field E = 2 i N/C in the region. Calculate the force F and torque τ experienced by the dipole.

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Official CBSE Marking Scheme · CBSE Board 2026 (Main), Set 55/1/1

(a)

Electric dipole with fields E+q, E–q and resultant E at an equatorial point P

[½] (figure)

The magnitude of the electric fields due to the two charges +q and –q are given by

E+q = q4πε0 1(r2 + a2) [½]

E–q = q4πε0 1(r2 + a2)

The components normal to the dipole axis cancel away. The components along the dipole axis add up. The total electric field is opposite to p.

E = –(E+q + E–q) cos θ p [1]

E = – 14πε0 2qa(r2 + a2)3/2 p [½]

At large distance r >> a

E = –2qa4πε0r3 p [½]

(b)

Charges q and –q on the x-axis at distances a and b from the origin

∵ F = F+q + F–q

Net force = [+q·2i – q·2i] = 0 N [½ + ½]

Torque τ = p × E

τ = p(–i) × 2i = 0 [½ + ½]

Alternatively:

τ = pE sin θ; angle between p and E is π, so τ = 0

Q42.2026 Main 55/3/1 · Q31 (OR option)5 marks

(i) Two infinitely long straight wires having linear charge densities − λ and 3λ are held vertically parallel to each other, distance r apart in free space. Find the nature and magnitude of the force/length exerted by one wire on the other.

(ii) A small hollow conducting sphere of radius r1 is given a charge Q. It is surrounded by a concentric conducting spherical shell of inner radius r2 and outer radius r3, having charge − 3q.

If a point charge 2q were kept at the centre, find :

(I) the electric flux through a concentric spherical Gaussian surface of radius x for (1) x < r1, and (2) r1 < x < r2.

(II) electric field at a point distant x from the centre for (1) x > r3, and (2) r1 < x < r2.

(III) surface charge density on the inner surface of (1) sphere, and (2) shell.

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Official CBSE Marking Scheme · CBSE Board 2026 (Main), Set 55/3/1

(i) Nature of force : Attractive [½]

dq = λdl

Electric field on wire 2 due to wire 1, E = −λ2πε0r [½]

Force on wire 2 = qE

Force/length = F1l = −λ × 3λ2πε0r [½]

= −3 λ22πε0r [½]

Similarly, F2l = −3λ22πε0r

(ii) (I) (1) x < r1 : φ = 2qε0 [½]

(2) r1 < x < r2 : φ = (Q + 2q)ε0 [½]

(II) (1) x > r3 : E = k (Q − q)x2 [½]

(2) r1 < x < r2 : E = k (Q + 2q)x2 [½]

(III) (1) σ = −2q4πr12 [½]

(2) σ = −(Q + 2q)4πr22 [½]

Q43.2025 Main 55/2/1 · Q31 (OR option)5 marks
  1. What is difference between an open surface and a closed surface ? Draw elementary surface vector dS for a spherical surface S.
  2. Define electric flux through a surface. Give the significance of a Gaussian surface. A charge outside a Gaussian surface does not contribute to total electric flux through the surface. Why ?
  3. A small spherical shell S1 has point charges q1 = –3 μC, q2 = –2 μC and q3 = 9 μC inside it. This shell is enclosed by another big spherical shell S2. A point charge Q is placed in between the two surfaces S1 and S2. If the electric flux through the surface S2 is four times the flux through surface S1, find charge Q.
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Q44.2025 Main 55/4/1 · Q31 (option 1)5 marks
  1. Two point charges + q and − q are held at (a, 0) and (− a, 0) in x-y plane. Obtain an expression for the net electric field due to the charges at a point (0, y). Hence, find electric field at a far off point (y >> a).
  2. Three point charges of − 2 nC, − 1 nC, and + 5 nC are kept at the vertices A, B and C of an equilateral triangle of side 0·2 m. Find the total amount of work done in shifting the charges from A to A1, B to B1 and C to C1. Here A1, B1 and C1 are the midpoints of sides AB, BC and CA, respectively.
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Q45.2025 Main 55/4/1 · Q31 (OR option)5 marks
  1. Show that Gauss’s theorem is consistent with Coulomb’s law. Using it, derive an expression for the electric field due to a uniformly charged thin spherical shell of radius r at a point at a distance y from the centre of the shell such that (I) y > r, and (II) y < r.
  2. A point charge of + 2 nC is kept at the origin of a three-dimensional coordinate system. Find the type and magnitude of the charge which should be kept at (0, 0, − 6m) so that the potential due to the system becomes zero at (0, 0, 2m).
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Q46.2025 Main 55/6/1 · Q33 (OR option)5 marks
  1. Obtain an expression for the electric field E due to a dipole of dipole moment p at a point on its equatorial plane and specify its direction. Hence, find the value of electric field :
    (I) at the centre of the dipole (r = 0), and
    (II) at a point r >> a,
    where 2a is the length of the dipole.
  2. An electric field E = (10x + 5) i N/C exists in a region in which a cube of side L is kept as shown in the figure. Here x and L are in metres. Calculate the net flux through the cube.
Cube OABCDA1C1D1 with axes X, Y, Z; field along +x
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Q47.2025 Supp 55/S/1 · Q32 (OR option)5 marks

(i) Write Coulomb’s law of electrostatics in vector form. Apply it to determine the electric field at a point due to a system of point charges.

(ii) ABC is an equilateral triangle of side l. Two point charges +2 μC each, are located at points B and C. Find the sign and magnitude of the point charge q to be kept at the midpoint M of the side BC, so that the net electric field at point A becomes zero.

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Official CBSE Marking Scheme · CBSE Board 2025 (Supplementary), Set 55/S/1

(i) F21 = 14πε0 q1q2r212 r21 [1]

Alternative : F12 = 14πε0 q1q2r122 r12

OR F = 14πε0 q1q2r2 r

For determining electric field due to a system of point charges

Point charges q1, q2, q3, q4 and the electric fields E1 to E4 at point P with their resultant

E1 = 14πε0 q1r1p2 r1p [½]

E2 = 14πε0 q2r2p2 r2p and so on [½]

By superposition principle

E = E1 + E2 + .............. + En [½]

E = 14πε0 Σi=1n qirip2 rip [½]

(ii)

Equilateral triangle ABC with +2 μC at B and C, charge q at M, and the fields E_AB, E_AC, E_BC and E_AM at A

[½] for figure

From the figure

|EAB| = 14πε0 · qr2 = 14πε0 · 2 × 10–6l2

|EAC| = 14πε0 · 2 × 10–6l2

∵ |EAB| = |EAC|

∴ Resultant of EAB & EAC

EBC = EAB + EAC

|EBC| = 2EAB cos 30° [½]

|EBC| = 14πε0 · 2 × 10–6l2 32

|EAM| = 14πε0 · q( l32 )2 [½]

∵ Net electric field at A is zero.

∴ EAM = –EBC

14πε0 · q( l32 )2 = –24πε0 · 2 × 10–6l2 32

q = –332 × 10–6 C or –332 μC [½]

Q48.2024 Main 55/1/1 · Q31 (option 1)5 marks

(i) Derive an expression for potential energy of an electric dipole p in an external uniform electric field E. When is the potential energy of the dipole (1) maximum, and (2) minimum ?

(ii) An electric dipole consists of point charges − 1·0 pC and + 1·0 pC located at (0, 0) and (3 mm, 4 mm) respectively in x − y plane. An electric field E = (1000 Vm) i is switched on in the region. Find the torque τ acting on the dipole.

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Official CBSE Marking Scheme · CBSE Board 2024 (Main), Set 55/1/1

(i) Deriving the expression for potential energy [2]; maximum and minimum values of potential energy [½ + ½]. (ii) Finding the torque [2]

Dipole of charges −q and +q at angle θ to a uniform field E along x, separation 2a

(i) The amount of work done in rotating the dipole from θ = θ0 to θ = θ1 by the external torque

W = ∫θ0θ1 τext dθ [½]

= ∫θ0θ1 pE sin θ dθ [½]

W = pE (cos θ0 − cos θ1) [½]

For θ0 = π2 and θ1 = θ

= pE (cos π2 − cos θ)

U(θ) = − pE cos θ = − p·E [½]

(1) Potential energy is maximum when p is antiparallel to E
Alternatively: θ = 180° or π radians [½]

(2) Potential energy is minimum when p is along E
Alternatively: θ = 0° [½]

(ii)

Dipole in the x-y plane with charge −1 pC at (0,0) and +1 pC at (3,4), field E along +x

τ = pE sin θ [½]

= (2aq) E sin θ [½]

= (5 × 10−3 × 1 × 10−12) 103 × 45 [½]

= 4 × 10−12 N m

Direction is along −ve Z direction. [½]

Q49.2024 Main 55/2/1 · Q31 (OR option)5 marks

(i) State Gauss’s Law in electrostatics. Apply this to obtain the electric field E at a point near a uniformly charged infinite plane sheet.

(ii) Two long straight wires 1 and 2 are kept as shown in the figure. The linear charge density of the two wires are λ1 = 10 µC/m and λ2 = − 20 µC/m. Find the net force F experienced by an electron held at point P.

Two parallel long wires 1 and 2, 30 cm apart, with point P 10 cm below wire 1; unit vectors i and j shown
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Official CBSE Marking Scheme · CBSE Board 2024 (Main), Set 55/2/1

(i) Electric flux through a closed surface is equal to qε0, where q is the total charge enclosed by the surface. φ = qε0 [1]

Alternatively

The surface integral of electric field over a closed surface is 1ε0 times the total charge enclosed by the surface.
∮ E·dS = qε0

(Award ½ mark for writing the formula only.)

Gaussian cuboid through a uniformly charged infinite plane sheet, with faces 1 and 2 at distance x on either side

(Gaussian surface can be cylindrical also) [½]

As seen from figure, only two faces 1 and 2 will contribute to the flux. Flux E·ds through both the surfaces is equal and add up. [½]

The charge enclosed by surface is σA, where σ is surface charge density.
According to Gauss’s theorem
2EA = σA/ε0 [½]
E = σ/2ε0

E = σ2ε0 n, where n is unit vector directed normally out of the plane [½]

(ii) E = λ2πε0r r

According to question
E1 (at point P) = λ12πε0r1
E = 10 × 10−62πε0(10 × 10−2) (−j) N/C [½]

E2 (at point P) = λ22πε0r2
E = 20 × 10−62πε0(20 × 10−2) (−j) N/C [½]

Enet = 10 × 10−62πε0(10·1 + 20·2) (−j) N/C
= 3·6 × 106 (−j) N/C [½]

Fnet = q × Enet
F = − 1·6 × 10−19 × 3·6 × 106 (−j) N
= 5·76 × 10−13 N (j) [½]

Q50.2024 Main 55/3/1 · Q31 (OR option)5 marks

(i) A charge + Q is placed on a thin conducting spherical shell of radius R. Use Gauss’s theorem to derive an expression for the electric field at a point lying (i) inside and (ii) outside the shell.

(ii) Show that the electric field for same charge density (σ) is twice in case of a conducting plate or surface than in a nonconducting sheet.

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Official CBSE Marking Scheme · CBSE Board 2024 (Main), Set 55/3/1

(i) Expression for electric field at a point lying (i) inside [1], (ii) outside [2]; (ii) Explanation [2]

(i) Field inside the shell

Spherical shell of radius R with surface charge density σ and a concentric Gaussian surface of radius r larger than R

The Flux through the Gaussian surface is = E × 4πR2 [½]

In this case Gaussian surface encloses no charge.

Hence E × 4πR2 = 0

E = 0 [½]

(Note: Award full credit of this part if a student writes directly E=0, mentioning as there is no charge enclosed by Gaussian surface)

(ii) Field outside the shell

Spherical shell of radius R with surface charge density σ and a concentric Gaussian surface of radius r smaller than R

[½]

Electric flux through Gaussian surface

E × 4πr2 = (σ 4πR2)ε0 [½]

Charge enclosed by the Gaussian surface

E × 4πr2 = (σ 4πR2)ε0

Using Gauss’s law: ∫ E · ds = Qε0 [½]

E × 4πr2 = (σ 4πR2)ε0

E = σε0 R2r2 = q4πε0r2 [½]

(ii) For conducting sheet,

Electric field due to a conducting sheet

Ec = σε0 [½]

Gaussian pillbox through a charged sheet with surface charge density σ, E arrows on both sides

For non-conducting sheet

Enc = σ2ε0 [½]

Since surface charge density is same. [½]

2Enc = Ec [½]

Q51.2024 Main 55/4/1 · Q32 (OR option)5 marks
  1. Using Gauss’s law, show that the electric field E at a point due to a uniformly charged infinite plane sheet is given by E = σ2ε0 n where symbols have their usual meanings.
  2. Electric field E in a region is given by
    E = (5x2 + 2) i
    where E is in N/C and x is in meters.
    A cube of side 10 cm is placed in the region as shown in figure.
    Cube of side 10 cm with one corner at origin, edges along x, y, z axes, field E along +x
    Calculate (1) the electric flux through the cube, and (2) the net charge enclosed by the cube.
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Official CBSE Marking Scheme · CBSE Board 2024 (Main), Set 55/4/1

(i)

Gaussian cylinder (pillbox) across an infinite plane sheet of surface charge density σ

[1]

∮ E·ds = ∫1 E·ds + ∫2 E·ds = 2EA [½]

From Gauss’s law ∮ E·ds = qε0 [½]

2EA = σAε0

E = σ2ε0 [½]

Vectorially E = σ2ε0 n [½]

Electric field is normally outward of the sheet.

(ii) (1) Electric flux through the cube

φ = φL + φR = ∫ EL·ds + ∫ ER·ds [½]

= – 2 × 100 × 10–4 + [5 × (10 × 10–2)2 + 2] × 100 × 10–4

φ = 5 × 10–4 N m2 C–1 [½]

(2) φ = qenε0 [½]

qen = φ · ε0 = 5 × 10–4 × 8.85 × 10–12 = 4.43 × 10–15 C [½]

Q52.2024 Main 55/5/1 · Q31 (OR option)5 marks
  1. A thin spherical shell of radius R has a uniform surface charge density σ. Using Gauss’ law, deduce an expression for electric field (i) outside and (ii) inside the shell.
  2. Two long straight thin wires AB and CD have linear charge densities 10 μC/m and –20 μC/m, respectively. They are kept parallel to each other at a distance 1 m. Find magnitude and direction of the net electric field at a point midway between them.
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Official CBSE Marking Scheme · CBSE Board 2024 (Main), Set 55/5/1

(i) Electric Field outside the shell

Charged spherical shell of radius R with a concentric Gaussian surface of radius r > R and point P on it

[½]

Electric flux through Gaussian surface

Φ = E × 4π r2

Charge enclosed by the Gaussian surface

Q = σ × 4πR2

Using Gauss’ law: ∫ E · ds = Qε0 [½]

E × 4π r2 = (σ 4πR2)ε0

∴ E = σR2ε0r2

E = σR2ε0r2 r [½]

(ii) Field inside the shell

Charged spherical shell of radius R with a concentric Gaussian surface of radius r < R and point P on it

[½]

Electric flux through Gaussian surface

Φ = E × 4π r2 (∵ r < R)

Charge enclosed by the Gaussian surface

Q = 0

By Gauss’ Law

E × 4π r2 = 0 [½]

i.e. E = 0 [½]

(Note: Award full credit of this part if a student writes directly E=0, mentioning as there is no charge enclosed by Gaussian surface)

(ii) Electric field due to a long straight charged wire of linear charged density λ

E = λ2πε0r [½]

Parallel wires AB and CD with fields E1 and E2 at the midpoint, wire separation 2r

Net electric field at the mid-point

Enet = E1 + E2

= λ12πε0r + λ22πε0r

Enet = 12πε0r[λ1 + λ2] [½]

= 2×9×1090.5[10 + 20]×10–6

= 1.08 × 106 NC-1 [½]

Enet is directed towards CD. [½]

Q53.2023 Main 55/1/1 · Q31 (option 1)5 marks

(i) State Coulomb’s law in electrostatics and write it in vector form, for two charges.

(ii) ‘Gauss’s law is based on the inverse-square dependence on distance contained in the Coulomb’s law.’ Explain.

(iii) Two charges A (charge q) and B (charge 2q) are located at points (0, 0) and (a, a) respectively. Let i and j be the unit vectors along x-axis and y-axis respectively. Find the force exerted by A on B, in terms of i and j.

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Official CBSE Marking Scheme · CBSE Board 2023 (Main), Set 55/1/1

i) Statement of Coulomb’s law and vector form [1 + 1]; ii) Explanation of Gauss’s law based on Coulomb’s law [1]; iii) Force exerted by charge A on charge B [2]

i) Force between two point charges varies inversely with the square of distance between the charges and is directly proportional to the product of magnitude of the two charges and acts along the line joining the two charges. [1]

F12 = 14πε0 q1q2r122 r12 [1]

Alternatively

F12 = 14πε0 q1q2r123 r12, where r12 is a vector from charge q2 to charge q1.

ii) In derivation of Gauss’s law, flux is calculated using Coulomb’s law and surface area. Here Coulomb’s law involves 1r2 factor and surface area involves r2 factor. When product is taken, the two factors cancel out and flux becomes independent of r. [1]

iii)

Charge q at origin A and charge 2q at B (a, a)

[½]

r = AB = ai + aj; r = |AB| = a2 + a2 = 2a [½]

F = 14πε0 q1q2r2 r

F = 14πε0 × q × 2q(2a)2 × ai + aj2a [½]

F = 14πε0 × 2q22a2 × (i + j)2

F = q242πε0a2 (i + j) [½]

Note: Award 1 mark if a student calculates the magnitude of force only. |F| = 14πε0 q2a2

Alternatively Give full credit if a student uses component method to solve the question.

Q54.2023 Main 55/1/1 · Q31 (OR option)5 marks

(i) Derive an expression for the electric field at a point on the equatorial plane of an electric dipole consisting of charges q and −q separated by a distance 2a.

(ii) The distance of a far off point on the equatorial plane of an electric dipole is halved. How will the electric field be affected for the dipole ?

(iii) Two identical electric dipoles are placed along the diagonals of a square ABCD of side 2 m as shown in the figure. Obtain the magnitude and direction of the net electric field at the centre (O) of the square.

Square ABCD with charges +q at A, −q at B, −q at C, +q at D and centre O; dipoles along the diagonals
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Official CBSE Marking Scheme · CBSE Board 2023 (Main), Set 55/1/1

i) Derivation of electric field [2]; ii) Effect on electric field [1]; iii) Finding magnitude and direction of electric field [2]

i)

Electric dipole with fields E(+q) and E(−q) at a point P on the equatorial plane; resultant E at P parallel to −p

[½]

E+q = q4πε0 × 1r2 + a2; E−q = q4πε0 × 1r2 + a2 [½]

The components normal to dipole axis cancel away. The components along the dipole axis add up. [½]

Total electric field is opposite to dipole moment.

E = −(E+q + E−q) cosθ p = −2qa4πε0(r2 + a2)3/2 p = −p4πε0(r2 + a2)3/2 [½ + ½]

Deduct ½ mark if the expression of electric field is not in vector form.

ii) At far off point r >> a, E = −p4πε0r3 [½]

When distance is halved, E = −p4πε0(r/2)3 = −8p4πε0r3

E becomes 8 times [½]

iii)

Dipole moments p1 (along OA), p2 (along OD), net dipole moment p_net and net field E_net at O

p1 = q × 2 C m (along OA); p2 = q × 2 C m (along OD)

pnet = p12 + p22 = 22 q C m [½]

Electric field at centre O: E = k pnet(r2 + a2)3/2 [½]

At point O, r = 0, a = 1 m: E = k × 22q13 = 22 kq = 22q4πε0 [½]

Along DC [½]

Alternatively

Fields E1 (along OB) and E2 (along OC) at O

E = kq/r2 [½]; AC = BD = 2 m; r = OA = OB = OC = OD = 1 m

Electric field at O due to charges at B and D: E1 = EB + ED = kq/12 + kq/12 = 2kq, along OB [½]

Electric field at O due to charges at A and C: E2 = EA + EC = kq/12 + kq/12 = 2kq, along OC [½]

Enet = E12 + E22 = 22 kq = 22q4πε0 [½]

Along DC

Alternatively

Fields E1 and E2 at O both pointing from O towards the side BC

Considering AB as dipole, electric field at O:

E1 = 2kq × a((1/2)2 + (1/2)2)3/2 = 2kqa(1/2 + 1/2)3/2 = 2kqa [½]

Similarly considering DC as another dipole, electric field at O: E2 = 2kqa [½]

Enet = E1 + E2 = 4kqa = 14πε0 × 4 × 12 × q = 22 kq = 22q4πε0 [½]

Along DC

Q55.2023 Main 55/2/1 · Q31 (option 1)5 marks
  1. Use Gauss’ law to obtain an expression for the electric field due to an infinitely long thin straight wire with uniform linear charge density λ.
  2. An infinitely long positively charged straight wire has a linear charge density λ. An electron is revolving in a circle with a constant speed v such that the wire passes through the centre, and is perpendicular to the plane, of the circle. Find the kinetic energy of the electron in terms of magnitudes of its charge and linear charge density λ on the wire.
  3. Draw a graph of kinetic energy as a function of linear charge density λ.
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Official CBSE Marking Scheme · CBSE Board 2023 (Main), Set 55/2/1

(i)

Cylindrical Gaussian surface of length l and radius r around an infinite line charge, field E at point P

[½]

Flux through the Gaussian surface φ = E·2πrl [½]

According to Gauss’s law

E·2πrl = qε0 [½]

∵ q = λl

E = λ2π ε0 r [½]

(ii) Electric field E = λ2π ε0 r [½]

Electron moving in a circle around a line of positive charge passing through the centre

mv2r = eE [½]

∴ Kinetic energy K = ½ mv2 = ½ eEr [½]

= ½ e λ·r2π ε0 r = eλ4π ε0 [½]

(iii) Kinetic energy K = eλ4π ε0, ∴ K ∝ λ [1]

Graph of kinetic energy K versus linear charge density λ: a straight line through the origin
Q56.2023 Main 55/3/1 · Q31 (option 1)5 marks

(i) Define electric flux and write its SI unit.

(ii) Use Gauss’ law to obtain the expression for the electric field due to a uniformly charged infinite plane sheet.

(iii) A cube of side L is kept in space, as shown in the figure. An electric field E = (Ax + B) i NC exists in the region. Find the net charge enclosed by the cube.

Cube of side L with one corner at the origin, edges along x, y and z axes
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Official CBSE Marking Scheme · CBSE Board 2023 (Main), Set 55/3/1

(i) Electric flux φ = E · A [½]

Alternatively: Electric flux is the number of electric field lines passing through an area normally.

S.I. unit of electric flux N m2/C or V-m. [½]

(ii) [diagram: Gaussian surface through the sheet] [½]

Gaussian pillbox crossing a uniformly charged infinite plane sheet of surface charge density σ

From Gauss’s law : φ = ∮ E · dA = qε0 [½]

2EA = σAε0 [½]

E = σ2ε0 [½]

Alternatively: If the shape of the Gaussian surface is taken cylindrical, full credit to be given.

(iii) φL = E ds cos 180° = –E ds = –BL2 [½ + ½]

φR = E ds cos 0° = E ds = (AL + B)L2 = AL3 + BL2

Net flux = φL + φR = (AL3 + BL2) – BL2 [½]

Net flux = AL3 = qε0 ⇒ q = AL3 ε0 [½]

Unit I · Electrostatics

2. Electrostatic Potential and Capacitance

Q1.2026 Main 55/1/1 · Q11 mark

In a region, the electric potential varies as V = 10 – 50x, where V is in volts and x in meters. The electric field in the region is :

  1. 10 N/C along +x
  2. 10 N/C along –x
  3. 50 N/C along +x
  4. 50 N/C along –x
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Official CBSE Marking Scheme · CBSE Board 2026 (Main), Set 55/1/1

(C) 50 N/C along +x

Q2.2026 Main 55/1/1 · Q21 mark

A conducting wire connects two charged metallic spheres A and B of radii r1 and r2 respectively. The distance between the spheres is very large compared to their radii. The ratio of electric fields, (EA/EB) at the surfaces of spheres A and B will be

  1. r1r2
  2. r2r1
  3. r12r22
  4. r22r12
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Official CBSE Marking Scheme · CBSE Board 2026 (Main), Set 55/1/1

(B) r2r1

Q3.2026 Main 55/2/1 · Q11 mark

Three point charges 2q, –2q and q are kept at the vertices of an equilateral triangle of side l. The potential energy of the system is

  1. zero
  2. –2q2πε0l
  3. q22πε0l
  4. – q2πε0l
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Official CBSE Marking Scheme · CBSE Board 2026 (Main), Set 55/2/1

(D) –q2πε0l

Q4.2026 Main 55/2/1 · Q21 mark

Two metal spheres of radii r1 and r2 (> r1) having charges q1 and q2 respectively kept in air, are brought in contact. Which of the following statements is not correct ?

  1. The total charge of the two spheres is conserved.
  2. Both spheres attain the same potential.
  3. The final potential of the system equals 14πε0 (q1 + q2)(r1 + r2)
  4. The final potential of the system equals 14πε0 (q1 + q2) (r1 + r2)r1 r2
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Official CBSE Marking Scheme · CBSE Board 2026 (Main), Set 55/2/1

(D) The final potential of the system equals 14πε0 (q1 + q2)(r1 + r2)r1r2

Q5.2026 Main 55/5/1 · Q21 mark

The figure represents the variation of the electric potential V at a point in a region of space as a function of its position along the x-axis. A charged particle will experience the maximum force at :

Graph of potential V versus x with points P, Q, R, S, T
  1. P
  2. Q
  3. R
  4. S
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Official CBSE Marking Scheme · CBSE Board 2026 (Main), Set 55/5/1

(D) S

Q6.2026 Supp 55/7/1 · Q11 mark

Assume that the surface charge density of the Earth is 1 electron charge/m2. The potential of the Earth’s surface (radius 6400 km) will be :

  1. – 0·12 V
  2. – 0·21 V
  3. + 2·1 V
  4. + 1·2 V
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Official CBSE Marking Scheme · CBSE Board 2026 (Supplementary), Set 55/7/1

(A) – 0.12 V

Q7.2025 Main 55/5/1 · Q11 mark

A metal sheet is inserted between the plates of a parallel plate capacitor of capacitance C. If the sheet partly occupies the space between the plates, the capacitance :

  1. remains C
  2. becomes greater than C
  3. becomes less than C
  4. becomes zero
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Q8.2025 Main 55/5/1 · Q21 mark

The electric field at a point in a region is given by E = α r|r|3, where α is a constant and r is the distance of the point from the origin. The magnitude of potential of the point is :

  1. αr
  2. α r22
  3. α2r2
  4. – αr
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Q9.2025 Main 55/7/1 · Q11 mark

Two horizontal plates, separated by 1 cm, are arranged one above the other. A particle of mass 5 mg and charge 2 nC is released in air between the plates. The potential difference that should be applied to the plates so that the particle remains suspended between them, is :

  1. 250 V
  2. 200 V
  3. 100 V
  4. 50 V
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Q10.2025 Supp 55/S/1 · Q131 mark

Questions number 13 to 16 are Assertion (A) and Reason (R) type questions. Two statements are given — one labelled Assertion (A) and the other labelled Reason (R). Select the correct answer from the codes (A), (B), (C) and (D) as given below.

Assertion (A) : Electrostatic force is a conservative force.

Reason (R) : In an electrostatic field, the work done between two points per unit positive charge depends on the path followed.

  1. Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of the Assertion (A).
  2. Both Assertion (A) and Reason (R) are true, but Reason (R) is not the correct explanation of the Assertion (A).
  3. Assertion (A) is true, but Reason (R) is false.
  4. Both Assertion (A) and Reason (R) are false.
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Official CBSE Marking Scheme · CBSE Board 2025 (Supplementary), Set 55/S/1

(C) Assertion (A) is true, but Reason (R) is false.

Q11.2024 Main 55/1/1 · Q21 mark

Ten capacitors, each of capacitance 1 µF, are connected in parallel to a source of 100 V. The total energy stored in the system is equal to :

  1. 10−2 J
  2. 10−3 J
  3. 0·5 × 10−3 J
  4. 5·0 × 10−2 J
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Official CBSE Marking Scheme · CBSE Board 2024 (Main), Set 55/1/1

(D) 5.0 × 10−2 J

Q12.2024 Main 55/3/1 · Q11 mark

Consider a group of charges q1, q2, q3 ... such that Σq ≠ 0. Then equipotentials at a large distance, due to this group are approximately :

  1. Plane
  2. Spherical surface
  3. Paraboloidal surface
  4. Ellipsoidal surface
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Official CBSE Marking Scheme · CBSE Board 2024 (Main), Set 55/3/1

(B) Spherical surface

Q13.2024 Main 55/3/1 · Q21 mark

A proton is taken from point P1 to point P2, both located in an electric field. The potentials at points P1 and P2 are – 5 V and + 5 V respectively. Assuming that kinetic energies of the proton at points P1 and P2 are zero, the work done on the proton is :

  1. – 1·6 × 10–18 J
  2. 1·6 × 10–18 J
  3. Zero
  4. 0·8 × 10–18 J
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Official CBSE Marking Scheme · CBSE Board 2024 (Main), Set 55/3/1

(B) 1.6 ×10-18 J

Q14.2024 Main 55/4/1 · Q11 mark

Two charges + q each are kept ‘2a’ distance apart. A third charge – 2q is placed midway between them. The potential energy of the system is –

  1. q28πε0a
  2. –6q28πε0a
  3. – 7q28πε0a
  4. 9q28πε0a
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Official CBSE Marking Scheme · CBSE Board 2024 (Main), Set 55/4/1

(C) –7q28πε0a

Q15.2024 Supp 55/S/1 · Q11 mark

Two particles A and B of the same mass but having charges q and 4q respectively, are accelerated from rest through different potential differences VA and VB such that they attain same kinetic energies. The value of (VAVB) is :

  1. 14
  2. 12
  3. 2
  4. 4
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Official CBSE Marking Scheme · CBSE Board 2024 (Supplementary), Set 55/S/1

(D) 4

Q16.2024 Supp 55/S/1 · Q121 mark

A parallel plate capacitor is charged by a battery. The battery is then disconnected and the plates of the charged capacitor are then moved farther apart. In the process :

  1. the charge on the capacitor increases.
  2. the potential difference across the plates decreases.
  3. the capacitance of the capacitor increases.
  4. the electrostatic energy stored in the capacitor increases.
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Official CBSE Marking Scheme · CBSE Board 2024 (Supplementary), Set 55/S/1

(D) The electrostatic energy stored in the capacitor increases

Q17.2023 Main 55/2/1 · Q21 mark

A point P lies at a distance x from the mid point of an electric dipole on its axis. The electric potential at point P is proportional to

  1. 1x2
  2. 1x3
  3. 1x4
  4. 1x1/2
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Official CBSE Marking Scheme · CBSE Board 2023 (Main), Set 55/2/1

(A) 1x2

Q18.2023 Main 55/2/1 · Q171 mark

Assertion (A) : Work done in moving a charge around a closed path, in an electric field is always zero.

Reason (R) : Electrostatic force is a conservative force.

  1. Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of Assertion (A).
  2. Both Assertion (A) and Reason (R) are true and Reason (R) is NOT the correct explanation of Assertion (A).
  3. Assertion (A) is true and Reason (R) is false.
  4. Assertion (A) is false and Reason (R) is also false.
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Official CBSE Marking Scheme · CBSE Board 2023 (Main), Set 55/2/1

(A) Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of Assertion (A).

Q19.2023 Main 55/5/1 · Q121 mark

The capacitors, each of 4 μF are to be connected in such a way that the effective capacitance of the combination is 6 μF. This can be achieved by connecting

  1. All three in parallel
  2. All three in series
  3. Two of them connected in series and the combination in parallel to the third.
  4. Two of them connected in parallel and the combination in series to the third.
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Official CBSE Marking Scheme · CBSE Board 2023 (Main), Set 55/5/1

(C) Two of them connected in series and the combination in parallel to the third

Q20.SQP 2025-26 · Q11 mark

If a charged hollow sphere and a solid sphere of aluminum and copper of equal radii are in electrostatic equilibrium, then which of the following statements is true?

  1. Both the spheres are having equal charges.
  2. The hollow sphere will have more charge than solid sphere at its surface.
  3. The aluminum sphere will have more charge on its surface than copper sphere.
  4. If hollow sphere is also made up of aluminum then it will have more charge.
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Official CBSE Marking Scheme · CBSE Sample Question Paper 2025-26

(A) Both are having equal charges

For two bodies to be in equilibrium, both should have same potential(V).

As V = Cq

Where C of sphere is 4πε0r. Which is independent of all the factors mentioned in options.

Q21.SQP 2026-27 · Q161 mark

Assertion (A): If a conductor is placed in an external electric field, the electric field inside the conductor becomes infinity in electrostatic equilibrium.

Reason (R): Free electrons inside the conductor rearrange themselves in such a way that the internal electric field is in the same direction as that of the external field.

  1. both Assertion and Reason are true and Reason is the correct explanation of Assertion.
  2. both Assertion and Reason are true but Reason is not the correct explanation of Assertion.
  3. Assertion is true but Reason is false.
  4. both Assertion and Reason are false.
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Official CBSE Marking Scheme · CBSE Sample Question Paper 2026-27

(D) both Assertion and Reason are false.

Electrons in a conductor move freely. When an external field is applied, charges rearrange until the internal field they create exactly cancels the external field. This is the reason the net internal field becomes zero.

Q22.2023 Main 55/4/1 · Q25 (option 1)2 marks

Obtain an expression for electrostatic potential energy of a system of three charges q, 2q and −3q placed at the vertices of an equilateral triangle of side a.

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Official CBSE Marking Scheme · CBSE Board 2023 (Main), Set 55/4/1
Equilateral triangle of side a with charges q, 2q and -3q at the vertices

(OR a similar diagram with different order of charges) [½]

U = 14π ε0 · q1q2r

U = 14π ε0 [ 2q2a − 6q2a − 3q2a ] [1]

U = 14π ε0 · (−7q2)a [½]

Q23.2023 Main 55/4/1 · Q25 (OR option)2 marks

Two small conducting balls A and B of radius r1 and r2 have charges q1 and q2 respectively. They are connected by a wire. Obtain the expression for charges on A and B, in equilibrium.

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Official CBSE Marking Scheme · CBSE Board 2023 (Main), Set 55/4/1

According to law of conservation of charge qi = qf

q1 + q2 = q1′ + q2′ = Q [½]

When two balls are connected with wire V1 = V2 [½]

kq1′r1 = kq2′r2 or q1′r1 = q2′r2

q1′ r2 = q2′ r1

q1′ r2 = (Q − q1′) r1

q1′ r2 = Q r1 − q1′ r1

q1′ (r1 + r2) = Q r1

q1′ = Q r1r1 + r2 = (q1 + q2) r1r1 + r2 [½]

q2′ = Q − q1′ = Q − Q r1r1 + r2 = Q r2r1 + r2 = (q1 + q2) r2r1 + r2 [½]

Note : Give full credit if done by any other method.

Q24.2023 Supp 55/C/1 · Q25 (option 1)2 marks

A uniform electric field E of 500 N/C is directed along +x axis. O, B and A are three points in the field having x and y coordinates (in cm) (0, 0), (4, 0) and (0, 3) respectively. Calculate the potential difference between the points (i) O and A, and (ii) O and B.

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Official CBSE Marking Scheme · CBSE Board 2023 (Supplementary), Set 55/C/1

(a) (i) VOA = E (x2 − x1) [½]

VOA = 500 × 0 = 0 volt [½]

(ii) VOB = −E (x2 − x1) [½]

VOB = −500 × (4 × 10−2) = −20 V [½]

Q25.2023 Supp 55/C/1 · Q25 (OR option)2 marks

Three point charges 1 μC, −1 μC and 2 μC are kept at the vertices A, B and C respectively of an equilateral triangle of side 1 m. A1, B1 and C1 are the midpoints of the sides AB, BC and CA respectively. Calculate the net amount of work done in displacing the charge from A to A1, from B to B1 and from C to C1.

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Official CBSE Marking Scheme · CBSE Board 2023 (Supplementary), Set 55/C/1

Initial electrostatic potential energy of the system

Ui = kr[1×(−1) + (−1)×2 + (1)×(2)] × 10−12

= 9×1091[−1 − 2 + 2] × 10−12

= −9 × 10−3 J [1]

Now A1B1 = B1C1 = A1C1 = 12 m

Final electrostatic potential energy of the system

Uf = −9×10−912 = −18 × 10−3 J [½]

Amount of work done W = Uf − Ui

W = −18 × 10−3 + 9 × 10−3 = −9 × 10−3 J [½]

Q26.SQP 2026-27 · Q192 marks

Two like charges +q and +q are placed at a distance d. What work must be done by an external agent to bring another charge +q from infinity to the midpoint of the line joining the two given charges?

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Official CBSE Marking Scheme · CBSE Sample Question Paper 2026-27

At the midpoint, distance from both charges is d/2 [½]

Potential due to one charge +q at the midpoint of the line: V1 = kqd/2 = 2kqd [½]

Potential due to both charges; V = 2 × 2kqd = 4kqd [½]

Work done by external agent: W = qV = q(4kqd) = 4kq2/d [½]

Q27.2026 Main 55/1/1 · Q233 marks

A parallel plate capacitor of capacitance C has a dielectric slab between its plates. It is charged to a potential difference V by connecting it across a battery. The battery is then disconnected. If the dielectric slab is now withdrawn from the capacitor, how will the following be affected ?

  1. Capacitance of the capacitor,
  2. Energy stored in the capacitor, and
  3. The potential difference between the plates of the capacitor.

Justify your answer in each case.

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Official CBSE Marking Scheme · CBSE Board 2026 (Main), Set 55/1/1

(a) C = Kε0Ad (capacitance of capacitor with dielectric slab) [½]

C′ = ε0Ad (capacitance of a capacitor without dielectric slab)

C′ = CK (Capacitance decreases) [½]

(b) As charge Q is constant

U = Q22C [½]

U′ = Q22C′ = KU (Energy stored in the capacitor increases) [½]

(c) V = QC [½]

V′ = QC′ = KV (Potential difference between the plates increases) [½]

Q28.2026 Main 55/2/1 · Q233 marks

(a) An electric field E = E0i exists in a region of space. Draw three equipotential surfaces in the region.

(b) Two point charges –q and +q are located at points (–a, 0, 0) and (a, 0, 0) respectively. Find the electrostatic potential at point (x, 0, 0) where x ≫ a.

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Official CBSE Marking Scheme · CBSE Board 2026 (Main), Set 55/2/1

(a) Three equipotential surfaces (planes perpendicular to the field, i.e. parallel to the yz-plane) [1]

Three equipotential planes perpendicular to uniform electric field lines

Alternatively :

Equipotential planes (dashed) perpendicular to field E along x-axis

(b) Potential at P(x, 0, 0) due to –q charge

V–q = – 14πε0 q(x + a) [½]

Potential at P due to +q charge

V+q = 14πε0 q(x – a) [½]

Net potential at P : V = V–q + V+q

V = –14πε0 q(x + a) + 14πε0 q(x – a)

V = 14πε0 · q(2a)(x2 – a2) [½]

For x ≫ a : V = 14πε0 q(2a)x2 [½]

Q29.2026 Main 55/5/1 · Q223 marks

Two parallel plate capacitors X and Y are connected in series to a 6 V battery. They have the same plate area and same plate separation but capacitor X has air between its plates, whereas capacitor Y contains a material of dielectric constant 4.

  1. Calculate the capacitances of X and Y, if the equivalent capacitance of the combination of X and Y is 4 μF.
  2. Calculate the potential difference across the plates of X and Y.
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Official CBSE Marking Scheme · CBSE Board 2026 (Main), Set 55/5/1

(a) Calculating the capacitance of X & Y [1½]; (b) Calculating the potential difference across plates of X & Y [1½]

(a) CX = C, CY = KC = 4C

1Ceq = 1CX + 1CY

Ceq = CXCYCX + CY [½]

4 = C·4CC + 4C

C = 5 μF [½]

CX = 5 μF

CY = 20 μF [½]

(b) Total charge Q = CeqV = 4 × 10–6 × 6 = 24 × 10–6 C [½]

VX = QCX = 24 × 10–65 × 10–6 = 4.8 V [½]

VY = QCY = 24 × 10–620 × 10–6 = 1.2 V [½]

Alternatively,

VY = V – VX = 6 – 4.8 V = 1.2 V

Q30.2026 Supp 55/7/1 · Q233 marks
  1. Briefly explain the process of charging a parallel plate capacitor, when it is connected across a dc source. [1]
  2. A capacitor, of capacitance C, plate area A and plate separation d is charged to V volts by a battery. The energy stored in it is U. After some time, the battery is disconnected and a slab of dielectric constant K, is inserted to fill the space between the plates of the capacitor. Let V′ and U′ now be the potential difference across the plates of the capacitor and the energy stored in it respectively. Find (V′V) and (U′U). [2]
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Official CBSE Marking Scheme · CBSE Board 2026 (Supplementary), Set 55/7/1

(a) When a battery is connected across the plates of a parallel plate capacitor, a potential difference between the plates is developed which sets up an electric field between the plates of the capacitor. The electric field continues to increase till the potential difference between the plates becomes equal to the applied potential by the battery. [1]

Alternatively: When a capacitor is connected to a battery, current will flow for a short time required to charge the capacitor. As charge accumulates on the capacitor plates, the voltage across them increases. When the capacitor is fully charged, the current in the circuit falls to zero.

(b) C′ = KC [½]

Since the battery is disconnected, the charge will remain conserved.

CV = C′V′ = KCV′

V′V = 1K [½]

U′U = ½C′V′2½CV2 = (C′C)(V′V)2 [½]

= K(1K)2

U′U = 1K [½]

Q31.2026 Supp 55/7/1 · Q28 (OR option)3 marks

Two charges of 8 μC and – 4 μC are located at points (–18 cm, 0, 0) and (18 cm, 0, 0) respectively.

  1. Calculate the electrostatic potential energy of the system of the two charges.
  2. The same system of charges is placed in a region where electrostatic potential V(r) = Ar, where A = 9 × 104 V.m. Calculate the potential energy of the system.
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Official CBSE Marking Scheme · CBSE Board 2026 (Supplementary), Set 55/7/1

(i) U = 14πε0 · q1q2r [½]

U = 9 × 109 (8 × 10-6)(–4 × 10-6)36 × 10-2

U = – 0.8 J [½]

(ii) U = V1q1 + V2q2 + 14πε0 · q1q2r [1]

= 9 × 10418 × 10-2 × 8 × 10-6 + 9 × 10418 × 10-2 × (–4 × 10-6) + (–0.8) [½]

= 4 – 2 – 0.8

U = 1.2 J [½]

Q32.2025 Main 55/5/1 · Q233 marks

A parallel plate capacitor has plate area A and plate separation d. Half of the space between the plates is filled with a material of dielectric constant K in two ways as shown in the figure.

Capacitor with dielectric K filling half the separation (a) and half the area (b)

Find the values of the capacitance of the capacitors in the two cases.

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Q33.2025 Main 55/5/1 · Q28 (option 1)3 marks

Two small solid metal balls A and B of radii R and 2R having charge densities 2σ and 3σ respectively are kept far apart. Find the charge densities on A and B after they are connected by a conducting wire.

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Q34.2024 Main 55/4/1 · Q233 marks

The electric field in a region is given by

E = (10x + 4) i

where x is in m and E is in N/C. Calculate the amount of work done in taking a unit charge from

  1. (5 m, 0) to (10 m, 0)
  2. (5 m, 0) to (5 m, 10 m)
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Official CBSE Marking Scheme · CBSE Board 2024 (Main), Set 55/4/1

(i) ΔV = – ∫x1x2 E dx [½]

ΔV = – ∫510 (10x + 4) dx = – [ 10x22 + 4x ]510 [½]

= – 395 V [½]

W = qΔV = – 395 × 1 = – 395 J [½]

(ii) ΔV = – ∫55 (10x + 4) dx = 0 [½]

W = q · ΔV = 0 [½]

Alternatively

If a student writes, displacement is perpendicular to electric field then ΔV = 0 [½]; W = q · ΔV = 0 [½]

Award full credit for part (ii)

Q35.2024 Supp 55/S/1 · Q233 marks

Two point charges of 10 µC and 20 µC are located at points (– 4 cm, 0, 0) and (5 cm, 0, 0) respectively, in a region with electric field E = Ar2, where A = 2 × 106 NC–1 m2 and r is the position vector of the point under consideration. Calculate the electrostatic potential energy of the system.

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Official CBSE Marking Scheme · CBSE Board 2024 (Supplementary), Set 55/S/1

Electrostatic potential energy of the system :

U = kq1q2r12 + q1V1 + q2V2 [1]

kq1q2r12 = 9×109 × 10×10–6 × 20×10–69×10–2 = 20 J [½]

q1V1 = q1Ar1 = 10×10–6 × 2×1064×10–2 = 500 J [½]

q2V2 = q2Ar2 = 20×10–6 × 2×1065×10–2 = 800 J [½]

U = (20 + 500 + 800) J [½]

U = 1320 J [½]

Q36.2023 Main 55/1/1 · Q263 marks

(a) Twelve negative charges of same magnitude are equally spaced and fixed on the circumference of a circle of radius R as shown in Fig. (i). Relative to potential being zero at infinity, find the electric potential and electric field at the centre C of the circle.

(b) If the charges are unequally spaced and fixed on an arc of 120° of radius R as shown in Fig. (ii), find electric potential at the centre C.

Fig. (i) twelve equally spaced charges on a circle of radius R; Fig. (ii) charges unequally spaced on a 120 degree arc of radius R
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Official CBSE Marking Scheme · CBSE Board 2023 (Main), Set 55/1/1

a) Finding electric potential at the centre [1]; Finding electric field at the centre [1]

b) Finding electric potential at the centre [1]

(a) Electric potential due to point charge V = kq/R [½]

Value of each charge = −q, Total charge = −12q

Total potential V = k(−12q)R = −12kqR = −12q4πε0R [½]

By symmetry the resultant of all electric field vectors becomes zero. So electric field is zero. [1]

(b) Electric potential is a scalar quantity and does not depend on placement of charges.

Therefore V = −12kqR = −12q4πε0R [1]

Q37.2023 Main 55/2/1 · Q26 (option 1)3 marks

Two charged conducting spheres of radii a and b are connected to each other by a wire. Find the ratio of the electric fields at their surfaces.

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Official CBSE Marking Scheme · CBSE Board 2023 (Main), Set 55/2/1

When connected by a conducting wire both spheres will be at the same potential. [½]

∴ kq1a = kq2b [½]

∴ q1q2 = ab [½]

E1E2 = k q1/a2k q2/b2 [1]

E1E2 = ba [½]

Q38.2023 Main 55/2/1 · Q26 (OR option)3 marks

A parallel plate capacitor (A) of capacitance C is charged by a battery to voltage V. The battery is disconnected and an uncharged capacitor (B) of capacitance 2C is connected across A. Find the ratio of

  1. final charges on A and B.
  2. total electrostatic energy stored in A and B finally and that stored in A initially.
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Official CBSE Marking Scheme · CBSE Board 2023 (Main), Set 55/2/1

i) Initially Q = CV

Finally qA = CAV1 & qB = CBV1 [½]

qAqB = CACB = 12 [½]

ii) qA + qB = Q [½]

∴ qA = Q3 & qB = 2Q3 [½]

UfUi = UA + UBUAi = qA2/2CA + qB2/2CBQ2/2CA [1]

= 13 [½]

Alternatively

Common potential

V1 = Q1 + Q2C1 + C2 = Q3C = V3 [∵ QC = V] [1]

UfUi = ½ CeqV12½ CAV2 [½]

= ½ × 3C × (V/3)2½ CV2 = 13 [½]

Q39.2026 Main 55/3/1 · Q294 marks

Capacitors are manufactured with certain standard capacitances and working voltages. However, these standard values may not be the ones that are actually needed in a particular application. Two or more capacitors can be grouped in series or in parallel to achieve desired capacitance and voltage. When connected in series, the total capacitance decreases while the voltage rating increases, whereas in parallel connections, the total capacitance increases and maintains the same voltage rating. A capacitor stores energy in the electric field between its plates and stored energy is proportional to the square of the voltage and capacitance U = 12CV2, where symbols have their usual meanings.

Circuit: 3 V cell, 5 Ω resistor, 6 μF and 3 μF capacitors in series between A and B with mid-point P, and 10 Ω resistor in parallel

Two capacitors, one of 3 μF and the other of 6 μF, are connected in series in the circuit as shown in the figure, for a long time.

(i) The total capacitance of the circuit is :

  1. 6 μF
  2. 3 μF
  3. 9 μF
  4. 2 μF

(ii) The current in the 10 Ω resistor is :

  1. 0·3 A
  2. 0·6 A
  3. 0·2 A
  4. 0

(iii) The potential difference between point A and B is :

  1. 2 V
  2. 0·3 V
  3. 0·2 V
  4. 3 V

(iv) Answer either (a) or (b) :

(a) The value of charge on the plates of the 6 μF capacitor is :

  1. 6 μC
  2. 4 μC
  3. 12 μC
  4. 8 μC

(b) The wire between two capacitors is cut at point P. The current in the circuit will :

  1. increase
  2. decrease
  3. remain the same
  4. first increase then become stable
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Official CBSE Marking Scheme · CBSE Board 2026 (Main), Set 55/3/1

(i) (D) 2 μF [1]

(ii) (C) 0·2 A [1]

(iii) (A) 2 V [1]

(iv) (a) (B) 4 μC OR (b) (C) remain the same [1]

Q40.2026 Main 55/5/1 · Q294 marks

The electric potential (V) and electric field (E) are closely related concepts in electrostatics. The electric field is a vector quantity that represents the force per unit charge at a given point in space, whereas electric potential is a scalar quantity that represents the potential energy per unit charge at a given point in space. Electric field and electric potential are related by the equations Er = –dVdr and E = Er r, i.e., electric field is the negative gradient of the electric potential. This means that electric field points in the direction of decreasing potential and its magnitude is the rate of change of potential with distance. The electric field is the force that drives a unit charge to move from higher potential region to lower potential region and electric potential difference between the two points determines the work done in moving a unit charge from one point to the other point.

Two parallel plates connected to a 200 V supply, 0.01 m apart, electron entering at 3×10^7 m/s, plate length 0.05 m, with axes

A pair of square conducting plates having sides of length 0·05 m are arranged parallel to each other in x-y plane. They are 0·01 m apart along z-axis and are connected to a 200 V power supply as shown in the figure. An electron enters with a speed of 3 × 107 ms–1 horizontally and symmetrically in the space between the two plates. Neglect the effect of gravity on the electron.

(i) The electric field E in the region between the plates is :

  1. (2 × 102 Vm) k
  2. – (2 × 102 Vm) k
  3. (2 × 104 Vm) k
  4. – (2 × 104 Vm) k

(ii) In the region between the plates, the electron moves with an acceleration a given by :

  1. – (3·5 × 1015 ms–2) k
  2. (3·5 × 1015 ms–2) k
  3. (3·5 × 1013 ms–2) i
  4. – (3·5 × 1013 ms–2) i

(iii) (a) Time interval during which an electron moves through the region between the plates is :

  1. 9·0 × 10–9 s
  2. 1·67 × 10–8 s
  3. 1·67 × 10–9 s
  4. 2·17 × 10–9 s

OR

(b) The vertical displacement of the electron which travels through the region between the plates is :

  1. 10 mm
  2. 4·9 mm
  3. 5·9 mm
  4. 3·0 mm

(iv) Which one of the following is the path traced by the electron in between the two plates ?

Electron entering between plates A and B with candidate paths a, b, c, d
  1. a
  2. b
  3. c
  4. d
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Official CBSE Marking Scheme · CBSE Board 2026 (Main), Set 55/5/1

(i) (D) (–2 × 104 V/m) k [1]

(ii) (B) (3.5 × 1015 m/s2) k [1]

(iii) (a) (C) 1.67 × 10–9 s

OR

(b) (B) 4.9 mm [1]

(iv) (B) b [1]

Q41.2025 Main 55/1/1 · Q294 marks

A circuit consisting of a capacitor C, a resistor of resistance R and an ideal battery of emf V, as shown in figure is known as RC series circuit.

RC series circuit with capacitor C, resistor R, battery V and keys S1 and S2

As soon as the circuit is completed by closing key S1 (keeping S2 open) charges begin to flow between the capacitor plates and the battery terminals. The charge on the capacitor increases and consequently the potential difference Vc (= q/C) across the capacitor also increases with time. When this potential difference equals the potential difference across the battery, the capacitor is fully charged (Q = VC). During this process of charging, the charge q on the capacitor changes with time t as q = Q[1 – e–t/RC]

The charging current can be obtained by differentiating it and using ddx (emx) = memx.

Consider the case when R = 20 kΩ, C = 500 μF and V = 10 V.

(i) The final charge on the capacitor, when key S1 is closed and S2 is open, is

  1. 5 μC
  2. 5 mC
  3. 25 mC
  4. 0.1 C

(ii) For sufficient time the key S1 is closed and S2 is open. Now key S2 is closed and S1 is open. What is the final charge on the capacitor ?

  1. Zero
  2. 5 mC
  3. 2.5 mC
  4. 5 μC

(iii) The dimensional formula for RC is

  1. [M L2 T–3 A–2]
  2. [M0 L0 T–1 A0]
  3. [M–1 L–2 T4 A2]
  4. [M0 L0 T A0]

(iv) The key S1 is closed and S2 is open. The value of current in the resistor after 5 seconds, is

  1. 12e mA
  2. e mA
  3. 1e mA
  4. 12e mA

Alternative to (iv) : (iv) The key S1 is closed and S2 is open. The initial value of charging current in the resistor, is

  1. 5 mA
  2. 0.5 mA
  3. 2 mA
  4. 1 mA
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Q42.2025 Main 55/2/1 · Q294 marks

A parallel plate capacitor has two parallel plates which are separated by an insulating medium like air, mica, etc. When the plates are connected to the terminals of a battery, they get equal and opposite charges and an electric field is set up in between them. This electric field between the two plates depends upon the potential difference applied, the separation of the plates and nature of the medium between the plates.

  1. The electric field between the plates of a parallel plate capacitor is E. Now the separation between the plates is doubled and simultaneously the applied potential difference between the plates is reduced to half of its initial value. The new value of the electric field between the plates will be :
    1. E
    2. 2E
    3. E4
    4. E2
  2. A constant electric field is to be maintained between the two plates of a capacitor whose separation d changes with time. Which of the graphs correctly depict the potential difference (V) to be applied between the plates as a function of separation between the plates (d) to maintain the constant electric field ?
    1. V versus d: increasing concave curve flattening out
    2. V versus d: straight line decreasing
    3. V versus d: increasing convex curve
    4. V versus d: straight line increasing from origin
  3. Parallel plates P (top) and Q (bottom, positive) with imaginary line MN perpendicular to the plates
    In the above figure P, Q are the two parallel plates of a capacitor. Plate Q is at positive potential with respect to plate P. MN is an imaginary line drawn perpendicular to the plates. Which of the graphs shows correctly the variations of the magnitude of electric field strength E along the line MN ?
    1. E versus position along MN: parabola-shaped arch
    2. E versus position along MN: rises, flat top, falls (trapezoid)
    3. E versus position along MN: constant
    4. E versus position along MN: falls, flat bottom, rises

(iv) Alternative 1 –

Three parallel plates are placed above each other with equal displacement d between neighbouring plates. The electric field between the first pair of the plates is E1 and the electric field between the second pair of the plates is E2. The potential difference between the third and the first plate is –

  1. (E1 + E2) · d
  2. (E1 – E2) · d
  3. (E2 – E1) · d
  4. d(E1 + E2)2

(iv) Alternative 2 –

A material of dielectric constant K is filled in a parallel plate capacitor of capacitance C. The new value of its capacitance becomes

  1. C
  2. CK
  3. CK
  4. C(1 + 1K)
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Q43.2025 Main 55/4/1 · Q294 marks

A parallel plate capacitor consists of two conducting plates kept generally parallel to each other at a distance. When the capacitor is charged, the charge resides on the inner surfaces of the plates and an electric field is set up between them. Thus, electrostatic energy is stored in the capacitor. The figure shows three large square metallic plates, each of side ‘L’ held parallel and equidistant from each other. The space between P1 and P2 and P2 and P3 is completely filled with mica sheets of dielectric constant ‘K’. The plate P2 is connected to point A and other plates P1 and P3 are connected to point B. Point A is maintained at a positive potential with respect to point B and the potential difference between A and B is V.

Three parallel plates P1, P2, P3 separated by d each; P2 connected to A, P1 and P3 joined to B
  1. The capacitance of the system between A and B will be :
    1. ε0KL2d
    2. ε0KL22d
    3. 2ε0KL2d
    4. 2ε0KdL2
  2. The charge on plate P1 is :
    1. ε0 VKL22d
    2. ε0 VKL2d
    3. 2ε0 VKL2d
    4. ε0 VKL24d
  3. The electric field in the region between P1 and P2 is :
    1. Vd
    2. 2Vd
    3. V2d
    4. dV
  4. (a) The separation between the plates of same area (L2) of a parallel plate air capacitor having capacitance equal to that of this system, will be :
    1. dK
    2. 2dK
    3. d2K
    4. d4K

    OR

    (b) If the source of potential difference applied between A and B is removed, and then A and B are connected by a conducting wire, the net charge on the system will be :

    1. ε0 VKL24d
    2. ε0 VKL22d
    3. ε0 VKL2d
    4. Zero
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Q44.2025 Main 55/6/1 · Q294 marks

A capacitor is a system of two conductors separated by an insulator. In practice, the two conductors have charges Q and – Q with potential difference V = V1 – V2 between them. The ratio QV is a constant, denoted by C and is called the capacitance of the capacitor. It is independent of Q or V. It depends only on the geometrical configuration (shape, size, separation) of the two conductors and the medium separating the conductors. When a parallel plate capacitor is charged, the electric field E0 is localised between the plates and is uniform throughout. When a slab of a dielectric is inserted between the charged plates (charge density σ), the dielectric is polarised by the field. Consequently opposite charges appear on the faces of the slab, near the plates, with surface charge density of magnitude σp. For a linear dielectric σp is proportional to E0. Introduction of a dielectric changes the electric field, and hence, the capacitance of a capacitor, and hence, the energy stored in the capacitor.

Like resistors, capacitors can also be arranged in series or in parallel or in a combination of series and parallel.

(i) Consider a capacitor of capacitance C, with plate area A and plate separation d, filled with air [Fig. (a)]. The distance between the plates is increased to 2d and one of the plates is shifted as shown in Fig. (b). The capacitance of the new system now is :

Fig (a): plates separated by d; Fig (b): plates separated by 2d with one plate shifted
  1. C4
  2. C2
  3. 2C
  4. 4C

(ii) A slab (area A and thickness d1) of a linear dielectric of dielectric constant K is inserted between charged plates (charge density σ) of a parallel plate capacitor [plate area A and plate separation d (> d1)] and opposite charges with charge density of magnitude σp appear on the faces of the slab. The dielectric constant K is given by :

  1. σ + σpσ
  2. σσ – σp
  3. σ + σpσp
  4. σσp

(iii) An electric field E is established between the plates of an air filled parallel plate capacitor, with charges Q and – Q. V is the volume of the space enclosed between the plates. The energy stored in the capacitor is :

  1. 12 ε0E2
  2. ε0Q2E
  3. 12 ε0E2V
  4. ε0EQV

(iv) (a) Three capacitors A, B and M, each of capacitance C are connected to a capacitor N of capacitance 2C and a battery as shown in the figure. If the charges on A and N are Q and Q′ respectively, then Q′Q is :

Capacitors A, B, M in series branch in parallel with N and battery V
  1. 16
  2. 13
  3. 3
  4. 6

(iv) (b) (alternative to (iv)(a)) A slab (area A and thickness d2) of dielectric constant K is inserted in a parallel plate capacitor of plate area A and plate separation d. If C and C0 are the capacitances of the capacitors with and without the dielectric, then CC0 is :

  1. K + 12K
  2. 2KK + 1
  3. KK – 1
  4. K – 1K
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Q45.2024 Main 55/3/1 · Q294 marks

The figure shows four pairs of parallel identical conducting plates, separated by the same distance 2·0 cm and arranged perpendicular to x-axis. The electric potential of each plate is mentioned. The electric field between a pair of plates is uniform and normal to the plates.

Four pairs of parallel plates I (–70 V, –50 V), II (–50 V, +150 V), III (–20 V, +200 V), IV (–100 V, –400 V) with x, y, z axes
  1. For which pair of the plates is the electric field E along i ?
    1. I
    2. II
    3. III
    4. IV
  2. An electron is released midway between the plates of pair IV. It will :
    1. move along i at constant speed
    2. move along – i at constant speed
    3. accelerate along i
    4. accelerate along – i
  3. Let V0 be the potential at the left plate of any set, taken to be at x = 0 m. Then potential V at any point (0 ≤ x ≤ 2 cm) between the plates of that set can be expressed as :
    1. V = V0 + αx
    2. V = V0 + αx2
    3. V = V0 + αx1/2
    4. V = V0 + αx3/2
    where α is a constant, positive or negative.
  4. (a) Let E1, E2, E3 and E4 be the magnitudes of the electric field between the pairs of plates, I, II, III and IV respectively. Then :
    1. E1 > E2 > E3 > E4
    2. E3 > E4 > E1 > E2
    3. E4 > E3 > E2 > E1
    4. E2 > E3 > E4 > E1

    OR

    (b) An electron is projected from the right plate of set I directly towards its left plate. It just comes to rest at the plate. The speed with which it was projected is about : (Take (e/m) = 1·76 × 1011 C/kg)
    1. 1·3 × 105 m/s
    2. 2·6 × 106 m/s
    3. 6·5 × 105 m/s
    4. 5·2 × 107 m/s
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Official CBSE Marking Scheme · CBSE Board 2024 (Main), Set 55/3/1

(i) (D) IV [1]

(ii) (D) accelerate along – i [1]

(iii) (A) V = V0 + αx [1]

(iv) (a) (C) E4 > E3 > E2 > E1

OR

    (b) (B) 2.6 × 106 m/s [1]

Q46.2024 Main 55/5/1 · Q29 (option 1)4 marks

Dielectrics play an important role in design of capacitors. The molecules of a dielectric may be polar or non-polar. When a dielectric slab is placed in an external electric field, opposite charges appear on the two surfaces of the slab perpendicular to electric field. Due to this an electric field is established inside the dielectric.

The capacitance of a capacitor is determined by the dielectric constant of the material that fills the space between the plates. Consequently, the energy storage capacity of a capacitor is also affected. Like resistors, capacitors can also be arranged in series and/or parallel.

  1. Which of the following is a polar molecule ?
    1. O2
    2. H2
    3. N2
    4. HCl
  2. Which of the following statements about dielectrics is correct ?
    1. A polar dielectric has a net dipole moment in absence of an external electric field which gets modified due to the induced dipoles.
    2. The net dipole moments of induced dipoles is along the direction of the applied electric field.
    3. Dielectrics contain free charges.
    4. The electric field produced due to induced surface charges inside a dielectric is along the external electric field.
  3. When a dielectric slab is inserted between the plates of an isolated charged capacitor, the energy stored in it :
    1. increases and the electric field inside it also increases.
    2. decreases and the electric field also decreases.
    3. decreases and the electric field increases.
    4. increases and the electric field decreases.
  4. An air-filled capacitor with plate area A and plate separation d has capacitance C0. A slab of dielectric constant K, area A and thickness (d5) is inserted between the plates. The capacitance of the capacitor will become
    1. [4K5K + 1]C0
    2. [K + 54]C0
    3. [5K4K + 1]C0
    4. [K + 45K]C0
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Official CBSE Marking Scheme · CBSE Board 2024 (Main), Set 55/5/1

(i) (D) HCl [1]

(ii) (B) The net dipole moment of induced dipoles is along the direction of the applied electric field. [1]

(iii) (B) decreases and the electric field also decreases. [1]

(iv) (a) (C) [5K4K + 1]C0 [1]

Q47.2024 Main 55/5/1 · Q29 (OR option)4 marks

Dielectrics play an important role in design of capacitors. The molecules of a dielectric may be polar or non-polar. When a dielectric slab is placed in an external electric field, opposite charges appear on the two surfaces of the slab perpendicular to electric field. Due to this an electric field is established inside the dielectric.

The capacitance of a capacitor is determined by the dielectric constant of the material that fills the space between the plates. Consequently, the energy storage capacity of a capacitor is also affected. Like resistors, capacitors can also be arranged in series and/or parallel.

  1. Which of the following is a polar molecule ?
    1. O2
    2. H2
    3. N2
    4. HCl
  2. Which of the following statements about dielectrics is correct ?
    1. A polar dielectric has a net dipole moment in absence of an external electric field which gets modified due to the induced dipoles.
    2. The net dipole moments of induced dipoles is along the direction of the applied electric field.
    3. Dielectrics contain free charges.
    4. The electric field produced due to induced surface charges inside a dielectric is along the external electric field.
  3. When a dielectric slab is inserted between the plates of an isolated charged capacitor, the energy stored in it :
    1. increases and the electric field inside it also increases.
    2. decreases and the electric field also decreases.
    3. decreases and the electric field increases.
    4. increases and the electric field decreases.
  4. Two capacitors of capacitances 2 C0 and 6 C0 are first connected in series and then in parallel across the same battery. The ratio of energies stored in series combination to that in parallel is
    1. 14
    2. 16
    3. 215
    4. 316
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Official CBSE Marking Scheme · CBSE Board 2024 (Main), Set 55/5/1

(i) (D) HCl [1]

(ii) (B) The net dipole moment of induced dipoles is along the direction of the applied electric field. [1]

(iii) (B) decreases and the electric field also decreases. [1]

(iv) (b) (D) 316 [1]

Q48.2023 Main 55/3/1 · Q344 marks

A beam of electrons moving horizontally with a velocity of 3 × 107 m/s enters a region between two plates as shown in the figure. A suitable potential difference is applied across the plates such that the electron beam just strikes the edge of the lower plate.

Electron beam entering between two parallel plates: plate length 3 cm, separation 1 cm

Answer the following questions based on the above :

(a) How long does an electron take to strike the edge ? [1]

(b) What is the shape of the path followed by the electron and why ? [1]

(c) Find the potential difference applied. [2]

Alternative to (c) :

(c) Find the magnitude and direction of the magnetic field which should be created in the space between the plates so that the electron beam goes straight undeviated. [2]

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Official CBSE Marking Scheme · CBSE Board 2023 (Main), Set 55/3/1

(a) Electron strikes the edge after travelling 3 cm horizontally (along x-axis). [½]

Sx = vx × t ; 3 × 10-2 = (3 × 107) × t

t = 10-9 s [½]

(b) Shape of the path is parabola. [½]

Reason : Force/acceleration is in a fixed direction perpendicular to the initial velocity. [½]

(c) Along y-direction

Sy = uyt + ½ ayt2

–0.5 × 10-2 = 0 + ½ ay(10-9)2

ay = –1016 m/s2 [½]

Magnitude of acceleration (ay) = eEm = em (Vl) [½]

V = 1016 × 9.1 × 10-31 × 1 × 10-21.6 × 10-19 [½]

V = 568.75 V [½]

Alternative to (c) :

(c) |qE| = |qvB| ; B = Ev = (m aye)(1v) [½]

Along y-direction

Sy = uyt + ½ ayt2

–0.5 × 10-2 = 0 + ½ ay(10-9)2

ay = –1016 m/s2 [½]

B = (9.1 × 10-31 × 10161.6 × 10-19) × (13 × 107)

B = 1.9 × 10-3 T [½]

Direction of magnetic field will be out of the plane of the paper. [½]

Q49.2023 Main 55/5/1 · Q35 (option 1)4 marks

A capacitor is a system of two conductors separated by an insulator. The two conductors have equal and opposite charges with a potential difference between them. The capacitance of a capacitor depends on the geometrical configuration (shape, size and separation) of the system and also on the nature of the insulator separating the two conductors. They are used to store charges. Like resistors, capacitors can be arranged in series or parallel or a combination of both to obtain desired value of capacitance.

  1. Find the equivalent capacitance between points A and B in the given diagram.
    Network of five capacitors of capacitance C between points A and B
  2. A dielectric slab is inserted between the plates of a parallel plate capacitor. The electric field between the plates decreases. Explain.
  3. A capacitor A of capacitance C, having charge Q is connected across another uncharged capacitor B of capacitance 2C. Find an expression for (a) the potential difference across the combination and (b) the charge lost by capacitor A.
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Official CBSE Marking Scheme · CBSE Board 2023 (Main), Set 55/5/1

(i)

Redrawn network: four C capacitors in a bridge in parallel with a fifth C between A and B

Cnet = C + C = 2C [½ + ½]

(Note: Give full credit to a student if 2C written directly)

(ii) Within the dielectric slab, induced electric field due to polarization, decreases the electric field. [1]

Alternatively: E = E0 – Ep

Alternatively: E = E0K

(iii) (a) V′ = QTotalCeqi = Q3C [½]

V′ = Q3C = V3 [½]

(b) QA′ = C × V3 = Q3, QA = CV = Q [½]

Charge lost by capacitor A is ΔQ = Q – Q3 = 2Q3 [½]

Q50.2023 Main 55/5/1 · Q35 (OR option)4 marks

A capacitor is a system of two conductors separated by an insulator. The two conductors have equal and opposite charges with a potential difference between them. The capacitance of a capacitor depends on the geometrical configuration (shape, size and separation) of the system and also on the nature of the insulator separating the two conductors. They are used to store charges. Like resistors, capacitors can be arranged in series or parallel or a combination of both to obtain desired value of capacitance.

  1. Find the equivalent capacitance between points A and B in the given diagram.
    Network of five capacitors of capacitance C between points A and B
  2. A dielectric slab is inserted between the plates of a parallel plate capacitor. The electric field between the plates decreases. Explain.
  3. Two slabs of dielectric constants 2K and K fill the space between the plates of a parallel plate capacitor of plate area A and plate separation d as shown in figure. Find an expression for capacitance of the system.
    Parallel plate capacitor: slab of dielectric constant 2K of thickness d/3 and slab of constant K of thickness 2d/3
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Official CBSE Marking Scheme · CBSE Board 2023 (Main), Set 55/5/1

(i)

Redrawn network: four C capacitors in a bridge in parallel with a fifth C between A and B

Cnet = C + C = 2C [½ + ½]

(Note: Give full credit to a student if 2C written directly)

(ii) Within the dielectric slab, induced electric field due to polarization, decreases the electric field. [1]

Alternatively: E = E0 – Ep

Alternatively: E = E0K

(iii) Capacitance of left portion, C1 = 6Kε0Ad [½]

Capacitance of right portion, C2 = 3Kε0A2d [½]

As the capacitors are in series 1Ceqi = 1C1 + 1C2 [½]

1Ceqi = d6Kε0A + 2d3Kε0A = 5d6Kε0A

Ceqi = 6Kε0A5d [½]

Q51.2023 Supp 55/C/1 · Q34 (option 1)4 marks
Parallel plate capacitor with dielectric slab between metal plates of area A, separation d, connected to a V volt battery

A parallel plate capacitor is an arrangement of two identical metal plates kept parallel, a small distance apart. The capacitance of a capacitor depends on the size and separation of the two plates and also on the dielectric constant of the medium between the plates. Like resistors, capacitors can also be arranged in series or parallel or a combination of both. By virtue of electric field between the plates, charged capacitors store energy.

(a) The capacitance of a parallel plate capacitor increases from 10 μF to 80 μF on introducing a dielectric medium between the plates. Find the dielectric constant of the medium. [1]

(b) n capacitors, each of capacitance C, are connected in series. Find the equivalent capacitance of the combination. [1]

(c) A capacitor is charged to a potential (V) by connecting it to a battery. After some time, the battery is disconnected and a dielectric is introduced between the plates. How will the potential difference between the plates, and the energy stored in it be affected ? Justify your answer. [2]

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Official CBSE Marking Scheme · CBSE Board 2023 (Supplementary), Set 55/C/1

(a) K = CC0 [½]

K = 80 μF10 μF = 8 [½]

(b) 1CS = 1C1 + 1C2 + ......... + 1Cn [½]

1CS = nC

CS = Cn [½]

(c) Charge is constant

Q1 = Q2 [½]

C2 = KC1

C1V1 = K C1V2

V2 = V1K Potential diff decreases by a factor (1/K) [½ + ½]

U2 = 12 Q2C2

= 12 Q2kC1 = 1k(Q22C1) [½]

U2 = U1K [½]

Energy reduces by a factor of 1/K. [½]

Q52.2023 Supp 55/C/1 · Q34 (OR option)4 marks
Parallel plate capacitor with dielectric slab between metal plates of area A, separation d, connected to a V volt battery

A parallel plate capacitor is an arrangement of two identical metal plates kept parallel, a small distance apart. The capacitance of a capacitor depends on the size and separation of the two plates and also on the dielectric constant of the medium between the plates. Like resistors, capacitors can also be arranged in series or parallel or a combination of both. By virtue of electric field between the plates, charged capacitors store energy.

(a) The capacitance of a parallel plate capacitor increases from 10 μF to 80 μF on introducing a dielectric medium between the plates. Find the dielectric constant of the medium. [1]

(b) n capacitors, each of capacitance C, are connected in series. Find the equivalent capacitance of the combination. [1]

(c) Find the equivalent capacitance between points A and B, if capacitance of each capacitor is C. [2]

Network of capacitors between A and B with junction points M, P, S, N
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Official CBSE Marking Scheme · CBSE Board 2023 (Supplementary), Set 55/C/1

(a) K = CC0 [½]

K = 80 μF10 μF = 8 [½]

(b) 1CS = 1C1 + 1C2 + ......... + 1Cn [½]

1CS = nC

CS = Cn [½]

(c) For calculating effective capacitance = 2 C. [2]

Q53.2026 Main 55/2/1 · Q32 (option 1)5 marks

(a) Derive an expression for the capacitance of a parallel plate capacitor of plate area A and plate separation d with air present between the plates.

(b) Two air-filled capacitors of capacitances C1 and C2 are connected in parallel with a dc battery. After the capacitors are fully charged, a slab of dielectric constant K is inserted between the plates of each capacitor. How will the (i) charge on each capacitor and (ii) energy stored in the capacitor affected after the slab is introduced.

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(a)

Parallel plate capacitor with plates of area A, surface charge densities +σ and –σ, separation d, field E between plates

Electric field between the plates : E = σε0 = qAε0 [1]

Potential difference : V = Ed = qdAε0 [1]

Capacitance : C = qV = Aε0d [1]

(b) (i) In parallel combination the potential difference is same across the capacitors. On insertion of dielectric slab capacitance of each capacitor becomes K (dielectric constant) times. [1]

Alternatively : C′ = KC. Hence final charge on each capacitor becomes K times.

Alternatively : q′ = C′V = KCV, q′ = Kq

Alternatively : q′1 = KC1V = Kq1, q′2 = KC2V = Kq2

(ii) Energy stored by capacitor, U = 12CV2. As the capacitance becomes K times, energy stored by each capacitor also increases by factor K. [1]

Alternatively : U′ = KU

Alternatively : U′1 = 12KC1V2 = KU1, U′2 = 12KC2V2 = KU2

Q54.2026 Main 55/3/1 · Q31 (option 1)5 marks

(i) In the figure, OA and OB show the variation of electric potential V at a point due to two point charges Q1 and Q2 with 1r respectively. Here r represents the distance of the point from the two point charges.

Graph of V against 1/r: line OA at 60° and line OB at 30° below the 1/r axis

(I) Identify the nature of the two charges Q1 and Q2.

(II) What is the value of (Q1Q2) ? Justify your answer.

(ii) Two point charges − 2 μC and 5 μC are placed at (− 30 cm, 0) and (30 cm, 0) respectively in an external electric field E = Ax2 i, where A = 9 × 105 Nm2 C−1. Find the electrostatic potential energy of this configuration.

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(i) (I) Q1 – Positive [½]

Q2 – Negative [½]

(II) Q1Q2 = Slope of ASlope of B [1]

= tan 60°tan 30°

= 31 [1]

(ii) V(r) = − ∫ E dx = − ∫ Ax2 dx [½]

= Ax [½]

U = q1V(x1) + q2V(x2) + q1q24πε0x12

= q1 Ax1 + q2 Ax2 + q1q24πε0x12

= −2 × 10−6 × 9 × 10530 × 10−2 + 5 × 10−6 × 9 × 10530 × 10−2 − 5 × 2 × 10−12 × 9 × 10960 × 10−2 [½]

= (−6 + 15 − 0·15) J

= 8·85 J [½]

Q55.2026 Main 55/4/1 · Q33 (option 1)5 marks

(a) (i) Explain the following statements giving reason :

  1. An equipotential surface through a point is normal to the electric field at that point.
  2. When a dielectric is placed in an external electric field, the electric field inside the dielectric is less than that outside it.
  3. The potential difference between the plates of a charged parallel plate capacitor decreases when its plates are brought closer.

(ii) Obtain an expression for the work done to dissociate the system of three charges q, − 4q and 2q placed at the vertices A, B and C respectively of an equilateral triangle of side ‘a’.

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Official CBSE Marking Scheme · CBSE Board 2026 (Main), Set 55/4/1

(i) (I) If the field were not normal to the equipotential surface, it would have non-zero component along the surface. To move a unit test charge against the direction of the component of the field, work would be done. This is in contradiction with the definition of equipotential surface. [1]

Alternatively : ∮ E·dl = VA − VB ; VA = VB for equipotential surface

∮ E·dl = 0 ⇒ cos θ = 0 as E ≠ 0 & dl ≠ 0 and θ = 90°

(II) When a dielectric is placed in an external electric field, due to polarization of molecules, an electric field is induced in a direction opposite to the external field, hence net electric field inside is reduced. [1]

Note : Full credit to be given if explained diagrammatically.

Alternatively : ε = ε0 − εP

(III) Potential difference (V) = QC

V = QdAε0

V ∝ d [1]

(ii) Work done = k[q1q2a + q1q3a + q2q3a] [½]

= ka[(q)(−4q) + (q)(2q) + (−4q)(2q)] [1]

= − k(10q2)a

Work done to dissociate is + k(10q2)a. [½]

Q56.2025 Main 55/1/1 · Q31 (option 1)5 marks

(i) Two point charges 5 μC and –1 μC are placed at points (–3 cm, 0, 0) and (3 cm, 0, 0) respectively. An external electric field E = Ar2 r where A = 3 × 105 Vm is switched on in the region. Calculate the change in electrostatic energy of the system due to the electric field.

(ii) A system of two conductors is placed in air and they have net charge of +80μC and –80μC which causes a potential difference of 16 V between them.

(1) Find the capacitance of the system.

(2) If the air between the capacitor is replaced by a dielectric medium of dielectric constant 3, what will be the potential difference between the two conductors ?

(3) If the charges on two conductors are changed to +160 μC and –160 μC, will the capacitance of the system change ? Give reason for your answer.

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Q57.2025 Main 55/1/1 · Q31 (OR option)5 marks

(i) Consider three metal spherical shells A, B and C, each of radius R. Each shell is having a concentric metal ball of radius R/10. The spherical shells A, B and C are given charges +6q, –4q, and 14q respectively. Their inner metal balls are also given charges –2q, +8q and –10q respectively. Compare the magnitude of the electric fields due to shells A, B and C at a distance 3R from their centres.

(ii) A charge –6 μC is placed at the centre B of a semicircle of radius 5 cm, as shown in the figure. An equal and opposite charge is placed at point D at a distance of 10 cm from B. A charge +5 μC is moved from point ‘C’ to point ‘A’ along the circumference. Calculate the work done on the charge.

Semicircle with centre B, diameter points C and A, charge +6q at D and –6q at B
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Q58.2025 Main 55/2/1 · Q31 (option 1)5 marks
  1. A small conducting sphere A of radius r charged to a potential V, is enclosed by a spherical conducting shell B of radius R. If A and B are connected by a thin wire, calculate the final potential on sphere A and shell B.
  2. Write two characteristics of equipotential surfaces. A uniform electric field of 50 NC-1 is set up in a region along +x axis. If the potential at the origin (0, 0) is 220 V, find the potential at a point (4m, 3m).
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Q59.2025 Main 55/6/1 · Q33 (option 1)5 marks
  1. The electric field in a region is given by E = 40x i N/C. Find the amount of work done in taking a unit positive charge from a point (0, 3m) to the point (5m, 0).
  2. A charge Q is distributed over two concentric hollow spheres of radii r and R (> r) such that their surface charge densities are equal. Find :
    (I) the electric field, and
    (II) the potential
    at their common centre.
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Q60.2025 Main 55/7/1 · Q31 (option 1)5 marks

(i) A parallel plate capacitor with plate area A and plate separation d has a capacitance C0. A slab of dielectric constant K having area A and thickness (d4) is inserted in the capacitor, parallel to the plates. Find the new value of its capacitance.

(ii) You are provided with a large number of 1 μF identical capacitors and a power supply of 1200 V. The dielectric medium used in each capacitor can withstand up to 200 V only. Find the minimum number of capacitors and their arrangement, required to build a capacitor system of equivalent capacitance of 2 μF for use with this supply.

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Q61.2025 Main 55/7/1 · Q31 (OR option)5 marks

(i) An electric dipole of dipole moment p consists of point charges q and – q, separated by 2a. Derive an expression for electric potential in terms of its dipole moment at a point at a distance x (>> a) from its centre and lying (I) along its axis, and (II) along its bisector line.

(ii) An electric dipole of dipole moment p = (0·8 i + 0·6 j) 10–29 Cm is placed in an electric field E = 1·0 × 107 k Vm. Calculate the magnitude of the torque acting on it and the angle it makes with the x-axis, at this instant.

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Q62.2025 Supp 55/S/1 · Q32 (option 1)5 marks

(i) A parallel plate capacitor A of capacitance C is charged by a battery to a potential ‘V’. The battery is disconnected and an uncharged identical capacitor B is connected across it. Calculate for the capacitor A the new value of the :

(I) charge

(II) potential difference

(III) energy stored

Justify your answers.

(ii) Draw the pattern of electric field lines due to :

(I) positively charged conducting sphere, and

(II) an electric dipole.

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Official CBSE Marking Scheme · CBSE Board 2025 (Supplementary), Set 55/S/1

(i) (I) Charge will become half [½]

Charge will flow from capacitors A (high potential) to capacitor B (low potential) till they achieve equilibrium.

Alternatively :

qA = CV   qB = 0

Vcommon = qA + qBCA + CB = CV2C = V2

q′A = CVcommon = CV2 = qA2 [½]

(II) Potential difference will become half. [½]

Vcommon = qA + qBCA + CB = CV + 02C = V2 [½]

(III) Energy stored will become one fourth [½]

UA = 12CV2

U′A = 12CV2common = 12C( V2 )2 = UA4 [½]

(ii) (I)

Electric field lines of a positively charged conducting sphere, radially outward

[1]

(II)

Electric field lines of an electric dipole (+q and –q)

[1]

Q63.2024 Main 55/1/1 · Q31 (OR option)5 marks

(i) An electric dipole (dipole moment p = p i), consisting of charges − q and q, separated by distance 2a, is placed along the x-axis, with its centre at the origin. Show that the potential V, due to this dipole, at a point x, (x >> a) is equal to 14πε0 · p·ix2.

(ii) Two isolated metallic spheres S1 and S2 of radii 1 cm and 3 cm respectively are charged such that both have the same charge density (2π × 10−9) C/m2. They are placed far away from each other and connected by a thin wire. Calculate the new charge on sphere S1.

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Official CBSE Marking Scheme · CBSE Board 2024 (Main), Set 55/1/1

(i) Deriving expression for potential [2½]; (ii) new charge on sphere S1 [2½]

(i)

Dipole with −q, centre O and +q on the x-axis, separation 2a, and point P at distance x from O

V = 14πε0 qr [½]

V = V+q − V−q [½]

V = 14πε0 [ q(x − a) − q(x + a) ] [½]

= q4πε0 [ x + a − x + a(x2 − a2) ]

V = q4πε0 2a(x2 − a2) = p4πε0(x2 − a2)

As p is along x-axis, so

V = 14πε0 p·i(x2 − a2) [½]

If x >> a

V = 14πε0 p·ix2 [½]

Alternatively:

Dipole with +q and −q separated by 2a, point P at distance r making angle θ with the dipole axis, distances r1 and r2 (figure appears rotated in the source)

V = 14πε0 (qr1 − qr2)   ----- (i) [½]

By geometry

r12 = r2 + a2 − 2ar cos θ

r22 = r2 + a2 + 2ar cos θ

r12 = r2 (1 − 2a cos θr + a2r2) ≅ r2 (1 − 2a cos θr) [½]

Similarly, r22 ≅ r2 (1 + 2a cos θr) [½]

Using binomial theorem & retaining terms upto the first order in ar ; we obtain

1r1 ≅ 1r (1 − 2a cos θr)−½ ≅ 1r (1 + ar cos θ)   ----- (ii)

1r2 ≅ 1r (1 + 2a cos θr)−½ ≅ 1r (1 − ar cos θ)   ----- (iii)

Using equations (i), (ii) & (iii) & p = 2qa

V = q4πε0 2a cos θr2 = p cos θ4πε0r2 [½]

p cos θ = p·r

As r is along the x-axis, p·r = p·i

⇒ V = 14πε0 p·ix2 [½]

(ii) Charge on sphere S1 :

Q1 = surface charge density × surface area = (2π × 10−9) × 4π (1 × 10−2)2 = 8 × 10−13 C [½]

Charge on sphere S2 :

Q2 = surface charge density × surface area = (2π × 10−9) × 4π (3 × 10−2)2 = 72 × 10−13 C [½]

When connected by a thin wire they acquire a common potential V and the charge remains conserved.

Q1 + Q2 = Q1′ + Q2′ = C1V + C2V [½]

Q1 + Q2 = (C1 + C2) V

Common potential (V) = Q1 + Q2C1 + C2

C1 = 4πε0r1 = 19 × 109 × 10−2 = 19 × 10−11 F

C2 = 4πε0r2 = 19 × 109 × 3 × 10−2 = 13 × 10−11 F

V = 80 × 10−13(19 + 13) × 10−11 = 1.8 V [½]

Q1′ = C1V = 19 × 10−11 × 1.8

Q1′ = 2 × 10−12 C [½]

Alternatively:

Q1 = 8 × 10−13 C [½]; Q2 = 72 × 10−13 C [½]

When connected by a thin wire they acquire a common potential V and the charge remains conserved.

Q1 + Q2 = Q1′ + Q2′ [½]

Q2′Q1′ = r2r1 [½]

On solving, Q1′ = 2 × 10−12 C [½]

Q64.2024 Main 55/2/1 · Q31 (option 1)5 marks

(i) Obtain an expression for the electric potential due to a small dipole of dipole moment p, at a point r from its centre, for much larger distances compared to the size of the dipole.

(ii) Three point charges q, 2q and nq are placed at the vertices of an equilateral triangle. If the potential energy of the system is zero, find the value of n.

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Official CBSE Marking Scheme · CBSE Board 2024 (Main), Set 55/2/1

(i)

Electric dipole of charges +q and −q separated by 2a, with point P at distance r making angle θ with the dipole axis

Potential due to the dipole is the sum of potentials due to charges q and −q [½]

V = 14πε0(qr1 − qr2) ……(1) [½]

By geometry
r12 = r2 + a2 − 2ar cos θ
r22 = r2 + a2 + 2ar cos θ

For r >> a, retaining terms only up to first order in a/r
r12 = r2(1 − 2a cos θr + a2r2) ≅ r2(1 − 2a cos θr)
Similarly r22 ≅ r2(1 + 2a cos θr) [½]

Using the binomial theorem and retaining terms up to the first order in a/r
1r1 ≅ 1r(1 + a cos θr) ……(2)
1r2 ≅ 1r(1 − a cos θr) ……(3) [½]

Using eqn. (1), (2), (3) and p = 2qa
V = q4πε0 2a cos θr2 = p cos θ4πε0r2 [½]

Alternatively –

Dipole AB with charges −q at A and +q at B, perpendiculars AC and BD drawn to the line OP

r2 = r + a cos θ [½]
r1 = r − a cos θ [½]

V = q4πε0(1r1 − 1r2) = q4πε0(1r − a cos θ − 1r + a cos θ) [½]

= q4πε0(2a cos θr2 − a2cos2θ) [½]

= p4πε0r2(cos θ1 − a2r2 cos2θ) [½]

For r >> a, neglecting a2r2
V = p cos θ4πε0r2 [½]

(ii) Consider the side of equilateral triangle as ‘a’.
Potential energy = U = kq1q2a + kq2q3a + kq1q3a [½]

According to question
U = k(q)(2q)a + k(2q)(nq)a + k(q)(nq)a = 0 [½]

= 2q2a + 2nq2a + nq2a = 0
2 + 2n + n = 0
3n = −2 [½]

n = − 23 [½]

Q65.2024 Main 55/3/1 · Q31 (option 1)5 marks

(i) Obtain the expression for the capacitance of a parallel plate capacitor with a dielectric medium between its plates.

(ii) A charge of 6 μC is given to a hollow metallic sphere of radius 0·2 m. Find the potential at (i) the surface and (ii) the centre of the sphere.

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Official CBSE Marking Scheme · CBSE Board 2024 (Main), Set 55/3/1

(i) Obtaining expression for the capacitance [3]; (ii) Finding the electric potential (i) at the surface, (ii) at the centre [2]

(i) When a dielectric slab is inserted between the plates of capacitor, there is induced charge density σP which opposes the original charge density (σ) on the plate of capacitance. [½]

Electric field with dielectric medium is

E = (σ – σP)ε0 [½]

V = E × d = (σ – σP)ε0 d [½]

(σ – σP) = σK [½]

V = σ dε0K = QdAε0K [½]

C = QV = Kε0Ad [½]

(ii) Electric potential due to a point charge

V = 14πε0 qr [½]

(i) At the surface

V = 14πε0 qr = 9×109×6×10–60.2 [½]

V = 2.7 × 105 V [½]

(ii) Since electric field inside the hollow sphere is zero, hence V is same as that of the surface and remains constant throughout the volume.

V = 2.7 × 105 V [½]

Q66.2024 Main 55/4/1 · Q32 (option 1)5 marks
  1. A dielectric slab of dielectric constant ‘K’ and thickness ‘t’ is inserted between plates of a parallel plate capacitor of plate separation d and plate area A. Obtain an expression for its capacitance.
  2. Two capacitors of different capacitances are connected first (1) in series and then (2) in parallel across a dc source of 100 V. If the total energy stored in the combination in the two cases are 40 mJ and 250 mJ respectively, find the capacitance of the capacitors.
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Official CBSE Marking Scheme · CBSE Board 2024 (Main), Set 55/4/1

(i) Electric field in air between plates

Parallel plate capacitor with dielectric slab of thickness t, plate separation d

E0 = σε0 [½]

Electric field inside the dielectric

E = σε0K [½]

Potential difference between the plates

V = E0(d – t) + Et [½]

V = σε0[ d – t + tK ]

V = qAε0[ d – t + tK ] [½]

Capacitance C = qV [½]

C = Aε0d – t + tK

C = Aε0d – t(1 – 1K) [½]

(ii) Total energy stored in series combination

12(C1C2C1 + C2)V2 = 40 × 10–3 J ………… (1) [½]

Energy stored in parallel combination

12(C1 + C2)V2 = 250 × 10–3 J ………… (2) [½]

Substituting value of V = 100 V in eq (1) and (2), on solving

C1 = 4 × 10–5 F or 40 μF [½]

C2 = 1 × 10–5 F or 10 μF [½]

Q67.2024 Main 55/5/1 · Q31 (option 1)5 marks
  1. Draw equipotential surfaces for an electric dipole.
  2. Two point charges q1 and q2 are located at r1 and r2 respectively in an external electric field E. Obtain an expression for the potential energy of the system.
  3. The dipole moment of a molecule is 10–30 Cm. It is placed in an electric field E of 105 V/m such that its axis is along the electric field. The direction of E is suddenly changed by 60° at an instant. Find the change in the potential energy of the dipole, at that instant.
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Official CBSE Marking Scheme · CBSE Board 2024 (Main), Set 55/5/1

(i) [1]

Equipotential surfaces of an electric dipole: concentric circles around the negative and positive charges

(ii) Work done in bringing a charge q1 from infinity to r1 :

W1 = q1V(r1) ------(1) [½]

Work done in bringing a charge q2 from infinity to r2 against the external field :

W2 = q2V(r2) ------(2) [½]

Work done on q2 against the field due to q1:

W12 = q1q24πε0r12 ------(3) [½]

Potential energy of the system = Total work done

= q1V(r1) + q2V(r2) + q1q24πε0r12 [½]

(iii) Change in Potential energy = Work done [1]

W = pE [cosθ0 – cosθ1]

W = 10-30 × 105 [cos0° – cos60°] [½]

W = 5.0 × 10-26 J [½]

Q68.2023 Main 55/2/1 · Q31 (OR option)5 marks
  1. Consider two identical point charges located at points (0, 0) and (a, 0).
    1. Is there a point on the line joining them at which the electric field is zero ?
    2. Is there a point on the line joining them at which the electric potential is zero ?
    Justify your answers for each case.
  2. State the significance of negative value of electrostatic potential energy of a system of charges.
    Three charges are placed at the corners of an equilateral triangle ABC of side 2.0 m as shown in figure. Calculate the electric potential energy of the system of three charges.
    Equilateral triangle ABC with charges +4.0 µC at A, −4.0 µC at B and +2.0 µC at C
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Official CBSE Marking Scheme · CBSE Board 2023 (Main), Set 55/2/1

(i) (1) Yes, electric field is zero at mid point. [½]

Electric field being a vector quantity, its resultant is zero. [½]

(2) No, potential cannot be zero on line joining the charges. [½]

Electric potential being a scalar quantity, the net potential due to two identical charges cannot be zero. [½]

(ii) Negative value of electrostatic potential energy of a system signifies that the system has attractive forces. [1]

Alternatively

Give full credit, if a candidate writes the system is stable / bound.

U = 14π ε0 × q1q2r [½]

U = 14π ε0[qAqBr + qBqCr + qCqAr] [½]

= 9×1092 [−16 − 8 + 8] × 10−12 [½]

= −7.2 × 10−2 J [½]

Q69.2023 Main 55/3/1 · Q31 (OR option)5 marks

(i) Define electric potential at a point and write its SI unit.

(ii) Two capacitors are connected in series. Derive an expression of the equivalent capacitance of the combination.

(iii) Two point charges + q and – q are located at points (3a, 0) and (0, 4a) respectively in x-y plane. A third charge Q is kept at the origin. Find the value of Q, in terms of q and a, so that the electrostatic potential energy of the system is zero.

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Official CBSE Marking Scheme · CBSE Board 2023 (Main), Set 55/3/1

(i) Electric Potential – Electrostatic potential at any point in a region with electrostatic field is the work done in bringing a unit positive charge (without acceleration) from infinity to that point. [½]

Alternatively: V = Work Doneq ; V = – ∫ E · dl

S.I. unit of electrostatic potential is volt. [½]

Alternatively: S.I. unit is J/C.

(ii) Series combination of two capacitors C1 and C2 [½]

Two capacitors C1 and C2 in series, each plate pair carrying charges +Q and -Q

V = V1 + V2 = QC1 + QC2 [½]

QCeq = Q (1C1 + 1C2) [½]

1Ceq = 1C1 + 1C2 [½]

(iii) Potential energy of the system = K [Q(–q)4a + Qq3a – q25a] [½]

Potential energy of the system = 0

⇒ K [–Qq4a + Qq3a – q25a] = 0 [½]

⇒ –Q4 + Q3 – q5 = 0

⇒ Q12 – q5 = 0

⇒ Q = + 12q5 [1]

Q70.SQP 2025-26 · Q31 (option 1)5 marks

(A) A dielectric slab of thickness t, is introduced between the plates of parallel plate capacitor of area A and separation d (where t<d). Find an expression for the capacitance with the dielectric slab. [2]

(B) A copper sphere of capacitor C is dropped in ocean. Will the capacitance of the sphere increase, decrease or remain same? Justify. [2]

(C) A capacitor is connected across a source of potential difference V and then the separation ‘d’ between the plates is increased using insulating stick. Plot ‘V’ vs ‘d’ graph for the given capacitor. [1]

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Official CBSE Marking Scheme · CBSE Sample Question Paper 2025-26

(A) In absence of dielectric slab, the capacitance of parallel plate capacitor is given by

C = Aε0d [½]

When a dielectric slab of thickness t (t < d) is introduced between the plates without touching the plates, the electric field in air

Eo = σε0 (σ is charge density given by qA)

but on account of polarisation of dielectric the electric field inside the dielectric changes to

E = EoK

If potential difference between the plates of capacitor be V. now, then clearly

V = Eo(d–t) + Et;

Or, V = Eo(d–t) + EoK t;

Or, V = Eo(d–t+tk) = σε0 (d–t+tk)

Or, V = qAε0 (d–t+tk) [1]

Parallel plate capacitor with dielectric slab of thickness t between plates separated by d, showing fields E0 and E = E0/K

[½]

(B) Capacitance of sphere will Increase.

Justification:

As C = qV

& V = ∫ E·dl

As, electric field will decrease, due to polarization of water. Resulting in decrease in potential.

Hence, capacitance of sphere will increase [1 + 1]

(C)

Straight-line graph of V against d through the origin

[1]

Q71.SQP 2025-26 · Q31 (OR option)5 marks

(A) If a charge of 1μC is placed at the origin and another charge of 3 μC placed at the point (20m,0m,0m) in an external uniform electric field of 40V/m i with the electric potential at origin to be zero. Find the electrical potential energy of system. [3]

(B) If one charge particle is moved from A to C to B and another charge particle of equal magnitude is moved from A to D to B, in uniform external magnetic field. Than for which charge particle more work will be needed? (use fig for reference) [1]

Uniform electric field E to the right with two paths from A to B: via C (curved) and via D

(C) Electrostatic potential is constant throughout the volume of conductor has the same value on its surface why? [1]

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Official CBSE Marking Scheme · CBSE Sample Question Paper 2025-26

(A) U = Kq1q2r12 + q1V(r1) + q2V(r2) [1]

Or, U = Kq1q2r12 + q1(E r1 – 0) + q2(E r2 – 0) [1]

Or, U = (9 × 109 × 10–6 × 3 × 10–620 + 0 + 3 × 10–6 × 40 × 20) J

Or, U = 37.5 × 10–4 J [1]

(B) Work done will be same for both paths, as electric field is conservative in nature. [1]

(C) As electric field inside the conductor is zero so there will be no work needed in moving unit positive charge inside or on the surface. [1]

Q72.SQP 2026-27 · Q31 (option 1)5 marks
  1. An isolated conductor cannot have a large capacitance. Why?
  2. How would you connect 6 µF, 9 µF and 18 µF capacitors to obtain-
    1. the minimum capacitance
    2. the maximum capacitance
  3. The capacitance of a parallel plate capacitor is increased as the area of the plates increases. What are the other two factors on which the capacitance depends?
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Official CBSE Marking Scheme · CBSE Sample Question Paper 2026-27

I. An isolated conductor cannot have a large capacitance because its capacitance depends directly on its size (radius). [½]

For an isolated sphere: C = 4πε0R [½]

Since ε0 is a constant, the only way to increase capacitance is by increasing R (the radius). [½]

A very large capacitance would require an extremely large-sized conductor, which is not practically possible. Therefore, an isolated conductor always has a small capacitance. [½]

II. (i) For minimum capacitance, the three capacitors must be connected in series, [½]

1Cmin = 16 + 19 + 118 = 618 = 13

Cmin = 3 µF [½]

(ii) For maximum capacitance, the three capacitors must be connected in parallel, [½]

Cmax = 6 + 9 + 18 = 33 µF [½]

III. The other two factors on which capacitance of a parallel plate capacitor depends are-

  1. Distance between the plates (C ∝ 1/d)
  2. Permittivity of the medium between the plates (C ∝ ε)

[½ × 2]

Q73.SQP 2026-27 · Q31 (OR option)5 marks
  1. Show that the electric field at any point is equal to the negative of the potential gradient at that point.
  2. A small metallic sphere is gradually charged in open air. It is observed that electrical breakdown of air occurs when the electric field at its surface reaches 2.5 x 106 V m-1. If the sphere is allowed to attain a maximum potential of 5 x 105 V without causing any discharge, determine the minimum radius the insulated sphere must have.
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Official CBSE Marking Scheme · CBSE Sample Question Paper 2026-27

I. Consider a small positive charge +q placed in an electric field E. Suppose it is displaced through a very small distance dx in the direction of the field. [½]

Force on the charge, F = qE [½]

Work done by the electric field in moving the charge through distance dr, dW = F · dr = qE.dr ……(1) [½]

Here, the change in electric potential, dV = − dWq [½]

From (1)- dV = − qE.drq

dV = − E.dr

⇒ E = − dVdr [½]

i.e. The electric field at any point is equal to the negative gradient of electric potential at that point. [½]

II. Potential gradient, dVdr = 2.5 × 106 V m-1 [½]

dV = 2.5 × 106 dr

V = 2.5 × 106 r [½]

r = V2.5 × 106 = 5 × 1052.5 × 106 [½]

r = 2 × 10-1 = 0.2 m [½]

Unit II · Current Electricity

3. Current Electricity

Q1.2026 Main 55/1/1 · Q111 mark

Two heaters rated as (P1, V) and (P2, V) are connected in series across a dc source of V2 volt. The power consumed by the combination will be –

  1. (P1 + P2)
  2. P1 + P22
  3. P1P22(P1 + P2)
  4. P1P24(P1 + P2)
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Official CBSE Marking Scheme · CBSE Board 2026 (Main), Set 55/1/1

(D) P1P24(P1 + P2)

Q2.2026 Main 55/2/1 · Q161 mark

Assertion (A) : Two electric heaters of power P1 and P2 (> P1) are joined in series across a dc source of voltage V. The power consumed by the combination will be less than that consumed by P1 when connected across the same source.

Reason (R) : The power consumed by a electric device when connected to a dc source of voltage V is proportional to its resistance.

  1. Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of Assertion (A).
  2. Both Assertion (A) and Reason (R) are true, but Reason (R) is not the correct explanation of Assertion (A).
  3. Assertion (A) is true, but Reason (R) is false.
  4. Assertion (A) is false and Reason (R) is also false.
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Official CBSE Marking Scheme · CBSE Board 2026 (Main), Set 55/2/1

(C) Assertion (A) is true but Reason (R) is false

Q3.2026 Main 55/3/1 · Q131 mark

Assertion (A) : In a Wheatstone bridge circuit, if we interchange the position of the cell and the galvanometer, the balance condition PQ = RS remains unchanged.

Reason (R) : PQ = RS ⇒ QS = PR so balance condition remains same.

  1. Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of the Assertion (A).
  2. Both Assertion (A) and Reason (R) are true, but Reason (R) is not the correct explanation of the Assertion (A).
  3. Assertion (A) is true, but Reason (R) is false.
  4. Both Assertion (A) and Reason (R) are false.
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Official CBSE Marking Scheme · CBSE Board 2026 (Main), Set 55/3/1

(A) Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of the Assertion (A)

Q4.2026 Main 55/4/1 · Q11 mark

Two wires of same length and same area of cross-section but made of different material of resistivities ρ1 and ρ2 are connected in series. The equivalent resistivity of the combination is :

  1. 12(ρ1 + ρ2)
  2. ρ1 + ρ2
  3. ρ1ρ2
  4. 2 (ρ1 + ρ2)
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Official CBSE Marking Scheme · CBSE Board 2026 (Main), Set 55/4/1

(A) (ρ1 + ρ2) / 2

Q5.2026 Main 55/4/1 · Q31 mark

An electron moves around the nucleus in a circular orbit of radius r and makes n revolutions per second. The value of equivalent current in the orbit is :

  1. en
  2. ne
  3. ner
  4. enr
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Official CBSE Marking Scheme · CBSE Board 2026 (Main), Set 55/4/1

(B) ne

Q6.2026 Main 55/4/1 · Q131 mark

Questions number 13 to 16 are Assertion (A) and Reason (R) type questions. Two statements are given — one labelled Assertion (A) and the other labelled Reason (R). Select the correct answer from the codes (A), (B), (C) and (D) as given below.

Assertion (A) : The work done, in taking a unit charge around a closed loop of an electric circuit involving cells and resistors in the loop, is zero.

Reason (R) : The potential at a point depends on the location of the point in the loop. After completing one round, the charge comes back to the point of start.

  1. Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of the Assertion (A).
  2. Both Assertion (A) and Reason (R) are true, but Reason (R) is not the correct explanation of the Assertion (A).
  3. Assertion (A) is true, but Reason (R) is false.
  4. Both Assertion (A) and Reason (R) are false.
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Official CBSE Marking Scheme · CBSE Board 2026 (Main), Set 55/4/1

(A) Both Assertion (A) and Reason (R) are true, and Reason (R) is the correct explanation of Assertion (A).

Q7.2026 Supp 55/7/1 · Q121 mark

Five resistors are joined as shown in the figure. Between the points A and B, a battery of 6 V is connected. How much current is flowing in arm FC ?

Five resistors arranged between C, D, E and F with battery terminals A and B
  1. 2·0 A
  2. 1·5 A
  3. 1·0 A
  4. 0·5 A
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Official CBSE Marking Scheme · CBSE Board 2026 (Supplementary), Set 55/7/1

(C) 1.0 A

Q8.2025 Main 55/1/1 · Q21 mark

Two wires P and Q are made of the same material. The wire Q has twice the diameter and half the length as that of wire P. If the resistance of wire P is R, the resistance of the wire Q will be

  1. R
  2. R2
  3. R8
  4. 2R
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Q9.2025 Main 55/2/1 · Q21 mark

A student has three resistors, each of resistance R. To obtain a resistance of 23R, she should connect

  1. all the three resistors in series.
  2. all the three resistors in parallel.
  3. two resistors in series and then this combination in parallel with the third resistor.
  4. two resistors in parallel and then this combination in series with the third resistor.
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Q10.2025 Main 55/4/1 · Q21 mark

A current flows through a cylindrical conductor of radius R. The current density at a point in the conductor is j = αr (along its axis), here α is a constant and r is distance from the axis of the conductor. The current flowing through the portion of the conductor from r = 0 to r = R2 is proportional to :

  1. R
  2. R2
  3. R3
  4. R4
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Q11.2025 Main 55/5/1 · Q31 mark

Four resistors, each of resistance R and a key K are connected as shown in the figure. The equivalent resistance between points A and B when key K is open, will be :

Circuit of four resistors R in a circle with key K between A and B
  1. 4R
  2. ∞
  3. R4
  4. 4R3
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Q12.2025 Main 55/6/1 · Q11 mark

The figure shows the voltage (V) versus the current (I) graphs for a wire at two temperatures T1 and T2. One can conclude that :

I versus V straight lines for T1 and T2, T1 steeper
  1. T2 = 2T1
  2. T1 > T2
  3. T1 = T2 / 3
  4. T1 < T2
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Q13.2025 Main 55/6/1 · Q21 mark

If Rs and Rp are the equivalent resistances of n resistors, each of value R, in series and parallel combinations respectively, then the value of (Rs – Rp) is :

  1. (n2 – 1n2)R
  2. (n2 + 1n2 – 1)R
  3. (n2 – 1n)R
  4. (n2 + 1)Rn2
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Q14.2025 Main 55/7/1 · Q21 mark

The effective resistance between points A and B in the given circuit is :

Resistor network between A and B with 2 Ω, 3 Ω, 4 Ω, 3 Ω, 7 Ω, 18 Ω, 6 Ω, 4 Ω and 12 Ω resistors
  1. 6 Ω
  2. 83 Ω
  3. 163 Ω
  4. 2 Ω
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Q15.2024 Main 55/1/1 · Q31 mark

Consider the circuit shown in the figure. The potential difference between points A and B is :

Two branches in parallel between A and B: a 12 V cell with 1 Ω resistor, and a 6 V cell with 0.5 Ω resistor
  1. 6 V
  2. 8 V
  3. 9 V
  4. 12 V
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Official CBSE Marking Scheme · CBSE Board 2024 (Main), Set 55/1/1

(B) 8 V

Q16.2024 Main 55/2/1 · Q21 mark

Electrons drift with speed vd in a conductor with potential difference V across its ends. If V is reduced to V2, their drift speed will become :

  1. vd2
  2. vd
  3. 2 vd
  4. 4 vd
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Official CBSE Marking Scheme · CBSE Board 2024 (Main), Set 55/2/1

(A) vd2

Q17.2024 Main 55/4/1 · Q141 mark

Assertion (A) : The temperature coefficient of resistance is positive for metals and negative for p-type semiconductors.

Reason (R) : The charge carriers in metals are negatively charged, whereas the majority charge carriers in p-type semiconductors are positively charged.

  1. If both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of Assertion (A).
  2. If both Assertion (A) and Reason (R) are true and Reason (R) is not the correct explanation of Assertion (A).
  3. If Assertion (A) is true and Reason (R) is false.
  4. If both Assertion (A) and Reason (R) are false.
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Official CBSE Marking Scheme · CBSE Board 2024 (Main), Set 55/4/1

(B) Both assertion (A) and Reason (R) are true and Reason (R) is not the correct explanation of Assertion (A).

Q18.2024 Main 55/4/1 · Q151 mark

Assertion (A) : When electrons drift in a conductor, it does not mean that all free electrons in the conductor are moving in the same direction.

Reason (R) : The drift velocity is superposed over large random velocities of electrons.

  1. If both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of Assertion (A).
  2. If both Assertion (A) and Reason (R) are true and Reason (R) is not the correct explanation of Assertion (A).
  3. If Assertion (A) is true and Reason (R) is false.
  4. If both Assertion (A) and Reason (R) are false.
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Official CBSE Marking Scheme · CBSE Board 2024 (Main), Set 55/4/1

(A) Both assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of Assertion (A).

Q19.2024 Main 55/5/1 · Q11 mark

A battery supplies 0.9 A current through a 2 Ω resistor and 0.3 A current through a 7 Ω resistor when connected one by one. The internal resistance of the battery is :

  1. 2 Ω
  2. 1.2 Ω
  3. 1 Ω
  4. 0.5 Ω
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Official CBSE Marking Scheme · CBSE Board 2024 (Main), Set 55/5/1

(D) 0.5 Ω

Q20.2024 Supp 55/S/1 · Q131 mark

Assertion (A) : The current density (J) at a point in a conducting wire is in the direction of electric field (E) at that point.

Reason (R) : A conducting wire obeys Ohm’s law.

  1. Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of the Assertion (A).
  2. Both Assertion (A) and Reason (R) are true, but Reason (R) is not the correct explanation of the Assertion (A).
  3. Assertion (A) is true, but Reason (R) is false.
  4. Both Assertion (A) and Reason (R) are false.
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Official CBSE Marking Scheme · CBSE Board 2024 (Supplementary), Set 55/S/1

(B) Both Assertion (A) and Reason (R) are true, but Reason (R) is not the correct explanation of the Assertion (A).

Q21.2023 Main 55/1/1 · Q21 mark

For a metallic conductor, the correct representation of variation of resistance R with temperature T is :

Four graphs (a)-(d) of resistance R versus temperature T
  1. Graph (a)
  2. Graph (b)
  3. Graph (c)
  4. Graph (d)
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Official CBSE Marking Scheme · CBSE Board 2023 (Main), Set 55/1/1

(D)

Resistance increasing non-linearly with temperature
Q22.2023 Main 55/1/1 · Q31 mark

The potential difference across a cell in an open circuit is 8 V. It falls to 4 V when a current of 4 A is drawn from it. The internal resistance of the cell is :

  1. 4 Ω
  2. 3 Ω
  3. 2 Ω
  4. 1 Ω
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Official CBSE Marking Scheme · CBSE Board 2023 (Main), Set 55/1/1

(D) 1 Ω

Q23.2023 Main 55/1/1 · Q41 mark

A steady current flows through a metallic wire whose area of cross-section (A) increases continuously from one end of the wire to the other. The magnitude of drift velocity (vd) of the free electrons as a function of ‘A’ can be shown by :

Four graphs (a)-(d) of drift velocity v_d versus area A
  1. Graph (a)
  2. Graph (b)
  3. Graph (c)
  4. Graph (d)
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Official CBSE Marking Scheme · CBSE Board 2023 (Main), Set 55/1/1

(A)

Drift velocity decreasing hyperbolically with area A
Q24.2023 Main 55/2/1 · Q31 mark

A current of 0.8 A flows in a conductor of 40 Ω for 1 minute. The heat produced in the conductor will be

  1. 1445 J
  2. 1536 J
  3. 1569 J
  4. 1640 J
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Official CBSE Marking Scheme · CBSE Board 2023 (Main), Set 55/2/1

(B) 1536 J

Q25.2023 Main 55/2/1 · Q41 mark

A cell of emf E is connected across an external resistance R. When current ‘I’ is drawn from the cell, the potential difference across the electrodes of the cell drops to V. The internal resistance ‘r’ of the cell is

  1. (E − VE)R
  2. (E − VR)
  3. (E − V)RI
  4. (E − VV)R
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Official CBSE Marking Scheme · CBSE Board 2023 (Main), Set 55/2/1

(D) (E − VV)R

Q26.2023 Main 55/3/1 · Q31 mark

The current density due to drift of electrons in a conductor is given by : (symbols have their usual meanings)

  1. n e A vd
  2. n A vde
  3. n vde A
  4. n e vd
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Official CBSE Marking Scheme · CBSE Board 2023 (Main), Set 55/3/1

(D) n e vd

Q27.2023 Main 55/3/1 · Q161 mark

Assertion (A) : The internal resistance of a cell is constant.

Reason (R) : Ionic concentration of the electrolyte remains same during use of a cell.

  1. Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of the Assertion (A).
  2. Both Assertion (A) and Reason (R) are true, but Reason (R) is not the correct explanation of the Assertion (A).
  3. Assertion (A) is true, but Reason (R) is false.
  4. Assertion (A) is false and Reason (R) is also false.
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Official CBSE Marking Scheme · CBSE Board 2023 (Main), Set 55/3/1

(D) Assertion (A) is false and Reason (R) is also false.

Q28.2023 Main 55/4/1 · Q161 mark

Questions 16 to 18 are Assertion (A) and Reason (R) type questions. Select the correct answer from the codes (a), (b), (c) and (d) given below.

Assertion (A) : When three electric bulbs of power 200 W, 100 W and 50 W are connected in series to a source, the power consumed by the 50 W bulb is maximum.

Reason (R) : In a series circuit, current is the same through each bulb, but the potential difference across each bulb is different.

  1. Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of the Assertion (A).
  2. Both Assertion (A) and Reason (R) are true, but Reason (R) is not the correct explanation of the Assertion (A).
  3. Assertion (A) is true, but Reason (R) is false.
  4. Assertion (A) is false and Reason (R) is also false.
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Official CBSE Marking Scheme · CBSE Board 2023 (Main), Set 55/4/1

(B) Both Assertion (A) and Reason (R) are true, but Reason (R) is not the correct explanation of the Assertion (A).

Q29.2023 Main 55/5/1 · Q171 mark

Assertion (A) : The equivalent resistance between points A and B in the given network is 2R.

Reason (R) : All the resistors are connected in parallel

Network of four 2R resistors between points A and B

Select the correct answer from the codes below :

  1. Both Assertion (A) and Reason (R) are true and (R) is the correct explanation of (A).
  2. Both Assertion (A) and Reason (R) are true and (R) is NOT the correct explanation of (A).
  3. Assertion (A) is true and Reason (R) is false.
  4. Assertion (A) is false and Reason (R) is also false.
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Official CBSE Marking Scheme · CBSE Board 2023 (Main), Set 55/5/1

(C) Assertion (A) is true and Reason (R) is false.

Q30.2023 Supp 55/C/1 · Q161 mark

Assertion (A) : The temperature coefficient of resistance is positive for metals and negative for semi-conductors.

Reason (R) : The charge carriers in metals are negatively charged whereas in semi-conductors they are positively charged.

  1. Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of the Assertion (A).
  2. Both Assertion (A) and Reason (R) are true, but Reason (R) is not the correct explanation of the Assertion (A).
  3. Assertion (A) is true, but Reason (R) is false.
  4. Assertion (A) is false and Reason (R) is also false.
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Official CBSE Marking Scheme · CBSE Board 2023 (Supplementary), Set 55/C/1

(C) Assertion (A) is true, but Reason (R) is false.

Q31.SQP 2025-26 · Q21 mark

A coil contains N turns of insulated copper wire of diameter d and resistivity ρ wound on a cylinder of diameter D. What is the total resistance between the two ends of the coil of copper wire? (given: D>>d)

  1. 4ρNDd2
  2. 8ρNDd2
  3. 2ρNDd2
  4. 12ρNDd2
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Official CBSE Marking Scheme · CBSE Sample Question Paper 2025-26

(A) 4ρNDd2

Diameter of copper wire d,

Diameter of cylindrical iron is D

No. of turns N, (D>>d)

Length = N × Circumference of cylinder

L = NπD

R = ρLA = ρNπDd2(π/4)

R = 4ρNDd2

Q32.SQP 2026-27 · Q21 mark

If two identical heaters each rated as (1000 W, 220 V) are connected in parallel to 220 V, then the total power consumed is-

  1. 200 W
  2. 250 W
  3. 2000 W
  4. 2500 W
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Official CBSE Marking Scheme · CBSE Sample Question Paper 2026-27

(C) 2000 W

For a parallel combination, P = P1 + P2 = 1000 + 1000 = 2000 W

Q33.SQP 2026-27 · Q91 mark

A bridge circuit is shown in the following figure. The total current flowing in the circuit is-

Bridge circuit between X and Y with 2 ohm, 4 ohm, 3 ohm, 6 ohm arms and 5 ohm central resistor, connected to an 18 V battery
  1. 18 A
  2. 5 A
  3. 5/18 A
  4. 18/5 A
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Official CBSE Marking Scheme · CBSE Sample Question Paper 2026-27

(B) 5 A

The Wheatstone bridge is balanced, because 2 Ω4 Ω = 3 Ω6 Ω.

The 5 Ω resistance is ineffective, we now have (2 Ω + 4 Ω) and (3 Ω + 6 Ω) resistances in parallel.

RAB = 6 × 96 + 9 = 185 Ω

Now, I = V/R

Or, I = 1818/5 A = 5 A

Q34.SQP 2026-27 · Q101 mark

Two electric bulbs, B1 and B2, are connected in series to an A.C. supply of 200 V. The power ratings printed on the bulbs are:

Bulbs B1 and B2 in series across a 200 V A.C. supply, with voltages V1 and V2 across them
  • B1: 200 V, 60 W
  • B2: 200 V, 100 W

Using their rated voltage–power information, the ratio of their resistances R1/R2 is-

  1. 5:3
  2. 3:5
  3. 3:4
  4. 4:3
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Official CBSE Marking Scheme · CBSE Sample Question Paper 2026-27

(A) 5:3

Here P1 = 60 W, P2 = 100 W

R1 = V2P1 and R2 = V2P2

∴ R1R2 = P2P1 = 10060 = 5 : 3

Q35.2026 Main 55/1/1 · Q202 marks

What is the order of magnitude of drift velocity of electrons in a conductor ? Deduce the relation between the current flowing through a conductor and drift velocity of electrons in it.

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Official CBSE Marking Scheme · CBSE Board 2026 (Main), Set 55/1/1

Magnitude of drift velocity in conductor is of the order of a few mm/s [½]

Conductor of cross-section A with electrons drifting opposite to E; length Δx = v_d Δt

For n number of free electrons per unit volume in a conductor of cross section area A, the total charge transported across the area in time Δt is

Q = – neAvdΔt [½]

As electron moves in a direction opposite to that of electric field

IΔt = neAvdΔt [½]

I = neAvd [½]

Q36.2026 Main 55/2/1 · Q202 marks

A heating element using nichrome is connected to a 220 V supply. Initially it draws a current of 2.9 A. After some time, the current attains a steady value of 2.5 A. Find the steady temperature of the heating element if the room temperature is 27 °C. The temperature coefficient of resistance of nichrome is 1.7 × 10–4 °C–1.

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Official CBSE Marking Scheme · CBSE Board 2026 (Main), Set 55/2/1

R1 = VI1 = 2202.9 Ω at temperature t1 [½]

R2 = VI2 = 2202.5 Ω at temperature t2 [½]

R2 = R1[1 + α(t2 – t1)] [½]

t2 – t1 = R2 – R1R1·α

t2 – 27 = (220/2.5 – 220/2.9)(220/2.9) × 1.7 × 10–4

t2 ≈ 968 °C [½]

Q37.2026 Main 55/3/1 · Q17 (option 1)2 marks

In the given figure, a steady current I flows through the circuit when points A and C are connected by a wire of negligible resistance. Find the potential difference between points B and C.

Circuit: A to B through cell E1 = 6 V and 1 Ω, B to C through cell E2 = 4 V and 3 Ω, current I
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Official CBSE Marking Scheme · CBSE Board 2026 (Main), Set 55/3/1

I = εnetRnet [½]

= 24 [½]

= 0·5 A

VBC = − ε2 − I r2 [½]

= − 4 − (0·5 × 3) [½]

= − 5·5 V

Q38.2026 Main 55/3/1 · Q17 (OR option)2 marks

A battery of emf 21 V and internal resistance 3 Ω is connected to a resistor. If the current in the circuit is 3 A, find :

  1. the resistance of the resistor.
  2. the terminal voltage of the battery.
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Official CBSE Marking Scheme · CBSE Board 2026 (Main), Set 55/3/1

(i) IR = ε − Ir

R = 21 − (3 × 3)3 [½]

= 4 Ω [½]

(ii) V = IR [½]

= 3 × 4

= 12 V [½]

Q39.2026 Main 55/5/1 · Q17 (option 1)2 marks

An electric iron rated 2·2 kW, 220 V is operated at 110 V supply. Find :

  1. its resistance, and
  2. heat produced by it in 10 minutes.
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Official CBSE Marking Scheme · CBSE Board 2026 (Main), Set 55/5/1

Calculating (i) Resistance [½]; (ii) heat produced in 10 min. [1½]

(i) R = V2P = (220)22.2 × 103 = 22 Ω [½]

(ii) H = I2Rt [½]

I = VappliedR = 11022 = 5 A [½]

H = 5 × 5 × 22 × 600 [½]

= 3.3 × 105 J [½]

Q40.2026 Main 55/5/1 · Q17 (OR option)2 marks

A current of 4·0 A flows through a wire of length 1 m and cross-sectional area 1·0 mm2, when potential difference of 2 V is applied across its ends.

Calculate the resistivity of the material of the wire.

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Official CBSE Marking Scheme · CBSE Board 2026 (Main), Set 55/5/1

Calculating the resistivity of the wire. [2]

ρ = RAl [½]

R = VI = 24 = 12 Ω [½]

ρ = 1 × 10–62 × 1 [½]

= 5 × 10–7 Ω m [½]

Q41.2026 Supp 55/7/1 · Q212 marks

Find the equivalent emf and the equivalent resistance between points A and B for two cells arranged as shown in the figure.

Two cells, 6 V 1 Ω and 3 V 0.8 Ω, connected in parallel between A and B with opposite polarity
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Official CBSE Marking Scheme · CBSE Board 2026 (Supplementary), Set 55/7/1

Eeq = E1r2 + E2r1r1 + r2 [½]

= 6 × 0.8 – 3 × 11 + 0.8 = 1 V [½]

1req = 1r1 + 1r2 = 11 + 10.8 [½]

req = 49 Ω or 0.44 Ω [½]

Q42.2025 Main 55/1/1 · Q172 marks

A battery of emf E and internal resistance r is connected to a rheostat. When a current of 2A is dawn from the battery, the potential difference across the rheostat is 5V. The potential difference becomes 4V when a current of 4A is drawn from the battery. Calculate the value of E and r.

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Q43.2025 Main 55/2/1 · Q17 (option 1)2 marks

Two wires of the same material and the same radius have their lengths in the ratio 2 : 3. They are connected in parallel to a battery which supplies a current of 15 A. Find the current through the wires.

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Q44.2025 Main 55/2/1 · Q17 (OR option)2 marks

In the circuit three ideal cells of e.m.f. V, V and 2V are connected to a resistor of resistance R, a capacitor of capacitance C and another resistor of resistance 2R as shown in figure. In the steady state find (i) the potential difference between P and Q and (ii) potential difference across capacitor C.

Circuit: three branches in parallel between P and Q: cell V with resistor R; cell V with capacitor C; cell 2V with resistor 2R
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Q45.2025 Main 55/4/1 · Q17 (option 1)2 marks

In the given figure, three identical bulbs P, Q and S are connected to a battery.

Bulb S in series with parallel bulbs P and Q; key K in series with P; battery
  1. Compare the brightness of bulbs P and Q with that of bulb S when key K is closed.
  2. Compare the brightness of the bulbs S and Q when the key K is opened.

Justify your answer in both cases.

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Q46.2025 Main 55/4/1 · Q17 (OR option)2 marks

Two cells of emf 10 V each, two resistors of 20 Ω and 10 Ω and a bulb B of 10 Ω resistance are connected together as shown in the figure. Find the current that flows through the bulb.

Two 10 V cells, 20 Ω and 10 Ω resistors and a 10 Ω bulb B connected in a network
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Q47.2025 Main 55/5/1 · Q212 marks

The resistance of a wire at 25°C is 10·0 Ω. When heated to 125°C, its resistance becomes 10·5 Ω. Find (i) the temperature coefficient of resistance of the wire, and (ii) the resistance of the wire at 425°C.

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Q48.2025 Main 55/6/1 · Q172 marks

Find the equivalent resistance between points A and B for the network shown in the figure.

Resistor network between A and B with 1, 5, 10, 25, 20, 5, 10, 15, 30 ohm resistors
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Q49.2025 Main 55/7/1 · Q172 marks

A cell of emf E and internal resistance r is connected across a resistor of variable resistance R. Show graphically the variation of

  1. the terminal voltage across the cell,
  2. the current supplied by the cell,

with R as it is increased from 0 to the maximum value.

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Q50.2025 Supp 55/S/1 · Q192 marks

The figure shows a network of resistors connected across a battery. Find the current supplied by the battery of emf 5 V and internal resistance 23 Ω.

Network of resistors between A and C: AB 8 Ω, BC 2 Ω, AD 4 Ω, DC 1 Ω, BD 3 Ω, with a 5 V battery across A and C
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Official CBSE Marking Scheme · CBSE Board 2025 (Supplementary), Set 55/S/1

RABRBC = RADRDC = 41 [½]

The bridge is balanced. [½]

Resistance of the circuit between points A and C :

1R = 1(8 + 2) + 1(4 + 1) [½]

R = 103 Ω

I = ER + r

I = 5103 + 23

I = 1.25 A [½]

Q51.2024 Main 55/1/1 · Q172 marks

Find the temperature at which the resistance of a wire made of silver will be twice its resistance at 20 °C. Take 20 °C as the reference temperature and temperature coefficient of resistance of silver at 20 °C = 4·0 × 10−3 K−1.

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Official CBSE Marking Scheme · CBSE Board 2024 (Main), Set 55/1/1

R = R0 [1 + α (T − T0)] [½]

R = 2R0 [Given] [½]

2R0 = R0 [1 + α (T − T0)]

On solving

T = T0 + 250

T = 270 °C or 543 K [1]

Q52.2024 Main 55/1/1 · Q212 marks

Plot a graph showing the variation of current with voltage for the material GaAs. On the graph, mark the region where :

(a) resistance is negative, and
(b) Ohm’s law is obeyed.

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Official CBSE Marking Scheme · CBSE Board 2024 (Main), Set 55/1/1

Plotting the graph [1]; marking the region where (a) resistance is negative [½] and (b) Ohm’s law is obeyed [½]

Current-voltage graph of GaAs with the Ohm's-law region near the origin and the negative-resistance region after the peak marked
Q53.2024 Main 55/2/1 · Q172 marks

Define resistivity of a conductor. How does the resistivity of a conductor depend upon the following :

  1. Number density of free electrons in the conductor (n)
  2. Their relaxation time (τ)
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Official CBSE Marking Scheme · CBSE Board 2024 (Main), Set 55/2/1

Resistance offered by a material of unit length and having unit cross-sectional area is called resistivity. [1]

ρ = mne2τ

(a) ρ ∝ 1n [½]

(b) ρ ∝ 1τ [½]

Q54.2024 Main 55/3/1 · Q17 (option 1)2 marks

What is meant by ‘relaxation time’ of free electrons in a conductor ? Show that the resistance of a conductor can be expressed by R = mlne2τA, where symbols have their usual meanings.

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Official CBSE Marking Scheme · CBSE Board 2024 (Main), Set 55/3/1

Meaning of relaxation time [½]; Derivation of R [1½]

Average time between two successive collisions of electron in presence of electric field [½]

Drift velocity of an electron

υd = eEm τ     ---(i) [½]

Current flowing through a conductor of length l and area of cross section A

I = neAυd     ---(ii) [½]

I = ne2AEτm = ne2AτVml

R = VI = mlne2τA [½]

Q55.2024 Main 55/3/1 · Q17 (OR option)2 marks

Draw the circuit diagram of a Wheatstone bridge. Obtain the condition when no current flows through the galvanometer in it.

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Official CBSE Marking Scheme · CBSE Board 2024 (Main), Set 55/3/1

Circuit diagram of Wheatstone bridge [½]; Obtaining the condition when no current flows through galvanometer [1½]

Wheatstone bridge circuit ABCD with galvanometer G between B and D and a cell

By applying Kirchhoff’s loop rule to closed loops ADBA and CBDC

– I1R1 + 0 + I2R2 = 0    -----(i)   [Ig = 0] [½]

I2R4 + 0 – I1R3 = 0    -----(ii) [½]

From eq (i) – I1I2 = R2R1

From eq (ii) – I1I2 = R4R3 [½]

Hence, R2R1 = R4R3 [½]

Q56.2024 Main 55/4/1 · Q202 marks

Two electric heaters have power ratings P1 and P2, at voltage V. They are connected in series to a dc source of voltage V. Find the power consumed by the combination. Will they consume the same power if connected in parallel across the same source ?

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Official CBSE Marking Scheme · CBSE Board 2024 (Main), Set 55/4/1

R1 = V2P1 & R2 = V2P2 [½]

Req = R1 + R2 = V2(1P1 + 1P2) [½]

Pseries = V2Req

Pseries = V2V2(1P1 + 1P2)

1Pseries = 1P1 + 1P2 [½]

No [½]

Q57.2024 Supp 55/S/1 · Q21 (option 1)2 marks

A cell is connected across an external resistance 12 Ω and supplies 0.25 A current. When the external resistance is increased by 4 Ω, the current reduces to 0.2 A. Calculate (i) the emf, and (ii) the internal resistance, of the cell.

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Official CBSE Marking Scheme · CBSE Board 2024 (Supplementary), Set 55/S/1

I = εR + r [½]

0.25 = ε12 + r ——— (1) [½]

0.2 = ε16 + r ——— (2)

On solving eq (1) and eq (2)

r = 4 Ω [½]

ε = 4 V [½]

Q58.2023 Main 55/4/1 · Q202 marks

Write two differences between the emf and terminal potential difference of a cell. What is the most important precaution that one should take while drawing current from a cell ?

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Official CBSE Marking Scheme · CBSE Board 2023 (Main), Set 55/4/1

Two differences

1. The potential difference across the electrodes in open circuit is e.m.f. (ε) and in closed circuit is terminal potential difference (V). [½]

2. V depends on r and ε is independent of r. [½]

Precaution

1. Some external resistance should be connected to cell in series.

2. Short circuiting should be avoided. [1]

Q59.2023 Main 55/5/1 · Q242 marks

The potential difference applied across a given conductor is doubled. How will this affect (i) the mobility of electrons and (ii) the current density in the conductor ? Justify your answers.

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Official CBSE Marking Scheme · CBSE Board 2023 (Main), Set 55/5/1

(i) No effect [½]

μ = vd lV = lenAR = Constant [½]

Alternatively: μ = vdE = eτm = Constant

(ii) As J ∝ V, current density gets doubled [½]

J = IA = VRA [½]

Q60.2023 Supp 55/C/1 · Q23 (option 1)2 marks

When is more power delivered to a light bulb — just after it is turned on and the glow of the filament is increasing or after the glow becomes steady ? Why ?

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Official CBSE Marking Scheme · CBSE Board 2023 (Supplementary), Set 55/C/1

(a) Power delivered just after it is turned on, will be more because resistance of the bulb is low. [1]

After some time temperature of the bulb increases and resistance also increases and therefore power (V2R) becomes low. [1]

Q61.2023 Supp 55/C/1 · Q23 (OR option)2 marks

A battery is connected first across the series combination and then across the parallel combination, of three resistances R, 2R and 3R. In which of the three resistances will power dissipated be maximum in the two cases ? Justify your answer.

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Official CBSE Marking Scheme · CBSE Board 2023 (Supplementary), Set 55/C/1

For series

Power dissipated will be maximum for 3R. [½]

Because current is same and power is proportional to resistance. [½]

For parallel

Power dissipated will be maximum for R. [½]

Because voltage is same and power is inversely proportional to resistance. [½]

Q62.SQP 2025-26 · Q182 marks

The following graph shows the potential difference across the terminals of a cell against its load current.

Straight-line graph of potential difference (V) against load current (A), from 6 V at zero current to zero at 12 A

Find,

(I) the emf of the cell and
(II) the internal resistance of the cell.

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Official CBSE Marking Scheme · CBSE Sample Question Paper 2025-26

(I) E.m.f of the cell is 6V, As when load current is zero potential difference becomes equal to emf of the cell. [1]

(II) Explanation: The internal resistance of a cell can be determined as the negative slope of its voltage–current graph.

First, we can determine the slope by choosing two points on the line:

Slope = 0 – 612 – 0 = – 0.5

This means that the internal resistance must be 0.50 ohm (Ω). [1]

Q63.SQP 2026-27 · Q182 marks

Write the nature of path of free electrons in a conductor in the

a) presence of electric field,
b) absence of electric field.

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Official CBSE Marking Scheme · CBSE Sample Question Paper 2026-27

The nature of the path of free electrons in a conductor differs depending on whether an electric field is present or not:

a. In the presence of an electric field [1]

  • Free electrons experience a net force opposite to the direction of the electric field.
  • This causes them to drift slowly in the direction opposite to the field (since electrons are negatively charged).
  • Despite this drift, electrons still undergo frequent collisions with atoms in the conductor, resulting in a zigzag or random motion superimposed with a net drift.
  • This net movement of electrons constitutes an electric current.

b. In the absence of an electric field [1]

  • Electrons move randomly due to thermal energy.
  • There is no net movement of electrons in any particular direction.
  • The motion is random and chaotic, similar to the movement of gas molecules.
  • As a result, no electric current flows in the conductor.
Q64.2026 Main 55/1/1 · Q22 (OR option)3 marks

(a) Obtain the condition of balance of a Wheatstone bridge.

(b) Find net resistance of the network of resistors connected between A and B, as shown in figure.

Network of resistors 2R, R, R, R, R, 3R between A and B with nodes M, O, P, N
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Official CBSE Marking Scheme · CBSE Board 2026 (Main), Set 55/1/1

(a)

Wheatstone bridge with resistors R1 to R4, galvanometer G and cell

[½] (figure)

By using Kirchhoff’s rule to closed loops ADBA and CBDC. The first loop gives

–I1R1 + 0 + I2R2 = 0 ………… (1) [∵ VB = VD, Ig = 0] [½]

Second loop gives

I4R4 + 0 – I3R3 = 0

∵ Ig = 0, hence I1 = I3 and I2 = I4

I2R4 – I1R3 = 0 ………… (2) [½]

From equations (1) and (2)

Hence R2R1 = R4R3 [½]

(b)

Due to balanced Wheatstone bridge

RMN = R [½]

Total resistance across AB

RAB = RAM + RMN + RNB

RAB = 2R + R + 3R

RAB = 6R Ω [½]

Q65.2026 Main 55/2/1 · Q273 marks

(a) Explain the statement : “Current is a scalar although we represent current with an arrow”.

(b) Use Kirchhoff’s rules to find the current through 3 Ω resistor in the circuit shown in the figure :

Circuit ABCDEFA: 3 V cell in AF, 5 V cell in BC, 3 ohm resistor in BE, 2 ohm in CD, 4 ohm in FE
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Official CBSE Marking Scheme · CBSE Board 2026 (Main), Set 55/2/1

(a) Electric current is a scalar quantity because it does not obey law of vector addition. The current with an arrow is only to signify the direction of conventional current. [1]

Alternatively : Current do not follow the law of vector addition. The current through an area of cross-section is given by the scalar product of two vectors. I = j · ΔS

(b)

Circuit with assumed current distribution: I in the outer branch, I1 through 3 ohm, (I – I1) through 4 ohm

In loop ABEFA : 3 + 3I1 – 4(I – I1) = 0   ⇒   4I – 7I1 = 3 …(1) [½ + ½]

In loop BCDEB : 5 – 2I – 3I1 = 0   ⇒   2I + 3I1 = 5 …(2) [½]

Solving equations (1) and (2), we get

Current through 3 Ω resistor is I1 = 713 A [½]

Note : Award full marks if a student solves by another appropriate current distribution in the circuit.

Q66.2026 Main 55/3/1 · Q223 marks

(a) Establish the relation between drift velocity of electrons (vd) and electric current (I) in a conductor.

(b) How is vd affected when the length of the conductor is doubled, keeping the voltage applied across the conductor constant ?

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Official CBSE Marking Scheme · CBSE Board 2026 (Main), Set 55/3/1

(a) For ‘n’ number of free electrons per unit volume in a conductor of cross section area A, the total charge transported across the area in time Δt

Q = − ne A vd Δt [1]

As electrons move in a direction opposite to that of electric field

I Δt = ne A vd Δt [½]

I = Anevd [½]

(b) vd = eVτml [½]

Length is doubled, drift velocity is halved. [½]

Q67.2026 Main 55/4/1 · Q223 marks

State Kirchhoff’s rules. Using these rules, find the current flowing through branch FC in the given circuit.

Circuit ABCDEF: 5 V, 1 Ω and 3 V in branch AB; 3 Ω in BC; 2 V and 1 Ω in branch FC; 4 Ω in ED
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Official CBSE Marking Scheme · CBSE Board 2026 (Main), Set 55/4/1

Junction Rule : At any junction, the sum of the currents entering the junction is equal to the sum of the currents leaving the junction. [½]

Loop Rule : The algebraic sum of changes in potential around any closed loop involving resistors and cells in the loop is zero. [½]

Note : Full credit to be given if stated mathematically.

Circuit with assumed currents i<sub>1</sub> in AB, i<sub>2</sub> in FC and i<sub>1</sub> + i<sub>2</sub> in ED, and loop directions

For loop AFCBA

3i1 + i1 − i2 = −3 + 5 − 2

4i1 − i2 = 0

4i1 = i2 ……(i) [½]

For loop FEDCF

i2 + 4(i1 + i2) = 2

4i1 + 5i2 = 2 ……(ii) [½]

Solving equations (i) and (ii),

i1 = 112 A

i2 = 13 A [1]

Note : Award full marks if a student solves using another appropriate current distribution in the above circuit.

Q68.2025 Main 55/1/1 · Q22 (option 1)3 marks

Two batteries of emf’s 3V & 6V and internal resistances 0.2 Ω & 0.4 Ω are connected in parallel. This combination is connected to a 4 Ω resistor. Find :

(i) the equivalent emf of the combination

(ii) the equivalent internal resistance of the combination

(iii) the current drawn from the combination

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Q69.2025 Main 55/1/1 · Q22 (OR option)3 marks

(i) A conductor of length l is connected across an ideal cell of emf E. Keeping the cell connected, the length of the conductor is increased to 2l by gradually stretching it. If R and R′ are initial and final values of resistance and vd and vd′ are initial and final values of drift velocity, find the relation between (i) R′ and R and (ii) vd′ and vd.

(ii) When electrons drift in a conductor from lower to higher potential, does it mean that all the ‘free electrons’ of the conductor are moving in the same direction ?

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Q70.2025 Main 55/2/1 · Q223 marks

(a) A cell of e.m.f. E and internal resistance r is connected with a variable external resistance R and a voltmeter showing potential drop V across R. Obtain the relationship between V, E, R and r.

(b) Draw the shape of the graph showing the variation of terminal voltage V of the cell as a function of current I drawn from it. How one can determine the e.m.f. of the cell and its internal resistance from this graph ?

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Q71.2025 Main 55/4/1 · Q22 (option 1)3 marks
  1. Derive an expression for the resistivity of a conductor in terms of number density of free electrons and relaxation time.
  2. The figure shows the plot of current through a cross-section of wire over two different time intervals. Compare the charges (Q1 and Q2) that pass through the cross-section during these time intervals.
Current vs time: rectangular pulse of 2.0 A from t = 1 s to 2 s (Q1); triangular ramp from 0 at t = 4 s to 1.5 A at t = 6 s (Q2)
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Q72.2025 Main 55/4/1 · Q22 (OR option)3 marks
  1. A battery of emf E and internal resistance r is connected to a variable external resistance R.
    1. Obtain the expression for current I in the circuit and the value of maximum current the battery can supply.
    2. Obtain the terminal voltage V across the battery and its maximum possible value.
  2. The above battery sends a current I1 when R = R1 and a current I2 when R = R2. Obtain the internal resistance of the battery in terms of I1, I2, R1 and R2.
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Q73.2025 Main 55/6/1 · Q223 marks

What is the difference between ‘emf’ and ‘terminal voltage’ of a cell ?

Two cells of emfs E1 and E2 and internal resistances r1 and r2 are connected in parallel. Derive an expression for the emf and internal resistance of the equivalent cell.

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Q74.2025 Main 55/7/1 · Q223 marks

Three cells A, B and C of emfs 2 V, 3 V and 5 V respectively are connected in parallel to each other. Their internal resistances are 5 Ω, 5 Ω and 1 Ω respectively. Calculate the currents flowing through the cells A, B and C.

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Q75.2025 Supp 55/S/1 · Q273 marks

Define the term ‘drift velocity’ of electrons in a current carrying conductor and derive, in its term, the expression for current flowing through the conductor.

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Official CBSE Marking Scheme · CBSE Board 2025 (Supplementary), Set 55/S/1

The drift velocity is defined as average velocity with which free electrons in a conductor get drifted under the influence of electric field applied across the conductor. [1]

Let us consider a conductor of length l and cross-sectional area A connected across the battery of voltage V.

Conductor of length l and area A connected to a battery, electrons drifting with velocity v_d

Let n be the electron density, current flowing through the conductor is given by

I = Total chargetime .... (1) [½]

i.e. Q = nAL × e .... (2) [½]

t = Lvd .... (3)

Using (2) and (3) in (1) [½]

I = Qt = nALeL/vd

I = neAvd [½]

Q76.2024 Main 55/1/1 · Q23 (option 1)3 marks

Define ‘current density’. Is it a scalar or a vector ? An electric field E is maintained in a metallic conductor. If n be the number of electrons (mass m, charge − e) per unit volume in the conductor and τ its relaxation time, show that the current density j = αE, where α = (ne2m) τ.

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Official CBSE Marking Scheme · CBSE Board 2024 (Main), Set 55/1/1

Current density is the amount of charge flowing per second per unit area normal to the flow. [½]

Alternatively: j = IA

It is a vector quantity. [½]

Cylindrical conductor of length Δx = v_d Δt with electric field E and electrons drifting

The amount of charge crossing the area A in time Δt is I Δt, where I is the magnitude of the current. Hence,

I Δt = ne A |vd| Δt [½]

I Δt = e2Am τ n Δt |E| [½]

I = |j|A [½]

|j| = ne2m τ |E|

j = α E [½]

Q77.2024 Main 55/1/1 · Q23 (OR option)3 marks

What is a Wheatstone bridge ? Obtain the necessary conditions under which the Wheatstone bridge is balanced.

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Official CBSE Marking Scheme · CBSE Board 2024 (Main), Set 55/1/1

Defining Wheatstone bridge [1]; obtaining balancing conditions [2]

Wheatstone bridge ABCD with resistors R1, R2, R3, R4, galvanometer G between B and D, and cell between A and C

Alternatively: If the figure is explained in words full credit to be given.

For loop ADBA: − I1R1 + I2R2 + IgG = 0   (1) [½]

For loop CBDC: I4R4 − I3R3 − IgG = 0   (2) [½]

For balanced Wheatstone bridge, Ig = 0

And by applying Kirchhoff’s junction rule to junction D and B,

I1 = I3 & I2 = I4 [½]

From eqn (1) and (2)

I1I2 = R2R1 and I1I2 = R4R3

⇒ R2R1 = R4R3 [½]

Q78.2024 Main 55/2/1 · Q223 marks

The figure shows a circuit with three ideal batteries. Find the magnitude and direction of currents in the branches AG, BF and CD.

Circuit with three batteries (3 V, 6 V, 6 V) and resistors, nodes A, B, C, D, F, G
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Official CBSE Marking Scheme · CBSE Board 2024 (Main), Set 55/2/1
Circuit with branch currents I3 (A to G), I2 (F to B) and I1 (D to C) marked

By Kirchhoff’s Laws (at point B)
I1 + I2 = I3 …….(1) [½]

In the closed loop AGFBA
3 + 2I3 − 6 + 4I2 + 2I3 = 0
I2 + I3 = 34 …….(2) [½]

From (1)
2I1 + I2 = 34 …….(3) [½]

In closed loop BFDCB
− 4I2 + 6 + 2I1 − 6 + 2I1 = 0
I2 − I1 = 0, i.e. I2 = I1 ……..(4) [½]

Putting in (3), I1 = 14 A [½]

From (4), I2 = 14 A

From (2), I3 = 12 A [½]

Currents : AG (I3) = 0·5 A, directed from A to G; BF (I2) = 0·25 A, directed from F to B; CD (I1) = 0·25 A, directed from D to C (as marked in the figure).

Q79.2024 Main 55/3/1 · Q223 marks

Determine the current in branches AB, AC and BC of the network shown in figure.

Network with nodes A, B, C, D: 4 Ω in AB, 4 Ω in AD, 10 V cell with 1 Ω in AC, 2 Ω in BC, 2 Ω in CD and a 5 V cell between B and D
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Official CBSE Marking Scheme · CBSE Board 2024 (Main), Set 55/3/1

Determination of current in branches AB, AC, BC [1+1+1]

Network with branch currents I1, I2, I3 marked

For closed loop ADCA,

10 – 4(I1 – I2) + 2(I2+I3–I1) – I1 = 0

7I1 – 6I2 – 2I3 = 10 ------(i) [½]

For closed loop ABCA,

10 – 4I2 – 2(I2+I3) – I1 = 0

I1 + 6I2 + 2I3 = 10 ------(ii) [½]

For closed loop BCDEB,

5 – 2(I2+I3) – 2(I2+I3–I1) = 0

2I1 – 4I2 – 4I3 = –5 -----(iii) [½]

Current in branch AB = I2 = 58 A [½]

Current in branch AC = I1 = 2.5 A [½]

Current in branch BC = I2 + I3 = 2.5 A [½]

Q80.2024 Main 55/4/1 · Q253 marks

Find the current in branch BM in the network shown :

Network with batteries 10E and 6E and resistors 3R, 4R, 4R, 2R, 4R, R, 2R, R between nodes A to N
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Official CBSE Marking Scheme · CBSE Board 2024 (Main), Set 55/4/1

Equivalent resistance across CH, RCH = 2R [½]

Equivalent circuit diagram [½]

Equivalent circuit with currents I, I1 and I – I1

In closed loop ABMNA

–3IR – 4I1R + 16E = 0 ………… (1) [½]

In closed loop BCHMB

–2R(I – I1) – 6E + 4I1R = 0 ………… (2) [½]

On solving equations (1) and (2)

I1 = 25E13R [1]

Q81.2024 Supp 55/S/1 · Q24 (option 1)3 marks

The radius of a conducting wire AB uniformly decreases from its one end A to another end B. It is connected across a battery. How will (i) electric field, (ii) current density, and (iii) mobility of electrons change from end A to end B ? Justify your answer in each case.

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Official CBSE Marking Scheme · CBSE Board 2024 (Supplementary), Set 55/S/1

With the decrease in area of cross-section,

(i) E = IAρ, electric field increases [½ + ½]

(ii) j = IA, current density increases [½ + ½]

(iii) μe = eτm, mobility remains same [½ + ½]

Q82.2023 Main 55/2/1 · Q273 marks

Define current density and relaxation time. Derive an expression for resistivity of a conductor in terms of number density of charge carriers in the conductor and relaxation time.

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Official CBSE Marking Scheme · CBSE Board 2023 (Main), Set 55/2/1

Current density is defined as the current flowing per unit area of cross section of a conductor. [½]

Alternatively

Give full credit if a candidate writes j = I/A in place of definition

Relaxation time is the average time interval between two successive collisions for drifting electrons in a conductor. [½]

From I = nAevd [½]

but vd = eEm τ [½]

∴ I = nAe·eEm τ [½]

j = IA = ne2Eτm [½]

But j = Eρ

∴ ρ = mne2τ [½]

Q83.2023 Main 55/3/1 · Q263 marks

A potential difference V is applied across a conductor of length l and uniform cross-section area A. How will the (i) electric field E, (ii) drift velocity vd, and (iii) current density j be affected when (a) V is doubled and (b) l is halved (keeping other factors constant) ?

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Official CBSE Marking Scheme · CBSE Board 2023 (Main), Set 55/3/1

(a) When V is doubled,

(i) Electric field is doubled. [½]

(ii) Drift velocity is doubled. [½]

(iii) Current density is doubled. [½]

(b) When l is halved,

(i) Electric field is doubled. [½]

(ii) Drift velocity is doubled. [½]

(iii) Current density is doubled. [½]

Alternatively:

(a) (i) E = Vl ; E′ = 2E

(ii) vd = eEm τ = eτm (Vl) ; vd ∝ V ; vd′ = 2vd

(iii) Current density J = IA = VRA ∴ J is also doubled.

(b) When l is halved

(i) E = Vl ; E′ = 2E

(ii) vd = eEm τ = em (Vl) τ ; ∴ vd′ = 2vd

(iii) J = IA = VRA = VA (Aρl) = Vρl ∴ J is also doubled.

Q84.SQP 2025-26 · Q233 marks

(I) The current I1 in a wire is getting divided in two wires with currents I2 and I3 at a junction in a circuit. The currents in the three wires are related by I1 = I2 + I3.

(A) State the fundamental law from which this relation is derived.
(B) Explain the validation of law of conservation of energy in Kirchhoff's voltage law?

(II) How the balancing condition gets affected if you are interchanging the galvanometer and the cell in the Wheat stone bridge?

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Official CBSE Marking Scheme · CBSE Sample Question Paper 2025-26

(I) (A) Conservation of electric charge [1]

(B) KVL is obeys law of conservation of energy as it is supplied voltage is equal to the voltage across each component in the loop. (OR) algebraic sum of voltages equal to zero. [1]

(II) No change in balancing condition is observed. [1]

Q85.SQP 2026-27 · Q233 marks

Explain-

  1. It is easier to start a car engine on a warm day than on a chilly day.
  2. The resistance of our body is so large, even then one experiences a strong shock (sometimes even fatal)) when one accidentally touches the live wire of, say 240 V supply.
  3. Heat is generated continuously in an electric heater but its temperature becomes constant after some time.
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Official CBSE Marking Scheme · CBSE Sample Question Paper 2026-27

I. The internal resistance of a car battery decreases with increase in temperature. Therefore, on a warm day a car battery gives large current which helps in starting the car engine. [1]

II. It is because our body is sensitive to minute currents even as low as a few mA. [1]

III. When the temperature of the heater becomes greater than the temperature of its surroundings, it begins to lose heat to surroundings. A stage is reached when the rate of production of heat becomes equal to the rate of loss of heat and the temperature becomes constant. [1]

Q86.2026 Main 55/4/1 · Q294 marks

Questions number 29 and 30 are Case Study-based questions. Read the following paragraphs and answer the questions that follow.

A student sets up the circuit as shown in the figure to find the value of unknown resistance X and records a set of readings of the voltmeter and the ammeter by using the rheostat.

Circuit with cell, key K, rheostat (terminals P and Q), ammeter A, unknown resistance X and voltmeter V across X

(i) If resistor X were made of manganin and readings for V and I are taken without switching off the circuit, the graph between V and I will be as :

  1. I versus V: straight line through origin
  2. I versus V: curve rising, flattening and dipping slightly
  3. I versus V: straight line then bending below the dotted straight line
  4. I versus V: straight line then bending upward above the dotted straight line

(ii) Error in the value of X obtained from different sets of voltmeter and ammeter readings, is :

  1. due to error in voltmeter reading only.
  2. due to error in ammeter reading only.
  3. equal to the sum of error in voltmeter reading and error in ammeter reading.
  4. equal to error in voltmeter reading divided by the error in ammeter reading.

(iii) If the movable end of rheostat is moved towards P, then :

  1. reading in ammeter decreases and reading in voltmeter increases.
  2. readings in both voltmeter and ammeter increase.
  3. reading in ammeter increases and reading in voltmeter decreases.
  4. readings in both voltmeter and ammeter decrease.

(iv) (a) Suppose the unknown resistance X is replaced by a wire made of the same metal. This wire consists of three parts, of the same length L but has radii r, r/3 and r/2 as shown in the figure.

Wire ACDB of three parts each of length L with radii r, r/3 and r/2

For a particular setting of the rheostat, let v1, v2 and v3 be the value of drift velocities in parts AC, CD and DB. Then :

  1. v1 > v2 > v3
  2. v2 > v3 > v1
  3. v3 > v2 > v1
  4. v1 = v2 = v3

(iv) (b) (alternative to (a)) Consider the same wire, as shown in figure in question (iv) (a) connected in place of X. For a particular setting of rheostat, let E1, E2 and E3 be the value of electric fields in part AC, CD and DB. Then :

  1. E1 = E2 = E3
  2. E3 > E2 > E1
  3. E2 > E3 > E1
  4. E1 > E2 > E3
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Official CBSE Marking Scheme · CBSE Board 2026 (Main), Set 55/4/1

(i) (C) [1]

Graph (C): I versus V, straight line then bending below the dotted line

(ii) (C) equal to the sum of error in voltmeter reading and the error in ammeter reading. [1]

(iii) (B) readings in both voltmeter and ammeter increase. [1]

(iv) (B) v2 > v3 > v1 [1]

Alternatively, for the (b) option : (C) E2 > E3 > E1 [1]

Q87.2025 Main 55/7/1 · Q294 marks

In a metallic conductor, an electron, moving due to thermal motion, suffers collisions with the heavy fixed ions but after collision, it will emerge out with the same speed but in random directions. If we consider all the electrons, their average velocity will be zero. When an electric field is applied, electrons move with an average velocity, known as drift velocity (vd). The average time between successive collisions is known as relaxation time (τ). The magnitude of drift velocity per unit electric field is called mobility (μ).

An expression for current through the conductor can be obtained in terms of drift velocity, number of electrons per unit volume (n), electronic charge (– e), and the cross-sectional area (A) of the conductor. This expression leads to an expression between current density (j) and the electric field (E). Hence, an expression for resistivity (ρ) of a metal is obtained. This expression helps us to understand increase in resistivity of a metal with increase in its temperature, in terms of change in the relaxation time (τ) and change in the number density of electrons (n).

(i) Consider two cylindrical conductors A and B, made of the same metal connected in series to a battery. The length and the radius of B are twice that of A. If μA and μB are the mobility of electrons in A and B respectively, then μAμB is :

  1. 12
  2. 14
  3. 2
  4. 1

(ii) A wire of length 0·5 m and cross-sectional area 1·0 × 10–7 m2 is connected to a battery of 2 V that maintains a current of 1·5 A in it. The conductivity of the material of the wire (in Ω–1 m–1) is :

  1. 2·5 × 104
  2. 3·0 × 105
  3. 3·75 × 106
  4. 5·0 × 107

(iii) The temperature coefficient of resistance of nichrome is 1·70 × 10–4 °C–1. In order to increase resistance of a nichrome wire by 8·5%, the temperature of the wire should be increased by :

  1. 250°C
  2. 500°C
  3. 850°C
  4. 1000°C

(iv) (a) Consider the contribution of the following two factors I and II in resistivity of a metal :

  1. Relaxation time of electrons
  2. Number of electrons per unit volume

The resistivity of a metal increases with increase in its temperature because :

  1. I decreases and II increases.
  2. I increases and II is almost constant.
  3. Both I and II increase.
  4. I decreases and II is almost constant.

Alternative to (iv)(a) : (b) A steady current flows in a copper wire of non-uniform cross-section. Consider the following three physical quantities :

  1. Electric field
  2. Current density
  3. Drift speed

Then at the different points along the wire :

  1. II and III change, but I is constant.
  2. I and II change, but III is constant.
  3. I and III change, but II is constant.
  4. All I, II and III change.
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The official marking scheme for this paper is not available in the source folder.

Q88.2024 Main 55/2/1 · Q294 marks

When the terminals of a cell are connected to a conductor of resistance R, an electric current flows through the circuit. The electrolyte of the cell also offers some resistance in the path of the current, like the conductor. This resistance offered by the electrolyte is called internal resistance of the cell (r). It depends upon the nature of the electrolyte, the area of the electrodes immersed in the electrolyte and the temperature. Due to internal resistance, a part of the energy supplied by the cell is wasted in the form of heat.

When no current is drawn from the cell, the potential difference between the two electrodes in known as emf of the cell (ε). With a current drawn from the cell, the potential difference between the two electrodes is termed as terminal potential difference (V).

(i) Choose the incorrect statement :

  1. The potential difference (V) between the two terminals of a cell in a closed circuit is always less than its emf (ε), during discharge of the cell.
  2. The internal resistance of a cell decreases with the decrease in temperature of the electrolyte.
  3. When current is drawn from the cell then V = ε − Ir.
  4. The graph between potential difference between the two terminals of the cell (V) and the current (I) through it is a straight line with a negative slope.

(ii) Two cells of emfs 2·0 V and 6·0 V and internal resistances 0·1 Ω and 0·4 Ω respectively, are connected in parallel. The equivalent emf of the combination will be :

  1. 2·0 V
  2. 2·8 V
  3. 6·0 V
  4. 8·0 V

(iii) Dipped in the solution, the electrode exchanges charges with the electrolyte. The positive electrode develops a potential V+ (V+ > 0), and the negative electrode develops a potential − (V−) (V− ≥ 0), relative to the electrolyte adjacent to it. When no current is drawn from the cell then :

  1. ε = V+ + V− > 0
  2. ε = V+ − V− > 0
  3. ε = V+ + V− < 0
  4. ε = V+ − V− = 0

(iv) (a) Five identical cells, each of emf 2 V and internal resistance 0·1 Ω are connected in parallel. This combination in turn is connected to an external resistor of 9·98 Ω. The current flowing through the resistor is :

  1. 0·05 A
  2. 0·1 A
  3. 0·15 A
  4. 0·2 A

(b) Potential difference across a cell in the open circuit is 6 V. It becomes 4 V when a current of 2 A is drawn from it. The internal resistance of the cell is :

  1. 1·0 Ω
  2. 1·5 Ω
  3. 2·0 Ω
  4. 2·5 Ω
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Official CBSE Marking Scheme · CBSE Board 2024 (Main), Set 55/2/1

(i) (B) The internal resistance of a cell decreases with the decrease in temperature of the electrolyte. [1]

(ii) (B) 2·8 V [1]

(iii) (A) ε = V+ + V− > 0 [1]

(iv) (a) (D) 0·2 A

(iv) (b) (A) 1·0 Ω [1]

Q89.2023 Main 55/1/1 · Q34 (option 1)4 marks

The following figure shows a circuit diagram. We can find the currents through and potential differences across different resistors using Kirchhoff’s rules.

Circuit: cell E = 6.0 V with r = 1.0 Ω in arm a-h, R1 = 10 Ω in arm b-g, R2 = 5.0 Ω and R3 = 5.0 Ω in series in arm c-f; points a to j marked

Answer the following questions based on the above :

(a) Which points are at the same potential in the circuit ? [1]

(b) What is the current through arm bg ? [1]

(c) Find the potential difference across resistance R3. [2]

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Official CBSE Marking Scheme · CBSE Board 2023 (Main), Set 55/1/1

a) Points (a, b, c); (d, e); (j, f, g, h) are at same potential [1]

Note: Give full credit if a student mentions any two points at same potential from the above.

b)

Circuit with currents i1 in arm ab, i2 in arm bg and i1 − i2 in arm bc

According to Kirchhoff’s loop rule, for closed loop abgha: −6 + 10 I2 + I1 = 0, i.e. I1 + 10 I2 = 6  (i)

for closed loop acfha: −6 + 10 (I1 − I2) + I1 = 0, i.e. 11 I1 − 10 I2 = 6  (ii) [½]

Adding (i) and (ii): 12 I1 = 12, I1 = 1 A, I2 = 0.5 A [½]

I2 = current through arm bg = 0.5 A

Note: Award 1 mark if a student calculates the current by any other method.

c) VR3 = (I1 − I2) × R3 [1] = 0.5 × 5 = 2.5 V [1]

Q90.2023 Main 55/1/1 · Q34 (OR option)4 marks

The following figure shows a circuit diagram. We can find the currents through and potential differences across different resistors using Kirchhoff’s rules.

Circuit: cell E = 6.0 V with r = 1.0 Ω in arm a-h, R1 = 10 Ω in arm b-g, R2 = 5.0 Ω and R3 = 5.0 Ω in series in arm c-f; points a to j marked

Answer the following questions based on the above :

(a) Which points are at the same potential in the circuit ? [1]

(b) What is the current through arm bg ? [1]

(c) What is the power dissipated in resistance R2 ? [2]

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Official CBSE Marking Scheme · CBSE Board 2023 (Main), Set 55/1/1

a) Points (a, b, c); (d, e); (j, f, g, h) are at same potential [1]

Note: Give full credit if a student mentions any two points at same potential from the above.

b)

Circuit with currents i1 in arm ab, i2 in arm bg and i1 − i2 in arm bc

According to Kirchhoff’s loop rule, for closed loop abgha: −6 + 10 I2 + I1 = 0, i.e. I1 + 10 I2 = 6  (i)

for closed loop acfha: −6 + 10 (I1 − I2) + I1 = 0, i.e. 11 I1 − 10 I2 = 6  (ii) [½]

Adding (i) and (ii): 12 I1 = 12, I1 = 1 A, I2 = 0.5 A [½]

I2 = current through arm bg = 0.5 A

Note: Award 1 mark if a student calculates the current by any other method.

c) P = (I1 − I2)2 × R2 = (0.5)2 × 5 [1] = 1.25 W [1]

Q91.2026 Main 55/1/1 · Q31 (OR option)5 marks

(a) Two cells of emf E1 and E2 with internal resistances r1 and r2 respectively, are connected in parallel by connecting their positive terminals together and negative terminals together. Deduce an expression for equivalent emf and equivalent internal resistance of the combination.

(b) A parallel combination, as stated in (a) above, of two cells of emfs E and 3E and internal resistances R each is connected across a resistance 2R. Find the current that flows through resistance 2R.

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Official CBSE Marking Scheme · CBSE Board 2026 (Main), Set 55/1/1

(a)

Two cells ε1, r1 and ε2, r2 in parallel between A and C, equivalent to a single cell ε_eq, r_eq

Let I1 and I2 are the currents leaving from the positive electrodes of the cells ε1 and ε2 respectively. Hence I = I1 + I2 [½]

Potential difference across the terminals of cell ε1 is V = ε1 – I1r1

Potential difference across the terminals of cell ε2 is V = ε2 – I2r2 [½]

I = I1 + I2

I = ε1 – Vr1 + ε2 – Vr2

I = (ε1r1 + ε2r2) – V(1r1 + 1r2)

V = ε1r2 + ε2r1r1 + r2 – I(r1r2r1 + r2) [½]

V = εeq – Ireq

εeq = (ε1r2 + ε2r1r1 + r2) [½]

req = (r1r2r1 + r2) [1]

(b)

Cells E and 3E, each with internal resistance R, in parallel and connected across 2R

Equivalent emf

Eeq = ε1r2 + ε2r1r1 + r2

Eeq = E × R + 3E × RR + R

Eeq = 4ER2R

Eeq = 2E [½]

Equivalent resistance

req = r1r2r1 + r2 = RRR + R = R2 [½]

I = Eeq2R + R/2

I = 4E5R A [½]

Q92.2026 Main 55/2/1 · Q32 (OR option)5 marks

(a) An electric field E is established across the ends of a cylindrical conductor of length L and area of cross-section A. Discuss how electrons attain an average velocity, independent of time. Hence, obtain a relation between current in the conductor and this ‘average velocity’ of electrons.

(b) (i) This ‘average velocity’ is found be few mm/s for currents in range of a few amperes. How then is current established almost the instant a circuit is closed ?

(ii) Two copper wires having their radii in the ratio of 3 : 2 are connected in series across a battery. Find the ratio of the drift velocities of the electrons in the wires.

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Official CBSE Marking Scheme · CBSE Board 2026 (Main), Set 55/2/1

(a) vd = –eEτm, where τ is the average time between two successive collisions, which is constant. Distance covered by electrons between two successive collisions is almost same. Hence S = 12at2 where a = constant, s = almost constant. So t is constant. [1]

Alternatively : At any instant some of the electrons would have spent time more than τ and some less than τ. Thus averaging over all the times we get a constant time. This way we get average velocity independent of time.

Let n be the number of free electrons per unit volume in the conductor of length L and cross-sectional area A. Total charge crossing area A in time Δt is q = –neA|vd|Δt [1]

Where negative sign indicates direction of flow of electrons is opposite to direction of electric field. By definition, electric current is opposite to flow of electrons.

∴ IΔt = neA|vd|Δt

I = neA|vd| [1]

(b) (i) Electric field is established throughout the circuit, almost instantly causing at every point a local electron drift. As a result the current is established instantaneously. [1]

(ii) vd = Ineπr2 [½]

vd1vd2 = (r2r1)2 = 49 [½]

Q93.2026 Main 55/4/1 · Q33 (OR option)5 marks

(b) (i) Answer the following giving reason :

  1. The electron drift speed is estimated to be only a few mm/s for currents in the range of a few amperes. How, then, is the current established almost the instant a circuit is closed ?
  2. A low voltage supply from which one needs high currents must have very low internal resistance. Why ?
  3. The assertion that V = IR is a statement of Ohm’s law is not true. Why ?

(ii) Two cells of emfs 12 V and 6 V are connected in parallel as shown in the figure. Their internal resistances are 1 Ω and 0·5 Ω respectively. Calculate the emf and internal resistance of the equivalent cell between points A and B.

Two cells, 12 V with 1 Ω and 6 V with 0.5 Ω, connected in parallel between points A and B
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Official CBSE Marking Scheme · CBSE Board 2026 (Main), Set 55/4/1

(i) (I) Electric field is established throughout the circuit, almost instantly (with the speed of light) causing at every point a local electron drift. Establishment of current does not have to wait for electrons from one end of the conductor travelling to the other end. However, it does take a little while for the current to reach its steady value. [1]

(II) I = εR + r

The maximum current that can be drawn from a cell is for R = 0 and it is Imax = εr [1]

(III) The Ohm’s law asserts that the plot of I vs V is linear i.e. R is independent of V. But this condition is not applicable for diodes. [1]

(ii) εAB = ε1r2 + ε2r1r1 + r2 [½]

= 12 × 0.5 + 6 × 11.5 = 8 V [½]

Alternatively (for emf only) : In loop PQRSP,

−6 − 0.5I + 12 − 1 × I = 0

−1.5I + 6 = 0

I = 61.5 = 4 A

VAB = ε − Ir = 12 − 4 × 1 = 8 V

Internal resistance = r1r2r1 + r2 [½]

= 0.5 × 11.5 = 13 Ω [½]

Q94.2026 Main 55/5/1 · Q31 (option 1)5 marks
  1. Derive the condition for which a Wheatstone Bridge is balanced.
  2. Determine the current in 3 Ω branch of a Wheatstone Bridge in the circuit shown in the figure.
Wheatstone bridge with 20, 2, 12, 1 and 3 ohm resistors and a 6 V battery
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Official CBSE Marking Scheme · CBSE Board 2026 (Main), Set 55/5/1

(i) Deriving the condition for balanced Wheatstone bridge [2½]; (ii) Determining the current in 3 Ω branch [2½]

(i)

Wheatstone bridge with resistors R1 to R4, galvanometer G, currents marked

[½]

Applying Kirchhoff’s rule to closed loop ADBA. For balanced condition Ig = 0. [½]

–I1R1 + 0 + I2R2 = 0 ……. (i) [½]

Applying Kirchhoff’s rule to closed loop CBDC using I3 = I1, I4 = I2

I2R4 + 0 – I1R3 = 0 ……. (ii)

From eq. (i): I1I2 = R2R1 [½]

and from eq. (ii): I1I2 = R4R3

hence R2R1 = R4R3 [½]

(ii)

Wheatstone bridge with loop currents I1, I1–I2 and I2 marked

[½]

Applying Kirchhoff’s rule to loop I (ABDA):

–20 I1 – 3 I2 + 12 (I – I1) = 0 ……. (i) [½]

Applying Kirchhoff’s rule to loop II (BCDB):

–2 (I1 – I2) + 1 (I – I1 + I2) + 3I2 = 0 ……. (ii) [½]

Applying Kirchhoff’s rule to loop III (EADCE):

–12 (I – I1) – 1 (I – I1 + I2) = –6 ……. (iii) [½]

After solving eq. (i), (ii), (iii)… [½]

I2 = 12/821 ≈ 0.014 A [½]

Note: Award full marks if student takes another appropriate current distribution in the circuit.

Q95.2026 Main 55/5/1 · Q31 (OR option)5 marks
  1. Consider a cylindrical conductor of length l and area of cross-section A. Current I is maintained in the conductor and electrons drift with velocity vd (|vd| = e|E|m τ), (where symbols have their usual meanings). Show that the conductivity σ of the material of the conductor is given by σ = ne2m τ.
  2. The resistance of a metal wire at 20°C is 1·05 Ω and at 100°C is 1·38 Ω. Determine the temperature coefficient of resistivity of this metal.
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Official CBSE Marking Scheme · CBSE Board 2026 (Main), Set 55/5/1

(i) Showing the conductivity of material σ = ne2τ/m [3]; (ii) Determining the temperature coefficient of resistivity [2]

(i)

Cylindrical conductor of area A with drift of electrons in field E

The amount of charge crossing the area in time Δt is

Q = Ne; Q = (nAl)e

I = QΔt = nAleΔt

I = nAe|vd| ……. (i) [1]

Substituting the value of |vd| in eq. (i)

IA = ne(e|E|τm) [½]

|J| = ne2τ|E|m ……. (ii) [½]

where J is current density and parallel to E, and

J = σE ……. (iii) [½]

From eq. (ii) and (iii): σ = ne2τ/m [½]

(ii)

α = (R2 – R1)[R1 (T2 – T1)] [½]

α = (1.38 – 1.05)[1.05 × (100 – 20)] [½]

α = 0.33(1.05 × 80) [½]

= 0.0039 °C–1 [½]

Alternatively,

R1 = R0(1 + α ΔT1) ……. (i)

R2 = R0(1 + α ΔT2) ……. (ii)

R1R2 = (1 + α ΔT1)(1 + α ΔT2)

1.051.38 = 1 + 20α1 + 100α

On solving, α = 112580 °C–1 = 0.0042 °C–1

Q96.2026 Supp 55/7/1 · Q31 (option 1)5 marks
  1. A potential difference V is applied across the ends of a cylindrical conductor of length l, area of cross-section A and resistance R. It is gradually stretched till its length is tripled. How will (i) the drift speed of electrons in the conductor, and (ii) the resistance of the conductor be affected ? Justify your answers. [2]
  2. In the given circuit diagram, find the potential difference between points B and D. [3]
    Rectangular circuit ABCD with cells 1 V 2 Ω, 4 V 3 Ω, 3 V 2 Ω, 2 V 1 Ω and a 2 Ω resistor along diagonal BD
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Official CBSE Marking Scheme · CBSE Board 2026 (Supplementary), Set 55/7/1

(a) (i) Drift speed vd will become one-third. [½]

vd = IneA = Vρlne

vd ∝ 1l [½]

(ii) Resistance will become 9 times of its previous value. [½]

R′ = ρl′A′ = ρ(3l)A/3 = 9R [½]

(b)

Current distribution in the circuit with loop currents I, I1 and I-I1

[1]

Using Kirchhoff’s Law in loop ABDA,

–1 – 2(I – I1) + 2I1 + 2 – (I – I1) = 0

–3I + 5I1 = –1 ……..(i) [½]

In loop DBCD,

–2I1 + 4 – 3I – 3 – 2I = 0

–5I – 2I1 = –1 ……..(ii) [½]

On solving equations (i) and (ii),

I1 = – 231 A [½]

|VAB| = I1 × 2 = 231 × 2

|VB| = 431 V [½]

Q97.2026 Supp 55/7/1 · Q31 (OR option)5 marks
  1. A cell of emf E and internal resistance r is connected across a resistor of variable resistance R. Draw a plot showing the variation of (i) the current (I) in the circuit with R, and (ii) the terminal potential difference of the cell with R. [2]
  2. The resistance of a heating coil is 100 Ω at 25°C and 120 Ω at 1025°C. (i) Find its resistance at 425°C. (ii) Draw a plot showing variation of its resistance as a function of increase in temperature. [3]
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Official CBSE Marking Scheme · CBSE Board 2026 (Supplementary), Set 55/7/1

(a) (i)

Graph of current I versus R, falling from E/r

[1]

(ii)

Graph of terminal potential difference V versus R, rising towards E

[1]

(b) α = R2 – R1R1(T2 – T1) = 120 – 100100(1025 – 25) = 20100 × 1000 [½]

α = 2 × 10-4 °C-1 [½]

R2 = R1[1 + α(T2 – T1)] = 100[1 + 2 × 10-4(425 – 25)] [½]

= 100 × 1.08

R2 = 108 Ω [½]

Graph of resistance R versus temperature T, a straight line with positive intercept and slope

[1]

Q98.2025 Main 55/5/1 · Q31 (option 1)5 marks

(i) Three batteries E1, E2 and E3 of emfs and internal resistances (4 V, 2 Ω), (2 V, 4 Ω) and (6 V, 2 Ω) respectively are connected as shown in the figure. Find the values of the currents passing through batteries E1, E2 and E3.

Three batteries E1, E2, E3 connected in parallel

(ii) The ends of six wires, each of resistance R (= 10 Ω) are joined as shown in the figure. The points A and B of the arrangement are connected in a circuit. Find the value of the effective resistance offered by it to the circuit.

Triangle with centre node: six resistors R between A and B
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Q99.2025 Main 55/5/1 · Q31 (OR option)5 marks

(i) Current I (= 1 A) is passing through a copper rod (n = 8·5 × 1028 m–3) of varying cross-sections as shown in the figure. The areas of cross-section at points A and B along its length are 1·0 × 10–7 m2 and 2·0 × 10–7 m2 respectively. Calculate :

Copper rod of increasing cross-section with points A and B

(I) the ratio of electric fields at points A and B.

(II) the drift velocity of free electrons at point B.

(ii) Two point charges q1 (= 16 μC) and q2 (= 1 μC) are placed at points r1 = (3 m)i and r2 = (4 m)j. Find the net electric field E at point r = (3 m)i + (4 m)j.

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Q100.2023 Main 55/4/1 · Q31 (option 1)5 marks

(i) Define mobility of electrons. Give its SI units.

(ii) A steady current flows through a wire AB, as shown in the figure. What happens to the electric field and the drift velocity along the wire ? Justify your answer.

Wire AB of increasing cross-section from A to B connected to a battery

(iii) Consider the circuit shown in the figure. Find the effective resistance of the circuit and the current drawn from the battery.

Circuit: 6 V battery with 10 Ω and 14 Ω in series branch, 10 Ω and 10 Ω in series branch, and 30 Ω branch, all in parallel
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Official CBSE Marking Scheme · CBSE Board 2023 (Main), Set 55/4/1

(i) Mobility : Mobility is defined as the magnitude of the drift velocity per unit electric field. [1]

S.I. Unit : m2V s or C skg [½]

(ii) Both electric field and the drift velocity decreases. [½ + ½]

Justification :

vd = IneA

As area increases across the wire, drift velocity decreases. [½]

vd = eEm τ

As drift velocity decreases, electric field decreases (since e, m and τ are constant). [½]

(iii) From the diagram

10 Ω and 14 Ω are in series R1 = 10 Ω + 14 Ω = 24 Ω

10 Ω and 10 Ω are in series R2 = 10 Ω + 10 Ω = 20 Ω

24 Ω, 20 Ω and 30 Ω are in parallel.

1R = 124 + 120 + 130 = 5 + 6 + 4120 = 15120

R = 8 Ω [½]

Electric current in the circuit I = VR = 68 = 34 A [1]

Note : Full credit of 1½ marks is to be awarded for part (iii) even if a student does not attempt this part.

Q101.2023 Main 55/4/1 · Q31 (OR option)5 marks

(i) Define electrical conductivity of a wire. Give its SI unit.

(ii) High current is to be drawn safely from (1) a low-voltage battery, and (2) a high-voltage battery. What can you say about the internal resistance of the two batteries ?

(iii) Calculate the total energy supplied by the batteries to the circuit shown in the figure, in one minute.

Series circuit with 10 V battery, 1 Ω, 4 V battery, 2 Ω, 6 Ω and 3 Ω resistors and current I
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Official CBSE Marking Scheme · CBSE Board 2023 (Main), Set 55/4/1

(i) Definition : Electrical conductivity is defined as the measure of a material’s ability to carry a current through it. [1]

Alternatively : It is the reciprocal of the resistivity.

Alternatively : It is defined as the current density per unit electric field.

S.I. Unit : (ohm)−1-m−1 or S-m−1 [½]

(ii) Low voltage battery – Internal resistance should be low. [1]

High voltage battery – Internal resistance should be high. [1]

(iii) Applying Kirchhoff’s loop rule

10 − I × 1 − 4 − 2I − 6I − 3I = 0 [½]

12 I = 6 ⟹ I = 0.5 A

Heat energy H = I2Rt [½]

H = 0.25 × 12 × 60 = 180 J [½]

Q102.2023 Main 55/5/1 · Q31 (option 1)5 marks

(i) Explain how free electrons in a metal at constant temperature attain an average velocity under the action of an electric field. Hence obtain an expression for it.

(ii) Consider two conducting wires A and B of the same diameter but made of different materials joined in series across a battery. The number density of electrons in A is 1.5 times that in B. Find the ratio of drift velocity of electrons in wire A to that in wire B.

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Official CBSE Marking Scheme · CBSE Board 2023 (Main), Set 55/5/1

(i) Under the effect of external field, an electron experiences a force F = –eE between collisions.

Due to this force the electron is accelerated and attains a velocity. This velocity is different for different electrons, which averaged over all electrons gives average drift velocity. This drift velocity is constant for a given temperature. [1]

Expression of average velocity:

Under the action of an electric field electrons get accelerated with a = –eEm [½]

Velocity of an electron at any instant of time is Vi = vi – eEm ti [½]

Average velocity of the electrons at time ‘t’ is the drift velocity vd = (Vi)average

(Vi)average = (vi)average – eEm (ti)average [½]

But (vi)average = 0 due to randomness

vd = 0 – eEmτ = –eEmτ [½]

(ii) vd = IenA = (4IeπD2) × 1n [½]

vd ∝ 1n for same diameter and current [½]

nA = 1.5 nB (Given)

vdAvdB = nBnA = 23 [1]

Q103.2023 Main 55/5/1 · Q31 (OR option)5 marks

(i) A cell emf of (E) and internal resistance (r) is connected across a variable load resistance (R). Draw plots showing the variation of terminal voltage V with (i) R and (ii) the current (I) in the load.

(ii) Three cells, each of emf E but internal resistances 2r, 3r and 6r are connected in parallel across a resistor R.

Obtain expressions for (i) current flowing in the circuit, and (ii) the terminal potential difference across the equivalent cell.

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Official CBSE Marking Scheme · CBSE Board 2023 (Main), Set 55/5/1

(i)

i. Plot showing variation of V with R [1]

Terminal voltage V versus R rising from 0 towards E

ii. Plot showing variation of V with I [1]

Terminal voltage V versus current I, straight line from E to E/r

(ii) 1req = 1r1 + 1r2 + 1r3 = 12r + 13r + 16r

req = r [½]

Given cells are of equal emf (E) and connected in parallel, so Eeqreq = E2r + E3r + E6r

Eeq = E [½]

Current flowing I = EeqR + req = ER + r [½]

The terminal potential difference V = Eeq – I req [½]

V = E – E(R + r) × r [½]

V = ERR + r [½]

Q104.2023 Supp 55/C/1 · Q32 (option 1)5 marks

(i) Derive the relation between the current and the drift velocity of free electrons in a conductor. Briefly explain the variation of resistance of a conductor with rise in temperature.

(ii) An ammeter, together with an unknown resistance in series is connected across two identical batteries, each of emf 1·5 V, connected (i) in series, and (ii) in parallel. If the current recorded in the two cases be (12) A and (13) A respectively, calculate the internal resistance of each battery.

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Official CBSE Marking Scheme · CBSE Board 2023 (Supplementary), Set 55/C/1

(i)

Conductor of cross-section A with electrons drifting, length Δx = v_d Δt, field E

[½]

Total charge transported across the area A in time Δt is

ΔQ = −neAvd Δt ------(1) [½]

Also the amount of charge crossing area ‘A’ in time Δt is

ΔQ = I Δt ------(2) [½]

Comparing equation (1) and (2)

I = neAvd [½]

With increase in temperature, average speed of electrons increases resulting in more frequent collisions

Hence relaxation time τ decreases

As R = mlne2τA

Resistance increases. [1]

(ii) For series I = ER + r

12 = 3R + 2r

R + 2r = 6 --------(1) [½]

For parallel 13 = 1.5R + r/2

2R + r = 9 ----------(2) [½]

After solving r = 1 Ω [1]

Q105.2023 Supp 55/C/1 · Q32 (OR option)5 marks

(i) State Kirchhoff’s rules. Use them to obtain the condition of balance for a Wheatstone Bridge.

(ii) Use Kirchhoff’s rule to determine the currents flowing through the branches MN, TO and SP in the circuit shown in the figure.

Circuit with branches MN (8 V cell and 2 ohm), TO (4 ohm), SP (4 V cell and 1 ohm)
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Official CBSE Marking Scheme · CBSE Board 2023 (Supplementary), Set 55/C/1

(i) Kirchhoff’s junction rule - at any junction, the sum of the current entering the junction is equal to the sum of currents leaving the junction. [½]

Kirchhoff second rule:

The algebraic sum of changes in potential around any closed loop involving resistors and cells in the loop is zero. [½]

Wheatstone bridge ABCD with arms R1, R2, R3, R4, galvanometer G and battery

[½]

In balanced bridge Ig = 0,

Hence I1 = I3 and I2 = I4

Using Kirchhoff’s loop rule for closed loops ADBA and CBDC

− I1R1 + 0 + I1R1 = 0 (Ig = 0) -----(1) [½]

In the second loop I3 = I1, I4 = I2

I2R4 + 0 − I1R3 = 0 -----(2) [½]

From equation (1) and (2)

I1I2 = R2R1 and I1I2 = R4R3

R2R1 = R4R3 [½]

This is the condition for balanced Wheatstone bridge

(ii) In loop MNOTM

2 I + 4 I1 = 8 ------(1) [½]

Loop OPSTO

− I + 5 I1 = − 4 ------(2) [½]

On solving

Current in MN, I = 4 A

Current in TO, I1 = 0 A

Current in SP, I − I1 = 4 A [1]

Unit III · Magnetic Effects of Current and Magnetism

4. Moving Charges and Magnetism

Q1.2026 Main 55/1/1 · Q31 mark

A long straight wire of circular cross-section (radius a) carries a steady current I. The current is uniformly distributed across this cross-section. The magnitude of the magnetic field produced at a point at a distance (a/2) from the axis of the wire will be

  1. Zero
  2. μ0I2πa
  3. μ0I4πa
  4. μ0I6πa
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Official CBSE Marking Scheme · CBSE Board 2026 (Main), Set 55/1/1

(C) μ0I4πa

Q2.2026 Main 55/2/1 · Q51 mark

A straight long wire lying along y-axis, carries a current of 1 A along –y direction. The magnetic field due to the conductor at a point (50 cm, 0, 0) will point along.

  1. z-axis
  2. –z-axis
  3. x-axis
  4. –x-axis
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Official CBSE Marking Scheme · CBSE Board 2026 (Main), Set 55/2/1

(A) z-axis

Q3.2026 Main 55/2/1 · Q71 mark

A galvanometer of resistance 27 Ω is converted into an ammeter of range (0 – 10 mA) using a resistance of 3 Ω. The galvanometer will show full scale deflection for a current of about –

  1. 10 mA
  2. 100 mA
  3. 1 mA
  4. 3 mA
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Official CBSE Marking Scheme · CBSE Board 2026 (Main), Set 55/2/1

(C) 1 mA

Q4.2026 Main 55/3/1 · Q141 mark

Assertion (A) : The cylindrical soft iron core in a moving coil galvanometer only makes the magnetic field radial and does not affect the strength of the magnetic field.

Reason (R) : In a moving coil galvanometer, the plane of the coil is always perpendicular to the magnetic field.

  1. Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of the Assertion (A).
  2. Both Assertion (A) and Reason (R) are true, but Reason (R) is not the correct explanation of the Assertion (A).
  3. Assertion (A) is true, but Reason (R) is false.
  4. Both Assertion (A) and Reason (R) are false.
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Official CBSE Marking Scheme · CBSE Board 2026 (Main), Set 55/3/1

(D) Both Assertion (A) and Reason (R) are false

Q5.2026 Main 55/4/1 · Q21 mark

Two particles of masses m1 and m2 having charges q1 and q2 respectively are projected with the same velocity in a region of uniform magnetic field B pointing vertically upward. If they describe circular paths as shown in the figure, one may conclude that :

Two circular paths 1 and 2 of particles m1 and m2 in a magnetic field pointing out of the page
  1. m1m2 > q1q2
  2. m1m2 > q2q1
  3. m1m2 < q1q2
  4. m1m2 < q2q1
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Official CBSE Marking Scheme · CBSE Board 2026 (Main), Set 55/4/1

(C) m1m2 < q1q2

Q6.2026 Main 55/5/1 · Q31 mark

Four long straight thin wires are held vertically at the corners A, B, C and D of a square of side ‘a’, kept on a table and carry equal current ‘I’. The wire at A carries current in upward direction whereas the current in the remaining wires flows in downward direction. The net magnetic field at the centre of the square will have the magnitude :

Square ABCD with wires at corners and centre O
  1. μ0Iπa and directed along OC
  2. μ0Iπa 2 and directed along OD
  3. μ0I 2πa and directed along OB
  4. 2 μ0Iπa and directed along OA
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Official CBSE Marking Scheme · CBSE Board 2026 (Main), Set 55/5/1

(C) μ0Iπa 2 and directed along OB

Q7.2026 Supp 55/7/1 · Q31 mark

Two infinitely long straight conductors lie along x-axis and y-axis in the plane of the page. If they carry current I each along +x-axis and +y-axis respectively, the net magnetic field at a point with position vector r = (3i + 4j) m will be :

  1. μ0I6πk
  2. – μ0I12πk
  3. – μ0I24πk
  4. μ0I48πk
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Official CBSE Marking Scheme · CBSE Board 2026 (Supplementary), Set 55/7/1

(C) – μ0I24πk

Q8.2025 Main 55/1/1 · Q31 mark

A 1 cm segment of a wire lying along x-axis carries current of 0.5 A along +x direction. A magnetic field B = (0.4 mT) j + (0.6 mT) k is switched on, in the region. The force acting on the segment is

  1. (2j + 3k) mN
  2. (–3j + 2k) μN
  3. (6j + 4k) mN
  4. (–4j + 6k) μN
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Q9.2025 Main 55/1/1 · Q141 mark

Assertion (A) : The deflection in a galvanometer is directly proportional to the current passing through it.

Reason (R) : The coil of a galvanometer is suspended in a uniform radial magnetic field.

  1. Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of Assertion (A).
  2. Both Assertion (A) and Reason (R) are true, but Reason (R) is not the correct explanation of Assertion (A).
  3. Assertion (A) is true, but Reason (R) is false.
  4. Assertion (A) is false and Reason (R) is also false.
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Q10.2025 Main 55/2/1 · Q31 mark

A 1 cm straight segment of a conductor carrying 1 A current in x direction lies symmetrically at origin of Cartesian coordinate system. The magnetic field due to this segment at point (1m, 1m, 0) is

  1. 1.0 × 10-9 k T
  2. –1.0 × 10-9 k T
  3. 5.02 × 10-10 k T
  4. –5.02 × 10-10 k T
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Q11.2025 Main 55/4/1 · Q31 mark

A particle having charge +q enters a uniform magnetic field B as shown in the figure. The particle will describe :

Uniform magnetic field into the page in the XY plane, charge +q moving along +Y with velocity v, axes X, Y, Z
  1. a circular path in XZ plane
  2. a semicircular path in XY plane
  3. a helical path with its axis parallel to Y-axis
  4. a semicircular path in YZ plane
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Q12.2025 Main 55/4/1 · Q121 mark

A long straight wire is held vertically and carries a steady current in upward direction. The shape of magnetic field lines produced by the current-carrying wire are :

  1. horizontal straight lines directed radially out from the wire.
  2. straight lines parallel to the current-carrying wire.
  3. concentric horizonal circles around the wire.
  4. coaxial helixes around the wire.
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Q13.2025 Main 55/5/1 · Q41 mark

A charged particle gains a speed of 106 ms–1, when accelerated from rest through a potential difference 10 kV. It enters a region of magnetic field of 0·4 T such that v ⊥ B. The radius of circular path described by it is :

  1. 2·5 cm
  2. 5 cm
  3. 8 cm
  4. 10 cm
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Q14.2025 Main 55/5/1 · Q51 mark

A current of (10π) A is maintained in a circular loop of radius 14 cm. The value of dipole moment associated with the loop is :

  1. 0·019 Am2
  2. 0·14 Am2
  3. 0·196 Am2
  4. 0·615 Am2
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Q15.2025 Main 55/6/1 · Q31 mark

The value of magnetic field at point O in the given figure is :

Semicircular wire of radius R carrying current I with straight leads, centre O
  1. μ0I2πR
  2. μ0IπR
  3. μ0I4R
  4. μ0IR
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Q16.2025 Main 55/6/1 · Q51 mark

A galvanometer can be converted into an ammeter of desired range by connecting a :

  1. small resistance in series
  2. large resistance in series
  3. small resistance in parallel
  4. large resistance in parallel
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Q17.2025 Main 55/6/1 · Q61 mark

A proton and an α-particle enter with the same velocity v in a uniform magnetic field B such that v ⊥ B. The ratio of the radii of their paths is :

  1. 2
  2. 12
  3. 14
  4. 4
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Q18.2025 Main 55/7/1 · Q131 mark

Assertion (A) : A charged particle is moving with velocity v in x-y plane, making an angle θ (0 < θ < π2) with x-axis. If a uniform magnetic field B is applied in the region, along y-axis, the particle will move in a helical path with its axis parallel to x-axis.

Reason (R) : The direction of the magnetic force acting on a charged particle moving in a magnetic field is along the velocity of the particle.

  1. Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of the Assertion (A).
  2. Both Assertion (A) and Reason (R) are true, but Reason (R) is not the correct explanation of the Assertion (A).
  3. Assertion (A) is true, but Reason (R) is false.
  4. Both Assertion (A) and Reason (R) are false.
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Q19.2025 Supp 55/S/1 · Q71 mark

A galvanometer having a coil resistance of 48 Ω shows full-scale deflection for a current of 2·0 A. It can be converted into an ammeter of range (0 – 10 A) by connecting :

  1. in series a resistance of 8 Ω
  2. in parallel a resistance of 8 Ω
  3. in parallel a resistance of 12 Ω
  4. in series a resistance of 12 Ω
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Official CBSE Marking Scheme · CBSE Board 2025 (Supplementary), Set 55/S/1

(C) In parallel a resistance of 12 Ω

Q20.2025 Supp 55/S/1 · Q111 mark

Two parallel beams of protons moving in the same direction will :

  1. attract each other
  2. repel each other
  3. be deflected normal to the plane containing the two beams
  4. not interact with each other
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Official CBSE Marking Scheme · CBSE Board 2025 (Supplementary), Set 55/S/1

(A) Attract each other

Q21.2024 Main 55/1/1 · Q41 mark

A loop carrying a current I clockwise is placed in x − y plane, in a uniform magnetic field directed along z-axis. The tendency of the loop will be to :

  1. move along x-axis
  2. move along y-axis
  3. shrink
  4. expand
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Official CBSE Marking Scheme · CBSE Board 2024 (Main), Set 55/1/1

(C) Shrink

Q22.2024 Main 55/1/1 · Q51 mark

A 10 cm long wire lies along y-axis. It carries a current of 1·0 A in positive y-direction. A magnetic field B = (5 mT) j − (8 mT) k exists in the region. The force on the wire is :

  1. (0·8 mN) i
  2. − (0·8 mN) i
  3. (80 mN) i
  4. − (80 mN) i
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Official CBSE Marking Scheme · CBSE Board 2024 (Main), Set 55/1/1

(B) (− 0.8 mN) i

Q23.2024 Main 55/1/1 · Q61 mark

A galvanometer of resistance G Ω is converted into an ammeter of range 0 to I A. If the current through the galvanometer is 0·1% of I A, the resistance of the ammeter is :

  1. G999 Ω
  2. G1000 Ω
  3. G1001 Ω
  4. G100·1 Ω
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Official CBSE Marking Scheme · CBSE Board 2024 (Main), Set 55/1/1

(B) G1000 Ω

Q24.2024 Main 55/1/1 · Q151 mark

Assertion (A) : Two long parallel wires, freely suspended and connected in series to a battery, move apart.

Reason (R) : Two wires carrying current in opposite directions repel each other.

  1. Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of the Assertion (A).
  2. Both Assertion (A) and Reason (R) are true, but Reason (R) is not the correct explanation of the Assertion (A).
  3. Assertion (A) is true, but Reason (R) is false.
  4. Assertion (A) is false and Reason (R) is also false.
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Official CBSE Marking Scheme · CBSE Board 2024 (Main), Set 55/1/1

(A) Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of the Assertion (A).

Q25.2024 Main 55/2/1 · Q31 mark

A wire of length 4·4 m is bent round in the shape of a circular loop and carries a current of 1·0 A. The magnetic moment of the loop will be :

  1. 0·7 Am2
  2. 1·54 Am2
  3. 2·10 Am2
  4. 3·5 Am2
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Official CBSE Marking Scheme · CBSE Board 2024 (Main), Set 55/2/1

(B) 1·54 Am2

Q26.2024 Main 55/2/1 · Q151 mark

Two statements are given — one labelled Assertion (A) and the other labelled Reason (R). Select the correct answer from the codes (A), (B), (C) and (D) as given below.

Assertion (A) : An electron and a proton enter with the same momentum p in a magnetic field B such that p ⊥ B. Then both describe a circular path of the same radius.

Reason (R) : The radius of the circular path described by the charged particle (charge q, mass m) moving in the magnetic field B is given by r = mvqB.

  1. Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of the Assertion (A).
  2. Both Assertion (A) and Reason (R) are true, but Reason (R) is not the correct explanation of the Assertion (A).
  3. Assertion (A) is true, but Reason (R) is false.
  4. Assertion (A) is false and Reason (R) is also false.
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Official CBSE Marking Scheme · CBSE Board 2024 (Main), Set 55/2/1

(A) Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of the Assertion (A)

Q27.2024 Main 55/3/1 · Q31 mark

A 2·0 cm segment of wire, carrying 5·0 A current in positive y-direction lies along y-axis, as shown in the figure. The magnetic field at a point (3 m, 4 m, 0) due to this segment (part of a circuit) is :

Wire segment PQ along the y-axis carrying 5.0 A in +y direction, with x, y and z axes
  1. (0·12 nT) j
  2. – (0·10 nT) j
  3. – (0·24 nT) k
  4. (0·24 nT) k
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Official CBSE Marking Scheme · CBSE Board 2024 (Main), Set 55/3/1

(C) –(0.24 nT) k

Q28.2024 Main 55/3/1 · Q41 mark

A circular loop of wire, carrying a current ‘I’ is lying in xy-plane with its centre coinciding with the origin. It is subjected to a uniform magnetic field pointing along + z-axis. The loop will :

  1. move along x-axis
  2. move along – y-axis
  3. move along z-axis
  4. remain stationary
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Official CBSE Marking Scheme · CBSE Board 2024 (Main), Set 55/3/1

(D) remain stationary

Q29.2024 Main 55/3/1 · Q51 mark

A current carrying circular loop of magnetic moment M is suspended in a vertical plane in an external magnetic field B such that its plane is normal to B. The work done in rotating this loop by 45° about an axis perpendicular to B is closest to :

  1. – 0·3 MB
  2. 0·3 MB
  3. – 1·7 MB
  4. 1·7 MB
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Official CBSE Marking Scheme · CBSE Board 2024 (Main), Set 55/3/1

(B) 0.3 MB

Q30.2024 Main 55/3/1 · Q151 mark

Assertion (A) : A proton and an electron enter a uniform magnetic field B with the same momentum p such that p is perpendicular to B. They describe circular paths of the same radius.

Reason (R) : In a magnetic field, orbital radius r is equal to pqB.

  1. Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of the Assertion (A).
  2. Both Assertion (A) and Reason (R) are true, but Reason (R) is not the correct explanation of the Assertion (A).
  3. Assertion (A) is true, but Reason (R) is false.
  4. Assertion (A) is false and Reason (R) is also false.
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Official CBSE Marking Scheme · CBSE Board 2024 (Main), Set 55/3/1

(A) Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of the Assertion (A).

Q31.2024 Main 55/4/1 · Q51 mark

A straight wire is kept horizontally along east-west direction. If a steady current flows in wire from east to west, the magnetic field at a point above the wire will point towards

  1. East
  2. West
  3. North
  4. South
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Official CBSE Marking Scheme · CBSE Board 2024 (Main), Set 55/4/1

(C) North

Q32.2024 Main 55/4/1 · Q71 mark

A galvanometer of resistance 100 Ω is converted into an ammeter of range (0 – 1 A) using a resistance of 0.1 Ω. The ammeter will show full scale deflection for a current of about

  1. 0.1 mA
  2. 1 mA
  3. 10 mA
  4. 0.1 A
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Official CBSE Marking Scheme · CBSE Board 2024 (Main), Set 55/4/1

(B) 1 mA

Q33.2024 Main 55/5/1 · Q21 mark

A particle of mass m and charge q describes a circular path of radius R in a magnetic field. If its mass and charge were 2 m and q2 respectively, the radius of its path would be

  1. R4
  2. R2
  3. 2 R
  4. 4 R
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Official CBSE Marking Scheme · CBSE Board 2024 (Main), Set 55/5/1

(D) 4R

Q34.2024 Main 55/5/1 · Q41 mark

A galvanometer of resistance 50 Ω is converted into a voltmeter of range (0 – 2V) using a resistor of 1.0 kΩ. If it is to be converted into a voltmeter of range (0 – 10 V), the resistance required will be

  1. 4.8 kΩ
  2. 5.0 kΩ
  3. 5.2 kΩ
  4. 5.4 kΩ
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Official CBSE Marking Scheme · CBSE Board 2024 (Main), Set 55/5/1

(C) 5.2 kΩ

Q35.2024 Main 55/5/1 · Q141 mark

Assertion (A) : The energy of a charged particle moving in a magnetic field does not change.

Reason (R) : It is because the work done by the magnetic force on the charge moving in a magnetic field is zero.

  1. If both Assertion (A) and Reason (R) are true and Reason (R) is correct explanation of Assertion (A).
  2. If both Assertion (A) and Reason (R) are true and Reason (R) is not the correct explanation of Assertion (A).
  3. If Assertion (A) is true but Reason (R) is false.
  4. If both Assertion (A) and Reason (R) are false.
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Official CBSE Marking Scheme · CBSE Board 2024 (Main), Set 55/5/1

(A) Both Assertion (A) and Reason (R) are true and Reason (R) is correct explanation of Assertion (A).

Q36.2024 Supp 55/S/1 · Q71 mark

A straight wire of length 1.0 m is placed along x-axis, in a region with magnetic field B = (3 i + 2 j) T. A current of 2.0 A flows in the wire along +x direction. The magnetic force acting on the wire is :

  1. 2.0 N, along z-axis
  2. 2.0 N, along –z-axis
  3. 4.0 N, along z-axis
  4. 4.0 N, along –z-axis
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Official CBSE Marking Scheme · CBSE Board 2024 (Supplementary), Set 55/S/1

(C) 4.0 N, along z-axis

Q37.2024 Supp 55/S/1 · Q141 mark

Assertion (A) : The torque acting on a current carrying coil is maximum when it is suspended in a radial magnetic field.

Reason (R) : The torque tends to rotate the coil on its own axis.

  1. Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of the Assertion (A).
  2. Both Assertion (A) and Reason (R) are true, but Reason (R) is not the correct explanation of the Assertion (A).
  3. Assertion (A) is true, but Reason (R) is false.
  4. Both Assertion (A) and Reason (R) are false.
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Official CBSE Marking Scheme · CBSE Board 2024 (Supplementary), Set 55/S/1

(B) Both Assertion (A) and Reason (R) are true, but Reason (R) is not the correct explanation of the Assertion (A).

Q38.2023 Main 55/2/1 · Q51 mark

Beams of electrons and protons move parallel to each other in the same direction. They

  1. attract each other.
  2. repel each other.
  3. neither attract nor repel.
  4. force of attraction or repulsion depends upon speed of beams.
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Official CBSE Marking Scheme · CBSE Board 2023 (Main), Set 55/2/1

(B) Repel each other

Q39.2023 Main 55/2/1 · Q61 mark

A long straight wire of radius ‘a’ carries a steady current ‘I’. The current is uniformly distributed across its area of cross-section. The ratio of magnitude of magnetic field B1 at a2 and B2 at distance 2a is

  1. 12
  2. 1
  3. 2
  4. 4
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Official CBSE Marking Scheme · CBSE Board 2023 (Main), Set 55/2/1

(B) 1

Q40.2023 Main 55/3/1 · Q41 mark

Which of the following graphs correctly represents the variation of the magnitude of the magnetic field outside a straight infinite current carrying wire of radius ‘a’, as a function of distance ‘r’ from the centre of the wire ?

Four graphs (a)-(d) of magnetic field B versus distance r, with r = a marked
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Official CBSE Marking Scheme · CBSE Board 2023 (Main), Set 55/3/1

(C) the graph labelled (c): B drops hyperbolically (∝ 1/r) for r > a, starting from the dashed line at r = a.

Q41.2023 Main 55/3/1 · Q51 mark

A particle of mass m and charge q moving with a uniform velocity v = v0x i + v0y j enters a region with a magnetic field B = B0 j. After some time, an electric field E = E0 j is also switched on in the region. The resulting path described by the particle will be :

  1. a circle in x-z plane
  2. a parabola in x-y plane
  3. a helix with constant pitch
  4. a helix with increasing pitch
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Official CBSE Marking Scheme · CBSE Board 2023 (Main), Set 55/3/1

(D) a helix with increasing pitch.

Q42.2023 Main 55/3/1 · Q171 mark

Assertion (A) : When radius of a circular loop carrying a steady current is doubled, its magnetic moment becomes four times.

Reason (R) : The magnetic moment of a circular loop carrying a steady current is proportional to the area of the loop.

  1. Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of the Assertion (A).
  2. Both Assertion (A) and Reason (R) are true, but Reason (R) is not the correct explanation of the Assertion (A).
  3. Assertion (A) is true, but Reason (R) is false.
  4. Assertion (A) is false and Reason (R) is also false.
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Official CBSE Marking Scheme · CBSE Board 2023 (Main), Set 55/3/1

(A) Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of the Assertion (A)

Q43.2023 Main 55/4/1 · Q21 mark

An electron enters a uniform magnetic field with speed v. It describes a semicircular path and comes out of the field. The final speed of the electron is :

  1. Zero
  2. v
  3. v2
  4. 2v
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Official CBSE Marking Scheme · CBSE Board 2023 (Main), Set 55/4/1

(B) v

Q44.2023 Main 55/4/1 · Q171 mark

Assertion (A) : A current carrying square loop made of a wire of length L is placed in a magnetic field. It experiences a torque which is greater than the torque on a circular loop made of the same wire carrying the same current in the same magnetic field.

Reason (R) : A square loop occupies more area than a circular loop, both made of wire of the same length.

  1. Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of the Assertion (A).
  2. Both Assertion (A) and Reason (R) are true, but Reason (R) is not the correct explanation of the Assertion (A).
  3. Assertion (A) is true, but Reason (R) is false.
  4. Assertion (A) is false and Reason (R) is also false.
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Official CBSE Marking Scheme · CBSE Board 2023 (Main), Set 55/4/1

(D) Assertion (A) is false and Reason (R) is also false.

Q45.2023 Main 55/5/1 · Q21 mark

Two long parallel wires kept 2 m apart carry 3A current each, in the same direction. The force per unit length on one wire due to the other is

  1. 4.5 × 10-5 Nm-1, attractive
  2. 4.5 × 10-7 N/m, repulsive
  3. 9 × 10-7 N/m, repulsive
  4. 9 × 10-5 N/m, attractive
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Official CBSE Marking Scheme · CBSE Board 2023 (Main), Set 55/5/1

No option is correct.

[Award one mark to each student]

Q46.2023 Main 55/5/1 · Q181 mark

Assertion (A) : The deflecting torque acting on a current carrying loop is zero when its plane is perpendicular to the direction of magnetic field.

Reason (R) : The deflecting torque acting on a loop of magnetic moment m in a magnetic field B is given by the dot product of m and B.

Select the correct answer from the codes below :

  1. Both Assertion (A) and Reason (R) are true and (R) is the correct explanation of (A).
  2. Both Assertion (A) and Reason (R) are true and (R) is NOT the correct explanation of (A).
  3. Assertion (A) is true and Reason (R) is false.
  4. Assertion (A) is false and Reason (R) is also false.
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Official CBSE Marking Scheme · CBSE Board 2023 (Main), Set 55/5/1

(C) Assertion (A) is true and Reason (R) is false.

Q47.2023 Supp 55/C/1 · Q21 mark

An electron with velocity v = (vx i + vy j) moves through a magnetic field B = (Bx i − By j). The force F on the electron is : (e is the magnitude of its charge)

  1. − e (vx By − vy Bx) k
  2. e (vx By − vy Bx) k
  3. − e (vx By + vy Bx) k
  4. e (vx By + vy Bx) k
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Official CBSE Marking Scheme · CBSE Board 2023 (Supplementary), Set 55/C/1

(D) e (vx By + vy Bx) k

Q48.2023 Supp 55/C/1 · Q171 mark

Assertion (A) : A planar loop of irregular shape carrying current is subjected to a magnetic field acting perpendicular to the plane of the loop. If the wire is flexible, the loop takes a circular shape.

Reason (R) : The force acting on each point of a current carrying loop, in a magnetic field perpendicular to its plane, is radially outward.

  1. Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of the Assertion (A).
  2. Both Assertion (A) and Reason (R) are true, but Reason (R) is not the correct explanation of the Assertion (A).
  3. Assertion (A) is true, but Reason (R) is false.
  4. Assertion (A) is false and Reason (R) is also false.
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Official CBSE Marking Scheme · CBSE Board 2023 (Supplementary), Set 55/C/1

(A) Both Assertion (A) and Reason (R) are true, but Reason (R) is the correct explanation of the Assertion (A)

Q49.SQP 2025-26 · Q81 mark

A charged particle is projected along the axis of a current carrying loop. Which of the following statements is true?

  1. The acceleration of the charged particle will depend on the velocity with which it is projected.
  2. The acceleration of the charged particle will depend on the magnitude of the current passing through the coil.
  3. The acceleration of the charged particle will depend on the radius of the coil.
  4. The charged particle will move with constant velocity.
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Official CBSE Marking Scheme · CBSE Sample Question Paper 2025-26

(D) The charged particle will move with constant velocity.

As charge particle is moving parallel to magnetic field, there will be no acceleration.

Q50.SQP 2026-27 · Q121 mark

A current of 5A is passing through a long wire which has semicircular loop of radius 10 cm as shown in the figure-

Long wire carrying 5 A with a semicircular loop of radius 10 cm

The magnetic field produced at the center of the loop is-

  1. 2 π µ T
  2. 4 π µ T
  3. 5 π µ T
  4. 8 π µ T
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Official CBSE Marking Scheme · CBSE Sample Question Paper 2026-27

(C) 5 π µ T

B = µ0I4R = 4π × 10−7 × 54 × 0.10 T = 5π × 10−6 T = 5 π µ T

Q51.2026 Main 55/1/1 · Q192 marks

A wire of length L is bent round into (i) a square coil having N turns and (ii) a circular coil having N turns. The coil in both cases is free to turn about a vertical axis coinciding with the plane of the coil, in a uniform, horizontal magnetic field and carry the same currents. Find the ratio of the maximum value of the torque acting on the square coil to that on the circular coil.

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Official CBSE Marking Scheme · CBSE Board 2026 (Main), Set 55/1/1

Side of square a = L4N and radius of circular coil r = L2πN [½]

τ = NIBA

τ1τ2 = NIBA1NIBA2 [½]

τ1τ2 = A1A2 [½]

τ1τ2 = (L/4N)2π(L/2πN)2

τ1τ2 = π4 [½]

Q52.2026 Supp 55/7/1 · Q192 marks

Two identical circular coils A and B of radius 10 cm each are placed concentrically and mutually perpendicular in XY and YZ planes respectively. Currents of 1 A and 3 A flow anticlockwise in the coils as seen from YZ plane. Find the magnitude of the net magnetic field at the common centre of the coils.

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Official CBSE Marking Scheme · CBSE Board 2026 (Supplementary), Set 55/7/1

B1 = μ0I2rk = 4π × 10-7 × 12 × 0.1k [½]

B1 = 2π × 10-6 k T

B2 = 4π × 10-7 × 32 × 0.1i [½]

B2 = 23 π × 10-6 i T

Bnet = B12 + B22 [½]

= 4π × 10-6 T

= 1.26 × 10-5 T [½]

Q53.2025 Main 55/5/1 · Q192 marks

A voltmeter of resistance 1000 Ω can measure up to 25 V. How will you convert it so that it can read up to 250 V ?

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The official marking scheme for this paper is not available in the source folder.

Q54.2024 Main 55/5/1 · Q182 marks

Derive an expression for magnetic force F acting on a straight conductor of length L carrying current I in an external magnetic field B. Is it valid when the conductor is in zig-zag form ? Justify.

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Official CBSE Marking Scheme · CBSE Board 2024 (Main), Set 55/5/1

Total number of mobile charge carriers in a conductor of length L, cross-sectional area A and number density of charge carriers n :

= nLA

Force acting on the charge carriers in external magnetic field B

F = (nAL) q vd × B ------(1) [½]

Where vd is the drift velocity of the charge carriers

Current flowing

I = vd q n A

IL = vd q n A L ------(2) [½]

On solving equation (1) and (2)

F = I(L × B) [½]

Yes, because this force can be calculated by considering zig-zag conductor as a collection of linear strips (dl) and summing them vectorially. [½]

Q55.2024 Supp 55/S/1 · Q182 marks

A long straight horizontal wire is carrying a current I. At an instant, an alpha particle at a distance r from it, is travelling with speed v parallel to the wire in a direction opposite to the current. Find the magnitude and direction of the force experienced by the particle at this instant.

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Official CBSE Marking Scheme · CBSE Board 2024 (Supplementary), Set 55/S/1

Magnetic field due to current carrying wire at distance r: B = μ0I2πr [½]

Charge on alpha particle, q = 2e [½]

Force acting on a charged particle in motion in a uniform magnetic field: F = q(v × B)

∵ θ = 90°

∴ F = 2ev(μ0I2πr) [½]

F = μ0evIπr [½]

Direction of the force on alpha particle is away from the wire.

Q56.2023 Main 55/1/1 · Q192 marks

An alpha particle is projected with velocity v = (3·0 × 105 m/s) i into a region in which magnetic field B = [(0·4 T) i + (0·3 T) j] exists. Calculate the acceleration of the particle in the region. i, j and k are unit vectors along x, y and z axis respectively and charge to mass ratio for alpha particle is 4·8 × 107 C/kg.

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Official CBSE Marking Scheme · CBSE Board 2023 (Main), Set 55/1/1

Calculation of acceleration of alpha particle [2]

F = q(v × B) [½]

= q (3 × 105 i × (0.4 i + 0.3 j)) N

F = q (0.9 × 105 k) N [½]

F = ma = q (0.9 × 105 k) N

a = qm (0.9 × 105 k) m s−2 [½]

= 4.8 × 107 × 0.9 × 105 k m s−2 [½]

= 4.32 × 1012 k m s−2

Note: Deduct ½ mark if a student does not mention the direction of acceleration.

Q57.2023 Main 55/2/1 · Q192 marks

Briefly explain why and how a galvanometer is converted into an ammeter.

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Official CBSE Marking Scheme · CBSE Board 2023 (Main), Set 55/2/1

Why: Due to very high sensitivity [1]

Alternatively

It has large resistance and hence will change the value of current in circuit.

How: A galvanometer can be converted into an ammeter of desired range by connecting a shunt of proper value across its coil. [1]

Q58.2023 Main 55/3/1 · Q192 marks

A wire of length l is in the form of a circular loop A of one turn. This loop is reshaped into loop B of three turns. Find the ratio of the magnetic fields at the centres of loop A and loop B for the same current through them.

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Official CBSE Marking Scheme · CBSE Board 2023 (Main), Set 55/3/1

B = μ0NI2r [½]

l = 2πr ; 2πrA = 3(2πrB)   ⇒ rB = rA3 [½]

BABB = μ0NAI2rA × 2rBμ0NBI [½]

BABB = 19 [½]

Q59.2023 Main 55/5/1 · Q23 (option 1)2 marks

Write the expression for the Lorentz force on a particle of charge q moving with a velocity v in a magnetic field B. When is the magnitude of this force maximum ? Show that no work is done by this force on the particle during its motion from a point r1 to point r2.

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Official CBSE Marking Scheme · CBSE Board 2023 (Main), Set 55/5/1

Fm = q(v × B) [½]

Fm = q v B sinθ

Force is maximum for θ = 90° (v ⊥ B) [½]

As magnetic force always acts in a direction perpendicular to the velocity vector, hence no work is done by this force on the particle during motion. [1]

Alternatively

W = F · (r2 – r1) = F |r2 – r1| cos 90° = 0

Q60.2023 Main 55/5/1 · Q23 (OR option)2 marks

A long straight wire AB carries a current I. A particle (mass m and charge q) moves with a velocity v, parallel to the wire, at a distance d from it as shown in the figure. Obtain the expression for the force experienced by the particle and mention its directions.

Wire AB with current I upward; particle at distance d moving with velocity v downward, parallel to wire
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Official CBSE Marking Scheme · CBSE Board 2023 (Main), Set 55/5/1

The magnetic field produced by current carrying conductor AB: B = μ0I2πd, directed into the plane of paper. [½]

Force experienced by charged particle Fm = q(v × B) [½]

∵ v ⊥ B

Fm = q v B sin 90°

Fm = q v × μ0I2πd [½]

Force is repulsive / acts towards right / away from the conductor. [½]

Q61.SQP 2025-26 · Q20 (option 1)2 marks

Write an expression for the magnetic force per unit length between two parallel thin current carrying wires. Hence define one ampere.

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Official CBSE Marking Scheme · CBSE Sample Question Paper 2025-26

FL = μ0I1I22πr (I1 is the current in first wire and I2 is the current in second wire) [1]

Thus we define ampere as the current flowing in each conductor separated by a unit distance so that one conductor applies a force of 2 × 10-7 N on a unit length of another parallel conductor. [1]

Q62.2026 Main 55/2/1 · Q25 (OR option)3 marks

Two long straight parallel conductors A and B carrying steady currents Ia and Ib in the same direction are separated by a distance d. Deduce the expressions for the force acting on length L of conductor B due to conductor A and show it in figure. Write the expression for the force acting on length L of conductor A due to conductor B and show that it follows Newton’s third law.

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Official CBSE Marking Scheme · CBSE Board 2026 (Main), Set 55/2/1
Parallel conductors A and B with currents Ia, Ib: fields B_a, B_b and forces F_AB, F_BA

[1] for figure

Note : Deduct ½ mark if direction of force on conductor B is not shown.

Magnetic field due to current carrying conductor A at all points along the conductor B : Ba = μ0Ia2πd (–k) [½]

Force experienced by a segment of length L of conductor B :

FBA = Ib L × Ba

FBA = IbL μ0Ia2πd [i × (–k)]

FBA = μ0IaIbL2πd j [½]

Similarly FAB = μ0IaIbL2πd (–j) [½]

FBA = –FAB [½]

It follows Newton’s third law.

Alternatively :

Alternative figure: parallel conductors A and B at separation d with forces F_AB and F_BA towards each other

Ba = μ0Ia2πd vertically inward

FBA = Ib L Ba = μ0IaIb2πd L directed towards A

FAB = Ia L Bb = μ0IaIbL2πd

It is equal in magnitude to FBA but directed towards the conductor ‘B’.

Thus FBA = –FAB. Thus, they follow Newton’s third law.

Note : Award 1 mark for the diagram, if the directions are not indicated in the diagram but written

Q63.2026 Main 55/2/1 · Q283 marks

Derive an expression for the magnetic field B, due to a circular coil of N turns, each of radius r carrying current I, at a distance ‘x’ from the centre along its axis.

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Official CBSE Marking Scheme · CBSE Board 2026 (Main), Set 55/2/1
Circular current loop of radius r with element dl, field dB at axial point P at distance x, and components dB_x, dB_perp

[½] for figure

The magnitude of magnetic field (dB) due to current element Idl is

dB = μ04π I|dl × a|a3 = μ04π I dl(r2 + x2) [½]

The components of the magnetic field perpendicular to the x-axis (dB⊥) are summed over and get cancelled. [½]

The components of the magnetic field along the x-axis contribute towards the net magnetic field

Bx = ∫ dB cosθ = ∫ μ04π I dl(x2 + r2) · rx2 + r2 [½]

B = Bxi = μ0Ir22(x2 + r2)3/2 (i) [½]

For N number of turns, net magnetic field

Bnet = μ0NIr22(x2 + r2)3/2 (i) [½]

Q64.2026 Main 55/3/1 · Q23 (option 1)3 marks

A circular coil of 30 turns and radius 8·0 cm carrying a current of 6 A is suspended vertically in a uniform horizontal magnetic field of 1·0 T. The field lines make an angle of 30° with the plane of the coil. Calculate the magnitude of the external torque that must be applied to prevent the coil from turning. What would happen if the circular coil is replaced by a planar coil of irregular shape that encloses the same area, keeping other parameters unchanged ?

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Official CBSE Marking Scheme · CBSE Board 2026 (Main), Set 55/3/1

τ = NIBA sinθ [½]

= 30 × 6 × 1 × 3·14 × (8 × 10−2)2 × sin 60° [½]

= 90 × 64 × 3·14 × 1·73 × 10−4 [½]

= 3·13 Nm [½]

Torque will remain the same. [1]

Q65.2026 Main 55/3/1 · Q23 (OR option)3 marks

An alpha particle (mass 6·4 × 10−27 kg and charge 3·2 × 10−19 C) having 8·0 MeV energy, enters a region of a uniform magnetic field of 0·5 T. If the field is directed perpendicular to the velocity of the particle, find the radius of the circular path described by the particle. Mention the condition under which the particle in this region (i) describes a helical path, and (ii) goes straight undeviated.

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Official CBSE Marking Scheme · CBSE Board 2026 (Main), Set 55/3/1

K.E. = q2B2r22m [½]

r = 2 × K.E. × mqB [½]

= 2 × 8 × 1·6 × 10−19 × 106 × 6·4 × 10−273·2 × 10−19 × 0·5 [½]

= 12·8 × 10−201·6 × 10−19 [½]

= 8 × 10−1 m

= 80 cm [½]

(i) Particle enters the magnetic field at angle 0° < θ < 90° [½]

(Alternatively) If a component of velocity is parallel or antiparallel to the magnetic field and another component is perpendicular to the magnetic field.

(ii) Particle moves parallel or antiparallel to the magnetic field. [½]

Q66.2026 Main 55/4/1 · Q23 (option 1)3 marks

(a) The figure given below shows three straight long parallel conductors ①, ② and ③ kept in x-y plane, carrying currents 2I, I and 3I respectively as shown in figure.

Three parallel conductors 1, 2, 3 in the x-y plane, separated by d, carrying 2I, I (both along +x) and 3I (along −x); axes x, y, z

Find the magnitude and direction of :

  1. net magnetic field at a point on conductor ① and
  2. net magnetic force acting on unit length of conductor ①, due to conductors ② and ③.
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Official CBSE Marking Scheme · CBSE Board 2026 (Main), Set 55/4/1

(i) Magnetic field due to conductor ② at a point on conductor ① is B1 = μ0I2πd along positive z-axis. [½]

Magnetic field due to conductor ③ at a point on conductor ① is B2 = 3μ0I4πd along negative z-axis. [½]

Magnitude of net magnetic field is B2 − B1 = 3μ0I4πd − μ0I2πd = μ0I4πd along negative z-axis.

Direction will be along negative z axis / into the plane of the paper. [½]

(ii) Force per unit length on conductor ① due to conductor ② = μ0I1I22πd = μ0I2πd (attractive) [½]

Force per unit length on conductor ① due to conductor ③ = 3μ0I22πd (repulsive)

Magnitude of net force per unit length on conductor ① due to ② & ③ = μ0I22πd [½]

Direction of the net force per unit length will be repulsive or along positive y-axis. [½]

Q67.2026 Main 55/5/1 · Q233 marks

Write the expression for the magnetic field due to a current element in vector form. Consider a 1 cm segment of a wire, centered at the origin, carrying a current of 10 A in positive x-direction. Calculate the magnetic field B at a point (1 m, 1 m, 0).

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Official CBSE Marking Scheme · CBSE Board 2026 (Main), Set 55/5/1

Writing the expression for magnetic field in vector form [1]; Calculating magnetic field B at a given point [2]

Magnetic field due to a current element (dl) at a distance r,

dB = μ04π I (dl × r)r3 [1]

dB = μ0 i dl sinθ4πr2 [½]

r = 2 m [½]

dB = 10–7 × 10 × 10–2 sin45°2 [½]

dB = 3.53 × 10–9 T

Direction : along +ve Z-axis [½]

Alternatively

r = (i + j) m, r = 2 m

dl = 10–2 m; I dl = 10 × 10–2 i Am

dB = μ0 I (dl × r)4πr3 = 4π × 10–74π × 10[(10–2) i × (i + j)](2)3

= 10–722 (0.1)(0 + k) = 52 × 10–9 k

dB = 3.53 × 10–9 k T

Q68.2026 Supp 55/7/1 · Q263 marks
  1. Derive the expression for the magnetic field (near its centre) due to an air-filled long solenoid carrying a current I and having n number of turns per unit length. [2]
  2. A charged particle moving with velocity v enters a region of uniform magnetic field B. The force acting on it is non-zero. Would the particle gain any energy ? Explain. [1]
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Official CBSE Marking Scheme · CBSE Board 2026 (Supplementary), Set 55/7/1
Rectangular Amperian loop abcd on a long solenoid

(a) Consider a rectangular Amperian loop abcd. Along cd, the magnetic field is zero. Along the transverse sections bc and ad, the field component is zero. [½]

Let n = number of turns per unit length, then the enclosed current is Ie = I(nh) [½]

From Ampere’s Circuital Law, BL = μ0Ie [½]

Bh = μ0I(nh)

B = μ0nI [½]

Note: 2 marks should be awarded to the students for writing the formula.

(b) The particle will not gain energy. [½]

Since the force on charged particle is non zero, it will describe a circular path. Work done by the magnetic force will be zero. [½]

Q69.2025 Main 55/1/1 · Q233 marks

Using Biot-Savart law, derive expression for the magnetic field (B) due to a circular current carrying loop at a point on its axis and hence at its centre.

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Q70.2025 Main 55/2/1 · Q23 (option 1)3 marks

In a region of a uniform electric field E, a negatively charged particle is moving with a constant velocity v = –v0 i near a long straight conductor coinciding with XX′ axis and carrying current I towards –X axis. The particle remains at a distance d from the conductor.

  1. Draw diagram showing direction of electric and magnetic fields.
  2. What are the various forces acting on the charged particle ?
  3. Find the value of v0 in terms of E, d and I.
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Q71.2025 Main 55/2/1 · Q23 (OR option)3 marks

Two infinitely long conductors kept along XX′ and YY′ axes are carrying current I1 and I2 along –X axis and –Y axis respectively. Find the magnitude and direction of the net magnetic field produced at point P(X, Y).

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Q72.2025 Main 55/4/1 · Q233 marks
  1. Write vector form of Biot-Savart law.
  2. Two insulated long straight wires, each carrying 2·0 A current are kept along xx′ and yy′ axis as shown in the figure. Find the magnitude and direction of resultant magnetic field at point P (4m, 5m).
Two long insulated wires along xx′ and yy′ axes, each carrying current I (towards +x and towards +y), point P at (4, 5), z-axis out of the page
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Q73.2025 Main 55/6/1 · Q233 marks

A rectangular loop carries a current of 1 A. A straight long wire carrying 2 A current is kept near the loop in the same plane as shown in the figure.

Rectangular loop 1 cm by 5 cm carrying 1 A, 1 cm from a long wire carrying 2 A

Find :

  1. the torque acting on the loop, and
  2. the magnitude and direction of the net force on the loop.
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Q74.2025 Main 55/7/1 · Q23 (option 1)3 marks

(i) Write Biot-Savart’s law in vector form.

(ii) Two identical circular coils A and B, each of radius R, carrying currents I and 3 I respectively, are placed concentrically in XY and YZ planes respectively. Find the magnitude and direction of the net magnetic field at their common centre.

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Q75.2025 Main 55/7/1 · Q23 (OR option)3 marks

(i) A rectangular loop of sides l and b carries a current I clockwise. Write the magnetic moment m of the loop and show its direction in a diagram.

(ii) The loop is placed in a uniform magnetic field B and is free to rotate about an axis which is perpendicular to B. Prove that the loop experiences no net force, but a torque τ = m × B.

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Q76.2025 Supp 55/S/1 · Q25 (OR option)3 marks

An electron (mass 9 × 10–31 kg and charge 1·6 × 10–19 C) is moving in a circle with a speed of 3·2 × 107 m/s in a magnetic field of 4·5 × 10–4 T perpendicular to it.

Calculate :

(i) the radius of the circular path,

(ii) its frequency of rotation, and

(iii) its energy.

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Official CBSE Marking Scheme · CBSE Board 2025 (Supplementary), Set 55/S/1

(i) r = mvqB [½]

= 9 × 10–31 × 3.2 × 1071.6 × 10–19 × 4.5 × 10–4

= 0.4 m [½]

(ii) v = v2πr [½]

= 3.2 × 1072π × 0.4

= 4π × 107 Hz [½]

(iii) E = 12 mv2 [½]

= 12 × 9 × 10–31 × 3.2 × 3.2 × 1014

= 4.608 × 10–16 J [½]

Q77.2025 Supp 55/S/1 · Q283 marks

A rectangular loop (4 cm × 5 cm) of wire carrying current of 0·25 A is placed in a magnetic field B = [(0·20 T) j + (0·50 T) k] such that outward unit vector normal to the plane of the loop is (0·80 i – 0·60 j).

Calculate :

(a) the magnetic moment m of the loop, and

(b) the torque τ acting on the loop.

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Official CBSE Marking Scheme · CBSE Board 2025 (Supplementary), Set 55/S/1

(a) m = IA = I |A| n [½]

= 0.25 A × (4 × 5 × 10–4 m2) [0.8i – 0.6j] [½]

= 5 × 10–4 (0.8i – 0.6j) A m2 [½]

(b) τ = m × B [½]

= 5 × 10–4(0.8i – 0.6j) × (0.2j + 0.5k) N m [½]

= 5 × 10–4(0.16k – 0.4j – 0.3i) N m

= 5 × 10–4(–0.3i – 0.4j + 0.16k) N m [½]

Q78.2024 Main 55/1/1 · Q243 marks

A proton with kinetic energy 1·3384 × 10−14 J moving horizontally from north to south, enters a uniform magnetic field B of 2·0 mT directed eastward. Calculate :

(a) the speed of the proton
(b) the magnitude of acceleration of the proton
(c) the radius of the path traced by the proton

[Take (q/m) for proton = 1·0 × 108 C/kg]

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Official CBSE Marking Scheme · CBSE Board 2024 (Main), Set 55/1/1

(a) v = 2 × K.E.m [½]

= 4 × 106 m/s [½]

(b) acceleration = qvB / m [½]

= 8 × 1011 m/s2 [½]

(c) r = mv / Bq [½]

= 20 m [½]

Q79.2024 Main 55/2/1 · Q25 (option 1)3 marks

Two long, straight, parallel conductors carry steady currents in opposite directions. Explain the nature of the force of interaction between them. Obtain an expression for the magnitude of the force between the two conductors. Hence define one ampere.

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Official CBSE Marking Scheme · CBSE Board 2024 (Main), Set 55/2/1

Nature of force is repulsive. [½]

Two parallel conductors a and b carrying opposite currents I_a and I_b a distance d apart, with field B and force F_ba

Magnetic field due to current Ia at all points of conductor b
Bab = µ0Ia2πd, directed downwards [½]

Force experienced by conductor b on its segment of length l
Fab = Ib l Bab = µ0IaIb2πd l, directed towards left [½]

Similarly
Force experienced by conductor a on its segment of length l
Fba = µ0IaIb2πd l, directed towards right [½]

One ampere is that steady current which when maintained in each of two very long straight parallel conductors of negligible cross-section, placed one metre apart in vacuum produces a force of 2 × 10−7 N/m on each conductor. [1]

Q80.2024 Main 55/2/1 · Q25 (OR option)3 marks

Obtain an expression for the torque τ acting on a current carrying loop in a uniform magnetic field B. Draw the necessary diagram.

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Official CBSE Marking Scheme · CBSE Board 2024 (Main), Set 55/2/1

Diagram [1]

Rectangular current loop ABCD in a uniform magnetic field B between N and S poles, with forces F1 and F2 forming a couple

Forces on arm BC and DA are equal and opposite and act along the axis of the coil. Being collinear they cancel each other. [½]

Forces on arms AB and CD are equal and opposite but not collinear. They form a couple.
F1 = F2 = IbB [½]

τ = F1 a2 sin θ + F2 a2 sin θ [½]

τ = IabB sin θ
τ = IAB sin θ (where A = ab & m = IA) [½]

τ = m × B [½]

Q81.2024 Main 55/3/1 · Q233 marks

Two long straight parallel conductors carrying currents, exert a force on each other. Why ? Derive an expression for the force per unit length between two long straight parallel conductors carrying currents in opposite directions. Explain the nature of the force between these conductors.

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Official CBSE Marking Scheme · CBSE Board 2024 (Main), Set 55/3/1

Reason for exerting force on straight parallel conductors [½]; Derivation for force per unit length [2]; Explanation of nature of force [½]

One conductor experiences a force due to magnetic field of the other conductor [½]

Two parallel conductors a and b carrying opposite currents I_a and I_b separated by d, with B and F_ba shown on b

Magnetic field produced by conductor ‘a’ at all points along the length of conductor ‘b’

Ba = μ0Ia2πd [½]

Force on conductor ‘b’ due to this magnetic field

Fba = IbLBa [½]

Fba = μ0IaIbL2πd

fba = FbaL = μ0IaIb2πd   directed away from a [½]

fab = FabL = μ0IaIb2πd   directed away from b

Repulsive, the forces acting on them are away from each other. [½]

Q82.2024 Main 55/4/1 · Q263 marks

A circular loop of radius 10 cm carrying current of 1.0 A lies in x-y plane. A long straight wire lies in the same plane parallel to x-axis at a distance of 20 cm as shown in figure.

Circular loop of radius 10 cm with 1.0 A current, centre O, and a straight wire 20 cm below O parallel to x-axis

Find the direction and value of current that has to be maintained in the wire so that the net magnetic field at O is zero.

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Official CBSE Marking Scheme · CBSE Board 2024 (Main), Set 55/4/1

Magnetic field due to circular current loop at its centre O.

B1 = μ0I12r [½]

= μ0 × 12 × 0.1 = 5μ0 T [½]

The magnetic field B1 is perpendicular to plane of loop and directed inwards.

Magnetic field due to long current carrying straight wire at O.

B2 = μ0I22πr [½]

B2 = μ0I22π × 0.2 T

For net magnetic field at O to be zero, B1 should be equal and opposite to B2. [½]

5μ0 = μ0I20.4π [½]

I2 = 2π A = 6.28 A [½]

Direction of current in the straight wire is along +ve x axis. [½]

Alternatively

Net magnetic field at O is zero.

Bloop = Bwire [½]

μ0I12r1 = μ0I22πr2 [1]

μ0 × 12 × 0.1 = μ0I22π × 0.2 [½]

I2 = 2π A = 6.28 A [½]

Direction of current is along + x-axis. [½]

Q83.2024 Main 55/5/1 · Q24 (OR option)3 marks
  1. State and explain Ampere’s circuital law.
  2. Two long straight parallel wires separated by 20 cm, carry 5 A and 10 A current respectively, in the same direction. Find the magnitude and direction of the net magnetic field at a point midway between them.
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Official CBSE Marking Scheme · CBSE Board 2024 (Main), Set 55/5/1

(i) Line integral of magnetic field over a closed loop in vacuum is equal to μ0 times the total current passing through the loop. [1]

Alternatively

∮ B · dl = μ0I

The integral in this expression is over a closed loop coinciding with the boundary of the surface.

(ii)

Two parallel wires carrying 5 A and 10 A in the same direction with point P midway between them

B = μ0I2πr [½]

Net magnetic field B = B2 – B1

B = μ0 × 10220π [10 – 5]

B = 4π × 10–7 × 102 × 520π [½]

B = 10–5 T [½]

Along the direction of magnetic field produced by the conductor carrying current 10A. [½]

Q84.2024 Main 55/5/1 · Q253 marks

An electron moving with a velocity v = (1.0 × 107 m/s)i + (0.5 × 107 m/s)j enters a region of uniform magnetic field B = (0.5 mT)j. Find the radius of the circular path described by it. While rotating; does the electron trace a linear path too ? If so, calculate the linear distance covered by it during the period of one revolution.

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Official CBSE Marking Scheme · CBSE Board 2024 (Main), Set 55/5/1

Radius of circular path

r = mvxeB [½]

r = 9.1×10–31 × 1×1071.6×10–19 × 0.5×10–3

= 11.38×10–2 m [½]

Yes, it traces a linear path too. [½]

Linear distance during period of one revolution

y = 2πmeB × vy [½]

= 2 × π × 9.1×10–31 × 0.5×1071.6×10–19 × 0.5×10–3 [½]

= 0.357 m

= 0.36 m [½]

Q85.2024 Supp 55/S/1 · Q273 marks

A galvanometer is converted into a voltmeter of range (0 – V) volt using a resistor of 9900 Ω. If a resistor of 4900 Ω is used, the range becomes half, i.e. (0 – V2) volt. Calculate :

(a) resistance of the galvanometer

(b) resistance required to convert it into a voltmeter of range (0 – 2V) volt.

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Official CBSE Marking Scheme · CBSE Board 2024 (Supplementary), Set 55/S/1

(a) V = Ig(R + G) [½]

V = Ig(9900 + G) ——— (1) [½]

V2 = Ig(4900 + G) ——— (2)

On solving above eq (1) and eq (2)

G = 100 Ω [1]

(b) 2V = Ig(R1 + G) ——— (3) [½]

From eq (1) and (3)

R1 = 19900 Ω [½]

Q86.2023 Main 55/4/1 · Q263 marks

Two circular loops A and B, each of radius 3 m, are placed coaxially at a distance of 4 m. They carry currents of 3 A and 2 A in opposite directions respectively. Find the net magnetic field at the centre of loop A.

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Official CBSE Marking Scheme · CBSE Board 2023 (Main), Set 55/4/1

B1 = μ0I2r

B1 = 4π × 10−7 × 32 × 3 = 2π × 10−7 T = 6.28 × 10−7 T [1]

B2 = μ0IR22(R2 + x2)3/2 = 2π × 10−7 × 2 × 9(32 + 42)3/2, opposite to B1

B2 = 36π × 10−7125 T = 0.9 × 10−7 T [1]

Bnet = B1 − B2 = 6.28 × 10−7 − 0.9 × 10−7

Bnet = 5.38 × 10−7 T [1]

Q87.2023 Main 55/5/1 · Q293 marks

(a) Briefly describe how the current sensitivity of a moving coil galvanometer can be increased.

(b) A galvanometer shows full scale deflection for current Ig. A resistance R1 is required to convert it into a voltmeter of range (0 – V) and a resistance R2 to convert it into a voltmeter of range (0 – 2V). Find the resistance of the galvanometer.

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Official CBSE Marking Scheme · CBSE Board 2023 (Main), Set 55/5/1

(a) Current sensitivity of moving coil galvanometer can be increased by

(1) increasing number of turns in the coil (Is ∝ N)

(2) increasing magnetic field strength (Is ∝ B)

(3) using a material having lesser value of torsional constant (Is ∝ 1/k)

(4) increasing area of cross-section of the coil

Note: Any one of the above [1]

(b) R = VIg – G [½]

For range (0 – V): R1 = VIg – G ——— (1) [½]

For range (0 – 2V): R2 = 2VIg – G ——— (2) [½]

On solving equations (1) and (2): G = R2 – 2R1 [½]

Q88.2023 Supp 55/C/1 · Q273 marks

(a) It is not advisable to use a galvanometer as such to measure current directly. Why ?

(b) Why should the value of resistance connected in parallel to a galvanometer be low ?

(c) Is the reading shown by an ammeter in a circuit less than or more than the actual value of current flowing in the circuit ? Why ?

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Official CBSE Marking Scheme · CBSE Board 2023 (Supplementary), Set 55/C/1

(a) It will not measure accurate value of current because its high resistance will affect the current in the circuit. [1]

(b) To reduce the galvanometer resistance a small resistance is connected in parallel. [1]

(c) It is less than the actual value of current because it has some resistance [½ + ½]

Q89.SQP 2025-26 · Q27 (option 1)3 marks

If a point sized object having charge 1C and mass 1g is projected with velocity of 2i m/s from a point (0, 2cm, 0) in the region of magnetic field –0.1k T which spreads in the first quadrant.

Uniform magnetic field into the page in the first quadrant, charge q at 2 cm on the Y axis moving along +X with velocity v

(A) What will be the shape of the path followed by the given charged particle?
(B) At what point it will cross the X-axis?
(C) What will be the kinetic energy of particle when it will enter in the fourth quadrant?

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Official CBSE Marking Scheme · CBSE Sample Question Paper 2025-26

As charge particle is moving perpendicular to magnetic field it will follow circular trajectory in clock wise direction. Magnetic force will act as centripetal force.

Given:

Q = 1C;

M = 10-3 kg;

v = 2 m/s &

B = –0.1Tk

Radius of trajectory is given by

R = mvqB = 2cm

(A) Quarter Circle [1]

(B) It will cross the X axis at 2cm. [1]

(C) As work done by B is on charge particle is zero it’s kinetic energy (K) will remain same

K = ½mv2

Or, K = ½ × 10-3 × 22 J = 2 × 10-3 J [1]

Q90.SQP 2026-27 · Q27 (option 1)3 marks
  1. An ammeter has an internal resistance of 13 Ω and is originally designed to measure currents only up to 100 A. To extend its range for high-current measurements, a shunt resistor is connected in parallel with it. After adding the shunt, the same meter can now read currents as high as 750 A without damage. Using the principle of current division in parallel circuits, determine the value of the shunt resistance required to increase the measuring range of the ammeter.
  2. The coil of a moving coil galvanometer is wound over a metal frame. Why?
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Official CBSE Marking Scheme · CBSE Sample Question Paper 2026-27

I. The original ammeter can be treated as the galvanometer. [1]

S = Ig GI − Ig [½]

= 100 × 13750 − 100 = 1300650 = 2 Ω [½]

II. To provide electromagnetic damping. The induced currents set up in the metal frame help to damp the oscillation of the coil in the magnetic field. [1]

Q91.2026 Main 55/1/1 · Q29 (option 1)4 marks

A galvanometer is used to detect or/and measure small currents in an electrical circuit. It essentially works on the fact that a current-carrying coil experiences a deflecting torque when placed in a magnetic field. This deflection in the coil can be measured and it is related to the current flowing in the coil, the number of turns in the coil, area of the coil and the magnetic field. A hair spring attached to the coil provides a counter torque and helps in measuring the deflection. A galvanometer can be converted to an ammeter or a voltmeter of desired range by using suitable resistances.

(I) The torque on the coil remains constant irrespective of the coil’s orientation during rotation due to

  1. use of soft iron core which increases the magnetic field.
  2. radial magnetic field
  3. hair spring which provides the counter torque
  4. eddy current in the iron core which causes damping.

(II) The best way to increase current sensitivity of a galvanometer is by

  1. increasing number of turns of the coil
  2. increasing area of coil and magnetic field strength
  3. decreasing area of coil and magnetic field strength
  4. increasing torsional constant of the hair spring

(III) A moving coil galvanometer has a coil with area of cross-section 4.0 × 10–3 m2 and number of turns 50. The coil is rotating in a magnetic field of 0.25 T. The torque acting on the coil when a current of 5 A passes through it is

  1. 1.0 N m
  2. 2.0 N m
  3. 0.50 N m
  4. 0.25 N m

(IV) A galvanometer with coil of resistance 20 Ω shows full scale deflection for a current of 5 mA. To convert it into an ammeter of range (0 – 10 A), a resistance of

  1. 0.05 Ω should be connected in series with it.
  2. 0.05 Ω should be connected in parallel with it.
  3. 0.01 Ω should be connected in parallel with it.
  4. 0.01 Ω should be connected in series with it.
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Official CBSE Marking Scheme · CBSE Board 2026 (Main), Set 55/1/1

(I) (B) radial magnetic field [1]

(II) (A) increasing number of turns of the coil [1]

(III) (D) 0.25 N m [1]

(IV) (C) 0.01 Ω should be connected in parallel with it. [1]

Q92.2026 Main 55/1/1 · Q29 (OR option)4 marks

A galvanometer is used to detect or/and measure small currents in an electrical circuit. It essentially works on the fact that a current-carrying coil experiences a deflecting torque when placed in a magnetic field. This deflection in the coil can be measured and it is related to the current flowing in the coil, the number of turns in the coil, area of the coil and the magnetic field. A hair spring attached to the coil provides a counter torque and helps in measuring the deflection. A galvanometer can be converted to an ammeter or a voltmeter of desired range by using suitable resistances.

(I) The torque on the coil remains constant irrespective of the coil’s orientation during rotation due to

  1. use of soft iron core which increases the magnetic field.
  2. radial magnetic field
  3. hair spring which provides the counter torque
  4. eddy current in the iron core which causes damping.

(II) The best way to increase current sensitivity of a galvanometer is by

  1. increasing number of turns of the coil
  2. increasing area of coil and magnetic field strength
  3. decreasing area of coil and magnetic field strength
  4. increasing torsional constant of the hair spring

(III) A galvanometer coil has a resistance of 15 Ω and the meter shows full scale deflection for a current of 3 mA. The value of resistance required to convert it into a voltmeter of range (0 – 12 V) is

  1. 4015 Ω
  2. 3985 Ω
  3. 415 Ω
  4. 385 Ω

(IV) A galvanometer with coil of resistance 20 Ω shows full scale deflection for a current of 5 mA. To convert it into an ammeter of range (0 – 10 A), a resistance of

  1. 0.05 Ω should be connected in series with it.
  2. 0.05 Ω should be connected in parallel with it.
  3. 0.01 Ω should be connected in parallel with it.
  4. 0.01 Ω should be connected in series with it.
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Official CBSE Marking Scheme · CBSE Board 2026 (Main), Set 55/1/1

(I) (B) radial magnetic field [1]

(II) (A) increasing number of turns of the coil [1]

(III) (B) 3985 Ω [1]

(IV) (C) 0.01 Ω should be connected in parallel with it. [1]

Q93.2026 Main 55/3/1 · Q304 marks

A charged particle +q in an electric field E experiences a force in the direction of the electric field. As a result, its kinetic energy changes. Similarly, the charged particle also experiences a force when it moves in a magnetic field B. But this magnetic force is perpendicular to both velocity v of the charged particle and the magnetic field B, so it cannot change the kinetic energy of the charged particle. Consider two charged particles 1 and 2 of masses m and m2 having charges −q and +2q respectively. They are accelerated from rest through the same potential difference V and acquire kinetic energy K1 and K2. Then they enter in a region of uniform magnetic field B perpendicular to their velocities.

(i) The ratio of their kinetic energies (K1K2) is :

  1. 12
  2. 14
  3. 4
  4. 1

(ii) The ratio of the radii of the circular paths described by them (r1r2) is :

  1. 12
  2. 2
  3. 12
  4. 2

(iii) Suppose particles 1 and 2 enter the magnetic field B = B0k with velocities v1 = v1i and v2 = v2i. Then :

  1. both particles revolve clockwise
  2. both particles revolve anticlockwise
  3. particle 1 revolves clockwise while particle 2 revolves anticlockwise
  4. particle 1 revolves anticlockwise while particle 2 revolves clockwise

(iv) Answer either (a) or (b) :

(a) If period of revolution for particle 1 is 4 s, then for particle 2, the period will be :

  1. 1 s
  2. 2 s
  3. 4 s
  4. 8 s

(b) If the value of momentum for particles 1 and 2 are p1 and p2, then :

  1. p1 = p22
  2. p1 = p2
  3. p1 = 2p2
  4. p1 = 4p2
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Official CBSE Marking Scheme · CBSE Board 2026 (Main), Set 55/3/1

(i) (A) 12 [1]

(ii) (D) 2 [1]

(iii) (D) Particle 1 revolves anticlockwise while particle 2 revolves clockwise [1]

(iv) (a) (A) 1 s OR (b) (B) p1 = p2 [1]

Q94.2026 Supp 55/7/1 · Q294 marks

A particle of mass m and charge q experiences a force Fe in an electric field E and it is directed along the field. When the particle is moving with velocity v in a magnetic field B, it experiences a force FB, that is perpendicular to both v and B.

In presence of both fields, the force acting on the particle is the vector sum of that due to E and that due to B.

A current-carrying conductor also experiences a force in presence of a magnetic field. The force dF acting on a current-element I dl is perpendicular to both the current element, and the magnetic field B. The net force acting on a conductor of a given length is the vector sum of all dFs.

  1. A negative ion enters a region of uniform magnetic field B as shown in the figure.
    Negative charge -q moving with velocity v to the right in a magnetic field B into the plane of the paper
    It will :
    1. go straight, undeviated
    2. describe a circular path
    3. describe a semicircular path
    4. describe a helical path
    1. An electron is moving in a circular path in a uniform magnetic field of 3·14 × 10-4 T. Its period of revolution will be approximately :
      1. 0·11 μs
      2. 0·28 μs
      3. 0·014 μs
      4. 0·024 μs
    2. OR An alpha particle (mass 6·4 × 10-27 kg) moves in a circular path with a speed of 6·28 × 105 m/s in a uniform magnetic field of 3·14 × 10-3 T. The radius of the path is :
      1. 1·00 m
      2. 2·00 m
      3. 3·00 m
      4. 4·00 m
  2. A charged particle enters with a velocity v = (6 × 106 m/s) i in a region of magnetic field B = (0·5 G) j and an electric field E. If the particle goes straight undeviated in the region, then E equals :
    1. (120 NC) k
    2. – (300 NC) k
    3. (600 NC) j
    4. – (150 NC) i
  3. A current-element (I dl) of mass per unit length (m) is held in air as shown in the figure. After withdrawing the support, it will remain suspended in its position, when subjected to an external magnetic field B equal to :
    Current element I dl along the x-axis, with z-axis vertical
    1. mgdlIk
    2. mgIj
    3. – mgI dli
    4. – (I dlmg) k
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Official CBSE Marking Scheme · CBSE Board 2026 (Supplementary), Set 55/7/1

(i) (B) describe a circular path [1]

(ii) (a) (A) 0.11 μs [1]

OR

(b) (D) 4.00 m [1]

(iii) (B) – (300 NC) k [1]

(iv) (B) mgIj [1]

Q95.2025 Main 55/5/1 · Q294 marks

A galvanometer is an instrument used to show the direction and strength of the current passing through it. In a galvanometer, a coil placed in a magnetic field experiences a torque and hence gets deflected when a current passes through it. The name is derived from the surname of Italian scientist L. Galvani, who in 1791 discovered that electric current makes a dead frog’s leg jerk. A spring attached with the coil provides a counter torque.

In equilibrium, the deflecting torque is balanced by the restoring torque of the spring and we have :

NBAI = kφ

where N is the total number of turns in the coil
A is the area of cross-section of each turn
B is the radial magnetic field
k is the torsional constant of the spring
φ is the angular deflection of the coil

As the current (Ig) which produces full scale deflection in the galvanometer is very small, the galvanometer cannot as such be used to measure current in electric circuits. A small resistance, called shunt, of a suitable value is connected with the galvanometer to convert it into an ammeter of desired range. By using a higher resistance, a galvanometer can also be converted into a voltmeter.

(i) The value of the current sensitivity of a galvanometer is given by :

  1. kNBA
  2. NBAk
  3. kBAN
  4. kNBA

(ii) A galvanometer of resistance 6 Ω shows full scale deflection for a current of 0·2 A. The value of shunt to be used with this galvanometer to convert it into an ammeter of range (0 – 5 A) is :

  1. 0·25 Ω
  2. 0·30 Ω
  3. 0·50 Ω
  4. 6·0 Ω

(iii) The value of resistance of the ammeter in case (ii) will be :

  1. 0·20 Ω
  2. 0·24 Ω
  3. 6·0 Ω
  4. 6·25 Ω

(iv)(a) A galvanometer is converted into a voltmeter of range (0 – V) by connecting with it, a resistance R1. If R1 is replaced by R2, the range becomes (0 – 2 V). The resistance of the galvanometer is :

  1. (R2 – 2R1)
  2. (R2 – R1)
  3. (R1 + R2)
  4. (R1 – 2R2)

(iv)(b) (alternative to (iv)(a)) A current of 5 mA flows through a galvanometer. Its coil has 100 turns, each of area of cross-section 18 cm2 and is suspended in a magnetic field 0·20 T. The deflecting torque acting on the coil will be :

  1. 3·6 × 10–3 Nm
  2. 1·8 × 10–4 Nm
  3. 2·4 × 10–3 Nm
  4. 1·2 × 10–4 Nm
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Q96.2023 Main 55/2/1 · Q34 (OR option)4 marks

Briefly explain various ways to increase the strength of magnetic field produced by a given solenoid.

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Official CBSE Marking Scheme · CBSE Board 2023 (Main), Set 55/2/1

Ways to increase strength of magnetic field produced by a solenoid.

• By inserting soft iron core inside the solenoid. [1]

• By increasing current in the solenoid. [1]

Q97.2026 Main 55/5/1 · Q32 (option 1)5 marks
  1. A rectangular loop of sides a and b carrying current I is placed in a magnetic field B such that its area vector A makes an angle θ with B. With the help of a suitable diagram, show that the torque τ acting on the loop is given by τ = m × B, where m (= I A) is the magnetic dipole moment of the loop.
  2. A circular coil of 100 turns and radius (10π) cm carrying current of 5·0 A is suspended vertically in a uniform horizontal magnetic field of 2·0 T. The field makes an angle 30° with the normal to the coil. Calculate :
    1. the magnetic dipole moment of the coil, and
    2. the magnitude of the counter torque that must be applied to prevent the coil from turning.
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(i) Showing the torque acting on the loop τ = m × B [3]; (ii)(I) Calculating magnetic dipole moment of the coil [1]; (II) Calculating the magnitude of counter torque [1]

(i)

Rectangular loop in magnetic field between poles N and S with forces F1 and F2 forming a couple

[½]

According to Right Hand Screw Rule, forces on BC & DA are equal and opposite in direction with same line of action, hence they cancel out each other. [½]

Force on arms AB & CD, are also equal and opposite in direction but there is a difference in line of action so, a torque will act on it. [½]

F1 = F2 = I b B

τ = Force × perpendicular distance [½]

τ = F1 (a/2) sinθ + F2 (a/2) sinθ = (IbB) a sinθ

τ = I A B sinθ

m = IA; τ = m B sinθ [½]

τ = m × B [½]

Alternatively,

Alternative diagram of rectangular loop PQRS in field B with forces F forming a couple

According to Fleming’s left-hand rule, the magnitude of force on sides PS and QR are equal and opposite to each other, so they cancel each other. The side PQ experiences a normal inward force equal to IbB while the side RS experiences an equal normal outward force. These two forces form a couple which exerts a torque given by:

τ = Force × perpendicular distance = IbB × a sinθ = IA B sinθ

But IA = m, the magnetic moment of the loop

τ = mB sinθ

τ = m × B

(ii)(I) m = NIA = 100 × 5 × π (10 × 10–2π)2 = 5 Am2 [½ + ½]

(II) τ = mB sinθ = 5 × 2 × sin30 [½] = 5 Nm [½]

Q98.2026 Main 55/5/1 · Q32 (OR option)5 marks
  1. Derive an expression for the force F acting on a conductor of length L and area of cross-section A carrying current I and placed in a magnetic field B.
  2. A part of a wire carrying 2·0 A current and bent at 90° at two points is placed in a region of uniform magnetic field B = – (0·50 T) k, as shown in the figure. Calculate the magnitude of the net force acting on the wire.
Wire carrying 2.0 A bent at 90 degrees twice in a region of field into the page; 20 cm and 50 cm marked; axes i, j, k
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(i) Deriving expression for the force acting on a conductor placed in magnetic field [2½]; (ii) Calculating the magnitude of net force acting on the wire [2½]

(i) Let the number density of the free electron in the conductor be n. So, the total charge of conductor, q = nALe [½]

Total force acting on q placed in field B, moving with velocity vd: [½]

F = q (v × B) [1]

F = n(AL)e(vd × B)

Here, ne vd is current density (J) and |ne vd|A is current in the conductor. Therefore

F = [J(AL)] × B; F = I(L × B) [½]

(ii) The force acting on the wire l1 (50 cm) placed along x-axis

F1 = I(l1 × B) = 2 (50 × 10–2) i × (–0.5) k = 0.5 j N [½ + ½]

The force acting on the wire l2 (20 cm) placed along y-axis

F2 = I(l2 × B) = 2 (20 × 10–2) (–j) × (–0.5) k = 0.2 i N [½]

As F1 and F2 are perpendicular to each other, the resultant force on the wire will be

F = F12 + F22 = 0.25 + 0.04 [½] = 0.29 N [½]

Alternatively,

F = F1 + F2 = I(lx × B) + I(ly × B) = 0.5 j + 0.2 i

F = F12 + F22 = 0.25 + 0.04 = 0.29 N

Q99.2025 Main 55/1/1 · Q32 (option 1)5 marks

(i) A proton moving with velocity V in a non-uniform magnetic field traces a path as shown in the figure.

Path of a proton in the plane of the paper passing through points P, Q and R

The path followed by the proton is always in the plane of the paper. What is the direction of the magnetic field in the region near points P, Q and R ? What can you say about relative magnitude of magnetic fields at these points ?

(ii) A current carrying circular loop of area A produces a magnetic field B at its centre. Show that the magnetic moment of the loop is 2 BAμ0 Aπ.

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Q100.2025 Main 55/1/1 · Q32 (OR option)5 marks

(i) Derive an expression for the torque acting on a rectangular current loop suspended in a uniform magnetic field.

(ii) A charged particle is moving in a circular path with velocity V in a uniform magnetic field B. It is made to pass through a sheet of lead and as a consequence, it looses one half of its kinetic energy without change in its direction. How will (1) the radius of its path (2) its time period of revolution change ?

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Q101.2025 Main 55/2/1 · Q32 (option 1)5 marks
  1. What is the source of force acting on a current-carrying conductor placed in a magnetic field ? Obtain the expression for force acting between two long straight parallel conductors carrying steady currents and hence define ‘ampere’.
  2. A point charge q is moving with velocity v in a uniform magnetic field B. Find the work done by the magnetic force on the charge.
  3. Explain the necessary conditions in which the trajectory of a charged particle is helical in a uniform magnetic field.
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Q102.2025 Main 55/2/1 · Q32 (OR option)5 marks
  1. A current carrying loop can be considered as a magnetic dipole placed along its axis. Explain.
  2. Obtain the relation for magnetic dipole moment M of current carrying coil. Give the direction of M.
  3. A current carrying coil is placed in an external uniform magnetic field. The coil is free to turn in the magnetic field. What is the net force acting on the coil ? Obtain the orientation of the coil in stable equilibrium. Show that in this orientation the flux of the total field (field produced by the loop + external field) through the coil is maximum.
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Q103.2025 Main 55/7/1 · Q32 (option 1)5 marks

(i) With the help of a labelled diagram, explain the principle of working of a moving coil galvanometer. Write the purpose of using (I) radial magnetic field, and (II) soft iron core, in it.

(ii) Define current sensitivity of a galvanometer. “Increasing the current sensitivity may not necessarily increase the voltage sensitivity.” Give reason.

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Q104.2025 Main 55/7/1 · Q32 (OR option)5 marks

(i) (I) Write Ampere’s circuital law in mathematical form and explain the terms used.

(II) As the current carrying solenoid is made longer, the magnetic field produced outside it approaches zero. Why ?

(III) A flexible loop of irregular shape carrying current when located in an external magnetic field, changes to a circular shape. Give reason.

(ii) A galvanometer of resistance G is converted into a voltmeter to measure up to V volts, by connecting a resistance R1 in series with the coil. If R1 is replaced by R2, then it can only measure up to V2 volt. Find the value of the resistance R3 (in terms of R1 and R2) needed to convert it into a voltmeter that can read up to 2 V.

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Q105.2024 Main 55/2/1 · Q32 (option 1)5 marks

(i) A particle of mass m and charge q is moving with a velocity v in a magnetic field B as shown in the figure. Show that it follows a helical path. Hence, obtain its frequency of revolution.

Velocity v at an angle to the field B (along x-axis) in x, y, z axes

(ii) In a hydrogen atom, the electron moves in an orbit of radius 2 Å making 8 × 1014 revolutions per second. Find the magnetic moment associated with the orbital motion of the electron.

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Helical path of a charged particle in a uniform magnetic field B along x, showing v⊥, v∥, pitch and radius

(i) [½]

v⊥ = v sin θ is perpendicular to B and v∥ = v cos θ is parallel to B.
Due to v⊥ the charge describes circular path and v∥ pushes it in the direction of B. Therefore under the combined effect of two components the charged particle describes helical path, as shown in the figure. [1]

The centripetal force mv⊥2r = B q v⊥ [½]

v⊥ = Bqrm (v⊥ = v sin θ) [½]

Time period T = 2πrv⊥ = 2πmBq
frequency ν = 1T = Bq2πm [½]

(ii) Magnetic moment m = I A [½]

I = eT = eν = 1·6 × 10−19 × 8 × 1014 = 1·28 × 10−4 A [½]

M = 1·28 × 10−4 × 3·14 × (2 × 10−10)2 [½]

= 5·12π × 10−24 Am2 = 1·6 × 10−23 Am2 [½]

Q106.2024 Main 55/2/1 · Q32 (OR option)5 marks

(i) What is current sensitivity of a galvanometer ? Show how the current sensitivity of a galvanometer may be increased. “Increasing the current sensitivity of a galvanometer may not necessarily increase its voltage sensitivity.” Explain.

(ii) A moving coil galvanometer has a resistance 15 Ω and takes 20 mA to produce full scale deflection. How can this galvanometer be converted into a voltmeter of range 0 to 100 V ?

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Official CBSE Marking Scheme · CBSE Board 2024 (Main), Set 55/2/1

(i) Deflection produced per unit current is called its current sensitivity. [1]

IS = θI = NBAK

Current sensitivity can be increased by
(a) increasing number of turns in coil
(b) increasing area of coil in magnetic field
(c) decreasing K (Torsional Constant)
(any one) [1]

Vs = θV = NBAKR

If current sensitivity is increased by increasing number of turns of the coil, the resistance of the galvanometer will also increase. Thus voltage sensitivity may not increase. [1]

(ii) V = IG(R + G) [½]

R = VIG − G = 10020 × 10−3 − 15 [½]

= 5000 − 15 = 4985 Ω [½]

By connecting 4985 Ω in series with galvanometer it is converted to voltmeter of range (0–100 V). [½]

Q107.2024 Main 55/3/1 · Q32 (option 1)5 marks

(i) (1) What is meant by current sensitivity of a galvanometer ? Mention the factors on which it depends.

(2) A galvanometer of resistance G is converted into a voltmeter of range (0 – V) by using a resistance R. Find the resistance, in terms of R and G, required to convert it into a voltmeter of range (0 – V2).

(ii) The magnetic flux through a coil of resistance 5 Ω increases with time as :

φ = (2·0 t3 + 5·0 t2 + 6·0 t) mWb

Find the magnitude of induced current through the coil at t = 2 s.

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Official CBSE Marking Scheme · CBSE Board 2024 (Main), Set 55/3/1

(i)(1) Meaning of current sensitivity, mentioning factors [2]; (2) Finding the required resistance [1½]; (ii) Finding the induced current [1½]

(i) (1) Current sensitivity of galvanometer is defined as the deflection per unit current. [1]

Alternatively, φI = NBAK

Factors

Number of turns in coil, Magnetic field intensity, Area of coil, Torsional Constant (Any two) [½ + ½]

(2) R = VI – G   for (0-V) Range [½]

R1 = V2I – G   for (0-V2) Range [½]

VI = R + G

R1 = (R+G2) – G

R1 = R – G2 [½]

(ii) φ = (2.0t3 + 5.0t2 + 6.0t) mWb

|ε| = dφdt = 50 × 10–3 V [½]

I = |ε|R [½]

I = 50×10–35 A = 10 mA [½]

Q108.2023 Main 55/1/1 · Q32 (option 1)5 marks

(i) State Biot-Savart’s law for the magnetic field due to a current carrying element. Use this law to obtain an expression for the magnetic field at the centre of a circular loop of radius ‘a’ and carrying a current ‘I’. Draw the magnetic field lines for a current loop indicating the direction of magnetic field.

(ii) An electron is revolving around the nucleus in a circular orbit with a speed of 107 m s−1. If the radius of the orbit is 10−10 m, find the current constituted by the revolving electron in the orbit.

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i) Statement of Biot-Savart’s law [1]; Expression for magnetic field [2]; Diagram for magnetic field lines [½]; ii) Finding current by revolving electron [1½]

(i) The magnetic field at a point due to a current carrying element is proportional to magnitude of current, element length and inversely proportional to the square of the distance from the element. [1]

dB = μ04π I dl × rr3; |dB| = μ04π I dl sinθr2

Current element Idl, position vector r and field dB at point P

Consider a circular coil of radius a carrying current I.

Circular loop of radius a carrying current I with element Idl at A

[½]

According to Biot-Savart’s law |dB| = μ04π I dl sinθr2 [½]

At point A, I dl ⊥ a, ∴ θ = 90°, sin 90° = 1. Hence dB = μ04π I dla2 [½]

Magnetic field at centre B = ∫ dB = ∫02πa μ04π I dla2 = μ04π Ia2 × 2πa [½]

B = μ0I2a [½]

Note: Give full credit of 2 marks if a student derives the expression for magnetic field at the axis of the loop and then puts the distance of point as 0 from the centre.

Magnetic field lines of a current loop

[½]

(ii) q = e, v = 107 m s−1, r = 10−10 m

i = qT = qv2πr = ev2πr [½]

= 1.6 × 10−19 × 1072 × π × 10−10 [½]

= 0.8π × 10−2 A = 0.255 × 10−2 A = 2.55 mA [½]

Q109.2023 Main 55/1/1 · Q32 (OR option)5 marks

(i) Derive an expression for the force acting on a current carrying straight conductor kept in a magnetic field. State the rule which is used to find the direction of this force. Give the condition under which this force is (1) maximum, and (2) minimum.

(ii) Two long parallel straight wires A and B are 2·5 cm apart in air. They carry 5·0 A and 2·5 A currents respectively in opposite directions. Calculate the magnitude of the force exerted by wire A on a 10 cm length of wire B.

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i) Derivation of expression for force [2]; Statement of Rule [½]; Conditions for maximum and minimum force [½ + ½]; ii) Calculation of magnitude of force [1½]

Consider a rod of uniform cross sectional area A and length l. Let the number density of mobile charge carriers in it be n. Thus the total number of mobile charge carriers in it is nlA. [½]

For steady current I, drift velocity of electrons vd, in the presence of external magnetic field B, the force on these carriers is [½]

F = nlAq(vd × B) [½]

= [jAl] × B = I(l × B) [½]

Where nqvd is current density (j) and |j|A is current (I)

Fleming’s left hand Rule: If forefinger, middle finger and thumb are stretched in mutually perpendicular directions, such that forefinger indicates the direction of magnetic field, middle finger indicates the direction of current in the conductor, then thumb indicates the direction of force on the conductor.

Alternatively

Right Hand Thumb Rule : If the fingers of right hand are made to rotate from l to B through angle θ, the thumb points in the direction of force on the current carrying conductor. [½]

Condition for maximum force θ = 90°, |F| = I l B sinθ = I l B [½]

Condition for minimum force θ = 0° or 180°, |F| = 0 [½]

ii)

Two parallel wires A (5 A) and B (2.5 A, opposite direction) separated by 2.5 cm, length 10 cm

F = μ04π 2I1I2d l [½]

= 10−7 × 2 × 5 × 2.52.5 × 10−2 × 10 × 10−2 N [½]

= 10−5 N [½]

Q110.2023 Main 55/3/1 · Q32 (option 1)5 marks

(i) Write the principle and explain the working of a moving coil galvanometer. A galvanometer as such cannot be used to measure the current in a circuit. Why ?

(ii) Why is the magnetic field made radial in a moving coil galvanometer ? How is it achieved ?

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Official CBSE Marking Scheme · CBSE Board 2023 (Main), Set 55/3/1

(i) Principle – When a rectangular loop carrying current I is placed in a uniform magnetic field, it experiences a torque. [1]

Working :– When a current flows through the coil of a galvanometer, a torque acts on it. [1]

τ = NiAB sin θ

For radial magnetic field ; sin θ = 1

The spring provides a counter or restoring torque kφ.

kφ = NiAB

In equilibrium ; φ = (NABk) i

Galvanometer cannot be used as such to measure current because : [1]

– It has large resistance and hence will change the value of current in the circuit.

– It is a sensitive device.

(Any one of the above)

(ii) The magnetic field is made radial in a moving coil galvanometer so that the magnetic dipole moment (m) is always perpendicular to the magnetic field (B). Hence, sin θ = 1 always. [1]

Alternatively: The magnetic field is made radial in a moving coil galvanometer to make the scale linear.

It is achieved by using curved magnetic poles. [1]

Alternatively: By using soft iron cylindrical core.

Q111.2023 Main 55/3/1 · Q32 (OR option)5 marks

(i) Derive an expression for magnetic field on the axis of a current carrying circular loop.

(ii) Write any two points of difference between a diamagnetic and a paramagnetic substance.

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(i) Diagram [½]

Circular current loop of radius R with point P on the axis at distance x; dB resolved into dB perpendicular and dB along x

dB = μ04π I |dl × r|r3 [½]

= μ0 i dl sin 90°4π (x2 + R2) [½]

dB⊥ cancels out.

Net B = ∫ dBx = ∫ dB cos θ [½]

= μ04π ∫ i dl(x2 + R2) × R(x2 + R2)1/2

= μ0 i R4π (x2 + R2)3/2 ∫ dl = μ0 i R4π (x2 + R2)3/2 (2πR) [½]

B = Bx i = μ0 I R22 (x2 + R2)3/2 i [½]

(ii) Differences [1 + 1]

Diamagnetic MaterialsParamagnetic Materials
(i) Susceptibility is between –1 and 0.(i) Susceptibility is a small positive number (slightly greater than zero).
(ii) Relative permeability is between 0 and 1.(ii) Relative permeability is slightly greater than 1.
(iii) μ < μ0(iii) μ > μ0
(iv) Tendency to move from stronger to weaker part of external magnetism.(iv) Tendency to move from region of weak to strong magnetic field.
(v) is repelled by a magnet.(v) is weakly attracted by a magnet.
(vi) Field inside the material is reduced.(vi) Field inside is slightly enhanced.
(vii) Field lines expelled from a diamagnetic specimen(vii) Field lines concentrated in a paramagnetic specimen

Any two of the above mentioned differences.

Q112.SQP 2025-26 · Q33 (option 1)5 marks

(A) State the working principle of a moving coil galvanometer? What modification is required in the galvanometer to make its scale linear? [2]

(B) If a galvanometer of resistance 49.5Ω has range of 0.05A. What will be the value of resistance needed to convert it in ammeter of range 5A? [2]

(C) How these two resistors should be connected to galvanometer in both cases? [1]

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Official CBSE Marking Scheme · CBSE Sample Question Paper 2025-26

(A) Torque due to current carrying coil. [1]

Modification in designing of galvanometer are

(i) Poles of magnet are made spherical [½]

(ii) Iron ore is placed inside the coil. [½]

(B) Given: Rg = 49.5Ω; Range = 0.05A

For ammeter let resistance needed be Ra.

As per requirement

Range × Rg = Ra(5 – 0.05) [1]

Ra = 0.5 × 49.54.95 = 0.5 Ω [1]

(C) Ra will be connected in series & Rv is connected in parallel. [1]

Unit III · Magnetic Effects of Current and Magnetism

5. Magnetism and Matter

Q1.2026 Main 55/1/1 · Q131 mark

Assertion (A) : All atoms have a net magnetic moment.

Reason (R) : A current loop does not always behave as a magnetic dipole.

  1. Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of the Assertion (A).
  2. Both Assertion (A) and Reason (R) are true, but Reason (R) is not the correct explanation of the Assertion (A).
  3. Assertion (A) is true, but Reason (R) is false.
  4. Both Assertion (A) and Reason (R) are false.
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(D) Both Assertion (A) and Reason (R) are false.

Q2.2026 Main 55/2/1 · Q61 mark

Which of the following materials has positive and small value of magnetic susceptibility ?

  1. Cu
  2. Al
  3. Bi
  4. Ni
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Official CBSE Marking Scheme · CBSE Board 2026 (Main), Set 55/2/1

(B) Al

Q3.2026 Main 55/4/1 · Q141 mark

Assertion (A) : When a ferromagnetic substance is heated to high temperature it becomes paramagnetic in nature.

Reason (R) : The disappearance of magnetisation of a ferromagnet is abrupt and not gradual.

  1. Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of the Assertion (A).
  2. Both Assertion (A) and Reason (R) are true, but Reason (R) is not the correct explanation of the Assertion (A).
  3. Assertion (A) is true, but Reason (R) is false.
  4. Both Assertion (A) and Reason (R) are false.
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Official CBSE Marking Scheme · CBSE Board 2026 (Main), Set 55/4/1

(C) Assertion (A) is true, but Reason (R) is false.

Q4.2026 Supp 55/7/1 · Q131 mark

Assertion (A) : If B be the magnetic field at a small patch of vector area dS, then the magnetic flux through the patch will be B . dS.

Reason (R) : The magnetic flux is a scalar quantity.

  1. Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of the Assertion (A).
  2. Both Assertion (A) and Reason (R) are true, but Reason (R) is not the correct explanation of the Assertion (A).
  3. Assertion (A) is true, but Reason (R) is false.
  4. Both Assertion (A) and Reason (R) are false.
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Official CBSE Marking Scheme · CBSE Board 2026 (Supplementary), Set 55/7/1

(B) Both Assertion (A) and Reason (R) are true, but Reason (R) is not the correct explanation of Assertion (A).

Q5.2025 Main 55/2/1 · Q41 mark

The magnetic field due to a small magnetic dipole of dipole moment ‘M’ at a distance ‘r’ from the centre along the axis of the dipole is given by

  1. μ04π × 2Mr3
  2. μ04π × Mr3
  3. μ04π × M2r3
  4. μ04π × 2Mr2
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Q6.2025 Main 55/4/1 · Q41 mark

A bar magnet is initially at right angles to a uniform magnetic field. The magnet is rotated till the torque acting on it becomes one-half of its initial value. The angle through which the bar magnet is rotated is :

  1. 30°
  2. 45°
  3. 60°
  4. 75°
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Q7.2025 Main 55/4/1 · Q51 mark

Which one out of the following materials is not paramagnetic ?

  1. Aluminium
  2. Sodium Chloride
  3. Calcium
  4. Copper Chloride
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Q8.2025 Main 55/6/1 · Q41 mark

A piece of a diamagnetic material, free to move when placed in a uniform magnetic field :

  1. moves along the field
  2. moves opposite to the field
  3. moves perpendicular to the field
  4. does not move at all
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Q9.2024 Main 55/4/1 · Q61 mark

The magnetic susceptibility for a diamagnetic material is

  1. small and negative
  2. small and positive
  3. large and negative
  4. large and positive
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Official CBSE Marking Scheme · CBSE Board 2024 (Main), Set 55/4/1

(A) Small and negative.

Q10.2024 Main 55/5/1 · Q31 mark

Which of the following pairs is that of paramagnetic materials ?

  1. Copper and Aluminium
  2. Sodium and Calcium
  3. Lead and Iron
  4. Nickel and Cobalt
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Official CBSE Marking Scheme · CBSE Board 2024 (Main), Set 55/5/1

(B) Sodium and Calcium

Q11.2024 Supp 55/S/1 · Q51 mark

Which of the following substances has relative magnetic permeability μr >> 1 ?

  1. Aluminium
  2. Copper chloride
  3. Nickel
  4. Sodium chloride
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Official CBSE Marking Scheme · CBSE Board 2024 (Supplementary), Set 55/S/1

(C) Nickel

Q12.2023 Main 55/1/1 · Q51 mark

A diamagnetic substance is brought near the north or south pole of a bar magnet. It will be :

  1. repelled by both the poles.
  2. attracted by both the poles.
  3. repelled by the north pole and attracted by the south pole.
  4. attracted by the north pole and repelled by the south pole.
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Official CBSE Marking Scheme · CBSE Board 2023 (Main), Set 55/1/1

(A) Repelled by both the poles.

Q13.2023 Main 55/1/1 · Q161 mark

Assertion (A) : When a bar of copper is placed in an external magnetic field, the field lines get concentrated inside the bar.

Reason (R) : Copper is a paramagnetic substance.

  1. Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of the Assertion (A).
  2. Both Assertion (A) and Reason (R) are true, but Reason (R) is not the correct explanation of the Assertion (A).
  3. Assertion (A) is true, but Reason (R) is false.
  4. Assertion (A) is false and Reason (R) is also false.
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Official CBSE Marking Scheme · CBSE Board 2023 (Main), Set 55/1/1

(D) Assertion (A) is false and Reason (R) is also false.

Q14.2023 Main 55/2/1 · Q161 mark

Note: In question numbers 16 to 18, two statements are given – one labelled Assertion (A) and the other labelled Reason (R). Select the correct answer to these questions from the codes (A), (B), (C) and (D) as given below:

Assertion (A) : Diamagnetic substances exhibit magnetism.

Reason (R) : Diamagnetic materials do not have permanent magnetic dipole moment.

  1. Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of Assertion (A).
  2. Both Assertion (A) and Reason (R) are true and Reason (R) is NOT the correct explanation of Assertion (A).
  3. Assertion (A) is true and Reason (R) is false.
  4. Assertion (A) is false and Reason (R) is also false.
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Official CBSE Marking Scheme · CBSE Board 2023 (Main), Set 55/2/1

(B) Both Assertion (A) and Reason (R) are true and Reason (R) is not the correct explanation of Assertion (A)

Q15.2023 Main 55/4/1 · Q31 mark

The magnetic field lines near a substance are as shown in the figure. The substance is :

Magnetic field lines bending away from (avoiding) a rectangular piece of substance
  1. Copper
  2. Iron
  3. Sodium
  4. Aluminium
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Official CBSE Marking Scheme · CBSE Board 2023 (Main), Set 55/4/1

(A) Copper

Q16.2023 Main 55/5/1 · Q31 mark

Which of the following has its permeability less than that of free space ?

  1. Copper
  2. Aluminium
  3. Copper chloride
  4. Nickel
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Official CBSE Marking Scheme · CBSE Board 2023 (Main), Set 55/5/1

(A) Copper

Q17.2023 Supp 55/C/1 · Q31 mark

A bar magnet is cut into two equal halves parallel to its magnetic axis. The physical quantity that remains unchanged is :

  1. pole strength
  2. magnitude of magnetisation
  3. moment of inertia
  4. magnetic moment
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Official CBSE Marking Scheme · CBSE Board 2023 (Supplementary), Set 55/C/1

(B) magnitude of magnetization

Q18.SQP 2025-26 · Q121 mark

When we move magnetic compass from point P to Q then which of the following statement is true

Current-carrying solenoid with point P on its axis and point Q on its perpendicular bisector
  1. The deflection of the magnetic needle at P and Q will be in the same direction.
  2. The deflection of the magnetic needle at P and Q will be in the opposite directions.
  3. The deflection of the magnetic needle at P and Q will be perpendicular to each other.
  4. The deflection of the magnetic needle at P and Q will be inclined at 45° with respect to each other.
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Official CBSE Marking Scheme · CBSE Sample Question Paper 2025-26

(B) The deflection of the magnetic needle at P and Q will be in the opposite directions.

As magnetic field at equator is antiparallel to magnetic field at pole.

Q19.SQP 2026-27 · Q81 mark

A ball of superconducting material is dipped in liquid nitrogen and placed near a bar magnet. The direction of the movement of the ball and the magnetic moment would be-

Direction of the movement of ballDirection of the magnetic moment
A.Towards the bar magnetOpposite to that of the bar magnet
B.Towards the bar magnetSame direction as that of the bar magnet
C.Away from the bar magnetSame direction as that of the bar magnet
D.Away from the bar magnetOpposite to that of the bar magnet
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Official CBSE Marking Scheme · CBSE Sample Question Paper 2026-27

(D) Away from the bar magnet and opposite to that of bar magnet

As the ball is dipped in liquid nitrogen, its temperature becomes less than the critical temperature and it becomes a superconductor. In this state, the ball becomes diamagnetic.

Q20.2023 Main 55/4/1 · Q212 marks

A small magnetised needle P is placed at the origin of x-y plane with its magnetic moment pointing along the y-axis. Another identical magnetised needle Q is placed in two positions, one by one.

Case 1 : at (a, 0) with its magnetic moment pointing along x-axis.
Case 2 : at (0, a) with its magnetic moment pointing along y-axis.

(a) In which case is the potential energy of P and Q minimum ?
(b) In which case is P and Q not in equilibrium ?
Justify your answers.

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Official CBSE Marking Scheme · CBSE Board 2023 (Main), Set 55/4/1

(a) Potential energy is minimum in case 2, since Q is placed along the direction of P / stable equilibrium. [½ + ½]

(b) P and Q are not in equilibrium in case 1. In this case Q is at the normal bisector of P / not in equilibrium. [½ + ½]

Alternatively :

(a) Since U = − MB cos θ and θ = 0° so P.E. is minimum. [½ + ½]

(b) Case 1, not in equilibrium, since τ = MB sin θ and θ = 90°, τ = MB. [½ + ½]

OR

(b) Since two needles are perpendicular they experience torque. [½ + ½]

Q21.2023 Supp 55/C/1 · Q202 marks

Two identical bars, one of a paramagnetic material and another of a diamagnetic material are kept in a uniform magnetic field. Show diagrammatically the modifications in the pattern of magnetic field in each case. How are the two materials affected by increase in temperature ?

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Official CBSE Marking Scheme · CBSE Board 2023 (Supplementary), Set 55/C/1

Showing the modifications in the pattern of magnetic field [½ + ½]

Magnetic field line patterns for (a) paramagnetic bar (lines concentrate into it) and (b) diamagnetic bar (lines expelled)

No effect in case of diamagnetic materials. [½]

Magnetization / magnetic susceptibility decrease in case of paramagnetic materials. [½]

Q22.SQP 2025-26 · Q20 (OR option)2 marks

Draw a diagram representing the behaviour of magnetic field lines for a

(A) diamagnetic &
(B) paramagnetic substance.

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Official CBSE Marking Scheme · CBSE Sample Question Paper 2025-26
Magnetic field lines (a) expelled from a diamagnetic substance and (b) concentrated in a paramagnetic substance

[1] for each diagram

Q23.SQP 2026-27 · Q20 (option 1)2 marks

A certain magnetic substance is found to have a relative permeability of 800. What type of magnetic material is it? Write any two characteristics that such a material exhibits.

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Official CBSE Marking Scheme · CBSE Sample Question Paper 2026-27

Given: Relative magnetic permeability, µr = 800

Since the relative permeability is much greater than 1, the material is a ferromagnetic material. [1]

Properties:

  1. Ferromagnetic materials are strongly attracted by a magnetic field because they have a very high value of magnetic permeability.
  2. They can retain magnetism even after the external magnetic field is removed, meaning they show hysteresis and can be made into permanent magnets.

[½ × 2]

Q24.SQP 2026-27 · Q20 (OR option)2 marks

State Gauss’s law of magnetism. Also discuss its significance.

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Official CBSE Marking Scheme · CBSE Sample Question Paper 2026-27

Gauss’s law for magnetism states that the net magnetic flux through any closed surface is always zero: ΦB = ∮ B · dS = 0

This means that the magnetic field lines neither begin nor end at any point. [1]

Significance:

  • Magnetic monopoles do not exist
  • Magnetic field lines always form continuous loops

[½ × 2]

Q25.2025 Main 55/5/1 · Q263 marks

An electron of mass m and charge – e is revolving anticlockwise around the nucleus of an atom.

(a) Obtain the expression for the magnetic dipole moment (μ) of the atom.

(b) If L is the angular momentum of electron, show that μ = – (e2m)L.

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Q26.2024 Main 55/4/1 · Q28 (OR option)3 marks

What are ferromagnetic materials ? Explain ferromagnetism with the help of suitable diagrams, using the concept of magnetic domain.

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Official CBSE Marking Scheme · CBSE Board 2024 (Main), Set 55/4/1

Ferromagnetic substances are those which get strongly magnetised when placed in an external magnetic field. [1]

Magnetic domains: (a) randomly oriented without field, (b) aligned along B0

Diagrams (a) and (b) [½ + ½]

In absence of external magnetic field, domains are randomly oriented and it exhibits weak magnetisation. [½]

In the presence of external magnetic field domains orient themselves in the direction of magnetic field and it exhibits strong magnetisation. [½]

Q27.SQP 2025-26 · Q27 (OR option)3 marks

A solenoid has a core of material with relative permeability 200. The windings of the solenoid are insulated from the core and carry a current of 1A. If the number of turns is 2000 per metre, calculate

(A) magnetic intensity,
(B) magnetic field &
(C) magnetisation

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Official CBSE Marking Scheme · CBSE Sample Question Paper 2025-26

Given:

μr = 200

I = 1A

N = 200 turn/m

(A) H = nI

Or, H = 2000/m × 1A = 2 × 103 A/m [1]

(B) B = μoμrH

Or, B = 200 × 4π × 10-7 × 2 × 103 A/m

Or, B = 0.50T [1]

(C) Magnetisation is given by

M = (μr – 1)H = 199 × 103 A/m

Or, M = 1.99 × 105 A/m [1]

Q28.SQP 2026-27 · Q27 (OR option)3 marks
  1. A small current loop behaves like a tiny magnet when placed in a magnetic field. Using this idea, explain what physical quantity determines the strength and orientation of such a magnetic dipole. Also state the SI unit of this quantity.
  2. A steel wire of length l has a magnetic moment m. It is bent into a semicircular arc. What is its new magnetic moment?
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Official CBSE Marking Scheme · CBSE Sample Question Paper 2026-27

I. Magnetic dipole moment [½]

SI unit: ampere metre2 (Am2) or joule per tesla (JT-1) [½]

II. Pole strength, qm = ml [½]

When the wire is bent into a semicircular arc, the separation between the poles changes from l to 2r, where r is the radius of the semicircular arc.

l = π r, r = lπ [½]

New magnetic moment = qm × 2r [½]

= ml × 2lπ [½]

= 2mπ [½]

Unit IV · Electromagnetic Induction and Alternating Currents

6. Electromagnetic Induction

Q1.2026 Main 55/2/1 · Q81 mark

The magnetic flux φ (in Wb) linked with a coil is related to time t (in s) as
φ = 5 At2 + Bt – 2C
The SI units of A and B are respectively

  1. Wbs2, Wbs
  2. Wb s–1, Wb
  3. Wbs–2, Wbs–1
  4. Wbs–1, Wbs–2
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Official CBSE Marking Scheme · CBSE Board 2026 (Main), Set 55/2/1

(C) Wbs–2, Wbs–1

Q2.2026 Main 55/2/1 · Q101 mark

A magnet held vertically, with its north pole down is dropped along the axis of a closed solenoid placed vertically on a table. If the observer looks down from the top,

  1. the induced current will flow in the anticlockwise direction.
  2. the induced current will flow in the clockwise direction.
  3. no induced current will flow in the solenoid.
  4. the magnet will fall with a constant velocity.
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Official CBSE Marking Scheme · CBSE Board 2026 (Main), Set 55/2/1

(A) The induced current will flow in the anticlockwise direction

Q3.2026 Main 55/3/1 · Q21 mark

A square loop of side 50 cm is placed in a uniform magnetic field of 3·0 T acting perpendicular to the plane of the loop. If the loop is rotated through an angle of 90° in 0·3 s, the value of emf induced in the loop would be :

  1. 0·25 V
  2. 0·50 V
  3. 0·75 V
  4. 1·0 V
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Official CBSE Marking Scheme · CBSE Board 2026 (Main), Set 55/3/1

No option is correct (Award full credit for attempting)

Q4.2026 Main 55/4/1 · Q41 mark

The dimensions of the rate of change of magnetic flux are :

  1. [M L T−3 A]
  2. [M L2 T−3 A−1]
  3. [M L2 T−2 A−1]
  4. [M L2 T−3 A−2]
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Official CBSE Marking Scheme · CBSE Board 2026 (Main), Set 55/4/1

(B) [M L2 T−3 A−1]

Q5.2026 Main 55/4/1 · Q51 mark

Two identical conductors 1 and 2 are placed on two frictionless conducting rails R and S in a uniform magnetic field directed vertically downward into the plane of the page. If conductor 1 is moved with a constant velocity in the direction as shown in figure, the force on conductor 2 will be along :

Conductors 1 and 2 on rails R and S in a field into the page; conductor 1 moves to the left; axes i, j, k shown
  1. −i
  2. −j
  3. k
  4. j
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Official CBSE Marking Scheme · CBSE Board 2026 (Main), Set 55/4/1

(A) −i

Q6.2026 Main 55/5/1 · Q41 mark

The magnetic flux through a loop placed in a magnetic field can be changed by changing :

  1. area of the loop only
  2. the value of magnetic field only
  3. orientation of the loop in the magnetic field only
  4. any one or more of the factors given in (A), (B) and (C)
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Official CBSE Marking Scheme · CBSE Board 2026 (Main), Set 55/5/1

(D) any one or more of the factors given in option (A), (B) and (C)

Q7.2026 Main 55/5/1 · Q131 mark

Assertion (A) : Induced emf produced in a coil will be more when the magnetic flux linked with the coil is more.

Reason (R) : Induced emf produced is directly proportional to the magnetic flux.

  1. Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of the Assertion (A).
  2. Both Assertion (A) and Reason (R) are true, but Reason (R) is not the correct explanation of the Assertion (A).
  3. Assertion (A) is true, but Reason (R) is false.
  4. Both Assertion (A) and Reason (R) are false.
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Official CBSE Marking Scheme · CBSE Board 2026 (Main), Set 55/5/1

(D) Both Assertion (A) and reason (R) are false.

Q8.2025 Main 55/1/1 · Q41 mark

A coil has 100 turns, each of area 0.05 m2 and total resistance 1.5 Ω. It is inserted at an instant in a magnetic field of 90 mT, with its axis parallel to the field. The charge induced in the coil at that instant is :

  1. 3.0 mC
  2. 0.30 C
  3. 0.45 C
  4. 1.5 C
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Q9.2025 Main 55/1/1 · Q51 mark

You are required to design an air-filled solenoid of inductance 0.016 H having a length 0.81 m and radius 0.02 m. The number of turns in the solenoid should be

  1. 2592
  2. 2866
  3. 2976
  4. 3140
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Q10.2025 Main 55/1/1 · Q131 mark

Assertion (A) : It is difficult to move a magnet into a coil of large number of turns when the circuit of the coil is closed.

Reason (R) : The direction of induced current in a coil with its circuit closed, due to motion of a magnet, is such that it opposes the cause.

  1. Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of Assertion (A).
  2. Both Assertion (A) and Reason (R) are true, but Reason (R) is not the correct explanation of Assertion (A).
  3. Assertion (A) is true, but Reason (R) is false.
  4. Assertion (A) is false and Reason (R) is also false.
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Q11.2025 Main 55/2/1 · Q51 mark

In the figure X is a coil wound over a hollow wooden pipe.

Coil X wound over a hollow wooden pipe with ammeter A connected across terminals Y and Z; bar magnet with N pole moving in from the right at speed v

A permanent magnet is pushed at a constant speed v from the right into the pipe and it comes out at the left end of the pipe. During the entry and the exit of the magnet, the current in the wire YZ will be from

  1. Y to Z and then Y to Z
  2. Z to Y and then Y to Z
  3. Y to Z and then Z to Y
  4. Z to Y and then Z to Y
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Q12.2025 Main 55/5/1 · Q61 mark

The magnetic flux linked with a coil changes with time t as φ = (8t2 + 5t + 7), where t is in seconds and φ is in Wb. The value of emf induced in the coil at t = 4 s is :

  1. 32 V
  2. 37 V
  3. 64 V
  4. 69 V
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Q13.2025 Main 55/6/1 · Q71 mark

A vertically held bar magnet is dropped along the axis of a copper ring having a cut as shown in the diagram. The acceleration of the falling magnet is :

Bar magnet (S top, N bottom) falling toward a copper ring with a cut
  1. zero
  2. less than g
  3. g
  4. greater than g
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Q14.2025 Main 55/7/1 · Q31 mark

A rectangular coil of area A is kept in a uniform magnetic field B such that the plane of the coil makes an angle α with B. The magnetic flux linked with the coil is :

  1. BA sin α
  2. BA cos α
  3. BA
  4. zero
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Q15.2025 Main 55/7/1 · Q61 mark

A metal rod of length 50 cm is held vertically and moved with a velocity of 10 m/s towards east. The horizontal component of the Earth’s magnetic field at the place is 0·4 G. The emf induced across the ends of the rod is :

  1. 0·1 mV
  2. 0·2 mV
  3. 0·8 mV
  4. 1·6 mV
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Q16.2025 Main 55/7/1 · Q71 mark

The dimensions of ‘self-inductance’ are :

  1. [M L T–2 A–2]
  2. [M L2 T–1 A–1]
  3. [M L–1 T–2 A–2]
  4. [M L2 T–2 A–2]
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Q17.2025 Supp 55/S/1 · Q51 mark

Whenever a magnet is moved either towards or away from a conducting coil, an emf is induced whose magnitude is independent of the :

  1. number of turns in the coil
  2. resistance of the coil
  3. speed with which the magnet is moved
  4. area of the coil
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Official CBSE Marking Scheme · CBSE Board 2025 (Supplementary), Set 55/S/1

(B) the resistance of the coil

Q18.2025 Supp 55/S/1 · Q81 mark

A rectangular loop of area vector A = [i + 3 j] × 10–2 m2 is placed in a magnetic field B = [4·0t2 + 2·0] i, where B is in tesla and t in seconds. At t = 1·0 s, the magnitude of induced emf in the loop is :

  1. 80 mV
  2. 0·11 V
  3. 0·22 V
  4. 0·8 V
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Official CBSE Marking Scheme · CBSE Board 2025 (Supplementary), Set 55/S/1

(A) 80 mV

Q19.2024 Main 55/1/1 · Q141 mark

Assertion (A) : The mutual inductance between two coils is maximum when the coils are wound on each other.

Reason (R) : The flux linkage between two coils is maximum when they are wound on each other.

  1. Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of the Assertion (A).
  2. Both Assertion (A) and Reason (R) are true, but Reason (R) is not the correct explanation of the Assertion (A).
  3. Assertion (A) is true, but Reason (R) is false.
  4. Assertion (A) is false and Reason (R) is also false.
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Official CBSE Marking Scheme · CBSE Board 2024 (Main), Set 55/1/1

(A) Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of the Assertion (A).

Q20.2024 Main 55/2/1 · Q41 mark

A circular coil of radius 10 cm is placed in a magnetic field B = (1·0 i + 0·5 j) mT such that the outward unit vector normal to the surface of the coil is (0·6 i + 0·8 j). The magnetic flux linked with the coil is :

  1. 0·314 µWb
  2. 3·14 µWb
  3. 31·4 µWb
  4. 1·256 µWb
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Official CBSE Marking Scheme · CBSE Board 2024 (Main), Set 55/2/1

(C) 31·4 µWb

Q21.2024 Main 55/2/1 · Q141 mark

Two statements are given — one labelled Assertion (A) and the other labelled Reason (R). Select the correct answer from the codes (A), (B), (C) and (D) as given below.

Assertion (A) : Lenz’s law is a consequence of the law of conservation of energy.

Reason (R) : There is no power loss in an ideal inductor.

  1. Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of the Assertion (A).
  2. Both Assertion (A) and Reason (R) are true, but Reason (R) is not the correct explanation of the Assertion (A).
  3. Assertion (A) is true, but Reason (R) is false.
  4. Assertion (A) is false and Reason (R) is also false.
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Official CBSE Marking Scheme · CBSE Board 2024 (Main), Set 55/2/1

(B) Both Assertion (A) and Reason (R) are true but Reason (R) is not the correct explanation of the Assertion (A)

Q22.2024 Main 55/3/1 · Q61 mark

The current in a coil of 15 mH increases uniformly from zero to 4 A in 0·004 s. The emf induced in the coil will be :

  1. 22·5 V
  2. 17·5 V
  3. 15·0 V
  4. 12·5 V
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Official CBSE Marking Scheme · CBSE Board 2024 (Main), Set 55/3/1

(C) 15.0 V

Q23.2024 Main 55/3/1 · Q71 mark

Consider a solenoid of length l and area of cross-section A with fixed number of turns. The self-inductance of the solenoid will increase if :

  1. both l and A are increased
  2. l is decreased and A is increased
  3. l is increased and A is decreased
  4. both l and A are decreased
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Official CBSE Marking Scheme · CBSE Board 2024 (Main), Set 55/3/1

(B) l is decreased and A is increased

Q24.2024 Main 55/4/1 · Q81 mark

A circular loop A of radius R carries a current I. Another circular loop B of radius r (= R20) is placed concentrically in the plane of A. The magnetic flux linked with loop B is proportional to

  1. R
  2. R
  3. R3/2
  4. R2
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Official CBSE Marking Scheme · CBSE Board 2024 (Main), Set 55/4/1

(A) R

Q25.2024 Main 55/4/1 · Q111 mark

A coil of N turns is placed in a magnetic field B such that B is perpendicular to the plane of the coil. B changes with time as B = B0 cos (2πTt) where T is time period. The magnitude of emf induced in the coil will be maximum at

  1. t = nT8
  2. t = nT4
  3. t = nT2
  4. t = nT

Here, n = 1, 2, 3, 4, ...

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Official CBSE Marking Scheme · CBSE Board 2024 (Main), Set 55/4/1

No option is correct, award 1 mark.

Q26.2024 Main 55/5/1 · Q51 mark

Two coils are placed near each other. When the current in one coil is changed at the rate of 5 A/s, an emf of 2 mV is induced in the other. The mutual inductance of the two coils is

  1. 0.4 mH
  2. 2.5 mH
  3. 10 mH
  4. 2.5 H
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Official CBSE Marking Scheme · CBSE Board 2024 (Main), Set 55/5/1

(A) 0.4 mH

Q27.2023 Main 55/1/1 · Q61 mark

A circular coil of radius 8·0 cm and 40 turns is rotated about its vertical diameter with an angular speed of 25π rad s−1 in a uniform horizontal magnetic field of magnitude 3·0 × 10−2 T. The maximum emf induced in the coil is :

  1. 0·12 V
  2. 0·15 V
  3. 0·19 V
  4. 0·22 V
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Official CBSE Marking Scheme · CBSE Board 2023 (Main), Set 55/1/1

(C) 0.19 V

Q28.2023 Main 55/1/1 · Q71 mark

Figure shows a rectangular conductor PSRQ in which movable arm PQ has a resistance ‘r’ and resistance of PSRQ is negligible. The magnitude of emf induced when PQ is moved with a velocity v does not depend on :

Rectangular conductor PSRQ in a uniform magnetic field into the page with movable arm PQ moving with velocity v
  1. magnetic field ( B )
  2. velocity ( v )
  3. resistance (r)
  4. length of PQ
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Official CBSE Marking Scheme · CBSE Board 2023 (Main), Set 55/1/1

(C) Resistance ( r )

Q29.2023 Main 55/4/1 · Q91 mark

The direction of induced current in the loop abc is :

Loop abc above a straight wire xy carrying current I from x to y
  1. along abc if I decreases
  2. along acb if I increases
  3. along abc if I is constant
  4. along abc if I increases
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Official CBSE Marking Scheme · CBSE Board 2023 (Main), Set 55/4/1

(D) along abc if I increases

Q30.2023 Main 55/5/1 · Q41 mark

A square shaped coil of side 10 cm, having 100 turns is placed perpendicular to a magnetic field which is increasing at 1 T/s. The induced emf in the coil is

  1. 0.1 V
  2. 0.5 V
  3. 0.75 V
  4. 1.0 V
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Official CBSE Marking Scheme · CBSE Board 2023 (Main), Set 55/5/1

(D) 1.0 V

Q31.2023 Supp 55/C/1 · Q141 mark

A planar loop is rotated in a magnetic field about an axis perpendicular to the field. The polarity of induced emf changes once in each :

  1. 1 revolution
  2. (12) revolution
  3. (14) revolution
  4. (34) revolution
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Official CBSE Marking Scheme · CBSE Board 2023 (Supplementary), Set 55/C/1

(B) (12) revolution

Q32.SQP 2025-26 · Q91 mark

Two small identical magnets are allowed to fall freely one through a vertical solenoid of 20 m made up of copper and another in air through the same vertical distance. The time taken by the two magnets to fall will be

  1. same in both the cases.
  2. more for the magnet falling in air.
  3. more for the magnet falling through the solenoid.
  4. infinite.
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Official CBSE Marking Scheme · CBSE Sample Question Paper 2025-26

(C) more for the magnet falling through the solenoid.

Emf will be induced in solenoid due to motion of magnet through it. As per Lenz’s law induced emf will oppose the motion of magnet.

Q33.SQP 2025-26 · Q101 mark

The emf generated by an AC generator is given by V = Vo sin ωt, where ω is angular frequency of armature of generator. What will be the emf if the angular frequency is doubled

  1. V = Vo sin 2ωt
  2. V = 2Vo sin ωt
  3. V = 2Vo sin 2ωt
  4. V = Vo sin ωt
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Official CBSE Marking Scheme · CBSE Sample Question Paper 2025-26

(C) V = 2Vo sin 2ωt

As V = NBAω sin ωt

Q34.SQP 2026-27 · Q31 mark

An infinitely long cylinder is kept parallel to a uniform magnetic field B directed along negative y –axis. What will be the direction of induced current as seen from the y axis?

  1. clockwise of the negative y-axis
  2. anticlockwise of the negative y-axis
  3. no current will be induced
  4. along the direction of the magnetic field
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Official CBSE Marking Scheme · CBSE Sample Question Paper 2026-27

(C) No current will be induced

As the cylinder is kept stationary and also the magnetic field is uniform, the flux linked with the cylinder is not changing, so no current is induced in the cylinder.

Q35.2026 Main 55/5/1 · Q182 marks

A plane circular coil is rotated about its vertical diameter with a constant angular speed ω in a uniform horizontal magnetic field. Initially the plane of the coil is parallel to the magnetic field. Draw plots showing the variation of the following physical quantities as the function of ωt, where t represents time elapsed :

  1. Magnetic flux φ linked with the coil, and
  2. emf induced in the coil.
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Official CBSE Marking Scheme · CBSE Board 2026 (Main), Set 55/5/1

Drawing plot showing variation of (a) magnetic flux linked with a coil [1]; (b) emf induced in the coil as a function of ωt [1]

Plots of flux φ = BA cos(ωt + π/2) and emf e = ωBA sin(ωt + π/2) versus ωt

Alternatively,

Alternative plots of flux and emf versus ωt

Note: Award full marks even if formula is not written by student.

Q36.2023 Main 55/5/1 · Q252 marks

Two coils C1 and C2 are placed close to each other. The magnetic flux φ2 linked with the coil C2 varies with the current I1 flowing in coil C1, as shown in the figure. Find

Graph of flux phi2 (wb) versus current I1 (A), a straight line through the origin reaching 15 wb at 6 A
  1. the mutual inductance of the arrangement, and
  2. the rate of change of current (dI1dt) that will induce an emf of 100 V in coil C2.
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Official CBSE Marking Scheme · CBSE Board 2023 (Main), Set 55/5/1

(i) φ2 = M I1 [½]

Slope of the graph, M = φ2I1 [½]

M = 2.5 H

(ii) |ε2| = M dI1dt [½]

dI1dt = |ε2|M = 1002.5 = 40 A s-1 [½]

Q37.2026 Main 55/2/1 · Q25 (option 1)3 marks

(i) Define mutual inductance of a pair of coils. Write its SI unit.

(ii) A long solenoid of radius R and length L has n turns per unit length. A circular loop of radius r(<R) is placed inside at the centre of the solenoid such that its axis coincides with the axis of the solenoid. Obtain the mutual inductance of the solenoid and the loop.

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Official CBSE Marking Scheme · CBSE Board 2026 (Main), Set 55/2/1

(i) It is the ratio of the magnetic flux linked with a coil to the current flowing in the neighbouring coil. [1]

Alternatively : It is the ratio of magnitude of induced emf in a coil to the rate of change of current in the neighbouring coil.

Note : Award (½) mark if a student writes M = φI OR M = |ε|di/dt

SI unit : henry (H) or Tm2A–1 or WbA–1 or VsA–1 [½]

(ii) If current I is set up through the solenoid, the magnetic field inside at its centre B = μ0nI [½]

Flux linked with the circular loop of radius r(< R) : φB = BA = (μ0nI)(πr2) [½]

Mutual inductance M = φBI [½]

M = μ0nπr2 [½]

Q38.2026 Main 55/4/1 · Q23 (OR option)3 marks

(b) A rectangular loop of sides l and b and resistance ‘R’ is kept in a region in which the magnetic field varies as B = B0 sin ωt.

  1. Derive expression for the emf induced in the loop.
  2. Find the effective value of current that flows in the loop.
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Official CBSE Marking Scheme · CBSE Board 2026 (Main), Set 55/4/1

(i) Induced emf = |ε| = dφ/dt [½]

= ddt(B·A) [½]

|ε| = ddt(B0 sin ωt × lb) [½]

= B0ωlb cos ωt [½]

(ii) Instantaneous value of current = |ε|R = B0ωlb cos ωtR [½]

Ieff = I02 = B0ωlbR2 [½]

Q39.2026 Main 55/4/1 · Q243 marks

The figure given below shows a square-shaped loop MNOP of side 25 cm placed horizontally in a uniform magnetic field B directed vertically downward. The position of the loop at t = 0 s is as shown in figure. The loop is pulled with a constant velocity of 25 cm/s till it goes out of the field.

Square loop MNOP of side 25 cm at the left edge of a 1 m wide field region (into the page), moving right at 25 cm/s
  1. What will be the direction of the induced current in the loop as it goes out of the field ? For how long would the current in the loop persist ?
  2. Plot graphs showing the variation of magnetic flux and magnitude of induced emf as a function of time.
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Official CBSE Marking Scheme · CBSE Board 2026 (Main), Set 55/4/1

(a) As the magnetic flux linked with the loop is decreasing, hence in accordance with Lenz’s Law, the induced current will be in the clockwise direction. [1]

t = 25 cm25 cm/s = 1 s [1]

(b) [½ + ½]

Graphs of magnetic flux (constant till t = 3 s, then decreasing linearly to zero at 4 s) and |emf| (zero till 3 s, constant between 3 s and 4 s) against time
Q40.2026 Main 55/5/1 · Q243 marks

A long solenoid of length L and radius r1 having N1 turns is surrounded symmetrically by a coil of radius r2 (> r1) having N2 turns (N2 << N1) around its mid-point. Derive an expression for the mutual inductance of solenoid and coil. Is M12 = M21 valid in this case ?

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Official CBSE Marking Scheme · CBSE Board 2026 (Main), Set 55/5/1

Deriving the expression for mutual inductance of solenoid and coil [2]; Validating M12 = M21 [1]

Long solenoid S1 of N1 turns surrounded by coil S2 of N2 turns

Let n1 and n2 be the total no. of turns per unit length of solenoid S1 and coil S2 respectively

N1 = n1L, N2 = n2L

When current I2 is set up in S2, flux linkage with S1 is:

N1φ1 = N1A1B2 ……. (i) [½]

N1φ1 = (n1L)(πr12)(μ0n2I2)

N1φ1 = (μ0n1n2πr12LI2) ……. (ii) [½]

N1φ1 = M12I2 ……. (iii) [½]

From (ii) and (iii)

M12 = (μ0n1n2πr12L) [½]

Considering the reverse case, when I1 current is set up in S1, flux linkage with S2 is

N2φ2 = N2A1B1 ……. (iv)

N2φ2 = (n2L)(πr12)(μ0n1I1) ……. (v)

N2φ2 = M21I1 ……. (vi)

From (v) and (vi)

M21 = (μ0n1n2πr12L) [½]

Therefore, M12 = M21 [½]

Alternatively,

M12 = M21 = (μ0N1N2πr12)L

Q41.2025 Main 55/1/1 · Q243 marks

(a) Show that the energy required to build up the current I in a coil of inductance L is 12 LI2.

(b) Considering the case of magnetic field produced by air-filled current carrying solenoid, show that the magnetic energy density of a magnetic field B is B22μ0.

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Q42.2025 Main 55/2/1 · Q243 marks

(a) State Lenz’s law.

(b) In the given figure :

AC generator: coil rotating between magnet poles with slip rings, brushes and voltmeter; parts labelled P, Q and R
  1. Identify the machine.
  2. Name the parts P and Q and R of the machine.
  3. Give the polarities of the magnetic poles.
  4. Write the two ways of increasing the output voltage.
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Q43.2025 Main 55/4/1 · Q243 marks

Two coils ‘1’ and ‘2’ are placed close to each other as shown in the figure. Find the direction of induced current in coil ‘1’ in each of the following situations, justifying your answers :

Coil 2 connected to a variable resistor R2 and a battery (M to N); coil 1 connected to resistor R1 (O to P)
  1. Coil ‘2’ is moving towards coil ‘1’.
  2. Coil ‘2’ is moving away from coil ‘1’.
  3. The resistance connected with coil ‘2’ is increased keeping both the coils stationary.
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Q44.2025 Main 55/6/1 · Q24 (option 1)3 marks

State Lenz’s law. A rod MN of length L is rotated about an axis passing through its end M perpendicular to its length, with a constant angular velocity ω in a uniform magnetic field B parallel to the axis. Obtain an expression for emf induced between its ends.

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Q45.2025 Main 55/6/1 · Q24 (OR option)3 marks

Define ‘self-inductance’ of a coil. Derive an expression for self-inductance of a long solenoid of cross-sectional area A and length l, having n turns per unit length.

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Q46.2025 Main 55/7/1 · Q243 marks

(a) State Faraday’s law of electromagnetic induction and explain the role of negative sign in its expression.

(b) Explain, with an example, that Lenz’s law is consistent with the law of conservation of energy.

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Q47.2024 Main 55/2/1 · Q243 marks

A 100-turn coil of radius 1·6 cm and resistance 5·0 Ω is co-axial with a solenoid of 250 turns/cm and radius 1·8 cm. The solenoid current drops from 1·5 A to zero in 25 ms. Calculate the current induced in the coil in this duration. (Take π2 = 10)

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Official CBSE Marking Scheme · CBSE Board 2024 (Main), Set 55/2/1

Induced emf (ε) = − N dφdt [½]

= − NA dBdt = − NA ddt(µ0nI) = − N µ0n(πr2) dIdt

ε = 100 × 4π × 10−7 × 250 × 102 × π × (1·6 × 10−2)2 × 1·525 × 10−3 [1]

= 0·1536 V [½]

I = εR [½]

= 0·03 A [½]

Alternatively

ε = − M dIdt [½]

M = µ0n1n2πr12l = µ0(n1l)n2πr12 [½]

= 4π × 10−7 × 100 × 250 × 102 × π × (1·6 × 10−2)2
= 2·56 × 10−3 H [½]

ε = − 2·56 × 10−3 × (0 − 1·5)25 × 10−3 = 0·1536 V [½]

I = εR = 0·15365 [½]

= 0·03 A [½]

Q48.2024 Main 55/4/1 · Q28 (option 1)3 marks

(i) Define mutual inductance. Write its SI unit.

(ii) Derive an expression for the mutual inductance of a system of two long coaxial solenoids of same length l, having turns N1 and N2 and of radii r1 and r2 (> r1).

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Official CBSE Marking Scheme · CBSE Board 2024 (Main), Set 55/4/1

(i) Mutual inductance between two coils is defined as the magnetic flux associated with a coil when unit current flows through neighbouring coil. [½]

Alternatively

Mutual inductance between two coils is defined as the magnitude of induced emf in a coil when the rate of change of current in neighbouring coil is unity.

SI unit of mutual inductance is henry (H). [½]

(ii)

Two long coaxial solenoids: inner with N1 turns and radius r1, outer with N2 turns and radius r2, length l

When current I2 flows in outer solenoid, the resulting flux linkage with inner solenoid.

N1φ1 = N1B2A1 [½]

N1φ1 = N1(μ0N2I2l)πr12

N1φ1 = μ0N1N2πr12I2l ………… (1) [½]

N1φ1 = M12I2 ………… (2) [½]

From equations (1) and (2)

M12 = μ0N1N2πr12l [½]

Q49.2024 Main 55/5/1 · Q24 (option 1)3 marks
  1. State Lenz’s Law. In a closed circuit, the induced current opposes the change in magnetic flux that produced it as per the law of conservation of energy. Justify.
  2. A metal rod of length 2 m is rotated with a frequency 60 rev/s about an axis passing through its centre and perpendicular to its length. A uniform magnetic field of 2T perpendicular to its plane of rotation is switched-on in the region. Calculate the e.m.f. induced between the centre and the end of the rod.
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Official CBSE Marking Scheme · CBSE Board 2024 (Main), Set 55/5/1

(i) The polarity of induced emf is such that it tends to produce a current which opposes the change in magnetic flux that produced it. [1]

In a closed loop, when the polarity of induced emf is such that, the induced current favours the change in magnetic flux then the magnetic flux and consequently the current will go on increasing without any external source of energy. This violates law of conservation of energy. [½]

(ii) ε = 12 Bl2ω [½]

= 12 × 2 × (2)2 × (2π × 60) [½]

= 480π V

= 1.51×103 V [½]

Q50.2024 Supp 55/S/1 · Q223 marks

A small circular loop of area 6π cm2 is placed inside a long solenoid at its centre such that its axis makes an angle of 60° with the axis of the solenoid. The number of turns per cm is 10 in the solenoid. The current in the solenoid changes uniformly from 5 A to zero in 10 ms. Calculate the emf induced in the loop.

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Official CBSE Marking Scheme · CBSE Board 2024 (Supplementary), Set 55/S/1

E.m.f induced in the circular loop:

|ε| = dφdt [½]

= ddt(BA cos θ)

= ddt(μ0nIA cos θ) [1]

|ε| = (μ0nA cos θ · dIdt) [1]

|ε| = 4π×10–7 × 1000 × 6π × 10–4 × cos 60° × 510–2

= 6×10–5 V [½]

Q51.2023 Main 55/1/1 · Q283 marks

A long solenoid of radius r consists of n turns per unit length. A current I = I0 sin ωt flows in the solenoid. A coil of N turns is wound tightly around it near its centre. What is :

(a) the induced emf in the coil ?

(b) the mutual inductance between the solenoid and the coil ?

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Official CBSE Marking Scheme · CBSE Board 2023 (Main), Set 55/1/1

Finding (a) Induced emf [2]; (b) Mutual inductance between solenoid and coil [1]

a) Magnetic field produced in the solenoid near the centre B = μ0nI [½]

Flux linked with the coil wound over solenoid φ = NBA = Nπr2B = Nπr2μ0nI [½]

Induced emf e = −dφdt = −πr2Nnμ0 dIdt   (i) [½]

= −μ0πr2nN I0ω cos ωt [½]

b) Comparing Eq. (i) with e = −M dIdt [½]

M = μ0πr2nN [½]

Q52.2023 Main 55/3/1 · Q273 marks

What is meant by the term ‘mutual inductance’ of a pair of coils ? Obtain an expression for the mutual inductance of two long coaxial solenoids, each of length l but having different number of turns N1 and N2 and radii r1 and r2 (r2 > r1).

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Official CBSE Marking Scheme · CBSE Board 2023 (Main), Set 55/3/1

Mutual Inductance: It is the magnetic flux in the secondary coil due to the flow of unit current in the primary coil. [1]

Alternatively: It is the emf induced in the secondary coil when rate of change of current in the primary coil is unity.

Expression for Mutual Inductance

Let N1 and N2 be the total number of turns of coils S1 & S2, respectively.

When a current I2 is set up through S2, it in turn sets up a magnetic flux through S1 ⇒ N1 φ1 = M12 I2   …(1) [½]

M12 is the mutual inductance of solenoid S1 w.r.t. S2.

Magnetic field due to I2 in S2 = μ0 (N2l) I2 [½]

N1 φ1 = (N1)(πr12)(μ0 N2l I2)   …(2) [½]

From equations (1) & (2) :

M12 = μ0 N1 N2 π r12l [½]

Q53.2023 Main 55/4/1 · Q273 marks

(a) The figure shows the variation of induced emf as a function of rate of change of current for two identical solenoids X and Y. One is air cored and the other is iron cored. Which one of them is iron cored ? Why ?

Induced emf versus dI/dt straight lines for solenoids X (steeper) and Y

(b) Obtain an expression for self-inductance of a long solenoid of length L and cross-sectional area A having N turns.

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Official CBSE Marking Scheme · CBSE Board 2023 (Main), Set 55/4/1

(a) X is iron cored. [½]

Reason – From the graph, slope of the graph = ε|dI/dt| = L [½]

Slope of X is more than that of Y. Hence X is iron cored because inductance of iron cored coil is more than that of air cored coil. [½]

(b) Magnetic field due to solenoid (axis) B = μ0NIl [½]

Magnetic flux through the solenoid ϕB = Nϕ = μ0N2AL I [½]

Since self-inductance = ϕBI = μ0N2AL [½]

Q54.2023 Main 55/5/1 · Q283 marks

State the basic principle behind the working of an ac generator. Briefly describe its working and obtain the expression for the instantaneous value of emf induced.

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Official CBSE Marking Scheme · CBSE Board 2023 (Main), Set 55/5/1

Principle: Whenever magnetic flux linked through a coil changes an emf is induced. [1]

Alternatively: ac generator is based on the phenomenon of electromagnetic induction.

Working: The coil (armature) is mechanically rotated in the uniform magnetic field by some external means. The rotation of the coil causes the magnetic flux through it to change. So an emf is induced in the coil. [1]

Expression for induced emf

Flux linked with the coil at any instant of time is – φB = BA cos ωt

ε = –N dφBdt [½]

ε = –NBA ddt(cos ωt)

Thus, instantaneous value of emf is ε = NBAω sin ωt [½]

Q55.2023 Supp 55/C/1 · Q26 (option 1)3 marks

State Lenz’s law. Determine the direction of the induced current when a rectangular conducting loop abcd in xy-plane is moved into a region of magnetic field which is directed along z-axis.

Rectangular loop abcd partly entering a region of magnetic field into the plane (crosses)
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Official CBSE Marking Scheme · CBSE Board 2023 (Supplementary), Set 55/C/1

Lenz’s law : “the polarity of induced e.m.f is such that it tends to produce a current which opposes the change in magnetic flux that produced it” [2]

Loop abcd entering magnetic field region

Direction of induced current is abcd / anticlockwise [1]

Q56.2023 Supp 55/C/1 · Q26 (OR option)3 marks

Two identical circular loops, one of copper and the other of aluminium are rotated about their diameters with the same angular speed in a magnetic field directed perpendicular to their axes of rotation. Compare (i) the emf induced, and (ii) the current in the two loops. Justify your answers.

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Official CBSE Marking Scheme · CBSE Board 2023 (Supplementary), Set 55/C/1

(i) Same e.m.f induced in both cases because they are identical and moving with same angular speed in same magnetic field. [½ + 1]

(ii) Induced current is more in copper loops, as its resistance is lesser than that of aluminum. [½ + 1]

Q57.SQP 2025-26 · Q283 marks

A conducting coil of 50 turns and area 5π cm2 is rotating along the axis of solenoid of length 50cm and 2000 turns, carrying current of 5 A. What will be the value of maximum emf generated?

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Official CBSE Marking Scheme · CBSE Sample Question Paper 2025-26

Given:

No of turns of coil Nc = 50

Area of coil = 5π cm2 = 5π × 10-4 m2

For solenoid:

Ns = 2000,

L = 0.5m,

n = N/L = 4000 turns/m,

I = 5A

Magnetic field due to solenoid ‘B’ = μonI

Or, B = 4000 × 4π × 10-7 × 5 T

Or, B = 8π × 10-2 T [1]

Flux linked to coil ΦB = NcB·A

Or, ΦB = NcBA cos ωt [1]

Emf ε = dΦBdt = NcBAω sin ωt

Or, εmax = NcBA

Or, εmax = 50 × 8π × 10-2 T × 5π × 10-4 m2

Or, εmax = 2 mV [1]

Q58.SQP 2026-27 · Q283 marks
  1. A bar magnet falls from a height through a metal ring. Its acceleration will be less than ‘g’. Explain.
  2. A coil of copper wire is wound uniformly on a wooden equilateral triangular frame. If each side of the triangle is increased to 5 times its original length, while keeping the number of turns per unit length unchanged, how does the self-inductance of the coil change?
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Official CBSE Marking Scheme · CBSE Sample Question Paper 2026-27

I. As the magnet falls, the magnetic flux linked with the copper ring increases. The current induced in the ring opposes the downward motion of the magnet. [1]

II. Self inductance, L ∝ N2Al [½]

Here N → 5N

Area as square of linear dimension, A → 52A = 25 A [½]

Magnetic path length, l → 5l [½]

L′L = (5N)2.25A/5lN2A/l = 25 × 255 = 125 [½]

L′ = 125 L [½]

Q59.2023 Main 55/2/1 · Q34 (option 1)4 marks

Consider the experimental set up shown in the figure. This jumping ring experiment is an outstanding demonstration of some simple laws of Physics. A conducting non-magnetic ring is placed over the vertical core of a solenoid. When current is passed through the solenoid, the ring is thrown off.

Jumping ring experiment: copper ring on an iron core inside a coil connected to a battery and switch

Answer the following questions :

  1. Explain the reason of jumping of the ring when the switch is closed in the circuit.
  2. What will happen if the terminals of the battery are reversed and the switch is closed ? Explain.
  3. Explain the two laws that help us understand this phenomenon.
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Official CBSE Marking Scheme · CBSE Board 2023 (Main), Set 55/2/1

(i) The direction of induced current in the ring is such that the polarity developed in the ring is same as that of the polarity on the face of the coil, hence it will jump up due to repulsive force. [1]

(ii) The polarity of the induced current in the ring will get reversed on changing the terminals of the battery, so the ring will jump again. [1]

(iii) Lenz’s law: It states that the polarity of induced emf is such that it tends to produce a current which opposes the change in magnetic flux that produces it. [1]

Faraday’s law of EMI: Whenever there is change in magnetic flux through a coil, an emf is induced. The magnitude of the induced emf in a coil is equal to the time rate of change of magnetic flux through the coil. [1]

Q60.2026 Main 55/1/1 · Q33 (option 1)5 marks

(a) State Faraday’s law of electromagnetic induction.

(b) Derive an expression for the self-inductance of an air-filled long solenoid of length l and cross-sectional area A having N turns.

(c) A conducting rod of length 50 cm, with one end pivoted, is rotated with angular speed of 60 rpm in a uniform magnetic field of 4.0 mT directed perpendicular to the plane of rotation of rod. Find the emf induced in the rod.

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Official CBSE Marking Scheme · CBSE Board 2026 (Main), Set 55/1/1

(a) The magnitude of the induced e.m.f. in a circuit is equal to the time rate of change of magnetic flux through the circuit. [1]

Alternatively:

ε = – dφBdt

(b) Magnetic field due to current carrying long solenoid of length l area of cross section A having n turns per unit length is B = μ0nI [½]

Total magnetic flux linked with the solenoid is

NφB = (nl)(μ0nI)(A) [½]

= μ0n2AlI

Where nl is total number of turns. Thus [½]

∴ Self inductance of the solenoid

L = NφBI

L = μ0n2Al [½]

Note: Award full marks for any other correct alternative method

(c) Induced e.m.f.

ε = 12Bl2ω [½]

= 12 × 4×10–3 × (50×10–2)2 × (2π × 1) [1]

= 3.14 mV [½]

Q61.2026 Main 55/2/1 · Q33 (OR option)5 marks

(i) With the help of a labelled diagram, explain the principle, construction and working of an a.c. generator.

(ii) Deduce an expression for the induced emf in the coil of the generator.

(iii) If T is the time period of the rotation of the coil, at what values of T in a cycle, the emf generator is maximum ?

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Official CBSE Marking Scheme · CBSE Board 2026 (Main), Set 55/2/1

(i)

AC generator: coil rotating between N and S poles with axle, slip rings, carbon brushes and alternating emf output

[1] for labelled diagram / construction

Principle : It is based on the principle of electromagnetic induction. When the magnetic flux passing through a coil changes continuously, emf is induced. [1]

Working : When coil is rotated in magnetic field, magnetic flux linked with it changes and emf is induced in the coil. [1]

(ii) Flux linked with a coil at any instant of time placed in magnetic field rotating with angular speed ω : φB = BA cosωt [½]

From Faraday’s law : ε = –N dφBdt [½]

ε = NBAω sinωt

(iii) EMF generated is maximum at T4 and 3T4 [1]

Q62.2026 Main 55/3/1 · Q32 (OR option)5 marks

(i) State Lenz’s law and explain that it follows the law of conservation of energy.

(ii) Write the dimensional formula for self-inductance.

The current in a coil changes from 8·0 A to 2·0 A in 0·6 s. If an average emf induced in the coil is 50 V, calculate the self-inductance of the coil.

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Official CBSE Marking Scheme · CBSE Board 2026 (Main), Set 55/3/1

(i) The polarity of induced emf is such that it tends to produce a current which opposes the change in magnetic flux that produces it. [1]

Consider north pole of a magnet being moved towards a coil. The magnetic flux through the coil increases, hence current is induced in counter clockwise direction making the side of the coil behave as north pole causing repulsion. If this is not the case then bringing magnet towards a coil will construct a perpetual motion machine by a suitable arrangement. Therefore production of induced emf in a direction which opposes the change in magnetic flux follows from law of conservation of energy. [1]

Alternatively

Magnet moved towards a coil with induced current direction

(Note: Award full credit for this part if a student draws the diagram and explains conservation of energy.)

(ii) [ML2T−2A−2] [1]

ε = − L dIdt [½]

L = −(50) × 0·6(2 − 8) [1]

= 5 H [½]

Q63.2026 Supp 55/7/1 · Q32 (OR option)5 marks
  1. An ac generator consists of a coil of 500 turns and size 25 cm × 100 cm. It is rotated at an angular speed of 60 rad/s in a uniform magnetic field B = 0·4 T between two fixed pole pieces. If the resistance of the circuit, including that of the coil, is 250 Ω, calculate :
    1. the maximum current drawn from the generator
    2. the flux through the coil when the current is zero
    [3]
  2. Give two examples of commercial generators. How is rotation of armature achieved in them ? Explain briefly. [2]
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Official CBSE Marking Scheme · CBSE Board 2026 (Supplementary), Set 55/7/1

(a) (i) N = 500, A = 25 × 100 × 10-4 m2, ω = 60 rad s-1, B = 0.4 T, R = 250 Ω [½]

I0 = NBAωR [½]

I0 = 500 × 0.4 × 25 × 100 × 10-4 × 60250 [½]

I0 = 12 A

(ii) When current through the coil = 0, flux is maximum. [½]

φmax = NBA [½]

φmax = 500 × 0.4 × 2500 × 10-4

φmax = 50 Wb [½]

(b) Hydroelectric Generator [½]

The mechanical energy required for rotation of the armature is provided by water falling from a height (dams). [½]

Thermal Generator [½]

Water is heated to produce steam using coal or other sources. The steam at high pressure produce the rotation. [½]

Nuclear Power Generator - Nuclear fuel is used to rotate the armature.

(Any two)

Q64.2025 Main 55/4/1 · Q32 (option 1)5 marks
  1. State Lenz’s law and explain how this law is a consequence of conservation of energy principle.
  2. A square shaped loop of side l2 is initially lying outside a region of uniform magnetic field B as shown in the figure. The loop is moved towards right with a constant velocity v till it goes out of the region of magnetic field.
    Square loop of side l/2 at distance l from a region of width l with uniform magnetic field B into the page, moving right with velocity v
    1. What will be the directions of induced current when the loop enters the field and when it leaves the field ?
    2. Draw the plots showing the variation of magnetic flux φ linked with the loop with time t and variation of induced emf E with time t. Mark the relevant values of E, φ and t on the graphs.
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Q65.2025 Main 55/5/1 · Q32 (option 1)5 marks

(i) Define self-inductance of a coil. Derive the expression for the energy required to build up a current I in a coil of self-inductance L.

(ii) The currents passing through two inductors of self-inductances 10 mH and 20 mH increase with time at the same rate.

Draw graphs showing the variation of :

(I) the magnitude of emf induced with the rate of change of current in each inductor.

(II) the energy stored in each inductor with the current flowing through it.

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Q66.2025 Main 55/5/1 · Q32 (OR option)5 marks

(i) Define the term mutual inductance. Deduce the expression for the mutual inductance of two long coaxial solenoids of the same length having different radii and different number of turns.

(ii) The current through an inductor is uniformly increased from zero to 2 A in 40 s. An emf of 5 mV is induced during this period. Find the flux linked with the inductor at t = 10 s.

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Q67.2025 Main 55/6/1 · Q31 (option 1)5 marks
  1. Write the principle of working of an ac generator. Draw its labelled diagram and explain its working.
  2. A resistor of 400 Ω, an inductor of (5π) H and a capacitor of (50π) μF are joined in series across an ac source v = 140 sin (100π)t V. Find the rms voltages across these three circuit elements. The algebraic sum of these voltages is more than the rms voltage of source. Explain.
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Q68.2024 Main 55/3/1 · Q32 (OR option)5 marks

(i) A rectangular coil of N turns and area of cross-section A is rotated at a steady angular speed ω in a uniform magnetic field. Obtain an expression for the emf induced in the coil at any instant of time.

(ii) Two coplanar and concentric circular loops L1 and L2 are placed coaxially with their centres coinciding. The radii of L1 and L2 are 1 cm and 100 cm respectively. Calculate the mutual inductance of the loops. (Take π2 = 10)

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Official CBSE Marking Scheme · CBSE Board 2024 (Main), Set 55/3/1

(i) Obtaining the expression of emf induced [3]; (ii) Calculation of mutual inductance [2]

AC generator: rectangular coil rotating between N and S poles with slip rings and carbon brushes

[1]

(i) The flux at any instant t is

φ = NBA cosθ = NBA cosωt [½]

From Faraday’s law

ε = – dφBdt [½]

= – NBA ddt (cosωt) [½]

ε = – NBA ω sinωt [½]

(ii) M = μ0πr122r2 = 4π×10–7×πr122r2 [½ + ½]

= 2×10–7×(10–2)2×10100×10–2 [½]

= 2 × 10–10 H [½]

Q69.2024 Main 55/4/1 · Q33 (OR option)5 marks
  1. With the help of a diagram, briefly explain the construction and working of ac generator.
  2. An electron is revolving around a proton in an orbit of radius r with a speed v. Obtain expression for magnetic moment associated with the electron.
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Official CBSE Marking Scheme · CBSE Board 2024 (Main), Set 55/4/1

(i) Diagram of ac generator [1]

AC generator: coil rotating between N and S poles, slip rings and carbon brushes

Construction – It consists of a coil placed in a magnetic field. The coil is mounted on a rotor shaft. The ends of the coil are connected to an external circuit by means of slip rings and brushes. [1]

Alternatively If a student draws only a labeled diagram of ac generator give 2 marks for construction and diagram.

Working – The coil is rotated in the uniform magnetic field by some external means. The rotation of the coil causes the magnetic flux through it to change, so an emf is induced in the coil. [1]

Alternatively If a student derives e = e0 sin ωt give one mark for working.

(ii) The equivalent current

I = qt = e2πrv = ev2πr [½]

Magnetic moment of revolving electron m = IA [½]

= ev2πr × πr2 [½]

= 12 evr [½]

Q70.2024 Supp 55/S/1 · Q32 (option 1)5 marks

(i) With the help of a labelled diagram, explain the working of an ac generator. Obtain the expression for the emf induced at an instant ‘t’.

(ii) A long, straight horizontal wire X′X is held stationary and carries a current of 3.0 A. A square loop MNPS of side 10 cm, carrying a current of 1.0 A is kept near the wire X′X as shown in the figure. Find the magnitude and direction of the net magnetic force acting on the loop due to the wire.

Straight wire X′X carrying 3 A and square loop MNPS carrying 1 A, with MN at 20 cm from the wire
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Official CBSE Marking Scheme · CBSE Board 2024 (Supplementary), Set 55/S/1

(i) Labelled diagram of ac generator [1]

AC generator: coil rotating between N and S poles with slip rings and carbon brushes

Working of ac generator

When coil is rotated in a uniform magnetic field with a constant angular speed ω, magnetic flux through it changes. As a result, an e.m.f is induced in the coil. [1]

Flux linked with the coil at any instant ‘t’ is

φB = BA cos ωt [½]

ε = – N dφBdt

ε = NBAω sin ωt [½]

(ii) F = μ0I1I2l2πr [½]

Force on arm MN of the loop

F1 = 4π×10–7 × 3 × 1 × 10×10–22π × 20×10–2

F1 = 3×10–7 N

Force is directed away from the wire [½]

Force on arm SP of the loop

F2 = 4π×10–7 × 3 × 1 × 10×10–22π × 30×10–2

F2 = 2×10–7 N

Force is directed towards the wire [½]

Net force on the loop

F = F1 – F2 = 10–7 N [½]

Net force on the loop is away from the wire. [½]

Q71.2024 Supp 55/S/1 · Q32 (OR option)5 marks

(i) State Faraday’s law of electromagnetic induction and mention the utility of Lenz’s law. Obtain an expression for self-inductance of a coil in terms of its geometry and permeability of the medium.

(ii) A resistance of 20 Ω, a capacitance of 80 µF and an inductor of 50 mH are connected in series. This combination is connected across a 220 V ac supply of variable frequency. When the frequency of supply equals the natural frequency of the circuit, calculate :

(1) angular frequency of supply

(2) impedance of the circuit

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Official CBSE Marking Scheme · CBSE Board 2024 (Supplementary), Set 55/S/1

(i) The magnitude of induced e.m.f in a circuit is equal to the time rate of change of magnetic flux through the circuit [½]

Utility of Lenz’s law

It give polarity of the induced e.m.f. [½]

Expression for self inductance

Consider a long solenoid of cross-sectional area A and length l, having n turns per unit length. If I is the current flowing in the solenoid, magnetic field inside the solenoid is

B = μ0nI [½]

Total magnetic flux linked with the solenoid is

NφB = (nl)(μ0nI)(A)

NφB = μ0n2AlI [½]

Self inductance

L = NφBI [½]

L = μ0n2Al

If solenoid is filled with a material of relative permeability μr, then

L = μrμ0n2Al [½]

(ii) (1) Resonant angular frequency is

ω0 = 1LC [½]

ω0 = 150×10–3 × 80×10–6 = 500 rad s–1 [½]

(2) When frequency of supply is equal to natural frequency of the circuit

Z = R [½]

Z = 20 Ω [½]

Q72.2023 Main 55/2/1 · Q32 (option 1)5 marks
  1. Define coefficient of self-induction. Obtain an expression for self-inductance of a long solenoid of length l, area of cross-section A having N turns.
  2. Calculate the self-inductance of a coil using the following data obtained when an AC source of frequency (200π) Hz and a DC source is applied across the coil.
AC Source
S.No.V (Volts)I (A)
13.00.5
26.01.0
39.01.5
DC Source
S.No.V (Volts)I (A)
14.01.0
26.01.5
38.02.0
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Official CBSE Marking Scheme · CBSE Board 2023 (Main), Set 55/2/1

(i) Coefficient of self induction is defined as the amount of magnetic flux associated with a coil when unit current flows through it. [1]

Alternatively

It is defined as the magnitude of emf induced in a coil when current changes at the rate of 1 A/s through it.

The magnetic field due to a current I flowing in solenoid is B = μ0NIl [½]

The total magnetic flux linked with solenoid

NφB = (N)(μ0NIl)(A) = μ0N2IAl [½]

The self inductance is L = NφBI [½]

L = μ0N2Al [½]

(ii) From the table, Z = 6 Ω, R = 4 Ω [½]

Z2 = R2 + XL2

XL2 = Z2 − R2 = 36 − 16 = 20

XL = 25 ≈ 4.5 Ω [½]

2πνL = 4.5 [½]

L = 4.52 × π × 200/π [½]

L = 1.1 × 10−2 H = 11 mH [½]

Note: Please do not deduct marks if a student writes answer as 0.55 × 10−2 H

Q73.2023 Main 55/2/1 · Q32 (OR option)5 marks
  1. With the help of a labelled diagram, describe the principle and working of an ac generator. Hence, obtain an expression for the instantaneous value of the emf generated.
  2. The coil of an ac generator consists of 100 turns of wire, each of area 0.5 m2. The resistance of the wire is 100 Ω. The coil is rotating in a magnetic field of 0.8 T perpendicular to its axis of rotation, at a constant angular speed of 60 radian per second. Calculate the maximum emf generated and power dissipated in the coil.
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Official CBSE Marking Scheme · CBSE Board 2023 (Main), Set 55/2/1

(i) Diagram [1]

AC generator: rectangular coil rotating between N and S poles with slip rings and carbon brushes

Principle – It is based on the principle of electromagnetic induction. Whenever there is a change in magnetic flux linked with a coil, an emf is induced in the coil. [½]

Alternatively

Give full credit if a candidate writes, it is based on the principle of electromagnetic induction.

Working – When a rectangular coil is rotated in a magnetic field, the magnetic flux changes continuously which induces an emf and the direction of current changes periodically. [½]

ε = −Ndφdt [½]

= −NBA ddt (cos ωt) [½]

ε = NBAω sin ωt [½]

(ii) ε0 = NBAω [½]

= 100 × 0.8 × 0.5 × 60 = 2400 V [½]

Power dissipated, P = εrms2R [½]

= (2400/2)2100 = 28.8 kW [½]

Alternatively

Give full credit if a candidate calculates power dissipated using formula εrmsIrms or Irms2R.

Unit IV · Electromagnetic Induction and Alternating Currents

7. Alternating Current

Q1.2026 Main 55/1/1 · Q61 mark

In a series LCR circuit, the voltage across the resistor, capacitor and inductor is 10 V each. If the capacitor is short circuited, the voltage across the inductor will be

  1. 10 V
  2. 52 V
  3. 52 V
  4. 102 V
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Official CBSE Marking Scheme · CBSE Board 2026 (Main), Set 55/1/1

(B) 52 V

Q2.2026 Main 55/2/1 · Q91 mark

The figure shows the variation of capacitive reactance (XC) of two ideal capacitors of capacitances C1 & C2 with the reciprocal of angular frequency (1/ω) of ac source. The value of C1/C2 is

Graph of X_C versus 1/ω: line C1 at 45° and line C2 at 30° to the 1/ω axis
  1. 12
  2. 2
  3. 3
  4. 13
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Official CBSE Marking Scheme · CBSE Board 2026 (Main), Set 55/2/1

(D) 13

Q3.2026 Main 55/4/1 · Q61 mark

An ac voltage is given as v = 14 sin (314t) V. The average and the effective value of the voltage (in V) over a cycle are respectively :

  1. 14 and 7
  2. 10 and 14
  3. 0 and 10
  4. 10 and 0
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Official CBSE Marking Scheme · CBSE Board 2026 (Main), Set 55/4/1

(C) 0 and 10

Q4.2026 Main 55/5/1 · Q51 mark

Which of the following statements is not true for electric energy in ac form compared to that in dc form ?

  1. Production of ac is economical.
  2. ac can be easily and efficiently converted from one voltage to the other.
  3. ac can be transmitted economically over long distances.
  4. ac is less dangerous.
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Official CBSE Marking Scheme · CBSE Board 2026 (Main), Set 55/5/1

(D) ac is less dangerous

Q5.2026 Main 55/5/1 · Q121 mark

The rms and the average value of an ac voltage V = V0 sin ωt volt over a cycle respectively will be :

  1. V02, V02
  2. V0π, V02
  3. V02, 0
  4. V0, V02
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Official CBSE Marking Scheme · CBSE Board 2026 (Main), Set 55/5/1

(C) V02, 0

Q6.2026 Supp 55/7/1 · Q61 mark

In a series LCR circuit connected to an ac source of voltage V and frequency ω, the power loss at resonance is :

  1. I2ωL
  2. V2/(ωC)
  3. V2/R
  4. V2/(ωL – 1ωC)
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Official CBSE Marking Scheme · CBSE Board 2026 (Supplementary), Set 55/7/1

(C) V2/R

Q7.2025 Main 55/1/1 · Q61 mark

A voltage v = v0 sin ωt applied to a circuit drives a current i = i0 sin (ωt + φ) in the circuit. The average power consumed in the circuit over a cycle is

  1. Zero
  2. i0 v0 cos φ
  3. i0 v02
  4. i0 v02 cos φ
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Q8.2025 Main 55/2/1 · Q61 mark

The alternating current I in an inductor is observed to vary with time t as shown in the graph for a cycle.

Triangular current waveform I versus t: rises linearly from 0 to peak at T/2 then falls linearly to 0 at T

Which one of the following graphs is the correct representation of wave form of voltage V with time t ?

  1. Triangular voltage waveform, positive, peak at T/2
  2. Triangular voltage waveform, positive then negative
  3. Square voltage waveform, positive then negative
  4. Sinusoidal voltage waveform
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Q9.2025 Main 55/2/1 · Q71 mark

A transformer is connected to a 200 V ac source. The transformer supplies 3000 V to a device. If the number of turns in the primary coil is 450, then the number of turns in its secondary coil is –

  1. 30
  2. 450
  3. 4500
  4. 6750
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Q10.2025 Main 55/2/1 · Q161 mark

For Questions 13 to 16, two statements are given – one labelled Assertion (A) and other labelled Reason (R). Select the correct answer to these questions from the codes (A), (B), (C) and (D) as given below :

Assertion (A) : In an ideal step-down transformer, the electrical energy is not lost.

Reason (R) : In a step-down transformer, voltage decreases but the current increases.

  1. If both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of Assertion (A).
  2. If both Assertion (A) and Reason (R) are true but Reason (R) is not the correct explanation of Assertion (A).
  3. If Assertion (A) is true but Reason (R) is false.
  4. If both Assertion (A) and Reason (R) are false.
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Q11.2025 Main 55/4/1 · Q61 mark

An ammeter connected in series in an ac circuit reads 10 A. The maximum value of current at any instant in the circuit is :

  1. 10 2 A
  2. 102 A
  3. 10π A
  4. 102 π A
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Q12.2025 Main 55/4/1 · Q141 mark

Assertion (A) : A series LCR circuit behaves as a pure resistive circuit at resonance.

Reason (R) : At resonance, XL = XC gives ω = 1LC.

  1. Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of the Assertion (A).
  2. Both Assertion (A) and Reason (R) are true, but Reason (R) is not the correct explanation of the Assertion (A).
  3. Assertion (A) is true, but Reason (R) is false.
  4. Both Assertion (A) and Reason (R) are false.
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Q13.2025 Main 55/6/1 · Q81 mark

An ac source is connected to a resistor and an inductor in series. The voltage across the resistor and inductor are 8 V and 6 V respectively. The voltage of the source is :

  1. 10 V
  2. 12 V
  3. 14 V
  4. 16 V
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Q14.2025 Main 55/7/1 · Q41 mark

An alternating current is given by I = I0 cos (100π)t. The least time the current takes to decrease from its maximum value to zero will be :

  1. (1200) s
  2. (1150) s
  3. (1100) s
  4. (150) s
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Q15.2025 Main 55/7/1 · Q51 mark

A capacitor and an inductor are connected in series across an ac source of voltage of variable frequency. The frequency is increased continuously. The nature of the circuit before and after the resonance will be :

  1. inductive only
  2. capacitive only
  3. capacitive and inductive respectively
  4. inductive and capacitive respectively
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Q16.2025 Supp 55/S/1 · Q101 mark

The average value of voltage of 220 V ac mains during its positive half cycle will be :

  1. 156 V
  2. 198 V
  3. 220 V
  4. zero
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Official CBSE Marking Scheme · CBSE Board 2025 (Supplementary), Set 55/S/1

(B) 198 V

Q17.2025 Supp 55/S/1 · Q121 mark

Which of the following assumptions has been used to obtain the relation vsvp = NsNp for a transformer ?

  1. The resistance of the primary coil is large.
  2. The same flux links both the primary and the secondary coils.
  3. The resistance of secondary coil is large.
  4. The transformer is 100% efficient.
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Official CBSE Marking Scheme · CBSE Board 2025 (Supplementary), Set 55/S/1

(B) The same flux links both the primary and secondary coils

Q18.2024 Main 55/1/1 · Q71 mark

The reactance of a capacitor of capacitance C connected to an ac source of frequency ω is ‘X’. If the capacitance of the capacitor is doubled and the frequency of the source is tripled, the reactance will become :

  1. X6
  2. 6X
  3. 23 X
  4. 32 X
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Official CBSE Marking Scheme · CBSE Board 2024 (Main), Set 55/1/1

(A) X6

Q19.2024 Main 55/2/1 · Q51 mark

Which of the following quantity/quantities remains same in primary and secondary coils of an ideal transformer ?
Current, Voltage, Power, Magnetic flux

  1. Current only
  2. Voltage only
  3. Power only
  4. Magnetic flux and Power both
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Official CBSE Marking Scheme · CBSE Board 2024 (Main), Set 55/2/1

(D) Magnetic flux and Power both

Q20.2024 Main 55/2/1 · Q61 mark

A resistor and an ideal inductor are connected in series to a 1002 V, 50 Hz ac source. When a voltmeter is connected across the resistor or the inductor, it shows the same reading. The reading of the voltmeter is :

  1. 1002 V
  2. 100 V
  3. 502 V
  4. 50 V
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Official CBSE Marking Scheme · CBSE Board 2024 (Main), Set 55/2/1

(B) 100 V

Q21.2024 Main 55/4/1 · Q91 mark

Figure shows the variation of inductive reactance XL of two ideal inductors of inductance L1 and L2, with angular frequency ω. The value of L1L2 is

Graph of X_L versus ω: line L1 at 30° and line L2 at 60° to the ω-axis
  1. 3
  2. 13
  3. 3
  4. 13
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Official CBSE Marking Scheme · CBSE Board 2024 (Main), Set 55/4/1

(D) 13

Q22.2024 Supp 55/S/1 · Q21 mark

A coil of resistance 20 Ω and self-inductance 10 mH is connected to an ac source of frequency 1000/π Hz. The phase difference between current in the circuit and the source voltage is :

  1. 30°
  2. 60°
  3. 75°
  4. 45°
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Official CBSE Marking Scheme · CBSE Board 2024 (Supplementary), Set 55/S/1

(D) 45°

Q23.2024 Supp 55/S/1 · Q41 mark

A bulb is rated (100 W, 110 V). It is operated by current of 1.0 A supplied by a step down transformer. If the input voltage and efficiency of the transformer are 220 V and 0.9 respectively, the input current drawn from the mains is :

  1. 12 A
  2. 38 A
  3. 59 A
  4. 47 A
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Official CBSE Marking Scheme · CBSE Board 2024 (Supplementary), Set 55/S/1

(C) 59 A

Q24.2023 Main 55/3/1 · Q61 mark

An inductor, a capacitor and a resistor are connected in series across an ac source of voltage. If the frequency of the source is decreased gradually, the reactance of :

  1. both the inductor and the capacitor decreases.
  2. inductor decreases and the capacitor increases.
  3. both the inductor and the capacitor increases.
  4. inductor increases and the capacitor decreases.
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Official CBSE Marking Scheme · CBSE Board 2023 (Main), Set 55/3/1

(B) inductor decreases and the capacitor increases

Q25.2023 Main 55/3/1 · Q151 mark

A voltage signal is described by :

v = V0   for 0 ≤ t ≤ T2
  = 0   for T2 ≤ t ≤ T

for a cycle. Its rms value is :

  1. V02
  2. V0
  3. V02
  4. 2 V0
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Official CBSE Marking Scheme · CBSE Board 2023 (Main), Set 55/3/1

(A) V02

Q26.2023 Main 55/4/1 · Q41 mark

The figure shows variation of current (I) with time (t) in four devices P, Q, R and S. The device in which an alternating current flows is :

Current versus time graphs for devices P, Q, R and S
  1. P
  2. Q
  3. R
  4. S
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Official CBSE Marking Scheme · CBSE Board 2023 (Main), Set 55/4/1

(D) S

Q27.2023 Main 55/4/1 · Q101 mark

An ac voltage v = v0 sin ωt is applied to a series combination of a resistor R and an element X. The instantaneous current in the circuit is I = I0 sin (ωt + π4). Then which of the following is correct ?

  1. X is a capacitor and XC = 2 R
  2. X is an inductor and XL = R
  3. X is an inductor and XL = 2 R
  4. X is a capacitor and XC = R
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Official CBSE Marking Scheme · CBSE Board 2023 (Main), Set 55/4/1

(D) X is a capacitor and XC = R

Q28.2023 Main 55/5/1 · Q131 mark

Which of the following statements about a series LCR circuit connected to an ac source is correct ?

  1. If the frequency of the source is increased, the impedance of the circuit first decreases and then increases.
  2. If the net reactance (XL – XC) of circuit becomes equal to its resistance, then the current leads the voltage by 45°.
  3. At resonance, the voltage drop across the inductor is more than that across the capacitor.
  4. At resonance, the voltage drop across the capacitor is more than that across the inductor.
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Official CBSE Marking Scheme · CBSE Board 2023 (Main), Set 55/5/1

(A) If the frequency of the source is increased, the impedance of the circuit first decreases and then increases.

Q29.2023 Supp 55/C/1 · Q41 mark

In a series LC circuit connected to an ac source, with the increase in the frequency of the source, the net reactance :

  1. increases linearly
  2. decreases linearly
  3. first increases to become maximum and then decreases to zero
  4. first decreases to become zero and then increases
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Official CBSE Marking Scheme · CBSE Board 2023 (Supplementary), Set 55/C/1

(D) first decreases to become zero and then increases

Q30.2023 Supp 55/C/1 · Q91 mark

The given figure shows a capacitor C and a resistor R connected in series to an ac source. V1 and V2 are voltmeters and A is an ammeter.

Series C and R across ac source with voltmeter V1 across C, voltmeter V2 across R, ammeter A

Which of the following statements is correct ?

  1. Current in the circuit lags in phase with voltage shown in V2.
  2. The voltage shown in V1 is ahead in phase with voltage shown in V2.
  3. The current in the circuit and the voltage shown in V1 are always in phase.
  4. The voltage shown in V1 lags behind in phase with the voltage shown in V2.
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Official CBSE Marking Scheme · CBSE Board 2023 (Supplementary), Set 55/C/1

(D) The voltage shown in V1 lags behind in phase with the voltage shown in V2.

Q31.SQP 2025-26 · Q31 mark

If the phasor diagram for a device connected to AC supply is as shown in the fig, then which of the following statements is true?

Phasor diagram showing voltage phasor V and current phasor I perpendicular to each other
  1. When the frequency of the AC source is increased than the impedance of the device decreases.
  2. This device behaves as conducting wire when connected across DC source.
  3. When the frequency of the AC source is decreased than the impedance of the device decreases.
  4. This device stores energy in the form of magnetic potential energy.
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Official CBSE Marking Scheme · CBSE Sample Question Paper 2025-26

(A) When the frequency of the AC source is increased than the impedance of the device decreases.

As in phasor diagram current leads the voltage, so given appliance is capacitor.

Q32.2026 Main 55/3/1 · Q243 marks

(a) Discuss the behaviour of an inductor connected to (i) a dc source, and (ii) a high frequency ac source.

(b) What is the phase relation between current and voltage in an ideal inductor connected to an ac source ? Draw a phasor diagram for the circuit.

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Official CBSE Marking Scheme · CBSE Board 2026 (Main), Set 55/3/1

(a) (i) Initially inductor opposes the change in current and gradually current becomes steady. [½]

(Alternatively) Inductor behaves like a conductor.

(ii) The magnetic field changes continuously around inductor and induces the back emf. At high frequency, the reactance becomes high, blocking the flow of current. [½]

(b) Current lags behind the voltage by an angle π2 radian. [1]

Phasor diagram [1]

Phasor diagram and waveforms showing current lagging voltage by π/2 in an inductor
Q33.2025 Supp 55/S/1 · Q25 (option 1)3 marks

A 20 μF capacitor is connected to a 220 V, 125π Hz ac source.

(i) Find the capacitive reactance and the peak value of current in the circuit.

(ii) If the frequency of the ac source is doubled, by what factor will the capacitive reactance and the current be changed ?

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Official CBSE Marking Scheme · CBSE Board 2025 (Supplementary), Set 55/S/1

(i) Xc = 12πvC [½]

= 12π × 125π × 20 × 10–6 [½]

= 200 Ω

I0 = V0XC [½]

= 220 × 1.414200

= 1.6 A or 1.12 A [½]

(ii) Xc becomes half [½]

Current becomes double. [½]

Q34.2024 Main 55/1/1 · Q253 marks

An inductor, a capacitor and a resistor are connected in series with an ac source v = vm sin ωt. Derive an expression for the average power dissipated in the circuit. Also obtain the expression for the resonant frequency of the circuit.

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Official CBSE Marking Scheme · CBSE Board 2024 (Main), Set 55/1/1

Deriving expression for average power in series LCR circuit [2]; obtaining expression for the resonant frequency [1]

v = vm sin ωt

i = im sin (ωt + φ)

Power, P = v i = (vm sin ωt) × [im sin (ωt + φ)] [½]

= vm im2 [cos φ − cos (2ωt + φ)]   (1) [½]

The average power over a cycle is given by the average of the two terms in RHS of eqn (1). It is only the 2nd term which is time dependent. Its average is zero. Therefore,

P = vm im2 cos φ [½]

P = V I cos φ
OR
P = I2 Z cos φ [½]

At resonance, XC = XL

1ωC = ωL [½]

ω = 1(LC)

⇒ υ = 12π(LC) [½]

Q35.2024 Main 55/3/1 · Q243 marks

A sinusoidal voltage is applied to an electric circuit containing a circuit element ‘X’ in which the current leads the voltage by π2.

  1. Identify the circuit element ‘X’ in the circuit.
  2. Write the formula for its reactance.
  3. Show graphically the variation of this reactance with frequency of ac voltage.
  4. Explain the behaviour of this element when it is used in (i) an ac circuit, and (ii) a dc circuit.
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Official CBSE Marking Scheme · CBSE Board 2024 (Main), Set 55/3/1

(a) Capacitor [½]

(b) Xc = 1ωC [½]

(c) [1]

Graph of capacitive reactance falling with increasing frequency

(d) (i) For ac Xc is finite and therefore allows the ac to pass. [½]

     (ii) For dc Xc is infinite and therefore does not allow the dc to pass. [½]

Q36.2023 Main 55/1/1 · Q27 (option 1)3 marks

How does the resistance differ from impedance ? With the help of a suitable phasor diagram, obtain an expression for impedance of a series LCR circuit, connected to a source v = vm sin ωt.

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Official CBSE Marking Scheme · CBSE Board 2023 (Main), Set 55/1/1

Difference between resistance and impedance [1]; Obtaining expression for impedance [2]

1. Resistance is the opposition offered to both alternating current and direct current while impedance is the opposition offered to alternating current only.

2. Resistance is independent of frequency of source while impedance depends on frequency.

3. Resistance is opposition offered by material of the conductor while impedance is combined opposition offered by different electrical components such as resistor, inductor or capacitor.

(Any One) [1]

(Note: Give credit of this part if a student writes any other correct answer.)

Phasor diagram of a series LCR circuit showing I, V_R, V_L and V_C

[½]

VR = imR, VC = imXC, VL = imXL [½]

im = Peak value of current in the circuit.

VL + VR + VC = Vm

(Vm)2 = VR2 + (VC − VL)2

= (imR)2 + (imXC − imXL)2 [½]

= im2 [R2 + (XC − XL)2]

im = VmR2 + (XC − XL)2

im = VmZ where Z = R2 + (XC − XL)2 = impedance [½]

Q37.2023 Main 55/1/1 · Q27 (OR option)3 marks

Find the condition for resonance in a series LCR circuit connected to a source v = vm sin ωt, where ω can be varied. Give the factors on which the resonant frequency of a series LCR circuit depends. Plot a graph showing the variation of electric current with frequency in a series LCR circuit.

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Official CBSE Marking Scheme · CBSE Board 2023 (Main), Set 55/1/1

Finding condition for resonance [1]; Factors affecting resonant frequency [1]; Graph [1]

Z = R2 + (XL − XC)2 [½]

For maximum current, Z should be minimum therefore to minimize Z, XL = XC [½]

Alternatively

XL = XC [½]

ωL = 1ωC

ωr = 1LC [½]

Resonant frequency depends on value of Inductance and Capacitance [½ + ½]

Current versus angular frequency curve for series LCR circuit, peaking at resonant frequency

[1]

Q38.2023 Main 55/2/1 · Q283 marks

A series CR circuit with R = 200 Ω and C = (50/π) μF is connected across an ac source of peak voltage ε0 = 100 V and frequency ν = 50 Hz. Calculate (a) impedance of the circuit (Z), (b) phase angle (φ), and (c) voltage across the resistor.

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Official CBSE Marking Scheme · CBSE Board 2023 (Main), Set 55/2/1

(a) Z = R2 + Xc2 = R2 + (12πνC)2 [½]

Xc = 12πνC = 12 × π × 50 × (50/π) × 10−6 = 200 Ω [½]

Z = (200)2 + (200)2 = 2002 Ω ≈ 282 Ω [½]

(b) tan φ = XcR = 200200 [½]

φ = 45° or π4 rad [½]

(c) Vrms = IrmsR = VrmsZ R [½]

= 1002 × 2002 × 200 = 50 V [½]

Q39.2023 Main 55/3/1 · Q28 (option 1)3 marks

An ac source v = vm sin ωt is connected across an ideal capacitor. Derive the expression for the (i) current flowing in the circuit, and (ii) reactance of the capacitor. Plot a graph of current i versus ωt.

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Official CBSE Marking Scheme · CBSE Board 2023 (Main), Set 55/3/1

(i) vm sin ωt = qC

I = dqdt = ddt(C vm sin ωt) [½]

I = ωC vm cos ωt [½]

Alternatively: I = vm1/ωC cos ωt = Im sin(ωt + π2)

(ii) I = vm1/ωC sin(ωt + π2) = Im sin(ωt + π2)

Comparing with Im = vm1/ωC

Reactance of the capacitor : Xc = 1ωC [1]

Graph of i versus ωt [1]

Voltage (solid) and current (dashed) versus ωt for a capacitor; current leads voltage by π/2
Q40.2023 Main 55/3/1 · Q28 (OR option)3 marks

A series combination of an inductor L, a capacitor C and a resistor R is connected across an ac source of voltage in a circuit. Obtain an expression for the average power consumed by the circuit. Find power factor for (i) purely inductive circuit, and (ii) purely resistive circuit.

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Official CBSE Marking Scheme · CBSE Board 2023 (Main), Set 55/3/1

Instantaneous Power :

P = VI = (Vm sin ωt) × Im sin(ωt + φ) [½]

P = Vm Im2 [cos φ – cos(2ωt + φ)]   …(1) [½]

The average power over a cycle is given by the average of the two terms in the R.H.S of equation (1). It is only the second term which is time dependent. Its average is zero (the positive half of the cosine cancels the negative half). Therefore, [½]

Pavg = Vm Im2 cos φ = Vm2 Im2 cos φ

Pavg = Vrms Irms cos φ [½]

Alternatively: If the expression is deduced using integration, then full credit to be given.

(i) Power factor for purely inductive circuit, φ = π2 ⇒ cos φ = 0 [½]

(ii) Power factor for purely resistive circuit, φ = 0 ⇒ cos φ = 1 [½]

Q41.2023 Main 55/4/1 · Q28 (option 1)3 marks

A resistor of 30 Ω and a capacitor of 250π μF are connected in series to a 200 V, 50 Hz ac source. Calculate (i) the current in the circuit, and (ii) voltage drops across the resistor and the capacitor. (iii) Is the algebraic sum of these voltages more than the source voltage ? If yes, solve the paradox.

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Official CBSE Marking Scheme · CBSE Board 2023 (Main), Set 55/4/1

XC = 1ωC; ω = 2πυ = 100π

XC = 1100π × 250/π × 10−6 = 40 Ω

Impedance of the circuit Z = XC2 + R2 = (40)2 + (30)2 = 50 Ω

(i) Current in the circuit Irms = VrmsZ = 20050 = 4 A [1]

(ii) Voltage drop across the capacitor, VC = Irms XC = 4 × 40 = 160 V [½]

Voltage drop across the resistor, VR = Irms × R = 4 × 30 = 120 V [½]

(iii) The algebraic sum of the two voltages VR and VC is 280 V, which is more than the source voltage of 200 V. [½] This paradox can be removed by considering impedance triangle because VR and VC are out of phase by 90°, therefore

V = VR2 + VC2 = (120)2 + (160)2 = 14400 + 25600 = 200 V

This is equal to the source voltage. [½]

Q42.2023 Main 55/4/1 · Q28 (OR option)3 marks

A series LCR circuit with R = 20 Ω, L = 2 H and C = 50 μF is connected to a 200 volts ac source of variable frequency. What is (i) the amplitude of the current, and (ii) the average power transferred to the circuit in one complete cycle, at resonance ? (iii) Calculate the potential drop across the capacitor.

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Official CBSE Marking Scheme · CBSE Board 2023 (Main), Set 55/4/1

(i) At resonance, Z = R

Irms = 20020 = 10 A [½]

Amplitude of the current I0 = 2 × Irms = 1.414 × 10 = 14.14 A [½]

(ii) Average power transferred to the circuit in one complete cycle at resonance

P = Irms2 R = (10)2 × 20 [½]

P = 2000 W [½]

(iii) Resonant frequency ωr = 1LC = 12 × 50 × 10−6 = 100 rad/s

XC = 1ωrC = 1100 × 50 × 10−6 [½]

VC = Irms XC = 10 × 1100 × 50 × 10−6 = 2000 V [½]

Q43.2023 Main 55/5/1 · Q273 marks

An alternating voltage of 220 V is applied across a device X. A current of 0.22 A flows in the circuit and it lags behind the applied voltage in phase by π/2 radian. When the same voltage is applied across another device Y, the current in the circuit remains the same and it is in phase with the applied voltage.

  1. Name the devices X and Y and,
  2. Calculate the current flowing in the circuit when the same voltage is applied across the series combination of X and Y.
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Official CBSE Marking Scheme · CBSE Board 2023 (Main), Set 55/5/1

(i) X is inductor [½]

Y is resistor [½]

(ii) XL = 2200.22 = 1000 Ω [½]

R = 2200.22 = 1000 Ω [½]

Irms = VrmsXL2 + R2 [½]

Irms = 2202 × 1000 = 0.222 = 0.11 × 2 A [½]

Irms = 0.16 A

Q44.2023 Supp 55/C/1 · Q283 marks

The figure shows a capacitor C, a variable resistor R and a bulb connected in series to the ac source of frequency (ν). The bulb glows with some brightness.

Series circuit of capacitor C, variable resistor R, bulb B and ac source

How will the glow of the bulb be affected, if the

(a) separation between the plates of the capacitor is doubled, keeping resistance R and frequency (ν) the same ?

(b) resistance R is decreased keeping the value of capacitance C and frequency (ν) same ?

(c) frequency of ac source is decreased for the same value of C and R ? Justify your answer in each case.

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Official CBSE Marking Scheme · CBSE Board 2023 (Supplementary), Set 55/C/1

(a) Glow of bulb will reduce. [½]

As capacitance reduces to half the net impedance of the circuit will increase and I decreases. [½]

(b) Glow will enhance [½]

as R is decreased, Z decreases and I increases [½]

(c) Glow of bulb will reduce [½]

When frequency is decreased, impedance increases which decreases current in the circuit. [½]

(Note: If a student attempt using relevant formulas and conclude correctly give full credit)

Q45.2026 Main 55/1/1 · Q33 (OR option)5 marks

(a) Draw a labelled diagram of a step-up transformer. State the principle on which it works and obtain the ratio of secondary voltage to primary voltage in terms of number of turns and currents in the two coils.

(b) The ratio of the number of turns in the primary to the secondary of an ideal transformer is 1 : 5. If 5 kW power at 200 V is supplied to the primary, find

  1. current in the primary, and
  2. output voltage.
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Official CBSE Marking Scheme · CBSE Board 2026 (Main), Set 55/1/1

(a) Fig [1]

Transformer: primary and secondary windings on a soft iron core

Alternatively:

Transformer: primary and secondary windings on a rectangular soft iron core

Principle: When an alternating voltage is applied to the primary, the resulting current produces an alternating magnetic flux which links the secondary and induces an e.m.f in it / mutual induction. [½]

Let φ be the flux in each turn in the core at time t due to current in the primary when a voltage vp is applied to it. The induced e.m.f in the secondary coil is

εs = –Nsdφdt ………… (1) [½]

Alternating flux φ also induces an e.m.f. called back e.m.f in the primary. This is

εp = –Npdφdt ………… (2)

But εp = vp

If this were not so, the primary current would be infinite since the primary has zero resistance. If the secondary is an open circuit or the current taken from it is small then εs = vs

Now equations (1) and (2) can be written as

∴ vs = –Nsdφdt ………… (3) [½]

vp = –Npdφdt ………… (4)

From equations (3) and (4)

vsvp = NsNp [½]

If the transformer is ideal the input power is equal to the power output

Since P = IV

IpVp = IsVs

VpVs = IsIp [½]

(b)

(i) P = VpIp

5000 = 200Ip

Ip = 25 A [½]

(ii) ∵ NpNs = VpVs

15 = 200Vs [½]

Vs = 1000 V [½]

Q46.2026 Main 55/2/1 · Q33 (option 1)5 marks

(a) A series combination of L, C and R is connected to an a.c. source. Using a phasor diagram, derive an expression for the impedance of the circuit and phase difference between V and I.

(b) Under what conditions the

(i) impedance of the circuit is minimum ?

(ii) Wattless current flows in the circuit ?

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Official CBSE Marking Scheme · CBSE Board 2026 (Main), Set 55/2/1

(a)

Phasor diagram of series LCR circuit: V_R, (V_C + V_L), resultant V and phase angle φ

[½] for phasor diagram

VL + VR + VC = V. Applying Pythagorean theorem, we get

vm2 = vRm2 + (vCm – vLm)2

vm2 = (imR)2 + (imXC – imXL)2 [1]

Im = vmR2 + (XC – XL)2 = vmZ

Where Z = R2 + (XC – XL)2 is impedance [½]

tan φ = VCm – VLmVRm = XC – XLR [1]

Or φ = tan–1(XC – XLR)

(b) (i) At resonance i.e., XC = XL [1]

(ii) When the average power consumed is zero. [1]

Alternatively : Pavg = VrmsIrms cosφ. For φ = π2, cosφ = 0, ∴ Pavg = 0

Q47.2026 Main 55/3/1 · Q32 (option 1)5 marks

(i) A light bulb and an open coil inductor are connected in series across an ac source of variable frequency. How will the glow of the bulb be affected when :

(I) an iron bar is inserted inside the coil, and

(II) the frequency of the source is decreased ?

Justify your answers. Assume that in each above case other factors remain unchanged.

(ii) An ac voltage V = 280 sin (100π t) volt is connected across a series LCR circuit in which R = 400 Ω, L = 5π H and C = 50π μF. Taking 2 = 1·4, calculate :

(I) impedance of the circuit.

(II) rms value of current that flows in the circuit.

(III) power factor of the circuit.

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Official CBSE Marking Scheme · CBSE Board 2026 (Main), Set 55/3/1

(i) (I) Brightness of the bulb decreases [½]

As iron bar is inserted, inductive reactance of the coil increases. Larger fraction of applied AC voltage appears across inductor, leaving less voltage across the bulb. [½]

(Alternatively) L′ = μrL, L increases, XL = ωL increases, current decreases.

(II) Brightness of the bulb increases [½]

As frequency of the source decreases, inductive reactance of coil decreases. Less fraction of applied AC voltage appears across inductor. Hence higher voltage appears across the bulb. [½]

(Alternatively) XL = ωL, ω decreases, XL decreases, current increases.

(ii) V = 280 sin (100πt)

XL = 100π × 5π = 500 Ω

XC = 106100π × 50π = 200 Ω

(I) Z = R2 + (XL − XC)2 [½]

= (400)2 + (300)2

= 500 Ω [½]

(II) Irms = (V0Z2) = 280500 × 1·4 [½]

= 0·4 A [½]

(III) Power factor = cos φ = R/Z [½]

= 400/500

= 0·8 [½]

Q48.2026 Main 55/4/1 · Q32 (option 1)5 marks

(a) (i) Define the terms (I) resonant frequency, and (II) power factor of a series LCR circuit. For what value of the power factor will the power dissipated in the circuit be maximum ?

(ii) An inductor of 5π H, a capacitor of 50π μF and a resistor of 400 Ω are connected in series across an ac voltage v = 140 sin (100π t) V. Calculate :

  1. impedance of the circuit, and
  2. rms value of current that flows in the circuit.

(Take 2 = 1·4)

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Official CBSE Marking Scheme · CBSE Board 2026 (Main), Set 55/4/1

(i) (I) For series LCR circuit, the frequency at which the current amplitude is maximum. [1]

Alternatively : The frequency at which the impedance is minimum (Z = R).

Alternatively : Frequency at which XL = XC

(II) Power factor : It is the ratio of resistance to impedance of the series LCR circuit. [1]

Alternatively : cos φ = RZ, where φ is the angle between the voltage and the current.

Alternatively : cos φ = VRV

P = VeffIeff cos φ, P is maximum when cos φ is unity. [1]

(ii) (I) Z = R2 + (ωL − 1ωC)2 [½]

= (400)2 + (100π × 5π − 1 × π100π × 50 × 10−6)2

= 160000 + 90000 = 250000

= 500 Ω [½]

(II) Irms = V0Z2 [½]

= 140500 × 1.4 = 0.2 A [½]

Q49.2026 Main 55/4/1 · Q32 (OR option)5 marks

(b) (i) Draw a labelled diagram of a step-up transformer. Obtain the ratio of secondary voltage to primary voltage in terms of number of turns in the two coils.

(ii) The number of turns in the primary and the secondary coil of an ideal transformer are 100 and 5000 respectively. If 3·3 kW power is supplied to the transformer at 220 V, find (I) current in the primary coil, and (II) output voltage.

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Official CBSE Marking Scheme · CBSE Board 2026 (Main), Set 55/4/1

(i) Labelled diagram [1]

Transformer with iron core, primary coil and secondary coil, voltages V<sub>p</sub> and V<sub>s</sub>

(Anyone of the above)

The induced emf or voltage across secondary with Ns turns is : εs = −Ns dφdt

The alternating φ flux also induces an emf, called back emf. This is : εP = −NP dφdt

εP = vP [∵ primary coil has zero resistance]

εs = vs [∵ secondary coil is an open circuit]

vs = −Ns dφdt ……(i) [½]

vP = −NP dφdt ……(ii) [½]

Dividing (i) by (ii) : vsvP = NsNP [½]

(ii) (I) P = VPIP [½]

3.3 × 103 = 220 × IP

IP = 3.3 × 103220 = 15 A [½]

(II) vsvP = NsNP [½]

220vs = 1005000

vs = 11000 V [½]

Q50.2026 Supp 55/7/1 · Q32 (option 1)5 marks
  1. Using a phasor diagram, derive an expression for the impedance of a series LCR circuit connected to an ac source. What is meant by resonance in this circuit ? Obtain expression for the resonant frequency of this circuit. [3]
  2. Write two disadvantages of transmitting ac over long distances at low voltage and high current. [2]
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Official CBSE Marking Scheme · CBSE Board 2026 (Supplementary), Set 55/7/1

(a)

Phasor diagram of series LCR circuit showing V_R, V_C + V_L and V

[½]

For : VC > VL

VRm = ImR, VCm = ImXC, VLm = ImXL

Phasor relation : VL + VR + VC = V [½]

Using Pythagoras’ theorem,

Vm2 = VRm2 + (VCm – VLm)2

Im2Z2 = (ImR)2 + (ImXC – ImXL)2

Im2Z2 = Im2[R2 + (XC – XL)2]

Z = R2 + (XC – XL)2 [½]

Resonance is achieved when : The voltage across the capacitor becomes out of phase with the voltage across the inductor. / The current amplitude becomes maximum. / The impedance becomes minimum. [½]

(Any one)

Expression for Resonant Frequency : For Z = minimum, XC = XL

1ω0C = ω0L

ω0 = 1LC or ν0 = 12πLC [½]

(b) (i) Heat loss (I2Rt) will be higher.

(ii) Voltage drop (IR) along the transmission line will be more.

(iii) Installation cost will be higher.

(Any two) [1 + 1]

Q51.2025 Main 55/4/1 · Q32 (OR option)5 marks
  1. Differentiate between peak and rms values of alternating current. How are they related ?
  2. A current element X is connected across an ac source of emf V = V0 sin 2πνt. It is found that the voltage leads the current in phase by π2 radian. If element X was replaced by element Y, the voltage lags behind the current in phase by π2 radian.
    1. Identify elements X and Y by drawing phasor diagrams.
    2. Obtain the condition of resonance when both elements X and Y are connected in series to the source and obtain expression for resonant frequency. What is the impedance value in this case ?
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The official marking scheme for this paper is not available in the source folder.

Q52.2025 Main 55/6/1 · Q31 (OR option)5 marks
  1. Write the principle of working of a transformer. With the help of a labelled diagram, explain the working of a step-up transformer.
  2. An ideal transformer is designed to convert 50 V into 250 V. It draws 200 W power from an ac source whose instantaneous voltage is given by vi = 20 sin (100π)t V.
    Find :
    (I) rms value of input current.
    (II) expression for instantaneous output voltage.
    (III) expression for instantaneous output current.
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The official marking scheme for this paper is not available in the source folder.

Q53.2025 Supp 55/S/1 · Q31 (option 1)5 marks

(i) A series combination of L, C and R is connected to an ac source V = Vm sin ωt. Obtain :

(I) the impedance of the circuit using phasor diagram,

(II) the expression for the instantaneous current I, and

(III) the phase relationship of current to the applied voltage.

(ii) Define power factor of an ac circuit. State the conditions under which it is :

(I) maximum,

(II) minimum.

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Official CBSE Marking Scheme · CBSE Board 2025 (Supplementary), Set 55/S/1

(i) (I) Let resistance R, inductor L and capacitor C be connected in series with an alternate e.m.f.

Series LCR circuit with voltage drops, and phasor diagram with V_R, V_L, V_C and phase angle φ

[½] for phasor diagram

ε = I2[R2 + (XL – Xc)2] [½]

Z = εI [½]

Z = R2 + (XL – XC)2 [½]

(II) Im = VmZ sin(ωt – φ) [½]

(III) Phase angle tan φ = VL – VCVR [½]

(ii) P = EVIV cos φ where P = true power and EVIV is the apparent power. φ is phase difference between current and voltage.

Power factor is defined as the ratio of true power to the apparent power. [1]

Power factor = True Power (P)Apparent power (EVIV) = cos φ

(I) Power factor is maximum when φ = 0 purely resistive circuit (It occurs when ω = 1LC at resonance in an LCR series circuit). [½]

(II) Power factor is minimum when φ = π2, in pure inductive or capacitive circuit. [½]

Q54.2025 Supp 55/S/1 · Q31 (OR option)5 marks

(i) Prove that the voltage is ahead of the current in phase by π2 rad in an ac circuit containing an ideal inductor.

(ii) The currents through two inductors of self-inductance 12 mH and 24 mH are increasing with time at the same rate. Draw graphs showing the variation of the :

(I) magnitude of emf induced with the rate of change of current in each inductor.

(II) energy stored in each inductor with the current flowing through it.

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Official CBSE Marking Scheme · CBSE Board 2025 (Supplementary), Set 55/S/1

(i) Let the voltage across the source be υ = υm sin ωt

AC source connected across an ideal inductor L

[½] for circuit

Using Kirchhoff’s loop rule

υ – L didt = 0

didt = υL = υmL sin ωt

di = υmL sin ωt dt [1]

Integrating

i = υmωL cos ωt [½]

i = im sin(ωt – π2) [½]

This show that current lags behind the voltage by π2 rad.

∴ Voltage is ahead of current in phase by π2 rad [½]

(ii) (I) Graph for induced e.m.f.

Graph of induced emf versus dI/dt: straight lines through the origin for 24 mH and 12 mH

[1]

(II) Graph for energy stored

Graph of energy stored U versus current I for the two inductors (curves labelled L2 = 30 mH and L1 = 12 mH in the marking scheme)

[1]

Q55.2024 Main 55/1/1 · Q32 (option 1)5 marks

(i) A resistor and a capacitor are connected in series to an ac source v = vm sin ωt. Derive an expression for the impedance of the circuit.

(ii) When does an inductor act as a conductor in a circuit ? Give reason for it.

(iii) An electric lamp is designed to operate at 110 V dc and 11 A current. If the lamp is operated on 220 V, 50 Hz ac source with a coil in series, then find the inductance of the coil.

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Official CBSE Marking Scheme · CBSE Board 2024 (Main), Set 55/1/1

(i) Deriving expression for impedance [2]; (ii) reason [1]; (iii) inductance of coil [2]

(i)

Phasor diagram of V_R, V_C and resultant V for a series RC circuit

VC + VR = V [½]

vm2 = vrm2 + vcm2

vrm = imR     vcm = imXc

vm2 = (imR)2 + (imXc)2 = im2 [R2 + Xc2] [½]

⇒ im = vmR2 + Xc2 [½]

⇒ Impedance Z = R2 + Xc2 [½]

(ii) For direct current (dc), an inductor behaves as a conductor.

As XL = ωL = 2πνL

For dc ν = 0 ⇒ XL = 0 [1]

Alternatively: Induced emf (ε) = − L dIdt. For dc; dI = 0 ⇒ ε = 0

(iii) R = 11011 = 10 Ω [½]

irms = vrmsR2 + XL2 = 220100 + XL2

11 = 220100 + XL2 [½]

100 + XL2 = 22011 = 20 Ω

Squaring both sides: ⇒ 100 + XL2 = 400

⇒ XL2 = 300 ⇒ XL = 103 Ω [½]

XL = 2πfL ⇒ 103 = 2π × 50 × L

L = 310π H [½]

Q56.2024 Main 55/1/1 · Q32 (OR option)5 marks

(i) Draw a labelled diagram of a step-up transformer and describe its working principle. Explain any three causes for energy losses in a real transformer.

(ii) A step-up transformer converts a low voltage into high voltage. Does it violate the principle of conservation of energy ? Explain.

(iii) A step-up transformer has 200 and 3000 turns in its primary and secondary coils respectively. The input voltage given to the primary coil is 90 V. Calculate :

(1) The output voltage across the secondary coil
(2) The current in the primary coil if the current in the secondary coil is 2·0 A.

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Official CBSE Marking Scheme · CBSE Board 2024 (Main), Set 55/1/1

(i) Labelled diagram of step-up transformer [1]; describing working principle [½]; three causes [1½]. (ii) Explanation [1]. (iii) (1) output voltage across secondary coil [½]; (2) current in primary coil [½]

(i)

Step-up transformer: primary and secondary windings on a soft iron core (core type)

OR

Step-up transformer: primary and secondary windings on a soft iron core (shell type)

The working principle of transformer is mutual induction. When an alternating voltage is applied to the primary, the resulting current produces an alternating magnetic flux which links the secondary and induces an emf in it. [½]

Causes of energy losses (Any three)
(a) Flux leakage
(b) Resistance of the windings
(c) Eddy currents
(d) Hysteresis [½ + ½ + ½]

(ii) No [½]

Current changes correspondingly. So, the input power is equal to the output power. [½]

(iii) (1) VsVP = NsNP

Vs = NsNP × VP = 3000200 × 90

Vs = 1350 V [½]

(2) IPIs = NsNP

IP = 3000200 × 2 = 30 A [½]

Q57.2024 Main 55/4/1 · Q33 (option 1)5 marks
  1. Mention the factors on which the resonant frequency of a series LCR circuit depends. Plot a graph showing variation of impedance of a series LCR circuit with the frequency of the applied a.c. source.
  2. With the help of a suitable diagram, explain the working of a step-up transformer.
  3. Write two causes of energy loss in a real transformer.
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Official CBSE Marking Scheme · CBSE Board 2024 (Main), Set 55/4/1

(i) Inductance [½]

Capacitance [½]

Alternatively ν0 = 12πLC

Graph of impedance (Z) versus frequency (ν) [1]

Impedance Z versus frequency: minimum value R at resonant frequency

(ii) Diagram of a transformer [1]

Transformer: primary and secondary coils wound on a soft iron core

Working – when an alternating voltage is applied to the primary, the resulting current produces an alternating magnetic flux which links the secondary and induces an emf in it. [1]

(iii) Causes of energy loss (any two) [½ + ½]

  • Flux leakage
  • Resistance of the windings
  • Hysteresis
  • Eddy currents
Q58.2024 Main 55/5/1 · Q32 (option 1)5 marks
  1. You are given three circuit elements X, Y and Z. They are connected one by one across a given ac source. It is found that V and I are in phase for element X. V leads I by (π4) for element Y while I leads V by (π4) for element Z. Identify elements X, Y and Z.
  2. Establish the expression for impedance of circuit when elements X, Y and Z are connected in series to an ac source. Show the variation of current in the circuit with the frequency of the applied ac source.
  3. In a series LCR circuit, obtain the conditions under which (i) impedance is minimum and (ii) wattless current flows in the circuit.
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Official CBSE Marking Scheme · CBSE Board 2024 (Main), Set 55/5/1

(i) X : Resistor [½]

Y : real inductor (such that its reactance is equal to its resistance) / Inductor [½]

Z : real capacitor (such that its reactance is equal to its resistance)/ Capacitor [½]

(ii) [½]

Phasor diagram of series LCR circuit showing V_R, V_C + V_L and resultant V at phase angle φ

From the fig.

Vm2 = VRm2 + (VCm – VLm)2 [½]

Vm2 = (im R)2 + (imXC – imXL)2 [½]

Impedance (Z) = VmIm = R2 + (XC – XL)2 [½]

Resonance curves: current amplitude i_m versus angular frequency ω with peak at ω0 for two resistances (i) and (ii)

[½]

(iii) Z = R2 + (XC – XL)2

For the minimum value of impedance

(i) XC = XL [½]

(ii) Average power consumed in A.C. circuit over a cycle

P = VI cos φ

For wattless current P = 0

Since V ≠ 0 , I ≠ 0

cos φ = 0

i.e. φ = π2 [½]

Q59.2024 Main 55/5/1 · Q32 (OR option)5 marks
  1. Describe the construction and working of a transformer and hence obtain the relation for (vsvp) in terms of number of turns of primary and secondary.
  2. Discuss four main causes of energy loss in a real transformer.
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Official CBSE Marking Scheme · CBSE Board 2024 (Main), Set 55/5/1

(i) Construction: A transformer consists of two sets of coils, insulated from each other. They are wound on a soft- iron core, either one on top of other or on separate limbs of the core.

Alternatively

Two transformer constructions: coils wound on a soft iron core, primary and secondary

[1]

Working: When an alternating voltage is applied to the primary, the resulting current produces an alternating magnetic flux which links with the secondary and induces an emf. in it. [1]

For an ideal transformer the induced emf (εp) in primary coil for applied alternating voltage (VP)

εp = VP = –NP dφdt ------(1) [½]

e.m.f. induced εs in the secondary coil

εs = VS = –NS dφdt ------(2)

From eq. (1) and (2)

VSVP = NSNP [½]

(ii) Any four energy losses [½ × 4]

1. Flux leakage.

2. Resistance of windings/ copper loss.

3. Eddy currents/iron loss.

4. Hysteresis.

5. Magnetostriction.

Q60.SQP 2025-26 · Q33 (OR option)5 marks

(A) An input potential Vin = 200 sin 100πt V is provided to an ideal transformer having 1000 turns in primary coil and 100 turns in secondary coil as shown in figure. The load circuit has a resistance of 4Ω, a capacitive reactance of 2Ω and an inductive reactance of 6Ω.

Transformer with AC source on the primary and a series R, L, C load on the secondary

Find:

(i) the output voltage across the load circuit
(ii) the current flowing through the load circuit
(iii) the power supplied to the load circuit by the transformer [3]

(B) State the working principle of a transformer and explain how it is a key component in the transfer of electrical power over long distances. [2]

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Official CBSE Marking Scheme · CBSE Sample Question Paper 2025-26

(A) Given:

In load circuit,

R = 4Ω,

Xc = 2 Ω,

Xl = 6 Ω,

Np = 1000,

Ns = 100,

Vin = 200V sin100πt

(i) Output voltage Across Load Circuit

VoutVin = NsNp = 0.1

Or, Vout = 0.1 × 200V sin100πt,

Or, Vout = 20V sin100πt. [1]

(ii) Current flowing through load circuit

As, I = Im sin(ωt + Φ)

Where,

Im = VmZ,

Z = R2 + (Xc2 – XL2)

Or, Z = 42 Ω, &

Im = 2042 A = 522 A;

Φ = tan-1Xc – XLR = tan-1 1 = π4

I = 522 A sin(100πt + π4) [1]

(iii) Find the Power supplied to load circuit By the transformer.

P = VmIm2 cosΦ

Where, cosΦ = cosπ4 = 12

P = 20V × 522 A × 12 = 50W [1]

(B) Ac transformer works on the principal of ‘Mutual Induction’ [1]

A.C transformer can increase output potential.

As P = V/I

So increase in output potential results in decrease in output current, resulting in significant decrease in power loss in transmission wires between power plants and Cities. In respective cities they are stepped down. [1]

Q61.SQP 2026-27 · Q33 (option 1)5 marks

A certain electrical device is used to convert an alternating voltage of smaller magnitude into a much larger alternating voltage without violating the law of conservation of energy.

  1. With the help of a neat labelled diagram, describe the principle on which this device operates and explain how it increases the voltage.
  2. State the basic working mechanism of the device and mention the condition required in its input signal for proper functioning.
  3. Give one reason why this device does not operate with 100% efficiency in practical situations.
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Official CBSE Marking Scheme · CBSE Sample Question Paper 2026-27

I. Device: Step–up transformer

Transformer: primary coil and secondary coil wound on a laminated core

[1] (diagram)

Principle – Mutual induction: When an alternating current flows in the primary coil, it produces a time-varying magnetic flux in the soft iron core. This changing flux links with the secondary coil and, by Faraday’s law of electromagnetic induction, an emf is induced in the secondary.

For an ideal transformer: VsVP = NSNP [½]

If Ns > Np, then Vs > Vp: the transformer steps up the voltage.

Because in an ideal transformer, Pin = Pout ⇒ VpIp = VsIs, [½]

The law of conservation of energy is not violated: when voltage increases, current decreases correspondingly.

II. Working mechanism:

  • The primary coil is supplied with an alternating voltage.
  • This produces an alternating magnetic field in the core.
  • The changing magnetic flux linked with the secondary coil induces an alternating emf in it.
  • Thus energy is transferred from primary to secondary coil through the changing magnetic field in the core, without any electrical contact between the two windings.

[1]

Condition for proper functioning: The input must be alternating current (a.c.), not direct current (d.c.). With d.c. there is no change of flux, hence no induced emf in the secondary coil. [1]

III. Any one of the following: [1]

  • Copper losses
  • Eddy current losses.
  • Hysteresis loss
  • Flux leakage

Because of these losses, output power is less than input power, so efficiency is less than 100%.

Q62.SQP 2026-27 · Q33 (OR option)5 marks
  1. What do you mean by the resonance condition of a series LCR–circuit?
  2. Write an expression for the resonant frequency.
  3. Describe the use of a series resonant circuit in the tuning of a radio receiver.
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Official CBSE Marking Scheme · CBSE Sample Question Paper 2026-27

I. Resonance condition in a series LCR circuit & resonant frequency

A series LCR circuit has a resistor (R), inductor (L) and capacitor (C) all in series with an AC source. [1]

Resonance condition: The circuit is said to be in resonance when its inductive reactance equals its capacitive reactance: XL = XC [½]

ωL = 1ωC, where ω is the angular frequency of the a.c. supply. [½]

At resonance:

  • The net reactance X = XL − XC = 0
  • The impedance Z = R (minimum)
  • The current in the circuit is maximum for the given applied voltage

Hence the circuit behaves as a purely resistive circuit at resonance. [1]

The resonant frequency, f0 = ω02π = 12πLC

II. Use of a series resonant circuit in tuning a radio receiver

A radio station transmits electromagnetic waves of a definite frequency. In a radio receiver, the input stage contains a tuneable series LCR circuit (or LC tank) whose resonance frequency can be changed.

  • The inductor (L) is usually fixed.
  • The capacitor (C) is made variable (variable capacitor).

[1]

By adjusting the variable capacitor C, the resonant frequency, f0 = 12πLC, is changed until it becomes equal to the frequency of the desired radio station.

At this condition:

  • The LCR circuit is at resonance with that station’s frequency.
  • Its impedance is minimum, so the current (and hence signal voltage) is maximum for that particular station.
  • Signals of other frequencies (other stations) are not at resonance, so their currents are much smaller and get largely rejected.

Thus, by using a series resonant circuit, the radio receiver can select (tune to) one particular station and reject others. [1]

Unit V · Electromagnetic Waves

8. Electromagnetic Waves

Q1.2026 Main 55/1/1 · Q51 mark

An electromagnetic wave passes from vacuum into a dielectric medium with relative electrical permittivity (3/2) and relative magnetic permeability (8/3). Then, its

  1. wavelength is doubled and frequency remains unchanged.
  2. wavelength is doubled and frequency is halved.
  3. wavelength is halved and frequency remains unchanged.
  4. wavelength and frequency both will remain unchanged.
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Official CBSE Marking Scheme · CBSE Board 2026 (Main), Set 55/1/1

(C) wavelength is halved and frequency remains unchanged.

Q2.2026 Main 55/1/1 · Q71 mark

Electromagnetic waves used in a diagnostic tool in medicine have a wavelength range

  1. 1 nm to 10–3 nm
  2. 400 nm to 1 nm
  3. 1 mm to 700 nm
  4. 0.1 m to 1 mm
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Official CBSE Marking Scheme · CBSE Board 2026 (Main), Set 55/1/1

(A) 1 nm to 10–3 nm

Q3.2026 Main 55/2/1 · Q111 mark

An electromagnetic wave is propagating along x-axis. At any instant, the phase difference (in radian) between the electric field (E) and the magnetic field (B) associated with the wave is

  1. zero
  2. π4
  3. π2
  4. π
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Official CBSE Marking Scheme · CBSE Board 2026 (Main), Set 55/2/1

(A) Zero

Q4.2026 Main 55/3/1 · Q31 mark

A plane electromagnetic wave travels through a medium and the magnetic field associated with it is given by

B = 5 × 10−8 sin (3 × 1010 t − 150 x) T

where x is in metres and t is in seconds.

The velocity of the wave is :

  1. 2·0 × 108 ms−1
  2. 4·5 × 107 ms−1
  3. 3·5 × 107 ms−1
  4. 2·5 × 108 ms−1
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Official CBSE Marking Scheme · CBSE Board 2026 (Main), Set 55/3/1

(A) 2·0 × 108 ms−1

Q5.2026 Main 55/4/1 · Q71 mark

The ratio of amplitude of electric field to the amplitude of the magnetic field associated with an electromagnetic wave propagating in glass (n = 1·5) is :

  1. 3 × 108 ms−1
  2. 2 × 108 ms−1
  3. 3·3 × 10−9 ms−1
  4. 5 × 10−9 ms−1
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Official CBSE Marking Scheme · CBSE Board 2026 (Main), Set 55/4/1

(B) 2×108 ms−1

Q6.2026 Main 55/5/1 · Q61 mark

The magnetic field in a plane electromagnetic wave travelling in glass (n = 1·5) is given by

By = (2 × 10–7 T) sin (αx + 1·5 × 1011 t)

where x is in metres and t is in seconds. The value of α is :

  1. 0·5 × 103 m–1
  2. 6·0 × 102 m–1
  3. 7·5 × 102 m–1
  4. 1·5 × 103 m–1
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Official CBSE Marking Scheme · CBSE Board 2026 (Main), Set 55/5/1

(C) 7.5 × 102 m–1

Q7.2026 Supp 55/7/1 · Q41 mark

In an electromagnetic wave, the electric field (E) and the magnetic field (B) oscillate along x-axis and y-axis respectively. The wave is propagating along :

  1. z-axis
  2. – z-axis
  3. x-axis
  4. – y-axis
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Official CBSE Marking Scheme · CBSE Board 2026 (Supplementary), Set 55/7/1

(A) z-axis

Q8.2026 Supp 55/7/1 · Q51 mark

Which of the following is also referred to as ‘heat waves’ ?

  1. Gamma rays
  2. X-rays
  3. UV rays
  4. Infrared rays
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Official CBSE Marking Scheme · CBSE Board 2026 (Supplementary), Set 55/7/1

(D) Infrared rays

Q9.2026 Supp 55/7/1 · Q71 mark

If νX, νG and νU are frequencies of X-rays, Gamma-rays and UV rays respectively, then :

  1. νX > νG > νU
  2. νG > νX > νU
  3. νU > νX > νG
  4. νX > νU > νG
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Official CBSE Marking Scheme · CBSE Board 2026 (Supplementary), Set 55/7/1

(B) νG > νX > νU

Q10.2025 Main 55/1/1 · Q71 mark

The given diagram exhibits the relationship between the wavelength of the electromagnetic waves and the energy of photon associated with them. The three points P, Q and R marked on the diagram may correspond respectively to :

Graph of photon energy versus wavelength with points P, R and Q
  1. X-rays, microwaves, UV radiation
  2. X-rays, UV radiation, microwaves
  3. UV radiation, microwaves, X-rays
  4. Microwaves, UV radiation, X-rays
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Q11.2025 Main 55/2/1 · Q151 mark

For Questions 13 to 16, two statements are given – one labelled Assertion (A) and other labelled Reason (R). Select the correct answer to these questions from the codes (A), (B), (C) and (D) as given below :

Assertion (A) : Out of Infrared and radio waves, the radio waves show more diffraction effect.

Reason (R) : Radio waves have greater frequency than infrared waves.

  1. If both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of Assertion (A).
  2. If both Assertion (A) and Reason (R) are true but Reason (R) is not the correct explanation of Assertion (A).
  3. If Assertion (A) is true but Reason (R) is false.
  4. If both Assertion (A) and Reason (R) are false.
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Q12.2025 Main 55/4/1 · Q71 mark

The amplitude of electric field in an electromagnetic wave in free space is 1000 Vm−1. The amplitude of the magnetic field in this electromagnetic wave is :

  1. 3·0 × 10−3 T
  2. 3·33 × 10−8 T
  3. 3·0 × 1011 T
  4. 3·33 × 10−6 T
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Q13.2025 Main 55/5/1 · Q71 mark

Which of the following rays coming from the Sun plays an important role in maintaining the Earth’s warmth ?

  1. Infrared rays
  2. γ rays
  3. UV rays
  4. Visible light rays
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Q14.2025 Main 55/5/1 · Q81 mark

The dimensions of (με)–1, where ε is permittivity and μ is permeability of a medium, are :

  1. [M0 L1 T–1]
  2. [M0 L2 T–2]
  3. [M1 L2 T–2]
  4. [M1 L–1 T1]
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Q15.2025 Main 55/5/1 · Q91 mark

Which of the following electromagnetic waves has photons of largest momentum ?

  1. X-rays
  2. AM radio waves
  3. Microwaves
  4. TV waves
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Q16.2025 Main 55/6/1 · Q151 mark

Assertion (A) : X-rays are produced when slow moving electrons are stopped by a metal target of high atomic number.

Reason (R) : X-rays consist of low-energy photons.

  1. Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of the Assertion (A).
  2. Both Assertion (A) and Reason (R) are true, but Reason (R) is not the correct explanation of the Assertion (A).
  3. Assertion (A) is true, but Reason (R) is false.
  4. Both Assertion (A) and Reason (R) are false.
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Q17.2025 Main 55/7/1 · Q151 mark

Assertion (A) : EM waves do not require a medium for their propagation.

Reason (R) : EM waves are transverse waves.

  1. Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of the Assertion (A).
  2. Both Assertion (A) and Reason (R) are true, but Reason (R) is not the correct explanation of the Assertion (A).
  3. Assertion (A) is true, but Reason (R) is false.
  4. Both Assertion (A) and Reason (R) are false.
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Q18.2025 Supp 55/S/1 · Q61 mark

Electromagnetic radiation used to kill germs in water purifiers is :

  1. Microwaves
  2. Ultraviolet rays
  3. Gamma rays
  4. Radio waves
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Official CBSE Marking Scheme · CBSE Board 2025 (Supplementary), Set 55/S/1

(B) Ultraviolet rays

Q19.2024 Main 55/1/1 · Q81 mark

In the four regions, I, II, III and IV, the electric fields are described as :

Region I : Ex = E0 sin (kz − ωt)
Region II : Ex = E0
Region III : Ex = E0 sin kz
Region IV : Ex = E0 cos kz

The displacement current will exist in the region :

  1. I
  2. IV
  3. II
  4. III
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Official CBSE Marking Scheme · CBSE Board 2024 (Main), Set 55/1/1

(A) I

Q20.2024 Main 55/2/1 · Q71 mark

Electromagnetic waves with wavelength 10 nm are called :

  1. Infrared waves
  2. Ultraviolet rays
  3. Gamma rays
  4. X-rays
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Official CBSE Marking Scheme · CBSE Board 2024 (Main), Set 55/2/1

(B) Ultraviolet rays

Q21.2024 Main 55/3/1 · Q81 mark

Which one of the following has the highest frequency ?

  1. Infrared rays
  2. Gamma rays
  3. Radio waves
  4. Microwaves
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Official CBSE Marking Scheme · CBSE Board 2024 (Main), Set 55/3/1

(B) Gamma rays

Q22.2024 Main 55/4/1 · Q101 mark

The phase difference between electric field E and magnetic field B in an electromagnetic wave propagating along z-axis is –

  1. zero
  2. π
  3. π2
  4. π4
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Official CBSE Marking Scheme · CBSE Board 2024 (Main), Set 55/4/1

(A) Zero

Q23.2024 Main 55/5/1 · Q61 mark

The electromagnetic waves used to purify water are

  1. Infrared rays
  2. Ultraviolet rays
  3. X-rays
  4. Gamma rays
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Official CBSE Marking Scheme · CBSE Board 2024 (Main), Set 55/5/1

(B) Ultraviolet rays

Q24.2024 Supp 55/S/1 · Q81 mark

The electric field E associated with an electromagnetic wave is represented by

Ey = E0 sin (kx – ωt)

Which of the following statements is correct ?

  1. The wave is propagating along +x-axis.
  2. The wave is propagating along +z-axis.
  3. The magnetic field B of the wave is acting along +y-axis.
  4. The magnetic field B of the wave is acting along –x-axis.
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Official CBSE Marking Scheme · CBSE Board 2024 (Supplementary), Set 55/S/1

(A) The wave is propagating along +x-axis.

Q25.2023 Main 55/1/1 · Q81 mark

In the process of charging of a capacitor, the current produced between the plates of the capacitor is :

  1. μ0 dφEdt
  2. 1μ0 dφEdt
  3. ε0 dφEdt
  4. 1ε0 dφEdt

where symbols have their usual meanings.

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Official CBSE Marking Scheme · CBSE Board 2023 (Main), Set 55/1/1

(C) ε0 dφEdt

Q26.2023 Main 55/2/1 · Q71 mark

E and B represent the electric and the magnetic field of an electromagnetic wave respectively. The direction of propagation of the wave is along

  1. B
  2. E
  3. E × B
  4. B × E
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Official CBSE Marking Scheme · CBSE Board 2023 (Main), Set 55/2/1

(C) E × B

Q27.2023 Main 55/3/1 · Q71 mark

The electromagnetic radiations used to kill germs in water purifiers are called :

  1. Infrared waves
  2. X-rays
  3. Gamma rays
  4. Ultraviolet rays
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Official CBSE Marking Scheme · CBSE Board 2023 (Main), Set 55/3/1

(D) ultraviolet

Q28.2023 Main 55/4/1 · Q51 mark

The electromagnetic waves used in radar systems are :

  1. Infrared waves
  2. Ultraviolet rays
  3. Microwaves
  4. X-rays
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Official CBSE Marking Scheme · CBSE Board 2023 (Main), Set 55/4/1

(C) Microwaves

Q29.2023 Main 55/5/1 · Q51 mark

Which one of the following electromagnetic radiation has the least wavelength ?

  1. Gamma rays
  2. Microwaves
  3. Visible light
  4. X-rays
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Official CBSE Marking Scheme · CBSE Board 2023 (Main), Set 55/5/1

(A) Gamma rays

Q30.2023 Supp 55/C/1 · Q51 mark

Which of the following physical quantities remain the same for X-ray, red light and radio waves when travelling through a medium ?

  1. Wavelength
  2. Speed
  3. Frequency
  4. Momentum
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Official CBSE Marking Scheme · CBSE Board 2023 (Supplementary), Set 55/C/1

(B) speed

Q31.SQP 2025-26 · Q41 mark

Which of the following statement is true for the radio waves and the gamma rays?

  1. The energy of gamma rays is lesser than that of the radio waves.
  2. The frequency of the radio waves is higher than that of gamma rays.
  3. The radio waves and the gamma rays have the same energy.
  4. The energy of radio waves is lesser than that of the gamma rays.
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Official CBSE Marking Scheme · CBSE Sample Question Paper 2025-26

(D) The energy of radio waves is lesser than that of the gamma rays.

Since the frequency of radio waves is less than gamma waves.

E = hν

Hence, energy of radio waves is less than gamma waves

Q32.SQP 2026-27 · Q41 mark

If electric field associated with an electromagnetic wave is given by E = 30 sin (0.2π × 10-4 m-1x − 0.6π × 104 Hz t) j then the direction of propagation of the wave is along,

  1. i
  2. -i
  3. k
  4. -k
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Official CBSE Marking Scheme · CBSE Sample Question Paper 2026-27

(A) i

The direction of propagation of electromagnetic waves is given by –ω/k and is along x axis as the equation of the wave is a function of x.

Q33.2023 Main 55/1/1 · Q202 marks

Consider an induced magnetic field due to changing electric field and an induced electric field due to changing magnetic field. Which one is more easily observed ? Justify your answer.

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Official CBSE Marking Scheme · CBSE Board 2023 (Main), Set 55/1/1

Identification [1], Justification [1]

Induced electric field due to changing magnetic field is easily observed. [1]

Induced electric field due to changing magnetic field can be easily produced by various ways like rotating/moving a coil in magnetic field, changing the shape of coil in magnetic field, bringing bar magnet near a coil etc. [1]

Q34.2023 Main 55/2/1 · Q20 (option 1)2 marks

How are infrared waves produced ? Why are these waves referred to as heat waves ? Give any two uses of infrared waves.

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Official CBSE Marking Scheme · CBSE Board 2023 (Main), Set 55/2/1

Infrared waves are produced by hot bodies and vibrations of molecules. [½]

They are referred as heat waves because they are readily absorbed by water molecules and increase their thermal energy and heat them. [½]

Uses [½ + ½]

1) Dehydration of fruits.

2) In greenhouse Effect.

3) In remote switches.

(any other relevant two uses)

Q35.2023 Main 55/2/1 · Q20 (OR option)2 marks

How are X-rays produced ? Give any two uses of these.

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Official CBSE Marking Scheme · CBSE Board 2023 (Main), Set 55/2/1

When fast moving electrons strike a heavy target like tungsten, X-rays are produced. [1]

Two uses – [½ + ½]

1. Used as a diagnostic tool in medicine,

2. Treatment for certain forms of cancer.

3. To study crystal structure.

(Any two uses from above or other uses)

Q36.2023 Main 55/3/1 · Q202 marks

What is meant by the term ‘displacement current’ ? Briefly explain how this current is different from a conduction current.

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Official CBSE Marking Scheme · CBSE Board 2023 (Main), Set 55/3/1

Displacement current: It is the current that arises due to the rate of change of electric field/flux. [1]

Alternatively: Id = ε0 dφEdt

Alternatively: The term with units of current to explain the continuity of current in a region.

Difference: [1]

Displacement current is due to change in electric flux.

Conduction current is due to flow of electrons.

Alternatively: Id = ε0 dφEdt ; Ic = dqdt

Q37.2023 Main 55/4/1 · Q22 (option 1)2 marks

What is a displacement current ? How is it different from a conduction current ?

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Official CBSE Marking Scheme · CBSE Board 2023 (Main), Set 55/4/1

Displacement current is the current produced due to changing electric field / electric flux in a region. [1]

Alternatively

id = ε0 dϕEdt & I = dqdt

Difference : Current carried by a conductor due to flow of charges is called conduction current. Displacement current is not due to flow of charges but due to changing electric field / electric flux. [1]

Q38.2023 Main 55/4/1 · Q22 (OR option)2 marks

Write any two characteristics of an electromagnetic wave. Why are microwaves used in radar systems ?

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Official CBSE Marking Scheme · CBSE Board 2023 (Main), Set 55/4/1

Any two characteristics :

1) No medium is required for their propagation.

2) Transverse in nature.

3) Consist of Electric and Magnetic field perpendicular to each other.

4) Energy is equally shared by electrical and magnetic field.

5) Travel with speed of light in vacuum. [½ + ½]

Reason : Short wavelength, do not diffract / unidirectional property. [1]

Q39.2023 Main 55/5/1 · Q212 marks

Identify the electromagnetic wave whose wavelengths range is from about

  1. 10-12 m to about 10-8 m.
  2. 10-3 m to about 10-1 m.

Write one use of each.

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Official CBSE Marking Scheme · CBSE Board 2023 (Main), Set 55/5/1

(a) X-rays [½]

Use: used as diagnostic tool in medicine. (Any one) [½]

Treatment for certain forms of cancer.

To study crystal structure

(Or any other one suitable use)

Alternatively

Gamma rays

Use: used in medicine to destroy the cancer cell.

(b) Microwaves [½]

Use: In radar system for aircraft navigation [½]

(Or any other one suitable use)

Q40.2023 Supp 55/C/1 · Q192 marks

(a) The electric field of an electromagnetic wave passing through vacuum is represented as Ex = E0 sin (kz − ωt). Identify the parameter which is related to the (i) wavelength, and (ii) the frequency of the wave in the above equation.

(b) Write two properties of a medium that determine the velocity of light in that medium.

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Official CBSE Marking Scheme · CBSE Board 2023 (Supplementary), Set 55/C/1

(a) Parameter relating wavelength is ‘k’ (= 2πλ) [½]

Parameter relating frequency is ‘ω’ (= 2πν) [½ + ½]

(b) 1. Electric properties of the medium

2. Magnetic properties of the medium [½ + ½]

Alternatively:

i. Permittivity (ε) of the medium

ii. Permeability (μ) of the medium

Q41.2022 Supp 55/6/1 · Q1 (option 1)2 marks
  1. Long distance radio broadcasts use short-wave bands. Why ? [1]
  2. The experimental demonstration of electromagnetic waves is possible only in the low frequency region (the radio wave region). Explain. [1]
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Official CBSE Marking Scheme · CBSE Board 2022 (Supplementary), Set 55/6/1

(i) Ionosphere reflects waves in these bands. [1]

(ii) The frequency that we get even with modern electronic circuits is hardly about 1011 Hz, this is why experimental demonstration of electromagnetic waves had to come in low frequency region. [1]

Q42.2022 Supp 55/6/1 · Q1 (OR option)2 marks
  1. Draw a diagram showing a linearly polarized electromagnetic wave, propagating in the z-direction depicting E and B. [1]
  2. The wavelength of an electromagnetic wave is 10-10 m. Identify the wave and mention its one use. [1]
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Official CBSE Marking Scheme · CBSE Board 2022 (Supplementary), Set 55/6/1

(i) Diagram [1]

Linearly polarised EM wave along z with E and B oscillating in perpendicular planes

[Note: Award full marks even if a student takes E and B on Y and X axis.]

(ii) X rays [½]

Use: As a diagnostic tool in medicine / treatment of cancer / or any other. [½]

Q43.SQP 2025-26 · Q172 marks

The amplitude of the magnetic field of a plane electromagnetic wave propagating along positive X axis in vacuum is 510 nTk and its angular frequency is 60 × 106 rad/sec. Write the expression for the electric field (E).

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Official CBSE Marking Scheme · CBSE Sample Question Paper 2025-26

Given, Bo = 510 nT = 510 × 10-9 T

ω = 60 × 106 rad/sec

Eo = cBo = 153 N/C

k = ω/c = 20 × 10-2 rad/m [1]

E = Eo sin (ωt – kz)

E = 153 sin (60 × 106 t – 20 × 10-2 x) N/C [1]

Q44.SQP 2026-27 · Q172 marks

The electric and magnetic fields associated with an electromagnetic radiation in a medium are given as follows:

E = 30 sin (2π × 1018 rad s-1 t + π × 1010 m-1 x) V m-1
B = 10-7 sin (2π × 1018 rad-1 t + π × 1010 m-1 x) T

  1. What is the refractive index of the medium through which the wave is propagating?
  2. Find the frequency of the wave and identify the electromagnetic wave.
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Official CBSE Marking Scheme · CBSE Sample Question Paper 2026-27

I. Refractive index of medium (µ) = cvmedium [½]

c = 3 × 108 m/s

vmedium = 2π × 1018π × 1010 = 2 × 108 m/s [½]

µ = 3 × 1082 × 108 = 1.5 [½]

II. Angular frequency (ω) = 2π × 1018 rad s-1

Or, 2πν = 2π × 1018 rad s-1

Or, ν = 1018 s-1

This frequency corresponds to X Ray [½]

Q45.2026 Main 55/2/1 · Q243 marks

Name the electromagnetic waves which are used for

  1. detection of fractures in bones
  2. physiotherapy
  3. radar systems

Also write their wavelength range.

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Official CBSE Marking Scheme · CBSE Board 2026 (Main), Set 55/2/1

(i) X-rays [½]
Wavelength range : 1 nm to 10–3 nm [½]

(ii) Infrared waves [½]
Wavelength range : 1 mm to 700 nm [½]

(iii) Microwaves [½]
Wavelength range : 0.1 m to 1 mm [½]

Q46.2026 Main 55/3/1 · Q253 marks

(a) Depict the variation of electric field (E) and magnetic field (B) with respect to the direction of propagation of an electromagnetic wave. Write their two important characteristics.

(b) Show that 1ε0 μ0 gives the velocity of an electromagnetic wave in free space.

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Official CBSE Marking Scheme · CBSE Board 2026 (Main), Set 55/3/1

(a) Variation of E and B along the direction of propagation [1]

Electric and magnetic field variation of an electromagnetic wave propagating along z

Characteristics of Electromagnetic wave : (any two) [½ + ½]

(i) Electric and magnetic fields are perpendicular to each other.

(ii) Speed of Electromagnetic wave in free space or vacuum is equal to 3 × 108 ms−1.

(iii) Electric and magnetic fields are in same phase.

(iv) Electromagnetic waves are transverse in nature.

(v) Electromagnetic waves can travel in vacuum.

(b) 1μ0ε0 [½]

= 14π × 10−7 × 14π × 9 × 109

= 9 × 1016

= 3 × 108 m/s = Velocity of electromagnetic wave in free space. [½]

Q47.2026 Main 55/4/1 · Q253 marks

What are infrared waves ? Why are these waves referred to as ‘heat waves’ ? Give any two uses of these waves.

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Official CBSE Marking Scheme · CBSE Board 2026 (Main), Set 55/4/1

Infrared waves are the waves produced by hot bodies and molecules. [1]

Alternatively : They are electromagnetic (em) waves whose wavelength lies between 1 mm to 700 nm.

They are referred to as heat waves because water molecules present in most materials readily absorb infrared waves. After absorption, their thermal motion increases, that is, they heat up and heat their surroundings. [1]

Alternatively : When infrared waves pass through a medium, they heat up the medium.

Uses (Any Two) [½ + ½]

  1. Infrared lamps are used in physical therapy.
  2. They help in maintaining Earth’s average temperature.
  3. They are used in satellites for military purpose.
  4. They are used to observe growth of crops.
  5. They are used in remote switches.
Q48.2026 Main 55/5/1 · Q253 marks

What is displacement current (id) ? Considering the case of charging of a capacitor, show that id = ε0 dφEdt. What is the value of id for a conductor across which a constant voltage is applied ?

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Official CBSE Marking Scheme · CBSE Board 2026 (Main), Set 55/5/1

Defining displacement current [1]; Showing id = ε0 dφedt [1½]; Writing value of id when constant voltage is applied [½]

Displacement current arises due to time varying electric field and hence the time varying electric flux in a region. [1]

The plates of the capacitor have an area A & a total charge Q, the magnitude of electric field between the plates:

E = Q/Aε0 ……. (i) [½]

φe = EA ……. (ii)

Substituting (i) in (ii)

φe = Qε0

dφedt = ddt(Qε0) = 1ε0 dQdt

ε0 dφedt = dQdt ……. (iii) [½]

If charge Q changes with time, the current id = dQdt

From eq. (iii) and (iv): id = ε0 dφedt [½]

For constant voltage E = constant, so φe = 0 [½]

∴ id = 0

Q49.2025 Main 55/1/1 · Q253 marks

(a) A parallel plate capacitor is charged by an ac source. Show that the sum of conduction current (Ic) and the displacement current (Id) has the same value at all points of the circuit.

(b) In case (a) above, is Kirchhoff’s first rule (junction rule) valid at each plate of the capacitor ? Explain.

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The official marking scheme for this paper is not available in the source folder.

Q50.2025 Main 55/2/1 · Q253 marks

(a) The electric field E of an electromagnetic wave propagating in north direction is oscillating in up and down direction. Describe the direction of magnetic field B of the wave.

(b) Are the wave length of radio waves and microwaves longer or shorter than those detectable by human eyes ?

(c) Write main use of each of the following in human life :
(i) Infrared waves (ii) Gamma rays

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The official marking scheme for this paper is not available in the source folder.

Q51.2025 Main 55/4/1 · Q253 marks
  1. State any three characteristics of electromagnetic waves.
  2. Briefly explain how and where the displacement current exists during the charging of a capacitor.
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The official marking scheme for this paper is not available in the source folder.

Q52.2025 Main 55/6/1 · Q253 marks

Name the electromagnetic wave used (i) in radar, (ii) in eye surgery and (iii) as diagnostic tool in medicine. Write their wavelength range also.

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The official marking scheme for this paper is not available in the source folder.

Q53.2025 Main 55/7/1 · Q253 marks

(a) Differentiate between ‘conduction current’ and ‘displacement current’, giving one similarity and one dissimilarity between them.

(b) Explain the existence of electromagnetic waves in free space, using the concept of displacement current.

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The official marking scheme for this paper is not available in the source folder.

Q54.2025 Supp 55/S/1 · Q233 marks

Depict the variation of electric and magnetic fields in an electromagnetic wave as it propagates along z-axis. In a plane electromagnetic wave in free space, the electric field oscillates sinusoidally at a frequency of 1·5 × 1010 Hz with amplitude 36 Vm–1. Find :

(a) the wavelength of the wave, and

(b) the amplitude of the associated magnetic field.

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Official CBSE Marking Scheme · CBSE Board 2025 (Supplementary), Set 55/S/1
Electric and magnetic fields of an electromagnetic wave along the z-axis (E along x, B along y)

[1] for the variation of E and B

(a) λ = cv [½]

= 3 × 1081.5 × 1010 = 2 × 10–2 = 0.02 m [½]

(b) B0 = E0c [½]

= 363 × 108

= 1.2 × 10–7 T [½]

Q55.2024 Main 55/1/1 · Q263 marks

(a) “The wavelength of the electromagnetic wave is often correlated with the characteristic size of the system that radiates.” Give two examples to justify this statement.

(b) (i) Long distance radio broadcasts use short-wave bands. Why ?
(ii) Optical and radio telescopes are built on the ground, but X-ray astronomy is possible only from satellites orbiting the Earth. Why ?

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Official CBSE Marking Scheme · CBSE Board 2024 (Main), Set 55/1/1

a) (Any Two)

• Gamma radiation having wavelength of 10−14 m to 10−15 m, typically originate from an atomic nucleus.

• X-rays are emitted from heavy atoms.

• Radio waves are produced by accelerating electrons in a circuit. A transmitting antenna can most efficiently radiate waves having a wavelength of about the same size as the antenna. [½ + ½]

b) (i) Ionosphere reflects waves in these bands [1]

(ii) Atmosphere absorbs x-rays, while visible and radio waves can penetrate it [1]

Note: Full credit to be given for part (b) for mere attempt.

Q56.2024 Main 55/2/1 · Q233 marks

(a) On what factors does the speed of an electromagnetic wave in a medium depend ?
(b) How is an electromagnetic wave produced ?
(c) Sketch a schematic diagram depicting the electric and magnetic fields for an electromagnetic wave propagating along z-axis.

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Official CBSE Marking Scheme · CBSE Board 2024 (Main), Set 55/2/1

(a) Speed of EM waves v = 1µε

Speed depends upon
(i) Permittivity (ε) of medium
(ii) Magnetic permeability (µ) of medium [½ + ½]

(b) Accelerated charges or oscillating charges produce electromagnetic waves [1]

(c) [1]

Electric field along x and magnetic field along y for an EM wave travelling along z
Q57.2024 Main 55/3/1 · Q253 marks

The electric field in an electromagnetic wave in vacuum is given by :

E = (6·3 N/C) [cos (1·5 rad/m) y + (4·5 × 108 rad/s) t] i

  1. Find the wavelength and frequency of the wave.
  2. What is the amplitude of the magnetic field of the wave ?
  3. Write an expression for the magnetic field of this wave.
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Official CBSE Marking Scheme · CBSE Board 2024 (Main), Set 55/3/1

(a) k = 2πλ [½]

λ = 2πK = 4π3 m = 4.18 m [½]

ω = 2πυ [½]

υ = ω2π = 4.5×1082π Hz = 94π × 108 Hz

υ = 7.16 × 10–1 Hz [½]

(b) B0 = E0c

B0 = 6.33×108 = 2.1 × 10–8 T [½]

(c) B = 2.1 × 10–8 [(cos 1.5 rad/m) y + (4.5 × 108 rad/s) t] k T [½]

Q58.2024 Main 55/4/1 · Q273 marks

Name the electromagnetic waves with their wavelength range which are used for

  1. FM radio broadcast
  2. detection of fracture in bones
  3. treatment of muscular strain
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Official CBSE Marking Scheme · CBSE Board 2024 (Main), Set 55/4/1
Electromagnetic wavesWavelength range
(i)Radio waves [½]> 0.1 m [½]
(ii)X-rays [½]1 nm – 10–3 nm [½]
(iii)Infrared waves [½]1 mm – 700 nm [½]
Q59.2024 Main 55/5/1 · Q263 marks
  1. Name the parts of the electromagnetic spectrum which are (i) also known as ‘heat waves’ and (ii) absorbed by ozone layer in the atmosphere.
  2. Write briefly one method each, of the production and detection of these radiations.
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Official CBSE Marking Scheme · CBSE Board 2024 (Main), Set 55/5/1

(a) (i) Infrared waves [½]

(ii) Ultraviolet Rays [½]

(b) Method of production

Infrared waves: Hot bodies / Vibration of atoms and molecules [½]

Ultraviolet Rays: Special UV lamps / Sun / Very hot bodies [½]

Method of detection

Infrared waves: Thermopiles / IR photographic film / Bolometer [½]

Ultraviolet Rays: Photocells / photographic film [½]

Q60.2024 Supp 55/S/1 · Q263 marks

Explain the following, giving proper reason :

(a) During charging of a capacitor, displacement current exists in the capacitor. But there is no displacement current when it gets fully charged.

(b) The frequency of microwaves in ovens matches with the resonant frequency of water molecules.

(c) Infrared waves are also known as heat waves.

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Official CBSE Marking Scheme · CBSE Board 2024 (Supplementary), Set 55/S/1

(a) id = ε0dφEdt

Displacement current is due to changing electric flux. During charging of capacitor, there is change in electric flux. When fully charged electric field hence electric flux does not change. Hence no displacement current. [1]

(b) The frequency of the microwaves is selected to match the resonant frequency of water molecules so that energy from the waves is transferred efficiently to the kinetic energy of the molecules. [1]

(c) Infrared waves are also known as heat waves, because water molecules present in most materials readily absorbs infrared waves and their thermal motion increases and heat up. [1]

Q61.2022 Main 55/1/1 · Q11 (option 1)3 marks
  1. Monochromatic light is incident on a surface separating two media. The frequency of the light after refraction remains unaffected but its wavelength changes. Why ?
  2. The frequency of an electromagnetic radiation is 1.0 × 1011 Hz. Identify the radiation and mention its two uses.
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Official CBSE Marking Scheme · CBSE Board 2022 (Main), Set 55/1/1

i) Refraction arises through interaction of incident light with the atomic constituents of matter. Atoms may be viewed as oscillators which take up the frequency of the external agency causing forced oscillations. Thus the frequency of refracted light equals the frequency of incident light. [1]

Alternatively

Frequency is the characteristic of the source of light. So it remains unaffected. But λ depends on refractive index (μ) of the medium as — λm = λoμ

ii) Infrared/ Microwaves/ Radio waves [1]

Uses of Infrared rays (any two) [½ + ½]

  • Remote control
  • Green house effect
  • Photography in foggy condition
  • To reveal secret writings
  • Infrared lamps

Uses of Microwaves (any two)

  • Radar System
  • Geostationary satellite
  • Microwave ovens

Uses of Radiowaves (any two)

  • TV transmission
  • Radio broadcast
  • Mobile transmission
Q62.2022 Main 55/2/1 · Q4 (option 1)3 marks
  1. Depict a plane electromagnetic wave propagating along the x-axis. Write the expressions for its oscillating electric and magnetic fields.
  2. Write three characteristics of electromagnetic waves.
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Official CBSE Marking Scheme · CBSE Board 2022 (Main), Set 55/2/1

(i) Depiction of plane EM wave [½]; Expression for electric field [½]; Expression for magnetic field [½]

Plane EM wave along x-axis with E along y and B along z

Ey = E0 sin (kx – ωt) [½]

Bz = B0 sin (kx – ωt) [½]

(ii) The three characteristics are: [½ + ½ + ½]

a) They travel with velocity of light.

b) They carry energy and momentum.

c) They are transverse in nature.

(Or any other characteristic given)

Q63.2022 Main 55/2/1 · Q4 (OR option)3 marks

Name the electromagnetic waves which are produced by the following :

  1. Radioactive decays of nucleus
  2. Welding arcs
  3. Hot bodies

Write one use each of these waves.

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Official CBSE Marking Scheme · CBSE Board 2022 (Main), Set 55/2/1

Naming of EM waves [½ + ½ + ½]; Uses of EM waves [½ + ½ + ½]

(a) Gamma Rays - Used for cancer treatment [½ + ½]

(b) Ultraviolet/Visible/Infrared (either) – Use of anyone of these three. [½ + ½]

(c) Infrared Rays – Used in night vision camera, bolometer & thermopiles [½ + ½]

(Note: Give full credit to any other use written.)

Q64.2022 Main 55/3/1 · Q11 (option 1)3 marks

Electromagnetic waves of wavelengths λ1, λ2 and λ3 are used in radar systems, in water purifiers and in remote switches of TV, respectively.

  1. Identify the electromagnetic waves, and
  2. Write one source of each of them.
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Official CBSE Marking Scheme · CBSE Board 2022 (Main), Set 55/3/1
EM WavesSources
λ1Microwavesmagnetron valve / klystron valve / gun diodes
λ2Ultra Violethigh voltage gas discharge tube
λ3Infra redhot bodies and molecules

[½ + ½] for each of the three

Q65.2022 Main 55/4/1 · Q11 (option 1)3 marks

Identify electromagnetic waves which :

  1. are used in radar system.
  2. affect a photographic plate.
  3. are used in surgery.

Write their frequency range.

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Official CBSE Marking Scheme · CBSE Board 2022 (Main), Set 55/4/1

(i) Microwave [½]

1010 Hz – 1012 Hz [½]

(ii) X-rays. [½]

1016 Hz – 1020 Hz [½]

Alternatively

Gamma Ray

1020 Hz – 1024 Hz

(iii) Gamma Ray [½]

1020 Hz – 1024 Hz [½]

Alternatively

Infrared Radiations

1012 Hz – 1014 Hz

Q66.2022 Main 55/5/1 · Q11 (option 1)3 marks
  1. Arrange the following electromagnetic radiation in the ascending order of their frequencies :
    X-rays, microwaves, gamma rays, radio waves
  2. Write two uses of any two of these radiation.
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Official CBSE Marking Scheme · CBSE Board 2022 (Main), Set 55/5/1

Arranging the e-m radiations in ascending order of frequency [1]; Uses of any two radiation [½ + ½ + ½ + ½]

(i) Radio waves < microwaves < X-rays < gamma rays [1]

(ii) Two uses each of any two of the radiation [2]

Radio waves –

  • TV transmission
  • Radio broadcast
  • Mobile communication
  • Radio telescope

Microwaves –

  • Microwave oven
  • Speed of automobiles
  • Radar
  • Air craft navigation

Gamma rays –

  • Treatment of cancer
  • Sterilisation and disinfection

X rays –

  • Diagnostic tool in medicine
  • Treatment for certain forms of cancer

(Two uses of any two of these radiations)

Unit VI · Optics

9. Ray Optics and Optical Instruments

Q1.2026 Main 55/1/1 · Q91 mark

A concave lens of focal length 10 cm is cut into two identical plano-concave lenses. The focal length of each lens will be

  1. 20 cm
  2. 30 cm
  3. 40 cm
  4. 5 cm
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Official CBSE Marking Scheme · CBSE Board 2026 (Main), Set 55/1/1

(A) 20 cm

Q2.2026 Main 55/3/1 · Q41 mark

A thin plano-convex lens and a thin equi-concave lens are kept coaxially in contact as shown in the figure. Assuming both the lenses are made of glass of refractive index μ, and R is the radius of curvature of each curved surface, the focal length of the combination is :

Plano-convex lens in contact with an equi-concave lens, both of refractive index μ
  1. Rμ − 1
  2. −Rμ − 1
  3. 2Rμ − 1
  4. −2Rμ − 1
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Official CBSE Marking Scheme · CBSE Board 2026 (Main), Set 55/3/1

(B) −Rμ − 1

Q3.2026 Main 55/3/1 · Q61 mark

A telescope has an objective lens of focal length 144 cm and an eyepiece of focal length 6·0 cm. The magnifying power and the length of telescope tube will be respectively :

  1. 24, 150 cm
  2. 42, 138 cm
  3. 24, 138 cm
  4. 42, 150 cm
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Official CBSE Marking Scheme · CBSE Board 2026 (Main), Set 55/3/1

(A) 24, 150 cm

Q4.2026 Main 55/3/1 · Q71 mark

In which of the following, total internal reflection does not occur ?

  1. Twinkling of stars
  2. Brilliance of diamonds
  3. Optical fibre
  4. Reflecting prism
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Official CBSE Marking Scheme · CBSE Board 2026 (Main), Set 55/3/1

(A) Twinkling of stars

Q5.2026 Main 55/4/1 · Q151 mark

Assertion (A) : When a convex lens made of glass is immersed in water, its converging power increases.

Reason (R) : The focal length of a lens depends only on the radii of curvature of its two faces.

  1. Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of the Assertion (A).
  2. Both Assertion (A) and Reason (R) are true, but Reason (R) is not the correct explanation of the Assertion (A).
  3. Assertion (A) is true, but Reason (R) is false.
  4. Both Assertion (A) and Reason (R) are false.
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Official CBSE Marking Scheme · CBSE Board 2026 (Main), Set 55/4/1

(D) Both assertion(A) and Reason (R) are false.

Q6.2026 Supp 55/7/1 · Q91 mark

A ray of light is incident at an angle of 60° on face AB of prism ABC with ∠ A = 30°. The ray emerging out of the prism makes an angle of 30° with the incident ray. The angle of emergence (∠ e) is equal to :

  1. 0°
  2. 30°
  3. 60°
  4. 90°
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Official CBSE Marking Scheme · CBSE Board 2026 (Supplementary), Set 55/7/1

(A) 0°

Q7.2026 Supp 55/7/1 · Q101 mark

A point source in air is placed in front of convex spherical glass surface (n = 1·5 and R = 10 cm). Its real image is formed at the same distance as the surface from the source. The distance of the source from the surface is :

  1. 25 cm
  2. 50 cm
  3. 75 cm
  4. 100 cm
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Official CBSE Marking Scheme · CBSE Board 2026 (Supplementary), Set 55/7/1

(B) 50 cm

Q8.2025 Main 55/1/1 · Q81 mark

A beaker is filled with water (refractive index 43) upto a height H. A coin is placed at its bottom. The depth of the coin, when viewed along the near normal direction, will be

  1. H4
  2. 3H4
  3. H
  4. 4H3
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Q9.2025 Main 55/2/1 · Q91 mark

A tub is filled with a transparent liquid to a height of 30.0 cm. The apparent depth of a coin lying at the bottom of the tub is found to be 16.0 cm. The speed of light in the liquid will be

  1. 1.6 × 108 m s-1
  2. 2.0 × 108 m s-1
  3. 3.0 × 108 m s-1
  4. 2.5 × 108 m s-1
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Q10.2025 Main 55/4/1 · Q81 mark

The magnification produced by a spherical mirror is − 2·0. The mirror used and the nature of the image formed will be

  1. Convex and virtual
  2. Concave and real
  3. Concave and virtual
  4. Convex and real
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Q11.2025 Main 55/5/1 · Q101 mark

A compound microscope has an objective and an eyepiece of focal lengths fo and fe, respectively. To obtain a large magnification of a small object, the microscope should have :

  1. fo and fe small, and fe > fo
  2. fo and fe small, and fo > fe
  3. fo and fe large, and fe > fo
  4. fo and fe large, and fo > fe
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Q12.2025 Main 55/6/1 · Q131 mark

Assertion (A) : In a reflecting telescope, the image does not have chromatic aberration.

Reason (R) : Chromatic aberration occurs only due to refraction of light through an optical medium.

  1. Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of the Assertion (A).
  2. Both Assertion (A) and Reason (R) are true, but Reason (R) is not the correct explanation of the Assertion (A).
  3. Assertion (A) is true, but Reason (R) is false.
  4. Both Assertion (A) and Reason (R) are false.
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Q13.2025 Main 55/7/1 · Q141 mark

Assertion (A) : A ray of light is incident normally on the face of a prism. The emergent ray will graze along the opposite face of the prism when the critical angle at glass-air interface is equal to the angle of the prism.

Reason (R) : The refractive index of a prism depends on angle of the prism.

  1. Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of the Assertion (A).
  2. Both Assertion (A) and Reason (R) are true, but Reason (R) is not the correct explanation of the Assertion (A).
  3. Assertion (A) is true, but Reason (R) is false.
  4. Both Assertion (A) and Reason (R) are false.
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Q14.2025 Supp 55/S/1 · Q141 mark

Assertion (A) : The image of an object placed between f and 2f from a convex lens can be seen on a screen. If the screen is removed, image is not formed.

Reason (R) : Rays from a given point on the object placed between f and 2f, after passing through a convex lens, do not converge on a point in space.

  1. Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of the Assertion (A).
  2. Both Assertion (A) and Reason (R) are true, but Reason (R) is not the correct explanation of the Assertion (A).
  3. Assertion (A) is true, but Reason (R) is false.
  4. Both Assertion (A) and Reason (R) are false.
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Official CBSE Marking Scheme · CBSE Board 2025 (Supplementary), Set 55/S/1

(D) Both Assertion (A) and Reason (R) are false.

Q15.2024 Main 55/1/1 · Q161 mark

Assertion (A) : Plane and convex mirrors cannot produce real images under any circumstance.

Reason (R) : A virtual image cannot serve as an object to produce a real image.

  1. Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of the Assertion (A).
  2. Both Assertion (A) and Reason (R) are true, but Reason (R) is not the correct explanation of the Assertion (A).
  3. Assertion (A) is true, but Reason (R) is false.
  4. Assertion (A) is false and Reason (R) is also false.
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Official CBSE Marking Scheme · CBSE Board 2024 (Main), Set 55/1/1

(D) Assertion (A) is false and reason (R) is also false.

Q16.2024 Main 55/2/1 · Q161 mark

Two statements are given — one labelled Assertion (A) and the other labelled Reason (R). Select the correct answer from the codes (A), (B), (C) and (D) as given below.

Assertion (A) : The magnifying power of a compound microscope is negative.

Reason (R) : The final image formed is erect with respect to the object.

  1. Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of the Assertion (A).
  2. Both Assertion (A) and Reason (R) are true, but Reason (R) is not the correct explanation of the Assertion (A).
  3. Assertion (A) is true, but Reason (R) is false.
  4. Assertion (A) is false and Reason (R) is also false.
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Official CBSE Marking Scheme · CBSE Board 2024 (Main), Set 55/2/1

(C) Assertion (A) is true, but Reason (R) is false

Q17.2024 Main 55/3/1 · Q161 mark

Assertion (A) : A convex lens, when immersed in a liquid, disappears.

Reason (R) : The refractive indices of material of the lens and the liquid are equal.

  1. Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of the Assertion (A).
  2. Both Assertion (A) and Reason (R) are true, but Reason (R) is not the correct explanation of the Assertion (A).
  3. Assertion (A) is true, but Reason (R) is false.
  4. Assertion (A) is false and Reason (R) is also false.
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Official CBSE Marking Scheme · CBSE Board 2024 (Main), Set 55/3/1

(A) Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of the Assertion (A).

Q18.2024 Main 55/5/1 · Q71 mark

The focal lengths of the objective and the eyepiece of a compound microscope are 1 cm and 2 cm respectively. If the tube length of the microscope is 10 cm, the magnification obtained by the microscope for most suitable viewing by relaxed eye is :

  1. 250
  2. 200
  3. 150
  4. 125
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Official CBSE Marking Scheme · CBSE Board 2024 (Main), Set 55/5/1

(D) 125

Q19.2024 Supp 55/S/1 · Q91 mark

A point object is placed in air at a distance of 4R on the principal axis of a convex spherical surface of radius of curvature R separating two mediums, air and glass. As the object is moved towards the surface, the image formed is :

  1. always real
  2. always virtual
  3. first virtual and then real
  4. first real and then virtual
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Official CBSE Marking Scheme · CBSE Board 2024 (Supplementary), Set 55/S/1

(D) First real and then virtual

Q20.2024 Supp 55/S/1 · Q151 mark

Assertion (A) : Although the surfaces of a goggle lens are curved, it does not have any power.

Reason (R) : In case of goggles, both the curved surfaces are curved on the same side and have equal radii of curvature.

  1. Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of the Assertion (A).
  2. Both Assertion (A) and Reason (R) are true, but Reason (R) is not the correct explanation of the Assertion (A).
  3. Assertion (A) is true, but Reason (R) is false.
  4. Both Assertion (A) and Reason (R) are false.
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Official CBSE Marking Scheme · CBSE Board 2024 (Supplementary), Set 55/S/1

(A) Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of the Assertion (A).

Q21.2023 Main 55/1/1 · Q91 mark

For a concave mirror of focal length ‘f’, the minimum distance between the object and its real image is :

  1. zero
  2. f
  3. 2f
  4. 4f
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Official CBSE Marking Scheme · CBSE Board 2023 (Main), Set 55/1/1

(A) Zero

Q22.2023 Main 55/2/1 · Q81 mark

A ray of monochromatic light propagating in air, is incident on the surface of water. Which of the following will be the same for the reflected and refracted rays ?

  1. Energy carried
  2. Speed
  3. Frequency
  4. Wavelength
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Official CBSE Marking Scheme · CBSE Board 2023 (Main), Set 55/2/1

(C) Frequency

Q23.2023 Main 55/2/1 · Q91 mark

A beam of light travels from air into a medium. Its speed and wavelength in the medium are 1.5 × 108 ms−1 and 230 nm respectively. The wavelength of light in air will be

  1. 230 nm
  2. 345 nm
  3. 460 nm
  4. 690 nm
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Official CBSE Marking Scheme · CBSE Board 2023 (Main), Set 55/2/1

(C) 460 nm

Q24.2023 Supp 55/C/1 · Q11 mark

A small object lies at the bottom of a vessel filled with water (refractive index 4/3) up to a height H. When viewed from a point above the surface of water, the object appears raised by n percent of H. The value of n is :

  1. 15
  2. 20
  3. 25
  4. 33
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Official CBSE Marking Scheme · CBSE Board 2023 (Supplementary), Set 55/C/1

(C) 25

Q25.SQP 2025-26 · Q51 mark

A glass prism has internal angles of 45°, 45° and 90°. The glass has a critical angle of 45°. Which of the following ray diagrams depicts the possible path the of light through the prism?

  1. Ray diagram A: ray entering face AB, reflected at BC and leaving
  2. Ray diagram B: two reflections inside the prism with the ray returning parallel to incident ray
  3. Ray diagram C: rays entering the hypotenuse side and reflecting back
  4. Ray diagram D: ray entering face AB, reflected and leaving through face AC
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Official CBSE Marking Scheme · CBSE Sample Question Paper 2025-26

(A) Diagram (A)

Total Internal reflection

Q26.SQP 2025-26 · Q141 mark

Assertion (A): The critical angle of light passing from glass to air is minimum for violet colour.

Reason (R): The wavelength of blue light is greater than the light of other colours.

  1. Both Assertion and Reason are true and Reason is the correct explanation of Assertion.
  2. Both Assertion and Reason are true but Reason is not the correct explanation of Assertion.
  3. Assertion is true but Reason is false.
  4. Both Assertion and Reason are false.
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Official CBSE Marking Scheme · CBSE Sample Question Paper 2025-26

(C) Assertion is true but Reason is false.

Q27.SQP 2026-27 · Q141 mark

A convex lens of focal length 20 cm is used to form a real image of a point object placed 30 cm from the lens. The lens is then immersed in water of refractive index 1.33 (the lens material has refractive index 1.5).

Assertion (A): The focal length of the lens increases when it is immersed in water.

Reason (R): The focal length of a lens depends on the relative refractive index of the lens material with respect to the surrounding medium, as given by the lens maker’s formula.

  1. both Assertion and Reason are true and Reason is the correct explanation of Assertion.
  2. both Assertion and Reason are true but Reason is not the correct explanation of Assertion.
  3. Assertion is true but Reason is false.
  4. both Assertion and Reason are false.
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Official CBSE Marking Scheme · CBSE Sample Question Paper 2026-27

(A) both Assertion and Reason are true and Reason is the correct explanation of Assertion.

Explanation: Both true and R explains A; lens maker's formula uses relative index (µlensµmedium − 1), so in water the relative refractive index decreases and f increases.

Q28.2026 Main 55/2/1 · Q192 marks

A ray of light in air is incident at angle ∠i on a face of an equilateral glass prism and is refracted through the prism. As ∠i is varied, it is observed that the ray undergoes minimum deviation, when the ∠i is three-fourth of the angle of the prism. Calculate the speed of light in the prism.

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Official CBSE Marking Scheme · CBSE Board 2026 (Main), Set 55/2/1

Angle of incidence i = 34A = 45° [½]

At minimum deviation r = A2 = 30° [½]

n = sin isin r = sin 45°sin 30° = 2 [½]

n = cv

v = 32 × 108 m/s = 2.12 × 108 m/s [½]

Q29.2026 Main 55/2/1 · Q21 (option 1)2 marks

A 5 cm long pencil is placed along the principal axis of a concave mirror of focal length 20 cm such that its nearest end is at a distance of 25 cm from the mirror. Calculate the length of the image of the pencil.

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Official CBSE Marking Scheme · CBSE Board 2026 (Main), Set 55/2/1

1f = 1v + 1u [½]

1v1 = 1f – 1u1 = 1(–20) – 1(–25) (for near end of the pencil)

v1 = –100 cm [½]

1v2 = 1(–20) – 1(–30) (for far end of the pencil)

v2 = –60 cm [½]

Length of the image L = |v1| – |v2| = 40 cm [½]

Q30.2026 Main 55/4/1 · Q192 marks

A small bulb is placed at the bottom of a tank, containing a transparent liquid of refractive index 2, to a depth of 1 m. Calculate the area of the surface of the liquid through which light from the bulb emerges.

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Official CBSE Marking Scheme · CBSE Board 2026 (Main), Set 55/4/1
Bulb at the bottom of a tank; light emerges through a circle of radius r on the liquid surface, limited by the critical angle

r = h × tan i [as i = ic] [½]

sin ic = 1μ = 12 ⇒ ic = 45° [½]

r = h [½]

Area = πr2 = 3.14 × (1)2

Area = 3.14 m2 [½]

Q31.2026 Main 55/5/1 · Q192 marks

A tank is filled with a liquid to a height of 12·5 m. The apparent depth of a needle lying at the bottom of the tank is measured to be 9·0 m. Calculate the speed of light in the liquid.

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Official CBSE Marking Scheme · CBSE Board 2026 (Main), Set 55/5/1

Calculating the speed of light in liquid. [2]

n = speed of light in vacuum (c)speed of light in medium (v) ……. (i) [½]

n = real depthapparent depth [½]

n = 12.59 ……. (ii)

From (i) and (ii)

v = 3 × 108 × 912.5 [½]

= 2.16 × 108 m/s [½]

Q32.2026 Main 55/5/1 · Q202 marks

Two thin lenses of focal length f1 and f2 are placed in contact with each other coaxially. Prove that the focal length f of the combination is given by f = f1 f2f1 + f2.

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Official CBSE Marking Scheme · CBSE Board 2026 (Main), Set 55/5/1

Proving the focal length of combination is f = f1f2f1 + f2 [2]

Two thin lenses A and B in contact forming image I1 then final image I

Ray diagram [½]

For the image formed by lens A: 1v1 – 1u = 1f1 ……. (i) [½]

For the image formed by second lens B: 1v – 1v1 = 1f2 ……. (ii) [½]

Adding (i) and (ii): 1v – 1u = 1f1 + 1f2

1f = 1f1 + 1f2

f = f1f2f1 + f2 [½]

Q33.2026 Supp 55/7/1 · Q18 (OR option)2 marks
  1. The lower half of a concave mirror’s reflecting surface is covered with an opaque non-reflecting material. What effect will this have on the image of an object placed in front of the mirror ?
  2. A lens made of glass of refractive index 1·5 disappears, when immersed in a trough of liquid. What is the refractive index of the liquid ?

Give reasons for your answer in each case.

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Official CBSE Marking Scheme · CBSE Board 2026 (Supplementary), Set 55/7/1

(i) The image will be formed that of the whole object. However, the intensity of the image will be low (or half) [1]

(ii) The refractive index of the liquid = 1.5 [½]

Since lens disappears inside the liquid which is possible only when both have same refractive indices. [½]

Q34.2025 Main 55/1/1 · Q192 marks

A transparent solid cylindrical rod (refractive index 23) is kept in air. A ray of light incident on its face travels along the surface of the rod, as shown in figure. Calculate the angle θ.

Ray entering the face of a cylindrical rod at angle θ to the axis and travelling along the curved surface
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Q35.2025 Main 55/2/1 · Q192 marks

Two concave lenses A and B, each of focal length 8.0 cm are arranged coaxially 16 cm apart as shown in figure. An object P is placed at a distance of 4.0 cm from A. Find the position and nature of the final image formed.

Object P on the axis to the left of concave lens A, with a second concave lens B to the right of A
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Q36.2025 Main 55/4/1 · Q192 marks

An equiconvex lens is made of glass of refractive index 1·55. If the focal length of the lens is 15·0 cm, calculate the radius of curvature of its surfaces.

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Q37.2025 Main 55/5/1 · Q18 (option 1)2 marks

A point object is placed in air at a distance R/3 in front of a convex surface of radius of curvature R, separating air from a medium of refractive index n (< 4). Find the nature and position of the image formed.

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Q38.2025 Main 55/6/1 · Q18 (OR option)2 marks

A point source of light in air is kept at a distance of 12 cm in front of a convex spherical surface of glass of refractive index 1·5 and radius of curvature 30 cm. Find the nature and position of the image formed.

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Q39.2025 Main 55/6/1 · Q202 marks

A right angled isosceles glass prism ABC is kept in contact with an equilateral triangular prism DBC as shown in the figure. Both prisms are made of the same glass of refractive index 1·6. Trace the path of the ray MN incident normally on face AB as it passes through the combination.

Right-angled isosceles prism ABC on equilateral prism DBC, ray MN normal to AB
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Q40.2025 Main 55/7/1 · Q18 (option 1)2 marks

Using the mirror equation and the formula of magnification, deduce that “the virtual image produced by a convex mirror is always diminished in size and is located between the pole and the focus.”

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Q41.2025 Main 55/7/1 · Q18 (OR option)2 marks

A convex lens of focal length 10 cm, a concave lens of focal length 15 cm and a third lens of unknown focal length are placed coaxially in contact. If the focal length of the combination is +12 cm, find the nature and focal length of the third lens, if all lenses are thin. Will the answer change if the lenses were thick ?

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Q42.2025 Supp 55/S/1 · Q18 (option 1)2 marks

A plano-convex lens of refractive index μ1 is placed coaxially in contact with a biconcave lens of refractive index μ2 as shown in the figure. All curved faces are of radius of curvature R each. Obtain the expression for the focal length of the combined lens.

Plano-convex lens (μ1) in contact with a biconcave lens (μ2), all curved faces of radius R
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Official CBSE Marking Scheme · CBSE Board 2025 (Supplementary), Set 55/S/1

1f = (μ – 1)( 1R1 – 1R2 ) [½]

For plano convex lens

1f1 = (μ1 – 1)( 1∞ – 1–R )

1f1 = (μ1 – 1)R [½]

For concave lens

1f2 = (μ2 – 1)( 1–R – 1R )

1f2 = –2 (μ2 – 1)R [½]

1f = 1f1 + 1f2

f = Rμ1 – 2μ2 + 1 [½]

Q43.2025 Supp 55/S/1 · Q18 (OR option)2 marks

A convex lens is kept coaxially on the left side of a concave mirror at a distance of 5f from it where f is focal length of each of them. An object is kept at a distance of 2f on the left side of the convex lens. Draw the ray diagram showing the formation of the image by the combination. Find the distance of the final image from the mirror.

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Official CBSE Marking Scheme · CBSE Board 2025 (Supplementary), Set 55/S/1
Ray diagram: object at 2f from convex lens, intermediate image I1 and final image I2 formed by the concave mirror

[1] for ray diagram

u = –3f [½]

1v + 1u = 1f

–1f = –13f + 1v

v = –32 f [½]

Q44.2024 Main 55/1/1 · Q18 (option 1)2 marks

Monochromatic light of frequency 5·0 × 1014 Hz passes from air into a medium of refractive index 1·5. Find the wavelength of the light (i) reflected, and (ii) refracted at the interface of the two media.

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Official CBSE Marking Scheme · CBSE Board 2024 (Main), Set 55/1/1

(i) v = υλ

3 × 108 = 5 × 1014 × λ

λ = 600 nm or 6 × 10−7 m [1]

(ii) λmedium = λairμ

λmedium = 600 nm1.5

= 400 nm or 4 × 10−7 m [1]

Q45.2024 Main 55/1/1 · Q18 (OR option)2 marks

A plano-convex lens of focal length 16 cm is made of a material of refractive index 1·4. Calculate the radius of the curved surface of the lens.

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Official CBSE Marking Scheme · CBSE Board 2024 (Main), Set 55/1/1

1f = (μ − 1) (1R1 − 1R2) [1]

116 = (1.4 − 1) (1R − 1∞)

116 = 0.4 × 1R

R = 16 × 0.4

R = 6.4 cm [1]

Q46.2024 Main 55/1/1 · Q192 marks

An object is placed 30 cm in front of a concave mirror of radius of curvature 40 cm. Find the (i) position of the image formed and (ii) magnification of the image.

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Official CBSE Marking Scheme · CBSE Board 2024 (Main), Set 55/1/1

(i) 1v + 1u = 1f [½]

1v + 1−30 = 1−20

On solving

v = − 60 cm [½]

(ii) m = − vu [½]

= − (−60−30)

= − 2 [½]

Q47.2024 Main 55/2/1 · Q192 marks

A convex lens (n = 1·52) has a focal length of 15·0 cm in air. Find its focal length when it is immersed in liquid of refractive index 1·65. What will be the nature of the lens ?

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Official CBSE Marking Scheme · CBSE Board 2024 (Main), Set 55/2/1

For convex lens in air

1fa = (ngna − 1)(1R1 − 1R2)

For convex lens in liquid

1fl = (ngnl − 1)(1R1 − 1R2) [½]

flfa = 1·52 − 11 ÷ 1·52 − 1·651·65 [½]

= − 6·6 [½]

fl = − 6·6 fa = − 99 cm [½]

Nature of the lens : Diverging / behaves like a concave lens. [½]

Q48.2024 Main 55/3/1 · Q182 marks

The magnifying power of an astronomical telescope is 24. In normal adjustment, distance between its two lenses is 150 cm. Find the focal length of the objective lens.

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Official CBSE Marking Scheme · CBSE Board 2024 (Main), Set 55/3/1

Magnifying power = 24, Distance between lenses = 150 cm

fofe = 24 [½]

fo + fe = 150 cm [½]

fe = 6 cm [½]

fo = 144 cm [½]

Q49.2024 Main 55/3/1 · Q192 marks

Explain the following :

  1. For a simple microscope, the angular size of the object equals the angular size of the image. Yet it offers magnification.
  2. Both plane and convex mirrors produce virtual images of objects. Can they produce real images under some circumstances ?
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Official CBSE Marking Scheme · CBSE Board 2024 (Main), Set 55/3/1

(a) Yes, it offers magnification. [½]

We can keep the small object much closer to the eye than 25 cm and hence have it subtend a large angle. [½]

(b) Yes, [½]

Rays converging to a point behind a plane or convex mirror are reflected to a point in front of the mirror on a screen [½]

Q50.2024 Main 55/4/1 · Q192 marks

A ray of light is incident normally on one face of an equilateral glass prism of refractive index μ. When the prism is completely immersed in a transparent medium, it is observed that the emergent ray just grazes the adjacent face. Find the refractive index of the medium.

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Official CBSE Marking Scheme · CBSE Board 2024 (Main), Set 55/4/1
Ray incident normally on one face of equilateral prism, emerging along the adjacent face with angle of incidence 60°

Ray diagram [½]

From Snell’s law, μ sin i = μm sin r [½]

μ sin 60° = μm sin 90° [½]

μm = μ · 32 [½]

Alternatively

μ = 1sin C [1]

μμm = 1sin 60° [½]

μm = 32 μ [½]

Q51.2024 Main 55/4/1 · Q21 (option 1)2 marks

An air bubble is trapped at point B (CB = 20 cm) in a glass sphere of radius 40 cm and refractive index 1.5 as shown in figure. Find the nature and position of the image of the bubble as seen by an observer at point P.

Glass sphere (μ = 1.5) with centre C, bubble at B and observer at P on the same line
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Official CBSE Marking Scheme · CBSE Board 2024 (Main), Set 55/4/1

Using refraction formula at spherical surface from denser to rarer medium

n1 = refractive index of rarer medium, n2 = refractive index of denser medium

n1v – n2u = n1 – n2R [½]

u = – 20 cm, R = – 40 cm, n1 = 1, n2 = 1.5 [½]

1v – 1.5(–20) = 1 – 1.5(–40) [½]

v = –16 cm

Nature of image is virtual. [½]

Q52.2024 Main 55/4/1 · Q21 (OR option)2 marks

In normal adjustment, for a refracting telescope, the distance between objective and eye piece lens is 1.00 m. If the magnifying power of the telescope is 19, find the focal length of the objective and the eyepiece lens.

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Official CBSE Marking Scheme · CBSE Board 2024 (Main), Set 55/4/1

Distance between objective and eyepiece: fo + fe = 1.00 m = 100 cm [½]

Magnifying power |m| = fofe = 19 [½]

On solving

fo = 95 cm = 0.95 m [½]

fe = 5 cm = 0.05 m [½]

Q53.2024 Main 55/5/1 · Q192 marks

A telescope has an objective lens of focal length 150 cm and an eyepiece of focal length 5 cm. Calculate its magnifying power in normal adjustment and the distance of the image formed by the objective.

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Official CBSE Marking Scheme · CBSE Board 2024 (Main), Set 55/5/1

|m| = f0fe [½]

= 1505 = 30 [½]

1f = 1v – 1u [½]

1150 = 1v – 1∞

v = 150 cm [½]

(Note: Award full credit of this part, if a student writes correct distance of image without calculation i.e. using object position at infinity.)

Q54.2024 Supp 55/S/1 · Q192 marks

A point light source rests on the bottom of a bucket filled with a liquid of refractive index μ = 1.25 up to height of 10 cm. Calculate :

(a) the critical angle for liquid-air interface

(b) radius of circular light patch formed on the surface by light emerging from the source.

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Official CBSE Marking Scheme · CBSE Board 2024 (Supplementary), Set 55/S/1

(a) sin ic = 1μ [½]

sin ic = 45   (∵ μ = 1.25 = 54) [½]

ic = sin–1(45) or ic = 53° [½]

(b) sin ic = rr2 + h2

Ray from source at bottom reaching the surface at critical angle, h = 10 cm, radius r

r2r2 + h2 = (45)2

25r2 = 16r2 + 16h2

9r2 = 1600

r = 403 cm [½]

Q55.2023 Main 55/1/1 · Q21 (OR option)2 marks

Why is a reflecting telescope preferred over a refracting telescope ? Justify your answer giving two reasons.

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Official CBSE Marking Scheme · CBSE Board 2023 (Main), Set 55/1/1

Reason for preferring reflecting type telescope over refracting telescope [½ + ½]; Justification [½ + ½]

1. No Chromatic Aberration – No refraction in mirrors

2. No Spherical Aberration – Due to use of parabolic reflector

3. Easy mechanical support required – Mirrors weigh less and can be supported over entire back surface.

4. High resolving power – Due to Mirror with large diameter.

5. Brighter image – Large mirrors gather more light waves.

(Any two) [1 + 1]

Q56.2023 Main 55/2/1 · Q212 marks

In the given figure the radius of curvature of curved face in the plano-convex and the plano-concave lens is 15 cm each. The refractive index of the material of the lenses is 1.5. Find the final position of the image formed.

Plano-convex lens and plano-concave lens 20 cm apart, parallel rays incident from the left
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Official CBSE Marking Scheme · CBSE Board 2023 (Main), Set 55/2/1

Using the formula 1f = (μ − 1)(1R1 − 1R2) [½]

Focal length of plano-convex lens = +30 cm

Focal length of plano-concave lens = −30 cm [½]

For plano-convex lens

As object is at ∞, its real image will be formed at its focus i.e. +30 cm

v1 = +30 cm

For plano-concave lens

u = +(30 − 20) cm = +10 cm

1f2 = 1v2 − 1u2 [½]

1−30 = 1v2 − 110

∴ v2 = 15 cm [½]

Q57.2023 Main 55/3/1 · Q222 marks

The refractive indices of two media A and B are 2 and 2 respectively. What is the critical angle for their interface ?

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Official CBSE Marking Scheme · CBSE Board 2023 (Main), Set 55/3/1

From Snell’s law :

μA sin ic = μB sin 90° [½]

2 × sin ic = 2 × 1 [½]

sin ic = 12 [½]

ic = 45° [½]

Alternatively:

sin ic = 1BμA [1]

sin ic = 12 [½]

ic = 45° [½]

Q58.2026 Main 55/1/1 · Q273 marks

(a) Draw the ray diagram to show the image formation by a refracting telescope and write the expression for angular magnification for the telescope in normal adjustment.

(b) Give two reasons to explain why a reflecting telescope is preferred over a refracting telescope.

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Official CBSE Marking Scheme · CBSE Board 2026 (Main), Set 55/1/1

(a)

Ray diagram of a refracting telescope with objective and eyepiece, final image at infinity

[1½]

|m| = fofe [½]

(b) Reasons: [½ + ½]

  • No chromatic aberration
  • Less mechanical support is required to install it.
  • Image formed is brighter and clear.
  • Cost efficient.
  • No / less spherical aberration

[Any two reasons]

Q59.2026 Main 55/1/1 · Q283 marks

(a) State the two conditions under which total internal reflection occurs.

(b) A transparent container contains layers of three immiscible transparent liquids A, B and C of refractive indices n, 34n and 23n, respectively. A laser beam is incident at the interface between A and B at an angle θ as shown in figure. Prove that the beam does not enter region C at all for sin θ ≥ 23.

Container with layers A (index n), B (3n/4) and C (2n/3) and a laser beam incident at angle θ on the A–B interface
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Official CBSE Marking Scheme · CBSE Board 2026 (Main), Set 55/1/1

(a) (i) Light must travel from denser medium to rarer medium. [½]

(ii) Angle of incidence must be greater than critical angle, i > ic. [½]

(b)

Ray refracting at the A–B interface (angle θ1 in B) and at the B–C interface

For refraction at AB interface

sin θsin θ1 = 3n / 4n

sin θ = 34 sin θ1 ………… (1) [½]

For refraction at BC interface

sin θ1sin r = 2n / 33n / 4

sin θ1 = 89 sin r ………… (2) [½]

From equations (1) and (2)

32 sin θ = sin r [½]

for sin θ = 23

sin r = 1 i.e. r = 90°

Ray grazes the surface

Hence for sin θ ≥ 23 does not enter the region C at all [½]

Q60.2026 Main 55/3/1 · Q263 marks

A ray of light is travelling through a rectangular glass slab (refractive index 32) and is incident on the horizontal glass-air surface at the critical angle for the two media. The slab is then brought in contact with water (refractive index 43) such that a thin horizontal layer of water is formed on the surface of the slab. Find the angle at which the ray will emerge into air from the water-air surface.

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Official CBSE Marking Scheme · CBSE Board 2026 (Main), Set 55/3/1

sin iC = 1n = 23 [1]

For glass – water surface

32 sin iC = 43 sin r [½]

32 × 23 = 43 sin r

sin r = 34 [½]

For water – air surface

43 sin r = sin e [½]

43 × 34 = sin e

sin e = 1

e = 90° [½]

Q61.2026 Supp 55/7/1 · Q243 marks

Derive an expression for refractive index of an equilateral prism in terms of the angle of minimum deviation and the angle of the prism. Plot a graph to show the variation of the angle of deviation as a function of angle of incidence.

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Official CBSE Marking Scheme · CBSE Board 2026 (Supplementary), Set 55/7/1
Ray passing through prism ABC with angles i, r1, r2, e and deviation δ

[½]

In quadrilateral AQNR, ∠A + ∠QNR = 180°

From triangle QNR, r1 + r2 + ∠QNR = 180°

Comparing the above equations, r1 + r2 = A [½]

The total deviation δ is the sum of deviations at the two faces:

δ = (i – r1) + (e – r2)

δ = i + e – A

At minimum deviation, δ = Dm, i = e which implies r1 = r2

r = A2

Dm = 2i – A or i = A + Dm2 [½]

The refractive index of the prism is

n = sin isin r = sin (A + Dm2)sin (A2) [½]

Graph of angle of deviation versus angle of incidence, with minimum at i = e

Graph [1]

Q62.2026 Supp 55/7/1 · Q273 marks
  1. Draw a labelled ray diagram of a refracting telescope in normal adjustment. Briefly explain its working. [2]
  2. Why is the eyepiece of a telescope of short focal length, while the objective is of large focal length ? Explain. [1]
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Official CBSE Marking Scheme · CBSE Board 2026 (Supplementary), Set 55/7/1

(a)

Ray diagram of a refracting telescope with objective, eyepiece and final image

Labelled ray diagram [1]

Working: When light from a distant object enters the objective lens, a real image is formed in the tube at its second focal point. The eyepiece magnifies this image, producing a final inverted image. [1]

(b) To increase the magnifying power or angular magnification. [1]

Q63.2025 Main 55/2/1 · Q263 marks

(a) When a parallel beam of light enters water surface obliquely at some angle, what is the effect on the width of the beam ?

(b) With the help of a ray diagram, show that a straw appears bent when it is partly dipped in water and explain it.

(c) Explain the transmission of optical signal through an optical fibre by a diagram.

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The official marking scheme for this paper is not available in the source folder.

Q64.2025 Main 55/5/1 · Q273 marks

A rectangular glass slab ABCD (refractive index 1·5) is surrounded by a transparent liquid (refractive index 1·25) as shown in the figure. A ray of light is incident on face AB at an angle i such that it is refracted out grazing the face AD. Find the value of angle i.

Glass slab ABCD surrounded by liquid, ray incident on AB at angle i and emerging along AD
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The official marking scheme for this paper is not available in the source folder.

Q65.2025 Main 55/6/1 · Q263 marks

Draw a ray diagram showing the image formation when a concave mirror produces a real, inverted and magnified image of an object and hence obtain the mirror formula.

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The official marking scheme for this paper is not available in the source folder.

Q66.2023 Main 55/2/1 · Q293 marks

Define critical angle for a given pair of media and total internal reflection. Obtain the relation between the critical angle and refractive index of the medium.

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Official CBSE Marking Scheme · CBSE Board 2023 (Main), Set 55/2/1

Critical angle – When a ray of light passes from a denser to a rarer medium, the value of angle of incidence for which the angle of refraction becomes 90° is called critical angle for that pair of media. [1]

Total internal reflection – When a ray of light passes from a denser to rarer medium and the angle of incidence exceeds the critical angle for pair of media, the ray undergoes reflection. This is called total internal reflection. [1]

From Snell’s law sin isin r = μrd [½]

When angle of incidence is equal to critical angle (∠i = ∠ic), ∠r = 90°

∴ sin icsin 90° = 1μdr

∴ μdr = 1sin ic [½]

Q67.2022 Main 55/1/1 · Q53 marks

A point source in air is kept 24 cm in front of a concave spherical glass surface (aμg = 1.5) and radius of curvature 60 cm. Find the nature of the image formed and its distance from the point source.

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Official CBSE Marking Scheme · CBSE Board 2022 (Main), Set 55/1/1

Relation for concave spherical surface

μ1−u + μ2v = μ2 − μ1R [½]

1−u + μv = μ − 1R [½]

1−(−24) + 1·5v = 1·5 − 1−60 [½]

∴ v = −30 cm [½]

Distance of image from point source = −24 − (−30) = 6 cm [½]

Nature of image = Virtual image [1]

Q68.2022 Main 55/1/1 · Q73 marks

Explain with the help of a suitable diagram, the phenomenon on which an optical fibre works. Mention any two uses of optical fibres.

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Official CBSE Marking Scheme · CBSE Board 2022 (Main), Set 55/1/1

An optical fibre works on the principle of Total internal reflection. [1]

Diagram of TIR [1]

Total internal reflection at water-air interface

Alternatively

Light guided by total internal reflection in an optical fibre

Uses of optical fibres (any two) [½ + ½]

  1. Medical and optical examination (endoscopy).
  2. Transmission and reception of signals
  3. Photometric sensors.
Q69.2022 Main 55/1/1 · Q8 (OR option)3 marks

A thin equiconvex lens of radius of curvature R made of material of refractive index μ1 is kept coaxially, in contact with an equiconcave lens of the same radius of curvature and refractive index μ2 (>μ1). Find :

  1. the ratio of their powers, and
  2. the power of the combination and its nature.
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Official CBSE Marking Scheme · CBSE Board 2022 (Main), Set 55/1/1

Finding the ratio of powers [1½]; Finding the power of combination and nature [1½]

(i) From P = (μ − 1)(1R1 − 1R2) [½]

P1 = Pconvex = (μ1 − 1)(1R − (−1R)) = (μ1 − 1)(2R) [½]

P2 = Pconcave = (μ2 − 1)(−1R − 1R) = −(μ2 − 1)(2R)

∴ P1P2 = (μ1 − 1)−(μ2 − 1) = (μ1 − 1)(1 − μ2) [½]

(ii) P = P1 + P2 [½]

= (μ1 − 1)(2R) + (−(μ2 − 1))(2R)

P = 2(μ1 − μ2)R [½]

As μ2 > μ1, P is negative

∴ Nature is diverging [½]

Q70.2022 Main 55/1/1 · Q11 (OR option)3 marks
  1. Trace the path of a ray of light PQ which is incident at an angle i on one face of a glass prism of angle A. It then emerges out from the other face at an angle e. Use the ray diagram to prove that the angle through which the ray is deviated is given by ∠δ = ∠i + ∠e − ∠A.
  2. What will be the minimum value of δ if the ray passes symmetrically through the prism ?
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Official CBSE Marking Scheme · CBSE Board 2022 (Main), Set 55/1/1

i) Diagram [1]

Ray PQRS through a prism ABC showing i, r1, r2, e and deviation δ

δ = (i − r1) + (e − r2) [½]

δ = (i + e) − (r1 + r2) [½]

In Quadrilateral AQOR
∠Q = ∠R = 90° ∴ ∠A + ∠O = 180° ---------(1)
In ΔQOR
∠O + r1 + r2 = 180° ---------(2)
Comparing (1) and (2)
∴ A = r1 + r2

∴ δ = (i + e) − A [½]

ii) If a ray passes symmetrically through a prism (parallel to base of prism), the value of angle of deviation is minimum. At this angle ∠i = ∠e and ∠r1 = ∠r2 [½]

Q71.2022 Main 55/2/1 · Q53 marks
  1. State the conditions for total internal reflection to take place.
  2. A tank is filled with a transparent liquid to height ‘H’. A coin suspended by a thread in the liquid is gradually lowered till it touches the bottom. The apparent depth is determined corresponding to different positions of the coin.
    1. Plot a graph showing variation of the apparent depth with the real depth of the coin.
    2. What is the physical significance of the slope of the graph ?
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Official CBSE Marking Scheme · CBSE Board 2022 (Main), Set 55/2/1

(a) The conditions for total internal reflection to take place are:

(i) The light should travel from denser to rarer medium. [½]

(ii) The angle of incidence should be greater than the critical angle for a given pair of media. [½]

(b) (i) [1]

Straight line graph of apparent depth versus real depth

(ii) The slope will give 1/μ where μ is refractive index as per the graph above. [1]

Q72.2022 Main 55/2/1 · Q6 (option 1)3 marks

Draw a labelled ray diagram showing the formation of an image by an astronomical refracting telescope in normal adjustment. Hence, obtain the expression for its magnifying power.

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Official CBSE Marking Scheme · CBSE Board 2022 (Main), Set 55/2/1

Labelled ray diagram of refracting telescope for normal arrangement [1½]

Ray diagram of astronomical refracting telescope in normal adjustment

m = tan βtan α ≈ βα (as α, β are small angles) [½]

≈ hfe × f0h [½]

= f0fe [½]

(Note: ½ mark to be deducted if arrows not shown or labelling is not done)

Q73.2022 Main 55/2/1 · Q73 marks

A converging lens made of glass ( μ = 1·5) has its spherical faces of radii of curvature 10 cm and 20 cm. Find its focal length

  1. in air, and
  2. when it is immersed in a liquid of refractive index 1·25.
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Official CBSE Marking Scheme · CBSE Board 2022 (Main), Set 55/2/1

(a) 1fa = (aμg – 1)(1R1 – 1R2) [½]

R1 = 10 cm & R2 = –20 cm, aμg = 1.5 [½]

1fa = (1.5 – 1)(110 – 1–20) [½]

fa = 403 = 13.3 cm [½]

(b) 1fl = (lμg – 1)(1R1 – 1R2) [½]

= (1.51.25 – 1)(110 – 1–20) [½]

fl = 1003 = 33.3 cm [½]

Q74.2022 Main 55/3/1 · Q8 (option 1)3 marks
  1. Write two necessary conditions for total internal reflection.
  2. Two prisms ABC and DBC are arranged as shown in figure.
    Two prisms ABC (angle A 45°) and DBC (angles 60°, 60°) sharing face BC, ray incident normally on AB
    The critical angles for the two prisms with respect to air are 41.1° and 45° respectively. Trace the path of the ray through the combination.
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Official CBSE Marking Scheme · CBSE Board 2022 (Main), Set 55/3/1

(a) Two conditions

(i) The light must travel from an optically denser medium to a rarer medium. [1]

(ii) Angle of incidence should be greater than the critical angle. [1]

(b) [1]

Ray enters AB normally, totally reflected at AC, passes down through BC, and emerges at BD side without deviation
Q75.2022 Main 55/3/1 · Q8 (OR option)3 marks
  1. An object is placed in front of a converging lens. Obtain the conditions under which the magnification produced by the lens is (i) negative and (ii) positive.
  2. A point object is placed at O in front of a glass sphere as shown in figure.
    Point object O at distance 3r from the surface of a glass sphere of radius r with centre C
    Show the formation of image by the sphere.
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Official CBSE Marking Scheme · CBSE Board 2022 (Main), Set 55/3/1

(a) Two conditions m = h′h = vu [½ + ½]

(i) For real images / when object is placed beyond F, As u is negative and v is positive. [½ + ½]

(ii) for virtual image / when object is kept between F and the optical centre of the lens. As u and v both are negative. [½ + ½]

(b) [1]

Ray diagram of image I formed by the glass sphere for object O at 3r
Q76.2022 Main 55/4/1 · Q8 (option 1)3 marks

(i) Draw a labelled ray diagram showing the formation of the image at infinity by an astronomical telescope.

(ii) A telescope consists of an objective of focal length 150 cm and an eyepiece of focal length 6·0 cm. If the final image is formed at infinity, then calculate :

  1. the length of the tube in this adjustment, and
  2. the magnification produced.
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Official CBSE Marking Scheme · CBSE Board 2022 (Main), Set 55/4/1

(i) Labelled ray diagram of astronomical telescope [1]

Ray diagram of astronomical telescope, final image at infinity

(ii) Given fo = 150 cm, fe = 6 cm

(I) Length of the tube L = fo + fe [½]

= 150 + 6

L = 156 cm [½]

(II) m = fofe [½]

= 1506 = 25 [½]

Q77.2022 Main 55/4/1 · Q8 (OR option)3 marks

(i) Draw a labelled ray diagram showing the formation of the image at least distance of distinct vision by a compound microscope.

(ii) A small object is placed at a distance of 3·0 cm from a magnifier of focal length 4·0 cm. Find :

  1. the position of the image formed, and
  2. the linear magnification produced.
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Official CBSE Marking Scheme · CBSE Board 2022 (Main), Set 55/4/1

(i) Labelled ray diagram of compound microscope [1]

Ray diagram of image formation by a compound microscope at least distance of distinct vision

(ii) Given u = −3 cm, f = 4 cm

(I) Using 1v − 1u = 1f [½]

1v = 1u + 1f = 1−3·0 + 14·0

1v = −4 + 312 = −112 v = −12 cm [½]

(II) Linear magnification m = vu [½]

m = −12−3 = 4 [½]

Q78.2022 Main 55/4/1 · Q103 marks

A ray of light is incident on a prism at an angle of 45° and passes symmetrically as shown in the figure. Calculate :

Prism ABC with apex angle 60°, ray incident at P at 45° passing symmetrically through to Q
  1. the angle of minimum deviation,
  2. the refractive index of the material of the prism, and
  3. the angle of refraction at the point P.
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Official CBSE Marking Scheme · CBSE Board 2022 (Main), Set 55/4/1

(a) As the ray passes symmetrically ∠i = ∠e and δ = δm

∴ i + e = A + δ

2i = A + δm

δm = 2i − A = 90° − 60° [½]

= 30° [½]

(b) μ = sin (A + δm)/2sin A/2 [½]

= sin 45°sin 30° = 2 [½]

(c) As δ = δm, r1 = r2 (Alternatively)

Hence, r1 (Angle of refraction at P)

1 × sin 45° = μ sin r1

1 × 12 = 2 sin r1 [½]

∴ ∠r1 = 30° [½]

Q79.2022 Main 55/5/1 · Q8 (option 1)3 marks
  1. Define SI unit of power of a lens.
  2. A plano convex lens is made of glass of refractive index 1.5. The radius of curvature of the convex surface is 25 cm.
    1. Calculate the focal length of the lens.
    2. If an object is placed 50 cm in front of the lens, find the nature and position of the image formed.
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Official CBSE Marking Scheme · CBSE Board 2022 (Main), Set 55/5/1

Definition [1]; Calculation of focal length [1]; Nature and position of image [1]

(i) Dioptre is the reciprocal of focal length of lens in metre. [1]

Alternatively: Dioptre = 1/focal length (m)

Alternatively: One dioptre is the power of a lens of focal length of one metre

(ii.i) 1f = (μ – 1)(1R1 – 1R2) [½]

R1 = ∞, R2 = –25 cm, μ = 1.5

1f = (μ – 1)(1∞ + 125)

or 1f = 0.5 × 125 or f = 50 cm [½]

(ii.ii) 1f = 1v – 1u

150 = 1v – 1–50 [½]

1v = 0

∴ v = ∞ [½]

Thus the image will be real and inverted

(Note: Award only ½ mark if a student draws ray diagram showing correct position of the image.)

Q80.2022 Main 55/5/1 · Q103 marks

A ray of light passes through a prism of refractive index 2 as shown in the figure. Find :

Ray passing through prism ABC (angle A = 60°, B = 60°, μ = √2), refracted at M, travelling to N with angles r<sub>1</sub>, r<sub>2</sub>, emerging along face AC
  1. The angle of incidence (∠r2) at face AC.
  2. The angle of minimum deviation for this prism.
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Official CBSE Marking Scheme · CBSE Board 2022 (Main), Set 55/5/1

Calculation of ∠r2 [1½]; Calculation of angle of minimum deviation [1½]

(i) As the emergent ray grazes along the side AC, therefore

12 = sin r2sin 90° [1]

∴ r2 = 45° [½]

(ii) μ = sin((A + δm)/2)sin(A/2) [½]

2 = sin((60° + δm)/2)sin 30° [½ + ½]

∴ δm = 30° [½]

Q81.2022 Main 55/5/1 · Q11 (OR option)3 marks

With the help of a ray diagram explain the working of a reflecting telescope. Mention two advantages of a reflecting telescope over a refracting telescope.

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Official CBSE Marking Scheme · CBSE Board 2022 (Main), Set 55/5/1

Ray diagram and explanation of working [2]; Advantages [½ + ½]

Ray diagram of reflecting telescope [1½]

Ray diagram of a reflecting telescope: objective mirror, secondary mirror, eyepiece

Working: Parallel beam of light gathered by objective mirror is reflected to the secondary mirror, which further forms the image in front of the eyepiece. [½]

(Note: Deduct ½ marks for not showing the direction of propagation of rays and give full credit for the ray diagram of Newtonian telescope)

Two Advantages (Any Two) [½ + ½]

  • High resolving power
  • No chromatic aberration
  • Reduced spherical aberration
  • Brighter image is formed
  • Easy mechanical support
  • Large magnifying power
Q82.2022 Supp 55/6/1 · Q4 (option 1)3 marks

Draw a ray diagram to show the formation of an image at the least distance of distinct vision, by a compound microscope. Hence, obtain an expression for its angular magnification. [3]

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Official CBSE Marking Scheme · CBSE Board 2022 (Supplementary), Set 55/6/1

Ray diagram [1]

Ray diagram of a compound microscope with final image at the near point D

Linear magnification due to objective (mo)

mo = h′h = Lfo ......(1) [½]

Linear magnification due to eye piece (me) when final image is formed at near point (D)

me = 1 + Dfe ......(2) [½]

Total magnification m = mo × me [½]

m = Lfo (1 + Dfe) [½]

Q83.2022 Supp 55/6/1 · Q4 (OR option)3 marks

Consider a convex spherical surface of radius of curvature R, separating two media of refractive indices n1 and n2 (> n1). A point object is placed at a distance u in front of the surface in medium of refractive index n1. Its real image is formed at a distance v. Obtain a relation between u and v in terms of n1, n2 and R. [3]

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Official CBSE Marking Scheme · CBSE Board 2022 (Supplementary), Set 55/6/1

Ray diagram [1]

Refraction at a convex spherical surface, object O, image I, centre C

For small angles,

tan ∠NOM = MNOM

tan ∠NCM = MNMC

tan ∠NIM = MNMI [½]

Now, for ΔNOC, ∠i is the exterior angle. Therefore, ∠i = ∠NOM + ∠NCM

i = MNOM + MNMC

Similarly, r = ∠NCM – ∠NIM

i.e. r = MNMC – MNMI [½]

By Snell’s law n1 sin i = n2 sin r

or for small angles, sin i ≈ i and sin r ≈ r

n1 i = n2 r

Substituting i and r, we get

n1OM + n2MI = n2 – n1MC [½]

Where OM, MI and MC represent magnitude of distances.

Applying the Cartesian sign convention OM = –u, MI = +v, MC = +R

Substituting these in equation, we get

n2v – n1u = n2 – n1R [½]

Q84.SQP 2025-26 · Q253 marks

A compound microscope consists of an objective lens of focal length 0.82 cm and an eyepiece lens of focal length 2.9 cm. An object is placed 0.91 cm from the objective lens. The image is formed at the near point (25 cm) from the eye.

(I) Calculate that the angular magnification of the microscope.
(II) Draw the ray diagram of compound microscope in normal adjustment.

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Official CBSE Marking Scheme · CBSE Sample Question Paper 2025-26

(I) 1v0 = 1f0 – 1u0

v0 = 8.3 cm [1]

Angular magnification M = mo × me

M = v0u0 (Dfe + 1)

M = – 8.30.91 × (252.9 + 1)

M = –87.7 [1]

(II)

Ray diagram of a compound microscope with objective, eyepiece and final image

[1]

Q85.SQP 2026-27 · Q253 marks

A diamond has a refractive index of 2.42.

  1. A ray of light inside the diamond strikes the surface at an angle of incidence of 300. State whether it will undergo refraction or total internal reflection.
  2. Explain why diamonds sparkle when cut properly.
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Official CBSE Marking Scheme · CBSE Sample Question Paper 2026-27

For diamond, refractive index µ = 2.42

Critical angle θc = sin-1 12.42

Or, θc < 30° [1]

I. Given incidence angle: 30°. Since 30° > θc, the ray exceeds the critical angle. Hence, the ray will undergo Total Internal Reflection (TIR) inside the diamond. [1]

II. A well-cut diamond has facets (the flat faces) angled just right so that light hitting them reflects back inward (TIR) instead of leaking out the sides or bottom. So the light gets trapped inside the piece of diamond. [1]

Q86.2026 Main 55/2/1 · Q304 marks

In an experiment with convex lens of focal length f, the screen is fixed at a distance D from the object. A student slowly moves the lens away from the object towards the screen and finds that she is able to form sharp image of the object for two positions of the lens. The distance between these two positions of the lens is d.

  1. The value of d is
    1. D(D – 4f)
    2. D(D – 2f)
    3. 2Df
    4. D(D – f)
  2. Compared to the size of the object, the images formed in the two positions of the lens are respectively
    1. reduced, enlarged
    2. reduced, reduced
    3. enlarged, enlarged
    4. enlarged, reduced
  3. If the distance between object and screen is 80.00 cm and the lens forms sharp images at two positions separated by 20.00 cm., the focal length of convex lens is
    1. 15.50 cm
    2. 18.75 cm
    3. 20.50 cm
    4. 22.75 cm
  4. Consider a convex lens of focal length 15 cms. For which of the following values of object-screen distance, two positions of the object can be found to obtain sharp image on the screen ?
    1. 45 cm
    2. 50 cm
    3. 55 cm
    4. 65 cm

    OR

    A thin convex lens of focal length 10 cm and another thin lens of focal length ‘f’ are placed coaxially in contact. If the power of their combination is 103 D, the value of ‘f’ is
    1. –15 cm
    2. –10 cm
    3. –20 cm
    4. –30 cm
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Official CBSE Marking Scheme · CBSE Board 2026 (Main), Set 55/2/1

(i) (A) D(D – 4f) [1]

(ii) (D) Enlarged, reduced [1]

(iii) (B) 18.75 cm [1]

(iv) (D) 65 cm [1]

OR (iv) (A) –15 cm [1]

Q87.2026 Main 55/4/1 · Q304 marks

An astronomical telescope consists of two converging lenses. One of them of large aperture and large focal length is called objective lens and the other one, of smaller focal length and smaller aperture is called the eyepiece. It is used to see distant objects which are not seen clearly with naked eyes. The image formed by the objective lens acts as an object for the eyepiece and the final image produced by the eyepiece is magnified.

(i) The images formed by the objective lens and the eyepiece are respectively :

  1. virtual, real
  2. real, virtual
  3. virtual, virtual
  4. real, real

(ii) The magnification produced by the telescope does not depend upon the :

  1. colour of light
  2. focal length of objective lens
  3. focal length of eyepiece
  4. apertures of objective lens and eyepiece

(iii) Which of the following statements is not correct for this telescope ?

  1. The focal length of objective lens (fo) is larger than the focal length of eyepiece (fe).
  2. Its magnifying power can be increased by increasing the focal length of objective lens (fo).
  3. The distance between two lenses is more than (fo + fe).
  4. The magnifying power can be decreased by increasing the focal length of eyepiece.

(iv) (a) An astronomical telescope has objective lens and eyepiece of focal lengths 80 cm and 4 cm respectively. To view the image in normal adjustment, the lenses must be separated by a distance of :

  1. 84 cm
  2. 76 cm
  3. 20 cm
  4. 320 cm

(iv) (b) (alternative to (a)) Consider the telescope described in question (iv) (a). Its magnifying power in normal adjustment will be :

  1. 320
  2. 84
  3. 76
  4. 20
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Official CBSE Marking Scheme · CBSE Board 2026 (Main), Set 55/4/1

(i) (B) real, virtual [1]

(ii) (D) apertures of objective lens and eyepiece [1]

(iii) (C) The distance between two lenses is more than (fo + fe) [1]

(iv) (A) 84 cm [1]

Alternatively, for the (b) option : (D) 20 [1]

Q88.2025 Main 55/1/1 · Q304 marks

A thin lens is a transparent optical medium bounded by two surfaces, at least one of which should be spherical. Applying the formula for image formation by a single spherical surface successively at the two surfaces of a lens, one can obtain the ‘lens maker formula’ and then the ‘lens formula’. A lens has two foci – called ‘first focal point’ and ‘second focal point’ of the lens, one on each side.

Optical bench with screen, convex lens and light box carrying a plate with an arrow and a line with a ball

Consider the arrangement shown in figure. A black vertical arrow and a horizontal thick line with a ball are painted on a glass plate. It serves as the object. When the plate is illuminated, its real image is formed on the screen.

(i) Which of the following correctly represents the image formed on the screen ?

  1. Upward arrow with line tilted up to the right ending in a ball on the right
  2. Upward arrow with line tilted up to the right and ball on the lower left
  3. Downward arrow with line and ball on the left
  4. Downward arrow with line tilted up to the right and ball on the upper right

(ii) Which of the following statements is incorrect ?

  1. For a convex mirror magnification is always negative.
  2. For all virtual images formed by a mirror magnification is positive.
  3. For a concave lens magnification is always positive.
  4. For real and inverted images, magnification is always negative.

(iii) A convex lens of focal length ‘f’ is cut into two equal parts perpendicular to the principal axis. The focal length of each part will be :

  1. f
  2. 2 f
  3. f2
  4. f4

Alternative to (iii) : (iii) If an object in case (i) above is 20 cm from the lens and the screen is 50 cm away from the object, the focal length of the lens used is

  1. 10 cm
  2. 12 cm
  3. 16 cm
  4. 20 cm

(iv) The distance of an object from first focal point of a biconvex lens is X1 and distance of the image from second focal point is X2. The focal length of the lens is

  1. X1 X2
  2. X1 + X2
  3. X1 X2
  4. X2X1
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Q89.2025 Main 55/7/1 · Q304 marks

When light travels from an optically denser medium to an optically rarer medium, at the interface it is partly reflected back into the same medium and partly refracted to the second medium. The angle of incidence corresponding to an angle of refraction 90° is called the critical angle (ic) for the given pair of media. This angle is related to the refractive index of medium 1 with respect to medium 2.

Refraction of light through a prism involves refraction at two plane interfaces. A relation for the refractive index of the material of the prism can be obtained in terms of the refracting angle of the prism and the angle of minimum deviation. For a thin prism, this relation reduces to a simple equation.

Laws of refraction are also valid for refraction of light at a spherical interface. When an object is placed in front of a spherical surface separating two media, its image is formed. A relation between object and image distance, in terms of refractive indices of two media and the radius of curvature of the spherical surface can be obtained. Using this relation for two surfaces of a lens, ‘lens maker formula’ is obtained.

(i) A small bulb is placed at the bottom of a tank containing a transparent liquid (refractive index n) to a depth H. The radius of the circular area of the surface of liquid, through which light from the bulb can emerge out, is R. Then (RH) is :

  1. 1n2 – 1
  2. n2 – 1
  3. 1n2 + 1
  4. n2 + 1

(ii) (a) A parallel beam of light is incident on a face of a prism with refracting angle 60°. The angle of minimum deviation is found to be 30°. The refractive index of the material of the prism is close to :

  1. 1·3
  2. 1·4
  3. 1·5
  4. 1·6

Alternative to (ii)(a) : (b) The angle of minimum deviation for a ray of light incident on a thin prism, made of crown glass (n = 1·52) is Dm. If the prism was made of dense flint glass (n = 1·62) instead of crown glass, the angle of minimum deviation will :

  1. decrease by 4%
  2. increase by 4%
  3. decrease by 19%
  4. increase by 19%

(iii) An object is placed in front of a convex spherical glass surface (n = 1·5 and radius of curvature R) at a distance of 4R from it. As the object is moved slowly close to the surface, the image formed is :

  1. always real
  2. always virtual
  3. first real and then virtual
  4. first virtual and then real

(iv) A double-convex lens, made of glass of refractive index 1·5, has focal length 10 cm. The radius of curvature of its each face, is :

  1. 10 cm
  2. 15 cm
  3. 20 cm
  4. 40 cm
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Q90.2025 Supp 55/S/1 · Q294 marks

A compound microscope is an optical instrument used for observing highly magnified images of tiny objects. It consists of two convex lenses. The lens near the object is called the objective and the lens near the eye is called the eyepiece. The magnifying power of a compound microscope is given by m = mome, where mo is the magnification produced by objective lens and me is the magnification produced by eyepiece. The expression for me depends on whether the final image is formed at the near point or at infinity (normal adjustment).

(i)(a) A compound microscope consists of an objective lens of focal length fo = 2·0 cm and an eyepiece of focal length fe = 6·25 cm separated by a distance of 15 cm. How far should an object be placed from the objective so that the final image is formed at infinity ?

  1. 3·45 cm
  2. 5 cm
  3. 1·29 cm
  4. 2·59 cm

OR

(b) A person with normal near point 25 cm, using a compound microscope with an objective of focal length 0·8 cm and eyepiece of focal length 2·5 cm can bring an object placed at 0·9 cm from the objective in sharp focus. If the image formed by the objective is at a distance of 2·27 cm from the eyepiece, the separation between the lenses is :

  1. 7·47 cm
  2. 8·47 cm
  3. 9·47 cm
  4. 10·47 cm

(ii) A compound microscope consists of an objective of focal length 2 cm and an eyepiece of focal length 10 cm. The eyepiece is kept 20 cm away from the objective. Find the magnifying power of the microscope if the final image is formed at infinity.

  1. 7
  2. 10
  3. 14
  4. 16

(iii) The intermediate image formed in a compound microscope is :

  1. real, inverted and magnified
  2. real, erect and magnified
  3. virtual, erect and magnified
  4. virtual, inverted and magnified

(iv) The focal lengths of the objective and the eyepiece of a compound microscope are fo and fe respectively. Which of the following statements is correct, in normal adjustment ?

  1. The object is at a distance less than fo from the objective.
  2. The object is at a distance more than fo and less than 2fo from the objective.
  3. The intermediate image is formed at a distance slightly more than fe from the eyepiece.
  4. The intermediate image is formed at 2fe from the eyepiece.
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Official CBSE Marking Scheme · CBSE Board 2025 (Supplementary), Set 55/S/1

(i) (a) (D) 2.59 cm [1]

OR

(b) (C) 9.47 cm [1]

(ii) (B) 10 [1]

(iii) (A) Real inverted and magnified. [1]

(iv) (B) The object is at a distance more than f0 and less than 2f0 from the objective. [1]

Q91.2024 Main 55/1/1 · Q294 marks

A lens is a transparent medium bounded by two surfaces, with one or both surfaces being spherical. The focal length of a lens is determined by the radii of curvature of its two surfaces and the refractive index of its medium with respect to that of the surrounding medium. The power of a lens is reciprocal of its focal length. If a number of lenses are kept in contact, the power of the combination is the algebraic sum of the powers of the individual lenses.

(i) A double-convex lens, with each face having same radius of curvature R, is made of glass of refractive index n. Its power is : [1]

  1. 2 (n − 1)R
  2. (2n − 1)R
  3. (n − 1)2R
  4. (2n − 1)2R

(ii) A double-convex lens of power P, with each face having same radius of curvature, is cut into two equal parts perpendicular to its principal axis. The power of one part of the lens will be : [1]

  1. 2P
  2. P
  3. 4P
  4. P2

(iii) The above two parts are kept in contact with each other as shown in the figure. The power of the combination will be : [1]

Two plano-convex halves of a lens placed in contact with their curved faces touching
  1. P2
  2. P
  3. 2P
  4. P4

(iv) Attempt any one of (a), (b). [1]

(a) A double-convex lens of power P, with each face having same radius of curvature, is cut along its principal axis. The two parts are arranged as shown in the figure. The power of the combination will be :

Two halves of a double-convex lens, cut along the principal axis, placed side by side above the axis
  1. Zero
  2. P
  3. 2P
  4. P2

(b) Two convex lenses of focal lengths 60 cm and 20 cm are held coaxially in contact with each other. The power of the combination is :

  1. 6·6 D
  2. 15 D
  3. 115 D
  4. 180 D
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Official CBSE Marking Scheme · CBSE Board 2024 (Main), Set 55/1/1

(i) (A) 2 (n − 1)R [1]

(ii) (D) P2 [1]

(iii) (B) P [1]

(iv) (a) (C) 2P

    (b) (A) 6.6 D [1]

Q92.2024 Main 55/2/1 · Q304 marks

When a ray of light propagates from a denser medium to a rarer medium, it bends away from the normal. When the incident angle is increased, the refracted ray deviates more from the normal. For a particular angle of incidence in the denser medium, the refracted ray just grazes the interface of the two surfaces. This angle of incidence is called the critical angle for the pair of media involved.

(i) For a ray incident at the critical angle, the angle of reflection is :

  1. 0°
  2. < 90°
  3. > 90°
  4. 90°

(ii) A ray of light of wavelength 600 nm is incident in water (n = 43) on the water-air interface at an angle less than the critical angle. The wavelength associated with the refracted ray is :

  1. 400 nm
  2. 450 nm
  3. 600 nm
  4. 800 nm

(iii) (a) The interface AB between the two media A and B is shown in the figure. In the denser medium A, the incident ray PQ makes an angle of 30° with the horizontal. The refracted ray is parallel to the interface. The refractive index of medium B w.r.t. medium A is :

Interface AB between media A and B with incident ray PQ at 30 degrees to the horizontal
  1. 32
  2. 52
  3. 43
  4. 23

(b) Two media A and B are separated by a plane boundary. The speed of light in medium A and B is 2 × 108 ms−1 and 2·5 × 108 ms−1 respectively. The critical angle for a ray of light going from medium A to medium B is :

  1. sin−1 12
  2. sin−1 45
  3. sin−1 35
  4. sin−1 25

(iv) The figure shows the path of a light ray through a triangular prism. In this phenomenon, the angle θ is given by :

Light ray entering a right-angled prism at angle theta, travelling parallel to the base inside, and emerging
  1. sin−1 n2 − 1
  2. sin−1 (n2 − 1)
  3. sin−1 [1n2 − 1]
  4. sin−1 [1(n2 − 1)]
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Official CBSE Marking Scheme · CBSE Board 2024 (Main), Set 55/2/1

(i) Since no option is correct, award 1 mark to all students. [1]

(ii) (D) 800 nm [1]

(iii) (a) (A) 32

(iii) (b) (B) sin−1 (45) [1]

(iv) (A) sin−1 n2 − 1 [1]

Q93.2024 Main 55/4/1 · Q304 marks

A lens is a transparent optical medium bounded by two surfaces; at least one of which should be spherical. Applying the formula of image formation by a single spherical surface successively at the two surfaces of a thin lens, a formula known as lens maker’s formula and hence the basic lens formula can be obtained. The focal length (or power) of a lens depends on the radii of its surfaces and the refractive index of its material with respect to the surrounding medium. The refractive index of a material depends on the wavelength of light used. Combination of lenses helps us to obtain diverging or converging lenses of desired power and magnification.

(i) A thin converging lens of focal length 20 cm and a thin diverging lens of focal length 15 cm are placed coaxially in contact. The power of the combination is

  1. –56 D
  2. –53 D
  3. 43 D
  4. 32 D

(ii) The radii of curvature of two surfaces of a convex lens are R and 2R. If the focal length of this lens is (43)R, the refractive index of the material of the lens is :

  1. 53
  2. 43
  3. 32
  4. 75

(iii) The focal length of an equiconvex lens

  1. increases when the lens is dipped in water.
  2. increases when the wavelength of incident light decreases.
  3. increases with decrease in radius of curvature of its surface.
  4. decreases when the lens is cut into two identical parts along its principal axis.

(iv) (a) A thin convex lens L of focal length 10 cm and a concave mirror M of focal length 15 cm are placed coaxially 40 cm apart as shown in figure. A beam of light coming parallel to the principal axis is incident on the lens. The final image will be formed at a distance of

Convex lens L and concave mirror M placed 40 cm apart on a common axis; parallel beam incident on the lens
  1. 10 cm, left of lens
  2. 10 cm, right of lens
  3. 20 cm, left of lens
  4. 20 cm, right of lens

or

(iv) (b) A beam of light coming parallel to the principal axis of a convex lens L1 of focal length 16 cm is incident on it. Another convex lens L2 of focal length 12 cm is placed coaxially at a distance 40 cm from L1. The nature and distance of the final image from L2 will be

  1. real, 24 cm
  2. virtual, 12 cm
  3. real, 32 cm
  4. virtual, 18 cm
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Official CBSE Marking Scheme · CBSE Board 2024 (Main), Set 55/4/1

(i) (B) –53 D [1]

(ii) (C) 32 [1]

(iii) (A) increases when a lens is dipped in water. [1]

(iv) (a) (B) 10 cm, right from lens. [1]

or (b) (A) real, 24 cm

Q94.2024 Main 55/5/1 · Q30 (option 1)4 marks

A prism is an optical medium bounded by three refracting plane surfaces. A ray of light suffers successive refractions on passing through its two surfaces and deviates by a certain angle from its original path. The refractive index of the material of the prism is given by μ = sin(A + δm2) / sin A2. If the angle of incidence on the second surface is greater than an angle called critical angle, the ray will not be refracted from the second surface and is totally internally reflected.

  1. The critical angle for glass is θ1 and that for water is θ2. The critical angle for glass-water surface would be (given aμg = 1.5, aμw = 1.33)
    1. less than θ2
    2. between θ1 and θ2
    3. greater than θ2
    4. less than θ1
  2. When a ray of light of wavelength λ and frequency υ is refracted into a denser medium
    1. λ and υ both increase.
    2. λ increases but υ is unchanged.
    3. λ decreases but υ is unchanged.
    4. λ and υ both decrease.
  3. The critical angle for a ray of light passing from glass to water is minimum for
    1. red colour
    2. blue colour
    3. yellow colour
    4. violet colour
  4. A ray of light is incident normally on a prism ABC of refractive index 2, as shown in figure. After it strikes face AC, it will
    Right-angled prism ABC with angle 60° at A and right angle at B; a ray enters normally on face AB
    1. go straight undeviated
    2. just graze along the face AC
    3. refract and go out of the prism
    4. undergo total internal reflection
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Official CBSE Marking Scheme · CBSE Board 2024 (Main), Set 55/5/1

(i) (C) greater than θ2 [1]

(ii) (C) λ decreases but ν is unchanged [1]

(iii) (a) (D) violet colour [1]

(iv) (D) undergo total internal reflection [1]

Q95.2024 Main 55/5/1 · Q30 (OR option)4 marks

A prism is an optical medium bounded by three refracting plane surfaces. A ray of light suffers successive refractions on passing through its two surfaces and deviates by a certain angle from its original path. The refractive index of the material of the prism is given by μ = sin(A + δm2) / sin A2. If the angle of incidence on the second surface is greater than an angle called critical angle, the ray will not be refracted from the second surface and is totally internally reflected.

  1. The critical angle for glass is θ1 and that for water is θ2. The critical angle for glass-water surface would be (given aμg = 1.5, aμw = 1.33)
    1. less than θ2
    2. between θ1 and θ2
    3. greater than θ2
    4. less than θ1
  2. When a ray of light of wavelength λ and frequency υ is refracted into a denser medium
    1. λ and υ both increase.
    2. λ increases but υ is unchanged.
    3. λ decreases but υ is unchanged.
    4. λ and υ both decrease.
  3. Three beams of red, yellow and violet colours are passed through a prism, one by one under the same condition. When the prism is in the position of minimum deviation, the angles of refraction from the second surface are rR, rY and rV respectively. Then
    1. rV < rY < rR
    2. rY < rR < rV
    3. rR < rY < rV
    4. rR = rY = rV
  4. A ray of light is incident normally on a prism ABC of refractive index 2, as shown in figure. After it strikes face AC, it will
    Right-angled prism ABC with angle 60° at A and right angle at B; a ray enters normally on face AB
    1. go straight undeviated
    2. just graze along the face AC
    3. refract and go out of the prism
    4. undergo total internal reflection
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Official CBSE Marking Scheme · CBSE Board 2024 (Main), Set 55/5/1

(i) (C) greater than θ2 [1]

(ii) (C) λ decreases but ν is unchanged [1]

(iii) (b) (C) rR < rY < rV [1]

(iv) (D) undergo total internal reflection [1]

Q96.2023 Main 55/1/1 · Q35 (option 1)4 marks

Strontium titanate is a rare oxide — a natural mineral found in Siberia. It is used as a substitute for diamond because its refractive index and critical angle are 2·41 and 24·5°, respectively, which are approximately equal to the refractive index and critical angle of diamond. It has all the properties of diamond. Even an expert jeweller is unable to differentiate between diamond and strontium titanate. A ray of light is incident normally on one face of an equilateral triangular prism ABC made of strontium titanate.

Equilateral prism ABC with apex angle 60 degrees at A and a ray incident on face AB

Answer the following questions based on the above :

(a) Trace the path of the ray showing its passage through the prism. [1]

(b) Find the velocity of light through the prism. [1]

(c) Briefly explain two applications of total internal reflection. [2]

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Official CBSE Marking Scheme · CBSE Board 2023 (Main), Set 55/1/1

a)

Ray enters face AB normally, strikes AC at 60 degrees (greater than critical angle) and is totally internally reflected towards BC

From fig. angle of incidence on second face i = 60°; critical angle ic = 24.5°; (i) > (ic), ∴ TIR takes place [1]

b) n = cv; v = cn = 3 × 1082.41 = 1.24 × 108 m/s [1]

c) Optical Fibre / Brilliance of diamond / mirage (any two) [1 + 1]

Note: Give full credit if students mention the names of applications only.

Q97.2023 Main 55/1/1 · Q35 (OR option)4 marks

Strontium titanate is a rare oxide — a natural mineral found in Siberia. It is used as a substitute for diamond because its refractive index and critical angle are 2·41 and 24·5°, respectively, which are approximately equal to the refractive index and critical angle of diamond. It has all the properties of diamond. Even an expert jeweller is unable to differentiate between diamond and strontium titanate. A ray of light is incident normally on one face of an equilateral triangular prism ABC made of strontium titanate.

Equilateral prism ABC with apex angle 60 degrees at A and a ray incident on face AB

Answer the following questions based on the above :

(a) Trace the path of the ray showing its passage through the prism. [1]

(b) Find the velocity of light through the prism. [1]

(c) Define total internal reflection of light. Give two conditions for it. [2]

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Official CBSE Marking Scheme · CBSE Board 2023 (Main), Set 55/1/1

a)

Ray enters face AB normally, strikes AC at 60 degrees (greater than critical angle) and is totally internally reflected towards BC

From fig. angle of incidence on second face i = 60°; critical angle ic = 24.5°; (i) > (ic), ∴ TIR takes place [1]

b) n = cv; v = cn = 3 × 1082.41 = 1.24 × 108 m/s [1]

c) When light travels from optically denser medium to rarer medium at an interface and gets reflected back into the same medium the phenomenon is called as total internal reflection. [1]

Conditions for TIR:

1. Light must travel from optically denser medium to rarer medium.

2. Angle of incidence at the interface must be greater than the critical angle for the pair of media. [½ + ½]

Q98.2023 Main 55/4/1 · Q35 (option 1)4 marks

The lens maker’s formula is useful to design lenses of desired focal lengths using surfaces of suitable radii of curvature. The focal length also depends on the refractive index of the material of the lens and the surrounding medium. The refractive index depends on the wavelength of the light used. The power of a lens is related to its focal length.

Answer the following questions based on the above :

(a) How will the power of a lens be affected with an increase of wavelength of light ?

(b) The radius of curvature of two surfaces of a convex lens is R each. For what value of μ of its material will its focal length become equal to R ?

(c) The focal length of a concave lens of μ = 1·5 is 20 cm in air. It is completely immersed in water of μ = 43. Calculate its focal length in water.

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Official CBSE Marking Scheme · CBSE Board 2023 (Main), Set 55/4/1

(a) P = 1f = (μ − 1) (1R1 − 1R2)

μ ∝ 1λ ⟹ P ∝ 1λ

So P decreases as λ increases. [1]

(b) 1f = (μ − 1) (1R1 − 1R2) [½]

1R = (μ − 1) 2R    (R1 = R, R2 = − R)

⟹ μ = 32 [½]

(c) 1f = (μ − 1) (1R1 − 1R2) [½]

120 = (32 − 1) (1R1 − 1R2) [½]

When the lens is immersed in water

1fwater = (98 − 1) (1R1 − 1R2) [½]

1/201/f = 1/21/8 ⟹ f = 80 cm [½]

Q99.2023 Main 55/4/1 · Q35 (OR option)4 marks

The lens maker’s formula is useful to design lenses of desired focal lengths using surfaces of suitable radii of curvature. The focal length also depends on the refractive index of the material of the lens and the surrounding medium. The refractive index depends on the wavelength of the light used. The power of a lens is related to its focal length.

Answer the following questions based on the above :

(a) How will the power of a lens be affected with an increase of wavelength of light ?

(b) The radius of curvature of two surfaces of a convex lens is R each. For what value of μ of its material will its focal length become equal to R ?

(c) An object is placed in front of a lens which forms its erect image of magnification 3. The power of the lens is 5 D. Calculate the distance of the object and the image from the lens.

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Official CBSE Marking Scheme · CBSE Board 2023 (Main), Set 55/4/1

(a) P = 1f = (μ − 1) (1R1 − 1R2)

μ ∝ 1λ ⟹ P ∝ 1λ

So P decreases as λ increases. [1]

(b) 1f = (μ − 1) (1R1 − 1R2) [½]

1R = (μ − 1) 2R    (R1 = R, R2 = − R)

⟹ μ = 32 [½]

(c) m = 3 = vu ⟹ v = 3u [½]

f = 100+5 = + 20 cm [½]

1v − 1u = 1f

13u − 1u = 120

1 − 33u = 120

u = − (40/3) cm [½]

v = − 40 cm [½]

Q100.2023 Main 55/5/1 · Q34 (option 1)4 marks

A lens is a transparent optical medium bounded by two surfaces; at least one of which should be spherical. Considering image formation by a single spherical surface successively at the two surfaces of a lens, lens maker’s formula is obtained. It is useful to design lenses of desired focal length using surfaces of suitable radii of curvature. This formula helps us obtain a relation between u, v and f for a lens. Lenses form images of objects and they are used in a number of optical devices, for example microscopes and telescopes.

  1. An object AB is kept in front of a composite convex lens, as shown in figure. Will the lens produce one image ? If not, explain.
    Object AB in front of a convex lens whose upper half has refractive index n1 and lower half n2
  2. A real image of an object formed by a convex lens is observed on a screen. If the screen is removed, will the image still be formed ? Explain.
  3. A double convex lens is made of glass of refractive index 1.55 with both faces of the same radius of curvature. Find the radius of curvature required if focal length is 20 cm.
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Official CBSE Marking Scheme · CBSE Board 2023 (Main), Set 55/5/1

(i) No, [½]

The lens is made up of two materials of different refractive indices. It has two focal lengths. [½]

(ii) Yes [½]

Rays are still intersecting / converging at the location of image. [½]

(iii) 1f = (n – 1)[1R1 – 1R2] [½]

R1 = +R, R2 = –R [½]

1f = (n – 1)[2R] [½]

R = 2(n – 1) f

R = 2 (1.55 – 1) × 20 = 22 cm [½]

Q101.2023 Main 55/5/1 · Q34 (OR option)4 marks

A lens is a transparent optical medium bounded by two surfaces; at least one of which should be spherical. Considering image formation by a single spherical surface successively at the two surfaces of a lens, lens maker’s formula is obtained. It is useful to design lenses of desired focal length using surfaces of suitable radii of curvature. This formula helps us obtain a relation between u, v and f for a lens. Lenses form images of objects and they are used in a number of optical devices, for example microscopes and telescopes.

  1. An object AB is kept in front of a composite convex lens, as shown in figure. Will the lens produce one image ? If not, explain.
    Object AB in front of a convex lens whose upper half has refractive index n1 and lower half n2
  2. A real image of an object formed by a convex lens is observed on a screen. If the screen is removed, will the image still be formed ? Explain.
  3. Two convex lenses A and B of focal lengths 15 cm and 10 cm respectively are placed coaxially ‘d’ distance apart. A point object is kept at a distance of 30 cm in front of lens A. Find the value of ‘d’ so that the rays emerging from lens B are parallel to its principal axis.
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Official CBSE Marking Scheme · CBSE Board 2023 (Main), Set 55/5/1

(i) No, [½]

The lens is made up of two materials of different refractive indices. It has two focal lengths. [½]

(ii) Yes [½]

Rays are still intersecting / converging at the location of image. [½]

(iii) For lens A (f = 15 cm), u = –30 cm

1v – 1u = 1f [½]

1v1 = 115 – 130 = 130

v1 = 30 cm [½]

For lens B: for rays to go parallel to principal axis out of lens B the image formed by lens A must lie at the focus of B. So d = 30 + 10 = 40 cm [1]

Alternatively: (Object is kept at 2f so image will also be formed at 2f on the other side of the lens i.e. at 30 cm. Now the final image is to be formed at infinity so the image formed must lie at the focus of the second lens (B). Thus separation is 30 + 10 = 40 cm)

Q102.2023 Supp 55/C/1 · Q35 (option 1)4 marks

A prism is a solid transparent medium bounded by three rectangular faces with a triangular base and a top. A ray of light incident at angle i on one face of a prism suffers two refractions on passing through a prism. Hence it deviates through a certain angle δ from its original path. The angle of deviation becomes minimum (δ = δm) for a certain value of angle i. In such a condition, the refracted ray inside the prism becomes parallel to its base. An expression for refractive index μ of the material of the prism can be obtained in terms of angle A and angle δm.

(a) Show in a figure the variation of angle δ with angle of incidence i. [1]

(b) Show that for a prism of small angle A, the refractive index μ of its material can be written as μ = 1 + δmA. [1]

(c) A ray of light passes through an equilateral prism such that both the angle of incidence and the angle of emergence are equal to the angle of prism A. Find the refractive index of the material of the prism, in terms of A. [2]

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(a) [1]

Graph of angle of deviation versus angle of incidence, U-shaped curve with minimum at i = e

(b) μ = sin[(A + δm)/2]sin A/2 [½]

For small angle

μ = (A + δm)/2A/2

μ = A + δmA

μ = 1 + δmA [½]

(c) i + e = A [½]

r = A/2 [½]

μ = sin isin r [½]

μ = sin Asin A/2 [½]

μ = 2 sin(A/2) cos(A/2)sin(A/2) = 2 cos(A/2) [½]

Q103.2023 Supp 55/C/1 · Q35 (OR option)4 marks

A prism is a solid transparent medium bounded by three rectangular faces with a triangular base and a top. A ray of light incident at angle i on one face of a prism suffers two refractions on passing through a prism. Hence it deviates through a certain angle δ from its original path. The angle of deviation becomes minimum (δ = δm) for a certain value of angle i. In such a condition, the refracted ray inside the prism becomes parallel to its base. An expression for refractive index μ of the material of the prism can be obtained in terms of angle A and angle δm.

(a) Show in a figure the variation of angle δ with angle of incidence i. [1]

(b) Show that for a prism of small angle A, the refractive index μ of its material can be written as μ = 1 + δmA. [1]

(c) A ray of light passes through a prism of angle 75°, as shown in the figure. The refractive index of the material of the prism, with respect to its surrounding is 2. Find the angle of incidence i. [2]

Ray passing through a prism of angle 75 degrees, refractive index root 2, travelling parallel to the base inside, angle of incidence i marked
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Official CBSE Marking Scheme · CBSE Board 2023 (Supplementary), Set 55/C/1

(a) [1]

Graph of angle of deviation versus angle of incidence, U-shaped curve with minimum at i = e

(b) μ = sin[(A + δm)/2]sin A/2 [½]

For small angle

μ = (A + δm)/2A/2

μ = A + δmA

μ = 1 + δmA [½]

(c) sin isin r = 2 [½]

sin r2sin 90° = 12

sin r2 = 12

r2 = 45° [½]

r1 + r2 = A

r1 + 45° = 75°

r1 = 30°

sin isin r1 = 2

sin isin 30° = 2

sin i = 2 × 12

sin i = 12

i = 45° [½ + ½]

Q104.2026 Main 55/1/1 · Q32 (option 1)5 marks

(a) Using the relation for refraction at a curved spherical surface, derive the expression for lens maker’s formula.

(b) Three lenses L1, L2 and L3, each of focal length 40 cm, are placed coaxially. The distance between L1 and L2 and between L2 and L3 are 120 cm and 20 cm respectively. An object is kept at a distance of 80 cm to the left of lens L1.

Find the distance of the final image formed from the object.

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Official CBSE Marking Scheme · CBSE Board 2026 (Main), Set 55/1/1

(a)

Image formation by a thin lens through two successive refracting surfaces: object O, intermediate image I1, final image I

[1] (figure)

The first refracting surface forms the image of the object O at I1

For refraction from first interface ABC

n1OB + n2BI1 = n2 – n1BC1 ………… (1) [½]

Similarly for refraction from second interface ADC. The image I1 acts as a virtual object for the second surface ADC

– n2DI1 + n1DI = n2 – n1DC2 ………… (2) [½]

For thin lens BI1 = DI1

By adding equation (1) and (2)

n1OB + n1DI = (n2 – n1)[1BC1 + 1DC2]

– n1u + n1v = (n2 – n1)[1R1 – 1R2]

1v – 1u = (n2n1 – 1)[1R1 – 1R2] [½]

If object is kept at ∞, image will form at focus hence

1f = (n2n1 – 1)(1R1 – 1R2) [½]

(b)

Three lenses L1, L2, L3 with object O at 80 cm from L1; separations 120 cm and 20 cm

For lens L1

1f = 1v – 1u

140 = 1v1 + 180

1v1 = 180

v1 = 80 cm [½]

For lens L2

u2 = 120 – 80 = 40 cm

140 = 1v2 + 140

1v2 = 0, v2 = ∞ [½]

For lens L3

u3 = ∞

140 = 1v3 + 1∞

v3 = 40 cm [½]

Distance between final image and object = 80 + 120 + 20 + 40 = 260 cm [½]

Alternatively:

Ray diagram: O, I1 after L1 (80 cm + 80 cm), then L2 and L3 forming final image I2; total 260 cm
Q105.2026 Main 55/1/1 · Q32 (OR option)5 marks

(a) Draw a ray diagram to show the image formation by a concave mirror when the object is kept between its focus and the centre of curvature. Using this diagram, derive the mirror formula.

(b) A concave mirror produces a two times magnified virtual image of an object kept 10 cm in front of it. Calculate the focal length of the mirror.

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Official CBSE Marking Scheme · CBSE Board 2026 (Main), Set 55/1/1

(a)

Concave mirror: object AB between F and C, real inverted enlarged image A′B′; points M, N, P marked

[1] (ray diagram)

ΔBAP and ΔB′A′P are similar then

BAB′A′ = APA′P ………… (1) [½]

∵ BA = MN

ΔMNF and ΔB′A′F are similar then

MNB′A′ = NFA′F

BAB′A′ = NFA′F = PFA′F (N is very close to P and BA = MN) ………… (2) [½]

From equation (1) and (2)

APA′P = PFA′F

–u–v = –f–v + f [½]

–uv + uf = –vf

–uvuvf + ufuvf = –vfuvf

–1f + 1v = 1–u

 1f = 1v + 1u  [½]

Note: Award full marks if student derives mirror formula by using any other justified method

(b) m = –vu [½]

m = 2

∵ v = –2u ⇒ v = –2(–10) = 20 cm [½]

1f = 1v + 1u

120 – 110 = 1f [½]

1 – 220 = 1f

f = –20 cm [½]

Q106.2026 Main 55/2/1 · Q31 (OR option)5 marks

(a) Draw a labelled ray diagram of a refracting telescope when it forms image of a distant object at infinity. Derive expression for its magnifying power.

(b) (i) In a telescope the objective has much larger aperture than the eye piece. Why ?

(ii) Write two advantages of reflecting telescope over refracting telescope.

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Official CBSE Marking Scheme · CBSE Board 2026 (Main), Set 55/2/1

(a)

Ray diagram of refracting telescope with objective, eyepiece, focal lengths f_o and f_e, image A′B′ at infinity

[1] for diagram

m = βα ≈ tan βtan α [1]

= h / feh / fo [½]

= fofe [½]

(b) (i) With larger aperture even the fainter objects can be observed clearly. [1]

Alternatively : With larger aperture, the light gathering power and its resolution / resolving power increases.

Alternatively : Large aperture helps in collecting more light from distant object which forms a sharper / brighter images.

(ii) (Any two) [½ + ½]

  • There is no chromatic aberration in a mirror.
  • Spherical aberration in case of parabolic reflectors is removed.
  • Requires less mechanical support
  • Cost effective
  • Brighter / sharper image is formed

(Any two from above or any other appropriate advantages)

Q107.2026 Main 55/3/1 · Q33 (option 1)5 marks

(i) A point object is kept in front of a convex spherical surface of radius of curvature R. Draw the ray diagram to show the formation of image and derive the relation between the object and image distance (u and v) in terms of refractive index n of the medium and R.

(ii) A convex lens of focal length of 20 cm is used to form the image of an object placed 30 cm away from the lens. Find the position and nature of the image formed.

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Official CBSE Marking Scheme · CBSE Board 2026 (Main), Set 55/3/1

(i) Ray diagram [1]

Refraction at a convex spherical surface: object O, image I, centre of curvature C, point N on the surface

For small angles

tan∠NOM = MNMO, tan∠NCM = MNMC, tan∠NIM = MNMI

i = ∠NOM + ∠NCM = MNMO + MNOC ------------(1) [½]

Similarly

r = ∠NCM − ∠NIM = MNMC − MNMI ------------(2) [½]

By Snell’s law for small angles

1 × i = n × r

substituting i and r from (1) and (2) :

1MO + nMI = n − 1MC ------------(3) [½]

Using cartesian sign conventions

MO = −u, MI = +v, MC = +R

Substituting in (3) :

nv − 1u = n − 1R [½]

(Note : Give full credit of this part if student derives the relation by forming virtual image.)

(ii) 1f = 1v − 1u [½]

120 = 1v − 1−30 [½]

v = 60 cm [½]

Nature of image : Real and inverted [½]

Q108.2026 Main 55/4/1 · Q31 (option 1)5 marks

(a) (i) Define refractive index of a medium in terms of speed of light.

(ii) Derive the relation for the refractive index (μ) of a prism in terms of angle of minimum deviation (δm) and angle of prism (A).

(iii) A ray of light QP is incident normally on the face BC of a triangular prism ABC of refractive index 1·5 kept in air, as shown in the figure. Trace the path of the ray as it passes through the prism and give relevant explanation.

Triangular prism ABC with ray QP incident normally on face BC at P
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Official CBSE Marking Scheme · CBSE Board 2026 (Main), Set 55/4/1

(i) Refractive index of a medium is the ratio of the speed of light in vacuum to the speed of light in the medium. [1]

Alternatively : n21 = cv

(ii) Figure [½]

Ray PQRS through a prism ABC at minimum deviation, with angles i, r<sub>1</sub>, r<sub>2</sub>, e and deviation δ

In the quadrilateral AQNR, two of the angles (at the vertices Q and R) are right angles. Therefore, the sum of the other angles of the quadrilateral is 180°.

∠A + ∠QNR = 180°

From the triangle QNR,

r1 + r2 + ∠QNR = 180°

Comparing these two equations, we get

r1 + r2 = A ………(i) [½]

The total deviation δ is the sum of the deviations at the two faces.

δ = i + e − A ………(ii) [½]

For δ = δm ; i = e which implies r1 = r2

From equation (i) and (ii)

r = A2 ; i = A + δm2

Refractive index of the prism

n21 = n2n1 = sin(A + δm)2sinA2 [½]

(iii) Path of the ray [1]

Ray QP enters BC undeviated, undergoes total internal reflection at face AB and emerges undeviated through face AC (A = 60°)

The ray QP strikes the face BC normally and hence goes undeviated. As i > ic, it undergoes TIR. It strikes the face AC normally and comes out undeviated. [1]

Alternatively :

Right-angled prism (A = 90°): ray QP undergoes TIR at both faces AB and AC and comes out of face BC undeviated

The ray QP strikes the surface BC normally and hence goes undeviated. As i > ic, it undergoes TIR. As at face AC, i > ic, it undergoes TIR. The ray comes out of the face BC undeviated.

(Award full credit of this part if the student takes the value of A other than 60° or 90° and traces the path correctly and according to the angle taken)

Q109.2026 Main 55/5/1 · Q33 (OR option)5 marks
  1. With the help of a ray diagram, describe the construction and working of a compound microscope.
    1. The real image of an object placed between f and 2f from a convex lens can be seen on a screen placed at the image location. If the screen is removed, is the image still there ? Explain.
    2. Plane and convex mirrors produce virtual images of objects. Can they produce real images under some circumstances ? Explain.
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Official CBSE Marking Scheme · CBSE Board 2026 (Main), Set 55/5/1

(i) Construction/labelling using ray diagram [1 + 1]; Working of a compound microscope with help of ray diagram [1]; (ii)(I) Explaining the formation of real image if the screen is removed [1]; (II) Explaining the circumstances for formation of real image [1]

(i)

Ray diagram of a compound microscope with objective, eyepiece and final virtual image at D

Construction: It consists of two convex lenses, objective lens with small aperture placed near the object and the eye piece with large aperture placed near the eye. Both lenses have small focal length. These lenses are placed coaxially in a narrow tube with rack and pinion arrangement. [1]

Working: The lens near to the object called objective forms a real, inverted & magnified image of the object. This serves as the object for eye piece which produces the final image, enlarged & virtual. [1]

(ii) I. Yes [½]. Real image is formed by actual intersection of rays. [½]

II. Yes [½]. If the object is virtual. [½]

Note: 1. If students do not write construction award full marks for well-labelled diagram representing objective lens, eye piece, along with image position at near point or normal adjustment.

2. Deduct ½ mark if arrows are not drawn.

Q110.2026 Supp 55/7/1 · Q33 (option 1)5 marks
  1. A small bulb is placed at the bottom of a tank containing water to a depth of 1·00 m. Find the area of the surface of water through which light from the bulb can emerge from. The refractive index of water is (4/3). [3]
  2. An object is placed at a distance of 15 cm from a convex lens of focal length 10 cm. On the other side of the lens, a convex mirror is placed coaxially at a distance of 10 cm from the lens. If the image formed by this combination coincides with the object itself, draw the ray diagram and find the focal length of the convex mirror. [2]
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Official CBSE Marking Scheme · CBSE Board 2026 (Supplementary), Set 55/7/1

(a)

Cone of light from a bulb at the bottom of a tank, with critical angle i_c, radius r and depth h

[1]

sin ic = 1n; tan ic = rh [½]

r = h tan ic & tan ic = 1n2 – 1 [½]

A = πr2 = πh2tan2ic = πh2n2 – 1 [½]

A = π × 1169 – 1 = 9π7 m2 = 9 × 3.147 = 4.04 m2 [½]

(b)

Ray diagram: object, convex lens, convex mirror 10 cm away, with centre of curvature C behind the mirror

[1]

1v – 1u = 1f

1v = 110 – 115 = 130

v = 30 cm [½]

Radius of curvature of the mirror, R = 30 – 10 = 20 cm

f = R2 = 10 cm [½]

Q111.2025 Main 55/1/1 · Q33 (OR option)5 marks

(i) Define a wavefront. An incident plane wave falls on a convex lens and gets refracted through it. Draw a diagram to show the incident and refracted wavefront.

(ii) A beam of light coming from a distant source is refracted by a spherical glass ball (refractive index 1.5) of radius 15 cm. Draw the ray diagram and obtain the position of the final image formed.

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Q112.2025 Main 55/2/1 · Q33 (option 1)5 marks
  1. A thin pencil of length (f/4) is placed coinciding with the principal axis of a mirror of focal length f. The image of the pencil is real and enlarged, just touches the pencil. Calculate the magnification produced by the mirror.
  2. A ray of light is incident on a refracting face AB of a prism ABC at an angle of 45°. The ray emerges from face AC and the angle of deviation is 15°. The angle of prism is 30°. Show that the emergent ray is normal to the face AC from which it emerges out. Find the refraction index of the material of the prism.
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Q113.2025 Main 55/4/1 · Q33 (option 1)5 marks
  1. An object is placed 30 cm from a thin convex lens of focal length 10 cm. The lens forms a sharp image on a screen. If a thin concave lens is placed in contact with the convex lens, the sharp image on the screen is formed when the screen is moved by 45 cm from its initial position. Calculate the focal length of the concave lens.
  2. Calculate the angle of minimum deviation of an equilateral prism. The refractive index of the prism is 3. Calculate the angle of incidence for this case of minimum deviation also.
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Q114.2025 Main 55/5/1 · Q33 (option 1)5 marks

(i) Draw a ray diagram of a reflecting telescope (Cassegrain) and explain the formation of image. State two important advantages that a reflecting telescope has over a refracting telescope.

(ii) In a refracting telescope, the focal length of the objective is 50 times the focal length of the eyepiece. When the final image is formed at infinity, the length of the tube is 102 cm. Find the focal lengths of the two lenses.

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Q115.2025 Main 55/5/1 · Q33 (OR option)5 marks

(i) Write any two advantages of a compound microscope over a simple microscope. Draw a ray diagram for the image formation at the near point by a compound microscope and explain it.

(ii) A thin planoconcave lens with its curved face of radius of curvature R is made of glass of refractive index n1. It is placed coaxially in contact with a thin equiconvex lens of same radius of curvature of refractive index n2. Obtain the power of the combination lens.

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Q116.2025 Main 55/6/1 · Q32 (option 1)5 marks
  1. Draw a ray diagram to show the image formation by a compound microscope. Obtain the expression for the total magnification of the microscope when the final image is formed at infinity.
  2. In a compound microscope, an object is placed at a distance of 1·5 cm from the objective of focal length 1·25 cm. The eyepiece has a focal length of 5 cm. The final image is formed at infinity. Calculate the distance between the objective and the eyepiece.
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Q117.2025 Main 55/7/1 · Q33 (option 1)5 marks

(i) Explain with the help of a labelled ray diagram the formation of final image by an astronomical telescope at infinity. Write the expression for its magnifying power.

(ii) The total magnification produced by a compound microscope is 20. The magnification produced by the eyepiece is 5. When the microscope is focussed on a certain object, the distance between the objective and eyepiece is observed to be 14 cm. Calculate the focal lengths of the objective and the eyepiece. (Given that the least distance of distinct vision = 25 cm)

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Q118.2025 Supp 55/S/1 · Q33 (option 1)5 marks

A ray of light is incident on one face of a triangular glass prism of refractive index n and is refracted out from the opposite face. Deduce the expression for refractive index of glass prism in terms of the angle of minimum deviation and angle of the prism. Write the condition on n for refraction to take place on the opposite face.

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Official CBSE Marking Scheme · CBSE Board 2025 (Supplementary), Set 55/S/1
Ray PQRS through a triangular prism ABC showing angles i, r1, r2, e and deviation δ

[1] for diagram

A + δ = i + e [½]

and A = r1 + r2 [½]

at angle of minimum deviation, δ = Dm

i = e, r1 = r2 = r

A + Dm = 2i [½]

i = A + Dm2

A = 2r ⇒ r = A2 [½]

From Snell’s law

n = sin isin r [½]

n = sin( A + Dm2 )sin( A2 ) [½]

For refraction to take place on opposite face

r2 < ic [½]

n = 1sin ic

ic = sin–1( 1n )

r2 < sin–1( 1n ) [½]

Q119.2024 Main 55/1/1 · Q33 (option 1)5 marks

(i) A ray of light passes through a triangular prism. Show graphically, how the angle of deviation varies with the angle of incidence ? Hence define the angle of minimum deviation.

(ii) A ray of light is incident normally on a refracting face of a prism of prism angle A and suffers a deviation of angle δ. Prove that the refractive index n of the material of the prism is given by n = sin (A + δ)sin A.

(iii) The refractive index of the material of a prism is 2. If the refracting angle of the prism is 60°, find the

(1) Angle of minimum deviation, and
(2) Angle of incidence.

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Official CBSE Marking Scheme · CBSE Board 2024 (Main), Set 55/1/1

(i) Graph showing variation of angle of deviation with angle of incidence [1]; defining angle of minimum deviation [1]. (ii) Proof [1]. (iii) (1) angle of minimum deviation [1]; (2) angle of incidence [1]

(i)

Graph of angle of deviation against angle of incidence: U-shaped curve with minimum deviation at i = e

Minimum deviation angle is defined as the angle at which angle of incidence is equal to the angle of emergence. [1]

Alternatively: At minimum deviation refracted ray inside the prism becomes parallel to the base of the prism.

(ii)

Ray incident normally on face XY of prism XYZ, refracted at face XZ at N with deviation δ

At the face XZ :

μ sin i = 1 × sin r   ----- (1)

r = i + δ   [from diagram]   ----- (2) [½]

In ΔXMN : A + (90 − i) + 90 = 180

⇒ A = i   ----- (3)

Putting eq. (3) & (2) in eq. (1)

μ sin A = sin (A + δ)

μ = sin (A + δ)sin A [½]

(iii) (1) μ = sin (A + δm2)sin A2

2 = sin (60 + δm2)sin 30° [½]

⇒ sin (60 + δm2) = 12 = sin 45°

60 + δm2 = 45° ⇒ δm = 30° [½]

(2) i = A + δm2 [½]

⇒ i = 60 + 302

i = 45° [½]

Q120.2024 Main 55/2/1 · Q33 (OR option)5 marks

(i) Draw a labelled ray diagram of a compound microscope showing image formation at least distance of distinct vision. Derive an expression for its magnifying power.

(ii) A telescope consists of two lenses of focal length 100 cm and 5 cm. Find the magnifying power when the final image is formed at infinity.

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Official CBSE Marking Scheme · CBSE Board 2024 (Main), Set 55/2/1

(i) Ray diagram [1]

Labelled ray diagram of a compound microscope with final image at the least distance of distinct vision D

The magnification obtained by eye-piece lens me = (1 + Dfe) [½]

The magnification obtained by objective lens m0 = v0−u0 [½]

Hence the total magnifying power is m = m0 × me [½]

= v0−u0(1 + Dfe) [½]

(ii) m = |f0fe| [1]

Identification of focal length of objective and eyepiece
f0 = 100 cm, fe = 5 cm [½]

m = |1005| = 20 [½]

Q121.2024 Main 55/3/1 · Q33 (option 1)5 marks

(i) Trace the path of a ray of light showing refraction through a triangular prism and hence obtain an expression for angle of deviation (δ) in terms of A, i and e, where symbols have their usual meanings. Draw a graph showing the variation of angle of deviation with the angle of incidence.

(ii) In the figure, a ray of light is incident on a transparent liquid contained in a thin glass box at an angle of 45° with its one face. The emergent ray passes along the face AB. Find the refractive index of the liquid.

Rectangular glass box ABCD with liquid; ray incident on face AD at 45 degrees and emerging along face AB
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Official CBSE Marking Scheme · CBSE Board 2024 (Main), Set 55/3/1

(i) Tracing the path of ray [½]; Obtaining an expression for angle of deviation [1½]; Drawing graph [1]; (ii) Finding the refractive index [2]

Ray PQRS refracted through prism ABC with angles i, r1, r2, e and deviation δ

[½]

For quadrilateral AQNR,

∠A + ∠QNR = 180°   --- (i) [½]

For triangle QNR

r1 + r2 + ∠QNR = 180°   ---- (ii)

comparing equation (i) and (ii)

r1 + r2 = A   ------ (iii) [½]

The angle of deviation

δ = (i – r1) + (e – r2)   ------ (iv)

from equation (iii) and (iv)

δ = i + e – A [½]

Graph [1]

Graph of angle of deviation against angle of incidence, minimum deviation at i = e

(ii)

Ray refracting at G on face AD and meeting face AB at E with angle 90° − θ

sin 45°sin θ = μ

12 = μ sin θ [½]

For second surface,

sin(90° – θ)sin 90° = 1μ [½]

12 cos θsin θ = 1

tan θ = 12 [½]

From the triangle GEF

sin θ = 13 [½]

μ = 32

Q122.2024 Main 55/4/1 · Q31 (option 1)5 marks
  1. Draw a ray diagram for the formation of the image of an object by a convex mirror. Hence, obtain the mirror equation.
  2. Why are multi-component lenses used for both the objective and the eyepiece in optical instruments ?
  3. The magnification of a small object produced by a compound microscope is 200. The focal length of the eyepiece is 2 cm and the final image is formed at infinity. Find the magnification produced by the objective.
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Official CBSE Marking Scheme · CBSE Board 2024 (Main), Set 55/4/1

(i) Ray diagram [1]

Image formation by a convex mirror: virtual, erect, diminished image A′B′ between P and F

For paraxial rays MP can be considered to be a straight line perpendicular to CP. Therefore right angled triangles A′B′F and MPF are similar

B′A′PM = B′FFP

Or B′A′BA = B′FFP (∵ PM = AB) ………… (1) [½]

Since ∠APB = ∠A′PB′, the right angled triangles A′PB′ and ABP are also similar

Therefore, B′A′BA = B′PBP ………… (2) [½]

Comparing eq (1) and (2), we get

B′FFP = B′PBP

PF – PB′FP = B′PBP

Using sign convention PF = f, PB′ = + v, PB = – u [½]

On solving 1v + 1u = 1f [½]

(ii) To improve image quality by minimizing various optical aberrations in lenses. [1]

(iii) Magnification produced by compound microscope m = mo × me [½]

mo = mme = m|Dfe|

mo = 200252 = 16 [½]

Q123.2024 Supp 55/S/1 · Q33 (option 1)5 marks

(i) What are the two main considerations for designing the objective and eyepiece lenses of an astronomical telescope ? Obtain the expression for magnifying power of the telescope when the final image is formed at infinity.

(ii) A ray of light is incident at an angle of 45° at one face of an equilateral triangular prism and passes symmetrically through the prism. Calculate :

(1) the angle of deviation produced by the prism

(2) the refractive index of the material of the prism

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Official CBSE Marking Scheme · CBSE Board 2024 (Supplementary), Set 55/S/1

Two main considerations

Objective should have

1. Larger diameter

2. Larger focal length [½]

Eye piece should have

1. Smaller diameter

2. Smaller focal length [½]

Ray diagram [1]

Ray diagram of an astronomical telescope with final image at infinity

Magnifying power of telescope

Magnifying power is the ratio of the angle β subtended at the eye by the final images to the angle α which the object subtends at the lens or eye

m ≈ βα ≈ hfe · f0h = f0fe [1]

(ii) i + e = D + A

at D = Dm, i = e

2i = Dm + A

2 × 45 = Dm + 60°

Dm = 30° [1]

μ = sin((A + Dm)/2)sin(A/2)

μ = sin((60° + 30°)/2)sin(60°/2)

= 2 [1]

Q124.2024 Supp 55/S/1 · Q33 (OR option)5 marks

(i) Describe a simple activity to observe diffraction pattern due to a single slit.

(ii) The figure below shows an equiconvex lens (of refractive index 1.50) in contact with a liquid layer on top of a plane mirror. A small needle with its tip on the principal axis is moved along the axis until its inverted image is found at the position of the needle. The distance of the needle from the lens is measured to be 45.0 cm. When the liquid is removed and the experiment is repeated, the new distance is 30.0 cm. Find the refractive index of the liquid.

Equiconvex lens on a liquid layer over a plane mirror, needle position P on the principal axis
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Official CBSE Marking Scheme · CBSE Board 2024 (Supplementary), Set 55/S/1

(i) We hold two razor blades in such a way that their edges are parallel and with a narrow slit in between. Keep the slit parallel to the filament of electric bulb, right in front of the eye. A diffraction is seen with its bright and dark bands. [2]

(ii) 1f = (μ – 1)[1R1 – 1R2] [½]

Focal length of convex lens, f1 = 30 cm

130 = (1.5 – 1)[1R – 1(–R)] [½]

R = 30 cm

focal length of combination, f = 45 cm

focal length of plane concave lens of liquid.

1f2 = 1f – 1f1 [½]

1f2 = 145 – 130

f2 = –90 cm [½]

Using lens maker’s formula

1–90 = (μl – 1)[1–30 – 1∞] [½]

μl = 43 [½]

Q125.2023 Main 55/1/1 · Q33 (OR option)5 marks

(i) A spherical surface of radius of curvature R separates two media of refractive indices n1 and n2. A point object is placed in front of the surface at distance u in medium of refractive index n1 and its image is formed by the surface at distance v, in the medium of refractive index n2. Derive a relation between u and v.

(ii) A solid glass sphere of radius 6·0 cm has a small air bubble trapped at a distance 3·0 cm from its centre C as shown in the figure. The refractive index of the material of the sphere is 1·5. Find the apparent position of this bubble when seen through the surface of the sphere from an outside point E in air.

Glass sphere with centre C, air bubble at O 3 cm from C, and outside point E on the line through O and C
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Official CBSE Marking Scheme · CBSE Board 2023 (Main), Set 55/1/1

i) Derivation of relation between u and v [3]; ii) Finding apparent position [2]

Refraction at a spherical surface: object O, image I, pole M, centre of curvature C, point N on surface

[1]

Assume that the aperture of the surface is small as compared to other distance involved, so that small angle approximation can be made.

For small angles, for ΔNOC, i is the exterior angle

∴ i = ∠NOM + ∠NCM; i = MNOM + MNMC  (i) [½]

Similarly r = ∠NCM − ∠NIM = MNMC − MNMI  (ii)

By Snell’s law n1 sin i = n2 sin r; for small angles n1i = n2r [½]

Substituting i and r from (i) and (ii) we get n1OM + n2MI = n2 − n1MC [½]

Applying Cartesian coordinates OM = −u, MI = +v, MC = +R: n2v − n1u = n2 − n1R [½]

(ii) n2v − n1u = n2 − n1R

R = −6 cm, u = −3 cm, n1 = 1.5, n2 = 1

1v + 1.53 = 1 − 1.5−6 [½]

1v = 0.56 − 1.53 [½]; 1v = 0.5 − 36 [½]; 1v = −2.56

v = −2.4 cm [½]

(i.e. 2.4 cm from the left surface, inside the sphere)

Q126.2023 Main 55/2/1 · Q33 (OR option)5 marks
  1. Draw a labelled ray diagram showing the image formation by a refracting telescope. Define its magnifying power. Write two limitations of a refracting telescope over a reflecting telescope.
  2. The focal lengths of the objective and the eye-piece of a compound microscope are 1.0 cm and 2.5 cm respectively. Find the tube length of the microscope for obtaining a magnification of 300.
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Official CBSE Marking Scheme · CBSE Board 2023 (Main), Set 55/2/1

(i) Labelled ray diagram [1½]

Ray diagram of a refracting telescope: objective, eyepiece, intermediate image A′B′ and final image

(Note deduct ½ mark if a student does not show the direction of propagation of the light.)

Alternatively

Give full credit for ray diagram if a candidate draws ray diagram for final image at the near point.

Magnifying power of a telescope – It is defined as the ratio the angle subtended at the eye by the final image to the angle subtended by the object at the lens or the eye. [1]

Two limitations of a refracting telescope over a reflecting telescope. [½ + ½]

(i) Less resolving power.

(ii) Difficult mechanical support.

(iii) Less bright image.

(iv) Suffers chromatic aberration.

(v) Image suffers with spherical aberration.

(Any two of the above)

(ii) fo = 1.0 cm, fe = 2.5 cm, m = 300, D = 25 cm, L = ? [½]

|m| = Lf0·Dfe [½]

300 = L1.0·252.5

L = 30 cm [½]

Q127.2023 Main 55/3/1 · Q33 (option 1)5 marks

(i) Draw a ray diagram showing the formation of a real image of an object placed at a distance ‘u’ in front of a concave mirror of radius of curvature ‘R’. Hence, obtain the relation for the image distance ‘v’ in terms of u and R.

(ii) A 1·8 m tall person stands in front of a convex lens of focal length 1 m, at a distance of 5 m. Find the position and height of the image formed.

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Official CBSE Marking Scheme · CBSE Board 2023 (Main), Set 55/3/1

(i) Ray diagram showing formation of real image in a concave mirror [1]

Ray diagram: real inverted image A′B′ of object AB formed by a concave mirror, with u, v and f marked

From the figure, the two right-angled triangles A′B′F and MPF are similar. (For paraxial rays, MP can be considered to be a straight line perpendicular to CP.) Therefore,

B′A′PM = B′FFP

or B′A′BA = B′FFP (∵ PM = AB)   …(i) [½]

Since ∠APB = ∠A′PB′, the right angled triangles A′B′P and ABP are also similar. Therefore,

B′A′BA = B′PBP   …(ii) [½]

Comparing equations (i) and (ii)

B′FFP = B′P – FPFP = B′PBP   …(iii) [½]

B′P = – v, FP = – f, BP = – u ;

Using these in Eq. (iii) we get 1v + 1u = 1f = 2R [½]

Alternatively: If the result derived by any other method, full credit to be given.

(ii) For lens : 1v – 1u = 1f [½]

u = –5 m ; f = +1 m

1v – 1–5 = 1(+1)

⇒ v = 54 m = 1.25 m [½]

m = IO = vu = (+5/4)(–5) [½]

I = (– 0.25) × (1.8) [½]

I = – 0.45 m [½]

Q128.2023 Main 55/3/1 · Q33 (OR option)5 marks

(i) Draw a ray diagram showing refraction of a ray of light through a triangular glass prism. Hence, obtain the relation for the refractive index (μ) in terms of angle of prism (A) and angle of minimum deviation (δm).

(ii) The radii of curvature of the two surfaces of a concave lens are 20 cm each. Find the refractive index of the material of the lens if its power is – 5·0 D.

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Official CBSE Marking Scheme · CBSE Board 2023 (Main), Set 55/3/1

(i) Ray diagram showing refraction of a ray of light through a prism [1]

Ray PQRS refracted through prism ABC with angles i, r1, r2, e and deviation δ marked

In the quadrilateral AQNR, two of the angles (at the vertices Q and R) are right angles. Therefore, the sum of the other angles of the quadrilateral is 180°.

∠A + ∠QNR = 180°

From the triangle QNR, r1 + r2 + ∠QNR = 180°

Comparing these two equations, we get

r1 + r2 = A   …(i) [½]

The total deviation δ is the sum of deviations at the two faces,

δ = (i – r1) + (e – r2) that is, δ = i + e – A   …(ii) [½]

When δ = δm ; i = e & r1 = r2 [½]

From (i) ; 2r = A or r = A/2

From (ii) ; δm = 2i – A or i = A + δm2

μ = sin isin r = sin(A + δm2)sin A/2 [½]

(ii) Given ; P = – 5 D

f (in cm) = 100(–5) = – 20 cm [½]

Using Lens Maker’s formula ; 1f = (μ – 1) [1R1 – 1R2] [½]

1(–20) = (μ – 1) [1(–20) – 1(+20)]

1(–20) = (μ – 1) [– 110] ; μ – 1 = 12 [½]

⇒ μ = 32 = 1.5 [½]

Q129.2023 Main 55/4/1 · Q32 (option 1)5 marks

(i) Draw a ray diagram to show how the final image is formed at infinity in an astronomical refracting telescope. Obtain an expression for its magnifying power.

(ii) Two thin lenses L1 and L2, L1 being a convex lens of focal length 24 cm and L2 a concave lens of focal length 18 cm are placed coaxially at a separation of 45 cm. A 1 cm tall object is placed in front of the lens L1 at a distance of 36 cm. Find the location and height of the image formed by the combination.

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Official CBSE Marking Scheme · CBSE Board 2023 (Main), Set 55/4/1

(i)

Ray diagram of an astronomical refracting telescope with final image at infinity

[1½]

Note : Deduct ½ mark, if the direction of propagation of light is not marked.

From the diagram β = hfe [½]

and α = hfo [½]

Magnifying Power = fofe [½]

(ii)

Convex lens f = 24 cm and concave lens f = -18 cm separated by 45 cm, object 1 cm tall at 36 cm from L1

For lens L1 :

1v − 1u = 1f [½]

1v − 1−36 = 124

1v = 124 − 136 = 3 − 272 = 172

v = 72 cm [½]

For lens L2 :

1v′ − 1u′ = 1f′

1v′ − 1(72 − 45) = 1−18

1v′ = 1−18 + 127 = −3 + 254 = −154

v′ = −54 cm [½]

Final distance v1′ = −54 − (−45) = −9 cm (to the left of convex lens)

Magnification hiho = v1′u

hi1 = −9−36 ⟹ hi = +14 cm [½]

Q130.2023 Main 55/4/1 · Q32 (OR option)5 marks

(i) Explain the working principle of an optical fibre with the help of a diagram. Mention one use of a light pipe.

(ii) A ray of light is incident at an angle of 60° on one face of a prism with the prism angle A = 60°. The ray passes symmetrically through the prism. Find the angle of minimum deviation (δm) and refractive index of the material of the prism. If the prism is immersed in water, how will δm be affected ? Justify your answer.

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Official CBSE Marking Scheme · CBSE Board 2023 (Main), Set 55/4/1

(i) Working Principle : Optical fibre uses the optical principle of total internal reflection to capture the light transmitted in an optical fibre and confine the light to the core of the fibre. [1]

Light undergoing total internal reflections inside an optical fibre core (high n) surrounded by cladding (low n)

[½]

Uses : Transmission of audio and video signal / Examination of internal organs / Endoscopy [½]

(ii) δm = i + e − A

δm = 2i − A [½]

δm = 60° [½]

Refractive Index

μ = sin ( A + δm2 )sin A / 2 [½]

μ = sin 120°2sin 60°2 = sin 60°sin 30° = 3 / 21 / 2

μ = 3 [½]

If the prism is immersed in water μ decreases and consequently angle of minimum deviation decreases. Since δm depends on μ through equation given above. [1]

Q131.2023 Main 55/5/1 · Q33 (option 1)5 marks

(i) Draw a ray diagram to show the working of a compound microscope. Obtain the expression for the total magnification for the final image to be formed at the near point.

(ii) In a compound microscope an object is placed at a distance of 1.5 cm from the objective of focal length 1.25 cm. If the eye-piece has a focal length of 5 cm and the final image is formed at the near point, find the magnifying power of the microscope.

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Official CBSE Marking Scheme · CBSE Board 2023 (Main), Set 55/5/1

(i) Ray diagram of compound microscope [1½]

Ray diagram of a compound microscope with final image at the near point

(Note: Give full credit of the diagram, if a student draws diagram for normal adjustment. Deduct ½ mark for not showing the direction of propagation of light.)

Linear magnification due to the objective is tan β = hf0 = h′L

m0 = h′h = Lf0 [½]

Here L is the distance between the second focal point of the objective and the first focal point of the eyepiece.

Linear magnification due to eyepiece is me = (1 + Dfe) [½]

Thus, the total magnification is, m = m0 × me

m = Lf0(1 + Dfe) [½]

Note: Full credit of the derivation should be given if a student derives m = v0u0(1 + Dfe)

(ii) Given: u0 = –1.5 cm, f0 = 1.25 cm, fe = 5 cm, D = 25 cm

1f0 = 1v0 – 1u0

1v0 = 11.25 – 11.5 = 215 [½]

v0 = 152 cm [½]

|m0| = v0u0 = 152 × 11.5 = 5

|me| = (1 + Dfe) = 1 + 255 = 6 [½]

m = m0 × me = 5 × 6 = 30 [½]

Q132.2023 Main 55/5/1 · Q33 (OR option)5 marks

(i) Draw a ray diagram for the formation of image of an object by an astronomical telescope, in normal adjustment. Obtain the expression for its magnifying power.

(ii) The magnifying power of an astronomical telescope in normal adjustment is 2.9 and the objective and the eyepiece are separated by a distance of 150 cm. Find the focal lengths of the two lenses.

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Official CBSE Marking Scheme · CBSE Board 2023 (Main), Set 55/5/1

(i) Ray diagram for image formation by astronomical telescope in normal adjustment [1½]

Ray diagram of an astronomical telescope in normal adjustment

Magnifying power (m)

m ≈ βα [½]

β = hfe [½]

α = hf0 [½]

m = f0fe

(ii) m = 2.9, d = 150 cm (Given)

m = fofe = 2910 [½]

f0 + fe = 150 [½]

fe = 38.5 cm [½]

f0 = 150 – 38.5 = 111.5 cm [½]

Q133.2023 Supp 55/C/1 · Q31 (option 1)5 marks

(i) Draw a labelled ray diagram of an astronomical telescope to show the image formation of a distant object by it in normal adjustment. What are the main considerations required in selecting the objective and eyepiece lenses so that the telescope has large magnifying power and high resolution ?

(ii) A biconvex lens of focal length 20 cm is immersed in water, whose refractive index is 43. Find the change, if any, in the nature and the focal length of the lens. Refractive index of the material of convex lens is 32.

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Official CBSE Marking Scheme · CBSE Board 2023 (Supplementary), Set 55/C/1

(i)

Ray diagram of astronomical telescope in normal adjustment

[1]

For large magnifying power

fo > fe [½]

For high resolution

Aperture of objective should be large [½]

(ii) 1fa = (ng − 1)[1R1 − 1R2] [½]

1fa = (ngnw − 1)[1R1 − 1R2] [½]

fwfa = (ng − 1)(ngnw − 1)

fw20 = (32 − 1)(32 × 34 − 1) = 1/21/8 = 4 [½]

fw = 80 cm [½]

Changes in focal length Δf = 80 − 60 = 20 cm [½]

Nature of the lense is converging because fw is positive [½]

Q134.2023 Supp 55/C/1 · Q31 (OR option)5 marks

(i) Draw a ray diagram showing refraction of light through a prism of angle A and obtain the relation between μ, A and the angle of minimum deviation δm.

(ii) An equiconvex lens of radius of curvature R and made of glass of refractive index μ is cut into two identical plano-convex lenses. Find the focal length of the plano-convex lenses.

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Official CBSE Marking Scheme · CBSE Board 2023 (Supplementary), Set 55/C/1

(i)

Ray diagram of refraction through a prism PQRS with angles marked

[1]

From the figure

∠A + ∠QNR = 180°

From △QNR

r1 + r2 + ∠QNR = 180°

comparing the above equations

r1 + r2 = A ------- (1) [½]

δ = (i − r1) + (e − r2)

δ = i + e − A ------- (2) [½]

At the minimum deviation Dm

δ = Dm, i = e and r1 = r2 [½]

From equation (1)

2r = A

r = A2

From equation (2)

Dm = 2i − A

i = A + Dm2

The refractive index of the prism is

n21 = sin isin r = sin[(A + Dm)/2]sin[A/2] [½]

(ii) 1f = (n − 1)[1R1 − 1R2] [½]

1f1 = (n − 1)[2R] [½]

1f2 = (n − 1)[1R] [½]

f2 = 2 f1 [½]

Q135.2022 Main 55/3/1 · Q125 marks

A compound microscope consists of two converging lenses. One of them, of smaller aperture and smaller focal length is called objective and the other of slightly larger aperture and slightly larger focal length is called eye-piece. Both the lenses are fitted in a tube with an arrangement to vary the distance between them. A tiny object is placed in front of the objective at a distance slightly greater than its focal length. The objective produces the image of the object which acts as an object for the eye-piece. The eye piece, in turn produces the final magnified image.

1 × 5 = 5

  1. In a compound microscope the images formed by the objective and the eye-piece are respectively
    1. virtual, real
    2. real, virtual
    3. virtual, virtual
    4. real, real
  2. The magnification due to a compound microscope does not depend upon
    1. the aperture of the objective and the eye-piece
    2. the focal length of the objective and the eye-piece
    3. the length of the tube
    4. the colour of the light used
  3. Which of the following is not correct in the context of a compound microscope ?
    1. Both the lenses are of short focal lengths.
    2. The magnifying power increases by decreasing the focal lengths of the two lenses.
    3. The distance between the two lenses is more than (fo + fe).
    4. The microscope can be used as a telescope by interchanging the two lenses.
  4. A compound microscope consists of an objective of 10X and an eye-piece of 20X. The magnification due to the microscope would be
    1. 2
    2. 10
    3. 30
    4. 200
  5. The focal lengths of objective and eye-piece of a compound microscope are 1.2 cm and 3.0 cm respectively. The object is placed at a distance of 1.25 cm from the objective. If the final image is formed at infinity, the magnifying power of the microscope would be
    1. 100
    2. 150
    3. 200
    4. 250
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Official CBSE Marking Scheme · CBSE Board 2022 (Main), Set 55/3/1

(I) (B) real, virtual [1]

(II) (A) The aperture of the objective and the eye piece [1]

(III) (D) The microscope can be used as a telescope by interchanging the two lenses. [1]

(IV) (D) 200 [1]

(V) (C) 200 [1]

Q136.2022 Main 55/4/1 · Q125 marks

Two transparent media of refractive indices n1 and n2 are separated by a spherical transparent surface. The rays of light incident on the surface get refracted into the medium on the other side. The laws of refraction are valid at each point of the spherical surface. A lens is a transparent optical medium bounded by two surfaces, at least one of which should be spherical. The focal length of a lens is determined by the radii of curvature (R1 and R2) of its two surfaces and the refractive index (n) of the medium of the lens with respect to the surrounding medium. Depending on R1 and R2, a lens behaves as a diverging or a converging lens. The ability of a lens to diverge or converge a beam of light incident on it defines its power.

(a) An object is placed at the point B as shown in the figure. The object distance (u) and the image distance (v) are related as

Spherical refracting surface between media n1 and n2 (n2 > n1), centre of curvature C, radius R, object at B at distance u
  1. 1v − 1u = (n2 − n1n1) 1R
  2. 1v − 1u = (n1 − n2n2) 1R
  3. n2v − n1u = (n2 − n1)R
  4. n1v − n2u = (n1 − n2)R

(b) A point object is placed in air at a distance ‘R’ in front of a convex spherical refracting surface of radius of curvature R. If the medium on the other side of the surface is glass, then the image is :

  1. real and formed in glass.
  2. real and formed in air.
  3. virtual and formed in glass.
  4. virtual and formed in air.

(c) An object is kept at 2F in front of an equiconvex lens. The image formed is :

  1. real and of the size of the object.
  2. virtual and of the size of the object.
  3. real and enlarged.
  4. virtual and diminished.

(d) A thin converging lens of focal length 10 cm and a thin diverging lens of focal length 20 cm are placed coaxially in contact. The power of the combination is :

  1. − 5 D
  2. + 5 D
  3. + 15 D
  4. − 15 D

(e) An equiconcave lens of focal length ‘f’ is cut into two identical parts along the dotted line as shown in the figure. The focal length of each part will be :

Equiconcave lens with radii R on both faces, cut along a horizontal dotted line
  1. f4
  2. f2
  3. f
  4. 2f
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Official CBSE Marking Scheme · CBSE Board 2022 (Main), Set 55/4/1

(a) (iv) n1v − n2u = (n1 − n2)R [1]

(b) (iv) virtual and formed in air [1]

(c) (i) real and of the size of the object [1]

(d) (ii) + 5 D [1]

(e) (iii) f [1]

Q137.2022 Main 55/5/1 · Q125 marks

A ray of light travels from a denser to a rarer medium. After refraction, it bends away from the normal. When we keep increasing the angle of incidence, the angle of refraction also increases till the refracted ray grazes along the interface of two media. The angle of incidence for which it happens is called critical angle. If the angle of incidence is increased further the ray will not emerge and it will be reflected back in the denser medium. This phenomenon is called total internal reflection of light.

  1. A ray of light travels from a medium into water at an angle of incidence of 18°. The refractive index of the medium is more than that of water and the critical angle for the interface between the two media is 20°. Which one of the following figures best represents the correct path of the ray of light ?
    Four ray diagrams (A)-(D) at a medium-water interface with angle of incidence 18°: (A) refracted ray away from normal, (B) reflected ray, (C) refracted ray bending away, (D) ray grazing the interface
  2. A point source of light is placed at the bottom of a tank filled with water, of refractive index μ, to a depth d. The area of the surface of water through which light from the source can emerge, is :
    1. πd22(μ2 – 1)
    2. πd2(μ2 – 1)
    3. πd22μ2 – 1
    4. 2πd2(μ2 – 1)
  3. For which of the following media, with respect to air, the value of critical angle is maximum ?
    1. Crown glass
    2. Flint glass
    3. Water
    4. Diamond
  4. The critical angle for a pair of two media A and B of refractive indices 2.0 and 1.0 respectively is :
    1. 0°
    2. 30°
    3. 45°
    4. 60°
  5. The critical angle of pair of a medium and air is 30°. The speed of light in the medium is :
    1. 1 × 108 m s–1
    2. 1.5 × 108 m s–1
    3. 2.2 × 108 m s–1
    4. 2.8 × 108 m s–1
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Official CBSE Marking Scheme · CBSE Board 2022 (Main), Set 55/5/1

(i) — (a) [1]

(ii) — (b) [1]

(iii) — (c) [1]

(iv) — (b) [1]

(v) — (b) [1]

Q138.2022 Supp 55/6/1 · Q125 marks

A lens is a portion of a transparent medium bounded by two surfaces and one of these surfaces is essentially spherical. It is used to converge or diverge the light incident on it. Power of a lens is the measure of its ability to converge or diverge the light incident on it. Power of a lens depends on the refractive index of the material of lens with respect to the surrounding medium and the radii of curvature of its two surfaces.

(a) The power of a convex lens

  1. increases when the lens is dipped in water.
  2. increases when the wavelength of incident light decreases.
  3. decreases when another convex lens is placed in contact with it.
  4. increases when the lens is cut into two identical plano-convex lenses.

(b) The focal length of a concave lens is 40 cm. The power of the lens is :

  1. 0·025 D
  2. 2·5 D
  3. – 0·025 D
  4. – 2·5 D

(c) The focal length of a concave lens (μg = 1·5) in air is 20 cm. What should be the refractive index of the surrounding medium so that the lens behaves as a converging lens of focal length 60 cm ?

  1. 1·4
  2. 1·8
  3. 1·7
  4. 1·2

(d) Beams of red light, blue light, yellow light and violet light are incident on a convex lens, one-by-one. Which one of them converges nearest to the lens ?

  1. Blue light
  2. Violet light
  3. Red light
  4. Yellow light

(e) A beam of light coming parallel to the principal axis of a convex lens L1 of focal length 15 cm is incident on it. Another convex lens L2 of focal length 25 cm is placed coaxially at a distance d from L1. For the final image to be formed at infinity, the value of d should be :

  1. 10 cm
  2. 15 cm
  3. 25 cm
  4. 40 cm

5×1=5

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Official CBSE Marking Scheme · CBSE Board 2022 (Supplementary), Set 55/6/1

(a) (ii) [1]

(b) (iv) [1]

(c) (ii) [1]

(d) (ii) [1]

(e) (iv) [1]

Q139.SQP 2025-26 · Q32 (option 1)5 marks

(A) Derive lens maker’s formula. [3]

(B) Equi-convex lenses are to be manufactured from a glass of refractive index 1.55, with both faces of the same radius of curvature. What is the radius of curvature required if the focal length is to be 10cm? [2]

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Official CBSE Marking Scheme · CBSE Sample Question Paper 2025-26

(A) Lens Maker’s Formula:

For refraction at LP1N,

μ1CO + μ2CI1 = μ2 – μ1CC1 [1]

(as if the image is formed in the denser medium)

For refraction at LP2N

μ2–CI1 + μ1CI = μ2 – μ1CC2 [1]

(as if the object is in the denser medium and the image is formed in the rarer medium)

Combining the refractions at both the surfaces.

μ1CO + μ2CI = (μ2 – μ1) (1CC1 + 1CC2)

Substituting the values with sign convections,

1–u + 1v = μ2 – μ1μ1 (1R1 – 1R2)

Ray diagram for a thin lens showing object O, image I, radii R1 and R2, and distances u and v

Since μ2μ1 = μ

1–u + 1v = μ2 – μ1μ1 (1R1 – 1R2)

(or)

1–u + 1v = (μ – 1) (1R1 – 1R2)

When the object is kept at infinity, the image is formed at the principal focus.

i.e. u = –∞, v = +f.

1f = (μ – 1) (1R1 – 1R2)

This equation is called ‘Lens Maker’s Formula’. [1]

(B) Refractive index of glass, μ = 1.55

Focal length of the convexo-concave lens, f = 10 cm

Radius of curvature of one face of the first Convex surface = R1

Radius of curvature of the other face of the second convex surface = –R1

Therefore, R1 = R and R2 = –R

The value of R can be calculated from Lens – Maker formula:

(1/f) = (μ – 1) [(1/ R1) – (1/ R2)]

(1/10) = (1.55 – 1) [(1/R) + (1/R)]

(1/10) = 0.55 × (2/R) [1]

Therefore R = (0.55 × 2 × 10)

= 11cm

Hence, the radius of curvature of the convexo-concave is 11cm [1]

Q140.SQP 2025-26 · Q32 (OR option)5 marks

(A) Define angle of deviation in a prism? [1]

(B) Obtain the relation A+δ=i+e for a prism where A is the angle of prism, δ is the angle of deviation, i is the angle of incidence and e is the angle of emergence. Write this relation for the minimum deviation? [3]

(C) Write the condition for minimum deviation. [1]

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Official CBSE Marking Scheme · CBSE Sample Question Paper 2025-26

(A) The angle of deviation represents the angle by which a light ray is deviated after passing through a prism. [1]

(B) Refraction of light through prism :

Ray refracted through a prism ABC with angles i, r1, r2, e, angle of prism A and deviation δ

In quadrilateral APOQ,

A + O = 180° …….(1)

In triangle OPQ,

r1 + r2 + O = 180° …….(2) [1]

In triangle DPQ

δ = (i – r1) + (e – r2)

δ = (i + e) – (r1 + r2) …….(3) [1]

From (1) and (2),

A = r1 + r2

From (3),

δ = (i + e) – (A)

i + e = A + δ

Sum of angle of incidence and angle of emergence is equal to the sum of angle of prism and angle of deviation. [1]

(C) When angle of incidence increases, the angle of deviation decreases. At a particular value of angle of incidence the angle of deviation becomes minimum and is called ‘angle of minimum deviation’.

At δm,

• i = e and r1 = r2 = r (say)

• At minimum deviation, refracted ray become parallel to incident ray.

(Award full marks if either of condition is mentioned) [1]

Q141.SQP 2026-27 · Q32 (option 1)5 marks

A compound microscope consists of an objective lens of focal length 2 cm and an eyepiece lens of focal length 5 cm. The object is placed 2.2 cm from the objective lens.

  1. Calculate the position of the image formed by the objective lens.
  2. The eyepiece is so placed that the final image is formed at infinity. Find the magnifying power of the microscope.
  3. Explain briefly why the compound microscope provides much higher magnification compared to a simple magnifier.
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Official CBSE Marking Scheme · CBSE Sample Question Paper 2026-27

I. Position of the image formed by the objective lens. Using the lens formula:

1fo = 1vo − 1uo

12 = 1vo − (1−2.2)

1vo = 12 − 12.2

1vo = 2.2 − 24.4 = 0.24.4

vo = 22 cm

Image formed by the objective lens is at 22 cm from it. [2]

II. Magnifying power of the microscope (final image at infinity)

Magnifying power of a compound microscope: M = (vouo)(Dfe)

where D = 25 cm (least distance of distinct vision)

M = (222.2)(255)

M = 10 × 5 = 50

Magnifying power of the microscope = 50 [2]

III. A compound microscope provides much higher magnification than a simple magnifier because a compound microscope uses two lenses:

  • The objective lens produces a highly magnified real image of a very small object.
  • The eyepiece lens further magnifies this image.

Thus, the total magnification is the product of the magnifications of the objective and the eyepiece, making it much greater than that of a simple magnifier, which uses only a single lens. [1]

Q142.SQP 2026-27 · Q32 (OR option)5 marks

An astronomical telescope uses an objective lens of focal length 100 cm and an eyepiece of focal length 5 cm.

  1. Calculate the magnifying power of the telescope when the final image is formed at infinity.
  2. If the least distance of distinct vision of the observer is 25 cm, calculate the magnifying power when the final image is formed at this distance.
  3. Explain why the objective lens of an astronomical telescope is made with a large focal length and aperture, while the eyepiece has a small focal length.
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Official CBSE Marking Scheme · CBSE Sample Question Paper 2026-27

Given: Focal length of objective lens, fo = 100 cm; focal length of eyepiece lens, fe = 5 cm; least distance of distinct vision, D = 25 cm

I. Magnifying power when final image is formed at infinity. For an astronomical telescope (normal adjustment):

M = fofe = 1005 = 20

Magnifying power = 20 [2]

II. Magnifying power when final image is formed at the least distance of distinct vision

M = fofe(1 + feD)

M = 1005(1 + 525)

M = 20(1 + 0.2) = 20 × 1.2 = 24

Magnifying power = 24 [2]

III. Reason for design of objective and eyepiece in an astronomical telescope

  • The objective lens is made with a large focal length to produce a highly magnified real image of distant celestial objects and with a large aperture to collect more light, increasing brightness and resolution.
  • The eyepiece lens has a small focal length so that it can further magnify the image formed by the objective.

This combination results in high magnifying power and a brighter, clearer image of distant astronomical objects. [1]

Unit VI · Optics

10. Wave Optics

Q1.2026 Main 55/1/1 · Q41 mark

The shape of the interference fringes in Young’s double-slit experiment, when the distance between the slit and the screen is very large as compared to the slit-separation, is nearly

  1. straight
  2. parabolic
  3. circular
  4. hyperbolic
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Official CBSE Marking Scheme · CBSE Board 2026 (Main), Set 55/1/1

(A) Straight

Q2.2026 Main 55/1/1 · Q101 mark

Four independent waves are expressed as

  1. y1 = A1 sin ωt
  2. y2 = A2 sin 2ωt
  3. y3 = A3 cos ωt
  4. y4 = A4 sin (ωt + π/3)

The interference between two of these waves is possible in

  1. (i) and (iii) only
  2. (iii) and (iv) only
  3. (i), (iii) and (iv) only
  4. All of them
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Official CBSE Marking Scheme · CBSE Board 2026 (Main), Set 55/1/1

(C) (i), (iii) and (iv) only

Q3.2026 Main 55/2/1 · Q151 mark

Assertion (A) : Light added to light can produce darkness.

Reason (R) : When two coherent light waves interfere, there is darkness at position of destructive interference.

  1. Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of Assertion (A).
  2. Both Assertion (A) and Reason (R) are true, but Reason (R) is not the correct explanation of Assertion (A).
  3. Assertion (A) is true, but Reason (R) is false.
  4. Assertion (A) is false and Reason (R) is also false.
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Official CBSE Marking Scheme · CBSE Board 2026 (Main), Set 55/2/1

(A) Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of Assertion (A)

Q4.2026 Main 55/3/1 · Q51 mark

The phase difference between the two superimposing waves that give rise to a bright spot in a Young’s double-slit experiment is (n is an integer) :

  1. 2nπ
  2. 2nπ + π4
  3. 2nπ + π2
  4. 2nπ + π
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Official CBSE Marking Scheme · CBSE Board 2026 (Main), Set 55/3/1

(A) 2nπ

Q5.2026 Main 55/5/1 · Q141 mark

Assertion (A) : In Young’s double-slit experiment, the fringe width for dark and bright fringes is the same.

Reason (R) : Fringe width is given by β = λDd, where symbols have their usual meanings.

  1. Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of the Assertion (A).
  2. Both Assertion (A) and Reason (R) are true, but Reason (R) is not the correct explanation of the Assertion (A).
  3. Assertion (A) is true, but Reason (R) is false.
  4. Both Assertion (A) and Reason (R) are false.
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Official CBSE Marking Scheme · CBSE Board 2026 (Main), Set 55/5/1

(A) Both Assertion (A) and reason (R) are true, reason (R) is the correct explanation of the assertion (A).

Q6.2025 Main 55/4/1 · Q151 mark

Assertion (A) : In double slit experiment if one slit is closed, diffraction pattern due to the other slit will appear on the screen.

Reason (R) : For interference, at least two waves are required.

  1. Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of the Assertion (A).
  2. Both Assertion (A) and Reason (R) are true, but Reason (R) is not the correct explanation of the Assertion (A).
  3. Assertion (A) is true, but Reason (R) is false.
  4. Both Assertion (A) and Reason (R) are false.
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The official marking scheme for this paper is not available in the source folder.

Q7.2025 Main 55/5/1 · Q111 mark

Two coherent light waves, each having amplitude ‘a’, superpose to produce an interference pattern on a screen. The intensity of light as seen on the screen varies between :

  1. 0 and 2a2
  2. 0 and 4a2
  3. a2 and 2a2
  4. 2a2 and 4a2
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The official marking scheme for this paper is not available in the source folder.

Q8.2025 Main 55/6/1 · Q91 mark

Two coherent waves, each of intensity I0, produce interference pattern on a screen. The average intensity of light on the screen is :

  1. zero
  2. I0
  3. 2I0
  4. 4I0
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The official marking scheme for this paper is not available in the source folder.

Q9.2024 Main 55/4/1 · Q161 mark

Assertion (A) : In interference and diffraction of light, light energy reduces in one region producing a dark fringe. It increases in another region and produces a bright fringe.

Reason (R) : This happens because energy is not conserved in the phenomena of interference and diffraction.

  1. If both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of Assertion (A).
  2. If both Assertion (A) and Reason (R) are true and Reason (R) is not the correct explanation of Assertion (A).
  3. If Assertion (A) is true and Reason (R) is false.
  4. If both Assertion (A) and Reason (R) are false.
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Official CBSE Marking Scheme · CBSE Board 2024 (Main), Set 55/4/1

(C) Assertion (A) is true and Reason (R) is false.

Q10.2024 Main 55/5/1 · Q101 mark

A Young’s double-slit experimental set up is kept in a medium of refractive index (43). Which maximum in this case will coincide with the 6th maximum obtained if the medium is replaced by air ?

  1. 4th
  2. 6th
  3. 8th
  4. 10th
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Official CBSE Marking Scheme · CBSE Board 2024 (Main), Set 55/5/1

(C) 8th

Q11.2024 Main 55/5/1 · Q151 mark

Assertion (A) : In a Young’s double-slit experiment, interference pattern is not observed when two coherent sources are infinitely close to each other.

Reason (R) : The fringe width is proportional to the separation between the two sources.

  1. If both Assertion (A) and Reason (R) are true and Reason (R) is correct explanation of Assertion (A).
  2. If both Assertion (A) and Reason (R) are true and Reason (R) is not the correct explanation of Assertion (A).
  3. If Assertion (A) is true but Reason (R) is false.
  4. If both Assertion (A) and Reason (R) are false.
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Official CBSE Marking Scheme · CBSE Board 2024 (Main), Set 55/5/1

(C) Assertion (A) is true but Reason (R) is false.

Q12.2023 Main 55/1/1 · Q171 mark

Assertion (A) : The phase difference between any two points on a wavefront is zero.

Reason (R) : All points on a wavefront are at the same distance from the source and thus oscillate in the same phase.

  1. Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of the Assertion (A).
  2. Both Assertion (A) and Reason (R) are true, but Reason (R) is not the correct explanation of the Assertion (A).
  3. Assertion (A) is true, but Reason (R) is false.
  4. Assertion (A) is false and Reason (R) is also false.
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Official CBSE Marking Scheme · CBSE Board 2023 (Main), Set 55/1/1

(A) Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of Assertion (A).

Q13.2023 Main 55/2/1 · Q181 mark

Assertion (A) : In Young’s double slit experiment all fringes are of equal width.

Reason (R) : The fringe width depends upon wavelength of light (λ) used, distance of screen from plane of slits (D) and slits separation (d).

  1. Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of Assertion (A).
  2. Both Assertion (A) and Reason (R) are true and Reason (R) is NOT the correct explanation of Assertion (A).
  3. Assertion (A) is true and Reason (R) is false.
  4. Assertion (A) is false and Reason (R) is also false.
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Official CBSE Marking Scheme · CBSE Board 2023 (Main), Set 55/2/1

(A) Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of Assertion (A).

Q14.2023 Main 55/3/1 · Q81 mark

In the wave picture of light, the intensity I of light is related to the amplitude A of the wave as :

  1. I ∝ A
  2. I ∝ A
  3. I ∝ A2
  4. I ∝ 1A2
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Official CBSE Marking Scheme · CBSE Board 2023 (Main), Set 55/3/1

(C) I ∝ A2

Q15.2023 Main 55/3/1 · Q91 mark

In a single-slit diffraction experiment, the width of the slit is halved. The width of the central maximum, in the diffraction pattern, will become :

  1. half
  2. twice
  3. four times
  4. one-fourth
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Official CBSE Marking Scheme · CBSE Board 2023 (Main), Set 55/3/1

(B) twice

Q16.2023 Main 55/4/1 · Q61 mark

In a Young’s double-slit experiment, the fringe width is found to be β. If the entire apparatus is immersed in a liquid of refractive index μ, the new fringe width will be :

  1. β
  2. μβ
  3. βμ
  4. βμ2
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Official CBSE Marking Scheme · CBSE Board 2023 (Main), Set 55/4/1

(C) βμ

Q17.2023 Main 55/4/1 · Q111 mark

A plane wavefront is incident on a concave mirror of radius of curvature R. The radius of the refracted wavefront will be :

  1. 2R
  2. R
  3. R2
  4. R4
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Official CBSE Marking Scheme · CBSE Board 2023 (Main), Set 55/4/1

(C) R2

Note: (C) R/2 is for students who opted to answer in Hindi medium only. English medium students: there is a misprint in the English version of the question as the word “reflected” appears as “refracted”. Therefore full mark to be awarded to each student who opted to answer the question in English medium.

Q18.2023 Main 55/5/1 · Q61 mark

In a Young’s double-slit experiment, the screen is moved away from the plane of the slits. What will be its effect on the following ?

  1. Angular separation of the fringes.
  2. Fringe-width.
  1. Both (i) and (ii) remain constant.
  2. (i) remains constant, but (ii) decreases.
  3. (i) remains constant, but (ii) increases.
  4. Both (i) and (ii) increase.
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Official CBSE Marking Scheme · CBSE Board 2023 (Main), Set 55/5/1

(C) (i) remains constant, but (ii) increases

Q19.2023 Main 55/5/1 · Q141 mark

According to Huygens principle, the amplitude of secondary wavelets is

  1. equal in both the forward and the backward directions.
  2. maximum in the forward direction and zero in the backward direction.
  3. large in the forward direction and small in the backward direction.
  4. small in the forward direction and large in the backward direction.
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Official CBSE Marking Scheme · CBSE Board 2023 (Main), Set 55/5/1

(B) Maximum in the forward direction and zero in the backward direction.

Q20.2023 Supp 55/C/1 · Q61 mark

In Young’s double-slit experiment, the intensity on the screen is I0 at a point where path difference is λ. The intensity at the point where path difference is λ4 is :

  1. I04
  2. I02
  3. I0
  4. zero
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Official CBSE Marking Scheme · CBSE Board 2023 (Supplementary), Set 55/C/1

(B) I02

Q21.2023 Supp 55/C/1 · Q131 mark

A plane wave is incident on a concave mirror of radius of curvature R. The reflected wave is a spherical wave of radius :

  1. R4
  2. R2
  3. R
  4. 2R
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Official CBSE Marking Scheme · CBSE Board 2023 (Supplementary), Set 55/C/1

(B) R2

Q22.SQP 2025-26 · Q61 mark

The light from a monochromatic source is incident on a single slit and the resulting diffraction pattern is viewed on a screen. The graph shows the variation of the intensity with the distance on the screen.

Single-slit diffraction intensity versus distance graph

The width of slit is increased keeping the intensity of the source the same. Which of the following graphs is correct? (The original curve is shown with a dashed line.)

  1. Narrower central peak of the same height as the dashed curve
  2. Wider central peak of the same height as the dashed curve
  3. Wider and taller central peak than the dashed curve
  4. Narrower and taller central peak than the dashed curve
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Official CBSE Marking Scheme · CBSE Sample Question Paper 2025-26

(D) Graph (D)

Slit width increases hence amplitude will increase, so intensity will also increase.

Q23.SQP 2025-26 · Q151 mark

Assertion (A): Two light sources emitting waves of similar wavelengths are coherent.

Reason (R): Two light sources emitting waves having zero or constant phase difference are known as coherent sources.

  1. Both Assertion and Reason are true and Reason is the correct explanation of Assertion.
  2. Both Assertion and Reason are true but Reason is not the correct explanation of Assertion.
  3. Assertion is true but Reason is false.
  4. Both Assertion and Reason are false.
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Official CBSE Marking Scheme · CBSE Sample Question Paper 2025-26

(D) Both Assertion and Reason are false.

Q24.SQP 2026-27 · Q51 mark

Which one of the following phenomenon is not explained by Huygens’ construction of wave front?

  1. Diffraction
  2. Refraction
  3. Reflection
  4. Origin of spectra
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Official CBSE Marking Scheme · CBSE Sample Question Paper 2026-27

(D) Origin of spectra

All three (diffraction, refraction, reflection) can be explained by Huygens’ construction of wave front but origin of spectra cannot.

Q25.SQP 2026-27 · Q61 mark

If wave front from star in nearby galaxy is detected by a telescope at the top of some mountain in India then the shape of the wavefront detected will be

  1. diverging spherical
  2. converging spherical
  3. plane
  4. cylindrical
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Official CBSE Marking Scheme · CBSE Sample Question Paper 2026-27

(C) Plane

As the star will act as a distant point source, hence the wavefront detected will be plane.

Q26.SQP 2026-27 · Q151 mark

In Young’s double slit experiment, the distance between the slits is 0.25mm and the screen is placed 1.0m away. A student uses monochromatic light of wavelength 500nm.

Assertion (A): The fringe width observed on the screen will be 2.0mm.

Reason (R): The fringe width in YDSE is directly proportional to the slit separation and inversely proportional to the wavelength of light.

  1. both Assertion and Reason are true and Reason is the correct explanation of Assertion.
  2. both Assertion and Reason are true but Reason is not the correct explanation of Assertion.
  3. Assertion is true but Reason is false.
  4. both Assertion and Reason are false.
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Official CBSE Marking Scheme · CBSE Sample Question Paper 2026-27

(C) Assertion is true but Reason is false

Fringe width in YDSE is inversely proportional to the slit separation and directly proportional to the wavelength of light.

Q27.2026 Main 55/1/1 · Q18 (option 1)2 marks

A beam of light consisting of two wavelengths 400 nm and 600 nm is used to illuminate a single slit of width 1 mm. Find the least distance of the point from the central maximum where the dark fringes due to both wavelengths coincide on the screen placed 1.5 m from the slit.

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Official CBSE Marking Scheme · CBSE Board 2026 (Main), Set 55/1/1

At the point of coincidence for dark fringes:

d sin θ = nλ2 = (n + 1)λ1

n × 600 = (n + 1) × 400 [½]

n = 2 [½]

Position of dark fringe from the central maxima

y = (n + 1)λ1Dd [½]

y = 3 × 400×10–9 × 1.51×10–3

y = 1.8 mm [½]

Alternatively:

y = nλ2Dd = 2 × 600×10–9 × 1.51×10–3 = 1.8 mm

Q28.2026 Main 55/1/1 · Q18 (OR option)2 marks

In a Young’s double-slit experimental set-up with slit separation 0.6 mm a beam of light consisting of two wavelengths 440 nm and 660 nm is used to obtain interference pattern on a screen kept 1.5 m in front of the slits. Find the least distance of the point from the central maximum where the bright fringes due to both the wavelengths coincide.

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Official CBSE Marking Scheme · CBSE Board 2026 (Main), Set 55/1/1

At the point of coincidence for bright fringes

nλ2Dd = (n + 1)λ1Dd

n × 660 = (n + 1) × 440 [½]

n = 2 [½]

Position of bright fringe from the central maxima

y = nλ2Dd [½]

y = 2 × 660×10–9 × 1.50.6×10–3

y = 3.3 mm [½]

Alternatively:

y = (n + 1)λ1Dd = 3.3 mm

Q29.2026 Main 55/2/1 · Q21 (OR option)2 marks

In a Young’s double-slit experiment, a beam of light consisting of two wavelengths 500 nm and 600 nm is used. The interference fringes are observed at a screen placed 1.8 m away from the plane of slits (slit separation 0.3 mm). Calculate the least distance from the central maximum where the bright fringes due to both the wavelengths coincide.

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Official CBSE Marking Scheme · CBSE Board 2026 (Main), Set 55/2/1

n1λ1 = n2λ2 [½]

n1n2 = λ2λ1 = 600500 = 65 [½]

y6max = n1 λ1Dd [½]

= 6 × 500 × 10–9 × 1.80.3 × 10–3 = 18 mm [½]

Alternatively : nλ1 = (n + 1)λ2, we get n = 5

y5max = n2λ2Dd = 5 × 600 × 10–9 × 1.80.3 × 10–3 = 18 mm

Q30.2026 Main 55/3/1 · Q182 marks

Explain why diffraction of sound is more common in daily experience than that of light.

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Official CBSE Marking Scheme · CBSE Board 2026 (Main), Set 55/3/1

For diffraction to take place, the size of obstacle has to be comparable to the wavelength of the wave. In case of sound wave, the wavelength is almost comparable to the size of the obstacles around us, while wavelength of light is much smaller than the dimensions of most obstacles around us. Hence diffraction of sound is more common in daily experience than that of light. [2]

Q31.2026 Main 55/4/1 · Q18 (option 1)2 marks

(a) State Huygens principle. How did Huygens justify the absence of the backwave on a spherical wavefront ?

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Official CBSE Marking Scheme · CBSE Board 2026 (Main), Set 55/4/1

Stating Huygens’ Principle [1]

According to Huygens’ principle, each point of the wavefront is the source of a secondary disturbance and the wavelets emanating from these points spread out in all directions with the speed of the wave. These wavelets emanating from the wavefront are usually referred to as secondary wavelets and if we draw a common tangent to all these spheres, we obtain the new position of the wavefront at a later time.

Justifying the absence of the back wave [1]

The amplitude of the secondary wavelets is maximum in the forward direction and zero in the backward direction.

Alternatively :

Amplitude (A) ∝ (1 + cos θ); for back wave θ = 180° ⇒ Amplitude = 0

Q32.2026 Main 55/4/1 · Q18 (OR option)2 marks

(b) In a single-slit diffraction experiment, light of wavelength λ illuminates the slit of width ‘a’. The diffraction pattern is observed on a screen kept at a distance D from the slits.

  1. Depict variation of intensity in the fringe pattern with the angular position of the fringes.
  2. How is the linear width of central maximum affected when separation between the slit and the screen is decreased ?
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Official CBSE Marking Scheme · CBSE Board 2026 (Main), Set 55/4/1

(i) Depicting the variation of intensity [1]

Single-slit diffraction: intensity versus angle, central maximum with minima at ±λ/a, ±2λ/a, ±3λ/a

(ii) The linear width of the central maximum will decrease. [1]

Note : Award ½ mark if the student gives only the formula.

Q33.2026 Supp 55/7/1 · Q18 (option 1)2 marks

A monochromatic beam of light is travelling from a rarer to a denser medium. Giving reasons, explain the following :

  1. Is the frequency of the reflected and the refracted light the same as the frequency of incident light ?
  2. Does the decrease in speed in denser medium imply a reduction in the energy carried by the light wave ?
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Official CBSE Marking Scheme · CBSE Board 2026 (Supplementary), Set 55/7/1

(i) Yes [½]

The atoms of the interface can be regarded as oscillators. The light incident on the interface forces the atomic oscillators to oscillate with the frequency of the incident light. Thus, the frequency of the reflected and the refracted light equals the frequency of incident light. [½]

(ii) No [½]

Energy carried by the light wave depends on the amplitude of the wave, not on the speed of wave propagation. [½]

Q34.2025 Main 55/1/1 · Q18 (option 1)2 marks

In a diffraction experiment, the slit is illuminated by light of wavelength 600 nm. The first minimum of the pattern falls at θ = 30°. Calculate the width of the slit.

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Q35.2025 Main 55/1/1 · Q18 (OR option)2 marks

In a Young’s double-slit experiment, two light waves, each of intensity Io, interfere at a point, having a path difference λ8 on the screen. Find the intensity at this point.

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Q36.2025 Main 55/2/1 · Q182 marks

In a double-slit experiment, 6th dark fringe is observed at a certain point of the screen. A transparent sheet of thickness t and refractive index n is now introduced in the path of one of the two interfering waves to increase its phase by 2π (n – 1) t/λ. The pattern is shifted and 8th bright fringe is observed at the same point. Find the relation for thickness t in terms of n and λ.

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Q37.2025 Main 55/4/1 · Q182 marks

Find the angle of diffraction (in degrees) for first secondary maximum of the pattern due to diffraction at a single slit. The width of the slit and wavelength of light used are 0·55 mm and 550 nm, respectively.

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Q38.2025 Main 55/5/1 · Q18 (OR option)2 marks

In Young’s double slit experimental set-up, the intensity of the central maximum is Io. Calculate the intensity at a point where the path difference between two interfering waves is λ/3.

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Q39.2025 Main 55/6/1 · Q18 (option 1)2 marks

Find the intensity at a point on the screen in Young’s double slit experiment, at which the interfering waves of intensity I0 each, have a path difference of (i) λ3, and (ii) λ2.

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Q40.2025 Main 55/7/1 · Q192 marks

Write two differences in the patterns of double-slit interference experiment and single-slit diffraction experiment.

Light waves from two pinholes illuminated by two sodium lamps do not produce interference patterns. Explain why.

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Q41.2025 Supp 55/S/1 · Q212 marks

Light of wavelength 640 nm incident on the slits in a Young’s double-slit experiment produces an interference pattern of fringe width 7·2 mm. Calculate the wavelength of light which will produce a pattern of fringe width 8·1 mm.

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Official CBSE Marking Scheme · CBSE Board 2025 (Supplementary), Set 55/S/1

β = Dλd [½]

β1β2 = λ1λ2 [½]

λ2 = β1β2 × λ1

λ2 = 8.17.2 × 640 [½]

λ2 = 720 nm [½]

Q42.2024 Main 55/2/1 · Q18 (option 1)2 marks

Two waves, each of amplitude ‘a’ and frequency ‘ω’ emanating from two coherent sources of light superpose at a point. If the phase difference between the two waves is φ, obtain an expression for the resultant intensity at that point.

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Official CBSE Marking Scheme · CBSE Board 2024 (Main), Set 55/2/1

x1 = a cos ωt
x2 = a cos(ωt + φ)
x = x1 + x2 [½]

= a (cos ωt + cos(ωt + φ))
= a (2 cos(ωt + φ2) cos φ2)
= 2a cos φ2 cos(ωt + φ2) [½]

Intensity I = K (amplitude)2, where K is a constant. [½]

= K (2a cos φ2)2
= 4I0 cos2 φ2 [½]

I0 = Ka2 = intensity of each incident wave.

(Note : Award full credit of this part for all other alternative correct methods)

Q43.2024 Main 55/2/1 · Q18 (OR option)2 marks

What is the effect on the interference pattern in Young’s double-slit experiment when (i) the source slit is moved closer to the plane of the slits, and (ii) the separation between the two slits is increased ? Justify your answers.

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Official CBSE Marking Scheme · CBSE Board 2024 (Main), Set 55/2/1

(i) Sharpness of interference pattern decreases [½]

sS < λd

As S decreases, interference patterns produced by different parts of the source overlap and finally fringes disappear. [½]

Alternatively

As the source slit is brought closer to the plane of the slits, the screen gets illuminated uniformly and fringes disappear.

Alternatively

Interference pattern is not formed.

(Note : Award full credit of this part if a student merely attempts this part.)

(ii) β = λDd [½]

As d increases, β decreases and fringes disappear. [½]

Q44.2024 Supp 55/S/1 · Q202 marks

State Huygens principle. Using it draw a diagram showing the details of passage of a plane wave from a denser into a rarer medium.

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Official CBSE Marking Scheme · CBSE Board 2024 (Supplementary), Set 55/S/1

Huygens Principle :

Each point of the wavefront is the source of a secondary disturbance and the wavelets emanating from these points spread out in all directions with the speed of the wave. These wavelets emanating from the wavefront are usually referred to as secondary wavelets and if we draw a common tangent to all these spheres, we obtain the new position of the wavefront at a later time. [1]

Diagram [1]

Refraction of a plane wavefront from medium 1 into medium 2 using Huygens construction (v2 > v1)
Q45.2023 Main 55/1/1 · Q21 (option 1)2 marks

Using Huygens’ principle, draw a ray diagram showing the propagation of a plane wave refracting at a plane surface separating two media. Also verify Snell’s law of refraction.

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Official CBSE Marking Scheme · CBSE Board 2023 (Main), Set 55/1/1

Ray diagram [1], Proof of Snell’s law of refraction [1]

Huygens construction of a plane wavefront AB refracting at a plane surface into medium 2, refracted wavefront CE

[1]

AB is incident wave front, incident at an angle i. Let τ be time taken by the wave front to travel distance BC.

BC = v1τ where v1 is speed of wave in medium 1.

To determine shape of refracted wave front, we draw a sphere of radius v2τ, where v2 is speed of wave in medium 2.

CE represents a tangent drawn from point C on sphere, CE is the refracted wave front.

sin i = BCAC = v1τAC; sin r = AEAC = v2τAC [½]

sin isin r = v1v2 = n21 [½]

Note: Give full credit if student derives Snell’s law by taking incident wavefront in denser medium.

Huygens construction with incident wavefront in the denser medium
Q46.2023 Main 55/1/1 · Q222 marks

Two coherent monochromatic light beams of intensities I and 4I superpose each other. Find the ratio of maximum and minimum intensities in the resulting beam.

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Official CBSE Marking Scheme · CBSE Board 2023 (Main), Set 55/1/1

Finding the ratio of maximum and minimum intensities [2]

ImaxImin = (I1 + I2)2(I1 − I2)2 [½]

= I1 + I2 + 2I1I2I1 + I2 − 2I1I2 [½]

= 5I + 4I5I − 4I [½]

= 91 [½]

Alternatively

I1I2 = a2b2 = 4II = 41 [½]

ab = 21 [½]

ImaxImin = (a + b)2(a − b)2 [½]

ImaxImin = (2 + 1)2(2 − 1)2 = 91 [½]

Q47.2023 Main 55/2/1 · Q222 marks

What happens to the interference pattern when two coherent sources are

  1. infinitely close, and
  2. far apart from each other
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Official CBSE Marking Scheme · CBSE Board 2023 (Main), Set 55/2/1

(a) When ‘d’ is very small, β ∝ 1d, β will be very large and a single patch will occupy the whole field of view hence pattern cannot be observed. [1]

Alternatively

Give full credit if a candidate writes that the fringe width will increase or the fringes will not be observed.

(b) When sources are far apart, i.e. d is very large, then fringe width will be so small that the fringes are not resolved and cannot be seen separately. [1]

Alternatively

Give full credit if a candidate writes that the fringe width will decrease or the fringes may not be observed.

Q48.2023 Main 55/3/1 · Q21 (option 1)2 marks

State Huygens’ principle. How did Huygens explain the absence of the backwave ?

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Official CBSE Marking Scheme · CBSE Board 2023 (Main), Set 55/3/1

Statement: Each point of the wavefront is the source of secondary disturbance in all directions. [½]

Common tangent to all the secondary wavelets gives new position of the wavefront. [½]

Explanation: Light energy cannot travel in backward direction. [1]

Alternatively: It was an adhoc assumption.

Alternatively: For back wave: I = ½ (1 + cos θ); at θ = 180°, contribution is zero.

Alternatively: Amplitude of secondary wavelets is maximum in forward direction and zero in backward direction.

Note: If any other relevant explanation given, give full credit.

Q49.2023 Main 55/3/1 · Q21 (OR option)2 marks

Use Huygens’ principle to show reflection/refraction of a plane wave by (i) concave mirror, and (ii) a convex lens.

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Official CBSE Marking Scheme · CBSE Board 2023 (Main), Set 55/3/1

(i) Diagram for concave mirror [1]

Plane wavefront reflected by a concave mirror becomes a spherical wavefront of radius R/2

(ii) Diagram for convex lens [1]

Plane wavefront refracted by a convex lens becomes a spherical wavefront of radius f
Q50.2025 Main 55/4/1 · Q263 marks

A double slit set-up was initially placed in a tank filled with water and the interference pattern was obtained using a laser light. When water is replaced by a transparent liquid of refractive index n > nwater, what will be the effect on the following ?

  1. Speed, frequency and wavelength of the light of laser beam.
  2. The fringe width, shape of interference fringes and shift in the position of central maximum.
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Q51.2025 Main 55/5/1 · Q243 marks

In Young’s double slit experiment, the separation between the two slits is 1·0 mm and the screen is 1·0 m away from the slits. A beam of light consisting of two wavelengths 500 nm and 600 nm is used to obtain interference fringes. Calculate :

(a) the distance between the first maxima for the two wavelengths.

(b) the least distance from the central maximum, where the bright fringes due to both the wavelengths coincide.

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Q52.2024 Supp 55/S/1 · Q283 marks

(a) A ray of light is incident on a surface separating air from a denser medium A of refractive index μ1. It is then made incident on the parallel surface of another medium B of refractive index μ2 at the same angle of incidence. If the angle of refraction in the two media are 30° and 35° respectively, then in which one of the two media (A or B) will light travel faster and why ?

(b) The intensity of the two interfering waves in Young’s double slit experiment is I0 each. Find the intensity at a point on the screen where path difference between the interfering waves is (i) λ2, and (ii) λ3.

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Official CBSE Marking Scheme · CBSE Board 2024 (Supplementary), Set 55/S/1

(a) Light travels faster in medium ‘B’ [½]

μ1 = sin isin r1,   μ2 = sin isin r2

∴ μ1μ2 = sin r2sin r1 = sin 35°sin 30° [½]

⇒ μ1μ2 = v2v1 > 1

∴ v2 > v1 [½]

(b) I = 4I0 cos2φ2 [½]

(i) φ = 2πλ × λ2 = π

I = 4I0(cos π2)2 = 0 [½]

(ii) φ = 2πλ × λ3 = 2π3 [½]

I = 4I0(cos π3)2

I = I0 [½]

Q53.2023 Main 55/4/1 · Q29 (option 1)3 marks

(i) In diffraction due to a single slit, the phase difference between light waves reaching a point on the screen is 5π. Explain whether a bright or a dark fringe will be formed at the point.

(ii) What should the width (a) of each slit be to obtain eight maxima of two double-slit patterns (slit separation d) within the central maximum of the single slit pattern ?

(iii) Draw the plot of intensity distribution in a diffraction pattern due to a single slit.

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Official CBSE Marking Scheme · CBSE Board 2023 (Main), Set 55/4/1

(i) For bright fringe, ϕ = (2n + 1)π = 5π for n = 2 [1]

Alternatively :

Δϕ = 2πλ × Δx

Δx = 52 λ

(ii) We want, aθ = λ, θ = λ/a

8 λd = 2 λa ⟹ a = d4 [1]

(iii) Intensity distribution [1]

Intensity versus phase difference plot for single slit diffraction pattern
Q54.2023 Main 55/4/1 · Q29 (OR option)3 marks

(i) In a Young’s double-slit experiment SS2 − SS1 = λ4, where S1 and S2 are the two slits as shown in the figure. Find the path difference (S2P − S1P) for constructive and destructive interference at P.

Source slit S, double slits S1 and S2 and screen with point P

(ii) What is the effect on the interference fringes in a Young’s double-slit experiment, if the monochromatic source S is replaced by a source of white light ?

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Official CBSE Marking Scheme · CBSE Board 2023 (Main), Set 55/4/1

(i) Net path difference between two waves reaching on the screen through two slits is given as

SS2P − SS1P = (SS2 − SS1) + (S2P − S1P)

Δx = λ4 + (S2P − S1P)

For constructive interference Δx = nλ

∴ (S2P − S1P) = nλ − λ4 = λ (n − 14) [1]

For destructive interference Δx = (2n − 1) λ2

Δx = λ4 + (S2P − S1P)

S2P − S1P = (2n − 1) λ2 − λ4 = λ4 [4n − 3] [1]

(ii) If monochromatic source is replaced by a source of white light the centre bright fringe is white. [1]

Q55.2023 Main 55/5/1 · Q26 (option 1)3 marks

A plane wave-front propagating in a medium of refractive index ‘μ1’ is incident on a plane surface making an angle of incidence (i). It enters into a medium of refractive index μ2 (μ2 > μ1).

Use Huygen’s construction of secondary wavelets to trace the retracted wave-front. Hence verify Snell’s law of refraction.

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Official CBSE Marking Scheme · CBSE Board 2023 (Main), Set 55/5/1

Tracing the refracted wavefront [1]

Huygens construction of refraction of a plane wavefront at a plane surface

AB is incident wavefront, incident at an angle i. Let τ be the time taken by wavefront to travel distance BC.

BC = v1τ where v1 is speed of wave in medium 1 [½]

To determine shape of refracted wavefront, we draw a sphere of radius v2τ, where v2 is speed of wave in medium 2. CE represents a tangent drawn from point C on sphere, CE is the refracted wavefront. [½]

sin i = BCAC = v1τAC and sin r = AEAC = v2τAC

Where i and r are the angles of incidence and refraction, respectively. [½]

sin isin r = v1v2 = μ2μ1

μ1 sin i = μ2 sin r

This is the Snell’s law of refraction. [½]

Q56.2023 Main 55/5/1 · Q26 (OR option)3 marks

Using Huygen’s construction, show how a plane wave is reflected from a surface. Hence verify the law of reflection.

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Official CBSE Marking Scheme · CBSE Board 2023 (Main), Set 55/5/1

Diagram showing the reflection of plane wave [1]

Huygens construction of reflection of a plane wavefront from surface MN

Consider a plane wave AB incident at an angle i on a reflecting surface MN. If v is the speed of the wave in the medium and if τ represents the time taken by the wavefront to advance from the point B to C then the distance BC = vτ [½]

In order to construct the reflected wavefront, draw a sphere of radius vτ from the point A. Let CE represent the tangent plane, drawn from the point C to this sphere. [½]

In ΔAEC and ΔABC

AE = BC = vτ

∠CEA = ∠ABC (90° each) [½]

AC is common side

ΔAEC ≅ ΔABC

∴ ∠i = ∠r [½]

This is the law of reflection.

Q57.2023 Supp 55/C/1 · Q29 (option 1)3 marks

(i) In which case is diffraction effect more dominant — slit formed by 2 blades or slit formed by two fingers ?

(ii) Yellow light (λ = 6000 Å) illuminates a single slit of width 1 × 10−4 m. Calculate (i) the distance between two dark lines on either side of central maximum, in the diffraction pattern observed on a screen kept 1·5 m away from the slit, and (ii) the angular spread of the first minimum.

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Official CBSE Marking Scheme · CBSE Board 2023 (Supplementary), Set 55/C/1

(i) slit formed by two blades [1]

(ii) (i) separation between two dark lines

= 2λDd [½]

= 2×6000×10−10×1.51×10−4 = 18 × 10−3 m [½]

(ii) φ = λa [½]

= 6000×10−101×10−4 = 6 × 10−3 rad [½]

Q58.2023 Supp 55/C/1 · Q29 (OR option)3 marks

(i) What will be the colour of the central bright fringe in Young’s double slit experiment if the monochromatic source is replaced by a source of white light ? Give reason for your answer.

(ii) In Young’s double slit experiment, the slits are separated by 0·3 mm and the screen is placed 1·5 m away from the slits. The distance between the central bright fringe and the sixth bright fringe is found to be 1·8 cm. Find the wavelength of light used in the experiment.

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Official CBSE Marking Scheme · CBSE Board 2023 (Supplementary), Set 55/C/1

(i) White. [½]

The interference pattern due to different component colors of white light overlap (incoherently). The central bright fringes for different colors are at the same position. Therefore the central fringe is white. [½]

(ii) λn = nλDd [½]

λ = λndnD [1]

λ = 1.8×10−2 × 0.3×10−36 × 1.5

λ = 6000 Å [½]

Q59.2022 Main 55/1/1 · Q8 (option 1)3 marks

A parallel beam of light of wavelength 600 nm is incident normally on a slit of width 0.2 mm. If the resulting diffraction pattern is observed on a screen 1 m away, find the distance of

  1. first minimum, and
  2. second maximum, from the central maximum.
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Official CBSE Marking Scheme · CBSE Board 2022 (Main), Set 55/1/1

(i) y = λDa [½]

= 600×10−9×10.2×10−3 [½]

= 3×10−3 m = 3 mm [½]

(ii) y = (n + ½) λDa [½]

y = (2 + ½) λDa [½]

y = 52 λDa = 52 × 600×10−9×10.2×10−3

= 7.5×10−3 m = 7·5 mm [½]

Q60.2022 Main 55/2/1 · Q6 (OR option)3 marks

A plane wavefront of light of wavelength ‘λ’ is incident normally on a narrow slit of width ‘a’ and a diffraction pattern is observed on a screen at a distance ‘D’ from the slit.

  1. Depict the intensity distribution in the pattern observed.
  2. Obtain the expression for the first maximum from the central maximum.
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Official CBSE Marking Scheme · CBSE Board 2022 (Main), Set 55/2/1

(i) Intensity distribution curve [1½]

Intensity distribution in single slit diffraction pattern

(ii) For maximum

a sin θ = (n + ½)λ [½]

For first maximum ; n = 1

For small θ; aθ = 3λ2 ⇒ θ = 3λ2a [½]

∴ xD = 32 λa

∴ x = 32 λDa [½]

Q61.2022 Main 55/3/1 · Q73 marks

In a diffraction pattern due to a single slit, how will the angular width of central maximum change, if

  1. Orange light is used in place of green light,
  2. the screen is moved closer to the slit,
  3. the slit width is decreased ?

Justify your answer in each case.

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Official CBSE Marking Scheme · CBSE Board 2022 (Main), Set 55/3/1

In diffraction pattern the angular width of central maximum = 2λa where a is the slit width and λ is the wavelength. [½]

(i) Increases [½]

As θ = 2λa and λorange > λgreen [½]

(ii) No change / no effect [½]

As θ does not depend upon the distance of the screen from the slit (D) [½]

(iii) Increases [½]

As θ is inversely proportional to the slit width (a). [½]

(Note :- Give ½ mark, if only the formula θ = 2λa is given.)

Q62.2022 Main 55/3/1 · Q103 marks

In a Young’s double slit experiment using light of wavelength 600 nm, the slit separation is 0.8 mm and the screen is kept 1.6 m from the plane of the slits. Calculate :

  1. the fringe width
  2. the distance of (a) third minimum and (b) fifth maximum, from the central maximum.
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Official CBSE Marking Scheme · CBSE Board 2022 (Main), Set 55/3/1

(i) β = λDd = 600 × 10-9 × 1.28 × 10-4 = 1.2 mm [½ + ½]

(ii) (a) x3 = 52 λDd = 52 × 1.2 mm = 3 mm [½ + ½]

(b) x5 = 5λDd = 5 × 1.2 mm = 6 mm [½ + ½]

Q63.2022 Main 55/3/1 · Q11 (OR option)3 marks
  1. State two conditions for two light sources to be coherent.
  2. Give two points of difference between an interference pattern due to a double – slit and a diffraction pattern due to a single slit.
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Official CBSE Marking Scheme · CBSE Board 2022 (Main), Set 55/3/1

(i) The two sources should emit the waves in the same phase or with a constant phase difference. [½]

The two sources must continuously emit light wave of the same wavelength. [½]

(ii) [1 + 1]

InterferenceDiffraction
1) Intensity of all bright bands are nearly same1) Intensity of bright bands decreases rapidly on both sides of central maximum.
2) All fringes are of same width.2) All the fringes are not of same width / the width of central maximum is double that of the width of secondary maxima.

Or any other two differences.

Q64.2022 Main 55/4/1 · Q73 marks

Briefly explain how bright and dark fringes are formed on the screen in Young’s double slit experiment. Hence, derive the expression for the fringe width.

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Official CBSE Marking Scheme · CBSE Board 2022 (Main), Set 55/4/1

We get bright fringes, when waves from two coherent sources meet at a point on the screen with a phase difference

Δφ = 2nπ (n = 1,2,3…….) [½]

We get dark fringes, when waves from two coherent sources meet at a point on the screen with a phase difference

Δφ = (2n + 1)π (n = 1,2,3…….) [½]

(Note: Give full credit for explaining using path difference)

YDSE geometry: slits S1 and S2 separated by d, screen GG' at distance D, point P at height x from O

For maxima, S2P − S1P = nλ n = 0,1,2,3........ (1) [½]

From figure, (S2P)2 − (S1P)2 = [D2 + (x + d2)2] − [D2 + (x − d2)2] = 2xd

S2P − S1P = 2xdS2P + S1P

S2P ≈ S1P ≈ D

S2P − S1P = 2xd2D = xdD (2) [½]

From equation (1) & (2): nλ = xdD

xn = nλDd for nth maxima [½]

Similarly for (n+1)th maxima: xn+1 = (n+1)λDd

∴ fringe width = xn+1 − xn = λDd [½]

Note- Give full credit, if a student derives this relation using condition of minimum intensity.

Q65.2022 Main 55/4/1 · Q11 (OR option)3 marks

A plane wavefront is propagating from a rarer into a denser medium. Use Huygens principle to show the refracted wavefront and verify Snell’s law.

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Official CBSE Marking Scheme · CBSE Board 2022 (Main), Set 55/4/1

Showing refracted wavefront [1]

Huygens construction of refraction of a plane wavefront at a plane surface, showing incident wavefront AB and refracted wavefront CE

We consider, refracted wavefront CE and triangles ABC & AEC. From the triangles we obtained

sin i = BCAC = v1τAC [½]

sin r = AEAC = v2τAC [½]

Thus sin isin r = v2v1 [½]

We know n = cv, So sin isin r = n2n1 [½]

which is the Snell’s law.

Q66.2022 Main 55/5/1 · Q73 marks

How can you differentiate whether a pattern is produced by a single slit or double slits ? Derive the expression for the angular position of (i) bright and (ii) dark fringes produced in a single slit diffraction.

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Official CBSE Marking Scheme · CBSE Board 2022 (Main), Set 55/5/1

Difference in the pattern of fringes due to single slit and double slits [1]; Derivation of angular position of (i) bright fringe and (ii) dark fringe [1 + 1]

In the pattern of fringes produced by a single slit, the central fringe (band) is brighter as compared to other fringes i.e intensity goes on decreasing as the order (n) of the maxima increases, while in the fringe pattern produced by double slits all bright fringes including central fringe are of same intensity. [1]

Alternatively: In the fringe pattern produced by single slit, the fringe at the centre is wider as compared to the width of other bright fringes, while in the fringe pattern produced by double slits all bright fringes are of equal width.

(Note : Give full credit, for the differentiation by drawing intensity distribution curves of the patterns produced by single slit and double slits)

Calculation of angular position

For the slit of width ‘a’ and angle of diffraction ‘θ’

Path difference (Δp) = a sin θ [½]

(i) Condition for Bright Fringe, Δp = (2n+1)λ2

∴ a sin θ = (2n+1)λ2 [½]

sin θ = (2n+1)λ2a (n = 1, 2 ….)

for small angle, sin θ ≈ θ

θ = (2n+1)λ2a [½]

(ii) Condition for dark fringe, Δp = nλ

a sin θ = nλ (n = 1, 2…) [½]

∴ sin θ = nλa

for small angle, sin θ ≈ θ

θ = nλa [½]

Q67.2022 Main 55/5/1 · Q8 (OR option)3 marks

A slit of width 0.6 mm is illuminated by a beam of light consisting of two wavelengths 600 nm and 480 nm. The diffraction pattern is observed on a screen 1.0 m from the slit. Find :

  1. The distance of the second bright fringe from the central maximum pertaining to light of 600 nm.
  2. The least distance from the central maximum at which bright fringes due to both the wavelengths coincide.
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Official CBSE Marking Scheme · CBSE Board 2022 (Main), Set 55/5/1

Calculation of (i) distance of second bright fringe [1½]; (ii) least distance [1½]

(i) sin θ ≃ θ = xD = 5λ2a [½]

∴ x = 5λD2a = 5 × 600 × 10–9 × 12 × 0.6 × 10–3 [½]

= 2.5 mm [½]

(ii) (2n+1)λ2 = (2(n+1)+1)λ′2 [½]

(2n+1) × 600 = (2n+3) × 480 [½]

∴ n = 3.5

xn = (2n+1)λD2d = (2 × 3.5 + 1) × 600 × 10–9 × 12 × 0.6 × 10–3

= 4 mm [½]

(Note: Give full credit for finding the position by taking n = 3 or n = 4; for n = 3, xn = 3.5 mm and for n = 4, xn = 4.5 mm)

Q68.2022 Supp 55/6/1 · Q53 marks
  1. In a Young’s double slit experimental set-up, the intensity of the light waves from two coherent sources are in the ratio of 9 : 1. Find the ratio of intensity of bright and dark fringes in the interference pattern. [2]
  2. Monochromatic light of wavelength 600 nm is incident from air on a water surface. If the refractive index of water is 43, then find the wavelength of the refracted light. [1]
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Official CBSE Marking Scheme · CBSE Board 2022 (Supplementary), Set 55/6/1

(a) r = A1A2 = I1I2 = 91 = 31 [1]

ImaxImin = (r + 1)2(r – 1)2 = (42)2 = 41 [1]

(b) Wavelength of refracted light

λ = λairn = 3 × 6004 = 18004 [½]

= 450 nm [½]

Q69.2022 Supp 55/6/1 · Q6 (option 1)3 marks

In single slit diffraction, explain why the maxima at θ = (n + 12)(λa) becomes weaker and weaker as n increases. State two important differences between interference and diffraction pattern. [3]

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Official CBSE Marking Scheme · CBSE Board 2022 (Supplementary), Set 55/6/1

Maximum intensity is obtained at a point on the screen when the path difference is (2n+1)λ2. With increasing ‘n’ only one-third, one-fifth, one-seventh etc. of the slit contribute, hence intensity of maxima decreases sharply. [1]

Alternatively

With the increase of order (n) the number of secondary wavelets responsible for the formation of secondary maxima decreases, resulting in sharp decrease of intensity.

Differences (any two) [1+1]

  1. Diffraction is a pattern formed as a result of superposition of waves from different portions of the same wave front. Interference is a pattern on a screen a result of superposition of single slit diffraction from two slits.
  2. In diffraction pattern width of central maximum is twice the width of secondary maxima. In interference pattern width of each maxima is same.
  3. In diffraction pattern intensity of maxima goes on decreasing as we move away from central maximum. In interference pattern intensity of all maxima is same.
  4. In diffraction pattern there is no absolute minima. In interference pattern absolute minima depends on amplitude of waves superposing.
Q70.2022 Supp 55/6/1 · Q6 (OR option)3 marks

A plane wavefront is incident on a surface separating two media of refractive indices n1 and n2 (> n1). With the help of a suitable diagram, explain its propagation from the rarer to denser medium. Hence, verify Snell’s law. [3]

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Official CBSE Marking Scheme · CBSE Board 2022 (Supplementary), Set 55/6/1

Diagram [1]

Refraction of a plane wavefront at a plane surface, wavefronts AB and CE

Consider the triangles ABC and AEC, we readily obtain [½]

sin i = BCAC = v1τAC

and

sin r = AEAC = v2τAC

where i and r are the angles of incidence and refraction, respectively.

Thus we obtain

sin isin r = v1v2 ......(1) [½]

If c represents the speed of light in vacuum, then,

n1 = cv1 and n2 = cv2 [½]

are known as the refractive indices of medium 1 and medium 2, respectively. In terms of the refractive indices, eq. (1) can be written as

n1 sin i = n2 sin r [½]

This is the Snell’s law of refraction.

Q71.SQP 2025-26 · Q263 marks

Draw the reflected wave front for a plane wave front incident on a plane reflecting surface. Hence verify the laws of reflection using Huygen’s principle.

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Official CBSE Marking Scheme · CBSE Sample Question Paper 2025-26
Huygens construction: incident wavefront AB and reflected wavefront CD at reflecting surface XY

[1]

If c be the speed of light, t be the time taken by light to go from B to C or A to D or E to G through F, then

t = EFc + FGc

t = AF sin ic + FC sin rc

t = AC sin r + AF(sin i – sin r)c [1]

For rays of light from different parts on the incident wavefront, the values of AF are different. But light from different points of the incident wavefront should take the same time to reach the corresponding points on the reflected wavefront.

So, t should not depend upon AF. This is possible only if sin i – sin r = 0.

i.e. sin i = sin r or i = r

Hence proved. [1]

Q72.SQP 2026-27 · Q263 marks

Derive Snell’s law of refraction using Huygens’ wave theory. What prediction of the wave theory was later experimentally verified?

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Official CBSE Marking Scheme · CBSE Sample Question Paper 2026-27

Huygens’ principle: Every point on a wavefront acts as a source of secondary wavelets, and the new wavefront at a later time is the common tangent (envelope) to these wavelets. [1]

Refraction of a plane wavefront AB at a boundary between medium 1 and medium 2 with refracted wavefront CE

Consider a plane wavefront AB incident on the boundary between two media at an angle of incidence i.

  • Velocity of light in medium 1 = v1
  • Velocity of light in medium 2 = v2
  • Angle of refraction = r

Let point A of the wavefront strike the boundary first. After time t:

  • Point B (still in medium 1) advances to C, where BC = v1t
  • From point A, a secondary wavelet grows into medium 2 with radius AE = v2t

The refracted wavefront is the tangent CE drawn to the wavelet from A.

From the geometry of the figure: sin i = BCAC ; sin r = AEAC

Or, sin isin r = BCAE = v1tv2t = v1v2 [1]

But the refractive index n of a medium is defined as n = cv, hence v1v2 = n1n2

Therefore, n1 sin i = n2 sin r

This is Snell’s law of refraction. [1]

Prediction of wave theory later experimentally verified: The wave theory of light (from Huygens’ construction) predicted that light travels more slowly in an optically denser medium than in a rarer medium: v ∝ 1n

This prediction was experimentally verified by Foucault (and later Fizeau), who showed that the speed of light in water is less than its speed in air—contradicting the earlier corpuscular (Newtonian) theory.

Q73.2026 Main 55/5/1 · Q304 marks

In a Young’s double-slit experiment, the two slits behave as coherent sources. When coherent light waves superpose over each other they create an interference pattern of successive bright and dark regions due to constructive and destructive interference.

Two slits 2 mm apart are illuminated by a source of monochromatic light and the interference pattern is observed on a screen 5·0 m away from the slits as shown in the figure.

Double slit with 2.0 mm separation, screen 5.0 m away; fringe pattern with scale in mm and point P

(i) What property of light does this interference experiment demonstrate ?

  1. Wave nature of light
  2. Particle nature of light
  3. Transverse nature of light
  4. Both wave nature and transverse nature of light

(ii) (a) The wavelength of light used in this experiment is :

  1. 720 nm
  2. 590 nm
  3. 480 nm
  4. 364 nm

OR

(b) The fringe width in the interference pattern formed on the screen is :

  1. 1·2 mm
  2. 0·2 mm
  3. 4·2 mm
  4. 6·8 mm

(iii) The path difference between the two waves meeting at point P, where there is a minimum in the interference pattern is :

  1. 8·1 × 10–7 m
  2. 7·2 × 10–7 m
  3. 6·5 × 10–7 m
  4. 6·0 × 10–7 m

(iv) When the experiment is performed in a liquid of refractive index greater than 1, then fringe pattern will :

  1. disappear
  2. become blurred
  3. be widened
  4. be compressed
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Official CBSE Marking Scheme · CBSE Board 2026 (Main), Set 55/5/1

(i) (A) wave nature of light. [1]

(ii) (a) (C) 480 nm

OR

(b) (A) 1.2 mm [1]

(iii) (B) 7.2 × 10–7 m [1]

(iv) (D) be compressed [1]

Q74.2024 Main 55/3/1 · Q304 marks

Diffraction and interference are closely related phenomena that occur together. Diffraction is the phenomenon of bending of light around the edges of the obstacle, while interference is the combination of waves that results in a new wave pattern. In order to get interference, there must be at least two waves that are diffracting. So while diffraction can occur without interference, interference cannot occur without diffraction.

Two slits of width 2 μm each in an opaque material are separated by a distance of 6 μm. Monochromatic light of wavelength 450 nm is incident normally on the slits. One finds a combined interference and diffraction pattern on the screen.

  1. The number of peaks of the interference fringes formed within the central peak of the envelope of the diffraction pattern will be :
    1. 2
    2. 3
    3. 4
    4. 6
  2. The number of peaks of the interference formed if the slit width is doubled while keeping the distance between the slits same will be :
    1. 1
    2. 2
    3. 3
    4. 4
  3. (a) If instead of 450 nm light, another light of wavelength 680 nm is used, number of peaks of the interference formed in the central peak of the envelope of the diffraction pattern will be :
    1. 2
    2. 4
    3. 6
    4. 9

    OR

    (b) Consider the diffraction of light by a single slit described in this case study. The first minimum falls at an angle θ equal to :
    1. sin–1 (0·12)
    2. sin–1 (0·225)
    3. sin–1 (0·32)
    4. sin–1 (0·45)
  4. The number of bright fringes formed due to interference on 1 m of screen placed at 43 m away from the slits is :
    1. 2
    2. 3
    3. 6
    4. 10
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Official CBSE Marking Scheme · CBSE Board 2024 (Main), Set 55/3/1

(i) (D) 6 [1]

(ii) (C) 3 [1]

(iii) (a) (C) 6

OR

    (b) sin–1(0.225) [1]

(iv) (D) 10 [1]

Q75.2023 Main 55/3/1 · Q354 marks

Diffraction of light is bending of light around the corners of an object whose size is comparable with the wavelength of light. Diffraction actually defines the limits of ray optics. This limit for optical instruments is set by the wavelength of light. An experimental arrangement is set up to observe the diffraction pattern due to a single slit.

Answer the following questions based on the above :

(a) How will the width of central maximum be affected if the wavelength of light is increased ? [1]

(b) Under what condition is the first minimum obtained ? [1]

(c) Write two points of difference between interference and diffraction patterns. [2]

Alternative to (c) :

(c) Two students are separated by a 7 m partition wall in a room 10 m high. If both light and sound waves can bend around obstacles, how is it that the students are unable to see each other even though they can converse easily ? [2]

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Official CBSE Marking Scheme · CBSE Board 2023 (Main), Set 55/3/1

(a) β0 ∝ λ

β0 will increase with increase in wavelength. [1]

(b) When path difference aθ = λ or at an angle θ ≈ λ/a. [1]

(c) Differences [1 + 1]

Interference PatternDiffraction Pattern
(i) All the maxima are equally spaced.(i) Width of central bright maxima is twice the width of the other maxima.
(ii) The dark fringe is having zero intensity.(ii) The dark fringe is not completely dark.
(iii) All the maxima are of the same intensity.(iii) There is a sharp decrease in the intensity of maxima after the central bright maxima.
(iv) Interference pattern: equal-intensity maxima versus path difference(iv) Diffraction pattern: intensity versus diffraction angle with broad central maximum
Alternatively: It is obtained by the superposition of two waves originating from two sources/slits.It is obtained by the superposition of waves from points on a single slit.

Any two of the above mentioned differences.

Alternative to (c) :

(c) The opening (slit) is 3 m, which is of the order of the wavelength of sound waves whereas it is very large compared to the wavelength of light. Hence, sound can bend around the obstacle while light cannot. [2]

Q76.2026 Main 55/2/1 · Q31 (option 1)5 marks

(a) What are coherent sources ? Why they are necessary for observing stable interference pattern ? Draw a graph showing the variation of intensity of light with the position on the screen in Young’s double-slit experiment.

(b) Find the intensity of light at a point on the screen when two interfering waves of the same intensity (I0) have a path difference of

(i) λ4 and (ii) λ3.

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Official CBSE Marking Scheme · CBSE Board 2026 (Main), Set 55/2/1

(a) Two sources of light emitting waves of same frequency and constant phase or zero phase difference are said to be coherent. [1]

If the sources are incoherent, interference pattern will change with time. Hence the positions of maxima and minima will also vary rapidly with time and we obtain ‘time averaged’ intensity distribution. [1]

Alternatively : If the two sources are coherent then the phase difference at any point will not change with time and we will have a stable interference pattern.

Graph of intensity versus position on the screen in YDSE, equally spaced maxima

[1] for graph

(b) (i) I = 4I0 cos2(φ2)

Δp = λ4; φ = 2πλ · λ4 = π2 [½]

I = 4I0 cos2 π4 = 2I0 [½]

(ii) Δp = λ3; φ = 2πλ · λ3 = 2π3 [½]

I = 4I0 cos2 π3 = I0 [½]

Note : Award (½) mark if the student writes I = 4I0 cos2(φ2)

Q77.2026 Main 55/3/1 · Q33 (OR option)5 marks

(i) Two thin converging lenses of focal length f1 and f2 are placed coaxially in contact. Derive expression for the focal length of the combination.

(ii) A beam of coherent light of wavelength 550 nm is incident normal to the plane of a pair of two slits S1 and S2 each of width 1·2 × 10−6 m separated by 1·1 mm. Dark and bright fringes are observed on a screen 2·2 m away from the plane of the slits.

Calculate :

(I) fringe width.

(II) distance of the second dark fringe from the central maximum.

(III) what will happen when the entire apparatus is immersed in water.

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Official CBSE Marking Scheme · CBSE Board 2026 (Main), Set 55/3/1

(i) Ray diagram [½]

Two thin lenses in contact: object O, first image I1, final image I

For image formed by first lens A : 1v1 − 1u = 1f1 [½]

For image formed by second lens B : 1v − 1v1 = 1f2 [½]

Adding and comparing with lens formula : 1f = 1f1 + 1f2 [½]

(ii) (I) β = λDd [½]

= 550 × 10−9 × 2·21·1 × 10−3 [½]

= 1·1 mm

(II) x = (2n − 1)λD2d [½]

n = 2

x = 1·65 mm [½]

(III) Fringe width decreases [1]

(Alternatively) β′ = βn

Q78.2026 Main 55/4/1 · Q31 (OR option)5 marks

(b) (i) What is the difference between a ray and a wavefront ?

(ii) A plane wave is incident on a reflecting surface. Using Huygens principle, show how it is reflected from the surface. Hence, verify the law of reflection.

(iii) Depict refraction of a plane wave by a convex lens.

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Official CBSE Marking Scheme · CBSE Board 2026 (Main), Set 55/4/1

(i) Ray – It is a straight line depicting the rectilinear propagation of light. [½]

Wavefront – It is the locus of points which oscillate in same phase. [½]

Alternatively :

Spherical wavefront with rays (left) and plane wavefronts with rays (right)

(ii) Reflected wavefront [1]

Huygens construction for reflection of a plane wave AB at a surface MN: incident wavefront AB, reflected wavefront EC, angles i and r

If v represents the speed of the wave in the medium and τ represents the time taken by the wavefront to advance from the point B to C then the distance

BC = vτ [½]

Let CE represent the tangent plane drawn from the point C to this sphere. Then,

AE = BC = vτ

If we now consider the triangles EAC and BAC we will find that they are congruent. [1]

Therefore, the angles i and r would be equal. [½]

(iii) Refraction of a plane wave by a convex lens [1]

Incident plane wave converted into a spherical wavefront converging to the focus P by a convex lens
Q79.2026 Main 55/5/1 · Q33 (option 1)5 marks
  1. A parallel beam of monochromatic light falls normally on a single slit of width ‘a’ and a diffraction pattern is observed on a screen placed at distance D from the slits. Explain :
    1. the formation of maxima and minima in the diffraction pattern, and
    2. why the maxima go on becoming weaker and weaker with its increasing number (n).
  2. Write any two points of difference between interference pattern due to double-slit and diffraction pattern due to single-slit.
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Official CBSE Marking Scheme · CBSE Board 2026 (Main), Set 55/5/1

(i)(I) Explaining the formation of maxima and minima in the diffraction pattern [2]; (II) Explaining the reason for weakening of maxima with increasing number (n) [1]; (ii) Writing two points of difference between interference and diffraction pattern [2]

(i)(I)

Single-slit diffraction geometry with slit LN and path difference NQ

[½]

Path difference between the waves originating from points L and N superimposed at P on the screen is, NQ = a sinθ [½]

Secondary minima: a sinθ = nλ, n = ±1, ±2, … [½]

Secondary maxima: a sinθ = (2n + 1) λ/2, n = ±1, ±2, … [½]

Alternatively,

The light coming from the slit superimposes at the center of screen forms central maxima with maximum intensity. For secondary maxima on screen light coming from a part (one-third, one-fifth, one-seventh….) of whole slit contribute for giving intensity on screen.

For secondary minima, contribution from two halves of whole slit cancel each other, so the intensity falls to zero.

(II) As n increases, maxima goes on becoming weaker, since the light coming from a fraction of the slit (one-third, one-fifth, one-seventh, ..) contributes to the intensity at a point. [1]

(ii) [(½) × 4]

InterferenceDiffraction
The width of bright and dark bands is equal.The width of central maxima is twice the width of secondary maxima or minima.
Intensity of all bright fringes is sameIntensity of bright fringe decreases with distance from central maxima.
There is a good contrast between bright and dark fringesThere is a poor contrast between bright and dark fringes.
Intensity distribution curves for interference (left) and diffraction (right)

Note: Award full marks for any two correct differences.

Q80.2026 Supp 55/7/1 · Q33 (OR option)5 marks
  1. Differentiate between interference and diffraction of light.
    Two coherent light waves, each of intensity I0 produce interference pattern on a screen. Find the intensity at a point at which the path difference between them is (i) λ3, and (ii) λ4. [3]
  2. In a Young’s double-slit experiment, fringes are obtained on a screen placed at a distance D from the slits. It is observed that if the screen is moved 5 cm towards the slits, the fringe width changes by 0·03 mm. If the distance between the slits is 1 mm, calculate the wavelength of the light used. [2]
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Official CBSE Marking Scheme · CBSE Board 2026 (Supplementary), Set 55/7/1

(a)

InterferenceDiffraction
1. The width of bright and dark bands is equal.1. The width of the central maxima is twice the width of secondary maxima or minima.
2. Intensity of all bright fringes is the same.2. Intensity of bright fringes decreases with distance from the central maxima.
3. There is good contrast between bright and dark fringes.3. There is poor contrast between bright and dark fringes.
4. Intensity distribution curve.
Intensity distribution curve for interference with equal maxima
4. Intensity distribution curve
Intensity distribution curve for diffraction with decreasing secondary maxima

(Any one correct difference) [1]

(i) I = 4I0cos2φ2 [½]

Δx = λ3; φ = 2πλ × λ3 = 2π3

I1 = 4I0cos2π3 = I0 [½]

(ii) Δx = λ4; φ = 2πλ × λ4 = π2 [½]

I2 = 4I0cos2π4 = 2I0 [½]

(b) Δβ = β1 – β2 [½]

0.03 = Dλd – (D – 5)λd [½]

0.03 = (D – D + 5)λ [½]

λ = 0.035 [½]

= 0.006 nm = 6 × 10-7 m

Q81.2025 Main 55/1/1 · Q33 (option 1)5 marks

(i) (1) What are coherent sources ? Why are they necessary for observing a sustained interference pattern ?

(2) Lights from two independent sources are not coherent. Explain.

(ii) Two slits 0.1 mm apart are arranged 1.20 m from a screen. Light of wavelength 600 nm from a distant source is incident on the slits.

(1) How far apart will adjacent bright interference fringes be on the screen ?

(2) Find the angular width (in degree) of the first bright fringe.

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Q82.2025 Main 55/2/1 · Q33 (OR option)5 marks
  1. Light consisting of two wavelengths 600 nm and 480 nm is used to obtain interference fringes in a double slit experiment. The screen is placed 1.0 m away from slits which are 1.0 nm apart.
    1. Calculate the distance of the third bright fringe on the screen from the central maximum for wavelength 600 nm.
    2. Find the least distance from the central maximum where the bright fringes due to both the wavelengths coincide.
    1. Draw the variation of intensity with angle of diffraction in single slit diffraction pattern. Write the expression for value of angle corresponding to zero intensity locations.
    2. In what way diffraction of light waves differs from diffraction of sound waves ?
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Q83.2025 Main 55/4/1 · Q33 (OR option)5 marks
  1. A physics teacher wants to demonstrate interference with the help of double slit experiment using a laser beam of 633 nm wavelength. Since the hall is large enough, interference pattern is formed on the wall 5·0 m from the slits. For clear and comfortable view by all the students they want the fringe width 5 mm.
    1. Find the slit separation for obtaining the desired interference pattern.
    2. How far will the first minimum be from the central maximum ?
  2. A parallel beam of light of wavelength 650 nm passes through a slit of width 0·6 mm. The diffraction pattern is obtained on a screen kept 60 cm away from the slit. Find the distance between first order minima on both sides of the central maximum.
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Q84.2025 Main 55/6/1 · Q32 (OR option)5 marks
  1. Using Huygens’ principle, explain the refraction of a plane wavefront, propagating in air, at a plane interface between air and glass. Hence verify Snell’s law.
  2. Use mirror formula to deduce that a convex mirror always produces a virtual image of an object kept in front of it.
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Q85.2025 Main 55/7/1 · Q33 (OR option)5 marks

(i) Two coherent light waves, each of intensity I0 superpose each other and produce interference pattern on a screen. Obtain the expression for the resultant intensity at a point where the phase difference between the waves is ϕ. Write its maximum and minimum possible values.

(ii) In a single slit diffraction experiment, the aperture of the slit is 3 mm and the separation between the slit and the screen is 1·5 m. A monochromatic light of wavelength 600 nm is normally incident on the slit. Calculate the distance of (I) first order minimum, and (II) second order maximum, from the centre of the screen.

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Q86.2025 Supp 55/S/1 · Q33 (OR option)5 marks

State Huygens principle. Using it, draw a diagram and discuss the case of refraction of plane wave of light from a rarer medium to a denser medium at their plane interface. Hence derive Snell’s law.

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Official CBSE Marking Scheme · CBSE Board 2025 (Supplementary), Set 55/S/1

Huygen’s Principle :

Each point of the wavefront is the source of a secondary disturbance and the wavelets emanating from these points spread out in all directions with the speed of the wave. These wavelets emanating from the wavefront are usually referred to as secondary wavelets and if we draw a common tangent to all these spheres, we obtain the new position of the wavefront at a later time. [1]

Refraction of a plane wavefront at a plane interface: incident wavefront AB, refracted wavefront CE, v2 < v1

[1½] for diagram

Discussion : In time τ distance BC travelled by the incident plane wavefront with the velocity v1 is v1τ. To determine the shape of the refracted wavefront, draw a sphere of radius v2τ (= AE) from point A and draw a tangent from C to E as shown in the diagram. [½]

From the above diagram, in △ABC and △AEC

sin i = BCAC = v1τAC [½]

and sin r = AEAC = v2τAC [½]

sin isin r = v1v2 ......................(1) [½]

As n1 = cv1 and n2 = cv2

∴ v1v2 = n2n1 ..................(2)

From equations (1) and (2)

n1 sin i = n2 sin r [½]

Q87.2024 Main 55/1/1 · Q33 (OR option)5 marks

(i) State Huygens’ principle. A plane wave is incident at an angle i on a reflecting surface. Construct the corresponding reflected wavefront. Using this diagram, prove that the angle of reflection is equal to the angle of incidence.

(ii) What are the coherent sources of light ? Can two independent sodium lamps act like coherent sources ? Explain.

(iii) A beam of light consisting of a known wavelength 520 nm and an unknown wavelength λ, used in Young’s double slit experiment produces two interference patterns such that the fourth bright fringe of unknown wavelength coincides with the fifth bright fringe of known wavelength. Find the value of λ.

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Official CBSE Marking Scheme · CBSE Board 2024 (Main), Set 55/1/1

(i) Statement of Huygens’ principle [½]; construction of reflected wavefront [½]; proof of angle of reflection equal to angle of incidence [1]. (ii) Definition of coherent sources [½]; explanation [1]. (iii) Finding the unknown wavelength [1½]

(i) Each point of the wavefront is the source of a secondary disturbance and the wavelets emanating from these points spread out in all directions with the speed of the wave. These wavelets emanating from the wavefront are usually referred to as secondary wavelets and if we draw a common tangent to all these spheres, we obtain the new position of the wavefront at a later time. [½]

Plane wavefront incident on reflecting surface MN, with incident wavefront BC and reflected wavefront AE constructed

ΔEAC is congruent to ΔBAC; so ∠i = ∠r [1]

(ii) Two sources are said to be coherent if the phase difference between them does not change with time. [½]

No, two independent sodium lamps cannot be coherent. [½]

Two independent sodium lamps cannot be coherent as the phase between them does not remain constant with time. [½]

(iii) 4β2 = 5β1 [½]

4 × λDd = 5 × λknownDd

⇒ λ = 54 × λknown = 54 × 520

= 650 nm [1]

Q88.2024 Main 55/2/1 · Q33 (option 1)5 marks

(i) Give any two differences between the interference pattern obtained in Young’s double-slit experiment and a diffraction pattern due to a single slit.

(ii) Draw an intensity distribution graph in case of a double-slit interference pattern.

(iii) In Young’s double-slit experiment using monochromatic light of wavelength λ, the intensity of light at a point on the screen, where path difference is λ, is K units. Find the intensity of light at a point on the screen where the path difference is λ6.

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Official CBSE Marking Scheme · CBSE Board 2024 (Main), Set 55/2/1

(i) [1 + 1]

InterferenceDiffraction
1Bands are equally spacedBands are not equally spaced.
2Intensity of bright bands is same.Intensity of maxima decreases on either side of central maxima.
3First maxima is at an angle λ/aFirst minima is at an angle λ/a

(ii) [1]

Intensity versus path difference for double-slit interference: equal-height fringes

(iii) Path difference (Δ) = λ
φ = 2πΔλ
φ = 2π [½]

I = 4I0 cos2 φ2
K = 4I0 cos2 π = 4I0 [½]

Path difference = λ6
φ = π/3 [½]

I = 4I0 cos2 π6 = 4I0 × 34 = 34 K [½]

Q89.2024 Main 55/3/1 · Q33 (OR option)5 marks

(i) The displacement of two light waves, each of amplitude ‘a’ and frequency ω, emanating from two coherent sources of light, are given by y1 = a cos ωt and y2 = a cos (ωt + φ). φ is the phase difference between the two waves. These light waves superpose at a point. Obtain the expression for the resultant intensity at that point.

(ii) In Young’s double slit experiment, find the ratio of intensities at two points on a screen when waves emanating from two slits reaching these points have path differences (i) λ6 and (ii) λ12.

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Official CBSE Marking Scheme · CBSE Board 2024 (Main), Set 55/3/1

(i) Expression for resultant intensity [3]; (ii) Ratio of intensities [2]

(i) y1 = a cos ωt

y2 = a cos(ωt + φ)

According to the principle of superposition

y = y1 + y2 [½]

y = a cos ωt + a cos(ωt + φ)

y = a cos ωt + a cos ωt cos φ – a sin ωt sin φ [½]

y = a cos ωt (1 + cos φ) – a sin φ sin ωt

Let,

a(1 + cos φ) = A cos θ   ------- (i)

a sin φ = A sin θ   --------(ii) [½]

Squaring and adding equation (i) and (ii)

A2 = a2(1 + cos φ)2 + a2 sin2φ

= a2(1 + cos2φ + 2 cos φ) + a2 sin2φ [½]

= 2a2(1 + cos φ)

= 4a2 cos2φ/2 [½]

I ∝ A2

I = kA2

where k is constant

I = 4ka2 cos2φ/2 [½]

[Award full credit for this part for any other alternative methods]

(ii) φ1 = 2πλ × λ6 = π/3 [½]

I1 = 4I0 cos2φ/2

= 4I0 cos2(π/6)

I1 = 3I0 [½]

φ2 = 2πλ × λ12 = π/6 [½]

I2 = 4I0 cos2(π/12)

I2 = 4I0 cos2 15° [½]

I1I2 = 34 cos2 15° [½]

Q90.2024 Main 55/4/1 · Q31 (OR option)5 marks
  1. Differentiate between a wavefront and a ray.
  2. State Huygen’s principle and verify laws of reflection using suitable diagram.
  3. In Young’s double slit experiment, the slits S1 and S2 are 3 mm apart and the screen is placed 1.0 m away from the slits. It is observed that the fourth bright fringe is at a distance of 5 mm from the second dark fringe. Find the wavelength of light used.
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Official CBSE Marking Scheme · CBSE Board 2024 (Main), Set 55/4/1

(i) Wavefront is a surface of constant phase. [½]

Alternatively Locus of points, which oscillate in phase

Ray – The straight line path along which light travels (or energy propagates). [½]

Alternatively – Ray is normal to wave front.

(ii) Huygens’ Principle Each point of the wave front is the source of secondary disturbance and the wavelets emanating from the points spread out in all directions with speed of wave. The wavelets emanating from wave front are usually referred to as secondary wavelets. A common tangent to all these spheres gives the new position of the wave front at a later time. [1]

Reflection of a plane wavefront at a surface MN: incident wavefront AB, reflected wavefront EC

[1]

Triangles EAC and BAC are congruent therefore ∠i = ∠r [½]

(iii) Position of 4th bright fringe

x4(bright) = 4 Dλd [½]

Position of 2nd dark fringe

x2(dark) = 32 Dλd [½]

x4(bright) – x2(dark) = 5 mm

4 Dλd – 32 Dλd = 5 × 10–3

λ = 6 × 10–6 m [½]

Q91.2024 Main 55/5/1 · Q33 (option 1)5 marks
  1. A plane light wave propagating from a rarer into a denser medium, is incident at an angle i on the surface separating two media. Using Huygen’s principle, draw the refracted wave and hence verify Snell’s law of refraction.
  2. In a Young’s double slit experiment, the slits are separated by 0.30 mm and the screen is kept 1.5 m away. The wavelength of light used is 600 nm. Calculate the distance between the central bright fringe and the 4th dark fringe.
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Official CBSE Marking Scheme · CBSE Board 2024 (Main), Set 55/5/1

(i) [1]

Huygens construction for refraction: incident wavefront AB, refracted wavefront EC at the surface PP′ between medium 1 and medium 2 with v2 < v1

Considering triangles ABC and AEC

sin i = BCAC = v1τAC ------(1) [½]

sin r = AEAC = v2τAC ------(2)

From equation (1) and equation (2)

sin isin r = v1v2 ------(3) [½]

If c represents the speed of light in vacuum, then

n1 = cv1 and n2 = cv2 [½]

In terms of refractive indices

n1 sin i = n2 sin r [½]

which is Snell’s law of refraction.

(ii)

X4 = (2n – 1)λD2d [½]

X4 = (2×4 – 1) × 600×10–9 × 1.52 × 0.3×10–3 [1]

= 1.05×10–2 m [½]

Q92.2024 Main 55/5/1 · Q33 (OR option)5 marks
  1. Discuss briefly diffraction of light from a single slit and draw the shape of the diffraction pattern.
  2. An object is placed between the pole and the focus of a concave mirror. Using mirror formula, prove mathematically that it produces a virtual and an enlarged image.
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Official CBSE Marking Scheme · CBSE Board 2024 (Main), Set 55/5/1

(i) A beam of light falls normally on a single slit and bends around its corners. This phenomenon is called diffraction. [1]

When a beam of light falls normally on a narrow single slit, then diffracted light goes on to meet on a screen. It is observed that at the center of the screen intensity is maximum and goes on decreasing as one move away from the center on either side of screen. [1]

Single-slit diffraction: incoming wave, slit, viewing screen with central maximum and intensity pattern

[1]

(ii)

1f = 1v + 1u

v = ufu – f

Following new cartesian sign convention

v = (–u)(–f)–u – (–f)

v = uff – u as f > u [1]

v is +ve, So image is virtual.

m = – vu = ff – u > 1 i.e. Enlarged image [1]

Q93.2023 Main 55/1/1 · Q33 (option 1)5 marks

(i) (1) Write two points of difference between an interference pattern and a diffraction pattern.

(2) Name any two factors on which the fringe width in a Young’s double-slit experiment depends.

(ii) In Young’s double-slit experiment, the two slits are separated by a distance equal to 100 times the wavelength of light that passes through the slits. Calculate :

(1) the angular separation in radians between the central maximum and the adjacent maximum.

(2) the distance between these two maxima on a screen 50 cm from the slits.

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Official CBSE Marking Scheme · CBSE Board 2023 (Main), Set 55/1/1

i) (1) Difference between interference pattern and diffraction pattern [1 + 1]; (2) Two factors affecting fringe width in Young’s double slit experiment [½ + ½]; ii) (1) calculation of angular separation [1]; (2) calculation of distance between two maxima [1]

(i) (1)

(a) The interference pattern has a number of equally spaced bright and dark bands while diffraction pattern has a central bright maximum which is twice as wide as the other maxima.

(b) Interference pattern is obtained by superposing two waves originating from two narrow slits, while diffraction pattern is a superposition of a continuous family of waves originating from each point on a single slit.

(c) The maxima in interference pattern is obtained at angle λ/a, while the first minima is obtained at same angle λ/a for diffraction pattern.

(d) In interference pattern the intensity of bright fringes remain same while in diffraction the intensity falls as we go to successive maxima away from the center on either side.

(any two) [1 + 1]

(2) Factors affecting fringe width: Wavelength (λ) / distance of screen from slits (D) / separation between slits (d). (any two) [½ + ½]

(ii) (1) d sinθ = nλ, n = 1, sinθ = λ/d [½]

For small angle sinθ ≈ θ = λ100λ = 1100 radian. [½]

(2) β = λDd = θD [½]

= 1100 × 50 × 10−2 = 50 × 10−4 m = 5 mm [½]

Q94.2023 Main 55/2/1 · Q33 (option 1)5 marks
  1. State Huygen’s principle. With the help of a diagram, show how a plane wave is reflected from a surface. Hence verify the law of reflection.
  2. A concave mirror of focal length 12 cm forms a three times magnified virtual image of an object. Find the distance of the object from the mirror.
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Official CBSE Marking Scheme · CBSE Board 2023 (Main), Set 55/2/1

(i) Huygen’s principle

Each point of the wavefront is the source of a secondary disturbance and the wavelets emanating from these points spread out in all directions with the speed of the wave. These wavelets emanating from the wavefront are usually referred to as secondary wavelets, a common tangent to all these spheres gives the new position of the wavefront at a later time. [1]

Plane wavefront reflected at surface MN: incident wavefront AB, reflected wavefront EC, angles i and r

Diagram [1]

Verification of law of reflection

In ΔAEC & ΔCBA

EC = AB (c × t each)

∠AEC = ∠CBA (90° each)

AC = AC (common side) [½]

By RHS congruency ΔAEC ≅ ΔCBA

⇒ ∠i = ∠r [½]

(ii) m = +3, f = −12 cm, u = ? [½]

m = −vu = 3 ⇒ v = −3u [½]

using mirror formula 1v + 1u = 1f [½]

1−3u + 1u = 1−12

u = −8 cm [½]

Q95.2022 Main 55/1/1 · Q125 marks

The principle of superposition is used to understand the phenomenon of interference of light waves. The principle states that at a particular point, the resultant displacement produced by a number of waves is the vector sum of the displacements produced by each wave. Light waves from two coherent sources produce interference pattern. Thomas Young devised a way to obtain two coherent sources using two identical pinholes (S1 and S2) illuminated by a single monochromatic pinhole source S. Using these sources in his experiment known as Young's double slit experiment, Young studied the interference pattern. The pattern consists of alternate bright and dark finges. The distance between two successive bright or dark fringes depends on the distance between S1 and S2, the distance of the screen from the plane of S1S2 and the wavelength of light used.

I. Consider the following waves :

(i) y1 = a sin ωt
(ii) y2 = a sin 2ωt
(iii) y3 = a sin (2ωt + φ)
(iv) y4 = a sin (4ωt + π2)

Which pair of the waves coming from two sources S1 and S2 will produce interference ?

  1. (i) and (ii)
  2. (ii) and (iii)
  3. (iii) and (iv)
  4. (iv) and (i)

II. Two light waves of the same intensity I0 each, having a path difference of λ/4, emanating from two coherent sources, meet at a point. The resultant intensity at the point will be

  1. Zero
  2. I0
  3. 2 I0
  4. 4 I0

III. Vandana performs Young's double slit experiment by using orange, green and red lights successively. If the fringe widths measured in the three cases are ω1, ω2 and ω3 respectively, then which of the following is correct ?

  1. ω2 > ω1 > ω3
  2. ω1 > ω2 > ω3
  3. ω2 > ω3 > ω1
  4. ω3 > ω1 > ω2

IV. In a Young's double slit experiment, the slit separation is 0.8 mm and the interference pattern is obtained on a screen kept 50 cm from the plane of the slits S1 and S2. If the first bright fringe is formed 0.4 mm from the central maximum, the wavelength of light used is

  1. 480 nm
  2. 560 nm
  3. 640 nm
  4. 680 nm

V. Consider the effect on the angular separation of the fringes in a Young's double slit experiment due to the following operations :

  1. the screen is moved away from the plane of the slits,
  2. the separation between the two slits is increased till fringes are observed.

Which of the following options is correct ?

  1. It remains constant in both cases.
  2. It decreases in both cases.
  3. It remains constant in (i) but decreases in (ii).
  4. It decreases in (i) but remains constant in (ii).
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I (B) (ii) and (iii) [1]

II (C) 2 I0 [1]

III (D) ω3 > ω1 > ω2 [1]

IV (C) 640 nm [1]

V (C) It remains constant in (i) but decreases in (ii). [1]

Q96.2022 Main 55/2/1 · Q125 marks

The British physicist Thomas Young explained the interference of light using the principle of superposition of waves. He observed the interference pattern on the screen, in his experimental set-up, known now as Young’s double slit experiment. The two slits S1 and S2 were illuminated by light from a slit S. The interference pattern consists of dark and bright bands of light. Such bands are called fringes. The distance between two consecutive bright and dark fringes is called fringe width.

  1. If the screen is moved closer to the plane of slits S1 and S2, then the fringe width :
    1. will decrease, but the intensity of bright fringe remains the same.
    2. will increase, but the intensity of bright fringe decreases.
    3. will decrease, but the intensity of bright fringe increases.
    4. and the intensity both remain the same.
  2. What will happen to the pattern on the screen, when the two slits S1 and S2 are replaced by two independent but identical sources ?
    1. The intensity of pattern will increase
    2. The intensity of pattern will decrease
    3. The number of fringes will become double
    4. No pattern will be observed on the screen
  3. Two sources of light are said to be coherent, when both emit light waves of :
    1. same amplitude and have a varying phase difference.
    2. same wavelength and a constant phase difference.
    3. different wavelengths and same intensity.
    4. different wavelengths and a constant phase difference.
  4. The fringe width in a Young’s double slit experiment is β. If the whole set-up is immersed in a liquid of refractive index ‘μ’, then the new fringe width will be :
    1. β
    2. βμ
    3. βμ
    4. βμ2
  5. The total path difference between two waves meeting at points P1 and P2 on the screen are 3λ2 and 2λ respectively. Then :
    1. bright fringes are formed at both points.
    2. dark fringes are formed at both points.
    3. a bright fringe is formed at P1 and a dark fringe is formed at P2.
    4. a bright fringe is formed at P2 and a dark fringe is formed at P1.
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Official CBSE Marking Scheme · CBSE Board 2022 (Main), Set 55/2/1

a) (iii) [1]

b) (iv) [1]

c) (ii) [1]

d) (iii) [1]

e) (iv) [1]

Unit VII · Dual Nature of Radiation and Matter

11. Dual Nature of Radiation and Matter

Q1.2026 Main 55/1/1 · Q141 mark

Assertion (A) : If accelerated electrons are passed through a narrow slit, a diffraction pattern is observed.

Reason (R) : Electrons behave as both particles and waves.

  1. Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of the Assertion (A).
  2. Both Assertion (A) and Reason (R) are true, but Reason (R) is not the correct explanation of the Assertion (A).
  3. Assertion (A) is true, but Reason (R) is false.
  4. Both Assertion (A) and Reason (R) are false.
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Official CBSE Marking Scheme · CBSE Board 2026 (Main), Set 55/1/1

(A) Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of the Assertion (A).

Q2.2026 Main 55/2/1 · Q31 mark

The maximum kinetic energy of the electrons emitted from a photo sensitive surface depends on

  1. work function of the surface ‘φ0’ only.
  2. frequency of the incident radiation ‘v’ only.
  3. intensity of the incident radiation ‘I’ only.
  4. Both φ0 and v.
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Official CBSE Marking Scheme · CBSE Board 2026 (Main), Set 55/2/1

(D) Both φ0 and v

Q3.2026 Main 55/2/1 · Q131 mark

Note : For question numbers 13 to 16, two statements are given – one labelled as Assertion (A) and the other labelled as Reason (R). Select the correct answer from the codes (A), (B), (C) and (D) as given below :

Assertion (A) : On increasing the intensity of incident light of frequency v(> v0) on a photosensitive surface, the photocurrent increases.

Reason (R) : The stopping potential for a photosensitive surface increases with increase of frequency v(> v0) of incident light.

  1. Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of Assertion (A).
  2. Both Assertion (A) and Reason (R) are true, but Reason (R) is not the correct explanation of Assertion (A).
  3. Assertion (A) is true, but Reason (R) is false.
  4. Assertion (A) is false and Reason (R) is also false.
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Official CBSE Marking Scheme · CBSE Board 2026 (Main), Set 55/2/1

(B) Both assertion (A) and Reason (R) are true, but Reason (R) is not the correct explanation of Assertion (A)

Q4.2026 Main 55/3/1 · Q81 mark

Radiation of wavelength 200 nm is incident on a photosensitive surface of work function 4·2 eV. The kinetic energy of fastest photoelectrons emitted from this surface will be close to :

  1. 3·5 eV
  2. 3·0 eV
  3. 2·5 eV
  4. 2·0 eV
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Official CBSE Marking Scheme · CBSE Board 2026 (Main), Set 55/3/1

(D) 2·0 eV

Q5.2026 Main 55/3/1 · Q91 mark

A proton and an alpha particle have equal momentum. The ratio of their kinetic energies (Ep/Ea) and de Broglie wavelengths associated with them (λp/λa) respectively are :

  1. 2, 1
  2. 1, 2
  3. 4, 1
  4. 1, 4
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(C) 4, 1

Q6.2026 Main 55/3/1 · Q151 mark

Assertion (A) : Photoelectric current depends upon the intensity of the incident radiation.

Reason (R) : Stopping potential is independent of the intensity of the incident radiation.

  1. Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of the Assertion (A).
  2. Both Assertion (A) and Reason (R) are true, but Reason (R) is not the correct explanation of the Assertion (A).
  3. Assertion (A) is true, but Reason (R) is false.
  4. Both Assertion (A) and Reason (R) are false.
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Official CBSE Marking Scheme · CBSE Board 2026 (Main), Set 55/3/1

(B) Both Assertion (A) and Reason (R) are true but Reason (R) is not the correct explanation of the Assertion (A)

Q7.2026 Main 55/4/1 · Q81 mark

While studying photoelectric emission from a given surface, the wavelength of the incident radiation is changed from 600 nm to 400 nm, keeping the intensity of radiation the same. Then :

  1. cut-off potential will decrease.
  2. cut-off potential will increase.
  3. saturation current will decrease.
  4. saturation current will increase.
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(B) cut-off potential will increase

Q8.2026 Main 55/4/1 · Q91 mark

Radiation of wavelength 331 nm irradiates the following metals :

MetalWork Function (eV)
Na1·92
K2·15
Ca3·20
Mo4·17

Which of the following statements is correct ?

  1. Only Na and K show photoelectric emission.
  2. Only Mo will not show photoelectric emission.
  3. All of the given metals show photoelectric emission.
  4. None of them show photoelectric emission.
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(B) Only Mo will not show photoelectric emission

Q9.2026 Main 55/4/1 · Q101 mark

The kinetic energy of a charged particle is increased to four times of its initial value. The de Broglie wavelength associated with the particle will :

  1. increase by 100% of its initial value.
  2. increase by 50% of its initial value.
  3. decrease by 25% of its initial value.
  4. decrease by 50% of its initial value.
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Official CBSE Marking Scheme · CBSE Board 2026 (Main), Set 55/4/1

(D) Decrease by 50% of its initial value

Q10.2026 Main 55/5/1 · Q71 mark

Light of which of the following colours will have the maximum energy in a photon associated with it ?

  1. Red light
  2. Yellow light
  3. Green light
  4. Blue light
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Official CBSE Marking Scheme · CBSE Board 2026 (Main), Set 55/5/1

(D) Blue light

Q11.2026 Main 55/5/1 · Q161 mark

Assertion (A) : Photoelectric effect is a spontaneous phenomenon.

Reason (R) : According to the wave picture of radiation, an electron would take hours/days to absorb sufficient energy to overcome the work function and come out from a metal surface.

  1. Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of the Assertion (A).
  2. Both Assertion (A) and Reason (R) are true, but Reason (R) is not the correct explanation of the Assertion (A).
  3. Assertion (A) is true, but Reason (R) is false.
  4. Both Assertion (A) and Reason (R) are false.
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(B) Both Assertion (A) and reason (R) are true, but reason (R) is not the correct explanation of the assertion (A).

Q12.2026 Supp 55/7/1 · Q161 mark

Assertion (A) : Electron emission from surface of zinc occurs when yellow light is incident on it.

Reason (R) : Energy of a photon of yellow light is more than the work function of zinc.

  1. Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of the Assertion (A).
  2. Both Assertion (A) and Reason (R) are true, but Reason (R) is not the correct explanation of the Assertion (A).
  3. Assertion (A) is true, but Reason (R) is false.
  4. Both Assertion (A) and Reason (R) are false.
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(D) Both Assertion (A) and Reason (R) are false.

Q13.2025 Main 55/1/1 · Q91 mark

The stopping potential V0 measured in a photoelectric experiment for a metal surface is plotted against frequency ν of the incident radiation. Let m be the slope of the straight line so obtained. Then the value of charge of an electron is given by (h is the Planck’s constant.)

  1. mh
  2. mh
  3. hm
  4. 1mh
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Q14.2025 Main 55/1/1 · Q101 mark

Let λe, λp and λd be the wavelengths associated with an electron, a proton and a deuteron, all moving with the same speed. Then the correct relation between them is

  1. λd > λp > λe
  2. λe > λp > λd
  3. λp > λe > λd
  4. λe = λp = λd
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Q15.2025 Main 55/2/1 · Q101 mark

Atomic spectral emission lines of hydrogen atom are incident on a zinc surface. The lines which can emit photoelectrons from the surface are members of

  1. Balmer series
  2. Paschen series
  3. Lyman series
  4. Neither Balmer, nor Paschen nor Lyman series
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Q16.2025 Main 55/4/1 · Q91 mark

Choose the correct statement :

  1. Photons of light show diffraction whereas electrons do not show diffraction.
  2. Electrons have momentum whereas photons do not have momentum.
  3. Photons of light and electrons both exhibit dual nature.
  4. All electromagnetic radiations do not have photons.
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Q17.2025 Main 55/4/1 · Q101 mark

A beam of red light and a beam of blue light have equal intensities. Which of the following statements is true ?

  1. The blue beam has more number of photons than the red beam.
  2. The red beam has more number of photons than the blue beam.
  3. Wavelength of red light is lesser than wavelength of blue light.
  4. The blue light beam has lesser energy per photon than that in the red light beam.
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Q18.2025 Main 55/4/1 · Q161 mark

Assertion (A) : For monochromatic incident radiation, the emitted photoelectrons from a given metal have speed ranging from zero to a certain maximum value.

Reason (R) : Each metal has a definite work function.

  1. Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of the Assertion (A).
  2. Both Assertion (A) and Reason (R) are true, but Reason (R) is not the correct explanation of the Assertion (A).
  3. Assertion (A) is true, but Reason (R) is false.
  4. Both Assertion (A) and Reason (R) are false.
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Q19.2025 Main 55/5/1 · Q121 mark

The kinetic energy of an alpha particle is four times the kinetic energy of a proton. The ratio (λαλp) of de Broglie wavelengths associated with them will be :

  1. 116
  2. 18
  3. 14
  4. 12
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Q20.2025 Main 55/6/1 · Q101 mark

The work function of a material is 2·21 eV. Which of the following cannot produce photoelectrons from it ?

  1. Red light
  2. Blue light
  3. Violet light
  4. Green light
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Q21.2025 Main 55/6/1 · Q111 mark

The momentum (in kg m/s) of a photon of frequency 6·0 × 1014 Hz is :

  1. 6·63 × 10–25
  2. 1·326 × 10–27
  3. 2·652 × 10–26
  4. 3·978 × 10–24
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Q22.2025 Main 55/7/1 · Q81 mark

The frequency of a photon of energy 1·326 eV is :

  1. 1·18 × 1014 Hz
  2. 3·20 × 1014 Hz
  3. 4·20 × 1015 Hz
  4. 4·80 × 1015 Hz
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Q23.2025 Main 55/7/1 · Q161 mark

Assertion (A) : The minimum negative potential applied to the anode in a photoelectric experiment at which photoelectric current becomes zero, is called cut-off voltage.

Reason (R) : The threshold frequency for a metal is the minimum frequency of incident radiation below which emission of photoelectrons does not take place.

  1. Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of the Assertion (A).
  2. Both Assertion (A) and Reason (R) are true, but Reason (R) is not the correct explanation of the Assertion (A).
  3. Assertion (A) is true, but Reason (R) is false.
  4. Both Assertion (A) and Reason (R) are false.
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The official marking scheme for this paper is not available in the source folder.

Q24.2025 Supp 55/S/1 · Q41 mark

The figure shows the variation of maximum kinetic energy (Kmax) of emitted electrons as a function of the frequency v of radiation incident on a photosensitive surface. The slope of the curve is :

Graph of K_max versus frequency v, a straight line starting from the frequency axis
  1. h
  2. hv
  3. he
  4. he
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Official CBSE Marking Scheme · CBSE Board 2025 (Supplementary), Set 55/S/1

(A) h

Q25.2025 Supp 55/S/1 · Q151 mark

Assertion (A) : In process of photoelectric emission using a monochromatic light, all emitted electrons do not have the same kinetic energy.

Reason (R) : If radiation falling on a photosensitive surface consists of different wavelengths, the energy of emitted photoelectrons by absorbing photons of different wavelengths, shall be different.

  1. Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of the Assertion (A).
  2. Both Assertion (A) and Reason (R) are true, but Reason (R) is not the correct explanation of the Assertion (A).
  3. Assertion (A) is true, but Reason (R) is false.
  4. Both Assertion (A) and Reason (R) are false.
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Official CBSE Marking Scheme · CBSE Board 2025 (Supplementary), Set 55/S/1

(B) Both Assertion (A) and Reason (R) are true, but Reason (R) is not the correct explanation of the Assertion (A).

Q26.2024 Main 55/1/1 · Q111 mark

Two beams, A and B whose photon energies are 3·3 eV and 11·3 eV respectively, illuminate a metallic surface (work function 2·3 eV) successively. The ratio of maximum speed of electrons emitted due to beam A to that due to beam B is :

  1. 3
  2. 9
  3. 13
  4. 19
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Official CBSE Marking Scheme · CBSE Board 2024 (Main), Set 55/1/1

(C) 13

Q27.2024 Main 55/1/1 · Q121 mark

The waves associated with a moving electron and a moving proton have the same wavelength λ. It implies that they have the same :

  1. momentum
  2. angular momentum
  3. speed
  4. energy
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Official CBSE Marking Scheme · CBSE Board 2024 (Main), Set 55/1/1

(A) momentum

Q28.2024 Main 55/1/1 · Q131 mark

Assertion (A) : In photoelectric effect, the kinetic energy of the emitted photoelectrons increases with increase in the intensity of the incident light.

Reason (R) : Photoelectric current depends on the wavelength of the incident light.

  1. Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of the Assertion (A).
  2. Both Assertion (A) and Reason (R) are true, but Reason (R) is not the correct explanation of the Assertion (A).
  3. Assertion (A) is true, but Reason (R) is false.
  4. Assertion (A) is false and Reason (R) is also false.
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Official CBSE Marking Scheme · CBSE Board 2024 (Main), Set 55/1/1

(D) Assertion (A) is false and reason (R) is also false.

Q29.2024 Main 55/2/1 · Q81 mark

The work function for a photosensitive surface is 3·315 eV. The cut-off wavelength for photoemission of electrons from this surface is :

  1. 150 nm
  2. 200 nm
  3. 375 nm
  4. 500 nm
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Official CBSE Marking Scheme · CBSE Board 2024 (Main), Set 55/2/1

(C) 375 nm

Q30.2024 Main 55/2/1 · Q131 mark

Two statements are given — one labelled Assertion (A) and the other labelled Reason (R). Select the correct answer from the codes (A), (B), (C) and (D) as given below.

Assertion (A) : Photoelectric current increases with an increase in intensity of incident radiation, for a given frequency of incident radiation and the accelerating potential.

Reason (R) : Increase in the intensity of incident radiation results in an increase in the number of photoelectrons emitted per second and hence an increase in the photocurrent.

  1. Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of the Assertion (A).
  2. Both Assertion (A) and Reason (R) are true, but Reason (R) is not the correct explanation of the Assertion (A).
  3. Assertion (A) is true, but Reason (R) is false.
  4. Assertion (A) is false and Reason (R) is also false.
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Official CBSE Marking Scheme · CBSE Board 2024 (Main), Set 55/2/1

(A) Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of the Assertion (A)

Q31.2024 Main 55/3/1 · Q101 mark

Which one of the following is the correct graph between the maximum kinetic energy (Km) of the emitted photoelectrons and the frequency of incident radiation (ν) for a given photosensitive surface ?

  1. Graph A: K_m decreasing linearly with frequency from a positive intercept
  2. Graph B: K_m increasing linearly from a positive intercept on the K_m axis
  3. Graph C: K_m zero up to a threshold frequency, then rising linearly
  4. Graph D: K_m increasing linearly from the origin
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Official CBSE Marking Scheme · CBSE Board 2024 (Main), Set 55/3/1

(C)

Graph C: K_m zero up to a threshold frequency, then rising linearly
Q32.2024 Main 55/3/1 · Q141 mark

Assertion (A) : Photoelectric effect demonstrates the particle nature of light.

Reason (R) : Photoelectric current is proportional to frequency of incident radiation.

  1. Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of the Assertion (A).
  2. Both Assertion (A) and Reason (R) are true, but Reason (R) is not the correct explanation of the Assertion (A).
  3. Assertion (A) is true, but Reason (R) is false.
  4. Assertion (A) is false and Reason (R) is also false.
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Official CBSE Marking Scheme · CBSE Board 2024 (Main), Set 55/3/1

(C) Assertion (A) is true but Reason (R) is false.

Q33.2024 Main 55/4/1 · Q31 mark

The quantum nature of light explains the observations on photoelectric effect as –

  1. there is a minimum frequency of incident radiation below which no electrons are emitted.
  2. the maximum kinetic energy of photoelectrons depends only on the frequency of incident radiation.
  3. when the metal surface is illuminated, electrons are ejected from the surface after sometime.
  4. the photoelectric current is independent of the intensity of incident radiation.
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Official CBSE Marking Scheme · CBSE Board 2024 (Main), Set 55/4/1

(A) There is a minimum frequency of incident radiation below which no electrons are emitted.

Q34.2024 Main 55/4/1 · Q131 mark

Assertion (A) : Electrons are ejected from the surface of zinc when it is irradiated by yellow light.

Reason (R) : Energy associated with a photon of yellow light is more than the work function of zinc.

  1. If both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of Assertion (A).
  2. If both Assertion (A) and Reason (R) are true and Reason (R) is not the correct explanation of Assertion (A).
  3. If Assertion (A) is true and Reason (R) is false.
  4. If both Assertion (A) and Reason (R) are false.
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Official CBSE Marking Scheme · CBSE Board 2024 (Main), Set 55/4/1

(D) Both Assertion (A) and Reason (R) are false.

Q35.2024 Main 55/5/1 · Q81 mark

The variation of the stopping potential (V0) with the frequency (υ) of the incident radiation for four metals A, B, C and D is shown in the figure. For the same frequency of incident radiation producing photo-electrons in all metals, the kinetic energy of photo-electrons will be maximum for metal

Graph of stopping potential V0 versus frequency υ: four parallel straight lines A, B, C, D with A leftmost and D rightmost
  1. A
  2. B
  3. C
  4. D
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Official CBSE Marking Scheme · CBSE Board 2024 (Main), Set 55/5/1

(A) A

Q36.2023 Main 55/1/1 · Q181 mark

Assertion (A) : Photoelectric effect demonstrates the particle nature of light.

Reason (R) : Photoelectric current is proportional to intensity of incident radiation for frequencies more than the threshold frequency.

  1. Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of the Assertion (A).
  2. Both Assertion (A) and Reason (R) are true, but Reason (R) is not the correct explanation of the Assertion (A).
  3. Assertion (A) is true, but Reason (R) is false.
  4. Assertion (A) is false and Reason (R) is also false.
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Official CBSE Marking Scheme · CBSE Board 2023 (Main), Set 55/1/1

(B) Both Assertion (A) and Reason (R) are true but Reason (R) is not the correct explanation of the Assertion (A).

Q37.2023 Main 55/2/1 · Q101 mark

Which one of the following metals does not exhibit emission of electrons from its surface when irradiated by visible light ?

  1. Rubidium
  2. Sodium
  3. Cadmium
  4. Caesium
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Official CBSE Marking Scheme · CBSE Board 2023 (Main), Set 55/2/1

(C) Cadmium

Q38.2023 Main 55/3/1 · Q101 mark

A graph is plotted between the stopping potential (on y-axis) and the frequency of incident radiation (on x-axis) for a metal. The product of the slope of the straight line obtained and the magnitude of charge on an electron is equal to :

  1. h
  2. hc
  3. 2hc
  4. h2c
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Official CBSE Marking Scheme · CBSE Board 2023 (Main), Set 55/3/1

(A) h

Q39.2023 Main 55/3/1 · Q111 mark

Light of frequency 6·4 × 1014 Hz is incident on a metal of work function 2·14 eV. The maximum kinetic energy of the emitted electrons is about :

  1. 0·25 eV
  2. 0·51 eV
  3. 1·02 eV
  4. 0·10 eV
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Official CBSE Marking Scheme · CBSE Board 2023 (Main), Set 55/3/1

(B) 0.51 eV

Q40.2023 Main 55/4/1 · Q71 mark

Photons of energy 3·2 eV are incident on a photosensitive surface. If the stopping potential for the emitted electrons is 1·5 V, the work function for the surface is :

  1. 1·5 eV
  2. 1·7 eV
  3. 3·2 eV
  4. 4·7 eV
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Official CBSE Marking Scheme · CBSE Board 2023 (Main), Set 55/4/1

(B) 1.7 eV

Q41.2023 Main 55/4/1 · Q121 mark

A proton and an alpha particle have the same kinetic energy. The ratio of de Broglie wavelengths associated with the proton to that with the alpha particle is :

  1. 1
  2. 2
  3. 2 2
  4. 12
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Official CBSE Marking Scheme · CBSE Board 2023 (Main), Set 55/4/1

(B) 2

Q42.2023 Main 55/5/1 · Q71 mark

The energy of a photon of wavelength λ is

  1. hc λ
  2. hc/λ
  3. λ/hc
  4. λh/c
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Official CBSE Marking Scheme · CBSE Board 2023 (Main), Set 55/5/1

(B) hc/λ

Q43.2023 Main 55/5/1 · Q111 mark

Which of the following graphs correctly represents the variation of a particle momentum with its associated de-Broglie wavelength ?

  1. Graph of momentum p versus wavelength λ, option A
  2. Graph of momentum p versus wavelength λ, option B
  3. Graph of momentum p versus wavelength λ, option C
  4. Graph of momentum p versus wavelength λ, option D
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Official CBSE Marking Scheme · CBSE Board 2023 (Main), Set 55/5/1

(D) Graph in which p decreases hyperbolically with increasing λ (p ∝ 1/λ)

Graph of p versus λ, a decreasing hyperbola
Q44.2023 Supp 55/C/1 · Q71 mark

Which of the following figures represents the variation of a particle’s momentum with the de Broglie wavelength associated with it ?

  1. P versus lambda: straight line through origin
  2. P versus lambda: decreasing hyperbola-like curve
  3. P versus lambda: straight line with negative slope
  4. P versus lambda: increasing convex curve
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Official CBSE Marking Scheme · CBSE Board 2023 (Supplementary), Set 55/C/1

(B)

P versus lambda: decreasing hyperbola-like curve
Q45.2023 Supp 55/C/1 · Q101 mark

Light of frequency 1·5 ν0 is incident on a photosensitive material of threshold frequency ν0. If the frequency of the incident radiation is kept constant and intensity is increased, the photo current will :

  1. increase
  2. decrease
  3. not change
  4. first decrease and then become zero
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Official CBSE Marking Scheme · CBSE Board 2023 (Supplementary), Set 55/C/1

(A) increase

Q46.2026 Main 55/1/1 · Q172 marks

In a photoelectric experiment, the emitter plate is irradiated with radiation of 200 nm. The photocurrent becomes zero when the collector plate potential is – 0.80 V. Calculate the work function (in eV) of the emitter.

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Official CBSE Marking Scheme · CBSE Board 2026 (Main), Set 55/1/1

hν = φ0 + KEmax [½]

hcλ = φ0 + eV0

φ0 = hcλ – eV0 [½]

φ0 = [6.63×10–34 × 3×1081.6×10–19 × 200×10–9 – 0.80] eV [½]

φ0 = 5.4 eV [½]

Q47.2026 Main 55/2/1 · Q182 marks

Find ratio (λa/λp) of the de Broglie wavelength λa and λp associated respectively with an alpha particle and a proton,

  1. if they are moving with the same kinetic energy.
  2. just after they are accelerated through the same potential difference.
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Official CBSE Marking Scheme · CBSE Board 2026 (Main), Set 55/2/1

(i) λ = h2mK [½]

λaλp = mpma = m4m = 12 [½]

(ii) λ = h2mqV [½]

λaλp = mpqpmaqa = m4m · e2e = 122 [½]

Q48.2026 Main 55/3/1 · Q202 marks

A particle of mass M at rest splits up into two particles of masses m1 and m2 having non-zero velocities. Calculate the ratio of the de Broglie wavelengths associated with the two particles.

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Official CBSE Marking Scheme · CBSE Board 2026 (Main), Set 55/3/1

According to law of conservation of momentum [½]

m1v1 − m2v2 = 0

m1v1 = m2v2

v1v2 = m2m1 ------------(1) [½]

λ1λ2 = hm1v1 × m2v2h [½]

= m2v2m1v1

= m2m1 × m1m2 from (1)

= 1 [½]

λ1 : λ2 = 1 : 1

Q49.2026 Supp 55/7/1 · Q172 marks

Calculate the kinetic energy of an electron which has the de Broglie wavelength of 600 nm associated with it.

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Official CBSE Marking Scheme · CBSE Board 2026 (Supplementary), Set 55/7/1

Using de-Broglie wavelength, λ = h2mK [½]

K = h22mλ2 [½]

= (6.63 × 10-34)22 × 9.1 × 10-31 × (600 × 10-9)2 [½]

= 6.7 × 10-25 J [½]

Q50.2025 Main 55/2/1 · Q202 marks

A light of wavelength 400 nm is incident on metal surface whose work function is 3.0 × 10-19 J. Calculate the speed of the fastest photoelectrons emitted.

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Q51.2025 Main 55/5/1 · Q172 marks

The threshold frequency for a given metal is 3·6 × 1014 Hz. If monochromatic radiations of frequency 6·8 × 1014 Hz are incident on this metal, find the cut-off potential for the photoelectrons.

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The official marking scheme for this paper is not available in the source folder.

Q52.2025 Main 55/6/1 · Q192 marks

A laser beam of frequency 3·0 × 1014 Hz produces average power of 9 mW. Find (i) the energy of photon of the beam, and (ii) the number of photons emitted per second on an average by the source.

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The official marking scheme for this paper is not available in the source folder.

Q53.2025 Supp 55/S/1 · Q172 marks

The work function of Caesium is 2·14 eV.

(a) Find the threshold frequency for Caesium.

(b) Find the wavelength of the incident light if the photocurrent is brought to zero by a stopping potential of 0·86 V.

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Official CBSE Marking Scheme · CBSE Board 2025 (Supplementary), Set 55/S/1

(a) v0 = φ0h [½]

= 2.14 × 1.6 × 10–196.63 × 10–34 [½]

= 5.16 × 1014 Hz

(b) hcλ = Kmax + φ0 [½]

λ = hcKmax + φ0

λ = 6.63 × 10–34 × 3 × 108(2.14 + 0.86) × 1.6 × 10–19 [½]

λ = 4.14 × 10–7 m [½]

Q54.2024 Main 55/1/1 · Q202 marks

Consider a neutron (mass m) of kinetic energy E and a photon of the same energy. Let λn and λp be the de Broglie wavelength of neutron and the wavelength of photon respectively. Obtain an expression for λnλp.

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Official CBSE Marking Scheme · CBSE Board 2024 (Main), Set 55/1/1

E = hcλp ⇒ λp = hcE [½]

λn = hp = h(2mE) [½]

λnλp = h(2mE) × Ehc [½]

λnλp = E2mc2 [½]

Q55.2024 Main 55/3/1 · Q202 marks

The minimum intensity of white light that our eyes can perceive is about 0·1 nWm–2. Calculate the number of photons of this light entering our pupil (area 0·4 cm2) per second.

(Take average wavelength of white light = 500 nm and Planck’s constant = 6·6 × 10–34 Js)

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Official CBSE Marking Scheme · CBSE Board 2024 (Main), Set 55/3/1

Total energy gained per second from photon = IA

E = Nhν [½]

IA = N × hcλ

N = [IA]λhc

N = [0.1×10–9 × 0.4×10–4] × 500×10–96.6×10–34 × 3×108 [1]

N = 1.01 × 104 [½]

Q56.2024 Main 55/4/1 · Q182 marks

A proton and α-particle are accelerated through different potentials V1 and V2 respectively so that they have the same de Broglie wavelengths. Find V1V2.

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Official CBSE Marking Scheme · CBSE Board 2024 (Main), Set 55/4/1

De Broglie wavelength of a proton

λp = h2mpqpV1 = h2meV1 [½]

De Broglie wavelength of an α particle

λα = h2mαqαV2 = h2(4m)(2e)V2 [½]

λp = λα

h2meV1 = h16meV2 [½]

V1V2 = 8 [½]

Q57.2023 Main 55/4/1 · Q232 marks

The figure shows vm2 versus 1λ graph for photoelectrons emitted from a surface where vm is the maximum speed of electrons and λ is the wavelength of incident radiation. Using this graph and Einstein’s photoelectric equation, obtain the expression for Planck’s constant and work function of the surface.

Straight line graph of v_m squared versus 1/lambda with positive slope and negative intercept on the v_m squared axis
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Official CBSE Marking Scheme · CBSE Board 2023 (Main), Set 55/4/1

12 m vm2 = hcλ − ϕ0

vm2 = (2hcm) 1λ − 2m ϕ0 [½]

According to this equation a plot of vm2 versus (1/λ) is a straight line.

Slope of the graph = 2hcm

Intercept = 2m ϕ0 [½]

Slope and intercept can be found from the graph

h = m2c × slope [½]

ϕ0 = m2 × intercept [½]

Q58.2023 Main 55/5/1 · Q20 (option 1)2 marks

How will the De Broglie wavelength associated with an electron be affected when the (i) velocity of the electron decreases ? and (ii) accelerating potential is increased ? Justify your answer.

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Official CBSE Marking Scheme · CBSE Board 2023 (Main), Set 55/5/1

(i) λ = h/mv [½]

With decrease in velocity, de-Broglie wavelength increases [½]

(ii) λ = h2meV [½]

With increase in accelerating potential, de-Broglie wavelength decreases [½]

Q59.2023 Main 55/5/1 · Q20 (OR option)2 marks

How would the stopping potential for a given photosensitive surface change if (i) the frequency of the incident radiation were increased ? and (ii) the intensity of incident radiation were decreased ? Justify your answer.

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Official CBSE Marking Scheme · CBSE Board 2023 (Main), Set 55/5/1

(i) With increase in frequency stopping potential will increase. [½]

V0 = heν – φ0e [½]

(ii) Stopping potential remains same. [½]

Energy of photoelectron ejected is independent of intensity of radiation [½]

Q60.2023 Supp 55/C/1 · Q212 marks

Light of wavelength 3500 Å is incident on two metals A and B. Which of them will yield photoelectrons, if their work functions are 4·2 eV and 1·9 eV respectively ? Make the necessary calculations to justify your answer.

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Official CBSE Marking Scheme · CBSE Board 2023 (Supplementary), Set 55/C/1

E = hcλ [½]

= 6.63×10−34 × 3×1083500×10−10 × 1.6×10−19 eV [½]

E = 3.55 eV

Metal B [½]

Since energy of incident light is more than the work function of metal ‘B’. [½]

(Note: Give full credit of one mark if student writes ‘Metal B’ only.)

Q61.2023 Supp 55/C/1 · Q242 marks

The figure shows the variation of stopping potential V0 with frequencies of incident monochromatic beam for a metal. Such a graph was first obtained by R.A. Millikan in 1916 for sodium.

Graph of stopping potential V0 versus frequency ν, straight line with positive intercept on ν axis

Explain how, using Einstein’s photoelectric equation and the graph, you can obtain the value of (i) Planck’s constant, and (ii) work function of the metal, given the value of charge on electron.

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Official CBSE Marking Scheme · CBSE Board 2023 (Supplementary), Set 55/C/1

According to Einstein’s equation

hν = Φo + eVo

eVo = hν − Φo [½]

Vo = heν − Φoe [½]

This is equation of straight line as shown in figure, where slope is he.

So from the value of slope from the graph and value of charge on electron value of Planck’s constant can be calculated. [½]

Extrapolating the graph we get y-intercept Φoe. From this we can find work function given value of e. [½]

Q62.2022 Main 55/1/1 · Q2 (OR option)2 marks

Plot suitable graphs to show the variation of photoelectric current with the collector plate potential for the incident radiation of

  1. the same intensity but different frequencies ν1, ν2 and ν3 (ν1 < ν2 < ν3)
  2. the same frequency but different intensities I1, I2 and I3 (I1 < I2 < I3)
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Official CBSE Marking Scheme · CBSE Board 2022 (Main), Set 55/1/1

(i) Same intensity different frequency [1]

(ii) Same frequency different intensity [1]

Photoelectric current vs collector plate potential: (i) different frequencies, (ii) different intensities
Q63.2022 Main 55/2/1 · Q2 (OR option)2 marks

Photoelectric emission occurs when a surface is irradiated with the radiation of frequency (i) v1, and (ii) v2. The maximum kinetic energy of the electrons emitted in the two cases are K and 2K respectively. Obtain the expression for the threshold frequency for the surface.

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Official CBSE Marking Scheme · CBSE Board 2022 (Main), Set 55/2/1

K = hν1 – φ0 and 2K = hν2 – φ0 [½ + ½]

⇒ 2(hν1 – φ0) = hν2 – φ0

⇒ 2hν1 – 2φ0 = hν2 – φ0 [½]

⇒ h(2ν1 – ν2) = φ0 = hν0

⇒ (2ν1 – ν2) = φ0 = ν0

⇒ ν0 = 2ν1 – ν2 [½]

Q64.2022 Main 55/3/1 · Q2 (OR option)2 marks
  1. Name the factors on which photoelectric emission from a surface depends.
  2. Define the term ‘threshold frequency’ for a photosensitive material.
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Official CBSE Marking Scheme · CBSE Board 2022 (Main), Set 55/3/1

(i) Factors

(a) Frequency of incident radiation [½]

(b) Work function of the surface [½]

(ii) The minimum frequency of the incident radiation below which photoelectric emission does not take place. [1]

Q65.2022 Main 55/4/1 · Q2 (OR option)2 marks

What are matter waves ? A proton and an alpha particle are accelerated through the same potential difference. Find the ratio of the de Broglie wavelength associated with the proton to that with the alpha particle.

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Official CBSE Marking Scheme · CBSE Board 2022 (Main), Set 55/4/1

Wave associated with moving material particle [½]

λ = h2qmV [½]

∴ λpλα = qαmαqpmp [½]

∴ λpλα = 2e · 4me · m

= 221 [½]

Q66.2022 Main 55/5/1 · Q2 (OR option)2 marks

Why it is the frequency and not the intensity of light source that determines whether emission of photoelectrons will occur or not ? Explain.

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Official CBSE Marking Scheme · CBSE Board 2022 (Main), Set 55/5/1

Explanation [2]

According to the photon picture of light the emission of photoelectrons depend on the energy of photon incident on the metal surface which is determined by the frequency not by the intensity.

Q67.SQP 2025-26 · Q21 (OR option)2 marks

Plot a graph showing variation of de-Broglie wavelength λ versus 1V, where V is accelerating potential for a particle of mass m and charge q. Obtain the slope of this graph.

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Official CBSE Marking Scheme · CBSE Sample Question Paper 2025-26
Straight line through the origin: de Broglie wavelength versus 1/√V

[1]

λ = hmv

λ = h2mqV, comparing this equation with y = mx

slope = h2mq [1]

Q68.2026 Main 55/4/1 · Q263 marks
  1. Explain how the dual aspect of matter is evident in the de Broglie relation.
  2. Radiation of wavelength λ is incident on a photosensitive surface. Find the de Broglie wavelength of electrons emitted from the surface. Assume that the work function of the surface is negligible.
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Official CBSE Marking Scheme · CBSE Board 2026 (Main), Set 55/4/1

(a) According to the de Broglie relation, wavelength λ associated with a particle of momentum p is given as : λ = hp = hmv [½]

λ is the characteristic of a wave whereas momentum p is a typical characteristic of a particle. The de Broglie relation connects the two characteristics. [½]

(b) Kinetic energy of the electron = 12mv2

12mv2 = hcλ

v = 2hcmλ [½]

De Broglie wavelength of the electron (λ′) = hmv [½]

λ′ = hm2hcmλ

λ′ = hλ2mc [1]

Q69.2026 Main 55/5/1 · Q273 marks

Photoemission of electrons occurs from a metal (φ0 = 1·96 eV) when light of frequency 6·4 × 1014 Hz is incident on it. Calculate :

  1. Energy of a photon in the incident light,
  2. The maximum kinetic energy of the emitted electrons, and
  3. The stopping potential.
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Official CBSE Marking Scheme · CBSE Board 2026 (Main), Set 55/5/1

Calculating (a) Energy of a photon in the incident light [1]; (b) Maximum kinetic energy emitted [1]; (c) Stopping potential [1]

(a) E = hν [½]

= 6.626 × 10–34 × 6.4 × 10141.6 × 10–19 [½]

= 2.64 eV or 42.24 × 10–20 J [½]

(b) Kmax = E – φ0 [½]

= 2.64 – 1.96 = 0.68 eV or (10.88 × 10–20 J) [½]

(c) eV0 = Kmax [½]

V0 = Kmax/e = 0.68 V

Q70.2025 Main 55/1/1 · Q263 marks

Answer the following giving reason :

(a) All the photo electrons do not eject with the same kinetic energy when monochromatic light is incident on a metal surface.

(b) The saturation current in case (a) is different for different intensity.

(c) If one goes on increasing the wavelength of light incident on a metal surface, keeping its intensity constant, emission of photo electrons stop at a certain wavelength for this metal.

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Q71.2025 Main 55/4/1 · Q273 marks

Explain the following observations using Einstein’s photoelectric equation :

  1. Photoelectric emission does not occur from a surface when the frequency of the light incident on it is less than a certain minimum value.
  2. It is the frequency, and not the intensity of the incident light which affects the maximum kinetic energy of the photoelectrons.
  3. The cut-off voltage (Vo) versus frequency (ν) of the incident light curve is a straight line with a slope he.
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Q72.2025 Main 55/7/1 · Q263 marks

(a) Define ‘work function’ of a metal. How can its value be determined from a graph between stopping potential and frequency of the incident radiation ?

(b) The work function of a metal is 2·4 eV. A stopping potential of 0·6 V is required to reduce the photocurrent to zero, in a photoelectric experiment. Calculate the wavelength of light used.

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Q73.2024 Main 55/2/1 · Q273 marks

de Broglie wavelength λ as a function of 1K, for two particles of masses m1 and m2 are shown in the figure. Here, K is the energy of the moving particles.

Graph of de Broglie wavelength versus 1/sqrt(K): two straight lines for masses m1 and m2

(a) What does the slope of a line represent ?
(b) Which of the two particles is heavier ?
(c) Is this graph also valid for a photon ?
Justify your answer in each case.

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Official CBSE Marking Scheme · CBSE Board 2024 (Main), Set 55/2/1

(a) λ = h2mK = h2m × 1K [½]

slope = h2m [½]

(b) slope ∝ 1m [½]

Slope of m2 is more than that of m1. Therefore, m1 is heavier. [½]

(c) No [½]

Momentum (p) = 2mK is not valid for a photon [½]

Q74.2024 Main 55/5/1 · Q223 marks
  1. Write Einstein’s photoelectric equation. How did Millikan prove the validity of this equation ?
  2. Explain the existence of threshold frequency of incident radiation for photoelectric emission from a given surface.
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Official CBSE Marking Scheme · CBSE Board 2024 (Main), Set 55/5/1

(a) hν = hν0 + Kmax = hν0 + eV0 [1½]

By finding the value of Planck’s constant using V0 versus υ straight line plot for sodium.

(b) Since Kmax must be non-negative therefore photo-electric emission is possible only when hν > hν0, which implies the existence of ν0. [1½]

Q75.2024 Supp 55/S/1 · Q253 marks

In photoelectric effect experiment, show the variation of

(a) photocurrent with collector plate potential for a given surface for different intensities of incident radiation. Do the curves meet at any point ? If so, why ?

(b) photocurrent with intensity of radiation incident on a surface keeping the frequency and plate potential fixed.

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Official CBSE Marking Scheme · CBSE Board 2024 (Supplementary), Set 55/S/1

(a) Graph [1]

Photocurrent versus collector plate potential for three intensities I1 < I2 < I3

Yes, these curves meet at stopping potential [½]

For any surface, as the energy of photons does not depend upon intensity of light, at the stopping potential current reduces to zero. [½]

(b) Graph [1]

Photoelectric current versus intensity of light: straight line through the origin
Q76.2023 Main 55/1/1 · Q293 marks

How does Einstein’s photoelectric equation explain the emission of electrons from a metal surface ? Explain briefly.

Plot the variation of photocurrent with :

(a) collector plate potential for different intensity of incident radiation, and

(b) intensity of incident radiation.

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Official CBSE Marking Scheme · CBSE Board 2023 (Main), Set 55/1/1

Explanation of emission of electron [1]; a) variation of photocurrent with collector plate potential for different intensity [1]; b) variation of photocurrent with intensity of incident radiation [1]

According to Einstein’s photoelectric equation, an electron absorbs a quantum of energy ‘hν’ of incident radiation. If the energy of absorbed quantum exceeds the minimum energy needed by the electron to escape from the metal surface (work function φ0), the electron is emitted. [1]

Kmax = hν − φ0

a)

Photocurrent versus collector plate potential for intensities I1 < I2 < I3, with stopping potential -V0

[1]

b)

Photoelectric current increasing linearly with intensity of light

[1]

Q77.2023 Main 55/3/1 · Q293 marks

Calculate the wavelength of de Broglie waves associated with a proton having (5001·673) eV energy. How will the wavelength be affected for an alpha particle having the same energy ?

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Official CBSE Marking Scheme · CBSE Board 2023 (Main), Set 55/3/1

λ = h2mK [½]

K = (5001.673) eV = 500 × 1.6 × 10-191.673 J [½]

λp = 6.63 × 10-342 × 1.673 × 10-27 × (500 × 1.6 × 10-191.673) [½ + ½]

= 1.65 × 10-12 m [½]

λ ∝ 1m [½]

For α particle ; mα = 4mp ⇒ λα = λp2 [½]

Q78.2022 Main 55/1/1 · Q93 marks

Photoelectrons are emitted from a metal surface when illuminated with UV light of wavelength 330 nm. The minimum amount of energy required to emit the electrons from the surface is 3.5 × 10−19 J. Calculate :

  1. the energy of the incident radiation, and
  2. the kinetic energy of the photoelectron.
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Official CBSE Marking Scheme · CBSE Board 2022 (Main), Set 55/1/1

i) Energy of incident radiation

E = hν = hcλ [½]

= 6·63×10−34 × 3×108330×10−9 [½]

= 6·027×10−19 J [½]

ii) Kinetic energy of photoelectron

K.E. = E − φ0 [½]

= (6·027×10−19 − 3·5×10−19) J [½]

= 2·527×10−19 J [½]

Q79.2022 Main 55/2/1 · Q103 marks

Find the ratio of the de Broglie wavelengths associated with an alpha particle and a proton, if both

  1. have the same speeds,
  2. have the same kinetic energy,
  3. are accelerated through the same potential difference.
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Official CBSE Marking Scheme · CBSE Board 2022 (Main), Set 55/2/1

(a) λ = hp [½]

λαλp = hmαvα × mpvph = 14 [½]

(b) p = 2m(K.E.) [½]

λαλp = h2mα(K.E.)α × 2mp(K.E.)ph = mpmα = 12 [½]

(c) v = 2qVm [½]

λαλp = hmαvα × mpvph = mpmα 2qpVmp × mα2qαV

= mpmα × mαmp × qpqα = 122 [½]

Q80.2022 Main 55/3/1 · Q93 marks

An electron is accelerated from rest through a potential difference of 100 V. Find :

  1. the wavelength associated with
  2. the momentum of and
  3. the velocity required by, the electron.
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Official CBSE Marking Scheme · CBSE Board 2022 (Main), Set 55/3/1

(i) λ = 1·23V nm = 1.23100 [½]

= 0·123 nm [½]

(ii) p = hλ [½]

= 6·63 × 10-340·123 × 10-9

= 5.4 × 10-24 kg m s-1 [½]

(iii) v = pm [½]

v = 5·4 × 10-249·1 × 10-31

= 5·9 × 106 m/s [½]

Alternative method

v = 2eVm

v = 2 × 1.6 × 10-19 × 1009.1 × 10-31

= 5.9 × 106 m/s

Q81.2022 Main 55/4/1 · Q93 marks

(a) Use Einstein’s photoelectric equation to depict the variation of the maximum kinetic energy (Ek) of electrons emitted, with the frequency (ν) of the incident radiation.

(b) A photosensitive surface is illuminated with a beam of (i) yellow light, and (ii) red light, both of the same intensity.

In which case will

  1. photoelectrons have more Ek ?
  2. more numbers of electrons be emitted ?

Justify your answer in each case.

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Official CBSE Marking Scheme · CBSE Board 2022 (Main), Set 55/4/1

(a) Graph of maximum kinetic energy versus frequency [1]

Graph of maximum kinetic energy Ek against frequency of incident radiation, straight line with intercept −φ0 and threshold frequency ν0

(b) (I) EK = hν − φ0 [½]

as νY > νR, (EK)Y > (EK)R [½]

For a given photosensitive surface, K.E. of photo-electrons will be more for yellow light.

(II) Since the number of photons incident on unit area per unit time for same intensity of two lights will be more in Red light (NR > NY) hence more number of photoelectrons will be ejected for Red light. [1]

Q82.2022 Main 55/5/1 · Q93 marks
  1. Calculate the energy and momentum of a photon in a monochromatic beam of wavelength 331.5 nm.
  2. How fast should a hydrogen atom travel in order to have the same momentum as that of the photon in part (a) ?
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Official CBSE Marking Scheme · CBSE Board 2022 (Main), Set 55/5/1

Calculation of energy and momentum [1 + 1]; Calculation of speed [1]

(a) Energy of photon, E = hν [½]

E = hcλ = 6.63 × 10–34 × 3 × 108331.5 × 10–9

E = 6 × 10–19 J [½]

momentum (p) = hλ [½]

= 6.63 × 10–34331.5 × 10–9

p = 2 × 10–27 kg m s–1 [½]

(b) Momentum of H atom = momentum of the photon = 2 × 10–27 kg m s–1

p = mu [½]

u = pm = 2 × 10–271.67 × 10–27 m s–1

u = 1.20 m s–1 [½]

Q83.2022 Supp 55/6/1 · Q73 marks
  1. A proton and an α particle are moving with the same speed. Which one of them has greater value of de Broglie wavelength associated with it ? Justify your answer. [1]
  2. Lights of wavelengths 430 nm, 450 nm and 660 nm are incident on a metal surface whose threshold wavelength is 600 nm. In which case/cases will photoemission take place and why ? Calculate the threshold frequency of the metal surface. [2]
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Official CBSE Marking Scheme · CBSE Board 2022 (Supplementary), Set 55/6/1

(a) de Broglie wavelength, λ = hmv

λαλp = mpmα [½]

As, mα > mp ⇒ λp > λα [½]

(b) For photoelectric emission wavelength of radiation must be lesser than the threshold wavelength. [½]

Thus lights of wavelength 430 nm and 450 nm can cause photoelectric emission [½]

Threshold frequency (νo) = cλo = 3 × 108600 × 10-9 = 5 × 1014 Hz [½ + ½]

Q84.2026 Main 55/1/1 · Q30 (option 1)4 marks

A researcher performs an experiment on photo-electric effect using two metals A and B with unknown work functions. She illuminates the surfaces of A and B with monochromatic radiation of various frequencies and records the value of corresponding stopping potentials (Vs). The graph shows the variation of stopping potential (Vs) with the frequency of incident radiation (ν) for metals A and B.

Stopping potential Vs versus frequency ν: two parallel straight lines A and B with threshold frequencies ν1 and ν2 and intercepts V1 and V2

Answer the following questions :

(I) From the graph, the work functions of A and B are (h is Planck’s constant and e value of charge on an electron)

  1. ν1 and ν2
  2. V1 and V2
  3. hν1 and hν2
  4. hν1e and hν2e

(II) For radiation of frequency ν > ν2 incident on the surfaces of A and B, the maximum kinetic energy of ejected electron is

  1. greater for metal A because it has a smaller work function.
  2. greater for metal B because it has a larger work function.
  3. greater for metal B because it has higher threshold frequency.
  4. the same for both metal A and metal B because it is independent of work functions of metals.

(III) If the intensity of the incident radiation for both metals A and B, is doubled keeping its frequency constant, then

  1. the slope of the parallel lines will increase.
  2. the slope of the parallel lines will decrease.
  3. the threshold frequencies for both A and B will decrease.
  4. the slope of the parallel lines will not change but more electrons will be emitted per second.

(IV) The threshold frequency for a metal surface is ν0. If the radiation of frequency 3ν0 illuminates the surface, the maximum kinetic energy (KE) of photoelectrons is E1. If the frequency were increased to 6ν0, the maximum KE of the photoelectrons becomes E2. Then (E1E2) equals

  1. 1/3
  2. 1/2
  3. 2/5
  4. 3/4
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Official CBSE Marking Scheme · CBSE Board 2026 (Main), Set 55/1/1

(I) (C) hν1 and hν2 [1]

(II) (A) greater for metal A because it has a smaller work function. [1]

(III) (D) the slope of the parallel lines will not change but more electrons will be emitted per second. [1]

(IV) (C) 25 [1]

Q85.2026 Main 55/1/1 · Q30 (OR option)4 marks

A researcher performs an experiment on photo-electric effect using two metals A and B with unknown work functions. She illuminates the surfaces of A and B with monochromatic radiation of various frequencies and records the value of corresponding stopping potentials (Vs). The graph shows the variation of stopping potential (Vs) with the frequency of incident radiation (ν) for metals A and B.

Stopping potential Vs versus frequency ν: two parallel straight lines A and B with threshold frequencies ν1 and ν2 and intercepts V1 and V2

Answer the following questions :

(I) From the graph, the work functions of A and B are (h is Planck’s constant and e value of charge on an electron)

  1. ν1 and ν2
  2. V1 and V2
  3. hν1 and hν2
  4. hν1e and hν2e

(II) For radiation of frequency ν > ν2 incident on the surfaces of A and B, the maximum kinetic energy of ejected electron is

  1. greater for metal A because it has a smaller work function.
  2. greater for metal B because it has a larger work function.
  3. greater for metal B because it has higher threshold frequency.
  4. the same for both metal A and metal B because it is independent of work functions of metals.

(III) If the intensity of the incident radiation for both metals A and B, is doubled keeping its frequency constant, then

  1. the slope of the parallel lines will increase.
  2. the slope of the parallel lines will decrease.
  3. the threshold frequencies for both A and B will decrease.
  4. the slope of the parallel lines will not change but more electrons will be emitted per second.

(IV) Let m be the slope of the graph line for metal B. If e is the value of electron charge, then Planck’s constant ‘h’ is given by

  1. me
  2. 1me
  3. me
  4. em
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Official CBSE Marking Scheme · CBSE Board 2026 (Main), Set 55/1/1

(I) (C) hν1 and hν2 [1]

(II) (A) greater for metal A because it has a smaller work function. [1]

(III) (D) the slope of the parallel lines will not change but more electrons will be emitted per second. [1]

(IV) (A) me [1]

Q86.2026 Supp 55/7/1 · Q304 marks

When radiation of suitable frequency is incident on a metal surface, electrons are emitted from the surface. This phenomenon is called photoelectric effect. Einstein explained this phenomenon considering the particle nature of radiation and using the law of conservation of energy. Wave theory of light could not explain this phenomenon.

    1. A photon of 3·8 eV energy is incident on a surface of work function 1·8 eV. The maximum kinetic energy of the electron emitted from the surface will be :
      1. 3·2 × 10-19 J
      2. 1·6 × 10-19 J
      3. 1·25 × 10-19 J
      4. 2·4 × 10-18 J
    2. OR The work function of a photosensitive surface is 2·21 eV. The value of threshold frequency for the surface is :
      1. 4·8 × 1014 Hz
      2. 5·3 × 1014 Hz
      3. 1·9 × 1015 Hz
      4. 1·2 × 1015 Hz
  1. The saturation current for a given photosensitive material depends on :
    1. work function of the material
    2. wavelength of the incident radiation
    3. intensity of the incident radiation
    4. collector plate potential
  2. The stopping potential of photoelectrons is found to be 0·40 V when light of wavelength 500 nm is incident on a surface. The work function of the surface is close to :
    1. 1·20 eV
    2. 1·48 eV
    3. 2·08 eV
    4. 6·48 eV
  3. A graph between the frequency of incident radiation (ν) and the corresponding stopping potential (V0) for a metal is a straight line. Let α and β be the value of the slope and the value of the intercept on the V0-axis respectively. Then, according to Einstein’s photoelectric equation (βα) is :
    1. ν0
    2. 1ν0
    3. eν0
    4. ν0h
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Official CBSE Marking Scheme · CBSE Board 2026 (Supplementary), Set 55/7/1

(i) (a) (A) 3.2 × 10-19 J [1]

OR

(b) (B) 5.3 × 1014 Hz [1]

(ii) (C) intensity of the incident radiation. [1]

(iii) (C) 2.08 eV [1]

(iv) (A) ν0 [1]

Q87.2025 Main 55/2/1 · Q304 marks

When a photon of suitable frequency is incident on a metal surface, photoelectron is emitted from it. If the frequency is below a threshold frequency (ν0) for the surface, no photoelectron is emitted. For a photon of frequency ν(ν > ν0), the kinetic energy of the emitted photoelectrons is h(ν – ν0). The photocurrent can be stopped by applying a potential V0 called ‘stopping potential’ on the anode. Thus maximum kinetic energy of photoelectrons Km = eV0 = h(ν – ν0). The experimental graph between V0 and ν for a metal is shown in figure. This is a straight line of slope m.

Stopping potential V0 versus frequency ν: straight line of slope m crossing the ν-axis at ν0, with dashed extension to negative V0
  1. The straight line graphs obtained for two metals
    1. coincide each other.
    2. are parallel to each other.
    3. are not parallel to each other and cross at a point on ν-axis.
    4. are not parallel to each other and do not cross at a point on ν-axis.
  2. The value of Planck’s constant for this metal is
    1. em
    2. 1me
    3. me
    4. me

(iii) Alternative 1 – The intercepts on ν-axis and V0-axis of the graph are respectively :

  1. ν0, hν0e
  2. ν0, hν0
  3. hν0e, ν0
  4. hν0, ν0

(iii) Alternative 2 – When the wavelength of a photon is doubled, how many times its wave number and frequency become, respectively ?

  1. 2, 12
  2. 12, 12
  3. 12, 2
  4. 2, 2

(iv) The momentum of a photon is 5.0 × 10-29 kg. m/s. Ignoring relativistic effects (if any), the wavelength of the photon is

  1. 1.33 μm
  2. 3.3 μm
  3. 16.6 μm
  4. 13.3 μm
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Q88.2025 Main 55/5/1 · Q304 marks

Einstein explained photoelectric effect on the basis of Planck’s quantum theory, where light travels in the form of small bundles of energy called photons. The energy of each photon is hν, where ν is the frequency of incident light and h is Planck’s constant. The number of photons in a beam of light determines the intensity of the incident light. A photon incident on a metal surface transfers its total energy hν to a free electron in the metal. A part of this energy is used in ejecting the electron from the metal and is called its work function. The rest of the energy is carried by the ejected electron as its kinetic energy.

(i) Which of the following graphs shows the variation of photoelectric current I with the intensity of light ?

Four graphs (A)-(D) of photoelectric current I versus intensity: decreasing line, constant, increasing line through origin, saturating curve

(ii) When the frequency of the incident light is increased without changing its intensity, the saturation current :

  1. increases linearly
  2. decreases
  3. increases non-linearly
  4. remains the same

(iii) Which of the following graphs can be used to obtain the value of Planck’s constant ?

  1. Photocurrent versus Intensity of incident light
  2. Photocurrent versus Frequency of incident light
  3. Cut-off potential versus Frequency of incident light
  4. Cut-off potential versus Intensity of incident light

(iv)(a) Red light, yellow light and blue light of the same intensity are incident on a metal surface successively. KR, KY and KB represent the maximum kinetic energy of photoelectrons respectively, then :

  1. KR > KY > KB
  2. KY > KB > KR
  3. KB > KY > KR
  4. KR > KB > KY

(iv)(b) (alternative to (iv)(a)) Which of the following metals exhibits photoelectric effect with visible light ?

  1. Caesium
  2. Zinc
  3. Cadmium
  4. Magnesium
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Q89.2023 Main 55/2/1 · Q35 (option 1)4 marks

Figure shows the variation of photoelectric current measured in a photo cell circuit as a function of the potential difference between the plates of the photo cell when light beams A, B, C and D of different wavelengths are incident on the photo cell. Examine the given figure and answer the following questions :

Photoelectric current versus potential difference for light beams A, B, C and D
  1. Which light beam has the highest frequency and why ?
  2. Which light beam has the longest wavelength and why ?
  3. Which light beam ejects photoelectrons with maximum momentum and why ?
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Official CBSE Marking Scheme · CBSE Board 2023 (Main), Set 55/2/1

(i) The light beam B because it requires maximum retarding potential to reduce the photoelectric current to zero. [½ + ½]

(ii) The light beam C because it requires minimum retarding potential to reduce photoelectric current to zero. [½ + ½]

(iii) The light beam B ejects photoelectrons with maximum momentum, because highest frequency light beam ejects photoelectrons with highest kinetic energy and hence highest momentum. [1 + 1]

Q90.2023 Main 55/2/1 · Q35 (OR option)4 marks

What is the effect on threshold frequency and stopping potential on increasing the frequency of incident beam of light ? Justify your answer.

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Official CBSE Marking Scheme · CBSE Board 2023 (Main), Set 55/2/1

There is no effect on threshold frequency since it is characteristic of the metal. [½ + ½]

With increase in frequency of incident beam of light, stopping potential increases because to stop the photoelectrons of higher kinetic energy, larger retarding potential is required. [½ + ½]

Alternatively

Give full credit if a candidate explains the effect of frequency on stopping potential using the following formula.

eV0 = h(ν − ν0)

Q91.SQP 2025-26 · Q304 marks

Photoelectric effect is phenomenon of the ejection of electrons when the radiation of suitable frequency is made to fall on the surface of a metal. When light of suitable wavelength falls on the emitter C given in the diagram, the photoelectrons are emitted. These photoelectrons are drawn to the collector A. The photoelectric current of the order of a few microamperes can be normally obtained from the device given in figure. The device given converts a change in intensity of illumination into a change in photocurrent. This current can be used to operate control systems and in light measuring devices. The devices are made up of metals with low ionization enthalpies, for example platinum whose work function is 6.35 eV.

Photocell with emitter C, collector A, incident light, battery with sliding terminal B and microammeter

(I) If infrared radiation of 3 × 1011 Hz is used as incident radiation, determine the reading of microammeter? Justify mathematically. [2]

(II) In the given diagram, if terminal B is shifted towards the left then how will it affect the reading of the microammeter? [1]

(III) Plot a graph showing this variation in reading of micrometre on shifting the terminal B towards the right. [1]

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Official CBSE Marking Scheme · CBSE Sample Question Paper 2025-26

(I) If infrared radiation is used as incident radiation, determine the reading

Wo = hνo

Threshold frequency, νo = Woh = 6.35 × 1.6 × 10–196.63 × 10–34 = 1.5 × 1015 Hz [1]

Frequency of infrared radiation < threshold frequency (νo), hence no emission of photoelectrons will take place, therefore reading of the microammeter = 0 [1]

(II) Photoelectric current decreases with decrease in potential. At some stage, for a certain potential of plate A, all the emitted electrons are stopped by the plate A and the photoelectric current becomes zero. [1]

(III)

Photocurrent versus collector plate potential graph, rising and saturating

[1]

Q92.SQP 2026-27 · Q304 marks

In 1905, Albert Einstein explained the photoelectric effect by proposing that light consists of discrete packets of energy called photons. Each photon carries an energy E=hν. When a photon strikes a metal surface, it can transfer its energy to an electron. If the photon’s energy exceeds the work function of the metal, the electron can be emitted with maximum kinetic energy Kmax=hν-φ0. This discovery challenged the classical wave theory, which predicted that electron emission should depend only on the intensity of light. Instead, experiments showed that emission of electrons depends on frequency too, proving the particle nature of light.

I. Define the term work function. [1]

II. Why the increase in the intensity of light does not increase the maximum kinetic energy of photoelectrons? [1]

III. A metal surface has a work function of 2 eV. The light of frequency 1.2 ×1015Hz falls on it. Which statement is correct? [1]

  1. No electron is emitted because the intensity is low.
  2. Electrons are emitted with maximum kinetic energy less than 2 eV.
  3. Electrons are emitted with maximum kinetic energy greater than 0 eV.
  4. Electron emission depends only on the intensity of light and not on the frequency.

IV. Which observation/s directly contradicts the classical wave theory of light? [1]

  1. Photoelectric current increases with intensity
  2. The maximum kinetic energy depends only on the frequency and not on the intensity.
  3. The electrons are emitted instantaneously.
  4. Both B and C.
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Official CBSE Marking Scheme · CBSE Sample Question Paper 2026-27

I. The work function (φ0) of a metal is the minimum energy required to remove an electron from the surface of the metal so that it just escapes with zero kinetic energy. [1]

II. Increasing light intensity increases the number of photons, not the energy of each photon. Since the energy of a photon depends only on its frequency (E = hν), the maximum kinetic energy of emitted electrons depends only on frequency, not intensity. Higher intensity only increases the number of emitted electrons (photoelectric current). [1]

III. (C) Electrons are emitted with maximum kinetic energy greater than 0 eV.

Photon energy: E = hν ≈ 4.14 × 10−15 × 1.2 × 1015 ≈ 5 eV

Maximum kinetic energy: Kmax = E − φ0 = 5 − 2 = 3 eV [1]

IV. (D) (Both B and C)

  • B: Maximum kinetic energy depends only on frequency, not intensity
  • C: Electrons are emitted instantaneously

Both observations cannot be explained by classical wave theory and support the particle nature of light. [1]

Q93.2024 Supp 55/S/1 · Q31 (option 1)5 marks

(i) What are matter waves ? A particle of mass m and charge q is accelerated from rest through a potential difference V. Obtain an expression for de Broglie wavelength associated with the particle.

(ii) Monochromatic light of frequency 5.0 × 1014 Hz is produced by a source of power output 3.315 mW. Calculate :

(1) energy of the photon in the beam

(2) number of photons emitted per second by the source

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Official CBSE Marking Scheme · CBSE Board 2024 (Supplementary), Set 55/S/1

(i) Wave associated with a mass in motion is called matter wave. [1]

Particle of mass m and charge q gains energy in the form of kinetic energy.

12mv2 = qV [½]

(mv)2 = 2qVm [½]

mv = 2mqV [½]

According to de Broglie relation

λ = hmv [½]

λ = h2mqV [½]

(ii) (1) E = hν [½]

= 6.63×10–34 × 5×1014

= 3.315×10–19 J [½]

(2) n = P/E [½]

= 3.315×10–33.315×10–19 = 1016 s–1 [½]

Unit VIII · Atoms and Nuclei

12. Atoms

Q1.2026 Main 55/1/1 · Q81 mark

The ‘distance of closest approach’ of an alpha-particle is ‘d’ when it moves with a velocity ν head-on towards the target nucleus. If the velocity of alpha particle is halved, the new ‘distance of closest approach’ will be –

  1. d2
  2. 2d
  3. d4
  4. 4d
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Official CBSE Marking Scheme · CBSE Board 2026 (Main), Set 55/1/1

(D) 4d

Q2.2026 Main 55/1/1 · Q161 mark

Assertion (A) : In Bohr model of hydrogen atom, the energy levels are discrete and quantised.

Reason (R) : In a hydrogen atom, the electrostatic force on the electron provides the necessary centripetal force to it to revolve around the nucleus.

  1. Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of the Assertion (A).
  2. Both Assertion (A) and Reason (R) are true, but Reason (R) is not the correct explanation of the Assertion (A).
  3. Assertion (A) is true, but Reason (R) is false.
  4. Both Assertion (A) and Reason (R) are false.
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Official CBSE Marking Scheme · CBSE Board 2026 (Main), Set 55/1/1

(B) Both Assertion (A) and Reason (R) are true, but Reason (R) is not the correct explanation of the Assertion (A).

Q3.2026 Main 55/2/1 · Q41 mark

In Bohr model of hydrogen atom, for large values of n, the distance between the consecutive orbits is proportional to

  1. n
  2. n
  3. n2
  4. n3
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Official CBSE Marking Scheme · CBSE Board 2026 (Main), Set 55/2/1

(B) n

Q4.2026 Main 55/2/1 · Q121 mark

In Bohr model of hydrogen atom, the electron makes a transition from n = 5 to n = 1 state. As a result, a photon of wavelength λ is emitted in the process. The wavelength of the photon emitted when an electron makes a transition from energy level n = 5 to n = 2 will be

  1. 87 λ
  2. 247 λ
  3. 167 λ
  4. 327 λ
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Official CBSE Marking Scheme · CBSE Board 2026 (Main), Set 55/2/1

(D) 327 λ

Q5.2026 Main 55/4/1 · Q111 mark

Paschen series in spectrum of hydrogen atom lies in :

  1. infrared region
  2. ultraviolet region
  3. visible region
  4. partly in ultraviolet region and partly in visible region
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Official CBSE Marking Scheme · CBSE Board 2026 (Main), Set 55/4/1

(A) Infrared region

Q6.2026 Main 55/5/1 · Q101 mark

The energy of an electron in an orbit in hydrogen atom is – 3·4 eV. Its angular momentum in the orbit will be :

  1. 3h2π
  2. 2hπ
  3. hπ
  4. h2π
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Official CBSE Marking Scheme · CBSE Board 2026 (Main), Set 55/5/1

(C) hπ

Q7.2026 Supp 55/7/1 · Q81 mark

The energy of an electron in an orbit in a hydrogen atom (Bohr model) is – 3·4 eV. Its angular momentum in this orbit is :

  1. hπ
  2. h2π
  3. h4π
  4. 2hπ
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Official CBSE Marking Scheme · CBSE Board 2026 (Supplementary), Set 55/7/1

(A) hπ

Q8.2026 Supp 55/7/1 · Q141 mark

Assertion (A) : The ratio of the speeds of the electron in the second to that in the third orbit in Bohr’s hydrogen atom is 32.

Reason (R) : The speed of the electron in nth permitted orbit in hydrogen atom varies proportional to 1n i.e. vn ∝ 1n.

  1. Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of the Assertion (A).
  2. Both Assertion (A) and Reason (R) are true, but Reason (R) is not the correct explanation of the Assertion (A).
  3. Assertion (A) is true, but Reason (R) is false.
  4. Both Assertion (A) and Reason (R) are false.
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Official CBSE Marking Scheme · CBSE Board 2026 (Supplementary), Set 55/7/1

(A) Both Assertion (A) and Reason (R) are true, and Reason (R) is the correct explanation of Assertion (A).

Q9.2025 Main 55/1/1 · Q161 mark

Assertion (A) : The potential energy of an electron revolving in any stationary orbit in a hydrogen atom is positive.

Reason (R) : The total energy of a charged particle is always positive.

  1. Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of Assertion (A).
  2. Both Assertion (A) and Reason (R) are true, but Reason (R) is not the correct explanation of Assertion (A).
  3. Assertion (A) is true, but Reason (R) is false.
  4. Assertion (A) is false and Reason (R) is also false.
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Q10.2025 Main 55/2/1 · Q111 mark

The energy of an electron in a hydrogen atom in ground state is –13.6 eV. Its energy in an orbit corresponding to quantum number n is –0.544 eV. The value of n is

  1. 2
  2. 3
  3. 4
  4. 5
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Q11.2025 Main 55/2/1 · Q141 mark

For Questions 13 to 16, two statements are given – one labelled Assertion (A) and other labelled Reason (R). Select the correct answer to these questions from the codes (A), (B), (C) and (D) as given below :

Assertion (A) : In Bohr model of hydrogen atom, the angular momentum of an electron in nth orbit is proportional to the square root of its orbit radius rn.

Reason (R) : According to Bohr model, electron can jump to its nearest orbits only.

  1. If both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of Assertion (A).
  2. If both Assertion (A) and Reason (R) are true but Reason (R) is not the correct explanation of Assertion (A).
  3. If Assertion (A) is true but Reason (R) is false.
  4. If both Assertion (A) and Reason (R) are false.
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Q12.2025 Main 55/5/1 · Q151 mark

Assertion (A) : The Balmer series in hydrogen atom spectrum is formed when the electron jumps from higher energy state to the ground state.

Reason (R) : In Bohr’s model of hydrogen atom, the electron can jump between successive orbits only.

  1. Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of the Assertion (A).
  2. Both Assertion (A) and Reason (R) are true, but Reason (R) is not the correct explanation of the Assertion (A).
  3. Assertion (A) is true, but Reason (R) is false.
  4. Both Assertion (A) and Reason (R) are false.
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Q13.2025 Main 55/5/1 · Q161 mark

Assertion (A) : In Rutherford’s alpha particle scattering experiment, the presence of only few alpha particles at angle of scattering π led him to the discovery of nucleus.

Reason (R) : The size of nucleus is approximately 10–5 times the size of an atom and therefore only few alpha particles are rebounded.

  1. Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of the Assertion (A).
  2. Both Assertion (A) and Reason (R) are true, but Reason (R) is not the correct explanation of the Assertion (A).
  3. Assertion (A) is true, but Reason (R) is false.
  4. Both Assertion (A) and Reason (R) are false.
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Q14.2025 Main 55/7/1 · Q101 mark

Out of the four options given, in which transition will the emitted photon have the maximum wavelength ?

  1. n = 4 to n = 3
  2. n = 3 to n = 2
  3. n = 2 to n = 1
  4. n = 3 to n = 1
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Q15.2025 Supp 55/S/1 · Q21 mark

In a Rutherford scattering experiment, when an alpha particle of mass m1 approaches a target nucleus of charge Ze and mass m2, the distance of the closest approach is d0. The energy of the particle is :

  1. directly proportional to Z
  2. inversely proportional to Z
  3. directly proportional to mass m2
  4. directly proportional to m1m2
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Official CBSE Marking Scheme · CBSE Board 2025 (Supplementary), Set 55/S/1

(A) directly proportional to Z

Q16.2024 Main 55/1/1 · Q91 mark

The transition of electron that gives rise to the formation of the second spectral line of the Balmer series in the spectrum of hydrogen atom corresponds to :

  1. nf = 2 and ni = 3
  2. nf = 3 and ni = 4
  3. nf = 2 and ni = 4
  4. nf = 2 and ni = ∞
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Official CBSE Marking Scheme · CBSE Board 2024 (Main), Set 55/1/1

(C) nf = 2 and ni = 4

Q17.2024 Main 55/2/1 · Q91 mark

Energy levels A, B and C of an atom correspond to increasing values of energy i.e. EA < EB < EC. Let λ1, λ2 and λ3 be the wavelengths of radiation corresponding to the transitions C to B, B to A and C to A, respectively. The correct relation between λ1, λ2 and λ3 is :

  1. λ12 + λ22 = λ32
  2. 1λ1 + 1λ2 = 1λ3
  3. λ1 + λ2 + λ3 = 0
  4. λ1 + λ2 = λ3
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Official CBSE Marking Scheme · CBSE Board 2024 (Main), Set 55/2/1

(B) 1λ1 + 1λ2 = 1λ3

Q18.2024 Main 55/2/1 · Q101 mark

An alpha particle approaches a gold nucleus in Geiger-Marsden experiment with kinetic energy K. It momentarily stops at a distance d from the nucleus and reverses its direction. Then d is proportional to :

  1. 1K
  2. K
  3. 1K
  4. K
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Official CBSE Marking Scheme · CBSE Board 2024 (Main), Set 55/2/1

(C) 1K

Q19.2024 Main 55/3/1 · Q91 mark

A proton and an alpha particle having equal velocities approach a target nucleus. They come momentarily to rest and then reverse their directions. The ratio of the distance of closest approach of the proton to that of the alpha particle will be :

  1. 12
  2. 2
  3. 14
  4. 4
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Official CBSE Marking Scheme · CBSE Board 2024 (Main), Set 55/3/1

(B) 2

Q20.2024 Main 55/3/1 · Q111 mark

An electron makes a transition from n = 2 level to n = 1 level in the Bohr model of a hydrogen atom. Its period of revolution :

  1. increases by 87·5%
  2. decreases by 87·5%
  3. increases by 43·75%
  4. decreases by 43·75%
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Official CBSE Marking Scheme · CBSE Board 2024 (Main), Set 55/3/1

(B) decreased by 87.5%

Q21.2024 Main 55/4/1 · Q41 mark

The radius (rn) of nth orbit in Bohr model of hydrogen atom varies with n as

  1. rn ∝ n
  2. rn ∝ 1n
  3. rn ∝ n2
  4. rn ∝ 1n2
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Official CBSE Marking Scheme · CBSE Board 2024 (Main), Set 55/4/1

(C) rn ∝ n2

Q22.2024 Main 55/4/1 · Q121 mark

In Balmer series of hydrogen atom, as the wavelength of spectral lines decreases, they appear

  1. equally spaced and equally intense.
  2. further apart and stronger in intensity.
  3. closer together and stronger in intensity.
  4. closer together and weaker in intensity.
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Official CBSE Marking Scheme · CBSE Board 2024 (Main), Set 55/4/1

(D) Closer together and weaker in intensity.

Q23.2024 Main 55/5/1 · Q91 mark

The energy of an electron in the ground state of hydrogen atom is –13.6 eV. The kinetic and potential energy of the electron in the first excited state will be

  1. –13.6 eV, 27.2 eV
  2. –6.8 eV, 13.6 eV
  3. 3.4 eV, –6.8 eV
  4. 6.8 eV, –3.4 eV
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Official CBSE Marking Scheme · CBSE Board 2024 (Main), Set 55/5/1

(C) 3.4 eV, –6.8 eV

Q24.2024 Main 55/5/1 · Q161 mark

Assertion (A) : An alpha particle is moving towards a gold nucleus. The impact parameter is maximum for the scattering angle of 180°.

Reason (R) : The impact parameter in an alpha particle scattering experiment does not depend upon the atomic number of the target nucleus.

  1. If both Assertion (A) and Reason (R) are true and Reason (R) is correct explanation of Assertion (A).
  2. If both Assertion (A) and Reason (R) are true and Reason (R) is not the correct explanation of Assertion (A).
  3. If Assertion (A) is true but Reason (R) is false.
  4. If both Assertion (A) and Reason (R) are false.
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Official CBSE Marking Scheme · CBSE Board 2024 (Main), Set 55/5/1

(D) Both Assertion (A) and Reason (R) are false.

Q25.2024 Supp 55/S/1 · Q61 mark

Which of the following statements is correct for alpha particle scattering experiment ?

  1. For angle of scattering θ ≈ 0, the impact parameter is small.
  2. For angle of scattering θ ≈ π, the impact parameter is large.
  3. The number of alpha particles undergoing head-on collision is small.
  4. The experiment provides an estimate of the upper limit to the size of target atom.
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Official CBSE Marking Scheme · CBSE Board 2024 (Supplementary), Set 55/S/1

(C) The number of alpha particles undergoing head on collision is small

Q26.2024 Supp 55/S/1 · Q101 mark

An electron makes a transition from orbit n = 2 to orbit n = 1, in Bohr’s model of hydrogen atom. Consider change in magnitudes of its kinetic energy (K) and potential energy (U).

  1. K increases and U decreases
  2. K decreases and U increases
  3. Both K and U decrease
  4. Both K and U increase
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Official CBSE Marking Scheme · CBSE Board 2024 (Supplementary), Set 55/S/1

(A) K increases and U decreases

Q27.2023 Main 55/1/1 · Q101 mark

The radius of the nth orbit in Bohr model of hydrogen atom is proportional to :

  1. 1n2
  2. 1n
  3. n2
  4. n
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Official CBSE Marking Scheme · CBSE Board 2023 (Main), Set 55/1/1

(C) n2

Q28.2023 Main 55/1/1 · Q111 mark

Hydrogen atom initially in the ground state, absorbs a photon which excites it to n = 5 level. The wavelength of the photon is :

  1. 975 nm
  2. 740 nm
  3. 523 nm
  4. 95 nm
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Official CBSE Marking Scheme · CBSE Board 2023 (Main), Set 55/1/1

(D) 95 nm

Q29.2023 Main 55/2/1 · Q111 mark

A hydrogen atom makes a transition from n = 5 to n = 1 orbit. The wavelength of photon emitted is λ. The wavelength of photon emitted when it makes a transition from n = 5 to n = 2 orbit is

  1. 87 λ
  2. 167 λ
  3. 247 λ
  4. 327 λ
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Official CBSE Marking Scheme · CBSE Board 2023 (Main), Set 55/2/1

(D) 327 λ

Q30.2023 Main 55/3/1 · Q121 mark

The ratio of maximum frequency and minimum frequency of light emitted in Balmer series of hydrogen spectrum, in Bohr’s model is :

  1. 119
  2. 95
  3. 117
  4. 167
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Official CBSE Marking Scheme · CBSE Board 2023 (Main), Set 55/3/1

(B) 95

Q31.2023 Main 55/4/1 · Q131 mark

The potential energy of an electron in the second excited state in hydrogen atom is :

  1. − 3·4 eV
  2. − 3·02 eV
  3. − 1·51 eV
  4. − 6·8 eV
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Official CBSE Marking Scheme · CBSE Board 2023 (Main), Set 55/4/1

(B) − 3.02 eV

Q32.2023 Main 55/5/1 · Q101 mark

The diagram shows four energy level of an electron in Bohr model of hydrogen atom. Identify the transition in which the emitted photon will have the highest energy.

Energy levels n=1 to n=4 with transitions I (n=3 to 1), II (n=4 to 3), III (n=2 to 1), IV (n=3 to 2)
  1. I
  2. II
  3. III
  4. IV
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Official CBSE Marking Scheme · CBSE Board 2023 (Main), Set 55/5/1

(A) I

Q33.2023 Main 55/5/1 · Q151 mark

The radius of the nth orbit in Bohr model of hydrogen atom is proportional to

  1. n2
  2. 1n2
  3. n
  4. 1n
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Official CBSE Marking Scheme · CBSE Board 2023 (Main), Set 55/5/1

(A) n2

Q34.2023 Supp 55/C/1 · Q111 mark

The impact parameter for an alpha particle approaching a target nucleus is maximum when the scattering angle (θ) is :

  1. 0°
  2. 90°
  3. 180°
  4. 45°
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Official CBSE Marking Scheme · CBSE Board 2023 (Supplementary), Set 55/C/1

(A) 0°

Q35.SQP 2025-26 · Q71 mark

Which of the following transitions corresponds to the emission of the radiation of the maximum wavelength?

Hydrogen energy level diagram with transitions I to VI marked between levels 1, 2, 3 and 4
  1. I
  2. III
  3. IV
  4. VI
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Official CBSE Marking Scheme · CBSE Sample Question Paper 2025-26

(C) IV

Transition III, V, VI corresponds to absorption of energy.

Maximum emitted wavelength corresponds minimum energy difference.

ΔEI > ΔEII > ΔEIV

Therefore, maximum emitted wavelength corresponds to transition IV.

Q36.SQP 2025-26 · Q131 mark

Assertion (A): Total energy of an electron in hydrogen atom is negative.

Reason (R): The centripetal force is provided by electrostatic force.

  1. Both Assertion and Reason are true and Reason is the correct explanation of Assertion.
  2. Both Assertion and Reason are true but Reason is not the correct explanation of Assertion.
  3. Assertion is true but Reason is false.
  4. Both Assertion and Reason are false.
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Official CBSE Marking Scheme · CBSE Sample Question Paper 2025-26

(B) Both Assertion and Reason are true but Reason is not the correct explanation of Assertion.

Q37.SQP 2026-27 · Q111 mark

If nucleus scaled to football size is approximately 22cm, then the radius of the orbit of electron will be approximately:

  1. 22 cm
  2. 22 m
  3. 22 km
  4. 22000 km
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Official CBSE Marking Scheme · CBSE Sample Question Paper 2026-27

(C) 22 km

Typical nuclear radius ≈ 10-15 m (1 femtometer). Atom radius (electron orbit, roughly Bohr radius) ~ 10-10 m. Ratio of atom size to nucleus size ~ 105.

Football (nucleus) radius ~ 22 cm = 0.22 m. Multiply by 105: 0.22 m × 105 = 22,000 m = 22 km.

Q38.SQP 2026-27 · Q131 mark

Assertion (A): In Rutherford’s scattering experiment, most of the alpha particles passed through the gold foil undeflected.

Reason (R): The electrons in the atom are very light and do not affect the path of the alpha particles significantly.

  1. both Assertion and Reason are true and Reason is the correct explanation of Assertion.
  2. both Assertion and Reason are true but Reason is not the correct explanation of Assertion.
  3. Assertion is true but Reason is false.
  4. both Assertion and Reason are false.
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Official CBSE Marking Scheme · CBSE Sample Question Paper 2026-27

(B) both Assertion and Reason are true but Reason is not the correct explanation of Assertion.

Explanation: Both are true; most of the space is empty (nucleus tiny), so alpha particles pass undeflected; electron mass has negligible deflecting effect on energetic alpha particles, as reason for statement is “most of the space is empty (nucleus tiny)”

Q39.2026 Main 55/4/1 · Q202 marks

Draw the graph showing variation of number of scattered alpha particles (N) detected with scattering angle (θ) in Geiger-Marsden experiment. Infer two conclusions from the graph.

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Official CBSE Marking Scheme · CBSE Board 2026 (Main), Set 55/4/1

Drawing graph [1]

Number of scattered particles detected decreases steeply with increasing scattering angle

Conclusions from the graph (Any Two) [½ + ½]

  1. Many of the α-particles pass through the foil, without suffering any collisions.
  2. Only a few incident α-particles scatter by more than 1°.
  3. About 1 in 8000 α-particles deflect by more than 90°.
Q40.2025 Main 55/1/1 · Q202 marks

Prove that, in Bohr model of hydrogen atom, the time period of revolution of an electron in nth orbit is proportional to n3.

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Q41.2024 Main 55/5/1 · Q202 marks
  1. Two energy levels of an electron in hydrogen atom are separated by 2.55 eV. Find the wavelength of radiation emitted when the electron makes transition from the higher energy level to the lower energy level.
  2. In which series of hydrogen spectrum this line shall fall ?
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Official CBSE Marking Scheme · CBSE Board 2024 (Main), Set 55/5/1

(a) E2 – E1 = hcλ [½]

Given

E2 – E1 = 2.55×1.6×10–19 J [½]

⇒ λ = 6.63×10–34 × 3×1082.55×1.6×10–19 = 487.5 nm [½]

(b) Balmer series [½]

Q42.2024 Main 55/5/1 · Q212 marks

The earth revolves around the sun in an orbit of radius 1.5 × 1011 m with orbital speed 30 km/s. Find the quantum number that characterises its revolution using Bohr’s model in this case (mass of earth = 6.0 × 1024 kg).

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Official CBSE Marking Scheme · CBSE Board 2024 (Main), Set 55/5/1

Using Bohr’s model

mvr = nh2π [1]

n = 2π × 6.0×1024 × 30×103 × 1.5×10116.63×10–34 [½]

n = 2.558×1074 [½]

Q43.2023 Main 55/1/1 · Q232 marks

The ground state energy of hydrogen atom is −13·6 eV. What is the potential energy and kinetic energy of an electron in the third excited state ?

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Official CBSE Marking Scheme · CBSE Board 2023 (Main), Set 55/1/1

Calculation of potential energy of electron [1]; Calculation of kinetic energy of electron [1]

En = −13.6n2 eV = Total energy

For third excited state n = 4

E4 = −13.642 = −13.616 = −0.85 eV [½]

Potential Energy = 2 × Total Energy = 2 × E4 = 2 × (−0.85) eV = −1.70 eV [½]

Kinetic energy = −(Total Energy) = −E4 = 0.85 eV [½ + ½]

Q44.2023 Main 55/2/1 · Q23 (option 1)2 marks

What is meant by ionisation energy ? Write its value for hydrogen atom.

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Official CBSE Marking Scheme · CBSE Board 2023 (Main), Set 55/2/1

Ionization energy is the minimum energy required to remove an electron from an isolated atom of an element. [1]

Alternatively

It is the energy required to excite an electron from energy level n = 1 to n = ∞ from an isolated atom of an element.

The ionization energy for hydrogen atom is 13.6 eV. [1]

Q45.2023 Main 55/3/1 · Q23 (OR option)2 marks

Using Bohr’s postulates, obtain the expression for radius of nth stable orbit in a hydrogen atom.

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Official CBSE Marking Scheme · CBSE Board 2023 (Main), Set 55/3/1

mv2rn = kq2rn2 = e24π ε0 rn2   …(1) [½]

mvrn = nh2π   …(2) [½]

Using equations (1) & (2) :

rn = n2h2 4π ε0m (2π)2 e2 = 0·53 × 10-10 n2 m [1]

Q46.2022 Main 55/1/1 · Q2 (option 1)2 marks

Draw the graph showing the variation of the number (N) of scattered alpha particles with scattering angle (θ) in Geiger – Marsden experiment. Infer two conclusions from the graph.

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Official CBSE Marking Scheme · CBSE Board 2022 (Main), Set 55/1/1

Graph [1]

Number of scattered alpha particles detected versus scattering angle

(Give full credit if axis are marked and values are not given)

Conclusions (any two) [½ + ½]

  • Most of the alpha particles pass undeviated through the gold foil.
  • A few alpha particles, get deflected through 90° or more.
  • Only about 0.14% of the incident alpha particles are reflected by large angle.
  • A very few alpha particles retrace their path.

Any other two conclusions

Q47.2022 Main 55/2/1 · Q2 (option 1)2 marks
  1. Define the terms : ‘impact parameter’ and ‘distance of closest approach’ for an α-particle in Geiger-Marsden scattering experiment.
  2. What will be the value of the impact parameter for scattering angle (I) θ = 0° and (II) θ = 180° ?
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Official CBSE Marking Scheme · CBSE Board 2022 (Main), Set 55/2/1

(i) Impact Parameter : It is the perpendicular distance of the initial velocity vector of the approaching α-particle from the centre of the nucleus. [½]

Distance of closest approach : It is the minimum distance of the approaching α-particle and the target gold nucleus [½]

d = 2Ze24πε0K ; where K is the kinetic energy

Alternatively : Distance of closest approach is the distance of the alpha particle from the centre of gold nucleus where its whole kinetic energy is converted into potential energy

(ii) θ = 0° ; b = maximum / almost of atomic size [½]

θ = 180° ; b = minimum = zero [½]

(Note: Allot ½ Mark for only formula.)

Q48.2022 Main 55/4/1 · Q2 (option 1)2 marks

Name the spectral series for a hydrogen atom which lies in the visible region. Find the ratio of the maximum to the minimum wavelengths of this series.

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Official CBSE Marking Scheme · CBSE Board 2022 (Main), Set 55/4/1

Balmer Series [½]

1λ = R(1nf2 − 1ni2) [½]

1λmax = R(122 − 132) (nf = 2, ni = 3)

λmax = 365R [½]

1λmin = R(122 − 1∞) (nf = 2, ni = ∞)

λmin = 4R

λmax / λmin = 95 [½]

Q49.2022 Main 55/5/1 · Q2 (option 1)2 marks

What results do you expect if α-particle scattering experiment is repeated using a thin sheet of hydrogen in place of a gold foil ? Explain. (Hydrogen is a solid at temperature below 14K)

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Official CBSE Marking Scheme · CBSE Board 2022 (Main), Set 55/5/1

Writing of the result [1]; Explanation [1]

There would be no large angle scattering / Size of nucleus can’t be determined. [1]

As hydrogen atom which is a target nucleus has only one proton whereas approaching α particle is more massive than the target nucleus. [1]

Alternatively: Repulsive force between target nucleus (Hydrogen) and α–particles will be very less.

(Note: Give full credit for other correct explanations.)

Q50.2022 Supp 55/6/1 · Q22 marks
  1. In Geiger-Marsden experiment, the distance of closest approach is considerably larger than the sum of the radii of the gold nucleus and the α-particle. Explain.
  2. The total energy of hydrogen atom in a state is – 3·4 eV. What are the kinetic and potential energies of the electron in this state ? [1+1]
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Official CBSE Marking Scheme · CBSE Board 2022 (Supplementary), Set 55/6/1

(a) Alpha particle reverses its direction of motion, due to strong repulsive force, exerted by positively charged nucleus without even actually touching the gold nucleus. [1]

(b) En = –3·4 eV (given)

Ek = –En = 3·4 eV [½]

U = 2En = –6·8 eV [½]

Q51.SQP 2025-26 · Q21 (option 1)2 marks

How does the impact parameter affect the trajectory of a α – particles scattered by a heavy nucleus? What is the value of impact parameter for head on collision of α – particles with the nucleus?

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Official CBSE Marking Scheme · CBSE Sample Question Paper 2025-26

Smaller is the impact parameter, larger is the angle at which α – particles scatters. [1]

Larger is the impact parameter, α – particles scatter less keeping its original trajectory.

For head on collision, the value of impact parameter is zero. [1]

Q52.SQP 2026-27 · Q21 (option 1)2 marks
  1. Why hydrogen spectral lines are considered to be a fingerprint of the element?
  2. If energy of an electron of the hydrogen atom in ground state is -13.6 eV, what should be the minimum energy of photon which can ionize the atom?
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Official CBSE Marking Scheme · CBSE Sample Question Paper 2026-27

I. Hydrogen spectral lines are considered to be a fingerprint of the element because they represent the unique set of colored lines emitted when hydrogen atoms transit between energy levels. This phenomenon, known as the line spectrum, is a result of the quantized energy levels of electrons in the hydrogen atom. Each spectral line corresponds to a specific energy transition, and the distinct pattern of these lines allows for the identification of hydrogen in various environments. [1]

II. The minimum energy of a photon that can ionize an electron in a hydrogen atom is 13.6 eV. This is the ionization energy required to remove the electron from its ground state, considering all the energy is transferred to the electron. [1]

Q53.SQP 2026-27 · Q21 (OR option)2 marks

State Bohr’s quantization condition for defining stationary orbit. How does de Broglie hypothesis explain the stationary orbit?

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Official CBSE Marking Scheme · CBSE Sample Question Paper 2026-27

Postulate: Bohr postulated that electrons revolve around the nucleus in certain discrete circular orbits without radiating energy. His quantization condition is: mvr = nh2π [1]

Where:

  • m = mass of the electron
  • v = velocity of the electron
  • r = radius of the orbit
  • n = principal quantum number

Explanation of stationary orbits using de Broglie hypothesis: According to de Broglie, a moving electron behaves like a matter wave with wavelength: λ = h/mv

For an orbit to be stable (stationary), the electron wave must form a standing wave around the nucleus. This is possible only when the circumference of the orbit is an integral multiple of the wavelength:

2πr = nλ

Substituting λ = h/mv

2πr = nh/mv

Or, mvr = nh/2π

This is exactly Bohr’s quantization condition.

Hence, de Broglie hypothesis explains Bohr’s stationary orbits as those orbits in which the electron’s matter wave forms a standing wave, leading to stable, non-radiating orbits. [1]

Q54.2026 Main 55/2/1 · Q263 marks

(a) State Bohr’s second postulate and mention its significance.

(b) Prove that, in Bohr model of hydrogen atom as principal quantum number n becomes large, the energy levels get closer and closer.

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Official CBSE Marking Scheme · CBSE Board 2026 (Main), Set 55/2/1

(a) The electron revolves around the nucleus only in those orbits for which angular momentum is some integral multiple of h2π where h is Planck’s constant. [1]

Alternatively : L = n(h2π)

It defines stable / stationary orbits [½]

Alternatively : Angular momentum is quantised

(b) En = –13.6n2 eV [½]

For any two consecutive energy levels, the energy difference is : ΔE = En – En–1

ΔE = 13.6 [1(n–1)2 – 1n2] [½]

ΔE = 13.6 (2n–1)n2(n–1)2

For large value of n : ΔE = 27.2n3 eV

Thus as n increases, ΔE decreases i.e., energy levels get closer and closer. [½]

Q55.2026 Main 55/5/1 · Q26 (OR option)3 marks
  1. Draw the number of scattered particles versus the scattering angle graph for scattering of alpha particles by a thin foil. Write two important conclusions that can be drawn from this plot.
  2. If Bohr’s quantization postulate (angular momentum = nh2π) is a basic law of nature, it should be equally valid for the case of planetary motion also. Why, then, do we never speak of quantization of orbits of planets around the Sun ? Explain.
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Official CBSE Marking Scheme · CBSE Board 2026 (Main), Set 55/5/1

(i) Drawing the graph between scattered particles & scattering angle [1]; writing two important conclusions [1]; (ii) Explaining invalidity of Bohr quantisation postulate for planetary motion [1]

(i) Graph [1]

Number of scattered alpha particles detected versus scattering angle

Important conclusions: [1]

1. Most of the space in the atom is empty.

2. Most part of mass of atom and its positive charge were tightly concentrated at its centre.

Note: No marks to be deducted for scale on the axis

(ii) According to Bohr quantisation postulate L = mvr = nh/2π

Since mass of planets is very large as compared to mass of atomic particles, angular momentum (L) will be very large.

Therefore, Principal quantum number (n) will be very large.

For large quantum number, the difference in successive energy levels will be negligible. Hence quantization fails. [1]

Q56.2025 Main 55/6/1 · Q273 marks

How is the necessary force provided to an electron to keep it moving in a circular orbit according to Bohr model of hydrogen atom ? Derive an expression for the total energy of an electron moving in an orbit of radius r in hydrogen atom. Give the significance of negative sign in this expression.

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Q57.2025 Main 55/7/1 · Q273 marks

Write the mathematical forms of three postulates of Bohr’s theory of the hydrogen atom. Using them prove that, for an electron revolving in the nth orbit,

  1. the radius of the orbit is proportional to n2, and
  2. the total energy of the atom is proportional to (1n2).
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Q58.2025 Supp 55/S/1 · Q243 marks

Write the three postulates of Bohr model of hydrogen atom. Show that, in this model, the frequency of revolution of an electron in its nth orbit is proportional to ( 1n3 ).

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Official CBSE Marking Scheme · CBSE Board 2025 (Supplementary), Set 55/S/1

Bohr’s Postulates

(i) An electron in an atom could revolve in certain stable orbits without the emission of radiant energy. [½]

(ii) Electron revolves around the nucleus only in those orbits for which the angular momentum is some integral multiple of h/2π. [½]

(iii) An electron might make a transition from one of its specified non-radiating orbits to another of lower energy. When it does so, a photon is emitted having energy equal to the energy difference between the initial and final states. The frequency of the emitted photon is then given by hν = Ei – Ef [½]

As frequency of revolution of electron is given by

v = v2πr ............(1) [½]

In Bohr’s model the velocity of the electron in the nth orbit is

vn = e22ε0nh

vn ∝ 1n ............(2) [½]

and radius of the electron in the nth orbit is

rn = ε0n2h2πme2

rn ∝ n2 ............(3) [½]

Using (3) and (2) in (1)

v ∝ 1n3 [½]

Q59.2024 Main 55/1/1 · Q273 marks

Write the drawbacks of Rutherford’s atomic model. How did Bohr remove them ? Show that different orbits in Bohr’s atom are not equally spaced.

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Official CBSE Marking Scheme · CBSE Board 2024 (Main), Set 55/1/1

Drawbacks:

i) According to classical electromagnetic theory, an accelerating charged particle emits radiation in the form of electromagnetic waves. The energy of an accelerating electron should therefore, continuously decrease. The electron would spiral inward and eventually fall into the nucleus. Thus, such an atom cannot be stable.

ii) As the electrons spiral inwards, their angular velocities and hence their frequencies would change continuously. Thus, they would emit a continuous spectrum, in contradiction to the line spectrum actually observed. [1]

Bohr postulated stable orbits in which electrons do not radiate energy [1]

Alternatively:
Bohr’s postulates (Any ONE of the three)
(i) An electron in an atom could revolve in certain stable orbits without the emission of radiant energy.
(ii) The electron revolves around the nucleus only in those orbits for which the angular momentum is some integral multiple of h/2π
(iii) An electron might make a transition from one of its specified non-radiating orbits to another of lower energy. When it does so, a photon is emitted having energy equal to the energy difference between the initial and final states.

The radius of the nth orbit is found as

rn = (n2m)(h2π)2 4πε0e2

rn ∝ n2 [1]

Alternatively:
Difference in radius of consecutive orbits is
rn+1 − rn = k [(n+1)2 − n2] = k (2n + 1) which depends on n, and is not a constant

Q60.2024 Main 55/2/1 · Q263 marks

Using Bohr’s postulates, derive the expression for the radius of the nth orbit of an electron in a hydrogen atom. Also find the numerical value of Bohr’s radius a0.

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Official CBSE Marking Scheme · CBSE Board 2024 (Main), Set 55/2/1

From Bohr’s second postulate
mvr = nh2π ……(1) [½]

Also mv2r = e24πε0r2 (Z = 1) [½]

v = e4πε0mr [½]

Substituting in (1) and simplifying
r = n2h2ε0πme2 [½]

For n = 1, r = a0 (Bohr’s radius)
a0 = (6·63 × 10−34)2 × 8·854 × 10−123·14 × 9·1 × 10−31 × (1·6 × 10−19)2 [½]

= 5·29 × 10−11 m = 0·53 Å [½]

Q61.2024 Main 55/3/1 · Q263 marks

State Bohr’s first and second postulates. Use them to derive an expression for the radius of the nth orbit in a hydrogen atom.

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Official CBSE Marking Scheme · CBSE Board 2024 (Main), Set 55/3/1

Statements of Bohr’s first and second postulates [½ + ½]; Derivation for radius of nth orbit [2]

  • Bohr’s first postulate
    An electron in an atom revolves in certain stable orbits without the emission of radiant energy. [½]
  • Bohr’s second postulate
    Electron revolves around the nucleus only in those orbits for which the angular momentum is integral multiple of h2π. [½]

Electrostatic force between revolving electron and nucleus provides requisite centripetal force

mvn2rn = 14πε0 e2rn2 [½]

vn = e4πε0mrn ------(i) [½]

mvnrn = nh2π ------(ii) [½]

using equations (i) and (ii)

rn = (nm)(h2π)2 4πε0e2 [½]

Q62.2024 Main 55/4/1 · Q243 marks

Draw the graph showing variation of scattered particles detected (N) with the scattering angle (θ) in Geiger-Marsden experiment. Write two conclusions that you can draw from this graph. Obtain the expression for the distance of closest approach in this experiment.

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Official CBSE Marking Scheme · CBSE Board 2024 (Main), Set 55/4/1

Graph showing variation of scattered particles detected (N) with scattering angle (θ) [1]

Number of scattered particles detected versus scattering angle: decreasing curve

Two conclusions [½ + ½]

(i) Most of an atom is empty space.

(ii) Almost entire mass and entire positive charge is concentrated in a very small region called nucleus.

At distance of closest approach

Ek = EP [½]

K = 14πε0 · (Ze)·(2e)d [½]

d = 14πε0 · 2Ze2K [½]

Q63.2023 Main 55/1/1 · Q30 (option 1)3 marks

Draw the energy level diagram for hydrogen atom. Mark the transitions corresponding to the series lying in the ultraviolet region, visible region and infrared region.

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Official CBSE Marking Scheme · CBSE Board 2023 (Main), Set 55/1/1

Energy level diagram for hydrogen atom [1½]; Transitions corresponding to ultraviolet region, visible region and infrared region [½ + ½ + ½]

Energy level diagram of hydrogen atom with Lyman (UV), Balmer (visible), Paschen and Brackett (IR) series marked

[1½ + ½ + ½ + ½]

Note: Award 1½ mark for energy level diagram if the student does not show the transitions.

Q64.2023 Main 55/3/1 · Q30 (OR option)3 marks

(i) Draw a graph to show the variation of the number of scattered particles detected (N) in Geiger-Marsden experiment as a function of scattering angle (θ).

(ii) Discuss briefly two conclusions that can be drawn from this graph and how they lead to the discovery of nucleus in an atom.

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Official CBSE Marking Scheme · CBSE Board 2023 (Main), Set 55/3/1

(i) Graph [1]

Number of scattered particles detected versus scattering angle (log scale), Geiger-Marsden experiment

(ii) – The entire positive charge and most of the mass of the atom are concentrated in a small space. [½]

– Many of the α-particles pass through the foil. It means that they do not suffer any collisions. [½]

To deflect the α-particle backwards, a large repulsive force is required, which is provided only if the greater part of the mass of the atom & its positive charge were concentrated tightly at its centre. This lead to the discovery of the nucleus in the atom. [1]

Q65.2023 Main 55/4/1 · Q303 marks

Briefly explain Geiger-Marsden experiment. Show the variation of the number of particles scattered (N) with scattering angle (θ) in this experiment. What is the main conclusion that can be inferred from this plot ?

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Official CBSE Marking Scheme · CBSE Board 2023 (Main), Set 55/4/1

Alpha-particles emitted by a radioactive source were allowed to fall on a thin foil of gold. The scattered alpha-particles were observed through a rotatable detector consisting of zinc sulphide screen and a microscope. The scattered alpha-particles on striking the screen produced brief light flashes or scintillations. [1]

Note : Give full credit of 1 mark if a student draws the labeled diagram of Geiger-Marsden experiment.

Graph [1]

Number of scattered particles detected versus scattering angle (degrees), decreasing curve

Note : Full marks to be given even if values are not marked.

Conclusion : The existence of positively charged nucleus inside an atom and provide an upper limit to the size of the nucleus. [1]

Q66.2023 Main 55/5/1 · Q30 (option 1)3 marks

(i) Differentiate between ‘distance of closest approach’ and ‘impact parameter’.

(ii) Determine the distance of closest approach when an alpha particle of kinetic energy 3.95 MeV approaches a nucleus of Z = 79, stops and reverses its directions.

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Official CBSE Marking Scheme · CBSE Board 2023 (Main), Set 55/5/1

(i) Distance of closest approach: It is the minimum distance of α-particle from the centre of nucleus at which its total kinetic energy gets converted into electrostatic potential energy. [1]

Alternatively: Distance of closest approach r0 = 14πε0 2Ze2EK

Impact parameter: Perpendicular distance of the initial velocity vector of the α-particle from the centre of the nucleus. [1]

Alternatively:

Alpha particle trajectories with impact parameter b from the target nucleus

(ii) r0 = 14πε0 2Ze2EK [½]

r0 = 9 × 109 × 2 × 79 × (1.6 × 10-19)23.95 × 1.6 × 10-13

= 57.6 × 10-15 m [½]

Q67.2023 Main 55/5/1 · Q30 (OR option)3 marks

(i) State three postulates of Bohr’s theory of hydrogen atom.

(ii) Find the angular momentum of an electron revolving in the second orbit in Bohr’s hydrogen atom.

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Official CBSE Marking Scheme · CBSE Board 2023 (Main), Set 55/5/1

(i)

(1) An electron in an atom revolves in certain stable orbit without the emission of radiant energy. [½]

(2) The electron revolves around the nucleus only in those orbits for which the angular momentum is integral multiple of h2π where h is the Planck’s constant. [½]

(3) When an electron makes a transition from one of its specified non-radiating orbits to another of lower energy orbit a photon is emitted having energy equal to the energy difference between the initial and final states. [½]

(ii) Angular momentum = nh2π [½]

For n = 2, angular momentum = 2h2π [½]

= 6.63 × 10-34π = 2.1 × 10-34 kg m2 s-1 [½]

Q68.2023 Supp 55/C/1 · Q30 (option 1)3 marks

(i) An electron in a hydrogen atom jumps from second excited state to the first excited state. Name the spectral series in the spectrum of hydrogen atom to which the emitted radiation belongs.

(ii) Find the ratio of the wavelengths of the “most energetic spectral” lines in the Balmer series to that in Paschen series of Hydrogen spectrum.

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Official CBSE Marking Scheme · CBSE Board 2023 (Supplementary), Set 55/C/1

(i) Balmer series [½]

(ii) 1λB = R[122 − 1∞] [½]

1λB = R4 [½]

1λp = R[132 − 1∞] [½]

1λp = R9 [½]

λBλp = 49 [½]

Q69.2023 Supp 55/C/1 · Q30 (OR option)3 marks

(i) An α-particle having kinetic energy K approaches a nucleus of atomic number Z. It gets close to the nucleus and then approaches a distance (d) and reverses its direction. Obtain an expression for the distance of closest approach (d) in terms of kinetic energy of the α-particle.

(ii) A proton and an alpha particle approach a target nucleus in head-on position, with equal velocities. Find the ratio of their distances of closest approach to the target nucleus.

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Official CBSE Marking Scheme · CBSE Board 2023 (Supplementary), Set 55/C/1

(i) At the distance of closest approach.

K = (Ze)2e4πεod [½]

d = 2Ze24πεoK [1]

d = 2(Ze)q12mv2 × 4πεo

d = (Ze)qmv2πεo [½]

d ∝ qm [½]

dpdα = e2e × 4mm = 21 [½]

Q70.2022 Main 55/1/1 · Q43 marks

Define the term – Distance of closest approach. How will it be affected, for an α - particle, if kinetic energy of the particle is doubled ?

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Official CBSE Marking Scheme · CBSE Board 2022 (Main), Set 55/1/1

The minimum distance up to which an alpha particle travel along the central line of the nucleus before it rebounds is called distance of closest approach. [1½]

Alternatively

An alpha particle travelling directly towards the centre of a nucleus slows down as it approaches the nucleus due to repulsive force. At a distance r0 from the nucleus, the α- particle stops and its total kinetic energy converts into electrostatic potential energy. This distance r0 is called distance of closest approach.

r0 = 2Ze24π ε0 K.E. [½]

i.e., r0 ∝ 1K.E. [½]

as K is doubled, r0 is halved. [½]

(Award full 1½ marks if a student writes r0 is halved without writing formula)

Q71.2022 Main 55/2/1 · Q83 marks

The energy of a hydrogen atom in the first excited state is – 3·4 eV. Find :

  1. the radius of this orbit. (Take Bohr radius = 0·53 Å)
  2. the angular momentum of the electron in the orbit.
  3. the kinetic and potential energy of the electron in the orbit.
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Official CBSE Marking Scheme · CBSE Board 2022 (Main), Set 55/2/1

a) r = n2 r0 = (2)2 × 0.53 = 2.12 Å [1]

b) L = nh2π = 2 × 6.63 × 10–342 × 3.14 = 2 × 10–34 kg m/s [1]

c) K.E = + 3.4 eV & P.E = – 6.8 eV [1]

Q72.2022 Main 55/3/1 · Q43 marks

State Bohr’s postulate to explain stable orbits in a hydrogen atom. Prove that the speed with which the electron revolves in nth orbit is proportional to (1/n).

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Official CBSE Marking Scheme · CBSE Board 2022 (Main), Set 55/3/1

(i) An electron can revolve around the nucleus in an orbit in which its angular momentum is an integral multiple of h2π. [1]

(ii) Proof

mv2r = 14πε0 e2r2 [½]

mvr = nh2π [½]

Eliminating r we get

v = e22ε0h · 1n [½]

∴ v ∝ 1n [½]

Q73.2022 Main 55/3/1 · Q63 marks

A narrow beam of protons, each having 4.1 MeV energy is approaching a sheet of lead (Z = 82). Calculate :

  1. the speed of a proton in the beam, and
  2. the distance of its closest approach
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Official CBSE Marking Scheme · CBSE Board 2022 (Main), Set 55/3/1

(i) ½mv2 = 4·1 × 1·6 × 10-13 J [½]

v = 2 × 4·1 × 1·6 × 10-131·673 × 10-27 [½]

= 2·8 × 107 m/s [½]

(ii) d = Ze24πε0 × Ek [½]

= 9 × 109 × 82 × 1·6 × 10-19 × 1·6 × 10-194·1 × 1·6 × 10-13 [½]

= 2.88 × 10-14 m [½]

Q74.2022 Main 55/4/1 · Q63 marks

(a) In Geiger-Marsden experiment, calculate the distance of closest approach for an alpha particle with energy 2·56 × 10−12 J. Consider that the particle approaches gold nucleus (Z = 79) in head-on position.

(b) If the above experiment is repeated with a proton of the same energy, then what will be the value of the distance of closest approach ?

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Official CBSE Marking Scheme · CBSE Board 2022 (Main), Set 55/4/1

(a) For α particles, distance of closest approach

rα = 14πε0 2Ze2Ek [½]

rα = 9 × 109 × 2 × 79 × (1.6 × 10−19)22.56 × 10−12 [½]

= 14·22 × 10−15 m

= 14.22 fm [½]

(b) For proton, distance of closest approach

rp = 14πε0 Ze2Ek [½]

rp = rα2 [½]

= 7·11 × 10−15 m [½]

Q75.2022 Main 55/5/1 · Q43 marks

Draw a graph showing the variation of number of particles scattered (N) with the scattering angle θ in Geiger-Marsden experiment. Why only a small fraction of the particles are scattered at θ > 90° ? Mention two limitations of Rutherford nuclear model of an atom.

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Official CBSE Marking Scheme · CBSE Board 2022 (Main), Set 55/5/1

Graph [1]

Number of scattered particles detected versus scattering angle θ (log scale)

(Note: Full credit if values on the axis are not mentioned)

Reason [1]

Due to the large empty space available in an atom, α particles heading towards the nucleus are less, therefore, small fraction of particles are scattered at θ > 90°

Alternatively:

For angle θ > 90°, the impact parameter is very small.

Limitations [½ + ½]

  • Could not explain the stability of the atom
  • Could not explain the existence of line spectra
Q76.2022 Supp 55/6/1 · Q83 marks
  1. How did de Broglie hypothesis provide an explanation for Bohr’s second postulate for quantisation of orbital angular momentum of the orbiting electron in hydrogen atom ? Discuss. [2]
  2. Identify the transition of electron in Bohr model which gives rise to (i) the maximum, and (ii) the minimum wavelength in Balmer series of hydrogen spectrum. [1]
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Official CBSE Marking Scheme · CBSE Board 2022 (Supplementary), Set 55/6/1
Standing de Broglie wave on a circular orbit around the nucleus

(a) For an electron moving in nth circular orbit of radius rn, the total distance is circumference of the orbit.

Thus 2π rn = nλ ........(1) where n = 1, 2, 3 ........ [½]

λ is de Broglie wavelength associated with the electron in the nth orbit.

Now λ = hp = hmvn ..........(2) [½]

From equation (1) and (2)

2π rn = nhmvn [½]

mvn rn = nh2π [½]

Which is quantum condition proposed by Bohr for the angular momentum of the electron.

(b) For Balmer series of hydrogen spectrum 1λ = R (122 – 1n2)

(i) For maximum wavelength: transition of electron is from n = 3 to n = 2 [½]

(ii) For minimum wavelength: transition of electron is from n = ∞ to n = 2 [½]

Q77.2025 Main 55/4/1 · Q304 marks

A hydrogen atom consists of an electron revolving in a circular orbit of radius r with certain velocity v around a proton located at the nucleus of the atom. The electrostatic force of attraction between the revolving electron and the proton provides the requisite centripetal force to keep it in the orbit. According to Bohr’s model, an electron can revolve only in certain stable orbits. The angular momentum of the electron in these orbits is some integral multiple of h2π, where h is the Planck’s constant.

Further, when an electron makes a transition from one orbit of higher energy to that of lower energy, a photon is emitted having energy equal to the difference between energies of the initial and final states. Assuming the mass and charge of an electron as m and e respectively, answer the following questions.

  1. The expression for the speed of electron v in terms of radius of the orbit (r) and physical constant (K = 14πε0) is :
    1. Ke2mr
    2. Ke2mr2
    3. Ke2mr
    4. Ke2mr2
  2. The total energy of the atom in terms of r and physical constant K is :
    1. Ke2r
    2. − Ke22r
    3. Ke22r
    4. 32 Ke2r
  3. A photon of wavelength 500 nm is emitted when an electron makes a transition from one state to the other state in an atom. The change in the total energy of the electron and change in its kinetic energy in eV as per Bohr’s model, respectively will be :
    1. 2·48, − 2·48
    2. − 1·24, 1·24
    3. − 2·48, 2·48
    4. 1·24, − 1·24
  4. (a) In Bohr’s model of hydrogen atom, the frequency of revolution of electron in its nth orbit is proportional to :
    1. n
    2. 1n
    3. 1n2
    4. 1n3

    OR

    (b) An electron makes a transition from − 3·4 eV state to the ground state in hydrogen atom. Its radius of orbit changes by : (radius of orbit of electron in ground state = 0·53 Å)

    1. 0·53 Å
    2. 1·06 Å
    3. 1·59 Å
    4. 2·12 Å
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Q78.2024 Supp 55/S/1 · Q31 (OR option)5 marks

(i) State Bohr’s postulates and derive an expression for the energy of electron in nth orbit in Bohr’s model of hydrogen atom.

(ii) Calculate binding energy per nucleon (in MeV) of 126C .

Given :

m(126C) = 12.000000 u

m(10n) = 1.008665 u

m(11H) = 1.007825 u

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Official CBSE Marking Scheme · CBSE Board 2024 (Supplementary), Set 55/S/1

Bohr’s Postulates

(a) Bohr’s first postulate was that an electron in an atom could revolve in certain stable orbits without the emission of radiant energy. [½]

(b) Bohr’s second postulate states that the electron revolves around the nucleus only in those orbits for which the angular momentum is some integral multiple of h/2π where h is the Planck’s constant. [½]

(c) Bohr’s third postulate states that an electron might make a transition from one of its specified non-radiating orbits to another of lower energy. When it does so, a photon is emitted having energy equal to the energy difference between the initial and final states. [½]

Derivation

Total energy of electron in the stationary state of hydrogen atoms is

E = – e28πε0rn ——— (1) [½]

Where rn is radius of nth orbit

rn = n2h2ε0πme2 ——— (2) [½]

Substituting eq (2) in eq (1)

En = – me48n2ε02h2 [1]

(ii) Mass defect, Δm = [6m(10n) + 6m(11H)] – m(126C) [½]

Δm = (6×1.008665 + 6×1.007825) – 12.000000

Δm = 0.09894 u [½]

B.E. = Δm × 931.5 MeV

= 92.16 MeV

Binding energy per nucleon, Ebn = EbA = 92.1612

= 7.68 MeV [½]

Unit VIII · Atoms and Nuclei

13. Nuclei

Q1.2026 Main 55/1/1 · Q151 mark

Assertion (A) : The mass of a nucleus is less than the sum of the masses of the constituent nucleons.

Reason (R) : Energy is absorbed when the nucleons are bound together to form a nucleus.

  1. Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of the Assertion (A).
  2. Both Assertion (A) and Reason (R) are true, but Reason (R) is not the correct explanation of the Assertion (A).
  3. Assertion (A) is true, but Reason (R) is false.
  4. Both Assertion (A) and Reason (R) are false.
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Official CBSE Marking Scheme · CBSE Board 2026 (Main), Set 55/1/1

(C) Assertion (A) is true, but Reason (R) is false.

Q2.2026 Main 55/3/1 · Q101 mark

If r1 and r2 are the radii of atomic nuclei of mass numbers 64 and 27 respectively, then the value of (r1r2) is :

  1. 1
  2. 43
  3. 34
  4. 2764
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Official CBSE Marking Scheme · CBSE Board 2026 (Main), Set 55/3/1

(B) 43

Q3.2026 Main 55/3/1 · Q161 mark

Assertion (A) : Nuclear forces are always attractive.

Reason (R) : The nuclear force between protons and neutrons in a nucleus is a weak force.

  1. Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of the Assertion (A).
  2. Both Assertion (A) and Reason (R) are true, but Reason (R) is not the correct explanation of the Assertion (A).
  3. Assertion (A) is true, but Reason (R) is false.
  4. Both Assertion (A) and Reason (R) are false.
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Official CBSE Marking Scheme · CBSE Board 2026 (Main), Set 55/3/1

(D) Both Assertion (A) and Reason (R) are false

Q4.2026 Main 55/5/1 · Q81 mark

Nuclides with the same number of neutrons are called :

  1. Isobars
  2. Isotones
  3. Isotopes
  4. Isomers
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Official CBSE Marking Scheme · CBSE Board 2026 (Main), Set 55/5/1

(B) Isotone

Q5.2026 Main 55/5/1 · Q91 mark

The radius of a nucleus of mass number 125 is

  1. 6·0 fm
  2. 30 fm
  3. 72 fm
  4. 150 fm
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Official CBSE Marking Scheme · CBSE Board 2026 (Main), Set 55/5/1

(A) 6.0 fm (Note: Award full marks for attempt)

Q6.2026 Main 55/5/1 · Q151 mark

Assertion (A) : Energy is released when heavy nuclei undergo fission or light nuclei undergo fusion.

Reason (R) : For heavy nuclei, binding energy per nucleon increases with increasing Z while for light nuclei, it decreases with increasing Z.

  1. Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of the Assertion (A).
  2. Both Assertion (A) and Reason (R) are true, but Reason (R) is not the correct explanation of the Assertion (A).
  3. Assertion (A) is true, but Reason (R) is false.
  4. Both Assertion (A) and Reason (R) are false.
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Official CBSE Marking Scheme · CBSE Board 2026 (Main), Set 55/5/1

(C) Assertion (A) is true, but reason (R) is false.

Q7.2026 Supp 55/7/1 · Q151 mark

Assertion (A) : A free neutron is not a stable particle.

Reason (R) : It decays into a proton and an electron.

  1. Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of the Assertion (A).
  2. Both Assertion (A) and Reason (R) are true, but Reason (R) is not the correct explanation of the Assertion (A).
  3. Assertion (A) is true, but Reason (R) is false.
  4. Both Assertion (A) and Reason (R) are false.
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Official CBSE Marking Scheme · CBSE Board 2026 (Supplementary), Set 55/7/1

(C) Assertion (A) is true, but Reason (R) is false.

Q8.2025 Main 55/1/1 · Q111 mark

Which of the following figures correctly represent the shape of curve of binding energy per nucleon as a function of mass number ?

  1. Curve of B.E./A versus A with peak at A = 56, gradual fall afterwards
  2. Curve of B.E./A versus A rising steeply to A = 56 then nearly flat
  3. Symmetric bell-shaped curve of B.E./A with peak at A = 80
  4. Curve of B.E./A rising to a peak at A = 80 then falling and rising slightly
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Q9.2025 Main 55/5/1 · Q141 mark

Assertion (A) : During formation of a nucleus, the mass defect produced is the source of the binding energy of the nucleus.

Reason (R) : For all nuclei, the value of binding energy per nucleon increases with mass number.

  1. Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of the Assertion (A).
  2. Both Assertion (A) and Reason (R) are true, but Reason (R) is not the correct explanation of the Assertion (A).
  3. Assertion (A) is true, but Reason (R) is false.
  4. Both Assertion (A) and Reason (R) are false.
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Q10.2025 Main 55/6/1 · Q121 mark

Inside a nucleus, the nuclear forces between proton and proton, proton and neutron, neutron and neutron are Fpp, Fpn and Fnn respectively. Then :

  1. Fpp > Fpn > Fnn
  2. Fpn > Fnn > Fpp
  3. Fnn > Fpp > Fpn
  4. Fpp = Fpn = Fnn
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Q11.2025 Main 55/6/1 · Q161 mark

Assertion (A) : The binding energy per nucleon is practically constant for mass number in the range (30 < A < 170).

Reason (R) : Nuclear forces between the nucleons for mass numbers in the range (30 < A < 170) are not short-range.

  1. Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of the Assertion (A).
  2. Both Assertion (A) and Reason (R) are true, but Reason (R) is not the correct explanation of the Assertion (A).
  3. Assertion (A) is true, but Reason (R) is false.
  4. Both Assertion (A) and Reason (R) are false.
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Q12.2025 Main 55/7/1 · Q121 mark

Isotones are the nuclides having :

  1. same mass numbers
  2. same atomic numbers
  3. same neutron number, but different atomic number
  4. different neutron number, and different mass number
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Q13.2025 Supp 55/S/1 · Q31 mark

The potential energy of a pair of nucleons is minimum when they are separated by about :

  1. 0·8 × 10–15 m
  2. 0·8 × 10–10 m
  3. 2·0 × 10–13 m
  4. 2·0 × 10–14 m
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Official CBSE Marking Scheme · CBSE Board 2025 (Supplementary), Set 55/S/1

(A) 0.8 × 10–15 m

Q14.2024 Main 55/5/1 · Q111 mark

The potential energy between two nucleons inside a nucleus is minimum at a distance of about

  1. 0.8 fm
  2. 1.6 fm
  3. 2.0 fm
  4. 2.8 fm
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Official CBSE Marking Scheme · CBSE Board 2024 (Main), Set 55/5/1

(A) 0.8 fm

Q15.2024 Supp 55/S/1 · Q31 mark

Isotones are nuclides having :

  1. same number of neutrons but different number of protons
  2. same number of protons but different number of neutrons
  3. same number of protons and also same number of neutrons
  4. different number of protons and also different number of neutrons
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Official CBSE Marking Scheme · CBSE Board 2024 (Supplementary), Set 55/S/1

(A) Same number of neutrons but different number of protons

Q16.2024 Supp 55/S/1 · Q161 mark

Assertion (A) : Nuclear fission reactions are responsible for energy generation in the Sun.

Reason (R) : Light nuclei fuse together in the nuclear fission reactions.

  1. Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of the Assertion (A).
  2. Both Assertion (A) and Reason (R) are true, but Reason (R) is not the correct explanation of the Assertion (A).
  3. Assertion (A) is true, but Reason (R) is false.
  4. Both Assertion (A) and Reason (R) are false.
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Official CBSE Marking Scheme · CBSE Board 2024 (Supplementary), Set 55/S/1

(D) Both Assertion (A) and Reason (R) are false.

Q17.2023 Main 55/1/1 · Q121 mark

The mass density of a nucleus of mass number A is :

  1. proportional to A1/3
  2. proportional to A2/3
  3. proportional to A3
  4. independent of A
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Official CBSE Marking Scheme · CBSE Board 2023 (Main), Set 55/1/1

(D) Independent of A

Q18.2023 Main 55/2/1 · Q121 mark

The curve of binding energy per nucleon as a function of atomic mass number has a sharp peak for helium nucleus. This implies that helium nucleus is

  1. radioactive
  2. unstable
  3. easily fissionable
  4. more stable nucleus than its neighbours
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Official CBSE Marking Scheme · CBSE Board 2023 (Main), Set 55/2/1

(D) More stable nucleus than its neighbours

Q19.2023 Main 55/3/1 · Q181 mark

Assertion (A) : The nucleus 73X is more stable than the nucleus 43Y.

Reason (R) : 73X contains more number of protons.

  1. Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of the Assertion (A).
  2. Both Assertion (A) and Reason (R) are true, but Reason (R) is not the correct explanation of the Assertion (A).
  3. Assertion (A) is true, but Reason (R) is false.
  4. Assertion (A) is false and Reason (R) is also false.
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Official CBSE Marking Scheme · CBSE Board 2023 (Main), Set 55/3/1

(C) Assertion (A) is true, but Reason (R) is false

Q20.2023 Main 55/4/1 · Q81 mark

Which of the following statements is not true for nuclear forces ?

  1. They are stronger than Coulomb forces.
  2. They have about the same magnitude for different pairs of nucleons.
  3. They are always attractive.
  4. They saturate as the separation between two nucleons increases.
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Official CBSE Marking Scheme · CBSE Board 2023 (Main), Set 55/4/1

(C) They are always attractive

Q21.2023 Main 55/4/1 · Q141 mark

The difference in mass of 7X nucleus and total mass of its constituent nucleons is 21·00 u. The binding energy per nucleon for this nucleus is equal to the energy equivalent of :

  1. 3 u
  2. 3·5 u
  3. 7 u
  4. 21 u
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Official CBSE Marking Scheme · CBSE Board 2023 (Main), Set 55/4/1

(A) 3 u

Q22.2023 Main 55/5/1 · Q81 mark

The ratio of the nuclear densities of two nuclei having mass numbers 64 and 125 is

  1. 64125
  2. 45
  3. 54
  4. 1
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Official CBSE Marking Scheme · CBSE Board 2023 (Main), Set 55/5/1

(D) 1

Q23.2023 Supp 55/C/1 · Q81 mark

In the nuclear reaction 7N14 + 2He4 → X + 1H1, X represents :

  1. 7O16
  2. 8N17
  3. 8O17
  4. 7N16
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Official CBSE Marking Scheme · CBSE Board 2023 (Supplementary), Set 55/C/1

(C) 8O17

Q24.2023 Supp 55/C/1 · Q121 mark

Two nuclei have their mass numbers in the ratio of 1 : 27. What is the ratio of their nuclear densities ?

  1. 1 : 27
  2. 1 : 1
  3. 1 : 9
  4. 1 : 3
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Official CBSE Marking Scheme · CBSE Board 2023 (Supplementary), Set 55/C/1

(B) 1 : 1

Q25.SQP 2025-26 · Q111 mark

The ratio of the nuclear densities of two nuclei having the mass numbers 8 and 27 is

  1. 8:27
  2. 3:2
  3. 2:3
  4. 1:1
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Official CBSE Marking Scheme · CBSE Sample Question Paper 2025-26

(D) 1:1

Nuclear density does not depend on mass number.

Q26.SQP 2026-27 · Q71 mark

The binding energy per nucleon curve shows a maximum around mass number A=56. What does this imply about nuclei near iron (Fe)?

  1. They are the least stable nuclei.
  2. They are the most tightly bound nuclei.
  3. They undergo spontaneous fission releasing maximum energy.
  4. They cannot participate in nuclear reactions.
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Official CBSE Marking Scheme · CBSE Sample Question Paper 2026-27

(B) They are the most tightly bound nuclei.

A maximum in binding energy per nucleon means these nuclei are the most stable and most tightly bound.

  • They do not undergo spontaneous fission (that’s typical for very heavy nuclei like uranium).
  • They are certainly not the least stable — in fact, they are the most stable.
  • They can still participate in nuclear reactions, but they are at the “energy valley,” so both fusion (lighter nuclei) and fission (heavier nuclei) tend to release energy by moving toward iron.
Q27.2026 Main 55/1/1 · Q212 marks

Draw the plot of potential energy of a pair of nucleons as a function of their separation. Write two important conclusions that can be drawn from this plot.

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Official CBSE Marking Scheme · CBSE Board 2026 (Main), Set 55/1/1

Graph [1]

Potential energy of a pair of nucleons versus separation r, with a minimum at r0

Important conclusions (any two) [½ + ½]

  • Potential energy is minimum at a distance r0 of about 0.8 fm
  • Force is attractive for distances larger than 0.8 fm.
  • Force is repulsive for distances smaller than 0.8 fm
Q28.2026 Main 55/3/1 · Q192 marks

Explain the terms mass defect and binding energy. How are they related ?

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Official CBSE Marking Scheme · CBSE Board 2026 (Main), Set 55/3/1

The difference in mass of a nucleus and its constituents is called mass defect. [½]

If a certain number of neutrons and protons are brought together to form a nucleus of a certain mass and charge, the energy released is called binding energy. [½]

Relation between ΔM and E : [1]

Eb = ΔMc2

= [[Zmp + (A − Z)mn] − M]c2

Q29.2026 Supp 55/7/1 · Q202 marks

Draw a plot showing the variation of potential energy of a pair of nucleons as a function of their separation. With the help of the plot, explain why the nuclear force is attractive for r > r0 and repulsive for r < r0.

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Official CBSE Marking Scheme · CBSE Board 2026 (Supplementary), Set 55/7/1
Potential energy of a pair of nucleons versus separation r, with minimum at r0

Plot [1]

If r > r0, the potential energy gradually decreases with increasing r which indicates attractive force. [½]

If r < r0, the potential energy increases rapidly with decreasing r which indicates repulsive force. [½]

Q30.2025 Main 55/4/1 · Q202 marks

Calculate the mass of an α-particle in atomic mass unit (u). Given,
Mass of a normal helium atom = 4·002603 u
Mass of carbon atom = 1·9926 × 10−26 kg

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Q31.2025 Main 55/5/1 · Q202 marks

When a neutron collides with 23592U, the nucleus gives 14054Xe and 9438Sr as fission products and two neutrons are ejected. Calculate the mass defect and the energy released (in MeV) in the process. Given :

m(23592U) = 235·04393 u, m(14054Xe) = 139·92164 u

m(9438Sr) = 93·91536 u, 10n = 1·00866 u

1 u = 931 MeV/c2

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Q32.2025 Main 55/7/1 · Q212 marks

Briefly explain how energy is produced in stars, giving two examples of the nuclear reactions involved.

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Q33.2024 Main 55/2/1 · Q202 marks

The carbon isotope 126C has a nuclear mass of 12·000000 u. Calculate the binding energy of its nucleus.
Given mp = 1·007825 u; mn = 1·008665 u.

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Official CBSE Marking Scheme · CBSE Board 2024 (Main), Set 55/2/1

Binding Energy = (Zmp + (A − Z)mn − MN) × 931·5 MeV [½]

B.E. = (6 × 1·007825 + 6 × 1·008665 − 12·000000) × 931·5 MeV [½]

= (0·09894) × 931·5 MeV [½]

B.E. = 92·16 MeV [½]

Q34.2023 Main 55/2/1 · Q23 (OR option)2 marks

Define the term, mass defect. How is it related to stability of the nucleus ?

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Official CBSE Marking Scheme · CBSE Board 2023 (Main), Set 55/2/1

Mass defect is the difference between the actual mass of the nucleus and the sum of the masses of its nucleons. [1]

Greater the mass defect, greater will be the binding energy and the nucleus will be more stable. [1]

Alternatively

Give full credit (1 mark) if a candidate writes, mass defect ∝ stability of the nucleus.

Q35.2023 Main 55/3/1 · Q23 (option 1)2 marks

Draw a graph showing the variation of binding energy per nucleon as a function of mass number A. The binding energy per nucleon for heavy nuclei (A > 170) decreases with the increase in mass number. Explain.

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Official CBSE Marking Scheme · CBSE Board 2023 (Main), Set 55/3/1

Graph of binding energy per nucleon as a function of mass number A [1]

Binding energy per nucleon (MeV) versus mass number A

Explanation: Nuclear forces are short range & show saturation, while the electrostatic force are neither short range nor show any saturation. Hence for heavier nuclei (A > 170) the electrostatic force of repulsion becomes predominant, decreasing the binding energy per nucleon. [1]

Alternatively: As the size of the nucleus increases, the nucleus becomes unstable.

Note: No deduction of marks if values of elements are not shown in the graph.

Q36.2023 Main 55/4/1 · Q242 marks

Draw the graph showing the variation of binding energy per nucleon with mass number A of nuclei (2 < A < 170). Use this graph to explain the release of energy in nuclear fission.

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Official CBSE Marking Scheme · CBSE Board 2023 (Main), Set 55/4/1

Graph [1]

Binding energy per nucleon (MeV) versus mass number A

Note : Full marks to be awarded even if values are not marked.

Explanation : A very heavy nucleus has lower binding energy per nucleon compared to that of lighter nuclei. Thus if a heavier nucleus breaks into two nuclei, nucleons get more tightly bound. This implies, energy would be released in the process. [1]

Alternatively

In nuclear fission, a heavy nucleus breaks, B.E/nucleon increases. So energy is released. [1]

Q37.2023 Main 55/5/1 · Q192 marks

Draw a graph showing the variation of potential energy of a pair of nucleons as a function of their separation. Indicate the region in which the nuclear force is (a) attractive and (b) repulsive.

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Official CBSE Marking Scheme · CBSE Board 2023 (Main), Set 55/5/1

Graph between potential energy and separation between the nucleons [1]

Potential energy versus separation of nucleons with minimum at r0

Repulsive nuclear force for r < r0 [½]

Attractive force for r > r0 [½]

Note: If a student draws a graph without indicating the region of attractive and repulsive force, award 1 mark only.

Q38.2022 Main 55/3/1 · Q2 (option 1)2 marks
  1. Distinguish between isotopes and isobars.
  2. Two nuclei have different mass numbers A1 and A2. Are these nuclei necessarily the isotopes of the same element ? Explain.
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Official CBSE Marking Scheme · CBSE Board 2022 (Main), Set 55/3/1

(i) Isotopes – These are the atoms having same atomic number (Z) but different atomic mass (A). [½]

Isobars – The atoms of different element having same atomic masses. [½]

(ii) No [½]

The mass number of a nucleus is the sum of number of proton (Z) and number of neutrons (N) / A = Z + N. Two nuclei with different mass numbers A1 and A2 may have, have different Z. [½]

Q39.2026 Main 55/2/1 · Q223 marks

(a) Consider the following nuclides :

126C, 19880Hg, 146C, 19779Au

Group them into isotopes and isotones.

(b) How does the size of a nucleus depend on its mass number A ? Hence prove that the density of nucleus is a constant, independent of A, for all nuclei.

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Official CBSE Marking Scheme · CBSE Board 2026 (Main), Set 55/2/1

(a) Isotopes : 126C, 146C [½]

Isotones : 19880Hg, 19779Au [½]

(b) Size of nucleus ∝ A1/3 [½]

Alternatively : R = R0A1/3

Suppose m is the average mass of a nucleon, density of nucleus

ρ = MassVolume [½]

ρ = mA4/3π(R0A1/3)3 [½]

ρ = 3m4πR03 = constant [½]

Q40.2026 Main 55/3/1 · Q273 marks

Differentiate between nuclear fission and nuclear fusion. Give one example for each with nuclear reaction.

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Official CBSE Marking Scheme · CBSE Board 2026 (Main), Set 55/3/1

Nuclear fission is decay of heavy nucleus into two or more intermediate mass fragments whereas in nuclear fusion, light nuclei fuse into a heavier nucleus. [1]

Nuclear Fission :

01n + 92235U → 56144Ba + 3689Kr + 3 01n [1]

Nuclear Fusion :

11H + 11H → 12H + e+ + ν + Q [1]

Q41.2026 Main 55/4/1 · Q273 marks

In the nuclear reaction :

21H + 21H → A2X + 10n

  1. Find the value of A.
  2. Calculate the amount of energy released in the reaction.

Given : m(21H) = 2·014102 u, m(A2X) = 3·016049 u
m(10n) = 1·008665 u, 1 u = 931·5 MeV/c2

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Official CBSE Marking Scheme · CBSE Board 2026 (Main), Set 55/4/1

(a) 21H + 21H → A2X + 10n

Mass number of reactants = Mass number of products

∴ A = 3 [1]

(b) Mass defect, ΔM = Mass of reactants − Mass of products

= [2 × m(21H)] − [m(A2X) + m(10n)] [½]

= [2 × 2.014102] − [3.016049 + 1.008665] [½]

ΔM = 0.00349 u [½]

Energy released = ΔM × 931.5 MeV/c2 = 0.00349 × 931.5

Energy released = 3.25 MeV [½]

Q42.2026 Main 55/5/1 · Q26 (option 1)3 marks
  1. Write any two features of nuclear forces.
  2. If both the number of protons and the neutrons are conserved in each nuclear reaction, in what way is mass converted into energy (or vice versa) in a nuclear reaction ? Explain.
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Official CBSE Marking Scheme · CBSE Board 2026 (Main), Set 55/5/1

(i) Writing two features of nuclear force [2]; (ii) Explaining how mass converted into energy (or vice versa) in a nuclear reaction [1]

(i) Features of nuclear force: [1 + 1]

  1. charge independent
  2. strongest force
  3. short-range force / saturated force
  4. does not follow inverse square law

Note: Award full marks for any two valid points.

(ii) The total mass of the nuclei on the left-hand side need not be same as that on right hand side of nuclear reaction. Since binding energy contributes to mass, the difference in these binding energies appear as energy released or absorbed in a nuclear reaction. [1]

Q43.2025 Main 55/1/1 · Q273 marks

(a) Define ‘Mass defect’ and ‘Binding energy’ of a nucleus. Describe ‘Fission process’ on the basis of binding energy per nucleon.

(b) A deuteron contains a proton and a neutron and has a mass of 2.013553 u. Calculate the mass defect for it in u and its energy equivalence in MeV. (mp = 1.007277 u, mn = 1.008665 u, 1u = 931.5 MeV/c2)

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Q44.2025 Main 55/2/1 · Q273 marks

(a) Show the variation of binding energy per nucleon with mass number. Write the significance of the binding energy curve.

(b) Two nuclei with lower binding energy per nucleon form a nuclei with more binding energy per nucleon.

  1. What type of nuclear reaction is it ?
  2. Whether the total mass of nuclei increases, decreases or remains unchanged ?
  3. Does the process require energy or produce energy ?
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Q45.2025 Main 55/6/1 · Q28 (option 1)3 marks

Consider the so-called ‘D-T reaction’ (Deuterium-Tritium reaction). In a thermonuclear fusion reactor, the following nuclear reaction occurs :

21H + 31H → 42He + 10n + Q

Find the amount of energy released in the reaction.

Given :

m(21H) = 2·014102 u
m(31H) = 3·016049 u
m(42He) = 4·002603 u
m(10n) = 1·008665 u
1 u = 931 MeV/c2

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Q46.2025 Main 55/6/1 · Q28 (OR option)3 marks

Show that the nuclear density is independent of mass number.

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Q47.2025 Supp 55/S/1 · Q263 marks

Define binding energy (BE) of a nucleus. An unstable nucleus has mass number A = 240 and BEA = 7·6 MeV. It splits in two fragments, each of A = 120 with BEA = 8·5 MeV. Calculate the energy released.

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Official CBSE Marking Scheme · CBSE Board 2025 (Supplementary), Set 55/S/1

Binding energy : Energy released in bringing all the nucleons of a nucleus together to form a nucleus. [1]

Alternatively : Energy required to separate the nucleons from the nucleus.

Binding Energy of X, B.EX = 240 × 7.6 = 1824 MeV [½]

Binding Energy of Y, B.EY = 120 × 8.5 = 1020 MeV [½]

Energy released E = 2(B.EY) – (B.EX) [½]

= (2 × 1020) – 1824

= 216 MeV [½]

Q48.2024 Main 55/1/1 · Q283 marks

(a) State any two properties of a nucleus.

(b) Why is the density of a nucleus much more than that of an atom ?

(c) Show that the density of the nuclear matter is the same for all nuclei.

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Official CBSE Marking Scheme · CBSE Board 2024 (Main), Set 55/1/1

a) (Any TWO)
(i) The nucleus is positively charged
(ii) The nucleus consists of protons and neutrons
(iii) The nuclear density is independent of mass number
(iv) The radius of the nucleus, R = R0 A1/3 [½ + ½]

b) Atoms have large amount of empty spaces. Mass is concentrated in nucleus. [1]

c) Density = Mass / Volume

= m A43πR3 = m A43πR03 A = m43πR03

So, density is independent of mass number [1]

Q49.2024 Main 55/3/1 · Q273 marks

(a) Define atomic mass unit (u).

(b) Calculate the energy required to separate a deuteron into its constituent parts (a proton and a neutron). Given :

m(D) = 2·014102 u

mH = 1·007825 u

mn = 1·008665 u

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Official CBSE Marking Scheme · CBSE Board 2024 (Main), Set 55/3/1

(a) Atomic mass unit (u) is defined as 1/12th of the mass of the carbon (12C) atom. [1]

(b) m(1H2) → m(1H1) + m(0n1) [½]

Q = (mR – mP) × 931.5 MeV

= (2.014102 – 1.007825 – 1.008665) × 931.5 MeV [½]

= – 0.002388 × 931.5 MeV [½]

= – 2.224 MeV

Hence energy required is 2.224 MeV [½]

Q50.2024 Main 55/4/1 · Q223 marks

(a) Differentiate between nuclear fission and fusion.

(b) The fission properties of 94Pu239 are very similar to those of 92U235. How much energy (in MeV), is released if all the atoms in 1 g of pure 94Pu239 undergo fission ? The average energy released per fission is 180 MeV.

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Official CBSE Marking Scheme · CBSE Board 2024 (Main), Set 55/4/1

(a) In nuclear fission, a heavy nucleus splits into two or more lighter nuclei and energy is released. [½]

In nuclear fusion, lighter nuclei combine together to form a heavy nucleus and larger amount of energy is released. [½]

(b) Number of atoms in 1 g of 94Pu239

= 6.023 × 1023239 = 2.5 × 1021 [1]

Energy released in fission of 1 g of 94Pu239,

E = 180 MeV × 2.5 × 1021

E = 4.5 × 1023 MeV [1]

Q51.2023 Main 55/1/1 · Q30 (OR option)3 marks

Draw a diagram to show the variation of binding energy per nucleon with mass number for different nuclei and mention its two features. Why do lighter nuclei usually undergo nuclear fusion ?

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Official CBSE Marking Scheme · CBSE Board 2023 (Main), Set 55/1/1

Diagram to show variation [1]; Two features of diagram [½ + ½]; Reason for nuclear fusion [1]

Binding energy per nucleon versus mass number curve

[1]

(Note: Award full credit even if a student does not mark so many elements and does not mention the values of Ebn.)

Features of diagram (any two)

1. Binding energy per nucleon is practically independent of atomic number for nuclei of middle mass number (30 < A < 170)

2. The curve has maximum of about 8.75 MeV for A = 56 and has a value of 7.6 MeV for A = 238

3. Binding energy per nucleon is lower for both light nuclei (A < 30) and heavy nuclei (A > 170) [½ + ½]

Two lighter nuclei fuse together to form heavier nuclei as the binding energy per nucleon of fused heavier nuclei is more than the binding energy per nucleon of the lighter nuclei. Thus the final system is more tightly bound than initial system. [1]

Alternatively

To attain the stability [1]

Q52.2023 Main 55/2/1 · Q30 (option 1)3 marks
  1. Distinguish between nuclear fission and fusion giving an example of each.
  2. Explain the release of energy in nuclear fission and fusion on the basis of binding energy per nucleon curve.
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Official CBSE Marking Scheme · CBSE Board 2023 (Main), Set 55/2/1

(i) Nuclear fission – It is a process in which a heavy nucleus when excited (say on bombarding by a slow moving neutron) splits into two lighter nuclei of nearly comparable masses with a release of large amount of energy. [½]

Example of nuclear fission [½]

01n + 92235U → 92236U → 56144Ba + 3689Kr + 3 01n + Q

Nuclear fusion – It is a process in which two lighter nuclei fuse (at extremely high temperature) to form a heavy nucleus and large amount of energy is released. [½]

Examples of nuclear fusion [½]

(i) 11H + 11H → 12H + e+ + ν + Q1

(ii) 12H + 12H → 23He + n + Q2

(iii) 12H + 12H → 13H + 11H + Q3

(any other possible reaction equation)

(ii) The binding energy per nucleon of the products in the nuclear reactions (nuclear fission and nuclear fusion) is greater than that of the reactants. [1]

Q53.2023 Main 55/2/1 · Q30 (OR option)3 marks
  1. How is the size of a nucleus found experimentally ? Write the relation between the radius and mass number of a nucleus.
  2. Prove that the density of a nucleus is independent of its mass number.
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Official CBSE Marking Scheme · CBSE Board 2023 (Main), Set 55/2/1

(i) Size of nucleus of an atom is determined by scattering experiments in which fast electrons are used to bombard targets. [½]

Relation between radius and mass number of nucleus: R = R0A1/3 [1]

(ii) Density of nucleus

ρ = massvolume [½]

ρ = m × A(4/3) πR3 = m A(4/3) π (R0A1/3)3 [½]

ρ = 3m4πR03

Hence, density of nucleus is independent of mass number (A). [½]

Q54.2023 Main 55/3/1 · Q30 (option 1)3 marks

(i) Prove that the nuclear density is same for all nuclei.

(ii) Draw a plot of potential energy of a pair of nucleons as a function of their separation. Draw two inferences from this plot.

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Official CBSE Marking Scheme · CBSE Board 2023 (Main), Set 55/3/1

(i) ρ = massvolume = mass number × mass of nucleonvolume of nucleus [½]

ρ = A × m4⁄3 π (R0A1/3)3 = 3m4π R03 [½]

Hence, density is independent of mass number.

(ii) Plot of potential energy U versus separation r [1]

Potential energy of a pair of nucleons versus separation, with minimum at r0

Inferences [½ + ½]

– The force is attractive for distances larger than r0.

– The force is repulsive for distance less than r0.

Alternatively: Any other relevant inference drawn from the graph should be given full credit.

Q55.2022 Main 55/1/1 · Q63 marks

Calculate the energy released in MeV in the following reaction :

21H + 31H → 42He + n

Given : m(21H) = 2.014102 u
m(31H) = 3.016049 u
m(42He) = 4.002603 u
mn = 1.008665 u

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Official CBSE Marking Scheme · CBSE Board 2022 (Main), Set 55/1/1

21H + 31H → 42He + n + Energy

Mass defect = mass of reactants − mass of products [½]

Δm = m(21H + 31H) − m(42He + 10n) [½]

Mass defect = (2·014102 + 3·016049) − (4·002603 + 1·008665) [½]

= 5·030151 − 5·011268 [½]

= 0·018883 u [½]

Energy released = Δm × 931.5 MeV = 0·018883 × 931.5 MeV [½]

= 17·58 MeV [½]

Q56.2022 Main 55/2/1 · Q93 marks
  1. Depict the variation of the potential energy of a pair of nucleons with the separation between them.
  2. Imagine the fission of a 5626Fe into two equal fragments of 2813Al nucleus. Is the fission energetically possible ? Justify your answer by working out Q value of the process.

Given : m(5626Fe) = 55·93494 u, m(2813Al) = 27·98191 u.

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Official CBSE Marking Scheme · CBSE Board 2022 (Main), Set 55/2/1

(a) Graph [1½]

Potential energy of a pair of nucleons versus separation

(b) Mass Difference = 55.93494 – 2 × 27.98191 = – 0.02442 u [1]

Fission not possible. [½]

Q57.2022 Main 55/4/1 · Q43 marks

(a) Differentiate between nuclear fission and nuclear fusion.

(b) Deuterium undergoes fusion as per the reaction :

21H + 21H ⟶ 32He + 10n + 3·27 MeV

Find the duration for which an electric bulb of 500 W can be kept glowing by the fusion of 100 g of deuterium.

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Official CBSE Marking Scheme · CBSE Board 2022 (Main), Set 55/4/1

(a) Nuclear fission – The process of breaking a very heavy nucleus into lighter nuclei, having mass number in the range of middle mass number (30 < A < 170). [½]

Nuclear Fusion – It is the process of joining of very light nuclei (A ≤ 10) to form a heavier nucleus. [½]

(b) No. of atoms in 100 g = 6·023 × 10232 × 100 = 3·0115 × 1025 [½]

Energy released/atom = 3·27 MeV2 = 1·635 MeV

Total energy released = 3·0115 × 1025 × 1·635 MeV

= 3·0115 × 1·635 × 1025 × 1·6 × 10−13

= 7·878 × 1012 J [½]

t = EP [½]

= 7·878 × 1012 J500 J/s = 1.5756 × 1010 s

= 1.5756 × 10103·15 × 107 ≃ 500 y [½]

Alternatively :

E = MQ2md [½]

= (0·1 kg) × (3·27 MeV)2(2·04) × (1·66 × 10−27 kg / 4)

= 0·0492 × 1027 = 4·92 × 1025 MeV [½]

t = EP [½]

= (4·92 × 1025) × (1·6 × 10−13)500 = 1.5756 × 1010 s

= 1.5756 × 10103·15 × 107 ≃ 500 y [½]

Q58.2022 Main 55/5/1 · Q63 marks

In a fission event of 23892U by fast moving neutrons, no neutrons are emitted and final products, after the beta decay of the primary fragments, are 14058Ce and 9944Ru. Calculate Q for this process. Neglect the masses of electrons/positrons emitted during the intermediate steps.

Given : m(23892U) = 238.05079u; m(14058Ce) = 139.90543u; m(9944Ru) = 98.90594u; m(10n) = 1.008665u

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Official CBSE Marking Scheme · CBSE Board 2022 (Main), Set 55/5/1

Calculation of mass defect [2]; Calculation of Q value [1]

Δm = total mass of the reactants – total mass of the products [½]

= [m(23892U) + mn – m(14058Ce) – m(9944Ru)] [½]

= [238.05079 + 1.008665 – 139.90543 – 98.90594] u [½]

= [239.059455 – 238.81137] u

= 0.248085 u [½]

Q-value = 0.248085 × 931.5 MeV

= 231.09 MeV [1]

(Note: Award this 1 mark even if Q-value is not calculated)

Q59.2022 Supp 55/6/1 · Q93 marks
  1. Determine from the given data, whether the following reaction is exothermic or endothermic : [2]

    11H + 31H → 21H + 21H

    Atomic masses : m(21H) = 2·014102 u, m(31H) = 3·016049 u, m(11H) = 1·007825 u

  2. What is the ratio of the nuclear density of the gold isotope 19779Au and the silver isotope 10747Ag. [1]
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Official CBSE Marking Scheme · CBSE Board 2022 (Supplementary), Set 55/6/1

(a) Mass of reactants = (1·007825 + 3·016049) u = 4·023874 u [½]

Mass of products = (2 × 2·014102) u = 4·028204 u [½]

Mass of reactants < Mass of products [½]

Hence, reaction is Endothermic [½]

(b) 1 : 1 [1]

Q60.SQP 2025-26 · Q243 marks

A fast-moving neutron collides with the nucleus of Plutonium (Pu), thereby producing Xenon (Xe) and Zirconium (Zr) along with neutrons.

(I) Write the nuclear fission reaction.
(II) Find the energy released in the above nuclear reaction.

Given atomic masses:
m (23994Pu) = 239.052157u,
m (10340Zr) = 102.926597u,
m (13454Xe) = 133.905040u &
m (10n) = 1.00866u.

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Official CBSE Marking Scheme · CBSE Sample Question Paper 2025-26

(I) Nuclear fission reaction.

23994Pu + 10n → 13454Xe + 10340Zr + 3 10n [1]

(II) Δm = [m (23994Pu) + m (10n)] – [m (13454Xe) + m (10340Zr) + 3 m (10n)]

= [239.052157 + 1.00866] – [133.905040 + 102.926597 + 3 × 1.00866] [1]

= 240.060817 – 239.857617

= 0.2032 amu

Q value = Δm c2

= 0.2032 × 931.5 MeV

= 189.2808 MeV [1]

Q61.SQP 2026-27 · Q243 marks

The atomic mass of 168O is 15.99493 u. Calculate the binding energy per nucleon of the oxygen nucleus.

(Given: mass of proton = 1.00727 u, mass of neutron = 1.00866 u, 1 u = 931.5 MeV/c2)

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Official CBSE Marking Scheme · CBSE Sample Question Paper 2026-27

Z = 8 protons, N = 8 neutrons.

Sum of nucleon masses: mpZ + mnN = 8(1.00727) + 8(1.00866) = 8.05816 + 8.06928 = 16.12744 u [1]

Mass defect: Δm = 16.12744 – 15.99493 = 0.132514 u

Binding energy: B = Δm × 931.5 = 0.13251 × 931.5 ≈ 123.4 MeV [1]

Binding energy per nucleon: BA = 123.416 ≈ 7.71 MeV [1]

Unit IX · Electronic Devices

14. Semiconductor Electronics: Materials, Devices and Simple Circuits

Q1.2026 Main 55/1/1 · Q121 mark

In an unbiased p-n junction, at equilibrium, which of the following statements is true ?

  1. Diffusion current is zero but drift current exists.
  2. Diffusion current exists but drift current is zero.
  3. Diffusion and drift currents are equal and opposite.
  4. Both the diffusion and drift currents exist but are unequal.
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Official CBSE Marking Scheme · CBSE Board 2026 (Main), Set 55/1/1

(C) Diffusion and drift currents are equal and opposite.

Q2.2026 Main 55/2/1 · Q141 mark

Assertion (A) : On forward biasing a p-n junction diode, the height of the barrier potential increases.

Reason (R) : In forward biasing of a p-n junction diode, the direction of the applied voltage is in the same direction as the built-in potential.

  1. Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of Assertion (A).
  2. Both Assertion (A) and Reason (R) are true, but Reason (R) is not the correct explanation of Assertion (A).
  3. Assertion (A) is true, but Reason (R) is false.
  4. Assertion (A) is false and Reason (R) is also false.
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Official CBSE Marking Scheme · CBSE Board 2026 (Main), Set 55/2/1

(D) Assertion (A) is false and Reason (R) is also false.

Q3.2026 Main 55/3/1 · Q111 mark

When the forward bias voltage in a semiconductor diode is changed from 0·8 V to 1·0 V, the forward current changes by 2·0 mA. The forward bias resistance of the diode will be :

  1. 200 Ω
  2. 175 Ω
  3. 100 Ω
  4. 125 Ω
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Official CBSE Marking Scheme · CBSE Board 2026 (Main), Set 55/3/1

(C) 100 Ω

Q4.2026 Main 55/3/1 · Q121 mark

The process named ‘minority carrier injection’ in a p-n junction diode occurs during :

  1. forward biasing
  2. reverse biasing
  3. no biasing at low temperature
  4. no biasing at high temperature
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Official CBSE Marking Scheme · CBSE Board 2026 (Main), Set 55/3/1

(A) forward biasing

Q5.2026 Main 55/4/1 · Q121 mark

In a reversed-biased p-n junction diode, the applied voltage mostly drops across :

  1. p-region only
  2. n-region only
  3. depletion region
  4. the diode
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Official CBSE Marking Scheme · CBSE Board 2026 (Main), Set 55/4/1

(C) Depletion region

Q6.2026 Main 55/4/1 · Q161 mark

Assertion (A) : The conductivity of an n-type semiconductor is higher than that of a p-type semiconductor at a given temperature.

Reason (R) : The electrons being in the conduction band in n-type semiconductor are more mobile than the holes in the valence band in p-type semiconductor.

  1. Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of the Assertion (A).
  2. Both Assertion (A) and Reason (R) are true, but Reason (R) is not the correct explanation of the Assertion (A).
  3. Assertion (A) is true, but Reason (R) is false.
  4. Both Assertion (A) and Reason (R) are false.
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Official CBSE Marking Scheme · CBSE Board 2026 (Main), Set 55/4/1

(A) Both Assertion (A) and Reason (R) are true, and Reason (R) is the correct explanation of Assertion (A).

Q7.2026 Main 55/5/1 · Q111 mark

A good diode checked by a multimeter should indicate :

  1. high resistance in reverse bias and a low resistance in forward bias
  2. high resistance in both forward bias and reverse bias
  3. low resistance in both reverse bias and forward bias
  4. high resistance in forward bias and low resistance in reverse bias
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Official CBSE Marking Scheme · CBSE Board 2026 (Main), Set 55/5/1

(A) high resistance in reverse bias and low resistance in forward bias

Q8.2026 Supp 55/7/1 · Q111 mark

In forward and reverse biased silicon p-n junction, the dominant mechanisms for the motion of charge carriers are :

  1. diffusion in forward bias, drift in reverse bias
  2. drift in both forward bias and in reverse bias
  3. drift in forward and diffusion in reverse bias
  4. diffusion in both forward and reverse bias
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Official CBSE Marking Scheme · CBSE Board 2026 (Supplementary), Set 55/7/1

(A) diffusion in forward bias, drift in reverse bias

Q9.2025 Main 55/1/1 · Q121 mark

When a p-n junction diode is forward biased

  1. the barrier height and the depletion layer width both increase.
  2. the barrier height increases and the depletion layer width decreases.
  3. the barrier height and the depletion layer width both decrease.
  4. the barrier height decreases and the depletion layer width increases.
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Q10.2025 Main 55/1/1 · Q151 mark

Assertion (A) : We cannot form a p-n junction diode by taking a slab of a p-type semiconductor and physically joining it to another slab of a n-type semiconductor.

Reason (R) : In a p-type semiconductor ηe >> ηh while in a n-type semiconductor ηh >> ηe.

  1. Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of Assertion (A).
  2. Both Assertion (A) and Reason (R) are true, but Reason (R) is not the correct explanation of Assertion (A).
  3. Assertion (A) is true, but Reason (R) is false.
  4. Assertion (A) is false and Reason (R) is also false.
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Q11.2025 Main 55/2/1 · Q121 mark

When the resistance measured between p and n ends of a p-n junction diode is high, it can act as a/an –

  1. resistor
  2. inductor
  3. capacitor
  4. switch
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Q12.2025 Main 55/2/1 · Q131 mark

For Questions 13 to 16, two statements are given – one labelled Assertion (A) and other labelled Reason (R). Select the correct answer to these questions from the codes (A), (B), (C) and (D) as given below :

Assertion (A) : In a semiconductor diode the thickness of depletion layer is not fixed.

Reason (R) : Thickness of depletion layer in a semiconductor device depends upon many factors such as biasing of the semiconductor.

  1. If both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of Assertion (A).
  2. If both Assertion (A) and Reason (R) are true but Reason (R) is not the correct explanation of Assertion (A).
  3. If Assertion (A) is true but Reason (R) is false.
  4. If both Assertion (A) and Reason (R) are false.
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Q13.2025 Main 55/4/1 · Q111 mark

Which of the following is an electrical conductor at room temperature ?

  1. Sn
  2. Mica
  3. Si
  4. C
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Q14.2025 Main 55/4/1 · Q131 mark

Questions number 13 to 16 are Assertion (A) and Reason (R) type questions. Two statements are given — one labelled Assertion (A) and the other labelled Reason (R). Select the correct answer from the codes (A), (B), (C) and (D) as given below.

Assertion (A) : n-type semiconductor is not negatively charged.

Reason (R) : Neutral pentavalent impurity atom doped in intrinsic semiconductor (neutral) donates its fifth unpaired electron to the crystal lattice and becomes a positive donor.

  1. Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of the Assertion (A).
  2. Both Assertion (A) and Reason (R) are true, but Reason (R) is not the correct explanation of the Assertion (A).
  3. Assertion (A) is true, but Reason (R) is false.
  4. Both Assertion (A) and Reason (R) are false.
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Q15.2025 Main 55/5/1 · Q131 mark

Assertion (A) : The impurities in p-type Si are not pentavalent atoms.

Reason (R) : The hole density in valance band in p-type semiconductor is almost equal to the acceptor density.

  1. Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of the Assertion (A).
  2. Both Assertion (A) and Reason (R) are true, but Reason (R) is not the correct explanation of the Assertion (A).
  3. Assertion (A) is true, but Reason (R) is false.
  4. Both Assertion (A) and Reason (R) are false.
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Q16.2025 Main 55/6/1 · Q141 mark

Assertion (A) : A hole is an apparent free particle with effective positive electronic charge.

Reason (R) : A hole is not necessarily a vacancy left behind by an electron in the valence band.

  1. Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of the Assertion (A).
  2. Both Assertion (A) and Reason (R) are true, but Reason (R) is not the correct explanation of the Assertion (A).
  3. Assertion (A) is true, but Reason (R) is false.
  4. Both Assertion (A) and Reason (R) are false.
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Q17.2025 Main 55/7/1 · Q91 mark

Germanium crystal is doped at room temperature with a minute quantity of boron. The charge carriers in the doped semiconductors will be :

  1. electrons only
  2. holes only
  3. holes and few electrons
  4. electrons and few holes
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Q18.2025 Main 55/7/1 · Q111 mark

A p-n junction diode is forward biased. As a result,

  1. both the potential barrier height and the width of depletion layer decrease.
  2. both the potential barrier height and the width of depletion layer increase.
  3. the potential barrier height decreases and the width of depletion layer increases.
  4. the potential barrier height increases and the width of depletion layer decreases.
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Q19.2025 Supp 55/S/1 · Q161 mark

Assertion (A) : The electrical conductivity of a pure Ge crystal increases with increase in its temperature.

Reason (R) : The number of electrons excited by thermal excitation from the valence band to the conduction band, in a semiconductor, increases with increase in temperature.

  1. Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of the Assertion (A).
  2. Both Assertion (A) and Reason (R) are true, but Reason (R) is not the correct explanation of the Assertion (A).
  3. Assertion (A) is true, but Reason (R) is false.
  4. Both Assertion (A) and Reason (R) are false.
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Official CBSE Marking Scheme · CBSE Board 2025 (Supplementary), Set 55/S/1

(A) Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of the Assertion (A).

Q20.2024 Main 55/1/1 · Q101 mark

Ge is doped with As. Due to doping,

  1. the structure of Ge lattice is distorted.
  2. the number of conduction electrons increases.
  3. the number of holes increases.
  4. the number of conduction electrons decreases.
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Official CBSE Marking Scheme · CBSE Board 2024 (Main), Set 55/1/1

(B) the number of conduction electrons increases

Q21.2024 Main 55/2/1 · Q111 mark

An n-type semiconducting Si is obtained by doping intrinsic Si with :

  1. Al
  2. B
  3. P
  4. In
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Official CBSE Marking Scheme · CBSE Board 2024 (Main), Set 55/2/1

(C) P

Q22.2024 Main 55/2/1 · Q121 mark

When a p-n junction diode is subjected to reverse biasing :

  1. the barrier height decreases and the depletion region widens.
  2. the barrier height increases and the depletion region widens.
  3. the barrier height decreases and the depletion region shrinks.
  4. the barrier height increases and the depletion region shrinks.
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Official CBSE Marking Scheme · CBSE Board 2024 (Main), Set 55/2/1

(B) The barrier height increases and the depletion region widens.

Q23.2024 Main 55/3/1 · Q121 mark

Si is doped with a pentavalent element. The energy required to set the additional electron free is about :

  1. 0·01 eV
  2. 0·05 eV
  3. 0·72 eV
  4. 1·1 eV
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Official CBSE Marking Scheme · CBSE Board 2024 (Main), Set 55/3/1

(B) 0.05 eV

Q24.2024 Main 55/3/1 · Q131 mark

Assertion (A) : In a semiconductor, the electrons in the conduction band have lesser energy than those in the valence band.

Reason (R) : Donor energy level is just above the valence band in a semiconductor.

  1. Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of the Assertion (A).
  2. Both Assertion (A) and Reason (R) are true, but Reason (R) is not the correct explanation of the Assertion (A).
  3. Assertion (A) is true, but Reason (R) is false.
  4. Assertion (A) is false and Reason (R) is also false.
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Official CBSE Marking Scheme · CBSE Board 2024 (Main), Set 55/3/1

(D) Assertion (A) is false and Reason (R) is also false.

Q25.2024 Main 55/5/1 · Q121 mark

A pure Si crystal having 5 × 1028 atoms m–3 is dopped with 1 ppm concentration of antimony. If the concentration of holes in the doped crystal is found to be 4.5 × 109 m–3, the concentration (in m–3) of intrinsic charge carriers in Si crystal is about

  1. 1.2 × 1015
  2. 1.5 × 1016
  3. 3.0 × 1015
  4. 2.0 × 1016
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Official CBSE Marking Scheme · CBSE Board 2024 (Main), Set 55/5/1

(B) 1.5×1016

Q26.2024 Supp 55/S/1 · Q111 mark

Which of the following statements is not true for a p-n junction diode under reverse bias ?

  1. The current is almost independent of the applied voltage.
  2. Holes flow from p-side to n-side.
  3. Electric field in the depletion region increases.
  4. n-side of the junction is connected to +ve terminal and p-side to –ve terminal of the battery.
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Official CBSE Marking Scheme · CBSE Board 2024 (Supplementary), Set 55/S/1

(B) Holes flow from p-side to n-side

Q27.2023 Main 55/1/1 · Q131 mark

An ac source of voltage is connected in series with a p-n junction diode and a load resistor. The correct option for output voltage across load resistance will be :

Four graphs (a)-(d) of output voltage V versus time t
  1. Graph (a)
  2. Graph (b)
  3. Graph (c)
  4. Graph (d)
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Official CBSE Marking Scheme · CBSE Board 2023 (Main), Set 55/1/1

(C)

Half-wave rectified output voltage
Q28.2023 Main 55/1/1 · Q141 mark

When an intrinsic semiconductor is doped with a small amount of trivalent impurity, then :

  1. its resistance increases.
  2. it becomes a p-type semiconductor.
  3. there will be more free electrons than holes in the semiconductor.
  4. dopant atoms become donor atoms.
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Official CBSE Marking Scheme · CBSE Board 2023 (Main), Set 55/1/1

(B) it becomes a p-type semiconductor

Q29.2023 Main 55/1/1 · Q151 mark

In the energy-band diagram of n-type Si, the gap between the bottom of the conduction band EC and the donor energy level ED is of the order of :

  1. 10 eV
  2. 1 eV
  3. 0·1 eV
  4. 0·01 eV
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Official CBSE Marking Scheme · CBSE Board 2023 (Main), Set 55/1/1

(D) 0.01 eV

Q30.2023 Main 55/2/1 · Q131 mark

In an extrinsic semiconductor, the number density of holes is 4 × 1020 m−3. If the number density of intrinsic carriers is 1.2 × 1015 m−3, the number density of electrons in it is

  1. 1.8 × 109 m−3
  2. 2.4 × 1010 m−3
  3. 3.6 × 109 m−3
  4. 3.2 × 1010 m−3
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Official CBSE Marking Scheme · CBSE Board 2023 (Main), Set 55/2/1

(C) 3.6 × 109 m−3

Q31.2023 Main 55/2/1 · Q141 mark

Pieces of copper and of silicon are initially at room temperature. Both are heated to temperature T. The conductivity of

  1. both increases.
  2. both decreases.
  3. copper increases and silicon decreases.
  4. copper decreases and silicon increases.
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Official CBSE Marking Scheme · CBSE Board 2023 (Main), Set 55/2/1

(D) Copper decreases and silicon increases

Q32.2023 Main 55/2/1 · Q151 mark

The formation of depletion region in a p-n junction diode is due to

  1. movement of dopant atoms
  2. diffusion of both electrons and holes
  3. drift of electrons only
  4. drift of holes only
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Official CBSE Marking Scheme · CBSE Board 2023 (Main), Set 55/2/1

(B) Diffusion of both electrons and holes

Q33.2023 Main 55/3/1 · Q131 mark

At a certain temperature in an intrinsic semiconductor, the electrons and holes concentration is 1·5 × 1016 m-3. When it is doped with a trivalent dopant, hole concentration increases to 4·5 × 1022 m-3. In the doped semiconductor, the concentration of electrons (ne) will be :

  1. 3 × 106 m-3
  2. 5 × 107 m-3
  3. 5 × 109 m-3
  4. 6·75 × 1038 m-3
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Official CBSE Marking Scheme · CBSE Board 2023 (Main), Set 55/3/1

(C) 5 × 109 m-3

Q34.2023 Main 55/3/1 · Q141 mark

If a p-n junction diode is reverse biased,

  1. the potential barrier is lowered.
  2. the potential barrier remains unaffected.
  3. the potential barrier is raised.
  4. the current is mainly due to majority carriers.
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Official CBSE Marking Scheme · CBSE Board 2023 (Main), Set 55/3/1

(C) the potential barrier is raised

Q35.2023 Main 55/4/1 · Q151 mark

The threshold voltage for a p-n junction diode used in the circuit is 0·7 V. The type of biasing and current in the circuit are :

Circuit with diode D, 100 Ω resistor and 0.5 V battery in series
  1. Forward biasing, 0 A
  2. Reverse biasing, 0 A
  3. Forward biasing, 5 mA
  4. Reverse biasing, 2 mA
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Official CBSE Marking Scheme · CBSE Board 2023 (Main), Set 55/4/1

(A) Forward biasing, 0 A

Q36.2023 Main 55/4/1 · Q181 mark

Assertion (A) : In ‘n’ type semiconductor, number density of electrons is greater than the number density of holes but the crystal maintains an overall charge neutrality.

Reason (R) : The charge of electrons donated by donor atoms is just equal and opposite to that of the ionised donor.

  1. Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of the Assertion (A).
  2. Both Assertion (A) and Reason (R) are true, but Reason (R) is not the correct explanation of the Assertion (A).
  3. Assertion (A) is true, but Reason (R) is false.
  4. Assertion (A) is false and Reason (R) is also false.
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Official CBSE Marking Scheme · CBSE Board 2023 (Main), Set 55/4/1

(A) Both Assertion (A) and Reason (R) are true, and Reason (R) is the correct explanation of the Assertion (A).

Q37.2023 Main 55/5/1 · Q91 mark

During the formation of a p-n junction :

  1. diffusion current keeps increasing.
  2. drift current remains constant.
  3. both the diffusion current and drift current remain constant.
  4. diffusion current remains almost constant but drift current increases till both currents become equal.
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Official CBSE Marking Scheme · CBSE Board 2023 (Main), Set 55/5/1

(D) Diffusion current remains almost constant but drift current increases till both currents become equal.

Q38.2023 Main 55/5/1 · Q161 mark

Assertion (A) : The resistance of an intrinsic semiconductor decreases with increase in its temperature.

Reason (R) : The number of conduction electrons as well as hole increase in an intrinsic semiconductor with rise in its temperature.

Select the correct answer from the codes below :

  1. Both Assertion (A) and Reason (R) are true and (R) is the correct explanation of (A).
  2. Both Assertion (A) and Reason (R) are true and (R) is NOT the correct explanation of (A).
  3. Assertion (A) is true and Reason (R) is false.
  4. Assertion (A) is false and Reason (R) is also false.
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Official CBSE Marking Scheme · CBSE Board 2023 (Main), Set 55/5/1

(A) Both Assertion (A) and Reason (R) are true and (R) is correct explanation of (A)

Note: In Hindi version none of the answer is correct, Therefore award one mark to the student who opted Hindi medium to write the answer.

Q39.2023 Supp 55/C/1 · Q151 mark

In which of the following diagrams is the capacitor ‘C’ connected correctly to provide smooth output of a half-wave rectifier ?

  1. Capacitor in series with the diode and load
  2. Capacitor across the source side before the diode
  3. Diode in series, capacitor in parallel with the load resistor
  4. Capacitor in series with the load resistor
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Official CBSE Marking Scheme · CBSE Board 2023 (Supplementary), Set 55/C/1

(C)

Diode in series, capacitor in parallel with the load resistor
Q40.2023 Supp 55/C/1 · Q181 mark

Assertion (A) : Silicon is preferred over germanium for making semiconductor devices.

Reason (R) : The energy gap for germanium is more than the energy gap for silicon.

  1. Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of the Assertion (A).
  2. Both Assertion (A) and Reason (R) are true, but Reason (R) is not the correct explanation of the Assertion (A).
  3. Assertion (A) is true, but Reason (R) is false.
  4. Assertion (A) is false and Reason (R) is also false.
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Official CBSE Marking Scheme · CBSE Board 2023 (Supplementary), Set 55/C/1

(C) Assertion (A) is true, but Reason (R) is false.

Q41.2026 Main 55/2/1 · Q172 marks

Write two points of difference between intrinsic and extrinsic semiconductors.

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Official CBSE Marking Scheme · CBSE Board 2026 (Main), Set 55/2/1
Intrinsic semiconductorsExtrinsic semiconductors
1. ne = nh1. ne ≠ nh (Alternatively ne ≫ nh or ne ≪ nh)
2. Low conductivity2. High conductivity
3. Semiconductor without any impurity atoms added / Pure semiconductor.3. Semiconductor with doping / Impure semiconductor.
4. Its conductivity depends only on temperature4. Its conductivity depends on temperature and concentration of dopants
5. They are naturally occurring semiconductors.5. They are manufactured semiconductors.

(Any two) or (Any other relevant differences) [1 + 1]

Q42.2026 Main 55/3/1 · Q212 marks

How are charge carriers created in an intrinsic semiconductor ? Explain.

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Official CBSE Marking Scheme · CBSE Board 2026 (Main), Set 55/3/1

At low temperatures, all the covalent bonds in an intrinsic semiconductor remain intact. As temperature increases, thermal energy becomes available and electrons may break away, becoming free electrons available for conduction. [1]

The neighbourhood from which the free electron has come out leaves a vacancy with an effective positive charge. This vacancy is called hole. [1]

Hence charge carriers are created.

Q43.2026 Main 55/4/1 · Q212 marks

The hole concentration in an intrinsic semiconductor is 5 × 108 m−3. When it is doped with certain impurity, the electron concentration becomes 4 × 1012 m−3. Find the new value of the hole concentration. Also identify the type of new semiconductor formed after doping.

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Official CBSE Marking Scheme · CBSE Board 2026 (Main), Set 55/4/1

nenh = ni2 [½]

Hole concentration (nh) = ni2ne [½]

= 5×108×5×1084×1012

nh = 6.25 × 104 m−3 [½]

The new semiconductor formed is n-type semiconductor. [½]

Q44.2026 Main 55/5/1 · Q212 marks

Suppose a pure Si crystal has 5 × 1028 atoms per m3. It is doped with 5 × 1022 atoms per m3 of Arsenic. Calculate the majority and minority carrier concentration in the doped silicon. (Given : ni = 1·5 × 1016 m–3)

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Official CBSE Marking Scheme · CBSE Board 2026 (Main), Set 55/5/1

Calculating majority carrier concentration [½]; minority carrier concentration [1½]

Majority carrier concentration = Doping concentration

∴ Majority carrier concentration (ne) = 5 × 1022 atoms per m3 [½]

ni2 = ne nh [½]

nh = ni2ne = (1.5 × 1016)25 × 1022 [½]

= 4.5 × 109 atoms per m3

Minority carrier concentration = 4.5 × 109 atoms per m3 [½]

Q45.2025 Main 55/1/1 · Q212 marks

A p-type Si semiconductor is made by doping an average of one dopant atom per 5 × 107 silicon atoms. If the number density of silicon atoms in the specimen is 5 × 1028 atoms m–3, find the number of holes created per cubic centimetre in the specimen due to doping. Also give one example of such dopants.

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The official marking scheme for this paper is not available in the source folder.

Q46.2025 Main 55/2/1 · Q212 marks

The threshold voltage of a silicon diode is 0.7 V. It is operated at this point by connecting the diode in series with a battery of V volt and a resistor of 1000 Ω. Find the value of V when the current drawn is 15 mA.

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The official marking scheme for this paper is not available in the source folder.

Q47.2025 Main 55/4/1 · Q212 marks

In an intrinsic semiconductor, carrier’s concentration is 5 × 108 m−3. On doping with impurity atoms, the hole concentration becomes 8 × 1012 m−3.

  1. Identify (i) the type of dopant and (ii) the extrinsic semiconductor so formed.
  2. Calculate the electron concentration in the extrinsic semiconductor.
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The official marking scheme for this paper is not available in the source folder.

Q48.2025 Main 55/6/1 · Q212 marks

In an n-type semiconductor electron-hole combination is a continuous process at room temperature. Yet the electron concentration is always greater than the hole concentration in it. Explain.

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The official marking scheme for this paper is not available in the source folder.

Q49.2025 Main 55/7/1 · Q202 marks

Draw energy band diagrams of n-type and p-type semiconductors at temperature T > 0 K. Show the donor/acceptor energy levels with the order of difference of their energies from the bands.

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The official marking scheme for this paper is not available in the source folder.

Q50.2025 Supp 55/S/1 · Q202 marks

With the help of a circuit diagram, describe the method to obtain p-n junction diode characteristic when it is forward biased. Draw the shape of the characteristic.

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Official CBSE Marking Scheme · CBSE Board 2025 (Supplementary), Set 55/S/1
Circuit for forward-bias characteristic of a p-n junction diode with battery, rheostat, milliammeter and voltmeter

[1] for circuit diagram

A very small current flows till the applied voltage reaches the barrier potential in the forward bias p-n junction diode. Diode current increases significantly when applied voltage becomes greater than barrier potential. [½]

Forward-bias characteristic: I (mA) versus V (volts)

[½] for shape of characteristic

Q51.2024 Main 55/2/1 · Q212 marks

How does the energy gap of an intrinsic semiconductor effectively change when doped with a (a) trivalent impurity, and (b) pentavalent impurity ? Justify your answer in each case.

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Official CBSE Marking Scheme · CBSE Board 2024 (Main), Set 55/2/1

(i) Decreases [½]

Justification : An acceptor energy level is formed just above the top of the valence band. [½]

(ii) Decreases [½]

Justification : A donor level is formed just below the bottom of conduction band. [½]

Alternatively

Energy band diagrams of n-type (donor level E_D) and p-type (acceptor level E_A) semiconductors

(Note : Award the credit of justification if a student draws band diagram)

Q52.2024 Main 55/3/1 · Q212 marks

Suppose a pure Si crystal has 5 × 1028 atoms m–3. It is doped by 1 ppm concentration of boron. Calculate the concentration of holes and electrons, given that ni = 1·5 × 1016 m–3. Is the doped crystal n-type or p-type ?

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Official CBSE Marking Scheme · CBSE Board 2024 (Main), Set 55/3/1

nenh = ni2 [½]

nh ≈ 5 × 1022 / m3

ne = ni2nh

ne = (1.5×1016)25×1022 [½]

ne = 4.5 × 109 / m3 [½]

nh > ne, it is a p-type crystal [½]

Q53.2024 Main 55/4/1 · Q172 marks

Draw the circuit diagram of a p-n junction diode in (i) forward biasing and (ii) reverse biasing. Also draw its I-V characteristics in the two cases.

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Official CBSE Marking Scheme · CBSE Board 2024 (Main), Set 55/4/1

Circuit diagram of p-n junction diode: (i) Forward bias [½] (ii) Reverse bias [½]

Circuit diagrams of a p-n junction diode in forward bias (a) and reverse bias (b)

I-V characteristics in forward and reverse bias [½ + ½]

I-V characteristics of a p-n junction diode: forward bias in mA, reverse bias in μA
Q54.2024 Supp 55/S/1 · Q172 marks

Two halves of a silicon crystal (A and B) are doped with arsenic and boron respectively, forming a p-n junction in it. A battery is connected across it as shown in the figure.

Silicon crystal halves A and B connected across a variable battery V

(a) Will the junction be forward biased or reverse biased ? Give reason.

(b) Draw V-I graph for this arrangement.

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Official CBSE Marking Scheme · CBSE Board 2024 (Supplementary), Set 55/S/1

(a) p-n junction is reverse biased. [½]

Crystal A is doped with arsenic which is pentavalent. Hence it is n-type. Crystal B is doped with boron which is trivalent. Hence it is p-type. [½]

(b) V-I graph [1]

Reverse-bias V-I characteristic of a p-n junction diode

Note : if the graph is drawn without values full credit will be given

Q55.2023 Main 55/1/1 · Q24 (option 1)2 marks

Differentiate between intrinsic and extrinsic semiconductors.

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Official CBSE Marking Scheme · CBSE Board 2023 (Main), Set 55/1/1

Difference between intrinsic and extrinsic semiconductor [2]

Intrinsic semiconductorExtrinsic semiconductor
1. Pure semiconductor.Semiconductor is doped with impurities.
2. Low conductivity at room temperature.High conductivity at room temperature.
3. ne = nhne ≠ nh

(Any one) [1 + 1]

Note: Give full credit if a student writes any other relevant correct answer.

Q56.2023 Main 55/1/1 · Q24 (OR option)2 marks

Draw the circuit arrangement for studying the V – I characteristics of a p-n junction diode in forward bias and reverse bias. Show the plot of V – I characteristic of a silicon diode.

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Official CBSE Marking Scheme · CBSE Board 2023 (Main), Set 55/1/1

Circuit diagram for forward and reverse biased p-n junction diode [½ + ½]; V-I characteristic (Forward and Reverse bias) [½ + ½]

Circuit arrangements for forward bias and reverse bias of a p-n junction diode

[½ + ½]

V-I characteristic of a silicon diode

[½ + ½]

Q57.2023 Main 55/1/1 · Q252 marks

Briefly explain how the diffusion and drift currents contribute to the formation of potential barrier in a p-n junction diode.

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Official CBSE Marking Scheme · CBSE Board 2023 (Main), Set 55/1/1

Formation of potential barrier [2]

The diffusion current due to concentration gradient at the junction forms a space charge region consisting of immobile charge carriers. Due to this an electric field is generated at the junction giving rise to drift current in a direction opposite to diffusion current.

The potential at which diffusion current becomes equal to drift current is called potential barrier. [2]

Q58.2023 Main 55/2/1 · Q242 marks

Draw energy band diagram for an n-type and p-type semiconductor at T > 0 K.

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Official CBSE Marking Scheme · CBSE Board 2023 (Main), Set 55/2/1

Energy band diagrams at T > 0 K: n-type semiconductor [1], p-type semiconductor [1]

Energy band diagrams of n-type (donor level E_D below E_C) and p-type (acceptor level E_A above E_V) semiconductors at T > 0 K
Q59.2023 Main 55/2/1 · Q252 marks

Answer the following giving reasons :

  1. A p-n junction diode is damaged by a strong current.
  2. Impurities are added in intrinsic semiconductors.
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Official CBSE Marking Scheme · CBSE Board 2023 (Main), Set 55/2/1

i) Due to strong current, a junction diode gets heated, consequently large number of covalent bonds are broken and the junction is damaged. [1]

ii) Deliberate addition of impurity atoms in intrinsic semiconductor increases its conductivity and is suitable for making electronic devices. [1]

Alternatively

Give full credit if a student writes that no electronic device can be developed using intrinsic semiconductor because of their low conductivity.

Q60.2023 Main 55/3/1 · Q242 marks

Explain the roles of diffusion current and drift current in the formation of the depletion layer in a p-n junction diode.

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Official CBSE Marking Scheme · CBSE Board 2023 (Main), Set 55/3/1

During the formation of p-n junction, and due to the concentration gradient across p- and n-sides, holes diffuse from p-side to n-side (p → n) and electrons diffuse from n-side to p-side (n → p). When an electron diffuses from (n → p), it leaves behind an ionized donor (positive charge) on n-side which is immobile. Similarly, when a hole diffuses from (p → n) due to the concentration gradient, it leaves behind an ionised acceptor (negative charge) which is immobile. This space–charge region on either side of the junction together is known as depletion region. [1]

As a result, an electric field is developed across the junction. Due to this field, an electron on p-side of the junction moves to n-side and a hole on n-side of the junction moves to p-side. The motion of charge carriers due to the electric field is called drift. Initially, diffusion current is large and drift current is small.

As the diffusion process continues, the electric field strength & hence drift current increases. This process continues till diffusion & drift current becomes equal. [1]

Q61.2023 Main 55/3/1 · Q252 marks

Explain the property of a p-n junction which makes it suitable for rectifying alternating voltages. Differentiate between a half-wave and a full-wave rectifier.

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Official CBSE Marking Scheme · CBSE Board 2023 (Main), Set 55/3/1

p-n junction conducts in forward bias. [1]

Alternatively: p-n junction is a uni-directional device.

The half-wave rectifier gives output only for half of the input cycle. The full-wave rectifier gives output for both the halves of the input cycles. [1]

Alternatively: If output waveform of both the rectifiers shown diagrammatically, then full credit to be given.

Q62.2022 Main 55/1/1 · Q12 marks

Draw energy band diagrams of n-type and p-type semiconductors at temperature T > 0K, depicting the donor and acceptor energy levels. Mention the significance of these levels.

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Official CBSE Marking Scheme · CBSE Board 2022 (Main), Set 55/1/1

Energy band diagram [½ + ½]

Energy band diagrams of n-type (donor level E_D) and p-type (acceptor level E_A) semiconductors

Significance [1]

n-type semiconductors – small energy gap between donor level and conduction band which can be easily covered by thermally excited electrons. [½]

p-type semiconductors – small energy gap between acceptor level and valence band which can be easily covered by thermally excited electrons. [½]

Alternatively

The conductivity of semiconductor is improved with the creation of donor and acceptor levels.

Q63.2022 Main 55/1/1 · Q32 marks

Write the characteristics of a p-n junction which make it suitable for rectification.

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Official CBSE Marking Scheme · CBSE Board 2022 (Main), Set 55/1/1

Explanation [2]

The unidirectional property of a diode makes it suitable for rectification.

Alternatively

The diode conducts when forward biased and does not conduct when reverse biased.

(Award 1 mark if a student draws the forward and reverse characteristics of a diode.)

Q64.2022 Main 55/2/1 · Q12 marks

Explain the formation of depletion region in a p-n junction.

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Official CBSE Marking Scheme · CBSE Board 2022 (Main), Set 55/2/1

When p-type semiconductor is chipped with n-type semiconductor, e– from the n-side diffuse towards p-side and holes from p-side diffuse towards n-side leaving behind a layer of immobile +ve ions on n-side and immobile –ve ions on p-side leading to formation of depletion layer. [2]

(Note : Award 1 mark, if a student draws a diagram showing depletion region)

Q65.2022 Main 55/2/1 · Q32 marksCheck syllabus: solar cell

How is an emf generated by a solar cell due to the three basic processes involved ? Explain.

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Official CBSE Marking Scheme · CBSE Board 2022 (Main), Set 55/2/1

The three basic processes for generation of e.m.f. by a solar cell are:

(i) Generation of e-h pairs due to light (with hν > Eg) close to the junction. [½]

(ii) Separation of e–s and holes due to electric field of the depletion region. [½]

(iii) Collection : e– reaching the n-side are collected by the front contact & holes reaching p-side are collected by back contact. [½]

Thus, p-side becomes positive and n-side becomes negative giving rise to photo voltage. [½]

Q66.2022 Main 55/3/1 · Q12 marks

What is meant by doping of an intrinsic semiconductor ? Name the two types of atoms used for doping of Ge/Si.

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Official CBSE Marking Scheme · CBSE Board 2022 (Main), Set 55/3/1

Doping is the process of adding some external impurity atoms in an intrinsic semiconductor to increase its conductivity. [1]

Dopant atoms may be pentavalent or trivalent. [½ + ½]

Q67.2022 Main 55/3/1 · Q32 marks

Explain the formation of the barrier potential in a p-n junction.

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Official CBSE Marking Scheme · CBSE Board 2022 (Main), Set 55/3/1

The diffusion of electrons from n-region to p-region and that of the holes from p-region to n-region creates positive charge on the n-side and negative charge on the p-side which causes a difference of potential across the junction. [1]

This potential, set up across the junction tends to prevent the movement of electrons from the n-region to p-region. This is called barrier potential. [1]

Q68.2022 Main 55/4/1 · Q12 marks

What is meant by energy band gap in a solid ? Draw the energy band diagrams for a conductor, an insulator and a semiconductor.

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Official CBSE Marking Scheme · CBSE Board 2022 (Main), Set 55/4/1

The gap between the top of the valence band and bottom of the conduction band is called the energy band gap. [½]

Energy band diagrams for insulator (Eg > 3 eV), conductor (overlapping bands, Eg ≈ 0) and semiconductor (Eg < 3 eV)

Energy band diagram for insulator [½]

Energy band diagram for conductor [½]

Energy band diagram for semiconductor [½]

Q69.2022 Main 55/4/1 · Q32 marksCheck syllabus: LED

Name the device which converts electrical energy into light energy. Write three advantages of the device.

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Official CBSE Marking Scheme · CBSE Board 2022 (Main), Set 55/4/1

Light emitting diode (LED) [½]

Advantages of LED (Any three of the following) [½ + ½ + ½]

(i) Low operational voltage and less power

(ii) Fast on-off switching capability

(iii) The bandwidth of emitted light is 100 Å to 500 Å or in other words it is nearly (but not exactly) monochromatic.

(iv) Long life and ruggedness.

(v) Fast action and no warm up time required

(Note- Give credit of half mark to the students for writing electric bulb or other relevant device, and further credit 1½ marks if they mention the advantages of written device only)

Q70.2022 Main 55/5/1 · Q12 marks

With the help of a circuit diagram, explain briefly how a p-n junction diode works as a half-wave rectifier.

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Official CBSE Marking Scheme · CBSE Board 2022 (Main), Set 55/5/1

Circuit diagram of p-n junction diode as half wave rectifier [1]

Half-wave rectifier circuit: transformer secondary, diode and load resistor R<sub>L</sub>

Explanation of its working [1]

  • When voltage at A is positive, diode is forward biased and it conducts and when voltage at A is negative, diode is reverse biased and it does not conduct, so output is zero.

Alternatively:

Input ac voltage and half-wave rectified output voltage across R<sub>L</sub>
Q71.2022 Main 55/5/1 · Q32 marksCheck syllabus: photodiode

Why a photo-diode is operated in reverse bias whereas current in the forward bias is much larger than that in the reverse bias ? Explain. Mention its two uses.

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Official CBSE Marking Scheme · CBSE Board 2022 (Main), Set 55/5/1

Explanation [1]; Two uses [½ + ½]

It is easier to observe the change in current with change in light intensity if a reverse bias is applied.

Alternatively:

The fractional change due to photo effect on the minority carrier dominated reverse bias current, is more readily measureable than the fractional change in the forward bias current.

Uses: (Any two uses) [½ + ½]

  • Smoke detector
  • Remote control
  • Medical devices
  • Optical signal detection

(Any other uses)

Q72.2022 Supp 55/6/1 · Q32 marks
  1. A p-type semiconductor is electrically neutral although it has holes as the majority carriers. Justify.
  2. In photon picture of light, what determines the intensity of light ? [1+1]
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Official CBSE Marking Scheme · CBSE Board 2022 (Supplementary), Set 55/6/1

(a) The crystal maintains an overall charge neutrality as the charge of additional charge carriers is just equal and opposite to that of the ionised cores in the lattice. [1]

(b) It is defined as the energy transmitting per unit area, per unit time.

Alternatively: It is proportional to number of photons per unit area per unit time. [1]

Q73.2026 Main 55/1/1 · Q253 marks
Two ideal diodes D1, D2 and three resistors 1 kΩ, 2 kΩ, 3 kΩ connected between points A, P, R, B with ac source Vi across A and B

An ac voltage Vi = 12 sin (100 πt) V is applied between points A and B in a network of two ideal diodes and three resistors as shown in figure.

During the positive half-cycle of the input voltage Vi supplied to the network.

  1. Identify which of the two diodes will conduct and why ?
  2. Redraw an equivalent circuit diagram to show the flow of current.
  3. Calculate the output voltage drops V0 across the three resistors when the input voltage attains its peak value.
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Official CBSE Marking Scheme · CBSE Board 2026 (Main), Set 55/1/1

(a) Diode D2 is in forward bias and it conducts. [½]

(b) [1]

Equivalent circuit with only D2 conducting: current flows A to R through D2, then through 1 kΩ and 2 kΩ in series in parallel with 3 kΩ to B

(c) Voltage drop across RP, PB and RB is

VRP = 4 V, VPB = 8 V, VRB = 12 V [½ + ½ + ½]

Alternatively:

If the student considers positive AC cycle reaching diode D1

(a) Diode D1 is in forward bias and it conducts.

(b)

Equivalent circuit with only D1 conducting

(c) VPB = 12 V, VPR = 3 V, VRB = 9 V

Q74.2026 Main 55/1/1 · Q263 marks

Briefly explain the two important processes that occur during the formation of a p-n junction.

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Official CBSE Marking Scheme · CBSE Board 2026 (Main), Set 55/1/1

Two important processes are in the formation of p-n junction. [½ + ½]

(i) Diffusion

(ii) Drift

Diffusion: During the formation of p-n junction, due to concentration gradient across p- and n-sides, holes diffuse from p-side to n-side (p → n) and electrons diffuse from n-side to p-side (n → p). This motion of charge carriers give rise to diffusion current across the junction. [1]

Drift: Due to the positive space charge region on n-side of the junction and negative space charge region on p-side of the junction, an electric field directed from positive charge towards negative charge develops. Due to this field an electron on p-side of junction moves to n-side and a hole on n-side of the junction moves to p-side. The motion of charge carriers due to the electric field is called drift and constitutes drift current. [1]

Q75.2026 Main 55/3/1 · Q283 marks

With the help of circuit diagrams, briefly explain the forward biasing and the reverse biasing of a p-n junction diode.

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Official CBSE Marking Scheme · CBSE Board 2026 (Main), Set 55/3/1

Forward biasing [1½]

Circuit for forward biasing of a p-n junction diode with voltmeter, milliammeter and battery

The p-side of the diode is connected to positive terminal and the n-side is connected to the negative terminal of the battery.

Reverse biasing [1½]

Circuit for reverse biasing of a p-n junction diode with voltmeter, microammeter and battery

The p-side of the diode is connected to negative terminal and n-side is connected to the positive terminal of the battery.

Q76.2026 Main 55/4/1 · Q283 marks

With the help of a circuit diagram, explain the working of a full wave rectifier. Depict the input and output waveforms.

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Official CBSE Marking Scheme · CBSE Board 2026 (Main), Set 55/4/1

Circuit diagram [1]

Full wave rectifier circuit: centre-tap transformer, diodes D<sub>1</sub> and D<sub>2</sub>, load resistance R<sub>L</sub>, output across X and Y

Working [1] : Let the input voltage A with respect to the centre tap at any instant is positive. At that instant, voltage at B being out of phase will be negative. So, diode D1 gets forward biased and conducts (while D2 being reverse biased is not conducting). Hence, during this positive half cycle we get an output current. When voltage at A becomes negative w.r.t. centre tap, the voltage at B would be positive. In this part of the cycle diode D1 would not conduct but diode D2 would be giving output current.

Input and output waveforms [½ + ½]

Input sinusoidal waveform and full wave rectified output waveform against time
Q77.2026 Main 55/5/1 · Q283 marks

Draw a circuit diagram of a full-wave rectifier using p-n junction diodes. Explain its working and show the input-output waveforms.

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Official CBSE Marking Scheme · CBSE Board 2026 (Main), Set 55/5/1

Drawing a circuit diagram [1]; Explaining its working [1]; Showing the input and output waveforms [1]

Centre-tap full-wave rectifier circuit with diodes D1 and D2 and load RL

Working: Suppose the input voltage to A w.r.t centre tap at any instant is positive. At that instant voltage at B being out of phase will be negative. So, diode D1 gets forward biased and conducts while D2 being reverse biased is not conducting. Hence, during this positive half cycle we get an output current and output voltage across the load. When the voltage at A becomes negative w.r.t centre tap, the voltage at B would be positive, the diode D1 would not conduct but D2 conducts and giving an output current and voltage across the load. [1]

Input and output wave form: [½ + ½]

Input waveforms at A and B and full-wave rectified output across RL
Q78.2026 Supp 55/7/1 · Q223 marks
  1. A diode is connected to 220 V (rms) ac source in series with a capacitor as shown in the diagram.
    Diode in series with a capacitor C across a 220 V rms ac source
    Find the voltage (rms) across the capacitor. [1]
  2. Find VAB in the circuit given below. [2]
    15 V battery with 5 Ω resistor and a 10 Ω resistor in series, with a diode D and another 10 Ω resistor in parallel branch between A and B
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Official CBSE Marking Scheme · CBSE Board 2026 (Supplementary), Set 55/7/1

(a) The diode will conduct for the positive half cycle of AC supply because it is forward biased. Negative half cycle of AC supply, this is reversed biased and does not conduct. So the output would be half wave rectified output.

Vrms = V02 = 22022 = 2202 volt [1]

Note : 1 mark should be awarded on attempting this part of the question.

(b) 1RAB = 1R1 + 1R2 = 110 + 110 [½]

RAB = 5 Ω [½]

Rnet = 5 + 5 = 10 Ω

I = 1510 = 1.5 A [½]

VAB = IRAB = 1.5 × 5 = 7.5 V [½]

Q79.2026 Supp 55/7/1 · Q253 marks

Draw a labelled diagram of a full wave rectifier circuit. State its working principle and show the input-output waveforms.

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Official CBSE Marking Scheme · CBSE Board 2026 (Supplementary), Set 55/7/1
Full wave rectifier with centre-tap transformer, two diodes D1 and D2 and load resistance R_L

Labelled diagram [1]

Working principle : A p-n junction diode allows current to pass only when it is forward biased. [1]

Alternatively: A p-n junction diode offers low resistance when forward biased.

Input sinusoidal voltage and full wave rectified output voltage waveforms

Input waveform [½]; output waveform [½]

Q80.2025 Main 55/1/1 · Q283 marks

(a) Draw circuit arrangement for studying V-I characteristics of a p-n junction diode.

(b) Show the shape of the characteristics of a diode.

(c) Mention two information that you can get from these characteristics.

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Q81.2025 Main 55/2/1 · Q283 marks

(a) What are majority and minority charge carriers in an extrinsic semiconductor ?

(b) A p-n junction is forward biased. Describe the movement of the charge carriers which produce current in it.

(c) The graph shows the variation of current with voltage for a p-n junction diode.

I-V characteristic of a p-n junction diode: forward bias with current in mA, reverse bias with voltage down to –12.0 V

Estimate the dynamic resistance of diode at V = –0.6 volt.

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Q82.2025 Main 55/4/1 · Q283 marks
  1. What are majority and minority charge carriers of p-type and n-type semiconductors ?
  2. Explain briefly the formation of diffusion current and drift current in a p-n junction diode.
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Q83.2025 Main 55/5/1 · Q223 marks

(a) Draw the energy-band diagrams for conductors, semiconductors and insulators at T = 0 K. How is an electron-hole pair formed in a semiconductor at room temperature ?

(b) Carbon and silicon both, are members of IV group of periodic table and have the same lattice structure. Carbon is an insulator whereas silicon is a semiconductor. Explain.

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Q84.2025 Main 55/5/1 · Q253 marks

Differentiate between half-wave and full-wave rectification. With the help of a circuit diagram, explain the working of a full-wave rectifier.

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Q85.2025 Main 55/7/1 · Q283 marks

Explain the process of formation of ‘depletion layer’ and ‘potential barrier’ in a p-n junction region of a diode, with the help of a suitable diagram. Which feature of junction diode makes it suitable for its use as a rectifier ?

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Q86.2024 Main 55/2/1 · Q283 marks

With the help of a circuit diagram, explain the working of a p-n junction diode as a full wave rectifier. Draw its input and output waveforms.

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Official CBSE Marking Scheme · CBSE Board 2024 (Main), Set 55/2/1

Circuit diagram [1]

Full wave rectifier circuit: centre-tap transformer, diodes D1 and D2, load resistance R_L

Working : When input voltage at A with respect to the centre tap at any instant is positive, at that instant voltage at B, being out of phase will be negative, during the positive half cycle diode D1 gets forward biased and conducts while diode D2 gets reverse biased and does not conduct. Hence during positive half cycle an output current and output voltage across RL is obtained. [½]

During second half of the cycle when voltage at A becomes negative with respect to centre tap, the voltage at B would be positive hence D1 would not conduct but D2 would be giving an output current and output voltage. We get output voltage in both positive and negative half cycles. [½]

Input and output waveforms [1]

Input waveforms at A and B and full-wave rectified output across R_L
Q87.2024 Main 55/3/1 · Q28 (option 1)3 marks

Draw the circuit diagrams for obtaining the V – I characteristics of a p-n junction diode. Explain briefly the salient features of the V – I characteristics in (i) forward biasing, and (ii) reverse biasing.

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Official CBSE Marking Scheme · CBSE Board 2024 (Main), Set 55/3/1

Drawing the circuit diagram for V-I characteristics [1]; Salient features of V-I characteristics in (i) forward biasing [1], (ii) reverse biasing [1]

Circuit diagrams for forward bias (milliammeter) and reverse bias (microammeter) of a p-n junction diode

[any one circuit diagram]

Salient features

(i) Forward biasing - After threshold voltage or cut in voltage diode current increases significantly (exponentially), even for a small increase in the diode bias voltage. [1]

(ii) Reverse biasing - Current is very small (~μA) and almost remains constant and it increases rapidly after breakdown voltage. [1]

Q88.2024 Main 55/3/1 · Q28 (OR option)3 marks

On the basis of energy band diagrams, distinguish between (i) an insulator, (ii) a semiconductor, and (iii) a conductor.

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Official CBSE Marking Scheme · CBSE Board 2024 (Main), Set 55/3/1

Energy band diagrams; Difference between (i) an insulator, (ii) a semiconductor, (iii) a metal [1+1+1]

Energy band diagrams: (i) insulator with Eg > 3 eV, (ii) semiconductor with Eg < 3 eV, (iii) conductor with overlapping bands
Q89.2024 Main 55/5/1 · Q273 marks
  1. Explain the characteristics of a p-n junction diode that makes it suitable for its use as a rectifier.
  2. With the help of a circuit diagram, explain the working of a full wave rectifier.
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Official CBSE Marking Scheme · CBSE Board 2024 (Main), Set 55/5/1

(a) p-n junction diode allows current to pass only when it is forward biased [1]

(b) Circuit diagram [1]

Full wave rectifier circuit with centre-tap transformer, diodes D1 and D2 and load resistor RL

When input voltage to A, with respect to the centre tap at any instant is positive, at that instant voltage at B, being out of phase will be negative, diode D1 gets forward biased and conducts while D2 being reverse biased does not conduct. Hence during this half cycle an output current and output voltage across RL is obtained. During second half of the cycle when voltage at A becomes negative with respect to centre tap, the voltage at B would be positive. Hence D1 would not conduct but D2 would be giving an output current and output voltage. Thus output voltage is obtained during both halves of the cycle. [1]

Q90.2024 Main 55/5/1 · Q283 marks

Explain the following, giving reasons :

  1. A doped semiconductor is electrically neutral.
  2. In a p-n junction under equilibrium, there is no net current.
  3. In a diode, the reverse current is practically not dependent on the applied voltage.
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Official CBSE Marking Scheme · CBSE Board 2024 (Main), Set 55/5/1

(a) Charge of additional charge carriers is just equal and opposite to that of the ionised cores in the lattice. [1]

(b) Under equilibrium, the diffusion current is equal to the drift current. [1]

(c) Reverse current is limited due to concentration of minority charge carriers on either side of the junction. [1]

Q91.2022 Main 55/1/1 · Q103 marksCheck syllabus: LED

State the working principle of an LED. Write any two important advantages and two disadvantages of LED.

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Official CBSE Marking Scheme · CBSE Board 2022 (Main), Set 55/1/1

When the diode is forward biased, electrons are sent from n region to p region and holes are sent from p region to n region. At the junction the concentration of minority carriers increases. Thus at the junction, the excess minority carriers recombine with majority carriers and energy is released in the form of photons. [1]

Advantages (any two): [½ + ½]

  1. Low operational voltage
  2. Less power consumption
  3. Fast action
  4. Long life and ruggedness

Disadvantages (any two): [½ + ½]

  1. High cost
  2. Can get damaged due to overheating
  3. Excess of voltage or current can damage LED

(Note: Award last 1 mark, even if disadvantages are not given.)

Q92.2022 Main 55/2/1 · Q113 marks

With the help of a circuit diagram, explain the working of a p-n junction diode as a full-wave rectifier. Also draw its input and output waveforms.

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Official CBSE Marking Scheme · CBSE Board 2022 (Main), Set 55/2/1

Explanation of working [1]; Circuit diagram of full wave rectifier [1]; Input & output waveform [½ + ½]

Centre-tap full-wave rectifier circuit with two diodes

Let input voltage at A w.r.t. the centre tap at any instant is positive, then voltage at B will be negative. So, diode D1 gets forward biased & D2 gets reverse biased. Hence output current is obtained. When voltage at A becomes –ve ; then voltage at B would be + ve , hence D1 gets reverse biased & D2 gets forward biased. So output current is again obtained as shown in the figure. [1]

Input and output waveforms of full-wave rectifier

[½ + ½]

(Note: If the student takes inverted input waveform full credit to be given.)

Q93.2022 Main 55/3/1 · Q53 marksCheck syllabus: solar cell

Briefly explain how emf is generated in a solar cell. Draw its I-V characteristics.

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Official CBSE Marking Scheme · CBSE Board 2022 (Main), Set 55/3/1

Three processes due to which emf is generated in a solar cell are:

(i) Generation of electron-hole pairs due to light incident close to the junction. [½]

(ii) Separation of electrons and holes due to electric field of the depletion region. Electrons swept to n-side and holes to p-side. [½]

(iii) The electrons reaching the n-side are collected by the front contact and the holes reaching the p-side are collected by the back contact. [½]

Thus p-side becomes positive and n-side becomes negative giving rise to photovoltage. [½]

I-V characteristic of a solar cell in the fourth quadrant, marking V_OC and I_SC

I-V characteristics [1]

Q94.2022 Main 55/4/1 · Q53 marksCheck syllabus: photodiode (part c)

Answer the following, giving reason :

(a) The resistance of a p-n junction is low when it is forward biased and is high when it is reversed biased.

(b) Doping of intrinsic semiconductors is a necessity for making electronic devices.

(c) Photodiodes are operated in reverse bias.

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Official CBSE Marking Scheme · CBSE Board 2022 (Main), Set 55/4/1

(a) When a p-n junction is forward biased, the junction width decreases and as a result, its resistance also decreases. [½]

On the other hand, when a p-n junction is reverse biased, the junction width increases hence resistance increases. [½]

(b) Conductivity of intrinsic semi-conductors is very low. Hence, no electronic device can be developed using them.

Dopping increases conductivity, hence makes intrinsic semiconductor suitable for making electronic devices. [1]

(c) It is easier to observe the change in the current with change in light intensity if a reverse bias is applied. [1]

Alternatively

The fractional change due to photo-effects on the minority charge carriers dominated reverse bias current, is more easily measurable than the fractional change in forward bias current.

Q95.2022 Main 55/5/1 · Q53 marks
  1. Draw V-I characteristics of a p-n Junction diode.
  2. Differentiate between the threshold voltage and the breakdown voltage for a diode.
  3. Write the property of a junction diode which makes it suitable for rectification of ac voltages.
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Official CBSE Marking Scheme · CBSE Board 2022 (Main), Set 55/5/1

(i) V-I characteristics [1]

V-I characteristics of a p-n junction diode: forward current in mA, reverse current in µA

(Note: Full credit if values on the axis are not mentioned)

(ii) Threshold Voltage: Forward bias voltage at which the current increases significantly (exponentially) even for a very small increase in voltage. [½]

Alternatively: Forward bias voltage at which the width depletion layer and barrier potential decreases significantly.

Alternatively: The voltage at which resistance of junction decreases significantly.

Break down voltage: Reverse bias voltage at which current increases suddenly [½]

Alternatively: Large number of covalent bonds present in the depletion layer break suddenly

(iii) Junction Diode conducts when it is forward biased and does not conduct when reverse biased. [1]

Q96.2022 Supp 55/6/1 · Q103 marks

Draw V – I characteristics of a p-n junction diode. Answer the following giving reasons : [3]

  1. Why is the reverse bias current almost independent of applied voltage up to breakdown voltage ?
  2. Why does the reverse current show a sudden increase at breakdown voltage ?
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Official CBSE Marking Scheme · CBSE Board 2022 (Supplementary), Set 55/6/1

V–I characteristics of p-n junction [1]

V-I characteristic of a p-n junction diode showing forward and reverse bias regions and breakdown voltage

[Note: Do not deduct marks for not writing the values on the axis.]

(a) The current is not limited by the magnitude of the applied voltage but is limited due to the concentration of the minority carriers on either side of the junction. [1]

(b) At the breakdown voltage, a large number of covalent bonds break, resulting in the increase of large number of charge carriers. Hence current increases suddenly. [1]

Q97.2022 Supp 55/6/1 · Q113 marks
  1. Explain the formation of a p-n junction.
  2. Can we take one slab of p-type semiconductor and physically join it to another n-type semiconductor to get a p-n junction ? Explain. [3]
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Official CBSE Marking Scheme · CBSE Board 2022 (Supplementary), Set 55/6/1

(a) Diagram [½]

p-n junction with depletion region, diffusion and drift of electrons and holes

During the formation of p-n junction due to concentration gradient across p and n sides, holes diffuse from p side to n side and electrons diffuse from n side to p side. This motion of charge carriers gives rise to diffusion current across the junction. [½]

Diffusion of electrons develops a layer of positive charge on n side of the junction and diffusion of holes develops a layer of negative charge on p side of the junction.

Due to this space charge region on either side of the junction an electric field is developed. This electric field drifts charge carriers across the junction and sets up drift current in a direction opposite to diffusion current. [½]

This process continues until the diffusion current is equal to drift current. Thus p-n junction is formed. [½]

(b) No. Any slab, howsoever flat, will have roughness much larger than the inter-atomic spacing (~2 to 3 Å) and hence continuous contact at the atomic level will not be possible. The junction will behave as a discontinuity for the flowing charge carriers. [1]

Q98.SQP 2025-26 · Q223 marks

With the help of circuit diagram explain working of the full wave rectifier.

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Official CBSE Marking Scheme · CBSE Sample Question Paper 2025-26

In the full wave rectifier: D1 and D2 are pn junction diode which allow current to pass only in forward biasing.

During odd half cycle the diode D1 will be forward biased hence potential at the Q will be more then Potential at P and during this cycle D2 will not permit current through it. [1]

During even half cycle the diode D2 will be forward biased hence potential at the Q will be more then Potential at P and during this cycle D1 will not permit current through it.

Hence we will get DC as output as shown in diagram. [2]

Full wave rectifier circuit with centre-tapped transformer, diodes D1 and D2, load R_L, and input and output waveforms
Q99.SQP 2026-27 · Q223 marks

Compare the energy band diagrams of metals, insulators, and semiconductors.

  1. Explain how the band gap determines conductivity.
  2. Apply this reasoning to justify why diamond (C) is an insulator while tin (Sn) is a conductor, though both are the elements of group IV.
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Official CBSE Marking Scheme · CBSE Sample Question Paper 2026-27

Metals: Overlapping bands or partially filled conduction band; free electrons yield high conductivity.

Insulators: Large band gap; negligible thermal excitation at room temperature; poor conductivity. [1]

Semiconductors: Moderate band gap; thermally/optically excited carriers; conductivity can be varied with doping. [1]

Diamond vs tin:

Diamond (C): Strong network, large gap (~ 5.4 eV); insulator at room temperature.

Tin (Sn): Metallic bonding with overlapping bands; conductor due to available states at Fermi level.

Conclusion: Band gap magnitude controls room-temperature carrier availability and hence conductivity. [1]

Q100.2026 Main 55/2/1 · Q294 marks

A p-type or n-type semiconductor can be converted into a p-n junction by doping it with suitable impurity. The motion of majority charge carriers causes diffusion current across the junction while the barrier electric field causes motion of minority carriers for drift current. In case of unbiased diode, the diffusion and drift currents are equal. This equilibrium is disturbed by the biasing batteries. Diodes, therefore, allow currents in one direction. This property of diode is used in making rectifiers.

  1. Silicon is doped with which of the following to obtain p-type semiconductor ?
    1. Phosphorus
    2. Arsenic
    3. Boron
    4. Antimony
  2. A semiconductor has an electron concentration of 5 × 1022 m–3. The concentration of holes is (given ni = 1.5 × 1016 m–3)
    1. 5 × 1022 m–3
    2. 1.5 × 106 m–3
    3. 9 × 108 m–3
    4. 4.5 × 109 m–3
  3. During forward biasing of a p-n junction diode, the
    1. current is mainly due to drifting of majority carriers.
    2. current is mainly due to drifting of minority carriers.
    3. diffusion and drift currents are equal.
    4. current is of the order of 1 A.
  4. The threshold voltage for silicon diode is about
    1. 0.2 V
    2. 0.5 V
    3. 0.7 V
    4. 1.5 V

    OR

    When we dope Ge with a pentavalent element, four of its electrons bond with four germanium neighbours but fifth electron remains weakly bound. The ionisation energy for this electron is about
    1. 0.01 eV
    2. 0.05 eV
    3. 0.1 eV
    4. 0.15 eV
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Official CBSE Marking Scheme · CBSE Board 2026 (Main), Set 55/2/1

(i) (C) Boron [1]

(ii) (D) 4.5 × 109 m–3 [1]

(iii) Award 1 mark to each student, who has attempted this part. [1]

(iv) (C) 0.7 V [1]

OR (iv) (A) 0.01 eV [1]

Q101.2025 Main 55/6/1 · Q304 marks

Extrinsic semiconductors are made by doping pure or intrinsic semiconductors with suitable impurity. There are two type of dopants used in doping, Si or Ge, and using them p-type and n-type semiconductors can be obtained. A p-n junction is the basic building block of many semiconductor devices. Two important processes occur during the formation of a p-n junction : diffusion and drift. When such a junction is formed, a ‘depletion layer’ is created consisting of immobile ion-cores. This is responsible for a junction potential barrier. The width of a depletion layer and the height of potential barrier changes when a junction is forward-biased or reverse-biased. A semiconductor diode is basically a p-n junction with metallic contacts provided at the ends for application of an external voltage. Using diodes, alternating voltages can be rectified.

(i) Which of the following is a donor impurity atom for Ge ?

  1. Boron
  2. Antimony
  3. Aluminium
  4. Indium

(ii) When a pentavalent atom occupies the position of an atom in the crystal lattice of Si, four of its electrons form covalent bonds with four silicon neighbours, while the fifth remains bound to the parent atom. The energy required to set this electron free is about :

  1. 0·5 eV
  2. 0·1 eV
  3. 0·05 eV
  4. 0·01 eV

(iii) During formation of a p-n junction :

  1. a layer of negative charge on n-side and a layer of positive charge on p-side appear.
  2. a layer of positive charge on n-side and a layer of negative charge on p-side appear.
  3. the electrons on p-side of the junction move to n-side initially.
  4. initially diffusion current is small and drift current is large.

(iv) (a) In reverse-biased p-n junction :

  1. the drift current is of the order of few mA.
  2. the applied voltage mostly drops across the depletion region.
  3. the depletion region width decreases.
  4. the current increases with increase in applied voltage.

(iv) (b) (alternative to (iv)(a)) The output frequency of a full-wave rectifier with 50 Hz as input frequency is :

  1. 25 Hz
  2. 50 Hz
  3. 100 Hz
  4. 200 Hz
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Q102.2025 Supp 55/S/1 · Q304 marks

According to the band theory of solids, a semiconductor has a valence band and a conduction band separated by a gap, known as energy band gap. Pure semiconductors are called intrinsic semiconductors. At room temperature, some electrons from the valence band can acquire enough energy to cross the band gap and enter the conduction band. The number of conduction electrons is equal to the number of holes is an intrinsic semiconductor. The number of charge carriers can be changed by doping of a suitable impurity in a pure semiconductor. Such semiconductors are known as extrinsic semiconductors. These are of two types (n-type and p-type).

A p-n junction is the basic building block of semiconductor devices. Two important processes occur during formation of a p-n junction : diffusion and drift. A ‘depletion layer’ is formed in a p-n junction. This is responsible for a junction potential barrier. A semiconductor diode is basically a p-n junction with metallic contacts provided at the ends for the application of an external voltage. A diode can be forward-biased or reverse-biased. The barrier height and the depletion layer width in a p-n junction changes depending on the nature of the biasing.

(i)(a) Which of the following statements is not true ?

  1. The resistance of an intrinsic semiconductor decreases with the increase of temperature.
  2. Doping pure Si with trivalent impurities gives p-type semiconductors.
  3. The majority charge carriers in n-type semiconductors are holes.
  4. A p-n junction can act as a semiconductor diode.

OR

(b) In a unbiased p-n junction :

  1. the diffusion current is zero everywhere.
  2. the drift current is zero everywhere.
  3. the electric potential is zero everywhere.
  4. the drift current and the diffusion current cancel each other.

(ii) The impurity atoms with which pure Ge should be doped to convert it into an n-type semiconductor, is :

  1. Boron
  2. Phosphorous
  3. Aluminium
  4. Indium

(iii) The energy band gap in Ge at 0 K is about :

  1. 0·72 eV
  2. 1·1 eV
  3. 3·0 eV
  4. 5·4 eV

(iv) In a p-n junction diode under reverse bias, the barrier height :

  1. is reduced and the depletion layer width decreases.
  2. is reduced and the depletion layer width increases.
  3. increases and the depletion layer width also increases.
  4. increases and the depletion layer width decreases.
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Official CBSE Marking Scheme · CBSE Board 2025 (Supplementary), Set 55/S/1

(i) (a) (C) Majority charge carrier in n type semiconductors are holes. [1]

OR

(b) (D) The drift current and the diffusion current cancel each other. [1]

(ii) (B) Phosphorus [1]

(iii) (A) 0.72 eV [1]

(iv) (C) Increases and the depletion layer width also increases. [1]

Q103.2024 Main 55/1/1 · Q304 marks

Junction Diode as a Rectifier :

The process of conversion of an ac voltage into a dc voltage is called rectification and the device which performs this conversion is called a rectifier. The characteristics of a p-n junction diode reveal that when a p-n junction diode is forward biased, it offers a low resistance and when it is reverse biased, it offers a high resistance. Hence, a p-n junction diode conducts only when it is forward biased. This property of a p-n junction diode makes it suitable for its use as a rectifier.

Thus, when an ac voltage is applied across a p-n junction, it conducts only during those alternate half cycles for which it is forward biased. A rectifier which rectifies only half cycle of an ac voltage is called a half-wave rectifier and one that rectifies both the half cycles is known as a full-wave rectifier.

(i) The root mean square value of an alternating voltage applied to a full-wave rectifier is V02. Then the root mean square value of the rectified output voltage is : [1]

  1. V02
  2. V022
  3. 2V02
  4. V022

(ii) In a full-wave rectifier, the current in each of the diodes flows for : [1]

  1. Complete cycle of the input signal
  2. Half cycle of the input signal
  3. Less than half cycle of the input signal
  4. Only for the positive half cycle of the input signal

(iii) In a full-wave rectifier : [1]

  1. Both diodes are forward biased at the same time.
  2. Both diodes are reverse biased at the same time.
  3. One is forward biased and the other is reverse biased at the same time.
  4. Both are forward biased in the first half of the cycle and reverse biased in the second half of the cycle.

(iv) Attempt any one of (a), (b). [1]

(a) An alternating voltage of frequency of 50 Hz is applied to a half-wave rectifier. Then the ripple frequency of the output will be :

  1. 100 Hz
  2. 50 Hz
  3. 25 Hz
  4. 150 Hz

(b) A signal, as shown in the figure, is applied to a p-n junction diode. Identify the output across resistance RL :

Input square pulse alternating between +5 V and −5 V, applied to a diode in series with load resistor R_L
  1. Output pulse between 0 and 10 V
  2. Output pulse inverted, at −10 V
  3. Output pulse inverted, at −5 V
  4. Output pulse between 0 and +5 V
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Official CBSE Marking Scheme · CBSE Board 2024 (Main), Set 55/1/1

(i) (A) V02 [1]

(ii) (B) half cycle of the input signal [1]

(iii) (C) One is forward biased and the other is reverse biased at the same time [1]

(iv) (a) (B) 50 Hz

    (b) (D) output pulse between 0 and +5 V (as in option D) [1]

Q104.2024 Main 55/4/1 · Q294 marks

A pure semiconductor like Ge or Si, when doped with a small amount of suitable impurity, becomes an extrinsic semiconductor. In thermal equilibrium, the electron and hole concentration in it are related to the concentration of intrinsic charge carriers. A p-type or n-type semiconductor can be converted into a p-n junction by doping it with suitable impurity. Two processes, diffusion and drift take place during formation of a p-n junction. A semiconductor diode is basically a p-n junction with metallic contacts provided at the ends for the application of an external voltage. A p-n junction diode allows currents to pass only in one direction when it is forward biased. Due to this property, a diode is widely used to rectify alternating voltages, in half-wave or full wave configuration.

(i) When Ge is doped with pentavalent impurity, the energy required to free the weakly bound electron from the dopant is about

  1. 0.001 eV
  2. 0.01 eV
  3. 0.72 eV
  4. 1.1 eV

(ii) At a given temperature, the number of intrinsic charge carriers in a semiconductor is 2.0 × 1010 cm–3. It is doped with pentavalent impurity atoms. As a result, the number of holes in it becomes 8 × 103 cm–3. The number of electrons in the semiconductor is

  1. 2 × 1024 m–3
  2. 4 × 1023 m–3
  3. 1 × 1022 m–3
  4. 5 × 1022 m–3

(iii) (a) During the formation of a p-n junction –

  1. electrons diffuse from p-region into n-region and holes diffuse from n-region into p-region.
  2. both electrons and holes diffuse from n-region into p-region.
  3. electrons diffuse from n-region into p-region and holes diffuse from p-region into n-region.
  4. both electrons and holes diffuse from p-region into n-region.

or

(iii) (b) Initially during the formation of a p-n junction –

  1. diffusion current is large and drift current is small.
  2. diffusion current is small and drift current is large.
  3. both the diffusion and the drift currents are large.
  4. both the diffusion and the drift currents are small.

(iv) An ac voltage V = 0.5 sin (100 πt) volt is applied, in turn, across a half-wave rectifier and a full-wave rectifier. The frequency of the output voltage across them respectively will be

  1. 25 Hz, 50 Hz
  2. 25 Hz, 100 Hz
  3. 50 Hz, 50 Hz
  4. 50 Hz, 100 Hz
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Official CBSE Marking Scheme · CBSE Board 2024 (Main), Set 55/4/1

(i) (B) 0.01 eV [1]

(ii) (D) 5 × 1022 m–3 [1]

(iii) (a) (C) Electrons diffuse from n-region into p-region and holes diffuse from p-region to n-region. [1]

or (b) (A) Diffusion current is large and drift current is small.

(iv) (D) 50 Hz, 100 Hz. [1]

Q105.2024 Supp 55/S/1 · Q304 marks

The process of converting ac into dc is called rectification and the device used is called a rectifier. When ac signal is fed to a junction diode during positive half cycle, the diode is forward biased and current flows through it. During the negative half cycle, the diode is reverse biased and it does not conduct. Thus the ac signal is rectified. The p-n junction diodes can be used as half-wave and full-wave rectifiers.

(i) Which bulb/bulbs will glow in the given circuit ? [1]

Circuit with diodes D1 and D2, a 9 V battery and bulbs B1 and B2
  1. B1 only
  2. B2 only
  3. Both B1 and B2
  4. Neither B1 nor B2

(ii) Answer either (a) or (b) :

(a) A full-wave rectifier circuit is shown in the figure. The contribution in output waveform from junction diode D1 is : [1]

Input waveform, full-wave rectifier circuit with diodes D1 and D2, and output waveform with humps A, B, C, D
  1. A, D
  2. A, C
  3. B, D
  4. B, C

(b) The output in a half-wave rectifier is : [1]

  1. unidirectional without ripple
  2. steady and continuous
  3. unidirectional with ripple
  4. steady but discontinuous

(iii) In a p-n junction diode, the majority charge carriers on p-side and on n-side are, respectively : [1]

  1. electrons, electrons
  2. electrons, holes
  3. holes, holes
  4. holes, electrons

(iv) If the frequency of the half-wave rectifier is 50 Hz, the frequency of full-wave rectifier is : [1]

  1. 25 Hz
  2. 50 Hz
  3. 100 Hz
  4. 200 Hz
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Official CBSE Marking Scheme · CBSE Board 2024 (Supplementary), Set 55/S/1

(i) (A) B1 only [1]

(ii) (a) (B) A, C [1]

OR

(ii) (b) (C) unidirectional with ripple [1]

(iii) (D) holes, electrons [1]

(iv) (C) 100 Hz [1]

Q106.SQP 2025-26 · Q294 marks

When an external voltage is applied across a semiconductor diode such that p-side is connected to the positive terminal of the battery and n-side to the negative terminal it is said to be forward biased. The applied voltage mostly drops across the depletion region and the voltage drop across the p-side and n-side of the junction is negligible. When an external voltage is applied across the diode such that n-side is positive and p-side is negative, it is said to be reverse biased. The applied voltage mostly drops across the depletion region.

(I) Ge and Si diodes start conducting at 0.3 V and 0.7 V respectively. In the following figure if Ge diode connection are reversed, the value of V0 changes by (assume that the Ge diode has large breakdown voltage)

Circuit with 12 V battery, Ge and Si diodes in parallel (opposite directions), and 5 kΩ resistor with output V0
  1. 0.2 V
  2. 0.4 V
  3. 0.6 V
  4. 0.8 V

(II) The value of V0 and Id for the network are :

12 V source, Si diode and Ge diode in series with a 5.6 kΩ resistor to ground, output V0
  1. 13 V, 2.32mA
  2. 11.7 V, 2.08mA
  3. 11.3V, 2.01mA
  4. 11V, 1.96mA

(III) The circuit shown below contains two ideal diodes, each with a forward resistance of 50Ω. If the battery voltage is 6 V, the current through the 100Ω resistance (in amperes) is

Circuit with diodes D1 (with 150 Ω) and D2 (with 75 Ω) in parallel branches, in series with a 100 Ω resistor and 6 V battery
  1. 0.036
  2. 0.020
  3. 0.030
  4. 0.027

(IV) In the figure, potential difference between A and B is

Circuit with 30 V battery, 10 kΩ resistor, 10 kΩ resistor between A and B, and a diode in series with a 10 kΩ resistor across A and B
  1. Zero
  2. 5 V
  3. 10 V
  4. 15 V
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Official CBSE Marking Scheme · CBSE Sample Question Paper 2025-26

(I) (B) [1]

Voltage drop across diode will change from 0.3 to 0.7 V.

Value of V0 changes by 0.4 V.

(II) (D) 11V, 1.96mA [1]

V0 = E – Vsi – VGe = 12.07 – 0.3 = 11V

Id = V0/R = 11/5.6 × 10–3 = 1.96 mA

(III) (B) [1]

I = 650 + 150 + 100 = 6300 A = 0.02 A

(IV) (C) [1]

Equivalent circuit with the forward-biased diode replaced by a wire: 30 V battery, 10 kΩ, and two 10 kΩ resistors in parallel across A and B

Here the diode is in forward bias. So we replace it by a connecting wire.

Va – Vb = I2 × 10

= 3015 × 2 × 10 V = 10 V

Q107.SQP 2026-27 · Q294 marks

In 2023, researchers developed flexible semiconductor films using organic polymers that could be printed like newspaper sheets. These films can bend, fold, and still conduct electricity efficiently. Unlike traditional silicon chips, polymer semiconductors are lightweight, biodegradable, and can be integrated into clothing or medical patches. Imagine a shirt that monitors your heartbeat or a bandage that tracks wound healing in real time. Such innovations extend the idea of semiconductors beyond rigid circuits, showing how manipulating band gaps and carrier mobility in new materials can revolutionize electronics for healthcare, sustainability, and wearable technology.

I. Which advantage of semiconductors makes them ideal for wearable devices like smart shirts? [1]

  1. High voltage operation (~100 V)
  2. Bulky design and short life
  3. Low power consumption and reliability
  4. Requirement of external heating

II. Carbon (diamond) has a band gap of 5.4 eV, while silicon has 1.1 eV. Why is silicon used in electronic devices but not the diamond? [1]

  1. Diamond has too many free electrons.
  2. Diamond’s band gap is too large for thermal excitation at room temperature.
  3. Silicon has weaker covalent bonds than diamond.
  4. Diamond cannot form covalent bonds.

III. In an intrinsic semiconductor at room temperature [1]

  1. only electrons contribute to conduction.
  2. only holes contribute to conduction.
  3. number of electrons equals number of holes.
  4. no conduction occurs at all.

IV. How will you use the concept of majority and minority carriers to explain the efficiency of polymer semiconductors in wearable technology? [1]

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Official CBSE Marking Scheme · CBSE Sample Question Paper 2026-27

I. (C) Low power consumption and reliability

Wearable devices need to operate continuously with minimal energy usage and dependable performance, which semiconductors provide. [1]

II. (B) Diamond’s band gap is too large for thermal excitation at room temperature.

Because of its large band gap, diamond does not allow sufficient charge carriers to be generated at room temperature, making it unsuitable as a semiconductor. [1]

III. (C) Number of electrons equals number of holes.

In intrinsic semiconductors, electrons and holes are generated in pairs, so their numbers are equal. [1]

IV. In polymer semiconductors, electrical conduction mainly occurs due to majority carriers (electrons in n-type or holes in p-type materials). By controlling the concentration and mobility of these carriers, polymer semiconductors can conduct electricity efficiently even when bent or stretched. Minority carriers play a role in recombination processes and stability. This balance allows polymer semiconductors to function reliably at low power, making them well-suited for flexible, lightweight, and wearable technologies such as smart clothing and medical patches. [1]

Q108.2023 Main 55/4/1 · Q33 (option 1)5 marks

(i) A germanium crystal is doped with antimony. With the help of energy-band diagram, explain how the conductivity of the doped crystal is affected.

(ii) Briefly explain the two processes involved in the formation of a p-n junction.

(iii) What will the effect of (1) forward biasing, and (2) reverse biasing be on the width of depletion layer in a p-n junction diode ?

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Official CBSE Marking Scheme · CBSE Board 2023 (Main), Set 55/4/1

(i) With proper level of doping, the number of conduction electrons can be made much larger than the number of holes. Due to this conductivity of the doped crystal increases. [1]

Energy band diagram of n-type semiconductor at T > 0 K showing donor level E_D just below conduction band

[1]

(ii) Two processes (a) Diffusion (b) drift

Diffusion : Due to concentration gradient majority charge carrier that is electron moves from n → p side and holes to p → n side. This movement of charges is called diffusion. [1]

Drift : Due to the junction field, an electron on p-side of the junction moves to n-side and a hole on n-side of the junction moves to p-side. The motion of the charge carrier due to electric field is called drift. [1]

(iii) (1) decreases [½]

(2) increases [½]

Q109.2023 Main 55/4/1 · Q33 (OR option)5 marks

(i) With the help of a circuit diagram, briefly explain the working of a full-wave rectifier using p-n junction diodes.

(ii) Draw V − I characteristics of a p-n junction diode. Explain how these characteristics make a diode suitable for rectification.

(iii) Carbon and silicon have the same lattice structure. Then why is carbon an insulator but silicon a semiconductor ?

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Official CBSE Marking Scheme · CBSE Board 2023 (Main), Set 55/4/1

(i)

Full-wave rectifier circuit with centre-tap transformer, diodes D1 and D2 and load resistor R_L

[1]

Working : Suppose the input voltage to A with respect to the centre tap at any instant is positive. At that instant voltage at B, being out of phase will be negative. So diode D1 gets forward biased and conducts, while D2 being reverse biased does not conduct. Similarly during second half of the cycle polarity get reversed so only D2 will conduct. [1]

(ii) V-I characteristics

V-I characteristic curve of a p-n junction diode

[½ + ½]

This diagram shows that the diode conducts when forward biased and does not conduct when reverse biased. This characteristics makes it suitable for use for rectification. [1]

(iii) The 4 bonding electrons of C and Si lie respectively, in the second and third orbit. Hence energy required to take out an electron from their atoms will be much less than that for C. Hence number of free e− for conduction in Si significant but negligibly small for C. [1]

Q110.2023 Main 55/5/1 · Q32 (option 1)5 marks

Draw the circuit arrangement for studying V-I characteristics of a p-n junction diode in (i) forward biasing and (ii) reverse biasing. Draw the typical V-I characteristics of a silicon diode. Describe briefly the following terms : (i) minority carrier injection in forward biasing and (ii) breakdown voltage in reverse biasing.

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Official CBSE Marking Scheme · CBSE Board 2023 (Main), Set 55/5/1

Circuit arrangement (i) forward bias [1]

Circuit for forward biasing a p-n junction diode with voltmeter, milliammeter and switch

(ii) reverse bias [1]

Circuit for reverse biasing a p-n junction diode with voltmeter, microammeter and switch

V-I characteristics [1]

V-I characteristics of a silicon p-n junction diode

(Note: Please do not deduct marks for not showing values.)

Minority carrier injection: Under forward bias electrons from n-side cross the depletion region and reach p-side. Similarly, holes from p-side cross the junction and reach the n-side. [1]

Breakdown voltage: It is the voltage under reverse bias for which reverse current increases sharply. [1]

Q111.2023 Main 55/5/1 · Q32 (OR option)5 marks

Name two important processes involved in the formation of a p-n junction diode. With the help of a circuit diagram, explain the working of junction diode as a full wave rectifier. Draw its input and output waveforms. State the characteristic property of a junction diode that makes it suitable for rectification.

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Official CBSE Marking Scheme · CBSE Board 2023 (Main), Set 55/5/1

(a) Diffusion [½]

Drift [½]

Circuit diagram [1]

Full wave rectifier circuit using a centre-tap transformer and two diodes D1, D2 with load RL

Suppose the input voltage to A with respect to the centre-tap at any instant is positive. At that instant, voltage at B being out of phase will be negative. So, diode D1 gets forward biased and conducts (while D2 being reverse biased is not conducting). Hence, during this positive half cycle we get an output current (and a output voltage across the load resistor RL). In the course of ac cycle when the voltage at A becomes negative with respect to centre tap, the voltage at B would be positive. In this part of the cycle diode D1 would not conduct but diode D2 would, giving an output current and output voltage (across RL) during the negative half cycle of the input ac. [1]

Input and output waveforms [½ + ½]

Input waveforms at A and B and full wave rectified output waveform across RL

Property: Junction diode allows current to pass only when it is forward biased. [1]

Q112.2023 Supp 55/C/1 · Q33 (option 1)5 marks

(i) Draw the circuit diagram used to study I − V characteristics of a p-n junction diode in conducting mode. Mark on the graph the threshold voltage of the diode. Explain the significance of this voltage.

(ii) In the circuit shown in the figure, the forward voltage drop across the diode is 0·3 V. Find the voltage difference between A and B.

Series circuit from A: 0.2 mA current, 5 K resistor, diode, 5 K resistor to B
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Official CBSE Marking Scheme · CBSE Board 2023 (Supplementary), Set 55/C/1

(i)

Circuit for I-V characteristics of forward biased p-n junction diode with voltmeter, milliammeter, switch and variable battery

[1]

I-V characteristic curve of a p-n junction diode

[1 + ½ (marking threshold voltage)]

Beyond threshold voltage in forward bias diode current increases significantly even for very small increases in diode bias voltage. [½]

(ii) VA − 5×103 × 0.2×10−3 − 0.3 − 5×103 × 0.2×10−3 − VB = 0

VA − VB = 2.3 volt [2]

Q113.2023 Supp 55/C/1 · Q33 (OR option)5 marks

(i) Briefly describe the classification of solids into metals, insulators and semi-conductors on the basis of energy level diagrams.

(ii) In a silicon diode, the current increases from 10 mA to 20 mA when the voltage changes from 0·6 V to 0·7 V. Calculate the dynamic resistance of the diode.

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Official CBSE Marking Scheme · CBSE Board 2023 (Supplementary), Set 55/C/1

(i)

Energy band diagram of a metal: overlapping conduction and valence bands, Eg approx 0
Energy band diagrams of an insulator (Eg greater than 3 eV) and a semiconductor (Eg less than 3 eV)

[1½]

For Eg > 3 eV material is insulator

For Eg < 3 eV material is semiconductor

For Eg = 0 or overlapping of conduction and valence band material is conductor. [½ + ½ + ½]

(ii) rd = ΔVΔI = 0.7 − 0.6(20 − 10)×10−3 [½ + ½]

rd = 10 Ω [1]