AIIMS 2005 Physics Question Paper with Answer and Solution

(A) $Parsec$ (parallax second) is a unit of length used to measure the large distances to astronomical objects outside the Solar System.
It is defined as the distance at which one astronomical unit $(AU)$ subtends an angle of one arcsecond.
$1 \ Parsec \approx 3.086 \times 10^{16} \ m$.
Therefore,the correct option is $A$.

(B) According to Ohm's law,$R = \frac{V}{I}$.
First,find the dimensions of potential difference $(V)$: $V = \frac{W}{q} = \frac{[M L^2 T^{-2}]}{[A T]} = [M L^2 T^{-3} A^{-1}]$.
Now,substitute the dimensions of $V$ and $I$ into the formula for $R$:
$R = \frac{[M L^2 T^{-3} A^{-1}]}{[A]} = [M L^2 T^{-3} A^{-2}]$.
Therefore,the correct option is $B$.

(A) The maximum height $h$ reached by a ball thrown vertically with initial velocity $u$ is given by the formula $h = \frac{u^2}{2g}$.
From this relation,we can see that $h \propto u^2$,which implies $u \propto \sqrt{h}$.
Let the initial velocity be $V_o$ for height $h$,and the new velocity be $V'$ for height $3h$.
Then,$\frac{V'}{V_o} = \sqrt{\frac{3h}{h}} = \sqrt{3}$.
Therefore,$V' = \sqrt{3} V_o$.

(B) The apparent weight $W'$ of a person in an elevator is given by $W' = m(g \pm a)$,where $m$ is the mass of the person,$g$ is the acceleration due to gravity,and $a$ is the acceleration of the elevator.
When the elevator moves downward with an acceleration $a$,the apparent weight is $W' = m(g - a)$.
Since $(g - a) < g$,the apparent weight $W'$ is less than the actual weight $W = mg$.
Therefore,the person feels lighter when the elevator moves downward with constant acceleration.

(C) According to Newton's First Law of Motion,a body remains in a state of rest or uniform motion unless acted upon by an external unbalanced force.
If a body is stationary,its acceleration is $0$.
By Newton's Second Law,$F_{net} = ma$.
Since $a = 0$,the net force $F_{net}$ acting on the body must be $0$.
This means that the vector sum of all individual forces acting on the body is zero,implying that the forces balance each other out.
Therefore,option $C$ is correct.

(C) The escape velocity $v_e$ from the surface of a spherical mass $m$ of radius $r$ is given by the formula: $v_e = \sqrt{\frac{2Gm}{r}}$.
For an object to be a black hole,not even light can escape its gravitational pull.
Therefore,the escape velocity must be greater than or equal to the speed of light $c$.
Thus,the condition is $\sqrt{\frac{2Gm}{r}} \ge c$ or $(2Gm/r)^{1/2} \ge c$.

(A) When the tube rotates with angular velocity $\omega$,the water in the horizontal arms experiences a centrifugal force directed outwards.
For a small element of water of mass $dm$ at a distance $r$ from the axis of rotation,the required centripetal force is provided by the pressure gradient: $dP = \rho \omega^2 r dr$.
Integrating this from the axis $(r=0)$ to a distance $r$,we get the pressure at distance $r$ as $P(r) = P_0 + \frac{1}{2} \rho \omega^2 r^2$,where $P_0$ is the pressure at the axis.
Since the pressure at the surface of the water in the vertical arms must be equal to the atmospheric pressure,the height $h$ of the water column in an arm at distance $r$ from the axis is given by $\rho gh = P(r) - P_{atm}$.
As $r$ increases,the pressure $P(r)$ increases,which means the height $h$ of the water column must increase to balance this pressure.
Since both arms $A$ and $B$ are at distances $L$ and $2L$ respectively from the axis,the water levels in both arms will rise compared to the level at the central axis to provide the necessary centripetal force for the rotation of the water mass.

(B) Let the density of the candle be $\rho_c$ and the density of the liquid be $\rho_l$. Let the total length of the candle be $L_0$ and the submerged length be $L_s$. According to the principle of floatation,the weight of the candle equals the weight of the displaced liquid:
$\rho_c \cdot A \cdot L_0 = \rho_l \cdot A \cdot L_s$
where $A$ is the cross-sectional area of the candle.
This implies $L_s = (\rho_c / \rho_l) L_0$. Let $k = \rho_c / \rho_l$. Then $L_s = k L_0$.
The height of the top of the candle above the liquid surface is $h = L_0 - L_s = L_0(1 - k)$.
As the candle burns,its total length $L_0$ decreases at a rate $v = dL_0/dt = 2 \text{ cm/hour}$.
The rate of change of the height of the top of the candle is $dh/dt = (dL_0/dt)(1 - k)$.
Assuming the candle is half-submerged (as suggested by the figure),$k = 0.5$.
Therefore,$dh/dt = 2 \text{ cm/hour} \times (1 - 0.5) = 1 \text{ cm/hour}$.
Since $dh/dt$ is positive,the top of the candle falls at a rate of $1 \text{ cm/hour}$.

(A) function represents simple harmonic motion $(S.H.M.)$ if it satisfies the differential equation $\frac{d^2x}{dt^2} + \omega^2 x = 0$.
Option $A$: $x = \sin \omega t - \cos \omega t$. This can be written as $x = \sqrt{2} [\frac{1}{\sqrt{2}} \sin \omega t - \frac{1}{\sqrt{2}} \cos \omega t] = \sqrt{2} \sin(\omega t - \frac{\pi}{4})$. This is a standard $S.H.M.$ equation.
Option $B$: $x = \sin^2 \omega t = \frac{1 - \cos 2\omega t}{2}$. This represents motion with a constant shift and frequency $2\omega$,which is not simple harmonic.
Options $C$ and $D$: These are superpositions of two different frequencies ($\omega$ and $2\omega$),which result in periodic motion but not simple harmonic motion.
Therefore,the correct option is $A$.

(A) Amorphous solids are those in which the constituent particles (atoms, molecules, or ions) are arranged in a completely irregular or random manner over long distances.
Among the given options, $\text{Glass}$ is a classic example of an amorphous solid, often referred to as a supercooled liquid.
Diamond, Salt $(NaCl)$, and Sugar are all examples of crystalline solids, which possess a long-range ordered structure.
Therefore, the correct option is $A$.

(A) Specific gravity (also known as relative density) is defined as the ratio of the density of a substance to the density of water at a standard temperature (usually $4^{\circ}C$).
$\text{Specific Gravity} = \frac{\rho_{\text{fluid}}}{\rho_{\text{water}}}$
Since it is a ratio of two identical physical quantities (density/density),the units cancel out,making it a dimensionless quantity.
Therefore,both the $Assertion$ and $Reason$ are correct,and the $Reason$ is the correct explanation for the $Assertion$.

(D) force is called conservative if the work done by or against it in moving a particle between two points is independent of the path taken.
Frictional force is a non-conservative force because the work done against friction depends on the path length.
Furthermore,work done against a non-conservative force like friction is dissipated as heat and cannot be recovered as potential energy.
Therefore,both the $Assertion$ and the $Reason$ are incorrect.

(B) According to the principle of conservation of angular momentum, since no external torque acts on the system, the angular momentum $L = I\omega$ remains constant.
When the viscous fluid is dropped at the centre and spreads out, the distribution of mass moves away from the axis of rotation, which increases the moment of inertia $I$ of the system.
Since $L = I\omega$ is constant, an increase in $I$ leads to a decrease in the angular velocity $\omega$.
As the fluid eventually falls off the platform, the mass of the system decreases, causing the moment of inertia $I$ to decrease back towards its original value.
Consequently, the angular velocity $\omega$ increases again as the fluid leaves the platform.

(A) To climb the inclined surface to a height $h$,the total initial kinetic energy of the rolling sphere must be at least equal to the potential energy gained at height $h$.
The total kinetic energy of a rolling body is the sum of its translational and rotational kinetic energies:
$K_{total} = K_{trans} + K_{rot} = \frac{1}{2}mv^2 + \frac{1}{2}I\omega^2$
For a solid sphere,the moment of inertia $I = \frac{2}{5}mr^2$ and for pure rolling,$\omega = \frac{v}{r}$.
Substituting these values:
$K_{total} = \frac{1}{2}mv^2 + \frac{1}{2}(\frac{2}{5}mr^2)(\frac{v}{r})^2$
$K_{total} = \frac{1}{2}mv^2 + \frac{1}{5}mv^2 = \frac{7}{10}mv^2$
By the law of conservation of energy,the total kinetic energy must be greater than or equal to the potential energy $mgh$ at the top:
$\frac{7}{10}mv^2 \ge mgh$
$v^2 \ge \frac{10}{7}gh$
$v \ge \sqrt{\frac{10}{7}gh}$

(B) Let $\ell$ be the length of the ladder and $(x, y)$ be the coordinates of its center of mass,which is the midpoint of the ladder.
From the geometry of the ladder leaning against the wall,the coordinates of the ends are $(2x, 0)$ and $(0, 2y)$.
Using the Pythagorean theorem for the length of the ladder $\ell$:
$(2x)^2 + (2y)^2 = \ell^2$
$4x^2 + 4y^2 = \ell^2$
$x^2 + y^2 = \left(\frac{\ell}{2}\right)^2$
This is the equation of a circle with radius $R = \frac{\ell}{2}$ centered at the origin $(0, 0)$.
As the ladder slips,the center of mass traces a circular arc. Therefore,the correct representation is a circular path.

(A) In a central force field,the force acting on each particle is directed along the line joining the center of force and the particle.
Since the force is radial,the position vector $\vec{r}$ and the force vector $\vec{F}$ are collinear.
The torque $\vec{\tau}$ is defined as $\vec{\tau} = \vec{r} \times \vec{F}$.
Since $\vec{r}$ and $\vec{F}$ are parallel,the cross product is zero,meaning the net torque acting on the system is zero.
According to the principle of conservation of angular momentum,if the net external torque acting on a system is zero,the total angular momentum $\vec{L}$ remains constant.
Therefore,both the $Assertion$ and $Reason$ are correct,and the $Reason$ is the correct explanation of the $Assertion$.

$(A)$ The Reynolds number $(Re)$ is defined as the ratio of inertial forces to viscous forces in a fluid flow: $Re = \frac{\text{Inertial Force}}{\text{Viscous Force}}$.
For $Re < 2000$, the flow is typically laminar, where viscous forces dominate.
For $Re > 2000$, the flow becomes turbulent, meaning inertial forces are dominant compared to viscous forces.
Therefore, the Assertion is correct, and the Reason correctly explains why the flow becomes turbulent at high Reynolds numbers.

(A) According to Kirchhoff's law of radiation,for an arbitrary body,the ratio of its emissive power $(e_{\lambda})$ to its absorptivity $(a_{\lambda})$ at a given wavelength and temperature is equal to the emissive power $(E_{\lambda})$ of a perfectly black body at the same wavelength and temperature.
Mathematically,$\frac{e_{\lambda}}{a_{\lambda}} = E_{\lambda}$.
Since $E_{\lambda}$ is a constant for a given wavelength and temperature,it follows that $e_{\lambda} \propto a_{\lambda}$.
This implies that a body with high emissivity (good radiator) must also have high absorptivity (good absorber) at that specific wavelength. Thus,both the Assertion and the Reason are correct,and the Reason is the correct explanation of the Assertion.

(A) According to the Clausius-Clapeyron equation,the slope of the melting curve is given by $\frac{dP}{dT} = \frac{L}{T(V_2 - V_1)}$,where $L$ is the latent heat of fusion,$T$ is the temperature,$V_2$ is the volume of the liquid phase,and $V_1$ is the volume of the solid phase.
For water,ice contracts upon melting,meaning the volume of water $(V_2)$ is less than the volume of ice $(V_1)$.
Therefore,$(V_2 - V_1) < 0$,which makes the slope $\frac{dP}{dT}$ negative.
Thus,both the Assertion and the Reason are correct,and the Reason is the correct explanation for the Assertion.

(C) According to Wien's displacement law, $\lambda_m T = \text{constant}$.
This implies that $\lambda_m \propto 1/T$. Therefore, as the temperature $T$ increases, the peak emission wavelength $\lambda_m$ decreases. Thus, the Assertion is correct.
According to the Stefan-Boltzmann law, the total energy radiated per unit area per unit time is $E = \sigma T^4$. This law relates energy to temperature, not the peak wavelength. The peak wavelength is governed by Wien's displacement law, not the fourth power of temperature. Thus, the Reason is incorrect.

(A) In any real-world process,some energy is inevitably converted into heat due to friction,viscosity,or other resistive forces,which is dissipative in nature.
Because of this energy loss,the system cannot be returned to its original state without leaving a change in the surroundings.
Therefore,all natural processes are irreversible,making perfectly reversible systems impossible to find in reality.
Thus,the $Assertion$ is correct,and the $Reason$ is the correct explanation for it.

(A) When air leaks out of a balloon quickly,the process happens so fast that there is no time for heat exchange with the surroundings,making it an adiabatic process.
During this process,the air expands against the external atmospheric pressure.
Since the air does work at the expense of its internal energy,its temperature decreases,causing the air to cool down.

(C) The electric field due to an infinitely large thin sheet of charge with surface charge density $\sigma$ is given by $E = \frac{\sigma}{2\varepsilon_0}$.
For two parallel plates with charge densities $+\sigma$ and $-\sigma$,the electric field from the positively charged plate points away from it,and the electric field from the negatively charged plate points towards it.
In the region between the plates,both electric fields point in the same direction (from the positive plate to the negative plate).
Therefore,the net electric field $E_{net}$ is the sum of the magnitudes of the individual fields:
$E_{net} = E_1 + E_2 = \frac{\sigma}{2\varepsilon_0} + \frac{|-\sigma|}{2\varepsilon_0} = \frac{\sigma}{2\varepsilon_0} + \frac{\sigma}{2\varepsilon_0} = \frac{\sigma}{\varepsilon_0} \text{ V/m}$.

(A) The four positive charges $(Q)$ at the corners of the square create an electric field directed towards the center of the square along the $Z$-axis.
For a negative charge $(-q)$ placed at a small distance $z$ on the $Z$-axis, the net electrostatic force exerted by the four charges will be directed towards the center of the square frame.
The magnitude of this force is proportional to the displacement $z$ for small $z$ (i.e., $F \propto -z$).
This acts as a restoring force. As the negative charge is attracted towards the center, it accelerates. When it reaches the center (the plane of the frame), the net force is zero, but due to its acquired velocity (inertia), it crosses the center and moves to the other side.
The restoring force then acts in the opposite direction, slowing it down until it stops and is pulled back again.
Thus, the negative charge performs simple harmonic motion (oscillates) along the $Z$-axis about the center of the frame.

(A) The electric field $E$ at a distance $x$ from the common centre is determined using Gauss's Law:
$1$. For $x < r_A$: The interior of a conducting shell has no electric field. Thus,$E = 0$.
$2$. For $r_A < x < r_B$: Only the charge $Q_A$ on shell $A$ contributes to the field. Thus,$E = \frac{k Q_A}{x^2}$. This is a positive value that decreases as $x$ increases.
$3$. For $x > r_B$: Both charges $Q_A$ and $-Q_B$ contribute to the field. The net charge is $(Q_A - Q_B)$. Since $|Q_B| > |Q_A|$,the net charge is negative. Thus,$E = \frac{k(Q_A - Q_B)}{x^2}$. This is a negative value that approaches zero as $x \to \infty$.
Comparing these characteristics with the given options,the graph shows $E=0$ for $x < r_A$,a positive $1/x^2$ curve for $r_A < x < r_B$,and a negative $1/x^2$ curve for $x > r_B$.

(C) Let the energy dissipated in each resistor be $H$. Since the resistors ${R_2}$ and ${R_3}$ are connected in parallel,the potential difference across them is the same. Therefore,the energy dissipated in them is given by $H = \frac{V^2}{R}t$. For the energy to be the same,we must have ${R_2} = {R_3}$.
Let the total current flowing through the circuit be $i$. The current ${R_1}$ is $i$. The current splits into ${i_1}$ and ${i_2}$ at the junction. Since ${R_2} = {R_3}$,the current splits equally,so ${i_1} = {i_2} = \frac{i}{2}$.
The energy dissipated in ${R_1}$ is $H = i^2 {R_1} t$.
The energy dissipated in ${R_2}$ is $H = i_1^2 {R_2} t = (\frac{i}{2})^2 {R_2} t = \frac{i^2}{4} {R_2} t$.
Since the energy dissipated in all resistors is the same,we equate the energy in ${R_1}$ and ${R_2}$:
$i^2 {R_1} t = \frac{i^2}{4} {R_2} t$
${R_1} = \frac{{R_2}}{4}$.

(B) The magnetic field is given by $B(t) = B_0 \sin(\omega t)$,where $B_0 = 0.01\,T$ and $\omega = 2\pi f = 2\pi \times 100 = 200\pi\,rad/s$.
Using Faraday's law of induction,the induced emf $\varepsilon$ is given by $\varepsilon = -\frac{d\phi}{dt} = -A \frac{dB}{dt}$.
The magnetic flux $\phi = B \cdot A = B_0 \sin(\omega t) \cdot \pi r^2$.
Thus,$\varepsilon = -\pi r^2 \frac{d}{dt}(B_0 \sin(\omega t)) = -\pi r^2 B_0 \omega \cos(\omega t)$.
The induced electric field $E$ along the circumference of the ring is given by $\oint E \cdot dl = \varepsilon$.
$E(2\pi r) = \pi r^2 \frac{dB}{dt} \implies E = \frac{r}{2} \frac{dB}{dt}$.
Since $\frac{dB}{dt} = B_0 \omega \cos(\omega t)$,the maximum induced electric field is $E_{max} = \frac{r}{2} B_0 \omega$.
Substituting the values: $E_{max} = \frac{1}{2} \times 0.01 \times 200\pi = \pi \approx 3.14\,V/m$.
However,considering the average induced emf approach provided in the context of the question: $\varepsilon_{avg} = \frac{\Delta \phi}{\Delta t} = \frac{B_0 A}{T/4} = 4 B_0 A f = 4 \times 0.01 \times \pi(1)^2 \times 100 = 4\pi$.
Then $E = \frac{\varepsilon}{2\pi r} = \frac{4\pi}{2\pi(1)} = 2\,V/m$.

(A) According to Faraday's law of electromagnetic induction,the induced $e.m.f.$ is given by $\varepsilon = -\frac{d\phi}{dt}$.
When the north pole of the magnet approaches the coil,the magnetic flux through the coil increases,inducing an $e.m.f.$ of one polarity (e.g.,positive) and a current that flows in an anticlockwise direction to oppose the motion.
As the magnet passes through the center of the coil,the rate of change of flux becomes zero momentarily,so the induced $e.m.f.$ becomes zero.
As the magnet moves away,the flux decreases,inducing an $e.m.f.$ of the opposite polarity (e.g.,negative) and a current in the clockwise direction.
When the magnet reaches the extreme end of its oscillation,it is momentarily at rest,so the $e.m.f.$ is zero.
As it returns,the process repeats in reverse,creating a symmetric variation in the $e.m.f.$ over one full cycle of oscillation. This results in a waveform that shows both positive and negative peaks,as depicted in the first option.

(B) According to Moseley's Law,the frequency $(f)$ of the characteristic $X-$rays emitted by an element is related to its atomic number $(Z)$ by the equation: $f = a(Z - b)^2$,where $a$ and $b$ are constants.
For high atomic numbers,$b$ is negligible,so $f \propto Z^2$.
Therefore,the frequency of characteristic $X-$rays varies with the square of the atomic number.

(B) For a hydrogen atom,the total energy $(E)$,kinetic energy $(K)$,and potential energy $(U)$ are related as follows:
$E = -K$
$U = 2E$
Given the ground state energy $E = -13.6 \, eV$.
Therefore,the potential energy $U = 2 \times (-13.6 \, eV) = -27.2 \, eV$.

(A) The magnetic moment $(\mu)$ of a revolving electron is given by the relation $\mu = \frac{e}{2m} L$,where $L$ is the angular momentum.
According to Bohr's quantization condition,the angular momentum $L$ of an electron in the $n$-th orbit is $L = \frac{nh}{2\pi}$.
Substituting this value into the expression for magnetic moment,we get $\mu = \frac{e}{2m} \left( \frac{nh}{2\pi} \right)$.
Rearranging the terms,we have $\mu = n \left( \frac{eh}{4\pi m} \right)$.
Since $e$,$h$,and $m$ are constants,the term $\frac{eh}{4\pi m}$ is also a constant (related to the Bohr magneton $\mu_B = \frac{eh}{4\pi m}$).
Therefore,$\mu \propto n$.

(C) The number of radioactive nuclei remaining after time $t$ is given by the formula: $N = N_0 \left( \frac{1}{2} \right)^{\frac{t}{T_{1/2}}}$.
Here,the initial amount is $N_0$,the half-life $T_{1/2} = 10$ days,and the total time $t = 30$ days.
The fraction of the material remaining is $\frac{N}{N_0} = \left( \frac{1}{2} \right)^{\frac{30}{10}}$.
$\frac{N}{N_0} = \left( \frac{1}{2} \right)^3 = \frac{1}{8}$.
$\frac{N}{N_0} = 0.125$.

(A) In a semiconductor,the mobility of charge carriers is defined as the drift velocity per unit electric field.
Electrons are lighter and move through the conduction band,whereas holes are essentially vacancies in the valence band that move through a process of successive electron jumps.
Due to their lower effective mass and the nature of their movement in the conduction band,electrons encounter less scattering and have a higher mobility compared to holes.
Therefore,the mobility of electrons is always greater than the mobility of holes,i.e.,$\mu_e > \mu_h$.

(B) Given: Current gain $\beta = 100$,Change in collector current $\Delta I_c = 1\, mA$.
We know that the current gain in common-emitter configuration is defined as $\beta = \frac{\Delta I_c}{\Delta I_b}$.
Therefore,the change in base current is $\Delta I_b = \frac{\Delta I_c}{\beta} = \frac{1\, mA}{100} = 0.01\, mA$.
The relationship between emitter,collector,and base currents is $\Delta I_e = \Delta I_c + \Delta I_b$.
Substituting the values,we get $\Delta I_e = 1\, mA + 0.01\, mA = 1.01\, mA$.
Thus,the change in emitter current is $1.01\, mA$.

(B) The given circuit is an inverting amplifier configuration using an operational amplifier (op-amp).
For an inverting amplifier,the voltage gain $A_v$ is given by the formula:
$A_v = -\frac{R_f}{R_i}$
Where $R_f$ is the feedback resistance and $R_i$ is the input resistance.
From the circuit diagram,we have:
$R_f = 100 \, k\Omega$
$R_i = 1 \, k\Omega$
Substituting these values into the formula,we get:
$A_v = -\frac{100 \, k\Omega}{1 \, k\Omega} = -100$
The magnitude of the voltage gain is $|A_v| = 100$.
Therefore,the correct option is $(b)$.

(B) The acceptance angle $\theta_a$ is defined as the maximum angle that a light ray can make with the axis of the optical fibre and still be guided through the core by total internal reflection.
Applying Snell's Law at the air-core interface: $1 \cdot \sin \theta_a = n_1 \cdot \sin \theta_r$,where $\theta_r$ is the angle of refraction.
At the core-cladding interface,for total internal reflection to occur,the angle of incidence $\theta_i$ must be at least the critical angle $\theta_c$,where $\sin \theta_c = n_2/n_1$.
Since $\theta_r + \theta_i = 90^\circ$,we have $\sin \theta_r = \cos \theta_i = \sqrt{1 - \sin^2 \theta_i} = \sqrt{1 - (n_2/n_1)^2} = \frac{\sqrt{n_1^2 - n_2^2}}{n_1}$.
Substituting this into the first equation: $\sin \theta_a = n_1 \cdot \frac{\sqrt{n_1^2 - n_2^2}}{n_1} = \sqrt{n_1^2 - n_2^2}$.
Therefore,$\theta_a = \sin^{-1}\sqrt{n_1^2 - n_2^2}$.

(A) The magnification produced by the objective lens of a telescope is given by the ratio of the height of the image $(I)$ to the height of the object $(O)$,which is also equal to the ratio of the focal length of the objective $(f_0)$ to the object distance $(u_0)$:
$m = \frac{I}{O} = \frac{f_0}{u_0}$
Given:
Height of the building $(O)$ = $50 \, m = 5000 \, cm$
Focal length of objective $(f_0)$ = $200 \, cm$
Distance of the building $(u_0)$ = $2 \, km = 2000 \, m = 200,000 \, cm$
Substituting the values into the formula:
$\frac{I}{5000} = \frac{200}{200000}$
$I = 5000 \times \frac{200}{200000}$
$I = 5000 \times \frac{1}{1000} = 5 \, cm$
Therefore,the height of the image formed by the objective lens is $5 \, cm$.

(B) The apparent depth $h'$ is related to the real depth $h$ by the formula $h' = h / \mu$,where $\mu = n_2/n_1$ is the refractive index of water with respect to air.
Given that the apparent depth is reducing at a rate of $x \, cm/min$,we have $\frac{dh'}{dt} = -x$.
Since $h' = h \cdot (n_1/n_2)$,differentiating with respect to time $t$ gives $\frac{dh'}{dt} = \frac{n_1}{n_2} \frac{dh}{dt}$.
Substituting the given rate: $-x = \frac{n_1}{n_2} \frac{dh}{dt}$,which implies $\frac{dh}{dt} = -x \cdot (n_2/n_1)$. Thus,the real depth is decreasing at a rate of $x \cdot (n_2/n_1) \, cm/min$.
The volume of water in the cylindrical tank is $V = \pi R^2 h$.
The rate of change of volume is $\frac{dV}{dt} = \pi R^2 \frac{dh}{dt}$.
Substituting the magnitude of $\frac{dh}{dt}$,the volume of water drained per minute is $\frac{dV}{dt} = \pi R^2 \cdot x \cdot (n_2/n_1) = x \pi R^2 (n_2/n_1)$.

(B) When light falls on a thin film of oil on water,it undergoes reflection and refraction at both the top and bottom surfaces of the oil film.
These reflected light waves from the upper and lower surfaces of the thin film interfere with each other.
Depending on the thickness of the film and the angle of incidence,some wavelengths of light undergo constructive interference (appearing bright) while others undergo destructive interference (appearing dark).
This phenomenon of superposition of light waves reflected from the two surfaces of a thin film is known as the interference of light,which results in the observation of brilliant colours.

(B) In an electromagnetic wave,the electric field vector $\vec{E}$ at any point in space is given by $\vec{E} = E_0 \sin(kx - \omega t) \hat{n}$.
Here,$E_0$ is the amplitude,$k$ is the wave number,$\omega$ is the angular frequency,and $t$ is time.
The magnitude of the electric field vector is $|\vec{E}| = |E_0 \sin(kx - \omega t)|$.
Since the sine function oscillates between $-1$ and $1$,the magnitude of the electric field vector varies periodically with time at any fixed position $x$.

(A) Hubble's law states that the recessional velocity $(v)$ of a galaxy is directly proportional to its distance $(r)$ from the observer.
Mathematically,this is expressed as $v = H_0 r$,where $H_0$ is the Hubble constant.
Since the redshift $(Z)$ is directly proportional to the recessional velocity $(v)$ for non-relativistic speeds,we have $Z \propto v$.
Substituting the relationship,we get $Z \propto r$.
Therefore,the redshift of a receding galaxy is directly proportional to its distance from Earth.

(A) Superconductors exhibit the Meissner effect,which means they expel magnetic fields from their interior when cooled below their critical temperature $(T_c)$.
Because of this,they act as perfect diamagnets.
When a magnet is placed near a superconductor,the superconductor generates surface currents that create a magnetic field opposing the magnet's field.
This results in a repulsive force that allows the magnet to levitate above the superconductor.
Therefore,the Assertion is correct because the magnet levitates,and the Reason is correct because superconductors exhibit a repulsive force towards magnets due to the Meissner effect.

(C) The Assertion is correct because roughening the surface of a glass sheet causes diffuse reflection or scattering of light,which reduces the amount of light transmitted directly through the glass,thereby reducing its transparency.
The Reason is incorrect because a rough surface does not necessarily absorb more light; rather,it scatters the incident light in various directions. The reduction in transparency is primarily due to scattering,not increased absorption.

(C) The Assertion is correct because diamond has a very high refractive index $(n \approx 2.42)$,which results in a very small critical angle $(C \approx 24.4^\circ)$.
When light enters a diamond,it undergoes multiple internal reflections due to this small critical angle,causing it to glitter brilliantly.
The Reason is incorrect because the glittering of a diamond is not related to the absorption of sunlight,but rather to the phenomenon of Total Internal Reflection $(TIR)$.

(C) The Assertion is correct because clouds appear white due to the scattering of light by water droplets.
The size of water droplets in clouds is much larger than the wavelength of visible light.
According to the theory of scattering, when the size of the scattering particle $(a)$ is much larger than the wavelength of light $(\lambda)$, i.e., $a \gg \lambda$, the scattering is independent of the wavelength.
Since all wavelengths of visible light are scattered equally, the combined effect is perceived as white light.
Diffraction is a phenomenon that occurs when light encounters an obstacle or aperture of a size comparable to its wavelength. Since cloud particles are much larger than the wavelength of light, they do not cause significant diffraction of visible light.
Therefore, the Reason is incorrect.

(B) The resolving power of a telescope is defined as the inverse of the smallest angular separation between two distant objects that can be just resolved by the telescope.
The formula for the resolving power of a telescope is given by: $\text{Resolving Power} = \frac{D}{1.22 \lambda}$,where $D$ is the diameter of the objective lens and $\lambda$ is the wavelength of light used.
From this formula,it is clear that the resolving power is directly proportional to the diameter $D$ of the objective lens. Thus,the Assertion is correct.
The Reason states that an objective lens of large diameter collects more light. This statement is physically true,as a larger aperture allows more light to enter the telescope,which increases its light-gathering power (brightness of the image). However,this is not the reason for the increase in resolving power. The resolving power depends on the diffraction limit,not the light-gathering capacity.
Therefore,both statements are correct,but the Reason is not the correct explanation for the Assertion.

(A) According to the theory of relativity,the energy of a particle is given by $E^2 = (pc)^2 + (m_0c^2)^2$.
For a photon,the rest mass $m_0$ is $0$.
Therefore,$E^2 = (pc)^2$,which simplifies to $E = pc$ or $p = E/c$.
This relationship is derived specifically from the particle nature of the photon,which allows it to possess momentum despite having no rest mass.
Thus,the Assertion is correct,and the Reason is the correct explanation for the Assertion.

(C) Nuclear fusion involves the combining of light nuclei to form a heavier,more stable nucleus,releasing energy in the process.
${}^{35}Cl$ is a relatively stable,medium-mass nucleus with a high binding energy per nucleon.
Because it is already stable,it does not undergo fusion to release energy.
The Assertion is correct because ${}^{35}Cl$ cannot be used as fuel for fusion.
The Reason is incorrect because the binding energy of ${}^{35}Cl$ is actually quite high,not small,which is precisely why it is stable and does not undergo fusion.

(D) $NOT$ gate is an inverter that requires an active component like a transistor to perform the inversion logic. $A$ diode is a passive,unidirectional device that allows current to flow in only one direction and cannot perform the logical inversion required for a $NOT$ gate.
Furthermore,a diode does not introduce a $180^o$ phase shift between the input and output voltages. Therefore,both the Assertion and the Reason are incorrect.

(A) According to the law of mass action,for a semiconductor,the product of the number density of electrons $(n_e)$ and holes $(n_h)$ is constant at a given temperature: $n_e n_h = n_i^2$,where $n_i$ is the intrinsic carrier concentration.
In a $p-$type semiconductor,trivalent impurity atoms are added,which significantly increases the number of holes $(n_h > n_i)$.
Since the product $n_e n_h$ must remain equal to $n_i^2$,an increase in $n_h$ leads to a decrease in the number of electrons $(n_e)$ such that $n_e = n_i^2 / n_h$. Since $n_h > n_i$,it follows that $n_e < n_i$.
Therefore,the number of electrons in a $p-$type semiconductor is less than the number of electrons in an intrinsic (pure) semiconductor at the same temperature.
Both the Assertion and the Reason are correct,and the Reason correctly explains the Assertion.

(C) In a common emitter transistor amplifier,the input current is the base current $(I_B)$ and the output current is the collector current $(I_C)$.
Since the current gain $\beta = I_C / I_B$ is typically much greater than $1$,it follows that $I_C \gg I_B$,meaning the input current is much less than the output current. Thus,the Assertion is correct.
However,a common emitter transistor amplifier is characterized by low input impedance,not high input impedance.
Therefore,the Reason is incorrect.

(D) universal gate is a logic gate that can be used to implement any other logic gate or Boolean function without the need for any other type of gate.
$NAND$ and $NOR$ gates are known as universal gates.
In the given options,$NAND$ is a universal gate.
Therefore,the correct option is $D$.