CBSE Class 12 Physics Sample Paper 2: marking scheme and solutions

Answer: (a) In the Huygens construction a crest from A reaches E as the crest from B reaches C, so v₁/λ₁ = v₂/λ₂: the frequency is unchanged, while speed and wavelength both increase in air and an oblique ray bends.

Answer: (c) ε₀μ₀ = (4πε₀)(μ₀/4π) = [1/(9 × 10⁹)] × 10⁻⁷ = 1/(3 × 10⁸)² = 1/c².

Answer: (b) The reverse current is a drift of minority carriers, only a few μA, so a microammeter is used; a milliammeter suits the forward current, which is in mA.

Answer: (b) ρ = 1/σ = 1/(2.0 × 10⁻³ S m⁻¹) = 500 Ω m, which lies in the semiconductor range 10⁻⁵ to 10⁶ Ω m; insulators start near 10¹¹ Ω m. Option (a) takes σ for ρ.

Answer: (b) E_b = ΔMc², so ΔM = 28.3 MeV ÷ 931.5 MeV/u = 0.0304 u. 0.0076 u is the mass defect per nucleon (÷ 4); 0.0152 u divides by 2; 0.304 u slips a decimal.

Answer: (d) M is the net magnetic moment per unit volume: (A m²)/m³ = A m⁻¹, the same unit as H.

Answer: (b) Each charge gives kQ/a² along its own side, the two fields 60° apart. Resultant = 2(kQ/a²)cos 30° = √3 kQ/a², along the bisector, away from the two charges.

Answer: (d) From R = ρl/A, ρ = RA/l, with unit Ω m² / m = Ω m. S m⁻¹ is the unit of conductivity.

Answer: (d) Energy supplied by a very strong electric field, of the order of 10⁸ V m⁻¹, gives field emission. Thermionic emission needs heating; photoelectric emission needs light of suitable frequency.

Answer: (a) For a pure inductor i = i_m sin(ωt − π/2): the current reaches its maximum a quarter period after the voltage, so it lags by π/2. A lead of π/2 is the capacitor's case; in phase is the resistor's.

Answer: (d) Infrared, of lower frequency than visible light, vibrates not only electrons but whole atoms and molecules (water in particular absorbs it readily), raising their internal energy and temperature. All electromagnetic waves travel at the same speed in vacuum; any hot body emits infrared.

Answer: (d) f = −10 cm. At u = −30 cm, 1/v = −1/10 + 1/30 = −1/15, v = −15 cm. At u = −20 cm, 1/v = −1/10 + 1/20 = −1/20, v = −20 cm. The image moves from 15 cm to 20 cm in front of the mirror: 5 cm away from it, as the object moves 10 cm towards it.

Answer: (b) Both are true, but the near-constancy comes from the short range of the nuclear force (saturation): an interior nucleon feels only its near neighbours, however large the nucleus. Charge independence does not explain it.

Answer: (a) Both true. Driving the current against the cell's emf adds the drop Ir to ε, so the terminal voltage V = ε + Ir exceeds ε; R explains A.

Answer: (a) In air (n₂/n₁ − 1) = 0.5 > 0 and the convex shape gives f > 0. In the liquid, 1.5/1.7 − 1 < 0 while (1/R₁ − 1/R₂) is still positive, so f < 0: the lens diverges. R is the relation that makes A true.

Answer: (a) Both are true. Because of the π/2 lag the product vi is a sin 2ωt term, whose average over a cycle is zero: R is exactly why A holds.

In the wave picture the energy of the radiation is spread continuously over the whole wavefront and shared by a very large number of electrons, so each absorbs only a little per unit time; calculations estimate that an electron could take hours or more to gather the work function. Observed: above the threshold, emission starts instantaneously, in about 10⁻⁹ s or less, even for very dim light.

• wave picture: energy shared over the wavefront, time lag of hours or more (1)
• observed: no time lag, ~10⁻⁹ s or less (1)

In the wave picture a greater intensity means a greater amplitude of the electric and magnetic fields, so each electron absorbs more energy, and Kmax should increase with intensity. Observed: for a given frequency, the stopping potential, and so Kmax, is independent of intensity; only the photocurrent rises.

• wave picture: Kmax should rise with intensity (1)
• observed: Kmax independent of intensity (1)

I = 4I₀ cos²(φ/2). (i) φ = π/2: I = 4I₀ cos²(π/4) = 4I₀ × 1/2 = 2I₀. (ii) φ = 5π/3: I = 4I₀ cos²(5π/6) = 4I₀ × 3/4 = 3I₀.

• I = 4I₀cos²(φ/2); (i) 2I₀ (1)
• (ii) 3I₀ (1)

β = λD/d = (600 × 10⁻⁹ × 1.5)/(0.30 × 10⁻³) = 3.0 mm. Moving one fringe width changes the path difference by λ, so the phase difference at distance x is φ = 2πx/β, and I = 4I₀ cos²(φ/2). (i) x = 0.50 mm: φ = 2π/6 = π/3, I = 4I₀ cos²(π/6) = 4I₀ × 3/4 = 3I₀. (ii) x = 4.5 mm = 1.5β: φ = 3π, I = 4I₀ cos²(3π/2) = 0, a dark point (the second dark fringe).

• β = 3.0 mm, φ = 2πx/β; (i) φ = π/3, I = 3I₀ (1)
• (ii) φ = 3π, I = 0 (1)

r = mv/qB, so at the same v and B, r ∝ m/q. r_α/r_p = (4m/2e)/(m/e) = 2.

• r = mv/qB, so r ∝ m/q at the same v and B (1)
• ratio r_α : r_p = 2 : 1 (1)

The charged comb produces a field that polarises the paper: it induces a dipole moment in the paper along the field. The comb's field is non-uniform, stronger nearer the comb, so the induced dipole feels a net force towards the stronger field, and the paper moves towards the comb.

• the comb's field polarises the paper, inducing a dipole along the field (1)
• field is non-uniform, so a net force acts towards the comb (1)

In the half-cycle when the diode is reverse biased it does not conduct, and the whole ac voltage of that half-cycle, up to its peak, appears across it. If the peak exceeded the breakdown voltage, the reverse current would rise sharply and the diode would conduct in the wrong half-cycle (and could be destroyed), spoiling the rectification.

• in the reverse half-cycle the full peak voltage appears across the diode (1)
• above breakdown it conducts sharply/is damaged, so rectification fails (1)

(a) Within each domain the atomic dipoles are aligned, but the magnetisation varies randomly from domain to domain, so the moments cancel and there is no bulk magnetisation. (b) In the magnet's field the domains turn towards the field and those already along it grow, so the field lines are highly concentrated; in the non-uniform field it moves towards high field, that is, it is attracted. (c) The nail is soft iron, a soft ferromagnet whose magnetisation disappears when the field is removed; lodestone is a hard ferromagnet whose magnetisation persists.

• domains randomly oriented, so moments cancel (1)
• domains align and grow in the field; attracted towards high field (1)
• soft ferromagnet loses magnetisation; lodestone is a hard ferromagnet (1)

(a) Each domain (about 1 mm, about 10¹¹ atoms) is magnetised, but the directions vary randomly from domain to domain: no bulk magnetisation. (b) The domains orient along B₀ and those along B₀ grow, until they amalgamate into a single giant domain; the field lines are highly concentrated. (c) In a hard ferromagnet (Alnico, lodestone) the alignment and magnetisation persist: a permanent magnet. In a soft ferromagnet (soft iron) the magnetisation disappears.

• random domains, no net M (1)
• domains align along B₀ and grow into a giant domain (1)
• hard keeps, soft loses its magnetisation (1)

(a) R_X = 10.0[1 + 5.0 × 10⁻³(T − 20)] = 15.0 Ω gives T − 20 = 100, so T = 120 °C. Y stays at 10.0 Ω because manganin's resistivity depends very little on temperature. (b) 15.0 × 10.0/25.0 = 6.0 Ω.

• R_T = R₀[1 + α(T − T₀)] set to 15.0 Ω (1)
• T = 120 °C (1)
• parallel resistance 6.0 Ω (1)

(a) Closing the switch makes the coil's current, and the flux through the ring, rise sharply (the iron core makes the flux large). By Lenz's law the induced current in the ring opposes this rise: its field is opposite to the coil's, so the ring is repelled and jumps. (b) Once the current is steady the flux stops changing, the induced current dies away, there is no repulsion and gravity pulls the ring back. (c) With a cut the ring is an open circuit: an emf appears across the cut but no current flows, so there is no force and the ring does not jump.

• rising flux induces a current opposing it, so the ring is repelled (1)
• steady current, no change of flux, no induced current (1)
• cut ring: emf only, no current, so no jump (1)

(a) Nₐ(Y) = 10 Nₐ(X), so nₕ(Y) = 10 nₕ(X). Since nₑnₕ = nᵢ² at a fixed temperature, nₑ = nᵢ²/Nₐ, so nₕ/nₑ = Nₐ²/nᵢ² ∝ Nₐ². Ten times the doping makes nₕ/nₑ larger by 10² = 100. (b) nₑ ∝ 1/Nₐ, so nₑ(Y)/nₑ(X) = 1/10. (c) The abundant majority holes give each thermally produced electron more chance of meeting a hole and recombining, so the minority concentration falls as the majority rises (nₑnₕ = nᵢ² stays fixed).

• nₕ ∝ Nₐ and nₑ = nᵢ²/Nₐ ∝ 1/Nₐ, so nₕ/nₑ ∝ Nₐ²: factor 100 (1)
• ratio 1/10 (1)
• more majority carriers, more recombination of minority carriers (1)

With σ = Q/A, each plate gives a field σ/2ε₀. Outside the plates the two fields are opposite: E = σ/2ε₀ − σ/2ε₀ = 0. Between the plates they add: E = σ/ε₀ = Q/ε₀A, uniform, from the positive to the negative plate. Then V = Ed = Qd/ε₀A, and C = Q/V = ε₀A/d.

• outside E = 0 (fields cancel) (1)
• inside E = σ/ε₀ = Q/ε₀A (1)
• V = Qd/ε₀A, so C = ε₀A/d (1)

Take a circular loop of radius r coaxial with the plates. No conduction current crosses it; the displacement current through it is the fraction r²/R² of the total i, since E is uniform over the plate area. So B(2πr) = μ₀ i r²/R², B = μ₀ i r/(2πR²). B is zero on the axis and rises linearly with r to μ₀i/(2πR) at the rim, equal to the field just outside the plates.

• displacement current through the loop = i r²/R² (1)
• B = μ₀ i r/(2πR²) (1)
• zero on the axis, linear rise to μ₀i/(2πR) at the rim (1)

(a) The gold nucleus, much heavier than the α-particle, is taken as stationary. Mechanical energy is conserved: at the turning point the α-particle is momentarily at rest, so all of K has become electric potential energy: K = (1/4πε₀)(2e)(Ze)/d, giving d = 2Ze²/(4πε₀K). (b) Setting d = 6 × 10⁻¹⁵ m: K = 9 × 10⁹ × 2 × 79 × (1.6 × 10⁻¹⁹)²/(6 × 10⁻¹⁵) J = 6.07 × 10⁻¹² J ≈ 37.9 MeV, about 38 MeV, roughly five times the energy of the α-particles that showed the nucleus to be smaller than 30 fm.

• nucleus stationary; K equals potential energy at turning point (1)
• d = 2Ze²/(4πε₀K) (1)
• K ≈ 38 MeV (1)

(a) A polaroid is a sheet whose long-chain molecules are aligned in one direction; it absorbs the electric vector along them and passes the component along the pass-axis, perpendicular to the molecules. Light through one polaroid P₁ is halved and does not change as P₁ turns; viewed through a second polaroid P₂, rotating one of them makes the light nearly vanish at one position and reach full brightness 90° from it. This happens only if the electric field is at right angles to the propagation: light is transverse. (b) After P₁: 32 W m⁻². After P₂: 32 cos²30° = 24 W m⁻². After P₃: 24 cos²30° = 18 W m⁻². Without P₂: 32 cos²60° = 8 W m⁻². Inserting a polaroid between the other two increases the light that gets through, because P₂ passes a component that has a larger projection on P₃'s pass-axis.

• polaroid: aligned molecules, passes E along the pass-axis (1)
• two-polaroid experiment: zero and maximum 90° apart, so transverse (1)
• 32 → 24 W m⁻² after P₂ (1)
• 18 W m⁻² after P₃ (1)
• without P₂: 8 W m⁻², so inserting P₂ increases the output (1)

(a) Let the electric field amplitude after P₁ be E, along P₁'s pass-axis. P₂ passes only its component along its own pass-axis, E cos θ. Intensity is proportional to the square of the amplitude, so I = I₀ cos²θ (Malus' law). With unpolarised light, P₁ passes 50%; P₂ then reduces it from that 50% (θ = 0) to zero (θ = 90°). (b) (i) After P₁: 24 W m⁻². One-eighth of 48 is 6 W m⁻², so cos²θ = 6/24 = 1/4, cos θ = 1/2, θ = 60°. (ii) P₃ is at 60° to P₂: 6 cos²60° = 6 × 1/4 = 1.5 W m⁻². (iii) No. Light from P₁ is already polarised along P₁'s pass-axis, so a polaroid parallel to P₁ passes all 24 W m⁻² and P₂ still gives 24 cos²60° = 6 W m⁻².

• component E cos θ passes; I ∝ amplitude², so I = I₀cos²θ (1)
• limits 50% to zero of the incident intensity (1)
• 24 W m⁻² after P₁; cos²θ = 1/4, θ = 60° (1)
• 1.5 W m⁻² from P₃ (1)
• no change: light from P₁ already polarised along P₁, so 6 W m⁻² (1)

(a) Due to B: F_B = 9 × 10⁹ × 2 × 10⁻⁶ × 3 × 10⁻⁶/(0.30)² = 0.60 N, repulsive, along BA (away from B). Due to C: F_C = 9 × 10⁹ × 2 × 10⁻⁶ × 8 × 10⁻⁶/(0.40)² = 0.90 N, attractive, along AC (towards C). (b) The two forces are at right angles: F = √(0.60² + 0.90²) = 1.08 N ≈ 1.1 N, at tan⁻¹(0.90/0.60) = 56° to the direction BA (extended beyond A), on the side of C. (c) E = F/q = 1.08/(2 × 10⁻⁶) = 5.4 × 10⁵ N C⁻¹, in the direction of the net force, since the charge at A is positive.

• F_B = 0.60 N along BA (repulsion) (1)
• F_C = 0.90 N along AC (attraction) (1)
• vector sum: F = 1.08 N (1)
• direction: 56° from BA, towards the side of C (1)
• E = F/q = 5.4 × 10⁵ N C⁻¹ along the net force (1)

(a) B and D are 0.30 m from C: each force = 9 × 10⁹ × 2 × 10⁻⁶ × 1 × 10⁻⁶/(0.30)² = 0.20 N, attractive, along CB and CD. A is 0.30√2 m away: force = 9 × 10⁹ × 2 × 10⁻¹²/0.18 = 0.10 N, along CA. (b) The forces along CB and CD are perpendicular and equal; their resultant is 0.20√2 = 0.283 N along CA. Adding the 0.10 N along CA: F = 0.38 N, along the diagonal CA, towards A. (c) E = F/q in magnitude = 0.383/(1 × 10⁻⁶) = 3.8 × 10⁵ N C⁻¹; since the test charge is negative, E points opposite to the force, along AC, away from A.

• 0.20 N each due to B and D, towards them (1)
• 0.10 N due to A, at distance 0.30√2 m (1)
• B and D combine to 0.28 N along CA (1)
• net 0.38 N along CA, towards A (1)
• E = 3.8 × 10⁵ N C⁻¹, along AC, away from A (1)

(a) The same current i = i_m sin(ωt + φ) flows through R, L and C. Take I as reference: V_R (length i_mR) along I, V_L (i_mX_L) π/2 ahead of I, V_C (i_mX_C) π/2 behind I. V_L and V_C are opposite, combining to i_m(X_L − X_C) along V_L. The source phasor V is the hypotenuse of the right triangle with sides V_R and (V_L + V_C): v_m² = (i_mR)² + [i_m(X_L − X_C)]², so i_m = v_m/Z with Z = √[R² + (X_C − X_L)²]. The angle between V and I gives tan φ = (X_C − X_L)/R. (b) Z = √[50² + (150 − 30)²] = √(2500 + 14 400) = 130 Ω, I = 130/130 = 1.0 A, cos φ = R/Z = 50/130 ≈ 0.38 (the current lags, since X_L > X_C).

• phasor diagram: V_R along I, V_L ahead and V_C behind by π/2 (1)
• v_m² = (i_mR)² + [i_m(X_L − X_C)]² (1)
• Z = √[R² + (X_C − X_L)²] and tan φ = (X_C − X_L)/R (1)
• Z = 130 Ω, I = 1.0 A (1)
• cos φ ≈ 0.38 (1)

(a) p = vi = v_m i_m sin ωt sin(ωt + φ) = (v_m i_m/2)[cos φ − cos(2ωt + φ)]. Over a cycle the average of cos(2ωt + φ) is zero, so P = (v_m i_m/2) cos φ = (v_m/√2)(i_m/√2) cos φ = VI cos φ. (b) Z = √[50² + (160 − 40)²] = 130 Ω, I = 260/130 = 2.0 A, P = I²R = 4.0 × 50 = 200 W. X_C > X_L, so φ > 0 and the current leads.

• p = vi expanded to (v_m i_m/2)[cos φ − cos(2ωt + φ)] (1)
• average of cos(2ωt + φ) is zero (1)
• P = (v_m i_m/2) cos φ = VI cos φ (1)
• Z = 130 Ω, I = 2.0 A, P = 200 W (1)
• current leads (1)