IIT JEE 2000 Physics Question Paper with Answer and Solution

(C) The resultant of $2$ vectors $\vec{A}$ and $\vec{B}$ is given by $\vec{R} = \vec{A} + \vec{B}$.
For the resultant to be zero,$\vec{A} + \vec{B} = 0$,which implies $\vec{A} = -\vec{B}$.
This means the two vectors must have the same magnitude $(|A| = |B|)$ but must be directed in opposite directions (an angle of $180^{\circ}$ or $\pi$ radians between them).
Therefore,the correct condition is that the $2$ vectors are same in magnitude but opposite in sense.

(A) For a ball dropped from height $d$,the velocity $v$ at any height $h$ is given by $v^2 = u^2 + 2a(s)$. Here,$u = 0$ and $a = -g$,so $v^2 = -2g(h - d) = 2g(d - h)$. This implies $v = \pm \sqrt{2g(d - h)}$.
$1$. During the downward motion from $h = d$ to $h = 0$,the velocity is negative (downward) and its magnitude increases as $h$ decreases. The relation $v = -\sqrt{2g(d - h)}$ represents a parabolic curve opening towards the positive $h$-axis.
$2$. At $h = 0$,the ball hits the ground. Just after the collision,it bounces up to a height $d/2$. The velocity becomes positive (upward) and its magnitude is determined by $v = \sqrt{2g(d/2 - h)}$.
$3$. As the ball moves upward from $h = 0$ to $h = d/2$,the velocity decreases from its maximum value to zero at $h = d/2$. This also follows a parabolic path.
Comparing these physical requirements with the given options,the graph that correctly represents the downward motion (negative velocity) and the subsequent upward motion (positive velocity) is option $(A)$.

(A) Let the bead start slipping after time $t$.
For the bead to slip,the net force acting on it must exceed the maximum static frictional force.
The forces acting on the bead in the rotating frame are the centrifugal force $F_c = m\omega^2 L$ (acting radially outward) and the tangential force $F_t = m a_t = m \alpha L$ (acting perpendicular to the rod).
The normal force $N$ is zero as gravity is neglected and there are no other vertical forces.
Wait,the friction force $f$ must balance the resultant of these forces. The maximum static friction is $f_{max} = \mu N$. Since there is no vertical force,$N$ is effectively provided by the constraint of the rod. However,in this specific problem context,the friction is assumed to be $\mu m a_t$ or similar. Re-evaluating: The bead slips when the resultant force $F_{net} = \sqrt{(m\omega^2 L)^2 + (m\alpha L)^2} = \mu N$. Assuming $N = m a_t = m \alpha L$ is not standard. Actually,the friction must oppose the tendency to move. The bead will slip when the net force required to keep it in circular motion exceeds the maximum available friction. The force required is $F = \sqrt{(m\omega^2 L)^2 + (m\alpha L)^2}$. Given the options,the intended logic is $m\omega^2 L = \mu m \alpha L$,which implies $\omega^2 = \mu \alpha$. Since $\omega = \alpha t$,we get $(\alpha t)^2 = \mu \alpha$,so $t^2 = \mu / \alpha$,or $t = \sqrt{\mu / \alpha}$.

(A) The process of heating ice from $-10^{\circ}C$ to steam at $100^{\circ}C$ involves several stages:
$1$. Heating ice from $-10^{\circ}C$ to $0^{\circ}C$: The temperature rises linearly as heat is supplied.
$2$. Melting ice at $0^{\circ}C$: The temperature remains constant at $0^{\circ}C$ while the phase changes from solid to liquid (latent heat of fusion).
$3$. Heating water from $0^{\circ}C$ to $100^{\circ}C$: The temperature rises linearly as heat is supplied.
$4$. Boiling water at $100^{\circ}C$: The temperature remains constant at $100^{\circ}C$ while the phase changes from liquid to gas (latent heat of vaporization).
Therefore, the heating curve must show two linear rising segments separated by a horizontal segment (melting) and followed by another horizontal segment (boiling). Option $A$ correctly represents this sequence.

(C) For an ideal gas at constant pressure,Charles's Law states that $V / T = \text{constant}.$
When the temperature changes from $T$ to $T + \Delta T,$ the volume changes from $V$ to $V + \Delta V.$
Thus,we have:
$\frac{V + \Delta V}{T + \Delta T} = \frac{V}{T}$
Cross-multiplying gives:
$T(V + \Delta V) = V(T + \Delta T)$
$VT + T \Delta V = VT + V \Delta T$
Subtracting $VT$ from both sides:
$T \Delta V = V \Delta T$
Rearranging to find the expression for $\delta$:
$\delta = \frac{\Delta V}{V \Delta T} = \frac{1}{T}$
Since $\delta = 1/T,$ the quantity $\delta$ is inversely proportional to the temperature $T.$ This relationship is represented by a rectangular hyperbola,which corresponds to the graph shown in option $C.$

(D) For an adiabatic process,the relationship between temperature $T$ and volume $V$ is given by $T V^{\gamma - 1} = \text{constant}$.
Thus,$T_1 V_1^{\gamma - 1} = T_2 V_2^{\gamma - 1}$,which implies $\frac{T_1}{T_2} = (\frac{V_2}{V_1})^{\gamma - 1}$.
Since the gas is monoatomic,the adiabatic index $\gamma = 5/3$. Therefore,$\gamma - 1 = 5/3 - 1 = 2/3$.
The volume of the gas in a cylinder is $V = A \times L$,where $A$ is the cross-sectional area and $L$ is the length of the gas column.
Substituting $V_1 = A L_1$ and $V_2 = A L_2$,we get $\frac{V_2}{V_1} = \frac{A L_2}{A L_1} = \frac{L_2}{L_1}$.
Substituting these into the temperature ratio equation: $\frac{T_1}{T_2} = (\frac{L_2}{L_1})^{2/3}$.

(B) According to Wien's displacement law,the wavelength corresponding to maximum intensity $(\lambda_m)$ is inversely proportional to the absolute temperature $(T)$,i.e.,$\lambda_m \propto \frac{1}{T}$.
From the given graph,we can observe the values of $\lambda_m$ for the three temperatures:
$(\lambda_m)_1 < (\lambda_m)_3 < (\lambda_m)_2$.
Since $\lambda_m$ is inversely proportional to $T$,a smaller $\lambda_m$ corresponds to a higher temperature.
Therefore,the relationship between the temperatures is $T_1 > T_3 > T_2$.

(A) The vehicle moves down a frictionless inclined plane with an acceleration $a = g\sin \alpha$.
In the frame of the vehicle,a pseudo force $F_p = ma = mg\sin \alpha$ acts on the bob of the pendulum in the upward direction along the incline.
The effective acceleration $g_{eff}$ is the vector sum of the acceleration due to gravity $\vec{g}$ and the negative of the vehicle's acceleration $-\vec{a}$.
Resolving the components,the component of $g$ perpendicular to the incline is $g\cos \alpha$ and the component of $g$ parallel to the incline is $g\sin \alpha$. The pseudo force cancels the component $g\sin \alpha$.
Thus,the effective acceleration is $g_{eff} = g\cos \alpha$.
The time period of the simple pendulum is given by $T = 2\pi \sqrt{\frac{L}{g_{eff}}} = 2\pi \sqrt{\frac{L}{g\cos \alpha}}$.

(B) The speed of sound $v$ in an ideal gas is given by the formula $v = \sqrt{\frac{\gamma RT}{M}}$,where $\gamma$ is the adiabatic index,$R$ is the universal gas constant,$T$ is the absolute temperature,and $M$ is the molar mass.
Since both gases are monoatomic,they have the same adiabatic index $\gamma = 5/3$.
Given that both gases are at the same temperature $T$,the speed of sound is inversely proportional to the square root of the molar mass: $v \propto \frac{1}{\sqrt{M}}$.
Therefore,the ratio of the speed of sound in gas $1$ $(v_1)$ to that in gas $2$ $(v_2)$ is $\frac{v_1}{v_2} = \sqrt{\frac{m_2}{m_1}}$.

(D) The fundamental frequency $n$ of a vibrating string is given by the formula: $n = \frac{1}{2l} \sqrt{\frac{T}{\mu}}$,where $\mu$ is the mass per unit length.
Since $\mu = \pi r^2 \rho$ (where $\rho$ is the density of the material),the frequency is $n = \frac{1}{2lr} \sqrt{\frac{T}{\pi \rho}}$.
Given that the material is the same,$\rho$ is constant. Since the tension $T$ is also the same,we have $n \propto \frac{1}{lr}$.
Therefore,the ratio of the frequencies is $\frac{n_1}{n_2} = \frac{l_2 r_2}{l_1 r_1}$.
Substituting the given values: $l_1 = L$,$l_2 = 2L$,$r_1 = 2r$,and $r_2 = r$.
$\frac{n_1}{n_2} = \frac{2L \times r}{L \times 2r} = \frac{2Lr}{2Lr} = 1$.

(D) According to the Doppler effect,when a source moves towards a stationary observer,the observed frequency $f'$ is given by $f' = f \left( \frac{v}{v - v_s} \right)$,where $v$ is the speed of sound and $v_s$ is the speed of the source.
For the first case,$v_s = 34 \ m/s$:
$f_1 = f \left( \frac{340}{340 - 34} \right) = f \left( \frac{340}{306} \right)$
For the second case,$v_s = 17 \ m/s$:
$f_2 = f \left( \frac{340}{340 - 17} \right) = f \left( \frac{340}{323} \right)$
Now,calculating the ratio $f_1/f_2$:
$\frac{f_1}{f_2} = \frac{f \left( \frac{340}{306} \right)}{f \left( \frac{340}{323} \right)} = \frac{323}{306}$
Dividing both numerator and denominator by $17$:
$\frac{323 \div 17}{306 \div 17} = \frac{19}{18}$
Thus,the ratio $f_1/f_2$ is $19/18$.

(C) To find the minimum force required to topple the block,we consider the torque about the edge of the block where it is about to rotate (the pivot point $P$).
At the critical condition for toppling,the normal reaction $N$ acts through the pivot point $P$.
The torque due to the gravitational force $mg$ about point $P$ is $\tau_{mg} = mg \times (L/2)$.
The torque due to the applied horizontal force $F$ about point $P$ is $\tau_{F} = F \times L$.
For the block to topple,the torque due to the applied force must be greater than the torque due to gravity:
$\tau_{F} > \tau_{mg}$
$F \times L > mg \times (L/2)$
$F > mg/2$
Thus,the minimum force required to topple the block is $F = mg/2$.

(D) The total mass of the wire is $M = \rho L$. Since the circumference of the loop is $L = 2\pi R$,the radius of the loop is $R = \frac{L}{2\pi}$.
The moment of inertia of a circular loop about its diameter is $I_{diam} = \frac{1}{2}MR^2$.
Using the parallel axis theorem,the moment of inertia about the tangent $XX'$ is $I = I_{cm} + Md^2$,where $I_{cm} = I_{diam} = \frac{1}{2}MR^2$ and $d = R$.
Thus,$I = \frac{1}{2}MR^2 + MR^2 = \frac{3}{2}MR^2$.
Substituting $M = \rho L$ and $R = \frac{L}{2\pi}$:
$I = \frac{3}{2}(\rho L)\left(\frac{L}{2\pi}\right)^2 = \frac{3}{2}\rho L \left(\frac{L^2}{4\pi^2}\right) = \frac{3\rho L^3}{8\pi^2}$.

(B) $1$. The system consists of the triangle and the two beads. Since there is no external torque acting on the system about the vertical axis $AO$,the total angular momentum $L$ of the system remains conserved.
$2$. As the beads slide down,the gravitational potential energy of the beads is converted into kinetic energy (both translational and rotational). Since there is no friction,the total mechanical energy (kinetic + potential) of the system is conserved.
$3$. The moment of inertia $I$ of the system about the axis $AO$ increases as the beads move away from the axis. Since $L = I\omega$ is constant,the angular velocity $\omega$ must decrease.
$4$. Therefore,the quantities that are conserved are the total angular momentum and the total energy.

(A) The volume flow rate (quantity of water per second) is given by $Q = A v$,where $A$ is the area of the hole and $v$ is the velocity of efflux.
According to Torricelli's law,the velocity of efflux at a depth $h$ is $v = \sqrt{2gh}$.
For the square hole: Area $A_1 = L^2$ and depth $h_1 = y$. So,$v_1 = \sqrt{2gy}$.
The flow rate $Q_1 = A_1 v_1 = L^2 \sqrt{2gy}$.
For the circular hole: Area $A_2 = \pi R^2$ and depth $h_2 = 4y$. So,$v_2 = \sqrt{2g(4y)} = 2\sqrt{2gy}$.
The flow rate $Q_2 = A_2 v_2 = \pi R^2 (2\sqrt{2gy})$.
Given that $Q_1 = Q_2$,we have:
$L^2 \sqrt{2gy} = 2\pi R^2 \sqrt{2gy}$.
Canceling $\sqrt{2gy}$ from both sides,we get $L^2 = 2\pi R^2$.
Therefore,$R^2 = \frac{L^2}{2\pi}$,which gives $R = \frac{L}{\sqrt{2\pi}}$.

(B) The capacitor can be viewed as a combination of three capacitors. The top half consists of two capacitors in parallel with areas $A/2$ and separation $d/2$,having capacitances $C_1 = \frac{k_1 \epsilon_0 (A/2)}{d/2} = \frac{k_1 \epsilon_0 A}{d}$ and $C_2 = \frac{k_2 \epsilon_0 (A/2)}{d/2} = \frac{k_2 \epsilon_0 A}{d}$.
These two are in parallel,so their equivalent capacitance is $C_{12} = C_1 + C_2 = \frac{\epsilon_0 A}{d} (k_1 + k_2)$.
The bottom half is a capacitor with area $A$ and separation $d/2$,having capacitance $C_3 = \frac{k_3 \epsilon_0 A}{d/2} = \frac{2k_3 \epsilon_0 A}{d}$.
Since $C_{12}$ and $C_3$ are in series,the total capacitance $C$ is given by $\frac{1}{C} = \frac{1}{C_{12}} + \frac{1}{C_3}$.
Substituting the values: $\frac{1}{C} = \frac{d}{\epsilon_0 A (k_1 + k_2)} + \frac{d}{2k_3 \epsilon_0 A} = \frac{d}{\epsilon_0 A} \left( \frac{1}{k_1 + k_2} + \frac{1}{2k_3} \right)$.
For a single dielectric $k$,$C = \frac{k \epsilon_0 A}{d}$,so $\frac{1}{C} = \frac{d}{k \epsilon_0 A}$.
Comparing the two expressions,we get $\frac{1}{k} = \frac{1}{k_1 + k_2} + \frac{1}{2k_3}$.

(C) The magnetic field at any point lying on the axis of a current-carrying straight conductor is zero.
Case $1$: The current $I$ flows through $PQ$ and $QR$. Point $M$ lies on the extension of $QR$. Therefore,the magnetic field at $M$ due to $QR$ is zero. The magnetic field $H_1$ at $M$ is only due to the segment $PQ$. Let $d$ be the perpendicular distance from $M$ to $PQ$. The magnetic field due to a semi-infinite wire is $H_1 = \frac{\mu_0 I}{4 \pi d}$.
Case $2$: The current in $PQ$ is $I$,in $QR$ is $I/2$,and in $QS$ is $I/2$. Point $M$ lies on the extension of $QR$,so the magnetic field due to $QR$ is still zero. The magnetic field at $M$ is now $H_2 = H_{PQ} + H_{QS}$.
Since the current in $PQ$ is unchanged,$H_{PQ} = H_1 = \frac{\mu_0 I}{4 \pi d}$.
The conductor $QS$ is perpendicular to $PQ$ and $M$ lies on the perpendicular bisector of the line passing through $QS$ if we consider the geometry. However,$M$ is at a distance $d$ from $Q$ along the line $QR$. The magnetic field at $M$ due to a semi-infinite wire $QS$ carrying current $I/2$ is $H_{QS} = \frac{\mu_0 (I/2)}{4 \pi d} = \frac{1}{2} H_1$.
Thus,$H_2 = H_1 + \frac{1}{2} H_1 = \frac{3}{2} H_1$.
The ratio $H_1/H_2 = H_1 / (\frac{3}{2} H_1) = 2/3 \approx 0.67$.

(C) The electric field $\vec{E}$ is along the $+x$ direction. Positive ions experience an electric force $\vec{F}_e = q\vec{E}$ along the $+x$ direction,while negative ions experience an electric force along the $-x$ direction.
As a result,positive ions acquire a velocity $\vec{v}$ in the $+x$ direction,and negative ions acquire a velocity $\vec{v}$ in the $-x$ direction.
The magnetic field $\vec{B}$ is along the $+z$ direction.
The magnetic force on a charged particle is given by $\vec{F}_m = q(\vec{v} \times \vec{B})$.
For positive ions: $\vec{F}_m = (+q)(v\hat{i} \times B\hat{k}) = -qvB\hat{j}$,which is in the $-y$ direction.
For negative ions: $\vec{F}_m = (-q)(-v\hat{i} \times B\hat{k}) = (-q)(-vB(-\hat{j})) = -qvB\hat{j}$,which is also in the $-y$ direction.
Thus,both types of ions deflect towards the $-y$ direction.

(B) If the current flows out of the paper,the magnetic field at points to the right of the wire will be upwards and to the left will be downward. Let the wires be at $A$ and $B$,and the midpoint be $C$. The magnetic field at $C$ is zero because the fields from $A$ and $B$ cancel each other out.
In the region to the right of $B$,the magnetic field is upwards $(+ve)$ because all points are to the right of both wires. Similarly,in the region to the left of $A$,the magnetic field is downwards $(-ve)$.
In the region $AC$,the points are closer to $A$ than to $B$,so the field due to $A$ dominates and is upwards $(+ve)$.
In the region $BC$,the points are closer to $B$ than to $A$,so the field due to $B$ dominates and is downwards $(-ve)$.
Graph $(b)$ correctly represents these variations in the magnetic field.

(D) Let the two magnets be placed such that point $P$ is at a distance $d$ from the center of each magnet.
For the first magnet,point $P$ lies on its axial line. The magnetic field due to this magnet at $P$ is $B_1 = \frac{\mu_0}{4\pi} \cdot \frac{2M}{d^3}$.
For the second magnet,point $P$ lies on its equatorial line. The magnetic field due to this magnet at $P$ is $B_2 = \frac{\mu_0}{4\pi} \cdot \frac{M}{d^3}$.
Since the axes are perpendicular,the fields $B_1$ and $B_2$ are perpendicular to each other.
The net magnetic field at $P$ is $B_{net} = \sqrt{B_1^2 + B_2^2}$.
Substituting the values: $B_{net} = \sqrt{\left(\frac{\mu_0}{4\pi} \cdot \frac{2M}{d^3}\right)^2 + \left(\frac{\mu_0}{4\pi} \cdot \frac{M}{d^3}\right)^2}$.
$B_{net} = \frac{\mu_0}{4\pi} \cdot \frac{M}{d^3} \sqrt{2^2 + 1^2} = \frac{\mu_0}{4\pi} \cdot \frac{\sqrt{5}M}{d^3}$.

(B) To find the induced electric field at a point $P$ outside the circular region $(r > a)$,we use Faraday's law of induction in integral form: $\oint \vec{E} \cdot d\vec{l} = -\frac{d\phi_B}{dt}$.
Consider a concentric circular path of radius $r$ passing through point $P$. Due to symmetry,the magnitude of the induced electric field $E$ is constant along this path,and $\vec{E}$ is tangential to the circle.
Thus,$\oint \vec{E} \cdot d\vec{l} = E(2\pi r)$.
The magnetic flux $\phi_B$ through this circular path is limited to the region of radius $a$ where the magnetic field exists: $\phi_B = B(t) \cdot (\pi a^2)$.
Applying Faraday's law:
$E(2\pi r) = \left| \frac{d}{dt} (B(t) \cdot \pi a^2) \right|$
$E(2\pi r) = \pi a^2 \left| \frac{dB}{dt} \right|$
$E = \frac{a^2}{2r} \left| \frac{dB}{dt} \right|$
Since $a$ and $\frac{dB}{dt}$ are constant,we have $E \propto \frac{1}{r}$.
Therefore,the magnitude of the induced electric field decreases as $\frac{1}{r}$.

(D) Let $k_1, k_2, k_3, k_4, k_5$ be constants.
The current in the $RL$ circuit is given by $I_1(t) = k_1(1 - e^{-t/\tau})$.
The magnetic field $B(t)$ at the axis of the coil is proportional to the current $I_1(t)$,so $B(t) = k_2 I_1(t) = k_2 k_1(1 - e^{-t/\tau})$.
The induced current $I_2(t)$ in the ring is proportional to the rate of change of magnetic flux,which is proportional to $dB(t)/dt$. Thus,$I_2(t) = k_3 \frac{dB(t)}{dt} = k_4 e^{-t/\tau}$.
Therefore,the product $I_2(t) B(t) = k_5 (1 - e^{-t/\tau}) e^{-t/\tau} = k_5 (e^{-t/\tau} - e^{-2t/\tau})$.
At $t = 0$,$I_2(t) B(t) = 0$. As $t \to \infty$,$I_2(t) B(t) \to 0$. Since the product is positive for $t > 0$,it must pass through a maximum.

(D) The minimum wavelength of the continuous $X$-ray spectrum is given by the formula $\lambda_{\min} = \frac{hc}{E}$.
Using $hc \approx 12400 \ eV \cdot \mathring{A}$ (or $12375 \ eV \cdot \mathring{A}$),we get $\lambda_{\min} = \frac{12375}{80 \times 10^3} \approx 0.155 \ \mathring{A}$.
Since the energy of the incident electrons $(80 \ keV)$ is greater than the ionization energy of the $K$-shell electrons $(72.5 \ keV)$,the incident electrons have sufficient energy to eject $K$-shell electrons from the tungsten atoms.
When an electron is ejected from the $K$-shell,electrons from higher energy levels transition to fill the vacancy,resulting in the emission of characteristic $X$-rays.
Therefore,the emitted $X$-rays consist of both a continuous spectrum (Bremsstrahlung) and the characteristic $X$-ray spectrum of tungsten.

(A) For a hydrogen atom,the total energy $E_n$,potential energy $U_n$,and kinetic energy $K_n$ are given by:
$E_n = -13.6 \frac{Z^2}{n^2} \text{ eV}$
$U_n = 2E_n = -27.2 \frac{Z^2}{n^2} \text{ eV}$
$K_n = |E_n| = 13.6 \frac{Z^2}{n^2} \text{ eV}$
When an electron transitions from an excited state $(n > 1)$ to the ground state $(n = 1)$,the value of $n$ decreases.
As $n$ decreases,$K_n$ increases because it is proportional to $1/n^2$.
Since $E_n$ and $U_n$ are negative and their magnitudes decrease as $n$ decreases,their values become more negative,meaning they decrease.
Therefore,kinetic energy increases,while potential and total energies decrease.

(C) In a hydrogen atom,the energy of the $n^{th}$ level is given by $E_n = -\frac{Rhc}{n^2}$.
Since the energy $E_n$ is directly proportional to the mass of the orbiting particle $(E_n \propto m)$,and the new particle has a mass $m' = 2m_e$,the energy levels for this hypothetical atom become $E'_n = -\frac{2Rhc}{n^2}$.
The longest wavelength photon corresponds to the minimum energy transition,which occurs between $n = 3$ and $n = 2$.
The energy difference is $\Delta E = E'_3 - E'_2 = -\frac{2Rhc}{3^2} - (-\frac{2Rhc}{2^2}) = 2Rhc \left( \frac{1}{4} - \frac{1}{9} \right)$.
Simplifying this,$\Delta E = 2Rhc \left( \frac{5}{36} \right) = \frac{5Rhc}{18}$.
Since $\Delta E = \frac{hc}{\lambda}$,we have $\frac{hc}{\lambda} = \frac{5Rhc}{18}$.
Therefore,$\lambda = \frac{18}{5R}$.

(D) Let $N_0$ be the initial number of radioactive atoms for both elements at $t = 0$.
According to the law of radioactive decay,the number of atoms remaining at time $t$ is given by $N(t) = N_0 e^{-\lambda t}$.
For element $X_1$ with decay constant $\lambda_1 = 10\lambda$,the number of atoms is $N_1 = N_0 e^{-10\lambda t}$.
For element $X_2$ with decay constant $\lambda_2 = \lambda$,the number of atoms is $N_2 = N_0 e^{-\lambda t}$.
We are given that the ratio $\frac{N_1}{N_2} = \frac{1}{e} = e^{-1}$.
Substituting the expressions for $N_1$ and $N_2$:
$\frac{N_0 e^{-10\lambda t}}{N_0 e^{-\lambda t}} = e^{-1}$
$e^{-10\lambda t + \lambda t} = e^{-1}$
$e^{-9\lambda t} = e^{-1}$
Comparing the exponents:
$-9\lambda t = -1$
$t = \frac{1}{9\lambda}$.

(B) Let the ground state energy of the hydrogen-like atom be $E_0 = -13.6 Z^2 \ eV$. The energy of an electron in state $k$ is given by $E_k = \frac{E_0}{k^2}$.
Given that the atom is in state $2n$ and emits a maximum energy photon of $204 \ eV$ to reach the ground state $(k=1)$:
$E_{2n} - E_1 = 204 \ eV$
$\frac{E_0}{(2n)^2} - E_0 = 204 \ eV$
$E_0 \left( \frac{1}{4n^2} - 1 \right) = 204 \ eV$ --- $(i)$
Given that the atom makes a transition from state $2n$ to state $n$ emitting a photon of $40.8 \ eV$:
$E_{2n} - E_n = 40.8 \ eV$
$\frac{E_0}{4n^2} - \frac{E_0}{n^2} = 40.8 \ eV$
$E_0 \left( \frac{1 - 4}{4n^2} \right) = 40.8 \ eV$
$E_0 \left( -\frac{3}{4n^2} \right) = 40.8 \ eV$ --- $(ii)$
Dividing equation $(i)$ by equation $(ii)$:
$\frac{E_0 \left( \frac{1 - 4n^2}{4n^2} \right)}{E_0 \left( -\frac{3}{4n^2} \right)} = \frac{204}{40.8}$
$\frac{1 - 4n^2}{-3} = 5$
$1 - 4n^2 = -15$
$4n^2 = 16$
$n^2 = 4$
$n = 2$

(C) In a compound microscope,the objective lens forms an image of the object placed just beyond its focal point.
This image is known as the intermediate image.
Since the light rays actually converge to form this image,it is real.
Because the object is placed beyond the focal point of the objective lens,the image formed is inverted and magnified relative to the object.
Therefore,the intermediate image is real,inverted,and magnified.

(B) Let the distance between the two mirrors be $h = 0.2 \ m$ and the length of the mirrors be $L = 2\sqrt{3} \ m$.
The angle of incidence is $i = 30^\circ$.
When the ray reflects between the two parallel mirrors,the horizontal distance $d$ covered by the ray between two consecutive reflections is given by:
$d = h \tan(i) = 0.2 \tan(30^\circ) = 0.2 \times \frac{1}{\sqrt{3}} \ m$.
The total horizontal distance covered by the ray before it emerges is $L = 2\sqrt{3} \ m$.
The number of reflections $n$ is given by the ratio of the total length to the horizontal distance per reflection:
$n = \frac{L}{d} = \frac{2\sqrt{3}}{0.2 / \sqrt{3}} = \frac{2 \times 3}{0.2} = \frac{6}{0.2} = 30$.
Thus,the maximum number of reflections is $30$.

(A) For the ray to emerge only from the surface $CD$,it must undergo total internal reflection at the surface $AD$.
Applying Snell's law at the surface $AB$:
$n_2 \sin \alpha_{max} = n_1 \sin r_1 \implies \alpha_{max} = \sin^{-1} \left( \frac{n_1}{n_2} \sin r_1 \right) \dots (i)$
At the surface $AD$,the angle of incidence is $r_2$. For total internal reflection to occur,$r_2$ must be at least the critical angle $C$,where $\sin C = \frac{n_2}{n_1}$.
Since the surface $AD$ is perpendicular to $AB$,we have $r_1 + r_2 = 90^\circ$,so $r_1 = 90^\circ - r_2$.
To ensure the ray reaches $CD$,we set $r_2 = C = \sin^{-1} \left( \frac{n_2}{n_1} \right)$.
Thus,$r_1 = 90^\circ - \sin^{-1} \left( \frac{n_2}{n_1} \right)$.
Substituting this into equation $(i)$:
$\alpha_{max} = \sin^{-1} \left[ \frac{n_1}{n_2} \sin \left( 90^\circ - \sin^{-1} \frac{n_2}{n_1} \right) \right]$
Using the identity $\sin(90^\circ - \theta) = \cos \theta$:
$\alpha_{max} = \sin^{-1} \left[ \frac{n_1}{n_2} \cos \left( \sin^{-1} \frac{n_2}{n_1} \right) \right]$.

(B) When a light ray passes through a glass slab with parallel faces,the emergent ray is parallel to the incident ray.
This phenomenon is known as lateral displacement.
Since the incident rays are diverging at an angle $\alpha$,and each ray undergoes the same lateral displacement without changing its direction relative to the normal,the emergent rays will also diverge at the same angle $\alpha$.
Therefore,the divergence angle of the emergent beam remains $\alpha$.

(D) The focal length $f$ of a lens is given by the Lens Maker's formula:
$\frac{1}{f} = \left( \frac{n_L}{n_m} - 1 \right) \left( \frac{1}{R_1} - \frac{1}{R_2} \right)$
where $n_L$ is the refractive index of the lens material and $n_m$ is the refractive index of the surrounding medium.
For a double concave lens,the term $\left( \frac{1}{R_1} - \frac{1}{R_2} \right)$ is negative.
For the lens to diverge light,it must act as a concave lens,meaning its focal length $f$ must be negative.
Since $\left( \frac{1}{R_1} - \frac{1}{R_2} \right) < 0$,the term $\left( \frac{n_L}{n_m} - 1 \right)$ must be positive for $f$ to be negative.
This implies $\frac{n_L}{n_m} > 1$,or $n_L > n_m$.
In option $(d)$,the lens is filled with liquid $L_2$ (refractive index $n_2$) and immersed in liquid $L_1$ (refractive index $n_1$). Since $n_2 > n_1$,the condition $n_L > n_m$ is satisfied,and the lens will diverge the light.

(A) In a standard Young's Double Slit Experiment $(YDSE)$ with equal slit widths, the amplitudes of the waves from both slits are equal, say $a$. The maximum intensity is $I_{max} \propto (a + a)^2 = 4a^2$ and the minimum intensity is $I_{min} \propto (a - a)^2 = 0$.
When one slit is made twice as wide as the other, the intensity of the light passing through it increases. Since intensity $I \propto \text{width}$, if the width of the first slit is $w$, the width of the second is $2w$. The amplitudes are related by $A_2 = \sqrt{2} A_1$. Let $A_1 = a$, then $A_2 = a\sqrt{2}$.
The new maximum intensity is $I'_{max} \propto (a + a\sqrt{2})^2 = a^2(1 + \sqrt{2})^2 \approx 5.83a^2$, which is greater than $4a^2$.
The new minimum intensity is $I'_{min} \propto (a\sqrt{2} - a)^2 = a^2(\sqrt{2} - 1)^2 \approx 0.17a^2$, which is greater than $0$.
Thus, the intensities of both the maxima and the minima increase.

(B) The expression $\frac{1}{2} \varepsilon_0 E^2$ represents the energy density of an electric field.
Energy density is defined as the energy per unit volume.
Dimensional formula of energy is $[M^1L^2T^{-2}]$.
Dimensional formula of volume is $[L^3]$.
Therefore,the dimensional formula of energy density is $\frac{[M^1L^2T^{-2}]}{[L^3]} = [M^1L^{-1}T^{-2}]$.
Thus,the dimension of $\frac{1}{2} \varepsilon_0 E^2$ is $[M^1L^{-1}T^{-2}]$.

(C) Let the mirror be placed along the $y$-axis from $y = -d/2$ to $y = d/2$. The source $S$ is at $(L, 0)$. The image $S'$ of the source $S$ is formed at $(-L, 0)$.
The man walks along the line $x = 2L$. The rays from the image $S'$ that reach the man must pass through the edges of the mirror.
The rays from $S'$ passing through the top edge $(0, d/2)$ and bottom edge $(0, -d/2)$ of the mirror define the field of view.
Using similar triangles,the height $h$ of the field of view at a distance $x = 2L$ from the mirror is given by the ratio of distances.
The distance from the image $S'$ to the mirror is $L$,and the distance from the image $S'$ to the man is $L + 2L = 3L$.
By similar triangles,the width of the field of view $h$ is given by $\frac{h}{d} = \frac{3L}{L} = 3$.
Therefore,$h = 3d$.

(D) The total electrostatic potential energy $U$ of a system of point charges is the sum of the potential energies of all distinct pairs of charges.
For the given configuration,the pairs are $(Q, +q)$,$(+q, +q)$,and $(Q, +q)$ with distances $l$,$l$,and $\sqrt{2}l$ respectively.
The total potential energy is given by:
$U = \frac{kQq}{l} + \frac{kq^2}{l} + \frac{kQq}{\sqrt{2}l} = 0$
Dividing by $k/l$ (assuming $k \neq 0$ and $l \neq 0$):
$Qq + q^2 + \frac{Qq}{\sqrt{2}} = 0$
$Qq(1 + \frac{1}{\sqrt{2}}) = -q^2$
$Qq(\frac{\sqrt{2}+1}{\sqrt{2}}) = -q^2$
$Q = -q^2 \cdot \frac{\sqrt{2}}{q(\sqrt{2}+1)}$
$Q = -\frac{\sqrt{2}q}{\sqrt{2}+1}$

(C) The magnetic moment $M$ of a particle of charge $q$ moving in a circular orbit is given by $M = IA$,where $I$ is the current and $A$ is the area of the orbit.
The current $I = \frac{q}{T} = \frac{q\omega}{2\pi}$.
The area $A = \pi r^2$.
Thus,$M = \left(\frac{q\omega}{2\pi}\right)(\pi r^2) = \frac{q\omega r^2}{2}$.
The angular momentum $L$ of the particle is $L = mvr = m(\omega r)r = m\omega r^2$.
The ratio of the magnetic moment to the angular momentum is $\frac{M}{L} = \frac{q\omega r^2 / 2}{m\omega r^2} = \frac{q}{2m}$.
This ratio is known as the gyromagnetic ratio and depends only on the charge '$q$' and mass '$m$' of the particle.