WBJEE 2014 Physics Question Paper with Answer and Solution

(B) For a satellite in a circular orbit of radius $r$ around the Earth of mass $M$,the potential energy $U$ is given by $U = -\frac{G M m}{r}$.
The kinetic energy $K$ of the satellite is given by $K = \frac{G M m}{2 r}$.
The total energy $E$ is the sum of potential and kinetic energy:
$E = U + K = -\frac{G M m}{r} + \frac{G M m}{2 r} = -\frac{G M m}{2 r}$.
Comparing the expressions for $U$ and $E$:
$U = -\frac{G M m}{r}$
$E = -\frac{G M m}{2 r}$
Therefore,$U = 2 \times (- \frac{G M m}{2 r}) = 2 E$.
Wait,let us re-evaluate: $U = -\frac{G M m}{r}$ and $E = -\frac{G M m}{2 r}$.
Thus,$U = 2 E$ is incorrect. The correct relation is $U = 2 E$ if $E$ were defined differently,but here $U = 2 E$ is not correct. Let's re-check: $U = -\frac{G M m}{r}$ and $E = -\frac{G M m}{2 r}$.
So,$U = 2 \times (-\frac{G M m}{2 r}) = 2 E$.
Actually,$U = 2 E$ is correct because $2 \times (- \frac{G M m}{2 r}) = - \frac{G M m}{r} = U$.

(A) For a smooth inclined plane,the acceleration is $a_s = g \sin \theta$. The time taken to cover distance $d$ is $t_s = \sqrt{2d / a_s} = \sqrt{2d / (g \sin \theta)}$.
For a rough inclined plane,the acceleration is $a_r = g(\sin \theta - \mu \cos \theta)$. The time taken to cover distance $d$ is $t_r = \sqrt{2d / a_r} = \sqrt{2d / (g(\sin \theta - \mu \cos \theta))}$.
Given $t_r = t$ and $t_s = t/2$,we have $t_r = 2t_s$.
Therefore,$\sqrt{2d / (g(\sin \theta - \mu \cos \theta))} = 2 \sqrt{2d / (g \sin \theta)}$.
Squaring both sides: $1 / (\sin \theta - \mu \cos \theta) = 4 / \sin \theta$.
$\sin \theta = 4 \sin \theta - 4 \mu \cos \theta$.
$4 \mu \cos \theta = 3 \sin \theta$.
$\mu = (3/4) \tan \theta$.
Substituting $\theta = 45^{\circ}$,$\mu = (3/4) \tan 45^{\circ} = 3/4 \times 1 = 3/4$.

(B) For a system of two masses connected by a string over a pulley,the acceleration of each mass is given by $a = \frac{m_{1} - m_{2}}{m_{1} + m_{2}} g$.
Since the two masses move in opposite directions,the acceleration of the center of mass $(a_{CM})$ is calculated as:
$a_{CM} = \frac{m_{1}a_{1} + m_{2}a_{2}}{m_{1} + m_{2}}$.
Taking the direction of $m_{1}$ as positive,$a_{1} = a$ and $a_{2} = -a$.
$a_{CM} = \frac{m_{1}a - m_{2}a}{m_{1} + m_{2}} = \left(\frac{m_{1} - m_{2}}{m_{1} + m_{2}}\right) a$.
Substituting the value of $a$:
$a_{CM} = \left(\frac{m_{1} - m_{2}}{m_{1} + m_{2}}\right) \times \left(\frac{m_{1} - m_{2}}{m_{1} + m_{2}}\right) g = \left(\frac{m_{1} - m_{2}}{m_{1} + m_{2}}\right)^{2} g$.
The total external force acting on the system is $F_{ext} = (m_{1} + m_{2}) a_{CM}$.
$F_{ext} = (m_{1} + m_{2}) \times \left(\frac{m_{1} - m_{2}}{m_{1} + m_{2}}\right)^{2} g = \frac{(m_{1} - m_{2})^{2}}{m_{1} + m_{2}} g$.

(C) The buoyant force acting on an immersed body is equal to the weight of the fluid displaced by it when the fluid is at rest. When the fluid is accelerating,the effective gravity $g_{\text{eff}}$ must be considered.
In the first case,the block is in equilibrium: $V_{\text{immersed}} \rho g = V_{\text{total}} \rho_b g$,where $\rho$ is the density of water and $\rho_b$ is the density of the block.
Given $40 \%$ is above the surface,$60 \%$ is immersed,so $V_{\text{immersed}} = 0.6 V_{\text{total}}$.
Thus,$0.6 V \rho g = V \rho_b g \implies \rho_b = 0.6 \rho$.
When the lift accelerates upward with $a = g/2$,the effective gravity is $g_{\text{eff}} = g + a = 1.5 g$.
The buoyant force becomes $F_B = V_{\text{immersed}} \rho (1.5 g)$.
The weight of the block in the accelerating frame is $W_{\text{eff}} = V_{\text{total}} \rho_b (1.5 g)$.
For equilibrium,$F_B = W_{\text{eff}} \implies V_{\text{immersed}} \rho (1.5 g) = V_{\text{total}} \rho_b (1.5 g)$.
Dividing by $1.5 g$,we get $V_{\text{immersed}} \rho = V_{\text{total}} \rho_b$.
Substituting $\rho_b = 0.6 \rho$,we get $V_{\text{immersed}} = 0.6 V_{\text{total}}$.
Therefore,the fraction of the volume immersed remains $60 \%$,and the fraction above the surface remains $40 \%$.

(C) For the rod to be in horizontal equilibrium,the net torque about the mid-point must be zero.
Initially,the torque balance is: $w \cdot l = w_{1} \cdot l_{1}$.
When the metal piece of weight $w$ is immersed in water,it experiences an upward buoyant force $F_{B}$. The effective weight becomes $w' = w - F_{B}$.
The buoyant force is $F_{B} = V \rho_{w} g$,where $V$ is the volume of the metal and $\rho_{w}$ is the density of water. Since $w = V \rho_{metal} g$,we have $F_{B} = w \cdot \frac{\rho_{w}}{\rho_{metal}} = \frac{w}{\sigma}$,where $\sigma$ is the specific gravity of the metal.
Thus,the new effective weight is $w' = w(1 - \frac{1}{\sigma})$.
For the new equilibrium,the torque balance is: $w' \cdot l = w_{1} \cdot l_{2}$.
Substituting $w'$,we get: $w(1 - \frac{1}{\sigma}) l = w_{1} l_{2}$.
From the initial condition,$w = \frac{w_{1} l_{1}}{l}$.
Substituting $w$ into the new equilibrium equation: $\frac{w_{1} l_{1}}{l} (1 - \frac{1}{\sigma}) l = w_{1} l_{2}$.
$l_{1} (1 - \frac{1}{\sigma}) = l_{2}$.
$1 - \frac{1}{\sigma} = \frac{l_{2}}{l_{1}}$.
$\frac{1}{\sigma} = 1 - \frac{l_{2}}{l_{1}} = \frac{l_{1} - l_{2}}{l_{1}}$.
Therefore,$\sigma = \frac{l_{1}}{l_{1} - l_{2}}$.

(A) Let $x$ be the weight of the first metal and $(w_1 - x)$ be the weight of the second metal in air.
Let $v_1$ and $v_2$ be the volumes of the two metals respectively.
$v_1 = \frac{x}{\rho_1}$ and $v_2 = \frac{w_1 - x}{\rho_2}$.
The total volume of the alloy is $V = v_1 + v_2 = \frac{x}{\rho_1} + \frac{w_1 - x}{\rho_2}$.
According to Archimedes' Principle,the loss in weight equals the weight of the displaced water:
$w_1 - w_2 = V \rho_w = \left( \frac{x}{\rho_1} + \frac{w_1 - x}{\rho_2} \right) \rho_w$.
Multiplying by $\rho_1 \rho_2$:
$(w_1 - w_2) \rho_1 \rho_2 = (x \rho_2 + (w_1 - x) \rho_1) \rho_w$.
$(w_1 - w_2) \rho_1 \rho_2 = (x \rho_2 + w_1 \rho_1 - x \rho_1) \rho_w$.
$(w_1 - w_2) \rho_1 \rho_2 = x(\rho_2 - \rho_1) \rho_w + w_1 \rho_1 \rho_w$.
$x(\rho_2 - \rho_1) \rho_w = w_1 \rho_1 \rho_2 - w_2 \rho_1 \rho_2 - w_1 \rho_1 \rho_w$.
$x(\rho_2 - \rho_1) \rho_w = w_1 \rho_1(\rho_2 - \rho_w) - w_2 \rho_1 \rho_2$.
$x = \frac{\rho_1}{\rho_w(\rho_2 - \rho_1)} [w_1(\rho_2 - \rho_w) - w_2 \rho_2]$.

(C) Let the thickness of the liquid layer be $x$.
The volume $V$ of the liquid is given by $V = A \times x$,where $A$ is the area of the layer.
Thus,$x = V / A$.
The liquid forms a thin film between two glass plates,creating a concave meniscus with radius of curvature $r$. For a thin film,the thickness $x$ is equal to the diameter of the meniscus,so $x = 2r$,which implies $r = x / 2 = V / (2A)$.
The pressure difference $\Delta P$ across the curved surface is given by $\Delta P = T / r$,where $T$ is the surface tension.
The force $F$ required to separate the plates is $F = \Delta P \times A$.
Substituting $\Delta P = T / r$ and $r = V / (2A)$,we get:
$F = (T / (V / (2A))) \times A = (2AT / V) \times A = (2A^{2}T) / V$.
Rearranging for $T$: $T = (F \times V) / (2A^{2})$.
Given: $F = 16 \times 10^{5} \ dyne$,$V = 0.04 \ cm^{3}$,$A = 20 \ cm^{2}$.
$T = (16 \times 10^{5} \times 0.04) / (2 \times 20^{2}) = (16 \times 10^{5} \times 0.04) / (2 \times 400) = (0.64 \times 10^{5}) / 800 = 64000 / 800 = 80 \ dyne \ cm^{-1}$.

(C) According to Stokes' Law,when a small sphere of radius $a$ moves with a velocity $v$ through a viscous fluid with a coefficient of viscosity $\eta$,it experiences a drag force (viscous force) that opposes its motion.
The formula for this viscous force $F$ is given by:
$F = 6 \pi \eta a v$
This force acts in the direction opposite to the velocity of the sphere.

(D) When a rod is fixed at both ends,its thermal expansion is prevented,resulting in thermal stress.
Thermal strain is given by $\epsilon = \frac{\Delta L}{L} = \alpha \cdot \Delta T$.
According to Hooke's Law,Stress $\sigma = Y \cdot \text{Strain}$.
Substituting the strain value,we get $\sigma = Y \cdot \alpha \cdot \Delta T$.
Since $Y$,$\alpha$,and $\Delta T$ are the factors determining the stress,the stress is independent of the length $L$ of the rod.
Therefore,the correct statement is that the stress is independent of $L$.

(A) The vertical displacement $y$ of a projectile at time $t$ is given by $y = (u \sin \theta)t - \frac{1}{2}gt^2$.
For heights $h_1$ and $h_2$ at times $t_1$ and $t_2$:
$h_1 = (u \sin \theta)t_1 - \frac{1}{2}gt_1^2 \implies \frac{h_1}{t_1} = u \sin \theta - \frac{1}{2}gt_1 \quad (1)$
$h_2 = (u \sin \theta)t_2 - \frac{1}{2}gt_2^2 \implies \frac{h_2}{t_2} = u \sin \theta - \frac{1}{2}gt_2 \quad (2)$
Subtracting $(2)$ from $(1)$:
$\frac{h_1}{t_1} - \frac{h_2}{t_2} = \frac{1}{2}g(t_2 - t_1) \implies \frac{h_1 t_2 - h_2 t_1}{t_1 t_2} = \frac{1}{2}g(t_2 - t_1)$
$\frac{g}{2} = \frac{h_1 t_2 - h_2 t_1}{t_1 t_2 (t_2 - t_1)} \quad (3)$
The time of flight $T$ is given by $T = \frac{2u \sin \theta}{g}$. From $(1)$,$u \sin \theta = \frac{h_1}{t_1} + \frac{1}{2}gt_1$.
$T = \frac{2}{g} \left( \frac{h_1}{t_1} + \frac{1}{2}gt_1 \right) = \frac{2h_1}{gt_1} + t_1$.
Substituting $\frac{g}{2}$ from $(3)$ into the expression for $T$:
$T = \frac{h_1}{t_1} \left( \frac{t_1 t_2 (t_2 - t_1)}{h_1 t_2 - h_2 t_1} \right) + t_1 = \frac{h_1 t_2 (t_2 - t_1) + t_1 (h_1 t_2 - h_2 t_1)}{h_1 t_2 - h_2 t_1}$
$T = \frac{h_1 t_2^2 - h_1 t_1 t_2 + h_1 t_1 t_2 - h_2 t_1^2}{h_1 t_2 - h_2 t_1} = \frac{h_1 t_2^2 - h_2 t_1^2}{h_1 t_2 - h_2 t_1}$.

(C) Given,$y = 4 \cos^{2}\left(\frac{t}{2}\right) \sin(1000 t)$.
Using the trigonometric identity $2 \cos^{2} \theta = 1 + \cos(2 \theta)$,we have $2 \cos^{2}\left(\frac{t}{2}\right) = 1 + \cos t$.
Substituting this into the equation:
$y = 2 \times [2 \cos^{2}\left(\frac{t}{2}\right)] \sin(1000 t)$
$y = 2(1 + \cos t) \sin(1000 t)$
$y = 2 \sin(1000 t) + 2 \sin(1000 t) \cos t$.
Using the product-to-sum formula $2 \sin A \cos B = \sin(A + B) + \sin(A - B)$:
$y = 2 \sin(1000 t) + [\sin(1000 t + t) + \sin(1000 t - t)]$
$y = 2 \sin(1000 t) + \sin(1001 t) + \sin(999 t)$.
This expression represents the superposition of $3$ independent harmonic oscillations with frequencies $1000, 1001,$ and $999$ rad/s.
Thus,$n = 3$.

$(B)$ The displacement of a particle executing $SHM$ is given by $y = a \sin(\omega t)$.
The velocity of the particle is $u = \frac{dy}{dt} = a\omega \cos(\omega t)$.
The kinetic energy $K$ is given by $K = \frac{1}{2}mu^2 = \frac{1}{2}m(a\omega \cos(\omega t))^2 = \frac{1}{2}m\omega^2 a^2 \cos^2(\omega t)$.
Using the trigonometric identity $\cos^2(\theta) = \frac{1 + \cos(2\theta)}{2}$, we get $K = \frac{1}{4}m\omega^2 a^2 (1 + \cos(2\omega t))$.
Since the frequency of the $SHM$ is $v = \frac{\omega}{2\pi}$, the frequency of the kinetic energy oscillation is determined by the term $\cos(2\omega t)$, which is $v' = \frac{2\omega}{2\pi} = 2v$.
Therefore, the kinetic energy changes periodically with a frequency of $2v$.

(C) Let $v$ be the final velocity of the centre of the sphere and $\omega$ be the final angular velocity when it starts rolling without slipping.
Since the friction force acts at the point of contact,the net torque about the point of contact is zero.
Therefore,the angular momentum about the point of contact remains conserved.
Initial angular momentum about the point of contact: $L_i = m v_0 r$
Final angular momentum about the point of contact: $L_f = mvr + I_{cm}\omega$
For a solid sphere,the moment of inertia about its centre is $I_{cm} = \frac{2}{5} mr^2$.
Since it rolls without slipping,the condition is $v = r\omega$,or $\omega = \frac{v}{r}$.
Equating initial and final angular momentum:
$mv_0 r = mvr + (\frac{2}{5} mr^2)(\frac{v}{r})$
$mv_0 r = mvr + \frac{2}{5} mvr$
$v_0 = v + \frac{2}{5} v$
$v_0 = \frac{7}{5} v$
$v = \frac{5}{7} v_0$

(B) The rotational kinetic energy $K$ is given by $K = \frac{1}{2} I \omega^{2}$.
Since the kinetic energies are equal,we have $I_{1} \omega_{1}^{2} = I_{2} \omega_{2}^{2} = I_{3} \omega_{3}^{2}$,which implies $\omega \propto \frac{1}{\sqrt{I}}$.
For a square plate of side $a$ and mass $M$:
$1$. For axis $1$ (passing through the center and parallel to sides),$I_{1} = \frac{Ma^{2}}{12}$.
$2$. For axis $2$ (passing through the center and parallel to the diagonal),$I_{2} = \frac{Ma^{2}}{12}$.
$3$. For axis $3$ (passing through the center and perpendicular to the plane of the plate),by the perpendicular axis theorem,$I_{3} = I_{x} + I_{y} = \frac{Ma^{2}}{12} + \frac{Ma^{2}}{12} = \frac{Ma^{2}}{6}$.
Thus,the moments of inertia are in the ratio $I_{1}: I_{2}: I_{3} = \frac{1}{12}: \frac{1}{12}: \frac{1}{6} = 1: 1: 2$.
The ratio of angular speeds is $\omega_{1}: \omega_{2}: \omega_{3} = \frac{1}{\sqrt{I_{1}}}: \frac{1}{\sqrt{I_{2}}}: \frac{1}{\sqrt{I_{3}}} = \frac{1}{\sqrt{1}}: \frac{1}{\sqrt{1}}: \frac{1}{\sqrt{2}} = 1: 1: \frac{1}{\sqrt{2}} = \sqrt{2}: \sqrt{2}: 1$.

(C) For a solid sphere rolling without slipping down an inclined plane,the total kinetic energy $K$ at the bottom is the sum of translational and rotational kinetic energy.
$K = K_{\text{trans}} + K_{\text{rot}} = \frac{1}{2}mv^2 + \frac{1}{2}I\omega^2$.
Since $I = \frac{2}{5}mR^2$ and $v = R\omega$,we have $K = \frac{1}{2}mv^2 + \frac{1}{2}(\frac{2}{5}mR^2)(\frac{v}{R})^2 = \frac{1}{2}mv^2 + \frac{1}{5}mv^2 = \frac{7}{10}mv^2$.
By conservation of energy,the potential energy at height $h$ equals the total kinetic energy at the bottom:
$mgh = \frac{7}{10}mv^2 \implies v^2 = \frac{10}{7}gh$.
When the ball is thrown vertically upwards with velocity $v$,it undergoes motion under gravity. At the maximum height $h'$,the final velocity is $0$. Using the kinematic equation $v_f^2 = v_i^2 - 2gh'$:
$0 = v^2 - 2gh' \implies h' = \frac{v^2}{2g}$.
Substituting $v^2 = \frac{10}{7}gh$:
$h' = \frac{10/7 gh}{2g} = \frac{5}{7}h$.

(A, C) Let the rod have length $L$ and mass $m$. The center of mass is at $L/2$ from $A$.
When the rod makes an angle $\theta$ with the horizontal,the center of mass has descended by $h = (L/2) \sin \theta$.
By conservation of energy,the loss in potential energy equals the gain in rotational kinetic energy:
$mg(L/2) \sin \theta = \frac{1}{2} I \omega^2$,where $I = mL^2/3$.
$mg(L/2) \sin \theta = \frac{1}{2} (mL^2/3) \omega^2 \Rightarrow \omega^2 \propto \sin \theta \Rightarrow \omega \propto \sqrt{\sin \theta}$.
Since the speed of end $B$ is $v = \omega L$,we have $v \propto \sqrt{\sin \theta}$. Thus,option $A$ is correct.
For angular acceleration $\alpha$,the torque about $A$ is $\tau = mg(L/2) \cos \theta$.
Using $\tau = I \alpha$,we get $mg(L/2) \cos \theta = (mL^2/3) \alpha$.
This implies $\alpha \propto \cos \theta$. Thus,option $C$ is correct.
Note: Both $A$ and $C$ are correct statements.

(A) The rate of heat flow through a material is given by the formula: $\frac{dQ}{dt} = \frac{KA(\Delta T)}{x}$,where $K$ is the thermal conductivity,$A$ is the surface area,$\Delta T$ is the temperature difference,and $x$ is the wall thickness.
Since the containers have the same size,shape,and wall thickness,$A$ and $x$ are constant. The surroundings are identical,so $\Delta T$ is also constant.
For a given mass of ice $m$,the total heat required to melt it is $Q = mL$,where $L$ is the latent heat of fusion. Thus,$Q$ is constant for both containers.
The rate of heat flow is inversely proportional to the time taken to melt the ice: $\frac{dQ}{dt} \propto \frac{1}{t}$.
Therefore,$K \propto \frac{1}{t}$,which implies $K_P t_1 = K_Q t_2$.
Rearranging this gives the ratio of thermal conductivities: $\frac{K_P}{K_Q} = \frac{t_2}{t_1}$.

(B) The heat supplied by the heater is absorbed by both the liquid and the container. The formula is: $(m_L s_L + m_C s_C) \Delta T = P \times t$.
For water: $m_w = 0.5 \ kg$,$s_w = 4200 \ J \ kg^{-1} \ K^{-1}$,$\Delta T = 3 \ K$,$t_1 = 15 \times 60 = 900 \ s$,$P = 10 \ W$.
$(0.5 \times 4200 + m_C s_C) \times 3 = 10 \times 900$
$(2100 + m_C s_C) \times 3 = 9000$
$2100 + m_C s_C = 3000 \implies m_C s_C = 900 \ J \ K^{-1}$.
For oil: $m_o = 2 \ kg$,$s_o = ?$,$\Delta T = 2 \ K$,$t_2 = 20 \times 60 = 1200 \ s$,$P = 10 \ W$.
$(2 \times s_o + m_C s_C) \times 2 = 10 \times 1200$
$(2 s_o + 900) \times 2 = 12000$
$2 s_o + 900 = 6000$
$2 s_o = 5100$
$s_o = 2550 \ J \ kg^{-1} \ K^{-1} = 2.55 \times 10^{3} \ J \ kg^{-1} \ K^{-1}$.

(D) Thermal radiation is the process of heat transfer in the form of electromagnetic waves.
These waves travel in straight lines at the speed of light $(c \approx 3 \times 10^8 \ m/s)$ and do not require a material medium for propagation.

(A) The energy density $u$ of black body radiation in a cavity at temperature $T$ is given by $u = aT^4$,where $a$ is the radiation constant.
The total energy $E$ contained in the box is the product of energy density $u$ and the volume $V$ of the box.
$E = u \times V = aT^4 \times V$.
Initially,the side length is $L$,so the volume $V_1 = L^3$. The energy is $E_1 = aT^4 L^3$.
Finally,the side length is doubled,so $L' = 2L$,and the new volume $V_2 = (2L)^3 = 8L^3$.
The temperature is halved,so $T' = T/2$.
The new total energy $E_2 = a(T')^4 V_2 = a(T/2)^4 (8L^3)$.
$E_2 = a(T^4 / 16) (8L^3) = (8/16) aT^4 L^3 = (1/2) aT^4 L^3$.
$E_2 = E_1 / 2$.
Therefore,the total energy halves.

(D) The ice point on the Celsius scale is $0^{\circ} C$ and the steam point is $100^{\circ} C$.
In the new scale,the ice point is $25 X$ and the steam point is $305 X$.
The temperature interval of $100^{\circ} C$ corresponds to $(305 - 25) X = 280 X$.
Therefore,a change of $1^{\circ} C$ is equivalent to a change of $\frac{280}{100} X = 2.8 X$.
This implies $1^{\circ} C = 2.8 X$,or $1 X = \frac{1}{2.8}^{\circ} C$.
The specific heat capacity of water is $c = 4200 \ J kg^{-1} (^{\circ} C)^{-1}$.
Substituting the relation $1^{\circ} C = 2.8 X$ into the unit:
$c = 4200 \ J kg^{-1} (2.8 X)^{-1} = \frac{4200}{2.8} J kg^{-1} X^{-1} = 1500 \ J kg^{-1} X^{-1}$.
Thus,$c = 1.5 \times 10^{3} J kg^{-1} X^{-1}$.

(A) For a cyclic process,the change in internal energy $\Delta U$ is zero. According to the first law of thermodynamics,$\Delta Q = \Delta U + \Delta W$. Since $\Delta U = 0$,the net heat absorbed $\Delta Q$ is equal to the net work done $\Delta W$ by the gas.
The net work done in a cyclic process is equal to the area enclosed by the cycle in the $p-V$ diagram.
The area of the triangle in the $p-V$ diagram is given by:
$\text{Area} = \frac{1}{2} \times \text{base} \times \text{height}$
Here,the base is $(V_{1}-V_{2})$ and the height is $(p_{1}-p_{2})$.
Therefore,the net heat absorbed $\Delta Q = \frac{1}{2}(p_{1}-p_{2})(V_{1}-V_{2})$.

(D) The change in internal energy $(\Delta U)$ for an ideal gas is given by the formula: $\Delta U = n C_{v} \Delta T$.
For a monoatomic gas,the molar heat capacity at constant volume is $C_{v} = \frac{3}{2} R$.
Given values:
$n = 1 \text{ mol}$
$\Delta T = T_{2} - T_{1} = (100 + 273) - (0 + 273) = 100 \text{ K}$
$R = 8.32 \text{ J mol}^{-1} \text{ K}^{-1}$
Substituting these values into the formula:
$\Delta U = 1 \times \left( \frac{3}{2} \times 8.32 \right) \times 100$
$\Delta U = 1.5 \times 8.32 \times 100$
$\Delta U = 1248 \text{ J}$
Rounding to the nearest significant figure provided in the options,we get $\Delta U \approx 1.25 \times 10^{3} \text{ J}$.

(A, C) For a diatomic gas,the degree of freedom $f = 5$. The internal energy is $U = \frac{n f R T}{2} = \frac{5 R}{2} (\alpha t + \frac{1}{2} \beta t^2)$.
The rate of increase in internal energy is $\frac{dU}{dt} = \frac{5 R}{2} (\alpha + \beta t)$.
Since the pressure is constant,the heat supplied is $dQ = n C_p dT$. For a diatomic gas,$C_p = \frac{7}{2} R$.
Thus,the power supplied by the heating element is $i^2 r = \frac{dQ}{dt} = n C_p \frac{dT}{dt} = 1 \times \frac{7}{2} R \times (\alpha + \beta t)$.
Therefore,the current $i = \sqrt{\frac{7 R}{2 r} (\alpha + \beta t)}$.
From the ideal gas law $PV = nRT$,since $P$ is constant,$V = \frac{nRT}{P} = \frac{R}{P} (\alpha t + \frac{1}{2} \beta t^2)$.
The position of the piston $x = \frac{V}{A} = \frac{R}{PA} (\alpha t + \frac{1}{2} \beta t^2)$.
The velocity $v = \frac{dx}{dt} = \frac{R}{PA} (\alpha + \beta t)$.
The acceleration $a = \frac{dv}{dt} = \frac{R \beta}{PA}$,which is constant. Thus,options $(a)$ and $(c)$ are correct.

(C) According to the Doppler effect, when an observer moves with velocity $v_0$ towards a stationary source, the observed frequency $N_{\text{approach}}$ is given by:
$N_{\text{approach}} = N \left( \frac{v + v_0}{v} \right)$
Given $N = 850 \text{ Hz}$, $v = 340 \text{ m/s}$, and $v_0 = 72 \text{ km/h} = 72 \times \frac{5}{18} = 20 \text{ m/s}$.
$N_{\text{approach}} = 850 \left( \frac{340 + 20}{340} \right) = 850 \left( \frac{360}{340} \right) = 900 \text{ Hz}$.
When the observer moves away from the source, the observed frequency $N_{\text{separation}}$ is given by:
$N_{\text{separation}} = N \left( \frac{v - v_0}{v} \right)$
$N_{\text{separation}} = 850 \left( \frac{340 - 20}{340} \right) = 850 \left( \frac{320}{340} \right) = 800 \text{ Hz}$.
The difference between the two frequencies is:
$\Delta N = N_{\text{approach}} - N_{\text{separation}} = 900 \text{ Hz} - 800 \text{ Hz} = 100 \text{ Hz}$.

(A) The path length of the straight route is $2r$ (the diameter of the semicircle).
The path length of the semicircular route is $\pi r$.
The path difference $\Delta x$ between the two routes is $\Delta x = \pi r - 2r = r(\pi - 2)$.
For maximum amplitude (constructive interference),the path difference must be an integral multiple of the wavelength $\lambda$,so $\Delta x = n\lambda$,where $n = 1, 2, 3, \dots$.
Substituting the path difference,we get $r(\pi - 2) = n\lambda$.
Since the velocity of sound is $v = f\lambda$,we have $\lambda = \frac{v}{f}$.
Substituting this into the equation: $r(\pi - 2) = n \frac{v}{f}$.
Rearranging for frequency $f$: $f = n \left[ \frac{v}{r(\pi - 2)} \right]$.
Thus,the frequencies of the resultant waves of maximum amplitude are integral multiples of $\frac{v}{r(\pi - 2)}$.

(B) For an open pipe,the fundamental frequency $f$ is given by the formula $f = \frac{v}{2l}$,where $v$ is the speed of sound and $l$ is the length of the pipe.
Given: $f = 5100 \ Hz$,$v = 340 \ ms^{-1}$.
Substituting the values into the formula:
$5100 = \frac{340}{2l}$
$l = \frac{340}{5100 \times 2}$
$l = \frac{340}{10200} = \frac{34}{1020} = \frac{1}{30} \ m$.
To convert the length into $cm$,multiply by $100$:
$l = \frac{1}{30} \times 100 = \frac{10}{3} \ cm$.

(B) When the capacitor is charged and then disconnected from the battery,the charge $Q$ on the plates remains constant because there is no path for the charge to flow.
The capacitance $C$ of a parallel plate capacitor is given by $C = \frac{A \varepsilon_0}{d}$.
When the plates are moved further apart,the distance $d$ increases,which causes the capacitance $C$ to decrease.
Since the charge $Q$ is constant and $Q = CV$,the voltage $V = \frac{Q}{C}$ must increase as $C$ decreases.
Therefore,the voltage across the capacitor increases.

(C) The energy of an electron in the $n$-th orbit of a hydrogen atom is given by $E_n = -\frac{13.6 \text{ eV}}{n^2}$.
For the ground state $(n_1 = 1)$,the energy is $E_1 = -13.6 \text{ eV}$.
For the first excited state $(n_2 = 2)$,the energy is $E_2 = -\frac{13.6 \text{ eV}}{2^2} = -\frac{13.6}{4} \text{ eV} = -3.4 \text{ eV}$.
The energy of the photon released during the transition from $n_2$ to $n_1$ is $\Delta E = E_2 - E_1$.
$\Delta E = -3.4 \text{ eV} - (-13.6 \text{ eV}) = 13.6 \text{ eV} - 3.4 \text{ eV} = 10.2 \text{ eV}$.

(B, C, D) The capacitor can be viewed as two capacitors in parallel,each with area $A/2$ and plate separation $d$.
The capacitance of the dielectric-filled part is $C_1 = \frac{K \varepsilon_0 (A/2)}{d} = \frac{K C_0}{2}$ and the air-filled part is $C_2 = \frac{\varepsilon_0 (A/2)}{d} = \frac{C_0}{2}$,where $C_0 = \frac{\varepsilon_0 A}{d}$.
The equivalent capacitance is $C_{eq} = C_1 + C_2 = \frac{C_0}{2}(K+1)$. Thus,option $(d)$ is correct.
Since the capacitors are in parallel,the potential difference $V$ across both is the same. The charges are $Q_1 = C_1 V$ and $Q_2 = C_2 V$.
The ratio of charges is $\frac{Q_1}{Q_2} = \frac{C_1}{C_2} = K$. Since $Q_1 + Q_2 = Q$,we have $Q_2 = \frac{Q}{K+1}$ and $Q_1 = \frac{KQ}{K+1}$. Thus,option $(c)$ is correct.
The charge densities are $\sigma_1 = Q_1 / (A/2)$ and $\sigma_2 = Q_2 / (A/2)$. Since $Q_1 \neq Q_2$,the charge densities are unequal. Thus,option $(b)$ is correct.
The electric fields are $E_1 = \frac{\sigma_1}{K \varepsilon_0}$ and $E_2 = \frac{\sigma_2}{\varepsilon_0}$. Substituting $\sigma_1 = K \sigma_2$,we get $E_1 = \frac{K \sigma_2}{K \varepsilon_0} = \frac{\sigma_2}{\varepsilon_0} = E_2$. The fields are equal,so $(a)$ is incorrect.

(D) For capacitors connected in series,the equivalent capacitance $C_{eq}$ is given by:
$\frac{1}{C_{eq}} = \frac{1}{C_1} + \frac{1}{C_2} + \frac{1}{C_3}$
$\frac{1}{C_{eq}} = \frac{1}{3} + \frac{1}{6} + \frac{1}{6} = \frac{2+1+1}{6} = \frac{4}{6} = \frac{2}{3} \mu F^{-1}$
So,$C_{eq} = 1.5 \mu F$.
The total charge $q$ flowing through the circuit is:
$q = C_{eq} \times V = 1.5 \mu F \times 120 V = 180 \mu C$.
In a series circuit,the charge on each capacitor is the same. Therefore,the charge on the $3 \mu F$ capacitor is $q = 180 \mu C$.
The potential difference $V_1$ across the $3 \mu F$ capacitor is:
$V_1 = \frac{q}{C_1} = \frac{180 \mu C}{3 \mu F} = 60 V$.

(A) When four cells of emf $E$ are connected in series,the total emf is $4E$.
If one cell is connected in reverse,its emf opposes the others.
Therefore,the effective emf of the circuit is $E_{eff} = E + E + E - E = 2E$.
The total internal resistance of the four cells in series is $r_{total} = r + r + r + r = 4r$.
The total resistance of the circuit is $R_{total} = 4r + R$.
Using Ohm's law,the current $I$ in the external circuit is given by $I = \frac{E_{eff}}{R_{total}} = \frac{2E}{4r + R}$.

(B) The batteries are connected in parallel. The equivalent emf $E_{eq}$ and equivalent internal resistance $r_{eq}$ of the combination are given by:
$E_{eq} = \frac{\frac{E_{1}}{r_{1}} + \frac{E_{2}}{r_{2}} + \frac{E_{3}}{r_{3}}}{\frac{1}{r_{1}} + \frac{1}{r_{2}} + \frac{1}{r_{3}}}$
$\frac{1}{r_{eq}} = \frac{1}{r_{1}} + \frac{1}{r_{2}} + \frac{1}{r_{3}}$
Given $E_{1} = 1 \text{ V}, r_{1} = 1 \Omega$; $E_{2} = 2 \text{ V}, r_{2} = 2 \Omega$; $E_{3} = 3 \text{ V}, r_{3} = 1 \Omega$.
First, calculate the equivalent internal resistance:
$\frac{1}{r_{eq}} = \frac{1}{1} + \frac{1}{2} + \frac{1}{1} = 1 + 0.5 + 1 = 2.5 \Omega^{-1} = \frac{5}{2} \Omega^{-1}$
$r_{eq} = \frac{2}{5} \Omega = 0.4 \Omega$
Now, calculate the equivalent emf:
$E_{eq} = r_{eq} \times \left( \frac{E_{1}}{r_{1}} + \frac{E_{2}}{r_{2}} + \frac{E_{3}}{r_{3}} \right)$
$E_{eq} = 0.4 \times \left( \frac{1}{1} + \frac{2}{2} + \frac{3}{1} \right) = 0.4 \times (1 + 1 + 3) = 0.4 \times 5 = 2.0 \text{ V}$
Thus, the potential difference between points $P$ and $Q$ is $2.0 \text{ V}$.

(A) To convert a galvanometer into a voltmeter,a high resistance $R$ must be connected in series with the galvanometer.
Given:
Internal resistance of galvanometer,$G = 10 \ \Omega$
Full-scale deflection current,$i_g = 0.01 \ A$
Required voltage range,$V = 120 \ V$
The formula for the series resistance is $R = \frac{V}{i_g} - G$.
Substituting the values:
$R = \frac{120}{0.01} - 10$
$R = 12000 - 10$
$R = 11990 \ \Omega$
Thus,a resistance of $11990 \ \Omega$ must be connected in series.

(C) The de-Broglie wavelength of an electron is given by $\lambda = \frac{h}{p} = \frac{h}{\sqrt{2mK_e}}$,where $K_e$ is the kinetic energy of the electron and $m$ is its mass.
Thus,$K_e = \frac{h^2}{2m\lambda^2}$.
The energy of a photon is given by $E_p = \frac{hc}{\lambda}$.
We are given that the wavelengths are equal,so $\lambda_e = \lambda_p = \lambda$.
The ratio of the energy of the photon to the kinetic energy of the electron is:
$\frac{E_p}{K_e} = \frac{hc/\lambda}{h^2/(2m\lambda^2)} = \frac{hc}{\lambda} \cdot \frac{2m\lambda^2}{h^2} = \frac{2mc\lambda}{h}$.
Since $E_p = \frac{hc}{\lambda}$,we have $\lambda = \frac{hc}{E_p}$.
Substituting this into the ratio:
$\frac{E_p}{K_e} = \frac{2mc}{h} \cdot \frac{hc}{E_p} = \frac{2mc^2}{E_p}$.
Given $mc^2 = 0.5 \ MeV = 500 \ keV$ and $E_p = 50 \ keV$:
$\frac{E_p}{K_e} = \frac{2 \times 500 \ keV}{50 \ keV} = \frac{1000}{50} = 20$.
Therefore,the ratio is $20:1$.

(A, B, D) According to the experimental observations of the photoelectric effect:
$1$. There is no significant time delay between the absorption of suitable radiation and the emission of electrons,even at very low intensities.
$2$. Einstein's photoelectric equation is $K_{max} = h\nu - \phi_0$,where $\phi_0$ is the work function. Electrons are emitted only if $\nu > \nu_0$ (threshold frequency). Thus,no electrons are emitted below the threshold frequency.
$3$. The maximum kinetic energy $K_{max}$ is a linear function of frequency $\nu$,but it is not directly proportional to it (due to the work function term $\phi_0$).
$4$. The maximum kinetic energy $K_{max}$ depends only on the frequency of incident radiation and the work function of the metal; it is independent of the intensity of the incident radiation.
Therefore,statements $(a)$,$(b)$,and $(d)$ are correct.

(C) The magnetic field $B$ at the center of the large loop of radius $b$ carrying current $I$ is given by $B = \frac{\mu_{0} I}{2 b}$.
Since the smaller loop of radius $a$ is very small,we assume the magnetic field is uniform over its area $A = \pi a^{2}$.
The magnetic flux $\phi$ through the smaller loop at time $t$ is $\phi = B A \cos(\theta)$,where $\theta = \omega t$ is the angle between the area vector of the small loop and the magnetic field.
$\phi = \left( \frac{\mu_{0} I}{2 b} \right) (\pi a^{2}) \cos(\omega t)$.
According to Faraday's law,the induced emf $\varepsilon$ is $\varepsilon = -\frac{d\phi}{dt}$.
$\varepsilon = -\frac{d}{dt} \left[ \frac{\mu_{0} I \pi a^{2}}{2 b} \cos(\omega t) \right]$.
$\varepsilon = -\frac{\mu_{0} I \pi a^{2}}{2 b} (-\omega \sin(\omega t)) = \frac{\pi a^{2} \mu_{0} I}{2 b} \omega \sin(\omega t)$.

(C) The energy of a photon is given by $E = h\nu = \frac{hc}{\lambda}$. Since the frequency $\nu$ of electromagnetic waves increases in the order: Visible light < $X$-rays < Gamma rays,the energy of the photons follows the same order.
$1$. For visible light,the energy range is approximately $1.6 \ eV$ to $3.2 \ eV$ $(E_{v} \approx 2.48 \ eV)$.
$2$. For $X$-rays,the energy range is approximately $100 \ eV$ to $100 \ keV$.
$3$. For gamma rays,the energy is typically greater than $100 \ keV$.
Comparing these ranges,we find that $E_{\gamma} > E_{X} > E_{v}$.

(A) The electric field due to an infinite sheet of charge with surface charge density $\sigma$ is given by $\vec{E} = \frac{\sigma}{2\varepsilon_{0}} \hat{k}$ for $z > 0$ and $\vec{E} = -\frac{\sigma}{2\varepsilon_{0}} \hat{k}$ for $z < 0$. Since the points $A$ and $B$ have positive $z$-coordinates ($3a$ and $6a$),the electric field is $\vec{E} = \frac{\sigma}{2\varepsilon_{0}} \hat{k}$.
Work done $W = -q \int_{A}^{B} \vec{E} \cdot d\vec{r} = -q \vec{E} \cdot (\vec{r}_{B} - \vec{r}_{A})$.
Given $\vec{r}_{A} = a(\hat{i} + 2\hat{j} + 3\hat{k})$ and $\vec{r}_{B} = a(\hat{i} - 2\hat{j} + 6\hat{k})$.
Displacement $\vec{d} = \vec{r}_{B} - \vec{r}_{A} = a(\hat{i} - \hat{i}) + a(-2\hat{j} - 2\hat{j}) + a(6\hat{k} - 3\hat{k}) = a(-4\hat{j} + 3\hat{k})$.
Work done $W = -q (\frac{\sigma}{2\varepsilon_{0}} \hat{k}) \cdot a(-4\hat{j} + 3\hat{k}) = -\frac{q \sigma a}{2\varepsilon_{0}} (\hat{k} \cdot -4\hat{j} + \hat{k} \cdot 3\hat{k}) = -\frac{q \sigma a}{2\varepsilon_{0}} (0 + 3) = -\frac{3q \sigma a}{2\varepsilon_{0}}$.
Note: The magnitude of work done is $\frac{3q \sigma a}{2\varepsilon_{0}}$.

(C) Let the charge on the inner shell be $q^{\prime}$.
Since the inner shell is grounded,its electric potential must be zero.
The potential at the surface of the inner shell is due to its own charge $q^{\prime}$ and the charge $q$ on the outer shell.
The potential $V_{1}$ at the inner shell is given by $V_{1} = \frac{k q^{\prime}}{r_{1}} + \frac{k q}{r_{2}}$.
Setting $V_{1} = 0$,we get $\frac{k q^{\prime}}{r_{1}} + \frac{k q}{r_{2}} = 0$.
Solving for $q^{\prime}$,we get $\frac{q^{\prime}}{r_{1}} = -\frac{q}{r_{2}}$.
Therefore,$q^{\prime} = -\left(\frac{r_{1}}{r_{2}}\right) q$.

(D) The magnetic field due to a finite straight wire at a perpendicular distance $r$ is given by $B = \frac{\mu_{0} I}{4 \pi r} (\sin \theta_{1} + \sin \theta_{2})$.
For the given geometry,the point $P$ is at a distance $d$ from the bend along the angle bisector. The perpendicular distance from $P$ to each wire segment is $r = d \sin(60^{\circ}) = \frac{d \sqrt{3}}{2}$.
The angles subtended by the ends of the wire segments at point $P$ are $\theta_{1} = 90^{\circ}$ and $\theta_{2} = 30^{\circ}$.
Since there are two identical segments,the total magnetic field is $B_{\text{net}} = 2 \times \frac{\mu_{0} I}{4 \pi r} (\sin 90^{\circ} + \sin 30^{\circ})$.
Substituting the values:
$B_{\text{net}} = 2 \times \frac{\mu_{0} I}{4 \pi (d \sqrt{3} / 2)} \times (1 + 1/2)$
$B_{\text{net}} = 2 \times \frac{\mu_{0} I}{2 \pi d \sqrt{3}} \times \frac{3}{2}$
$B_{\text{net}} = \frac{3 \mu_{0} I}{2 \pi d \sqrt{3}} = \frac{\sqrt{3} \mu_{0} I}{2 \pi d}$.

(D) Given:
Radius $r = 0.05 \ nm = 0.05 \times 10^{-9} \ m$
Frequency $f = 10^{16} \ Hz$ (revolutions per second)
Charge of electron $e = 1.6 \times 10^{-19} \ C$
The magnetic moment $M$ of a current loop is given by $M = I \times A$,where $I$ is the current and $A$ is the area of the loop.
The current $I$ due to the revolving electron is $I = e \times f$.
The area of the circular orbit is $A = \pi r^{2}$.
Substituting these into the formula for magnetic moment:
$M = (e \times f) \times (\pi r^{2})$
$M = (1.6 \times 10^{-19} \ C) \times (10^{16} \ s^{-1}) \times (3.14) \times (0.05 \times 10^{-9} \ m)^{2}$
$M = 1.6 \times 10^{-19} \times 10^{16} \times 3.14 \times 0.0025 \times 10^{-18}$
$M = 1.6 \times 3.14 \times 0.0025 \times 10^{-21} \ m^{2} \cdot C/s$
$M = 0.01256 \times 10^{-21} \ A \ m^{2}$
$M = 1.26 \times 10^{-23} \ A \ m^{2}$

(B) Given: Kinetic energy $= E$,Mass $= m$,Magnetic field $= B$,Charge $= q$.
When a charged particle moves perpendicular to a uniform magnetic field,the magnetic force provides the necessary centripetal force for circular motion.
$F_m = qvB$ and $F_c = \frac{mv^2}{r}$.
Equating these,we get $qvB = \frac{mv^2}{r}$,which simplifies to $r = \frac{mv}{qB}$.
We know that kinetic energy $E = \frac{1}{2}mv^2$,so $v = \sqrt{\frac{2E}{m}}$.
Substituting the value of $v$ into the radius formula:
$r = \frac{m}{qB} \sqrt{\frac{2E}{m}} = \frac{\sqrt{m^2} \cdot \sqrt{2E}}{\sqrt{m} \cdot qB} = \frac{\sqrt{2Em}}{qB}$.

(A) For a charged particle moving with velocity $v$ in a region with electric field $E$ and magnetic field $B$,the Lorentz force is $F = q(E + v \times B)$.
For the particle to pass undeflected,the net force must be zero,i.e.,$qE = -q(v \times B)$.
For a proton $(q > 0)$: The electric force $F_E = qE$ is downwards. The magnetic force $F_B = q(v \times B)$ is upwards (using the right-hand rule,$v$ is right,$B$ is out,so $v \times B$ is up). For no deflection,$qE = qvB$,which gives $v = E/B$.
For an electron $(q < 0)$: The electric force $F_E = qE$ is upwards. The magnetic force $F_B = q(v \times B)$ is downwards (since $q$ is negative,the direction of $F_B$ reverses). For no deflection,$|q|E = |q|vB$,which also gives $v = E/B$.
However,the question asks for the behavior of the stream. Since both require the same speed $v = E/B$ to pass undeflected,option $B$ is incorrect as it implies electrons would not pass. Option $A$ is the correct statement.

(A) The intensity of magnetization $I$ is defined as the magnetic moment $M$ per unit volume $V$,i.e.,$I = M/V$.
Therefore,the magnetic moment is given by $M = I \times V$.
The volume $V$ of the bar magnet is the product of its magnetic length $l$ and its cross-sectional area $A$.
Given:
$I = 5.0 \times 10^{4} \text{ A m}^{-1}$
$l = 12 \text{ cm} = 0.12 \text{ m}$
$A = 1 \text{ cm}^{2} = 1 \times 10^{-4} \text{ m}^{2}$
Volume $V = l \times A = 0.12 \times 10^{-4} \text{ m}^{3} = 1.2 \times 10^{-5} \text{ m}^{3}$.
Now,calculating the magnetic moment $M$:
$M = (5.0 \times 10^{4} \text{ A m}^{-1}) \times (1.2 \times 10^{-5} \text{ m}^{3})$
$M = 6.0 \times 10^{-1} \text{ A m}^{2} = 0.6 \text{ A m}^{2}$.

(B) In $\alpha$ decay,the atomic number $Z$ decreases by $2$ and the mass number $A$ decreases by $4$. In $\beta^-$ decay,$Z$ increases by $1$ and $A$ remains constant. In $\beta^+$ decay,$Z$ decreases by $1$ and $A$ remains constant.
An isotope has the same atomic number $Z$ but a different mass number $A$. Since $\alpha$ decay changes $Z$ and $\beta$ decay changes $Z$,neither process can produce an isotope of the original nucleus. Thus,the correct answer is that an isotope of the original nucleus cannot be produced.

(D) For the first condition,using the lens formula $\frac{1}{f} = \frac{1}{v} - \frac{1}{u}$:
Given $f_1 = 10 \ cm$ and $u = -30 \ cm$.
$\frac{1}{10} = \frac{1}{v} - \frac{1}{-30} \Rightarrow \frac{1}{v} = \frac{1}{10} - \frac{1}{30} = \frac{3-1}{30} = \frac{2}{30} = \frac{1}{15}$.
So,$v = 15 \ cm$.
For the second condition,when the concave lens is placed in contact,the new image distance $v' = v + 45 \ cm = 15 + 45 = 60 \ cm$.
The object distance $u$ remains $-30 \ cm$.
Let $F$ be the focal length of the combination: $\frac{1}{F} = \frac{1}{v'} - \frac{1}{u} = \frac{1}{60} - \frac{1}{-30} = \frac{1}{60} + \frac{2}{60} = \frac{3}{60} = \frac{1}{20}$.
So,$F = 20 \ cm$.
Using the combination formula $\frac{1}{F} = \frac{1}{f_1} + \frac{1}{f_2}$:
$\frac{1}{20} = \frac{1}{10} + \frac{1}{f_2} \Rightarrow \frac{1}{f_2} = \frac{1}{20} - \frac{1}{10} = \frac{1-2}{20} = -\frac{1}{20}$.
Thus,$f_2 = -20 \ cm$.
The magnitude of the focal length of the concave lens is $20 \ cm$.

(D) For a convex lens to form a real image of an object on a screen separated by a fixed distance $d$,the condition for the existence of the image is given by the displacement method.
Let $u$ be the object distance and $v$ be the image distance. Then $u + v = d$.
For a real image,the lens formula is $\frac{1}{v} - \frac{1}{u} = \frac{1}{f}$.
Substituting $u = -(d - v)$,we get $\frac{1}{v} + \frac{1}{d - v} = \frac{1}{f}$.
This simplifies to the quadratic equation $v^2 - dv + df = 0$.
For $v$ to have real roots,the discriminant must be non-negative: $D = d^2 - 4df \geq 0$.
This implies $d^2 \geq 4df$,or $f \leq \frac{d}{4}$.
Thus,the maximum possible focal length of the convex lens is $\frac{d}{4}$.

(A) In a compound microscope,the objective lens forms an intermediate image of the object.
Since the object is placed just beyond the focal point of the objective lens,the lens creates a real,inverted,and magnified image.
This image then acts as an object for the eyepiece,which further magnifies it to produce the final virtual image.

(D) According to Snell's law,for a series of parallel layers,the product of the refractive index and the sine of the angle of incidence remains constant at every interface.
Let $\mu_0$ be the refractive index of the $0^{\text{th}}$ layer and $i_0$ be the angle of incidence at the first interface.
$\mu_0 \sin i_0 = \mu_m \sin r_m$
Given:
$\mu_0 = \mu = 1.5$
$i_0 = 30^{\circ}$
$\mu_m = \mu - m \Delta \mu = 1.5 - m(0.015)$
Since the ray emerges parallel to the interface after the $m^{\text{th}}$ refraction,the angle of refraction $r_m = 90^{\circ}$.
Substituting the values:
$1.5 \sin 30^{\circ} = (1.5 - m \times 0.015) \sin 90^{\circ}$
$1.5 \times 0.5 = (1.5 - 0.015m) \times 1$
$0.75 = 1.5 - 0.015m$
$0.015m = 1.5 - 0.75$
$0.015m = 0.75$
$m = \frac{0.75}{0.015} = \frac{750}{15} = 50$
Thus,the value of $m$ is $50$.

(B) For an ideal diode,it conducts when the potential at the anode is greater than or equal to the potential at the cathode. Here,the cathode is at a constant potential of $3 \ V$.
Case $1$: When $V_{i} = 2 \ V$,the potential at the anode $(2 \ V)$ is less than the potential at the cathode $(3 \ V)$. Thus,the diode is reverse-biased and acts as an open circuit. Therefore,the initial current $I_{initial} = 0 \ A$.
Case $2$: When $V_{i} = 6 \ V$,the potential at the anode $(6 \ V)$ is greater than the potential at the cathode $(3 \ V)$. Thus,the diode is forward-biased and acts as a short circuit (ideal). The current $I_{final}$ is given by Ohm's law:
$I_{final} = \frac{V_{i} - V_{cathode}}{R} = \frac{6 \ V - 3 \ V}{150 \ \Omega} = \frac{3 \ V}{150 \ \Omega} = 0.02 \ A = 20 \ mA$.
The change in current is $\Delta I = I_{final} - I_{initial} = 20 \ mA - 0 \ mA = 20 \ mA$.

(D) The classification of materials based on energy band theory is as follows:
$1$. Conductors: The valence band and conduction band overlap,or the energy gap is negligible.
$2$. Semiconductors: The energy band gap is small,typically around $1 \ eV$ (e.g.,$Si \approx 1.1 \ eV$,$Ge \approx 0.7 \ eV$).
$3$. Insulators: The energy band gap is very large,typically greater than $3 \ eV$.
Since the given bandgap is $5.0 \ eV$,which is significantly greater than $3 \ eV$,the material is classified as an insulator.

(B) In a common emitter configuration of a transistor,the base current $I_{B}$ is very small and is typically in the order of microamperes $(\mu A)$.
Since the collector current $I_{C} = \beta I_{B}$ and the current gain $\beta$ is typically large (around $100$),the collector current $I_{C}$ is in the order of milliamperes $(mA)$.
The collector-emitter voltage $V_{CE}$ is typically maintained in the order of volts $(V)$ to keep the transistor in the active region.
Therefore,the correct order of magnitude is $I_{B}$ in $\mu A$,$I_{C}$ in $mA$,and $V_{CE}$ in volts.

(B) The input $A$ passes through a $NOT$ gate,resulting in $\bar{A}$.
This $\bar{A}$ is fed into an $AND$ gate along with input $B$,producing an output of $\bar{A} \cdot B$.
This result $(\bar{A} \cdot B)$ and the original $\bar{A}$ are then fed into an $OR$ gate.
Therefore,the final output $Y$ is given by:
$Y = \bar{A} + (\bar{A} \cdot B)$
Using the absorption law of Boolean algebra,which states that $X + (X \cdot Y) = X$,we can simplify the expression:
$Y = \bar{A} \cdot (1 + B)$
Since $(1 + B) = 1$,we get:
$Y = \bar{A} \cdot 1 = \bar{A}$

(D) The maximum intensity for the superposition of two coherent waves is given by:
$I_{\max} = (\sqrt{I_1} + \sqrt{I_2})^2$
Given $I_1 = I$ and $I_2 = 4I$:
$I_{\max} = (\sqrt{I} + \sqrt{4I})^2 = (\sqrt{I} + 2\sqrt{I})^2 = (3\sqrt{I})^2 = 9I$
The minimum intensity for the superposition of two coherent waves is given by:
$I_{\min} = (\sqrt{I_1} - \sqrt{I_2})^2$
$I_{\min} = (\sqrt{I} - \sqrt{4I})^2 = (\sqrt{I} - 2\sqrt{I})^2 = (-\sqrt{I})^2 = I$
Thus,the maximum and minimum intensities are $9I$ and $I$ respectively.