AIPMT 2007 Physics Question Paper with Answer and Solution

(C) According to Ohm's law,$V = RI$,which implies $R = \frac{V}{I}$.
Dimensions of potential difference $V$ are given by $V = \frac{W}{q}$,where $W$ is work and $q$ is charge.
Dimensions of work $W = [ML^2T^{-2}]$ and charge $q = [IT]$.
Therefore,dimensions of $V = \frac{[ML^2T^{-2}]}{[IT]} = [ML^2T^{-3}I^{-1}]$.
Now,substituting this into the expression for resistance: $R = \frac{[ML^2T^{-3}I^{-1}]}{[I]} = [ML^2T^{-3}I^{-2}]$.

(A) Given the position equation: $x = 9t^2 - t^3$.
Velocity $v$ is the derivative of position with respect to time: $v = \frac{dx}{dt} = \frac{d}{dt}(9t^2 - t^3) = 18t - 3t^2$.
To find the time at which speed is maximum,we differentiate velocity with respect to time and set it to zero: $\frac{dv}{dt} = 18 - 6t = 0$.
Solving for $t$,we get $t = 3 \ s$.
Now,substitute $t = 3 \ s$ into the position equation to find the position at that instant: $x = 9(3)^2 - (3)^3 = 9(9) - 27 = 81 - 27 = 54 \ m$.

(C) Average speed is defined as the ratio of total distance traveled to the total time taken.
Let the distance between $X$ and $Y$ be $d$.
Time taken to travel from $X$ to $Y$ is $t_1 = \frac{d}{v_1}$.
Time taken to travel from $Y$ to $X$ is $t_2 = \frac{d}{v_2}$.
Total distance $= d + d = 2d$.
Total time $= t_1 + t_2 = \frac{d}{v_1} + \frac{d}{v_2}$.
Average speed $\bar v = \frac{\text{Total distance}}{\text{Total time}} = \frac{2d}{\frac{d}{v_1} + \frac{d}{v_2}}$.
Canceling $d$ from the numerator and denominator,we get $\bar v = \frac{2}{\frac{1}{v_1} + \frac{1}{v_2}}$.
Rearranging this gives $\frac{2}{\bar v} = \frac{1}{v_1} + \frac{1}{v_2}$.

(C) Given: At time $t = 0$,velocity $v = 0$.
Acceleration $f = f_0(1 - t/T)$.
At the instant when $f = 0$,we have $0 = f_0(1 - t/T)$. Since $f_0$ is a constant,$1 - t/T = 0$,which implies $t = T$.
We know that acceleration $f = \frac{dv}{dt}$,so $dv = f dt$.
Integrating both sides from $t = 0$ to $t = T$:
$\int_{0}^{v_x} dv = \int_{0}^{T} f_0(1 - t/T) dt$
$v_x = f_0 \left[ t - \frac{t^2}{2T} \right]_{0}^{T}$
$v_x = f_0 \left( T - \frac{T^2}{2T} \right) = f_0 \left( T - \frac{T}{2} \right) = \frac{1}{2} f_0 T$.

(B) Let $\theta$ be the angle that the particle's path makes with the $x$-axis.
From the given coordinates $(x, y) = (\sqrt{3}, 3)$,the slope of the line is given by $\tan \theta = \frac{y}{x}$.
Substituting the values,we get $\tan \theta = \frac{3}{\sqrt{3}}$.
Simplifying this,$\tan \theta = \sqrt{3}$.
Therefore,$\theta = \tan^{-1}(\sqrt{3}) = 60^o$.

(A) Given the equation: $|\vec A \times \vec B| = \sqrt{3}(\vec A \cdot \vec B)$.
We know that the magnitude of the cross product is $|\vec A \times \vec B| = AB \sin \theta$ and the dot product is $\vec A \cdot \vec B = AB \cos \theta$.
Substituting these into the given equation:
$AB \sin \theta = \sqrt{3} AB \cos \theta$.
Dividing both sides by $AB \cos \theta$ (assuming $A, B \neq 0$ and $\cos \theta \neq 0$):
$\frac{\sin \theta}{\cos \theta} = \sqrt{3}$.
$\tan \theta = \sqrt{3}$.
Therefore,$\theta = \tan^{-1}(\sqrt{3}) = 60^\circ$.

(D) Given: Initial velocity $u = V$,final velocity $v = 0$.
The force of kinetic friction acting on the block is $f = \mu R = \mu mg$,where $m$ is the mass of the block and $g$ is the acceleration due to gravity.
According to Newton's second law,the retardation $a$ produced by this friction is:
$a = \frac{f}{m} = \frac{\mu mg}{m} = \mu g$.
Using the first equation of motion,$v = u - at$ (where $a$ is retardation):
$0 = V - (\mu g)t$
$V = \mu gt$
$t = \frac{V}{\mu g}$.

(A) According to the work-energy theorem,the net work done on the system is equal to the change in kinetic energy of the ball.
Let the initial position of the ball be at height $h$ above the spring. The final position is when the spring is compressed by distance $d$.
The total vertical displacement of the ball is $(h + d)$.
The forces acting on the ball are gravity (downward) and the spring force (upward).
The work done by gravity is $W_g = mg(h + d)$.
The work done by the spring force is $W_s = -\int_0^d kx \, dx = -\frac{1}{2}kd^2$.
The net work done on the ball is $W_{net} = W_g + W_s = mg(h + d) - \frac{1}{2}kd^2$.
Since the ball starts from rest and comes to rest momentarily at the maximum compression $d$,the change in kinetic energy is zero,which implies $W_{net} = 0$ for the entire process. However,the question asks for the net work done by the external forces (gravity and spring) during the displacement $d$,which is $mg(h + d) - \frac{1}{2}kd^2$.

(A) According to Stefan-Boltzmann law,the rate of energy radiated $(E)$ by a black body is directly proportional to the fourth power of its absolute temperature ($T$ in Kelvin).
The formula is given by $E \propto T^4$.
Given the temperature in Celsius is $727^\circ C$,we must convert it to Kelvin:
$T = 727 + 273 = 1000 \ K$.
Substituting this value into the proportionality relation:
$E \propto (1000)^4$.
Therefore,the rate of energy emission is proportional to $(1000)^4$.

(C) The total power $P$ radiated by the sun,acting as a black body with radius $r$ and absolute temperature $T = (t + 273) \ K$,is given by the Stefan-Boltzmann law: $P = \sigma A T^4 = \sigma (4\pi r^2) (t + 273)^4$.
At a distance $R$ from the center of the sun,this power is distributed over a spherical surface area of $4\pi R^2$.
The power received per unit area (intensity $S$) by a surface normal to the incident rays is given by: $S = \frac{P}{4\pi R^2}$.
Substituting the value of $P$: $S = \frac{\sigma (4\pi r^2) (t + 273)^4}{4\pi R^2} = \frac{r^2 \sigma (t + 273)^4}{R^2}$.

(C) The torque $\tau$ about point $A$ is due to the gravitational force acting at the center of mass of the rod,which is at a distance $l/2$ from $A.$
$\tau = mg \times \frac{l}{2} = \frac{mgl}{2}$
Using the relation between torque and angular acceleration,$\tau = I\alpha$,where $I$ is the moment of inertia about $A$ and $\alpha$ is the angular acceleration.
Given $I = \frac{ml^2}{3}$,we have:
$\alpha = \frac{\tau}{I} = \frac{mgl/2}{ml^2/3} = \frac{mgl}{2} \times \frac{3}{ml^2} = \frac{3g}{2l}$

(A) The angular momentum $L$ of a particle with respect to an origin $O$ is given by the formula $L = r \times p = r \times (mv),$ where $r$ is the position vector of the particle and $p$ is its linear momentum.
This can also be expressed as $L = m v d,$ where $d$ is the perpendicular distance from the origin $O$ to the line of motion of the particle.
Since the particle is moving along a straight line $AB,$ the perpendicular distance $d$ from the origin $O$ to this line remains constant at all points on the line.
Therefore,the angular momentum $L$ is constant for all points on the line $AB.$
Thus,$L_A = L_B.$

(D) The efficiency 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 in Kelvin.
Given $\eta_1 = 1/6$,so $1/6 = 1 - \frac{T_2}{T_1}$,which implies $\frac{T_2}{T_1} = 5/6$ or $T_2 = \frac{5}{6}T_1$ $...(i)$.
When the sink temperature is reduced by $62^{\circ}C$ (which is equivalent to a change of $62 \ K$),the new efficiency $\eta_2 = 2 \times \eta_1 = 2 \times (1/6) = 1/3$.
The new sink temperature is $T_2' = T_2 - 62$.
Using the efficiency formula: $1/3 = 1 - \frac{T_2 - 62}{T_1}$.
Rearranging gives $\frac{T_2 - 62}{T_1} = 1 - 1/3 = 2/3$.
Substituting $T_2 = \frac{5}{6}T_1$ from equation $(i)$:
$\frac{\frac{5}{6}T_1 - 62}{T_1} = 2/3$.
$\frac{5}{6} - \frac{62}{T_1} = 2/3$.
$\frac{62}{T_1} = \frac{5}{6} - \frac{4}{6} = 1/6$.
$T_1 = 62 \times 6 = 372 \ K$.
To convert to Celsius: $T(^{\circ}C) = 372 - 273 = 99^{\circ}C$.

(A) The kinetic energy of a particle in simple harmonic motion is given by $K = K_0 \cos^2(\omega t)$.
Since the maximum value of $\cos^2(\omega t)$ is $1$,the maximum kinetic energy is $K_{max} = K_0$.
In simple harmonic motion,the total energy $E$ remains constant and is equal to the maximum kinetic energy or the maximum potential energy.
Thus,$E = K_{max} = K_0$.
Since the total energy $E = K + U$,where $U$ is the potential energy,we have $U = E - K = K_0 - K_0 \cos^2(\omega t) = K_0 \sin^2(\omega t)$.
The maximum value of potential energy is $U_{max} = K_0$.
Therefore,the maximum potential energy is $K_0$ and the total energy is $K_0$.

(C) Let the displacement of a particle executing simple harmonic motion be $y = A \sin(\omega t)$.
The instantaneous velocity $v$ is given by the derivative of displacement with respect to time:
$v = \frac{dy}{dt} = A\omega \cos(\omega t) = A\omega \sin(\omega t + \frac{\pi}{2})$.
The instantaneous acceleration $a$ is given by the derivative of velocity with respect to time:
$a = \frac{dv}{dt} = -A\omega^2 \sin(\omega t) = A\omega^2 \sin(\omega t + \pi)$.
The phase of velocity is $(\omega t + \frac{\pi}{2})$ and the phase of acceleration is $(\omega t + \pi)$.
The phase difference between acceleration and velocity is $(\omega t + \pi) - (\omega t + \frac{\pi}{2}) = \frac{\pi}{2}$ or $0.5\pi$.

(A) The spring has a natural length $l$. When mass $m$ is placed on it,the spring compresses by $x_0$ to reach the equilibrium position $O'$.
At equilibrium,the spring force balances the weight of the mass: $k x_0 = m g$.
$x_0 = \frac{m g}{k} = \frac{2.0 \times 10}{200} = 0.10\, m = 10\, cm$.
When the mass oscillates with amplitude $A$,the maximum upward acceleration of the mass is $a_{max} = A \omega^2$,where $\omega^2 = \frac{k}{m}$.
The mass will detach from the pan when the downward acceleration of the mass exceeds the acceleration due to gravity $g$ at the highest point of its motion,or equivalently,when the normal force becomes zero.
The condition for the mass to lose contact with the pan is $A \omega^2 \ge g$.
Substituting $\omega^2 = \frac{k}{m}$,we get $A (\frac{k}{m}) \ge g$.
$A \ge \frac{m g}{k} = x_0$.
Therefore,the minimum amplitude required for the mass to detach is $A = 10\, cm$.

(B) The displacement of a particle in simple harmonic motion starting from the equilibrium position is given by $x(t) = a \sin(\omega t)$.
We need to find the time $t$ when the displacement $x(t) = a/2$.
Substituting the values,we get: $a/2 = a \sin(\omega t)$.
This simplifies to $\sin(\omega t) = 1/2$.
Since $\sin(\pi/6) = 1/2$,we have $\omega t = \pi/6$.
Substituting $\omega = 2\pi/T$,we get $(2\pi/T) \cdot t = \pi/6$.
Solving for $t$,we find $t = T/12$.

(B) Given:
Initial angular speed,$\omega_0 = 2.00\, rad/s$
Angular acceleration,$\alpha = 3.0\, rad/s^2$
Time,$t = 2\, s$
Using the kinematic equation for angular displacement:
$\theta = \omega_0 t + \frac{1}{2} \alpha t^2$
Substituting the given values:
$\theta = (2.00)(2) + \frac{1}{2}(3.0)(2)^2$
$\theta = 4 + \frac{1}{2}(3.0)(4)$
$\theta = 4 + 6 = 10\, radians$
Thus,the wheel has rotated through an angle of $10\, radians$.

(C) The orbital speed of a satellite is given by $v = \sqrt{\frac{GM}{r}}$,where $G$ is the gravitational constant,$M$ is the mass of the earth,and $r$ is the radius of the orbit.
Since both satellites are in the same orbit,$r$ is the same for both.
Therefore,the orbital speed $v$ is independent of the mass of the satellite.
Thus,$S_{1}$ and $S_{2}$ move with the same speed.
The time period is given by $T = \frac{2\pi r}{v}$,which is also independent of the mass of the satellite.
Potential energy $U = -\frac{GMm}{r}$ and kinetic energy $K = \frac{GMm}{2r}$ both depend on the mass $m$ of the satellite,so they are not equal for $S_{1}$ and $S_{2}$.

(C) In an electromagnetic wave,the oscillating electric field vector $\vec{E}$ and the magnetic field vector $\vec{B}$ are always mutually perpendicular to each other and also perpendicular to the direction of wave propagation.
Furthermore,these fields are in phase,which means they reach their maximum and minimum values at the same time and at the same spatial location.
Therefore,the correct option is $C$.

(C) From the figure,$AC = L$,$BC = L$. Since $C$ is the midpoint of $AB$ and $CD$ is a semicircle with diameter $CD$,the distance $BD = L$ (as $C, B, D$ are collinear and $CD$ is the diameter of the semicircle,$CB = BD = L$).
Potential at $C$ is given by:
$V_C = \frac{1}{4\pi\varepsilon_0} \left[ \frac{q}{AC} + \frac{-q}{BC} \right] = \frac{1}{4\pi\varepsilon_0} \left[ \frac{q}{L} - \frac{q}{L} \right] = 0$
Potential at $D$ is given by:
$AD = AB + BD = 2L + L = 3L$
$V_D = \frac{1}{4\pi\varepsilon_0} \left[ \frac{q}{AD} + \frac{-q}{BD} \right] = \frac{1}{4\pi\varepsilon_0} \left[ \frac{q}{3L} - \frac{q}{L} \right]$
$V_D = \frac{q}{4\pi\varepsilon_0 L} \left[ \frac{1}{3} - 1 \right] = \frac{q}{4\pi\varepsilon_0 L} \left( -\frac{2}{3} \right) = -\frac{q}{6\pi\varepsilon_0 L}$
Work done in moving charge $+Q$ from $C$ to $D$ is:
$W = Q(V_D - V_C) = Q \left( -\frac{q}{6\pi\varepsilon_0 L} - 0 \right) = -\frac{qQ}{6\pi\varepsilon_0 L}$

(B) The capacitors are connected in parallel to the battery of potential $V$.
The equivalent capacitance of the parallel combination is $C_{eq} = C + \frac{C}{2} = \frac{3}{2} C$.
The work done $W$ in charging a capacitor is equal to the energy stored in the capacitor,which is given by $W = \frac{1}{2} C_{eq} V^2$.
Substituting the value of $C_{eq}$:
$W = \frac{1}{2} \times \left( \frac{3}{2} C \right) V^2 = \frac{3}{4} C V^2$.

(D) Let $\phi_A, \phi_B,$ and $\phi_C$ be the electric flux linked with the surfaces $A, B,$ and $C$ respectively.
According to Gauss's Law,the total electric flux through the closed surface is $\phi_{total} = \phi_A + \phi_B + \phi_C = \frac{q}{\varepsilon_0}$.
Due to the symmetry of the cylinder,the flux linked with the two plane surfaces $A$ and $C$ is equal,so $\phi_A = \phi_C$.
Substituting this into the Gauss's Law equation,we get $2\phi_A + \phi_B = \frac{q}{\varepsilon_0}$.
Given that the flux through the curved surface $B$ is $\phi_B = \phi$,we have $2\phi_A + \phi = \frac{q}{\varepsilon_0}$.
Rearranging for $\phi_A$,we get $2\phi_A = \frac{q}{\varepsilon_0} - \phi$.
Therefore,$\phi_A = \frac{1}{2}\left(\frac{q}{\varepsilon_0} - \phi\right)$.

(B) In the given circuit,the resistors $6 \,\Omega$ and $3 \,\Omega$ are connected in parallel. Their equivalent resistance $R_p$ is given by:
$\frac{1}{R_p} = \frac{1}{6} + \frac{1}{3} = \frac{1+2}{6} = \frac{3}{6} = \frac{1}{2} \implies R_p = 2 \,\Omega$.
The equivalent circuit consists of this $R_p = 2 \,\Omega$ resistor in series with the $4 \,\Omega$ resistor and the $18 \,V$ battery.
The total resistance of the circuit is $R_{eq} = R_p + 4 \,\Omega = 2 \,\Omega + 4 \,\Omega = 6 \,\Omega$.
The total current $I$ flowing through the circuit is given by Ohm's law:
$I = \frac{V}{R_{eq}} = \frac{18 \,V}{6 \,\Omega} = 3 \,A$.
The total power dissipated in the circuit is given by $P = I^2 R_{eq}$ or $P = VI_{total}$.
Using $P = VI_{total} = 18 \,V \times 3 \,A = 54 \,W$.

(A) Let $X$ be the equivalent resistance of $S$ and $6 \,\Omega$ connected in parallel.
For a parallel combination,the equivalent resistance $X$ is given by:
$\frac{1}{X} = \frac{1}{S} + \frac{1}{6} \quad \dots(i)$
For a balanced Wheatstone bridge,the condition is $\frac{P}{Q} = \frac{R}{X}$.
Given $P = Q = R = 2 \,\Omega$,we substitute these values:
$\frac{2}{2} = \frac{2}{X} \implies 1 = \frac{2}{X} \implies X = 2 \,\Omega$.
Now,substitute $X = 2 \,\Omega$ into equation $(i)$:
$\frac{1}{2} = \frac{1}{S} + \frac{1}{6}$
$\frac{1}{S} = \frac{1}{2} - \frac{1}{6}$
$\frac{1}{S} = \frac{3 - 1}{6} = \frac{2}{6} = \frac{1}{3}$
Therefore,$S = 3 \,\Omega$.

(A) Let the shunt resistance be $S$.
Given:
Total current to be measured,$I = 750\, A$
Full-scale deflection current of the ammeter,$I_g = 100\, A$
Resistance of the ammeter,$R_G = 13\, \Omega$
When a shunt $S$ is connected in parallel with the ammeter,the potential difference across the ammeter and the shunt must be equal:
$I_g R_G = (I - I_g) S$
Substituting the given values:
$100 \times 13 = (750 - 100) \times S$
$1300 = 650 \times S$
Solving for $S$:
$S = \frac{1300}{650} = 2\, \Omega$
Thus,the value of the shunt resistance is $2\, \Omega$.

(D) The magnetic moment $\mu$ of a current loop is given by $\mu = I A$.
Here,the current $I$ produced by the revolving charge $q$ is $I = \frac{q}{T}$,where $T$ is the time period of revolution.
The time period $T$ for a particle moving in a circle of radius $R$ with speed $v$ is $T = \frac{2 \pi R}{v}$.
Substituting $T$ into the expression for $I$,we get $I = \frac{q}{2 \pi R / v} = \frac{qv}{2 \pi R}$.
The area $A$ of the circular path is $A = \pi R^2$.
Therefore,the magnetic moment is $\mu = I A = \left( \frac{qv}{2 \pi R} \right) (\pi R^2) = \frac{qvR}{2}$.

(B) The refractive index $\mu$ is defined as the ratio of the speed of light in a vacuum $(c)$ to the speed of light in the medium $(v)$: $\mu = \frac{c}{v}$.
The speed of light in the medium is given by $v = f \lambda$,where $f$ is the frequency and $\lambda$ is the wavelength.
Given: $f = 2 \times 10^{14} \ Hz$ and $\lambda = 5000 \ \mathring{A} = 5000 \times 10^{-10} \ m = 5 \times 10^{-7} \ m$.
Calculating the velocity in the medium: $v = (2 \times 10^{14}) \times (5 \times 10^{-7}) = 10 \times 10^{7} = 10^{8} \ m/s$.
Using $c = 3 \times 10^{8} \ m/s$,the refractive index is: $\mu = \frac{3 \times 10^{8}}{10^{8}} = 3$.

(D) From the figure,the light ray travels from the coin to the surface at the critical angle $C$. The base of the triangle formed is $3 \, cm$ and the height is $4 \, cm$. The hypotenuse is $\sqrt{3^2 + 4^2} = 5 \, cm$.
By definition of the critical angle,$\sin C = \frac{\text{opposite}}{\text{hypotenuse}} = \frac{3}{5}$.
Also,the refractive index of the liquid with respect to air is $\mu = \frac{1}{\sin C} = \frac{1}{3/5} = \frac{5}{3}$.
We know that $\mu = \frac{c}{v}$,where $c = 3 \times 10^8 \, m/s$ is the speed of light in air and $v$ is the speed of light in the liquid.
Therefore,$v = \frac{c}{\mu} = \frac{3 \times 10^8}{5/3} = \frac{9 \times 10^8}{5} = 1.8 \times 10^8 \, m/s$.

(D) The power $P$ of a monochromatic light beam is given by $P = N h \nu$,where $N$ is the number of photons emitted per second,$h$ is Planck's constant,and $\nu$ is the frequency.
Given:
Power $P = 2 \times 10^{-3} \text{ W}$
Frequency $\nu = 6.0 \times 10^{14} \text{ Hz}$
Planck's constant $h \approx 6.63 \times 10^{-34} \text{ J s}$
Energy of one photon $E = h \nu = (6.63 \times 10^{-34}) \times (6.0 \times 10^{14}) \text{ J} \approx 3.978 \times 10^{-19} \text{ J}$.
Number of photons emitted per second $N = P / E$:
$N = \frac{2 \times 10^{-3}}{3.978 \times 10^{-19}} \approx 0.5027 \times 10^{16} \approx 5 \times 10^{15}$ photons per second.
Thus,the correct option is $D$.

(B) Given: Output power $P_{out} = 100\,W$.
Primary voltage $V_{p} = 220\,V$.
Primary current $I_{p} = 0.5\,A$.
The input power is calculated as $P_{in} = V_{p} \times I_{p} = 220\,V \times 0.5\,A = 110\,W$.
The efficiency $\eta$ is defined as the ratio of output power to input power: $\eta = (P_{out} / P_{in}) \times 100$.
Substituting the values: $\eta = (100 / 110) \times 100 \approx 90.9\%$.
Rounding to the nearest given option, the efficiency is approximately $90\%$.

(A) Given: Number of turns in the primary coil $N_p = 50$. Number of turns in the secondary coil $N_s = 1500$. The magnetic flux linked with the primary coil is $\phi = \phi_0 + 4t$.
According to Faraday's law of electromagnetic induction,the induced voltage $(EMF)$ is given by $V = \frac{d\phi}{dt}$.
Therefore,the voltage across the primary coil is $V_p = \frac{d}{dt}(\phi_0 + 4t) = 4\, V$.
Using the transformer ratio formula,$\frac{V_s}{V_p} = \frac{N_s}{N_p}$.
Substituting the values,$V_s = V_p \times \frac{N_s}{N_p} = 4 \times \frac{1500}{50} = 4 \times 30 = 120\, V$.

(C) In a series $LCR$ circuit,the current is maximum at resonance.
At resonance,the inductive reactance equals the capacitive reactance,i.e.,$\omega L = \frac{1}{\omega C}$.
Therefore,the resonant frequency is given by $\omega = \frac{1}{\sqrt{LC}}$.
Squaring both sides,we get $\omega^2 = \frac{1}{LC}$,which implies $L = \frac{1}{\omega^2 C}$.
Given values: $\omega = 1000 \, rad/sec$ and $C = 10 \, \mu F = 10 \times 10^{-6} \, F = 10^{-5} \, F$.
Substituting these values into the formula:
$L = \frac{1}{(1000)^2 \times 10^{-5}} = \frac{1}{10^6 \times 10^{-5}} = \frac{1}{10^1} = 0.1 \, H$.
Converting to $mH$:
$0.1 \, H = 0.1 \times 1000 \, mH = 100 \, mH$.

(D) The total energy of an electron in the $n^{\text{th}}$ orbit of a hydrogen atom is given by $E_n = \frac{-13.6}{n^2} \text{ eV}$.
For the ground state,$n = 1$,so $E_1 = -13.6 \text{ eV}$.
For the first excited state,$n = 2$.
Therefore,the total energy in the first excited state is $E_2 = \frac{-13.6}{2^2} = \frac{-13.6}{4} = -3.4 \text{ eV}$.
In any orbit,the kinetic energy $K$ is equal to the negative of the total energy $E$,i.e.,$K = -E$.
Thus,the kinetic energy of the electron in the first excited state is $K = -(-3.4 \text{ eV}) = 3.4 \text{ eV}$.

(C) In beta minus decay $(\beta^{-})$, a neutron is transformed into a proton, and an electron is emitted from the nucleus along with an antineutrino.
The process is represented by the equation: $n \rightarrow p + e^{-} + \bar{\nu}$
where $n$ is the neutron, $p$ is the proton, $e^{-}$ is the emitted electron ($\beta$-particle), and $\bar{\nu}$ is the antineutrino.

(D) The nuclear radius $R$ is given by the formula $R = R_0 A^{1/3}$,where $A$ is the mass number and $R_0$ is a constant.
For the given nuclei:
For ${}_{13}^{27}Al$,$A_1 = 27$ and $R_1 = 3.6 \, fm$.
For ${}_{52}^{125}Te$,$A_2 = 125$ and we need to find $R_2$.
Taking the ratio:
$\frac{R_2}{R_1} = \left( \frac{A_2}{A_1} \right)^{1/3}$
Substituting the values:
$\frac{R_2}{3.6} = \left( \frac{125}{27} \right)^{1/3}$
$\frac{R_2}{3.6} = \frac{5}{3}$
$R_2 = \frac{5}{3} \times 3.6 = 5 \times 1.2 = 6 \, fm$.
Thus,the radius of ${}_{52}^{125}Te$ is $6 \, fm$.

(A) The binding energy $(B.E.)$ of a nucleus is the energy equivalent of the mass defect $(\Delta m)$.
The mass defect is defined as the difference between the sum of the masses of individual nucleons (protons and neutrons) and the actual mass of the nucleus.
Number of protons = $Z$
Number of neutrons = $A - Z$
Mass defect $\Delta m = [Z M_p + (A - Z) M_n - M(A, Z)]$
According to Einstein's mass-energy equivalence principle, $B.E. = \Delta m c^2$.
Therefore, $B.E. = [Z M_p + (A - Z) M_n - M(A, Z)] c^2$.

(C) Given: Decay constants $\lambda_{A} = 5\lambda$ and $\lambda_{B} = \lambda$.
At $t=0$,the initial number of nuclei are equal,i.e.,$(N_{0})_{A} = (N_{0})_{B}$.
We are given the ratio $\frac{N_{A}}{N_{B}} = (\frac{1}{e})^{2} = e^{-2}$.
According to the law of radioactive decay,$N = N_{0}e^{-\lambda t}$.
For material $A$,$N_{A} = (N_{0})_{A} e^{-5\lambda t}$.
For material $B$,$N_{B} = (N_{0})_{B} e^{-\lambda t}$.
Dividing the two equations:
$\frac{N_{A}}{N_{B}} = \frac{(N_{0})_{A} e^{-5\lambda t}}{(N_{0})_{B} e^{-\lambda t}} = e^{-(5\lambda - \lambda)t} = e^{-4\lambda t}$.
Equating the ratios:
$e^{-4\lambda t} = e^{-2}$.
Comparing the exponents:
$4\lambda t = 2$.
Solving for $t$:
$t = \frac{2}{4\lambda} = \frac{1}{2\lambda}$.

(C) The circuit consists of a $NOR$ gate followed by a $NOR$ gate acting as a $NOT$ gate (since both inputs are tied together).
Let the output of the first $NOR$ gate be $Y^{\prime} = \overline{A + B}$.
The second gate is a $NOR$ gate with both inputs as $Y^{\prime}$,so its output is $Y = \overline{Y^{\prime} + Y^{\prime}} = \overline{Y^{\prime}} = \overline{\overline{A + B}} = A + B$.
This is the Boolean expression for an $OR$ gate.
The truth table for $Y = A + B$ is:

$A$	$B$	$Y$
$0$	$0$	$0$
$0$	$1$	$1$
$1$	$0$	$1$
$1$	$1$	$1$

Comparing this with the given options,option $C$ is correct.

(B) The voltage gain $(A_v)$ of an amplifier is given by the product of the current gain $(\beta)$ and the ratio of output impedance $(R_{\text{out}})$ to input impedance $(R_{\text{in}})$:
$A_v = \beta \times \frac{R_{\text{out}}}{R_{\text{in}}}$
Given $A_v = 50$,$R_{\text{in}} = 100\; \Omega$,and $R_{\text{out}} = 200\; \Omega$,we can find the current gain $(\beta)$:
$50 = \beta \times \frac{200}{100}$
$50 = \beta \times 2$
$\beta = 25$
The power gain $(A_p)$ is defined as the product of the current gain $(\beta)$ and the voltage gain $(A_v)$:
$A_p = \beta \times A_v$
$A_p = 25 \times 50 = 1250$

(D) In the provided energy band diagram,the open circles represent holes in the valence band $(E_v)$,and the filled circles represent electrons in the conduction band $(E_c)$.
By observing the diagram,we can see that the number of holes in the valence band is significantly greater than the number of electrons in the conduction band.
In a semiconductor,if the majority charge carriers are holes,it is classified as a $p-$ type semiconductor.
Therefore,the material is a $p-$ type semiconductor.

(B) When a charged particle enters a magnetic field with a velocity perpendicular to the magnetic field,it experiences a magnetic Lorentz force $F = q(v \times B)$.
Since the force is always perpendicular to the velocity,it acts as a centripetal force.
This centripetal force causes the particle to move in a circular path with a constant speed.
Therefore,when the electric field is switched off,the electrons will move in a circular orbit.

(A) The intensity of light $I$ at a distance $r$ from a point source is given by $I = \frac{P_0}{4 \pi r^2}$, where $P_0$ is the power of the source.
Since the number of photoelectrons liberated per second is directly proportional to the number of photons incident on the surface, and the number of photons is proportional to the intensity of light, we have $N \propto I$.
Therefore, $N \propto \frac{1}{r^2}$.
Given $r_1 = 0.5\; m$ and $r_2 = 1.0\; m$, the ratio of the number of photoelectrons is:
$\frac{N_2}{N_1} = \frac{r_1^2}{r_2^2} = \left( \frac{0.5}{1.0} \right)^2 = \left( \frac{1}{2} \right)^2 = \frac{1}{4}$.
Thus, the number of photoelectrons liberated is reduced by a factor of $4$.

(B) The given charge assembly can be represented using the coordinate axes $x$ and $y$ as shown in the figure.
The charge $-2q$ is placed at the origin $O(0, 0, 0)$. One $+q$ charge is placed at $(a, 0, 0)$ and the other $+q$ charge is placed at $(0, a, 0)$.
This system can be viewed as two electric dipoles: one along the $x$-axis with dipole moment $\vec{p}_1 = q a \hat{i}$ and another along the $y$-axis with dipole moment $\vec{p}_2 = q a \hat{j}$.
The resultant dipole moment $\vec{P}_R$ is the vector sum of these two dipoles:
$\vec{P}_R = \vec{p}_1 + \vec{p}_2 = qa \hat{i} + qa \hat{j}$.
The magnitude of the resultant dipole moment is:
$P_R = \sqrt{(qa)^2 + (qa)^2} = \sqrt{2} qa$.
The direction of the resultant dipole moment is along the vector $\hat{i} + \hat{j}$,which is the line joining the origin $(0, 0, 0)$ and the point $(a, a, 0)$.

(C) When a magnetic field is perpendicular to the motion of a charged particle,the magnetic force provides the necessary centripetal force.
$F_{c} = F_{m}$
$\frac{m v^{2}}{R} = B q v$
From this,the radius of the circular path is given by:
$R = \frac{m v}{B q}$
The time period $T$ of the circular motion is the time taken to complete one full circumference:
$T = \frac{2 \pi R}{v}$
Substituting the expression for $R$:
$T = \frac{2 \pi}{v} \left( \frac{m v}{B q} \right)$
$T = \frac{2 \pi m}{B q}$
Since $T$ depends only on the mass $m$,charge $q$,and magnetic field $B$,it is independent of both the radius $R$ and the speed $v$.

(A) Nickel exhibits ferromagnetism due to a quantum physical effect known as exchange coupling,where the electron spins of one atom interact with those of neighboring atoms.
This interaction results in the alignment of the magnetic dipole moments of the atoms,overcoming the randomizing tendency of thermal collisions.
This persistent alignment is responsible for the permanent magnetism in ferromagnetic materials.
When the temperature of a ferromagnetic material is raised above a specific critical value,known as the Curie temperature $(T_C)$,the exchange coupling is no longer effective.
Consequently,the material transitions from being ferromagnetic to paramagnetic.
In the paramagnetic state,the dipoles still tend to align with an external magnetic field,but this alignment is much weaker,and thermal agitation can easily disrupt it.