Class 12 (CBSE) · Board Sample Papers (CBSE)

Class 12 Physics — CBSE Board Sample Paper

16 practice questions with full step-by-step solutions — free, no sign-up.

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Class 12 Physics — CBSE Board Sample Paper — solved practice questions

8 Class 12 Board Sample Papers (CBSE) questions with step-by-step solutions. Attempt each, then reveal the answer.

  1. Q1medium

    A uniform electric field pointing in positive X-direction exists in a region. Let A be the origin, B be the point on the X-axis at x = +1 cm and C be the point on the Y-axis at y = +1 cm. Then the potential at points A, B and C satisfy.

    • AV_A < V_B
    • BV_A > V_B
    • CV_A < V_C
    • DV_A > V_C
    Show answer & solution

    Correct answer: (B) V_A > V_B

    Potential decreases along the direction of the electric field, so V_A > V_B. Since C is perpendicular to the field (on the Y-axis), V_A = V_C.

  2. Q2medium

    A conducting wire connects two charged conducting spheres of radii r1 and r2 such that they attain equilibrium with respect to each other. The distance of separation between the two spheres is very large as compared to either of their radii. The ratio of the magnitudes of the electric fields at the surfaces of the spheres of radii r1 and r2 is

    • Ar1 / r2
    • Br2 / r1
    • Cr2^2 / r1^2
    • Dr1^2 / r2^2
    Show answer & solution

    Correct answer: (B) r2 / r1

    At equilibrium the spheres share a common potential, giving q1/q2 = r1/r2. Since E = kq/r^2, the ratio E1/E2 = (q1/q2)(r2^2/r1^2) = (r1/r2)(r2^2/r1^2) = r2/r1.

  3. Q3hard

    A long straight wire of circular cross section of radius a carries a steady current I. The current is uniformly distributed across its cross section. The ratio of magnitudes of the magnetic field at a point a/2 above the surface of wire to that of a point a/2 below its surface is

    • A4 : 1
    • B1 : 1
    • C4 : 3
    • D3 : 4
    Show answer & solution

    Correct answer: (C) 4 : 3

    Outside at distance 3a/2: B = mu0 I / (2 pi (3a/2)) = mu0 I / (3 pi a). Inside at distance a/2: B = mu0 I (a/2) / (2 pi a^2) = mu0 I / (4 pi a). The ratio is 4 : 3.

  4. Q4easy

    The diffraction effect can be observed in

    • Asound waves only
    • Blight waves only
    • Cultrasonic waves only
    • Dsound waves as well as light waves
    Show answer & solution

    Correct answer: (D) sound waves as well as light waves

    Diffraction is a general wave phenomenon, so it is observed in both sound waves and light waves.

  5. Q5medium

    A capacitor consists of two parallel plates, with an area of cross-section of 0.001 m^2, separated by a distance of 0.0001 m. If the voltage across the plates varies at the rate of 10^8 V/s, then the value of displacement current through the capacitor is

    • A8.85 x 10^-3 A
    • B8.85 x 10^-4 A
    • C7.85 x 10^-3 A
    • D9.85 x 10^-3 A
    Show answer & solution

    Correct answer: (A) 8.85 x 10^-3 A

    I_d = epsilon0 (A/d) dV/dt = 8.854e-12 x (0.001/0.0001) x 10^8 = 8.85 x 10^-3 A.

  6. Q6medium

    In a series LCR circuit, the voltage across the resistance, capacitance and inductance is 10 V each. If the capacitance is short circuited the voltage across the inductance will be

    • A10 V
    • B10*sqrt(2) V
    • C10/sqrt(2) V
    • D20 V
    Show answer & solution

    Correct answer: (C) 10/sqrt(2) V

    Initially X_C = X_L = R and V_S = 10 V. With the capacitor short-circuited, Z = sqrt(R^2 + X_L^2) = R*sqrt(2), so current I = 10/(R*sqrt(2)) and V_L = I*X_L = 10/sqrt(2) V.

  7. Q7medium

    Correct match of column I with column II is. Column I (waves): (1) Infra-red (2) Radio (3) Light (4) Microwave ; Column II (Production): P. Rapid vibration of electrons in aerials (Q) Electrons in atoms emit light when they move from higher to lower energy level (R) Klystron valve (S) Vibration of atoms and molecules

    • A1-P, 2-R, 3-S, 4-Q
    • B1-S, 2-P, 3-Q, 4-R
    • C1-Q, 2-P, 3-S, 4-R
    • D1-S, 2-R, 3-P, 4-Q
    Show answer & solution

    Correct answer: (B) 1-S, 2-P, 3-Q, 4-R

    Infra-red is produced by vibration of atoms and molecules (S), radio waves by rapid vibration of electrons in aerials (P), light by atomic electron transitions (Q), and microwaves by a klystron valve (R).

  8. Q8medium

    The distance of closest approach of an alpha particle is d when it moves with a speed V towards a nucleus. Another alpha particle is projected with higher energy such that the new distance of the closest approach is d/2. What is the speed of projection of the alpha particle in this case?

    • AV/sqrt(2)
    • Bsqrt(2) V
    • C2 V
    • D4 V
    Show answer & solution

    Correct answer: (B) sqrt(2) V

    The distance of closest approach d is inversely proportional to the kinetic energy, i.e. d is proportional to 1/V^2. Halving d requires doubling the kinetic energy, so V increases by sqrt(2), giving sqrt(2) V.

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