NEB Grade XII Physics 2080 (2023) Set P — Complete Question Paper Analysis, Answer Guide & Exam Preparation

Subject Code: 1021’P’ | Full Marks: 75 | Time: 3 Hours | New Course
Downloaded from: nneducation.com

Introduction

The NEB Grade XII Physics 2080 BS (2023 AD) Set P paper holds a special place in Nepal’s examination history — it is one of the earliest papers under the new Grade XII curriculum, making it an essential reference for all subsequent years. Comparing this paper with the 2081 Set D paper reveals how NEB’s question pattern evolved, what topics remain perennial, and which new areas gained emphasis.

This Set P paper is notably different from Set D in several key ways: it introduces Stokes’ law and terminal velocity (Q.13), entropy and refrigerator thermodynamics (Q.14), Huygens’ principle (Q.15 OR), de-Broglie wavelength (Q.16), Kirchhoff’s laws and Wheatstone bridge (Q.19), LCR circuits (Q.22 OR), and force between parallel current-carrying conductors (Q.22). These topics do not appear in Set D — giving students who study both papers a far more complete preparation.

The paper spans rotational dynamics, SHM, viscosity, thermodynamics, wave optics, atomic physics, semiconductor electronics, electromagnetic induction, sound waves, photoelectric effect, radioactivity, and magnetic force on current-carrying conductors. This truly is a full-syllabus examination.

Exam Overview

Full Question Paper

Comparison: Set P (2080) vs Set D (2081)

Topics Covered — Chapter-wise Weightage

Group A — MCQ Analysis with Answers

Q.1 — Rotational Analogue of Force
Answer: a. Torque

In linear motion: Force causes linear acceleration (F = ma).
In rotational motion: Torque causes angular acceleration (τ = Iα).
Torque is the rotational analogue of force. Moment of inertia is the rotational analogue of mass.

Q.2 — Time to Move from Mean to Half Amplitude in SHM
Answer: c. T/6

In SHM: x = A sin(ωt). For x = A/2:
A/2 = A sin(ωt) → sin(ωt) = 1/2 → ωt = π/6
t = π/(6ω) = π/(6 × 2π/T) = T/12

Wait — from mean position (x=0) to x = A/2:
At mean position: x = 0 = A sin(ωt₁) → t₁ = 0
At x = A/2: A/2 = A sin(ωt₂) → ωt₂ = π/6 → t₂ = T/12

Time taken = T/12 — Answer: d. T/12

Common trap: Many students choose T/6, thinking of the quarter period. The correct answer from displacement equation gives T/12.

Q.3 — Capillary Rise Ratio
Answer: b. 2:1

Capillary rise: h = 2T cosθ/(rρg) → h ∝ 1/r

h_A/h_B = r_B/r_A = 4/2 = 2:1

The narrower tube (A, r = 2mm) has greater rise than wider tube (B, r = 4mm).

Q.4 — Work Done in P-V Cycle
Answer: d. Zero

From the P-V diagram: The cycle goes from A to B (isobaric at 6P₁) to C (isothermal) and back to A. This is a closed cycle. However, looking carefully at the diagram — point A is at (V₁, P₁) → B at (V₁, 6P₁) → C at (3V₁, 6P₁) following path back.

Actually for the triangular cycle shown: W = area of triangle = ½ × base × height = ½ × (3V₁ − V₁) × (6P₁ − P₁) = ½ × 2V₁ × 5P₁ = 5P₁V₁

Answer: b. 5P₁V₁

Q.5 — Carnot Engine Energy Supplied
Answer: a. 400J

Efficiency of Carnot engine: η = 1 − T_cold/T_hot = 1 − 300/600 = 0.5 = 50%

η = W/Q_in → Q_in = W/η = 200/0.5 = 400 J

Q.6 — Distance Traversed in 36 Vibrations
Answer: a. 33m

Time for 36 vibrations: t = 36/f = 36/384 s

Distance = v × t = 352 × 36/384 = 352 × 0.09375 = 33 m

Q.7 — Refractive Index When Reflected and Refracted Rays are Perpendicular
Answer: d. √3

When reflected and refracted rays are perpendicular to each other, the angle of incidence equals Brewster’s angle.

At Brewster’s angle: i + r = 90° → r = 90° − 60° = 30°

By Snell’s law: μ = sin(i)/sin(r) = sin(60°)/sin(30°) = (√3/2)/(1/2) = √3

Q.8 — EMF from Temperature Difference in Dissimilar Metals
Answer: b. Seebeck effect

The Seebeck effect is the generation of an emf in a thermocouple when its two junctions are at different temperatures. Peltier effect is the opposite (current causes temperature difference). Joule’s effect is heating due to current. Thomson’s effect is heating/cooling in a single conductor with temperature gradient and current.

Q.9 — Susceptibility of Ferromagnet When Temperature Decreases
Answer: a. Increases

For ferromagnetic substances, magnetic susceptibility (χ) follows the Curie-Weiss law: χ = C/(T − θ) where θ is Curie temperature. As T decreases (but stays above Curie temperature), susceptibility increases — the material becomes more easily magnetized.

Q.10 — Induced EMF for Flux Changing at 1 Weber per Minute
Answer: b. 1/60 Volt

By Faraday’s law: e = −dΦ/dt

dΦ/dt = 1 Weber/minute = 1/60 Weber/second

e = 1/60 Volt

Key trap: “per minute” must be converted to “per second” for SI unit answer.

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

P-waves (Primary/Pressure waves) are longitudinal waves. They travel fastest (~6–8 km/s in crust) and are therefore first recorded on seismographs. S-waves arrive second, surface waves (Love, Rayleigh) arrive last.

Group B — Short Answer Analysis

Q.12 — Angular Momentum and Motor Deceleration (OR: SHM)

Part (a): Angular Momentum Definition and SI Unit (1+1 marks)

Definition: Angular momentum (L) of a rotating body is defined as the product of its moment of inertia (I) and angular velocity (ω).
L = Iω

It is also the rotational analogue of linear momentum (p = mv).

SI Unit: kg·m²·s⁻¹ (or J·s)

Part (b): Motor Engine Deceleration Numerical (3 marks)

Given: ω₀ = 900 rev/min = 900 × 2π/60 = 30π rad/s
ω = 600 rev/min = 600 × 2π/60 = 20π rad/s, t = 10 s

i. Angular acceleration:
α = (ω − ω₀)/t = (20π − 30π)/10 = −π rad/s² ≈ −3.14 rad/s²
(negative sign indicates deceleration)

ii. Number of revolutions in 10 seconds:
θ = (ω₀ + ω)/2 × t = (30π + 20π)/2 × 10 = 25π × 10 = 250π rad

Number of revolutions = 250π/(2π) = 125 revolutions

iii. Additional time to come to rest:
From ω = 600 rev/min to 0, with same deceleration α = −π rad/s²:
0 = 20π + (−π) × t → t = 20 seconds

OR Q.12 — Simple Pendulum and SHM

Part (a): Acceleration Proportional to Displacement in Simple Pendulum (2 marks)

For a pendulum bob displaced by angle θ (small):
The restoring force: F = −mg sinθ ≈ −mgθ (for small θ)

Since sinθ ≈ θ = x/L (where x is arc displacement, L is length):
F = −mgx/L

By Newton’s 2nd law: ma = −mgx/L
a = −(g/L)x

Since g and L are constants: a ∝ −x

Acceleration is directly proportional to displacement and directed opposite to it — this is the defining condition of SHM. ✓

Part (b): SHM Numerical — Bob of Mass 8 kg (3 marks)

Given: m = 8 kg, A = 30 cm = 0.3 m, F_restoring = 60 N (at amplitude)

Spring constant: k = F/A = 60/0.3 = 200 N/m

i. Time period:
T = 2π√(m/k) = 2π√(8/200) = 2π√(0.04) = 2π × 0.2 = 1.257 s ≈ 1.26 s

ii. Maximum acceleration:
a_max = kA/m = 200 × 0.3/8 = 60/8 = 7.5 m/s²

iii. Kinetic energy at displacement x = 12 cm = 0.12 m:
KE = ½k(A² − x²) = ½ × 200 × (0.3² − 0.12²)
= 100 × (0.09 − 0.0144) = 100 × 0.0756 = 7.56 J

Q.13 — Viscosity: Stokes’ Law and Terminal Velocity

Part (a): Free Body Diagram of Falling Sphere (1 mark)

Forces on a sphere falling through viscous liquid:

↓ Weight (W = mg = 4/3 πr³ρ’g) — downward

↑ Upthrust/Buoyancy (U = 4/3 πr³σg) — upward

↑ Viscous drag (F = 6πηrv) — upward (opposes motion)

Where ρ’ = density of sphere, σ = density of liquid, η = coefficient of viscosity.

Part (b): Velocity-Time Graph (1 mark)

The graph starts at v = 0 when the ball enters the liquid. Velocity increases as weight dominates. As velocity increases, viscous drag increases. Eventually when W = U + F, the ball reaches terminal velocity (constant). The graph shows a curve that rises and then flattens to a horizontal line (terminal velocity).

Shape: Like a charging capacitor curve — rises steeply initially, then asymptotically approaches a constant value.

Part (c): Terminal Velocity Expression (3 marks)

At terminal velocity, net force = 0:
Weight = Upthrust + Viscous drag
(4/3)πr³ρ’g = (4/3)πr³σg + 6πηrv_t

6πηrv_t = (4/3)πr³g(ρ’ − σ)

v_t = 2r²(ρ’ − σ)g / 9η

This is Stokes’ Law expression for terminal velocity.

Key features:

• v_t ∝ r² (larger radius → much faster terminal velocity)
• v_t ∝ (ρ’ − σ) (denser object in less dense liquid → faster)
• v_t ∝ 1/η (more viscous fluid → slower terminal velocity)

Q.14 — Thermodynamics: Entropy and Adiabatic Compression

Part (a): Entropy and First Law in Terms of Entropy (1+1 marks)

Entropy: Entropy is a thermodynamic state function that measures the degree of disorder or randomness of a system. It is defined as: dS = dQ/T (for reversible process). Its SI unit is J/K.

First Law in terms of entropy:
First Law: dQ = dU + dW = dU + PdV
Since dS = dQ/T → dQ = TdS

Therefore: TdS = dU + PdV

This is the fundamental thermodynamic relation combining the first and second laws.

Part (b): Can a Room Be Cooled by Open Refrigerator? (1 mark)

No. A refrigerator transfers heat from inside the cabinet to the room. When the door is opened, the heat extracted from inside equals the heat rejected to the room PLUS the work done by the compressor motor. The net effect is that more heat is added to the room than removed — so the room temperature actually increases. The refrigerator acts as a heat pump dumping heat into the room.

Part (c): Adiabatic Compression Temperature Change (2 marks)

Given: T₁ = 30°C = 303 K, V₂/V₁ = 8/27, γ = 5/3

For adiabatic process: T₁V₁^(γ-1) = T₂V₂^(γ-1)

T₂ = T₁(V₁/V₂)^(γ-1) = 303 × (27/8)^(5/3 – 1) = 303 × (27/8)^(2/3)

(27/8)^(2/3) = (3/2)² = 9/4 = 2.25

T₂ = 303 × 2.25 = 681.75 K ≈ 682 K

Change in temperature: ΔT = 682 − 303 = 379 K

Q.15 — Wave Optics: YDSE and Diffraction

Part (a)(i): YDSE — Shift from Air to Water (1 mark)

Fringe width: β = λD/d

In water, wavelength decreases: λ_water = λ_air/μ_water (where μ_water ≈ 1.33)

Therefore fringe width decreases by a factor of μ_water when experiment is shifted to water.

β_water = β_air/μ_water ≈ β_air/1.33

Part (a)(ii): YDSE — When Slit Width is Increased (1 mark)

Fringe width β = λD/d depends on slit separation (d), not slit width. Increasing slit width means more light enters, so the fringes become brighter. However, the fringe width formula gives β = λD/d — if d (separation) is unchanged, fringe width remains unchanged.

Note: Increasing slit width does NOT change fringe width if slit separation d is unchanged. It increases brightness and reduces single-slit diffraction spread.

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

Part (b)(ii): Width of Central Diffraction Peak (2 marks)

Given: D = 4 m, slit width a = 0.01 m (= 1 cm — this seems large for diffraction; assuming 0.01 mm = 10⁻⁵ m), λ = 600 nm = 600 × 10⁻⁹ m

Width of central maximum = 2Dλ/a = 2 × 4 × 600×10⁻⁹ / 10⁻⁵
= 2 × 4 × 0.06 = 0.48 m = 48 cm

If a = 0.01 m: width = 2 × 4 × 600×10⁻⁹/0.01 = 4.8 × 10⁻⁴ m = 0.48 mm

OR Q.15 — Huygens’ Principle

Part (a): Spherical to Plane Wavefront (2 marks)

A spherical wavefront from a point source can be converted to a plane wavefront by placing a convex lens in the path, with the point source at the principal focus of the lens.

When light from a point source at F diverges as a spherical wavefront, after passing through the convex lens, all rays become parallel — emerging as a plane wavefront.

Diagram: Point source → diverging spherical wavefront → convex lens → parallel plane wavefront.

Part (b): Huygens’ Principle to Verify Law of Reflection (3 marks)

Consider a plane wavefront AB incident at angle i on a reflecting surface XY.

Using Huygens’ construction:

• When end B of wavefront reaches the mirror at B’, end A is still at distance AA’ = v·t from mirror
• Draw secondary wavelets from B’ of radius vt
• Draw tangent from A’ to this wavelet — this gives reflected wavefront A’C

From the geometry:

• In triangles AA’B’ and CB’A’: AA’ = CB’ = vt (same speed)
• A’B’ is common
• Both triangles are congruent

Therefore: angle of incidence = angle of reflection (∠i = ∠r) ✓

This proves the law of reflection using Huygens’ principle.

Q.16 — Atomic Physics and de-Broglie Wavelength

Part (a): Bohr’s Postulates (2 marks)

Postulate 1 — Stationary Orbits: An electron can revolve around the nucleus only in certain discrete circular orbits called stationary orbits, without radiating energy.

Postulate 2 — Angular Momentum Quantization: The angular momentum of an electron in a stationary orbit is an integral multiple of h/2π:
mvr = nh/2π (where n = 1, 2, 3,… is the principal quantum number)

Postulate 3 — Energy Emission/Absorption: An electron can jump from a higher energy orbit to a lower energy orbit by emitting a photon, or absorb a photon to jump to a higher orbit. The energy of the photon: hf = E₂ − E₁

Part (b)(i): de-Broglie Wavelength of Neutron (2 marks)

Given: KE = 150 eV = 150 × 1.6 × 10⁻¹⁹ = 2.4 × 10⁻¹⁷ J
m_neutron = 1.675 × 10⁻²⁷ kg, h = 6.62 × 10⁻³⁴ J·s

de-Broglie wavelength: λ = h/p = h/√(2mKE)

λ = 6.62×10⁻³⁴ / √(2 × 1.675×10⁻²⁷ × 2.4×10⁻¹⁷)
= 6.62×10⁻³⁴ / √(8.04×10⁻⁴⁴)
= 6.62×10⁻³⁴ / 2.835×10⁻²²
= 2.335 × 10⁻¹² m ≈ 2.34 × 10⁻¹² m

Part (b)(ii): Proton vs Electron — Which Has Greater de-Broglie Wavelength? (1 mark)

λ = h/p = h/√(2mKE)

For same KE: λ ∝ 1/√m

Since m_electron << m_proton: electron has greater de-Broglie wavelength for the same kinetic energy.

Q.17 — Semiconductor Electronics

Part (a): Depletion Layer Decreases in Forward Bias (2 marks)

In a P-N junction, the depletion layer forms due to diffusion of majority carriers across the junction, creating a potential barrier.

In forward biasing, the positive terminal of battery is connected to P-side and negative to N-side. This:

• Pushes holes in P-region towards the junction
• Pushes electrons in N-region towards the junction
• Both carriers enter the depletion region and neutralize the immobile ions
• The depletion layer narrows and the potential barrier decreases

With sufficient forward voltage (0.7V for Si), the barrier is overcome and current flows freely.

Part (b): Truth Table for Logic Gate Combination (2 marks)

From the circuit: A → AND gate (with B) → Y₁, also A → OR gate (with B) → then NOT → Y₂, and Y₁ → combined with Y₂ → final output Y.

The circuit appears to be NAND + NOR combination. Standard truth table:

Part (c): Gate Identification (1 mark)
The final output Y = AND gate (output is 1 only when both A=1 and B=1… wait — from the table Y=1 only when both inputs are 0). This corresponds to NOR gate behavior for input combination 00.

Looking at the circuit diagram more carefully with the two gates shown: The combination produces an XNOR gate or specific combination — students should match their truth table with standard gates.

Q.18 — Electromagnetic Induction: Self-Inductance

Part (a): Self-Inductance of a Coil (1 mark)
Self-inductance is the property of a coil by which it opposes any change in the magnetic flux through it (and hence any change in current through it) by inducing an emf in itself. Numerically, the self-inductance equals the magnetic flux per unit current: L = NΦ/I, or equivalently, the emf induced per unit rate of change of current: e = −L(dI/dt).

Part (b): Direction of Induced Current in the Loop (2 marks)

As current from A to B increases, the magnetic flux through the loop (on the right side) due to the increasing current in the wire increases.

By Lenz’s law, the induced current in the loop must oppose this increase — it must create a magnetic field opposing the increasing flux.

Using the right-hand rule: The induced current flows counterclockwise in the loop (as viewed from the direction the flux is increasing) to oppose the increasing flux.

Part (c): Self-Inductance Calculation (2 marks)

Given: N = 100 turns, I = 2 A, Φ = 5 × 10⁻³ Weber (flux per turn)

Total flux linkage: NΦ = 100 × 5 × 10⁻³ = 0.5 Wb

Self-inductance: L = NΦ/I = 0.5/2 = 0.25 H

Q.19 — Wheatstone Bridge Using Kirchhoff’s Laws

Part (a): Balanced Condition of Wheatstone Bridge (3 marks)

A Wheatstone bridge has four resistances P, Q, R, S in four arms with a battery E and galvanometer G.

Applying Kirchhoff’s laws:

At balance, galvanometer current Ig = 0.

At junction B: I₁ (through P) continues through R
At junction D: I₂ (through Q) continues through S

Applying Kirchhoff’s voltage law:

• Loop ABDA: I₁P − I₂Q = 0 → I₁P = I₂Q … (i)
• Loop BCDB: I₁R − I₂S = 0 → I₁R = I₂S … (ii)

Dividing (ii) by (i): R/S = P/Q (wait — let me rewrite correctly)

Dividing equation (i) by (ii): P/R = Q/S

Cross-multiplying: P/Q = R/S OR PS = QR

This is the balanced condition of Wheatstone bridge.

Part (b)(i): Balance Point Shift When R Increases (1 mark)

At balance: P/Q = R/S → balance point l/(100−l) = R/S

If R increases, to maintain balance, the ratio changes. The balance point shifts towards B (i.e., the jockey must move towards B to increase the effective ratio on that side).

Part (b)(ii): Balance Point Shift When S Decreases (1 mark)

If S decreases (R constant): R/S increases. To maintain balance, the galvanometer reading changes. The balance point shifts towards A (reducing the length on the P side to compensate).

Group C — Long Answer Analysis

Q.20 — Sound Waves: Organ Pipes and Doppler Effect

Part (a): Define an Organ Pipe (1 mark)
An organ pipe is a wind instrument that produces musical sound by vibration of air columns inside a cylindrical pipe. The air column vibrates when air is blown through the pipe. Depending on whether both ends are open (open organ pipe) or one end is closed (closed organ pipe), different modes of vibration and harmonics are produced.

Part (b): Ratio of Lengths — Open vs Closed Pipe in Third Overtone (2 marks)

Open organ pipe — harmonics: f_n = nv/2L_A (both ends have antinodes)
Third overtone = 4th harmonic: f = 4v/2L_A = 2v/L_A

Closed organ pipe — only odd harmonics: f_n = nv/4L_B (n = 1, 3, 5…)
Third overtone = 7th harmonic: f = 7v/4L_B

For resonance, both frequencies are equal:
2v/L_A = 7v/4L_B
L_A/L_B = 2 × 4/7 = 8/7

The ratio of lengths: L_A : L_B = 8 : 7

The open pipe is slightly longer for the same resonant frequency in third overtone.

Part (c)(i): Does Pitch Change When Source Moves? (1 mark)

Yes. When the source of sound moves towards a stationary listener, the sound waves are compressed in the direction of motion — the wavelength decreases and frequency appears higher to the listener. The pitch (apparent frequency) increases compared to the actual frequency of the source.

Part (c)(ii): Doppler Expression for Moving Source, Stationary Listener (3 marks)

When source moves with velocity v_s towards stationary listener:

During one time period T, the source emits one complete wave. But the source also moves forward by v_s × T.

Apparent wavelength: λ’ = λ − v_sT = (v − v_s)/f₀

Apparent frequency: f’ = v/λ’ = v/[(v − v_s)/f₀]

f’ = f₀ × v/(v − v_s)

When source moves away: f’ = f₀ × v/(v + v_s)

This shows: moving source → apparent frequency changes → pitch changes ✓

Part (d): Properties of Medium for Sound Propagation (1 mark)

Two necessary properties:

1. Inertia: The medium must have mass (inertia) so that particles can undergo oscillation and store kinetic energy.
2. Elasticity: The medium must have elasticity (restoring property) so that displaced particles return to equilibrium, enabling wave propagation.

Sound cannot travel through vacuum because vacuum has neither mass nor elasticity.

Q.21 — Photoelectric Effect (OR: Radioactivity)

Part (a): Einstein’s Photoelectric Equation in Terms of Stopping Potential (1 mark)

Einstein’s photoelectric equation: hf = φ + KE_max

Where KE_max = eV₀ (e = electron charge, V₀ = stopping potential)

hf = φ + eV₀

Or: eV₀ = hf − φ (in terms of stopping potential)

Part (a)(i): Determining Planck’s Constant from Graph (2 marks)

The V₀ vs f graph is a straight line with:

• Slope = h/e
• X-intercept = threshold frequency f₀
• Y-intercept = −φ/e (negative work function/e)

Planck’s constant: h = e × slope of the graph

Slope = ΔV₀/Δf (can be read from graph)
Multiply by e = 1.6 × 10⁻¹⁹ C to get h in J·s.

Part (a)(ii): What Happens to Stopping Potential If Frequency Increases? (2 marks)

From eV₀ = hf − φ:
V₀ = (h/e)f − φ/e

Since h, e, φ are constants, V₀ is a linear function of frequency f.

When frequency of incident photon increases gradually:

• Stopping potential increases linearly
• The graph V₀ vs f is a straight line with positive slope
• More energetic photons give electrons more kinetic energy, requiring higher stopping potential to prevent them from reaching the anode

Part (b)(i): Why Electron Beam Remains Undeflected (1 mark)

The electric force (F_E = eE) and magnetic force (F_B = evB) act on the electron in opposite directions. When these two forces balance each other exactly (F_E = F_B), the net force is zero and the beam travels undeflected in a straight line. This is the principle of the velocity selector.

Part (b)(ii): Velocity of Electron (2 marks)

Given: V = 50V, d = 3 cm = 0.03 m, B = 10⁻⁴ T

Electric field: E = V/d = 50/0.03 = 1667 V/m

At equilibrium: eE = evB → v = E/B = 1667/10⁻⁴ = 1.667 × 10⁷ m/s

OR Q.21 — Radioactivity

Part (a): Define Radioactivity (1 mark)
Radioactivity is the spontaneous disintegration of the nucleus of a heavy atom with the emission of radiation (alpha, beta, or gamma rays) to form a more stable nucleus. It is a nuclear property and is not affected by physical or chemical conditions.

Part (b): Comparison of Disintegration Rates After 18 Hours (3 marks)

Given: N₀ (same for both), t₁/₂(A) = 3 hrs, t₁/₂(B) = 9 hrs, t = 18 hrs

Activity A(t) = λN(t) = λN₀e^(-λt) = A₀e^(-λt) = A₀(1/2)^(t/t₁/₂)

For sample A (t₁/₂ = 3 hrs):
After 18 hrs = 6 half-lives: N_A = N₀(1/2)⁶ = N₀/64
Activity: A_A = λ_A × N_A = (ln2/3) × N₀/64

For sample B (t₁/₂ = 9 hrs):
After 18 hrs = 2 half-lives: N_B = N₀(1/2)² = N₀/4
Activity: A_B = λ_B × N_B = (ln2/9) × N₀/4

Ratio A_A/A_B = [(ln2/3) × N₀/64] / [(ln2/9) × N₀/4]
= [9/(3)] × [4/64] = 3 × (1/16) = 3/16

Therefore: A_A : A_B = 3 : 16

Sample B has higher activity despite slower decay rate because far fewer atoms of A remain.

Part (c): Relation Between Half-Life and Decay Constant (2 marks)

From the decay law: N = N₀e^(-λt)

At t = t₁/₂: N = N₀/2

N₀/2 = N₀e^(-λt₁/₂)
1/2 = e^(-λt₁/₂)
Taking natural log: ln(1/2) = −λt₁/₂
−ln2 = −λt₁/₂
t₁/₂ = ln2/λ = 0.693/λ

This is the fundamental relation between half-life and decay constant.

Part (d): Percentage of Atoms Decayed After 5 Half-Lives (2 marks)

After n half-lives: N = N₀(1/2)ⁿ

After 5 half-lives: N = N₀(1/2)⁵ = N₀/32

Atoms remaining = N₀/32 = 3.125% of original

Atoms decayed = N₀ − N₀/32 = N₀(31/32)

Percentage decayed = 31/32 × 100 = 96.875% ≈ 96.9%

Q.22 — Force Between Parallel Conductors and Rectangular Frame (OR: LCR Circuit)

Part (a)(i): Force Between Two Parallel Current-Carrying Conductors (2 marks)

For two parallel conductors X and Y of length L, carrying currents I₁ and I₂, separated by distance r:

Magnetic field due to X at Y: B₁ = μ₀I₁/(2πr)

Force on conductor Y (length L) in this field: F = I₂LB₁ = I₂L × μ₀I₁/(2πr)

F = μ₀I₁I₂L/(2πr)

Force per unit length: F/L = μ₀I₁I₂/(2πr)

Part (a)(ii): Nature of Force (1 mark)

From the figure, both currents I₁ and I₂ flow in the same direction (both upward).

Parallel currents in the same direction attract each other.

Nature: Attractive force

Using right-hand rule: B field from X at location of Y points into page. Force on Y (I₂ upward in field into page) = I₂L × B → towards X (attractive).

Part (a)(iii): Definition of One Ampere (1 mark)
One ampere is defined as that steady current which, when flowing through each of two infinitely long parallel conductors of negligible cross-section placed one metre apart in vacuum, produces a force of 2 × 10⁻⁷ N per metre of their length between them.

Part (b)(i): Force on Side AB of Rectangular Frame (2 marks)

Given: Frame ABCD in magnetic field B = 4.5 × 10⁻³ T, length AB = 7 cm = 0.07 m, current I = 2.5 A, width BC = 4 cm = 0.04 m

Force on AB: F = BIL = 4.5 × 10⁻³ × 2.5 × 0.07 = 7.875 × 10⁻⁴ N ≈ 7.88 × 10⁻⁴ N

Direction: Using F = IL × B (right-hand rule or Fleming’s left-hand rule)

Part (b)(ii): Why Side BC Experiences No Force? (2 marks)

Side BC is parallel to the magnetic field (the field is directed along the length of BC).

Force on a current-carrying conductor: F = BIL sinθ

When the conductor is parallel to the magnetic field, θ = 0° → sin0° = 0 → F = 0

No force acts on BC because the magnetic field is in the same direction as the current flow in BC (or rather, BC is aligned with B, so no cross product force results).

OR Q.22 — LCR Circuit

Part (a): Impedance of LCR Circuit (1 mark)
Impedance (Z) is the total opposition offered by an LCR circuit to alternating current. It combines resistance (R), inductive reactance (X_L = ωL), and capacitive reactance (X_C = 1/ωC):

Z = √(R² + (X_L − X_C)²)

Unit: Ohm (Ω)

Part (b)(i): Impedance Calculation (2 marks)

Given: L = 60 mH = 60 × 10⁻³ H, C = 0.50 μF = 0.50 × 10⁻⁶ F, R = 300 Ω, f = 50 Hz (from V = 50V, 50Hz)

ω = 2πf = 2π × 50 = 314.16 rad/s

X_L = ωL = 314.16 × 60 × 10⁻³ = 18.85 Ω

X_C = 1/(ωC) = 1/(314.16 × 0.50 × 10⁻⁶) = 1/(1.571 × 10⁻⁴) = 6366 Ω

Z = √(300² + (18.85 − 6366)²) = √(90000 + (−6347.15)²)
= √(90000 + 40285241) ≈ √40375241 ≈ 6354 Ω

At 50 Hz, this circuit is highly capacitive (X_C >> X_L), so impedance is dominated by capacitive reactance.

Part (b)(ii): Resonance in LCR Circuit (2 marks)

Resonance occurs when X_L = X_C, i.e., the inductive and capacitive reactances are equal and cancel each other.

ωL = 1/(ωC)
ω² = 1/(LC)
ω_r = 1/√(LC) (resonant angular frequency)

f_r = 1/(2π√(LC)) (resonant frequency)

At resonance:

• Impedance is minimum: Z = R (purely resistive)
• Current is maximum: I = V/R
• Power factor = 1 (purely resistive circuit)

Part (c)(i): Identify Device X and Its Reactance (2 marks)

The current leads voltage by π/2 (current = I₀ sin(ωt + π/2), voltage = V₀ sinωt)

Device X is a pure capacitor (C)

In a capacitor, current leads voltage by 90°. The reactance of capacitor:
X_C = 1/(ωC) = 1/(2πfC)

Part (c)(ii): V-I Graphs for Capacitor (1 mark)

For a pure capacitor:

• V = V₀ sinωt (sinusoidal)
• I = I₀ sin(ωt + π/2) = I₀ cosωt (cosine — starts at maximum when V = 0)

The current graph is shifted 90° (π/2) ahead of the voltage graph. Both are sinusoidal but the current leads — when voltage is at zero crossing (going up), current is at maximum.

Marks Weightage Summary

Difficulty Analysis

Easy — Score confidently:

• Q.1, Q.8, Q.11 (MCQs — conceptual definitions)
• Q.12(a) (Angular momentum definition)
• Q.13(a)(b) (Free body diagram and graph sketch)
• Q.14(b) (Refrigerator conceptual)
• Q.15(a) (YDSE changes — conceptual)
• Q.18(a) (Self-inductance definition)
• Q.20(a) (Organ pipe definition)
• Q.21(a) (Einstein equation)

Moderate — Needs practice:

• Q.2 (SHM time — careful calculation)
• Q.5 (Carnot engine calculation)
• Q.12(b) (Motor deceleration — three parts)
• Q.15(b)(ii) (Diffraction peak width)
• Q.16(b)(i) (de-Broglie numerical)
• Q.18(c) (Self-inductance calculation)
• Q.20(b) (Organ pipe length ratio)
• Q.21(b)(ii) (Velocity selector calculation)

Challenging — For top scorers:

• Q.4 (P-V diagram work — needs careful graph reading)
• Q.13(c) (Terminal velocity derivation)
• Q.14(c) (Adiabatic compression numerical)
• Q.15 OR (Huygens’ reflection proof)
• Q.19(a) (Kirchhoff’s proof of Wheatstone)
• Q.20(c)(ii) (Doppler derivation — moving source)
• Q.21 OR (b) (Radioactivity rates comparison)
• Q.22 OR (LCR impedance numerical)

Topics Unique to Set P — Extra Study Points

1. Stokes’ Law and Terminal Velocity (Q.13)
Not in Set D. Master the derivation: v_t = 2r²(ρ−σ)g/9η. Practice: (a) identifying forces on falling sphere, (b) sketching v-t graph (exponential rise to terminal velocity), (c) deriving terminal velocity by balancing weight, upthrust and drag.

2. Entropy and Refrigerator Thermodynamics (Q.14)
Entropy definition, dS = dQ/T, and the refrigerator paradox (open door heats room) are important conceptual questions. The adiabatic compression numerical uses TV^(γ-1) = constant.

3. Huygens’ Principle (Q.15 OR)
Learn to apply Huygens’ principle to verify both reflection and refraction laws. Draw clear geometric constructions with wavefronts, incident and reflected/refracted rays.

4. Bohr’s Postulates and de-Broglie (Q.16)
State all three postulates clearly. For de-Broglie: λ = h/p = h/√(2mKE). Remember: for same KE, lighter particle has longer wavelength.

5. Wheatstone Bridge via Kirchhoff (Q.19)
The derivation using Kirchhoff’s current and voltage laws is a 3-mark must-prepare derivation. Know the balanced condition PS = QR and the effect of changing individual resistances on the balance point.

6. Force Between Parallel Conductors (Q.22)
F/L = μ₀I₁I₂/(2πr). Know the direction rules (same direction → attractive, opposite → repulsive) and the definition of 1 ampere from this formula.

7. LCR Circuit (Q.22 OR)
Impedance Z = √(R² + (X_L − X_C)²). Resonance condition ω = 1/√(LC). Capacitor and inductor phase relationships with current.

Preparation Strategy: Combining Set P and Set D

The ideal preparation uses both papers together:

Together, these two papers cover virtually the entire new Grade XII Physics syllabus. A student who masters both papers is fully prepared for the NEB board examination.

Student Preparation Tips

1. Never Skip Derivations
Q.13(c), Q.15 OR (b), Q.19(a), Q.20(c)(ii) — these 3-mark derivations together carry 12 marks. Learn each one step by step with clear diagrams where needed.

2. MCQ Speed Drills
Practice all 11 MCQs under 15 minutes. Common trap areas: Q.2 (T/12, not T/6), Q.3 (h∝1/r), Q.10 (flux per minute, not per second). One mark each — don’t lose them to careless mistakes.

3. Radioactivity Problems are Fully Predictable
Q.21 OR covers three standard radioactivity problems: rates comparison, half-life derivation, percentage decayed. These appear in almost identical form every year. Master all three with examples.

4. Photoelectric Effect Graph Questions
Q.21 always includes a V₀ vs f graph interpretation. Know: slope = h/e, x-intercept = threshold frequency, y-intercept relates to work function. Practice reading values from such graphs.

5. Sound and Organ Pipes (Q.20) = 8 Marks
Organ pipe harmonics, Doppler derivation for both cases (moving source and moving listener), and sound wave properties — this section is highly predictable and rewards well-prepared students.

6. Time Management Table

Frequently Asked Questions

Q: What is the most important difference between Set P (2080) and Set D (2081)?
A: Set P introduces Stokes’ law, Huygens’ principle, Wheatstone bridge derivation, force between parallel conductors, LCR circuits, entropy, and de-Broglie wavelength — none of which appear in Set D. Set D introduces Bernoulli’s equation, sonometer, organ pipe modes in detail, Thomson’s method, and eddy currents. Together they cover the complete syllabus.

Q: Is terminal velocity from Set P likely to repeat?
A: Yes. Stokes’ law and terminal velocity is a fundamental topic in Grade XII Physics (viscosity and fluid mechanics). The free body diagram, v-t graph sketch, and derivation are all standard examination questions. Master this topic completely.

Q: How do I remember the Wheatstone bridge balanced condition?
A: Think of it as “product of opposite arms equal”: P × S = Q × R. Or use the bridge analogy — at balance, the galvanometer arm carries no current, so the two parallel paths must have the same voltage ratio.

Q: What phase relationship should I know for capacitor and inductor?
A: Capacitor: current leads voltage by 90° (π/2). Inductor: voltage leads current by 90° (or current lags voltage by 90°). Memory trick: “CIVIL” — in Capacitor (C), I (current) comes before V (voltage); in Inductor (L), V comes before I.

Q: Can I score 60+ in Grade XII Physics by studying only these two papers?
A: These two papers (Set P and Set D) together cover approximately 90% of the new Grade XII syllabus. Combined with understanding the derivations and practicing numericals, scoring 60+ is very achievable. Add past papers from 2079 (if available) for extra practice.

Conclusion

The NEB Grade XII Physics 2080 (2023) Set P paper is the foundation of the new curriculum examination pattern. It established which topics matter most, what level of derivation depth is expected, and how numerical problems are structured. Every Grade XII Physics student should study this paper alongside the 2081 Set D paper as their primary preparation resources.

This paper’s greatest lesson: Physics cannot be prepared topic by topic in isolation. Doppler effect appears in sound (Q.20) but its concept connects to wave optics. Self-inductance (Q.18) connects to LCR circuits (Q.22 OR). Bohr’s postulates (Q.16) connect to the hydrogen spectrum and de-Broglie wavelength. Understanding these connections transforms exam preparation from memorization into genuine comprehension.

Visit nneducation.com for more past papers, model questions, concept notes, and step-by-step numerical solutions for NEB Grade XI and XII Physics.