CBSE Class 12 Physics Board Paper 5 — 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

   A steady current of 3.2 A flows through a conductor. The number of electrons crossing any cross-section of it each second is (e = 1.6×10⁻¹⁹ C):

   A.2×10¹⁸

   B.2×10¹⁹✓ Correct

   C.5×10¹⁸

   D.1.6×10¹⁹

   Answer: B. 2×10¹⁹

   Explanation: Current is charge per second, so n = I/e = 3.2/(1.6×10⁻¹⁹) = 2×10¹⁹ electrons per second. Dividing the electronic charge by the current instead, or slipping one power of ten in 3.2/1.6 = 2, is what produces the neighbouring values.

2. Q2Easy

   The north pole of a bar magnet is pushed rapidly towards a closed conducting loop. Viewed from the side on which the magnet approaches, the induced current in the loop flows:

   A.clockwise, making the near face a south pole that attracts the magnet

   B.anticlockwise, making the near face a south pole

   C.anticlockwise, making the near face a north pole that opposes the magnet✓ Correct

   D.only after the magnet touches the loop

   Answer: C. anticlockwise, making the near face a north pole that opposes the magnet

   Explanation: By Lenz's law the induced current must OPPOSE the approach, so the face turned towards the magnet must become a north pole and repel it; seen from the magnet's side, a north face means anticlockwise current. A south face would attract the magnet, accelerating it and creating energy from nothing, which is why Lenz's law is a statement of energy conservation.

3. Q3Easy

   In a Young's double-slit experiment one of the two slits is covered by an opaque card. The pattern on the screen then shows:

   A.the same fringes, but with twice the fringe width

   B.the same fringes, but brighter

   C.the same fringes at the same spacing, with half the intensity

   D.no interference fringes, only a broad single-slit diffraction pattern✓ Correct

   Answer: D. no interference fringes, only a broad single-slit diffraction pattern

   Explanation: Interference requires two coherent beams to superpose; block one and there is nothing for the other to interfere with, so the fringe system vanishes. What remains is the diffraction pattern of the single open slit — a wide bright central band with faint side maxima. Fringe width λD/d depends on the slit SEPARATION, which becomes meaningless once only one slit is open.

4. Q4Medium

   Three equal point charges q are fixed at the corners of an equilateral triangle of side a. The magnitude of the net electrostatic force on any one of them is:

   A.kq²/a²

   B.√3 kq²/a²✓ Correct

   C.2kq²/a²

   D.√2 kq²/a²

   Answer: B. √3 kq²/a²

   Explanation: Each of the other two charges pushes with force F = kq²/a², and these two forces are separated by 60°. Their resultant is 2F cos30° = √3 F = √3 kq²/a². Adding the magnitudes directly to get 2F would be right only if the two forces were parallel, and the √2 combination applies to forces at 90°.

5. Q5Medium

   A potentiometer wire carries a steady current from a driver cell, giving a potential gradient of 0.20 V per metre along the wire. A cell of unknown emf, connected in the secondary circuit, gives a null point at 250 cm from the starting end. The emf of that cell is:

   A.2 V

   B.0.8 V

   C.0.25 V

   D.0.5 V✓ Correct

   Answer: D. 0.5 V

   Explanation: At the null point no current is drawn from the unknown cell, so the length balances its full emf: E = (potential gradient) × (balancing length) = 0.20 × 2.50 = 0.50 V. The length must first be converted to metres to match the gradient's units; leaving it as 250 cm inflates the answer a hundredfold.

6. Q6Medium

   A galvanometer of resistance 50 Ω gives full-scale deflection for a current of 2 mA. To convert it into an ammeter reading up to 1 A, one must connect a shunt of nearly:

   A.25 Ω in parallel

   B.0.1 Ω in parallel✓ Correct

   C.0.5 Ω in series

   D.50 Ω in parallel

   Answer: B. 0.1 Ω in parallel

   Explanation: The shunt must carry all but 2 mA of the 1 A, so S = I_gG/(I − I_g) = (0.002 × 50)/0.998 ≈ 0.1 Ω. A very small parallel resistance is what makes an ammeter low-resistance, so that inserting it barely disturbs the circuit; a series resistance would raise the total resistance and make a voltmeter instead.

7. Q7Hard

   A hollow conducting sphere of radius 10 cm carries a charge of 5 μC. The electric field at a point 5 cm from its centre and at a point 20 cm from its centre are respectively (1/4πε₀ = 9×10⁹ N·m²/C²):

   A.1.8×10⁷ N/C and 1.125×10⁶ N/C

   B.zero and 4.5×10⁶ N/C

   C.zero and 1.125×10⁶ N/C✓ Correct

   D.1.125×10⁶ N/C and zero

   Answer: C. zero and 1.125×10⁶ N/C

   Explanation: A Gaussian sphere of radius 5 cm lies inside the shell and encloses no charge, so the field there is exactly zero — the charge resides entirely on the outer surface of a conductor. Outside, the shell behaves like a point charge at its centre: E = 9×10⁹ × 5×10⁻⁶/(0.20)² = 1.125×10⁶ N/C. Using the shell's own radius 10 cm rather than the field point's 20 cm gives the four-times-larger 4.5×10⁶ N/C.

8. Q8Hard

   Capacitors of 2 μF, 3 μF and 6 μF are joined in series across a 100 V supply. The charge on the 3 μF capacitor and the potential difference across it are:

   A.100 μC and 33.3 V✓ Correct

   B.300 μC and 100 V

   C.100 μC and 50 V

   D.33.3 μC and 33.3 V

   Answer: A. 100 μC and 33.3 V

   Explanation: In series 1/C = 1/2 + 1/3 + 1/6 = 1, so C_eq = 1 μF and the total charge is Q = 1 × 100 = 100 μC. That SAME charge sits on every capacitor in a series chain, so V across the 3 μF unit is Q/C = 100/3 ≈ 33.3 V. Computing the charge as 3 μF × 100 V = 300 μC wrongly assumes each capacitor gets the full supply voltage.

More Class 12 Physics board papers

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

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