MHT CET 2024 Physics Question Paper with Answer and Solution

(A) For a particle in simple harmonic motion $(SHM)$ starting from the equilibrium position,the displacement is given by $x = A \sin(\omega t)$.
Given $t = \frac{T}{12}$ and $\omega = \frac{2\pi}{T}$,the displacement is $x = A \sin(\frac{2\pi}{T} \cdot \frac{T}{12}) = A \sin(\frac{\pi}{6}) = \frac{A}{2}$.
The potential energy ($P$.$E$.) is given by $U = \frac{1}{2} k x^2 = \frac{1}{2} m \omega^2 x^2$.
Substituting $x = \frac{A}{2}$,we get $U = \frac{1}{2} m \omega^2 (\frac{A^2}{4}) = \frac{1}{8} m \omega^2 A^2$.
The kinetic energy ($K$.$E$.) is given by $K = \frac{1}{2} m \omega^2 (A^2 - x^2)$.
Substituting $x = \frac{A}{2}$,we get $K = \frac{1}{2} m \omega^2 (A^2 - \frac{A^2}{4}) = \frac{1}{2} m \omega^2 (\frac{3A^2}{4}) = \frac{3}{8} m \omega^2 A^2$.
The ratio of potential energy to kinetic energy is $\frac{U}{K} = \frac{\frac{1}{8} m \omega^2 A^2}{\frac{3}{8} m \omega^2 A^2} = \frac{1}{3}$.

(D) For a simple harmonic oscillator,the potential energy $PE$ at displacement $x$ is given by $PE = \frac{1}{2} k x^2$,where $k$ is the force constant.
At the extreme position,the displacement $x = A$,so $PE = \frac{1}{2} k A^2$.
For a simple pendulum,the restoring force is $F = -\frac{Mg}{L} x$,so the force constant $k = \frac{Mg}{L}$.
Substituting the value of $k$ into the potential energy formula:
$PE = \frac{1}{2} \left( \frac{Mg}{L} \right) A^2$.
$PE = \frac{MgA^2}{2L}$.

(B) The kinetic energy $(KE)$ of a particle in simple harmonic motion is given by $KE = \frac{1}{2} k A^2 \cos^2(\omega t)$,where $A$ is the amplitude and $\omega$ is the angular frequency.
Given that the body starts from the mean position,the displacement is $x = A \sin(\omega t)$,and the velocity is $v = A \omega \cos(\omega t)$.
The total energy $(TE)$ is $\frac{1}{2} k A^2$.
We are given $KE = 75 \% \text{ of } TE = \frac{3}{4} TE$.
So,$\frac{1}{2} k (A \cos(\omega t))^2 = \frac{3}{4} (\frac{1}{2} k A^2)$.
$\cos^2(\omega t) = \frac{3}{4}$.
$\cos(\omega t) = \frac{\sqrt{3}}{2}$.
Since $\cos(30^{\circ}) = \frac{\sqrt{3}}{2}$,we have $\omega t = 30^{\circ} = \frac{\pi}{6} \text{ radians}$.
Substituting $\omega = \frac{2\pi}{T}$,we get $\frac{2\pi}{T} t = \frac{\pi}{6}$.
Solving for $t$,we get $t = \frac{T}{12}$.

(D) The potential energy $(P.E.)$ of a simple harmonic oscillator at its extreme position is given by the formula:
$P.E. = \frac{1}{2} m \omega^2 A^2$
For a simple pendulum,the angular frequency $\omega$ is given by $\omega = \sqrt{\frac{g}{L}}$.
Substituting this value into the potential energy equation:
$P.E. = \frac{1}{2} m \left(\sqrt{\frac{g}{L}}\right)^2 A^2$
$P.E. = \frac{1}{2} m \left(\frac{g}{L}\right) A^2$
$P.E. = \frac{mgA^2}{2L}$

(A) Given: Kinetic Energy $(K.E.)$ at mean position $= 16 \ J$,Amplitude $(A)$ $= 25 \ cm = 0.25 \ m$,Mass $(m)$ $= 5.12 \ kg$.
The kinetic energy at the mean position is equal to the total energy $(E)$ of the particle in simple harmonic motion.
The formula for total energy is $E = \frac{1}{2} m \omega^2 A^2$.
Substituting the values: $16 = \frac{1}{2} \times 5.12 \times \omega^2 \times (0.25)^2$.
Solving for $\omega^2$: $\omega^2 = \frac{16 \times 2}{5.12 \times 0.0625} = \frac{32}{0.32} = 100$.
Therefore,$\omega = \sqrt{100} = 10 \ rad/s$.
The period of oscillation $(T)$ is given by $T = \frac{2 \pi}{\omega}$.
Substituting $\omega$: $T = \frac{2 \pi}{10} = \frac{\pi}{5} \ s$.

(D) For a particle performing uniform circular motion, the centripetal acceleration is given by $a = \omega^2 r$.
Given, diameter $d = 1 \ m$, so the radius $r = d/2 = 0.5 \ m$.
The frequency $f = 4 \ Hz$.
The angular velocity $\omega = 2 \pi f = 2 \pi (4) = 8 \pi \ rad/s$.
Substituting these values into the acceleration formula:
$a = (8 \pi)^2 \times 0.5$
$a = 64 \pi^2 \times 0.5$
$a = 32 \pi^2 \ m/s^2$.

(A) Given the displacement equations:
$y_1 = a_1 \sin \left(\frac{2 \pi x}{\lambda} - \omega t\right) \quad \dots(i)$
$y_2 = a_2 \cos \left(\frac{2 \pi x}{\lambda} - \omega t + \phi\right) \quad \dots(ii)$
To compare the phases,we convert the cosine function into a sine function using the identity $\cos(\theta) = \sin(\theta + \frac{\pi}{2})$:
$y_2 = a_2 \sin \left(\frac{2 \pi x}{\lambda} - \omega t + \phi + \frac{\pi}{2}\right) \quad \dots(iii)$
Comparing the phase of $y_1$ (which is $\frac{2 \pi x}{\lambda} - \omega t$) with the phase of $y_2$ (which is $\frac{2 \pi x}{\lambda} - \omega t + \phi + \frac{\pi}{2}$),the phase difference is:
$\Delta \phi = \left(\frac{2 \pi x}{\lambda} - \omega t + \phi + \frac{\pi}{2}\right) - \left(\frac{2 \pi x}{\lambda} - \omega t\right)$
$\Delta \phi = \phi + \frac{\pi}{2}$

(C) For a damped harmonic oscillator,the equation of motion is $m \frac{d^2x}{dt^2} + b \frac{dx}{dt} + kx = 0$.
Defining the natural angular frequency $\omega_0 = \sqrt{\frac{k}{m}}$ and the damping constant $r = \frac{b}{2m}$,the angular frequency of the damped oscillations $\omega^{\prime}$ is given by:
$\omega^{\prime} = \sqrt{\omega_0^2 - r^2}$
Substituting the values:
$\omega^{\prime} = \sqrt{\frac{k}{m} - \left(\frac{b}{2m}\right)^2}$

(C) The displacement equation is $y = A \cos(\pi t + \pi \phi)$.
Taking the derivative with respect to time,the velocity is $v = \frac{dy}{dt} = -A\pi \sin(\pi t + \pi \phi)$.
At $t = 0$,we have $y_0 = A \cos(\pi \phi)$ and $v_0 = -A\pi \sin(\pi \phi)$.
From these,we get $\cos(\pi \phi) = \frac{y_0}{A}$ and $\sin(\pi \phi) = -\frac{v_0}{A\pi}$.
Using the identity $\cos^2(\pi \phi) + \sin^2(\pi \phi) = 1$,we get $\left(\frac{y_0}{A}\right)^2 + \left(-\frac{v_0}{A\pi}\right)^2 = 1$.
This simplifies to $y_0^2 + \frac{v_0^2}{\pi^2} = A^2$.
Substituting the given values $y_0 = 2 \text{ cm}$ and $v_0 = 2\pi \text{ cm/s}$:
$2^2 + \frac{(2\pi)^2}{\pi^2} = A^2$.
$4 + 4 = A^2$.
$A^2 = 8$.
$A = \sqrt{8} = 2\sqrt{2} \text{ cm}$.

(D) In $S.H.M.$,one complete oscillation corresponds to a phase change of $2 \pi \ rad$.
Therefore,for $n$ oscillations,the total phase change is given by $\Delta \phi = n \times 2 \pi \ rad$.
Given that the particle completes $n = 2$ oscillations.
Thus,the phase change $= 2 \times 2 \pi \ rad = 4 \pi \ rad$.

(B) The frequency of oscillation is $n = 10 \ Hz$. The angular frequency is $\omega = 2 \pi n = 20 \pi \ rad/s$.
For a vertical spring-mass system,the equilibrium position is at a distance $x_0 = \frac{mg}{k}$ below the unstretched position.
Since the spring is unstretched at the highest point,the amplitude $A$ of the oscillation is equal to the displacement of the equilibrium position from the unstretched position,so $A = x_0 = \frac{mg}{k}$.
We know that $\omega^2 = \frac{k}{m}$,so $k = m \omega^2$.
Substituting $k$ into the expression for $A$: $A = \frac{mg}{m \omega^2} = \frac{g}{\omega^2}$.
The maximum speed is $v_{\max} = \omega A = \omega \left( \frac{g}{\omega^2} \right) = \frac{g}{\omega}$.
Substituting the values: $v_{\max} = \frac{10}{20 \pi} = \frac{1}{2 \pi} \ m/s$.

(C) For $S.H.M.$,the maximum velocity is given by $\alpha = A \omega$,where $A$ is the amplitude and $\omega$ is the angular frequency.
From this,we get $\omega = \frac{\alpha}{A}$ ... $(i)$
The maximum acceleration is given by $\beta = A \omega^2$.
Substituting the value of $\omega$ from equation $(i)$ into the acceleration formula:
$\beta = A \left( \frac{\alpha}{A} \right)^2 = A \left( \frac{\alpha^2}{A^2} \right) = \frac{\alpha^2}{A}$
Rearranging for amplitude $A$,we get $A = \frac{\alpha^2}{\beta}$.
The path length of a particle in $S.H.M.$ is equal to the total distance between the two extreme positions,which is $2A$.
Therefore,Path length $= 2A = \frac{2 \alpha^2}{\beta}$.

(C) At the mean position, the velocity is maximum, given by $v_{\max} = A\omega$.
Given $A = 4 \,cm$ and $v_{\max} = 12 \,cm/s$.
$\omega = \frac{v_{\max}}{A} = \frac{12}{4} = 3 \,rad/s$.
The velocity $v$ at any displacement $x$ is given by $v = \omega \sqrt{A^2 - x^2}$.
Squaring both sides: $v^2 = \omega^2 (A^2 - x^2)$.
Rearranging for $x$: $x^2 = A^2 - \frac{v^2}{\omega^2}$.
Substituting the values $A = 4$, $v = 6$, and $\omega = 3$:
$x^2 = 4^2 - \frac{6^2}{3^2} = 16 - \frac{36}{9} = 16 - 4 = 12$.
$x = \sqrt{12} = 2\sqrt{3} \,cm$.

(A) Given: Displacement $Y = A \cos 30^{\circ} = 40 \times \frac{\sqrt{3}}{2} = 20\sqrt{3} \ cm = 0.2\sqrt{3} \ m$.
Amplitude $A = 40 \ cm = 0.4 \ m$.
Kinetic Energy $K.E. = \frac{1}{2} k(A^2 - Y^2)$.
Substituting the values: $200 = \frac{1}{2} k((0.4)^2 - (0.2\sqrt{3})^2)$.
$200 = \frac{1}{2} k(0.16 - 0.12)$.
$200 = \frac{1}{2} k(0.04)$.
$200 = k(0.02)$.
$k = \frac{200}{0.02} = 10000 \ N/m = 1 \times 10^4 \ N/m$.
Comparing with $1 \times 10^x \ N/m$,we get $x = 4$.

(C) For a particle in Simple Harmonic Motion $(SHM)$,the velocity $V$ at a displacement $x$ is given by the formula: $V = \omega \sqrt{A^2 - x^2}$,where $A$ is the amplitude and $\omega$ is the angular frequency. Squaring both sides,we get $V^2 = \omega^2(A^2 - x^2)$.
Applying this to the given conditions:
$V_1^2 = \omega^2(A^2 - x_1^2)$ --- $(1)$
$V_2^2 = \omega^2(A^2 - x_2^2)$ --- $(2)$
Dividing equation $(1)$ by equation $(2)$:
$\frac{V_1^2}{V_2^2} = \frac{A^2 - x_1^2}{A^2 - x_2^2}$
Cross-multiplying gives:
$V_1^2(A^2 - x_2^2) = V_2^2(A^2 - x_1^2)$
$V_1^2 A^2 - V_1^2 x_2^2 = V_2^2 A^2 - V_2^2 x_1^2$
Rearranging terms to solve for $A^2$:
$A^2(V_1^2 - V_2^2) = V_1^2 x_2^2 - V_2^2 x_1^2$
$A^2 = \frac{V_1^2 x_2^2 - V_2^2 x_1^2}{V_1^2 - V_2^2}$
Taking the square root,we find the amplitude:
$A = \sqrt{\frac{V_1^2 x_2^2 - V_2^2 x_1^2}{V_1^2 - V_2^2}}$

(C) Given:
- The initial maximum velocity is $V = A\omega$.
- The amplitude doubles: $A' = 2A$.
- The period becomes one-third: $T' = \frac{T}{3}$.
Step $1$: Calculate the new angular frequency $\omega'$.
Since $\omega = \frac{2\pi}{T}$,the new angular frequency is:
$\omega' = \frac{2\pi}{T'} = \frac{2\pi}{T/3} = 3 \left(\frac{2\pi}{T}\right) = 3\omega$.
Step $2$: Calculate the new maximum velocity $V'$.
The formula for maximum velocity is $V_{\max} = A\omega$.
$V' = A' \omega' = (2A)(3\omega) = 6(A\omega)$.
Since $V = A\omega$,we get:
$V' = 6V$.

(D) Let the positions of the particle be $x_1$ and $x_2$ at the two instants.
The acceleration in $SHM$ is given by $a = -\omega^2 x$. Considering magnitudes,$a_1 = \omega^2 |x_1|$ and $a_2 = \omega^2 |x_2|$.
The velocity in $SHM$ is given by $v^2 = \omega^2 (A^2 - x^2)$.
For the first instant: $u^2 = \omega^2 A^2 - \omega^2 x_1^2$ . . . $(i)$
For the second instant: $V^2 = \omega^2 A^2 - \omega^2 x_2^2$ . . . $(ii)$
Subtracting $(ii)$ from $(i)$: $u^2 - V^2 = \omega^2 (x_2^2 - x_1^2) = \omega^2 (x_2 - x_1)(x_2 + x_1)$ . . . $(iii)$
From the acceleration equations: $a_2 - a_1 = \omega^2 (x_2 - x_1)$ (assuming $x_2 > x_1$).
However,the distance between positions is $|x_2 - x_1|$.
Using $a_1 = \omega^2 x_1$ and $a_2 = \omega^2 x_2$,we have $x_1 = a_1/\omega^2$ and $x_2 = a_2/\omega^2$.
Substituting into the velocity difference: $u^2 - V^2 = \omega^2 (x_2^2 - x_1^2) = \omega^2 (a_2^2/\omega^4 - a_1^2/\omega^4) = \frac{1}{\omega^2} (a_2^2 - a_1^2)$.
This implies $\omega^2 = \frac{a_2^2 - a_1^2}{u^2 - V^2}$.
The distance $d = |x_2 - x_1| = |\frac{a_2 - a_1}{\omega^2}| = |\frac{a_2 - a_1}{(a_2^2 - a_1^2)/(u^2 - V^2)}| = |\frac{u^2 - V^2}{a_2 + a_1}|$.
Thus,the distance is $\frac{u^2 - V^2}{a_1 + a_2}$.

(B) The speed $V$ of a particle performing $S.H.M.$ at a displacement $x$ from the mean position is given by the formula: $V = \omega \sqrt{a^2 - x^2}$.
Given,angular frequency $\omega = \frac{2 \pi}{T}$ and displacement $x = \frac{a}{3}$.
Substituting these values into the formula:
$V = \frac{2 \pi}{T} \sqrt{a^2 - (\frac{a}{3})^2}$
$V = \frac{2 \pi}{T} \sqrt{a^2 - \frac{a^2}{9}}$
$V = \frac{2 \pi}{T} \sqrt{\frac{8 a^2}{9}}$
$V = \frac{2 \pi}{T} \times \frac{2 \sqrt{2} a}{3}$
$V = \frac{4 \sqrt{2} \pi a}{3 T}$

(D) The given wave equation is $Y = 3 \sin 2 \pi \left( \frac{t}{0.04} - \frac{x}{0.01} \right)$.
Comparing this with the standard wave equation $Y = A \sin 2 \pi \left( ft - \frac{x}{\lambda} \right)$,we get the frequency $f = \frac{1}{0.04} = 25 \ Hz$.
The angular frequency is $\omega = 2 \pi f = 2 \pi \times 25 = 50 \pi \ rad/s$.
The maximum acceleration $a_{\max}$ is given by $a_{\max} = \omega^2 A$.
Substituting the values: $a_{\max} = (50 \pi)^2 \times 3 = 2500 \times \pi^2 \times 3$.
Given $\pi^2 = 10$,we have $a_{\max} = 2500 \times 10 \times 3 = 75000 \ cm/s^2 = 7.5 \times 10^4 \ cm/s^2$.

(C) The velocity of a particle in $S.H.M.$ at displacement $x$ is given by $V = \omega \sqrt{A^2 - x^2}$.
Maximum velocity is $V_{\max} = \omega A$.
Given that the speed $V = \frac{V_{\max}}{3} = \frac{\omega A}{3}$.
Equating the two expressions: $\frac{\omega A}{3} = \omega \sqrt{A^2 - x^2}$.
Squaring both sides: $\frac{A^2}{9} = A^2 - x^2$.
Rearranging for $x^2$: $x^2 = A^2 - \frac{A^2}{9} = \frac{8A^2}{9}$.
Taking the square root: $x = \sqrt{\frac{8}{9} A^2} = \frac{2\sqrt{2}}{3} A$.

(A) Let the positions of the particle be $x_1$ and $x_2$ at the two instants.
In $S.H.M.$,acceleration $a = -\omega^2 x$. Considering magnitudes,$\alpha = \omega^2 |x_1|$ and $\beta = \omega^2 |x_2|$.
Thus,$|x_1| = \frac{\alpha}{\omega^2}$ and $|x_2| = \frac{\beta}{\omega^2}$.
The velocity in $S.H.M.$ is given by $v^2 = \omega^2(A^2 - x^2)$.
For the first instant: $u^2 = \omega^2 A^2 - \omega^2 x_1^2$ . . . $(i)$
For the second instant: $v^2 = \omega^2 A^2 - \omega^2 x_2^2$ . . . $(ii)$
Subtracting $(ii)$ from $(i)$: $u^2 - v^2 = \omega^2(x_2^2 - x_1^2) = \omega^2(x_2 - x_1)(x_2 + x_1)$.
Since $\alpha = \omega^2 x_1$ and $\beta = \omega^2 x_2$,we have $\alpha + \beta = \omega^2(x_1 + x_2)$.
Substituting this into the equation: $u^2 - v^2 = \omega^2(x_2 - x_1) \cdot \frac{\alpha + \beta}{\omega^2}$.
Therefore,the distance between the two positions is $|x_2 - x_1| = \frac{u^2 - v^2}{\alpha + \beta}$.

(A) The time period of a simple pendulum is given by the formula $T = 2 \pi \sqrt{\frac{l}{g_{eff}}}$,where $l$ is the length of the pendulum and $g_{eff}$ is the effective acceleration due to gravity.
When the point of suspension moves vertically upward with an acceleration '$a$',the effective acceleration due to gravity becomes $g_{eff} = g + a$.
Substituting this into the formula,we get $T' = 2 \pi \sqrt{\frac{l}{g + a}}$.
Since $g + a > g$,the denominator increases,which causes the time period $T'$ to decrease compared to the original time period $T$.

(C) The frequency of a stretched string is given by the formula $n = \frac{1}{2L} \sqrt{\frac{T}{\mu}}$,where $L$ is the length,$T$ is the tension,and $\mu$ is the mass per unit length.
Initially,the frequency is $n_1 = \frac{1}{2L} \sqrt{\frac{T_1}{\mu}}$.
When the tension is increased by $44\%$,the new tension $T_2$ becomes $T_2 = T_1 + 0.44T_1 = 1.44T_1$.
The new frequency $n_2$ is given by $n_2 = \frac{1}{2L} \sqrt{\frac{1.44T_1}{\mu}}$.
Taking the ratio of $n_2$ to $n_1$:
$\frac{n_2}{n_1} = \frac{\frac{1}{2L} \sqrt{\frac{1.44T_1}{\mu}}}{\frac{1}{2L} \sqrt{\frac{T_1}{\mu}}} = \sqrt{\frac{1.44T_1}{T_1}} = \sqrt{1.44} = 1.2$.
Converting $1.2$ to a fraction,we get $1.2 = \frac{12}{10} = \frac{6}{5}$.
Thus,the ratio $n_2:n_1$ is $6:5$.

(B) We know that 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 $T \propto \sqrt{l}$.
Let the initial time period be $T_1$ with length $l_1$,and the final time period be $T_2$ with length $l_2$.
We want the period to be doubled,so $T_2 = 2 T_1$.
Using the ratio:
$\frac{T_2}{T_1} = \sqrt{\frac{l_2}{l_1}}$
Substituting $T_2 = 2 T_1$:
$2 = \sqrt{\frac{l_2}{l_1}}$
Squaring both sides:
$4 = \frac{l_2}{l_1} \Rightarrow l_2 = 4 l_1$
Therefore,the period of a simple pendulum is doubled when its length is made four times the original length.

(C) For a simple pendulum,the time period is given by $T = 2 \pi \sqrt{\frac{l}{g}}$.
Thus,$T_1 = 2 \pi \sqrt{\frac{l_1}{g}}$ and $T_2 = 2 \pi \sqrt{\frac{l_2}{g}}$.
Squaring both sides,we get $T_1^2 = 4 \pi^2 \frac{l_1}{g}$ and $T_2^2 = 4 \pi^2 \frac{l_2}{g}$.
From these,we can express the lengths as $l_1 = \frac{T_1^2 g}{4 \pi^2}$ and $l_2 = \frac{T_2^2 g}{4 \pi^2}$.
Now,for a pendulum of length $(l_1 - l_2)$,the time period $T$ is given by $T = 2 \pi \sqrt{\frac{l_1 - l_2}{g}}$.
Substituting the expressions for $l_1$ and $l_2$:
$T = 2 \pi \sqrt{\frac{1}{g} \left( \frac{T_1^2 g}{4 \pi^2} - \frac{T_2^2 g}{4 \pi^2} \right)}$.
$T = 2 \pi \sqrt{\frac{g}{g \cdot 4 \pi^2} (T_1^2 - T_2^2)}$.
$T = 2 \pi \sqrt{\frac{1}{4 \pi^2} (T_1^2 - T_2^2)}$.
$T = 2 \pi \cdot \frac{1}{2 \pi} \sqrt{T_1^2 - T_2^2}$.
$T = \sqrt{T_1^2 - T_2^2}$.

(D) Given the equation for velocity in $S.H.M.$: $4V^2 = 50 - x^2$.
Rearranging this,we get $V^2 = \frac{1}{4}(50 - x^2) = \frac{1}{4}(50 - x^2)$.
Comparing this with the standard $S.H.M.$ velocity equation $V^2 = \omega^2(A^2 - x^2)$,we can rewrite the given equation as $V^2 = \frac{1}{4} \cdot 50 - \frac{1}{4}x^2$.
This implies $\omega^2 = \frac{1}{4}$,so $\omega = \frac{1}{2} \text{ rad/s}$.
The time period $T$ is given by $T = \frac{2\pi}{\omega} = \frac{2\pi}{1/2} = 4\pi$.
Substituting $\pi = \frac{22}{7}$,we get $T = 4 \times \frac{22}{7} = \frac{88}{7} \text{ seconds}$.
Comparing this with the given time period $\frac{x}{7}$,we find $x = 88$.

(C) The time period of a simple pendulum is given by the formula $T = 2\pi \sqrt{\frac{l}{g}}$.
From this formula,we can see that the time period $T$ is directly proportional to the square root of the length $l$,i.e.,$T \propto \sqrt{l}$.
The time period of a simple pendulum is independent of the mass,size,or density of the bob.
Let the initial length be $l_1 = l$ and the initial time period be $T_1 = T$.
Let the new length be $l_2$ and the new time period be $T_2 = 2T$.
Using the proportionality $T \propto \sqrt{l}$,we have $\frac{T_2}{T_1} = \sqrt{\frac{l_2}{l_1}}$.
Substituting the given values: $\frac{2T}{T} = \sqrt{\frac{l_2}{l}}$.
$2 = \sqrt{\frac{l_2}{l}}$.
Squaring both sides,we get $4 = \frac{l_2}{l}$.
Therefore,the new length $l_2 = 4l$.

(B) The velocity of a particle in $S.H.M.$ is given by $v = \omega \sqrt{A^2 - x^2}$.
Substituting the given values:
For $x_1 = 3 \ cm$,$v_1 = 8 \ cm/s$: $8 = \omega \sqrt{A^2 - 3^2} \implies 8 = \omega \sqrt{A^2 - 9}$ ... $(i)$
For $x_2 = 4 \ cm$,$v_2 = 6 \ cm/s$: $6 = \omega \sqrt{A^2 - 4^2} \implies 6 = \omega \sqrt{A^2 - 16}$ ... (ii)
Dividing equation $(i)$ by (ii):
$\frac{8}{6} = \frac{\sqrt{A^2 - 9}}{\sqrt{A^2 - 16}} \implies \frac{4}{3} = \frac{\sqrt{A^2 - 9}}{\sqrt{A^2 - 16}}$
Squaring both sides:
$\frac{16}{9} = \frac{A^2 - 9}{A^2 - 16} \implies 16(A^2 - 16) = 9(A^2 - 9)$
$16A^2 - 256 = 9A^2 - 81 \implies 7A^2 = 175 \implies A^2 = 25 \ cm^2$.
Substituting $A^2 = 25$ into equation (ii):
$6 = \omega \sqrt{25 - 16} = \omega \sqrt{9} = 3\omega \implies \omega = 2 \ rad/s$.
The periodic time $T$ is given by $T = \frac{2\pi}{\omega} = \frac{2\pi}{2} = \pi \ s$.

(B) The displacement of a particle performing $S.H.M.$ starting from the equilibrium position is given by $x = A \sin(\omega t)$.
Velocity of the particle is $v = \frac{dx}{dt} = A \omega \cos(\omega t) \dots (i)$.
Given: $v = \pi \ m/s$,$T = 12 \ s$,and $t = 2 \ s$.
Angular frequency $\omega = \frac{2 \pi}{T} = \frac{2 \pi}{12} = \frac{\pi}{6} \ rad/s$.
Substituting these values into equation $(i)$:
$\pi = A \times \frac{\pi}{6} \times \cos\left(\frac{\pi}{6} \times 2\right)$
$\pi = A \times \frac{\pi}{6} \times \cos\left(\frac{\pi}{3}\right)$
Since $\cos(60^{\circ}) = 0.5 = \frac{1}{2}$,we have:
$1 = \frac{A}{6} \times \frac{1}{2}$
$1 = \frac{A}{12}$
$A = 12 \ m$.

(D) The time period of $S.H.M.$ for small vertical oscillations of a floating body is given by $T = 2 \pi \sqrt{\frac{h'}{g}}$,where $h'$ is the depth of the object submerged in the liquid.
For a floating body,the weight of the object equals the weight of the displaced liquid.
Let $V$ be the volume of the wood,$V = a \times b \times c$.
Mass of wood $= V \times d \times \rho_w = (abc) \times d \times \rho_w$ (where $\rho_w$ is the density of water).
Volume of displaced water $V_{disp} = b \times c \times h'$.
Weight of displaced water $= (bc h') \times \rho_w \times g$.
Equating the two: $(abc) \times d \times \rho_w \times g = (bc h') \times \rho_w \times g$.
Solving for $h'$,we get $h' = ad$.
Substituting this into the time period formula: $T = 2 \pi \sqrt{\frac{ad}{g}}$.

(D) The maximum velocity $(V_{\max})$ of a particle in Simple Harmonic Motion ($S$.$H$.$M$.) is given by the formula: $V_{\max} = A \omega$,where $A$ is the amplitude and $\omega$ is the angular frequency.
Given: Amplitude $A = 7 \ mm = 7 \times 10^{-3} \ m$,Maximum velocity $V_{\max} = 4.4 \ ms^{-1}$,and $\pi = \frac{22}{7}$.
We know that $\omega = \frac{2\pi}{T}$,where $T$ is the time period.
Substituting the values into the formula: $4.4 = (7 \times 10^{-3}) \times \left(\frac{2 \times 22/7}{T}\right)$.
Simplifying the expression: $4.4 = (7 \times 10^{-3}) \times \left(\frac{44}{7T}\right)$.
$4.4 = \frac{44 \times 10^{-3}}{T}$.
$T = \frac{44 \times 10^{-3}}{4.4} = 10 \times 10^{-3} = 0.01 \ s$.

(C) For the object to remain on the platform, the normal force $N$ must be greater than or equal to zero. The equation of motion for the object of mass $m$ is given by $mg - N = ma$, where $a$ is the acceleration of the platform.
For the object to just lose contact with the surface, the normal force $N$ becomes $0$.
Thus, $mg = ma$, which implies $a = g$.
The acceleration of a particle in $S.H.M.$ is given by $a = A\omega^2$, where $A$ is the amplitude and $\omega$ is the angular frequency.
To avoid detachment, the maximum downward acceleration of the platform must not exceed $g$.
Therefore, $A\omega^2 \leq g$.
Substituting the given values, $A = 40 \text{ cm} = 0.4 \text{ m}$ and $g = 10 \text{ m/s}^2$:
$0.4 \omega^2 = 10$
$\omega^2 = \frac{10}{0.4} = 25$
$\omega = 5 \text{ rad/s}$.
Since $\omega = \frac{2\pi}{T}$, we have:
$T = \frac{2\pi}{\omega} = \frac{2\pi}{5} = 0.4\pi \text{ s}$.
Thus, the least period of oscillation is $0.4\pi \text{ s}$.

(B) Given the equation of motion for the particle is $a = -bx$.
Comparing this with the standard equation for Simple Harmonic Motion $(SHM)$,which is $a = -\omega^2 x$,where $\omega$ is the angular frequency.
Equating the two expressions,we get $\omega^2 = b$,which implies $\omega = \sqrt{b}$.
The time period $T$ of an $SHM$ is given by the formula $T = \frac{2 \pi}{\omega}$.
Substituting the value of $\omega$,we get $T = \frac{2 \pi}{\sqrt{b}}$.

(C) The velocity of a particle in $S.H.M.$ at a distance $x$ from the mean position is given by $V = \omega \sqrt{A^2 - x^2}$,where $A$ is the amplitude and $\omega$ is the angular frequency.
Squaring both sides,we get $V^2 = \omega^2(A^2 - x^2)$.
For the given conditions:
$V_1^2 = \omega^2(A^2 - x_1^2)$ $(i)$
$V_2^2 = \omega^2(A^2 - x_2^2)$ $(ii)$
Subtracting equation $(i)$ from $(ii)$:
$V_2^2 - V_1^2 = \omega^2(A^2 - x_2^2 - A^2 + x_1^2)$
$V_2^2 - V_1^2 = \omega^2(x_1^2 - x_2^2)$
$\omega^2 = \frac{V_2^2 - V_1^2}{x_1^2 - x_2^2}$
$\omega = \sqrt{\frac{V_2^2 - V_1^2}{x_1^2 - x_2^2}}$
Since the frequency $f = \frac{\omega}{2\pi}$,we have:
$f = \frac{1}{2\pi} \sqrt{\frac{V_2^2 - V_1^2}{x_1^2 - x_2^2}}$

(B) The angular momentum $L$ is given by the formula $L = I \omega$,where $I$ is the moment of inertia and $\omega$ is the angular velocity.
For a solid sphere,the moment of inertia about its axis of rotation is $I = \frac{2}{5} MR^2$.
The angular velocity $\omega$ in terms of the period of rotation $T$ is given by $\omega = \frac{2 \pi}{T}$.
Substituting these values into the angular momentum formula:
$L = I \omega = \left( \frac{2}{5} MR^2 \right) \left( \frac{2 \pi}{T} \right)$.
Therefore,$L = \frac{4 \pi MR^2}{5 T}$.

(A) The angular momentum $L$ of a rotating body is given by $L = I\omega$,where $I$ is the moment of inertia and $\omega$ is the angular velocity.
Kinetic energy $K$ is given by $K = \frac{1}{2}I\omega^2$.
We can express $L$ in terms of $K$ and $\omega$ as $L = \frac{2K}{\omega}$.
Given that the frequency $f$ is doubled,the angular velocity $\omega = 2\pi f$ is also doubled. Let the initial state be $(L_1, K_1, \omega_1)$ and the final state be $(L_2, K_2, \omega_2)$.
We have $K_2 = \frac{K_1}{2}$ and $\omega_2 = 2\omega_1$.
Using the relation $\frac{L_2}{L_1} = \frac{K_2}{K_1} \times \frac{\omega_1}{\omega_2}$,we get:
$\frac{L_2}{L} = \frac{K_1/2}{K_1} \times \frac{\omega_1}{2\omega_1} = \frac{1}{2} \times \frac{1}{2} = \frac{1}{4}$.
Therefore,$L_2 = \frac{L}{4}$.

(B) The angular momentum of the system is conserved because no external torque acts on the system.
Initial angular momentum $L_i = I_1 \omega_1$,where $I_1 = \frac{1}{2} M R^2$ and $\omega_1 = \omega$.
Final moment of inertia $I_2 = I_{\text{disc1}} + I_{\text{disc2}} = \frac{1}{2} M R^2 + \frac{1}{2} (\frac{M}{3}) R^2 = \frac{1}{2} R^2 (M + \frac{M}{3}) = \frac{1}{2} R^2 (\frac{4M}{3}) = \frac{2}{3} M R^2$.
By the law of conservation of angular momentum,$I_1 \omega_1 = I_2 \omega_2$.
Substituting the values: $(\frac{1}{2} M R^2) \omega = (\frac{2}{3} M R^2) \omega_2$.
Solving for $\omega_2$: $\omega_2 = \frac{1/2}{2/3} \omega = \frac{3}{4} \omega$.

(C) For a particle of mass $m$ moving in a circular path of radius $r$ with velocity $v$,the angular momentum $L$ is given by $L = mvr$.
From this,we can express velocity as $v = \frac{L}{mr}$.
The centripetal force $F$ acting on the particle is given by $F = \frac{mv^2}{r}$.
Substituting the expression for $v$ into the force equation:
$F = \frac{m}{r} \left( \frac{L}{mr} \right)^2$
$F = \frac{m}{r} \cdot \frac{L^2}{m^2 r^2}$
$F = \frac{L^2}{mr^3}$
Rearranging the terms,we get $\frac{L^2}{m} = Fr^3$.

(A) The kinetic energy $E$ of a particle in rotational motion is given by $E = \frac{1}{2} I \omega^2$.
Angular momentum $L$ is defined as $L = I \omega$,which implies $L^2 = I^2 \omega^2$.
Substituting $\omega^2 = \frac{L^2}{I^2}$ into the kinetic energy equation,we get $E = \frac{1}{2} I \left( \frac{L^2}{I^2} \right) = \frac{L^2}{2I}$.
For a particle of mass '$m$' moving in a circle of radius '$r$',the moment of inertia about the axis passing through the center is $I = mr^2$.
Substituting $I = mr^2$ into the expression for kinetic energy,we get $E = \frac{L^2}{2(mr^2)} = \frac{L^2}{2 mr^2}$.

(C) The rotational kinetic energy $(KE)$ of a body is given by $KE = \frac{1}{2} I \omega^2$,where $I$ is the moment of inertia and $\omega$ is the angular velocity.
Since $I$ remains constant,$KE \propto \omega^2$.
Given that the angular velocity increases by $20 \%$,the new angular velocity $\omega_2$ is:
$\omega_2 = \omega_1 + 0.20 \omega_1 = 1.2 \omega_1$.
The ratio of the new kinetic energy $(KE_2)$ to the initial kinetic energy $(KE_1)$ is:
$\frac{KE_2}{KE_1} = \frac{\omega_2^2}{\omega_1^2} = \frac{(1.2 \omega_1)^2}{\omega_1^2} = (1.2)^2 = 1.44$.
Therefore,$KE_2 = 1.44 KE_1$.
The percentage increase in kinetic energy is:
$\text{Percentage Increase} = \left( \frac{KE_2 - KE_1}{KE_1} \right) \times 100 = (1.44 - 1) \times 100 = 0.44 \times 100 = 44 \%$.

(B) The moment of inertia $I$ of an annular ring (or hollow disk) about an axis passing through its centre and perpendicular to its plane is given by the formula:
$I = \frac{1}{2} M(R_1^2 + R_2^2)$
where $M$ is the mass,$R_1$ is the inner radius,and $R_2$ is the outer radius.
Given:
$M = 10 \ kg$
$R_1 = 5 \ m$
$R_2 = 10 \ m$
Substituting the values:
$I = \frac{1}{2} \times 10 \times (5^2 + 10^2)$
$I = 5 \times (25 + 100)$
$I = 5 \times 125 = 625 \ kg \cdot m^2$

(A) The moment of inertia of a solid sphere about its center of mass is $I = \frac{2}{5} m R^2$.
The moment of inertia of a disc of mass $m$ and radius $r$ about an axis passing through its rim and perpendicular to its plane is given by the parallel axis theorem: $I_{rim} = I_{cm} + m r^2 = \frac{1}{2} m r^2 + m r^2 = \frac{3}{2} m r^2$.
Given that the moment of inertia remains the same $(I_{sphere} = I_{disc})$:
$\frac{2}{5} m R^2 = \frac{3}{2} m r^2$.
Solving for $r$:
$r^2 = \frac{2}{5} \times \frac{2}{3} R^2 = \frac{4}{15} R^2$.
Therefore,$r = \sqrt{\frac{4}{15}} R = \frac{2 R}{\sqrt{15}}$.

(B) The two loops are formed from the same uniform wire,so the linear mass density $\lambda$ is constant.
Mass of loop $P$ is $M_P = \lambda \times (2\pi r) = 2\pi r\lambda$.
Mass of loop $Q$ is $M_Q = \lambda \times (2\pi nr) = 2\pi nr\lambda$.
The moment of inertia of a circular loop about its axis is $I = Mr^2$.
For loop $P$,$I_P = M_P r^2 = (2\pi r\lambda) r^2 = 2\pi r^3 \lambda$.
For loop $Q$,$I_Q = M_Q (nr)^2 = (2\pi nr\lambda) (nr)^2 = 2\pi n^3 r^3 \lambda$.
Given that $I_Q = 4 I_P$,we have:
$2\pi n^3 r^3 \lambda = 4 \times (2\pi r^3 \lambda)$.
$n^3 = 4$.
$n = (4)^{1/3} = (2^2)^{1/3} = (2)^{2/3}$.

(A) Let $M$ be the mass and $R$ be the radius of both the circular disc and the circular ring.
For a circular disc,the moment of inertia about its central axis is $I_d = \frac{1}{2} MR^2$.
If $K_d$ is the radius of gyration for the disc,then $I_d = MK_d^2$.
Equating the two,we get $MK_d^2 = \frac{1}{2} MR^2$,which simplifies to $K_d = \frac{R}{\sqrt{2}}$.
For a circular ring,the moment of inertia about its central axis is $I_r = MR^2$.
If $K_r$ is the radius of gyration for the ring,then $I_r = MK_r^2$.
Equating the two,we get $MK_r^2 = MR^2$,which simplifies to $K_r = R$.
The ratio of the radius of gyration of the disc to that of the ring is $\frac{K_d}{K_r} = \frac{R/\sqrt{2}}{R} = \frac{1}{\sqrt{2}}$.
Thus,the ratio is $1 : \sqrt{2}$.

(C) The moment of inertia of a loop (ring) about its central axis is given by $I = MR^2$.
Thus,$I_A = M_1 R_1^2$ and $I_B = M_2 R_2^2$ ... $(i)$
Since the loops are made from a uniform wire,the mass $M$ is proportional to the circumference $(2 \pi R)$.
Let $m$ be the mass per unit length. Then $M_1 = 2 \pi R_1 m$ and $M_2 = 2 \pi R_2 m$.
Therefore,$\frac{M_1}{M_2} = \frac{R_1}{R_2}$ ... (ii)
Substituting (ii) into the ratio of moments of inertia:
$\frac{I_A}{I_B} = \frac{M_1 R_1^2}{M_2 R_2^2} = \left(\frac{M_1}{M_2}\right) \left(\frac{R_1}{R_2}\right)^2 = \left(\frac{R_1}{R_2}\right) \left(\frac{R_1}{R_2}\right)^2 = \left(\frac{R_1}{R_2}\right)^3$.
Given $\frac{I_A}{I_B} = 27$,we have $\left(\frac{R_1}{R_2}\right)^3 = 27$.
Taking the cube root on both sides,$\frac{R_1}{R_2} = 3$.
Therefore,$\frac{R_2}{R_1} = \frac{1}{3}$.

(C) For a disc of mass $M$ and radius $R$,the moment of inertia about its diameter $(I_D)$ is given by:
$I_D = \frac{MR^2}{4}$
For a ring of mass $M$ and radius $R$,the moment of inertia about a tangent in its plane $(I_T)$ is calculated using the parallel axis theorem:
$I_T = I_{CM} + MR^2$
Since the moment of inertia of a ring about its diameter is $I_{CM} = \frac{MR^2}{2}$,we have:
$I_T = \frac{MR^2}{2} + MR^2 = \frac{3}{2} MR^2$
The ratio of the moment of inertia of the disc about its diameter to that of the ring about a tangent in its plane is:
$\frac{I_D}{I_T} = \frac{\frac{MR^2}{4}}{\frac{3}{2} MR^2} = \frac{1}{4} \times \frac{2}{3} = \frac{1}{6}$
Thus,the ratio is $1: 6$.

(B) Mass $=$ Volume $\times$ Density.
Let the masses be $M_A$ and $M_B$,and radii be $R_A$ and $R_B$.
Since $M_A = M_B = M$,we have:
$M = \frac{4}{3} \pi R_A^3 \rho_A = \frac{4}{3} \pi R_B^3 \rho_B$.
From this,$\frac{R_B^3}{R_A^3} = \frac{\rho_A}{\rho_B}$,which implies $\frac{R_B}{R_A} = (\frac{\rho_A}{\rho_B})^{1/3}$.
The moment of inertia of a solid sphere about its diameter is $I = \frac{2}{5} MR^2$.
Therefore,$\frac{I_B}{I_A} = \frac{\frac{2}{5} M R_B^2}{\frac{2}{5} M R_A^2} = \frac{R_B^2}{R_A^2}$.
Substituting the ratio of radii:
$\frac{I_B}{I_A} = ((\frac{\rho_A}{\rho_B})^{1/3})^2 = (\frac{\rho_A}{\rho_B})^{2/3}$.

(C) The moment of inertia of a disc about its central axis is $I = \frac{1}{2} M R_d^2$.
Since the mass $M$ remains constant,we equate the volumes:
$V_{\text{disc}} = V_{\text{sphere}}$
$\pi R_d^2 \times (R_d/8) = \frac{4}{3} \pi R_s^3$
$\frac{R_d^3}{8} = \frac{4}{3} R_s^3 \implies R_s^3 = \frac{3}{32} R_d^3 \implies R_s^2 = \left(\frac{3}{32}\right)^{2/3} R_d^2$.
The moment of inertia of a solid sphere about its diameter is $I_{\text{sphere}} = \frac{2}{5} M R_s^2$.
Substituting $R_s^2$:
$I_{\text{sphere}} = \frac{2}{5} M \left(\frac{3}{32}\right)^{2/3} R_d^2$.
Since $M R_d^2 = 2I$,we have:
$I_{\text{sphere}} = \frac{2}{5} (2I) \left(\frac{3}{32}\right)^{2/3} = \frac{4I}{5} \left(\frac{9}{1024}\right)^{1/3} \approx \frac{I}{5}$.

(D) Let $I$ be the moment of inertia of the square plate about an axis passing through the centre '$O$' and perpendicular to the plane of the plate.
According to the perpendicular axis theorem,for any two mutually perpendicular axes in the plane of the plate intersecting at the centre,the sum of the moments of inertia about these axes equals the moment of inertia about the axis perpendicular to the plane.
For axes $1$ and $2$ (diagonals),$I = I_1 + I_2$.
For axes $3$ and $4$ (mid-lines),$I = I_3 + I_4$.
Since the plate is a square,by symmetry,$I_1 = I_2$ and $I_3 = I_4$.
Thus,$I = 2I_1$ and $I = 2I_3$,which implies $I_1 = I_3$.
Therefore,the moment of inertia about the perpendicular axis is $I = I_1 + I_3$ (or $I_2 + I_4$,or $I_1 + I_4$,etc.).
Comparing with the given options,the correct expression is $I_1 + I_3$.

(A) The total moment of inertia of the system about the $x$-axis is the sum of the moments of inertia of the three individual rods about the $x$-axis: $I_{\text{total}} = I_x + I_y + I_z$.
For a thin rod of mass $m$ and length $L$ rotating about an axis passing through one end and perpendicular to the rod,the moment of inertia is $I = \frac{mL^2}{3}$.
$1$. For the rod along the $x$-axis: Since the rod lies on the $x$-axis,its distance from the $x$-axis is zero for all points. Thus,$I_x = 0$.
$2$. For the rod along the $y$-axis: The axis of rotation ($x$-axis) is perpendicular to the rod and passes through one end (the origin). Here,$m = 2M$,so $I_y = \frac{(2M)L^2}{3} = \frac{2ML^2}{3}$.
$3$. For the rod along the $z$-axis: Similarly,the $x$-axis is perpendicular to this rod and passes through one end. Here,$m = 2M$,so $I_z = \frac{(2M)L^2}{3} = \frac{2ML^2}{3}$.
Summing these values: $I_{\text{total}} = 0 + \frac{2ML^2}{3} + \frac{2ML^2}{3} = \frac{4ML^2}{3}$.

(A) According to Einstein's photoelectric equation,the maximum kinetic energy is given by $K.E_{\max} = E - \phi$.
For the first case,$E_1 = 3\phi$,so $K.E_1 = 3\phi - \phi = 2\phi$.
For the second case,$E_2 = 9\phi$,so $K.E_2 = 9\phi - \phi = 8\phi$.
Since $K.E = \frac{1}{2}mv^2$,we have the ratio $\frac{K.E_1}{K.E_2} = \frac{v_1^2}{v_2^2}$.
Substituting the values,$\frac{2\phi}{8\phi} = \frac{1}{4} = \frac{v_1^2}{v_2^2}$.
Taking the square root,$\frac{v_1}{v_2} = \frac{1}{2}$,which implies $v_2 = 2v_1$.
Given $v_1 = 6 \times 10^6 \ m/s$,then $v_2 = 2 \times 6 \times 10^6 = 12 \times 10^6 \ m/s$.

(D) According to Einstein's photoelectric equation,the maximum kinetic energy of emitted photoelectrons is given by $K.E. = h n - \phi$,where $\phi$ is the work function of the metal.
For the first photocathode: $\frac{1}{2}mV_1^2 = hn_1 - \phi$ $(1)$
For the second photocathode: $\frac{1}{2}mV_2^2 = hn_2 - \phi$ $(2)$
Subtracting equation $(2)$ from equation $(1)$:
$\frac{1}{2}mV_1^2 - \frac{1}{2}mV_2^2 = (hn_1 - \phi) - (hn_2 - \phi)$
$\frac{1}{2}m(V_1^2 - V_2^2) = h(n_1 - n_2)$
$V_1^2 - V_2^2 = \frac{2h}{m}(n_1 - n_2)$

(A) According to Einstein's photoelectric equation,the stopping potential $V_0$ is related to the frequency $f$ of the incident radiation by the equation: $eV_0 = hf - \Phi$,where $e$ is the charge of an electron,$h$ is Planck's constant,and $\Phi$ is the work function of the metal.
From the graph,the stopping potentials are $|V_0^1| < |V_0^2| < |V_0^3| < |V_0^4|$.
Since the stopping potential $V_0$ is directly proportional to the frequency $f$ for a given metal,a larger magnitude of stopping potential corresponds to a higher frequency of incident radiation.
Therefore,the order of frequencies is $f_a > f_b > f_c > f_d$.

(A) According to Einstein's photoelectric equation,the maximum kinetic energy of emitted photoelectrons is given by: $K_{max} = hv - \phi$,where $h$ is Planck's constant,$v$ is the frequency of incident radiation,and $\phi$ is the work function of the metal.
Comparing this equation with the equation of a straight line,$y = mx + c$,we have $y = K_{max}$,$x = v$,slope $m = h$,and intercept $c = -\phi$.
Since the slope $h$ is positive and the intercept $-\phi$ is negative,the graph is a straight line starting from a threshold frequency $v_0$ (where $K_{max} = 0$) and increasing linearly with frequency $v$.
This corresponds to graph $(1)$.

(C) The photocurrent is independent of the frequency of the incident photon,provided the frequency is above the threshold frequency.
However,when the intensity of the incident light increases,the number of photons incident per unit area per unit time increases.
This leads to an increase in the number of photoelectrons emitted from the metal surface,thereby increasing the photocurrent.

(C) According to Einstein's photoelectric equation,the kinetic energy $E$ of an emitted photoelectron is given by $E = \frac{hc}{\lambda} - W_0$,where $W_0$ is the work function.
For wavelength $\lambda_1$,$E_1 = \frac{hc}{\lambda_1} - W_0 \implies E_1 \lambda_1 = hc - W_0 \lambda_1 \implies hc = E_1 \lambda_1 + W_0 \lambda_1$ ... $(i)$
For wavelength $\lambda_2$,$E_2 = \frac{hc}{\lambda_2} - W_0 \implies E_2 \lambda_2 = hc - W_0 \lambda_2 \implies hc = E_2 \lambda_2 + W_0 \lambda_2$ ... (ii)
Equating $(i)$ and (ii):
$E_1 \lambda_1 + W_0 \lambda_1 = E_2 \lambda_2 + W_0 \lambda_2$
$E_1 \lambda_1 - E_2 \lambda_2 = W_0 \lambda_2 - W_0 \lambda_1$
$E_1 \lambda_1 - E_2 \lambda_2 = W_0 (\lambda_2 - \lambda_1)$
$W_0 = \frac{E_1 \lambda_1 - E_2 \lambda_2}{\lambda_2 - \lambda_1}$

(C) According to Einstein's photoelectric equation,the maximum kinetic energy $E_k$ is given by $E_k = \frac{hc}{\lambda} - \phi$,where $\phi$ is the work function.
For the first case with wavelength $\lambda$:
$E_1 = \frac{hc}{\lambda} - \phi$ ... $(i)$
For the second case with wavelength $\frac{\lambda}{2}$:
$E_2 = \frac{hc}{\lambda/2} - \phi = \frac{2hc}{\lambda} - \phi$ ... (ii)
Given that $E_1 = \frac{1}{4} E_2$,which implies $4E_1 = E_2$ ... (iii)
Substituting $(i)$ and (ii) into (iii):
$4\left(\frac{hc}{\lambda} - \phi\right) = \frac{2hc}{\lambda} - \phi$
$\frac{4hc}{\lambda} - 4\phi = \frac{2hc}{\lambda} - \phi$
$\frac{4hc}{\lambda} - \frac{2hc}{\lambda} = 4\phi - \phi$
$\frac{2hc}{\lambda} = 3\phi$
$\phi = \frac{2hc}{3\lambda}$

(A) According to Einstein's photoelectric equation,the maximum kinetic energy of emitted photoelectrons is given by $K_{max} = hv - W_0$,where $h$ is Planck's constant,$v$ is the frequency of incident radiation,and $W_0$ is the work function of the metal.
Since the stopping potential $V_s$ is related to the maximum kinetic energy by $eV_s = K_{max}$,we have $eV_s = hv - W_0$.
Rearranging for $V_s$,we get $V_s = \frac{h}{e}v - \frac{W_0}{e}$.
From this equation,it is clear that the stopping potential $V_s$ is directly proportional to the frequency $v$ of the incident radiation.
Therefore,if the frequency $v$ is increased,the stopping potential $V_s$ will also increase.

(C) According to Einstein's photoelectric equation,the maximum kinetic energy $K_{max}$ is given by $K_{max} = \frac{hc}{\lambda} - \phi$,where $\phi$ is the work function.
For the first case with wavelength $\lambda$:
$K_1 = \frac{hc}{\lambda} - \phi$ ... $(i)$
For the second case with wavelength $\lambda / 3$:
$K_2 = \frac{hc}{\lambda / 3} - \phi = \frac{3hc}{\lambda} - \phi$ ... $(ii)$
Given that $K_2 = 4K_1$,we substitute the expressions:
$\frac{3hc}{\lambda} - \phi = 4 \left( \frac{hc}{\lambda} - \phi \right)$
Expanding the equation:
$\frac{3hc}{\lambda} - \phi = \frac{4hc}{\lambda} - 4\phi$
Rearranging to solve for $\phi$:
$4\phi - \phi = \frac{4hc}{\lambda} - \frac{3hc}{\lambda}$
$3\phi = \frac{hc}{\lambda}$
$\phi = \frac{hc}{3\lambda}$

(B) According to Einstein's photoelectric equation,the maximum kinetic energy $(K_{max})$ is given by: $K_{max} = h\nu - \Phi_0$,where $\Phi_0$ is the work function.
Since $K_{max} = eV_0$,we can write: $eV_0 = h\nu - \Phi_0$.
Dividing by $e$,we get: $V_0 = (\frac{h}{e})\nu - \frac{\Phi_0}{e}$.
Comparing this with the equation of a straight line $y = mx + c$,where $y = V_0$ and $x = \nu$,the slope $(m)$ is $\frac{h}{e}$.
Multiplying the slope by the charge of the electron $(e)$,we get: $m \times e = (\frac{h}{e}) \times e = h$.
Thus,the product gives the Planck's constant $(h)$.

(A) According to Einstein's photoelectric equation,the maximum kinetic energy is given by $K.E_{\max} = E - \phi_0$,where $E$ is the energy of the incident photon and $\phi_0$ is the work function.
For the first case,$E_1 = 2\phi_0$,so $K.E_1 = 2\phi_0 - \phi_0 = \phi_0$.
For the second case,$E_2 = 3\phi_0$,so $K.E_2 = 3\phi_0 - \phi_0 = 2\phi_0$.
Since $K.E. = \frac{1}{2}mv^2$,we have $\frac{K.E_1}{K.E_2} = \frac{v_1^2}{v_2^2}$.
Substituting the values,$\frac{v_1^2}{v_2^2} = \frac{\phi_0}{2\phi_0} = \frac{1}{2}$.
Therefore,the ratio of maximum velocities is $\frac{v_1}{v_2} = \sqrt{\frac{1}{2}} = \frac{1}{\sqrt{2}}$.

(B) According to Einstein's photoelectric equation, the stopping potential $V_s$ is given by:
$e V_s = h \nu - \Phi_0$
$V_s = \frac{h}{e} \nu - \frac{\Phi_0}{e}$
where $\Phi_0$ is the work function and $\nu_0$ is the threshold frequency, such that $\Phi_0 = h \nu_0$.
From the given graph, the intercept on the frequency axis represents the threshold frequency $\nu_0$.
Since the threshold frequency $\nu_{0,A}$ for surface $A$ is less than the threshold frequency $\nu_{0,B}$ for surface $B$ ( $\nu_{0,A} < \nu_{0,B}$ ),
it follows that the work function $\Phi_{0,A} = h \nu_{0,A}$ is smaller than the work function $\Phi_{0,B} = h \nu_{0,B}$.
Therefore, the work function of $A$ is smaller than that of $B$.

(A) Using Einstein's photoelectric equation,$h \nu = \phi_0 + KE_{\text{max}}$.
Since $KE_{\text{max}} = eV_s$,we have $\frac{hc}{\lambda} = \phi_0 + e(4V_0)$ ....$(i)$
Similarly,for wavelength $3\lambda$,we have $\frac{hc}{3\lambda} = \phi_0 + eV_0$ ....(ii)
Multiply equation (ii) by $4$:
$\frac{4hc}{3\lambda} = 4\phi_0 + 4eV_0$ ....(iii)
Subtract equation $(i)$ from equation (iii):
$\frac{4hc}{3\lambda} - \frac{hc}{\lambda} = (4\phi_0 + 4eV_0) - (\phi_0 + 4eV_0)$
$\frac{hc}{3\lambda} = 3\phi_0$
$\phi_0 = \frac{hc}{9\lambda}$
Since the work function $\phi_0 = \frac{hc}{\lambda_0}$,where $\lambda_0$ is the threshold wavelength,we get $\lambda_0 = 9\lambda$.
Thus,option $(A)$ is correct.

(C) In the photoelectric effect,the stopping potential depends only on the frequency of the incident radiation and the work function of the metal surface. Since all four curves intersect the potential axis at the same point (the same stopping potential),the frequency of all four radiations must be the same. Therefore,$f_a = f_b = f_c = f_d$.
The saturation photocurrent is directly proportional to the intensity of the incident radiation. From the graph,the saturation current values are $I_a < I_b < I_c < I_d$. Therefore,the intensities follow the same order: $I_a < I_b < I_c < I_d$.

(C) The stopping potential $(V_0)$ in the photoelectric effect is determined by the maximum kinetic energy of the emitted photoelectrons, which depends solely on the frequency of the incident light and the work function of the metal surface, as given by Einstein's photoelectric equation: $eV_0 = h\nu - \Phi$.
Moving the point source of light farther away changes the intensity of the light incident on the metal surface, but it does not change the frequency of the incident photons.
Since the stopping potential is independent of the intensity of the incident light, it will remain constant.

(A) The brightness of the bulb depends on the current $I$ flowing through the circuit,where $I = \frac{V}{Z}$. The impedance $Z$ of the $LR$ circuit is given by $Z = \sqrt{R^2 + X_L^2}$,where $X_L = \omega L = 2\pi f L$. For the brightness to decrease,the current $I$ must decrease,which means the impedance $Z$ must increase.
$1$. When an iron rod is inserted into the coil,the self-inductance $L$ increases due to the increase in permeability. Consequently,$X_L$ increases,$Z$ increases,and the current $I$ decreases,leading to a decrease in brightness.
$2$. If the frequency $f$ is decreased,$X_L$ decreases,$Z$ decreases,and the current $I$ increases,leading to an increase in brightness.
$3$. If the number of turns $N$ is reduced,$L$ decreases (since $L \propto N^2$),$X_L$ decreases,$Z$ decreases,and the current $I$ increases,leading to an increase in brightness.
$4$. Adding a capacitor such that the net reactance becomes zero (resonance) would increase the current,not decrease it.
Therefore,the correct option is $A$.

(B) The energy stored in the inductor is given by $U = \frac{1}{2} LI^2$.
Substituting the given values,$U = \frac{1}{2} \times 1 \times 1^2 = 0.5 \text{ J}$.
When the circuit is broken,this energy is transferred to the capacitor to prevent sparking.
The energy stored in a capacitor is $U = \frac{1}{2} CV^2$.
Equating the two energies: $\frac{1}{2} CV^2 = \frac{1}{2} LI^2$.
Solving for $C$: $C = L \left( \frac{I}{V} \right)^2$.
Substituting the values: $C = 1 \times \left( \frac{1}{500} \right)^2 = \frac{1}{250000} \text{ F}$.
$C = 4 \times 10^{-6} \text{ F} = 4 \mu \text{ F}$.

(C) The magnetic energy $U$ stored in an inductor is given by the formula: $U = \frac{1}{2} LI^2$.
Given: Inductance $L = 5 \mu H = 5 \times 10^{-6} \ H$ and current $I = 2 \ A$.
Substituting the values: $U = \frac{1}{2} \times (5 \times 10^{-6}) \times (2)^2$.
$U = \frac{1}{2} \times 5 \times 10^{-6} \times 4$.
$U = 10 \times 10^{-6} \ J = 10 \mu J$.

(A) The magnetic energy $U_B$ in an inductor is given by $U_B = \frac{1}{2} L I^2$. Since the current $I$ in an $A.C.$ circuit varies as $I = I_0 \sin(\omega t)$,the energy varies as $U_B \propto \sin^2(\omega t)$.
Energy changes from maximum to minimum in $\frac{1}{4}$ of a time period $T$.
Given,$\frac{T}{4} = 5 \ ms = 5 \times 10^{-3} \ s$.
Therefore,$T = 20 \times 10^{-3} \ s = 0.02 \ s$.
The frequency $f$ is given by $f = \frac{1}{T} = \frac{1}{0.02} = 50 \ Hz$.

(D) The circuit consists of two inductors,each of inductance $L/2$,connected in parallel,which are then connected in series with an inductor of inductance $3L/4$.
First,calculate the equivalent inductance $L_p$ of the two inductors in parallel:
$\frac{1}{L_p} = \frac{1}{L/2} + \frac{1}{L/2} = \frac{2}{L} + \frac{2}{L} = \frac{4}{L}$
$\therefore L_p = \frac{L}{4}$
Now,calculate the total equivalent inductance $L_{eq}$ by adding the series inductor:
$L_{eq} = L_p + \frac{3L}{4} = \frac{L}{4} + \frac{3L}{4} = \frac{4L}{4} = L$

(C) The energy stored in an inductor is given by the formula: $E = \frac{1}{2} LI^2$.
Given: Energy $E = 64 \times 10^{-3} \ J$ and current $I = 80 \ mA = 80 \times 10^{-3} \ A$.
Rearranging the formula to solve for inductance $L$: $L = \frac{2E}{I^2}$.
Substituting the values: $L = \frac{2 \times 64 \times 10^{-3}}{(80 \times 10^{-3})^2}$.
$L = \frac{128 \times 10^{-3}}{6400 \times 10^{-6}} = \frac{128 \times 10^{-3}}{6.4 \times 10^{-3}} = \frac{128}{6.4} = 20 \ H$.

(D) The inductance of a coil is directly proportional to its length $(L \propto l)$.
When the coil is divided into two equal parts,each part has an inductance of $L_1 = L_2 = \frac{L}{2}$.
When these two inductors are connected in parallel,the equivalent inductance $L_{eq}$ is given by the formula:
$\frac{1}{L_{eq}} = \frac{1}{L_1} + \frac{1}{L_2}$
Substituting the values:
$\frac{1}{L_{eq}} = \frac{1}{L/2} + \frac{1}{L/2} = \frac{2}{L} + \frac{2}{L} = \frac{4}{L}$
Therefore,$L_{eq} = \frac{L}{4}$.

(B) Given: Current $I = 4 \ A$,Number of turns $N = 400$,Magnetic flux $\phi = 3 \times 10^{-3} \ Wb$.
The magnetic flux linkage is given by $N\phi = LI$,where $L$ is the self-inductance of the coil.
First,calculate the self-inductance $L$:
$L = \frac{N\phi}{I} = \frac{400 \times 3 \times 10^{-3}}{4} = 100 \times 3 \times 10^{-3} = 0.3 \ H$.
The energy stored in an inductor is given by the formula $U = \frac{1}{2} LI^2$.
Substituting the values:
$U = \frac{1}{2} \times 0.3 \times (4)^2$
$U = \frac{1}{2} \times 0.3 \times 16$
$U = 0.3 \times 8 = 2.4 \ J$.

(A) The reactance of an inductor in an $A.C.$ circuit is given by $X_L = \omega L = 2\pi f L$,where $f$ is the frequency of the $A.C.$ source.
For a $D.C.$ source,the frequency $f = 0$.
Therefore,the inductive reactance in a $D.C.$ circuit is $X_L = 2\pi (0) L = 0$.
The ratio of reactance in an $A.C.$ source to that in a $D.C.$ source is $\frac{X_{L(ac)}}{X_{L(dc)}} = \frac{\omega L}{0} = \infty$.

(A) According to Faraday's law of electromagnetic induction,the induced electromotive force $(e)$ is given by $e = -N \frac{d\phi}{dt}$.
Since the coil and the magnet are moving in the same direction with the same speed $V$,their relative velocity is zero.
As a result,the magnetic flux $(\phi)$ linked with the coil remains constant over time.
Therefore,$\frac{d\phi}{dt} = 0$.
Consequently,the induced e.m.f. is zero.

(D) The induced e.m.f. in the first coil is given by $e_1 = M \frac{dI_2}{dt}$.
Given $e_1 = 20 \,mV = 20 \times 10^{-3} \,V$ and $\frac{dI_2}{dt} = 10 \,A/s$.
Substituting these values, we get $M = \frac{e_1}{dI_2/dt} = \frac{20 \times 10^{-3}}{10} = 2 \times 10^{-3} \,H$.
The flux linkage in the second coil when current $I_1$ flows through the first coil is given by $\phi_2 = M I_1$.
Given $I_1 = 3.6 \,A$, we have $\phi_2 = (2 \times 10^{-3} \,H) \times (3.6 \,A) = 7.2 \times 10^{-3} \,Wb$.

(A) According to Faraday's law of electromagnetic induction, the magnitude of induced e.m.f. is given by $|e| = N \frac{|\Delta \phi|}{\Delta t}$.
Here, $N = 50$, $B_1 = 2 \times 10^{-2} \,T$, $B_2 = 0 \,T$, $A = 100 \,cm^2 = 100 \times 10^{-4} \,m^2 = 10^{-2} \,m^2$, and $|e| = 0.1 \,V$.
The change in magnetic flux is $\Delta \phi = (B_2 - B_1) A \cos 0^{\circ} = (0 - 2 \times 10^{-2}) \times 10^{-2} = -2 \times 10^{-4} \,Wb$.
Substituting the values into the formula:
$0.1 = 50 \times \frac{2 \times 10^{-4}}{t}$.
$t = \frac{50 \times 2 \times 10^{-4}}{0.1} = \frac{100 \times 10^{-4}}{0.1} = \frac{10^{-2}}{10^{-1}} = 0.1 \,s$.

(D) According to Faraday's law of electromagnetic induction,the magnitude of induced electromotive force $(EMF)$ is given by $|e| = \frac{\Delta \phi}{\Delta t}$.
Since the coil has resistance $R$,the induced current $I$ is given by $I = \frac{|e|}{R} = \left(\frac{\Delta \phi}{\Delta t}\right) \frac{1}{R}$.
The total induced charge $Q$ passing through the circuit is given by $Q = I \times \Delta t$.
Substituting the value of $I$,we get $Q = \left(\frac{\Delta \phi}{\Delta t} \cdot \frac{1}{R}\right) \times \Delta t = \frac{\Delta \phi}{R}$.
Thus,the induced current is $\left(\frac{\Delta \phi}{\Delta t}\right) \frac{1}{R}$ and the induced charge is $\frac{\Delta \phi}{R}$.

(D) The induced e.m.f. in the second coil is given by the formula: $e = M \frac{dI_1}{dt} \dots (i)$
Given $I_1 = I_0 \sin \omega t$, we differentiate with respect to time $t$:
$\frac{dI_1}{dt} = I_0 \omega \cos \omega t$
Substituting this into equation $(i)$, we get:
$e = M I_0 \omega \cos \omega t$
The maximum value of e.m.f. $(e_{max})$ occurs when $\cos \omega t = 1$:
$e_{max} = M I_0 \omega$
Given values: $M = 5 \times 10^{-3} \text{ H}$, $I_0 = 10 \text{ A}$, and $\omega = 100 \pi \text{ rad/s}$.
Substituting these values:
$e_{max} = (5 \times 10^{-3}) \times 10 \times 100 \pi$
$e_{max} = 5 \times 10^{-3} \times 10^3 \pi$
$e_{max} = 5 \pi \text{ V}$

(D) The instantaneous current in the primary coil is given by $I = I_0 \sin(\omega t)$.
Given peak current $I_0 = \frac{2}{\pi} \text{ A}$ and frequency $v = 50 \text{ Hz}$.
The angular frequency is $\omega = 2 \pi v = 2 \pi(50) = 100 \pi \text{ rad/s}$.
The rate of change of current is $\frac{dI}{dt} = I_0 \omega \cos(\omega t)$.
The maximum value of the rate of change of current is $\left(\frac{dI}{dt}\right)_{\text{max}} = I_0 \omega$.
Substituting the values: $\left(\frac{dI}{dt}\right)_{\text{max}} = \left(\frac{2}{\pi}\right) \times (100 \pi) = 200 \text{ A/s}$.
The induced e.m.f. in the secondary coil is $E = M \left|\frac{dI}{dt}\right|$,where $M = 1 \text{ H}$.
Therefore,the peak induced e.m.f. is $E_0 = 1 \times 200 = 200 \text{ V}$.

(D) The induced electromotive force (e.m.f.) is given by Faraday's law: $|e| = \frac{d\phi}{dt}$.
Given the magnetic flux $\phi = 50t^2 + 7$.
Taking the derivative with respect to time $t$,we get $|e| = \frac{d}{dt}(50t^2 + 7) = 100t$.
At time $t = 4 \ s$,the induced e.m.f. is $|e| = 100(4) = 400 \ V$.
The current $I$ in the coil is given by Ohm's law: $I = \frac{e}{R}$.
Substituting the values,$I = \frac{400 \ V}{250 \ \Omega} = 1.6 \ A$.

(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):
$f = \frac{e}{B \pi l^2}$

(A) The angular velocity $\omega$ in radians per second is given by $\omega = 2 \pi f$,where $f$ is the frequency in revolutions per second. Since $F$ is in r.p.m.,$f = \frac{F}{60}$.
Thus,$\omega = 2 \pi \frac{F}{60} = \frac{\pi F}{30}$.
Consider a small element $dr$ at a distance $r$ from the center. The motional e.m.f. $dE$ induced across this element is $dE = B v dr = B (r \omega) dr$.
Integrating from $r = 0$ to $r = R$:
$E = \int_{0}^{R} B \omega r dr = B \omega \left[ \frac{r^2}{2} \right]_{0}^{R} = \frac{1}{2} B \omega R^2$.
Substituting $\omega = \frac{2 \pi F}{60}$:
$E = \frac{1}{2} B \left( \frac{2 \pi F}{60} \right) R^2 = \frac{B \pi F R^2}{60}$.
Note: If $F$ is treated as revolutions per second (rps) in the context of standard physics problems of this type,the answer is $\frac{1}{2} B \pi F R^2$. Given the options,the intended answer is $A$.

(B) The magnetic flux linked with the coil is given by $\phi_B = B A \cos \theta$,where $\theta$ is the angle between the normal to the plane of the coil and the magnetic field vector $B$.
When the plane of the coil is perpendicular to the magnetic field,the normal to the coil is parallel to the magnetic field,so $\theta = 0^{\circ}$.
Thus,the magnetic flux $\phi_B = B A \cos(0^{\circ}) = B A$,which is the maximum value.
The induced $e.m.f.$ is given by $\varepsilon = -\frac{d\phi_B}{dt} = B A \omega \sin(\omega t)$.
At $\theta = 0^{\circ}$,$\omega t = 0$,so $\varepsilon = B A \omega \sin(0^{\circ}) = 0$.
Therefore,the magnetic flux is maximum and the induced $e.m.f.$ is zero.

(B) Consider a small element of length $dr$ at a distance $r$ from the fixed end of the rod.
The velocity of this element is $v = r \omega$.
The motional e.m.f. induced across this small element is $de = B v dr = B (r \omega) dr$.
To find the total induced e.m.f. across the entire length of the rod,we integrate from $r = 0$ to $r = l$:
$e = \int_{0}^{l} B \omega r dr$
$e = B \omega \left[ \frac{r^2}{2} \right]_{0}^{l}$
$e = \frac{1}{2} B l^2 \omega = 0.5 B l^2 \omega$.

(A) The induced electromotive force $(e)$ in a rotating coil is given by $e = NAB \omega \sin(\omega t)$.
The peak value of the induced $EMF$ is $e_0 = NAB \omega$.
The peak current is $i_0 = \frac{e_0}{R} = \frac{NAB \omega}{R}$.
The average power dissipated in an $AC$ circuit is given by $P_{av} = \frac{e_0 i_0}{2}$ or $P_{av} = \frac{e_0^2}{2R}$.
Substituting the value of $e_0$:
$P_{av} = \frac{(NAB \omega)^2}{2R} = \frac{N^2 A^2 B^2 \omega^2}{2R}$.

(B) When a conductor of length $L$ moves with velocity $V$ in a uniform magnetic field $B$ perpendicular to it,the motional electromotive force $(e.m.f.)$ induced is given by $e = BLV$.
Since the loop is moving in a uniform magnetic field,the magnetic flux linked with the loop remains constant if the entire loop is within the field. However,if the loop is entering or leaving the field,an $e.m.f.$ is induced across the side cutting the field lines.
The rate of production of heat energy (power dissipated) in the loop is given by $P = \frac{e^2}{R}$.
Substituting the value of $e = BLV$,we get:
$P = \frac{(BLV)^2}{R} = \frac{B^2 L^2 V^2}{R}$.

(D) The induced electromotive force $(e)$ in a conductor moving through a magnetic field is given by the formula: $e = B \cdot v \cdot l$, where $B$ is the magnetic field, $v$ is the velocity, and $l$ is the length of the rod.
Given:
$B = 3.6 \times 10^{-5} \text{ T}$
$v = 2.00 \text{ m/s}$
$l = 2 \text{ m}$
Substituting these values into the formula:
$e = (3.6 \times 10^{-5}) \times 2.00 \times 2$
$e = 14.4 \times 10^{-5} \text{ V}$
$e = 0.144 \times 10^{-3} \text{ V}$
$e = 0.144 \text{ mV}$

(D) Consider a small radial element of length $dr$ at a distance $r$ from the centre of the disc.
As the disc rotates,this element moves with a linear velocity $v = r\omega$ perpendicular to the magnetic field $B$.
The motional e.m.f. $de$ induced across this small element is given by $de = Bv dr = B(r\omega) dr$.
To find the total induced e.m.f. $e$ between the centre (axis) and the rim (radius $R$),we integrate the expression from $r = 0$ to $r = R$:
$e = \int_{0}^{R} B\omega r dr$
$e = B\omega \left[ \frac{r^2}{2} \right]_{0}^{R}$
$e = \frac{1}{2} B\omega R^2$.

(A) The magnetic flux through the coil at any time $t$ is given by $\phi = N A B \cos(\omega t)$.
According to Faraday's law,the induced electromotive force $(EMF)$ is $e = -\frac{d\phi}{dt} = N A B \omega \sin(\omega t)$.
The peak $EMF$ is $e_0 = N A B \omega$.
The peak current is $i_0 = \frac{e_0}{R} = \frac{N A B \omega}{R}$.
The average power dissipated in a complete cycle is given by $P_{avg} = \frac{1}{2} e_0 i_0$.
Substituting the values,$P_{avg} = \frac{1}{2} (N A B \omega) \left( \frac{N A B \omega}{R} \right) = \frac{N^2 A^2 B^2 \omega^2}{2 R}$.

(A) The magnetic flux $\phi$ linked with an inductor is given by $\phi = LI$,where $L$ is the self-inductance of the inductor.
Comparing this equation with the equation of a straight line $y = mx$,where $y = \phi$ and $x = I$,we get the slope $m = L$.
Since the slope of the graph represents the self-inductance $L$,the inductor with the smallest slope will have the smallest value of self-inductance.
Looking at the graph,the line $D$ makes the smallest angle with the current axis ($I$-axis),meaning it has the lowest slope.
Therefore,inductor $D$ has the smallest self-inductance.

(A) The formula for self-inductance $L$ is given by $L = \frac{N \phi}{i}$,where $N$ is the number of turns,$\phi$ is the magnetic flux per turn,and $i$ is the current.
Given:
$N = 400$
$\phi = 3 \times 10^{-3} \ Wb$
$i = 0.5 \ A$
Substituting the values into the formula:
$L = \frac{400 \times 3 \times 10^{-3}}{0.5}$
$L = \frac{1200 \times 10^{-3}}{0.5}$
$L = \frac{1.2}{0.5} = 2.4 \ H$
Therefore,the self-inductance of the solenoid is $2.4 \ H$.

(D) The magnetic field at the center of a circular loop of radius $r_1$ carrying current $I$ is given by $B = \frac{\mu_0 I}{2 r_1}$.
Since the rings are concentric and coplanar,the magnetic field $B$ is uniform over the area of the smaller ring (radius $r_2$).
The magnetic flux $\phi$ linked with the smaller ring is $\phi = B \times A_2$,where $A_2 = \pi r_2^2$ is the area of the smaller ring.
$\phi = \left( \frac{\mu_0 I}{2 r_1} \right) \times \pi r_2^2 = \frac{\mu_0 \pi r_2^2}{2 r_1} I$.
The mutual inductance $M$ is defined as $M = \frac{\phi}{I}$.
Therefore,$M = \frac{\mu_0 \pi r_2^2}{2 r_1}$.

(C) The mutual inductance $M$ between two coils with self-inductances $L_1$ and $L_2$ is given by the formula $M = k \sqrt{L_1 L_2}$,where $k$ is the coefficient of coupling.
Since the flux in one coil is completely linked with the other,the coupling is perfect,meaning $k = 1$.
Given $L_1 = 25 \ mH$ and $L_2 = 9 \ mH$.
Substituting these values into the formula:
$M = \sqrt{25 \ mH \times 9 \ mH} = \sqrt{225 \ mH^2} = 15 \ mH$.

(B) The induced e.m.f. in the second coil is given by the formula: $|e_s| = M \left| \frac{dI_p}{dt} \right|$.
Given $I_p = I_0 \sin \omega t$, we differentiate with respect to time $t$:
$\frac{dI_p}{dt} = I_0 \omega \cos \omega t$.
Substituting this into the e.m.f. equation:
$|e_s| = M I_0 \omega \cos \omega t$.
The maximum value of e.m.f. occurs when $\cos \omega t = 1$:
$|e_s|_{\max} = M I_0 \omega$.
Substituting the given values $M = 0.003 \ H$, $I_0 = 8 \ A$, and $\omega = 100 \pi \ rad \ s^{-1}$:
$|e_s|_{\max} = 0.003 \times 8 \times 100 \pi = 2.4 \pi \ V$.

(D) The magnetic field at the center of a circular loop of radius $r_1$ carrying current $I$ is given by $B = \frac{\mu_0 I}{2 r_1}$.
Since $r_2 \ll r_1$,the magnetic field $B$ is approximately uniform over the area of the smaller loop.
The magnetic flux $\phi$ linked with the smaller loop of radius $r_2$ is $\phi = B \times A_2$,where $A_2 = \pi r_2^2$ is the area of the smaller loop.
Substituting the values,we get $\phi = \left( \frac{\mu_0 I}{2 r_1} \right) \times (\pi r_2^2) = \frac{\mu_0 \pi r_2^2}{2 r_1} I$.
The mutual inductance $M$ is defined by the relation $\phi = M I$.
Therefore,$M = \frac{\phi}{I} = \frac{\mu_0 \pi r_2^2}{2 r_1}$.

(A) The formula for induced e.m.f. in the secondary coil due to mutual induction is given by:
$e_s = M \cdot \frac{dI_p}{dt}$
Where:
$M = 2 \ H$ (coefficient of mutual induction)
$e_s = 2 \ kV = 2 \times 10^3 \ V$
$dI_p = 6 \ A - 3 \ A = 3 \ A$
Rearranging the formula to solve for time $(dt)$:
$dt = M \cdot \frac{dI_p}{e_s}$
Substituting the values:
$dt = 2 \times \frac{3}{2 \times 10^3} \ s$
$dt = 3 \times 10^{-3} \ s$

(A) The self-inductance $L$ of a solenoid is given by the formula:
$L = \frac{\mu_0 N^2 A}{l}$
where $N$ is the total number of turns,$A$ is the cross-sectional area,and $l$ is the length of the coil.
From this formula,we can see that $L \propto N^2$ when $A$ and $l$ remain constant.
Given that the number of turns is doubled,i.e.,$N_2 = 2N_1$.
Therefore,the new self-inductance $L_2$ is:
$L_2 = L_1 \times \left(\frac{N_2}{N_1}\right)^2$
$L_2 = L_1 \times (2)^2$
$L_2 = 4 L_1$
Thus,the self-inductance becomes $4$ times the original value.

(A) Given: $N = 100$, $A = 1 \,cm^2 = 1 \times 10^{-4} \,m^2$, $L = 1 \,mH = 1 \times 10^{-3} \,H$, and $I = 2 \,A$.
Self-inductance of a solenoid is given by $L = \frac{\mu_0 N^2 A}{l}$, where $l$ is the length of the coil.
From this, the length $l$ is:
$l = \frac{\mu_0 N^2 A}{L} = \frac{(4 \pi \times 10^{-7}) \times (100)^2 \times (1 \times 10^{-4})}{1 \times 10^{-3}} = 4 \pi \times 10^{-3} \,m$.
The magnetic induction $B$ at the centre of the solenoid is given by $B = \frac{\mu_0 N I}{l}$.
Substituting the value of $l$:
$B = \frac{(4 \pi \times 10^{-7}) \times 100 \times 2}{4 \pi \times 10^{-3}} = 10^{-7} \times 200 \times 10^3 = 0.2 \,Wb/m^2$.

(C) The self-inductance $L$ of a solenoid is given by the formula $L = \mu_0 n^2 A l$,where $n$ is the number of turns per unit length,$A$ is the cross-sectional area,and $l$ is the length of the solenoid.
Since $L \propto n^2$,if the number of turns per unit length $n$ is tripled $(n' = 3n)$,the new self-inductance $L'$ will be:
$L' = \mu_0 (3n)^2 A l = 9 (\mu_0 n^2 A l) = 9L$.
Therefore,the self-inductance becomes $9$ times the original value.