Class 12 Physics Test Paper Chapter 3 and 4 | CBSE Board | Detailed Solutions & Marking Scheme

VEDANT SKILL ASSESSMENT SERIES – ACADEMIC YEAR 2026–27

CLASS XII – PHYSICS (MATHEMATICAL STRUCTURES)
Chapters: 3 & 4
Time Allowed: 1 Hour
Maximum Marks: 30

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GENERAL INSTRUCTIONS

1. Read all questions carefully before answering.

2. Show proper steps wherever required.

3. Write neatly and clearly.

4. Manage your time wisely.

5. All questions are compulsory.

6. Use SI units throughout the paper wherever applicable.

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SECTION A (7 MARKS)

Q1. A steady current I flows through a metallic conductor of non-uniform cross-section. Which of the following quantities remains constant along the entire length of the conductor? [1]

(A) Current density

(B) Drift speed

(C) Electric field

(D) Current

Q2. Two wires of equal lengths are made of copper and manganin respectively and have the same electrical resistance. The ratio of their cross-sectional areas (A_copper/A_manganin) is: [1]

(A) Equal to 1

(B) Greater than 1

(C) Less than 1

(D) Depends on the applied voltage

Q3. A charged particle moves through a region of uniform magnetic field B with velocity v perpendicular to B. The kinetic energy of the particle after a time interval t will: [1]

(A) Increase linearly with time

(B) Decrease exponentially with time

(C) Remain unchanged

(D) Double its initial value

Q4. A long straight wire carries a current of 5 A. An electron moves parallel to the wire at a distance of 10 cm with velocity 1 × 10⁶ m/s in the direction of current. The magnitude of the magnetic force acting on the electron is: [1]

(A) 1.6 × 10⁻¹⁸ N

(B) 3.2 × 10⁻¹⁸ N

(C) 1.6 × 10⁻¹⁹ N

(D) Zero

Q5. To convert a galvanometer into a voltmeter of high range, one must connect a: [1]

(A) Low resistance in parallel

(B) High resistance in series

(C) Low resistance in series

(D) High resistance in parallel

Directions for Q6 and Q7:

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

(B) Both A and R are true, but R is not the correct explanation of A.

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

(D) A is false, but R is true.

Q6.

Assertion (A): The terminal voltage of a cell is always lower than its emf when current is being drawn from it. [1]

Reason (R): Every real cell possesses a non-zero internal resistance which causes an internal potential drop given by:

V = E − Ir

Q7.

Assertion (A): A magnetic field cannot change the kinetic energy of a moving charged particle. [1]

Reason (R): The magnetic force acting on a moving charged particle is always perpendicular to its velocity vector. [1]

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SECTION B (8 MARKS)

Q8. Establish the relation between current density J and drift velocity v_d of free electrons in a metallic conductor. State the assumptions made, if any. [2]

Q9. A potential difference V is applied across a copper conductor of length L and diameter D.

Discuss mathematically how the drift velocity v_d changes if:

(a) The length L is doubled while V remains constant.

(b) The applied voltage V is doubled while L remains constant.

[2]

Q10. State the Biot-Savart law in vector form. Clearly define each term and symbol used in the equation. [2]

Q11. Two very long, straight parallel wires carry currents I₁ and I₂ in opposite directions.

Show mathematically that the magnetic force per unit length acting between them is repulsive. [2]

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SECTION C (6 MARKS)

Q12. State Kirchhoff's two laws for electrical networks.

Write the mathematical equations for both laws and identify the conservation principle on which each law is based. [3]

Q13. Using Ampere's circuital law, derive the expression for the magnetic field B inside a long solenoid carrying current I.

Draw a neat labelled diagram showing the Amperian loop used in the derivation. [3]

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SECTION D (5 MARKS)

Q14. Derive the expression for the magnetic field B at a point on the axis of a circular current-carrying loop of radius R at a distance x from its centre.

Hence show that at the centre of the loop:

B = μ₀I/(2R)

[5]

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SECTION E (4 MARKS)

Q15. Case Study: Real-Life Engineering of Shunts in Ammeters

A moving-coil galvanometer cannot be directly connected in a high-current circuit because of its high sensitivity and low resistance.

To measure larger currents, a low resistance called a shunt (S) is connected in parallel with the galvanometer.

Given:

Galvanometer resistance = G

Full-scale deflection current = I_g

Required ammeter range = I

(a) Write the formula for shunt resistance S required to convert the galvanometer into an ammeter. [1]

(b) What is the effective resistance of the modified ammeter? Will it be larger or smaller than the shunt resistance S? [1]

(c) A galvanometer has resistance G = 60 Ω and full-scale deflection current I_g = 5 mA.

Calculate the shunt resistance required to convert it into an ammeter of range 6 A. [2]

------------------- END OF QUESTION PAPER -------------------

ANSWER KEY WITH MARKING SCHEME

SECTION A (7 MARKS)

Q1. Answer: (D) Current

Current remains constant throughout the conductor due to conservation of charge.

Current density J, drift velocity v_d, and electric field E vary with cross-sectional area.

[1 Mark]

Q2. Answer: (C) Less than 1

R = ρL/A

For equal R and L:

A ∝ ρ

Since:

ρ_copper < ρ_manganin

Therefore:

A_copper < A_manganin

Hence:

A_copper/A_manganin < 1

[1 Mark]

Q3. Answer: (C) Remain unchanged

Magnetic force is always perpendicular to velocity.

Hence, no work is done:

W = F·s cos 90° = 0

Therefore, kinetic energy remains constant.

[1 Mark]

Q4.

Magnetic field due to wire:

B = μ₀I/(2πr)

= (4π × 10⁻⁷ × 5)/(2π × 0.1)

= 1 × 10⁻⁵ T

Magnetic force:

F = evB sin 90°

= (1.6 × 10⁻¹⁹)(1 × 10⁶)(1 × 10⁻⁵)

= 1.6 × 10⁻¹⁸ N

Answer: (A)

[1 Mark]

Q5. Answer: (B) High resistance in series

[1 Mark]

Q6. Answer: (A)

Both A and R are true, and R correctly explains A.

V = E − Ir

Since r > 0 and I > 0:

V < E

[1 Mark]

Q7. Answer: (A)

Both A and R are true, and R correctly explains A.

Magnetic force:

F = q(v × B)

Hence:

F ⊥ v

No work is done by magnetic force.

[1 Mark]

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SECTION B (8 MARKS)

Q8.

Consider a conductor having:

n = number density of free electrons

A = cross-sectional area

v_d = drift velocity

Charge crossing area A in time dt:

dq = nA(v_d dt)e

Current:

I = dq/dt

I = nAev_d

Current density:

J = I/A

Therefore:

J = nev_d

Vector form:

J = −ne(v_d)

Magnitude:

J = nev_d

Assumptions:

1. Electron density is uniform.

2. Drift velocity is constant.

3. Steady current flows through the conductor.

[2 Marks]

Q9.

Drift velocity:

v_d = (eVτ)/(mL)

(a) If L becomes 2L:

v_d' = (eVτ)/(2mL)

Therefore:

v_d' = v_d/2

[1 Mark]

(b) If V becomes 2V:

v_d' = (2eVτ)/(mL)

Therefore:

v_d' = 2v_d

[1 Mark]

Q10.

Biot-Savart law:

dB = (μ₀/4π) × [I(dℓ × r̂)]/r²

where:

dB = magnetic field due to current element

μ₀ = permeability of free space

I = current in conductor

dℓ = current element vector

r = distance from current element to observation point

r̂ = unit vector from source to observation point

Magnitude form:

dB = (μ₀/4π)(Idℓ sinθ)/r²

[2 Marks]

Q11.

Magnetic field due to wire 1 at wire 2:

B₁ = μ₀I₁/(2πd)

Force on wire 2:

F₂ = I₂LB₁

F₂/L = μ₀I₁I₂/(2πd)

For opposite currents, Fleming's left-hand rule shows that the force acts away from each wire.

Hence, the force is repulsive.

[2 Marks]

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SECTION C (6 MARKS)

Q12.

Kirchhoff's Current Law (KCL):

The algebraic sum of currents at a junction is zero.

ΣI = 0

Based on conservation of charge.

[1.5 Marks]

Kirchhoff's Voltage Law (KVL):

The algebraic sum of potential differences around any closed loop is zero.

ΣV = 0

or

ΣIR = ΣE

Based on conservation of energy.

[1.5 Marks]

Q13.

Ampere's circuital law:

∮ B·dl = μ₀I_enclosed

For a long solenoid having:

n = number of turns per unit length

Current = I

Consider an Amperian loop of length l inside the solenoid.

Magnetic field outside is approximately zero.

Hence:

Bl = μ₀(nl)I

Therefore:

B = μ₀nI

[3 Marks]

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SECTION D (5 MARKS)

Q14.

Consider a circular loop of radius R carrying current I.

At point P on the axis at distance x:

Using Biot-Savart law:

dB = (μ₀/4π)(Idℓ sin90°)/r²

where:

r = √(R² + x²)

Only axial components add.

Net magnetic field:

B = (μ₀IR²)/(2(R² + x²)³ᐟ²)

At centre:

x = 0

B = (μ₀IR²)/(2R³)

B = μ₀I/(2R)

[5 Marks]

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SECTION E (4 MARKS)

Q15.

(a)

Voltage across galvanometer equals voltage across shunt:

I_gG = (I − I_g)S

Therefore:

S = (I_gG)/(I − I_g)

[1 Mark]

(b)

Effective resistance:

R_A = GS/(G + S)

Since S is connected in parallel:

R_A < S < G

Hence, effective resistance is smaller than shunt resistance.

[1 Mark]

(c)

Given:

G = 60 Ω

I_g = 5 mA = 0.005 A

I = 6 A

S = (0.005 × 60)/(6 − 0.005)

= 0.3/5.995

= 0.05004 Ω

Therefore:

S ≈ 0.050 Ω

[2 Marks]

------------------- TOTAL MARKS: 30 -------------------