CBSE Class 12 Physics Board Paper 2 — 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

   The electric field at a point 30 cm from an isolated point charge in air is 900 N/C, directed away from the charge. The magnitude of the charge is (1/4πε₀ = 9×10⁹ N·m²/C²):

   A.3 nC

   B.9 nC✓ Correct

   C.27 nC

   D.0.9 nC

   Answer: B. 9 nC

   Explanation: From E = kq/r², q = Er²/k = 900 × (0.30)²/(9×10⁹) = 900 × 0.09/9×10⁹ = 9×10⁻⁹ C = 9 nC. The distance must be squared and converted to metres first; using r = 30 without squaring, or in centimetres, throws the answer off by orders of magnitude.

2. Q2Easy

   Resistors of 4 Ω and 6 Ω are connected in parallel across a 12 V battery of negligible internal resistance. The current drawn from the battery is:

   A.1.2 A

   B.2 A

   C.5 A✓ Correct

   D.12 A

   Answer: C. 5 A

   Explanation: Each branch carries its own current across the same 12 V: 12/4 = 3 A and 12/6 = 2 A, so the battery supplies 5 A. Equivalently the parallel combination is (4×6)/(4+6) = 2.4 Ω and 12/2.4 = 5 A. Adding the resistances as if in series gives 10 Ω and the too-small 1.2 A.

3. Q3Easy

   An alternating voltage has a peak value of 311 V. The reading shown by an AC voltmeter connected across the source is nearly:

   A.220 V✓ Correct

   B.440 V

   C.155.5 V

   D.311 V

   Answer: A. 220 V

   Explanation: AC meters are calibrated to read the rms value, and V_rms = V₀/√2 = 311/1.414 ≈ 220 V — which is why domestic mains quoted as 220 V actually peaks at about 311 V. Halving the peak instead of dividing by √2 gives 155.5 V, a common shortcut that is simply wrong for a sinusoid.

4. Q4Medium

   A 2 μF capacitor charged to 100 V is connected, positive plate to positive plate, across a 4 μF capacitor charged to 50 V. The common potential difference across the pair is nearly:

   A.75 V

   B.66.7 V✓ Correct

   C.50 V

   D.150 V

   Answer: B. 66.7 V

   Explanation: Charge is conserved and redistributes until both capacitors share one potential: V = (C₁V₁ + C₂V₂)/(C₁ + C₂) = (200 + 200) μC / 6 μF ≈ 66.7 V. Simply averaging the two voltages to get 75 V ignores that the 4 μF capacitor carries more weight in the sharing.

5. Q5Medium

   Two long straight parallel wires 10 cm apart carry currents of 5 A and 10 A in the SAME direction. The force per unit length between them is (μ₀/4π = 10⁻⁷ T·m/A):

   A.1×10⁻⁴ N/m, attractive✓ Correct

   B.1×10⁻⁴ N/m, repulsive

   C.1×10⁻⁵ N/m, attractive

   D.5×10⁻⁴ N/m, repulsive

   Answer: A. 1×10⁻⁴ N/m, attractive

   Explanation: F/l = μ₀I₁I₂/(2πd) = 2×10⁻⁷ × (5×10)/0.10 = 1×10⁻⁴ N/m. Parallel currents in the same direction ATTRACT — this is the definition-of-the-ampere case — so a repulsive answer reverses the rule that applies to antiparallel currents.

6. Q6Medium

   At a certain place the horizontal component of the earth's magnetic field is 0.30 G and the angle of dip is 60°. The total magnetic field of the earth at that place is:

   A.0.15 G

   B.0.35 G

   C.0.52 G

   D.0.60 G✓ Correct

   Answer: D. 0.60 G

   Explanation: The horizontal component is the projection of the total field onto the horizontal, so H = B cosδ and B = H/cosδ = 0.30/cos60° = 0.30/0.5 = 0.60 G. The value 0.52 G is H tanδ, which is the VERTICAL component, and the total field must always be larger than either component.

7. Q7Hard

   A point charge q is placed exactly at one corner of a cube. The total electric flux passing through the cube is:

   A.q/6ε₀

   B.q/24ε₀

   C.q/8ε₀✓ Correct

   D.q/2ε₀

   Answer: C. q/8ε₀

   Explanation: Imagine eight identical cubes stacked around the corner so that the charge sits at the shared centre; those eight cubes together enclose the charge completely and carry total flux q/ε₀. By symmetry each cube takes one eighth, so the flux is q/8ε₀. The value q/24ε₀ is the flux through just ONE of the three faces that do not touch the charge, not through the whole cube.

8. Q8Hard

   A uniform wire of resistance R is cut into five equal pieces, and these five pieces are then connected in parallel. The resistance of this combination compared with R is:

   A.1 : 5

   B.5 : 1

   C.1 : 10

   D.1 : 25✓ Correct

   Answer: D. 1 : 25

   Explanation: Cutting the wire into five equal pieces gives each a resistance R/5, since resistance is proportional to length. Putting five equal resistances in parallel divides again by five: (R/5)/5 = R/25, so the ratio is 1 : 25. Applying only one of the two factors of five gives the incomplete 1 : 5.

More Class 12 Physics board papers

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