MHT CET 2025 Physics Question Paper with Answer and Solution

(D) According to the principle of conservation of angular momentum,the total angular momentum of the system remains constant when the two discs are coupled.
Initial angular momentum $L_i = I_1 \omega_1 + I_2 \omega_2$.
When the discs rotate together,they form a single system with a combined moment of inertia $I = I_1 + I_2$ and a common angular velocity $\omega$.
The final angular momentum is $L_f = (I_1 + I_2) \omega$.
Equating $L_i = L_f$,we get $\omega = \frac{I_1 \omega_1 + I_2 \omega_2}{I_1 + I_2}$.
The rotational kinetic energy $K$ of the system is given by $K = \frac{1}{2} I \omega^2$.
Substituting the values of $I$ and $\omega$:
$K = \frac{1}{2} (I_1 + I_2) \left( \frac{I_1 \omega_1 + I_2 \omega_2}{I_1 + I_2} \right)^2$.
$K = \frac{(I_1 \omega_1 + I_2 \omega_2)^2}{2(I_1 + I_2)}$.

(A) Let the initial moment of inertia be $I_1$ and the final moment of inertia be $I_2 = 0.75 I_1 = \frac{3}{4} I_1$.
Since no external torque acts on the system,the angular momentum $L$ is conserved,so $L = I_1 \omega_1 = I_2 \omega_2$.
Thus,$\omega_2 = \frac{I_1}{I_2} \omega_1 = \frac{I_1}{0.75 I_1} \omega_1 = \frac{4}{3} \omega_1$.
The initial rotational kinetic energy is $K_1 = \frac{L^2}{2 I_1}$.
The final rotational kinetic energy is $K_2 = \frac{L^2}{2 I_2} = \frac{L^2}{2 (0.75 I_1)} = \frac{L^2}{1.5 I_1} = \frac{4}{3} K_1$.
The percentage increase in kinetic energy is $\frac{K_2 - K_1}{K_1} \times 100 = (\frac{4}{3} - 1) \times 100 = \frac{1}{3} \times 100 \approx 33.3 \%$.

(A) The angular momentum $L$ of a rotating body is given by $L = I\omega$,where $I$ is the moment of inertia and $\omega$ is the angular velocity. The kinetic energy $K$ is given by $K = \frac{1}{2} I\omega^2 = \frac{L^2}{2I}$.
Given: Initial state is $L_1 = L$,$K_1 = K$,$\omega_1 = \omega$,$I_1 = I$.
New state: Frequency $f_2 = 3f_1$,which implies $\omega_2 = 3\omega_1 = 3\omega$. Kinetic energy $K_2 = \frac{1}{3} K_1 = \frac{K}{3}$.
Using $K = \frac{L^2}{2I}$,we have $I = \frac{L^2}{2K}$.
For the new state: $I_2 = \frac{L_2^2}{2K_2}$.
Also,$L_2 = I_2 \omega_2 = I_2 (3\omega)$.
From $K_2 = \frac{1}{2} I_2 \omega_2^2$,we have $\frac{K}{3} = \frac{1}{2} I_2 (3\omega)^2 = \frac{9}{2} I_2 \omega^2$.
Since $K = \frac{1}{2} I \omega^2$,we substitute this: $\frac{1}{3} (\frac{1}{2} I \omega^2) = \frac{9}{2} I_2 \omega^2$.
Solving for $I_2$: $I_2 = \frac{I}{27}$.
Now,calculate the new angular momentum $L_2 = I_2 \omega_2 = (\frac{I}{27}) (3\omega) = \frac{I\omega}{9} = \frac{L}{9}$.

(B) Let the two spheres be $S_1$ and $S_2$. The axis of rotation passes through the centre of $S_1$ and is perpendicular to the rod.
For sphere $S_1$,the moment of inertia about its own centre is $I_1 = \frac{2}{5} MR^2$.
For sphere $S_2$,the distance of its centre from the axis of rotation is $d = R + 4R + R = 6R$ is incorrect based on the diagram. The rod length is $4R$ between the surfaces,or the distance between centres is $4R$. Given the diagram shows the rod length $4R$ between the spheres,the distance between the centres is $d = R + 4R + R = 6R$. However,if the rod length $4R$ is the distance between centres,then $d = 4R$.
Assuming the distance between centres is $4R$:
For $S_1$,$I_1 = \frac{2}{5} MR^2$.
For $S_2$,using the parallel axis theorem,$I_2 = I_{cm} + Md^2 = \frac{2}{5} MR^2 + M(4R)^2 = \frac{2}{5} MR^2 + 16MR^2 = \frac{82}{5} MR^2$.
Total moment of inertia $I = I_1 + I_2 = \frac{2}{5} MR^2 + \frac{82}{5} MR^2 = \frac{84}{5} MR^2$.

(C) The moment of inertia $I$ of a disc about its central axis is given by $I = \frac{1}{2} M R^2$.
Since the discs are made of the same material and have the same thickness $t$,their mass $M$ is given by $M = \text{density} \times \text{volume} = \rho \times (\pi R^2 t)$.
Substituting $M$ into the formula for $I$,we get $I = \frac{1}{2} (\rho \pi R^2 t) R^2 = \frac{1}{2} \rho \pi t R^4$.
Since $\rho$,$\pi$,and $t$ are constant for both discs,$I \propto R^4$.
Therefore,the ratio of the moments of inertia is $\frac{I_A}{I_B} = \left( \frac{R_A}{R_B} \right)^4$.
Given $R_A = R$ and $R_B = 3R$,we have $\frac{I_A}{I_B} = \left( \frac{R}{3R} \right)^4 = \left( \frac{1}{3} \right)^4 = \frac{1}{81}$.
Thus,the ratio is $1: 81$.

(C) The moment of inertia of a solid sphere of mass $m$ and radius $r$ about its diameter is $I_{cm} = \frac{2}{5} mr^2$.
For sphere $1$,the axis $YY^1$ passes through its center (diameter). Thus,$I_1 = \frac{2}{5} mr^2$.
For spheres $2$ and $3$,the axis $YY^1$ is tangent to them. The distance of the center of these spheres from the axis $YY^1$ is $r$. Using the parallel axis theorem,$I = I_{cm} + md^2$,where $d = r$.
So,$I_2 = I_3 = \frac{2}{5} mr^2 + mr^2 = \frac{7}{5} mr^2$.
The total moment of inertia of the system is $I_{total} = I_1 + I_2 + I_3 = \frac{2}{5} mr^2 + \frac{7}{5} mr^2 + \frac{7}{5} mr^2 = \frac{16}{5} mr^2$.

(B) The moment of inertia $I$ of a circular loop of mass $M$ and radius $R$ about its central axis is given by $I = MR^2$.
Let the linear mass density of the wire be $\lambda$.
The mass of loop $P$ is $M_P = \lambda \cdot (2\pi r)$ and its radius is $R_P = r$.
Thus,$I_P = M_P R_P^2 = (2\pi r \lambda) r^2 = 2\pi \lambda r^3$.
The mass of loop $Q$ is $M_Q = \lambda \cdot (2\pi nr)$ and its radius is $R_Q = nr$.
Thus,$I_Q = M_Q R_Q^2 = (2\pi nr \lambda) (nr)^2 = 2\pi \lambda n^3 r^3$.
Given that $I_Q = 4 I_P$,we have $2\pi \lambda n^3 r^3 = 4(2\pi \lambda r^3)$.
Canceling $2\pi \lambda r^3$ from both sides,we get $n^3 = 4$.
Therefore,$n = 4^{1/3} = (2^2)^{1/3} = 2^{2/3}$.

(B) $1$. For a thin uniform rod of mass $M$ and length $L$,the moment of inertia about the perpendicular axis passing through its centre is $I = \frac{ML^2}{12}$.
$2$. When the rod is bent into a ring of radius $R$,the circumference of the ring equals the length of the rod: $2\pi R = L$,so $R = \frac{L}{2\pi}$.
$3$. The moment of inertia of a ring of mass $M$ and radius $R$ about its diameter is $I_1 = \frac{1}{2}MR^2$.
$4$. Substituting $R = \frac{L}{2\pi}$ into the expression for $I_1$: $I_1 = \frac{1}{2}M(\frac{L}{2\pi})^2 = \frac{1}{2}M(\frac{L^2}{4\pi^2}) = \frac{ML^2}{8\pi^2}$.
$5$. Given $I_1 = xI$,we have $\frac{ML^2}{8\pi^2} = x(\frac{ML^2}{12})$.
$6$. Solving for $x$: $x = \frac{12}{8\pi^2} = \frac{3}{2\pi^2}$.

(C) The moment of inertia of a solid sphere about its diameter is given by $I_{s} = \frac{2}{5}MR^{2}$.
The moment of inertia of a thin-walled hollow sphere about its diameter is given by $I_{h} = \frac{2}{3}MR^{2}$.
Since both spheres have the same mass $M$ and the same material,they must have the same volume if they were the same size,but for a solid and hollow sphere of the same mass,the radius $R$ of the hollow sphere must be larger than the radius of the solid sphere to maintain the same mass density distribution.
Even if we assume the radii are the same,comparing the coefficients: $\frac{2}{3} \approx 0.667$ and $\frac{2}{5} = 0.4$.
Since $0.667 > 0.4$,it follows that $I_{h} > I_{s}$.

(C) $1$. The length of the wire is $L$. Since it is bent into a semicircular ring,the circumference of a full circle would be $2 \pi R = 2L$,where $R$ is the radius of the semicircle. Thus,$\pi R = L$,which gives $R = \frac{L}{\pi}$.
$2$. The moment of inertia of a complete circular ring of mass $M'$ and radius $R$ about its diameter is $I_{diam} = \frac{1}{2} M' R^2$.
$3$. For a semicircular ring of mass $M$ and radius $R$,the moment of inertia about the axis passing through its ends (the diameter) is the same as that of a full ring of the same radius and mass $M$ about its diameter. Therefore,$I = \frac{1}{2} M R^2$.
$4$. Substituting $R = \frac{L}{\pi}$ into the formula,we get $I = \frac{1}{2} M \left(\frac{L}{\pi}\right)^2 = \frac{M L^2}{2 \pi^2}$.

(B) The moment of inertia $I$ of a system of particles about an axis is given by $I = \sum m_i r_i^2$.
For a square of side $L$,the distance of each corner from the centre is $r = \frac{L}{\sqrt{2}}$.
Since there are four particles of mass $M$,the total moment of inertia about an axis perpendicular to the plane and passing through the centre is $I = 4 \times M \times r^2 = 4 \times M \times (\frac{L}{\sqrt{2}})^2 = 4 \times M \times \frac{L^2}{2} = 2ML^2$.
The radius of gyration $k$ is defined by $I = Mk_{total} k^2$,where $M_{total} = 4M$.
So,$2ML^2 = (4M)k^2$.
$k^2 = \frac{2ML^2}{4M} = \frac{L^2}{2}$.
$k = \frac{L}{\sqrt{2}}$.

(C) The moment of inertia of a solid sphere about its diametric axis is given by $I = \frac{2}{5} mR^2$.
According to the parallel axis theorem,the moment of inertia about a tangent is $I_{tangent} = I_{cm} + mR^2$.
Substituting the value of $I_{cm} = I = \frac{2}{5} mR^2$,we get:
$I_{tangent} = \frac{2}{5} mR^2 + mR^2 = \frac{7}{5} mR^2$.
Since $I = \frac{2}{5} mR^2$,we can write $mR^2 = \frac{5}{2} I$.
Substituting this into the expression for $I_{tangent}$:
$I_{tangent} = \frac{7}{5} \times (\frac{5}{2} I) = \frac{7}{2} I = 3.5 I$.

(B) Let the mass of the rod be $M$ and its length be $L$. The moment of inertia of the rod about an axis passing through its center and perpendicular to its length is $I_1 = \frac{ML^2}{12}$.
When the rod is bent into a ring of radius $R$,the circumference of the ring is equal to the length of the rod,so $2\pi R = L$,which implies $R = \frac{L}{2\pi}$.
The moment of inertia of a ring of mass $M$ and radius $R$ about its diameter is $I_2 = \frac{1}{2}MR^2$.
Substituting $R = \frac{L}{2\pi}$ into the expression for $I_2$,we get $I_2 = \frac{1}{2}M(\frac{L}{2\pi})^2 = \frac{1}{2}M(\frac{L^2}{4\pi^2}) = \frac{ML^2}{8\pi^2}$.
Now,calculating the ratio $I_1 / I_2 = \frac{ML^2/12}{ML^2/8\pi^2} = \frac{8\pi^2}{12} = \frac{2\pi^2}{3}$.

(D) The moment of inertia of a solid sphere of mass $M$ and radius $R$ about its diameter is given by $I = \frac{2}{5}MR^2$.
Let the radius of the large sphere be $R$ and its mass be $M$.
When it is recast into $27$ small spheres of radius $r$,the volume remains constant: $\frac{4}{3}\pi R^3 = 27 \times \frac{4}{3}\pi r^3$.
This implies $R^3 = 27r^3$,so $R = 3r$ or $r = \frac{R}{3}$.
The mass of each small sphere $m$ is $\frac{M}{27}$.
The moment of inertia of each small sphere about its diameter is $I' = \frac{2}{5}mr^2$.
Substituting $m = \frac{M}{27}$ and $r = \frac{R}{3}$:
$I' = \frac{2}{5} \left( \frac{M}{27} \right) \left( \frac{R}{3} \right)^2 = \frac{2}{5} \left( \frac{M}{27} \right) \left( \frac{R^2}{9} \right) = \frac{1}{243} \left( \frac{2}{5}MR^2 \right)$.
Since $I = \frac{2}{5}MR^2$,we get $I' = \frac{I}{243}$.

(A) $1$. The total mass $M$ of the wire is $M = \lambda L$.
$2$. The circumference of the ring is $2 \pi R = L$,so the radius $R = \frac{L}{2 \pi}$.
$3$. The moment of inertia of a ring about its central axis (perpendicular to the plane) is $I_{cm} = MR^2$.
$4$. By the perpendicular axis theorem,the moment of inertia about a diameter is $I_{diameter} = \frac{1}{2} MR^2$.
$5$. Using the parallel axis theorem,the moment of inertia about a tangential axis in the plane is $I = I_{diameter} + MR^2$.
$6$. $I = \frac{1}{2} MR^2 + MR^2 = \frac{3}{2} MR^2$.
$7$. Substituting $M = \lambda L$ and $R = \frac{L}{2 \pi}$:
$I = \frac{3}{2} (\lambda L) (\frac{L}{2 \pi})^2 = \frac{3}{2} \lambda L \cdot \frac{L^2}{4 \pi^2} = \frac{3 \lambda L^3}{8 \pi^2}$.

(C) $1$. The moment of inertia of the disc about its central axis is $I_{cm} = \frac{1}{2}MR^2$.
$2$. Using the Parallel Axis Theorem,the moment of inertia of the disc about an axis passing through a point on its rim (perpendicular to the plane) is $I_{disc} = I_{cm} + MR^2 = \frac{1}{2}MR^2 + MR^2 = \frac{3}{2}MR^2$.
$3$. Let the four particles be at positions $(R, 0), (0, R), (-R, 0), (0, -R)$. Let the axis pass through the particle at $(R, 0)$.
$4$. The distances of the four particles from the axis are: $r_1 = 0$,$r_2 = \sqrt{R^2 + R^2} = R\sqrt{2}$,$r_3 = \sqrt{(2R)^2 + 0} = 2R$,$r_4 = \sqrt{R^2 + R^2} = R\sqrt{2}$.
$5$. The moment of inertia of the particles is $I_{particles} = m(r_1^2 + r_2^2 + r_3^2 + r_4^2) = m(0 + 2R^2 + 4R^2 + 2R^2) = 8mR^2$.
$6$. Total moment of inertia $I = I_{disc} + I_{particles} = \frac{3}{2}MR^2 + 8mR^2 = (\frac{M}{2} + 8m)R^2$.

(D) The moment of inertia of a thin uniform rod about an axis passing through its center of mass and perpendicular to its length is $I_{cm} = \frac{ML^2}{12}$.
According to the parallel axis theorem,$I = I_{cm} + Md^2$,where $d$ is the distance between the center of mass and the given axis.
The center of mass of the rod is at a distance $L/2$ from one end.
The given axis is at a distance $L/4$ from the same end.
Therefore,the distance $d$ between the center of mass and the axis is $d = |L/2 - L/4| = L/4$.
Substituting these values into the parallel axis theorem:
$I = \frac{ML^2}{12} + M(L/4)^2$
$I = \frac{ML^2}{12} + \frac{ML^2}{16}$
Taking the least common multiple of $12$ and $16$,which is $48$:
$I = \frac{4ML^2 + 3ML^2}{48} = \frac{7ML^2}{48}$.

(A) For a body sliding down a smooth inclined plane of height $h$,the potential energy is converted into kinetic energy: $mgh = \frac{1}{2}mV^2$,which gives $V = \sqrt{2gh}$.
For a ring rolling down the same inclined plane,the potential energy is converted into both translational and rotational kinetic energy: $mgh = \frac{1}{2}mv^2 + \frac{1}{2}I\omega^2$.
For a ring,the moment of inertia $I = mr^2$ and the rolling condition is $v = r\omega$,so $\omega = v/r$.
Substituting these into the energy equation: $mgh = \frac{1}{2}mv^2 + \frac{1}{2}(mr^2)(v/r)^2 = \frac{1}{2}mv^2 + \frac{1}{2}mv^2 = mv^2$.
Thus,$v^2 = gh$,which means $v = \sqrt{gh}$.
Comparing the two velocities: $v = \sqrt{gh} = \frac{\sqrt{2gh}}{\sqrt{2}} = \frac{V}{\sqrt{2}}$.

(B) The linear acceleration $a$ of a body rolling down an inclined plane without slipping is given by the formula: $a = \frac{g \sin \theta}{1 + \frac{I}{MR^2}}$.
For a solid sphere,the moment of inertia $I$ about its center of mass is $\frac{2}{5}MR^2$.
Substituting this into the formula: $a = \frac{g \sin \theta}{1 + \frac{2/5 MR^2}{MR^2}} = \frac{g \sin \theta}{1 + 2/5} = \frac{g \sin \theta}{7/5} = \frac{5}{7} g \sin \theta$.
Given $\theta = 30^{\circ}$ and $\sin 30^{\circ} = 0.5$,we have:
$a = \frac{5}{7} g (0.5) = \frac{5}{7} g \left(\frac{1}{2}\right) = \frac{5g}{14}$.
Therefore,the correct option is $B$.

(A) For a solid sphere rolling without slipping, the total kinetic energy $K$ is the sum of translational and rotational kinetic energy: $K = \frac{1}{2}mv^2 + \frac{1}{2}I\omega^2$.
Since $I = \frac{2}{5}mr^2$ and $v = r\omega$, we have $K = \frac{1}{2}mv^2 + \frac{1}{2}(\frac{2}{5}mr^2)(\frac{v}{r})^2 = \frac{1}{2}mv^2 + \frac{1}{5}mv^2 = \frac{7}{10}mv^2$.
When the sphere moves up the ramp, this kinetic energy is converted into potential energy $U = mgh$, where $h = d \sin \theta$ and $d$ is the distance along the ramp.
Equating $K = U$: $\frac{7}{10}mv^2 = mgd \sin \theta$.
Solving for $d$: $d = \frac{7v^2}{10g \sin \theta}$.
Substituting the given values: $v = 2 \,m/s$, $g = 10 \,m/s^2$, $\theta = 30^{\circ}$:
$d = \frac{7 \times (2)^2}{10 \times 10 \times (1/2)} = \frac{7 \times 4}{100 \times 0.5} = \frac{28}{50} = 0.56 \,m$.

(B) For a body rolling down an inclined plane without slipping,mechanical energy is conserved.
Initial potential energy at the top is $PE = mgh$.
Initial kinetic energy at the top is $KE = 0$ (since it starts from rest).
At the bottom,the total kinetic energy is the sum of translational and rotational kinetic energy: $KE_{total} = \frac{1}{2}mv^2 + \frac{1}{2}I\omega^2$.
For a solid sphere,the moment of inertia $I = \frac{2}{5}mr^2$.
Since it rolls without slipping,$v = r\omega$,so $\omega = v/r$.
Substituting these into the kinetic energy equation: $KE_{total} = \frac{1}{2}mv^2 + \frac{1}{2}(\frac{2}{5}mr^2)(\frac{v}{r})^2 = \frac{1}{2}mv^2 + \frac{1}{5}mv^2 = \frac{7}{10}mv^2$.
By conservation of energy,$mgh = \frac{7}{10}mv^2$.
Solving for $v$,we get $v^2 = \frac{10gh}{7}$,which implies $v = \left(\frac{10gh}{7}\right)^{1/2}$.

(A) Let the initial angular velocity be $\omega_0$ and the final angular velocity be $\omega$. The equation of rotational motion is $\omega^2 = \omega_0^2 + 2\alpha\theta$.
For the first part,the velocity becomes $\frac{\omega_0}{3}$ after $\theta_1 = 24$ rotations.
$(\frac{\omega_0}{3})^2 = \omega_0^2 + 2\alpha(24 \times 2\pi) \implies \frac{\omega_0^2}{9} - \omega_0^2 = 48\pi\alpha \implies -\frac{8}{9}\omega_0^2 = 48\pi\alpha \implies \alpha = -\frac{\omega_0^2}{54\pi}$.
Now,for the motion from $\frac{\omega_0}{3}$ to $0$ (rest),let the additional rotations be $\theta_2$.
$0^2 = (\frac{\omega_0}{3})^2 + 2\alpha(\theta_2 \times 2\pi) \implies 0 = \frac{\omega_0^2}{9} + 2(-\frac{\omega_0^2}{54\pi})(2\pi\theta_2)$.
$0 = \frac{\omega_0^2}{9} - \frac{\omega_0^2}{13.5}\theta_2 \implies \frac{\omega_0^2}{9} = \frac{\omega_0^2}{13.5}\theta_2 \implies \theta_2 = \frac{13.5}{9} = 3$.
Thus,the fan makes $3$ more rotations.

(A) For a disc rolling down an inclined plane,the total potential energy at the top is converted into translational and rotational kinetic energy at the bottom.
By the law of conservation of energy: $mgh = K_{trans} + K_{rot}$.
For a disc,the moment of inertia $I = \frac{1}{2}mr^2$.
Since the disc rolls without slipping,$v = r\omega$,so $\omega = \frac{v}{r}$.
The translational kinetic energy is $K_{trans} = \frac{1}{2}mv^2$.
The rotational kinetic energy is $K_{rot} = \frac{1}{2}I\omega^2 = \frac{1}{2}(\frac{1}{2}mr^2)(\frac{v}{r})^2 = \frac{1}{4}mv^2$.
Total energy $mgh = \frac{1}{2}mv^2 + \frac{1}{4}mv^2 = \frac{3}{4}mv^2$.
Thus,$\frac{1}{2}mv^2 = \frac{2}{3}mgh$.
Substituting this into the expression for rotational kinetic energy: $K_{rot} = \frac{1}{4}mv^2 = \frac{1}{2}(\frac{1}{2}mv^2) = \frac{1}{2}(\frac{2}{3}mgh) = \frac{mgh}{3}$.

(D) The angular displacement $\theta$ for an object starting from rest with constant angular acceleration $\alpha$ is given by $\theta = \frac{1}{2} \alpha t^2$.
For the first $2 \ s$ $(t_1 = 2 \ s)$: $\theta_1 = \frac{1}{2} \alpha (2)^2 = 2\alpha$.
For the total time of $4 \ s$ $(t_2 = 2 + 2 = 4 \ s)$: $\theta_{total} = \frac{1}{2} \alpha (4)^2 = 8\alpha$.
The angle rotated in the next $2 \ s$ is $\theta_2 = \theta_{total} - \theta_1 = 8\alpha - 2\alpha = 6\alpha$.
Therefore,the ratio $\theta_1 : \theta_2 = 2\alpha : 6\alpha = 1 : 3$.

(B) The rotational kinetic energy $K$ of a body rotating with angular velocity $\omega$ and moment of inertia $I$ is given by $K = \frac{1}{2} I \omega^2$.
The angular momentum $L$ is given by $L = I \omega$.
We can express the kinetic energy in terms of angular momentum as:
$K = \frac{1}{2} I \left( \frac{L}{I} \right)^2 = \frac{L^2}{2I}$.
Given $K = x$ and $L = y$,we have $x = \frac{y^2}{2I}$.
Rearranging for the moment of inertia $I$,we get $I = \frac{y^2}{2x}$.
Therefore,the correct option is $B$.

(B) The moment of inertia of a solid sphere about its diameter is $I_s = \frac{2}{5} mR^2$.
The moment of inertia of a solid cylinder about its geometrical axis is $I_c = \frac{1}{2} mR^2$.
Let the angular speed of the sphere be $\omega$. Then the angular speed of the cylinder is $2\omega$.
The rotational kinetic energy is given by $K = \frac{1}{2} I \omega^2$.
Kinetic energy of the sphere: $K_s = \frac{1}{2} I_s \omega^2 = \frac{1}{2} (\frac{2}{5} mR^2) \omega^2 = \frac{1}{5} mR^2 \omega^2$.
Kinetic energy of the cylinder: $K_c = \frac{1}{2} I_c (2\omega)^2 = \frac{1}{2} (\frac{1}{2} mR^2) (4\omega^2) = mR^2 \omega^2$.
The ratio of kinetic energy of the sphere to the cylinder is $\frac{K_s}{K_c} = \frac{\frac{1}{5} mR^2 \omega^2}{mR^2 \omega^2} = \frac{1}{5}$.
Thus,the ratio is $1: 5$.

(C) The rotational kinetic energy is given by $K = \frac{1}{2} I \omega^2$.
For a solid cylinder rotating about its geometrical axis,the moment of inertia is $I_c = \frac{1}{2} M R^2$.
Let the angular speed of the cylinder be $\omega_c = \omega$.
So,$K_c = \frac{1}{2} (\frac{1}{2} M R^2) \omega^2 = \frac{1}{4} M R^2 \omega^2$.
For a solid sphere rotating about its diameter,the moment of inertia is $I_s = \frac{2}{5} M R^2$.
The angular speed of the sphere is $\omega_s = \frac{\omega}{2}$.
So,$K_s = \frac{1}{2} (\frac{2}{5} M R^2) (\frac{\omega}{2})^2 = \frac{1}{5} M R^2 (\frac{\omega^2}{4}) = \frac{1}{20} M R^2 \omega^2$.
The ratio of the kinetic energy of the sphere to that of the cylinder is $\frac{K_s}{K_c} = \frac{\frac{1}{20} M R^2 \omega^2}{\frac{1}{4} M R^2 \omega^2} = \frac{1}{20} \times \frac{4}{1} = \frac{4}{20} = \frac{1}{5}$.

(A) The linear velocity $\vec{v}$ is given by the cross product of angular velocity $\vec{\omega}$ and position vector $\vec{r}$:
$\vec{v} = \vec{\omega} \times \vec{r}$
$\vec{v} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 3 & -4 & 1 \\ 5 & -6 & 6 \end{vmatrix}$
Expanding the determinant:
$\vec{v} = \hat{i} [(-4)(6) - (1)(-6)] - \hat{j} [(3)(6) - (1)(5)] + \hat{k} [(3)(-6) - (-4)(5)]$
$\vec{v} = \hat{i} [-24 + 6] - \hat{j} [18 - 5] + \hat{k} [-18 + 20]$
$\vec{v} = -18 \hat{i} - 13 \hat{j} + 2 \hat{k}$

(D) The rod rotates about an axis passing through its end. The moment of inertia of the rod about this axis is $I = \frac{1}{3}ML^2$.
The rotational kinetic energy of the rod at its lowest point (maximum angular speed) is $K = \frac{1}{2}I\omega^2 = \frac{1}{2}(\frac{1}{3}ML^2)\omega^2 = \frac{1}{6}ML^2\omega^2$.
At the maximum height $h$,the centre of mass of the rod rises by $h$. The potential energy gained by the rod is $U = Mgh$.
By the law of conservation of energy,the rotational kinetic energy at the bottom equals the potential energy at the maximum height:
$\frac{1}{6}ML^2\omega^2 = Mgh$.
Solving for $h$:
$h = \frac{L^2\omega^2}{6g}$.

(A) Let the angular velocity of the disc be $\omega$ and the linear velocity of the bob be $v$. Since there is no slipping,$v = R\omega$.
Applying the law of conservation of energy: The loss in potential energy of the bob equals the gain in kinetic energy of the bob plus the gain in rotational kinetic energy of the disc.
$mgh = \frac{1}{2}mv^2 + \frac{1}{2}I\omega^2$
For a disc,the moment of inertia $I = \frac{1}{2}mR^2$.
Substituting $I$ and $v = R\omega$ into the equation:
$mgh = \frac{1}{2}m(R\omega)^2 + \frac{1}{2}(\frac{1}{2}mR^2)\omega^2$
$mgh = \frac{1}{2}mR^2\omega^2 + \frac{1}{4}mR^2\omega^2$
$mgh = \frac{3}{4}mR^2\omega^2$
$gh = \frac{3}{4}R^2\omega^2$
$\omega^2 = \frac{4gh}{3R^2}$
$\omega = \frac{2}{R} \sqrt{\frac{gh}{3}}$.

(C) Given the angular velocity $\omega(t) = \alpha - \beta t$.
At time $t = 0$,$\omega = \alpha$.
The body comes to rest when $\omega(t) = 0$.
$\alpha - \beta t = 0 \implies t = \frac{\alpha}{\beta}$.
The angular displacement $\theta$ is given by the integral of angular velocity with respect to time:
$\theta = \int_{0}^{t} \omega(t) dt = \int_{0}^{\alpha/\beta} (\alpha - \beta t) dt$.
$\theta = [\alpha t - \frac{1}{2} \beta t^2]_{0}^{\alpha/\beta}$.
Substituting the limits:
$\theta = \alpha(\frac{\alpha}{\beta}) - \frac{1}{2} \beta (\frac{\alpha}{\beta})^2$.
$\theta = \frac{\alpha^2}{\beta} - \frac{1}{2} \frac{\alpha^2}{\beta} = \frac{\alpha^2}{2 \beta}$.

(C) The angular momentum of a rotating body is given by $L = I\omega$,where $I$ is the moment of inertia and $\omega$ is the angular velocity.
The rotational kinetic energy is given by $K = \frac{1}{2} I \omega^2$.
The ratio of angular momentum to kinetic energy is $\frac{L}{K} = \frac{I\omega}{\frac{1}{2} I \omega^2} = \frac{2}{\omega}$.
Given the ratio $\frac{L}{K} = \frac{\pi}{22}$,we equate the two expressions:
$\frac{2}{\omega} = \frac{\pi}{22}$.
Substituting $\pi = \frac{22}{7}$ into the equation:
$\frac{2}{\omega} = \frac{22/7}{22} = \frac{1}{7}$.
Solving for $\omega$:
$\omega = 2 \times 7 = 14 \ rad/s$.
Thus,the angular velocity of the sphere is $14 \ rad/s$.

(A) The rod is pivoted at one end. The moment of inertia of the rod about the pivot is $I = \frac{mL^2}{3}$.
The gravitational force $mg$ acts at the center of mass of the rod,which is at a distance $L/2$ from the pivot.
The torque $\tau$ about the pivot due to gravity is $\tau = mg \cdot (L/2) \sin \theta$.
Using the rotational analog of Newton's second law,$\tau = I \alpha$,where $\alpha$ is the angular acceleration:
$mg \frac{L}{2} \sin \theta = \left( \frac{mL^2}{3} \right) \alpha$
Solving for $\alpha$:
$\alpha = \frac{mg(L/2) \sin \theta}{mL^2/3} = \frac{mgL/2}{mL^2/3} \sin \theta = \frac{3g}{2L} \sin \theta$.
Thus,the angular acceleration is $\frac{3g \sin \theta}{2L}$.

(B) The torque $\vec{\tau}$ is given by the cross product of the position vector $\vec{r}$ and the force vector $\vec{F}$:
$\vec{\tau} = \vec{r} \times \vec{F}$
Given $\vec{r} = (4 \hat{i} - 3 \hat{j})$ and $\vec{F} = (5 \hat{i} - 10 \hat{j})$:
$\vec{\tau} = (4 \hat{i} - 3 \hat{j}) \times (5 \hat{i} - 10 \hat{j})$
Using the distributive property of the cross product:
$\vec{\tau} = 4 \hat{i} \times (5 \hat{i}) - 4 \hat{i} \times (10 \hat{j}) - 3 \hat{j} \times (5 \hat{i}) + 3 \hat{j} \times (10 \hat{j})$
Since $\hat{i} \times \hat{i} = 0$ and $\hat{j} \times \hat{j} = 0$,and knowing $\hat{i} \times \hat{j} = \hat{k}$ and $\hat{j} \times \hat{i} = -\hat{k}$:
$\vec{\tau} = 0 - 40(\hat{k}) - 15(-\hat{k}) + 0$
$\vec{\tau} = -40 \hat{k} + 15 \hat{k}$
$\vec{\tau} = -25 \hat{k} \text{ N m}$

(C) Given:
Mass $M = 25 \ kg$,Radius $R = 0.2 \ m$,Initial angular velocity $\omega_0 = 240 \ r.p.m. = 240 \times \frac{2\pi}{60} \ rad/s = 8\pi \ rad/s$.
Final angular velocity $\omega = 0 \ rad/s$,Time $t = 20 \ s$.
The moment of inertia of the disc is $I = \frac{1}{2}MR^2 = \frac{1}{2} \times 25 \times (0.2)^2 = 0.5 \ kg \cdot m^2$.
The angular acceleration $\alpha$ is given by $\alpha = \frac{\omega - \omega_0}{t} = \frac{0 - 8\pi}{20} = -0.4\pi \ rad/s^2$.
The retarding torque is $\tau = I|\alpha| = 0.5 \times 0.4\pi = 0.2\pi \ N \cdot m$.
Since the torque is applied tangentially,$\tau = F \times R$,so $F = \frac{\tau}{R} = \frac{0.2\pi}{0.2} = \pi \ N$.

(C) The torque $\tau$ is related to the moment of inertia $I$ and angular acceleration $\alpha$ by the equation $\tau = I\alpha$.
Since the torque $\tau$ is the same for both,the angular acceleration is given by $\alpha = \frac{\tau}{I}$.
The moment of inertia of a ring is $I_{\text{ring}} = MR^2$ and for a disc is $I_{\text{disc}} = \frac{1}{2}MR^2$.
Since $I_{\text{ring}} > I_{\text{disc}}$,the angular acceleration of the disc will be greater than that of the ring $(\alpha_{\text{disc}} > \alpha_{\text{ring}})$.
Consequently,the disc will rotate with a greater angular frequency (or angular velocity) compared to the ring after the same time interval.

(C) The potential energy of the object at height $h$ is converted into heat energy upon impact.
Given: Mass $m = 3.5 \ kg$,height $h = 2000 \ m$,$g = 10 \ m/s^2$,Latent heat $L = 3.5 \times 10^5 \ J/kg$.
The potential energy $PE = mgh = 3.5 \times 10 \times 2000 = 70,000 \ J$.
The heat required to melt a mass $m'$ of ice is $Q = m'L$.
Since the object comes to rest,all potential energy is converted into heat: $mgh = m'L$.
$70,000 = m' \times 3.5 \times 10^5$.
$m' = \frac{70,000}{3.5 \times 10^5} = \frac{7 \times 10^4}{3.5 \times 10^5} = 2 \times 10^{-1} \ kg = 0.2 \ kg$.
Converting to grams: $0.2 \ kg = 200 \ g$.

(C) According to Newton's Law of Cooling,the rate of heat loss is proportional to the temperature difference between the body and the surroundings: $\frac{dQ}{dt} = k(T - T_s)$.
Here,the heat lost by the system (calorimeter + water) is $\Delta Q = (m_w + w)c \Delta T$,where $w$ is the water equivalent of the calorimeter.
The rate of cooling is $\frac{\Delta Q}{\Delta t} = \frac{(m_w + w)c \Delta T}{\Delta t} = k(T_{avg} - T_s)$.
Since the temperature change $\Delta T = 5^{\circ}C$ and the average temperature $T_{avg} = 17.5^{\circ}C$ are the same in both cases,the rate of heat loss is constant.
Thus,$\frac{(m_1 + w)c \Delta T}{t_1} = \frac{(m_2 + w)c \Delta T}{t_2}$.
This simplifies to $\frac{m_1 + w}{t_1} = \frac{m_2 + w}{t_2}$.
Given $m_1 = 10 \ g, t_1 = 10 \ min$ and $m_2 = 20 \ g, t_2 = 15 \ min$.
Substituting the values: $\frac{10 + w}{10} = \frac{20 + w}{15}$.
$15(10 + w) = 10(20 + w) \implies 150 + 15w = 200 + 10w$.
$5w = 50 \implies w = 10 \ g$.

(D) In a steady state,the rate of heat flow $(H)$ through materials in series is the same.
For the first material: $H = \frac{KA(T_2 - T)}{x}$,where $T$ is the interface temperature.
For the second material: $H = \frac{(2K)A(T - T_1)}{4x} = \frac{KA(T - T_1)}{2x}$.
Equating the two expressions: $\frac{KA(T_2 - T)}{x} = \frac{KA(T - T_1)}{2x}$.
$2(T_2 - T) = T - T_1 \implies 2T_2 - 2T = T - T_1 \implies 3T = 2T_2 + T_1 \implies T = \frac{2T_2 + T_1}{3}$.
Substituting $T$ back into the first expression: $H = \frac{KA}{x} (T_2 - \frac{2T_2 + T_1}{3}) = \frac{KA}{x} (\frac{3T_2 - 2T_2 - T_1}{3}) = \frac{KA(T_2 - T_1)}{3x}$.
Comparing this with the given expression $\left[\frac{A(T_2 - T_1)K}{x}\right] f$,we find $f = 1/3$.

(B) The rate of heat transfer $H$ through a rod by conduction is given by the formula:
$H = \frac{Q}{t} = \frac{KA(T_1 - T_2)}{x}$
Where:
$Q/t$ is the rate of heat transfer,
$K$ is the coefficient of thermal conductivity,
$A$ is the cross-sectional area,
$x$ is the length of the rod,
$(T_1 - T_2)$ is the temperature difference.
Rearranging the formula to solve for $K$:
$K = \frac{(Q/t) \cdot x}{A(T_1 - T_2)}$
$K = \frac{xQ}{tA(T_1 - T_2)}$
Thus,the correct option is $B$.

(D) According to the Stefan-Boltzmann Law,the power $P$ radiated by a black body of surface area $A$ at temperature $T$ is given by $P = \sigma A T^4$.
For a sphere of radius $R$,the surface area is $A = 4 \pi R^2$.
Thus,the initial power is $P_1 = \sigma (4 \pi R^2) T^4$.
When the radius is doubled $(R' = 2R)$ and the temperature is doubled $(T' = 2T)$,the new power $P_2$ is:
$P_2 = \sigma (4 \pi (2R)^2) (2T)^4$
$P_2 = \sigma (4 \pi \cdot 4R^2) (16T^4)$
$P_2 = 16 \cdot 4 \cdot \sigma (4 \pi R^2) T^4$
$P_2 = 64 P_1$.
Therefore,the new power radiated is $64 P$.

(C) For any surface,the sum of the coefficient of absorption $(a)$,the coefficient of reflection $(r)$,and the coefficient of transmission $(t)$ is equal to $1$.
$a + r + t = 1$
Given: $a = 0.77$,$r = 0.17$.
Substituting these values: $0.77 + 0.17 + t = 1$
$0.94 + t = 1$
$t = 1 - 0.94 = 0.06$
The quantity of heat transmitted is given by the product of the incident heat $(Q)$ and the coefficient of transmission $(t)$.
$Q_{\text{transmitted}} = Q \times t$
$Q_{\text{transmitted}} = 250 \ kcal \times 0.06$
$Q_{\text{transmitted}} = 15 \ kcal$
Therefore,the correct option is $C$.

(B) According to Newton's law of cooling,the rate of cooling is given by $\frac{dT}{dt} = -k(T - T_s)$,where $T_s$ is the temperature of the surroundings.
For the first interval: $\frac{60 - 40}{6} = k \left( \frac{60 + 40}{2} - 10 \right) \implies \frac{20}{6} = k(50 - 10) \implies \frac{10}{3} = 40k \implies k = \frac{1}{12}$.
For the second interval,let the final temperature be $T_f$: $\frac{40 - T_f}{6} = k \left( \frac{40 + T_f}{2} - 10 \right)$.
Substituting $k = \frac{1}{12}$: $\frac{40 - T_f}{6} = \frac{1}{12} \left( 20 + \frac{T_f}{2} - 10 \right) \implies \frac{40 - T_f}{6} = \frac{1}{12} \left( 10 + \frac{T_f}{2} \right)$.
Multiplying by $12$: $2(40 - T_f) = 10 + 0.5T_f \implies 80 - 2T_f = 10 + 0.5T_f \implies 70 = 2.5T_f \implies T_f = \frac{70}{2.5} = 28^{\circ} C$.

(B) perfectly black body is defined as an object that absorbs all incident radiation of any wavelength. According to Kirchhoff's law of thermal radiation,for an arbitrary body in thermal equilibrium with its surroundings,the emissivity is equal to its absorptivity. Since a perfectly black body has an absorptivity of $1$,its emissivity (coefficient of emission) must also be $1$,which is referred to as unity.

(A) According to Newton's Law of Cooling,the rate of cooling is given by: $\frac{d\theta}{dt} = K(\theta - \theta_0)$,where $\theta$ is the temperature of the body and $\theta_0$ is the temperature of the surroundings.
For the first case:
$1.5 = K(80 - \theta_0)$ --- (Equation $1$)
For the second case:
$0.3 = K(40 - \theta_0)$ --- (Equation $2$)
Dividing Equation $1$ by Equation $2$:
$\frac{1.5}{0.3} = \frac{K(80 - \theta_0)}{K(40 - \theta_0)}$
$5 = \frac{80 - \theta_0}{40 - \theta_0}$
$5(40 - \theta_0) = 80 - \theta_0$
$200 - 5\theta_0 = 80 - \theta_0$
$200 - 80 = 5\theta_0 - \theta_0$
$120 = 4\theta_0$
$\theta_0 = 30^{\circ} C$
Thus,the temperature of the surrounding is $30^{\circ} C$.

(D) The graph shows the spectral emissive power $(E_{\lambda})$ on the y-axis and wavelength $(\lambda)$ on the x-axis.
The area under the curve is given by the integral $\int_{0}^{\infty} E_{\lambda} d\lambda$.
According to the definition of spectral emissive power,this integral represents the total radiant energy emitted per unit time per unit surface area of the black body across all possible wavelengths.
This is consistent with the Stefan-Boltzmann Law,which states that the total power radiated per unit area is proportional to the fourth power of the absolute temperature $(E = \sigma T^4)$.

(B) According to Stefan-Boltzmann Law,the energy emitted per second (power) $P$ is given by $P = \sigma A T^4$,where $\sigma$ is the Stefan-Boltzmann constant,$A$ is the surface area,and $T$ is the absolute temperature in Kelvin.
Initial state: $P_1 = E = \sigma A T_1^4$,where $T_1 = 27 + 273 = 300 \ K$.
Final state: Length and breadth are halved,so the new area $A' = (L/2) \times (B/2) = A/4$. The new temperature $T_2 = 327 + 273 = 600 \ K$.
The new power $P_2 = \sigma A' T_2^4 = \sigma (A/4) (600)^4$.
Taking the ratio: $P_2 / P_1 = [\sigma (A/4) (600)^4] / [\sigma A (300)^4] = (1/4) \times (600/300)^4 = (1/4) \times 2^4 = (1/4) \times 16 = 4$.
Therefore,$P_2 = 4 E$.

(C) According to Stefan-Boltzmann law,the rate of cooling $R$ of a body is given by $R \propto (T^4 - T_0^4)$,where $T$ is the temperature of the body and $T_0$ is the temperature of the surroundings.
For the first case: $R_0 = k(400^4 - 300^4) = k(256 \times 10^8 - 81 \times 10^8) = k(175 \times 10^8)$.
For the second case: $R' = k(900^4 - 300^4) = k(6561 \times 10^8 - 81 \times 10^8) = k(6480 \times 10^8)$.
Taking the ratio: $\frac{R'}{R_0} = \frac{6480}{175} \approx 37.02$.
Comparing this with the given options,the closest value is $36 R_0$.

(A) According to Newton's Law of Cooling,the rate of cooling is given by $\frac{dT}{dt} = -k(T - T_0)$,where $T$ is the temperature of the body,$T_0$ is the room temperature,and $k$ is a constant.
For the first interval: $\frac{80 - 60}{1} = k \left( \frac{80 + 60}{2} - 30 \right) \implies 20 = k(70 - 30) \implies 20 = 40k \implies k = 0.5 \ \text{min}^{-1}$.
For the second interval: $\frac{60 - 50}{t} = k \left( \frac{60 + 50}{2} - 30 \right) \implies \frac{10}{t} = 0.5(55 - 30) \implies \frac{10}{t} = 0.5(25) \implies \frac{10}{t} = 12.5$.
Solving for $t$: $t = \frac{10}{12.5} = 0.8 \ \text{minutes}$.
Converting to seconds: $0.8 \times 60 = 48 \ \text{s}$.

(C) Let $Q$ be the total incident thermal energy,$Q_a$ be the absorbed energy,$Q_t$ be the transmitted energy,and $Q_r$ be the reflected energy.
Given: $Q = 120 \ J$,$Q_t = 12 \ J$,and coefficient of absorption $a = 0.6$.
We know that the coefficient of absorption $a = \frac{Q_a}{Q}$.
Therefore,$Q_a = a \times Q = 0.6 \times 120 \ J = 72 \ J$.
According to the law of conservation of energy,the total incident energy is the sum of absorbed,transmitted,and reflected energy:
$Q = Q_a + Q_t + Q_r$
$120 \ J = 72 \ J + 12 \ J + Q_r$
$120 \ J = 84 \ J + Q_r$
$Q_r = 120 \ J - 84 \ J = 36 \ J$.
Thus,the amount of heat reflected is $36 \ J$.

(C) When two coils of inductance $L_1$ and $L_2$ are connected in series,the equivalent inductance $L_{eq}$ is given by $L_{eq} = L_1 + L_2 \pm 2M$,where $M$ is the mutual inductance between the coils.
Since the coils are identical,$L_1 = L_2 = L$.
Because the winding directions are exactly opposite,the magnetic flux produced by one coil opposes the flux produced by the other,resulting in a negative mutual inductance effect.
For two identical coils placed very close to each other,the mutual inductance $M$ is equal to the self-inductance $L$ (i.e.,$M = L$).
Substituting these values into the formula: $L_{eq} = L + L - 2M = 2L - 2L = 0$.

(C) The coefficient of mutual inductance $M$ is defined by the relation $\Delta \phi = M \Delta I$,where $\Delta \phi$ is the change in magnetic flux and $\Delta I$ is the change in current.
Given:
Initial flux $\phi_1 = 6.5 \times 10^{-2} \ Wb$
Final flux $\phi_2 = 11 \times 10^{-2} \ Wb$
Change in flux $\Delta \phi = \phi_2 - \phi_1 = (11 - 6.5) \times 10^{-2} \ Wb = 4.5 \times 10^{-2} \ Wb$
Change in current $\Delta I = 0.03 \ A = 3 \times 10^{-2} \ A$
Using the formula $M = \frac{\Delta \phi}{\Delta I}$:
$M = \frac{4.5 \times 10^{-2}}{3 \times 10^{-2}} = \frac{4.5}{3} = 1.5 \ H$
Therefore,the coefficient of mutual inductance is $1.5 \ H$.

(D) The induced emf in coil $P$ due to the changing current in coil $Q$ is given by $\epsilon_P = M \frac{dI_Q}{dt}$,where $M$ is the mutual inductance.
Given $\epsilon_P = 12 \ mV = 12 \times 10^{-3} \ V$ and $\frac{dI_Q}{dt} = 10 \ A/s$.
$12 \times 10^{-3} = M \times 10 \implies M = 1.2 \times 10^{-3} \ H = 1.2 \ mH$.
Now,the magnetic flux $\phi_Q$ linked with coil $Q$ when current $I_P = 1.5 \ A$ flows through coil $P$ is given by $\phi_Q = M \times I_P$.
$\phi_Q = 1.2 \ mH \times 1.5 \ A = 1.8 \ mWb$.
Thus,the correct option is $D$.

(C) The magnetic flux $\phi$ linked with a coil is given by the relation $\phi = LI$,where $L$ is the self-inductance of the coil.
This equation represents a straight line passing through the origin,where the slope of the $\phi-I$ graph is equal to the self-inductance $L$ (i.e.,$L = \frac{\phi}{I} = \tan \theta$).
From the given figure,the slope of the line for $L_1$ is greater than the slope of the line for $L_2$ (since the angle $\theta_1 > \theta_2$).
Therefore,the self-inductance $L_1$ is greater than the self-inductance $L_2$.

(B) The displacement current $I_d$ is defined as $I_d = \epsilon_0 \frac{d\Phi_E}{dt}$,where $\Phi_E$ is the electric flux through the surface.
For a parallel plate capacitor,the electric field $E$ between the plates is uniform and given by $E = \frac{q}{\epsilon_0 A}$,where $q$ is the charge on the plates.
The electric flux $\Phi_E$ through an area $A'$ parallel to the plates is $\Phi_E = E \cdot A' = \frac{q}{\epsilon_0 A} \cdot A'$.
Substituting $A' = A/2$,we get $\Phi_E = \frac{q}{\epsilon_0 A} \cdot \frac{A}{2} = \frac{q}{2\epsilon_0}$.
The displacement current is $I_d = \epsilon_0 \frac{d}{dt} \left( \frac{q}{2\epsilon_0} \right) = \frac{1}{2} \frac{dq}{dt}$.
Since the charging current $I = \frac{dq}{dt}$,we have $I_d = \frac{I}{2}$.

(A) Let the proton (charge $+e$) be placed at a distance $x$ from $q_1$ on the $x$-axis. Since $q_2$ is at distance $L$ from $q_1$,the distance of the proton from $q_2$ is $(x - L)$.
For the proton to be in equilibrium,the net electrostatic force on it must be zero.
The force due to $q_1$ is $F_1 = \frac{k q_1 e}{x^2}$ (repulsive,towards the right).
The force due to $q_2$ is $F_2 = \frac{k |q_2| e}{(x - L)^2}$ (attractive,towards the left).
Equating the magnitudes: $\frac{k (6q) e}{x^2} = \frac{k (3q) e}{(x - L)^2}$.
Simplifying: $\frac{6}{x^2} = \frac{3}{(x - L)^2} \implies \frac{2}{x^2} = \frac{1}{(x - L)^2}$.
Taking the square root on both sides: $\frac{\sqrt{2}}{x} = \frac{1}{x - L}$.
$\sqrt{2}(x - L) = x \implies \sqrt{2}x - \sqrt{2}L = x$.
$x(\sqrt{2} - 1) = \sqrt{2}L$.
$x = \left(\frac{\sqrt{2}}{\sqrt{2} - 1}\right) L$.

(C) Let the side length of the square be $a$. The charges are $q_1 = q_2 = q_3 = q$.
According to Coulomb's Law,the force between two charges $q_i$ and $q_j$ separated by a distance $r$ is $F = \frac{k q_i q_j}{r^2}$.
The distance between $q_1$ and $q_2$ is $r_{12} = a$. Thus,$F_{12} = \frac{k q^2}{a^2}$.
The distance between $q_1$ and $q_3$ is the diagonal of the square,$r_{13} = a\sqrt{2}$. Thus,$F_{13} = \frac{k q^2}{(a\sqrt{2})^2} = \frac{k q^2}{2a^2}$.
The ratio of $F_{13}$ to $F_{12}$ is $\frac{F_{13}}{F_{12}} = \frac{k q^2 / 2a^2}{k q^2 / a^2} = \frac{1}{2}$.

(B) Let the charges be $q_1 = +3q$ at $x = 0$,$q_2 = Q$ at $x = \frac{\ell}{2}$,and $q_3 = +q$ at $x = \ell$.
For the net force on the charge $+q$ at $x = \ell$ to be zero,the sum of the electrostatic forces exerted by $q_1$ and $q_2$ must be zero.
Using Coulomb's Law,$F = \frac{k q_1 q_2}{r^2}$.
The force exerted by $q_1$ on $q_3$ is $F_1 = \frac{k(3q)(q)}{\ell^2}$.
The force exerted by $q_2$ on $q_3$ is $F_2 = \frac{k(Q)(q)}{(\ell/2)^2} = \frac{kQq}{\ell^2/4} = \frac{4kQq}{\ell^2}$.
For the net force to be zero,$F_1 + F_2 = 0$,so $\frac{3kq^2}{\ell^2} + \frac{4kQq}{\ell^2} = 0$.
Dividing by $\frac{kq}{\ell^2}$,we get $3q + 4Q = 0$.
Therefore,$4Q = -3q$,which gives $Q = -\frac{3}{4}q$.
Comparing this with $Q = xq$,we find $x = -\frac{3}{4}$.

(C) According to Coulomb's Law, the electrostatic force between two point charges is given by $F = k \frac{|q_1 q_2|}{l^2}$.
Initially, the force is $F_1 = k \frac{q_1 q_2}{l^2}$.
When one charge is doubled $(q_1' = 2q_1)$ and the distance is halved $(l' = l/2)$, the new force $F_2$ is:
$F_2 = k \frac{(2q_1)(q_2)}{(l/2)^2} = k \frac{2q_1 q_2}{l^2 / 4} = 8 \left( k \frac{q_1 q_2}{l^2} \right)$.
Therefore, $F_2 = 8 F_1$.
Comparing this with $F_2 = n F_1$, we get $n = 8$.

(B) Let the charges on the two balls be $q$ and $q$,and the distance between them be $r$. The initial force is $F = k \frac{q^2}{r^2}$.
When an uncharged ball touches one of the charged balls,the charge $q$ is shared equally between them. Thus,the charge on the first ball becomes $q/2$ and the charge on the third ball becomes $q/2$.
The third ball is placed at the midpoint,so its distance from both balls is $r/2$.
The force on the third ball due to the first ball is $F_1 = k \frac{(q/2)(q/2)}{(r/2)^2} = k \frac{q^2/4}{r^2/4} = k \frac{q^2}{r^2} = F$ (directed away from the first ball).
The force on the third ball due to the second ball is $F_2 = k \frac{(q/2)(q)}{(r/2)^2} = k \frac{q^2/2}{r^2/4} = 2k \frac{q^2}{r^2} = 2F$ (directed away from the second ball).
Since these forces are in opposite directions,the net force on the third ball is $F_{net} = |F_2 - F_1| = |2F - F| = F$.

(A) According to Coulomb's Law,the force between two charges $q_1$ and $q_2$ separated by distance $r$ is given by $F = k \frac{|q_1 q_2|}{r^2}$.
Initially,$q_1 = +2 \ C$ and $q_2 = +6 \ C$. The force is $F_1 = k \frac{(2)(6)}{r^2} = 18 \ N$.
This implies $k \frac{12}{r^2} = 18$,so $k/r^2 = 18/12 = 1.5$.
Now,$-4 \ C$ is added to each charge:
New charge $q_1' = +2 \ C - 4 \ C = -2 \ C$.
New charge $q_2' = +6 \ C - 4 \ C = +2 \ C$.
The new force is $F_2 = k \frac{|q_1' q_2'|}{r^2} = k \frac{|(-2)(2)|}{r^2} = k \frac{4}{r^2}$.
Substituting $k/r^2 = 1.5$,we get $F_2 = 1.5 \times 4 = 6 \ N$.
Since the charges are of opposite signs ($-2 \ C$ and $+2 \ C$),the force is attractive.

(A) The potential energy $U$ of an electric dipole in an external electric field is given by $U = -pE \cos \theta$,where $\theta$ is the angle between the dipole moment $p$ and the electric field $E$.
Initially,the dipole is aligned parallel to the electric field,so $\theta_1 = 0^{\circ}$.
The initial potential energy is $U_1 = -pE \cos 0^{\circ} = -pE(1) = -pE$.
Finally,the dipole is rotated by $90^{\circ}$,so $\theta_2 = 90^{\circ}$.
The final potential energy is $U_2 = -pE \cos 90^{\circ} = -pE(0) = 0$.
The work done (energy required) to rotate the dipole is equal to the change in potential energy: $W = \Delta U = U_2 - U_1$.
$W = 0 - (-pE) = pE$.

(B) The torque $\tau$ acting on an electric dipole in an electric field $E$ is given by $\tau = pE \sin \theta$,where $p = q \times (2a)$ is the dipole moment and $(2a)$ is the length of the dipole.
For maximum torque,$\sin \theta = 1$,so $\tau_{max} = pE = q(2a)E$.
Given: $q = 2 \mu C = 2 \times 10^{-6} \ C$,$E = 8 \times 10^{4} \ N/C$,and $\tau_{max} = 4 \times 10^{-3} \ N \cdot m$.
Substituting the values: $4 \times 10^{-3} = (2 \times 10^{-6}) \times (2a) \times (8 \times 10^{4})$.
$4 \times 10^{-3} = 16 \times 10^{-2} \times (2a)$.
$2a = \frac{4 \times 10^{-3}}{16 \times 10^{-2}} = \frac{4}{16} \times 10^{-1} = 0.25 \times 10^{-1} \ m$.
$2a = 0.025 \ m = 25 \ mm$.
Thus,the length of the dipole is $25 \ mm$.

(B) polar molecule is one that possesses a permanent electric dipole moment. This occurs when there is an asymmetric distribution of charge within the molecule.
$(a)$ $H_2O$ (Water): It has a bent geometry. The bond dipoles of the two $O-H$ bonds do not cancel each other out,resulting in a net dipole moment. Thus,it is a polar molecule.
$(b)$ $N_2$: It is a homonuclear diatomic molecule with a linear structure,so its dipole moment is zero.
$(c)$ $CO_2$: It has a linear structure where the two $C=O$ bond dipoles are equal and opposite,canceling each other out. Thus,it is non-polar.
$(d)$ $H_2$: It is a homonuclear diatomic molecule,so its dipole moment is zero.
Therefore,the correct option is $(a)$.

(B) The torque $\tau$ experienced by an electric dipole in a uniform electric field is given by the formula: $\tau = pE \sin \theta$,where $p = q \times (2a)$ is the dipole moment,$E$ is the electric field,and $\theta$ is the angle between the dipole axis and the electric field.
Given:
Length of dipole $(2a) = 2 \ cm = 0.02 \ m = 2 \times 10^{-2} \ m$
Electric field $(E) = 10^{5} \ N/C$
Angle $(\theta) = 60^{\circ}$
Torque $(\tau) = 8 \sqrt{3} \ Nm$
Substituting the values into the formula:
$8 \sqrt{3} = (q \times 2 \times 10^{-2}) \times 10^{5} \times \sin 60^{\circ}$
$8 \sqrt{3} = q \times 2 \times 10^{3} \times \frac{\sqrt{3}}{2}$
$8 \sqrt{3} = q \times \sqrt{3} \times 10^{3}$
$q = \frac{8 \sqrt{3}}{\sqrt{3} \times 10^{3}} = 8 \times 10^{-3} \ C$.
Thus,the magnitude of the charge is $8 \times 10^{-3} \ C$.

(B) The force acting on the charged particle in a uniform electric field $E$ is given by $F = qE$.
According to Newton's second law,the acceleration $a$ of the particle is $a = \frac{F}{m} = \frac{qE}{m}$.
Since the particle starts from rest,its initial velocity $u = 0$. After time $t$,the velocity $v$ of the particle is given by $v = u + at = 0 + (\frac{qE}{m})t = \frac{qEt}{m}$.
The kinetic energy $K$ of the particle is given by $K = \frac{1}{2}mv^2$.
Substituting the value of $v$,we get $K = \frac{1}{2}m(\frac{qEt}{m})^2 = \frac{1}{2}m(\frac{q^2E^2t^2}{m^2}) = \frac{q^2E^2t^2}{2m}$.

(A) For a conducting sphere,the charge $Q$ resides entirely on its outer surface.
Inside a conducting sphere,the electric field $E$ is zero everywhere because there is no enclosed charge $(q_{enclosed} = 0)$ according to Gauss's Law.
Since the electric field is zero inside the sphere,the potential $V$ is constant throughout the interior and is equal to the potential at the surface.
The potential at the surface (and thus at the centre) is given by $V = \frac{Q}{4 \pi \epsilon_0 R}$.
Therefore,the electric field at the centre is $0$ and the electric potential at the centre is $\frac{Q}{4 \pi \epsilon_0 R}$.

(C) Electric field lines are imaginary lines used to represent the electric field in a region.
$1$. They originate from positive charges and terminate on negative charges.
$2$. They never intersect each other because if they did,the electric field at the point of intersection would have two directions,which is physically impossible.
$3$. The density of field lines is proportional to the magnitude of the electric field intensity.
$4$. Electric field lines do not pass through the interior of a conductor in electrostatic equilibrium because the electric field inside a conductor is zero.
Therefore,the statement that they pass through the conductor is incorrect.

(A) Let the distance from the centre $O$ to each vertex be $r$. The electric field due to a charge $q$ at distance $r$ is $E = \frac{kq}{r^2}$.
Let $\vec{E}_A, \vec{E}_B, \vec{E}_C, \vec{E}_D, \vec{E}_E, \vec{E}_F$ be the electric fields at $O$ due to charges at $A, B, C, D, E, F$ respectively.
The charges are: $A(+q), B(-q), C(-q), D(+q), E(+Q), F(-q)$.
Electric fields at $O$:
$\vec{E}_A$ is directed away from $A$ (towards $D$).
$\vec{E}_D$ is directed away from $D$ (towards $A$).
Since $q_A = q_D = +q$,$\vec{E}_A + \vec{E}_D = 0$.
Similarly,$\vec{E}_B$ is directed towards $B$ (away from $E$),and $\vec{E}_E$ is directed away from $E$ (towards $B$).
$\vec{E}_C$ is directed towards $C$ (away from $F$),and $\vec{E}_F$ is directed towards $F$ (away from $C$).
Let $\vec{E}_0$ be the resultant field due to charges at $A, B, C, D, F$.
$\vec{E}_0 = \vec{E}_A + \vec{E}_B + \vec{E}_C + \vec{E}_D + \vec{E}_F$.
Since $\vec{E}_A + \vec{E}_D = 0$,we have $\vec{E}_0 = \vec{E}_B + \vec{E}_C + \vec{E}_F$.
All these are directed towards the respective vertices (as they are negative charges $-q$): $\vec{E}_B$ towards $B$,$\vec{E}_C$ towards $C$,$\vec{E}_F$ towards $F$.
By symmetry,the resultant of these three fields is a vector of magnitude $\frac{kq}{r^2}$ directed towards $E$.
Given: $|\vec{E}_0| = 2 |\vec{E}_E|$,where $\vec{E}_E$ is the field due to $+Q$ at $E$.
$\frac{kq}{r^2} = 2 \frac{kQ}{r^2} \implies Q = \frac{q}{2}$.

(D) The surface charge density $\sigma = 20 \ \mu C \ m^{-2} = 20 \times 10^{-6} \ C \ m^{-2}$.
The diameter of the sphere $d = 3.5 \ cm$,so the radius $r = 1.75 \ cm = 1.75 \times 10^{-2} \ m$.
The surface area of the sphere $A = 4 \pi r^2 = 4 \times 3.14 \times (1.75 \times 10^{-2})^2 \ m^2$.
$A = 12.56 \times 3.0625 \times 10^{-4} \approx 3.848 \times 10^{-3} \ m^2$.
The total charge $q = \sigma \times A = (20 \times 10^{-6}) \times (3.848 \times 10^{-3}) \approx 7.696 \times 10^{-8} \ C$.
According to Gauss's Law,the total electric flux $\phi = q / \epsilon_0$.
$\phi = (7.696 \times 10^{-8}) / (8.85 \times 10^{-12}) \approx 0.8696 \times 10^4 \approx 8.7 \times 10^3 \ N \cdot m^2 / C$.

(D) According to Gauss's Law,the total electric flux $\phi_{total}$ through a closed surface is given by $\phi_{total} = \frac{q}{\epsilon_0}$.
For the hollow cylinder,the total flux is the sum of the flux through the two plane surfaces ($A$ and $C$) and the curved surface $(B)$.
Let $\phi_A$,$\phi_C$,and $\phi_B$ be the flux through surfaces $A$,$C$,and $B$ respectively.
Thus,$\phi_A + \phi_C + \phi_B = \frac{q}{\epsilon_0}$.
Given that $\phi_B = \phi$,we have $\phi_A + \phi_C + \phi = \frac{q}{\epsilon_0}$.
Due to the symmetry of the cylinder,the flux through the two plane ends $A$ and $C$ must be equal,so $\phi_A = \phi_C$.
Substituting this into the equation: $2\phi_A + \phi = \frac{q}{\epsilon_0}$.
Solving for $\phi_A$: $2\phi_A = \frac{q}{\epsilon_0} - \phi$.
Therefore,$\phi_A = \frac{1}{2}\left(\frac{q}{\epsilon_0} - \phi\right)$.

(C) According to Gauss's Law, the total electric flux $\phi_{total}$ through a closed surface is given by $\phi_{total} = \frac{q}{\epsilon_0}$.
For the hollow cylinder, the total flux is the sum of the flux through the two plane surfaces ($A$ and $B$) and the curved surface $(C)$: $\phi_{total} = \phi_A + \phi_B + \phi_C$.
Given that $\phi_C = \phi$ and by symmetry, the flux through the two plane surfaces is equal, i.e., $\phi_A = \phi_B$.
Substituting these into the Gauss's Law equation: $\frac{q}{\epsilon_0} = \phi_A + \phi_A + \phi$.
$\frac{q}{\epsilon_0} - \phi = 2\phi_A$.
Therefore, the flux linked with the plane surface $A$ is $\phi_A = \frac{1}{2}\left(\frac{q}{\epsilon_0} - \phi\right)$.

(D) The total charge $Q$ of a solid sphere of radius $r$ with uniform volume charge density $\rho$ is given by $Q = \rho \times V = \rho \times (\frac{4}{3} \pi r^3)$.
According to Gauss's Law,the electric field $E$ at a distance $r$ from the center of a charged sphere is $E = \frac{1}{4 \pi \epsilon_0} \frac{Q}{r^2}$.
Substituting the value of $Q$ into the formula,we get $E = \frac{1}{4 \pi \epsilon_0} \frac{\rho (\frac{4}{3} \pi r^3)}{r^2}$.
Simplifying the expression,we get $E = \frac{\rho r}{3 \epsilon_0}$.

(A) According to Gauss's Law,the net electric flux $\phi$ through a closed surface is given by $\phi = \frac{q_{enclosed}}{\epsilon_0}$.
Here,the enclosed charges are $q_1 = 2 \mu C, q_2 = -3 \mu C, q_3 = 4 \mu C, q_4 = -4 \mu C$,and $q_5 = -1 \mu C$.
The net enclosed charge $q_{net} = q_1 + q_2 + q_3 + q_4 + q_5$.
$q_{net} = (2 - 3 + 4 - 4 - 1) \mu C = -2 \mu C$.
Therefore,the net outward flux $\phi = \frac{-2 \mu C}{\epsilon_0} = -\frac{2}{\epsilon_0} \mu V-m$.
Note: Since the question asks for the magnitude or the value based on the provided options,and the sum is $-2 \mu C$,there might be a typo in the question's provided options. However,based on the calculation,the net flux is $-\frac{2}{\epsilon_0}$.

(B) According to Gauss's Law,the total electric flux $\phi_{total}$ through a closed surface is given by $\phi_{total} = \frac{Q_{enclosed}}{\epsilon_0}$.
Since the charge $Q$ is placed at the centre of the cube,the flux is distributed equally among all $6$ faces of the cube due to symmetry.
Therefore,the flux through one face is $\phi_{face} = \frac{1}{6} \phi_{total} = \frac{Q}{6 \epsilon_0}$.
The question asks for the flux through two opposite faces.
Thus,the flux through two opposite faces is $2 \times \phi_{face} = 2 \times \frac{Q}{6 \epsilon_0} = \frac{Q}{3 \epsilon_0}$.

(C) According to Gauss's Law,the electric field $E$ near the surface of a charged conductor is given by $E = \frac{\sigma}{\epsilon_0}$.
This electric field is always directed perpendicular (normal) to the surface of the conductor.
If the field were not normal,there would be a component of the electric field parallel to the surface,which would cause the free charges on the conductor to move,contradicting the assumption of electrostatic equilibrium.
Therefore,the correct expression is $\frac{\sigma}{\epsilon_0}$ and it is normal to the surface.

(C) According to Gauss's Law,the total electric flux $\Phi_E$ through any closed surface is given by $\Phi_E = \frac{q_{enclosed}}{\epsilon_0}$,where $q_{enclosed}$ is the total charge enclosed by the surface and $\epsilon_0$ is the permittivity of free space.
In this problem,the charge $q$ is distributed on the surface of the balloon. As the balloon is blown up,the total charge $q$ enclosed by the surface remains constant.
Since the enclosed charge $q$ does not change,the total electric flux $\Phi_E = \frac{q}{\epsilon_0}$ also remains constant.
Therefore,the electric flux remains unchanged.

(C) According to Gauss's Law,the total electric flux $\phi_E$ through any closed surface is equal to the net charge $q_{enclosed}$ enclosed by the surface divided by the permittivity of free space $\epsilon_0$.
Mathematically,$\phi_E = \frac{q_{enclosed}}{\epsilon_0}$.
In all four figures $(a)$,$(b)$,$(c)$,and $(d)$,the charge enclosed by the surface is the same,which is $+q$.
Since the enclosed charge is identical for all surfaces and $\epsilon_0$ is a constant,the electric flux $\phi_E$ through each surface must be equal.
Therefore,the electric flux is the same for all the figures.

(D) The work done $W$ in moving a charge $q$ from point $A$ to point $B$ is given by the formula:
$W = q(V_B - V_A)$
Here,the potential at point $A$ is $V_A = -5 \ V$ and the potential at point $B$ is $V_B = V$.
Substituting these values into the formula:
$W = q(V - (-5))$
$W = q(V + 5)$
Now,solve for $V$:
$\frac{W}{q} = V + 5$
$V = \frac{W}{q} - 5$
Therefore,the correct option is $D$.

(B) The relationship between electric field $E$ and electric potential $V$ is given by the formula $E = -\frac{dV}{dx}$.
Given $V = 4x^2 + 8x - 3$.
Differentiating $V$ with respect to $x$:
$\frac{dV}{dx} = \frac{d}{dx}(4x^2 + 8x - 3) = 8x + 8$.
Substituting this into the electric field formula:
$E = -(8x + 8) = -8x - 8$.
Now,evaluate the electric field at $x = 0.5 \ m$:
$E = -8(0.5) - 8 = -4 - 8 = -12 \ V/m$.
Therefore,the correct option is $B$.

(D) Let the radius of each small drop be ' $r$ ' and the charge on each be ' $q$ '. The potential of a small drop is given by $V = \frac{kq}{r}$.
When ' $n$ ' drops coalesce to form a big drop of radius ' $R$ ',the volume remains conserved: $\frac{4}{3} \pi R^3 = n \cdot \frac{4}{3} \pi r^3$,which implies $R = n^{1/3} r$.
The total charge on the big drop is $Q = n \cdot q$.
The potential of the big drop ' $V_{big}$ ' is given by $V_{big} = \frac{kQ}{R} = \frac{k(nq)}{n^{1/3}r} = n^{1 - 1/3} \cdot \frac{kq}{r} = n^{2/3} \cdot V$.

(A) The electrical potential energy $U$ of a system of point charges is given by the sum of the potential energies of all pairs of charges: $U = k \left( \frac{q_1 q_2}{r_{12}} + \frac{q_2 q_3}{r_{23}} + \frac{q_3 q_1}{r_{31}} \right)$.
For an equilateral triangle with side length $a$,the distance between any two charges is $a$.
Given charges are $q_1 = Q$,$q_2 = 2Q$,and $q_3 = q$.
The total potential energy is $U = \frac{k}{a} (Q \cdot 2Q + 2Q \cdot q + q \cdot Q)$.
Setting $U = 0$ for the system to have zero potential energy:
$2Q^2 + 2Qq + qQ = 0$.
$2Q^2 + 3Qq = 0$.
$Q(2Q + 3q) = 0$.
Since $Q \neq 0$,we have $2Q + 3q = 0$.
Therefore,$q = -\frac{2}{3} Q$.

(A) The surface charge densities are $\sigma_A = +\sigma$,$\sigma_B = -\sigma$,and $\sigma_C = +\sigma$. The charges on the shells are $Q_A = 4\pi a^2 \sigma$,$Q_B = 4\pi b^2 (-\sigma) = -4\pi b^2 \sigma$,and $Q_C = 4\pi c^2 \sigma$.
The potential at the surface of shell $A$ is the sum of potentials due to all three shells: $V_A = V_{A,A} + V_{A,B} + V_{A,C}$.
Since $A$ is inside $B$ and $C$,the potential due to $B$ and $C$ at the surface of $A$ is equal to the potential at their respective surfaces: $V_A = \frac{1}{4\pi\epsilon_0} \left( \frac{Q_A}{a} + \frac{Q_B}{b} + \frac{Q_C}{c} \right)$.
Substituting the values: $V_A = \frac{1}{4\pi\epsilon_0} \left( \frac{4\pi a^2 \sigma}{a} + \frac{-4\pi b^2 \sigma}{b} + \frac{4\pi c^2 \sigma}{c} \right)$.
$V_A = \frac{1}{4\pi\epsilon_0} (4\pi a \sigma - 4\pi b \sigma + 4\pi c \sigma)$.
$V_A = \frac{\sigma}{\epsilon_0} (a - b + c)$.

(B) The electric potential $V$ at a point due to a charge $q$ at a distance $r$ is given by $V = \frac{1}{4\pi\epsilon_0} \frac{q}{r}$.
Point $A$ is at the midpoint of the side with charges $+2q$ and $+2q$. The distance of $A$ from each of these charges is $L$.
The distance of $A$ from each of the charges $-2q$ at the opposite corners can be found using the Pythagorean theorem. The horizontal distance is $2L$ and the vertical distance is $L$, so the distance $r = \sqrt{(2L)^2 + L^2} = \sqrt{5L^2} = L\sqrt{5}$.
The total potential at $A$ is the sum of potentials due to all four charges:
$V_A = \frac{1}{4\pi\epsilon_0} \left[ \frac{2q}{L} + \frac{2q}{L} + \frac{-2q}{L\sqrt{5}} + \frac{-2q}{L\sqrt{5}} \right]$
$V_A = \frac{1}{4\pi\epsilon_0} \left[ \frac{4q}{L} - \frac{4q}{L\sqrt{5}} \right]$
$V_A = \frac{4q}{4\pi\epsilon_0 L} \left[ 1 - \frac{1}{\sqrt{5}} \right] = \frac{q}{\pi\epsilon_0 L} \left[ 1 - \frac{1}{\sqrt{5}} \right]$.

(C) The potential energy $U$ of a system of point charges is given by $U = \sum \frac{1}{4 \pi \epsilon_0} \frac{q_i q_j}{r_{ij}}$.
For three identical charges $q$ placed at the vertices of an equilateral triangle of side $r$,the total potential energy is $U = 3 \times \left( \frac{1}{4 \pi \epsilon_0} \frac{q^2}{r} \right)$.
Given: $q = 3 \mu C = 3 \times 10^{-6} \ C$,$r = 6 \ cm = 0.06 \ m$,and $\frac{1}{4 \pi \epsilon_0} = 9 \times 10^9 \ N \cdot m^2/C^2$.
Substituting the values:
$U = 3 \times \left( 9 \times 10^9 \times \frac{(3 \times 10^{-6})^2}{0.06} \right)$
$U = 3 \times \left( 9 \times 10^9 \times \frac{9 \times 10^{-12}}{0.06} \right)$
$U = 3 \times \left( \frac{81 \times 10^{-3}}{0.06} \right)$
$U = 3 \times 1.35 = 4.05 \ J$.
Rounding to the nearest value,we get $4.1 \ J$.

(B) In a regular hexagon,the distance from the centre to each vertex is equal to the side length of the hexagon. Given side $a = 6 \ cm = 0.06 \ m$.
Since there are $6$ vertices,each with a charge $q = 2 \ \mu C = 2 \times 10^{-6} \ C$,the total potential $V$ at the centre is the sum of the potentials due to each charge.
$V = 6 \times \left( \frac{1}{4 \pi \epsilon_0} \cdot \frac{q}{a} \right)$
Substituting the values:
$V = 6 \times 9 \times 10^9 \times \frac{2 \times 10^{-6}}{0.06}$
$V = 54 \times 10^9 \times \frac{2 \times 10^{-6}}{6 \times 10^{-2}}$
$V = 54 \times 10^9 \times \frac{1}{3} \times 10^{-4}$
$V = 18 \times 10^5 \ V = 1.8 \times 10^6 \ V$.

(B) The work done $W$ to change the distance between two point charges $q_1$ and $q_2$ from $r_1$ to $r_2$ is given by the change in electrostatic potential energy: $W = U_f - U_i = \frac{1}{4 \pi \epsilon_0} q_1 q_2 \left( \frac{1}{r_2} - \frac{1}{r_1} \right)$.
Given: $q_1 = 10 \times 10^{-6} \ C$,$q_2 = 4 \times 10^{-6} \ C$,$r_1 = 10 \times 10^{-2} \ m$,$r_2 = (10 - 2) \times 10^{-2} = 8 \times 10^{-2} \ m$.
Substituting the values:
$W = (9 \times 10^9) \times (10 \times 10^{-6}) \times (4 \times 10^{-6}) \times \left( \frac{1}{8 \times 10^{-2}} - \frac{1}{10 \times 10^{-2}} \right)$.
$W = 360 \times 10^{-3} \times \left( \frac{100}{8} - \frac{100}{10} \right) \times 10^{-2} = 0.36 \times (12.5 - 10) = 0.36 \times 2.5 = 0.9 \ J$.

(D) Let the radius of each small drop be $r$ and the charge on each be $q$. The potential of a small drop is given by $v = \frac{kq}{r}$.
When $n$ small drops combine to form a big drop of radius $R$,the volume remains conserved: $\frac{4}{3} \pi R^3 = n \cdot \frac{4}{3} \pi r^3$,which gives $R = n^{1/3} r$.
The total charge on the big drop is $Q = nq$.
The potential of the big drop is $V = \frac{kQ}{R} = \frac{k(nq)}{n^{1/3} r} = n^{1 - 1/3} \cdot \frac{kq}{r} = n^{2/3} v$.

(C) The electrostatic potential energy $U$ of a system of point charges is given by $U = \sum \frac{k q_i q_j}{r_{ij}}$.
For the given system,the charges are $q_1 = Q$,$q_2 = -2q$,and $q_3 = -2q$. The distances between them are $r_{12} = l$,$r_{13} = l$,and $r_{23} = \sqrt{l^2 + l^2} = l\sqrt{2}$.
The total potential energy is:
$U = \frac{k Q(-2q)}{l} + \frac{k Q(-2q)}{l} + \frac{k (-2q)(-2q)}{l\sqrt{2}} = 0$
Dividing by $k$ and simplifying:
$-\frac{2Qq}{l} - \frac{2Qq}{l} + \frac{4q^2}{l\sqrt{2}} = 0$
$-\frac{4Qq}{l} + \frac{4q^2}{l\sqrt{2}} = 0$
$\frac{4Qq}{l} = \frac{4q^2}{l\sqrt{2}}$
$Q = \frac{q}{\sqrt{2}}$
Thus,the correct option is $C$.

(B) Let $r$ be the radius of each small drop and $q$ be the charge on each small drop.
The potential of a small drop is given by $V = \frac{kq}{r} = 20 \ V$.
When $n = 27$ drops combine to form a big drop of radius $R$ and charge $Q$,the volume remains constant:
$\frac{4}{3} \pi R^3 = n \cdot \frac{4}{3} \pi r^3 \implies R^3 = n r^3 \implies R = n^{1/3} r$.
For $n = 27$,$R = (27)^{1/3} r = 3r$.
The total charge on the big drop is $Q = nq = 27q$.
The potential of the big drop is $V' = \frac{kQ}{R} = \frac{k(nq)}{n^{1/3}r} = n^{2/3} \left( \frac{kq}{r} \right) = n^{2/3} V$.
Substituting the values: $V' = (27)^{2/3} \times 20 = (3^3)^{2/3} \times 20 = 3^2 \times 20 = 9 \times 20 = 180 \ V$.

(A) Let the point where the net potential is zero be at a distance $x$ from charge $A$ $(2 \mu C)$ along the line joining $A$ and $B$.
The potential $V$ due to a point charge $q$ at a distance $r$ is given by $V = \frac{kq}{r}$.
For the net potential to be zero,the sum of potentials from both charges must be zero:
$V_A + V_B = 0$
$\frac{k(2 \times 10^{-6})}{x} + \frac{k(-3 \times 10^{-6})}{1 - x} = 0$
$\frac{2}{x} = \frac{3}{1 - x}$
$2(1 - x) = 3x$
$2 - 2x = 3x$
$2 = 5x$
$x = \frac{2}{5} = 0.4 \ m$.
Thus,the distance from $A$ is $0.4 \ m$.

(D) The potential $V$ of a conducting sphere of radius $r$ carrying a charge $q$ is given by the formula $V = \frac{1}{4\pi\epsilon_0} \frac{q}{r}$.
Given that both spheres have the same charge,let $q_1 = q_2 = q$.
The potential of the first sphere is $V_1 = \frac{1}{4\pi\epsilon_0} \frac{q}{r_1}$.
The potential of the second sphere is $V_2 = \frac{1}{4\pi\epsilon_0} \frac{q}{r_2}$.
The ratio of their potentials is $\frac{V_1}{V_2} = \frac{\frac{1}{4\pi\epsilon_0} \frac{q}{r_1}}{\frac{1}{4\pi\epsilon_0} \frac{q}{r_2}} = \frac{r_2}{r_1}$.
Therefore,the correct option is $D$.

(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,$R = 2r_1 - r_1 - r_2 = r_1 - r_2$.

(A) The magnetic field at the centre $O$ is the sum of the magnetic fields due to the three parts: straight wire $PQ$, curved part $QRS$, and straight wire $ST$.
$1$. For the straight wire $PQ$: The point $O$ lies on the axis of the wire, so the magnetic field $B_{PQ} = 0$.
$2$. For the curved part $QRS$: The angle subtended at the centre is $\theta = 270^\circ = \frac{3\pi}{2} \text{ radians}$. The magnetic field is $B_{QRS} = \frac{\mu_0 I}{4\pi r} \theta = \frac{\mu_0 I}{4\pi r} \left(\frac{3\pi}{2}\right)$. Using the right-hand rule, the direction is into the page $(-\widehat{k})$.
$3$. For the straight wire $ST$: The point $O$ is at a perpendicular distance $r$ from the wire. The wire extends from $S$ to infinity. The magnetic field is $B_{ST} = \frac{\mu_0 I}{4\pi r} (\sin 90^\circ + \sin 0^\circ) = \frac{\mu_0 I}{4\pi r}$. Using the right-hand rule, the direction is into the page $(-\widehat{k})$.
Total magnetic field $B = B_{PQ} + B_{QRS} + B_{ST} = 0 + \frac{\mu_0 I}{4\pi r} \left(\frac{3\pi}{2}\right) + \frac{\mu_0 I}{4\pi r} = \frac{\mu_0 I}{4\pi r} \left(\frac{3\pi}{2} + 1\right) (-\widehat{k})$.

(C) According to the Biot-Savart Law,the magnetic field $d\vec{B}$ due to a current element $I d\vec{\ell}$ is given by:
$d\vec{B} = \frac{\mu_0}{4\pi} \frac{I (d\vec{\ell} \times \vec{r})}{r^3}$
Here,$I = 10 \ A$,$d\vec{\ell} = \Delta x \hat{i} = 10^{-2} \hat{i} \ m$,and the position vector $\vec{r} = 0.5 \hat{j} \ m$.
The cross product is $(d\vec{\ell} \times \vec{r}) = (10^{-2} \hat{i}) \times (0.5 \hat{j}) = 0.5 \times 10^{-2} (\hat{i} \times \hat{j}) = 0.005 \hat{k} \ m^2$.
The magnitude of the magnetic field is:
$|d\vec{B}| = \frac{\mu_0}{4\pi} \frac{I |d\vec{\ell} \times \vec{r}|}{r^3} = 10^{-7} \times \frac{10 \times 0.005}{(0.5)^3}$
$|d\vec{B}| = 10^{-7} \times \frac{0.05}{0.125} = 10^{-7} \times 0.4 = 4 \times 10^{-8} \ T$.

(C) The magnetic force per unit length between two parallel wires carrying currents $I_1$ and $I_2$ separated by distance $d$ is given by $f = \frac{\mu_0 I_1 I_2}{2 \pi d}$.
For length $L$,the force is $F = \frac{\mu_0 I^2 L}{2 \pi d}$.
In the new condition,the distance $d' = \frac{d}{2}$ and the currents $I' = 2I$.
The new force $F_2$ is given by $F_2 = \frac{\mu_0 (2I)(2I) L}{2 \pi (d/2)}$.
Simplifying this,we get $F_2 = \frac{\mu_0 (4I^2) L}{2 \pi (d/2)} = 8 \times \left( \frac{\mu_0 I^2 L}{2 \pi d} \right)$.
Therefore,$F_2 = 8F$.

(A) The force per unit length between two long parallel conductors is given by $F/L = \frac{\mu_0 I_1 I_2}{2 \pi d}$.
Initially,$F = \frac{\mu_0 I_1 I_2}{2 \pi d}$.
After changing the distance to $3d$,the new force $F'$ is given as $2/3 F$ (repulsive).
Since the original force was attractive (same direction currents),a repulsive force implies the direction of one current must be reversed.
Let the new currents be $I_1$ and $I_2'$. Then $F' = \frac{\mu_0 I_1 I_2'}{2 \pi (3d)} = \frac{2}{3} F$.
Substituting $F = \frac{\mu_0 I_1 I_2}{2 \pi d}$,we get $\frac{\mu_0 I_1 I_2'}{6 \pi d} = \frac{2}{3} \left( \frac{\mu_0 I_1 I_2}{2 \pi d} \right)$.
Simplifying,$\frac{I_2'}{3} = \frac{2}{3} \cdot \frac{I_2}{2} = \frac{I_2}{3}$.
Thus,$I_2' = I_2$. The magnitude remains the same,but the direction is reversed.

(D) The magnetic force on a current-carrying wire is given by the formula $\vec{F} = I(\vec{L} \times \vec{B})$.
Here,the wire is along the $x$-axis,so the length vector is $\vec{L} = L\hat{i}$.
The magnetic field is $\vec{B} = B_0(\hat{i} - \hat{j} - \hat{k})$.
Substituting these into the force equation:
$\vec{F} = I(L\hat{i}) \times B_0(\hat{i} - \hat{j} - \hat{k})$
$\vec{F} = ILB_0 [(\hat{i} \times \hat{i}) - (\hat{i} \times \hat{j}) - (\hat{i} \times \hat{k})]$
Using the cross product rules ($\hat{i} \times \hat{i} = 0$,$\hat{i} \times \hat{j} = \hat{k}$,$\hat{i} \times \hat{k} = -\hat{j}$):
$\vec{F} = ILB_0 [0 - \hat{k} - (-\hat{j})]$
$\vec{F} = ILB_0 (\hat{j} - \hat{k})$
The magnitude of the force is $|\vec{F}| = ILB_0 \sqrt{(1)^2 + (-1)^2} = ILB_0 \sqrt{1 + 1} = \sqrt{2} ILB_0$.

(A) Let the length of the wire be $L$. For the square loop, the perimeter is $4a = L$, so the side length $a = L/4$. The area $A_s = a^2 = (L/4)^2 = L^2/16$.
For the circular loop, the circumference is $2\pi r = L$, so the radius $r = L/(2\pi)$. The area $A_c = \pi r^2 = \pi (L/(2\pi))^2 = L^2/(4\pi)$.
Since $\pi \approx 3.14$, $4\pi \approx 12.56$, which is less than $16$. Therefore, $A_c > A_s$.
The torque experienced by a current-carrying loop in a magnetic field is given by $\tau = NIAB \sin \theta$. Since $N$, $I$, $B$, and $\theta$ are the same for both loops, the torque is directly proportional to the area $A$.
Since $A_c > A_s$, the torque experienced by the circular loop is greater.

(D) The magnetic field $B$ due to the straight wire at a distance $r$ is $B = \frac{\mu_0 I_2}{2 \pi r}$.
For side $AB$ (distance $r_1 = L/3$),the force is $F_{AB} = I_1 L B_1 = I_1 L \left( \frac{\mu_0 I_2}{2 \pi (L/3)} \right) = \frac{3 \mu_0 I_1 I_2}{2 \pi}$ (Attractive,towards the wire).
For side $CD$ (distance $r_2 = L/3 + L = 4L/3$),the force is $F_{CD} = I_1 L B_2 = I_1 L \left( \frac{\mu_0 I_2}{2 \pi (4L/3)} \right) = \frac{3 \mu_0 I_1 I_2}{8 \pi}$ (Repulsive,away from the wire).
The sides $BC$ and $AD$ are perpendicular to the wire,and the forces on them cancel each other out due to symmetry.
The net force $F_{net} = F_{AB} - F_{CD} = \frac{3 \mu_0 I_1 I_2}{2 \pi} - \frac{3 \mu_0 I_1 I_2}{8 \pi} = \frac{12 \mu_0 I_1 I_2 - 3 \mu_0 I_1 I_2}{8 \pi} = \frac{9 \mu_0 I_1 I_2}{8 \pi}$.