NEB Grade XII Physics Model Question 2078 — Complete Analysis, Answer Guide & Exam Preparation

Subject Code: 102 | Full Marks: 75 (11 Obj + 64 Sub) | Time: 3 Hours
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Introduction

The NEB Grade XII Physics Model Question 2078 (School Leaving Certificate Examination) is a landmark paper — it represents the old curriculum’s final model question format and serves as a foundational reference for understanding how Grade XII Physics questions are structured in Nepal. Unlike the new curriculum papers (2080–2081), this paper carries Subject Code 102 rather than 1021, and its marking scheme differs slightly: 11 marks objective + 64 marks subjective.

What makes this paper uniquely valuable for students is its depth of conceptual questioning. Rather than straightforward numerical problems, it frequently asks students to explain, justify, prove, and derive — skills that separate truly understanding students from those who merely memorize. Questions about why ball P falls slower than ball Q through a magnetic field (Q.17), whether a refrigerator violates the second law (Q.15), and what happens when an oil drop is struck by an alpha particle (Q.22) — these are the kinds of physically rich questions that build lasting understanding.

This paper also introduces topics less commonly tested in newer papers: Millikan’s oil drop experiment, potentiometer circuit analysis, LRC circuit phasor diagrams, radioactive decay graph interpretation, and SHM applied to earthquake mechanics — making it essential preparation material for any serious Grade XII Physics student.

Exam Overview

Full Question Paper

Key Differences: 2078 Model vs 2080/2081 New Curriculum

Topics Covered — Chapter-wise Weightage

Group A — MCQ Analysis with Answers

Q.1 — Radius of Gyration Formula
Answer: (a) [latex]k^2 = \frac{I}{m}[/latex]

The radius of gyration k is defined such that [latex]I = mk^2[/latex], where I = moment of inertia and m = total mass. Therefore [latex]k^2 = \frac{I}{m}[/latex] and [latex]k = \sqrt{\frac{I}{m}}[/latex]. Option (b) gives k = I/m which has wrong dimensions. Option (c) gives m/I. Option (d) gives (I/m)² which would mean [latex]k = \frac{I}{m}[/latex] — also wrong.

Q.2 — Effect of Air Stream Under Pan
Answer: (b) goes down

When air is blown horizontally under the pan, by Bernoulli’s principle the air pressure below the pan decreases (fast-moving air = low pressure). The atmospheric pressure above the pan is now greater than the pressure below. This net downward force pushes the pan downward.

This is a beautiful application of Bernoulli’s principle — the same reason why airplane wings generate lift, but here reversed because the air flows beneath.

Q.3 — Capillary Height on Moon
Answer: (c) 6h

Capillary rise formula: [latex]h = \frac{2T\cos\theta}{r\rho g}[/latex]

On the Moon, [latex]g_{moon} = \frac{g_{earth}}{6}[/latex]. All other factors (T, θ, r, ρ) remain the same.

[latex]h_{moon} = \frac{2T\cos\theta}{r\rho g_{moon}} = \frac{2T\cos\theta}{r\rho \cdot \frac{g}{6}} = 6 \times \frac{2T\cos\theta}{r\rho g} = 6h[/latex]

The capillary rise on the Moon is 6 times greater than on Earth.

Q.4 — COP of Ideal Refrigerator (Ice Point to Room Temperature)
Answer: (d) 10

Ice point = 0°C = 273 K, Room temperature = 27°C = 300 K

COP of ideal refrigerator: [latex]\beta = \frac{T_{cold}}{T_{hot} – T_{cold}} = \frac{273}{300 – 273} = \frac{273}{27} \approx 10.1 \approx 10[/latex]

Q.5 — Work Done in P-V Cycle
Answer: (c) ACBDA

The total work done by the system in one complete cycle equals the area enclosed by the closed curve in the P-V diagram. The system goes A→B via C (upper path, gas expands = positive work) and returns B→A via D (lower path, gas compressed = negative work). The net work = area enclosed = area of region ACBDA.

Q.6 — What Affects Quality of Sound
Answer: (d) wave form

The “quality” or “timbre” of sound is what distinguishes two sounds of the same pitch (frequency) and loudness (intensity) produced by different instruments. It is determined entirely by the waveform — specifically by the number and relative intensities of harmonics present.

Q.7 — Diffraction Pattern: Red Light Replaced by White Light
Answer: (d) All bright fringes, except the central one, become colourful

The central bright fringe (zero-order maximum) occurs at the same position for all wavelengths — it remains white. All other maxima positions depend on wavelength: [latex]y_n = \frac{n\lambda D}{d}[/latex]. Different wavelengths in white light produce maxima at different positions — so all other bright fringes become spread out and colourful (with violet innermost and red outermost).

Q.8 — Currents Satisfying I₁ – I₂ = I₃ – I₄

Rearranging: I₁ + I₃ = I₂ + I₄

By Kirchhoff’s Current Law at a junction: sum of currents entering = sum leaving. The correct diagram is the one where I₁ and I₃ flow into the junction and I₂ and I₄ flow out (or equivalent). From the four circuit diagrams shown, the answer is (a) — where I₁ and I₃ are incoming and I₂ and I₄ are outgoing, satisfying I₁ + I₃ = I₂ + I₄, i.e., I₁ – I₂ = I₃ – I₄… wait — rearranging differently: I₁ – I₂ = I₄ – I₃ — students must identify the junction where this current equation applies.

Q.9 — Product NIA Represents
Answer: (c) magnetic moment of the coil

Magnetic moment of a current loop: [latex]m = NIA[/latex]

where N = number of turns, I = current, A = cross-sectional area. This is the fundamental definition of magnetic moment of a coil.

• Magnetic flux = BA (not NIA)
• Magnetic flux density = B (not NIA)
• Magnetic susceptibility = dimensionless ratio (not NIA)

Q.10 — Neutral Temperature When Cold Junction is Decreased
Answer: (c) remains the same

The neutral temperature [latex]T_n[/latex] of a thermocouple is the temperature of the hot junction at which thermo-emf is maximum. It depends only on the material of the thermocouple, not on the temperature of the cold junction. Therefore, decreasing the cold junction temperature does not change the neutral temperature — it remains the same.

Note: The inversion temperature changes when cold junction temperature changes: [latex]T_i = 2T_n – T_c[/latex]. If [latex]T_c[/latex] decreases, [latex]T_i[/latex] increases.

Q.11 — Point Where Seismic Waves Start
Answer: (b) hypocentre

The hypocentre (also called the focus) is the actual point inside the Earth where an earthquake originates — where seismic waves start. The epicentre is the point on the Earth’s surface directly above the hypocentre. Metacentre relates to floating bodies.

Group B — Short Answer Analysis

Q.12 — Surface Tension and Terminal Velocity of Merged Drops

Part (i): Define Surface Tension (1 mark)
Surface tension is the property of a liquid surface by which it behaves like a stretched elastic membrane, tending to contract to the minimum possible area. It is defined as the force per unit length acting along the surface of the liquid at right angles to any line drawn on the surface.

[latex]T = \frac{F}{l}[/latex] (SI unit: N/m)

Part (ii): Relation Between Surface Tension and Surface Energy (2 marks)

Consider a soap film on a wire frame with a movable side of length l. When the side is pulled outward by distance dx against surface tension:

Work done = Force × distance = [latex](T \times 2l) \times dx[/latex] (factor 2 for two surfaces of soap film)

Increase in surface area = [latex]2l \times dx[/latex]

Surface energy = Work done per unit increase in area:

[latex]E = \frac{T \times 2l \times dx}{2l \times dx} = T[/latex]

Therefore: Surface tension = Surface energy per unit area

[latex]T = \frac{\text{Work done}}{\text{Increase in surface area}}[/latex] (SI unit: J/m² = N/m)

Part (iii): Terminal Velocity of Merged Drops (2 marks)

For a single drop of radius r, terminal velocity: [latex]v_t = \frac{2r^2(\rho – \sigma)g}{9\eta}[/latex]

So [latex]v_t \propto r^2[/latex]

When two drops of radius r merge, volume is conserved:
[latex]\frac{4}{3}\pi R^3 = 2 \times \frac{4}{3}\pi r^3[/latex]

[latex]R^3 = 2r^3 \Rightarrow R = 2^{1/3} r[/latex]

New terminal velocity:

[latex]\frac{v_{new}}{v_{original}} = \frac{R^2}{r^2} = \frac{(2^{1/3}r)^2}{r^2} = 2^{2/3} \approx 1.587[/latex]

The new terminal velocity is [latex]2^{2/3}[/latex] ≈ 1.59 times greater than the original. The merged drop falls faster because its radius (and hence the ratio of weight to drag) is larger.

Q.13 — Rotational Dynamics: Moment of Inertia and Conservation of Angular Momentum

Part (i): Define Moment of Inertia (1 mark)
Moment of inertia of a body about a given axis is the sum of the products of the mass of each particle and the square of its perpendicular distance from the axis of rotation.

[latex]I = \sum m_i r_i^2[/latex]

It is the rotational analogue of mass. SI unit: kg·m²

Part (ii): Why Earth’s Angular Velocity Increases When Closer to Sun (2 marks)

When Earth moves closer to the Sun in its elliptical orbit, its distance from the Sun (r) decreases. By conservation of angular momentum (no external torque in the Sun-Earth system):

[latex]L = I\omega = mr^2\omega = \text{constant}[/latex]

When r decreases, [latex]\omega[/latex] must increase to keep L constant. Therefore Earth’s angular velocity increases when it comes closer to the Sun — this is consistent with Kepler’s second law (equal areas in equal times).

Part (iii): If Earth Shrinks — What Happens to Length of Day? (2 marks)

If Earth shrinks suddenly, its radius R decreases while mass M remains constant. Moment of inertia of Earth: [latex]I = \frac{2}{5}MR^2[/latex]

If R decreases, I decreases. By conservation of angular momentum:
[latex]I\omega = I’\omega'[/latex]

[latex]\frac{2}{5}MR^2 \cdot \omega = \frac{2}{5}MR’^2 \cdot \omega'[/latex]

Since R’ < R → I’ < I → ω’ > ω

Greater angular velocity means shorter period of rotation: [latex]T = \frac{2\pi}{\omega}[/latex]

Length of the day decreases — Earth would rotate faster.

OR Q.13 — Bernoulli’s Principle

Part (i): Bernoulli’s Principle (1 mark)
Bernoulli’s principle states that for a steady, non-viscous, incompressible flow of a 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 gh = \text{constant}[/latex]

Part (ii): Derive Bernoulli’s Equation (2 marks)

Using the work-energy theorem for fluid flow between two cross-sections 1 and 2:

Work done by pressure forces: [latex]W = P_1A_1\Delta x_1 – P_2A_2\Delta x_2 = (P_1 – P_2)\Delta V[/latex]

Change in KE: [latex]\Delta KE = \frac{1}{2}\rho\Delta V(v_2^2 – v_1^2)[/latex]

Change in PE: [latex]\Delta PE = \rho\Delta V \cdot g(h_2 – h_1)[/latex]

By work-energy theorem: W = ΔKE + ΔPE

[latex](P_1 – P_2)\Delta V = \frac{1}{2}\rho\Delta V(v_2^2 – v_1^2) + \rho\Delta Vg(h_2 – h_1)[/latex]

Rearranging: [latex]\boxed{P_1 + \frac{1}{2}\rho v_1^2 + \rho gh_1 = P_2 + \frac{1}{2}\rho v_2^2 + \rho gh_2}[/latex]

Part (iii): Garden Hose — Thumb Partially Covering Opening (2 marks)

By the continuity equation: [latex]A_1v_1 = A_2v_2[/latex]

When you partially cover the opening with your thumb, the cross-sectional area A₂ decreases. Since flow rate is conserved, v₂ must increase significantly.

By Bernoulli’s equation, the water exits with higher velocity and therefore travels a much greater horizontal distance — the same principle used in sprinkler systems. The kinetic energy at exit = [latex]\frac{1}{2}\rho v_2^2[/latex] is larger, enabling greater range.

Q.14 — Sound: Harmonics, Flute Frequency, and Doppler Effect

Part (i): Define Harmonics (1 mark)
Harmonics are musical tones whose frequencies are whole-number multiples of a fundamental frequency. The fundamental (lowest frequency) is the first harmonic, the tone with twice the fundamental frequency is the second harmonic, and so on. They contribute to the quality (timbre) of a musical note.

Part (ii): Frequency of 30 cm Flute in First Harmonic (2 marks)

Given: L = 30 cm = 0.30 m, v = 330 m/s, open at both ends (open organ pipe)

For an open organ pipe, fundamental frequency (first harmonic):

[latex]f_1 = \frac{v}{2L} = \frac{330}{2 \times 0.30} = \frac{330}{0.60} = \mathbf{550 \text{ Hz}}[/latex]

Part (iii): Change in Pitch When Bus Moves Away from Person (2 marks)

When the bus moves uniformly away from the stationary person at the bus park, the flute (source) moves away from the listener. By the Doppler effect for a moving source:

[latex]f’ = f_0 \cdot \frac{v}{v + v_s}[/latex]

where [latex]v_s[/latex] = speed of bus (source), v = speed of sound.

Since [latex]v + v_s > v[/latex], we have [latex]f’ < f_0[/latex].

The person at the bus park hears a lower pitch (lower frequency) than the actual frequency of the flute. The sound appears to have a lower tone as the bus moves away — this is the familiar effect of a receding vehicle.

Q.15 — Thermodynamics: Second Law and Heat Engine

Part (i): Second Law of Thermodynamics (1 mark)

Kelvin-Planck statement: It is impossible to construct a heat engine that operates in a cycle and converts all the heat absorbed from a single reservoir completely into work without any other effect.

Clausius statement: It is impossible for a self-acting device to transfer heat from a colder body to a hotter body without the expenditure of work.

Part (ii): Does Refrigerator Violate Second Law? (2 marks)

No, a refrigerator does NOT violate the second law of thermodynamics.

The Clausius statement says heat cannot flow spontaneously from cold to hot — but a refrigerator is not spontaneous. A refrigerator uses external electrical work (from the compressor motor) to transfer heat from the cold interior to the hot room. The total entropy of the universe still increases. The refrigerator satisfies energy conservation: [latex]Q_1 = Q_2 + W[/latex] (heat rejected to room = heat absorbed from cold body + work done by motor).

Part (iii.a): Expression for Efficiency of Heat Engine (1 mark)

[latex]\eta = \frac{W}{Q_1} = \frac{Q_1 – Q_2}{Q_1} = 1 – \frac{Q_2}{Q_1}[/latex]

For an ideal (Carnot) engine: [latex]\eta = 1 – \frac{T_2}{T_1}[/latex]

Part (iii.b): Condition for Zero Efficiency (1 mark)

Efficiency becomes zero when [latex]W = 0[/latex], i.e., when [latex]Q_1 = Q_2[/latex].

For a Carnot engine: [latex]\eta = 0[/latex] when [latex]T_2 = T_1[/latex]

If both reservoirs are at the same temperature, no work can be extracted — this would require a reversible engine operating between identical temperatures, which produces no net output.

Q.16 — Magnetic Flux Density Measurement

Part (i): Magnetic Flux Density (1 mark)
Magnetic flux density (B) is defined as the force per unit length per unit current on a long straight conductor placed perpendicular to the magnetic field.

[latex]B = \frac{F}{IL}[/latex]

or: it is the number of magnetic field lines passing through unit area perpendicular to the field. SI unit: Tesla (T) = Wb/m²

Part (ii): Why Hall Probe is Not Suitable Here (1 mark)

The Hall probe measures magnetic field by detecting the Hall voltage across a semiconductor, which requires the field to be perpendicular to the probe’s flat surface. For the arrangement shown (field between two bar magnets on a yoke), the field direction is non-uniform and not easily aligned with the Hall probe’s sensitive face. Additionally, the field is horizontal between the poles, making it difficult to position the Hall probe correctly without disturbing the setup. A Hall probe requires precise angular alignment that is impractical here.

Part (iii): Determining Flux Density Using Current-Carrying Wire (3 marks)

Method:

1. Insert a straight wire of known length l between the poles of the magnet, perpendicular to the field
2. Connect the wire to a circuit with an ammeter and variable resistor (rheostat)
3. Attach a sensitive balance or force sensor to the wire support
4. Pass current I through the wire and measure the force F acting on it using the balance

Using F = BIL:

[latex]B = \frac{F}{IL}[/latex]

By varying current I and measuring corresponding force F, plot F vs I — the slope gives BL. Knowing l, calculate B. The force direction is given by Fleming’s Left-Hand Rule — the wire experiences a force perpendicular to both the current direction and field direction.

Additional apparatus needed: Ammeter, rheostat (variable resistor), sensitive spring balance or force sensor.

Q.17 — Electromagnetic Induction: Faraday’s Law and Copper Ball

Part (i): Symbols in [latex]\epsilon = -N\frac{d\phi}{dt}[/latex] (2 marks)

• ε (epsilon): The induced electromotive force (emf) in the coil, measured in Volts (V)
• N: The number of turns in the coil (dimensionless)
• [latex]\frac{d\phi}{dt}[/latex]: The rate of change of magnetic flux through one turn of the coil, measured in Webers per second (Wb/s) or Volts

The equation states that the induced emf equals N times the rate of change of flux through each turn.

Part (ii): Significance of Negative Sign (1 mark)

The negative sign represents Lenz’s Law — the induced emf always acts in a direction such that the current it drives would oppose the change in magnetic flux that caused it. In other words, the induced effect always opposes its cause. This is a consequence of conservation of energy — if the induced emf aided the change, it would create energy from nothing.

Part (iii): Why Ball P Takes Longer Than Ball Q (2 marks)

Ball P passes through a region of uniform horizontal magnetic field B. As the conducting copper ball moves downward through the field, the magnetic flux through the ball changes — by Faraday’s law, this induces an emf and hence eddy currents within the copper ball.

By Lenz’s law, these eddy currents create a magnetic force that opposes the downward motion of ball P. This opposing (upward) force reduces the net downward force on ball P, causing it to have a smaller acceleration than ball Q (which experiences no field and falls freely under gravity alone).

Result: Ball P accelerates more slowly and takes longer to reach the ground. Ball Q falls freely and arrives first.

Q.18 — Photoelectric Effect: Aluminium Plate

Part (i): Show Maximum Speed = 8.6 × 10⁵ m/s (3 marks)

Given: f = 1.5 × 10¹⁵ Hz, φ = 6.6 × 10⁻¹⁹ J, h = 6.63 × 10⁻³⁴ J·s, m_e = 9.1 × 10⁻³¹ kg

Einstein’s photoelectric equation:

[latex]hf = \phi + \frac{1}{2}mv_{max}^2[/latex]

[latex]\frac{1}{2}mv_{max}^2 = hf – \phi[/latex]

[latex]= 6.63 \times 10^{-34} \times 1.5 \times 10^{15} – 6.6 \times 10^{-19}[/latex]

[latex]= 9.945 \times 10^{-19} – 6.6 \times 10^{-19}[/latex]

[latex]= 3.345 \times 10^{-19} \text{ J}[/latex]

[latex]v_{max}^2 = \frac{2 \times 3.345 \times 10^{-19}}{9.1 \times 10^{-31}} = \frac{6.69 \times 10^{-19}}{9.1 \times 10^{-31}} = 7.352 \times 10^{11}[/latex]

[latex]v_{max} = \sqrt{7.352 \times 10^{11}} = 8.57 \times 10^5 \approx \mathbf{8.6 \times 10^5 \text{ m/s}} \checkmark[/latex]

Part (ii): Effect of Increasing Intensity on Maximum Speed (2 marks)

No change in maximum speed. The maximum kinetic energy (and hence maximum speed) of emitted electrons depends only on the frequency of incident radiation, not on its intensity.

By Einstein’s equation: [latex]KE_{max} = hf – \phi[/latex]

Intensity determines the number of photons hitting per second — increasing intensity means more photons, which means more electrons are emitted (greater photocurrent), but each individual photon still has the same energy hf. Since each emitted electron is ejected by a single photon, the maximum speed remains [latex]8.6 \times 10^5[/latex] m/s regardless of intensity.

Q.19 — Bohr’s Postulates and Lyman Series

Part (i): Bohr’s Postulates (3 marks)

Postulate 1 — Stationary Orbits:
An electron can revolve around the nucleus only in certain discrete circular orbits called stationary states or allowed orbits, without emitting any radiation.

Postulate 2 — Quantization of Angular Momentum:
The angular momentum of an electron in a stationary orbit is an integral multiple of [latex]\frac{h}{2\pi}[/latex]:

[latex]mvr = \frac{nh}{2\pi}[/latex]

where n = 1, 2, 3, … is the principal quantum number.

Postulate 3 — Quantum Jump and Radiation:
An electron can jump from a higher energy orbit (E₂) to a lower energy orbit (E₁) by emitting a photon of frequency:

[latex]hf = E_2 – E_1[/latex]

Conversely, a photon of this frequency can be absorbed to cause the reverse transition.

Part (ii): Frequency of Photon from Second Excited State to Ground Level (2 marks)

Second excited state = n = 3, energy E₃ = -1.51 eV
Ground level = n = 1, energy E₁ = -13.6 eV (not shown in figure; from figure: ground level is lowest)

Wait — from the Lyman series energy diagram given: The levels shown are E = 0, -0.54 eV, -0.85 eV, -1.51 eV, -3.40 eV and ground state.

Second excited state (n=3): E₃ = -1.51 eV
Ground state (n=1): E₁ = -13.6 eV (or from figure context)

Actually using the values from the figure:
From -1.51 eV to ground (-13.6 eV):

[latex]\Delta E = -1.51 – (-13.6) = 12.09 \text{ eV} = 12.09 \times 1.6 \times 10^{-19} = 1.934 \times 10^{-18} \text{ J}[/latex]

[latex]f = \frac{\Delta E}{h} = \frac{1.934 \times 10^{-18}}{6.63 \times 10^{-34}} = \mathbf{2.92 \times 10^{15} \text{ Hz}}[/latex]

This is in the Lyman series (UV region), consistent with a transition to n=1.

OR Q.19 — P-N Junction Diode and Bridge Rectifier

Part (i): P-N Junction Diode Symbol (1 mark)

Symbol: A →|— K (anode → triangle → vertical bar → cathode)

The arrowhead points from P-type (anode) to N-type (cathode), indicating conventional current flow direction when forward biased.

Part (ii): Bridge Rectifier with Four Diodes (2 marks)

In a bridge rectifier, four diodes are arranged in a bridge configuration:

• During positive half-cycle: current flows through D₁ and D₂
• During negative half-cycle: current flows through D₃ and D₄
• Both half-cycles drive current in the same direction through load R_L

The diodes D₁, D₂, D₃, D₄ are placed so that during each half-cycle, two diodes conduct and two are reverse biased.

Part (iii): One Diode Damaged — Effect on Output (2 marks)

If one diode (say D₁) is damaged and stops conducting:

• During the positive half-cycle (which required D₁): current cannot flow → no output
• During the negative half-cycle (which uses D₃ and D₄): current flows normally → normal half-cycle output

Result: The bridge rectifier becomes a half-wave rectifier. Only every alternate half-cycle appears across R_L. The output pd vs time graph shows: one normal half-cycle pulse, then zero, then one normal pulse, alternating — with gaps between pulses (unlike the original full-wave output).

Group C — Long Answer Analysis

Q.20 — SHM and Earthquake Mechanics

Part (i): Prove F = -Ax Causes SHM (2 marks)

Given: [latex]F = -Ax[/latex], where x is displacement and A is a positive constant.

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

[latex]ma = -Ax[/latex]

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

Let [latex]\omega^2 = \frac{A}{m}[/latex] (a positive constant since A, m > 0):

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

This shows acceleration is directly proportional to displacement and directed opposite to displacement — this is the defining condition of Simple Harmonic Motion. ✓

Angular frequency: [latex]\omega = \sqrt{\frac{A}{m}}[/latex], Time period: [latex]T = 2\pi\sqrt{\frac{m}{A}}[/latex]

Part (ii): Derive [latex]v = \pm\omega\sqrt{A^2 – x^2}[/latex] (2 marks)

For SHM: [latex]a = \frac{dv}{dt} = v\frac{dv}{dx} = -\omega^2 x[/latex]

[latex]v,dv = -\omega^2 x,dx[/latex]

Integrating both sides:

[latex]\frac{v^2}{2} = -\frac{\omega^2 x^2}{2} + C[/latex]

At x = A (amplitude), v = 0:

[latex]0 = -\frac{\omega^2 A^2}{2} + C \Rightarrow C = \frac{\omega^2 A^2}{2}[/latex]

Substituting back:

[latex]\frac{v^2}{2} = \frac{\omega^2}{2}(A^2 – x^2)[/latex]

[latex]\boxed{v = \pm\omega\sqrt{A^2 – x^2}}[/latex] ✓

(+ for motion toward equilibrium in one direction, – for other direction)

Part (iii): Maximum Speed of Building in S-waves (2 marks)

Given: f = 21 Hz, amplitude A = 0.05 m

Maximum speed occurs at [latex]x = 0[/latex] (equilibrium position):

[latex]v_{max} = \omega A = 2\pi f A[/latex]

[latex]v_{max} = 2\pi \times 21 \times 0.05 = 2\pi \times 1.05 = 6.597 \approx \mathbf{6.6 \text{ m/s}}[/latex]

Part (iv): Why Tall Buildings More Susceptible to Low-Frequency S-waves (2 marks)

Tall buildings have a low natural frequency of vibration due to their large mass and height — they tend to sway at low frequencies (1–5 Hz typically). Low-frequency S-waves have frequencies comparable to the natural frequencies of tall buildings.

When the forcing frequency (S-wave) matches the natural frequency of the building, resonance occurs. At resonance, the amplitude of oscillation builds up dramatically — far beyond what the structural materials can withstand. This causes catastrophic failure. Low-rise buildings have higher natural frequencies and are less affected by low-frequency S-waves.

Q.21 — LRC Circuit (OR: Potentiometer)

Part (i): Phase Constant — Positive or Negative? (2 marks)

From the graph, the emf (voltage) curve reaches its maximum before the current curve — the current lags the voltage.

In an LRC circuit: [latex]I = I_0\sin(\omega t – \phi)[/latex] when voltage [latex]V = V_0\sin\omega t[/latex]

If current lags voltage, [latex]\phi > 0[/latex] — the phase constant is positive.

Current lagging voltage indicates the circuit is predominantly inductive (X_L > X_C).

Part (ii): Phasor Diagram (2 marks)

For an inductive circuit (I lags V by angle φ):

Draw:

• Voltage phasor V₀ along reference direction (e.g., horizontal)
• Current phasor I₀ rotated clockwise by angle φ from voltage
• V_R (resistive component) in phase with I
• V_L (inductive component) leading I by 90° (vertical, upward)
• V_C (capacitive component) lagging I by 90° (vertical, downward)
• Net reactive voltage = V_L – V_C (net upward since inductive)
• Resultant V₀ = √(V_R² + (V_L – V_C)²)

The phase angle: [latex]\tan\phi = \frac{V_L – V_C}{V_R} = \frac{X_L – X_C}{R}[/latex]

Part (iii): Is Circuit More Inductive or Capacitive? (2 marks)

From the graph: voltage leads current (current lags voltage).

In a series LRC circuit:

• If X_L > X_C → voltage leads current → circuit is inductive
• If X_C > X_L → current leads voltage → circuit is capacitive

Since current lags voltage in this case, the circuit is predominantly inductive (X_L > X_C).

Part (iv): Should Inductance Be Increased or Decreased? (2 marks)

Power delivered to resistive load: [latex]P = V_{rms}I_{rms}\cos\phi[/latex]

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

To maximize power transfer: maximize cos φ → minimize (X_L – X_C) → approach resonance condition X_L = X_C.

Currently X_L > X_C (inductive circuit). To reduce X_L toward X_C value:
Inductance L should be decreased (since X_L = ωL).

Alternatively, at resonance (X_L = X_C), Z = R (minimum impedance), cos φ = 1, and power transfer is maximum.

OR Q.21 — Potentiometer Circuit

Part (a): Why No Balance Point Found and Required Change (2 marks)

The student cannot find a balance point because the test cell’s emf is greater than the emf of the driving cell. In a potentiometer, the balance condition requires the pd across some length of wire to equal the emf of the test cell. If the driving cell’s emf is smaller than the test cell’s emf, the pd across even the full length of wire is insufficient to balance the test cell — no null point exists.

Required change: Replace the driving cell with one of larger emf (greater than the test cell’s emf), or reduce the current by increasing the series resistance so that the pd per unit length decreases and the full wire length provides a pd range that encompasses the test cell emf.

Part (b): How to Recognize Balance Point (2 marks)

The balance point (null point) is recognized when the galvanometer reads zero — no current flows through the galvanometer. At this position, the pd across the potentiometer wire from A to the jockey position exactly equals the emf of the test cell. The student should press the jockey at various points and note where the galvanometer deflects to zero. At balance, pressing or releasing the jockey makes no change to the galvanometer reading.

Part (c): Calculate emf of Test Cell (2 marks)

For a potentiometer: [latex]\frac{E_{test}}{E_{standard}} = \frac{l_{test}}{l_{standard}}[/latex]

Given: [latex]E_{standard}[/latex] = 1.50 V, [latex]l_{standard}[/latex] = 37.5 cm, [latex]l_{test}[/latex] = 25.0 cm

[latex]E_{test} = E_{standard} \times \frac{l_{test}}{l_{standard}} = 1.50 \times \frac{25.0}{37.5} = 1.50 \times \frac{2}{3} = \mathbf{1.00 \text{ V}}[/latex]

Part (d): Why Potentiometer is More Precise Than Voltmeter (2 marks)

An ordinary voltmeter measures emf by drawing current from the test cell, which causes a voltage drop across the cell’s internal resistance — the voltmeter reads the terminal voltage (V = E – Ir), not the true emf.

A potentiometer measures emf at the null point — when galvanometer reads zero, no current is drawn from the test cell. Therefore the test cell is in an open-circuit condition and its terminal voltage equals its true emf. There is no error due to internal resistance.

This makes the potentiometer measurement fundamentally more accurate — it is a null method that does not disturb the quantity being measured.

Q.22 — Millikan’s Oil Drop Experiment and Charge Quantization (OR: Radioactive Decay)

Part (a): Quantization of Charge (2 marks)

Electric charge is quantized — it exists only in discrete, indivisible units. The smallest possible charge is the charge of an electron:

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

Any observable charge Q is always an integer multiple of e:

[latex]Q = ne[/latex] where n = ±1, ±2, ±3, …

Charge cannot be fractional in ordinary matter. This was demonstrated by Millikan’s oil drop experiment, which showed that all measured charges were integral multiples of [latex]1.6 \times 10^{-19}[/latex] C.

Part (b)(i): Condition for Stationary Drop and Force Diagram (2 marks)

For the oil drop to remain stationary, the electric force must exactly balance the gravitational force (weight):

[latex]qE = mg[/latex] (upward electric force = downward weight)

Forces acting on the stationary drop:

↓ Weight [latex]W = mg = 1.5 \times 10^{-14}[/latex] N (downward)
↑ Electric force [latex]F_E = qE[/latex] (upward, if charge is negative and field points downward, or positive charge with field upward)

The electric field E = V/d points from positive plate to negative plate. For the drop to be stationary: [latex]qE = mg[/latex]

Part (b)(ii): Calculate Charge on Oil Drop (2 marks)

Given: W = mg = 1.5 × 10⁻¹⁴ N, d = 10 mm = 0.01 m, V = 470 V

Electric field: [latex]E = \frac{V}{d} = \frac{470}{0.01} = 47000 \text{ V/m}[/latex]

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

[latex]q = \frac{mg}{E} = \frac{1.5 \times 10^{-14}}{47000} = 3.19 \times 10^{-19} \text{ C} \approx 3.2 \times 10^{-19} \text{ C}[/latex]

This equals [latex]2e[/latex] (twice the electron charge) — consistent with charge quantization (n = 2).

Part (b)(iii): Effect of Alpha Particle Strike (2 marks)

An alpha particle carries a charge of +2e. When it strikes the oil drop (which has charge q = −2e or +2e):

If the oil drop is negatively charged (−2e) and the alpha particle (+2e) attaches to it:
New charge = [latex]-2e + 2e = 0[/latex]

The drop becomes neutral — the electric force disappears (F_E = qE = 0). Only gravity acts, so the drop falls downward and is no longer stationary.

If the oil drop is positively charged: new charge = [latex]+2e + 2e = +4e[/latex] — electric force doubles, drop moves upward.

In either case, the equilibrium is destroyed and the drop moves. This demonstrates that charge is transferred in discrete units (quantized).

OR Q.22 — Radioactive Decay

Part (a): Derive N = N₀e^(-λt) (3 marks)

The rate of decay of a radioactive sample is proportional to the number of undecayed nuclei N present at that instant:

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

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

where λ = decay constant (probability of decay per nucleus per unit time).

Separating variables:

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

Integrating both sides:

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

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

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

[latex]\ln\left(\frac{N}{N_0}\right) = -\lambda t[/latex]

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

This shows that radioactive decay follows an exponential law — the number of undecayed nuclei decreases exponentially with time.

Part (b)(i): Why Data is Scattered (1 mark)

Radioactive decay is a random statistical process — it is impossible to predict exactly when any specific nucleus will decay. The actual number of decays in any given time interval fluctuates randomly around the average value. Therefore, when measured count rates are plotted, they scatter around the smooth theoretical exponential curve. With larger samples and longer counting times, the scatter decreases but never disappears entirely.

Part (b)(ii): Determine Half-Life from Graph (2 marks)

From the activity vs time graph:

• Read the initial count rate: approximately 40 counts per second (at t = 0)
• Find the time at which activity = half of initial = 20 counts/s
• Reading from graph: this occurs at approximately t = 80 s

Half-life ≈ 80 s

(Students should draw construction lines on the graph: horizontal at 20 counts/s, vertical down to time axis.)

Part (b)(iii): Shape of Curve With Larger Decay Constant (2 marks)

A larger decay constant λ means the sample decays faster. From [latex]t_{1/2} = \frac{\ln 2}{\lambda} = \frac{0.693}{\lambda}[/latex]:

A larger λ → shorter half-life → the curve falls more steeply and reaches low count rates much sooner.

The new curve would:

• Start at the same initial count rate (same N₀)
• Decrease more rapidly (steeper slope)
• Reach half the initial activity in a shorter time
• The exponential curve is “compressed” horizontally compared to the original

Marks Weightage Summary

Difficulty Analysis

Easy — Reliable marks:

• Q.1 (radius of gyration — direct formula)
• Q.6 (quality of sound — recall)
• Q.9 (NIA = magnetic moment — direct)
• Q.11 (hypocentre — definition)
• Q.12(i) (surface tension definition)
• Q.13(i) (moment of inertia definition)
• Q.14(i) (harmonics definition)
• Q.15(i) (second law statement)
• Q.17(i) (symbols in Faraday’s equation)
• Q.19(i) (Bohr’s postulates)

Moderate — Needs good understanding:

• Q.3 (Moon capillary — calculation)
• Q.4 (refrigerator COP — numerical)
• Q.12(iii) (merged drops terminal velocity)
• Q.13(ii)(iii) (angular momentum conservation applications)
• Q.14(ii)(iii) (flute frequency + Doppler)
• Q.16(iii) (flux density with current wire method)
• Q.18(i) (photoelectric maximum speed)
• Q.20(iii) (S-wave maximum speed)

Challenging — For top scorers:

• Q.2 (Bernoulli + beam balance — conceptual application)
• Q.5 (P-V diagram work — needs careful graph reading)
• Q.7 (diffraction + white light — understanding of dispersion)
• Q.13 OR (Bernoulli derivation)
• Q.17(iii) (copper ball in magnetic field — eddy currents explanation)
• Q.20(i)(ii) (SHM proof + velocity derivation)
• Q.21 (LRC phasor diagram — all four parts)
• Q.21 OR (d) (potentiometer vs voltmeter precision argument)
• Q.22(b)(iii) (alpha particle striking oil drop — conceptual)

Student Preparation Tips

1. This Paper Tests “Why” More Than “What”
Unlike newer papers, the 2078 model frequently asks students to explain, justify, give reason. For every formula you learn, also learn the physical reason behind it. Why does ball P fall slower? Why can’t a refrigerator cool a room? Why does the potentiometer give a more precise reading? These conceptual explanations are worth 2 marks each.

2. Master the Mathematical Form of SHM (Q.20 is 8 marks)
The SHM group C question — proving F = -Ax gives SHM, deriving velocity expression, and calculating maximum speed — follows a completely predictable pattern. Master these three: (a) F = ma → a = -ω²x → SHM, (b) v·dv = -ω²x·dx → integrate → [latex]v = \pm\omega\sqrt{A^2 – x^2}[/latex], (c) v_max = ωA = 2πfA. Practice these until they take less than 5 minutes.

3. LRC Circuit Questions Need Both Graph Reading and Phasor Skills
Q.21 is 8 marks and covers phase constant determination from graphs, phasor diagram drawing, inductive vs capacitive circuit identification, and optimization of power transfer. Practice: (a) reading phase from I-V graphs, (b) drawing phasor diagrams with correct angle directions, (c) knowing that resonance maximizes power.

4. Millikan’s Oil Drop Experiment is Completely Learnable
Q.22 covers charge quantization, force balance condition, charge calculation, and effect of alpha particle strike. The numerical is straightforward (q = mg/E). The conceptual parts are logical. This entire question can be mastered in one focused study session.

5. Radioactive Decay Graph Interpretation
For Q.22 OR — reading half-life from a scattered activity-time graph, understanding why data scatters (statistical nature of decay), and predicting how the curve changes with larger λ — these are conceptual skills that don’t require calculation practice, just understanding.

6. Time Management for This Paper
Group A (25 min separate, 11 MCQs): Average 2.25 min per MCQ — don’t exceed this.
Group B (8 × 5 = 40 marks): ~50 min total, about 6 min per question.
Group C (3 × 8 = 24 marks): ~55 min total, about 18 min per question.
Buffer: 10 min for checking.

Frequently Asked Questions

Q: Is this 2078 model paper still relevant for NEB 2082/2083 board exams?
A: Yes — the core physics concepts never change. Surface tension, rotational dynamics, SHM, thermodynamics, electromagnetic induction, photoelectric effect, atomic physics, and radioactivity are all part of the new curriculum too. The 2078 paper tests these at a deeper conceptual level. Students who can answer 2078’s explanatory questions will handle 2080/2081 numerical questions with ease.

Q: What is the difference between Subject Code 102 and 1021’D’/’P’?
A: Subject Code 102 was the old curriculum code for Grade XII Physics. The new curriculum uses codes like 1021’D’ and 1021’P’. The syllabi overlap significantly but the new curriculum adds rotational dynamics (in more detail), SHM systems, wave optics derivations, de-Broglie wavelength, LCR circuits, and digital electronics.

Q: The Millikan’s oil drop experiment doesn’t appear in newer papers — should I still study it?
A: Yes, for two reasons. First, NEB can reintroduce any topic from the syllabus in any year. Second, understanding Millikan’s experiment builds a complete picture of charge quantization, which connects to atomic physics, photoelectric effect, and modern physics — all of which appear in newer papers.

Q: How do I draw a proper phasor diagram for an LRC circuit?
A: Start with the current phasor I₀ horizontally. Since V_R is in phase with I, draw V_R in the same direction. V_L leads I by 90° — draw it vertically upward. V_C lags I by 90° — draw it vertically downward. The net vertical component is (V_L – V_C). The total voltage phasor V₀ is the vector sum of V_R (horizontal) and (V_L – V_C) (vertical). The phase angle φ = arctan[(V_L – V_C)/V_R].

Q: Why is [latex]v_{max} = \omega A[/latex] in SHM?
A: From the velocity expression [latex]v = \omega\sqrt{A^2 – x^2}[/latex], maximum velocity occurs when x = 0 (at the equilibrium position). At x = 0: [latex]v_{max} = \omega\sqrt{A^2 – 0} = \omega A[/latex]. This makes physical sense — at the equilibrium position, all potential energy has been converted to kinetic energy.

Where to Find the Full Question Paper

The complete original question paper with all figures — including the beam balance diagram (Q.2), the P-V cycle diagram (Q.5), the junction current diagrams (Q.8), the heat engine figure (Q.15), the steel yoke magnet figure (Q.16), the copper ball magnetic field figure (Q.17), the Lyman series energy diagram (Q.19), the bridge rectifier figure (Q.19 OR), the LRC graph (Q.21), the potentiometer circuit (Q.21 OR), and the radioactive decay graph (Q.22 OR) — is available as a PDF above.

Conclusion

The NEB Grade XII Physics Model Question 2078 is one of the most conceptually rich Physics papers in Nepal’s examination history. Its questions about why the Earth spins faster when closer to the Sun, why a potentiometer beats a voltmeter, why an alpha particle strike destroys the oil drop’s equilibrium, and why tall buildings resonate with low-frequency S-waves — these are not just exam questions. They are windows into the deep physical principles that govern our universe.

Students who study this paper carefully do not just prepare for an examination — they develop the genuine physical intuition that separates a physicist from a formula-memorizer. Every “why” and “explain” and “justify” in this paper is an invitation to think more deeply.

Study this paper alongside the 2080 Set P and 2081 Set D papers, and you will have covered virtually every important concept in the NEB Grade XII Physics syllabus.

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