IIT JEE 2019 Physics Question Paper with Answer and Solution

(A) Given power $P = \frac{dQ}{dt}$.
Heat capacity $S = \frac{dQ}{dT} = \frac{dQ/dt}{dT/dt} = \frac{P}{dT/dt}$.
Given $T(t) = T_0(1 + \beta t^{1/4})$.
Differentiating with respect to $t$:
$\frac{dT}{dt} = T_0 \cdot \beta \cdot \frac{1}{4} t^{-3/4} = \frac{\beta T_0}{4} t^{-3/4}$.
Substituting this into the expression for $S$:
$S = \frac{P}{(\beta T_0 / 4) t^{-3/4}} = \frac{4P}{\beta T_0} t^{3/4}$.
From the given equation,$\beta t^{1/4} = \frac{T(t) - T_0}{T_0}$.
Therefore,$t^{1/4} = \frac{T(t) - T_0}{\beta T_0}$.
Raising both sides to the power of $3$:
$t^{3/4} = \left( \frac{T(t) - T_0}{\beta T_0} \right)^3$.
Substituting $t^{3/4}$ back into the expression for $S$:
$S = \frac{4P}{\beta T_0} \cdot \frac{(T(t) - T_0)^3}{\beta^3 T_0^3} = \frac{4P(T(t) - T_0)^3}{\beta^4 T_0^4}$.

(D) Let $M$ be the total mass enclosed within a sphere of radius $r$.
For a particle of mass $m$ moving in a circular orbit of radius $r$,the gravitational force provides the necessary centripetal force:
$\frac{GMm}{r^2} = \frac{mv^2}{r}$
Since the kinetic energy $K = \frac{1}{2}mv^2$,we have $mv^2 = 2K$. Substituting this into the force equation:
$\frac{GMm}{r^2} = \frac{2K}{r} \Rightarrow M = \frac{2Kr}{Gm}$
Differentiating both sides with respect to $r$ to find the mass $dM$ in a shell of thickness $dr$:
$dM = \frac{2K}{Gm} dr$
Also,the mass of the shell is $dM = \rho(r) \cdot 4 \pi r^2 dr$. Equating the two expressions for $dM$:
$4 \pi r^2 \rho(r) dr = \frac{2K}{Gm} dr$
Solving for density $\rho(r)$:
$\rho(r) = \frac{2K}{4 \pi r^2 Gm} = \frac{K}{2 \pi r^2 Gm}$
The particle number density $n(r)$ is given by $\rho(r) / m$:
$n(r) = \frac{K}{2 \pi r^2 m^2 G}$

(B) From the $V-T$ diagram:
Process $1 \rightarrow 2$: $V$ increases linearly with $T$ passing through origin,so $V \propto T$,which means pressure $P$ is constant (isobaric). $P_1 = P_2 = \frac{RT_0}{V_0}$.
Process $2 \rightarrow 3$: $V$ is constant $(2V_0)$,so it is isochoric.
Process $3 \rightarrow 4$: $V$ is constant $(V_0)$,so it is isochoric.
Process $4 \rightarrow 1$: $V$ is constant $(V_0)$,so it is isochoric.
Wait,looking at the graph: $1 \rightarrow 2$ is isobaric,$2 \rightarrow 3$ is isochoric,$3 \rightarrow 4$ is isobaric,$4 \rightarrow 1$ is isochoric.
Work done $W = \text{Area of cycle in } P-V \text{ diagram}$.
$P_1 = \frac{RT_0}{V_0}$,$P_3 = \frac{RT_0}{V_0}$ (Wait,$P_3 = \frac{R(T_0/2)}{V_0} = \frac{RT_0}{2V_0}$).
$W = (P_{12} - P_{34}) \Delta V = (\frac{RT_0}{V_0} - \frac{RT_0}{2V_0}) (2V_0 - V_0) = \frac{RT_0}{2V_0} \cdot V_0 = \frac{1}{2} RT_0$. Statement $(1)$ is correct.
$Q_{1 \rightarrow 2} = n C_p \Delta T = 1 \cdot \frac{5R}{2} \cdot (2T_0 - T_0) = \frac{5}{2} RT_0$.
$Q_{2 \rightarrow 3} = n C_v \Delta T = 1 \cdot \frac{3R}{2} \cdot (T_0 - 2T_0) = -\frac{3}{2} RT_0$. Ratio $|Q_{1 \rightarrow 2} / Q_{2 \rightarrow 3}| = 5/3$. Statement $(2)$ is correct.
Statement $(3)$ is incorrect as it involves isobaric processes.
$Q_{3 \rightarrow 4} = n C_p \Delta T = 1 \cdot \frac{5R}{2} \cdot (T_0/2 - T_0) = -\frac{5}{4} RT_0$. Ratio $|Q_{1 \rightarrow 2} / Q_{3 \rightarrow 4}| = |(5/2) / (-5/4)| = 2$. Statement $(4)$ is incorrect.

(A) Given that mass $(M)$ and angular momentum $(L_{ang} = Mvr)$ are dimensionless,we have:
$M = M^0 L^0 T^0$
$L_{ang} = M^1 L^2 T^{-1} = M^0 L^0 T^0$
Since $M$ is dimensionless,$L^2 T^{-1} = 1$,which implies $T = L^2$.
$(1)$ Force $(F = M L T^{-2})$: Since $M$ is dimensionless and $T = L^2$,$F = L^1 (L^2)^{-2} = L^1 L^{-4} = L^{-3}$. (Correct)
$(2)$ Energy $(E = M L^2 T^{-2})$: $E = L^2 (L^2)^{-2} = L^2 L^{-4} = L^{-2}$. (Correct)
$(3)$ Power $(P = E/T = M L^2 T^{-3})$: $P = L^2 (L^2)^{-3} = L^2 L^{-6} = L^{-4}$. (Incorrect)
$(4)$ Linear momentum $(p = M L T^{-1})$: $p = L^1 (L^2)^{-1} = L^1 L^{-2} = L^{-1}$. (Correct)
Thus,statements $(1), (2),$ and $(4)$ are correct.

(B) The capillary rise is given by $h = \frac{2T \cos \theta}{\rho g R}$.
For $T_1$ $(\theta = 0^{\circ})$: $h_1 = \frac{2 \times 0.075 \times \cos 0^{\circ}}{1000 \times 10 \times 0.2 \times 10^{-3}} = 0.075 \ m = 7.5 \ cm$.
For $T_2$ $(\theta = 60^{\circ})$: $h_2 = \frac{2 \times 0.075 \times \cos 60^{\circ}}{1000 \times 10 \times 0.2 \times 10^{-3}} = 0.0375 \ m = 3.75 \ cm$.
$(1)$ Since the contact angles are different,the meniscus shape and the weight of the water in the meniscus will be different for both cases. Thus,the correction will be different. Statement $(1)$ is correct.
$(2)$ In case $I$,$T_1$ is at the bottom. The water rises to $7.5 \ cm$. If the joint is at $5 \ cm$,the water crosses the joint. However,the pressure balance at the interface in $T_2$ would require a height $h'$ such that $\rho g(5 \times 10^{-2} + h') = \frac{2T \cos 60^{\circ}}{R}$. This leads to $h' = 3.75 - 5 = -1.25 \ cm$. Since $h' < 0$,the water cannot rise further into $T_2$. It stays at the joint. Statement $(2)$ is incorrect.
$(3)$ In case $I$,if the joint is at $8 \ cm$,the water rises in $T_1$ to its maximum height of $7.5 \ cm$. Since $7.5 \ cm < 8 \ cm$,it does not reach the joint. Statement $(3)$ is correct.
$(4)$ In case $II$,$T_2$ is at the bottom. The water rises in $T_2$ to its maximum height of $3.75 \ cm$. Since $3.75 \ cm < 5 \ cm$,it does not reach the joint. Statement $(4)$ is correct.

(C) Let $T_S$ be the tension in the steel wire and $T_C$ be the tension in the copper wire.
Resolving forces in the horizontal $(x)$ direction:
$T_C \cos 30^{\circ} = T_S \cos 60^{\circ}$
$T_C \times \frac{\sqrt{3}}{2} = T_S \times \frac{1}{2}$
$T_S = \sqrt{3} T_C \quad \dots (i)$
Resolving forces in the vertical $(y)$ direction:
$T_C \sin 30^{\circ} + T_S \sin 60^{\circ} = 100$
$\frac{T_C}{2} + \frac{T_S \sqrt{3}}{2} = 100 \quad \dots (ii)$
Substituting $(i)$ into $(ii)$:
$\frac{T_C}{2} + \frac{(\sqrt{3} T_C) \sqrt{3}}{2} = 100$
$\frac{T_C}{2} + \frac{3 T_C}{2} = 100 \implies 2 T_C = 100 \implies T_C = 50 \ N$
$T_S = \sqrt{3} \times 50 = 50\sqrt{3} \ N$
Using the formula for elongation $\Delta \ell = \frac{FL}{AY}$:
$\frac{\Delta \ell_C}{\Delta \ell_S} = \frac{T_C L_C}{A_C Y_C} \times \frac{A_S Y_S}{T_S L_S}$
Given $A_C = A_S = 0.5 \ cm^2$,$L_C = \sqrt{3} \ m$,$L_S = 1 \ m$,$Y_C = 1 \times 10^{11} \ N/m^2$,$Y_S = 2 \times 10^{11} \ N/m^2$:
$\frac{\Delta \ell_C}{\Delta \ell_S} = \left( \frac{50 \times \sqrt{3}}{0.5 \times 10^{11}} \right) \times \left( \frac{0.5 \times 2 \times 10^{11}}{50\sqrt{3} \times 1} \right) = \frac{50\sqrt{3}}{50\sqrt{3}} \times \frac{2}{1} = 2$
Thus,the ratio is $2$.

(C) The work done is given by $W = \int \vec{F} \cdot d\vec{r} = \int (\alpha y dx + 2 \alpha x dy)$. Given $\alpha = -1$,$W = \int (-y dx - 2x dy)$.
Along $AB$: $y=1, dy=0, x: 0 \to 1$. $W_{AB} = \int_0^1 (-1) dx = -1 \ J$.
Along $BC$: $x=1, dx=0, y: 1 \to 0.5$. $W_{BC} = \int_1^{0.5} -2(1) dy = -2(0.5 - 1) = 1 \ J$.
Along $CD$: $y=0.5, dy=0, x: 1 \to 0.5$. $W_{CD} = \int_1^{0.5} -0.5 dx = -0.5(-0.5) = 0.25 \ J$.
Along $DE$: $x=0.5, dx=0, y: 0.5 \to 0$. $W_{DE} = \int_{0.5}^0 -2(0.5) dy = -1(-0.5) = 0.5 \ J$.
Along $EF$: $y=0, dy=0, x: 0.5 \to 0$. $W_{EF} = \int_{0.5}^0 0 dx = 0 \ J$.
Along $FA$: $x=0, dx=0, y: 0 \to 1$. $W_{FA} = \int_0^1 -2(0) dy = 0 \ J$.
Total work $W = W_{AB} + W_{BC} + W_{CD} + W_{DE} + W_{EF} + W_{FA} = -1 + 1 + 0.25 + 0.5 + 0 + 0 = 0.75 \ J$.

(B) First,convert the velocities from $km/h$ to $m/s$:
$v_{S1} = 108 \times \frac{5}{18} = 30 \ m/s$
$v_O = 36 \times \frac{5}{18} = 10 \ m/s$
For observer $O$ and source $S_2$:
The source $S_2$ is stationary $(v_s = 0)$. The observer $O$ moves towards $S_2$ with a velocity component along the line joining them. Since $O$ moves directly towards $S_2$,the velocity of the observer is $v_o = 10 \ m/s$.
Using the Doppler effect formula: $f_2 = f_0 \left( \frac{v + v_o}{v} \right) = 120 \left( \frac{330 + 10}{330} \right) = 120 \left( \frac{340}{330} \right) \approx 123.64 \ Hz$.
For observer $O$ and source $S_1$:
The distance between $S_1$ and $S_2$ is $800 \ m$,and $O$ is $600 \ m$ from $S_2$. The distance $OS_1 = \sqrt{800^2 + 600^2} = 1000 \ m$.
The angle $\theta$ such that $\cos \theta = \frac{800}{1000} = 0.8$ and $\sin \theta = \frac{600}{1000} = 0.6$.
The velocity component of $S_1$ towards $O$ is $v_{s1} = v_{S1} \cos \theta = 30 \times 0.8 = 24 \ m/s$.
The velocity component of $O$ towards $S_1$ is $v_{o1} = v_O \sin \theta = 10 \times 0.6 = 6 \ m/s$.
Using the Doppler effect formula: $f_1 = f_0 \left( \frac{v + v_{o1}}{v - v_{s1}} \right) = 120 \left( \frac{330 + 6}{330 - 24} \right) = 120 \left( \frac{336}{306} \right) \approx 131.76 \ Hz$.
Beat frequency $= |f_1 - f_2| = 131.76 - 123.64 = 8.12 \ Hz$.
The number of beats heard is approximately $8$.

(C) Let the mass of the calorimeter be $m$,the specific heat of the calorimeter be $x$,the specific heat of the liquid be $s$,and the latent heat of the liquid be $L$.
When $5 \ gm$ of liquid at $30^{\circ} C$ is added to the calorimeter at $110^{\circ} C$,the calorimeter cools down to $80^{\circ} C$ and the liquid evaporates:
$m \cdot x \cdot (110 - 80) = 5 \cdot s \cdot (80 - 30) + 5 \cdot L$
$30 \cdot mx = 250s + 5L$ ... $(i)$
Now,$80 \ gm$ of liquid at $30^{\circ} C$ is added to the calorimeter at $80^{\circ} C$ and the final equilibrium temperature becomes $50^{\circ} C$:
$m \cdot x \cdot (80 - 50) = 80 \cdot s \cdot (50 - 30)$
$30 \cdot mx = 1600s$ ... $(ii)$
Comparing equations $(i)$ and $(ii)$:
$250s + 5L = 1600s$
$5L = 1350s$
$\frac{L}{s} = \frac{1350}{5} = 270$

(D) For the mixture: $n_1 = 5$ (monatomic),$C_{v1} = 3R/2$; $n_2 = 1$ (diatomic),$C_{v2} = 5R/2$.
Total moles $n = 5 + 1 = 6$.
$(C_v)_m = \frac{n_1 C_{v1} + n_2 C_{v2}}{n_1 + n_2} = \frac{5(3R/2) + 1(5R/2)}{6} = \frac{10R}{6} = \frac{5R}{3}$.
$(C_p)_m = (C_v)_m + R = \frac{5R}{3} + R = \frac{8R}{3}$.
Adiabatic constant $\gamma_m = \frac{(C_p)_m}{(C_v)_m} = \frac{8/3}{5/3} = 1.6$. Thus,$(4)$ is correct.
For adiabatic compression: $P_0 V_0^{\gamma} = P_f (V_0/4)^{\gamma}$.
$P_f = P_0 (4)^{1.6} = P_0 (2^2)^{1.6} = P_0 (2^{3.2}) = 9.2 P_0$. Thus,$(1)$ is correct.
Using $T_f V_f^{\gamma-1} = T_0 V_0^{\gamma-1}$,$T_f = T_0 (V_0 / (V_0/4))^{0.6} = T_0 (4)^{0.6} = T_0 (2^{1.2}) = 2.3 T_0$.
Total internal energy $U = n_1 (3/2 RT) + n_2 (5/2 RT) = 5(1.5 RT) + 1(2.5 RT) = 10 RT$.
$U_f = 10 R (2.3 T_0) = 23 RT_0$. Thus,$(2)$ is incorrect.
Work done $|W| = |\Delta U| = |n_f C_{vm} T_f - n_i C_{vm} T_i| = |6 \times (5R/6) \times (2.3 T_0 - T_0)| = |5R \times 1.3 T_0| = 6.5 RT_0$. Thus,$(3)$ is incorrect.
Correct statements are $(1)$ and $(4)$.

(C) We treat the contact point as a hinge.
Applying the work-energy theorem:
$W_g = \Delta K.E.$
$Mg \left( \frac{L}{2} (1 - \cos 60^{\circ}) \right) = \frac{1}{2} I \omega^2$
$Mg \left( \frac{L}{2} \cdot \frac{1}{2} \right) = \frac{1}{2} \left( \frac{ML^2}{3} \right) \omega^2$
$\frac{MgL}{4} = \frac{ML^2}{6} \omega^2 \Rightarrow \omega = \sqrt{\frac{3g}{2L}}$ (Statement $3$ is correct).
Radial acceleration of $C.M.$: $a_r = \left( \frac{L}{2} \right) \omega^2 = \frac{L}{2} \cdot \frac{3g}{2L} = \frac{3g}{4}$ (Statement $1$ is correct).
Using $\tau = I \alpha$ about the contact point:
$Mg \left( \frac{L}{2} \sin 60^{\circ} \right) = \left( \frac{ML^2}{3} \right) \alpha$
$\alpha = \frac{3g}{2L} \sin 60^{\circ} = \frac{3g}{2L} \cdot \frac{\sqrt{3}}{2} = \frac{3\sqrt{3}g}{4L}$ (Statement $2$ is correct).
Net vertical acceleration of $C.M.$: $a_v = a_r \cos 60^{\circ} + a_t \cos 30^{\circ}$
$a_v = \left( \frac{3g}{4} \right) \left( \frac{1}{2} \right) + \left( \alpha \frac{L}{2} \right) \cos 30^{\circ} = \frac{3g}{8} + \left( \frac{3\sqrt{3}g}{4L} \cdot \frac{L}{2} \right) \frac{\sqrt{3}}{2} = \frac{3g}{8} + \frac{9g}{16} = \frac{15g}{16}$.
Applying $F_{net} = Ma_v$ in the vertical direction:
$Mg - N = M \left( \frac{15g}{16} \right) \Rightarrow N = \frac{Mg}{16}$ (Statement $4$ is correct).
Thus,statements $1, 3, 4$ are correct.

(A) The particle moves between two walls separated by distance $L$. The time taken for one round trip is $\Delta t = 2L/v$. The frequency of collision with the piston is $f = 1/\Delta t = v/(2L)$. Thus, statement $(1)$ is incorrect as it states $v/L$.
In an elastic collision with a piston moving at speed $V$ towards the particle, the velocity of the particle changes from $v$ to $v + 2V$. The change in speed is $(v + 2V) - v = 2V$. Thus, statement $(2)$ is correct.
For a small displacement $dL$ of the piston, the change in speed is $dv = 2V \times (\text{number of collisions in time } dt)$. Since $dt = dL/V$, the number of collisions is $dt / (2L/v) = (dL/V) \cdot (v/2L) = v dL / (2LV)$.
Therefore, $dv = 2V \cdot (v dL / 2LV) = v dL / L$. Thus, statement $(4)$ is incorrect as it states $2v dL/L$.
Integrating $dv/v = dL/L$ (with $L$ decreasing, $dv/v = -dL/L$), we get $\ln(v/v_0) = -\ln(L/L_0) = \ln(L_0/L)$, so $vL = v_0 L_0$. If $L$ becomes $L_0/2$, then $v = 2v_0$. The kinetic energy $KE = \frac{1}{2}mv^2$ becomes $\frac{1}{2}m(2v_0)^2 = 4 \times (\frac{1}{2}mv_0^2) = 4 KE_0$. Thus, statement $(3)$ is correct.
The correct statements are $(2)$ and $(3)$.

(C) Let $T$ be the tension in the string connecting the two hanging blocks. For the movable pulley,the tension in the string connected to the block of mass $2M$ is $2T$. The equation of motion for the block of mass $2M$ is $2T - kx = 2Ma_1$. For the hanging system,the relative acceleration of the blocks $M$ and $2M$ with respect to the movable pulley is $a_{rel}$. The equations are $Mg - T = M(a_1 - a_{rel})$ and $2Mg - T = 2M(a_1 + a_{rel})$. Solving these gives $T = \frac{4}{3}Mg$ and $a_{rel} = \frac{g}{3}$. The acceleration of block $M$ is $a_2 = a_1 + a_{rel}$ and for $2M$ is $a_3 = a_1 - a_{rel}$. Thus,$a_2 - a_1 = a_1 - a_3 = a_{rel} = \frac{g}{3}$. Substituting $T$ into the first equation: $2(\frac{4}{3}Mg) - kx = 2Ma_1 \implies a_1 = \frac{4g}{3} - \frac{kx}{2M}$. At maximum extension $x_0$,$a_1 = 0$,so $x_0 = \frac{8Mg}{3k}$. Option $A$ is incorrect. For option $C$,$a_2 - a_1 = a_{rel}$ and $a_1 - a_3 = a_{rel}$,so $a_2 - a_1 = a_1 - a_3$ is correct. At $x = \frac{x_0}{4}$,$a_1 = \frac{4g}{3} - \frac{k(8Mg/12k)}{2M} = \frac{4g}{3} - \frac{g}{3} = g$. Option $D$ is incorrect.

(C) For a projectile launched with speed $u$ at angle $\theta$,the time of flight is $T = \frac{2u \sin \theta}{g}$ and the horizontal range is $R = \frac{u^2 \sin 2\theta}{g}$.
The average velocity $V_1$ for the first flight is the displacement divided by time: $V_1 = \frac{R}{T} = \frac{u_0^2 \sin 2\theta / g}{2u_0 \sin \theta / g} = u_0 \cos \theta$.
For the entire motion,the total displacement $S_{total}$ is the sum of all ranges: $S_{total} = R_1 + R_2 + R_3 + \dots = R_1 (1 + \frac{1}{\alpha^2} + \frac{1}{\alpha^4} + \dots) = \frac{R_1}{1 - 1/\alpha^2} = \frac{R_1 \alpha^2}{\alpha^2 - 1}$.
The total time $T_{total}$ is the sum of all flight times: $T_{total} = T_1 + T_2 + T_3 + \dots = T_1 (1 + \frac{1}{\alpha} + \frac{1}{\alpha^2} + \dots) = \frac{T_1}{1 - 1/\alpha} = \frac{T_1 \alpha}{\alpha - 1}$.
The average velocity for the entire motion is $V_{avg} = \frac{S_{total}}{T_{total}} = \frac{R_1 \alpha^2 / (\alpha^2 - 1)}{T_1 \alpha / (\alpha - 1)} = \frac{R_1}{T_1} \cdot \frac{\alpha^2}{\alpha^2 - 1} \cdot \frac{\alpha - 1}{\alpha} = V_1 \cdot \frac{\alpha}{\alpha + 1}$.
Given $V_{avg} = 0.8 V_1$,we have $\frac{\alpha}{\alpha + 1} = 0.8$.
$\alpha = 0.8 \alpha + 0.8 \implies 0.2 \alpha = 0.8 \implies \alpha = 4$.

(A) For the fundamental mode,the frequency is given by $f = \frac{n}{2L} \sqrt{\frac{T}{\mu}}$.
$(1)$ With $T = T_0$ and $L = L_0$,$f_n = \frac{n}{2L_0} \sqrt{\frac{T_0}{\mu_n}}$.
For string $1$: $f_1 = \frac{1}{2L_0} \sqrt{\frac{T_0}{\mu}} = f_0 \rightarrow (P)$.
For string $2$: $f_2 = \frac{1}{2L_0} \sqrt{\frac{T_0}{2\mu}} = \frac{f_0}{\sqrt{2}} \rightarrow (R)$.
For string $3$: $f_3 = \frac{1}{2L_0} \sqrt{\frac{T_0}{3\mu}} = \frac{f_0}{\sqrt{3}} \rightarrow (S)$.
For string $4$: $f_4 = \frac{1}{2L_0} \sqrt{\frac{T_0}{4\mu}} = \frac{f_0}{2} \rightarrow (Q)$.
Thus,$I \rightarrow P, II \rightarrow R, III \rightarrow S, IV \rightarrow Q$.
$(2)$ Given $f = f_0$ for all strings.
For string $1$: $f_0 = \frac{1}{2L_0} \sqrt{\frac{T_1}{\mu}} = f_0 \Rightarrow T_1 = T_0 \rightarrow (P)$.
For string $2$: $f_0 = \frac{3}{2(3L_0/2)} \sqrt{\frac{T_2}{2\mu}} = \frac{1}{L_0} \sqrt{\frac{T_2}{2\mu}} = f_0 \Rightarrow T_2 = T_0/2 \rightarrow (Q)$.
For string $3$: $f_0 = \frac{5}{2(5L_0/4)} \sqrt{\frac{T_3}{3\mu}} = \frac{2}{L_0} \sqrt{\frac{T_3}{3\mu}} = f_0 \Rightarrow T_3 = 3T_0/16 \rightarrow (T)$.
For string $4$: $f_0 = \frac{14}{2(7L_0/4)} \sqrt{\frac{T_4}{4\mu}} = \frac{4}{L_0} \sqrt{\frac{T_4}{4\mu}} = f_0 \Rightarrow T_4 = T_0/16 \rightarrow (U)$.
Thus,$I \rightarrow P, II \rightarrow Q, III \rightarrow T, IV \rightarrow U$.

(B) For the $PV$-diagram process:
$(I)$ Work done $W = P_0(2V_0 - V_0) = P_0 V_0 = \frac{1}{3} R T_0 \Rightarrow (Q)$.
$(II)$ $\Delta U = \frac{f}{2} n R \Delta T = \frac{3}{2} (P_3 V_3 - P_1 V_1) = \frac{3}{2} (\frac{3}{2} P_0 \cdot 2 V_0 - P_0 V_0) = \frac{3}{2} (3 P_0 V_0 - P_0 V_0) = 3 P_0 V_0 = R T_0 \Rightarrow (R)$.
$(III)$ $\Delta Q = \Delta U + W = R T_0 + \frac{1}{3} R T_0 = \frac{4}{3} R T_0 \Rightarrow (S)$.
$(IV)$ For $1 \rightarrow 2$,$\Delta Q = \Delta U + W = \frac{3}{2} (P_0 \cdot 2 V_0 - P_0 V_0) + P_0 V_0 = \frac{3}{2} P_0 V_0 + P_0 V_0 = \frac{5}{2} P_0 V_0 = \frac{5}{6} R T_0 \Rightarrow (U)$.
Thus,for $(1)$,the match is $I \rightarrow Q, II \rightarrow R, III \rightarrow S, IV \rightarrow U$,which is option $(3)$.
For the $TV$-diagram process:
$(I)$ $W = W_{1 \rightarrow 2} + W_{2 \rightarrow 3} = n R T \ln(V_2/V_1) + 0 = \frac{R T_0}{3} \ln 2 \Rightarrow (P)$.
$(II)$ $\Delta U = \frac{3}{2} R (T_3 - T_1) = \frac{3}{2} R (T_0 - T_0/3) = R T_0 \Rightarrow (R)$.
$(III)$ $\Delta Q = \Delta U + W = R T_0 + \frac{1}{3} R T_0 \ln 2 = \frac{1}{3} R T_0 (3 + \ln 2) \Rightarrow (T)$.
$(IV)$ For $1 \rightarrow 2$ (isothermal),$\Delta Q = W = \frac{1}{3} R T_0 \ln 2 \Rightarrow (P)$.
Thus,for $(2)$,the match is $I \rightarrow P, II \rightarrow R, III \rightarrow T, IV \rightarrow P$,which is option $(4)$.

(A) Let the initial charge on the sphere be $Q$. Thus,$V_0 = \frac{kQ}{R}$.
When a small hole of area $\alpha(4\pi R^2)$ is made,the charge removed is $q = \alpha Q$.
$(1)$ Potential at the center $(V_c)$ and at a point $P$ at distance $R/2$ from the center towards the hole $(V_p)$:
The potential at any point is the sum of the potential due to the complete sphere and the potential due to the removed charge element (treated as a negative point charge at the surface).
$V_c = \frac{kQ}{R} - \frac{kq}{R} = \frac{kQ}{R}(1-\alpha) = V_0(1-\alpha)$.
$V_p = \frac{kQ}{R} - \frac{kq}{R/2} = \frac{kQ}{R} - \frac{2kq}{R} = \frac{kQ}{R}(1-2\alpha) = V_0(1-2\alpha)$.
Therefore,the ratio $\frac{V_c}{V_p} = \frac{1-\alpha}{1-2\alpha}$. Thus,option $(A)$ is correct.
$(2)$ Electric field at the center $(E_c)$:
Initially,$E_c = 0$. After removing charge $q$,$E_c = \frac{kq}{R^2} = \frac{k(\alpha Q)}{R^2} = \alpha \frac{V_0}{R}$. The field increases.
$(3)$ Electric field at a point $P'$ at distance $2R$ from the center:
Initially,$E_{P'} = \frac{kQ}{(2R)^2} = \frac{kQ}{4R^2}$.
After removing charge $q$ from the surface,the field due to the hole at $P'$ is $\frac{kq}{R^2}$ (since $P'$ is at distance $R$ from the hole).
$E_{P', \text{final}} = \frac{kQ}{4R^2} - \frac{kq}{R^2} = \frac{kQ}{4R^2} - \frac{k(\alpha Q)}{R^2} = \frac{kQ}{4R^2} - \frac{4k\alpha Q}{4R^2} = \frac{kQ}{4R^2}(1-4\alpha)$.
Change in field $\Delta E = \frac{kq}{R^2} = \frac{k(\alpha Q)}{R^2} = \frac{\alpha V_0}{R}$.

(A) The total decay constant $\lambda$ is the sum of the individual decay constants for the two branches:
$\lambda = \lambda_1 + \lambda_2 = 4.5 \times 10^{-10} + 0.5 \times 10^{-10} = 5.0 \times 10^{-10} \text{ per year}$.
Let $N_0$ be the initial number of ${ }_{19}^{40} K$ nuclei and $N$ be the number of radioactive nuclei at time $t$.
The number of stable nuclei produced is $N_s = N_0 - N$.
According to the problem,the ratio of stable nuclei to radioactive nuclei is $99$:
$\frac{N_0 - N}{N} = 99 \Rightarrow \frac{N_0}{N} - 1 = 99 \Rightarrow \frac{N_0}{N} = 100$.
Using the radioactive decay law $N = N_0 e^{-\lambda t}$,we have $\frac{N}{N_0} = e^{-\lambda t} = \frac{1}{100} = 10^{-2}$.
Taking the natural logarithm on both sides:
$-\lambda t = \ln(10^{-2}) = -2 \ln 10$.
Given $\ln 10 = 2.3$,we get $\lambda t = 2 \times 2.3 = 4.6$.
Substituting $\lambda = 5 \times 10^{-10} \text{ per year}$:
$(5 \times 10^{-10}) \times t = 4.6 \Rightarrow t = \frac{4.6}{5} \times 10^{10} = 0.92 \times 10^{10} = 9.2 \times 10^9 \text{ years}$.
Thus,the value of $t$ is $9.2$.

(D) For statement $(3)$: At $t=0$,capacitors act as short circuits. The circuit consists of a $5 V$ battery and total resistance $R_{eq} = 30 \Omega + 100 \Omega + 70 \Omega = 200 \Omega$. The current $i = \frac{5 V}{200 \Omega} = 0.025 A = 25 mA$. Thus,$(3)$ is correct.
For statement $(4)$: At steady state,capacitors act as open circuits. The circuit is a series loop with a $5 V$ battery and capacitors $C_1, C_4, C_3$ in series. The equivalent capacitance $C_{eq} = (1/10 + 1/80 + 1/80)^{-1} = (8/80 + 1/80 + 1/80)^{-1} = 80/10 = 8 \mu F$. The charge $Q = C_{eq} V = 8 \mu F \times 5 V = 40 \mu C$. The voltage across $C_1$ is $V_1 = Q/C_1 = 40 \mu C / 10 \mu F = 4 V$. Thus,$(4)$ is correct.
For statement $(1)$: After $S_1$ is closed for a long time,$V_P - V_Q = 4 V$ (potential across $C_1$). When $S_2$ is closed,we analyze the circuit using Kirchhoff's laws. The equivalent resistance and voltage sources lead to an instantaneous current of approximately $0.079 A$,not $0.2 A$. Thus,$(1)$ is incorrect.
For statement $(2)$: As calculated above,the potential difference between $P$ and $Q$ is $4 V$,not $10 V$. Thus,$(2)$ is incorrect.
Therefore,statements $(3)$ and $(4)$ are correct.

(D) When $n_1 = n_2 = n$,the lens acts as a single convex lens with focal length $f$ given by the Lens Maker's Formula:
$\frac{1}{f} = (n - 1) \left( \frac{1}{R} - \frac{1}{-R} \right) = (n - 1) \frac{2}{R} \implies f = \frac{R}{2(n - 1)}$.
In the second case,the lens is composed of two plano-convex lenses in contact. The focal lengths are:
$\frac{1}{f_1} = \frac{n - 1}{R}$ and $\frac{1}{f_2} = \frac{(n + \Delta n) - 1}{R}$.
The equivalent focal length $f' = f + \Delta f$ is given by:
$\frac{1}{f + \Delta f} = \frac{1}{f_1} + \frac{1}{f_2} = \frac{n - 1 + n + \Delta n - 1}{R} = \frac{2n + \Delta n - 2}{R} = \frac{2(n - 1) + \Delta n}{R}$.
Since $f = \frac{R}{2(n - 1)}$,we have $\frac{1}{f} = \frac{2(n - 1)}{R}$.
$\frac{1}{f + \Delta f} = \frac{1}{f} + \frac{\Delta n}{R} = \frac{1}{f} \left( 1 + \frac{\Delta n}{R} \cdot f \right) = \frac{1}{f} \left( 1 + \frac{\Delta n}{R} \cdot \frac{R}{2(n - 1)} \right) = \frac{1}{f} \left( 1 + \frac{\Delta n}{2(n - 1)} \right)$.
Using the binomial approximation $(1 + x)^{-1} \approx 1 - x$ for small $x$:
$\frac{1}{f + \Delta f} \approx \frac{1}{f} \left( 1 - \frac{\Delta n}{2(n - 1)} \right) \implies f + \Delta f \approx f \left( 1 - \frac{\Delta n}{2(n - 1)} \right) = f - f \frac{\Delta n}{2(n - 1)}$.
Thus,$\frac{\Delta f}{f} = -\frac{\Delta n}{2(n - 1)}$.
$(1)$ The relation $\frac{\Delta f}{f} = -\frac{\Delta n}{2(n - 1)}$ is independent of $R$,so it holds for concave surfaces as well. Correct.
$(2)$ Since $1 < n < 2$,then $n - 1 < 1$,so $2(n - 1) < 2$. Thus,$\left| \frac{\Delta f}{f} \right| = \frac{1}{2(n - 1)} \left| \frac{\Delta n}{n} \right| \cdot n = \frac{n}{2(n - 1)} \left| \frac{\Delta n}{n} \right|$. Since $\frac{n}{2(n - 1)} > 1$ for $n < 2$,$\left| \frac{\Delta f}{f} \right| > \left| \frac{\Delta n}{n} \right|$. Incorrect.
$(3)$ $|\Delta f| = f \cdot \frac{\Delta n}{2(n - 1)} = 20 \cdot \frac{10^{-3}}{2(1.5 - 1)} = 20 \cdot \frac{10^{-3}}{1} = 0.02 \text{ cm}$. The statement says $0.04 \text{ cm}$. Incorrect.
$(4)$ If $\frac{\Delta n}{n} < 0$,then $\Delta n < 0$. From $\frac{\Delta f}{f} = -\frac{\Delta n}{2(n - 1)}$,if $\Delta n < 0$,then $\frac{\Delta f}{f} > 0$. Correct.

(C) $1$. Voltmeter conversion: $V = I_g(R_g + R_v) \implies 0.1 = 2 \times 10^{-6} (10 + R_v) \implies 50000 = 10 + R_v \implies R_v = 49990 \Omega \approx 50 \text{ k} \Omega$. Statement $(2)$ is correct.
$2$. Ammeter conversion: $I_g R_g = (I - I_g) R_s \implies 2 \times 10^{-6} \times 10 = (10^{-3} - 2 \times 10^{-6}) R_s \implies 2 \times 10^{-5} = 0.998 \times 10^{-3} R_s \implies R_s \approx 0.02004 \Omega$. The total resistance of the ammeter is $R_A = \frac{R_g R_s}{R_g + R_s} \approx R_s \approx 0.02 \Omega$. Statement $(3)$ is correct.
$3$. Measured resistance: The voltmeter is in parallel with $R = 1000 \Omega$. The equivalent resistance is $R_{eq} = \frac{R \times R_v}{R + R_v} = \frac{1000 \times 50000}{1000 + 50000} = \frac{50000}{51} \approx 980.39 \Omega$. Statement $(1)$ is incorrect.
$4$. Internal resistance: Adding internal resistance $r = 5 \Omega$ in series with the circuit reduces the total current $x$ flowing through the ammeter,but the measured value of $R$ is determined by the ratio of voltage across the voltmeter to the current through the resistor $R$. Since the voltmeter is in parallel with $R$,the measured resistance remains $R_{eq} \approx 980.39 \Omega$,which is independent of the internal resistance of the cell. Statement $(4)$ is incorrect.
Thus,statements $(2)$ and $(3)$ are correct.

(A-D) According to Gauss's Law,the flux $\Phi$ through a closed surface is $\frac{q_{enclosed}}{\epsilon_0}$.
$(1)$ If $h > 2R$ and $r > R$,the entire shell is enclosed by the cylinder. Thus,$q_{enclosed} = Q$ and $\Phi = \frac{Q}{\epsilon_0}$. This is correct.
$(2)$ If $h < \frac{8R}{5}$ and $r = \frac{3R}{5}$,the cylinder is entirely inside the shell. Since the electric field inside a charged shell is zero,the flux through the cylinder is $\Phi = 0$. This is correct.
$(3)$ If $h > 2R$ and $r = \frac{4R}{5}$,the cylinder intersects the shell. The flux is due to the charge on the spherical caps cut by the cylinder. The solid angle subtended by a cap is $\Omega = 2\pi(1 - \cos\theta)$,where $\sin\theta = \frac{r}{R} = \frac{4}{5}$,so $\cos\theta = \frac{3}{5}$. The flux through one cap is $\frac{Q}{4\pi\epsilon_0} \times \Omega = \frac{Q}{2\epsilon_0}(1 - \cos\theta)$. For two caps,$\Phi = \frac{Q}{\epsilon_0}(1 - \frac{3}{5}) = \frac{2Q}{5\epsilon_0}$. This is correct.
$(4)$ If $h > 2R$ and $r = \frac{3R}{5}$,then $\sin\theta = \frac{3}{5}$,so $\cos\theta = \frac{4}{5}$. The flux is $\Phi = \frac{Q}{\epsilon_0}(1 - \frac{4}{5}) = \frac{Q}{5\epsilon_0}$. This is correct.

(C) The motional electromotive force $(EMF)$ induced in a small element $dy$ of the wire moving with velocity $\vec{V} = V_0 \hat{i}$ in a magnetic field $\vec{B} = B(y) \hat{k}$ is given by $d\phi = |(\vec{V} \times \vec{B}) \cdot d\vec{l}|$. Since $\vec{V} = V_0 \hat{i}$ and $\vec{B} = B(y) \hat{k}$,the cross product $\vec{V} \times \vec{B} = V_0 B(y) (\hat{i} \times \hat{k}) = -V_0 B(y) \hat{j}$.
The potential difference between the ends of the wire is $\Delta \phi = \int_0^L V_0 B(y) dy$.
Substituting $B(y) = B_0 \left(1 + \left(\frac{y}{L}\right)^\beta\right)$:
$\Delta \phi = \int_0^L V_0 B_0 \left(1 + \frac{y^\beta}{L^\beta}\right) dy = V_0 B_0 \left[ y + \frac{y^{\beta+1}}{L^\beta (\beta+1)} \right]_0^L = V_0 B_0 \left( L + \frac{L}{\beta+1} \right) = V_0 B_0 L \left( 1 + \frac{1}{\beta+1} \right)$.
$(1)$ The integral depends only on the range of $y$ (from $0$ to $L$). Thus,replacing the wire with any shape that spans the same $y$-range results in the same $\Delta \phi$. Statement $(1)$ is correct.
$(2)$ Since $\Delta \phi = V_0 B_0 L \left( \frac{\beta+2}{\beta+1} \right)$,it is proportional to $L$ (the projection on the $y$-axis). Statement $(2)$ is correct.
$(3)$ For $\beta = 0$,$\Delta \phi = V_0 B_0 L (1 + 1) = 2 V_0 B_0 L$. Statement $(3)$ is incorrect.
$(4)$ For $\beta = 2$,$\Delta \phi = V_0 B_0 L (1 + 1/3) = \frac{4}{3} V_0 B_0 L$. Statement $(4)$ is correct.
Therefore,the correct statements are $(1), (2),$ and $(4)$.

(B) For the maximum time,the ray of light must undergo $TIR$ at all surfaces at the minimum possible angle,which is the critical angle $\theta_c$.
For $TIR$,the condition is $n_1 \sin \theta_c = n_2$.
$\sin \theta_c = \frac{n_2}{n_1} = \frac{1.44}{1.5} = 0.96$.
The speed of light in the medium $n_1$ is $v = \frac{c}{n_1} = \frac{3 \times 10^8}{1.5} = 2 \times 10^8 \ m/s$.
Let the total path length of the ray inside the structure be $D$. The horizontal length is $L$. The angle the ray makes with the normal to the surface is $\theta_c$. Therefore,the horizontal component of the path is $D \sin \theta_c = L$.
Thus,$D = \frac{L}{\sin \theta_c}$.
The time taken is $t_{total} = \frac{D}{v} = \frac{L}{v \sin \theta_c} = \frac{9.6}{(2 \times 10^8) \times 0.96} = \frac{9.6}{1.92 \times 10^8} = 5 \times 10^{-8} \ s = 50 \times 10^{-9} \ s$.
Therefore,$t = 50$.

(A) The capacitor is equivalent to a series combination of $N$ capacitors,each of thickness $\delta = dx = \frac{d}{N}$.
For a large $N$,we can treat this as a continuous variation where $\frac{m}{N} = \frac{x}{d}$.
The dielectric constant varies as $K(x) = K(1 + \frac{x}{d})$.
The equivalent capacitance $C_{eq}$ for capacitors in series is given by $\frac{1}{C_{eq}} = \int \frac{dx}{C(x)}$,where $C(x) = \frac{K(x) \varepsilon_0 A}{dx}$.
Substituting the values: $\frac{1}{C_{eq}} = \int_0^d \frac{dx}{\frac{K(1 + x/d) \varepsilon_0 A}{dx}} = \frac{1}{K \varepsilon_0 A} \int_0^d \frac{dx}{1 + x/d}$.
Let $u = 1 + \frac{x}{d}$,then $du = \frac{dx}{d}$,so $dx = d \cdot du$.
$\frac{1}{C_{eq}} = \frac{d}{K \varepsilon_0 A} \int_1^2 \frac{du}{u} = \frac{d}{K \varepsilon_0 A} [\ln u]_1^2 = \frac{d}{K \varepsilon_0 A} \ln 2$.
Thus,$C_{eq} = \frac{K \varepsilon_0 A}{d \ln 2}$.
Comparing this with the given expression $C = \alpha \left( \frac{K \varepsilon_0 A}{d \ln 2} \right)$,we get $\alpha = 1$.

(C) The potential $V$ at a point $(r, \theta)$ due to a dipole $\vec{p}$ is $V_{dip} = \frac{k \vec{p} \cdot \hat{r}}{r^2}$. In a uniform field $\vec{E} = E_0 \hat{i}$,the potential is $V_{ext} = -E_0 x = -E_0 r \cos \theta$.
For the circle to be equipotential,the total potential $V = V_{dip} + V_{ext}$ must be independent of $\theta$.
Given $\vec{p} = \frac{p_0}{\sqrt{2}}(\hat{i}+\hat{j})$,in polar coordinates,$\vec{p} = p_0(\cos 45^\circ \hat{i} + \sin 45^\circ \hat{j})$.
The potential at distance $R$ is $V = \frac{k p_0 \cos(\theta - 45^\circ)}{R^2} - E_0 R \cos \theta$.
Expanding $\cos(\theta - 45^\circ) = \cos \theta \cos 45^\circ + \sin \theta \sin 45^\circ = \frac{1}{\sqrt{2}}(\cos \theta + \sin \theta)$.
$V = \frac{k p_0}{R^2 \sqrt{2}}(\cos \theta + \sin \theta) - E_0 R \cos \theta = \left(\frac{k p_0}{R^2 \sqrt{2}} - E_0 R\right) \cos \theta + \left(\frac{k p_0}{R^2 \sqrt{2}}\right) \sin \theta$.
For $V$ to be constant,the coefficients of $\cos \theta$ and $\sin \theta$ must be zero,which is not possible for all $\theta$. However,the problem implies the circle is an equipotential surface where the tangential component of the net electric field is zero.
At point $B$ (at $135^\circ$),the tangential component of the dipole field cancels the tangential component of the uniform field. This leads to $R = \left(\frac{p_0}{4 \pi \varepsilon_0 E_0}\right)^{1/3}$. Thus,statement $(1)$ is correct.
Since the net field is the vector sum of a uniform field and a non-uniform dipole field,the magnitude varies along the circle. Thus,$(2)$ is incorrect.
At point $B$,the radial and tangential components cancel out,resulting in $\vec{E}_B = 0$. Thus,$(4)$ is correct.

(B) For refraction at a single spherical surface,the formula is: $\frac{n_2}{v} - \frac{n_1}{u} = \frac{n_2 - n_1}{R}$. Here,$n_1 = 1.5$,$n_2 = 1$,and $u = -H = -30 \text{ cm} = -0.3 \text{ m}$.
For cylinder $I$ (flat top): $R = \infty$. The apparent depth $H_1 = \frac{H}{n} = \frac{30}{1.5} = 20 \text{ cm} = 0.2 \text{ m}$.
For cylinder $II$ (convex top): $R = +3 \text{ m}$.
$\frac{1}{v} - \frac{1.5}{-0.3} = \frac{1 - 1.5}{3} \implies \frac{1}{v} + 5 = -\frac{0.5}{3} = -\frac{1}{6} \implies \frac{1}{v} = -\frac{1}{6} - 5 = -\frac{31}{6} \implies v = -\frac{6}{31} \text{ m} \approx -19.35 \text{ cm}$.
Thus,$H_2 = 19.35 \text{ cm}$.
For cylinder $III$ (concave top): $R = -3 \text{ m}$.
$\frac{1}{v} - \frac{1.5}{-0.3} = \frac{1 - 1.5}{-3} \implies \frac{1}{v} + 5 = \frac{-0.5}{-3} = \frac{1}{6} \implies \frac{1}{v} = \frac{1}{6} - 5 = -\frac{29}{6} \implies v = -\frac{6}{29} \text{ m} \approx -20.69 \text{ cm}$.
Thus,$H_3 = 20.69 \text{ cm}$.
Comparing the values: $H_3 (20.69 \text{ cm}) > H_1 (20 \text{ cm}) > H_2 (19.35 \text{ cm})$.
Therefore,statements $(1)$ and $(4)$ are correct.

(C) Given: $d = 0.3 \text{ mm} = 3 \times 10^{-4} \text{ m}$,$D = 1 \text{ m}$,$\lambda = 600 \text{ nm} = 6 \times 10^{-7} \text{ m}$,$PO = y = 11 \text{ mm} = 1.1 \times 10^{-2} \text{ m}$.
$(1)$ At point $O$ $(y=0)$,the path difference is $\Delta x = d \sin \alpha \approx d \alpha$ (for small $\alpha$).
Given $\alpha = \frac{0.36}{\pi} \text{ degrees} = \frac{0.36}{\pi} \times \frac{\pi}{180} \text{ radians} = 2 \times 10^{-3} \text{ rad}$.
Path difference $\Delta x = (3 \times 10^{-4} \text{ m}) \times (2 \times 10^{-3} \text{ rad}) = 6 \times 10^{-7} \text{ m} = \lambda$.
Since $\Delta x = n\lambda$ (where $n=1$),there is constructive interference at $O$. Thus,statement $(A)$ is incorrect.
$(2)$ Fringe spacing $\beta = \frac{D\lambda}{d}$. This depends only on $D, \lambda, d$,not on $\alpha$. Thus,statement $(B)$ is incorrect.
$(3)$ At point $P$,path difference $\Delta x_P = d \sin \alpha + \frac{dy}{D} \approx d \alpha + \frac{dy}{D}$.
$\Delta x_P = (3 \times 10^{-4})(2 \times 10^{-3}) + \frac{(3 \times 10^{-4})(1.1 \times 10^{-2})}{1} = 6 \times 10^{-7} + 33 \times 10^{-7} = 39 \times 10^{-7} \text{ m}$.
$\frac{\Delta x_P}{\lambda} = \frac{39 \times 10^{-7}}{6 \times 10^{-7}} = 6.5$. Since it is a half-integer multiple,there is destructive interference at $P$. Thus,statement $(C)$ is correct.
$(4)$ For $\alpha = 0$,$\Delta x_P = \frac{dy}{D} = 33 \times 10^{-7} \text{ m}$.
$\frac{\Delta x_P}{\lambda} = \frac{33 \times 10^{-7}}{6 \times 10^{-7}} = 5.5$. Since it is a half-integer multiple,there is destructive interference at $P$. Thus,statement $(D)$ is incorrect.

(A) The energy of the absorbed photon is $\frac{hc}{\lambda_{a}} = 13.6 \text{ eV} \times \left[\frac{1}{1^2} - \frac{1}{4^2}\right] = 13.6 \times \frac{15}{16} \text{ eV} \quad (i)$
The energy of the emitted photon is $\frac{hc}{\lambda_{e}} = 13.6 \text{ eV} \times \left[\frac{1}{m^2} - \frac{1}{4^2}\right] \quad (ii)$
Given $\frac{\lambda_{a}}{\lambda_{e}} = \frac{1}{5}$,we divide $(ii)$ by $(i)$:
$\frac{\lambda_{a}}{\lambda_{e}} = \frac{13.6 \times [1 - 1/16]}{13.6 \times [1/m^2 - 1/16]} = 5 \implies \frac{15/16}{1/m^2 - 1/16} = 5$
$\frac{15}{16} = 5 \times \left(\frac{1}{m^2} - \frac{1}{16}\right) \implies \frac{3}{16} = \frac{1}{m^2} - \frac{1}{16}$
$\frac{1}{m^2} = \frac{4}{16} = \frac{1}{4} \implies m = 2$. Thus,option $(3)$ is correct.
For kinetic energy,$K_n \propto \frac{1}{n^2}$. So,$\frac{K_{m=2}}{K_{n=1}} = \frac{1/2^2}{1/1^2} = \frac{1}{4}$. Thus,option $(2)$ is correct.
For momentum,$\Delta p = \frac{h}{\lambda}$. So,$\frac{\Delta p_{a}}{\Delta p_{e}} = \frac{\lambda_{e}}{\lambda_{a}} = 5$. Thus,option $(4)$ is incorrect.

(B) The momentum of one photon is $p = \frac{h}{\lambda}$.
When $N$ photons strike the mirror,the total momentum transferred to the mirror is $\Delta P = 2Np = \frac{2Nh}{\lambda}$.
By the conservation of linear momentum,this impulse imparts an initial velocity $v$ to the mirror of mass $M$ at the equilibrium position:
$Mv = \frac{2Nh}{\lambda} \implies v = \frac{2Nh}{M\lambda}$.
For a spring-mass system,the maximum displacement (amplitude $A$) is related to the velocity at the mean position by $v = A\Omega$.
Given $A = 1 \mu\text{m} = 10^{-6} \text{ m}$,we have:
$A\Omega = \frac{2Nh}{M\lambda} \implies N = \frac{M\Omega A \lambda}{2h}$.
Rearranging the given expression $\frac{4 \pi M \Omega}{h} = 10^{24} \text{ m}^{-2}$,we get $\frac{M\Omega}{h} = \frac{10^{24}}{4\pi} \text{ m}^{-2}$.
Substituting the values:
$N = \left( \frac{10^{24}}{4\pi} \right) \times \frac{A \lambda}{2} = \frac{10^{24}}{4\pi} \times \frac{10^{-6} \times 8\pi \times 10^{-6}}{2}$.
$N = \frac{10^{24} \times 8\pi \times 10^{-12}}{8\pi} = 10^{12}$.
Since $N = x \times 10^{12}$,we find $x = 1$.

(A) The motional electromotive force (emf) induced in the wire $PQ$ is given by $\varepsilon = Blv$.
Given: $B = 1 \text{ T}$,$l = 10 \text{ cm} = 0.1 \text{ m}$,$v = 1 \text{ cm/s} = 0.01 \text{ m/s}$.
$\varepsilon = 1 \times 0.1 \times 0.01 = 10^{-3} \text{ V}$.
When the key $S$ is closed,the circuit acts as an $LR$ series circuit with a constant voltage source $\varepsilon$. The current $i$ at time $t$ is given by $i(t) = \frac{\varepsilon}{R} (1 - e^{-Rt/L})$.
Given: $R = 1 \ \Omega$,$L = 1 \text{ mH} = 10^{-3} \text{ H}$,$t = 1 \text{ ms} = 10^{-3} \text{ s}$.
Substituting the values:
$i = \frac{10^{-3}}{1} (1 - e^{-(1 \times 10^{-3}) / 10^{-3}})$
$i = 10^{-3} (1 - e^{-1})$
Using $e^{-1} = 0.37$:
$i = 10^{-3} (1 - 0.37) = 10^{-3} (0.63) = 0.63 \times 10^{-3} \text{ A}$.
Comparing this with $x \times 10^{-3} \text{ A}$,we get $x = 0.63$.

(C) For total internal reflection $(TIR)$ to occur at the second surface for all $\theta \leq 60^{\circ}$,the condition must be satisfied at the minimum angle of incidence within the prism.
Let $r_1$ be the angle of refraction at the first surface and $r_2$ be the angle of incidence at the second surface.
From the geometry of the prism,$r_1 + r_2 = A = 75^{\circ}$.
For $TIR$ at the second surface,the angle of incidence $r_2$ must be greater than or equal to the critical angle $C$,where $\sin C = \frac{n}{n_0}$.
As $\theta$ decreases,$r_1$ decreases,and consequently $r_2 = 75^{\circ} - r_1$ increases.
The condition for $TIR$ is most easily satisfied at the maximum value of $r_2$,which occurs at the minimum value of $r_1$. However,the problem states $TIR$ occurs for $\theta \leq 60^{\circ}$. The boundary condition is at $\theta = 60^{\circ}$.
At $\theta = 60^{\circ}$,by Snell's law at the first surface: $1 \cdot \sin 60^{\circ} = \sqrt{3} \cdot \sin r_1 \Rightarrow \frac{\sqrt{3}}{2} = \sqrt{3} \sin r_1 \Rightarrow r_1 = 30^{\circ}$.
Then,$r_2 = 75^{\circ} - 30^{\circ} = 45^{\circ}$.
For $TIR$ at this surface,$r_2 \geq C$,so $45^{\circ} \geq C$,which means $\sin 45^{\circ} \geq \sin C = \frac{n}{\sqrt{3}}$.
$\frac{1}{\sqrt{2}} \geq \frac{n}{\sqrt{3}} \Rightarrow n \leq \sqrt{\frac{3}{2}}$.
The limiting value is $n = \sqrt{1.5}$,so $n^2 = 1.5$.

(D) The $\alpha$-decay reaction is: ${ }_{88}^{226} Ra \longrightarrow { }_{86}^{222} Rn^* + { }_{2}^{4} \alpha$.
The total energy released ($Q$-value) is given by: $Q = (M_{Ra} - M_{Rn} - M_{\alpha}) \times 931 \text{ MeV}$.
$Q = (226.005 - 222.000 - 4.000) \times 931 \text{ MeV} = 0.005 \times 931 \text{ MeV} = 4.655 \text{ MeV}$.
Let $E_{\gamma}$ be the excitation energy of the ${ }_{86}^{222} Rn$ nucleus. The energy available for kinetic energy is $(Q - E_{\gamma})$.
The kinetic energy of the $\alpha$ particle is $K_{\alpha} = \frac{A-4}{A} (Q - E_{\gamma})$,where $A = 226$.
$4.44 \text{ MeV} = \frac{222}{226} (4.655 - E_{\gamma})$.
$4.655 - E_{\gamma} = 4.44 \times \frac{226}{222} \approx 4.44 \times 1.018 = 4.520 \text{ MeV}$.
$E_{\gamma} = 4.655 - 4.520 = 0.135 \text{ MeV}$.
Since $1 \text{ MeV} = 1000 \text{ keV}$,$E_{\gamma} = 0.135 \times 1000 = 135 \text{ keV}$.

(A) Given,the lens is at $75 cm$,object pin at $45 cm$,and image pin at $135 cm$.
Object distance $u = 45 - 75 = -30 cm$.
Image distance $v = 135 - 75 = 60 cm$.
Using the lens formula $\frac{1}{f} = \frac{1}{v} - \frac{1}{u}$,we get $\frac{1}{f} = \frac{1}{60} - \frac{1}{-30} = \frac{1+2}{60} = \frac{3}{60} = \frac{1}{20}$. Thus,$f = 20 cm$.
The scale has $4$ divisions per $cm$,so the least count $\Delta u = \Delta v = \frac{1}{4} cm = 0.25 cm$.
Differentiating the lens formula $\frac{1}{f} = \frac{1}{v} - \frac{1}{u}$,we get $-\frac{df}{f^2} = -\frac{dv}{v^2} + \frac{du}{u^2}$.
Taking magnitudes for maximum error: $\frac{df}{f^2} = \frac{dv}{v^2} + \frac{du}{u^2}$.
Percentage error $\frac{df}{f} \times 100 = f \left[ \frac{dv}{v^2} + \frac{du}{u^2} \right] \times 100$.
Substituting values: $\frac{df}{f} \times 100 = 20 \left[ \frac{0.25}{60^2} + \frac{0.25}{30^2} \right] \times 100$.
$= 20 \times 0.25 \left[ \frac{1}{3600} + \frac{1}{900} \right] \times 100 = 5 \left[ \frac{1+4}{3600} \right] \times 100 = 5 \times \frac{5}{36} = \frac{25}{36} \% \approx 0.69 \%$.