IIT JEE 2018 Physics Question Paper with Answer and Solution

(B) The potential energy of the particle is given by $V(r) = \frac{kr^2}{2}$.
The force acting on the particle is $F = -\frac{dV}{dr} = -\frac{d}{dr}(\frac{kr^2}{2}) = -kr$.
The magnitude of the force is $F = kr$. This force acts as the centripetal force required for circular motion.
Equating the force to the centripetal force: $kr = \frac{mv^2}{r}$.
At $r = R$,we have $kR = \frac{mv^2}{R}$,which gives $v^2 = \frac{kR^2}{m}$,so $v = \sqrt{\frac{k}{m}} R$. Thus,statement $(B)$ is correct.
The angular momentum $L$ is given by $L = mvr$.
Substituting $v = \sqrt{\frac{k}{m}} R$ and $r = R$,we get $L = m \left( \sqrt{\frac{k}{m}} R \right) R = \sqrt{mk} R^2$. Thus,statement $(C)$ is correct.
Therefore,the correct statements are $(B)$ and $(C)$.

(A) The force applied on the body is $\overrightarrow{F}=(\alpha t) \hat{i}+\beta \hat{j}$.
Given $\alpha=1.0 \ Ns^{-1}$ and $\beta=1.0 \ N$,we have $\overrightarrow{F}=t \hat{i}+\hat{j}$.
Using Newton's second law,$m \frac{d\overrightarrow{v}}{dt} = \overrightarrow{F}$,with $m=1.0 \ kg$,we get $\frac{d\overrightarrow{v}}{dt} = t \hat{i}+\hat{j}$.
Integrating with respect to time $t$ (with $\overrightarrow{v}=0$ at $t=0$),we get $\overrightarrow{v} = \int_{0}^{t} (t \hat{i}+\hat{j}) dt = \frac{t^2}{2} \hat{i} + t \hat{j}$.
At $t=1 \ s$,$\overrightarrow{v} = \frac{1}{2} \hat{i} + \hat{j} = \frac{1}{2}(\hat{i} + 2\hat{j}) \ ms^{-1}$. Thus,statement $(C)$ is true.
Integrating velocity to find position $\overrightarrow{r}$ (with $\overrightarrow{r}=0$ at $t=0$),we get $\overrightarrow{r} = \int_{0}^{t} (\frac{t^2}{2} \hat{i} + t \hat{j}) dt = \frac{t^3}{6} \hat{i} + \frac{t^2}{2} \hat{j}$.
At $t=1 \ s$,$\overrightarrow{r} = \frac{1}{6} \hat{i} + \frac{1}{2} \hat{j}$.
The displacement is $\overrightarrow{s} = \overrightarrow{r}(1) - \overrightarrow{r}(0) = \frac{1}{6} \hat{i} + \frac{1}{2} \hat{j}$.
The magnitude of displacement is $|\overrightarrow{s}| = \sqrt{(\frac{1}{6})^2 + (\frac{1}{2})^2} = \sqrt{\frac{1}{36} + \frac{9}{36}} = \sqrt{\frac{10}{36}} = \frac{\sqrt{10}}{6} \ m$. Thus,statement $(D)$ is false.
The torque is $\vec{\tau} = \overrightarrow{r} \times \overrightarrow{F} = (\frac{t^3}{6} \hat{i} + \frac{t^2}{2} \hat{j}) \times (t \hat{i} + \hat{j})$.
At $t=1 \ s$,$\vec{\tau} = (\frac{1}{6} \hat{i} + \frac{1}{2} \hat{j}) \times (1 \hat{i} + 1 \hat{j}) = (\frac{1}{6} - \frac{1}{2}) \hat{k} = -\frac{1}{3} \hat{k} \ Nm$.
The magnitude is $|\vec{\tau}| = \frac{1}{3} \ Nm$. Thus,statement $(A)$ is true and $(B)$ is false.

(B) The pressure balance at the meniscus is given by $\frac{2 \sigma}{R} = \rho g h$,where $R$ is the radius of the meniscus.
From the geometry of the meniscus,$R = \frac{r}{\cos \theta}$,where $r$ is the radius of the capillary and $\theta$ is the angle of contact.
Substituting $R$ into the height equation: $h = \frac{2 \sigma \cos \theta}{\rho g r}$.
$(A)$ For a given material,$\theta$ is constant,so $h \propto \frac{1}{r}$. Thus,$h$ decreases as $r$ increases. This is true.
$(B)$ From the formula,$h \propto \sigma$,so $h$ depends on $\sigma$. This is false.
$(C)$ If the lift accelerates upward with acceleration $a$,the effective gravity becomes $g_{\text{eff}} = g + a$. The new height is $h' = \frac{2 \sigma \cos \theta}{\rho (g+a) r}$. Since $g+a > g$,$h'$ decreases. This is true.
$(D)$ From the formula,$h \propto \cos \theta$,not $\theta$. This is false.
Therefore,statements $(A)$ and $(C)$ are true.

(C) In the given $T-V$ diagram:
Process $I$: $V$ is changing,so it is not isochoric. Statement $(A)$ is false.
Process $II$: This is an isothermal expansion ($T$ is constant,$V$ increases). Since $W > 0$ and $\Delta U = 0$,the gas absorbs heat $(\Delta Q = W > 0)$. Statement $(B)$ is true.
Process $III$: This is an isothermal compression ($T$ is constant,$V$ decreases). Since $W < 0$ and $\Delta U = 0$,the gas releases heat $(\Delta Q = W < 0)$. Statement $(C)$ is true.
Process $IV$: This is an isobaric compression ($T$ is constant,$V$ decreases). Wait,looking at the graph,$IV$ is a horizontal line at constant $T$,which is isothermal. Let's re-evaluate: Process $I$ and $III$ are curves,not linear,so they are not isobaric. Statement $(D)$ is true.
Therefore,statements $(B), (C),$ and $(D)$ are true.

(B) Given $\vec{A} = a \hat{i}$ and $\vec{B} = a \cos \omega t \hat{i} + a \sin \omega t \hat{j}$.
$\vec{A} + \vec{B} = a(1 + \cos \omega t) \hat{i} + a \sin \omega t \hat{j}$.
$|\vec{A} + \vec{B}|^2 = a^2(1 + \cos \omega t)^2 + a^2 \sin^2 \omega t = a^2(1 + 2 \cos \omega t + \cos^2 \omega t + \sin^2 \omega t) = a^2(2 + 2 \cos \omega t) = 2a^2(1 + \cos \omega t) = 4a^2 \cos^2(\omega t / 2)$.
So,$|\vec{A} + \vec{B}| = 2a \cos(\omega t / 2)$.
Similarly,$\vec{A} - \vec{B} = a(1 - \cos \omega t) \hat{i} - a \sin \omega t \hat{j}$.
$|\vec{A} - \vec{B}|^2 = a^2(1 - \cos \omega t)^2 + a^2 \sin^2 \omega t = a^2(1 - 2 \cos \omega t + \cos^2 \omega t + \sin^2 \omega t) = a^2(2 - 2 \cos \omega t) = 4a^2 \sin^2(\omega t / 2)$.
So,$|\vec{A} - \vec{B}| = 2a \sin(\omega t / 2)$.
Given $|\vec{A} + \vec{B}| = \sqrt{3}|\vec{A} - \vec{B}|$,we have $2a \cos(\omega t / 2) = \sqrt{3} \cdot 2a \sin(\omega t / 2)$.
$\tan(\omega t / 2) = 1 / \sqrt{3}$.
$\omega t / 2 = \pi / 6 \implies \omega t = \pi / 3$.
Since $\omega = \pi / 6 \text{ rad s}^{-1}$,we get $(\pi / 6) \cdot t = \pi / 3$.
Therefore,$t = 2 \text{ s}$.

(A) Let the stationary man be at point $C$ and the horizontal line be $AB$. Let $O$ be the point on the line $AB$ directly below $C$. Given $CO = 12 \,m$. The men are at $A$ and $B$ such that $AO = OB = 5 \,m$. The distance $AC = BC = \sqrt{12^2 + 5^2} = 13 \,m$. The angle $\theta$ between the line $AC$ (or $BC$) and the vertical $CO$ is given by $\cos \theta = \frac{12}{13}$ and $\sin \theta = \frac{5}{13}$.
The man at $A$ (behind) is moving towards $O$ with velocity $v_A = 2.0 \,m \,s^{-1}$. The component of his velocity along the line $AC$ is $v_{A, \text{eff}} = v_A \sin \theta = 2.0 \times \frac{5}{13} \,m \,s^{-1}$. Since he is moving towards $C$,the observed frequency is $f_A = f \left( \frac{v}{v - v_A \sin \theta} \right) = 1430 \left( \frac{330}{330 - 2 \times \frac{5}{13}} \right) \approx 1430 \left( 1 + \frac{10}{13 \times 330} \right)$.
The man at $B$ (front) is moving away from $O$ with velocity $v_B = 1.0 \,m \,s^{-1}$. The component of his velocity along the line $BC$ is $v_{B, \text{eff}} = v_B \sin \theta = 1.0 \times \frac{5}{13} \,m \,s^{-1}$. Since he is moving away from $C$,the observed frequency is $f_B = f \left( \frac{v}{v + v_B \sin \theta} \right) = 1430 \left( \frac{330}{330 + 1 \times \frac{5}{13}} \right) \approx 1430 \left( 1 - \frac{5}{13 \times 330} \right)$.
The beat frequency is $\Delta f = |f_A - f_B| = 1430 \left( \frac{10}{13 \times 330} + \frac{5}{13 \times 330} \right) = 1430 \left( \frac{15}{13 \times 330} \right) = \frac{1430}{330} \times \frac{15}{13} = \frac{13}{3} \times \frac{15}{13} = 5 \,Hz$.

(C) The acceleration of a body rolling down an inclined plane is given by $a = \frac{g \sin \theta}{1 + \frac{k^2}{R^2}}$.
For a ring, $k^2 = R^2$, so $a_R = \frac{g \sin \theta}{1 + 1} = \frac{g \sin \theta}{2}$.
For a disc, $k^2 = \frac{R^2}{2}$, so $a_D = \frac{g \sin \theta}{1 + 0.5} = \frac{2g \sin \theta}{3}$.
The distance traveled along the incline is $d = \frac{h}{\sin \theta}$.
Using $d = \frac{1}{2} a t^2$, we have $t = \sqrt{\frac{2d}{a}} = \sqrt{\frac{2h}{a \sin \theta}}$.
For the ring: $t_R = \sqrt{\frac{2h}{(g \sin \theta / 2) \sin \theta}} = \sqrt{\frac{4h}{g \sin^2 \theta}}$.
For the disc: $t_D = \sqrt{\frac{2h}{(2g \sin \theta / 3) \sin \theta}} = \sqrt{\frac{3h}{g \sin^2 \theta}}$.
Given $\theta = 60^{\circ}$, $\sin \theta = \frac{\sqrt{3}}{2}$, so $\sin^2 \theta = \frac{3}{4}$.
$t_R = \sqrt{\frac{4h}{g(3/4)}} = \sqrt{\frac{16h}{3g}}$ and $t_D = \sqrt{\frac{3h}{g(3/4)}} = \sqrt{\frac{4h}{g}}$.
Given $t_R - t_D = \frac{2-\sqrt{3}}{\sqrt{10}}$, we have $\sqrt{\frac{h}{g}} \left( \frac{4}{\sqrt{3}} - 2 \right) = \frac{2-\sqrt{3}}{\sqrt{10}}$.
$\sqrt{\frac{h}{10}} \left( \frac{4-2\sqrt{3}}{\sqrt{3}} \right) = \frac{2-\sqrt{3}}{\sqrt{10}}$.
$\sqrt{h} \left( \frac{2(2-\sqrt{3})}{\sqrt{3}} \right) = 2-\sqrt{3} \Rightarrow \sqrt{h} = \frac{\sqrt{3}}{2} \Rightarrow h = \frac{3}{4} = 0.75 \,m$.

(C) Let $m_1 = 1.0 \,kg$ and $m_2 = 2.0 \,kg$. Initial velocity of $m_1$ is $u_1 = 2.0 \,m \,s^{-1}$ and $m_2$ is $u_2 = 0$.
By conservation of linear momentum: $m_1 u_1 = m_1 v_1 + m_2 v_2 \implies 1(2) = 1(v_1) + 2(v_2) \implies v_1 + 2v_2 = 2$ $(i)$
For an elastic collision, the coefficient of restitution $e = 1$: $v_2 - v_1 = e(u_1 - u_2) = 1(2 - 0) = 2$ $(ii)$
Solving $(i)$ and $(ii)$: Adding them gives $3v_2 = 4 \implies v_2 = \frac{4}{3} \,m \,s^{-1}$. Substituting into $(ii)$, $v_1 = v_2 - 2 = \frac{4}{3} - 2 = -\frac{2}{3} \,m \,s^{-1}$.
The $2.0 \,kg$ block attached to the spring undergoes simple harmonic motion. The time taken for the spring to return to its unstretched position is half the time period: $\Delta t = \frac{T}{2} = \pi \sqrt{\frac{m_2}{k}} = \pi \sqrt{\frac{2}{2}} = \pi \,s$.
During this time, the $1.0 \,kg$ block moves with constant velocity $v_1 = -\frac{2}{3} \,m \,s^{-1}$ (moving left). The $2.0 \,kg$ block moves to the right, compresses the spring, and returns to the equilibrium position.
Displacement of $1.0 \,kg$ block: $s_1 = v_1 \Delta t = -\frac{2}{3} \pi \,m$.
Displacement of $2.0 \,kg$ block: $s_2 = 0$ (as it returns to the starting position).
The distance between the blocks is $|s_2 - s_1| = |0 - (- \frac{2}{3} \pi)| = \frac{2}{3} \pi = \frac{2 \times 3.14159}{3} \approx 2.094 \,m$. Rounded to two decimal places, the distance is $2.09 \,m$.

(A) In the steady state,the rate of heat flow through both cylinders is the same.
Let $r_1$ be the radius of the smaller cylinder and $r_2$ be the radius of the larger cylinder.
Given $r_2 = 2 r_1$,the cross-sectional areas are $A_1 = \pi r_1^2$ and $A_2 = \pi r_2^2 = \pi (2 r_1)^2 = 4 \pi r_1^2 = 4 A_1$.
The rate of heat flow is given by $\frac{dQ}{dt} = \frac{K A \Delta T}{L}$.
For the first cylinder: $\frac{dQ}{dt} = \frac{K_1 A_1 (300 - 200)}{L} = \frac{K_1 A_1 (100)}{L}$.
For the second cylinder: $\frac{dQ}{dt} = \frac{K_2 A_2 (200 - 100)}{L} = \frac{K_2 (4 A_1) (100)}{L}$.
Equating the two rates of heat flow:
$\frac{K_1 A_1 (100)}{L} = \frac{K_2 (4 A_1) (100)}{L}$.
$K_1 = 4 K_2$.
Therefore,$\frac{K_1}{K_2} = 4$.

(C) From the $V-T$ graph:
$1$. Process $AB$: Temperature is constant $(T_1)$,so it is an isothermal process. For an ideal gas,$\Delta U = 0$.
$2$. Process $AC$: Volume is constant $(V_1)$,so it is an isochoric process. Heat exchanged $q_{AC} = \Delta U_{AC} = nC_v(T_2 - T_1)$.
$3$. Process $BC$: Pressure is constant $(P_2)$,so it is an isobaric process. Heat exchanged $q_{BC} = \Delta H_{BC} = nC_p(T_2 - T_1)$.
Analyzing the options:
- For process $AC$ (isochoric),$q_{AC} = \Delta U_{AC} = nC_v(T_2 - T_1)$.
- For process $BC$ (isobaric),$\Delta U_{BC} = nC_v(T_1 - T_2) = -nC_v(T_2 - T_1)$. Thus,$q_{AC} = -\Delta U_{BC}$ is not generally true.
- For process $CA$ (isochoric),$\Delta H_{CA} = nC_p(T_1 - T_2)$ and $\Delta U_{CA} = nC_v(T_1 - T_2)$. Since $C_p > C_v$ and $(T_1 - T_2) < 0$,we have $\Delta H_{CA} < \Delta U_{CA}$.
- For process $BC$ (isobaric),$W_{BC} = P_2(V_1 - V_2) = -P_2(V_2 - V_1)$.
Given the standard interpretation of such problems,option $(C)$ is correct as $\Delta H_{CA} < \Delta U_{CA}$ holds true.

(B) Let the stationary man be at point $C$ and the two moving men be at $A$ and $B$. The distance $CO = 12 \ m$. Since the men are $10 \ m$ apart and the stationary man is equidistant,$AO = OB = 5 \ m$.
The distance $AC = BC = \sqrt{12^2 + 5^2} = 13 \ m$.
The angle $\theta$ between the line of motion and the line joining the observer to the source is given by $\cos \theta = \frac{5}{13}$.
For the man behind $(A)$ moving towards $O$ with speed $v_s = 2.0 \ m \ s^{-1}$,the component of velocity along the line $AC$ is $v_s \cos \theta$ towards $C$. The observed frequency is $f_A = f \left( \frac{v}{v - v_s \cos \theta} \right) = 1430 \left( \frac{330}{330 - 2 \cos \theta} \right) \approx 1430 \left( 1 + \frac{2 \cos \theta}{330} \right)$.
For the man in front $(B)$ moving away from $O$ with speed $v_s = 1.0 \ m \ s^{-1}$,the component of velocity along the line $BC$ is $v_s \cos \theta$ away from $C$. The observed frequency is $f_B = f \left( \frac{v}{v + v_s \cos \theta} \right) = 1430 \left( \frac{330}{330 + 1 \cos \theta} \right) \approx 1430 \left( 1 - \frac{\cos \theta}{330} \right)$.
The beat frequency is $\Delta f = f_A - f_B = 1430 \left( \frac{2 \cos \theta + \cos \theta}{330} \right) = 1430 \left( \frac{3 \cos \theta}{330} \right) = 13 \cos \theta$.
Substituting $\cos \theta = \frac{5}{13}$,we get $\Delta f = 13 \times \frac{5}{13} = 5 \ Hz$.

(B) The acceleration $a$ of a body rolling down an inclined plane is given by $a = \frac{g \sin \theta}{1 + \frac{k^2}{R^2}}$.
For a ring,$k^2 = R^2$,so $a_R = \frac{g \sin \theta}{1+1} = \frac{g \sin \theta}{2}$.
For a disc,$k^2 = R^2/2$,so $a_D = \frac{g \sin \theta}{1+0.5} = \frac{2g \sin \theta}{3}$.
The distance to be covered along the incline is $d = \frac{h}{\sin \theta}$.
Using $d = \frac{1}{2} a t^2$,we have $t = \sqrt{\frac{2d}{a}} = \sqrt{\frac{2h}{a \sin \theta}}$.
For the ring: $t_R = \sqrt{\frac{2h}{(g \sin \theta / 2) \sin \theta}} = \sqrt{\frac{4h}{g \sin^2 \theta}} = \sqrt{\frac{4h}{g (3/4)}} = \sqrt{\frac{16h}{3g}}$.
For the disc: $t_D = \sqrt{\frac{2h}{(2g \sin \theta / 3) \sin \theta}} = \sqrt{\frac{3h}{g \sin^2 \theta}} = \sqrt{\frac{3h}{g (3/4)}} = \sqrt{\frac{4h}{g}}$.
Given $t_R - t_D = \frac{2-\sqrt{3}}{\sqrt{10}}$,we have $\sqrt{\frac{h}{g}} \left( \sqrt{\frac{16}{3}} - 2 \right) = \frac{2-\sqrt{3}}{\sqrt{10}}$.
$\sqrt{\frac{h}{10}} \left( \frac{4 - 2\sqrt{3}}{\sqrt{3}} \right) = \frac{2-\sqrt{3}}{\sqrt{10}}$.
$\sqrt{h} \left( \frac{2(2-\sqrt{3})}{\sqrt{3}} \right) = 2-\sqrt{3}$.
$\sqrt{h} = \frac{\sqrt{3}}{2} \implies h = \frac{3}{4} = 0.75 \ m$.

(C) Let $m_1 = 1.0 \ kg$ and $m_2 = 2.0 \ kg$. Initial velocity of $m_1$ is $u_1 = 2.0 \ m \ s^{-1}$ and $m_2$ is $u_2 = 0$. Let $v_1$ and $v_2$ be their velocities after the elastic collision.
Using conservation of linear momentum: $m_1 u_1 = m_1 v_1 + m_2 v_2 \implies 1(2) = 1(v_1) + 2(v_2) \implies v_1 + 2v_2 = 2$ . . . . . $(1)$
Using the coefficient of restitution for an elastic collision $(e=1)$: $v_2 - v_1 = e(u_1 - u_2) = 1(2 - 0) = 2$ . . . . . $(2)$
Solving equations $(1)$ and $(2)$,we get $v_2 = 4/3 \ m \ s^{-1}$ and $v_1 = -2/3 \ m \ s^{-1}$.
The $2.0 \ kg$ block attached to the spring undergoes simple harmonic motion. The time period is $T = 2\pi \sqrt{m_2/k} = 2\pi \sqrt{2/2} = 2\pi \ s$.
The spring returns to its unstretched position after half the time period,i.e.,$\Delta t = T/2 = \pi \ s$.
During this time,the $1.0 \ kg$ block moves with constant velocity $v_1 = -2/3 \ m \ s^{-1}$ (away from the spring). Its displacement is $\Delta x_1 = v_1 \Delta t = (-2/3) \times \pi = -2\pi/3 \ m$.
The $2.0 \ kg$ block moves from $x=0$ to $x_{max}$ and back to $x=0$. Its displacement is $\Delta x_2 = 0$.
The distance between the blocks is $|\Delta x_1 - \Delta x_2| = |-2\pi/3 - 0| = 2\pi/3 \approx 2(3.14)/3 = 6.28/3 \approx 2.09 \ m$.

(C) In the steady state,the rate of heat flow through both cylinders is the same.
Let $r_1$ be the radius of the smaller cylinder and $r_2$ be the radius of the larger cylinder.
Given $r_2 = 2r_1$.
The cross-sectional areas are $A_1 = \pi r_1^2$ and $A_2 = \pi r_2^2 = \pi (2r_1)^2 = 4\pi r_1^2 = 4A_1$.
The rate of heat flow is given by $\frac{dQ}{dt} = \frac{KA(T_{high} - T_{low})}{L}$.
For the first cylinder: $\frac{dQ}{dt} = \frac{K_1 A_1 (300 - 200)}{L} = \frac{K_1 A_1 (100)}{L}$.
For the second cylinder: $\frac{dQ}{dt} = \frac{K_2 A_2 (200 - 100)}{L} = \frac{K_2 (4A_1) (100)}{L}$.
Equating the rates of heat flow:
$\frac{K_1 A_1 (100)}{L} = \frac{K_2 (4A_1) (100)}{L}$.
$K_1 = 4K_2$.
Therefore,$K_1 / K_2 = 4$.

$(C, D)$ $(1)$ The Lorentz force on a charge $q$ moving with velocity $v$ is given by $F = qE + q(v \times B)$. For the dimensions to be consistent, the magnitude of the electric force $qE$ must be equal to the magnitude of the magnetic force $qvB$.
Thus, $qE = qvB$, which implies $E = vB$.
Since the dimension of velocity $v$ is $[L][T]^{-1}$, we have $[E] = [L][T]^{-1}[B]$. This matches option $(C)$.
$(2)$ The speed of light $c$ in free space is related to permittivity $\varepsilon_0$ and permeability $\mu_0$ by the relation $c = \frac{1}{\sqrt{\mu_0 \varepsilon_0}}$.
Squaring both sides, we get $c^2 = \frac{1}{\mu_0 \varepsilon_0}$, or $\mu_0 = \frac{1}{\varepsilon_0 c^2}$.
Since the dimension of $c$ is $[L][T]^{-1}$, the dimension of $c^2$ is $[L]^2[T]^{-2}$.
Therefore, $[\mu_0] = [\varepsilon_0]^{-1} ([L]^2[T]^{-2})^{-1} = [\varepsilon_0]^{-1}[L]^{-2}[T]^2$. This matches option $(D)$.

(D) From the $V-T$ graph:
$AB$: $T$ is constant $(T_1)$,so it is an isothermal process. Thus,$\Delta U_{AB} = 0$ and $W_{AB} = nRT_1 \ln(V_2/V_1)$.
$AC$: $V$ is constant $(V_1)$,so it is an isochoric process. Thus,$W_{AC} = 0$ and $q_{AC} = \Delta U_{AC} = nC_v(T_2 - T_1)$.
$BC$: $P$ is constant $(P_2)$,so it is an isobaric process. Thus,$W_{BC} = P_2(V_1 - V_2)$ and $q_{BC} = \Delta H_{BC} = nC_p(T_1 - T_2)$.
For process $CA$: $\Delta H_{CA} = nC_p(T_1 - T_2)$ and $\Delta U_{CA} = nC_v(T_1 - T_2)$. Since $C_p > C_v$ and $(T_1 - T_2) < 0$,we have $\Delta H_{CA} < \Delta U_{CA}$.
Therefore,options $(B)$ and $(D)$ are correct.

(B) Given $\frac{dK}{dt} = \gamma t$. Since $K = \frac{1}{2}mv^2$,we have $\frac{dK}{dt} = mv \frac{dv}{dt} = \gamma t$.
Thus,$v \frac{dv}{dt} = \frac{\gamma t}{m}$.
Integrating both sides,$\int v dv = \int \frac{\gamma}{m} t dt$,which gives $\frac{v^2}{2} = \frac{\gamma t^2}{2m}$.
Therefore,$v = \sqrt{\frac{\gamma}{m}} t$. This shows that speed is proportional to time,so $(B)$ is true.
The acceleration $a = \frac{dv}{dt} = \sqrt{\frac{\gamma}{m}}$,which is a constant. Since $F = ma$,the force $F = \sqrt{\gamma m}$ is also constant,so $(A)$ is true.
Since $v = \frac{dx}{dt} = \sqrt{\frac{\gamma}{m}} t$,integrating gives $x = \int \sqrt{\frac{\gamma}{m}} t dt = \frac{1}{2} \sqrt{\frac{\gamma}{m}} t^2$. The distance increases quadratically with time,not linearly,so $(C)$ is false.
Since the force is constant,it is a conservative force,so $(D)$ is true.
Thus,statements $(A), (B),$ and $(D)$ are true.

(C) According to Newton's law of viscosity,the viscous force $F$ acting on a plate of area $A$ moving with velocity $u_0$ over a liquid layer of thickness $h$ is given by:
$F = \eta A \frac{dv}{dx} = \eta A \frac{u_0}{h}$
From this expression:
$1$. The resistive force $F$ is inversely proportional to $h$ $(F \propto 1/h)$. Thus,statement $(A)$ is true.
$2$. The resistive force $F$ is directly proportional to the area $A$ $(F \propto A)$. Thus,statement $(B)$ is false.
$3$. The tangential (shear) stress $\tau$ is defined as $\tau = F/A = \eta \frac{u_0}{h}$.
$4$. Since $\tau = \eta \frac{u_0}{h}$,the shear stress on the floor (or plate) is directly proportional to $u_0$. Thus,statement $(C)$ is true.
$5$. Since $\tau = \eta \frac{u_0}{h}$,the shear stress is directly proportional to the viscosity $\eta$. Thus,statement $(D)$ is true.
Therefore,statements $(A), (C),$ and $(D)$ are true.

(B) For two successive resonances at lengths $L_1$ and $L_2$,the half-wavelength is given by $\lambda/2 = L_2 - L_1$.
$\lambda/2 = 83.9 \text{ cm} - 50.7 \text{ cm} = 33.2 \text{ cm}$.
Therefore,the wavelength $\lambda = 2 \times 33.2 \text{ cm} = 66.4 \text{ cm}$. (Statement $C$ is true).
Speed of sound $v = f \lambda = 500 \text{ Hz} \times 0.664 \text{ m} = 332 \text{ m s}^{-1}$. (Statement $A$ is true).
For the first resonance,$L_1 + e = \lambda/4$,where $e$ is the end correction.
$50.7 \text{ cm} + e = 66.4 \text{ cm} / 4 = 16.6 \text{ cm}$.
$e = 16.6 \text{ cm} - 50.7 \text{ cm} = -34.1 \text{ cm}$.
Note: The problem setup implies $L_1$ and $L_2$ are the $2^{\text{nd}}$ and $3^{\text{rd}}$ harmonics (or higher) because $L_1$ is quite large. The end correction $e$ is typically small and positive. Given the options,$A$ and $C$ are correct.

(B) Given,$m = 0.4 \ kg$,Impulse $J = 1.0 \ N \ s$,$\tau = 4 \ s$,and $v(t) = v_0 e^{-t/\tau}$.
At $t = 0$,the velocity is $v(0) = v_0 e^0 = v_0$.
From the impulse-momentum theorem,$J = \Delta p = m \Delta v = m(v(0) - 0) = m v_0$.
Therefore,$v_0 = J/m = 1.0 / 0.4 = 2.5 \ m/s$.
The displacement $S$ at $t = \tau$ is given by the integral of velocity:
$S = \int_0^{\tau} v(t) \ dt = \int_0^{\tau} v_0 e^{-t/\tau} \ dt$.
$S = v_0 \left[ -\tau e^{-t/\tau} \right]_0^{\tau} = v_0 \tau (1 - e^{-1})$.
Substituting the values: $S = (2.5) \times (4) \times (1 - 0.37)$.
$S = 10 \times 0.63 = 6.3 \ m$.

(B) Let the initial velocity be $u$. The maximum height $H$ is given by $H = \frac{u^2 \sin^2 45^{\circ}}{2g} = 120 \ m$.
Since $\sin 45^{\circ} = \frac{1}{\sqrt{2}}$, we have $H = \frac{u^2 (1/2)}{2g} = \frac{u^2}{4g} = 120 \ m$, so $u^2 = 480g$.
The kinetic energy before the bounce is $K_i = \frac{1}{2}mu^2$.
After the bounce, the kinetic energy is $K_f = \frac{1}{2}K_i = \frac{1}{4}mu^2 = \frac{1}{2}mv^2$, where $v$ is the velocity after the bounce.
Thus, $v^2 = \frac{u^2}{2} = \frac{480g}{2} = 240g$.
The new maximum height $h$ is given by $h = \frac{v^2 \sin^2 30^{\circ}}{2g}$.
Substituting $v^2 = 240g$ and $\sin 30^{\circ} = \frac{1}{2}$, we get $h = \frac{240g \times (1/2)^2}{2g} = \frac{240g \times (1/4)}{2g} = \frac{60}{2} = 30 \ m$.

(A) The change in length $\Delta L$ is given by the formula $\Delta L = \frac{FL}{AY}$.
Given: $F = mg = 1.2 \times 10 = 12 \text{ N}$,$L = 1.0 \text{ m}$,$d = 0.5 \times 10^{-3} \text{ m}$,$Y = 2 \times 10^{11} \text{ N m}^{-2}$,and $\pi = 3.2$.
Area $A = \pi r^2 = \pi \left(\frac{d}{2}\right)^2 = 3.2 \times \frac{(0.5 \times 10^{-3})^2}{4} = 0.8 \times 0.25 \times 10^{-6} = 0.2 \times 10^{-6} \text{ m}^2$.
Calculating $\Delta L$: $\Delta L = \frac{12 \times 1.0}{0.2 \times 10^{-6} \times 2 \times 10^{11}} = \frac{12}{0.4 \times 10^5} = 30 \times 10^{-5} \text{ m} = 0.3 \text{ mm}$.
The Least Count $(LC)$ of the vernier scale is $LC = \text{Main Scale Division} - \text{Vernier Scale Division} = 1.0 \text{ mm} - \frac{9}{10} \times 1.0 \text{ mm} = 0.1 \text{ mm}$.
The reading on the vernier scale is given by $\Delta L = n \times LC$,where $n$ is the vernier division coinciding with the main scale.
$0.3 \text{ mm} = n \times 0.1 \text{ mm} \implies n = 3$.

(C) For an adiabatic process,the relation between temperature and volume is $T_1 V_1^{\gamma-1} = T_2 V_2^{\gamma-1}$.
For a monatomic gas,the adiabatic index $\gamma = \frac{5}{3}$,so $\gamma - 1 = \frac{2}{3}$.
Given $V_2 = 8 V_1$ and $T_1 = 100 K$.
$T_2 = T_1 \left(\frac{V_1}{V_2}\right)^{\gamma-1} = 100 \times \left(\frac{1}{8}\right)^{2/3} = 100 \times \left(\frac{1}{4}\right) = 25 K$.
The change in internal energy is given by $\Delta U = n C_v \Delta T$.
For a monatomic gas,$C_v = \frac{3}{2} R$.
$\Delta U = 1 \times \frac{3}{2} \times 8.0 \times (25 - 100) = 12 \times (-75) = -900 J$.
The decrease in internal energy is $|\Delta U| = 900 J$.

(B) Given: $\frac{m_1}{m_2} = 2$ and $\frac{R_1}{R_2} = \frac{1}{4}$.
$(P)$ For orbital speed $v = \sqrt{\frac{GM}{R}}$:
$\frac{v_1}{v_2} = \sqrt{\frac{R_2}{R_1}} = \sqrt{4} = 2$. Thus,$P \rightarrow 3$.
$(Q)$ For angular momentum $L = mvr = m\sqrt{GMR}$:
$\frac{L_1}{L_2} = \frac{m_1}{m_2} \sqrt{\frac{R_1}{R_2}} = 2 \times \sqrt{\frac{1}{4}} = 2 \times \frac{1}{2} = 1$. Thus,$Q \rightarrow 2$.
$(R)$ For kinetic energy $K = \frac{GMm}{2R}$:
$\frac{K_1}{K_2} = \frac{m_1}{m_2} \times \frac{R_2}{R_1} = 2 \times 4 = 8$. Thus,$R \rightarrow 4$.
$(S)$ For time period $T = 2\pi \sqrt{\frac{R^3}{GM}}$:
$\frac{T_1}{T_2} = \left(\frac{R_1}{R_2}\right)^{3/2} = \left(\frac{1}{4}\right)^{3/2} = \frac{1}{8}$. Thus,$S \rightarrow 1$.
Therefore,the correct matching is $P \rightarrow 3, Q \rightarrow 2, R \rightarrow 4, S \rightarrow 1$.

(A) For path $P$: $\vec{r} = \alpha t \hat{i} + \beta t \hat{j}$. Velocity $\vec{v} = \alpha \hat{i} + \beta \hat{j}$ (constant), so acceleration $\vec{a} = 0$ and force $\vec{F} = 0$. Linear momentum $\vec{p} = m\vec{v}$ is constant. Angular momentum $\vec{L} = \vec{r} \times \vec{p} = m(\alpha t \hat{i} + \beta t \hat{j}) \times (\alpha \hat{i} + \beta \hat{j}) = m(\alpha\beta t - \beta\alpha t)\hat{k} = 0$ (constant). Since $\vec{F}=0$, $K$, $U$, and $E$ are constant. Thus, $P \rightarrow 1, 2, 3, 4, 5$.
For path $Q$: $\vec{r} = \alpha \cos \omega t \hat{i} + \beta \sin \omega t \hat{j}$. This is an elliptical path. The force is central, so angular momentum $\vec{L}$ is conserved. Since the force is conservative, total energy $E$ is conserved. Thus, $Q \rightarrow 2, 5$.
For path $R$: $\vec{r} = \alpha(\cos \omega t \hat{i} + \sin \omega t \hat{j})$. This is a circular path with constant speed $v = \alpha\omega$. Thus, $K$ is constant. Since the force is central, $\vec{L}$ is conserved. For a circular path in a central potential, $U$ and $E$ are also constant. Thus, $R \rightarrow 2, 3, 4, 5$.
For path $S$: $\vec{r} = \alpha t \hat{i} + \frac{\beta}{2} t^2 \hat{j}$. Velocity $\vec{v} = \alpha \hat{i} + \beta t \hat{j}$ (time-dependent), so $\vec{p}$ is not constant. Acceleration $\vec{a} = \beta \hat{j}$ (constant force). $\vec{L} = m(\alpha t \hat{i} + \frac{\beta}{2} t^2 \hat{j}) \times (\alpha \hat{i} + \beta t \hat{j}) = m(\alpha\beta t^2 - \frac{\alpha\beta}{2} t^2)\hat{k} = \frac{m\alpha\beta t^2}{2}\hat{k}$ (time-dependent). Only total energy $E$ is conserved. Thus, $S \rightarrow 5$.

(D) Let $I_1$ be the current in the left loop and $I_2$ be the current in the right loop. The current $I$ in the middle wire is $I = I_2 - I_1$.
For the left loop: $V - I_1 R - L \frac{dI_1}{dt} = 0 \implies I_1(t) = \frac{V}{R}(1 - e^{-(R/L)t})$.
For the right loop: $V - I_2 R - 2L \frac{dI_2}{dt} = 0 \implies I_2(t) = \frac{V}{R}(1 - e^{-(R/2L)t})$.
The current in the middle wire is $I(t) = I_2(t) - I_1(t) = \frac{V}{R} [e^{-(R/L)t} - e^{-(R/2L)t}]$.
To find the maximum current,set $\frac{dI}{dt} = 0$:
$\frac{dI}{dt} = \frac{V}{R} [-\frac{R}{L} e^{-(R/L)t} + \frac{R}{2L} e^{-(R/2L)t}] = 0$.
$\frac{1}{L} e^{-(R/L)t} = \frac{1}{2L} e^{-(R/2L)t} \implies e^{-(R/2L)t} = \frac{1}{2}$.
Taking the natural logarithm on both sides: $-\frac{R}{2L} T = \ln(1/2) = -\ln 2 \implies T = \frac{2L}{R} \ln 2$.
Substituting $T$ back into the expression for $I(t)$:
$I_{\max} = \frac{V}{R} [e^{-(R/L) \cdot (2L/R) \ln 2} - e^{-(R/2L) \cdot (2L/R) \ln 2}] = \frac{V}{R} [e^{-2 \ln 2} - e^{-\ln 2}] = \frac{V}{R} [\frac{1}{4} - \frac{1}{2}] = -\frac{V}{4R}$.
The magnitude is $I_{\max} = \frac{V}{4R}$.
Thus,statements $(B)$ and $(D)$ are true.

(A) The magnetic field at the origin due to the wire at $x=R$ is $\vec{B}_1 = \frac{\mu_0 I_1}{2 \pi R} (\hat{k})$ and due to the wire at $x=-R$ is $\vec{B}_2 = \frac{\mu_0 I_2}{2 \pi R} (-\hat{k})$.
$(A)$ If $I_1=I_2$, the magnetic field due to the wires at the origin is $\vec{B}_{wires} = \vec{B}_1 + \vec{B}_2 = 0$. However, the circular loop creates a non-zero magnetic field at the origin. Thus, the net magnetic field $\vec{B}_{net} \neq 0$. Statement $(A)$ is true.
$(B)$ If $I_1 > 0$ and $I_2 < 0$, the magnetic field due to the wires at the origin is $\vec{B}_{wires} = \frac{\mu_0}{2 \pi R} (I_1 - I_2) \hat{k}$. Since $I_1 > 0$ and $I_2 < 0$, $(I_1 - I_2) > 0$, so $\vec{B}_{wires}$ is along $+\hat{k}$. The magnetic field due to the loop at the origin is along $-\hat{k}$. Since both fields are along the $z$-axis, they can cancel each other out. Statement $(B)$ is true.
$(C)$ If $I_1 < 0$ and $I_2 > 0$, $\vec{B}_{wires}$ is along $-\hat{k}$. The magnetic field due to the loop at the origin is also along $-\hat{k}$. They cannot cancel each other out. Statement $(C)$ is false.
$(D)$ The magnetic field due to the wires at the center of the loop $(0, 0, \sqrt{3}R)$ is purely along the $x$-axis. The $z$-component of the magnetic field at the center is solely due to the loop, which is $\left(-\frac{\mu_0 I}{2 R}\right)$. Statement $(D)$ is true.

(A) $1$. Initially,$S_1$ is closed and $S_2$ is open. Capacitor $C_3$ is connected directly across the battery of emf $V_0 = 8 \,V$.
$2$. The charge on $C_3$ becomes $Q_3 = C_3 V_0 = (1.0 \mu F)(8 \,V) = 8 \mu C$.
$3$. When $S_1$ is opened and $S_2$ is closed,$C_3$ is connected in parallel with the series combination of $C_1$ (with dielectric $\varepsilon_r$) and $C_2$.
$4$. Let the final charge on $C_3$ be $Q_3' = 5 \mu C$. The potential difference across $C_3$ is $V = \frac{Q_3'}{C_3} = \frac{5 \mu C}{1.0 \mu F} = 5 \,V$.
$5$. The charge lost by $C_3$ is $8 \mu C - 5 \mu C = 3 \mu C$. This charge flows to the series combination of $C_1$ and $C_2$.
$6$. The capacitance of $C_1$ with dielectric is $C_1' = \varepsilon_r C_1 = \varepsilon_r (1.0 \mu F)$.
$7$. The equivalent capacitance of $C_1'$ and $C_2$ in series is $C_{eq} = \frac{C_1' C_2}{C_1' + C_2} = \frac{\varepsilon_r (1.0)}{\varepsilon_r + 1.0} \mu F$.
$8$. The charge on this series combination is $3 \mu C$. Thus,$V = \frac{Q}{C_{eq}} \implies 5 \,V = \frac{3 \mu C}{\frac{\varepsilon_r}{\varepsilon_r + 1} \mu F}$.
$9$. $5 = \frac{3(\varepsilon_r + 1)}{\varepsilon_r} \implies 5\varepsilon_r = 3\varepsilon_r + 3 \implies 2\varepsilon_r = 3 \implies \varepsilon_r = 1.50$.

(C) The radius of the circular path of a charged particle in a magnetic field is given by $R = \frac{mv}{qB}$.
For the region $y > 0$, the radius is $R_1 = \frac{mv_0}{qB_1}$.
For the region $y < 0$, the radius is $R_2 = \frac{mv_0}{qB_2}$.
Given $B_2 = 4B_1$, we have $R_2 = \frac{mv_0}{q(4B_1)} = \frac{R_1}{4}$.
The particle moves in a semicircle in the region $y > 0$ and then in a semicircle in the region $y < 0$ before crossing the $x$-axis again.
The total displacement along the $x$-axis is $\Delta x = 2R_1 + 2R_2 = 2R_1 + 2(\frac{R_1}{4}) = 2R_1 + \frac{R_1}{2} = \frac{5R_1}{2}$.
The time taken to complete a semicircle in region $y > 0$ is $t_1 = \frac{\pi m}{qB_1}$.
The time taken to complete a semicircle in region $y < 0$ is $t_2 = \frac{\pi m}{qB_2} = \frac{\pi m}{q(4B_1)} = \frac{t_1}{4}$.
The total time taken is $T = t_1 + t_2 = t_1 + \frac{t_1}{4} = \frac{5t_1}{4}$.
The average speed along the $x$-axis is $v_{avg} = \frac{\Delta x}{T} = \frac{5R_1/2}{5t_1/4} = \frac{5R_1}{2} \times \frac{4}{5t_1} = \frac{2R_1}{t_1}$.
Substituting $R_1 = \frac{mv_0}{qB_1}$ and $t_1 = \frac{\pi m}{qB_1}$, we get $v_{avg} = \frac{2(mv_0/qB_1)}{\pi m/qB_1} = \frac{2v_0}{\pi}$.
Given $v_0 = \pi \text{ m s}^{-1}$, the average speed is $v_{avg} = \frac{2(\pi)}{\pi} = 2 \text{ m s}^{-1}$.

(B) For $(1)$: The force on a charge $q$ moving with velocity $v$ in an electric field $E$ and magnetic field $B$ is given by the Lorentz force law: $F = qE + q(v \times B)$. For the dimensions to be consistent, the units of $E$ and $vB$ must be the same. Thus, $E = vB$. Since the dimension of velocity $v$ is $[L][T]^{-1}$, we have $[E] = [L][T]^{-1}[B]$. This matches option $(A)$.
For $(2)$: The speed of light $c$ in free space is related to permittivity $\varepsilon_0$ and permeability $\mu_0$ by the relation $c = \frac{1}{\sqrt{\mu_0 \varepsilon_0}}$. Squaring both sides, we get $c^2 = \frac{1}{\mu_0 \varepsilon_0}$, which implies $\mu_0 = \frac{1}{\varepsilon_0 c^2}$. Since the dimension of $c$ is $[L][T]^{-1}$, the dimension of $c^2$ is $[L]^2[T]^{-2}$. Therefore, $[\mu_0] = [\varepsilon_0]^{-1} ([L]^2[T]^{-2})^{-1} = [\varepsilon_0]^{-1}[L]^{-2}[T]^2$. This matches option $(B)$.

(C) $(1)$ Given $r = \frac{1-a}{1+a}$. Taking the natural logarithm on both sides: $\ln r = \ln(1-a) - \ln(1+a)$. Differentiating both sides: $\frac{dr}{r} = \frac{-da}{1-a} - \frac{da}{1+a}$. Since errors are always added,$\frac{\Delta r}{r} = \frac{\Delta a}{1-a} + \frac{\Delta a}{1+a} = \Delta a \frac{(1+a) + (1-a)}{(1-a)(1+a)} = \frac{2 \Delta a}{1-a^2}$. Thus,$\Delta r = r \cdot \frac{2 \Delta a}{1-a^2} = \frac{1-a}{1+a} \cdot \frac{2 \Delta a}{(1-a)(1+a)} = \frac{2 \Delta a}{(1+a)^2}$. The correct option is $B$.
$(2)$ The number of nuclei remaining is $N = N_0 - N_{decayed} = 3000 - 1000 = 2000$. The decay law is $N = N_0 e^{-\lambda t}$,so $\ln N = \ln N_0 - \lambda t$. Differentiating,$\frac{dN}{N} = -t \cdot d\lambda$. Considering magnitudes for errors,$\Delta \lambda = \frac{\Delta N}{N \cdot t}$. Here $\Delta N = 40$,$N = 2000$,and $t = 1.0 \ s$. Therefore,$\Delta \lambda = \frac{40}{2000 \times 1} = 0.02 \ s^{-1}$. The correct option is $C$.

(D) $1$. Initially,$S_1$ is closed and $S_2$ is open. $C_3$ is connected directly to the battery $V_0 = 8 \ V$. The charge on $C_3$ becomes $Q_3 = C_3 V_0 = (1.0 \mu F)(8 \ V) = 8 \mu C$.
$2$. When $S_1$ is opened and $S_2$ is closed,the charge $8 \mu C$ on $C_3$ redistributes among the capacitors $C_1, C_2$ and $C_3$. Let the final charge on $C_3$ be $Q_3' = 5 \mu C$. Since $C_3$ is in parallel with the series combination of $C_1$ and $C_2$,the voltage across $C_3$ must equal the sum of voltages across $C_1$ and $C_2$.
$3$. The voltage across $C_3$ is $V_3 = \frac{Q_3'}{C_3} = \frac{5 \mu C}{1 \mu F} = 5 \ V$.
$4$. The charge remaining for the series combination of $C_1$ and $C_2$ is $Q_{12} = Q_{initial} - Q_3' = 8 \mu C - 5 \mu C = 3 \mu C$. Thus,$Q_1 = Q_2 = 3 \mu C$.
$5$. The voltage across $C_1$ is $V_1 = \frac{Q_1}{C_1'} = \frac{3 \mu C}{\varepsilon_r (1 \mu F)} = \frac{3}{\varepsilon_r} \ V$,where $C_1' = \varepsilon_r C_1$.
$6$. The voltage across $C_2$ is $V_2 = \frac{Q_2}{C_2} = \frac{3 \mu C}{1 \mu F} = 3 \ V$.
$7$. Applying the loop rule: $V_3 = V_1 + V_2 \implies 5 = \frac{3}{\varepsilon_r} + 3$.
$8$. Solving for $\varepsilon_r$: $2 = \frac{3}{\varepsilon_r} \implies \varepsilon_r = \frac{3}{2} = 1.50$.

(A) The radius of the circular path of a charged particle in a magnetic field is given by $R = \frac{mv}{qB}$.
For the region $y > 0$,the radius is $R_1 = \frac{mv_0}{qB_1}$.
For the region $y < 0$,the radius is $R_2 = \frac{mv_0}{qB_2}$.
Given $B_2 = 4B_1$,we have $R_2 = \frac{mv_0}{q(4B_1)} = \frac{R_1}{4}$.
The particle moves in a semicircle of radius $R_1$ in the region $y > 0$ and then in a semicircle of radius $R_2$ in the region $y < 0$.
The total distance traveled along the $x$-axis is $\Delta x = 2R_1 + 2R_2 = 2R_1 + 2(\frac{R_1}{4}) = 2R_1 + \frac{R_1}{2} = \frac{5R_1}{2}$.
The time taken to complete a semicircle in region $y > 0$ is $t_1 = \frac{\pi m}{qB_1}$.
The time taken to complete a semicircle in region $y < 0$ is $t_2 = \frac{\pi m}{qB_2} = \frac{\pi m}{q(4B_1)} = \frac{t_1}{4}$.
The total time $T = t_1 + t_2 = t_1 + \frac{t_1}{4} = \frac{5t_1}{4}$.
The average speed along the $x$-axis is $v_{avg} = \frac{\Delta x}{T} = \frac{5R_1/2}{5t_1/4} = \frac{5R_1}{2} \times \frac{4}{5t_1} = 2 \frac{R_1}{t_1}$.
Substituting $R_1 = \frac{mv_0}{qB_1}$ and $t_1 = \frac{\pi m}{qB_1}$,we get $v_{avg} = 2 \frac{mv_0/qB_1}{\pi m/qB_1} = 2 \frac{v_0}{\pi}$.
Given $v_0 = \pi \text{ m s}^{-1}$,the average speed is $v_{avg} = 2 \frac{\pi}{\pi} = 2 \text{ m s}^{-1}$.

(B) Given: Intensity of incident light $I_0 = 1.3 \text{ kW m}^{-2}$,focal length $f = 20 \text{ cm}$.
Let $R$ be the radius of the lens aperture and $r$ be the radius of the light beam at a distance $d = 22 \text{ cm}$ from the lens.
The distance of this plane from the focus $F$ is $x = d - f = 22 \text{ cm} - 20 \text{ cm} = 2 \text{ cm}$.
By similar triangles $\Delta ABF$ and $\Delta PQF$ (where $AB$ is the diameter of the lens and $PQ$ is the diameter of the beam at the given plane):
$\frac{r}{R} = \frac{x}{f} = \frac{2 \text{ cm}}{20 \text{ cm}} = \frac{1}{10}$.
The area of the lens is $A = \pi R^2$ and the area of the beam at the given plane is $a = \pi r^2$.
Thus,$\frac{a}{A} = \left(\frac{r}{R}\right)^2 = \left(\frac{1}{10}\right)^2 = \frac{1}{100}$.
Since the total power $P$ incident on the lens is conserved and passes through the area $a$,the intensity $I$ at this plane is given by $I \times a = I_0 \times A$.
Therefore,$I = I_0 \times \left(\frac{A}{a}\right) = 1.3 \times 100 = 130 \text{ kW m}^{-2}$.

(B, C) $(1)$ Given $r = \frac{1 - a}{1 + a}$.
Taking the natural logarithm: $\ln r = \ln(1 - a) - \ln(1 + a)$.
Differentiating: $\frac{dr}{r} = \frac{-da}{1 - a} - \frac{da}{1 + a}$.
Taking magnitudes for maximum error: $\frac{\Delta r}{r} = \frac{\Delta a}{1 - a} + \frac{\Delta a}{1 + a} = \Delta a \left( \frac{1 + a + 1 - a}{1 - a^2} \right) = \frac{2 \Delta a}{1 - a^2}$.
Substituting $r = \frac{1 - a}{1 + a}$: $\Delta r = r \left( \frac{2 \Delta a}{1 - a^2} \right) = \left( \frac{1 - a}{1 + a} \right) \left( \frac{2 \Delta a}{(1 - a)(1 + a)} \right) = \frac{2 \Delta a}{(1 + a)^2}$.
$(2)$ Given $N = N_0 e^{-\lambda t}$,where $N_0 = 3000$.
Number of decayed nuclei is $1000 \pm 40$,so remaining nuclei $N = 3000 - 1000 = 2000$. The error in $N$ is $\Delta N = 40$.
Taking $\ln$: $\ln N = \ln N_0 - \lambda t$.
Differentiating: $\frac{dN}{N} = -d(\lambda t) \implies \Delta \lambda = \frac{\Delta N}{N \cdot t}$.
Substituting values: $\Delta \lambda = \frac{40}{2000 \times 1.0} = 0.02 \ s^{-1}$.

(A) According to Gauss's law,the electric flux $\phi$ through a closed surface is given by $\phi = \frac{Q_{\text{enclosed}}}{\epsilon_0}$.
The length of the wire segment $PQ$ inside the spherical shell is $L = 2R \sin(120^{\circ}/2) = 2R \sin(60^{\circ}) = 2R \times \frac{\sqrt{3}}{2} = R\sqrt{3}$.
The charge enclosed by the shell is $Q_{\text{enclosed}} = \lambda L = \lambda R \sqrt{3}$.
Therefore,the electric flux through the shell is $\phi = \frac{\lambda R \sqrt{3}}{\epsilon_0}$. Thus,statement $(A)$ is true and $(C)$ is false.
Since the wire is parallel to the $z$-axis,the electric field lines are directed radially outward from the wire in the $xy$-plane. Therefore,the electric field has no component along the $z$-axis. Thus,statement $(B)$ is true.
The electric field is only normal to the surface of the shell if the charge distribution is spherically symmetric,which is not the case here. Thus,statement $(D)$ is false.
Therefore,the correct statements are $(A)$ and $(B)$.

(B) The wire is placed between $f/2$ and $f$. For a concave mirror,the mirror formula is $\frac{1}{v} + \frac{1}{u} = \frac{1}{f}$.
Let a point $P$ on the hypotenuse be at a distance $x$ from the focus $f$,so $u = -(f-x)$.
Substituting this into the mirror formula: $\frac{1}{v} - \frac{1}{f-x} = -\frac{1}{f}$.
$\frac{1}{v} = \frac{1}{f-x} - \frac{1}{f} = \frac{x}{f(f-x)}$.
Thus,$v = \frac{f(f-x)}{x}$.
The magnification $M = -\frac{v}{u} = -\frac{f(f-x)/x}{-(f-x)} = \frac{f}{x}$.
For the vertical leg of the triangle at $u = -f/2$,the image is at $v = -f$ (real,inverted,magnified by $2$).
For the point at the focus $(u = -f)$,the image is at infinity $(v \rightarrow \infty)$.
As $x$ increases from $0$ to $f/2$,the magnification $M = f/x$ decreases from $\infty$ to $2$. The image of the hypotenuse is a curve that extends to infinity,consistent with option $B$.

(B) The decay process is represented as: ${ }_{90}^{232} Th \rightarrow { }_{82}^{212} Pb + N_{\alpha} { }_{2}^{4} He + N_{\beta} { }_{-1}^{0} e$.
First,consider the change in mass number $(\Delta A)$:
$\Delta A = 232 - 212 = 20$.
Since each $\alpha$-particle has a mass number of $4$,the number of $\alpha$-particles emitted is $N_{\alpha} = \frac{20}{4} = 5$.
Next,consider the change in atomic number $(\Delta Z)$:
$\Delta Z = 90 - 82 = 8$.
In the decay,the change in atomic number is given by $2N_{\alpha} - N_{\beta} = \Delta Z$.
Substituting the values: $2(5) - N_{\beta} = 8$.
$10 - N_{\beta} = 8$,which gives $N_{\beta} = 2$.
Thus,$N_{\alpha} = 5$ and $N_{\beta} = 2$. The correct statements are $(A)$ and $(C)$.

(A) The force on the particle is given by $F = qE = qE_0 \sin(\omega t)$.
Using Newton's second law,$F = ma$,we have $a = \frac{qE_0}{m} \sin(\omega t)$.
Since $a = \frac{dv}{dt}$,we integrate with respect to time:
$v(t) = \int_0^t \frac{qE_0}{m} \sin(\omega t') dt' = \frac{qE_0}{m\omega} [-\cos(\omega t')]_0^t = \frac{qE_0}{m\omega} (1 - \cos(\omega t))$.
The speed is maximum when $\cos(\omega t) = -1$,which occurs at $\omega t = \pi, 3\pi, \dots$.
At these times,$v_{\max} = \frac{qE_0}{m\omega} (1 - (-1)) = \frac{2qE_0}{m\omega}$.
Substituting the given values: $q = 1.0 \ C$,$E_0 = 1.0 \ N \ C^{-1}$,$m = 10^{-3} \ kg$,and $\omega = 10^3 \ rad \ s^{-1}$.
$v_{\max} = \frac{2 \times 1.0 \times 1.0}{10^{-3} \times 10^3} = \frac{2}{1} = 2 \ m \ s^{-1}$.

(C) Given:
Number of turns $n = 50$
Area $A = 2 \times 10^{-4} \ m^2$
Magnetic field $B = 0.02 \ T$
Torsional constant $C = 10^{-4} \ N \ m \ rad^{-1}$
Full-scale deflection angle $\theta = 0.2 \ rad$
Galvanometer resistance $G = 50 \ \Omega$
The torque balance equation for a moving coil galvanometer is given by:
$C \theta = n i_g A B$
Where $i_g$ is the full-scale deflection current.
$i_g = \frac{C \theta}{n A B} = \frac{10^{-4} \times 0.2}{50 \times 2 \times 10^{-4} \times 0.02} = \frac{0.2 \times 10^{-4}}{200 \times 10^{-4} \times 0.02} = \frac{0.2}{4} = 0.05 \ A$
Wait,recalculating: $i_g = \frac{0.2 \times 10^{-4}}{50 \times 2 \times 10^{-4} \times 0.02} = \frac{0.2}{50 \times 2 \times 0.02} = \frac{0.2}{2} = 0.1 \ A$
To convert the galvanometer into an ammeter of range $I = 1.0 \ A$,a shunt resistance $S$ is connected in parallel.
The current through the shunt is $I_s = I - i_g = 1.0 - 0.1 = 0.9 \ A$.
Since the galvanometer and shunt are in parallel,the voltage across them is equal:
$i_g G = I_s S$
$0.1 \times 50 = 0.9 \times S$
$S = \frac{5}{0.9} = \frac{50}{9} \approx 5.56 \ \Omega$

(B) Given power $P = 200 \ W$ and work function $\phi = 6.25 \ eV$.
Since the frequency is just above the threshold frequency,the initial kinetic energy of photoelectrons is zero.
Energy of one photon $E_1 = h\nu = \phi = 6.25 \ eV = 6.25 \times 1.6 \times 10^{-19} \ J = 10^{-18} \ J$.
Number of photons incident per second $N = P / E_1 = 200 / 10^{-18} = 2 \times 10^{20} \ s^{-1}$.
Since efficiency is $100 \%$,the number of electrons emitted per second is also $N = 2 \times 10^{20} \ s^{-1}$.
When these electrons are accelerated by a potential $V = 500 \ V$,their final kinetic energy $K = eV = 1.6 \times 10^{-19} \times 500 = 8 \times 10^{-17} \ J$.
The momentum of each electron upon impact is $p = \sqrt{2m_eK} = \sqrt{2 \times 9 \times 10^{-31} \times 8 \times 10^{-17}} = \sqrt{144 \times 10^{-48}} = 12 \times 10^{-24} \ kg \ m/s$.
The force exerted on the anode is $F = N \times p = (2 \times 10^{20}) \times (12 \times 10^{-24}) = 24 \times 10^{-4} \ N$.
Comparing with $F = n \times 10^{-4} \ N$,we get $n = 24$.

(A) The energy of a photon emitted during a transition from $n_2$ to $n_1$ in a hydrogen-like atom is given by $\Delta E = 13.6 Z^2 \left[ \frac{1}{n_1^2} - \frac{1}{n_2^2} \right] \ eV$.
For the transition $n = 2$ to $n = 1$:
$E_1 = 13.6 Z^2 \left[ \frac{1}{1^2} - \frac{1}{2^2} \right] = 13.6 Z^2 \left( 1 - \frac{1}{4} \right) = 13.6 Z^2 \left( \frac{3}{4} \right)$.
For the transition $n = 3$ to $n = 2$:
$E_2 = 13.6 Z^2 \left[ \frac{1}{2^2} - \frac{1}{3^2} \right] = 13.6 Z^2 \left( \frac{1}{4} - \frac{1}{9} \right) = 13.6 Z^2 \left( \frac{5}{36} \right)$.
According to the problem,$E_1 - E_2 = 74.8 \ eV$:
$13.6 Z^2 \left( \frac{3}{4} - \frac{5}{36} \right) = 74.8$.
Calculating the term in the bracket:
$\frac{3}{4} - \frac{5}{36} = \frac{27 - 5}{36} = \frac{22}{36} = \frac{11}{18}$.
Substituting back:
$13.6 Z^2 \times \frac{11}{18} = 74.8$.
$Z^2 = \frac{74.8 \times 18}{13.6 \times 11} = 5.5 \times \frac{18}{11} = 0.5 \times 18 = 9$.
$Z = 3$.

(B) $(1)$ Electric field due to a point charge at the origin is $E = \frac{kQ}{d^2}$,so $E \propto \frac{1}{d^2}$.
$(2)$ Electric field at any point on the axis of a dipole is $E = \frac{2kp}{d^3} = \frac{4kQl}{d^3}$,so $E \propto \frac{1}{d^3}$.
$(3)$ Electric field due to an infinite long line charge is $E = \frac{2k\lambda}{d}$,so $E \propto \frac{1}{d}$.
$(4)$ Electric field due to two infinite long wires is $\vec{E} = \vec{E}_1 + \vec{E}_2 = \frac{2k\lambda}{d-l} - \frac{2k\lambda}{d+l} = \frac{4k\lambda l}{d^2-l^2}$. If $d \gg l$,then $E = \frac{4k\lambda l}{d^2}$,so $E \propto \frac{1}{d^2}$.
$(5)$ Electric field due to an infinite plane charge is $E = \frac{\sigma}{2\epsilon_0}$,which is independent of $d$.
Matching: $P \rightarrow 5$,$Q \rightarrow 3$,$R \rightarrow 1, 4$,$S \rightarrow 2$. Thus,option $B$ is correct.