HISSAN Central Examination 2082 (2026) — 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 (Higher Secondary Schools’ Association of Nepal) Central Examination is one of Nepal’s most prestigious and widely-administered pre-board examinations for Grade XII students. Unlike school-level tests or generic model papers, HISSAN papers are centrally set and administered simultaneously across hundreds of affiliated private schools nationwide — making them one of the most reliable benchmarks for how students will actually perform in the NEB board examination.

The 2082 (2026) HISSAN Physics Set D2 paper follows the standard NEB new curriculum structure: 11 marks MCQ in Group A, 40 marks short answer in Group B, and 24 marks long answer in Group C. What makes this particular paper stand out is its remarkable range — it tests rotational dynamics, viscosity and fluid flow, thermodynamics, open organ pipe theory, wave equation analysis, photoelectric effect, P-N junction and digital logic, thermoelectricity, LCR circuits, wave optics, Millikan’s oil drop, Bragg’s law, radioactivity, potentiometer theory, meter bridge calculation, moving coil galvanometer, and Biot-Savart law — all in a single 3-hour sitting.

For Grade XII students across Nepal preparing for the NEB board exam, this HISSAN paper is essential practice material. The difficulty level, question style, and topic coverage closely mirror what NEB will ask. This complete guide breaks down every question with full solutions, LaTeX-formatted mathematics, and targeted preparation strategies.

What Makes HISSAN Different from NEB Board Exams

Students who perform well in HISSAN consistently perform well in NEB. Treating HISSAN as seriously as the board exam is the right approach.

Full Question Paper

Exam Overview

Topics Covered — Chapter-wise Weightage

Group A — MCQ Analysis with Answers

Q.1 — Rotational Kinetic Energy of Fan
Answer: d) 20.00 J

Given: [latex]I = 0.30 \text{ kgm}^2[/latex], [latex]n = 110 \text{ rpm}[/latex]

Convert to rad/s: [latex]\omega = \frac{2\pi \times 110}{60} = \frac{220\pi}{60} = \frac{11\pi}{3} \approx 11.52 \text{ rad/s}[/latex]

Rotational KE: [latex]KE = \frac{1}{2}I\omega^2 = \frac{1}{2} \times 0.30 \times (11.52)^2 = 0.15 \times 132.7 \approx 19.9 \approx \mathbf{20 \text{ J}}[/latex]

Q.2 — Capillary Height with One-Third Radius
Answer: c) 9 mm

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

If new radius [latex]r’ = \frac{r}{3}[/latex]:

[latex]h’ = h \times \frac{r}{r’} = 3 \times 3 = \mathbf{9 \text{ mm}}[/latex]

Q.3 — Frequency and Amplitude from Displacement-Time Graph
Answer: d) 5 Hz, 1 cm

From the graph: amplitude = 1.0 cm (maximum displacement from equilibrium).
Time period appears to be T = 0.2 s (reading from the graph scale showing t in seconds with markings at 0.2).
Frequency: [latex]f = \frac{1}{T} = \frac{1}{0.2} = \mathbf{5 \text{ Hz}}[/latex]

Q.4 — Process Where Work Done is Zero
Answer: d) isochoric process

Work done: [latex]W = \int P,dV[/latex]

In an isochoric (constant volume) process, [latex]dV = 0[/latex], so [latex]W = 0[/latex].

In isothermal: W ≠ 0 (gas expands/contracts). In adiabatic: W ≠ 0. In isobaric: W = PΔV ≠ 0.

Q.5 — “Heat Cannot Flow from Cold to Hot” — Consequence Of
Answer: b) second law of thermodynamics

This is the Clausius statement of the second law of thermodynamics: Heat cannot spontaneously flow from a colder body to a hotter body. It is not the first law (energy conservation) or conservation of mass.

Q.6 — Maximum Speed of Building Shaken by S-Waves
Answer: c) 80 m/s

Given: f = 20 Hz, λ = 4 m

Wave speed: [latex]v = f\lambda = 20 \times 4 = 80 \text{ m/s}[/latex]

Maximum speed of vibration = wave speed = 80 m/s

Note: For a particle undergoing SHM in a wave: [latex]v_{max} = \omega A = 2\pi f A[/latex]. But if the question treats wave speed as maximum particle speed (as in some contexts), v = fλ = 80 m/s.

Q.7 — Refractive Index When Reflected Light is Completely Polarized
Answer: b) 1.6 (or c) 1.7)

The reflected light is completely polarized at Brewster’s angle. From the figure, polarizing angle [latex]i_p = 63°[/latex].

By Brewster’s Law: [latex]\mu = \tan(i_p) = \tan(63°) = 1.963 \approx \mathbf{1.96}[/latex]

Closest option: c) 1.7 — Note: if angle is 59°, tan(59°) ≈ 1.66 ≈ 1.6. Students should use the exact angle from the figure. With 63°: μ ≈ 2.0, but closest answer is c) 1.7 given the options.

Use Brewster’s Law: [latex]\mu = \tan(i_B)[/latex] directly with the polarizing angle shown in the figure.

Q.8 — Curie Law: Susceptibility vs Temperature
Answer: a) [latex]T^{-1}[/latex]

Curie’s Law: [latex]\chi = \frac{C}{T}[/latex]

Magnetic susceptibility is inversely proportional to absolute temperature: [latex]\chi \propto T^{-1}[/latex]

Q.9 — Galvanometer Converted to Voltmeter
Answer: c) a high resistance in series with the galvanometer

A voltmeter must have very high resistance so it draws negligible current from the circuit. A galvanometer is converted to a voltmeter by connecting a high resistance (multiplier) in series. (A low resistance in parallel converts it to an ammeter.)

Q.10 — Rectangular Loop Near Long Wire with Current
Answer: a) move towards the wire

When current flows downward in the long wire (as shown by arrow in figure), the magnetic field to the right of the wire points into the page. The rectangular loop carrying current I will experience:

• Attractive force on the closer side (stronger field)
• Repulsive force on the farther side (weaker field)

Net force = towards the wire. The loop moves towards the wire.

By Lenz’s law: The induced effect in the loop opposes any tendency to increase flux — here the loop moves to increase the flux through it from the wire, meaning it moves toward the wire.

Q.11 — [latex]E_k[/latex] vs Frequency Graph for Photoelectric Effect
Answer: (a)

From Einstein’s photoelectric equation:

[latex]E_k = hf – \phi = hf – hf_0[/latex]

This is a straight line with:

• Positive slope h (Planck’s constant)
• X-intercept at threshold frequency [latex]f_0[/latex] (no emission below [latex]f_0[/latex])
• Zero [latex]E_k[/latex] for [latex]f < f_0[/latex] (no emission)
• Linear increase for [latex]f > f_0[/latex]

Graph (a) shows: no emission below [latex]f_0[/latex], then linear increase — this is correct.

Group B — Short Answer Analysis

Q.12 — Conservation of Angular Momentum and Moment of Inertia (OR: SHM)

Part (a): Conservation of Angular Momentum — State and Prove (2 marks)

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

Proof:
Angular momentum: [latex]L = I\omega[/latex]

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

If external torque [latex]\tau = 0[/latex]:

[latex]\frac{dL}{dt} = 0 \Rightarrow L = \text{constant}[/latex]

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

If moment of inertia changes from [latex]I_1[/latex] to [latex]I_2[/latex]: [latex]I_1\omega_1 = I_2\omega_2[/latex] ✓

Part (b): Ice Melting in Polar Region — Effect on Length of Day (1 mark)

When polar ice melts, water spreads away from the poles toward the equator. This increases the average distance of mass from Earth’s rotation axis, increasing Earth’s moment of inertia [latex]I[/latex].

By conservation of angular momentum: [latex]I\omega = \text{constant}[/latex]

If I increases → ω decreases → Time period [latex]T = \frac{2\pi}{\omega}[/latex] increases.

Length of day increases — Earth rotates more slowly.

Part (c): Moment of Inertia of Uniform Rod (2 marks)

Given: L = 2.5 m, m = 25 kg, rotating about centre.

For a uniform rod rotating about its centre:

[latex]I = \frac{mL^2}{12} = \frac{25 \times (2.5)^2}{12} = \frac{25 \times 6.25}{12} = \frac{156.25}{12} = \mathbf{13.02 \text{ kgm}^2}[/latex]

OR Q.12 — Simple Harmonic Motion

Part (a): Two Characteristics of SHM (1 mark)

1. The restoring force (and hence acceleration) is directly proportional to the displacement from equilibrium: [latex]F = -kx[/latex] or [latex]a = -\omega^2 x[/latex]
2. The restoring force always directs toward the equilibrium (mean) position — opposite to displacement.

Part (b): Total Energy of SHM is Independent of Time (2 marks)

For a body in SHM with displacement [latex]x = A\sin(\omega t)[/latex]:

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

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

Total Energy:

[latex]E = KE + PE = \frac{1}{2}m\omega^2 A^2[\cos^2(\omega t) + \sin^2(\omega t)] = \frac{1}{2}m\omega^2 A^2[/latex]

Since [latex]\frac{1}{2}m\omega^2 A^2[/latex] is constant (no time dependence), total energy is independent of time. ✓

Part (c): Spring-Mass System (2 marks)

Given: m = 2 kg, x = 0.1 m (extension at equilibrium)

At equilibrium: [latex]mg = kx[/latex]

[latex]k = \frac{mg}{x} = \frac{2 \times 10}{0.1} = \mathbf{200 \text{ N/m}}[/latex]

Time period: [latex]T = 2\pi\sqrt{\frac{m}{k}} = 2\pi\sqrt{\frac{2}{200}} = 2\pi\sqrt{0.01} = 2\pi \times 0.1 = 0.2\pi \approx \mathbf{0.628 \text{ s}}[/latex]

Q.13 — Viscosity and Fluid Flow

Part (a): What is Viscosity? (1 mark)
Viscosity is the property of a fluid (liquid or gas) by which it resists the relative motion between adjacent layers. It is the internal friction between fluid layers moving at different velocities. High viscosity fluids (like honey) flow slowly; low viscosity fluids (like water) flow easily.

Part (b): Newton’s Formula for Coefficient of Viscosity (2 marks)

According to Newton’s law of viscosity, the viscous force F between two adjacent fluid layers is:

• Proportional to the area A of the layers: [latex]F \propto A[/latex]
• Proportional to the velocity gradient [latex]\frac{dv}{dy}[/latex] (rate of change of velocity with distance): [latex]F \propto \frac{dv}{dy}[/latex]

Combining: [latex]F = \eta A \frac{dv}{dy}[/latex]

Therefore: [latex]\eta = \frac{F}{A \cdot \frac{dv}{dy}}[/latex]

where [latex]\eta[/latex] = coefficient of viscosity. SI unit: Pa·s (Pascal-second) or N/m²·s.

Part (c): Bernoulli’s Equation — Pressure Calculation (2 marks)

Given: [latex]P_1 = 15[/latex] Pa, [latex]v_1 = 0.35[/latex] m/s, [latex]v_2 = 0.25[/latex] m/s, horizontal pipe so [latex]h_1 = h_2[/latex]

Applying Bernoulli’s equation (horizontal pipe):

[latex]P_1 + \frac{1}{2}\rho v_1^2 = P_2 + \frac{1}{2}\rho v_2^2[/latex]

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

Using [latex]\rho_{water} = 1000[/latex] kg/m³:
[latex]P_2 = 15 + \frac{1}{2} \times 1000 \times [(0.35)^2 – (0.25)^2][/latex]
[latex]= 15 + 500 \times [0.1225 – 0.0625][/latex]

[latex]= 15 + 500 \times 0.06 = 15 + 30 = \mathbf{45 \text{ Pa}}[/latex]

Q.14 — Thermodynamics: First Law and Molar Heat Capacities

Part (a): First Law of Thermodynamics (1 mark)

The first law of thermodynamics states that the total energy of a system is conserved. The heat energy (Q) supplied to a system equals the increase in internal energy (ΔU) plus the work done (W) by the system:

[latex]\mathbf{Q = \Delta U + W}[/latex]

or: [latex]dQ = dU + PdV[/latex]

Part (b): Is Efficiency of Heat Engine Always Less Than 100%? (1 mark)

Yes, he is correct. By the second law of thermodynamics, no heat engine can convert all the heat absorbed from a source completely into work. Some heat is always rejected to the sink. For a Carnot engine:

[latex]\eta = 1 – \frac{T_2}{T_1}[/latex]

For η = 100%, we need [latex]T_2 = 0[/latex] K (absolute zero) or [latex]T_1 = \infty[/latex] — both are physically impossible. Therefore efficiency is always less than 100%.

Part (c): Relation Between [latex]C_p[/latex] and [latex]C_v[/latex] — Mayer’s Relation (3 marks)

Consider N moles of ideal gas:

At constant volume (no work done): [latex]Q_v = NC_v\Delta T[/latex], all heat goes to internal energy:
[latex]\Delta U = NC_v\Delta T[/latex] … (i)

At constant pressure: [latex]Q_p = NC_p\Delta T[/latex]

By first law: [latex]Q_p = \Delta U + W = NC_v\Delta T + P\Delta V[/latex]

For ideal gas at constant pressure: [latex]P\Delta V = NR\Delta T[/latex] … (ii)

Therefore: [latex]NC_p\Delta T = NC_v\Delta T + NR\Delta T[/latex]

Dividing by [latex]N\Delta T[/latex]:

[latex]\boxed{C_p – C_v = R}[/latex]

This is Mayer’s relation — the difference between molar heat capacities at constant pressure and constant volume equals the universal gas constant R = 8.314 J/mol/K.

Q.15 — Open Organ Pipe and Wave Equation (OR: Newton-Laplace)

Part (a): Odd and Even Harmonics in Open Organ Pipe (3 marks)

In an open organ pipe (both ends open), both ends are antinodes and the air column vibrates freely at both ends.

Fundamental mode (1st harmonic):

[latex]L = \frac{\lambda_1}{2} \Rightarrow \lambda_1 = 2L \Rightarrow f_1 = \frac{v}{2L}[/latex]

1st overtone (2nd harmonic):

[latex]L = \lambda_2 \Rightarrow \lambda_2 = L \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 pattern: [latex]f_n = \frac{nv}{2L}[/latex] for n = 1, 2, 3, 4, …

Since all positive integer values of n are allowed, both odd (n = 1, 3, 5…) and even (n = 2, 4, 6…) harmonics are present. An open organ pipe produces all harmonics. ✓

Part (b): Wave Equation — Amplitude and Frequency (2 marks)

Given: [latex]y = 0.01\sin(4\pi t – 0.02\pi x)[/latex]

Comparing with standard form [latex]y = A\sin(\omega t – kx)[/latex]:

• Amplitude: [latex]A = 0.01 \text{ m} = \mathbf{1 \text{ cm}}[/latex]
• Angular frequency: [latex]\omega = 4\pi[/latex] rad/s
• Frequency: [latex]f = \frac{\omega}{2\pi} = \frac{4\pi}{2\pi} = \mathbf{2 \text{ Hz}}[/latex]
• Wave number: [latex]k = 0.02\pi[/latex] rad/m
• Wavelength: [latex]\lambda = \frac{2\pi}{k} = \frac{2\pi}{0.02\pi} = 100 \text{ m}[/latex]
• Wave speed: [latex]v = f\lambda = 2 \times 100 = 200 \text{ m/s}[/latex]

OR Q.15 — Newton vs Laplace: Velocity of Sound

Part (a): Why Newton’s Assumption Was Wrong (1 mark)

Newton assumed sound propagation in air is an isothermal process — that compressions and rarefactions occur so slowly that heat has time to flow between compressed and rarefied regions, maintaining constant temperature. This gave:

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

This is wrong because the actual speed of sound in air (~332 m/s at 0°C) is significantly higher. Newton’s assumption was incorrect — sound propagation is too rapid for heat exchange, so the process is actually adiabatic, not isothermal.

Part (b): Laplace’s Correction (2 marks)

Laplace corrected Newton by recognizing that sound propagation is an adiabatic process — compressions and rarefactions happen so rapidly that no heat exchange occurs with the surroundings.

For adiabatic process: [latex]PV^\gamma = \text{constant}[/latex], giving adiabatic bulk modulus = [latex]\gamma P[/latex]

Laplace’s formula: [latex]v = \sqrt{\frac{\gamma P}{\rho}}[/latex]

For air: [latex]\gamma = \frac{C_p}{C_v} = 1.4[/latex]

[latex]v_{Laplace} = \sqrt{1.4} \times v_{Newton} = \sqrt{1.4} \times 280 \approx 331 \text{ m/s}[/latex]

This matches the experimentally measured value. ✓

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

Given: f = 480 Hz, λ = 700 mm = 0.7 m at 25°C

Velocity at 25°C:

[latex]v_{25} = f\lambda = 480 \times 0.7 = \mathbf{336 \text{ m/s}}[/latex]

Velocity at 0°C:

[latex]v_T = v_0\sqrt{\frac{T}{273}}[/latex]

[latex]336 = v_0\sqrt{\frac{298}{273}} \Rightarrow v_0 = 336 \times \sqrt{\frac{273}{298}} = 336 \times 0.957 = \mathbf{321.6 \text{ m/s}}[/latex]

Wavelength at 0°C (frequency unchanged, only v changes):

[latex]\lambda_0 = \frac{v_0}{f} = \frac{321.6}{480} = \mathbf{0.67 \text{ m} = 670 \text{ mm}}[/latex]

Q.16 — Photoelectric Effect

Part (a): Einstein Photoelectric Equation and Work Function (2 marks)

Einstein’s Photoelectric Equation:

[latex]hf = \phi + KE_{max} = \phi + \frac{1}{2}mv_{max}^2[/latex]

or equivalently: [latex]hf = hf_0 + eV_0[/latex]

where h = Planck’s constant, f = frequency of incident photon, [latex]\phi[/latex] = work function, [latex]f_0[/latex] = threshold frequency, [latex]V_0[/latex] = stopping potential.

Physical significance of work function:
Work function latex[/latex] is the minimum energy required to liberate one electron from the surface of a metal. It depends on the nature of the metal surface (not on the frequency or intensity of incident light). Metals with low work functions (like cesium, potassium) are more photosensitive. It represents the energy barrier that holds surface electrons bound to the metal.

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

Given: f = 4.95 × 10¹⁴ Hz, KE = 2.25 × 10⁻¹⁹ J, e = 1.6 × 10⁻¹⁹ C, h = 6.62 × 10⁻³⁴ J·s

Energy of incident photon:

[latex]E = hf = 6.62 \times 10^{-34} \times 4.95 \times 10^{14} = 32.769 \times 10^{-20} = 3.277 \times 10^{-19} \text{ J}[/latex]

Work function:

[latex]\phi = hf – KE_{max} = 3.277 \times 10^{-19} – 2.25 \times 10^{-19} = \mathbf{1.027 \times 10^{-19} \text{ J}}[/latex]

Converting to eV: [latex]\phi = \frac{1.027 \times 10^{-19}}{1.6 \times 10^{-19}} = \mathbf{0.642 \text{ eV}}[/latex]

Stopping potential:
[latex]eV_0 = KE_{max}[/latex]

[latex]V_0 = \frac{KE_{max}}{e} = \frac{2.25 \times 10^{-19}}{1.6 \times 10^{-19}} = \mathbf{1.406 \text{ V} \approx 1.41 \text{ V}}[/latex]

Q.17 — P-N Junction and NOR Gate

Part (a): How P-N Junction Diode is Formed (1 mark)

A P-N junction is formed when a P-type semiconductor (doped with acceptor impurities like Boron, creating holes as majority carriers) is brought into metallurgical contact with an N-type semiconductor (doped with donor impurities like Phosphorus, creating electrons as majority carriers). At the junction, electrons from the N-side diffuse to the P-side and holes from the P-side diffuse to the N-side, creating a depletion region with an internal electric field (potential barrier ~0.7V for Si, ~0.3V for Ge).

Part (b): Characteristics of Junction Diode (2 marks)

Forward bias characteristics:
When positive terminal of battery connects to P-side: depletion layer narrows, current increases exponentially once forward voltage exceeds threshold (~0.7V for Si). The graph shows a rapid current rise after threshold voltage.

Reverse bias characteristics:
When positive terminal connects to N-side: depletion layer widens, only tiny reverse saturation current ([latex]I_0[/latex] ~μA) flows due to minority carriers. At very high reverse voltage, breakdown occurs (Zener/avalanche), current increases rapidly.

The V-I graph has: large forward current for small forward voltage (beyond threshold), nearly zero reverse current up to breakdown voltage, then sharp reverse current increase at breakdown.

Part (c): NOR Gate — Working with Truth Table (2 marks)

A NOR gate is an OR gate followed by a NOT gate. Output Y = NOT(A OR B) = [latex]\overline{A+B}[/latex]

Truth Table:

Working: Output is HIGH (1) only when both inputs are LOW (0). If either or both inputs are HIGH, output is LOW. NOR is called a universal gate because any logic circuit can be built using NOR gates alone.

Q.18 — Thermoelectricity

Part (a): Seebeck Effect vs Peltier Effect (2 marks)

The Seebeck effect is the basis of thermocouples — temperature difference generates emf. The Peltier effect is used in thermoelectric coolers — current flow creates temperature difference (one junction heats, other cools).

Part (b): [latex]\theta_1, \theta_2, \theta_3[/latex] in Thermo-emf Graph (2 marks)

From the thermo-emf vs temperature graph (parabolic shape with maximum):

• [latex]\theta_1[/latex] = Cold junction temperature (temperature of cold junction, often kept at 0°C)
• [latex]\theta_2[/latex] = Neutral temperature (temperature of hot junction at which thermo-emf is maximum)
• [latex]\theta_3[/latex] = Temperature of inversion (temperature at which thermo-emf returns to zero; beyond this it reverses)

Relation between them:

[latex]\theta_2 = \frac{\theta_1 + \theta_3}{2}[/latex]

i.e., neutral temperature is the arithmetic mean of cold junction temperature and inversion temperature.

Or equivalently: [latex]\theta_3 – \theta_2 = \theta_2 – \theta_1[/latex]

Part (c): Two Uses of Thermocouple (1 mark)

1. Temperature measurement — Thermocouples measure temperatures ranging from very low (cryogenic) to very high (furnaces, up to 1600°C) — far beyond mercury thermometer range
2. Thermoelectric generator — Converting waste heat directly into electricity in industrial processes, spacecraft (RTG — Radioisotope Thermoelectric Generator)

Q.19 — AC Circuits and LCR Impedance

Part (a): RMS Value of AC (1 mark)

The RMS (Root Mean Square) value of alternating current is the equivalent DC current that produces the same heating effect in a resistance as the alternating current does in the same time period.

[latex]I_{rms} = \frac{I_0}{\sqrt{2}} \approx 0.707 \times I_0[/latex]

where [latex]I_0[/latex] = peak current. Similarly: [latex]V_{rms} = \frac{V_0}{\sqrt{2}}[/latex]

Part (b): Impedance of LCR Circuit — Derivation (2 marks)

In a series LCR circuit:

• Voltage across R: [latex]V_R = IR[/latex] (in phase with current)
• Voltage across L: [latex]V_L = IX_L = I\omega L[/latex] (leads current by 90°)
• Voltage across C: [latex]V_C = IX_C = \frac{I}{\omega C}[/latex] (lags current by 90°)

Since [latex]V_L[/latex] and [latex]V_C[/latex] are 180° out of phase with each other, net reactive voltage = [latex](V_L – V_C)[/latex]

Using phasor addition (V_R perpendicular to net reactive voltage):

[latex]V_{total} = \sqrt{V_R^2 + (V_L – V_C)^2} = I\sqrt{R^2 + (X_L – X_C)^2}[/latex]

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

Phase angle: [latex]\tan\phi = \frac{X_L – X_C}{R} = \frac{\omega L – \frac{1}{\omega C}}{R}[/latex]

Part (c): LCR Circuit Current Calculation (2 marks)

Given: L = 2H, [latex]R_{coil}[/latex] = 50Ω, [latex]R_{series}[/latex] = 950Ω, V = 220V, f = 50 Hz

Total resistance: [latex]R = 50 + 950 = 1000 \text{ Ω}[/latex]

Inductive reactance: [latex]X_L = \omega L = 2\pi \times 50 \times 2 = 200\pi = 628.3 \text{ Ω}[/latex]

(No capacitor given, so only R and X_L)

Impedance: [latex]Z = \sqrt{R^2 + X_L^2} = \sqrt{1000^2 + 628.3^2} = \sqrt{10^6 + 3.947 \times 10^5} = \sqrt{1.3947 \times 10^6} = 1181 \text{ Ω}[/latex]

Current: [latex]I = \frac{V}{Z} = \frac{220}{1181} = \mathbf{0.186 \text{ A} \approx 0.19 \text{ A}}[/latex]

Group C — Long Answer Analysis

Q.20 — Wave Optics: Interference and Diffraction

Part (a): Differences Between Interference and Diffraction (2 marks)

Part (b): Single Slit Diffraction — Secondary Maxima and Minima Conditions (3 marks)

When monochromatic light of wavelength λ passes through a single slit of width a, the intensity pattern on a screen shows a broad central maximum and secondary maxima on either side.

Condition for minima (dark fringes):

Divide the slit into equal zones. For destructive interference (minima):

[latex]a\sin\theta = m\lambda[/latex] where m = ±1, ±2, ±3, …

Position on screen: [latex]y_m = \frac{m\lambda D}{a}[/latex]

Condition for secondary maxima:

Secondary maxima occur approximately midway between minima:

[latex]a\sin\theta = \left(m + \frac{1}{2}\right)\lambda[/latex] where m = ±1, ±2, ±3, …

i.e., [latex]a\sin\theta = \frac{3\lambda}{2}, \frac{5\lambda}{2}, \frac{7\lambda}{2}, …[/latex]

Width of central maximum:

[latex]w = \frac{2\lambda D}{a}[/latex]

Secondary maxima are half as wide as the central maximum. Intensity of secondary maxima decreases rapidly: first secondary is ~4.5% of central maximum intensity.

Part (c): Wavelength from Dark Fringe Spacing in YDSE (3 marks)

Given: slit separation d = 0.3 mm = 3 × 10⁻⁴ m, D = 50 cm = 0.5 m

Distance between 2nd and 3rd dark fringes = fringe width β = 1 mm = 10⁻³ m

For dark fringes in YDSE:

[latex]y_n = \frac{(2n-1)\lambda D}{2d}[/latex]

Distance between consecutive dark fringes = fringe width:

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

Therefore:

[latex]\lambda = \frac{\beta d}{D} = \frac{10^{-3} \times 3 \times 10^{-4}}{0.5} = \frac{3 \times 10^{-7}}{0.5} = \mathbf{6 \times 10^{-7} \text{ m} = 600 \text{ nm}}[/latex]

This corresponds to orange-red visible light. ✓

Q.21 — Millikan’s Oil Drop, Electron Paths, and Bragg’s Law (OR: Radioactivity)

Part (a): Principle Used in Millikan’s Oil Drop Experiment (1 mark)

Millikan’s oil drop experiment uses the principle of equilibrium of forces — specifically the balance between:

1. Gravitational force (downward): [latex]W = mg[/latex]
2. Electric force (upward): [latex]F_E = qE[/latex]
3. Viscous drag (upward/downward): [latex]F_v = 6\pi\eta r v[/latex] (Stokes’ law)

When the drop is stationary: electric force = gravitational force → [latex]qE = mg[/latex]. This allows calculation of the charge q on the drop.

Part (b): Millikan’s Oil Drop Diagram and Function of X-rays (2 marks)

Diagram components:

• Two parallel metal plates (upper positive, lower negative for levitating negative drops)
• Atomizer/spray nozzle to introduce oil drops
• Microscope for observation
• X-ray source (or radioactive source)
• High voltage source with voltmeter
• Illumination source

Function of X-rays in the experiment:
X-rays ionize the air and the oil droplets — they strip electrons from or add electrons to the droplets, changing the charge on the drop. This allows studying drops with different amounts of charge (different numbers of electrons) and verifying that all charges are integral multiples of the elementary charge e. X-rays make it possible to change the drop’s charge while observing it, without needing to replace the drop.

Part (c): Circular vs Parabolic Path of Electron (2 marks)

In magnetic field (circular path):
When an electron moves perpendicular to a uniform magnetic field B, the magnetic force [latex]F = qvB[/latex] acts always perpendicular to the velocity. This means the force only changes the direction, never the speed. The result is uniform circular motion — the electron traces a circle.

[latex]qvB = \frac{mv^2}{r} \Rightarrow r = \frac{mv}{qB}[/latex]

In electric field (parabolic path):
When an electron moves perpendicular to a uniform electric field E, the electric force [latex]F = qE[/latex] acts parallel to the field (constant in direction and magnitude). This is analogous to projectile motion under gravity — constant acceleration in one direction, uniform motion in the perpendicular direction. The result is parabolic motion (just like a projectile).

Part (d): Bragg’s Law — Spacing of Atomic Planes (3 marks)

Given: λ = 0.36 Å = 0.36 × 10⁻¹⁰ m, first order (n = 1), glancing angle θ = 4.8°

Bragg’s Law: [latex]2d\sin\theta = n\lambda[/latex]

[latex]d = \frac{n\lambda}{2\sin\theta} = \frac{1 \times 0.36 \times 10^{-10}}{2 \times \sin(4.8°)}[/latex]

[latex]\sin(4.8°) = 0.0837[/latex]

[latex]d = \frac{0.36 \times 10^{-10}}{2 \times 0.0837} = \frac{0.36 \times 10^{-10}}{0.1674} = 2.15 \times 10^{-10} \text{ m} = \mathbf{2.15 \text{ Å}}[/latex]

This is typical of interplanar spacing in crystals — consistent with crystal lattice dimensions. ✓

OR Q.21 — Radioactivity

Part (a): Activity of Radioactive Substance (1 mark)
Activity of a radioactive substance is the number of disintegrations (nuclear decays) occurring per unit time. It is also called the decay rate.

[latex]A = -\frac{dN}{dt} = \lambda N[/latex]

SI unit: Becquerel (Bq) = 1 disintegration per second. Older unit: Curie (1 Ci = 3.7 × 10¹⁰ Bq).

Part (b): Decay Law and Decay Equation (1+2 marks)

Decay Law (Statement): The rate of radioactive disintegration of a substance is directly proportional to the number of undecayed nuclei present at that instant.

[latex]-\frac{dN}{dt} \propto N \Rightarrow -\frac{dN}{dt} = \lambda N[/latex]

where λ = decay constant.

Derivation:

[latex]\frac{dN}{N} = -\lambda,dt[/latex]

Integrating: [latex]\int_{N_0}^{N}\frac{dN}{N} = -\lambda\int_0^t dt[/latex]

[latex]\ln\frac{N}{N_0} = -\lambda t[/latex]

[latex]\boxed{N = N_0 e^{-\lambda t}}[/latex]

Activity at time t: [latex]A = A_0 e^{-\lambda t}[/latex]

Part (c): Two Applications of Radioactivity (1 mark)

1. Carbon-14 dating — Determining age of ancient organic materials (fossils, archaeological objects) using the decay of C-14
2. Cancer treatment (radiotherapy) — Using gamma rays from radioactive sources (e.g., Co-60) to destroy cancerous cells

Part (d): Time for Radium Activity to Reduce to 1/10 (3 marks)

Given: [latex]t_{1/2} = 1620[/latex] years, activity reduced to [latex]\frac{A_0}{10}[/latex]

[latex]A = A_0 e^{-\lambda t}[/latex]

[latex]\frac{A_0}{10} = A_0 e^{-\lambda t}[/latex]

[latex]e^{-\lambda t} = \frac{1}{10}[/latex]

[latex]\lambda t = \ln 10 = 2.3026[/latex]

[latex]t = \frac{2.3026}{\lambda} = \frac{2.3026 \times t_{1/2}}{\ln 2} = \frac{2.3026 \times 1620}{0.6931}[/latex]

[latex]t = \frac{3730.2}{0.6931} = \mathbf{5382 \text{ years}}[/latex]

Q.22 — Potentiometer and Meter Bridge

Part (a): Why Prefer Potentiometer to Measure EMF? (2 marks)

A potentiometer is preferred over a voltmeter to measure EMF because:

1. No current flows at balance: At null point, no current is drawn from the cell being tested — it is an open circuit condition. Therefore terminal voltage = true EMF. A voltmeter always draws current, causing a drop across internal resistance: V = E − Ir < E.
2. Null method — highly accurate: The potentiometer uses a null deflection method (galvanometer reads zero at balance). At null, the reading is independent of the galvanometer’s own resistance. This eliminates systematic errors and gives much higher precision than any deflection instrument.
3. Can measure very small EMFs and compare EMFs of cells with high precision.

Part (b): Potentiometer to Measure Internal Resistance of Cell (3 marks)

Circuit setup:

1. Connect driver cell through rheostat across potentiometer wire AB
2. Connect test cell (EMF = E, internal resistance = r) through key K₁ to jockey — find balance length [latex]l_1[/latex] (K₁ open → no current through cell)
3. Connect known external resistance R across cell terminals, close K₁ → find new balance length [latex]l_2[/latex]

Derivation:

Without R (open circuit): [latex]E \propto l_1[/latex] → [latex]E = kl_1[/latex] … (i)

With R (current flows): Terminal voltage [latex]V = E – Ir = \frac{ER}{R+r}[/latex]

[latex]V = kl_2[/latex] → [latex]\frac{ER}{R+r} = kl_2[/latex] … (ii)

Dividing (i) by (ii):

[latex]\frac{E}{\frac{ER}{R+r}} = \frac{l_1}{l_2}[/latex]

[latex]\frac{R+r}{R} = \frac{l_1}{l_2}[/latex]

[latex]\boxed{r = R\left(\frac{l_1}{l_2} – 1\right) = R\left(\frac{l_1 – l_2}{l_2}\right)}[/latex]

Part (c): Meter Bridge — Find R and S (3 marks)

Case 1: Null point at 40 cm from A. R in one gap, S in other.

By Wheatstone bridge balance: [latex]\frac{R}{S} = \frac{l}{100-l} = \frac{40}{60} = \frac{2}{3}[/latex]

[latex]3R = 2S[/latex] → [latex]S = \frac{3R}{2}[/latex] … (i)

Case 2: 12Ω connected in parallel with S. New balance at 50 cm from A.

Parallel combination of S and 12Ω: [latex]S’ = \frac{12S}{12+S}[/latex]

New balance: [latex]\frac{R}{S’} = \frac{50}{50} = 1[/latex]

Therefore: [latex]R = S’ = \frac{12S}{12+S}[/latex] … (ii)

From (i): [latex]S = \frac{3R}{2}[/latex]

Substituting in (ii): [latex]R = \frac{12 \times \frac{3R}{2}}{12 + \frac{3R}{2}} = \frac{18R}{12 + \frac{3R}{2}} = \frac{18R}{\frac{24+3R}{2}} = \frac{36R}{24+3R}[/latex]

[latex]R(24 + 3R) = 36R[/latex]

[latex]24 + 3R = 36[/latex]

[latex]3R = 12[/latex]

[latex]\mathbf{R = 4 \text{ Ω}}[/latex]

From (i): [latex]S = \frac{3 \times 4}{2} = \mathbf{6 \text{ Ω}}[/latex]

Verification: 12 || 6 = [latex]\frac{12 \times 6}{12+6} = 4 \text{ Ω} = R[/latex] → balance at 50 cm ✓

OR Q.22 — Moving Coil Galvanometer and Solenoid

Part (a): Principle of Moving Coil Galvanometer (1 mark)

A moving coil galvanometer works on the principle that a current-carrying coil placed in a magnetic field experiences a torque. The torque [latex](\tau = NBIA)[/latex] due to the magnetic force deflects the coil, and this deflection is opposed by the restoring torque of the suspension (phosphor-bronze strip). At equilibrium: deflection is proportional to current.

Part (b): How Radial Field is Created (1 mark)

A radial (uniform) magnetic field is created in a moving coil galvanometer by:

1. Using curved (concave) pole pieces of the permanent magnet — shaped to follow the curvature of the cylindrical coil
2. Placing a soft iron cylinder (core) inside the coil

Together, these ensure that the magnetic field is always parallel to the plane of the coil (radial) regardless of the coil’s position. This makes the torque [latex]NBIA\sin\theta \approx NBIA[/latex] (since field is always perpendicular to coil normal), ensuring a linear scale — deflection is directly proportional to current.

Part (c): Biot-Savart Law and Magnetic Field at Centre of Circular Coil (1+2 marks)

Biot-Savart Law:
The magnetic field [latex]d\vec{B}[/latex] at a point P due to a small current element [latex]Id\vec{l}[/latex] at distance r is:

[latex]d\vec{B} = \frac{\mu_0}{4\pi} \frac{Id\vec{l} \times \hat{r}}{r^2}[/latex]

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

Field at Centre of Circular Coil:

For a circular loop of radius R with current I, every element [latex]dl[/latex] is perpendicular to the radius vector (θ = 90°, sin θ = 1):

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

All contributions point in the same direction (perpendicular to the plane of the loop, by right-hand rule).

Integrating around the full loop ([latex]\oint dl = 2\pi R[/latex]):

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

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

Part (d): Magnetic Flux Density in Solenoid (3 marks)

Given: L = 1.96 m, N = 2500 turns, I = 2.6 A, radius = 3 cm (not needed for B), [latex]\mu_0 = 4\pi \times 10^{-7}[/latex] H/m

Number of turns per unit length: [latex]n = \frac{N}{L} = \frac{2500}{1.96} = 1275.5[/latex] turns/m

(a) Flux density at centre:
[latex]B_{centre} = \mu_0 nI = 4\pi \times 10^{-7} \times 1275.5 \times 2.6[/latex]
[latex]= 4\pi \times 10^{-7} \times 3316.3[/latex]
[latex]= 4 \times 3.1416 \times 10^{-7} \times 3316.3[/latex]
[latex]= 12.566 \times 10^{-7} \times 3316.3[/latex]

[latex]= 4.17 \times 10^{-3} \text{ T} = \mathbf{4.17 \text{ mT}}[/latex]

(b) Flux density at one end:
At the end of a long solenoid, the field is exactly half the central value:

[latex]B_{end} = \frac{B_{centre}}{2} = \frac{4.17 \times 10^{-3}}{2} = \mathbf{2.08 \times 10^{-3} \text{ T} = 2.08 \text{ mT}}[/latex]

Marks Weightage Summary

Difficulty Analysis

Easy — Reliable marks for all students:

• Q.1 (Rotational KE — direct formula)
• Q.4 (Isochoric process — recall)
• Q.5 (Second law — conceptual)
• Q.8 (Curie law — direct)
• Q.9 (Galvanometer conversion — recall)
• Q.12(b) (Ice melting — conceptual)
• Q.14(a) (First law statement)
• Q.15(b) (Wave equation reading)
• Q.17(a) (P-N junction formation)
• Q.18(c) (Thermocouple uses)
• Q.19(a) (RMS definition)

Moderate — Needs regular practice:

• Q.2 (Capillary height ratio)
• Q.3 (SHM graph reading)
• Q.12(c) (Moment of inertia of rod)
• Q.13(c) (Bernoulli numerical)
• Q.14(b) (Heat engine efficiency explanation)
• Q.15(a) (Open organ pipe harmonics proof)
• Q.16(a) (Photoelectric equation + work function)
• Q.17(b)(c) (Diode characteristics + NOR gate)
• Q.18(a)(b) (Seebeck-Peltier + temperatures)
• Q.20(a) (Interference vs diffraction)

Challenging — For top scorers:

• Q.6 (Maximum speed of building — conceptual)
• Q.10 (Current loop near wire — force direction)
• Q.14(c) (Mayer’s relation derivation — 3 marks)
• Q.16(b) (Photoelectric numerical — work function + stopping potential)
• Q.19(b)(c) (LCR impedance derivation + numerical)
• Q.20(b)(c) (Diffraction conditions + YDSE wavelength)
• Q.21(c)(d) (Circular vs parabolic paths + Bragg’s law)
• Q.22(b)(c) (Internal resistance by potentiometer + meter bridge calculation)
• Q.22 OR (c)(d) (Biot-Savart + solenoid flux density)

Student Preparation Tips

1. The Meter Bridge Calculation (Q.22c) is a Must-Master
The meter bridge problem with parallel resistance is one of the most common and trickiest numerical problems in NEB/HISSAN exams. Key steps: (a) write Wheatstone balance for Case 1, (b) compute parallel combination for Case 2, (c) write new balance, (d) solve simultaneous equations. Practice at least 5 such problems.

2. Open Organ Pipe (Q.15a) = 3 Marks — Learn it Once, Use it Forever
The proof that open organ pipes produce all harmonics (both odd and even) is completely predictable. Set up the boundary conditions (antinodes at both ends), derive [latex]f_n = \frac{nv}{2L}[/latex] for all n, and conclude. This takes less than 3 minutes if well-prepared.

3. Mayer’s Relation (Q.14c) is a Core Derivation
[latex]C_p – C_v = R[/latex] — this derivation appears in almost every HISSAN and NEB paper. Master the three-step proof: constant volume (Q_v = NC_vΔT = ΔU), constant pressure (Q_p = NC_pΔT = ΔU + W), use PΔV = NRΔTand subtract.

4. Bragg’s Law Numericals are Always Straightforward
[latex]2d\sin\theta = n\lambda[/latex] — given two of the three (λ, θ, d), find the third. Always check units (Å vs nm vs m) and use the correct order n. These 3-mark numericals are completely predictable.

5. Radioactivity Numericals — Time for Activity Reduction
[latex]t = \frac{t_{1/2} \times \ln(A_0/A)}{\ln 2}[/latex] — memorize this formula. For Q.21 OR(d): time for activity to become 1/10 = [latex]\frac{1620 \times \ln 10}{\ln 2} = \frac{1620 \times 2.303}{0.693} \approx 5382[/latex] years.

6. Prepare Both Options in Every OR Question
Q.12 (Rotational/SHM), Q.15 (Organ pipe/Newton-Laplace), Q.21 (Millikan-Bragg/Radioactivity), Q.22 (Potentiometer/Galvanometer-Solenoid) — all have OR options. Enter the exam hall having prepared all options. Choose the easier one on exam day.

7. Time Management
Group A (11 MCQs): 15 minutes. Group B (Q.12–Q.19): 80 minutes (~10 min each). Group C (Q.20–Q.22): 60 minutes (~20 min each). Buffer: 25 minutes for revision.

Frequently Asked Questions

Q: How does HISSAN 2082 Set D2 compare to NEB 2081 Set D in difficulty?
A: HISSAN Set D2 is slightly more comprehensive — it includes topics like the Seebeck-Peltier comparison (Q.18), Newton-Laplace correction (Q.15 OR), Bragg’s law numerical (Q.21d), and internal resistance by potentiometer (Q.22b) that require deeper understanding. The NEB 2081 Set D focused more heavily on electrostatics derivations (Gauss law, capacitor). Both papers are at similar overall difficulty levels.

Q: What is the most frequently tested topic across HISSAN and NEB papers?
A: Based on this paper and past papers: Rotational dynamics (conservation of angular momentum, moment of inertia), thermodynamics (Mayer’s relation, heat engine efficiency), wave optics (YDSE fringe width, single slit diffraction), photoelectric effect (Einstein equation, numerical), and potentiometer/meter bridge — these five topic areas appear in virtually every HISSAN and NEB Grade XII Physics paper.

Q: For Q.22(c) meter bridge problem — what’s the most common mistake?
A: Students often forget to compute the parallel combination correctly for Case 2. The parallel combination of S and 12Ω is [latex]\frac{12S}{12+S}[/latex], NOT just one of them. Then set this equal to R (since balance is at 50 cm, ratio = 1). Always verify your answer by checking both cases.

Q: The LCR current calculation (Q.19c) — why is the answer so large?
A: The iron-cored coil has L = 2H, giving X_L = 2π × 50 × 2 ≈ 628Ω. With total R = 1000Ω, Z ≈ 1181Ω, giving I = 220/1181 ≈ 0.186A. This is reasonable for a 220V AC circuit with 1kΩ total resistance.

Q: What is the difference between Seebeck and Peltier effects in simple terms?
A: Seebeck: heat difference → electricity (thermocouple measures temperature). Peltier: electricity → heat difference (thermoelectric cooler/heater). One converts thermal energy to electrical, the other converts electrical energy to thermal. They are inverse effects of each other.

Where to Find the Full Question Paper

The complete original question paper with all figures — including the SHM displacement-time graph (Q.3), the Brewster’s angle diagram (Q.7), the rectangular loop near wire diagram (Q.10), the photoelectric [latex]E_k[/latex] vs f graphs (Q.11), the thermo-emf parabola graph (Q.18), and the meter bridge circuit diagram (Q.22c) — is available as a PDF download from above.

Conclusion

The HISSAN Central Examination 2082 Set D2 Physics paper is a well-constructed, comprehensive pre-board assessment that closely mirrors the NEB board examination format. It tests every major chapter of the Grade XII Physics new curriculum — from the classical mechanics of rotating fans to the quantum physics of photoelectric emission, from the thermoelectric behavior of junction pairs to the crystallographic measurement of atomic spacings using Bragg’s law.

What sets this paper apart is the presence of several questions that demand not just formula recall but genuine physical reasoning: explaining why electrons follow circular paths in magnetic fields but parabolic paths in electric fields (Q.21c), justifying why potentiometers beat voltmeters (Q.22a), proving that open organ pipes produce all harmonics (Q.15a), and deriving Mayer’s relation from thermodynamic first principles (Q.14c).

Students who work through this paper completely — attempting every question, checking every numerical, and understanding every derivation — will be genuinely well-prepared for the NEB board examination. The topics tested here are exactly what NEB tests, at exactly the right level of depth.

Keep visiting nneducation.com for more HISSAN papers, NEB past papers, model questions, concept notes, and step-by-step numerical solutions for Grade XI and XII Physics, Chemistry, Mathematics, and Nepali.