HISSAN Central Examination 2081 (2025) — Grade XII Physics Set D1 | Complete Analysis, Answer Guide & Exam Preparation

Subject Code: 1021 D1 | Full Marks: 75 (11 Obj. + 64 Subj.) | Time: 3 Hours
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Introduction

The HISSAN Central Examination 2081 (2025) Set D1 is the companion paper to Set D2 of the same year, and together they form the most comprehensive pre-board Physics examination set for Grade XII students in Nepal. While Set D2 focused heavily on Bragg’s law, Heisenberg uncertainty, and AC generator theory, Set D1 brings a completely different flavor — emphasizing topics that students often underestimate.

This paper stands out for several remarkable questions. The “God particle” (Higgs Boson, Q.11) reflects awareness of cutting-edge particle physics. The ball dropped through Earth’s diameter exhibiting SHM (Q.2) is a classic thought experiment connecting gravitational physics with oscillatory motion. The two pulses meeting on a stretched string (Q.7) is a beautiful wave superposition question testing energy understanding. The flywheel energy problem (Q.12c) requires angular deceleration calculation and combined use of rotational kinematics equations.

Group C is equally rich: Brewster’s law derivation (Q.20d), Doppler effect for approaching and receding source (Q.20 OR c), hypothetical atom energy level transitions (Q.22a,b), Bohr’s model kinetic energy derivation (Q.22c), electron deflection between capacitor plates (Q.19 OR b), and Kirchhoff’s laws applied to a complex PQRS network (Q.21 OR b) — a 3-mark problem that requires systematic application of both KCL and KVL.

Studying this paper alongside HISSAN 2081 D2 and both 2082 papers gives the most complete possible preparation for any NEB Grade XII Physics board examination.

HISSAN 2081 Set D1 vs Set D2 — Key Differences

Full Question Paper

Exam Overview

Topics Covered — Chapter-wise Weightage

Group A — MCQ Analysis with Answers

Q.1 — When Torque is Zero, What Remains Constant?
Answer: c. Angular momentum

Torque: [latex]\tau = \frac{dL}{dt}[/latex]

If [latex]\tau = 0[/latex], then [latex]\frac{dL}{dt} = 0[/latex] → [latex]L = \text{constant}[/latex]

Angular momentum [latex]L = I\omega[/latex] is conserved when no external torque acts. Angular velocity is NOT necessarily constant (I can change while L stays constant, as in the spinning skater).

Q.2 — Ball Dropped Through Earth’s Diameter
Answer: c. it exhibits SHM

If a ball is dropped through a hole along Earth’s diameter, the gravitational force at distance r from centre is [latex]F = -\frac{GMm}{R^3}r[/latex] — directed toward the centre and proportional to r. This is exactly the SHM restoring force condition: [latex]F = -kr[/latex]. The ball oscillates back and forth through the centre with SHM, emerging from the other end momentarily before returning.

Period of this SHM ≈ 84 minutes — same as a low-Earth orbit satellite.

Q.3 — Capillary Height on Moon
Answer: c. 6h

[latex]h = \frac{2T\cos\theta}{r\rho g} \propto \frac{1}{g}[/latex]

Moon’s gravity: [latex]g_{moon} = g_{earth}/6[/latex]

[latex]h_{moon} = h_{earth} \times \frac{g_{earth}}{g_{moon}} = h \times 6 = \mathbf{6h}[/latex]

Q.4 — Efficiency of Carnot Engine Between Steam Point and Ice Point
Answer: c. 36.8%

Steam point: [latex]T_1 = 373[/latex] K (100°C). Ice point: [latex]T_2 = 273[/latex] K (0°C).

[latex]\eta = 1 – \frac{T_2}{T_1} = 1 – \frac{273}{373} = 1 – 0.732 = 0.268 \approx \mathbf{26.8%}[/latex]

Closest answer: b. 26.81% → Answer: b. 26.81%

Q.5 — Work Done Order: Isobaric, Isothermal, Adiabatic, Isochoric
Answer: c. [latex]W_1 > W_2 > W_3 > W_4[/latex]

For the same initial state expanding to the same final volume:

• Isobaric (W₁): Maximum work — P stays highest throughout
• Isothermal (W₂): Less than isobaric — P decreases as V increases
• Adiabatic (W₃): Less than isothermal — temperature drops, reducing pressure more steeply
• Isochoric (W₄): Zero — no volume change

[latex]W_1 > W_2 > W_3 > W_4 = 0[/latex] → Answer: c

Q.6 — Phase Difference Between Velocity and Acceleration in SHM
Answer: c. π/2

In SHM:
[latex]x = A\sin(\omega t)[/latex]
[latex]v = A\omega\cos(\omega t) = A\omega\sin\left(\omega t + \frac{\pi}{2}\right)[/latex]

[latex]a = -A\omega^2\sin(\omega t) = A\omega^2\sin\left(\omega t + \pi\right)[/latex]

Phase of velocity: [latex]\omega t + \pi/2[/latex]. Phase of acceleration: [latex]\omega t + \pi[/latex].

Phase difference between velocity and acceleration: [latex]\pi – \pi/2 = \mathbf{\pi/2}[/latex]

Q.7 — Two Pulses Meeting on String After 2 Seconds
Answer: a. Zero

Two pulses of same amplitude but opposite shapes (one positive, one negative — “same amplitude” in opposite directions). They are 8 cm apart and each moves at 2 cm/s toward each other. After 2 seconds: each has moved 4 cm → they are exactly overlapping.

When they completely overlap at equal and opposite amplitudes → complete destructive interference → total displacement = 0 at that instant.

At the moment of complete cancellation: all energy is purely kinetic (the string elements are moving through zero displacement). Wait — actually: when displacement is zero, all energy is kinetic.

Answer: b. Purely kinetic

At the instant of complete cancellation, displacement = 0 everywhere but the string is moving — all stored energy is kinetic (not potential). Total energy is conserved.

Q.8 — Magnetic Field Intensity at Centre of Long Solenoid
Answer: a. [latex]\mu_0 nI[/latex]

For a long solenoid with n turns per unit length carrying current I:

[latex]B = \mu_0 nI[/latex]

At the centre of a long solenoid, the field is uniform and equals [latex]\mu_0 nI[/latex]. At the end, [latex]B = \mu_0 nI/2[/latex]. For a single circular coil of radius r: [latex]B = \mu_0 I/2r[/latex].

Q.9 — Inversion Temperature of Thermocouple
Answer: b. 570°C

Given: [latex]\theta_n = 300°C[/latex], [latex]\theta_c = 30°C[/latex]

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

[latex]300 = \frac{30 + \theta_i}{2}[/latex]

[latex]600 = 30 + \theta_i[/latex]

[latex]\theta_i = \mathbf{570°C}[/latex]

Q.10 — Curie Temperature is the Temperature Above Which
Answer: a. Ferromagnetic substance becomes paramagnetic

Above the Curie temperature, the thermal energy disrupts the magnetic domains of a ferromagnetic material — it loses its ferromagnetic ordering and becomes paramagnetic (weakly attracted to magnetic field). Below the Curie temperature, the material is ferromagnetic (strongly attracted).

Q.11 — “God Particle”
Answer: b. Higgs Boson

The Higgs Boson is nicknamed the “God particle” — a term popularized by physicist Leon Lederman. It is the quantum of the Higgs field, which gives mass to other fundamental particles. It was discovered at CERN’s Large Hadron Collider (LHC) in 2012, confirming the Standard Model of particle physics. Its discovery was awarded the Nobel Prize in Physics 2013.

Group B — Short Answer Analysis

Q.12 — Rotational Dynamics: Flywheel (OR: SHM Effects of Double Amplitude)

Part (a): Define Moment of Inertia (1 mark)
Moment of inertia of a body about a given axis is 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 measures resistance to rotational acceleration. SI unit: kg·m².

Part (b): Ballet Dancer Stretches Arms — Why Rotation Slows (1 mark)

When a ballet dancer stretches her arms outward, the mass moves farther from the rotation axis — moment of inertia I increases.

By conservation of angular momentum (no external torque):

[latex]L = I\omega = \text{constant}[/latex]

If I increases → ω decreases. The dancer’s rotation slows down. Conversely, pulling arms in reduces I, increases ω — she spins faster.

Part (c): Flywheel Stopping — Revolutions and Time (3 marks)

Given: Initial KE = 200 J, opposing torque τ = 5 N·m, I = 4 kgm²

Step 1 — Number of revolutions before stopping:

Using work-energy theorem for rotation:
[latex]W = \Delta KE[/latex]
[latex]\tau \times \theta = 200 \text{ J}[/latex] (work done by opposing torque = initial KE)

[latex]\theta = \frac{200}{\tau} = \frac{200}{5} = 40 \text{ rad}[/latex]

Number of revolutions: [latex]N = \frac{\theta}{2\pi} = \frac{40}{2\pi} = \frac{20}{\pi} \approx \mathbf{6.37 \text{ revolutions}}[/latex]

Step 2 — Initial angular velocity:

[latex]KE = \frac{1}{2}I\omega_0^2[/latex]

[latex]200 = \frac{1}{2} \times 4 \times \omega_0^2 \Rightarrow \omega_0^2 = 100 \Rightarrow \omega_0 = 10 \text{ rad/s}[/latex]

Step 3 — Angular deceleration:

[latex]\alpha = \frac{\tau}{I} = \frac{5}{4} = 1.25 \text{ rad/s}^2[/latex] (deceleration)

Step 4 — Time to stop:

[latex]\omega = \omega_0 – \alpha t \Rightarrow 0 = 10 – 1.25t[/latex]

[latex]t = \frac{10}{1.25} = \mathbf{8 \text{ seconds}}[/latex]

OR Q.12 — SHM: Effect of Doubling Amplitude

Part (a): Doubling Amplitude — Effect on Period, Total Energy, Maximum Velocity (3 marks)

For SHM: [latex]T = 2\pi\sqrt{\frac{m}{k}}[/latex], [latex]E = \frac{1}{2}kA^2[/latex], [latex]v_{max} = \omega A[/latex]

If amplitude doubles (A → 2A):

i. Period: [latex]T = 2\pi\sqrt{m/k}[/latex] — independent of A. Period remains unchanged. ✓

ii. Total energy: [latex]E’ = \frac{1}{2}k(2A)^2 = 4 \times \frac{1}{2}kA^2 = 4E[/latex]. Total energy becomes 4 times. ✓

iii. Maximum velocity: [latex]v_{max}’ = \omega(2A) = 2\omega A = 2v_{max}[/latex]. Maximum velocity doubles. ✓

Part (b): SHM — Force, Maximum Velocity and Period (2 marks)

Given: m = 0.2 kg, A = 20 mm = 0.02 m, [latex]F_{max} = 0.064[/latex] N

[latex]F_{max} = m \cdot a_{max} = m\omega^2 A[/latex]

[latex]\omega^2 = \frac{F_{max}}{mA} = \frac{0.064}{0.2 \times 0.02} = \frac{0.064}{0.004} = 16 \text{ rad}^2/\text{s}^2[/latex]

[latex]\omega = 4 \text{ rad/s}[/latex]

Maximum velocity: [latex]v_{max} = \omega A = 4 \times 0.02 = \mathbf{0.08 \text{ m/s} = 8 \text{ cm/s}}[/latex]

Period: [latex]T = \frac{2\pi}{\omega} = \frac{2\pi}{4} = \frac{\pi}{2} \approx \mathbf{1.571 \text{ s}}[/latex]

Q.13 — Viscosity and Terminal Velocity: Coalescing Drops

Part (a)(i): Speed vs Time Graph for Steel Ball in Viscous Liquid (1 mark)

The graph shows speed starting from zero, increasing rapidly (when weight dominates over drag), then gradually leveling off to a constant terminal velocity [latex]v_t[/latex].

Shape: Exponential approach to terminal velocity — like a charging capacitor curve. Starts steep, then flattens asymptotically at [latex]v_t[/latex].

Part (a)(ii): Can the Velocity Be Zero? (1 mark)

No. Once the ball starts falling (v > 0), the viscous drag acts upward. The ball never comes to rest inside the liquid during its downward fall — it either continues at terminal velocity or approaches it asymptotically from below.

Velocity can only be zero at the initial moment (t = 0 when dropped from rest). After that, gravity always maintains a net downward force until terminal velocity is reached, at which point speed stays constant.

Part (b): Eight Drops Coalescing — New Terminal Speed (3 marks)

Given: 8 drops, each radius r, terminal speed [latex]v = 5[/latex] cm/s

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

Finding new radius:
Volume of 8 drops = Volume of 1 big drop:

[latex]8 \times \frac{4}{3}\pi r^3 = \frac{4}{3}\pi R^3[/latex]

[latex]R^3 = 8r^3 \Rightarrow R = 2r[/latex]

New terminal velocity:

[latex]\frac{V_{new}}{v} = \frac{R^2}{r^2} = \frac{(2r)^2}{r^2} = 4[/latex]

[latex]V_{new} = 4v = 4 \times 5 = \mathbf{20 \text{ cm/s}}[/latex]

Q.14 — Thermodynamics: Entropy and P-V Work

Part (a): Entropy and First Law in Terms of Entropy (2 marks)

Entropy (S): Entropy is a thermodynamic state function that measures the degree of disorder or irreversibility of a system. For a reversible process:

[latex]dS = \frac{dQ_{rev}}{T}[/latex]

SI unit: J/K. Entropy always increases or remains constant for any natural (spontaneous) process.

First Law in Terms of Entropy:
First Law: [latex]dQ = dU + PdV[/latex]

Since [latex]dQ = TdS[/latex] for reversible process:

[latex]\boxed{TdS = dU + PdV}[/latex]

Yes — the first law can be expressed in terms of entropy: [latex]dU = TdS – PdV[/latex]. This is the fundamental thermodynamic relation.

Part (b): Work Done from D to E to F on P-V Diagram (3 marks)

From the P-V diagram:

• D is at (2.0 cc, 600 atm)
• E is at (5.0 cc, 300 atm)
• F is at (2.0 cc, 300 atm)

Work from D to E (linear process):
A linear process on P-V diagram means the path is a straight line from D(2, 600) to E(5, 300).

Work = Area under D→E on P-V diagram (trapezoid):

[latex]W_{DE} = \frac{(P_D + P_E)}{2} \times (V_E – V_D) = \frac{(600 + 300)}{2} \times (5 – 2)[/latex]

Converting: P in atm = P × 101325 Pa, V in cc = V × 10⁻⁶ m³

[latex]W_{DE} = \frac{900}{2} \times 3 = 1350 \text{ atm·cc}[/latex]

[latex]= 1350 \times 101325 \times 10^{-6} = 1350 \times 0.101325 = \mathbf{136.8 \text{ J}}[/latex]

Work from E to F (isobaric at P = 300 atm):

[latex]W_{EF} = P \times (V_F – V_E) = 300 \times (2 – 5) = -900 \text{ atm·cc}[/latex]

[latex]= -900 \times 0.101325 = \mathbf{-91.2 \text{ J}}[/latex] (negative — compression)

Total work D→E→F:

[latex]W_{total} = W_{DE} + W_{EF} = 136.8 + (-91.2) = \mathbf{45.6 \text{ J}}[/latex]

Q.15 — Sound: Laplace Formula, Sound in Solids, and Velocity at 0°C

Part (a): Laplace Corrected Formula for Velocity of Sound in Air (2 marks)

Newton assumed isothermal propagation: [latex]v = \sqrt{P/\rho}[/latex] — gave ~280 m/s (too low).

Laplace corrected this by recognizing the process is adiabatic — compressions and rarefactions occur so rapidly that no heat exchange occurs. For adiabatic process: effective bulk modulus = [latex]\gamma P[/latex].

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

where [latex]\gamma = C_p/C_v = 1.4[/latex] for air.

[latex]v = \sqrt{1.4} \times 280 \approx \mathbf{331.5 \text{ m/s at 0°C}}[/latex] — matches experiment. ✓

Part (b): Why Velocity of Sound Greater in Solid Despite Higher Density (1 mark)

Velocity of sound: [latex]v = \sqrt{\frac{E}{\rho}}[/latex] where E = elastic modulus (bulk/Young’s modulus).

For solids, the elastic modulus E is enormously larger than for gases/liquids — it dominates over the higher density. The ratio E/ρ is much larger for solids than gases, resulting in higher sound velocity.

Example: Steel — E ≈ 2×10¹¹ Pa, ρ ≈ 7800 kg/m³ → v ≈ 5000 m/s. Air — E ≈ 1.4×10⁵ Pa, ρ ≈ 1.29 kg/m³ → v ≈ 330 m/s. Despite steel being ~6000× denser, its modulus is ~10⁶× larger.

Part (c): Velocity of Sound at 0°C (2 marks)

Given: f = 512 Hz, λ = 64.5 cm = 0.645 m at 20°C

Velocity at 20°C: [latex]v_{20} = f\lambda = 512 \times 0.645 = 330.24 \text{ m/s}[/latex]

Using [latex]v \propto \sqrt{T}[/latex]:

[latex]\frac{v_0}{v_{20}} = \sqrt{\frac{T_0}{T_{20}}} = \sqrt{\frac{273}{293}}[/latex]

[latex]v_0 = 330.24 \times \sqrt{\frac{273}{293}} = 330.24 \times \sqrt{0.9317} = 330.24 \times 0.9653[/latex]

[latex]v_0 = \mathbf{318.8 \text{ m/s} \approx 319 \text{ m/s}}[/latex]

Q.16 — Electromagnetic Induction: Bar Magnet in Two Coils and Mutual Inductance

Part (a): Induced Current Directions in Coils PQ and CD (2 marks)

From the figure: Bar magnet moves between coils PQ (left) and CD (right), with North pole facing right (toward CD) and South pole facing left (toward PQ). The magnet moves in the direction of the arrow (rightward, from N toward S end in the figure).

Coil PQ (magnet’s South pole approaching): Flux through PQ increases (flux from N to S through PQ increases as magnet moves right). By Lenz’s law, induced current opposes increase — creates a magnetic field opposing the approaching South pole (repulsion) → in coil PQ, using right-hand rule: current flows from A to P (counterclockwise when viewed from the approaching south pole side).

Coil CD (magnet’s North pole moving away): Flux through CD decreases as magnet moves right (N pole moving away reduces flux through CD). Lenz’s law: induced current tries to maintain flux → creates field in same direction as the retreating N pole → in coil CD, current flows to attract the leaving magnet: from D to A (or as determined by the right-hand rule applied to the specific geometry).

Students should apply Lenz’s law carefully to each coil separately, considering the direction and polarity of flux change.

Part (b): Mutual Inductance Definition and Changes (3 marks)

Mutual inductance (M): Mutual inductance between two coils is defined as the ratio of the EMF induced in the secondary coil to the rate of change of current in the primary coil:

[latex]M = \frac{\varepsilon_2}{-dI_1/dt}[/latex]

Or: M = flux linkage in secondary per unit current in primary: [latex]M = \frac{N_2\Phi_{12}}{I_1}[/latex]

SI unit: Henry (H). One Henry = 1 V·s/A.

i. Distance between coils increased:
As distance increases, less flux from the primary coil links with the secondary coil. Mutual inductance decreases — fewer field lines pass through the secondary.

ii. Number of turns increased:
If turns in either coil increase, more flux is produced by primary (more turns × same B) or more flux is linked by secondary (more turns × same Φ per turn). Mutual inductance increases proportionally.

[latex]M = \frac{\mu_0 N_1 N_2 A}{l}[/latex] — M is proportional to N₁ and N₂ (for solenoid geometry).

Q.17 — Ampere’s Law, Force Between Conductors, and 1 Ampere Definition

Part (a): Ampere’s Circuital Law (1 mark)

Ampere’s circuital law states that the line integral of the magnetic field [latex]\vec{B}[/latex] around any closed loop is equal to [latex]\mu_0[/latex] times the total current enclosed by the loop:

[latex]\oint \vec{B} \cdot d\vec{l} = \mu_0 I_{enclosed}[/latex]

Part (b): Force Per Unit Length Between Two Parallel Conductors (3 marks)

Two long parallel conductors separated by distance d, carrying currents [latex]I_1[/latex] and [latex]I_2[/latex]:

Step 1 — Magnetic field due to conductor 1 at the position of conductor 2:

[latex]B_1 = \frac{\mu_0 I_1}{2\pi d}[/latex]

Step 2 — Force on length L of conductor 2:

[latex]F = I_2 L B_1 = I_2 L \times \frac{\mu_0 I_1}{2\pi d}[/latex]

Step 3 — Force per unit length:

[latex]\boxed{\frac{F}{L} = \frac{\mu_0 I_1 I_2}{2\pi d}}[/latex]

Direction: Parallel currents (same direction) → attractive. Antiparallel currents → repulsive.

Part (c): Definition of 1 Ampere (1 mark)

One Ampere is defined as that steady current which, when flowing through each of two infinitely long parallel wires of negligible cross-section separated by 1 metre in vacuum, produces between them a force of [latex]2 \times 10^{-7}[/latex] N per metre of their length.

Proof from formula: [latex]\frac{F}{L} = \frac{\mu_0 I_1 I_2}{2\pi d} = \frac{4\pi \times 10^{-7} \times 1 \times 1}{2\pi \times 1} = 2 \times 10^{-7}[/latex] N/m ✓

Q.18 — Logic Gates and Full-Wave Rectifier

Part (a): Truth Table and AND Gate Truth Table (1+1 marks)

Definition of Truth Table: A truth table is a table that lists all possible combinations of input values and shows the corresponding output for a logic gate or logic circuit.

AND Gate Truth Table: [latex]Y = A \cdot B[/latex]

Output is HIGH only when both inputs are HIGH.

Part (b): Full-Wave Rectifier — Working, Input and Output Waveforms (3 marks)

Construction: Four diodes D₁, D₂, D₃, D₄ in bridge configuration. AC input connected to two opposite corners. Load [latex]R_L[/latex] and DC output at the other two corners.

Working:

Positive half-cycle (top terminal positive): D₁ and D₃ forward biased → conduct. Current flows: top → D₁ → [latex]R_L[/latex] → D₃ → bottom. Current through [latex]R_L[/latex]: top to bottom.

Negative half-cycle (bottom terminal positive): D₂ and D₄ forward biased → conduct. Current: bottom → D₂ → [latex]R_L[/latex] → D₄ → top. Current through [latex]R_L[/latex]: still top to bottom.

Result: Both half-cycles produce current in the same direction through [latex]R_L[/latex].

Input waveform: Sinusoidal AC — alternating between positive and negative.
Output waveform: Pulsating DC — all pulses positive (each original half-cycle appears as a positive pulse), with ripple frequency = 2f. A smoothing capacitor reduces ripple to near-steady DC.

Q.19 — Radioactivity: Half-Life and Activity Calculation (OR: Photoelectric Effect)

Part (a): Half-Life of Radon (3.8 days) — Meaning and Activity Plot (2 marks)

Meaning of half-life 3.8 days:
The half-life of Radon-222 is 3.8 days, which means that every 3.8 days, exactly half of the remaining Radon atoms undergo radioactive decay. Starting with N₀ atoms:

• After 3.8 days: N₀/2 remain
• After 7.6 days: N₀/4 remain
• After 11.4 days: N₀/8 remain

Activity vs time plot:
The graph is an exponential decay curve: [latex]A = A_0 e^{-\lambda t}[/latex]

Starting at [latex]A_0[/latex], halving every 3.8 days. The curve starts steeply and flattens asymptotically toward zero. X-axis: time (in units of 3.8 days), Y-axis: activity. Half-life marked at each A/2 point.

Part (b): Activity After 260 Days and Total Energy Released (3 marks)

Given: [latex]t_{1/2} = 130[/latex] days, [latex]N_0 = 1 \times 10^{20}[/latex] atoms, energy per disintegration = 5 MeV, t = 260 days

Number of half-lives: [latex]n = \frac{260}{130} = 2[/latex]

Atoms remaining after 260 days:

[latex]N = N_0 \left(\frac{1}{2}\right)^2 = 10^{20} \times \frac{1}{4} = 2.5 \times 10^{19} \text{ atoms}[/latex]

Activity after 260 days:

[latex]\lambda = \frac{\ln 2}{t_{1/2}} = \frac{0.693}{130 \times 24 \times 3600} = \frac{0.693}{1.123 \times 10^7} = 6.17 \times 10^{-8} \text{ s}^{-1}[/latex]

[latex]A = \lambda N = 6.17 \times 10^{-8} \times 2.5 \times 10^{19} = \mathbf{1.54 \times 10^{12} \text{ Bq} = 1.54 \text{ TBq}}[/latex]

Atoms decayed in 260 days:

[latex]\Delta N = N_0 – N = 10^{20} – 2.5 \times 10^{19} = 7.5 \times 10^{19} \text{ atoms}[/latex]

Total energy released:

[latex]E = \Delta N \times 5 \text{ MeV} = 7.5 \times 10^{19} \times 5 \times 10^6 \times 1.6 \times 10^{-19} \text{ J}[/latex]

[latex]E = 7.5 \times 10^{19} \times 8 \times 10^{-13} = \mathbf{6 \times 10^{7} \text{ J} = 60 \text{ MJ}}[/latex]

OR Q.19 — Stopping Potential and Electron Deflection

Part (a): Stopping Potential — Definition and V₀ vs f Graph (2 marks)

Stopping potential (V₀): The minimum negative potential applied to the anode plate that completely stops even the most energetic photoelectrons from reaching it, reducing photocurrent to zero.

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

V₀ vs f graph:
A straight line with:

• Positive slope = h/e (Planck’s constant/electron charge)
• X-intercept = threshold frequency [latex]f_0[/latex] (V₀ = 0)
• Y-intercept = −φ/e (negative of work function/e)
• V₀ = 0 for f ≤ f₀; V₀ increases linearly for f > f₀

Part (b): Vertical Deflection of Electron Between Capacitor Plates (3 marks)

Given: Plate length L = 4 cm = 0.04 m, plate separation d = 3 cm = 0.03 m, V = 300 V, [latex]v_x = 10^7[/latex] m/s, e = [latex]1.6 \times 10^{-19}[/latex] C, [latex]m_e = 9.1 \times 10^{-31}[/latex] kg

Electric field between plates:

[latex]E = \frac{V}{d} = \frac{300}{0.03} = 10^4 \text{ V/m}[/latex]

Vertical acceleration:

[latex]a = \frac{eE}{m_e} = \frac{1.6 \times 10^{-19} \times 10^4}{9.1 \times 10^{-31}} = \frac{1.6 \times 10^{-15}}{9.1 \times 10^{-31}} = 1.758 \times 10^{15} \text{ m/s}^2[/latex]

Time in the field:

[latex]t = \frac{L}{v_x} = \frac{0.04}{10^7} = 4 \times 10^{-9} \text{ s}[/latex]

Vertical deflection:

[latex]y = \frac{1}{2}at^2 = \frac{1}{2} \times 1.758 \times 10^{15} \times (4 \times 10^{-9})^2[/latex]

[latex]= \frac{1}{2} \times 1.758 \times 10^{15} \times 16 \times 10^{-18}[/latex]

[latex]= \frac{1}{2} \times 28.13 \times 10^{-3} = \mathbf{1.406 \times 10^{-2} \text{ m} = 1.41 \text{ cm}}[/latex]

Group C — Long Answer Analysis

Q.20 — Wave Optics: YDSE, Diffraction, and Brewster’s Law

Part (a)(i): YDSE Immersed in Liquid of Refractive Index 1.33 (1 mark)

Fringe width: [latex]\beta = \frac{\lambda D}{d}[/latex]

In liquid: wavelength reduces to [latex]\lambda’ = \lambda/\mu[/latex]

New fringe width: [latex]\beta’ = \frac{\lambda’ D}{d} = \frac{\lambda D}{\mu d} = \frac{\beta}{\mu} = \frac{\beta}{1.33}[/latex]

Fringe width decreases by a factor of 1.33 — fringes become narrower and closer together.

Part (a)(ii): Slit Width Increased (1 mark)

Fringe width [latex]\beta = \frac{\lambda D}{d}[/latex] depends on slit separation d (distance between slits), not slit width. If slit width increases but separation d is unchanged: fringe width remains unchanged. However, the intensity of fringes increases (more light enters) and the single-slit diffraction envelope becomes narrower (fewer bright fringes visible).

Part (b): What is Diffraction of Light? (1 mark)
Diffraction is the phenomenon of bending or spreading of light waves around the edges of an obstacle or through a narrow opening (slit), whose size is comparable to the wavelength of light. It is evidence of the wave nature of light.

Part (c): Width of Central Diffraction Peak (2 marks)

Given: D = 4 m, slit width a = 0.01 m (=1 cm), λ = 600 nm = [latex]600 \times 10^{-9}[/latex] m

If a = 0.01 mm = [latex]10^{-5}[/latex] m (more physically reasonable for diffraction):

Width of central maximum: [latex]w = \frac{2\lambda D}{a} = \frac{2 \times 600 \times 10^{-9} \times 4}{10^{-5}}[/latex]

[latex]= \frac{4.8 \times 10^{-6}}{10^{-5}} = \mathbf{0.48 \text{ m} = 48 \text{ cm}}[/latex]

If a = 0.01 m literally: w = 2 × 600×10⁻⁹ × 4 / 0.01 = 4.8×10⁻⁴ m = 0.48 mm — very narrow, consistent with large slit.

Part (d): Brewster’s Law — State and Prove (3 marks)

Statement: When unpolarized light is incident at a special angle [latex]i_B[/latex] (Brewster’s angle or polarizing angle) on the boundary between two media, the reflected light is completely plane polarized. The polarizing angle satisfies:

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

where μ = refractive index of the second medium.

Proof using Snell’s Law:

At Brewster’s angle, the reflected and refracted rays are perpendicular to each other:

[latex]i_B + r = 90° \Rightarrow r = 90° – i_B[/latex]

Applying Snell’s law at the interface:

[latex]\mu_1 \sin i_B = \mu_2 \sin r = \mu_2 \sin(90° – i_B) = \mu_2 \cos i_B[/latex]

If medium 1 is air ([latex]\mu_1 = 1[/latex]):

[latex]\sin i_B = \mu_2 \cos i_B[/latex]

[latex]\frac{\sin i_B}{\cos i_B} = \mu_2[/latex]

[latex]\boxed{\tan i_B = \mu}[/latex] ✓

Physical explanation: At Brewster’s angle, the refracted and reflected rays are perpendicular. The oscillating dipoles in the medium that would radiate in the reflected direction cannot vibrate along that direction — so the component parallel to the plane of incidence cannot be reflected, leaving only perpendicular polarization in the reflected beam → completely polarized.

OR Q.20 — Progressive vs Stationary Waves, Organ Pipe, Doppler Effect

Part (a): Progressive vs Stationary Waves (2 marks)

Part (b): All Harmonics Present in Open Organ Pipe (3 marks)

Open pipe (both ends open): antinodes at both ends.

For length L, the standing wave condition:

[latex]L = \frac{n\lambda_n}{2} \Rightarrow \lambda_n = \frac{2L}{n}[/latex]

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

All integers n are permitted → frequencies [latex]f_1, 2f_1, 3f_1, 4f_1…[/latex]

Both odd and even harmonics are present — open organ pipe produces a complete harmonic series. ✓

Part (c): Doppler Effect — Car Passing Observer (3 marks)

Given: Source frequency f₀ = 500 Hz, car speed [latex]v_s[/latex] = 20 m/s, v_sound = 340 m/s

Frequency when approaching:

[latex]f_{app} = f_0 \times \frac{v}{v – v_s} = 500 \times \frac{340}{340 – 20} = 500 \times \frac{340}{320} = 500 \times 1.0625 = \mathbf{531.25 \text{ Hz}}[/latex]

Frequency when receding:

[latex]f_{rec} = f_0 \times \frac{v}{v + v_s} = 500 \times \frac{340}{340 + 20} = 500 \times \frac{340}{360} = 500 \times 0.9444 = \mathbf{472.2 \text{ Hz}}[/latex]

Change in frequency heard:

[latex]\Delta f = f_{app} – f_{rec} = 531.25 – 472.2 = \mathbf{59.1 \text{ Hz}}[/latex]

The observer hears a sudden drop of ~59 Hz as the car passes.

Q.21 — AC Circuits: Power Factor, Current, and Phasor Diagram

Part (a): Power Factor of AC Circuit (1 mark)

Power factor of an AC circuit is the cosine of the phase angle (φ) between the voltage and current:

[latex]\text{Power factor} = \cos\phi = \frac{R}{Z}[/latex]

It represents the ratio of actual power (P = VI cosφ) to apparent power (VI). A power factor of 1 means purely resistive (maximum power transfer); 0 means purely reactive (no power dissipated).

Part (b): Current in [latex]I = 2\sin(50t)[/latex] (3 marks)

i. Peak value of current:
Comparing [latex]I = 2\sin(50t)[/latex] with [latex]I = I_0\sin(\omega t)[/latex]:
[latex]I_0 = 2[/latex] A

ii. Frequency of supply:

[latex]\omega = 50 \text{ rad/s} = 2\pi f[/latex]

[latex]f = \frac{50}{2\pi} = \frac{25}{\pi} \approx \mathbf{7.96 \text{ Hz}}[/latex]

iii. RMS value of current:

[latex]I_{rms} = \frac{I_0}{\sqrt{2}} = \frac{2}{\sqrt{2}} = \sqrt{2} \approx \mathbf{1.414 \text{ A}}[/latex]

Part (c): Phase Relationship in Pure Inductive AC Circuit (2 marks)

In a pure inductor: [latex]V = V_0\sin(\omega t)[/latex], induced back-EMF opposes change in current.

The current lags the voltage by 90°:

[latex]I = I_0\sin\left(\omega t – \frac{\pi}{2}\right) = -I_0\cos(\omega t)[/latex]

Phasor diagram:
Draw voltage phasor [latex]V_0[/latex] along reference direction (rightward, horizontal). Current phasor [latex]I_0[/latex] is drawn 90° clockwise (lagging) from voltage — i.e., downward. The phase difference is π/2, with current lagging.

Physical reason: The inductor opposes changes in current — when voltage is maximum (pushing hardest), current hasn’t yet reached its maximum (still building up).

Part (d): RL Circuit — Impedance and Phase Angle (2 marks)

Given: L = 100 mH = 0.1 H, R = 50 Ω, V = 230V, f = 50 Hz

Inductive reactance: [latex]X_L = \omega L = 2\pi \times 50 \times 0.1 = 10\pi = 31.42[/latex] Ω

Impedance: [latex]Z = \sqrt{R^2 + X_L^2} = \sqrt{50^2 + 31.42^2} = \sqrt{2500 + 987.2} = \sqrt{3487.2} = \mathbf{59.05 \text{ Ω}}[/latex]

Phase angle: [latex]\tan\phi = \frac{X_L}{R} = \frac{31.42}{50} = 0.6284[/latex]

[latex]\phi = \tan^{-1}(0.6284) = \mathbf{32.1°}[/latex] (current lags voltage by 32.1°)

OR Q.21 — Wheatstone Bridge, Network, Biot-Savart Circular Coil, Magnetic Hysteresis

Part (a): Balanced Condition of Wheatstone Bridge (1 mark)

At balance (galvanometer reads zero):
[latex]\frac{P}{Q} = \frac{R}{S}[/latex] or equivalently [latex]PS = QR[/latex]

Part (b): PQRS Network — Current in Milliammeter (3 marks)

Given: Battery 4V between Q and P (bottom), Battery 5V between S and P (left), S = 200Ω (top), R = 60Ω (right), milliammeter 20Ω between P and R.

This is a complex network requiring Kirchhoff’s laws.

Let currents:

• [latex]I_1[/latex] through 5V battery and 200Ω (loop SPQS top-left)
• [latex]I_2[/latex] through 60Ω (right branch SR)
• [latex]I_m[/latex] through milliammeter (PR)
• 4V battery between Q and P (bottom)

Applying KCL and KVL systematically:

At junction R: [latex]I_2 = I_m + I_3[/latex] (where [latex]I_3[/latex] goes through QR path)

Loop containing 5V, 200Ω, milliammeter (20Ω):
[latex]5 = 200I_1 + 20I_m[/latex] … (i)

Loop containing 4V, 60Ω:
[latex]4 = 60I_2[/latex] → [latex]I_2 = 4/60 = 0.0667[/latex] A … (ii)

Loop PQRS with both batteries:
Applying KVL around the outer loop:
[latex]5 – 4 = 200I_1 – 60I_2[/latex]
[latex]1 = 200I_1 – 60 \times 0.0667[/latex]
[latex]1 = 200I_1 – 4[/latex]
[latex]200I_1 = 5 \Rightarrow I_1 = 0.025[/latex] A

From (i): [latex]I_m = \frac{5 – 200 \times 0.025}{20} = \frac{5 – 5}{20} = \mathbf{0 \text{ mA}}[/latex]

Milliammeter reads 0 mA — the bridge is balanced under these conditions.

Part (c): Magnetic Field at Centre of Current Carrying Circular Coil Using Biot-Savart (3 marks)

For a circular coil of radius R, N turns, current I:

Each element dl is perpendicular to the radius vector r = R (θ = 90°):

[latex]dB = \frac{\mu_0 I,dl\sin 90°}{4\pi R^2} = \frac{\mu_0 I,dl}{4\pi R^2}[/latex]

All elements contribute in the same direction (perpendicular to coil plane):

[latex]B = \oint dB = \frac{\mu_0 I}{4\pi R^2}\oint dl = \frac{\mu_0 I}{4\pi R^2} \times 2\pi R = \frac{\mu_0 I}{2R}[/latex]

For N turns: [latex]\boxed{B = \frac{\mu_0 NI}{2R}}[/latex]

Part (d): Magnetic Hysteresis (1 mark)

Magnetic hysteresis is the phenomenon by which the magnetization of a ferromagnetic material lags behind the applied magnetizing field. When the magnetizing field H is cycled, the B-H graph traces a closed loop called the hysteresis loop. The area of the hysteresis loop represents the energy dissipated as heat per unit volume per magnetization cycle. It is caused by the irreversible motion of magnetic domain walls during magnetization and demagnetization.

Q.22 — Atomic Physics: Energy Levels, Bohr’s Model, and X-rays

Part (a): Transition Emitting Photon of Wavelength 275 nm (2 marks)

From the energy level diagram: levels at 0, −2, −4.5, −10 eV.

Transitions shown: A (0 to −2 eV), B (0 to −4.5 eV), C (−2 to −4.5 eV), D (−4.5 to −10 eV)

Energy of photon with λ = 275 nm:

[latex]E = \frac{hc}{\lambda} = \frac{6.63 \times 10^{-34} \times 3 \times 10^8}{275 \times 10^{-9}} = \frac{1.989 \times 10^{-25}}{2.75 \times 10^{-7}} = 7.23 \times 10^{-19} \text{ J}[/latex]

[latex]E = \frac{7.23 \times 10^{-19}}{1.6 \times 10^{-19}} = 4.52 \text{ eV} \approx 4.5 \text{ eV}[/latex]

Which transition gives ΔE = 4.5 eV?

• A: 0 − (−2) = 2 eV ✗
• B: 0 − (−4.5) = 4.5 eV ✓
• C: −2 − (−4.5) = 2.5 eV ✗
• D: −4.5 − (−10) = 5.5 eV ✗

Transition B (from 0 eV to −4.5 eV) emits a photon of 275 nm.

Part (b): Maximum and Minimum Wavelength Transitions (1+2 marks)

i. Maximum wavelength (minimum energy):
[latex]E = hc/\lambda[/latex] → minimum energy → maximum wavelength.

Energy differences:

• A: 0 − (−2) = 2 eV
• B: 0 − (−4.5) = 4.5 eV
• C: −2 − (−4.5) = 2.5 eV
• D: −4.5 − (−10) = 5.5 eV

Minimum ΔE = 2 eV → Transition A gives maximum wavelength.

[latex]\lambda_{max} = \frac{hc}{\Delta E} = \frac{6.63 \times 10^{-34} \times 3 \times 10^8}{2 \times 1.6 \times 10^{-19}} = \frac{1.989 \times 10^{-25}}{3.2 \times 10^{-19}} = \mathbf{621.6 \text{ nm}}[/latex] (red light)

ii. Minimum wavelength (maximum energy):
Maximum ΔE = 5.5 eV → Transition D gives minimum wavelength.

[latex]\lambda_{min} = \frac{hc}{\Delta E} = \frac{1.989 \times 10^{-25}}{5.5 \times 1.6 \times 10^{-19}} = \frac{1.989 \times 10^{-25}}{8.8 \times 10^{-19}} = \mathbf{226 \text{ nm}}[/latex] (UV)

Part (c): KE of Electron in Stationary State Using Bohr’s Postulates (2 marks)

For an electron in circular orbit of radius r:
Coulomb force = Centripetal force:

[latex]\frac{ke^2}{r^2} = \frac{mv^2}{r}[/latex]

[latex]mv^2 = \frac{ke^2}{r}[/latex]

Kinetic energy:

[latex]KE = \frac{1}{2}mv^2 = \frac{ke^2}{2r}[/latex]

Using Bohr’s quantization condition: [latex]mvr = \frac{nh}{2\pi}[/latex]

Solving for r from both equations:
[latex]r_n = \frac{n^2h^2\epsilon_0}{\pi me^2}[/latex] (Bohr radius for nth orbit)

Substituting:

[latex]KE = \frac{me^4}{8\epsilon_0^2 h^2 n^2} = \frac{13.6}{n^2} \text{ eV}[/latex]

[latex]\boxed{KE = \frac{13.6}{n^2} \text{ eV}}[/latex]

KE is positive and decreases with increasing n — outer orbits have less kinetic energy.

Part (d): Work Function of Cesium is 1.9 eV — Explain (1 mark)

The work function of cesium is 1.9 eV means that a minimum energy of 1.9 eV (= [latex]1.9 \times 1.6 \times 10^{-19}[/latex] J = [latex]3.04 \times 10^{-19}[/latex] J) is required to completely remove one electron from the surface of cesium metal.

This corresponds to a threshold frequency [latex]f_0 = \phi/h = 3.04 \times 10^{-19}/6.63 \times 10^{-34} = 4.59 \times 10^{14}[/latex] Hz (near visible light range). This is why cesium is highly photosensitive — visible light can cause photoelectric emission from cesium, making it useful in photoelectric cells and light detectors.

Part (e): Why X-rays are NOT Deflected by Electric and Magnetic Fields (1 mark)

X-rays are electromagnetic waves — they have no electric charge. Only charged particles experience forces from electric and magnetic fields ([latex]F_E = qE[/latex] and [latex]F_B = qvB[/latex]).

Since X-rays carry no charge (q = 0), neither electric nor magnetic fields exert any force on them. They travel in straight lines undeflected — just like all other electromagnetic radiation (light, gamma rays, radio waves). This is a fundamental distinction from charged particles like electrons and protons, which are deflected by fields.

Marks Weightage Summary

Difficulty Analysis

Easy — Reliable marks:

• Q.1 (Torque zero → angular momentum constant)
• Q.3 (Capillary height on moon)
• Q.6 (Phase difference v and a in SHM — π/2)
• Q.8 (Solenoid field at centre)
• Q.10 (Curie temperature — ferromagnetic → paramagnetic)
• Q.12(a) (Moment of inertia definition)
• Q.12(b) (Ballet dancer explanation)
• Q.13(a)(i) (Speed-time graph for ball)
• Q.15(b) (Sound faster in solid)
• Q.17(a) (Ampere’s law statement)
• Q.17(c) (Definition of 1 ampere)
• Q.18(a) (AND gate truth table)
• Q.19(a) (Half-life meaning and plot)
• Q.21(a) (Power factor definition)
• Q.22(d)(e) (Cesium work function + X-rays not deflected)

Moderate — Needs regular practice:

• Q.2 (Ball through Earth — SHM connection)
• Q.5 (Work done order comparison)
• Q.9 (Inversion temperature calculation)
• Q.12(c) (Flywheel stopping — three parts)
• Q.13(b) (Eight drops coalescing)
• Q.14(a) (Entropy + first law in entropy form)
• Q.15(a)(c) (Laplace formula + velocity at 0°C)
• Q.16(a) (Bar magnet in two coils — Lenz’s law)
• Q.20(a)(b) (YDSE changes + diffraction definition)
• Q.21(b)(c)(d) (AC current analysis + phasor + RL impedance)

Challenging — For top scorers:

• Q.4 (Carnot efficiency calculation)
• Q.7 (Two pulses meeting — energy analysis)
• Q.11 (Higgs Boson — modern physics awareness)
• Q.12 OR (a) (SHM doubling amplitude — all three effects)
• Q.14(b) (P-V diagram D→E→F work calculation)
• Q.16(b) (Mutual inductance — changes)
• Q.17(b) (Force per unit length derivation — full derivation)
• Q.19(b) (Activity + total energy after 260 days)
• Q.19 OR (b) (Electron deflection between plates)
• Q.20(d) (Brewster’s law — proof with Snell’s law)
• Q.21 OR (b) (PQRS network via Kirchhoff — complex)
• Q.21 OR (c) (Biot-Savart circular coil derivation)
• Q.22(a)(b) (Hypothetical atom transitions — energy level calculations)
• Q.22(c) (Bohr model KE derivation)

Student Preparation Tips

1. The Hypothetical Atom Problem (Q.22a,b) is Pure Energy Level Arithmetic
Don’t be intimidated by “hypothetical” atoms. Calculate E = hc/λ for given wavelength, convert to eV, identify which transition (ΔE) matches. For max/min wavelength: find max/min ΔE from the diagram. Practice with hydrogen energy levels (−13.6/n² eV) and any given energy level diagram.

2. Brewster’s Law Proof (Q.20d) is 3 Marks — Learn It Once
Key steps: (a) State that reflected and refracted rays are perpendicular at Brewster’s angle, (b) write [latex]i_B + r = 90°[/latex], (c) apply Snell’s law with sin(90° − i_B) = cos(i_B), (d) get tan(i_B) = μ. Five clear lines earns full 3 marks.

3. Doppler Effect (Q.20 OR c) Requires Both Approaching and Receding Formulas
[latex]f_{app} = f_0 \frac{v}{v – v_s}[/latex] and [latex]f_{rec} = f_0 \frac{v}{v + v_s}[/latex]. The “change in frequency” = f_app − f_rec. Always state which formula applies to which case before substituting.

4. Flywheel Problem (Q.12c) = Three Results from Two Methods
(i) Revolutions: use W = τθ to find θ, then N = θ/2π. (ii) Initial ω: use KE = ½Iω². (iii) Time: use ω = ω₀ − αt with α = τ/I. Practice combining rotational work-energy theorem with kinematics.

5. Mutual Inductance (Q.16b) — Qualitative But Marks-Rich
State definition, then explain each factor logically. Distance increases → less flux linkage → M decreases. More turns → more flux produced AND more turns to link with → M increases (M ∝ N₁N₂). Clear physical reasoning earns full 3 marks.

6. The PQRS Network (Q.21 OR b) Needs Systematic Kirchhoff
Complex network problems require: (1) label all branch currents with assumed directions, (2) apply KCL at junctions, (3) apply KVL to independent loops, (4) solve simultaneous equations. Always check by substituting back. In this case, the clever result (zero milliammeter reading) indicates bridge balance.

FAQs

Q: In Q.7 (two pulses), why is the energy purely kinetic when they meet, not zero?
A: Energy is always conserved in wave superposition. At the instant of complete cancellation (displacement = 0 everywhere), the string elements are passing through zero displacement with maximum velocity — all energy is kinetic. Total energy = KE + PE. PE (∝ displacement²) = 0 when displacement = 0. So all energy is kinetic at that instant. Answer is b (purely kinetic), not a (zero).

Q: Q.4 gives 26.81% but I calculated 26.8% — which is exact?
A: η = 1 − 273/373 = 100/373 = 0.26810… = 26.81%. The correct answer is b. 26.81%. The calculation 273/373 = 0.73188, so η = 1 − 0.73188 = 0.26812 ≈ 26.81%. Always carry enough decimal places in Carnot efficiency calculations.

Q: For Q.22(c) Bohr’s kinetic energy — should I derive it fully or just state the result?
A: The question says “obtain the expression” — so derive it. Start from Coulomb force = centripetal force, get mv² = ke²/r, then KE = ke²/2r. Then optionally use the quantization condition to get KE = 13.6/n² eV. Both routes are acceptable — the derivation path shows more understanding.

Q: What exactly is the “God particle” and why does it matter for students?
A: The Higgs Boson, discovered at CERN in 2012, explains why other fundamental particles have mass. Without the Higgs field permeating all of space, particles like the W and Z bosons would be massless. This question tests modern physics awareness. The answer is simply: Higgs Boson (b) — one easy mark that rewards students who follow current physics news.

Q: For Q.14(b) P-V diagram work — are the pressures in atm and volumes in cc giving J directly?
A: No — you must convert. 1 atm·cc = 101325 Pa × 10⁻⁶ m³ = 0.101325 J. So 1350 atm·cc × 0.101325 = 136.8 J. Always convert to SI units before calculating work in Joules.

Combining All Four HISSAN Papers — The Complete Picture

A student who masters all four papers is comprehensively prepared for any NEB Grade XII Physics board examination.

Where to Find the Full Question Paper

The complete original question paper with all figures — including the two pulses on string (Q.7), the P-V diagram DEF (Q.14), the two coils with bar magnet (Q.16), the hypothetical atom energy level diagram (Q.22), the PQRS network circuit (Q.21 OR), and the electron deflection setup (Q.19 OR) — is available as a PDF above.

Conclusion

The HISSAN Central Examination 2081 (2025) Set D1 Physics paper rewards students who have built a genuine understanding of physics across its full breadth. From the elegant simplicity of “why can’t X-rays be deflected?” (Q.22e — one mark for understanding a fundamental principle) to the computational depth of the radioactivity activity-and-energy calculation (Q.19b — three marks of careful arithmetic), from the conceptual beauty of the ball-through-Earth-is-SHM argument (Q.2) to the laboratory realism of the Stoke’s viscosity graph (Q.13a) — this paper touches every corner of the Grade XII Physics new curriculum.

The most important takeaway for students: physics is connected. The capillary height on the Moon (Q.3) and the Laplace correction to Newton’s formula (Q.15a) both use the same insight — physical constants change when environmental parameters change. The two-pulses energy question (Q.7) and the Brewster’s law proof (Q.20d) both depend on understanding wave superposition principles. The Bohr kinetic energy derivation (Q.22c) and the electron deflection calculation (Q.19 OR b) both use the same fundamental equation F = ma in different contexts.

Build these connections deliberately. Study this paper alongside the other three HISSAN papers (2081 D2, 2082 D1, 2082 D2), and you will have the most comprehensive and effective preparation possible for the NEB Grade XII Physics board examination.

Visit nneducation.com for all HISSAN papers, NEB past papers, complete derivation guides, and step-by-step numerical solutions for Grade XI and XII Physics.