TS EAMCET 2022 Physics Question Paper with Answer and Solution

(A) The coordinates of the masses arranged on the circle of radius $R = 1 \ m$ are:
$M$ at $(1, 0)$
$2M$ at $(0, 1)$
$3M$ at $(-1, 0)$
$4M$ at $(0, -1)$
The $x$-coordinate of the center of mass $(X_{cm})$ is given by:
$X_{cm} = \frac{M(1) + 2M(0) + 3M(-1) + 4M(0)}{M + 2M + 3M + 4M} = \frac{M - 3M}{10M} = \frac{-2M}{10M} = -\frac{1}{5} \ m$
The $y$-coordinate of the center of mass $(Y_{cm})$ is given by:
$Y_{cm} = \frac{M(0) + 2M(1) + 3M(0) + 4M(-1)}{M + 2M + 3M + 4M} = \frac{2M - 4M}{10M} = \frac{-2M}{10M} = -\frac{1}{5} \ m$
Thus,the position vector of the center of mass is $-\frac{1}{5} \hat{i} - \frac{1}{5} \hat{j}$.

(B) Let the mass of the moving particle be $m$ and its initial velocity be $u$. The mass of the stationary particle is $m' = \frac{m}{n}$.
Assuming a perfectly elastic head-on collision,we apply the conservation of linear momentum:
$mu = mv_1 + \frac{m}{n}v_2 \Rightarrow u = v_1 + \frac{v_2}{n}$ (Equation $1$)
Using the coefficient of restitution $e=1$ for an elastic collision:
$v_2 - v_1 = u \Rightarrow v_1 = v_2 - u$ (Equation $2$)
Substituting Equation $2$ into Equation $1$:
$u = (v_2 - u) + \frac{v_2}{n} \Rightarrow 2u = v_2(1 + \frac{1}{n}) = v_2(\frac{n+1}{n})$
$v_2 = \frac{2nu}{n+1}$
The kinetic energy transferred to the stationary particle is $K' = \frac{1}{2} m' v_2^2 = \frac{1}{2} (\frac{m}{n}) (\frac{2nu}{n+1})^2 = \frac{1}{2} \frac{m}{n} \frac{4n^2 u^2}{(n+1)^2} = \frac{2mnu^2}{(n+1)^2}$.
The initial kinetic energy is $K = \frac{1}{2} mu^2$.
The fraction of kinetic energy transferred is $\frac{K'}{K} = \frac{\frac{2mnu^2}{(n+1)^2}}{\frac{1}{2} mu^2} = \frac{4n}{(n+1)^2}$.

(A) Let the mass of both particles be $m$. Let $V_1$ and $V_2$ be the velocities of particles $A$ and $B$ after the collision,respectively.
By the law of conservation of momentum:
$m \times 10 + 0 = m V_1 + m V_2 \Rightarrow V_1 + V_2 = 10$ ... $(i)$
Given that the kinetic energy of the system decreases by $1 \%$,the final kinetic energy $K_f = 0.99 K_i$.
$\frac{1}{2} m V_1^2 + \frac{1}{2} m V_2^2 = 0.99 \times (\frac{1}{2} m \times 10^2)$
$V_1^2 + V_2^2 = 0.99 \times 100 = 99$ ... (ii)
We know that $(V_1 + V_2)^2 = V_1^2 + V_2^2 + 2 V_1 V_2$.
Substituting the values: $10^2 = 99 + 2 V_1 V_2 \Rightarrow 100 = 99 + 2 V_1 V_2 \Rightarrow 2 V_1 V_2 = 1 \Rightarrow V_1 V_2 = 0.5$.
Now,$(V_1 - V_2)^2 = (V_1 + V_2)^2 - 4 V_1 V_2 = 100 - 4(0.5) = 100 - 2 = 98$.
$V_1 - V_2 = \sqrt{98} = 7\sqrt{2} \approx 9.899 \ m/s$.
Adding $(i)$ and (ii): $2 V_1 = 10 + 9.899 = 19.899 \Rightarrow V_1 \approx 9.95 \ m/s$.

(B) The impulse imparted to the ball is equal to the area under the $F-t$ graph.
Impulse = Area of the triangle = $\frac{1}{2} \times \text{base} \times \text{height}$
Impulse = $\frac{1}{2} \times (2 \times 10^{-3} \ s) \times (20 \times 10^3 \ N)$
Impulse = $20 \ N \ s = 20 \ kg \ m \ s^{-1}$
According to the impulse-momentum theorem,impulse is equal to the change in momentum $(\Delta p)$:
$\Delta p = m v - m u = \text{Impulse}$
Given: mass $m = 100 \ g = 0.1 \ kg$,initial velocity $u = 10 \ m \ s^{-1}$.
$0.1 \times v - 0.1 \times 10 = 20$
$0.1 \times v - 1 = 20$
$0.1 \times v = 21$
$v = \frac{21}{0.1} = 210 \ m \ s^{-1}$

(C) In an elastic collision,both the total linear momentum and the total kinetic energy of the system are conserved.
Assertion $(A)$ is true because by definition,an elastic collision is one in which there is no loss of kinetic energy.
Reason $(R)$ is false because,during an elastic collision,there is indeed an exchange of energy between the colliding bodies (momentum and kinetic energy are transferred between them),even though the total kinetic energy of the system remains constant.
Therefore,$(A)$ is true but $(R)$ is false.

(C) Let the mass of the original sphere $A$ be $M$. The force $F$ on particle $B$ of mass $m$ at distance $2R$ is $F = \frac{GMm}{(2R)^2} = \frac{GMm}{4R^2}$.
The mass of the removed spherical portion of radius $r = R/2$ is $M^{\prime} = \rho \cdot \frac{4}{3} \pi (R/2)^3 = \frac{M}{\frac{4}{3} \pi R^3} \cdot \frac{4}{3} \pi \frac{R^3}{8} = \frac{M}{8}$.
The distance of the centre of the removed portion from the particle $B$ is $d = 2R - R/2 = 3R/2$.
The force exerted by the removed portion on $B$ is $F_{removed} = \frac{G M^{\prime} m}{d^2} = \frac{G (M/8) m}{(3R/2)^2} = \frac{GMm}{8 \cdot (9R^2/4)} = \frac{GMm}{18R^2}$.
Since $F = \frac{GMm}{4R^2}$,we have $\frac{GMm}{R^2} = 4F$. Thus,$F_{removed} = \frac{4F}{18} = \frac{2}{9} F$.
The new force $F^{\prime}$ is $F - F_{removed} = F - \frac{2}{9} F = \frac{7}{9} F$.

(B) Using the principle of conservation of mechanical energy, the total energy at the surface equals the total energy at the maximum height $h$ reached by the rocket.
Total energy at surface = Total energy at height $h$
$-\frac{GMm}{R} + \frac{1}{2}mu^2 = -\frac{GMm}{R+h} + 0$
Dividing by $m$ and using $GM = gR^2$:
$-\frac{gR^2}{R} + \frac{1}{2}u^2 = -\frac{gR^2}{R+h}$
$-gR + \frac{1}{2}u^2 = -\frac{gR^2}{R+h}$
Given $u = 4000 \ m/s$, $R = 6.4 \times 10^6 \ m$, and $g = 10 \ m/s^2$:
$-(10)(6.4 \times 10^6) + \frac{1}{2}(4000)^2 = -\frac{10 \times (6.4 \times 10^6)^2}{6.4 \times 10^6 + h}$
$-6.4 \times 10^7 + 8 \times 10^6 = -\frac{4.096 \times 10^{14}}{6.4 \times 10^6 + h}$
$-5.6 \times 10^7 = -\frac{4.096 \times 10^{14}}{6.4 \times 10^6 + h}$
$6.4 \times 10^6 + h = \frac{4.096 \times 10^{14}}{5.6 \times 10^7} = 0.7314 \times 10^7 = 7.314 \times 10^6 \ m$
$h = 7.314 \times 10^6 - 6.4 \times 10^6 = 0.914 \times 10^6 \ m = 914 \ km$.
Thus, the correct option is $914.28 \ km$.

(B) The gravitational potential energy $U$ of a system of particles is given by the sum of the potential energies of all possible pairs: $U = -\sum \frac{Gm_i m_j}{r_{ij}}$.
For a rectangle with sides $a = 3 l_o$ and $b = 4 l_o$,the diagonal $d = \sqrt{(3 l_o)^2 + (4 l_o)^2} = 5 l_o$.
There are $4$ pairs with side length $3 l_o$ (two sides),$4 l_o$ (two sides),and $5 l_o$ (two diagonals).
$U = -\left[ \frac{Gm^2}{3 l_o} \times 2 + \frac{Gm^2}{4 l_o} \times 2 + \frac{Gm^2}{5 l_o} \times 2 \right]$
$U = -\frac{Gm^2}{l_o} \left[ \frac{2}{3} + \frac{2}{4} + \frac{2}{5} \right]$
$U = -\frac{Gm^2}{l_o} \left[ \frac{40 + 30 + 24}{60} \right] = -\frac{94}{60} \frac{Gm^2}{l_o} = -\frac{47}{30} \frac{Gm^2}{l_o}$.
The magnitude is $|U| = \frac{47}{30} \frac{Gm^2}{l_o}$.

(A) By the law of conservation of mechanical energy,the total energy at the surface of the earth must equal the total energy at a point far away from the earth (where potential energy is zero).
Let $m$ be the mass of the object and $M$ be the mass of the earth.
The escape speed $V_0$ is given by $V_0 = \sqrt{\frac{2GM}{R}}$. Thus,$V_0^2 = \frac{2GM}{R}$.
At the surface: $E_i = \frac{1}{2} m(5V_0)^2 - \frac{GMm}{R}$.
Far away: $E_f = \frac{1}{2} mV^2 + 0$.
Equating $E_i = E_f$:
$\frac{1}{2} m(25V_0^2) - \frac{GMm}{R} = \frac{1}{2} mV^2$.
Substitute $\frac{GM}{R} = \frac{V_0^2}{2}$:
$\frac{25}{2} mV_0^2 - m(\frac{V_0^2}{2}) = \frac{1}{2} mV^2$.
$12 mV_0^2 = \frac{1}{2} mV^2$.
$V^2 = 24 V_0^2$.
$V = \sqrt{24} V_0 = 2\sqrt{6} V_0$.

(C) According to the Shell Theorem,the gravitational field $E$ inside a uniform hollow spherical shell is $0$. Therefore,no net gravitational force acts on a point mass placed inside it. Thus,Statement $(I)$ is false.
For any point outside a uniform hollow spherical shell,the shell behaves as if its entire mass were concentrated at its center. Thus,the gravitational force is calculated using Newton's Law of Gravitation as $F = G M m / r^2$,where $r$ is the distance from the center. Thus,Statement $(II)$ is correct.

(C) As the particles move under gravitational influence,the gravitational force provides the necessary centripetal force.
The gravitational force between two bodies of mass $M$ separated by distance $L$ is $F = \frac{G M^2}{L^2}$.
The net force on a particle at vertex $A$ due to particles at $B$ and $C$ is the vector sum of the forces $\vec{F}_{AB}$ and $\vec{F}_{AC}$.
Since the angle between these forces is $60^\circ$,the magnitude of the resultant force is $F_{net} = 2 F \cos(30^\circ) = 2 \left( \frac{G M^2}{L^2} \right) \frac{\sqrt{3}}{2} = \frac{\sqrt{3} G M^2}{L^2}$.
The particles move in a circle of radius $R$. For an equilateral triangle,the distance from the centroid to a vertex is $R = \frac{L}{\sqrt{3}}$.
Equating the net force to the centripetal force,$\frac{M v^2}{R} = F_{net}$.
Substituting the values,$\frac{M v^2}{L/\sqrt{3}} = \frac{\sqrt{3} G M^2}{L^2}$.
Solving for $v$,$v^2 = \frac{\sqrt{3} G M^2}{L^2} \cdot \frac{L}{\sqrt{3} M} = \frac{G M}{L}$.
Therefore,$v = \sqrt{\frac{G M}{L}}$.

(A) Since the container is placed in a reservoir at a constant temperature $T$,the process is isothermal.
According to Boyle's Law for an ideal gas at constant temperature,$PV = \text{constant}$.
Therefore,$P_1 V_1 = P_2 V_2$.
Given $P_1 = P$,$V_1 = V$,and $V_2 = 2V$.
Substituting these values: $P \times V = P^{\prime} \times (2V)$.
Solving for $P^{\prime}$: $P^{\prime} = \frac{PV}{2V} = \frac{1}{2} P$.

(C) Statement $(I)$ is false because gas thermometers are more sensitive than liquid thermometers due to the higher coefficient of thermal expansion of gases.
Statement $(II)$ is true. Boltzmann's constant $k_B$ is defined as the ratio of the universal gas constant $R$ to Avogadro's number $N_A$,i.e.,$k_B = \frac{R}{N_A}$.
Statement $(III)$ is true. From the ideal gas equation $PV = nRT$,we have $PV = \frac{m}{M}RT$,where $m$ is mass and $M$ is molar mass. Since density $\rho = \frac{m}{V}$,we get $P = \frac{\rho RT}{M}$. At constant pressure $P$,$\rho \propto \frac{1}{T}$.
Therefore,statements $(II)$ and $(III)$ are true,but statement $(I)$ is false.

(D) Let $M$ be the mass of chamber $B$ and $m$ be the mass of each ball. The total downward force exerted by the chamber and the balls on the gas is $F = (M + nm)g$.
If $A$ is the cross-sectional area of the chamber,the pressure $P$ of the gas is given by $P = F/A = (M + nm)g / A$.
When one ball is removed,the new force is $F^{\prime} = (M + (n-1)m)g$.
The new pressure $P^{\prime}$ is $P^{\prime} = F^{\prime} / A = (M + (n-1)m)g / A$.
The difference in pressure is $P - P^{\prime} = (M + nm)g/A - (M + nm - m)g/A = mg/A$.
Thus,$(P - P^{\prime}) / P = (mg/A) / ((M + nm)g/A) = m / (M + nm)$.
Assuming the mass of the chamber $M$ is negligible compared to the total mass of the balls (or that the problem implies the weight is primarily due to the balls),we have $M \approx 0$.
Then,$(P - P^{\prime}) / P = m / (nm) = 1/n$.

(A) The root mean square (rms) speed of gas molecules is given by the formula $v_{rms} = \sqrt{\frac{3RT}{M}}$, where $R$ is the universal gas constant, $T$ is the absolute temperature, and $M$ is the molar mass of the gas.
For a given temperature $T$, the rms speed is inversely proportional to the square root of the molar mass: $v_{rms} \propto \frac{1}{\sqrt{M}}$.
Therefore, the ratio of the rms speeds of Helium $(He)$ and Nitrogen $(N_2)$ is $\frac{v_{He}}{v_{N_2}} = \sqrt{\frac{M_{N_2}}{M_{He}}}$.
The molar mass of Nitrogen $(N_2)$ is $M_{N_2} = 28 \ g \ mol^{-1}$ and the molar mass of Helium $(He)$ is $M_{He} = 4 \ g \ mol^{-1}$.
Given $v_{N_2} = 100 \ m \ s^{-1}$, we substitute the values: $\frac{v_{He}}{100} = \sqrt{\frac{28}{4}} = \sqrt{7}$.
Thus, $v_{He} = 100 \sqrt{7} \ m \ s^{-1}$.

(B) The root mean square (r.m.s.) velocity of a gas molecule is given by the formula: $v_{rms} = \sqrt{\frac{3RT}{M}}$.
Here,$R$ is the universal gas constant,$T$ is the absolute temperature,and $M$ is the molar mass of the gas.
For hydrogen $(H_2)$,the molar mass $M_{H_2} = 2 \times 10^{-3} \ kg/mol$ and temperature $T_{H_2} = 20 \ K$.
For oxygen $(O_2)$,the molar mass $M_{O_2} = 32 \times 10^{-3} \ kg/mol$.
We are given that the r.m.s. velocities are equal: $v_{rms(H_2)} = v_{rms(O_2)}$.
Therefore,$\sqrt{\frac{3RT_{H_2}}{M_{H_2}}} = \sqrt{\frac{3RT_{O_2}}{M_{O_2}}}$.
Squaring both sides and simplifying: $\frac{T_{H_2}}{M_{H_2}} = \frac{T_{O_2}}{M_{O_2}}$.
Substituting the values: $\frac{20}{2} = \frac{T_{O_2}}{32}$.
$10 = \frac{T_{O_2}}{32} \implies T_{O_2} = 320 \ K$.

(C) The root mean square speed of a gas is given by the formula $V_{rms} = \sqrt{\frac{3RT}{M}}$,where $R$ is the universal gas constant,$T$ is the temperature in Kelvin,and $M$ is the molar mass.
Given,$(V_{rms})_{Ne} = (V_{rms})_{He}$.
Substituting the formula: $\sqrt{\frac{3RT_{Ne}}{M_{Ne}}} = \sqrt{\frac{3RT_{He}}{M_{He}}}$.
Squaring both sides and canceling common terms: $\frac{T_{Ne}}{M_{Ne}} = \frac{T_{He}}{M_{He}}$.
Given $T_{He} = -33^{\circ} C = (-33 + 273) \ K = 240 \ K$,$M_{Ne} = 20.2 \ u$,and $M_{He} = 4.0 \ u$.
Substituting the values: $\frac{T_{Ne}}{20.2} = \frac{240}{4.0}$.
$\frac{T_{Ne}}{20.2} = 60$.
$T_{Ne} = 60 \times 20.2 = 1212 \ K$.

(B) The frictional force acting on both blocks is kinetic in nature.
For the $1 \ kg$ block:
$f_1 = \mu m_1 g \cos 45^{\circ} = 0.4 \times 1 \times 10 \times \frac{1}{\sqrt{2}} = \frac{4}{\sqrt{2}} = 2\sqrt{2} \ N$
For the $2 \ kg$ block:
$f_2 = \mu m_2 g \cos 45^{\circ} = 0.4 \times 2 \times 10 \times \frac{1}{\sqrt{2}} = \frac{8}{\sqrt{2}} = 4\sqrt{2} \ N$
The net force along the incline in the downward direction provides the acceleration $a = \alpha \sqrt{2}$:
$(m_1 + m_2)g \sin 45^{\circ} - (f_1 + f_2) = (m_1 + m_2)a$
$(1 + 2) \times 10 \times \frac{1}{\sqrt{2}} - (2\sqrt{2} + 4\sqrt{2}) = (1 + 2) \times (\alpha \sqrt{2})$
$30 \times \frac{1}{\sqrt{2}} - 6\sqrt{2} = 3\alpha \sqrt{2}$
Multiply throughout by $\sqrt{2}$:
$30 - 6 \times 2 = 3\alpha \times 2$
$30 - 12 = 6\alpha$
$18 = 6\alpha$
$\alpha = 3$

(B) The block will start slipping when the slope of the curve at that point is equal to the tangent of the angle of repose,$\tan \theta = \mu_s$.
The slope of the curve at any point $x$ is given by the derivative:
$m = \frac{dy}{dx} = \frac{d}{dx} \left( \frac{x^2}{20} \right) = \frac{2x}{20} = \frac{x}{10}$.
For the block to remain in equilibrium without slipping,the slope must satisfy:
$\frac{dy}{dx} \leq \mu_s$
At the point of slipping,$\frac{x}{10} = 0.5$.
Solving for $x$:
$x = 0.5 \times 10 = 5 \ m$.
Now,substitute $x = 5 \ m$ into the equation of the parabola to find the maximum height $h$:
$h = y = \frac{x^2}{20} = \frac{5^2}{20} = \frac{25}{20} = 1.25 \ m$.

(D) Given: initial velocity $u = 7 \ m/s$,final velocity $v = 0 \ m/s$,distance $s = 10 \ m$,and $g = 9.8 \ m/s^2$.
Using the equation of motion $v^2 = u^2 + 2as$:
$0^2 = (7)^2 + 2 \cdot a \cdot 10$
$0 = 49 + 20a$
$a = -\frac{49}{20} = -2.45 \ m/s^2$.
We know that $g = 9.8 \ m/s^2$,so $a = -\frac{9.8}{4} = -\frac{g}{4}$.
According to Newton's second law,the resistance force $R = ma$.
Substituting $a = -\frac{g}{4}$:
$R = m \left(-\frac{g}{4}\right) = -\frac{mg}{4}$.
Since weight $W = mg$,we get $R = -\frac{W}{4}$.

(A) First,calculate the mass of block $A$ using Newton's second law: $F = m_A a \Rightarrow 10 = m_A \times 20 \Rightarrow m_A = 0.5 \,kg$.
When block $A$ is placed against block $B$ and a force $F^{\prime} = 20 \,N$ is applied,both blocks move together with a common acceleration $a$.
The equation of motion for block $A$ is: $F^{\prime} - N = m_A a \Rightarrow 20 - N = 0.5 a$ $(i)$.
The equation of motion for block $B$ is: $N = m_B a \Rightarrow N = 1.5 a$ $(ii)$.
Adding equations $(i)$ and $(ii)$: $20 = 2 a \Rightarrow a = 10 \,m/s^2$.
Substituting $a$ into equation $(ii)$: $N = 1.5 \times 10 = 15 \,N$.
Thus,the force on block $B$ is $15 \,N$.

(B) Let at $t=t_0$,the body leaves the plane. Then,at $t=t_0$,the normal force $N=0$.
The vertical component of the force is $F_y = F \sin 45^{\circ} = \alpha t_0 \sin 45^{\circ}$.
For the body to break off the plane,the vertical component of the force must balance the weight of the body:
$N + \alpha t_0 \sin 45^{\circ} = mg$
Since $N=0$ at the moment of breaking off,we have:
$\alpha t_0 \sin 45^{\circ} = mg$
Given $m = 1 \text{ kg}$ and $g = 10 \text{ m/s}^2$:
$\alpha t_0 \left(\frac{1}{\sqrt{2}}\right) = 1 \times 10$
$t_0 = \frac{10 \sqrt{2}}{\alpha} \text{ s}$.
The horizontal component of the force is $F_x = F \cos 45^{\circ} = \alpha t \cos 45^{\circ}$.
The acceleration of the body in the horizontal direction is $a = \frac{F_x}{m} = \frac{\alpha t \cos 45^{\circ}}{1} = \frac{\alpha t}{\sqrt{2}}$.
The velocity $V$ at time $t_0$ is given by:
$V = \int_0^{t_0} a \, dt = \int_0^{t_0} \frac{\alpha t}{\sqrt{2}} \, dt = \frac{\alpha}{\sqrt{2}} \left[ \frac{t^2}{2} \right]_0^{t_0} = \frac{\alpha t_0^2}{2 \sqrt{2}}$.
Substituting $t_0 = \frac{10 \sqrt{2}}{\alpha}$:
$V = \frac{\alpha}{2 \sqrt{2}} \left( \frac{10 \sqrt{2}}{\alpha} \right)^2 = \frac{\alpha}{2 \sqrt{2}} \times \frac{100 \times 2}{\alpha^2} = \frac{100}{\sqrt{2} \alpha} = \frac{50 \sqrt{2}}{\alpha} \text{ m/s}$.

(C) Statement $(I)$ is incorrect because the resultant velocity magnitude is given by the vector addition formula: $|\vec{v}| = \sqrt{|\vec{v_1}|^2 + |\vec{v_2}|^2 + 2|\vec{v_1}| |\vec{v_2}| \cos \theta}$. It is only equal to the sum of magnitudes if the vectors are in the same direction $(\theta = 0^\circ)$.
Statement $(II)$ is correct because displacement is the shortest distance between two points,while path length is the total distance covered. Thus,displacement $\leq$ distance.
Statement $(III)$ is correct by definition. Instantaneous acceleration is defined as $\vec{a} = \lim_{\Delta t \rightarrow 0} \frac{\Delta \vec{v}}{\Delta t}$.

(C) Given the relation $S = \frac{a^{1/2} b}{c^3 d^4}$.
To find the maximum relative error in $S$,we use the formula for propagation of errors:
$\frac{\Delta S}{S} = \frac{1}{2} \frac{\Delta a}{a} + \frac{\Delta b}{b} + 3 \frac{\Delta c}{c} + 4 \frac{\Delta d}{d}$.
Substituting the given percentage errors:
$\left( \frac{\Delta S}{S} \times 100 \right) = \frac{1}{2} \times (2 \%) + (1 \%) + 3 \times (1 \%) + 4 \times (1 \%)$.
Calculating the values:
$= 1 \% + 1 \% + 3 \% + 4 \% = 9 \%$.
Therefore,the percentage error in $S$ is $9 \%$.

(A) The four fundamental forces in nature are gravitational force, weak nuclear force, electromagnetic force, and strong nuclear force.
$1$. The range of the weak nuclear force is approximately $10^{-18} \,m$, which is the shortest among all four fundamental forces.
$2$. The range of gravitational and electromagnetic forces is infinite.
$3$. The relative strengths of the forces are: Strong nuclear force $(1)$ > Electromagnetic force $(10^{-2})$ > Weak nuclear force $(10^{-13})$ > Gravitational force $(10^{-39})$.
Therefore, the statement that the range for the weak nuclear force is the shortest is correct.

(A) The ant's displacement vector $\vec{a}$ is given by the movement along the $x$ and $y$ axes:
$\vec{a} = 10 \hat{i} + 20 \hat{j}$
The position vector $\vec{b}$ of the point with magnitude $r = \sqrt{2} \ cm$ at an angle $\theta = 45^{\circ}$ with the $x$-axis is:
$\vec{b} = r \cos \theta \hat{i} + r \sin \theta \hat{j}$
$\vec{b} = \sqrt{2} \cos 45^{\circ} \hat{i} + \sqrt{2} \sin 45^{\circ} \hat{j}$
$\vec{b} = \sqrt{2} \left( \frac{1}{\sqrt{2}} \right) \hat{i} + \sqrt{2} \left( \frac{1}{\sqrt{2}} \right) \hat{j} = \hat{i} + \hat{j}$
The dot product $\vec{a} \cdot \vec{b}$ is:
$\vec{a} \cdot \vec{b} = (10 \hat{i} + 20 \hat{j}) \cdot (\hat{i} + \hat{j})$
$= (10 \times 1) + (20 \times 1) = 10 + 20 = 30 \ cm^2$

(B) The volume of the material of the hollow sphere is $V_S = \frac{4}{3} \pi (R^3 - r^3)$,where $R = 4 \ cm$ and $r = 2 \ cm$.
For a floating body,the buoyant force equals the weight of the body.
$B = W$
$\rho_L V_{sub} g = \rho_S V_S g$
Since the sphere is half submerged,the submerged volume $V_{sub} = \frac{1}{2} \times ( \frac{4}{3} \pi R^3 ) = \frac{2}{3} \pi R^3$.
Substituting the values:
$2.0 \times \frac{2}{3} \pi (4)^3 = \rho_S \times \frac{4}{3} \pi (4^3 - 2^3)$
$2.0 \times 64 = \rho_S \times (64 - 8)$
$128 = \rho_S \times 56$
$\rho_S = \frac{128}{56} \approx 2.28 \ g \ cm^{-3}$.
Wait,re-evaluating the calculation: $V_S = \frac{4}{3} \pi (64 - 8) = \frac{4}{3} \pi (56) = \frac{224}{3} \pi$.
$V_{sub} = \frac{1}{2} \times \frac{4}{3} \pi (4^3) = \frac{128}{3} \pi$.
Equating forces: $2.0 \times (\frac{128}{3} \pi) = \rho_S \times (\frac{224}{3} \pi)$.
$\rho_S = \frac{256}{224} = 1.1428 \approx 1.14 \ g \ cm^{-3}$.

(A) The apparent weight of an object in a liquid is given by $W_{app} = W_{air} - F_B$,where $F_B = V \rho g$ is the buoyant force.
At $T_1 = 32^{\circ}C$: $W_{app1} = 39 \ gm$,$W_{air} = 49 \ gm$,$\rho_1 = 1.2 \times 10^3 \ kg/m^3$.
$V_1 = \frac{W_{air} - W_{app1}}{\rho_1} = \frac{(49 - 39) \times 10^{-3} \ kg}{1.2 \times 10^3 \ kg/m^3} = \frac{10 \times 10^{-3}}{1.2 \times 10^3} = \frac{1}{12} \times 10^{-5} \ m^3 \approx 8.33 \times 10^{-6} \ m^3$.
At $T_2 = 42^{\circ}C$: $W_{app2} = 40 \ gm$,$W_{air} = 49 \ gm$,$\rho_2 = 1.0 \times 10^3 \ kg/m^3$.
$V_2 = \frac{W_{air} - W_{app2}}{\rho_2} = \frac{(49 - 40) \times 10^{-3} \ kg}{1.0 \times 10^3 \ kg/m^3} = 9 \times 10^{-6} \ m^3$.
The change in volume is $\Delta V = V_2 - V_1 = V_1 (3\alpha \Delta T)$.
$9 \times 10^{-6} - 8.33 \times 10^{-6} = 8.33 \times 10^{-6} \times 3 \alpha \times (42 - 32)$.
$0.67 \times 10^{-6} = 8.33 \times 10^{-6} \times 3 \alpha \times 10$.
$\alpha = \frac{0.67}{8.33 \times 30} \approx 2.67 \times 10^{-3} /^{\circ}C = \frac{8}{3} \times 10^{-3} /^{\circ}C$.

(C) The velocity of efflux $(v)$ for an orifice at depth $h$ is given by Torricelli's Law:
$v = \sqrt{2gh}$
Given $h = 20.0 \ m$ and $g = 10 \ m s^{-2}$:
$v = \sqrt{2 \times 10 \times 20} = \sqrt{400} = 20 \ m s^{-1}$
The volume flow rate $(Q)$ is the product of the area of the orifice $(A)$ and the velocity of efflux $(v)$:
$Q = A \times v$
$3.08 \times 10^{-5} = \left( \frac{\pi d^2}{4} \right) \times 20$
Rearranging for $d^2$:
$d^2 = \frac{4 \times 3.08 \times 10^{-5}}{20 \times \pi} = \frac{12.32 \times 10^{-5}}{62.83} \approx 1.96 \times 10^{-6} \ m^2$
Taking the square root:
$d = \sqrt{1.96 \times 10^{-6}} = 1.4 \times 10^{-3} \ m = 1.4 \ mm$

(B) According to the equation of continuity for an incompressible fluid,the product of the cross-sectional area and the velocity of the fluid remains constant at all points along the flow:
$A_A v_A = A_B v_B$
Given that the pipe is cylindrical,the cross-sectional area $A = \pi r^2$.
At point $A$,the radius is $2R$,so $A_A = \pi (2R)^2 = 4\pi R^2$.
At point $B$,the radius is $R$,so $A_B = \pi R^2$.
The velocity at point $B$ is given as $v_B = 4v$.
Substituting these values into the continuity equation:
$4\pi R^2 \times v_A = \pi R^2 \times 4v$
Dividing both sides by $4\pi R^2$:
$v_A = v$

(B) Given:
Diameter of pipe $D_1 = 4 \,cm$,radius $r_1 = 2 \,cm$.
Diameter of throat $D_2 = 2 \,cm$,radius $r_2 = 1 \,cm$.
Velocity in pipe $V_1 = 10 \,m/s$.
Density of water $\rho = 1000 \,kg/m^3$.
Step $1$: Using the equation of continuity $A_1 V_1 = A_2 V_2$:
$\pi r_1^2 V_1 = \pi r_2^2 V_2$
$(2)^2 \times 10 = (1)^2 \times V_2$
$V_2 = 40 \,m/s$.
Step $2$: Using Bernoulli's theorem for horizontal flow:
$P_1 + \frac{1}{2} \rho V_1^2 = P_2 + \frac{1}{2} \rho V_2^2$
$P_1 - P_2 = \frac{1}{2} \rho (V_2^2 - V_1^2)$
$P_1 - P_2 = \frac{1}{2} \times 1000 \times (40^2 - 10^2)$
$P_1 - P_2 = 500 \times (1600 - 100)$
$P_1 - P_2 = 500 \times 1500 = 7.5 \times 10^5 \,Pa$.

(B) The excess pressure inside an air bubble at a depth $h$ in a liquid is given by the sum of the pressure due to the liquid column and the excess pressure due to surface tension.
For an air bubble in a liquid,there is only one free surface.
Excess pressure due to surface tension is given by $\Delta P_s = \frac{2S}{R}$.
Pressure due to depth is given by $\Delta P_h = \rho g h$.
Total excess pressure $\Delta P = \frac{2S}{R} + \rho g h$.
Given: $S = 0.1 \ N \ m^{-1}$,$R = 1 \ mm = 1 \times 10^{-3} \ m$,$\rho = 2000 \ kg \ m^{-3}$,$h = 8 \ cm = 8 \times 10^{-2} \ m$,and $g = 10 \ m \ s^{-2}$.
Substituting the values:
$\Delta P = \frac{2 \times 0.1}{1 \times 10^{-3}} + (2000 \times 10 \times 8 \times 10^{-2})$
$\Delta P = 200 + 1600 = 1800 \ N \ m^{-2}$.

(D) soap bubble has two surfaces (inner and outer),so the surface area is $2 \times 4 \pi r^2 = 8 \pi r^2$.
Initial radius $r_i = R$.
Initial surface energy $U_i = T \times (8 \pi R^2) = 8 \pi R^2 T$.
Final diameter is doubled,so final radius $r_f = 2R$.
Final surface energy $U_f = T \times (8 \pi (2R)^2) = T \times (8 \pi \times 4R^2) = 32 \pi R^2 T$.
The surface energy needed is $\Delta U = U_f - U_i$.
$\Delta U = 32 \pi R^2 T - 8 \pi R^2 T = 24 \pi R^2 T$.

(C) Let $H = 50 \ cm$ be the total height of the vessel. Let the hole be made at a height $y$ from the bottom. Then the depth of the hole from the free surface is $h = H - y = 50 - y$.
The velocity of the water jet is $v = \sqrt{2gh} = \sqrt{2g(50-y)}$.
The time taken for the water to reach the table is $t = \sqrt{\frac{2y}{g}}$.
The horizontal range $R$ is given by $R = v \times t = \sqrt{2g(50-y)} \times \sqrt{\frac{2y}{g}} = 2\sqrt{y(50-y)}$.
To maximize the range $R$,we differentiate $R$ with respect to $y$ and set it to zero:
$\frac{dR}{dy} = 2 \times \frac{1}{2\sqrt{y(50-y)}} \times (50 - 2y) = 0$.
This gives $50 - 2y = 0$,so $y = 25 \ cm$.
Thus,the hole should be made at a height of $25 \ cm$ from the bottom.

(A) Terminal velocity $(v_t)$ is given by the formula derived from Stokes' Law: $v_t = \frac{2r^2(\rho - \sigma)g}{9\eta}$.
Given: Radius $r = 0.02 \ mm = 2 \times 10^{-5} \ m$,viscosity $\eta = 1.8 \times 10^{-5} \ N \ s \ m^{-2}$,density of water $\rho = 1000 \ kg \ m^{-3}$,density of air $\sigma \approx 0$,and $g = 10 \ m \ s^{-2}$.
Substituting the values: $v_t = \frac{2 \times (2 \times 10^{-5})^2 \times 1000 \times 10}{9 \times 1.8 \times 10^{-5}}$.
$v_t = \frac{2 \times 4 \times 10^{-10} \times 10^4}{16.2 \times 10^{-5}} = \frac{8 \times 10^{-6}}{16.2 \times 10^{-5}} = \frac{80}{16.2} \times 10^{-2} \approx 4.938 \times 10^{-2} \ m \ s^{-1}$.
Converting to $cm \ s^{-1}$: $v_t \approx 4.938 \times 10^{-2} \times 100 \ cm \ s^{-1} = 4.938 \ cm \ s^{-1}$.
Rounding to the nearest option,we get $4.9 \ cm \ s^{-1}$.

$(A)$ Given:
Side of the cube, $L = 10 \,cm = 0.1 \,m$.
Area of the base, $A = L^2 = (0.1 \,m)^2 = 0.01 \,m^2$.
Thickness of the liquid film, $dx = 0.2 \,mm = 0.2 \times 10^{-3} \,m$.
Applied force, $F = 0.1 \,N$.
Constant velocity, $v = 0.08 \,m/s$.
Since the cube moves with a constant speed, the net force is zero, meaning the applied force equals the viscous drag force: $F = F_{drag}$.
According to Newton's law of viscosity, $F = \eta A \frac{dv}{dx}$.
Substituting the values: $0.1 = \eta \times 0.01 \times \frac{0.08}{0.2 \times 10^{-3}}$.
$0.1 = \eta \times 0.01 \times 400$.
$0.1 = \eta \times 4$.
$\eta = \frac{0.1}{4} = 0.025 \,Ns/m^2 = 2.5 \times 10^{-2} \,Ns/m^2$.

(D) According to Poiseuille's law,the volume flow rate $Q$ is given by:
$Q = \frac{\pi \Delta P r^4}{8 L \eta}$
where $r$ is the radius,$\eta$ is the viscosity,and $\Delta P$ is the pressure difference.
Since the flow rate $Q$ does not depend on the density of blood,the change in density does not affect the flow rate.
Taking the logarithmic differentiation of the formula:
$\frac{\Delta Q}{Q} = 4 \frac{\Delta r}{r} - \frac{\Delta \eta}{\eta}$
Given that the radius $r$ increases by $1 \%$ $(\frac{\Delta r}{r} = 0.01)$ and the viscosity $\eta$ increases by $1 \%$ $(\frac{\Delta \eta}{\eta} = 0.01)$:
$\frac{\Delta Q}{Q} = 4(0.01) - 0.01 = 0.04 - 0.01 = 0.03$
Therefore,the percentage change in the flow rate is $0.03 \times 100 = 3 \%$.

(C) The initial surface area of the drop is $A_i = 4 \pi r^2$. The initial surface energy is $U_i = S \times 4 \pi r^2$.
Since the volume remains constant,the volume of the large drop equals the sum of the volumes of the $8$ small droplets: $\frac{4}{3} \pi r^3 = 8 \times \frac{4}{3} \pi (r')^3$.
This simplifies to $r^3 = 8(r')^3$,so $r = 2r'$,which means $r' = r/2$.
The final surface area of $8$ droplets is $A_f = 8 \times 4 \pi (r')^2 = 8 \times 4 \pi (r/2)^2 = 8 \times 4 \pi (r^2/4) = 8 \pi r^2$.
The final surface energy is $U_f = S \times 8 \pi r^2$.
The work done is equal to the change in surface energy: $W = \Delta U = U_f - U_i = S(8 \pi r^2 - 4 \pi r^2) = 4 \pi r^2 S$.

(D) The increase in pressure at the bottom of the pool due to the water column is given by $\Delta P = \rho g h$.
Substituting the given values: $\Delta P = 1000 \times 10 \times 22 = 2.2 \times 10^5 \ N \ m^{-2}$.
The bulk modulus $B$ is defined as $B = -\frac{\Delta P}{\Delta V / V}$.
Therefore,the fractional change in volume is $\left| \frac{\Delta V}{V} \right| = \frac{\Delta P}{B}$.
Substituting the values: $\left| \frac{\Delta V}{V} \right| = \frac{2.2 \times 10^5}{2.2 \times 10^9} = 10^{-4}$.

(B) Shear modulus is the ratio of shearing stress to shearing strain,representing the resistance to change against deformation force. Thus,$A \rightarrow V$.
$(B)$ Shearing stress is the force applied tangentially to the surface,also known as tangential stress. Thus,$B \rightarrow III$.
$(C)$ Elastic fatigue is the temporary loss of elastic properties of a material due to repeated alternating deforming forces. Thus,$C \rightarrow IV$.
$(D)$ Modulus of elasticity is the proportionality constant between stress and strain within the elastic limit. Thus,$D \rightarrow II$.
Therefore,the correct match is $A-V, B-III, C-IV, D-II$.

(B) At the lowest point of the vertical circle, the tension $T$ in the wire provides the necessary centripetal force and balances the weight of the object.
The net force $F$ acting on the wire is given by $F = mg + m\omega^2l$.
Given values: $m = 15 \,kg$, $l = 1.0 \,m$, $\omega = 4 \,rad/s$, $A = 0.05 \,cm^2 = 0.05 \times 10^{-4} \,m^2$, $Y = 2 \times 10^{11} \,N/m^2$, and $g = 10 \,m/s^2$.
Calculating the force $F$:
$F = (15 \times 10) + (15 \times 4^2 \times 1) = 150 + 240 = 390 \,N$.
Using the formula for Young's modulus $Y = \frac{F/A}{\Delta l/l}$, the elongation $\Delta l$ is:
$\Delta l = \frac{Fl}{AY} = \frac{390 \times 1.0}{(0.05 \times 10^{-4}) \times (2 \times 10^{11})}$
$\Delta l = \frac{390}{0.1 \times 10^7} = \frac{390}{10^6} = 390 \times 10^{-6} \,m = 0.39 \,mm$.
Thus, the correct option is $B$.

(B) The formula for extension $\Delta l$ is given by $\Delta l = \frac{F l}{Y A} = \frac{m g l}{Y \pi r^2}$.
Since $m, g, l$ are the same for both wires,we have $\Delta l \propto \frac{1}{Y r^2}$.
Therefore,$\frac{\Delta l_1}{\Delta l_2} = \frac{Y_2}{Y_1} \times \left( \frac{r_2}{r_1} \right)^2$.
Given $\Delta l_2 = 2 \Delta l_1$,so $\frac{\Delta l_1}{\Delta l_2} = \frac{1}{2}$.
Substituting the values: $\frac{1}{2} = \frac{Y_2}{Y_1} \times \left( \frac{1.5}{2} \right)^2$.
$\frac{1}{2} = \frac{Y_2}{Y_1} \times \left( \frac{3}{4} \right)^2 = \frac{Y_2}{Y_1} \times \frac{9}{16}$.
$\frac{Y_1}{Y_2} = \frac{9}{16} \times 2 = \frac{18}{16} = \frac{9}{8}$.

(A) Given: Radius $r = 10.0 \ mm = 0.01 \ m$,Length $L = 50.0 \ cm = 0.5 \ m$,Force $F = 10.0 \times \pi \ kN = 10^4 \pi \ N$,Young's modulus $Y = 2.0 \times 10^{11} \ N/m^2$.
The formula for change in length is $\Delta L = \frac{FL}{AY}$,where $A = \pi r^2$.
Substituting the values: $A = \pi \times (0.01)^2 = \pi \times 10^{-4} \ m^2$.
$\Delta L = \frac{(10^4 \pi) \times 0.5}{(\pi \times 10^{-4}) \times (2.0 \times 10^{11})}$.
$\Delta L = \frac{0.5 \times 10^4}{2.0 \times 10^7} = 0.25 \times 10^{-3} \ m = 0.25 \ mm$.

(B) The work done in stretching a wire is given by the formula for elastic potential energy stored in the wire:
$W = \frac{1}{2} \times \text{Stress} \times \text{Strain} \times \text{Volume}$
$W = \frac{1}{2} \times \left( Y \frac{\Delta \ell}{\ell} \right) \times \left( \frac{\Delta \ell}{\ell} \right) \times (A \ell)$
$W = \frac{1}{2} \frac{Y A}{\ell} (\Delta \ell)^2$
Given:
$Y = 2 \times 10^{11} \ N/m^2$
$A = 10^{-6} \ m^2$
$\ell = 2 \ m$
$\Delta \ell = 2.004 - 2 = 0.004 \ m = 4 \times 10^{-3} \ m$
Substituting the values:
$W = \frac{1}{2} \times \frac{2 \times 10^{11} \times 10^{-6}}{2} \times (4 \times 10^{-3})^2$
$W = \frac{1}{2} \times 10^5 \times 16 \times 10^{-6}$
$W = 0.5 \times 1.6 = 0.8 \ J$

(B) The particle starts from rest,so its initial velocity $u = 0 \ m \ s^{-1}$.
In an acceleration-time $(a-t)$ graph,the change in velocity $(\Delta v)$ is equal to the area under the curve.
Since the acceleration is positive throughout the interval from $t = 0 \ s$ to $t = 15 \ s$,the velocity of the particle increases continuously.
Therefore,the maximum speed is attained at $t = 15 \ s$.
The area under the $a-t$ graph is the area of the right-angled triangle with base $15 \ s$ and height $10 \ m \ s^{-2}$.
$\Delta v = \text{Area} = \frac{1}{2} \times \text{base} \times \text{height}$
$\Delta v = \frac{1}{2} \times 15 \ s \times 10 \ m \ s^{-2} = 75 \ m \ s^{-1}$.
Since $v_{max} = u + \Delta v = 0 + 75 \ m \ s^{-1} = 75 \ m \ s^{-1}$.

(C) Step $1$: Calculate the velocity and height at the end of the powered flight $(t = 5 \,s)$.
The initial velocity $u = 0$, acceleration $a = 20 \,m/s^2$, and time $t = 5 \,s$.
The velocity at $t = 5 \,s$ is $V_{max} = u + at = 0 + 20 \times 5 = 100 \,m/s$.
The height reached at $t = 5 \,s$ is $S = ut + \frac{1}{2}at^2 = 0 + \frac{1}{2} \times 20 \times 5^2 = 250 \,m$.
Step $2$: Calculate the final speed when it hits the ground.
After $t = 5 \,s$, the rocket is in free fall under gravity $(g = 10 \,m/s^2)$.
Using the work-energy theorem or kinematic equations, the total energy at the peak height is conserved, or we can use $V^2 = u^2 + 2aS$.
Here, the rocket starts from height $S = 250 \,m$ with an initial upward velocity $V_{max} = 100 \,m/s$.
When it hits the ground, the final velocity $V$ is given by $V^2 = V_{max}^2 + 2gS$.
$V^2 = (100)^2 + 2 \times 10 \times 250$.
$V^2 = 10000 + 5000 = 15000$.
$V = \sqrt{15000} = \sqrt{2500 \times 6} = 50 \sqrt{6} \,m/s$.

(C) The equation of motion for height $H$ at time $t$ is given by $H = ut - \frac{1}{2}gt^2$.
Since the ball passes the same height $H$ at $t_1 = 2 \,s$ and $t_2 = 10 \,s$,we have:
$H = u(2) - \frac{1}{2}g(2)^2 = 2u - 2g$
$H = u(10) - \frac{1}{2}g(10)^2 = 10u - 50g$
Equating the two expressions for $H$:
$2u - 2g = 10u - 50g$
$8u = 48g$
$u = 6g = 6 \times 9.8 = 58.8 \,m/s$
Now,substitute $u$ back into the expression for $H$:
$H = 2(58.8) - 2(9.8) = 117.6 - 19.6 = 98 \,m$.

(C) Let the initial velocity of the bullet be $u$.
After travelling a distance $x$,the velocity becomes $v_1 = u - \frac{1}{3}u = \frac{2u}{3}$.
Using the equation of motion $v^2 = u^2 + 2as$,where $a$ is the retardation (deceleration) inside the target:
$(\frac{2u}{3})^2 = u^2 - 2ax$
$\frac{4u^2}{9} = u^2 - 2ax$
$2ax = u^2 - \frac{4u^2}{9} = \frac{5u^2}{9} \quad \dots (1)$
Now,for the second part of the motion,the bullet starts with velocity $\frac{2u}{3}$ and comes to rest (final velocity $v_2 = 0$) after travelling a further distance $x^{\prime}$:
$0^2 = (\frac{2u}{3})^2 - 2ax^{\prime}$
$2ax^{\prime} = \frac{4u^2}{9} \quad \dots (2)$
Dividing equation $(2)$ by equation $(1)$:
$\frac{2ax^{\prime}}{2ax} = \frac{4u^2/9}{5u^2/9}$
$\frac{x^{\prime}}{x} = \frac{4}{5}$.

(A) Given: Initial velocity $u = 0 \ m \ s^{-1}$,Final velocity $v = 10 \ m \ s^{-1}$,Time $t = 2 \ s$.
For constant acceleration,we use the first equation of motion:
$v = u + at$
$10 = 0 + a \times 2$
$a = \frac{10}{2} = 5 \ m \ s^{-2}$.
Now,for the distance travelled $(s)$,we use the second equation of motion:
$s = ut + \frac{1}{2}at^2$
$s = 0 \times 2 + \frac{1}{2} \times 5 \times (2)^2$
$s = 0 + \frac{1}{2} \times 5 \times 4$
$s = 10 \ m$.
Thus,the acceleration is $5 \ m \ s^{-2}$ and the distance travelled is $10 \ m$.

(D) When a body reverses its direction,it comes to rest for a moment,for example,a body thrown upward at the topmost point. Hence,the reason statement is correct.
In the same example,at the topmost point,the velocity of the particle is $0$,but the acceleration of the particle is equal to the acceleration due to gravity $(g = 9.8 \ m/s^2)$,which is non-zero.
Therefore,the assertion statement is false.

(C) The circuit consists of two $NOT$ gates connected to the inputs $A$ and $B$,followed by a $NAND$ gate.
$1$. The inputs to the $NAND$ gate are $\overline{A}$ and $\overline{B}$.
$2$. The output $Y$ of the $NAND$ gate is given by $Y = \overline{\overline{A} \cdot \overline{B}}$.
$3$. According to De Morgan's law,$\overline{X \cdot Y} = \overline{X} + \overline{Y}$.
$4$. Applying this to our expression: $Y = \overline{\overline{A}} + \overline{\overline{B}} = A + B$.
$5$. The expression $Y = A + B$ represents the $OR$ logic operation.

(A) Given: Resistance $R = 100 \Omega$,$AC$ source $\varepsilon = 10 \sin (250 \pi t)$.
Comparing with $\varepsilon = \varepsilon_0 \sin (\omega t)$,we get $\varepsilon_0 = 10 \text{ V}$ and $\omega = 250 \pi \text{ rad/s}$.
The energy dissipated as heat $H$ is given by $H = \int_0^{t} \frac{\varepsilon^2}{R} dt$.
$H = \frac{1}{R} \int_0^{10^{-3}} (10 \sin (250 \pi t))^2 dt = \frac{100}{100} \int_0^{10^{-3}} \sin^2 (250 \pi t) dt$.
Using $\sin^2 \theta = \frac{1 - \cos 2\theta}{2}$,we get $H = \int_0^{10^{-3}} \frac{1 - \cos (500 \pi t)}{2} dt$.
$H = \frac{1}{2} \left[ t - \frac{\sin (500 \pi t)}{500 \pi} \right]_0^{10^{-3}}$.
$H = \frac{1}{2} \left[ 10^{-3} - \frac{\sin (500 \pi \times 10^{-3})}{500 \pi} \right] = \frac{1}{2} \left[ 10^{-3} - \frac{\sin (0.5 \pi)}{500 \pi} \right]$.
Since $\sin (0.5 \pi) = 1$,$H = \frac{1}{2} \left[ \frac{1}{1000} - \frac{1}{500 \pi} \right] = \frac{1}{2} \left[ \frac{\pi - 2}{1000 \pi} \right] = \frac{\pi - 2}{2000 \pi} \text{ J}$.
Approximating $\pi \approx 3.14$,$\pi - 2 \approx 1.14$. So $H \approx \frac{1.14}{2000 \pi} \text{ J} = \frac{0.57}{\pi} \times 10^{-3} \text{ J} = \frac{0.57}{\pi} \text{ mJ}$.

(A) The $Q$-factor (quality factor) of a series $LCR$ circuit is given by the formula:
$Q = \frac{1}{R} \sqrt{\frac{L}{C}}$
Given values are:
$L = 2 \ H$
$C = 32 \ \mu F = 32 \times 10^{-6} \ F$
$R = 20 \ \Omega$
Substituting these values into the formula:
$Q = \frac{1}{20} \sqrt{\frac{2}{32 \times 10^{-6}}}$
$Q = \frac{1}{20} \sqrt{\frac{1}{16 \times 10^{-6}}}$
$Q = \frac{1}{20} \times \frac{1}{4 \times 10^{-3}}$
$Q = \frac{1}{20} \times \frac{1000}{4}$
$Q = \frac{250}{20} = 12.5$
Thus,the $Q$-value is $12.5$.

(C) Given: $C = 100 \mu F = 10^{-4} F$,$R = 20 \Omega$,$L = 12.5 mH = 12.5 \times 10^{-3} H$,$V_{rms} = 220 V$,$f = \frac{200}{\pi} Hz$.
First,calculate the angular frequency: $\omega = 2 \pi f = 2 \pi \times \frac{200}{\pi} = 400 rad/s$.
Calculate inductive reactance: $X_L = \omega L = 400 \times 12.5 \times 10^{-3} = 5 \Omega$.
Calculate capacitive reactance: $X_C = \frac{1}{\omega C} = \frac{1}{400 \times 100 \times 10^{-6}} = \frac{1}{0.04} = 25 \Omega$.
Calculate impedance: $Z = \sqrt{R^2 + (X_L - X_C)^2} = \sqrt{20^2 + (5 - 25)^2} = \sqrt{400 + (-20)^2} = \sqrt{400 + 400} = \sqrt{800} = 20\sqrt{2} \Omega$.
The peak voltage is $V_0 = V_{rms} \sqrt{2} = 220\sqrt{2} V$.
The maximum current is $I_0 = \frac{V_0}{Z} = \frac{220\sqrt{2}}{20\sqrt{2}} = 11 A$.

(B) The standard equation for an alternating current is given by $I(t) = I_0 \sin(\omega t)$,where $I_0$ is the peak current and $\omega$ is the angular frequency.
Comparing this with the given equation $I(t) = 50 \sin(200 \pi t)$,we find the peak current $I_0 = 50 \text{ A}$ and the angular frequency $\omega = 200 \pi \text{ rad/s}$.
The $RMS$ value of the current is calculated as $I_{RMS} = \frac{I_0}{\sqrt{2}} = \frac{50}{\sqrt{2}} = 25 \sqrt{2} \text{ A}$.
The frequency $f$ is related to the angular frequency by the formula $\omega = 2 \pi f$.
Substituting the value of $\omega$,we get $200 \pi = 2 \pi f$,which simplifies to $f = 100 \text{ Hz}$.
Thus,the frequency is $100 \text{ Hz}$ and the $RMS$ value is $25 \sqrt{2} \text{ A}$.

(B) The power produced by the generator is $P = V \times I = 4000 \, V \times 100 \, A = 4 \times 10^5 \, W$.
Since the transformer is ideal, the power remains constant. The current in the transmission line $(I')$ is given by $P = V' \times I'$, where $V' = 2 \times 10^5 \, V$.
$I' = P / V' = (4 \times 10^5 \, W) / (2 \times 10^5 \, V) = 2 \, A$.
The power loss in the transmission line is $P_{loss} = (I')^2 \times R = (2 \, A)^2 \times 50 \, \Omega = 4 \times 50 = 200 \, W$.
The percentage of power loss is $(P_{loss} / P) \times 100 = (200 / 4 \times 10^5) \times 100 = (2 \times 10^2 / 4 \times 10^5) \times 10^2 = 0.5 \times 10^{-1} \times 10^2 = 0.05 \%$.

(C) The energy of the emitted photon during a transition from $n_i = 3$ to $n_f = 2$ is given by the Rydberg formula:
$E = 13.6 \left[ \frac{1}{n_f^2} - \frac{1}{n_i^2} \right] \text{ eV}$
Substituting the values $n_f = 2$ and $n_i = 3$:
$E = 13.6 \left[ \frac{1}{2^2} - \frac{1}{3^2} \right] \text{ eV}$
$E = 13.6 \left[ \frac{1}{4} - \frac{1}{9} \right] \text{ eV}$
$E = 13.6 \left[ \frac{9 - 4}{36} \right] \text{ eV}$
$E = 13.6 \times \frac{5}{36} \text{ eV} \approx 1.89 \text{ eV}$
For the photoelectric effect to occur, the energy of the incident photon must be greater than or equal to the work function $(\Phi)$ of the metal.
Therefore, the maximum work function is $\Phi_{\text{max}} = 1.89 \text{ eV}$.

(C) The frequency $\nu$ of emitted radiation for a transition in a hydrogen atom is given by $\nu = cR \left[ \frac{1}{n_1^2} - \frac{1}{n_2^2} \right]$.
For the series limit,the transition occurs from $n_2 = \infty$ to $n_1$.
For the Balmer series,$n_1 = 2$,so $\nu_{B} = cR \left[ \frac{1}{2^2} - \frac{1}{\infty^2} \right] = \frac{cR}{4}$.
For the Brackett series,$n_1 = 4$,so $\nu' = cR \left[ \frac{1}{4^2} - \frac{1}{\infty^2} \right] = \frac{cR}{16}$.
Dividing the two expressions: $\frac{\nu'}{\nu_{B}} = \frac{cR/16}{cR/4} = \frac{4}{16} = \frac{1}{4}$.
Therefore,$\nu' = \frac{\nu_{B}}{4}$.

(B) The Rydberg formula for the Balmer series is given by $\frac{1}{\lambda} = R_H \left( \frac{1}{2^2} - \frac{1}{n^2} \right)$,where $n = 3, 4, 5, \dots$
For the maximum wavelength (minimum energy),we take $n = 3$:
$\frac{1}{\lambda_{max}} = 10^7 \left( \frac{1}{4} - \frac{1}{9} \right) = 10^7 \left( \frac{5}{36} \right) \implies \lambda_{max} = \frac{36}{5} \times 10^{-7} \ m = 7.2 \times 10^{-7} \ m = 7200 \ Å$.
For the minimum wavelength (maximum energy),we take $n = \infty$:
$\frac{1}{\lambda_{min}} = 10^7 \left( \frac{1}{4} - 0 \right) = 10^7 \left( \frac{1}{4} \right) \implies \lambda_{min} = 4 \times 10^{-7} \ m = 4000 \ Å$.
The difference in wavelength is $\Delta \lambda = \lambda_{max} - \lambda_{min} = 7200 \ Å - 4000 \ Å = 3200 \ Å$.

(A) The radius of a nucleus is given by the relation $R = R_0 A^{1/3}$,where $R_0$ is a constant and $A$ is the mass number.
For nucleus $1$: $R_1 = R_0 A_1^{1/3}$.
For nucleus $2$: $R_2 = R_0 A_2^{1/3}$.
Taking the ratio: $\frac{R_2}{R_1} = \left( \frac{A_2}{A_1} \right)^{1/3}$.
Given that $A_2$ is $2\%$ larger than $A_1$,we have $A_2 = A_1(1 + 0.02) = 1.02 A_1$,so $\frac{A_2}{A_1} = 1.02$.
Using the binomial approximation $(1 + x)^n \approx 1 + nx$ for small $x$:
$\frac{R_2}{R_1} = (1.02)^{1/3} = (1 + 0.02)^{1/3} \approx 1 + \frac{1}{3}(0.02) = 1 + 0.00666...$
Thus,$\frac{R_2}{R_1} \approx 1 + \frac{2}{300} = 1 + \frac{2}{3} \%$.
Therefore,$R_2$ is larger than $R_1$ by $\frac{2}{3} \%$.

(A) According to the Bohr model of the hydrogen atom,the velocity $V$ of an electron in the $n^{\text{th}}$ orbit is given by the relation $V_n = \frac{V_0}{n}$,where $V_0$ is a constant.
This implies that the velocity is inversely proportional to the principal quantum number $n$,i.e.,$V \propto \frac{1}{n}$.
To find the ratio of the velocity in the $4^{\text{th}}$ orbit $(V_4)$ to the velocity in the $9^{\text{th}}$ orbit $(V_9)$:
$\frac{V_4}{V_9} = \frac{1/4}{1/9} = \frac{9}{4}$.
Therefore,the ratio is $9 : 4$.

(D) The energy of an electron in the $n^{th}$ orbit of a hydrogen atom is given by the formula $E_n = \frac{-13.6}{n^2} eV$.
For the ground state,$n = 1$.
The first excited state is $n = 2$,the second excited state is $n = 3$,the third excited state is $n = 4$,and the fourth excited state is $n = 5$.
Substituting $n = 5$ into the formula:
$E_5 = \frac{-13.6}{5^2} eV = \frac{-13.6}{25} eV = -0.544 eV$.

(C) Initial capacitance $C = \frac{\varepsilon_0 A}{d} = \frac{\varepsilon_0 \times 10 \times 10^{-4}}{3 \times 10^{-3}} = \frac{\varepsilon_0}{3} \text{ F}$.
Initial charge $Q = CV = \frac{\varepsilon_0}{3} \times 12 = 4 \varepsilon_0 \text{ C}$.
Initial energy $U_i = \frac{Q^2}{2C} = \frac{(4 \varepsilon_0)^2}{2(\varepsilon_0/3)} = \frac{16 \varepsilon_0^2}{2 \varepsilon_0 / 3} = 24 \varepsilon_0 \text{ J}$.
After inserting the dielectric,the new capacitance $C' = KC = 3 \times \frac{\varepsilon_0}{3} = \varepsilon_0 \text{ F}$.
Since the battery is disconnected,the charge $Q$ remains constant.
Final energy $U_f = \frac{Q^2}{2C'} = \frac{(4 \varepsilon_0)^2}{2 \varepsilon_0} = \frac{16 \varepsilon_0^2}{2 \varepsilon_0} = 8 \varepsilon_0 \text{ J}$.
Work done $W = U_f - U_i = 8 \varepsilon_0 - 24 \varepsilon_0 = -16 \varepsilon_0 \text{ J}$.
The magnitude of work done on the system is $16 \varepsilon_0$. Thus,$\alpha = 16$.

(C) Initially,when the switch is thrown to the right,capacitor $C_1$ is connected to the $9 \ V$ battery. The charge on $C_1$ is given by $Q_0 = C_1 \times V = 1 \ \mu F \times 9 \ V = 9 \ \mu C$.
When the switch is thrown to the left,$C_1$ is connected in parallel with the series combination of $C_2$ and $C_3$. Let $Q$ be the final charge on $C_1$ and $q$ be the final charge on $C_2$ and $C_3$ (since they are in series,they have the same charge).
The potential difference across $C_1$ must be equal to the potential difference across the series combination of $C_2$ and $C_3$:
$\frac{Q}{C_1} = \frac{q}{C_2} + \frac{q}{C_3}$
Since $C_1 = C_2 = C_3 = 1 \ \mu F$,we have:
$\frac{Q}{1} = \frac{q}{1} + \frac{q}{1} \Rightarrow Q = 2q$.
By the law of conservation of charge,the total charge remains constant:
$Q + q = Q_0 = 9 \ \mu C$.
Substituting $Q = 2q$ into the equation:
$2q + q = 9 \ \mu C \Rightarrow 3q = 9 \ \mu C \Rightarrow q = 3 \ \mu C$.
Thus,the final charge on capacitor $C_2$ is $3 \ \mu C$.

(A) From the given circuit diagram,the capacitors $3 C$ and $2 C$ are connected in parallel between the same two nodes.
Therefore,their equivalent capacitance $C_p$ is given by:
$C_p = 3 C + 2 C = 5 C$
Now,this equivalent capacitor $C_p = 5 C$ is in series with the capacitor $C$.
The total equivalent capacitance $C_{AB}$ between points $A$ and $B$ is given by the series formula:
$C_{AB} = \frac{C \times C_p}{C + C_p} = \frac{C \times 5 C}{C + 5 C} = \frac{5 C^2}{6 C} = \frac{5}{6} C$

(B) The frequencies of the side bands in amplitude modulation are given by the carrier frequency $\nu_c$ plus and minus the message signal frequency $f_m$.
Given: $f_m = 15 kHz$,Upper Side Band $(USB)$ $= 1515 kHz$,Lower Side Band $(LSB)$ $= 1485 kHz$.
The side bands are $\nu_c + f_m$ and $\nu_c - f_m$.
Therefore,$\nu_c + 15 kHz = 1515 kHz$ or $\nu_c - 15 kHz = 1485 kHz$.
Solving for $\nu_c$: $\nu_c = 1515 kHz - 15 kHz = 1500 kHz$.
Converting to $MHz$: $1500 kHz = 1.5 MHz$.

(A) Standard $AM$ (Amplitude Modulation) broadcast radio operates in the Medium Frequency $(MF)$ band.
The frequency range allocated for standard $AM$ broadcasting is $540 \text{ kHz}$ to $1600 \text{ kHz}$.
Therefore,the correct option is $A$.

(B) The power $P$ radiated by a linear antenna is given by the relation $P \propto \frac{l^2}{\lambda^2}$,where $l$ is the length of the antenna and $\lambda$ is the wavelength.
Since $P \propto l^2$,statement $A$ is true.
Since $\lambda = \frac{c}{f}$,we have $P \propto \frac{l^2}{(c/f)^2} \propto f^2$. This means power radiated increases with the square of the frequency.
Therefore,statement $B$ is false because it claims power decreases with increasing frequency.
For efficient radiation,the antenna size $l$ must be comparable to the wavelength $\lambda$ (typically $l \geq \frac{\lambda}{4}$),so statement $C$ is true.
Since long wavelength signals correspond to low frequencies,the radiated power $P \propto f^2$ is very small,making statement $D$ true.

(A) The modulation index $m$ is defined as the ratio of the peak voltage of the modulating signal $(A_m)$ to the peak voltage of the carrier wave $(A_c)$.
Given:
Carrier wave peak voltage,$A_c = 10 \,V$
Modulation index,$m = 80 \% = 0.8$
Formula:
$m = \frac{A_m}{A_c}$
Substituting the values:
$0.8 = \frac{A_m}{10 \,V}$
$A_m = 0.8 \times 10 \,V = 8 \,V$
Therefore,the peak voltage of the modulating signal should be $8 \,V$.

(B) The modulation index $m$ for an amplitude modulated wave is given by the formula:
$m = \frac{A_{\max} - A_{\min}}{A_{\max} + A_{\min}}$
Given $m = 0.5$ and $A_{\max} = 10.0 \ V$.
Substituting the values into the formula:
$0.5 = \frac{10 - A_{\min}}{10 + A_{\min}}$
$0.5(10 + A_{\min}) = 10 - A_{\min}$
$5 + 0.5 A_{\min} = 10 - A_{\min}$
$1.5 A_{\min} = 5$
$A_{\min} = \frac{5}{1.5} = \frac{50}{15} = 3.33 \ V$.

(B) The range $(d)$ of a $TV$ transmission antenna of height $h$ is given by the formula: $d = \sqrt{2hR}$,where $R$ is the radius of the Earth.
Given: $h = 40 \ m$ and $R = 6400 \ km = 6400 \times 10^3 \ m$.
The service area covered by the antenna is the area of a circle with radius $d$,given by: $Area = \pi d^2$.
Substituting the value of $d^2 = 2hR$ into the area formula:
$Area = \pi (2hR)$
$Area = \pi \times 2 \times 40 \times (6400 \times 10^3)$
$Area = \pi \times 80 \times 6400 \times 10^3$
$Area = 512000 \times 10^3 \pi \ m^2$
$Area = 512 \times 10^6 \pi \ m^2$.

(B) Let the potential at point $A$ be $V_A = 0 \ V$.
Moving from $A$ to $C$ through the $2 \ V$ battery,the potential at $C$ is $V_C = 0 + 2 = 2 \ V$.
Moving from $A$ to $B$ is not direct,but looking at the bottom branch,if we assume the potential at $B$ is $0 \ V$,then moving from $B$ to $D$ through the $2 \ V$ battery gives $V_D = 0 + 2 = 2 \ V$.
The potential difference across the first resistor $R$ is $V_{CD} = V_C - V_D = 2 \ V - 2 \ V = 0 \ V$.
Thus,the current $i_1 = \frac{V_{CD}}{R} = 0$.
Similarly,for the second branch,the potential at $E$ is $V_E = V_C + 2 - 2 = 2 \ V$ and the potential at $F$ is $V_F = V_D + 2 - 2 = 2 \ V$.
Thus,$V_{EF} = 0 \ V$,so $i_2 = 0$.
Finally,for the third branch,$V_G = V_E + 2 - 2 = 2 \ V$ and $V_H = V_F + 2 - 2 = 2 \ V$.
Thus,$V_{GH} = 0 \ V$,so $i_3 = 0$.
Therefore,$i_1 = i_2 = i_3 = 0$.

(A) The expression for electrical resistivity $\rho$ is given by $\rho = \frac{m}{ne^2 \tau}$.
Rearranging the formula to solve for the relaxation time $\tau$:
$\tau = \frac{m}{ne^2 \rho}$
Given values:
$m = 9 \times 10^{-31} \ kg$
$n = 9 \times 10^{28} \ m^{-3}$
$e = 1.6 \times 10^{-19} \ C$
$\rho = 1 \times 10^{-8} \ \Omega \cdot m$
Substituting these values into the equation:
$\tau = \frac{9 \times 10^{-31}}{(9 \times 10^{28}) \times (1.6 \times 10^{-19})^2 \times (1 \times 10^{-8})}$
$\tau = \frac{9 \times 10^{-31}}{9 \times 10^{28} \times 2.56 \times 10^{-38} \times 10^{-8}}$
$\tau = \frac{9 \times 10^{-31}}{23.04 \times 10^{-18}}$
$\tau \approx 0.39 \times 10^{-13} \ s = 3.9 \times 10^{-14} \ s$
Rounding to the nearest value, we get $\tau \approx 4 \times 10^{-14} \ s$.

(B) The power dissipated in a resistor is given by $P = I^2 R$, where $I$ is the current and $R$ is the resistance.
Resistance $R$ is given by $R = \rho \frac{l}{A}$, where $\rho = 10^{-6} \, \Omega \cdot m$, $l = 0.04 \, m$, and $A = \pi r^2 = \pi (7 \times 10^{-3})^2 \, m^2$.
Substituting $R$ in the power equation: $P = I^2 \left( \frac{\rho l}{A} \right) \Rightarrow I = \sqrt{\frac{P A}{\rho l}}$.
Current density $J$ is defined as $J = \frac{I}{A}$.
Substituting $I$: $J = \frac{1}{A} \sqrt{\frac{P A}{\rho l}} = \sqrt{\frac{P}{\rho l A}}$.
Substituting the values: $J = \sqrt{\frac{1.54}{10^{-6} \times 0.04 \times \pi \times (7 \times 10^{-3})^2}}$.
$J = \sqrt{\frac{1.54}{10^{-6} \times 0.04 \times \pi \times 49 \times 10^{-6}}} = \sqrt{\frac{1.54}{1.96 \times 10^{-9} \times \pi}} \approx 5 \times 10^5 \, A/m^2$.

(A) Statement $(I)$: For most pure metals, the resistance increases with temperature, making the temperature coefficient of resistance positive. Thus, $(I)$ is true.
Statement $(II)$: Given diameter $d = 2 \ mm$, radius $r = 1 \ mm = 10^{-3} \ m$. Area $A = \pi r^2 = \pi \times (10^{-3})^2 = \pi \times 10^{-6} \ m^2$. Charge $q = 360 \pi \ C$, time $t = 2 \ hours = 7200 \ s$. Current $i = q/t = 360 \pi / 7200 = \pi / 20 \ A$. Electron density $n = 5 \times 10^{22} \ cm^{-3} = 5 \times 10^{28} \ m^{-3}$. Drift velocity $v_d = i / (neA) = (\pi / 20) / (5 \times 10^{28} \times 1.6 \times 10^{-19} \times \pi \times 10^{-6}) = 1 / (20 \times 5 \times 1.6 \times 10^3) = 1 / (160000) = 6.25 \times 10^{-6} \ m/s$. Thus, $(II)$ is true.
Statement $(III)$: Semiconductors are non-ohmic conductors; they do not follow a linear $V-I$ relationship for all electric field values. Thus, $(III)$ is true.

(B) The mobility $\mu$ of an electron is defined as the ratio of drift velocity $V_d$ to the electric field $E$,given by the formula: $\mu = \frac{V_d}{E} = \frac{e \tau}{m}$.
Given values are:
Charge of electron $e = 1.6 \times 10^{-19} \,C$
Mass of electron $m = 9.1 \times 10^{-31} \,kg$
Average collision time (relaxation time) $\tau = 9.1 \times 10^{-15} \,s$.
Substituting these values into the formula:
$\mu = \frac{(1.6 \times 10^{-19} \,C) \times (9.1 \times 10^{-15} \,s)}{9.1 \times 10^{-31} \,kg}$
$\mu = \frac{1.6 \times 10^{-19} \times 9.1 \times 10^{-15}}{9.1 \times 10^{-31}}$
$\mu = 1.6 \times 10^{-19 - 15 + 31} \,m^2/V \cdot s$
$\mu = 1.6 \times 10^{-3} \,m^2/V \cdot s$.

(B) Let the potential gradient of the potentiometer wire be $\alpha$.
In the first case,the cell is in an open circuit,so the balancing length $l_1 = 60 \ cm = 0.6 \ m$ corresponds to the $EMF$ $\varepsilon$ of the cell:
$\varepsilon = \alpha \times 0.6 \quad \dots(1)$
In the second case,the cell is shunted with an external resistance $R = 4 \ \Omega$. The terminal potential difference $V$ is balanced,where $V = \varepsilon - Ir = \varepsilon \left( \frac{R}{R+r} \right)$.
The new balancing length is $l_2 = 40 \ cm = 0.4 \ m$,so:
$V = \alpha \times 0.4 \quad \dots(2)$
Dividing equation $(2)$ by equation $(1)$:
$\frac{V}{\varepsilon} = \frac{\alpha \times 0.4}{\alpha \times 0.6} = \frac{2}{3}$
Since $\frac{V}{\varepsilon} = \frac{R}{R+r}$,we have:
$\frac{4}{4+r} = \frac{2}{3}$
$12 = 8 + 2r$
$2r = 4$
$r = 2 \ \Omega$
Thus,the internal resistance of the cell is $2 \ \Omega$.

(A) The resistance of a wire is given by the formula $R = \rho \frac{L}{A}$,where $\rho$ is the resistivity,$L$ is the length,and $A$ is the cross-sectional area.
Since the cross-section is circular,$A = \pi r^2$.
For the first wire: $R = \rho \frac{L}{\pi r^2} \quad (1)$
For the second wire: $L_2 = 2L$ and $r_2 = 3r$.
The new resistance $R_2 = \rho \frac{2L}{\pi (3r)^2} = \rho \frac{2L}{9 \pi r^2} \quad (2)$
Dividing equation $(2)$ by equation $(1)$:
$\frac{R_2}{R} = \frac{\rho \frac{2L}{9 \pi r^2}}{\rho \frac{L}{\pi r^2}} = \frac{2}{9}$
Therefore,$R_2 = \frac{2}{9} R$.

(C) The resistance of a uniform wire is given by $R = \frac{\rho L}{A}$.
Since the volume $V = AL$ remains constant during stretching,we have $A = \frac{V}{L}$.
Substituting this into the resistance formula,we get $R = \frac{\rho L}{(V/L)} = \left(\frac{\rho}{V}\right) L^2$.
Since $\rho$ and $V$ are constant,$R \propto L^2$.
For small percentage changes,we can use the relation $\frac{\Delta R}{R} = 2 \frac{\Delta L}{L}$.
Given $\frac{\Delta R}{R} \times 100 = 6\%$,we substitute this into the equation:
$6\% = 2 \times \left(\frac{\Delta L}{L} \times 100\right)$.
Therefore,the percentage increase in length is $\frac{\Delta L}{L} \times 100 = \frac{6\%}{2} = 3\%$.

(C) Specific resistance (resistivity) depends on both the temperature and the nature of the material. Therefore,Statement $(I)$ is incorrect.
When a wire is stretched to $n$ times its original length,its new resistance $R'$ becomes $n^2 R$. Here,$n = 4$ and $R = 6 \ \Omega$,so $R' = 4^2 \times 6 = 16 \times 6 = 96 \ \Omega$. Therefore,Statement $(II)$ is incorrect.
Drift velocity is defined as the average velocity with which free electrons get drifted in the direction opposite to the applied electric field. Therefore,Statement $(III)$ is correct.
Thus,only Statement $(III)$ is true.

(B) The total current $I$ flowing through the wire is given by the integral of current density $J$ over the cross-sectional area $A$: $I = \int J \, dA$.
Since the wire is circular,$dA = 2 \pi r \, dr$.
$I = \int_0^{R} J(r) \cdot 2 \pi r \, dr$,where $R = 2 \times 10^{-3} \text{ m}$.
$I = \int_0^{2 \times 10^{-3}} (10^5 r) \cdot (2 \pi r) \, dr = 2 \pi \times 10^5 \int_0^{2 \times 10^{-3}} r^2 \, dr$.
$I = 2 \pi \times 10^5 \left[ \frac{r^3}{3} \right]_0^{2 \times 10^{-3}} = 2 \pi \times 10^5 \times \frac{8 \times 10^{-9}}{3} = \frac{16 \pi}{3} \times 10^{-4} \text{ A}$.
The energy consumed $E$ is given by $E = V \cdot I \cdot t$.
$E = 70 \text{ V} \times \left( \frac{16 \pi}{3} \times 10^{-4} \text{ A} \right) \times 1000 \text{ s}$.
$E = 70 \times \frac{16 \pi}{3} \times 10^{-1} = \frac{1120 \pi}{30} = \frac{112}{3} \pi \approx 37.33 \pi \text{ J}$.
Given the options,the closest value is $37 \pi \text{ J}$ (Note: The unit in the provided options appears to be in Joules,not kJ).

(B) The de-Broglie wavelength $\lambda$ of an electron accelerated through a potential difference $V$ is given by the formula:
$\lambda = \frac{h}{\sqrt{2meV}}$
Given:
$h = 6.6 \times 10^{-34} \ J \cdot s$
$m = 9 \times 10^{-31} \ kg$
$e = 1.6 \times 10^{-19} \ C$
$V = 121 \ V$
Substituting the values:
$\lambda = \frac{6.6 \times 10^{-34}}{\sqrt{2 \times 9 \times 10^{-31} \times 1.6 \times 10^{-19} \times 121}}$
$\lambda = \frac{6.6 \times 10^{-34}}{\sqrt{3484.8 \times 10^{-50}}}$
$\lambda = \frac{6.6 \times 10^{-34}}{59.03 \times 10^{-25}}$
$\lambda \approx 0.1118 \times 10^{-9} \ m = 0.112 \ nm$
Thus,the correct option is $B$.

(D) The de Broglie wavelength is given by the formula $\lambda = \frac{h}{mv}$,where $h$ is Planck's constant,$m$ is the mass,and $v$ is the velocity.
Since $\lambda \propto \frac{1}{mv}$,the particle with the smallest momentum $(mv)$ will have the largest de Broglie wavelength.
Calculating momentum $(p = mv)$ for each case:
$A$: $p = 0.02 \ kg \times 1000 \ m/s = 20 \ kg \cdot m/s$
$B$: $p = 0.06 \ kg \times 10 \ m/s = 0.6 \ kg \cdot m/s$
$C$: $p = 0.01 \ kg \times 100 \ m/s = 1 \ kg \cdot m/s$
$D$: $p = 0.03 \ kg \times 1 \ m/s = 0.03 \ kg \cdot m/s$
Comparing the values,the momentum in option $D$ is the smallest $(0.03 \ kg \cdot m/s)$.
Therefore,the ball in option $D$ has the largest de Broglie wavelength.

(A) For the shortest wavelength in the Balmer series, the transition occurs from $n_{i} = \infty$ to $n_{f} = 2$.
Using the Rydberg formula: $\frac{1}{\lambda} = R_{H} \left( \frac{1}{n_{f}^2} - \frac{1}{n_{i}^2} \right)$.
Substituting the values: $\frac{1}{\lambda} = 1.097 \times 10^7 \left( \frac{1}{2^2} - \frac{1}{\infty^2} \right)$.
$\frac{1}{\lambda} = 1.097 \times 10^7 \times \frac{1}{4} = 2742500 \ \text{m}^{-1}$.
$\lambda = \frac{1}{2742500} \ \text{m} \approx 3.646 \times 10^{-7} \ \text{m}$.
Converting to $\text{Å}$: $\lambda = 3.646 \times 10^{-7} \times 10^{10} \ \text{Å} = 3646 \ \text{Å}$.

(C) photon is a quantum of electromagnetic radiation. According to the quantum theory of light, the energy of a photon is given by $E = h\nu$, where $h$ is Planck's constant and $\nu$ is the frequency of the radiation.
Since the energy depends only on the frequency, statement $C$ is incorrect because it claims energy depends on intensity.
Photons are massless particles that travel at the speed of light $(c \approx 3 \times 10^8 \ m/s)$.
Because photons are electrically neutral, they do not experience any force in electric or magnetic fields, meaning they are not deflected by them.

(C) According to Einstein's photoelectric equation,the maximum kinetic energy $(K_{\max})$ of emitted photoelectrons is given by $K_{\max} = h\nu - \phi$,where $h$ is Planck's constant,$\nu$ is the frequency of incident light,and $\phi$ is the work function of the metal.
$1$. Statement $I$ is incorrect because $K_{\max}$ depends directly on the frequency $\nu$.
$2$. Statement $II$ is incorrect because the photoelectric effect depends on the frequency of light being greater than the threshold frequency $(\nu_0)$,regardless of intensity.
$3$. Statement $III$ is correct because $K_{\max}$ depends only on the frequency of incident light,not on its intensity.
$4$. Statement $IV$ is correct because as frequency $\nu$ increases,$K_{\max} = h\nu - \phi$ also increases.
Thus,statements $III$ and $IV$ are correct.

(C) From Einstein's photoelectric equation,the stopping potential $V_s$ is related to the frequency $\nu$ as:
$e V_s = h \nu - \phi$
Rearranging for $V_s$,we get:
$V_s = \left( \frac{h}{e} \right) \nu - \frac{\phi}{e}$
This is a linear equation of the form $y = mx + c$,where the slope $m = \frac{h}{e}$.
Given that the slope is $4 \times 10^{-15} \ V \ s$,we have:
$\frac{h}{e} = 4 \times 10^{-15} \ V \ s$
$h = (4 \times 10^{-15} \ V \ s) \times (1.6 \times 10^{-19} \ C)$
$h = 6.4 \times 10^{-34} \ J \ s$
Thus,the value of Planck's constant is $6.4 \times 10^{-34} \ J \ s$.

(D) According to Einstein's photoelectric equation, the maximum kinetic energy $(K.E._{max})$ of the emitted electrons is given by:
$K.E._{max} = E - W$
where $E$ is the energy of the incident photon and $W$ is the work function of the metal.
The energy of the incident photon is:
$E = \frac{hc}{\lambda} = \frac{1240 \, eV \cdot nm}{248 \, nm} = 5.0 \, eV$
The maximum kinetic energy is related to the stopping potential $(V_s)$ by:
$K.E._{max} = e V_s = 2.8 \, eV$
Substituting these values into the photoelectric equation:
$2.8 \, eV = 5.0 \, eV - W$
$W = 5.0 \, eV - 2.8 \, eV$
$W = 2.2 \, eV$
Thus, the work function of the metal is $2.2 \, eV$.

(B) The total power of the bulb is $P_{total} = 100 \ W$.
Since $20 \%$ of the power is converted to visible radiation,the power of visible radiation $P_{vis}$ is:
$P_{vis} = 100 \ W \times \frac{20}{100} = 20 \ W$.
The intensity $I$ of radiation emitted isotropically at a distance $r$ from a point source is given by the formula:
$I = \frac{P_{vis}}{4 \pi r^2}$.
Given $r = 5 \ m$,we substitute the values:
$I = \frac{20}{4 \pi \times (5)^2} = \frac{20}{4 \pi \times 25} = \frac{5}{25 \pi} \ W/m^2$.
Comparing this with the given expression $\frac{\alpha}{25 \pi} \ W/m^2$,we find that $\alpha = 5$.

(C) The de-Broglie wavelength $\lambda$ is given by the formula: $\lambda = \frac{h}{p} = \frac{h}{\sqrt{2 m_{e} K}}$, where $K$ is the kinetic energy.
Given: $h = 6.0 \times 10^{-34} \text{ J s}$, $m_{e} = 9.0 \times 10^{-31} \text{ kg}$, $K = 320 \text{ eV} = 320 \times 1.6 \times 10^{-19} \text{ J}$.
Substituting the values:
$\lambda = \frac{6.0 \times 10^{-34}}{\sqrt{2 \times 9.0 \times 10^{-31} \times 320 \times 1.6 \times 10^{-19}}}$
$\lambda = \frac{6.0 \times 10^{-34}}{\sqrt{18 \times 10^{-31} \times 512 \times 10^{-19}}}$
$\lambda = \frac{6.0 \times 10^{-34}}{\sqrt{9216 \times 10^{-50}}}$
$\lambda = \frac{6.0 \times 10^{-34}}{96 \times 10^{-25}}$
$\lambda = 0.0625 \times 10^{-9} \text{ m} = 62.5 \times 10^{-12} \text{ m} = 62.5 \text{ pm}$.

(A) The photoelectric equation is given by $eV_S = \frac{hc}{\lambda} - \Phi$, where $\Phi = \frac{hc}{\lambda_0}$ is the work function of the metal.
Thus, the stopping potential is $V_S = \frac{hc}{e} \left( \frac{1}{\lambda} - \frac{1}{\lambda_0} \right)$.
For the first wavelength $\lambda_1 = 200 \, nm$, the stopping potential is $V_{S1} = \frac{1240}{e} \left( \frac{1}{200} - \frac{1}{\lambda_0} \right)$.
For the second wavelength $\lambda_2 = 400 \, nm$, the stopping potential is $V_{S2} = \frac{1240}{e} \left( \frac{1}{400} - \frac{1}{\lambda_0} \right)$.
The decrease in stopping potential is $\Delta V_S = V_{S1} - V_{S2}$.
$\Delta V_S = \frac{1240}{e} \left( \frac{1}{200} - \frac{1}{\lambda_0} - \left( \frac{1}{400} - \frac{1}{\lambda_0} \right) \right)$.
$\Delta V_S = \frac{1240}{e} \left( \frac{1}{200} - \frac{1}{400} \right) = \frac{1240}{e} \left( \frac{2-1}{400} \right) = \frac{1240}{400} = 3.1 \, V$.

(D) Statement $(I)$ is incorrect because the photocurrent increases with potential difference only until it reaches a saturation value,after which it remains constant.
Statement $(II)$ is incorrect because the maximum kinetic energy is given by $K_{max} = E - \phi$. For photon $A$,$K_A = 2.5 \ eV - 2.0 \ eV = 0.5 \ eV$. For photon $B$,$K_B = 3.5 \ eV - 2.0 \ eV = 1.5 \ eV$. The ratio $K_A / K_B = 0.5 / 1.5 = 1/3$,not $3$.
Statement $(III)$ is incorrect because the work function is defined as the $MINIMUM$ energy required to eject an electron from the metal surface,not the maximum.

(D) The intensity $I$ at a distance $r = 5 \ m$ from the source is given by $I = \frac{P}{4 \pi r^2}$.
Substituting the values: $I = \frac{942}{4 \times 3.14 \times 5^2} = \frac{942}{12.56 \times 25} = \frac{942}{314} = 3 \ W/m^2$.
The energy of a single photon is $E = \frac{hc}{\lambda} = \frac{6.6 \times 10^{-34} \times 3 \times 10^8}{660 \times 10^{-9}} = \frac{19.8 \times 10^{-26}}{6.6 \times 10^{-7}} = 3 \times 10^{-19} \ J$.
The photon flux $\phi$ is the number of photons per unit area per unit time,given by $\phi = \frac{I}{E}$.
$\phi = \frac{3}{3 \times 10^{-19}} = 1 \times 10^{19} \ \frac{\text{photon}}{m^2 \cdot s}$.

(A) The maximum kinetic energy $(K.E.)_{\max}$ of the emitted electron is given by the formula:
$(K.E.)_{\max} = \frac{1}{2} m v_{\max}^2$
Given $m = 9 \times 10^{-31} \ kg$ and $v_{\max} = 1.6 \times 10^6 \ m/s$,we have:
$(K.E.)_{\max} = \frac{1}{2} \times (9 \times 10^{-31}) \times (1.6 \times 10^6)^2$
$(K.E.)_{\max} = 0.5 \times 9 \times 10^{-31} \times 2.56 \times 10^{12}$
$(K.E.)_{\max} = 11.52 \times 10^{-19} \ J$
The stopping potential $V_s$ is related to the maximum kinetic energy by the equation:
$e V_s = (K.E.)_{\max}$
$V_s = \frac{(K.E.)_{\max}}{e} = \frac{11.52 \times 10^{-19} \ J}{1.6 \times 10^{-19} \ C}$
$V_s = 7.2 \ V$
Therefore,the correct option is $A$.

(B) The magnetic field $B$ inside a long solenoid is given by $B = \mu_0 n I$,where $n$ is the number of turns per unit length.
Given $n = 20 \text{ turns/cm} = 2000 \text{ turns/m}$.
$B = (4\pi \times 10^{-7} \text{ T m/A}) \times (2000 \text{ m}^{-1}) \times I = 8\pi \times 10^{-4} I \text{ T}$.
The magnetic flux $\phi$ through the loop of area $A = \frac{4}{\pi} \text{ cm}^2 = \frac{4}{\pi} \times 10^{-4} \text{ m}^2$ is $\phi = B \cdot A$.
The induced emf $\varepsilon$ is given by Faraday's Law: $\varepsilon = \left| \frac{\Delta \phi}{\Delta t} \right| = A \left| \frac{\Delta B}{\Delta t} \right| = A \mu_0 n \left| \frac{\Delta I}{\Delta t} \right|$.
Substituting the values: $\Delta I = 3.0 \text{ A} - 1.0 \text{ A} = 2.0 \text{ A}$,$\Delta t = 0.2 \text{ s}$.
$\varepsilon = \left( \frac{4}{\pi} \times 10^{-4} \text{ m}^2 \right) \times (4\pi \times 10^{-7} \text{ T m/A}) \times (2000 \text{ m}^{-1}) \times \left( \frac{2.0 \text{ A}}{0.2 \text{ s}} \right)$.
$\varepsilon = (4 \times 10^{-4}) \times (4 \times 10^{-7}) \times (2000) \times (10) \text{ V}$.
$\varepsilon = 32 \times 10^{-7} \text{ V} = 3.2 \times 10^{-6} \text{ V} = 3.2 \mu \text{V}$.

(C) According to Faraday's Law of Electromagnetic Induction,the induced electromotive force $(EMF)$ in a coil is given by the rate of change of magnetic flux linkage.
$|\varepsilon| = N \left| \frac{d\phi}{dt} \right| = N A \left| \frac{dB}{dt} \right|$
Given:
Number of turns $N = 100$
Radius $r = 10 \ cm = 0.1 \ m$
Area $A = \pi r^2 = \pi \times (0.1)^2 = 0.01 \pi \ m^2$
Rate of change of magnetic field $\frac{dB}{dt} = 0.1 \ T \ s^{-1}$
Substituting the values:
$|\varepsilon| = 100 \times (0.01 \pi) \times 0.1$
$|\varepsilon| = 1 \times 0.1 \pi = 0.1 \pi \ V = \frac{\pi}{10} \ V$

(B) The magnetic flux $\phi$ is given by the formula $\phi = \vec{B} \cdot \vec{A} = BA \cos \theta$.
Here,the magnetic field $B = 2 \text{ T}$ is perpendicular to the area,so the angle $\theta = 0^{\circ}$ and $\cos 0^{\circ} = 1$.
The area $A$ of the right-angled triangle is $\frac{1}{2} \times \text{base} \times \text{height}$.
Given base $= 2 \text{ cm} = 2 \times 10^{-2} \text{ m}$ and height $= 1 \text{ cm} = 1 \times 10^{-2} \text{ m}$.
$A = \frac{1}{2} \times (2 \times 10^{-2} \text{ m}) \times (1 \times 10^{-2} \text{ m}) = 1 \times 10^{-4} \text{ m}^2$.
Now,calculating the flux: $\phi = 2 \text{ T} \times (1 \times 10^{-4} \text{ m}^2) \times 1 = 2 \times 10^{-4} \text{ Wb}$.

(B) The induced emf $\varepsilon$ is given by Faraday's law of induction: $|\varepsilon| = |\frac{d\phi}{dt}| = |\frac{d(B \cdot A)}{dt}|$.
Since the area $A$ is constant,$|\varepsilon| = A \cdot |\frac{dB}{dt}|$.
Given $A = 0.2 \, m^2$,initial $B = 0.25 \, T$,final $B = 0 \, T$,and $\Delta t = 10^{-4} \, s$.
$|\varepsilon| = 0.2 \cdot \frac{0.25 - 0}{10^{-4}} = 0.2 \cdot 2500 = 500 \, V$.
The induced current $I$ is given by Ohm's law: $I = \frac{\varepsilon}{R}$.
Given $R = 20 \, \Omega$,$I = \frac{500}{20} = 25 \, A$.

(B) Consider a small radial element of length $dr$ at a distance $r$ from the centre of the disc. As the disc rotates,this element moves perpendicular to the magnetic field $B$.
The motional electromotive force $(de)$ induced across this small element is given by $de = Bv \cdot dr$,where $v = r\omega$.
Substituting $v$,we get $de = B(r\omega)dr$.
To find the total potential difference $(e)$ between the centre $(r=0)$ and the rim $(r=R)$,we integrate the expression:
$e = \int_0^R B\omega r \, dr = B\omega \left[ \frac{r^2}{2} \right]_0^R = \frac{1}{2} B\omega R^2$.
Given values: $B = 4 \ mT = 4 \times 10^{-3} \ T$,$\omega = 100 \ rad/s$,and $R = 30 \ cm = 0.3 \ m$.
Substituting these values:
$e = \frac{1}{2} \times (4 \times 10^{-3}) \times 100 \times (0.3)^2$
$e = 2 \times 10^{-3} \times 100 \times 0.09$
$e = 0.2 \times 0.09 = 0.018 \ V = 18 \ mV$.

(B) The capacitor is charged to a potential difference of $V = 50 \ V$. When the battery is removed and the capacitor is connected to the inductor,the circuit forms an $LC$ oscillator.
At $t=0$,the charge on the capacitor is maximum,so the potential difference across it is $V_{max} = 50 \ V$.
The voltage across the inductor is given by $V_L = L \frac{dI}{dt}$.
In an $LC$ circuit,the sum of potential differences is zero: $V_C + V_L = 0$,which implies $|V_L| = |V_C|$.
At $t=0$,the current $I=0$,so the entire potential of the capacitor appears across the inductor.
Thus,$L \left( \frac{dI}{dt} \right)_{max} = V_{max}$.
Given $L = 10 \ mH = 10 \times 10^{-3} \ H$ and $V_{max} = 50 \ V$.
$\left( \frac{dI}{dt} \right)_{max} = \frac{50}{10 \times 10^{-3}} = \frac{50}{0.01} = 5000 \ A s^{-1}$.