MHT CET 2022 Physics Question Paper with Answer and Solution

(D) When the liquid column is displaced by a small distance $y$ from its equilibrium position,a restoring force is generated due to the difference in the heights of the liquid levels in the two arms.
The difference in height between the two arms becomes $2y$.
The restoring force $F$ is equal to the weight of the excess liquid column of height $2y$: $F = -(A \cdot 2y \cdot \rho \cdot g)$,where $A$ is the cross-sectional area and $\rho$ is the density.
The mass of the liquid column is $m = A \cdot L \cdot \rho$.
Using Newton's second law,$F = ma$:
$A \cdot L \cdot \rho \cdot \frac{d^2 y}{dt^2} = -2y \cdot A \cdot \rho \cdot g$
$\frac{d^2 y}{dt^2} = -(\frac{2g}{L})y$
Comparing this with the standard $SHM$ equation $\frac{d^2 y}{dt^2} = -\omega^2 y$,we get $\omega^2 = \frac{2g}{L}$.
Therefore,the time period $T = \frac{2\pi}{\omega} = 2\pi \sqrt{\frac{L}{2g}}$.

(C) The time period $T$ of a simple pendulum is given by the formula:
$T = 2 \pi \sqrt{\frac{L}{g}}$
From this relation,we can see that $T \propto \sqrt{L}$.
If a pendulum clock is running slow,it means the time period $T$ is greater than the required value.
To correct the time,we need to decrease the time period $T$.
Since $T$ is directly proportional to the square root of the length $L$,decreasing the length $L$ will decrease the time period $T$.
Therefore,we should reduce the length of the pendulum.

(C) For a simple pendulum,the restoring torque is given by $\tau = -mgL \sin \theta$.
For small oscillations,$\sin \theta \approx \theta$,so $\tau \approx -mgL \theta$.
The angular acceleration is $\alpha = \frac{\tau}{I} = \frac{-mgL \theta}{mL^2} = -\frac{g}{L} \theta$.
Comparing this with the $SHM$ equation $\alpha = -\omega^2 \theta$,we get $\omega^2 = \frac{g}{L}$.
The force constant $k$ for an oscillating system is defined as $k = m \omega^2$.
Substituting $\omega^2$,we get $k = m \left(\frac{g}{L}\right)$.
Thus,the force constant $k$ is directly proportional to the mass of the bob $m$ and inversely proportional to the length $L$ of the pendulum.
Among the given options,the most appropriate relationship is that it is directly proportional to the mass of the bob.

(D) The potential energy $P$ of a body executing $S$.$H$.$M$. at displacement $x$ is given by $P = \frac{1}{2} k x^2$,where $k = m \omega^2$.
Given,$P_1 = \frac{1}{2} k x^2 \implies \sqrt{P_1} = x \sqrt{\frac{k}{2}}$ ---$(1)$
And $P_2 = \frac{1}{2} k y^2 \implies \sqrt{P_2} = y \sqrt{\frac{k}{2}}$ ---$(2)$
Let the potential energy at displacement $(x+y)$ be $P$. Then,
$P = \frac{1}{2} k (x+y)^2$
Taking the square root of both sides:
$\sqrt{P} = (x+y) \sqrt{\frac{k}{2}}$
$\sqrt{P} = x \sqrt{\frac{k}{2}} + y \sqrt{\frac{k}{2}}$
Substituting equations $(1)$ and $(2)$ into this expression:
$\sqrt{P} = \sqrt{P_1} + \sqrt{P_2}$

(B) The total energy $E$ of a body executing simple harmonic motion $(SHM)$ is given by $E = \frac{1}{2} m \omega^2 a^2$,where $a$ is the amplitude.
At any displacement $y$,the potential energy $U$ is given by $U = \frac{1}{2} m \omega^2 y^2$.
The kinetic energy $K$ is the difference between total energy and potential energy: $K = E - U$.
Substituting the expressions,$K = \frac{1}{2} m \omega^2 a^2 - \frac{1}{2} m \omega^2 y^2 = \frac{1}{2} m \omega^2 (a^2 - y^2)$.
Given that the displacement $y = \frac{a}{2}$,we substitute this into the kinetic energy formula:
$K = \frac{1}{2} m \omega^2 (a^2 - (\frac{a}{2})^2) = \frac{1}{2} m \omega^2 (a^2 - \frac{a^2}{4}) = \frac{1}{2} m \omega^2 (\frac{3a^2}{4})$.
Since $E = \frac{1}{2} m \omega^2 a^2$,we can write $K = \frac{3}{4} (\frac{1}{2} m \omega^2 a^2) = \frac{3E}{4}$.

(A) The displacement of a particle performing $SHM$ starting from the mean position is given by $x = A \sin(\omega t)$,where $\omega = \frac{2\pi}{T}$.
At $t = \frac{T}{12}$,the displacement is $x = A \sin\left(\frac{2\pi}{T} \cdot \frac{T}{12}\right) = A \sin\left(\frac{\pi}{6}\right) = \frac{A}{2}$.
The potential energy $U$ is given by $U = \frac{1}{2}m\omega^2x^2 = \frac{1}{2}m\omega^2\left(\frac{A}{2}\right)^2 = \frac{1}{8}m\omega^2A^2$.
The kinetic energy $K$ is given by $K = \frac{1}{2}m\omega^2(A^2 - x^2) = \frac{1}{2}m\omega^2\left(A^2 - \frac{A^2}{4}\right) = \frac{1}{2}m\omega^2\left(\frac{3A^2}{4}\right) = \frac{3}{8}m\omega^2A^2$.
The ratio of potential energy to kinetic energy is $\frac{U}{K} = \frac{\frac{1}{8}m\omega^2A^2}{\frac{3}{8}m\omega^2A^2} = \frac{1}{3}$.
Thus,the ratio is $1: 3$.

(B) The correct option is $(B)$.
Concept: For $SHM$,the acceleration is given by $a = -\omega^2 x$.
The displacement is $x = A \sin(\omega t + \phi)$.
Velocity is the first derivative of displacement: $v = \frac{dx}{dt} = A\omega \cos(\omega t + \phi)$.
Acceleration is the derivative of velocity: $a = \frac{dv}{dt} = -A\omega^2 \sin(\omega t + \phi)$.
At the extreme positions,displacement $x = \pm A$,so acceleration $a = \mp A\omega^2$ is maximum (in magnitude),and velocity $v = 0$.
At the mean position,displacement $x = 0$,so acceleration $a = 0$,and velocity $v = \pm A\omega$ is maximum.
Therefore,when '$a$' is maximum,'$v$' is zero.

(D) The correct option is $D$.
Concept: For $S.H.M.$,the displacement is given by $x = A \sin(\omega t + \phi)$.
The velocity is $v = \frac{dx}{dt} = A \omega \cos(\omega t + \phi)$.
The maximum speed is $V = A \omega$.
In one complete time period $T$,the particle travels a total distance of $4A$.
The average speed is defined as the total distance divided by the total time.
$\text{Average speed} = \frac{\text{Total distance}}{\text{Total time}} = \frac{4A}{T}$.
Since $T = \frac{2\pi}{\omega}$ and $\omega = \frac{V}{A}$,we have $T = \frac{2\pi A}{V}$.
Substituting $T$ into the average speed formula:
$\text{Average speed} = \frac{4A}{\left(\frac{2\pi A}{V}\right)} = \frac{4A \cdot V}{2\pi A} = \frac{2V}{\pi}$.

(B) The general equation of a simple harmonic progressive wave is $y = A \sin(\omega t - kx)$.
Comparing this with the given equation $y = A \sin(100 \pi t - 3x)$,we get the propagation constant $k = 3 \text{ rad/m}$.
The relationship between phase difference $(\Delta \phi)$ and path difference $(\Delta x)$ is given by $\Delta \phi = k \cdot \Delta x$.
Given the phase difference $\Delta \phi = \frac{\pi}{18}$.
Substituting the values,we get $\frac{\pi}{18} = 3 \cdot \Delta x$.
Therefore,the distance between the two particles is $\Delta x = \frac{\pi}{18 \cdot 3} = \frac{\pi}{54} \text{ m}$.
Thus,the correct option is $B$.

(B) The displacement of a particle starting from the equilibrium position is given by $x = a \sin(\omega t)$.
Given $t = \frac{T}{12}$ and $\omega = \frac{2\pi}{T}$,the displacement is $x = a \sin\left(\frac{2\pi}{T} \cdot \frac{T}{12}\right) = a \sin\left(\frac{\pi}{6}\right) = \frac{a}{2}$.
The kinetic energy $(K.E.)$ is given by $\frac{1}{2}k(a^2 - x^2)$ and the potential energy $(P.E.)$ is given by $\frac{1}{2}kx^2$.
The ratio of kinetic energy to potential energy is $\frac{K.E.}{P.E.} = \frac{a^2 - x^2}{x^2}$.
Substituting $x = \frac{a}{2}$,we get $\frac{K.E.}{P.E.} = \frac{a^2 - (a/2)^2}{(a/2)^2} = \frac{a^2 - a^2/4}{a^2/4} = \frac{3a^2/4}{a^2/4} = \frac{3}{1}$.
Thus,the ratio is $3: 1$.

(A) For a particle in $S.H.M.$ suspended from a vertical spring,the equilibrium position is where the spring force balances the gravitational force,i.e.,$kx = mg$.
At the highest point of oscillation,the spring is unstretched,meaning the extension $x = 0$.
Since the equilibrium position is at a distance $A$ (amplitude) below the highest point,the extension at equilibrium is $x = A$.
Therefore,$kA = mg$,which gives the amplitude $A = \frac{mg}{k} = \frac{g}{\omega^2}$.
Given frequency $f = 5 \ Hz$,the angular frequency is $\omega = 2 \pi f = 2 \pi \times 5 = 10 \pi \ rad/s$.
Substituting the values,$A = \frac{10}{(10 \pi)^2} = \frac{10}{100 \pi^2} = \frac{1}{10 \pi^2} \ m$.
The maximum speed is $V_{\max} = A \omega$.
$V_{\max} = \left( \frac{1}{10 \pi^2} \right) \times (10 \pi) = \frac{1}{\pi} \ m/s$.

(A) The total distance covered by a particle in one complete oscillation of linear $S.H.M.$ is equal to $4A$,where $A$ is the amplitude.
The time period $T$ of the oscillation is related to the frequency $n$ by the formula $T = \frac{1}{n}$.
Average speed is defined as the total distance divided by the total time taken.
Therefore,average speed $v_{av} = \frac{\text{Total Distance}}{\text{Total Time}} = \frac{4A}{T} = \frac{4A}{1/n} = 4nA$.

(C) The standard equation for linear $S.H.M$ is given by $x = a \sin(\omega t + \phi)$.
Comparing the given equation $x = 5 \sin(4t - \frac{\pi}{6})$ with the standard form,we get the amplitude $a = 5 \ cm$ and angular frequency $\omega = 4 \ rad/s$.
The velocity $v$ of a particle in $S.H.M$ at a displacement $x$ is given by the formula:
$v = \omega \sqrt{a^2 - x^2}$
Substituting the given values $a = 5 \ cm$,$x = 3 \ cm$,and $\omega = 4 \ rad/s$:
$v = 4 \sqrt{5^2 - 3^2}$
$v = 4 \sqrt{25 - 9}$
$v = 4 \sqrt{16}$
$v = 4 \times 4 = 16 \ cm/s$.
Therefore,the velocity of the particle is $16 \ cm/s$.

(D) The equations of motion for the particles starting from the mean position are:
$X_A = A_1 \sin(\omega_A t) = A_1 \sin(\frac{2\pi}{T} t)$
$X_B = A_2 \sin(\omega_B t) = A_2 \sin(\frac{2\pi}{3T/2} t) = A_2 \sin(\frac{4\pi}{3T} t)$
The phase of particle '$A$' is $\phi_A = \frac{2\pi}{T} t$ and the phase of particle '$B$' is $\phi_B = \frac{4\pi}{3T} t$.
When particle '$A$' completes one oscillation,the time elapsed is $t = T$.
At $t = T$,the phase of particle '$A$' is $\phi_A = \frac{2\pi}{T} \times T = 2\pi$.
At $t = T$,the phase of particle '$B$' is $\phi_B = \frac{4\pi}{3T} \times T = \frac{4\pi}{3}$.
The phase difference $\Delta \phi = |\phi_A - \phi_B| = |2\pi - \frac{4\pi}{3}| = \frac{2\pi}{3}$.

(D) In $S.H.M.$,the displacement is given by $x = A \sin(\omega t + \phi)$.
Velocity is given by $v = \frac{dx}{dt} = A\omega \cos(\omega t + \phi)$. At the mean position $(x = 0)$,the velocity is maximum $(v_{max} = A\omega)$.
Acceleration is given by $a = -\omega^2 x$. At the mean position $(x = 0)$,acceleration is minimum $(a = 0)$.
The restoring force $F = -kx$ is always directed towards the mean position.
The total energy $E = K.E. + P.E. = \frac{1}{2}kA^2$ remains constant at all times.
Therefore,statement $D$ is incorrect because the velocity is maximum,not minimum,at the mean position.

(D) Given,$y_1 = 10 \sin \omega t$ and $y_2 = 10 \sin \omega t + 5 \cos \omega t$.
For $y_1$,the amplitude $A_1 = 10$.
For $y_2$,the expression is of the form $A \sin \omega t + B \cos \omega t$,where the resultant amplitude $A_2 = \sqrt{A^2 + B^2}$.
Here,$A = 10$ and $B = 5$,so $A_2 = \sqrt{10^2 + 5^2} = \sqrt{100 + 25} = \sqrt{125} = 5\sqrt{5}$.
The ratio of the amplitudes is $\frac{A_1}{A_2} = \frac{10}{5\sqrt{5}} = \frac{2}{\sqrt{5}}$.
Thus,the ratio is $2 : \sqrt{5}$.

(D) Given,$x = A \cos \left(\omega t + \frac{\pi}{4}\right)$.
Velocity $v = \frac{dx}{dt} = -A \omega \sin \left(\omega t + \frac{\pi}{4}\right)$.
The velocity is maximum when the magnitude of the sine function is maximum,i.e.,$\sin \left(\omega t + \frac{\pi}{4}\right) = -1$.
This occurs when the phase angle $\left(\omega t + \frac{\pi}{4}\right) = \frac{3\pi}{2}$.
$\omega t + \frac{\pi}{4} = \frac{3\pi}{2} \implies \omega t = \frac{3\pi}{2} - \frac{\pi}{4} = \frac{5\pi}{4}$.
$t = \frac{5\pi}{4\omega}$.
However,checking the options provided,the question implies the time when the particle reaches the equilibrium position $(x=0)$ moving in the negative direction,where velocity magnitude is maximum. Setting $\omega t + \frac{\pi}{4} = \frac{\pi}{2}$ gives $\omega t = \frac{\pi}{4}$,so $t = \frac{\pi}{4\omega}$.

(D) For a particle starting from the positive extreme position $(x = +A)$,the displacement equation is given by:
$x = A \cos(\omega t)$
Given: $A = 6 \ cm$,$T = 3 \ s$.
The angular frequency is $\omega = \frac{2\pi}{T} = \frac{2\pi}{3} \ rad/s$.
We need to find the time $t$ when the particle has traveled a distance of $3 \ cm$ from the positive extreme position. This means the new position is $x = A - 3 = 6 - 3 = 3 \ cm$.
Substituting the values into the displacement equation:
$3 = 6 \cos(\omega t)$
$\cos(\omega t) = \frac{3}{6} = \frac{1}{2}$
Since $\cos(60^{\circ}) = \frac{1}{2}$,we have:
$\omega t = 60^{\circ} = \frac{\pi}{3} \ rad$
Substituting $\omega = \frac{2\pi}{3}$:
$(\frac{2\pi}{3}) t = \frac{\pi}{3}$
$t = \frac{\pi}{3} \times \frac{3}{2\pi} = 0.5 \ s$
Therefore,the time required is $0.5 \ s$.

(C) Concept: The wooden cube will leave the plank at the top extreme position if the downward acceleration of the plank exceeds the acceleration due to gravity.
For the block not to leave the plank,the normal force $N$ must be greater than or equal to zero.
From the free body diagram of the block,the equation of motion is: $mg - N = ma$.
For the block to remain in contact,$N \geq 0$,which implies $mg \geq ma$,or $a \leq g$.
In $S$.$H$.$M$.,the maximum acceleration is given by $a_{max} = \omega^2 A$.
Therefore,the condition for the block not to leave the plank is $\omega^2 A \leq g$,or $A \leq \frac{g}{\omega^2}$.
Given frequency $f = \frac{3}{\pi} \text{ Hz}$.
Angular frequency $\omega = 2\pi f = 2\pi \left( \frac{3}{\pi} \right) = 6 \text{ rad/s}$.
Substituting the values: $A \leq \frac{10}{6^2} = \frac{10}{36} = \frac{5}{18} \text{ m}$.
Thus,the maximum amplitude is $\frac{5}{18} \text{ m}$.

(B) The displacement equation for a particle executing simple harmonic motion $(SHM)$ starting from the mean position is given by $x = A \sin(\omega t)$,where $A$ is the amplitude and $\omega = \frac{2\pi}{T}$ is the angular frequency.
We want to find the time $t$ when the displacement $x = \frac{A}{2}$.
Substituting this into the equation: $\frac{A}{2} = A \sin(\omega t)$.
This simplifies to $\sin(\omega t) = \frac{1}{2}$.
Since $\sin(30^\circ) = \frac{1}{2}$,we have $\omega t = \frac{\pi}{6}$.
Substituting $\omega = \frac{2\pi}{T}$,we get $\left(\frac{2\pi}{T}\right) t = \frac{\pi}{6}$.
Solving for $t$,we find $t = \frac{T}{12}$.

(B) The path length of oscillation is the total distance between the two extreme positions, which is equal to $2a$, where $a$ is the amplitude.
Given, $2a = 16 \,cm$, so the amplitude $a = 8 \,cm$.
The length of the pendulum is $l = 1 \,m$.
The angular frequency of a simple pendulum is given by $\omega = \sqrt{\frac{g}{l}}$.
Substituting $g = \pi^2 \,m/s^2$ and $l = 1 \,m$, we get $\omega = \sqrt{\frac{\pi^2}{1}} = \pi \,rad/s$.
The maximum velocity $v_{max}$ in simple harmonic motion is given by $v_{max} = a\omega$.
Substituting the values, $v_{max} = 8 \,cm \times \pi \,rad/s = 8\pi \,cm/s$.

(B) The resultant amplitude $R$ of two SHMs with phase difference $\delta$ is given by $R=\sqrt{A_1^2+A_2^2+2 A_1 A_2 \cos \delta}$.
$A$. For $A_1=A_2=A$ and $\delta=0^{\circ}$:
$R=\sqrt{A^2+A^2+2A^2 \cos 0^{\circ}} = \sqrt{4A^2} = 2A$. Thus,$A-III$.
$B$. For $A_1 \neq A_2$ and $\delta=0^{\circ}$:
$R=\sqrt{A_1^2+A_2^2+2A_1 A_2 \cos 0^{\circ}} = \sqrt{(A_1+A_2)^2} = A_1+A_2$. Thus,$B-I$.
$C$. For $A_1=A_2=A$ and $\delta=90^{\circ}$:
$R=\sqrt{A^2+A^2+2A^2 \cos 90^{\circ}} = \sqrt{2A^2} = A\sqrt{2}$. Thus,$C-IV$.
$D$. For $A_1=A_2=A$ and $\delta=180^{\circ}$:
$R=\sqrt{A^2+A^2+2A^2 \cos 180^{\circ}} = \sqrt{2A^2-2A^2} = 0$. Thus,$D-II$.

(D) The correct option is $D$.
Concept: The standard differential equation for $S.H.M.$ is given by $\frac{d^2 x}{dt^2} + \omega^2 x = 0$,where $\omega$ is the angular frequency.
Comparing the given equation $\frac{d^2 x}{dt^2} + \alpha x = 0$ with the standard equation,we get $\omega^2 = \alpha$,which implies $\omega = \sqrt{\alpha}$.
The time period $T$ of the motion is defined as $T = \frac{2 \pi}{\omega}$.
Substituting the value of $\omega$,we get $T = \frac{2 \pi}{\sqrt{\alpha}}$.

(D) For $S.H.M.$,the acceleration $a$ is proportional to the displacement $x$ and is directed towards the mean position of the particle:
$a = -\omega^2 x = -px$
where $\omega$ is the angular frequency of the $S.H.M.$
Comparing the two expressions,we get $\omega^2 = p$,which implies $\omega = \sqrt{p}$.
The time period of oscillation $T$ is given by the formula:
$T = \frac{2 \pi}{\omega}$
Substituting the value of $\omega$:
$T = \frac{2 \pi}{\sqrt{p}}$

(B) When the sphere is displaced by a small angle $\theta$,the center of the sphere moves along a circular arc of radius $(R-r)$.
The restoring force is provided by the component of gravity acting along the tangent to the path.
The restoring torque about the center of curvature $O$ is given by:
$\tau = -mg(R-r) \sin \theta$
For small oscillations,$\sin \theta \approx \theta$,so $\tau \approx -mg(R-r) \theta$.
Using the rotational analog of Newton's second law,$\tau = I \alpha$,where $I$ is the moment of inertia about the axis of rotation passing through $O$.
Since the sphere is rolling without slipping,the effective moment of inertia about $O$ is $I = I_{cm} + m(R-r)^2 = \frac{2}{5}mr^2 + m(R-r)^2$.
However,for a simple oscillation problem where we consider the motion of the center of mass,we use the equation of motion for the center of mass: $F_{restoring} = -mg \sin \theta = m a_{cm}$.
Since $a_{cm} = (R-r) \alpha$,we have $m(R-r) \alpha = -mg \theta$.
$\alpha = -\left(\frac{g}{R-r}\right) \theta$.
Comparing this with the $S$.$H$.$M$. equation $\alpha = -\omega^2 \theta$,we get $\omega^2 = \frac{g}{R-r}$.
The time period is $T = \frac{2 \pi}{\omega} = 2 \pi \sqrt{\frac{R-r}{g}}$.

(B) The standard differential equation for simple harmonic motion $(SHM)$ is given by $\frac{d^2 x}{d t^2} + \omega^2 x = 0$,which can be rewritten as $\frac{d^2 x}{d t^2} = -\omega^2 x$.
Given equation: $\alpha \frac{d^2 x}{d t^2} + \beta x = 0$.
Rearranging the given equation: $\frac{d^2 x}{d t^2} = -\frac{\beta}{\alpha} x$.
Comparing this with the standard form $\frac{d^2 x}{d t^2} = -\omega^2 x$,we get $\omega^2 = \frac{\beta}{\alpha}$,which implies $\omega = \sqrt{\frac{\beta}{\alpha}}$.
The time period $T$ is given by $T = \frac{2 \pi}{\omega}$.
Substituting the value of $\omega$: $T = \frac{2 \pi}{\sqrt{\frac{\beta}{\alpha}}} = 2 \pi \sqrt{\frac{\alpha}{\beta}}$.

(A) The gravitational acceleration on the surface of a planet with mass $M$ and radius $R$ is given by $g = \frac{GM}{R^2}$.
For the given planet,the mass $M_p = 2M_e$ and the radius $R_p = 2R_e$.
Therefore,the gravitational acceleration on the planet $g_p$ is:
$g_p = \frac{G(2M_e)}{(2R_e)^2} = \frac{2GM_e}{4R_e^2} = \frac{1}{2} g_e$.
The time period of a pendulum is given by $T = 2\pi \sqrt{\frac{l}{g}}$.
For a second's pendulum on Earth,$T_e = 2 \ s$ with acceleration $g_e$.
On the planet,the new time period $T_p$ is:
$T_p = 2\pi \sqrt{\frac{l}{g_p}} = 2\pi \sqrt{\frac{l}{g_e/2}} = \sqrt{2} \times (2\pi \sqrt{\frac{l}{g_e}}) = \sqrt{2} \times T_e$.
Substituting $T_e = 2 \ s$,we get $T_p = 2\sqrt{2} \ s$.

(B) The correct option is $B$.
Concept: Angular momentum $L$ of a particle of mass $m$ moving in a circle of radius $r$ with angular velocity $\omega$ is given by $L = I\omega = mr^2\omega$.
Given that the mass $m$ and angular velocity $\omega$ remain constant.
Initially,$L = m\omega r^2$.
When the length of the string is halved,the new radius becomes $r' = \frac{r}{2}$.
The new angular momentum $L'$ is given by:
$L' = m\omega(r')^2 = m\omega\left(\frac{r}{2}\right)^2$
$L' = m\omega\left(\frac{r^2}{4}\right) = \frac{1}{4}(mr^2\omega)$
$L' = \frac{L}{4}$.

(C) The impulse $J$ applied at one end provides both linear and angular momentum to the system.
$1$. Linear impulse equation: $J = M_{total} v_{cm} \Rightarrow MV = (2M) v_{cm} \Rightarrow v_{cm} = \frac{V}{2}$.
$2$. Angular impulse equation about the center of mass $(COM)$: $J \times r = I_{cm} \omega$.
Here,$r = \frac{L}{2}$ and $I_{cm} = M(\frac{L}{2})^2 + M(\frac{L}{2})^2 = \frac{ML^2}{2}$.
Substituting the values: $(MV) \times \frac{L}{2} = (\frac{ML^2}{2}) \omega$.
Solving for $\omega$: $\frac{MVL}{2} = \frac{ML^2 \omega}{2} \Rightarrow \omega = \frac{V}{L}$.

(A) Concept: The angular impulse applied to the rod is equal to the change in its angular momentum.
Angular impulse about the center of mass $O$ is given by $J_{\theta} = P \cdot \frac{l}{2}$.
The angular momentum of the rod about its center of mass is $L = I \omega$,where $I = \frac{m l^2}{12}$ is the moment of inertia of the rod about its center.
Equating angular impulse to change in angular momentum:
$P \cdot \frac{l}{2} = I \omega$
$P \cdot \frac{l}{2} = \left( \frac{m l^2}{12} \right) \omega$
Solving for angular velocity $\omega$:
$\omega = \frac{P \cdot l}{2} \cdot \frac{12}{m l^2} = \frac{6 P}{m l}$
The rod rotates with a constant angular velocity $\omega$. The time $\Delta t$ taken to turn through an angle $\Delta \theta = \frac{\pi}{2}$ is:
$\Delta t = \frac{\Delta \theta}{\omega} = \frac{\pi / 2}{6 P / (m l)} = \frac{\pi}{2} \cdot \frac{m l}{6 P} = \frac{\pi m l}{12 P}$

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

(D) The rotational kinetic energy of a body is given by $x = \frac{1}{2} I \omega^2$.
The angular momentum of a body is given by $y = I \omega$.
We can express the kinetic energy in terms of angular momentum as:
$x = \frac{(I \omega)^2}{2 I} = \frac{y^2}{2 I}$.
Rearranging this equation to solve for the moment of inertia $I$,we get:
$I = \frac{y^2}{2 x}$.

(A) The moment of inertia of a ring about its central axis perpendicular to its plane is $I_{cm} = MR^2 = 4 \,kg \,m^2$.
By the perpendicular axis theorem, the moment of inertia about a diameter in its plane is $I_d = \frac{1}{2} MR^2 = \frac{1}{2} \times 4 = 2 \,kg \,m^2$.
Using the parallel axis theorem, the moment of inertia about a tangent in the plane is $I_t = I_d + MR^2$.
Substituting the values: $I_t = 2 \,kg \,m^2 + 4 \,kg \,m^2 = 6 \,kg \,m^2$.

(D) The moment of inertia of a solid sphere about its diameter is $I_{cm} = \frac{2}{5} m r^2$.
For the three spheres whose centers lie on the axis of rotation $A-A'$,the moment of inertia of each about this axis is $I_1 = \frac{2}{5} m r^2$.
For the two spheres whose centers are at a distance $r$ from the axis of rotation $A-A'$,we use the parallel axis theorem: $I = I_{cm} + m d^2$,where $d = r$.
Thus,$I_2 = \frac{2}{5} m r^2 + m r^2 = \frac{7}{5} m r^2$.
The total moment of inertia of the system is $I_{total} = 3 \times I_1 + 2 \times I_2$.
$I_{total} = 3 \left( \frac{2}{5} m r^2 \right) + 2 \left( \frac{7}{5} m r^2 \right) = \frac{6}{5} m r^2 + \frac{14}{5} m r^2 = \frac{20}{5} m r^2 = 4 m r^2$.

(D) The moment of inertia of a solid sphere about an axis passing through its centre of mass is $I_{cm} = \frac{2}{5} MR^2$.
According to the parallel axis theorem,the moment of inertia $I$ about an axis at a distance $d$ from the centre is given by $I = I_{cm} + Md^2$.
Here,$d = \frac{R}{2}$.
Substituting the values,we get $I = \frac{2}{5} MR^2 + M\left(\frac{R}{2}\right)^2$.
$I = \frac{2}{5} MR^2 + M\left(\frac{R^2}{4}\right)$.
$I = \left(\frac{2}{5} + \frac{1}{4}\right) MR^2$.
$I = \left(\frac{8 + 5}{20}\right) MR^2 = \frac{13}{20} MR^2$.

(A) The radius of gyration $k$ is related to the moment of inertia $I$ by the formula $I = mk^2$,where $m$ is the mass of the body.
For each case:
$(A)$ Solid sphere about its tangent: $I = \frac{2}{5}mR^2 + mR^2 = \frac{7}{5}mR^2$. Thus,$k = \sqrt{\frac{7}{5}}R$. Matches $(R)$.
$(B)$ Ring about its tangent perpendicular to its plane: $I = mR^2 + mR^2 = 2mR^2$. Thus,$k = \sqrt{2}R$. Matches $(P)$.
$(C)$ Uniform solid right circular cone about its central axis: $I = \frac{3}{10}mR^2$. Thus,$k = \sqrt{\frac{3}{10}}R$. Matches $(S)$.
$(D)$ Uniform disc about its diameter: $I = \frac{1}{2}(\frac{1}{2}mR^2) = \frac{1}{4}mR^2$. Thus,$k = \frac{R}{2}$. Matches $(Q)$.
Therefore,the correct matching is $(A)-(R), (B)-(P), (C)-(S), (D)-(Q)$.

(A) The radius of gyration $K$ is given by $K = \sqrt{\frac{I}{m}}$.
For a ring of mass $m$ and radius $R$,the moment of inertia about its central axis perpendicular to the plane is $I_{\text{ring, center}} = mR^2$.
For a disc of mass $m$ and radius $R$,the moment of inertia about its central axis perpendicular to the plane is $I_{\text{disc, center}} = \frac{1}{2}mR^2$.
Using the parallel axis theorem,the moment of inertia $I'$ about a tangential axis perpendicular to the plane is $I' = I_{\text{center}} + mR^2$.
For the ring: $I'_{\text{ring}} = mR^2 + mR^2 = 2mR^2$. Thus,$K_{\text{ring}} = \sqrt{\frac{2mR^2}{m}} = \sqrt{2}R$.
For the disc: $I'_{\text{disc}} = \frac{1}{2}mR^2 + mR^2 = \frac{3}{2}mR^2$. Thus,$K_{\text{disc}} = \sqrt{\frac{3}{2}}R$.
The ratio of the radius of gyration of the ring to that of the disc is $\frac{K_{\text{ring}}}{K_{\text{disc}}} = \frac{\sqrt{2}R}{\sqrt{3/2}R} = \frac{\sqrt{2}}{\sqrt{3}/\sqrt{2}} = \frac{2}{\sqrt{3}}$.

(D) The moment of inertia of a disc of mass $m$ and radius $r$ about an axis passing through its center and perpendicular to its plane is $I_{cm} = \frac{1}{2} m r^2$.
For disc $A$,the axis passes through its center,so its moment of inertia is $I_A = \frac{1}{2} m r^2$.
For disc $B$,the axis is at a distance $d = 2r$ from its center. Using the parallel axis theorem,$I_B = I_{cm} + m d^2 = \frac{1}{2} m r^2 + m(2r)^2 = \frac{1}{2} m r^2 + 4 m r^2 = \frac{9}{2} m r^2$.
The total moment of inertia of the system is $I_{total} = I_A + I_B = \frac{1}{2} m r^2 + \frac{9}{2} m r^2 = \frac{10}{2} m r^2 = 5 m r^2$.

(D) Concept: Application of the parallel axis theorem.
The moment of inertia of a solid sphere about an axis passing through its centre is:
$I_{\text{cm}} = \frac{2}{5} MR^2$
According to the parallel axis theorem,the moment of inertia $I$ about an axis parallel to the one passing through the centre of mass at a distance $d$ is given by:
$I = I_{\text{cm}} + Md^2$
Here,the distance $d = \frac{R}{2}$.
Substituting the values:
$I = \frac{2}{5} MR^2 + M\left(\frac{R}{2}\right)^2$
$I = \frac{2}{5} MR^2 + \frac{1}{4} MR^2$
$I = \left(\frac{8 + 5}{20}\right) MR^2$
$I = \frac{13}{20} MR^2$

(A) The moment of inertia of a thin rod of mass $M$ and length $L$ about an axis passing through its centre and perpendicular to its length is given by $I = \frac{1}{12} M L^2$.
When the rod is cut into four equal parts,each part has a mass $M' = \frac{M}{4}$ and a length $L' = \frac{L}{4}$.
The moment of inertia $I'$ of each part about an axis passing through its own centre and perpendicular to its length is given by:
$I' = \frac{1}{12} M' (L')^2$
Substituting the values of $M'$ and $L'$:
$I' = \frac{1}{12} \left( \frac{M}{4} \right) \left( \frac{L}{4} \right)^2$
$I' = \frac{1}{12} \left( \frac{M}{4} \right) \left( \frac{L^2}{16} \right)$
$I' = \frac{1}{64} \left( \frac{1}{12} M L^2 \right)$
Since $I = \frac{1}{12} M L^2$,we get:
$I' = \frac{I}{64}$

(C) The formulas for the moment of inertia about their central axes are as follows:
$I_{\text{Ring}} = M R^2 = 1.0 M R^2$
$I_{\text{Sphere}} = \frac{2}{5} M R^2 = 0.4 M R^2$
$I_{\text{Disc}} = \frac{1}{2} M R^2 = 0.5 M R^2$
Comparing the coefficients,we see that $1.0 > 0.5 > 0.4$.
Therefore,the ring has the largest moment of inertia because its mass is distributed at the maximum distance $(R)$ from the axis of rotation.

(C) The correct option is $C$.
Concept: The angular speed is given by $\omega = \frac{\Delta \theta}{\Delta t}$.
Relative angular speed is given by $\omega_{rel} = \omega_{s} - \omega_{h}$.
The angular speed of the second hand is $\omega_{s} = \frac{2 \pi}{60} = \frac{\pi}{30} \ rad/s$.
The angular speed of the hour hand is $\omega_{h} = \frac{2 \pi}{12 \times 3600} = \frac{2 \pi}{43200} = \frac{\pi}{21600} \ rad/s$.
Therefore,the relative angular speed is $\omega_{rel} = \omega_{s} - \omega_{h} = \frac{\pi}{30} - \frac{\pi}{21600}$.
Taking the common denominator $21600$: $\omega_{rel} = \frac{720 \pi - \pi}{21600} = \frac{719 \pi}{21600} \ rad/s$.

(D) particle moves along a circular path of radius $r$ with uniform speed $v$.
The angular velocity $\omega$ of the particle is given by the relation $\omega = \frac{v}{r}$.
The angle $\theta$ described by the particle in time $t$ is given by $\theta = \omega t$.
For $t = 1 \ s$,the angle described is $\theta = \omega \times 1 = \frac{v}{r}$ radians.

(B) The angular speed of the hour hand $(\omega_h)$ is given by the angle covered in one full rotation ($2 \pi$ radians) divided by the time taken $(12 \text{ hours} = 12 \times 3600 \text{ seconds})$: $\omega_h = \frac{2 \pi}{12 \times 3600} \text{ rad/s}$.
The angular speed of the minute hand $(\omega_m)$ is given by the angle covered in one full rotation ($2 \pi$ radians) divided by the time taken $(1 \text{ hour} = 3600 \text{ seconds})$: $\omega_m = \frac{2 \pi}{3600} \text{ rad/s}$.
The relative angular speed is the difference between the angular speeds of the two hands: $\Delta \omega = \omega_m - \omega_h$.
Substituting the values: $\Delta \omega = \frac{2 \pi}{3600} - \frac{2 \pi}{12 \times 3600} = \frac{2 \pi}{3600} \left(1 - \frac{1}{12}\right) = \frac{2 \pi}{3600} \left(\frac{11}{12}\right) = \frac{22 \pi}{43200} = \frac{11 \pi}{21600} \text{ rad/s}$.

(B) The angular velocity $\omega$ is defined as the angle described per unit time.
For the minute hand of a clock,it completes one full revolution $(360^{\circ})$ in $60$ minutes.
Since $1$ minute $= 60$ seconds,the time taken for one full revolution is $60 \times 60 = 3600 \ s$.
Therefore,the angular velocity $\omega = \frac{360^{\circ}}{3600 \ s} = 0.1^{\circ}/s$.

(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) Before impact,the bullet is moving with velocity $v$. The initial angular momentum of the system about the hinge point $O$ is $J = m v L$.
After the bullet gets embedded in the rod,suppose the system attains angular velocity $\omega$. The moment of inertia of the bullet-rod system about the axis through $O$ is:
$I = I_{\text{bullet}} + I_{\text{rod}} = m L^2 + \frac{1}{3} M L^2 = \left( \frac{M + 3m}{3} \right) L^2$.
By the principle of conservation of angular momentum,the initial angular momentum equals the final angular momentum:
$J = J' \implies m v L = I \omega$
$m v L = \left( \frac{M + 3m}{3} \right) L^2 \omega$
Solving for $\omega$:
$\omega = \frac{3 m v}{(M + 3m) L}$.

(D) The initial angular velocity is $\omega_0 = 240 \ r.p.m. = \frac{240 \times 2\pi}{60} \ rad/s = 8\pi \ rad/s$.
Using the rotational kinematic equation $\omega = \omega_0 - \alpha t$,where the final angular velocity $\omega = 0$ at $t = 20 \ s$:
$0 = 8\pi - \alpha(20) \Rightarrow \alpha = \frac{8\pi}{20} = 0.4\pi \ rad/s^2$.
The torque $\tau$ is given by $\tau = I\alpha$,where $I = \frac{1}{2}MR^2$ for a disc.
Given $M = 25 \ kg$ and $R = 0.2 \ m$,$I = \frac{1}{2} \times 25 \times (0.2)^2 = 0.5 \ kg \cdot m^2$.
The torque is also $\tau = F \cdot R$,where $F$ is the tangential force.
Equating the two expressions for torque: $F \cdot R = I \alpha
\Rightarrow F = \frac{I \alpha}{R} = \frac{0.5 \times 0.4\pi}{0.2} = \frac{0.2\pi}{0.2} = \pi \ N$.

(C) The torque $\tau$ required to open or close the door remains constant.
According to the principle of moments, $\tau = r_1 \times F_1 = r_2 \times F_2$.
Given $r_1 = 1.2 \,m$, $F_1 = 1 \,N$, and $r_2 = 0.2 \,m$.
Substituting these values: $1.2 \,m \times 1 \,N = 0.2 \,m \times F_2$.
$F_2 = \frac{1.2}{0.2} \,N = 6 \,N$.

(C) Let $T^{\circ} C$ be the final equilibrium temperature of the mixture.
Since the container is closed and isolated,the heat lost by the hotter gas $(O_2)$ equals the heat gained by the cooler gas $(CO_2)$.
For $CO_2$ (a non-linear triatomic gas),the molar heat capacity at constant volume is $C_{V, CO_2} = 3R$.
For $O_2$ (a diatomic gas),the molar heat capacity at constant volume is $C_{V, O_2} = \frac{5}{2}R$.
The number of moles are: $\mu_{CO_2} = \frac{22}{44} = 0.5 \ mol$ and $\mu_{O_2} = \frac{16}{32} = 0.5 \ mol$.
Applying the principle of calorimetry: $\mu_{CO_2} C_{V, CO_2} (T - 27) = \mu_{O_2} C_{V, O_2} (37 - T)$.
$0.5 \times 3R \times (T - 27) = 0.5 \times \frac{5}{2}R \times (37 - T)$.
$3(T - 27) = 2.5(37 - T)$.
$3T - 81 = 92.5 - 2.5T$.
$5.5T = 173.5$.
$T = \frac{173.5}{5.5} \approx 31.5^{\circ} C$.

(B) Let the total current entering at point $P$ be $I = 2.1 \text{ A}$.
Let $i$ be the current flowing through the upper branch ($PQ$ and $QR$).
Then,the current flowing through the lower branch ($PS$ and $SR$) is $(I - i) = (2.1 - i) \text{ A}$.
Since the potential at $Q$ and $S$ is the same if the bridge were balanced,but here it is not,we use Kirchhoff's laws or potential division.
The potential difference between $P$ and $R$ is the same along both paths:
$V_{PR} = i(R_{PQ} + R_{QR}) = (I - i)(R_{PS} + R_{SR})$
$V_{PR} = i(5 + 1) = (2.1 - i)(12.5 + 2.5)$
$6i = (2.1 - i)(15)$
$6i = 31.5 - 15i$
$21i = 31.5$
$i = \frac{31.5}{21} = 1.5 \text{ A}$.
Thus,the current through the $1 \Omega$ resistor is $1.5 \text{ A}$.

(A) Let the two resistances be $R_1$ and $R_2$. The meter bridge balance condition is $\frac{R_1}{R_2} = \frac{l}{100-l}$.
Given $l = 20 \ cm$,so $\frac{R_1}{R_2} = \frac{20}{80} = \frac{1}{4}$,which implies $R_2 = 4R_1$.
Since $R_2 = 4R_1$,$R_1$ is the smaller resistance.
When $15 \ \Omega$ is connected in series with $R_1$,the new resistance is $R_1' = R_1 + 15$.
The new null point is $l' = 40 \ cm$.
Using the balance condition again: $\frac{R_1 + 15}{R_2} = \frac{40}{100-40} = \frac{40}{60} = \frac{2}{3}$.
Substituting $R_2 = 4R_1$: $\frac{R_1 + 15}{4R_1} = \frac{2}{3}$.
Cross-multiplying: $3(R_1 + 15) = 8R_1 \Rightarrow 3R_1 + 45 = 8R_1 \Rightarrow 5R_1 = 45 \Rightarrow R_1 = 9 \ \Omega$.

(B) The de-Broglie wavelength is given by $\lambda = \frac{h}{p}$,where $h$ is Planck's constant and $p$ is the momentum of the electron.
Since $p = mv$ and kinetic energy $E_K = \frac{p^2}{2m}$,we have $p = \sqrt{2mE_K}$.
In a hydrogen atom,the kinetic energy of an electron in the $n^{th}$ orbit is proportional to $\frac{1}{n^2}$,i.e.,$E_K \propto \frac{1}{n^2}$.
Therefore,the momentum $p = \sqrt{2mE_K} \propto \sqrt{\frac{1}{n^2}} = \frac{1}{n}$.
Since $\lambda = \frac{h}{p}$,it follows that $\lambda \propto n$.
When the electron jumps from the third excited state $(n = 4)$ to the ground state $(n = 1)$,the ratio of the new wavelength $\lambda_{final}$ to the initial wavelength $\lambda_{initial}$ is $\frac{\lambda_{final}}{\lambda_{initial}} = \frac{n_{final}}{n_{initial}} = \frac{1}{4}$.
Thus,the de-Broglie wavelength becomes $\frac{1}{4}$ of its initial value.

(B) According to the de-Broglie hypothesis,the wavelength $\lambda$ of a particle is given by $\lambda = \frac{h}{mv}$,where $h$ is Planck's constant,$m$ is the mass,and $v$ is the velocity of the electron.
Rearranging this,we get $mv = \frac{h}{\lambda} \quad --- (1)$.
According to Bohr's postulate for the quantization of angular momentum,the angular momentum $L$ of an electron in the $n^{\text{th}}$ orbit is given by $L = mvr = \frac{nh}{2\pi}$.
Rearranging this for $mv$,we get $mv = \frac{nh}{2\pi r} \quad --- (2)$.
Equating the expressions for $mv$ from equation $(1)$ and $(2)$:
$\frac{h}{\lambda} = \frac{nh}{2\pi r}$.
Canceling $h$ from both sides,we get $\frac{1}{\lambda} = \frac{n}{2\pi r}$.
Therefore,the de-Broglie wavelength is $\lambda = \frac{2\pi r}{n}$.

(B) The de-Broglie wavelength of a particle is given by $\lambda = \frac{h}{p}$.
For an electron with energy $E$ and mass $m$,the momentum $p_e = \sqrt{2mE}$. Thus,$\lambda_e = \frac{h}{\sqrt{2mE}}$.
For a photon with energy $E$,the momentum $p_p = \frac{E}{c}$. Thus,$\lambda_p = \frac{h}{p_p} = \frac{hc}{E}$.
The ratio of the wavelengths is $\frac{\lambda_e}{\lambda_p} = \frac{h}{\sqrt{2mE}} \times \frac{E}{hc} = \frac{1}{c} \frac{E}{\sqrt{2mE}} = \frac{1}{c} \sqrt{\frac{E}{2m}} = \frac{1}{c} \left[\frac{E}{2m}\right]^{1/2}$.

(A) The de-Broglie wavelength of an electron is given by $\lambda = \frac{h}{p}$,where $p$ is the momentum of the electron.
Since the kinetic energy $E = \frac{p^2}{2m}$,we have $p = \sqrt{2mE}$.
Substituting this into the de-Broglie equation: $\lambda = \frac{h}{\sqrt{2mE}}$.
Squaring both sides: $\lambda^2 = \frac{h^2}{2mE}$,which gives $E = \frac{h^2}{2m\lambda^2}$.
The cut-off wavelength $\lambda_0$ of the emitted $X$-rays corresponds to the maximum energy of the photon,which is equal to the kinetic energy $E$ of the incident electron: $E = \frac{hc}{\lambda_0}$.
Equating the two expressions for $E$: $\frac{hc}{\lambda_0} = \frac{h^2}{2m\lambda^2}$.
Solving for $\lambda_0$: $\lambda_0 = \frac{hc \cdot 2m\lambda^2}{h^2} = \frac{2mc\lambda^2}{h}$.

(C) The de-Broglie wavelength $\lambda$ of an electron accelerated through a potential difference $V$ is given by the formula: $\lambda = \frac{h}{\sqrt{2mqV}}$.
From this relation,we can see that $\lambda \propto \frac{1}{\sqrt{V}}$.
Let the initial potential difference be $V_1 = V$ and the final potential difference be $V_2 = 2V$.
The ratio of the wavelengths is $\frac{\lambda_2}{\lambda_1} = \sqrt{\frac{V_1}{V_2}} = \sqrt{\frac{V}{2V}} = \frac{1}{\sqrt{2}}$.
Therefore,the de-Broglie wavelength decreases by a factor of $\frac{1}{\sqrt{2}}$.

(B) The de-Broglie wavelength of a charged particle accelerated through a potential difference '$V$' is given by:
$\lambda = \frac{h}{\sqrt{2mqV}}$
For an electron of mass '$m$' and charge '$q$':
$\lambda = \frac{h}{\sqrt{2mqV}}$ ---$(1)$
For a proton of mass '$M$' and charge '$q$' (since the magnitude of charge on an electron and a proton is the same):
$\lambda_p = \frac{h}{\sqrt{2Mq(9V)}}$ ---$(2)$
Dividing equation $(1)$ by equation $(2)$:
$\frac{\lambda}{\lambda_p} = \frac{\frac{h}{\sqrt{2mqV}}}{\frac{h}{\sqrt{2Mq(9V)}}} = \sqrt{\frac{2Mq(9V)}{2mqV}} = \sqrt{\frac{9M}{m}} = 3\sqrt{\frac{M}{m}}$
Therefore,the wavelength associated with the proton is:
$\lambda_p = \frac{\lambda}{3} \sqrt{\frac{m}{M}}$

(D) The relation between kinetic energy $K$ and momentum $P$ is given by $P = \sqrt{2mK}$.
If the kinetic energy is tripled,the new kinetic energy $K' = 3K$.
The new momentum $P'$ is given by $P' = \sqrt{2m(3K)} = \sqrt{3} \sqrt{2mK} = \sqrt{3}P$.
The de-Broglie wavelength $\lambda$ is given by $\lambda = h/P$.
Therefore,the new wavelength $\lambda'$ is $\lambda' = h/P' = h/(\sqrt{3}P) = \lambda / \sqrt{3}$.
Thus,the wavelength changes by a factor of $1/\sqrt{3}$.

(B) The de-Broglie wavelength $\lambda$ is given by the formula $\lambda = \frac{h}{p}$,where $h$ is Planck's constant and $p$ is the momentum of the electron.
Since kinetic energy $K = \frac{p^2}{2m}$,we can express momentum as $p = \sqrt{2mK}$.
Substituting this into the wavelength formula,we get $\lambda = \frac{h}{\sqrt{2mK}}$.
This shows that $\lambda \propto \frac{1}{\sqrt{K}}$.
If the kinetic energy $K$ is doubled to $K' = 2K$,the new wavelength $\lambda'$ becomes $\lambda' = \frac{h}{\sqrt{2m(2K)}} = \frac{1}{\sqrt{2}} \cdot \frac{h}{\sqrt{2mK}} = \frac{\lambda}{\sqrt{2}}$.
Therefore,the wavelength changes by a factor of $\frac{1}{\sqrt{2}}$.

(A) The de-Broglie wavelength $\lambda$ is given by $\lambda = \frac{h}{p}$.
Since the kinetic energy $K = \frac{p^2}{2m} = qV$,the momentum is $p = \sqrt{2mqV}$.
Substituting this into the wavelength formula,we get $\lambda = \frac{h}{\sqrt{2mqV}} = \left( \frac{h}{\sqrt{2mq}} \right) \left( \frac{1}{\sqrt{V}} \right)$.
Comparing this with the equation of a straight line $y = mx$,where $y = \lambda$ and $x = \frac{1}{\sqrt{V}}$,the slope is given by $\text{slope} = \frac{h}{\sqrt{2mq}}$.
Since $h$ and $q$ are constants,the slope is inversely proportional to the square root of the mass,i.e.,$\text{slope} \propto \frac{1}{\sqrt{m}}$.
Therefore,a larger slope corresponds to a smaller mass.
From the graph,the line corresponding to $m_4$ has the largest slope.
Thus,$m_4$ represents the particle with the smallest mass.

(B) Concept: Photoelectric effect and the motion of a charge in a direction perpendicular to a magnetic field.
According to Einstein's photoelectric equation,the maximum kinetic energy $K$ of a photoelectron is given by:
$K = E - W_0$
Since $K = \frac{1}{2} m v^2$,we have:
$v = \sqrt{\frac{2(E - W_0)}{m}}$
When a charge $e$ moves with velocity $v$ perpendicular to a magnetic field $B$,it experiences a magnetic Lorentz force that acts as a centripetal force,causing it to move in a circular path of radius $r$:
$e v B = \frac{m v^2}{r}$
Solving for $r$:
$r = \frac{m v}{e B}$
Substituting the expression for $v$:
$r = \frac{m}{e B} \sqrt{\frac{2(E - W_0)}{m}} = \frac{\sqrt{m^2 \cdot \frac{2(E - W_0)}{m}}}{e B} = \frac{\sqrt{2m(E - W_0)}}{e B}$

(B) According to Einstein's photoelectric equation,the maximum kinetic energy $K$ is given by $K = \frac{hc}{\lambda} - \phi$,where $\phi$ is the work function.
For the first case with wavelength $\lambda$: $K_1 = \frac{hc}{\lambda} - \phi$.
For the second case with wavelength $\frac{\lambda}{2}$: $K_2 = \frac{hc}{\lambda/2} - \phi = \frac{2hc}{\lambda} - \phi$.
Given that $K_1 = \frac{1}{3} K_2$,we have $3K_1 = K_2$.
Substituting the expressions: $3(\frac{hc}{\lambda} - \phi) = \frac{2hc}{\lambda} - \phi$.
$\frac{3hc}{\lambda} - 3\phi = \frac{2hc}{\lambda} - \phi$.
$\frac{3hc}{\lambda} - \frac{2hc}{\lambda} = 3\phi - \phi$.
$\frac{hc}{\lambda} = 2\phi$.
Therefore,$\phi = \frac{hc}{2\lambda}$.

(B) According to Einstein's photoelectric equation,the maximum kinetic energy of emitted photoelectrons is given by $K_{max} = \frac{1}{2} m v^2 = E_p - \phi$,where $E_p$ is the energy of the incident photon and $\phi$ is the work function of the metal.
Given $\phi = 0.8 eV$.
For the first photon with energy $E_1 = 1.3 eV$:
$\frac{1}{2} m v_1^2 = 1.3 eV - 0.8 eV = 0.5 eV$ --- $(1)$
For the second photon with energy $E_2 = 2.8 eV$:
$\frac{1}{2} m v_2^2 = 2.8 eV - 0.8 eV = 2.0 eV$ --- $(2)$
Dividing equation $(1)$ by equation $(2)$:
$\frac{\frac{1}{2} m v_1^2}{\frac{1}{2} m v_2^2} = \frac{0.5 eV}{2.0 eV}$
$\frac{v_1^2}{v_2^2} = \frac{0.5}{2.0} = \frac{1}{4}$
Taking the square root on both sides:
$\frac{v_1}{v_2} = \sqrt{\frac{1}{4}} = \frac{1}{2}$
Therefore,the ratio of maximum speeds is $1: 2$.

(C) Concept: The stopping potential $V$ is related to the maximum kinetic energy $K_{\max}$ of the photoelectrons via Einstein's photoelectric effect equation:
$eV = K_{\max} = hv - hv_0$
Therefore,
$V = \left(\frac{h}{e}\right)v - \left(\frac{h}{e}\right)v_0$
This equation is of the form $y = mx + c$,which represents a straight line.
Here,the slope $m = \frac{h}{e}$ is positive,and the intercept on the $V$-axis is $-\left(\frac{h}{e}\right)v_0$.
At $v = v_0$,the stopping potential $V = 0$.
Thus,the graph is a straight line starting from $v = v_0$ on the frequency axis and increasing linearly with frequency.
Graph $(A)$ correctly represents this relationship.

(C) According to Einstein's photoelectric equation,the maximum kinetic energy $(K.E._{max})$ of emitted electrons is given by:
$K.E._{max} = E - \phi$
Where $E$ is the energy of the incident photon and $\phi$ is the work function of the metal.
Given that the cutoff frequency (threshold frequency) is $v$,the work function is $\phi = hv$.
The energy of the incident radiation with frequency $2v$ is $E = h(2v) = 2hv$.
Substituting these values into the equation:
$\frac{1}{2}mv_{max}^2 = 2hv - hv$
$\frac{1}{2}mv_{max}^2 = hv$
$v_{max}^2 = \frac{2hv}{m}$
$v_{max} = \sqrt{\frac{2hv}{m}}$

(B) Einstein's photoelectric equation is given by: $e V = \frac{h c}{\lambda} - \phi$,which can be written as $\frac{h c}{\lambda} = \phi + e V$.
For incident wavelength $\lambda$,the stopping potential is $V_1$:
$\frac{h c}{\lambda} = \phi + e V_1$ --- $(1)$
For incident wavelength $\frac{\lambda}{2}$,the stopping potential is $V_2$:
$\frac{h c}{\lambda / 2} = \phi + e V_2 \Rightarrow \frac{2 h c}{\lambda} = \phi + e V_2$ --- $(2)$
Substitute the value of $\frac{h c}{\lambda}$ from equation $(1)$ into equation $(2)$:
$2(\phi + e V_1) = \phi + e V_2$
$2 \phi + 2 e V_1 = \phi + e V_2$
$e V_2 = 2 e V_1 + \phi$
Dividing by $e$,we get:
$V_2 = 2 V_1 + \frac{\phi}{e}$

(C) According to Einstein's photoelectric equation,the maximum kinetic energy $K_{max}$ of the emitted photoelectrons is given by $K_{max} = h v - h v_0$,where $v_0$ is the threshold frequency.
Since the stopping potential $V_s$ is related to the maximum kinetic energy by $K_{max} = e V_s$,we have $e V_s = h v - h v_0$ ... $(1)$.
For the second case,with frequency $v/2$ and stopping potential $V_s/4$,we have $e(V_s/4) = h(v/2) - h v_0$ ... $(2)$.
From equation $(1)$,$e V_s = h v - h v_0$. Substituting this into equation $(2)$ multiplied by $4$:
$e V_s = 4(h v/2 - h v_0) = 2h v - 4h v_0$.
Equating the two expressions for $e V_s$:
$h v - h v_0 = 2h v - 4h v_0$.
Rearranging the terms: $3h v_0 = h v$.
Therefore,$v_0 = v/3$.

(D) The energy of a photon is given by the formula $E = h\nu$.
Given frequency $\nu = 10^{12} \text{ MHz} = 10^{12} \times 10^6 \text{ Hz} = 10^{18} \text{ Hz}$.
Planck's constant $h = 6.63 \times 10^{-34} \text{ Js}$.
To convert the energy from Joules to electron-volts (eV),we divide by the elementary charge $e = 1.6 \times 10^{-19} \text{ C}$.
$E(\text{eV}) = \frac{h\nu}{e} = \frac{6.63 \times 10^{-34} \times 10^{18}}{1.6 \times 10^{-19}}$.
$E(\text{eV}) = \frac{6.63}{1.6} \times 10^{-34 + 18 + 19} = 4.14375 \times 10^3 \text{ eV}$.
Rounding to significant figures,$E \approx 4.14 \times 10^3 \text{ eV}$.

(D) The energy of the incident photon is given by $E = \frac{hc}{\lambda}$.
Given $h = 6.63 \times 10^{-34} J s$, $c = 3 \times 10^8 m/s$, and $\lambda = 4100 \times 10^{-10} m$.
$E = \frac{6.63 \times 10^{-34} \times 3 \times 10^8}{4100 \times 10^{-10}} J = 4.85 \times 10^{-19} J$.
To convert this energy into electron-volts $(eV)$, divide by $e = 1.6 \times 10^{-19} C$:
$E = \frac{4.85 \times 10^{-19}}{1.6 \times 10^{-19}} eV \approx 3.03 eV$.
Photoelectric emission occurs if the incident photon energy is greater than the work function $(\Phi)$ of the metal.
For metal $A$: $\Phi_A = 1.92 eV < 3.03 eV$ (Emission occurs).
For metal $B$: $\Phi_B = 2.0 eV < 3.03 eV$ (Emission occurs).
For metal $C$: $\Phi_C = 5 eV > 3.03 eV$ (No emission).
Therefore, metals $A$ and $B$ will emit photoelectrons.

(A) According to Einstein's photoelectric equation:
$eV = \frac{hc}{\lambda} - \Phi$ --- $(1)$
where $V$ is the stopping potential,$\lambda$ is the incident wavelength,and $\Phi$ is the work function of the metal.
When the incident wavelength is changed to $2\lambda$,let the new stopping potential be $V'$. The equation becomes:
$eV' = \frac{hc}{2\lambda} - \Phi$ --- $(2)$
Subtracting equation $(2)$ from equation $(1)$:
$eV - eV' = \left( \frac{hc}{\lambda} - \Phi \right) - \left( \frac{hc}{2\lambda} - \Phi \right)$
$e(V - V') = \frac{hc}{\lambda} - \frac{hc}{2\lambda} = \frac{hc}{2\lambda}$
$V - V' = \frac{hc}{2e\lambda}$
$V' = V - \frac{hc}{2e\lambda}$

(A) According to Einstein's photoelectric equation, the maximum kinetic energy $K_{max}$ of the emitted photoelectron is given by:
$K_{max} = h\nu - \phi$
where $\phi$ is the work function of the metal.
Given that the threshold frequency is $\nu$, the work function is $\phi = h\nu$.
When radiation of frequency $2\nu$ is incident on the metal, the maximum kinetic energy is:
$K_{max} = h(2\nu) - h\nu = h\nu$
Since $K_{max} = \frac{1}{2}mv_{max}^2$, we have:
$\frac{1}{2}mv_{max}^2 = h\nu$
$v_{max}^2 = \frac{2h\nu}{m}$
$v_{max} = \sqrt{\frac{2h\nu}{m}}$
Therefore, option $A$ is correct.

(C) The threshold frequency of the metal is given as $f_0 = 15 \times 10^{14} \, Hz$.
The frequency of the incident light with wavelength $\lambda = 6000 \, \text{Å} = 6000 \times 10^{-10} \, m$ is calculated as:
$f = \frac{c}{\lambda} = \frac{3 \times 10^8 \, m/s}{6000 \times 10^{-10} \, m} = \frac{3 \times 10^8}{6 \times 10^{-7}} \, Hz = 0.5 \times 10^{15} \, Hz = 5 \times 10^{14} \, Hz$.
Since the frequency of the incident light $(f = 5 \times 10^{14} \, Hz)$ is less than the threshold frequency $(f_0 = 15 \times 10^{14} \, Hz)$, the energy of the incident photons is insufficient to overcome the work function of the metal.
Therefore, no photoelectric emission takes place.

(B) The kinetic energy of the fastest emitted photoelectron is given by $K_{max} = \frac{1}{2} m v^2$.
The stopping potential $V$ is the potential required to stop these fastest electrons,such that the work done by the electric field equals the maximum kinetic energy:
$e V = K_{max}$
$e V = \frac{1}{2} m v^2$
Solving for $V$,we get:
$V = \frac{m v^2}{2 e}$

(C) According to Einstein's photoelectric equation, the maximum kinetic energy $(K_{max})$ of an emitted photoelectron is given by $K_{max} = h\nu - \Phi$, where $h$ is Planck's constant, $\nu$ is the frequency of the incident light, and $\Phi$ is the work function of the metal surface.
Since $h$ and $\Phi$ are constants for a given metal, the kinetic energy $K_{max}$ depends directly on the frequency $\nu$ of the incident light.
Intensity of light affects the number of photoelectrons emitted per second, but not their individual kinetic energy.

(C) From Faraday's law of electromagnetic induction,the induced electromotive force (emf) $e$ in the circuit is given by:
$e = -\frac{\Delta \phi}{\Delta t}$
Given the resistance of the coil is $R$,the induced current $I$ is:
$I = \frac{|e|}{R} = \frac{\Delta \phi}{R \cdot \Delta t}$
The total charge $Q$ that flows through the circuit in time $\Delta t$ is given by:
$Q = I \cdot \Delta t$
Substituting the expression for $I$:
$Q = \left( \frac{\Delta \phi}{R \cdot \Delta t} \right) \cdot \Delta t$
$Q = \frac{\Delta \phi}{R}$
Thus,the total quantity of induced electric charge is $\frac{\Delta \phi}{R}$.

(C) The energy stored in an inductor (coil) is given by the formula $E = \frac{1}{2} L I^2$,where $L$ is the self-inductance and $I$ is the current.
Let the initial current be $I_1 = I$ and the initial energy be $E_1 = \frac{1}{2} L I^2$.
If the current is reduced by $50 \%$,the new current becomes $I_2 = I - 0.5 I = 0.5 I = \frac{I}{2}$.
The new energy stored in the coil is $E_2 = \frac{1}{2} L (I_2)^2 = \frac{1}{2} L (\frac{I}{2})^2 = \frac{1}{2} L (\frac{I^2}{4}) = \frac{1}{4} E_1$.
The change in energy is $\Delta E = E_1 - E_2 = E_1 - \frac{1}{4} E_1 = \frac{3}{4} E_1$.
Since $\frac{3}{4} = 0.75$,the energy decreases by $75 \%$.

(B) The energy stored in an inductor (coil) is given by the formula $E = \frac{1}{2} L I^2$,where $L$ is the self-inductance of the coil.
From this formula,we can see that $E \propto I^2$.
Initially,when the current is $I$,the energy is $E_1 = \frac{1}{2} L I^2$.
When the current is reduced to $I' = I/2$,the new energy $E_2$ is given by $E_2 = \frac{1}{2} L (I/2)^2 = \frac{1}{2} L (I^2/4) = \frac{1}{4} E_1$.
The change in energy is $\Delta E = E_1 - E_2$.
Substituting the value of $E_2$,we get $\Delta E = E_1 - \frac{E_1}{4} = \frac{3E_1}{4}$.

(A) Given: Current $i = 5 \, A$, Resistance $R = 1 \, \Omega$, Inductance $L = 5 \, mH = 5 \times 10^{-3} \, H$, Electromotive force $E = 15 \, V$.
The current is decreasing, so $\frac{di}{dt} = -10^3 \, A/s$.
Applying Kirchhoff's Voltage Law from point $A$ to $B$:
$V_A - iR + E - L\left(\frac{di}{dt}\right) = V_B$
$V_A - V_B = iR + L\left(\frac{di}{dt}\right) - E$
Substituting the values:
$V_A - V_B = (5 \, A \times 1 \, \Omega) + (5 \times 10^{-3} \, H \times -10^3 \, A/s) - 15 \, V$
$V_A - V_B = 5 - 5 - 15 = -15 \, V$
Therefore, $V_B - V_A = 15 \, V$.

(B) When $S_1$ is closed,the capacitor $C$ charges to a potential difference $V$.
When $S_1$ is opened and $S_2$ is closed,the capacitor $C$ and inductor $L$ form an $LC$ oscillating circuit.
The energy initially stored in the capacitor is $U_E = \frac{1}{2} C V^2$.
As the capacitor discharges,the energy is transferred to the inductor as magnetic energy $U_B = \frac{1}{2} L I^2$.
By the law of conservation of energy,the maximum current $I_{max}$ occurs when all electrostatic energy is converted to magnetic energy:
$\frac{1}{2} C V^2 = \frac{1}{2} L I_{max}^2$
$I_{max}^2 = \frac{C V^2}{L}$
$I_{max} = V \sqrt{\frac{C}{L}}$.
Thus,the instantaneous current in the circuit can reach this maximum value.

(B) The induced e.m.f. in coil $S$ due to a change in current in coil $P$ is given by Faraday's law of induction: $E = -M \frac{dI}{dt}$.
Given,mutual inductance $M = 3 \times 10^{-3} \ H$ and current $I = 20 \sin(50 \pi t) \ A$.
Differentiating the current with respect to time $t$: $\frac{dI}{dt} = 20 \times 50 \pi \cos(50 \pi t) = 1000 \pi \cos(50 \pi t) \ A/s$.
Substituting these values into the e.m.f. equation: $E = -(3 \times 10^{-3}) \times (1000 \pi \cos(50 \pi t)) = -3 \pi \cos(50 \pi t) \ V$.
The maximum value of the induced e.m.f. is $|E_{max}| = 3 \pi \ V$.
Using $\pi \approx 3.14$,we get $|E_{max}| = 3 \times 3.14 = 9.42 \ V$.

(A) The total charge $q$ flowing through a circuit due to a change in magnetic flux $\Delta \phi$ is given by the formula $q = \frac{\Delta \phi}{R_{total}}$.
Here,$R_{total} = R_{galvanometer} + R_{coil} = 200 \ \Omega + 400 \ \Omega = 600 \ \Omega$.
The change in flux is $\Delta \phi = N A \Delta B$,where $N = 1000$ and $A = \pi r^2$.
The diameter is $20 \ mm$,so the radius $r = 10 \ mm = 0.01 \ m$.
Thus,$A = \pi \times (0.01)^2 = \pi \times 10^{-4} \ m^2$.
The change in magnetic field $\Delta B = |0 - 0.012| = 0.012 \ T$.
Substituting these values into the charge formula:
$q = \frac{1000 \times \pi \times 10^{-4} \times 0.012}{600}$
$q = \frac{1000 \times 3.14159 \times 10^{-4} \times 0.012}{600}$
$q = \frac{3.14159 \times 0.012}{600} \times 1000 = \frac{0.037699}{600} \times 1000 \approx 0.0628 \times 10^{-3} \ C = 62.8 \ \mu C$.
Rounding to the nearest value,we get $q \approx 63 \ \mu C$.

(D) The magnetic field produced inside a long solenoid is given by $B = \mu_0 n I$.
Here, $n$ is the number of turns per unit length and $I$ is the current.
The magnetic flux $\phi$ linked with the loop of area $A$ placed normal to the axis is $\phi = B A = \mu_0 n I A$.
The induced emf $e$ is given by Faraday's law: $e = -\frac{d\phi}{dt} = -\mu_0 n A \frac{dI}{dt}$.
Given values:
$n = 15 \text{ turns/cm} = 1500 \text{ turns/m}$.
$A = 2.0 \text{ cm}^2 = 2.0 \times 10^{-4} \text{ m}^2$.
$\frac{dI}{dt} = \frac{4.0 - 2.0}{0.1} = 20 \text{ A/s}$.
$\mu_0 = 4\pi \times 10^{-7} \text{ T m/A}$.
Substituting these values:
$|e| = (4 \times 3.14 \times 10^{-7}) \times 1500 \times (2.0 \times 10^{-4}) \times 20$.
$|e| = 12.56 \times 10^{-7} \times 1500 \times 2.0 \times 10^{-4} \times 20$.
$|e| = 7.536 \times 10^{-6} \text{ V} \approx 7.54 \times 10^{-6} \text{ V}$.

(C) The magnetic field lines produced by a circular coil are perpendicular to the plane of the coil at its center.
Since the two coils $A$ and $B$ are placed mutually perpendicular,the magnetic field lines produced by coil $A$ will lie in the plane of coil $B$,and vice-versa.
The magnetic flux $\Phi$ is given by $\Phi = \vec{B} \cdot \vec{A} = BA \cos \theta$,where $\theta$ is the angle between the magnetic field vector $\vec{B}$ and the area vector $\vec{A}$ (which is perpendicular to the plane of the coil).
Since the magnetic field lines are parallel to the plane of the other coil,they are perpendicular to the area vector of that coil. Thus,$\theta = 90^\circ$ and $\cos 90^\circ = 0$.
Therefore,the magnetic flux linked with the other coil is zero.
According to Faraday's law of electromagnetic induction,the induced $EMF$ is $\varepsilon = -d\Phi/dt$. Since $\Phi = 0$ at all times,the induced $EMF$ and consequently the induced current will be zero.

(B) The magnetic flux $\phi$ through the coil is given by $\phi = B \cdot A \cdot \cos(\theta)$. Since the plane of the coil is normal to the magnetic field,$\theta = 0^\circ$ and $\cos(0^\circ) = 1$,so $\phi = BA$.
The initial magnetic flux is $\phi_i = B \cdot A$.
The magnetic induction is reduced to $25 \%$ of its initial value,so the final magnetic induction is $B_f = 0.25 B = \frac{B}{4}$.
The final magnetic flux is $\phi_f = \frac{B}{4} \cdot A = \frac{BA}{4}$.
The change in magnetic flux is $\Delta \phi = \phi_i - \phi_f = BA - \frac{BA}{4} = \frac{3BA}{4}$.
According to Faraday's law of induction,the magnitude of the induced e.m.f. is $e = \left| \frac{\Delta \phi}{\Delta t} \right|$.
Given $\Delta t = 1 \ s$,we have $e = \frac{3BA/4}{1} = \frac{3AB}{4} \ V$.

(B) The induced e.m.f. $e$ in a rotating rod is given by the rate of change of magnetic flux $\phi$ linked with the area swept by the rod.
$e = \frac{d\phi}{dt} = B \frac{dA}{dt}$
In one complete revolution,the rod sweeps an area $A = \pi l^2$.
If the rod makes $f$ revolutions per second,the area swept per unit time is $\frac{dA}{dt} = f \cdot A = f \cdot \pi l^2$.
Substituting this into the e.m.f. equation:
$e = B \cdot (f \cdot \pi l^2)$
Rearranging to solve for the frequency $f$ (number of revolutions per second):
$f = \frac{e}{B \pi l^2}$

(D) The magnetic flux linked with the coil is given by $\Phi = 4t^2 + 3t + 7$.
According to Faraday's law of electromagnetic induction,the magnitude of the induced e.m.f. $(e)$ is given by $e = |\frac{d\Phi}{dt}|$.
Calculating the derivative of $\Phi$ with respect to time $t$:
$\frac{d\Phi}{dt} = \frac{d}{dt}(4t^2 + 3t + 7) = 8t + 3$.
Therefore,the magnitude of the induced e.m.f. is $e = |8t + 3|$.
At $t = 2 \ s$,substituting the value of $t$ into the expression:
$e = 8(2) + 3 = 16 + 3 = 19 \ V$.
Thus,the magnitude of the induced e.m.f. at $t = 2 \ s$ is $19 \ V$.

(B) According to Faraday's Law of electromagnetic induction,the induced e.m.f. $\varepsilon$ is given by $\varepsilon = -N \frac{d\phi}{dt}$.
Here,$N$ is the number of turns,and $\frac{d\phi}{dt}$ is the rate of change of magnetic flux.
The rate of change of flux $\frac{d\phi}{dt}$ depends on the speed of the magnet,the strength of the magnetic field,and the geometry of the coil.
The induced e.m.f. $\varepsilon$ does not depend on the resistance of the coil $(R)$.
Note: While the induced current $I = \frac{\varepsilon}{R}$ depends on the resistance,the induced e.m.f. itself is independent of the resistance of the circuit.
Therefore,the correct option is $B$.

(D) The correct option is $D$.
We can imagine the disc to be a collection of thin rods connected in parallel between the center of the disc and the rim. So,if we calculate the induced e.m.f. on a thin rod rotating about its axis,it should be equal to that of the disc.
The tiny motional e.m.f. developed across the element $dr$ at a distance $r$ from the center can be written as:
$dE = Bv dr$
Taking the linear velocity $v = \omega r$,we get:
$dE = B \omega r dr$
On integrating across the rod from $r = 0$ to $r = R$:
$E = \int_0^{R} B \omega r dr = B \omega \left[ \frac{r^2}{2} \right]_0^{R} = \frac{B \omega R^2}{2}$

(B) The motional emf induced in the loop is given by: $V = B v l$.
The current $i$ in the loop is given by Ohm's law: $i = \frac{V}{R} = \frac{B v l}{R}$.
As the loop falls,it experiences a magnetic force $F_m = i l B$ acting upwards. For the loop to fall with a constant terminal velocity $v$,the magnetic force must balance the gravitational force $mg$ acting downwards.
Therefore,$i l B = m g$.
Substituting the expression for $i$:
$B \left( \frac{B v l}{R} \right) l = m g$
$\frac{B^2 l^2 v}{R} = m g$
$v = \frac{m g R}{B^2 l^2}$.

(C) The correct option is $C$.
Concept: The magnetic flux $\phi$ associated with a coil is given by $\phi = B A$,and the coefficient of self-inductance $L$ is defined as $L = \frac{N\phi}{I}$,where $I$ is the current flowing through the coil.
The magnetic field $B$ at the center of a circular coil with radius $r$ carrying current $I$ is given by $B = \frac{\mu_0 I}{2r}$.
For a coil with $N$ turns,the total magnetic field is $B_N = N \left( \frac{\mu_0 I}{2r} \right)$.
The magnetic flux linked with each turn is $\phi_1 = B_N A = N \left( \frac{\mu_0 I}{2r} \right) A$.
The total flux linkage for $N$ turns is $\Phi = N \phi_1 = N^2 \left( \frac{\mu_0 I A}{2r} \right)$.
Using the definition $L = \frac{\Phi}{I}$,we get $L = N^2 \left( \frac{\mu_0 A}{2r} \right)$.
Therefore,the self-inductance $L$ is proportional to the square of the number of turns,i.e.,$L \propto N^2$.

(D) The magnetic field '$B$' inside a long solenoid is given by $B = \mu_0 n i$,where '$n$' is the number of turns per unit length and '$i$' is the current.
The cross-sectional area '$A$' of the solenoid with diameter '$d$' is $A = \pi r^2 = \pi (d/2)^2 = \frac{\pi d^2}{4}$.
The magnetic flux '$\phi$' through a single turn is $\phi = B \cdot A = (\mu_0 n i) \left( \frac{\pi d^2}{4} \right)$.
For a length '$l$' of the solenoid,the total number of turns is $N = n \cdot l$.
The total flux '$\Phi$' linked with length '$l$' is $\Phi = N \cdot \phi = (n l) \left( \mu_0 n i \frac{\pi d^2}{4} \right) = \mu_0 n^2 i l \frac{\pi d^2}{4}$.
Since $\Phi = L \cdot i$,the total inductance '$L$' for length '$l$' is $L = \frac{\mu_0 n^2 i l \pi d^2}{4 i} = \frac{\mu_0 n^2 l \pi d^2}{4}$.
Therefore,the inductance per unit length is $\frac{L}{l} = \frac{\mu_0 \pi n^2 d^2}{4}$.

(D) The magnetic flux $\Phi$ linked with coil $B$ due to current $i_1$ in coil $A$ is given by the relation: $\Phi = M i_1$.
Here, $M$ is the coefficient of mutual inductance.
Given that the change in current $\Delta i_1 = 0.8 \,A$ results in a change in magnetic flux $\Delta \Phi = 0.16 \,Wb$.
The mutual inductance $M$ is calculated as:
$M = \frac{\Delta \Phi}{\Delta i_1} = \frac{0.16 \,Wb}{0.8 \,A} = 0.2 \,H$.

(B) The self-inductance $L$ of a solenoid of length $\lambda$ and area of cross-section $A$ with a fixed number of turns $N$ is given by the formula:
$L = \frac{\mu_0 N^2 A}{\lambda}$
From this expression,we can see that $L$ is directly proportional to the area $A$ and inversely proportional to the length $\lambda$.
Therefore,for $L$ to increase,the area $A$ must increase and the length $\lambda$ must decrease.

(C) When two coils with self-inductances $L_1$ and $L_2$ and mutual inductance $M$ are connected in series,the equivalent inductance $L_{eq}$ is given by the formula:
$L_{eq} = L_1 + L_2 + 2M$
Given:
$L_1 = 1 H$
$L_2 = 3 H$
$M = 5 H$
Substituting these values into the formula:
$L_{eq} = 1 H + 3 H + 2(5 H)$
$L_{eq} = 1 H + 3 H + 10 H$
$L_{eq} = 14 H$
Therefore,the equivalent self-inductance of the combination is $14 H$.

(C) The change in current in the primary coil is given by $di_1 = (8 - 4) \,A = 4 \,A$.
The time interval is $dt = 0.6 \,s$.
The induced e.m.f. in the secondary coil is $E_2 = 50 \,mV = 50 \times 10^{-3} \,V$.
The formula for mutual inductance $M$ is given by $E_2 = M \cdot \frac{di_1}{dt}$.
Rearranging for $M$, we get $M = \frac{E_2 \cdot dt}{di_1}$.
Substituting the values: $M = \frac{50 \times 10^{-3} \,V \times 0.6 \,s}{4 \,A}$.
$M = \frac{30 \times 10^{-3}}{4} \,H = 7.5 \times 10^{-3} \,H = 7.5 \,mH$.

(B) The formula for induced emf in a coil due to self-inductance is given by $e = L \left| \frac{dI}{dt} \right|$.
Given:
Induced emf $e = 0.5 \,V$
Initial current $I_1 = 10 \,A$
Final current $I_2 = -10 \,A$ (opposite direction)
Change in current $\Delta I = I_2 - I_1 = -10 \,A - 10 \,A = -20 \,A$
Time interval $\Delta t = 0.8 \,s$
Magnitude of rate of change of current $|\frac{\Delta I}{\Delta t}| = \frac{|-20 \,A|}{0.8 \,s} = \frac{20}{0.8} \,A/s = 25 \,A/s$.
Substituting these values into the formula:
$0.5 = L \times 25$
$L = \frac{0.5}{25} \,H = \frac{1}{50} \,H = 0.02 \,H$.
Converting to millihenry: $0.02 \,H = 0.02 \times 1000 \,mH = 20 \,mH$.

(D) The coefficient of self-induction $L$ is defined as $L = \frac{\phi}{I}$,where $\phi$ is the total magnetic flux linked with the coil and $I$ is the current flowing through it.
For a toroid with $N$ turns,the total flux $\phi$ is given by $\phi = N \cdot A \cdot B$,where $A$ is the cross-sectional area of the toroid and $B$ is the magnetic field inside it.
The cross-sectional area $A = \pi r^2$.
The magnetic field inside a toroid is $B = \mu_0 n I$,where $n = \frac{N}{2 \pi R}$ is the number of turns per unit length.
Substituting these into the flux equation:
$\phi = N (\pi r^2) (\mu_0 \frac{N}{2 \pi R} I) = \frac{\mu_0 N^2 r^2 I}{2 R}$.
Finally,the self-induction $L = \frac{\phi}{I} = \frac{\mu_0 N^2 r^2}{2 R}$.

(C) The magnetic field $B$ at the center of a solenoid is given by $B = \mu_0 (N/l) I$,where $N$ is the number of turns,$l$ is the length,and $I$ is the current.
Total magnetic flux $\phi$ linked with the solenoid is $\phi = B A N = \mu_0 (N^2/l) A I$,where $A$ is the cross-sectional area.
By the definition of self-induction,$\phi = L I$.
Comparing the two expressions,the self-inductance $L$ is given by $L = \frac{\mu_0 N^2 A}{l}$.
From this formula,it is clear that $L$ depends on the number of turns $N$,cross-sectional area $A$,and length $l$,but it is independent of the current $I$ flowing through the solenoid.

(B) The magnetic field $B$ at the center of the solenoid is given by $B = \mu_0 (N/l) i$.
The magnetic flux $\Phi$ linked with the solenoid is $\Phi = N B A = N (\mu_0 N i A / l) = (\mu_0 N^2 A / l) i$.
The self-inductance $L$ is defined as $L = \Phi / i = \mu_0 N^2 A / l$.
Substituting the given values: $N = 500$,$l = 0.3 \ m$,$A = 25 \times 10^{-4} \ m^2$,$\mu_0 = 4\pi \times 10^{-7} \ T \cdot m/A$.
$L = \frac{4\pi \times 10^{-7} \times (500)^2 \times 25 \times 10^{-4}}{0.3} \approx 2.618 \times 10^{-3} \ H$.
The average back e.m.f. is $e = L (\Delta i / \Delta t)$.
Given $\Delta i = 2.5 \ A$ and $\Delta t = 10^{-3} \ s$.
$e = (2.618 \times 10^{-3}) \times (2.5 / 10^{-3}) = 2.618 \times 2.5 \approx 6.545 \ V$.
Rounding to the nearest value,$e \approx 6.5 \ V$.