A 4 μF capacitor charged to 200 V is disconnected from its battery and then connected across an identical uncharged 4 μF capacitor. The energy lost in the process is:

zero, since charge is conserved

CBSE Class 12 Physics Board Paper 3 — MCQs with Answers

Original paper · CBSE patternWritten by our subject team, not a reprint of an official sample paper.

Questions: 20
Time: 30 min
Marking: +1, no negative marking
Mix: 4 Easy · 8 Medium · 8 Hard

Board Paper 3 — CBSE Class 12 Physics is a full 20-question objective paper for Class 12 Physics, written in the CBSE board pattern and marked the way the board marks: +1 for a correct answer and 0 for a wrong one — no negative marking, exactly like the real board paper. It follows Section A of a real CBSE paper. Q1–Q16 are MCQs with one correct answer. Q17–Q18 are case-based questions on a short source. Q19–Q20 are assertion–reason questions that use the board's own four answer options.

It is set a little harder than the real board paper on purpose (4 Easy · 8 Medium · 8 Hard), because a paper you find easy tells you nothing about exam day. You get 30 minutes, about 90 seconds a question, which is the speed the real paper needs. 8 of the questions are solved in full below, each with the correct option and a worked explanation. Sign in free to attempt the other 12, timed and scored.

Take it like a real exam, with your books closed. Your score on a timed full-length paper is the only honest way to know if your Physics revision holds up across all the chapters. That is exactly what the board tests, and chapter-wise practice cannot tell you.

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.

Q1Easy
Charges of +2 μC and −2 μC are fixed at the two ends of a 20 cm line. The electric potential at the midpoint of the line is:
A.3.6×10⁵ V
B.1.8×10⁵ V
C.3.6×10⁴ V
D.zero✓ Correct
Answer: D. zero
Explanation: Potential is a scalar and adds with sign, so at the midpoint the two contributions are +kq/0.1 and −kq/0.1, which cancel exactly to zero. Note that the electric FIELD there is not zero — both charges push it in the same direction — so a non-zero value like 3.6×10⁵ V confuses the vector quantity with the scalar one.
Q2Easy
An electric bulb is rated 100 W, 220 V. The resistance of its filament while operating at its rated values is:
A.484 Ω✓ Correct
B.220 Ω
C.2.2 Ω
D.44 Ω
Answer: A. 484 Ω
Explanation: From P = V²/R, R = V²/P = (220)²/100 = 48400/100 = 484 Ω. Dividing V by P instead of V² by P gives 2.2 Ω, a resistance so low it would draw 100 A from the mains — a quick sanity check that catches the slip.
Q3Easy
A lens has a focal length of −20 cm. Its power and nature are:
A.−0.05 D, diverging
B.−5 D, diverging✓ Correct
C.+5 D, converging
D.−20 D, diverging
Answer: B. −5 D, diverging
Explanation: Power is the reciprocal of the focal length expressed in METRES: P = 1/(−0.20) = −5 D. A negative focal length under the Cartesian convention means the lens is concave and diverges the beam. Leaving the focal length in centimetres gives −0.05 D, which is the classic unit error here.
Q4Medium
An electric dipole of dipole moment 4×10⁻⁹ C·m is placed in a uniform electric field of 5×10⁴ N/C, its axis making an angle of 30° with the field. The torque acting on the dipole is:
A.1×10⁻⁴ N·m✓ Correct
B.2×10⁻⁴ N·m
C.1.73×10⁻⁴ N·m
D.1×10⁻⁵ N·m
Answer: A. 1×10⁻⁴ N·m
Explanation: Torque on a dipole is τ = pE sinθ = 4×10⁻⁹ × 5×10⁴ × sin30° = 2×10⁻⁴ × 0.5 = 1×10⁻⁴ N·m. Omitting the sine gives the maximum torque 2×10⁻⁴ N·m, which applies only at 90°, and using cos30° gives 1.73×10⁻⁴ N·m, an expression belonging to the potential energy instead.
Q5Medium
In a metre-bridge experiment an unknown resistance X is placed in the left gap and a 6 Ω standard resistance in the right gap. The balance point is obtained at 40 cm from the left end. The value of X is:
A.9 Ω
B.4 Ω✓ Correct
C.6 Ω
D.2.4 Ω
Answer: B. 4 Ω
Explanation: At balance X/R = l/(100 − l), so X = 6 × 40/60 = 4 Ω. The unknown must be SMALLER than the standard because its balance length is shorter than 50 cm; inverting the ratio to 6 × 60/40 gives 9 Ω and contradicts that check.
Q6Medium
A magnetising field of 1600 A/m produces a magnetisation of 4.8×10⁵ A/m in a certain material. Its magnetic susceptibility and relative permeability are:
A.300 and 301✓ Correct
B.300 and 300
C.3.0×10⁻³ and 1.003
D.480 and 481
Answer: A. 300 and 301
Explanation: Susceptibility is χ = M/H = 4.8×10⁵/1600 = 300, and relative permeability is μ_r = 1 + χ = 301. The large positive susceptibility marks this as a ferromagnetic material. Forgetting the +1 makes χ and μ_r identical, which is only a good approximation when χ is huge, not an exact relation.
Q7Hard
An infinitely long straight wire carries a uniform linear charge density of 2×10⁻⁸ C/m. The electric field at a point 20 cm from the wire is (1/4πε₀ = 9×10⁹ N·m²/C²):
A.900 N/C
B.450 N/C
C.3600 N/C
D.1800 N/C✓ Correct
Answer: D. 1800 N/C
Explanation: A cylindrical Gaussian surface gives E = λ/(2πε₀r) = 2kλ/r = 2 × 9×10⁹ × 2×10⁻⁸ / 0.20 = 360/0.20 = 1800 N/C. Note the field falls as 1/r for a line, not 1/r² as for a point charge; dropping the factor of 2 gives the 900 N/C trap.
Q8Hard
A 4 μF capacitor charged to 200 V is disconnected from its battery and then connected across an identical uncharged 4 μF capacitor. The energy lost in the process is:
A.0.08 J
B.zero, since charge is conserved
C.0.04 J✓ Correct
D.0.02 J
Answer: C. 0.04 J
Explanation: Initially U = ½ × 4×10⁻⁶ × 200² = 0.08 J. Charge is conserved but redistributes, so the common potential is 100 V and the final energy is ½ × 8×10⁻⁶ × 100² = 0.04 J, leaving 0.04 J lost. Conservation of CHARGE does not imply conservation of energy — the difference is dissipated as heat in the connecting wires and as radiation.

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Is this an official CBSE sample paper?

No, and that is on purpose. This is an original paper written by our subject team in the CBSE board pattern. It is not a copy of CBSE's official Sample Question Paper. So you get fresh questions you have not already seen on a dozen other sites, with instant scoring and worked solutions instead of a PDF.

How is this Class 12 Physics paper marked?

+1 for a correct answer and 0 for a wrong one — no negative marking, exactly like the real board paper. So attempt every question. A board paper takes nothing away for a wrong answer, and leaving a question blank can only cost you.

How long should this paper take?

30 minutes for 20 questions, about 90 seconds each. The timer keeps running whether you watch it or not, and that is the point. Students lose board marks to poor timing at least as often as to gaps in what they know.

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