HISSAN Central Examination 2081 (2025) — Grade XII Physics Set D2 | Complete Analysis, Answer Guide & Exam Preparation

Subject Code: 1021 D2 | Full Marks: 75 (11 Obj. + 64 Subj.) | Time: 3 Hours
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Introduction

The HISSAN Central Examination 2081 (2025) Set D2 is the previous year’s companion to the 2082 papers, making it invaluable for understanding how HISSAN question patterns evolve year to year. A student who studies both 2081 and 2082 HISSAN papers develops a complete picture of every high-priority topic and question style used in pre-board and board examinations.

This 2081 Set D2 paper stands out for several reasons. It features one of the most intellectually demanding sets of Group C questions seen in any HISSAN paper: Bragg’s law with crystal plane spacing (Q.21b), Heisenberg uncertainty principle applied to nuclear physics (Q.21c), the mathematical proof that an electron’s path in an electric field is parabolic (Q.21 OR a), and Planck’s constant determination from a stopping potential graph (Q.21 OR c). The AC generator question (Q.22) is unusually comprehensive — covering working principle, EMF calculation, waveform sketch, Lenz’s law, transformer calculations, all in one question.

Other highlights include: a Carnot engine driving a Carnot refrigerator problem (Q.14b — requiring three separate heat flow calculations), the Biot-Savart law applied to magnetic forces on a current-carrying square frame in a magnetic field (Q.16b), displacement amplitude from pressure amplitude (Q.17 OR b), and the justification of why diffraction grating is superior to Young’s double slit for wavelength measurement (Q.20d).

This paper rewards genuine physical understanding at every level — it is exactly the kind of paper that separates well-prepared students from those who have only surface-level knowledge.

HISSAN 2081 Set D2 vs HISSAN 2082 Set D2 — Key Differences

Full Question Paper

Exam Overview

Topics Covered — Chapter-wise Weightage

Group A — MCQ Analysis with Answers

Q.1 — Ratio of Terminal Velocities of Drops of Radii R and R/2
Answer: d. 4:1

Terminal velocity: [latex]v_t = \frac{2r^2(\rho – \sigma)g}{9\eta} \Rightarrow v_t \propto r^2[/latex]

[latex]\frac{v_R}{v_{R/2}} = \frac{R^2}{(R/2)^2} = \frac{R^2}{R^2/4} = \mathbf{4:1}[/latex]

Q.2 — Kinetic Energy vs Time Graph for SHM Starting at Maximum Displacement
Answer: A

At t = 0, particle has maximum displacement → velocity = 0 → KE = 0.

For displacement [latex]x = A\cos(\omega t)[/latex] (starts at amplitude):

[latex]KE = \frac{1}{2}m\omega^2(A^2 – x^2) = \frac{1}{2}m\omega^2 A^2\sin^2(\omega t)[/latex]

KE is zero at t = 0, rises to maximum at T/4, back to zero at T/2, maximum at 3T/4, zero at T. This is a [latex]\sin^2[/latex] curve — always non-negative, with period T/2 (two complete oscillations per period T). Graph A shows this correctly: starting at zero, two symmetric peaks per period.

Q.3 — Radius of Gyration of Uniform Rod
Answer: a. [latex]\frac{L}{2\sqrt{3}}[/latex]

Moment of inertia of uniform rod about centre: [latex]I = \frac{mL^2}{12}[/latex]

Radius of gyration: [latex]I = mk^2 \Rightarrow k^2 = \frac{L^2}{12}[/latex]

[latex]k = \frac{L}{\sqrt{12}} = \frac{L}{2\sqrt{3}}[/latex]

Q.4 — Work Done in Cyclic Process ABCDA
Answer: d. [latex]200\pi[/latex] J

From the P-V diagram: The part ABC is semicircular. The semicircle lies between:

• P range: 30 to 50 N/m² → radius in P direction = 10 N/m²
• V range: 20 to 60 m³ → radius in V direction = 20 m³

Area of semicircle ABC = [latex]\frac{1}{2}\pi r_P r_V = \frac{1}{2}\pi \times 10 \times 20 = 100\pi[/latex] J

The DA portion is a straight horizontal line (isobaric at P = 30). Work done in DA = P × ΔV = 30 × (60-20) = 1200 J

Net work = area enclosed = [latex]\pi \times 10 \times 20 = 200\pi \approx 628[/latex] J (full ellipse area for the cycle)

Answer: d. [latex]200\pi[/latex] J — the net enclosed area of the ABCDA cycle.

Q.5 — Gas Receives Heat Q, Internal Energy Increases, Does Work W
Answer: b. Isobaric

From first law: Q = ΔU + W. Given Q > 0, ΔU > 0, W > 0.

• Isochoric: W = 0 (not matching)
• Isothermal: ΔU = 0 for ideal gas (not matching)
• Adiabatic: Q = 0 (not matching)
• Isobaric: Q = ΔU + PΔV. Both ΔU > 0 and W = PΔV > 0 possible ✓

Answer: Isobaric

Q.6 — Second Mode Frequency of Closed Organ Pipe with Fundamental 330 Hz
Answer: d. 990 Hz

For a closed organ pipe, only odd harmonics are present:

• Fundamental (1st harmonic): [latex]f_1 = \frac{v}{4L} = 330[/latex] Hz
• 1st overtone (2nd mode = 3rd harmonic): [latex]f_3 = 3f_1 = 3 \times 330 = \mathbf{990 \text{ Hz}}[/latex]

Note: “Second mode” means the second mode of vibration possible in the closed pipe — which is the 3rd harmonic (1st overtone).

Q.7 — Polarizing Angle for Glass with μ = 1.5
Answer: b. 56.3°

By Brewster’s Law: [latex]\tan i_B = \mu = 1.5[/latex]

[latex]i_B = \tan^{-1}(1.5) = 56.31° \approx \mathbf{56.3°}[/latex]

Q.8 — Temperature of Cold Junction
Answer: a. 40°C

Given: [latex]\theta_i = 600°C[/latex], [latex]\theta_n = 320°C[/latex]

Relation: [latex]\theta_n = \frac{\theta_c + \theta_i}{2}[/latex]

[latex]320 = \frac{\theta_c + 600}{2}[/latex]

[latex]640 = \theta_c + 600[/latex]

[latex]\theta_c = \mathbf{40°C}[/latex]

Q.9 — RMS Voltage When Frequency is Doubled in AC Generator
Answer: b. [latex]\frac{V}{\sqrt{2}}[/latex]

The EMF in an AC generator: [latex]e = NBA\omega\sin(\omega t)[/latex]

Peak EMF: [latex]V_0 = NBA\omega[/latex]

When frequency is doubled, [latex]\omega[/latex] doubles → [latex]V_0[/latex] doubles → new peak = 2V… wait.

But the question says the generator produces peak value V originally. When frequency doubles: new [latex]V_0′ = 2V[/latex].

New RMS: [latex]V_{rms}’ = \frac{V_0′}{\sqrt{2}} = \frac{2V}{\sqrt{2}} = V\sqrt{2}[/latex]

Answer: c. [latex]V\sqrt{2}[/latex]

However, if peak value V is fixed regardless of frequency (just the frequency changes): [latex]V_{rms} = V/\sqrt{2}[/latex] — answer b. The question is ambiguous — if the peak voltage stays V, rms = V/√2.

Q.10 — Electric Field to Move Oil Drop Upward at Same Terminal Speed
Answer: d. [latex]\frac{2mgv}{q}[/latex]

At terminal velocity v (falling): Weight = Upthrust + Viscous drag
[latex]mg = F_{upthrust} + F_{viscous}[/latex] (but at terminal velocity with no electric field: [latex]mg – F_{upthrust} = F_{viscous} = kv[/latex])

For upward motion at same speed v: Electric force must overcome weight + viscous drag (now both opposing upward motion):

[latex]qE = mg + F_{viscous} = mg + kv[/latex]

Since at terminal velocity (falling): [latex]kv = mg – F_{upthrust} \approx mg[/latex] (ignoring upthrust for simplicity):

[latex]qE = mg + mg = 2mg[/latex]

[latex]\boxed{E = \frac{2mg}{q}}[/latex]

Answer: b. [latex]\frac{2mg}{q}[/latex]

Explanation: Moving up at speed v means viscous drag now acts downward (opposing upward motion). So E must overcome both weight and drag, each equal to mg at terminal velocity.

Q.11 — Region of Initiation of Seismic Energy Within Earth
Answer: b. Hypocenter

The hypocenter (or focus) is the actual point within the Earth where seismic energy originates — where the fault rupture begins. The epicenter is the point on Earth’s surface directly above the hypocenter.

Group B — Short Answer Analysis

Q.12 — Rotational Dynamics and Angular Momentum (OR: SHM)

Part (a): Define Moment of Inertia (1 mark)
Moment of inertia of a body about a given axis is defined as the sum of products of the mass of each particle and the square of its perpendicular distance from the axis:

[latex]I = \sum m_i r_i^2[/latex]

It is the rotational analogue of mass. SI unit: kg·m². It depends on mass distribution and axis position.

Part (b): Conservation of Angular Momentum — State and Explain (2 marks)

Statement: The total angular momentum of a system remains constant when no external torque acts on it.

Mathematical form: [latex]\frac{dL}{dt} = \tau_{ext}[/latex]. If [latex]\tau_{ext} = 0 \Rightarrow L = I\omega = \text{constant}[/latex]

Explanation: When a spinning figure skater pulls her arms inward, her moment of inertia I decreases. Since [latex]I\omega[/latex] = constant, her angular velocity ω increases — she spins faster. This happens because no external torque acts on her (friction is negligible). The total angular momentum before and after arm-pulling is the same.

Part (c): Power Output of Electric Motor (2 marks)

Given: n = 4800 rpm, τ = 4.30 N·m

[latex]\omega = \frac{2\pi n}{60} = \frac{2\pi \times 4800}{60} = \frac{9600\pi}{60} = 160\pi \text{ rad/s}[/latex]

Power: [latex]P = \tau\omega = 4.30 \times 160\pi = 688\pi = \mathbf{2160.7 \text{ W} \approx 2.16 \text{ kW}}[/latex]

OR Q.12 — Simple Harmonic Motion

Part (a): Define SHM (1 mark)
Simple Harmonic Motion is a type of periodic oscillatory motion where the restoring force is directly proportional to the displacement from the mean position and directed opposite to it: [latex]F = -kx[/latex] or [latex]a = -\omega^2 x[/latex].

Part (b): Velocity vs Time Graph — Particle at Mean Position at t = 0 (1 mark)

For a particle starting at mean position (x = 0) at t = 0:
[latex]x = A\sin(\omega t)[/latex] → [latex]v = A\omega\cos(\omega t)[/latex]

The velocity-time graph is a cosine curve:

• At t = 0: v = Aω (maximum, positive)
• At t = T/4: v = 0 (at maximum displacement)
• At t = T/2: v = −Aω (maximum negative, returning through mean)
• At t = 3T/4: v = 0 again
• At t = T: v = Aω (back to start)

The graph is a smooth cosine wave oscillating between +Aω and −Aω.

Part (c): Maximum Velocity and Acceleration from Two Position-Velocity Data (3 marks)

Given: v₁ = 8 cm/s at x₁ = 12 cm; v₂ = 12 cm/s at x₂ = 8 cm

Using SHM velocity formula: [latex]v = \omega\sqrt{A^2 – x^2}[/latex]

[latex]v_1^2 = \omega^2(A^2 – x_1^2) \Rightarrow 64 = \omega^2(A^2 – 144)[/latex] … (i)

[latex]v_2^2 = \omega^2(A^2 – x_2^2) \Rightarrow 144 = \omega^2(A^2 – 64)[/latex] … (ii)

Dividing (ii) by (i):

[latex]\frac{144}{64} = \frac{A^2 – 64}{A^2 – 144}[/latex]

[latex]2.25(A^2 – 144) = A^2 – 64[/latex]

[latex]2.25A^2 – 324 = A^2 – 64[/latex]

[latex]1.25A^2 = 260 \Rightarrow A^2 = 208[/latex]

[latex]A = \sqrt{208} = 4\sqrt{13} \approx 14.42 \text{ cm}[/latex]

From (i): [latex]\omega^2 = \frac{64}{208 – 144} = \frac{64}{64} = 1[/latex]

[latex]\omega = 1 \text{ rad/s}[/latex]

Maximum velocity: [latex]v_{max} = \omega A = 1 \times 4\sqrt{13} = \mathbf{4\sqrt{13} \approx 14.42 \text{ cm/s}}[/latex]

Maximum acceleration: [latex]a_{max} = \omega^2 A = 1 \times 4\sqrt{13} = \mathbf{4\sqrt{13} \approx 14.42 \text{ cm/s}^2}[/latex]

Q.13 — Fluid Mechanics: Meniscus, Viscosity, and Flow Rate

Part (a): Water Concave, Mercury Convex Meniscus in Glass (1 mark)

Water in glass (concave meniscus): The adhesive force between water molecules and glass is greater than the cohesive force between water molecules themselves. Water is attracted to glass — it wets glass. The water surface curves upward at the edges (concave), pulling water up.

Mercury in glass (convex meniscus): The cohesive force between mercury molecules is much greater than the adhesive force between mercury and glass. Mercury does not wet glass — it is repelled. The mercury surface curves downward at the edges (convex), pushing mercury down.

Part (b): How Viscosity of Air Changes with Temperature (1 mark)

Viscosity of air increases with increasing temperature.

Unlike liquids (where viscosity decreases with temperature due to weakening cohesive bonds), gases gain viscosity with temperature because warmer gas molecules move faster and exchange momentum more rapidly between layers — increasing the internal friction between layers. For gases: [latex]\eta \propto \sqrt{T}[/latex].

Part (c): Rate of Flow of Glycerin Through Conical Pipe (3 marks)

Given: [latex]\rho = 1.25 \times 10^3[/latex] kg/m³, r₁ = 0.1 m, r₂ = 0.04 m, pressure drop [latex]\Delta P = 10[/latex] N/m²

Using Bernoulli’s equation (horizontal pipe) and continuity:

Continuity: [latex]A_1v_1 = A_2v_2[/latex]

[latex]\pi r_1^2 v_1 = \pi r_2^2 v_2 \Rightarrow v_2 = v_1\frac{r_1^2}{r_2^2} = v_1\frac{(0.1)^2}{(0.04)^2} = v_1 \times 6.25[/latex]

Bernoulli (horizontal): [latex]P_1 + \frac{1}{2}\rho v_1^2 = P_2 + \frac{1}{2}\rho v_2^2[/latex]

[latex]\Delta P = P_1 – P_2 = \frac{1}{2}\rho(v_2^2 – v_1^2) = \frac{1}{2}\rho v_1^2(6.25^2 – 1) = \frac{1}{2}\rho v_1^2(39.0625 – 1)[/latex]

[latex]10 = \frac{1}{2} \times 1250 \times v_1^2 \times 38.0625[/latex]

[latex]v_1^2 = \frac{10}{625 \times 38.0625} = \frac{10}{23789} = 4.204 \times 10^{-4}[/latex]

[latex]v_1 = 0.02050 \text{ m/s}[/latex]

Volume flow rate: [latex]Q = A_1 v_1 = \pi r_1^2 v_1 = \pi \times (0.1)^2 \times 0.02050[/latex]

[latex]Q = \pi \times 0.01 \times 0.02050 = \pi \times 2.05 \times 10^{-4} = \mathbf{6.44 \times 10^{-4} \text{ m}^3/\text{s}}[/latex]

Q.14 — Thermodynamics: First Law and Carnot Engine-Refrigerator Chain

Part (a): First Law and Adiabatic Cooling (1+1 marks)

First Law: [latex]Q = \Delta U + W[/latex]

Heat supplied = increase in internal energy + work done by the system.

Why cooling in adiabatic expansion:
In adiabatic expansion, no heat is exchanged with surroundings: [latex]Q = 0[/latex].

From first law: [latex]0 = \Delta U + W[/latex] → [latex]\Delta U = -W[/latex]

The gas does positive work (W > 0) during expansion. This work comes entirely from the internal energy of the gas. Internal energy decreases → temperature decreases → cooling effect observed. ✓

Part (b): Carnot Engine Driving Carnot Refrigerator — Find Q₂, Q₃, Q₄ (3 marks)

From the figure: Carnot Heat Engine operates between 600K (source) and 300K (sink). It receives [latex]Q_1 = 1000[/latex] J from 600K source and rejects [latex]Q_2[/latex] to 300K sink, doing work [latex]W_{out}[/latex].

Carnot Refrigerator operates between 500K (hot) and 400K (cold). It receives work [latex]W_{in} = W_{out}[/latex] from engine, absorbs [latex]Q_4[/latex] from 400K cold reservoir, rejects [latex]Q_3[/latex] to 500K hot reservoir.

For Carnot Engine (600K → 300K):

[latex]\eta = 1 – \frac{T_{cold}}{T_{hot}} = 1 – \frac{300}{600} = \frac{1}{2}[/latex]

[latex]W_{out} = \eta Q_1 = \frac{1}{2} \times 1000 = 500 \text{ J}[/latex]

[latex]Q_2 = Q_1 – W_{out} = 1000 – 500 = \mathbf{500 \text{ J}}[/latex]

For Carnot Refrigerator (400K → 500K):

[latex]COP = \frac{T_{cold}}{T_{hot} – T_{cold}} = \frac{400}{500 – 400} = \frac{400}{100} = 4[/latex]

[latex]W_{in} = W_{out} = 500 \text{ J}[/latex]

[latex]COP = \frac{Q_4}{W_{in}} \Rightarrow Q_4 = COP \times W_{in} = 4 \times 500 = \mathbf{2000 \text{ J}}[/latex]

[latex]Q_3 = Q_4 + W_{in} = 2000 + 500 = \mathbf{2500 \text{ J}}[/latex]

Q.15 — Wheatstone Bridge and Potentiometer Gradient

Part (a): Balanced Condition of Wheatstone Bridge Using Kirchhoff’s Laws (3 marks)

Consider four resistances P, Q, R, S in a bridge, with battery in one diagonal and galvanometer in the other. Let currents through P, R be [latex]I_1[/latex] and through Q, S be [latex]I_2[/latex].

At balance: Galvanometer current [latex]I_g = 0[/latex]

Apply KCL at junction B:
[latex]I_1 = I_{P} = I_{R}[/latex] (no current branches to galvanometer)

Apply KCL at junction D:

[latex]I_2 = I_Q = I_S[/latex]

Apply KVL to loop ABD (left loop):
[latex]I_1 P – I_2 Q = 0 \Rightarrow I_1 P = I_2 Q[/latex] … (i)

Apply KVL to loop BCD (right loop):
[latex]I_1 R – I_2 S = 0 \Rightarrow I_1 R = I_2 S[/latex] … (ii)

Dividing (i) by (ii):

[latex]\frac{P}{R} = \frac{Q}{S}[/latex]

Cross-multiplying: [latex]\boxed{PS = QR}[/latex] or equivalently [latex]\frac{P}{Q} = \frac{R}{S}[/latex]

This is the balanced condition of the Wheatstone bridge. ✓

Part (b): Potential Gradient Along Potentiometer Wire (2 marks)

Given: Wire length L = 4 m, wire resistance [latex]R_w[/latex] = 10 Ω, battery EMF = 2V, series resistance [latex]R_s[/latex] = 5Ω

Total circuit resistance = [latex]R_w + R_s = 10 + 5 = 15[/latex] Ω

Current in circuit: [latex]I = \frac{V}{R_{total}} = \frac{2}{15} \text{ A}[/latex]

Voltage across potentiometer wire: [latex]V_w = I \times R_w = \frac{2}{15} \times 10 = \frac{20}{15} = \frac{4}{3} \text{ V}[/latex]

Potential gradient: [latex]k = \frac{V_w}{L} = \frac{4/3}{4} = \mathbf{\frac{1}{3} \text{ V/m} \approx 0.333 \text{ V/m}}[/latex]

Q.16 — Biot-Savart Law and Square Frame in Magnetic Field

Part (a): Biot-Savart Law — Magnetic Field Due to Long Straight Conductor (3 marks)

Biot-Savart Law:

[latex]dB = \frac{\mu_0 I,dl\sin\theta}{4\pi r^2}[/latex]

For a long straight conductor carrying current I, the field at perpendicular distance a:

Consider a small element dl at angle φ from the perpendicular. Setting up the integral:

[latex]B = \frac{\mu_0 I}{4\pi a}\int_{-\pi/2}^{\pi/2}\cos\phi,d\phi[/latex]

(where we substitute [latex]l = a\tan\phi[/latex], [latex]dl = a\sec^2\phi,d\phi[/latex], [latex]r = a\sec\phi[/latex], [latex]\sin\theta = \cos\phi[/latex])

[latex]B = \frac{\mu_0 I}{4\pi a}[\sin\phi]_{-\pi/2}^{\pi/2} = \frac{\mu_0 I}{4\pi a}[1-(-1)] = \frac{\mu_0 I}{4\pi a} \times 2[/latex]

[latex]\boxed{B = \frac{\mu_0 I}{2\pi a}}[/latex]

Direction: Using right-hand rule — field circles around the conductor.

Part (b): Magnetic Forces on Four Sides of Square Frame (2 marks)

Given: Square side = 20 cm = 0.2 m, I = 2A (enters at ‘a’, leaves at ‘c’ — opposite corner), B = 0.25 T (perpendicular to plane, out of page from dots in figure)

Since current enters at ‘a’ and leaves at ‘c’, the current splits into two paths:

• Path a→b→c (two sides)
• Path a→d→c (two sides)

Each path carries I/2 = 1A (by symmetry, since both paths have equal resistance — same length, same wire).

Force on each side: [latex]F = BIL = 0.25 \times 1 \times 0.2 = 0.05 \text{ N}[/latex]

Side ‘ab’ (current flows from a to b, say leftward): [latex]F = BIL = 0.05[/latex] N, direction by Fleming’s left-hand rule: downward

Side ‘bc’ (current flows from b to c, say downward): F = 0.05 N, direction: rightward

Side ‘ad’ (current flows from a to d, say rightward): F = 0.05 N, direction: upward

Side ‘dc’ (current flows from d to c, say downward): F = 0.05 N, direction: leftward

The four forces are equal in magnitude (0.05 N each) and directed outward from the centre — the frame experiences a tendency to expand in the magnetic field. Each force = 0.05 N.

Q.17 — Organ Pipes and Stretched String (OR: Laplace and Displacement Amplitude)

Part (a): Open Pipe First Harmonic = Closed Pipe First Overtone (2 marks)

Open pipe 1st harmonic: [latex]f_1^{open} = \frac{v}{2L_o}[/latex]

Closed pipe 1st overtone (3rd harmonic): [latex]f_{1st \text{ overtone}}^{closed} = \frac{3v}{4L_c}[/latex]

Both produce same note: [latex]f_1^{open} = f_{1st \text{ overtone}}^{closed}[/latex]

[latex]\frac{v}{2L_o} = \frac{3v}{4L_c}[/latex]

[latex]4L_c = 6L_o \Rightarrow \frac{L_o}{L_c} = \frac{4}{6} = \frac{2}{3}[/latex]

[latex]\boxed{L_o : L_c = 2 : 3}[/latex]

The closed pipe is 1.5 times longer than the open pipe for the same note.

Part (b): End Correction (1 mark)
End correction is the small additional length that must be added to the actual length of an organ pipe to account for the fact that the antinode at the open end forms slightly outside the pipe (not exactly at the rim). For a pipe of radius r, end correction ≈ 0.6r for each open end. For an open pipe: effective length = L + 2(0.6r). For a closed pipe: effective length = L + 0.6r.

Part (c): Modes of Vibration in a Stretched String (2 marks)

A stretched string fixed at both ends forms standing waves with nodes at both ends.

Fundamental (1st harmonic):

[latex]L = \frac{\lambda_1}{2} \Rightarrow f_1 = \frac{v}{2L} = \frac{1}{2L}\sqrt{\frac{T}{\mu}}[/latex]

1st overtone (2nd harmonic):

[latex]L = \lambda_2 \Rightarrow f_2 = \frac{v}{L} = 2f_1[/latex]

2nd overtone (3rd harmonic):

[latex]L = \frac{3\lambda_3}{2} \Rightarrow f_3 = \frac{3v}{2L} = 3f_1[/latex]

General: [latex]f_n = \frac{nv}{2L} = nf_1[/latex] for n = 1, 2, 3,…

A stretched string produces all harmonics (both odd and even). The fundamental gives the pitch; the presence of overtones determines the quality (timbre) of the sound.

OR Q.17 — Newton vs Laplace and Displacement Amplitude

Part (a): Laplace’s Correction to Newton’s Formula (3 marks)

Newton’s formula and why it was wrong:
Newton assumed sound propagation is isothermal (T = constant), giving:

[latex]v_{Newton} = \sqrt{\frac{P}{\rho}} \approx 280 \text{ m/s at 0°C}[/latex]

This is wrong because sound travels too rapidly for heat to flow between adjacent compressed and rarefied layers. The actual speed (~332 m/s) is much higher.

Laplace’s correction:
Laplace recognized the process is adiabatic ([latex]PV^\gamma = \text{constant}[/latex]), giving effective bulk modulus = [latex]\gamma P[/latex]:

[latex]\boxed{v_{Laplace} = \sqrt{\frac{\gamma P}{\rho}}}[/latex]

For air: [latex]\gamma = 1.4[/latex], so [latex]v = \sqrt{1.4} \times 280 = 1.183 \times 280 = 331.2[/latex] m/s — matches experiment. ✓

Part (b): Displacement Amplitude from Pressure Amplitude (2 marks)

Given: f = 100 Hz, pressure amplitude [latex]\Delta P_0 = 10[/latex] Pa, v = 340 m/s, [latex]\rho = 1.29[/latex] kg/m³

Relation between displacement amplitude A and pressure amplitude [latex]\Delta P_0[/latex]:

[latex]\Delta P_0 = \rho v \omega A = \rho v (2\pi f) A[/latex]

[latex]A = \frac{\Delta P_0}{\rho v (2\pi f)} = \frac{10}{1.29 \times 340 \times 2\pi \times 100}[/latex]

[latex]= \frac{10}{1.29 \times 340 \times 628.3}[/latex]

[latex]= \frac{10}{275,820} = \mathbf{3.63 \times 10^{-5} \text{ m} = 36.3 \text{ μm}}[/latex]

Q.18 — P-N Junction, Full-Wave Rectifier, and NAND Gate

Part (a): Property Enabling P-N Diode as Rectifier (1 mark)
The key property is unidirectional current flow (rectifying property). A P-N junction diode conducts current easily in forward bias (low resistance) but blocks current in reverse bias (very high resistance). This asymmetric behavior — acting as a one-way valve for current — makes it ideal for converting AC (bidirectional) into DC (unidirectional).

Part (b): Full-Wave Rectifier — Construction and Working (3 marks)

Construction (Bridge Rectifier):
Four diodes D₁, D₂, D₃, D₄ arranged in a bridge configuration between the AC input and load [latex]R_L[/latex].

Working:

Positive half-cycle (A positive, B negative):
Current path: A → D₁ → [latex]R_L[/latex] → D₃ → B. Diodes D₁ and D₃ conduct; D₂ and D₄ are reverse biased.

Negative half-cycle (B positive, A negative):
Current path: B → D₂ → [latex]R_L[/latex] → D₄ → A. Diodes D₂ and D₄ conduct; D₁ and D₃ are reverse biased.

Result: Current through [latex]R_L[/latex] flows in the same direction during both half-cycles → full-wave rectification. Output is pulsating DC with ripple frequency = 2f. A smoothing capacitor reduces the ripple further.

Part (c): NAND Gate Truth Table (1 mark)

NAND = NOT(AND): [latex]Y = \overline{A \cdot B}[/latex]

NAND is a universal gate — any logic circuit can be built from NAND gates alone.

Q.19 — Radioactivity: Curie, Half-Life, and Percentage Remaining

Part (a): Define One Curie (2 marks)
One Curie (Ci) is the unit of activity of a radioactive substance, defined as the quantity of radioactive material in which [latex]3.7 \times 10^{10}[/latex] disintegrations occur per second (equal to the activity of 1 gram of Radium-226).

[latex]1 \text{ Ci} = 3.7 \times 10^{10} \text{ Bq}[/latex]

The SI unit of radioactivity is the Becquerel (Bq) = 1 disintegration per second. The Curie is the older but still widely used unit in medicine and industry.

Part (c): Relation Between Half-Life and Decay Constant (2 marks)

From the decay equation [latex]N = N_0 e^{-\lambda t}[/latex]:

At [latex]t = t_{1/2}[/latex]: [latex]N = \frac{N_0}{2}[/latex]

[latex]\frac{N_0}{2} = N_0 e^{-\lambda t_{1/2}}[/latex]

[latex]\frac{1}{2} = e^{-\lambda t_{1/2}}[/latex]

Taking natural log: [latex]\ln\frac{1}{2} = -\lambda t_{1/2}[/latex]

[latex]-\ln 2 = -\lambda t_{1/2}[/latex]

[latex]\boxed{t_{1/2} = \frac{\ln 2}{\lambda} = \frac{0.693}{\lambda}}[/latex]

Part (d): Percentage Remaining After 25 Days (2 marks)

Given: [latex]t_{1/2} = 10[/latex] days, t = 25 days

Number of half-lives: [latex]n = \frac{t}{t_{1/2}} = \frac{25}{10} = 2.5[/latex]

Fraction remaining: [latex]\frac{N}{N_0} = \left(\frac{1}{2}\right)^n = \left(\frac{1}{2}\right)^{2.5} = \frac{1}{2^{2.5}} = \frac{1}{4\sqrt{2}} = \frac{1}{5.657}[/latex]

Percentage remaining: [latex]\frac{N}{N_0} \times 100 = \frac{100}{5.657} = \mathbf{17.68% \approx 17.7%}[/latex]

Group C — Long Answer Analysis

Q.20 — Wave Optics: Polarization, YDSE, and Diffraction Grating

Part (a): Why Sound Waves Cannot be Plane Polarized (1 mark)

Sound waves are longitudinal waves — the particle oscillations are parallel to the direction of wave propagation. Polarization requires that the transverse oscillations be restricted to one plane.

Since longitudinal waves oscillate only along the direction of travel (not transversely), there is no transverse component to polarize — sound cannot be plane polarized.

Only transverse waves (like light, microwaves) can be polarized.

Part (b): YDSE — Proof that Dark and Bright Fringes are Equally Spaced (3 marks)

Setup: Two slits [latex]S_1[/latex] and [latex]S_2[/latex] separated by d, screen at distance D, point P at height y from centre.

Path difference: [latex]\Delta = \frac{yd}{D}[/latex]

Bright fringes (constructive): [latex]\Delta = n\lambda[/latex]
[latex]y_n^{bright} = \frac{n\lambda D}{d}[/latex] for n = 0, 1, 2, 3, …

Dark fringes (destructive): [latex]\Delta = (2n-1)\frac{\lambda}{2}[/latex]
[latex]y_n^{dark} = \frac{(2n-1)\lambda D}{2d}[/latex] for n = 1, 2, 3, …

Fringe width (consecutive brights):

[latex]\beta^{bright} = y_{n+1} – y_n = \frac{(n+1)\lambda D}{d} – \frac{n\lambda D}{d} = \frac{\lambda D}{d}[/latex]

Fringe width (consecutive darks):

[latex]\beta^{dark} = y_{n+1}^{dark} – y_n^{dark} = \frac{(2n+1)\lambda D}{2d} – \frac{(2n-1)\lambda D}{2d} = \frac{2\lambda D}{2d} = \frac{\lambda D}{d}[/latex]

Since [latex]\beta^{bright} = \beta^{dark} = \frac{\lambda D}{d}[/latex], bright and dark fringes are equally spaced. ✓

Part (c): Polarizing Angle of Water with Critical Angle 48° (2 marks)

From critical angle: [latex]\sin\theta_c = \frac{1}{\mu} \Rightarrow \mu = \frac{1}{\sin 48°} = \frac{1}{0.7431} = 1.346[/latex]

By Brewster’s Law:

[latex]\tan i_B = \mu = 1.346[/latex]

[latex]i_B = \tan^{-1}(1.346) = \mathbf{53.4°}[/latex]

Part (d): Why Diffraction Grating is Better than YDSE for Wavelength Measurement (2 marks)

1. Much sharper maxima: A diffraction grating has thousands of slits (N ~ 400-2000 per mm). Maxima have width proportional to 1/N — extremely narrow and bright. YDSE with 2 slits produces broad, less distinct maxima.
2. Higher angular dispersion: Grating equation [latex]d\sin\theta = n\lambda[/latex] with very small d (grating spacing) gives large θ for a given λ. Small differences in wavelength produce measurable angular separations. YDSE has large d (mm scale), giving very small angular separation — hard to measure precisely.
3. Higher resolving power: Grating resolving power = nN (order × number of slits), enabling wavelengths very close together to be resolved. YDSE cannot resolve close wavelengths.
4. More precise measurement: The angle of diffraction maximum can be measured to high precision with a spectrometer, giving highly accurate λ values.

Q.21 — Bragg’s Law, Heisenberg, and Bohr (OR: Photoelectric Effect and Electron Path)

Part (a): Bragg’s Law — State and Explain (2 marks)

Statement: When X-rays of wavelength λ are incident on crystal planes separated by spacing d at a glancing angle θ, constructive interference (diffraction maximum) occurs when:

[latex]\boxed{2d\sin\theta = n\lambda}[/latex]

where n = 1, 2, 3, … is the order of diffraction.

Explanation: X-rays reflect from successive parallel crystal planes. The path difference between rays reflected from adjacent planes = 2d sinθ. For constructive interference, this path difference must equal a whole number of wavelengths. This allows measurement of crystal interplanar spacing d if λ and θ are known, or measurement of λ if d is known.

Part (b): Crystal Plane Spacing Calculation (2 marks)

Given: λ = [latex]4.00 \times 10^{-11}[/latex] m, scattering angle = 26.4°

Note: Bragg’s law uses the glancing angle (angle from the crystal plane). The scattering angle in the diagram is 2θ (angle between incident and scattered beam), so:

Glancing angle: [latex]\theta = \frac{26.4°}{2} = 13.2°[/latex]

For first order (n = 1): [latex]2d\sin\theta = \lambda[/latex]

[latex]d = \frac{\lambda}{2\sin\theta} = \frac{4.00 \times 10^{-11}}{2 \times \sin(13.2°)} = \frac{4.00 \times 10^{-11}}{2 \times 0.2284} = \frac{4.00 \times 10^{-11}}{0.4568}[/latex]

[latex]d = \mathbf{8.76 \times 10^{-11} \text{ m} = 0.876 \text{ Å}}[/latex]

Part (c): Why Electron Cannot Exist in Nucleus — Heisenberg Uncertainty Principle (2 marks)

The nucleus has diameter [latex]\approx 10^{-14}[/latex] m. If an electron existed inside the nucleus, its position uncertainty would be at most [latex]\Delta x \approx 10^{-14}[/latex] m.

By Heisenberg’s Uncertainty Principle:

[latex]\Delta x \cdot \Delta p \geq \frac{h}{4\pi}[/latex]

[latex]\Delta p \geq \frac{h}{4\pi \Delta x} = \frac{6.63 \times 10^{-34}}{4\pi \times 10^{-14}} = \frac{6.63 \times 10^{-34}}{1.257 \times 10^{-13}} = 5.27 \times 10^{-21} \text{ kg·m/s}[/latex]

Minimum kinetic energy:

[latex]KE = \frac{p^2}{2m_e} = \frac{(5.27 \times 10^{-21})^2}{2 \times 9.1 \times 10^{-31}} = \frac{2.78 \times 10^{-41}}{1.82 \times 10^{-30}} = 1.53 \times 10^{-11} \text{ J} \approx 95 \text{ MeV}[/latex]

But the nuclear binding energy per nucleon is only ~8 MeV. An electron with 95 MeV kinetic energy cannot be bound inside the nucleus — it would immediately escape. Therefore, electrons cannot exist within the nucleus. ✓

Part (d): Bohr’s Postulates (2 marks)

Postulate 1: Electrons revolve in fixed circular orbits (stationary states) without emitting radiation.

Postulate 2: Angular momentum is quantized: [latex]mvr = \frac{nh}{2\pi}[/latex] (n = 1, 2, 3, …)

Postulate 3: Electrons emit or absorb a photon of frequency [latex]f = \frac{E_2 – E_1}{h}[/latex] when jumping between orbits.

OR Q.21 — Electron in Electric Field and Photoelectric Graph

Part (a): Proof That Electron Path in Electric Field is Parabolic (3 marks)

From the figure: An electron beam enters the electric field region horizontally with velocity [latex]v_x = v[/latex]. The electric field E is directed vertically (perpendicular to initial motion).

Horizontal motion (no force in x-direction):
[latex]x = v \cdot t \Rightarrow t = \frac{x}{v}[/latex] … (i)

Vertical motion (constant electric force [latex]F = eE[/latex]):
Acceleration: [latex]a = \frac{eE}{m_e}[/latex]

Vertical displacement:
[latex]y = \frac{1}{2}at^2 = \frac{1}{2} \cdot \frac{eE}{m_e} \cdot t^2[/latex] … (ii)

Substituting (i) into (ii):

[latex]y = \frac{1}{2} \cdot \frac{eE}{m_e} \cdot \frac{x^2}{v^2} = \left(\frac{eE}{2m_e v^2}\right)x^2[/latex]

This is of the form [latex]y = kx^2[/latex] where [latex]k = \frac{eE}{2m_e v^2}[/latex] is a constant.

This is the equation of a parabola — confirming the path is parabolic. ✓

This is analogous to projectile motion under gravity — constant transverse acceleration produces parabolic trajectory.

Part (b): Photoelectric Effect, Work Function, and Stopping Potential (3 marks)

Photoelectric effect: When light of sufficient frequency falls on a metal surface, electrons are emitted from the surface. This phenomenon is called the photoelectric effect. It demonstrates the particle (photon) nature of light.

Work function (φ): The minimum energy required to liberate one electron from the metal surface. It is characteristic of the metal. [latex]\phi = hf_0[/latex] where [latex]f_0[/latex] = threshold frequency. Metals with low work function (like Cs, K) are more photosensitive.

Stopping potential (V₀): The minimum negative potential applied to the collector that stops even the most energetic photoelectrons and reduces photocurrent to zero. Related to maximum KE:

[latex]eV_0 = KE_{max} = hf – \phi[/latex]

V₀ increases with increasing frequency; it is independent of intensity.

Part (c): From V₀ vs f Graph — Find Planck’s Constant and Work Function (1+1 marks)

From the graph: [latex]V_0[/latex] (in volts) vs f (in [latex]\times 10^{14}[/latex] Hz)

Reading from graph:

• Line passes through approximately (3, 0.2) and (5, 1.0)… let’s identify threshold frequency and slope.
• X-intercept (threshold frequency): approximately [latex]f_0 \approx 2 \times 10^{14}[/latex] Hz (where V₀ = 0)
• The line appears to go from (0, −0.8) to (5, 1.0) approximately

Slope:

[latex]\text{slope} = \frac{\Delta V_0}{\Delta f} = \frac{1.0 – 0}{(5-2) \times 10^{14}} \approx \frac{1.0}{3 \times 10^{14}} = 3.33 \times 10^{-15} \text{ V/Hz}[/latex]

(i) Planck’s constant:

[latex]h = e \times \text{slope} = 1.6 \times 10^{-19} \times 3.33 \times 10^{-15} = \mathbf{5.33 \times 10^{-34} \text{ J·s}}[/latex]

(Accepted value: [latex]6.63 \times 10^{-34}[/latex] J·s — actual reading from graph gives a value close to this depending on the scale)

(ii) Work function:
At threshold ([latex]V_0 = 0[/latex]): [latex]\phi = hf_0 = 5.33 \times 10^{-34} \times 2 \times 10^{14} = 1.07 \times 10^{-19} \text{ J} \approx \mathbf{0.67 \text{ eV}}[/latex]

Q.22 — AC Generator and Transformer (OR: LCR Circuit)

Part (a): Working Principle of AC Generator (1 mark)

An AC generator works on the principle of electromagnetic induction (Faraday’s law): when a conducting coil rotates in a magnetic field, the magnetic flux through the coil changes continuously, inducing an alternating EMF in the coil.

[latex]\varepsilon = -N\frac{d\Phi}{dt}[/latex]

Part (b): Quantitative Value of Induced EMF (2 marks)

For a rectangular coil of N turns, area A, rotating with angular velocity ω in field B:

Flux: [latex]\Phi = NBA\cos(\omega t)[/latex]

Induced EMF: [latex]\varepsilon = -N\frac{d\Phi}{dt} = NBA\omega\sin(\omega t) = \varepsilon_0\sin(\omega t)[/latex]

Peak EMF: [latex]\varepsilon_0 = NBA\omega = NBA(2\pi f)[/latex]

RMS EMF: [latex]\varepsilon_{rms} = \frac{\varepsilon_0}{\sqrt{2}} = \frac{NBA\omega}{\sqrt{2}}[/latex]

Part (c): Waveform Sketch (1 mark)

The EMF waveform is a sinusoidal curve: [latex]\varepsilon = \varepsilon_0\sin(\omega t)[/latex]

The graph shows:

• Sinusoidal shape oscillating between [latex]+\varepsilon_0[/latex] and [latex]-\varepsilon_0[/latex]
• Period T = 2π/ω
• Zero at t = 0, maximum at t = T/4, zero at t = T/2, minimum at t = 3T/4, zero at t = T

Part (d): Lenz’s Law and Conservation of Energy (2 marks)

Lenz’s Law: The direction of the induced EMF and hence the induced current is always such as to oppose the change in magnetic flux that caused it.

Lenz’s Law and Conservation of Energy:
If the induced current aided the change in flux (instead of opposing it), it would create a positive feedback loop — more flux → more induced current → more flux — generating energy from nothing. This would violate conservation of energy.

By opposing the cause, the induced current creates a force that opposes the motion causing the flux change. Mechanical work must be done against this opposing force. This mechanical energy is converted to electrical energy in the coil. Energy is conserved — not created. Therefore Lenz’s law is fully consistent with conservation of energy. ✓

Part (e): Transformer — Output Voltage and Current (2 marks)

Given: [latex]N_p = 500[/latex], [latex]N_s = 100[/latex], [latex]V_p = 4000[/latex] V

Output voltage:

[latex]\frac{V_s}{V_p} = \frac{N_s}{N_p} = \frac{100}{500} = \frac{1}{5}[/latex]

[latex]V_s = \frac{4000}{5} = \mathbf{800 \text{ V}}[/latex]

This is a step-down transformer (N_s < N_p).

Output current (ideal transformer — power conserved):

[latex]V_p I_p = V_s I_s[/latex]

[latex]I_s = \frac{V_p I_p}{V_s} = \frac{4000 \times I_p}{800} = 5I_p[/latex]

Output current is 5 times the input current. If input current [latex]I_p[/latex] is not given, state: [latex]I_s = \frac{N_p}{N_s} \times I_p = 5I_p[/latex]

OR Q.22 — LCR Series Circuit

Part (a): Current Expression and Impedance (2 marks)

For series LCR with [latex]V = V_0\sin\omega t[/latex] and [latex]X_L > X_C[/latex] (inductive circuit):

Phasor combination: [latex]V = \sqrt{V_R^2 + (V_L – V_C)^2}[/latex]

[latex]I = \frac{V}{Z}[/latex] where [latex]Z = \sqrt{R^2 + (X_L – X_C)^2} = \sqrt{R^2 + \left(\omega L – \frac{1}{\omega C}\right)^2}[/latex]

Current: [latex]I = \frac{V_0}{Z}\sin(\omega t – \phi)[/latex]

Phase angle: [latex]\tan\phi = \frac{X_L – X_C}{R}[/latex] (current lags voltage when inductive)

Part (b): Iron Core Inserted in Inductor — Effect on Current (1 mark)

When an iron core is inserted inside the inductor:

• Iron has high magnetic permeability → inductance L increases significantly
• Inductive reactance [latex]X_L = \omega L[/latex] increases
• Impedance Z increases
• Current [latex]I = V/Z[/latex] decreases

The iron core makes the inductor more effective at opposing AC current.

Part (c): Inductor as Open Circuit, Capacitor as Perfect Conductor (1 mark)

Inductor as open circuit: At very high frequency ([latex]\omega \to \infty[/latex]): [latex]X_L = \omega L \to \infty[/latex]. The inductor blocks very high frequency AC — behaves like an open circuit.

Capacitor as perfect conductor: At very high frequency ([latex]\omega \to \infty[/latex]): [latex]X_C = \frac{1}{\omega C} \to 0[/latex]. The capacitor passes very high frequency AC without opposition — behaves like a short circuit (perfect conductor).

Part (d): Resonance and Resonant Frequency (2 marks)

Resonance in a series LCR circuit occurs when the inductive reactance equals the capacitive reactance ([latex]X_L = X_C[/latex]), resulting in minimum impedance (Z = R) and maximum current.

Derivation:
[latex]X_L = X_C[/latex]
[latex]\omega L = \frac{1}{\omega C}[/latex]
[latex]\omega^2 = \frac{1}{LC}[/latex]

[latex]\omega_r = \frac{1}{\sqrt{LC}}[/latex]

[latex]\boxed{f_r = \frac{1}{2\pi\sqrt{LC}}}[/latex]

At resonance: Z = R, cos φ = 1, power factor = 1, maximum power transfer to resistive load.

Part (e): LR Circuit — Current and Phase Shift (2 marks)

Given: V = 100V, f = 50 Hz, [latex]X_L = 31.4[/latex] Ω, R = 25 Ω (no capacitor — RL circuit)

Impedance: [latex]Z = \sqrt{R^2 + X_L^2} = \sqrt{25^2 + 31.4^2} = \sqrt{625 + 985.96} = \sqrt{1610.96} = 40.14[/latex] Ω

Current: [latex]I = \frac{V}{Z} = \frac{100}{40.14} = \mathbf{2.49 \text{ A}}[/latex]

Phase shift: [latex]\tan\phi = \frac{X_L}{R} = \frac{31.4}{25} = 1.256[/latex]

[latex]\phi = \tan^{-1}(1.256) = \mathbf{51.5°}[/latex]

Current lags voltage by 51.5°.

Marks Weightage Summary

Difficulty Analysis

Easy — Reliable marks:

• Q.6 (Closed pipe 2nd mode — pattern recognition)
• Q.7 (Brewster’s angle — direct formula)
• Q.11 (Hypocenter — definition)
• Q.12(a) (Moment of inertia definition)
• Q.13(a) (Concave/convex meniscus)
• Q.14(a) (First law + adiabatic cooling)
• Q.15(b) (Potential gradient)
• Q.17(b) (End correction definition)
• Q.18(a) (Rectifier property of diode)
• Q.18(c) (NAND truth table)
• Q.19(a) (One curie definition)
• Q.22(a) (Generator working principle)

Moderate — Needs regular practice:

• Q.1 (Terminal velocity ratio — formula recall)
• Q.3 (Radius of gyration — moment of inertia)
• Q.5 (Identifying isobaric from Q = ΔU + W)
• Q.8 (Cold junction temperature calculation)
• Q.12(c) (Motor power calculation)
• Q.13(b) (Viscosity of air with temperature)
• Q.17(a) (Organ pipe resonance length ratio)
• Q.17(c) (Stretched string modes)
• Q.18(b) (Full-wave rectifier)
• Q.19(c)(d) (Half-life relation + percentage)
• Q.20(a)(b) (Sound polarization + YDSE proof)

Challenging — For top scorers:

• Q.2 (KE-time graph for SHM starting at maximum)
• Q.4 (Cyclic process work — reading P-V diagram)
• Q.10 (Electric field for upward motion at terminal speed)
• Q.12 OR (c) (SHM: amplitude and ω from two velocity points)
• Q.13(c) (Glycerin flow rate — Bernoulli + continuity)
• Q.14(b) (Carnot engine driving refrigerator — Q₂, Q₃, Q₄)
• Q.15(a) (Wheatstone bridge via Kirchhoff — full derivation)
• Q.16 (Biot-Savart + long conductor derivation + square frame forces)
• Q.20(c)(d) (Brewster from critical angle + grating superiority)
• Q.21(b)(c) (Bragg’s law numerical + Heisenberg in nucleus)
• Q.21 OR (a) (Mathematical proof of parabolic path)
• Q.22(e) (Transformer output voltage + current)

Student Preparation Tips

1. The Carnot Engine → Refrigerator Problem (Q.14b) Requires Systematic Thinking
This is a 3-mark chain problem. Step 1: Find engine efficiency → Work output. Step 2: Use this work as input to refrigerator. Step 3: Find refrigerator COP → Q₄ → Q₃. Always draw the energy flow diagram before calculating. Practice similar cascaded thermodynamic problems.

2. The Heisenberg Nuclear Problem (Q.21c) is Conceptually Powerful
To show electrons can’t be in the nucleus: (a) use nuclear diameter as Δx, (b) apply ΔxΔp ≥ h/4π to find minimum Δp, (c) calculate KE = Δp²/2mₑ, (d) compare with ~8 MeV binding energy. This proves electrons can’t be confined there. Practice this full calculation.

3. The Parabolic Path Proof (Q.21 OR a) is Standard Projectile Motion
Treat horizontal velocity as constant (vₓ = v), vertical as uniformly accelerated (a = eE/mₑ). Show t = x/v, substitute into y = ½at², get y = (eE/2mₑv²)x² = kx². This is a parabola. Very similar to how a positive charge behaves in a uniform electric field.

4. Master the Wheatstone Bridge via Kirchhoff (Q.15a)
This 3-mark derivation appears in virtually every HISSAN/NEB paper. Apply KCL at junctions + KVL to two loops at balanced condition (Ig = 0). Result: PS = QR. Know every step — it’s worth 3 marks and takes less than 5 minutes if well-prepared.

5. YDSE Proof (Q.20b) is Essential
Show both fringe widths (bright and dark) = λD/d. Set up path difference = yd/D. Find consecutive bright positions (nλD/d and (n+1)λD/d), subtract → β = λD/d. Do same for dark → same result. Equal spacing proven in ~6 lines.

6. LCR Circuit (Q.22 OR) Covers 5 Different Sub-concepts
Prepare: (a) impedance derivation, (b) iron core effect on current, (c) inductor/capacitor at high frequency, (d) resonance and resonant frequency, (e) RL numerical. Each is a separate concept — don’t confuse them.

FAQs

Q: The Q.2 KE-time graph — why is it graph A and not D?
A: At t = 0, maximum displacement → v = 0 → KE = 0. Then KE increases as particle moves to mean position. The pattern is [latex]KE = \frac{1}{2}m\omega^2 A^2\sin^2(\omega t)[/latex] — a [latex]\sin^2[/latex] curve starting at zero. Graph A starts at zero and shows two symmetric peaks per period T. Graph D starts at a maximum — that would be for a particle starting at the mean position (x = 0 at t = 0).

Q: For Q.10 (oil drop upward), why is the answer 2mg/q and not mg/q?
A: Moving downward at terminal velocity: [latex]mg = F_{upthrust} + kv \approx kv[/latex] (at terminal speed, weight ≈ drag). Moving upward at same speed v: drag now acts downward (opposes motion) + weight downward = [latex]mg + kv = mg + mg = 2mg[/latex]. So electric force qE = 2mg → E = 2mg/q.

Q: For the Carnot engine-refrigerator chain (Q.14b), the COP of refrigerator seems very high (COP = 4). Is that correct?
A: Yes — COP = [latex]T_{cold}/(T_{hot} – T_{cold}) = 400/100 = 4[/latex]. A refrigerator COP > 1 is normal and expected — it pumps more heat than the work input. A Carnot refrigerator between 400K and 500K has COP = 4, meaning for every joule of work, it moves 4 joules of heat. The total Q₃ = 2500J = Q₄ + W = 2000 + 500, which correctly satisfies energy conservation.

Q: For Q.21(b) Bragg’s law — the scattering angle shown is θ in the figure, but is it the glancing angle or twice the glancing angle?
A: The “scattering angle” in Bragg diffraction usually refers to the angle between the incident beam and the scattered beam, which equals 2θ (twice the glancing angle). If the given angle (26.4°) is the scattering angle, glancing angle = 13.2°. If it’s already the glancing angle, use it directly. Check the figure carefully — from the description “scattering angle of 26.4°”, use θ = 13.2° in Bragg’s equation.

Where to Find the Full Question Paper

The complete original question paper with all figures — including the SHM KE vs time graphs (Q.2), the ABCDA cyclic process P-V diagram (Q.4), the Carnot engine-refrigerator chain diagram (Q.14b), the square frame in magnetic field (Q.16b), the X-ray crystal scattering diagram (Q.21), the electron in electric field diagram (Q.21 OR a), the stopping potential vs frequency graph (Q.21 OR c), and the AC generator diagram (Q.22) — is available as a PDF from above.

Conclusion

The HISSAN Central Examination 2081 (2025) Set D2 Physics paper is one of the most intellectually stimulating pre-board papers available for Grade XII students. It challenges students at every level — from reading a KE-time graph in SHM (Q.2) to proving that electrons cannot exist in an atomic nucleus using Heisenberg’s uncertainty principle (Q.21c), from calculating the heat flows in a Carnot engine driving a Carnot refrigerator (Q.14b) to finding Planck’s constant from a stopping potential graph (Q.21 OR c).

What makes this paper particularly valuable is how it integrates topics that are often studied in isolation. The Bragg’s law question connects wave optics with crystal physics. The electron-in-electric-field proof connects classical kinematics with modern physics. The Carnot engine-refrigerator chain connects engine efficiency with refrigerator COP in a single coherent problem.

Students who study this paper alongside HISSAN 2082 Set D1 and D2 will have encountered virtually every important concept in the Grade XII Physics new curriculum, tested at exactly the depth and style of the actual NEB board examination.

Visit nneducation.com for all HISSAN papers (2081 and 2082, both D1 and D2), complete NEB past papers, model questions, derivation guides, and step-by-step numerical solutions for Grade XI and XII Physics.