AIPMT 2002 Physics Question Paper with Answer and Solution

(C) Stefan's law is given by $E = \sigma T^4$,where $E$ is the power radiated per unit area.
Rearranging for $\sigma$,we get $\sigma = \frac{E}{T^4}$.
The unit of $E$ (power per unit area) is $\frac{\text{Watt}}{\text{m}^2} = W\,m^{-2}$.
The unit of temperature $T$ is Kelvin $(K)$.
Therefore,the unit of $\sigma$ is $\frac{W\,m^{-2}}{K^4} = W\,m^{-2}\,K^{-4}$.
Thus,the correct option is $C$.

(A) For both particles,the vertical motion is governed by the equation of motion $h = ut + \frac{1}{2}gt^2$.
Since both particles are released from the same height $h$ and their initial vertical velocity component $u_y$ is $0$ in both cases,the time taken to reach the ground is given by $t = \sqrt{\frac{2h}{g}}$.
Since $h$ and $g$ are the same for both,the time taken $t$ is identical for both particles.
Therefore,both particles will reach the ground simultaneously.

(C) The forces acting on the lift are the tension $T$ acting upwards and the weight $mg$ acting downwards.
Since the lift is accelerating upwards with acceleration $a$,the net force is given by Newton's second law: $F_{net} = T - mg = ma$.
Rearranging the equation to solve for tension: $T = m(g + a)$.
Given values: $m = 1000\,kg$,$g = 9.8\,m/s^2$,and $a = 1\,m/s^2$.
Substituting these values into the formula: $T = 1000(9.8 + 1) = 1000(10.8) = 10800\,N$.
Therefore,the tension in the cable is $10800\,N$.

(B) Given: Mass $m = 10 \, kg$,coefficient of friction $\mu = 0.5$,applied force $F = 100 \, N$,and acceleration due to gravity $g = 10 \, m/s^2$.
First,calculate the limiting friction force: $f_k = \mu \cdot m \cdot g = 0.5 \times 10 \times 10 = 50 \, N$.
Since the applied force $F = 100 \, N$ is greater than the limiting friction $f_k = 50 \, N$,the block will move.
The net force acting on the block is $F_{net} = F - f_k = 100 - 50 = 50 \, N$.
Using Newton's second law,$F_{net} = m \cdot a$,the acceleration $a = \frac{F_{net}}{m} = \frac{50}{10} = 5 \, m/s^2$.

(A) Let the initial kinetic energy be $E_1 = E$.
Final kinetic energy $E_2 = E + 300\% \text{ of } E = E + 3E = 4E$.
We know that the relationship between momentum $P$ and kinetic energy $E$ is $P = \sqrt{2mE}$,which implies $P \propto \sqrt{E}$.
Therefore,the ratio of final momentum $P_2$ to initial momentum $P_1$ is $\frac{P_2}{P_1} = \sqrt{\frac{E_2}{E_1}} = \sqrt{\frac{4E}{E}} = \sqrt{4} = 2$.
This means $P_2 = 2P_1$.
The percentage change in momentum is given by $\frac{P_2 - P_1}{P_1} \times 100\% = \frac{2P_1 - P_1}{P_1} \times 100\% = 100\%$.
Thus,the momentum increases by $100\%$.

(B) The gravitational potential energy $U$ at a distance $r$ from the center of the earth is given by $U = -\frac{GMm}{r}$.
At the earth's surface,$r_1 = R$,so $U_1 = -\frac{GMm}{R}$.
At a height $h = 3R$,the distance from the center is $r_2 = R + h = R + 3R = 4R$.
So,$U_2 = -\frac{GMm}{4R}$.
The change in gravitational potential energy is $\Delta U = U_2 - U_1 = -\frac{GMm}{4R} - (-\frac{GMm}{R}) = \frac{GMm}{R} - \frac{GMm}{4R} = \frac{3GMm}{4R}$.
Since the acceleration due to gravity at the surface is $g = \frac{GM}{R^2}$,we have $GM = gR^2$.
Substituting this into the expression,$\Delta U = \frac{3(gR^2)m}{4R} = \frac{3}{4}mgR$.

(B) The efficiency $\eta$ of a Carnot engine is given by $\eta = 1 - \frac{T_2}{T_1}$,where $T_1$ is the source temperature and $T_2$ is the sink temperature.
Initially,$\eta_1 = 0.5$ and $T_2 = 500 \ K$.
$0.5 = 1 - \frac{500}{T_1} \implies \frac{500}{T_1} = 0.5 \implies T_1 = 1000 \ K$.
Now,the efficiency is increased to $\eta_2 = 0.6$ while keeping $T_1$ constant at $1000 \ K$.
$0.6 = 1 - \frac{T_2'}{1000} \implies \frac{T_2'}{1000} = 1 - 0.6 = 0.4$.
$T_2' = 0.4 \times 1000 = 400 \ K$.
Therefore,the required temperature of the sink is $400 \ K$.

(B) The rate of heat flow through a rod by conduction is given by the formula:
$\frac{dQ}{dt} = \frac{KA(T_{1} - T_{2})}{l}$
For the two rods,the rates of heat loss are:
$(\frac{dQ}{dt})_{1} = \frac{K_{1}A_{1}(T_{1} - T_{2})}{l}$
$(\frac{dQ}{dt})_{2} = \frac{K_{2}A_{2}(T_{1} - T_{2})}{l}$
Given that the rate of loss of heat is equal for both rods:
$(\frac{dQ}{dt})_{1} = (\frac{dQ}{dt})_{2}$
Substituting the expressions:
$\frac{K_{1}A_{1}(T_{1} - T_{2})}{l} = \frac{K_{2}A_{2}(T_{1} - T_{2})}{l}$
Since the lengths $l$ and the temperature differences $(T_{1} - T_{2})$ are the same for both rods,we can cancel these terms from both sides:
$K_{1}A_{1} = K_{2}A_{2}$

(C) In $S.H.M.$,the potential energy $U$ is given by $U = \frac{1}{2} k y^2$,where $y$ is the displacement.
Maximum potential energy occurs at the extreme positions,where $y = \pm a$.
Kinetic energy $K$ is given by $K = \frac{1}{2} k (a^2 - y^2)$.
Maximum kinetic energy occurs at the mean position,where $y = 0$.
The displacement between the position of maximum potential energy $(y = \pm a)$ and the position of maximum kinetic energy $(y = 0)$ is $| \pm a - 0 | = a$.

(D) For a spring-mass system,the time period is given by $t = 2\pi \sqrt{\frac{m}{K}}$.
For the individual springs,we have $t_1 = 2\pi \sqrt{\frac{m}{K_1}}$ and $t_2 = 2\pi \sqrt{\frac{m}{K_2}}$.
Squaring these,we get $t_1^2 = 4\pi^2 \frac{m}{K_1} \implies K_1 = \frac{4\pi^2 m}{t_1^2}$ and $t_2^2 = 4\pi^2 \frac{m}{K_2} \implies K_2 = \frac{4\pi^2 m}{t_2^2}$.
In the given figure,the springs are in parallel. The equivalent spring constant is $K_{eq} = K_1 + K_2$.
The time period for the combined system is $t = 2\pi \sqrt{\frac{m}{K_1 + K_2}}$.
Squaring this,$t^2 = 4\pi^2 \frac{m}{K_1 + K_2} \implies \frac{1}{t^2} = \frac{K_1 + K_2}{4\pi^2 m} = \frac{K_1}{4\pi^2 m} + \frac{K_2}{4\pi^2 m}$.
Substituting the expressions for $K_1$ and $K_2$,we get $\frac{1}{t^2} = \frac{1}{t_1^2} + \frac{1}{t_2^2}$,which can be written as $t^{-2} = t_1^{-2} + t_2^{-2}$.

(C) wave travelling in the positive $x-$direction is represented by the equation:
$y = A \sin \left( \frac{2\pi}{\lambda} (vt - x) \right)$
Given values are $A = 0.2\;m,$ $v = 360\;m/s,$ and $\lambda = 60\;m.$
Substituting these values into the equation:
$y = 0.2 \sin \left( \frac{2\pi}{60} (360t - x) \right)$
$y = 0.2 \sin \left( 2\pi \left( \frac{360}{60}t - \frac{x}{60} \right) \right)$
$y = 0.2 \sin \left[ 2\pi \left( 6t - \frac{x}{60} \right) \right]$
Thus,the correct expression is $y = 0.2 \sin \left[ 2\pi \left( 6t - \frac{x}{60} \right) \right].$

(B) The linear speed of the whistle is given by $v_s = r\omega.$
Given $r = 50 \; cm = 0.5 \; m$ and $\omega = 20 \; rad/s.$
So,$v_s = 0.5 \times 20 = 10 \; m/s.$
The minimum frequency is heard when the source is moving directly away from the observer.
The formula for the observed frequency is $n_{\min} = n \left( \frac{v}{v + v_s} \right).$
Substituting the values: $n_{\min} = 385 \left( \frac{340}{340 + 10} \right) = 385 \left( \frac{340}{350} \right) = 385 \times \frac{34}{35} = 11 \times 34 = 374 \; Hz.$

(B) The moment of inertia $I$ of a body about an axis is given by $I = \int r^2 dm$,where $dm$ is the mass element at a distance $r$ from the axis.
To maximize the moment of inertia for a fixed total mass and radius,we need to place as much mass as possible at the largest possible distance $r$ from the axis.
Since iron has a higher density than aluminium,placing iron at the outer periphery (surrounding the interior) increases the mass distribution at larger radii.
Therefore,placing aluminium at the interior and iron at the exterior (surrounding it) results in a higher moment of inertia compared to other configurations.
Thus,option $B$ is correct.

(D) According to the principle of conservation of angular momentum,if the net external torque acting on a system is zero,the total angular momentum of the system remains constant.
When a child sits on a rotating disc,the weight of the child acts vertically downwards,passing through the axis of rotation or creating no torque about the axis of rotation.
Since there is no external torque applied to the system (disc + child),the angular momentum of the system remains conserved.

(D) An ideal black body is defined as a body that absorbs all incident radiation falling on it,regardless of the frequency or angle of incidence.
$A$ cavity maintained at a constant temperature,often called a Fery's black body,is the closest physical approximation to an ideal black body.
In this setup,any radiation entering the small hole of the cavity undergoes multiple internal reflections. At each reflection,a portion of the radiation is absorbed by the walls. After many such reflections,almost all of the incident radiation is absorbed,making it an excellent black body.

(B) According to Stefan-Boltzmann Law,the net power radiated by a black body is given by $P = \sigma A (T^4 - T_0^4)$,where $T$ is the temperature of the body and $T_0$ is the temperature of the surroundings.
Given:
$T_1 = 727^{\circ} C = 727 + 273 = 1000 \; K$
$T_0 = 227^{\circ} C = 227 + 273 = 500 \; K$
$P_1 = 60 \; W$
$P_1 = k(T_1^4 - T_0^4) \Rightarrow 60 = k(1000^4 - 500^4) \quad \dots(1)$
Now,$T_2 = 1227^{\circ} C = 1227 + 273 = 1500 \; K$
$P_2 = k(T_2^4 - T_0^4) \Rightarrow P_2 = k(1500^4 - 500^4) \quad \dots(2)$
Dividing equation $(2)$ by $(1)$:
$\frac{P_2}{60} = \frac{1500^4 - 500^4}{1000^4 - 500^4} = \frac{500^4 (3^4 - 1^4)}{500^4 (2^4 - 1^4)}$
$\frac{P_2}{60} = \frac{81 - 1}{16 - 1} = \frac{80}{15} = \frac{16}{3}$
$P_2 = 60 \times \frac{16}{3} = 20 \times 16 = 320 \; W$.

(B) When a wheel of radius $R$ rolls without slipping,the horizontal distance covered by the center of the wheel in half a revolution is equal to $\pi R$.
The vertical displacement of the point $P$ (which was initially at the contact point $A$ and moves to the topmost point $A'$) is equal to the diameter of the wheel,which is $2R$.
The total displacement is the vector sum of the horizontal and vertical displacements,which is the hypotenuse of a right-angled triangle with sides $\pi R$ and $2R$.
Displacement $= \sqrt{(\pi R)^2 + (2R)^2} = R\sqrt{\pi^2 + 4}$.
Given $R = 1 \ m$,the displacement $= 1 \times \sqrt{\pi^2 + 4} = \sqrt{\pi^2 + 4} \ m$.

(D) Wien's displacement law states that the product of the absolute temperature $(T)$ and the wavelength $(\lambda_{\max})$ at which the spectral emissive power of a black body is maximum is a constant.
The mathematical expression is given by:
$\lambda_{\max} T = b$
where $b$ is Wien's constant.
Therefore, the law expresses the relationship between the wavelength corresponding to maximum energy and the absolute temperature of the body.

(B) In damped oscillations,the amplitude decays exponentially with time,given by $a = a_0 e^{-bt}$,where $b$ is the damping constant and $t$ is the time.
Let $T$ be the time period of one oscillation. After $n$ oscillations,the time elapsed is $t = nT$.
Initially,after $100$ oscillations,the amplitude is $a = \frac{a_0}{3}$.
So,$\frac{a_0}{3} = a_0 e^{-b(100T)}$,which implies $e^{-100bT} = \frac{1}{3}$.
After $200$ oscillations,the time elapsed is $t = 200T$.
The amplitude $a'$ will be $a' = a_0 e^{-b(200T)}$.
This can be written as $a' = a_0 (e^{-100bT})^2$.
Substituting the value from the first step: $a' = a_0 (\frac{1}{3})^2 = \frac{a_0}{9}$.
Thus,the amplitude becomes $\frac{1}{9}$ of the initial value.

(B) The surface is smooth,so there is no friction at the contact surface.
According to Newton's second law,the linear acceleration $a$ of the centre of mass is given by $a = \frac{F}{m}$,where $F$ is the applied force and $m$ is the mass of the sphere.
Since the force $F$ and the mass $m$ are constant,the acceleration $a$ is independent of the height $h$ at which the force is applied.
Therefore,there is no specific relation between $h$ and $R$ required to maximize the acceleration; it remains constant for any $h$.

(A) Given mass $m = 3\,kg$ and force $\vec{F} = (6t^2\hat{i} + 4t\hat{j})\,N$.
Using Newton's second law,$\vec{a} = \frac{\vec{F}}{m} = \frac{6t^2\hat{i} + 4t\hat{j}}{3} = (2t^2\hat{i} + \frac{4}{3}t\hat{j})\,m/s^2$.
Since the object starts from rest,$\vec{v}(0) = 0$. The velocity $\vec{v}$ at time $t$ is given by $\vec{v} = \int_{0}^{t} \vec{a} dt$.
$\vec{v} = \int_{0}^{3} (2t^2\hat{i} + \frac{4}{3}t\hat{j}) dt$.
$\vec{v} = [\frac{2t^3}{3}\hat{i} + \frac{4t^2}{6}\hat{j}]_{0}^{3}$.
$\vec{v} = [\frac{2(3)^3}{3}\hat{i} + \frac{4(3)^2}{6}\hat{j}] = [\frac{54}{3}\hat{i} + \frac{36}{6}\hat{j}] = 18\hat{i} + 6\hat{j}\,m/s$.

(D) Let the linear mass density be $\rho = kx$,where $k$ is a constant.
The mass of a small element of length $dx$ is $dm = \rho \cdot dx = kx \cdot dx$.
The position of the centre of mass $x_{cm}$ is given by the formula:
$x_{cm} = \frac{\int x \cdot dm}{\int dm}$
Substituting the values:
$x_{cm} = \frac{\int_{0}^{3} x(kx \cdot dx)}{\int_{0}^{3} kx \cdot dx} = \frac{\int_{0}^{3} x^2 \cdot dx}{\int_{0}^{3} x \cdot dx}$
Evaluating the integrals:
$x_{cm} = \frac{[x^3/3]_{0}^{3}}{[x^2/2]_{0}^{3}} = \frac{27/3}{9/2} = \frac{9}{4.5} = 2 \; m$.

(D) The length of the body diagonal of a cube with side $b$ is $\sqrt{3}b$.
The distance of the center of the cube from each vertex is $r = \frac{\sqrt{3}b}{2}$.
The electric potential energy $U$ of a charge $q$ at the center due to $8$ charges of $-q$ at the corners is given by the sum of individual potential energies:
$U = \sum_{i=1}^{8} \frac{1}{4\pi\varepsilon_0} \frac{(-q)(q)}{r}$
$U = 8 \times \left( \frac{1}{4\pi\varepsilon_0} \cdot \frac{-q^2}{\sqrt{3}b/2} \right)$
$U = 8 \times \left( \frac{-2q^2}{4\pi\varepsilon_0\sqrt{3}b} \right)$
$U = \frac{-16q^2}{4\pi\varepsilon_0\sqrt{3}b} = \frac{-4q^2}{\sqrt{3}\pi\varepsilon_0b}$

(C) When two capacitors are connected,charge flows from the higher potential to the lower potential until they reach a common potential $V'$.
By the principle of conservation of charge,the total charge before connection equals the total charge after connection.
Initial charge $Q_{total} = C_1 V + C_2 (0) = C_1 V$.
Total capacitance after connection $C_{total} = C_1 + C_2$.
Common potential $V' = \frac{Q_{total}}{C_{total}} = \frac{C_1 V}{C_1 + C_2}$.

(A) galvanometer has a low internal resistance,but a voltmeter must have a very high resistance to ensure it draws minimal current from the circuit it is measuring.
To convert a galvanometer into a voltmeter,we need to increase its effective resistance. This is achieved by connecting a high resistance in series with the galvanometer coil.
As shown in the diagram,the combination of the galvanometer and the series resistor acts as a voltmeter.
Note: The resistance of an ideal voltmeter should be infinite.

(C) The magnetic field at the center of a circular coil of $n$ turns carrying current $I$ and having radius $r$ is given by $B = \frac{\mu_0 n I}{2r}$.
If the same wire of length $L$ is used,then for $n=1$,$L = 2\pi r_1$,so $r_1 = \frac{L}{2\pi}$. Thus,$B = \frac{\mu_0 I}{2(L/2\pi)} = \frac{\mu_0 I \pi}{L}$.
For $n=2$,the new radius $r_2$ is given by $L = 2(2\pi r_2)$,so $r_2 = \frac{L}{4\pi} = \frac{r_1}{2}$.
The new magnetic field $B'$ is $B' = \frac{\mu_0 (2) I}{2r_2} = \frac{\mu_0 I}{r_2} = \frac{\mu_0 I}{r_1/2} = 2 \left( \frac{\mu_0 I}{r_1} \right) = 4 \left( \frac{\mu_0 I}{2r_1} \right) = 4B$.
Therefore,the new magnetic field is $4B$.

(A) When a charge $q$ moves with velocity $\overrightarrow{v}$ in a region containing both an electric field $\overrightarrow{E}$ and a magnetic field $\overrightarrow{B}$,it experiences two forces:
$1$. The electric force: $\overrightarrow{F_e} = q\overrightarrow{E}$
$2$. The magnetic force (Lorentz force): $\overrightarrow{F_m} = q(\overrightarrow{v} \times \overrightarrow{B})$
The total force,known as the Lorentz force,is the vector sum of these two forces:
$\overrightarrow{F} = \overrightarrow{F_e} + \overrightarrow{F_m} = q\overrightarrow{E} + q(\overrightarrow{v} \times \overrightarrow{B})$
Therefore,the correct option is $A$.

(D) Cathode rays consist of a stream of negatively charged particles (electrons). Because they carry a charge,they are deflected by both electric and magnetic fields. Therefore,the statement that they do not deflect in an electric field is incorrect.

(C) The de-Broglie wavelength $\lambda$ is given by the formula $\lambda = \frac{h}{mv}$,where $h$ is Planck's constant,$m$ is the mass of the particle,and $v$ is its velocity.
Given that the velocity $v$ is the same for all particles,the relationship becomes $\lambda \propto \frac{1}{m}$.
This means that the particle with the smallest mass will have the maximum de-Broglie wavelength.
Comparing the masses: $m_{\beta} < m_{proton} \approx m_{neutron} < m_{\alpha}$.
Since the $\beta$-particle (an electron) has the smallest mass among the given options,it will have the maximum de-Broglie wavelength.

(A) The photoelectric effect occurs when the incident radiation has a frequency greater than the threshold frequency of the metal, or equivalently, a wavelength shorter than the threshold wavelength.
Since ultraviolet $(UV)$ rays do not cause the photoelectric effect, the incident radiation must have a higher energy (shorter wavelength) than $UV$ rays.
Among the given options, $X$-rays have a shorter wavelength $(\lambda_{X-ray} < \lambda_{UV-ray})$ and thus higher energy than $UV$ rays.
Therefore, $X$-rays will cause the photoelectric effect.

(B) The nuclei of light elements have a lower binding energy per nucleon than that for the elements of intermediate mass. They are therefore less stable. Consequently,the fusion of light elements results in a more stable nucleus with a higher binding energy per nucleon,releasing a large amount of energy.

(D) The nuclear reaction can be represented as: $_8O^{16} + _1H^2 \to _Z^AX + _2He^4$.
Applying the law of conservation of mass number: $16 + 2 = A + 4$,which gives $A = 14$.
Applying the law of conservation of atomic number: $8 + 1 = Z + 2$,which gives $Z = 7$.
Thus,the product nucleus is $_7N^{14}$.

(B) When a $P$-type semiconductor is joined with an $N$-type semiconductor to form a $PN$-junction,electrons diffuse from the $N$-region to the $P$-region,and holes diffuse from the $P$-region to the $N$-region.
This diffusion creates a depletion layer near the junction.
Due to the migration of charge carriers,the $N$-side becomes positively charged relative to the $P$-side,creating an electric field directed from the $N$-side to the $P$-side.
Consequently,the $P$-side is at a lower potential and the $N$-side is at a higher potential. This potential difference is known as the barrier potential.

(A) In forward biasing,an ideal $PN$ junction diode acts as a closed switch with zero resistance. Therefore,the entire applied voltage $V$ appears across the resistor $R$.
In reverse biasing,an ideal $PN$ junction diode acts as an open switch with infinite resistance. Therefore,no current flows through the circuit,and the voltage across the resistor $R$ is $0$.

(B) In a common base configuration,the current gain $\alpha$ is defined as the ratio of collector current $I_C$ to emitter current $I_E$.
Given: $\alpha = \frac{I_C}{I_E} = 0.96$.
The current gain in common emitter configuration,denoted by $\beta$,is related to $\alpha$ by the formula:
$\beta = \frac{\alpha}{1 - \alpha}$.
Substituting the given value:
$\beta = \frac{0.96}{1 - 0.96} = \frac{0.96}{0.04}$.
$\beta = 24$.
Therefore,the maximum current gain in common emitter configuration is $24$.

(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) The greenhouse effect is primarily caused by infrared radiations.
Solar energy reaches the Earth's surface,which then re-emits this energy as infrared radiation.
Greenhouse gases in the atmosphere absorb these infrared rays and reflect them back towards the Earth's surface,thereby trapping heat and keeping the planet warm.

(C) The electromagnetic wave propagates in a direction perpendicular to both the electric field vector $\overrightarrow E$ and the magnetic field vector $\overrightarrow B$.
According to the properties of electromagnetic waves,the direction of wave propagation is given by the direction of the Poynting vector $\vec S$,which is proportional to $\overrightarrow E \times \overrightarrow B$.
Therefore,the direction of propagation is along $\overrightarrow E \times \overrightarrow B$.

(B) The electromagnetic spectrum is ordered by wavelength. The approximate wavelength ranges are as follows:
$1$. Cosmic rays: $10^{-14} \ m$ to $10^{-12} \ m$
$2$. $\gamma$-rays: $10^{-12} \ m$ to $10^{-10} \ m$
$3$. $X$-rays: $10^{-10} \ m$ to $10^{-8} \ m$
$4$. Ultraviolet rays: $10^{-8} \ m$ to $4 \times 10^{-7} \ m$
Comparing these values,cosmic rays have the shortest (minimum) wavelength. Therefore,the correct option is $B$.

(C) When a charge is given to a conductor,it resides entirely on its outer surface due to electrostatic repulsion between like charges.
Inside the conductor,the electric field $E$ is zero.
Since the electric field $E = -dV/dr$,if $E = 0$,then the potential $V$ must be constant throughout the volume of the conductor.
Therefore,the potential at every point inside and on the surface of the conductor is the same,making the conductor an equipotential volume.

(B) Specific resistance, also known as resistivity $(\rho)$, is an intrinsic property of a material.
It depends on the nature of the material and the temperature.
For metallic conductors, resistivity increases with an increase in temperature due to increased scattering of electrons by vibrating lattice ions.
Resistivity does not depend on the physical dimensions of the conductor, such as its length or cross-sectional area.

(A) Given: Electromotive force $(E)$ = $2.2\;V$,Terminal voltage $(V)$ = $1.8\;V$,External resistance $(R)$ = $5\;\Omega$.
The formula for internal resistance $(r)$ is given by:
$r = \left( \frac{E}{V} - 1 \right) R$
Substituting the given values:
$r = \left( \frac{2.2}{1.8} - 1 \right) \times 5$
$r = \left( \frac{22}{18} - 1 \right) \times 5$
$r = \left( \frac{11}{9} - 1 \right) \times 5$
$r = \left( \frac{11 - 9}{9} \right) \times 5$
$r = \left( \frac{2}{9} \right) \times 5$
$r = \frac{10}{9}\;\Omega$

(C) The impedance $Z$ of a series $LCR$ circuit is given by $Z = \sqrt{R^{2} + (X_{L} - X_{C})^{2}}$.
At resonance,the inductive reactance $X_{L}$ is equal to the capacitive reactance $X_{C}$,i.e.,$X_{L} = X_{C}$.
Substituting this into the impedance formula,we get $Z = \sqrt{R^{2} + 0} = R$.
The power loss in an $AC$ circuit is given by $P = I^{2} R$,where $I$ is the $R.M.S.$ current.
Since $Z = R$ at resonance,the circuit behaves as a purely resistive circuit,and the power loss is $I^{2} R$.

(A) The time period of oscillation for a combination of two magnets in a magnetic field $B_{H}$ is given by $T = 2\pi \sqrt{\frac{I_{total}}{M_{total} B_{H}}}$.
Case $1$: When similar poles are on the same side,the effective magnetic moment is $M_{eff} = M + 2M = 3M$. The moment of inertia is $I = I_{1} + I_{2}$.
Thus,$T_{1} = 2\pi \sqrt{\frac{I_{1} + I_{2}}{3M B_{H}}}$.
Case $2$: When the polarity of one magnet is reversed,the effective magnetic moment is $M_{eff} = |2M - M| = M$. The moment of inertia remains $I = I_{1} + I_{2}$.
Thus,$T_{2} = 2\pi \sqrt{\frac{I_{1} + I_{2}}{M B_{H}}}$.
Comparing $T_{1}$ and $T_{2}$,since the denominator in $T_{2}$ is smaller than in $T_{1}$,we have $T_{1} < T_{2}$.

(C) The number of active nuclei present in the original sample is $N_0 = 4 \times 10^{16}$.
The half-life of the element is $T_{1/2} = 10$ days.
The number of half-lives in $30$ days is $n = \frac{30}{10} = 3$.
The number of active nuclei remaining after $30$ days is $N = \frac{N_0}{2^n} = \frac{4 \times 10^{16}}{2^3} = \frac{4 \times 10^{16}}{8} = 0.5 \times 10^{16}$.
The number of decayed nuclei is given by $N_{decayed} = N_0 - N$.
$N_{decayed} = 4 \times 10^{16} - 0.5 \times 10^{16} = 3.5 \times 10^{16}$.

(C) The limit of resolution for an aperture is given by the formula $\theta = \frac{y}{D} = 1.22 \frac{\lambda}{d}$.
For small angles,we consider the Rayleigh criterion $\frac{y}{D} \approx \frac{\lambda}{d}$.
Given:
Diameter of lens $d = 2\,mm = 2 \times 10^{-3}\,m$.
Distance $D = 50\,m$.
Wavelength $\lambda = 5000\,\mathring{A} = 5 \times 10^{-7}\,m$.
Substituting the values:
$y = \frac{\lambda D}{d} = \frac{5 \times 10^{-7} \times 50}{2 \times 10^{-3}}$.
$y = \frac{250 \times 10^{-7}}{2 \times 10^{-3}} = 125 \times 10^{-4}\,m$.
$y = 1.25 \times 10^{-2}\,m = 1.25\,cm$.

(D) At point $A$,by Snell's law:
$1 \cdot \sin 45^{\circ} = \mu \cdot \sin r$
$\sin r = \frac{1}{\mu \sqrt{2}}$ ..... $(i)$
At point $B$,for total internal reflection to occur at the vertical face,the angle of incidence $i_1$ must be greater than or equal to the critical angle $C$.
$\sin i_1 = \frac{1}{\mu}$
From the geometry of the triangle,$i_1 = 90^{\circ} - r$.
Therefore,$\sin(90^{\circ} - r) = \frac{1}{\mu} \Rightarrow \cos r = \frac{1}{\mu}$ ..... $(ii)$
Using the identity $\sin^2 r + \cos^2 r = 1$:
$\left(\frac{1}{\mu \sqrt{2}}\right)^2 + \left(\frac{1}{\mu}\right)^2 = 1$
$\frac{1}{2\mu^2} + \frac{1}{\mu^2} = 1$
$\frac{1 + 2}{2\mu^2} = 1$
$\frac{3}{2\mu^2} = 1 \Rightarrow \mu^2 = \frac{3}{2}$
$\mu = \sqrt{\frac{3}{2}}$

(C) Let the distance between the two parallel walls be $D$. The bulb is on the first wall and the image is formed on the second wall.
For a convex lens to form a real image of equal size (magnification $m = -1$),the object distance $u$ and image distance $v$ must satisfy $u = v = 2f$.
The total distance between the object and the screen is $D = u + v = 2f + 2f = 4f$.
In this problem,the lens is placed at a distance $d$ from the second wall,which means the image distance $v = d$.
Since the image is of equal size,the object distance $u$ must also be equal to $d$ (because $u = v$ for $m = -1$).
Therefore,the total distance between the walls is $D = u + v = d + d = 2d$.
Using the relation $D = 4f$,we get $2d = 4f$.
Solving for $f$,we find $f = 2d / 4 = d / 2$.

(B) Planck's constant $(h)$ relates the energy of a photon to its frequency according to the equation $E = h \nu$.
Since $E$ is measured in Joules $(J)$ and frequency $\nu$ is in $s^{-1}$,the unit of $h$ is $J \cdot s$.
Alternatively,since $E = \text{Force} \times \text{Distance} = (kg \cdot m/s^2) \times m = kg \cdot m^2/s^2$,then $h = E/\nu = (kg \cdot m^2/s^2) / (1/s) = kg \cdot m^2/s$.
The numerical value of Planck's constant is approximately $6.63 \times 10^{-34} \; J \cdot s$ or $6.63 \times 10^{-34} \; kg \cdot m^2/s$.
Comparing this with the given options,option $B$ represents the correct value and units.