NEB Class 12 Science Physics Question Paper 2082 (Set H) Nepal

For a rod of mass $M$ and length $L$: about the centre $I_{c}=\frac{ML^{2}}{12}$ and about one end $I_{e}=\frac{ML^{2}}{3}$. The difference is $I_{e}-I_{c}=\frac{ML^{2}}{3}-\frac{ML^{2}}{12}=\frac{4ML^{2}-ML^{2}}{12}=\frac{3ML^{2}}{12}=\frac{ML^{2}}{4}$.

From the graph, amplitude $A=0.20\,\text{m}$ and period $T=2.0\,\text{s}$, so $\omega=\frac{2\pi}{T}=\pi\,\text{rad/s}$. Maximum velocity (at the mean position) is $v_{max}=A\omega=0.20\times\pi=0.2\pi\,\text{m/s}$.

Terminal velocity $v\propto r^{2}$. Combining two equal drops: $2\cdot\frac{4}{3}\pi r^{3}=\frac{4}{3}\pi R^{3}$, so $R=2^{1/3}r$. Then $v'=v\left(\frac{R}{r}\right)^{2}=0.15\times 2^{2/3}=0.15\times 1.587\approx 0.238\approx 0.24\,\text{m/s}$.

In an adiabatic process there is no heat exchange between the system and surroundings ($Q=0$), i.e. the system is insulated from the surroundings. Volume, pressure and temperature all change.

Steam point $T_{1}=373\,\text{K}$, ice point $T_{2}=273\,\text{K}$. Efficiency $\eta=1-\frac{T_{2}}{T_{1}}=1-\frac{273}{373}=\frac{100}{373}=0.268=26.8\%$.

Sound in air and waves inside water are longitudinal. A wave produced in the stem (prong) of a vibrating tuning fork and waves in a stretched string are transverse, but the standard textbook answer here is the wave produced in the stretched string, which is the classic example of a transverse mechanical wave.

Huygens' wavefront construction explains reflection, refraction and diffraction (wave phenomena) but does not explain polarization, which requires the transverse nature of light, established later. Hence polarization is not explained by Huygens' construction.

Magnetic field at the centre of a circular coil (single turn): $B=\frac{\mu_{0}I}{2R}$.

Relative permeability is related to susceptibility by $\mu_{r}=1+\chi$.

Flux $\phi=BA\cos 0^{\circ}=10^{3}\times 10^{-2}=10\,\text{Weber}$.

The point inside the earth where seismic waves originate (the focus) is called the hypocenter. The point on the surface directly above it is the epicenter.

By conservation of angular momentum $L=I\omega=\text{constant}$ (no external torque). When the dancer folds her arms her moment of inertia $I$ decreases, so her angular velocity $\omega$ increases and she spins faster; when she stretches her arms $I$ increases and $\omega$ decreases, so she spins slower. This lets her control her rate of rotation.

Or (SHM): For a particle of mass $m$, displacement $y=A\sin\omega t$. Kinetic energy $K=\frac{1}{2}m\omega^{2}(A^{2}-y^{2})$ and potential energy $U=\frac{1}{2}m\omega^{2}y^{2}$. Total energy $E=K+U=\frac{1}{2}m\omega^{2}(A^{2}-y^{2})+\frac{1}{2}m\omega^{2}y^{2}=\frac{1}{2}m\omega^{2}A^{2}=\text{constant}$, independent of $y$.

For a rigid body, each particle of mass $m_{i}$ at distance $r_{i}$ has tangential acceleration $a_{i}=r_{i}\alpha$. The torque on it is $\tau_{i}=m_{i}a_{i}r_{i}=m_{i}r_{i}^{2}\alpha$. Total torque $\tau=\sum m_{i}r_{i}^{2}\,\alpha=I\alpha$, where $I=\sum m_{i}r_{i}^{2}$ is the moment of inertia. Hence $\tau=I\alpha$.

Or: The potential energy $U=\frac{1}{2}m\omega^{2}y^{2}$ varies parabolically with displacement $y$ — an upward-opening parabola symmetric about $y=0$, minimum (zero) at the mean position and maximum at the extreme positions.

Torque $\tau=F\times r$. A longer arm $r$ gives a larger torque (turning effect) for the same applied force, so less force is needed to loosen/tighten a bolt. Hence a longer wrench is preferred.

Or: $T=2\pi\sqrt{\frac{l}{g}}$, so $T\propto\frac{1}{\sqrt{g}}$. On the Moon $g_{m}=\frac{g}{6}$, so $T_{m}=T\sqrt{\frac{g}{g_{m}}}=T\sqrt{6}\approx 2.45\,T$.

Laminar (streamline) flow: the liquid flows in smooth parallel layers, every particle following the path of the one ahead, with velocity below the critical velocity; the layers do not mix. Turbulent flow: the flow is irregular and chaotic with eddies and mixing of layers, occurring when the velocity exceeds the critical velocity (high Reynolds number).

By continuity, $A_{1}v_{1}=A_{2}v_{2}$: $v_{2}=\frac{A_{1}v_{1}}{A_{2}}=\frac{200\times 2}{50}=8\,\text{m/s}$.

By Bernoulli's equation (horizontal pipe): $P_{1}+\frac{1}{2}\rho v_{1}^{2}=P_{2}+\frac{1}{2}\rho v_{2}^{2}$.

$P_{2}=P_{1}+\frac{1}{2}\rho(v_{1}^{2}-v_{2}^{2})=4\times10^{4}+\frac{1}{2}\times1000\times(2^{2}-8^{2})$

$=4\times10^{4}+500\times(4-64)=4\times10^{4}-30000=10000\,\text{N/m}^{2}=1\times10^{4}\,\text{Nm}^{-2}$.

Sink $T_{2}=9+273=282\,\text{K}$.

With $\eta_{1}=0.25$: $0.25=1-\frac{T_{2}}{T_{1}}\Rightarrow\frac{T_{2}}{T_{1}}=0.75\Rightarrow T_{1}=\frac{282}{0.75}=376\,\text{K}$.

With $\eta_{2}=0.50$: $0.50=1-\frac{T_{2}}{T_{1}'}\Rightarrow T_{1}'=\frac{282}{0.50}=564\,\text{K}$.

Increase in source temperature $=T_{1}'-T_{1}=564-376=188\,\text{K}$ (i.e. $188^{\circ}\text{C}$).

First law of thermodynamics: the heat supplied to a system equals the increase in its internal energy plus the work done by the system, $dQ=dU+dW$.

Yes, it follows the principle of conservation of energy: the heat energy given to the system is not lost but is accounted for partly as an increase in internal energy ($dU$) and partly as work done ($dW$). Energy is merely transformed, not created or destroyed.

Velocity of sound in a gas $v=\sqrt{\frac{\gamma P}{\rho}}$.

Effect of pressure: at constant temperature, increasing pressure increases density proportionally ($\frac{P}{\rho}$ remains constant by Boyle's law), so velocity of sound is independent of pressure.

Effect of temperature: $v\propto\sqrt{T}$ (in kelvin), so velocity increases with increase in temperature (about $0.61\,\text{m/s}$ rise per $1^{\circ}\text{C}$ in air).

The cliff first receives the sound (source moving toward stationary cliff): $f_{1}=f\frac{v}{v-v_{s}}=600\times\frac{340}{340-10}=600\times\frac{340}{330}$. The cliff reflects this, and the driver (now a moving observer approaching the stationary reflected source) hears: $f'=f_{1}\frac{v+v_{o}}{v}=600\times\frac{340}{330}\times\frac{340+10}{340}=600\times\frac{350}{330}=636.4\,\text{Hz}$.

So the echo frequency heard by the driver is approximately $636\,\text{Hz}$.

Kirchhoff's first law (junction/current law): the algebraic sum of currents meeting at a junction is zero, $\sum I=0$ (sum of currents entering a junction equals sum of currents leaving). It follows from conservation of charge.

Kirchhoff's second law (loop/voltage law): in any closed loop, the algebraic sum of the products of current and resistance (emf drops) equals the algebraic sum of the emfs, $\sum IR=\sum E$. It follows from conservation of energy.

Let the four arms have resistances $P,Q,R,S$ with a galvanometer $G$ across BD. At balance, no current flows through the galvanometer ($I_{g}=0$), so the current through $P$ also flows through $Q$ ($I_{1}$) and the current through $R$ also flows through $S$ ($I_{2}$).

Applying Kirchhoff's voltage law to loop ABDA: $I_{1}P-I_{2}R=0\Rightarrow I_{1}P=I_{2}R$. Applying it to loop BCDB: $I_{1}Q-I_{2}S=0\Rightarrow I_{1}Q=I_{2}S$.

Dividing: $\frac{I_{1}P}{I_{1}Q}=\frac{I_{2}R}{I_{2}S}\Rightarrow\frac{P}{Q}=\frac{R}{S}$. This is the balanced condition of the Wheatstone bridge.

Lenz's law: the direction of the induced current (or induced emf) is always such that it opposes the change in magnetic flux that produces it. It is a consequence of the conservation of energy.

The induced emf is $e=e_{0}\sin\omega t$, where $e_{0}=NBA\omega$. The graph of $e$ versus $t$ is a sine curve (sinusoidal), oscillating between $+e_{0}$ and $-e_{0}$ with period $T=\frac{2\pi}{\omega}$.

When current $i$ grows in an inductor of inductance $L$, the back emf is $e=-L\frac{di}{dt}$. The work done against this emf in time $dt$ is $dW=ei\,dt=Li\,di$.

Total work (energy stored) as current rises from $0$ to $I$: $W=\int_{0}^{I}Li\,di=\frac{1}{2}LI^{2}$.

Hence the energy stored in the magnetic field of the inductor is $U=\frac{1}{2}LI^{2}$.

The NAND gate is an AND gate followed by a NOT (a small circle at the output of the AND-gate symbol). Output $Y=\overline{A\cdot B}$.

Truth table:

The output is 0 only when both inputs are 1; otherwise it is 1.

In a centre-tapped full-wave rectifier, the secondary of a transformer has its centre tapped, and two diodes $D_{1}$ and $D_{2}$ are connected to its two ends; the load is connected between the junction of the diode cathodes and the centre tap.

During the positive half cycle the upper end is positive, $D_{1}$ is forward biased and conducts while $D_{2}$ is reverse biased; current flows through the load in one direction. During the negative half cycle the lower end is positive, $D_{2}$ conducts while $D_{1}$ is off; current again flows through the load in the same direction. Thus both halves of the AC input drive current through the load in the same direction, giving a pulsating DC output (full-wave rectification).

Mass $m=3\times10^{-11}\,\text{g}=3\times10^{-14}\,\text{kg}$, radius $r=2\times10^{-4}\,\text{cm}=2\times10^{-6}\,\text{m}$, $\eta=1.8\times10^{-5}$, charge $q=10\times1.6\times10^{-19}=1.6\times10^{-18}\,\text{C}$.

i) Without field, weight is balanced by viscous drag at terminal velocity: $mg=6\pi\eta r v$. $v=\frac{mg}{6\pi\eta r}=\frac{3\times10^{-14}\times 9.8}{6\pi\times1.8\times10^{-5}\times2\times10^{-6}}=\frac{2.94\times10^{-13}}{6.786\times10^{-10}}\approx 4.33\times10^{-4}\,\text{m/s}$ (downward).

ii) With a downward field $E$ on negative charge, electric force $qE$ acts upward. Net downward force $=mg-qE$. $qE=1.6\times10^{-18}\times3\times10^{5}=4.8\times10^{-13}\,\text{N}$; $mg=2.94\times10^{-13}\,\text{N}$. Here $qE>mg$, so the net force is $4.8\times10^{-13}-2.94\times10^{-13}=1.86\times10^{-13}\,\text{N}$ upward, and the drop moves upward with terminal speed $v'=\frac{|qE-mg|}{6\pi\eta r}=\frac{1.86\times10^{-13}}{6.786\times10^{-10}}\approx 2.74\times10^{-4}\,\text{m/s}$ (upward).

b) In Thomson's $e/m$ experiment the electric and magnetic fields are kept mutually perpendicular (and both perpendicular to the beam) so that the electric force ($eE$) and the magnetic force ($Bev$) act on the electron beam in exactly opposite directions along the same line. They can then be balanced ($eE=Bev$) to give the velocity $v=E/B$ of the electrons in a straight (undeflected) path, which is essential to determine the specific charge $e/m$.

X-rays are used to ionise the air inside the chamber so that the oil drops can pick up (or change) electric charge; this lets the drop acquire excess electrons whose charge can then be measured.

Or: (a) Decay constant ($\lambda$): the fraction of the radioactive nuclei that decay per unit time, i.e. $\lambda=\frac{-\,dN/dt}{N}$. Half life ($T_{1/2}$): the time in which half the nuclei present in a sample decay; $T_{1/2}=\frac{0.693}{\lambda}$. (b) The rate of decay is proportional to the number of undecayed nuclei: $\frac{dN}{dt}=-\lambda N$. Separating variables, $\frac{dN}{N}=-\lambda\,dt$. Integrating from $N_{0}$ at $t=0$ to $N$ at $t$: $\ln\frac{N}{N_{0}}=-\lambda t$, hence $N=N_{0}e^{-\lambda t}$.

Stationary (standing) waves are formed by the superposition of two identical progressive waves of the same amplitude and frequency travelling in opposite directions; the resultant pattern does not advance but has fixed nodes (zero amplitude) and antinodes (maximum amplitude).

Resultant: $y=2a\cos\frac{2\pi x}{\lambda}\sin\frac{2\pi vt}{\lambda}$ (taking one form). Nodes occur where the amplitude $2a\cos\frac{2\pi x}{\lambda}=0$, i.e. $\frac{2\pi x}{\lambda}=(2n+1)\frac{\pi}{2}$, giving $x=(2n+1)\frac{\lambda}{4}$. Consecutive nodes correspond to $n$ and $n+1$: $x_{n+1}-x_{n}=\frac{\lambda}{4}\big[(2n+3)-(2n+1)\big]=\frac{\lambda}{4}\times 2=\frac{\lambda}{2}$.

Hence the distance between two consecutive nodes is $\frac{\lambda}{2}$.

Or (YDSE): Fringe positions are bright at $y_{n}=\frac{n\lambda D}{d}$ and dark at $y_{m}=\frac{(2m+1)\lambda D}{2d}$. The fringe width $\beta=y_{n+1}-y_{n}=\frac{\lambda D}{d}$ is the same constant for both bright and dark fringes, so they are equally spaced.

Strain $=1\%=0.01$.

i) Stress $=Y\times\text{strain}=2\times10^{11}\times0.01=2\times10^{9}\,\text{N/m}^{2}$.

ii) Wave speed in string $v=\sqrt{\frac{\text{stress}}{\rho}}=\sqrt{\frac{2\times10^{9}}{7800}}=\sqrt{2.564\times10^{5}}\approx 506.4\,\text{m/s}$. For the second mode (first overtone) of a string fixed at both ends, $f_{2}=\frac{2v}{2L}=\frac{v}{L}=\frac{506.4}{1.5}\approx 337.6\,\text{Hz}$.

Or (grating): $d=\frac{1}{500}\,\text{mm}=\frac{10^{-3}}{500}\,\text{m}=2\times10^{-6}\,\text{m}$. For maxima $d\sin\theta=n\lambda$ with $n=2$, $\theta=30^{\circ}$: $\lambda=\frac{d\sin\theta}{n}=\frac{2\times10^{-6}\times0.5}{2}=5\times10^{-7}\,\text{m}=500\,\text{nm}$.

i) The figure shows three loops (antinodes at both open ends with two internal nodes), which corresponds to the third harmonic / second overtone of the open organ pipe.

ii) For an open pipe the harmonics are $f_{n}=\frac{nv}{2L}$. For this mode $n=3$, so $f_{3}=\frac{3v}{2L}$, where $v$ is the speed of sound in air and $L$ the length of the pipe.

Or (Brewster's law): When unpolarised light is incident on a transparent medium at the polarising angle $\theta_{p}$, the reflected light is completely plane polarised, and the refractive index equals the tangent of the polarising angle: $\mu=\tan\theta_{p}$. Proof: at the polarising angle the reflected and refracted rays are perpendicular, so $\theta_{p}+r=90^{\circ}\Rightarrow r=90^{\circ}-\theta_{p}$. By Snell's law $\mu=\frac{\sin\theta_{p}}{\sin r}=\frac{\sin\theta_{p}}{\sin(90^{\circ}-\theta_{p})}=\frac{\sin\theta_{p}}{\cos\theta_{p}}=\tan\theta_{p}$.

$\omega=2\pi f=2\pi\times\frac{500}{\pi}=1000\,\text{rad/s}$.

Inductive reactance $X_{L}=\omega L=1000\times0.1=100\,\Omega$. Capacitive reactance $X_{C}=\frac{1}{\omega C}=\frac{1}{1000\times2\times10^{-6}}=\frac{1}{2\times10^{-3}}=500\,\Omega$.

i) $\tan\phi=\frac{X_{L}-X_{C}}{R}=\frac{100-500}{400}=\frac{-400}{400}=-1$, so $\phi=-45^{\circ}$ (magnitude $45^{\circ}$).

ii) Since $X_{C}>X_{L}$ the circuit is net capacitive, so the current leads the voltage (by $45^{\circ}$).

Impedance ($Z$) is the total effective opposition offered by an ac circuit (combination of resistance and reactance) to the flow of alternating current. It is the ratio of the rms voltage to the rms current, $Z=\frac{V_{rms}}{I_{rms}}$, measured in ohms; for a series LCR circuit $Z=\sqrt{R^{2}+(X_{L}-X_{C})^{2}}$.

i) Wire P produces magnetic field at Q: $B_{1}=\frac{\mu_{0}I_{1}}{2\pi r}$. The force on length $l$ of Q is $F=B_{1}I_{2}l=\frac{\mu_{0}I_{1}I_{2}l}{2\pi r}$. So force per unit length $\frac{F}{l}=\frac{\mu_{0}I_{1}I_{2}}{2\pi r}$.

ii) As shown, both currents are in the same direction (parallel), so the force between the wires is attractive. (Anti-parallel currents would repel.)

iii) $\frac{F}{l}\propto\frac{1}{r}$. Halving the distance ($r\to r/2$) doubles the force per unit length: $\frac{F'}{l}=\frac{\mu_{0}I_{1}I_{2}}{2\pi(r/2)}=\frac{\mu_{0}I_{1}I_{2}}{\pi r}=2\times\frac{F}{l}$.

a) As the temperature of the hot junction is increased (cold junction fixed), the thermo-emf first increases, reaches a maximum at the neutral temperature ($\theta_{n}$), then decreases, becomes zero at the temperature of inversion ($\theta_{i}$), and reverses (becomes negative) on further heating. The variation is parabolic: $E=\alpha\theta+\frac{1}{2}\beta\theta^{2}$, where $\theta$ is the temperature difference between the junctions.

b) Driver current $I=\frac{V}{R_{B}+R_{wire}}=\frac{3}{290+10}=\frac{3}{300}=0.01\,\text{A}$. Resistance per metre of wire $=\frac{10}{4}=2.5\,\Omega/\text{m}$, so potential gradient $k=I\times2.5=0.01\times2.5=0.025\,\text{V/m}$. Balance length $=240\,\text{cm}=2.4\,\text{m}$. Unknown emf $E=k\times l=0.025\times2.4=0.06\,\text{V}$.

c) Ampere's circuital law: $\oint\vec{B}\cdot d\vec{l}=\mu_{0}I$. For a long straight conductor, take a circular Amperian loop of radius $r$ centred on the wire; by symmetry $B$ is constant along it and parallel to $d\vec{l}$, so $\oint B\,dl=B(2\pi r)=\mu_{0}I$, giving $B=\frac{\mu_{0}I}{2\pi r}$.

Threshold frequency is the minimum frequency of incident radiation that can just eject photoelectrons from the surface of a given metal. Below this frequency no photoemission occurs, however intense the light. It is related to the work function by $\phi=h f_{0}$.

Work function $\phi=2\,\text{eV}=2\times1.6\times10^{-19}=3.2\times10^{-19}\,\text{J}$.

Energy of incident photon $E=\frac{hc}{\lambda}=\frac{6.62\times10^{-34}\times3\times10^{8}}{150\times10^{-9}}=\frac{1.986\times10^{-25}}{1.5\times10^{-7}}=1.324\times10^{-18}\,\text{J}$.

Maximum kinetic energy $K_{max}=E-\phi=1.324\times10^{-18}-3.2\times10^{-19}=1.004\times10^{-18}\,\text{J}$.

$\frac{1}{2}mv^{2}=K_{max}\Rightarrow v=\sqrt{\frac{2K_{max}}{m}}=\sqrt{\frac{2\times1.004\times10^{-18}}{9.1\times10^{-31}}}=\sqrt{2.207\times10^{12}}\approx 1.49\times10^{6}\,\text{m/s}$.

Least (threshold) frequency $f_{0}=\frac{\phi}{h}=\frac{3.2\times10^{-19}}{6.62\times10^{-34}}\approx 4.83\times10^{14}\,\text{Hz}$.

Heisenberg's uncertainty principle states that it is impossible to simultaneously determine both the exact position and the exact momentum of a particle; the product of the uncertainties is at least of the order of $\frac{h}{4\pi}$: $\Delta x\cdot\Delta p\ge\frac{h}{4\pi}$.

In Bohr's model, the Coulomb attraction provides the centripetal force for the electron in the $n^{th}$ orbit of radius $r_{n}$: $\frac{1}{4\pi\varepsilon_{0}}\frac{e^{2}}{r_{n}^{2}}=\frac{mv_{n}^{2}}{r_{n}}\quad(1)$

Bohr's quantisation of angular momentum: $mv_{n}r_{n}=\frac{nh}{2\pi}\Rightarrow r_{n}=\frac{nh}{2\pi mv_{n}}\quad(2)$

From (1): $\frac{e^{2}}{4\pi\varepsilon_{0}r_{n}}=mv_{n}^{2}\Rightarrow r_{n}=\frac{e^{2}}{4\pi\varepsilon_{0}mv_{n}^{2}}$. Equating with (2): $\frac{nh}{2\pi mv_{n}}=\frac{e^{2}}{4\pi\varepsilon_{0}mv_{n}^{2}}$

Solving for $v_{n}$: $v_{n}=\frac{e^{2}}{2\varepsilon_{0}nh}=\frac{e^{2}}{2nh\varepsilon_{0}}$.

Thus the velocity in the $n^{th}$ orbit is $v_{n}=\frac{e^{2}}{2\varepsilon_{0}nh}$, i.e. $v_{n}\propto\frac{1}{n}$.