CBSE Class 12 Physics Board Paper 1 — MCQs with Answers

8 solved questions from this paper

Answer and worked explanation shown for each. The remaining 12 are timed and scored when you sign in free.

1. Q1Easy

   Two identical conducting spheres carrying charges +8 μC and −2 μC are brought into contact and then separated to a distance of 10 cm. The force between them is now (1/4πε₀ = 9×10⁹ N·m²/C²):

   A.14.4 N, attractive

   B.2.025 N, repulsive

   C.8.1 N, repulsive✓ Correct

   D.4.05 N, attractive

   Answer: C. 8.1 N, repulsive

   Explanation: Identical spheres in contact share the TOTAL charge equally: (8 − 2)/2 = +3 μC each. Then F = 9×10⁹ × (3×10⁻⁶)² / (0.1)² = 8.1 N, and since both are now positive the force is repulsive. Using the original charges gives 14.4 N attractive, which ignores the redistribution caused by contact.

2. Q2Easy

   Three capacitors, each of 6 μF, are connected so that two are in series and this combination is joined in parallel with the third. The equivalent capacitance is:

   A.9 μF✓ Correct

   B.18 μF

   C.4 μF

   D.2 μF

   Answer: A. 9 μF

   Explanation: Two 6 μF capacitors in series give 6/2 = 3 μF, because in series the reciprocals add. That 3 μF sits in parallel with the third capacitor, and parallel capacitances simply add: 3 + 6 = 9 μF. Adding all three directly gives 18 μF, which forgets that the series pair is weaker than a single capacitor.

3. Q3Easy

   A copper wire of cross-sectional area 1×10⁻⁶ m² carries a steady current of 2 A. If the free-electron density is 8×10²⁸ m⁻³ and e = 1.6×10⁻¹⁹ C, the drift speed of the electrons is about:

   A.1.6×10⁻³ m/s

   B.2.5×10⁻⁴ m/s

   C.1.6×10⁻⁵ m/s

   D.1.6×10⁻⁴ m/s✓ Correct

   Answer: D. 1.6×10⁻⁴ m/s

   Explanation: From I = neAv_d, v_d = I/(neA) = 2 / (8×10²⁸ × 1.6×10⁻¹⁹ × 1×10⁻⁶). The denominator is 8×1.6 = 12.8 times 10²⁸⁻¹⁹⁻⁶ = 10³, so v_d = 2/12800 ≈ 1.6×10⁻⁴ m/s — a few centimetres per minute, which is why the exponent must be tracked carefully rather than guessed.

4. Q4Medium

   A parallel plate capacitor of capacitance 10 pF has a dielectric slab of dielectric constant 5 and thickness equal to HALF the plate separation inserted between its plates. The new capacitance is nearly:

   A.16.7 pF✓ Correct

   B.50 pF

   C.30 pF

   D.12 pF

   Answer: A. 16.7 pF

   Explanation: A partly-filled gap behaves as two capacitors in series, so the effective gap is (d/2) + (d/2)/K = (d/2)(1 + 1/5) = 0.6d. Capacitance scales inversely with the effective gap: C = C₀/0.6 = 10/0.6 ≈ 16.7 pF. Multiplying by K to get 50 pF would only be right if the slab filled the whole gap.

5. Q5Medium

   A solenoid 50 cm long has 500 turns and carries a current of 2 A. The magnetic field near its middle, well inside the solenoid, is (μ₀ = 4π×10⁻⁷ T·m/A):

   A.1.26×10⁻³ T

   B.2.51×10⁻³ T✓ Correct

   C.5.03×10⁻³ T

   D.1.26×10⁻⁶ T

   Answer: B. 2.51×10⁻³ T

   Explanation: For a long solenoid B = μ₀nI where n is turns per METRE, not total turns: n = 500/0.5 = 1000 m⁻¹. So B = 4π×10⁻⁷ × 1000 × 2 = 8π×10⁻⁴ ≈ 2.51×10⁻³ T. Substituting N = 500 straight into the formula halves the answer to 1.26×10⁻³ T, which is the commonest slip here.

6. Q6Medium

   A bar magnet of magnetic moment 0.40 A·m² is held at 30° to a uniform magnetic field of 0.16 T. The torque acting on it is:

   A.0.032 N·m✓ Correct

   B.0.055 N·m

   C.0.064 N·m

   D.0.016 N·m

   Answer: A. 0.032 N·m

   Explanation: Torque on a magnetic dipole is τ = mB sinθ, with θ measured between the magnet's axis and the field: τ = 0.40 × 0.16 × sin30° = 0.064 × 0.5 = 0.032 N·m. Using cos30° gives 0.055 N·m — that expression belongs to the potential energy −mB cosθ, not the torque.

7. Q7Hard

   A cell is connected across a potentiometer wire and gives a balance length of 560 cm. When a 5 Ω resistor is connected across the terminals of the same cell, the balance length falls to 400 cm. The internal resistance of the cell is:

   A.1 Ω

   B.2.5 Ω

   C.2 Ω✓ Correct

   D.3 Ω

   Answer: C. 2 Ω

   Explanation: The open-circuit balance length measures the emf and the loaded balance length measures the terminal voltage, so r = R(l₁ − l₂)/l₂ = 5 × (560 − 400)/400 = 5 × 0.4 = 2 Ω. The division must be by the SHORTER (loaded) length, since that is the one proportional to the terminal voltage; dividing by 560 instead would give the too-small 1.43 Ω.

8. Q8Hard

   A proton and an alpha particle, starting from rest, are accelerated through the SAME potential difference and then enter a uniform magnetic field at right angles to it. The ratio of the radius of the alpha particle's path to that of the proton is:

   A.1 : 1

   B.1 : √2

   C.2 : 1

   D.√2 : 1✓ Correct

   Answer: D. √2 : 1

   Explanation: Acceleration through V gives momentum p = √(2mqV), and in the field r = p/qB = √(2mV/q)/B, so r ∝ √(m/q). The alpha particle has 4 times the proton mass and 2 times the charge, so the ratio is √(4/2 ÷ 1/1) = √2. Treating r ∝ m/q instead of the square root would wrongly give 2 : 1.

More Class 12 Physics board papers

• Board Paper 1 — you are here
• Board Paper 2 — 20 questions
• Board Paper 3 — 20 questions
• Board Paper 4 — 20 questions
• Board Paper 5 — 20 questions

Prefer chapter-by-chapter revision first? Class 12 Physics notes & chapter MCQs →

Questions students ask about this paper