NEB Class 12 Science Physics Question Paper 2080 (Set A) Nepal

The relation between moment of inertia $I$, mass $M$, and radius of gyration $k$ is given by $I = Mk^2$. Solving for $k$ gives $k = \sqrt{\frac{I}{M}}$. Therefore, the correct option is (C).

Maximum velocity $v_{\text{max}} = \omega A = 4$ and maximum acceleration $a_{\text{max}} = \omega^2 A = 2$. Dividing $a_{\text{max}}$ by $v_{\text{max}}$ gives $\omega = \frac{2}{4} = 0.5~\text{rad/s}$. The time period $T = \frac{2\pi}{\omega} = \frac{2\pi}{0.5} = 4\pi~\text{sec}$. Therefore, the correct option is (D).

In a PV diagram, work done is the area under the curve. For expansion between the same initial and final volumes, the pressure remains highest during an isobaric process, thus enclosing the largest area and yielding maximum work done. Therefore, the correct option is (C).

A refrigerator is a heat engine operating in reverse; it extracts heat from the inside and releases it outside into the room along with the work done by the compressor compressor. If the door is left open, it removes heat from the room and discharges that same heat plus the extra heat generated by electrical work back into the room, causing the temperature to increase. Therefore, the correct option is (A).

In a stationary or standing wave, the distance between two consecutive nodes or two consecutive antinodes is equal to half of the wavelength ($\lambda/2$). Therefore, the correct option is (B).

We know that refractive index $\mu = \frac{1}{\sin C}$. Given $C = \sin^{-1}(\frac{3}{5}) \implies \sin C = \frac{3}{5}$, so $\mu = \frac{5}{3}$. By Brewster's law, $\mu = \tan i_p \implies i_p = \tan^{-1}(\mu) = \tan^{-1}(\frac{5}{3})$. Therefore, the correct option is (D).

In the magnetic hysteresis loop, the value of the reverse magnetizing field $H$ required to reduce the residual magnetic flux density $B$ to zero is defined as coercivity. Therefore, the correct option is (C).

The specified standard AC supply voltage of $220~\text{V}$ represents the root-mean-square voltage ($V_{\text{rms}}$). The relation between peak voltage ($V_0$) and $V_{\text{rms}}$ is $V_0 = \sqrt{2} \times V_{\text{rms}} = 1.414 \times 220 \approx 311~\text{V}$. The closest option provided is $310~\text{V}$. Therefore, the correct option is (A).

The magnetic field inside a long solenoid is given by $B = \mu_0 n I$. When the new current $I' = 2I$ and new turns per unit length $n' = n/2$, the new magnetic field becomes $B' = \mu_0 (n/2) (2I) = \mu_0 n I = B$. Therefore, the correct option is (A).

Primary waves (P-waves) are longitudinal waves that travel the fastest through the Earth's crust and thus arrive first and are recorded first on a seismograph. Therefore, the correct option is (B).

a) Definition of Simple Harmonic Motion (SHM)

Simple harmonic motion is the to-and-fro periodic motion of a particle in which the restoring force (or acceleration) is directly proportional to the displacement from the mean position and is always directed towards that mean position.

b) Total Mechanical Energy in SHM

For a particle of mass $m$ executing SHM of amplitude $A$ and angular frequency $\omega$, the displacement is $x = A\sin\omega t$ and velocity is $v = A\omega\cos\omega t = \omega\sqrt{A^2 - x^2}$.

Kinetic Energy (KE):

Potential Energy (PE): The restoring force is $F = -m\omega^2 x$, so

Total Energy (E):

The total mechanical energy is constant and independent of displacement, confirming conservation of energy.

c) Graphical Variation with Displacement

Plotting energy (y-axis) against displacement $x$ (x-axis), from $-A$ to $+A$:

• Potential Energy is an upward parabola $PE = \tfrac{1}{2}m\omega^2 x^2$, zero at $x=0$ (mean) and maximum at $x=\pm A$.
• Kinetic Energy is an inverted (downward) parabola, maximum at $x=0$ and zero at $x=\pm A$.
• Total Energy is a horizontal straight line at $E = \tfrac{1}{2}m\omega^2 A^2$, constant for all $x$.

The KE and PE curves intersect at $x = \pm A/\sqrt{2}$, where each equals half the total energy.

a) Surface Energy and its Relation with Surface Tension

The surface energy of a liquid is the extra potential energy possessed by the molecules in the surface film due to the work done in bringing them from the interior to the surface against the inward cohesive force, i.e. the work required to increase the surface area by unit amount at constant temperature.

If surface tension is $T$ and the area is increased by $\Delta A$, the work done (= increase in surface energy) is

Hence surface tension equals surface energy per unit area:

Numerically, surface tension (N/m) is equal to the surface energy per unit area (J/m²).

b) Coefficient of Viscosity (Rising Air Bubble)

An air bubble rising at steady (terminal) speed obeys Stokes' law. Since the density of air is neglected, the upward effective body force is the buoyant weight of the displaced liquid, balanced by the viscous drag.

Given: radius $r = 1~\text{cm} = 0.01~\text{m}$, terminal velocity $v = 5~\text{mm/s} = 5\times10^{-3}~\text{m/s}$, liquid density $\rho = 0.8~\text{g/cm}^3 = 800~\text{kg/m}^3$, $g = 9.8~\text{m/s}^2$.

For a bubble (density of air negligible), terminal velocity:

a) PV Diagram and Working of a Petrol Engine (Otto Cycle)

The petrol engine works on the Otto cycle, consisting of two adiabatics and two isochorics (plus intake/exhaust strokes). On a PV diagram the closed loop has the following stages:

• Intake (5→1): Air–petrol mixture is drawn in at constant pressure (horizontal line at the bottom).
• Compression (1→2): The mixture is compressed adiabatically; pressure and temperature rise steeply (steep curve upward-left).
• Power/explosion (2→3): The spark ignites the mixture; heat is added at constant volume, raising pressure sharply (vertical line up).
• Working/expansion (3→4): Hot gases expand adiabatically pushing the piston and doing work (curve downward-right).
• Exhaust valve opens (4→1): Heat is rejected at constant volume; pressure falls (vertical line down).
• Exhaust (1→5): Burnt gases are pushed out at constant pressure.

The enclosed area of the loop represents the net work done per cycle. The cycle is four-stroke: suction, compression, power and exhaust.

b) Why $C_p > C_v$

• $C_v$ is the molar heat capacity at constant volume; all heat supplied raises only the internal energy (temperature) since no external work is done ($\Delta V = 0$).
• $C_p$ is the molar heat capacity at constant pressure; when heated, the gas expands and does external work against the surroundings in addition to raising its internal energy.

Thus to raise the temperature of one mole by 1 K at constant pressure, extra heat is needed to perform the work of expansion. By the first law and Mayer's relation:

Hence $C_p$ is always greater than $C_v$.

a) Doppler's Effect and Apparent Frequency (Source Approaching)

Doppler's effect is the apparent change in the frequency (or pitch) of a wave observed due to the relative motion between the source of the wave and the observer.

Source moving towards a stationary observer: Let the true frequency be $f$, velocity of sound $v$, and source speed $v_s$. In one period $T$ the source advances a distance $v_s T$ towards the observer, so the wavelength ahead is compressed:

The apparent frequency heard is

Since $v - v_s < v$, $f' > f$, i.e. the pitch increases as the source approaches.

b) Sound Waves as Pressure Waves

Sound is a longitudinal wave that propagates through a medium as alternate compressions (regions of high density and high pressure) and rarefactions (regions of low density and low pressure). As the wave travels, each layer of the medium undergoes periodic variation in pressure about the normal atmospheric value. Because the disturbance is described most directly by this oscillating pressure variation, sound waves are also called pressure waves. The pressure variation is maximum at displacement nodes and minimum at displacement antinodes, i.e. the pressure wave is $90°$ out of phase with the displacement wave.

a) Lenz's Law and Conservation of Energy

Lenz's law: The direction of an induced current (or induced emf) is always such that it opposes the very cause (the change in magnetic flux) that produces it.

Does it violate energy conservation? No. Lenz's law is in fact a consequence of the conservation of energy. Because the induced current opposes the change in flux, work must be done by the external agent (e.g. pushing the magnet) against this opposing force. This mechanical work done against the opposition is what gets converted into the electrical energy of the induced current. If the induced current aided the change instead, the motion would accelerate on its own and produce energy from nothing, violating conservation. Hence the opposition demanded by Lenz's law guarantees that energy is conserved.

b) Direction of Induced Current when Magnet is Withdrawn

Yes, the direction reverses. When the magnet moves towards the solenoid, the flux through the coil increases, so by Lenz's law the induced current flows in a direction that makes the near face of the coil behave like the same pole as the approaching pole, repelling the magnet (current in one sense through the galvanometer).

When the magnet moves away, the flux now decreases, so the induced current flows in the opposite sense to oppose this decrease, making the near face behave like the opposite pole and attracting the magnet. Since the induced current now opposes the withdrawal instead of the approach, its direction is reversed and the galvanometer deflects to the opposite side.

a) Definition of One Ampere

One ampere is that steady current which, when maintained in each of two infinitely long straight parallel conductors of negligible cross-section placed 1 metre apart in vacuum, produces between them a force of $2\times10^{-7}$ newton per metre of length.

b) Contraction of a Solenoidal Coil Carrying Current

When current flows through a solenoid, adjacent turns carry current in the same direction. Two parallel conductors carrying currents in the same direction attract each other. Therefore the neighbouring turns of the solenoid attract one another, pulling the turns closer together, and the coil tends to contract (shorten) along its length.

c) Potential Difference for a Self-Supporting Wire

Given: mass $m = 0.12~\text{g} = 0.12\times10^{-3}~\text{kg}$, length $L = 10~\text{cm} = 0.10~\text{m}$, $B = 0.6~\text{T}$, resistance per unit length $= 3.8~\Omega\,\text{m}^{-1}$, $g = 9.8~\text{m/s}^2$.

For the wire to be just self-supporting, the magnetic force must balance its weight:

Resistance of the wire:

Required potential difference:

a) Seebeck Effect

The Seebeck effect is the phenomenon in which an electromotive force (thermo-emf) is set up and an electric current flows in a closed circuit made of two different metals when their two junctions are maintained at different temperatures.

b) Variation of Thermo-emf with Temperature of Junction

Keeping the cold junction at a fixed temperature and gradually raising the hot-junction temperature $\theta$, the thermo-emf $E$ varies parabolically:

• As $\theta$ increases, $E$ first increases, reaches a maximum value at the neutral temperature $\theta_n$.
• On heating further, $E$ decreases, becomes zero at the temperature of inversion $\theta_i$, and then reverses direction beyond it.

The neutral temperature is independent of the cold-junction temperature, whereas the temperature of inversion depends on it. They are symmetrically placed about the neutral temperature.

c) Temperature of Inversion

The neutral temperature is the mean of the cold-junction and inversion temperatures:

Given: $\theta_c = 25^{\circ}\text{C}$, $\theta_n = 270^{\circ}\text{C}$.

a) Specific Charge of an Electron

The specific charge of a particle is the ratio of its charge to its mass, $e/m$. For an electron its value is $1.76\times10^{11}~\text{C/kg}$.

Why constant for cathode rays but not for positive rays: Cathode rays always consist of the same particle — the electron — regardless of the gas used, so $e/m$ has a single fixed value. Positive rays, however, are positive ions produced from the residual gas; the nature (and hence mass) of these ions depends on the gas in the tube, and they may carry single or multiple positive charges. Therefore their charge-to-mass ratio varies from gas to gas and is not constant.

b) Electron Beam in Crossed Fields

Given: $V = 1.36\times10^4~\text{V}$ across plates $d = 4~\text{cm} = 0.04~\text{m}$, $B = 2\times10^{-3}~\text{Wb/m}^2$, $e = 1.6\times10^{-19}~\text{C}$, $m = 9.1\times10^{-31}~\text{kg}$.

Electric field between plates:

i) Velocity for no deflection (crossed fields balance):

ii) Radius when electric field is removed (only magnetic field): The magnetic force provides the centripetal force:

a) Half-life and Decay Constant

• Half-life ($T_{1/2}$): The time taken for half the radioactive nuclei in a given sample to disintegrate (decay), i.e. for the number of undecayed nuclei to reduce to one-half of its initial value.
• Decay constant ($\lambda$): The fraction of the total number of radioactive nuclei that decays per unit time; equivalently, the reciprocal of the mean (average) life. It is related to half-life by $T_{1/2} = \dfrac{0.693}{\lambda}$.

b) Time for Equal Amounts of A and B

Given: Initial ratio $N_{0A} : N_{0B} = 1 : 4$. Half-lives $T_A = 100$ years, $T_B = 50$ years.

Using $N = N_0\left(\tfrac{1}{2}\right)^{t/T_{1/2}}$:

Setting $N_A = N_B$:

a) Characteristic Used in Rectification

A junction diode conducts appreciable current only when forward biased and blocks current when reverse biased; this unidirectional (one-way) conduction property is used for rectification.

b) Junction Diode as a Full-Wave Rectifier

A full-wave rectifier (centre-tap type) uses two diodes $D_1$ and $D_2$ fed from the two halves of a centre-tapped transformer secondary, with the load $R_L$ between the centre tap and the joined diode outputs.

• During the positive half-cycle of the AC input, $D_1$ is forward biased (conducts) while $D_2$ is reverse biased; current flows through $R_L$ in a fixed direction.
• During the negative half-cycle, $D_2$ becomes forward biased and $D_1$ reverse biased; current again flows through $R_L$ in the same direction.

Thus both halves of the AC cycle drive current through the load in one direction, giving a pulsating DC output for the complete cycle.

c) AND Gate Followed by NOT Gate (NAND)

Feeding the output of a two-input AND gate into a NOT gate gives a NAND gate.

Logic symbol: A standard AND-gate symbol (D-shape) with a small circle (bubble) at its output, equivalent to an AND followed by a NOT.

Boolean expression: $Y = \overline{A \cdot B}$

Truth table:

a) Significance of Negative Energy of an Orbiting Electron

The negative energy of an electron in an orbit signifies that the electron is bound to the nucleus. The zero of energy is taken when the electron is free at infinity; the negative value means energy must be supplied (work done) to remove the electron from the atom and free it, i.e. the electron cannot escape on its own.

b) Energy Level and Photon Momentum

Given: transition $n=4 \to n=2$, $E_2 = -13.6~\text{eV}$, wavelength $\lambda = 121.9~\text{nm} = 121.9\times10^{-9}~\text{m}$.

Photon energy:

i) Energy of $n=4$ level: $E_4 - E_2 = E_{photon}$

ii) Momentum of the photon:

c) Photoelectric Effect on Silver ($\phi = 4.7~\text{eV}$)

i) Work function: The minimum energy required to just eject (liberate) an electron from the surface of a metal.

ii) Maximum kinetic energy (Einstein's equation $KE_{max} = E - \phi$):

iii) Stopping potential $V_0$: $eV_0 = KE_{max}$

d) de-Broglie Wavelength of an Electron (2 kV)

For an electron accelerated through $V = 2000~\text{V}$:

a) Pattern from Two Slits

The pattern obtained is an interference pattern (Young's double-slit fringes). Light from the single monochromatic source reaches the two slits, which act as two coherent sources. The secondary wavelets from the two slits superpose on the screen; where they arrive in phase (path difference $= n\lambda$) constructive interference gives bright fringes, and where they arrive out of phase (path difference $= (n+\tfrac{1}{2})\lambda$) destructive interference gives dark fringes, producing equally spaced bright and dark bands.

b) Pattern with One Slit Covered

With only a single slit open, the pattern obtained is a single-slit diffraction pattern — a broad central bright band with much fainter secondary maxima on either side.

c) Difference on the Basis of Huygens' Principle

In interference (two slits), superposition occurs between wavelets coming from two separate coherent sources (slits); the fringes are of nearly equal width and equal intensity. In diffraction (single slit), superposition occurs between the infinitely many secondary wavelets originating from different points of the same single wavefront within one slit; this gives a bright central maximum that is much wider and brighter than the rapidly fading side maxima of unequal width.

d) Condition for First Minimum in Single-Slit Diffraction

Consider a slit of width $d$ illuminated normally by light of wavelength $\lambda$, diffracted at angle $\theta$. Imagine the slit divided into two equal halves of width $d/2$. The path difference between wavelets from corresponding points of the two halves is $\tfrac{d}{2}\sin\theta$.

For these to cancel in pairs (destructive interference, first minimum), this path difference must equal $\lambda/2$:

This is the condition for the first minimum of intensity.

e) Wavelength from a Diffraction Grating

Given: 500 lines per mm, order $n = 2$, angle $\theta = 30^{\circ}$.

Grating element:

Grating equation $d\sin\theta = n\lambda$:

a) Open Organ Pipe

An open organ pipe is a cylindrical air column (pipe) that is open at both ends, in which longitudinal stationary sound waves are set up with antinodes at both open ends.

b) Modes of Vibration of an Open Organ Pipe

In an open pipe both ends are antinodes. Let pipe length be $L$ and speed of sound $v$.

• Fundamental (first harmonic): One node at the centre, antinodes at both ends. $L = \dfrac{\lambda_1}{2} \Rightarrow \lambda_1 = 2L$, so $f_1 = \dfrac{v}{2L}$.
• Second harmonic (first overtone): $L = \lambda_2 \Rightarrow \lambda_2 = L$, so $f_2 = \dfrac{v}{L} = 2f_1$.
• Third harmonic (second overtone): $L = \dfrac{3\lambda_3}{2} \Rightarrow \lambda_3 = \dfrac{2L}{3}$, so $f_3 = \dfrac{3v}{2L} = 3f_1$.

Thus the frequencies are in the ratio $f_1 : f_2 : f_3 = 1 : 2 : 3$, i.e. an open pipe produces both even and odd harmonics, giving a richer musical quality.

c) Why a Musician Turns the Screws

The frequency (pitch) of a stretched string is $f = \dfrac{1}{2L}\sqrt{\dfrac{T}{\mu}}$, so it depends on the tension $T$. By turning the tuning screws (pegs), the musician changes the tension in the strings, thereby changing their fundamental frequencies until each string sounds the correct pitch — this is how the guitar is tuned.

d) Fundamental Frequency of a Stretched Steel Wire

Given: $L = 1~\text{m}$, mass $= 80~\text{g} = 0.08~\text{kg}$, tension $T = 1000~\text{N}$.

Mass per unit length:

Fundamental frequency:

a) Kirchhoff's Laws

1. Kirchhoff's Current Law (Junction rule): The algebraic sum of currents meeting at any junction in a circuit is zero, i.e. the total current entering a junction equals the total current leaving it ($\sum I = 0$). It is based on conservation of charge.

2. Kirchhoff's Voltage Law (Loop rule): In any closed loop of a circuit, the algebraic sum of the emfs equals the algebraic sum of the products of current and resistance ($\sum E = \sum IR$). It is based on conservation of energy.

b) Wheatstone Bridge

i) Circuit diagram: Four resistors $P$, $Q$, $R$, $S$ form the four arms of a diamond-shaped quadrilateral ABCD. A cell with key is connected across one diagonal (A–C) and a galvanometer $G$ across the other diagonal (B–D).

        B
       / \
     P/   \Q
     /     \
    A---G---C   (battery across A-C)
     \     /
     R\   /S
       \ /
        D

ii) Principle: When the bridge is balanced, no current flows through the galvanometer, and the four resistances are related by

iii) Derivation using Kirchhoff's laws: At balance, galvanometer current $I_g = 0$. Let current $I_1$ flow through $P$ and $R$, and $I_2$ through $Q$ and $S$.

Since $I_g = 0$, points B and D are at the same potential.

• Loop ABDA: potential drop across $P$ = drop across $Q$: $I_1 P = I_2 Q$ ...(1)
• Loop BCDB: drop across $R$ = drop across $S$: $I_1 R = I_2 S$ ...(2)

Dividing (1) by (2):

This is the balanced condition of the Wheatstone bridge.

c) Thermoelectric Effect and Variation of Thermo-emf

Thermoelectric (Seebeck) effect: When two different metals are joined to form a closed circuit and their two junctions are kept at different temperatures, an emf (thermo-emf) is developed and a current flows in the circuit.

Variation with hot-junction temperature $\theta$ (cold junction fixed):

• As $\theta$ increases, $E$ rises, reaching a maximum at the neutral temperature $\theta_n$.
• On further heating, $E$ decreases, falls to zero at the temperature of inversion $\theta_i$, and then reverses direction.

The graph of $E$ versus $\theta$ is a downward parabola, symmetric about $\theta_n$, with $\theta_n = \tfrac{1}{2}(\theta_c + \theta_i)$.

a) Electrical Resonance of a Series LCR Circuit

Electrical resonance is the condition in a series LCR circuit when the inductive reactance equals the capacitive reactance ($X_L = X_C$). The net reactance becomes zero, the impedance is minimum (= $R$), and the current is maximum and in phase with the applied voltage.

b) Resonant Frequency

At resonance $X_L = X_C$:

Since $\omega = 2\pi f_r$:

c) LCR Circuit Calculations

Given: $R = 8~\Omega$, $C = 800~\mu\text{F} = 800\times10^{-6}~\text{F}$, $L = 300~\text{mH} = 0.3~\text{H}$, source $220~\text{V}$, $50~\text{Hz}$.

(The inductance is taken as $300~\text{mH}$, the standard value for this NEB problem.)

Angular frequency: $\omega = 2\pi f = 2\pi \times 50 = 314.2~\text{rad/s}$.

Reactances:

i) Impedance:

ii) Current in the circuit:

iii) Voltage across resistor R:

d) Wattless Current

Wattless current is the component of the AC current that is $90^{\circ}$ out of phase (quadrature) with the applied voltage, so it consumes no net (average) power over a cycle. It occurs in a purely reactive (inductive or capacitive) circuit where the average power $P = V_{rms}I_{rms}\cos\phi = 0$ since $\cos 90^{\circ} = 0$.