NEET 2016 Physics Question Paper with Answer and Solution

(C) Let the dimension of length $l$ be expressed as $l \propto h^p c^q G^r$.
Substituting the dimensions of each constant:
$[L] = [ML^2T^{-1}]^p [LT^{-1}]^q [M^{-1}L^3T^{-2}]^r$
$[M^0 L^1 T^0] = M^{p-r} L^{2p+q+3r} T^{-p-q-2r}$
Comparing the powers of $M, L,$ and $T$ on both sides:
$p - r = 0 \implies p = r$
$2p + q + 3r = 1$
$-p - q - 2r = 0$
Substituting $p = r$ into the third equation: $-r - q - 2r = 0 \implies q = -3r$.
Substituting $p = r$ and $q = -3r$ into the second equation: $2r - 3r + 3r = 1 \implies 2r = 1 \implies r = 1/2$.
Thus,$p = 1/2$ and $q = -3/2$.
Therefore,$l \propto h^{1/2} c^{-3/2} G^{1/2} = \sqrt{\frac{hG}{c^3}} = \frac{\sqrt{hG}}{c^{3/2}}$.

(B) The position of car $P$ at any time $t$ is given by $x_P(t) = at + bt^2$.
The velocity of car $P$ is $v_P(t) = \frac{dx_P(t)}{dt} = a + 2bt$ ... $(i)$.
Similarly,for car $Q$,the position is $x_Q(t) = ft - t^2$.
The velocity of car $Q$ is $v_Q(t) = \frac{dx_Q(t)}{dt} = f - 2t$ ... $(ii)$.
Given that the cars have the same velocity,$v_P(t) = v_Q(t)$.
Therefore,$a + 2bt = f - 2t$.
Rearranging the terms to solve for $t$: $2bt + 2t = f - a$.
$2t(b + 1) = f - a$.
Thus,$t = \frac{f - a}{2(1 + b)}$.

(B) The velocity of the particle is given by $v = At + Bt^2$.
Since $v = \frac{ds}{dt}$,we have $ds = (At + Bt^2) dt$.
To find the distance travelled between $t = 1 \ s$ and $t = 2 \ s$,we integrate the velocity with respect to time:
$s = \int_{1}^{2} (At + Bt^2) dt$
$s = \left[ \frac{At^2}{2} + \frac{Bt^3}{3} \right]_{1}^{2}$
$s = \left( \frac{A(2)^2}{2} + \frac{B(2)^3}{3} \right) - \left( \frac{A(1)^2}{2} + \frac{B(1)^3}{3} \right)$
$s = \left( 2A + \frac{8B}{3} \right) - \left( \frac{A}{2} + \frac{B}{3} \right)$
$s = (2A - \frac{A}{2}) + (\frac{8B}{3} - \frac{B}{3})$
$s = \frac{3}{2}A + \frac{7}{3}B$
Since the velocity does not change sign in the interval $[1, 2]$,the distance is equal to the magnitude of displacement.

(A) Given:
Total acceleration $a = 15 \, m s^{-2}$
Radius $R = 2.5 \, m$
Angle between total acceleration and centripetal acceleration is $30^{\circ}$.
The centripetal acceleration $a_c$ is the component of the total acceleration $a$ directed towards the center of the circle.
From the figure,$a_c = a \cos(30^{\circ})$.
$a_c = 15 \times \frac{\sqrt{3}}{2} \approx 15 \times 0.866 = 12.99 \, m s^{-2}$.
We know that the centripetal acceleration is given by $a_c = \frac{v^2}{R}$,where $v$ is the speed of the particle.
Therefore,$v = \sqrt{a_c R}$.
$v = \sqrt{12.99 \times 2.5} = \sqrt{32.475} \approx 5.698 \, m/s$.
Rounding to one decimal place,we get $v \approx 5.7 \, m/s$.
Thus,the correct option is $A$.

(B) Given the position vector: $\vec{r} = \cos(\omega t) \hat{i} + \sin(\omega t) \hat{j}$.
$1$. Velocity $\vec{v}$ is the derivative of position with respect to time:
$\vec{v} = \frac{d\vec{r}}{dt} = -\omega \sin(\omega t) \hat{i} + \omega \cos(\omega t) \hat{j}$.
$2$. Acceleration $\vec{a}$ is the derivative of velocity with respect to time:
$\vec{a} = \frac{d\vec{v}}{dt} = -\omega^2 \cos(\omega t) \hat{i} - \omega^2 \sin(\omega t) \hat{j} = -\omega^2 \vec{r}$.
$3$. Since $\vec{a} = -\omega^2 \vec{r}$,the acceleration is directed towards the origin (centripetal acceleration).
$4$. To check the relationship between $\vec{r}$ and $\vec{v}$,calculate the dot product:
$\vec{r} \cdot \vec{v} = (\cos(\omega t))(-\omega \sin(\omega t)) + (\sin(\omega t))(\omega \cos(\omega t)) = -\omega \sin(\omega t) \cos(\omega t) + \omega \sin(\omega t) \cos(\omega t) = 0$.
Since the dot product is $0$,the velocity is perpendicular to the position vector $\vec{r}$.

(A) Let the two vectors be $\vec{A}$ and $\vec{B}$.
The magnitude of the sum of $\vec{A}$ and $\vec{B}$ is given by: $|\vec{A} + \vec{B}| = \sqrt{A^2 + B^2 + 2AB \cos \theta}$.
The magnitude of the difference of $\vec{A}$ and $\vec{B}$ is given by: $|\vec{A} - \vec{B}| = \sqrt{A^2 + B^2 - 2AB \cos \theta}$.
Given that $|\vec{A} + \vec{B}| = |\vec{A} - \vec{B}|$,we square both sides:
$A^2 + B^2 + 2AB \cos \theta = A^2 + B^2 - 2AB \cos \theta$.
Subtracting $A^2 + B^2$ from both sides,we get:
$2AB \cos \theta = -2AB \cos \theta$.
$4AB \cos \theta = 0$.
Since $A$ and $B$ are non-zero vectors,$4AB \neq 0$,therefore $\cos \theta = 0$.
This implies $\theta = 90^o$.

(C) The impulse imparted by the wall on the ball is equal to the change in momentum of the ball.
Let the initial velocity be $\vec{v}_i$ and the final velocity be $\vec{v}_f$.
The initial momentum is $\vec{p}_i = m\vec{v}_i$ and the final momentum is $\vec{p}_f = m\vec{v}_f$.
Taking the normal to the wall as the $x$-axis,the initial velocity components are $v_{ix} = v \cos 60^\circ$ (towards the wall) and $v_{iy} = v \sin 60^\circ$ (parallel to the wall).
The final velocity components are $v_{fx} = -v \cos 60^\circ$ (away from the wall) and $v_{fy} = v \sin 60^\circ$ (parallel to the wall).
The change in momentum $\Delta \vec{p} = \vec{p}_f - \vec{p}_i = m(\vec{v}_f - \vec{v}_i)$.
$\Delta p_x = m(v_{fx} - v_{ix}) = m(-v \cos 60^\circ - v \cos 60^\circ) = -2mv \cos 60^\circ = -2mv(0.5) = -mv$.
$\Delta p_y = m(v_{fy} - v_{iy}) = m(v \sin 60^\circ - v \sin 60^\circ) = 0$.
The magnitude of the impulse is $|\Delta \vec{p}| = |-mv| = mv$.

(A) For vertical equilibrium on the road:
$N \cos\theta = mg + f \sin\theta$
$mg = N \cos\theta - f \sin\theta$ ... $(i)$
For safe turning,the centripetal force is provided by the horizontal components of the normal force and friction:
$N \sin\theta + f \cos\theta = \frac{mv^2}{R}$ ... $(ii)$
Dividing equation $(ii)$ by equation $(i)$:
$\frac{v^2}{Rg} = \frac{N \sin\theta + f \cos\theta}{N \cos\theta - f \sin\theta}$
At maximum velocity,the friction force $f$ reaches its limiting value,$f = \mu_s N$. Substituting this into the equation:
$\frac{v_{\max}^2}{Rg} = \frac{N \sin\theta + \mu_s N \cos\theta}{N \cos\theta - \mu_s N \sin\theta}$
Dividing the numerator and denominator by $N \cos\theta$:
$\frac{v_{\max}^2}{Rg} = \frac{\tan\theta + \mu_s}{1 - \mu_s \tan\theta}$
Therefore,the maximum safe velocity is:
$v_{\max} = \sqrt{gR \left( \frac{\mu_s + \tan\theta}{1 - \mu_s \tan\theta} \right)}$

(C) Given: Force $\overrightarrow{F} = (2t\hat{i} + 3t^2\hat{j})\, N$ and mass $m = 1\, kg$.
Acceleration of the body is given by $\overrightarrow{a} = \frac{\overrightarrow{F}}{m} = \frac{2t\hat{i} + 3t^2\hat{j}}{1} = (2t\hat{i} + 3t^2\hat{j})\, m/s^2$.
Velocity $\overrightarrow{v}$ at time $t$ is obtained by integrating acceleration with respect to time: $\overrightarrow{v} = \int \overrightarrow{a} dt = \int (2t\hat{i} + 3t^2\hat{j}) dt = t^2\hat{i} + t^3\hat{j}\, m/s$.
Power $P$ developed by the force is the dot product of force and velocity: $P = \overrightarrow{F} \cdot \overrightarrow{v}$.
$P = (2t\hat{i} + 3t^2\hat{j}) \cdot (t^2\hat{i} + t^3\hat{j}) = (2t)(t^2) + (3t^2)(t^3) = 2t^3 + 3t^5\, W$.

(B) To complete a vertical loop,the body must maintain a minimum velocity at the highest point $C$ such that the tension $T_C$ in the string (or normal force) is at least zero.
At the highest point $C$,the forces acting on the body are gravity $(mg)$ acting downwards and tension $(T_C)$ acting downwards. The centripetal force is provided by the sum of these forces:
$T_C + mg = \frac{mv_C^2}{R}$
For the minimum velocity,we set $T_C = 0$,which gives:
$mg = \frac{mv_C^2}{R} \Rightarrow v_C = \sqrt{gR}$
Now,we apply the law of conservation of mechanical energy between the lowest point $A$ and the highest point $C$. Let $v_0$ be the velocity at point $A$:
Total Energy at $A$ = Total Energy at $C$
$\frac{1}{2}mv_0^2 = \frac{1}{2}mv_C^2 + mg(2R)$
Substituting $v_C^2 = gR$:
$\frac{1}{2}mv_0^2 = \frac{1}{2}m(gR) + 2mgR$
$\frac{1}{2}mv_0^2 = \frac{5}{2}mgR$
$v_0^2 = 5gR$
$v_0 = \sqrt{5gR}$

(B) Given:
Mass of bullet,$m = 10\, g = 0.01\, kg$
Initial speed of bullet,$u = 400\, m s^{-1}$
Mass of block,$M = 2\, kg$
Vertical rise of block,$h = 10\, cm = 0.1\, m$
Let $v_1$ be the speed of the block immediately after the collision and $v$ be the speed of the bullet after emerging from the block.
Applying the principle of conservation of mechanical energy to the block after the collision:
$\frac{1}{2} M v_1^2 = Mgh$
$v_1 = \sqrt{2gh} = \sqrt{2 \times 9.8 \times 0.1} = \sqrt{1.96} = 1.4\, m s^{-1}$
(Using $g = 9.8\, m s^{-2}$ for precision).
Applying the principle of conservation of linear momentum for the system (bullet + block) during the collision:
$mu = Mv_1 + mv$
$0.01 \times 400 = 2 \times 1.4 + 0.01 \times v$
$4 = 2.8 + 0.01v$
$0.01v = 1.2$
$v = 120\, m s^{-1}$

(C) According to Wien's displacement law,the wavelength corresponding to maximum spectral emissive power is given by:
$\lambda_m = \frac{b}{T} = \frac{2.88 \times 10^6 \ nm \ K}{5760 \ K} = 500 \ nm$
This means the peak of the black body radiation curve occurs at $\lambda = 500 \ nm$,where the energy emitted is $U_2$.
Comparing the values at different wavelengths from the spectral distribution curve:
At $\lambda = 250 \ nm$,the energy is $U_1$.
At $\lambda = 500 \ nm$,the energy is $U_2$ (maximum).
At $\lambda = 1000 \ nm$,the energy is $U_3$.
From the graph,it is clear that $U_2$ is the maximum value,so $U_2 > U_1$ and $U_2 > U_3$. Also,comparing $U_1$ and $U_3$ from the graph,$U_3 > U_1$. Thus,the correct relation is $U_2 > U_1$.

(D) According to Newton's law of cooling,the rate of cooling is proportional to the temperature difference between the body and the surroundings:
$\frac{T_1 - T_2}{t} = K \left( \frac{T_1 + T_2}{2} - T_s \right)$
For the first interval of $10$ minutes:
$T_1 = 3T, T_2 = 2T, T_s = T, t = 10$
$\frac{3T - 2T}{10} = K \left( \frac{3T + 2T}{2} - T \right)$
$\frac{T}{10} = K \left( 2.5T - T \right) = K(1.5T) \implies K = \frac{1}{15} \dots (i)$
For the next interval of $10$ minutes:
Let the final temperature be $T'$.
$T_1 = 2T, T_2 = T', T_s = T, t = 10$
$\frac{2T - T'}{10} = K \left( \frac{2T + T'}{2} - T \right)$
$\frac{2T - T'}{10} = K \left( \frac{T'}{2} \right) \dots (ii)$
Substituting $K = \frac{1}{15}$ into equation $(ii)$:
$\frac{2T - T'}{10} = \frac{1}{15} \left( \frac{T'}{2} \right)$
$\frac{2T - T'}{10} = \frac{T'}{30}$
$3(2T - T') = T'$
$6T - 3T' = T'$
$6T = 4T' \implies T' = \frac{6}{4}T = \frac{3}{2}T$

(D) The ideal gas equation is given by $PV = nRT$,where $n$ is the number of moles.
We know that $n = \frac{\text{mass of gas}}{M}$,where $M$ is the molar mass.
Also,the molar mass $M = m N_A$,where $m$ is the mass of one molecule and $N_A$ is Avogadro's number.
Substituting these into the ideal gas equation: $PV = \left(\frac{\text{mass}}{m N_A}\right) RT$.
Rearranging for density $\rho = \frac{\text{mass}}{V}$,we get $\rho = \frac{P m N_A}{RT}$.
Since the Boltzmann constant $k = \frac{R}{N_A}$,we have $R = N_A k$.
Substituting $R$ in the density equation: $\rho = \frac{P m N_A}{(N_A k) T} = \frac{Pm}{kT}$.

(A) The $r.m.s.$ velocity of gas molecules is given by the formula $v_{rms} = \sqrt{\frac{3RT}{M}}$.
This shows that $v_{rms} \propto \sqrt{T}$.
The pressure does not affect the $r.m.s.$ velocity of an ideal gas.
Given $T_1 = 27^o C = 300 \,K$ and $v_1 = 200 \,m s^{-1}$.
Given $T_2 = 127^o C = 400 \,K$.
Using the ratio: $\frac{v_2}{v_1} = \sqrt{\frac{T_2}{T_1}}$.
$\frac{v_2}{200} = \sqrt{\frac{400}{300}} = \sqrt{\frac{4}{3}} = \frac{2}{\sqrt{3}}$.
Therefore,$v_2 = 200 \times \frac{2}{\sqrt{3}} = \frac{400}{\sqrt{3}} \,m s^{-1}$.

(A) The rotational kinetic energy $K$ is related to the angular momentum $L$ and moment of inertia $I$ by the formula $K = \frac{L^2}{2I}$.
Given that the kinetic energies are equal,$K_A = K_B$.
Therefore,$\frac{L_A^2}{2I_A} = \frac{L_B^2}{2I_B}$.
This implies $\frac{L_A^2}{L_B^2} = \frac{I_A}{I_B}$.
Taking the square root on both sides,we get $\frac{L_A}{L_B} = \sqrt{\frac{I_A}{I_B}}$.
Since it is given that $I_B > I_A$,it follows that $\frac{I_A}{I_B} < 1$.
Thus,$\frac{L_A}{L_B} < 1$,which means $L_A < L_B$ or $L_B > L_A$.

(D) The rotational kinetic energy is given by $E = \frac{1}{2} I \omega^2$.
For the solid sphere rotating about its diameter,the moment of inertia is $I_s = \frac{2}{5} m R^2$.
For the solid cylinder rotating about its geometrical axis,the moment of inertia is $I_c = \frac{1}{2} m R^2$.
Given that the angular speed of the cylinder is twice that of the sphere,we have $\omega_c = 2 \omega_s$.
The ratio of their kinetic energies is:
$\frac{E_{sphere}}{E_{cylinder}} = \frac{\frac{1}{2} I_s \omega_s^2}{\frac{1}{2} I_c \omega_c^2} = \frac{I_s \omega_s^2}{I_c (2 \omega_s)^2} = \frac{I_s}{4 I_c}$.
Substituting the values of $I_s$ and $I_c$:
$\frac{E_{sphere}}{E_{cylinder}} = \frac{\frac{2}{5} m R^2}{4 \times \frac{1}{2} m R^2} = \frac{2/5}{2} = \frac{1}{5}$.
Thus,the ratio is $1:5$.

(C) Let the distances of masses $m_1$ and $m_2$ from the center of mass be $l_1$ and $l_2$ respectively.
Given that the total length of the rod is $l$,so $l_1 + l_2 = l$.
By the definition of the center of mass,$m_1 l_1 = m_2 l_2$.
Substituting $l_2 = l - l_1$,we get $m_1 l_1 = m_2 (l - l_1)$,which simplifies to $l_1 = \frac{m_2 l}{m_1 + m_2}$.
Similarly,$l_2 = \frac{m_1 l}{m_1 + m_2}$.
The moment of inertia $I$ about an axis passing through the center of mass is given by $I = m_1 l_1^2 + m_2 l_2^2$.
Substituting the values of $l_1$ and $l_2$:
$I = m_1 \left( \frac{m_2 l}{m_1 + m_2} \right)^2 + m_2 \left( \frac{m_1 l}{m_1 + m_2} \right)^2$
$I = \frac{m_1 m_2^2 l^2 + m_2 m_1^2 l^2}{(m_1 + m_2)^2}$
$I = \frac{m_1 m_2 l^2 (m_2 + m_1)}{(m_1 + m_2)^2}$
$I = \frac{m_1 m_2}{m_1 + m_2} l^2$.

(A) Mass per unit area of the disc is $\sigma = \frac{M}{\pi R^2}$.
The radius of the removed circular hole is $r = \frac{R}{2}$.
The mass of the removed portion is $M' = \sigma \times \pi r^2 = \frac{M}{\pi R^2} \times \pi \left(\frac{R}{2}\right)^2 = \frac{M}{4}$.
The moment of inertia of the removed portion about its own central axis (perpendicular to the plane) is $I_{cm} = \frac{1}{2} M' r^2 = \frac{1}{2} \left(\frac{M}{4}\right) \left(\frac{R}{2}\right)^2 = \frac{MR^2}{32}$.
Using the parallel axis theorem,the moment of inertia of the removed portion about the axis passing through the center $O$ of the original disc is $I'_{0} = I_{cm} + M' d^2$,where $d = \frac{R}{2}$ is the distance between the center of the disc and the center of the hole.
$I'_{0} = \frac{MR^2}{32} + \left(\frac{M}{4}\right) \left(\frac{R}{2}\right)^2 = \frac{MR^2}{32} + \frac{MR^2}{16} = \frac{3MR^2}{32}$.
The moment of inertia of the complete disc about the center $O$ is $I_{0} = \frac{1}{2} MR^2$.
The moment of inertia of the remaining part is $I = I_{0} - I'_{0} = \frac{1}{2} MR^2 - \frac{3MR^2}{32} = \frac{16MR^2 - 3MR^2}{32} = \frac{13MR^2}{32}$.

(A) The time $t$ taken by a body to roll down an inclined plane of length $l$ and inclination $\theta$ without slipping is given by:
$t = \sqrt{\frac{2l(1 + \frac{k^2}{R^2})}{g \sin \theta}}$
where $k$ is the radius of gyration and $R$ is the radius of the object.
Since $l$,$\theta$,and $g$ are the same for both,the time depends on the factor $(1 + \frac{k^2}{R^2})$.
For a disc,$I = \frac{1}{2}MR^2$,so $k^2 = \frac{1}{2}R^2$,which gives $(1 + \frac{k^2}{R^2}) = 1 + 0.5 = 1.5$.
For a solid sphere,$I = \frac{2}{5}MR^2$,so $k^2 = \frac{2}{5}R^2$,which gives $(1 + \frac{k^2}{R^2}) = 1 + 0.4 = 1.4$.
Since $1.4 < 1.5$,the time taken by the sphere is less than the time taken by the disc $(t_s < t_d)$.
Therefore,the sphere reaches the bottom first.

(D) Given: Radius $r = 50\, cm = 0.5\, m$,angular acceleration $\alpha = 2.0\, rad\, s^{-2}$,initial angular velocity $\omega_0 = 0$,and time $t = 2.0\, s$.
First,calculate the angular velocity $\omega$ at $t = 2.0\, s$ using $\omega = \omega_0 + \alpha t = 0 + (2.0)(2.0) = 4.0\, rad\, s^{-1}$.
The tangential acceleration is $a_t = r\alpha = 0.5 \times 2.0 = 1.0\, m\, s^{-2}$.
The radial (centripetal) acceleration is $a_r = \omega^2 r = (4.0)^2 \times 0.5 = 16 \times 0.5 = 8.0\, m\, s^{-2}$.
The net acceleration $a$ is given by $a = \sqrt{a_t^2 + a_r^2} = \sqrt{1.0^2 + 8.0^2} = \sqrt{1 + 64} = \sqrt{65} \approx 8.06\, m\, s^{-2}$.
Rounding to the nearest value,the net acceleration is approximately $8.0\, m\, s^{-2}$.

(A) The formula for escape velocity is $v = \sqrt{\frac{2GM}{R}}$.
Since mass $M = \rho \times V = \rho \times \frac{4}{3}\pi R^3$,we substitute this into the formula:
$v = \sqrt{\frac{2G}{R} \times \frac{4}{3}\pi R^3 \rho} = R \sqrt{\frac{8\pi G \rho}{3}}$.
Thus,the escape velocity is proportional to $R\sqrt{\rho}$,i.e.,$v \propto R\sqrt{\rho}$.
Given for the planet: $R_p = 2R_e$ and $\rho_p = 2\rho_e$.
Therefore,the ratio is:
$\frac{v_e}{v_p} = \frac{R_e}{R_p} \times \sqrt{\frac{\rho_e}{\rho_p}} = \frac{R_e}{2R_e} \times \sqrt{\frac{\rho_e}{2\rho_e}} = \frac{1}{2} \times \frac{1}{\sqrt{2}} = \frac{1}{2\sqrt{2}}$.
Hence,the ratio is $1 : 2\sqrt{2}$.

(D) The gravitational potential at a height $h$ from the surface of the Earth is given by $V_h = -\frac{GM}{R+h} = -5.4 \times 10^7\, J kg^{-1}$.
The acceleration due to gravity at the same height is $g_h = \frac{GM}{(R+h)^2} = 6.0\, m s^{-2}$.
We can relate these two expressions as $g_h = \frac{GM}{(R+h)^2} = \frac{1}{R+h} \left( \frac{GM}{R+h} \right) = \frac{-V_h}{R+h}$.
Rearranging for $(R+h)$,we get $R+h = \frac{-V_h}{g_h}$.
Substituting the given values: $R+h = \frac{-(-5.4 \times 10^7)}{6.0} = \frac{5.4 \times 10^7}{6.0} = 0.9 \times 10^7 = 9.0 \times 10^6\, m$.
Converting the radius of the Earth to meters: $R = 6400\, km = 6.4 \times 10^6\, m$.
Now,$h = (R+h) - R = 9.0 \times 10^6 - 6.4 \times 10^6 = 2.6 \times 10^6\, m$.
Converting back to kilometers: $h = 2600\, km$.

(D) The total energy $E$ of a satellite is the sum of its potential energy $PE$ and kinetic energy $KE$.
$E = PE + KE = -\frac{GMm}{R + h} + \frac{1}{2}mv^2$
For a satellite in a circular orbit,the gravitational force provides the necessary centripetal force:
$\frac{mv^2}{R + h} = \frac{GMm}{(R + h)^2} \implies v^2 = \frac{GM}{R + h}$
Substituting $v^2$ into the energy equation:
$E = -\frac{GMm}{R + h} + \frac{1}{2}m\left(\frac{GM}{R + h}\right) = -\frac{GMm}{2(R + h)}$
Since the acceleration due to gravity at the surface is $g_0 = \frac{GM}{R^2}$,we have $GM = g_0 R^2$.
Substituting this into the expression for $E$:
$E = -\frac{m g_0 R^2}{2(R + h)}$

(D) The height of capillary rise is given by $h = \frac{2T \cos \theta}{r \rho g}$.
For a given value of $T$ and $r$,we have $h \propto \frac{\cos \theta}{\rho}$.
Since the liquids rise to the same height $(h_1 = h_2 = h_3)$,we have $\frac{\cos \theta_1}{\rho_1} = \frac{\cos \theta_2}{\rho_2} = \frac{\cos \theta_3}{\rho_3}$.
Given $\rho_1 > \rho_2 > \rho_3$,it follows that $\cos \theta_1 > \cos \theta_2 > \cos \theta_3$.
Since the cosine function is a decreasing function in the interval $[0, \frac{\pi}{2}]$,we have $\theta_1 < \theta_2 < \theta_3$.
Therefore,$0 \le \theta_1 < \theta_2 < \theta_3 < \frac{\pi}{2}$.

(C) Let $d$ be the density of the cylinder and $A$ be the area of the cross-section of the cylinder.
According to the law of floatation,the weight of the cylinder is equal to the total upthrust exerted by the two liquids.
Weight of cylinder = $L \times A \times d \times g$
Upthrust by denser liquid (density $n\rho$) = $(pL \times A) \times n\rho \times g$
Upthrust by lighter liquid (density $\rho$) = $((L - pL) \times A) \times \rho \times g$
Equating weight to total upthrust:
$L \times A \times d \times g = (pL \times A) \times n\rho \times g + ((L - pL) \times A) \times \rho \times g$
Dividing both sides by $A \times L \times g$:
$d = p \times n\rho + (1 - p) \times \rho$
$d = np\rho + \rho - p\rho$
$d = [1 + (n - 1)p]\rho$

(D) The work done in stretching a liquid film is given by $W = T \times \Delta A_{total}$.
Since a liquid film has two surfaces,the change in area is $\Delta A_{total} = 2 \times (A_2 - A_1)$.
Initial area $A_1 = 4 \, cm \times 2 \, cm = 8 \, cm^2 = 8 \times 10^{-4} \, m^2$.
Final area $A_2 = 5 \, cm \times 4 \, cm = 20 \, cm^2 = 20 \times 10^{-4} \, m^2$.
Change in area $\Delta A = A_2 - A_1 = 12 \, cm^2 = 12 \times 10^{-4} \, m^2$.
Total change in area $\Delta A_{total} = 2 \times 12 \times 10^{-4} \, m^2 = 24 \times 10^{-4} \, m^2$.
Given work done $W = 3 \times 10^{-4} \, J$.
Using $W = T \times \Delta A_{total}$,we get $T = \frac{W}{\Delta A_{total}} = \frac{3 \times 10^{-4}}{24 \times 10^{-4}} = \frac{1}{8} = 0.125 \, N \, m^{-1}$.

(D) The temperature inside the refrigerator is $T_2 = (t_2 + 273) \, K$ and the room temperature is $T_1 = (t_1 + 273) \, K$.
For an ideal refrigerator (Carnot cycle),the ratio of heat rejected to the room $(Q_1)$ to the heat absorbed from the cold reservoir $(Q_2)$ is given by $\frac{Q_1}{Q_2} = \frac{T_1}{T_2}$.
From the first law of thermodynamics,the work done $W$ is $W = Q_1 - Q_2$,which implies $Q_2 = Q_1 - W$.
Substituting this into the ratio: $\frac{Q_1}{Q_1 - W} = \frac{T_1}{T_2}$.
Rearranging for $\frac{Q_1}{W}$: $\frac{Q_1 - W}{Q_1} = \frac{T_2}{T_1} \Rightarrow 1 - \frac{W}{Q_1} = \frac{T_2}{T_1}$.
$\frac{W}{Q_1} = 1 - \frac{T_2}{T_1} = \frac{T_1 - T_2}{T_1}$.
Therefore,the heat delivered to the room per unit of work done is $\frac{Q_1}{W} = \frac{T_1}{T_1 - T_2}$.
Substituting $T_1 = t_1 + 273$ and $T_2 = t_2 + 273$,we get $\frac{Q_1}{W} = \frac{t_1 + 273}{(t_1 + 273) - (t_2 + 273)} = \frac{t_1 + 273}{t_1 - t_2}$.

(B) The given process is described by the equation $PV^3 = \text{constant}$.
This is a polytropic process of the form $PV^n = \text{constant}$, where $n = 3$.
For an ideal gas, the molar heat capacity $C$ for a polytropic process is given by the formula $C = C_v + \frac{R}{1 - n}$.
For a monatomic gas, the molar heat capacity at constant volume is $C_v = \frac{3}{2}R$.
Substituting the values $C_v = \frac{3}{2}R$ and $n = 3$ into the formula:
$C = \frac{3}{2}R + \frac{R}{1 - 3}$
$C = \frac{3}{2}R + \frac{R}{-2}$
$C = \frac{3}{2}R - \frac{1}{2}R = R$.
Therefore, the heat capacity of the gas during this process is $R$.

(D) Let the heat capacity of the bodies be $C(T)$,where $C(T)$ is an increasing function of temperature $T$.
When the two bodies are brought into contact,heat flows from the hotter body $(100^{\circ} C)$ to the colder body $(0^{\circ} C)$ until they reach a common final temperature $T_f$.
According to the principle of calorimetry,the heat lost by the hotter body equals the heat gained by the colder body:
$\int_{T_f}^{100} C(T) dT = \int_{0}^{T_f} C(T) dT$.
Since $C(T)$ increases with temperature,the heat capacity of the hotter body is higher than that of the colder body throughout the process of reaching equilibrium.
Because the hotter body has a higher heat capacity,it requires more energy to change its temperature by a certain amount compared to the colder body.
Therefore,the final equilibrium temperature $T_f$ will be closer to the initial temperature of the body with the higher heat capacity.
Thus,$T_f > 50^{\circ} C$.

(C) Let the lengths of the brass and steel rods at temperature $T$ be $L_1$ and $L_2$ respectively.
After a change in temperature $\Delta T$,the new lengths are:
$L_1' = l_1(1 + \alpha_1 \Delta T)$
$L_2' = l_2(1 + \alpha_2 \Delta T)$
Given that the difference $(L_2' - L_1')$ remains constant and equal to $(l_2 - l_1)$ at all temperatures:
$L_2' - L_1' = l_2 - l_1$
$l_2(1 + \alpha_2 \Delta T) - l_1(1 + \alpha_1 \Delta T) = l_2 - l_1$
$l_2 + l_2 \alpha_2 \Delta T - l_1 - l_1 \alpha_1 \Delta T = l_2 - l_1$
$(l_2 - l_1) + \Delta T(l_2 \alpha_2 - l_1 \alpha_1) = l_2 - l_1$
For this to hold for any $\Delta T$,the coefficient of $\Delta T$ must be zero:
$l_2 \alpha_2 - l_1 \alpha_1 = 0$
$\alpha_1 l_1 = \alpha_2 l_2$

(A) Let the initial volume be $V_1 = V$ and the final volume be $V_2 = V/2$.
In a $P-V$ diagram,the work done during a compression process is represented by the area under the curve.
For a given change in volume,the adiabatic curve is steeper than the isothermal curve.
Since the adiabatic curve lies above the isothermal curve for the same volume range during compression,the area under the adiabatic curve is greater than the area under the isothermal curve.
Therefore,compressing the gas through an adiabatic process requires more work to be done than compressing it isothermally.

(B) The potential energy of the ice at height $h$ is $PE = mgh$.
According to the problem, all this energy is converted into heat during the fall.
Only one-quarter of this heat is absorbed by the ice to melt it completely.
The heat required to melt a mass $m$ of ice is $Q = mL$, where $L$ is the latent heat of fusion.
Therefore, the energy balance equation is $\frac{1}{4} (mgh) = mL$.
Canceling $m$ from both sides, we get $\frac{gh}{4} = L$.
Solving for $h$, we get $h = \frac{4L}{g}$.
Substituting the given values: $h = \frac{4 \times 3.4 \times 10^{5}}{10} \text{ m}$.
$h = 4 \times 3.4 \times 10^{4} \text{ m} = 13.6 \times 10^{4} \text{ m} = 136,000 \text{ m}$.
Converting to kilometers, $h = 136 \text{ km}$.

(B) Given: Cold reservoir temperature $T_2 = 4^{\circ}C = 277 \, K$,Hot reservoir temperature $T_1 = 30^{\circ}C = 303 \, K$.
Heat removed per second $Q_2 = 600 \, cal/s$.
Coefficient of performance $\alpha = \frac{T_2}{T_1 - T_2}$.
$\alpha = \frac{277}{303 - 277} = \frac{277}{26}$.
We know that $\alpha = \frac{Q_2}{W}$,where $W$ is the work done per second (power).
Therefore,$W = \frac{Q_2}{\alpha} = \frac{600 \times 26}{277} \, cal/s$.
Converting to Joules: $W = \frac{600 \times 26}{277} \times 4.2 \, J/s$.
$W \approx 236.5 \, W$.

(B) The time period of a spring-block system is given by $T = 2 \pi \sqrt{\frac{m}{k}}$.
For a given spring,$T \propto \sqrt{m}$.
Therefore,$\frac{T_{1}}{T_{2}} = \sqrt{\frac{m_{1}}{m_{2}}}$.
Here,$T_{1} = 3 \ s$,$m_{1} = m$,$T_{2} = 5 \ s$,and $m_{2} = m + 1$.
Substituting the values: $\frac{3}{5} = \sqrt{\frac{m}{m+1}}$.
Squaring both sides: $\frac{9}{25} = \frac{m}{m+1}$.
Cross-multiplying: $9(m + 1) = 25m \Rightarrow 9m + 9 = 25m$.
$16m = 9 \Rightarrow m = \frac{9}{16} \ kg$.

(D) For an open organ pipe of length $L'$,the frequency of the $n^{th}$ harmonic is given by $f_n = n \frac{v}{2L'}$,where $n = 1, 2, 3, \dots$. The second overtone corresponds to the $3^{rd}$ harmonic $(n=3)$. Thus,$f_{open} = 3 \frac{v}{2L'}$.
For a closed organ pipe of length $L$,the frequency of the $n^{th}$ harmonic is given by $f_n = n \frac{v}{4L}$,where $n = 1, 3, 5, \dots$. The first overtone corresponds to the $3^{rd}$ harmonic $(n=3)$. Thus,$f_{closed} = 3 \frac{v}{4L}$.
According to the problem,the frequencies are equal:
$3 \frac{v}{2L'} = 3 \frac{v}{4L}$
Canceling $3v$ from both sides:
$\frac{1}{2L'} = \frac{1}{4L}$
Solving for $L'$:
$2L' = 4L \implies L' = 2L \; m$.

(B) The frequencies of the three sound waves are $f_1 = n - 1$,$f_2 = n$,and $f_3 = n + 1$.
Beats are produced by the superposition of waves with different frequencies.
The beat frequencies between pairs of waves are:
$|f_2 - f_1| = |n - (n - 1)| = 1 \text{ Hz}$
$|f_3 - f_2| = |(n + 1) - n| = 1 \text{ Hz}$
$|f_3 - f_1| = |(n + 1) - (n - 1)| = 2 \text{ Hz}$
The resultant beat frequency is the greatest common divisor or the effective frequency of the superposition,which is determined by the difference between the extreme frequencies.
The number of beats produced per second is the difference between the maximum and minimum frequencies: $(n + 1) - (n - 1) = 2 \text{ Hz}$.

(B) The frequency of the sound emitted by the siren is $f_0 = 800 \, Hz$.
The speed of the source (siren) is $v_s = 15 \, m s^{-1}$.
The speed of sound in air is $v = 330 \, m s^{-1}$.
Since the siren is moving towards the cliff,the cliff acts as an observer receiving the sound. The frequency $f'$ received by the cliff is given by the Doppler effect formula for a moving source and stationary observer:
$f' = f_0 \left( \frac{v}{v - v_s} \right) = 800 \left( \frac{330}{330 - 15} \right) = 800 \left( \frac{330}{315} \right) \approx 838.09 \, Hz$.
The cliff reflects this sound back to the original observer. Since the cliff is stationary,it acts as a stationary source emitting sound at frequency $f'$. The original observer is stationary relative to the cliff,so they hear the reflected sound at the same frequency $f'$.
Therefore,the frequency heard by the observer is approximately $838 \, Hz$.

(B) For an air column closed at one end,the resonance occurs at odd harmonics.
The fundamental frequency (first harmonic) corresponds to the smallest length $l_1$:
$l_1 = \frac{\lambda}{4} = 50\, cm$
The next resonance (third harmonic) occurs at length $l_2$:
$l_2 = \frac{3\lambda}{4}$
Substituting $\lambda = 4 \times 50\, cm = 200\, cm$:
$l_2 = 3 \times \left(\frac{200}{4}\right) = 3 \times 50 = 150\, cm$.

(A) The speed of a transverse wave on a string is given by $v = \sqrt{\frac{T}{\mu}}$,where $T$ is the tension and $\mu$ is the linear mass density.
Since the frequency $f$ of the pulse remains constant,the wavelength $\lambda = \frac{v}{f}$ is directly proportional to the velocity $v$. Therefore,$\lambda \propto \sqrt{T}$.
At the lower end of the rope (where the block $m_2$ is attached),the tension $T_1$ is due to the weight of the block: $T_1 = m_2 g$.
At the top of the rope,the tension $T_2$ is due to the weight of both the rope and the block: $T_2 = (m_1 + m_2) g$.
Thus,the ratio of the wavelengths is:
$\frac{\lambda_2}{\lambda_1} = \sqrt{\frac{T_2}{T_1}} = \sqrt{\frac{(m_1 + m_2) g}{m_2 g}} = \sqrt{\frac{m_1 + m_2}{m_2}}$.

(D) In an elastic collision between two bodies of equal mass,the velocities of the bodies are interchanged after the collision.
Given: Initial velocity of ball $A$ $(u_A)$ = $0.5 \, m s^{-1}$ and initial velocity of ball $B$ $(u_B)$ = $-0.3 \, m s^{-1}$.
Since the masses are identical and the collision is elastic,the final velocity of ball $A$ $(v_A)$ will be equal to $u_B$,and the final velocity of ball $B$ $(v_B)$ will be equal to $u_A$.
Therefore,$v_A = -0.3 \, m s^{-1}$ and $v_B = 0.5 \, m s^{-1}$.
The question asks for the velocities of $B$ and $A$ respectively,which are $v_B$ and $v_A$.
Thus,the velocities are $0.5 \, m s^{-1}$ and $-0.3 \, m s^{-1}$.

(A) The work done $W$ by a constant force $\vec{F}$ during a displacement $\vec{d}$ is given by the dot product: $W = \vec{F} \cdot \vec{d}$.
Here,the initial position is $\vec{r}_1 = -2\hat i + 5\hat j$ and the final position is $\vec{r}_2 = 4\hat j + 3\hat k$.
The displacement vector $\vec{d} = \vec{r}_2 - \vec{r}_1 = (4\hat j + 3\hat k) - (-2\hat i + 5\hat j) = 2\hat i - \hat j + 3\hat k$.
The force is $\vec{F} = 4\hat i + 3\hat j$.
Therefore,$W = (4\hat i + 3\hat j) \cdot (2\hat i - \hat j + 3\hat k) = (4 \times 2) + (3 \times -1) + (0 \times 3) = 8 - 3 = 5 \ J$.

(D) Given: Mass $m = 10\, g = 10^{-2}\, kg$,Radius $R = 6.4\, cm = 6.4 \times 10^{-2}\, m$,Final kinetic energy $K_f = 8 \times 10^{-4}\, J$,Initial kinetic energy $K_i = 0$.
The work done by the tangential force is equal to the change in kinetic energy,as the centripetal force does no work.
Work done $W = F_t \times d$,where $F_t = m a_t$ and $d$ is the distance covered in two revolutions.
Distance $d = 2 \times (2\pi R) = 4\pi R$.
Using the work-energy theorem: $W = K_f - K_i = K_f$.
$m a_t (4\pi R) = K_f$.
$a_t = \frac{K_f}{4\pi R m}$.
Substituting the values: $a_t = \frac{8 \times 10^{-4}}{4 \times 3.14159 \times 6.4 \times 10^{-2} \times 10^{-2}}$.
$a_t = \frac{8 \times 10^{-4}}{4 \times 3.14159 \times 6.4 \times 10^{-4}} = \frac{8}{25.6 \times 3.14159} \approx \frac{8}{80.42} \approx 0.0995\, m/s^2$.
Rounding to the nearest value,$a_t \approx 0.1\, m/s^2$.

(D) The variation of acceleration due to gravity $g$ with distance $d$ from the center of the earth is given by:
$1$. Inside the earth $(d < R)$:
$g = \frac{GM}{R^3} d$
Since $G, M,$ and $R$ are constants,$g \propto d$. This represents a straight line passing through the origin.
$2$. At the surface of the earth $(d = R)$:
$g = \frac{GM}{R^2} = g_s$ (maximum value).
$3$. Outside the earth $(d > R)$:
$g = \frac{GM}{d^2}$
Here,$g \propto \frac{1}{d^2}$. This represents a rectangular hyperbola.
Combining these,the graph shows a linear increase from the center to the surface,followed by a hyperbolic decrease as distance increases beyond the surface. This matches the graph in option $D$.

(C) The given circuit consists of an $OR$ gate followed by an $AND$ gate. The output of the $OR$ gate is $(A + B)$. This output is then fed into an $AND$ gate along with input $C$. Thus,the final Boolean expression for the output $Y$ is $Y = (A + B) \cdot C$.
To get an output $Y = 1$,both inputs to the $AND$ gate must be $1$. This means $(A + B) = 1$ and $C = 1$.
For $(A + B) = 1$,at least one of $A$ or $B$ must be $1$.
Checking the options:
- For option $A$: $A=0, B=1, C=0 \implies Y = (0+1) \cdot 0 = 0$.
- For option $B$: $A=1, B=0, C=0 \implies Y = (1+0) \cdot 0 = 0$.
- For option $C$: $A=1, B=0, C=1 \implies Y = (1+0) \cdot 1 = 1$.
- For option $D$: $A=1, B=1, C=0 \implies Y = (1+1) \cdot 0 = 0$.
Thus,the correct input is $A = 1, B = 0, C = 1$.

(A) The capacitor can be modeled as a combination of four smaller capacitors. The top portion of thickness $d/2$ is divided into three parts,each of area $A/3$. These three capacitors $(C_1, C_2, C_3)$ are in parallel.
$C_1 = \frac{\epsilon_0 K_1 (A/3)}{d/2} = \frac{2\epsilon_0 K_1 A}{3d}$
$C_2 = \frac{2\epsilon_0 K_2 A}{3d}$
$C_3 = \frac{2\epsilon_0 K_3 A}{3d}$
Their equivalent capacitance $C_p = C_1 + C_2 + C_3 = \frac{2\epsilon_0 A}{3d}(K_1 + K_2 + K_3)$.
This parallel combination is in series with the bottom capacitor $C_4$ of thickness $d/2$ and area $A$.
$C_4 = \frac{\epsilon_0 K_4 A}{d/2} = \frac{2\epsilon_0 K_4 A}{d}$.
The total equivalent capacitance $C$ is given by $\frac{1}{C} = \frac{1}{C_p} + \frac{1}{C_4}$.
Substituting the values: $\frac{1}{C} = \frac{3d}{2\epsilon_0 A(K_1 + K_2 + K_3)} + \frac{d}{2\epsilon_0 K_4 A}$.
For a single dielectric $K$,$C = \frac{K\epsilon_0 A}{d}$,so $\frac{1}{C} = \frac{d}{K\epsilon_0 A}$.
Equating the two: $\frac{d}{K\epsilon_0 A} = \frac{d}{\epsilon_0 A} [\frac{3}{2(K_1 + K_2 + K_3)} + \frac{1}{2K_4}]$.
$\frac{1}{K} = \frac{3}{2(K_1 + K_2 + K_3)} + \frac{1}{2K_4}$.
Multiplying by $2$: $\frac{2}{K} = \frac{3}{K_1 + K_2 + K_3} + \frac{1}{K_4}$.

(D) Given: $\theta = 30^\circ$,$E = 2 \times 10^5 \, \text{NC}^{-1}$,$\tau = 4 \, \text{Nm}$,and $l = 2 \, \text{cm} = 0.02 \, \text{m}$.
The torque experienced by an electric dipole in an electric field is given by the formula: $\tau = pE \sin \theta$,where $p = ql$.
Substituting $p = ql$ into the torque equation: $\tau = qlE \sin \theta$.
Rearranging to solve for $q$: $q = \frac{\tau}{El \sin \theta}$.
Substituting the given values: $q = \frac{4}{2 \times 10^5 \times 0.02 \times \sin(30^\circ)}$.
Since $\sin(30^\circ) = 0.5$: $q = \frac{4}{2 \times 10^5 \times 0.02 \times 0.5} = \frac{4}{2 \times 10^3} = 2 \times 10^{-3} \, \text{C}$.
Therefore,$q = 2 \, \text{mC}$.

(B) From the equilibrium condition of the spheres,the forces acting are tension $T$,weight $mg$,and electrostatic repulsion $F_e = \frac{kq^2}{x^2}$.
Resolving forces: $T \cos \theta = mg$ and $T \sin \theta = \frac{kq^2}{x^2}$.
Dividing the equations,we get $\tan \theta = \frac{kq^2}{x^2 mg}$.
Since $\theta$ is small,$\tan \theta \approx \sin \theta = \frac{x/2}{l} = \frac{x}{2l}$.
Equating the two expressions for $\tan \theta$: $\frac{x}{2l} = \frac{kq^2}{x^2 mg} \implies q^2 = \frac{mg}{2lk} x^3 \implies q \propto x^{3/2}$.
Differentiating with respect to time $t$: $\frac{dq}{dt} \propto \frac{3}{2} x^{1/2} \frac{dx}{dt}$.
Given that $\frac{dq}{dt}$ is constant,we have $1 \propto x^{1/2} v$,which implies $v \propto x^{-1/2}$.

(C) Initially,the energy stored in the $2\,\mu F$ capacitor is:
$U_{i} = \frac{1}{2} C V^{2} = \frac{1}{2} (2 \times 10^{-6}) V^{2} = V^{2} \times 10^{-6} \, J$
Initially,the charge stored in the $2\,\mu F$ capacitor is:
$Q_{i} = C V = (2 \times 10^{-6}) V = 2V \times 10^{-6} \, C$
When the switch $S$ is turned to position $2,$ the charge flows and both capacitors share charges until a common potential $V_{c}$ is reached.
$V_{c} = \frac{\text{Total Charge}}{\text{Total Capacitance}} = \frac{2V \times 10^{-6}}{(2 + 8) \times 10^{-6}} = \frac{V}{5} \, V$
Finally,the energy stored in both capacitors is:
$U_{f} = \frac{1}{2} (C_{1} + C_{2}) V_{c}^{2} = \frac{1}{2} (10 \times 10^{-6}) \left(\frac{V}{5}\right)^{2} = 5 \times 10^{-6} \times \frac{V^{2}}{25} = \frac{V^{2}}{5} \times 10^{-6} \, J$
Percentage loss of energy is given by:
$\Delta U \% = \frac{U_{i} - U_{f}}{U_{i}} \times 100 \%$
$\Delta U \% = \frac{V^{2} \times 10^{-6} - \frac{V^{2}}{5} \times 10^{-6}}{V^{2} \times 10^{-6}} \times 100 \% = \left(1 - \frac{1}{5}\right) \times 100 \% = \frac{4}{5} \times 100 \% = 80 \%$

(B) To find the potential difference $(V_A - V_B)$,we apply Kirchhoff's voltage law along the path from $A$ to $B$.
Starting from point $A$ and moving towards $B$ in the direction of the current $I = 2 \, A$:
$V_A - I R_1 - E - I R_2 = V_B$
Where $R_1 = 2 \, \Omega$,$E = 3 \, V$,and $R_2 = 1 \, \Omega$.
Substituting the values:
$V_A - (2 \, A \times 2 \, \Omega) - 3 \, V - (2 \, A \times 1 \, \Omega) = V_B$
$V_A - 4 \, V - 3 \, V - 2 \, V = V_B$
$V_A - 9 \, V = V_B$
Therefore,$V_A - V_B = 9 \, V$.

(A) First,calculate the resistance of the bulb $(R_B)$:
$R_B = \frac{V^2}{P} = \frac{(100)^2}{500} = \frac{10000}{500} = 20 \, \Omega$.
Next,calculate the current $(I)$ flowing through the bulb when it operates at its rated power:
$I = \frac{P}{V} = \frac{500}{100} = 5 \, A$.
Since the bulb and resistor $R$ are in series,the same current $I = 5 \, A$ flows through the resistor $R$. The voltage drop across the resistor $(V_R)$ is the difference between the supply voltage and the bulb's rated voltage:
$V_R = V_{supply} - V_{bulb} = 230 \, V - 100 \, V = 130 \, V$.
Using Ohm's law for the resistor $R$:
$V_R = I \times R$
$130 = 5 \times R$
$R = \frac{130}{5} = 26 \, \Omega$.

(C) Let the emfs of the two cells be $\varepsilon_{1}$ and $\varepsilon_{2}$ (where $\varepsilon_{1} > \varepsilon_{2}$).
Let $k$ be the potential gradient (potential difference per unit length) of the potentiometer wire.
When the cells are connected in series to support each other,the total emf is $\varepsilon_{1} + \varepsilon_{2}$. The balance point is at $l_{1} = 50 \, cm$.
So,$\varepsilon_{1} + \varepsilon_{2} = k \cdot l_{1} = 50k$ .....$(i)$
When the cells are connected in opposite directions,the total emf is $\varepsilon_{1} - \varepsilon_{2}$. The balance point is at $l_{2} = 10 \, cm$.
So,$\varepsilon_{1} - \varepsilon_{2} = k \cdot l_{2} = 10k$ .....$(ii)$
Adding equations $(i)$ and $(ii)$:
$2\varepsilon_{1} = 60k \Rightarrow \varepsilon_{1} = 30k$
Subtracting equation $(ii)$ from $(i)$:
$2\varepsilon_{2} = 40k \Rightarrow \varepsilon_{2} = 20k$
The ratio of the emfs is $\frac{\varepsilon_{1}}{\varepsilon_{2}} = \frac{30k}{20k} = \frac{3}{2}$.

(D) Given,$Q = at - bt^2$.
The current $I$ is given by $I = \frac{dQ}{dt} = a - 2bt$.
The current becomes zero when $a - 2bt = 0$,which gives $t = \frac{a}{2b}$.
The total heat $H$ produced in the resistance $R$ is given by $H = \int_0^{a/2b} I^2 R dt$.
Substituting $I = a - 2bt$:
$H = R \int_0^{a/2b} (a - 2bt)^2 dt = R \int_0^{a/2b} (a^2 + 4b^2 t^2 - 4abt) dt$.
Integrating with respect to $t$:
$H = R \left[ a^2 t + \frac{4b^2 t^3}{3} - 2abt^2 \right]_0^{a/2b}$.
Substituting the limits:
$H = R \left[ a^2 \left( \frac{a}{2b} \right) + \frac{4b^2}{3} \left( \frac{a^3}{8b^3} \right) - 2ab \left( \frac{a^2}{4b^2} \right) \right]$.
$H = R \left[ \frac{a^3}{2b} + \frac{a^3}{6b} - \frac{a^3}{2b} \right] = \frac{a^3 R}{6b}$.

(C) Given: Magnetic field $B = 3.57 \times 10^{-2} \, T$ and specific charge $\frac{e}{m} = 1.76 \times 10^{11} \, C/kg$.
The frequency of revolution $f$ of a charged particle in a magnetic field is given by the formula:
$f = \frac{1}{T} = \frac{qB}{2 \pi m}$
Substituting the given values:
$f = \frac{1}{2 \times 3.14} \times (1.76 \times 10^{11}) \times (3.57 \times 10^{-2})$
$f = \frac{1}{6.28} \times 6.2832 \times 10^9$
$f \approx 1 \times 10^9 \, Hz = 1 \, GHz$.

(B) The force on a current-carrying wire near a long straight conductor is given by $F = \frac{{\mu _0}IiL}{2\pi r}$.
For the side of the loop nearest to the wire (distance $r_1 = L/2$),the current flows in the same direction as the wire $XY$. The force $F_1$ is attractive:
$F_1 = \frac{{\mu _0}IiL}{2\pi (L/2)} = \frac{{\mu _0}Ii}{\pi}$.
For the side of the loop farthest from the wire (distance $r_2 = L/2 + L = 3L/2$),the current flows in the opposite direction to the wire $XY$. The force $F_2$ is repulsive:
$F_2 = \frac{{\mu _0}IiL}{2\pi (3L/2)} = \frac{{\mu _0}Ii}{3\pi}$.
The forces on the top and bottom sides of the loop are equal and opposite,so they cancel out.
The net force is $F_{net} = F_1 - F_2 = \frac{{\mu _0}Ii}{\pi} - \frac{{\mu _0}Ii}{3\pi} = \frac{{\mu _0}Ii}{\pi} (1 - 1/3) = \frac{2{\mu _0}Ii}{3\pi}$.

(B) For a long straight wire of radius $a$ carrying a steady current $I$ uniformly distributed over its cross-section:
$1$. The magnetic field $B$ at a distance $r$ inside the wire $(r < a)$ is given by $B = \frac{\mu_0 I r}{2 \pi a^2}$.
For $r = \frac{a}{2}$,we have $B = \frac{\mu_0 I (a/2)}{2 \pi a^2} = \frac{\mu_0 I}{4 \pi a}$.
$2$. The magnetic field $B'$ at a distance $r$ outside the wire $(r > a)$ is given by $B' = \frac{\mu_0 I}{2 \pi r}$.
For $r = 2a$,we have $B' = \frac{\mu_0 I}{2 \pi (2a)} = \frac{\mu_0 I}{4 \pi a}$.
$3$. The ratio of the magnetic fields is $\frac{B}{B'} = \frac{\frac{\mu_0 I}{4 \pi a}}{\frac{\mu_0 I}{4 \pi a}} = 1$.

(D) At equilibrium,the initial potential energy of the dipole is $U_{i} = -MB_{H} \cos 0^{\circ} = -MB_{H}$.
The final potential energy of the dipole after rotating by $60^{\circ}$ is $U_{f} = -MB_{H} \cos 60^{\circ} = -\frac{MB_{H}}{2}$.
The work done $W$ is the change in potential energy: $W = U_{f} - U_{i} = -\frac{MB_{H}}{2} - (-MB_{H}) = \frac{MB_{H}}{2}$.
From this,we get $MB_{H} = 2W$.
The torque required to keep the magnet in this new position is $\tau = MB_{H} \sin 60^{\circ}$.
Substituting $MB_{H} = 2W$,we get $\tau = (2W) \times \frac{\sqrt{3}}{2} = \sqrt{3} W$.

(D) Magnetic susceptibility,denoted by $\chi$,is a measure of how much a material will become magnetized in an applied magnetic field.
For diamagnetic materials,the magnetic susceptibility $\chi$ is always negative,which means they are weakly repelled by a magnetic field.
For paramagnetic materials,$\chi$ is small and positive.
For ferromagnetic materials,$\chi$ is large and positive.
Therefore,the magnetic susceptibility is negative only for diamagnetic materials.

(A) Given: Angle of incidence $i = 45^o$,Angle of prism $A = 60^o$.
Since the ray undergoes minimum deviation,the angle of emergence $e$ is equal to the angle of incidence $i$,so $e = 45^o$.
The angle of minimum deviation $\delta_m$ is given by the formula: $\delta_m = i + e - A$.
Substituting the values: $\delta_m = 45^o + 45^o - 60^o = 30^o$.
The refractive index $\mu$ is given by: $\mu = \frac{\sin((A + \delta_m)/2)}{\sin(A/2)}$.
Substituting the values: $\mu = \frac{\sin((60^o + 30^o)/2)}{\sin(60^o/2)} = \frac{\sin(45^o)}{\sin(30^o)}$.
Calculating the values: $\mu = \frac{1/\sqrt{2}}{1/2} = \frac{2}{\sqrt{2}} = \sqrt{2}$.
Thus,the angle of minimum deviation is $30^o$ and the refractive index is $\sqrt{2}$.

(C) Given: Focal length of objective $f_o = 40\, cm$,focal length of eyepiece $f_e = 4\, cm$,and object distance $u_o = -200\, cm$.
For the objective lens,using the lens formula $\frac{1}{v_o} - \frac{1}{u_o} = \frac{1}{f_o}$:
$\frac{1}{v_o} - \frac{1}{-200} = \frac{1}{40}$
$\frac{1}{v_o} = \frac{1}{40} - \frac{1}{200} = \frac{5-1}{200} = \frac{4}{200} = \frac{1}{50}$
Thus,$v_o = 50\, cm$.
For the final image to be formed at infinity (normal adjustment),the image formed by the objective must lie at the principal focus of the eyepiece. Therefore,the distance between the lenses is $l = v_o + f_e$.
$l = 50\, cm + 4\, cm = 54\, cm$.

(D) The magnification for a spherical mirror is given by $m = -v/u$.
$1$. For $m = -2$: Since $m$ is negative,the image is real. $A$ real image is always inverted and formed by a concave mirror. Thus,$(1-b, c)$.
$2$. For $m = -1/2$: Since $m$ is negative,the image is real. $A$ real image is formed by a concave mirror. Thus,$(2-b, c)$.
$3$. For $m = +2$: Since $m$ is positive,the image is virtual. $A$ virtual image with magnification greater than $1$ is formed by a concave mirror. Thus,$(3-b, d)$.
$4$. For $m = +1/2$: Since $m$ is positive,the image is virtual. $A$ virtual image with magnification less than $1$ is formed by a convex mirror. Thus,$(4-a, d)$.
Therefore,the correct matching is $(1-b, c), (2-b, c), (3-b, d), (4-a, d)$.

(B) Given: Refractive index of glass $\mu_g = 3/2$,refractive index of water $\mu_w = 4/3$.
The focal length $f$ of an equiconvex glass lens is given by the lens maker's formula:
$\frac{1}{f} = (\mu_g - 1) \left( \frac{1}{R} - \frac{1}{-R} \right) = (\frac{3}{2} - 1) \frac{2}{R} = \frac{1}{2} \cdot \frac{2}{R} = \frac{1}{R}$.
Thus,$f = R$.
The water lens formed between the two convex lenses is a biconcave lens with radii of curvature $R$. Its focal length $f_w$ is given by:
$\frac{1}{f_w} = (\mu_w - 1) \left( -\frac{1}{R} - \frac{1}{R} \right) = (\frac{4}{3} - 1) \left( -\frac{2}{R} \right) = \frac{1}{3} \cdot (-\frac{2}{R}) = -\frac{2}{3R}$.
Since $R = f$,we have $\frac{1}{f_w} = -\frac{2}{3f}$.
The combination consists of two convex lenses and one concave water lens in contact. The equivalent focal length $f_{eq}$ is:
$\frac{1}{f_{eq}} = \frac{1}{f} + \frac{1}{f_w} + \frac{1}{f} = \frac{1}{f} - \frac{2}{3f} + \frac{1}{f} = \frac{3 - 2 + 3}{3f} = \frac{4}{3f}$.
Therefore,$f_{eq} = \frac{3f}{4}$.

(A) Let the thickness of the glass slab be $t$.
Let the air bubble be at a distance $x$ from one surface. Then its distance from the opposite surface is $(t - x)$.
Given the refractive index of the glass slab is $\mu = 1.5$.
The apparent depth of the bubble from the first surface is $d_1 = \frac{x}{\mu} = 5\, cm$.
So,$x = 5 \times 1.5 = 7.5\, cm$.
The apparent depth of the bubble from the second surface is $d_2 = \frac{t - x}{\mu} = 3\, cm$.
So,$t - x = 3 \times 1.5 = 4.5\, cm$.
Adding the two equations: $x + (t - x) = 7.5 + 4.5$.
$t = 12\, cm$.
Alternatively,the sum of apparent depths is $\frac{x}{\mu} + \frac{t - x}{\mu} = \frac{t}{\mu} = 5 + 3 = 8$.
Therefore,$t = 8 \times 1.5 = 12\, cm$.

(A) The person suffers from myopia (nearsightedness) because they cannot see objects beyond $400\, cm$.
To correct this, we need to place a lens such that an object at infinity $(\infty)$ forms a virtual image at the person's far point $(400\, cm)$.
Given: Object distance $u = -\infty$, Image distance $v = -400\, cm = -4\, m$.
Using the lens formula: $\frac{1}{f} = \frac{1}{v} - \frac{1}{u}$.
Substituting the values: $\frac{1}{f} = \frac{1}{-4} - \frac{1}{-\infty} = -0.25\, m^{-1}$.
Since the focal length $f$ is negative, the lens must be concave.
The power of the lens $P = \frac{1}{f} = -0.25\, D$.

(C) The kinetic energy $K$ of an electron with de-Broglie wavelength $\lambda$ is given by $K = \frac{p^2}{2m}$.
Since $p = \frac{h}{\lambda}$,we have $K = \frac{(h/\lambda)^2}{2m} = \frac{h^2}{2m\lambda^2}$.
In an $X$-ray tube,the maximum energy of an emitted $X$-ray photon is equal to the kinetic energy of the incident electron.
Therefore,$\frac{hc}{\lambda_0} = K = \frac{h^2}{2m\lambda^2}$.
Solving for $\lambda_0$,we get $\lambda_0 = \frac{hc \cdot 2m\lambda^2}{h^2} = \frac{2mc\lambda^2}{h}$.

(B) According to Einstein's photoelectric equation,the maximum kinetic energy of photoelectrons is given by:
$KE_{\max} = E - \phi$
For the first case,$E = 5\, eV$ and $KE_{\max} = 2\, eV$:
$2 = 5 - \phi \implies \phi = 3\, eV$
When photons of energy $E' = 6\, eV$ are incident,the new maximum kinetic energy is:
$KE'_{\max} = E' - \phi = 6 - 3 = 3\, eV$
For no photoelectrons to reach the anode $A,$ the potential of the anode relative to the cathode $(V_A - V_C)$ must be equal to the negative of the stopping potential $(V_s)$.
The stopping potential $V_s$ is defined by $e V_s = KE'_{\max} = 3\, eV,$ so $V_s = 3\, V$.
Since the anode must be at a negative potential relative to the cathode to stop the electrons,the potential difference $V_A - V_C = -3\, V$.

(D) For an electron of energy $E$,the de-Broglie wavelength is given by $\lambda_{e} = \frac{h}{p} = \frac{h}{\sqrt{2mE}}$.
For a photon of energy $E$,the relation is $E = \frac{hc}{\lambda_{p}}$,which implies $\lambda_{p} = \frac{hc}{E}$.
Taking the ratio of the wavelengths:
$\frac{\lambda_{e}}{\lambda_{p}} = \frac{h}{\sqrt{2mE}} \times \frac{E}{hc} = \frac{1}{c} \times \frac{E^{1}}{\sqrt{E}} \times \frac{1}{\sqrt{2m}} = \frac{1}{c} \sqrt{\frac{E}{2m}} = \frac{1}{C}(\frac{E}{2m})^{1/2}$.

(C) According to Einstein's photoelectric equation,the stopping potential $V_s$ is given by:
$e V_s = \frac{hc}{\lambda} - \frac{hc}{\lambda_0}$
where $\lambda_0$ is the threshold wavelength.
For the first case:
$eV = \frac{hc}{\lambda} - \frac{hc}{\lambda_0}$ ..... $(i)$
For the second case:
$e(\frac{V}{4}) = \frac{hc}{2\lambda} - \frac{hc}{\lambda_0}$ ..... $(ii)$
Multiply equation $(ii)$ by $4$:
$eV = \frac{4hc}{2\lambda} - \frac{4hc}{\lambda_0} = \frac{2hc}{\lambda} - \frac{4hc}{\lambda_0}$ ..... $(iii)$
Equating $(i)$ and $(iii)$:
$\frac{hc}{\lambda} - \frac{hc}{\lambda_0} = \frac{2hc}{\lambda} - \frac{4hc}{\lambda_0}$
$\frac{4hc}{\lambda_0} - \frac{hc}{\lambda_0} = \frac{2hc}{\lambda} - \frac{hc}{\lambda}$
$\frac{3hc}{\lambda_0} = \frac{hc}{\lambda}$
$\lambda_0 = 3\lambda$

(C) Given: Number of turns $N = 1000$,Current $I = 4\, A$,Magnetic flux per turn $\phi_{0} = 4 \times 10^{-3}\, Wb$.
The total magnetic flux $\phi$ linked with the solenoid is given by $\phi = N \phi_{0}$.
Substituting the values,$\phi = 1000 \times 4 \times 10^{-3}\, Wb = 4\, Wb$.
We know that the relation between flux and self-inductance $L$ is $\phi = L I$.
Therefore,the self-inductance $L = \frac{\phi}{I} = \frac{4\, Wb}{4\, A} = 1\, H$.

(B) According to Faraday's law of electromagnetic induction,the induced $e.m.f.$ is given by $\varepsilon = -\frac{d\phi}{dt}$.
For loop $1$ (radius $R > r$): The magnetic flux $\phi_1$ is linked only with the region of radius $r$ where the magnetic field exists. Thus,$\phi_1 = B \cdot A = B(\pi r^2)$.
The induced $e.m.f.$ is $\varepsilon_1 = -\frac{d}{dt}(B \pi r^2) = -\pi r^2 \frac{dB}{dt}$.
For loop $2$ (radius $R$): The loop is completely outside the region where the magnetic field exists. Therefore,the magnetic flux linked with loop $2$ is $\phi_2 = 0$.
The induced $e.m.f.$ is $\varepsilon_2 = -\frac{d\phi_2}{dt} = 0$.

(A) Given: $V_{R} = 80 \, V$,$V_{C} = 40 \, V$,$V_{L} = 100 \, V$.
The power factor of an $L-C-R$ circuit is given by $\cos \phi = \frac{R}{Z}$.
Since $V = IZ$ and $V_{R} = IR$,we can write $\cos \phi = \frac{V_{R}}{V}$.
The total voltage $V$ in an $L-C-R$ circuit is $V = \sqrt{V_{R}^{2} + (V_{L} - V_{C})^{2}}$.
Substituting the given values:
$V = \sqrt{80^{2} + (100 - 40)^{2}}$
$V = \sqrt{80^{2} + 60^{2}}$
$V = \sqrt{6400 + 3600} = \sqrt{10000} = 100 \, V$.
Now,calculate the power factor:
$\cos \phi = \frac{V_{R}}{V} = \frac{80}{100} = 0.8$.

(A) When an ideal capacitor is connected to an $AC$ voltage source,the current $I(t)$ leads the voltage $V(t)$ by a phase angle of $90^\circ$.
Since the energy stored in the capacitor during the charging phase is returned to the circuit during the discharging phase,the net energy consumed by an ideal capacitor over a complete cycle is zero.
Therefore,statement $A$ is the correct statement.

(D) Given: $L = 20 \, mH = 20 \times 10^{-3} \, H$,$C = 50 \, \mu F = 50 \times 10^{-6} \, F$,$R = 40 \, \Omega$,$V = 10 \, \sin \, 340t$.
Comparing with $V = V_0 \sin \omega t$,we get $V_0 = 10 \, V$ and $\omega = 340 \, rad/s$.
Inductive reactance $X_L = \omega L = 340 \times 20 \times 10^{-3} = 6.8 \, \Omega$.
Capacitive reactance $X_C = \frac{1}{\omega C} = \frac{1}{340 \times 50 \times 10^{-6}} = \frac{10^6}{17000} \approx 58.82 \, \Omega$.
Impedance $Z = \sqrt{R^2 + (X_C - X_L)^2} = \sqrt{40^2 + (58.82 - 6.8)^2} = \sqrt{1600 + (52.02)^2} = \sqrt{1600 + 2706.08} = \sqrt{4306.08} \approx 65.62 \, \Omega$.
Peak current $I_0 = \frac{V_0}{Z} = \frac{10}{65.62} \, A$.
Power factor $\cos \phi = \frac{R}{Z} = \frac{40}{65.62}$.
Power loss $P = V_{rms} I_{rms} \cos \phi = \frac{1}{2} V_0 I_0 \cos \phi = \frac{1}{2} \times 10 \times \frac{10}{65.62} \times \frac{40}{65.62} = \frac{2000}{4306} \approx 0.46 \, W$.

(C) Given: Resistance $R = 100 \, \Omega$,Capacitive reactance $X_C = 100 \, \Omega$,and $RMS$ voltage $V_{rms} = 220 \, V$.
The total impedance of the $RC$ series circuit is $Z = \sqrt{R^2 + X_C^2} = \sqrt{100^2 + 100^2} = 100\sqrt{2} \, \Omega$.
The peak voltage of the source is $V_0 = V_{rms} \sqrt{2} = 220\sqrt{2} \, V$.
The peak conduction current in the circuit is $I_0 = \frac{V_0}{Z} = \frac{220\sqrt{2}}{100\sqrt{2}} = 2.2 \, A$.
According to Maxwell's modification of Ampere's law,the displacement current $I_d$ between the plates of a capacitor is equal to the conduction current $I_c$ in the connecting wires.
Therefore,the peak value of the displacement current is equal to the peak value of the conduction current,which is $2.2 \, A$.

(C) According to Maxwell's theory,an accelerating charge produces time-varying electric and magnetic fields. These oscillating fields propagate through space as an electromagnetic wave. $A$ stationary charge produces only a static electric field,and a charge moving at a constant velocity produces a steady magnetic field along with an electric field,but neither generates a propagating electromagnetic wave.

(B) Given:
Width of aperture $a = 0.02\, cm = 2 \times 10^{-4}\, m$
Wavelength $\lambda = 5 \times 10^{-5}\, cm = 5 \times 10^{-7}\, m$
Focal length of lens $f = 60\, cm = 0.6\, m$. Since the screen is at the focal plane,the distance $D = f = 0.6\, m$.
For Fraunhofer diffraction at a single slit,the condition for the $n^{th}$ dark band (minima) is given by $a \sin \theta = n\lambda$.
For the first dark band,$n = 1$,so $\sin \theta \approx \theta = \frac{\lambda}{a}$.
The distance from the centre of the screen is $y_1 = D \theta = \frac{D\lambda}{a}$.
Substituting the values:
$y_1 = \frac{0.6 \times 5 \times 10^{-7}}{2 \times 10^{-4}} = \frac{3 \times 10^{-7}}{2 \times 10^{-4}} = 1.5 \times 10^{-3}\, m = 0.15\, cm$.

(C) For the first minimum in a single slit diffraction pattern,the condition is given by $a \sin \theta = n \lambda$. For the first minimum,$n = 1$,so $a \sin \theta = \lambda$.
Given $\theta = 30^{\circ}$,we have $\sin 30^{\circ} = \frac{1}{2}$.
Thus,$a (\frac{1}{2}) = \lambda$,which implies $a = 2 \lambda$ ..... $(i)$.
For the first secondary maximum,the condition is $a \sin \theta' = (n + \frac{1}{2}) \lambda$ where $n = 1$. So,$a \sin \theta' = \frac{3}{2} \lambda$.
Substituting $a = 2 \lambda$ from equation $(i)$ into this expression:
$(2 \lambda) \sin \theta' = \frac{3}{2} \lambda$.
Dividing both sides by $2 \lambda$,we get $\sin \theta' = \frac{3}{4}$.
Therefore,$\theta' = \sin^{-1} \left( \frac{3}{4} \right)$.

(C) Given: $d = 5\lambda$,$D = 10d$,and the position in front of one slit is $y = \frac{d}{2}$.
The path difference $\Delta x$ at position $y$ is given by $\Delta x = d \sin \theta \approx d \tan \theta = d \left( \frac{y}{D} \right)$.
Substituting the values: $\Delta x = d \left( \frac{d/2}{10d} \right) = \frac{d}{20} = \frac{5\lambda}{20} = \frac{\lambda}{4}$.
The phase difference $\phi$ is $\phi = \frac{2\pi}{\lambda} \Delta x = \frac{2\pi}{\lambda} \left( \frac{\lambda}{4} \right) = \frac{\pi}{2}$.
In a Young's double slit experiment,the resultant intensity $I_y$ is given by $I_y = I_1 + I_2 + 2\sqrt{I_1 I_2} \cos \phi$.
Since $I_1 = I_2 = I$,we have $I_{max} = I + I + 2\sqrt{I^2} = 4I = I_0$,so $I = \frac{I_0}{4}$.
Substituting $\phi = \frac{\pi}{2}$ into the intensity formula:
$I_y = I + I + 2I \cos(\frac{\pi}{2}) = 2I + 0 = 2I$.
Since $I = \frac{I_0}{4}$,the intensity $I_y = 2 \left( \frac{I_0}{4} \right) = \frac{I_0}{2}$.

(A) Let the intensities of the two sources be $I_1$ and $I_2$ such that $\frac{I_1}{I_2} = \alpha$. Since $I \propto A^2$,we have $\frac{A_1}{A_2} = \sqrt{\alpha}$.
The maximum and minimum intensities in an interference pattern are given by $I_{max} = (A_1 + A_2)^2$ and $I_{min} = (A_1 - A_2)^2$.
We need to calculate the value of $\frac{I_{max} - I_{min}}{I_{max} + I_{min}}$.
Substituting the expressions for $I_{max}$ and $I_{min}$:
$\frac{I_{max} - I_{min}}{I_{max} + I_{min}} = \frac{(A_1 + A_2)^2 - (A_1 - A_2)^2}{(A_1 + A_2)^2 + (A_1 - A_2)^2}$
Expanding the squares:
$= \frac{(A_1^2 + A_2^2 + 2A_1A_2) - (A_1^2 + A_2^2 - 2A_1A_2)}{(A_1^2 + A_2^2 + 2A_1A_2) + (A_1^2 + A_2^2 - 2A_1A_2)}$
$= \frac{4A_1A_2}{2(A_1^2 + A_2^2)} = \frac{2A_1A_2}{A_1^2 + A_2^2}$
Dividing the numerator and denominator by $A_2^2$:
$= \frac{2(A_1/A_2)}{(A_1/A_2)^2 + 1}$
Since $\frac{A_1}{A_2} = \sqrt{\alpha}$,we substitute this into the expression:
$= \frac{2\sqrt{\alpha}}{(\sqrt{\alpha})^2 + 1} = \frac{2\sqrt{\alpha}}{\alpha + 1}$.

(B) The wave number $\bar{\nu}$ for the hydrogen spectrum is given by the Rydberg formula: $\bar{\nu} = R \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right)$.
For the Balmer series,the transition ends at $n_1 = 2$.
The last line of the Balmer series corresponds to the transition from $n_2 = \infty$ to $n_1 = 2$.
Given $R = 10^7 \, m^{-1}$,we substitute the values:
$\bar{\nu} = 10^7 \left( \frac{1}{2^2} - \frac{1}{\infty^2} \right) = 10^7 \left( \frac{1}{4} - 0 \right) = \frac{10^7}{4} = 0.25 \times 10^7 \, m^{-1}$.

(A) The Rydberg formula for the wavelength of emitted radiation is given by $\frac{1}{\lambda} = R \left( \frac{1}{n_{f}^{2}} - \frac{1}{n_{i}^{2}} \right)$.
For the transition from $n_i = 3$ to $n_f = 2$:
$\frac{1}{\lambda} = R \left( \frac{1}{2^{2}} - \frac{1}{3^{2}} \right) = R \left( \frac{1}{4} - \frac{1}{9} \right) = R \left( \frac{9-4}{36} \right) = \frac{5R}{36}$.
For the transition from $n_i = 4$ to $n_f = 3$:
$\frac{1}{\lambda'} = R \left( \frac{1}{3^{2}} - \frac{1}{4^{2}} \right) = R \left( \frac{1}{9} - \frac{1}{16} \right) = R \left( \frac{16-9}{144} \right) = \frac{7R}{144}$.
Taking the ratio of the two equations:
$\frac{\lambda'}{\lambda} = \frac{5R/36}{7R/144} = \frac{5}{36} \times \frac{144}{7} = \frac{5 \times 4}{7} = \frac{20}{7}$.
Therefore,$\lambda' = \frac{20}{7}\lambda$.

(C) The distance of closest approach $(r_0)$ is the distance at which the initial kinetic energy of the $\alpha$-particle is completely converted into electrostatic potential energy.
Equating kinetic energy to potential energy:
$\frac{1}{2}mv^2 = \frac{1}{4\pi\varepsilon_0} \cdot \frac{(Ze)(2e)}{r_0}$
Solving for $r_0$:
$r_0 = \frac{2Ze^2}{4\pi\varepsilon_0 \cdot \frac{1}{2}mv^2} = \frac{Ze^2}{\pi\varepsilon_0 mv^2}$
Since $Z$,$e$,$\varepsilon_0$,and $v$ are constants for a given experiment,we have:
$r_0 \propto \frac{1}{m}$

(B) Let $N_{0}$ be the initial number of nuclei at time $t=0$.
$N_{1}$ is the number of remaining nuclei after $40\%$ decay:
$N_{1} = (1 - 0.40) N_{0} = 0.6 N_{0}$
$N_{2}$ is the number of remaining nuclei after $85\%$ decay:
$N_{2} = (1 - 0.85) N_{0} = 0.15 N_{0}$
Now,find the ratio of remaining nuclei:
$\frac{N_{2}}{N_{1}} = \frac{0.15 N_{0}}{0.6 N_{0}} = \frac{1}{4} = \left(\frac{1}{2}\right)^{2}$
Since $\frac{N_{2}}{N_{1}} = \left(\frac{1}{2}\right)^{n}$,where $n$ is the number of half-lives,we have $n = 2$.
Therefore,the time taken is $t = n \times T_{1/2} = 2 \times 30 \text{ minutes} = 60 \text{ minutes}$.

(D) Given: Collector resistance $R_C = 2 \, k\Omega = 2000 \, \Omega$,Output voltage $V_0 = 4 \, V$,Current amplification factor $\beta = 100$,Base resistance $R_B = 1 \, k\Omega = 1000 \, \Omega$.
The voltage gain $A_v$ of a $CE$ amplifier is given by the formula: $A_v = \beta \times \frac{R_C}{R_B}$.
Substituting the values: $A_v = 100 \times \frac{2000}{1000} = 200$.
We know that voltage gain $A_v = \frac{V_0}{V_i}$,where $V_i$ is the input signal voltage.
Therefore,$V_i = \frac{V_0}{A_v} = \frac{4 \, V}{200} = 0.02 \, V$.
Converting to millivolts $(mV)$: $V_i = 0.02 \times 1000 \, mV = 20 \, mV$.

(C) In the given circuit,diode $D_1$ is reverse biased because its p-terminal is at a lower potential than its n-terminal,so it acts as an open circuit and blocks the current.
Diode $D_2$ is forward biased because its p-terminal is at a higher potential than its n-terminal,so it acts as a closed switch (ideal diode).
Hence,the equivalent circuit consists of the $10 \ V$ battery,resistance $R_1 = 2 \ \Omega$,and resistance $R_3 = 2 \ \Omega$ in series.
The total resistance of the circuit is $R_{eq} = R_1 + R_3 = 2 \ \Omega + 2 \ \Omega = 4 \ \Omega$.
The current flowing through the resistance $R_1$ is given by $I = \frac{V}{R_{eq}} = \frac{10 \ V}{4 \ \Omega} = 2.5 \ A$.

(A) The $p-n$ junction diode is forward biased because the $p$-terminal is at a higher potential $(+4\;V)$ than the $n$-terminal $(-6\;V)$.
An ideal diode in forward bias acts as a short circuit (zero resistance).
Therefore,the current $I_{AB}$ flowing through the circuit is given by Ohm's law:
$I_{AB} = \frac{V_A - V_B}{R} = \frac{4\;V - (-6\;V)}{1\;k\Omega} = \frac{10\;V}{1000\;\Omega} = 10^{-2}\;A$.

(D) Given: Load resistance $R_L = 800 \,\,\Omega$,Input resistance $R_i = 192 \,\,\Omega$,Current amplification factor $\beta = 0.96$.
The voltage gain $(A_v)$ for a common emitter amplifier is given by the formula:
$A_v = \beta \times \frac{R_L}{R_i}$
Substituting the values:
$A_v = 0.96 \times \frac{800}{192} = 0.96 \times 4.166... = 4$.
The power gain $(A_p)$ is given by the product of current gain and voltage gain:
$A_p = \beta \times A_v$
Substituting the values:
$A_p = 0.96 \times 4 = 3.84$.
Thus,the voltage gain is $4$ and the power gain is $3.84$.

(C) The given circuit consists of two $NAND$ gates. Let the output of the first $NAND$ gate be $X$. The inputs to this gate are $A$ and $B$. Thus, $X = \overline{A \cdot B}$.
The second $NAND$ gate takes $X$ and $C$ as inputs. Thus, the final output $Y = \overline{X \cdot C} = \overline{(\overline{A \cdot B}) \cdot C}$.
Case $1$: When $A = 0, B = 0, C = 0$:
$X = \overline{0 \cdot 0} = \overline{0} = 1$.
$Y = \overline{1 \cdot 0} = \overline{0} = 1$.
Case $2$: When $A = 1, B = 1, C = 1$:
$X = \overline{1 \cdot 1} = \overline{1} = 0$.
$Y = \overline{0 \cdot 1} = \overline{0} = 1$.
Wait, re-evaluating the circuit diagram: The first gate is a $NAND$ gate, and the second gate is also a $NAND$ gate. Let's re-check the logic.
For $A=0, B=0, C=0$: $X = \overline{0 \cdot 0} = 1$. Then $Y = \overline{1 \cdot 0} = 1$.
For $A=1, B=1, C=1$: $X = \overline{1 \cdot 1} = 0$. Then $Y = \overline{0 \cdot 1} = 1$.
Looking at the provided solution image, the first gate is actually an $AND$ gate. Let's re-read the diagram. The first gate is an $AND$ gate, and the second is a $NAND$ gate.
If first is $AND$: $X = A \cdot B$. Then $Y = \overline{X \cdot C} = \overline{(A \cdot B) \cdot C}$.
Case $1$: $A=0, B=0, C=0 \implies X = 0 \cdot 0 = 0 \implies Y = \overline{0 \cdot 0} = 1$.
Case $2$: $A=1, B=1, C=1 \implies X = 1 \cdot 1 = 1 \implies Y = \overline{1 \cdot 1} = 0$.
Thus, the outputs are $1, 0$. The correct option is $C$.

(D) For better tuning of an $L-C-R$ circuit used in communication,the quality factor $Q$ must be high.
The formula for the quality factor is $Q = \frac{1}{R} \sqrt{\frac{L}{C}}$.
To maximize $Q$,we need a small resistance $R$,a large inductance $L$,and a small capacitance $C$.
Comparing the given options:
$A: R=20, L=1.5, C=35$
$B: R=25, L=2.5, C=45$
$C: R=25, L=1.5, C=45$
$D: R=15, L=3.5, C=30$
Option $D$ has the minimum resistance $(15 \Omega)$,the maximum inductance $(3.5 \text{ H})$,and the minimum capacitance $(30 \mu\text{F})$.
Therefore,the combination in option $D$ provides the highest quality factor,resulting in better tuning.