Class 12 CBSE Physics Practice Paper Set 5 with Solutions

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Class 12 Physics · Practice Paper SET-5 with Solutions

CBSE PRACTICE PAPER 2026–27 · PHYSICS (SET – 5)

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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. 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 statements about electric field lines is correct?

(a) Field lines of static charges form closed loops(b) Field lines are always straight lines(c) Two electric field lines can never intersect each other(d) Field lines start from negative charges and end on positive charges

2. A body carries a charge of −3.2 × 10⁻¹⁹ C. It has

(a) an excess of 2 electrons(b) an excess of 2 protons(c) an excess of 1 electron(d) an excess of 4 electrons

3. The energy stored per unit volume in an electric field E in vacuum is

(a) ε₀E²(b) ½ ε₀E(c) E²/2μ₀(d) ½ ε₀E²

4. Two 2 Ω resistors connected in parallel are joined in series with a third 2 Ω resistor. The equivalent resistance is

(a) 6 Ω(b) 3 Ω(c) 1 Ω(d) 4/3 Ω

5. The magnetic field outside an ideal toroid is

(a) μ₀NI/2πr(b) μ₀nI(c) zero(d) μ₀NI/2r

6. The SI unit of magnetic flux, the weber, is equivalent to

(a) T m(b) T m²(c) T/m²(d) N/A m²

7. The quality factor of a series LCR circuit at resonance is

(a) R/ω₀L(b) ω₀R/L(c) LR/ω₀(d) ω₀L/R

8. Electromagnetic waves with wavelengths just longer than those of visible light are

(a) infrared rays(b) ultraviolet rays(c) X-rays(d) γ-rays

9. In Rayleigh scattering the ratio of the intensities of light of wavelengths 400 nm and 800 nm scattered by air molecules is

(a) 1 : 16(b) 4 : 1(c) 2 : 1(d) 16 : 1

10. In Young’s double-slit experiment each slit alone produces an intensity I₀ on the screen. The intensity at a bright fringe is

(a) 2I₀(b) I₀(c) 4I₀(d) I₀/2

11. The de Broglie wavelength of a 0.1 kg ball moving at 10 m/s is

(a) 6.6 × 10⁻³³ m(b) 6.6 × 10⁻³⁴ m(c) 6.6 × 10⁻³¹ m(d) 6.6 × 10⁻²⁵ m

12. The energy equivalent of a mass defect of 0.02 u is

(a) 18.63 MeV(b) 9.3 MeV(c) 1.86 MeV(d) 186.3 MeV

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): Electric field lines are perpendicular to the surface of a conductor in electrostatics.
Reason (R): Electric field lines start from positive charges and end on negative charges.

14. Assertion (A): A freely suspended bar magnet comes to rest along the north–south direction.
Reason (R): The magnetic south pole of the earth lies near the geographic south pole.

15. Assertion (A): The sky appears blue.
Reason (R): The scattering of light by air molecules varies inversely as the fourth power of the wavelength.

16. Assertion (A): An n-type semiconductor is negatively charged.
Reason (R): In an n-type semiconductor the majority charge carriers are electrons.

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

17. Define electric flux and write its SI unit. A charge of 8.85 μC is enclosed in a closed surface. Find the net electric flux through the surface.(2)

18. Two resistors of 3 Ω and 6 Ω are connected in series across a 9 V battery of negligible internal resistance. Find the potential drop across each resistor.(2)

19. Write the expression for the magnetic field inside a toroid. A toroid of 1000 turns and mean radius 10 cm carries a current of 2 A. Find the magnetic field inside it.(2)

20. Define the quality factor (Q-factor) of a series LCR circuit and write its expression. What does a high value of Q signify?(2)

21. A 3.3 W lamp emits monochromatic light of wavelength 600 nm. Find the number of photons emitted per second.(2)

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

22. Using Gauss’s law, derive the expression for the electric field at a distance r from an infinitely long, uniformly charged straight wire of linear charge density λ.(3)

23. Derive the expression for the energy stored in a parallel-plate capacitor and hence show that the energy density in the electric field is ½ε₀E².(3)

24. A cell of emf E and internal resistance r is connected to an external resistance R. Show that the power delivered to R is maximum when R = r and find this maximum power.(3)

25. Use Ampere’s circuital law to derive the expression for the magnetic field inside a toroid. What is the field outside the toroid and in the open space enclosed by it?(3)

26. Describe Faraday’s experiments which led to the discovery of electromagnetic induction. State Faraday’s laws.(3)

27. Derive the relation n₂/v − n₁/u = (n₂ − n₁)/R for refraction at a convex spherical surface separating two media (light travelling from the rarer medium n₁ to the denser medium n₂).(3)

OR

Two thin lenses of focal lengths f₁ and f₂ are placed in contact. Derive the expression for the focal length and the power of the combination.

28. Name the series of lines in the hydrogen spectrum with the final orbits n = 1, 2, 3, 4 and 5. Write the formula for the wavelength of the lines and state the region of the electromagnetic spectrum in which the first three series lie.(3)

OR

Calculate the energy released in the fusion reaction ₁²H + ₁²H → ₂³He + ₀¹n. Given: m(²H) = 2.0141 u, m(³He) = 3.0160 u, m(n) = 1.0087 u, 1 u = 931.5 MeV/c².

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

29. Case Study – Tuning a radio receiver. A radio receiver uses a series LCR circuit with L = 1 mH, C = 1 μF and R = 10 Ω, connected to a 100 V (rms) source of variable frequency. The circuit is tuned to resonance to select a particular station.

(i) Find the resonant angular frequency. (1)
(ii) Find the resonant frequency in Hz. (1)
(iii) Find the impedance and the rms current at resonance. (1)
(iv) Calculate the quality factor of the circuit. (1)

OR

(iv) How will the current at resonance and the sharpness of resonance change if R is increased? (1)

30. Case Study – Nuclear reactor. In a nuclear reactor, slow neutrons cause fission of uranium-235. Each fission releases about 200 MeV of energy. A moderator and control rods are used to run the chain reaction in a controlled manner.

(i) What is the function of a moderator? Name one moderator. (1)
(ii) What is the function of control rods? Name one material used for them. (1)
(iii) Calculate the number of fissions per second needed for a power output of 1 MW. (1)
(iv) Why is nuclear fusion difficult to achieve on Earth? (1)

OR

(iv) What is the function of the coolant in a nuclear reactor? (1)

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

31. (a) Derive the expression for the potential energy of an electric dipole placed in a uniform electric field. (3)
(b) A dipole of moment 3 × 10⁻⁸ C m is placed in a uniform field of 2 × 10⁴ N/C. Find the work done in rotating it from the position θ = 0° to θ = 180°. (2)(5)

OR

(a) Derive the expression for the electric potential at a point on the axial line of an electric dipole. (3)
(b) Find the potential at a point on the axis of a dipole of moment 4 × 10⁻⁹ C m, at a distance of 20 cm from its centre. (2)

32. (a) Derive the expressions for the radius and the time period of the circular path of a charged particle moving perpendicular to a uniform magnetic field. What is the path if the velocity has a component along the field? (3)
(b) An electron moves at 3 × 10⁶ m/s perpendicular to a magnetic field of 0.01 T. Find the radius of its path and the time period. (2)(5)

OR

(a) Derive the expression for the magnetic field at a point on the equatorial line of a short bar magnet. (3)
(b) Find the field at a point on the equatorial line 0.5 m from the centre of a bar magnet of magnetic moment 2 A m². (2)

33. (a) Explain the formation of energy bands in solids. On this basis distinguish between conductors, semiconductors and insulators. (3)
(b) The energy gap of silicon is 1.1 eV. Find the longest wavelength of light that can create an electron–hole pair in it (use hc = 1240 eV nm). (2)(5)

OR

(a) State three experimental observations of the photoelectric effect that cannot be explained by the wave theory of light and explain how Einstein’s photon picture accounts for them. (3)
(b) The threshold wavelength of a metal is 500 nm. Find the maximum kinetic energy of photoelectrons and the stopping potential for light of wavelength 400 nm (use hc = 1240 eV nm). (2)

ANSWER KEY & SOLUTIONS · PHYSICS PRACTICE PAPER (SET – 5)

www.careerinformationportal.in

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

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

SECTION A — Answer Key (1 mark each)

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

Hints for Q.1–16. (1) Two field lines cannot intersect, otherwise the field would have two directions at the point of intersection; lines of static charges start on positive and end on negative charges and are not closed loops. (2) n = 3.2×10⁻¹⁹/1.6×10⁻¹⁹ = 2 electrons in excess (negative charge). (3) u = ½ε₀E². (4) 2 ∥ 2 = 1 Ω; 1 + 2 = 3 Ω. (5) The field outside an ideal toroid is zero (net enclosed current is zero). (6) 1 Wb = 1 T m². (7) Q = ω₀L/R = 1/(ω₀CR). (8) λ_infrared > λ_visible. (9) I ∝ 1/λ⁴ ⇒ I₄₀₀/I₈₀₀ = (800/400)⁴ = 16 : 1. (10) Amplitudes add: A = 2A₀ ⇒ I = 4I₀. (11) λ = h/mv = 6.6×10⁻³⁴/(0.1 × 10) = 6.6×10⁻³⁴ m (too small to observe). (12) 0.02 × 931.5 = 18.63 MeV. (13) (b): both true; the perpendicularity follows from the field inside a conductor being zero in electrostatics, not from the origin of lines. (14) (c): A is true; R is false — the magnetic south pole of the earth lies near the geographic north. (15) (a): blue light (shorter λ) is scattered more strongly, so the sky looks blue. (16) (d): A is false — an n-type semiconductor is electrically neutral (donor ions balance the extra electrons); R is true.

SECTION B — Solutions (2 marks each)

17. Electric flux through a surface is Φ = ∮E·dA, the number of field lines crossing it; SI unit N m² C⁻¹ (or V m) (1). By Gauss’s law, Φ = q/ε₀ = (8.85×10⁻⁶)/(8.85×10⁻¹²) = 10⁶ N m² C⁻¹ (1).

18. I = V/(R₁ + R₂) = 9/9 = 1 A (1). V_3Ω = 1 × 3 = 3 V; V_6Ω = 1 × 6 = 6 V (sum = 9 V) (1).

19. Inside a toroid B = μ₀NI/2πr (1). B = (4π×10⁻⁷ × 1000 × 2)/(2π × 0.1) = 4 × 10⁻³ T (1).

20. The Q-factor is the ratio of the voltage across the inductor (or capacitor) at resonance to the applied voltage; Q = ω₀L/R = (1/R)√(L/C) (also Q = ω₀/2Δω) (1). A high Q means a sharp resonance: the circuit is highly selective and responds only to a narrow band of frequencies around ω₀ (1).

21. Energy of one photon E = hc/λ = (6.6×10⁻³⁴ × 3×10⁸)/(6×10⁻⁷) = 3.3 × 10⁻¹⁹ J (1). Number per second n = P/E = 3.3/(3.3×10⁻¹⁹) = 10¹⁹ photons/s (1).

SECTION C — Solutions (3 marks each)

22. Take a coaxial cylindrical Gaussian surface of radius r and length l around the wire. By symmetry E is radial and has the same magnitude everywhere on the curved surface (1). Flux through the curved surface = E(2πrl); flux through the two flat ends = 0 (1). Charge enclosed = λl. Gauss’s law: E(2πrl) = λl/ε₀ ⇒ E = λ/2πε₀r, directed radially outward for λ > 0 (1).

23. Work done in charging the capacitor from 0 to Q: W = ∫ (q/C) dq = Q²/2C, so U = ½CV² (1). For a parallel-plate capacitor C = ε₀A/d and V = Ed: U = ½(ε₀A/d)(Ed)² = ½ε₀E²(Ad) (1). The field occupies the volume Ad, so energy density u = U/Ad = ½ε₀E² (with a dielectric, ½Kε₀E²) (1).

24. Current I = E/(R + r), so the power in R is P = I²R = E²R/(R + r)² (1). For maximum P, dP/dR = 0: [(R + r)² − 2R(R + r)]·E²/(R + r)⁴ = 0 ⇒ (R + r) = 2R ⇒ R = r (1). Then P_max = E²r/(2r)² = E²/4r (the efficiency at that point is 50%) (1).

25. A toroid is a hollow circular ring with N turns carrying current I. Take a circular Amperian loop of radius r inside the toroid, along its mean circumference (1). By symmetry B is tangential and constant on the loop: ∮B·dl = B(2πr). Current enclosed = NI. Ampere’s law ⇒ B(2πr) = μ₀NI ⇒ B = μ₀NI/2πr = μ₀nI with n = N/2πr (1). For a loop outside the toroid (or in the open space enclosed by it) the net current enclosed is zero, so B = 0 there (1).

26. (i) A bar magnet is moved towards/away from a coil connected to a galvanometer: the galvanometer deflects (in opposite directions for approach and withdrawal); the deflection is larger for faster motion and zero when the magnet is at rest (1). (ii) Two coils are placed near each other: a galvanometer connected to the secondary coil shows a momentary deflection when the current in the primary is switched on, switched off or varied (e.g., with a rheostat) (1). Faraday’s laws: (1) an emf is induced whenever the magnetic flux linked with a circuit changes; (2) the magnitude of the induced emf equals the rate of change of flux, ε = −N dΦ/dt (1).

27. Let a convex surface of centre of curvature C and radius R separate media n₁ and n₂ (n₂ > n₁). A point object O on the axis sends a ray OM which meets the surface at M and refracts to the image I; N is the foot of the perpendicular from M on the axis. Let ∠MOC = α, ∠MCN = γ, ∠MIC = β (1). In △OMC the exterior angle gives i = α + γ; in △MCI, γ = r + β ⇒ r = γ − β. For small angles Snell’s law gives n₁i = n₂r, i.e. n₁(α + γ) = n₂(γ − β) (1). With α ≈ MN/NO, β ≈ MN/NI, γ ≈ MN/NC: n₁/NO + n₂/NI = (n₂ − n₁)/NC. Using the sign convention NO = −u, NI = +v, NC = +R: n₂/v − n₁/u = (n₂ − n₁)/R (1).

OR

Let an object O be placed at distance u from the first lens L₁. It forms an image I₁ at v₁: 1/v₁ − 1/u = 1/f₁ (1). I₁ acts as the object for L₂ (lenses in contact, thickness negligible), which forms the final image at v: 1/v − 1/v₁ = 1/f₂ (1). Adding: 1/v − 1/u = 1/f₁ + 1/f₂ = 1/F, so 1/F = 1/f₁ + 1/f₂ and, since P = 1/f, P = P₁ + P₂ (1).

28. Formula (Rydberg): 1/λ = R(1/n_f² − 1/n_i²), n_i > n_f, R = 1.097×10⁷ m⁻¹ (1). Series: n_f = 1 Lyman; n_f = 2 Balmer; n_f = 3 Paschen; n_f = 4 Brackett; n_f = 5 Pfund (1). Regions: Lyman — ultraviolet; Balmer — visible (partly near-UV); Paschen — infrared (1).

OR

Mass of reactants = 2 × 2.0141 = 4.0282 u; mass of products = 3.0160 + 1.0087 = 4.0247 u (1). Mass defect Δm = 4.0282 − 4.0247 = 0.0035 u (1). Energy released Q = 0.0035 × 931.5 ≈ 3.26 MeV (1).

SECTION D — Solutions (4 marks each)

29. (i) ω₀ = 1/√(LC) = 1/√(10⁻³ × 10⁻⁶) = 1/√(10⁻⁹) ≈ 3.16 × 10⁴ rad/s (1). (ii) f₀ = ω₀/2π ≈ 3.16×10⁴/6.28 ≈ 5.0 × 10³ Hz (1). (iii) At resonance X_L = X_C, so Z = R = 10 Ω and I_rms = V/R = 100/10 = 10 A (1). (iv) Q = ω₀L/R = (3.16×10⁴ × 10⁻³)/10 ≈ 3.16 (1). OR (iv): on increasing R, the resonant current V/R decreases and the resonance becomes less sharp (Q decreases, bandwidth increases) (1).

30. (i) A moderator slows down the fast neutrons produced in fission to thermal energies so that they can cause further fission of U-235; e.g. heavy water, graphite (1). (ii) Control rods absorb excess neutrons to control the rate of the chain reaction (multiplication factor ≈ 1); e.g. cadmium, boron (1). (iii) Energy per fission = 200 MeV = 200 × 1.6×10⁻¹³ = 3.2×10⁻¹¹ J. Fissions/s = 10⁶/(3.2×10⁻¹¹) ≈ 3.1 × 10¹⁶ (1). (iv) Fusion needs the nuclei to overcome their Coulomb repulsion, which requires extremely high temperature (≈ 10⁷–10⁸ K) and pressure, difficult to attain and confine (1). OR (iv): the coolant (water, heavy water, liquid sodium) carries away the heat produced in the core to the steam generator and prevents overheating (1).

SECTION E — Solutions (5 marks each)

31. (a) A dipole of moment p in a uniform field E at angle θ experiences torque τ = pE sin θ (1). The work done in rotating it from θ₀ to θ against the torque is W = ∫_θ0^θ pE sin θ dθ = pE(cos θ₀ − cos θ) (1). Taking the reference orientation θ₀ = 90° (U = 0), the potential energy at angle θ is U = −pE cos θ = −p·E (1). (b) W = U(180°) − U(0°) = pE − (−pE) = 2pE = 2 × 3×10⁻⁸ × 2×10⁴ = 1.2 × 10⁻³ J (1+1).

OR

(a) Dipole: −q at A, +q at B, AB = 2a, centre O. For a point P on the axis at distance r from O (on the +q side) the potentials add (scalars): V = q/4πε₀(r − a) − q/4πε₀(r + a) (1). V = (q/4πε₀)·2a/(r² − a²) = p/[4πε₀(r² − a²)] (1); for r ≫ a, V = p/4πε₀r² (1). (b) V = (9×10⁹ × 4×10⁻⁹)/(0.2)² = 36/0.04 = 900 V (1+1).

32. (a) The magnetic force qvB is perpendicular to v, so it does no work and acts as the centripetal force: qvB = mv²/r ⇒ r = mv/qB (1). Time period T = 2πr/v = 2πm/qB, independent of the speed and radius (1). If v has a component v_∥ along B, that component is unaffected, and the path is a helix of radius mv_⊥/qB and pitch v_∥T (1). (b) r = mv/eB = (9.1×10⁻³¹ × 3×10⁶)/(1.6×10⁻¹⁹ × 0.01) ≈ 1.7 × 10⁻³ m (1); T = 2πm/eB = (2π × 9.1×10⁻³¹)/(1.6×10⁻²¹) ≈ 3.6 × 10⁻⁹ s (1).

OR

(a) Represent the magnet by poles ±q_m at distance l on either side of O (m = 2q_ml). A point P on the equatorial line at distance r is at distance √(r² + l²) from each pole, so the two fields have equal magnitude (μ₀/4π)q_m/(r² + l²) (1). The components perpendicular to the axis cancel; those along the axis add and point from N to S (opposite to m): B = 2(μ₀/4π)·q_m/(r² + l²) · l/√(r² + l²) = (μ₀/4π)·m/(r² + l²)^3/2 (1). For a short magnet (r ≫ l): B = (μ₀/4π)·m/r³, directed opposite to m (1). (b) B = 10⁻⁷ × 2/(0.5)³ = 2×10⁻⁷/0.125 = 1.6 × 10⁻⁶ T (1+1).

33. (a) In an isolated atom the electron energy levels are discrete. In a solid, the atoms are close together and the levels of the outer electrons split into closely spaced levels that form continuous energy bands: the filled valence band, the conduction band, separated by a forbidden gap E_g (1). Conductors: the conduction and valence bands overlap or the conduction band is partly filled, so there are plenty of free electrons (E_g = 0). Insulators: large gap (E_g > 3 eV, e.g. diamond 5.4 eV), valence band full, conduction band empty, so no conduction (1). Semiconductors: small gap (Si 1.1 eV, Ge 0.7 eV); at 0 K they are insulators but at room temperature some electrons are thermally excited to the conduction band, leaving holes in the valence band (1). (b) λ_max = hc/E_g = 1240/1.1 ≈ 1127 nm (≈ 1.13 μm, near infrared) (1+1).

OR

(a) (i) Threshold frequency: no emission below a certain frequency, however intense the light; wave theory predicts emission at any frequency if the intensity is high enough. (ii) Instantaneous emission: there is no time lag (< 10⁻⁹ s), whereas wave theory predicts a delay to accumulate energy. (iii) K_max depends on frequency, not on intensity, whereas wave theory predicts K_max should increase with intensity (1.5). Einstein’s explanation: light consists of photons of energy hν; one photon is absorbed by one electron, so emission occurs only if hν ≥ φ (threshold), immediately, and K_max = hν − φ is independent of the number of photons (intensity), which only decides the number of emitted electrons (1.5). (b) φ = 1240/500 = 2.48 eV; E = 1240/400 = 3.1 eV; K_max = 3.1 − 2.48 = 0.62 eV, so V₀ = 0.62 V (1+1).