HISSAN Central Examination 2082 (2026) — 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
Downloaded from: nneducation.com

Introduction

The HISSAN Central Examination 2082 Set D1 is the companion paper to Set D2, administered simultaneously to different groups of Grade XII students across HISSAN-affiliated schools in Nepal. While both papers share the same structure and difficulty level, Set D1 covers a distinctly different selection of topics and question styles — making studying both papers together the most comprehensive preparation strategy for the NEB board examination.

Set D1 is particularly notable for several reasons. It introduces gravitational waves (Q.11) — a cutting-edge astrophysics topic that reflects modern physics awareness. It tests the diesel engine P-V diagram (Q.15) — a thermodynamics application not commonly seen in model papers. It has a beautiful shower head fluid flow problem (Q.13c) requiring the continuity equation across 30 circular openings. It also features the falling magnet through a copper ring (Q.17c) — an electromagnetic induction conceptual problem that perfectly illustrates Lenz’s law in action.

The Group C section is equally impressive: solenoid field using Ampere’s theorem (Q.20), circular coil magnetic field calculation (Q.20b), diffraction grating numerical (Q.21c), electron circular motion with potential difference (Q.22c), and the path of neutron in magnetic field (Q.22d) — a conceptual question that tests the fundamental understanding of magnetic force on neutral particles.

Together, HISSAN 2082 Set D1 and Set D2 form a complete, near-comprehensive review of the entire Grade XII Physics new curriculum.

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 — Rotational Kinetic Energy of Wheel
Answer: b) 2220 J

Given: [latex]I = 0.5 \text{ kgm}^2[/latex], [latex]f = 15 \text{ Hz}[/latex]

[latex]\omega = 2\pi f = 2\pi \times 15 = 30\pi = 94.25 \text{ rad/s}[/latex]

[latex]KE = \frac{1}{2}I\omega^2 = \frac{1}{2} \times 0.5 \times (30\pi)^2 = 0.25 \times 900\pi^2[/latex]

[latex]= 0.25 \times 900 \times 9.8696 = 0.25 \times 8882.6 = \mathbf{2220.7 \approx 2220 \text{ J}}[/latex]

Q.2 — Viscosity of Liquid with Increasing Temperature
Answer: b) it decreases

For liquids, viscosity decreases with increasing temperature. This is because rising temperature provides more thermal energy, breaking the intermolecular cohesive forces that cause viscous resistance — molecules slide past each other more easily. (Note: For gases, viscosity increases with temperature — opposite behavior.)

Q.3 — Time Period and Amplitude from SHM Graph
Answer: c) 8 s, 1 cm

From the displacement-time graph: amplitude = maximum displacement = 1 cm (reading the y-axis scale). One complete cycle spans from 0 to 8 s on the time axis → T = 8 s.

Q.4 — Correct Relation for Isothermal Process
Answer: c) PV = constant

For an isothermal process, temperature T is constant. By Boyle’s Law for an ideal gas: PV = constant (Boyle’s Law).

• [latex]PV^\gamma = \text{constant}[/latex] → adiabatic process
• [latex]TV^{\gamma-1} = \text{constant}[/latex] → adiabatic process
• [latex]\frac{T^\gamma}{P^{\gamma-1}} = \text{constant}[/latex] → also adiabatic

Only PV = constant is isothermal.

Q.5 — Work Done by Carnot Engine
Answer: d) 5.33 J

Given: [latex]T_1 = 900[/latex] K, [latex]T_2 = 300[/latex] K, [latex]Q_1 = 8[/latex] J

Efficiency: [latex]\eta = 1 – \frac{T_2}{T_1} = 1 – \frac{300}{900} = 1 – \frac{1}{3} = \frac{2}{3}[/latex]

Work done: [latex]W = \eta Q_1 = \frac{2}{3} \times 8 = \frac{16}{3} = \mathbf{5.33 \text{ J}}[/latex]

Q.6 — Speed of Sound When Temperature Doubles and Pressure Halves
Answer: c) [latex]330\sqrt{2}[/latex] m/s

Speed of sound: [latex]v = \sqrt{\frac{\gamma P}{\rho}}[/latex]

For an ideal gas: [latex]P = \rho RT/M[/latex], so [latex]v = \sqrt{\frac{\gamma RT}{M}} \propto \sqrt{T}[/latex]

Speed is independent of pressure (at constant temperature — when pressure halves and density halves together, [latex]P/\rho[/latex] remains constant). Only temperature change matters.

NTP: T₁ = 273 K. New temperature: T₂ = 2 × 273 = 546 K

[latex]\frac{v_2}{v_1} = \sqrt{\frac{T_2}{T_1}} = \sqrt{\frac{546}{273}} = \sqrt{2}[/latex]

[latex]v_2 = 330\sqrt{2} \approx \mathbf{466.7 \text{ m/s}}[/latex]

Q.7 — Length of Pipe in Figure
Answer: c) [latex]l = \frac{3\lambda}{4}[/latex]

From the figure showing a closed organ pipe (one end closed, one end open) with two nodes and two antinodes visible — this represents the first overtone (3rd harmonic) of a closed pipe.

For a closed organ pipe at first overtone: [latex]l = \frac{3\lambda}{4}[/latex]

The figure shows the pipe with crossing lines (nodes) at both ends and a middle node — indicating 3/4 wavelength fits in the pipe length.

Q.8 — Temperature of Inversion
Answer: c) 1210°C

Given: Neutral temperature [latex]\theta_n = 600°C[/latex], cold junction [latex]\theta_c = -10°C[/latex]

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

[latex]600 = \frac{-10 + \theta_i}{2}[/latex]

[latex]1200 = -10 + \theta_i[/latex]

[latex]\theta_i = \mathbf{1210°C}[/latex]

Q.9 — Ratio of X to Y in Meter Bridge at 30 cm
Answer: a) 3:7

At null point 30 cm from A (out of 100 cm total):

By Wheatstone bridge principle:

[latex]\frac{X}{Y} = \frac{l}{100-l} = \frac{30}{70} = \frac{3}{7}[/latex]

X:Y = 3:7

Q.10 — Substance with Negative Susceptibility
Answer: c) diamagnetic

Magnetic susceptibility [latex]\chi = \frac{M}{H}[/latex]:

• Diamagnetic: [latex]\chi < 0[/latex] (small negative) — repelled by magnetic field, e.g., bismuth, copper, gold
• Paramagnetic: [latex]\chi > 0[/latex] (small positive) — weakly attracted
• Ferromagnetic: [latex]\chi >> 0[/latex] (large positive) — strongly attracted

Q.11 — Gravitational Waves are Due To
Answer: b) ripple in space time curvature

Gravitational waves are disturbances in the fabric of spacetime caused by accelerating massive objects (such as merging black holes or neutron stars). They propagate as ripples in spacetime curvature at the speed of light, first predicted by Einstein’s General Theory of Relativity and first directly detected by LIGO in 2015.

This is a cutting-edge modern physics question — its inclusion in HISSAN reflects the updated awareness of contemporary physics in Nepal’s curriculum.

Group B — Short Answer Analysis

Q.12 — Simple Harmonic Motion and Simple Pendulum

Part (a): What is Simple Harmonic Motion? (1 mark)
Simple Harmonic Motion (SHM) is a type of periodic motion in which the restoring force acting on the body is directly proportional to its displacement from the equilibrium (mean) position and is always directed toward the equilibrium position.

Mathematically: [latex]F = -kx[/latex] or equivalently [latex]a = -\omega^2 x[/latex]

The negative sign indicates the force/acceleration is opposite to displacement.

Part (b): Spring Pendulum is Simple Harmonic — Justify (2 marks)

For a mass m attached to a spring of force constant k, displaced by x from equilibrium:

Restoring force: [latex]F = -kx[/latex]

By Newton’s second law: [latex]ma = -kx[/latex]

[latex]a = -\frac{k}{m}x[/latex]

Let [latex]\omega^2 = \frac{k}{m}[/latex]:

[latex]a = -\omega^2 x[/latex]

Since acceleration is directly proportional to displacement and directed opposite to it, the motion satisfies the definition of SHM. ✓

The time period: [latex]T = 2\pi\sqrt{\frac{m}{k}}[/latex], frequency: [latex]f = \frac{1}{2\pi}\sqrt{\frac{k}{m}}[/latex]

Part (c): Length of Simple Pendulum with T = 6 s (2 marks)

Given: T = 6 s, g = 10 m/s²

For a simple pendulum: [latex]T = 2\pi\sqrt{\frac{L}{g}}[/latex]

[latex]6 = 2\pi\sqrt{\frac{L}{10}}[/latex]

[latex]\sqrt{\frac{L}{10}} = \frac{6}{2\pi} = \frac{3}{\pi}[/latex]

[latex]\frac{L}{10} = \frac{9}{\pi^2}[/latex]

[latex]L = \frac{90}{\pi^2} = \frac{90}{9.8696} = \mathbf{9.12 \text{ m}}[/latex]

Q.13 — Bernoulli’s Principle and Shower Head (OR: Rotational Dynamics)

Part (a): Bernoulli’s Principle (1 mark)
For a steady, non-viscous, incompressible fluid flow along a streamline, the sum of pressure energy, kinetic energy, and potential energy per unit volume remains constant:

[latex]P + \frac{1}{2}\rho v^2 + \rho gh = \text{constant}[/latex]

Part (b): Lifting of Aeroplane During Takeoff (2 marks)

An aeroplane wing (airfoil) is designed with a curved upper surface and a flatter lower surface. When the plane moves forward:

• Air travels a longer path over the curved upper surface → higher speed
• Air travels a shorter path under the flat lower surface → lower speed

By Bernoulli’s principle: Higher speed → Lower pressure (upper surface). Lower speed → Higher pressure (lower surface).

This pressure difference creates a net upward force — the aerodynamic lift [latex](L = \Delta P \times A)[/latex]. When lift exceeds weight, the aeroplane rises. During takeoff, the pilot increases speed and angle of attack to maximize this pressure difference.

Part (c): Shower Head Speed Calculation (2 marks)

Given: Pipe radius R = 0.90 cm = 0.009 m, number of openings N = 30, radius of each opening r = 2 mm = 0.002 m, pipe speed v₁ = 4 m/s

By continuity equation (conservation of mass flow):

[latex]A_{pipe} \times v_1 = N \times A_{opening} \times v_2[/latex]

[latex]\pi R^2 \times v_1 = N \times \pi r^2 \times v_2[/latex]

[latex]v_2 = \frac{R^2 v_1}{N r^2} = \frac{(0.009)^2 \times 4}{30 \times (0.002)^2}[/latex]

[latex]= \frac{8.1 \times 10^{-5} \times 4}{30 \times 4 \times 10^{-6}} = \frac{3.24 \times 10^{-4}}{1.2 \times 10^{-4}} = \mathbf{2.7 \text{ m/s}}[/latex]

OR Q.13 — Angular Momentum and Rotational Kinetic Energy

Part (a): Angular Momentum and Relation with Moment of Inertia (2 marks)

Angular momentum (L) is the rotational analogue of linear momentum. It is defined as the product of moment of inertia and angular velocity:

[latex]\vec{L} = I\vec{\omega}[/latex]

Alternatively: [latex]L = mvr[/latex] for a particle of mass m moving with velocity v at perpendicular distance r from the axis.

Relation: [latex]L = I\omega[/latex], where I = moment of inertia (kg·m²) and ω = angular velocity (rad/s). SI unit of L: kg·m²·s⁻¹ (or J·s).

Torque is the rate of change of angular momentum: [latex]\tau = \frac{dL}{dt}[/latex]

Part (b): Kinetic Energy Gained After 20 Revolutions (3 marks)

Given: τ = 300 N·m, I = 150 kgm², θ = 20 revolutions = 20 × 2π = 40π rad

Angular acceleration:

[latex]\alpha = \frac{\tau}{I} = \frac{300}{150} = 2 \text{ rad/s}^2[/latex]

Starting from rest (ω₀ = 0), using rotational kinematics:

[latex]\omega^2 = \omega_0^2 + 2\alpha\theta = 0 + 2 \times 2 \times 40\pi = 160\pi[/latex]

Kinetic energy gained:

[latex]KE = \frac{1}{2}I\omega^2 = \frac{1}{2} \times 150 \times 160\pi = 75 \times 160\pi[/latex]

[latex]= 12000\pi = 12000 \times 3.1416 = \mathbf{37699 \text{ J} \approx 37.7 \text{ kJ}}[/latex]

Alternatively using work-energy theorem: [latex]KE = \tau \times \theta = 300 \times 40\pi = 12000\pi \approx 37699 \text{ J}[/latex] ✓

Q.14 — Waves, Stationary Waves, and Flute Frequency

What are waves? Wavelength-Frequency Relation (1 mark)
Waves are disturbances that transfer energy from one place to another without permanent displacement of the medium’s particles. The relation between wavelength (λ) and frequency (f):

[latex]v = f\lambda[/latex]

where v = wave speed, f = frequency (Hz), λ = wavelength (m).

Part (a): Formation of Stationary Waves — Musical Instrument Example (2 marks)

Stationary (standing) waves are formed by the superposition of two identical waves travelling in opposite directions in the same medium. When an incident wave and its reflected wave interfere, nodes (zero displacement) and antinodes (maximum displacement) form at fixed positions — the wave pattern appears “stationary.”

Mathematically: Two waves [latex]y_1 = A\sin(\omega t – kx)[/latex] and [latex]y_2 = A\sin(\omega t + kx)[/latex] superpose to give:

[latex]y = 2A\cos(kx)\sin(\omega t)[/latex]

Example in musical instrument: In a flute (open pipe), sound waves travel down the tube and reflect from the open ends. The incident and reflected waves superpose to form stationary waves. Antinodes form at both open ends, and the air column vibrates in harmonics.

Part (b): Frequency of 25 cm Open Flute in Fundamental Mode (2 marks)

Given: L = 25 cm = 0.25 m, v = 342 m/s, fundamental mode (1st harmonic)

For open pipe at fundamental: [latex]L = \frac{\lambda}{2} \Rightarrow \lambda = 2L = 0.50 \text{ m}[/latex]

[latex]f = \frac{v}{\lambda} = \frac{342}{0.50} = \mathbf{684 \text{ Hz}}[/latex]

Q.15 — Thermodynamics: Diesel Engine and Adiabatic Compression

Part (a): P-V Diagram of Diesel Engine (1 mark)

The diesel engine cycle (Diesel cycle) consists of:

• A→B: Adiabatic compression (steep curve, volume decreases sharply)
• B→C: Isobaric (constant pressure) heat addition — fuel burns at constant pressure
• C→D: Adiabatic expansion
• D→A: Isochoric (constant volume) heat rejection

Key difference from Otto cycle: Heat addition is at constant pressure (not constant volume).

Part (b): Net Work Done from P-V Diagram (1 mark)

The net work done by a heat engine in one complete cycle equals the area enclosed by the closed loop in the P-V diagram. Going clockwise around the loop: the area represents net positive work output. If the cycle is traced counterclockwise, the enclosed area represents net work input (refrigerator cycle).

[latex]W_{net} = \oint P,dV = \text{area enclosed by the P-V cycle}[/latex]

Part (c): Adiabatic Compression — Final Pressure and Temperature (3 marks)

Given: Compression ratio V₁/V₂ = 10, [latex]P_1 = 1.01 \times 10^5[/latex] Pa, [latex]T_1 = 27°C = 300[/latex] K, [latex]\gamma = 1.40[/latex]

Final Pressure:
For adiabatic process: [latex]P_1 V_1^\gamma = P_2 V_2^\gamma[/latex]

[latex]P_2 = P_1\left(\frac{V_1}{V_2}\right)^\gamma = 1.01 \times 10^5 \times (10)^{1.40}[/latex]

latex^{1.40} = 10^{1.4} = 25.12[/latex]

[latex]P_2 = 1.01 \times 10^5 \times 25.12 = \mathbf{2.537 \times 10^6 \text{ Pa} \approx 25.1 \text{ atm}}[/latex]

Final Temperature:
For adiabatic process: [latex]T_1 V_1^{\gamma-1} = T_2 V_2^{\gamma-1}[/latex]

[latex]T_2 = T_1\left(\frac{V_1}{V_2}\right)^{\gamma-1} = 300 \times (10)^{0.40}[/latex]

latex^{0.4} = 2.512[/latex]

[latex]T_2 = 300 \times 2.512 = \mathbf{753.6 \text{ K} = 480.6°C \approx 481°C}[/latex]

Q.16 — Kirchhoff’s Laws and Wheatstone Bridge (OR: Potentiometer and Shunt)

Part (a): Kirchhoff’s Laws (2 marks)

Kirchhoff’s Current Law (KCL) — Junction Rule:
The algebraic sum of all currents at any junction (node) in a circuit is zero:
[latex]\sum I = 0[/latex]
i.e., the sum of currents entering a junction = sum of currents leaving it. Based on conservation of charge.

Kirchhoff’s Voltage Law (KVL) — Loop Rule:
The algebraic sum of all potential differences (EMFs and voltage drops) around any closed loop in a circuit is zero:
[latex]\sum V = 0[/latex]
i.e., [latex]\sum \varepsilon = \sum IR[/latex]. Based on conservation of energy.

Part (b): Wheatstone Bridge Balance Condition — Galvanometer and Cell Interchanged (2 marks)

Balanced condition: [latex]\frac{P}{Q} = \frac{R}{S}[/latex] or [latex]PS = QR[/latex]

If galvanometer and cell positions are interchanged:
The balanced condition remains the same — [latex]PS = QR[/latex]. This is because the Wheatstone bridge balance is a reciprocal condition — it doesn’t matter which diagonal carries the galvanometer and which carries the cell. The balance point (null deflection) depends only on the ratio of resistances, not on the source-detector configuration.

This is a consequence of the reciprocity theorem in linear circuit theory.

Part (c): What is a Superconductor? (1 mark)
A superconductor is a material that exhibits zero electrical resistance below a critical temperature (transition temperature [latex]T_c[/latex]). Below [latex]T_c[/latex], electrical current flows without any energy loss. Additionally, superconductors expel magnetic fields (Meissner effect). Examples: Mercury below 4.2 K, YBCO below 93 K. Applications: MRI magnets, maglev trains, particle accelerators.

OR Q.16 — Potentiometer and Shunt

Part (a): Why Potentiometer is Superior to Voltmeter (2 marks)

A potentiometer is superior to a voltmeter for measuring EMF because:

1. No current drawn at balance: The potentiometer works on the null method — at balance, no current flows through the cell being measured. Therefore it measures the true open-circuit EMF. A voltmeter always draws current, measuring only terminal voltage (V = E − Ir), not true EMF.
2. Infinite input impedance at null: At the balance point, the potentiometer presents effectively infinite resistance to the cell — no loading effect. A voltmeter has finite resistance and always loads the circuit.
3. Higher accuracy and sensitivity: The null method eliminates errors from meter resistance and gives results repeatable to very high precision.

Part (b): What is Shunt? (1 mark)
A shunt is a low resistance connected in parallel with a galvanometer to convert it into an ammeter. Most of the current bypasses through the shunt, allowing only a small safe current through the galvanometer. This extends the measurable current range without damaging the galvanometer.

Part (c): Maximum Measurable Current from Ammeter (2 marks)

Given: Shunt [latex]S = 0.002[/latex] Ω, galvanometer resistance [latex]G = 5[/latex] Ω, maximum galvanometer current [latex]I_g = 2[/latex] mA = [latex]2 \times 10^{-3}[/latex] A

For parallel connection: [latex]I_g \times G = (I – I_g) \times S[/latex]

[latex]I – I_g = \frac{I_g \times G}{S} = \frac{2 \times 10^{-3} \times 5}{0.002} = \frac{10^{-2}}{2 \times 10^{-3}} = 5 \text{ A}[/latex]

[latex]I = I_g + 5 = 0.002 + 5 = \mathbf{5.002 \text{ A} \approx 5 \text{ A}}[/latex]

The ammeter can measure up to 5 A of current.

Q.17 — Lenz’s Law and Falling Magnet

Part (a): State Lenz’s Law (1 mark)
Lenz’s Law states that the direction of the induced EMF (and hence the induced current) in a circuit is always such that it opposes the change in magnetic flux that caused it. In other words, the induced effect always acts to oppose its own cause.

Mathematically: [latex]\varepsilon = -\frac{d\Phi}{dt}[/latex] (the negative sign represents Lenz’s law)

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

Lenz’s law is a consequence of the conservation of energy:

If the induced current had aided the change in flux (instead of opposing it), the process would be self-reinforcing — the induced current would increase the flux, which would increase the induced current, which would increase the flux further, without any energy input. This would violate conservation of energy by creating energy from nothing.

By opposing the cause, Lenz’s law ensures that mechanical work must be done against the opposing magnetic force to maintain the change in flux. This mechanical energy is converted to electrical energy — energy is conserved, not created.

Example: When you push a magnet into a coil, the induced current creates a force opposing the push. You must do work against this force — that work becomes the electrical energy in the induced current. ✓

Part (c): Acceleration of Falling Magnet Through Copper Ring (2 marks)

When a bar magnet falls through a horizontally held copper ring:

As the magnet approaches the ring, increasing flux induces a current in the ring. By Lenz’s law, this induced current creates a magnetic field that opposes the increasing flux — i.e., the ring acts as a magnet with its north pole facing the approaching north pole of the bar magnet, creating a repulsive force that slows the magnet’s fall.

As the magnet passes through and moves away, decreasing flux induces a current creating an attractive force — again opposing the change (slowing the exit).

At all times, the induced electromagnetic force opposes the motion. Therefore the magnet experiences:

• Gravity (downward) = mg
• Induced electromagnetic braking force (upward)

Net downward force < mg → acceleration a < g

The magnet falls with acceleration less than g (free fall acceleration). It reaches a terminal velocity when the electromagnetic braking force equals gravity. The magnet never falls in true free fall while passing through the ring — energy is continuously dissipated as heat in the copper ring (eddy current heating).

Q.18 — Quantum Nature of Light and Photoelectric Effect

Part (a): Which Law Deals with Particle Nature of Light? (1 mark)

Photoelectric law (Einstein’s Photoelectric Law) deals with the particle nature of light.

Statement: When light of frequency f falls on a metal surface, the maximum kinetic energy of the emitted photoelectrons is:

[latex]KE_{max} = hf – \phi[/latex]

where h = Planck’s constant, f = frequency of incident light, φ = work function of the metal. Each photon interacts with one electron — demonstrating that light behaves as particles (photons) of energy hf.

Part (b): Work Function and Stopping Potential (2 marks)

Work function (φ):
The minimum energy required to completely remove an electron from the surface of a metal is called its work function. It represents the energy barrier that binds surface electrons to the metal. It depends on the nature of the metal, not on the incident light. Unit: eV or joule. [latex]\phi = hf_0[/latex] where [latex]f_0[/latex] = threshold frequency.

Stopping potential (V₀):
The minimum negative potential applied to the anode (with respect to the emitter) that brings the photoelectric current to zero is called stopping potential. It measures the maximum kinetic energy of emitted photoelectrons:

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

Stopping potential depends on the frequency of incident light, not on its intensity.

Part (c): Work Function Calculation (2 marks)

Given: f = [latex]5 \times 10^{14}[/latex] Hz, [latex]KE_{max} = 2.3 \times 10^{-19}[/latex] J, [latex]h = 6.6 \times 10^{-34}[/latex] J·s

Energy of incident photon:

[latex]E = hf = 6.6 \times 10^{-34} \times 5 \times 10^{14} = 33 \times 10^{-20} = 3.3 \times 10^{-19} \text{ J}[/latex]

Work function:

[latex]\phi = hf – KE_{max} = 3.3 \times 10^{-19} – 2.3 \times 10^{-19} = \mathbf{1.0 \times 10^{-19} \text{ J}}[/latex]

In eV: [latex]\phi = \frac{1.0 \times 10^{-19}}{1.6 \times 10^{-19}} = \mathbf{0.625 \text{ eV}}[/latex]

Q.19 — P-N Junction Diode and Full-Wave Rectifier

Part (a): P-N Junction Diode (1 mark)
A P-N junction diode is a semiconductor device formed by joining a P-type semiconductor (with holes as majority carriers) and an N-type semiconductor (with electrons as majority carriers). At the junction, a depletion region forms due to diffusion of carriers, creating a potential barrier. The diode allows current to flow easily in one direction (forward bias) and blocks it in the other (reverse bias) — making it a one-way valve for current.

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

Forward bias V-I characteristics:
When P-side is connected to positive terminal of battery, depletion layer narrows. No significant current flows until the forward voltage exceeds the threshold voltage (~0.7V for Si, ~0.3V for Ge). Beyond threshold, current rises steeply and exponentially.

Reverse bias V-I characteristics:
When N-side is connected to positive terminal, depletion layer widens. Only a tiny reverse saturation current ([latex]I_0[/latex] ~μA) flows due to minority carriers. At very high reverse voltage, breakdown occurs and current increases sharply.

The V-I graph is asymmetric: steep exponential rise in forward direction, nearly zero (then breakdown) in reverse direction.

Part (c): Full-Wave Rectifier Using Junction Diodes (2 marks)

A bridge rectifier (full-wave) uses four P-N junction diodes arranged in a bridge. This converts both half-cycles of AC into pulsating DC across the load.

Working:

During positive half-cycle: Current flows through D₁ → Load → D₃ (D₂ and D₄ are reverse biased)

During negative half-cycle: Current flows through D₂ → Load → D₄ (D₁ and D₃ are reverse biased)

In both cases, current flows in the same direction through the load — producing full-wave rectified output. The output is pulsating DC with ripple frequency = 2f (twice the input frequency). A smoothing capacitor further reduces ripple.

Group C — Long Answer Analysis

Q.20 — Magnetic Field: Solenoid, Circular Coil, and Force Between Conductors

Part (a): Magnetic Field Inside Long Solenoid Using Ampere’s Theorem (3 marks)

Consider a long solenoid with n turns per unit length carrying current I.

Apply Ampere’s circuital law to a rectangular Amperian loop ABCD of length l:

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

Contributions from each side:

• AB (inside solenoid, parallel to B): [latex]B \cdot l[/latex]
• BC (perpendicular to B): [latex]0[/latex]
• CD (outside solenoid, B ≈ 0): [latex]0[/latex]
• DA (perpendicular to B): [latex]0[/latex]

Total: [latex]Bl = \mu_0 \times (nl) \times I[/latex]

(enclosed current = n turns per unit length × length l × current I)

[latex]\boxed{B = \mu_0 nI}[/latex]

For N total turns in length L: [latex]n = N/L[/latex] → [latex]B = \frac{\mu_0 NI}{L}[/latex]

This field is uniform inside the solenoid and directed along the axis.

Part (b): Magnetic Field at Centre of Circular Coil (3 marks)

Given: Wire length = 30 m, radius of coil r = 5 cm = 0.05 m, I = 2 A, [latex]\mu_0 = 4\pi \times 10^{-7}[/latex] H/m

Number of turns:
Circumference of one turn = [latex]2\pi r = 2\pi \times 0.05 = 0.1\pi = 0.3142[/latex] m

[latex]N = \frac{\text{total length}}{\text{circumference}} = \frac{30}{0.1\pi} = \frac{30}{0.3142} = 95.49 \approx \mathbf{95 \text{ turns}}[/latex]

Magnetic field at centre:

[latex]B = \frac{\mu_0 NI}{2r} = \frac{4\pi \times 10^{-7} \times 95.49 \times 2}{2 \times 0.05}[/latex]

[latex]= \frac{4\pi \times 10^{-7} \times 190.98}{0.10}[/latex]

[latex]= \frac{4 \times 3.1416 \times 10^{-7} \times 190.98}{0.10}[/latex]

[latex]= \frac{2.4 \times 10^{-3}}{0.10} = \mathbf{2.4 \times 10^{-2} \text{ T} = 24 \text{ mT}}[/latex]

(Using N = 30/0.1π more precisely: N = 300/π ≈ 95.5)

Part (c): Force Per Unit Length Between Two Parallel Conductors (2 marks)

Consider two long parallel conductors separated by distance d, carrying currents [latex]I_1[/latex] and [latex]I_2[/latex].

Magnetic field due to conductor 1 at the location of conductor 2:

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

Force on length L of conductor 2 in this field:

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

Force per unit length:

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

Nature: If currents are in the same direction → force is attractive. If opposite directions → force is repulsive.

This formula defines the SI unit of current: 1 Ampere = current that produces force of [latex]2 \times 10^{-7}[/latex] N/m between two parallel wires 1 m apart.

OR Q.20 — LCR Circuit: Impedance and AC vs DC Safety

Part (a): Impedance and Phase Angle of LCR Circuit (3 marks)

In a series LCR circuit with AC voltage [latex]V = V_0\sin\omega t[/latex]:

Voltages: [latex]V_R = IR[/latex] (in phase), [latex]V_L = IX_L[/latex] (leads by 90°), [latex]V_C = IX_C[/latex] (lags by 90°)

Using phasor addition:

[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 = \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]

• If [latex]X_L > X_C[/latex]: circuit is inductive, current lags voltage (φ > 0)
• If [latex]X_L < X_C[/latex]: circuit is capacitive, current leads voltage (φ < 0)
• At resonance [latex]X_L = X_C[/latex]: Z = R (minimum), [latex]\phi = 0[/latex], maximum current

Part (b): Why 220 V AC is More Dangerous Than 220 V DC (2 marks)

220 V AC has a peak voltage of [latex]V_0 = V_{rms}\sqrt{2} = 220\sqrt{2} \approx 311[/latex] V.

The AC voltage oscillates between +311V and −311V. Even though the RMS value is 220V (equivalent heating to 220V DC), the peak voltage is 311V — much higher than 220V DC.

This higher peak voltage can:

• Cause more severe electric shock (muscles contract more violently at peak)
• More easily break down insulation
• Create dangerous capacitive/inductive effects

Additionally, AC frequency (50 Hz) is close to the resonant frequency of the human heart’s nerve impulses, making it more likely to cause ventricular fibrillation — the deadliest cardiac response to electric shock. DC of the same voltage tends to cause muscle contraction (which can be released), whereas AC causes tetanic (sustained) muscle contraction, preventing the victim from letting go.

Part (c): LCR Circuit — Current and Voltage Across Coil (3 marks)

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

Total resistance: R = 50 + 950 = 1000 Ω

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

Impedance: [latex]Z = \sqrt{R^2 + X_L^2} = \sqrt{1000^2 + 628.3^2} = \sqrt{10^6 + 3.948 \times 10^5} = \sqrt{1.3948 \times 10^6} = 1181.0[/latex] Ω

(i) Current: [latex]I = \frac{V}{Z} = \frac{220}{1181.0} = \mathbf{0.186 \text{ A}}[/latex]

(ii) Voltage across the coil (coil has both resistance [latex]R_c[/latex] = 50Ω and inductance [latex]X_L[/latex] = 628.3Ω):

[latex]Z_{coil} = \sqrt{R_c^2 + X_L^2} = \sqrt{50^2 + 628.3^2} = \sqrt{2500 + 394750} = \sqrt{397250} = 630.3 \text{ Ω}[/latex]

[latex]V_{coil} = I \times Z_{coil} = 0.186 \times 630.3 = \mathbf{117.2 \text{ V} \approx 117 \text{ V}}[/latex]

Q.21 — Wave Optics: Interference and Diffraction Grating

Part (a): Wavefront and Conditions for Sustainable Interference (2 marks)

Wavefront: A wavefront is the locus of all points in a wave that are in the same phase of vibration at any given instant. It is a surface of constant phase. Types: spherical (from point source), cylindrical (from line source), plane (from distant source).

Two conditions for sustainable (stable) interference:

1. Coherence: The two sources must be coherent — they must emit waves of the same frequency with a constant (or zero) phase difference. Incoherent sources produce randomly varying phase differences, washing out the interference pattern.
2. Nearly equal amplitudes: The amplitudes of the interfering waves should be nearly equal for a high contrast interference pattern (dark fringes approach zero intensity). If amplitudes are very different, minima are not dark and the pattern is indistinct.

Additional condition: The two sources must emit light of the same wavelength (monochromatic or narrow bandwidth).

Part (b): Young’s Double Slit Experiment — Fringe Width (3 marks)

In Young’s double slit experiment, two coherent slits [latex]S_1[/latex] and [latex]S_2[/latex] are separated by distance d. A screen is placed at distance D.

Path difference at point P (distance y from centre):
[latex]\Delta = \frac{yd}{D}[/latex] (for small angles)

Bright fringes (constructive interference): [latex]\Delta = n\lambda[/latex] → [latex]y_n = \frac{n\lambda D}{d}[/latex]

Dark fringes (destructive): [latex]\Delta = (2n-1)\frac{\lambda}{2}[/latex] → [latex]y_n = \frac{(2n-1)\lambda D}{2d}[/latex]

Fringe width (β) = distance between consecutive bright or dark fringes:

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

Fringe width is constant (uniform spacing), independent of order n, and increases with larger D, larger λ, and smaller d.

Part (c): Diffraction Grating — Grating Spacing and First-Order Angle (3 marks)

Given: 400 lines/mm, λ = 600 nm = [latex]600 \times 10^{-9}[/latex] m

(i) Grating spacing:

[latex]d = \frac{1}{400 \text{ lines/mm}} = \frac{1}{400 \times 10^3 \text{ lines/m}} = 2.5 \times 10^{-6} \text{ m} = \mathbf{2.5 \times 10^{-3} \text{ mm} = 2500 \text{ nm}}[/latex]

(ii) Angle for first-order maximum:
Grating equation: [latex]d\sin\theta = n\lambda[/latex]

For n = 1: [latex]\sin\theta = \frac{\lambda}{d} = \frac{600 \times 10^{-9}}{2.5 \times 10^{-6}} = \frac{600}{2500} = 0.24[/latex]

[latex]\theta = \sin^{-1}(0.24) = \mathbf{13.88° \approx 13.9°}[/latex]

Q.22 — Atomic Physics: Bohr’s Postulates, de Broglie, and Electron in Magnetic Field

Part (a): Bohr’s Postulates (2 marks)

Postulate 1: Electrons revolve around the nucleus in discrete circular orbits (stationary states) without emitting radiation.

Postulate 2: The angular momentum of the electron in a stationary orbit is quantized: [latex]mvr = \frac{nh}{2\pi}[/latex] (n = 1, 2, 3, …)

Postulate 3: An electron emits or absorbs a photon only when it transitions between two allowed orbits. The energy of the photon equals the difference in energy between the orbits:

[latex]hf = E_2 – E_1[/latex]

Part (b): Matter Wave and de Broglie Wavelength (2 marks)

Matter wave: According to de Broglie’s hypothesis, every moving particle has wave-like properties. These are called matter waves or de Broglie waves. The wave nature of particles was confirmed by Davisson-Germer electron diffraction experiment.

de Broglie wavelength expression:
From Einstein’s photon energy: [latex]E = pc = hf[/latex], momentum [latex]p = h/\lambda[/latex]

For a particle of mass m and velocity v:

[latex]\boxed{\lambda = \frac{h}{mv} = \frac{h}{p}}[/latex]

where h = Planck’s constant, m = mass, v = velocity, p = momentum.

For kinetic energy KE: [latex]\lambda = \frac{h}{\sqrt{2mKE}}[/latex]

Part (c): Electron in Circular Path — Speed and Potential Difference (3 marks)

Given: r = 20 cm = 0.20 m, B = [latex]2 \times 10^{-3}[/latex] T, e = [latex]1.6 \times 10^{-19}[/latex] C, [latex]m_e = 9.1 \times 10^{-31}[/latex] kg

Speed of electron:
For circular motion in magnetic field: [latex]evB = \frac{m_e v^2}{r}[/latex]

[latex]v = \frac{eBr}{m_e} = \frac{1.6 \times 10^{-19} \times 2 \times 10^{-3} \times 0.20}{9.1 \times 10^{-31}}[/latex]

[latex]= \frac{6.4 \times 10^{-23}}{9.1 \times 10^{-31}} = 7.03 \times 10^7 \text{ m/s}[/latex]

[latex]\mathbf{v \approx 7.03 \times 10^7 \text{ m/s}}[/latex]

Potential difference to accelerate electron to this speed:

[latex]eV = \frac{1}{2}m_e v^2[/latex]

[latex]V = \frac{m_e v^2}{2e} = \frac{9.1 \times 10^{-31} \times (7.03 \times 10^7)^2}{2 \times 1.6 \times 10^{-19}}[/latex]

[latex]= \frac{9.1 \times 10^{-31} \times 4.942 \times 10^{15}}{3.2 \times 10^{-19}}[/latex]

[latex]= \frac{4.497 \times 10^{-15}}{3.2 \times 10^{-19}} = \mathbf{14054 \text{ V} \approx 14.1 \text{ kV}}[/latex]

Part (d): Path of Neutron in Magnetic Field (1 mark)

A neutron is electrically neutral (charge = 0). The magnetic force on a moving charged particle is [latex]F = qvB\sin\theta[/latex]. Since q = 0 for a neutron, F = 0 — no magnetic force acts on it.

Therefore, a neutron entering a magnetic field follows a straight line path — completely unaffected by the magnetic field, regardless of its speed or direction of entry.

(Note: A neutron has a magnetic moment and experiences a much weaker force from a non-uniform magnetic field, but for a uniform magnetic field, the path is straight.)

OR Q.22 — Radioactivity

Part (a): Radioactivity and Its Units (2 marks)

Radioactivity: Radioactivity is the spontaneous disintegration (decay) of the nucleus of a heavy, unstable atom with the emission of radiation — alpha (α), beta (β), or gamma (γ) rays. It is a purely nuclear phenomenon, unaffected by temperature, pressure, or chemical state.

Units of radioactivity (activity):

• Becquerel (Bq): SI unit. 1 Bq = 1 disintegration per second
• Curie (Ci): Older unit. [latex]1 \text{ Ci} = 3.7 \times 10^{10}[/latex] disintegrations per second = [latex]3.7 \times 10^{10}[/latex] Bq
• Rutherford (Rd): [latex]1 \text{ Rd} = 10^6[/latex] disintegrations per second

Part (b): Laws of Radioactive Disintegration and Decay Equation (3 marks)

Radioactive Decay Law:
The rate of disintegration of a radioactive substance at any instant is directly proportional to the number of undecayed atoms present at that instant:

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

where λ = decay constant (probability of decay per nucleus per unit time, unit: s⁻¹).

Derivation of decay equation:

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

Integrating both sides:

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

[latex]\ln N – \ln N_0 = -\lambda t[/latex]

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

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

This exponential decay law shows the number of undecayed nuclei decreases exponentially with time.

Half-life: [latex]t_{1/2} = \frac{\ln 2}{\lambda} = \frac{0.693}{\lambda}[/latex]

Part (c): Scheme of Beta Decay (1 mark)

In beta (β⁻) decay, a neutron in the nucleus converts to a proton, emitting an electron (β particle) and an antineutrino:

[latex]n \rightarrow p + e^- + \bar{\nu}_e[/latex]

General nuclear equation:

[latex]^A_Z X \rightarrow ^A_{Z+1}Y + ^0_{-1}e + \bar{\nu}_e[/latex]

Mass number A unchanged, atomic number Z increases by 1.

Part (d)(i): Half-Life from Graph (1 mark)

From the radioactive decay graph (N vs time in minutes):

Initial number N₀ = 100 (at t = 0). Half of initial = 50 atoms.

Reading from graph: N = 50 at approximately t = 4 minutes.

Half-life ≈ 4 minutes

Part (d)(ii): Decay Constant (1 mark)

[latex]\lambda = \frac{0.693}{t_{1/2}} = \frac{0.693}{4 \text{ min}} = 0.173 \text{ min}^{-1}[/latex]

Converting: [latex]\lambda = \frac{0.173}{60} = \mathbf{2.89 \times 10^{-3} \text{ s}^{-1}}[/latex]

Marks Weightage Summary

Difficulty Analysis

Easy — Reliable marks:

• Q.2 (Viscosity decrease — recall)
• Q.4 (Isothermal PV = constant)
• Q.10 (Diamagnetic susceptibility)
• Q.12(a) (SHM definition)
• Q.13(a) (Bernoulli’s principle statement)
• Q.14 (Waves + wavelength-frequency relation)
• Q.16(a) (Kirchhoff’s laws)
• Q.16(c) (Superconductor definition)
• Q.17(a) (Lenz’s law statement)
• Q.19(a) (P-N junction definition)
• Q.22(a) (Bohr’s postulates)
• Q.22(d) (Neutron path in B-field)

Moderate — Needs regular practice:

• Q.1 (Rotational KE calculation)
• Q.5 (Carnot engine work)
• Q.8 (Inversion temperature)
• Q.9 (Meter bridge ratio)
• Q.12(b)(c) (Spring SHM proof + pendulum length)
• Q.14(b) (Open flute frequency)
• Q.15(b) (PV diagram work explanation)
• Q.18(a)(b) (Photoelectric law + definitions)
• Q.19(b)(c) (Diode characteristics + full-wave rectifier)
• Q.21(a) (Wavefront + interference conditions)

Challenging — For top scorers:

• Q.3 (SHM graph reading — time period)
• Q.6 (Sound velocity with doubled T and halved P)
• Q.7 (Pipe length and wavelength — closed pipe)
• Q.11 (Gravitational waves — modern physics)
• Q.13(c) (Shower head continuity equation — 30 openings)
• Q.15(c) (Adiabatic compression — two unknowns)
• Q.16(c) shunt numerical
• Q.17(b)(c) (Lenz’s law + energy conservation + falling magnet)
• Q.20 (Solenoid + circular coil + force formula)
• Q.21(b)(c) (YDSE derivation + diffraction grating numerical)
• Q.22(c) (Electron speed + potential difference)

Student Preparation Tips

1. The Shower Head Problem (Q.13c) — Understand Continuity
This 2-mark problem requires applying the continuity equation across a branching pipe system: [latex]A_{pipe}v_1 = N \times A_{opening}v_2[/latex]. The key insight is that 30 openings each carry 1/30 of the total flow. Always compute the total exit area (N × π r²) and use [latex]v_2 = \frac{A_{pipe}}{N \times A_{opening}} \times v_1[/latex].

2. Adiabatic Compression Has Two Separate Formulas
For Q.15(c): final pressure uses [latex]P_2 = P_1(V_1/V_2)^\gamma[/latex] and final temperature uses [latex]T_2 = T_1(V_1/V_2)^{\gamma-1}[/latex]. These are different formulas — don’t confuse them. Practice both separately.

3. Gravitational Waves (Q.11) — One Easy Mark
Gravitational waves are ripples in spacetime curvature caused by accelerating masses. First predicted by Einstein (1915), first detected by LIGO (2015) from merging black holes. This is directly answer (b) — don’t overthink it.

4. Diffraction Grating Numerical is Completely Predictable
For Q.21(c): (i) grating spacing d = 1/N (where N is lines per meter), (ii) first order angle from [latex]d\sin\theta = \lambda[/latex]. The only trick is unit conversion — lines/mm to lines/m. Practice 3–4 such problems and this becomes automatic.

5. de Broglie + Electron in B-field (Q.22b,c) — Two Standard Results
de Broglie wavelength: [latex]\lambda = h/mv[/latex]. For electron in circular orbit in B-field: [latex]v = eBr/m_e[/latex]. Then find KE and hence accelerating voltage from [latex]eV = \frac{1}{2}m_ev^2[/latex]. These are standard sequences — practice the full chain once.

6. Kirchhoff’s Laws Must Be Stated Precisely
Q.16(a) gives 2 marks for stating both KCL (sum of currents at junction = 0) and KVL (sum of EMFs = sum of IR drops around loop). State the physical basis (charge conservation for KCL, energy conservation for KVL) for full marks.

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

Frequently Asked Questions

Q: In Q.6, why doesn’t halving the pressure change the speed of sound?
A: Speed of sound [latex]v = \sqrt{\gamma P/\rho}[/latex]. When pressure halves at constant temperature, density also halves (ideal gas: P/ρ = RT/M = constant at constant T). So P/ρ is unchanged. Speed only depends on temperature: [latex]v \propto \sqrt{T}[/latex].

Q: For Q.7 (pipe length = 3λ/4), how do we identify it from the figure?
A: The figure shows a closed organ pipe with crossing-line nodes. Count the pattern: node at closed end, antinode at open end. If there are two nodes and two antinodes, it’s the first overtone (3rd harmonic) of a closed pipe, giving l = 3λ/4. Alternatively: 3/4 of a wavelength fills the pipe → l = 3λ/4.

Q: The falling magnet problem (Q.17c) — does the magnet reach terminal velocity?
A: Yes. As the magnet accelerates, the induced EMF and hence the braking force increase (braking force ∝ v). Eventually braking force equals gravitational force and the magnet falls at constant terminal velocity. For a short copper ring this happens quickly. For a long solenoid, the magnet can visibly slow to near-terminal velocity.

Q: In Q.22(b), why is de Broglie wavelength important for Bohr’s model?
A: de Broglie provided the physical justification for Bohr’s angular momentum quantization rule. For standing wave condition of electron waves around the orbit: 2πr = nλ = n(h/mv) → mvr = nh/2π — exactly Bohr’s quantization condition. de Broglie waves explain why only certain orbits are allowed.

Q: For Q.21(c) diffraction grating — what’s the maximum order visible?
A: Maximum order when [latex]\sin\theta = 1[/latex]: [latex]n_{max} = d/\lambda = 2500/600 = 4.17[/latex]. So maximum 4 orders are visible. This is a common follow-up that HISSAN/NEB sometimes ask.

Where to Find the Full Question Paper

The complete original question paper with all figures — including the SHM displacement-time graph (Q.3), the closed pipe resonance figure (Q.7), the meter bridge circuit (Q.9), the shower head diagram (Q.13c), the diesel P-V diagram (Q.15), the Wheatstone bridge (Q.16), the falling magnet through ring (Q.17), and the radioactive decay graph (Q.22 OR) — is available as a PDF from above.

Studying D1 and D2 Together — The Complete Coverage Strategy

A student who masters both papers covers the complete Grade XII Physics new curriculum for NEB.

Conclusion

The HISSAN Central Examination 2082 Set D1 Physics paper is a thoughtfully constructed, curriculum-aligned pre-board examination that tests Grade XII students at exactly the right depth and breadth. Its inclusion of gravitational waves (Q.11), the shower head fluid dynamics problem (Q.13c), the diesel engine cycle (Q.15), the falling magnet through copper ring (Q.17c), and the neutron path in magnetic field (Q.22d) demonstrates that modern HISSAN papers go beyond rote testing — they reward students who truly understand physics.

Studied alongside Set D2, these two papers provide the most comprehensive pre-board preparation available. The topics they cover together form a near-complete picture of the Grade XII Physics new curriculum, with the same question styles, difficulty levels, and marking patterns that students will encounter in the actual NEB board examination.

The most important lesson from both papers: Physics rewards understanding over memorization. The student who understands why the copper ring slows the magnet, why 220V AC is more dangerous than 220V DC, why the shower water speeds up despite 30 openings, and why the neutron travels straight through a magnetic field — that student will answer every question confidently, regardless of which specific numerical or diagram appears on exam day.

Visit nneducation.com for more HISSAN papers (D1 and D2), NEB past papers, model questions, complete derivation guides, and step-by-step numerical solutions for Grade XI and XII Physics, Chemistry, Mathematics, and Nepali.