TS EAMCET 2025 Physics Question Paper with Answer and Solution

(C) Let the centre of the sphere be the origin $(0, 0)$.
The radius of the sphere is $R_s = 10 \ cm$.
The sphere is in contact with the cylinder,so the centre of the cylinder is at a distance of $R_s + L/2$ from the centre of the sphere,where $L = 40 \ cm$ is the length of the cylinder.
Distance of the centre of the cylinder from the centre of the sphere,$x_c = 10 \ cm + 20 \ cm = 30 \ cm$.
Mass of the sphere,$m_s = 0.5 \ kg$.
Mass of the cylinder,$m_c = 2 \ kg$.
The centre of mass $X_{cm}$ of the system from the centre of the sphere is given by:
$X_{cm} = \frac{m_s \cdot x_s + m_c \cdot x_c}{m_s + m_c}$
$X_{cm} = \frac{0.5 \cdot 0 + 2 \cdot 30}{0.5 + 2}$
$X_{cm} = \frac{60}{2.5} = 24 \ cm$.

(D) Let $v_1$ and $v_2$ be the velocities of the bodies of mass $m$ and $2m$ respectively after the collision.
By the law of conservation of linear momentum: $mv + (2m)(0) = mv_1 + 2mv_2$,which simplifies to $v = v_1 + 2v_2$ (Equation $1$).
The coefficient of restitution $e$ is defined as $e = \frac{v_2 - v_1}{u_1 - u_2}$.
Given $u_1 = v$ and $u_2 = 0$,we have $e = \frac{v_2 - v_1}{v}$,which implies $v_1 = v_2 - ev$ (Equation $2$).
Substitute Equation $2$ into Equation $1$: $v = (v_2 - ev) + 2v_2$.
$v(1+e) = 3v_2$,so $v_2 = \frac{v(1+e)}{3}$.
Now,substitute $v_2$ back into Equation $2$: $v_1 = \frac{v(1+e)}{3} - ev = \frac{v + ev - 3ev}{3} = \frac{v(1-2e)}{3}$.
The ratio of the velocities $v_1/v_2$ is $\frac{v(1-2e)/3}{v(1+e)/3} = \frac{1-2e}{1+e}$.

(D) When a ball is dropped from a height $H$ and the coefficient of restitution is $e$,the height reached after the first bounce is $h_1 = e^2 H$,after the second bounce is $h_2 = e^4 H$,and so on.
The total distance $D$ travelled by the ball is given by the sum of the initial fall and the infinite series of bounces (up and down):
$D = H + 2h_1 + 2h_2 + 2h_3 + ...$
$D = H + 2(e^2 H + e^4 H + e^6 H + ...)$
$D = H + 2e^2 H (1 + e^2 + e^4 + ...)$
Using the sum of an infinite geometric progression $S = \frac{a}{1-r}$,where $a = 1$ and $r = e^2$:
$D = H + 2e^2 H \left( \frac{1}{1 - e^2} \right)$
$D = H \left( 1 + \frac{2e^2}{1 - e^2} \right) = H \left( \frac{1 - e^2 + 2e^2}{1 - e^2} \right) = H \left( \frac{1 + e^2}{1 - e^2} \right)$
Given $H = 42 \ m$ and $e = 0.4$:
$e^2 = (0.4)^2 = 0.16$
$D = 42 \times \left( \frac{1 + 0.16}{1 - 0.16} \right) = 42 \times \left( \frac{1.16}{0.84} \right)$
$D = 42 \times \frac{116}{84} = 42 \times \frac{116}{2 \times 42} = \frac{116}{2} = 58 \ m$.

(B) The correct matches are as follows:
$A$. Steam engine operates on the $\text{Laws}$ $\text{of}$ $\text{thermodynamics}$ $(II)$.
$B$. Electron microscope utilizes the $\text{Wave}$ $\text{nature}$ $\text{of}$ $\text{electrons}$ $(III)$.
$C$. Non-reflecting coatings are based on the $\text{Interference}$ $\text{of}$ $\text{light}$ $(IV)$.
$D$. Tokamak uses $\text{Magnetic}$ $\text{confinement}$ $\text{of}$ $\text{plasma}$ $(I)$.
Therefore, the correct sequence is $A-II, B-III, C-IV, D-I$.

(C) The time period of a simple pendulum is given by $T = 2\pi \sqrt{\frac{L}{g'}}$.
Here,$g'$ is the acceleration due to gravity at a height $h$ from the surface of the earth,given by $g' = g (\frac{R_E}{R_E + h} )^2$.
Therefore,$T \propto \frac{1}{\sqrt{g'}} \propto \frac{R_E + h}{R_E}$.
At height $h_1 = 2 R_E$,the time period is $T_1 \propto \frac{R_E + 2 R_E}{R_E} = \frac{3 R_E}{R_E} = 3$.
At height $h_2 = 3 R_E$,the time period is $T_2 \propto \frac{R_E + 3 R_E}{R_E} = \frac{4 R_E}{R_E} = 4$.
The ratio of the time periods is $\frac{T_1}{T_2} = \frac{3}{4}$ or $3: 4$.

(C) Using the law of conservation of energy, the total energy at the surface equals the total energy at the maximum height $h$.
At the surface: $E_i = K_i + U_i = \frac{1}{2}mv^2 - \frac{GMm}{R}$
At maximum height: $E_f = K_f + U_f = 0 - \frac{GMm}{R+h}$
Equating $E_i = E_f$: $\frac{1}{2}mv^2 - \frac{GMm}{R} = - \frac{GMm}{R+h}$
Dividing by $m$ and using $GM = gR^2$: $\frac{v^2}{2} - gR = - \frac{gR^2}{R+h}$
Rearranging: $\frac{gR^2}{R+h} = gR - \frac{v^2}{2} = gR(1 - \frac{v^2}{2gR})$
$\frac{R}{R+h} = 1 - \frac{v^2}{2gR} = 1 - \frac{(8000)^2}{2 \times 10 \times 6400 \times 10^3} = 1 - \frac{64 \times 10^6}{128 \times 10^6} = 1 - 0.5 = 0.5$
$\frac{R}{R+h} = 0.5 \implies R+h = 2R \implies h = R$
Since $R = 6400 \,km$, the maximum height $h = 6400 \,km$.

(A) According to the law of conservation of energy, the total energy at the surface of the earth must equal the total energy at infinity.
Let $m$ be the mass of the body and $M$ be the mass of the earth.
The initial speed is $v = \sqrt{5} V_{e}$, where $V_{e} = \sqrt{\frac{2GM}{R}}$.
The total initial energy $E_{i} = \frac{1}{2}mv^2 - \frac{GMm}{R}$.
Substituting $v = \sqrt{5} V_{e}$, we get $E_{i} = \frac{1}{2}m(5 V_{e}^2) - \frac{GMm}{R} = \frac{5}{2}m(\frac{2GM}{R}) - \frac{GMm}{R} = \frac{5GMm}{R} - \frac{GMm}{R} = \frac{4GMm}{R}$.
At infinity, the potential energy is $0$. Let the final speed be $v_{f}$.
The total final energy $E_{f} = \frac{1}{2}mv_{f}^2 + 0$.
Equating $E_{i} = E_{f}$, we get $\frac{4GMm}{R} = \frac{1}{2}mv_{f}^2$.
$v_{f}^2 = \frac{8GM}{R} = 4 \times (\frac{2GM}{R}) = 4 V_{e}^2$.
Therefore, $v_{f} = 2 V_{e}$.

(A) According to the law of conservation of energy,the total energy of the meteor at infinity is equal to the total energy at the earth's surface.
At infinity,the potential energy is $0$ and the kinetic energy is $\frac{1}{2}mv^2$.
At the earth's surface,the potential energy is $-\frac{GMm}{R}$ and the kinetic energy is $\frac{1}{2}mv_f^2$,where $v_f$ is the final speed.
So,$\frac{1}{2}mv^2 + 0 = \frac{1}{2}mv_f^2 - \frac{GMm}{R}$.
We know that the escape speed $v_e = \sqrt{\frac{2GM}{R}}$,which implies $v_e^2 = \frac{2GM}{R}$,or $\frac{GM}{R} = \frac{v_e^2}{2}$.
Substituting this into the energy equation: $\frac{1}{2}mv^2 = \frac{1}{2}mv_f^2 - m(\frac{v_e^2}{2})$.
Dividing by $\frac{m}{2}$ gives $v^2 = v_f^2 - v_e^2$.
Therefore,$v_f^2 = v^2 + v_e^2$,which means $v_f = \sqrt{v^2 + v_e^2}$.

(A) The orbital speed $v_0$ of a satellite at a distance $r$ from the center of the earth is given by $v_0 = \sqrt{\frac{GM}{r}}$.
For a body near the earth's surface,$r_1 = R = 6400 \ km$,so $v_1 = \sqrt{\frac{GM}{R}} = 8 \ km s^{-1}$.
For a body at a height $h = 19,200 \ km$,the distance from the center is $r_2 = R + h = 6400 + 19,200 = 25,600 \ km$.
The orbital speed at this height is $v_2 = \sqrt{\frac{GM}{r_2}}$.
Taking the ratio: $\frac{v_2}{v_1} = \sqrt{\frac{R}{r_2}} = \sqrt{\frac{6400}{25600}} = \sqrt{\frac{1}{4}} = \frac{1}{2}$.
Therefore,$v_2 = \frac{v_1}{2} = \frac{8 \ km s^{-1}}{2} = 4 \ km s^{-1}$.

(A) According to Newton's Law of Universal Gravitation,every particle of matter in the universe attracts every other particle with a force that is directly proportional to the product of their masses and inversely proportional to the square of the distance between their centers.
This force of mutual attraction that exists between any two objects due to their masses is known as the Gravitational force.

(C) The gravitational force between two point masses is given by Newton's Law of Gravitation,$F = G \frac{m_1 m_2}{r^2}$.
$1$. It is a conservative force,meaning the work done by it is independent of the path.
$2$. It is always an attractive force between two masses.
$3$. It is a central force,as it acts along the line joining the centers of the two bodies.
$4$. It is a non-contact force (or action-at-a-distance force),meaning it does not require physical contact between the bodies.
Therefore,the statement that it is 'Not a central force' is incorrect.

(B) For a monoatomic gas,the degrees of freedom $f = 3$.
The molar specific heat at constant volume is given by $C_V = \frac{f}{2}R = \frac{3}{2}R$.
The molar specific heat at constant pressure is given by $C_P = C_V + R = \frac{3}{2}R + R = \frac{5}{2}R$.
Given $R = 8.3 \,J \,mol^{-1} \,K^{-1}$.
Substituting the value of $R$,we get $C_P = \frac{5}{2} \times 8.3 = 2.5 \times 8.3 = 20.75 \,J \,mol^{-1} \,K^{-1}$.

(D) $1$. Calculate the number of moles $(n)$ for each gas:
$n_{O_2} = \frac{64 \ g}{32 \ g/mol} = 2 \ mol$
$n_{N_2} = \frac{28 \ g}{28 \ g/mol} = 1 \ mol$
$n_{CO_2} = \frac{132 \ g}{44 \ g/mol} = 3 \ mol$
$2$. The total number of moles initially is $n_{total} = 2 + 1 + 3 = 6 \ mol$.
$3$. According to the ideal gas law,$PV = nRT$. Since $V, R, T$ are constant,$P \propto n$.
$4$. Therefore,$P = k \times 6$,where $k$ is a constant.
$5$. After removing oxygen,the remaining moles are $n_{remaining} = 1 (N_2) + 3 (CO_2) = 4 \ mol$.
$6$. The new pressure $P'$ is $P' = k \times 4$.
$7$. Taking the ratio: $\frac{P'}{P} = \frac{4}{6} = \frac{2}{3}$.
$8$. Thus,$P' = \frac{2 P}{3}$.

(C) The average translational kinetic energy $(K_{avg})$ of a gas molecule is given by the formula: $K_{avg} = \frac{3}{2} k_B T$,where $k_B$ is the Boltzmann constant and $T$ is the absolute temperature.
Since the temperature $T$ is the same for both hydrogen and oxygen,the average translational kinetic energy depends only on the temperature.
Therefore,the ratio of the average translational kinetic energies of hydrogen and oxygen is $1: 1$.

(C) The rms speed of a gas is given by $v_{rms} = \sqrt{\frac{3RT}{M}}$.
Since $v_{rms} \propto \sqrt{T}$,if the speed increases from $v$ to $\sqrt{3}v$,the temperature increases from $T_1$ to $T_2$ such that $\frac{v_{rms2}}{v_{rms1}} = \sqrt{\frac{T_2}{T_1}} = \sqrt{3}$.
Thus,$\frac{T_2}{T_1} = 3$,which implies $T_2 = 3T_1$.
The heat required for a diatomic gas at constant volume is given by $Q = n C_v \Delta T$.
For a diatomic gas,the degrees of freedom $f = 5$,so $C_v = \frac{5}{2}R$.
Given $n = 4 \ mol$,$Q = 83.1 \ kJ = 83100 \ J$,and $R = 8.31 \ J \ mol^{-1} \ K^{-1}$.
Substituting the values: $83100 = 4 \times \frac{5}{2} \times 8.31 \times (3T_1 - T_1)$.
$83100 = 10 \times 8.31 \times 2T_1$.
$83100 = 166.2 \times T_1$.
$T_1 = \frac{83100}{166.2} = 500 \ K$.
Converting to Celsius: $T_1(^{\circ}C) = 500 - 273 = 227^{\circ}C$.

(C) The root mean square (rms) speed of gas molecules is given by the formula $v_{rms} = \sqrt{\frac{3RT}{M}}$.
This implies that $v_{rms} \propto \sqrt{T}$, where $T$ is the absolute temperature in Kelvin.
Given $T_1 = 77^{\circ} C = 77 + 273 = 350 \,K$ and $v_1 = 50 \,ms^{-1}$.
Given $T_2 = 150.5^{\circ} C = 150.5 + 273 = 423.5 \,K$.
Using the ratio $\frac{v_2}{v_1} = \sqrt{\frac{T_2}{T_1}}$, we get:
$v_2 = v_1 \times \sqrt{\frac{T_2}{T_1}} = 50 \times \sqrt{\frac{423.5}{350}}$.
$v_2 = 50 \times \sqrt{1.21} = 50 \times 1.1 = 55 \,ms^{-1}$.
Thus, the correct option is $C$.

(B) The root mean square (rms) speed of gas molecules is given by the formula $v_{rms} = \sqrt{\frac{3RT}{M}}$,where $T$ is the absolute temperature in Kelvin.
Since $v_{rms} \propto \sqrt{T}$,the ratio of the final rms speed $(v_2)$ to the initial rms speed $(v_1)$ is $\frac{v_2}{v_1} = \sqrt{\frac{T_2}{T_1}}$.
Initial temperature $T_1 = 127^{\circ} C = 127 + 273 = 400 \ K$.
Final temperature $T_2 = 527^{\circ} C = 527 + 273 = 800 \ K$.
Substituting these values into the ratio formula: $\frac{v_2}{v_1} = \sqrt{\frac{800}{400}} = \sqrt{2}$.
Therefore,the rms speed becomes $\sqrt{2}$ times the initial speed.

(D) The apparent weight $W'$ of a person in a lift is given by the formula $W' = m(g + a)$,where $a$ is the acceleration of the lift.
When the lift is moving up with a retardation,the acceleration $a$ is negative.
Given: mass $m = 60 \ kg$,acceleration due to gravity $g = 9.8 \ ms^{-2}$,and retardation $a = -2.8 \ ms^{-2}$.
Substituting these values into the formula:
$W' = 60 \times (9.8 - 2.8)$
$W' = 60 \times 7.0$
$W' = 420 \ N$.
Therefore,the apparent weight of the man is $420 \ N$.

(C) Total mass of the system $M = 8 \text{ ton} + 2 \text{ ton} = 10 \text{ ton} = 10000 \text{ kg}$.
Breaking force $F = 25000 \text{ N}$.
Deceleration of the truck $a = F / M = 25000 / 10000 = 2.5 \text{ m/s}^2$.
The block of mass $m = 2000 \text{ kg}$ experiences a pseudo force $F_p = m \times a = 2000 \times 2.5 = 5000 \text{ N}$ in the forward direction.
The maximum static frictional force available is $f_{max} = \mu \times m \times g = 0.3 \times 2000 \times 10 = 6000 \text{ N}$.
Since the required force to keep the block stationary $(5000 \text{ N})$ is less than the maximum static friction $(6000 \text{ N})$, the block will not slide.
Therefore, the frictional force acting on the block is equal to the pseudo force, which is $5000 \text{ N}$.

(C) The block is on a horizontal surface. The forces acting on the block are:
$1$. Weight $mg$ acting downwards.
$2$. Normal reaction $N$ acting upwards.
$3$. Applied force $F$ at $45^{\circ}$ to the horizontal.
Resolving $F$ into components: $F_x = F \cos 45^{\circ}$ and $F_y = F \sin 45^{\circ}$.
For vertical equilibrium: $N + F \sin 45^{\circ} = mg$.
So, $N = mg - F \sin 45^{\circ} = \sqrt{2} \times 10 - F \times \frac{1}{\sqrt{2}} = 10\sqrt{2} - \frac{F}{\sqrt{2}}$.
The block starts to move when the horizontal component of the force equals the limiting friction: $F \cos 45^{\circ} = \mu N$.
Substituting the values: $F \times \frac{1}{\sqrt{2}} = 0.25 \times (10\sqrt{2} - \frac{F}{\sqrt{2}})$.
Multiply by $\sqrt{2}$: $F = 0.25 \times (10 \times 2 - F) = 0.25 \times (20 - F)$.
$F = 5 - 0.25F$.
$1.25F = 5$.
$F = \frac{5}{1.25} = 4 \,N$.

(B) The power delivered by a force is given by $P = \vec{F} \cdot \vec{v}$.
Since the force $\vec{F}$ is proportional to acceleration $\vec{a}$ (by Newton's second law,$\vec{F} = m\vec{a}$),the condition that velocity $\vec{v}$ and acceleration $\vec{a}$ are perpendicular implies that $\vec{F} \cdot \vec{v} = 0$.
Therefore,the power delivered to the particle is zero $(P = 0)$.
Since power is the rate of change of kinetic energy $(P = \frac{dK}{dt})$,$P = 0$ implies that the kinetic energy $K$ is constant.
Thus,the kinetic energy of the particle remains constant throughout the motion.

(B) Let the masses of the two blocks be $M_1 = m$ and $M_2 = n$. Assume $m > n$.
The acceleration $a$ of the blocks in an Atwood machine is given by $a = \frac{|M_1 - M_2|}{M_1 + M_2} g = \frac{m-n}{m+n} g$.
The acceleration of block $1$ is $a_1 = a$ (downwards) and the acceleration of block $2$ is $a_2 = a$ (upwards).
Taking the downward direction as positive,the acceleration of the centre of mass $a_{cm}$ is given by:
$a_{cm} = \frac{M_1 a_1 + M_2 a_2}{M_1 + M_2} = \frac{m(a) + n(-a)}{m+n} = \frac{(m-n)a}{m+n}$.
Substituting the value of $a$:
$a_{cm} = \frac{(m-n)}{m+n} \cdot \left( \frac{m-n}{m+n} g \right) = \left( \frac{m-n}{m+n} \right)^2 g$.

(D) Let the masses be $m_1 = 4 \ kg$ (left side),$m_2 = 3 \ kg$ (upper right),and $m_3 = 3 \ kg$ (lower right).
The total mass on the right side is $M_R = m_2 + m_3 = 3 \ kg + 3 \ kg = 6 \ kg$.
The mass on the left side is $M_L = 4 \ kg$.
Since $M_R > M_L$,the system accelerates towards the right with acceleration $a$.
The acceleration $a$ is given by $a = \frac{(M_R - M_L)g}{M_R + M_L} = \frac{(6 - 4)g}{6 + 4} = \frac{2g}{10} = 0.2g$.
For the lower $3 \ kg$ block,the equation of motion is $T_1 - m_3g = m_3a$.
$T_1 = m_3(g + a) = 3(g + 0.2g) = 3(1.2g) = 3.6g$.
For the entire system,the tension $T_2$ in the string passing over the pulley is $T_2 = \frac{2 M_L M_R g}{M_L + M_R} = \frac{2(4)(6)g}{4 + 6} = \frac{48g}{10} = 4.8g$.
The ratio of tensions is $\frac{T_1}{T_2} = \frac{3.6g}{4.8g} = \frac{36}{48} = \frac{3}{4}$.

(A) Let the force be $F$. According to Newton's second law,$F = ma$,so $m = F/a$.
For mass $P$,$P = F/18$.
For mass $Q$,$Q = F/9$.
For mass $R$,$R = F/6$.
When the same force $F$ is applied to the combined mass $(P+Q+R)$,the acceleration $a'$ is given by $a' = F / (P+Q+R)$.
Substituting the values of $P, Q,$ and $R$:
$a' = F / (F/18 + F/9 + F/6) = F / [F(1/18 + 2/18 + 3/18)]$.
$a' = 1 / (6/18) = 1 / (1/3) = 3 \ m/s^2$.
Thus,the acceleration is $3 \ m/s^2$.

(B) For a body falling under gravity with air resistance,the acceleration is not constant. However,we can use the average velocity concept for motion with constant acceleration or approximate the motion.
Given: Initial velocity $u = 0 \ m/s$,final velocity $v = 18 \ m/s$,and displacement $s = 20 \ m$.
The average velocity $v_{avg}$ is given by $v_{avg} = \frac{u + v}{2} = \frac{0 + 18}{2} = 9 \ m/s$.
The time taken $t$ is given by $t = \frac{s}{v_{avg}} = \frac{20 \ m}{9 \ m/s} \approx 2.22 \ s$.
Rounding to the nearest value,the time taken is approximately $2.2 \ s$.

(C) The maximum speed $v_{max}$ on a banked road with friction is given by the formula: $v_{max} = \sqrt{rg \left( \frac{\tan \theta + \mu}{1 - \mu \tan \theta} \right)}$.
Given: Radius $r = 75 \ m$,angle $\tan \theta = 0.2$,coefficient of friction $\mu = 0.1$,and acceleration due to gravity $g = 10 \ m/s^2$.
Substituting the values:
$v_{max} = \sqrt{75 \times 10 \times \left( \frac{0.2 + 0.1}{1 - (0.1 \times 0.2)} \right)}$
$v_{max} = \sqrt{750 \times \left( \frac{0.3}{1 - 0.02} \right)}$
$v_{max} = \sqrt{750 \times \left( \frac{0.3}{0.98} \right)}$
$v_{max} = \sqrt{750 \times 0.3061} \approx \sqrt{229.57} \approx 15.15 \ m/s$.
Rounding to the nearest option,the maximum permissible speed is $15 \ m/s$.

(C) The component of vector $\vec{A}$ along $\vec{B}$ is given by $\frac{\vec{A} \cdot \vec{B}}{|\vec{B}|}$.
The component of vector $\vec{B}$ along $\vec{A}$ is given by $\frac{\vec{A} \cdot \vec{B}}{|\vec{A}|}$.
According to the problem,the component of $\vec{A}$ along $\vec{B}$ is twice the component of $\vec{B}$ along $\vec{A}$:
$\frac{\vec{A} \cdot \vec{B}}{|\vec{B}|} = 2 \left( \frac{\vec{A} \cdot \vec{B}}{|\vec{A}|} \right)$.
Assuming $\vec{A} \cdot \vec{B} \neq 0$,we can divide both sides by $(\vec{A} \cdot \vec{B})$:
$\frac{1}{|\vec{B}|} = \frac{2}{|\vec{A}|}$.
Rearranging the terms to find the ratio of magnitudes $\frac{|\vec{A}|}{|\vec{B}|}$:
$\frac{|\vec{A}|}{|\vec{B}|} = 2$.
Thus,the ratio is $2: 1$.

(C) Let the outer volume of the block be $V = 1000 \text{ ml} = 10^{-3} \text{ m}^3$.
The specific gravity of the wood is $\rho_w / \rho_{water} = 0.75$.
Thus,the density of the wood material is $\rho_w = 0.75 \times 1000 \text{ kg/m}^3 = 750 \text{ kg/m}^3$.
Let $V_c$ be the volume of the cavity. The volume of the actual wooden material is $V_m = V - V_c$.
The mass of the block is $M = \rho_w \times V_m = 750 \times (10^{-3} - V_c)$.
According to the law of floatation,the weight of the block equals the weight of the water displaced.
The block floats with half its volume immersed,so $V_{displaced} = \frac{V}{2} = 500 \text{ ml} = 5 \times 10^{-4} \text{ m}^3$.
Weight of displaced water = $\rho_{water} \times V_{displaced} \times g = 1000 \times 5 \times 10^{-4} \times g = 0.5g$.
Weight of the block = $M \times g = 750 \times (10^{-3} - V_c) \times g$.
Equating the two: $0.5g = 750 \times (10^{-3} - V_c) \times g$.
$0.5 = 0.75 - 750 \times V_c$.
$750 \times V_c = 0.25$.
$V_c = \frac{0.25}{750} = \frac{1}{3000} \text{ m}^3$.
$V_c = \frac{1}{3000} \times 10^6 \text{ ml} = 333.33 \text{ ml}$.

(C) Let the side of the cube be $a = 40 \ cm = 0.4 \ m$. The volume of the cube is $V = a^3 = (0.4)^3 = 0.064 \ m^3$.
Density of water $\rho_w = 1000 \ kg/m^3$.
Initially,the cube floats with $\frac{1}{4}$ of its volume immersed. By the law of floatation,the weight of the cube equals the weight of the displaced water:
$m_{cube} \cdot g = \rho_w \cdot V_{immersed} \cdot g$
$m_{cube} = 1000 \cdot (\frac{1}{4} \cdot 0.064) = 1000 \cdot 0.016 = 16 \ kg$.
When the disc of mass $M$ is placed on the cube,the total weight is $(m_{cube} + M)g$. The new volume immersed is $\frac{2}{5}$ of $V$:
$(m_{cube} + M)g = \rho_w \cdot (\frac{2}{5} \cdot V) \cdot g$
$16 + M = 1000 \cdot (\frac{2}{5} \cdot 0.064)$
$16 + M = 1000 \cdot 0.0256 = 25.6$
$M = 25.6 - 16 = 9.6 \ kg$.

(B) The velocity of efflux $v$ is given by Torricelli's law: $v = \sqrt{2gh}$.
Given: $h = 20 \ cm = 0.2 \ m$,$g = 10 \ m \ s^{-2}$.
$v = \sqrt{2 \times 10 \times 0.2} = \sqrt{4} = 2 \ m \ s^{-1}$.
The volume of water flowing out per second is $Q = A \times v$,where $A = 1 \ mm^2 = 1 \times 10^{-6} \ m^2$.
$Q = 1 \times 10^{-6} \ m^2 \times 2 \ m \ s^{-1} = 2 \times 10^{-6} \ m^3 \ s^{-1}$.
The mass of water flowing out in time $t = 0.6 \ s$ is $m = \rho \times Q \times t$.
$m = 1000 \ kg \ m^{-3} \times 2 \times 10^{-6} \ m^3 \ s^{-1} \times 0.6 \ s$.
$m = 1000 \times 1.2 \times 10^{-6} \ kg = 1.2 \times 10^{-3} \ kg = 1.2 \ g$.

(D) Let the surface area of the bubble at the bottom be $A_1$ and at the top be $A_2$. Given $A_2 = A_1 + 1.25 A_1 = 2.25 A_1$.
Since the bubble is spherical,$A = 4 \pi r^2$,which implies $r \propto \sqrt{A}$. Thus,$r_2 = \sqrt{2.25} r_1 = 1.5 r_1$.
The volume $V = \frac{4}{3} \pi r^3$,so $V_2 = (1.5)^3 V_1 = 3.375 V_1$.
Since the temperature is uniform,Boyle's Law applies: $P_1 V_1 = P_2 V_2$.
Here,$P_2 = P_{atm} = 10 \ m$ of water column.
$P_1 = P_{atm} + h = 10 + h$,where $h$ is the depth of the tank.
Substituting the values: $(10 + h) V_1 = 10 \times (3.375 V_1)$.
$10 + h = 33.75$.
$h = 33.75 - 10 = 23.75 \ m$.

(D) The work done $W$ in blowing a soap bubble of radius $r$ is given by the formula $W = T \times \Delta A$,where $T$ is the surface tension and $\Delta A$ is the change in surface area.
Since a soap bubble has two surfaces (inner and outer),the change in surface area is $\Delta A = 2 \times (4 \pi r^2) = 8 \pi r^2$.
Given: Diameter $d = 3 \ cm$,so radius $r = 1.5 \ cm = 1.5 \times 10^{-2} \ m$.
Surface tension $T = 0.035 \ N/m$.
Substituting the values:
$W = 0.035 \times 8 \times \pi \times (1.5 \times 10^{-2})^2$
$W = 0.035 \times 8 \times 3.14159 \times 2.25 \times 10^{-4}$
$W = 0.28 \times 3.14159 \times 2.25 \times 10^{-4}$
$W \approx 1.979 \times 10^{-4} \ J$
$W \approx 198 \times 10^{-6} \ J = 198 \ \mu J$.
Thus,the correct option is $D$.

(A) The work done in increasing the radius of a soap bubble from $r_1$ to $r_2$ is given by the formula $W = T \times \Delta A$,where $T$ is the surface tension and $\Delta A$ is the change in surface area. Since a soap bubble has two surfaces,the change in area is $\Delta A = 2 \times (4\pi r_2^2 - 4\pi r_1^2) = 8\pi(r_2^2 - r_1^2)$.
For $W_1$ (from $r$ to $2r$): $W_1 = 8\pi T ((2r)^2 - r^2) = 8\pi T (4r^2 - r^2) = 8\pi T (3r^2) = 24\pi T r^2$.
For $W_2$ (from $2r$ to $3r$): $W_2 = 8\pi T ((3r)^2 - (2r)^2) = 8\pi T (9r^2 - 4r^2) = 8\pi T (5r^2) = 40\pi T r^2$.
Therefore,the ratio $W_1: W_2 = (24\pi T r^2) : (40\pi T r^2) = 24:40 = 3:5$.

(B) The work done in increasing the area of a film is given by $W = T \times \Delta A$,where $T$ is the surface tension and $\Delta A$ is the change in surface area.
Since a thin film has two surfaces,the total change in area is $\Delta A = 2 \times l \times (d_2 - d_1)$.
Given: $l = 8 \ cm = 0.08 \ m$,$d_1 = 0.6 \ cm = 0.006 \ m$,$d_2 = 0.8 \ cm = 0.008 \ m$,and $T = 0.07 \ N/m$.
Change in distance $\Delta d = d_2 - d_1 = 0.8 \ cm - 0.6 \ cm = 0.2 \ cm = 0.002 \ m$.
$\Delta A = 2 \times 0.08 \ m \times 0.002 \ m = 0.00032 \ m^2$.
Work done $W = 0.07 \ N/m \times 0.00032 \ m^2 = 0.0000224 \ J$.
$W = 22.4 \times 10^{-6} \ J = 22.4 \ \mu J$.

(B) Let the radius of each small drop be $r$ and the radius of the big drop be $R$. Since the total volume remains constant,we have $n \times (4/3) \pi r^3 = (4/3) \pi R^3$,which gives $R = n^{1/3} r$.
The initial surface area of $n$ drops is $A_i = n \times 4 \pi r^2$.
The final surface area of the big drop is $A_f = 4 \pi R^2 = 4 \pi (n^{1/3} r)^2 = 4 \pi n^{2/3} r^2$.
Since $n^{2/3} < n$ for $n > 1$,the final surface area $A_f$ is less than the initial surface area $A_i$. Thus,the surface area decreases.
Surface energy is given by $U = T \times A$,where $T$ is the surface tension. Since the surface area decreases,the surface energy of the system decreases.
According to the law of conservation of energy,the decrease in surface energy is released as heat. Therefore,surface area decreases and heat is released.

(D) For a capillary tube open at both ends,the water column is supported by surface tension forces at the meniscus.
When a water column of length $h$ is held in a capillary tube,the pressure difference due to surface tension must balance the hydrostatic pressure of the water column.
For a tube open at both ends,the water will not be able to be supported if the tube is vertical because the pressure at the top and bottom meniscus would be atmospheric,and gravity would pull the water down.
However,if we consider the capillary rise formula $h = \frac{2T \cos \theta}{r \rho g}$,this applies to a tube dipping in a reservoir.
In a tube open at both ends,the water column cannot be supported against gravity unless there is a pressure difference.
Since the tube is open at both ends,the pressure at both ends is atmospheric pressure $(P_0)$.
Thus,the water column will fall out due to gravity.
Therefore,the maximum length of the water column that can stay without falling is $0 \ cm$.

(C) The terminal velocity $v_t$ of a sphere of radius $r$ and density $\rho$ falling through a liquid of density $\sigma$ and viscosity $\eta$ is given by the formula: $v_t = \frac{2}{9} \frac{r^2 g (\rho - \sigma)}{\eta}$.
Since both spheres are made of the same metal and fall through the same liquid,$\rho, \sigma, g,$ and $\eta$ are constant. Thus,$v_t \propto r^2$.
Given the mass $m = \frac{4}{3} \pi r^3 \rho$,we have $r^3 \propto m$,which implies $r \propto m^{1/3}$.
Substituting this into the proportionality for $v_t$,we get $v_t \propto (m^{1/3})^2 = m^{2/3}$.
Let $v_1 = 3 \ cm s^{-1}$ for $m_1 = 8 \ g$ and $v_2$ be the terminal velocity for $m_2 = 64 \ g$.
Then,$\frac{v_2}{v_1} = \left( \frac{m_2}{m_1} \right)^{2/3} = \left( \frac{64}{8} \right)^{2/3} = (8)^{2/3} = (2^3)^{2/3} = 2^2 = 4$.
Therefore,$v_2 = 4 \times v_1 = 4 \times 3 \ cm s^{-1} = 12 \ cm s^{-1}$.

$(A)$ The Reynolds number $(Re)$ is a dimensionless quantity used to predict the flow regime of a fluid.
For flow in a pipe, the criteria are generally defined as follows:
$1$. If $Re < 2000$, the flow is streamlined or laminar.
$2$. If $2000 < Re < 3000$, the flow is in a transition state.
$3$. If $Re > 3000$, the flow is turbulent.
Comparing the given options with the criteria for streamlined flow $(Re < 2000)$:
- Option $A$: $900 < 2000$ (Streamlined)
- Option $B$: $2100$ (Transition)
- Option $C$: $2900$ (Transition)
- Option $D$: $4000$ (Turbulent)
Therefore, the correct option is $A$.

(B) According to Stokes' Law,the viscous force $F$ acting on a spherical object of radius $r$ moving with a terminal velocity $v$ in a fluid of viscosity $\eta$ is given by $F = 6 \pi \eta r v$.
Given:
Diameter $d = 1 \ mm = 10^{-3} \ m$,so radius $r = 0.5 \times 10^{-3} \ m$.
Terminal velocity $v = 0.7 \ ms^{-1}$.
Coefficient of viscosity $\eta = 2 \times 10^{-5} \ Pa \cdot s$.
Substituting these values into the formula:
$F = 6 \times 3.14 \times (2 \times 10^{-5}) \times (0.5 \times 10^{-3}) \times 0.7$
$F = 6 \times 3.14 \times 10^{-5} \times 0.5 \times 10^{-3} \times 0.7$
$F = 6 \times 0.5 \times 0.7 \times 3.14 \times 10^{-8}$
$F = 2.1 \times 3.14 \times 10^{-8}$
$F = 6.594 \times 10^{-8} \ N \approx 6.6 \times 10^{-8} \ N$.
Thus,the correct option is $B$.

(B) The Bulk modulus $B$ is defined as $B = -\frac{\Delta P}{\Delta V/V}$.
Given: $B = 60 \ GPa = 60 \times 10^9 \ Pa$,$\Delta P = 10^7 \ Pa$,$r = 36 \ cm = 0.36 \ m$.
The volume of a sphere is $V = \frac{4}{3}\pi r^3$,so the fractional change in volume is $\frac{\Delta V}{V} = 3 \frac{\Delta r}{r}$.
Substituting this into the Bulk modulus formula: $B = -\frac{\Delta P}{3 \Delta r / r}$.
Rearranging for $\Delta r$: $\Delta r = -\frac{\Delta P \cdot r}{3B}$.
Taking the magnitude: $|\Delta r| = \frac{10^7 \times 0.36}{3 \times 60 \times 10^9}$.
$|\Delta r| = \frac{3.6 \times 10^6}{180 \times 10^9} = \frac{3.6}{180} \times 10^{-3} = 0.02 \times 10^{-3} \ m = 2 \times 10^{-5} \ m$.
Converting to cm: $2 \times 10^{-5} \times 10^2 \ cm = 2 \times 10^{-3} \ cm$.

(C) Let the original radius of the wire be $r$ and its length be $L$. The area of cross-section is $A = \pi r^2$.
Taking the derivative,$\Delta A = 2 \pi r \Delta r$.
Poisson's ratio $\sigma$ is defined as $\sigma = -\frac{\Delta r / r}{\Delta L / L}$,so $\frac{\Delta r}{r} = -\sigma \frac{\Delta L}{L}$.
Substituting this into the area change equation: $\Delta A = 2 \pi r^2 (-\sigma \frac{\Delta L}{L}) = -2 A \sigma \frac{\Delta L}{L}$.
Taking the magnitude,$|\Delta A| = 2 A \sigma \frac{\Delta L}{L}$,which gives $\frac{\Delta L}{L} = \frac{\Delta A}{2 A \sigma}$.
Young's modulus $Y$ is defined as $Y = \frac{F/A}{\Delta L/L}$,so $F = Y A \frac{\Delta L}{L}$.
Substituting $\frac{\Delta L}{L}$: $F = Y A (\frac{\Delta A}{2 A \sigma}) = \frac{Y \Delta A}{2 \sigma}$.

(C) The energy stored per unit volume $(u)$ in a stretched rod is given by the formula: $u = \frac{1}{2} \times \text{stress} \times \text{strain}$.
We know that $\text{stress} = \frac{F}{A}$ and $\text{strain} = \frac{\text{stress}}{Y} = \frac{F}{AY}$.
Substituting these into the formula, we get $u = \frac{1}{2} \times \frac{F}{A} \times \frac{F}{AY} = \frac{F^2}{2A^2Y}$.
Given: $F = 9 \times 10^4 \,N$, $A = 3 \,cm^2 = 3 \times 10^{-4} \,m^2$, and $Y = 2 \times 10^{11} \,Nm^{-2}$.
Calculating stress: $\sigma = \frac{9 \times 10^4}{3 \times 10^{-4}} = 3 \times 10^8 \,Nm^{-2}$.
Now, $u = \frac{1}{2} \times \sigma \times \frac{\sigma}{Y} = \frac{\sigma^2}{2Y}$.
$u = \frac{(3 \times 10^8)^2}{2 \times 2 \times 10^{11}} = \frac{9 \times 10^{16}}{4 \times 10^{11}} = 2.25 \times 10^5 \,Jm^{-3}$.

(A) The work done in stretching a wire is given by the formula: $W = \frac{1}{2} \times \text{Stress} \times \text{Strain} \times \text{Volume}$.
Given: Strain $(\epsilon) = 10^{-3}$,Young's modulus $(Y) = 2 \times 10^{11} \ Nm^{-2}$,Mass $(m) = 2.96 \ kg$,Density $(\rho) = 7.4 \ g \ cm^{-3} = 7400 \ kg \ m^{-3}$.
First,calculate the volume $(V)$ of the wire: $V = \frac{m}{\rho} = \frac{2.96}{7400} = 4 \times 10^{-4} \ m^3$.
Next,calculate the stress: $\text{Stress} = Y \times \epsilon = (2 \times 10^{11}) \times (10^{-3}) = 2 \times 10^8 \ Nm^{-2}$.
Now,calculate the work done: $W = \frac{1}{2} \times (2 \times 10^8) \times (10^{-3}) \times (4 \times 10^{-4})$.
$W = 10^8 \times 10^{-3} \times 4 \times 10^{-4} = 4 \times 10^1 = 40 \ J = 0.04 \ kJ$.

(B) The elongation $\Delta L$ of a wire is given by $\Delta L = \frac{FL}{AY}$,where $F$ is the force,$L$ is the length,$A$ is the area of cross-section,and $Y$ is Young's modulus.
Since the wires are made of the same material,$Y_A = Y_B = Y$. The force $F$ is also the same.
We know that mass $m = \rho A L$,where $\rho$ is the density. Since the material is the same,$\rho_A = \rho_B = \rho$.
Thus,$L = \frac{m}{\rho A}$.
Substituting this into the elongation formula: $\Delta L = \frac{F}{\rho A^2 Y} \cdot m$.
For the ratio of elongations: $\frac{\Delta L_A}{\Delta L_B} = \frac{m_A}{m_B} \cdot \left( \frac{A_B}{A_A} \right)^2$.
Given $\frac{m_A}{m_B} = \frac{2}{3}$ and $\frac{A_A}{A_B} = \frac{1}{2}$,we have $\frac{A_B}{A_A} = 2$.
Therefore,$\frac{\Delta L_A}{\Delta L_B} = \frac{2}{3} \cdot (2)^2 = \frac{2}{3} \cdot 4 = \frac{8}{3}$.

(D) Let the length of both rods be $L$,density be $\rho$,and Young's modulus be $Y$.
Since the masses are equal,$m_1 = m_2 \implies \rho A_1 L = \rho A_2 L \implies A_1 = A_2$.
For the circular rod,$A_1 = \pi r^2 = \pi (d/2)^2 = \pi (0.5)^2 = \frac{\pi}{4} \ cm^2$.
For the square rod,$A_2 = s^2 = 1^2 = 1 \ cm^2$.
Wait,the problem states they have equal mass and equal length (implied by rods),so $A_1 = A_2$. However,the dimensions given ($d=1 \ cm$ and $s=1 \ cm$) lead to $A_1 = \frac{\pi}{4}$ and $A_2 = 1$. This implies the lengths must be different to keep mass equal.
Let $L_1$ and $L_2$ be the lengths. $m = \rho A_1 L_1 = \rho A_2 L_2 \implies L_1 A_1 = L_2 A_2$.
Elongation $\Delta L = \frac{FL}{AY}$.
Ratio $\frac{\Delta L_1}{\Delta L_2} = \frac{L_1}{A_1} \times \frac{A_2}{L_2} = \frac{L_1}{L_2} \times \frac{A_2}{A_1}$.
Since $\frac{L_1}{L_2} = \frac{A_2}{A_1}$,the ratio is $(\frac{A_2}{A_1})^2$.
$A_1 = \frac{\pi}{4}$,$A_2 = 1$.
Ratio $= (\frac{1}{\pi/4})^2 = (\frac{4}{\pi})^2 = \frac{16}{\pi^2}$.

(A) Given: Initial velocity $u = 20 \,ms^{-1}$, final velocity $v = 0 \,ms^{-1}$ at maximum height, and acceleration $a = -g = -10 \,ms^{-2}$.
Using the third equation of motion: $v^2 = u^2 + 2as$.
Substituting the values: $0^2 = (20)^2 + 2(-10)h$.
$0 = 400 - 20h$.
$20h = 400$.
$h = 20 \,m$.
Note: The value $18 \,ms^{-1}$ provided in the question is irrelevant to finding the maximum height, as the final velocity at the peak of a vertical projection is always $0 \,ms^{-1}$.

(B) For a body falling freely from rest,the distance $s$ covered in time $t$ is given by $s = \frac{1}{2}gt^2$.
Thus,the time taken to cover a distance $s$ is $t = \sqrt{\frac{2s}{g}}$.
Let $t_1$ be the time to travel the first $5 \ m$ $(s_1 = 5 \ m)$. Then $t_1 = \sqrt{\frac{2(5)}{g}} = \sqrt{\frac{10}{g}}$.
Let $t_2$ be the time to travel the first $10 \ m$ $(s_2 = 10 \ m)$. Then $t_2 = \sqrt{\frac{2(10)}{g}} = \sqrt{\frac{20}{g}}$.
Let $t_3$ be the time to travel the first $15 \ m$ $(s_3 = 15 \ m)$. Then $t_3 = \sqrt{\frac{2(15)}{g}} = \sqrt{\frac{30}{g}}$.
The time taken to travel the first $5 \ m$ is $T_1 = t_1 = \sqrt{\frac{10}{g}}$.
The time taken to travel the second $5 \ m$ is $T_2 = t_2 - t_1 = \sqrt{\frac{20}{g}} - \sqrt{\frac{10}{g}} = \sqrt{\frac{10}{g}}(\sqrt{2} - 1)$.
The time taken to travel the third $5 \ m$ is $T_3 = t_3 - t_2 = \sqrt{\frac{30}{g}} - \sqrt{\frac{20}{g}} = \sqrt{\frac{10}{g}}(\sqrt{3} - \sqrt{2})$.
The ratio $T_1 : T_2 : T_3$ is $\sqrt{\frac{10}{g}} : \sqrt{\frac{10}{g}}(\sqrt{2} - 1) : \sqrt{\frac{10}{g}}(\sqrt{3} - \sqrt{2})$.
Simplifying,we get $1 : (\sqrt{2} - 1) : (\sqrt{3} - \sqrt{2})$.

(C) Step $1$: Calculate the time taken for free fall $(t_1)$ and velocity attained $(v)$.
For free fall,initial velocity $u = 0$,distance $s_1 = 20 \ m$,and $g = 10 \ m/s^2$.
Using $v^2 = u^2 + 2gs_1$,we get $v^2 = 0 + 2 \times 10 \times 20 = 400$,so $v = 20 \ m/s$.
Using $v = u + gt_1$,we get $20 = 0 + 10t_1$,so $t_1 = 2 \ s$.
Step $2$: Calculate the time taken for uniform motion $(t_2)$.
The remaining distance is $s_2 = 2000 \ m - 20 \ m = 1980 \ m$.
The velocity is constant at $v = 20 \ m/s$.
Time $t_2 = s_2 / v = 1980 / 20 = 99 \ s$.
Step $3$: Calculate total time.
Total time $T = t_1 + t_2 = 2 \ s + 99 \ s = 101 \ s$.

(B) Given the relation: $t = 2x^2 + 3x$.
To find velocity $v$,differentiate $t$ with respect to $x$: $\frac{dt}{dx} = 4x + 3$.
Since $v = \frac{dx}{dt}$,we have $v = \frac{1}{4x + 3} = (4x + 3)^{-1}$.
To find acceleration $a$,differentiate $v$ with respect to $t$: $a = \frac{dv}{dt} = \frac{dv}{dx} \cdot \frac{dx}{dt} = \frac{dv}{dx} \cdot v$.
$\frac{dv}{dx} = -1(4x + 3)^{-2} \cdot 4 = -\frac{4}{(4x + 3)^2}$.
Thus,$a = -\frac{4}{(4x + 3)^2} \cdot \frac{1}{4x + 3} = -\frac{4}{(4x + 3)^3}$.
Given displacement $x = 25 \ cm = 0.25 \ m = \frac{1}{4} \ m$.
Substitute $x = \frac{1}{4}$ into the acceleration formula: $a = -\frac{4}{(4(1/4) + 3)^3} = -\frac{4}{(1 + 3)^3} = -\frac{4}{4^3} = -\frac{4}{64} = -\frac{1}{16} \ ms^{-2}$.

(D) The displacement of a particle in the $n^{th}$ second is given by the formula: $S_n = u + \frac{a}{2}(2n - 1)$,where $u$ is the initial velocity and $a$ is the constant acceleration.
For the $3^{rd}$ second $(n=3)$: $10 = u + \frac{a}{2}(2(3) - 1) \implies 10 = u + 2.5a$ --- (Equation $1$)
For the $5^{th}$ second $(n=5)$: $18 = u + \frac{a}{2}(2(5) - 1) \implies 18 = u + 4.5a$ --- (Equation $2$)
Subtracting Equation $1$ from Equation $2$: $(18 - 10) = (u + 4.5a) - (u + 2.5a) \implies 8 = 2a \implies a = 4 \ ms^{-2}$.
Substituting $a = 4$ into Equation $1$: $10 = u + 2.5(4) \implies 10 = u + 10 \implies u = 0 \ ms^{-1}$.
The velocity at time $t$ is given by $v = u + at$.
At $t = 4 \ s$: $v = 0 + (4)(4) = 16 \ ms^{-1}$.

(C) When an iron rod is inserted into the inductor,the self-inductance $L$ of the coil increases because the permeability of the core increases.
The inductive reactance of the circuit is given by $X_L = \omega L$.
As $L$ increases,$X_L$ increases.
The total impedance of the series circuit is $Z = \sqrt{R^2 + X_L^2}$,where $R$ is the resistance of the bulb.
Since $X_L$ increases,the total impedance $Z$ of the circuit increases.
The current in the circuit is given by $I = V/Z$.
As $Z$ increases,the current $I$ flowing through the bulb decreases.
Since the power dissipated in the bulb is $P = I^2 R$,a decrease in current leads to a decrease in the power,and thus the glow of the bulb decreases.

(B) Given: Resistance $R = 100 \sqrt{3} \Omega$,Voltage $V = 100 \sin(200t) \text{ V}$,Phase difference $\phi = 30^{\circ}$.
Comparing the voltage equation with $V = V_m \sin(\omega t)$,we get angular frequency $\omega = 200 \text{ rad/s}$.
In an $RC$ series circuit,the phase difference $\phi$ is given by $\tan \phi = \frac{X_C}{R}$,where $X_C = \frac{1}{\omega C}$.
Substituting the values: $\tan 30^{\circ} = \frac{1}{\omega C R}$.
Since $\tan 30^{\circ} = \frac{1}{\sqrt{3}}$,we have $\frac{1}{\sqrt{3}} = \frac{1}{200 \times C \times 100 \sqrt{3}}$.
Canceling $\sqrt{3}$ from both sides: $1 = \frac{1}{200 \times 100 \times C}$.
$C = \frac{1}{20000} \text{ F} = 0.5 \times 10^{-4} \text{ F} = 50 \times 10^{-6} \text{ F} = 50 \mu \text{F}$.
Thus,the capacitance is $50 \mu \text{F}$.

(B) In an $LCR$ series circuit,the total impedance $Z$ is given by $Z = \sqrt{R^2 + (X_L - X_C)^2}$.
For the circuit to be purely resistive,the net reactance must be zero,which means the inductive reactance must equal the capacitive reactance.
Therefore,$X_L - X_C = 0$,which implies $X_L = X_C$.
At this condition,the circuit is said to be in resonance,and the impedance $Z$ becomes equal to $R$.

(B) The instantaneous voltage $V$ is given by $V = V_0 \sin(\omega t)$,where $V_0$ is the peak voltage and $\omega = 2\pi f$.
Given $f = 50 \ Hz$,so $\omega = 2 \times \pi \times 50 = 100\pi \ rad/s$.
We want to find the time $t$ when $V = \frac{V_0}{2}$.
Substituting this into the equation: $\frac{V_0}{2} = V_0 \sin(100\pi t)$.
$\sin(100\pi t) = \frac{1}{2}$.
Since $\sin(30^\circ) = \frac{1}{2}$,we have $100\pi t = \frac{\pi}{6}$.
$t = \frac{\pi}{6 \times 100\pi} = \frac{1}{600} \ s$.

(A) In an $LR$ series circuit,the voltage across the inductor $(V_L)$ and the voltage across the resistor $(V_R)$ are out of phase by $90^{\circ}$.
The total voltage $(V)$ of the ac supply is given by the phasor sum of the individual voltages:
$V = \sqrt{V_L^2 + V_R^2}$
Given:
$V_L = 180 \ V$
$V_R = 240 \ V$
Substituting the values:
$V = \sqrt{(180)^2 + (240)^2}$
$V = \sqrt{32400 + 57600}$
$V = \sqrt{90000}$
$V = 300 \ V$
Therefore,the voltage of the ac supply is $300 \ V$.

(C) Eddy currents are induced currents that circulate within the core of a transformer when it is subjected to a changing magnetic field,leading to energy loss in the form of heat.
To minimize these losses,the core is constructed using thin,insulated sheets of metal stacked together,known as a laminated core.
This lamination increases the electrical resistance of the path for eddy currents,thereby significantly reducing their magnitude and the associated energy dissipation.

(C) The frequency of a spectral line in the hydrogen atom is given by $f = R c \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right)$,where $R$ is the Rydberg constant and $c$ is the speed of light.
For the Lyman series,$n_1 = 1$. The first line is $n_2 = 2$ and the second line is $n_2 = 3$.
$f_1 = R c \left( \frac{1}{1^2} - \frac{1}{2^2} \right) = R c \left( 1 - \frac{1}{4} \right) = \frac{3}{4} R c$
$f_2 = R c \left( \frac{1}{1^2} - \frac{1}{3^2} \right) = R c \left( 1 - \frac{1}{9} \right) = \frac{8}{9} R c$
The difference is $f = f_2 - f_1 = R c \left( \frac{8}{9} - \frac{3}{4} \right) = R c \left( \frac{32 - 27}{36} \right) = \frac{5}{36} R c$. Thus,$R c = \frac{36 f}{5}$.
For the Balmer series,$n_1 = 2$. The first line is $n_2 = 3$ and the second line is $n_2 = 4$.
$f'_1 = R c \left( \frac{1}{2^2} - \frac{1}{3^2} \right) = R c \left( \frac{1}{4} - \frac{1}{9} \right) = \frac{5}{36} R c$
$f'_2 = R c \left( \frac{1}{2^2} - \frac{1}{4^2} \right) = R c \left( \frac{1}{4} - \frac{1}{16} \right) = \frac{3}{16} R c$
The difference is $f' = f'_2 - f'_1 = R c \left( \frac{3}{16} - \frac{5}{36} \right) = R c \left( \frac{27 - 20}{144} \right) = \frac{7}{144} R c$.
Substituting $R c = \frac{36 f}{5}$,we get $f' = \frac{7}{144} \times \frac{36 f}{5} = \frac{7 f}{4 \times 5} = \frac{7 f}{20}$.

(C) The wavelength $\lambda$ for a transition in a hydrogen atom is given by the Rydberg formula: $\frac{1}{\lambda} = R \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right)$,where $R$ is the Rydberg constant.
For the second line of the Balmer series,$n_1 = 2$ and $n_2 = 4$. Thus,$\frac{1}{\lambda_B} = R \left( \frac{1}{2^2} - \frac{1}{4^2} \right) = R \left( \frac{1}{4} - \frac{1}{16} \right) = R \left( \frac{3}{16} \right)$. So,$\lambda_B = \frac{16}{3R}$.
For the first line of the Lyman series,$n_1 = 1$ and $n_2 = 2$. Thus,$\frac{1}{\lambda_L} = R \left( \frac{1}{1^2} - \frac{1}{2^2} \right) = R \left( 1 - \frac{1}{4} \right) = R \left( \frac{3}{4} \right)$. So,$\lambda_L = \frac{4}{3R}$.
The ratio of the wavelengths is $\frac{\lambda_B}{\lambda_L} = \frac{16/3R}{4/3R} = \frac{16}{4} = 4:1$.

(C) The kinetic energy $(K)$ of an electron in the $n^{th}$ orbit of a hydrogen atom is given by $K_n = \frac{13.6 \text{ eV}}{n^2}$.
For the third excited state,the principal quantum number is $n_1 = 3 + 1 = 4$.
For the fourth excited state,the principal quantum number is $n_2 = 4 + 1 = 5$.
The kinetic energy in the third excited state is $K_4 = \frac{13.6}{4^2} = \frac{13.6}{16}$.
The kinetic energy in the fourth excited state is $K_5 = \frac{13.6}{5^2} = \frac{13.6}{25}$.
The ratio of the kinetic energies is $\frac{K_4}{K_5} = \frac{13.6/16}{13.6/25} = \frac{25}{16}$.
Thus,the ratio is $25: 16$.

(C) The total energy $E$ of an electron in an orbit is given by the sum of its kinetic energy $K$ and potential energy $U$. For a bound system,such as an electron in an atom,the total energy is negative $(E < 0)$.
If the total energy $E$ is positive $(E > 0)$,it implies that the kinetic energy is greater than the magnitude of the potential energy. In such a case,the electron is not bound to the nucleus and will escape the influence of the electrostatic force.
Therefore,the electron will not follow a closed orbit and will move away to infinity.

(C) The energy required to ionize a hydrogen atom from its ground state $(n=1)$ is equal to the ionization energy, which is $E = 13.6 \, eV$.
To find the maximum wavelength $(\lambda_{max})$ of the incident radiation, we use the relation $E = \frac{hc}{\lambda}$.
Substituting the values: $13.6 \, eV = \frac{1240 \, eV \cdot nm}{\lambda}$.
$\lambda = \frac{1240}{13.6} \, nm \approx 91.17 \, nm$.
Converting this to $\mathring{A}$ $(Å)$: $91.17 \, nm = 911.7 \, Å \approx 912 \, Å$.
Thus, the maximum wavelength required is nearly $912 \, Å$.

(B) $1$. Initial state: The capacitor has a dielectric of thickness $d/2$ and constant $K$. The capacitance $C_i$ is given by the series combination of two capacitors: one with dielectric $(C_1 = \frac{2K\epsilon_0 A}{d})$ and one with air $(C_2 = \frac{2\epsilon_0 A}{d})$.
$C_i = \frac{C_1 C_2}{C_1 + C_2} = \frac{\frac{2K\epsilon_0 A}{d} \cdot \frac{2\epsilon_0 A}{d}}{\frac{2\epsilon_0 A}{d}(K+1)} = \frac{2K\epsilon_0 A}{d(K+1)}$.
$2$. Charge on the capacitor: $Q = C_i V = \frac{2K\epsilon_0 A V}{d(K+1)}$.
$3$. After disconnecting the battery,the charge $Q$ remains constant.
$4$. Final state: The dielectric is removed,so the capacitor becomes an air-filled parallel plate capacitor with capacitance $C_f = \frac{\epsilon_0 A}{d}$.
$5$. Final potential difference $V_f = \frac{Q}{C_f} = \frac{2K\epsilon_0 A V}{d(K+1)} \cdot \frac{d}{\epsilon_0 A} = V \left( \frac{2K}{K+1} \right)$.

(D) Initial capacitance with air $C_0 = 4 \mu F$.
After filling with dielectric $(K = 5)$,the new capacitance is $C = K C_0 = 5 \times 4 \mu F = 20 \mu F$.
The charge on the capacitor is $Q = C V = 20 \mu F \times 100 \ V = 2000 \mu C = 2 \times 10^{-3} \ C$.
Since the capacitor is disconnected from the battery,the charge $Q$ remains constant.
Initial energy stored with dielectric: $U_i = \frac{Q^2}{2C} = \frac{(2 \times 10^{-3})^2}{2 \times 20 \times 10^{-6}} = \frac{4 \times 10^{-6}}{40 \times 10^{-6}} = 0.1 \ J$.
Final capacitance after removing the dielectric: $C_f = C_0 = 4 \mu F$.
Final energy stored: $U_f = \frac{Q^2}{2C_f} = \frac{(2 \times 10^{-3})^2}{2 \times 4 \times 10^{-6}} = \frac{4 \times 10^{-6}}{8 \times 10^{-6}} = 0.5 \ J$.
Work done by external agent $W = U_f - U_i = 0.5 \ J - 0.1 \ J = 0.4 \ J$.

(A) Initial energy stored in the first capacitor: $U_i = \frac{1}{2} C_1 V^2 = \frac{1}{2} \times 10 \times 10^{-6} \times (220)^2 = 5 \times 10^{-6} \times 48400 = 0.242 \text{ J} = 242 \text{ mJ}$.
When connected to the second capacitor,the charge $Q = C_1 V = 10 \times 10^{-6} \times 220 = 2.2 \times 10^{-3} \text{ C}$ is redistributed.
The common potential $V'$ is given by $V' = \frac{Q}{C_1 + C_2} = \frac{2.2 \times 10^{-3}}{10 \times 10^{-6} + 12 \times 10^{-6}} = \frac{2.2 \times 10^{-3}}{22 \times 10^{-6}} = 100 \text{ V}$.
Final energy stored in the system: $U_f = \frac{1}{2} (C_1 + C_2) (V')^2 = \frac{1}{2} \times 22 \times 10^{-6} \times (100)^2 = 11 \times 10^{-6} \times 10000 = 0.11 \text{ J} = 110 \text{ mJ}$.
Loss of energy: $\Delta U = U_i - U_f = 242 \text{ mJ} - 110 \text{ mJ} = 132 \text{ mJ}$.

(B) Speech signals are human voice signals which typically have a frequency range from $20 \ Hz$ to $20 \ kHz$. However,for commercial telephonic communication,it is not necessary to transmit the entire audible range. To ensure clarity and efficiency in transmission,the frequency range is restricted. The standard frequency range adequate for speech signals in commercial telephony is $300 \ Hz$ to $3100 \ Hz$.

(C) coaxial cable is a type of transmission line that consists of an inner conductor surrounded by a tubular insulating layer,surrounded by a tubular conducting shield.
It is widely used for cable television,broadband internet,and other high-frequency signal transmissions.
The frequency bandwidth of a coaxial cable is typically in the range of up to $750 \text{ MHz}$.
Therefore,the correct option is $C$.

(A) The frequency of the carrier wave is $f_c = 1 \text{ MHz} = 1000 \text{ kHz}$.
The frequency of the message signal is $f_m = 28 \text{ kHz}$.
The frequencies of the side bands are given by the upper side band $(USB)$ and lower side band $(LSB)$ formulas:
$USB = f_c + f_m = 1000 \text{ kHz} + 28 \text{ kHz} = 1028 \text{ kHz}$.
$LSB = f_c - f_m = 1000 \text{ kHz} - 28 \text{ kHz} = 972 \text{ kHz}$.
Therefore,the side band frequencies are $1028 \text{ kHz}$ and $972 \text{ kHz}$.

(B) The range $d$ of a transmitting antenna of height $h$ is given by the formula $d = \sqrt{2Rh}$,where $R$ is the radius of the Earth.
Given: $h = 980 \ m = 0.98 \ km$ and $R = 6400 \ km$.
Substituting the values into the formula:
$d = \sqrt{2 \times 6400 \times 0.98}$
$d = \sqrt{12800 \times 0.98}$
$d = \sqrt{12544}$
$d = 112 \ km$.
Therefore,the range of the transmitting antenna is $112 \ km$.

(D) The layer of the atmosphere that reflects radio waves,including high-frequency waves,is the ionosphere,which is a part of the thermosphere.
During the night,the ionosphere becomes more stable and effective at reflecting these waves,allowing for long-distance communication.
Therefore,the correct layer is the thermosphere.

(C) The radio horizon distance $d$ for a transmitting antenna of height $h$ is given by the formula $d = \sqrt{2Rh}$,where $R$ is the radius of the Earth.
Given: $h = 39.2 \ m = 39.2 \times 10^{-3} \ km$ and $R = 6400 \ km$.
Substituting the values into the formula:
$d = \sqrt{2 \times 6400 \times 39.2 \times 10^{-3}}$
$d = \sqrt{12800 \times 0.0392}$
$d = \sqrt{501.76}$
$d = 22.4 \ km$.
Therefore,the correct option is $C$.

(A) The total resistance of the wire is $R = 36 \Omega$.
When the wire is bent into a semi-circular loop,the wire is divided into two equal halves along the diameter.
Each half has a resistance of $R' = \frac{R}{2} = \frac{36}{2} = 18 \Omega$.
These two halves are connected in parallel between the ends of the diameter.
The equivalent resistance $R_{eq}$ is given by the formula for parallel resistors: $\frac{1}{R_{eq}} = \frac{1}{R'} + \frac{1}{R'}$.
$\frac{1}{R_{eq}} = \frac{1}{18} + \frac{1}{18} = \frac{2}{18} = \frac{1}{9}$.
Therefore,$R_{eq} = 9 \Omega$.

(D) Let the electromotive force of the cell be $E$ and its internal resistance be $r$. The current flowing through the circuit is $I = 2 \ A$.
When the current flows in the normal direction,the terminal potential difference is given by $V_1 = E - Ir$.
Substituting the given values: $20 = E - 2r$ --- (Equation $1$).
When the direction of the current is reversed,the cell is being charged,so the terminal potential difference is given by $V_2 = E + Ir$.
Substituting the given values: $30 = E + 2r$ --- (Equation $2$).
Subtracting Equation $1$ from Equation $2$: $(E + 2r) - (E - 2r) = 30 - 20$.
$4r = 10$.
$r = 2.5 \ \Omega$.
Thus,the internal resistance of the cell is $2.5 \ \Omega$.

(A) The mobility $\mu$ of charge carriers is defined as the ratio of drift velocity $v_d$ to the applied electric field $E$.
Formula: $\mu = \frac{v_d}{E}$
Given:
Drift velocity $v_d = 0.3 \ m s^{-1}$
Electric field $E = 2 \ V m^{-1}$
Calculation:
$\mu = \frac{0.3}{2} = 0.15 \ m^2 V^{-1} s^{-1}$
Therefore,the correct option is $A$.

(C) The drift velocity $v_d$ is given by the formula $v_d = \frac{I}{neA}$,where $I$ is the current,$n$ is the electron density,$e$ is the charge of an electron,and $A$ is the cross-sectional area.
Since the wires are connected in parallel to the same battery,the potential difference $V$ across both wires is the same.
Using Ohm's law,$I = \frac{V}{R}$,where $R = \rho \frac{L}{A}$. Thus,$I = \frac{VA}{\rho L}$.
Substituting this into the drift velocity formula: $v_d = \frac{VA}{neA\rho L} = \frac{V}{ne\rho L}$.
Since both wires are made of the same material,$n$,$e$,and $\rho$ are constant. Also,$V$ is constant.
Therefore,$v_d \propto \frac{1}{L}$.
The ratio of lengths is $L_1 : L_2 = 2 : 3$.
Thus,the ratio of drift velocities is $\frac{v_{d1}}{v_{d2}} = \frac{L_2}{L_1} = \frac{3}{2}$.

(A) The resistance per unit length of the potentiometer wire is $\lambda = \frac{R}{L} = \frac{5 \ \Omega}{10 \ m} = 0.5 \ \Omega/m$.
The resistance of the wire segment of length $6.6 \ m$ $(660 \ cm)$ is $R' = 0.5 \ \Omega/m \times 6.6 \ m = 3.3 \ \Omega$.
The potential difference across this segment is $V' = I \times R'$,where $I$ is the current in the potentiometer wire.
Given $V' = 1.1 \ V$,we have $1.1 = I \times 3.3$,which gives $I = \frac{1.1}{3.3} = \frac{1}{3} \ A$.
The current in the circuit is given by $I = \frac{E}{R + r}$,where $E = 2.2 \ V$,$R = 5 \ \Omega$,and $r$ is the internal resistance.
Substituting the values: $\frac{1}{3} = \frac{2.2}{5 + r}$.
$5 + r = 3 \times 2.2 = 6.6$.
$r = 6.6 - 5 = 1.6 \ \Omega$.

(C) The total resistance in the primary circuit is $R_{total} = R_{wire} + R_{series} = 8 \ \Omega + 242 \ \Omega = 250 \ \Omega$.
The current flowing through the potentiometer wire is $I = \frac{V}{R_{total}} = \frac{2.5 \ V}{250 \ \Omega} = 0.01 \ A$.
The potential drop across the entire wire is $V_{wire} = I \times R_{wire} = 0.01 \ A \times 8 \ \Omega = 0.08 \ V$.
The potential gradient $k$ is the potential drop per unit length: $k = \frac{V_{wire}}{L} = \frac{0.08 \ V}{2.5 \ m} = 0.032 \ V/m$.
The potential difference across a length $l = 20 \ cm = 0.2 \ m$ is $V' = k \times l = 0.032 \ V/m \times 0.2 \ m = 0.0064 \ V = 6.4 \ mV$.

(A) The voltage sensitivity $(V_s)$ of a moving coil galvanometer is given by $V_s = \frac{NBA}{kR}$,where $N$ is the number of turns,$B$ is the magnetic field,$A$ is the area,$k$ is the torsional constant,and $R$ is the resistance.
Given the ratio of voltage sensitivities: $\frac{V_{sA}}{V_{sB}} = \frac{4}{3}$.
Given the ratio of resistances: $\frac{R_A}{R_B} = \frac{3}{4}$.
Given the ratio of areas: $\frac{A_A}{A_B} = \frac{1}{2}$.
Since $B$ and $k$ are constant,we have $\frac{V_{sA}}{V_{sB}} = \frac{N_A A_A R_B}{N_B A_B R_A}$.
Substituting the given values: $\frac{4}{3} = \frac{N_A}{N_B} \times (\frac{1}{2}) \times (\frac{4}{3})$.
$\frac{4}{3} = \frac{N_A}{N_B} \times \frac{2}{3}$.
$\frac{N_A}{N_B} = \frac{4}{3} \times \frac{3}{2} = 2$.
Given $N_A = 200$,then $N_B = \frac{N_A}{2} = \frac{200}{2} = 100$.

(A) In a potentiometer,the balancing length $l$ is directly proportional to the terminal potential difference $V$ across the cell.
Initially,when the cell is in an open circuit,the balancing length is $l_1 = kE$,where $E$ is the $EMF$ of the cell and $k$ is the potential gradient.
When an external resistance $R$ is connected in parallel,the terminal potential difference becomes $V = E \left( \frac{R}{R + r} \right)$,where $r$ is the internal resistance.
The new balancing length is $l_2 = kV = kE \left( \frac{R}{R + r} \right)$.
Given that the balancing length decreases by $10 \%$,we have $l_2 = l_1 - 0.1l_1 = 0.9l_1$.
Substituting the expressions for $l_1$ and $l_2$: $kE \left( \frac{R}{R + r} \right) = 0.9 kE$.
Dividing both sides by $kE$: $\frac{R}{R + r} = 0.9 = \frac{9}{10}$.
Cross-multiplying gives $10R = 9(R + r) = 9R + 9r$.
Therefore,$R = 9r$,which implies $r = \frac{R}{9}$.

(D) The power $P$ of an electric motor is given by $P = \frac{V^2}{R}$,where $V$ is the voltage and $R$ is the resistance of the motor.
First,we calculate the resistance $R$ using the initial conditions: $P_1 = 242 \ W$ and $V_1 = 220 \ V$.
$R = \frac{V_1^2}{P_1} = \frac{220 \times 220}{242} = \frac{48400}{242} = 200 \ \Omega$.
Now,when the motor is operated at $V_2 = 200 \ V$,the current $I$ drawn by the motor is given by Ohm's law: $I = \frac{V_2}{R}$.
$I = \frac{200 \ V}{200 \ \Omega} = 1 \ A$.
Thus,the current drawn is $1 \ A$.

(A) To find the potential difference between points $C$ and $D$,we first need to determine the current flowing through the resistor between $C$ and $D$.
Applying Kirchhoff's Current Law $(KCL)$ at junction $C$:
The sum of currents entering the junction must equal the sum of currents leaving the junction.
Let $I_{BC}$ be the current flowing from $B$ to $C$,$I_{CF}$ be the current flowing from $C$ to $F$,$I_{CG}$ be the current flowing from $C$ to $G$,and $I_{CD}$ be the current flowing from $C$ to $D$.
First,find $I_{BC}$ using $KCL$ at junction $B$:
$I_{AB} + I_{EB} = I_{BC}$
$4 \ A + 1.8 \ A = I_{BC}$
$I_{BC} = 5.8 \ A$
Now,apply $KCL$ at junction $C$:
$I_{BC} = I_{CF} + I_{CG} + I_{CD}$
$5.8 \ A = 1.3 \ A + 1 \ A + I_{CD}$
$5.8 \ A = 2.3 \ A + I_{CD}$
$I_{CD} = 5.8 \ A - 2.3 \ A = 3.5 \ A$
The potential difference between $C$ and $D$ is given by Ohm's Law:
$V_{CD} = I_{CD} \times R_{CD}$
$V_{CD} = 3.5 \ A \times 8 \ \Omega = 28 \ V$.

(A) Let the potential at junction $A$ be $V_A = 3 \text{ V}$ and at junction $C$ be $V_C = 0 \text{ V}$.
Let the potential at junction $B$ be $V_B$.
Applying Kirchhoff's Current Law $(KCL)$ at junction $B$:
$\frac{V_B - V_A}{6} + \frac{V_B - V_C}{12} + \frac{V_B - V_C}{12} = 0$
$\frac{V_B - 3}{6} + \frac{V_B}{12} + \frac{V_B}{12} = 0$
Multiplying by $12$:
$2(V_B - 3) + V_B + V_B = 0$
$2V_B - 6 + 2V_B = 0$
$4V_B = 6 \implies V_B = 1.5 \text{ V}$.
The current through the $6 \Omega$ resistor between $A$ and $B$ is $I = \frac{V_A - V_B}{6} = \frac{3 - 1.5}{6} = \frac{1.5}{6} = 0.25 \text{ A}$.

(B) Given:
Area of cross-section,$A = 6 \times 10^{-7} \ m^2$
Potential gradient,$x = \frac{V}{L} = 0.15 \ Vm^{-1}$
Current,$I = 0.3 \ A$
From Ohm's law,the potential difference across a length $L$ is $V = I \times R$,where $R = \rho \frac{L}{A}$.
Substituting $R$ in the equation for $V$,we get $V = I \times \rho \frac{L}{A}$.
Rearranging for potential gradient: $\frac{V}{L} = \frac{I \times \rho}{A}$.
Substituting the given values: $0.15 = \frac{0.3 \times \rho}{6 \times 10^{-7}}$.
Solving for $\rho$: $\rho = \frac{0.15 \times 6 \times 10^{-7}}{0.3}$.
$\rho = 0.5 \times 6 \times 10^{-7} = 3 \times 10^{-7} \ \Omega \ m$.

(D) Let the resistance in the left gap be $R_L$ and the resistance in the right gap be $R_R$.
In the first case,the right gap has two equal resistors $r$ in series,so $R_R = r + r = 2r$.
The balancing point is at $l_1 = 50 \ cm$.
Using the meter bridge formula: $\frac{R_L}{R_R} = \frac{l_1}{100 - l_1} \implies \frac{R_L}{2r} = \frac{50}{50} = 1 \implies R_L = 2r$.
In the second case,one resistor $r$ is removed from the right gap,so the new right resistance is $R_R' = r$.
This removed resistor $r$ is connected in parallel to $R_L$.
The new left resistance is $R_L' = \frac{R_L \cdot r}{R_L + r} = \frac{2r \cdot r}{2r + r} = \frac{2r^2}{3r} = \frac{2}{3}r$.
Let the new balancing point be $l_2$.
Then $\frac{R_L'}{R_R'} = \frac{l_2}{100 - l_2} \implies \frac{(2/3)r}{r} = \frac{l_2}{100 - l_2} \implies \frac{2}{3} = \frac{l_2}{100 - l_2}$.
$2(100 - l_2) = 3l_2 \implies 200 - 2l_2 = 3l_2 \implies 5l_2 = 200 \implies l_2 = 40 \ cm$.

(A) According to Einstein's photoelectric equation,the maximum kinetic energy $K_{max}$ of the emitted photoelectrons is given by $K_{max} = E - \Phi$,where $E$ is the energy of the incident photon and $\Phi$ is the work function of the material.
Given $E = 4.5 \ eV$ and $\Phi = 3 \ eV$,we have $K_{max} = 4.5 \ eV - 3 \ eV = 1.5 \ eV$.
Converting this energy to Joules: $K_{max} = 1.5 \times 1.6 \times 10^{-19} \ J = 2.4 \times 10^{-19} \ J$.
The de Broglie wavelength $\lambda$ is given by $\lambda = \frac{h}{\sqrt{2mK_{max}}}$,where $h = 6.63 \times 10^{-34} \ J \cdot s$ and $m = 9.1 \times 10^{-31} \ kg$.
Substituting the values: $\lambda = \frac{6.63 \times 10^{-34}}{\sqrt{2 \times 9.1 \times 10^{-31} \times 2.4 \times 10^{-19}}} = \frac{6.63 \times 10^{-34}}{\sqrt{43.68 \times 10^{-50}}} = \frac{6.63 \times 10^{-34}}{6.61 \times 10^{-25}} \approx 1.0 \times 10^{-9} \ m$.
Since $1 \ Å = 10^{-10} \ m$,we have $\lambda \approx 10 \ Å$.

(B) The potential energy $U$ of an electron in the $n^{th}$ orbit of a hydrogen atom is given by $U = -27.2 / n^2 \text{ eV}$.
Given $U = -6.8 \text{ eV}$,we have $-27.2 / n^2 = -6.8$,which implies $n^2 = 27.2 / 6.8 = 4$,so $n = 2$.
The radius of the $n^{th}$ orbit is given by $r_n = n^2 r_0$. For $n = 2$,$r_2 = 2^2 r_0 = 4 r_0$.
According to Bohr's quantization condition,the circumference of the orbit is an integer multiple of the de Broglie wavelength: $2 \pi r_n = n \lambda$.
Substituting the values,$2 \pi (4 r_0) = 2 \lambda$.
Solving for $\lambda$,we get $\lambda = (8 \pi r_0) / 2 = 4 \pi r_0$.

(C) The de Broglie wavelength $\lambda$ is given by $\lambda = \frac{h}{p}$,where $p = \sqrt{2mK}$ and $K$ is the kinetic energy.
Thus,$\lambda = \frac{h}{\sqrt{2mK}}$,which implies $K = \frac{h^2}{2m\lambda^2}$.
Given: $h = 6.6 \times 10^{-34} \text{ J s}$,$m = 9 \times 10^{-31} \text{ kg}$,and $\lambda = 6600 \times 10^{-10} \text{ m} = 6.6 \times 10^{-7} \text{ m}$.
Substituting the values:
$K = \frac{(6.6 \times 10^{-34})^2}{2 \times 9 \times 10^{-31} \times (6.6 \times 10^{-7})^2}$
$K = \frac{6.6^2 \times 10^{-68}}{18 \times 10^{-31} \times 6.6^2 \times 10^{-14}}$
$K = \frac{10^{-68}}{18 \times 10^{-45}} = \frac{1}{18} \times 10^{-23} \approx 0.0556 \times 10^{-23} \text{ J} = 5.56 \times 10^{-25} \text{ J}$.
Since the work done equals the kinetic energy gained,the work done is $5.56 \times 10^{-25} \text{ J}$.

(C) According to Einstein's photoelectric equation,the maximum kinetic energy of photoelectrons is given by $K_{max} = E - \phi$,where $E$ is the energy of the incident photon and $\phi$ is the work function of the material.
Since $K_{max} = e V_s$,where $V_s$ is the stopping potential,we have $e V_s = E - \phi$.
For the first case: $e(1.7 \ V) = 3.1 \ eV - \phi$,which implies $\phi = 3.1 \ eV - 1.7 \ eV = 1.4 \ eV$.
For the second case: $e V_s' = 2.5 \ eV - \phi$.
Substituting the value of $\phi$: $e V_s' = 2.5 \ eV - 1.4 \ eV = 1.1 \ eV$.
Therefore,the stopping potential $V_s' = 1.1 \ V$.

(A) The maximum kinetic energy $(K_{max})$ of the emitted photoelectrons is given by the formula: $K_{max} = \frac{1}{2} m v_{max}^2$.
Given $m = 9 \times 10^{-31} \text{ kg}$ and $v_{max} = 8 \times 10^5 \text{ m/s}$.
$K_{max} = \frac{1}{2} \times (9 \times 10^{-31}) \times (8 \times 10^5)^2$.
$K_{max} = 0.5 \times 9 \times 10^{-31} \times 64 \times 10^{10} = 288 \times 10^{-21} \text{ J}$.
To convert this energy into electron-volts (eV),divide by the charge of an electron $(e = 1.6 \times 10^{-19} \text{ C})$:
$K_{max} \text{ (in eV)} = \frac{288 \times 10^{-21}}{1.6 \times 10^{-19}} = 180 \times 10^{-2} = 1.8 \text{ eV}$.
The stopping potential $(V_s)$ is related to the maximum kinetic energy by the equation: $K_{max} = e V_s$.
Therefore,$V_s = \frac{K_{max}}{e} = 1.8 \text{ V}$.

(C) According to Einstein's photoelectric equation, the maximum kinetic energy of emitted photoelectrons is given by $K_{max} = h\nu - \phi_0$, where $h$ is Planck's constant, $\nu$ is the frequency of incident radiation, and $\phi_0$ is the work function.
Since $K_{max} = eV_s$, where $e$ is the charge of an electron and $V_s$ is the stopping potential, we can write:
$eV_s = h\nu - \phi_0$
$V_s = (\frac{h}{e})\nu - \frac{\phi_0}{e}$
This equation is in the form of a straight line $y = mx + c$, where $y = V_s$, $x = \nu$, and the slope $m = \frac{h}{e}$.
Given $h = 6.6 \times 10^{-34} \text{ Js}$ and $e = 1.6 \times 10^{-19} \text{ C}$.
Slope $m = \frac{6.6 \times 10^{-34}}{1.6 \times 10^{-19}} = 4.125 \times 10^{-15} \text{ JsC}^{-1}$.
Thus, the correct option is $C$.

(C) The magnetic flux through the coil is given by $\phi = 2t^2 + 3t - 2$.
According to Faraday's law of electromagnetic induction,the induced electromotive force $(e)$ is given by $e = -\frac{d\phi}{dt}$.
$e = -\frac{d}{dt}(2t^2 + 3t - 2) = -(4t + 3)$.
The magnitude of the induced $EMF$ is $|e| = 4t + 3$.
At time $t = 3 \ s$,the induced $EMF$ is $|e| = 4(3) + 3 = 12 + 3 = 15 \ V$.
The induced current in the coil is $I = \frac{|e|}{R} = \frac{15 \ V}{7.5 \ \Omega} = 2 \ A$.
The induced power $(P)$ in the coil is given by $P = I^2 R = (2)^2 \times 7.5 = 4 \times 7.5 = 30 \ W$.

(B) The magnetic flux $\phi$ through a coil is given by $\phi = NBA \cos \theta$,where $\theta$ is the angle between the area vector and the magnetic field.
Given that the plane of the coil makes an angle of $\frac{\pi}{6}$ with the magnetic field,the angle $\theta_1$ between the area vector and the magnetic field is $\theta_1 = \frac{\pi}{2} - \frac{\pi}{6} = \frac{\pi}{3} = 60^\circ$.
Initial flux $\phi_1 = NBA \cos(60^\circ) = 250 \times 2 \times (120 \times 10^{-4}) \times 0.5 = 3 \ Wb$.
When the plane of the coil is parallel to the magnetic field,the area vector is perpendicular to the magnetic field,so $\theta_2 = 90^\circ$.
Final flux $\phi_2 = NBA \cos(90^\circ) = 0 \ Wb$.
The change in flux is $\Delta \phi = |\phi_2 - \phi_1| = 3 \ Wb$.
The induced $EMF$ is $\varepsilon = \frac{\Delta \phi}{\Delta t} = \frac{3}{100 \times 10^{-3}} = 30 \ V$.
The induced current is $I = \frac{\varepsilon}{R} = \frac{30}{8} = 3.75 \ A$.

(A) Given: Area $A = 3 \times 10^{-2} \, m^2$, Number of turns $N = 900$, Resistance $R = 1.8 \, \Omega$, Magnetic field $B = 3.5 \times 10^{-5} \, T$, Time $t = 0.5 \, s$.
Initial magnetic flux $\phi_i = N B A \cos(0^{\circ}) = N B A$.
Final magnetic flux $\phi_f = N B A \cos(180^{\circ}) = -N B A$.
Change in flux $\Delta \phi = \phi_f - \phi_i = -2 N B A$.
Induced $EMF$ $\varepsilon = -\frac{\Delta \phi}{\Delta t} = \frac{2 N B A}{t}$.
Induced current $I = \frac{\varepsilon}{R} = \frac{2 N B A}{R t}$.
Substituting the values: $I = \frac{2 \times 900 \times 3.5 \times 10^{-5} \times 3 \times 10^{-2}}{1.8 \times 0.5}$.
$I = \frac{1800 \times 10.5 \times 10^{-7}}{0.9} = 2000 \times 10.5 \times 10^{-7} = 21000 \times 10^{-7} = 2.1 \times 10^{-3} \, A = 2.1 \, mA$.

(D) The magnetic flux $\phi$ linked with a coil is given by the formula $\phi = N \vec{B} \cdot \vec{A} = N B A \cos \theta$,where $\theta$ is the angle between the magnetic field vector $\vec{B}$ and the area vector $\vec{A}$.
The area vector $\vec{A}$ is always perpendicular to the plane of the coil.
Given that the plane of the coil is parallel to the magnetic field $\vec{B}$,the angle between the area vector $\vec{A}$ and the magnetic field $\vec{B}$ is $90^\circ$.
Therefore,$\theta = 90^\circ$.
Substituting this into the formula: $\phi = N B A \cos(90^\circ) = N B A (0) = 0$.
Thus,the magnetic flux linked with the coil is zero.

(A) According to Faraday's law of electromagnetic induction, the induced electromotive force $(EMF)$ is given by $\varepsilon = -\frac{d\phi}{dt}$.
Given $\phi = (5t^2 + 4t + 2) \times 10^{-3} \text{ Wb}$.
Differentiating with respect to $t$:
$\frac{d\phi}{dt} = (10t + 4) \times 10^{-3} \text{ Wb/s}$.
At $t = 6 \text{ s}$, the magnitude of induced $EMF$ is:
$|\varepsilon| = |10(6) + 4| \times 10^{-3} = 64 \times 10^{-3} \text{ V}$.
The induced current $I$ is given by $I = \frac{|\varepsilon|}{R}$.
Given $R = 16 \Omega$, we have:
$I = \frac{64 \times 10^{-3}}{16} = 4 \times 10^{-3} \text{ A} = 4 \text{ mA}$.
Thus, the correct option is $A$.

(A) The induced emf $(e)$ in a coil due to self-inductance $(L)$ is given by the formula: $e = -L \frac{di}{dt}$.
Given values are:
Induced emf,$e = 2.8 \ mV = 2.8 \times 10^{-3} \ V$.
Change in current,$di = 8 \ A - 3 \ A = 5 \ A$.
Time interval,$dt = 0.2 \ s$.
Substituting these values into the formula (ignoring the negative sign as we are calculating the magnitude of inductance):
$2.8 \times 10^{-3} = L \times \frac{5}{0.2}$.
$2.8 \times 10^{-3} = L \times 25$.
$L = \frac{2.8 \times 10^{-3}}{25}$.
$L = 0.112 \times 10^{-3} \ H$.
$L = 112 \times 10^{-6} \ H = 112 \ \mu H$.
Therefore,the self-inductance of the loop is $112 \ \mu H$.

(D) The speed of an electromagnetic wave in a medium is given by $v = \frac{1}{\sqrt{\mu \epsilon}} = \frac{1}{\sqrt{\mu_0 \mu_r \epsilon_0 \epsilon_r}}$.
Since $c = \frac{1}{\sqrt{\mu_0 \epsilon_0}} = 3 \times 10^8 \text{ m/s}$,we can write $v = \frac{c}{\sqrt{\mu_r \epsilon_r}}$.
Given $\epsilon_r = 8$ and $\mu_r = 200$,the refractive index $n = \sqrt{\mu_r \epsilon_r} = \sqrt{200 \times 8} = \sqrt{1600} = 40$.
Thus,the speed $v = \frac{3 \times 10^8}{40} = 0.075 \times 10^8 = 7.5 \times 10^6 \text{ m/s}$.
The wavelength $\lambda$ is given by $\lambda = \frac{v}{f}$.
Given $f = 100 \text{ MHz} = 10^8 \text{ Hz}$.
Therefore,$\lambda = \frac{7.5 \times 10^6}{10^8} = 7.5 \times 10^{-2} \text{ m} = 7.5 \text{ cm}$.

(D) The force $F$ exerted by electromagnetic waves on a non-reflecting (perfectly absorbing) surface is given by the formula $F = \frac{P}{c}$,where $P$ is the power of the waves and $c$ is the speed of light in vacuum.
Given,power $P = 600 \ W$.
The speed of light $c = 3 \times 10^8 \ m/s$.
Substituting the values,we get $F = \frac{600}{3 \times 10^8} \ N$.
$F = 200 \times 10^{-8} \ N$.
$F = 2 \times 10^{-6} \ N$.
Therefore,the correct option is $D$.

(A) The displacement current $I_d$ is given by the formula $I_d = \epsilon_0 \frac{d\Phi_E}{dt}$,where $\epsilon_0$ is the permittivity of free space and $\frac{d\Phi_E}{dt}$ is the rate of change of electric flux.
Given $\epsilon_0 = 8.854 \times 10^{-12} \text{ F/m}$ and $\frac{d\Phi_E}{dt} = 9 \pi \times 10^3 \text{ Vm s}^{-1}$.
Substituting the values: $I_d = (8.854 \times 10^{-12}) \times (9 \times 3.14159 \times 10^3)$.
$I_d \approx 8.854 \times 10^{-12} \times 28.27 \times 10^3 \approx 250.3 \times 10^{-9} \text{ A} = 0.25 \mu \text{A}$.

(C) The intensity $I$ of an electromagnetic wave is related to the amplitude of the electric field $E_0$ by the formula: $I = \frac{1}{2} c \epsilon_0 E_0^2$.
Given: $I = \frac{15}{\pi} \text{ W m}^{-2}$, $c = 3 \times 10^8 \text{ m s}^{-1}$, and $\epsilon_0 = \frac{1}{36\pi} \times 10^{-9} \text{ F m}^{-1}$.
Substituting the values: $\frac{15}{\pi} = \frac{1}{2} \times (3 \times 10^8) \times (\frac{1}{36\pi} \times 10^{-9}) \times E_0^2$.
$\frac{15}{\pi} = \frac{3 \times 10^{-1}}{72\pi} \times E_0^2$.
$\frac{15}{\pi} = \frac{1}{240\pi} \times E_0^2$.
$E_0^2 = 15 \times 240 = 3600$.
$E_0 = \sqrt{3600} = 60 \text{ N C}^{-1}$.

(D) The displacement current $I$ in a parallel plate capacitor is given by the formula $I = \varepsilon_0 \frac{d\phi_E}{dt}$,where $\phi_E$ is the electric flux.
Since the electric flux $\phi_E$ through the area $A$ of the plates is given by $\phi_E = E_{field} \cdot A$,where $E_{field}$ is the electric field between the plates.
Substituting this into the displacement current formula,we get $I = \varepsilon_0 \frac{d}{dt}(E_{field} \cdot A)$.
Since the area $A$ is constant,we have $I = \varepsilon_0 A \frac{dE_{field}}{dt}$.
Given that the rate of change of the electric field is $E$,we substitute $\frac{dE_{field}}{dt} = E$.
Thus,$I = \varepsilon_0 A E$.
Solving for the area $A$,we get $A = \frac{I}{\varepsilon_0 E}$.