JEE Main 2024 Physics Question Paper with Answer and Solution

(B) Given: Main Scale Reading $(MSR)$ = $1 \,mm$, Circular Scale Reading $(CSR)$ = $42$ divisions, Pitch = $1 \,mm$, Number of circular scale divisions $(n)$ = $100$.
First, calculate the Least Count $(LC)$ of the screw gauge:
$LC = \frac{\text{Pitch}}{n} = \frac{1 \,mm}{100} = 0.01 \,mm$.
The total diameter is given by the formula:
$\text{Diameter} = MSR + (LC \times CSR)$.
Substituting the values:
$\text{Diameter} = 1 \,mm + (0.01 \,mm \times 42) = 1 + 0.42 = 1.42 \,mm$.
According to the problem, the diameter is $\frac{x}{50} \,mm$:
$1.42 = \frac{x}{50}$.
Solving for $x$:
$x = 1.42 \times 50 = 71$.

(D) From the first law of thermodynamics,$dQ = du + dW$.
For a process,$C dT = C_V dT + P dV$,where $C$ is the molar heat capacity.
Dividing by $dT$,we get $C = C_V + P \frac{dV}{dT}$.
Given the equation of state $PV^2 = RT$.
Differentiating with respect to $T$ at constant pressure $P$:
$P(2V) \frac{dV}{dT} = R$.
Therefore,$P \frac{dV}{dT} = \frac{R}{2}$.
Substituting this into the expression for $C$:
$C = C_V + \frac{R}{2}$.

(B) $1$. Torque $(\tau) = r \times F$. Dimensional formula: $[M^1 L^2 T^{-2}]$. Thus,$A-IV$.
$2$. Magnetic field $(B) = F / (qv)$. Dimensional formula: $[M^1 T^{-2} A^{-1}]$. Thus,$B-III$.
$3$. Magnetic moment $(M) = I \times A$. Dimensional formula: $[L^2 A^1]$. Thus,$C-II$.
$4$. Permeability of free space $(\mu_0) = [B \cdot r^2 / (I \cdot l)]$. Dimensional formula: $[M^1 L^1 T^{-2} A^{-2}]$. Thus,$D-I$.
Therefore,the correct matching is $A-IV, B-III, C-II, D-I$.

(B) The root mean square speed $(V_{rms})$ of a gas is given by the formula: $V_{rms} = \sqrt{\frac{3RT}{M_w}}$.
Since the gases are in the same sample,they are at the same temperature $(T)$.
Therefore,$V_{rms} \propto \frac{1}{\sqrt{M_w}}$.
The ratio of the root mean square speed of helium $(He)$ to oxygen $(O_2)$ is:
$\frac{V_{He}}{V_{O_2}} = \sqrt{\frac{M_{w, O_2}}{M_{w, He}}}$.
Given the molar mass of helium $(M_{w, He} = 4 \ g/mol)$ and oxygen $(M_{w, O_2} = 32 \ g/mol)$:
$\frac{V_{He}}{V_{O_2}} = \sqrt{\frac{32}{4}} = \sqrt{8} = 2\sqrt{2}$.
Thus,the ratio is $\frac{2\sqrt{2}}{1}$.

(A) For an Atwood machine with masses $m_1$ and $m_2$ $(m_2 > m_1)$,the acceleration $a$ is given by:
$a = \left( \frac{m_2 - m_1}{m_1 + m_2} \right) g$
Given $a = \frac{g}{\sqrt{2}}$,we substitute this into the equation:
$\frac{g}{\sqrt{2}} = \left( \frac{m_2 - m_1}{m_1 + m_2} \right) g$
$\frac{1}{\sqrt{2}} = \frac{m_2 - m_1}{m_1 + m_2}$
$m_1 + m_2 = \sqrt{2} m_2 - \sqrt{2} m_1$
$m_1 + \sqrt{2} m_1 = \sqrt{2} m_2 - m_2$
$m_1(1 + \sqrt{2}) = m_2(\sqrt{2} - 1)$
$\frac{m_1}{m_2} = \frac{\sqrt{2} - 1}{\sqrt{2} + 1}$

(C) The kinetic energy $KE$ of a particle is related to its momentum $p$ and mass $m$ by the formula: $KE = \frac{p^2}{2m}$.
Since all four particles have the same momentum $p$,the kinetic energy is inversely proportional to the mass $(KE \propto \frac{1}{m})$.
Therefore,the particle with the smallest mass will have the maximum kinetic energy.
Comparing the masses: $\frac{m}{2} < m < 2m < 4m$.
The particle $A$ has the smallest mass,$\frac{m}{2}$.
Thus,particle $A$ has the maximum kinetic energy.

(B) The train starts from rest,so initial velocity $u = 0$. It accelerates uniformly to $v = 80 \ km/h$ in time $t$.
Distance covered during acceleration $(d_1)$ = $\text{Average velocity} \times \text{time} = \frac{0 + 80}{2} \times t = 40t \ km$.
Then,it moves with a constant speed of $80 \ km/h$ for time $3t$.
Distance covered during constant speed $(d_2)$ = $80 \times 3t = 240t \ km$.
Total distance = $d_1 + d_2 = 40t + 240t = 280t \ km$.
Total time = $t + 3t = 4t$.
Average speed = $\frac{\text{Total distance}}{\text{Total time}} = \frac{280t}{4t} = 70 \ km/h$.

(D) When the ball falls with a constant velocity (terminal velocity),the net force acting on it is zero.
The forces acting on the ball are:
$1$. Weight of the ball $(W = mg)$ acting downwards.
$2$. Buoyant force $(F_B)$ acting upwards.
$3$. Viscous force $(F_v)$ acting upwards.
According to the condition of constant velocity,the acceleration $a = 0$.
Therefore,$mg - F_B - F_v = 0$,which implies $F_v = mg - F_B$.
The buoyant force $F_B$ is equal to the weight of the displaced liquid: $F_B = V \rho_0 g$,where $V$ is the volume of the ball.
Since the volume $V = \frac{m}{\rho}$,we have $F_B = \frac{m}{\rho} \rho_0 g$.
Substituting this into the equation for $F_v$:
$F_v = mg - \frac{m}{\rho} \rho_0 g$
$F_v = mg \left(1 - \frac{\rho_0}{\rho}\right)$.

(A) By the law of conservation of angular momentum,the angular momentum $L = I\omega$ remains constant.
Since $I = \frac{2}{5}MR^2$ and $\omega = \frac{2\pi}{T}$,we have $I_1\omega_1 = I_2\omega_2$.
Substituting the values:
$\left(\frac{2}{5}MR^2\right) \frac{2\pi}{T_1} = \left(\frac{2}{5}M(\frac{3}{4}R)^2\right) \frac{2\pi}{T_2}$
$R^2 \cdot \frac{1}{T_1} = (\frac{3}{4}R)^2 \cdot \frac{1}{T_2}$
$\frac{1}{T_1} = \frac{9}{16} \cdot \frac{1}{T_2}$
$T_2 = \frac{9}{16} \cdot T_1$
Given $T_1 = 24$ hours:
$T_2 = \frac{9}{16} \times 24 = \frac{9 \times 3}{2} = \frac{27}{2} = 13.5$ hours.
Thus,the duration of the day will be $13$ hours $30$ minutes.

(C) Let $r$ be the radius of each small droplet and $R$ be the radius of the big drop.
Since the volume is conserved,the volume of $1000$ small droplets equals the volume of the big drop:
$1000 \times \frac{4}{3} \pi r^3 = \frac{4}{3} \pi R^3$
$1000 r^3 = R^3$
$R = 10r$
The surface energy of a drop is given by $E = T \times A$,where $T$ is surface tension and $A$ is the surface area.
Surface energy of $1000$ small droplets $(E_s)$ = $1000 \times (4 \pi r^2 T) = 4000 \pi r^2 T$
Surface energy of the big drop $(E_b)$ = $4 \pi R^2 T = 4 \pi (10r)^2 T = 400 \pi r^2 T$
The ratio of surface energy of $1000$ droplets to that of the big drop is:
$\frac{E_s}{E_b} = \frac{4000 \pi r^2 T}{400 \pi r^2 T} = 10$
Given that the ratio is $\frac{10}{x}$,we have:
$10 = \frac{10}{x}$
Therefore,$x = 1$.

(A) The maximum velocity $(V_{\max})$ of a particle in simple harmonic motion is given by the formula: $V_{\max} = \omega A$.
Here, $\omega$ is the angular frequency and $A$ is the amplitude.
We know that $\omega = \frac{2\pi}{T}$, where $T$ is the time period.
Given: $A = 0.06 \,m$ and $T = 3.14 \,s \approx \pi \,s$.
Substituting the values:
$V_{\max} = \left(\frac{2\pi}{\pi}\right) \times 0.06 = 2 \times 0.06 = 0.12 \,m/s$.
To convert the velocity into $cm/s$, we multiply by $100$:
$V_{\max} = 0.12 \times 100 = 12 \,cm/s$.

(B) The scalar triple product $\vec{A} \cdot (\vec{B} \times \vec{C}) = 0$ implies that the three vectors are coplanar.
This can be calculated using the determinant of the components:
$\begin{vmatrix} -x & -6 & -2 \\ -1 & 4 & 3 \\ -8 & -1 & 3 \end{vmatrix} = 0$
Expanding the determinant along the first row:
$-x(4(3) - 3(-1)) - (-6)((-1)(3) - 3(-8)) + (-2)((-1)(-1) - 4(-8)) = 0$
$-x(12 + 3) + 6(-3 + 24) - 2(1 + 32) = 0$
$-15x + 6(21) - 2(33) = 0$
$-15x + 126 - 66 = 0$
$-15x + 60 = 0$
$15x = 60$
$x = 4$

(D) According to the $1^{\text{st}}$ law of thermodynamics:
$\Delta Q = \Delta U + W$
Here,$\Delta Q = 48 \,J$,$n = 1 \,mol$,and $\Delta T = 2 \,K$ (since a change of $2^{\circ}C$ is equal to a change of $2 \,K$).
For a monatomic gas like helium,the molar heat capacity at constant volume is $C_v = \frac{3}{2}R$.
The change in internal energy is $\Delta U = n C_v \Delta T = (1) \left(\frac{3}{2} R\right) (2) = 3R$.
Substituting the values into the first law equation:
$48 = 3R + W$
$W = 48 - 3(8.3)$
$W = 48 - 24.9$
$W = 23.1 \,J$.

(D) At the surface of the earth, the weight of the body is given by $W_s = mg_s = 300 \,N$.
The acceleration due to gravity at a depth $d$ below the surface of the earth is given by the formula $g_d = g_s \left(1 - \frac{d}{R}\right)$, where $g_s$ is the acceleration due to gravity at the surface and $R$ is the radius of the earth.
Given the depth $d = R/4$, we substitute this into the formula:
$g_d = g_s \left(1 - \frac{R/4}{R}\right)$
$g_d = g_s \left(1 - \frac{1}{4}\right)$
$g_d = g_s \left(\frac{3}{4}\right)$
The weight of the body at depth $d$ is $W_d = mg_d$.
Substituting $g_d$ into the weight equation:
$W_d = m \times \left(\frac{3}{4} g_s\right)$
$W_d = \frac{3}{4} \times (mg_s)$
Since $mg_s = 300 \,N$, we have:
$W_d = \frac{3}{4} \times 300 \,N = 225 \,N$.

(B) Given: Mass $m = 800 \,kg$, Radius $r = 300 \,m$, Angle of banking $\theta = 30^{\circ}$, Coefficient of static friction $\mu = 0.2$, Acceleration due to gravity $g = 10 \,m/s^2$.
The formula for the maximum safe speed on a banked road with friction is given by:
$V_{\max} = \sqrt{rg \left[ \frac{\tan \theta + \mu}{1 - \mu \tan \theta} \right]}$
Substitute the values:
$V_{\max} = \sqrt{300 \times 10 \times \left[ \frac{\tan 30^{\circ} + 0.2}{1 - 0.2 \times \tan 30^{\circ}} \right]}$
Since $\tan 30^{\circ} = \frac{1}{\sqrt{3}} \approx \frac{1}{1.73} \approx 0.577$
$V_{\max} = \sqrt{3000 \times \left[ \frac{0.577 + 0.2}{1 - 0.2 \times 0.577} \right]}$
$V_{\max} = \sqrt{3000 \times \left[ \frac{0.777}{0.8846} \right]}$
$V_{\max} = \sqrt{3000 \times 0.8783} \approx \sqrt{2635} \approx 51.33 \,m/s$
Rounding to the nearest value, $V_{\max} \approx 51.4 \,m/s$.

(D) For a non-rigid diatomic molecule, the degrees of freedom $(f)$ include $3$ translational, $2$ rotational, and $2$ vibrational modes.
Thus, $f = 3 + 2 + 2 = 7$.
The energy of one molecule is given by $U = \frac{f}{2} K_B T$.
Substituting $f = 7$, the energy of one molecule is $U = \frac{7}{2} K_B T$.
For $10$ such molecules, the total energy is $E = 10 \times \frac{7}{2} K_B T = 35 K_B T$.

(B) The system consisting of the body and the chain is in equilibrium.
The total downward force acting on the branch is the sum of the weight of the body and the weight of the chain.
Weight of the body,$W_b = 200 \,N$.
Weight of the chain,$W_c = m \times g = 10 \,kg \times 10 \,m/s^2 = 100 \,N$.
Since the system is in equilibrium,the tension $T$ in the chain at the point where it is attached to the branch must balance the total weight.
$T = W_b + W_c = 200 \,N + 100 \,N = 300 \,N$.
Therefore,the branch pulls the chain with a force of $300 \,N$.

(A) The relationship between kinetic energy $K$ and momentum $P$ is given by $K = \frac{P^2}{2m}$,where $m$ is the mass of the body.
From this,we can write $P = \sqrt{2mK}$.
Let the initial kinetic energy be $K_i$ and the final kinetic energy be $K_f = 36 K_i$.
The initial momentum is $P_i = \sqrt{2mK_i}$ and the final momentum is $P_f = \sqrt{2mK_f} = \sqrt{2m(36K_i)} = 6\sqrt{2mK_i} = 6P_i$.
The percentage increase in momentum is given by $\frac{P_f - P_i}{P_i} \times 100 \%$.
Substituting the values,we get $\frac{6P_i - P_i}{P_i} \times 100 \% = \frac{5P_i}{P_i} \times 100 \% = 500 \%$.

(A) soap bubble has two liquid-air surfaces: one on the inside and one on the outside.
For a single spherical surface,the excess pressure is given by $\Delta P = \frac{2 S}{R}$.
Since a soap bubble has two surfaces,the total excess pressure is $\Delta P = 2 \times \left( \frac{2 S}{R} \right)$.
Therefore,the pressure inside the soap bubble is greater than the pressure outside by $\Delta P = \frac{4 S}{R}$.

(C) The zero error is given as $0.04 \text{ mm} = 0.004 \text{ cm}$. Since the zero of the vernier scale shifts to the left,the zero error is negative.
The formula for zero error is: $\text{Zero Error} = -(\text{n} \times \text{Least Count})$,where $n$ is the coinciding division.
Given $50 \text{ VSD} = 49 \text{ MSD}$,the Least Count $(LC)$ is $1 \text{ MSD} - 1 \text{ VSD} = 1 \text{ MSD} - \frac{49}{50} \text{ MSD} = \frac{1}{50} \text{ MSD}$.
Let $1 \text{ MSD} = x \text{ cm}$. Then $LC = \frac{x}{50} \text{ cm}$.
The $4^{\text{th}}$ division coincides,so the zero error is $4 \times LC = 4 \times \frac{x}{50} = 0.004 \text{ cm}$.
Solving for $x$: $\frac{4x}{50} = 0.004 \implies 4x = 0.2 \implies x = 0.05 \text{ cm}$.
The number of main scale divisions in $1 \text{ cm}$ is $N = \frac{1 \text{ cm}}{x} = \frac{1}{0.05} = 20$.

(C) The formula for heat energy is $\Delta Q = mS \Delta T$,where $S$ is the specific heat capacity.
Therefore,$S = \frac{\Delta Q}{m \Delta T}$.
The dimensions are $[S] = \frac{[ML^2 T^{-2}]}{[M][K]} = [L^2 T^{-2} K^{-1}]$.
Thus,Statement $(I)$ is correct.
The ideal gas equation is $PV = nRT$,where $R$ is the universal gas constant.
Therefore,$R = \frac{PV}{nT}$.
The dimensions are $[R] = \frac{[ML^{-1} T^{-2}][L^3]}{[mol][K]} = [ML^2 T^{-2} mol^{-1} K^{-1}]$.
Comparing this with the given statement,Statement $(II)$ is incorrect.

(A) Let the height of the tower be $h$ and the initial speed be $u$. Taking the downward direction as positive,the equation of motion is $h = ut + \frac{1}{2}gt^2$.
For upward projection,the initial velocity is $-u$: $h = -ut_1 + \frac{1}{2}gt_1^2 \Rightarrow \frac{1}{2}gt_1^2 - ut_1 - h = 0$. Solving for $t_1$ (taking positive root): $t_1 = \frac{u + \sqrt{u^2 + 2gh}}{g}$.
For downward projection,the initial velocity is $+u$: $h = ut_2 + \frac{1}{2}gt_2^2 \Rightarrow \frac{1}{2}gt_2^2 + ut_2 - h = 0$. Solving for $t_2$ (taking positive root): $t_2 = \frac{-u + \sqrt{u^2 + 2gh}}{g}$.
If the body is dropped,$u = 0$,so $h = \frac{1}{2}gt^2 \Rightarrow t = \sqrt{\frac{2h}{g}}$.
Multiplying $t_1$ and $t_2$: $t_1 t_2 = \left(\frac{\sqrt{u^2 + 2gh} + u}{g}\right) \left(\frac{\sqrt{u^2 + 2gh} - u}{g}\right) = \frac{(u^2 + 2gh) - u^2}{g^2} = \frac{2gh}{g^2} = \frac{2h}{g} = t^2$.
Therefore,$t = \sqrt{t_1 t_2}$.

(C) Given the displacement equation: $x^2 = 1 + t^2$.
Differentiating both sides with respect to $t$: $2x \frac{dx}{dt} = 2t$,which simplifies to $x v = t$,where $v$ is the velocity.
Differentiating $x v = t$ with respect to $t$ using the product rule: $x \frac{dv}{dt} + v \frac{dx}{dt} = 1$.
Since $\frac{dv}{dt} = a$ (acceleration) and $\frac{dx}{dt} = v$,we get $x a + v^2 = 1$.
Substituting $v = \frac{t}{x}$ into the equation: $x a + (\frac{t}{x})^2 = 1$.
$x a = 1 - \frac{t^2}{x^2} = \frac{x^2 - t^2}{x^2}$.
Since $x^2 - t^2 = 1$ from the original equation,we have $x a = \frac{1}{x^2}$.
Therefore,$a = \frac{1}{x^3} = x^{-3}$.
Comparing this with $x^{-n}$,we find $n = 3$.

(D) The distance $r$ from the centroid to the midpoint of each side of an equilateral triangle of side length $a = 2 \,m$ is given by $r = \frac{a}{2\sqrt{3}} = \frac{2}{2\sqrt{3}} = \frac{1}{\sqrt{3}} \,m$.
The moment of inertia $I$ of the system about an axis passing through the centroid and perpendicular to the plane is the sum of the moments of inertia of individual masses: $I = \sum m_i r_i^2$.
Since all masses are at the same distance $r$ from the centroid,$I = (m_1 + m_2 + m_3) r^2$.
Substituting the values: $I = (2 + 4 + 6) \times (\frac{1}{\sqrt{3}})^2$.
$I = 12 \times \frac{1}{3} = 4 \,kg \,m^2$.

(C) For the equilibrium of the mass in the vertical direction,the vertical components of the tension $T$ in the wire must balance the weight of the mass:
$2T \sin \theta = mg$
Given $m = 2 \text{ kg}$,$g = 10 \text{ m/s}^2$,and $\theta = \frac{1}{100} \text{ rad}$. Using the small angle approximation $\sin \theta \approx \theta$:
$2T \theta = 20$
$T = \frac{10}{\theta} = \frac{10}{1/100} = 1000 \text{ N}$
The original length of the wire is $2L = 2 \text{ m}$,so $L = 1 \text{ m}$.
The new length of the wire is $2 \sqrt{L^2 + x^2}$,where $x = L \tan \theta \approx L \theta$.
The change in length $\Delta L$ is:
$\Delta L = 2 \sqrt{L^2 + x^2} - 2L = 2L \left( \sqrt{1 + \tan^2 \theta} - 1 \right) \approx 2L \left( 1 + \frac{\tan^2 \theta}{2} - 1 \right) = L \tan^2 \theta \approx L \theta^2$
$\Delta L = 1 \times (1/100)^2 = 10^{-4} \text{ m}$
Using Young's modulus $Y = \frac{T/A}{\Delta L / (2L)}$:
$2 \times 10^{11} = \frac{1000 / A}{10^{-4} / 2}$
$2 \times 10^{11} = \frac{2000}{A \times 10^{-4}}$
$A = \frac{2000}{2 \times 10^{11} \times 10^{-4}} = \frac{1000}{10^7} = 10^{-4} \text{ m}^2$
Thus,the value of $A$ is $1 \times 10^{-4} \text{ m}^2$.

(A) The frequency of an open organ pipe is given by $f_n = \frac{n v}{2 L}$, where $n$ is the harmonic number, $v$ is the velocity of sound, and $L$ is the length of the pipe.
For the first pipe: $L_1 = 0.6 \,m$, $n_1 = 6$.
$f_1 = \frac{6 \times 333}{2 \times 0.6} = \frac{1998}{1.2} = 1665 \,Hz$.
For the second pipe: $L_2 = 0.9 \,m$, $n_2 = 5$.
$f_2 = \frac{5 \times 333}{2 \times 0.9} = \frac{1665}{1.8} = 925 \,Hz$.
The difference in frequencies is $\Delta f = |f_1 - f_2| = |1665 - 925| = 740 \,Hz$.

(A) The kinetic energy $KE$ of a body is related to its linear momentum $P$ and mass $m$ by the formula $KE = \frac{P^2}{2m}$.
Since the kinetic energies are equal, we have $P^2 = 2m(KE)$, which implies $P = \sqrt{2m(KE)}$.
Since $2$ and $KE$ are constants, the linear momentum is directly proportional to the square root of the mass: $P \propto \sqrt{m}$.
Given masses are $m_A = 400 \,g = 0.4 \,kg$, $m_B = 1.2 \,kg$, and $m_C = 1.6 \,kg$.
The ratio of their momenta is $P_A : P_B : P_C = \sqrt{m_A} : \sqrt{m_B} : \sqrt{m_C}$.
Substituting the values: $P_A : P_B : P_C = \sqrt{0.4} : \sqrt{1.2} : \sqrt{1.6}$.
Multiplying by $\sqrt{10}$ to simplify: $\sqrt{4} : \sqrt{12} : \sqrt{16} = 2 : 2\sqrt{3} : 4$.
Dividing by $2$, we get $1 : \sqrt{3} : 2$.

(C) For a monoatomic gas,the molar specific heat at constant volume is $(C_v)_{\text{mono}} = \frac{3}{2}R$.
For a rigid diatomic gas,the molar specific heat at constant volume is $(C_v)_{\text{dia}} = \frac{5}{2}R$.
The ratio of the specific heat of these gases at constant volume is given by:
$\frac{(C_v)_{\text{mono}}}{(C_v)_{\text{dia}}} = \frac{\frac{3}{2}R}{\frac{5}{2}R} = \frac{3}{5}$.

(A) The order of magnitude of a number expressed as $a \times 10^b$ is determined by the value of $a$.
If $1 \leq a \leq 5$,the order of magnitude is $b$.
If $5 < a < 10$,the order of magnitude is $b + 1$.
Therefore,$b$ is the order of magnitude when $a \leq 5$.

(C) The length of the second hand is $r_s = 75 \ cm = 0.75 \ m$. The length of the minute hand is $r_m = 60 \ cm = 0.60 \ m$.
In $30$ minutes,the second hand completes $30$ full revolutions. The distance traveled by the tip of the second hand is $d_s = 30 \times (2 \pi r_s) = 30 \times 2 \times 3.14 \times 0.75 = 141.3 \ m$.
In $30$ minutes,the minute hand completes $0.5$ revolutions (half a circle). The distance traveled by the tip of the minute hand is $d_m = 0.5 \times (2 \pi r_m) = \pi r_m = 3.14 \times 0.60 = 1.884 \ m$.
The difference in distance is $x = d_s - d_m = 141.3 - 1.884 = 139.416 \ m$.
Rounding to one decimal place,$x \approx 139.4 \ m$.

(C) The formula for Young's modulus is $Y = 49000 \frac{M}{\ell}$.
The relative error in $Y$ is given by $\frac{\Delta Y}{Y} = \frac{\Delta M}{M} + \frac{\Delta \ell}{\ell}$.
Given values are $M = 500 \text{ g}$,$\Delta M = 5 \text{ g}$,$\ell = 2 \text{ cm}$,and $\Delta \ell = 0.02 \text{ cm}$.
Substituting these values into the relative error formula:
$\frac{\Delta Y}{Y} = \frac{5}{500} + \frac{0.02}{2}$.
Calculating each term:
$\frac{\Delta Y}{Y} = 0.01 + 0.01 = 0.02$.
To find the percentage error:
$\% \text{ error} = \frac{\Delta Y}{Y} \times 100 = 0.02 \times 100 = 2 \%$.

(C) For an adiabatic process,the relation between temperature $T$ and volume $V$ is given by $TV^{\gamma-1} = \text{constant}$.
For the adiabatic path connecting points $a$ and $d$,we have:
$T_a V_a^{\gamma-1} = T_d V_d^{\gamma-1}$
$\left(\frac{V_a}{V_d}\right)^{\gamma-1} = \frac{T_d}{T_a}$
For the adiabatic path connecting points $b$ and $c$,we have:
$T_b V_b^{\gamma-1} = T_c V_c^{\gamma-1}$
$\left(\frac{V_b}{V_c}\right)^{\gamma-1} = \frac{T_c}{T_b}$
Since points $a$ and $b$ lie on the same isothermal curve,$T_a = T_b$. Similarly,points $d$ and $c$ lie on the same isothermal curve,so $T_d = T_c$.
Substituting these into the equations,we get:
$\frac{T_d}{T_a} = \frac{T_c}{T_b}$
Therefore,$\left(\frac{V_a}{V_d}\right)^{\gamma-1} = \left(\frac{V_b}{V_c}\right)^{\gamma-1}$
This implies $\frac{V_a}{V_d} = \frac{V_b}{V_c}$.

(C) For a planet in a circular orbit,the angular momentum $L$ is given by $L = mvr = m(r\omega)r = mr^2\left(\frac{2\pi}{T}\right)$.
Thus,$\frac{L}{m} = \frac{2\pi r^2}{T}$,which implies $\frac{T}{r^2} = \frac{2\pi m}{L}$.
For planet $A$: $\frac{T_A}{r_1^2} = \frac{2\pi m_1}{L}$.
For planet $B$: $\frac{T_B}{r_2^2} = \frac{2\pi m_2}{3L}$.
Dividing the two equations: $\frac{T_A}{T_B} \cdot \left(\frac{r_2}{r_1}\right)^2 = \frac{m_1}{m_2} \cdot 3 \implies \frac{T_A}{T_B} = 3 \left(\frac{m_1}{m_2}\right) \left(\frac{r_1}{r_2}\right)^2$.
From Kepler's Third Law,$T^2 \propto r^3$,so $\left(\frac{T_A}{T_B}\right)^2 = \left(\frac{r_1}{r_2}\right)^3$,which means $\left(\frac{r_1}{r_2}\right)^2 = \left(\frac{T_A}{T_B}\right)^{4/3}$.
Substituting this into the expression: $\frac{T_A}{T_B} = 3 \left(\frac{m_1}{m_2}\right) \left(\frac{T_A}{T_B}\right)^{4/3}$.
Rearranging: $\left(\frac{T_A}{T_B}\right)^{-1/3} = 3 \frac{m_1}{m_2} \implies \frac{T_A}{T_B} = \left(3 \frac{m_1}{m_2}\right)^{-3} = \frac{1}{27} \left(\frac{m_2}{m_1}\right)^3$.

(B) Bernoulli's principle states that for an incompressible,non-viscous,and streamline flow of a fluid,the sum of pressure energy,potential energy per unit volume,and kinetic energy per unit volume remains constant along a streamline.
Mathematically,this is expressed as: $P + \rho gh + \frac{1}{2}\rho v^2 = \text{constant}$.
Here,$P$ is the pressure,$\rho$ is the density of the fluid,$g$ is the acceleration due to gravity,$h$ is the height,and $v$ is the velocity of the fluid.

(C) Given: Mass $m = 150 \ g = 0.15 \ kg$,Initial velocity $u = 20 \ m/s$,Final velocity $v = 0 \ m/s$,Time interval $\Delta t = 0.1 \ s$.
According to Newton's second law of motion,the force $F$ exerted is equal to the rate of change of momentum:
$F = \frac{\Delta p}{\Delta t} = \frac{m(v - u)}{\Delta t}$
$F = \frac{0.15 \times (0 - 20)}{0.1}$
$F = \frac{0.15 \times (-20)}{0.1} = \frac{-3}{0.1} = -30 \ N$
The magnitude of the force is $|F| = 30 \ N$.

(B) Since the particle is initially stationary,the initial momentum is $0$.
By the law of conservation of linear momentum,the final momentum must also be $0$.
Therefore,the magnitudes of the momenta of the two parts must be equal: $|P_A| = |P_B|$,which implies $m_A v_A = m_B v_B$.
The kinetic energy of a particle is given by $K = \frac{P^2}{2m}$.
Thus,the ratio of kinetic energies is $\frac{K_B}{K_A} = \frac{P_B^2 / 2m_B}{P_A^2 / 2m_A}$.
Since $|P_A| = |P_B|$,this simplifies to $\frac{K_B}{K_A} = \frac{2m_A}{2m_B} = \frac{m_A}{m_B}$.

(D) Given: $9 \,MSD = 10 \,VSD$.
Mass $= 8.635 \,g$.
Least Count $(LC)$ $= 1 \,MSD - 1 \,VSD = 1 \,MSD - 0.9 \,MSD = 0.1 \,MSD$.
Since $1 \,MSD = 1 \,mm = 0.1 \,cm$, $LC = 0.1 \times 0.1 \,cm = 0.01 \,cm$.
Diameter $= MSR + (VSR \times LC) = 2.0 \,cm + (2 \times 0.01 \,cm) = 2.02 \,cm$.
Radius $r = \frac{d}{2} = 1.01 \,cm$.
Volume $V = \frac{4}{3} \pi r^3 = \frac{4}{3} \times 3.1416 \times (1.01)^3 \approx 4.318 \,cm^3$.
Density $\rho = \frac{\text{mass}}{\text{volume}} = \frac{8.635}{4.318} \approx 1.9997 \,g/cm^3 \approx 2.0 \,g/cm^3$.

(A) Let the surface mass density be $\sigma$. The total area of the plate is $A = (3 \times 2) - (1 \times 1) = 5 \text{ units}^2$. Given mass $M = 10 \ kg$,so $\sigma = \frac{10}{5} = 2 \ kg/\text{unit}^2$.
We can consider the plate as a large rectangle of $3 \times 2$ (mass $M_{total} = 3 \times 2 \times 2 = 12 \ kg$) minus a small square of $1 \times 1$ (mass $m_{cut} = 1 \times 1 \times 2 = 2 \ kg$).
The center of mass of the large rectangle is at $(1.5, 1.0)$.
The center of mass of the cut-out square is at $(1.5, 1.5)$.
Let the center of mass of the remaining plate be $(x, y)$. Using the principle of moments:
$M_{total} X_{CM} = M_{plate} x + m_{cut} x_{cut} \Rightarrow 12(1.5) = 10(x) + 2(1.5)$
$18 = 10x + 3 \Rightarrow 10x = 15 \Rightarrow x = 1.5$.
$M_{total} Y_{CM} = M_{plate} y + m_{cut} y_{cut} \Rightarrow 12(1.0) = 10(y) + 2(1.5)$
$12 = 10y + 3 \Rightarrow 10y = 9 \Rightarrow y = 0.9$.
The ratio $\frac{x}{y} = \frac{1.5}{0.9} = \frac{15}{9}$.
Comparing with $\frac{n}{9}$,we get $n = 15$.

(B) The excess pressure inside a soap bubble is given by $\Delta P = \frac{4S}{R}$, where $S$ is the surface tension and $R$ is the radius of the bubble.
This pressure is balanced by the hydrostatic pressure of the liquid column: $\Delta P = \rho g h$.
Equating the two: $\rho g h = \frac{4S}{R}$.
Given: $\rho = 8 \times 10^3 \,kg \,m^{-3}$, $g = 10 \,m \,s^{-2}$, $h = 0.04 \,cm = 4 \times 10^{-4} \,m$, and $S = 0.28 \,N \,m^{-1}$.
Substituting the values: $(8 \times 10^3) \times 10 \times (4 \times 10^{-4}) = \frac{4 \times 0.28}{R}$.
$32 = \frac{1.12}{R}$.
$R = \frac{1.12}{32} \,m = 0.035 \,m = 3.5 \,cm$.
The diameter $D = 2R = 2 \times 3.5 \,cm = 7 \,cm$.

(C) For a closed organ pipe,the frequency of the $n$-th overtone is given by $f_{c} = (2n + 1) \frac{v}{4\ell}$,where $n$ is the overtone number.
For the $7$-th overtone $(n=7)$,$f_{c} = (2 \times 7 + 1) \frac{v}{4\ell} = \frac{15v}{4\ell}$.
For an open organ pipe,the frequency of the $n$-th overtone is given by $f_{o} = (n + 1) \frac{v}{2\ell}$.
For the $7$-th overtone $(n=7)$,$f_{o} = (7 + 1) \frac{v}{2\ell} = \frac{8v}{2\ell} = \frac{4v}{\ell} = \frac{16v}{4\ell}$.
The ratio of the frequencies is $\frac{f_{c}}{f_{o}} = \frac{15v/4\ell}{16v/4\ell} = \frac{15}{16}$.
Given the ratio is $\left(\frac{a-1}{a}\right) = \frac{15}{16}$,comparing the terms,we get $a = 16$.

(D) Let the vector $\overrightarrow{OP}$ be along the $x$-axis,$\overrightarrow{OP} = A \hat{i}$.
Vector $\overrightarrow{OQ}$ is at $90^\circ$ to $\overrightarrow{OP}$,so $\overrightarrow{OQ} = A \hat{j}$.
Vector $\overrightarrow{OR}$ is at $45^\circ$ below the $x$-axis,so $\overrightarrow{OR} = A \cos(45^\circ) \hat{i} - A \sin(45^\circ) \hat{j} = \frac{A}{\sqrt{2}} \hat{i} - \frac{A}{\sqrt{2}} \hat{j}$.
The resultant vector $\vec{R} = \overrightarrow{OP} + \overrightarrow{OQ} + \overrightarrow{OR} = (A + \frac{A}{\sqrt{2}}) \hat{i} + (A - \frac{A}{\sqrt{2}}) \hat{j}$.
The magnitude of the resultant is $|\vec{R}| = \sqrt{(A + \frac{A}{\sqrt{2}})^2 + (A - \frac{A}{\sqrt{2}})^2}$.
$|\vec{R}| = \sqrt{A^2(1 + \frac{1}{\sqrt{2}})^2 + A^2(1 - \frac{1}{\sqrt{2}})^2} = A \sqrt{(1 + \frac{1}{2} + \sqrt{2}) + (1 + \frac{1}{2} - \sqrt{2})} = A \sqrt{1 + 0.5 + 1 + 0.5} = A \sqrt{3}$.
Comparing $A \sqrt{3}$ with $A \sqrt{x}$,we get $x = 3$.

(C) Let the block be released from height $h = L \sin(30^{\circ}) = 10 \times 0.5 = 5 \text{ m}$.
Using the Work-Energy Theorem: $W_{\text{gravity}} + W_{\text{friction}} + W_{\text{spring}} = \Delta KE = 0$.
Work done by gravity: $W_g = mgh = 5 \times 10 \times 5 = 250 \text{ J}$.
Work done by friction on the horizontal surface: $W_f = -\mu mg(d + x) = -0.5 \times 5 \times 10 \times (2 + x) = -25(2 + x) = -50 - 25x$.
Work done by the spring: $W_s = -\frac{1}{2} kx^2 = -\frac{1}{2} \times 100 \times x^2 = -50x^2$.
Equating the total work to zero: $250 - 50 - 25x - 50x^2 = 0$.
$200 - 25x - 50x^2 = 0$.
Dividing by $25$: $8 - x - 2x^2 = 0$,or $2x^2 + x - 8 = 0$.
Using the quadratic formula $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} = \frac{-1 \pm \sqrt{1 - 4(2)(-8)}}{4} = \frac{-1 \pm \sqrt{65}}{4}$.
Since $x > 0$,$x = \frac{-1 + 8.06}{4} \approx 1.76 \text{ m}$.
Re-evaluating the provided options,if we assume the friction acts only on the $2 \text{ m}$ part and the spring compression is $x$,the equation is $250 - (0.5 \times 5 \times 10 \times 2) - (0.5 \times 5 \times 10 \times x) - 50x^2 = 0$.
$250 - 50 - 25x - 50x^2 = 0 \Rightarrow 200 - 25x - 50x^2 = 0 \Rightarrow 2x^2 + x - 8 = 0$.
Given the options,if $x=1$,$2(1)^2 + 1 - 8 = -5 \neq 0$. If $x=2$,$2(4) + 2 - 8 = 2 \neq 0$. There is a discrepancy in the provided options vs the physics model. Assuming the intended answer based on standard problem types is $x=1$ or $x=2$ depending on friction parameters,but mathematically $x \approx 1.76$.

(B) The energy density $u$ of an electric field is given by the formula $u = \frac{1}{2} \varepsilon_0 E^2$.
Energy density is defined as energy per unit volume.
The dimensions of energy are $[M L^2 T^{-2}]$ and the dimensions of volume are $[L^3]$.
Therefore,the dimensions of energy density are $\frac{[M L^2 T^{-2}]}{[L^3]} = [M L^{-1} T^{-2}]$.
Since $\varepsilon_0 E^2$ is proportional to energy density,it shares the same dimensions.
Thus,the dimensions of $\varepsilon_0 E^2$ are $[M L^{-1} T^{-2}]$.

(B) The given equation of the plane progressive wave is $y = 2 \cos 2 \pi (330 t - x) \ m$.
The standard equation for a plane progressive wave is $y = A \cos (\omega t - kx)$.
By comparing the given equation with the standard form,we have:
$y = 2 \cos (2 \pi \times 330 t - 2 \pi x)$
Here,the angular frequency $\omega = 2 \pi \times 330 \ rad/s$.
We know that $\omega = 2 \pi f$,where $f$ is the frequency of the wave.
Equating the two expressions for $\omega$:
$2 \pi f = 2 \pi \times 330$
$f = 330 \ Hz$.
Therefore,the frequency of the wave is $330 \ Hz$.

(C) The initial moment of inertia of the first disc about the central axis is $I_1 = \frac{1}{2} MR^2$.
The initial angular momentum of the system is $L_i = I_1 \omega = \frac{1}{2} MR^2 \omega$.
When the second disc of mass $M/2$ and radius $R$ is placed co-axially,the new moment of inertia of the system becomes $I_2 = I_1 + I_{disc2} = \frac{1}{2} MR^2 + \frac{1}{2} (M/2) R^2 = \frac{1}{2} MR^2 + \frac{1}{4} MR^2 = \frac{3}{4} MR^2$.
According to the law of conservation of angular momentum,$L_i = L_f$,which implies $I_1 \omega = I_2 \omega_2$.
Substituting the values: $(\frac{1}{2} MR^2) \omega = (\frac{3}{4} MR^2) \omega_2$.
Solving for the new angular velocity $\omega_2$: $\omega_2 = \frac{1/2}{3/4} \omega = \frac{2}{3} \omega$.

(D) Let $V_w$ be the volume of ice immersed in water and $V_k$ be the volume of ice immersed in kerosene oil.
According to the principle of floatation, the weight of the ice cube is equal to the total buoyant force exerted by both liquids.
Weight of ice $= (V_w + V_k) \rho_{ice} g$
Buoyant force $= V_w \rho_w g + V_k \rho_k g$
Equating the two: $(V_w + V_k) \rho_{ice} g = V_w \rho_w g + V_k \rho_k g$
Dividing by $\rho_w g$ (where $\rho_w = 1 \text{ g/cm}^3$):
$(V_w + V_k) \times 0.9 = V_w \times 1 + V_k \times 0.8$
$0.9 V_w + 0.9 V_k = V_w + 0.8 V_k$
$0.9 V_k - 0.8 V_k = V_w - 0.9 V_w$
$0.1 V_k = 0.1 V_w$
Therefore, $V_w / V_k = 1 / 1$ or $1: 1$.

(D) The mean free path $\lambda$ of gas molecules is given by the formula $\lambda = \frac{k_B T}{\sqrt{2} \pi d^2 P}$,where $d$ is the molecular diameter. Since $\lambda \propto \frac{1}{d^2}$,Statement $(I)$ is true.
The average kinetic energy of gas molecules is given by $KE_{avg} = \frac{3}{2} k_B T$ (for monatomic gases). Since $KE_{avg} \propto T$,the average kinetic energy is directly proportional to the absolute temperature. Thus,Statement $(II)$ is true.
Therefore,both statements are true.

(C) The orbital speed of a satellite at a distance $r$ from the center of a planet of mass $M$ is given by $v = \sqrt{\frac{GM}{r}}$.
Given the radii of the orbits are $R_A = 4R$ and $R_B = R$.
The ratio of the speeds is $\frac{v_A}{v_B} = \sqrt{\frac{R_B}{R_A}} = \sqrt{\frac{R}{4R}} = \sqrt{\frac{1}{4}} = \frac{1}{2}$.
Given $v_A = 3v$,we have $\frac{3v}{v_B} = \frac{1}{2}$.
Therefore,$v_B = 2 \times 3v = 6v$.

(B) The formula for horizontal range $R$ is $R = \frac{u^2 \sin 2\theta}{g} = \frac{2u^2 \sin \theta \cos \theta}{g}$.
The formula for maximum height $H$ is $H = \frac{u^2 \sin^2 \theta}{2g}$.
Given that the range is equal to the maximum height,we set $R = H$:
$\frac{2u^2 \sin \theta \cos \theta}{g} = \frac{u^2 \sin^2 \theta}{2g}$.
Canceling common terms $u^2/g$ from both sides:
$2 \sin \theta \cos \theta = \frac{\sin^2 \theta}{2}$.
Dividing both sides by $\sin \theta$ (assuming $\sin \theta \neq 0$):
$2 \cos \theta = \frac{\sin \theta}{2}$.
Rearranging to solve for $\tan \theta = \frac{\sin \theta}{\cos \theta}$:
$4 = \frac{\sin \theta}{\cos \theta} = \tan \theta$.
Therefore,$\theta = \tan^{-1}(4)$.

(B) The least count $(LC)$ of the screw gauge is calculated as: $LC = \frac{\text{Pitch}}{\text{Total circular scale divisions}} = \frac{1 \,mm}{100} = 0.01 \,mm$.
The zero error is positive because the zero of the circular scale is below the reference line: $\text{Zero Error} = +5 \times LC = +5 \times 0.01 \,mm = +0.05 \,mm$.
The observed reading is: $\text{Observed Reading} = \text{Main Scale Reading} + (\text{Circular Scale Reading} \times LC) = 4 \,mm + (60 \times 0.01 \,mm) = 4.60 \,mm$.
The corrected diameter is: $\text{Diameter} = \text{Observed Reading} - \text{Zero Error} = 4.60 \,mm - 0.05 \,mm = 4.55 \,mm$.

(D) The displacement current $(I_d)$ in a capacitor is equal to the conduction current $(I_c)$ flowing through the circuit.
The capacitive reactance $(X_C)$ is given by the formula:
$X_C = \frac{1}{\omega C} = \frac{1}{2 \pi f C}$
Given:
$V_{max} = 40 \, V$
$f = 4 \, kHz = 4 \times 10^3 \, Hz$
$C = 12 \, \mu F = 12 \times 10^{-6} \, F$
Calculating $X_C$:
$X_C = \frac{1}{2 \times 3.1416 \times 4 \times 10^3 \times 12 \times 10^{-6}}$
$X_C = \frac{1}{8 \pi \times 12 \times 10^{-3}} = \frac{1}{0.3016} \approx 3.317 \, \Omega$
The maximum current $(I_{max})$ is:
$I_{max} = \frac{V_{max}}{X_C} = \frac{40}{3.317} \approx 12.06 \, A$
Thus, the maximum displacement current is nearly $12 \, A$.

(A) The fringe width $\beta$ is given by the formula $\beta = \frac{\lambda D}{d}$.
Given: $\lambda = 5000 \ \mathring{A} = 5 \times 10^{-7} \ \text{m}$,$D = 200 \ \text{cm} = 2 \ \text{m}$,and $d = 0.3 \ \text{mm} = 3 \times 10^{-4} \ \text{m}$.
Substituting these values,we get $\beta = \frac{5 \times 10^{-7} \times 2}{3 \times 10^{-4}} = \frac{10 \times 10^{-3}}{3} \ \text{m}$.
The position of the $n^{\text{th}}$ maxima is given by $y_n = n \beta$.
For the third maxima $(n=3)$,$y_3 = 3 \times \left( \frac{10 \times 10^{-3}}{3} \right) \ \text{m} = 10 \times 10^{-3} \ \text{m} = 10 \ \text{mm}$.

(B) The resistance of the wire is given by $R = \frac{\rho \ell}{A}$.
Given $R = 1 \, \Omega$, $\rho = 2 \times 10^{-6} \, \Omega m$, and $A = 10 \, mm^2 = 10 \times 10^{-6} \, m^2 = 10^{-5} \, m^2$.
Substituting these values: $1 = \frac{2 \times 10^{-6} \times \ell}{10^{-5}} \Rightarrow 1 = 0.2 \times \ell \Rightarrow \ell = 5 \, m$.
For the wire to be suspended in mid-air, the magnetic force must balance the gravitational force: $F_m = F_g$.
$Bi\ell = mg$.
Given $i = 2 \, A$, $m = 500 \, g = 0.5 \, kg$, $g = 10 \, m/s^2$, and $\ell = 5 \, m$.
$B \times 2 \times 5 = 0.5 \times 10$.
$10B = 5$.
$B = 0.5 \, T = 5 \times 10^{-1} \, T$.
Thus, the magnitude is $5$.

(C) The electric field $E$ between the plates is proportional to the potential difference $V$ across the capacitor $(E = V/d)$.
Given that the electric field drops to one-third of its initial value,the potential difference also drops to one-third: $V = V_0 / 3$.
The discharging equation for a capacitor is $V = V_0 e^{-t/\tau}$,where $\tau = RC$ is the time constant.
Substituting the given values: $V_0 / 3 = V_0 e^{-t/\tau} \Rightarrow 1/3 = e^{-t/\tau}$.
Taking the natural logarithm on both sides: $\ln(3) = t/\tau$.
Given $\ln(3) = 1.1$,$t = 6.6 \times 10^{-6} \ s$,and $C = 1.5 \times 10^{-6} \ F$.
$1.1 = (6.6 \times 10^{-6}) / (R \times 1.5 \times 10^{-6})$.
$1.1 = 6.6 / (1.5 \times R)$.
$R = 6.6 / (1.5 \times 1.1) = 6.6 / 1.65 = 4 \ \Omega$.

(C) The nuclear reaction is: $3 \, _2^4\text{He} \longrightarrow _6^{12}\text{C} + Q$.
The mass of three helium nuclei is $3 \times 4.002603 \text{ u} = 12.007809 \text{ u}$.
The mass of one carbon nucleus is $12.000000 \text{ u}$.
The mass defect $\Delta m$ is calculated as: $\Delta m = (3 \times m_{\text{He}}) - m_{\text{C}} = 12.007809 \text{ u} - 12.000000 \text{ u} = 0.007809 \text{ u}$.
The energy released $Q$ is given by: $Q = \Delta m \times 931 \text{ MeV/u}$.
$Q = 0.007809 \times 931 \text{ MeV} \approx 7.270179 \text{ MeV}$.
Expressing this in terms of $10^{-2} \text{ MeV}$,we get $727.0179 \times 10^{-2} \text{ MeV} \approx 727 \times 10^{-2} \text{ MeV}$.

(D) Given: $L = 1 \text{ H}$,$C = 20 \mu F = 20 \times 10^{-6} \text{ F}$,$R = 300 \Omega$,$V = 50 \sqrt{2} \sin 100 t$.
$1$. The angular frequency $\omega = 100 \text{ rad/s}$.
$2$. The $RMS$ voltage $V_{\text{rms}} = \frac{V_0}{\sqrt{2}} = \frac{50 \sqrt{2}}{\sqrt{2}} = 50 \text{ V}$.
$3$. Inductive reactance $X_L = \omega L = 100 \times 1 = 100 \Omega$.
$4$. Capacitive reactance $X_C = \frac{1}{\omega C} = \frac{1}{100 \times 20 \times 10^{-6}} = \frac{1}{2 \times 10^{-3}} = 500 \Omega$.
$5$. Impedance $Z = \sqrt{R^2 + (X_L - X_C)^2} = \sqrt{300^2 + (100 - 500)^2} = \sqrt{300^2 + (-400)^2} = \sqrt{90000 + 160000} = \sqrt{250000} = 500 \Omega$.
$6$. $RMS$ current $I_{\text{rms}} = \frac{V_{\text{rms}}}{Z} = \frac{50}{500} = 0.1 \text{ A}$.
$7$. $RMS$ potential difference across the capacitor $V_C = I_{\text{rms}} \times X_C = 0.1 \times 500 = 50 \text{ V}$.

(A) The current $I$ in the galvanometer is given by $I = K \theta$,where $K$ is the figure of merit.
From the circuit,the current is $I = \frac{E}{G+R}$,where $E = 2 \text{ V}$ is the emf of the source,$G$ is the galvanometer resistance,and $R$ is the external resistance.
Equating the two expressions: $K \theta = \frac{E}{G+R} \Rightarrow \frac{1}{\theta} = \frac{G+R}{E} = \frac{1}{E} R + \frac{G}{E}$.
Comparing this with the equation of a straight line $y = mx + c$,we have the slope $m = \frac{1}{E}$.
From the graph,the slope $m = \frac{y_2 - y_1}{x_2 - x_1} = \frac{2 - 1}{6 - 2} = \frac{1}{4} \text{ } \Omega^{-1}$.
Since $m = \frac{K}{E}$,we have $\frac{K}{2} = \frac{1}{4} \Rightarrow K = 0.5 \text{ A/division}$.
Expressing $K$ in terms of $10^{-1}$,we get $K = 5 \times 10^{-1} \text{ A/division}$.
Thus,the correct option is $A$.

(B) In parallel combination,the potential difference $V$ is the same across all capacitors. The equivalent capacitance is $C_p = C_1 + C_2 + C_3 = (25 + 30 + 45) \mu F = 100 \mu F$.
The energy stored is $E = \frac{1}{2} C_p V^2 = \frac{1}{2} \times 100 \times 10^{-6} \times (100)^2 = 0.5 \ J$.
In series combination,the equivalent capacitance $C_s$ is given by $\frac{1}{C_s} = \frac{1}{C_1} + \frac{1}{C_2} + \frac{1}{C_3} = \frac{1}{25} + \frac{1}{30} + \frac{1}{45} = \frac{18 + 15 + 10}{450} = \frac{43}{450} \mu F^{-1}$.
Thus,$C_s = \frac{450}{43} \mu F$.
The energy stored in series is $E' = \frac{1}{2} C_s V^2 = \frac{1}{2} \times \frac{450}{43} \times 10^{-6} \times (100)^2 = \frac{1}{2} \times \frac{450}{43} \times 10^{-2} = \frac{4.5}{86} \ J$.
Given $E' = \frac{9}{x} E$,we have $\frac{4.5}{86} = \frac{9}{x} \times 0.5$.
$\frac{4.5}{86} = \frac{4.5}{x} \implies x = 86$.

(A) According to Cauchy's equation,the refractive index $\mu$ of a medium depends on the wavelength $\lambda$ as $\mu(\lambda) = A + B/\lambda^2 + ...$.
Since $\lambda_{\text{red}} > \lambda_{\text{yellow}} > \lambda_{\text{violet}}$,the refractive index for red light is the smallest,and for violet light,it is the largest.
For a thin prism,the angle of deviation $\delta = (\mu - 1)A$.
Since $\mu_{\text{red}} < \mu_{\text{yellow}} < \mu_{\text{violet}}$,the deviation for red light is the least,and for violet light,it is the most.
Thus,Statement $I$ is true.
Statement $II$ is also true because the refractive index of a dispersive medium varies with wavelength,which is the fundamental cause of dispersion.

(C) According to Einstein's photoelectric equation,the maximum kinetic energy of emitted photoelectrons is given by $KE_{\max} = h\nu - \phi_0$,where $h$ is Planck's constant,$\nu$ is the frequency of incident light,and $\phi_0$ is the work function of the material.
Since the stopping potential $V_0$ is defined as the potential required to stop the most energetic electrons,we have $eV_0 = KE_{\max}$.
Therefore,$V_0 = \frac{KE_{\max}}{e} = \frac{h\nu - \phi_0}{e}$.
$1$. $V_0$ depends on the frequency $\nu$ of incident light (Option $B$ is true).
$2$. $V_0$ depends on the work function $\phi_0$,which depends on the nature of the emitter material (Option $A$ is true).
$3$. $V_0$ is $1/e$ times the maximum kinetic energy (Option $D$ is true).
$4$. $V_0$ does not depend on the intensity of the incident light,as intensity only affects the number of photoelectrons emitted,not their individual kinetic energy (Option $C$ is false).

(A) The centripetal force required for the electron to orbit the nucleus is provided by the electrostatic force of attraction: $F_{C} = F_{e}$.
Using the formula for centripetal force $F_{C} = \frac{mv^2}{r}$ and Coulomb's law $F_{e} = \frac{kZe^2}{r^2}$,we have: $\frac{mv^2}{r} = \frac{kZe^2}{r^2}$.
Multiplying both sides by $mr^2$,we get: $m^2v^2r^2 = mkZe^2r$.
Since angular momentum $L = mvr$,we can write: $L^2 = mkZe^2r$.
Taking the square root of both sides,we find: $L = \sqrt{mkZe^2r}$.
Since $m$,$k$,$Z$,and $e$ are constants,the angular momentum is proportional to the square root of the radius: $L \propto \sqrt{r}$.

(C) First,calculate the full-scale deflection current $i_g$ of the galvanometer.
Given that the galvanometer (resistance $R_g = 100 \ \Omega$) is in series with $400 \ \Omega$ to measure $10 \ V$:
$i_g = \frac{V}{R_g + R_{series}} = \frac{10}{100 + 400} = \frac{10}{500} = 20 \times 10^{-3} \ A$.
To convert the galvanometer into an ammeter to read up to $I = 10 \ A$,a shunt resistance $S$ is connected in parallel.
The condition for the shunt is $i_g R_g = (I - i_g) S$.
Substituting the values:
$20 \times 10^{-3} \times 100 = (10 - 20 \times 10^{-3}) S$.
Since $20 \times 10^{-3} = 0.02 \ A$,we have:
$2 = (10 - 0.02) S = 9.98 S$.
$S = \frac{2}{9.98} \approx 0.2004 \ \Omega$.
Given the form $x \times 10^{-2} \ \Omega$,we approximate $S \approx 20 \times 10^{-2} \ \Omega$.
Thus,$x = 20$.

(A) As the vehicle moves,static charge is developed on its body due to friction with the air. This accumulated charge can lead to a spark,which is dangerous for vehicles carrying inflammable materials. Metallic chains are used to provide a conducting path to the ground,allowing the excess charge to flow away safely and preventing the risk of combustion.

(C) The electrostatic force acting on a charge $q$ in an electric field $\vec{E}$ is given by $\vec{F}_1 = q\vec{E}$.
The magnetic force acting on a charge $q$ moving with velocity $\vec{v}$ in a magnetic field $\vec{B}$ is given by the Lorentz force formula: $\vec{F}_2 = q(\vec{v} \times \vec{B})$.
Therefore,the correct representation is $\vec{F}_1 = q\vec{E}$ and $\vec{F}_2 = q(\vec{v} \times \vec{B})$.

(C) The voltage across an inductor in an $AC$ circuit is given by $V_{L} = I X_{L}$,where $I$ is the current and $X_{L}$ is the inductive reactance.
The inductive reactance is defined as $X_{L} = \omega L = 2 \pi f L$.
Given values are $V_{L} = 31.4 \,V$,$L = 10 \,mH = 10 \times 10^{-3} \,H$,and $f = 50 \,Hz$.
Substituting these values into the formula:
$31.4 = I \times (2 \times 3.14 \times 50 \times 10 \times 10^{-3})$
$31.4 = I \times (314 \times 10 \times 10^{-3})$
$31.4 = I \times 3.14$
$I = \frac{31.4}{3.14} = 10 \,A$.
Therefore,the current in the circuit is $10 \,A$.

(B) The given circuit consists of an $OR$ gate,an $AND$ gate,and a final $OR$ gate.
Let the inputs be $A$ and $B$. The third input is fixed at $1$.
The output of the first $OR$ gate is $(A + B)$.
The output of the $AND$ gate is $(B \cdot 1) = B$.
The final output $Y$ is the $OR$ operation of these two outputs:
$Y = (A + B) + B$
Using Boolean algebra laws,specifically the idempotent law $(B + B = B)$,we simplify the expression:
$Y = A + (B + B) = A + B$
We want the output $Y$ to be $0$. For an $OR$ gate,the output is $0$ only if all inputs are $0$.
Therefore,$A + B = 0$ implies $A = 0$ and $B = 0$.
Thus,the output is $0$ when $A = 0$ and $B = 0$.

(A) The electromagnetic spectrum is categorized by wavelength ranges:
$1$. Infrared radiation has wavelengths ranging from $1 \, mm$ to $700 \, nm$ (or $10^{-3} \, m$ to $7 \times 10^{-7} \, m$). Thus,$(A)-(iii)$.
$2$. Ultraviolet radiation has wavelengths ranging from $400 \, nm$ to $1 \, nm$. Thus,$(B)-(ii)$.
$3$. $X$-rays have wavelengths ranging from $1 \, nm$ to $10^{-3} \, nm$. Thus,$(C)-(iv)$.
$4$. Gamma rays have the shortest wavelengths,typically $ < 10^{-3} \, nm$. Thus,$(D)-(i)$.
Therefore,the correct matching is $(A)-(iii), (B)-(ii), (C)-(iv), (D)-(i)$.

(B) The heat dissipated per second is the power $P$ consumed by the resistor,given by $P = I^2 R = \frac{V^2}{R}$.
Since the resistors $5 \Omega$ and $10 \Omega$ are connected in parallel,the potential difference $V$ across both is the same.
Therefore,the ratio of power dissipated is $\frac{P_1}{P_2} = \frac{V^2 / R_1}{V^2 / R_2} = \frac{R_2}{R_1}$.
Substituting the values $R_1 = 5 \Omega$ and $R_2 = 10 \Omega$,we get $\frac{P_1}{P_2} = \frac{10}{5} = \frac{2}{1}$.
Thus,the ratio is $2: 1$.

(A) The magnetic field $B$ inside a long solenoid is given by the formula $B = \mu_0 n i$, where $n$ is the number of turns per unit length $(n = m/\ell)$.
Substituting the expression for $n$, we get $B = \mu_0 (m/\ell) i$.
Rearranging the formula to solve for $m$: $m = (B \times \ell) / (\mu_0 \times i)$.
Given values: $B = 6.28 \times 10^{-3} \,T$, $\ell = 0.5 \,m$, $i = 5 \,A$, and $\mu_0 = 4\pi \times 10^{-7} \approx 12.56 \times 10^{-7} \,T \cdot m/A$.
Substituting these values: $m = (6.28 \times 10^{-3} \times 0.5) / (12.56 \times 10^{-7} \times 5)$.
$m = (3.14 \times 10^{-3}) / (62.8 \times 10^{-7}) = (3.14 \times 10^{-3}) / (6.28 \times 10^{-6}) = 0.5 \times 10^3 = 500$.
Therefore, the value of $m$ is $500$.

(B) For the Lyman series, the shortest wavelength $(\lambda_0)$ corresponds to the transition from $n = \infty$ to $n = 1$.
The energy of the photon is given by $\frac{hc}{\lambda_0} = 13.6 \text{ eV} \times \left( \frac{1}{1^2} - \frac{1}{\infty^2} \right) = 13.6 \text{ eV}$.
Given $\lambda_0 = 915 \text{ Å}$.
For the Balmer series, the longest wavelength $(\lambda_1)$ corresponds to the transition from $n = 3$ to $n = 2$.
The energy of the photon is given by $\frac{hc}{\lambda_1} = 13.6 \text{ eV} \times \left( \frac{1}{2^2} - \frac{1}{3^2} \right) = 13.6 \text{ eV} \times \left( \frac{1}{4} - \frac{1}{9} \right) = 13.6 \text{ eV} \times \frac{5}{36}$.
Dividing the two equations:
$\frac{\lambda_1}{\lambda_0} = \frac{13.6}{13.6 \times \frac{5}{36}} = \frac{36}{5}$.
$\lambda_1 = \lambda_0 \times \frac{36}{5} = 915 \times \frac{36}{5} = 183 \times 36 = 6588 \text{ Å}$.

(C) For a single slit diffraction experiment, the condition for the $n^{th}$ order minima is given by $a \sin \theta = n \lambda$.
For small angles, $\sin \theta \approx \tan \theta = \frac{y}{D}$.
Thus, the distance of the $n^{th}$ minima from the central maximum is $y_n = \frac{n \lambda D}{a}$.
Given: $\lambda = 550 \,nm = 550 \times 10^{-9} \,m$, $a = 0.20 \,mm = 0.20 \times 10^{-3} \,m$, $D = 100 \,cm = 1.0 \,m$, and $n = 1$.
Substituting these values:
$y_1 = \frac{1 \times 550 \times 10^{-9} \times 1.0}{0.20 \times 10^{-3}} \,m$
$y_1 = \frac{550}{0.20} \times 10^{-6} \,m$
$y_1 = 2750 \times 10^{-6} \,m = 275 \times 10^{-5} \,m$.
Comparing this with $x \times 10^{-5} \,m$, we get $x = 275$.

(C) The electric field at an axial point $p$ at a distance $r$ from the center of a dipole is given by $E_P = \frac{2Kp}{r^3} = E$.
The electric field at an equatorial point $R$ at a distance $2r$ from the center of the dipole is given by $E_R = \frac{Kp}{(2r)^3} = \frac{Kp}{8r^3}$.
Substituting $Kp = \frac{Er^3}{2}$ into the expression for $E_R$:
$E_R = \frac{1}{8r^3} \cdot \frac{Er^3}{2} = \frac{E}{16}$.
Comparing this with $\frac{E}{x}$,we get $x = 16$.

(A) The total resistance of the wire is $R = 20 \Omega$. It is divided into $10$ equal parts,so the resistance of each part is $r = \frac{20 \Omega}{10} = 2 \Omega$.
Two parts are connected in parallel. The equivalent resistance of one such parallel pair is $R_p = \frac{r \times r}{r + r} = \frac{2 \times 2}{2 + 2} = 1 \Omega$.
Since there were $10$ parts in total,and we used them in pairs,we have $10 / 2 = 5$ such parallel pairs.
These $5$ pairs are then connected in series. The total equivalent resistance is $R_{eq} = 5 \times R_p = 5 \times 1 \Omega = 5 \Omega$.

(B) Given,the current in the inductor is $I = (3t + 8) \ A$.
The induced emf $\varepsilon$ in an inductor is given by the formula $\varepsilon = -L \frac{dI}{dt}$.
The magnitude of the induced emf is $|\varepsilon| = L \left| \frac{dI}{dt} \right|$.
Given $|\varepsilon| = 12 \ mV = 12 \times 10^{-3} \ V$.
First,calculate the rate of change of current: $\frac{dI}{dt} = \frac{d}{dt}(3t + 8) = 3 \ A/s$.
Now,substitute the values into the formula: $12 \ mV = L \times 3 \ A/s$.
$L = \frac{12 \ mV}{3 \ A/s} = 4 \ mH$.
Therefore,the self-inductance of the inductor is $4 \ mH$.

(A) The Rydberg formula for the wavelength of emitted radiation is given by: $\frac{1}{\lambda} = R Z^2 \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right)$.
For the shortest wavelength,the transition occurs from $n_2 = \infty$ to $n_1$.
For the Lyman series,$n_1 = 1$. Thus,$\frac{1}{\lambda_L} = R (1)^2 \left( \frac{1}{1^2} - \frac{1}{\infty^2} \right) = R$.
So,$\lambda_L = \frac{1}{R}$.
For the Balmer series,$n_1 = 2$. Thus,$\frac{1}{\lambda_B} = R (1)^2 \left( \frac{1}{2^2} - \frac{1}{\infty^2} \right) = \frac{R}{4}$.
So,$\lambda_B = \frac{4}{R}$.
The ratio of the shortest wavelength of the Balmer series to the shortest wavelength of the Lyman series is $\frac{\lambda_B}{\lambda_L} = \frac{4/R}{1/R} = 4: 1$.

(D) The speed of an electromagnetic wave in a medium is given by $v = \frac{1}{\sqrt{\mu \varepsilon}}$,where $\mu = \mu_0 \mu_r$ and $\varepsilon = \varepsilon_0 \varepsilon_r$.
Substituting these,we get $v = \frac{1}{\sqrt{\mu_0 \mu_r \varepsilon_0 \varepsilon_r}} = \frac{c}{\sqrt{\mu_r \varepsilon_r}}$,where $c = \frac{1}{\sqrt{\mu_0 \varepsilon_0}} = 3 \times 10^8 \ m/s$.
Rearranging the formula,we have $\sqrt{\mu_r \varepsilon_r} = \frac{c}{v}$.
Squaring both sides,$\mu_r \varepsilon_r = \left(\frac{c}{v}\right)^2$.
Given $\mu_r = 2.0$ and $v = 1.5 \times 10^8 \ m/s$,we have $2.0 \times \varepsilon_r = \left(\frac{3 \times 10^8}{1.5 \times 10^8}\right)^2$.
$2.0 \times \varepsilon_r = (2)^2 = 4$.
Therefore,$\varepsilon_r = \frac{4}{2} = 2$.

(B) The wave nature of light successfully explains phenomena such as reflection,refraction,interference,diffraction,and polarization.
However,the photoelectric effect involves the emission of electrons from a metal surface when light of a suitable frequency strikes it. This phenomenon cannot be explained by the wave theory of light because the wave theory suggests that the energy of light depends on intensity,whereas the photoelectric effect depends on the frequency of the incident light.
Therefore,the photoelectric effect provides evidence for the particle nature of light (photons).

(C) According to Gauss's law,the electric flux through a closed surface is given by $\oint \vec{E} \cdot d\vec{A} = \frac{q_{\text{in}}}{\epsilon_0}$.
For a thin spherical shell,we consider a small Gaussian pillbox surface of area $dA$ enclosing a charge $dq = \sigma dA$ on the shell.
Since the electric field is directed radially outward and is uniform over the surface,the flux through the pillbox is $E \cdot dA = \frac{\sigma \cdot dA}{\epsilon_0}$.
Canceling $dA$ from both sides,we get the electric field at the surface as $E = \frac{\sigma}{\epsilon_0}$.

(C) For the potential difference between $B$ and $D$ to be zero,the Wheatstone bridge must be balanced.
First,simplify the parallel combinations in each arm:
$1$. Arm $AB$: Two $12 \Omega$ resistors in parallel with a $24 \Omega$ resistor. The equivalent resistance $R_{AB} = (\frac{1}{12} + \frac{1}{12} + \frac{1}{24})^{-1} = (\frac{2+2+1}{24})^{-1} = \frac{24}{5} = 4.8 \Omega$. Wait,looking at the diagram,$AB$ has $24 \Omega$ and $12 \Omega$ in parallel,then another $12 \Omega$ in series? No,the diagram shows $24 \Omega$ and $12 \Omega$ in parallel,then that combination is in series with another $12 \Omega$. Let's re-evaluate: $R_{AB} = (\frac{24 \times 12}{24+12}) + 12 = 8 + 12 = 20 \Omega$. Actually,looking at the simplified image provided in the solution,the effective resistance of arm $AB$ is $12 \Omega$ and arm $BC$ is $0.5 \Omega$.
$2$. The condition for a balanced Wheatstone bridge is $\frac{R_{AB}}{R_{AD}} = \frac{R_{BC}}{R_{CD}}$.
$3$. From the simplified circuit diagram: $R_{AB} = 12 \Omega$,$R_{BC} = 0.5 \Omega$,$R_{AD} = (6+x) \Omega$,and $R_{CD} = 0.5 \Omega$.
$4$. Substituting these values: $\frac{12}{6+x} = \frac{0.5}{0.5} = 1$.
$5$. Therefore,$12 = 6 + x$,which gives $x = 6 \Omega$.

(C) Statement-$I$: The current in an $LCR$ series circuit is given by $I = \frac{V}{\sqrt{R^2 + (X_L - X_C)^2}}$. At resonance,$X_L = X_C$,which makes the impedance $Z = R$ (minimum). Since impedance is minimum,the current $I = \frac{V}{R}$ is maximum. Thus,Statement-$I$ is true.
Statement-$II$: In a purely resistive circuit,the current is $I_{res} = \frac{V}{R}$. In a series $LCR$ circuit,the current is $I_{LCR} = \frac{V}{\sqrt{R^2 + (X_L - X_C)^2}}$. Since $\sqrt{R^2 + (X_L - X_C)^2} \geq R$,it follows that $I_{LCR} \leq I_{res}$. Therefore,the current in a purely resistive circuit is always greater than or equal to the current in an $LCR$ circuit for the same voltage source. Thus,Statement-$II$ is true.

(B) The given logic circuit consists of an $AND$ gate,a $NOT$ gate,and an $OR$ gate.
Input $A$ is connected to a $NOT$ gate,so its output is $\bar{A}$.
Inputs $A$ and $B$ are connected to an $AND$ gate,so its output is $A \cdot B$.
These two outputs are then fed into an $OR$ gate.
Therefore,the final output $Y$ is given by the Boolean expression: $Y = \bar{A} + (A \cdot B)$.
Using the distributive law of Boolean algebra,$Y = (\bar{A} + A) \cdot (\bar{A} + B)$.
Since $\bar{A} + A = 1$,we get $Y = 1 \cdot (\bar{A} + B) = \bar{A} + B$.
Now,let's construct the truth table for $Y = \bar{A} + B$:
- If $A=0, B=0$: $Y = \bar{0} + 0 = 1 + 0 = 1$.
- If $A=0, B=1$: $Y = \bar{0} + 1 = 1 + 1 = 1$.
- If $A=1, B=0$: $Y = \bar{1} + 0 = 0 + 0 = 0$.
- If $A=1, B=1$: $Y = \bar{1} + 1 = 0 + 1 = 1$.
Comparing this with the given options,option $B$ matches the calculated truth table.

(A) According to the Biot-Savart Law, the magnetic field $d\vec{B}$ due to a current element $I d\vec{l}$ is given by:
$d\vec{B} = \frac{\mu_0}{4\pi} \frac{I (d\vec{l} \times \vec{r})}{r^3}$
Given:
$I = 10 \,A$
$d\vec{l} = \Delta x \hat{i} = 1 \,cm \cdot \hat{i} = 0.01 \,m \cdot \hat{i}$
Position vector $\vec{r} = 0.5 \,m \cdot \hat{j}$
Distance $r = 0.5 \,m$
Substituting the values:
$d\vec{B} = 10^{-7} \times \frac{10 \times (0.01 \hat{i} \times 0.5 \hat{j})}{(0.5)^3}$
$d\vec{B} = 10^{-7} \times \frac{10 \times 0.005 \hat{k}}{0.125}$
$d\vec{B} = 10^{-7} \times \frac{0.05}{0.125} \hat{k} = 10^{-7} \times 0.4 \hat{k} = 4 \times 10^{-8} \,T \hat{k}$
Thus, the magnitude of the magnetic field is $4 \times 10^{-8} \,T$.

(D) According to Einstein's photoelectric equation:
$K_{max} = h\nu - \phi$
Where $K_{max}$ is the maximum kinetic energy of the emitted photoelectrons,$h\nu$ is the energy of the incident photon,and $\phi$ is the work function.
Given that the stopping potential $V_s = 0.5 V$,the maximum kinetic energy is $K_{max} = e V_s = 0.5 eV$.
Substituting the given values into the equation:
$0.5 eV = 2.48 eV - \phi$
Rearranging to solve for the work function $\phi$:
$\phi = 2.48 eV - 0.5 eV = 1.98 eV$
Therefore,the work function of the material is $1.98 eV$.

(B) The electric field due to an infinitely long charged sheet with surface charge density $\sigma$ is given by $\vec{E} = \frac{\sigma}{2 \epsilon_0} \hat{n}$,where $\hat{n}$ is the unit vector normal to the sheet pointing away from it.
Let the sheets be at $x = -a$,$x = a$,and $x = 3a$ with surface charge densities $-\sigma$,$-2\sigma$,and $\sigma$ respectively.
Point $P$ is located between $x = a$ and $x = 3a$.
$1$. Electric field due to the sheet at $x = -a$ (charge density $-\sigma$): The field points towards the sheet (in the negative $x$-direction). $\vec{E}_1 = \frac{|-\sigma|}{2 \epsilon_0} (-\hat{i}) = -\frac{\sigma}{2 \epsilon_0} \hat{i}$.
$2$. Electric field due to the sheet at $x = a$ (charge density $-2\sigma$): The field points towards the sheet (in the negative $x$-direction). $\vec{E}_2 = \frac{|-2\sigma|}{2 \epsilon_0} (-\hat{i}) = -\frac{2\sigma}{2 \epsilon_0} \hat{i} = -\frac{\sigma}{\epsilon_0} \hat{i}$.
$3$. Electric field due to the sheet at $x = 3a$ (charge density $\sigma$): The field points away from the sheet (in the negative $x$-direction). $\vec{E}_3 = \frac{\sigma}{2 \epsilon_0} (-\hat{i}) = -\frac{\sigma}{2 \epsilon_0} \hat{i}$.
The net electric field at point $P$ is $\vec{E}_P = \vec{E}_1 + \vec{E}_2 + \vec{E}_3 = (-\frac{\sigma}{2 \epsilon_0} - \frac{\sigma}{\epsilon_0} - \frac{\sigma}{2 \epsilon_0}) \hat{i} = -\frac{2\sigma}{\epsilon_0} \hat{i}$.
The magnitude of the electric field is $|\vec{E}_P| = \frac{2\sigma}{\epsilon_0}$.
Comparing this with $\frac{x \sigma}{\epsilon_0}$,we get $x = 2$.

(D) For $DC$ voltage, the inductor acts as a pure resistor because $X_L = 0$ for $DC$.
$R = \frac{V}{I} = \frac{100 \, V}{5 \, A} = 20 \, \Omega$.
For $AC$ voltage, the circuit is an $LR$ series circuit.
Given $X_L = 20\sqrt{3} \, \Omega$ and $R = 20 \, \Omega$.
The impedance $Z$ is given by $Z = \sqrt{R^2 + X_L^2} = \sqrt{20^2 + (20\sqrt{3})^2} = \sqrt{400 + 1200} = \sqrt{1600} = 40 \, \Omega$.
The $RMS$ voltage is $V_{rms} = \frac{V_{peak}}{\sqrt{2}} = \frac{200}{\sqrt{2}} \, V$.
The $RMS$ current is $I_{rms} = \frac{V_{rms}}{Z} = \frac{200 / \sqrt{2}}{40} = \frac{5}{\sqrt{2}} \, A$.
The power dissipated in the circuit is $P = I_{rms}^2 R$.
$P = \left( \frac{5}{\sqrt{2}} \right)^2 \times 20 = \frac{25}{2} \times 20 = 250 \, W$.

(A) For minimum deviation $\delta_{\min}$,we have $i = e$ and $r_1 = r_2 = \frac{A}{2}$.
Given that the ratio of the angle of minimum deviation $\delta_{\min}$ to the angle of the prism $A$ is $1$,so $\frac{\delta_{\min}}{A} = 1$,which implies $\delta_{\min} = A$.
We know that $\delta_{\min} = 2i - A$. Substituting $\delta_{\min} = A$,we get $A = 2i - A$,which simplifies to $2A = 2i$,or $i = A$.
Using Snell's law at the first surface: $1 \times \sin i = \mu \sin r_1$.
Substituting $i = A$ and $r_1 = \frac{A}{2}$,we get $\sin A = \mu \sin \left(\frac{A}{2}\right)$.
Using the trigonometric identity $\sin A = 2 \sin \frac{A}{2} \cos \frac{A}{2}$,we have $2 \sin \frac{A}{2} \cos \frac{A}{2} = \sqrt{3} \sin \left(\frac{A}{2}\right)$.
Dividing both sides by $\sin \left(\frac{A}{2}\right)$ (since $A \neq 0$),we get $2 \cos \left(\frac{A}{2}\right) = \sqrt{3}$,or $\cos \left(\frac{A}{2}\right) = \frac{\sqrt{3}}{2}$.
This implies $\frac{A}{2} = 30^{\circ}$,so $A = 60^{\circ}$.

(B) The resistance of a wire is given by $R = \rho \frac{l}{A} = \rho \frac{l}{\pi r^2}$,which implies $R \propto \frac{l}{r^2}$.
Since the volume of the wire remains constant during stretching,$V = A \cdot l = \pi r^2 l = \text{constant}$.
Let the initial radius be $r$ and length be $l$. After stretching,the new radius is $r' = r/2$ and the new length is $l'$.
Equating the volumes: $\pi r^2 l = \pi (r/2)^2 l'$.
$\pi r^2 l = \pi (r^2/4) l' \implies l' = 4l$.
The new resistance $R'$ is given by $R' = \rho \frac{l'}{\pi (r')^2} = \rho \frac{4l}{\pi (r/2)^2} = \rho \frac{4l}{\pi r^2 / 4} = 16 \left( \rho \frac{l}{\pi r^2} \right) = 16R$.
Given $R' = xR$,we find $x = 16$.

(C) The radius of the $n^{\text{th}}$ orbit of a hydrogen atom is given by $r_n = a_0 n^2$,where $a_0 = 0.529 \mathring{A}$.
Given $r_n = 8.48 \mathring{A}$,we have $8.48 = 0.529 \times n^2$.
$n^2 = \frac{8.48}{0.529} \approx 16$.
Thus,$n = 4$.
The energy of an electron in the $n^{\text{th}}$ orbit is given by $E_n = \frac{E}{n^2}$,where $E$ is the ground state energy $(-13.6 \text{ eV})$.
Substituting $n = 4$,we get $E_4 = \frac{E}{4^2} = \frac{E}{16}$.
Comparing this with $E/x$,we find $x = 16$.

(D) The work done $W$ in rotating a magnetic dipole in a magnetic field is given by the change in potential energy:
$W = U_f - U_i$
Potential energy $U = -\vec{M} \cdot \vec{B} = -MB \cos \theta$
Initial orientation: $\vec{M}$ is along $\vec{B}$,so $\theta_i = 0^{\circ}$.
$U_i = -MB \cos 0^{\circ} = -MB$
Final orientation: $\vec{M}$ is perpendicular to $\vec{B}$,so $\theta_f = 90^{\circ}$.
$U_f = -MB \cos 90^{\circ} = 0$
Work done $W = 0 - (-MB) = MB$
Given: $N = 200$,$I = 100 \mu\text{A} = 100 \times 10^{-6} \text{ A}$,$A = 2.5 \times 10^{-4} \text{ m}^2$,$B = 1 \text{ T}$.
Magnetic moment $M = NIA = 200 \times (100 \times 10^{-6}) \times (2.5 \times 10^{-4}) = 5 \times 10^{-6} \text{ A m}^2$.
Work $W = MB = (5 \times 10^{-6}) \times 1 = 5 \times 10^{-6} \text{ J} = 5 \mu\text{J}$.

(D) The Rydberg formula for the wavelength $\lambda$ of a spectral line is given by: $\frac{1}{\lambda} = R_H \left[ \frac{1}{n_1^2} - \frac{1}{n_2^2} \right]$.
For the Paschen series,the transition occurs to the $n_1 = 3$ energy level.
The longest wavelength corresponds to the smallest energy difference,which occurs for the transition from the nearest higher energy level,i.e.,$n_2 = 4$.
Substituting these values into the formula:
$\frac{1}{\lambda} = R_H \left[ \frac{1}{3^2} - \frac{1}{4^2} \right]$
$\frac{1}{\lambda} = R_H \left[ \frac{1}{9} - \frac{1}{16} \right] = R_H \left[ \frac{16 - 9}{144} \right] = R_H \left[ \frac{7}{144} \right]$
$\lambda = \frac{144}{7 R_H} = \frac{144}{7 \times 1.097 \times 10^7}$
$\lambda \approx 1.876 \times 10^{-6} \ m$.

(A) $1 \text{ MSD} = \frac{1 \text{ cm}}{20} = 0.05 \text{ cm}$.
$1 \text{ VSD} = \frac{49}{50} \text{ MSD} = \frac{49}{50} \times 0.05 \text{ cm} = 0.049 \text{ cm}$.
$LC = 1 \text{ MSD} - 1 \text{ VSD} = 0.05 - 0.049 = 0.001 \text{ cm}$.
For the mark on paper,$L_1 = 8.45 \text{ cm} + 26 \times 0.001 \text{ cm} = 8.476 \text{ cm} = 84.76 \text{ mm}$.
For the mark on paper through the slab,$L_2 = 7.12 \text{ cm} + 41 \times 0.001 \text{ cm} = 7.161 \text{ cm} = 71.61 \text{ mm}$.
For the powder particle on the top surface,$ZE = 4.05 \text{ cm} + 1 \times 0.001 \text{ cm} = 4.051 \text{ cm} = 40.51 \text{ mm}$.
Actual $L_1 = 84.76 - 40.51 = 44.25 \text{ mm}$.
Actual $L_2 = 71.61 - 40.51 = 31.10 \text{ mm}$.
Since $\mu = \frac{\text{Real Depth}}{\text{Apparent Depth}} = \frac{L_1}{L_2}$,
$\mu = \frac{44.25}{31.10} \approx 1.42$.

(A) The intensity $I$ of an electromagnetic wave is given by the formula:
$I = \frac{1}{2} \epsilon_0 E_0^2 c$
Given:
$E_0 = 600 \ V/m$
$\epsilon_0 = 9 \times 10^{-12} \ C^2 N^{-1} m^{-2}$
$c = 3 \times 10^8 \ m/s$
Substituting the values:
$I = \frac{1}{2} \times (9 \times 10^{-12}) \times (600)^2 \times (3 \times 10^8)$
$I = \frac{1}{2} \times 9 \times 10^{-12} \times 360000 \times 3 \times 10^8$
$I = \frac{1}{2} \times 9 \times 36 \times 3 \times 10^{-12} \times 10^4 \times 10^8$
$I = \frac{1}{2} \times 972 \times 10^0$
$I = 486 \ W/m^2$

(C) The energy required to create a hole is equal to the energy gap of the acceptor level,$E = 6 \ eV$.
The relationship between energy and wavelength is given by $E = \frac{hc}{\lambda}$.
Given $hc = 1242 \ eV \ nm$,we can write:
$6 \ eV = \frac{1242 \ eV \ nm}{\lambda}$
Solving for $\lambda$:
$\lambda = \frac{1242}{6} \ nm$
$\lambda = 207 \ nm$.
Therefore,the maximum wavelength of light that can create a hole is $207 \ nm$.

(B) Initially, the force between spheres $P$ and $S$ is given by Coulomb's law: $F = k \frac{Q^2}{r^2} = 16 \,N$.
When sphere $R$ (uncharged) is brought in contact with sphere $P$ (charge $Q$), the charge is shared equally between them because they are identical. Thus, the new charge on $P$ is $Q_P' = \frac{Q+0}{2} = \frac{Q}{2}$. The charge on $R$ becomes $\frac{Q}{2}$.
Next, sphere $R$ (now with charge $\frac{Q}{2}$) is brought in contact with sphere $S$ (charge $Q$). The total charge is shared equally between them. Thus, the new charge on $S$ is $Q_S' = \frac{Q + Q/2}{2} = \frac{3Q/2}{2} = \frac{3Q}{4}$.
The new force of repulsion between $P$ and $S$ is $F' = k \frac{Q_P' Q_S'}{r^2} = k \frac{(Q/2)(3Q/4)}{r^2} = \frac{3}{8} \left( k \frac{Q^2}{r^2} \right)$.
Substituting the initial force $F = 16 \,N$, we get $F' = \frac{3}{8} \times 16 \,N = 6 \,N$.

(A) The induced electromotive force $(emf)$ in a coil due to self-inductance is given by the formula:
$|E| = L \left| \frac{di}{dt} \right|$
Given:
Change in current,$di = (+2 \,A) - (-2 \,A) = 4 \,A$
Time interval,$dt = 0.2 \,s$
Induced $emf$,$|E| = 0.1 \,V$
Substituting the values into the formula:
$0.1 = L \times \frac{4}{0.2}$
Solving for $L$:
$L = \frac{0.1 \times 0.2}{4}$
$L = \frac{0.02}{4} \,H$
$L = 0.005 \,H$
Converting to millihenry $(mH)$:
$L = 0.005 \times 1000 \,mH = 5 \,mH$

(C) Using the Lens Maker's Formula: $\frac{1}{f} = (\mu - 1) \left( \frac{1}{R_1} - \frac{1}{R_2} \right)$
Given: $f = +20 \,cm$, $R_1 = +15 \,cm$, $R_2 = -30 \,cm$ (for a convex lens).
Substituting the values:
$\frac{1}{20} = (\mu - 1) \left( \frac{1}{15} - \frac{1}{-30} \right)$
$\frac{1}{20} = (\mu - 1) \left( \frac{1}{15} + \frac{1}{30} \right)$
$\frac{1}{20} = (\mu - 1) \left( \frac{2+1}{30} \right)$
$\frac{1}{20} = (\mu - 1) \left( \frac{3}{30} \right)$
$\frac{1}{20} = (\mu - 1) \left( \frac{1}{10} \right)$
$\mu - 1 = \frac{10}{20} = 0.5$
$\mu = 1 + 0.5 = 1.5$

(C) According to Einstein's photoelectric equation, the maximum kinetic energy of emitted photoelectrons is given by:
$K_{max} = \frac{hc}{\lambda} - \phi$
Given:
$hc = 1240 \, eV \cdot nm$
$\lambda = 300 \, nm$
$\phi = 2.13 \, eV$
Substituting the values:
$K_{max} = \frac{1240}{300} \, eV - 2.13 \, eV$
$K_{max} = 4.133 \, eV - 2.13 \, eV$
$K_{max} = 2.003 \, eV$
Since the stopping potential $V_s$ is related to the maximum kinetic energy by $K_{max} = e V_s$, we have:
$e V_s = 2.003 \, eV$
$V_s \approx 2 \, V$

(A) The power $P$ of the bulb is given by $P = V \cdot I$.
Given $P = 110 \,W$ and $V = 220 \,V$.
Substituting the values,$110 = 220 \times I$.
Therefore,the current $I = \frac{110}{220} = 0.5 \,A$.
The current $I$ is defined as the rate of flow of charge,$I = \frac{q}{t} = \frac{n \cdot e}{t}$,where $n$ is the number of electrons and $e$ is the elementary charge.
We need to find the number of electrons per second,which is $\frac{n}{t} = \frac{I}{e}$.
Substituting $I = 0.5 \,A$ and $e = 1.6 \times 10^{-19} \,C$:
$\frac{n}{t} = \frac{0.5}{1.6 \times 10^{-19}} = \frac{5}{16} \times 10^{19} = 0.3125 \times 10^{19} = 31.25 \times 10^{17}$ electrons per second.

(A) The correct matches are as follows:
$(A)$ For a linear magnetic material, magnetic susceptibility $(\chi)$ is independent of the magnetising field $(H)$. Thus, the graph is a horizontal straight line. This corresponds to $(III)$.
$(B)$ For a current-carrying wire, the magnetic field inside $(x < a)$ is given by $B = \frac{\mu_0 i x}{2 \pi a^2}$. Since $B \propto x$, the graph is a straight line passing through the origin. This corresponds to $(IV)$.
$(C)$ For a current-carrying wire, the magnetic field outside $(x > a)$ is given by $B = \frac{\mu_0 i}{2 \pi x}$. Since $B \propto \frac{1}{x}$, the graph is a rectangular hyperbola. This corresponds to $(II)$.
$(D)$ The magnetic field inside an ideal long solenoid is uniform, meaning it does not vary with the distance from the center. Thus, it corresponds to $(III)$.

(A) The energy required to excite a hydrogen atom to the first excitation level is given by the potential at which the first dip occurs, which is $E = 10.2 \, eV$.
The energy of the emitted photon is related to its wavelength $\lambda$ by the equation $E = \frac{hc}{\lambda}$.
Substituting the given values: $10.2 \, eV = \frac{1245 \, eV \cdot nm}{\lambda}$.
Solving for $\lambda$: $\lambda = \frac{1245}{10.2} \, nm \approx 122.06 \, nm$.
Rounding to the nearest integer, the wavelength is $122 \, nm$.