MHT CET 2020 Physics Question Paper with Answer and Solution

(A) Let the vertices of the equilateral triangle be $A$,$B$,and $C$. Let the axis pass through vertex $A$ and be parallel to the side $BC$.
The mass at vertex $A$ lies on the axis,so its perpendicular distance from the axis is $0$. Its contribution to the moment of inertia is $m(0)^2 = 0$.
The masses at vertices $B$ and $C$ are at a perpendicular distance $h$ from the axis,where $h$ is the altitude of the equilateral triangle.
The altitude $h$ is given by $h = \ell \sin 60^{\circ} = \ell \times \frac{\sqrt{3}}{2}$.
The moment of inertia $I$ of the system is the sum of the moments of inertia of the individual masses:
$I = m(0)^2 + m(h)^2 + m(h)^2 = 2mh^2$.
Substituting the value of $h$:
$I = 2m \left( \ell \times \frac{\sqrt{3}}{2} \right)^2 = 2m \left( \frac{3}{4} \ell^2 \right) = \frac{3}{2} m \ell^2$.

(B) Let $m$ be the mass per unit length of the wire.
The mass of loop $A$ is $M_A = (2 \pi R)m$.
The mass of loop $B$ is $M_B = (2 \pi NR)m$.
Thus,the ratio of masses is $\frac{M_B}{M_A} = \frac{2 \pi NRm}{2 \pi Rm} = N$.
The moment of inertia of a circular loop about its central axis is $I = MR^2$.
Therefore,$I_A = M_A R^2$ and $I_B = M_B (NR)^2 = M_B N^2 R^2$.
Substituting $M_B = N M_A$,we get $I_B = (N M_A) N^2 R^2 = N^3 M_A R^2 = N^3 I_A$.
Given that $I_B = 3 I_A$,we have $N^3 = 3$.
Therefore,$N = (3)^{1/3}$.

(D) The moment of inertia of a ring of mass $m$ and radius $r$ about an axis passing through its centre and perpendicular to its plane is $I = mr^2$.
Since both rings are made from the same wire,the mass $m$ is proportional to the circumference $(2 \pi r)$,so $m \propto r$.
Therefore,$m_A = k(2 \pi nR)$ and $m_B = k(2 \pi R)$,where $k$ is the mass per unit length.
Thus,$\frac{m_A}{m_B} = \frac{nR}{R} = n$.
The moment of inertia of ring $A$ is $I_A = m_A (nR)^2 = (n m_B) (n^2 R^2) = n^3 (m_B R^2)$.
Since $I_B = m_B R^2$,we have $I_A = n^3 I_B$.
Given $I_A = 64 I_B$,we get $n^3 = 64$.
Therefore,$n = \sqrt[3]{64} = 4$.

(C) Let the vertices of the equilateral triangle be $A, B,$ and $C$. Let the axis of rotation be along the side $BC$.
The masses at $B$ and $C$ lie on the axis of rotation,so their perpendicular distance from the axis is $0$. Thus,their moment of inertia is $I_{B} = m(0)^{2} = 0$ and $I_{C} = m(0)^{2} = 0$.
The mass at $A$ is at a perpendicular distance $h$ from the side $BC$. In an equilateral triangle of side $\ell$,the height $h$ is given by $h = \ell \sin(60^{\circ}) = \frac{\sqrt{3}}{2} \ell$.
The moment of inertia of the mass at $A$ about the axis $BC$ is $I_{A} = m h^{2} = m \left( \frac{\sqrt{3}}{2} \ell \right)^{2} = m \left( \frac{3}{4} \ell^{2} \right) = \frac{3}{4} m \ell^{2}$.
The total moment of inertia of the system is $I = I_{A} + I_{B} + I_{C} = \frac{3}{4} m \ell^{2} + 0 + 0 = \frac{3}{4} m \ell^{2}$.

(A) The moment of inertia of a disc about an axis passing through its centre and perpendicular to its plane is given by $I = \frac{1}{2} M R^2$.
Since both discs have the same mass $M$,we have $I_1 = \frac{1}{2} M R_1^2$ and $I_2 = \frac{1}{2} M R_2^2$.
Therefore,the ratio is $\frac{I_1}{I_2} = \frac{R_1^2}{R_2^2}$.
Since both discs are made of the same material,they have the same density $\rho$. The mass is given by $M = \text{Volume} \times \rho = (\pi R^2 t) \rho$.
Since $M$ and $\rho$ are the same for both,we have $\pi R_1^2 t_1 \rho = \pi R_2^2 t_2 \rho$,which simplifies to $R_1^2 t_1 = R_2^2 t_2$.
From this,we get $\frac{R_1^2}{R_2^2} = \frac{t_2}{t_1}$.
Substituting this into the ratio of moments of inertia,we get $\frac{I_1}{I_2} = \frac{t_2}{t_1}$.
Cross-multiplying gives $I_1 t_1 = I_2 t_2$.

(D) Given,the moment of inertia about the center $O$ is $I_{o} = \frac{Ma^{2}}{6}$.
To find the moment of inertia about an axis passing through a corner (say $A$),we use the parallel axis theorem: $I_{A} = I_{o} + Mh^{2}$,where $h$ is the distance between the center and the corner.
The diagonal of the square is $d = \sqrt{a^{2} + a^{2}} = a\sqrt{2}$.
The distance $h$ from the center to the corner is half the diagonal: $h = \frac{a\sqrt{2}}{2} = \frac{a}{\sqrt{2}}$.
Substituting the values into the theorem:
$I_{A} = \frac{Ma^{2}}{6} + M\left(\frac{a}{\sqrt{2}}\right)^{2}$
$I_{A} = \frac{Ma^{2}}{6} + \frac{Ma^{2}}{2}$
$I_{A} = \frac{Ma^{2} + 3Ma^{2}}{6} = \frac{4Ma^{2}}{6} = \frac{2Ma^{2}}{3}$.

(B) The moment of inertia of a solid sphere about its diameter is given by $I = \frac{2}{5} MR^{2}$.
Initially,the moment of inertia is $I_{1} = \frac{2}{5} MR^{2}$.
After the planet shrinks to half its size,the new radius is $R' = \frac{R}{2}$ and the new mass is $M' = \frac{M}{2}$.
The new moment of inertia $I_{2}$ is calculated as:
$I_{2} = \frac{2}{5} M' (R')^{2}$
$I_{2} = \frac{2}{5} \left(\frac{M}{2}\right) \left(\frac{R}{2}\right)^{2}$
$I_{2} = \frac{2}{5} \times \frac{M}{2} \times \frac{R^{2}}{4}$
$I_{2} = \frac{2}{40} MR^{2} = \frac{MR^{2}}{20}$.

(A) For a solid cylinder rolling down an inclined plane,the total energy at the top is potential energy $PE = Mgh$.
At the bottom,the total energy is the sum of translational kinetic energy and rotational kinetic energy: $KE = \frac{1}{2} Mv^2 + \frac{1}{2} I\omega^2$.
For a solid cylinder,the moment of inertia $I = \frac{1}{2} MR^2$ and the rolling condition is $v = R\omega$,so $\omega = \frac{v}{R}$.
Substituting these into the energy equation: $Mgh = \frac{1}{2} Mv^2 + \frac{1}{2} (\frac{1}{2} MR^2) (\frac{v}{R})^2$.
$Mgh = \frac{1}{2} Mv^2 + \frac{1}{4} Mv^2$.
$Mgh = \frac{3}{4} Mv^2$.
Solving for height $h$: $h = \frac{3 v^2}{4 g}$.

(A) Let the radius of the wheel be $R = 2 \ cm$.
Initially,point $P$ is at the origin $(0, 0)$.
After half a rotation,the center of the wheel moves forward by a distance equal to half the circumference,which is $\pi R$.
The new position of the center is $(\pi R, R)$.
After half a rotation,point $P$ moves from the bottom to the top of the wheel.
The new coordinates of point $P$ will be $(\pi R, 2R)$.
The displacement $d$ is the distance between the initial position $(0, 0)$ and the final position $(\pi R, 2R)$.
Using the distance formula: $d = \sqrt{(\pi R - 0)^2 + (2R - 0)^2} = \sqrt{\pi^2 R^2 + 4R^2} = R\sqrt{\pi^2 + 4}$.
Substituting $R = 2 \ cm$: $d = 2\sqrt{\pi^2 + 4} \ cm$ or $2(\pi^2 + 4)^{1/2} \ cm$.

(A) For a solid cylinder rolling down an inclined plane,the total potential energy $Mgh$ is converted into translational kinetic energy and rotational kinetic energy.
Total energy $E = Mgh = \frac{1}{2} Mv^2 + \frac{1}{2} I\omega^2$.
For a solid cylinder,the moment of inertia $I = \frac{1}{2} MR^2$ and the rolling condition is $v = R\omega$.
Substituting these into the energy equation:
$Mgh = \frac{1}{2} Mv^2 + \frac{1}{2} (\frac{1}{2} MR^2) (\frac{v}{R})^2$
$Mgh = \frac{1}{2} Mv^2 + \frac{1}{4} Mv^2 = \frac{3}{4} Mv^2$.
Thus,$Mv^2 = \frac{4}{3} Mgh$.
The rotational kinetic energy $K_{rot} = \frac{1}{2} I\omega^2 = \frac{1}{2} (\frac{1}{2} MR^2) (\frac{v}{R})^2 = \frac{1}{4} Mv^2$.
Substituting $Mv^2 = \frac{4}{3} Mgh$ into the expression for $K_{rot}$:
$K_{rot} = \frac{1}{4} (\frac{4}{3} Mgh) = \frac{Mgh}{3}$.

(A) Given that the wheel is initially at rest,the initial angular velocity $\omega_0 = 0$.
The angular acceleration $\alpha$ produced by a constant torque $\tau$ is given by $\alpha = \frac{\tau}{I}$.
After time $t$,the angular velocity $\omega$ is given by $\omega = \omega_0 + \alpha t = 0 + \left(\frac{\tau}{I}\right)t = \frac{\tau t}{I}$.
The rotational kinetic energy $K$ is given by $K = \frac{1}{2} I \omega^2$.
Substituting the value of $\omega$,we get $K = \frac{1}{2} I \left(\frac{\tau t}{I}\right)^2 = \frac{1}{2} I \cdot \frac{\tau^2 t^2}{I^2} = \frac{\tau^2 t^2}{2 I}$.

(A) The rotational kinetic energy is given by $K = \frac{1}{2} I \omega^2$.
For a solid sphere rotating about its diameter,the moment of inertia is $I_s = \frac{2}{5} MR^2$.
Let the angular speed of the sphere be $\omega_s = \omega$.
Thus,the kinetic energy of the sphere is $K_s = \frac{1}{2} (\frac{2}{5} MR^2) \omega^2 = \frac{1}{5} MR^2 \omega^2$.
For a disc rotating about an axis passing through its centre and perpendicular to its plane,the moment of inertia is $I_d = \frac{1}{2} MR^2$.
The angular speed of the disc is given as $\omega_d = 2\omega$.
Thus,the kinetic energy of the disc is $K_d = \frac{1}{2} (\frac{1}{2} MR^2) (2\omega)^2 = \frac{1}{4} MR^2 (4\omega^2) = MR^2 \omega^2$.
The ratio of the kinetic energy of the disc to that of the sphere is $\frac{K_d}{K_s} = \frac{MR^2 \omega^2}{\frac{1}{5} MR^2 \omega^2} = 5$.
Therefore,the ratio is $5: 1$.

(A) The initial moment of inertia of the ring about the transverse axis is $I = M R^2$. The initial angular momentum is $L_i = I \omega = M R^2 \omega$.
When two objects of mass $m$ are attached to the opposite ends of a diameter,the new moment of inertia becomes $I' = M R^2 + m R^2 + m R^2 = (M + 2m) R^2$.
Since no external torque acts on the system,the angular momentum is conserved,so $L_i = L_f$.
$M R^2 \omega = (M + 2m) R^2 \omega'$
$\omega' = \frac{M R^2 \omega}{(M + 2m) R^2} = \frac{M \omega}{M + 2m}$.

(B) The angular speed of the hour hand is $\omega_{h} = \frac{2 \pi}{T_{h}} = \frac{2 \pi}{12 \times 3600} \text{ rad/s}$.
The angular speed of the second hand is $\omega_{s} = \frac{2 \pi}{T_{s}} = \frac{2 \pi}{60} \text{ rad/s}$.
The relative angular speed is given by $\omega_{rel} = |\omega_{s} - \omega_{h}|$.
$\omega_{rel} = \left| \frac{2 \pi}{60} - \frac{2 \pi}{43200} \right| = 2 \pi \left( \frac{720 - 1}{43200} \right)$.
$\omega_{rel} = 2 \pi \left( \frac{719}{43200} \right) = \frac{719 \pi}{21600} \text{ rad/s}$.

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

(B) The torque $\tau$ applied by the force $F$ at a distance $r$ from the axis is given by $\tau = F \times r$.
Substituting the values, $\tau = 25 \,N \times 0.4 \,m = 10 \,Nm$.
The moment of inertia $I$ of a solid cylinder about its own axis is $I = \frac{1}{2} M R^2$.
Substituting the values, $I = \frac{1}{2} \times 1 \,kg \times (0.4 \,m)^2 = 0.5 \times 0.16 = 0.08 \,kg \cdot m^2$.
Using the relation $\tau = I \alpha$, the angular acceleration $\alpha$ is $\alpha = \frac{\tau}{I}$.
$\alpha = \frac{10 \,Nm}{0.08 \,kg \cdot m^2} = 125 \,rad/s^2$.

(A) Initial angular velocity $\omega_i = 120 \ rpm = \frac{120 \times 2\pi}{60} = 4\pi \ rad/s$.
Final angular velocity $\omega_f = 0 \ rad/s$.
Time taken $t = 10 \ s$.
Angular acceleration $\alpha = \frac{\omega_f - \omega_i}{t} = \frac{0 - 4\pi}{10} = -0.4\pi \ rad/s^2$.
Moment of inertia of the disc $I = \frac{1}{2}MR^2 = \frac{1}{2} \times 10 \times (0.1)^2 = 0.05 \ kg \ m^2$.
Retarding torque $\tau = I|\alpha| = 0.05 \times 0.4\pi = 0.02\pi \ Nm$.
Since $\tau = F \times R$, the force $F = \frac{\tau}{R} = \frac{0.02\pi}{0.1} = 0.2\pi \ N$.

(A) The relationship between torque $\tau$ and the change in angular momentum $\Delta L$ is given by the impulse-momentum theorem for rotation: $\Delta L = \int \tau \ dt$.
Since the torque $\tau = 50 \ Nm$ is constant and acts for a time interval $\Delta t = 8 \ s$,the change in angular momentum is:
$\Delta L = \tau \times \Delta t$
$\Delta L = 50 \ Nm \times 8 \ s$
$\Delta L = 400 \ kg \cdot m^2/s$.
Therefore,the change in angular momentum is $400 \ kg \cdot m^2/s$.

(A) The relationship between torque $\tau$ and the rate of change of angular momentum $L$ is given by Newton's second law for rotation: $\tau = \frac{dL}{dt}$.
Given initial angular momentum $L_1 = 1 \, J \cdot s$ and final angular momentum $L_2 = 4 \, J \cdot s$.
The change in angular momentum is $\Delta L = L_2 - L_1 = 4 - 1 = 3 \, J \cdot s$.
The time interval is $\Delta t = 4 \, s$.
Therefore, the torque is $\tau = \frac{\Delta L}{\Delta t} = \frac{3}{4} = 0.75 \, N \cdot m$.

(A) The point $P$ is given as $(1, -1)$,so its position vector is $\vec{r}_P = \hat{i} - \hat{j}$.
Since the force acts at the origin $O(0, 0)$,the position vector of the point of application relative to point $P$ is $\vec{r} = \vec{r}_O - \vec{r}_P = 0 - (\hat{i} - \hat{j}) = -\hat{i} + \hat{j}$.
The torque (moment of force) about point $P$ is given by $\vec{\tau} = \vec{r} \times \vec{F}$.
Substituting the values: $\vec{\tau} = (-\hat{i} + \hat{j}) \times (-F\hat{k})$.
Using the distributive property of cross product: $\vec{\tau} = F[(\hat{i} \times \hat{k}) - (\hat{j} \times \hat{k})]$.
Using the cross product relations $\hat{i} \times \hat{k} = -\hat{j}$ and $\hat{j} \times \hat{k} = \hat{i}$:
$\vec{\tau} = F[-\hat{j} - \hat{i}] = -F(\hat{i} + \hat{j})$.

(D) The correct option is $D$.
Concept:
According to the Stefan-Boltzmann law,the emissive power $(E)$ of a body is given by $E = e \sigma T^4$,where $e$ is the emissivity,$\sigma$ is the Stefan-Boltzmann constant,and $T$ is the absolute temperature.
Given that the emissive power of both spheres is the same,we have $E_1 = E_2$.
Therefore,$e_1 \sigma T_1^4 = e_2 \sigma T_2^4$.
This implies $\frac{T_1^4}{T_2^4} = \frac{e_2}{e_1}$.
Given the ratio of emissivity $e_1 : e_2 = 1 : 4$,we have $\frac{e_2}{e_1} = \frac{4}{1} = 4$.
Thus,$\left(\frac{T_1}{T_2}\right)^4 = 4$.
Taking the fourth root on both sides,$\frac{T_1}{T_2} = (4)^{1/4} = (2^2)^{1/4} = 2^{1/2} = \sqrt{2}$.
So,the ratio $T_1 : T_2 = \sqrt{2} : 1$.

(B) According to Wien's displacement law,$\lambda_{max} T = b$ (constant).
Let $\lambda_A$ and $\lambda_B$ be the wavelengths of maximum energy for bodies $A$ and $B$,and $T_A$ and $T_B$ be their absolute temperatures.
Given: $T_A = 3 T_B$ and $\lambda_B - \lambda_A = 4 \mu m$ (since $T_A > T_B$,$\lambda_A < \lambda_B$).
From Wien's law: $\lambda_A T_A = \lambda_B T_B$.
Substituting $T_A = 3 T_B$: $\lambda_A (3 T_B) = \lambda_B T_B$,which gives $\lambda_B = 3 \lambda_A$.
Substitute this into the difference equation: $3 \lambda_A - \lambda_A = 4 \mu m$.
$2 \lambda_A = 4 \mu m \implies \lambda_A = 2 \mu m$.
Therefore,$\lambda_B = 3 \times 2 \mu m = 6 \mu m$.

(A) black body is an ideal body that emits and absorbs all frequencies of radiation. According to Planck's law of black body radiation,the intensity of radiation emitted by a black body depends on the wavelength and temperature. The intensity distribution is not uniform for all wavelengths; it follows a specific curve (Planck's curve) where intensity varies with wavelength. Therefore,the statement that 'intensity is same for all wavelengths' is incorrect. Additionally,the intensity does not strictly increase or decrease monotonically for all ranges; it peaks at a specific wavelength determined by Wien's displacement law. However,among the given options,the statement that intensity is the same for all wavelengths is fundamentally wrong.

(B) According to Wien's displacement law,$\lambda_m T = b$ (constant),so $T \propto 1/\lambda_m$.
Given the ratio of wavelengths $\lambda_P : \lambda_Q = 3:4$,the ratio of temperatures is $T_P : T_Q = 4:3$.
According to Stefan-Boltzmann law,the power radiated by a black body is $P = \sigma A T^4 = \sigma (4\pi r^2) T^4$.
Thus,the ratio of power radiated is $\frac{P_P}{P_Q} = \left( \frac{r_P}{r_Q} \right)^2 \left( \frac{T_P}{T_Q} \right)^4$.
Substituting the given values: $\frac{P_P}{P_Q} = \left( \frac{3}{2} \right)^2 \times \left( \frac{4}{3} \right)^4$.
$\frac{P_P}{P_Q} = \frac{9}{4} \times \frac{256}{81} = \frac{1}{1} \times \frac{64}{9} = \frac{64}{9}$.

(D) According to Wien's displacement law,the product of the wavelength of maximum intensity $\lambda_{m}$ and the absolute temperature $T$ is a constant.
$\lambda_{m} T = \text{constant}$
Given:
$T_{1} = 2200 \ K$
$T_{2} = 3300 \ K$
Let the new wavelength be $\lambda_{m}^{\prime}$.
Using the relation $\lambda_{m} T_{1} = \lambda_{m}^{\prime} T_{2}$:
$\lambda_{m}^{\prime} = \lambda_{m} \times \frac{T_{1}}{T_{2}}$
$\lambda_{m}^{\prime} = \lambda_{m} \times \frac{2200}{3300}$
$\lambda_{m}^{\prime} = \frac{2}{3} \lambda_{m}$

(D) According to the Stefan-Boltzmann Law, the rate of radiation $P$ (or $E$) is given by $P = e \sigma A T^{4}$, where $A = 4 \pi r^{2}$.
For sphere $S_{1}$: $E = e \sigma (4 \pi R^{2}) T^{4}$.
For sphere $S_{2}$: $E' = e \sigma (4 \pi (3R)^{2}) (T/3)^{4}$.
$E' = e \sigma (4 \pi \cdot 9R^{2}) (T^{4} / 81)$.
$E' = e \sigma (4 \pi R^{2}) T^{4} \cdot (9 / 81)$.
$E' = E \cdot (1 / 9) = E/9$.

(C) According to the Stefan-Boltzmann law,the energy emitted per second is given by $E = \sigma A T^4$.
Here,$\sigma$ is the Stefan-Boltzmann constant,$A$ is the surface area,and $T$ is the absolute temperature in Kelvin.
Let the initial state be $E_1 = \sigma A_1 T_1^4$ and the final state be $E_2 = \sigma A_2 T_2^4$.
Given that the length and breadth are reduced to $1/3$ of their initial values,the new area $A_2 = (L/3) \times (B/3) = A_1 / 9$.
Initial temperature $T_1 = 27 + 273 = 300 \ K$.
Final temperature $T_2 = 327 + 273 = 600 \ K$.
Taking the ratio: $\frac{E_2}{E_1} = \frac{A_2}{A_1} \times \left(\frac{T_2}{T_1}\right)^4$.
Substituting the values: $\frac{E_2}{E_1} = \frac{1}{9} \times \left(\frac{600}{300}\right)^4 = \frac{1}{9} \times (2)^4 = \frac{16}{9}$.
Therefore,$E_2 = \frac{16}{9} E$.

(A) According to the Stefan-Boltzmann Law,the power $P$ radiated by a spherical black body of radius $r$ and temperature $T$ is given by $P = \sigma A T^{4}$,where $A = 4 \pi r^{2}$ is the surface area.
Since both bodies radiate the same power,we have $P_{1} = P_{2}$.
Substituting the formula,we get $4 \pi r_{1}^{2} \sigma T_{1}^{4} = 4 \pi r_{2}^{2} \sigma T_{2}^{4}$.
Canceling the common terms $4 \pi \sigma$,we get $r_{1}^{2} T_{1}^{4} = r_{2}^{2} T_{2}^{4}$.
Rearranging the terms to find the ratio $\frac{r_{1}^{2}}{r_{2}^{2}}$,we get $\frac{r_{1}^{2}}{r_{2}^{2}} = \frac{T_{2}^{4}}{T_{1}^{4}}$.
Taking the square root on both sides,we get $\frac{r_{1}}{r_{2}} = \frac{T_{2}^{2}}{T_{1}^{2}} = \left(\frac{T_{2}}{T_{1}}\right)^{2}$.

(B) The coefficient of transmission $(T)$ is defined as the ratio of the radiant energy transmitted through a substance to the total radiant energy incident upon it.
An athermanous substance is a material that does not allow thermal radiation to pass through it.
Since no radiant energy is transmitted through an athermanous substance,the transmitted energy is $0$.
Therefore,the coefficient of transmission $T = 0$.

(D) According to Wien's displacement law,$\lambda_{m} T = b$ (constant),so $T \propto \frac{1}{\lambda_{m}}$.
According to Stefan-Boltzmann law,the total emissive power $E$ of a black body is given by $E = \sigma T^{4}$.
Substituting $T \propto \frac{1}{\lambda_{m}}$,we get $E \propto \left(\frac{1}{\lambda_{m}}\right)^{4} = \frac{1}{\lambda_{m}^{4}}$.
Note that the emissive power $E$ is independent of the surface area $A$ of the black body.
Therefore,the ratio of emissive powers is $Ex : Ey : Ez = \frac{1}{\lambda_{x}^{4}} : \frac{1}{\lambda_{y}^{4}} : \frac{1}{\lambda_{z}^{4}}$.
Substituting the given values: $Ex : Ey : Ez = \frac{1}{(200)^{4}} : \frac{1}{(300)^{4}} : \frac{1}{(400)^{4}}$.
Since $200 < 300 < 400$,it follows that $\frac{1}{200^{4}} > \frac{1}{300^{4}} > \frac{1}{400^{4}}$.
Thus,$Ex > Ey > Ez$.

(B) According to Wien's displacement law, $\lambda_{\max} T = \text{constant}$, so $\lambda_{1} T_{1} = \lambda_{2} T_{2}$.
Given $\lambda_{1} = \lambda$ and $\lambda_{2} = \frac{2 \lambda}{3}$.
Thus, $T_{2} = \frac{\lambda_{1} T_{1}}{\lambda_{2}} = \frac{\lambda T_{1}}{2 \lambda / 3} = \frac{3}{2} T_{1}$.
According to Stefan-Boltzmann law, the emissive power $E$ is proportional to $T^{4}$, i.e., $E = \sigma A T^{4}$.
Therefore, $\frac{E_{2}}{E_{1}} = \left( \frac{T_{2}}{T_{1}} \right)^{4}$.
Substituting the value of $T_{2}$, we get $\frac{E_{2}}{E} = \left( \frac{3/2 T_{1}}{T_{1}} \right)^{4} = \left( \frac{3}{2} \right)^{4} = \frac{81}{16}$.
Hence, $E_{2} = \frac{81 E}{16}$.

(C) The thermal contraction that would occur due to cooling is $\Delta L = L \alpha \Delta T$.
To prevent this contraction, the tensile force $F$ applied by the mass $m$ must produce an equal extension: $\Delta L = \frac{FL}{AY}$.
Equating the two expressions: $L \alpha \Delta T = \frac{FL}{AY}$.
Thus, the force required is $F = AY \alpha \Delta T$.
Since $F = Mg$, we have $Mg = AY \alpha \Delta T$.
Substituting the given values: $M = \frac{AY \alpha \Delta T}{g}$.
$M = \frac{(3 \times 10^{-6} \,m^{2}) \times (10^{11} \,N/m^{2}) \times (10^{-5} /K) \times (100 - 0) \,K}{10 \,m/s^{2}}$.
$M = \frac{3 \times 10^{0} \times 100}{10} = \frac{300}{10} = 30 \,kg$.

(C) The thermal expansion of the rod is given by $\Delta L = L \alpha \Delta \theta$.
To prevent this expansion,a compressive force $F$ must be applied such that the compression equals the thermal expansion.
From the definition of Young's modulus,$Y = \frac{\text{Stress}}{\text{Strain}} = \frac{F/A}{\Delta L/L}$.
Thus,the stress $\sigma = \frac{F}{A} = Y \frac{\Delta L}{L}$.
Substituting $\Delta L = L \alpha \Delta \theta$ into the stress equation:
$\sigma = Y \frac{L \alpha \Delta \theta}{L} = Y \alpha \Delta \theta$.

(C) The thermal expansion of the rod is given by $\Delta L = L \alpha T$.
To prevent this expansion,we must apply a compressive force $F$ such that the compressive strain equals the thermal strain.
The stress developed is $\sigma = Y \times \text{strain} = Y \times \frac{\Delta L}{L_{final}}$.
The final length of the rod after heating is $L_{final} = L(1 + \alpha T)$.
Thus,the force $F = \text{Stress} \times A = Y \times \frac{\Delta L}{L_{final}} \times A$.
Substituting the values,$F = Y \times \frac{L \alpha T}{L(1 + \alpha T)} \times A = \frac{YA \alpha T}{1 + \alpha T}$.

(B) The thermal strain produced in the wire due to a change in temperature $\Delta T = t$ is given by $\epsilon = \frac{\Delta L}{L} = \alpha \Delta T = \alpha t$.
By the definition of Young's modulus,$Y = \frac{\text{Stress}}{\text{Strain}} = \frac{F/A}{\Delta L/L}$.
Substituting the expression for strain,we get $Y = \frac{F/A}{\alpha t}$.
Rearranging for the tension $F$,we obtain $F = YA \alpha t$.
Thus,the increase in tension when the temperature falls by $t^{\circ} C$ is $YA \alpha t$.

(C) For an ideal gas,the heat supplied at constant pressure is given by $\Delta Q = n C_p \Delta T$.
The change in internal energy is given by $\Delta U = n C_v \Delta T$.
The work done at constant pressure is $\Delta W = P \Delta V = n R \Delta T$.
Therefore,the ratio is $\Delta W : \Delta U : \Delta Q = n R \Delta T : n C_v \Delta T : n C_p \Delta T = R : C_v : C_p$.
For a diatomic gas,the degrees of freedom $f = 5$.
Thus,$C_v = \frac{f}{2} R = \frac{5}{2} R$.
Using Mayer's relation,$C_p = C_v + R = \frac{5}{2} R + R = \frac{7}{2} R$.
Substituting these values into the ratio: $R : \frac{5}{2} R : \frac{7}{2} R = 1 : \frac{5}{2} : \frac{7}{2}$.
Multiplying by $2$,we get the ratio $2 : 5 : 7$.

(B) For a process at constant pressure,the heat supplied is given by $\Delta Q = n C_p \Delta T$.
The change in internal energy is given by $\Delta U = n C_v \Delta T$.
The work done is given by $\Delta W = P \Delta V = n R \Delta T$.
Thus,the ratio $\Delta Q : \Delta U : \Delta W = C_p : C_v : R$.
For a rigid diatomic gas,the degrees of freedom $f = 5$.
Therefore,$C_v = \frac{f}{2} R = \frac{5}{2} R$ and $C_p = C_v + R = \frac{7}{2} R$.
Substituting these values,we get $\Delta Q : \Delta U : \Delta W = \frac{7}{2} R : \frac{5}{2} R : R$.
Multiplying by $\frac{2}{R}$,we obtain the ratio $7 : 5 : 2$.

(D) Given: $\gamma = \frac{5}{3}$.
Step $1$: Isothermal expansion from volume $V$ to $2V$.
For an isothermal process,$P_1 V_1 = P_2 V_2$.
$P \times V = P_2 \times 2V$.
Therefore,$P_2 = \frac{P}{2}$.
Step $2$: Adiabatic expansion from volume $2V$ to $16V$.
For an adiabatic process,$P_2 V_2^{\gamma} = P_3 V_3^{\gamma}$.
$P_3 = P_2 \left( \frac{V_2}{V_3} \right)^{\gamma}$.
$P_3 = \frac{P}{2} \left( \frac{2V}{16V} \right)^{5/3} = \frac{P}{2} \left( \frac{1}{8} \right)^{5/3}$.
$P_3 = \frac{P}{2} \left( (2^{-3})^{5/3} \right) = \frac{P}{2} \times 2^{-5} = \frac{P}{2 \times 32} = \frac{P}{64}$.

(C) For an adiabatic process, the relation between pressure $P$ and volume $V$ is given by $PV^{\gamma} = \text{constant}$.
From the ideal gas equation, $PV = RT$, we have $V = \frac{RT}{P}$.
Substituting $V$ in the adiabatic equation:
$P \left( \frac{RT}{P} \right)^{\gamma} = \text{constant}$
$P \cdot \frac{T^{\gamma}}{P^{\gamma}} = \text{constant}'$
$P^{1-\gamma} T^{\gamma} = \text{constant}'$
$P^{1-\gamma} = \frac{\text{constant}'}{T^{\gamma}}$
$P = \text{constant}'' \cdot T^{\frac{\gamma}{\gamma-1}}$
Comparing this with $P \propto T^{x}$, we get $x = \frac{\gamma}{\gamma-1}$.
For a diatomic gas, the adiabatic exponent $\gamma = 1.4$ (or $\frac{7}{5}$).
Substituting the value of $\gamma$:
$x = \frac{1.4}{1.4 - 1} = \frac{1.4}{0.4} = 3.5$.
Therefore, the correct option is $C$.

(C) The dimensions of a physical quantity are given as $[L^{a} M^{b} T^{c}]$.
Let us check the dimensions of the given options:
$1$. Velocity: $[M^{0} L^{1} T^{-1}]$. Here,$a=1, b=0, c=-1$. Option $A$ is incorrect.
$2$. Force: $[M^{1} L^{1} T^{-2}]$. Here,$a=1, b=1, c=-2$. Option $B$ is incorrect.
$3$. Pressure: Pressure is defined as $\text{Force} / \text{Area} = [M^{1} L^{1} T^{-2}] / [L^{2}] = [M^{1} L^{-1} T^{-2}]$. Here,$a=-1, b=1, c=-2$. This matches option $C$.
$4$. Acceleration: $[M^{0} L^{1} T^{-2}]$. Here,$a=1, b=0, c=-2$. Option $D$ is incorrect.
Therefore,the correct option is $C$.

(A) The dimensions of the given constants are:
$G = [M^{-1} L^{3} T^{-2}]$
$h = [M L^{2} T^{-1}]$
$c = [L T^{-1}]$
Let the expression be $T = G^{a} h^{b} c^{d}$.
Substituting the dimensions:
$[T^{1}] = [M^{-1} L^{3} T^{-2}]^{a} [M L^{2} T^{-1}]^{b} [L T^{-1}]^{d}$
$[M^{0} L^{0} T^{1}] = [M^{-a+b} L^{3a+2b+d} T^{-2a-b-d}]$
Comparing the powers on both sides:
For $M$: $-a + b = 0 \implies a = b$
For $L$: $3a + 2b + d = 0 \implies 3a + 2a + d = 0 \implies d = -5a$
For $T$: $-2a - b - d = 1 \implies -2a - a - (-5a) = 1 \implies 2a = 1 \implies a = 1/2$
Thus,$a = 1/2, b = 1/2, d = -5/2$.
The expression is $G^{1/2} h^{1/2} c^{-5/2} = \left[\frac{G h}{c^{5}}\right]^{\frac{1}{2}}$.

(C) The correct option is $C$.
From the relation $E = h \nu$,where $E$ is energy,$h$ is Planck's constant,and $\nu$ is frequency.
$h = \frac{E}{\nu}$.
The dimension of energy $E$ is $[ML^2 T^{-2}]$.
The dimension of frequency $\nu$ is $[T^{-1}]$.
Therefore,the dimension of $h = \frac{[ML^2 T^{-2}]}{[T^{-1}]} = [ML^2 T^{-1}]$.
Now,let us check the dimensions of the product of force,displacement,and time:
Dimension of force $F = [MLT^{-2}]$.
Dimension of displacement $d = [L]$.
Dimension of time $t = [T]$.
Product dimension $= [MLT^{-2}] \times [L] \times [T] = [ML^2 T^{-1}]$.
This matches the dimension of Planck's constant.

(A) The dimensions of the given physical quantities are:
$[E] = [M L^{2} T^{-2}]$
$[M] = [M]$
$[L] = [M L^{2} T^{-1}]$
$[G] = [M^{-1} L^{3} T^{-2}]$
Now,substitute these into the expression $\left(\frac{E L^{2}}{G^{2} M^{5}}\right)$:
Dimensions $= \frac{[M L^{2} T^{-2}] \cdot [M L^{2} T^{-1}]^{2}}{[M^{-1} L^{3} T^{-2}]^{2} \cdot [M]^{5}}$
$= \frac{[M L^{2} T^{-2}] \cdot [M^{2} L^{4} T^{-2}]}{[M^{-2} L^{6} T^{-4}] \cdot [M^{5}]}$
$= \frac{[M^{3} L^{6} T^{-4}]}{[M^{3} L^{6} T^{-4}]}$
$= [M^{0} L^{0} T^{0}]$
Since the resulting dimension is dimensionless,it corresponds to the dimensions of an angle (which is a dimensionless quantity).

(B) The refractive index $\mu$ is a dimensionless quantity,so its dimension is $[M^0 L^0 T^0]$.
According to the principle of homogeneity of dimensions,the dimensions of each term in an equation must be the same.
Therefore,the dimensions of $\frac{B}{\lambda^2}$ must be equal to the dimensions of $\mu$ (which is dimensionless).
$\left[ \frac{B}{\lambda^2} \right] = [M^0 L^0 T^0]$
$[B] = [\lambda^2] \times [M^0 L^0 T^0]$
Since $\lambda$ is wavelength,its dimension is $[L]$.
$[B] = [L^2]$
The dimension $[L^2]$ corresponds to the dimension of area.

(A) The time constant of an $LR$ circuit is given by $\tau = \frac{L}{R}$.
Dimensional formula of inductance $L$ is $[M^{1} L^{2} T^{-2} A^{-2}]$.
Dimensional formula of resistance $R$ is $[M^{1} L^{2} T^{-3} A^{-2}]$.
Therefore,the dimensions of $\frac{L}{R}$ are:
$\frac{[M^{1} L^{2} T^{-2} A^{-2}]}{[M^{1} L^{2} T^{-3} A^{-2}]} = M^{1-1} L^{2-2} T^{-2-(-3)} A^{-2-(-2)} = M^{0} L^{0} T^{1} A^{0}$.
Thus,the dimensions of $\left(\frac{L}{R}\right)$ are $[L^{0} M^{0} T^{1}]$.

(B) In the given equation $F = P \cos(Ax) + Q \sin(Bt)$,the arguments of trigonometric functions must be dimensionless.
Therefore,the dimensions of $(Ax)$ must be $[M^0 L^0 T^0]$.
Since $[x] = [L]$,we have $[A][L] = [M^0 L^0 T^0]$,which implies $[A] = [L^{-1}]$.
Similarly,the dimensions of $(Bt)$ must be $[M^0 L^0 T^0]$.
Since $[t] = [T]$,we have $[B][T] = [M^0 L^0 T^0]$,which implies $[B] = [T^{-1}]$.
Now,calculating the dimensions of the ratio $\left[\frac{B}{A}\right]$:
$\left[\frac{B}{A}\right] = \frac{[T^{-1}]}{[L^{-1}]} = [L T^{-1}]$.
The dimension $[L T^{-1}]$ corresponds to the physical quantity of velocity.

(D) The argument of a trigonometric function must be dimensionless.
Given $F = A \sin(Ct) + B \cos(Dx)$.
For the term $\sin(Ct)$,the argument $Ct$ must be dimensionless.
$[Ct] = [M^0 L^0 T^0] \implies [C][T] = [T^0] \implies [C] = [T^{-1}]$.
For the term $\cos(Dx)$,the argument $Dx$ must be dimensionless.
$[Dx] = [M^0 L^0 T^0] \implies [D][L] = [L^0] \implies [D] = [L^{-1}]$.
Now,the dimensions of $\frac{C}{D}$ are:
$\left[ \frac{C}{D} \right] = \frac{[T^{-1}]}{[L^{-1}]} = [L T^{-1}]$.
These are the dimensions of velocity.
Therefore,the correct option is $D$.

(D) The dimension of Stefan's constant $\sigma$ is given by the formula $P = \sigma A T^4$,where $P$ is power,$A$ is area,and $T$ is temperature.
$[\sigma] = \frac{[P]}{[A][T]^4} = \frac{[ML^2 T^{-3}]}{[L^2][K^4]} = [MT^{-3} K^{-4}]$.
The dimension of Wien's constant $b$ is given by $\lambda_{max} T = b$,where $\lambda$ is wavelength and $T$ is temperature.
$[b] = [L][K]$.
Therefore,the dimensions of $\sigma b$ are:
$[\sigma b] = [MT^{-3} K^{-4}] \times [LK] = [L^1 M^1 T^{-3} K^{-3}]$.

(A) Given the equation $P = \frac{c - t^{2}}{DS}$.
According to the principle of homogeneity of dimensions,the dimensions of $c$ must be equal to the dimensions of $t^{2}$.
Therefore,$[c] = [t^{2}] = [T^{2}]$.
Now,the dimension of pressure $P$ is $[P] = [M L^{-1} T^{-2}]$.
Given $S$ is distance,so $[S] = [L]$.
Substituting these into the equation $P = \frac{c}{DS} - \frac{t^{2}}{DS}$,we consider the term $\frac{c}{DS}$:
$[P] = \frac{[c]}{[D][S]} \Rightarrow [M L^{-1} T^{-2}] = \frac{[T^{2}]}{[D][L]}$.
Solving for $[D]$:
$[D] = \frac{[T^{2}]}{[M L^{-1} T^{-2}][L]} = \frac{[T^{2}]}{[M T^{-2}]} = [M^{-1} T^{4}]$.
We need to find the dimensions of $\left(\frac{D}{c}\right)$:
$\left[\frac{D}{c}\right] = \frac{[M^{-1} T^{4}]}{[T^{2}]} = [M^{-1} T^{2}] = [L^{0} M^{-1} T^{2}]$.

(A) According to the principle of homogeneity of dimensions,only quantities with the same dimensions can be added or subtracted.
In the expression $A = B + \frac{C}{D+E}$,since $B$ is added to the term $\frac{C}{D+E}$,the dimensions of $A$ must be equal to the dimensions of $B$.
Given $[B] = [L^{1} M^{0} T^{-1}]$,therefore $[A] = [L^{1} M^{0} T^{-1}]$.
Also,in the denominator $(D+E)$,$D$ and $E$ must have the same dimensions,so $[D] = [E]$.
The dimension of the entire term $\frac{C}{D+E}$ must be equal to the dimension of $B$.
$[B] = \frac{[C]}{[D+E]} \implies [D+E] = \frac{[C]}{[B]}$.
Substituting the given dimensions: $[D+E] = \frac{[L^{1} M^{0} T^{0}]}{[L^{1} M^{0} T^{-1}]} = [T^{1}]$.
Since $[D] = [E]$,we have $[D] = [T^{1}]$ and $[E] = [T^{1}]$.

(C) For a thin prism with a small angle '$A$',the relationship between the angle of incidence '$i$',the angle of refraction '$r_1$',and the angle of emergence '$r_2$' is given by $A = r_1 + r_2$.
Since the ray emerges normally from the opposite surface,the angle of emergence is $0$,which implies $r_2 = 0$.
Therefore,$A = r_1 + 0$,so $r_1 = A$.
According to Snell's Law,$\mu = \frac{\sin i}{\sin r_1}$. For small angles,$\sin i \approx i$ and $\sin r_1 \approx r_1$.
Thus,$\mu = \frac{i}{r_1}$.
Substituting $r_1 = A$,we get $\mu = \frac{i}{A}$,which simplifies to $i = \mu A$.

(D) For an equilateral prism,the angle of the prism $A = 60^{\circ}$.
Given that the angle of incidence $i$ is equal to the angle of emergence $e$,and $i = e = \frac{3}{4}A$.
Substituting the value of $A$:
$i = e = \frac{3}{4} \times 60^{\circ} = 45^{\circ}$.
The formula for the angle of deviation $\delta$ is given by $\delta = i + e - A$.
Substituting the values:
$\delta = 45^{\circ} + 45^{\circ} - 60^{\circ}$.
$\delta = 90^{\circ} - 60^{\circ} = 30^{\circ}$.
Therefore,the angle of deviation is $30^{\circ}$.

(A) For a prism with a small angle '$A$',the angle of deviation '$\delta$' is given by $\delta = (\mu - 1)A$.
We know the relation for a prism: $i + e = A + \delta$.
Given that the ray emerges normally from the other surface,the angle of emergence '$e$' is $0$.
Substituting $e = 0$ and $\delta = (\mu - 1)A$ into the relation:
$i + 0 = A + (\mu - 1)A$
$i = A + \mu A - A$
$i = \mu A$.
Therefore,the angle of incidence is $\mu A$.

(D) For the light to retrace its path,it must strike the silvered surface normally.
Since the second surface is silvered,the angle of refraction at the second surface $r_2$ must be $0^{\circ}$.
Using the prism formula $A = r_1 + r_2$,where $A = 30^{\circ}$ is the refracting angle and $r_2 = 0^{\circ}$,we get $r_1 = 30^{\circ}$.
Applying Snell's Law at the first surface: $n = \frac{\sin i}{\sin r_1}$.
Given $n = \sqrt{2}$ and $r_1 = 30^{\circ}$,we have $\sqrt{2} = \frac{\sin i}{\sin 30^{\circ}}$.
$\sin i = \sqrt{2} \times \sin 30^{\circ} = \sqrt{2} \times \frac{1}{2} = \frac{1}{\sqrt{2}}$.
Therefore,$i = \sin^{-1}\left(\frac{1}{\sqrt{2}}\right)$.

(A) For a thin prism,the relationship between the prism angle '$A$' and the angles of refraction '$r_1$' and '$r_2$' is given by: $A = r_1 + r_2$.
Since the ray emerges normally from the other face,the angle of emergence '$e = 0$',which implies that the second angle of refraction '$r_2 = 0$'.
Therefore,$A = r_1$.
According to Snell's Law at the first surface,$n = \frac{\sin i}{\sin r_1}$.
For a thin prism,the angles '$i$' and '$r_1$' are very small,so we can approximate $\sin i \approx i$ and $\sin r_1 \approx r_1$.
Thus,$n = \frac{i}{r_1}$.
Substituting $r_1 = A$,we get $n = \frac{i}{A}$,which implies $i = An$.

(A) When a light ray travels from one medium to another,its frequency '$v$' remains unchanged because it depends only on the source of the light.
The speed of light in a medium is given by $v = \frac{c}{n}$,where '$c$' is the speed of light in vacuum and '$n$' is the refractive index.
The wavelength in the medium is given by $\lambda' = \frac{v}{f} = \frac{c/n}{f} = \frac{\lambda}{n}$.
Given the refractive index $n = \frac{3}{2}$,the new wavelength is $\lambda' = \frac{\lambda}{3/2} = \left(\frac{2}{3}\right) \lambda$.
Therefore,the frequency remains '$v$' and the wavelength becomes $\left(\frac{2}{3}\right) \lambda$.

(C) When light enters glass from vacuum,the wavelength of light decreases.
This occurs because the speed of light in glass is less than the speed of light in a vacuum.
The refractive index $\mu$ of a medium is defined as $\mu = \frac{c}{v}$,where $c$ is the speed of light in a vacuum and $v$ is the speed of light in the medium.
Since the refractive index of glass is greater than that of a vacuum,the speed of light $v$ decreases.
Given the relationship $v = f \lambda$,where $f$ is the frequency (which remains constant) and $\lambda$ is the wavelength,the speed $v$ is directly proportional to the wavelength $\lambda$.
Therefore,as the speed of light decreases,the wavelength of light also decreases.

(C) Let $n_{aw}$ be the refractive index of water with respect to air,$n_{gw}$ be the refractive index of glass with respect to water,and $n_{ag}$ be the refractive index of air with respect to glass. Given: $x = n_{aw}$,$y = n_{gw}$,and $z = n_{ag}$.
By the principle of reversibility and the definition of relative refractive indices,we have:
$x = \frac{n_w}{n_a}$
$y = \frac{n_g}{n_w}$
$z = \frac{n_a}{n_g}$
Multiplying these three expressions:
$x \cdot y \cdot z = \left(\frac{n_w}{n_a}\right) \cdot \left(\frac{n_g}{n_w}\right) \cdot \left(\frac{n_a}{n_g}\right) = 1$
Therefore,$xyz = 1$.

(A) The refractive index $\mu$ of a medium is defined as the ratio of the speed of light in vacuum $(c)$ to the speed of light in that medium $(v)$.
Thus,for the two media,we have:
$\mu_{1} = \frac{c}{v_{1}}$ and $\mu_{2} = \frac{c}{v_{2}}$
From these equations,we can write:
$c = \mu_{1} v_{1}$ and $c = \mu_{2} v_{2}$
Since both expressions are equal to the speed of light in vacuum $(c)$,we have:
$\mu_{1} v_{1} = \mu_{2} v_{2}$

(B) When an object is placed in a medium,it becomes visible due to the reflection and refraction of light at its boundaries.
If the refractive index of the object is exactly the same as the refractive index of the surrounding fluid $(\mu_{object} = \mu_{fluid} = \mu)$,there will be no change in the speed or direction of light as it passes from the fluid into the object.
Consequently,no refraction or reflection occurs at the interface between the object and the fluid.
As a result,the light rays pass through the object as if it were not there,making the object invisible to an observer outside the fluid.

(C) The refractive index $\mu$ of a medium is defined as the ratio of the speed of light in a vacuum $(c)$ to the speed of light in the medium $(v)$: $\mu = \frac{c}{v}$.
Since the frequency $(f)$ of light remains constant when it travels from one medium to another,we can express the speeds as $c = f \lambda_0$ and $v = f \lambda$,where $\lambda_0$ is the wavelength in a vacuum and $\lambda$ is the wavelength in the medium.
Substituting these into the refractive index formula: $\mu = \frac{f \lambda_0}{f \lambda} = \frac{\lambda_0}{\lambda}$.
Since $\lambda_0$ (wavelength in vacuum) is a constant,it follows that $\mu \propto \frac{1}{\lambda}$.
Therefore,the correct option is $C$.

(B) According to Snell's Law,$\mu_1 \sin i = \mu_2 \sin r$.
Here,$\mu_1 = \sqrt{2}$ (refractive index of glass),$i = 45^{\circ}$ (angle of incidence),and $\mu_2 = 1$ (refractive index of air).
Substituting the values: $\sqrt{2} \sin 45^{\circ} = 1 \cdot \sin r$.
Since $\sin 45^{\circ} = \frac{1}{\sqrt{2}}$,we get: $\sqrt{2} \cdot \frac{1}{\sqrt{2}} = \sin r$.
$1 = \sin r$.
Since $\sin 90^{\circ} = 1$,the angle of refraction $r = 90^{\circ}$.

(D) When a light ray passes through a series of parallel media and finally emerges into the original medium,the product of the relative refractive indices is equal to $1$.
For the given sequence (air $\rightarrow$ water $\rightarrow$ glass $\rightarrow$ air),the principle of reversibility and the property of parallel slabs imply:
$_{a}n_{w} \times _{w}n_{g} \times _{g}n_{a} = 1$.
We know that $_{g}n_{a} = \frac{1}{_{a}n_{g}}$.
Substituting this into the equation:
$_{a}n_{w} \times _{w}n_{g} \times \frac{1}{_{a}n_{g}} = 1$.
Rearranging the terms to solve for $_{w}n_{g}$:
$_{w}n_{g} = \frac{_{a}n_{g}}{_{a}n_{w}}$.
Thus,option $D$ is the correct relation.

(C) The convex lens receives a parallel beam of light from infinity and forms an image at its focal point,which is at a distance $f_{2}$ from the lens.
For the light rays to retrace their path after reflecting from the concave mirror,the rays must strike the mirror normally.
This occurs if the rays are directed towards the center of curvature of the concave mirror.
The center of curvature of a concave mirror is at a distance $2f_{1}$ from its pole.
Therefore,the distance $d$ between the lens and the mirror must be the sum of the focal length of the lens and the radius of curvature of the mirror.
Thus,$d = f_{2} + 2f_{1}$.

(A) Brewster's law states that the refractive index $n$ of the medium is given by $n = \tan(i_B)$,where $i_B$ is the Brewster's angle.
Given $i_B = 54.74^{\circ}$,we have $n = \tan(54.74^{\circ}) = \sqrt{2}$.
According to Snell's law,$n = \frac{\sin i}{\sin r}$,where $i$ is the angle of incidence and $r$ is the angle of refraction.
Given $i = 45^{\circ}$ and $n = \sqrt{2}$,we substitute these values:
$\sqrt{2} = \frac{\sin 45^{\circ}}{\sin r}$
$\sqrt{2} = \frac{1/\sqrt{2}}{\sin r}$
$\sin r = \frac{1}{\sqrt{2} \cdot \sqrt{2}} = \frac{1}{2} = 0.5$
Therefore,the angle of refraction $r = \sin^{-1}(0.5)$.

(B) The refractive index of the second medium with respect to the first medium is defined as the ratio of the speed of light in the first medium to the speed of light in the second medium.
Mathematically,it is expressed as: $n_{21} = \frac{\mu_{2}}{\mu_{1}} = \frac{C_{1}}{C_{2}} = \frac{\lambda_{1}}{\lambda_{2}}$.
However,the question asks for the refractive index of the second medium with respect to the first medium,which is denoted by the ratio of the refractive indices themselves,i.e.,$\frac{\mu_{2}}{\mu_{1}}$.

(A) The velocity of light in a medium is given by $v = \frac{C}{\mu_{w}}$,where $C$ is the speed of light in air and $\mu_{w}$ is the refractive index of water.
Time taken $(t)$ to travel a distance $h$ is given by the formula: $t = \frac{\text{distance}}{\text{velocity}}$.
Substituting the values,we get $t = \frac{h}{v}$.
Since $v = \frac{C}{\mu_{w}}$,we substitute this into the time equation:
$t = \frac{h}{(C / \mu_{w})} = \frac{h \mu_{w}}{C}$.
Therefore,the correct option is $A$.

(A) The length of the wire is $L$. When it is bent into a circular coil of $n$ turns,the circumference of one turn is $2\pi r = L/n$,where $r$ is the radius of the coil.
Thus,the radius $r = \frac{L}{2\pi n}$.
The area of one turn is $A = \pi r^2 = \pi \left(\frac{L}{2\pi n}\right)^2 = \frac{\pi L^2}{4\pi^2 n^2} = \frac{L^2}{4\pi n^2}$.
The total magnetic moment of the coil with $n$ turns is $M = nIA = nI \left(\frac{L^2}{4\pi n^2}\right) = \frac{IL^2}{4\pi n}$.
The maximum torque acting on a coil in a magnetic field is given by $\tau_{max} = MB$.
Substituting the value of $M$,we get $\tau_{max} = \left(\frac{IL^2}{4\pi n}\right)B = \frac{BIL^2}{4\pi n}$.

(B) In an unbiased $p-n$ junction,the diffusion of electrons from the $n$-region to the $p$-region and holes from the $p$-region to the $n$-region occurs near the junction.
When these charge carriers cross the junction,they recombine and neutralize each other.
As a result,the region near the junction becomes devoid of mobile charge carriers (free electrons and holes).
This region is called the depletion layer or depletion region.
Therefore,the layer is depleted of mobile charge carriers,which are electrons and holes.

(D) In the given circuit,the two diodes $D_{1}$ and $D_{2}$ are connected in series with a $DC$ voltage source.
For the circuit to conduct,both diodes must be forward-biased.
However,looking at the orientation of the diodes,$D_{1}$ is forward-biased while $D_{2}$ is reverse-biased.
In a series circuit,the current $I$ is the same through all components.
Since $D_{2}$ is reverse-biased,it offers a very high resistance,effectively acting as an open switch.
Consequently,the entire potential difference of the battery appears across the reverse-biased diode $D_{2}$,while the potential difference across the forward-biased diode $D_{1}$ is negligible (approximately $0.7 \ V$ for Silicon).
If the potential differences across $D_{1}$ and $D_{2}$ are stated to be equal,it implies that the circuit configuration or the assumption of identical characteristics must be re-evaluated,or the diodes are operating in a specific breakdown region. Given the standard interpretation,the potential difference across a reverse-biased diode is much larger than that across a forward-biased one.

(D) In a $PN$ junction diode,the diode is forward biased if the potential at the anode (p-side) is higher than the potential at the cathode (n-side). It is reverse biased if the potential at the anode is lower than the potential at the cathode.
In diagram $(A)$,the anode is at $-4 \ V$ and the cathode is at $-3 \ V$. Since $-4 \ V < -3 \ V$,the anode is at a lower potential than the cathode,so diagram $(A)$ is reverse biased.
In diagram $(B)$,assuming the standard configuration where the anode is at $-2 \ V$ and the cathode is at $-4 \ V$ (based on typical textbook problems of this type),the anode is at a higher potential than the cathode,so diagram $(B)$ is forward biased.
Therefore,diagram $(A)$ is reverse biased and diagram $(B)$ is forward biased.

(B) In the given circuit,the $p$-side of the diode is connected to $+2 V$ and the $n$-side is connected to $-2 V$ through a $400 \Omega$ resistor.
Since the potential of the $p$-side is higher than the potential of the $n$-side,the diode is forward biased.
For an ideal diode,the forward resistance is $0 \Omega$.
Therefore,the total resistance in the circuit is $R = 400 \Omega$.
The potential difference across the circuit is $V = 2 V - (-2 V) = 4 V$.
Using Ohm's law,the current $I$ is given by $I = V / R$.
$I = 4 V / 400 \Omega = 1 / 100 A = 0.01 A$.
Converting to milliamperes,$I = 0.01 \times 1000 mA = 10 mA$.

(D) $(i)$ When a $p-n$ junction is forward biased,the external electric field opposes the internal electric field of the depletion region. This reduces the potential barrier and consequently decreases the width of the depletion layer.
$(ii)$ When a $p-n$ junction is reverse biased,the external electric field supports the internal electric field of the depletion region. This increases the potential barrier and consequently increases the width of the depletion layer.

(D) When a $p-n$ junction diode is reverse biased,the positive terminal of the battery is connected to the $n$-region and the negative terminal to the $p$-region.
This configuration pulls the majority charge carriers away from the junction.
As a result,the width of the depletion layer increases.
Consequently,the potential barrier also increases,which opposes the flow of current,making electrical conduction negligible.

(C) The $I-V$ characteristics of a device operating in the fourth quadrant (where voltage is positive and current is negative,or vice versa depending on convention) represent a device that generates power rather than consuming it.
In a $photocell$,the $I-V$ characteristics are typically shown in the fourth quadrant because it acts as a source of electrical energy when illuminated.
Therefore,the correct option is $C$.

(A) In a $p-n$ junction diode,when reverse bias is applied,the positive terminal of the battery is connected to the $n$-region and the negative terminal to the $p$-region.
This causes the majority charge carriers (electrons in $n$-region and holes in $p$-region) to move away from the junction.
As a result,the depletion layer widens,and the barrier potential increases.
Due to the increased width of the depletion layer,the flow of majority charge carriers is blocked,which means the diode offers high resistance to the current.

(A) In conductors,the valence band and the conduction band overlap each other. This overlap allows electrons to move freely from the valence band to the conduction band even at low temperatures,which is why conductors have high electrical conductivity.

(A) In insulators,the valence band is completely filled with electrons,and the conduction band is completely empty. The energy gap (forbidden energy gap) between the valence band and the conduction band is very large (typically $> 3 \ eV$). Therefore,electrons cannot jump from the valence band to the conduction band even at room temperature,making them poor conductors of electricity. Thus,the band gap is very high and the conduction band is empty.

(C) For sustained oscillations in an oscillator circuit,the Barkhausen criterion must be satisfied.
This criterion states that the loop gain of the feedback system must be equal to unity,which is expressed as $|A\beta| = 1$.
Here,$A$ is the open-loop gain of the amplifier and $\beta$ is the feedback factor.
Therefore,the condition for sustained oscillations is $A\beta = 1$.

(D) In a common-emitter transistor amplifier configuration,the current gain (denoted by $\beta$) is defined as the ratio of the change in collector current $(\Delta I_{C})$ to the change in base current $(\Delta I_{B})$ while keeping the collector-emitter voltage $(V_{CE})$ constant.
Mathematically,$\beta = \frac{\Delta I_{C}}{\Delta I_{B}}$.
Therefore,the correct option is $D$.

(D) Given: $\beta = 20$, $I_{E} = 7 \,mA$.
We know that the current gain in common emitter mode is defined as $\beta = \frac{I_{C}}{I_{B}}$.
Also, the relationship between emitter current, collector current, and base current is $I_{E} = I_{C} + I_{B}$, which implies $I_{B} = I_{E} - I_{C}$.
Substituting $I_{B}$ in the gain formula: $\beta = \frac{I_{C}}{I_{E} - I_{C}}$.
Rearranging the equation: $20 = \frac{I_{C}}{7 - I_{C}}$.
$20(7 - I_{C}) = I_{C}$.
$140 - 20I_{C} = I_{C}$.
$140 = 21I_{C}$.
$I_{C} = \frac{140}{21} = \frac{20}{3} \,mA$.

(B) Given: Input resistance $R_{i} = 1000 \Omega$,Input signal voltage $v_{i} = 5 mV = 5 \times 10^{-3} V$,and current gain $\beta = 60$.
First,calculate the input current $I_{b}$ using Ohm's law: $I_{b} = \frac{v_{i}}{R_{i}} = \frac{5 \times 10^{-3} V}{1000 \Omega} = 5 \times 10^{-6} A$.
The output current in a common emitter amplifier is the collector current $I_{c}$.
Using the relation $I_{c} = \beta \times I_{b}$,we get: $I_{c} = 60 \times 5 \times 10^{-6} A = 300 \times 10^{-6} A$.
Therefore,the peak value of the output current is $3 \times 10^{-4} A$.

(B) The voltage gain of an amplifier with feedback is given by $A_f = \frac{A}{1 - A\beta}$,where $A$ is the open-loop voltage gain and $\beta$ is the feedback factor.
For an oscillator to produce sustained oscillations,the Barkhausen criterion must be satisfied.
The Barkhausen criterion states that the loop gain must be equal to unity,which is expressed as $A\beta = 1$.

(C) Given: $\alpha = 0.8$ and $\Delta I_B = 6 \text{ mA}$.
First, we calculate the current gain $\beta$ for the common-emitter configuration using the relation $\beta = \frac{\alpha}{1 - \alpha}$.
Substituting the value of $\alpha$: $\beta = \frac{0.8}{1 - 0.8} = \frac{0.8}{0.2} = 4$.
The relationship between the change in collector current $\Delta I_C$ and the change in base current $\Delta I_B$ is given by $\Delta I_C = \beta \times \Delta I_B$.
Substituting the values: $\Delta I_C = 4 \times 6 \text{ mA} = 24 \text{ mA}$.
Therefore, the change in collector current is $24 \text{ mA}$.

(D) In a transistor,the emitter current is the sum of the base current and the collector current: $I_{e} = I_{b} + I_{c}$.
Since $I_{b}$ and $I_{c}$ are both positive currents,it follows that $I_{c} < I_{e}$ and $I_{b} < I_{e}$.
The current gain in common emitter configuration is defined as $\beta = \frac{I_{c}}{I_{b}}$.
For a transistor,$I_{c}$ is typically much larger than $I_{b}$ (since the base region is very thin and lightly doped),which makes $\beta > 1$.
Therefore,the condition $I_{c} > I_{b}$ is the fundamental reason why the current gain $\beta$ is greater than $1$.

(D) Given: The ratio of collector current $(i_c)$ to emitter current $(i_e)$ is $\alpha = \frac{i_c}{i_e} = 0.98$.
The collector current is $i_c = 3 \text{ mA}$.
We know that the emitter current is the sum of collector current and base current: $i_e = i_c + i_b$.
Substituting this into the ratio: $\frac{i_c}{i_c + i_b} = 0.98$.
Rearranging the equation: $\frac{i_c + i_b}{i_c} = \frac{1}{0.98} = \frac{100}{98}$.
$1 + \frac{i_b}{i_c} = \frac{100}{98} \implies \frac{i_b}{i_c} = \frac{100}{98} - 1 = \frac{2}{98} = \frac{1}{49}$.
Therefore,$i_b = \frac{i_c}{49} = \frac{3 \text{ mA}}{49} = \frac{3000 \mu\text{A}}{49} \approx 61.22 \mu\text{A}$.
Rounding to the nearest given option,the base current is approximately $60 \mu\text{A}$.

(D) The given circuit consists of two $NOT$ gates ($G_1$ and $G_2$) and one $NOR$ gate $(G_3)$.
$1$. The input $A$ passes through the $NOT$ gate $G_1$,so the output at $C$ is $C = \overline{A}$.
$2$. The input $B$ passes through the $NOT$ gate $G_2$,so the output at $D$ is $D = \overline{B}$.
$3$. These outputs $C$ and $D$ are fed into the $NOR$ gate $G_3$. The output $Y$ of a $NOR$ gate is the complement of the $OR$ of its inputs.
$4$. Therefore,$Y = \overline{C + D}$.
$5$. Substituting the values of $C$ and $D$,we get $Y = \overline{\overline{A} + \overline{B}}$.
$6$. According to De Morgan's theorem,$\overline{\overline{A} + \overline{B}} = \overline{\overline{A}} \cdot \overline{\overline{B}} = A \cdot B$.
$7$. The Boolean expression $Y = A \cdot B$ represents an $AND$ gate.
Thus,the resultant gate is an $AND$ gate with the expression $A \cdot B$.

(A) The logic gate for which the output is high $(1)$ if and only if all inputs are high $(1)$ is the $AND$ gate.
For an $AND$ gate with inputs $A$ and $B$,the output $Y$ is given by $Y = A \cdot B$.
If $A = 1$ and $B = 1$,then $Y = 1 \cdot 1 = 1$.
For any other combination of inputs,the output is low $(0)$.

(C) When a pentavalent impurity (donor impurity) is added to an intrinsic semiconductor (like $Si$ or $Ge$),each donor atom provides an extra electron to the conduction band.
Since the number of free electrons increases significantly,the semiconductor becomes an $n$-type semiconductor.
In an $n$-type semiconductor,electrons are the majority charge carriers and holes are the minority charge carriers.

(A) The conductivity of a semiconductor is given by $\sigma = n_{e} e \mu_{e} + n_{h} e \mu_{h}$.
Here,$n_{e}$ and $n_{h}$ are the number of electrons and holes per unit volume,and $\mu_{e}$ and $\mu_{h}$ are the mobilities of electrons and holes,respectively.
When impurity atoms are added to a semiconductor (a process known as doping),the concentration of charge carriers ($n_{e}$ or $n_{h}$) increases significantly.
Since conductivity $\sigma$ is directly proportional to the carrier concentration,the conductivity increases.
As resistivity $\rho$ is the reciprocal of conductivity $(\rho = 1/\sigma)$,an increase in conductivity leads to a decrease in resistivity.

(C)
At absolute zero temperature $(T = 0 \ K)$,there is no thermal energy available to excite electrons from the valence band to the conduction band in a pure semiconductor like silicon.
Consequently,the valence band is completely filled and the conduction band is completely empty.
Due to the absence of free charge carriers,the material cannot conduct electricity.
Therefore,pure silicon behaves as an insulator at absolute zero temperature.

(B) In an intrinsic (pure) semiconductor,charge carriers are generated only due to the thermal excitation of electrons from the valence band to the conduction band.
When an electron is excited to the conduction band,it leaves behind a vacancy in the valence band,which is known as a hole.
Since each electron-hole pair is created simultaneously,the number of electrons in the conduction band $(n_{e})$ must be equal to the number of holes in the valence band $(n_{h})$.
Therefore,the correct relation is $n_{e} = n_{h}$.

(D) photodiode is a special type of $p-n$ junction diode that is designed to operate under reverse bias conditions. It is sensitive to light. The symbol for a photodiode consists of a standard $p-n$ junction diode symbol with two arrows pointing towards the diode,indicating that light is incident on it. Since the provided image shows a Zener diode symbol (which has '$Z$' shaped ends on the cathode),it does not represent a photodiode. However,based on standard textbook representations,a photodiode is characterized by incoming light rays.

(D) photodiode is a special type of $p-n$ junction diode that is fabricated with a transparent window to allow light to fall on the diode.
It is operated under reverse bias conditions.
When light with energy $h
u$ greater than the energy gap $E_g$ of the semiconductor is incident on the diode,electron-hole pairs are generated.
Due to the electric field at the junction,these charge carriers are separated before they can recombine,resulting in a current flow in the external circuit.
The magnitude of this photo-current depends on the intensity of the incident light,not significantly on the reverse bias voltage once a certain threshold is reached.
Therefore,the correct statement is that it is always operated in reverse bias.

(C) light emitting diode $(LED)$ is a heavily doped $p-n$ junction diode.
When the diode is forward-biased,electrons from the $n$-region and holes from the $p$-region are injected into the junction region.
In the junction region,these charge carriers undergo recombination.
During the process of recombination,the energy released is in the form of photons.
If the semiconductor material has a suitable band gap,the energy of the emitted photons corresponds to the visible light spectrum.
Therefore,the light emission is due to the recombination of holes and electrons.

(C) Magnetic intensity $(H)$ is defined as the magnetizing force per unit length.
The formula for magnetic intensity is $H = \frac{nI}{L}$,where $n$ is the number of turns,$I$ is the current,and $L$ is the length.
Since the number of turns $(n)$ is a dimensionless quantity,the dimensions of $H$ are given by the dimensions of $\frac{I}{L}$.
The dimension of current $(I)$ is $[I^1]$ and the dimension of length $(L)$ is $[L^1]$.
Therefore,the dimensions of magnetic intensity are $[L^{-1} M^0 T^0 I^1]$.

(C) The magnetic dipole moment $M$ is defined by the relation $\tau = M \times B$,where $\tau$ is torque and $B$ is magnetic field.
From this,$M = \tau / B$.
The unit of torque $\tau$ is $N m$ (Newton-meter) and the unit of magnetic field $B$ is $T$ (Tesla).
Therefore,the unit of $M$ is $N m / T$.
Since $1 \ T = 1 \ Wb / m^2$,we can write $M = (N m) / (Wb / m^2) = N m^3 / Wb$.
Also,$M = I A$,where $I$ is current $(A)$ and $A$ is area $(m^2)$,so the unit is $A m^2$.
Since $1 \ J = 1 \ N m$,the unit $N m / T$ is equivalent to $J / T$.
Comparing these with the options,$J-T$ (Option $C$) is dimensionally incorrect as it represents energy multiplied by magnetic field,not magnetic dipole moment.

(B) The angular width of the central maximum in a single-slit diffraction pattern is given by the formula $\theta = \frac{2\lambda}{a}$,where $\lambda$ is the wavelength of the light and $a$ is the width of the slit.
From this relation,it is clear that the angular width $\theta$ is inversely proportional to the slit width $a$,i.e.,$\theta \propto \frac{1}{a}$.
Therefore,as the slit width $a$ increases,the angular width of the central maximum decreases.

(B) The angular width of the central maximum in a single-slit diffraction experiment is given by the formula $\theta = \frac{2 \lambda}{a}$,where $\lambda$ is the wavelength of the light used and $a$ is the width of the slit.
From this formula,it is clear that the angular width depends on the wavelength $\lambda$ and the slit width $a$.
It does not depend on the distance $D$ between the slit and the screen.
Therefore,the correct option is $B$.

(A) The fringe width $(z)$ in a biprism experiment (or Young's Double Slit Experiment) is given by the formula:
$z = \frac{\lambda D}{d}$
where $\lambda$ is the wavelength of light,$D$ is the distance between the slit and the eyepiece (screen),and $d$ is the distance between the two coherent sources.
Given that the wavelength $\lambda$ and the distance between the coherent sources $d$ are kept constant,we can write:
$z \propto D$
This implies that the ratio $\frac{z}{D}$ is a constant.
Therefore,$\frac{z_{1}}{D_{1}} = \frac{z_{2}}{D_{2}} = \frac{z_{3}}{D_{3}} = \frac{z_{4}}{D_{4}}$.
Thus,option $A$ is correct.