NEB Grade XII Physics 2079 (2023) Model Question Paper — Complete Analysis & Guide

The National Examinations Board (NEB) Grade XII Physics examination is one of the most challenging yet rewarding papers that science students face in Nepal. Physics at the +2 level demands not just memory but genuine conceptual understanding, numerical skill, and the ability to interpret graphs and diagrams under time pressure. The 2079 B.S. (2023 A.D.) model question paper — carrying Subject Code 1021 — reflects these demands very well and serves as an excellent benchmark for students preparing for their board exams.

This article provides a thorough, teacher-guided analysis of the NEB XII Physics 2079 model question paper. Whether you are a student trying to understand which chapters carry the most weight, a teacher looking for a structured discussion of the syllabus coverage, or a self-study learner who wants authentic answers alongside the questions, this resource has everything you need. We go beyond simply reproducing the paper: we analyze the question pattern, estimate chapter-wise weightage, discuss the difficulty level of individual questions, and suggest the most important topics to focus on for future exams.

The 2079 paper is a 75-mark, 3-hour examination divided into three groups: Group A (11 MCQs × 1 mark), Group B (8 questions × 5 marks), and Group C (3 questions × 8 marks). The paper covers a wide range of topics from rotational mechanics and SHM all the way to modern physics, optics, and AC circuits. Notably, several questions require multi-step numerical solutions, making time management a critical skill.

At NN Education, our goal is to make quality educational content accessible to every NEB student across Nepal. This article is part of our comprehensive NEB Physics past paper series covering all years from 2070 to the most recent examinations. We encourage students to read through the complete analysis, attempt the questions independently first, and then verify using the answer key and discussions provided below.

Exam Overview

Full Question Paper

Question Pattern Analysis

The NEB XII Physics 2079 paper follows the standard three-group format introduced after the curriculum reform. Understanding this structure is the first step toward scoring well.

Group A — Multiple Choice Questions (MCQ)

Group A contains 11 questions, each carrying 1 mark, for a total of 11 marks. All questions are compulsory. The MCQs in this paper cover a diverse spread of topics — from rotational dynamics and oscillations to thermodynamics, seismology, and AC circuits. Importantly, some MCQs (like Q2 on pendulum kinetic energy and Q9 on the thermocouple) are numerically based, meaning students cannot simply recall a fact but must quickly perform a calculation to select the correct option. This is a characteristic feature of NEB XII Physics MCQs that students often underestimate.

Group B — Short Answer Questions

Group B has 8 questions (Q12–Q19), each worth 5 marks, totalling 40 marks. All 8 questions are compulsory, but Q17 and Q19 each offer an OR alternative, giving students a choice between two sub-topics. Each question is subdivided into parts (a, b, c) testing definition/statement, derivation or diagram, and numerical calculation in a single question. This integrated format tests depth of understanding.

Group C — Long Answer Questions

Group C has 3 questions (Q20–Q22), each worth 8 marks, totalling 24 marks. Q21 and Q22 each offer OR alternatives. These questions demand extended answers, multi-part numerical solutions, derivations, and graph/diagram interpretation — the hallmarks of higher-order thinking in physics.

Topics and Chapters Covered

The 2079 paper covers the following chapters from the NEB XII Physics curriculum:

• Rotational Dynamics (Moment of Inertia, Angular Momentum)
• Simple Harmonic Motion (SHM) — Definition, Spring-Mass System, Satellite
• Fluid Mechanics — Bernoulli’s Principle
• Thermodynamics — Petrol Engine PV Diagram, Diesel Engine, Internal Energy
• Seismology (basic concept)
• Waves — Superposition, Wave Pulses, Sonometer, Standing Waves
• Sound — Properties, Doppler Effect, Pressure Waves
• Heat (Thermocouple — neutral and inversion temperature)
• Radioactivity — Half-life
• Electromagnetism — Lenz’s Law, Electromagnetic Induction, Potentiometer, Meter Bridge, Kirchhoff’s Laws
• AC Circuits — Power Dissipation, Choke Coil, LCR Circuits, Effect of Frequency
• Transformers (Laminated Cores)
• Photoelectric Effect — Einstein’s Equation, Work Function, Planck’s Constant
• Millikan’s Oil Drop Experiment
• Semiconductor Devices — Diode Rectifier, Logic Gates (NOR, AND, NAND)
• Atomic Structure — Bohr’s Model, Energy Levels, Hydrogen Spectrum, Balmer Series
• Wave Optics — Young’s Double Slit, Diffraction Grating, Polarization, Brewster’s Law
• Magnetism — Force on Current-Carrying Conductor, Torque on Coil

Chapter-wise Weightage Estimation

Note: Where OR options exist, only one set of marks is counted. The table reflects the main (non-OR) questions.

Quick MCQ Answer Key — Group A

Note on Q9: Upon careful calculation, the correct answer is a. 2%. The change in inversion temperature is 20°C out of the original 1000°C, which is 2%.

Complete Question Paper — NEB XII Physics 2079

Subject Code: 1021  |  Full Marks: 75  |  Time: 3 hrs

Candidates are required to give answers in their own words as far as practicable. Figures in the margin indicate full marks.

Group – ‘A’   [11×1 = 11]

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

1. The product of moment of inertia and angular velocity gives
   • a. Force
   • b. Torque
   • c. Linear momentum
   • d. Angular momentum ✓
2. The bob of a simple pendulum has a mass of 0.40 kg. The pendulum oscillates with a period of 2.0 s and an amplitude of 0.15 m. At an extreme point in its cycle, it has a potential energy of 0.044 J. What is the kinetic energy of the pendulum bob at its mean point?
   • a. 0.022 J
   • b. 0.044 J ✓
   • c. 0.011 J
   • d. 0.033 J
3. What causes earthquakes?
   • a. The flow of magma
   • b. The expansion of the earth’s crust
   • c. The rubbing together of earth’s plates ✓
   • d. Tsunami
4. What percentage of original radioactive atoms is left after 4 half-lives?
   • a. 1%
   • b. 6% ✓
   • c. 10%
   • d. 20%
5. Two wave pulses travel toward each other. Which diagram represents the superposition of the pulses when they meet?
   • a., b., c., d. (diagram-based — see original paper)
   • Correct: b ✓
6. Which one of the following properties of sound is affected by the change in air temperature?
   • a. Amplitude
   • b. Frequency
   • c. Wavelength ✓
   • d. Intensity
7. Internal energy of an ideal gas depends on
   • a. Volume only
   • b. Pressure only
   • c. Temperature only ✓
   • d. Both pressure and volume
8. In which of the following processes of the gas, work done is the maximum?
   • a. Isothermal
   • b. Isobaric ✓
   • c. Adiabatic
   • d. Isochoric
9. The neutral temperature of a thermocouple is 500°C when cold junction is 0°C. Percentage change in temperature of inversion when cold junction is 20°C is
   • a. 2% ✓
   • b. 3%
   • c. 4%
   • d. 5%
10. In which circuit is maximum power dissipation observed?
    • a. Inductor and resistor in series
    • b. Pure resistive circuit ✓
    • c. Pure inductive circuit
    • d. Pure capacitive circuit
11. Why are laminated cores placed in transformers?
    • a. To reduce hysteresis loss
    • b. To reduce eddy current ✓
    • c. To reduce the magnetic effect
    • d. To increase coercivity

Group – ‘B’   [8×5 = 40]

Q.12 — Simple Harmonic Motion

a. Define simple harmonic motion. [1]

Answer: Simple Harmonic Motion (SHM) is a type of periodic motion in which the restoring force acting on the particle is directly proportional to its displacement from the equilibrium position and always directed toward that equilibrium position. Mathematically: [latex]F = -kx[/latex], where [latex]k[/latex] is the force constant and [latex]x[/latex] is displacement.

b. Derive an expression for the time period of oscillation for a mass m attached to a vertical spring of spring constant k. [3]

Answer:

Consider a mass [latex]m[/latex] attached to a vertical spring of spring constant [latex]k[/latex]. At equilibrium, the spring is extended by [latex]x_0[/latex] such that [latex]mg = kx_0[/latex].

If the mass is displaced by [latex]x[/latex] below equilibrium and released, the net restoring force is:

[latex]F = mg – k(x_0 + x) = mg – kx_0 – kx = -kx[/latex]

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

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

Comparing with [latex]a = -\omega^2 x[/latex], we get [latex]\omega^2 = \frac{k}{m}[/latex]

Time period: [latex]T = \frac{2\pi}{\omega} = 2\pi\sqrt{\frac{m}{k}}[/latex]

c. What will be the time period of this system if taken inside a satellite? [1]

Answer: The time period [latex]T = 2\pi\sqrt{\frac{m}{k}}[/latex] does not depend on gravitational acceleration [latex]g[/latex]. Therefore, the time period remains the same inside a satellite as it would on Earth’s surface.

Q.13 — Bernoulli’s Principle & Fluid Mechanics

a. State Bernoulli’s principle. [1]

Answer: Bernoulli’s principle states that for a steady, incompressible, non-viscous flow of fluid, the sum of pressure energy, kinetic energy, and potential energy per unit volume remains constant along a streamline:

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

b. Numerical (Bernoulli’s Application): [2+2]

Given: Density [latex]\rho = 1200\ \text{kg m}^{-3}[/latex], [latex]A_A = 1.0\ \text{cm}^2 = 1.0 \times 10^{-4}\ \text{m}^2[/latex], [latex]A_B = 20\ \text{mm}^2 = 20 \times 10^{-6}\ \text{m}^2[/latex], [latex]v_A = 10\ \text{cm/s} = 0.1\ \text{m/s}[/latex], A and B at the same horizontal level.

(i) Speed at B:

By continuity equation: [latex]A_A v_A = A_B v_B[/latex]

[latex]v_B = \frac{A_A v_A}{A_B} = \frac{1.0 \times 10^{-4} \times 0.1}{20 \times 10^{-6}} = \frac{1.0 \times 10^{-5}}{2.0 \times 10^{-5}} = 0.5\ \text{m/s}[/latex]

(ii) Pressure difference at A and B:

Since A and B are at the same height ([latex]h_A = h_B[/latex]), Bernoulli’s equation gives:

[latex]P_A + \frac{1}{2}\rho v_A^2 = P_B + \frac{1}{2}\rho v_B^2[/latex]

[latex]P_A – P_B = \frac{1}{2}\rho(v_B^2 – v_A^2)[/latex]

[latex]P_A – P_B = \frac{1}{2} \times 1200 \times (0.5^2 – 0.1^2)[/latex]

[latex]P_A – P_B = 600 \times (0.25 – 0.01) = 600 \times 0.24 = 144\ \text{Pa}[/latex]

Q.14 — Thermodynamics: Petrol Engine

a. Draw a PV diagram of a petrol engine and explain its working. [3]

Answer: The petrol engine works on the Otto cycle. The PV diagram consists of the following processes:

• Process 1→2 (Adiabatic Compression): The air-fuel mixture is compressed adiabatically. Volume decreases, pressure and temperature increase.
• Process 2→3 (Isochoric Heat Addition): The spark plug ignites the mixture. Heat is added at constant volume. Pressure rises sharply.
• Process 3→4 (Adiabatic Expansion / Power Stroke): Gases expand adiabatically, doing work on the piston. Pressure and temperature fall.
• Process 4→1 (Isochoric Heat Rejection): The exhaust valve opens. Heat is rejected at constant volume. Pressure drops to initial value.

Efficiency of the Otto (petrol) engine: [latex]\eta = 1 – \frac{1}{r^{\gamma-1}}[/latex], where [latex]r[/latex] is the compression ratio and [latex]\gamma = C_p/C_v[/latex].

b. Compare efficiency of petrol and diesel engines based on compression ratios. [2]

Answer: The efficiency of both engines follows [latex]\eta = 1 – \frac{1}{r^{\gamma-1}}[/latex]. The petrol engine typically operates with a compression ratio of about 6–10, while diesel engines operate at 14–22. Since efficiency increases with compression ratio, the diesel engine has a higher theoretical efficiency than the petrol engine. However, petrol engines cannot use very high compression ratios because the air-fuel mixture would auto-ignite (knocking). Diesel engines avoid this by injecting fuel only after compression.

Q.15 — Waves: Sonometer and Standing Waves

a. Sonometer frequency calculation: [3]

Given: Original length [latex]L_1 = 75\ \text{cm}[/latex], shortened length [latex]L_2 = 74.5\ \text{cm}[/latex], beats per second = 3.

Frequency is inversely proportional to length (at constant tension and linear density):

[latex]f \propto \frac{1}{L}[/latex], so [latex]\frac{f_2}{f_1} = \frac{L_1}{L_2}[/latex]

Let frequency of tuning fork = [latex]f[/latex]. Since shortening the wire increases its frequency, [latex]f_2 > f_1 = f[/latex]. The beats = [latex]f_2 – f = 3[/latex].

[latex]f_2 = f \times \frac{75}{74.5}[/latex]

[latex]f_2 – f = f\left(\frac{75}{74.5} – 1\right) = f \times \frac{0.5}{74.5} = 3[/latex]

[latex]f = \frac{3 \times 74.5}{0.5} = \frac{223.5}{0.5} = 447\ \text{Hz}[/latex]

b. Speed of progressive wave in string: [2]

From the diagram, the standing wave pattern over 75 cm shows 3 loops (3 antinodes visible), which means [latex]\frac{3\lambda}{2} = 75\ \text{cm}[/latex].

[latex]\lambda = \frac{2 \times 75}{3} = 50\ \text{cm} = 0.5\ \text{m}[/latex]

Speed: [latex]v = f\lambda = 120 \times 0.5 = 60\ \text{m/s}[/latex]

Q.16 — Electromagnetic Induction: Lenz’s Law

a. State Lenz’s Law and justify it supports conservation of energy. [1+2]

Answer: Lenz’s Law: The direction of the induced current is such that it opposes the cause that produces it (i.e., opposes the change in magnetic flux).

Justification: When a magnet is moved toward a coil, the induced current creates a magnetic field that repels the magnet (opposing its motion). To continue moving the magnet, external work must be done. This mechanical work is converted into electrical energy in the coil. If Lenz’s law were reversed (induced current aiding the magnet), the coil would attract the magnet, accelerating it indefinitely — creating energy from nothing, violating conservation of energy. Therefore, Lenz’s law is fully consistent with the principle of conservation of energy.

b. Direction of induced current in coils C₁ and C₂ when magnet moves from C₁ toward C₂ (N-pole leading): [2]

Answer:

• Coil C₁ (magnet approaching with N-pole): The flux through C₁ is increasing. By Lenz’s law, the induced current must oppose this increase, so C₁ face must become N-pole. Using the right-hand rule, the induced current in C₁ flows anticlockwise when viewed from the magnet’s side (i.e., from the front face of C₁). The galvanometer in C₁ deflects in one direction (say, towards A).
• Coil C₂ (magnet receding with S-pole): The flux through C₂ is increasing (S-pole approaching). C₂ face must attract (become S-pole to oppose). Current flows clockwise when viewed from the magnet’s side. The galvanometer in C₂ deflects toward A as well (both attract).

Q.17 — Potentiometer and Meter Bridge

a. Principle of potentiometer; why it is called a voltmeter of infinite resistance. [1+1]

Answer: Principle: The potential difference across any portion of a wire of uniform cross-section carrying a steady current is directly proportional to the length of that portion: [latex]V \propto l[/latex].

Infinite resistance: A potentiometer measures EMF at the null point (balance point) where no current flows through the galvanometer branch connected to the unknown EMF source. Since no current is drawn from the source at balance, it behaves as if the resistance of the measuring instrument is infinite — hence it is called a voltmeter of infinite resistance.

b(i). New balance point when R and X are doubled and interchanged: [2]

Original balance: [latex]\frac{R}{X} = \frac{l}{100-l} = \frac{20}{80} = \frac{1}{4}[/latex], so [latex]X = 4R[/latex].

After doubling: [latex]R’ = 2R[/latex], [latex]X’ = 2X = 8R[/latex]. After interchanging: [latex]R” = 8R[/latex] (in R position), [latex]X” = 2R[/latex] (in X position).

New balance: [latex]\frac{R”}{X”} = \frac{8R}{2R} = 4 = \frac{l’}{100-l’}[/latex]

[latex]4(100-l’) = l’ \Rightarrow 400 = 5l’ \Rightarrow l’ = 80\ \text{cm}[/latex]

b(ii). Effect of interchanging galvanometer and battery at balance: [1]

Answer: The balance point is not affected. The Wheatstone bridge is balanced, meaning [latex]P/Q = R/S[/latex], which is independent of the positions of the galvanometer and battery. The bridge remains balanced at the same point.

OR — Kirchhoff’s Laws and Meter Bridge

a. Kirchhoff’s Laws: [2]

• KCL (Junction Rule): The algebraic sum of currents at any junction is zero: [latex]\sum I = 0[/latex]. (Charge conservation)
• KVL (Loop Rule): The algebraic sum of all EMFs and potential drops around any closed loop is zero: [latex]\sum E = \sum IR[/latex]. (Energy conservation)

b. Meter Bridge — finding R and S: [3]

Case 1: Null point at [latex]l_1 = 60\ \text{cm}[/latex]

[latex]\frac{R}{S} = \frac{60}{40} = \frac{3}{2}[/latex] → [latex]R = \frac{3S}{2}[/latex] … (i)

Case 2: 5 Ω added in series with S; null point at [latex]l_2 = 50\ \text{cm}[/latex]

[latex]\frac{R}{S+5} = \frac{50}{50} = 1[/latex] → [latex]R = S + 5[/latex] … (ii)

From (i) and (ii): [latex]\frac{3S}{2} = S + 5 \Rightarrow 3S = 2S + 10 \Rightarrow S = 10\ \Omega[/latex]

[latex]R = 10 + 5 = 15\ \Omega[/latex]

Q.18 — Photoelectric Effect

a. Threshold frequency from graph: [1]

From the graph, the line intersects the x-axis (E = 0) at approximately [latex]f_0 = 4.5 \times 10^{14}\ \text{Hz}[/latex]. This is the maximum frequency that still gives zero kinetic energy — i.e., the threshold frequency.

b. Work function for sodium: [2]

[latex]\phi = hf_0 = h \times 4.5 \times 10^{14}[/latex]

Using [latex]h \approx 6.63 \times 10^{-34}\ \text{Js}[/latex]: [latex]\phi = 6.63 \times 10^{-34} \times 4.5 \times 10^{14} \approx 2.98 \times 10^{-19}\ \text{J} \approx 1.86\ \text{eV}[/latex]

c. Planck’s constant from graph: [2]

The slope of the [latex]E_{k,max}[/latex] vs [latex]f[/latex] graph equals Planck’s constant [latex]h[/latex].

From the graph, taking two points, e.g., at [latex]f = 8 \times 10^{14}\ \text{Hz}[/latex], [latex]E = 2.0\ \text{eV}[/latex] and at [latex]f = 4.5 \times 10^{14}[/latex] (threshold), [latex]E = 0[/latex]:

[latex]h = \frac{\Delta E}{\Delta f} = \frac{2.0 \times 1.6 \times 10^{-19}}{(8.0 – 4.5) \times 10^{14}} = \frac{3.2 \times 10^{-19}}{3.5 \times 10^{14}} \approx 9.14 \times 10^{-34}\ \text{Js}[/latex]

(Exact value depends on the precise graph reading; the accepted value of h is [latex]6.63 \times 10^{-34}\ \text{Js}[/latex].)

Q.19 — Millikan’s Oil Drop Experiment

a. Expression for charge of oil drop moving at constant velocity downward: [2]

At constant (terminal) velocity downward, net force = 0:

Forces: Weight [latex]W = mg[/latex] downward, Buoyancy [latex]F_b[/latex] upward, Viscous drag [latex]F_d = 6\pi\eta r v[/latex] upward (Stokes’ law), Electric force [latex]qE[/latex] upward (for drop moving downward — E field points up to slow it or there may be no field in this scenario).

If the drop moves downward at constant velocity with no electric field (to find terminal velocity first):

[latex]mg = F_b + 6\pi\eta r v[/latex]

[latex]\frac{4}{3}\pi r^3 \rho g = \frac{4}{3}\pi r^3 \sigma g + 6\pi\eta r v[/latex]

When electric field E is applied upward with drop still moving down at constant velocity:

[latex]mg = qE + F_b + 6\pi\eta r v'[/latex]

[latex]q = \frac{mg – F_b – 6\pi\eta r v’}{E}[/latex]

For the standard case where drop is in equilibrium (no motion): [latex]qE = mg – F_b = \frac{4}{3}\pi r^3(\rho – \sigma)g[/latex], so [latex]q = \frac{4\pi r^3(\rho-\sigma)g}{3E}[/latex]

b. Expression when electric force is greater than weight: [1]

If [latex]qE > mg[/latex], the drop moves upward. At constant upward velocity, viscous drag acts downward:

[latex]qE = mg + F_b’ + 6\pi\eta r v[/latex]

[latex]q = \frac{mg + 6\pi\eta r v}{E}[/latex] (simplified, neglecting buoyancy in air)

c. Electric field to balance oil drop with 4 electrons, mass [latex]1.3 \times 10^{-14}\ \text{kg}[/latex]: [2]

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

[latex]q = 4e = 4 \times 1.6 \times 10^{-19} = 6.4 \times 10^{-19}\ \text{C}[/latex]

[latex]E = \frac{mg}{q} = \frac{1.3 \times 10^{-14} \times 9.8}{6.4 \times 10^{-19}} = \frac{1.274 \times 10^{-13}}{6.4 \times 10^{-19}} \approx 1.99 \times 10^{5}\ \text{V/m} \approx 2.0 \times 10^5\ \text{V/m}[/latex]

Group – ‘C’   [3×8 = 24]

Q.20 — Atomic Physics & Hydrogen Spectrum

a. Significance of negative energy of electron in an orbit: [1]

Answer: The negative energy of an electron in an orbit signifies that the electron is in a bound state — it is bound to the nucleus. Energy must be supplied to the electron to remove it from the atom. The more negative the energy, the more tightly bound the electron is. When the energy is zero, the electron is free (unbound).

b. Transition giving wavelength of 275 nm: [3]

Energy of photon: [latex]E = \frac{hc}{\lambda} = \frac{6.64 \times 10^{-34} \times 3.0 \times 10^8}{275 \times 10^{-9}}[/latex]

[latex]E = \frac{1.992 \times 10^{-25}}{2.75 \times 10^{-7}} = 7.244 \times 10^{-19}\ \text{J}[/latex]

Converting to eV: [latex]E = \frac{7.244 \times 10^{-19}}{1.6 \times 10^{-19}} \approx 4.5\ \text{eV}[/latex]

From the energy level diagram: [latex]-10\ \text{eV}[/latex] (ground), [latex]-4.5\ \text{eV}[/latex], [latex]-2\ \text{eV}[/latex], [latex]0\ \text{eV}[/latex].

Transition from [latex]-4.5\ \text{eV}[/latex] to [latex]-10\ \text{eV}[/latex]: [latex]\Delta E = 10 – 4.5 = 5.5\ \text{eV}[/latex] — too large.

Transition from [latex]0[/latex] to [latex]-4.5\ \text{eV}[/latex]: [latex]\Delta E = 4.5\ \text{eV}[/latex] ✓

Therefore, the transition is from [latex]0\ \text{eV}[/latex] level to [latex]-4.5\ \text{eV}[/latex] level (i.e., transition B or D in the figure, from the highest level to the [latex]-4.5\ \text{eV}[/latex] level). This corresponds to transition A (from 0 eV down to −4.5 eV).

c. Expression for wavelength of radiation from hydrogen atom: [2]

When an electron jumps from [latex]n_2[/latex] to [latex]n_1[/latex] ([latex]n_2 > n_1[/latex]):

[latex]E_{n} = -\frac{13.6}{n^2}\ \text{eV}[/latex]

Energy of emitted photon: [latex]E = E_{n_2} – E_{n_1} = 13.6\left(\frac{1}{n_1^2} – \frac{1}{n_2^2}\right)\ \text{eV}[/latex]

Since [latex]E = \frac{hc}{\lambda}[/latex]:

[latex]\frac{1}{\lambda} = R\left(\frac{1}{n_1^2} – \frac{1}{n_2^2}\right)[/latex]

where [latex]R = 1.097 \times 10^7\ \text{m}^{-1}[/latex] is the Rydberg constant. Therefore:

[latex]\lambda = \frac{1}{R\left(\frac{1}{n_1^2} – \frac{1}{n_2^2}\right)}[/latex]

d. Wavelength of second Balmer series line: [2]

Second Balmer line: [latex]n_1 = 2[/latex], [latex]n_2 = 4[/latex]

[latex]\frac{1}{\lambda} = R\left(\frac{1}{2^2} – \frac{1}{4^2}\right) = 1.09 \times 10^7\left(\frac{1}{4} – \frac{1}{16}\right) = 1.09 \times 10^7 \times \frac{3}{16}[/latex]

[latex]\frac{1}{\lambda} = 1.09 \times 10^7 \times 0.1875 = 2.044 \times 10^6\ \text{m}^{-1}[/latex]

[latex]\lambda = \frac{1}{2.044 \times 10^6} \approx 4.89 \times 10^{-7}\ \text{m} = 489\ \text{nm}[/latex]

Q.21 — Wave Optics

a(i). Width of ten fringes in double slit experiment: [2]

[latex]\lambda = 546\ \text{nm} = 546 \times 10^{-9}\ \text{m}[/latex], [latex]d = 0.50\ \text{mm} = 5.0 \times 10^{-4}\ \text{m}[/latex], [latex]D = 0.80\ \text{m}[/latex]

Fringe width: [latex]\beta = \frac{\lambda D}{d} = \frac{546 \times 10^{-9} \times 0.80}{5.0 \times 10^{-4}} = \frac{4.368 \times 10^{-7}}{5.0 \times 10^{-4}} = 8.736 \times 10^{-4}\ \text{m} \approx 0.874\ \text{mm}[/latex]

Width of 10 fringes: [latex]= 10\beta = 10 \times 0.874 = 8.74\ \text{mm}[/latex]

a(ii). Angle of second-order maximum with diffraction grating: [2]

Grating: 3000 lines/cm → [latex]d = \frac{1}{3000}\ \text{cm} = \frac{10^{-2}}{3000}\ \text{m} = 3.33 \times 10^{-6}\ \text{m}[/latex]

For second order: [latex]d\sin\theta = m\lambda[/latex], [latex]m = 2[/latex]

[latex]\sin\theta = \frac{2\lambda}{d} = \frac{2 \times 546 \times 10^{-9}}{3.33 \times 10^{-6}} = \frac{1.092 \times 10^{-6}}{3.33 \times 10^{-6}} = 0.3279[/latex]

[latex]\theta = \sin^{-1}(0.328) \approx 19.1°[/latex]

a(iii). Which experiment gives more accurate measurement? [1]

Answer: The diffraction grating experiment gives a more accurate measurement of wavelength. This is because the grating has many more slits (3000 lines/cm), producing very sharp and bright maxima at precise angles. The wider angular dispersion allows more precise angle measurement. In contrast, the double slit fringes are closely spaced and harder to measure precisely.

b. Distinguishing polaroid from non-polaroid sunglasses: [1]

Answer: Take both sunglasses and look through one at a light source. Now rotate the other in front of it. If the intensity of light seen through the combination changes (alternately bright and dark) as one is rotated, at least one pair is polaroid. Alternatively, look at the sky (source of partially polarized light) through each pair and rotate them — a polaroid pair will show a change in intensity as it is rotated; a non-polaroid pair will not.

c. Angle of incidence for complete polarization by reflection from water (μ = 1.3): [2]

By Brewster’s Law: [latex]\tan \theta_B = \mu[/latex]

[latex]\tan \theta_B = 1.3[/latex]

[latex]\theta_B = \tan^{-1}(1.3) \approx 52.4°[/latex]

OR — Sound Waves

a. Change in intensity when distance becomes 4 times: [1]

[latex]I \propto \frac{1}{r^2}[/latex]. If [latex]r’ = 4r[/latex], then [latex]I’ \propto \frac{1}{(4r)^2} = \frac{1}{16r^2}[/latex]. Intensity becomes [latex]\frac{1}{16}[/latex] of the original — it decreases 16 times.

b. Doppler Effect — frequency heard by passenger on second train: [2+2]

Train 1 speed [latex]v_s = 30\ \text{m/s}[/latex], frequency [latex]f_0 = 262\ \text{Hz}[/latex], Train 2 speed [latex]v_o = 18\ \text{m/s}[/latex], speed of sound [latex]v = 340\ \text{m/s}[/latex]. Trains move in opposite directions.

(i) Approaching each other:

[latex]f = f_0 \times \frac{v + v_o}{v – v_s} = 262 \times \frac{340 + 18}{340 – 30} = 262 \times \frac{358}{310}[/latex]

[latex]f = 262 \times 1.1548 \approx 302.6\ \text{Hz} \approx 303\ \text{Hz}[/latex]

(ii) Receding from each other:

[latex]f = f_0 \times \frac{v – v_o}{v + v_s} = 262 \times \frac{340 – 18}{340 + 30} = 262 \times \frac{322}{370}[/latex]

[latex]f = 262 \times 0.8703 \approx 228\ \text{Hz}[/latex]

c. Maximum displacement in sound wave: [3]

Given: Pressure amplitude [latex]\Delta P_{max} = 3.0 \times 10^{-2}\ \text{Pa}[/latex], [latex]f = 1000\ \text{Hz}[/latex], [latex]v = 344\ \text{m/s}[/latex], Bulk modulus [latex]B = 1.42 \times 10^5\ \text{Pa}[/latex]

Relation: [latex]\Delta P_{max} = B \cdot k \cdot s_{max}[/latex], where [latex]k = \frac{2\pi f}{v} = \frac{\omega}{v}[/latex]

[latex]k = \frac{2\pi \times 1000}{344} = \frac{6283.2}{344} = 18.26\ \text{m}^{-1}[/latex]

[latex]s_{max} = \frac{\Delta P_{max}}{B \cdot k} = \frac{3.0 \times 10^{-2}}{1.42 \times 10^5 \times 18.26} = \frac{3.0 \times 10^{-2}}{2.593 \times 10^6}[/latex]

[latex]s_{max} \approx 1.16 \times 10^{-8}\ \text{m} \approx 11.6\ \text{nm}[/latex]

Q.22 — AC Circuits and Choke Coil

a. What is a choke coil? [1]

Answer: A choke coil is a coil of high inductance and low resistance used in AC circuits to reduce or control the current without significant energy loss. It consists of a large number of turns of insulated copper wire wound around a laminated soft iron core.

b. Why is it preferred over a resistor in AC circuits? [2]

Answer: A choke coil is preferred over a resistor because:

• A resistor dissipates electrical energy as heat ([latex]P = I^2R[/latex]), leading to energy loss.
• A choke coil has negligible resistance and offers inductive reactance [latex]X_L = \omega L[/latex] to AC. Energy stored in the magnetic field during one half-cycle is returned to the circuit in the next half-cycle — so there is no net energy dissipation (power factor ≈ 0).

Therefore, a choke coil controls current efficiently without wasting energy.

c(i). Effect of increasing frequency on currents in circuits (a) R, (b) L, (c) C: [3]

• Circuit (a) — Pure Resistor R: The resistance [latex]R[/latex] is independent of frequency. Therefore, increasing frequency has no effect on the current; it remains constant.
• Circuit (b) — Pure Inductor L: Inductive reactance [latex]X_L = 2\pi f L[/latex] increases with frequency. By [latex]I = V/X_L[/latex], the current decreases as frequency increases.
• Circuit (c) — Pure Capacitor C: Capacitive reactance [latex]X_C = \frac{1}{2\pi f C}[/latex] decreases with frequency. By [latex]I = V/X_C[/latex], the current increases as frequency increases.

c(ii). Opposition in circuits (a) R and (b) L when AC is replaced by equivalent DC: [2]

• Circuit (a) — Resistor: A resistor offers the same resistance [latex]R[/latex] to both AC and DC. No change in opposition.
• Circuit (b) — Inductor: For DC ([latex]f = 0[/latex]), [latex]X_L = 2\pi \times 0 \times L = 0[/latex]. The inductor offers zero opposition to DC (behaves like a pure conductor with only its coil resistance). For AC, it offers [latex]X_L = \omega L[/latex]. Therefore, the opposition reduces to practically zero when DC replaces AC.

OR — Magnetism and Force on Conductors

a. Definition of 1 Ampere: [1]

Answer: One ampere is that constant current which, when flowing through two infinitely long parallel straight conductors of negligible circular cross-section placed 1 metre apart in vacuum, produces a force of [latex]2 \times 10^{-7}\ \text{N}[/latex] per metre length between them.

b. Contraction of solenoid when current flows: [2]

Answer: When current passes through a solenoid, each turn carries current in the same direction. Adjacent turns of the solenoid act like two parallel wires carrying current in the same direction. By Ampere’s law, parallel wires carrying current in the same direction attract each other. Therefore, consecutive turns attract one another, causing the solenoid to contract along its length.

c(i). Current in suspended conductor when tension is zero: [3]

Given: Linear mass density [latex]\lambda = 0.2\ \text{g m}^{-1} = 0.2 \times 10^{-3}\ \text{kg m}^{-1}[/latex], [latex]B = 1\ \text{T}[/latex] (into page), tension = 0.

For tension to be zero, the magnetic force must support the weight:

[latex]BIl = mg \Rightarrow BI = \frac{m}{l}g = \lambda g[/latex]

[latex]I = \frac{\lambda g}{B} = \frac{0.2 \times 10^{-3} \times 9.8}{1} = 1.96 \times 10^{-3}\ \text{A} \approx 1.96\ \text{mA}[/latex]

Direction: For upward force with B into the page, using [latex]\vec{F} = I\vec{l} \times \vec{B}[/latex], the current must flow from right to left (by right-hand/Fleming’s left-hand rule).

c(ii). Torque on 100-turn coil: [2]

Given: [latex]N = 100[/latex], [latex]A = 100\ \text{cm}^2 = 100 \times 10^{-4}\ \text{m}^2 = 10^{-2}\ \text{m}^2[/latex], [latex]B = 10\ \text{T}[/latex], axis perpendicular to B (plane of coil parallel to field → [latex]\theta = 90°[/latex]).

[latex]\tau = NIAB\sin\theta = 100 \times 1.96 \times 10^{-3} \times 10^{-2} \times 10 \times \sin 90°[/latex]

[latex]\tau = 100 \times 1.96 \times 10^{-3} \times 10^{-2} \times 10 \times 1 = 1.96 \times 10^{-2}\ \text{N\cdot m}[/latex]

Chapter-wise Discussion and Common Student Mistakes

SHM and Spring-Mass System

This is one of the most predictable topics in NEB XII Physics. The derivation of [latex]T = 2\pi\sqrt{m/k}[/latex] appears almost every year in some form. Students commonly confuse vertical and horizontal spring systems — remember that for a vertical spring, gravity cancels out at equilibrium and the formula remains [latex]T = 2\pi\sqrt{m/k}[/latex], independent of [latex]g[/latex]. The satellite sub-question (Q12c) catches many students off guard, but since [latex]T[/latex] doesn’t depend on [latex]g[/latex], the answer is simply “the same.”

Bernoulli’s Principle (Fluid Mechanics)

Q13 is a standard numerical. The most common mistake is unit conversion — students forget that [latex]1\ \text{cm}^2 = 10^{-4}\ \text{m}^2[/latex] and [latex]1\ \text{mm}^2 = 10^{-6}\ \text{m}^2[/latex]. Always convert all quantities to SI units first.

Thermodynamics

The MCQs on internal energy (Q7) and isobaric work (Q8) are conceptual. Many students confuse the isothermal process (which does give maximum work in Carnot cycle comparisons) with the isobaric process (which gives maximum work among the four standard processes at the same pressure). For a given pressure and volume change, isobaric gives [latex]W = P\Delta V[/latex], which is the largest since temperature rises in isobaric expansion and no work is done in isochoric.

Electromagnetic Induction and Lenz’s Law

Lenz’s law justification in terms of energy conservation (Q16a) is a standard 2-mark question. Write it clearly: “if induced current aided the cause, perpetual motion would result — violating energy conservation.” For Q16b, always draw the coil faces, identify flux change, and apply the right-hand rule systematically.

Modern Physics: Photoelectric Effect

Q18 is a graph-based question testing Einstein’s photoelectric equation [latex]E_{k,max} = hf – \phi[/latex]. Reading the threshold frequency from the x-intercept of the graph is the key skill. Then [latex]h[/latex] = slope of the graph. Many students attempt to use a single point rather than the slope — always use two well-separated points for accuracy.

Wave Optics

Q21 covers double slit and diffraction grating — both must be practiced numerically. The grating equation [latex]d\sin\theta = m\lambda[/latex] is straightforward but students often forget to convert lines/cm to metres for [latex]d[/latex]. Brewster’s Law ([latex]\tan\theta_B = \mu[/latex]) gives the polarizing angle — a quick 2-mark calculation.

Difficulty Level Analysis

Student Preparation Tips for NEB XII Physics

1. Master the standard derivations. The NEB XII syllabus has about 20–25 derivations that appear repeatedly: spring-mass period, Bernoulli’s equation, photoelectric equation, Rydberg formula, Biot-Savart law, transformer equation, etc. Write each one out at least three times from memory.

2. Practice numerical problems by topic. For every chapter, solve at least 10 numericals covering different variations of the same formula. For example, in fluid mechanics, practice Bernoulli problems with different configurations (horizontal, inclined, Venturi meter).

3. Focus on unit conversions. A large number of marks are lost due to unit errors. Make it a habit to write units at every step and convert to SI before substituting into formulas.

4. Understand OR questions strategically. In Group B (Q17, Q19) and Group C (Q21, Q22), you have OR choices. Study both options but in the exam, choose the one where you can answer all sub-parts correctly. Attempting parts of both ORs is not allowed.

5. Draw clear diagrams. Many questions explicitly ask for diagrams (PV diagram, circuit diagrams, free body diagrams). Labels, arrows, and clear axes earn partial marks even if the numerical answer is wrong.

6. Revise graph-based questions. Photoelectric effect graphs, PV diagrams, standing wave diagrams, and AC circuit phasors appear regularly. Practice interpreting and drawing them.

7. Allocate time wisely. With 75 marks in 3 hours, you have about 2.4 minutes per mark. MCQs should take no more than 15 minutes total. Group B averages 5 minutes per 5-mark question. Group C deserves 10–12 minutes per 8-mark question.

8. Solve past papers under timed conditions. This is the single most effective exam preparation strategy. The 2079 model question paper, along with papers from 2075, 2076, 2077, and 2078, gives you an excellent picture of recurring topics.

Important Questions Likely to Appear in Future NEB XII Physics Exams

Based on the pattern observed in the 2079 paper and previous years:

• Derivation of time period of simple pendulum and spring-mass system
• Bernoulli’s principle — statement, derivation, and numerical
• Explanation of working of a heat engine with PV diagram (both petrol and diesel)
• Lenz’s Law — statement, proof using energy conservation, and application
• Meter Bridge / Wheatstone Bridge — balance condition and numerical
• Kirchhoff’s Laws — statement and circuit problems
• Einstein’s photoelectric equation — derivation and graph-based problems
• Millikan’s oil drop experiment — derivation of charge expression
• Bohr’s model of hydrogen atom — energy levels, wavelength formula
• Balmer, Lyman, Paschen series calculations using Rydberg formula
• Young’s double slit experiment — fringe width formula and numerical
• Diffraction grating — angle of maxima numerical
• Brewster’s Law — polarizing angle calculation
• Doppler effect in sound — both approaching and receding cases
• Choke coil — definition, construction, and comparison with resistor
• Effect of frequency on LCR circuit components
• Force on current-carrying conductor in magnetic field
• Torque on current-carrying coil in uniform magnetic field
• Logic gates — truth tables for combination circuits (NOR, NAND, AND, OR)
• Radioactive decay — half-life problems

Frequently Asked Questions (FAQs)

Q1. How many marks does Group A (MCQ) carry in NEB XII Physics?
Group A carries 11 marks — one mark for each of the 11 MCQs. All are compulsory and there is no negative marking.

Q2. Are there optional questions in NEB XII Physics 2079?
Yes. In Group B, Questions 17 and 19 each have an OR alternative. In Group C, Questions 21 and 22 each have an OR alternative. You must attempt only one of the two options for each such question.

Q3. What is the passing marks for NEB XII Physics?
The pass mark is 27 out of 75 (approximately 36%). However, students aiming for distinction should target 60+ marks.

Q4. Which chapters have the highest weightage in NEB XII Physics?
Based on the 2079 paper, Waves and Sound (~13 marks), Electromagnetism and Current Electricity (~10 marks), Modern Physics (~10 marks), and AC Circuits (~9 marks) carry the highest combined weightage.

Q5. Is the formula [latex]T = 2\pi\sqrt{m/k}[/latex] independent of gravity?
Yes. The time period of a spring-mass system does not depend on gravitational acceleration [latex]g[/latex]. This is why the time period remains the same inside a satellite (where effective [latex]g = 0[/latex]).

Q6. Why does the diesel engine have higher efficiency than the petrol engine?
Because the diesel engine operates at a much higher compression ratio (14–22) compared to the petrol engine (6–10). Since efficiency [latex]\eta = 1 – r^{1-\gamma}[/latex] increases with compression ratio, the diesel engine is inherently more efficient.

Q7. What does the x-intercept of the Ek vs frequency graph in the photoelectric effect represent?
The x-intercept represents the threshold frequency [latex]f_0[/latex] — the minimum frequency of incident light below which no photoelectrons are emitted, regardless of intensity.

Q8. What is Brewster’s angle and how is it calculated?
Brewster’s angle [latex]\theta_B[/latex] is the angle of incidence at which reflected light is completely plane-polarized. It is calculated using Brewster’s Law: [latex]\tan\theta_B = \mu[/latex], where [latex]\mu[/latex] is the refractive index of the medium. For water ([latex]\mu = 1.3[/latex]): [latex]\theta_B = \tan^{-1}(1.3) \approx 52.4°[/latex].

Q9. In the sonometer problem, why does shortening the wire increase its frequency?
The fundamental frequency of a string is [latex]f = \frac{1}{2L}\sqrt{T/\mu}[/latex]. Since [latex]f \propto \frac{1}{L}[/latex] at constant tension and mass per unit length, a shorter wire vibrates at a higher frequency.

Q10. Where can I find more NEB XII Physics past papers with solutions?
You can find complete NEB XII Physics past papers with solutions at nneducation.com — our dedicated NEB study portal for Grade XI and XII students.

Conclusion

The NEB XII Physics 2079 model question paper is a well-balanced examination that tests students across the full breadth of the Grade 12 Physics curriculum. From MCQs on angular momentum and radioactive decay to long-answer questions involving atomic spectra, wave optics, and AC circuits, the paper demands both conceptual clarity and computational skill.

The key takeaway from this analysis is that NEB XII Physics is very predictable in terms of topic coverage. The same core chapters — SHM, fluid mechanics, thermodynamics, electromagnetic induction, modern physics, optics, and AC circuits — appear year after year. Students who master these topics thoroughly, practice past papers under timed conditions, and strengthen their numerical problem-solving skills are very well positioned to score high.

At NN Education, we remain committed to providing high-quality, teacher-created study materials for every NEB student in Nepal. We encourage you to bookmark this page, share it with your classmates, and explore our other resources for NEB Grade XI and XII across Physics, Chemistry, Mathematics, Biology, and more.

Best of luck with your NEB examinations!