MHT CET 2023 Physics Question Paper with Answer and Solution

(B) The time period of a spring-mass system is given by $T = 2\pi \sqrt{m/k}$.
For the initial spring with constant $k$,the period is $T_1 = 2\pi \sqrt{m/k}$.
When a spring of length $l$ and spring constant $k$ is cut into two equal halves,the spring constant of each half becomes $k' = 2k$ because $k \propto 1/l$.
For the new system with the same mass $m$ and new spring constant $k' = 2k$,the period is $T_2 = 2\pi \sqrt{m/(2k)}$.
Taking the ratio: $T_1 / T_2 = \frac{2\pi \sqrt{m/k}}{2\pi \sqrt{m/(2k)}} = \sqrt{\frac{m/k}{m/(2k)}} = \sqrt{2}$.
Thus,the ratio $T_1 / T_2 = \sqrt{2}$.

(B) The time period of a simple pendulum is given by $T = 2 \pi \sqrt{\frac{l}{g_{eff}}}$.
In a stationary lift,the effective acceleration is $g_{eff} = g$,so $T = 2 \pi \sqrt{\frac{l}{g}}$.
When the lift accelerates upwards with an acceleration $a = \frac{g}{3}$,the effective acceleration experienced by the pendulum is $g_{eff} = g + a$.
Substituting the value of $a$,we get $g_{eff} = g + \frac{g}{3} = \frac{4g}{3}$.
The new time period $T^{\prime}$ is given by $T^{\prime} = 2 \pi \sqrt{\frac{l}{g_{eff}}} = 2 \pi \sqrt{\frac{l}{4g/3}}$.
Simplifying this,$T^{\prime} = 2 \pi \sqrt{\frac{3l}{4g}} = \frac{\sqrt{3}}{2} \left( 2 \pi \sqrt{\frac{l}{g}} \right)$.
Since $T = 2 \pi \sqrt{\frac{l}{g}}$,we have $T^{\prime} = \frac{\sqrt{3}}{2} T$.

(D) The formula for the time period of a spring-mass system is $T = 2 \pi \sqrt{\frac{m}{k}}$.
For the initial mass $m$,the time period is $T = 3 \text{ s}$,so $3 = 2 \pi \sqrt{\frac{m}{k}}$.
For the increased mass $m+1$,the time period is $T' = 5 \text{ s}$,so $5 = 2 \pi \sqrt{\frac{m+1}{k}}$.
Taking the ratio of the two equations:
$\frac{T}{T'} = \frac{2 \pi \sqrt{\frac{m}{k}}}{2 \pi \sqrt{\frac{m+1}{k}}} = \sqrt{\frac{m}{m+1}}$.
Substituting the given values:
$\frac{3}{5} = \sqrt{\frac{m}{m+1}}$.
Squaring both sides:
$\frac{9}{25} = \frac{m}{m+1}$.
Cross-multiplying gives:
$9(m+1) = 25m \implies 9m + 9 = 25m$.
$16m = 9 \implies m = \frac{9}{16} \text{ kg}$.

(A) When the block is displaced vertically by a small distance $x$,the additional buoyant force acting on it is $F = - (A x \rho) g$.
This force acts as a restoring force,so $F = -kx$,where $k = A \rho g$ is the effective spring constant.
The angular frequency of oscillation is $\omega = \sqrt{\frac{k}{m}} = \sqrt{\frac{A \rho g}{m}}$.
The frequency $n$ is given by $n = \frac{\omega}{2 \pi}$.
Therefore,$n = \frac{1}{2 \pi} \sqrt{\frac{A \rho g}{m}}$.

(B) Given mass $m = 0.04 \text{ kg}$.
From the force-displacement graph,the force $F$ is related to displacement $x$ by $F = -kx$.
The slope of the graph gives the force constant $k$.
$k = |\frac{F}{x}| = |\frac{-8}{2}| = 4 \text{ N/m}$.
The time period $T$ for simple harmonic motion is given by the formula:
$T = 2 \pi \sqrt{\frac{m}{k}}$
Substituting the values:
$T = 2 \pi \sqrt{\frac{0.04}{4}}$
$T = 2 \pi \sqrt{0.01}$
$T = 2 \pi \times 0.1$
$T = 0.2 \pi \text{ s}$.

(A) For a body oscillating under a restoring force $F$,we have $F = kx$,where $k = m\omega^2$.
Thus,$\omega^2 = \frac{F}{mx}$.
For the first force $F_1$,$\omega_1^2 = \frac{F_1}{mx}$.
For the second force $F_2$,$\omega_2^2 = \frac{F_2}{mx}$.
When both forces act simultaneously,the resultant force is $F_{res} = F_1 + F_2$.
The resultant angular frequency $\omega$ is given by $\omega^2 = \frac{F_1 + F_2}{mx} = \frac{F_1}{mx} + \frac{F_2}{mx} = \omega_1^2 + \omega_2^2$.
Since $\omega = \frac{2\pi}{T}$,we have $\left(\frac{2\pi}{T}\right)^2 = \left(\frac{2\pi}{T_1}\right)^2 + \left(\frac{2\pi}{T_2}\right)^2$.
Dividing by $4\pi^2$,we get $\frac{1}{T^2} = \frac{1}{T_1^2} + \frac{1}{T_2^2}$.
$\frac{1}{T^2} = \frac{T_1^2 + T_2^2}{T_1^2 T_2^2}$.
Therefore,$T = \sqrt{\frac{T_1^2 T_2^2}{T_1^2 + T_2^2}}$.

(A) The angular velocity $\omega$ of a particle moving in a circular path is related to its time period $T$ by the formula: $\omega = \frac{2\pi}{T}$.
Since the time period $T$ is the same for both particles $(T_1 = T_2 = T)$,their angular velocities are:
$\omega_1 = \frac{2\pi}{T_1}$ and $\omega_2 = \frac{2\pi}{T_2}$.
Therefore,$\omega_1 = \omega_2$.
The ratio of the angular velocity of the first particle to the second particle is $\frac{\omega_1}{\omega_2} = \frac{1}{1}$.

(A) The angular momentum $L$ of a particle with respect to the origin is given by the cross product of its position vector $r$ and its linear momentum $p = Mv$.
$L = r \times (Mv)$
Since the mass $M$ is moving with a constant velocity $v$ parallel to the $X$-axis, we can write its position as $r = xi + yj$, where $y$ is the constant perpendicular distance from the $X$-axis.
The velocity vector is $v = vi$.
The angular momentum is $L = (xi + yj) \times (Mvi) = Mv(xi \times i) + Mv(yj \times i)$.
Since $i \times i = 0$ and $j \times i = -k$, we get $L = -Mvyk$.
As $M$, $v$, and $y$ are all constants, the magnitude and direction of the angular momentum $L$ remain constant over time.

(B) The moment of inertia of a disc about its central axis is $I = \frac{1}{2} M R^2$.
Using the rotational analog of Newton's second law,$\tau = I \alpha$,where $\tau$ is the torque and $\alpha$ is the angular acceleration.
The tangential force $F$ applied at the rim provides a torque $\tau = F \times R$.
The angular acceleration is given by $\alpha = \frac{\omega}{t}$.
Substituting these into the torque equation: $F \times R = (\frac{1}{2} M R^2) \times (\frac{\omega}{t})$.
Solving for $F$: $F = \frac{M R \omega}{2 t}$.

(D) Let $m$ be the mass,$v$ be the velocity of the center of mass,$R$ be the radius,and $I$ be the moment of inertia of the body.
The translational kinetic energy is $KE_{\text{trans}} = \frac{1}{2} mv^2$.
The rotational kinetic energy is $KE_{\text{rot}} = \frac{1}{2} I \omega^2$.
Since the body is rolling without slipping,$v = \omega R$,so $\omega = \frac{v}{R}$.
Substituting this into the rotational kinetic energy formula: $KE_{\text{rot}} = \frac{1}{2} I \left(\frac{v}{R}\right)^2 = \frac{1}{2} \left(\frac{I}{R^2}\right) v^2$.
Given $KE_{\text{trans}} = KE_{\text{rot}}$,we have $\frac{1}{2} mv^2 = \frac{1}{2} \left(\frac{I}{R^2}\right) v^2$.
This implies $m = \frac{I}{R^2}$,or $I = mR^2$.
The moment of inertia $I = mR^2$ corresponds to a ring (or a thin hollow cylinder) about its central axis.

(B) The moment of inertia of a disc of mass $M$ and radius $R$ about an axis passing through its centre and perpendicular to its plane is $I = \frac{1}{2} M R^2$.
Since the mass $M$ is the same for both discs,the ratio is $\frac{I_1}{I_2} = \frac{R_1^2}{R_2^2}$.
The mass of a disc is given by $M = \text{Volume} \times \text{Density} = (\pi R^2 t) d$.
Since $M$ and $t$ are constant,we have $R^2 \propto \frac{1}{d}$,which implies $R^2 d = \text{constant}$.
Therefore,$R_1^2 d_1 = R_2^2 d_2$,which gives $\frac{R_1^2}{R_2^2} = \frac{d_2}{d_1}$.
Substituting this into the ratio of moments of inertia,we get $\frac{I_1}{I_2} = \frac{d_2}{d_1}$.

(B) The moment of inertia of a square lamina of mass $M$ and side $b$ about an axis perpendicular to its plane and passing through its centre is given by $I_{\text{lamina}} = \frac{M(b^2 + b^2)}{12} = \frac{Mb^2}{6}$.
The moment of inertia of a disc of mass $M$ and radius $R$ about an axis perpendicular to its plane and passing through its centre is given by $I_{\text{disc}} = \frac{MR^2}{2}$.
According to the problem,the moments of inertia are equal:
$\frac{Mb^2}{6} = \frac{MR^2}{2}$.
Canceling $M$ from both sides,we get:
$\frac{b^2}{6} = \frac{R^2}{2}$.
Rearranging the terms to find the ratio $\frac{b^2}{R^2}$:
$\frac{b^2}{R^2} = \frac{6}{2} = 3$.
Taking the square root of both sides:
$\frac{b}{R} = \sqrt{3} = \frac{\sqrt{3}}{1}$.

(D) The moment of inertia $(I)$ of a single circular disc about an axis passing through its center and perpendicular to its plane is $I_{cm} = \frac{MR^2}{2}$.
For the system of seven discs,the total moment of inertia about the axis passing through the center of the central disc (point $O$) and normal to the plane is the sum of the moment of inertia of the central disc and the six surrounding discs.
$1$. For the central disc: The axis passes through its center,so its moment of inertia is $I_1 = \frac{MR^2}{2}$.
$2$. For the six surrounding discs: The distance between the center of the central disc and the center of any surrounding disc is $d = 2R$. Using the parallel axis theorem,the moment of inertia of each surrounding disc about the central axis is $I_2 = I_{cm} + Md^2 = \frac{MR^2}{2} + M(2R)^2 = \frac{MR^2}{2} + 4MR^2 = \frac{9MR^2}{2}$.
Since there are six such surrounding discs,their total moment of inertia is $6 \times I_2 = 6 \times \frac{9MR^2}{2} = 27MR^2$.
Total moment of inertia $I = I_1 + 6I_2 = \frac{MR^2}{2} + 27MR^2 = \frac{MR^2 + 54MR^2}{2} = \frac{55MR^2}{2}$.

(B) Let $I_0$ be the moment of inertia about the axis passing through the center $O$ and perpendicular to the plane,given by $I_0 = \frac{Ma^2}{6}$.
The distance $h$ from the center $O$ to any corner $A$ is half the length of the diagonal of the square.
The diagonal of the square is $\sqrt{a^2 + a^2} = \sqrt{2}a$.
Therefore,the distance $h = \frac{\sqrt{2}a}{2} = \frac{a}{\sqrt{2}}$.
Using the parallel axis theorem,the moment of inertia $I_A$ about an axis passing through the corner $A$ and parallel to the axis through the center is:
$I_A = I_0 + Mh^2$
$I_A = \frac{Ma^2}{6} + M\left(\frac{a}{\sqrt{2}}\right)^2$
$I_A = \frac{Ma^2}{6} + \frac{Ma^2}{2}$
$I_A = \frac{Ma^2 + 3Ma^2}{6} = \frac{4Ma^2}{6} = \frac{2}{3}Ma^2$.

(D) Let the axis pass through the centre of sphere $1$. The distance between the centres of the two spheres is $2R$.
Using the parallel axis theorem for sphere $1$,the moment of inertia about its own centre is $I_{c1} = \frac{2}{5} M (\frac{R}{2})^2 = \frac{1}{10} MR^2$.
For sphere $2$,the axis is at a distance $d = 2R$ from its centre. Using the parallel axis theorem,$I_2 = I_{c2} + Md^2 = \frac{2}{5} M (\frac{R}{2})^2 + M(2R)^2 = \frac{1}{10} MR^2 + 4MR^2 = \frac{41}{10} MR^2$.
The rod is massless,so its moment of inertia is $0$.
The total moment of inertia $I = I_{c1} + I_2 = \frac{1}{10} MR^2 + \frac{41}{10} MR^2 = \frac{42}{10} MR^2 = \frac{21}{5} MR^2$.

(A) Let the mass of the disc be $M$ and its radius be $R$.
The moment of inertia of a circular disc about an axis passing through its centre and perpendicular to its plane is $I_c = \frac{1}{2}MR^2$.
The radius of gyration $K_c$ is given by $I_c = MK_c^2$,so $MK_c^2 = \frac{1}{2}MR^2$,which gives $K_c = \frac{R}{\sqrt{2}}$.
The moment of inertia of the disc about its diameter is $I_d = \frac{1}{4}MR^2$.
The radius of gyration $K_d$ is given by $I_d = MK_d^2$,so $MK_d^2 = \frac{1}{4}MR^2$,which gives $K_d = \frac{R}{2}$.
The ratio $K_c : K_d = \frac{R/\sqrt{2}}{R/2} = \frac{2}{\sqrt{2}} = \sqrt{2} : 1$.

(A) The moment of inertia of the remaining part of the disc $(I_r)$ is given by the principle of superposition:
$I_r = I_{\text{disc}} - I_{\text{hole}}$
where $I_{\text{disc}}$ is the moment of inertia of the original disc about the central axis,and $I_{\text{hole}}$ is the moment of inertia of the removed circular part about the same axis.
$1$. Moment of inertia of the original disc:
$I_{\text{disc}} = \frac{1}{2} MR^2$
$2$. Properties of the removed hole:
Radius of the hole $r = \frac{R}{2}$.
Since the surface mass density $\sigma = \frac{M}{\pi R^2}$ is uniform,the mass of the hole $M_h$ is:
$M_h = \sigma \cdot \pi r^2 = \left(\frac{M}{\pi R^2}\right) \cdot \pi \left(\frac{R}{2}\right)^2 = \frac{M}{4}$.
$3$. Moment of inertia of the hole about the central axis of the original disc:
Using the parallel axis theorem,$I_{\text{hole}} = I_{\text{cm}} + M_h d^2$,where $I_{\text{cm}} = \frac{1}{2} M_h r^2$ and $d = \frac{R}{2}$ is the distance between the center of the hole and the center of the original disc.
$I_{\text{hole}} = \frac{1}{2} \left(\frac{M}{4}\right) \left(\frac{R}{2}\right)^2 + \left(\frac{M}{4}\right) \left(\frac{R}{2}\right)^2$
$I_{\text{hole}} = \frac{MR^2}{32} + \frac{MR^2}{16} = \frac{MR^2 + 2MR^2}{32} = \frac{3MR^2}{32}$.
$4$. Moment of inertia of the remaining part:
$I_r = \frac{1}{2} MR^2 - \frac{3MR^2}{32} = \frac{16MR^2 - 3MR^2}{32} = \frac{13MR^2}{32}$.

(A) The moment of inertia of the first disc about the central axis is $I_1 = \frac{1}{2} MR^2$.
Since no external torque acts on the system,the angular momentum is conserved: $L_i = L_f$.
Initially,$L_i = I_1 \omega = \frac{1}{2} MR^2 \omega$.
When the second disc of mass $\frac{M}{2}$ is placed on the first,the total moment of inertia becomes $I_2 = I_1 + I_{\text{disc2}} = \frac{1}{2} MR^2 + \frac{1}{2} (\frac{M}{2}) R^2 = \frac{1}{2} MR^2 + \frac{1}{4} MR^2 = \frac{3}{4} MR^2$.
Using conservation of angular momentum: $I_1 \omega = I_2 \omega_2$.
$\frac{1}{2} MR^2 \omega = \frac{3}{4} MR^2 \omega_2$.
Solving for $\omega_2$: $\omega_2 = (\frac{1}{2} \times \frac{4}{3}) \omega = \frac{2}{3} \omega$.

(D) Using the parallel axis theorem,$I_2 = I_{CM} + Mh^2$.
For a circular disc,the moment of inertia about an axis passing through its centre and perpendicular to its plane is $I_1 = I_{CM} = \frac{1}{2}MR^2$.
The distance between the parallel axes is $h = \frac{2R}{3}$.
Substituting these values into the parallel axis theorem:
$I_2 = \frac{1}{2}MR^2 + M\left(\frac{2R}{3}\right)^2$
$I_2 = \frac{1}{2}MR^2 + M\left(\frac{4R^2}{9}\right)$
$I_2 = MR^2 \left(\frac{1}{2} + \frac{4}{9}\right) = MR^2 \left(\frac{9 + 8}{18}\right) = \frac{17}{18}MR^2$.
Now,find the ratio $\frac{I_2}{I_1}$:
$\frac{I_2}{I_1} = \frac{\frac{17}{18}MR^2}{\frac{1}{2}MR^2} = \frac{17}{18} \times 2 = \frac{17}{9}$.
Comparing this with the given ratio $\frac{x}{9}$,we get $x = 17$.

(A) The total mass of the wire is $M = \lambda L$. Since the wire is bent into a circular coil of radius $R$,the circumference is $2 \pi R = L$,so $R = \frac{L}{2 \pi}$.
The moment of inertia of a circular ring about its diameter is $I_{diam} = \frac{1}{2} M R^2$.
Using the parallel axis theorem,the moment of inertia about a tangential axis in the plane of the coil is $I = I_{diam} + M R^2$.
$I = \frac{1}{2} M R^2 + M R^2 = \frac{3}{2} M R^2$.
Substituting $M = \lambda L$ and $R = \frac{L}{2 \pi}$:
$I = \frac{3}{2} (\lambda L) \left( \frac{L}{2 \pi} \right)^2 = \frac{3}{2} \lambda L \left( \frac{L^2}{4 \pi^2} \right) = \frac{3 \lambda L^3}{8 \pi^2}$.

(C) The moment of inertia of a solid sphere about an axis passing through its center is given by $I_{cm} = \frac{2}{5}MR^2$.
In this problem,the axis of rotation for each sphere passes through its own center and is perpendicular to the plane of the spheres.
Since all four spheres are identical (same mass '$M$' and radius '$R$') and the axis of rotation for each sphere is the same (passing through its own center),the moment of inertia for each sphere is independent of the positions of the other spheres.
Therefore,$I_{A} = I_{B} = I_{C} = I_{D} = \frac{2}{5}MR^2$.
However,if the question implies the moment of inertia of the *entire system* about an axis passing through a specific point,or if the axes are different,the interpretation changes. Given the standard interpretation of such problems where the axis passes through the center of each individual sphere,all moments of inertia are equal. If the question implies the moment of inertia of the *entire system* about an axis passing through the center of each sphere,then $I_A = I_B = I_C = I_D$.

(B) The moment of inertia of a single solid sphere about its own central axis is $I_{cm} = \frac{2}{5} MR^2$. Using the parallel axis theorem,$I = I_{cm} + Md^2$,where $d$ is the distance from the axis of rotation.
For the axis passing through the center of sphere $A$:
$I_A = I_A' + I_B' + I_C' + I_D'$
$I_A = \frac{2}{5} MR^2 + (\frac{2}{5} MR^2 + M(2R)^2) + (\frac{2}{5} MR^2 + M(4R)^2) + (\frac{2}{5} MR^2 + M(6R)^2)$
$I_A = \frac{8}{5} MR^2 + M(4R^2 + 16R^2 + 36R^2) = 1.6 MR^2 + 56 MR^2 = 57.6 MR^2$.
For the axis passing through the center of sphere $B$:
$I_B = I_A' + I_B' + I_C' + I_D'$
$I_B = (\frac{2}{5} MR^2 + M(2R)^2) + \frac{2}{5} MR^2 + (\frac{2}{5} MR^2 + M(2R)^2) + (\frac{2}{5} MR^2 + M(4R)^2)$
$I_B = \frac{8}{5} MR^2 + M(4R^2 + 4R^2 + 16R^2) = 1.6 MR^2 + 24 MR^2 = 25.6 MR^2$.
The difference is $|I_A - I_B| = 57.6 MR^2 - 25.6 MR^2 = 32 MR^2$.

(C) Given: Length of pendulum $l = 2 \ m$,angular displacement $\theta = 60^{\circ}$,mass $m = 200 \ g = 0.2 \ kg$,acceleration due to gravity $g = 10 \ m/s^2$.
When the pendulum bob moves in a horizontal circle,the radius of the circular path is $r = l \sin \theta$.
For the bob to move in a horizontal circle,the forces acting on it are tension $T$ and weight $mg$.
The vertical component of tension balances the weight: $T \cos \theta = mg$ ... $(i)$
The horizontal component of tension provides the necessary centripetal force: $T \sin \theta = m r \omega^2 = m (l \sin \theta) \omega^2$ ... (ii)
Dividing equation (ii) by equation $(i)$:
$\frac{T \sin \theta}{T \cos \theta} = \frac{m l \sin \theta \omega^2}{mg}$
$\tan \theta = \frac{l \omega^2 \sin \theta}{g}$
Since $\tan \theta = \frac{\sin \theta}{\cos \theta}$,we get:
$\frac{1}{\cos \theta} = \frac{l \omega^2}{g}$
$\omega^2 = \frac{g}{l \cos \theta}$
Substituting the values: $\omega^2 = \frac{10}{2 \times \cos 60^{\circ}} = \frac{10}{2 \times 0.5} = \frac{10}{1} = 10$
$\omega = \sqrt{10} = \sqrt{4 \times 2.5} = 2 \sqrt{2.5} \ rad/s$.

(D) The centripetal force $F$ acting on a particle of mass $m$ moving in a circular path of radius $r$ with angular velocity $\omega$ is given by $F = m \omega^2 r$.
We know that angular momentum $L = I \omega$,where $I = mr^2$ is the moment of inertia of the particle.
Therefore,$\omega = \frac{L}{I} = \frac{L}{mr^2}$.
Substituting the value of $\omega$ into the centripetal force equation:
$F = m \left( \frac{L}{mr^2} \right)^2 r$
$F = m \left( \frac{L^2}{m^2 r^4} \right) r$
$F = \frac{L^2}{mr^3}$.

(C) The angular displacement $\theta_n$ covered in the $n^{\text{th}}$ second for rotational motion is given by the formula: $\theta_n = \omega_0 + \frac{\alpha}{2}(2n - 1)$.
Given that the particle starts from rest,the initial angular velocity $\omega_0 = 0$.
The angular displacement in the $P^{\text{th}}$ second is given as $\beta$.
Substituting these values into the formula: $\beta = 0 + \frac{\alpha}{2}(2P - 1)$.
Rearranging the equation to solve for the angular acceleration $\alpha$: $\alpha = \frac{2 \beta}{(2P - 1)}$.

(C) Density is given by $\rho = \frac{M}{V}$. Since all objects are made of the same material,$\rho$ is the same. Given that the mass $M$ is also the same,the volume $V$ of all objects must be the same.
For a constant volume,the surface area $A$ of a thin circular plate is the maximum among a sphere,a cube,and a plate.
According to Stefan-Boltzmann law,the rate of heat loss is $\frac{dQ}{dt} \propto A(T^4 - T_0^4)$.
Since the surface area $A_{\text{plate}}$ is the largest,the rate of cooling is highest for the plate.
Therefore,the circular plate will reach room temperature the fastest.

(B) The total incident heat energy rate is $P_i = 1000 \ J \ min^{-1}$.
According to the law of conservation of energy for heat radiation,the sum of the coefficients of reflection $(r)$,absorption $(a)$,and transmission $(t)$ is equal to $1$,i.e.,$r + a + t = 1$.
Given $r = 0.1$ and $a = 0.8$,we can find the coefficient of transmission $(t)$:
$t = 1 - (r + a) = 1 - (0.1 + 0.8) = 1 - 0.9 = 0.1$.
The rate of transmitted heat energy is $P_t = t \times P_i = 0.1 \times 1000 \ J \ min^{-1} = 100 \ J \ min^{-1}$.
For a time duration of $5 \ minutes$,the total transmitted heat energy $(Q_t)$ is:
$Q_t = P_t \times \text{time} = 100 \ J \ min^{-1} \times 5 \ min = 500 \ J$.

(B) Convection is a mode of heat transfer that occurs in fluids (liquids and gases) due to the actual movement of the molecules of the medium.
In the case of heating a room by a heater,the air near the heater gets warm,becomes less dense,and rises,while the cooler,denser air moves in to take its place. This continuous circulation of air,known as convection currents,heats the entire room.
Heating of a copper utensil and an iron rod are examples of conduction,while heat transfer from the sun to the earth occurs via radiation.
Therefore,the correct option is $B$.

(B) According to Wien's Displacement Law,$\lambda_{\max} T = b$,where $b$ is Wien's constant. Therefore,$T = \frac{b}{\lambda_{\max}}$.
According to the Stefan-Boltzmann Law,the total emissive power $E$ of a black body is proportional to the fourth power of its absolute temperature: $E = \sigma T^4$.
Substituting the expression for $T$,we get $E = \sigma \left( \frac{b}{\lambda_{\max}} \right)^4$.
Let the initial wavelength be $\lambda_1 = \lambda$ and the final wavelength be $\lambda_2 = \frac{2 \lambda}{3}$.
Since $E \propto \frac{1}{\lambda_{\max}^4}$,we can write the ratio of the emissive powers as:
$\frac{E'}{E} = \left( \frac{\lambda_1}{\lambda_2} \right)^4$.
Substituting the values: $\frac{E'}{E} = \left( \frac{\lambda}{\frac{2 \lambda}{3}} \right)^4 = \left( \frac{3}{2} \right)^4$.
Calculating the value: $\frac{E'}{E} = \frac{81}{16}$.
Therefore,the new emissive power is $E' = \frac{81}{16} E$.

(D) According to Stefan-Boltzmann law,the power radiated by a black body is proportional to the fourth power of its absolute temperature: $E \propto T^4$.
Let the initial temperature be $T_1 = T$ and the final temperature be $T_2 = T + 0.5T = 1.5T$.
The initial radiation is $E_1 = \sigma T^4$.
The final radiation is $E_2 = \sigma (1.5T)^4 = \sigma (5.0625) T^4 = 5.0625 E_1$.
The percentage increase in radiation is given by $\frac{E_2 - E_1}{E_1} \times 100$.
Substituting the values: $\frac{5.0625 E_1 - E_1}{E_1} \times 100 = 4.0625 \times 100 = 406.25 \%$.
Rounding to the nearest given option,the increase is approximately $400 \%$.

(B) The power radiated by a black body is given by Stefan-Boltzmann Law: $P = \sigma A T^4$,where $A = 4\pi r^2$ is the surface area of the sphere.
For the first black body: $P_1 = \sigma (4\pi r_1^2) T_1^4$.
For the second black body: $P_2 = \sigma (4\pi r_2^2) T_2^4$.
The ratio of the powers is given as $\frac{P_1}{P_2} = \frac{1}{2}$.
Substituting the expressions: $\frac{1}{2} = \frac{\sigma 4\pi r_1^2 T_1^4}{\sigma 4\pi r_2^2 T_2^4} = \frac{r_1^2 T_1^4}{r_2^2 T_2^4}$.
Rearranging for the ratio of radii: $\frac{r_1^2}{r_2^2} = \frac{1}{2} \left(\frac{T_2}{T_1}\right)^4$.
Taking the square root on both sides: $\frac{r_1}{r_2} = \frac{1}{\sqrt{2}} \left(\frac{T_2}{T_1}\right)^2$.

(B) According to the Stefan-Boltzmann law,the heat radiated per unit area per unit time is given by $E = e \sigma T^4$,where $e$ is the emissivity,$\sigma$ is the Stefan-Boltzmann constant,and $T$ is the absolute temperature.
Given that the heat radiated per unit area per unit time is the same for both bodies,we have $e_1 \sigma T_1^4 = e_2 \sigma T_2^4$.
Since $\sigma$ is a constant,this simplifies to $e_1 T_1^4 = e_2 T_2^4$.
Given the ratio of emissivities $e_1 : e_2 = 1 : 3$,we have $\frac{e_1}{e_2} = \frac{1}{3}$.
Rearranging the equation for the ratio of temperatures: $\left(\frac{T_1}{T_2}\right)^4 = \frac{e_2}{e_1} = \frac{3}{1}$.
Taking the fourth root on both sides,we get $\frac{T_1}{T_2} = \left(\frac{3}{1}\right)^{1/4} = 3^{1/4} : 1$.

(A) black body is an ideal body that absorbs and emits all incident electromagnetic radiation. According to Planck's Law of black body radiation,the intensity of radiation emitted by a black body depends on the wavelength and temperature.
$1$. The intensity is not the same for all wavelengths; it follows a specific distribution curve (Planck's curve) that peaks at a particular wavelength depending on the temperature.
$2$. Therefore,the statement 'For all wavelengths,intensity is same' is incorrect.
$3$. As temperature increases,the peak intensity shifts towards shorter wavelengths,but the intensity varies continuously across the spectrum.

(D) According to the Stefan-Boltzmann law,the rate of radiation $E$ from a black body is given by $E = A \sigma T^4$,where $A$ is the surface area,$\sigma$ is the Stefan-Boltzmann constant,and $T$ is the absolute temperature.
For a sphere,the surface area $A = 4 \pi R^2$.
Thus,$E = (4 \pi R^2) \sigma T^4$,which implies $E \propto R^2 T^4$.
Let the initial state be $E_1 = E$,$R_1 = R$,and $T_1 = T$.
Let the final state be $E_2$,$R_2 = R/3$,and $T_2 = 3T$.
Taking the ratio: $\frac{E_2}{E_1} = \left( \frac{R_2}{R_1} \right)^2 \left( \frac{T_2}{T_1} \right)^4$.
Substituting the values: $\frac{E_2}{E} = \left( \frac{R/3}{R} \right)^2 \left( \frac{3T}{T} \right)^4$.
$\frac{E_2}{E} = \left( \frac{1}{3} \right)^2 \times (3)^4 = \frac{1}{9} \times 81 = 9$.
Therefore,$E_2 = 9E$.

(C) According to Wien's displacement law,$\lambda T = b$ (constant),so $T \propto \frac{1}{\lambda}$.
Since the new wavelength is $\lambda' = \frac{\lambda}{3}$,the new temperature $T'$ becomes $T' = 3T$.
The emissive power of a black body is given by Stefan-Boltzmann law: $E = \sigma T^4$.
Therefore,the new emissive power $E'$ is:
$E' = \sigma (T')^4 = \sigma (3T)^4 = 81 \sigma T^4$.
Since $E = \sigma T^4$,we get $E' = 81 E$.

(A) According to Stefan-Boltzmann's Law,the rate of heat radiation $E$ is proportional to the fourth power of the absolute temperature $T$:
$E = \sigma T^4$
$\Rightarrow E \propto T^4$
Given:
$T_1 = 127^{\circ} C = 127 + 273 = 400 \ K$
$E_1 = 5 \ cal / cm^2 \ s$
$T_2 = 927^{\circ} C = 927 + 273 = 1200 \ K$
Using the ratio:
$\frac{E_2}{E_1} = \left(\frac{T_2}{T_1}\right)^4$
$\frac{E_2}{5} = \left(\frac{1200}{400}\right)^4$
$\frac{E_2}{5} = (3)^4$
$\frac{E_2}{5} = 81$
$E_2 = 81 \times 5 = 405 \ cal / cm^2 \ s$

(A) For any body,the sum of the coefficients of absorption $(a)$,reflection $(r)$,and transmission $(t)$ is always equal to $1$,i.e.,$a + r + t = 1$.
An opaque body is defined as a body that does not allow any radiation to pass through it.
Therefore,the transmission coefficient $t$ for an opaque body is $0$.
Substituting $t=0$ into the equation $a + r + t = 1$,we get $a + r = 1$.
Thus,for an opaque body,$t=0$ and $a+r=1$.

(C) The change in volume $\Delta V$ for a solid due to thermal expansion is given by $\Delta V = V \gamma \Delta T$,where $V$ is the initial volume,$\gamma$ is the coefficient of volume expansion,and $\Delta T$ is the change in temperature.
Since both rods are made of brass,$\gamma$ is the same for both. Given that $\Delta T$ is also the same,the ratio of the increase in volume is equal to the ratio of the initial volumes.
Initial volume of rod $A$: $V_A = \pi (2r)^2 l = 4 \pi r^2 l$.
Initial volume of rod $B$: $V_B = \pi (r)^2 (2l) = 2 \pi r^2 l$.
The ratio of the increase in volume is $\frac{\Delta V_A}{\Delta V_B} = \frac{V_A}{V_B} = \frac{4 \pi r^2 l}{2 \pi r^2 l} = \frac{2}{1}$.

(B) The initial volume of the cube is $V = L^3 = (1 \ m)^3 = 1 \ m^3$.
The change in temperature is $\Delta T = 100^{\circ} C - 0^{\circ} C = 100^{\circ} C$.
The coefficient of volume expansion $\gamma$ is related to the coefficient of linear expansion $\alpha$ by $\gamma = 3\alpha$.
Given $\alpha = 18 \times 10^{-6} /^{\circ} C$,we have $\gamma = 3 \times 18 \times 10^{-6} = 54 \times 10^{-6} /^{\circ} C$.
The change in volume is given by $\Delta V = V \gamma \Delta T$.
Substituting the values: $\Delta V = 1 \times (54 \times 10^{-6}) \times 100 = 54 \times 10^{-4} \ m^3$.

(C) The work done to blow a soap bubble of radius $r$ 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 soap bubble has two surfaces,$\Delta A = 2 \times (4 \pi r^2) = 8 \pi r^2$.
For the first bubble of radius $R$: $W_1 = 8 \pi R^2 T_1$.
For the second bubble of radius $2R$: $W_2 = 8 \pi (2R)^2 T_2 = 32 \pi R^2 T_2$.
If the temperature remained constant $(T_1 = T_2)$,the work $W_2$ would be exactly $4 W_1$.
However,heating the solution decreases the surface tension $(T_2 < T_1)$.
Therefore,$W_2 = 4 W_1 \times (T_2 / T_1)$. Since $T_2 < T_1$,it follows that $W_2 < 4 W_1$.

$(A)$ The formula for the coefficient of linear expansion is given by $\alpha = \frac{\Delta L}{L_1 \Delta T}$.
Given:
Initial length $L_1 = 2 \,m$.
Change in length $\Delta L = 1.6 \,mm = 1.6 \times 10^{-3} \,m$.
Change in temperature $\Delta T = 60^{\circ} C - 0^{\circ} C = 60^{\circ} C$.
Substituting these values into the formula:
$\alpha = \frac{1.6 \times 10^{-3}}{2 \times 60}$
$\alpha = \frac{1.6 \times 10^{-3}}{120}$
$\alpha = \frac{1.6}{120} \times 10^{-3} = 0.01333 \times 10^{-3} = 1.33 \times 10^{-5} /{ }^{\circ} C$.
Therefore, the correct option is $A$.

(A) Given:
Rate of flow of heat (conduction rate) $P_{\text{cond}} = 1600 \text{ cal/s}$.
Temperature difference $\Delta T = 40^{\circ} C$.
The formula for thermal resistance $R_T$ is given by the ratio of temperature difference to the rate of heat flow:
$R_T = \frac{\Delta T}{P_{\text{cond}}}$
Substituting the given values:
$R_T = \frac{40}{1600}$
$R_T = \frac{1}{40} = 0.025^{\circ} C s/cal$.
Thus,the thermal resistance is $0.025^{\circ} C s/cal$.

(B) Given: The difference in length $\Delta l = l_{A} - l_{B} = 60 \text{ cm}$ is constant at all temperatures.
This implies that the change in length for both rods must be equal for any temperature change $\Delta T$.
$\Delta l_{A} = \Delta l_{B}$
$l_{A} \alpha_{A} \Delta T = l_{B} \alpha_{B} \Delta T$
$l_{A} \alpha_{A} = l_{B} \alpha_{B}$
Substituting the given values:
$l_{A} (18 \times 10^{-6}) = l_{B} (27 \times 10^{-6})$
$l_{A} (18) = l_{B} (27)$
$l_{A} = \frac{27}{18} l_{B} = 1.5 l_{B}$
We know $l_{A} - l_{B} = 60 \text{ cm}$.
Substituting $l_{A} = 1.5 l_{B}$:
$1.5 l_{B} - l_{B} = 60 \text{ cm}$
$0.5 l_{B} = 60 \text{ cm}$
$l_{B} = 120 \text{ cm}$
Then,$l_{A} = 1.5 \times 120 = 180 \text{ cm}$.

(C) The fractional change in volume is given by $\frac{\Delta V}{V} = \frac{0.225}{100} = 0.00225$.
The change in temperature is $\Delta T = 30^{\circ} C$.
We know that the coefficient of volume expansion $\gamma$ is related to the coefficient of linear expansion $\alpha$ by the relation $\gamma = 3\alpha$.
The formula for volume expansion is $\Delta V = V \gamma \Delta T$,which can be rewritten as $\frac{\Delta V}{V} = \gamma \Delta T$.
Substituting $\gamma = 3\alpha$,we get $\frac{\Delta V}{V} = 3\alpha \Delta T$.
Substituting the given values: $0.00225 = 3 \times \alpha \times 30$.
Solving for $\alpha$: $\alpha = \frac{0.00225}{90} = 2.5 \times 10^{-5} /{ }^{\circ} C$.

(B) According to the kinetic theory of gases,the pressure $P$ exerted by an ideal gas is given by the formula: $P = \frac{1}{3} \rho v_{rms}^2$,where $\rho$ is the density and $v_{rms}$ is the root mean square speed.
Since $\rho = \frac{M}{V}$,we can write $P = \frac{1}{3} \frac{N m}{V} v_{rms}^2$.
Also,the mean square speed is defined as $\langle v^2 \rangle = \frac{1}{N} \sum v_i^2 = v_{rms}^2$.
Therefore,the pressure $P$ is directly proportional to the mean of the square of the speeds of the gas molecules,which is $\langle v^2 \rangle$.

(D) The thermal resistance $R$ of a slab is given by $R = \frac{L}{KA}$,where $L$ is the thickness,$K$ is the thermal conductivity,and $A$ is the area.
For the first material: $R_1 = \frac{x}{KA}$.
For the second material: $R_2 = \frac{4x}{(2K)A} = \frac{2x}{KA}$.
Since the slabs are in series,the equivalent thermal resistance $R_{eq}$ is:
$R_{eq} = R_1 + R_2 = \frac{x}{KA} + \frac{2x}{KA} = \frac{3x}{KA}$.
The rate of heat transfer $\frac{dQ}{dt}$ is given by:
$\frac{dQ}{dt} = \frac{T_2 - T_1}{R_{eq}} = \frac{T_2 - T_1}{\frac{3x}{KA}} = \frac{KA(T_2 - T_1)}{3x}$.
Comparing this with the given expression $\left[\frac{A(T_2 - T_1)K}{x}\right] \cdot f$,we get:
$f = \frac{1}{3}$.

(B) The temperature gradient is defined as the rate of change of temperature with respect to distance,given by $T_g = \frac{T_h - T_c}{L}$,where $T_h$ is the temperature of the hotter end,$T_c$ is the temperature of the cooler end,and $L$ is the length of the rod.
Given:
Length $L = 75 \ cm = 0.75 \ m$
Temperature gradient $T_g = 40^{\circ} C/m$
Cooler end temperature $T_c = 10^{\circ} C$
Substituting the values into the formula:
$40 = \frac{T_h - 10}{0.75}$
$T_h - 10 = 40 \times 0.75$
$T_h - 10 = 30$
$T_h = 30 + 10 = 40^{\circ} C$
Therefore,the temperature of the hotter end is $40^{\circ} C$.

(C) According to Stefan-Boltzmann law,the rate of loss of heat $dQ/dt$ is given by $dQ/dt = e \sigma A (T^4 - T_0^4)$.
Since both objects are black,made of the same material,and cooling in the same atmosphere at the same temperature,the rate of loss of heat is directly proportional to the surface area $A$.
Let $V$ be the volume of both objects. For a sphere of radius $r$,$V = \frac{4}{3} \pi r^3$. For a cube of side $a$,$V = a^3$.
Since $V$ is equal,$a^3 = \frac{4}{3} \pi r^3$,which implies $a = (\frac{4}{3} \pi r^3)^{1/3}$.
The surface area of the sphere is $A_s = 4 \pi r^2$.
The surface area of the cube is $A_c = 6 a^2 = 6 (\frac{4}{3} \pi r^3)^{2/3}$.
The ratio of the rate of loss of heat is $\frac{A_s}{A_c} = \frac{4 \pi r^2}{6 (\frac{4}{3} \pi r^3)^{2/3}}$.
Simplifying this,$\frac{A_s}{A_c} = \frac{4 \pi r^2}{6 (\frac{4}{3})^{2/3} \pi^{2/3} r^2} = \frac{4^{1/3} \pi^{1/3}}{6 (\frac{1}{3})^{2/3}} = \frac{4^{1/3} \pi^{1/3}}{6 \cdot 3^{-2/3}} = \frac{4^{1/3} \pi^{1/3} \cdot 3^{2/3}}{6} = \frac{(4 \cdot 9)^{1/3} \pi^{1/3}}{6} = \frac{(36)^{1/3} \pi^{1/3}}{6} = (\frac{36 \pi}{216})^{1/3} = (\frac{\pi}{6})^{1/3}$.

(C) According to Newton's law of cooling,the rate of cooling $R$ is directly proportional to the temperature difference between the body and its surroundings: $R = k(\theta - \theta_0)$.
Given:
Case $1$: $R_1 = 4^{\circ}C/min$,$\theta_1 = 90^{\circ}C$
Case $2$: $R_2 = 1^{\circ}C/min$,$\theta_2 = 30^{\circ}C$
Taking the ratio of the two rates:
$\frac{R_1}{R_2} = \frac{\theta_1 - \theta_0}{\theta_2 - \theta_0}$
$\frac{4}{1} = \frac{90 - \theta_0}{30 - \theta_0}$
$4(30 - \theta_0) = 90 - \theta_0$
$120 - 4\theta_0 = 90 - \theta_0$
$120 - 90 = 4\theta_0 - \theta_0$
$30 = 3\theta_0$
$\theta_0 = 10^{\circ}C$
Thus,the temperature of the surroundings is $10^{\circ}C$.

$(A)$ According to the first law of thermodynamics, $Q = \Delta U + W$, where $W = P \Delta V$.
Given: Pressure $P = 50 \,N/m^2$, initial volume $V_1 = 10 \,m^3$, final volume $V_2 = 4 \,m^3$, and heat added $Q = 100 \,J$.
The change in volume is $\Delta V = V_2 - V_1 = 4 - 10 = -6 \,m^3$.
The work done on the gas is $W = P \Delta V = 50 \times (-6) = -300 \,J$.
Using the first law: $Q = \Delta U + W$, we have $100 = \Delta U + (-300)$.
Therefore, $\Delta U = 100 + 300 = 400 \,J$.
Since $\Delta U$ is positive, the internal energy is increased by $400 \,J$.

(A) The potential at any point $P$ due to a system of charges is given by $V = \sum \frac{1}{4 \pi \varepsilon_0} \frac{q_i}{r_i}$.
In the isosceles triangle $ABC$,charges $+q$ are placed at $A, B,$ and $C$. $D$ is the midpoint of $AB$ and $E$ is the midpoint of $AC$.
By symmetry,the distance of $D$ from $A$ is $a$,from $B$ is $a$,and from $C$ is $\sqrt{AC^2 + AD^2 - 2(AC)(AD) \cos A} = \sqrt{(2a)^2 + a^2 - 2(2a)(a) \cos A}$.
Similarly,the distance of $E$ from $A$ is $a$,from $C$ is $a$,and from $B$ is $\sqrt{AB^2 + AE^2 - 2(AB)(AE) \cos A} = \sqrt{(2a)^2 + a^2 - 2(2a)(a) \cos A}$.
Since the configuration is symmetric with respect to the angle $A$,the potential at $D$ $(V_D)$ and the potential at $E$ $(V_E)$ are equal.
$V_D = \frac{1}{4 \pi \varepsilon_0} \left( \frac{q}{AD} + \frac{q}{BD} + \frac{q}{CD} \right) = \frac{1}{4 \pi \varepsilon_0} \left( \frac{q}{a} + \frac{q}{a} + \frac{q}{CD} \right)$.
$V_E = \frac{1}{4 \pi \varepsilon_0} \left( \frac{q}{AE} + \frac{q}{CE} + \frac{q}{BE} \right) = \frac{1}{4 \pi \varepsilon_0} \left( \frac{q}{a} + \frac{q}{a} + \frac{q}{BE} \right)$.
Since $CD = BE$ by symmetry,$V_D = V_E$.
The work done $W = Q(V_E - V_D) = Q(0) = 0$.

(D) The electric potential $V$ at a distance $r$ from a point charge $Q$ is given by $V = \frac{1}{4 \pi \varepsilon_0} \frac{Q}{r}$.
Since both points $A$ and $B$ are at the same distance $r$ from the charge $Q$,they lie on the same equipotential surface.
The potential at $A$ $(V_A)$ is equal to the potential at $B$ $(V_B)$.
The work done in moving a charge $q$ from $A$ to $B$ is given by $W = q(V_B - V_A)$.
Since $V_B = V_A$,the work done $W = q(0) = 0$.

(C) The total charge $q$ on the surface of the conducting sphere is given by $q = \sigma A$,where $\sigma$ is the surface charge density and $A = 4 \pi R^2$ is the surface area of the sphere.
Given: $\sigma = 1.8 \times 10^{-6} \ C/m^2$,$R = 0.1 \ m$.
$q = (1.8 \times 10^{-6}) \times (4 \pi \times (0.1)^2) = 1.8 \times 10^{-6} \times 4 \pi \times 0.01 = 7.2 \pi \times 10^{-8} \ C$.
The electric field $E$ at a distance $d = 0.2 \ m$ from the center is given by $E = \frac{1}{4 \pi \varepsilon_0} \frac{q}{d^2}$.
Substituting the values: $E = \frac{1}{4 \pi \varepsilon_0} \times \frac{7.2 \pi \times 10^{-8}}{(0.2)^2}$.
$E = \frac{1}{4 \pi \varepsilon_0} \times \frac{7.2 \pi \times 10^{-8}}{0.04} = \frac{1.8 \pi \times 10^{-8}}{\pi \varepsilon_0} = \frac{1.8 \times 10^{-8}}{\varepsilon_0} \ Vm^{-1}$.
Rounding to the nearest provided option,we get $E \approx \frac{2 \times 10^{-7}}{\varepsilon_0} \ Vm^{-1}$ (Note: The calculation yields $1.8 \times 10^{-8}$,but based on the options provided,the closest magnitude is selected).

(C) The dipole moment $p$ is given by the product of the magnitude of one charge $q$ and the separation distance $2l$.
Given: $q = 2 \times 10^{-6} \text{ C}$,$2l = 3 \text{ cm} = 3 \times 10^{-2} \text{ m}$.
$p = q \times 2l = (2 \times 10^{-6} \text{ C}) \times (3 \times 10^{-2} \text{ m}) = 6 \times 10^{-8} \text{ C-m}$.
The torque $\tau$ on a dipole in an electric field $E$ is given by $\tau = pE \sin \theta$.
For maximum torque,$\theta = 90^{\circ}$,so $\sin 90^{\circ} = 1$.
$\tau_{\max} = pE = (6 \times 10^{-8} \text{ C-m}) \times (2 \times 10^5 \text{ N/C}) = 12 \times 10^{-3} \text{ N-m}$.

(D) The work done in rotating an electric dipole in an electric field is given by $W = pE(\cos \theta_1 - \cos \theta_2)$.
Initially,the dipole is parallel to the field,so $\theta_1 = 0^{\circ}$.
For the first case,rotating through $180^{\circ}$ means $\theta_2 = 180^{\circ}$.
$w = pE(\cos 0^{\circ} - \cos 180^{\circ}) = pE(1 - (-1)) = 2pE$.
Therefore,$pE = \frac{w}{2}$.
For the second case,rotating through $60^{\circ}$ means $\theta_2 = 60^{\circ}$.
$W' = pE(\cos 0^{\circ} - \cos 60^{\circ}) = pE(1 - \frac{1}{2}) = pE(\frac{1}{2})$.
Substituting $pE = \frac{w}{2}$ into the equation:
$W' = (\frac{w}{2}) \times (\frac{1}{2}) = \frac{w}{4}$.

(D) For a short electric dipole of dipole moment $p$,the electric field intensity at a point on the axial line at a distance $r$ from the centre is given by $E_{a} = \frac{1}{4\pi\epsilon_{0}} \frac{2p}{r^{3}}$.
The electric field intensity at a point on the equatorial line at the same distance $r$ from the centre is given by $E_{q} = \frac{1}{4\pi\epsilon_{0}} \frac{p}{r^{3}}$.
Comparing the two expressions,we get $E_{a} = 2 \times (\frac{1}{4\pi\epsilon_{0}} \frac{p}{r^{3}}) = 2E_{q}$.
Therefore,the correct relation is $E_{a} = 2E_{q}$.

(C) To find the electric field at a distance $R$ where $A < R < B$, we use Gauss's Law.
Consider a Gaussian surface as a sphere of radius $R$ centered at the origin.
The total charge enclosed by this Gaussian surface is only the charge on the inner solid sphere, which is $+3Q$.
According to Gauss's Law, the electric field $E$ at a distance $R$ is given by:
$E = \frac{1}{4 \pi \varepsilon_0} \frac{q_{\text{enclosed}}}{R^2}$
Substituting $q_{\text{enclosed}} = 3Q$, we get:
$E = \frac{1}{4 \pi \varepsilon_0} \frac{3Q}{R^2} = \frac{3Q}{4 \pi \varepsilon_0 R^2}$

(D) According to Coulomb's Law,the force between two charges in a vacuum or air is given by $F = \frac{1}{4\pi\epsilon_0} \frac{q_1 q_2}{d^2}$.
When the spheres are immersed in a dielectric medium with dielectric constant $k$,the force becomes $F' = \frac{1}{4\pi\epsilon_0 k} \frac{q_1 q_2}{d^2}$.
Therefore,the relationship between the new force $F'$ and the original force $F$ is $F' = \frac{F}{k}$.
Given that the dielectric constant $k = 5$,the new force is $F' = \frac{F}{5}$.

(A) The electric field intensity due to a short dipole on the axial line is given by $E_{\text{axial}} = \frac{1}{4\pi\epsilon_0} \frac{2p}{r_1^3}$.
The electric field intensity due to a short dipole on the equatorial line is given by $E_{\text{equatorial}} = \frac{1}{4\pi\epsilon_0} \frac{p}{r_2^3}$.
Given that $E_{\text{axial}} = E_{\text{equatorial}}$,we have:
$\frac{1}{4\pi\epsilon_0} \frac{2p}{r_1^3} = \frac{1}{4\pi\epsilon_0} \frac{p}{r_2^3}$.
Simplifying the equation:
$\frac{2}{r_1^3} = \frac{1}{r_2^3}$.
Rearranging for the ratio $\frac{r_1^3}{r_2^3} = 2$.
Taking the cube root on both sides,we get $\frac{r_1}{r_2} = 2^{1/3} = \sqrt[3]{2}$.
Therefore,$r_1: r_2 = \sqrt[3]{2}: 1$.

(C) For a conductor in electrostatic equilibrium,the electric field just outside the surface must be perpendicular to the surface at every point. If it had a tangential component,charges would experience a force and move along the surface,which contradicts the static condition. Therefore,the statement that the field must be tangential is '$WRONG$'.

(A) Consider a small element of length $dl = r d\theta$ at an angle $\theta$ with the vertical axis. The charge on this element is $dq = \lambda dl = \lambda r d\theta$.
The electric field $dE$ at the centre due to this element is $dE = \frac{1}{4 \pi \varepsilon_0} \frac{dq}{r^2} = \frac{\lambda d\theta}{4 \pi \varepsilon_0 r}$.
By symmetry,the horizontal components of the electric field cancel out. We only need to integrate the vertical component $dE \cos \theta$ from $-\pi/2$ to $\pi/2$.
$E = \int_{-\pi/2}^{\pi/2} \frac{\lambda \cos \theta d\theta}{4 \pi \varepsilon_0 r} = \frac{\lambda}{4 \pi \varepsilon_0 r} [\sin \theta]_{-\pi/2}^{\pi/2}$.
$E = \frac{\lambda}{4 \pi \varepsilon_0 r} (1 - (-1)) = \frac{2 \lambda}{4 \pi \varepsilon_0 r} = \frac{\lambda}{2 \pi \varepsilon_0 r}$.

(A) According to Gauss's theorem,the electric flux through a closed surface is given by $\phi = E \cdot A = \frac{q_{enc}}{\varepsilon_0}$.
For a solid sphere of radius $r$ with uniform volume charge density $\rho$,the total enclosed charge $q_{enc}$ is the product of density and volume:
$q_{enc} = \rho \times V = \rho \left( \frac{4}{3} \pi r^3 \right)$.
The surface area of the sphere is $A = 4 \pi r^2$.
Substituting these into Gauss's law:
$E (4 \pi r^2) = \frac{\rho (\frac{4}{3} \pi r^3)}{\varepsilon_0}$.
Dividing both sides by $4 \pi r^2$,we get:
$E = \frac{\rho r}{3 \varepsilon_0}$.

(A) The surface charge density $\sigma$ is given by the ratio of charge $q$ to area $A$: $\sigma = \frac{q}{A} = \frac{17.7 \times 10^{-4} \ C}{200 \ m^2} = 8.85 \times 10^{-6} \ C/m^2$.
For a large non-conducting sheet,the electric field intensity $E$ at any point near it is independent of the distance and is given by the formula: $E = \frac{\sigma}{2\varepsilon_0}$.
Substituting the values: $E = \frac{8.85 \times 10^{-6}}{2 \times 8.85 \times 10^{-12}} = \frac{10^{-6}}{2 \times 10^{-12}} = 0.5 \times 10^6 = 5 \times 10^5 \ N/C$.
Thus,the correct option is $A$.

(A) The electric force $F$ on a charge $q$ in an electric field $E$ is given by $F = qE$.
Thus, the electric field strength is $E = F/q = 3000 \, N / 3 \, C = 1000 \, N/C$.
The potential difference $V$ between two points separated by a distance $d$ in a uniform electric field is given by $V = Ed$.
Given $d = 1 \, cm = 10^{-2} \, m$.
Substituting the values, $V = 1000 \, N/C \times 10^{-2} \, m = 10 \, V$.

(A) The relationship between electric field $\vec{E}$ and electric potential $V$ is given by the relation: $dV = -\vec{E} \cdot d\vec{r}$.
Given $\vec{E} = 30x^2 \hat{i}$ and $d\vec{r} = dx \hat{i}$,we have $dV = -30x^2 dx$.
To find the potential difference $(V_A - V_0)$,we integrate from $x = 0$ to $x = 2$:
$\int_{V_0}^{V_A} dV = -\int_{0}^{2} 30x^2 dx$.
$V_A - V_0 = -[10x^3]_{0}^{2}$.
$V_A - V_0 = -(10(2)^3 - 10(0)^3) = -(10 \times 8) = -80 \ V$.

(C) The energy stored in a capacitor is given by the formula $U = \frac{1}{2} CV^2$.
Given for the first capacitor: $C_1 = 900 \mu F$ and $V_1 = 100 \ V$.
The energy stored in the first capacitor is $U_1 = \frac{1}{2} C_1 V_1^2$.
When this energy is transferred to a second capacitor of capacity $C_2 = 100 \mu F$, let the new potential be $V_u$.
Since the total energy is transferred, $U_1 = U_2$, where $U_2 = \frac{1}{2} C_2 V_u^2$.
Equating the energies: $\frac{1}{2} C_1 V_1^2 = \frac{1}{2} C_2 V_u^2$.
Substituting the values: $\frac{1}{2} \times 900 \times 10^{-6} \times (100)^2 = \frac{1}{2} \times 100 \times 10^{-6} \times V_u^2$.
Simplifying the equation: $900 \times 100^2 = 100 \times V_u^2$.
$V_u^2 = \frac{900 \times 10000}{100} = 90000$.
$V_u = \sqrt{90000} = 300 \ V$.

(A) The potential energy $U$ of a system of point charges is given by the sum of the potential energies of all pairs of charges: $U = \sum \frac{k q_i q_j}{r_{ij}}$.
For an equilateral triangle with side length $r$, the potential energy of the system is:
$U = \frac{k(q)(2q)}{r} + \frac{k(q)(Q)}{r} + \frac{k(2q)(Q)}{r} = 0$
Dividing by $\frac{k}{r}$ (assuming $k \neq 0$ and $r \neq 0$):
$2q^2 + qQ + 2qQ = 0$
$2q^2 + 3qQ = 0$
$3qQ = -2q^2$
$Q = -\frac{2q}{3}$

(C) The circuit consists of two parallel branches connected between points $P$ and $R$. The total current entering at $P$ is $I = 5 \, A$.
Branch $1$ (upper) has resistors $6 \, \Omega$ and $3 \, \Omega$ in series. Total resistance $R_1 = 6 + 3 = 9 \, \Omega$.
Branch $2$ (lower) has resistors $8 \, \Omega$ and $4 \, \Omega$ in series. Total resistance $R_2 = 8 + 4 = 12 \, \Omega$.
Using the current divider rule, the current $I_1$ flowing through the upper branch is:
$I_1 = I \times \frac{R_2}{R_1 + R_2} = 5 \times \frac{12}{9 + 12} = 5 \times \frac{12}{21} = 5 \times \frac{4}{7} \approx 2.857 \, A$.
The potential at $P$ relative to $Q$ is the voltage drop across the $6 \, \Omega$ resistor:
$V_{PQ} = I_1 \times 6 = 2.857 \times 6 \approx 17.14 \, V$.
Thus, the potential difference is nearly $17 \, V$.

(D) The potential energy $U$ of a system of $n$ point charges is given by the formula:
$U = \frac{1}{4 \pi \varepsilon_0} \sum_{\text{all pairs}} \frac{q_j q_k}{r_{jk}}$
For a system of three charges $q_1 = q$,$q_2 = -2q$,and $q_3 = q$ placed at positions $x = 0$,$x = r$,and $x = 2r$ respectively:
$U = \frac{1}{4 \pi \varepsilon_0} \left[ \frac{q_1 q_2}{r_{12}} + \frac{q_2 q_3}{r_{23}} + \frac{q_1 q_3}{r_{13}} \right]$
Substituting the values:
$U = \frac{1}{4 \pi \varepsilon_0} \left[ \frac{(q)(-2q)}{r} + \frac{(-2q)(q)}{r} + \frac{(q)(q)}{2r} \right]$
$U = \frac{1}{4 \pi \varepsilon_0} \left[ -\frac{2q^2}{r} - \frac{2q^2}{r} + \frac{q^2}{2r} \right]$
$U = \frac{1}{4 \pi \varepsilon_0} \left[ -\frac{4q^2}{r} + \frac{q^2}{2r} \right] = \frac{1}{4 \pi \varepsilon_0} \left[ \frac{-8q^2 + q^2}{2r} \right]$
$U = \frac{1}{4 \pi \varepsilon_0} \left( -\frac{7q^2}{2r} \right) = -\frac{7q^2}{8 \pi \varepsilon_0 r}$

(A) Let the total current entering at $P$ be $I = 5 \,A$. The circuit is a bridge network. Let the potential at $P$ be $V_P$ and at the output node be $V_R = 0 \,V$.
We need to find the potential at $Q$ $(V_Q)$.
The upper branch has resistors $6 \,\Omega$ and $3 \,\Omega$ in series. The lower branch has resistors $8 \,\Omega$ and $4 \,\Omega$ in series.
Let $I_1$ be the current in the upper branch and $I_2$ be the current in the lower branch.
$I_1 = I \times \frac{R_{lower}}{R_{upper} + R_{lower}} = 5 \times \frac{8+4}{(6+3) + (8+4)} = 5 \times \frac{12}{9+12} = 5 \times \frac{12}{21} = 5 \times \frac{4}{7} \approx 2.857 \,A$.
The potential at $Q$ relative to $P$ is the voltage drop across the $6 \,\Omega$ resistor.
$V_P - V_Q = I_1 \times 6 = 2.857 \times 6 = 17.14 \,V$.
Thus,the potential difference between $P$ and $Q$ is approximately $17 \,V$.

(A) The net electrostatic potential energy $U$ of a system of point charges is given by the sum of the potential energies of all pairs of charges: $U = \sum \frac{k q_i q_j}{r_{ij}}$.
For the given configuration,the charges are $+Q$,$+2q$,and $+q$. The distances between them are $a$,$a$,and $\sqrt{a^2 + a^2} = a\sqrt{2}$.
The total potential energy is:
$U = \frac{1}{4 \pi \varepsilon_0} \left( \frac{Q(2q)}{a} + \frac{Q(q)}{a} + \frac{(2q)(q)}{a\sqrt{2}} \right) = 0$
Since $\frac{1}{4 \pi \varepsilon_0} \neq 0$,we have:
$\frac{2Qq}{a} + \frac{Qq}{a} + \frac{2q^2}{a\sqrt{2}} = 0$
$\frac{3Qq}{a} + \frac{\sqrt{2}q^2}{a} = 0$
$3Qq + \sqrt{2}q^2 = 0$
$3Q = -\sqrt{2}q$
$Q = -\frac{\sqrt{2}}{3} q$

(C) The electrostatic potential energy $U$ of a system of two point charges $q_1$ and $q_2$ separated by a distance $r$ is given by $U = \frac{K q_1 q_2}{r}$.
Initial potential energy $U_i$ at distance $d$ is $U_i = \frac{K q_1 q_2}{d}$.
When charge $q_2$ is moved towards $q_1$ by a distance $x$,the new separation becomes $r' = d - x$.
The final potential energy $U_f$ is $U_f = \frac{K q_1 q_2}{d - x}$.
The increase in potential energy $\Delta U$ is given by $\Delta U = U_f - U_i$.
Substituting the values:
$\Delta U = \frac{K q_1 q_2}{d - x} - \frac{K q_1 q_2}{d}$
$\Delta U = K q_1 q_2 \left( \frac{1}{d - x} - \frac{1}{d} \right)$
$\Delta U = K q_1 q_2 \left( \frac{d - (d - x)}{d(d - x)} \right)$
$\Delta U = \frac{K q_1 q_2 x}{d(d - x)}$.

(A) The total $EMF$ of the series combination is $E_{eq} = E + E = 2E$.
The total resistance of the circuit is $R_{total} = r_1 + r_2 + R$.
The current $I$ flowing through the circuit is given by $I = \frac{2E}{r_1 + r_2 + R}$.
The potential difference across cell $E_1$ is given by $V_1 = E - Ir_1$.
Given that $V_1 = 0$,we have $E - Ir_1 = 0$,which implies $E = Ir_1$.
Substituting the value of $I$,we get $E = \left( \frac{2E}{r_1 + r_2 + R} \right) r_1$.
Dividing both sides by $E$,we get $1 = \frac{2r_1}{r_1 + r_2 + R}$.
Rearranging the terms,$r_1 + r_2 + R = 2r_1$.
Therefore,$R = 2r_1 - r_1 - r_2 = r_1 - r_2$.

(C) $1$. Gravitational force is much weaker than electrostatic force. Thus,option $A$ is incorrect.
$2$. Gravitational force is always attractive,but electrostatic force can be attractive or repulsive depending on the charges. Thus,option $B$ is incorrect.
$3$. Both gravitational force (Newton's Law of Gravitation) and electrostatic force (Coulomb's Law) are central forces,meaning they act along the line joining the centers of the two objects. Thus,option $C$ is correct.
$4$. Both gravitational force and electrostatic force follow the inverse square law $(F \propto \frac{1}{r^2})$. Thus,option $D$ is incorrect.

(B) Area of the square loop $A = 25 \, cm^2 = 25 \times 10^{-4} \, m^2$.
Side length $l = \sqrt{A} = 5 \, cm = 0.05 \, m$.
Velocity $v = \frac{l}{t} = \frac{0.05 \, m}{1 \, s} = 0.05 \, m/s$.
Induced electromotive force $(EMF)$ $e = B l v$.
Induced current $I = \frac{e}{R} = \frac{B l v}{R}$.
Substituting the values: $I = \frac{40 \times 0.05 \times 0.05}{10} = 0.01 \, A$.
The magnetic force on the conductor is $F = B I l$.
$F = 40 \times 0.01 \times 0.05 = 0.02 \, N$.
Work done $W = F \times l = 0.02 \, N \times 0.05 \, m = 1 \times 10^{-3} \, J$.

(D) Given: Power $P = 1 \text{ kW} = 1000 \text{ W}$,Voltage $V = 100 \text{ V}$,Distance $a = 2 \text{ mm} = 2 \times 10^{-3} \text{ m}$.
First,calculate the current $I$ flowing through the wires using $P = VI$:
$I = \frac{P}{V} = \frac{1000}{100} = 10 \text{ A}$.
The force per unit length $f$ between two parallel wires carrying current $I$ is given by the formula:
$f = \frac{\mu_0 I^2}{2\pi a}$.
Substituting the values:
$f = \frac{(4\pi \times 10^{-7}) \times (10)^2}{2\pi \times (2 \times 10^{-3})}$.
$f = \frac{4\pi \times 10^{-7} \times 100}{4\pi \times 10^{-3}}$.
$f = 10^{-7} \times 10^2 \times 10^3 = 10^{-2} \text{ N/m}$.
Thus,the force per metre is $10^{-2} \text{ N}$.

(C) The force per unit length between two parallel long conductors is given by the formula: $F = \frac{\mu_0 I_1 I_2}{2 \pi d}$.
Initially,the force is $F = \frac{\mu_0 I_1 I_2}{2 \pi d}$.
When the current in one conductor is doubled $(I_1' = 2I_1)$ and its direction is reversed,the new current becomes $-2I_1$. The distance is increased to $d' = 3d$.
The new force $F'$ is given by: $F' = \frac{\mu_0 (-2I_1) I_2}{2 \pi (3d)}$.
Simplifying this,we get: $F' = -\frac{2}{3} \left( \frac{\mu_0 I_1 I_2}{2 \pi d} \right)$.
Substituting the initial force $F$,we get: $F' = -\frac{2F}{3}$.

(B) The force per unit length between two parallel current-carrying wires is given by $F = \frac{\mu_0 I_1 I_2}{2 \pi r}$.
For wire $A$ and $B$: Both carry current $I$ in the same direction. The force $F_1$ exerted by $B$ on $A$ is attractive,with magnitude $F_1 = \frac{\mu_0 I^2}{2 \pi d}$.
For wire $A$ and $C$: Wire $A$ carries current $I$ upwards and wire $C$ carries current $2I$ downwards. Since the currents are in opposite directions,the force $F_2$ exerted by $C$ on $A$ is repulsive,with magnitude $F_2 = \frac{\mu_0 I (2I)}{2 \pi (2d)} = \frac{\mu_0 I^2}{2 \pi d}$.
Since the forces $F_1$ and $F_2$ have equal magnitudes but act in opposite directions (one is attractive and the other is repulsive),we have $F_1 = -F_2$.

(C) The magnetic field $B$ inside a long solenoid is given by $B = \mu_0 n I_s$, where $n = \frac{N_s}{L}$.
Given: $N_s = 500$, $L = 0.4 \ m$, $I_s = 3 \ A$.
$B = (4\pi \times 10^{-7}) \times (500 / 0.4) \times 3 = 4\pi \times 10^{-7} \times 1250 \times 3 = 15000\pi \times 10^{-7} = 1.5\pi \times 10^{-3} \ T$.
The torque $\tau$ on a coil in a magnetic field is $\tau = N_c I_c A B \sin \theta$.
Given: $N_c = 10$, $I_c = 0.4 \ A$, $r = 0.1 \ m$, $\theta = 90^{\circ}$.
Area $A = \pi r^2 = \pi (0.1)^2 = 0.01\pi \ m^2$.
$\tau = 10 \times 0.4 \times (0.01\pi) \times (1.5\pi \times 10^{-3}) \times \sin 90^{\circ}$.
$\tau = 4 \times 0.01 \times 1.5 \times \pi^2 \times 10^{-3} \times 1$.
Using $\pi^2 = 10$:
$\tau = 0.06 \times 10 \times 10^{-3} = 0.6 \times 10^{-3} = 6 \times 10^{-4} \ Nm$.

(B) The induced electromotive force $(EMF)$ in a wire moving through a magnetic field is given by $e = B_H \cdot l \cdot v$.
Since the wire falls freely from a height $h = 10 \ m$,its velocity $v$ just before hitting the ground is given by $v = \sqrt{2gh}$.
Substituting the values: $v = \sqrt{2 \times 10 \times 10} = \sqrt{200} = 10\sqrt{2} \ m/s$.
Now,calculate the induced $EMF$: $e = (2 \times 10^{-5} \ T) \times (2500 \ \text{m}) \times (10\sqrt{2} \ m/s)$.
$e = 50000 \times 10^{-5} \times \sqrt{2} = 0.5 \sqrt{2} \ V$.
The induced current $I$ is given by $I = \frac{e}{R}$.
$I = \frac{0.5 \sqrt{2}}{25 \sqrt{2}} = \frac{0.5}{25} = 0.02 \ A$.

(D) The magnetic field $B$ at a distance $R$ from a long straight current-carrying wire is given by the formula:
$B = \frac{\mu_0 I}{2\pi R}$
Rearranging the formula to solve for $R$:
$R = \frac{\mu_0 I}{2\pi B}$
Given values:
$I = 12 \ A$
$B = 3 \times 10^{-5} \ Wb/m^2$
$\mu_0 = 4\pi \times 10^{-7} \ Wb/Am$
Substituting the values into the equation:
$R = \frac{4\pi \times 10^{-7} \times 12}{2\pi \times 3 \times 10^{-5}}$
$R = \frac{2 \times 10^{-7} \times 12}{3 \times 10^{-5}}$
$R = \frac{24 \times 10^{-7}}{3 \times 10^{-5}}$
$R = 8 \times 10^{-2} \ m$
To convert meters to millimeters $(mm)$,multiply by $1000$:
$R = 8 \times 10^{-2} \times 10^3 \ mm = 80 \ mm$

(D) The motional electromotive force $(emf)$ induced in a conductor of length $L$ moving with velocity $V$ in a magnetic field $B$ is given by $e = BLV$.
The current $I$ flowing through the circuit with resistance $R$ is $I = \frac{e}{R} = \frac{BLV}{R}$.
The rate of production of heat energy (power dissipated) in the loop is given by $P = I^2 R$.
Substituting the value of $I$:
$P = \left( \frac{BLV}{R} \right)^2 R = \frac{B^2 L^2 V^2}{R^2} \cdot R = \frac{B^2 L^2 V^2}{R}$.
Alternatively,the mechanical power required to pull the loop is $P = F \cdot V$,where $F = BIL$ is the magnetic force.
$P = (B \cdot I \cdot L) \cdot V = B \cdot \left( \frac{BLV}{R} \right) \cdot L \cdot V = \frac{B^2 L^2 V^2}{R}$.

(A) The magnetic field produced by a long straight wire at a distance $X$ is given by $B = \frac{\mu_0 I}{2 \pi X}$.
Let $X$ be the distance from point $P$ to each wire. Since the lines joining $P$ to the wires are perpendicular to each other,the triangle formed by the two wires and point $P$ is a right-angled isosceles triangle with hypotenuse $7 \,cm$.
Using the Pythagorean theorem: $X^2 + X^2 = 7^2 \implies 2X^2 = 49 \implies X = \frac{7}{\sqrt{2}} \,cm = \frac{7}{1.4} \times 10^{-2} \,m = 5 \times 10^{-2} \,m$.
The magnetic fields due to the two wires are $B_1 = \frac{\mu_0 I_1}{2 \pi X}$ and $B_2 = \frac{\mu_0 I_2}{2 \pi X}$.
Since the directions of the currents are opposite,the magnetic field vectors at $P$ are perpendicular to each other. Thus,the net magnetic field is $B_{\text{net}} = \sqrt{B_1^2 + B_2^2} = \frac{\mu_0}{2 \pi X} \sqrt{I_1^2 + I_2^2}$.
Substituting the values: $B_{\text{net}} = \frac{4 \pi \times 10^{-7}}{2 \pi \times 5 \times 10^{-2}} \sqrt{15^2 + 8^2} = \frac{2 \times 10^{-7}}{5 \times 10^{-2}} \sqrt{225 + 64} = \frac{2 \times 10^{-5}}{5} \sqrt{289} = 0.4 \times 10^{-5} \times 17 = 6.8 \times 10^{-5} \,T = 68 \times 10^{-6} \,T$.

(B) The energy $E$ stored in an inductor is given by the formula $E = \frac{1}{2} LI^2$.
Given: Inductance $L = 100 mH = 100 \times 10^{-3} H = 0.1 H$ and Current $I = 1 A$.
Substituting the values:
$E = \frac{1}{2} \times 0.1 \times (1)^2$
$E = 0.05 J$.

(C) Given: $n = 500 \ turns/m$,$I = 3 \ A$,$\mu_r = 5001$.
First,calculate the magnetic field intensity $H$ inside the solenoid:
$H = nI = 500 \times 3 = 1500 \ A/m$.
Next,calculate the magnetic susceptibility $\chi_m$:
$\chi_m = \mu_r - 1 = 5001 - 1 = 5000$.
The magnitude of magnetization $M$ is given by the formula:
$M = \chi_m H$.
Substituting the values:
$M = 5000 \times 1500 = 7.5 \times 10^6 \ Am^{-1}$.

(D) The magnetic field at the centre of a circular coil is given by the formula $B = \frac{\mu_0 I}{2R}$.
For the first coil: $B_1 = \frac{\mu_0 \times 2}{2R} = \frac{\mu_0}{R}$.
For the second coil: $B_2 = \frac{\mu_0 \times 4}{2(3R)} = \frac{2\mu_0}{3R}$.
Taking the ratio: $\frac{B_1}{B_2} = \frac{\mu_0 / R}{2\mu_0 / 3R} = \frac{\mu_0}{R} \times \frac{3R}{2\mu_0} = \frac{3}{2}$.
Thus,the ratio $B_1: B_2$ is $3: 2$.

(D) The magnetic field $B$ at a distance $r$ from a long straight wire carrying current $I$ is given by $B = \frac{\mu_0 I}{2 \pi r}$.
Given the distance between the wires is $d = 3 \ m$,the midpoint is at a distance $r = 1.5 \ m$ from each wire.
For the first wire,$I_1 = 3 \ A$. The magnetic field is $B_1 = \frac{\mu_0 \times 3}{2 \pi \times 1.5} = \frac{2 \mu_0}{2 \pi}$.
For the second wire,$I_2 = 4.5 \ A$. The magnetic field is $B_2 = \frac{\mu_0 \times 4.5}{2 \pi \times 1.5} = \frac{3 \mu_0}{2 \pi}$.
Since the currents are in opposite directions,by the right-hand rule,the magnetic fields produced by both wires at the midpoint point in the same direction.
Therefore,the net magnetic field is $B = B_1 + B_2 = \frac{2 \mu_0}{2 \pi} + \frac{3 \mu_0}{2 \pi} = \frac{5 \mu_0}{2 \pi}$.

(B) The inductance of a solenoid is given by $L = \frac{\mu_0 N^2 A}{l}$, where $l$ is the length of the solenoid and $A = \pi r^2$ is the cross-sectional area.
Let $x$ be the total length of the wire required. The wire is wound into $N$ turns, so $x = (2 \pi r) N$, which implies $N = \frac{x}{2 \pi r}$.
Substituting $N$ into the inductance formula:
$L = \frac{\mu_0 (x / 2 \pi r)^2 \times \pi r^2}{l} = \frac{\mu_0 x^2}{4 \pi l}$.
Rearranging for $x$:
$x^2 = \frac{4 \pi L l}{\mu_0} \implies x = \sqrt{\frac{4 \pi L l}{\mu_0}}$.
Given $L = 1 \,mH = 10^{-3} \,H$, $l = 1 \,m$, and $\mu_0 = 4 \pi \times 10^{-7} \,T \cdot m/A$:
$x = \sqrt{\frac{4 \pi \times 10^{-3} \times 1}{4 \pi \times 10^{-7}}} = \sqrt{10^4} = 100 \,m$.
Converting to kilometers: $100 \,m = 0.10 \,km$.

(B) The magnetic force on a current-carrying wire is given by $\vec{F} = I(\vec{L} \times \vec{B})$, where $L$ is the length vector of the wire.
Let the sides of the triangle be $EF = 5 \,cm$, $EG = 12 \,cm$, and $GF = 13 \,cm$.
The magnetic field $\vec{B}$ is parallel to the side $GF$. Let $\theta$ be the angle between the side $EF$ and the side $GF$.
From the geometry of the triangle, $\sin \theta = \frac{EG}{GF} = \frac{12}{13}$.
The force on the side $EF$ is $F_{EF} = I \cdot L_{EF} \cdot B \sin \theta$.
Given $I = 2 \,A$, $L_{EF} = 0.05 \,m$, $B = 0.75 \,T$, and $\sin \theta = \frac{12}{13}$.
$F_{EF} = 2 \times 0.05 \times 0.75 \times \frac{12}{13} = 0.1 \times 0.75 \times \frac{12}{13} = 0.075 \times \frac{12}{13} = \frac{0.9}{13} = \frac{9}{130} \,N$.
Comparing this with $\frac{x}{130} \,N$, we get $x = 9$.

(D) The magnetic field at the centre of a circular coil of radius $R$ carrying current $I$ is given by $B = \frac{\mu_0 I}{2 R}$.
For the first coil with current $I$,the magnetic field is $B_1 = \frac{\mu_0 I}{2 R}$.
For the second coil with current $2I$,the magnetic field is $B_2 = \frac{\mu_0 (2I)}{2 R} = \frac{\mu_0 I}{R}$.
Since the planes of the two coils are at right angles to each other,the magnetic fields $B_1$ and $B_2$ are perpendicular to each other.
The resultant magnetic field $B$ at the centre is given by $B = \sqrt{B_1^2 + B_2^2}$.
Substituting the values:
$B = \sqrt{\left(\frac{\mu_0 I}{2 R}\right)^2 + \left(\frac{\mu_0 I}{R}\right)^2}$
$B = \sqrt{\left(\frac{\mu_0 I}{2 R}\right)^2 + \left(\frac{2 \mu_0 I}{2 R}\right)^2}$
$B = \frac{\mu_0 I}{2 R} \sqrt{1^2 + 2^2}$
$B = \frac{\sqrt{5} \mu_0 I}{2 R}$.

(A) The inductance $L$ of a long solenoid is given by the formula:
$L = \frac{\mu_0 N^2 A}{l}$
Where $N$ is the total number of turns,$A$ is the cross-sectional area,and $l$ is the length.
Since the cross-sectional area $A = \pi r^2 = \pi (d/2)^2 = \frac{\pi d^2}{4}$,we substitute this into the equation:
$L = \mu_0 \left(\frac{N}{l}\right)^2 \times l \times \frac{\pi d^2}{4}$
Given that $n = N/l$ is the number of turns per unit length,the inductance per unit length is:
$\frac{L}{l} = \mu_0 n^2 \left(\frac{\pi d^2}{4}\right)$
$\frac{L}{l} = \mu_0 \pi \left(\frac{nd}{2}\right)^2$

(A) The current $I$ produced by an electron moving in a circle is given by $I = \frac{q}{T}$,where $q$ is the charge of the electron and $T$ is the time period.
Given $q = 1.6 \times 10^{-19} \ C$ and $T = 1 \ s$,we have $I = \frac{1.6 \times 10^{-19}}{1} = 1.6 \times 10^{-19} \ A$.
The magnetic field $B$ at the centre of a circular current loop is given by $B = \frac{\mu_0 I}{2r}$.
Substituting the values: $B = \frac{4 \pi \times 10^{-7} \times 1.6 \times 10^{-19}}{2 \times 0.8}$.
$B = \frac{4 \pi \times 10^{-7} \times 1.6 \times 10^{-19}}{1.6} = 4 \pi \times 10^{-26} \ T$.

(B) The magnetic field at the center of a circular coil is given by $B = \frac{\mu_0 n I}{2r}$.
Since the currents are in opposite directions, the net magnetic field is the difference between the fields produced by each coil: $B_{\text{net}} = |B_1 - B_2|$.
Given: $n_1 = n_2 = 10$, $r_1 = 0.2 \, m$, $r_2 = 0.4 \, m$, $I_1 = 0.2 \, A$, $I_2 = 0.3 \, A$.
$B_1 = \frac{\mu_0 \times 10 \times 0.2}{2 \times 0.2} = \frac{10 \mu_0}{2} = 5 \mu_0$.
$B_2 = \frac{\mu_0 \times 10 \times 0.3}{2 \times 0.4} = \frac{3 \mu_0}{0.8} = 3.75 \mu_0$.
$B_{\text{net}} = 5 \mu_0 - 3.75 \mu_0 = 1.25 \mu_0$.
Substituting $\mu_0 = 4 \pi \times 10^{-7} \, T \cdot m/A$:
$B_{\text{net}} = 1.25 \times 4 \pi \times 10^{-7} = 5 \pi \times 10^{-7} \, T$.

(C) The magnetic field at the centre of a circular coil is given by $B = \frac{\mu_0 I}{2R}$.
For the net magnetic field at the common centre to be zero,the magnetic fields produced by coils $A$ and $B$ must be equal in magnitude and opposite in direction.
Since coil $A$ has a current in the anticlockwise direction,its magnetic field points out of the plane. Therefore,coil $B$ must produce a magnetic field pointing into the plane,which requires a clockwise current.
Let $I_A = 0.5 \text{ A}$,$R_A = 0.2 \text{ m}$,$R_B = 0.1 \text{ m}$,and $I_B$ be the current in coil $B$.
Setting the magnitudes equal: $\frac{\mu_0 I_A}{2 R_A} = \frac{\mu_0 I_B}{2 R_B}$.
Substituting the values: $\frac{0.5}{0.2} = \frac{I_B}{0.1}$.
$I_B = \frac{0.5 \times 0.1}{0.2} = 0.25 \text{ A}$.
Thus,the current in coil $B$ is $0.25 \text{ A}$ in the clockwise direction.

(C) The magnetic field at a distance '$h$' along the axis of a circular coil is given by:
$B_{a} = \frac{\mu_0 n I r^2}{2(r^2 + h^2)^{3/2}}$
The magnetic field at the centre of the coil $(h=0)$ is:
$B_{c} = \frac{\mu_0 n I}{2r}$
Taking the ratio of $B_c$ to $B_a$:
$\frac{B_c}{B_a} = \frac{\mu_0 n I / 2r}{\mu_0 n I r^2 / 2(r^2 + h^2)^{3/2}}$
$\frac{B_c}{B_a} = \frac{(r^2 + h^2)^{3/2}}{r^3} = \left(\frac{r^2 + h^2}{r^2}\right)^{3/2} = \left(1 + \frac{h^2}{r^2}\right)^{3/2}$
Therefore,$B_c = B_a \left(1 + \frac{h^2}{r^2}\right)^{3/2}$.

(B) The magnetic induction at the centre of a circular current-carrying coil is given by $B_{C} = \frac{\mu_{0} I}{2 R}$.
The magnetic induction at a distance $r$ from the centre along the axis is given by $B_{A} = \frac{\mu_{0} I R^{2}}{2(R^{2} + r^{2})^{3/2}}$.
Given that $r = \sqrt{3} R$,we substitute this into the formula for $B_{A}$:
$B_{A} = \frac{\mu_{0} I R^{2}}{2(R^{2} + (\sqrt{3} R)^{2})^{3/2}} = \frac{\mu_{0} I R^{2}}{2(R^{2} + 3 R^{2})^{3/2}} = \frac{\mu_{0} I R^{2}}{2(4 R^{2})^{3/2}}$.
Simplifying the denominator: $(4 R^{2})^{3/2} = (2^{2} R^{2})^{3/2} = (2 R)^{3} = 8 R^{3}$.
So,$B_{A} = \frac{\mu_{0} I R^{2}}{2(8 R^{3})} = \frac{\mu_{0} I}{16 R}$.
Now,calculating the ratio $B_{A} : B_{C}$:
$\frac{B_{A}}{B_{C}} = \frac{\frac{\mu_{0} I}{16 R}}{\frac{\mu_{0} I}{2 R}} = \frac{2 R}{16 R} = \frac{1}{8}$.
Thus,the ratio is $1: 8$.

(D) The radius of the $n^{\text{th}}$ Bohr orbit is given by $r_n \propto n^2$.
The velocity of the electron in the $n^{\text{th}}$ orbit is $v_n \propto \frac{1}{n}$.
The angular velocity is $\omega_n = \frac{v_n}{r_n} \propto \frac{1/n}{n^2} = \frac{1}{n^3}$.
The current $I_n$ produced by the revolving electron is $I_n = \frac{e}{T_n} = \frac{e \omega_n}{2 \pi} \propto \omega_n \propto \frac{1}{n^3}$.
The magnetic field at the centre of the orbit is $B_n = \frac{\mu_0 I_n}{2 r_n}$.
Substituting the proportionalities: $B_n \propto \frac{I_n}{r_n} \propto \frac{1/n^3}{n^2} = \frac{1}{n^5}$.
Therefore,$B_n \propto n^{-5}$.

(C) The magnetic field at the center of a circular coil is given by $B = \frac{\mu_0 I}{2r}$.
Given that the magnetic fields at the centers are equal,we have $\frac{\mu_0 I_1}{2 r_1} = \frac{\mu_0 I_2}{2 r_2}$.
This implies $\frac{I_1}{I_2} = \frac{r_1}{r_2}$. Since $r_1 = 2r_2$,we get $\frac{I_1}{I_2} = 2$ (Equation $i$).
The resistance of a wire is given by $R = \rho \frac{l}{A}$. Since the coils are made from the same wire,$\rho$ and $A$ are constant,so $R \propto l$.
The length of the wire for a circular coil is $l = 2\pi r$,so $R \propto r$.
Therefore,$\frac{R_1}{R_2} = \frac{r_1}{r_2} = 2$ (Equation $ii$).
The potential difference is $V = IR$. The ratio of potential differences is $\frac{V_1}{V_2} = \frac{I_1 R_1}{I_2 R_2}$.
Substituting the ratios from (Equation $i$) and (Equation $ii$),we get $\frac{V_1}{V_2} = 2 \times 2 = 4$.

(C) The magnetic field at the centre of a circular coil with $N$ turns,radius $r$,and current $I$ is given by $B = \frac{\mu_0 NI}{2r}$.
For the first case,$N_1 = 1$,radius is $r_1$,so $B_1 = \frac{\mu_0 I}{2r_1}$.
For the second case,$N_2 = n$,radius is $r_2$,so $B_2 = \frac{\mu_0 nI}{2r_2}$.
The total length of the wire $L$ remains constant. Thus,$L = 2\pi r_1 = n(2\pi r_2)$.
This implies $r_1 = n r_2$,or $\frac{r_2}{r_1} = \frac{1}{n}$.
Taking the ratio of the magnetic fields:
$\frac{B_1}{B_2} = \frac{\mu_0 I / 2r_1}{\mu_0 nI / 2r_2} = \frac{1}{n} \cdot \frac{r_2}{r_1} = \frac{1}{n} \cdot \frac{1}{n} = \frac{1}{n^2}$.
Therefore,the ratio is $1:n^2$.

(A) The magnetic field at the centre of a semicircular arc of radius $R$ carrying current $I$ is given by $B = \frac{\mu_0 I}{4R}$.
In the given figure,the currents in the two semicircular arcs flow in the same sense (clockwise) relative to the centre $O$.
Therefore,the magnetic fields produced by both arcs at the centre $O$ are in the same direction (into the plane of the paper).
The total magnetic field at the centre $O$ is the sum of the individual magnetic fields:
$B_{\text{net}} = B_1 + B_2$
$B_{\text{net}} = \frac{\mu_0 I}{4R_1} + \frac{\mu_0 I}{4R_2}$
$B_{\text{net}} = \frac{\mu_0 I}{4} \left( \frac{1}{R_1} + \frac{1}{R_2} \right)$