CBSE Class 12 Physics Sample Paper 1: marking scheme and solutions

Answer: (c) A p-type dopant must be trivalent (third group) so that it leaves a hole in one bond; In is trivalent. Sb, As and P are pentavalent donors and give n-type material.

Answer: (a) V = (1/4πε₀) q/r = 9 × 10⁹ × 5.0 × 10⁻⁹ / 0.15 = 45/0.15 = 300 V. Dividing by r² gives 2000 V (the field's form, not the potential's); leaving out r gives 45 V; taking r as 1.5 cm gives 3000 V.

Answer: (a) The magnetic force is q(v × B), of magnitude qvB sin θ. It vanishes when θ = 0° or 180°, that is, when v is parallel or anti-parallel to B.

Answer: (a) Manganin, like nichrome and constantan, has a resistivity that depends very little on temperature, so it is used for wire-bound standard resistors. Copper and aluminium change appreciably; silicon is a semiconductor.

Answer: (b) L = μ_r μ₀ n² A l, so L is multiplied by μ_r: 400 × 2.0 × 10⁻³ H = 0.80 H. 0.0050 H divides by μ_r; 0.040 H uses √μ_r; 320 H uses μ_r².

Answer: (a) Volume ∝ r³, so the fraction is (10⁻⁴)³ = 10⁻¹²: the atom is almost empty, though the nucleus holds more than 99.9% of its mass.

Answer: (b) p̄ = ½ i_m² R = ½ × (2.0)² × 50 = 100 W. Using i_m² R without the ½ gives 200 W; taking I = i_m/2 gives 50 W; the average current is zero but i² is never negative, so the power is not zero.

Answer: (b) Tanning is the skin's response to ultraviolet: exposure induces more melanin. Ordinary glass absorbs ultraviolet, so the radiation that would tan her does not reach her; visible light clearly passes, since the room is sunlit.

Answer: (c) Reverse bias acts in the same direction as the barrier, so the barrier becomes V₀ + V = 0.60 + 2.0 = 2.6 V and the depletion region widens. (b) is the forward-bias result V₀ − V.

Answer: (d) The two forces cancel when 2/x² = 8/(12 − x)², i.e. (12 − x)/x = 2, so x = 4 cm, nearer the smaller charge. 2.4 cm comes from dividing in the ratio of the charges instead of the square root of that ratio.

Answer: (d) For a projectile of charge q, K = (1/4πε₀)(q)(Ze)/d, so d = qZe/(4πε₀K) ∝ q at the same K and Z. dα/dp = 2e/e = 2. The mass of the projectile does not enter, since K is given; 4 squares the charge ratio and 1/2 inverts it.

Answer: (a) n21 is a property of the pair of media and depends on the wavelength of light, not on the angle of incidence. Because it differs for different colours, white light splits into its colours (dispersion).

Answer: (b) Both statements are true. A mirror forms images by reflection, which does not depend on wavelength, so it has no chromatic aberration. Support over the back surface is a separate advantage, about weight and mechanical strength; it does not explain the absence of chromatic aberration.

Answer: (a) Both are true. By Lenz's law the induced current opposes the rising flux, so the near face becomes a north pole and repels the magnet; the person pushing does work against this force, and that work appears as Joule heat. R is the reason for A.

Answer: (b) Both are true. The crowding follows from E_n ∝ 1/n²: the gaps E_n+1 − E_n shrink as n grows. The sign of the energy only says the electron is bound; a negative energy alone does not fix how the levels are spaced, so R does not explain A.

Answer: (b) Both are true statements from the chapter. A rests on ultraviolet's ability to destroy germs; the absorption of ultraviolet by glass is a separate property and does not explain A.

Each piece has a quarter of the length and the same cross-section, so R ∝ l gives 32/4 = 8.0 Ω each. Four equal resistors in parallel: R = 8.0/4 = 2.0 Ω.

• resistance of each piece, 8.0 Ω, from R ∝ l (1)
• four in parallel: 2.0 Ω (1)

R = ρl/A with A = πd²/4, so R ∝ l/d². R_A/R_B = (l/d²)/(3l/4d²) = 4/3, so R_A : R_B = 4 : 3.

• R ∝ l/d² from R = ρl/A (1)
• R_A : R_B = 4 : 3 (1)

Required power P = 1/f = 1/(0.50 m) = +2 D. Powers in contact add: P = P1 + P2, so P2 = 2 − 6 = −4 D. f2 = 1/P2 = −0.25 m = −25 cm. The power is negative, so the second lens is a concave (diverging) lens.

• P = 1/f = +2 D and P2 = P − P1 = −4 D (1)
• f2 = −25 cm, concave lens (1)

A diverging combination of focal length 50 cm has f = −0.50 m, so P = 1/f = −2 D. P = P1 + P2 gives P2 = −2 − 5 = −7 D. f2 = 1/P2 = −0.143 m ≈ −14.3 cm. The power is negative, so the second lens is a concave (diverging) lens.

• P = 1/f = −2 D and P2 = P − P1 = −7 D (1)
• f2 ≈ −14.3 cm, concave lens (1)

Any two: eddy currents in the core, reduced by a laminated core; hysteresis of the core, reduced by a core of low hysteresis loss; resistance of the windings (I²R heating), reduced by thick wire for the high-current winding; flux leakage, reduced by winding the primary and secondary one over the other.

• first loss with its remedy (1)
• second loss with its remedy (1)

Energy bands are delocalised in space, not at any particular place in the solid; the diagram's vertical axis is energy, not position, and the energies are overall averages. E_C stands for the lowest energy level of the conduction band (the bottom of the band).

• bands are space delocalised; the line marks an energy, not a place (1)
• E_C is the bottom (lowest level) of the conduction band (1)

(i) Same A = 40, different Z: isobars. (ii) N = 13 − 6 = 7 and N = 14 − 7 = 7, different Z: isotones. (iii) Same Z = 17, different A: isotopes. (iv) Both have Z = 1 (deuterium and tritium), different A: isotopes.

• (i) isobars (0.5)
• (ii) isotones, with N = 7 for both (0.5)
• (iii) isotopes (0.5)
• (iv) isotopes (0.5)

(a) Q₀ = CV = 10 × 10⁻⁶ × 100 = 1.0 × 10⁻³ C. With the battery connected V is fixed and C becomes 30 μF, so Q = 3.0 × 10⁻³ C: an extra 2.0 × 10⁻³ C flows. (b) U₀ = ½ × 10 × 10⁻⁶ × 100² = 0.050 J; U = ½ × 30 × 10⁻⁶ × 100² = 0.15 J; increase 0.10 J. (c) The battery moves 2.0 × 10⁻³ C through a potential difference of 100 V: W = qV = 0.20 J, twice the increase in stored energy; the other 0.10 J does not stay in the capacitor's field.

• extra charge 2.0 × 10⁻³ C (1)
• stored energy rises from 0.050 J to 0.15 J, by 0.10 J (1)
• battery work 0.20 J, twice the increase (1)

(a) Axial: E = 2kp/r³, so p = Er³/2k = 3.2 × 10³ × (0.10)³/(2 × 9 × 10⁹) = 1.8 × 10⁻¹⁰ C m. (b) Equatorial: E = kp/r³ = 9 × 10⁹ × 1.78 × 10⁻¹⁰/(0.20)³ = 2.0 × 10² N C⁻¹, directed opposite to p. Equivalently, E_eq(2r) = E_ax(r)/(2 × 2³) = 3.2 × 10³/16 = 200 N C⁻¹.

• p = Er³/2k = 1.8 × 10⁻¹⁰ C m (1)
• equatorial E = kp/r³ = 2.0 × 10² N C⁻¹ (1)
• direction opposite to p (1)

(a) Equatorial: E = kp/r³, so p = Er³/k = 8.0 × 10³ × (0.050)³/(9 × 10⁹) = 1.1 × 10⁻¹⁰ C m. (b) Axial: E = 2kp/r³ = 2 × 9 × 10⁹ × 1.11 × 10⁻¹⁰/(0.10)³ = 2.0 × 10³ N C⁻¹, directed along p.

• p = Er³/k = 1.1 × 10⁻¹⁰ C m (1)
• axial E = 2kp/r³ = 2.0 × 10³ N C⁻¹ (1)
• direction along p (1)

(a) The p-side is connected towards the positive terminal, so the diode is forward biased. V_R = 6.0 − 0.7 = 5.3 V; I = 5.3/530 = 0.010 A = 10 mA. (b) Reversed, the diode is reverse biased; only the reverse saturation current of a few μA flows, practically zero. Then the resistor drops almost nothing and nearly the full 6.0 V appears across the diode.

• V_R = 5.3 V (1)
• I = 10 mA (1)
• reversed: current only of the order of μA (practically zero); almost all 6.0 V across the diode (1)

(i) n = c/v = 3.00/2.25 = 1.33 (4/3). (ii) sin i_c = n₂/n₁ = 1/(4/3) = 0.75, i_c ≈ 48.6°. (iii) Going into the rarer medium, v₂ > v₁, so the wavelet from A in air has radius v₂τ greater than BC = v₁τ. The refracted wavefront is the tangent from C to this wavelet, with sin r = (v₂/v₁) sin i. When sin i > v₁/v₂, the wavelet's radius v₂τ exceeds AC, so C lies inside it and no tangent plane can be drawn from C (sin r would exceed 1): there is no refracted wave, and the light is totally reflected.

• n = c/v = 4/3 (1)
• sin i_c = 3/4, i_c ≈ 48.6° (1)
• for i > i_c the wavelet radius v₂τ exceeds AC; no tangent from C, so no refracted wavefront (1)

(a) n = I/e = 0.40 × 10⁻⁶/1.6 × 10⁻¹⁹ = 2.5 × 10¹² per second. (b) Photons per second N = 1000n = 2.5 × 10¹⁵. (c) Energy of each photon E = hc/λ = (6.63 × 10⁻³⁴ × 3 × 10⁸)/(589 × 10⁻⁹) ≈ 3.38 × 10⁻¹⁹ J; P = NE = 2.5 × 10¹⁵ × 3.38 × 10⁻¹⁹ ≈ 8.4 × 10⁻⁴ W, about 0.84 mW.

• n = I/e = 2.5 × 10¹² s⁻¹ (1)
• N = 2.5 × 10¹⁵ s⁻¹ (1)
• E ≈ 3.38 × 10⁻¹⁹ J, P ≈ 0.84 mW (1)

(i) Iron filings (or a field-line plot) show very similar patterns round a bar magnet and round a current-carrying finite solenoid. (ii) A small compass needle moved near each is deflected in the same way. Cutting either in half gives two smaller magnets (two smaller solenoids), each with its own north and south pole and a smaller magnetic moment; the field lines remain continuous, leaving one face and entering the other.

• similar field-line or iron-filing patterns (1)
• similar compass deflections (1)
• cut: two smaller magnets/solenoids, each with N and S (1)

(a) The coil and the shunt are in parallel, so they share one potential difference, and the shunt carries I − I_g. With V = IR, I_g·R_G = (I − I_g)·r_s, so r_s = (4.0 × 10⁻³ × 30)/(1.0 − 0.004) = 0.12 Ω. (b) R_Gr_s/(R_G + r_s) = 30 × 0.12/30.12 ≈ 0.12 Ω, close to r_s since R_G ≫ r_s. (c) In series one current I_g flows through the coil and R, and their potential differences add: V = I_g(R_G + R), so R = V/I_g − R_G = 6.0/(4.0 × 10⁻³) − 30 = 1470 Ω.

• shunt r_s = 0.12 Ω (1)
• ammeter resistance ≈ 0.12 Ω (1)
• series resistance 1470 Ω (1)

(a) A current I₂ in the outer solenoid gives B = μ₀n₂I₂ inside it (edge effects neglected). The flux linked with the inner solenoid is N₁Φ₁ = (n₁l)(πr₁²)(μ₀n₂I₂), so M₁₂ = μ₀n₁n₂πr₁²l. (b) M₂₁ = μ₀n₁n₂πr₁²l as well (the inner solenoid's flux lies within area πr₁²), so M₁₂ = M₂₁ = M. (c) M = 4π × 10⁻⁷ × 2000 × 1000 × π × (0.015)² × 0.60 = 1.07 × 10⁻³ H. dI/dt = 3.0/0.15 = 20 A s⁻¹, so |ε| = 1.07 × 10⁻³ × 20 = 2.1 × 10⁻² V = 21 mV. (d) M becomes μ_r times larger: M = μ_rμ₀n₁n₂πr₁²l.

• B = μ₀n₂I₂ and flux linkage (n₁l)(πr₁²)(μ₀n₂I₂) (1)
• M₁₂ = μ₀n₁n₂πr₁²l (1)
• M₂₁ the same, M₁₂ = M₂₁ (1)
• M = 1.07 mH and ε = 21 mV (1)
• M multiplied by μ_r (1)

(a) A current I₂ in the outer solenoid gives B = μ₀n₂I₂ inside; flux linkage of the inner one N₁Φ₁ = (n₁l)(πr₁²)(μ₀n₂I₂), so M₁₂ = μ₀n₁n₂πr₁²l. (b) A current I₁ in the inner one gives B = μ₀n₁I₁ confined to its area πr₁², so each of the n₂l turns of the outer solenoid links flux μ₀n₁I₁πr₁²; N₂Φ₂ = μ₀n₁n₂πr₁²l I₁, the same M. (c) M = 4π × 10⁻⁷ × 1200 × 600 × π × (0.020)² × 0.80 = 9.1 × 10⁻⁴ H. dI/dt = 4.0/0.020 = 200 A s⁻¹; |ε| = 9.1 × 10⁻⁴ × 200 = 0.18 V. (d) M = 200 × 9.1 × 10⁻⁴ = 0.18 H.

• B = μ₀n₂I₂ and flux linkage of the inner solenoid (1)
• M₁₂ = μ₀n₁n₂πr₁²l (1)
• inner flux confined to πr₁², so M₂₁ is the same (1)
• M = 0.91 mH and ε = 0.18 V (1)
• M = 0.18 H with the core (1)

(a) The ray meets face AB at i, refracts at r1, meets AC at r2 and emerges at e. The normals at the two faces meet at N; in the quadrilateral AQNR, angles A and QNR add to 180°, and in triangle QNR, r1 + r2 + ∠QNR = 180°, so r1 + r2 = A. The deviation is (i − r1) + (e − r2) = i + e − A. At minimum deviation i = e, so r1 = r2 = r = A/2 and Dm = 2i − A, i = (A + Dm)/2. n = sin i/sin r = sin[(A + Dm)/2]/sin(A/2). (b) The graph falls to a minimum Dm and rises again (a U-shaped curve). Since i and e enter δ = i + e − A symmetrically, the ray path can be reversed with the same deviation: any δ other than Dm occurs once with incidence i and once with incidence e. (c) sin r1 = sin 60°/√3 = 1/2, r1 = 30°; r2 = A − r1 = 30° = r1, so the ray inside is symmetric and e = i = 60°: this is minimum deviation. Dm = i + e − A = 60°.

• ray diagram; r1 + r2 = A from the geometry (1)
• δ = i + e − A; at minimum i = e, r = A/2 (1)
• n = sin[(A + Dm)/2]/sin(A/2) obtained (1)
• δ–i curve with a minimum; two values of i by symmetry of i and e (1)
• r1 = r2 = 30°, e = 60°: Dm = 60° (1)

(a) The ray meets face AB at i, refracts at r1, meets AC at r2, emerges at e. The normals at the faces meet at N; in quadrilateral AQNR, A + ∠QNR = 180°; in triangle QNR, r1 + r2 + ∠QNR = 180°; so r1 + r2 = A. The deviation is the sum of the deviations at the two faces: (i − r1) + (e − r2) = i + e − A. (b) The path can be reversed with the same deviation, so a deviation is generally reached at two angles of incidence, i and e; the two coincide only at the minimum, so there i = e, r1 = r2 = A/2 and Dm = 2i − A. Then n = sin i/sin r = sin[(A + Dm)/2]/sin(A/2). (c) r = 30°, sin i = √2 × sin 30° = 1/√2, i = 45°. Dm = 2 × 45° − 60° = 30°.

• ray diagram; r1 + r2 = A from the geometry (1)
• δ = i + e − A (1)
• reversibility: two i for a δ, coincide at minimum, i = e (1)
• n = sin[(A + Dm)/2]/sin(A/2) (1)
• i = 45°, Dm = 30° (1)

(a) With I_g = 0, the junction rule at B and D gives the same current I₁ in P and Q and the same current I₂ in R and S. Loop ABDA: −I₁P + I₂R = 0; loop BCDB: −I₁Q + I₂S = 0. Dividing, P/Q = R/S. (b) S = RQ/P = 8.0 × 18/12 = 12 Ω. (c) With the cell across B and D and the galvanometer across A and C, the arms pair as P with R and Q with S; balance needs P/R = Q/S, i.e. PS = QR, the same condition, so it still holds.

• junction rule at B and D with I_g = 0 (1)
• two loop equations (1)
• P/Q = R/S (1)
• S = 12 Ω (1)
• PS = QR again after interchange (1)

(a) P/Q = R/S; the unknown is found from the setting for zero deflection, so the reading does not depend on calibrating the galvanometer. (b) P/Q = R/S = 2, balanced, so the galvanometer carries no current and can be ignored: ABC is 9.0 Ω in parallel with ADC, 12 Ω, R_AC = 36/7 Ω, I = 9.0 × 7/36 = 1.75 A. (c) Taking C as zero, V_B = 9.0 × 3.0/9.0 = 3.0 V and V_D = 9.0 × 6.0/14 ≈ 3.9 V (with the galvanometer removed). D is at the higher potential, so current flows through the galvanometer from D to B. (With the galvanometer in place the two potentials shift to 3.16 V and 3.58 V; their order is unchanged.)

• P/Q = R/S and why it is a null method (1)
• bridge balanced, galvanometer ignored (1)
• I = 1.75 A (1)
• V_B = 3.0 V, V_D ≈ 3.9 V (1)
• current from D to B (1)