TS EAMCET 2023 Physics Question Paper with Answer and Solution

(A) Let the length of each step be $d$. The initial position is $(0, 0)$.
$1$. Moves $2$ steps East: Position becomes $(2d, 0)$.
$2$. Turns right (South) and moves $4$ steps: Position becomes $(2d, -4d)$.
$3$. Turns right (West) and moves $6$ steps: Position becomes $(2d - 6d, -4d) = (-4d, -4d)$.
$4$. Turns right (North) and moves the remaining steps. Total steps taken so far $= 2 + 4 + 6 = 12$. Remaining steps $= 20 - 12 = 8$ steps North.
Final position $= (-4d, -4d + 8d) = (-4d, 4d)$.
The final position is in the second quadrant (West and North). Since the $x$ and $y$ coordinates have equal magnitude $(|-4d| = |4d|)$,the position is exactly at an angle of $45^{\circ}$ North of West,which is North-West.

(B) The velocity of the bird is given by $v(t) = t - 2$.
Since distance is the integral of the magnitude of velocity, we calculate:
$\text{Distance} = \int_{0}^{4} |v(t)| \text{ dt} = \int_{0}^{4} |t - 2| \text{ dt}$.
We split the integral at $t = 2 \text{ s}$ where the velocity changes sign:
$\text{Distance} = \int_{0}^{2} -(t - 2) \text{ dt} + \int_{2}^{4} (t - 2) \text{ dt}$.
Evaluating the first part:
$\int_{0}^{2} (2 - t) \text{ dt} = [2t - \frac{t^2}{2}]_{0}^{2} = (4 - 2) - 0 = 2 \text{ m}$.
Evaluating the second part:
$\int_{2}^{4} (t - 2) \text{ dt} = [\frac{t^2}{2} - 2t]_{2}^{4} = (8 - 8) - (2 - 4) = 0 - (-2) = 2 \text{ m}$.
Total distance $= 2 \text{ m} + 2 \text{ m} = 4 \text{ m}$.

(D) Given that the body starts from rest,so initial velocity $u = 0$.
Let the uniform acceleration be $a$.
The velocity after $n$ seconds is given by $v = u + at$.
Substituting the values,$V = 0 + a(n)$,which gives $a = \frac{V}{n}$.
The displacement $S$ in the last $t = 2 \text{ s}$ can be calculated using the formula $S = v_{final}t - \frac{1}{2}at^2$,where $v_{final} = V$.
$S = V(2) - \frac{1}{2} \left(\frac{V}{n}\right)(2)^2$.
$S = 2V - \frac{1}{2} \left(\frac{V}{n}\right)(4)$.
$S = 2V - \frac{2V}{n}$.
$S = 2V \left(1 - \frac{1}{n}\right) = \frac{2V(n-1)}{n}$.

(B) Given: Velocity of the aeroplane,$u = 60 \ km/h = 60 \times \frac{5}{18} \ m/s = \frac{50}{3} \ m/s$.
Height of the aeroplane,$h = 490 \ m$.
Acceleration due to gravity,$g = 9.8 \ m/s^2$.
The time taken by the bomb to reach the ground is given by $h = \frac{1}{2}gt^2$,so $t = \sqrt{\frac{2h}{g}}$.
$t = \sqrt{\frac{2 \times 490}{9.8}} = \sqrt{\frac{980}{9.8}} = \sqrt{100} = 10 \ s$.
The horizontal distance covered by the bomb in this time is $R = u \times t$.
$R = \left(\frac{50}{3}\right) \times 10 = \frac{500}{3} \ m$.

(C) Initial velocity $u = 50 \ m/s$,angle $\theta = 30^{\circ}$,$g = 10 \ m/s^2$.
Horizontal component of velocity $u_x = u \cos \theta = 50 \cos 30^{\circ} = 50 \times \frac{\sqrt{3}}{2} = 25\sqrt{3} \ m/s$.
Vertical component of velocity $u_y = u \sin \theta = 50 \sin 30^{\circ} = 50 \times 0.5 = 25 \ m/s$.
Total range $R = \frac{u^2 \sin 2\theta}{g} = \frac{50^2 \sin 60^{\circ}}{10} = \frac{2500 \times \sqrt{3}}{20} = 125\sqrt{3} \ m \approx 216.5 \ m$.
Horizontal distance covered in $t = 3 \ s$ is $x = u_x \times t = 25\sqrt{3} \times 3 = 75\sqrt{3} \ m \approx 129.9 \ m$.
The distance beyond the wall where the stone strikes the ground is $R - x = 125\sqrt{3} - 75\sqrt{3} = 50\sqrt{3} \ m$.
$50 \times 1.732 = 86.6 \ m$.

(D) The velocity vector $\vec{v}$ can be resolved into its rectangular components along the $X$ and $Y$ axes.
Given magnitude $v = 10 \ m/s$ and angle $\theta = 60^{\circ}$.
The components are given by:
$v_x = v \cos \theta = 10 \cos 60^{\circ} = 10 \times \frac{1}{2} = 5 \ m/s$
$v_y = v \sin \theta = 10 \sin 60^{\circ} = 10 \times \frac{\sqrt{3}}{2} = 5 \sqrt{3} \ m/s$
Thus,the velocity vector is $\vec{v} = v_x \hat{i} + v_y \hat{j} = 5 \hat{i} + 5 \sqrt{3} \hat{j} \ m/s$.

(D) The maximum horizontal range $R_{\max}$ for a projectile is given by $R_{\max} = \frac{u^2}{g}$.
Given $R_{\max} = 80 \,m$, we have $80 = \frac{u^2}{g}$, which implies $u^2 = 80g$.
When the ball is thrown vertically upward with the same velocity $u$, the maximum height $H$ reached is given by the kinematic equation $v^2 = u^2 - 2gH$.
At maximum height, the final velocity $v = 0$.
Substituting the values: $0 = u^2 - 2gH$.
$2gH = u^2$.
Since $u^2 = 80g$, we get $2gH = 80g$.
$H = \frac{80g}{2g} = 40 \,m$.

(B) The Cartesian equation of the path of a projectile is given by $y = x \tan \theta - \frac{g x^2}{2 u^2 \cos^2 \theta}$.
Given the initial velocity vector $\vec{u} = \hat{i} + 2 \hat{j} \,ms^{-1}$.
Here, the horizontal component $u_x = 1 \,ms^{-1}$ and the vertical component $u_y = 2 \,ms^{-1}$.
The horizontal displacement is $x = u_x t = 1 \cdot t$, so $t = x$.
The vertical displacement is $y = u_y t - \frac{1}{2} g t^2$.
Substituting $t = x$ into the equation for $y$:
$y = 2(x) - \frac{1}{2} (10) (x)^2$.
$y = 2x - 5x^2$.

(A) The centripetal acceleration $a_c$ of a particle moving in a circular path of radius $R$ with angular speed $\omega$ is given by $a_c = R \omega^2$.
Let the initial angular speed be $\omega_1 = \omega$ and the final angular speed be $\omega_2 = 2\omega$.
The initial centripetal acceleration is $a_1 = R \omega^2$.
The final centripetal acceleration is $a_2 = R \omega_2^2 = R (2\omega)^2 = 4 R \omega^2$.
Comparing the two,we get $a_2 = 4 a_1$.
Therefore,the centripetal acceleration becomes $4$ times the initial value.

(A) When a cylindrical vessel containing liquid is rotated with an angular velocity $\omega$ about its vertical axis, the surface of the liquid takes the shape of a paraboloid.
Let $r$ be the radius of the vessel and $h$ be the height difference between the liquid level at the center and at the sides.
For a particle of liquid at the surface at a distance $r$ from the axis, the forces acting in the rotating frame are the centrifugal force $(m r \omega^2)$ and the gravitational force $(mg)$.
The slope of the liquid surface is given by $\frac{dh}{dr} = \frac{r \omega^2}{g}$.
Integrating this from $r=0$ to $r=R$:
$\int_0^h dh = \int_0^R \frac{\omega^2}{g} r dr$
$h = \frac{\omega^2 R^2}{2g}$
Given $R = 0.05 \,m$, $\omega = 10 \,rad \,s^{-1}$, and $g = 10 \,m \,s^{-2}$:
$h = \frac{(10)^2 \times (0.05)^2}{2 \times 10}$
$h = \frac{100 \times 0.0025}{20} = \frac{0.25}{20} = 0.0125 \,m$
$h = 125 \times 10^{-4} \,m$.

(B) The car experiences two types of acceleration while moving on a circular path with increasing speed:
$1$. Radial (centripetal) acceleration: $a_r = \frac{V^2}{r}$,which is directed towards the center of the circular path.
$2$. Tangential acceleration: $a_t = a$,which is directed along the tangent to the path.
Since these two accelerations are perpendicular to each other,the resultant acceleration $a_R$ is given by the vector sum:
$a_R = \sqrt{a_r^2 + a_t^2}$
Substituting the values:
$a_R = \sqrt{(\frac{V^2}{r})^2 + a^2}$
$a_R = \sqrt{\frac{V^4}{r^2} + a^2}$

(B) The formula for centripetal force is $F = \frac{m v^2}{R}$.
Given that the mass $m$ and the centripetal force $F$ are constant for both particles.
Rearranging the formula for speed $v$,we get $v = \sqrt{\frac{F R}{m}}$.
Since $F$ and $m$ are constant,the relationship between speed and radius is $v \propto \sqrt{R}$.
Given the ratio of radii is $\frac{R_1}{R_2} = \frac{1}{2}$.
Therefore,the ratio of their speeds is $\frac{v_1}{v_2} = \sqrt{\frac{R_1}{R_2}} = \sqrt{\frac{1}{2}} = \frac{1}{\sqrt{2}}$.
Thus,the ratio of their speeds is $1 : \sqrt{2}$.

(A) The time period $T$ of a spring-block system is given by $T = 2\pi \sqrt{\frac{m}{k}}$.
If the mass $m$ is replaced by $4m$,the new time period $T'$ becomes $T' = 2\pi \sqrt{\frac{4m}{k}} = 2 \times (2\pi \sqrt{\frac{m}{k}}) = 2T$.
Since the time period doubles,the clock takes twice as long to complete one oscillation.
This means the clock runs slow. For every $1 \text{ s}$ of actual time,the clock only records $0.5 \text{ s}$,meaning it loses $0.5 \text{ s}$ for every $1 \text{ s}$ elapsed.

(B) The time period of a simple pendulum in air is given by $T = 2 \pi \sqrt{\frac{L}{g}}$.
When the pendulum oscillates in a liquid of density $\sigma$,the effective acceleration due to gravity $g'$ is given by $g' = g \left(1 - \frac{\sigma}{\rho}\right)$,where $\rho$ is the density of the bob.
The time period in water is $T' = 2 \pi \sqrt{\frac{L}{g'}}$.
Given $T' = \sqrt{2} T$,we have $\sqrt{2} = \frac{T'}{T} = \sqrt{\frac{g}{g'}} = \sqrt{\frac{g}{g(1 - \sigma/\rho)}} = \frac{1}{\sqrt{1 - \sigma/\rho}}$.
Squaring both sides,$2 = \frac{1}{1 - \sigma/\rho}$,which implies $1 - \frac{\sigma}{\rho} = \frac{1}{2}$.
Thus,$\frac{\sigma}{\rho} = \frac{1}{2}$.
Since the density of water $\sigma = 1 \text{ g/cm}^3$,the relative density of the bob is $\rho = 2$.

(B) Given: Force $F = 6.4 \,N$, extension $x = 0.1 \,m$, and time period $T = \frac{\pi}{4} \,s$.
First, calculate the spring constant $k$ using Hooke's Law: $F = kx$.
$6.4 = k \times 0.1 \implies k = 64 \,N/m$.
The formula for the time period of a spring-mass system is $T = 2\pi \sqrt{\frac{m}{k}}$.
Substituting the values: $\frac{\pi}{4} = 2\pi \sqrt{\frac{m}{64}}$.
Dividing both sides by $\pi$: $\frac{1}{4} = 2 \sqrt{\frac{m}{64}}$.
Squaring both sides: $\frac{1}{16} = 4 \times \frac{m}{64}$.
$\frac{1}{16} = \frac{m}{16}$.
Therefore, $m = 1 \,kg$.

(C) The average energy $E$ of a one-dimensional harmonic oscillator in thermal equilibrium at temperature $T$ is given by the equipartition theorem as $E = k_B T$.
However,for a quantum harmonic oscillator,the average energy is given by $E = \frac{h\nu}{e^{h\nu/k_B T} - 1} + \frac{1}{2}h\nu$.
In the classical limit where $k_B T \gg h\nu$,the average energy is $E = k_B T$.
Assuming the question refers to the classical average energy per degree of freedom for a simple harmonic oscillator (which involves both kinetic and potential energy),the total average energy is $E = k_B T$.
Given $k_B = 1.38 \times 10^{-23} \ J K^{-1}$ and $T = 300 \ K$:
$E = (1.38 \times 10^{-23}) \times 300 = 4.14 \times 10^{-21} \ J$.

(A) The kinetic energy $K$ of a particle in simple harmonic motion at a displacement $x$ is given by $K = \frac{1}{2} k(A^2 - x^2)$,where $A$ is the amplitude.
Maximum kinetic energy $K_{max} = \frac{1}{2} kA^2$.
Given that at $x = 4 \,cm$,$K = \frac{1}{3} K_{max}$.
Substituting the expressions: $\frac{1}{2} k(A^2 - 4^2) = \frac{1}{3} (\frac{1}{2} kA^2)$.
Dividing both sides by $\frac{1}{2} k$: $A^2 - 16 = \frac{A^2}{3}$.
Rearranging the terms: $A^2 - \frac{A^2}{3} = 16$.
$\frac{2A^2}{3} = 16$.
$A^2 = \frac{16 \times 3}{2} = 24$.
$A = \sqrt{24} = 2\sqrt{6} \,cm$.

(A) The frequency of the $n^{th}$ harmonic of a string is given by:
$f = \frac{n}{2\ell} \sqrt{\frac{T}{\mu}}$
where $T$ is the tension and $\mu$ is the linear mass density.
The frequency of the first harmonic $(n=1)$ of string $AB$ is:
$f_A = \frac{1}{2\ell} \sqrt{\frac{T_A}{\mu}}$
The frequency of the second harmonic $(n=2)$ of string $CD$ is:
$f_C = \frac{2}{2\ell} \sqrt{\frac{T_C}{\mu}} = \frac{1}{\ell} \sqrt{\frac{T_C}{\mu}}$
Given that $f_A = f_C$:
$\frac{1}{2\ell} \sqrt{\frac{T_A}{\mu}} = \frac{1}{\ell} \sqrt{\frac{T_C}{\mu}}$
$\frac{1}{2} \sqrt{T_A} = \sqrt{T_C}$
Squaring both sides:
$\frac{1}{4} T_A = T_C \implies T_A = 4T_C$
For the rod to be in rotational equilibrium,the net torque about point $O$ must be zero:
$T_A \cdot x = T_C \cdot (L - x)$
Substituting $T_A = 4T_C$:
$4T_C \cdot x = T_C \cdot (L - x)$
$4x = L - x$
$5x = L$
$x = \frac{L}{5}$

(C) The given displacement equation is $x = 4(\cos \pi t + \sin \pi t)$.
To find the amplitude,we express the equation in the form $x = A \sin(\omega t + \phi)$.
Multiply and divide by $\sqrt{1^2 + 1^2} = \sqrt{2}$:
$x = 4 \sqrt{2} \left( \frac{1}{\sqrt{2}} \cos \pi t + \frac{1}{\sqrt{2}} \sin \pi t \right)$.
Using the trigonometric identity $\sin(A + B) = \sin A \cos B + \cos A \sin B$,where $\sin \frac{\pi}{4} = \frac{1}{\sqrt{2}}$ and $\cos \frac{\pi}{4} = \frac{1}{\sqrt{2}}$:
$x = 4 \sqrt{2} \left( \sin \frac{\pi}{4} \cos \pi t + \cos \frac{\pi}{4} \sin \pi t \right)$.
$x = 4 \sqrt{2} \sin \left( \pi t + \frac{\pi}{4} \right)$.
Comparing this with the standard $SHM$ equation $x = A \sin(\omega t + \phi)$,the amplitude $A$ is $4 \sqrt{2}$.

(B) The general equation for simple harmonic motion is given by $x = A \cos(\omega t)$.
Comparing the given equation $x = 2 \cos(t)$ with the general equation,we find the angular frequency $\omega = 1 \text{ rad/s}$.
The relationship between the time period $T$ and angular frequency $\omega$ is given by $T = \frac{2 \pi}{\omega}$.
Substituting the value of $\omega$:
$T = \frac{2 \pi}{1} = 2 \pi \text{ s}$.
Therefore,the time period of the particle is $2 \pi \text{ seconds}$.

(C) The spinning of the 'ground chakkar' is a rotational motion where no external torque acts on the system to change its state of rotation.
According to Newton's second law for rotation,the torque $\tau$ is equal to the rate of change of angular momentum $L$,given by $\tau = \frac{dL}{dt}$.
Since there is no external torque acting on the 'ground chakkar' $(\tau = 0)$,the rate of change of angular momentum is zero,which implies $\frac{dL}{dt} = 0$.
Therefore,the angular momentum $L$ remains constant.
This phenomenon is based on the law of conservation of angular momentum.

(C) The equation of the line is $y = x + a$. Comparing this with the slope-intercept form $y = mx + c$,we get the slope $m_{slope} = 1$. Since $\tan \theta = m_{slope} = 1$,the angle $\theta = 45^{\circ}$.
The velocity vector $\vec{v}$ makes an angle of $45^{\circ}$ with the positive $x$-axis. Thus,$\vec{v} = v \cos 45^{\circ} \hat{i} + v \sin 45^{\circ} \hat{j} = \frac{v}{\sqrt{2}} (\hat{i} + \hat{j})$.
The angular momentum $\vec{L}$ about the origin is given by $\vec{L} = \vec{r} \times \vec{p} = m (\vec{r} \times \vec{v})$.
The perpendicular distance $d$ from the origin $(0, 0)$ to the line $x - y + a = 0$ is given by $d = \frac{|(1)(0) - (1)(0) + a|}{\sqrt{1^2 + (-1)^2}} = \frac{a}{\sqrt{2}}$.
The magnitude of angular momentum is $L = mvd = m v \left( \frac{a}{\sqrt{2}} \right) = \frac{mva}{\sqrt{2}}$.

(D) The angular momentum of a particle in circular motion is given by $L = I\omega$,where $I$ is the moment of inertia and $\omega$ is the angular velocity.
The kinetic energy is given by $K = \frac{1}{2}I\omega^2$.
We can express kinetic energy in terms of angular momentum as $K = \frac{1}{2}L\omega$.
From this,the angular momentum is $L = \frac{2K}{\omega}$.
Since the frequency $f$ is doubled,the angular velocity $\omega = 2\pi f$ is also doubled,so $\omega_2 = 2\omega_1$.
The kinetic energy is halved,so $K_2 = \frac{K_1}{2}$.
Now,calculating the new angular momentum $L_2$:
$L_2 = \frac{2K_2}{\omega_2} = \frac{2(K_1/2)}{2\omega_1} = \frac{K_1}{2\omega_1} = \frac{1}{4} \left( \frac{2K_1}{\omega_1} \right) = \frac{L_1}{4}$.
Therefore,the new angular momentum is $\frac{L}{4}$.

(A) The moment of inertia $I$ of a solid sphere of mass $M$ and radius $R$ about its diameter is given by the formula $I = \frac{2}{5} MR^2$.
Since the mass $M$ is the same for both spheres,the moment of inertia is directly proportional to the square of the radius,i.e.,$I \propto R^2$.
Given the ratio of the radii is $\frac{R_1}{R_2} = \frac{2}{3}$.
Therefore,the ratio of the moments of inertia is $\frac{I_1}{I_2} = \left(\frac{R_1}{R_2}\right)^2 = \left(\frac{2}{3}\right)^2 = \frac{4}{9}$.
Thus,the ratio is $4:9$.

(D) For rolling without slipping,the velocity is $v = r \omega$.
Total kinetic energy $E = K_{rot} + K_{trans} = \frac{1}{2} I \omega^2 + \frac{1}{2} m v^2$.
Substituting $v = r \omega$,we get $E = \frac{1}{2} I \omega^2 + \frac{1}{2} m r^2 \omega^2 = \frac{1}{2} (I + m r^2) \omega^2$.
Rotational kinetic energy is $K_{rot} = \frac{1}{2} I \omega^2$.
Given that $K_{rot} = 50 \% \text{ of } E$,so $K_{rot} = \frac{1}{2} E$.
Substituting the expressions: $\frac{1}{2} I \omega^2 = \frac{1}{2} [\frac{1}{2} (I + m r^2) \omega^2]$.
$I = \frac{1}{2} I + \frac{1}{2} m r^2$.
$\frac{1}{2} I = \frac{1}{2} m r^2$,which implies $I = m r^2$.
The moment of inertia $I = m r^2$ corresponds to a thin circular ring.

(C) The angular momentum of the Earth remains conserved because no external torque acts on it.
$L = I \omega$
Since $I = \frac{2}{5} MR^2$ and $\omega = \frac{2 \pi}{T}$,we have:
$I_1 \omega_1 = I_2 \omega_2$
$\frac{2}{5} MR^2 \times \frac{2 \pi}{T_1} = \frac{2}{5} M(xR)^2 \times \frac{2 \pi}{T_2}$
Canceling common terms:
$\frac{R^2}{T_1} = \frac{x^2 R^2}{T_2}$
$T_2 = T_1 x^2$
Given the present period of rotation $T_1 = 24 \text{ hours}$,the new period is:
$T_2 = 24 x^2 \text{ hours}$.

(A) The relationship between torque $\tau$ and angular momentum $L$ is given by the equation $\tau = \frac{dL}{dt}$.
Given the initial angular momentum $L_i = A_0$ and the final angular momentum $L_f = 4 \,A_0$.
The time interval $\Delta t = 4 \,s$.
The change in angular momentum is $\Delta L = L_f - L_i = 4 \,A_0 - A_0 = 3 \,A_0$.
Therefore, the magnitude of the torque is $\tau = \frac{\Delta L}{\Delta t} = \frac{3 \,A_0}{4}$.

(C) Let $c$ be the specific heat capacity of the material. The heat supplied is given by the formula $Q = mc\Delta T$.
For the object of mass $2m$,the initial temperature is $T$ and the final temperature is $2T$. The change in temperature is $\Delta T_1 = 2T - T = T$.
Thus,the heat supplied is $Q = (2m)c(T) = 2mcT$.
For the object of mass $m$,the initial temperature is $2T$. Let the final temperature be $T_f$. The change in temperature is $\Delta T_2 = T_f - 2T$.
Since the same heat $Q$ is supplied,we have $Q = m c (T_f - 2T)$.
Equating the two expressions for $Q$: $2mcT = mc(T_f - 2T)$.
Dividing both sides by $mc$: $2T = T_f - 2T$.
Solving for $T_f$: $T_f = 2T + 2T = 4T$.

(D) According to the principle of calorimetry, the heat lost by the steam equals the heat gained by the water.
Heat gained by water: $Q_{gain} = m_w c_w \Delta T_w = 100 \,g \times 1 \,cal/g^{\circ}C \times (90^{\circ}C - 24^{\circ}C) = 100 \times 66 = 6600 \,cal$.
Heat lost by steam: $Q_{lost} = m_s L_v + m_s c_w \Delta T_s = m_s(540) + m_s(1)(100^{\circ}C - 90^{\circ}C) = 540 m_s + 10 m_s = 550 m_s$.
Equating the two: $550 m_s = 6600$.
$m_s = \frac{6600}{550} = 12 \,g$.

(A) Mass of ice,$m = 10 \,kg = 10^4 \,g$.
Specific heat capacity of ice,$c = 0.5 \,cal \,g^{-1} \,^{\circ}C^{-1} = 2093.4 \,J \,kg^{-1} \,K^{-1}$.
Latent heat of fusion of ice,$L = 80 \,cal \,g^{-1} = 334944 \,J \,kg^{-1}$.
The total heat energy $Q$ required is the sum of heat to raise the temperature of ice to $0^{\circ}C$ and the heat to melt the ice at $0^{\circ}C$.
$Q = m c \Delta T + m L$
$Q = 10 \,kg \times 2093.4 \,J \,kg^{-1} \,K^{-1} \times (0 - (-10)) \,K + 10 \,kg \times 334944 \,J \,kg^{-1}$
$Q = 10 \times 2093.4 \times 10 + 3349440$
$Q = 209340 + 3349440$
$Q = 3558780 \,J \approx 357 \times 10^4 \,J$.

(C) The heat energy required to raise the temperature of a substance is given by $\Delta Q = m c \Delta T$.
Since mass $m = d V$ (where $d$ is density and $V$ is volume),we have $\Delta Q = d V c \Delta T$.
The heat energy required per unit volume is $\frac{\Delta Q}{V} = d c \Delta T$.
Given the ratio of densities $\frac{d_1}{d_2} = \frac{5}{6}$ and the ratio of specific heat capacities $\frac{c_1}{c_2} = \frac{3}{5}$.
For the same temperature rise,$\Delta T_1 = \Delta T_2 = \Delta T$.
The ratio of heat energies required per unit volume is $\frac{(\Delta Q/V)_1}{(\Delta Q/V)_2} = \frac{d_1 c_1 \Delta T}{d_2 c_2 \Delta T} = \frac{d_1}{d_2} \times \frac{c_1}{c_2}$.
Substituting the given values: $\frac{5}{6} \times \frac{3}{5} = \frac{15}{30} = \frac{1}{2}$.

(A) According to the principle of calorimetry, Heat gained by water $=$ Heat lost by steam.
Let $m$ be the mass of steam added.
Heat gained by water: $Q_1 = m_w s_w \Delta T_w = 150 \times 1 \times (40 - 20) = 150 \times 20 = 3000 \,cal$.
Heat lost by steam: $Q_2 = m L_v + m s_w \Delta T_s = m \times 540 + m \times 1 \times (100 - 40) = m(540 + 60) = 600m \,cal$.
Equating the two: $3000 = 600m$.
$m = \frac{3000}{600} = 5 \,g$.
The total mass of water at $40^{\circ} C$ is the initial mass of water plus the mass of condensed steam: $150 \,g + 5 \,g = 155 \,g$.

(D) Change in length of the first rod is given by $\Delta \ell_1 = L(2 \alpha) \Delta T = 2 \,L \alpha \Delta T$.
Change in length of the second rod is given by $\Delta \ell_2 = (2 \,L)(\alpha)(\Delta T) = 2 \,L \alpha \Delta T$.
Total change in length of the composite rod is $\Delta l = \Delta l_1 + \Delta l_2 = 2 \,L \alpha \Delta T + 2 \,L \alpha \Delta T = 4 \,L \alpha \Delta T$.
For the composite rod of length $3 \,L$,the change in length is $\Delta l = (3 \,L) \alpha_C \Delta T$.
Equating the two expressions for $\Delta l$: $4 \,L \alpha \Delta T = 3 \,L \alpha_C \Delta T$.
Solving for $\alpha_C$,we get $\alpha_C = \frac{4 \alpha}{3}$.

(D) The thermal stress $\sigma$ developed in a wire when its expansion is prevented is given by the formula: $\sigma = Y \alpha \Delta T$,where $Y$ is Young's modulus,$\alpha$ is the coefficient of linear expansion,and $\Delta T$ is the change in temperature.
Given that $Y$,$L$,and $\Delta T$ are the same for both wires $A$ and $B$,the thermal stress is directly proportional to the coefficient of linear expansion: $\sigma \propto \alpha$.
Given $\alpha_A = \frac{3}{2} \alpha_B$.
Therefore,the ratio of thermal stresses is $\frac{\sigma_A}{\sigma_B} = \frac{\alpha_A}{\alpha_B} = \frac{\frac{3}{2} \alpha_B}{\alpha_B} = \frac{3}{2}$.
Thus,the ratio is $3: 2$.

(D) For an ideal gas,the equation of state is $PV = nRT$.
At constant pressure $P$,differentiating with respect to temperature $T$,we get $P \frac{dV}{dT} = nR$,which implies $\frac{dV}{dT} = \frac{nR}{P}$.
The coefficient of volume expansion $\gamma$ is defined as $\gamma = \frac{1}{V} \frac{dV}{dT}$.
Substituting $\frac{dV}{dT} = \frac{nR}{P}$,we get $\gamma = \frac{1}{V} \left( \frac{nR}{P} \right) = \frac{nR}{PV}$.
Since $PV = nRT$,we have $\gamma = \frac{nR}{nRT} = \frac{1}{T}$.
Given temperature $T = 27^{\circ} C = 27 + 273 = 300 \ K$.
Therefore,$\gamma = \frac{1}{300} \ K^{-1} \approx 0.00333 \ K^{-1} = 33 \times 10^{-4} \ K^{-1}$.

(D) The relationship between the Fahrenheit scale $(F)$ and the Celsius scale $(C)$ is given by the formula: $\frac{F-32}{9} = \frac{C}{5}$.
Given that the reading in Fahrenheit is twice the reading in Celsius,we have $F = 2C$,which implies $C = \frac{F}{2}$.
Substituting $C = \frac{F}{2}$ into the temperature conversion formula:
$\frac{F-32}{9} = \frac{F/2}{5}$
$\frac{F-32}{9} = \frac{F}{10}$
Cross-multiplying gives: $10(F - 32) = 9F$.
$10F - 320 = 9F$.
$10F - 9F = 320$.
$F = 320^{\circ} F$.

(C) The formula for the coefficient of linear expansion $\alpha$ is given by:
$\alpha = \frac{\Delta L}{L \Delta T}$
Given:
Initial length $L = 30 \ cm$
Change in length $\Delta L = 0.027 \ cm$
Initial temperature $T_1 = 30^{\circ} C$
Final temperature $T_2 = 105^{\circ} C$
Change in temperature $\Delta T = T_2 - T_1 = 105^{\circ} C - 30^{\circ} C = 75^{\circ} C$
Substituting the values into the formula:
$\alpha = \frac{0.027}{30 \times 75}$
$\alpha = \frac{0.027}{2250}$
$\alpha = 0.000012 /{ }^{\circ} C$
$\alpha = 12 \times 10^{-6} /{ }^{\circ} C$

(C) The relationship between the Fahrenheit $(F)$ and Kelvin $(K)$ scales is given by the formula:
$\frac{F - 32}{9} = \frac{K - 273.15}{5}$
Let the temperature where both scales have the same numerical value be $X$.
Substituting $F = X$ and $K = X$ into the equation:
$\frac{X - 32}{9} = \frac{X - 273.15}{5}$
Cross-multiplying gives:
$5(X - 32) = 9(X - 273.15)$
$5X - 160 = 9X - 2458.35$
Rearranging the terms to solve for $X$:
$9X - 5X = 2458.35 - 160$
$4X = 2298.35$
$X = 574.5875 \approx 574.6$
Thus,the temperature is $574.6^{\circ} F$.

(B) The diameter of the iron ring $d_i = 5 \ m$ at $T_1 = 27^{\circ} C$.
The diameter of the wooden wheel $d_w = 5.012 \ m$.
We need the iron ring to expand to a diameter of $5.012 \ m$.
Using the formula for linear expansion: $d_w = d_i(1 + \alpha \Delta T)$.
$\Delta T = \frac{d_w - d_i}{d_i \alpha} = \frac{5.012 - 5}{5 \times 1.2 \times 10^{-5}} = \frac{0.012}{6 \times 10^{-5}} = \frac{1.2 \times 10^{-2}}{6 \times 10^{-5}} = 0.2 \times 10^3 = 200^{\circ} C$.
The final temperature $T_2 = T_1 + \Delta T = 27^{\circ} C + 200^{\circ} C = 227^{\circ} C$.

(B) The relationship between two temperature scales is given by the formula:
$\frac{\text{Reading on scale} - \text{LFP}}{\text{UFP} - \text{LFP}} = \text{constant}$
For the Celsius scale,the Lower Fixed Point $(LFP)$ is $0^{\circ} C$ and the Upper Fixed Point $(UFP)$ is $100^{\circ} C$.
For the new scale $X$,the $LFP$ is $20^{\circ} X$ and the $UFP$ is $110^{\circ} X$.
Let the temperature on the new scale be $z^{\circ} X$ corresponding to $40^{\circ} C$.
$\frac{z - 20}{110 - 20} = \frac{40 - 0}{100 - 0}$
$\frac{z - 20}{90} = \frac{40}{100}$
$\frac{z - 20}{90} = 0.4$
$z - 20 = 0.4 \times 90$
$z - 20 = 36$
$z = 36 + 20 = 56^{\circ} X$
Therefore,$40^{\circ} C$ corresponds to $56^{\circ} X$.

(B) Given: $n = 2 \text{ moles}$,$\gamma = 4/3$,$T_1 = 327^{\circ} C = 600 \text{ K}$.
Step $1$: Adiabatic expansion.
$T_1 V_1^{\gamma-1} = T_2 V_2^{\gamma-1}$.
Given $V_2 = 8 V_1$,so $T_2 = T_1 (V_1/V_2)^{\gamma-1} = 600 \times (1/8)^{(4/3 - 1)} = 600 \times (1/8)^{1/3} = 600 \times (1/2) = 300 \text{ K}$.
Work done in adiabatic process: $W_1 = \frac{nR(T_1 - T_2)}{\gamma - 1} = \frac{2R(600 - 300)}{4/3 - 1} = \frac{2R(300)}{1/3} = 1800 R$.
Step $2$: Isochoric process.
In an isochoric process,volume is constant,so work done $W_2 = 0$.
Total work done $W = W_1 + W_2 = 1800 R + 0 = 1800 R$.

(C) For an isobaric process, the pressure remains constant $(P = \text{constant})$.
From the first law of thermodynamics, $\Delta Q = \Delta U + \Delta W$.
For a diatomic gas, the molar heat capacities are $C_p = \frac{7}{2}R$ and $C_v = \frac{5}{2}R$.
The heat supplied is $\Delta Q = n C_p \Delta T$.
The work done is $\Delta W = n R \Delta T$.
The fraction of heat converted into work is $\frac{\Delta W}{\Delta Q} = \frac{n R \Delta T}{n C_p \Delta T} = \frac{R}{C_p} = \frac{R}{\frac{7}{2}R} = \frac{2}{7}$.
Converting this to a percentage: $\frac{2}{7} \times 100 \approx 28.6 \%$.
Therefore, the percentage of heat converted into work is $28.6 \%$.

(B) For an isothermal process,the ideal gas law gives $PV = nRT$. Since $T$ is constant,$P = \frac{nRT}{V}$.
For container $A$,the change in pressure is $\Delta P_A = P_{initial} - P_{final} = \frac{n_A RT}{V} - \frac{n_A RT}{2V} = \frac{n_A RT}{2V}$.
Given $\Delta P_A = 2\Delta P$,so $2\Delta P = \frac{n_A RT}{2V}$ ... $(1)$.
For container $B$,the change in pressure is $\Delta P_B = P_{initial} - P_{final} = \frac{n_B RT}{V} - \frac{n_B RT}{2V} = \frac{n_B RT}{2V}$.
Given $\Delta P_B = 3\Delta P$,so $3\Delta P = \frac{n_B RT}{2V}$ ... $(2)$.
Dividing equation $(1)$ by $(2)$:
$\frac{2\Delta P}{3\Delta P} = \frac{n_A RT / 2V}{n_B RT / 2V} = \frac{n_A}{n_B}$.
Thus,$\frac{n_A}{n_B} = \frac{2}{3}$.
Since $n = \frac{m}{M}$ (where $M$ is molar mass),$\frac{m_A / M}{m_B / M} = \frac{2}{3}$,which implies $\frac{m_A}{m_B} = \frac{2}{3}$.
Therefore,$3m_A = 2m_B$.

(A) For a monatomic gas,the molar heat capacity at constant volume is $C_v = \frac{3}{2}R$ and at constant pressure is $C_p = \frac{5}{2}R$.
When heat $Q$ is supplied at constant pressure,the total heat supplied is $Q = n C_p \Delta T$.
The change in internal energy is $\Delta U = n C_v \Delta T$.
The fraction of heat used for increasing internal energy is $\frac{\Delta U}{Q} = \frac{n C_v \Delta T}{n C_p \Delta T} = \frac{C_v}{C_p}$.
Given $\gamma = \frac{C_p}{C_v} = \frac{5}{3}$,therefore $\frac{C_v}{C_p} = \frac{3}{5}$.
Percentage of heat energy used = $\frac{3}{5} \times 100 = 60\%$.

(B) According to the first law of thermodynamics,the change in internal energy $dU$ is given by $dU = dQ - dW$,where $dQ$ is the heat supplied to the system and $dW$ is the work done by the system.
Given that the work done by the system is equal to the decrease in its internal energy,we have $dW = -dU$.
Substituting this into the first law equation:
$dU = dQ - (-dU)$
$dU = dQ + dU$
$dQ = 0$
Since the heat exchange $dQ$ is zero,the process is an adiabatic process.

(C) The dimension of the spring constant $k$ is given by $k = \frac{F}{x}$,so its unit is $\frac{N}{m} = \frac{kg \cdot m/s^2}{m} = kg \cdot s^{-2}$. The dimensional formula is $[M L^0 T^{-2}]$.
Surface energy is defined as energy per unit area: $\text{Surface energy} = \frac{\text{Energy}}{\text{Area}} = \frac{J}{m^2} = \frac{N \cdot m}{m^2} = \frac{N}{m}$.
Thus,the unit of surface energy is also $\frac{N}{m}$,which corresponds to the dimensional formula $[M L^0 T^{-2}]$.
Since both have the same dimensional formula,option $C$ is correct.

(A) Given: $v_2 = \frac{n}{m^2} v_1$ and $a_2 = \frac{a_1}{mn}$.
We know that $a = \frac{v}{t}$,so $\frac{a_2}{a_1} = \frac{v_2}{v_1} \times \frac{t_1}{t_2}$.
Substituting the given ratios: $\frac{1}{mn} = \frac{n}{m^2} \times \frac{t_1}{t_2}$.
Rearranging for $t_2$: $t_2 = \frac{n}{m^2} \times mn \times t_1 = \frac{n^2}{m} t_1$.
Now,we know $v = \frac{S}{t}$,so $\frac{v_2}{v_1} = \frac{S_2}{S_1} \times \frac{t_1}{t_2}$.
Substituting the known values: $\frac{n}{m^2} = \frac{S_2}{S_1} \times \frac{t_1}{(n^2/m)t_1} = \frac{S_2}{S_1} \times \frac{m}{n^2}$.
Solving for $S_2$: $S_2 = S_1 \times \frac{n}{m^2} \times \frac{n^2}{m} = S_1 \times \frac{n^3}{m^3} = \left(\frac{n}{m}\right)^3 S_1$.

(D) Efficiency $\eta$ is a dimensionless quantity. The argument of the logarithmic function must also be dimensionless,so $\frac{\beta x}{kT} = 1$ (dimensionless).
Since $\eta$ is dimensionless,$[\eta] = [M^0 L^0 T^0]$.
From $\frac{\beta x}{kT} = 1$,we get $\beta = \frac{kT}{x}$.
Given $k = [M L^2 T^{-2} K^{-1}]$ and $T = [K]$,we have $\beta = \frac{[M L^2 T^{-2} K^{-1}][K]}{[L]} = [M L T^{-2}]$.
This is the dimension of force,so option $A$ is correct.
Since $\eta = \frac{\alpha \beta}{\sin \theta} \cdot \text{constant}$,and $\eta$ and $\sin \theta$ are dimensionless,$[\alpha \beta] = [1]$,which implies $[\alpha] = [\beta^{-1}] = [M^{-1} L^{-1} T^2]$.
Option $D$ states $[\alpha] = [\beta]$,which is incorrect.
For option $B$,$[\alpha^{-1} x] = [\beta x] = [M L T^{-2}][L] = [M L^2 T^{-2}]$,which is the dimension of energy. Thus,$B$ is correct.
For option $C$,$[\eta^{-1} \sin \theta] = [1] = [\alpha \beta]$,which is correct.

(B) The relative strengths of the fundamental forces in nature are as follows:
$1$. Strong Nuclear Force: $1$
$2$. Electromagnetic Force: $10^{-2}$
$3$. Weak Nuclear Force: $10^{-13}$
$4$. Gravitational Force: $10^{-39}$
Given that $F_1$ is the gravitational force,$F_2$ is the weak nuclear force,and $F_3$ is the electromagnetic force:
$F_1 = 10^{-39}$
$F_2 = 10^{-13}$
$F_3 = 10^{-2}$
Comparing these values,we get $10^{-39} < 10^{-13} < 10^{-2}$,which implies $F_1 < F_2 < F_3$.

(D) The relative strength of the gravitational force is $F_1 \approx 10^{-39}$.
The relative strength of the weak nuclear force is $F_2 \approx 10^{-13}$.
To find the ratio $\frac{F_2}{F_1}$,we calculate:
$\frac{F_2}{F_1} = \frac{10^{-13}}{10^{-39}} = 10^{-13 - (-39)} = 10^{26}$.
Therefore,the ratio is $10^{26}$.

(B) The correct matching is as follows:
$A$) $X$-rays are used for the study of crystal structure because their wavelength is comparable to the interatomic spacing in crystals. Thus,$A \rightarrow R$.
$B$) $UV$-rays are used in forensic laboratories for detecting finger prints and analyzing documents. Thus,$B \rightarrow Q$.
$C$) Radio waves are widely used in $TV$ and radio communication systems. Thus,$C \rightarrow S$.
$D$) $IR$-rays (Infrared rays) are used in remote switches for electronic devices like $TV$ remotes. Thus,$D \rightarrow P$.
Therefore,the correct sequence is $A \rightarrow R, B \rightarrow Q, C \rightarrow S, D \rightarrow P$.

(A) Given frequency, $f = 3 \text{ GHz} = 3 \times 10^9 \text{ Hz}$.
Speed of light, $c = 3 \times 10^8 \text{ m/s}$.
The relationship between wavelength $\lambda$, frequency $f$, and speed of light $c$ is given by $\lambda = \frac{c}{f}$.
Substituting the values: $\lambda = \frac{3 \times 10^8}{3 \times 10^9} = 10^{-1} \text{ m} = 0.1 \text{ m}$.

(D) is incorrect because electromagnetic $(EM)$ waves do not require any medium to travel; thus,they can travel in a vacuum.
$B$ is incorrect because $EM$ waves are transverse waves,not longitudinal.
$C$ is incorrect because $EM$ waves are produced by an accelerating charge,not by charges moving with uniform velocity.
$D$ is correct because $EM$ waves carry both energy and momentum as they propagate through space.

(C) The speed of an electromagnetic wave in a medium is given by $v = \frac{c}{\sqrt{\mu_r \epsilon_r}}$,where $c$ is the speed of light in vacuum,$\mu_r$ is the relative permeability,and $\epsilon_r$ is the relative permittivity.
Given: $v = 1.5 \times 10^8 \ m/s$,$c = 3 \times 10^8 \ m/s$,and $\epsilon_r = 2$.
Substituting the values: $1.5 \times 10^8 = \frac{3 \times 10^8}{\sqrt{\mu_r \times 2}}$.
Squaring both sides: $(1.5)^2 = \frac{3^2}{2\mu_r} \Rightarrow 2.25 = \frac{9}{2\mu_r}$.
Solving for $\mu_r$: $2\mu_r = \frac{9}{2.25} = 4 \Rightarrow \mu_r = 2$.
The magnetic susceptibility $\chi_m$ is related to relative permeability by $\mu_r = 1 + \chi_m$.
Therefore,$\chi_m = \mu_r - 1 = 2 - 1 = 1$.

(B) The electric field $E$ due to a charge $Q$ at distance $r$ is given by $E = \frac{k|Q|}{r^2}$. Since all charges are at the same distance $r$ from the centre,the electric field due to a charge $nq$ is proportional to $n$.
Let the electric field due to charge $-q$ at position $1$ be $\vec{E}_1$ directed towards the charge (since it is negative).
The net electric field $\vec{E}_{net}$ is the vector sum of all fields: $\vec{E}_{net} = \sum_{n=1}^{12} \vec{E}_n$.
Note that the charge at position $n$ is $-nq$. The field $\vec{E}_n$ is directed from the centre towards the position $n$.
We can pair opposite charges: $(n)$ and $(n+6)$. The net field from these two is $\vec{E}_{net, n} = \vec{E}_n + \vec{E}_{n+6} = \frac{k}{r^2} [(-n\hat{r}_n) + (-(n+6)\hat{r}_{n+6})]$.
Since $\hat{r}_{n+6} = -\hat{r}_n$,we have $\vec{E}_{net, n} = \frac{k}{r^2} [-n\hat{r}_n + (n+6)\hat{r}_n] = \frac{6k}{r^2} \hat{r}_n$.
This means the net field from each pair $(n, n+6)$ is directed towards the position $n+6$.
There are $6$ such pairs: $(1,7), (2,8), (3,9), (4,10), (5,11), (6,12)$.
The net field is $\vec{E}_{net} = \frac{6k}{r^2} [\hat{r}_7 + \hat{r}_8 + \hat{r}_9 + \hat{r}_{10} + \hat{r}_{11} + \hat{r}_{12}]$.
These unit vectors point towards $7, 8, 9, 10, 11, 12$ on the clock. The resultant of these vectors points towards the direction between $9$ and $10$,which corresponds to $9:30$.

(D) The electric field $E$ near the surface of a conducting sphere with surface charge density $\sigma$ is given by the formula $E = \frac{\sigma}{\epsilon_0}$.
Since both spheres are charged to the same surface charge density,we have $\sigma_1 = \sigma_2 = \sigma$.
Therefore,the electric field near the surface of the first sphere is $E_1 = \frac{\sigma}{\epsilon_0}$.
Similarly,the electric field near the surface of the second sphere is $E_2 = \frac{\sigma}{\epsilon_0}$.
Comparing the two,we find $E_1 = E_2$.
Thus,the ratio of the electric fields is $\frac{E_1}{E_2} = \frac{1}{1}$.

(A) The electric field is given by $\overrightarrow{E} = 24 \hat{i} + 30 \hat{j} + 28 \hat{k} \text{ NC}^{-1}$.
Since the area lies on the $yz$ plane,its area vector $\vec{A}$ is directed along the $x$-axis.
Thus,$\vec{A} = 20 \hat{i} \text{ m}^2$.
The electric flux $\Phi_E$ is defined as the dot product of the electric field and the area vector: $\Phi_E = \overrightarrow{E} \cdot \vec{A}$.
Substituting the values: $\Phi_E = (24 \hat{i} + 30 \hat{j} + 28 \hat{k}) \cdot (20 \hat{i})$.
Using the dot product property $\hat{i} \cdot \hat{i} = 1$ and $\hat{i} \cdot \hat{j} = \hat{i} \cdot \hat{k} = 0$,we get:
$\Phi_E = 24 \times 20 = 480 \text{ Nm}^2 \text{C}^{-1}$.

(D) The total charge $q$ on the hollow spherical shell is given by the product of surface charge density $\sigma$ and the surface area of the sphere $(4 \pi r^2)$: $q = \sigma \times 4 \pi r^2$.
According to Gauss's law,the total electric flux $\phi_{total}$ through the closed surface of the cube is $\frac{q}{\varepsilon_0}$.
Since the cube is a symmetric closed surface and the charge is at its center,the flux is distributed equally through all $6$ faces of the cube.
Therefore,the electric flux $\phi'$ through one face of the cube is $\phi' = \frac{1}{6} \times \frac{q}{\varepsilon_0}$.
Substituting the value of $q$: $\phi' = \frac{1}{6} \times \frac{\sigma \times 4 \pi r^2}{\varepsilon_0} = \frac{2 \pi r^2 \sigma}{3 \varepsilon_0}$.

(C) The electric field $\vec{E}$ is related to the electric potential $V$ by the relation $\vec{E} = -\nabla V = -\left( \frac{\partial V}{\partial x} \hat{i} + \frac{\partial V}{\partial y} \hat{j} \right)$.
Given $V = \frac{1}{2}y^2 - 2x$.
Calculating the partial derivatives:
$\frac{\partial V}{\partial x} = \frac{\partial}{\partial x} (\frac{1}{2}y^2 - 2x) = -2$.
$\frac{\partial V}{\partial y} = \frac{\partial}{\partial y} (\frac{1}{2}y^2 - 2x) = y$.
Thus,$\vec{E} = -(-2 \hat{i} + y \hat{j}) = 2 \hat{i} - y \hat{j}$.
At the point $(x = 1 \text{ m}, y = 1 \text{ m})$:
$\vec{E} = 2 \hat{i} - (1) \hat{j} = 2 \hat{i} - \hat{j} \text{ V m}^{-1}$.

(C) Given: Electric field $E = 500 \ NC^{-1}$,Electric potential $V = 30 \ V$.
The relationship between electric field and potential for a point charge is $E = \frac{V}{r}$.
Therefore,the distance $r$ is $r = \frac{V}{E} = \frac{30}{500} = 0.06 \ m$.
The formula for electric potential is $V = \frac{kq}{r}$,where $k = 9 \times 10^9 \ Nm^2C^{-2}$.
Rearranging for charge $q$: $q = \frac{V \cdot r}{k}$.
Substituting the values: $q = \frac{30 \times 0.06}{9 \times 10^9} = \frac{1.8}{9 \times 10^9} = 0.2 \times 10^{-9} \ C$.
Thus,$q = 2 \times 10^{-10} \ C$.

(D) Given: $Q_1 = +2 \times 10^{-6} \ C$, $Q_2 = -4 \times 10^{-6} \ C$, distance $d = 3 \ m$. Let $P$ be at a distance $r_1$ from $Q_1$ and $r_2$ from $Q_2$. Since $P$ is between them, $r_1 + r_2 = 3 \ m$. Let $r_2 = x$, then $r_1 = 3 - x$.
Electric potential $V = \frac{1}{4 \pi \epsilon_0} \left( \frac{Q_1}{r_1} + \frac{Q_2}{r_2} \right) = 0$.
$\frac{2 \times 10^{-6}}{3 - x} + \frac{-4 \times 10^{-6}}{x} = 0 \Rightarrow \frac{2}{3 - x} = \frac{4}{x} \Rightarrow x = 6 - 2x \Rightarrow 3x = 6 \Rightarrow x = 2 \ m$.
So, $r_2 = 2 \ m$ and $r_1 = 1 \ m$.
The electric field $E$ at $P$ due to $Q_1$ is $E_1 = \frac{k |Q_1|}{r_1^2} = \frac{9 \times 10^9 \times 2 \times 10^{-6}}{1^2} = 18000 \ N/C$ (directed towards $Q_2$).
The electric field $E$ at $P$ due to $Q_2$ is $E_2 = \frac{k |Q_2|}{r_2^2} = \frac{9 \times 10^9 \times 4 \times 10^{-6}}{2^2} = 9000 \ N/C$ (directed towards $Q_2$).
Since both fields point in the same direction (towards $Q_2$), the net electric field $E = E_1 + E_2 = 18000 + 9000 = 27000 \ N/C$.

(A) For a hollow conducting sphere,the electric charge resides entirely on its outer surface.
Inside the sphere,the electric field is zero $(E = 0)$.
Since the electric field is the negative gradient of the potential $(E = -\frac{dV}{dr})$,if $E = 0$,then the potential $V$ must be constant throughout the interior of the sphere.
Therefore,the potential at any point inside the sphere is equal to the potential at its surface.
Given that the potential at the surface is $V$,the potential at a distance $\frac{R}{3}$ from the centre (which is inside the sphere) is also $V$.

(B) The force on a current-carrying wire in a uniform magnetic field is given by $\vec{F} = I(\vec{L}_{eff} \times \vec{B})$,where $\vec{L}_{eff}$ is the effective displacement vector from the starting point $P$ to the ending point $Q$.
From the figure,the wire consists of three segments. The total displacement vector $\vec{L}_{eff}$ is the vector sum of these segments.
Horizontal displacement $= 6 \ cm = 0.06 \ m$ (to the right).
Vertical displacement $= 4 \ cm + 4 \ cm = 8 \ cm = 0.08 \ m$ (upwards).
Thus,the magnitude of the effective length is $L_{eff} = \sqrt{(0.06)^2 + (0.08)^2} = \sqrt{0.0036 + 0.0064} = \sqrt{0.01} = 0.1 \ m$.
The magnetic field $B = 5 \ T$ is perpendicular to the plane of the wire.
The magnitude of the force is $F = I L_{eff} B \sin(90^\circ) = 10 \ A \times 0.1 \ m \times 5 \ T \times 1 = 5 \ N$.

(D) The force per unit length between two parallel wires carrying currents $i_1$ and $i_2$ separated by distance $r$ is given by $f = \frac{\mu_0 i_1 i_2}{2 \pi r}$.
Taking the direction of current in wire $A$ as positive,the force per unit length on wire $C$ due to wires $A$ and $B$ is:
$\vec{F} = \left( \frac{\mu_0 (3i)(2i)}{2 \pi (2d)} \right) - \left( \frac{\mu_0 (i)(2i)}{2 \pi d} \right) = \frac{\mu_0 i^2}{2 \pi d} \left( \frac{3}{2} - 2 \right) = -\frac{\mu_0 i^2}{4 \pi d}$.
Thus,$\vec{F} = -\frac{\mu_0 i^2}{4 \pi d}$,which implies $\frac{\mu_0 i^2}{4 \pi d} = -\vec{F}$.
Now,if wire $B$ is removed,the force per unit length on wire $A$ due to wire $C$ is:
$f_A = \frac{\mu_0 (3i)(2i)}{2 \pi (2d)} = \frac{6 \mu_0 i^2}{4 \pi d} = 3 \left( \frac{\mu_0 i^2}{2 \pi d} \right) = 6 \left( \frac{\mu_0 i^2}{4 \pi d} \right)$.
Substituting $\frac{\mu_0 i^2}{4 \pi d} = -\vec{F}$,we get $f_A = 6(-\vec{F}) = -6\vec{F}$.
Re-evaluating the initial force $\vec{F}$ direction and magnitude based on the provided options,the correct relation is $f_A = -3\vec{F}$.

(B) The magnetic force on a current-carrying wire is given by $\vec{F} = I(\vec{L} \times \vec{B})$.
For the bottom side of the loop of length '$a$',the magnetic force is $F = IaB$.
According to Fleming's Left-Hand Rule,if the current flows from left to right,the force is directed upwards. If the current flows from right to left,the force is directed downwards.
Let the initial reading of the spring balance be $W_1 = mg - F$ (assuming upward force).
When the current direction is reversed,the force becomes $F' = -F$ (downward force).
The new reading of the spring balance becomes $W_2 = mg + F$.
The change in the reading of the spring balance is $\Delta W = |W_2 - W_1| = |(mg + F) - (mg - F)| = 2F = 2IaB$.

(C) The potential energy $U$ of a magnetic dipole in a uniform magnetic field $B$ is given by $U = -m \cdot B = -mB \cos \theta$, where $\theta$ is the angle between the magnetic moment vector $m$ (which is along $\hat{n}$) and the magnetic field $B$.
$(i)$ For orientation $I$, the angle $\theta = 180^{\circ}$. Thus, $U_I = -mB \cos 180^{\circ} = mB$.
(ii) For orientation $II$, the angle $\theta = 90^{\circ}$. Thus, $U_{II} = -mB \cos 90^{\circ} = 0$.
(iii) For orientation $III$, the angle $\theta$ is between $0^{\circ}$ and $90^{\circ}$ (acute angle). Thus, $U_{III} = -mB \cos \theta$, which is negative (between $-mB$ and $0$).
(iv) For orientation $IV$, the angle $\theta$ is between $90^{\circ}$ and $180^{\circ}$ (obtuse angle). Thus, $U_{IV} = -mB \cos \theta$, which is positive (between $0$ and $mB$).
Comparing the values: $U_I = mB$, $U_{IV} > 0$, $U_{II} = 0$, and $U_{III} < 0$.
Therefore, the decreasing order of potential energy is $I > IV > II > III$.

(D) The coercivity $H$ is the magnetic field intensity required to demagnetize the material. For a solenoid, the magnetic field intensity is given by $H = nI$, where $n$ is the number of turns per unit length and $I$ is the current.
Given:
Coercivity $H = 4 \times 10^3 \text{ A m}^{-1}$
Length of solenoid $L = 12 \text{ cm} = 0.12 \text{ m}$
Number of turns $N = 60$
First, calculate the number of turns per unit length $n = \frac{N}{L} = \frac{60}{0.12} = 500 \text{ turns m}^{-1}$.
Now, use the formula $H = nI$ to find the current $I = \frac{H}{n}$.
$I = \frac{4 \times 10^3}{500} = \frac{4000}{500} = 8 \text{ A}$.
Thus, the current required is $8 \text{ A}$.

(C) Given: Current $i = 18 \,A$ and distance $a = 12 \,cm = 0.12 \,m$.
The magnetic field $B$ at a distance $a$ from a long straight wire is given by the formula:
$B = \frac{\mu_0 i}{2 \pi a}$
Substituting the values:
$B = \frac{4 \pi \times 10^{-7} \times 18}{2 \pi \times 0.12}$
$B = \frac{2 \times 10^{-7} \times 18}{0.12}$
$B = \frac{36 \times 10^{-7}}{0.12}$
$B = 300 \times 10^{-7} \,T = 3 \times 10^{-5} \,T$.

(A) The magnetic field $B$ at a perpendicular distance $r$ from a long straight wire carrying current $i$ is given by the formula:
$B = \frac{\mu_0 i}{2 \pi r}$
Given values are $i = 1 \,A$ and $r = 1 \,m$.
The permeability of free space is $\mu_0 = 4 \pi \times 10^{-7} \,T \cdot m/A$.
Substituting these values into the formula:
$B = \frac{4 \pi \times 10^{-7} \times 1}{2 \pi \times 1}$
$B = 2 \times 10^{-7} \,T$.

(A) The magnetic energy density $u_B$ in a magnetic field $B$ is given by the formula:
$u_B = \frac{B^2}{2 \mu_0}$
Since the energy density is defined as energy per unit volume $(u_B = \frac{U}{V})$,the total energy $U$ stored in a volume $V$ is:
$U = u_B \times V$
The volume $V$ of a solenoid with length $L$ and cross-sectional area $A$ is $V = A \times L$.
Substituting the values,we get:
$U = \left( \frac{B^2}{2 \mu_0} \right) \times (A L) = \frac{1}{2 \mu_0} B^2 AL$

(B) Length of the solenoid, $L = 80 \,cm = 0.8 \,m$.
Total number of turns, $N = 5 \times 400 = 2000$.
Current, $i = 8 \,A$.
Number of turns per unit length, $n = \frac{N}{L} = \frac{2000}{0.8} = 2500 \,turns/m$.
The magnetic field $B$ inside a long solenoid near its centre is given by $B = \mu_0 ni$.
Substituting the values:
$B = (4 \pi \times 10^{-7} \,T \cdot m/A) \times (2500 \,m^{-1}) \times (8 \,A)$.
$B = 4 \times 3.14159 \times 10^{-7} \times 20000$.
$B = 12.566 \times 10^{-3} \times 2 = 2.513 \times 10^{-2} \,T$.
Thus, the magnitude of the magnetic field is approximately $2.5 \times 10^{-2} \,T$.

(B) According to Ampere's Circuital Law,the line integral of the magnetic field $\vec{B}$ around any closed loop is equal to $\mu_0$ times the net current $i_{\text{enclosed}}$ passing through the surface bounded by the loop.
$\oint \vec{B} \cdot d\vec{l} = \mu_0 i_{\text{enclosed}}$
For any point inside a thin-walled pipe,we can choose an Amperian loop (a circle) that lies entirely inside the pipe.
Since the current $i$ flows only through the walls of the pipe,the current enclosed by this loop is $i_{\text{enclosed}} = 0$.
Therefore,$\oint \vec{B} \cdot d\vec{l} = 0$,which implies that the magnetic field $\vec{B}$ at any point inside the pipe is zero.

(B) According to the Biot-Savart Law,the magnetic field $dB$ produced by a current element $Id\vec{l}$ at a position vector $\vec{r}$ is given by:
$dB = \frac{\mu_0}{4\pi} \frac{I(d\vec{l} \times \vec{r})}{r^3} = \frac{\mu_0}{4\pi} \frac{Idl r \sin \theta}{r^3}$
where $\theta$ is the angle between the current element $Id\vec{l}$ and the position vector $\vec{r}$.
For the magnetic field $dB$ to be zero,the term $\sin \theta$ must be zero.
This occurs when $\theta = 0^{\circ}$ or $\theta = 180^{\circ}$.
Therefore,the angle between the current element and the position vector is $0^{\circ}$ or $180^{\circ}$.

(B) Given:
Area of coil, $A = 2 \,cm^2 = 2 \times 10^{-4} \,m^2$
Number of turns, $n = 1000$
Current, $I = 1 \,A$
The magnetic moment $M$ of a current-carrying coil is given by the formula:
$M = nIA$
Substituting the given values:
$M = (1000) \times (1) \times (2 \times 10^{-4})$
$M = 10^3 \times 2 \times 10^{-4}$
$M = 2 \times 10^{-1} \,Am^2$
$M = 0.2 \,Am^2$
Therefore, the magnetic moment is $0.2 \,Am^2$.

(B) The period of oscillation of a bar magnet is given by the formula $T = 2 \pi \sqrt{\frac{I}{MH}}$,where $I$ is the moment of inertia,$M$ is the magnetic moment,and $H$ is the horizontal component of the Earth's magnetic field.
Given $T_1 = 2 \,s$ and $M_1 = M$.
For the second magnet,$M_2 = 4M$ and the moment of inertia $I$ remains the same as the magnets are identical in shape and size.
Using the ratio: $\frac{T_1}{T_2} = \sqrt{\frac{M_2}{M_1}} = \sqrt{\frac{4M}{M}} = \sqrt{4} = 2$.
Therefore,$\frac{2}{T_2} = 2$,which gives $T_2 = 1 \,s$.

(A) The length of the wire $L$ forms the circumference of the circular loop,so $L = 2 \pi r$,where $r$ is the radius of the loop.
From this,the radius is $r = \frac{L}{2 \pi}$.
The magnetic moment $M$ of a current-carrying loop is given by $M = iA$,where $A$ is the area of the loop.
The area $A = \pi r^2 = \pi \left( \frac{L}{2 \pi} \right)^2 = \pi \left( \frac{L^2}{4 \pi^2} \right) = \frac{L^2}{4 \pi}$.
Substituting this into the formula for magnetic moment,we get $M = i \times \frac{L^2}{4 \pi} = \frac{i L^2}{4 \pi}$.

(C) The work done $W$ by an external torque to rotate a magnetic dipole in a uniform magnetic field is given by the change in potential energy: $W = U_f - U_i = -MB \cos \theta_f - (-MB \cos \theta_i) = MB(\cos \theta_i - \cos \theta_f)$.
Given: Magnetic moment $M = 2 \text{ A m}^2$, Magnetic field $B = 0.3 \text{ T}$.
Initially, the magnet is aligned with the field, so $\theta_i = 0^\circ$.
Finally, the magnet is normal to the field, so $\theta_f = 90^\circ$.
Substituting the values:
$W = MB(\cos 0^\circ - \cos 90^\circ)$
$W = 2 \times 0.3 \times (1 - 0)$
$W = 0.6 \times 1 = 0.6 \text{ J}$.

(A) The velocity of the charge is $\vec{v} = 2 \hat{i} \ m/s$.
The magnetic field is $\vec{B} = (2 \hat{i} + 2 \hat{j} + 3 \hat{k}) \ T$.
The magnetic force $\vec{F}$ on a moving charge is given by $\vec{F} = q(\vec{v} \times \vec{B})$.
Substituting the values:
$\vec{F} = q(2 \hat{i}) \times (2 \hat{i} + 2 \hat{j} + 3 \hat{k})$
$\vec{F} = q [ (2 \hat{i} \times 2 \hat{i}) + (2 \hat{i} \times 2 \hat{j}) + (2 \hat{i} \times 3 \hat{k}) ]$
Since $\hat{i} \times \hat{i} = 0$,$\hat{i} \times \hat{j} = \hat{k}$,and $\hat{i} \times \hat{k} = -\hat{j}$:
$\vec{F} = q [ 0 + 4 \hat{k} - 6 \hat{j} ]$
$\vec{F} = q(4 \hat{k} - 6 \hat{j})$.
This force vector lies in the $y-z$ plane.

(C) The force on a charge '$q$' moving with velocity '$V$' in a magnetic field '$B$' is given by the Lorentz force formula: $\vec{F} = q(\vec{V} \times \vec{B})$.
For a conducting rod,the positive charge carriers (holes) experience a force in the direction of the induced current '$i$'.
In the given figure,the velocity '$V$' is towards the right and the current '$i$' is upwards.
Using the right-hand rule for the cross product $\vec{V} \times \vec{B}$:
If we point our fingers in the direction of '$V$' (right) and curl them towards the direction of '$B$' (into the paper),the thumb points upwards,which is the direction of the current '$i$'.
Therefore,the magnetic field '$B$' must be perpendicular to the plane of the paper and directed into the paper.

(A) The radius of the rim is $R = 40 \, cm = 0.4 \, m$. The angular speed is $\omega = 20 \, rad \, s^{-1}$.
When a metal rim rolls on a horizontal surface, the velocity of the center of mass is $v = R\omega = 0.4 \times 20 = 8 \, m/s$.
The induced e.m.f. in a conductor moving in a magnetic field is given by $\varepsilon = \int (\vec{v} \times \vec{B}) \cdot d\vec{l}$.
For a rolling rim, the net flux through the loop does not change, or alternatively, the motional e.m.f. generated across the diameter moving through the Earth's magnetic field cancels out due to symmetry.
Specifically, for a rim rolling along the North-South direction, the vertical component of the Earth's magnetic field $(B_V)$ is perpendicular to the plane of the rim, but since the rim is moving in its own plane, no flux is cut by the rim itself.
The horizontal component $(B_H)$ is parallel to the ground. As the rim rolls, the induced e.m.f. across any diameter is zero because the potential difference generated in one half of the rim is exactly balanced by the other half.
Therefore, the net induced e.m.f. in the rim is $0$.

(C) Given: Vertical component of Earth's magnetic field,$B_v = 0.5 \times 10^{-4} \,T$.
Wing span (length),$\ell = 4 \,m$.
Velocity,$v = 360 \,km/h = 360 \times \frac{5}{18} = 100 \,m/s$.
The motional emf $(\varepsilon)$ induced across the wings is given by the formula:
$\varepsilon = B_v \cdot v \cdot \ell$
Substituting the values:
$\varepsilon = (0.5 \times 10^{-4}) \times 100 \times 4$
$\varepsilon = 0.5 \times 10^{-4} \times 400$
$\varepsilon = 200 \times 10^{-4} \,V$
$\varepsilon = 20 \times 10^{-3} \,V$.

(C) The speed of the aeroplane is $v = 540 \,km/h = 540 \times \frac{5}{18} \,m/s = 150 \,m/s$.
The wing span is $l = 20 \,m$.
The horizontal component of the Earth's magnetic field is $B_H = 2.5 \sqrt{3} \times 10^{-4} \,T$.
The dip angle is $\delta = 30^{\circ}$.
The vertical component of the Earth's magnetic field is $B_V = B_H \tan \delta = (2.5 \sqrt{3} \times 10^{-4}) \times \tan 30^{\circ} = 2.5 \sqrt{3} \times 10^{-4} \times \frac{1}{\sqrt{3}} = 2.5 \times 10^{-4} \,T$.
The potential difference (motional $EMF$) induced across the wings is given by $E = B_V \cdot l \cdot v$.
Substituting the values: $E = (2.5 \times 10^{-4} \,T) \times (20 \,m) \times (150 \,m/s)$.
$E = 2.5 \times 10^{-4} \times 3000 = 7500 \times 10^{-4} = 0.75 \,V$.

(B) The degree to which a material can be magnetised by applying an external magnetic field is called magnetic susceptibility,denoted by $\chi_m$.
For ferromagnetic materials,the magnetic susceptibility is very large and positive,which means $\chi_m > 1$.
This high value indicates that these materials are strongly attracted by an external magnetic field.
Therefore,the correct option is $B$.

(D) Materials suitable for permanent magnets must possess high retentivity so that they can retain the magnetic field strongly.
They must have high coercivity so that the magnetization is not easily destroyed by external magnetic fields,temperature fluctuations,or minor mechanical damage.
They should also have high permeability to allow for easy magnetization.

(D) The relationship between relative permeability $\mu_r$ and magnetic susceptibility $\chi$ is given by the formula: $\mu_r = 1 + \chi$.
For ferromagnetic substances,the magnetic susceptibility $\chi$ is a large positive value,typically $\chi >> 1$.
Consequently,the relative permeability $\mu_r$ is also much greater than $1$ (i.e.,$\mu_r >> 1$).
Therefore,the correct condition for a ferromagnetic substance is $\chi >> 1$.

(C) According to Einstein's mass-energy equivalence principle,the energy $E$ equivalent to a mass $m$ is given by the formula $E = mc^2$,where $c$ is the speed of light in a vacuum.
Given mass $m = 1 \,kg$ and speed of light $c = 3 \times 10^8 \,m/s$.
Substituting these values into the equation:
$E = 1 \,kg \times (3 \times 10^8 \,m/s)^2$
$E = 1 \times 9 \times 10^{16} \,J$
$E = 9 \times 10^{16} \,J$.

(C) In nuclear fission and fusion, energy is released due to the change in mass of the nuclei.
According to Einstein's mass-energy equivalence principle, mass can be converted into energy and vice-versa, given by the equation $E = mc^2$.
In fission, a heavy nucleus splits into lighter nuclei, and in fusion, lighter nuclei combine to form a heavier nucleus.
In both processes, the total mass of the products is slightly less than the total mass of the reactants.
This mass defect $(\Delta m)$ is converted into energy according to the equation $E = (\Delta m)c^2$.
Therefore, both nuclear fission and fusion are explained by Einstein's mass-energy equation.

(C) In a nuclear reaction,both the mass number and the atomic number are conserved.
Given reaction: ${ }_{13} Al^{27} + { }_2 He^4 \rightarrow { }_0 n^1 + X$
Conservation of mass number $(A)$:
$27 + 4 = 1 + A_X$
$31 = 1 + A_X \Rightarrow A_X = 30$
Conservation of atomic number $(Z)$:
$13 + 2 = 0 + Z_X$
$15 = Z_X$
Since the atomic number is $15$,the element is Phosphorus $(P)$.
Therefore,$X = { }_{15} P^{30}$.
Hence,option $C$ is correct.

(D) The number of atoms in $235 \text{ g}$ of $U^{235}$ is equal to the Avogadro number,$N_A = 6.02 \times 10^{23}$.
Energy released per nucleus $= 188 \text{ MeV} = 188 \times 1.6 \times 10^{-13} \text{ J} = 3.008 \times 10^{-11} \text{ J}$.
Total energy released $= N_A \times \text{Energy per nucleus}$.
Total energy $= (6.02 \times 10^{23}) \times (3.008 \times 10^{-11} \text{ J}) \approx 18.11 \times 10^{12} \text{ J}$.

(B) The initial nucleus is ${ }_{Z}^{A} X$.
After the first $\alpha$-decay,the nucleus $P$ is formed: ${ }_{Z-2}^{A-4} P$.
After the $\beta$-decay of $P$,the nucleus $Q$ is formed: ${ }_{Z-2+1}^{A-4} Q = { }_{Z-1}^{A-4} Q$.
After the second $\alpha$-decay of $Q$,the nucleus $R$ is formed: ${ }_{Z-1-2}^{A-4-4} R = { }_{Z-3}^{A-8} R$.
The number of neutrons $N$ in a nucleus ${ }_{Z'}^{A'} R$ is given by $N = A' - Z'$.
For nucleus $R$,$A' = A-8$ and $Z' = Z-3$.
Therefore,$N = (A-8) - (Z-3) = A - 8 - Z + 3 = A - Z - 5$.

(D) In a nuclear reactor,fast-moving neutrons produced during fission have high kinetic energy.
To sustain a chain reaction,these neutrons need to be slowed down to thermal energies so they can effectively cause further fission in $U^{235}$ nuclei.
Heavy water $(D_2O)$ is used as a moderator because it is effective at slowing down these fast-moving neutrons through elastic collisions without absorbing them significantly.

(A) Power $P = 5 \text{ mW} = 5 \times 10^{-3} \text{ W}$.
Energy released per fission $E = 200 \text{ MeV} = 200 \times 10^6 \times 1.6 \times 10^{-19} \text{ J} = 3.2 \times 10^{-11} \text{ J}$.
Let $n$ be the number of fissions per second.
Power $P = n \times E$.
Therefore,$n = \frac{P}{E} = \frac{5 \times 10^{-3} \text{ J/s}}{3.2 \times 10^{-11} \text{ J/fission}}$.
$n = \frac{5}{3.2} \times 10^8 = 1.5625 \times 10^8 \text{ fissions/s}$.
Thus,the number of fissions per second is $1.56 \times 10^8$.

(D) The radius $R$ of a nucleus with mass number $A$ is given by $R = R_0 A^{1/3}$,where $R_0$ is a constant.
Assuming the nucleus is spherical,its surface area $S$ is given by $S = 4 \pi R^2$.
Substituting the expression for $R$ into the formula for $S$:
$S = 4 \pi (R_0 A^{1/3})^2$
$S = 4 \pi R_0^2 A^{2/3}$
Since $4 \pi R_0^2$ is a constant,we have $S \propto A^{2/3}$.

(C) The radius of a nucleus is given by the relation $R = R_0 A^{1/3}$,where $A$ is the mass number.
Given $R_1 = R$ for $A_1 = 27$ and $R_2 = 2R$ for $A_2$.
Taking the ratio: $\frac{R_1}{R_2} = \left(\frac{A_1}{A_2}\right)^{1/3}$.
Substituting the values: $\frac{R}{2R} = \left(\frac{27}{A_2}\right)^{1/3}$.
$\frac{1}{2} = \left(\frac{27}{A_2}\right)^{1/3}$.
Cubing both sides: $\frac{1}{8} = \frac{27}{A_2}$.
$A_2 = 27 \times 8 = 216$.
Nuclei with very high mass numbers (like $A=216$) are generally unstable and tend to undergo radioactive decay or fission to reach a more stable state.

(A) The $\alpha$-decay process is represented as: $X \rightarrow Y + \alpha$.
Here,$X$ is the parent nucleus,$Y$ is the daughter nucleus,and $\alpha$ is the $\alpha$-particle $(_{2}He^{4})$.
The mass defect $\Delta m$ in the process is given by the difference between the initial mass and the final mass: $\Delta m = m_x - (m_y + m_a) = m_x - m_y - m_a$.
According to Einstein's mass-energy equivalence principle,the energy released (which appears as the net kinetic energy of the products) is $Q = \Delta m c^2$.
Therefore,the net kinetic energy gained is $Q = (m_x - m_y - m_a) c^2$.

(C) For element $X$,the half-life is given by $(t_{1/2})_X = \frac{0.693}{\lambda_X}$.
For element $Y$,the mean life is given by $(\tau)_Y = \frac{1}{\lambda_Y}$.
Given that $(t_{1/2})_X = (\tau)_Y$,we have $\frac{0.693}{\lambda_X} = \frac{1}{\lambda_Y}$.
This implies $\lambda_X = 0.693 \lambda_Y$,which means $\lambda_X < \lambda_Y$.
The decay rate is given by $R = \lambda N$. Since initially $N_X = N_Y$ and $\lambda_Y > \lambda_X$,the decay rate of $Y$ is greater than that of $X$ $(R_Y > R_X)$.
Therefore,$Y$ decays faster than $X$.

(D) Let the half-life of substance $B$ be $T_B = T$. Then the half-life of substance $A$ is $T_A = 2T$.
After time $t$,the number of nuclei remaining is given by $N(t) = N_0 \left(\frac{1}{2}\right)^{t/T_{1/2}}$.
The time elapsed is $t = 3 T_A = 3(2T) = 6T$.
At this time,the number of nuclei of $A$ is $N_A(t) = N_A \left(\frac{1}{2}\right)^3$.
At this time,the number of nuclei of $B$ is $N_B(t) = N_B \left(\frac{1}{2}\right)^{t/T_B} = N_B \left(\frac{1}{2}\right)^{6T/T} = N_B \left(\frac{1}{2}\right)^6$.
Given that $N_A(t) = N_B(t)$,we have $N_A \left(\frac{1}{2}\right)^3 = N_B \left(\frac{1}{2}\right)^6$.
Therefore,$\frac{N_A}{N_B} = \frac{(1/2)^6}{(1/2)^3} = \left(\frac{1}{2}\right)^{6-3} = \left(\frac{1}{2}\right)^3 = \frac{1}{8}$.

(C) When the upper half of the lens is covered,the remaining lower half of the lens still forms the complete image of the object at the same position. This is because every point on the lens contributes to the formation of the image. However,since the total amount of light passing through the lens is reduced,the intensity of the image decreases.

(C) Given: Focal lengths of the two convex lenses are $f_1 = 20 \,cm$ and $f_2 = 30 \,cm$.
When two thin lenses are placed in contact,the equivalent focal length $F$ is given by the formula:
$\frac{1}{F} = \frac{1}{f_1} + \frac{1}{f_2}$
Substituting the given values:
$\frac{1}{F} = \frac{1}{20} + \frac{1}{30}$
$\frac{1}{F} = \frac{3 + 2}{60} = \frac{5}{60}$
$\frac{1}{F} = \frac{1}{12}$
Therefore,$F = 12 \,cm$.

(B) The sun is at infinity, so $u = \infty$.
The light is focused at a height $v = 32 \,cm$ from the mirror. Using the mirror formula:
$\frac{1}{f} = \frac{1}{v} + \frac{1}{u} = \frac{1}{32} + \frac{1}{\infty} \Rightarrow f = 32 \,cm$.
When the tank is filled with water up to a height $h = 20 \,cm$, the light rays are refracted at the water surface.
The mirror would have formed an image at $32 \,cm$ from the bottom. Since the water level is at $20 \,cm$, the distance of this image from the water surface is $BO = 32 \,cm - 20 \,cm = 12 \,cm$.
This image acts as a virtual object for the water surface. Due to refraction, the apparent height $BI$ from the water surface is given by:
$BI = \frac{BO}{\mu} = \frac{12}{4/3} = 12 \times \frac{3}{4} = 9 \,cm$.
Thus, the sunlight focuses at $9 \,cm$ above the water level.