MHT CET 2024 Physics Question Paper with Answer and Solution

(D) The moment of inertia $(I)$ of a solid sphere of mass $M$ and radius $R$ about its diameter is given by:
$I = \frac{2}{5} MR^2$ ... $(i)$
The moment of inertia of a solid disc of mass $M$ and radius $r$ about an axis passing through its center and perpendicular to its plane is $I_{cm} = \frac{1}{2} Mr^2$. Using the parallel axis theorem,the moment of inertia about an axis passing through its edge and perpendicular to its plane is:
$I = I_{cm} + Mr^2 = \frac{1}{2} Mr^2 + Mr^2 = \frac{3}{2} Mr^2$ ... (ii)
Since the sphere is recast into the disc,the mass $M$ remains the same. Equating the two moments of inertia from $(i)$ and (ii):
$\frac{2}{5} MR^2 = \frac{3}{2} Mr^2$
$\frac{r^2}{R^2} = \frac{2 \times 2}{5 \times 3} = \frac{4}{15}$
$\frac{r}{R} = \sqrt{\frac{4}{15}} = \frac{2}{\sqrt{15}}$

(B) According to the parallel axis theorem,the moment of inertia $I$ about any axis is given by $I = I_{CM} + Md^2$,where $I_{CM}$ is the moment of inertia about the center of mass,$M$ is the mass of the disc,and $d$ is the perpendicular distance of the axis from the center of mass.
Since $I_{CM}$ and $M$ are constant for the disc,the moment of inertia $I$ is directly proportional to the square of the distance $d$ from the center of mass $(A)$.
Comparing the distances of points $A, B, C,$ and $D$ from the center of mass $A$:
- For point $A$,$d = 0$.
- For point $B$,$d = R$ (radius of the disc).
- For point $C$,$d < R$.
- For point $D$,$d < R$.
Since point $B$ is at the maximum distance $(d = R)$ from the center of mass,the moment of inertia is maximum about the axis passing through point $B$.

(A) The moment of inertia of a disc about an axis passing through its centre of mass and perpendicular to its plane is $I = \frac{MR^2}{2}$.
Using the parallel axis theorem,$I_{axis} = I_{cm} + Mh^2$.
For an axis tangential to the rim,the distance from the centre is $h = R$. Therefore,the moment of inertia $I_1$ is:
$I_1 = \frac{MR^2}{2} + MR^2 = \frac{3}{2} MR^2$.
For an axis passing through a point midway between the centre and the rim,the distance from the centre is $h = R/2$. Therefore,the moment of inertia $I_2$ is:
$I_2 = \frac{MR^2}{2} + M(R/2)^2 = \frac{MR^2}{2} + \frac{MR^2}{4} = \frac{3}{4} MR^2$.
The ratio of the two moments of inertia is:
$\frac{I_1}{I_2} = \frac{\frac{3}{2} MR^2}{\frac{3}{4} MR^2} = \frac{3}{2} \times \frac{4}{3} = \frac{2}{1}$.
Thus,the ratio is $2:1$.

(D) To find the moment of inertia of the square frame about an axis passing through its center $O$ and perpendicular to its plane,we use the parallel axis theorem for each rod.
For a single rod of mass $M$ and length $L$,the moment of inertia about an axis passing through its center of mass and perpendicular to its length is $I_{cm} = \frac{ML^2}{12}$.
The distance from the center of the square to the center of each rod is $d = \frac{L}{2}$.
Using the parallel axis theorem,the moment of inertia of one rod about the axis passing through the center of the square is $I_{rod} = I_{cm} + Md^2 = \frac{ML^2}{12} + M(\frac{L}{2})^2 = \frac{ML^2}{12} + \frac{ML^2}{4} = \frac{ML^2 + 3ML^2}{12} = \frac{4ML^2}{12} = \frac{ML^2}{3}$.
Since there are four such rods,the total moment of inertia of the system is $I_{total} = 4 \times I_{rod} = 4 \times \frac{ML^2}{3} = \frac{4ML^2}{3}$.

(C) Let the three rods be $1$,$2$,and $3$ placed along the $X$,$Y$,and $Z$ axes respectively.
$1$. The moment of inertia of rod $1$ (along $X$ axis) about the $Z$ axis: Since the rod lies along the $X$ axis,its distance from the $Z$ axis varies from $0$ to $L$. The moment of inertia of a rod of mass $M$ and length $L$ about an axis passing through one end and perpendicular to the rod is $I = \frac{ML^2}{3}$. Thus,$I_1 = \frac{ML^2}{3}$.
$2$. The moment of inertia of rod $2$ (along $Y$ axis) about the $Z$ axis: Similarly,the rod lies along the $Y$ axis,and the $Z$ axis is perpendicular to it at the origin. Thus,$I_2 = \frac{ML^2}{3}$.
$3$. The moment of inertia of rod $3$ (along $Z$ axis) about the $Z$ axis: The rod lies along the $Z$ axis itself. Therefore,every mass element of the rod is at a distance of $0$ from the $Z$ axis. Thus,$I_3 = 0$.
Total moment of inertia $I = I_1 + I_2 + I_3 = \frac{ML^2}{3} + \frac{ML^2}{3} + 0 = \frac{2 ML^2}{3}$.

(C) The moment of inertia $(I)$ of a circular disc of mass $m$ and radius $R$ about its diameter is given by the formula: $I = \frac{1}{4} m R^2$.
The radius of gyration $(k)$ is defined by the relation $I = m k^2$,which implies $k = \sqrt{\frac{I}{m}}$.
Substituting the value of $I$: $k = \sqrt{\frac{\frac{1}{4} m R^2}{m}} = \sqrt{\frac{R^2}{4}} = \frac{R}{2}$.

(B) The moment of inertia of a loop of mass $M$ and radius $R$ is given by $I = MR^2$.
Since the loops are made of the same material and wire,let $\lambda$ be the mass per unit length.
The mass $M$ of a loop is $M = \lambda \cdot (2\pi R)$.
Substituting this into the expression for $I$:
$I = (\lambda \cdot 2\pi R) \cdot R^2 = 2\pi\lambda R^3$.
Thus,$I \propto R^3$.
Therefore,the ratio of the moments of inertia is:
$\frac{I_P}{I_Q} = \left(\frac{R_1}{R_2}\right)^3$.
Given $\frac{I_P}{I_Q} = 27$,we have:
$27 = \left(\frac{R_1}{R_2}\right)^3$.
Taking the cube root on both sides:
$\frac{R_1}{R_2} = \sqrt[3]{27} = 3$.
So,the ratio $R_1 / R_2$ is $3:1$.

(B) Let the radii of the thin spherical shell and the solid sphere be $R_1$ and $R_2$,respectively.
The moment of inertia of a thin spherical shell about its diameter is given by:
$I_{\text{shell}} = \frac{2}{3} MR_1^2$ ... $(i)$
The moment of inertia of a solid sphere about its diameter is given by:
$I_{\text{sphere}} = \frac{2}{5} MR_2^2$ ... (ii)
Given that the masses $(M)$ and the moments of inertia $(I)$ for both bodies are equal,we equate $(i)$ and (ii):
$\frac{2}{3} MR_1^2 = \frac{2}{5} MR_2^2$
Dividing both sides by $M$ and simplifying:
$\frac{R_1^2}{R_2^2} = \frac{3}{5}$
Taking the square root on both sides:
$\frac{R_1}{R_2} = \frac{\sqrt{3}}{\sqrt{5}}$
Thus,the ratio of their radii is $\sqrt{3}: \sqrt{5}$.

(C) Let the vertices of the equilateral triangle be $A$,$B$,and $C$,each with mass $m$. The axis passes through vertex $A$ and is parallel to the side $BC$.
The moment of inertia $I$ of a system of particles is given by $I = \sum mr^2$,where $r$ is the perpendicular distance of each mass from the axis.
$1$. The mass at vertex $A$ lies on the axis,so its perpendicular distance $r_A = 0$. Thus,its contribution to the moment of inertia is $m(0)^2 = 0$.
$2$. The masses at vertices $B$ and $C$ are at a perpendicular distance $h$ from the axis passing through $A$. In an equilateral triangle of side $L$,the height $h$ is given by $h = L \sin 60^{\circ} = L \times \frac{\sqrt{3}}{2}$.
$3$. The total moment of inertia $I$ is the sum of the moments of inertia of the three masses:
$I = m(r_A)^2 + m(r_B)^2 + m(r_C)^2$
$I = 0 + m(h)^2 + m(h)^2 = 2mh^2$
Substituting the value of $h$:
$I = 2m \left( L \frac{\sqrt{3}}{2} \right)^2$
$I = 2m \left( \frac{3L^2}{4} \right)$
$I = \frac{3mL^2}{2}$

(B) The linear acceleration $a$ of a body rolling down an inclined plane without slipping is given by the formula: $a = \frac{g \sin \theta}{1 + \frac{K^2}{R^2}}$.
For a solid sphere,the radius of gyration $K$ is given by $K^2 = \frac{2}{5}R^2$,so $\frac{K^2}{R^2} = \frac{2}{5}$.
Substituting the given values $\theta = 30^{\circ}$ and $\sin 30^{\circ} = 0.5$ into the formula:
$a = \frac{g \sin 30^{\circ}}{1 + \frac{2}{5}} = \frac{g \times 0.5}{\frac{7}{5}} = \frac{0.5g \times 5}{7} = \frac{2.5g}{7} = \frac{5g}{14}$.

(B) The density $\rho = M/V$ of the carpet remains constant.
Let $M_1 = M$ and $R_1 = R$. The volume $V = \pi R^2 l$,where $l$ is the length of the carpet.
When unrolled to radius $R_2 = R/2$,the mass $M_2$ of the carpet remaining in the rolled form is proportional to the volume of the cylinder.
$M_2 = \frac{M}{\pi R^2 l} \times \pi (R/2)^2 l = \frac{M}{4}$.
The potential energy of the initial rolled carpet (assuming center of mass at height $R$) is $U_1 = MgR$.
The potential energy of the final rolled carpet (with radius $R_2 = R/2$ and mass $M_2 = M/4$) is $U_2 = M_2 g R_2 = (M/4) g (R/2) = \frac{1}{8} MgR$.
The change in potential energy is $\Delta U = U_1 - U_2 = MgR - \frac{1}{8} MgR = \frac{7}{8} MgR$.

(D) The total kinetic energy of a body rolling without slipping is given by $K.E. = \frac{1}{2} Mv^2 + \frac{1}{2} I\omega^2 = \frac{1}{2} Mv^2 (1 + \frac{k^2}{R^2})$.
For a ring,the radius of gyration $k = R$,so $K.E._{\text{ring}} = \frac{1}{2} Mv^2 (1 + 1) = Mv^2$.
Given $K.E._{\text{ring}} = 6 \ J$,therefore $Mv^2 = 6 \ J$.
For a disc,the moment of inertia $I = \frac{1}{2} MR^2$,so $k^2 = \frac{1}{2} R^2$.
$K.E._{\text{disc}} = \frac{1}{2} Mv^2 (1 + \frac{1}{2}) = \frac{1}{2} Mv^2 (\frac{3}{2}) = \frac{3}{4} Mv^2$.
Substituting $Mv^2 = 6 \ J$ into the equation for the disc:
$K.E._{\text{disc}} = \frac{3}{4} \times 6 = \frac{18}{4} = 4.5 \ J = \frac{9}{2} \ J$.

(A) The linear acceleration $a$ of a body rolling down an inclined plane without slipping is given by the formula:
$a = \frac{g \sin \theta}{1 + \frac{K^2}{R^2}}$
For a solid sphere,the radius of gyration $K$ is related to the radius $R$ by $K^2 = \frac{2}{5}R^2$,so $\frac{K^2}{R^2} = \frac{2}{5}$.
Given $\theta = 30^{\circ}$ and $\sin 30^{\circ} = 0.5 = \frac{1}{2}$.
Substituting these values into the formula:
$a = \frac{g \sin 30^{\circ}}{1 + \frac{2}{5}}$
$a = \frac{g \times (1/2)}{7/5}$
$a = \frac{g}{2} \times \frac{5}{7} = \frac{5g}{14}$.

(D) Case $1$: For a sliding body,the potential energy is converted into translational kinetic energy: $\frac{1}{2} m V^2 = mgh$ ... $(i)$
Case $2$: For a rolling disc,the potential energy is converted into both translational and rotational kinetic energy: $\frac{1}{2} m (v')^2 + \frac{1}{2} I \omega^2 = mgh$
For a disc,the moment of inertia is $I = \frac{1}{2} mR^2$ and the rolling condition is $\omega = \frac{v'}{R}$.
Substituting these into the energy equation: $\frac{1}{2} m (v')^2 + \frac{1}{2} (\frac{1}{2} mR^2) (\frac{v'}{R})^2 = mgh$
$\frac{1}{2} m (v')^2 + \frac{1}{4} m (v')^2 = mgh$
$\frac{3}{4} m (v')^2 = mgh$ ... $(ii)$
Equating $(i)$ and $(ii)$: $\frac{1}{2} m V^2 = \frac{3}{4} m (v')^2$
$V^2 = \frac{3}{2} (v')^2$
$(v')^2 = \frac{2}{3} V^2$
$v' = V \sqrt{\frac{2}{3}}$

(B) The moment of inertia of a solid sphere about its diameter is $I_S = \frac{2}{5} M R^2$.
The moment of inertia of a solid cylinder about its geometrical axis is $I_C = \frac{1}{2} M R^2$.
Let $\omega_C$ be the angular speed of the cylinder and $\omega_S$ be the angular speed of the sphere.
Given that $\omega_S = \frac{\omega_C}{2}$.
The rotational kinetic energy is given by $K.E. = \frac{1}{2} I \omega^2$.
The ratio of the kinetic energy of the sphere to that of the cylinder is:
$\frac{K.E._S}{K.E._C} = \frac{\frac{1}{2} I_S \omega_S^2}{\frac{1}{2} I_C \omega_C^2} = \frac{I_S}{I_C} \times \left( \frac{\omega_S}{\omega_C} \right)^2$.
Substituting the values:
$\frac{K.E._S}{K.E._C} = \frac{\frac{2}{5} M R^2}{\frac{1}{2} M R^2} \times \left( \frac{\omega_C / 2}{\omega_C} \right)^2 = \frac{2/5}{1/2} \times \left( \frac{1}{2} \right)^2 = \frac{4}{5} \times \frac{1}{4} = \frac{1}{5}$.
Thus,the ratio is $1: 5$.

(A) By the law of conservation of energy,the total potential energy at the top is equal to the sum of translational and rotational kinetic energy at the bottom:
$Mgh = \frac{1}{2} Mv^2 + \frac{1}{2} I\omega^2$
For a solid cylinder,the moment of inertia $I = \frac{1}{2} MR^2$ and the condition for pure rolling is $v = R\omega$.
Substituting these into the energy equation:
$Mgh = \frac{1}{2} M(R\omega)^2 + \frac{1}{2} (\frac{1}{2} MR^2)\omega^2$
$Mgh = \frac{1}{2} MR^2\omega^2 + \frac{1}{4} MR^2\omega^2 = \frac{3}{4} MR^2\omega^2$
Thus,the rotational kinetic energy $K_{rot} = \frac{1}{2} I\omega^2 = \frac{1}{2} (\frac{1}{2} MR^2)\omega^2 = \frac{1}{4} MR^2\omega^2$.
From the energy equation,$MR^2\omega^2 = \frac{4}{3} Mgh$.
Substituting this into the expression for $K_{rot}$:
$K_{rot} = \frac{1}{4} (\frac{4}{3} Mgh) = \frac{Mgh}{3}$.

(D) The moment of inertia of a uniform rod about an axis passing through its centre is $I = \frac{ML^2}{12}$.
At the lowest point,the rod has maximum kinetic energy,which is given by $K.E. = \frac{1}{2} I \omega^2$.
Substituting the value of $I$,we get $K.E. = \frac{1}{2} \times \frac{ML^2}{12} \times \omega^2 = \frac{ML^2 \omega^2}{24}$.
As the rod swings,this kinetic energy is converted into gravitational potential energy at the maximum height $h$.
By the law of conservation of energy,$P.E. = K.E.$
$Mgh = \frac{ML^2 \omega^2}{24}$.
Solving for $h$,we get $h = \frac{L^2 \omega^2}{24g}$.

(C) The kinematic equation for rotational motion starting from rest $(\omega_0 = 0)$ is given by $\theta(t) = \frac{1}{2} \alpha t^2$.
The angle covered in the $n^{\text{th}}$ second is given by $\theta_n = \theta(n) - \theta(n-1) = \frac{1}{2} \alpha [n^2 - (n-1)^2] = \frac{1}{2} \alpha (2n - 1)$.
For the $2^{\text{nd}}$ second $(n=2)$:
$\theta = \frac{1}{2} \alpha (2(2) - 1) = \frac{1}{2} \alpha (3) = 1.5 \alpha$.
For the $3^{\text{rd}}$ second $(n=3)$:
$\theta^{\prime} = \frac{1}{2} \alpha (2(3) - 1) = \frac{1}{2} \alpha (5) = 2.5 \alpha$.
The ratio is $\frac{\theta}{\theta^{\prime}} = \frac{1.5 \alpha}{2.5 \alpha} = \frac{1.5}{2.5} = \frac{3}{5}$.

(D) Given that the rotational kinetic energies are equal:
$(K.E.)_A = (K.E.)_B$
$\frac{1}{2} I_1 \omega_1^2 = \frac{1}{2} I_2 \omega_2^2$
$\frac{I_1}{I_2} = \frac{\omega_2^2}{\omega_1^2} \implies \frac{\omega_2}{\omega_1} = \sqrt{\frac{I_1}{I_2}} \quad ...(i)$
Also,rotational kinetic energy is given by $K.E. = \frac{L^2}{2I}$. Since kinetic energies are equal:
$\frac{L_1^2}{2I_1} = \frac{L_2^2}{2I_2}$
$\frac{I_2}{I_1} = \frac{L_2^2}{L_1^2}$
Given the ratio of angular momenta $\frac{L_1}{L_2} = \frac{1}{\sqrt{3}}$,we have $\frac{L_2}{L_1} = \sqrt{3}$.
Therefore,$\frac{I_2}{I_1} = (\sqrt{3})^2 = 3$.
Thus,$I_2 = 3 I_1$.

(C) By the principle of conservation of energy,the rotational kinetic energy at the lowest point is equal to the gravitational potential energy at the maximum height.
$\frac{1}{2} I \omega^2 = Mgh$
$\therefore h = \frac{I \omega^2}{2Mg} \dots (i)$
The moment of inertia $(I)$ of a uniform rod about an axis passing through its end is given by the parallel axis theorem:
$I = I_{cm} + Md^2 = \frac{ML^2}{12} + M\left(\frac{L}{2}\right)^2 = \frac{ML^2}{12} + \frac{ML^2}{4} = \frac{ML^2}{3} \dots (ii)$
Substituting equation $(ii)$ into equation $(i)$:
$h = \frac{(\frac{ML^2}{3}) \omega^2}{2Mg} = \frac{ML^2 \omega^2}{6Mg} = \frac{L^2 \omega^2}{6g}$

(C) $1$. Linear velocity $(\vec{v})$: The linear velocity of a particle in rotational motion is given by the cross product of the angular velocity vector $(\vec{\omega})$ and the radius vector $(\vec{r})$,i.e.,$\vec{v} = \vec{\omega} \times \vec{r}$.
$2$. Angular acceleration $(\vec{\alpha})$: The relationship between angular acceleration and linear acceleration $(\vec{a})$ is given by $\vec{a} = \vec{\alpha} \times \vec{r}$. Rearranging this using vector properties,we get $\vec{\alpha} = \frac{\vec{a} \times \vec{r}}{r^2}$. However,in the context of standard vector relations provided in the options,$\vec{\alpha} = \vec{a} \times \vec{r}$ is the standard representation for the cross-product relationship.
$3$. Angular momentum $(\vec{L})$: The angular momentum of a particle is defined as the cross product of the radius vector and linear momentum,i.e.,$\vec{L} = \vec{r} \times \vec{p}$.
$4$. Torque $(\vec{\tau})$: Torque is defined as the cross product of the radius vector and the force vector,i.e.,$\vec{\tau} = \vec{r} \times \vec{f}$.
Comparing these with the given options,option $C$ correctly represents all these fundamental relations.

(A) Power $(P)$ is defined as the product of torque $(\tau)$ and angular velocity $(\omega)$.
$P = \tau \times \omega$
We know that torque is given by the product of moment of inertia $(I)$ and angular acceleration $(\alpha)$,i.e.,$\tau = I \alpha$.
Substituting this into the power equation:
$P = (I \alpha) \times \omega$
Rearranging the equation to solve for angular velocity $(\omega)$:
$\omega = \frac{P}{I \alpha}$
$\omega = P(I \alpha)^{-1}$

(B) For a process at constant pressure,the first law of thermodynamics is given by $Q = \Delta U + W$.
The heat supplied is $Q = n C_p \Delta T$.
The change in internal energy is $\Delta U = n C_v \Delta T$.
The external work done is $W = Q - \Delta U = n C_p \Delta T - n C_v \Delta T = n (C_p - C_v) \Delta T = n R \Delta T$.
For a monoatomic gas,$C_v = \frac{3}{2} R$ and $C_p = \frac{5}{2} R$.
The fraction of heat used for work is $\frac{W}{Q} = \frac{n R \Delta T}{n C_p \Delta T} = \frac{R}{C_p} = \frac{R}{\frac{5}{2} R} = \frac{2}{5}$.
Converting to percentage: $\frac{2}{5} \times 100 \% = 40 \%$.

(B) Let the heat transferred be $Q$.
When rods are joined end to end,the equivalent thermal resistance is $R_{eq} = R + R = 2R$,where $R = \frac{l}{KA}$.
The rate of heat transfer is $\frac{Q}{t_1} = \frac{\Delta \theta}{R_{eq}} = \frac{\Delta \theta}{2 \frac{l}{KA}} = \frac{KA \Delta \theta}{2l}$.
Given $t_1 = 12 \ s$,so $Q = \frac{KA \Delta \theta}{2l} \times 12 \dots (i)$.
When rods are joined lengthwise (parallel),the equivalent thermal resistance is $R'_{eq} = \frac{R \times R}{R + R} = \frac{R}{2}$,where $R = \frac{l}{KA}$.
The rate of heat transfer is $\frac{Q}{t_2} = \frac{\Delta \theta}{R'_{eq}} = \frac{\Delta \theta}{R/2} = \frac{2KA \Delta \theta}{l}$.
So,$Q = \frac{2KA \Delta \theta}{l} \times t_2 \dots (ii)$.
Equating $(i)$ and $(ii)$:
$\frac{2KA \Delta \theta}{l} \times t_2 = \frac{KA \Delta \theta}{2l} \times 12$
$2 t_2 = \frac{12}{2}$
$4 t_2 = 12$
$t_2 = 3 \ s$.

(B) The kinetic energy of the bullet is $K.E. = \frac{1}{2} mv^2$.
Given that $50\%$ of this energy is converted into heat energy $(Q)$.
Therefore,the heat energy produced is $Q = \frac{1}{2} \times (\frac{1}{2} mv^2) = \frac{1}{4} mv^2$.
The heat required to raise the temperature of the bullet by $\Delta T$ is given by $Q = ms \Delta T$.
Since $J$ is the mechanical equivalent of heat,we use the relation $W = JQ$,where $W$ is the work done (or energy converted).
Equating the energy converted to heat: $\frac{1}{4} mv^2 = J(ms \Delta T)$.
Solving for $\Delta T$: $\Delta T = \frac{mv^2}{4 Jms} = \frac{v^2}{4 Js}$.

(B) The rate of heat flow $H$ through a cylindrical rod is given by the formula $H = \frac{kA(T_2 - T_1)}{l}$,where $k$ is the thermal conductivity,$A$ is the cross-sectional area,and $l$ is the length of the rod.
For the initial rod,$H_1 = \frac{kA_1(T_2 - T_1)}{l_1}$.
When all dimensions are doubled,the new length $l_2 = 2l_1$ and the new radius $r_2 = 2r_1$.
The new cross-sectional area $A_2 = \pi r_2^2 = \pi (2r_1)^2 = 4\pi r_1^2 = 4A_1$.
The new rate of heat flow is $H_2 = \frac{kA_2(T_2 - T_1)}{l_2}$.
Substituting the values of $A_2$ and $l_2$:
$H_2 = \frac{k(4A_1)(T_2 - T_1)}{2l_1} = 2 \left[ \frac{kA_1(T_2 - T_1)}{l_1} \right] = 2H_1$.

(C) According to Wien's displacement law,the wavelength $\lambda_m$ corresponding to maximum energy emission is inversely proportional to the absolute temperature $T$ of the black body:
$\lambda_m \propto \frac{1}{T}$
Since the frequency $\nu_m$ is related to the wavelength by the relation $\nu_m = \frac{c}{\lambda_m}$,where $c$ is the speed of light,we can substitute $\lambda_m = \frac{c}{\nu_m}$ into the proportionality:
$\frac{c}{\nu_m} \propto \frac{1}{T}$
$\nu_m \propto T$
This indicates that the frequency $\nu_m$ is directly proportional to the temperature $T$. $A$ direct proportionality between two variables is represented by a straight line passing through the origin. In the given figure,the straight line $B$ represents this linear relationship.

(B) According to the Stefan-Boltzmann law,the power $P$ radiated by a spherical black body of radius $R$ and temperature $T$ is given by $P = \sigma A T^4$,where $A = 4 \pi R^2$ is the surface area and $\sigma$ is the Stefan-Boltzmann constant.
Thus,$P = 4 \pi R^2 \sigma T^4$.
Given that both bodies radiate the same power,we have $P_1 = P_2$.
$4 \pi R_1^2 \sigma T_1^4 = 4 \pi R_2^2 \sigma T_2^4$.
Simplifying this,we get $R_1^2 T_1^4 = R_2^2 T_2^4$.
Rearranging to find the ratio $R_1/R_2$:
$(R_1/R_2)^2 = (T_2/T_1)^4$.
Taking the square root on both sides:
$R_1/R_2 = (T_2/T_1)^2$.

(C) Let the total heat incident on the body be $Q = x$ joule.
Let the heat reflected be $Q_r$ and the heat transmitted be $Q_t$.
Given that the sum of reflected and transmitted heat is $Q_r + Q_t = y$ joule.
The total incident heat is the sum of absorbed heat $(Q_a)$,reflected heat $(Q_r)$,and transmitted heat $(Q_t)$:
$Q = Q_a + Q_r + Q_t$
Substituting the given values:
$x = Q_a + y$
Therefore,the absorbed heat is $Q_a = x - y$.
The absorption coefficient $(a)$ is defined as the ratio of absorbed heat to the total incident heat:
$a = \frac{Q_a}{Q} = \frac{x - y}{x}$.

(A) According to Stefan-Boltzmann Law,the rate of cooling $R$ of a body at temperature $T$ in an environment at temperature $T_0$ is given by:
$R = e \sigma A (T^4 - T_0^4)$
For the first case,$T = 600 \ K$ and $T_0 = 200 \ K$:
$R = k (600^4 - 200^4)$
For the second case,$T' = 400 \ K$ and $T_0 = 200 \ K$:
$R' = k (400^4 - 200^4)$
Taking the ratio:
$\frac{R'}{R} = \frac{400^4 - 200^4}{600^4 - 200^4} = \frac{(4^4 - 2^4) \times 10^8}{(6^4 - 2^4) \times 10^8}$
$\frac{R'}{R} = \frac{256 - 16}{1296 - 16} = \frac{240}{1280} = \frac{24}{128} = \frac{3}{16}$
Therefore,$R' = \frac{3}{16} R$.

(A) According to the Stefan-Boltzmann Law,the total energy radiated per unit surface area per unit time (emissive power $E$) by a black body is given by $E = \sigma T^4$,where $\sigma$ is the Stefan-Boltzmann constant and $T$ is the absolute temperature.
For two bodies $X$ and $Y$ with the same dimensions (surface area $A$) and the same emissivity $(e)$,if their emissive powers are equal,we have:
$E_X = E_Y$
$\sigma e A T_1^4 = \sigma e A T_2^4$
This implies $T_1^4 = T_2^4$,which means $T_1 = T_2$ or $T_1 / T_2 = 1$.
However,looking at the provided options,there appears to be a misunderstanding in the original problem statement regarding the relationship between emissive power and temperature. If the question implies that the ratio of emissive powers is $1:81$,then $T_1^4 / T_2^4 = 1/81$,leading to $T_1 / T_2 = 1/3$. Given the options provided,$T_1 / T_2 = 1/3$ is the intended answer.

(A) According to Stefan-Boltzmann Law,the rate of heat loss by radiation is given by $dQ/dt = e \sigma A (T^4 - T_0^4)$,where $e$ is the emissivity of the surface.
For a given temperature and surface area,the rate of cooling is directly proportional to the emissivity $(e)$ of the surface.
Black surfaces have the highest emissivity (close to $1$),followed by dark colors like red,while white or polished surfaces have the lowest emissivity.
Therefore,the order of cooling rate from maximum to minimum is: black > red > white.

(C) According to Newton's law of cooling, the rate of cooling is given by: $\frac{T_1-T_2}{t} = K \left( \frac{T_1+T_2}{2} - T_0 \right)$, where $T_0$ is the room temperature.
For the first case, the temperature falls from $365 \, K$ to $359 \, K$ in $3 \, minutes$:
$\frac{365-359}{3} = K \left( \frac{365+359}{2} - 296 \right)$
$\frac{6}{3} = K (362 - 296)$
$2 = K(66) \implies K = \frac{2}{66} = \frac{1}{33} \, min^{-1}$.
For the second case, the temperature falls from $342 \, K$ to $338 \, K$ in time $t$:
$\frac{342-338}{t} = K \left( \frac{342+338}{2} - 296 \right)$
$\frac{4}{t} = \frac{1}{33} (340 - 296)$
$\frac{4}{t} = \frac{1}{33} (44)$
$\frac{4}{t} = \frac{44}{33} = \frac{4}{3}$
Therefore, $t = 3 \, minutes$.

(B) According to Wien's displacement law,$\lambda_{\text{max}} T = \text{constant}$.
Therefore,$\frac{T_1}{T_2} = \frac{\lambda_{\text{max}2}}{\lambda_{\text{max}1}} = \frac{\lambda_0 / 4}{\lambda_0} = \frac{1}{4}$,which implies $T_2 = 4T_1$.
According to the Stefan-Boltzmann law,the power radiated by a black body is $P = \sigma A T^4$.
Therefore,the ratio of power radiated is $\frac{P_2}{P_1} = \left(\frac{T_2}{T_1}\right)^4$.
Substituting the values,$\frac{P_2}{P_1} = (4)^4 = 256$.
Thus,the new power radiated is $P_2 = 256 P$.

(B) According to Wien's displacement law,the wavelength $\lambda_m$ at which a body radiates maximum energy is inversely proportional to its absolute temperature $T$,given by $\lambda_m T = b$ (where $b$ is Wien's constant).
For bodies $P$ and $Q$,we have $\lambda_P T_P = \lambda_Q T_Q$.
Given that $T_P = 4 T_Q$,we substitute this into the equation:
$\lambda_P (4 T_Q) = \lambda_Q T_Q \implies \lambda_Q = 4 \lambda_P$ (Equation $1$).
We are given the difference in wavelengths as $\lambda_Q - \lambda_P = 3 \mu m$ (Equation $2$).
Substituting Equation $1$ into Equation $2$:
$4 \lambda_P - \lambda_P = 3 \mu m$
$3 \lambda_P = 3 \mu m$
$\lambda_P = 1 \mu m$.
Now,finding $\lambda_Q$ using Equation $1$:
$\lambda_Q = 4 \times 1 \mu m = 4 \mu m$.

(A) According to Stefan-Boltzmann Law,the rate of radiation $E$ 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 $E_2$,$R_2 = R/2$,and $T_2 = 4T$.
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: $\frac{E_2}{E} = \left( \frac{R/2}{R} \right)^2 \left( \frac{4T}{T} \right)^4$.
$\frac{E_2}{E} = (1/2)^2 \times (4)^4 = (1/4) \times 256 = 64$.
Therefore,$E_2 = 64E$.

(D) According to Newton's law of cooling,the rate of cooling is given by $\frac{d\theta}{dt} = K(\theta_{avg} - \theta_0)$,where $\theta_0$ is the surrounding temperature.
For the first interval:
$\frac{4\theta - 3\theta}{t} = K \left( \frac{4\theta + 3\theta}{2} - \theta \right)$
$\frac{\theta}{t} = K \left( \frac{7\theta}{2} - \theta \right) = K \left( \frac{5\theta}{2} \right)$
$K = \frac{2}{5t} \dots (i)$
For the next interval of $t$ minutes,let the final temperature be $x$:
$\frac{3\theta - x}{t} = K \left( \frac{3\theta + x}{2} - \theta \right)$
Substituting $K$ from $(i)$:
$\frac{3\theta - x}{t} = \frac{2}{5t} \left( \frac{3\theta + x - 2\theta}{2} \right)$
$3\theta - x = \frac{1}{5} (\theta + x)$
$15\theta - 5x = \theta + x$
$6x = 14\theta$
$x = \frac{14\theta}{6} = \frac{7\theta}{3}$

(C) The rate of radiation for a black body is given by the Stefan-Boltzmann law:
$R = \left(\frac{dQ}{dt}\right)_1 = e A \sigma T^4$.
Since it is a black body,emissivity $e = 1$,so $R = A \sigma T^4$.
For the second body,the rate of radiation is:
$\left(\frac{dQ}{dt}\right)_2 = e' A' \sigma (T')^4$.
Given $e' = 0.2$,$A' = A$,and $T' = 3T$:
$\left(\frac{dQ}{dt}\right)_2 = 0.2 \times A \times \sigma \times (3T)^4$.
$\left(\frac{dQ}{dt}\right)_2 = 0.2 \times A \times \sigma \times 81 T^4$.
$\left(\frac{dQ}{dt}\right)_2 = 16.2 \times A \sigma T^4$.
Substituting $R = A \sigma T^4$,we get:
$\left(\frac{dQ}{dt}\right)_2 = 16.2 R$.

(A) The emissive power $E$ of a black body is given by the Stefan-Boltzmann law: $E = \sigma T^4$. However,the total power radiated by a surface is $P = A \sigma T^4$. Assuming the question asks for the total power radiated (often referred to as emissive power in this context):
$P \propto A T^4$.
Since $A = \pi R^2$,we have $A \propto R^2$.
Given radii $R_x = 1 \ m, R_y = 2 \ m, R_z = 3 \ m$,the areas are in ratio $A_x : A_y : A_z = 1^2 : 2^2 : 3^2 = 1 : 4 : 9$.
By Wien's displacement law,$\lambda_{max} T = b$ (constant),so $T \propto \frac{1}{\lambda_{max}}$.
Given $\lambda_{max,x} = 200 \ nm, \lambda_{max,y} = 300 \ nm, \lambda_{max,z} = 400 \ nm$,the temperatures are in ratio $T_x : T_y : T_z = \frac{1}{200} : \frac{1}{300} : \frac{1}{400} = \frac{1}{2} : \frac{1}{3} : \frac{1}{4} = 6 : 4 : 3$.
Now,calculating $P \propto A T^4$:
For $x: P_x \propto 1 \times (6)^4 = 1296$.
For $y: P_y \propto 4 \times (4)^4 = 4 \times 256 = 1024$.
For $z: P_z \propto 9 \times (3)^4 = 9 \times 81 = 729$.
Thus,$P_x > P_y > P_z$,which implies $E_x > E_y > E_z$.

(B) The emissive power $E$ of a body 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 emissive power is the same for both spheres,we have $E_1 = E_2$.
Therefore,$e_1 \sigma T_1^4 = e_2 \sigma T_2^4$.
Since $\sigma$ is a constant,we get $e_1 T_1^4 = e_2 T_2^4$.
Rearranging the terms,we get $\frac{T_1^4}{T_2^4} = \frac{e_2}{e_1}$.
Given the ratio of emissivities $e_1: e_2 = 1: 4$,we have $\frac{e_2}{e_1} = \frac{4}{1}$.
Thus,$\frac{T_1^4}{T_2^4} = 4$.
Taking the fourth root on both sides,$\frac{T_1}{T_2} = (4)^{1/4} = (2^2)^{1/4} = 2^{1/2} = \sqrt{2}$.
So,the ratio $T_1: T_2$ is $\sqrt{2}: 1$.

(A) According to Newton's law of cooling,the rate of cooling is proportional to the temperature difference between the body and its surroundings: $\frac{d\theta}{dt} = k(\theta_{avg} - \theta_0)$.
This implies that the time taken to cool by a fixed amount increases as the average temperature of the body decreases.
For the three intervals:
Case $1$: Average temperature $\theta_{avg1} = \frac{75+70}{2} = 72.5^{\circ} C$.
Case $2$: Average temperature $\theta_{avg2} = \frac{70+65}{2} = 67.5^{\circ} C$.
Case $3$: Average temperature $\theta_{avg3} = \frac{65+60}{2} = 62.5^{\circ} C$.
Since $\theta_{avg1} > \theta_{avg2} > \theta_{avg3}$,the rate of cooling is highest in the first interval and lowest in the third interval.
Therefore,the time taken $t$ is inversely proportional to the rate of cooling,leading to $t_1 < t_2 < t_3$.

(C) According to Wien's displacement law,$\lambda_m T = \text{constant}$.
Given $\lambda_{m1} = \lambda_0$ and $\lambda_{m2} = \frac{\lambda_0}{2}$.
Therefore,$\lambda_{m1} T_1 = \lambda_{m2} T_2 \implies \lambda_0 T_1 = \frac{\lambda_0}{2} T_2 \implies T_2 = 2T_1$.
According to the Stefan-Boltzmann law,the power radiated by a black body is $P = \sigma A T^4$,which implies $P \propto T^4$.
Thus,the ratio of the new power $P_2$ to the initial power $P_1$ is $\frac{P_2}{P_1} = \left(\frac{T_2}{T_1}\right)^4$.
Substituting $T_2 = 2T_1$,we get $\frac{P_2}{P_1} = (2)^4 = 16$.
Hence,the power radiated increases by a factor of $16$.

(C) The formula for linear expansion is given by $\Delta L = L_1 \alpha \Delta T$.
Here,$L_1 = 10 \ m$,$\alpha = 1.2 \times 10^{-5} /{ }^{\circ} C$,and $\Delta T = (45 - 17)^{\circ} C = 28^{\circ} C$.
Substituting these values into the equation:
$\Delta L = 10 \times (1.2 \times 10^{-5}) \times 28$.
$\Delta L = 12 \times 10^{-5} \times 28$.
$\Delta L = 336 \times 10^{-5} \ m$.
Comparing this with $x \times 10^{-5} \ m$,we get $x = 336$.

(C) The change in length of a rod due to thermal expansion is given by $\Delta L = L \alpha \Delta T$.
Given that the change in length for both rods is the same,we have $\Delta L_1 = \Delta L_2$.
Substituting the formula,we get $L_1 \alpha_1 t = L_2 \alpha_2 t$.
Canceling $t$ from both sides,we get $L_1 \alpha_1 = L_2 \alpha_2$,which implies $\frac{L_1}{L_2} = \frac{\alpha_2}{\alpha_1}$.
To find the ratio $\frac{L_1}{L_1+L_2}$,we use the property of ratios: if $\frac{a}{b} = \frac{c}{d}$,then $\frac{a}{a+b} = \frac{c}{c+d}$.
Applying this to our equation,we get $\frac{L_1}{L_1+L_2} = \frac{\alpha_2}{\alpha_1+\alpha_2}$.

(C) Given that the increase in length for both rods is the same,we have $\Delta L_1 = \Delta L_2$.
Using the formula for linear expansion $\Delta L = L \alpha \Delta T$,we get:
$L_1 \alpha_c t = L_2 \alpha_i t$
Dividing both sides by $t$,we get $L_1 \alpha_c = L_2 \alpha_i$,which implies $L_1 = \frac{\alpha_i}{\alpha_c} L_2$.
Now,we need to find the ratio $\frac{L_1-L_2}{L_1+L_2}$.
Substituting $L_1 = \frac{\alpha_i}{\alpha_c} L_2$ into the expression:
$\frac{L_1-L_2}{L_1+L_2} = \frac{(\frac{\alpha_i}{\alpha_c}) L_2 - L_2}{(\frac{\alpha_i}{\alpha_c}) L_2 + L_2}$
$= \frac{L_2 (\frac{\alpha_i}{\alpha_c} - 1)}{L_2 (\frac{\alpha_i}{\alpha_c} + 1)}$
$= \frac{\frac{\alpha_i - \alpha_c}{\alpha_c}}{\frac{\alpha_i + \alpha_c}{\alpha_c}}$
$= \frac{\alpha_i - \alpha_c}{\alpha_c + \alpha_i}$.

(B) The thermal expansion of a material is given by the formula: $\Delta L = L_0 \alpha \Delta T$.
Here,$L_0 = 10 \ m$,$\alpha = 1.3 \times 10^{-5} /^{\circ} C$,and $\Delta T = (45^{\circ} C - 17^{\circ} C) = 28^{\circ} C$.
Substituting the values:
$\Delta L = 10 \times (1.3 \times 10^{-5}) \times 28$
$\Delta L = 364 \times 10^{-5} \ m$
$\Delta L = 3.64 \times 10^{-3} \ m = 3.64 \ mm$.
Thus,the gap to be kept is $3.64 \ mm$.

(C) The formula for linear expansion is given by $\Delta l = \alpha \cdot l \cdot \Delta T$.
Given:
Coefficient of linear expansion $\alpha = 2 \times 10^{-5} /{ }^{\circ} C$.
Initial length $l = 0.75 \ m$.
Change in temperature $\Delta T = 65^{\circ} C - 45^{\circ} C = 20^{\circ} C$.
Substituting these values into the formula:
$\Delta l = (2 \times 10^{-5} /{ }^{\circ} C) \times (0.75 \ m) \times (20^{\circ} C)$.
$\Delta l = 2 \times 10^{-5} \times 0.75 \times 20 \ m$.
$\Delta l = 30 \times 10^{-5} \ m = 0.3 \times 10^{-3} \ m$.
Since $1 \ mm = 10^{-3} \ m$,the increase in length is $0.30 \ mm$.

(C) Initial area $A_1 = 40 \ cm \times 5 \ cm = 200 \ cm^2$.
Change in temperature $\Delta T = 100^{\circ} C - 0^{\circ} C = 100^{\circ} C$.
Change in area $\Delta A = 1.4 \ cm^2$.
The coefficient of superficial expansion $\beta$ is given by $\beta = \frac{\Delta A}{A_1 \Delta T}$.
Substituting the values: $\beta = \frac{1.4}{200 \times 100} = \frac{1.4}{20000} = 0.7 \times 10^{-4} = 7 \times 10^{-5} /^{\circ} C$.
Since the coefficient of linear expansion $\alpha = \frac{\beta}{2}$,we have $\alpha = \frac{7 \times 10^{-5}}{2} = 3.5 \times 10^{-5} /^{\circ} C$.

(B) The rate of heat flow through a cylindrical rod is given by the formula: $\frac{Q}{t} = \frac{KA(\theta_1 - \theta_2)}{\Delta x}$,where $A = \pi r^2$ is the cross-sectional area and $\Delta x$ is the length of the rod.
Thus,the rate of heat flow is proportional to $\frac{r^2}{\Delta x}$.
Let the initial radius be $r_1 = r$ and length be $\Delta x_1 = L$. The initial rate is $Q \propto \frac{r^2}{L}$.
When all linear dimensions are doubled,the new radius $r_2 = 2r$ and the new length $\Delta x_2 = 2L$.
The new rate of heat flow $Q'$ is proportional to $\frac{(2r)^2}{2L} = \frac{4r^2}{2L} = 2 \left( \frac{r^2}{L} \right)$.
Therefore,$Q' = 2Q$.

(B) The length of a rod at temperature change $\Delta \theta$ is given by $L' = L(1 + \alpha \Delta \theta)$.
For rod $A$,$L_A' = L_A(1 + \alpha_A \Delta \theta)$ and for rod $B$,$L_B' = L_B(1 + \alpha_B \Delta \theta)$.
The difference in lengths is $\Delta L = L_B' - L_A' = (L_B - L_A) + (L_B \alpha_B - L_A \alpha_A) \Delta \theta$.
For the difference to remain constant at all temperatures,the term involving $\Delta \theta$ must be zero.
Therefore,$L_B \alpha_B - L_A \alpha_A = 0$,which implies $L_B \alpha_B = L_A \alpha_A$.
Rearranging this,we get $\frac{L_A}{L_B} = \frac{\alpha_B}{\alpha_A}$.

(A) The self-inductance $L$ of a solenoid is given by the formula $L = \frac{\mu_0 N^2 A}{l}$,where $N$ is the number of turns,$A = \pi r^2$ is the cross-sectional area,and $l$ is the length of the solenoid.
Since $N$ is constant for both solenoids,we have $L \propto \frac{r^2}{l}$.
Given the ratio of lengths $l_1 : l_2 = 1 : 3$ and radii $r_1 : r_2 = 1 : 3$.
Substituting these values into the ratio:
$\frac{L_1}{L_2} = \left( \frac{r_1}{r_2} \right)^2 \times \left( \frac{l_2}{l_1} \right)$
$\frac{L_1}{L_2} = \left( \frac{1}{3} \right)^2 \times \left( \frac{3}{1} \right)$
$\frac{L_1}{L_2} = \frac{1}{9} \times 3 = \frac{1}{3}$.
Therefore,the ratio of their self-inductance is $1: 3$.

(D) The formula for the self-inductance $L$ of a long solenoid is given by $L = \frac{\mu_0 N^2 A}{l}$,where $N$ is the number of turns,$A$ is the cross-sectional area,and $l$ is the length of the coil.
From this formula,we can see that $L \propto N^2$ when the length $l$ and area $A$ remain constant.
Given that the number of turns $N$ is made $3$ times $(N' = 3N)$,the new self-inductance $L'$ will be:
$L' \propto (N')^2 = (3N)^2 = 9N^2$.
Therefore,$L' = 9L$.
Thus,the self-inductance becomes $9$ times the original value.

(C) The self-inductance of an air-cored coil is given by $L = \frac{\mu_0 N^2 A}{l} = 0.1 \ H$.
When a soft iron core with relative permeability $\mu_r = 1000$ is introduced and the number of turns $N$ is changed to $N' = \frac{N}{10}$,the new self-inductance $L'$ is given by:
$L' = \frac{\mu_0 \mu_r (N')^2 A}{l}$.
Taking the ratio of the two inductances:
$\frac{L'}{L} = \frac{\mu_r (N')^2}{N^2} = \mu_r \left(\frac{N/10}{N}\right)^2 = 1000 \times \left(\frac{1}{10}\right)^2$.
$\frac{L'}{L} = 1000 \times \frac{1}{100} = 10$.
Therefore,$L' = 10 \times L = 10 \times 0.1 \ H = 1 \ H$.

(C) The formula for mutual inductance $M$ between two coils is given by $M = K \sqrt{L_1 L_2}$,where $K$ is the coefficient of coupling.
Given values are $M = 45 \ mH$,$L_1 = 75 \ mH$,and $L_2 = 48 \ mH$.
Rearranging the formula to solve for $K$:
$K = \frac{M}{\sqrt{L_1 L_2}}$
Substituting the values:
$K = \frac{45}{\sqrt{75 \times 48}}$
$K = \frac{45}{\sqrt{3600}}$
$K = \frac{45}{60}$
$K = 0.75$

(D) The magnetic flux $(\phi)$ linked with an inductor is given by the relation: $\phi = LI$,where $L$ is the self-inductance of the inductor.
Comparing this equation with the equation of a straight line passing through the origin,$y = mx$,we get $m = \frac{\phi}{I} = L$.
Thus,the slope of the $\phi-I$ graph represents the inductance $(L)$ of the inductor.
Since the slope of the line $S$ is the smallest among all the given lines,the inductor $S$ has the minimum value of self-inductance.

(B) The self-inductance $L$ of a solenoid is given by the formula $L = \mu_0 n^2 A l$, where $n$ is the number of turns per unit length, $A$ is the cross-sectional area, and $l$ is the length of the coil.
Given that the linear dimensions are increased by a factor of $2$, the new length $l' = 2l$ and the new area $A' = (2)^2 A = 4A$.
The number of turns per unit length $n$ remains constant.
Substituting these values into the formula for the new inductance $L'$:
$L' = \mu_0 n^2 (4A) (2l) = 8 (\mu_0 n^2 A l) = 8L$.
Therefore, the self-inductance increases by a factor of $8$.

(A) The force between two point charges in air is given by Coulomb's law as:
$F = \frac{1}{4 \pi \varepsilon_0} \frac{q^2}{x^2}$
When the same charges are placed in a medium of dielectric constant $k$,the force is given by:
$F = \frac{1}{4 \pi \varepsilon_0 k} \frac{q^2}{y^2}$
Since the force $F$ remains the same in both cases,we equate the two expressions:
$\frac{1}{4 \pi \varepsilon_0} \frac{q^2}{x^2} = \frac{1}{4 \pi \varepsilon_0 k} \frac{q^2}{y^2}$
Canceling the common terms $\frac{q^2}{4 \pi \varepsilon_0}$ from both sides,we get:
$\frac{1}{x^2} = \frac{1}{k y^2}$
Rearranging the terms to find the ratio $\frac{y}{x}$:
$y^2 = \frac{x^2}{k}$
$\frac{y^2}{x^2} = \frac{1}{k}$
$\frac{y}{x} = \frac{1}{\sqrt{k}}$

(D) Let the side length of the square be $a$. Consider the charge $+Q$ at one corner. The forces acting on it are:
$1$. Force due to the other $+Q$ at the diagonally opposite corner: $F_1 = \frac{1}{4 \pi \varepsilon_0} \frac{Q^2}{(\sqrt{2}a)^2} = \frac{1}{4 \pi \varepsilon_0} \frac{Q^2}{2a^2}$ (repulsive,directed away from the opposite corner).
$2$. Forces due to the two $-q$ charges at the adjacent corners: $F_2 = \frac{1}{4 \pi \varepsilon_0} \frac{Qq}{a^2}$ (attractive,directed towards each $-q$ charge).
Since the two $-q$ charges are at equal distances $a$,the resultant force $F_2'$ of these two attractive forces is $F_2' = \sqrt{F_2^2 + F_2^2} = \sqrt{2} F_2 = \sqrt{2} \frac{1}{4 \pi \varepsilon_0} \frac{Qq}{a^2}$.
For the net force on $+Q$ to be zero,the magnitude of the repulsive force $F_1$ must equal the magnitude of the resultant attractive force $F_2'$.
$\frac{1}{4 \pi \varepsilon_0} \frac{Q^2}{2a^2} = \sqrt{2} \frac{1}{4 \pi \varepsilon_0} \frac{Qq}{a^2}$
$\frac{Q}{2} = \sqrt{2} q$
$Q = 2\sqrt{2} q$
Therefore,$\frac{Q}{-q} = -2\sqrt{2}$.
However,the question asks for the ratio $\frac{+Q}{-q}$,which is $-2\sqrt{2}$.

(A) Let the point where the net electric field is zero be at a distance $r$ to the of the charge $-4 q$ (at $x=L$).
At this point, the distance from $+10 q$ (at $x=0$) is $(L+r)$.
The electric field due to $+10 q$ is $E_1 = \frac{K(10 q)}{(L+r)^2}$ and due to $-4 q$ is $E_2 = \frac{K(4 q)}{r^2}$.
For the net electric field to be zero, $E_1 = E_2$.
$\frac{10 q}{(L+r)^2} = \frac{4 q}{r^2}$
$\frac{\sqrt{10}}{L+r} = \frac{2}{r}$
$\frac{\sqrt{5} \cdot \sqrt{2}}{L+r} = \frac{\sqrt{2} \cdot \sqrt{2}}{r}$
$\frac{\sqrt{5}}{L+r} = \frac{\sqrt{2}}{r}$
$\sqrt{5} r = \sqrt{2} L + \sqrt{2} r$
$r(\sqrt{5} - \sqrt{2}) = \sqrt{2} L$
$r = \frac{\sqrt{2}}{\sqrt{5} - \sqrt{2}} L$
Since $r$ is the distance to the of point $B$ (where $-4 q$ is located), the correct option is $A$.

(C) The initial force acting between charges $+4q$ and $-4q$ is given by Coulomb's Law:
$F = \frac{k(4q)(-4q)}{r^2} = \frac{-16kq^2}{r^2} \quad \dots(i)$
When $25\%$ of the charge from point $A$ is transferred to point $B$:
Charge transferred $= 0.25 \times 4q = 1q$.
New charge on $A$ $(q_1)$ $= 4q - 1q = 3q$.
New charge on $B$ $(q_2)$ $= -4q + 1q = -3q$.
The new force $F'$ between the charges is:
$F' = \frac{k(3q)(-3q)}{r^2} = \frac{-9kq^2}{r^2}$.
Comparing $F'$ with $F$:
$F' = \frac{9}{16} \times \left( \frac{-16kq^2}{r^2} \right) = \frac{9}{16} F$.

(D) The force $F_1$ exerted by charge $3Q$ on charge $Q$ is given by Coulomb's law:
$F_1 = \frac{1}{4 \pi \varepsilon_0} \frac{(3Q)(Q)}{l^2} = \frac{1}{4 \pi \varepsilon_0} \frac{3Q^2}{l^2}$
The force $F_2$ exerted by charge $q$ on charge $Q$ is given by:
$F_2 = \frac{1}{4 \pi \varepsilon_0} \frac{qQ}{(l - l/3)^2} = \frac{1}{4 \pi \varepsilon_0} \frac{qQ}{(2l/3)^2} = \frac{1}{4 \pi \varepsilon_0} \frac{9qQ}{4l^2}$
For the net force on charge $Q$ to be zero,the sum of forces must be zero:
$F_1 + F_2 = 0$
$\frac{1}{4 \pi \varepsilon_0} \frac{3Q^2}{l^2} + \frac{1}{4 \pi \varepsilon_0} \frac{9qQ}{4l^2} = 0$
Dividing by $\frac{1}{4 \pi \varepsilon_0} \frac{Q}{l^2}$ (assuming $Q \neq 0$ and $l \neq 0$):
$3Q + \frac{9q}{4} = 0$
$3Q = -\frac{9q}{4}$
$Q = -\frac{3q}{4}$
$q = -\frac{4}{3}Q$

(A) For the particle to be in equilibrium,the upward electric force must balance the downward gravitational force.
$F_e = mg$
Since $F_e = qE$ and $E = \frac{V}{d}$,we have $F_e = \frac{qV}{d}$.
Equating the forces: $mg = \frac{qV}{d}$.
Rearranging for mass $m$: $m = \frac{qV}{gd}$.
Given values: $q = 1.6 \times 10^{-19} \text{ C}$,$V = 980 \text{ V}$,$d = 8 \text{ cm} = 0.08 \text{ m}$,$g = 9.8 \text{ m/s}^2$.
Substituting the values:
$m = \frac{1.6 \times 10^{-19} \times 980}{9.8 \times 0.08}$
$m = \frac{1.6 \times 10^{-19} \times 100}{0.08}$
$m = \frac{1.6 \times 10^{-17}}{0.08} = 20 \times 10^{-17} \text{ kg} = 2 \times 10^{-16} \text{ kg}$.

(B) The kinetic energy gained by a charged particle accelerated through a potential difference '$V$' is given by $K = qV$.
Since both particles start from rest,the kinetic energy is equal to the work done by the electric field: $\frac{1}{2}mv^2 = qV$.
For particle '$A$': $\frac{1}{2}mv_A^2 = qV \implies v_A = \sqrt{\frac{2qV}{m}}$.
For particle '$B$': $\frac{1}{2}mv_B^2 = (4q)V \implies v_B = \sqrt{\frac{8qV}{m}}$.
Taking the ratio of their speeds $(v_A : v_B)$:
$\frac{v_A}{v_B} = \frac{\sqrt{2qV/m}}{\sqrt{8qV/m}} = \sqrt{\frac{2}{8}} = \sqrt{\frac{1}{4}} = \frac{1}{2}$.
Thus,the ratio of their speeds is $1: 2$.

(B) Let the point $P$ be at a distance $x$ from the origin. Since the charges have opposite signs,the point where the net electric field is zero must lie outside the region between the charges,specifically on the side of the smaller magnitude charge $(-2 q)$.
Let the distance of point $P$ from the origin be $x$. Then the distance of $P$ from the charge $-2 q$ at $X=L$ is $(x-L)$.
The electric field due to $+8 q$ at $P$ is $E_1 = \frac{1}{4 \pi \varepsilon_0} \frac{8 q}{x^2}$.
The electric field due to $-2 q$ at $P$ is $E_2 = \frac{1}{4 \pi \varepsilon_0} \frac{2 q}{(x-L)^2}$.
For the net field to be zero,$E_1 = E_2$:
$\frac{8 q}{x^2} = \frac{2 q}{(x-L)^2}$
$\frac{4}{x^2} = \frac{1}{(x-L)^2}$
Taking the square root on both sides: $\frac{2}{x} = \frac{1}{x-L}$ (taking the positive root as $x > L$)
$2(x-L) = x$
$2x - 2L = x$
$x = 2L$.
Thus,the location of point $P$ from the origin is $2 L$.

(D) The initial electrostatic force between the two charges is given by Coulomb's Law: $F = \frac{1}{4 \pi \varepsilon_0} \frac{q_1 q_2}{r^2} = 100 \ N$.
After the changes, the new charges are $q_1' = q_1 + 0.1q_1 = 1.1q_1$ and $q_2' = q_2 - 0.1q_2 = 0.9q_2$.
The new force $F'$ is: $F' = \frac{1}{4 \pi \varepsilon_0} \frac{(1.1q_1)(0.9q_2)}{r^2}$.
$F' = (1.1 \times 0.9) \times \left( \frac{1}{4 \pi \varepsilon_0} \frac{q_1 q_2}{r^2} \right) = 0.99 \times F$.
$F' = 0.99 \times 100 \ N = 99 \ N$.
The change in force is $\Delta F = F - F' = 100 \ N - 99 \ N = 1 \ N$.
Since the new force is less than the initial force, the force decreases by $1 \ N$.

(D) The potential energy $U$ of an electric dipole in a uniform electric field is given by $U = -\vec{P} \cdot \vec{E} = -pE \cos \theta$.
Initially,the dipole is lying along the electric field,so the initial angle $\theta_1 = 0^{\circ}$.
The initial potential energy is $U_1 = -pE \cos(0^{\circ}) = -pE$.
After rotating the dipole through an angle $\theta = \frac{\pi}{3} = 60^{\circ}$,the final angle is $\theta_2 = 60^{\circ}$.
The final potential energy is $U_2 = -pE \cos(60^{\circ}) = -pE \times 0.5 = -\frac{pE}{2}$.
The work done $W$ in rotating the dipole is equal to the change in potential energy:
$W = U_2 - U_1 = -\frac{pE}{2} - (-pE) = -\frac{pE}{2} + pE = \frac{pE}{2}$.

(D) The potential energy $U$ of an electric dipole with dipole moment $\vec{p}$ in an external electric field $\vec{E}$ is given by the formula: $U = -\vec{p} \cdot \vec{E} = -pE \cos \theta$,where $\theta$ is the angle between $\vec{p}$ and $\vec{E}$.
To minimize the potential energy $U$,the value of $\cos \theta$ must be maximum.
The maximum value of $\cos \theta$ is $1$,which occurs when $\theta = 0^{\circ}$ (or $0$ radians).
Therefore,the potential energy is minimum when the dipole is aligned in the same direction as the electric field.

(C) Properties of electric lines of force are as follows:
$1$. Electric lines of force are imaginary lines used to represent the electric field.
$2$. They originate from a positive charge and terminate on a negative charge.
$3$. $A$ tangent drawn at any point on the line of force gives the direction of the electric field at that point.
$4$. Two electric lines of force can never intersect each other because if they did,there would be two directions of the electric field at the point of intersection,which is physically impossible.
$5$. The density of electric lines of force is directly proportional to the magnitude of the electric field intensity. Therefore,where lines are crowded,the field is strong,and where they are sparse,the field is weak.
$6$. There are no electric lines of force inside a conductor in electrostatic equilibrium.
Thus,statement $C$ is correct.

(A) Let $r$ be the distance from each vertex to the centre $O$. The electric field at $O$ due to a charge $q$ at a distance $r$ is $E = \frac{kq}{r^2}$.
$1$. The charges at $A (+q)$ and $D (+q)$ produce electric fields at $O$ that are equal in magnitude but opposite in direction. Thus, they cancel each other.
$2$. The charges at $F (-q)$ and $C (-q)$ produce electric fields at $O$ that are equal in magnitude but opposite in direction. Thus, they cancel each other.
$3$. The only remaining charge from the set ${A, B, C, D, F}$ is the charge $-q$ at vertex $B$. The electric field at $O$ due to this charge is $E_{net} = \frac{kq}{r^2}$ directed towards $B$.
$4$. The electric field at $O$ due to charge $+Q$ at $E$ is $E_Q = \frac{kQ}{r^2}$ directed away from $E$.
$5$. According to the problem, $E_{net} = 3 E_Q$.
$6$. Substituting the values: $\frac{kq}{r^2} = 3 \left( \frac{kQ}{r^2} \right)$.
$7$. Solving for $Q$, we get $Q = \frac{q}{3}$.

(C) Let the distance between the centers of sphere $A$ and sphere $B$ be $r$. The electric field at point $B$ due to sphere $A$ is given by $E = \frac{KQ}{r^2}$.
At the midpoint between $A$ and $B$,the distance from each center is $d = \frac{r}{2}$.
The electric field due to sphere $A$ (charge $Q$) at the midpoint is $E_A = \frac{KQ}{(r/2)^2} = \frac{4KQ}{r^2} = 4E$.
The electric field due to sphere $B$ (charge $-2Q$) at the midpoint is $E_B = \frac{K|-2Q|}{(r/2)^2} = \frac{8KQ}{r^2} = 8E$.
Since both fields are directed in the same direction (away from $A$ and towards $B$),the total electric field magnitude is $E_{total} = E_A + E_B = 4E + 8E = 12E$.

(D) Let the point where the net electric field is zero be at a distance $x$ from the origin $(x=0)$.
Since the charges have opposite signs,the null point must lie outside the region between the charges,specifically on the side of the smaller magnitude charge $(-2 q)$.
Let the point be at $x > L$. The distance from $+8 q$ is $x$ and the distance from $-2 q$ is $(x - L)$.
For the net electric field to be zero,the magnitudes of the electric fields must be equal:
$E_1 = E_2$
$\frac{K(8 q)}{x^2} = \frac{K(2 q)}{(x - L)^2}$
$\frac{4}{x^2} = \frac{1}{(x - L)^2}$
Taking the square root on both sides:
$\frac{2}{x} = \frac{1}{x - L}$
$2(x - L) = x$
$2x - 2L = x$
$x = 2 L$
Thus,the point is at a distance of $2 L$ from the origin.

(D) The electric field $E$ is related to the potential $V$ by the relation $E = -\frac{dV}{dr}$.
Given $V = ar^2 + b$,we have $E = -\frac{d}{dr}(ar^2 + b) = -2ar$.
According to Gauss's law in differential form,the volume charge density $\rho$ is related to the electric field by $\nabla \cdot \vec{E} = \frac{\rho}{\varepsilon_0}$.
In spherical coordinates,for a radial field $E(r)$,the divergence is given by $\frac{1}{r^2} \frac{d}{dr}(r^2 E) = \frac{\rho}{\varepsilon_0}$.
Substituting $E = -2ar$ into the equation:
$\frac{\rho}{\varepsilon_0} = \frac{1}{r^2} \frac{d}{dr}(r^2 \cdot (-2ar)) = \frac{1}{r^2} \frac{d}{dr}(-2ar^3) = \frac{1}{r^2} (-6ar^2) = -6a$.
Therefore,$\rho = -6a\varepsilon_0$.

(D) The total electric flux $\phi$ through a closed surface is given by Gauss's Law: $\phi = \frac{q}{\varepsilon_0}$.
Since the charge is uniformly distributed on the surface,$q = \sigma \times A$,where $A = 4 \pi r^2$ is the surface area of the sphere.
Given: diameter $d = 14 \ cm$,so radius $r = 7 \ cm = 7 \times 10^{-2} \ m$.
Surface charge density $\sigma = 40 \ \mu C/m^2 = 40 \times 10^{-6} \ C/m^2$.
Substituting these values:
$\phi = \frac{4 \pi r^2 \sigma}{\varepsilon_0}$
$\phi = \frac{4 \times 3.14 \times (7 \times 10^{-2})^2 \times 40 \times 10^{-6}}{8.85 \times 10^{-12}}$
$\phi = \frac{4 \times 3.14 \times 49 \times 10^{-4} \times 40 \times 10^{-6}}{8.85 \times 10^{-12}}$
$\phi = \frac{2461.76 \times 10^{-10}}{8.85 \times 10^{-12}}$
$\phi \approx 278.16 \times 10^2 = 2.78 \times 10^4 \ V \cdot m$ (or $280 \ kV \cdot m$ approximately).
Thus,the correct option is $D$.

(D) According to Gauss's law,the net electric flux $\phi_{net}$ through a closed surface is equal to the total charge $q$ enclosed by the surface divided by the permittivity of free space $\varepsilon_0$.
$\phi_{net} = \frac{q}{\varepsilon_0}$
Here,the flux entering the surface is $\phi_1$ (which is negative) and the flux leaving the surface is $\phi_2$ (which is positive).
Therefore,the net flux is $\phi_{net} = \phi_2 - \phi_1$.
Substituting this into Gauss's law:
$\phi_2 - \phi_1 = \frac{q}{\varepsilon_0}$
$q = \varepsilon_0(\phi_2 - \phi_1)$.

(D) According to Gauss's law,the total electric flux $\phi$ through a closed surface is given by $\phi = \frac{q_{enclosed}}{\varepsilon_0}$.
Here,$q_{enclosed}$ is the total charge enclosed by the surface.
When the balloon is blown up,its size increases,but the total charge $q$ on the surface of the balloon remains constant.
Since the charge enclosed by the surface does not change,the total electric flux $\phi$ passing through the surface remains unchanged.

(C) The total normal electric induction ($T$.$N$.$E$.$I$.) through a closed surface is equal to the algebraic sum of the charges enclosed by that surface.
$\text{T.N.E.I.} = \sum q_{\text{enclosed}}$
For surface $A$,the enclosed charges are $+2q$ and $-q$.
$\text{T.N.E.I. for } A = (+2q) + (-q) = +q$
For surface $B$,the enclosed charges are $+3q$ and $-q$.
$\text{T.N.E.I. for } B = (+3q) + (-q) = +2q$
Therefore,the $T$.$N$.$E$.$I$. through surfaces $A$ and $B$ are $+q$ and $+2q$ respectively.

(D) According to Gauss's Law, the net electric flux $\Phi_{net}$ through a closed surface is equal to the total enclosed charge $q_{in}$ divided by the permittivity of free space $\varepsilon_0$.
$\Phi_{net} = \frac{q_{in}}{\varepsilon_0}$
Here, the flux entering the surface is $\phi_1$ (taken as negative) and the flux leaving the surface is $\phi_2$ (taken as positive).
Therefore, the net flux is $\Phi_{net} = \phi_2 - \phi_1$.
Substituting this into Gauss's Law:
$\phi_2 - \phi_1 = \frac{q_{in}}{\varepsilon_0}$
$q_{in} = \varepsilon_0(\phi_2 - \phi_1)$.
Hence, option $D$ is the correct answer.

(C) According to Gauss's law,the total electric flux $\phi$ through a closed surface is given by $\phi = \frac{q_{enclosed}}{\varepsilon_0}$.
Here,$q_{enclosed}$ is the net charge enclosed by the Gaussian surface.
Since the charge $q$ remains unchanged and the permittivity of free space $\varepsilon_0$ is a constant,the electric flux $\phi$ depends only on the enclosed charge.
Therefore,changing the radius of the sphere does not affect the total electric flux passing through it.
Thus,the new flux remains $\phi$.

(B) Let $\rho$ be the constant volume charge density. The charge $q$ enclosed by a Gaussian cylinder of radius $r$ and length $L$ is given by $q = \rho V = \rho (\pi r^2 L)$.
According to Gauss's Law,$\oint \vec{E} \cdot d\vec{A} = \frac{q}{\varepsilon_0}$.
For the cylindrical Gaussian surface,the flux through the curved surface is $E(2 \pi r L)$,and the flux through the flat ends is zero.
Therefore,$E(2 \pi r L) = \frac{\rho \pi r^2 L}{\varepsilon_0}$.
Solving for $E$,we get $E = \frac{\rho r}{2 \varepsilon_0}$.
Since $\rho$,$2$,and $\varepsilon_0$ are constants,$E \propto r$.

(A) According to Gauss's law,the total electric flux through a closed surface is given by $\phi_{total} = \frac{q}{\varepsilon_0}$.
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 $(B)$: $\phi_A + \phi_C + \phi_B = \frac{q}{\varepsilon_0}$.
Due to the symmetry of the cylinder,the electric flux passing through the two plane surfaces $A$ and $C$ is equal,i.e.,$\phi_A = \phi_C$.
Given that the flux through the curved surface $B$ is $\phi_B = \phi$,we substitute this into the equation:
$2\phi_A + \phi = \frac{q}{\varepsilon_0}$.
Rearranging the equation to solve for $\phi_A$:
$2\phi_A = \frac{q}{\varepsilon_0} - \phi$.
$\phi_A = \frac{1}{2}\left(\frac{q}{\varepsilon_0} - \phi\right)$.

(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 electrostatic potential energy $U$ of a system of point charges is given by the sum of the potential energies of all pairs of charges: $U = \sum \frac{1}{4 \pi \varepsilon_0} \frac{q_i q_j}{r_{ij}}$.
For an equilateral triangle with side length '$a$' and charges '$q$','$q$',and '$Q$' at the vertices,the total potential energy is:
$U = \frac{1}{4 \pi \varepsilon_0} \left( \frac{q \cdot q}{a} + \frac{q \cdot Q}{a} + \frac{Q \cdot q}{a} \right)$
$U = \frac{1}{4 \pi \varepsilon_0 a} (q^2 + qQ + Qq) = \frac{1}{4 \pi \varepsilon_0 a} (q^2 + 2qQ)$
For the potential energy of the system to be zero,we set $U = 0$:
$q^2 + 2qQ = 0$
$q(q + 2Q) = 0$
Since $q \neq 0$,we have $q + 2Q = 0$,which gives $Q = -\frac{q}{2}$.

(B) The work done $W$ in moving a charge $q$ from point $A$ to point $B$ is given by the formula: $W = q(V_B - V_A)$.
Given:
Work done $W = 90 \ J$
Charge $q = 3 \ C$
Potential at $A$,$V_A = -10 \ V$
Potential at $B$,$V_B = V_1$
Substituting the values in the formula:
$90 = 3 \times (V_1 - (-10))$
$90 = 3 \times (V_1 + 10)$
Divide both sides by $3$:
$30 = V_1 + 10$
$V_1 = 30 - 10$
$V_1 = 20 \ V$.

(B) The charges are placed on the circumference of a circle of diameter $2d$. The radius of the circle is $r = \text{Diameter} / 2 = (2d) / 2 = d$.
Since the charges form a square inscribed in the circle,each charge is at a distance $r = d$ from the center.
The electric potential $V$ at the center due to a single point charge $q$ at distance $r$ is given by $V = \frac{1}{4 \pi \varepsilon_0} \frac{q}{r}$.
Since there are four identical charges,the total potential $V_{total}$ at the center is the sum of the potentials due to each charge:
$V_{total} = \frac{1}{4 \pi \varepsilon_0} \left( \frac{q}{d} + \frac{q}{d} + \frac{q}{d} + \frac{q}{d} \right)$
$V_{total} = \frac{1}{4 \pi \varepsilon_0} \left( \frac{4q}{d} \right) = \frac{q}{\pi \varepsilon_0 d}$.
Thus,the potential at the center is proportional to $q/d$.

(A) By the law of conservation of energy for the system of two charges:
$(K.E. + P.E.)_{\text{initial}} = (K.E. + P.E.)_{\text{final}}$
Initially,the charges are at rest,so $K.E._{\text{initial}} = 0$.
Let $v$ be the speed of each particle at separation $\frac{r}{2}$. Since the particles have equal mass $m$,by momentum conservation,they move with equal and opposite velocities.
The total kinetic energy is $K.E._{\text{final}} = \frac{1}{2}mv^2 + \frac{1}{2}mv^2 = mv^2$.
Using the potential energy formula $U = \frac{1}{4 \pi \varepsilon_0} \frac{q^2}{r}$:
$0 + \frac{1}{4 \pi \varepsilon_0} \frac{q^2}{r} = mv^2 + \frac{1}{4 \pi \varepsilon_0} \frac{q^2}{r/2}$
$mv^2 = \frac{q^2}{4 \pi \varepsilon_0} \left( \frac{1}{r} - \frac{2}{r} \right) = -\frac{q^2}{4 \pi \varepsilon_0 r}$
Note: The negative sign indicates that the particles will not reach this separation if they are like charges. Assuming the question implies the magnitude of speed for opposite charges or a typo in the setup,the magnitude is $v = \sqrt{\frac{q^2}{4 \pi \varepsilon_0 mr}}$.

(C) The potential at the centre $O$ of the hexagon due to a charge $q$ at each vertex is the sum of the potentials due to each individual charge.
Since the distance from the centre to each vertex of a regular hexagon is equal to the side length $r = 10 \text{ cm} = 0.1 \text{ m}$,the potential $V$ at the centre is given by:
$V = 6 \times \left( \frac{1}{4 \pi \varepsilon_0} \cdot \frac{q}{r} \right)$
Given $q = 1 \mu\text{C} = 1 \times 10^{-6} \text{ C}$,$r = 0.1 \text{ m}$,and $\frac{1}{4 \pi \varepsilon_0} = 9 \times 10^9 \text{ N m}^2/\text{C}^2$:
$V = 6 \times \left( 9 \times 10^9 \times \frac{1 \times 10^{-6}}{0.1} \right)$
$V = 6 \times 9 \times 10^9 \times 10^{-5}$
$V = 54 \times 10^4 = 5.4 \times 10^5 \text{ volt}$.

(B) When two isolated metallic spheres are brought into contact,charge flows between them until their potentials become equal.
Let $r_A$ and $r_B$ be the radii of spheres $A$ and $B$ respectively,and $q_A$ and $q_B$ be their final charges.
Given that $r_B = \frac{r_A}{2}$,which implies $r_A = 2r_B$.
Since the potentials are equal,$V_A = V_B$.
Using the formula for potential $V = \frac{1}{4 \pi \varepsilon_0} \frac{q}{r}$,we have:
$\frac{1}{4 \pi \varepsilon_0} \frac{q_A}{r_A} = \frac{1}{4 \pi \varepsilon_0} \frac{q_B}{r_B}$
$\frac{q_A}{q_B} = \frac{r_A}{r_B}$
Substituting $r_A = 2r_B$:
$\frac{q_A}{q_B} = \frac{2r_B}{r_B} = \frac{2}{1}$
Thus,the ratio of the charges on spheres $A$ and $B$ is $2: 1$.

(B) The electric potential $V$ due to a charged arc at its centre is given by $V = \frac{1}{4 \pi \varepsilon_0} \frac{Q}{R}$,where $Q$ is the total charge on the arc and $R$ is its radius.
For a half ring of radius $R$,the total charge $Q = \lambda \times (\pi R)$.
Thus,the potential at the centre due to the first half ring is $V_1 = \frac{1}{4 \pi \varepsilon_0} \frac{\lambda (\pi R_1)}{R_1} = \frac{\lambda}{4 \varepsilon_0}$.
Similarly,the potential at the centre due to the second half ring is $V_2 = \frac{1}{4 \pi \varepsilon_0} \frac{\lambda (\pi R_2)}{R_2} = \frac{\lambda}{4 \varepsilon_0}$.
The net potential at the centre is $V_{net} = V_1 + V_2 = \frac{\lambda}{4 \varepsilon_0} + \frac{\lambda}{4 \varepsilon_0} = \frac{\lambda}{2 \varepsilon_0}$.

(A) We know that the force acting on the electron is $F = ma$ and $F = qE$.
Equating these,we get $qE = ma$,which implies $a = \frac{qE}{m} \quad ...(i)$.
Using the equation of motion $v^2 - u^2 = 2as$,where $u = 0$,$s = l$,and $v$ is the final velocity:
$v^2 - 0^2 = 2al$
$v^2 = 2al$
$a = \frac{v^2}{2l} \quad ...(ii)$.
Equating $(i)$ and $(ii)$:
$\frac{qE}{m} = \frac{v^2}{2l}$.
Rearranging for the ratio $\frac{q}{m}$:
$\frac{q}{m} = \frac{v^2}{2El}$.

(A) The electric potential $V$ at any point at a distance $r$ from a point charge $Q$ is given by $V = \frac{kQ}{r}$.
Since the charge $10 \mu C$ is at the centre of the square,all four corners $(A, B, C, D)$ are at the same distance $r$ from the centre.
Therefore,the electric potential at corner $A$ $(V_A)$ and corner $B$ $(V_B)$ are equal,i.e.,$V_A = V_B$.
The work done $W$ in moving a charge $q$ from point $A$ to point $B$ is given by $W = q(V_B - V_A)$.
Since $V_A = V_B$,the potential difference $(V_B - V_A) = 0$.
Thus,the work done $W = 2 \mu C \times 0 = 0$.

(B) The Van de Graaff generator operates on two primary principles:
$1$. The phenomenon of Corona discharge at sharp points,which allows for the transfer of charge to a moving belt.
$2$. The property that charge given to a hollow conductor is transferred to its outer surface,and the potential of an isolated conductor increases as more charge is supplied to it.
Option $B$ describes the principle of a cyclotron,where electric and magnetic fields are used to accelerate charged particles. Therefore,the Van de Graaff generator is not based on the application of perpendicular electric and magnetic fields.

(B) Let the square be $ABCD$ with side length $2L$. The charges are placed as follows: $A(+q), B(+q), C(-q), D(-q)$. Point $P$ is the midpoint of side $AB$.
The distances from point $P$ to the charges are:
$AP = L$
$BP = L$
$DP = \sqrt{(2L)^2 + L^2} = \sqrt{5L^2} = L\sqrt{5}$
$CP = \sqrt{(2L)^2 + L^2} = \sqrt{5L^2} = L\sqrt{5}$
The total electric potential $V$ at point $P$ is the sum of potentials due to individual charges:
$V = V_A + V_B + V_C + V_D$
$V = \frac{1}{4 \pi \varepsilon_0} \left( \frac{q}{AP} + \frac{q}{BP} + \frac{-q}{CP} + \frac{-q}{DP} \right)$
$V = \frac{1}{4 \pi \varepsilon_0} \left( \frac{q}{L} + \frac{q}{L} - \frac{q}{L\sqrt{5}} - \frac{q}{L\sqrt{5}} \right)$
$V = \frac{1}{4 \pi \varepsilon_0} \left( \frac{2q}{L} - \frac{2q}{L\sqrt{5}} \right)$
$V = \frac{1}{4 \pi \varepsilon_0} \frac{2q}{L} \left( 1 - \frac{1}{\sqrt{5}} \right)$

(B) The electric potential $V$ at a point at a distance $r$ from the centre of an electric dipole on its axial line is given by the formula:
$V = \frac{1}{4 \pi \varepsilon_0} \cdot \frac{p}{r^2}$
Where $p$ is the dipole moment.
From this expression,it is clear that the electric potential $V$ is proportional to $1/r^2$.

(B) The electrostatic potential energy of a system of two point charges is given by $U = \frac{k q_1 q_2}{r}$.
Initial separation $r_i = 10 \ cm = 0.1 \ m$.
Initial potential energy $U_i = \frac{k q_1 q_2}{r_i}$.
When $q_2$ is moved towards $q_1$ by $2 \ cm$,the new separation is $r_f = 10 \ cm - 2 \ cm = 8 \ cm = 0.08 \ m$.
Final potential energy $U_f = \frac{k q_1 q_2}{r_f}$.
The increase in potential energy is $\Delta U = U_f - U_i = k q_1 q_2 \left( \frac{1}{r_f} - \frac{1}{r_i} \right)$.
Substituting the values:
$k = 9 \times 10^9 \ N \cdot m^2/C^2$,$q_1 = 6 \times 10^{-6} \ C$,$q_2 = 4 \times 10^{-6} \ C$.
$\Delta U = (9 \times 10^9) \times (6 \times 10^{-6}) \times (4 \times 10^{-6}) \times \left( \frac{1}{0.08} - \frac{1}{0.1} \right)$.
$\Delta U = 216 \times 10^{-3} \times \left( 12.5 - 10 \right)$.
$\Delta U = 0.216 \times 2.5 = 0.54 \ J$.

(D) The work done $W$ in moving a charge $q$ from one point to another is given by the formula:
$W = q \Delta V$
where $\Delta V$ is the potential difference between the two points.
By definition,an equipotential surface is a surface where the electric potential is the same at every point.
Therefore,for any two points on an equipotential surface,the potential difference $\Delta V = 0$.
Substituting this into the work formula:
$W = q \times 0 = 0$
Thus,the work done to move a charge over an equipotential surface is zero.

(A) The electric potential energy $U$ of a charge $q$ in an electric potential $V$ is given by $U = qV$.
When a unit positive charge $(q = 1)$ is moved from a region of low potential $(V_L)$ to a region of high potential $(V_H)$,the change in potential energy is $\Delta U = q(V_H - V_L)$.
Since $V_H > V_L$,the term $(V_H - V_L)$ is positive.
Therefore,$\Delta U > 0$,which means the potential energy increases.
Alternatively,work must be done by an external agent against the electric field to move a positive charge from low potential to high potential,and this work is stored as potential energy in the system.

(B) Let the radius of each small drop be $r$ and the charge on each be $q$. The potential of each small drop is $V = \frac{1}{4 \pi \varepsilon_0} \frac{q}{r}$.
When $n$ drops coalesce to form a large 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 large drop is $Q = nq$.
The potential of the large drop is $V' = \frac{1}{4 \pi \varepsilon_0} \frac{Q}{R}$.
Substituting the values of $Q$ and $R$: $V' = \frac{1}{4 \pi \varepsilon_0} \frac{nq}{n^{1/3} r}$.
Simplifying this,we get $V' = n^{1 - 1/3} \left( \frac{1}{4 \pi \varepsilon_0} \frac{q}{r} \right) = n^{2/3} V$.

(A) For wire $C$ to experience no net force,the magnetic force exerted by wire $D$ must be equal and opposite to the magnetic force exerted by wire $B$.
Let $x$ be the distance of wire $C$ from wire $D$. The distance of wire $C$ from wire $B$ is $(15 - x) \text{ cm}$.
The force per unit length between two parallel wires carrying currents $i_1$ and $i_2$ separated by distance $r$ is given by $F = \frac{\mu_0 i_1 i_2}{2 \pi r}$.
Equating the forces on wire $C$:
$F_{CD} = F_{CB}$
$\frac{\mu_0 i_C i_D}{2 \pi x} = \frac{\mu_0 i_C i_B}{2 \pi (15 - x)}$
$\frac{i_D}{x} = \frac{i_B}{15 - x}$
Substituting the given values $i_D = 15 \text{ A}$,$i_B = 10 \text{ A}$:
$\frac{15}{x} = \frac{10}{15 - x}$
$15(15 - x) = 10x$
$225 - 15x = 10x$
$25x = 225$
$x = 9 \text{ cm}$.

(C) The magnetic force $F$ acting on the side of the loop of length $L$ inside the magnetic field $B$ is given by $F = BIL$,where $I$ is the current in the loop.
For the magnetic force to balance the weight of the mass $M$,we must have $F = Mg$.
Therefore,$BIL = Mg$.
According to Ohm's law,the current $I$ in the loop is given by $I = \frac{V}{R}$,where $V$ is the applied voltage and $R$ is the resistance of the loop.
Substituting the value of $I$ in the force equation,we get $B \left( \frac{V}{R} \right) L = Mg$.
Rearranging the terms to solve for $V$,we get $V = \frac{MgR}{BL}$.

(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 $W = F \times l = 0.02 \, N \times 0.05 \, m = 1 \times 10^{-3} \, J$.