CBSE Class 12 Physics Board Paper 4 — 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 point charges of +3 μC and −2 μC are held 30 cm apart in air. The electrostatic potential energy of this pair is (1/4πε₀ = 9×10⁹ N·m²/C²):

   A.+0.18 J

   B.−0.054 J

   C.−0.18 J✓ Correct

   D.−0.6 J

   Answer: C. −0.18 J

   Explanation: U = kq₁q₂/r = 9×10⁹ × (3×10⁻⁶)(−2×10⁻⁶)/0.30 = −0.054/0.30 = −0.18 J. Unlike force, potential energy carries a single power of r in the denominator, so dividing by r² would give −0.6 J. The negative sign says the pair is bound: work must be done to pull the charges apart.

2. Q2Easy

   A charged particle moves with velocity v exactly parallel to a uniform magnetic field B. The magnetic force acting on it is:

   A.qvB, directed along the field

   B.qvB, perpendicular to the field

   C.qvB/2, perpendicular to the velocity

   D.zero✓ Correct

   Answer: D. zero

   Explanation: The magnetic force is F = qvB sinθ, and for motion parallel to the field θ = 0 so sinθ = 0 and the force vanishes — the particle travels in a straight line at constant speed. The force is maximum only when the velocity is perpendicular to the field, and it is never directed along the field because F = q(v × B) is perpendicular to both.

3. Q3Easy

   The number of neutrons in the nucleus of ²³⁸U, for which the atomic number is 92, is:

   A.146✓ Correct

   B.92

   C.238

   D.330

   Answer: A. 146

   Explanation: The mass number A counts protons plus neutrons, so the neutron number is N = A − Z = 238 − 92 = 146. Adding rather than subtracting gives 330, a figure larger than the mass number itself, which is impossible for any nucleus.

4. Q4Medium

   Charges of +4q and +q are fixed a distance L apart. The point on the line joining them at which the resultant electric field is zero lies at:

   A.L/3 from the larger charge

   B.L/2 from the larger charge

   C.L/3 from the smaller charge✓ Correct

   D.2L/3 from the smaller charge

   Answer: C. L/3 from the smaller charge

   Explanation: The null point lies between two like charges. Setting 4q/(L − x)² = q/x², with x measured from the smaller charge, gives 2x = L − x, so x = L/3 from the smaller charge and 2L/3 from the larger one. The point must sit CLOSER to the weaker charge, since only there can its field grow enough to match the stronger one.

5. Q5Medium

   A battery of emf 12 V has an internal resistance of 2 Ω. The maximum power it can deliver to an external resistance, and the value of that resistance, are:

   A.36 W at R = 2 Ω

   B.18 W at R = 4 Ω

   C.72 W at R = 0

   D.18 W at R = 2 Ω✓ Correct

   Answer: D. 18 W at R = 2 Ω

   Explanation: External power P = E²R/(R + r)² is maximised when R = r, giving P_max = E²/4r = 144/8 = 18 W. At that setting exactly half the power generated goes to the load and half is wasted inside the battery. Short-circuiting (R = 0) maximises the CURRENT but delivers zero external power, since all of it is dissipated in r.

6. Q6Medium

   The core of an electromagnet, which must switch on and off rapidly, should be made of a material having:

   A.high retentivity and high coercivity, so that it stays magnetised

   B.low permeability and high coercivity

   C.high permeability and low coercivity, with a narrow hysteresis loop✓ Correct

   D.high permeability and high retentivity, with a broad hysteresis loop

   Answer: C. high permeability and low coercivity, with a narrow hysteresis loop

   Explanation: High permeability gives a strong field for a modest current, while low coercivity and low retentivity let the core lose its magnetism the moment the current stops. A narrow hysteresis loop also means little energy is wasted as heat per cycle, which is why soft iron is used. The high-retentivity, broad-loop combination describes a PERMANENT magnet material such as steel or alnico.

7. Q7Hard

   A parallel plate capacitor remains CONNECTED to a battery while a dielectric slab of dielectric constant 3 is slid in to fill the space between its plates completely. Which statement describes the result correctly?

   A.Capacitance and charge each become three times, while the electric field between the plates is unchanged✓ Correct

   B.Capacitance becomes three times while the charge on the plates stays the same

   C.Charge becomes three times while the potential difference falls to one third

   D.Capacitance becomes three times while the stored energy falls to one third

   Answer: A. Capacitance and charge each become three times, while the electric field between the plates is unchanged

   Explanation: With the battery still attached, V is held fixed, so C = KC₀ triples and Q = CV triples with it, while the field E = V/d is unchanged because both V and d are unchanged. Energy U = ½CV² also triples, supplied by the battery. The answers with a falling voltage or a fixed charge belong to the DISCONNECTED case, where Q is what stays constant.

8. Q8Hard

   The resistance of a metallic wire is 20 Ω at 20 °C and 60 Ω at 500 °C. Its temperature coefficient of resistance, referred to 20 °C, is:

   A.2.08×10⁻³ °C⁻¹

   B.4.17×10⁻³ °C⁻¹✓ Correct

   C.4.17×10⁻² °C⁻¹

   D.8.33×10⁻³ °C⁻¹

   Answer: B. 4.17×10⁻³ °C⁻¹

   Explanation: From R = R₀[1 + α(T − T₀)], α = (R − R₀)/[R₀(T − T₀)] = (60 − 20)/(20 × 480) = 40/9600 ≈ 4.17×10⁻³ °C⁻¹. The temperature INTERVAL is 480 °C, not 500 °C; using 500 in the denominator gives 4.0×10⁻³ and dividing by the final resistance instead of the initial one gives 1.39×10⁻³.

More Class 12 Physics board papers

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• Board Paper 2 — 20 questions
• Board Paper 3 — 20 questions
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• Board Paper 5 — 20 questions

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