CBSE Class 12 Physics Board Paper 2026 (55/1/1): Solutions

Potential in a region is V = 10 − 50x (V in volts, x in metres). Find the electric field.

1. The electric field is given by the negative gradient of the potential, so E = -dV/dx.
2. Differentiating V = 10 - 50x with respect to x gives dV/dx = -50 V/m.
3. Substituting this into the field equation yields E = -(-50) = +50 N/C, which corresponds to option (C).

Answer: E = −dV/dx = +50, so the field is 50 N/C along +x — option (C).

Two far-apart charged metal spheres A and B (radii r1, r2) are joined by a conducting wire. Find E_A/E_B at their surfaces.

1. When connected by a wire, both spheres attain the same potential V.
2. Since V = kq/r, the charge is proportional to radius (q ∝ r).
3. Surface field E = kq/r² is proportional to 1/r.
4. Thus E_A/E_B = r2/r1.

Answer: Both reach the same potential, so q ∝ r and surface field E ∝ q/r² ∝ 1/r; E_A/E_B = r2/r1 — option (B).

A long wire of radius a carries current I spread uniformly over its cross-section. Find the magnetic field magnitude at distance a/2 from its axis.

1. Apply Ampere's law to a circular path of radius r inside the wire, where the enclosed current is I_enc = I(r²/a²).
2. The magnetic field is given by B = μ₀I_enc/(2πr) = μ₀Ir/(2πa²).
3. Substitute r = a/2 to get B = μ₀I(a/2)/(2πa²) = μ₀I/(4πa).

Answer: Inside, B = μ₀Ir/(2πa²); at r = a/2, B = μ₀I/(4πa) — option (C).

In Young's double-slit set-up with screen distance much larger than slit separation, what shape are the fringes nearly?

1. Points of constant path difference from the two slits lie on hyperbolas, so the bright and dark bands follow these curves.
2. On a screen far away (D ≫ d), the part of each hyperbola near the centre is almost straight, so the fringes look nearly like straight lines — option (A).

Answer: Nearly straight lines — option (A).

An EM wave goes from vacuum into a medium with relative permittivity 3/2 and relative permeability 8/3. What happens to its wavelength and frequency?

1. Calculate the refractive index using the formula n = √(μr × εr) = √(8/3 × 3/2) = √4 = 2.
2. Since the refractive index is 2, the speed and wavelength are halved, while the frequency remains unchanged, making option (C) correct.

Answer: Refractive index = √(3/2 × 8/3) = 2, so speed and wavelength are halved while frequency stays the same — option (C).

In a series LCR circuit, the resistor, capacitor and inductor each have 10 V across them. If the capacitor is shorted, what is the inductor voltage?

1. Initially, V_L = V_C = 10 V, so the net reactive voltage is zero and the supply voltage equals the resistor voltage, V = 10 V.
2. Since V_L = V_C, the inductive and capacitive reactances are equal (X_L = X_C), and since V_L = V_R, the resistance equals the reactance (R = X_L).
3. Shorting the capacitor removes it from the circuit, leaving a series LR circuit with impedance Z = √(R² + X_L²) = √(2R²) = R√2.
4. The current becomes I = V/Z = 10/(R√2), so the inductor voltage is V_L' = I × X_L = (10/R√2) × R = 10/√2 = 5√2 V ≈ 7.07 V.

Answer: Supply = 10 V (V_L and V_C cancel) and X_L = R, so V_L = 10/√2 = 5√2 V ≈ 7.07 V — option (B).

What wavelength range do EM waves used as a medical diagnostic tool (X-rays) have?

1. X-rays are electromagnetic waves used in medical diagnostics to image internal body structures.
2. The wavelength range of X-rays is approximately 1 nm to 10⁻³ nm.

Answer: About 1 nm to 10⁻³ nm (X-rays) — option (A).

An alpha particle moving head-on at speed v has distance of closest approach d. What is it if the speed is halved?

1. At the distance of closest approach, the kinetic energy of the alpha particle is fully converted into electrostatic potential energy, so (1/2)mv² = kQq/d.
2. This implies d is proportional to 1/v², so halving the speed v results in the distance increasing by a factor of 4, giving 4d.

Answer: d ∝ 1/v², so halving v makes it 4d — option (D).

A diverging lens of focal length 10 cm is sliced into two equal plano-concave pieces. What is the focal length of each piece?

1. The original lens has f = −10 cm, so 1/f = −0.1 cm⁻¹. Its two curved surfaces contribute equally to this.
2. Cutting it into two plano-concave halves makes one surface of each piece flat, so each piece keeps only half: 1/f' = −0.05 cm⁻¹.
3. So f' = 1/(−0.05 cm⁻¹) = −20 cm: each piece is still diverging, with focal length 20 cm — option (A).

Answer: One surface becomes flat, so the power halves and each half has focal length 20 cm (diverging) — option (A).

Waves: y1 = A1 sin ωt, y2 = A2 sin 2ωt, y3 = A3 cos ωt, y4 = A4 sin(ωt + π/3). Which of them can interfere with each other?

1. Interference requires waves to have the same frequency and a constant phase difference.
2. y1 and y3 have frequency ω, while y2 has frequency 2ω, so y2 cannot interfere with the others.
3. y1 and y4 have frequency ω and a constant phase difference of π/3, so they can interfere.
4. y1 and y3 have frequency ω and a constant phase difference of π/2, so they can interfere.
5. Thus, waves (i), (iii), and (iv) can interfere with each other, corresponding to option (C).

Answer: Only waves of the same frequency with a constant phase difference: (i), (iii) and (iv) — option (C).

Heaters rated (P1, V) and (P2, V) are joined in series across a dc supply of V/2. Total power drawn?

1. Each heater's resistance from its rating: R1 = V²/P1 and R2 = V²/P2.
2. In series, R1 + R2 = V²/P1 + V²/P2 = V²(P1 + P2)/(P1P2).
3. Power drawn at V/2: P = (V/2)²/(R1 + R2) = (V²/4) × P1P2/[V²(P1 + P2)] = P1P2/[4(P1 + P2)] — option (D).

Answer: R = V²/P for each; P = (V/2)²/(R1 + R2) = P1P2/[4(P1 + P2)] — option (D).

For an unbiased p-n junction in equilibrium, how are diffusion and drift currents related?

1. When the device is not connected to a battery, the flow of majority carriers due to concentration difference is exactly counteracted by the flow of minority carriers due to the built-in electric field.
2. These two opposing flows are equal in magnitude, resulting in a total current of zero, which corresponds to option (C).

Answer: They are equal in size and opposite in direction, so net current is zero — option (C).

Assertion: every atom has a net magnetic moment. Reason: a current loop is not always a magnetic dipole.

1. Assertion is false because atoms with paired electrons, such as diamagnetic materials, have a net magnetic moment of zero.
2. Reason is false because any current loop, regardless of its shape, always acts as a magnetic dipole.
3. Since both the assertion and the reason are incorrect, the correct option is (D).

Answer: Both are false: atoms with paired electrons (diamagnetic) have zero net moment, and every current loop acts as a magnetic dipole — option (D).

Assertion: accelerated electrons sent through a narrow slit give a diffraction pattern. Reason: electrons show both particle and wave nature.

1. The assertion is true because accelerated electrons passing through a narrow slit exhibit diffraction, which is a characteristic wave phenomenon.
2. The reason is true because electrons possess dual nature, behaving as both particles and waves.
3. Since diffraction is a wave property, the wave nature of electrons (Reason) correctly explains why they form a diffraction pattern (Assertion).

Answer: Both true and the reason explains the assertion — option (A).

Assertion: a nucleus weighs less than its separate nucleons. Reason: energy is absorbed when nucleons bind into a nucleus.

1. The assertion is true because the total mass of a bound nucleus is lower than the sum of the masses of its free protons and neutrons.
2. The reason is false because energy is released during the formation of a nucleus, not absorbed, which corresponds to the mass defect.
3. Thus, the correct option is (C) as the assertion is correct but the reason is incorrect.

Answer: Assertion true (mass defect); reason false — binding energy is released, not absorbed — option (C).

Note: The paper prints 'Reason (A)' for the first statement; it is the Assertion.

Assertion: in Bohr's hydrogen model, energy levels are discrete. Reason: the Coulomb force on the electron supplies its centripetal force.

1. The assertion is true because Bohr's postulate of quantized angular momentum restricts the electron to specific orbits, resulting in discrete energy levels.
2. The reason is true because the electrostatic attraction between the proton and electron provides the necessary centripetal force for circular motion.
3. Option (B) is correct because while the Coulomb force explains the orbital mechanics, it does not account for the quantization of energy levels, which arises from the angular momentum postulate.

Answer: Both true, but the Coulomb force alone does not explain quantisation (that comes from the angular-momentum postulate) — option (B).

Note: The paper prints 'Reason (A)' for the first statement; it is the Assertion.

Light of 200 nm falls on an emitter; photocurrent stops at a collector potential of −0.80 V. Find the work function in eV.

1. Photon energy: E = hc/λ = (6.63 × 10⁻³⁴ × 3 × 10⁸)/(200 × 10⁻⁹) ≈ 9.95 × 10⁻¹⁹ J; dividing by 1.6 × 10⁻¹⁹ J/eV gives about 6.22 eV.
2. The stopping potential is 0.80 V, so the fastest electrons have K_max = eV₀ = 0.80 eV.
3. Einstein's equation: φ = E − eV₀ = 6.22 − 0.80 = 5.42 eV ≈ 5.4 eV.

Answer: φ = hc/(λe) − V₀ = 6.22 − 0.80 ≈ 5.4 eV.

Note: Using hc ≈ 1240 eV nm gives 6.20 eV for the photon, and the work function still rounds to about 5.4 eV.

Light of 400 nm and 600 nm falls on a 1 mm wide single slit; screen 1.5 m away. Least distance from the central maximum where minima of both coincide?

1. Condition for single-slit minima is a sin θ = nλ, where a is slit width and n is the order number.
2. Minima coincide when the path difference nλ₁ equals mλ₂, so n × 600 nm = m × 400 nm.
3. The least common multiple of 600 and 400 is 1200 nm, corresponding to the 2nd minimum of 600 nm and 3rd minimum of 400 nm.
4. Using the small angle approximation y = nλD/a, substitute values: y = 1200 × 10⁻⁹ m × 1.5 m / 10⁻³ m.
5. Calculate the result: y = 1.8 × 10⁻³ m = 1.8 mm.

Answer: Minima coincide when n×600 = m×400, first at path 1200 nm (3rd minimum of 400 nm, 2nd of 600 nm): y = 1200 nm × 1.5 m / 1 mm = 1.8 mm.

Double-slit set-up: slits 0.6 mm apart, screen 1.5 m away, light of 440 nm and 660 nm. Least distance from the central maximum where bright fringes of both coincide?

1. A bright fringe of order n lies at y = nλD/d. Two bright fringes coincide when n × 660 nm = m × 440 nm.
2. So n/m = 440/660 = 2/3; the smallest whole numbers are n = 2 (for 660 nm) and m = 3 (for 440 nm), a path difference of 2 × 660 = 1320 nm.
3. y = (1320 × 10⁻⁹ m × 1.5 m)/(0.6 × 10⁻³ m) = 3.3 × 10⁻³ m = 3.3 mm.

Answer: Coincide when n×660 = m×440, first at 1320 nm (2nd bright of 660 nm, 3rd of 440 nm): y = 1320 nm × 1.5 m / 0.6 mm = 3.3 mm.

A wire of length L is wound into an N-turn square coil, and again into an N-turn circular coil; same current, same uniform field. Ratio of maximum torques (square : circle)?

1. Maximum torque is given by τ = NIAB, where A is the area of the coil.
2. Area of square coil: side a = L/(4N), so A_sq = (L/4N)².
3. Area of circular coil: radius r = L/(2πN), so A_circ = π(L/2πN)².
4. Ratio τ_sq/τ_circ = A_sq/A_circ = (L/4N)² / [π(L/2πN)²] = π/4 ≈ 0.785.

Answer: τmax = NIAB; side = L/(4N), radius = L/(2πN), so τ_sq/τ_circ = (L/4N)² / [π(L/2πN)²] = π/4 ≈ 0.785.

Give the typical order of the electron drift speed in a conductor and derive how current depends on it.

1. Drift speed of free electrons in a conductor is very small, of the order of 10⁻³ m/s (about a mm per second).
2. Take a conductor of cross-section A with n free electrons per unit volume, each of charge e, drifting at speed v_d.
3. In time Δt the electrons within a length v_dΔt cross the area A; that cylinder holds nAv_dΔt electrons, so the charge crossing is Q = neAv_dΔt.
4. Current I = Q/Δt = neAv_d, so current is directly proportional to drift speed.

Answer: Drift speed is tiny, of the order of 10⁻³ m/s (about a millimetre or a few millimetres per second). In time Δt, electrons in a length v_dΔt cross area A, carrying charge neAv_dΔt, so I = neAv_d.

Note: NCERT's worked example gives a drift speed of about 1 mm/s and CBSE's marking scheme says a few mm/s: both are of the order of 10⁻³ m/s.

Sketch potential energy of two nucleons versus their separation and give two conclusions from it.

1. Sketch a graph with separation r on the x-axis and potential energy U on the y-axis.
2. Draw the curve starting from a large positive value at very small r, dropping sharply to a minimum near r = 0.8 fm, and then rising to approach zero as r increases beyond a few fm.
3. State that the steep rise at small r indicates a strong repulsive force, while the dip indicates an attractive force beyond 0.8 fm.
4. Conclude that the nuclear force is short-ranged because the potential energy becomes negligible beyond a few femtometers.

Answer: The curve dips to a minimum near r₀ ≈ 0.8 fm, rises steeply at smaller r and tends to zero beyond a few fm. Conclusions: force is attractive beyond about 0.8 fm and strongly repulsive closer than that; it is short-ranged.

(a) Use Gauss's law to get the field of a uniformly charged infinite thin sheet. (b) Two large parallel sheets, each with charge density σ, are close together in air: field (i) between and (ii) outside them?

1. Consider a Gaussian surface in the form of a cylinder perpendicular to the sheet, with area A on each end face.
2. The electric field is normal to the sheet, so flux through the curved surface is zero and total flux is 2EA.
3. By Gauss's law, total flux equals charge enclosed divided by ε₀, which is σA/ε₀.
4. Equating fluxes gives 2EA = σA/ε₀, so the field is E = σ/(2ε₀) directed normal to the sheet.
5. For two parallel sheets with equal charge density σ, the field between them is the difference of the two fields: σ/(2ε₀) - σ/(2ε₀) = 0.
6. Outside the sheets, the fields add up: σ/(2ε₀) + σ/(2ε₀) = σ/ε₀.

Answer: (a) A cylinder through the sheet has flux only through its two end faces: 2EA = σA/ε₀, so E = σ/(2ε₀), normal to the sheet. (b) With equal like charges: (i) between them the fields cancel, E = 0; (ii) outside, E = σ/ε₀.

(a) Derive the balance condition of a Wheatstone bridge. (b) Find the resistance between A and B of the network in the figure.

Uses a figure printed in CBSE's paper — what the figure shows: A to M: 2R. Between M and N: M–O is R, M–P (upper branch) is R, O–P is R, O–N (lower branch) is R, P–N is R. N to B: 3R.

1. Take a bridge with R1 and R2 in series in one branch, R3 and R4 in series in the other, the battery across the two ends and the galvanometer joining the R1–R2 junction to the R3–R4 junction.
2. At balance the galvanometer current is zero, so the same current I1 flows through R1 and R2, and the same current I2 through R3 and R4.
3. Loop rule on the loop with R1, galvanometer and R3: I1R1 + 0 − I2R3 = 0, so I1R1 = I2R3. On the loop with R2, galvanometer and R4: I1R2 = I2R4.
4. Dividing the second result by the first gives R2/R1 = R4/R3, the balance condition.
5. In the network, M–O, O–N, M–P and P–N are all R, so R(MO)/R(ON) = R(MP)/R(PN): the bridge is balanced and the O–P resistor carries no current.
6. So between M and N there are two paths of 2R each in parallel: R_MN = (2R × 2R)/(4R) = R.
7. Series total: R_AB = 2R + R + 3R = 6R.

Answer: (a) With zero galvanometer current, Kirchhoff's loop rule on the two loops gives R2/R1 = R4/R3 (ratio of resistances in adjacent arms equal). (b) Between M and N the five R's form a balanced bridge (the O–P resistor carries no current), so R_MN = R; R_AB = 2R + R + 3R = 6R.

A capacitor C with a dielectric slab is charged to V, then the battery is removed and the slab pulled out. Effect on (a) capacitance, (b) stored energy, (c) voltage? Justify.

1. Since the battery is removed, the charge Q on the capacitor remains constant.
2. Removing the dielectric slab reduces the capacitance from C to C/K.
3. The stored energy is given by U = Q²/2C; substituting the new capacitance gives U' = Q²/2(C/K) = K(Q²/2C), so the energy increases by a factor of K.
4. The voltage is given by V = Q/C; substituting the new capacitance gives V' = Q/(C/K) = K(Q/C) = KV, so the voltage increases by a factor of K.

Answer: Charge Q stays fixed. (a) Capacitance falls to C/K. (b) Energy Q²/2C rises K times. (c) Voltage Q/C rises to KV.

Electrons enter midway between two horizontal plates 2 cm apart at 3 × 10⁷ m/s. Plates are 3 cm long. What plate voltage makes the beam just hit the far end of the lower plate?

Uses a figure printed in CBSE's paper — what the figure shows: Beam enters along the central line between upper plate P1P1' and lower plate P2P2'; it bends down to strike the end P2' of the lower plate.

1. The electron enters midway between the plates, so the vertical deflection required to hit the lower plate is half the plate separation: y = 2 cm / 2 = 1 cm = 0.01 m.
2. The time taken to travel the length of the plates is t = L / u, where L = 3 cm = 0.03 m and u = 3 × 10⁷ m/s.
3. The vertical acceleration is a = eE / m = eV / (md), where d = 2 cm = 0.02 m is the plate separation.
4. Using the kinematic equation for vertical displacement y = ½ a t², we substitute a and t to get y = ½ (eV / md) (L / u)².
5. Rearranging for voltage V gives V = 2 y m d u² / (e L²).
6. Substituting the values: V = 2 × 0.01 × 9.1 × 10⁻³¹ × 0.02 × (3 × 10⁷)² / (1.6 × 10⁻¹⁹ × (0.03)²).
7. Calculating the result: V = 2275 V.

Answer: Required deflection y = 1 cm in time x/u; y = ½(eV/md)(x/u)², so V = 2ymdu²/(ex²) = 2275 V.

An ac input Vi = 12 sin(100πt) V acts across A and B of the diode–resistor network in the figure. In the positive half-cycle: (a) which diode conducts and why, (b) redraw the circuit, (c) voltage across each resistor at peak input.

Uses a figure printed in CBSE's paper — what the figure shows: Diamond A (top), P (left), R (right), B (bottom). D1 between P and A with anode at P (conducts P→A). D2 between A and R with anode at A (conducts A→R). 1 kΩ between P and R, 2 kΩ between P and B, 3 kΩ between R and B. Source Vi across A and B.

1. (a) In the positive half-cycle, terminal A is positive relative to B. Diode D2 has its anode at A and cathode at R, so it is forward biased and conducts. Diode D1 has its anode at P and cathode at A; since A is at a higher potential than P, D1 is reverse biased and acts as an open circuit.
2. (b) With D1 open and D2 shorted, the circuit consists of the source Vi connected to node R. From node R, the current splits into two parallel branches connected to node B: one branch is the 3 kΩ resistor, and the other branch is the series combination of the 1 kΩ and 2 kΩ resistors.
3. (c) The peak input voltage is 12 V. The 3 kΩ resistor is connected directly across the source, so the voltage across it is V(RB) = 12 V. The series branch (1 kΩ + 2 kΩ = 3 kΩ) is also connected across the source, so the total voltage across this branch is 12 V.
4. Using the voltage divider rule for the series branch, the voltage across the 1 kΩ resistor is V(RP) = 12 × (1 / (1 + 2)) = 4 V. The voltage across the 2 kΩ resistor is V(PB) = 12 × (2 / (1 + 2)) = 8 V.

Answer: (a) D2 conducts: A is positive and D2 points from A to R, so it is forward biased; D1 is reverse biased. (b) Current goes A → D2 → R, then splits: R → 3 kΩ → B, and R → 1 kΩ → P → 2 kΩ → B. (c) Both branches carry 12 V: V(RP) = 4 V, V(PB) = 8 V, V(RB) = 12 V.

Note: CBSE's marking scheme also accepts the reverse reading (D1 conducting): 12 V across 2 kΩ, 3 V across 1 kΩ and 9 V across 3 kΩ.

Briefly explain the two key processes in forming a p-n junction.

1. Diffusion occurs because of the concentration difference, causing holes to move from the p-side to the n-side and electrons from the n-side to the p-side.
2. This movement creates a diffusion current and leaves behind charged ions, forming a depletion layer.
3. Drift occurs because the electric field set up by these ions (positive on the n-side, negative on the p-side) pushes minority carriers back across the junction.
4. This creates a drift current, and equilibrium is reached when the magnitude of the diffusion current equals the drift current.

Answer: Diffusion: due to the concentration difference, holes move from p to n and electrons from n to p, giving a diffusion current and leaving charged ions (depletion layer). Drift: the field set up by these ions (n side +, p side −) pushes minority carriers back across, giving a drift current; equilibrium comes when the two are equal.

(a) Ray diagram of a refracting telescope and its angular magnification in normal adjustment. (b) Two reasons a reflecting telescope is preferred.

1. Draw the ray diagram: parallel rays from a distant object enter the objective lens and converge to form a real, inverted image at its focal point.
2. Show that this real image coincides with the focal point of the eyepiece, causing the rays to emerge parallel, which means the final image is formed at infinity.
3. State that the angular magnification in normal adjustment is the ratio of the focal length of the objective to the focal length of the eyepiece, given by m = f₀/fₑ.
4. State that a reflecting telescope is preferred because it uses mirrors, which eliminates chromatic aberration.
5. State that a reflecting telescope is preferred because it uses a paraboloid mirror, which minimizes spherical aberration.

Answer: (a) Objective forms a real image at its focus, which is also the eyepiece focus, so the final image is at infinity; m = f₀/fₑ. (b) Any two: no chromatic aberration; little or no spherical aberration (paraboloid mirror); a mirror is lighter and easier to support; brighter image with large aperture; cheaper to make.

(a) Two conditions for total internal reflection. (b) Liquids A, B, C (indices n, 3n/4, 2n/3) are layered; a laser in A meets the A–B boundary at angle θ. Show it never enters C when sin θ ≥ 2/3.

Uses a figure printed in CBSE's paper — what the figure shows: Bottom layer A (index n), middle B (3n/4), top C (2n/3); the beam rises through A and meets the A–B boundary at angle θ to the normal.

1. For total internal reflection, light must travel from a denser to a rarer medium.
2. The angle of incidence must be greater than the critical angle.
3. Since the layers are parallel, the angle of refraction in B equals the angle of incidence at the B-C boundary.
4. Apply Snell's law across the parallel interfaces: n sin θ = (2n/3) sin r.
5. Solve for sin r: sin r = (3/2) sin θ.
6. Substitute sin θ = 2/3 to get sin r = 1, which means r = 90°.
7. For sin θ > 2/3, sin r > 1, so total internal reflection occurs and the ray never enters C.

Answer: (a) Light must go from a denser to a rarer medium, and the angle of incidence must exceed the critical angle. (b) The boundaries are parallel, so n sin θ = (3n/4) sin θ₁ = (2n/3) sin r. Hence sin r = (3/2) sin θ; at sin θ = 2/3, r = 90° (grazing), and for larger θ sin r would exceed 1, so the beam is totally reflected and never enters C.

Why does the torque on a galvanometer coil stay the same at every orientation?

1. Torque on the coil is τ = NIAB sin θ, where θ is the angle between B and the coil's normal.
2. The concave pole pieces and soft-iron core make the field radial, so the plane of the coil always lies along the field lines and θ = 90° at every position.
3. So sin θ = 1 always and τ = NIAB stays the same — option (B).

Answer: Because the field is radial, so the coil's plane is always along the field — option (B).

Best practical way to raise a galvanometer's current sensitivity?

1. The current sensitivity of a galvanometer is given by the formula S = NAB/k, where N is the number of turns, A is the area, B is the magnetic field, and k is the spring constant.
2. Increasing the number of turns N directly increases the sensitivity, making it the best practical method to raise the current sensitivity.

Answer: Increase the number of turns N (sensitivity = NAB/k) — option (A).

A 50-turn coil of area 4.0 × 10⁻³ m² in a 0.25 T field carries 5 A. Torque on it?

1. Use the formula for torque on a current-carrying coil: τ = NIAB sin θ.
2. Assume the maximum torque case where sin θ = 1, so τ = NIAB.
3. Substitute the given values: N = 50, I = 5 A, A = 4.0 × 10⁻³ m², B = 0.25 T.
4. Calculate the product: τ = 50 × 5 × 4.0 × 10⁻³ × 0.25 = 0.25 N m.

Answer: τ = NIAB = 0.25 N m — option (D).

A 15 Ω galvanometer gives full deflection at 3 mA. Series resistance needed to make a 0–12 V voltmeter?

1. Calculate the total resistance required for the voltmeter using R_total = V / Ig = 12 / 0.003 = 4000 Ω.
2. Subtract the galvanometer resistance to find the series resistance: R = R_total - G = 4000 - 15 = 3985 Ω.

Answer: R = V/Ig − G = 4000 − 15 = 3985 Ω in series — option (B).

A 20 Ω galvanometer gives full deflection at 5 mA. What shunt makes it a 0–10 A ammeter, and how is it connected?

1. Given G = 20 Ω, Ig = 5 mA = 0.005 A, and total current I = 10 A.
2. Current through the shunt is Is = I − Ig = 10 − 0.005 = 9.995 A.
3. Using the shunt formula S = IgG / (I − Ig), we get S = (0.005 × 20) / 9.995 ≈ 0.01 Ω.
4. The shunt is connected in parallel with the galvanometer to divert the excess current, making it an ammeter.

Answer: S = IgG/(I − Ig) ≈ 0.01 Ω, connected in parallel — option (C).

From the stopping-potential versus frequency graph, what are the work functions of metals A and B?

Uses a figure printed in CBSE's paper — what the figure shows: Two parallel straight lines of Vs against ν: A cuts the ν-axis at ν₁, B at ν₂ (ν₂ > ν₁); extended back, they cut the Vs-axis at −V₁ and −V₂.

1. Einstein's equation: eV_s = hν − φ, so V_s = (h/e)ν − φ/e, a straight line.
2. Where a line cuts the ν-axis, V_s = 0 and ν equals the threshold frequency, so φ = hν₀.
3. A cuts the axis at ν₁ and B at ν₂, so the work functions are hν₁ for A and hν₂ for B — option (C).

Answer: φ = h × threshold frequency: hν₁ for A and hν₂ for B — option (C).

For frequency ν > ν₂ falling on both metals, which gives electrons of higher maximum KE?

Uses a figure printed in CBSE's paper — what the figure shows: Same graph: threshold ν₁ for A is lower than ν₂ for B.

1. The maximum kinetic energy of photoelectrons is given by the equation KEmax = hν − φ, where φ is the work function.
2. Since the threshold frequency for metal A (ν₁) is lower than for metal B (ν₂), the work function of A is smaller than that of B (φA < φB).
3. For the same incident frequency ν, subtracting the smaller work function of A results in a higher KEmax, so metal A gives electrons of higher maximum KE.

Answer: Metal A, since its work function is smaller (KEmax = hν − φ) — option (A).

If the light intensity on both metals is doubled at fixed frequency, what changes?

1. The slope of the stopping potential versus frequency graph is h/e, which depends only on Planck's constant and electron charge, so it remains unchanged.
2. Doubling the light intensity increases the number of photons incident per second, leading to more photoelectrons emitted per second without changing their maximum kinetic energy.

Answer: Slopes (h/e) stay the same; only more electrons are emitted per second — option (D).

Threshold frequency is ν₀. Max KE is E1 at frequency 3ν₀ and E2 at 6ν₀. Find E1/E2.

1. Using the photoelectric equation K.E. = hν - hν₀, for frequency 3ν₀, E1 = h(3ν₀) - hν₀ = 2hν₀.
2. For frequency 6ν₀, E2 = h(6ν₀) - hν₀ = 5hν₀.
3. Calculating the ratio E1/E2 = 2hν₀ / 5hν₀ = 2/5.

Answer: E1 = 2hν₀, E2 = 5hν₀, so E1/E2 = 2/5 — option (C).

If m is the slope of metal B's Vs–ν line, express h in terms of m and e.

1. The photoelectric equation is hν = φ + eVs, which rearranges to Vs = (h/e)ν − φ/e.
2. Comparing this with the straight-line equation y = mx + c, the slope m equals h/e.
3. Therefore, h = me.

Answer: Vs = (h/e)ν − φ/e, so slope m = h/e and h = me — option (A).

(a) Charges q and −q are 2a apart. Derive the field at distance r on the equatorial plane, and its form for r ≫ a. (b) q sits at x = a and −q at x = b; field E = 2î N/C. Find the force and torque on the dipole.

1. The distance from each charge to the equatorial point is R = √(r² + a²), and the field magnitude from each charge is E₁ = q / [4πε₀(r² + a²)].
2. The perpendicular components of the fields cancel, while the axial components add, giving a net field E = 2E₁(a/R).
3. Substituting E₁ and R yields E = 2qa / [4πε₀(r² + a²)^(3/2)], which is opposite to the dipole moment p = 2qa.
4. For r ≫ a, the term (r² + a²)^(3/2) approximates to r³, resulting in the far-field form E = -p / (4πε₀r³).
5. In part (b), the net force on the dipole is F = qE - qE = 0 because the external field is uniform.
6. The dipole moment p lies along the x-axis, parallel to the field E, so the torque τ = p × E = 0.

Answer: (a) Each charge gives q/[4πε₀(r² + a²)]; perpendicular parts cancel, axial parts add: E = −p/[4πε₀(r² + a²)^(3/2)] with p = 2qa (opposite to p). For r ≫ a, E = −p/(4πε₀r³). (b) Net force = q(2î) − q(2î) = 0; p lies along the x-axis, parallel/antiparallel to E, so torque p × E = 0.

(a) Cells (E1, r1) and (E2, r2) are joined in parallel, like terminals together. Derive the equivalent emf and internal resistance. (b) Such a pair with emfs E and 3E, each of internal resistance R, feeds a 2R resistor. Current in 2R?

1. Let the common terminal voltage be V and the total current I = I1 + I2. For each cell V = E1 − I1r1 = E2 − I2r2.
2. So I1 = (E1 − V)/r1 and I2 = (E2 − V)/r2; adding, I = (E1/r1 + E2/r2) − V(1/r1 + 1/r2).
3. Solving for V: V = (E1r2 + E2r1)/(r1 + r2) − I × r1r2/(r1 + r2).
4. Comparing with V = E_eq − I r_eq: E_eq = (E1r2 + E2r1)/(r1 + r2) and r_eq = r1r2/(r1 + r2).
5. For (b), E1 = E, E2 = 3E, r1 = r2 = R: E_eq = (E × R + 3E × R)/(2R) = 2E and r_eq = R²/(2R) = R/2.
6. Current in 2R: I = E_eq/(2R + r_eq) = 2E/(2R + R/2) = 2E/(5R/2) = 4E/(5R).

Answer: (a) E_eq = (E1r2 + E2r1)/(r1 + r2), r_eq = r1r2/(r1 + r2). (b) E_eq = (E·R + 3E·R)/2R = 2E, r_eq = R/2, so I = 2E/(2R + R/2) = 4E/(5R).

(a) Derive the lens maker's formula from refraction at a spherical surface. (b) Lenses L1, L2, L3 (each f = 40 cm) are coaxial; L1–L2 gap 120 cm, L2–L3 gap 20 cm. Object 80 cm left of L1. How far is the final image from the object?

1. At the first surface (radius R1), an object O in medium n1 gives an image I1 in glass: n1/OB + n2/BI1 = (n2 − n1)/BC1.
2. I1 acts as the object for the second surface (radius R2), which forms the final image I in medium n1: −n2/DI1 + n1/DI = (n2 − n1)/DC2.
3. For a thin lens BI1 ≈ DI1, so adding the two equations cancels the n2 terms: n1/OB + n1/DI = (n2 − n1)(1/BC1 + 1/DC2).
4. With the sign convention OB = −u, DI = +v, BC1 = R1, DC2 = −R2, and n21 = n2/n1, this gives 1/v − 1/u = (n21 − 1)(1/R1 − 1/R2); for u = −∞, v = f, so this equals 1/f.
5. (b) L1: 1/v = 1/40 + 1/(−80) = 1/80, so v = 80 cm. This image is 120 − 80 = 40 cm before L2, at its focus, so light leaves L2 parallel.
6. Parallel light entering L3 is focused at its focal point, 40 cm beyond L3.
7. Distance from the object to the final image = 80 + 120 + 20 + 40 = 260 cm.

Answer: (a) Applying n1/OB + n2/BI1 = (n2 − n1)/BC1 at the first surface and the same at the second (with I1 as virtual object), then adding for a thin lens gives 1/v − 1/u = (n21 − 1)(1/R1 − 1/R2) = 1/f. (b) L1: u = −80 cm gives v = 80 cm; this image is 40 cm before L2, i.e. at its focus, so light leaves L2 parallel and L3 focuses it 40 cm beyond L3. Distance from object = 80 + 120 + 20 + 40 = 260 cm.

Note: The paper does not say what kind of lenses these are; like CBSE's marking scheme, take all three as convex (f = +40 cm).

(a) Ray diagram for a concave mirror with the object between F and C; derive the mirror formula from it. (b) A concave mirror gives a virtual image twice the size of an object 10 cm in front. Focal length?

1. Ray diagram: object AB stands on the axis at B, between F and C. A ray from A parallel to the axis reflects through F; a ray from A through F reflects parallel. They meet beyond C, forming a real, inverted, enlarged image A'B'.
2. Triangles A'B'F and MPF are similar (M is where the parallel ray hits the mirror, PM ≈ AB for a small mirror): B'A'/BA = B'F/FP.
3. Triangles A'B'P and ABP are similar: B'A'/BA = B'P/BP. So B'F/FP = B'P/BP, i.e. (B'P − FP)/FP = B'P/BP.
4. With signs, B'P = −v, FP = −f, BP = −u: (−v + f)/(−f) = v/u, so v/f − 1 = v/u. Dividing by v gives 1/v + 1/u = 1/f.
5. (b) The image is virtual and erect, so m = −v/u = +2. With u = −10 cm, v = +20 cm (behind the mirror).
6. 1/f = 1/v + 1/u = 1/20 − 1/10 = −1/20, so f = −20 cm: the concave mirror has focal length 20 cm.

Answer: (a) Similar triangles in the ray diagram give B'A'/BA = A'P/AP and B'A'/BA = A'F/FP; equating and using u, v, f with sign convention gives 1/v + 1/u = 1/f. (b) m = −v/u = 2 with u = −10 cm, so v = +20 cm; 1/f = 1/20 − 1/10, f = -20 cm (focal length 20 cm, concave).

(a) State Faraday's law of induction. (b) Derive the self-inductance of a long air-core solenoid (length l, area A, N turns). (c) A 50 cm rod pivoted at one end spins at 60 rpm in a 4.0 mT field normal to its rotation plane. Induced emf?

1. Faraday's law states that the induced emf is equal to the negative rate of change of magnetic flux linkage, ε = −dΦ/dt.
2. For a long solenoid, the magnetic field is B = μ₀nI, where n = N/l is the number of turns per unit length.
3. The total magnetic flux linkage is N times the flux through one turn, which is N(BA) = N(μ₀nI)A = μ₀n²AlI.
4. Self-inductance L is defined as flux linkage per unit current, so L = μ₀n²Al = μ₀N²A/l.
5. Convert the rotation speed to angular velocity: ω = 2π × (60/60) = 2π rad/s.
6. The induced emf in a rotating rod is given by ε = ½Bωl².
7. Substitute the values: ε = ½ × 4.0 × 10⁻³ × 2π × (0.5)² = 3.14 × 10⁻³ V ≈ 3.14 mV.

Answer: (a) Induced emf equals the rate of change of magnetic flux linkage, ε = −dΦ/dt. (b) B = μ₀nI with n = N/l; flux linkage NBA = μ₀n²AlI, so L = μ₀n²Al = μ₀N²A/l. (c) ω = 2π rad/s; ε = ½Bωl² = ½ × 4×10⁻³ × 2π × 0.5² ≈ 3.14 mV.

(a) Draw and label a step-up transformer, give its working principle, and Vs/Vp in terms of turns and currents. (b) Ideal transformer, turns ratio primary : secondary = 1 : 5, input 5 kW at 200 V. Find (i) primary current, (ii) output voltage.

1. Draw a diagram of a step-up transformer with a laminated iron core, a primary coil with Np turns, and a secondary coil with Ns turns.
2. State that the working principle is mutual induction, where alternating current in the primary coil produces a changing magnetic flux in the core.
3. Explain that this changing flux links with the secondary coil, inducing an alternating emf in it according to Faraday's law of electromagnetic induction.
4. Write the transformer equation for an ideal transformer: Vs/Vp = Ns/Np = Ip/Is.
5. For part (b), calculate the primary current using the power formula: Ip = P/Vp = 5000 W / 200 V = 25 A.
6. Calculate the output voltage using the turns ratio: Vs/Vp = Ns/Np = 5/1, so Vs = 5 × 200 V = 1000 V.

Answer: (a) Works on mutual induction; the same flux links both coils so Vs/Vp = Ns/Np = Ip/Is (ideal). (b) Ip = P/Vp = 25 A; output voltage 1000 V.