NEB Grade XII Physics 2082 (2025) — Set H Complete Guide with Question Paper, Analysis & Preparation Tips

Subject: Physics | Code: 1021 ‘H’ | Year: 2082 B.S. (2025 A.D.) | Full Marks: 75 | Time: 3 Hours

Introduction

The NEB Grade XII Physics 2082 B.S. (2025 A.D.) examination, Set H, bearing subject code 1021 ‘H’, is a rigorous test of conceptual understanding, mathematical ability, and applied physics across the complete Grade XII New Course syllabus. Physics is one of the most marks-intensive and conceptually demanding subjects in the NEB +2 curriculum, and performing well requires both deep understanding and systematic examination technique.

This paper, like all NEB Physics papers, is structured into three groups. Group A presents 11 multiple choice questions worth 1 mark each, testing rapid conceptual recognition across a broad range of topics. Group B contains 8 short answer questions at 5 marks each, with OR options for most questions, combining theory, derivation, and numerical calculation. Group C has 3 long answer questions at 8 marks each, demanding extended derivations, multi-part numerical problems, and diagram-based analysis.

The 2082 Set H paper covers an impressive breadth of the Grade XII Physics curriculum: rotational mechanics and SHM, fluid mechanics and Bernoulli’s equation, thermodynamics and Carnot engines, wave motion and the Doppler effect, DC circuits and Kirchhoff’s laws, electromagnetic induction and inductors, electronics (NAND gates and rectifiers), atomic physics (Thomson’s experiment, Millikan’s oil drop, photoelectric effect), nuclear physics (radioactive decay), stationary waves and diffraction, LCR circuits, and the hydrogen atom model. This breadth is typical of NEB Physics, which demands that students study every chapter thoroughly.

For NEB Physics students across Nepal — whether sitting for the first time or revisiting for grade improvement — this article provides a complete breakdown of the paper, worked solutions to every numerical, answers to all MCQs with reasoning, and targeted preparation strategies based on what this paper reveals about NEB’s examining priorities.

Exam Overview

Full Question Paper

Question Pattern Analysis

Strategic note: Group B carries the most total marks (40) and offers OR options for most questions — this is where exam strategy matters most. Students who know their strong chapters should select OR options wisely. Group C’s 8-mark questions each combine a derivation sub-part, a numerical sub-part, and a conceptual sub-part. Time management is critical: allocate roughly 15 minutes to Group A, 60 minutes to Group B, and 45 minutes to Group C.

Chapter-wise Weightage Analysis

Key insight: Waves, Electromagnetic Induction, and Atomic/Nuclear Physics together account for roughly 42 marks — more than half the paper. Students must prioritize these chapters without neglecting the others.

Group A — MCQ Solutions with Explanations

Q1. Difference in moment of inertia of a rod about its end vs. centre

• Moment of inertia about centre: [latex]I_{centre} = \frac{ML^2}{12}[/latex]
• Moment of inertia about one end: [latex]I_{end} = \frac{ML^2}{3}[/latex]
• Difference: [latex]I_{end} – I_{centre} = \frac{ML^2}{3} – \frac{ML^2}{12} = \frac{4ML^2 – ML^2}{12} = \frac{3ML^2}{12} = \frac{ML^2}{4}[/latex]

Answer: (B) [latex]\frac{ML^2}{4}[/latex]

Q2. Velocity of oscillating object at mean position (from graph)

From the displacement-time graph: amplitude [latex]A = 0.20[/latex] m, time period [latex]T = 2.0[/latex] s.

Angular frequency: [latex]\omega = \frac{2\pi}{T} = \frac{2\pi}{2} = \pi \text{ rad/s}[/latex]

Maximum velocity (at mean position): [latex]v_{max} = A\omega = 0.20 \times \pi = 0.2\pi \text{ m/s}[/latex]

Answer: (A) [latex]0.2\pi[/latex] m/s

Q3. Terminal velocity of combined raindrop

When two identical drops combine, volume is conserved:

[latex]\frac{4}{3}\pi R’^3 = 2 \times \frac{4}{3}\pi r^3 \Rightarrow R’ = 2^{1/3} r[/latex]

Terminal velocity is proportional to [latex]r^2[/latex]:

[latex]\frac{v’}{v} = \left(\frac{R’}{r}\right)^2 = (2^{1/3})^2 = 2^{2/3} \approx 1.587[/latex]

[latex]v’ = 0.15 \times 2^{2/3} \approx 0.15 \times 1.587 \approx 0.238 \approx 0.24 \text{ m/s}[/latex]

Answer: (B) 0.24 m/s

Q4. Adiabatic process

In an adiabatic process, no heat exchange occurs between the system and surroundings — the system is thermally insulated.

Answer: (D) The system is insulated from the surroundings

Q5. Efficiency of Carnot engine between steam point and ice point

Steam point [latex]T_H = 373[/latex] K, Ice point [latex]T_C = 273[/latex] K

[latex]\eta = 1 – \frac{T_C}{T_H} = 1 – \frac{273}{373} = \frac{100}{373} \approx 0.268 = 26.8%[/latex]

Answer: (A) 26.8%

Q6. Example of transverse wave

In transverse waves, particles oscillate perpendicular to the direction of propagation. Sound in air and waves in the tuning fork stem are longitudinal. Waves inside water are surface waves (partially transverse). Waves in a stretched string are purely transverse.

Answer: (D) Wave produced in the stretched string

Q7. Phenomenon not explained by Huygens’ wave construction

Huygens’ principle explains reflection, refraction, and diffraction through wavefront construction. However, it cannot explain polarization, because Huygens originally proposed a longitudinal wave model — polarization is a transverse wave phenomenon that requires additional treatment beyond the basic Huygens construction.

Answer: (C) Polarization

Q8. Magnetic field at centre of circular coil

The standard formula for magnetic field at the centre of a circular coil of radius [latex]R[/latex] carrying current [latex]I[/latex] is:

[latex]B = \frac{\mu_0 I}{2R}[/latex]

Answer: (B) [latex]\frac{\mu_0 I}{2R}[/latex]

Q9. Relative permeability in terms of magnetic susceptibility

By definition: [latex]\mu_r = 1 + \chi[/latex]

Answer: (A) [latex]\chi + 1[/latex]

Note: Option (D) lists [latex]1 + \chi^2[/latex] which is incorrect. The correct relationship is linear: [latex]\mu_r = 1 + \chi[/latex].

Q10. Magnetic flux through coil

[latex]\Phi = BA\cos\theta[/latex]

The coil is perpendicular to the field, so [latex]\theta = 0°[/latex], [latex]\cos 0° = 1[/latex].

[latex]\Phi = 10^3 \times 10^{-2} \times 1 = 10 \text{ Weber}[/latex]

Answer: (B) 10 Weber

Q11. Point where seismic waves originate

The hypocenter (also called the focus) is the underground point where an earthquake originates. The epicenter is the point on the Earth’s surface directly above the hypocenter.

Answer: (C) Hypocenter

Group B — Short Answer Solutions

Q12 (Main): Rotational Mechanics

a) Why does a ballet dancer stretch and fold her arms?

A ballet dancer uses the principle of conservation of angular momentum. When she stretches her arms outward, her moment of inertia [latex]I[/latex] increases, so her angular velocity [latex]\omega[/latex] decreases (she spins slowly). When she folds her arms inward, [latex]I[/latex] decreases, causing [latex]\omega[/latex] to increase (she spins faster). Since no external torque acts, [latex]L = I\omega = \text{constant}[/latex] throughout her spin.

b) Relation between torque and moment of inertia

Consider a rigid body rotating about a fixed axis. Let a tangential force [latex]F[/latex] act on a particle of mass [latex]m[/latex] at distance [latex]r[/latex] from the axis.

By Newton’s second law: [latex]F = ma_t = m r\alpha[/latex]

Torque on this particle: [latex]\tau = Fr = mr^2\alpha[/latex]

For the entire rigid body, summing over all particles:

[latex]\tau_{total} = \left(\sum m_i r_i^2\right)\alpha = I\alpha[/latex]

Therefore: [latex]\tau = I\alpha[/latex] where [latex]I[/latex] is the moment of inertia and [latex]\alpha[/latex] is angular acceleration.

c) Why prefer a longer arm wrench?

Torque [latex]\tau = F \times d[/latex], where [latex]d[/latex] is the length of the arm. For the same applied force [latex]F[/latex], a longer arm produces greater torque, making it easier to turn a bolt or nut. Less effort is needed to achieve the same rotational effect.

Q12 (OR): Simple Harmonic Motion

a) Total energy in SHM is constant

For a particle of mass [latex]m[/latex] executing SHM with amplitude [latex]A[/latex] and angular frequency [latex]\omega[/latex]:

• Displacement: [latex]x = A\sin(\omega t)[/latex]
• Velocity: [latex]v = A\omega\cos(\omega t)[/latex]

Kinetic energy:

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

Potential energy:

[latex]PE = \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 – x^2) + \frac{1}{2}m\omega^2 x^2 = \frac{1}{2}m\omega^2 A^2[/latex]

Since [latex]m[/latex], [latex]\omega[/latex], and [latex]A[/latex] are all constants, total energy is constant and independent of position.

b) Potential energy vs. displacement graph: PE is a parabola symmetric about [latex]x = 0[/latex], with minimum PE = 0 at [latex]x = 0[/latex] and maximum PE = [latex]\frac{1}{2}m\omega^2 A^2[/latex] at [latex]x = \pm A[/latex].

c) Time period on Moon’s surface

[latex]T = 2\pi\sqrt{\frac{L}{g}}[/latex]

On the Moon, [latex]g_{moon} = \frac{g}{6}[/latex], so:

[latex]T_{moon} = 2\pi\sqrt{\frac{L}{g/6}} = 2\pi\sqrt{\frac{6L}{g}} = \sqrt{6} \cdot T \approx 2.45T[/latex]

The time period on the Moon is [latex]\sqrt{6}[/latex] times the time period on Earth.

Q13: Fluid Mechanics

a) Laminar and turbulent flow

Laminar flow: A smooth, orderly flow in which fluid particles move in parallel layers (streamlines) without mixing. Each layer slides smoothly past adjacent layers. Occurs at low velocities.

Turbulent flow: An irregular, chaotic flow in which fluid particles move in disorderly paths with mixing between layers. Occurs at high velocities when the Reynolds number exceeds approximately 4000.

b) Pressure at reduced cross section — Bernoulli’s equation

Given:

• [latex]P_1 = 4 \times 10^4 \text{ N/m}^2[/latex], [latex]v_1 = 2 \text{ m/s}[/latex], [latex]A_1 = 200 \text{ cm}^2 = 200 \times 10^{-4} \text{ m}^2[/latex]
• [latex]A_2 = 50 \text{ cm}^2 = 50 \times 10^{-4} \text{ m}^2[/latex], [latex]\rho = 1000 \text{ kg/m}^3[/latex]
• Horizontal pipe (no height difference)

By continuity equation: [latex]A_1 v_1 = A_2 v_2[/latex]

[latex]v_2 = \frac{A_1 v_1}{A_2} = \frac{200 \times 2}{50} = 8 \text{ m/s}[/latex]

By Bernoulli’s equation:

[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]

[latex]P_2 = 4 \times 10^4 + \frac{1}{2} \times 1000 \times (4 – 64)[/latex]

[latex]P_2 = 40000 + 500 \times (-60) = 40000 – 30000 = 10000 \text{ N/m}^2[/latex]

[latex]P_2 = 1 \times 10^4 \text{ N/m}^2[/latex]

Q14: Thermodynamics

a) Temperature increase to raise Carnot efficiency from 25% to 50%

Given: Sink temperature [latex]T_C = 9°C = 282 \text{ K}[/latex]

At 25% efficiency:

[latex]0.25 = 1 – \frac{282}{T_{H1}} \Rightarrow \frac{282}{T_{H1}} = 0.75 \Rightarrow T_{H1} = \frac{282}{0.75} = 376 \text{ K}[/latex]

At 50% efficiency:

[latex]0.50 = 1 – \frac{282}{T_{H2}} \Rightarrow \frac{282}{T_{H2}} = 0.50 \Rightarrow T_{H2} = \frac{282}{0.50} = 564 \text{ K}[/latex]

Increase in temperature:

[latex]\Delta T = T_{H2} – T_{H1} = 564 – 376 = 188 \text{ K} = 188°C[/latex]

The source temperature must be increased by 188°C (or 188 K).

b) First law of thermodynamics and conservation of energy

Statement: The heat supplied to a system equals the sum of the increase in internal energy of the system and the work done by the system:

[latex]Q = \Delta U + W[/latex]

Yes, it follows the principle of conservation of energy. The first law is essentially an expression of energy conservation: heat energy added to the system is not destroyed but is accounted for — partly stored as internal energy and partly converted to mechanical work. No energy is created or lost; it only changes form.

Q15: Waves and Doppler Effect

a) Effect of temperature and pressure on velocity of sound in a gas

Effect of temperature: [latex]v = \sqrt{\frac{\gamma RT}{M}}[/latex]

Velocity of sound is directly proportional to the square root of absolute temperature: [latex]v \propto \sqrt{T}[/latex]. As temperature increases, velocity increases. At 0°C, velocity in air is about 331 m/s; it increases by approximately 0.61 m/s for every 1°C rise.

Effect of pressure: At constant temperature, pressure changes do not affect the velocity of sound in a gas. This is because when pressure increases, density also increases proportionally, and their ratio ([latex]P/\rho[/latex]) remains constant. Since [latex]v = \sqrt{\frac{\gamma P}{\rho}}[/latex], the velocity remains unchanged.

b) Frequency of echo heard by train driver — Doppler Effect

The train (source and observer) approaches a cliff (reflector) at [latex]v_s = 10 \text{ m/s}[/latex].

Step 1 — Frequency received by cliff (stationary observer, moving source):

[latex]f_{cliff} = f_0 \cdot \frac{v}{v – v_s} = 600 \times \frac{340}{340 – 10} = 600 \times \frac{340}{330} = \frac{204000}{330} \approx 618.18 \text{ Hz}[/latex]

Step 2 — The cliff now acts as a source of frequency 618.18 Hz; the driver (moving observer) approaches it:

[latex]f_{echo} = f_{cliff} \cdot \frac{v + v_o}{v} = 618.18 \times \frac{340 + 10}{340} = 618.18 \times \frac{350}{340}[/latex]

[latex]f_{echo} = 618.18 \times 1.02941 \approx 636.4 \text{ Hz}[/latex]

Alternatively, using the combined formula directly:

[latex]f_{echo} = f_0 \cdot \frac{v + v_s}{v – v_s} = 600 \times \frac{340 + 10}{340 – 10} = 600 \times \frac{350}{330} = 600 \times 1.0606 \approx \mathbf{636.4 \text{ Hz}}[/latex]

The frequency of echo as heard by the driver is approximately 636.4 Hz.

Q16: Kirchhoff’s Laws and Wheatstone Bridge

a) Kirchhoff’s Laws

Junction Law (KCL): The algebraic sum of all currents at any junction in a circuit is zero. In other words, total current flowing into a junction equals total current flowing out: [latex]\sum I = 0[/latex]

Loop Law (KVL): The algebraic sum of all potential differences (emf sources and resistor voltage drops) around any closed loop in a circuit is zero: [latex]\sum V = 0[/latex]

b) Balanced condition of Wheatstone bridge

In a Wheatstone bridge, four resistors P, Q, R, and S are connected in a diamond (bridge) configuration with a battery across one diagonal and a galvanometer across the other.

Applying KCL: Let current [latex]I_1[/latex] flow through P and Q (upper arm), and [latex]I_2[/latex] through R and S (lower arm). At balance, galvanometer current [latex]I_g = 0[/latex].

Applying KVL to upper loop (A-B-G-A):

[latex]I_1 P – I_2 R = 0 \Rightarrow I_1 P = I_2 R \quad \cdots (1)[/latex]

Applying KVL to lower loop (B-C-G-B):

[latex]I_1 Q – I_2 S = 0 \Rightarrow I_1 Q = I_2 S \quad \cdots (2)[/latex]

Dividing (1) by (2):

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

This is the balanced condition of the Wheatstone bridge. When this condition is satisfied, no current flows through the galvanometer.

Q17: Electromagnetic Induction

a) Lenz’s Law

The direction of the induced emf (and hence induced current) in a conductor is always such that it opposes the change in magnetic flux that caused it.

b) Graph of induced emf vs. time

As a coil rotates in a uniform magnetic field, the induced emf varies sinusoidally: [latex]\varepsilon = \varepsilon_0 \sin(\omega t)[/latex]. The graph is a sine wave with amplitude [latex]\varepsilon_0 = NBA\omega[/latex] and period [latex]T = \frac{2\pi}{\omega}[/latex], passing through zero at [latex]t = 0[/latex] and repeating periodically.

c) Energy stored in an inductor

When current [latex]I[/latex] flows through an inductor of inductance [latex]L[/latex], the induced back emf is [latex]\varepsilon = L\frac{dI}{dt}[/latex].

The work done against this back emf in a small time [latex]dt[/latex]:

[latex]dW = \varepsilon I , dt = L\frac{dI}{dt} \cdot I , dt = LI , dI[/latex]

Total work done (energy stored) as current increases from 0 to [latex]I[/latex]:

[latex]U = \int_0^I LI , dI = L \cdot \frac{I^2}{2}[/latex]

[latex]\boxed{U = \frac{1}{2}LI^2}[/latex]

This energy is stored in the magnetic field of the inductor.

Q18: Electronics

a) NAND Gate — symbol and truth table

The NAND gate is the complement of the AND gate. Its Boolean expression is: [latex]Y = \overline{A \cdot B}[/latex]

The symbol is an AND gate symbol with a small circle (bubble) at the output.

Truth Table:

b) Full Wave Rectification using two P-N junction diodes

A full wave rectifier uses two diodes (D₁ and D₂) and a centre-tapped transformer:

Circuit: The secondary of the transformer has a centre tap (C). D₁ is connected to one end (A) and D₂ to the other end (B). Both diodes share a common load resistor [latex]R_L[/latex] connected between the output and the centre tap.

Working:

• Positive half cycle: Terminal A is positive. D₁ is forward biased and conducts; D₂ is reverse biased and does not conduct. Current flows through [latex]R_L[/latex] in one direction.
• Negative half cycle: Terminal B is positive. D₂ is forward biased and conducts; D₁ is reverse biased. Current flows through [latex]R_L[/latex] in the same direction as before.

Result: Both halves of the AC input produce output in the same direction across [latex]R_L[/latex] — giving a unidirectional (DC) pulsating output. The output frequency is twice the input frequency.

Q19 (Main): Atomic Physics — Millikan’s Oil Drop and Thomson’s Experiment

a) Terminal velocity of oil drop

Given: mass [latex]m = 3 \times 10^{-11} \text{ g} = 3 \times 10^{-14} \text{ kg}[/latex], radius [latex]r = 2 \times 10^{-4} \text{ cm} = 2 \times 10^{-6} \text{ m}[/latex], charge on drop [latex]q = 10e = 10 \times 1.6 \times 10^{-19} = 1.6 \times 10^{-18} \text{ C}[/latex], [latex]\eta = 1.8 \times 10^{-5} \text{ Ns/m}^2[/latex]

i) No electric field — terminal velocity (downward)

At terminal velocity, drag force = weight (buoyancy neglected for small drops):

[latex]6\pi\eta r v_t = mg[/latex]

[latex]v_t = \frac{mg}{6\pi\eta r} = \frac{3 \times 10^{-14} \times 10}{6\pi \times 1.8 \times 10^{-5} \times 2 \times 10^{-6}}[/latex]

[latex]v_t = \frac{3 \times 10^{-13}}{6\pi \times 3.6 \times 10^{-11}} = \frac{3 \times 10^{-13}}{6.786 \times 10^{-10}} \approx 4.42 \times 10^{-4} \text{ m/s}[/latex]

[latex]v_t \approx 4.42 \times 10^{-4} \text{ m/s}[/latex] (downward)

ii) Electric field directed downward = [latex]3 \times 10^5 \text{ V/m}[/latex]

Electric force on drop (downward, since electrons give negative charge but field is downward — force on negative charge is upward if field is downward… wait: field directed downward means [latex]\vec{E}[/latex] points down. Force on charge [latex]q = -10e[/latex] is [latex]\vec{F} = q\vec{E}[/latex] which is upward for negative charge):

Actually: the drop carries 10 excess electrons, so charge is negative: [latex]q = -1.6 \times 10^{-18} \text{ C}[/latex]. Field is directed downward. Electric force [latex]F_E = qE[/latex] — for negative charge with downward field, force is upward (opposing weight).

Net downward force = Weight – Electric force upward:

[latex]F_{net} = mg – qE = 3 \times 10^{-13} – 1.6 \times 10^{-18} \times 3 \times 10^5[/latex]

[latex]= 3 \times 10^{-13} – 4.8 \times 10^{-13} = -1.8 \times 10^{-13} \text{ N}[/latex]

The negative result means the electric force exceeds gravity — the drop moves upward at terminal velocity.

[latex]v_t = \frac{|F_{net}|}{6\pi\eta r} = \frac{1.8 \times 10^{-13}}{6\pi \times 1.8 \times 10^{-5} \times 2 \times 10^{-6}} = \frac{1.8 \times 10^{-13}}{6.786 \times 10^{-10}} \approx 2.65 \times 10^{-4} \text{ m/s (upward)}[/latex]

b) Why is E field perpendicular to B field in Thomson’s experiment?

In Thomson’s experiment, the electron beam passes through crossed electric ([latex]E[/latex]) and magnetic ([latex]B[/latex]) fields. When they are perpendicular to each other and to the beam, the electric force ([latex]eE[/latex]) acts in one direction while the magnetic force ([latex]evB[/latex]) acts in the opposite direction. By adjusting the fields until the beam is undeflected, we get [latex]eE = evB[/latex], which gives the velocity [latex]v = E/B[/latex]. This velocity selector arrangement is only possible when E and B are mutually perpendicular — if they were parallel, their forces would not oppose each other along the same axis.

c) Use of X-rays in Millikan’s oil drop experiment

X-rays are used to ionize the air between the plates of the Millikan apparatus. When X-rays pass through the air gap, they knock electrons off air molecules, producing ions. These ions then attach to the oil droplets, giving them specific (measurable) charges. By controlling the X-ray exposure, the experimenter can change the charge on the droplet during the experiment to observe how it moves under different charge conditions.

Q19 (OR): Nuclear Physics — Radioactive Decay

a) Decay constant and half-life

Decay constant ([latex]\lambda[/latex]): The probability of decay of a radioactive nucleus per unit time. It is the fraction of the total number of radioactive atoms that decay per unit time. Unit: s⁻¹ (or per second).

Half-life ([latex]T_{1/2}[/latex]): The time required for half the radioactive atoms in a sample to decay. It is related to the decay constant by: [latex]T_{1/2} = \frac{\ln 2}{\lambda} = \frac{0.693}{\lambda}[/latex]

b) Derivation of [latex]N = N_0 e^{-\lambda t}[/latex]

The rate of decay of radioactive atoms is proportional to the number of atoms present at that instant:

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

(The negative sign indicates N is decreasing.)

Separating variables:

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

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

where [latex]N_0[/latex] = initial number of radioactive atoms, [latex]N[/latex] = number remaining after time [latex]t[/latex], and [latex]\lambda[/latex] = decay constant.

Group C — Long Answer Solutions

Q20 (Main): Stationary Waves and Guitar String

a) Stationary waves and distance between nodes

Stationary (standing) waves are formed by the superposition of two identical waves (same amplitude, frequency, and speed) travelling in opposite directions. Unlike progressive waves, stationary waves do not transfer energy — they have fixed nodes (zero amplitude) and antinodes (maximum amplitude).

Proof that consecutive nodes are [latex]\lambda/2[/latex] apart:

Let two waves travelling in opposite directions be: [latex]y_1 = A\sin(kx – \omega t)[/latex] and [latex]y_2 = A\sin(kx + \omega t)[/latex]

By superposition:

[latex]y = y_1 + y_2 = A[\sin(kx – \omega t) + \sin(kx + \omega t)][/latex]

Using the identity [latex]\sin C + \sin D = 2\sin\frac{C+D}{2}\cos\frac{C-D}{2}[/latex]:

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

Nodes occur where amplitude = 0, i.e., where [latex]\sin(kx) = 0[/latex]:

[latex]kx = n\pi \Rightarrow \frac{2\pi}{\lambda}x = n\pi \Rightarrow x = \frac{n\lambda}{2}[/latex]

Distance between consecutive nodes ([latex]n[/latex] and [latex]n+1[/latex]):

[latex]\Delta x = \frac{(n+1)\lambda}{2} – \frac{n\lambda}{2} = \frac{\lambda}{2}[/latex]

Proved: distance between two consecutive nodes = [latex]\frac{\lambda}{2}[/latex]

b) Guitar string — stress and frequency

Given: Length [latex]L = 1.5 \text{ m}[/latex], density [latex]\rho = 7800 \text{ kg/m}^3[/latex], Young’s modulus [latex]Y = 2 \times 10^{11} \text{ N/m}^2[/latex], elastic strain = 1% = 0.01

i) Stress:

[latex]\text{Stress} = Y \times \text{Strain} = 2 \times 10^{11} \times 0.01 = 2 \times 10^9 \text{ N/m}^2[/latex]

ii) Frequency of second mode (first overtone) for a string fixed at both ends:

Velocity of wave in string: [latex]v = \sqrt{\frac{T}{\mu}}[/latex]

Tension per unit area = Stress, so [latex]T = \text{Stress} \times A[/latex] and [latex]\mu = \rho A[/latex]:

[latex]v = \sqrt{\frac{\text{Stress} \times A}{\rho A}} = \sqrt{\frac{\text{Stress}}{\rho}} = \sqrt{\frac{2 \times 10^9}{7800}}[/latex]

[latex]v = \sqrt{2.564 \times 10^5} \approx 506.4 \text{ m/s}[/latex]

For a string fixed at both ends, frequency of [latex]n[/latex]th mode: [latex]f_n = \frac{nv}{2L}[/latex]

Second mode ([latex]n = 2[/latex]):

[latex]f_2 = \frac{2 \times 506.4}{2 \times 1.5} = \frac{1012.8}{3} \approx 337.6 \text{ Hz}[/latex]

Frequency of second mode ≈ 337.6 Hz

c) Open organ pipe — mode of vibration and frequency

From the figure showing a standing wave in an open organ pipe with the given pattern (the figure in the paper shows 3 antinodes and 2 nodes inside, which is the second overtone / third harmonic for an open pipe):

i) The figure shows second overtone (third harmonic) — third mode of vibration for an open pipe.

ii) For an open pipe in the [latex]n[/latex]th mode: [latex]f_n = \frac{nv}{2L}[/latex]. For third harmonic, [latex]n = 3[/latex]: [latex]f_3 = \frac{3v}{2L}[/latex]. (Numerical value requires the pipe length, which is not given in this sub-part — express in terms of [latex]v[/latex] and [latex]L[/latex].)

Q20 (OR): Wave Optics — YDSE and Diffraction Grating

a) Equal spacing of fringes in Young’s Double Slit Experiment

In YDSE, let [latex]d[/latex] = slit separation, [latex]D[/latex] = distance to screen, [latex]\lambda[/latex] = wavelength.

Bright fringe positions: [latex]y_n = \frac{n\lambda D}{d}[/latex]

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

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

Fringe width (dark): [latex]\beta = \frac{\lambda D}{d}[/latex] (same calculation)

Since fringe width [latex]\beta = \frac{\lambda D}{d}[/latex] is independent of [latex]n[/latex], all bright and dark fringes are equally spaced.

b) Wavelength from diffraction grating

Given: grating has 500 lines/mm, so grating element [latex]d = \frac{1}{500} \text{ mm} = 2 \times 10^{-3} \text{ mm} = 2 \times 10^{-6} \text{ m}[/latex]

Order [latex]n = 2[/latex], angle [latex]\theta = 30°[/latex]

Grating equation: [latex]d\sin\theta = n\lambda[/latex]

[latex]\lambda = \frac{d\sin\theta}{n} = \frac{2 \times 10^{-6} \times \sin 30°}{2} = \frac{2 \times 10^{-6} \times 0.5}{2} = \frac{10^{-6}}{2} = 5 \times 10^{-7} \text{ m}[/latex]

[latex]\lambda = 500 \text{ nm}[/latex] (green light)

c) Brewster’s Law

Statement: When light is incident on a surface at the polarizing angle (Brewster’s angle) [latex]i_B[/latex], the reflected light is completely plane polarized, and the reflected and refracted rays are perpendicular to each other.

Mathematical form: [latex]\tan i_B = n[/latex] (refractive index of the medium)

Proof: At Brewster’s angle, reflected ray ⊥ refracted ray, so:

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

By Snell’s law: [latex]n = \frac{\sin i_B}{\sin r} = \frac{\sin i_B}{\sin(90° – i_B)} = \frac{\sin i_B}{\cos i_B} = \tan i_B[/latex]

Therefore: [latex]\tan i_B = n[/latex] — Brewster’s Law proved.

Q21 (Main): LCR Circuit and Parallel Wires

a) Series LCR circuit: [latex]L = 100 \text{ mH}[/latex], [latex]C = 2 \mu\text{F}[/latex], [latex]R = 400 \Omega[/latex], [latex]f = \frac{500}{\pi} \text{ Hz}[/latex]

[latex]\omega = 2\pi f = 2\pi \times \frac{500}{\pi} = 1000 \text{ rad/s}[/latex]

[latex]X_L = \omega L = 1000 \times 100 \times 10^{-3} = 100 \Omega[/latex]

[latex]X_C = \frac{1}{\omega C} = \frac{1}{1000 \times 2 \times 10^{-6}} = \frac{1}{2 \times 10^{-3}} = 500 \Omega[/latex]

i) Phase angle:

[latex]\tan\phi = \frac{X_C – X_L}{R} = \frac{500 – 100}{400} = \frac{400}{400} = 1[/latex]

[latex]\phi = \tan^{-1}(1) = 45°[/latex]

ii) Since [latex]X_C > X_L[/latex], the circuit is capacitively reactive — voltage lags current (or current leads voltage). Voltage leads in an inductive circuit; here the circuit is capacitive, so current leads the voltage by 45°.

b) Impedance of an AC circuit

Impedance ([latex]Z[/latex]) is the total opposition offered by an AC circuit to the flow of alternating current. It combines resistance ([latex]R[/latex]) and reactance (inductive [latex]X_L[/latex] or capacitive [latex]X_C[/latex]) as a phasor sum:

[latex]Z = \sqrt{R^2 + (X_L – X_C)^2}[/latex]

Unit: Ohm (Ω). For the LCR circuit above: [latex]Z = \sqrt{400^2 + (100 – 500)^2} = \sqrt{160000 + 160000} = \sqrt{320000} \approx 565.7 \Omega[/latex]

c) Force between two parallel current-carrying wires

i) Expression for force per unit length:

Magnetic field due to wire P at the location of wire Q:

[latex]B_P = \frac{\mu_0 I_1}{2\pi r}[/latex]

Force per unit length on wire Q due to this field:

[latex]\frac{F}{l} = I_2 B_P = \frac{\mu_0 I_1 I_2}{2\pi r}[/latex]

ii) Nature of force: Since both wires carry currents in the same direction (as shown in the figure — both currents pointing upward), the force between them is attractive.

iii) If distance r is halved: [latex]\frac{F}{l} \propto \frac{1}{r}[/latex], so halving r doubles the force per unit length.

New force per unit length = [latex]\frac{2\mu_0 I_1 I_2}{2\pi r} = \frac{\mu_0 I_1 I_2}{\pi r}[/latex] (twice the original)

Q21 (OR): Thermocouple and Potentiometer

a) Variation of thermoelectric emf with hot junction temperature

In a thermocouple, when the hot junction temperature [latex]T[/latex] is varied while the cold junction is kept at constant temperature, the thermoelectric emf varies as:

[latex]\varepsilon = aT + \frac{1}{2}bT^2[/latex]

where [latex]a[/latex] and [latex]b[/latex] are constants. This is a parabolic relationship. The emf first increases with temperature, reaching a maximum at the neutral temperature [latex]T_n = -\frac{a}{b}[/latex]. Beyond this, the emf decreases. At the temperature of inversion [latex]T_i = -\frac{2a}{b} = 2T_n – T_c[/latex] (where [latex]T_c[/latex] = cold junction temperature), the emf becomes zero and then reverses sign.

b) Potentiometer — unknown emf

Given: Total wire length [latex]L = 4 \text{ m}[/latex], total resistance [latex]R_w = 10 \Omega[/latex], main battery [latex]V = 3 \text{ V}[/latex], series resistance [latex]R = 290 \Omega[/latex], balance length [latex]l = 240 \text{ cm} = 2.4 \text{ m}[/latex]

Current through potentiometer wire:

[latex]I = \frac{V}{R + R_w} = \frac{3}{290 + 10} = \frac{3}{300} = 0.01 \text{ A}[/latex]

Resistance per unit length of wire: [latex]\frac{R_w}{L} = \frac{10}{4} = 2.5 \text{ Ω/m}[/latex]

Potential gradient: [latex]k = I \times \frac{R_w}{L} = 0.01 \times 2.5 = 0.025 \text{ V/m}[/latex]

Unknown emf:

[latex]E = k \times l = 0.025 \times 2.4 = 0.06 \text{ V}[/latex]

[latex]E = 0.06 \text{ V} = 60 \text{ mV}[/latex]

c) Magnetic field due to straight current-carrying conductor using Ampere’s Circuital Law

Consider a long straight conductor carrying current [latex]I[/latex]. By symmetry, the magnetic field at a perpendicular distance [latex]r[/latex] is tangential and uniform in magnitude on a circular Amperian loop of radius [latex]r[/latex].

Applying Ampere’s circuital law:

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

[latex]B \times 2\pi r = \mu_0 I[/latex]

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

The field is directed tangentially (by right-hand rule) and decreases inversely with distance from the conductor.

Q22: Photoelectric Effect and Hydrogen Atom

a) Threshold wavelength

The threshold wavelength ([latex]\lambda_0[/latex]) is the maximum wavelength of incident radiation that can cause the photoelectric effect for a given metal. At this wavelength, the photon energy exactly equals the work function of the metal: [latex]\phi = \frac{hc}{\lambda_0}[/latex]. Radiation with wavelength greater than [latex]\lambda_0[/latex] cannot eject photoelectrons regardless of intensity.

b) Maximum speed of photoelectrons and threshold frequency

Given: Work function [latex]\phi = 2 \text{ eV} = 2 \times 1.6 \times 10^{-19} = 3.2 \times 10^{-19} \text{ J}[/latex], [latex]\lambda = 150 \text{ nm} = 150 \times 10^{-9} \text{ m}[/latex]

Photon energy:

[latex]E = \frac{hc}{\lambda} = \frac{6.62 \times 10^{-34} \times 3 \times 10^8}{150 \times 10^{-9}} = \frac{19.86 \times 10^{-26}}{150 \times 10^{-9}} = 1.324 \times 10^{-18} \text{ J}[/latex]

Maximum KE of photoelectrons:

[latex]KE_{max} = E – \phi = 1.324 \times 10^{-18} – 3.2 \times 10^{-19} = 1.004 \times 10^{-18} \text{ J}[/latex]

Maximum speed:

[latex]\frac{1}{2}mv_{max}^2 = 1.004 \times 10^{-18}[/latex]

[latex]v_{max} = \sqrt{\frac{2 \times 1.004 \times 10^{-18}}{9.1 \times 10^{-31}}} = \sqrt{\frac{2.008 \times 10^{-18}}{9.1 \times 10^{-31}}} = \sqrt{2.207 \times 10^{12}} \approx 1.486 \times 10^6 \text{ m/s}[/latex]

[latex]v_{max} \approx 1.49 \times 10^6 \text{ m/s}[/latex]

Threshold frequency:

[latex]\phi = hf_0 \Rightarrow f_0 = \frac{\phi}{h} = \frac{3.2 \times 10^{-19}}{6.62 \times 10^{-34}} = 4.83 \times 10^{14} \text{ Hz}[/latex]

[latex]f_0 \approx 4.83 \times 10^{14} \text{ Hz}[/latex]

c) Heisenberg’s Uncertainty Principle

It is impossible to simultaneously determine both the exact position and exact momentum (or velocity) of a particle with arbitrary precision. If [latex]\Delta x[/latex] is the uncertainty in position and [latex]\Delta p[/latex] is the uncertainty in momentum:

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

Similarly for energy and time: [latex]\Delta E \cdot \Delta t \geq \frac{h}{4\pi}[/latex]

This is not a measurement limitation but a fundamental property of quantum systems.

d) Velocity of electron in nth orbit of hydrogen atom

In Bohr’s model, an electron moves in a circular orbit. The electrostatic force provides centripetal force:

[latex]\frac{mv^2}{r} = \frac{ke^2}{r^2} \Rightarrow mv^2 = \frac{ke^2}{r} \quad \cdots(1)[/latex]

Bohr’s quantization condition:

[latex]mvr = \frac{nh}{2\pi} \Rightarrow v = \frac{nh}{2\pi mr} \quad \cdots(2)[/latex]

Bohr radius for nth orbit: [latex]r_n = \frac{n^2 h^2 \varepsilon_0}{\pi m e^2}[/latex] (or equivalently [latex]r_n = n^2 a_0[/latex] where [latex]a_0 = 0.529 \text{ Å}[/latex])

Substituting [latex]r_n[/latex] into (2):

[latex]v_n = \frac{nh}{2\pi m} \times \frac{\pi m e^2}{n^2 h^2 \varepsilon_0} = \frac{e^2}{2\varepsilon_0 nh}[/latex]

[latex]\boxed{v_n = \frac{e^2}{2\varepsilon_0 nh} = \frac{2.18 \times 10^6}{n} \text{ m/s}}[/latex]

The velocity decreases as [latex]n[/latex] increases — electrons in higher orbits move more slowly.

Difficulty Analysis

Student Preparation Tips

1. Practise All Standard Derivations

NEB Group B and C questions repeatedly ask for derivations: energy in SHM, torque-inertia relation, Wheatstone bridge balanced condition, energy in inductor, stationary wave node spacing, fringe width in YDSE, Brewster’s law, radioactive decay law, velocity in nth Bohr orbit. These are predictable — a student who can reproduce all of them perfectly secures 20+ marks from derivations alone.

2. Master the Doppler Effect for Echo Problems

The echo problem (Q15b) is a two-step Doppler application. The train is both the source and the observer. Many students apply the formula only once — this gives the wrong answer. Always apply it in two stages: (1) cliff receives the whistle frequency as a moving source, (2) driver receives the echo from the stationary cliff as a moving observer.

3. Group A MCQs — Learn the Formula Patterns

Most MCQ errors come from formula confusion. Key formulas to know instantly:

• [latex]I_{rod, centre} = \frac{ML^2}{12}[/latex], [latex]I_{rod, end} = \frac{ML^2}{3}[/latex]
• [latex]B = \frac{\mu_0 I}{2R}[/latex] (circular coil centre)
• [latex]\mu_r = 1 + \chi[/latex]
• [latex]\Phi = BA\cos\theta[/latex]
• Terminal velocity [latex]\propto r^2[/latex] (two drops combining → [latex]v’ = 2^{2/3}v[/latex])

4. Numerical Problems — Show Every Step

NEB awards step marks. Even if your final answer is wrong, correct substitution, correct formula, and correct unit conversion each earn partial credit. Never skip steps in Group B and C numericals.

5. Know Your OR Options Before the Exam

Each Group B question has an OR option. Decide in advance — based on your preparation — which OR options you will choose. Don’t waste time in the exam reading both options and then deciding. Weak areas in rotational mechanics → choose SHM OR. Weak in Millikan experiment → choose radioactive decay OR.

6. Electronics and Logic Gates Are Easy Marks

Q18 (NAND gate and full wave rectifier) is among the most predictable and formulaic questions in NEB Physics. Study the truth tables of all basic gates (AND, OR, NOT, NAND, NOR, XOR) and the working of half-wave and full-wave rectifiers. These 5 marks are among the most reliably scored in the paper.

Frequently Asked Questions (FAQs)

Q1. What is the Sub Code for NEB Grade XII Physics 2082 Set H? The sub code is 1021 ‘H’. Different sets (A through H or beyond) are issued for different student registration number ranges.

Q2. What is the answer to MCQ Q1 about moment of inertia? The difference between [latex]I_{end}[/latex] and [latex]I_{centre}[/latex] for a rod is [latex]\frac{ML^2}{3} – \frac{ML^2}{12} = \frac{ML^2}{4}[/latex]. Answer: (B).

Q3. What is the answer to Q5 — efficiency of Carnot engine between steam and ice points? [latex]\eta = 1 – \frac{273}{373} = \frac{100}{373} \approx 26.8%[/latex]. Answer: (A) 26.8%.

Q4. Why is polarization not explained by Huygens’ principle? Huygens’ principle treats light as a wave but does not specify the direction of oscillation. Polarization requires the transverse nature of light — where vibrations are restricted to one plane — which is beyond what Huygens’ construction alone can explain. Answer to Q7: (C) Polarization.

Q5. What is the terminal velocity of the combined raindrop in Q3? Using [latex]v’ = v \times 2^{2/3} = 0.15 \times 1.587 \approx 0.24 \text{ m/s}[/latex]. Answer: (B) 0.24 m/s.

Q6. What is the frequency of the echo in Q15b? Using [latex]f_{echo} = f_0 \times \frac{v + v_s}{v – v_s} = 600 \times \frac{350}{330} \approx 636.4 \text{ Hz}[/latex].

Q7. What is the unknown emf in the potentiometer problem (Q21 OR)? [latex]E = k \times l = 0.025 \times 2.4 = 0.06 \text{ V} = 60 \text{ mV}[/latex].

Q8. What is the maximum speed of photoelectrons in Q22b? [latex]v_{max} \approx 1.49 \times 10^6 \text{ m/s}[/latex], and threshold frequency [latex]f_0 \approx 4.83 \times 10^{14} \text{ Hz}[/latex].

Q9. How do I calculate the wavelength of light in the diffraction grating problem? [latex]\lambda = \frac{d\sin\theta}{n} = \frac{2 \times 10^{-6} \times 0.5}{2} = 5 \times 10^{-7} \text{ m} = 500 \text{ nm}[/latex].

Q10. What chapters carry the most marks in NEB Physics 2082 Set H? Waves & Sound, Electromagnetic Induction, and Atomic/Nuclear Physics together account for approximately 42 out of 75 marks. These three chapters are the highest priority for exam preparation.

Important Questions for Future NEB Physics Exams

Chapter: Rotational Mechanics / SHM

• Derive [latex]\tau = I\alpha[/latex]; prove parallel and perpendicular axis theorems
• Show total energy in SHM is constant; derive [latex]E = \frac{1}{2}m\omega^2 A^2[/latex]
• Numerical: moment of inertia of composite bodies; time period changes with gravity

Chapter: Fluid Mechanics

• Bernoulli’s theorem derivation and applications (venturimeter, pitot tube)
• Continuity equation derivation
• Numerical: Stokes’ law terminal velocity; Bernoulli-based pressure calculations

Chapter: Thermodynamics

• Carnot cycle — P-V diagram, derivation of efficiency
• First and second laws with applications
• Numerical: efficiency calculations; source/sink temperature problems

Chapter: Waves and Sound

• Stationary waves — derivation, nodes/antinodes spacing
• Doppler effect — all four cases (source/observer moving towards/away from each other)
• Echo problems (train + cliff)

Chapter: Optics

• YDSE — derive fringe width; prove equal spacing
• Diffraction grating — grating equation and numericals
• Brewster’s law — statement and proof

Chapter: Electricity and Magnetism

• Kirchhoff’s laws; Wheatstone bridge balanced condition
• Biot-Savart law; Ampere’s circuital law applications
• Force between parallel wires; potentiometer theory and numericals

Chapter: Electromagnetic Induction

• Lenz’s law; Faraday’s laws
• Energy stored in inductor: [latex]U = \frac{1}{2}LI^2[/latex]
• LCR circuits — impedance, resonance, phase angle

Chapter: Electronics

• Logic gates truth tables (AND, OR, NOT, NAND, NOR)
• Half-wave and full-wave rectification
• Common applications of diodes and transistors

Chapter: Atomic and Nuclear Physics

• Bohr’s model — radius, velocity, energy of nth orbit
• Photoelectric effect — Einstein’s equation, numerical
• Radioactive decay law; half-life and decay constant
• Millikan’s oil drop experiment; Thomson’s specific charge experiment

Complete Question Paper

NEB — Grade XII | 2082 (2025)

Physics | Sub. Code: 1021 ‘H’

Attempt all the questions.

Group ‘A’ — Multiple Choice Questions [11×1=11]

Rewrite the correct option of each question in your answer sheet.

1. What is the difference between moment of inertia of a rod about an axis passing through its one end and through its centre perpendicular to its length? (A) [latex]\frac{ML^2}{3}[/latex] (B) [latex]\frac{ML^2}{4}[/latex] (C) [latex]\frac{ML^2}{6}[/latex] (D) [latex]\frac{ML^2}{12}[/latex]
2. For an oscillating object, the displacement versus time curve shown in figure. What is the velocity of the object in the mean position? (A) [latex]0.2\pi[/latex] m/s (B) [latex]0.4\pi[/latex] m/s (C) [latex]2\pi[/latex] m/s (D) [latex]4\pi[/latex] m/s (From graph: amplitude = 0.20 m, time period = 2.0 s)
3. Two spherical rain drops of each size are falling vertically downward with terminal velocity of 0.15 m/s. What would be the terminal velocity if these drops were combined to form a larger drop? (A) 0.15 m/s (B) 0.24 m/s (C) 0.31 m/s (D) 0.48 m/s
4. What happens in an adiabatic process? (A) Volume remains constant (B) Pressure remains constant (C) Temperature remains constant (D) The system is insulated from the surroundings
5. The efficiency of Carnot engine operating between steam point and ice point is: (A) 26.8% (B) 50% (C) 73.2% (D) 100%
6. Which of the following is an example of transverse wave? (A) Sound wave in air (B) Waves inside the water (C) Wave produced in the stem of vibrating tuning fork (D) Wave produced in the stretched string
7. Which one of the following phenomena is not explained by Huygen’s construction of wave front? (A) Refraction (B) Reflection (C) Polarization (D) Diffraction
8. The magnetic field intensity at the center of a circular coil of radius R and carrying current I is: (A) [latex]\frac{\mu_0 I}{R}[/latex] (B) [latex]\frac{\mu_0 I}{2R}[/latex] (C) [latex]\frac{\mu_0 I}{2\pi R}[/latex] (D) [latex]\frac{2\mu_0 I}{R}[/latex]
9. If [latex]\chi[/latex] is the magnetic susceptibility of a material then its relative permeability ([latex]\mu_r[/latex]) is: (A) [latex]\chi + 1[/latex] (B) [latex]\chi – 1[/latex] (C) [latex]\frac{1}{\chi}[/latex] (D) [latex]1 + \chi^2[/latex]
10. A square coil of area [latex]10^{-2}[/latex] m² is placed perpendicular to a uniform magnetic field of intensity [latex]10^3[/latex] T. What is the magnetic flux through the coil? (A) 100 Weber (B) 10 Weber (C) [latex]10^{-3}[/latex] Weber (D) [latex]10^{-5}[/latex] Weber
11. What is the point where the Seismic waves start called? (A) Epicenter (B) Metacenter (C) Hypocenter (D) Seismic center

Group ‘B’ — Short Answer Questions [8×5=40]

12. a) A ballet dancer sometimes stretches and sometimes folds her arms during her performance, why? Justify. [2] b) Establish a relation between torque and moment of inertia for a rigid body. [2] c) Why do we prefer a wrench of longer arm over a wrench of shorter arm? [1]

OR

a) Show that the total energy of the particles executing SHM is constant. [3] b) Draw a graph showing the variation of potential energy of particle in SHM with displacement. [1] c) The time period of simple pendulum on the surface of earth is T. What will be its time period on the Moon’s surface? [1]

13. a) Define laminar flow and turbulent flow of liquid. [2] b) Water flows steadily through a horizontal pipe of non-uniform cross section. If the pressure of water is [latex]4 \times 10^4[/latex] Nm⁻² at a point where the velocity of flow is 2 m/s and cross section is 200 cm². Calculate the pressure at a point where cross section reduces to 50 cm². (density of water = 1000 kg/m³) [3]
14. a) A Carnot’s engine has 25% efficiency with a sink at 9°C. By how many degrees should the temperature of the source be increased in order to raise the efficiency to 50%? [3] b) State first law of thermodynamics. Does it follow the principle of conservation of energy? Explain. [2]
15. a) Discuss the effect of temperature and pressure on velocity of sound in a gas. [2] b) A train is approaching a cliff at 10 m/s. The driver sounds a whistle of frequency 600 Hz. What will be the frequency of echo as heard by the driver? (Velocity of sound in air = 340 ms⁻¹) [3]
16. a) State the two Kirchhoff’s laws of electrical circuit. [2] b) Obtain an expression for the balanced condition of Wheatstone bridge using Kirchhoff’s laws. [3]
17. a) State Lenz’s law. [1] b) Plot a graph to show the variation of induced emf in a coil rotating in a uniform magnetic field with time. [1] c) Derive an expression for the energy stored in an inductor. [3]
18. a) Write the symbol and truth table of NAND gate. [2] b) Establish the full wave rectification using two P-N junction diodes. [3]
19. a) An oil drop of mass [latex]3 \times 10^{-11}[/latex] g and of radius [latex]2 \times 10^{-4}[/latex] cm carries 10 excess electrons. What is its terminal velocity when: i) falling in a region in which there is no electric field? [1] ii) falling in an electric field of [latex]3 \times 10^5[/latex] Vm⁻¹ directed downward? (Viscosity of air = [latex]1.8 \times 10^{-5}[/latex] NSm⁻²). [1] b) In Thomson’s method for determining specific charge of an electron, why is the electric field kept perpendicular to magnetic field? Justify. [2] c) What is the use of X-rays in Millikan’s oil drop experiment? [1]

OR

a) Define decay constant and half-life. [2] b) Derive the equation: [latex]N = N_0 e^{-\lambda t}[/latex] for radioactive process where symbols have their usual meanings. [3]

Group ‘C’ — Long Answer Questions [3×8=24]

20. a) What are stationary waves? Prove that the distance between two consecutive nodes in it is equal to half of the wavelength. [3] b) Guitar string of length 1.5 m is made of steel of density 7800 kg/m³ and Young’s modulus [latex]2 \times 10^{11}[/latex] N/m². It produces an elastic strain of 1% in the string. Calculate: [3] i) stress developed in the string. ii) frequency of second mode of vibration. c) The given figure shows standing wave formed in an open organ pipe. i) Which mode of vibration is shown? [1] ii) Obtain the frequency of vibration for this mode. [1]

OR

a) Prove that bright and dark fringes are equally spaced in Young’s double slit experiment. [3] b) A parallel beam of monochromatic light is incident on diffraction grating having 500 lines/mm to get second order maxima at 30°. Calculate the wavelength of light used. [3] c) State and prove Brewster’s law. [2]

21. a) The figure below shows a series LCR circuit (L = 100 mH, C = 2μF, R = 400Ω). If the frequency of an ac source is [latex]\frac{500}{\pi}[/latex] Hz then, i) Calculate the phase angle between current and voltage. [2] ii) Which one leads in phase, current or voltage? Justify. [1] b) What is meant by impedance of an ac circuit? [1] c) Two parallel wire P and Q carrying currents [latex]I_1[/latex] and [latex]I_2[/latex] are placed in the air at a distance r apart (both currents in same direction). i) Obtain an expression for the magnitude of force per unit length between them. [2] ii) What is the nature of force in this case? [1] iii) What will be the force per unit length between two wires if distance between wires is halved, keeping all other quantities constant? [1]

OR

a) Explain the variation of thermoelectric emf with the temperature of hot junction of a thermocouple. [3] b) A potentiometer circuit: total wire length 4 m, resistance 10Ω, main battery 3V, series resistance 290Ω, balance point at 240 cm. Calculate the value of unknown emf E. [3] c) Use Ampere’s circuital law to find the magnetic field due to a straight current carrying conductor. [2]

22. a) Define threshold wavelength. [1] b) A metal has a work function of 2 eV. Calculate the maximum speed of the emitted photoelectrons when it is illuminated by radiation of wavelength 150 nm. What is the least frequency of the radiation for which electrons are emitted? (h = 6.62 × 10⁻³⁴ Js, m = 9.1 × 10⁻³¹ kg, e = 1.6 × 10⁻¹⁹ C) [3] c) State Heisenberg’s Uncertainty principle. [1] d) Obtain an expression for velocity of an electron in nth orbit of hydrogen atom. [3]

-0-

Quick Answer Key

Group A — MCQ

1. (B) | 2. (A) | 3. (B) | 4. (D) | 5. (A) | 6. (D) | 7. (C) | 8. (B) | 9. (A) | 10. (B) | 11. (C)

Key Numerical Answers

• Q13b: [latex]P_2 = 1 \times 10^4 \text{ N/m}^2[/latex]
• Q14a: Temperature increase = 188°C
• Q15b: Echo frequency ≈ 636.4 Hz
• Q19a(i): [latex]v_t \approx 4.42 \times 10^{-4} \text{ m/s}[/latex]
• Q20b(i): Stress = [latex]2 \times 10^9 \text{ N/m}^2[/latex]
• Q20b(ii): [latex]f_2 \approx 337.6 \text{ Hz}[/latex]
• Q20(OR)b: [latex]\lambda = 500 \text{ nm}[/latex]
• Q21a(i): Phase angle = 45°
• Q21(OR)b: [latex]E = 0.06 \text{ V}[/latex]
• Q22b: [latex]v_{max} \approx 1.49 \times 10^6 \text{ m/s}[/latex], [latex]f_0 \approx 4.83 \times 10^{14} \text{ Hz}[/latex]

Conclusion

The NEB Grade XII Physics 2082 Set H paper is a comprehensive, well-designed examination that genuinely tests both conceptual understanding and mathematical ability across the full scope of the Grade XII New Course Physics curriculum. No chapter has been ignored — from rotational mechanics to quantum physics, every major topic contributes to the paper’s 75 marks.

For students, the most important lesson from this paper is that NEB Physics rewards systematic preparation over selective study. The derivations, the numerical problem types, and even many specific values (Carnot efficiency between ice and steam points, terminal velocity of combined drops, Doppler echo problems) are patterns that recur across years and sets. A student who masters these patterns — not just memorizes them but truly understands the physics behind them — can approach any NEB Physics paper with confidence.

Work through every numerical in this article with pen and paper. Check each step. Understand why the echo frequency formula uses [latex]\frac{v+v_s}{v-v_s}[/latex] rather than a single application. Understand why [latex]\mu_r = 1 + \chi[/latex] and not [latex]1 + \chi^2[/latex]. These distinctions are exactly what NEB tests — and what this article is designed to help you master.

Article prepared for educational purposes by NN Education (nneducation.com). All question paper content belongs to the National Examinations Board (NEB), Nepal. All mathematical expressions are wrapped in [latex][/latex] tags for proper rendering.