CBSE Class 12 Physics Board Paper 2025 (55/1/1): Solutions

A graph of force F against 1/r² for charge pairs (q1, q2) and (q2, q3) is given. q2 is positive and has the smallest magnitude. Order the magnitudes of q1, q2, q3.

Uses a figure printed in CBSE's paper — what the figure shows: In the figure, the (q1, q2) line rises steeply with positive slope (like charges, repulsion); the (q2, q3) line has a smaller negative slope (q3 opposite in sign to q2).

1. The force between two charges is given by F = k q1 q2 / r², so a graph of F versus 1/r² is a straight line with slope m = k q1 q2.
2. Since the slope of the (q1, q2) line is steeper than that of the (q2, q3) line, the magnitude of the product |q1 q2| is greater than |q2 q3|.
3. Dividing both sides by the positive quantity |q2| gives |q1| > |q3|.
4. Given that q2 has the smallest magnitude, the order of magnitudes is |q2| < |q3| < |q1|, which corresponds to option (A).

Answer: Slope of F vs 1/r² is kq·q; the (q1, q2) line is steeper than the (q2, q3) line, so |q1| > |q3| > q2, i.e. q2 < q3 < q1 — option (A).

Wires P and Q are of the same material. Q has double the diameter and half the length of P. If P has resistance R, find Q's resistance.

1. Resistance is given by R = ρl/A, and since A = πd²/4, R is proportional to l/d².
2. Substituting the ratios for Q (half length, double diameter) gives R_Q = R × (1/2)/(2²) = R/8.

Answer: R ∝ l/d², so R_Q = R × (1/2)/(2²) = R/8 — option (C).

A 1 cm wire segment on the x-axis carries 0.5 A in the +x direction in a field B = (0.4 mT) ĵ + (0.6 mT) k̂. Find the force on it.

1. Identify the current element vector as I L = 0.5 A × 0.01 m î = 0.005 A m î.
2. Calculate the cross product I L × B = 0.005 î × (0.4 × 10⁻³ ĵ + 0.6 × 10⁻³ k̂) T.
3. Apply unit vector relations î × ĵ = k̂ and î × k̂ = -ĵ to find F = (0.005 × 0.6 × 10⁻³)(-ĵ) + (0.005 × 0.4 × 10⁻³)(k̂) N.
4. Compute the components to get F = -3 × 10⁻⁶ ĵ + 2 × 10⁻⁶ k̂ N, which is (-3ĵ + 2k̂) μN.

Answer: F = I L × B = 0.5 × 0.01 î × (0.4ĵ + 0.6k̂) mT = (-3ĵ + 2k̂) μN — option (B).

A 100-turn coil (each turn 0.05 m², total resistance 1.5 Ω) is suddenly put into a 90 mT field with its axis along the field. Find the charge that flows.

1. The total change in flux linkage is NΔΦ = NBA = 100 × 90 × 10⁻³ T × 0.05 m² = 0.45 Wb.
2. The charge that flows is Q = NΔΦ/R = 0.45 Wb / 1.5 Ω = 0.3 C.

Answer: Q = NΔΦ/R = NBA/R = 0.3 C — option (B).

An air-cored solenoid must have L = 0.016 H, length 0.81 m and radius 0.02 m. How many turns does it need?

1. For an air-cored solenoid L = μ₀N²πr²/l, so N = √(Ll/(μ₀πr²)).
2. Numerator: Ll = 0.016 × 0.81 = 0.01296.
3. Denominator with π ≈ 3.14: μ₀πr² = (4 × 3.14 × 10⁻⁷) × 3.14 × (0.02)² ≈ 1.578 × 10⁻⁹.
4. N² = 0.01296 ÷ (1.578 × 10⁻⁹) ≈ 8.215 × 10⁶, so N ≈ 2866 turns — option (B).

Answer: L = μ₀N²πr²/l gives N = √(Ll/(μ₀πr²)) ≈ 2866 turns (with π ≈ 3.14) — option (B).

Note: With π = 3.1416 the answer comes out as about 2865 turns; the option 2866 uses π ≈ 3.14, so choose (B).

Voltage v = v₀ sin ωt drives current i = i₀ sin(ωt + φ). Find the average power over one cycle.

1. The instantaneous power is p = vi = v₀i₀ sin(ωt) sin(ωt + φ).
2. The average power over one cycle is P_avg = (1/T) ∫₀ᵀ p dt = (v₀i₀/2) cos φ.
3. This equals V_rms I_rms cos φ, so option (D) is correct.

Answer: P_avg = V_rms I_rms cos φ = (i₀v₀/2) cos φ — option (D).

On a falling curve of photon energy against wavelength, P is at short wavelength (high energy), R in the middle and Q at long wavelength. Which radiations could P, Q, R be?

Uses a figure printed in CBSE's paper — what the figure shows: In the figure, energy falls with wavelength; P is near the top-left (shortest λ), R is in the middle and Q is far right (longest λ).

1. Photon energy is inversely proportional to wavelength, so shorter wavelength corresponds to higher energy.
2. P has the shortest wavelength and highest energy, corresponding to X-rays.
3. Q has the longest wavelength and lowest energy, corresponding to microwaves.
4. R is in the middle, corresponding to UV radiation, making option (A) correct.

Answer: Highest energy P = X-rays, lowest Q = microwaves, middle R = UV — option (A).

A coin lies at the bottom of water (n = 4/3) of depth H. Its apparent depth when seen nearly normally?

1. The formula for apparent depth when viewed nearly normally is Apparent depth = Real depth / Refractive index.
2. Substituting the given values, Apparent depth = H / (4/3) = 3H/4.

Answer: Apparent depth = H/n = 3H/4 — option (B).

The graph of stopping potential V₀ against frequency ν has slope m. Express the electron's charge using m and Planck's constant h.

1. The photoelectric equation is eV₀ = hν − φ₀.
2. Comparing with the straight-line equation y = mx + c, the slope m = h/e.
3. Therefore, the charge of the electron is e = h/m.

Answer: eV₀ = hν − φ₀, so slope m = h/e and e = h/m — option (C).

An electron, a proton and a deuteron move at equal speeds. Compare their de Broglie wavelengths λe, λp, λd.

1. The de Broglie wavelength is given by the formula λ = h / (mv), where h is Planck's constant, m is mass, and v is speed.
2. Since the speeds are equal, the wavelength is inversely proportional to the mass (λ ∝ 1/m).
3. The masses satisfy m_e < m_p < m_d, which implies the wavelengths satisfy λe > λp > λd.

Answer: λ = h/(mv) and m_e < m_p < m_d, so λe > λp > λd — option (B).

Pick the correct shape of the binding energy per nucleon vs mass number A curve.

Uses a figure printed in CBSE's paper — what the figure shows: Option (A) shows a rapid rise, a flat peak at A = 56 and a slow decline; (B) has the peak at 56 but stays flat; (C) and (D) peak at A = 80 with a sharp hump.

1. The binding energy per nucleon curve rises steeply for light nuclei and reaches a broad maximum near A = 56 (iron-56).
2. Beyond A = 56, the curve declines slowly for heavier nuclei, which matches the shape described in option (A).

Answer: It rises steeply for light nuclei, peaks broadly near A ≈ 56 and then falls slowly — the graph shown as option (A).

What happens to barrier height and depletion width when a p-n junction is forward biased?

1. In forward bias the applied voltage V opposes the built-in barrier potential V₀, so the effective barrier falls to V₀ − V: the barrier height decreases.
2. A smaller barrier needs fewer uncovered ions on each side, so the depletion layer becomes narrower. Both decrease — option (C).

Answer: Both the barrier height and the depletion layer width decrease — option (C).

A: Pushing a magnet into a closed coil with many turns is hard. R: The induced current in the closed coil opposes the magnet's motion that causes it.

1. Assertion A is true because the induced current creates a magnetic field that opposes the insertion of the magnet.
2. Reason R is true as it correctly states Lenz's law, which dictates that the induced current opposes the change causing it.
3. Since R provides the physical explanation for why pushing the magnet is difficult, R is the correct explanation for A.

Answer: Both true; by Lenz's law the induced current opposes the motion, which is why it is hard — R explains A — option (A).

A: Galvanometer deflection is proportional to current. R: Its coil hangs in a uniform radial magnetic field.

1. In a radial field the plane of the coil always lies along the field lines, so the angle between the coil's normal and B stays 90°.
2. So the deflecting torque is always NIAB, and at equilibrium NIAB = kφ, giving φ = (NAB/k) I, i.e. φ ∝ I. A is true.
3. R is true (the pole pieces are curved and a soft-iron core makes the field radial), and it is exactly why A holds — option (A).

Answer: Both true; the radial field keeps the coil plane along B so torque NIAB balances kφ, giving φ ∝ I — R explains A — option (A).

A: Simply pressing a p-type slab against an n-type slab does not make a p-n junction. R: Electrons outnumber holes in p-type, and holes outnumber electrons in n-type.

1. Assertion A is true because simply pressing the slabs leaves surface roughness and air gaps, preventing the continuous contact required for a p-n junction.
2. Reason R is false because p-type semiconductors have holes as majority carriers, while n-type semiconductors have electrons as majority carriers.
3. Since A is true and R is false, the correct option is (C).

Answer: A is true (surface roughness and gaps prevent a continuous junction); R is false since holes dominate in p-type and electrons in n-type — option (C).

A: The electron's potential energy in any stationary orbit of hydrogen is positive. R: A charged particle always has positive total energy.

1. In a hydrogen atom, the electron is bound to the proton, so its potential energy is negative, given by U = -ke²/r, making Assertion A false.
2. The total energy of a bound electron in a stationary orbit is also negative, given by E = -ke²/2r, making Reason R false.
3. Since both the assertion and the reason are false, the correct option is (D).

Answer: Both false: the potential energy (−ke²/r) and the total energy of the bound electron are negative — option (D).

A cell (emf E, internal resistance r) feeds a rheostat. The rheostat voltage is 5 V at 2 A and 4 V at 4 A. Find E and r.

1. Apply the terminal voltage formula for a cell: V = E − Ir.
2. Substitute the first set of values (V = 5 V, I = 2 A) to get the equation: 5 = E − 2r.
3. Substitute the second set of values (V = 4 V, I = 4 A) to get the equation: 4 = E − 4r.
4. Subtract the second equation from the first to eliminate E: 1 = 2r, which gives r = 0.5 Ω.
5. Substitute r = 0.5 Ω into the first equation to find E: E = 5 + 2(0.5) = 6 V.

Answer: V = E − Ir: E = 5 + 2r and E = 4 + 4r, so r = 0.5 Ω and E = 6 V.

Light of 600 nm passes a single slit; the first minimum is at θ = 30°. Find the slit width.

1. For the first minimum in single-slit diffraction, the condition is a sin θ = λ.
2. Substitute the given values: a sin 30° = 600 nm.
3. Since sin 30° = 0.5, the equation becomes a × 0.5 = 600 nm.
4. Solving for a gives a = 600 nm / 0.5 = 1200 nm.
5. Converting to micrometers, a = 1.2 μm (or 1.2 × 10⁻⁶ m).

Answer: a sin θ = λ for the first minimum, so a = 600 nm / 0.5 = 1.2 μm (1.2 × 10⁻⁶ m).

Two waves, each of intensity I₀, meet at a point with path difference λ/8 in a double-slit set-up. Find the resultant intensity there.

1. Calculate the phase difference using the formula φ = (2π/λ) × path difference, which gives φ = (2π/λ) × (λ/8) = π/4.
2. Substitute the phase difference into the resultant intensity formula I = 4I₀cos²(φ/2) to get I = 4I₀cos²(π/8).
3. Evaluate the expression using the half-angle identity cos²(π/8) = (1 + cos(π/4))/2 = (1 + 1/√2)/2, resulting in I = 4I₀ × (1 + 1/√2)/2 = (2 + √2)I₀ ≈ 3.414 I₀.

Answer: Phase difference = (2π/λ)(λ/8) = π/4; I = 4I₀cos²(π/8) = (2 + √2)I₀ ≈ 3.414 I₀.

A ray enters the flat end face of a cylindrical rod (n = 2/√3, in air) at angle θ to the axis and then just grazes along the curved surface. Find θ.

Uses a figure printed in CBSE's paper — what the figure shows: In the figure, the ray meets the end face at angle θ to the rod's axis (the normal to that face), refracts and then travels along the curved surface.

1. The ray just grazes the curved surface, so the angle of incidence at the curved surface equals the critical angle C.
2. Using the critical angle formula sin C = 1/n, we have sin C = 1/(2/√3) = √3/2, which gives C = 60°.
3. From the geometry of the cylindrical rod, the angle of refraction r at the end face and the critical angle C are complementary, so r = 90° − C = 90° − 60° = 30°.
4. Applying Snell's law at the end face, n_air sin θ = n_rod sin r, so 1 × sin θ = (2/√3) sin 30°.
5. Calculating the value, sin θ = (2/√3) × (1/2) = 1/√3, therefore θ = sin⁻¹(1/√3) ≈ 35.3°.

Answer: sin C = √3/2 so C = 60°; refraction angle at the end face r = 90° − 60° = 30°; sin θ = (2/√3) sin 30° = 1/√3, so θ = sin⁻¹(1/√3) ≈ 35.3°.

Show that in Bohr's hydrogen model the period of the electron in the nth orbit varies as n³.

1. The period in the nth orbit is T = 2πr/v.
2. Bohr's radius r = n²h²ε₀/(πme²), so r ∝ n².
3. From mvr = nh/2π, v = nh/(2πmr); with r ∝ n², v ∝ n/n² = 1/n.
4. So T ∝ n² ÷ (1/n) = n³.

Answer: T = 2πr/v; with r ∝ n² and v ∝ 1/n (from mvr = nh/2π and r = n²h²ε₀/(πme²)), T ∝ n²/(1/n) = n³.

Si is doped with one acceptor atom per 5 × 10⁷ Si atoms; Si atom density is 5 × 10²⁸ m⁻³. Find holes per cm³ due to doping and name one such dopant.

1. Calculate the number of acceptor atoms per cubic meter by dividing the total silicon atom density by the doping ratio: 5 × 10²⁸ / 5 × 10⁷ = 10²¹ m⁻³.
2. Convert the hole concentration from per cubic meter to per cubic centimeter by dividing by 10⁶: 10²¹ / 10⁶ = 10¹⁵ cm⁻³.
3. Identify the dopant as a trivalent element, such as boron, aluminium, indium, or gallium.

Answer: Holes = 5 × 10²⁸ / 5 × 10⁷ = 10²¹ per m³ = 10^15 per cm³. Dopant: a trivalent atom such as aluminium, indium, gallium or boron.

Cells of 3 V (0.2 Ω) and 6 V (0.4 Ω) are joined in parallel and connected to a 4 Ω resistor. Find (i) the equivalent emf, (ii) the equivalent internal resistance, (iii) the current drawn.

1. Given E1 = 3 V, r1 = 0.2 Ω, E2 = 6 V, r2 = 0.4 Ω, and external resistance R = 4 Ω.
2. (i) The equivalent emf for cells in parallel is E_eq = (E1r2 + E2r1) / (r1 + r2).
3. Substituting values: E_eq = (3 × 0.4 + 6 × 0.2) / (0.2 + 0.4) = (1.2 + 1.2) / 0.6 = 2.4 / 0.6 = 4 V.
4. (ii) The equivalent internal resistance for parallel cells is r_eq = (r1 × r2) / (r1 + r2).
5. Substituting values: r_eq = (0.2 × 0.4) / (0.2 + 0.4) = 0.08 / 0.6 = 2/15 ≈ 0.133 Ω.
6. (iii) The total current drawn is I = E_eq / (R + r_eq).
7. Substituting values: I = 4 / (4 + 0.133) = 4 / 4.133 ≈ 0.97 A.

Answer: (i) E_eq = (E₁r₂ + E₂r₁)/(r₁ + r₂) = (1.2 + 1.2)/0.6 = 4 V. (ii) r_eq = r₁r₂/(r₁ + r₂) = 0.08/0.6 = 2/15 ≈ 0.13 Ω. (iii) I = 4/(4 + 0.133) = 0.97 A.

Note: CBSE's marking scheme writes the current as about 0.9 A; exactly it is 4 ÷ (4 + 2/15) = 30/31 A ≈ 0.97 A.

(i) A wire of length l across an ideal cell (emf E) is slowly stretched to 2l with the cell still connected. Relate the final resistance R' to R and the final drift speed v_d' to v_d. (ii) If electrons drift from lower to higher potential, are all free electrons moving the same way?

1. Since the volume of the wire remains constant during stretching, the cross-sectional area A halves to A/2 when the length doubles to 2l.
2. Using the resistance formula R = ρl/A, the new resistance is R' = ρ(2l)/(A/2) = 4(ρl/A) = 4R.
3. The electric field inside the wire is E_field = V/l = E/l, so the drift speed is v_d = eτ(E/l)/m.
4. Since the emf E is constant and length doubles, the new drift speed is v_d' = eτ(E/2l)/m = v_d/2.
5. No, free electrons do not all move in the same direction; they undergo random thermal motion in all directions with only a small net drift superposed.

Answer: (i) Volume fixed, so area halves: R' = ρ(2l)/(A/2) = 4R. v_d = eτE/(m l) since the field is E/l, so doubling the length halves it: v_d' = v_d/2. (ii) No — free electrons still move randomly in all directions; only a small net drift is superposed on this motion.

Use the Biot–Savart law to find B on the axis of a current loop, and then at its centre.

1. Take a current element I dl on the loop (radius R) and a point P on the axis at distance x from the centre; the distance from dl to P is r = √(x² + R²), and dl is perpendicular to r.
2. Biot–Savart law: dB = (μ₀/4π) I dl/r², directed perpendicular to the plane of dl and r.
3. Resolve dB: the components perpendicular to the axis from opposite elements cancel; the axial components add. The axial part is dB × (R/r).
4. B = ∮ (μ₀/4π)(I dl/r²)(R/r) = (μ₀/4π)(IR/r³)(2πR) = μ₀IR²/(2r³).
5. With r = √(x² + R²): B = μ₀IR²/[2(x² + R²)^(3/2)], along the axis.
6. At the centre x = 0, so B = μ₀IR²/(2R³) = μ₀I/(2R).

Answer: Perpendicular components of dB cancel; axial parts add: B = μ₀IR²/[2(x² + R²)^(3/2)] along the axis. At the centre (x = 0), B = μ₀I/(2R).

(a) Show that setting up current I in an inductor L needs energy LI²/2. (b) Using an air-cored solenoid, show that magnetic energy density is B²/(2μ₀).

1. (a) As the current grows, the back emf is ε = −L dI/dt. The source must do work against it at the rate P = −εI = LI dI/dt.
2. So dW = LI dI, and W = ∫₀ᴵ LI dI = LI²/2.
3. (b) For a long air-cored solenoid, L = μ₀n²Al and B = μ₀nI, so I = B/(μ₀n).
4. U = LI²/2 = (μ₀n²Al) × B²/(μ₀²n²) ÷ 2 = B²Al/(2μ₀).
5. The field fills volume Al, so energy density u = U/(Al) = B²/(2μ₀).

Answer: (a) Work against back emf: dW = L I dI, so W = ∫₀ᴵ LI dI = LI²/2. (b) For a solenoid L = μ₀n²Al and B = μ₀nI, so U = LI²/2 = B²Al/(2μ₀); dividing by volume Al gives u = B²/(2μ₀).

(a) A parallel-plate capacitor is charged by an ac source. Show that I_c + I_d is the same everywhere in the circuit. (b) Does Kirchhoff's junction rule hold at each plate? Explain.

1. (a) In the connecting wires charge flows, so the current is the conduction current I_c, and there is no changing electric flux there, so I_d = 0.
2. Between the plates no charge crosses the gap, so I_c = 0 there.
3. Between the plates E = Q/(ε₀A), so the flux is Φ_E = EA = Q/ε₀ and I_d = ε₀ dΦ_E/dt = dQ/dt.
4. dQ/dt is exactly the conduction current in the wires, so I_c + I_d has the same value everywhere in the circuit.
5. (b) Yes. At each plate the conduction current coming in equals the displacement current going out across the gap, so the total current into the plate equals the total current out.

Answer: (a) In the wires I_d = 0 so I = I_c; between the plates I_c = 0 and I_d = ε₀ dΦ_E/dt = ε₀ A d(Q/ε₀A)/dt = dQ/dt = I_c, so the sum is the same everywhere. (b) Yes — the conduction current into a plate equals the displacement current leaving it between the plates.

Give reasons: (a) photoelectrons from monochromatic light have a spread of kinetic energies; (b) saturation current changes with intensity; (c) at fixed intensity, raising the wavelength eventually stops emission.

1. (a) Electrons in the metal are bound with different energies; those deeper inside lose more energy escaping, so only the least bound electrons achieve the maximum kinetic energy.
2. (b) The number of photoelectrons emitted per second is directly proportional to the intensity of incident light, so the saturation current increases with intensity.
3. (c) Increasing the wavelength decreases the frequency and photon energy; when the frequency falls below the threshold frequency, the photon energy becomes less than the work function, stopping emission.

Answer: (a) Electrons are bound with different energies; deeper ones lose more energy escaping, so only the least bound get the maximum KE. (b) Number of photoelectrons per second, hence saturation current, is proportional to intensity. (c) Longer λ means lower ν and photon energy; beyond the threshold wavelength (ν < ν₀) hν is less than the work function, so no emission.

(a) Define mass defect and binding energy, and explain fission using binding energy per nucleon. (b) A deuteron (one p, one n) has mass 2.013553 u. With m_p = 1.007277 u, m_n = 1.008665 u and 1 u = 931.5 MeV/c², find its mass defect and energy equivalent.

1. Mass defect is the difference between the sum of the masses of individual protons and neutrons and the actual mass of the nucleus.
2. Binding energy is the energy required to disassemble a nucleus into its constituent protons and neutrons, calculated as the mass defect multiplied by c squared.
3. In fission, a heavy nucleus splits into lighter nuclei with higher binding energy per nucleon, releasing energy due to the mass difference.
4. Calculate the total mass of the separate nucleons: 1.007277 u + 1.008665 u = 2.015942 u.
5. Calculate the mass defect: 2.015942 u - 2.013553 u = 0.002389 u.
6. Calculate the energy equivalent: 0.002389 u × 931.5 MeV/u = 2.225 MeV.

Answer: (a) Mass defect = mass of separate nucleons minus nuclear mass; binding energy = energy needed to pull the nucleus apart into nucleons (Δm c²). In fission a heavy nucleus splits into middle-mass nuclei with higher BE per nucleon, so energy is released. (b) Δm = 2.015942 − 2.013553 = 0.002389 u; energy = 0.002389 × 931.5 ≈ 2.225 MeV.

Note: CBSE's marking scheme computes 2.2253 MeV but writes ≈ 2.22 MeV; rounded correctly it is 2.225 MeV (≈ 2.23 MeV to three figures).

(a) Draw the circuit to study a p-n diode's V-I characteristics. (b) Sketch the characteristic curves. (c) State two things learnt from them.

1. (a) Forward bias: battery with a rheostat used as a potential divider, the diode (p-side to +), and a milliammeter in series; a voltmeter across the diode.
2. Reverse bias: the battery is reversed (n-side to +) and a microammeter replaces the milliammeter, since the current is tiny.
3. (b) Forward curve: almost no current until the threshold (knee) voltage, about 0.7 V for Si, then the current rises steeply. Reverse curve: a very small, nearly constant current (μA) until the breakdown voltage, then a sudden large rise.
4. (c) Any two: the diode conducts only above the knee voltage; forward resistance is low; reverse current is very small (reverse resistance very high); breakdown occurs at a definite reverse voltage.

Answer: (a) Battery, rheostat (potential divider), diode, milliammeter in series and voltmeter across the diode for forward bias; reversed battery with microammeter for reverse bias. (b) Forward: tiny current until the knee (~0.7 V for Si), then a steep rise; reverse: very small nearly constant current until breakdown, then a sudden rise. (c) Any two: knee/threshold voltage, reverse saturation current, breakdown voltage, low forward resistance, very high reverse resistance.

RC circuit with R = 20 kΩ, C = 500 μF, battery V = 10 V. Closing S1 (S2 open) charges C. What is its final charge?

Uses a figure printed in CBSE's paper — what the figure shows: In the figure, S1 connects the battery so it charges C through R; closing S2 (with S1 open) joins C and R in a loop without the battery.

1. Upon closing S1, the capacitor charges until the potential difference across it equals the battery voltage, so Vc = 10 V.
2. Calculate the final charge using the formula Q = CV, which gives Q = 500 μF × 10 V = 5 mC.

Answer: Q = CV = 500 μF × 10 V = 5 mC — option (B).

In the same RC circuit, after C is fully charged, S2 is closed and S1 opened. What is the final charge on C?

Uses a figure printed in CBSE's paper — what the figure shows: Closing S2 with S1 open puts C and R in a closed loop with no battery.

1. When S2 is closed and S1 is opened, the capacitor C and resistor R form a closed loop without any battery.
2. The capacitor discharges through the resistor until all stored energy is dissipated as heat, leaving the final charge on C as zero.

Answer: C discharges completely through R, so the final charge is zero — option (A).

Give the dimensional formula of the product RC.

1. Resistance R has dimensions [M L² T⁻³ A⁻²] and Capacitance C has dimensions [M⁻¹ L⁻² T⁴ A²].
2. Multiplying these gives [M L² T⁻³ A⁻²] × [M⁻¹ L⁻² T⁴ A²] = [M⁰ L⁰ T¹ A⁰], which corresponds to the dimension of time.

Answer: RC is the time constant, so its dimensions are [M⁰L⁰T A⁰] — option (D).

With S1 closed and S2 open (R = 20 kΩ, C = 500 μF, V = 10 V), find the current in R 5 s after closing.

1. Calculate the time constant τ = RC = (20 × 10³ Ω)(500 × 10⁻⁶ F) = 10 s.
2. Calculate the initial current I₀ = V/R = 10 V / (20 × 10³ Ω) = 0.5 mA.
3. Substitute t = 5 s into the decay formula I = I₀e^(−t/τ) to get I = 0.5 × e^(−5/10) = 0.5 × e^(−0.5) mA.
4. Evaluate the expression: I = 0.5 / √e ≈ 0.5 / 1.648 ≈ 0.303 mA, which matches option (A).

Answer: I = (V/R)e^(−t/RC); V/R = 0.5 mA and RC = 10 s, so I = 0.5e^(−0.5) mA = 1/(2√e) mA ≈ 0.303 mA — option (A).

Note: CBSE's marking scheme prints this option as 1/(2e) mA, which drops the root sign; the correct value is 1/(2√e) mA ≈ 0.303 mA. The scheme gave this mark to every student who attempted it.

With S1 closed and S2 open (R = 20 kΩ, V = 10 V), find the charging current in R just after closing.

1. At t = 0, the capacitor acts as a short circuit because it has no initial voltage.
2. Using Ohm's law, the initial current is I₀ = V/R = 10/20000 A = 0.5 mA.

Answer: At t = 0 the capacitor has no voltage, so I₀ = V/R = 10/20000 A = 0.5 mA — option (B).

A glass plate painted with an upward arrow and a crossing line ending in a ball on the right is lit, and a convex lens forms its real image on a screen. Which picture shows the image?

Uses a figure printed in CBSE's paper — what the figure shows: Object: vertical arrow pointing up, with a slightly slanted horizontal line through it ending in a ball on the right. Option (C) shows the arrow pointing down with the ball at the left end of the line.

1. A real image formed by a convex lens is inverted both vertically and horizontally.
2. The upward arrow becomes downward, and the ball on the right moves to the left, matching option (C).

Answer: A real image is inverted both up-down and left-right: the arrow points down and the ball is on the left — option (C).

Four statements give the sign of magnification for a convex mirror, virtual mirror images, a concave lens and real inverted images. Which one is wrong?

1. A convex mirror always forms a virtual and erect image, which means the magnification is always positive.
2. Statement (A) claims the magnification is negative, which contradicts the property of convex mirrors, making it the incorrect statement.

Answer: A convex mirror always gives a virtual erect image, so its magnification is positive; statement (A) is the incorrect one — option (A).

A convex lens of focal length f is split into two equal halves by a cut perpendicular to its principal axis. Focal length of each half?

1. A convex lens of focal length f is split into two equal halves by a cut perpendicular to its principal axis.
2. Each half becomes a plano-convex lens with one curved surface and one flat surface.
3. The power of each half is half the power of the original lens, so its focal length becomes 2f.

Answer: Each half is plano-convex with one curved surface, so its power halves and focal length becomes 2f — option (B).

In the set-up of part (i), the object is 20 cm from the lens and the screen is 50 cm from the object. Find the focal length.

1. Given object distance u = -20 cm and total distance between object and screen is 50 cm, the image distance v is 50 - 20 = 30 cm.
2. Using the lens formula 1/f = 1/v - 1/u, substitute the values to get 1/f = 1/30 - (-1/20) = 1/30 + 1/20.
3. Solving the equation gives 1/f = 5/60, so f = 12 cm, which corresponds to option (B).

Answer: u = −20 cm, v = +30 cm; 1/f = 1/30 + 1/20, so f = 12 cm — option (B).

For a biconvex lens, the object is X₁ from the first focal point and the image is X₂ from the second focal point. Find the focal length.

1. Apply Newton's form of the lens equation, which relates the distances of the object and image from their respective focal points: X₁X₂ = f².
2. Solve for the focal length f by taking the square root of both sides, yielding f = √(X₁X₂).

Answer: Newton's form of the lens equation gives X₁X₂ = f², so f = √(X₁X₂) — option (C).

(i) Charges 5 μC at (−3 cm, 0, 0) and −1 μC at (3 cm, 0, 0) are placed in an external field E = (A/r²) r̂ with A = 3 × 10⁵ V m. Find the change in the system's electrostatic energy due to this field. (ii) Two conductors in air carry +80 μC and −80 μC with 16 V between them. (1) Capacitance? (2) New voltage if a dielectric of K = 3 fills the space? (3) Does the capacitance change if the charges become ±160 μC? Why?

1. The potential of the external field is V = A/r, and both charges are at r = 3 cm = 0.03 m.
2. The change in electrostatic energy is ΔU = (5 − 1) × 10⁻⁶ × (3 × 10⁵/0.03) = 40 J.
3. For the capacitor, capacitance is C = Q/V = 80 μC / 16 V = 5 μF.
4. With dielectric K = 3, the new capacitance is C' = K × C = 3 × 5 μF = 15 μF.
5. The new voltage is V' = Q/C' = 80 μC / 15 μF ≈ 5.33 V.
6. The capacitance does not change if the charge becomes ±160 μC because capacitance depends only on the geometry and the medium, not on the amount of charge.

Answer: (i) The field's potential is V = A/r, so each charge gains qV(r) with r = 3 cm: ΔU = (5 − 1) × 10⁻⁶ × 3 × 10⁵/0.03 = 40 J. (ii)(1) C = Q/V = 80 μC/16 V = 5 μF. (2) C' = 3 × 5 = 15 μF and V = 80/15 ≈ 5.33 V. (3) No — capacitance depends only on the geometry and medium, not on the charge.

(i) Shells A, B, C (radius R) each hold a concentric metal ball of radius R/10. Shell charges are +6q, −4q, 14q and ball charges −2q, +8q, −10q. Compare the field magnitudes due to A, B, C at 3R from their centres. (ii) −6 μC sits at the centre B of a semicircle of radius 5 cm and +6 μC at D, 10 cm from B on the diameter line beyond C. Find the work done in taking +5 μC from C to A along the arc.

Uses a figure printed in CBSE's paper — what the figure shows: In the figure, D (+6 μC), C, B (−6 μC) and A lie on one straight line in that order; C and A are the ends of the semicircle centred at B (radius 5 cm), so DC = 5 cm and DA = 15 cm.

1. (i) Outside a spherical charge distribution the field depends only on the total charge enclosed: E = kQ/r².
2. Total charge for each shell plus ball: A: 6q − 2q = 4q; B: −4q + 8q = 4q; C: 14q − 10q = 4q.
3. At r = 3R each field is k(4q)/(3R)² = 4kq/(9R²), so E_A : E_B : E_C = 1 : 1 : 1.
4. (ii) Distances: C is 5 cm from B and 5 cm from D; A is 5 cm from B and 15 cm from D. Use k × 6 μC = 9 × 10⁹ × 6 × 10⁻⁶ = 5.4 × 10⁴ V m.
5. V_C = −5.4 × 10⁴/0.05 + 5.4 × 10⁴/0.05 = 0.
6. V_A = −5.4 × 10⁴/0.05 + 5.4 × 10⁴/0.15 = −1.08 × 10⁶ + 3.6 × 10⁵ = −7.2 × 10⁵ V.
7. Work by the external agent W = q(V_A − V_C) = 5 × 10⁻⁶ × (−7.2 × 10⁵) = −3.6 J; the work does not depend on the path along the arc (the field itself does +3.6 J).

Answer: (i) Net enclosed charge is 4q for each (6q − 2q, −4q + 8q, 14q − 10q), so E = k(4q)/(3R)² for all: E_A : E_B : E_C = 1 : 1 : 1. (ii) V_C = k(−6 μC)/0.05 + k(6 μC)/0.05 = 0; V_A = k(−6 μC)/0.05 + k(6 μC)/0.15 = −7.2 × 10⁵ V. Work by the external agent W = q(V_A − V_C) = 5 μC × (−7.2 × 10⁵ V) = -3.6 J (the field does +3.6 J).

Note: The figure labels the charges ±6q, but the question text gives ±6 μC; the text values are used, as in CBSE's marking scheme.

(i) A proton moving right reaches P and then follows a curved path through Q up to a crest at R, staying in the paper's plane. Give the field direction near P, Q, R and compare field strengths there. (ii) A circular loop of area A gives field B at its centre. Show its magnetic moment is (2BA/μ₀)√(A/π).

Uses a figure printed in CBSE's paper — what the figure shows: In the figure, the proton enters horizontally at P, the path rises with an upward (anticlockwise) bend through Q, and turns over at a crest R, bending downward (clockwise) there.

1. Part (i): The magnetic force F = qv × B points toward the center of curvature of the path.
2. Near P and Q, the path bends upward, so the force is upward; for a proton moving right, the field B is into the paper.
3. Near R, the path bends downward at the crest, so the force is downward; for a proton moving right, the field B is out of the paper.
4. Using r = mv/(qB), we see B is inversely proportional to the radius of curvature r.
5. Since the bend is sharpest (smallest r) at R and flattest (largest r) at Q, the field strengths are B_Q < B_P < B_R.
6. Part (ii): The field at the center is B = μ₀I/(2r), so the current is I = 2Br/μ₀.
7. Substituting r = √(A/π) into m = IA gives m = (2BA/μ₀)√(A/π).

Answer: (i) Force F = q v × B points to the centre of curvature: near P and Q the path bends upward, so B is into the paper; near R it bends downward, so B is out of the paper. Since r = mv/(qB), B ∝ 1/r: B_Q < B_P < B_R (sharpest bend at R). (ii) B = μ₀I/(2r) gives I = 2Br/μ₀; with r = √(A/π), m = IA = (2BA/μ₀)√(A/π).

(i) Derive the torque on a rectangular current loop in a uniform magnetic field. (ii) A charged particle circling in a uniform field passes through a lead sheet and loses half its kinetic energy without changing direction. How do (1) its path radius and (2) its period change?

1. Consider a rectangular loop of sides a and b with current I in a uniform field B, where the normal makes an angle θ with B.
2. The forces on the sides of length a are F = IaB, acting in opposite directions to form a couple.
3. The perpendicular lever arm between these forces is b sin θ.
4. The torque is τ = F × arm = (IaB)(b sin θ) = IabB sin θ.
5. Since area A = ab, the torque is τ = IAB sin θ, which equals the vector product m × B with m = IA.
6. The radius of the particle's path is r = mv/qB = √(2mK)/qB, so r is proportional to √K.
7. Since kinetic energy halves, the radius becomes r/√2, while the period T = 2πm/qB remains unchanged as it is independent of speed.

Answer: (i) Forces on the two sides parallel to the axis of rotation are IbB each, equal and opposite, forming a couple with arm a sin θ: τ = IabB sin θ = IAB sin θ, i.e. τ = m × B with m = IA. (ii)(1) r = mv/(qB) = √(2mK)/(qB) ∝ √K, so r becomes r/√2. (2) T = 2πm/(qB) does not depend on speed or energy, so the period is unchanged.

(i) (1) What are coherent sources and why are they needed for steady interference? (2) Why are two independent lamps not coherent? (ii) Slits 0.1 mm apart, screen 1.20 m away, λ = 600 nm. Find (1) the fringe spacing, (2) the angular width of a bright fringe in degrees.

1. Coherent sources are defined as sources that emit waves of the same frequency and maintain a constant phase difference between them.
2. A constant phase difference is required for steady interference because it ensures that the positions of bright and dark fringes on the screen remain fixed over time.
3. Two independent lamps are not coherent because they emit light with random and rapidly changing phases, resulting in a phase difference that is not constant.
4. The fringe spacing is calculated using the formula β = λD/d, which gives β = (600 × 10⁻⁹ × 1.20) / (0.1 × 10⁻³) = 7.2 × 10⁻³ m or 7.2 mm.
5. The angular width of a bright fringe is given by θ = λ/d, which equals 600 × 10⁻⁹ / (0.1 × 10⁻³) = 6 × 10⁻³ rad.
6. Converting the angular width to degrees using the factor 180/π, we get θ ≈ 6 × 10⁻³ × 57.3 = 0.34°.

Answer: (i)(1) Sources with the same frequency and a constant phase difference; a constant phase difference keeps the bright and dark positions fixed. (2) Independent sources emit random, rapidly changing phases, so their phase difference is not constant. (ii)(1) β = λD/d = 600 × 10⁻⁹ × 1.2/(0.1 × 10⁻³) = 7.2 mm. (2) Angular width = λ/d = 6 × 10⁻³ rad ≈ 0.34°.

(i) Define a wavefront and sketch a plane wavefront refracted by a convex lens. (ii) Light from a distant source passes through a glass sphere (n = 1.5) of radius 15 cm. Draw the ray diagram and locate the final image.

1. Define a wavefront as the locus of points vibrating in the same phase and sketch a plane wavefront becoming a converging spherical wavefront after a convex lens.
2. For the first surface of the sphere, use the refraction formula with u = infinity, R = +15 cm, and n = 1.5 to find the image distance v = 45 cm.
3. Calculate the object distance for the second surface by subtracting the sphere diameter from the first image position, giving u = 45 - 30 = 15 cm.
4. Apply the refraction formula for the second surface with n1 = 1.5, n2 = 1, u = 15 cm, and R = -15 cm to set up the equation 1/v - 1.5/15 = (1 - 1.5)/(-15).
5. Solve the equation to find the final image distance v = 7.5 cm beyond the second surface.
6. State that the final image is located 22.5 cm from the center of the sphere, which is 15 cm from the center plus 7.5 cm from the surface.

Answer: (i) A wavefront is the locus of points vibrating in the same phase; a plane wavefront becomes a spherical wavefront converging to the focus after the convex lens. (ii) First surface (u = ∞, R = +15 cm): 1.5/v = 0.5/15 gives v = 45 cm, i.e. 15 cm beyond the second surface. Second surface (n 1.5 → 1, u = +15 cm, R = −15 cm): 1/v − 1.5/15 = (1 − 1.5)/(−15), giving v = 7.5 cm beyond the second surface (22.5 cm from the centre).