CBSE Class 12 Physics Sample Paper 2026 (with Solutions) — Free PDF
This is a full-length CBSE Class 12 Physics paper built to the current board blueprint: 33 questions across five sections for 70 marks in 3 hours, the 30 r
TL;DR: This is a full-length CBSE Class 12 Physics paper built to the current board blueprint: 33 questions across five sections for 70 marks in 3 hours, the…
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This is a full-length CBSE Class 12 Physics paper built to the current board blueprint: 33 questions across five sections for 70 marks in 3 hours, the 30 remaining marks being the practical component assessed in school. Section A is objective, Section D is case-based and Section E carries the long derivations. Every question is followed by a complete worked solution with the derivation set out step by step.
Board: CBSE · Class: Class 12 · Subject: Physics · Total: 70 marks · Time: 3 hours
Section A — Objective (1 mark each) — 16 marks
Questions 1–14 are multiple choice. Questions 15 and 16 are Assertion–Reason: mark (a) if both A and R are true and R explains A, (b) if both are true but R does not explain A, (c) if A is true and R is false, (d) if A is false and R is true.
Q1. Two point charges each of +2 μC are placed 30 cm apart in air. The force between them is (1 mark)
Answer: 0.4 N, repulsive. F = kq₁q₂/r² = (9 × 10⁹)(2 × 10⁻⁶)²/(0.3)² = (9 × 10⁹)(4 × 10⁻¹²)/0.09 = 0.4 N.
Q2. The electric field inside a charged hollow conductor in electrostatic equilibrium is (1 mark)
Answer: Zero. Charge resides entirely on the outer surface and the field inside the conducting material and cavity is zero, which is the basis of electrostatic shielding.
Q3. The capacitance of a parallel plate capacitor is doubled when (1 mark)
Answer: The separation between the plates is halved. C = ε₀A/d, so halving d doubles C (equivalently, doubling the plate area also doubles it).
Q4. The drift velocity of electrons in a conductor is of the order of (1 mark)
Answer: 10⁻⁴ m/s — less than a millimetre per second, even though the electrical signal itself travels at nearly the speed of light.
Q5. The resistance of a wire is R. If it is stretched to twice its original length, its new resistance is (1 mark)
Answer: 4R. Stretching keeps the volume constant, so doubling the length halves the area; R = ρl/A becomes ρ(2l)/(A/2) = 4ρl/A = 4R.
Q6. A charged particle moving parallel to a uniform magnetic field experiences a force of magnitude (1 mark)
Answer: Zero. F = qvB sin θ and θ = 0°, so sin θ = 0 and the force vanishes.
Q7. The magnetic field at the centre of a circular coil of n turns, radius r, carrying current I is (1 mark)
Answer: B = μ₀nI/2r, directed along the axis of the coil as given by the right-hand rule.
Q8. Lenz's law is a consequence of the law of conservation of (1 mark)
Answer: Energy. The induced current always opposes the change producing it, so work must be done against that opposition — otherwise energy would be created from nothing.
Q9. In a purely inductive AC circuit, the current (1 mark)
Answer: Lags the voltage by π/2 (90°). The inductor opposes the change in current, so the current peaks a quarter cycle after the voltage.
Q10. Electromagnetic waves are produced by (1 mark)
Answer: Accelerated charges. A charge at rest gives only an electrostatic field and one in uniform motion only a steady magnetic field; acceleration is required to radiate.
Q11. The refractive index of a medium for which the critical angle is 30° is (1 mark)
Answer: 2. n = 1/sin C = 1/sin 30° = 1/0.5 = 2.
Q12. In Young's double slit experiment, the fringe width is doubled when (1 mark)
Answer: The distance between the slits is halved. β = λD/d, so halving d doubles β (as would doubling D).
Q13. The stopping potential in a photoelectric experiment depends on (1 mark)
Answer: The frequency of the incident light and the work function of the metal — not on the intensity, which affects only the number of photoelectrons.
Q14. In a p-n junction diode under forward bias, the width of the depletion layer (1 mark)
Answer: Decreases. Forward bias opposes the built-in potential, so the depletion region narrows and the current rises steeply.
Q15. Assertion (A): The nucleus of an atom contains most of its mass. Reason (R): Protons and neutrons are much more massive than electrons. (1 mark)
Answer: (a) Both true and R explains A. A proton or neutron is about 1836 times the mass of an electron, so nearly all the mass sits in the nucleus.
Q16. Assertion (A): A convex lens behaves as a diverging lens when immersed in a liquid of higher refractive index. Reason (R): The focal length of a lens depends on the refractive index of the surrounding medium. (1 mark)
Answer: (a) Both true and R explains A. By the lens maker's formula 1/f = (n_lens/n_medium − 1)(1/R₁ − 1/R₂), if n_medium exceeds n_lens the bracket changes sign and the lens diverges.
Section B — Very Short Answer (2 marks each) — 10 marks
Answer all 5 questions. Each carries 2 marks.
Q1. Define electric flux and write its SI unit. (2 marks)
Answer: Electric flux through a surface is the total number of electric field lines crossing it, defined as Φ = ∮E·dA, the surface integral of the electric field over the area. Its SI unit is the newton metre squared per coulomb (N m² C⁻¹), equivalently the volt metre.
Q2. State Kirchhoff's two rules for electrical networks. (2 marks)
Answer: Junction rule: the algebraic sum of currents meeting at any junction is zero, which expresses conservation of charge. Loop rule: the algebraic sum of the changes in potential around any closed loop is zero, which expresses conservation of energy.
Q3. Write the expression for the force per unit length between two long parallel current-carrying conductors and state when it is attractive. (2 marks)
Answer: F/l = μ₀I₁I₂/(2πd), where d is the separation. The force is attractive when the currents flow in the same direction and repulsive when they are antiparallel.
Q4. Distinguish between the resistance and the impedance of an AC circuit. (2 marks)
Answer: Resistance opposes current in a purely resistive circuit and dissipates energy, and it is independent of frequency. Impedance is the total opposition of a circuit containing resistance together with inductance or capacitance, Z = √(R² + (X_L − X_C)²); it depends on frequency and its reactive part stores rather than dissipates energy.
Q5. Define the term 'mass defect' and relate it to binding energy. (2 marks)
Answer: Mass defect Δm is the difference between the sum of the masses of the free nucleons and the actual mass of the nucleus they form. This missing mass appears as the binding energy that holds the nucleus together, given by E = Δm·c².
Section C — Short Answer (3 marks each) — 21 marks
Answer all 7 questions. Each carries 3 marks.
Q1. Derive an expression for the electric field on the axis of an electric dipole at a distance r from its centre, for r far greater than the dipole length. (3 marks)
Answer: Take charges −q at −a and +q at +a on the axis, so the dipole moment is p = q(2a). At an axial point distance r from the centre, the field from +q is kq/(r − a)² directed outward and from −q is kq/(r + a)² directed inward. The resultant is E = kq[1/(r − a)² − 1/(r + a)²] = kq·4ar/(r² − a²)². For r ≫ a the denominator becomes r⁴, giving E = 4kqar/r⁴ = 2kp/r³, directed along the dipole moment.
Q2. Two capacitors of 3 μF and 6 μF are connected first in series and then in parallel across a 12 V supply. Find the equivalent capacitance and the total charge in each case. (3 marks)
Answer: In series: 1/C = 1/3 + 1/6 = 1/2, so C = 2 μF and Q = CV = 2 × 12 = 24 μC (the same charge on each capacitor). In parallel: C = 3 + 6 = 9 μF and Q = 9 × 12 = 108 μC total, shared as 36 μC and 72 μC. The parallel combination therefore stores four and a half times the charge.
Q3. State the principle of a potentiometer and explain why it is preferred over a voltmeter for measuring emf. (3 marks)
Answer: A potentiometer works on the principle that when a steady current flows through a wire of uniform cross-section, the potential difference across any portion is directly proportional to its length: V ∝ l. At the balance point no current is drawn from the cell being measured, so the potentiometer measures the true emf rather than the terminal voltage. A voltmeter, having finite resistance, always draws some current, so the reading falls short of the emf by the drop across the cell's internal resistance.
Q4. Derive the expression for the torque on a rectangular current-carrying loop placed in a uniform magnetic field. (3 marks)
Answer: Consider a rectangular loop of sides l and b carrying current I in a uniform field B, with the normal to the loop at angle θ to B. The forces on the two sides of length b are equal, opposite and along the same line, so they cancel. The forces on the two sides of length l are F = BIl each, opposite in direction but separated by a perpendicular distance b sin θ, so they form a couple. The torque is τ = BIl × b sin θ = BIA sin θ, where A = lb; for n turns τ = nBIA sin θ, or in vector form τ = m × B with m = nIA.
Q5. A coil of 100 turns and area 0.05 m² is placed in a magnetic field which falls uniformly from 0.8 T to 0.2 T in 0.1 s. Calculate the induced emf. (3 marks)
Answer: The change in flux per turn is ΔΦ = A·ΔB = 0.05 × (0.8 − 0.2) = 0.03 Wb. By Faraday's law the magnitude of the induced emf is ε = N·ΔΦ/Δt = 100 × 0.03/0.1 = 30 V. By Lenz's law the induced current flows so as to oppose the decrease, that is, to maintain the original flux.
Q6. Draw a labelled ray diagram of a compound microscope in normal adjustment and write its magnifying power. (3 marks)
Answer: The object is placed just beyond the focus of the short-focal-length objective, which forms a real, inverted, magnified image inside the focal length of the eyepiece; the eyepiece then acts as a simple magnifier and the final image is virtual, inverted relative to the object and highly magnified, formed at the least distance of distinct vision. The magnifying power is m = m₀ × mₑ = (v₀/u₀)(1 + D/fₑ), where D is 25 cm.
Q7. Explain the photoelectric effect and state two observations that the wave theory of light fails to explain. (3 marks)
Answer: The photoelectric effect is the emission of electrons from a metal surface when light of sufficiently high frequency falls on it; Einstein explained it by treating light as photons of energy hν, giving hν = φ₀ + K_max. Wave theory fails on two counts. First, there is a threshold frequency below which no emission occurs however intense the light, whereas waves should eventually supply enough energy at any frequency. Second, emission is instantaneous, whereas wave theory predicts a measurable time lag while the electron absorbs sufficient energy.
Section D — Case-based questions (4 marks each) — 8 marks
Answer both case-based questions. Each has sub-parts totalling 4 marks.
Q1. CASE STUDY 1 — A step-down transformer is used to run a 12 V, 24 W lamp from a 240 V AC mains supply. Assume the transformer is ideal. (i) Find the turns ratio Np : Ns. (ii) Find the current in the secondary. (iii) Find the current drawn from the mains. (iv) Why are transformer cores laminated? (4 marks)
Answer: (i) For an ideal transformer Np/Ns = Vp/Vs = 240/12 = 20, so the ratio is 20 : 1. (ii) Secondary current Is = P/Vs = 24/12 = 2 A. (iii) An ideal transformer conserves power, so Ip = P/Vp = 24/240 = 0.1 A. (iv) The core is laminated, with thin sheets separated by insulating varnish, to break up the paths available to eddy currents; this greatly reduces the eddy-current heat losses and improves efficiency.
Q2. CASE STUDY 2 — Light of wavelength 600 nm falls on a double slit with slit separation 0.2 mm, and the interference pattern is observed on a screen 1.0 m away. (i) Calculate the fringe width. (ii) Where is the central maximum located and why is it bright? (iii) What happens to the pattern if the whole apparatus is immersed in water of refractive index 1.33? (4 marks)
Answer: (i) β = λD/d = (600 × 10⁻⁹ × 1.0)/(0.2 × 10⁻³) = 3 × 10⁻³ m = 3 mm. (ii) The central maximum lies on the perpendicular bisector of the two slits, equidistant from both; the path difference there is zero, so the waves arrive in phase and interfere constructively. (iii) In water the wavelength shortens to λ/n, so the fringe width becomes β/1.33 ≈ 2.26 mm — the pattern contracts, with the fringes closer together.
Section E — Long Answer (5 marks each) — 15 marks
Answer all 3 questions. Each carries 5 marks.
Q1. (a) Using Gauss's law, derive the electric field due to an infinitely long uniformly charged straight wire of linear charge density λ. (b) A wire carries λ = 4 × 10⁻⁶ C/m; find the field 20 cm from it. (5 marks)
Answer: (a) By symmetry the field is radial and equal in magnitude at every point of a coaxial cylinder of radius r and length l. Choose that cylinder as the Gaussian surface. No flux crosses the flat ends because the field is parallel to them, so the flux is E × 2πrl through the curved surface. The charge enclosed is λl, so Gauss's law gives E × 2πrl = λl/ε₀, hence E = λ/(2πε₀r), directed away from the wire for positive λ. (b) E = (2 × 9 × 10⁹ × 4 × 10⁻⁶)/0.2 = 72 × 10³/0.2 = 3.6 × 10⁵ N/C, directed radially outward.
Q2. (a) Derive the lens maker's formula for a thin lens. (b) A biconvex lens of refractive index 1.5 has both radii of curvature equal to 20 cm. Find its focal length and power. (5 marks)
Answer: (a) Refraction at the first surface takes an object at u to an image at v₁: n₂/v₁ − n₁/u = (n₂ − n₁)/R₁. This image acts as the object for the second surface: n₁/v − n₂/v₁ = (n₁ − n₂)/R₂. Adding the two equations eliminates v₁ and gives n₁(1/v − 1/u) = (n₂ − n₁)(1/R₁ − 1/R₂). Dividing by n₁ and using 1/v − 1/u = 1/f yields 1/f = (n − 1)(1/R₁ − 1/R₂) with n = n₂/n₁. (b) For a biconvex lens R₁ = +20 cm and R₂ = −20 cm, so 1/f = (1.5 − 1)(1/20 − 1/(−20)) = 0.5 × (2/20) = 1/20, giving f = 20 cm = 0.2 m. Power P = 1/f in metres = +5 D.
Q3. (a) Draw a labelled circuit diagram of a full-wave rectifier using two diodes and explain its working. (b) Sketch the input and output waveforms. (c) State one advantage over a half-wave rectifier. (5 marks)
Answer: (a) A centre-tapped transformer feeds two diodes whose outputs join at a common load resistor. During the positive half cycle the upper diode is forward biased and conducts while the lower is reverse biased; during the negative half cycle the roles reverse. In both halves the current through the load flows in the same direction, so both halves of the input are used. (b) The input is a full sine wave alternating about zero; the output is a series of positive humps, one for each half cycle, touching zero between them. (c) A full-wave rectifier uses both half cycles, so its output frequency is twice the input and the ripple is far easier to smooth, and its rectification efficiency is about double that of a half-wave rectifier.
FAQs
What is the total marks for Class 12 Physics board exam?
70 marks for theory and 30 marks for practicals; total 100 marks.
Which topics have the highest weightage?
Electrostatics, magnetism, electromagnetic induction, and optics carry significant weightage.
Are derivations important for the exam?
Yes, derivations of important laws and formulas are frequently asked and carry 3-4 marks.
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