MHT CET 2020 Physics Question Paper with Answer and Solution

(C) The total energy $(E)$ of a particle executing simple harmonic motion is given by the sum of its potential energy and kinetic energy.
The formula for total energy is $E = \frac{1}{2} k A^{2}$,where $k$ is the force constant.
We know that the angular frequency $\omega = \frac{2 \pi}{T}$.
Also,the force constant $k = m \omega^{2}$.
Substituting the value of $\omega$ into the expression for $k$,we get $k = m \left( \frac{2 \pi}{T} \right)^{2} = \frac{4 \pi^{2} m}{T^{2}}$.
Now,substitute $k$ into the total energy formula: $E = \frac{1}{2} \left( \frac{4 \pi^{2} m}{T^{2}} \right) A^{2}$.
Simplifying this,we get $E = \frac{2 \pi^{2} m A^{2}}{T^{2}}$.

(B) Given: Displacement $= x$,Potential Energy $(P.E.) = E$,Restoring Force $= F$.
For a particle in $S.H.M.$,the restoring force is given by $F = -kx$,where $k$ is the force constant.
The potential energy is given by $E = \frac{1}{2}kx^2$.
From the force equation,we have $k = -\frac{F}{x}$.
Substituting this value of $k$ into the potential energy equation:
$E = \frac{1}{2} \left(-\frac{F}{x}\right) x^2$
$E = -\frac{1}{2} Fx$
Multiplying by $2$:
$2E = -Fx$
Rearranging the terms:
$2E + Fx = 0$
Dividing by $F$:
$\frac{2E}{F} + x = 0$.

(A) For a simple pendulum undergoing small oscillations,the restoring force is $F = -\frac{mg}{L} x$.
The potential energy $PE$ of a simple harmonic oscillator is given by $PE = \frac{1}{2} k x^{2}$.
Comparing the restoring force $F = -kx$ with $F = -\frac{mg}{L} x$,we get the spring constant $k = \frac{mg}{L}$.
At the extreme position,the displacement $x$ is equal to the amplitude $A$.
Substituting these values into the potential energy formula: $PE = \frac{1}{2} \left( \frac{mg}{L} \right) A^{2} = \frac{mgA^{2}}{2L}$.

(D) The potential energy of a body in simple harmonic motion is given by $P = \frac{1}{2} k x^2$,where $k$ is the force constant.
Given $P_{1} = \frac{1}{2} k x_{1}^2$ and $P_{2} = \frac{1}{2} k x_{2}^2$.
We need to find the potential energy $P$ at displacement $(x_{1} + x_{2})$:
$P = \frac{1}{2} k (x_{1} + x_{2})^2$
$P = \frac{1}{2} k (x_{1}^2 + x_{2}^2 + 2 x_{1} x_{2})$
$P = \frac{1}{2} k x_{1}^2 + \frac{1}{2} k x_{2}^2 + 2 \left( \sqrt{\frac{1}{2} k x_{1}^2} \right) \left( \sqrt{\frac{1}{2} k x_{2}^2} \right)$
$P = P_{1} + P_{2} + 2 \sqrt{P_{1} P_{2}}$.

(D) The time period of a simple pendulum is given by $T = 2 \pi \sqrt{\frac{\ell}{g}}$.
Since $\omega = \frac{2 \pi}{T}$,we have $\omega = \sqrt{\frac{g}{\ell}}$,which implies $\omega \propto \frac{1}{\sqrt{\ell}}$.
Given the new length $\ell_2 = \frac{\ell_1}{3}$,the new angular frequency $\omega_2$ is related to the initial frequency $\omega_1$ by $\frac{\omega_2}{\omega_1} = \sqrt{\frac{\ell_1}{\ell_2}} = \sqrt{\frac{\ell_1}{\ell_1/3}} = \sqrt{3}$.
The total energy of a simple harmonic oscillator is given by $E = \frac{1}{2} m \omega^2 A^2$.
Since the mass $(m)$ and amplitude $(A)$ remain constant,$E \propto \omega^2$.
Therefore,$\frac{E_2}{E_1} = \left( \frac{\omega_2}{\omega_1} \right)^2 = (\sqrt{3})^2 = 3$.
Thus,the new total energy is $E_2 = 3 E_1$.

(A) Given,period $T = 6 \ s$. The kinetic energy $(K.E.)$ is given by $K.E. = \frac{1}{2} m \omega^2 (A^2 - x^2)$ and total energy $(T.E.)$ is $T.E. = \frac{1}{2} m \omega^2 A^2$.
We are given that $K.E. = 50\% \text{ of } T.E.$,so $K.E. = \frac{1}{2} T.E.$.
Substituting the expressions: $\frac{1}{2} m \omega^2 (A^2 - x^2) = \frac{1}{2} (\frac{1}{2} m \omega^2 A^2)$.
This simplifies to $A^2 - x^2 = \frac{A^2}{2}$,which gives $x^2 = \frac{A^2}{2}$ or $x = \frac{A}{\sqrt{2}}$.
Since the particle starts from the mean position,the displacement equation is $x = A \sin(\omega t)$.
Substituting $x = \frac{A}{\sqrt{2}}$,we get $\frac{A}{\sqrt{2}} = A \sin(\frac{2\pi}{T} t)$.
$\frac{1}{\sqrt{2}} = \sin(\frac{2\pi}{6} t) = \sin(\frac{\pi}{3} t)$.
Since $\sin(45^{\circ}) = \frac{1}{\sqrt{2}}$,we have $\frac{\pi}{4} = \frac{\pi}{3} t$.
Solving for $t$: $t = \frac{3}{4} = 0.75 \ s$.

(A) In simple harmonic motion,the acceleration $a$ is given by the formula $a = -\omega^2 x$,where $\omega$ is the angular frequency and $x$ is the displacement from the mean position.
At the mean position,the displacement $x = 0$.
Substituting $x = 0$ into the formula,we get $a = -\omega^2 (0) = 0$.
Therefore,the acceleration of the weight is zero at the mean position.

(D) Given: Length $L = 0.4 \ m$,maximum speed $v_{max} = 4 \ m/s$ at the lowest point.
When the string makes an angle of $30^{\circ}$ with the horizontal,it makes an angle of $\theta = 60^{\circ}$ with the vertical.
The height $h$ of the pendulum bob above the lowest point is given by $h = L - L \cos \theta = L(1 - \cos 60^{\circ})$.
$h = 0.4(1 - 0.5) = 0.4 \times 0.5 = 0.2 \ m$.
Using the law of conservation of energy between the lowest point and the point at angle $\theta$:
$\frac{1}{2} m v_{max}^{2} = \frac{1}{2} m v^{2} + mgh$
$v^{2} = v_{max}^{2} - 2gh$
$v^{2} = (4)^{2} - 2 \times 10 \times 0.2$
$v^{2} = 16 - 4 = 12$
$v = \sqrt{12} = 2 \sqrt{3} \ m/s$.

(C) Given,$y_{1} = 2 \sin (10 t + \theta)$.
The velocity $V_{1} = \frac{dy_{1}}{dt} = 2 \times 10 \cos (10 t + \theta) = 20 \cos (10 t + \theta)$.
Given,$y_{2} = 3 \cos 10 t = 3 \sin (10 t + \frac{\pi}{2})$.
The velocity $V_{2} = \frac{dy_{2}}{dt} = 3 \times 10 \cos (10 t + \frac{\pi}{2}) = 30 \cos (10 t + \frac{\pi}{2})$.
The phase of $V_{1}$ is $\phi_{1} = 10 t + \theta$.
The phase of $V_{2}$ is $\phi_{2} = 10 t + \frac{\pi}{2}$.
The phase difference $\Delta \phi = \phi_{1} - \phi_{2} = (10 t + \theta) - (10 t + \frac{\pi}{2}) = \theta - \frac{\pi}{2}$.

(B) The velocity of a particle in $S.H.M.$ at displacement $x$ is given by $v = \omega \sqrt{A^2 - x^2}$.
At $x = \frac{2A}{3}$,the velocity is $v = \omega \sqrt{A^2 - (\frac{2A}{3})^2} = \omega \sqrt{A^2 - \frac{4A^2}{9}} = \omega \sqrt{\frac{5A^2}{9}} = \frac{\sqrt{5}}{3} \omega A$.
When the speed is tripled,the new velocity $v' = 3v = 3 \times \frac{\sqrt{5}}{3} \omega A = \sqrt{5} \omega A$.
The total energy of the $S.H.M.$ remains constant at any point,so $E' = K' + U$.
The new total energy is $E' = \frac{1}{2} m \omega^2 A'^2$.
The potential energy at $x = \frac{2A}{3}$ is $U = \frac{1}{2} m \omega^2 x^2 = \frac{1}{2} m \omega^2 (\frac{2A}{3})^2 = \frac{1}{2} m \omega^2 \frac{4A^2}{9}$.
The new kinetic energy is $K' = \frac{1}{2} m v'^2 = \frac{1}{2} m (\sqrt{5} \omega A)^2 = \frac{1}{2} m \omega^2 (5A^2)$.
Equating the total energy: $\frac{1}{2} m \omega^2 A'^2 = \frac{1}{2} m \omega^2 (\frac{4A^2}{9}) + \frac{1}{2} m \omega^2 (5A^2)$.
$A'^2 = \frac{4A^2}{9} + 5A^2 = A^2 (\frac{4+45}{9}) = \frac{49A^2}{9}$.
Taking the square root,$A' = \frac{7A}{3}$.

(A) For a particle performing $S.H.M.$,the displacement $x$ as a function of time $t$ is given by $x = A \sin(\omega t)$,where $A$ is the amplitude and $\omega$ is the angular frequency.
According to Hooke's Law for $S.H.M.$,the restoring force $F$ is given by $F = -kx$,where $k$ is the force constant.
Substituting the expression for $x$,we get $F = -k(A \sin(\omega t)) = -kA \sin(\omega t)$.
This equation shows that the force $F$ is proportional to the negative of the displacement $x$.
Therefore,the force-time graph will be an inverted version of the displacement-time graph. If the displacement graph starts from the origin and goes positive,the force graph must start from the origin and go negative.

(D) The total energy of a particle in $S.H.M.$ is given by $E = \frac{1}{2} m \omega^2 A^2$.
At any displacement $x$,the potential energy is $U = \frac{1}{2} m \omega^2 x^2$.
The kinetic energy is $K = E - U = \frac{1}{2} m \omega^2 (A^2 - x^2)$.
Given that $K = \frac{3E}{4}$,the potential energy $U$ must be $E - \frac{3E}{4} = \frac{E}{4}$.
Substituting the expressions for $U$ and $E$:
$\frac{1}{2} m \omega^2 x^2 = \frac{1}{4} (\frac{1}{2} m \omega^2 A^2)$.
Simplifying,we get $x^2 = \frac{A^2}{4}$.
Taking the square root,$x = \frac{A}{2}$.

(A) The velocity $v$ of a particle in $S.H.M.$ is given by $v = \omega \sqrt{A^2 - x^2}$,where $\omega = \sqrt{\frac{k}{m}}$.
Given: $m = 10 \ kg$,$k = 10 \ N/m$,$A = 0.5 \ m$,and $v = 40 \ cm/s = 0.4 \ m/s$.
First,calculate the angular frequency $\omega = \sqrt{\frac{10}{10}} = 1 \ rad/s$.
Now,substitute the values into the velocity equation: $0.4 = 1 \cdot \sqrt{(0.5)^2 - x^2}$.
Squaring both sides: $0.16 = 0.25 - x^2$.
Rearranging for $x^2$: $x^2 = 0.25 - 0.16 = 0.09$.
Taking the square root: $x = 0.3 \ m$.

(A) When mass $m_{1}$ is attached to the spring,the total energy of the $S.H.M.$ is $E = \frac{1}{2} k A_{1}^{2}$.
At the mean position,the velocity $v_{1}$ of the mass $m_{1}$ is maximum,given by $v_{1} = \omega_{1} A_{1} = \sqrt{\frac{k}{m_{1}}} A_{1}$.
When mass $m_{2}$ is placed on $m_{1}$ at the mean position,the momentum of the system is conserved because there is no external horizontal force.
Initial momentum $p_{i} = m_{1} v_{1}$.
Final momentum $p_{f} = (m_{1} + m_{2}) v_{2}$,where $v_{2}$ is the new velocity at the mean position.
Since $p_{i} = p_{f}$,we have $m_{1} v_{1} = (m_{1} + m_{2}) v_{2}$.
$v_{2} = \frac{m_{1}}{m_{1} + m_{2}} v_{1}$.
The new angular frequency is $\omega_{2} = \sqrt{\frac{k}{m_{1} + m_{2}}}$.
Since $v_{2} = \omega_{2} A_{2}$,we have $A_{2} = \frac{v_{2}}{\omega_{2}} = \frac{m_{1} v_{1}}{(m_{1} + m_{2})} \sqrt{\frac{m_{1} + m_{2}}{k}}$.
Substituting $v_{1} = \sqrt{\frac{k}{m_{1}}} A_{1}$,we get $A_{2} = \frac{m_{1}}{(m_{1} + m_{2})} \sqrt{\frac{k}{m_{1}}} A_{1} \sqrt{\frac{m_{1} + m_{2}}{k}} = A_{1} \sqrt{\frac{m_{1}}{m_{1} + m_{2}}}$.
Therefore,$\frac{A_{2}}{A_{1}} = \left[\frac{m_{1}}{m_{1} + m_{2}}\right]^{1/2}$.

(D) The given wave equation is $Y = A \sin 2 \pi (n t - \frac{x}{\lambda})$.
Comparing this with the standard wave equation $Y = A \sin (\omega t - kx)$,we have $\omega = 2 \pi n$ and $k = \frac{2 \pi}{\lambda}$.
The maximum particle velocity is given by $v_{p, \text{max}} = A \omega = A (2 \pi n)$.
The wave velocity is given by $v_w = \frac{\omega}{k} = \frac{2 \pi n}{2 \pi / \lambda} = n \lambda$.
According to the problem,$v_{p, \text{max}} = 4 v_w$.
Substituting the expressions,we get $A (2 \pi n) = 4 (n \lambda)$.
Simplifying the equation,$2 \pi A n = 4 n \lambda$.
Dividing both sides by $2n$,we get $\lambda = \pi A$.

(A) The equation for displacement in simple harmonic motion starting from the mean position is given by $y = A \sin(\omega t)$.
Given,the period $T = 3 \ s$,so the angular frequency $\omega = \frac{2\pi}{T} = \frac{2\pi}{3} \ rad/s$.
We need to find the time $t$ when the displacement $y = \frac{A}{2}$.
Substituting these values into the equation: $\frac{A}{2} = A \sin\left(\frac{2\pi}{3} t\right)$.
$\frac{1}{2} = \sin\left(\frac{2\pi}{3} t\right)$.
Since $\sin 30^{\circ} = 0.5$,we have $\sin\left(\frac{\pi}{6}\right) = \sin\left(\frac{2\pi}{3} t\right)$.
Equating the angles: $\frac{\pi}{6} = \frac{2\pi}{3} t$.
Solving for $t$: $t = \frac{\pi}{6} \times \frac{3}{2\pi} = \frac{3}{12} = 0.25 \ s = \frac{1}{4} \ s$.

(B) The damping force $F$ is given by the relation $F = -bv$,where $b$ is the constant of proportionality (damping constant) and $v$ is the velocity.
To find the unit of $b$,we rearrange the formula: $b = \frac{F}{v}$.
The $SI$ unit of force $F$ is Newton $(N)$,which is equivalent to $kg \cdot m \cdot s^{-2}$.
The $SI$ unit of velocity $v$ is $m \cdot s^{-1}$.
Substituting these units into the formula for $b$:
$b = \frac{kg \cdot m \cdot s^{-2}}{m \cdot s^{-1}} = kg \cdot s^{-1}$.
Therefore,the unit of the constant of proportionality is $kg \cdot s^{-1}$.

(A) The coin rests on the plate and moves with it. The acceleration of the plate in $S.H.M.$ is given by $a = \omega^{2} x$,where $x$ is the displacement from the mean position.
As the plate moves downward,its acceleration is directed upwards. As the plate moves upward,its acceleration is directed downwards.
The coin loses contact with the plate when the downward acceleration of the plate exceeds the acceleration due to gravity $(g)$.
At the topmost point of the oscillation,the downward acceleration is maximum,given by $a_{max} = \omega^{2} A$.
For the coin to lose contact,the condition is $a_{max} \geq g$.
Therefore,the minimum amplitude at which contact is lost is $A = \frac{g}{\omega^{2}}$.

(D) The span of oscillation $L$ is the total distance between the extreme positions,which is equal to $2A$,where $A$ is the amplitude of oscillation.
Therefore,$A = \frac{L}{2}$.
The angular frequency $\omega$ is given by $\omega = 2 \pi f$.
The acceleration $a$ of a particle in $S.H.M.$ is given by $a = -\omega^{2} x$.
At the top of the oscillation (extreme position),the displacement $x = A = \frac{L}{2}$.
Substituting the values,the magnitude of acceleration is $|a| = \omega^{2} A = (2 \pi f)^{2} \times \frac{L}{2}$.
$|a| = 4 \pi^{2} f^{2} \times \frac{L}{2} = 2 \pi^{2} f^{2} L$.
Thus,the correct option is $D$.

(D) Let $a$ be the amplitude of the $S$.$H$.$M$.
Given that the particle is at a distance $\frac{a}{3}$ from the extreme position,its displacement $x$ from the mean position is $x = a - \frac{a}{3} = \frac{2a}{3}$.
The magnitude of acceleration is $a_p = \omega^2 x = \omega^2 \left( \frac{2a}{3} \right)$.
The magnitude of velocity is $v_p = \omega \sqrt{a^2 - x^2} = \omega \sqrt{a^2 - \left( \frac{2a}{3} \right)^2} = \omega \sqrt{a^2 - \frac{4a^2}{9}} = \omega \sqrt{\frac{5a^2}{9}} = \frac{\omega a \sqrt{5}}{3}$.
According to the problem,$v_p = \frac{1}{3} a_p$.
Substituting the expressions for $v_p$ and $a_p$:
$\frac{\omega a \sqrt{5}}{3} = \frac{1}{3} \left( \omega^2 \frac{2a}{3} \right)$.
$\frac{\omega a \sqrt{5}}{3} = \frac{2 \omega^2 a}{9}$.
Dividing both sides by $\omega a$ (assuming $\omega, a \neq 0$):
$\frac{\sqrt{5}}{3} = \frac{2 \omega}{9}$.
$\omega = \frac{9 \sqrt{5}}{3 \times 2} = \frac{3 \sqrt{5}}{2}$.
Since the period $T = \frac{2 \pi}{\omega}$,we have:
$T = \frac{2 \pi}{\frac{3 \sqrt{5}}{2}} = \frac{4 \pi}{3 \sqrt{5}} \text{ s}$.

(D) The velocity $v$ of a particle in simple harmonic motion at a displacement $x$ is given by the formula: $v = \omega \sqrt{A^2 - x^2}$.
Given that the particle is halfway between the mean position $(x = 0)$ and the extreme position $(x = A)$,the displacement is $x = \frac{A}{2}$.
The angular frequency $\omega$ is related to the period $T$ by $\omega = \frac{2 \pi}{T}$.
Substituting these values into the velocity formula:
$v = \frac{2 \pi}{T} \sqrt{A^2 - (\frac{A}{2})^2}$
$v = \frac{2 \pi}{T} \sqrt{A^2 - \frac{A^2}{4}}$
$v = \frac{2 \pi}{T} \sqrt{\frac{3 A^2}{4}}$
$v = \frac{2 \pi}{T} \times \frac{\sqrt{3} A}{2}$
$v = \frac{\sqrt{3} \pi A}{T}$

(D) The time period of a mass-spring system is given by $T = 2\pi \sqrt{\frac{M}{k}}$.
When the mass is increased by $m$,the new time period $T'$ is given as $T' = \frac{5T}{3}$.
Thus,$\frac{5T}{3} = 2\pi \sqrt{\frac{M+m}{k}}$.
Substituting $T = 2\pi \sqrt{\frac{M}{k}}$ into the equation:
$\frac{5}{3} \times 2\pi \sqrt{\frac{M}{k}} = 2\pi \sqrt{\frac{M+m}{k}}$.
Squaring both sides:
$\frac{25}{9} \times \frac{M}{k} = \frac{M+m}{k}$.
$\frac{25}{9}M = M + m$.
Dividing by $M$:
$\frac{25}{9} = 1 + \frac{m}{M}$.
$\frac{m}{M} = \frac{25}{9} - 1 = \frac{16}{9}$.
Therefore,the ratio $\frac{M}{m} = \frac{9}{16}$.

(C) The time period of a simple pendulum is given by $T = 2\pi \sqrt{\frac{L}{g}}$.
For a seconds pendulum on Earth,$T = 2 \text{ s}$.
The acceleration due to gravity on a planet is $g' = \frac{GM'}{R'^2}$.
Given $M' = 3M$ and $R' = 3R$,we have:
$\frac{g'}{g} = \frac{M'}{M} \cdot \frac{R^2}{R'^2} = 3 \cdot \frac{1}{3^2} = \frac{3}{9} = \frac{1}{3}$.
The time period on the planet is $T' = 2\pi \sqrt{\frac{L}{g'}}$.
Thus,$\frac{T'}{T} = \sqrt{\frac{g}{g'}} = \sqrt{3}$.
$T' = T \cdot \sqrt{3} = 2 \sqrt{3} \text{ seconds}$.

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

(A) When a mass $m$ is suspended from a wire,the wire acts like a spring with a spring constant $k$.
From Hooke's Law,the tension $T$ in the wire for an extension $x$ is given by $T = \frac{YA}{L} x$.
Comparing this with the spring force equation $F = kx$,we find the effective spring constant $k = \frac{YA}{L}$.
The frequency of oscillation $f$ for a mass-spring system is given by $f = \frac{1}{2 \pi} \sqrt{\frac{k}{m}}$.
Substituting the value of $k$,we get $f = \frac{1}{2 \pi} \sqrt{\frac{YA}{mL}}$.

(A) The frequency $f$ of a simple pendulum is given by the formula $f = \frac{1}{2 \pi} \sqrt{\frac{g}{\ell}}$.
From this,we can see that $f \propto \frac{1}{\sqrt{\ell}}$,which implies $\frac{f_1}{f_2} = \sqrt{\frac{\ell_2}{\ell_1}}$.
Given the ratio of frequencies $\frac{f_1}{f_2} = \frac{3}{4}$,we substitute this into the equation:
$\frac{3}{4} = \sqrt{\frac{\ell_2}{\ell_1}}$.
Squaring both sides,we get $\frac{9}{16} = \frac{\ell_2}{\ell_1}$.
Therefore,the ratio of their lengths is $\frac{\ell_1}{\ell_2} = \frac{16}{9}$.

(B) The standard equation for linear $S.H.M.$ is given by $x = A \sin(\omega t + \phi)$.
Comparing the given equation $x = 0.25 \sin(11t + 0.5)$ with the standard equation,we identify the angular frequency $\omega = 11 \ rad/s$.
The time period $T$ of $S.H.M.$ is related to angular frequency by the formula $T = \frac{2\pi}{\omega}$.
Substituting the given values $\pi = \frac{22}{7}$ and $\omega = 11$:
$T = \frac{2 \times (22/7)}{11} = \frac{2 \times 22}{11 \times 7} = \frac{44}{77} = \frac{4}{7} \ s$.
Thus,the period of $S.H.M.$ is $\frac{4}{7} \ s$.

(B) At equilibrium,the gravitational force is balanced by the spring force: $mg = kx$.
From this,the spring constant is $k = \frac{mg}{x}$.
The angular frequency of a mass-spring system is given by $\omega = \sqrt{\frac{k}{m}}$.
The time period $T$ is given by $T = \frac{2\pi}{\omega} = 2\pi\sqrt{\frac{m}{k}}$.
Substituting the value of $k$ from the equilibrium condition: $T = 2\pi\sqrt{\frac{m}{(mg/x)}} = 2\pi\sqrt{\frac{mx}{mg}}$.
Simplifying the expression,we get $T = 2\pi\sqrt{\frac{x}{g}}$.

(D) The standard equation of a simple harmonic wave is given by $y = A \sin(\omega t - kx)$.
Comparing the given equation $y = 5 \sin \frac{\pi}{2}(100t - x)$ with the standard form:
$y = 5 \sin(50\pi t - \frac{\pi}{2}x)$.
Here,the angular frequency $\omega = 50\pi \ rad/s$.
We know that the time period $T$ is related to angular frequency by the formula $T = \frac{2\pi}{\omega}$.
Substituting the value of $\omega$:
$T = \frac{2\pi}{50\pi} = \frac{1}{25} \ s$.
$T = 0.04 \ s$.
Therefore,the correct option is $D$.

(B) The time period of a simple pendulum in a stationary lift is given by $T = 2\pi \sqrt{\frac{\ell}{g}} = \sqrt{3} \ s$.
When the lift moves upward with an acceleration $a = \frac{g}{3}$,the effective acceleration due to gravity becomes $g_{eff} = g + a = g + \frac{g}{3} = \frac{4g}{3}$.
The new time period $T^{\prime}$ is given by $T^{\prime} = 2\pi \sqrt{\frac{\ell}{g_{eff}}} = 2\pi \sqrt{\frac{\ell}{4g/3}}$.
Dividing $T^{\prime}$ by $T$,we get $\frac{T^{\prime}}{T} = \sqrt{\frac{g}{4g/3}} = \sqrt{\frac{3}{4}} = \frac{\sqrt{3}}{2}$.
Substituting the value of $T = \sqrt{3} \ s$,we get $T^{\prime} = \sqrt{3} \times \frac{\sqrt{3}}{2} = \frac{3}{2} = 1.5 \ s$.

(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,$L_2 = \frac{L}{4}$.

(C) The angular momentum $L$ of a particle in circular motion is given by $L = mvr$,where $m$ is the mass of the earth,$v$ is its orbital velocity,and $r$ is the radius of the orbit $R$.
For a planet revolving around the sun,the centripetal force is provided by the gravitational force:
$\frac{mv^2}{R} = \frac{GMm}{R^2}$
Solving for $v$,we get $v = \sqrt{\frac{GM}{R}}$.
Substituting this into the expression for angular momentum:
$L = m \times \sqrt{\frac{GM}{R}} \times R$
$L = m \sqrt{GM} \times \frac{R}{\sqrt{R}}$
$L = m \sqrt{GM} \times \sqrt{R}$
Since $m$,$G$,and $M$ are constants,we find that $L \propto \sqrt{R}$.

(C) The kinetic energy of the satellite is given by $E = \frac{1}{2}mv^{2}$.
From this,we can express the velocity $v$ as $v = \sqrt{\frac{2E}{m}}$.
The angular momentum $L$ of a particle moving in a circular orbit of radius $r$ is defined as $L = mvr$.
Substituting the expression for $v$ into the formula for $L$:
$L = m \cdot \sqrt{\frac{2E}{m}} \cdot r$
$L = \sqrt{m^{2} \cdot \frac{2E}{m} \cdot r^{2}}$
$L = \sqrt{2mEr^{2}}$
$L = (2mEr^{2})^{\frac{1}{2}}$.

(A) The relationship between torque $(\tau)$ and the change in angular momentum $(\Delta L)$ is given by the angular impulse-momentum theorem: $\Delta L = \int \tau dt$.
Since the torque is constant, the change in angular momentum is simply the product of torque and time: $\Delta L = \tau \times \Delta t$.
Given:
Torque $(\tau)$ = $200 \, N-m$
Time $(\Delta t)$ = $4 \, s$
Calculation:
$\Delta L = 200 \, N-m \times 4 \, s = 800 \, kg-m^{2}/s$.
Therefore, the change in angular momentum is $800 \, kg-m^{2}/s$.

(D) The rotational kinetic energy is given by $K = \frac{1}{2} I \omega^2$ and angular momentum is $L = I \omega$.
Given that the angular frequency is doubled,$\omega' = 2\omega$.
The new rotational kinetic energy is $K' = \frac{K}{2}$.
Substituting the expressions: $\frac{1}{2} I' \omega'^2 = \frac{1}{2} (\frac{1}{2} I \omega^2)$.
$\frac{1}{2} I' (2\omega)^2 = \frac{1}{4} I \omega^2$.
$2 I' \omega^2 = \frac{1}{4} I \omega^2 \implies I' = \frac{I}{8}$.
The new angular momentum $L' = I' \omega' = (\frac{I}{8}) (2\omega) = \frac{I \omega}{4} = \frac{L}{4}$.

(B) The orbital velocity $v$ of a satellite revolving around the Earth at a distance $r$ is given by $v = \sqrt{\frac{GM}{r}}$.
Angular momentum $L$ is defined as $L = mvr$.
Substituting the value of $v$ into the equation for $L$:
$L = m \left(\sqrt{\frac{GM}{r}}\right) r$
$L = m \sqrt{GM} \cdot \sqrt{r}$
$L = m \sqrt{GMr}$
$L = m(GMr)^{1/2}$.

(A) The moment of inertia of the remaining part of the disc $(I_r)$ is given by the principle of superposition:
$I_r = I_{\text{disc}} - I_{\text{hole}}$
where $I_{\text{disc}}$ is the moment of inertia of the original disc about the central axis,and $I_{\text{hole}}$ is the moment of inertia of the removed circular part about the same axis.
$1$. Moment of inertia of the original disc:
$I_{\text{disc}} = \frac{1}{2} MR^2$
$2$. Properties of the removed hole:
Radius of the hole $r = \frac{R}{2}$.
Since the surface mass density $\sigma = \frac{M}{\pi R^2}$ is uniform,the mass of the hole $M_h$ is:
$M_h = \sigma \cdot \pi r^2 = \left(\frac{M}{\pi R^2}\right) \cdot \pi \left(\frac{R}{2}\right)^2 = \frac{M}{4}$.
$3$. Moment of inertia of the hole about the central axis of the original disc:
Using the parallel axis theorem,$I_{\text{hole}} = I_{\text{cm}} + M_h d^2$,where $I_{\text{cm}} = \frac{1}{2} M_h r^2$ and $d = \frac{R}{2}$ is the distance between the center of the hole and the center of the original disc.
$I_{\text{hole}} = \frac{1}{2} \left(\frac{M}{4}\right) \left(\frac{R}{2}\right)^2 + \left(\frac{M}{4}\right) \left(\frac{R}{2}\right)^2$
$I_{\text{hole}} = \frac{MR^2}{32} + \frac{MR^2}{16} = \frac{MR^2 + 2MR^2}{32} = \frac{3MR^2}{32}$.
$4$. Moment of inertia of the remaining part:
$I_r = \frac{1}{2} MR^2 - \frac{3MR^2}{32} = \frac{16MR^2 - 3MR^2}{32} = \frac{13MR^2}{32}$.

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

(A) The moment of inertia of a solid sphere of mass $M$ and radius $R$ about its diameter is given by $I = \frac{2}{5}MR^2$.
Since the volume remains constant,the volume of the large sphere equals the sum of the volumes of the $27$ small spheres: $\frac{4}{3}\pi R^3 = 27 \times \frac{4}{3}\pi r^3$,where $r$ is the radius of each small sphere.
This simplifies to $R^3 = 27r^3$,so $R = 3r$ or $r = R/3$.
The mass of each small sphere $m$ is $M/27$ because the density is uniform.
The moment of inertia of each small sphere $I'$ is $\frac{2}{5}mr^2$.
Substituting $m = M/27$ and $r = R/3$:
$I' = \frac{2}{5} \times (M/27) \times (R/3)^2 = \frac{2}{5} \times \frac{M}{27} \times \frac{R^2}{9} = \frac{2}{5}MR^2 \times \frac{1}{27 \times 9} = I \times \frac{1}{243} = \frac{I}{243}$.

(A) Let the mass of both the ring and the disc be $M$ and the radius be $R$.
For a ring,the moment of inertia about its center of mass is $I_{cm} = MR^2$. By the parallel axis theorem,the moment of inertia about a tangent in its plane is $I = I_{cm} + MR^2 = MR^2 + MR^2 = \frac{3}{2}MR^2$ is incorrect; the correct formula for a tangent in the plane is $I = I_{cm} + MR^2 = MR^2 + MR^2 = 2MR^2$ is also incorrect. Let's re-evaluate: The moment of inertia of a ring about its diameter is $\frac{1}{2}MR^2$. By the parallel axis theorem,the moment of inertia about a tangent in its plane is $I = I_{diameter} + MR^2 = \frac{1}{2}MR^2 + MR^2 = \frac{3}{2}MR^2$.
For a disc,the moment of inertia about its diameter is $I' = \frac{1}{4}MR^2$.
The ratio is $\frac{I}{I'} = \frac{\frac{3}{2}MR^2}{\frac{1}{4}MR^2} = \frac{3}{2} \times 4 = 6$.
Thus,the ratio is $6: 1$.

(A) The mass of the original disc is $M_{total} = 9M$ and its radius is $R$. The moment of inertia of the complete disc about an axis passing through its centre and perpendicular to its plane is $I_{1} = \frac{1}{2} M_{total} R^{2} = \frac{1}{2} (9M) R^{2} = \frac{9 MR^{2}}{2}$.
Since the mass is proportional to the area,the mass of the removed disc $m$ is given by $m = M_{total} \times \frac{\pi (R/3)^{2}}{\pi R^{2}} = 9M \times \frac{1}{9} = M$.
The moment of inertia of the removed disc about the same axis is calculated using the parallel axis theorem: $I^{\prime} = I_{cm} + m d^{2}$,where $I_{cm} = \frac{1}{2} m (R/3)^{2}$ and $d = \frac{2R}{3}$.
$I^{\prime} = \frac{1}{2} M (R/3)^{2} + M (2R/3)^{2} = \frac{MR^{2}}{18} + \frac{4MR^{2}}{9} = \frac{MR^{2} + 8MR^{2}}{18} = \frac{9MR^{2}}{18} = \frac{MR^{2}}{2}$.
The moment of inertia of the remaining disc is $I_{remaining} = I_{1} - I^{\prime} = \frac{9 MR^{2}}{2} - \frac{MR^{2}}{2} = \frac{8 MR^{2}}{2} = 4 MR^{2}$.

(A) The moment of inertia of a thin uniform rod of mass $M$ and length $L$ about an axis passing through its center and perpendicular to its length is $I_1 = \frac{ML^2}{12}$.
By definition,$I_1 = MK_1^2$,where $K_1$ is the radius of gyration.
So,$MK_1^2 = \frac{ML^2}{12} \implies K_1 = \frac{L}{\sqrt{12}} = \frac{L}{2\sqrt{3}}$.
The moment of inertia about an axis passing through one end and perpendicular to its length is $I_2 = \frac{ML^2}{3}$.
By definition,$I_2 = MK_2^2$,where $K_2$ is the radius of gyration.
So,$MK_2^2 = \frac{ML^2}{3} \implies K_2 = \frac{L}{\sqrt{3}}$.
The ratio of the radius of gyration in the first case to the second case is $\frac{K_1}{K_2} = \frac{L / (2\sqrt{3})}{L / \sqrt{3}} = \frac{1}{2}$.
Thus,the ratio is $1:2$.

(A) Let $M$ be the mass of the rod and $L$ be its length. The moment of inertia of the rod about an axis passing through one end and perpendicular to its length is $I = \frac{ML^2}{3}$.
When the rod is bent into a ring of radius $r$,the circumference of the ring is equal to the length of the rod: $2\pi r = L$,which implies $r = \frac{L}{2\pi}$.
The moment of inertia of a ring about its diameter is given by $I_{1} = \frac{Mr^2}{2}$.
Substituting the value of $r$: $I_{1} = \frac{M}{2} \left( \frac{L}{2\pi} \right)^2 = \frac{M}{2} \cdot \frac{L^2}{4\pi^2} = \frac{ML^2}{8\pi^2}$.
Now,calculating the ratio $\frac{I}{I_{1}}$:
$\frac{I}{I_{1}} = \frac{ML^2/3}{ML^2/8\pi^2} = \frac{ML^2}{3} \cdot \frac{8\pi^2}{ML^2} = \frac{8\pi^2}{3}$.

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

(A) The rod is bent at the middle point $O$ into two segments,each of length $l = \frac{L}{2}$ and mass $m = \frac{M}{2}$.
Each segment acts as a rod of length $l$ rotating about one of its ends.
The moment of inertia of a uniform rod of mass $m$ and length $l$ about an axis passing through one of its ends and perpendicular to its length is given by $I = \frac{ml^2}{3}$.
For each segment,$m = \frac{M}{2}$ and $l = \frac{L}{2}$.
Therefore,the moment of inertia of one segment about point $O$ is:
$I_1 = \frac{(\frac{M}{2})(\frac{L}{2})^2}{3} = \frac{(\frac{M}{2})(\frac{L^2}{4})}{3} = \frac{ML^2}{24}$.
Since the system consists of two such segments,the total moment of inertia $I$ about the axis passing through $O$ and perpendicular to the plane is:
$I = I_1 + I_1 = 2 \times \frac{ML^2}{24} = \frac{ML^2}{12}$.

(A) Let the two rings be $Ring_1$ and $Ring_2$.
For $Ring_1$,the axis passing through its center and perpendicular to its plane is the central axis. Its moment of inertia is $I_1 = MR^2$.
For $Ring_2$,the axis passing through the common center and perpendicular to $Ring_1$ lies in the plane of $Ring_2$. This axis is a diameter of $Ring_2$. Its moment of inertia is $I_2 = \frac{MR^2}{2}$.
The total moment of inertia of the system about this axis is $I = I_1 + I_2 = MR^2 + \frac{MR^2}{2} = \frac{3 MR^2}{2}$.

(A) The rod swings about a horizontal axis passing through its end. By the principle of conservation of energy,the maximum rotational kinetic energy at the lowest point is equal to the maximum gravitational potential energy gained by the centre of mass at the highest point.
$1$. The rotational kinetic energy is given by $K = \frac{1}{2} I \omega^{2}$.
$2$. The moment of inertia of a rod of mass $M$ and length $L$ about an axis passing through its end is $I = \frac{ML^{2}}{3}$.
$3$. The potential energy gained by the centre of mass is $U = Mgh$,where $h$ is the maximum height reached by the centre of mass.
$4$. Equating kinetic energy and potential energy: $Mgh = \frac{1}{2} I \omega^{2}$.
$5$. Substituting the value of $I$: $Mgh = \frac{1}{2} \left( \frac{ML^{2}}{3} \right) \omega^{2}$.
$6$. Simplifying the equation: $Mgh = \frac{ML^{2} \omega^{2}}{6}$.
$7$. Solving for $h$: $h = \frac{L^{2} \omega^{2}}{6g}$.

(C) To find the moment of inertia about an axis passing through a point at a distance $\frac{L}{3}$ from one end,we use the parallel axis theorem: $I = I_{cm} + Mh^2$.
Here,$I_{cm}$ is the moment of inertia about the center of mass,which is $\frac{ML^2}{12}$.
The distance $h$ between the center of mass (at $\frac{L}{2}$ from the end) and the given axis (at $\frac{L}{3}$ from the end) is $h = |\frac{L}{2} - \frac{L}{3}| = \frac{L}{6}$.
Substituting these values into the theorem:
$I = \frac{ML^2}{12} + M(\frac{L}{6})^2$
$I = \frac{ML^2}{12} + \frac{ML^2}{36}$
$I = \frac{3ML^2 + ML^2}{36} = \frac{4ML^2}{36} = \frac{ML^2}{9}$.

(A) The moment of inertia of a body is given by $I = Mk^2$,where $k$ is the radius of gyration.
For a ring of mass $M$ and radius $R$ about a tangential axis perpendicular to its plane,using the parallel axis theorem: $I_{\text{ring}} = I_{\text{cm}} + MR^2 = MR^2 + MR^2 = 2MR^2$.
Thus,$Mk_{\text{ring}}^2 = 2MR^2 \implies k_{\text{ring}} = \sqrt{2}R$.
For a disc of mass $M$ and radius $R$ about a tangential axis perpendicular to its plane,using the parallel axis theorem: $I_{\text{disc}} = I_{\text{cm}} + MR^2 = \frac{1}{2}MR^2 + MR^2 = \frac{3}{2}MR^2$.
Thus,$Mk_{\text{disc}}^2 = \frac{3}{2}MR^2 \implies k_{\text{disc}} = \sqrt{\frac{3}{2}}R$.
The ratio of the radii of gyration 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 rod of mass $M$ and length $L$ about an axis passing through its center and perpendicular to its length is given by $I = \frac{ML^2}{12}$.
When the rod is cut into $4$ 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 small part about an axis passing through its own center and perpendicular to its length is $I' = \frac{M'(L')^2}{12}$.
Substituting the values of $M'$ and $L'$: $I' = \frac{(\frac{M}{4}) \cdot (\frac{L}{4})^2}{12} = \frac{M \cdot \frac{L^2}{16}}{4 \cdot 12} = \frac{ML^2}{12 \cdot 64}$.
Since $I = \frac{ML^2}{12}$,we get $I' = \frac{I}{64}$.

(A) The resonance condition for a cyclotron is that the frequency of the oscillator '$v$' must be equal to the cyclotron frequency: $v = \frac{eB}{2 \pi m}$.
From this,we can express the magnetic field as $B = \frac{2 \pi m v}{e}$.
The radius of the path of the proton in the cyclotron is given by $R = \frac{mv}{eB}$,where '$v_{p}$' is the velocity of the proton.
Rearranging for velocity: $v_{p} = \frac{eBR}{m}$.
Substituting the expression for '$B$': $v_{p} = \frac{e}{m} \times \left( \frac{2 \pi m v}{e} \right) \times R = 2 \pi v R$.
The kinetic energy $(K.E.)$ of the proton is given by $K.E. = \frac{1}{2} m v_{p}^{2}$.
Substituting '$v_{p}$': $K.E. = \frac{1}{2} m (2 \pi v R)^{2} = \frac{1}{2} m (4 \pi^{2} v^{2} R^{2}) = 2 m \pi^{2} v^{2} R^{2}$.

(D) The horizontal component of the Earth's magnetic field is given by the formula $H = B_e \cos \delta$,where $B_e$ is the total magnetic field of the Earth and $\delta$ is the angle of dip.
At place $A$,the angle of dip $\delta_A = 30^{\circ}$. Therefore,the horizontal component $H_A = B_e \cos 30^{\circ}$.
At place $B$,the angle of dip $\delta_B = 45^{\circ}$. Therefore,the horizontal component $H_B = B_e \cos 45^{\circ}$.
The ratio of the horizontal component at $A$ to that at $B$ is given by:
$\frac{H_A}{H_B} = \frac{B_e \cos 30^{\circ}}{B_e \cos 45^{\circ}} = \frac{\cos 30^{\circ}}{\cos 45^{\circ}}$
Substituting the values: $\frac{H_A}{H_B} = \frac{\sqrt{3}/2}{1/\sqrt{2}} = \frac{\sqrt{3}}{2} \times \sqrt{2} = \frac{\sqrt{3}}{\sqrt{2}}$.
Thus,the ratio is $\sqrt{3}: \sqrt{2}$.

(A) The torque $\tau$ acting on a magnetic dipole in a magnetic field is given by $\tau = MB \sin \theta$.
Given,$\tau_1 = 1.732 \times 10^{-5} \text{ Nm}$ at $\theta_1 = 90^{\circ}$.
So,$\tau_1 = MB \sin 90^{\circ} = MB(1) = MB$.
Therefore,$MB = 1.732 \times 10^{-5} \text{ Nm}$.
Now,for $\theta_2 = 60^{\circ}$,the torque $\tau_2$ is:
$\tau_2 = MB \sin 60^{\circ} = (1.732 \times 10^{-5}) \times \frac{\sqrt{3}}{2}$.
Since $\sqrt{3} = 1.732$,we have:
$\tau_2 = 1.732 \times 10^{-5} \times \frac{1.732}{2} = 1.732 \times 10^{-5} \times 0.866 = 1.4999 \times 10^{-5} \approx 1.5 \times 10^{-5} \text{ Nm}$.

(B) According to the Curie-Weiss Law,for a ferromagnetic material,the magnetic susceptibility $\chi$ above the Curie temperature $T_C$ is given by the relation: $\chi = \frac{C}{T - T_C}$,where $C$ is the Curie constant and $T$ is the absolute temperature.
For $T > T_C$,the susceptibility is inversely proportional to the difference between the absolute temperature and the Curie temperature. In many simplified contexts,it is stated that the susceptibility varies inversely as the absolute temperature $(T)$ as the material behaves paramagnetically.

(B) Magnetic susceptibility $(\chi)$ is a measure of how much a material will become magnetized in an applied magnetic field.
For diamagnetic materials, the magnetic susceptibility $(\chi)$ is small and negative $(-1 \le \chi < 0)$.
For paramagnetic materials, the magnetic susceptibility $(\chi)$ is small and positive $(\chi > 0)$.
For ferromagnetic materials, the magnetic susceptibility $(\chi)$ is large and positive $(\chi \gg 0)$.
Therefore, materials with negative magnetic susceptibility are diamagnetic.

(A) diamagnetic material is weakly repelled by a magnetic field.
When a thin rod of a diamagnetic substance is placed in a uniform external magnetic field,it experiences a torque that tends to align the rod in a position of minimum potential energy.
For a diamagnetic rod,the potential energy is minimum when the rod is oriented perpendicular to the direction of the external magnetic field.
Therefore,the rod will align itself perpendicular to the magnetic field to minimize its magnetic potential energy.

(B) The volume of the cube is $V = L^3 = (1 \mu m)^3 = (10^{-6} \ m)^3 = 10^{-18} \ m^3$.
The total magnetic moment $M_{total}$ is the product of the number of atoms and the dipole moment per atom:
$M_{total} = (8 \times 10^{10}) \times (9 \times 10^{-24} \ A \cdot m^2) = 72 \times 10^{-14} \ A \cdot m^2 = 7.2 \times 10^{-13} \ A \cdot m^2$.
The magnetisation $I$ (or $M$) is defined as the magnetic moment per unit volume:
$I = \frac{M_{total}}{V} = \frac{7.2 \times 10^{-13} \ A \cdot m^2}{10^{-18} \ m^3} = 7.2 \times 10^5 \ A/m$.

(C) The magnetic susceptibility $(\chi)$ of a substance is a measure of how easily it can be magnetized when placed in an external magnetic field.
For paramagnetic substances, the atoms or molecules possess a permanent magnetic dipole moment.
When placed in an external magnetic field, these dipoles align weakly in the direction of the field.
Consequently, paramagnetic substances exhibit a small and positive magnetic susceptibility ($0 < \chi < \epsilon$, where $\epsilon$ is a small positive value).
Therefore, the correct option is $C$.

(B) The magnetic properties of materials arise from the magnetic moments associated with the electrons in the atoms.
These magnetic moments are primarily due to two types of motions of the electrons:
$1$. The orbital motion of the electron around the nucleus,which creates an orbital magnetic moment.
$2$. The intrinsic spin motion of the electron about its own axis,which creates a spin magnetic moment.
Therefore,the net magnetic moment of an atom is the vector sum of the orbital and spin magnetic moments of all its electrons.

(A) Tungsten is a paramagnetic material. According to Curie's Law,the magnetic susceptibility $\chi$ of a paramagnetic material is inversely proportional to its absolute temperature $T$,i.e.,$\chi \propto \frac{1}{T}$.
Therefore,$\chi_1 T_1 = \chi_2 T_2$.
Given: $\chi_1 = 6.8 \times 10^{-5}$,$T_1 = 300 \ K$,$T_2 = 400 \ K$.
Substituting the values: $(6.8 \times 10^{-5}) \times 300 = \chi_2 \times 400$.
$\chi_2 = \frac{6.8 \times 10^{-5} \times 300}{400} = 6.8 \times 10^{-5} \times 0.75 = 5.1 \times 10^{-5}$.

(C) The magnetic moment of a bar magnet is given by $M = m \times 2l$,where $m$ is the pole strength and $2l$ is the length of the magnet.
When the magnet is cut into two equal parts perpendicular to its length,the length of each part becomes $l$ and the pole strength remains $m$.
Thus,the magnetic moment of each part is $M' = m \times l = \frac{M}{2}$.
Let the original magnet have poles $A$ (say North) and $B$ (South). After cutting,the first part has poles $A$ and $C_1$,and the second part has poles $C_2$ and $B$.
When the second part is placed over the first such that $C_2$ is above $C_1$,the poles $C_1$ and $C_2$ will have opposite polarities. If $C_1$ is South,then $C_2$ will be North.
In this configuration,the magnetic moments of the two parts are in opposite directions.
The net magnetic moment of the combination is $M_{net} = M' - M' = 0$.

(A) Magnetic field lines are continuous closed loops. Outside a bar magnet,the magnetic field lines emerge from the $N$-pole and enter the $S$-pole. Inside the bar magnet,to complete the closed loop,the magnetic field lines travel from the $S$-pole to the $N$-pole. Therefore,the correct option is $A$.

(A) Magnetization $(M)$ is defined as the net magnetic dipole moment $(m_{net})$ per unit volume $(V)$ of the material.
Mathematically,it is expressed as: $M = \frac{m_{net}}{V}$.
It represents the degree to which a substance is magnetized when placed in a magnetic field.

(D) The volume of the magnet $V = \text{length} \times \text{area} = 5 \ cm \times 4 \ cm^{2} = 20 \ cm^{3}$.
Converting to $SI$ units: $V = 20 \times 10^{-6} \ m^{3} = 2 \times 10^{-5} \ m^{3}$.
The magnetization $M$ is defined as the magnetic moment per unit volume: $M = \frac{m}{V} = \frac{2 \ Am^{2}}{2 \times 10^{-5} \ m^{3}} = 10^{5} \ A/m$.
The relation between magnetization $M$,magnetic intensity $H$,and magnetic susceptibility $\chi$ is $M = \chi H$.
Therefore,$H = \frac{M}{\chi} = \frac{10^{5}}{5 \times 10^{-6}} = 0.2 \times 10^{11} \ A/m = 2 \times 10^{10} \ A/m$.

(C) The relationship between relative permeability $\mu_r$ and magnetic susceptibility $\chi$ is given by the formula: $\mu_r = 1 + \chi$.
Given that the relative permeability $\mu_r = 5500$.
Therefore,the magnetic susceptibility $\chi = \mu_r - 1$.
Substituting the value: $\chi = 5500 - 1 = 5499$.
Thus,the magnetic susceptibility of iron is $5499$.

(D) The volume $V$ of the sample is given by the ratio of mass to density: $V = \frac{\text{mass}}{\text{density}} = \frac{2 \text{ g}}{4 \text{ g/cm}^3} = 0.5 \text{ cm}^3$.
Converting the volume to $SI$ units: $V = 0.5 \times 10^{-6} \text{ m}^3$.
Magnetization $M$ is defined as the magnetic moment $m$ per unit volume $V$: $M = \frac{m}{V}$.
Substituting the given values: $M = \frac{8 \times 10^{-7} \text{ A} \cdot \text{m}^2}{0.5 \times 10^{-6} \text{ m}^3} = 1.6 \text{ A/m}$.

(D) The magnetic field of a short bar magnet at an axial point at distance '$r$' is given by $B_{axial} = \frac{\mu_0}{4\pi} \frac{2M}{r^3}$.
At the equatorial point at the same distance '$r$',the magnetic field is given by $B_{equatorial} = \frac{\mu_0}{4\pi} \frac{M}{r^3}$.
As the point moves from the axial position to the equatorial position along a circular path of radius '$r$',the angle $\theta$ with the magnetic moment vector changes from $0^\circ$ to $90^\circ$.
The general formula for the magnetic field at any point $(r, \theta)$ is $B = \frac{\mu_0}{4\pi} \frac{M}{r^3} \sqrt{1 + 3\cos^2\theta}$.
As $\theta$ increases from $0^\circ$ to $90^\circ$,$\cos^2\theta$ decreases from $1$ to $0$,and therefore the magnitude of the magnetic field '$B$' will go on decreasing.

(B) For a short bar magnet of magnetic moment $M$,the magnetic field at a distance $r$ on the axial (longitudinal) line is given by $B_{axial} = \frac{\mu_{0}}{4\pi} \frac{2M}{r^{3}}$.
For the same distance $r$ on the equatorial (transverse) line,the magnetic field is given by $B_{equatorial} = \frac{\mu_{0}}{4\pi} \frac{M}{r^{3}}$.
Taking the ratio of the magnetic field in the longitudinal position to the transverse position:
$\frac{B_{axial}}{B_{equatorial}} = \frac{\frac{\mu_{0}}{4\pi} \frac{2M}{r^{3}}}{\frac{\mu_{0}}{4\pi} \frac{M}{r^{3}}} = \frac{2}{1}$.
Therefore,the ratio is $2:1$.

(A) The total magnetic field $(B)$ inside a material is given by the sum of the magnetic field due to external current $(B_{0} = \mu_{0}H)$ and the magnetic field due to magnetization $(B_{m} = \mu_{0}M)$.
Thus,$B = \mu_{0}(H + M)$.
We know that magnetization $(M)$ is related to magnetic intensity $(H)$ by the susceptibility $(\chi)$ as $M = \chi H$.
Substituting this into the equation for $B$:
$B = \mu_{0}(H + \chi H) = \mu_{0}H(1 + \chi)$.
Rearranging the terms to find the ratio $\frac{B}{H}$:
$\frac{B}{H} = \mu_{0}(1 + \chi)$.

(C) The magnetic permeability $\mu$ is defined as the ratio of magnetic induction $B$ to the magnetic field intensity $H$,given by $\mu = \frac{B}{H}$.
Since magnetic flux $\phi = B \cdot A$,we have $B = \frac{\phi}{A}$.
Substituting this into the permeability formula: $\mu = \frac{\phi}{A \cdot H}$.
Given values:
$\phi = 6 \times 10^{-4} \text{ Wb}$
$A = 3 \text{ cm}^2 = 3 \times 10^{-4} \text{ m}^2$
$H = 2000 \text{ A/m} = 2 \times 10^3 \text{ A/m}$
Calculating $\mu$:
$\mu = \frac{6 \times 10^{-4}}{(3 \times 10^{-4}) \times (2 \times 10^3)} = \frac{6 \times 10^{-4}}{6 \times 10^{-1}} = 10^{-3} \text{ Wb/(A} \cdot \text{m)}$.

(C) According to Curie's Law,the magnetic susceptibility $\chi$ of a paramagnetic material is inversely proportional to its absolute temperature $T$,i.e.,$\chi \propto \frac{1}{T}$.
Therefore,$\frac{\chi_1}{\chi_2} = \frac{T_2}{T_1}$.
Given:
$T_1 = -73^{\circ} C = 273 - 73 = 200 \ K$
$T_2 = -173^{\circ} C = 273 - 173 = 100 \ K$
$\chi_1 = 0.0075$
Substituting the values:
$\chi_2 = \chi_1 \times \frac{T_1}{T_2} = 0.0075 \times \frac{200}{100} = 0.0075 \times 2 = 0.0150$.
Thus,the correct option is $C$.

(B) Gold is a diamagnetic substance.
Diamagnetic substances are weakly repelled by a magnetic field.
Therefore,when a sphere of gold is brought towards a powerful magnet,it experiences a weak repulsive force.
Thus,the correct option is $B$.

(D) The torque (couple) acting on a bar magnet in a uniform magnetic field is given by $T = MB \sin \theta$,where $M$ is the magnetic moment,$B$ is the magnetic field,and $\theta$ is the angle between the magnet and the field.
Initially,the magnet is perpendicular to the field,so $\theta = \frac{\pi}{2} = 90^{\circ}$.
The initial torque is $T = MB \sin(90^{\circ}) = MB$.
We want the new torque $T'$ to be half of the initial torque,so $T' = \frac{T}{2} = \frac{MB}{2}$.
Let the new angle be $\theta'$. Then $T' = MB \sin \theta' = \frac{MB}{2}$.
Dividing both sides by $MB$,we get $\sin \theta' = \frac{1}{2} = 0.5$.
Therefore,the angle $\theta'$ is $\sin^{-1}(0.5) = 30^{\circ}$.

(D) The intensity of magnetisation $(I)$ is related to the magnetic field intensity $(H)$ by the relation $I = \chi H$,where $\chi$ is the magnetic susceptibility.
For diamagnetic materials,$\chi$ is small and negative,so $I$ is negative for positive $H$. This corresponds to line $OX$.
For paramagnetic materials,$\chi$ is small and positive,so $I$ is positive and small for a given $H$. This corresponds to line $OZ$.
For ferromagnetic materials,$\chi$ is large and positive,so $I$ is positive and large for a given $H$. This corresponds to line $OY$.
Therefore,$X$ is diamagnetic,$Y$ is ferromagnetic,and $Z$ is paramagnetic.

(B) The magnetic susceptibility $\chi$ of a material defines its magnetic behavior.
For diamagnetic materials, $\chi$ is negative and small $(-1 \le \chi < 0)$.
For paramagnetic materials, $\chi$ is positive and small $(\chi > 0)$.
For ferromagnetic materials, $\chi$ is positive and very large $(\chi \gg 1)$.
Since the given material has a positive and small susceptibility, it is classified as paramagnetic.

(C) The electrostatic pressure (stress) on the surface of a charged conductor is given by $P = \frac{\sigma^{2}}{2 \epsilon_{0}}$.
Here,the surface charge density $\sigma = \frac{q}{4 \pi r^{2}} = \frac{10^{-2}}{4 \pi (1)^{2}} = \frac{10^{-2}}{4 \pi} \ C/m^{2}$.
Substituting $\sigma$ into the pressure formula: $P = \frac{1}{2 \epsilon_{0}} \left( \frac{10^{-2}}{4 \pi} \right)^{2} = \frac{10^{-4}}{32 \pi^{2} \epsilon_{0}}$.
Volume strain is defined as $\text{strain} = \frac{\text{stress}}{B}$,where $B$ is the bulk modulus.
Given $B = \frac{10^{11}}{4 \pi^{2}}$.
Therefore,$\text{strain} = \frac{10^{-4}}{32 \pi^{2} \epsilon_{0}} \times \frac{4 \pi^{2}}{10^{11}} = \frac{10^{-4}}{8 \epsilon_{0} \times 10^{11}} = \frac{10^{-15}}{8 \epsilon_{0}}$.

(B) The resistance $R$ of a wire is given by the formula $R = \rho \frac{L}{A}$,where $\rho$ is the resistivity,$L$ is the length,and $A$ is the cross-sectional area.
Since the wire is reshaped,its volume $V = A \times L$ remains constant.
We can express the area $A$ in terms of diameter $D$ as $A = \pi \left(\frac{D}{2}\right)^2 = \frac{\pi D^2}{4}$.
Substituting $A$ into the resistance formula: $R = \rho \frac{L}{\pi D^2 / 4} = \frac{4 \rho L}{\pi D^2}$.
Since the volume $V = A \times L = \frac{\pi D^2 L}{4}$ is constant,we have $L = \frac{4V}{\pi D^2}$.
Substituting this into the resistance formula: $R = \frac{4 \rho}{\pi D^2} \times \left(\frac{4V}{\pi D^2}\right) = \frac{16 \rho V}{\pi^2 D^4}$.
To minimize $R$,we need to maximize $D$. Since volume is constant,increasing $D$ will result in a decrease in $L$. Therefore,we should decrease $L$ and increase $D$.

(C) The activity $A$ of a radioactive sample is given by the formula $A = \lambda N$,where $\lambda$ is the decay constant and $N$ is the number of undecayed atoms.
The decay constant $\lambda$ is related to the half-life $T$ by the relation $\lambda = \frac{\ln 2}{T}$.
For the two radioactive elements,the activities are:
$A_1 = \lambda_1 N_1 = \frac{\ln 2}{T_1} N_1$
$A_2 = \lambda_2 N_2 = \frac{\ln 2}{T_2} N_2$
The ratio of their activities is:
$\frac{A_1}{A_2} = \frac{(\frac{\ln 2}{T_1}) N_1}{(\frac{\ln 2}{T_2}) N_2}$
Simplifying the expression:
$\frac{A_1}{A_2} = \frac{N_1}{T_1} \times \frac{T_2}{N_2} = \frac{N_1 T_2}{N_2 T_1}$

(C) According to the law of radioactive decay,the rate of decay is given by the expression:
$-\frac{dN}{dt} = \lambda N$
where $\lambda$ is the decay constant and $N$ is the number of active nuclei present at time $t$.
This can be rewritten as:
$\frac{dN}{dt} = -\lambda N$
This equation represents a linear relationship between the decay rate $(R = -\frac{dN}{dt})$ and the number of active nuclei $(N)$,where the slope is negative $(-\lambda)$.
Comparing this to the equation of a straight line passing through the origin,$y = mx$,where $y = \frac{dN}{dt}$ and $x = N$,the slope $m = -\lambda$.
Therefore,the graph of $\frac{dN}{dt}$ versus $N$ is a straight line passing through the origin with a negative slope.
Looking at the provided options,graph $A$ represents a straight line passing through the origin with a negative slope.
Thus,the correct option is $A$.

(D) The activity $A$ of a radioactive substance at any time $t$ is given by $A = A_0 e^{-\lambda t}$,where $\lambda$ is the decay constant.
The decay constant $\lambda$ is related to the half-life $T$ by the relation $\lambda = \frac{\ln 2}{T}$.
The rate of change of activity is given by $\frac{dA}{dt} = \frac{d}{dt}(A_0 e^{-\lambda t}) = -\lambda A_0 e^{-\lambda t} = -\lambda A$.
The magnitude of the rate of change of activity is $|\frac{dA}{dt}| = \lambda A$.
Since $\lambda = \frac{\ln 2}{T}$,we have $|\frac{dA}{dt}| = \frac{\ln 2}{T} A$.
Thus,the rate of change of activity is proportional to $T^{-1}$. However,looking at the instantaneous rate of change of activity $\frac{dA}{dt} = -\lambda^2 N_0 e^{-\lambda t}$,where $N = N_0 e^{-\lambda t}$.
Since $A = \lambda N$,then $\frac{dA}{dt} = -\lambda A = -\lambda (\lambda N) = -\lambda^2 N$.
Substituting $\lambda = \frac{\ln 2}{T}$,we get $\frac{dA}{dt} \propto \lambda^2 \propto (\frac{1}{T})^2 = T^{-2}$.
Therefore,the rate of change of activity is proportional to $T^{-2}$.

(B) Let the initial nucleus be $^A_Z X$.
Emission of one $\alpha$ particle $(^4_2 He)$ decreases the mass number by $4$ and atomic number by $2$.
Emission of $4 \alpha$ particles results in a change: $A' = A - (4 \times 4) = A - 16$ and $Z' = Z - (4 \times 2) = Z - 8$.
Emission of one $\beta$ particle $(^0_{-1} e)$ increases the atomic number by $1$ and leaves the mass number unchanged.
Emission of $7 \beta$ particles results in: $A_{final} = A - 16$ and $Z_{final} = Z - 8 + 7 = Z - 1$.
The number of protons in the final nucleus is $P = Z_{final} = Z - 1$.
The number of neutrons is $N = A_{final} - Z_{final} = (A - 16) - (Z - 1) = A - Z - 15$.
The ratio of neutrons to protons is $\frac{N}{P} = \frac{A - Z - 15}{Z - 1}$.

(D) The radius of a charged particle moving in a uniform magnetic field is given by $r = \frac{mv}{qB} = \frac{\sqrt{2mK}}{qB}$,where $K$ is the kinetic energy.
Since the path $(r)$ and the magnetic field $(B)$ are the same for both particles,we have $\frac{\sqrt{2m_{\alpha}K_{\alpha}}}{q_{\alpha}B} = \frac{\sqrt{2m_{p}K_{p}}}{q_{p}B}$.
Squaring both sides and rearranging,we get $\frac{2m_{\alpha}K_{\alpha}}{q_{\alpha}^2} = \frac{2m_{p}K_{p}}{q_{p}^2}$,which implies $K_{p} = K_{\alpha} \cdot \left(\frac{m_{\alpha}}{m_{p}}\right) \cdot \left(\frac{q_{p}}{q_{\alpha}}\right)^2$.
Given $m_{\alpha} = 4m_{p}$ and $q_{\alpha} = 2q_{p}$,we substitute these values:
$K_{p} = 10 \ eV \cdot \left(\frac{4m_{p}}{m_{p}}\right) \cdot \left(\frac{q_{p}}{2q_{p}}\right)^2 = 10 \ eV \cdot 4 \cdot \frac{1}{4} = 10 \ eV$.

(A) For a biconvex lens,the lens maker's formula is given by: $\frac{1}{f} = (n-1) \left( \frac{1}{R_{1}} - \frac{1}{R_{2}} \right)$.
Given $n = 1.5$,$R_{1} = 20 \ cm$,and $R_{2} = -20 \ cm$.
Substituting these values: $\frac{1}{f} = (1.5 - 1) \left( \frac{1}{20} - \frac{1}{-20} \right) = 0.5 \left( \frac{1}{20} + \frac{1}{20} \right) = 0.5 \left( \frac{2}{20} \right) = 0.5 \times 0.1 = 0.05$.
Thus,$f = \frac{1}{0.05} = 20 \ cm$.
The focal length of the concave mirror is equal to the focal length of the lens,so $f_{mirror} = 20 \ cm$.
For a concave mirror,the focal length is negative,so $f = -20 \ cm$.
The radius of curvature $R$ is given by $R = 2f = 2(-20 \ cm) = -40 \ cm$.

(A) According to the Lens Maker's Formula,the focal length $f$ of a lens is given by $\frac{1}{f} = (\mu - 1) \left( \frac{1}{R_1} - \frac{1}{R_2} \right)$.
For an equiconvex lens,$R_1 = R$ and $R_2 = -R$,so $\frac{1}{f} = (\mu - 1) \left( \frac{1}{R} + \frac{1}{R} \right) = \frac{2(\mu - 1)}{R}$.
When the lens is cut along the axis $AB$ (perpendicular to the principal axis),the radius of curvature of the surfaces remains the same for each half.
For each half,the lens becomes a plano-convex lens where one surface has radius $R$ and the other is flat (infinite radius,$R' = \infty$).
Applying the formula for one half: $\frac{1}{f'} = (\mu - 1) \left( \frac{1}{R} - \frac{1}{\infty} \right) = \frac{\mu - 1}{R}$.
Comparing this with the original focal length,we see that $\frac{1}{f'} = \frac{1}{2} \times \frac{2(\mu - 1)}{R} = \frac{1}{2} \times \frac{1}{f}$.
Therefore,$f' = 2f$.

(A) The critical angle $c$ for total internal reflection $(TIR)$ when light travels from a medium of refractive index $n_1$ to a medium of lower refractive index $n_2$ is given by $\sin c = \frac{n_2}{n_1}$,where $n_1 > n_2$.
The smaller the ratio $\frac{n_2}{n_1}$,the smaller $\sin c$,and hence the smaller the critical angle $c$.
From the options,the valid scenarios for $TIR$ are only those in which the first medium has a higher index than the second.
For glass to air $(n_{\text{glass}} \approx 1.5, n_{\text{air}} \approx 1.0)$: $\sin c = \frac{1.0}{1.5} = \frac{2}{3} \approx 0.667$,which gives $c \approx 41.8^{\circ}$.
For glass to water $(n_{\text{glass}} \approx 1.5, n_{\text{water}} \approx 1.33)$: $\sin c = \frac{1.33}{1.5} \approx 0.887$,which gives $c \approx 62.5^{\circ}$.
Clearly,the critical angle is smaller for glass-to-air.
Hence,the critical angle is minimum when light travels from glass to air.

(D) The refractive index of glass is $\mu_g = \frac{3}{2}$ and the refractive index of water is $\mu_w = \frac{4}{3}$.
When light travels from a denser medium (glass) to a rarer medium (water),the critical angle $C$ is given by the formula $\sin C = \frac{\mu_w}{\mu_g}$.
Substituting the given values:
$\sin C = \frac{4/3}{3/2} = \frac{4}{3} \times \frac{2}{3} = \frac{8}{9}$.
Therefore,the critical angle $C = \sin^{-1}\left(\frac{8}{9}\right)$.

(A) The refractive index of a medium is inversely proportional to the speed of light in that medium,given by $\mu = \frac{c}{v}$.
Given speeds are $v_{A} = 1.8 \times 10^{8} \ m/s$ and $v_{B} = 2.7 \times 10^{8} \ m/s$.
The relative refractive index of medium $A$ with respect to medium $B$ is ${}_{B}\mu_{A} = \frac{\mu_{A}}{\mu_{B}} = \frac{v_{B}}{v_{A}}$.
Substituting the values: ${}_{B}\mu_{A} = \frac{2.7 \times 10^{8}}{1.8 \times 10^{8}} = \frac{27}{18} = \frac{3}{2}$.
The formula for the critical angle $C$ is $\sin C = \frac{1}{{}_{B}\mu_{A}}$.
Therefore,$\sin C = \frac{1}{3/2} = \frac{2}{3}$.
Thus,the critical angle is $C = \sin^{-1}\left(\frac{2}{3}\right)$.

(A) The image is real and hence inverted.
Therefore,the magnification $m = \frac{v}{u} = -n$,which implies $u = -\frac{v}{n}$.
Using the lens formula,$\frac{1}{v} - \frac{1}{u} = \frac{1}{f}$.
Substituting the value of $u$,we get $\frac{1}{v} - (-\frac{n}{v}) = \frac{1}{f}$.
This simplifies to $\frac{1+n}{v} = \frac{1}{f}$.
Therefore,the image distance $v = f(n+1)$.

(D) For a spherical mirror,the sign convention states that the direction of incident light is taken as positive.
When an image is formed on the same side as the object,the image distance $v$ is negative,which corresponds to a real image.
When an image is formed on the opposite side of the object (behind the mirror),the image distance $v$ is positive,which corresponds to a virtual image.
Therefore,images formed on the side of the object are real,and images formed on the opposite side are virtual.

(C) The magnifying power $(m)$ of an astronomical telescope in normal adjustment is given by the formula: $m = \frac{f_o}{f_e}$,where $f_o$ is the focal length of the objective lens and $f_e$ is the focal length of the eyepiece.
If the focal length of the eyepiece is doubled,the new focal length becomes $f_e' = 2f_e$.
The new magnifying power $(m')$ will be: $m' = \frac{f_o}{f_e'} = \frac{f_o}{2f_e} = \frac{1}{2} \left( \frac{f_o}{f_e} \right) = \frac{m}{2}$.
Therefore,the magnifying power becomes half of the original value.

(C) The magnifying power $(M)$ of a simple microscope when the image is formed at the near point (distance of distinct vision,$D = 25 \ cm$) is given by the formula:
$M = 1 + \frac{D}{f}$
Given,focal length $f = 5 \ cm$ and $D = 25 \ cm$.
Substituting the values:
$M = 1 + \frac{25}{5}$
$M = 1 + 5 = 6$
Therefore,the magnifying power is $6$.

(B) The magnifying power $(M)$ of a simple microscope is given by the formula: $M = 1 + \frac{D}{f}$, where $D$ is the least distance of distinct vision and $f$ is the focal length of the lens.
According to Cauchy's formula, the refractive index $(\mu)$ of a material depends on the wavelength $(\lambda)$ as $\mu \propto \frac{1}{\lambda^2}$. Since the wavelength of red light $(\lambda_r)$ is greater than the wavelength of blue light $(\lambda_b)$, the refractive index for red light is less than that for blue light $(\mu_r < \mu_b)$.
From the lens maker's formula, $\frac{1}{f} = (\mu - 1) \left( \frac{1}{R_1} - \frac{1}{R_2} \right)$, we see that $f \propto \frac{1}{\mu - 1}$. Since $\mu_r < \mu_b$, it follows that $f_r > f_b$.
As the magnifying power $M = 1 + \frac{D}{f}$, an increase in focal length $(f)$ leads to a decrease in the magnifying power. Therefore, when changing from blue light to red light, the magnifying power decreases.

(C) The numerical aperture $(NA)$ of a microscope objective is defined by the formula $NA = \mu \sin \alpha$,where $\mu$ is the refractive index of the medium between the object and the objective lens,and $\alpha$ is the semi-vertical angle of the cone of light entering the objective lens from the object. Therefore,the correct representation is $\mu \sin \alpha$.

(A) The magnifying power $(M)$ of an astronomical telescope is given by the formula $M = \frac{f_o}{f_e}$,where $f_o$ is the focal length of the objective lens and $f_e$ is the focal length of the eyepiece lens.
To obtain a high magnifying power,the numerator $(f_o)$ must be large and the denominator $(f_e)$ must be small.
Therefore,the objective should have a large focal length and the eyepiece should have a small focal length.

(C) For a telescope,the length $L$ is given by $L = f_{o} + f_{e} = 10 \ cm$.
The magnifying power $M$ is given by $M = \frac{f_{o}}{f_{e}} = 4$.
From the second equation,$f_{o} = 4f_{e}$.
Substituting this into the first equation: $4f_{e} + f_{e} = 10 \ cm$,which gives $5f_{e} = 10 \ cm$,so $f_{e} = 2 \ cm$.
Then,$f_{o} = 4 \times 2 \ cm = 8 \ cm$.
Thus,the objective lens has a focal length of $8 \ cm$ $(L_{4})$ and the eye lens has a focal length of $2 \ cm$ $(L_{1})$.
Therefore,the objective and eye lenses are $L_{4}$ and $L_{1}$ respectively.

(D) The resolving power $(RP)$ of an optical instrument is inversely proportional to the wavelength $(\lambda)$ of the light used, i.e., $RP \propto \frac{1}{\lambda}$.
Given: $\lambda_A = 4500 \, \text{Å}$ and $\lambda_B = 6000 \, \text{Å}$.
The ratio of the resolving power of $A$ to $B$ is given by:
$\frac{RP_A}{RP_B} = \frac{\lambda_B}{\lambda_A} = \frac{6000 \, \text{Å}}{4500 \, \text{Å}} = \frac{60}{45} = \frac{4}{3}$.
Therefore, the ratio is $4:3$.

(D) For an astronomical telescope in normal adjustment,the length of the tube $L$ is given by the sum of the focal lengths of the objective lens $(f_o)$ and the eyepiece $(f_e)$:
$L = f_o + f_e$.
Given $L = 50 \ cm$,we have $f_o + f_e = 50 \ cm$.
For an astronomical telescope,both the objective and the eyepiece are convex lenses,meaning their focal lengths are positive.
Additionally,for high magnification,the objective focal length $f_o$ is typically much larger than the eyepiece focal length $f_e$.
Comparing the options,$f_o = 45 \ cm$ and $f_e = 5 \ cm$ satisfies $f_o + f_e = 50 \ cm$ and both are positive.

(A) The resolving power $(RP)$ of an astronomical telescope is given by the formula: $RP = \frac{a}{1.22 \lambda}$,where $a$ is the aperture (diameter) of the objective lens and $\lambda$ is the wavelength of light.
$1$. The wavelength of light $(\lambda)$ depends on the source and is independent of the aperture of the objective lens. Thus,it is not affected.
$2$. The focal length $(f)$ of the objective lens is a property determined by its curvature and refractive index,which does not change by simply increasing the aperture (diameter). Thus,it is not affected.
$3$. From the formula $RP \propto a$,we see that the resolving power is directly proportional to the aperture $a$. Therefore,increasing the aperture increases the resolving power.
Hence,the correct sequence is: is not affected,is not affected,increases.

(A) simple microscope consists of a convex lens of short focal length $(f)$.
To obtain a magnified,virtual,and erect image,the object must be placed between the optical centre $(O)$ and the principal focus $(F)$ of the convex lens.
When the image is formed at the distance of distinct vision $(D)$,the magnification is given by $m = 1 + D/f$.
Therefore,the correct position for the object is between the principal focus and the optical centre.

(D) For a thin prism,the angle of deviation is given by $\delta = A(n - 1)$.
For dispersion without deviation,the net deviation must be zero,which means the deviation produced by the first prism must be equal to the deviation produced by the second prism in the opposite direction.
Therefore,$\delta_{1} = \delta_{2}$.
Substituting the formula,we get $A_{1}(n_{1} - 1) = A_{2}(n_{2} - 1)$.
Given $A_{1} = 4^{\circ}$,$n_{1} = 1.54$,and $n_{2} = 1.72$.
$4(1.54 - 1) = A_{2}(1.72 - 1)$.
$4 \times 0.54 = A_{2} \times 0.72$.
$A_{2} = \frac{4 \times 0.54}{0.72} = \frac{2.16}{0.72} = 3^{\circ}$.