NEB Grade XII Physics 2081 (2024) Set D — Complete Question Paper Analysis, Answer Guide & Exam Preparation

Subject Code: 1021’D’ | Full Marks: 75 | Time: 3 Hours | New Course
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Introduction

The NEB Grade XII Physics paper of 2081 BS (2024 AD) represents a significant milestone — it is one of the first board examinations under the New Course for Grade XII, covering topics that were not tested in the old curriculum. This Set D paper spans an impressive range of modern physics concepts alongside classical mechanics, thermodynamics, waves, and electromagnetism.

What makes this paper particularly challenging and rewarding is its breadth. In a single three-hour sitting, students must handle rotational dynamics, fluid mechanics, thermodynamics, wave optics, AC circuits, semiconductor electronics, atomic physics, radioactivity, electromagnetic induction, and the photoelectric effect. This is not a paper you can prepare for by studying two or three chapters — it demands comprehensive mastery of the entire Grade XII Physics syllabus.

This guide breaks down every question, explains the underlying concepts, provides worked solutions to numericals, and gives targeted preparation strategies to help students score maximum marks.

Exam Overview

Full Question Paper

Question Pattern Analysis

Group A — MCQs (11 marks)
Eleven one-mark questions covering surface tension/capillarity, SHM, rotational dynamics, thermodynamics, waves, optics, thermoelectricity, magnetism, electromagnetism, and seismology. Speed and concept clarity are essential — no working space needed.

Group B — Short Answer (40 marks)
Eight questions (Q.12–Q.19), each 5 marks, with sub-parts. Most have OR alternatives. Topics: rotational motion/SHM, fluid mechanics, thermodynamics, waves/sonometer, electrical measurements, AC circuits, semiconductor electronics, and atomic/nuclear physics.

Group C — Long Answer (24 marks)
Three questions (Q.20–Q.22), each 8 marks. Topics: Sound/Wave optics (Q.20), Electromagnetic induction/Thermoelectricity (Q.21), and Thomson’s method/Photoelectric effect (Q.22).

Topics Covered — Chapter-wise Weightage

Key observation: Sound and waves (Q.15, Q.20) and electromagnetic topics (Q.17, Q.21) together carry the highest marks in this paper.

Group A — MCQ Analysis with Answers

Q.1 — Angle of Contact for Plane Meniscus
Answer: d. equal to 90°

When the meniscus is perfectly plane (flat), the liquid neither wets nor avoids the glass — the angle of contact is exactly 90°. When concave (wetting liquid like water in glass), angle < 90°. When convex (non-wetting like mercury), angle > 90°.

Q.2 — Period of Spring-Mass System
Answer: c. 4 sec

T = 2π√(m/k). Original: T₁ = 2π√(M/k) = 2 sec.
New mass = 4M: T₂ = 2π√(4M/k) = 2 × 2π√(M/k) = 2 × 2 = 4 sec

Key formula: T ∝ √m. Doubling mass → period increases by √2. Quadrupling mass → period doubles.

Q.3 — New Moment of Inertia After Wax Drop
Answer: c. I + mr²

When wax of mass m is added at distance r from the axis, it adds mr² to the existing moment of inertia I. Moment of inertia is additive. The new moment of inertia = I + mr²

Q.4 — Efficiency of Heat Engine After Raising Both Temperatures
Answer: c. less than η

Efficiency η = 1 − T₂/T₁. When both T₁ and T₂ increase by 50K, the ratio T₂/T₁ increases (since adding same amount to smaller number increases the ratio). Therefore efficiency decreases and becomes less than η.

This is a commonly misunderstood MCQ. The intuition that “more heat = more efficiency” is wrong here.

Q.5 — Work Done by Gas in a Cycle (P-V Diagram)
Answer: b. 12 × 10⁻⁶ J

Work done in a thermodynamic cycle = area enclosed in the P-V diagram.
From the diagram: rectangular cycle with ΔP = (6−2) Pa = 4 Pa and ΔV = (6−3) cm³ = 3 × 10⁻⁶ m³

W = ΔP × ΔV = 4 × 3 × 10⁻⁶ = 12 × 10⁻⁶ J

Q.6 — Speed of Sound Highest In
Answer: d. Steel

Speed of sound = √(E/ρ) where E = elastic modulus. Solids have far greater elastic modulus than liquids or gases. Steel has v ≈ 5100 m/s, water ≈ 1480 m/s, hydrogen ≈ 1270 m/s, vacuum = 0 (sound cannot travel).

Q.7 — Polarizing Angle for Glass with Critical Angle 42°
Answer: c. 56°

From critical angle: sin(42°) = 1/μ → μ = 1/sin(42°) = 1/0.669 = 1.494

Brewster’s Law: tan(ip) = μ = 1.494
ip = tan⁻¹(1.494) ≈ 56°

Note: ip + ic ≠ 90°. Use Brewster’s law separately.

Q.8 — Thermoelectric Series: Least emf Combination
Answer: d. Zn and Mn

Thermoelectric series: Sb, Fe, Zn, Pb, Mn, Cu, Bi (in order of decreasing thermoelectric power). The farther apart two metals are in the series, the greater the thermo-emf. Metals closest together in the series give the least emf. Zn and Mn are adjacent in the series → least emf.

Q.9 — Area of Hysteresis Curve
Answer: b. loss of energy per cycle

The area enclosed by the hysteresis loop represents the energy lost as heat per unit volume per magnetization cycle. Retentivity = residual magnetism at zero field. Coercivity = field needed to demagnetize. Curie’s temperature = temperature at which ferromagnetism vanishes.

Q.10 — Magnetic Force on Moving Charge
Answer: b. 2N

F = qvB sinθ = 2 × 0.5 × 4 × sin30° = 2 × 0.5 × 4 × 0.5 = 2 N

Q.11 — First Earthquake Wave Recorded
Answer: b. P-waves

P-waves (Primary/Pressure waves) are longitudinal waves that travel fastest through both solid and liquid layers of Earth. They arrive first at seismographs. S-waves (Secondary/Shear waves) arrive second. Love and Rayleigh waves are surface waves that arrive last.

Group B — Short Answer Analysis

Q.12 — Rotational Dynamics (OR: SHM)

Part (a): Moment of Inertia (1 mark)
Moment of inertia (I) is the rotational analogue of mass in linear motion. It is defined as the sum of products of mass of each particle and the square of its distance from the axis of rotation: I = Σmr². It depends on the mass, distribution of mass, and the position of the axis of rotation.

Part (b): Conservation of Angular Momentum (1+1 marks)

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

Mathematical form: If τ_ext = 0, then L = Iω = constant

Example: A figure skater spinning on ice — when she pulls her arms inward, her moment of inertia decreases, so her angular velocity increases to keep L = Iω constant.

Part (c): Wheel Rotation Numerical (2 marks)

Given: ω₀ = 0 (starts from rest), ω = 8 rev/s = 8 × 2π = 16π rad/s, t = 5 s

i. Angular acceleration:
α = (ω − ω₀)/t = 16π/5 = 10.05 rad/s²

ii. Angle rotated in 3 seconds:
θ = ω₀t + ½αt² = 0 + ½ × (16π/5) × 3² = ½ × 10.05 × 9 = 45.24 rad ≈ 7.2 revolutions

OR Q.12 — Simple Harmonic Motion

Part (a): Definition of SHM (1 mark)
Simple harmonic motion is a type of periodic motion in which the restoring force is directly proportional to the displacement from the equilibrium position and always directed towards it.
Mathematically: F = −kx or a = −ω²x

Part (b): Frequency of Vertical Mass-Spring System (2 marks)

For a mass m on a spring of constant k displaced vertically by x:

• Restoring force: F = −kx
• From Newton’s 2nd law: ma = −kx → a = −(k/m)x
• Comparing with SHM: ω² = k/m
• Angular frequency: ω = √(k/m)
• Frequency: f = (1/2π)√(k/m)

Part (c): Velocity of Simple Pendulum at Lowest Point (2 marks)

Given: L = 4 m, amplitude A = 0.2 m, g = 9.8 m/s²

At the lowest point, all PE converts to KE → maximum velocity.

Using energy conservation:
v_max = Aω = A√(g/L) = 0.2 × √(9.8/4) = 0.2 × √2.45 = 0.2 × 1.565 = 0.313 m/s

Q.13 — Fluid Mechanics: Capillarity and Bernoulli’s Equation

Part (a): Capillarity with Examples (2 marks)

Capillarity (or capillary action) is the phenomenon by which a liquid rises or falls in a narrow tube (capillary tube) due to surface tension and adhesive forces.

Examples:

1. Water rises in a thin glass tube when dipped in it
2. Ink rises through the pores of blotting paper
3. Oil rises through the wick of a lamp
4. Water absorbed by the roots of plants through soil pores

Part (b): Bernoulli’s Equation Numerical (3 marks)

Given:

• P₁ = 4 × 10⁴ N/m², v₁ = 2 m/s, A₁ = 0.02 m²
• A₂ = 0.01 m², ρ = 1000 kg/m³

Step 1 — Find v₂ using continuity equation:
A₁v₁ = A₂v₂
0.02 × 2 = 0.01 × v₂
v₂ = 4 m/s

Step 2 — Apply Bernoulli’s equation:
P₁ + ½ρv₁² = P₂ + ½ρv₂²
P₂ = P₁ + ½ρ(v₁² − v₂²)
P₂ = 4×10⁴ + ½ × 1000 × (4 − 16)
P₂ = 4×10⁴ + 500 × (−12)
P₂ = 40000 − 6000 = 34000 N/m² = 3.4 × 10⁴ N/m²

Note: Pressure decreases as velocity increases — this is the Venturi effect.

Q.14 — Thermodynamics: Adiabatic Process and Refrigerator

Part (a)(i): Why is Adiabatic Curve Steeper than Isothermal? (1 mark)

For an adiabatic process: PV^γ = constant → slope of P-V curve = −γP/V

For an isothermal process: PV = constant → slope = −P/V

Since γ > 1 for all real gases, the adiabatic slope is steeper (larger negative magnitude) than the isothermal slope at any given point.

Physical reason: In an isothermal expansion, heat flows in to maintain temperature, so pressure doesn’t fall as rapidly. In adiabatic expansion, no heat input — the gas cools, causing a sharper pressure drop.

Part (a)(ii): Workdone in Adiabatic Process Derivation (2 marks)

For an adiabatic process: PV^γ = K (constant), so P = K/V^γ

W = ∫PdV from V₁ to V₂ = K∫V^(-γ)dV = K[V^(1-γ)/(1-γ)] from V₁ to V₂

= (P₁V₁^γ × V₂^(1-γ) − P₁V₁)/(1-γ) = (P₁V₁ − P₂V₂)/(γ-1)

Using PV = nRT: W = nR(T₁ − T₂)/(γ−1)

Part (b): Refrigerator Numerical (2 marks)

Given: β (COP) = 1.95, Q₂ = 3 × 10⁴ J (heat absorbed from cold reservoir)

COP formula: β = Q₂/(Q₁ − Q₂)
1.95 = 3×10⁴/(Q₁ − 3×10⁴)
Q₁ − 3×10⁴ = 3×10⁴/1.95 = 1.538 × 10⁴
Q₁ = 3×10⁴ + 1.538×10⁴ = 4.538 × 10⁴ J ≈ 4.54 × 10⁴ J

Q.15 — Sonometer and Organ Pipe

Part (a): Does Frequency Depend on Hanging Weight? (1 mark)

Yes. Fundamental frequency of a sonometer wire: f = (1/2L)√(T/μ)

The tension T = Mg (weight of hanging mass). Since f ∝ √T, frequency depends on the hanging weight. Increasing the weight increases tension and hence increases the fundamental frequency.

Part (b): Frequency When Weight Immersed in Liquid (1 mark)

When the hanging weight is immersed in a liquid with upthrust U, the effective tension becomes:
T’ = Mg − U

New fundamental frequency: f’ = (1/2L)√((Mg−U)/μ)

Since T’ < T, the new frequency is less than the original frequency.

Part (c): Modes of Vibration in Closed Organ Pipe (3 marks)

A closed organ pipe has one end closed and one end open. At the closed end there is always a node, and at the open end there is always an antinode.

Fundamental mode (1st harmonic):
Length L = λ₁/4 → λ₁ = 4L → f₁ = v/4L

First overtone (3rd harmonic):
L = 3λ₃/4 → λ₃ = 4L/3 → f₃ = 3v/4L = 3f₁

Second overtone (5th harmonic):
L = 5λ₅/4 → λ₅ = 4L/5 → f₅ = 5v/4L = 5f₁

Key property: Closed organ pipe produces only odd harmonics (1st, 3rd, 5th…). Even harmonics are absent. This gives it a hollow, woody tone quality compared to open pipes.

Q.16 — Electrical Measurements: Potential Gradient and Meter Bridge

Part (a): Potential Gradient (2 marks)

Definition: Potential gradient is the fall of potential per unit length of the potentiometer wire.
K = V/L (in V/m)

In terms of specific resistance:
If wire has length L, cross-sectional area A, specific resistance ρ, and current I flows:
Resistance of wire: R = ρL/A
Potential across wire: V = IR = IρL/A
Potential gradient: K = V/L = Iρ/A

Part (b): Meter Bridge Numerical (3 marks)

Case 1: R₁ in series with 10Ω in one gap, R₂ in other gap. Balance at 50 cm.
(R₁ + 10)/R₂ = 50/50 = 1
→ R₁ + 10 = R₂ … (i)

Case 2: 10Ω removed, only R₁ in first gap. Balance at 40 cm.
R₁/R₂ = 40/60 = 2/3
→ 3R₁ = 2R₂ … (ii)

From (i): R₂ = R₁ + 10
Substituting in (ii): 3R₁ = 2(R₁ + 10)
3R₁ = 2R₁ + 20
R₁ = 20 Ω

And R₂ = 20 + 10 = 30 Ω

Q.17 — AC Circuits

Part (a): Graph-Based Questions (3 marks)

From the AC current vs time graph (I vs t):

i. Current at t = 5 ms:
From the graph, at t = 5 ms the current appears to be at its maximum — reading the graph: I = 3 A (peak value based on graph scale showing I(A) axis at 3)

ii. Period T:
The graph shows one complete cycle between t = 0 and t = 20 ms.
T = 20 ms = 0.02 s

iii. Frequency:
f = 1/T = 1/0.02 = 50 Hz

Part (b): Peak Value and RMS Value (2 marks)

Peak value (I₀): The maximum value of the alternating current in either direction during one complete cycle. It is the amplitude of the AC waveform.

Relationship with RMS (I_rms):
I_rms = I₀/√2

This means the RMS value is approximately 0.707 times the peak value. The RMS value represents the equivalent DC current that produces the same heating effect.

Similarly for voltage: V_rms = V₀/√2

Q.18 — Semiconductor Electronics

Part (a): Depletion Layer and Potential Barrier (2 marks)

Depletion Layer: When a P-N junction is formed, electrons from the N-side diffuse to the P-side and holes from the P-side diffuse to the N-side. This leaves a region near the junction depleted of free charge carriers. This region is called the depletion layer (width ~10⁻⁶ m).

Potential Barrier: The diffusion of charge carriers creates positive ions on the N-side and negative ions on the P-side near the junction. This charge separation creates an electric field that opposes further diffusion. The corresponding potential difference is called the potential barrier (~0.3V for Ge, ~0.7V for Si).

Part (b): Truth Table for the Given Digital Circuit (2 marks)

The circuit shows an OR gate followed by a NOT gate (NOR gate combination), and another element producing Y₂.

From the circuit: A and B → OR gate → output Y (A+B) → NOT gate → Y₁ = NOT(A+B) = NOR

Part (c): Gate Identification (1 mark)
The combination produces a NOR gate output at Y₁ (NOT of OR).

Q.19 — Atomic Physics

Part (a): Ionization Potential (1 mark)
Ionization potential is the minimum potential that must be applied to accelerate an electron so that it gains enough energy to completely remove another electron from an atom in its ground state.

Part (b)(i): Potential Energy in First Excited State (1 mark)

For hydrogen atom: Total Energy (E) = KE + PE
Also: KE = −E (for circular orbit) and PE = 2E

Given E = −3.4 eV:
PE = 2E = 2 × (−3.4) = −6.8 eV

Note: KE = +3.4 eV, PE = −6.8 eV, Total E = −3.4 eV. This satisfies E = KE + PE.

Part (b)(ii): Wavelength of Emitted Radiation (3 marks)

Electron jumps from n = 2 (first excited state) to n = 1 (ground state).

Using Rydberg formula:
1/λ = R(1/n₁² − 1/n₂²) = 1.097 × 10⁷ × (1/1² − 1/2²)
= 1.097 × 10⁷ × (1 − 1/4)
= 1.097 × 10⁷ × 3/4
= 8.228 × 10⁶ m⁻¹

λ = 1/8.228 × 10⁶ = 1.215 × 10⁻⁷ m = 121.5 nm (Lyman series, UV)

OR Q.19 — Radioactivity

Part (a)(i): Decay Constant (1 mark)
Decay constant (λ) is the fraction of atoms of a radioactive substance that disintegrates per unit time. It is the probability of decay of a single nucleus per unit time. Unit: s⁻¹

Part (a)(ii): Fraction of Activity After Fourth Half-Life (1 mark)

After each half-life, activity reduces to half. After n half-lives: A = A₀(1/2)ⁿ

After 4th half-life: A/A₀ = (1/2)⁴ = 1/16

Part (b): GM Counter Numerical (3 marks)

Given: N₀ = 895 counts/min, N = 327 counts/min, t = 10 min

Using decay law: N = N₀e^(-λt)
327 = 895 × e^(-10λ)
e^(-10λ) = 327/895 = 0.3654
−10λ = ln(0.3654) = −1.006
λ = 0.1006 min⁻¹ ≈ 0.1 min⁻¹

Half-life: t₁/₂ = ln2/λ = 0.693/0.1006 = 6.89 minutes ≈ 6.9 min

Group C — Long Answer Analysis

Q.20 — Sound Waves and Doppler Effect (OR: Wave Optics)

Part (a): Sound Waves as Pressure Waves (1 mark)
Sound waves are called pressure waves because they propagate through a medium as alternating regions of compression (high pressure) and rarefaction (low pressure). The particles of the medium oscillate to and fro about their equilibrium positions, causing periodic pressure variations that travel as the wave.

Part (b): One Bel and Threshold of Hearing (2 marks)

One Bel: The sound level is said to be one bel when the intensity of sound is 10 times the intensity at the threshold of hearing.
Mathematically: β(bel) = log₁₀(I/I₀)
1 decibel = 1/10 bel, so β(dB) = 10 log₁₀(I/I₀)

Threshold of Hearing: The minimum intensity of sound that a normal human ear can hear is called the threshold of hearing. Its value is: I₀ = 10⁻¹² W/m² at frequency 1000 Hz.

Part (c): Doppler Effect — Listener Moving Towards Stationary Source (2 marks)

When listener moves towards a stationary source with velocity v_L:

The listener encounters more wave crests per second. The apparent wavelength remains λ = v/f₀, but the relative speed of waves with respect to listener is (v + v_L).

Apparent frequency: f’ = f₀(v + v_L)/v

where v = speed of sound, f₀ = original frequency, v_L = speed of listener.

Part (d): Echo Frequency Numerical (3 marks)

Given: Speed of car v_s = 20 m/s, original frequency f₀ = 800 Hz, v_sound = 350 m/s

This problem involves two Doppler shifts:

Step 1 — Frequency of echo striking the cliff (car as source, cliff as observer):
f₁ = f₀(v)/(v − v_s) = 800 × 350/(350 − 20) = 800 × 350/330 = 848.5 Hz

Step 2 — Frequency heard by car driver (cliff as source, car as moving observer):
f₂ = f₁(v + v_s)/v = 848.5 × (350 + 20)/350 = 848.5 × 370/350 = 897.7 Hz ≈ 898 Hz

OR Q.20 — Wave Optics

Part (a): Interference and Conservation of Energy (1 mark)
Yes, interference of light follows conservation of energy. In constructive interference, the intensity is maximum (4I₀), and in destructive interference it is zero. The energy is simply redistributed — not created or destroyed. The average intensity remains 2I₀, equal to what two independent sources would give.

Part (b): Position of nth Order Maxima in Young’s Double Slit Experiment (2 marks)

For constructive interference (maxima), path difference = nλ

If d = slit separation, D = distance to screen, and y = distance of nth maxima from center:
Path difference = d sinθ ≈ d × y/D (for small angles)

For maxima: dy/D = nλ
y_n = nλD/d

where n = 0, ±1, ±2, … (order of maxima)

Part (c): Central Diffraction Peak Width (3 marks)

Given: D = 3.5 m, slit width a = 0.01 mm = 10⁻⁵ m, λ = 500 nm = 500 × 10⁻⁹ m

The first minima of single slit diffraction occur at: sinθ = ±λ/a

Position of first minima: y = ±Dλ/a = ±3.5 × 500×10⁻⁹ / 10⁻⁵ = ±0.175 m

Width of central peak = 2y = 2 × 0.175 = 0.35 m = 35 cm

Part (d): Brewster’s Law (2 marks)

Statement: When unpolarized light is incident at a specific angle (polarizing angle ip) on a surface, the reflected light is completely plane polarized. The polarizing angle is related to the refractive index by: tan(ip) = μ

At Brewster’s angle, the reflected and refracted rays are perpendicular to each other: ip + r = 90°

Proof: From Snell’s law: sin(ip)/sin(r) = μ
Since r = 90° − ip: sin(r) = cos(ip)
Therefore: sin(ip)/cos(ip) = μ → tan(ip) = μ ✓

Q.21 — Electromagnetic Induction and Eddy Currents

Part (a): EMF Induced in Rotating Rectangular Coil (3 marks)

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

At angle θ = ωt from initial position:
Magnetic flux: Φ = NBA cos(ωt)

By Faraday’s law: e = −dΦ/dt = NBAω sin(ωt)

Maximum emf (e₀) = NBAω when ωt = 90°

e = e₀ sin(ωt) = NBAω sin(ωt)

This sinusoidal variation is the basis of AC generation.

Part (b): Eddy Currents (1 mark)
Eddy currents are circulating currents induced in a solid conductor when it is placed in a changing magnetic field. They are called “eddy” currents because they flow in closed loops within the conductor, like eddies in water. They cause energy loss as heat (Joule heating). Applications: induction heating, electromagnetic braking in speedometers. They are minimized by using laminated cores.

Part (c): Current Through Wire in Magnetic Field (3 marks)

Given: Length L = 50 cm = 0.5 m, mass m = 20 g = 0.02 kg, B = 0.6 T

For zero tension in the springs, the magnetic force must balance gravity:
BIL = mg
I = mg/(BL) = (0.02 × 10)/(0.6 × 0.5) = 0.2/0.3 = 0.667 A ≈ 0.67 A

OR Q.21 — Thermoelectricity and Magnetism

Part (a): Seebeck Effect (1 mark)
The Seebeck effect is the phenomenon in which an electromotive force (emf) is generated in a thermocouple when its two junctions are maintained at different temperatures. The emf produced is called thermo-emf and the current produced is called thermoelectric current.

Part (b): Variation of Thermo-emf with Temperature (2 marks)

When cold junction is at 0°C and hot junction temperature (T) increases:

• Thermo-emf increases linearly at first: e = aT + bT²
• Reaches a maximum at neutral temperature (Tₙ)
• Beyond Tₙ, emf decreases
• At temperature of inversion (Ti = 2Tₙ), emf returns to zero
• For T > Ti, emf reverses direction

The graph is a parabola with maximum at Tₙ.

Part (c): Magnetic Field Inside Solenoid Using Ampere’s Law (3 marks)

Consider a long solenoid with n turns per unit length carrying current I. Apply Ampere’s circuital law to a rectangular loop ABCD of length l:

∮B·dl = μ₀I_enclosed

• Along AB (inside, parallel to B): B·l
• Along BC and DA (perpendicular to B): 0
• Along CD (outside, B≈0): 0

Total: B·l = μ₀ × (nl) × I
B = μ₀nI

For N turns in length L: n = N/L → B = μ₀NI/L

Part (d): Alpha Particle Acceleration (2 marks)

Given: m = 6.65 × 10⁻²⁷ kg, v = 6 × 10⁶ m/s, B = 0.2 T, charge of alpha = 2e = 3.2 × 10⁻¹⁹ C

Magnetic force (centripetal): F = qvB = 3.2×10⁻¹⁹ × 6×10⁶ × 0.2 = 3.84 × 10⁻¹³ N

Acceleration: a = F/m = 3.84×10⁻¹³ / 6.65×10⁻²⁷ = 5.77 × 10¹³ m/s²

Q.22 — Thomson’s Method and Photoelectric Effect

Part (a)(i): Why Electric Field Perpendicular to Magnetic Field in Thomson’s Method? (1 mark)
In Thomson’s method, the electric and magnetic fields are kept perpendicular to each other so that when both fields are applied simultaneously, the deflections they cause on the electron beam are in opposite directions. By adjusting the fields until the beam is undeflected (E and B forces balance), the condition qE = qvB gives the velocity v = E/B directly without knowing the charge.

Part (a)(ii): Ratio of Electric to Magnetic Field (1 mark)
When the beam travels undeflected: Electric force = Magnetic force
qE = qvB → E/B = v (velocity of electrons)

The ratio E/B gives the speed of the electron beam.

Part (b): Speed and Period of Electron in Circular Path (3 marks)

Given: r = 20 cm = 0.2 m, B = 2 × 10⁻³ T, m = 9.1 × 10⁻³¹ kg, e = 1.6 × 10⁻¹⁹ C

For circular motion in magnetic field: qvB = mv²/r → v = qBr/m

v = (1.6×10⁻¹⁹ × 2×10⁻³ × 0.2)/(9.1×10⁻³¹)
v = (6.4×10⁻²³)/(9.1×10⁻³¹)
v = 7.03 × 10⁷ m/s

Period: T = 2πr/v = 2π × 0.2/(7.03×10⁷) = 1.787 × 10⁻⁸ s ≈ 1.79 × 10⁻⁸ s

Part (c): Stopping Potential vs Frequency Graph (1 mark)

The graph shows two straight parallel lines for materials A and B:

• Both lines have the same slope (h/e — Planck’s constant over electron charge)
• Material A has threshold frequency f₀ > f₀’ (material B’s threshold)
• Material A line is shifted right (higher threshold = higher work function)

i. Which has higher work function? (1 mark)
Material A has higher work function φ = hf₀. Since f₀ > f₀’, φ_A > φ_B.

ii. What does intercept of Graph A on potential axis give? (1 mark)
The intercept on the stopping potential (V₀) axis represents −φ_A/e — the negative of the work function of material A divided by electron charge. This is the stopping potential axis intercept and gives the work function information.

Marks Weightage Summary

Difficulty Analysis

Easy — Score confidently:

• Q.1, Q.3, Q.6, Q.9, Q.11 (MCQs with clear concepts)
• Q.12(a)(b) (definitions)
• Q.14(a)(i) (conceptual)
• Q.15(a)(b) (1-mark justifications)
• Q.17(a) (graph reading)
• Q.18(a) (definitions)
• Q.20(a)(b) (definitions)

Moderate — Needs regular practice:

• Q.4, Q.7, Q.8 (MCQs requiring calculation)
• Q.12(c) (rotational kinematics)
• Q.13(b) (Bernoulli numerical)
• Q.16(b) (meter bridge numerical)
• Q.21(a) (EMF derivation)
• Q.22(a) (Thomson’s method concepts)

Challenging — For top scorers:

• Q.5 (P-V diagram work calculation)
• Q.14(a)(ii) (adiabatic work derivation)
• Q.15(c) (organ pipe modes — full derivation)
• Q.19(b)(ii) (Rydberg formula wavelength)
• Q.19 OR (b) (GM counter decay constant)
• Q.20(d) (double Doppler echo)
• Q.21 OR (c) (solenoid field derivation)
• Q.22(b) (electron circular motion)

Student Preparation Tips

1. Master the Derivations — They’re Worth 2–3 Marks Each
Key derivations that appear repeatedly across NEB Grade XII papers: EMF in rotating coil, frequency of mass-spring system, adiabatic work expression, Bernoulli’s equation, position of nth maxima in YDSE, magnetic field in solenoid, and Rydberg formula applications. Learn them with understanding — not memorization.

2. Group A is Pure Profit
11 marks at 1 mark each — this section alone can be the difference between passing and failing. Practice all MCQ types: Doppler effect calculations, spring period scaling, moment of inertia addition, efficiency of heat engines, Brewster’s angle calculations, and magnetic force calculations. Aim for 9/11 minimum.

3. Waves and Sound (Q.15, Q.20) = 14 Marks Combined
This is the single largest topic block in this paper. Master: organ pipe harmonics (open and closed), sonometer frequency, Doppler effect derivations, sound level in decibels, and echo calculations. Practice at least 10 numerical problems from each.

4. Electromagnetic Induction (Q.17, Q.21) = 15 Marks Combined
AC circuits, EMF induction, eddy currents, and magnetic force on current-carrying wire — this block is equally critical. Learn the EMF derivation perfectly and practice numerical problems on meter bridge, Wheatstone bridge, and magnetic force.

5. Always Show Working in Numericals
NEB awards partial marks for correct method even if the final answer is wrong. Always write the formula, substitute values with units, and circle the final answer. Never skip intermediate steps.

6. Prepare Both Options
Q.12 (Rotational/SHM), Q.20 (Sound/Optics), Q.21 (Induction/Thermoelectricity), Q.19 (Atomic/Radioactivity) — all have OR options. Prepare both for each, then choose the one you’re more confident with on exam day.

7. Modern Physics is Predictable
Q.22 on Thomson’s method and photoelectric effect appears in virtually identical form across years. Learn: specific charge of electron, E/B = v condition, stopping potential vs frequency graph interpretation, and circular motion of charged particle in magnetic field.

8. Time Management
Group A: 15 minutes. Group B: 80 minutes (10 min per question). Group C: 60 minutes (20 min per question). Revision: 25 minutes.

Frequently Asked Questions (FAQs)

Q: What chapters are most important for NEB Grade XII Physics?
A: Based on this paper and past trends, the highest-priority chapters are: Rotational Dynamics, Thermodynamics, Waves and Acoustics, Wave Optics, Electromagnetic Induction, AC Circuits, Semiconductor Electronics, Atomic Physics, and Radioactivity. Together these cover over 85% of the marks.

Q: How many derivations should I prepare for Grade XII Physics?
A: At minimum, prepare these 15 key derivations: (1) EMF in rotating coil, (2) mass-spring SHM frequency, (3) adiabatic work expression, (4) Bernoulli’s equation, (5) position of nth maxima in YDSE, (6) diffraction peak width, (7) magnetic field in solenoid using Ampere’s law, (8) Doppler formula for moving listener/source, (9) expression for equivalent capacitance in series/parallel, (10) expression for equivalent resistance in series/parallel, (11) decay law N = N₀e^(-λt), (12) Rydberg formula application, (13) thermo-emf variation with temperature, (14) frequency in closed/open organ pipe, and (15) sonometer frequency formula.

Q: What is the difference between Grade XI and Grade XII Physics difficulty?
A: Grade XII Physics (New Course) adds significant new chapters — rotational dynamics, thermodynamics beyond basics, wave optics, electromagnetic induction, AC circuits, semiconductor electronics with digital logic, atomic physics, radioactivity, and solid state physics. The mathematical complexity is also higher, with more multi-step derivations and numericals.

Q: How do I handle the digital electronics questions (Q.18)?
A: For truth tables, systematically try all 4 input combinations (00, 01, 10, 11) for two inputs A and B. For gate identification, compare your truth table with standard gates: AND, OR, NAND, NOR, XOR, XNOR. Practice drawing and recognizing all basic logic gates.

Q: The Doppler echo problem in Q.20(d) seems complicated — is there a simpler approach?
A: Break it into two steps: (1) Treat the car as a source and cliff as a stationary observer — find the frequency hitting the cliff. (2) Treat the cliff as a stationary source and the car as a moving observer — find the frequency heard. Apply the correct Doppler formula for each step. Practice at least 5 echo problems.

Q: Can I score above 60 in this paper with two months of preparation?
A: Yes — with focused preparation. Week 1–2: Master MCQ concepts and short definitions. Week 3–4: Learn all derivations and short answer numerical methods. Week 5–6: Practice long answer questions and past papers. Week 7–8: Full paper practice under timed conditions, identify and strengthen weak areas.

Conclusion

The NEB Grade XII Physics 2081 (2024) Set D paper is a comprehensive, well-balanced examination that tests students across the full breadth of the new curriculum. Its strength lies in the diversity of topics — from the classical mechanics of rotating discs to the quantum physics of hydrogen atom energy levels, from sound waves bouncing off cliffs to electrons spinning in magnetic fields.

The paper rewards students who have built genuine conceptual understanding rather than those who have merely memorized formulas. The MCQs test concept clarity, the Group B questions test application, and the Group C questions test depth of understanding and ability to connect ideas.

Three things that separate high scorers from average scorers in this paper: mastery of derivations, accuracy in numerical calculations, and speed in MCQs. Build all three systematically, and 60+ marks in Grade XII Physics is very achievable.

Keep visiting nneducation.com for more past papers, model questions, concept notes, and numerical practice sets for NEB Grade XI and XII Physics.