Class 12 Physics Practice Paper Set 6 with Solutions

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Class 12 Physics · Practice Paper SET-6 with Solutions

CBSE PRACTICE PAPER 2026–27 · PHYSICS (SET – 6)

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CLASS : XIISUBJECT : PHYSICSROLL NO. : ..................

Time : 3 Hrs.Session : 2026–27M.M. : 70

General Instructions

1. There are 33 questions in all. All questions are compulsory.
2. This question paper has five sections: Section A, B, C, D and E.
3. Section A: Q. 1 to 16 – 1 mark each (Q. 1–12 MCQ, Q. 13–16 Assertion–Reason).
4. Section B: Q. 17 to 21 – 2 marks each. Section C: Q. 22 to 28 – 3 marks each (internal choice in Q. 27 and Q. 28).
5. Section D: Q. 29 and 30 – case-based questions of 4 marks each (internal choice in one part).
6. Section E: Q. 31 to 33 – 5 marks each (internal choice in all three).
7. There is no overall choice. Use of calculator is not permitted. Draw neat labelled diagrams wherever required.
8. The detailed Answer Key & Solutions are given at the end of this paper.

Useful constants: c = 3 × 10⁸ m/s; h = 6.6 × 10⁻³⁴ J s; e = 1.6 × 10⁻¹⁹ C; μ₀ = 4π × 10⁻⁷ T m/A; ε₀ = 8.85 × 10⁻¹² C²/N m²; 1 u = 931.5 MeV/c²; 1 eV = 1.6 × 10⁻¹⁹ J. Take π = 3.14 where required. Mass of electron = 9.1 × 10⁻³¹ kg; 1/4πε₀ = 9 × 10⁹ N m²/C².

SECTION A (16 × 1 = 16 Marks)

Q. 1 to Q. 12 : Multiple Choice Questions (choose the correct option)

1. Which of the following is a non-polar molecule?

(a) H₂O(b) CO₂(c) HCl(d) NH₃

2. The electric potential in a region is V = 4x² volt (x in metres). The magnitude of the electric field at x = 1 m is

(a) 4 V/m(b) 16 V/m(c) 8 V/m(d) 2 V/m

3. Kirchhoff’s junction rule is a consequence of the conservation of

(a) charge(b) energy(c) momentum(d) mass

4. A cell of emf 2 V and internal resistance 0.5 Ω supplies a current of 2 A to a circuit. The terminal voltage is

(a) 1.5 V(b) 1 V(c) 2 V(d) 0.5 V

5. A charged particle passes undeflected through a region having perpendicular electric and magnetic fields E = 2 × 10⁴ V/m and B = 0.1 T. Its speed is

(a) 2 × 10³ m/s(b) 2 × 10⁴ m/s(c) 5 × 10⁻⁶ m/s(d) 2 × 10⁵ m/s

6. The ratio of the magnetic moment to the angular momentum of an electron revolving in an orbit is

(a) e/m(b) 2e/m(c) e/2m(d) m/e

7. The magnetic flux linked with a coil of resistance 5 Ω changes by 0.05 Wb. The charge that flows through the coil is

(a) 0.25 C(b) 0.1 C(c) 0.05 C(d) 0.01 C

8. A series LR circuit has R = 30 Ω and X_L = 40 Ω. Its impedance is

(a) 50 Ω(b) 70 Ω(c) 10 Ω(d) 35 Ω

9. Maxwell modified Ampère’s circuital law by adding

(a) conduction current(b) displacement current(c) eddy current(d) induced charge

10. Two sources of light are coherent if they have

(a) the same intensity(b) different frequencies(c) a constant phase difference(d) a random phase difference

11. The slope of the graph of stopping potential versus frequency of incident light in a photoelectric experiment is

(a) h(b) e/h(c) the work function φ(d) h/e

12. The mean life τ of a radioactive substance is related to its half-life T_1/2 by

(a) τ = 1.44 T_1/2(b) τ = 0.693 T_1/2(c) τ = 2 T_1/2(d) τ = T_1/2/2

Q. 13 to Q. 16 : Assertion–Reason Questions

Options: (a) Both A and R are true and R is the correct explanation of A. (b) Both A and R are true but R is not the correct explanation of A. (c) A is true but R is false. (d) A is false but R is true.

13. Assertion (A): Equipotential surfaces are always perpendicular to the electric field lines.
Reason (R): No work is done in moving a charge along an equipotential surface.

14. Assertion (A): A voltmeter is connected in parallel with the circuit element and has a very high resistance.
Reason (R): A voltmeter is made by connecting a low resistance in parallel with a galvanometer.

15. Assertion (A): Electromagnetic waves are longitudinal in nature.
Reason (R): In an electromagnetic wave the electric and magnetic fields are perpendicular to the direction of propagation.

16. Assertion (A): Doping increases the conductivity of a semiconductor.
Reason (R): The forbidden energy gap of pure silicon is about 1.1 eV.

SECTION B (5 × 2 = 10 Marks) — Very Short Answer Questions

17. What is polarisation of a dielectric? A dielectric slab of dielectric constant 5 is placed in a uniform electric field E₀. What is the electric field inside the slab?(2)

18. State Joule’s law of heating. A wire of resistance 20 Ω carries a current of 3 A for 10 s. Calculate the heat produced.(2)

19. A galvanometer of resistance 50 Ω gives full-scale deflection for a current of 2 mA. How can it be converted into a voltmeter of range 0–10 V?(2)

20. Show that the charge flowing through a closed circuit of resistance R due to a change in magnetic flux is independent of the time taken. Find it for R = 10 Ω and a flux change from 0.08 Wb to 0.03 Wb.(2)

21. Write any four properties of the nuclear force.(2)

SECTION C (7 × 3 = 21 Marks) — Short Answer Questions

22. Explain the behaviour of polar and non-polar dielectrics in an external electric field. Define dielectric constant in terms of the electric field.(3)

23. Derive the relation between the electric field and the electric potential. Hence show that the electric field is perpendicular to an equipotential surface.(3)

24. Explain why the resistivity of a metal increases but that of a semiconductor decreases with increase in temperature.(3)

25. Explain the principle of a velocity selector. Show that a charged particle moves undeflected in crossed electric and magnetic fields only if its speed is v = E/B. Find v for E = 2 × 10⁴ V/m and B = 0.1 T.(3)

26. Derive the expression for the magnetic moment of an electron revolving in a circular orbit in terms of its orbital angular momentum. Define and write the value of the Bohr magneton.(3)

27. Use Ampere’s circuital law to derive the expression for the magnetic field inside a long solenoid.(3)

OR

Derive the relation between the rms value and the peak value of an alternating current.

28. Draw and explain the binding energy per nucleon versus mass number curve. How does it explain the release of energy in nuclear fission and fusion?(3)

OR

Calculate the energy released in the α-decay ₉₂²³⁸U → ₉₀²³⁴Th + ₂⁴He. Given: m(U) = 238.0508 u, m(Th) = 234.0436 u, m(He) = 4.0026 u, 1 u = 931.5 MeV/c².

SECTION D (2 × 4 = 8 Marks) — Case-Based Questions

29. Case Study – Electromagnetic wave. A plane electromagnetic wave travels in vacuum. The amplitude of its electric field is E₀ = 300 V/m.

(i) Find the amplitude of the magnetic field. (1)
(ii) Find the average energy density of the wave. (1)
(iii) Find the intensity of the wave. (1)
(iv) State the direction of propagation of the wave with respect to E and B. (1)

OR

(iv) Name two electromagnetic waves whose frequencies are higher than that of ultraviolet rays. (1)

30. Case Study – Photoelectric effect graph. In an experiment on a metal surface, the stopping potential is zero at a frequency of 5.0 × 10¹⁴ Hz and 2.07 V at a frequency of 10.0 × 10¹⁴ Hz. The V₀–ν graph is a straight line. (Take e = 1.6 × 10⁻¹⁹ C.)

(i) What is the threshold frequency of the metal? (1)
(ii) Find the slope of the V₀–ν graph. (1)
(iii) Hence calculate the value of Planck’s constant. (1)
(iv) Find the work function of the metal in eV. (1)

OR

(iv) What changes in the graph and in the photocurrent if the intensity of light is doubled at the same frequency? (1)

SECTION E (3 × 5 = 15 Marks) — Long Answer Questions

31. (a) Derive the expression for the capacitance of an isolated spherical conductor of radius R. (3)
(b) Two metal spheres of radii 10 cm and 20 cm are connected by a thin wire. A total charge of 3 μC is shared between them. Find the final charge on each sphere and the common potential. (2)(5)

OR

(a) A dielectric slab of thickness t and dielectric constant K is introduced between the plates (separation d) of a parallel-plate capacitor of area A. Derive the expression for the capacitance. (3)
(b) Find the capacitance for A = 0.01 m², d = 5 mm, t = 3 mm and K = 3. (2)

32. (a) Show that a current-carrying circular loop behaves as a magnetic dipole and find its magnetic moment. (3)
(b) A circular loop of radius 2 cm carries a current of 5 A. Find its magnetic moment and the magnetic field on its axis at a distance of 50 cm from the centre. (2)(5)

OR

(a) Derive the expression for the frequency of oscillation of charge in an LC circuit and explain the analogy with a spring–mass system. (3)
(b) Find the frequency of oscillation for L = 0.1 H and C = 10 μF. (2)

33. (a) With a ray diagram, explain the construction and working of a reflecting (Cassegrain) telescope. State two advantages over a refracting telescope. (3)
(b) A refracting telescope has an objective of focal length 100 cm and an eyepiece of focal length 5 cm. Find its magnifying power and tube length in normal adjustment. (2)(5)

OR

(a) Explain the working of a half-wave rectifier with a circuit diagram. Draw the input and output waveforms. (3)
(b) A half-wave rectifier is fed with ac of peak value 10 V and frequency 50 Hz. Find the output frequency and the average dc voltage (ideal diode). (2)

ANSWER KEY & SOLUTIONS · PHYSICS PRACTICE PAPER (SET – 6)

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CLASS : XIISUBJECT : PHYSICSROLL NO. : ..................

Time : 3 Hrs.Class XII · Session : 2026–27M.M. : 70

SECTION A — Answer Key (1 mark each)

Q.1	(b)	Q.2	(c)	Q.3	(a)	Q.4	(b)
Q.5	(d)	Q.6	(c)	Q.7	(d)	Q.8	(a)
Q.9	(b)	Q.10	(c)	Q.11	(d)	Q.12	(a)
Q.13	(a)	Q.14	(c)	Q.15	(d)	Q.16	(b)

Hints for Q.1–16. (1) CO₂ is linear and symmetric, so the bond dipoles cancel; H₂O, HCl, NH₃ are polar. (2) E = −dV/dx = −8x ⇒ |E| = 8 V/m at x = 1 m. (3) The junction rule (ΣI = 0) expresses charge conservation. (4) V = E − Ir = 2 − 2 × 0.5 = 1 V. (5) qE = qvB ⇒ v = E/B = 2×10⁴/0.1 = 2 × 10⁵ m/s. (6) μ = (e/2m)L. (7) q = ΔΦ/R = 0.05/5 = 0.01 C. (8) Z = √(30² + 40²) = 50 Ω. (9) Displacement current ε₀dΦ_E/dt. (10) Coherent sources keep a constant phase difference (and have the same frequency). (11) eV₀ = hν − φ ⇒ V₀ = (h/e)ν − φ/e; slope = h/e. (12) τ = 1/λ = T_1/2/0.693 = 1.44 T_1/2. (13) (a): E has no component along an equipotential surface (otherwise work would be done along it), so E ⟂ surface; R explains A. (14) (c): A is true; R is false — a voltmeter is made by connecting a high resistance in series with the galvanometer. (15) (d): A is false — EM waves are transverse; R is true. (16) (b): both true, but the forbidden gap of pure silicon does not itself explain the increase in conductivity on doping (extra carriers from impurities do).

SECTION B — Solutions (2 marks each)

17. In an external field a dielectric develops a net dipole moment per unit volume (surface bound charges appear); this is polarisation (1). The induced field opposes the external field, so inside the slab E = E₀/K = E₀/5 (1).

18. Joule’s law: the heat produced in a resistor is proportional to the square of the current, the resistance and the time, H = I²Rt (1). H = (3)² × 20 × 10 = 1800 J (1).

19. Connect a high resistance R in series with the galvanometer (1): V = I_g(G + R) ⇒ R = V/I_g − G = 10/(2×10⁻³) − 50 = 5000 − 50 = 4950 Ω (1).

20. Induced emf ε = −dΦ/dt, so I = ε/R and q = ∫I dt = −(1/R)∫dΦ ⇒ |q| = ΔΦ/R, which depends only on the flux change and R, not on the time taken (1). q = (0.08 − 0.03)/10 = 5 × 10⁻³ C (1).

21. Any four: (i) it is the strongest force (≈ 100 times the electromagnetic force); (ii) it is short-ranged (a few fm, ≈ 10⁻¹⁵ m); (iii) it is charge independent (p–p, n–n, p–n forces are nearly equal); (iv) it is saturated (a nucleon interacts only with its neighbours); (v) it is spin dependent and non-central; (vi) it is attractive at distances ≈ 1 fm and repulsive below ≈ 0.8 fm (2).

SECTION C — Solutions (3 marks each)

22. Non-polar dielectrics (e.g. N₂, CO₂): the centres of positive and negative charge coincide, so there is no permanent dipole moment. An external field E₀ displaces the charges and induces a dipole moment in each molecule along E₀ (1). Polar dielectrics (e.g. H₂O, HCl): molecules have permanent dipole moments, randomly oriented. E₀ exerts a torque that partially aligns them along E₀ (1). In both cases a net dipole moment per unit volume (polarisation) appears along E₀, and bound surface charges produce a field E_p opposite to E₀, so E = E₀ − E_p. Dielectric constant K = E₀/E (= ε/ε₀ = C/C₀) (1).

23. The potential difference between two close points separated by dr along the field is dV = −E·dr, because the work done per unit charge by the field in moving it by dr is E·dr and this equals the fall in potential (1). In one dimension E = −dV/dr: the field points in the direction in which the potential decreases most rapidly and its magnitude is the rate of fall of potential per unit distance (1). For two points on an equipotential surface dV = 0, so E·dr = 0 for every displacement dr along the surface, i.e. E ⟂ the equipotential surface (1).

24. ρ = m/ne²τ (1). Metal: the number density n of free electrons is almost constant; on heating the lattice ions vibrate with larger amplitude, collisions become more frequent and the relaxation time τ decreases, so ρ increases (1). Semiconductor: on heating more covalent bonds break, so n increases rapidly (exponentially); this increase far outweighs the small decrease in τ, so ρ decreases (negative temperature coefficient) (1).

25. In a velocity selector uniform E and B are applied perpendicular to each other and to the velocity of the charged particles. The electric force qE and the magnetic force qvB act in opposite directions (1). The particle is undeflected when qE = qvB ⇒ v = E/B, independent of the charge and mass; particles with any other speed are deflected and cannot pass through the exit slit (1). v = (2×10⁴)/0.1 = 2 × 10⁵ m/s (1).

26. Let the electron (charge −e, mass m) move with speed v in a circle of radius r. The current I = e/T = ev/2πr (1). Magnetic moment μ = IA = (ev/2πr)·πr² = evr/2. The orbital angular momentum is L = mvr, so μ = (e/2m)L (directed opposite to L for the negatively charged electron: μ = −eL/2m) (1). For the first Bohr orbit L = h/2π, giving the Bohr magneton μ_B = eh/4πm ≈ 9.27 × 10⁻²⁴ A m², the smallest unit of atomic magnetic moment (1).

27. Take a long solenoid with n turns per unit length carrying current I. Inside, the field is uniform and parallel to the axis; outside it is negligible. Choose a rectangular Amperian loop abcd with side ab = h inside the solenoid, parallel to the axis, and the other sides perpendicular to it or outside (1). ∮B·dl = Bh (only the side ab contributes). The loop encloses nh turns, i.e. current nhI (1). Ampere’s law: Bh = μ₀nhI ⇒ B = μ₀nI (1).

OR

The rms value is the square root of the mean of i² over a cycle (1). For i = I₀ sin ωt: ⟨i²⟩ = (1/T)∫₀^T I₀² sin²ωt dt = I₀²/2, since the mean of sin²ωt over a cycle is ½ (1). Hence I_rms = I₀/√2 ≈ 0.707 I₀; similarly V_rms = V₀/√2 (1).

28. The curve (drawn: B.E. per nucleon on the y-axis from 0–9 MeV, mass number A on the x-axis) rises steeply for light nuclei (e.g. ≈ 1.1 MeV for ²H), reaches a maximum of ≈ 8.75 MeV near A ≈ 56 (iron) and then falls slowly to ≈ 7.6 MeV for uranium (A ≈ 238) (1). Fusion: light nuclei (A < 56) combine into a heavier nucleus with a larger B.E. per nucleon, so the product is more tightly bound and energy is released (1). Fission: a heavy nucleus (A > 56) splits into two medium-mass nuclei with a larger B.E. per nucleon (≈ 8.5 MeV vs ≈ 7.6 MeV), so the difference (≈ 200 MeV per fission) is released (1).

OR

Mass of products = 234.0436 + 4.0026 = 238.0462 u (1). Mass defect Δm = 238.0508 − 238.0462 = 0.0046 u (1). Energy released Q = 0.0046 × 931.5 ≈ 4.28 MeV (shared mainly as the kinetic energy of the α-particle) (1).

SECTION D — Solutions (4 marks each)

29. (i) B₀ = E₀/c = 300/(3×10⁸) = 1 × 10⁻⁶ T (1). (ii) u_av = ½ε₀E₀² = ½ × 8.85×10⁻¹² × (300)² ≈ 3.98 × 10⁻⁷ J/m³ (1). (iii) I = u_avc = 3.98×10⁻⁷ × 3×10⁸ ≈ 119 W/m² (≈ 1.2 × 10² W/m²) (1). (iv) The wave travels along the direction of E × B, perpendicular to both E and B (transverse wave) (1). OR (iv): X-rays and γ-rays (1).

30. (i) ν₀ (V₀ = 0) = 5.0 × 10¹⁴ Hz (1). (ii) Slope = ΔV₀/Δν = 2.07/[(10 − 5) × 10¹⁴] = 4.14 × 10⁻¹⁵ V s (1). (iii) Slope = h/e ⇒ h = 4.14×10⁻¹⁵ × 1.6×10⁻¹⁹ ≈ 6.6 × 10⁻³⁴ J s (1). (iv) φ = hν₀ = 6.63×10⁻³⁴ × 5×10¹⁴ = 3.31×10⁻¹⁹ J = 2.07 eV (1). OR (iv): the graph does not change (V₀ depends only on frequency); the photocurrent (and saturation current) doubles (1).

SECTION E — Solutions (5 marks each)

31. (a) Let an isolated conducting sphere of radius R carry charge Q; its charge may be taken to be concentrated at the centre for points outside (1). Its potential (with V = 0 at infinity) is V = Q/4πε₀R (1). Capacitance C = Q/V = 4πε₀R (depends only on R; for the earth ≈ 711 μF) (1). (b) On connection both spheres reach the same potential: kQ₁/R₁ = kQ₂/R₂ ⇒ Q₁/Q₂ = R₁/R₂ = 1/2, and Q₁ + Q₂ = 3 μC ⇒ Q₁ = 1 μC, Q₂ = 2 μC (1). V = (9×10⁹ × 1×10⁻⁶)/0.1 = 9 × 10⁴ V (1).

OR

(a) Let the plates carry ±Q, σ = Q/A. In the air gap (d − t) the field is E₀ = σ/ε₀; inside the slab E = E₀/K (1). V = E₀(d − t) + E₀t/K = (σ/ε₀)[(d − t) + t/K] (1). C = Q/V = ε₀A / [d − t + t/K]; for t = d it reduces to Kε₀A/d (1). (b) d − t + t/K = 5 − 3 + 3/3 = 3 mm; C = (8.85×10⁻¹² × 0.01)/(3×10⁻³) ≈ 2.95 × 10⁻¹¹ F (≈ 29.5 pF) (1+1).

32. (a) On the axis of a loop of radius R carrying current I, at a distance x from the centre, B = μ₀IR²/[2(R² + x²)^3/2] (from the Biot–Savart law) (1). For x ≫ R: B ≈ μ₀IR²/2x³ = (μ₀/4π)·2(I·πR²)/x³ (1). This has the same form as the axial field of a bar magnet (μ₀/4π)·2m/x³, so the loop behaves as a magnetic dipole of moment m = IA (N turns: m = NIA), along the axis in the direction given by the right-hand rule (1). (b) m = IA = 5 × 3.14 × (0.02)² ≈ 6.28 × 10⁻³ A m² (1). B = 10⁻⁷ × 2 × 6.28×10⁻³/(0.5)³ ≈ 1.0 × 10⁻⁸ T (1).

OR

(a) A capacitor with charge q₀ is connected to an inductor L. At any instant the loop rule gives L di/dt + q/C = 0, with i = dq/dt (1). So d²q/dt² + q/(LC) = 0, an SHM equation: q = q₀ cos ω₀t with ω₀ = 1/√(LC), i.e. f = 1/(2π√LC) (1). Analogy: q ↔ x, i ↔ v, L ↔ m (inertia), 1/C ↔ k (spring constant); electrical energy q²/2C ↔ potential energy ½kx², magnetic energy ½Li² ↔ kinetic energy ½mv²; the total energy stays constant (in the absence of resistance) (1). (b) ω₀ = 1/√(0.1 × 10×10⁻⁶) = 1/√(10⁻⁶) = 1000 rad/s (1); f = ω₀/2π = 1000/6.28 ≈ 159 Hz (1).

33. (a) Construction/working (ray diagram): parallel rays from a distant object fall on a large concave parabolic primary mirror with a central hole. Before they converge, a small convex secondary mirror reflects them back through the hole in the primary, where a real image is formed and viewed through the eyepiece (1). The final image is magnified by the eyepiece, which acts as a simple microscope (1). Advantages (any two): no chromatic aberration; a parabolic mirror removes spherical aberration; a large mirror can be supported from the back, so a large aperture gives more light-gathering power and higher resolving power; the folded design makes the tube shorter (1). (b) m = f_o/f_e = 100/5 = 20 (1); tube length L = f_o + f_e = 105 cm (1).

OR

(a) Circuit: the ac source (through a transformer) is connected in series with a single p-n diode and the load resistance R_L; the output is taken across R_L (1). Working: in the positive half-cycle the diode is forward biased and conducts, so a current flows through R_L; in the negative half-cycle it is reverse biased and no current flows (1). The input is a full sine wave; the output is a series of positive half-sine pulses with gaps (one pulse per cycle), a pulsating unidirectional dc of the same frequency as the input (1). (b) Output frequency = 50 Hz (1); V_dc = V_m/π = 10/3.14 ≈ 3.18 V (1).