CBSE Class 12 Physics Board Paper 2024 (55/1/1): Solutions

A thin plastic ring of radius R carries a uniform linear charge density λ. Find the electric field magnitude at its centre.

1. Consider two small charge elements on opposite sides of the ring.
2. The electric field vectors produced by these opposite elements at the center are equal in magnitude and opposite in direction.
3. Since the ring has uniform charge density, these fields cancel each other out in pairs, resulting in a net electric field of zero at the center.

Answer: Opposite elements of the ring cancel each other's fields in pairs, so the field at the centre is zero — option (B).

Ten 1 μF capacitors are joined in parallel across a 100 V supply. Find the total stored energy.

1. For capacitors in parallel, the total capacitance is the sum of individual capacitances: C = 10 × 1 μF = 10 μF = 10 × 10⁻⁶ F.
2. The total energy stored is calculated using U = ½CV² = ½ × 10 × 10⁻⁶ × 100² = 0.05 J = 5.0 × 10⁻² J.

Answer: C = 10 μF, so U = ½CV² = ½ × 10 × 10⁻⁶ × 100² = 0.05 J = 5.0 × 10⁻² J — option (D).

Between points A and B, a 12 V cell with 1 Ω and a 6 V cell with 0.5 Ω are connected in parallel (same polarity). Find the potential difference between A and B.

Uses a figure printed in CBSE's paper — what the figure shows: In the figure, the upper branch is a 12 V cell in series with 1 Ω and the lower branch a 6 V cell in series with 0.5 Ω; both cells face the same way and the two branches join at A and at B.

1. Calculate the equivalent emf using the formula for parallel cells: E_eq = (ε₁/r₁ + ε₂/r₂) / (1/r₁ + 1/r₂).
2. Substitute the given values: E_eq = (12/1 + 6/0.5) / (1/1 + 1/0.5) = (12 + 12) / (1 + 2) = 24 / 3 = 8 V.

Answer: Equivalent emf = (ε₁/r₁ + ε₂/r₂)/(1/r₁ + 1/r₂) = (12 + 12)/3 = 8 V — option (B).

A loop in the x-y plane carries current I clockwise (seen from +z) in a uniform magnetic field along the z-axis. What does the loop tend to do?

1. The magnetic force on a current element is given by dF = I dl × B.
2. Since the current is clockwise and B is along +z, the cross product dl × B points radially inward for every element of the loop.
3. Thus, the net force acts inward, causing the loop to tend to shrink.

Answer: For a clockwise current with B along +z, the force on every element I dl × B points inward, so the loop tends to shrink — option (C).

A 10 cm wire on the y-axis carries 1.0 A in the +y direction in a field B = (5 mT) ĵ − (8 mT) k̂. Find the force on the wire.

1. Calculate the cross product of the current direction and magnetic field: ĵ × (5ĵ − 8k̂) = 5(ĵ × ĵ) − 8(ĵ × k̂) = 0 − 8î = −8î.
2. Compute the force magnitude and direction: F = I L (−8î) mT = 1.0 A × 0.1 m × (−8 mT) î = −0.8 mN î.

Answer: F = I L × B = 1.0 × 0.1 ĵ × (5ĵ − 8k̂) mT; ĵ × ĵ = 0 and ĵ × k̂ = î, so F = (-0.8 mN) î — option (B).

A galvanometer of resistance G Ω is made into a 0 to I A ammeter, with 0.1% of I passing through the galvanometer. Find the ammeter's resistance.

1. The current through the galvanometer is Ig = 0.1% of I = 0.001I, and the current through the shunt is Is = I - 0.001I = 0.999I.
2. Since the galvanometer and shunt are in parallel, the potential drop is equal: IgG = IsS, so S = (0.001I × G) / (0.999I) = G/999.
3. The ammeter resistance is the parallel combination of G and S: R = (G × S) / (G + S) = (G × G/999) / (G + G/999) = G/1000 Ω.

Answer: Shunt S = G × 0.001I/(0.999I) = G/999; ammeter resistance = GS/(G + S) = G/1000 Ω — option (B).

A capacitor has reactance X at angular frequency ω. Its capacitance is doubled and the frequency tripled. New reactance?

1. The capacitive reactance is given by the formula X_C = 1/(ωC).
2. Substituting the new values C' = 2C and ω' = 3ω, the new reactance is X' = 1/(3ω × 2C) = X/6.

Answer: X_C = 1/(ωC), so it becomes X/(2 × 3) = X/6 — option (A).

Electric fields in four regions: I: Ex = E₀ sin(kz − ωt); II: Ex = E₀; III: Ex = E₀ sin kz; IV: Ex = E₀ cos kz. In which region is there a displacement current?

1. Displacement current is defined as Id = ε₀ dΦE/dt, which requires a time-varying electric field.
2. In regions II, III, and IV, the electric field Ex is independent of time t, so dΦE/dt = 0 and Id = 0.
3. In region I, Ex = E₀ sin(kz − ωt) depends on time t, so dΦE/dt ≠ 0 and a displacement current exists.
4. Therefore, the correct option is (A).

Answer: Displacement current needs a time-varying field (ε₀ dΦ/dt ≠ 0); only region I's field changes with time — option (A).

Which electron transition in hydrogen produces the second line of the Balmer series?

1. The Balmer series consists of transitions where the electron falls to the final energy level n_f = 2.
2. The first line of the series corresponds to the transition from n_i = 3 to n_f = 2.
3. Therefore, the second line of the Balmer series corresponds to the transition from n_i = 4 to n_f = 2.

Answer: Balmer lines end on n = 2; the first line is 3 → 2, so the second is from n_i = 4 to n_f = 2 — option (C).

What happens when germanium is doped with arsenic?

1. Arsenic is a pentavalent element with five valence electrons, while germanium is tetravalent with four.
2. When doped, four arsenic electrons form covalent bonds with germanium, leaving one extra electron free to move.
3. This extra electron acts as a majority charge carrier, creating an n-type semiconductor where the number of conduction electrons increases.

Answer: Arsenic is pentavalent, so each atom donates an extra electron: the number of conduction electrons increases (n-type) — option (B).

Photons of 3.3 eV (beam A) and 11.3 eV (beam B) fall in turn on a metal with work function 2.3 eV. Find the ratio of maximum photoelectron speeds, A to B.

1. Calculate the maximum kinetic energy for beam A using K_A = 3.3 eV - 2.3 eV = 1 eV.
2. Calculate the maximum kinetic energy for beam B using K_B = 11.3 eV - 2.3 eV = 9 eV.
3. Use the relation v ∝ √K to find the ratio v_A/v_B = √(1/9) = 1/3.

Answer: K_max = 1 eV for A and 9 eV for B; v ∝ √K, so v_A : v_B = 1/3 — option (C).

An electron and a proton in motion have equal de Broglie wavelength λ. What quantity must they share?

1. According to the de Broglie relation, the wavelength λ is given by λ = h/p, where h is Planck's constant and p is the momentum.
2. Since both particles have the same wavelength λ and h is a universal constant, their momenta p must be equal.

Answer: λ = h/p, so equal λ means equal momentum — option (A).

Assertion: In the photoelectric effect, photoelectron kinetic energy rises with the intensity of light. Reason: Photocurrent depends on the wavelength of the light.

1. Assertion is false because the maximum kinetic energy of photoelectrons depends on the frequency of incident light, not its intensity.
2. Reason is false because the photocurrent depends on the intensity of light (provided frequency is above threshold), not on the wavelength.
3. Since both Assertion and Reason are false, the correct option is (D).

Answer: Assertion is false: maximum kinetic energy depends on frequency, not intensity. Reason is false: photocurrent depends on intensity (for frequency above threshold), not on wavelength — option (D).

Assertion: Two coils have the greatest mutual inductance when one is wound over the other. Reason: Their flux linkage is greatest in that arrangement.

1. Mutual inductance M = N₂Φ₂/I₁: it is the flux linked with one coil per unit current in the other.
2. When one coil is wound over the other, almost all the flux of one passes through the other, so the flux linkage and hence M is greatest; both statements are true and the Reason explains the Assertion — option (A).

Answer: Both are true and the Reason explains the Assertion, since M = NΦ/I grows with flux linkage — option (A).

Assertion: Two long, freely hung parallel wires joined in series to a battery move away from each other. Reason: Wires with oppositely directed currents repel.

1. In a series circuit, the current flows in opposite directions through the two parallel wires.
2. According to the principle of magnetic force between conductors, parallel wires carrying currents in opposite directions repel each other.
3. Since the wires repel, the Assertion is true, and the Reason correctly explains this phenomenon, making option (A) the correct choice.

Answer: In series the same current runs in opposite directions in the two wires, so they repel; both true and the Reason explains it — option (A).

Assertion: A plane or convex mirror can never form a real image. Reason: A virtual image can never act as the object for a real image.

1. Assertion is false because a converging beam (virtual object) falling on a plane or convex mirror can form a real image.
2. Reason is false because a virtual image can act as the object for another lens or mirror to form a real image.
3. Since both the assertion and the reason are false, the correct option is (D).

Answer: Both are false: a converging beam (virtual object) falling on a plane or convex mirror can give a real image, and a virtual image can act as an object for another lens or mirror — option (D).

Silver has α = 4.0 × 10⁻³ K⁻¹ at the reference temperature 20°C. At what temperature does a silver wire's resistance become double its value at 20°C?

1. The resistance of a conductor at temperature T is given by R = R₀[1 + α(T − T₀)], where R₀ is the resistance at reference temperature T₀.
2. Given that the resistance doubles, we set R = 2R₀ and substitute the values: 2R₀ = R₀[1 + 4.0 × 10⁻³(T − 20)].
3. Dividing both sides by R₀ and simplifying gives 2 = 1 + 4.0 × 10⁻³(T − 20), which leads to 1 = 4.0 × 10⁻³(T − 20).
4. Solving for the temperature difference, T − 20 = 1 / (4.0 × 10⁻³) = 250 °C.
5. Adding the reference temperature, T = 20 + 250 = 270 °C, which is equivalent to 270 + 273 = 543 K.

Answer: R = R₀[1 + α(T − 20)] with R = 2R₀ gives T − 20 = 1/α = 250, so T = 270°C (543 K).

Light of frequency 5.0 × 10¹⁴ Hz goes from air into a medium with n = 1.5. Find the wavelength of (i) the reflected light and (ii) the refracted light.

1. The frequency of light remains unchanged during reflection and refraction, so ν = 5.0 × 10¹⁴ Hz for both cases.
2. For the reflected light, it stays in air, so its wavelength is λ = c/ν = (3 × 10⁸) / (5 × 10¹⁴) = 6 × 10⁻⁷ m = 600 nm.
3. For the refracted light, the wavelength in the medium is given by λ' = λ/n, where n = 1.5.
4. Substituting the values, λ' = 600 / 1.5 = 400 nm.

Answer: (i) Reflected light stays in air: λ = c/ν = 3 × 10⁸/5 × 10¹⁴ = 600 nm. (ii) Refracted: λ/n = 600/1.5 = 400 nm (frequency is unchanged).

A plano-convex lens has focal length 16 cm and refractive index 1.4. Find the radius of its curved face.

1. Use the lens maker's formula for a plano-convex lens: 1/f = (n − 1)(1/R − 1/∞).
2. Since 1/∞ = 0, the formula simplifies to 1/f = (n − 1)/R.
3. Rearrange the equation to solve for R: R = (n − 1)f.
4. Substitute the given values n = 1.4 and f = 16 cm: R = (1.4 − 1) × 16.
5. Calculate the result: R = 0.4 × 16 = 6.4 cm.

Answer: 1/f = (n − 1)(1/R − 1/∞), so R = (n − 1)f = 0.4 × 16 = 6.4 cm.

A concave mirror (R = 40 cm) has an object 30 cm before it. Find (i) the image position and (ii) the magnification.

1. Given radius of curvature R = 40 cm, so focal length f = R/2 = 40/2 = 20 cm. Using sign convention, f = -20 cm and object distance u = -30 cm.
2. Apply mirror formula: 1/v = 1/f - 1/u = 1/(-20) - 1/(-30) = -1/20 + 1/30 = (-3 + 2)/60 = -1/60, so v = -60 cm.
3. Calculate magnification m = -v/u = -(-60)/(-30) = -60/30 = -2. The negative sign indicates the image is real and inverted, located 60 cm in front of the mirror with twice the object size.

Answer: f = −20 cm, u = −30 cm; 1/v = 1/f − 1/u gives v = -60 cm, m = -2 (real, inverted image 60 cm in front of the mirror, twice the size).

A neutron (mass m) and a photon both have energy E. With λn the neutron's de Broglie wavelength and λp the photon's wavelength, find λn/λp.

1. For a non-relativistic neutron with energy E, the momentum is p = √(2mE), so its de Broglie wavelength is λn = h/p = h/√(2mE).
2. For a photon with energy E, the wavelength is given by λp = hc/E.
3. Dividing the two expressions, λn/λp = (h/√(2mE)) / (hc/E) = E/(c√(2mE)).
4. Simplifying the ratio gives λn/λp = (1/c)√(E/2m).

Answer: λn = h/√(2mE) and λp = hc/E, so λn/λp = E/(c√(2mE)) = (1/c)√(E/2m).

Sketch the current-voltage curve of GaAs and mark (a) the negative-resistance region and (b) the region obeying Ohm's law.

1. Draw the current-voltage graph with voltage on the x-axis and current on the y-axis.
2. Sketch a curve starting from the origin with a positive slope, indicating the region where current is directly proportional to voltage (Ohm's law region).
3. Mark the point where the current reaches its maximum value.
4. Draw the curve bending downwards as voltage increases further, indicating a decrease in current; label this section as the negative-resistance region where dI/dV is negative.
5. Show the curve rising again at higher voltages to complete the characteristic shape of the GaAs device.

Answer: Current rises linearly from the origin at first (Ohm's law region), reaches a peak, then falls as voltage keeps rising (negative resistance region: slope dI/dV < 0), and finally rises again.

A cube of edge 0.1 m sits in a field E = 500x î N/C (x in m), placed as in the figure. Find (a) the net flux through the cube and (b) the charge inside it.

Uses a figure printed in CBSE's paper — what the figure shows: In the figure, the cube's edges are parallel to the axes; its face nearest the origin lies in the plane x = 0.1 m and the far face in x = 0.2 m.

1. Identify that the electric field is non-uniform and varies with x, so flux is calculated separately for each face.
2. Calculate the flux through the face at x = 0.1 m: Φ₁ = E(0.1) × Area × cos(180°) = (500 × 0.1) × (0.1)² × (−1) = −0.5 N m²/C.
3. Calculate the flux through the face at x = 0.2 m: Φ₂ = E(0.2) × Area × cos(0°) = (500 × 0.2) × (0.1)² × 1 = 1.0 N m²/C.
4. Sum the fluxes to find the net flux: Φ_net = Φ₁ + Φ₂ = −0.5 + 1.0 = 0.5 N m²/C.
5. Use Gauss's law to find the charge: q = ε₀ × Φ_net = 8.854 × 10⁻¹² × 0.5 = 4.43 × 10⁻¹² C.

Answer: Only the faces normal to x carry flux. Face at x = 0.1 m: −500 × 0.1 × 0.01 = −0.5; face at x = 0.2 m: 500 × 0.2 × 0.01 = 1. (a) net flux = 0.5 N m²/C; (b) q = ε₀Φ = 8.854 × 10⁻¹² × 0.5, so q = 4.43 × 10⁻¹² C (≈ 4.4 × 10⁻¹² C).

Define current density and say whether it is scalar or vector. For a metal in a field E, with electron density n, electron mass m, charge −e and relaxation time τ, show that j = αE with α = ne²τ/m.

1. Current density is defined as the current flowing per unit area normal to the direction of flow, expressed as j = I/A.
2. Since current density has both magnitude and direction, it is a vector quantity.
3. The drift velocity of electrons in an electric field E is given by v_d = eEτ/m.
4. The total current is the product of charge density, area, and drift velocity: I = neAv_d.
5. Substituting the expression for v_d into the current equation gives I = (ne²τ/m)AE.
6. Dividing by area A yields j = (ne²τ/m)E, which matches the form j = αE with α = ne²τ/m.

Answer: Current density is the current per unit area normal to the flow (j = I/A); it is a vector. Drift velocity v_d = eEτ/m, so I = neAv_d = (ne²τ/m)AE, giving j = (ne²τ/m)E, i.e. α = ne²τ/m.

What is a Wheatstone bridge? Derive the condition for it to be balanced.

1. A Wheatstone bridge has four resistors in a closed loop: R₁ and R₃ in series in one path, R₂ and R₄ in series in the other; a cell is joined across the two ends shared by both paths and a galvanometer between the junction of R₁, R₃ and the junction of R₂, R₄.
2. Balance means no current flows through the galvanometer, so its two ends are at the same potential.
3. With no current in the galvanometer, Kirchhoff's junction rule gives I₁ = I₃ (R₁ and R₃ carry the same current) and I₂ = I₄.
4. Applying the loop rule to the loop with R₁, R₂ and the galvanometer: I₁R₁ = I₂R₂. Applying it to the loop with R₃, R₄ and the galvanometer: I₃R₃ = I₄R₄.
5. Dividing, I₁R₁/(I₃R₃) = I₂R₂/(I₄R₄); with I₁ = I₃ and I₂ = I₄ this gives R₁/R₃ = R₂/R₄, i.e. the balance condition R₂/R₁ = R₄/R₃.

Answer: Four resistors R₁, R₂, R₃, R₄ in a closed loop with a cell across one diagonal and a galvanometer across the other. With zero galvanometer current, Kirchhoff's rules give I₁ = I₃, I₂ = I₄ and equal drops across R₁, R₂ and across R₃, R₄, so R₂/R₁ = R₄/R₃.

A proton of KE 1.3384 × 10⁻¹⁴ J travels horizontally north to south into a uniform 2.0 mT field pointing east. Taking proton mass 1.673 × 10⁻²⁷ kg and q/m = 1.0 × 10⁸ C/kg, find (a) its speed, (b) its acceleration and (c) its path radius.

1. (a) K = ½mv², so v = √(2K/m) = √(2 × 1.3384 × 10⁻¹⁴ / 1.673 × 10⁻²⁷) = √(1.6 × 10¹³) = 4 × 10⁶ m/s.
2. (b) The velocity (south) is perpendicular to B (east), so the force is qvB, directed vertically upward (along v × B). Then a = qvB/m = (q/m)vB = 1.0 × 10⁸ × 4 × 10⁶ × 2.0 × 10⁻³ = 8 × 10¹¹ m/s².
3. (c) The magnetic force provides the centripetal force: mv²/r = qvB, so r = v/((q/m)B) = 4 × 10⁶ / (1.0 × 10⁸ × 2.0 × 10⁻³) = 4 × 10⁶ / 2 × 10⁵ = 20 m.

Answer: (a) v = √(2K/m) with m = 1.673 × 10⁻²⁷ kg: v = 4 × 10⁶ m/s. (b) a = (q/m)vB = 8 × 10¹¹ m/s². (c) r = v/((q/m)B) = 20 m.

Note: The paper's q/m = 1.0 × 10⁸ C/kg does not quite match the proton mass it gives (e/m = 9.58 × 10⁷ C/kg). Using that exact value gives a ≈ 7.7 × 10¹¹ m/s² and r ≈ 20.9 m; follow the paper's q/m in the exam, as CBSE's marking scheme does.

An L, a C and an R are in series with v = vm sin ωt. Derive the average power dissipated and the resonant frequency.

1. Write the instantaneous current as i = im sin(ωt + φ), where φ is the phase difference between voltage and current.
2. Calculate instantaneous power p = vi = vm im sin(ωt) sin(ωt + φ).
3. Use the identity 2 sin A sin B = cos(A − B) − cos(A + B) to expand p = (vm im/2)[cos φ − cos(2ωt + φ)].
4. State that the average of cos(2ωt + φ) over a full cycle is zero, leaving average power P = (vm im/2) cos φ.
5. Express P in terms of RMS values as P = V I cos φ, noting that cos φ is the power factor.
6. Define resonance as the condition where inductive reactance equals capacitive reactance, so ωL = 1/(ωC).
7. Solve for the resonant frequency to get ω₀ = 1/√(LC), or ν₀ = 1/(2π√(LC)).

Answer: With i = im sin(ωt + φ), p = vi = (vm im/2)[cos φ − cos(2ωt + φ)]; the second term averages to zero, so P = (vm im/2) cos φ = V I cos φ (= I²Z cos φ). Resonance when X_L = X_C, i.e. ω₀L = 1/(ω₀C), so ω₀ = 1/√(LC) (ν₀ = 1/(2π√(LC))).

(a) Give two examples showing that an EM wave's wavelength matches the size of the system emitting it. (b)(i) Why are short-wave bands chosen for radio broadcasting over long distances? (ii) Why is X-ray astronomy possible only from satellites, while optical and radio telescopes work on the ground?

1. For part (a), state that gamma rays have wavelengths of 10⁻¹⁴ to 10⁻¹⁵ m, which matches the size of atomic nuclei.
2. For part (a), state that X-rays originate from heavy atoms, where the wavelength is comparable to the atomic scale.
3. For part (b)(i), explain that short-wave bands are chosen because the ionosphere reflects them back to Earth, enabling long-distance communication.
4. For part (b)(ii), explain that X-ray astronomy requires satellites because the Earth's atmosphere absorbs X-rays, whereas it allows visible light and radio waves to pass through.

Answer: (a) Any two: gamma rays (10⁻¹⁴–10⁻¹⁵ m) come from nuclei; X-rays come from heavy atoms; radio waves come from accelerated charges in circuits and an antenna radiates best at wavelengths near its own size. (b)(i) The ionosphere reflects short-wave bands back to Earth. (ii) The atmosphere absorbs X-rays but lets visible light and radio waves through.

State the shortcomings of Rutherford's atom model, how Bohr overcame them, and show that Bohr orbits are not evenly spaced.

1. Rutherford's model fails because an accelerating electron in orbit should radiate energy, causing it to spiral into the nucleus and making the atom unstable.
2. It also predicts a continuous spectrum of radiation, which contradicts the observed discrete line spectra of atoms.
3. Bohr postulated that electrons move in specific stationary orbits without radiating energy, defined by the angular momentum quantization condition L = nh/2π.
4. Energy is emitted only as a photon when an electron jumps from a higher to a lower orbit, explaining the discrete spectral lines.
5. Since the radius of the nth orbit is proportional to n² (rₙ ∝ n²), the difference between consecutive radii is rₙ₊₁ − rₙ ∝ (n+1)² − n² = 2n + 1.
6. As the value (2n + 1) increases with n, the spacing between successive orbits is not constant, proving they are not evenly spaced.

Answer: Drawbacks: an orbiting (accelerating) electron should radiate, spiral in and collapse, so the atom would be unstable; and it would emit a continuous spectrum, not the observed lines. Bohr: electrons move in certain stationary orbits (L = nh/2π) without radiating and emit a photon only on jumping between orbits. Since rₙ ∝ n², rₙ₊₁ − rₙ ∝ (2n + 1), which grows with n, so orbits are not equally spaced.

(a) Name two nuclear properties. (b) Why is a nucleus far denser than an atom? (c) Show that nuclear density does not depend on the nucleus.

1. (a) Two nuclear properties are: the nucleus is positively charged and it is composed of protons and neutrons.
2. (b) A nucleus is far denser than an atom because the atom is mostly empty space while nearly all its mass is concentrated in the tiny nucleus.
3. (c) The mass of a nucleus is given by M = mA, where m is the mass of a nucleon and A is the mass number.
4. The radius of the nucleus is given by R = R₀A^(1/3), so the volume is V = (4/3)πR³ = (4/3)πR₀³A.
5. Density is calculated as ρ = M/V = mA / ((4/3)πR₀³A).
6. Simplifying the expression, the mass number A cancels out, resulting in ρ = 3m / (4πR₀³).
7. Since the final expression contains no A, the nuclear density is independent of the nucleus.

Answer: (a) Any two: it is positively charged; it is made of protons and neutrons; R = R₀A^(1/3); its density is independent of A. (b) An atom is mostly empty space while nearly all its mass sits in the tiny nucleus. (c) ρ = mA/((4/3)πR₀³A) = 3m/(4πR₀³), where m is the nucleon mass; A cancels, so density is the same for all nuclei.

Case study on lenses. A double-convex lens of glass with index n has both radii equal to R. Its power?

1. Apply the lens maker's formula: P = (n − 1)(1/R₁ − 1/R₂).
2. For a double-convex lens, R₁ = R and R₂ = −R, so P = (n − 1)(1/R + 1/R) = 2(n − 1)/R.

Answer: P = (n − 1)(1/R + 1/R) = 2(n − 1)/R — option (A).

An equiconvex lens of power P is cut into two equal halves by a plane perpendicular to the principal axis. Power of each half?

1. Using the lens maker's formula for the original equiconvex lens, P = (n − 1)(1/R + 1/R) = 2(n − 1)/R.
2. Cutting the lens in half creates two plano-convex lenses, each with one curved surface of radius R and one flat surface (infinite radius).
3. The power of each half is P' = (n − 1)(1/R + 0) = (n − 1)/R, which is equal to P/2.

Answer: Each half is plano-convex with P' = (n − 1)/R = P/2 — option (D).

The two plano-convex halves from (ii) are put in contact as in the figure. Power of the combination?

Uses a figure printed in CBSE's paper — what the figure shows: In the figure, the two plano-convex halves are placed coaxially, touching each other on the principal axis.

1. Since the two plano-convex halves are placed in contact, the total power of the combination is the sum of their individual powers.
2. Each half has a power of P/2, so the total power is P/2 + P/2 = P, which corresponds to option (B).

Answer: P/2 + P/2 = P — option (B).

An equiconvex lens of power P is cut along its principal axis and the two halves are placed one after the other on the axis as in the figure. Power of the combination?

Uses a figure printed in CBSE's paper — what the figure shows: In the figure, the two half-lenses (each a half of the double-convex lens, flat cut face on the axis line) stand side by side in contact along the principal axis.

1. Cutting an equiconvex lens along its principal axis does not change the radius of curvature of its surfaces, so each half retains the original power P.
2. When two lenses of power P are placed in contact, the total power is the sum of individual powers, which is P + P = 2P.

Answer: Cutting along the axis leaves each half with power P; two in contact give 2P — option (C).

Two convex lenses (f = 60 cm and f = 20 cm) touch each other coaxially. Power of the pair?

1. Power in dioptres is P = 100/f with f in cm: P₁ = 100/60 ≈ 1.67 D and P₂ = 100/20 = 5 D.
2. Lenses in contact add their powers: P = 1.67 + 5 = 6.67 D, which is the value printed as 6.6 D in option (A).

Answer: P = 100/60 + 100/20 = 6.67 D, printed as 6.6 D — option (A).

Note: The exact power is 6.67 D (≈ 6.7 D); the paper's option (A) prints it as 6.6 D.

Case study on rectifiers. An ac input of rms value V₀/√2 feeds a full-wave rectifier. RMS value of the rectified output?

1. Full-wave rectification inverts the negative half-cycles of the AC input to make them positive, but the magnitude of the voltage at any instant remains unchanged.
2. Since the square of the voltage (v²) is identical for both the original AC and the rectified waveform, the root mean square value remains V₀/√2.

Answer: Full-wave rectification only flips the negative halves, so v² is unchanged and the rms stays V₀/√2 — option (A).

For how long does current flow through each diode of a full-wave rectifier?

1. In a full-wave rectifier, the two diodes are connected such that one conducts during the positive half-cycle and the other during the negative half-cycle of the AC input.
2. Therefore, each individual diode conducts for exactly half of the total input cycle, making option (B) the correct answer.

Answer: Each diode conducts for half of the input cycle — option (B).

In a full-wave rectifier, how are the two diodes biased at any instant?

1. During the positive half-cycle, the diode connected to the positive terminal conducts current while the other is cut off.
2. During the negative half-cycle, the roles reverse so that the other diode conducts while the first is cut off.
3. Thus, at any instant, one diode allows current flow while the other blocks it, corresponding to option (C).

Answer: At any moment one diode is forward biased while the other is reverse biased — option (C).

A 50 Hz ac voltage feeds a half-wave rectifier. Ripple frequency of the output?

1. A half-wave rectifier allows current to flow during only one half of the AC input cycle.
2. Therefore, it produces one output pulse per input cycle, making the ripple frequency equal to the input frequency of 50 Hz.

Answer: A half-wave rectifier gives one output pulse per cycle, so the ripple frequency equals the input frequency, 50 Hz — option (B).

A square wave switching between +5 V and −5 V is applied to a diode in series with a load R_L (output taken across R_L). What output appears?

Uses a figure printed in CBSE's paper — what the figure shows: In the figure, the input is a square pulse from −5 V up to +5 V and back; the diode (anode at input) is in series with R_L, output across R_L. Options: (A) a 10 V positive pulse, (B) a −10 V negative pulse, (C) a −5 V negative pulse, (D) a +5 V positive pulse.

1. When the input is +5 V, the diode is forward biased and conducts, allowing current to flow through R_L, so the output voltage is +5 V.
2. When the input is −5 V, the diode is reverse biased and acts as an open switch, so no current flows and the output voltage is 0 V.
3. Thus, the output is a +5 V positive pulse, which corresponds to option (D).

Answer: The diode conducts only on the +5 V half, so the output is a +5 V pulse with zero output during the −5 V half — option (D).

(i) Derive the potential energy of a dipole p in a uniform field E; when is it maximum and minimum? (ii) Charges −1.0 pC at (0, 0) and +1.0 pC at (3 mm, 4 mm) in the x-y plane form a dipole in a field E = (1000 V/m) î. Find the torque on it.

1. The work done in rotating the dipole from angle θ₀ to θ is W = ∫ pE sin θ dθ.
2. Integrating from π/2 to θ gives U(θ) = -pE cos θ, which is the potential energy.
3. In vector form, the potential energy is U = -p · E.
4. The potential energy is maximum when θ = 180° (p is antiparallel to E) and minimum when θ = 0° (p is parallel to E).
5. The dipole moment p is calculated as q × d = 10⁻¹² C × (3î + 4ĵ) × 10⁻³ m = (3î + 4ĵ) × 10⁻¹⁵ C m.
6. The torque is τ = p × E = (3î + 4ĵ) × 10⁻¹⁵ × (1000î) = 4 × 10⁻¹² (ĵ × î).
7. Using the cross product identity ĵ × î = -k̂, the torque is τ = -4 × 10⁻¹² k̂ N m.

Answer: (i) Work to turn the dipole from θ₀ = π/2 to θ is ∫pE sin θ dθ, so U(θ) = −pE cos θ = −p·E; maximum when p is antiparallel to E (θ = 180°), minimum when parallel (θ = 0°). (ii) p = 1 × 10⁻¹² × (3î + 4ĵ) × 10⁻³ C m; τ = p × E = (-4 × 10⁻¹² N m) k̂, i.e. 4 × 10⁻¹² N m along −z.

(i) A dipole p = p î (charges −q and q, 2a apart) lies on the x-axis centred at the origin. Show that at a point x with x >> a, V = (1/4πε₀)(p·î)/x². (ii) Isolated metal spheres S₁ and S₂ of radii 1 cm and 3 cm both have surface charge density (2/π) × 10⁻⁹ C/m². They are far apart and joined by a thin wire. Find the new charge on S₁.

1. The potential at x is V = (1/4πε₀)[q/(x-a) - q/(x+a)] = (1/4πε₀)(2qa)/(x² - a²).
2. Since p = 2qa î, p·î = 2qa, so V = (1/4πε₀)(p·î)/(x² - a²).
3. For x >> a, x² - a² ≈ x², giving V = (1/4πε₀)(p·î)/x².
4. Initial charge on S1 is Q1 = σ4πr1² = (2/π)×10⁻⁹ × 4π × (10⁻²)² = 8 × 10⁻¹³ C.
5. Initial charge on S2 is Q2 = σ4πr2² = (2/π)×10⁻⁹ × 4π × (3×10⁻²)² = 72 × 10⁻¹³ C.
6. Total charge Q = Q1 + Q2 = 80 × 10⁻¹³ C = 8 × 10⁻¹² C.
7. At common potential, Q1'/Q2' = r1/r2 = 1/3, so Q1' = (1/4)Q = (1/4)(8 × 10⁻¹²) = 2 × 10⁻¹² C.

Answer: (i) V = (q/4πε₀)[1/(x − a) − 1/(x + a)] = (1/4πε₀)(2qa)/(x² − a²); for x >> a this is (1/4πε₀)(p·î)/x². (ii) Q₁ = σ4πr₁² = 8 × 10⁻¹³ C, Q₂ = 72 × 10⁻¹³ C, total 8 × 10⁻¹² C. At a common potential charge divides in the ratio of radii (C ∝ r), so Q₁' = 8 × 10⁻¹² × 1/4 = 2 × 10⁻¹² C.

(i) Derive the impedance of a series RC circuit on v = vm sin ωt. (ii) When does an inductor behave as a plain conductor, and why? (iii) A lamp rated 110 V dc, 11 A is run from a 220 V, 50 Hz supply with a coil in series. Find the coil's inductance.

1. Part (i): Since V_R and V_C are 90° apart, vm² = (imR)² + (imX_C)², which gives Z = √(R² + X_C²) with X_C = 1/(ωC).
2. Part (ii): With DC, ω = 0, so X_L = ωL = 0, meaning no induced emf opposes the steady current.
3. Part (iii): Calculate the lamp resistance: R = 110/11 = 10 Ω.
4. Calculate the total impedance: Z = 220/11 = 20 Ω.
5. Find the inductive reactance: X_L = √(20² − 10²) = √300 = 10√3 Ω.
6. Calculate the inductance: L = X_L/(2πf) = 10√3/(2π × 50) = √3/(10π) H ≈ 0.055 H.

Answer: (i) V_R and V_C are 90° apart, so vm² = (imR)² + (imX_C)², giving Z = √(R² + X_C²) with X_C = 1/(ωC). (ii) With dc: ω = 0, so X_L = ωL = 0 (no changing current, no induced emf). (iii) R = 110/11 = 10 Ω, Z = 220/11 = 20 Ω, X_L = √(400 − 100) = 10√3 Ω, L = X_L/(2π × 50) = √3/(10π) H ≈ 0.055 H.

Note: Part (iii) treats the coil as a pure inductor with no resistance of its own, as CBSE's marking scheme does.

(i) Draw a labelled step-up transformer, state its working principle and three causes of energy loss in a real one. (ii) Does stepping up voltage violate energy conservation? Explain. (iii) Primary 200 turns, secondary 3000 turns, input 90 V. Find (1) the secondary voltage and (2) the primary current when the secondary current is 2.0 A.

1. Draw a transformer with a laminated soft-iron core, a primary coil with fewer turns, and a secondary coil with more turns.
2. State that the working principle is mutual induction, where an alternating current in the primary creates a changing magnetic flux that induces an emf in the secondary.
3. List three causes of energy loss: flux leakage, resistance of the windings, and eddy currents.
4. Explain that stepping up voltage does not violate energy conservation because the secondary current decreases in the same ratio that the voltage increases, keeping output power equal to input power.
5. Calculate the secondary voltage using the turns ratio: Vs = 90 × (3000/200) = 1350 V.
6. Calculate the primary current using the power balance: Ip = 2.0 × (3000/200) = 30 A.

Answer: (i) Primary (few turns) and secondary (more turns) on a laminated soft-iron core; works on mutual induction—alternating primary current gives changing flux that induces emf in the secondary. Losses (any three): flux leakage, resistance of windings, eddy currents, hysteresis. (ii) No: current falls in the same ratio as voltage rises, so output power equals input power (ideally). (iii) Vs = 90 × 3000/200 and Ip = 2.0 × 3000/200 give 1350 V and 30 A.

(i) Sketch how deviation through a prism varies with angle of incidence and define minimum deviation. (ii) A ray falls normally on one face of a prism of angle A and is deviated by δ. Prove n = sin(A + δ)/sin A. (iii) A prism of index √2 has refracting angle 60°. Find (1) the minimum deviation and (2) the angle of incidence.

1. (i) A graph of deviation δ against angle of incidence i is U-shaped: δ first falls to a least value δm, then rises.
2. At minimum deviation δm the angle of incidence equals the angle of emergence and the ray inside the prism runs parallel to the base.
3. (ii) For normal incidence on the first face r₁ = 0, and since r₁ + r₂ = A, r₂ = A.
4. Deviation δ = i + e − A with i = 0, so the angle of emergence is e = A + δ.
5. Snell's law at the second face: n sin r₂ = sin e, so n = sin(A + δ)/sin A.
6. (iii) n = sin((A + δm)/2)/sin(A/2): √2 = sin((60° + δm)/2)/sin 30°, so sin((60° + δm)/2) = √2 × 0.5 = 1/√2 and (60° + δm)/2 = 45°.
7. Hence δm = 90° − 60° = 30°, and the angle of incidence i = (A + δm)/2 = 45°.

Answer: (i) Deviation first falls, reaches a minimum δm, then rises (U-shaped curve); at δm the angle of incidence equals the angle of emergence (ray inside is parallel to the base). (ii) At the first face r₁ = 0, so r₂ = A; emergence angle e = A + δ, so n = sin e/sin r₂ = sin(A + δ)/sin A. (iii) n = sin((A + δm)/2)/sin(A/2) gives (60 + δm)/2 = 45°: δm = 30°, i = 45°.

(i) State Huygens' principle; construct the reflected wavefront for a plane wave meeting a mirror at angle i and prove reflection angle = incidence angle. (ii) Define coherent sources; can two separate sodium lamps be coherent? (iii) In YDSE with light of 520 nm and an unknown λ, the 4th bright fringe of λ falls on the 5th bright fringe of 520 nm. Find λ.

1. State Huygens' principle: every point on a wavefront acts as a source of secondary wavelets moving at the wave speed, and their forward envelope forms the new wavefront.
2. Construct the reflected wavefront by drawing secondary wavelets from points on the incident wavefront; the two right triangles formed share a hypotenuse and have equal sides vτ, making them congruent.
3. Since the triangles are congruent, the angle of reflection r equals the angle of incidence i.
4. Define coherent sources as those maintaining a constant phase difference over time; two separate sodium lamps are not coherent because their phases change randomly.
5. In YDSE, the position of the 4th bright fringe for λ is 4λD/d, and for the 5th bright fringe of 520 nm is 5 × 520 × D/d.
6. Equate the positions: 4λD/d = 5 × 520 × D/d, which simplifies to 4λ = 2600.
7. Solve for λ: λ = 2600 / 4 = 650 nm.

Answer: (i) Every point of a wavefront acts as a source of secondary wavelets moving at the wave speed; their forward envelope gives the new wavefront. In the construction the two right triangles share the hypotenuse and have equal sides vτ, so they are congruent and r = i. (ii) Coherent sources keep a constant phase difference in time; two independent sodium lamps cannot, since their phases change randomly. (iii) 4λD/d = 5 × 520 × D/d gives λ = 650 nm.