AIPMT 2001 Physics Question Paper with Answer and Solution

(C) The magnitude of the resultant of two vectors $\vec A$ and $\vec B$ is given by the formula:
$|\vec R| = |\vec A + \vec B| = \sqrt{A^2 + B^2 + 2AB \cos \theta}$,where $\theta$ is the angle between the two vectors.
Given the condition $|\vec A + \vec B| = |\vec A| + |\vec B|$,we substitute the magnitude formula:
$\sqrt{A^2 + B^2 + 2AB \cos \theta} = A + B$
Squaring both sides:
$A^2 + B^2 + 2AB \cos \theta = (A + B)^2$
$A^2 + B^2 + 2AB \cos \theta = A^2 + B^2 + 2AB$
$2AB \cos \theta = 2AB$
$\cos \theta = 1$
Therefore,$\theta = 0^o$.

(D) The energy of a photon is given by $E = h\nu$,where $h$ is Planck's constant and $\nu$ is the frequency.
Thus,the dimension of $h$ is $[h] = [E] / [\nu]$.
$[E] = [M L^2 T^{-2}]$ and $[\nu] = [T^{-1}]$.
So,$[h] = [M L^2 T^{-2}] / [T^{-1}] = [M L^2 T^{-1}]$.
Angular momentum $L$ is defined as $L = mvr$,where $m$ is mass,$v$ is velocity,and $r$ is radius.
$[L] = [M] [L T^{-1}] [L] = [M L^2 T^{-1}]$.
Therefore,the dimension of Planck's constant is equal to that of angular momentum.

(B) Let the initial velocity be $u$ and the maximum height be $H$. The maximum height is given by $H = \frac{u^2}{2g}$.
At height $h = \frac{H}{2}$,the velocity $v = 10 \; m/s$.
Using the equation of motion $v^2 = u^2 - 2gh$:
$(10)^2 = u^2 - 2g \left( \frac{H}{2} \right)$
$100 = u^2 - gH$
Substitute $H = \frac{u^2}{2g}$ into the equation:
$100 = u^2 - g \left( \frac{u^2}{2g} \right)$
$100 = u^2 - \frac{u^2}{2} = \frac{u^2}{2}$
$u^2 = 200 \; m^2/s^2$.
Now,calculate the maximum height $H$:
$H = \frac{u^2}{2g} = \frac{200}{2 \times 10} = \frac{200}{20} = 10 \; m$.

(B) The initial kinetic energy of the ball is $E = \frac{1}{2}mv^2$,where $v$ is the initial velocity.
At the highest point of the trajectory,the vertical component of the velocity becomes zero,while the horizontal component remains constant.
The horizontal component of velocity is $v_x = v \cos \theta$.
At the highest point,the velocity of the ball is $v_h = v \cos \theta$.
The kinetic energy at the highest point is $E' = \frac{1}{2}m(v_h)^2 = \frac{1}{2}m(v \cos \theta)^2$.
$E' = (\frac{1}{2}mv^2) \cos^2 \theta = E \cos^2 \theta$.
Given $\theta = 45^\circ$,we have $\cos 45^\circ = \frac{1}{\sqrt{2}}$.
Therefore,$E' = E (\frac{1}{\sqrt{2}})^2 = E (\frac{1}{2}) = \frac{E}{2}$.

(A) The limiting friction force is given by $F_l = \mu mg = 0.6 \times 1 \times 9.8 = 5.88 \; N$.
The pseudo force acting on the block in the frame of the truck is $F_p = ma = 1 \times 5 = 5 \; N$.
Since the pseudo force $(5 \; N)$ is less than the limiting friction $(5.88 \; N)$,the block does not slide relative to the truck.
Therefore,the static frictional force acting on the block is equal to the applied pseudo force,which is $5 \; N$.

(A) The maximum theoretical efficiency of a heat engine is given by the Carnot efficiency formula: $\eta_{\max} = 1 - \frac{T_2}{T_1}$.
First,convert the temperatures from Celsius to Kelvin: $T_1 = 127 + 273 = 400 \text{ K}$ and $T_2 = 27 + 273 = 300 \text{ K}$.
Substituting these values into the formula: $\eta_{\max} = 1 - \frac{300}{400} = 1 - 0.75 = 0.25$ or $25\%$.
Since the second law of thermodynamics states that no heat engine can have an efficiency greater than the Carnot efficiency operating between the same two temperatures,an efficiency of $26\%$ is impossible.

(B) The rate of heat flow through a cylindrical rod is given by the formula: $\frac{dQ}{dt} = \frac{kA(T_1 - T_2)}{L}$,where $A = \pi r^2$.
Thus,the rate of heat flow is proportional to $\frac{r^2}{L}$.
Let the initial radius be $r_1$ and length be $L_1$. Then $Q_1 \propto \frac{r_1^2}{L_1}$.
When all linear dimensions are doubled,the new radius $r_2 = 2r_1$ and the new length $L_2 = 2L_1$.
The new rate of heat flow $Q_2 \propto \frac{r_2^2}{L_2} = \frac{(2r_1)^2}{2L_1} = \frac{4r_1^2}{2L_1} = 2 \left( \frac{r_1^2}{L_1} \right)$.
Therefore,$Q_2 = 2Q_1$.

(B) The total energy $E$ of a particle performing Simple Harmonic Motion $(S.H.M.)$ is given by the formula:
$E = \frac{1}{2} K a^2$
Where:
$K$ is the force constant (or spring constant),
$a$ is the amplitude of the oscillation.
Since the total energy is the sum of kinetic and potential energy at any point,it remains constant throughout the motion and is independent of the instantaneous displacement $x$.
Therefore,the total energy depends only on the force constant $K$ and the amplitude $a$.

(D) The standard wave equation is given by $y = A \sin(\omega t - kx)$.
Comparing this with the given equation $y = 10^{-4} \sin(100t - x/10)$,we get:
Angular frequency $\omega = 100 \, rad/s$.
Wave number $k = 1/10 \, m^{-1}$.
The velocity of the wave $v$ is given by the formula $v = \frac{\omega}{k}$.
Substituting the values,$v = \frac{100}{1/10} = 1000 \, m/s$.

(B) The resultant amplitude $A$ of two superimposing waves with equal amplitude $a$ is given by the formula: $A^2 = a^2 + a^2 + 2a^2 \cos \phi$,where $\phi$ is the phase difference.
Given that the resultant amplitude $A$ is equal to the amplitude of the individual waves,$A = a$.
Substituting this into the equation: $a^2 = a^2 + a^2 + 2a^2 \cos \phi$.
$a^2 = 2a^2 + 2a^2 \cos \phi$.
$-a^2 = 2a^2 \cos \phi$.
$\cos \phi = -\frac{1}{2}$.
Since $\cos(120^\circ) = -\frac{1}{2}$,the phase difference $\phi = \frac{2\pi}{3}$ radians.

(C) The fundamental frequency $n$ of a sonometer wire is given by the formula: $n = \frac{1}{2l} \sqrt{\frac{T}{\mu}}$,where $\mu = \pi r^2 \rho$ is the linear mass density.
Thus,$n \propto \sqrt{\frac{T}{r^2 \rho}}$.
Let the initial values be $T_1 = T$,$r_1 = r$,and $\rho_1 = \rho$. The new values are $T_2 = 2T$,$r_2 = 2r$,and $\rho_2 = \frac{\rho}{2}$.
The ratio of the frequencies is given by:
$\frac{n_2}{n_1} = \sqrt{\frac{T_2}{T_1} \cdot \left(\frac{r_1}{r_2}\right)^2 \cdot \frac{\rho_1}{\rho_2}}$
Substituting the values:
$\frac{n_2}{n_1} = \sqrt{\frac{2T}{T} \cdot \left(\frac{r}{2r}\right)^2 \cdot \frac{\rho}{\rho/2}} = \sqrt{2 \cdot \frac{1}{4} \cdot 2} = \sqrt{1} = 1$.
Therefore,$n_2 = n_1 = n$.

(D) The correct answer is $D$.
According to Stefan-Boltzmann law,the total energy $E$ radiated per unit surface area of a black body per unit time is directly proportional to the fourth power of its absolute temperature $T$.
The formula is given by $E = \sigma T^4$,where $\sigma$ is the Stefan-Boltzmann constant.
By measuring the total energy radiated by the sun,we can determine its surface temperature using this law.

(D) Let the total mass be $M = 1 \; kg$. The masses of the three parts are $m_1 = 0.2 \; kg$,$m_2 = 0.2 \; kg$,and $m_3 = 0.6 \; kg$ (ratio $1:1:3$).
Since the bomb is initially stationary,the initial momentum is $0$.
By the law of conservation of linear momentum,the final momentum must also be $0$.
Let the velocities of the two smaller parts be $\vec{v}_1 = 30\hat{i} \; m/s$ and $\vec{v}_2 = 30\hat{j} \; m/s$.
The momentum of these two parts is $\vec{p}_{12} = m_1\vec{v}_1 + m_2\vec{v}_2 = 0.2(30\hat{i} + 30\hat{j}) = 6\hat{i} + 6\hat{j} \; kg \cdot m/s$.
The magnitude of this momentum is $p_{12} = \sqrt{6^2 + 6^2} = 6\sqrt{2} \; kg \cdot m/s$.
For the total momentum to be zero,the third part must have momentum $\vec{p}_3 = -\vec{p}_{12}$.
Thus,$m_3 v_3 = 6\sqrt{2}$.
$0.6 \cdot v_3 = 6\sqrt{2}$.
$v_3 = \frac{6\sqrt{2}}{0.6} = 10\sqrt{2} \; m/s$.

(D) For a disc rolling without slipping on a horizontal surface, the velocity of any point on the disc is the vector sum of the velocity of the centre of mass $(v_{cm})$ and the tangential velocity due to rotation $(v_{rot} = R\omega)$.
Since the disc is rolling without slipping, the velocity of the point of contact with the ground is zero, which implies $v_{cm} = R\omega$.
At the highest point, the velocity is $v_{top} = v_{cm} + R\omega = v_{cm} + v_{cm} = 2v_{cm}$.
At the point of contact, the velocity is $v_{contact} = v_{cm} - R\omega = v_{cm} - v_{cm} = 0$.
Therefore, the velocity of the highest point is $2v_{cm}$ and the velocity of the point of contact is zero.

(A) The angular velocity $\omega$ of a particle moving in a circular path is related to its time period $T$ by the formula: $\omega = \frac{2 \pi}{T}$.
Since the time periods of both particles are the same $(T_1 = T_2 = T)$,the angular velocity for both particles is given by $\omega_1 = \frac{2 \pi}{T}$ and $\omega_2 = \frac{2 \pi}{T}$.
Therefore,the ratio of their angular velocities is $\frac{\omega_1}{\omega_2} = \frac{2 \pi / T}{2 \pi / T} = 1$.
Thus,the ratio is $1$.

(A) The efficiency $\eta$ of a pulley system is defined as the ratio of output work to input work.
$\text{Input Work} = F \times d = 250 \; N \times 12 \; m = 3000 \; J$
$\text{Output Work} = m \times g \times h = 75 \; kg \times 10 \; m/s^2 \times 3 \; m = 2250 \; J$
$\eta = \frac{\text{Output Work}}{\text{Input Work}} \times 100\%$
$\eta = \frac{2250}{3000} \times 100\% = 0.75 \times 100\% = 75\%$

(A) According to the law of conservation of mechanical energy,the loss in potential energy is equal to the gain in kinetic energy at the lowest point.
Let $m$ be the mass of the child,$g$ be the acceleration due to gravity $(g = 10\,m/s^2)$,$h_{max} = 2\,m$,and $h_{min} = 0.75\,m$.
Loss in potential energy = $mg(h_{max} - h_{min})$
Gain in kinetic energy = $\frac{1}{2}mv^2$
Equating the two:
$mg(2 - 0.75) = \frac{1}{2}mv^2$
$g(1.25) = \frac{1}{2}v^2$
$v^2 = 2 \times 10 \times 1.25$
$v^2 = 25$
$v = 5\,m/s$.

(A) According to the law of conservation of energy,the total energy at the surface of the Earth is equal to the total energy at the maximum height $h$.
Initial energy at the surface: $E_i = K_i + U_i = \frac{1}{2}mv^2 - \frac{GMm}{R}$
At maximum height $h$,the velocity is zero,so the final energy is: $E_f = K_f + U_f = 0 - \frac{GMm}{R+h}$
Equating $E_i = E_f$:
$\frac{1}{2}mv^2 - \frac{GMm}{R} = - \frac{GMm}{R+h}$
$\frac{1}{2}v^2 = GM \left( \frac{1}{R} - \frac{1}{R+h} \right) = GM \left( \frac{R+h-R}{R(R+h)} \right) = \frac{GMh}{R(R+h)}$
Given $h = R$,substitute this into the equation:
$\frac{1}{2}v^2 = \frac{GMR}{R(R+R)} = \frac{GM}{2R}$
$v^2 = \frac{GM}{R}$
$v = \sqrt{\frac{GM}{R}} = \left( \frac{GM}{R} \right)^{1/2}$

(A) The potential energy $E$ stored in a spring with spring constant $K$ when stretched by a force $F$ is given by the formula $E = \frac{F^2}{2K}$.
Since the applied force $F$ is the same for both springs,the energy stored is inversely proportional to the spring constant: $E \propto \frac{1}{K}$.
Therefore,the ratio of energies is $\frac{E_{A}}{E_{B}} = \frac{K_{B}}{K_{A}}$.
Given that $K_{A} = 2 K_{B}$,we have $\frac{K_{B}}{K_{A}} = \frac{1}{2}$.
Substituting this into the ratio,we get $\frac{E_{A}}{E_{B}} = \frac{1}{2}$.
Thus,$E_{B} = 2 E_{A}$.

(C) Given: Mass $m = 150 \ g = 0.15 \ kg$,Initial velocity $u = 20 \ m/s$,Final velocity $v = 0 \ m/s$,Time interval $\Delta t = 0.1 \ s$.
According to Newton's second law of motion,the force $F$ exerted is equal to the rate of change of momentum:
$F = \frac{\Delta p}{\Delta t} = \frac{m(v - u)}{\Delta t}$
$F = \frac{0.15 \times (0 - 20)}{0.1}$
$F = \frac{0.15 \times (-20)}{0.1} = \frac{-3}{0.1} = -30 \ N$
The magnitude of the force is $|F| = 30 \ N$.

(B) According to Wien's displacement law,the product of the wavelength of maximum emission and the absolute temperature is a constant:
$\lambda_m T = \text{constant}$
Therefore,$\lambda_m \propto \frac{1}{T}$.
Given $T_1 = 2000 \ K$ and $T_2 = 3000 \ K$,and the initial wavelength is $\lambda_m$:
$\frac{\lambda_2}{\lambda_m} = \frac{T_1}{T_2}$
$\frac{\lambda_2}{\lambda_m} = \frac{2000}{3000} = \frac{2}{3}$
$\lambda_2 = \frac{2}{3} \lambda_m$
Thus,the corresponding wavelength at $3000 \ K$ is $\frac{2}{3} \lambda_m$.

(D) According to Gauss's Law,the total electric flux $\varphi_{net}$ through a closed surface is equal to the total charge enclosed divided by the permittivity of free space $\varepsilon_{0}$.
For a cube with a charge $Q \; \mu C$ placed at its centre,the total flux is $\varphi_{net} = \frac{Q \times 10^{-6}}{\varepsilon_{0}}$.
Since a cube has $6$ identical faces and the charge is at the centre,the flux is distributed equally among all faces.
Therefore,the flux through any one face is $\varphi_{face} = \frac{\varphi_{net}}{6} = \frac{Q \times 10^{-6}}{6 \varepsilon_{0}} = \frac{Q}{6 \varepsilon_{0}} \times 10^{-6}$.

(C) The potential gradient $(x)$ is defined as the potential drop per unit length of the wire, given by $x = \frac{V}{L}$.
Since $V = IR$ and $R = \frac{\rho L}{A}$, we can substitute these into the expression for the potential gradient:
$x = \frac{I \cdot (\rho L / A)}{L} = \frac{I \rho}{A}$.
Given values are:
Current $(I) = 0.1 \, A$,
Resistivity $(\rho) = 10^{-7} \, \Omega \cdot m$,
Area $(A) = 10^{-6} \, m^2$.
Substituting these values into the formula:
$x = \frac{0.1 \times 10^{-7}}{10^{-6}} = \frac{10^{-1} \times 10^{-7}}{10^{-6}} = \frac{10^{-8}}{10^{-6}} = 10^{-2} \, V/m$.

(D) The magnetic dipole moment $(M)$ of a current-carrying coil is defined as the product of the number of turns $(N)$,the current $(I)$ flowing through the coil,and the area $(A)$ of the coil.
Mathematically,it is expressed as:
$M = NIA$
Therefore,the correct option is $D$.

(C) When a current $I$ is passed through the coil of a tangent galvanometer,a magnetic field $B$ is produced at right angles to the plane of the coil,i.e.,at right angles to the horizontal component of the Earth's magnetic field $H$.
Under the influence of these two crossed magnetic fields $B$ and $H$,the magnetic needle of the galvanometer undergoes a deflection $\theta$,which is governed by the tangent law.
According to the tangent law,the relationship is given by $B = H \tan \theta$.
Since $B \propto I$,we get $I = k \tan \theta$,where $k$ is the reduction factor of the galvanometer.
This clearly indicates that a tangent galvanometer is an instrument used for the measurement of electric current in a circuit.

(C) The magnetic susceptibility $(\chi)$ of diamagnetic substances is independent of temperature.
For paramagnetic substances, $\chi$ follows Curie's Law, $\chi \propto 1/T$.
For ferromagnetic and ferrimagnetic substances, $\chi$ depends on temperature according to the Curie-Weiss Law, $\chi \propto 1/(T - T_c)$, where $T_c$ is the Curie temperature.
Therefore, diamagnetic substances are the only ones among the options whose magnetic susceptibility does not depend on temperature.

(D) The phenomena of refraction,interference,and polarization are explained by the wave theory of light,thus demonstrating the wave nature of light.
The photoelectric effect involves the emission of electrons when light of a suitable frequency strikes a metal surface. This phenomenon cannot be explained by wave theory and is successfully explained by Einstein's quantum theory,which treats light as consisting of discrete packets of energy called photons. Therefore,the photoelectric effect demonstrates the particle nature of light.

(D) The intensity of light $I$ at a distance $d$ from a point source is given by $I \propto \frac{1}{d^2}$.
Since the number of photoelectrons emitted per second is directly proportional to the intensity of the incident light,we have $N \propto I$.
Therefore,$N \propto \frac{1}{d^2}$.
When the distance is changed from $d$ to $\frac{d}{2}$,the new number of photoelectrons $N'$ is given by $N' \propto \frac{1}{(\frac{d}{2})^2} = \frac{4}{d^2}$.
Comparing $N'$ with $N$,we get $N' = 4N$.
Thus,the number of electrons emitted increases by a factor of $4$.

(A) The energy of an electron in the $n^{th}$ orbit of a hydrogen-like atom is given by the formula: ${E_n} = -13.6 \times \frac{Z^2}{n^2} \text{ eV}$.
For a hydrogen atom,the atomic number $Z_H = 1$,so the energy is ${E_n} = -13.6 \times \frac{1^2}{n^2} = -\frac{13.6}{n^2} \text{ eV}$.
For a singly ionized helium atom $(He^+)$,the atomic number $Z_{He} = 2$. The energy in the $n^{th}$ orbit is given by: ${E'_{n}} = -13.6 \times \frac{Z_{He}^2}{n^2} = -13.6 \times \frac{2^2}{n^2} = 4 \times \left( -\frac{13.6}{n^2} \right)$.
Substituting the expression for the hydrogen atom energy ${E_n}$,we get: ${E'_{n}} = 4{E_n}$.

(A) The actual mass of a nucleus $(M)$ is always less than the sum of the individual masses of its constituent nucleons (protons and neutrons).
This difference in mass is known as the mass defect $(\Delta m)$, which is converted into binding energy according to Einstein's mass-energy equivalence principle $(E = \Delta m c^2)$.
Therefore, the mass of the nucleus $M$ is related to the sum of the masses of $N$ neutrons and $Z$ protons as:
$M < (N M_n + Z M_p)$.

(A) The given nuclear reaction is $X(n, \alpha) {_3Li^7}$.
This can be written as: $_ZX^A + _0n^1 \to _3Li^7 + _2He^4$.
According to the law of conservation of mass number and atomic number:
For atomic number $(Z)$: $Z + 0 = 3 + 2 \implies Z = 5$.
For mass number $(A)$: $A + 1 = 7 + 4 \implies A + 1 = 11 \implies A = 10$.
Thus,the nucleus $X$ is $_5X^{10}$,which corresponds to Boron-$10$ $(_{5}B^{10})$.

(D) For a $NAND$ gate,the output $C$ is given by the Boolean expression $C = \overline{A \cdot B}$.
Checking the values:
$1$. For $A = 0, B = 0$: $C = \overline{0 \cdot 0} = \overline{0} = 1$.
$2$. For $A = 0, B = 1$: $C = \overline{0 \cdot 1} = \overline{0} = 1$.
$3$. For $A = 1, B = 0$: $C = \overline{1 \cdot 0} = \overline{0} = 1$.
$4$. For $A = 1, B = 1$: $C = \overline{1 \cdot 1} = \overline{1} = 0$.
Comparing these results with the given table,the output matches the $NAND$ gate logic.

(B) In the given circuit,the battery of $5 \text{ V}$ is connected in series with a $20 \text{ } \Omega$ resistor and two parallel branches.
The upper branch contains a $20 \text{ } \Omega$ resistor and a diode. The diode is connected such that its p-side is towards the right,making it reverse-biased with respect to the positive terminal of the battery.
The middle branch contains a diode and a $30 \text{ } \Omega$ resistor. The diode is connected such that its p-side is towards the left,making it forward-biased with respect to the positive terminal of the battery.
Since the upper branch is reverse-biased,it acts as an open circuit (no current flows through it).
The total resistance of the circuit is the sum of the $20 \text{ } \Omega$ resistor in the main line and the $30 \text{ } \Omega$ resistor in the active middle branch: $R_{eq} = 20 \text{ } \Omega + 30 \text{ } \Omega = 50 \text{ } \Omega$.
Using Ohm's law,the current $i$ is:
$i = \frac{V}{R_{eq}} = \frac{5 \text{ V}}{50 \text{ } \Omega} = \frac{5}{50} \text{ A}$.

(D) The correct answer is $(d)$.
Total internal reflection is a powerful process since it can be used to confine light. One of the most common applications of total internal reflection is in fibre optics.
An optical fibre is a thin,transparent fibre,usually made of glass or plastic,used for transmitting light.
If light is incident on a cable end with an angle of incidence greater than the critical angle,then the light will remain trapped inside the glass strand due to repeated total internal reflections.
In this way,light travels very quickly down the length of the cable over a very long distance (tens of kilometers) without significant loss of intensity.

(B) The light from the source will be cut off if the disc covers the area corresponding to the critical angle $\theta_c$.
For a point source at depth $h$,the radius $r$ of the disc is given by $r = h \tan \theta_c$.
Since $\sin \theta_c = \frac{1}{\mu}$,we have $\tan \theta_c = \frac{1}{\sqrt{\mu^2 - 1}}$.
Substituting the given values $h = 4 \; m$ and $\mu = 5/3$:
$r = \frac{4}{\sqrt{(5/3)^2 - 1}} = \frac{4}{\sqrt{25/9 - 1}} = \frac{4}{\sqrt{16/9}} = \frac{4}{4/3} = 3 \; m$.
The minimum diameter of the disc is $D = 2r = 2 \times 3 = 6 \; m$.

(A) Ozone $(O_3)$ is present in the stratosphere region of the atmosphere.
It protects life on Earth by absorbing harmful ultraviolet $(UV)$ radiation emitted by the sun.
Exposure to high levels of $UV$ radiation causes various health problems in humans,such as skin cancer and cataracts,and damages plants and other animals,leading to ecosystem disruption.
Plants cannot grow effectively under intense ultraviolet radiation.
Therefore,the primary biological importance of the ozone layer is the absorption of these harmful rays.
Hence,option $A$ is correct.

(A) In a common base configuration,the current gain $\alpha$ is defined as the ratio of collector current to emitter current: $\alpha = \frac{I_C}{I_E} = 0.98$.
The current gain for a common emitter circuit,denoted by $\beta$,is related to $\alpha$ by the formula: $\beta = \frac{\alpha}{1 - \alpha}$.
Substituting the given value of $\alpha$ into the formula:
$\beta = \frac{0.98}{1 - 0.98} = \frac{0.98}{0.02} = 49$.
Therefore,the current gain for the common emitter circuit is $49$.

(A) The specific resistance or resistivity $\rho$ of a material depends on temperature.
For metals like copper,the resistivity decreases as the temperature decreases because the scattering of electrons by lattice vibrations reduces.
For semiconductors like silicon,the resistivity increases as the temperature decreases because the number of charge carriers (electrons and holes) decreases significantly due to the freezing out of carriers.
Therefore,when cooling from $300 \ K$ to $60 \ K$,the specific resistance of copper decreases,while the specific resistance of silicon increases.

(D) The capacitive reactance $X$ is given by the formula $X = \frac{1}{2 \pi f C}$,where $f$ is the frequency and $C$ is the capacitance.
Given that the new capacitance $C^{\prime} = 2C$ and the new frequency $f^{\prime} = 2f$.
The new reactance $X^{\prime}$ is calculated as:
$X^{\prime} = \frac{1}{2 \pi f^{\prime} C^{\prime}} = \frac{1}{2 \pi (2f) (2C)} = \frac{1}{4 (2 \pi f C)}$.
Since $X = \frac{1}{2 \pi f C}$,we substitute this into the equation:
$X^{\prime} = \frac{X}{4}$.

(A) In a nuclear process,energy is released if the binding energy per nucleon of the daughter products increases.
In the nuclear fission process,the total binding energy of the fragments formed is greater than the binding energy of the parent fissionable nucleus.
This difference in binding energy is released as energy,consistent with the mass-energy equivalence principle where the mass defect is converted into energy.

(C) Wheatstone bridge is balanced when the ratio of resistances in opposite arms is equal. Here, $\frac{P}{Q} = \frac{R}{S} = \frac{10}{10} = 1$.
In a balanced Wheatstone bridge, no current flows through the galvanometer arm.
However, the problem states that a resistance of $10 \, \Omega$ is connected in series with the galvanometer. Since no current flows through the galvanometer branch in a balanced state, this additional resistance does not affect the equivalent resistance of the circuit.
The circuit effectively consists of two parallel branches, each containing two $10 \, \Omega$ resistors in series.
Resistance of each branch = $10 \, \Omega + 10 \, \Omega = 20 \, \Omega$.
Since these two branches are in parallel, the equivalent resistance $R_{eq}$ is given by:
$\frac{1}{R_{eq}} = \frac{1}{20} + \frac{1}{20} = \frac{2}{20} = \frac{1}{10}$.
Therefore, $R_{eq} = 10 \, \Omega$.

(B) When a light ray travels from one medium to another,its frequency $(n)$ depends only on the source of light and remains unchanged regardless of the medium.
However,the wavelength $(\lambda)$ and velocity $(v)$ change because they depend on the refractive index of the medium.
Intensity $(I)$ also changes due to partial reflection at the interface.
Therefore,the correct relation is $n = n'$.

(B) When a charged particle enters a magnetic field perpendicularly,it undergoes circular motion.
The magnetic force provides the necessary centripetal force:
$F_m = F_c$
$qvB = \frac{mv^2}{r}$
From this,the angular velocity $\omega$ is given by:
$\omega = \frac{v}{r} = \frac{qB}{m}$
Since the frequency $f$ is related to angular velocity by $\omega = 2\pi f$,we have:
$f = \frac{\omega}{2\pi} = \frac{qB}{2\pi m}$
For an electron,the charge $q = e$,so the frequency is:
$f = \frac{eB}{2\pi m}$

(A) $\alpha -$ particles are doubly ionized helium nuclei $(He^{2+})$. They carry a positive charge and are deflected towards negatively charged plates in an electric field.
$\beta -$ particles are high-speed electrons $(e^-)$,which carry a negative charge.
$\gamma -$ rays are high-energy electromagnetic waves and are electrically neutral.
$X-$ rays are also electromagnetic waves and are electrically neutral.
Therefore,the ray containing positively charged particles is the $\alpha -$ ray.

(B) The formula for radioactive decay is given by $M = M_{0} \left(\frac{1}{2}\right)^{n}$,where $n = \frac{t}{T_{1/2}}$ is the number of half-lives.
Given: $M = 1 \; g$,$M_{0} = 256 \; g$,and $T_{1/2} = 12.5 \; h$.
Substituting the values: $1 = 256 \left(\frac{1}{2}\right)^{n}$.
This simplifies to $\frac{1}{256} = \left(\frac{1}{2}\right)^{n}$.
Since $256 = 2^{8}$,we have $\left(\frac{1}{2}\right)^{8} = \left(\frac{1}{2}\right)^{n}$.
Therefore,$n = 8$.
Using $n = \frac{t}{T_{1/2}}$,we get $t = n \times T_{1/2} = 8 \times 12.5 \; h = 100 \; h$.

(B) The dipole moment of the dipole is $\vec{p}$ and the uniform electric field is $\vec{E}$.
When an electric dipole is placed in a uniform electric field,each charge $q$ experiences a force $\vec{F} = q\vec{E}$ and $\vec{F} = -q\vec{E}$.
These two equal and opposite forces form a couple,which exerts a torque on the dipole.
The magnitude of the torque is given by the product of the magnitude of one of the forces and the perpendicular distance between them.
$\tau = F \times (d \sin \theta) = (qE) \times (a \sin \theta) = (qa) E \sin \theta$.
Since the dipole moment $p = qa$,we have $\tau = pE \sin \theta$.
In vector form,this is expressed as $\vec{\tau} = \vec{p} \times \vec{E}$.

(C) Given: Self-inductance $L = 2 \, mH$ and current $i = t^2 e^{-t}$.
The induced $emf$ $(E)$ in an inductor is given by the formula $E = -L \frac{di}{dt}$.
First,calculate the derivative of current with respect to time:
$\frac{di}{dt} = \frac{d}{dt} (t^2 e^{-t}) = (2t) e^{-t} + t^2 (-e^{-t}) = e^{-t} (2t - t^2)$.
Now,substitute this into the $emf$ equation:
$E = -L [e^{-t} (2t - t^2)]$.
For the $emf$ to be zero $(E = 0)$,the term inside the bracket must be zero:
$e^{-t} (2t - t^2) = 0$.
Since $e^{-t}$ is never zero for finite $t$,we have:
$2t - t^2 = 0$
$t(2 - t) = 0$.
This gives $t = 0 \, s$ or $t = 2 \, s$. Excluding the initial state at $t = 0$,the $emf$ is zero at $t = 2 \, s$.

(A) The energy density $u$ (energy per unit volume) of a capacitor is given by the formula $u = \frac{1}{2} \varepsilon_{0} E^{2}$.
Since the electric field $E$ between the plates of a capacitor is related to the potential difference $V$ and separation $d$ by $E = \frac{V}{d}$,
Substituting this into the energy density formula,we get $u = \frac{1}{2} \varepsilon_{0} (\frac{V}{d})^{2} = \frac{1}{2} \varepsilon_{0} \frac{V^{2}}{d^{2}}$.