MHT CET 2021 Physics Question Paper with Answer and Solution

(B) Let the displacements of particles $P$ and $Q$ be $x_1 = a \sin(\omega t + \phi_1)$ and $x_2 = a \sin(\omega t + \phi_2)$.
The distance between them is $d = |x_1 - x_2| = |a \sin(\omega t + \phi_1) - a \sin(\omega t + \phi_2)|$.
Using the formula $\sin C - \sin D = 2 \sin(\frac{C-D}{2}) \cos(\frac{C+D}{2})$,we get $d = |2a \sin(\frac{\phi_1 - \phi_2}{2}) \cos(\omega t + \frac{\phi_1 + \phi_2}{2})|$.
The maximum distance occurs when $|\cos(\omega t + \frac{\phi_1 + \phi_2}{2})| = 1$,so $d_{max} = |2a \sin(\frac{\Delta \phi}{2})|$,where $\Delta \phi = \phi_1 - \phi_2$.
Given $d_{max} = \sqrt{2} a$,we have $\sqrt{2} a = 2a \sin(\frac{\Delta \phi}{2})$.
$\sin(\frac{\Delta \phi}{2}) = \frac{\sqrt{2}}{2} = \frac{1}{\sqrt{2}}$.
$\frac{\Delta \phi}{2} = \frac{\pi}{4}$,which gives $\Delta \phi = \frac{\pi}{2}$.

(A) Given: Mass $m = 5 \,g = 5 \times 10^{-3} \,kg$,Amplitude $A = 0.3 \,m$,Time period $T = \frac{\pi}{5} \,s$.
Angular frequency $\omega = \frac{2\pi}{T} = \frac{2\pi}{\pi/5} = 10 \,rad/s$.
The maximum force acting on a particle in $S.H.M.$ is given by $F_{max} = m\omega^2A$.
Substituting the values: $F_{max} = (5 \times 10^{-3} \,kg) \times (10 \,rad/s)^2 \times (0.3 \,m)$.
$F_{max} = 5 \times 10^{-3} \times 100 \times 0.3 = 5 \times 10^{-1} \times 0.3 = 0.5 \times 0.3 = 0.15 \,N$.

(A) The maximum velocity of a particle performing $S.H.M.$ is given by the formula: $V_{max} = \omega A = \frac{2 \pi}{T} A$.
Here,$A$ is the amplitude and $T$ is the periodic time.
Given that the initial maximum velocity is $v = \frac{2 \pi A}{T}$.
If the amplitude is tripled $(A' = 3A)$ and the periodic time is doubled $(T' = 2T)$,the new maximum velocity $V'_{max}$ is:
$V'_{max} = \frac{2 \pi A'}{T'} = \frac{2 \pi (3A)}{2T} = \frac{3}{2} \left( \frac{2 \pi A}{T} \right) = 1.5 v$.
Therefore,the new maximum velocity will be $1.5 v$.

(D) The period $T$ of a simple harmonic motion is given by $T = 2\pi \sqrt{\frac{m}{k}}$,which shows that the period is independent of the amplitude $A$.
Therefore,the period will not change.
The total energy $E$ of a particle in $S.H.M.$ is given by $E = \frac{1}{2} m \omega^2 A^2$.
Since $E \propto A^2$,if the amplitude $A$ is decreased,the total energy $E$ will also decrease.
Thus,the period remains constant while the total energy decreases.

(B) Given: $m = 0.4 \,kg$, $f = \frac{16}{\pi} \,Hz$, $K.E. = 2 \,J$, $P.E. = 1.2 \,J$.
Angular frequency $\omega = 2 \pi f = 2 \pi \times \frac{16}{\pi} = 32 \,rad/s$.
Total energy $T.E. = K.E. + P.E. = 2 + 1.2 = 3.2 \,J$.
The formula for total energy in $S.H.M.$ is $T.E. = \frac{1}{2} m \omega^2 A^2$.
Substituting the values: $3.2 = \frac{1}{2} \times 0.4 \times (32)^2 \times A^2$.
$3.2 = 0.2 \times 1024 \times A^2$.
$3.2 = 204.8 \times A^2$.
$A^2 = \frac{3.2}{204.8} = \frac{32}{2048} = \frac{1}{64}$.
$A = \sqrt{\frac{1}{64}} = 0.125 \,m$.

(B) The acceleration $a$ of a particle in $S.H.M.$ is given by $a = \omega^2 x$,where $x$ is the displacement.
Given $a = 0.5 \,m/s^2$ and $x = 0.02 \,m$.
$\omega^2 = \frac{a}{x} = \frac{0.5}{0.02} = 25 \,rad^2/s^2$.
Taking the square root,we get $\omega = 5 \,rad/s$.
The maximum velocity $V_{\max}$ is given by $V_{\max} = A \omega$,where $A$ is the amplitude.
Given $A = 0.1 \,m$.
$V_{\max} = 0.1 \times 5 = 0.5 \,m/s$.

(D) The velocity $V$ of a particle in $S.H.M.$ at displacement $x$ is given by $V = \omega \sqrt{A^2 - x^2}$.
Maximum speed $V_{\max}$ occurs at the mean position and is given by $V_{\max} = A\omega$.
According to the problem,the speed $V$ is one-fourth of the maximum speed:
$V = \frac{1}{4} V_{\max}$
$\omega \sqrt{A^2 - x^2} = \frac{1}{4} (A\omega)$
$\sqrt{A^2 - x^2} = \frac{A}{4}$
Squaring both sides:
$A^2 - x^2 = \frac{A^2}{16}$
$x^2 = A^2 - \frac{A^2}{16} = \frac{15A^2}{16}$
$x = \frac{A\sqrt{15}}{4}$

(C) The given equation for displacement is $x = 5 \sin (3t + 3)$.
Comparing this with the standard equation of $S.H.M.$,$x = A \sin (\omega t + \phi)$,we get:
Amplitude $A = 5 \ cm$
Angular frequency $\omega = 3 \ rad/s$
The formula for the maximum acceleration of a particle in $S.H.M.$ is $a_{max} = A \omega^2$.
Substituting the values,we get $a_{max} = 5 \times (3)^2$.
$a_{max} = 5 \times 9 = 45 \ cm \ s^{-2}$.

(C) For a body performing $S.H.M.$ under a restoring force $F = kx$,the time period is given by $T = 2\pi \sqrt{\frac{m}{k}}$,which implies $k = \frac{4\pi^2 m}{T^2}$.
Given $F_1 = k_1 x$ and $F_2 = k_2 x$,we have $k_1 = \frac{4\pi^2 m}{T_1^2}$ and $k_2 = \frac{4\pi^2 m}{T_2^2}$.
When both forces act simultaneously in the same direction,the effective force constant is $k_{eff} = k_1 + k_2$.
The new time period $T$ is given by $T = 2\pi \sqrt{\frac{m}{k_1 + k_2}}$.
Squaring both sides,$T^2 = 4\pi^2 \frac{m}{k_1 + k_2}$,so $\frac{1}{T^2} = \frac{k_1 + k_2}{4\pi^2 m} = \frac{k_1}{4\pi^2 m} + \frac{k_2}{4\pi^2 m}$.
Substituting the expressions for $k_1$ and $k_2$,we get $\frac{1}{T^2} = \frac{1}{T_1^2} + \frac{1}{T_2^2} = \frac{T_1^2 + T_2^2}{T_1^2 T_2^2}$.
Therefore,$T = \frac{T_1 T_2}{\sqrt{T_1^2 + T_2^2}}$.

(C) The time period of a mass-spring system is given by $T = 2\pi \sqrt{\frac{m}{k}}$.
For the initial mass $m_1$,the time period is $T_1 = 2\pi \sqrt{\frac{m_1}{k}}$.
For the new mass $(m_1 + m_2)$,the time period is $T_2 = 2\pi \sqrt{\frac{m_1 + m_2}{k}}$.
Dividing the two equations: $\frac{T_2}{T_1} = \sqrt{\frac{m_1 + m_2}{m_1}}$.
Squaring both sides: $\frac{T_2^2}{T_1^2} = \frac{m_1 + m_2}{m_1} = 1 + \frac{m_2}{m_1}$.
Rearranging to find the ratio: $\frac{m_2}{m_1} = \frac{T_2^2}{T_1^2} - 1 = \frac{T_2^2 - T_1^2}{T_1^2}$.

(C) The time period $T$ of a simple pendulum is given by the formula:
$T = 2\pi \sqrt{\frac{\ell}{g}}$
From this relation,we can see that $T \propto \sqrt{\ell}$.
If a clock is running fast,it means its time period $T$ is too small (it completes oscillations too quickly).
To correct the time,we need to increase the time period $T$.
Since $T$ is directly proportional to the square root of the length $\ell$,increasing the length $\ell$ will increase the time period $T$.
The time period is independent of the mass of the bob and the amplitude of oscillation (for small angles).

(A) The time period of a particle of mass $m$ attached to a spring of constant $k$ is given by $T = 2\pi \sqrt{\frac{m}{k}}$.
For the first spring,$T_1 = 2\pi \sqrt{\frac{m}{k_1}}$,so $T_1^2 = 4\pi^2 \frac{m}{k_1}$.
For the second spring,$T_2 = 2\pi \sqrt{\frac{m}{k_2}}$,so $T_2^2 = 4\pi^2 \frac{m}{k_2}$.
When two springs are connected in series,the effective spring constant $k$ is given by $\frac{1}{k} = \frac{1}{k_1} + \frac{1}{k_2}$.
The time period $T$ for the series combination is $T = 2\pi \sqrt{\frac{m}{k}}$.
Squaring both sides,$T^2 = 4\pi^2 \frac{m}{k} = 4\pi^2 m \left( \frac{1}{k_1} + \frac{1}{k_2} \right)$.
Substituting the expressions for $T_1^2$ and $T_2^2$,we get $T^2 = 4\pi^2 \frac{m}{k_1} + 4\pi^2 \frac{m}{k_2} = T_1^2 + T_2^2$.
Therefore,$T = \sqrt{T_1^2 + T_2^2}$.

(A) In the series combination,the effective spring constant $k_s$ is given by:
$k_s = \frac{K \cdot K}{K + K} = \frac{K}{2}$
In the parallel combination,the effective spring constant $k_p$ is given by:
$k_p = K + K = 2K$
The frequency of oscillation for a mass-spring system is $f = \frac{1}{2\pi} \sqrt{\frac{k}{M}}$.
For the series combination,the frequency $f_s$ is:
$f_s = \frac{1}{2\pi} \sqrt{\frac{K/2}{M}} = \frac{1}{2\pi} \sqrt{\frac{K}{2M}}$
For the parallel combination,the frequency $f_p$ is:
$f_p = \frac{1}{2\pi} \sqrt{\frac{2K}{M}}$
The ratio of the frequencies in series to parallel combination is:
$\frac{f_s}{f_p} = \frac{\frac{1}{2\pi} \sqrt{\frac{K}{2M}}}{\frac{1}{2\pi} \sqrt{\frac{2K}{M}}} = \sqrt{\frac{K}{2M} \cdot \frac{M}{2K}} = \sqrt{\frac{1}{4}} = \frac{1}{2}$
Thus,the ratio is $1:2$.

(D) For a body of mass $m$ executing $SHM$,the force is $F = kx$,where $k$ is the force constant. The time period is given by $T = 2\pi \sqrt{\frac{m}{k}}$,which implies $k = \frac{4\pi^2 m}{T^2}$.
For force $F_1$,$k_1 = \frac{4\pi^2 m}{T_1^2}$.
For force $F_2$,$k_2 = \frac{4\pi^2 m}{T_2^2}$.
When both forces act simultaneously in the same direction,the effective force constant is $k_{eff} = k_1 + k_2$.
The new time period $T$ is given by $T = 2\pi \sqrt{\frac{m}{k_1 + k_2}}$.
Squaring both sides,$T^2 = \frac{4\pi^2 m}{k_1 + k_2}$.
Taking the reciprocal,$\frac{1}{T^2} = \frac{k_1 + k_2}{4\pi^2 m} = \frac{k_1}{4\pi^2 m} + \frac{k_2}{4\pi^2 m}$.
Substituting the expressions for $k_1$ and $k_2$,we get $\frac{1}{T^2} = \frac{1}{T_1^2} + \frac{1}{T_2^2} = \frac{T_1^2 + T_2^2}{T_1^2 T_2^2}$.
Therefore,$T = \frac{T_1 T_2}{\sqrt{T_1^2 + T_2^2}}$.

(C) Consider a mass $m$ at corner $A$. The gravitational forces exerted on it by the other two masses at $B$ and $C$ are $F_1$ and $F_2$,where $F_1 = F_2 = G \frac{m^2}{L^2}$.
The angle between these two forces is $60^{\circ}$. The resultant force $F$ is given by:
$F = \sqrt{F_1^2 + F_2^2 + 2F_1 F_2 \cos 60^{\circ}} = \sqrt{3} F_1 = \sqrt{3} G \frac{m^2}{L^2}$.
The distance of each mass from the center of the triangle is $r = \frac{L}{\sqrt{3}}$.
For uniform circular motion,the gravitational force provides the necessary centripetal force:
$mr \omega^2 = F$
$m \left( \frac{L}{\sqrt{3}} \right) \omega^2 = \sqrt{3} G \frac{m^2}{L^2}$
$\omega^2 = 3 G \frac{m}{L^3}$
Since $\omega = \frac{2\pi}{T}$,we have:
$\left( \frac{2\pi}{T} \right)^2 = \frac{3Gm}{L^3}$
$T^2 = \frac{4\pi^2 L^3}{3Gm}$
$T \propto L^{3/2}$.

(A) The frequency of a simple pendulum is given by $n = \frac{1}{2\pi} \sqrt{\frac{g}{l}}$.
At the surface of the Earth,$n = \frac{1}{2\pi} \sqrt{\frac{g}{l}}$.
At a depth $d$ below the surface,the acceleration due to gravity $g'$ is given by $g' = g(1 - \frac{d}{R})$.
Given $d = \frac{R}{2}$,we have $g' = g(1 - \frac{R/2}{R}) = g(1 - \frac{1}{2}) = \frac{g}{2}$.
The new frequency $n'$ at depth $d$ is $n' = \frac{1}{2\pi} \sqrt{\frac{g'}{l}}$.
Substituting $g' = \frac{g}{2}$,we get $n' = \frac{1}{2\pi} \sqrt{\frac{g/2}{l}} = \frac{1}{\sqrt{2}} \left( \frac{1}{2\pi} \sqrt{\frac{g}{l}} \right)$.
Therefore,$n' = \frac{n}{\sqrt{2}}$.

(B) The rotational kinetic energy $E$ is given by $E = \frac{L^2}{2I}$,where $L$ is the angular momentum and $I$ is the moment of inertia.
From this,we can write $L = \sqrt{2IE}$.
Given $I_1 = \frac{I_2}{3} \implies I_2 = 3I_1$ and $E_1 = 27E_2$.
The ratio of angular momenta is $\frac{L_1}{L_2} = \frac{\sqrt{2I_1E_1}}{\sqrt{2I_2E_2}}$.
Substituting the given values:
$\frac{L_1}{L_2} = \sqrt{\frac{I_1}{I_2} \cdot \frac{E_1}{E_2}} = \sqrt{\frac{I_1}{3I_1} \cdot \frac{27E_2}{E_2}}$.
$\frac{L_1}{L_2} = \sqrt{\frac{1}{3} \cdot 27} = \sqrt{9} = 3$.
Thus,the ratio $L_1 : L_2$ is $3: 1$.

(C) The kinetic energy of a rotating body is given by $K = \frac{1}{2} I \omega^2$,where $I$ is the moment of inertia and $\omega$ is the angular velocity.
Since $\omega = 2\pi f$,where $f$ is the frequency,we have $K \propto I f^2$.
Given that $f_2 = 2f_1$ and $K_2 = \frac{K_1}{4}$,we can write:
$\frac{K_2}{K_1} = \frac{I_2 f_2^2}{I_1 f_1^2}$
$\frac{1}{4} = \frac{I_2}{I_1} \times (2)^2$
$\frac{1}{4} = \frac{I_2}{I_1} \times 4$
$\frac{I_2}{I_1} = \frac{1}{16}$
Angular momentum is given by $L = I \omega = I(2\pi f)$,so $L \propto I f$.
$\frac{L_2}{L_1} = \frac{I_2 f_2}{I_1 f_1} = \left(\frac{I_2}{I_1}\right) \times \left(\frac{f_2}{f_1}\right)$
$\frac{L_2}{L_1} = \frac{1}{16} \times 2 = \frac{1}{8}$
Therefore,$L_2 = \frac{L}{8}$.

(D) The rotational kinetic energy $K$ is given by the formula $K = \frac{L^2}{2I}$,where $L$ is the angular momentum and $I$ is the moment of inertia.
Since the kinetic energies are equal,we have $K_P = K_Q$.
Therefore,$\frac{L_P^2}{2I_P} = \frac{L_Q^2}{2I_Q}$.
Rearranging the terms,we get $\frac{L_Q^2}{L_P^2} = \frac{I_Q}{I_P}$.
Taking the square root on both sides,$\frac{L_Q}{L_P} = \sqrt{\frac{I_Q}{I_P}}$.
Given that $I_Q > I_P$,it follows that $\frac{I_Q}{I_P} > 1$.
Thus,$\sqrt{\frac{I_Q}{I_P}} > 1$,which implies $\frac{L_Q}{L_P} > 1$.
Therefore,$L_Q > L_P$.

(B) The angular momentum of a particle is defined by the cross product of its position vector and linear momentum: $\overrightarrow{L} = \overrightarrow{r} \times \overrightarrow{p}$.
By the properties of the cross product,the vector $\overrightarrow{L}$ is perpendicular to both $\overrightarrow{r}$ and $\overrightarrow{p}$.
The magnitude of the angular momentum is given by $L = rp \sin \theta$,where $\theta$ is the angle between $\overrightarrow{r}$ and $\overrightarrow{p}$.
For the magnitude $L$ to be maximum,$\sin \theta$ must be maximum,which occurs when $\theta = 90^{\circ}$.
Therefore,$\overrightarrow{L}$ is maximum when $\overrightarrow{p}$ is perpendicular to $\overrightarrow{r}$.

(A) The initial kinetic energy of the system is given by $K = \frac{1}{2} I \omega^2$,where $I$ is the initial moment of inertia and $\omega$ is the initial angular velocity.
According to the law of conservation of angular momentum,since no external torque acts on the system,the angular momentum $L$ remains constant: $L = I \omega = I^{\prime} \omega^{\prime}$.
Given that the new moment of inertia is $I^{\prime} = 2I$,we have $2I \omega^{\prime} = I \omega$,which implies $\omega^{\prime} = \frac{\omega}{2}$.
The new kinetic energy $K^{\prime}$ is given by $K^{\prime} = \frac{1}{2} I^{\prime} \omega^{\prime 2}$.
Substituting the values,$K^{\prime} = \frac{1}{2} (2I) \left( \frac{\omega}{2} \right)^2 = I \left( \frac{\omega^2}{4} \right) = \frac{1}{2} \left( \frac{1}{2} I \omega^2 \right) = \frac{K}{2}$.

(C) The radius of gyration $K$ is defined by the relation $I = MK^2$,where $I$ is the moment of inertia and $M$ is the mass.
For a circular ring of mass $M$ and radius $R$,the moment of inertia about an axis passing through its centre and perpendicular to its plane is $I_{r} = MR^2$.
Thus,$MR^2 = MK_{r}^2$,which gives $K_{r} = R$.
For a circular disc of mass $M$ and radius $R$,the moment of inertia about an axis passing through its centre and perpendicular to its plane is $I_{d} = \frac{1}{2}MR^2$.
Thus,$\frac{1}{2}MR^2 = MK_{d}^2$,which gives $K_{d} = \frac{R}{\sqrt{2}}$.
The ratio of the radii of gyration is $\frac{K_{r}}{K_{d}} = \frac{R}{R/\sqrt{2}} = \sqrt{2} : 1$.

(C) The moment of inertia of a solid sphere about its diameter is given by $I = \frac{2}{5} MR^2$.
When the sphere is recast into a disc of radius $R'$ and mass $M$,the moment of inertia of the disc about an axis passing through its edge and perpendicular to its plane is given by the parallel axis theorem: $I' = I_{cm} + Md^2 = \frac{1}{2} MR'^2 + MR'^2 = \frac{3}{2} MR'^2$.
Given that $I' = I$,we equate the two expressions:
$\frac{3}{2} MR'^2 = \frac{2}{5} MR^2$.
Canceling $M$ from both sides,we get $R'^2 = \frac{2}{5} \times \frac{2}{3} R^2 = \frac{4}{15} R^2$.
Taking the square root,we find $R' = \sqrt{\frac{4}{15}} R = \frac{2}{\sqrt{15}} R$.

(A) The moment of inertia of a disc about an axis passing through its center of mass and perpendicular to its plane is given by $I_{cm} = \frac{1}{2} MR^2$.
Given $M = 1 \,kg$ and $R = 2 \,m$,$I_{cm} = \frac{1}{2} \times 1 \times (2)^2 = 2 \,kg \cdot m^2$.
According to the parallel axis theorem,the moment of inertia about an axis parallel to the $XY$ axis and passing through the edge of the disc is $I' = I_{cm} + Md^2$,where $d = R$ is the distance between the two axes.
Substituting the values,$I' = 2 + (1 \times 2^2) = 2 + 4 = 6 \,kg \cdot m^2$.

(B) The axis $YY'$ passes through the center of the top ring and is a diameter of that ring. The moment of inertia of the top ring about its diameter is $I_1 = \frac{1}{2} MR^2$.
For the two lower rings,the axis $YY'$ is a tangent to each ring in its own plane. According to the parallel axis theorem,the moment of inertia of a ring about a tangent in its plane is $I_{tangent} = I_{cm} + MR^2 = \frac{1}{2} MR^2 + MR^2 = \frac{3}{2} MR^2$.
Since there are two such lower rings,their combined moment of inertia is $I_2 + I_3 = \frac{3}{2} MR^2 + \frac{3}{2} MR^2 = 3 MR^2$.
The total moment of inertia of the system about the axis $YY'$ is $I = I_1 + I_2 + I_3 = \frac{1}{2} MR^2 + \frac{3}{2} MR^2 + \frac{3}{2} MR^2 = \frac{7}{2} MR^2$.

(A) The moment of inertia of a ring of mass $M$ and radius $R$ about an axis passing through its center and perpendicular to its plane is $I = MR^2$.
Given mass per unit length is $\lambda$. The mass of the first ring is $M_1 = \lambda(2\pi R)$ and its radius is $R_1 = R$.
So,$I_1 = M_1 R_1^2 = (2\pi R \lambda) R^2 = 2\pi \lambda R^3$.
The mass of the second ring is $M_2 = \lambda(2\pi nR)$ and its radius is $R_2 = nR$.
So,$I_2 = M_2 R_2^2 = (2\pi nR \lambda) (nR)^2 = 2\pi \lambda n^3 R^3$.
The ratio is given as $\frac{I_1}{I_2} = \frac{1}{8}$.
Substituting the expressions: $\frac{2\pi \lambda R^3}{2\pi \lambda n^3 R^3} = \frac{1}{n^3} = \frac{1}{8}$.
Therefore,$n^3 = 8$,which gives $n = 2$.

(A) The moment of inertia $I$ of a body in terms of its mass $m$ and radius of gyration $K$ is given by $I = mK^2$.
Angular momentum $L$ is related to moment of inertia $I$ and angular velocity $\omega$ by the formula $L = I\omega$.
Substituting the expression for $I$ into the angular momentum formula,we get $L = (mK^2)\omega$.
Rearranging this to solve for angular velocity $\omega$,we get $\omega = \frac{L}{mK^2}$.

(A) The moment of inertia of the upper sphere about the axis $YY'$ passing through its center is $I_1 = \frac{2}{5} MR^2$.
For each of the two lower spheres,the axis $YY'$ is at a distance $R$ from their centers. Using the parallel axis theorem,the moment of inertia for each lower sphere is $I_2 = I_{cm} + MR^2 = \frac{2}{5} MR^2 + MR^2 = \frac{7}{5} MR^2$.
Therefore,the total moment of inertia of the system is $I = I_1 + 2 \times I_2 = \frac{2}{5} MR^2 + 2 \times \frac{7}{5} MR^2 = \frac{2}{5} MR^2 + \frac{14}{5} MR^2 = \frac{16}{5} MR^2$.

(C) The moment of inertia of a body about an axis is given by $I = \sum m_i r_i^2$,where $r_i$ is the perpendicular distance of the $i$-th particle from the axis of rotation.
For a given mass distribution,the moment of inertia is larger if the mass is distributed at a greater average distance from the axis of rotation.
In the given triangular lamina $PQR$,the axis $QR$ is the side of the triangle. The entire mass of the lamina is distributed such that most of the area (and thus mass) is at a relatively smaller distance from the axis $PQ$ or $PR$ compared to the axis $QR$ which is the base.
However,considering the perpendicular distance of the vertices from the axes,the axis $QR$ allows the vertex $P$ to be at the maximum perpendicular distance compared to other axes like $PQ$ or $PR$ where the vertices are closer to the axis.
Therefore,the moment of inertia is maximum about the axis $QR$.

(C) The moment of inertia of a circular loop of mass $M$ and radius $R$ about its central axis is $I = MR^2$.
Let $m$ be the mass per unit length of the wire. Then the masses of the loops are $M_P = 2\pi R_1 m$ and $M_Q = 2\pi R_2 m$.
The moments of inertia are $I_P = M_P R_1^2 = (2\pi R_1 m) R_1^2 = 2\pi m R_1^3$ and $I_Q = M_Q R_2^2 = (2\pi R_2 m) R_2^2 = 2\pi m R_2^3$.
Taking the ratio: $\frac{I_P}{I_Q} = \frac{2\pi m R_1^3}{2\pi m R_2^3} = \left(\frac{R_1}{R_2}\right)^3$.
Given $\frac{I_P}{I_Q} = \frac{1}{8}$,we have $\left(\frac{R_1}{R_2}\right)^3 = \frac{1}{8}$.
Taking the cube root on both sides: $\frac{R_1}{R_2} = \frac{1}{2}$.
Therefore,$\frac{R_2}{R_1} = 2$.

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

(D) The moment of inertia of a solid sphere about its diameter is $I_{\text{sphere}} = \frac{2}{5} M R^2$. The rotational kinetic energy is $K_{\text{sphere}} = \frac{1}{2} I_{\text{sphere}} \omega_{\text{sphere}}^2 = \frac{1}{2} \times \frac{2}{5} M R^2 \omega_{\text{sphere}}^2 = \frac{1}{5} M R^2 \omega_{\text{sphere}}^2$.
The moment of inertia of a solid cylinder about its geometrical axis is $I_{\text{cylinder}} = \frac{1}{2} M R^2$. Given the angular speed $\omega_{\text{cylinder}} = 2 \omega_{\text{sphere}}$,the rotational kinetic energy is $K_{\text{cylinder}} = \frac{1}{2} I_{\text{cylinder}} \omega_{\text{cylinder}}^2 = \frac{1}{2} \times \frac{1}{2} M R^2 (2 \omega_{\text{sphere}})^2 = \frac{1}{4} M R^2 (4 \omega_{\text{sphere}}^2) = M R^2 \omega_{\text{sphere}}^2$.
The ratio of their kinetic energies is $\frac{K_{\text{sphere}}}{K_{\text{cylinder}}} = \frac{\frac{1}{5} M R^2 \omega_{\text{sphere}}^2}{M R^2 \omega_{\text{sphere}}^2} = \frac{1}{5}$.

(C) Given: Angular acceleration $\alpha = 4 \ rad/s^2$,initial angular velocity $\omega_0 = 0$.
Centripetal acceleration is given by $a_c = r\omega^2$.
Tangential acceleration is given by $a_t = r\alpha$.
We are given that $a_c = a_t$,so $r\omega^2 = r\alpha$.
This simplifies to $\omega^2 = \alpha$,which means $\omega = \sqrt{\alpha} = \sqrt{4} = 2 \ rad/s$.
Using the kinematic equation for rotational motion,$\omega = \omega_0 + \alpha t$.
Substituting the known values: $2 = 0 + 4t$.
Solving for $t$: $t = 2/4 = 1/2 \ s$.

(A) Given: Radius $R = 0.4 \,m$,Mass $M = 1 \,kg$,Angular acceleration $\alpha = 10 \,rad \,s^{-2}$.
The moment of inertia of a disc about its central axis is $I = \frac{1}{2} MR^2$.
$I = \frac{1}{2} \times 1 \,kg \times (0.4 \,m)^2 = 0.5 \times 0.16 = 0.08 \,kg \,m^2$.
The torque $\tau$ is given by $\tau = I \alpha$.
$\tau = 0.08 \,kg \,m^2 \times 10 \,rad \,s^{-2} = 0.8 \,N \,m$.
The tangential force $F$ applied at the rim is related to torque by $\tau = F \times R$.
Therefore,$F = \frac{\tau}{R} = \frac{0.8 \,N \,m}{0.4 \,m} = 2 \,N$.

(C) Given: Radius $R = 0.4 \,m$,Mass $M = 1 \,kg$,Angular acceleration $\alpha = 10 \,rad/s^2$.
The moment of inertia $I$ of a disc about an axis passing through its center and perpendicular to its plane is given by $I = \frac{MR^2}{2}$.
Substituting the values: $I = \frac{1 \times (0.4)^2}{2} = \frac{0.16}{2} = 0.08 \,kg \cdot m^2$.
The torque $\tau$ is given by $\tau = I\alpha$. Also,for a tangential force $F$ applied at the rim,$\tau = RF$.
Equating the two: $RF = I\alpha$.
Solving for $F$: $F = \frac{I\alpha}{R} = \frac{0.08 \times 10}{0.4} = \frac{0.8}{0.4} = 2 \,N$.

(C) The moment of inertia $I$ of the system about the axis passing through the midpoint is given by $I = m(\frac{d}{2})^2 + m(\frac{d}{2})^2 = 2m(\frac{d^2}{4}) = \frac{md^2}{2}$.
Rotational kinetic energy is given by $K = \frac{1}{2} I \omega^2$.
Substituting the value of $I$,we get $K = \frac{1}{2} (\frac{md^2}{2}) \omega^2 = \frac{md^2 \omega^2}{4}$.
Rearranging for $\omega^2$,we get $\omega^2 = \frac{4K}{md^2}$.
Taking the square root on both sides,we get $\omega = \sqrt{\frac{4K}{md^2}} = \frac{2}{d} \sqrt{\frac{K}{m}}$.

(A) The rotational kinetic energy $E$ of a rigid body is given by $E = \frac{1}{2} I \omega^2$,where $I$ is the moment of inertia and $\omega$ is the angular frequency.
For a molecule consisting of two atoms of mass $m$ separated by a distance $d$,the moment of inertia about an axis passing through the center of mass and perpendicular to the line joining the atoms is $I = m(\frac{d}{2})^2 + m(\frac{d}{2})^2 = 2m(\frac{d^2}{4}) = \frac{md^2}{2}$.
Substituting $I$ into the energy equation: $E = \frac{1}{2} (\frac{md^2}{2}) \omega^2 = \frac{md^2}{4} \omega^2$.
Solving for $\omega^2$: $\omega^2 = \frac{4E}{md^2}$.
Taking the square root: $\omega = \sqrt{\frac{4E}{md^2}} = \frac{2}{d} \sqrt{\frac{E}{m}}$.

(C) The rotational kinetic energy $(K)$ of a body is given by the formula: $K = \frac{1}{2} I \omega^2$.
Given that the angular velocity $\omega = 1 \ rad/s$,we substitute this into the formula: $K = \frac{1}{2} I (1)^2 = \frac{1}{2} I$.
According to the problem,the moment of inertia $(I)$ is numerically equal to '$P$' times the rotational kinetic energy $(K)$: $I = P \cdot K$.
Substituting the expression for $K$: $I = P \cdot (\frac{1}{2} I)$.
Dividing both sides by $I$ (assuming $I \neq 0$): $1 = P \cdot \frac{1}{2}$.
Therefore,$P = 2$.

(D) body becomes weightless at the equator when the centrifugal force equals the gravitational force,which is expressed as $R \omega^2 = g$,or $\omega^2 = \frac{g}{R}$.
The rotational kinetic energy of the earth is given by $K = \frac{1}{2} I \omega^2$.
For a solid sphere (assuming the earth is a solid sphere),the moment of inertia is $I = \frac{2}{5} MR^2$.
Substituting the values of $I$ and $\omega^2$ into the kinetic energy formula:
$K = \frac{1}{2} \times (\frac{2}{5} MR^2) \times (\frac{g}{R})$.
Simplifying the expression:
$K = \frac{1}{2} \times \frac{2}{5} \times M \times R \times g = \frac{1}{5} MgR$.

(A) The centre of gravity of a metre scale is at the $50 \text{ cm}$ mark. The wedge is placed at this point.
The distance of the weight '$W$' from the wedge is $50 \text{ cm} - 20 \text{ cm} = 30 \text{ cm}$.
The distance of the $25 \text{ g-wt}$ weight from the wedge is $74 \text{ cm} - 50 \text{ cm} = 24 \text{ cm}$.
For the scale to remain horizontal, the clockwise moment must equal the anticlockwise moment about the pivot (wedge):
$W \times 30 \text{ cm} = 25 \text{ g-wt} \times 24 \text{ cm}$
$W = \frac{25 \times 24}{30} \text{ g-wt}$
$W = \frac{600}{30} \text{ g-wt} = 20 \text{ g-wt}$

(B) The rate of heat flow through the rod is given by the formula: $Q = \frac{KA(\Delta \theta)t}{\ell}$.
This heat is used to melt the ice,so $Q = mL$,where $m$ is the mass of ice melted and $L$ is the latent heat of fusion.
Equating the two expressions: $mL = \frac{KA(\Delta \theta)t}{\ell}$.
Given values: $K = 96 \,cal/(s \cdot m \cdot ^{\circ}C)$,$A = 10^{-3} \,m^2$,$\Delta \theta = 100^{\circ}C - 0^{\circ}C = 100^{\circ}C$,$t = 60 \,s$,$\ell = 1 \,m$,and $L = 8 \times 10^4 \,cal/kg$.
Substituting these values into the equation:
$m = \frac{96 \times 10^{-3} \times 100 \times 60}{1 \times 8 \times 10^4}$.
$m = \frac{96 \times 10^{-1} \times 60}{8 \times 10^4} = \frac{576}{8 \times 10^5} = 72 \times 10^{-4} \,kg = 7.2 \times 10^{-3} \,kg$.

(C) The rate of heat flow through a conductor is given by the formula: $\frac{Q}{t} = \frac{kA \Delta \theta}{d}$.
Rearranging for thermal conductivity $k$: $k = \frac{Q}{tA} \cdot \frac{d}{\Delta \theta}$.
Given: Heat flux $\frac{Q}{tA} = 900 \ kcal / (min \cdot m^2) = \frac{900}{60} \ kcal / (s \cdot m^2) = 15 \ kcal / (s \cdot m^2)$.
Thickness $d = 3 \ cm = 3 \times 10^{-2} \ m$.
Temperature difference $\Delta \theta = 15^{\circ} C$.
Substituting the values: $k = \frac{15 \times 3 \times 10^{-2}}{15} = 3 \times 10^{-2} \ kcal / (m \cdot s \cdot ^{\circ}C)$.

(B) Let $R$ be the thermal resistance of each rod. When connected in series,the total thermal resistance is $R_{s} = R + R = 2R$.
The heat transferred in series is $Q = \frac{\Delta T}{R_{s}} \times t_{s} = \frac{\Delta T}{2R} \times 8 = \frac{4 \Delta T}{R}$.
When connected in parallel,the equivalent thermal resistance is $R_{p} = \frac{R \times R}{R + R} = \frac{R}{2}$.
The heat transferred in parallel is $Q = \frac{\Delta T}{R_{p}} \times t_{p} = \frac{\Delta T}{R/2} \times t_{p} = \frac{2 \Delta T}{R} \times t_{p}$.
Since the heat transferred $Q$ is the same in both cases,we equate the expressions:
$\frac{4 \Delta T}{R} = \frac{2 \Delta T}{R} \times t_{p}$.
Solving for $t_{p}$,we get $t_{p} = \frac{4}{2} = 2 \ s$.

(C) The rate of heat flow through a rod is given by $Q = \frac{kA(\theta_1 - \theta_2)}{\ell}$,where $k$ is thermal conductivity,$A$ is the cross-sectional area,and $\ell$ is the length of the rod.
When all linear dimensions are doubled,the radius $r$ becomes $2r$ and the length $\ell$ becomes $2\ell$.
The new area $A' = \pi(2r)^2 = 4\pi r^2 = 4A$.
The new length $\ell' = 2\ell$.
The new rate of heat flow $Q'$ is given by $Q' = \frac{kA'(\theta_1 - \theta_2)}{\ell'}$.
Substituting the new values: $Q' = \frac{k(4A)(\theta_1 - \theta_2)}{2\ell} = 2 \left( \frac{kA(\theta_1 - \theta_2)}{\ell} \right) = 2Q$.

(B) The thermal resistance $(R_{th})$ is defined as the ratio of the temperature difference $(\Delta T)$ to the rate of heat flow ($H$ or thermal current).
Formula: $R_{th} = \frac{\Delta T}{H}$
Given:
Temperature difference $\Delta T = 28^{\circ} C$
Rate of heat flow $H = 1400 \ cal s^{-1}$
Calculation:
$R_{th} = \frac{28}{1400} \ ^{\circ} C s cal^{-1}$
$R_{th} = 0.02 \ ^{\circ} C s cal^{-1}$
Therefore, the thermal resistance is $0.02 \ ^{\circ} C s cal^{-1}$.

(A) The rate of heat flow per unit area is given by the formula: $\frac{Q}{At} = \frac{k \Delta \theta}{d}$.
Given:
Heat flux $\frac{Q}{At} = 10 \ kcal / (s \cdot m^2)$
Thickness $d = 1.8 \ cm = 1.8 \times 10^{-2} \ m$
Temperature difference $\Delta \theta = 9^{\circ} C$
Substituting the values into the formula:
$10 = k \times \frac{9}{1.8 \times 10^{-2}}$
$10 = k \times \frac{9}{0.018}$
$10 = k \times 500$
$k = \frac{10}{500} = \frac{1}{50} = 0.02 \ kcal / (m \cdot s \cdot ^{\circ} C)$.

(D) The rate of heat flow through a cylindrical rod is given by the formula: $Q = \frac{kA(T_1 - T_2)}{L}$,where $A = \pi r^2$.
Thus,$Q_1 = \frac{k \pi r_1^2 (T_1 - T_2)}{L_1}$.
When the length and radius are doubled,we have $L_2 = 2L_1$ and $r_2 = 2r_1$.
The new rate of heat flow is $Q_2 = \frac{k \pi r_2^2 (T_1 - T_2)}{L_2}$.
Taking the ratio: $\frac{Q_2}{Q_1} = \left( \frac{r_2}{r_1} \right)^2 \cdot \left( \frac{L_1}{L_2} \right)$.
Substituting the values: $\frac{Q_2}{Q_1} = (2)^2 \cdot \left( \frac{1}{2} \right) = 4 \cdot \frac{1}{2} = 2$.
Therefore,$Q_2 = 2 Q_1$.

(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.

(D) According to Wien's displacement law,the product of the wavelength of maximum emission $(\lambda_{m})$ and the absolute temperature $(T)$ is a constant,i.e.,$\lambda_{m} T = b$.
This implies that $T \propto \frac{1}{\lambda_{m}}$.
The wavelength of blue light is shorter than the wavelength of yellow light $(\lambda_{\text{blue}} < \lambda_{\text{yellow}})$.
Since the star '$Q$' emits blue light and star '$P$' emits yellow light,we have $\lambda_{Q} < \lambda_{P}$.
Therefore,$T_{Q} > T_{P}$,which can also be written as $T_{P} < T_{Q}$.

(C) According to Stefan-Boltzmann Law,the energy emitted per second by a black body is given by $E = \sigma A T^4$.
Initially,$E = \sigma A T^4$,where $T = 27^{\circ}C = 300 \ K$.
When the length and breadth are reduced to $1/3$ of their initial values,the new area $A' = (L/3) \times (B/3) = A/9$.
The new temperature $T' = 327^{\circ}C = 600 \ K$.
The new energy emitted per second is $E' = \sigma A' (T')^4$.
Taking the ratio: $\frac{E'}{E} = \frac{A'}{A} \times \left(\frac{T'}{T}\right)^4$.
Substituting the values: $\frac{E'}{E} = \frac{1}{9} \times \left(\frac{600}{300}\right)^4 = \frac{1}{9} \times (2)^4 = \frac{16}{9}$.
Therefore,$E' = \frac{16 E}{9}$.

(C) In the photoelectric effect,the photoelectric current is directly proportional to the intensity of the incident light,provided the frequency is above the threshold frequency. Since the frequency and accelerating potential are kept constant,increasing the intensity of the incident light increases the number of photons incident per unit time,which in turn increases the number of photoelectrons emitted per unit time. Therefore,the photoelectric current increases.

(B) According to Einstein's photoelectric equation,the kinetic energy $K$ is given by $K = \frac{hc}{\lambda} - W_0$,where $W_0$ is the work function.
For wavelength $\lambda_1$,the kinetic energy is $K_1 = \frac{hc}{\lambda_1} - W_0$.
For wavelength $\lambda_2$,the kinetic energy is $K_2 = \frac{hc}{\lambda_2} - W_0$.
Given that $K_2 = 3K_1$,we substitute the expressions:
$\frac{hc}{\lambda_2} - W_0 = 3\left(\frac{hc}{\lambda_1} - W_0\right)$.
$\frac{hc}{\lambda_2} - W_0 = \frac{3hc}{\lambda_1} - 3W_0$.
Rearranging the terms to solve for $W_0$:
$3W_0 - W_0 = \frac{3hc}{\lambda_1} - \frac{hc}{\lambda_2}$.
$2W_0 = hc\left(\frac{3}{\lambda_1} - \frac{1}{\lambda_2}\right)$.
$2W_0 = hc\left(\frac{3\lambda_2 - \lambda_1}{\lambda_1\lambda_2}\right)$.
Therefore,$W_0 = \frac{hc}{2}\left(\frac{3\lambda_2 - \lambda_1}{\lambda_1\lambda_2}\right)$.

(A) In the photoelectric effect,the photocurrent is directly proportional to the number of photoelectrons emitted per unit time.
Since each incident photon ejects at most one photoelectron,the number of photoelectrons emitted depends solely on the number of incident photons per unit time,which is determined by the intensity of the incident light.
The frequency of the incident light determines the kinetic energy of the emitted photoelectrons,provided the frequency is above the threshold frequency,but it does not affect the magnitude of the photocurrent.

(B) According to Einstein's photoelectric equation,the maximum kinetic energy is given by:
$(KE)_{\max} = h\nu - \phi_0$
where $\phi_0 = h\nu_0$ is the work function.
Comparing this with the equation of a straight line $y = mx + c$:
Here,$y = (KE)_{\max}$,$x = \nu$,$m = h$ (slope),and $c = -h\nu_0$ (y-intercept).
$1$. The $X$-intercept $(A)$ occurs when $(KE)_{\max} = 0$:
$0 = h\nu - h\nu_0 \implies \nu = \nu_0 = A$.
$2$. The $Y$-intercept $(B)$ occurs when $\nu = 0$:
$(KE)_{\max} = -h\nu_0 = B$. Since $B$ represents the magnitude of the intercept,we take $|B| = h\nu_0$.
Therefore,the ratio of the magnitude of the $Y$-intercept to the $X$-intercept is:
$\frac{|B|}{A} = \frac{h\nu_0}{\nu_0} = h$.
Thus,Planck's constant $h$ is given by $\frac{B}{A}$ (considering magnitudes).

(D) According to Einstein's photoelectric equation,the kinetic energy $E$ of an emitted photoelectron is given by $E = \frac{hc}{\lambda} - \phi$,where $\phi$ is the work function of the material. This shows that $E$ depends only on the wavelength $\lambda$ of the incident light,specifically $E \propto \frac{1}{\lambda}$.
Intensity $I$ of light is proportional to the number of photons incident per unit area per unit time. Since each photon interacts with one electron to cause emission,the number of emitted photoelectrons $N$ is directly proportional to the intensity $I$ of the incident light $(N \propto I)$.

(C) According to Faraday's law of induction,the induced electromotive force $(e)$ is given by $e = -\frac{\Delta \phi}{\Delta t}$.
Using Ohm's law,the induced current $(i)$ is $i = \frac{e}{R} = \frac{1}{R} \cdot \frac{\Delta \phi}{\Delta t}$.
Since current is the rate of flow of charge,we have $i = \frac{\Delta q}{\Delta t}$.
Equating the two expressions for current: $\frac{\Delta q}{\Delta t} = \frac{1}{R} \cdot \frac{\Delta \phi}{\Delta t}$.
Therefore,the total induced charge is $\Delta q = \frac{\Delta \phi}{R}$.
This shows that the total induced charge depends only on the total change in magnetic flux $(\Delta \phi)$ and the resistance $(R)$.

(C) In a $DC$ circuit at steady state,the inductor acts as a simple wire with zero resistance (ideal inductor) or just a conductor with its inherent resistance $R$.
Since the coil is broken into two identical parts,the resistance of each part becomes $R' = \frac{R}{2}$.
When these two parts are connected in parallel,the equivalent resistance $R_{eq}$ is given by $\frac{1}{R_{eq}} = \frac{1}{R'} + \frac{1}{R'} = \frac{2}{R'} = \frac{2}{R/2} = \frac{4}{R}$.
Thus,$R_{eq} = \frac{R}{4}$.
The current $I$ through the battery is given by Ohm's law: $I = \frac{E}{R_{eq}} = \frac{E}{R/4} = \frac{4 E}{R}$.

(D) The magnetic potential energy $U$ stored in an inductor is given by the formula: $U = \frac{1}{2} LI^2$.
Given:
$U = 25 \ mJ = 25 \times 10^{-3} \ J$
$I = 50 \ mA = 50 \times 10^{-3} \ A$
Rearranging the formula to solve for inductance $L$:
$L = \frac{2U}{I^2}$
Substituting the values:
$L = \frac{2 \times 25 \times 10^{-3}}{(50 \times 10^{-3})^2}$
$L = \frac{50 \times 10^{-3}}{2500 \times 10^{-6}}$
$L = \frac{50 \times 10^{-3}}{2.5 \times 10^{-3}}$
$L = 20 \ H$.

(D) In the given circuit,all three inductors have their left terminals connected to point $P$ and their right terminals connected to point $Q$.
This means the three inductors are connected in parallel.
The formula for the equivalent inductance $L_{eq}$ of inductors connected in parallel is given by:
$\frac{1}{L_{eq}} = \frac{1}{L_1} + \frac{1}{L_2} + \frac{1}{L_3}$
Given $L_1 = L_2 = L_3 = 6 \ H$,we have:
$\frac{1}{L_{eq}} = \frac{1}{6} + \frac{1}{6} + \frac{1}{6} = \frac{3}{6} = \frac{1}{2} \ H^{-1}$
Therefore,$L_{eq} = 2 \ H$.

(A) The energy $E$ stored in a coil with self-inductance $L$ carrying current $I$ is given by the formula: $E = \frac{1}{2} LI^2$.
Let the initial current be $I_1$ and the initial energy be $E_1 = \frac{1}{2} LI_1^2$.
Let the new current be $I_2 = \frac{I_1}{2}$ and the new energy be $E_2 = \frac{1}{2} LI_2^2$.
Taking the ratio of the two energies:
$\frac{E_2}{E_1} = \frac{\frac{1}{2} LI_2^2}{\frac{1}{2} LI_1^2} = \left( \frac{I_2}{I_1} \right)^2$.
Substituting $I_2 = \frac{I_1}{2}$:
$\frac{E_2}{E_1} = \left( \frac{I_1 / 2}{I_1} \right)^2 = \left( \frac{1}{2} \right)^2 = \frac{1}{4}$.
Therefore,$E_2 = \frac{E_1}{4}$.

(C) The energy stored in the inductor is given by $U = \frac{1}{2} LI^2$.
When the circuit is switched off,this energy must be transferred to the capacitor to prevent sparking.
The energy stored by the capacitor is given by $W = \frac{1}{2} CV^2$.
Equating the two energies: $\frac{1}{2} CV^2 = \frac{1}{2} LI^2$.
Solving for $C$: $C = \frac{LI^2}{V^2} = L\left(\frac{I}{V}\right)^2$.

(B) The motional electromotive force (emf) induced in a conductor moving through a magnetic field is given by $e = B_{H} l v$.
Given:
Length of wire $l = 2500 \ m$
Speed of wire $v = 10 \ m/s$
Horizontal component of Earth's magnetic field $B_{H} = 2 \times 10^{-5} \ T$
Resistance of wire $R = 25 \ \Omega$
Calculating the induced emf:
$e = (2 \times 10^{-5} \ T) \times (2500 \ m) \times (10 \ m/s) = 0.5 \ V$
Now,calculating the induced current $I$ using Ohm's law:
$I = e / R = 0.5 \ V / 25 \ \Omega = 0.02 \ A$
Therefore,the induced current is $0.02 \ A$.

(D) The induced electromotive force $(e)$ in a conductor of length $\ell$ rotating with angular velocity $\omega$ in a magnetic field $B$ is given by the formula:
$e = \frac{1}{2} B \omega \ell^2$
Given values:
$B = 0.2 \times 10^{-4} \ T$
$\omega = 5 \ rad/s$
$\ell = 1 \ m$
Substituting these values into the formula:
$e = \frac{1}{2} \times (0.2 \times 10^{-4}) \times 5 \times (1)^2$
$e = 0.5 \times 10^{-4} \ V$
$e = 50 \times 10^{-6} \ V = 50 \ \mu V$

(B) The induced electromotive force (e.m.f.) in a conductor moving through a magnetic field is given by the formula:
$E = B \ell v$
Where:
$B = 1.2 \, Wb \cdot m^{-2}$ (Magnetic field induction)
$\ell = 0.6 \, m$ (Length of the conductor)
$v = 10 \, ms^{-1}$ (Speed of the conductor)
Substituting the values into the formula:
$E = 1.2 \times 0.6 \times 10$
$E = 7.2 \, V$
Therefore, the induced e.m.f. across the conductor is $7.2 \, V$.

(B) According to Lenz's Law,the induced current in a loop always opposes the change in magnetic flux that produces it.
As the current in the outer loop is clockwise and increasing,the magnetic field lines passing through the inner loop (directed into the plane) are increasing.
To oppose this increase in inward magnetic flux,the inner loop must generate an outward magnetic field.
By the right-hand thumb rule,an outward magnetic field is produced by an anticlockwise current.
Therefore,the induced current in the inner loop is anticlockwise.

(A) When a wire falls freely from a height $\ell$ under gravity,its velocity $v$ at the moment it reaches the ground is given by the equation of motion $v^2 = u^2 + 2g\ell$. Since the initial velocity $u = 0$,we have $v = \sqrt{2g\ell}$.
The motional electromotive force (emf) induced in a conductor of length $L$ moving with velocity $v$ perpendicular to a magnetic field $B$ is given by the formula $E = BLv$.
Substituting the value of $v$ into the emf equation,we get $E = BL \sqrt{2g\ell}$.

(A) According to Lenz's law,the induced current in a closed circuit always flows in a direction such that it opposes the change in magnetic flux that produces it.
When the north pole of a bar magnet is brought towards the coil,the magnetic flux linked with the coil increases.
To oppose this increase in flux,the coil must behave like a north pole on the side facing the magnet.
According to the right-hand rule,a face behaving as a north pole has current flowing in an anticlockwise direction.
Therefore,the induced current in the coil flows in an anticlockwise direction.

(D) The magnetic field inside a long solenoid is given by $B = \frac{\mu_0 NI}{L}$.
Since magnetic flux $\phi = B \cdot A$, we have $B = \frac{\phi}{A}$.
Equating the two expressions for $B$: $\frac{\phi}{A} = \frac{\mu_0 NI}{L}$.
Rearranging to find the magnetic moment $M = NIA$: $NIA = \frac{\phi L}{\mu_0}$.
Given $\phi = 1.57 \times 10^{-6} \, Wb$, $L = 0.6 \, m$, and $\mu_0 = 4 \pi \times 10^{-7} \, T \cdot m/A$.
Substituting the values: $M = \frac{1.57 \times 10^{-6} \times 0.6}{4 \pi \times 10^{-7}}$.
Using $\pi \approx 3.14$, $4 \pi \approx 12.56$.
$M = \frac{1.57 \times 10^{-6} \times 0.6}{12.56 \times 10^{-7}} = \frac{0.942 \times 10^{-6}}{12.56 \times 10^{-7}} = \frac{9.42}{12.56} \approx 0.75 \, Am^2$.

(B) The induced electromotive force (emf) $e$ in the moving wire is given by $e = B \ell v$.
Substituting the given values: $e = 0.5 \, T \times 1 \, m \times 2 \, m/s = 1 \, V$.
The rate at which work is done to keep the wire moving at a constant speed is equal to the power dissipated in the resistance $R$.
The power $P$ is given by $P = \frac{e^2}{R}$.
Substituting the values: $P = \frac{(1 \, V)^2}{6 \, \Omega} = \frac{1}{6} \, W$.

(C) Given: $\phi = 5t^2 - 6t + 9$ and $R = 20 \ \Omega$.
According to Faraday's law of induction,the induced electromotive force $(e)$ is given by $e = -\frac{d\phi}{dt}$.
$e = -\frac{d}{dt}(5t^2 - 6t + 9) = -(10t - 6)$.
The magnitude of the induced electromotive force is $|e| = |10t - 6|$.
At $t = 0.2 \ s$,$|e| = |10(0.2) - 6| = |2 - 6| = |-4| = 4 \ V$.
The induced current $i$ is given by $i = \frac{|e|}{R}$.
$i = \frac{4 \ V}{20 \ \Omega} = 0.2 \ A$.

(B) The self-inductance $L$ of a coil with $N$ turns is defined by the relation $N\phi = LI$,where $\phi$ is the magnetic flux through each turn.
Thus,the inductance is $L = \frac{N\phi}{I}$.
The magnetic energy $U$ stored in an inductor is given by the formula $U = \frac{1}{2}LI^2$.
Substituting the expression for $L$ into the energy formula:
$U = \frac{1}{2} \left( \frac{N\phi}{I} \right) I^2$.
Simplifying this,we get $U = \frac{1}{2} N\phi I$ or $U = \frac{N\phi I}{2}$.

(D) The induced electromotive force (e.m.f) is given by Faraday's law of induction: $e = -N \frac{d\phi}{dt} = -N \frac{\phi_2 - \phi_1}{t}$.
Given: Area $A = 0.06 \ m^2$,Number of turns $N = 600$,Magnetic field $B = 5 \times 10^{-5} \ Wb/m^2$,Time $t = 0.2 \ s$.
The magnetic flux is $\phi = BA \cos \theta$.
Initially,the coil is perpendicular to the magnetic field,so $\theta_1 = 0^{\circ}$ and $\phi_1 = BA \cos 0^{\circ} = BA$.
After rotating by $90^{\circ}$,the coil is parallel to the magnetic field,so $\theta_2 = 90^{\circ}$ and $\phi_2 = BA \cos 90^{\circ} = 0$.
The magnitude of the average induced e.m.f is:
$|e| = N \frac{|\phi_2 - \phi_1|}{t} = N \frac{|0 - BA|}{t} = \frac{NBA}{t}$.
Substituting the values:
$|e| = \frac{600 \times 5 \times 10^{-5} \times 0.06}{0.2} = \frac{1.8 \times 10^{-3}}{0.2} = 9 \times 10^{-3} \ V$.

(D) According to Ohm's law, the induced electromotive force $(e)$ is given by $e = I \times R$.
Given: Current $(I)$ = $1.5 \, mA = 1.5 \times 10^{-3} \, A$ and Resistance $(R)$ = $5 \, \Omega$.
Substituting the values, we get $e = (1.5 \times 10^{-3} \, A) \times (5 \, \Omega) = 7.5 \times 10^{-3} \, V$.
According to Faraday's law of electromagnetic induction, the induced emf is equal to the rate of change of magnetic flux, i.e., $e = \frac{d\phi}{dt}$.
Therefore, the rate at which the conductor cuts the magnetic flux is $\frac{d\phi}{dt} = 7.5 \times 10^{-3} \, Wb/s$.

(B) According to Faraday's law of electromagnetic induction, the magnitude of the induced e.m.f. $(|e|)$ is given by the rate of change of magnetic flux $(\phi)$ with respect to time $(t)$:
$|e| = \left| \frac{d\phi}{dt} \right|$
Given $\phi = 3t^2 + 4t + 7$.
Differentiating with respect to $t$:
$\frac{d\phi}{dt} = \frac{d}{dt}(3t^2 + 4t + 7) = 6t + 4$
At $t = 2 \ s$:
$|e| = 6(2) + 4 = 12 + 4 = 16 \ V$
Therefore, the magnitude of the induced e.m.f. is $16 \ V$.

(A) Given:
Magnetic field $B = 0.1 \, T$
Length of the arm $PQ$, $\ell = 12 \, cm = 0.12 \, m$
Velocity of the arm, $v = 20 \, ms^{-1}$
Resistance of the loop, $R = 2 \, \Omega$
The motional electromotive force (emf) induced in the moving arm is given by:
$e = B \ell v$
$e = 0.1 \, T \times 0.12 \, m \times 20 \, ms^{-1}$
$e = 0.24 \, V$
The induced current $I$ in the loop is given by Ohm's law:
$I = \frac{e}{R}$
$I = \frac{0.24 \, V}{2 \, \Omega} = 0.12 \, A$
Therefore, the current induced in the loop is $0.12 \, A$.

(C) The magnetic field at the center of a circular coil carrying current $I$ is given by $B = \frac{\mu_0 NI}{2R}$.
The total magnetic flux $\phi$ linked with the coil is the product of the number of turns $N$, the magnetic field $B$, and the area of the coil $A = \pi R^2$.
$\phi = N \cdot B \cdot A = N \cdot \left( \frac{\mu_0 NI}{2R} \right) \cdot \pi R^2 = \frac{\mu_0 N^2 \pi R I}{2}$.
The coefficient of self-induction $L$ is defined as $L = \frac{\phi}{I}$.
Substituting the value of $\phi$, we get $L = \frac{\mu_0 N^2 \pi R}{2}$.

(B) Given:
Length of solenoid,$\ell = 31.4 \ cm = 0.314 \ m$
Area of cross-section,$A = 10^{-3} \ m^2$
Total number of turns,$N = 500$
Permeability of free space,$\mu_0 = 4 \pi \times 10^{-7} \ T \cdot m/A$
The formula for self-inductance of a solenoid is given by:
$L = \frac{\mu_0 N^2 A}{\ell}$
Substituting the values:
$L = \frac{(4 \times 3.14 \times 10^{-7}) \times (500)^2 \times 10^{-3}}{0.314}$
$L = \frac{4 \times 3.14 \times 10^{-7} \times 250000 \times 10^{-3}}{0.314}$
$L = \frac{4 \times 3.14 \times 10^{-7} \times 2.5 \times 10^5 \times 10^{-3}}{0.314}$
$L = \frac{4 \times 3.14 \times 2.5 \times 10^{-5}}{0.314}$
$L = 4 \times 10 \times 2.5 \times 10^{-5} = 10 \times 10^{-4} = 10^{-3} \ H$

(A) The induced emf $e$ in the neighboring coil is given by $e = M \frac{dI}{dt}$.
Given $I = 10 \sin(100 \pi t)$, we find the rate of change of current:
$\frac{dI}{dt} = 10 \times 100 \pi \cos(100 \pi t) = 1000 \pi \cos(100 \pi t)$.
Substituting this into the emf equation:
$e = M \times 1000 \pi \cos(100 \pi t)$.
The maximum value of the induced emf $e_0$ occurs when $\cos(100 \pi t) = 1$, so $e_0 = 1000 \pi M$.
Given $e_0 = 5 \pi \text{ V}$, we have $1000 \pi M = 5 \pi$.
Solving for $M$: $M = \frac{5 \pi}{1000 \pi} = 0.005 \text{ H} = 5 \text{ mH}$.

(D) The total magnetic flux linkage in a coil with $n$ turns is given by $N\phi_{total} = L I$,where $N$ is the number of turns,$\phi_{total}$ is the flux per turn,$L$ is the self-inductance,and $I$ is the current.
Given that the flux per turn is $\phi$ and the number of turns is $n$,the total flux linkage is $n\phi$.
Therefore,the relation is $n\phi = LI$.
Rearranging for $\phi$,we get $\phi = \frac{LI}{n}$.

(D) The magnetic field $B$ at the center of a large coil of radius $R$ carrying current $I$ is given by $B = \frac{\mu_{0} I}{2R}$.
Since $R \gg r$,we can assume the magnetic field is uniform over the area of the smaller coil.
The magnetic flux $\phi$ passing through the smaller coil of radius $r$ is $\phi = B \cdot A = B \cdot (\pi r^{2})$.
Substituting the value of $B$,we get $\phi = \left( \frac{\mu_{0} I}{2R} \right) \cdot (\pi r^{2})$.
The mutual inductance $M$ is defined as $M = \frac{\phi}{I}$.
Therefore,$M = \frac{\mu_{0} \pi r^{2}}{2R}$.
Since $\mu_{0}$,$\pi$,and $2$ are constants,we have $M \propto \frac{r^{2}}{R}$.

(B) According to Coulomb's Law,the force $F$ between two charges $q_1$ and $q_2$ separated by a distance $r$ is given by $F = k \frac{q_1 q_2}{r^2}$.
Given that the particles are identical,let $q_1 = q_2 = q$. Initially,the distance is $r_1 = Y$,so $F_1 = k \frac{q^2}{Y^2} = F$.
When the distance is reduced to half,the new distance is $r_2 = \frac{Y}{2}$.
The new force $F_2$ is given by $F_2 = k \frac{q^2}{(Y/2)^2} = k \frac{q^2}{Y^2 / 4} = 4 \left( k \frac{q^2}{Y^2} \right)$.
Substituting $F_1 = F$,we get $F_2 = 4F$.

(D) The charge on the polythene is given by $q = 4 \times 10^{-7} \ C$.
According to the quantization of charge,the total charge $q$ is given by $q = Ne$,where $N$ is the number of electrons and $e$ is the elementary charge.
Given $e = 1.6 \times 10^{-19} \ C$.
To find the number of electrons $N$,we use the formula $N = \frac{q}{e}$.
Substituting the values: $N = \frac{4 \times 10^{-7}}{1.6 \times 10^{-19}}$.
$N = \frac{4}{1.6} \times 10^{-7 - (-19)} = 2.5 \times 10^{12}$.
Thus,the number of electrons transferred is $2.5 \times 10^{12}$.

(B) For the system to be in equilibrium,the net force on each charge must be zero.
Let the distance between the two charges $+q$ be $r$. The charge $Q$ is placed at the center,so its distance from each $+q$ charge is $r/2$.
The force on the central charge $Q$ due to the two $+q$ charges is equal and opposite,so it is already in equilibrium.
For the system to be in equilibrium,the net force on one of the $+q$ charges must be zero.
The force on a $+q$ charge due to the other $+q$ charge is $F_1 = \frac{1}{4 \pi \varepsilon_0} \cdot \frac{q^2}{r^2}$ (repulsive).
The force on this $+q$ charge due to the central charge $Q$ is $F_2 = \frac{1}{4 \pi \varepsilon_0} \cdot \frac{qQ}{(r/2)^2} = \frac{1}{4 \pi \varepsilon_0} \cdot \frac{4qQ}{r^2}$.
For equilibrium,$F_1 + F_2 = 0$,which implies $F_1 = -F_2$.
$\frac{1}{4 \pi \varepsilon_0} \cdot \frac{q^2}{r^2} = -\left( \frac{1}{4 \pi \varepsilon_0} \cdot \frac{4qQ}{r^2} \right)$.
$q^2 = -4qQ$.
$Q = -\frac{q}{4}$.

(D) According to Coulomb's law,the force of repulsion between two charges $q$ separated by a distance $d$ is given by:
$F = \frac{1}{4 \pi \varepsilon_0} \cdot \frac{q^2}{d^2}$
Since the ions are positive,the charge $q$ on each ion is due to the loss of $n$ electrons,so $q = ne$.
Substituting $q = ne$ into the force equation:
$F = \frac{1}{4 \pi \varepsilon_0} \cdot \frac{(ne)^2}{d^2} = \frac{1}{4 \pi \varepsilon_0} \cdot \frac{n^2 e^2}{d^2}$
Now,solving for $n$:
$n^2 = \frac{F \cdot 4 \pi \varepsilon_0 \cdot d^2}{e^2}$
$n = \sqrt{\frac{4 \pi \varepsilon_0 F d^2}{e^2}}$

(A) Initial charges are $q_1 = 3 \mu C$ and $q_2 = 8 \mu C$. The force between them is given by Coulomb's Law: $F = k \frac{q_1 q_2}{r^2} = 40 \ N$.
When a charge of $-5 \mu C$ is added to each,the new charges become:
$q_1' = 3 \mu C - 5 \mu C = -2 \mu C$
$q_2' = 8 \mu C - 5 \mu C = 3 \mu C$
The new force $F'$ is:
$F' = k \frac{q_1' q_2'}{r^2} = k \frac{(-2)(3)}{r^2} = -6k \frac{1}{r^2}$
Dividing the new force by the initial force:
$\frac{F'}{40} = \frac{k \frac{(-2)(3)}{r^2}}{k \frac{(3)(8)}{r^2}} = \frac{-6}{24} = -\frac{1}{4}$
$F' = 40 \times (-\frac{1}{4}) = -10 \ N$.

(D) Let the three charges be $q$ at the vertices of an equilateral triangle.
Consider one charge at a vertex. It experiences two repulsive forces from the other two charges, each of magnitude $F$.
The angle between these two forces is $60^{\circ}$ (the interior angle of an equilateral triangle).
The resultant force $R$ is given by the vector addition formula:
$R = \sqrt{F_1^2 + F_2^2 + 2 F_1 F_2 \cos \theta}$
Substituting $F_1 = F$, $F_2 = F$, and $\theta = 60^{\circ}$:
$R = \sqrt{F^2 + F^2 + 2 F^2 \cos 60^{\circ}}$
Since $\cos 60^{\circ} = 0.5$:
$R = \sqrt{2 F^2 + 2 F^2 (0.5)} = \sqrt{2 F^2 + F^2} = \sqrt{3 F^2} = \sqrt{3} F$.

(D) An electric dipole consists of two equal and opposite charges,$+q$ and $-q$,separated by a distance $2\ell$.
In a uniform electric field $E$,the force on the positive charge is $F_+ = qE$ in the direction of the field,and the force on the negative charge is $F_- = -qE$ opposite to the direction of the field.
The net force on the dipole is $F_{net} = F_+ + F_- = qE - qE = 0$.
The potential energy $U$ of an electric dipole in an electric field is given by $U = -P \cdot E = -PE \cos \theta$.
When the dipole moment $P$ is along the direction of the field,the angle $\theta = 0^{\circ}$.
Thus,$U = -PE \cos(0^{\circ}) = -PE$,which is the minimum possible value (stable equilibrium).

(A) The electric field $E$ due to a charged arc of radius $r$ and linear charge density $\lambda$ at its center is given by the formula $E = \frac{2 k \lambda}{r} \sin \alpha$,where $2\alpha$ is the total angle subtended by the arc at the center.
For a semicircular arc,the total angle subtended is $180^{\circ}$,so $2\alpha = 180^{\circ}$,which means $\alpha = 90^{\circ}$.
Substituting $k = \frac{1}{4 \pi \epsilon_0}$ and $\alpha = 90^{\circ}$ into the formula:
$E = \frac{2 (\frac{1}{4 \pi \epsilon_0}) \lambda}{r} \sin(90^{\circ})$
Since $\sin(90^{\circ}) = 1$,we get:
$E = \frac{2 \lambda}{4 \pi \epsilon_0 r} = \frac{\lambda}{2 \pi \epsilon_0 r}$.

(C) The force acting on the electron in a uniform electric field is given by $F = qE$.
Since the electron is moving in the direction of the force,the work done by the electric field over a distance $L$ is $W = F \times L = qEL$.
According to the work-energy theorem,the work done is equal to the change in kinetic energy of the electron.
Since the electron starts from rest,the initial kinetic energy is $0$.
Therefore,$\frac{1}{2} mv^2 = qEL$.
Solving for the velocity $v$,we get $v^2 = \frac{2qEL}{m}$,which implies $v = \sqrt{\frac{2qEL}{m}}$.

(B) The electric field intensity $E$ at a distance $r$ from the center of a charged sphere is given by Coulomb's law as:
$E = \frac{1}{4 \pi \varepsilon_0} \frac{q}{r^2} \quad (1)$
For a solid sphere of radius $r$ with uniform volume charge density $\rho$,the total charge $q$ is given by:
$q = \rho \times \text{Volume} = \rho \times \left( \frac{4}{3} \pi r^3 \right)$
Substituting this value of $q$ into Eq. $(1)$:
$E = \frac{1}{4 \pi \varepsilon_0} \frac{\rho \times \frac{4}{3} \pi r^3}{r^2}$
Simplifying the expression:
$E = \frac{\rho r}{3 \varepsilon_0}$

(B) Since the spheres are joined by a metal wire,their potentials will be the same.
Let the radius of the smaller sphere be $r_1$ and the radius of the larger sphere be $r_2 = 4r_1$.
Since the potentials are equal,$V = \frac{1}{4 \pi \varepsilon_0} \cdot \frac{q_1}{r_1} = \frac{1}{4 \pi \varepsilon_0} \cdot \frac{q_2}{r_2}$.
Therefore,$\frac{q_2}{q_1} = \frac{r_2}{r_1} = 4$.
The electric field near the surface of the spheres is given by $E = \frac{1}{4 \pi \varepsilon_0} \cdot \frac{q}{r^2}$.
Thus,$\frac{E_2}{E_1} = \frac{q_2}{q_1} \cdot \left(\frac{r_1}{r_2}\right)^2 = \left(\frac{r_2}{r_1}\right) \cdot \left(\frac{r_1}{r_2}\right)^2 = \frac{r_1}{r_2}$.
Substituting $r_2 = 4r_1$,we get $\frac{E_2}{E_1} = \frac{r_1}{4r_1} = \frac{1}{4}$.
So,the electric field near the larger sphere is a quarter of that near the smaller sphere.

(A) $1$. According to the principle of electrostatic induction,when a charge $-q$ is placed at the center of a conducting shell,an equal and opposite charge $+q$ is induced on the inner surface of the shell to ensure the electric field inside the conductor is zero.
$2$. The inner surface of the shell now has a charge of $+q$. The surface charge density on the inner surface is given by $\sigma_{inner} = \frac{q}{4 \pi r_1^2}$.
$3$. Since the shell has a total charge $Q$ and the inner surface has a charge $+q$,the charge on the outer surface must be $Q_{outer} = Q - q$ to conserve the total charge.
$4$. The surface charge density on the outer surface is given by $\sigma_{outer} = \frac{Q-q}{4 \pi r_2^2}$.
$5$. Therefore,the surface charge densities are $\frac{q}{4 \pi r_1^2}$ and $\frac{Q-q}{4 \pi r_2^2}$.

(B) The potential at the surface of the sphere is $V_s = \frac{1}{4\pi\epsilon_0} \frac{q}{R}$.
The potential at a distance $5R$ from the centre is $V_p = \frac{1}{4\pi\epsilon_0} \frac{q}{5R}$.
The potential difference $V$ is given by $V = V_s - V_p = \frac{q}{4\pi\epsilon_0} (\frac{1}{R} - \frac{1}{5R}) = \frac{q}{4\pi\epsilon_0} (\frac{4}{5R}) = \frac{q}{5\pi\epsilon_0 R}$.
From this,we can express the charge $q$ as $q = \frac{5\pi\epsilon_0 RV}{1}$.
The electric field intensity $E$ at a distance $r = 5R$ is given by $E = \frac{1}{4\pi\epsilon_0} \frac{q}{r^2} = \frac{1}{4\pi\epsilon_0} \frac{q}{(5R)^2} = \frac{q}{100\pi\epsilon_0 R^2}$.
Substituting the value of $q$ in terms of $V$: $E = \frac{1}{100\pi\epsilon_0 R^2} \cdot (5\pi\epsilon_0 RV) = \frac{5}{100} \frac{V}{R} = \frac{V}{20R}$.

(D) The electric potential $dV$ due to a small charge element $dq$ at the centre of the ring is given by $dV = \frac{1}{4 \pi \epsilon_0} \cdot \frac{dq}{R}$.
Since the ring is a half-ring,the total charge $q$ is the product of the linear charge density $\sigma$ and the length of the half-ring $(L = \pi R)$.
Thus,$q = \sigma \cdot \pi R$.
The total potential $V$ at the centre is the integral of $dV$ over the entire half-ring:
$V = \int dV = \int \frac{1}{4 \pi \epsilon_0} \cdot \frac{dq}{R} = \frac{1}{4 \pi \epsilon_0 R} \int dq$.
Substituting the total charge $q = \sigma \pi R$:
$V = \frac{1}{4 \pi \epsilon_0 R} \cdot (\sigma \pi R) = \frac{\sigma}{4 \epsilon_0}$.

(D) Let the charge on the sphere be $Q$.
The potential at the surface of the sphere is $V_{surface} = \frac{kQ}{r}$.
The potential at a distance $3r$ from the center is $V_{3r} = \frac{kQ}{3r}$.
The potential difference $V$ is given by:
$V = V_{surface} - V_{3r} = \frac{kQ}{r} - \frac{kQ}{3r} = \frac{2kQ}{3r}$.
From this,we find $kQ = \frac{3Vr}{2}$.
The electric intensity $E$ at a distance $3r$ from the center is given by:
$E = \frac{kQ}{(3r)^2} = \frac{kQ}{9r^2}$.
Substituting the value of $kQ$:
$E = \frac{1}{9r^2} \cdot \frac{3Vr}{2} = \frac{3V}{18r} = \frac{V}{6r}$.

(A) Given that the two spherical conductors are at the same potential $V$.
For a spherical conductor of radius $r$ and charge $q$,the potential is given by $V = \frac{1}{4 \pi \varepsilon_0} \cdot \frac{q}{r}$.
Since $V_1 = V_2$,we have $\frac{q_1}{r_1} = \frac{q_2}{r_2}$,which implies $\frac{q_1}{q_2} = \frac{r_1}{r_2}$.
The surface charge density $\sigma$ is defined as $\sigma = \frac{q}{4 \pi r^2}$.
Therefore,the ratio of surface charge densities is $\frac{\sigma_1}{\sigma_2} = \frac{q_1 / (4 \pi r_1^2)}{q_2 / (4 \pi r_2^2)} = \frac{q_1}{q_2} \cdot \frac{r_2^2}{r_1^2}$.
Substituting $\frac{q_1}{q_2} = \frac{r_1}{r_2}$,we get $\frac{\sigma_1}{\sigma_2} = \frac{r_1}{r_2} \cdot \frac{r_2^2}{r_1^2} = \frac{r_2}{r_1}$.
Given $r_1 = 4 \ cm$ and $r_2 = 5 \ cm$,the ratio is $\frac{\sigma_1}{\sigma_2} = \frac{5}{4}$.

(A) The kinetic energy gained by a particle of charge $q$ and mass $m$ falling through a potential difference $V$ is given by $K = qV = \frac{1}{2}mv^2$.
For particle $A$ with charge $q$:
$\frac{1}{2}mv_A^2 = qV$ (Equation $1$)
For particle $B$ with charge $4q$:
$\frac{1}{2}mv_B^2 = 4qV$ (Equation $2$)
Dividing Equation $1$ by Equation $2$:
$\frac{v_A^2}{v_B^2} = \frac{qV}{4qV} = \frac{1}{4}$
Taking the square root on both sides:
$\frac{v_A}{v_B} = \sqrt{\frac{1}{4}} = \frac{1}{2}$
Therefore,the ratio of their speeds $v_A : v_B$ is $1:2$.

(C) The electric potential $V$ in an electric field $\overrightarrow{E}$ is given by the relation $dV = -\overrightarrow{E} \cdot d\overrightarrow{r}$.
This implies that the electric potential decreases as we move in the direction of the electric field.
In the given figure,point $B$ is at the leftmost position,meaning it is at the highest potential among the three points.
Points $A$ and $C$ are further along the direction of the electric field compared to point $B$,so their potentials will be lower than that of point $B$.
Therefore,the electric potential is maximum at point $B$.

(D) The force $F$ experienced by a charge $q$ in an electric field $E$ is given by the formula $F = qE$.
Here,the charge of the electron is $q = 1.6 \times 10^{-19} \ C$ and the electric field intensity due to the proton is $E = 5 \times 10^{11} \ NC^{-1}$.
Substituting these values into the formula:
$F = (1.6 \times 10^{-19} \ C) \times (5 \times 10^{11} \ NC^{-1})$
$F = 8.0 \times 10^{-8} \ N$.
Thus,the magnitude of the force between the electron and the proton is $8 \times 10^{-8} \ N$.

(D) The potential at the surface of the conductor is $V_s = \frac{q}{4\pi\varepsilon_0 r}$.
The potential at a distance $3r$ from the center is $V_p = \frac{q}{4\pi\varepsilon_0 (3r)}$.
The potential difference $V$ is given by $V = V_s - V_p = \frac{q}{4\pi\varepsilon_0 r} - \frac{q}{12\pi\varepsilon_0 r} = \frac{q}{4\pi\varepsilon_0 r} \left(1 - \frac{1}{3}\right) = \frac{2q}{12\pi\varepsilon_0 r} = \frac{q}{6\pi\varepsilon_0 r}$.
Thus,$\frac{q}{4\pi\varepsilon_0} = \frac{3}{2} Vr$.
The electric intensity $E$ at a distance $3r$ is $E = \frac{q}{4\pi\varepsilon_0 (3r)^2} = \frac{q}{4\pi\varepsilon_0 (9r^2)}$.
Substituting the value of $\frac{q}{4\pi\varepsilon_0}$,we get $E = \frac{3Vr}{2(9r^2)} = \frac{V}{6r}$.