AIEEE 2005 Physics Question Paper with Answer and Solution

(B) Initial velocity $\vec{v}_i = 5\hat{i}\,m/s$.
Final velocity $\vec{v}_f = 5\hat{j}\,m/s$.
Change in velocity $\Delta \vec{v} = \vec{v}_f - \vec{v}_i = 5\hat{j} - 5\hat{i}$.
Magnitude of change in velocity $|\Delta \vec{v}| = \sqrt{(-5)^2 + 5^2} = \sqrt{25 + 25} = \sqrt{50} = 5\sqrt{2}\,m/s$.
The direction of $\Delta \vec{v}$ is North-West (since it points in the direction of $-\hat{i} + \hat{j}$).
Average acceleration $\vec{a}_{avg} = \frac{\Delta \vec{v}}{\Delta t} = \frac{5\sqrt{2}}{10} = \frac{\sqrt{2}}{2} = \frac{1}{\sqrt{2}}\,m/s^2$.
Thus,the average acceleration is $\frac{1}{\sqrt{2}}\,m/s^2$ in the North-West direction.

(A) To determine which pair does not have identical dimensions,we analyze the dimensional formulas of each:
$1$. Moment of inertia $(I = MR^2)$ has dimensions $[ML^2T^0]$. Moment of force (Torque,$\tau = r \times F$) has dimensions $[ML^2T^{-2}]$. These are not identical.
$2$. Work $(W = F \cdot d)$ has dimensions $[ML^2T^{-2}]$. Torque $(\tau = r \times F)$ has dimensions $[ML^2T^{-2}]$. These are identical.
$3$. Angular momentum $(L = mvr)$ has dimensions $[ML^2T^{-1}]$. Planck's constant $(h = E/f)$ has dimensions $[ML^2T^{-1}]$. These are identical.
$4$. Impulse $(J = F \Delta t)$ has dimensions $[MLT^{-1}]$. Momentum $(p = mv)$ has dimensions $[MLT^{-1}]$. These are identical.
Therefore,the pair that does not have identical dimensions is Moment of inertia and moment of force.

(B) Let the initial velocity of the bullet be $u$.
After penetrating $3\,cm$,its velocity becomes $u/2$.
Using the equation of motion $v^2 = u^2 - 2as$:
$(u/2)^2 = u^2 - 2a(3)$
$u^2/4 = u^2 - 6a$
$6a = 3u^2/4$
$a = u^2/8$
Let the bullet penetrate an additional distance $x$ before coming to rest.
For this motion,initial velocity is $u/2$,final velocity is $0$,and acceleration is $-a = -u^2/8$.
Using $v^2 = u^2 - 2as$:
$0^2 = (u/2)^2 - 2(u^2/8)x$
$0 = u^2/4 - (u^2/4)x$
$u^2/4 = (u^2/4)x$
$x = 1\,cm$.

(A) Given the relation: $t = \alpha x^2 + \beta x$
Differentiating with respect to $x$:
$\frac{dt}{dx} = 2\alpha x + \beta$
Since velocity $v = \frac{dx}{dt}$,we have:
$v = \frac{1}{2\alpha x + \beta} \implies 2\alpha x + \beta = \frac{1}{v}$
Acceleration $a$ is given by $a = v \frac{dv}{dx}$.
First,find $\frac{dv}{dx}$ by differentiating $v = (2\alpha x + \beta)^{-1}$ with respect to $x$:
$\frac{dv}{dx} = -1(2\alpha x + \beta)^{-2} \cdot (2\alpha) = -\frac{2\alpha}{(2\alpha x + \beta)^2}$
Substitute $(2\alpha x + \beta) = \frac{1}{v}$ into the expression:
$\frac{dv}{dx} = -2\alpha \cdot v^2$
Now,calculate acceleration:
$a = v \cdot (-2\alpha v^2) = -2\alpha v^3$
Retardation is the negative of acceleration:
$\text{Retardation} = -a = 2\alpha v^3$

(C) Let the car start from point $A$ from rest and move up to point $B$ with acceleration $f$ over distance $S$.
The velocity of the car at point $B$ is $v = \sqrt{2fS}$ (using $v^2 = u^2 + 2as$).
The car moves distance $BC = x$ with this constant velocity $v$ in time $t$,so $x = vt = \sqrt{2fS} \cdot t$ ... $(i)$.
At point $C$,the car begins to decelerate at rate $a' = \frac{f}{2}$ until it comes to rest at point $D$. Let the distance $CD = y$.
Using $v^2 = u^2 - 2a'y$,where final velocity is $0$: $0 = v^2 - 2(\frac{f}{2})y \implies y = \frac{v^2}{f} = \frac{2fS}{f} = 2S$ ... (ii).
The total distance $AD = AB + BC + CD = S + x + 2S = 15S$.
$3S + x = 15S \implies x = 12S$.
Substituting $x = 12S$ into equation $(i)$: $12S = \sqrt{2fS} \cdot t$.
Squaring both sides: $144S^2 = 2fS \cdot t^2$.
Dividing by $2S$: $72S = f \cdot t^2 \implies S = \frac{1}{72}ft^2$.

(A) $1$. Free fall phase (from point $A$ to $B$): The parachutist falls freely under gravity for a distance $s_1 = 50\, m$. Initial velocity $u_1 = 0$,acceleration $a_1 = 9.8\, m/s^2$.
The velocity $v$ at point $B$ is given by $v^2 = u_1^2 + 2a_1s_1 = 0 + 2 \times 9.8 \times 50 = 980$.
So,$v = \sqrt{980}\, m/s$.
$2$. Deceleration phase (from point $B$ to $C$): The parachute opens,and the parachutist decelerates at $a_2 = -2\, m/s^2$. The final velocity at the ground is $v_f = 3\, m/s$. Let the distance covered be $h$.
Using $v_f^2 = v^2 + 2a_2h$:
$(3)^2 = 980 + 2(-2)h$
$9 = 980 - 4h$
$4h = 980 - 9 = 971$
$h = 971 / 4 = 242.75\, m$.
$3$. Total height: The total height from which he bailed out is $H = s_1 + h = 50 + 242.75 = 292.75\, m \approx 293\, m$.

(B) For a given velocity $u$,the range $R$ is the same for two angles of projection,$\theta$ and $(90^\circ - \theta)$.
The time of flight for angle $\theta$ is $t_1 = \frac{2u \sin \theta}{g}$.
The time of flight for angle $(90^\circ - \theta)$ is $t_2 = \frac{2u \sin(90^\circ - \theta)}{g} = \frac{2u \cos \theta}{g}$.
Multiplying the two times of flight:
$t_1 t_2 = \left( \frac{2u \sin \theta}{g} \right) \left( \frac{2u \cos \theta}{g} \right) = \frac{4u^2 \sin \theta \cos \theta}{g^2}$.
Using the identity $\sin 2\theta = 2 \sin \theta \cos \theta$,we get:
$t_1 t_2 = \frac{2u^2 (2 \sin \theta \cos \theta)}{g^2} = \frac{2(u^2 \sin 2\theta)}{g^2}$.
Since the range $R = \frac{u^2 \sin 2\theta}{g}$,we can substitute this into the equation:
$t_1 t_2 = \frac{2R}{g}$.
Since $g$ is constant,we conclude that $t_1 t_2 \propto R$.

(B) Given: Mass $m = 0.3 \, kg$,force constant $k = 15 \, N/m$,and displacement $x = 20 \, cm = 0.2 \, m$.
According to Hooke's Law,the magnitude of the force is $F = kx$.
Substituting the values: $F = 15 \times 0.2 = 3 \, N$.
Using Newton's Second Law,$F = ma$,the acceleration $a$ is given by $a = \frac{F}{m}$.
Substituting the values: $a = \frac{3}{0.3} = 10 \, m/s^2$.
Therefore,the initial acceleration is $10 \, m/s^2$.

(B) Let the mass of the block be $m$. To keep the block stationary relative to the inclined surface,we analyze the forces in the frame of the incline.
$1$. The forces acting on the block are its weight $(mg)$ acting downwards,the normal force $(N)$ perpendicular to the surface,and the pseudo force $(ma)$ acting horizontally in the direction opposite to the acceleration of the incline.
$2$. Resolving the forces along the inclined surface:
- The component of the gravitational force down the incline is $mg \sin \alpha$.
- The component of the pseudo force up the incline is $ma \cos \alpha$.
$3$. For the block to remain stationary relative to the incline,these two forces must balance each other:
$ma \cos \alpha = mg \sin \alpha$
$4$. Dividing both sides by $m \cos \alpha$:
$a = g \frac{\sin \alpha}{\cos \alpha}$
$a = g \tan \alpha$

(D) Given: Initial velocity $u = 100\, m/s$,Final velocity $v = 0\, m/s$,Coefficient of kinetic friction $\mu_k = 0.5$,Acceleration due to gravity $g = 10\, m/s^2$.
Using the work-energy theorem,the work done by the frictional force is equal to the change in kinetic energy:
$W = \Delta K$
$-f_k \cdot s = 0 - \frac{1}{2} m u^2$
$-(\mu_k m g) s = -\frac{1}{2} m u^2$
$s = \frac{u^2}{2 \mu_k g}$
Substituting the values:
$s = \frac{(100)^2}{2 \times 0.5 \times 10} = \frac{10000}{10} = 1000\, m$.

(A) For a smooth inclined plane,the acceleration is $a_s = g \sin \theta$. The time taken to cover distance $s$ is $t_s = \sqrt{\frac{2s}{g \sin \theta}}$.
For a rough inclined plane,the acceleration is $a_r = g(\sin \theta - \mu \cos \theta)$. The time taken is $t_r = \sqrt{\frac{2s}{g(\sin \theta - \mu \cos \theta)}}$.
Given $t_r = n t_s$,so $t_r^2 = n^2 t_s^2$.
$\frac{2s}{g(\sin \theta - \mu \cos \theta)} = n^2 \frac{2s}{g \sin \theta}$.
$\sin \theta - \mu \cos \theta = \frac{\sin \theta}{n^2}$.
$\mu \cos \theta = \sin \theta \left( 1 - \frac{1}{n^2} \right)$.
$\mu = \tan \theta \left( 1 - \frac{1}{n^2} \right)$.
Since $\theta = 45^{\circ}$,$\tan 45^{\circ} = 1$,so $\mu = 1 - \frac{1}{n^2}$.

(D) Let the total length of the inclined plane be $l$. The upper half has length $l/2$ and is smooth,while the lower half has length $l/2$ and is rough with coefficient of friction $\mu$.
For the upper half (smooth):
The acceleration is $a_1 = g \sin \theta$. Starting from rest $(u=0)$,the velocity $v$ at the midpoint is given by:
$v^2 = u^2 + 2 a_1 (l/2) = 0 + 2(g \sin \theta)(l/2) = gl \sin \theta$.
For the lower half (rough):
The acceleration is $a_2 = g(\sin \theta - \mu \cos \theta)$. The body starts with velocity $v$ and comes to rest $(v_f = 0)$ at the bottom after covering distance $l/2$:
$v_f^2 = v^2 + 2 a_2 (l/2) = 0$.
Substituting $v^2 = gl \sin \theta$ and $a_2 = g(\sin \theta - \mu \cos \theta)$:
$gl \sin \theta + 2[g(\sin \theta - \mu \cos \theta)](l/2) = 0$.
$gl \sin \theta + gl(\sin \theta - \mu \cos \theta) = 0$.
$2 \sin \theta - \mu \cos \theta = 0$.
$\mu \cos \theta = 2 \sin \theta$.
$\mu = 2 \tan \theta$.

(C) Since the surface is smooth,mechanical energy is conserved throughout the motion.
Let the initial height be $h_1 = 100 \, m$ and the final height be $h_2 = 20 \, m$.
The initial velocity $u = 0 \, m/s$.
According to the law of conservation of energy:
$PE_{initial} + KE_{initial} = PE_{final} + KE_{final}$
$mgh_1 + 0 = mgh_2 + \frac{1}{2}mv^2$
$mg(h_1 - h_2) = \frac{1}{2}mv^2$
$v = \sqrt{2g(h_1 - h_2)}$
Substituting the values $(g = 10 \, m/s^2)$:
$v = \sqrt{2 \times 10 \times (100 - 20)}$
$v = \sqrt{20 \times 80}$
$v = \sqrt{1600}$
$v = 40 \, m/s$.

(C) When a block of mass $M$ moving with velocity $v$ collides with a spring,its kinetic energy is converted into the elastic potential energy of the spring at the point of maximum compression $L$.
By the law of conservation of energy:
$\frac{1}{2} M v^2 = \frac{1}{2} K L^2$
Solving for $v$:
$v^2 = \frac{K}{M} L^2 \implies v = L \sqrt{\frac{K}{M}}$
The momentum $P$ of the block is given by $P = Mv$.
Substituting the value of $v$:
$P = M \left( L \sqrt{\frac{K}{M}} \right) = \sqrt{M^2 \cdot \frac{K}{M}} L = \sqrt{MK} L$
Thus,the maximum momentum of the block is $\sqrt{MK} L$.

(A) Let mass $A$ move with velocity $v$ and collide inelastically with mass $B$,which is initially at rest.
According to the problem,mass $A$ moves in a perpendicular direction after the collision with velocity $\frac{v}{\sqrt{3}}$. Let mass $B$ move at an angle $\theta$ with the horizontal with velocity $V$.
Initial horizontal momentum of the system (before collision) $= mv$.
Final horizontal momentum of the system (after collision) $= m \left( \frac{v}{\sqrt{3}} \right) \cos(90^{\circ}) + mV \cos \theta = mV \cos \theta$.
From the conservation of horizontal linear momentum: $mv = mV \cos \theta \implies v = V \cos \theta$ $...(i)$.
Initial vertical momentum of the system (before collision) $= 0$.
Final vertical momentum of the system (after collision) $= m \left( \frac{v}{\sqrt{3}} \right) - mV \sin \theta$.
From the conservation of vertical linear momentum: $m \left( \frac{v}{\sqrt{3}} \right) - mV \sin \theta = 0 \implies \frac{v}{\sqrt{3}} = V \sin \theta$ $...(ii)$.
Squaring and adding $(i)$ and $(ii)$:
$v^2 + \frac{v^2}{3} = V^2 (\cos^2 \theta + \sin^2 \theta)$.
$\frac{4v^2}{3} = V^2$.
$V = \frac{2}{\sqrt{3}}v$.

(B) The elastic potential energy stored per unit volume $(u)$ in a material is given by the formula:
$u = \frac{1}{2} \times \text{stress} \times \text{strain}$
We know that Young's modulus $(Y)$ is defined as the ratio of stress $(S)$ to strain $(\epsilon)$:
$Y = \frac{S}{\epsilon} \implies \epsilon = \frac{S}{Y}$
Substituting the expression for strain into the energy formula:
$u = \frac{1}{2} \times S \times \left( \frac{S}{Y} \right)$
$u = \frac{S^2}{2Y}$
Thus,the correct option is $B$.

(C) The height of the water column in a capillary tube is given by the formula $h = \frac{2T \cos \theta}{r \rho g}$.
In a freely falling elevator,the effective acceleration due to gravity $g_{eff}$ becomes $0$ because the elevator is in a state of weightlessness.
As $g_{eff} \to 0$,the height of the water column $h$ tends to infinity $(h \propto 1/g_{eff})$.
However,the water column cannot exceed the physical length of the capillary tube.
Therefore,the water will rise until it fills the entire length of the capillary tube,which is $20 \,cm$.

(D) Number of moles of Helium $(n_1)$ = $\frac{16}{4} = 4 \, \text{mol}$.
Number of moles of Oxygen $(n_2)$ = $\frac{16}{32} = 0.5 \, \text{mol}$.
For Helium (monatomic),$\gamma_1 = \frac{5}{3}$ and degrees of freedom $f_1 = 3$.
For Oxygen (diatomic),$\gamma_2 = \frac{7}{5}$ and degrees of freedom $f_2 = 5$.
The equivalent $\gamma$ for the mixture is given by $\frac{n_1 + n_2}{\gamma - 1} = \frac{n_1}{\gamma_1 - 1} + \frac{n_2}{\gamma_2 - 1}$.
Substituting the values: $\frac{4 + 0.5}{\gamma - 1} = \frac{4}{\frac{5}{3} - 1} + \frac{0.5}{\frac{7}{5} - 1}$.
$\frac{4.5}{\gamma - 1} = \frac{4}{2/3} + \frac{0.5}{2/5} = 6 + 1.25 = 7.25$.
$\gamma - 1 = \frac{4.5}{7.25} \approx 0.62$.
$\gamma \approx 1.62$.

(B) The first law of thermodynamics is based on the law of conservation of energy and introduces the concept of internal energy $(U)$. It states that $\Delta Q = \Delta U + \Delta W$. It is applicable to all processes,including cyclic processes. However,the concept of entropy is introduced by the second law of thermodynamics,not the first. Therefore,statement $(b)$ is incorrect.

(C) Internal energy $(U)$ is a state function,which means its value depends only on the state of the system (defined by variables like pressure,volume,and temperature) and not on the path taken to reach that state.
Since both processes $I$ and $II$ start at the same initial state $A$ and end at the same final state $B$,the change in internal energy for both processes must be identical.
Therefore,$\Delta U_I = \Delta U_{II}$.

(A) The area under the $T-S$ curve represents the heat exchanged in a process.
For the given cycle, the heat absorbed $(Q_{in})$ is the area under the top slanted line from $S_0$ to $2S_0$.
The area is a rectangle plus a triangle: $Q_{in} = (T_0 \times S_0) + \frac{1}{2} \times (2T_0 - T_0) \times (2S_0 - S_0) = T_0 S_0 + \frac{1}{2} T_0 S_0 = 1.5 T_0 S_0$.
The heat rejected $(Q_{out})$ is the area under the bottom horizontal line from $2S_0$ to $S_0$ at temperature $T_0$: $Q_{out} = T_0 \times (2S_0 - S_0) = T_0 S_0$.
The efficiency $\eta$ is given by $\eta = 1 - \frac{Q_{out}}{Q_{in}}$.
Substituting the values: $\eta = 1 - \frac{T_0 S_0}{1.5 T_0 S_0} = 1 - \frac{1}{1.5} = 1 - \frac{2}{3} = \frac{1}{3} \approx 0.33$.

(A) Consider a concentric spherical shell of radius $r$ and thickness $dr$ as shown in the figure.
In a steady state,the radial rate of flow of heat $H$ through this shell is given by Fourier's law of heat conduction:
$H = \frac{{dQ}}{{dt}} = - KA\frac{{dT}}{{dr}}$
Since the area of the spherical shell is $A = 4\pi {r^2}$,we have:
$H = - K(4\pi {r^2})\frac{{dT}}{{dr}}$
Rearranging the terms to integrate:
$\frac{{dr}}{{{r^2}}} = - \frac{{4\pi K}}{H}dT$
Integrating from $r_1$ to $r_2$ and $T_1$ to $T_2$:
$\int_{{r_1}}^{{r_2}} \frac{{dr}}{{{r^2}}} = - \frac{{4\pi K}}{H} \int_{{T_1}}^{{T_2}} dT$
$\left[ - \frac{1}{r} \right]_{{r_1}}^{{r_2}} = - \frac{{4\pi K}}{H} (T_2 - T_1)$
$\left( \frac{1}{{{r_1}}} - \frac{1}{{{r_2}}} \right) = \frac{{4\pi K}}{H} (T_1 - T_2)$
$\frac{{{r_2} - {r_1}}}{{{r_1}{r_2}}} = \frac{{4\pi K}}{H} (T_1 - T_2)$
Solving for $H$:
$H = \frac{{4\pi K{r_1}{r_2}({T_1} - {T_2})}}{{{r_2} - {r_1}}}$
Thus,the rate of heat flow is proportional to $\frac{{{r_1}{r_2}}}{{{r_2} - {r_1}}}$.

(C) The velocity of the first particle is given by the derivative of its displacement: ${v_1} = \frac{d{y_1}}{dt} = 0.1 \times 100\pi \cos(100\pi t + \frac{\pi}{3}) = 10\pi \cos(100\pi t + \frac{\pi}{3})$.
The velocity of the second particle is given by the derivative of its displacement: ${v_2} = \frac{d{y_2}}{dt} = -0.1\pi \sin(\pi t) = 0.1\pi \cos(\pi t + \frac{\pi}{2})$.
The phase of the velocity of the first particle is ${\phi_1} = 100\pi t + \frac{\pi}{3}$,and the phase of the velocity of the second particle is ${\phi_2} = \pi t + \frac{\pi}{2}$.
However,the question asks for the phase difference between the two velocities. Since the frequencies are different ($100\pi$ vs $\pi$),the phase difference is time-dependent. Assuming the question implies the difference in initial phases at $t=0$:
$\Delta \phi = \phi_1(0) - \phi_2(0) = \frac{\pi}{3} - \frac{\pi}{2} = \frac{2\pi - 3\pi}{6} = -\frac{\pi}{6}$.

(C) The given equation of motion is $\frac{d^2y}{dt^2} + Ky = 0$.
Comparing this with the standard equation of Simple Harmonic Motion $(SHM)$,which is $\frac{d^2y}{dt^2} + \omega^2 y = 0$,we get $\omega^2 = K$.
Therefore,the angular frequency is $\omega = \sqrt{K}$.
We know that the time period $T$ is related to the angular frequency $\omega$ by the formula $T = \frac{2\pi}{\omega}$.
Substituting the value of $\omega$,we get $T = \frac{2\pi}{\sqrt{K}}$.

(D) Given function is $y = \sin^2(\omega t)$.
Using the trigonometric identity $\sin^2 \theta = \frac{1 - \cos(2\theta)}{2}$,we can rewrite the function as:
$y = \frac{1}{2} - \frac{1}{2}\cos(2\omega t)$.
The period $T$ of a function $\cos(kt)$ is given by $T = \frac{2\pi}{k}$.
Here,$k = 2\omega$,so the period is $T = \frac{2\pi}{2\omega} = \frac{\pi}{\omega}$.
Simple Harmonic Motion $(S.H.M.)$ must satisfy the differential equation $\frac{d^2y}{dt^2} = -\Omega^2 y$. The given function represents a constant offset plus a cosine term,which does not satisfy the condition for $S.H.M.$ because the equilibrium position is shifted and it is not a pure sinusoidal oscillation about the origin. Therefore,it is a periodic motion but not $S.H.M.$

(D) The given system acts as a simple pendulum, where the effective length $(l)$ is the distance between the point of suspension and the center of gravity $(C.G.)$ of the oscillating body.
Initially, when the sphere is full, the $C.G.$ is at the center of the sphere. As water flows out, the center of mass of the remaining water shifts downwards, causing the resultant $C.G.$ of the system to move downwards. This increases the effective length $(l)$, and since the time period $T = 2\pi \sqrt{l/g}$, the time period $T$ increases.
As more water flows out, the weight of the remaining water becomes less than the weight of the empty sphere. The resultant $C.G.$ starts shifting back upwards towards the center of the sphere. Consequently, the effective length $(l)$ decreases, which causes the time period $T$ to decrease.
Finally, when the sphere is completely empty, the $C.G.$ returns to the center of the sphere, making the effective length equal to its initial value. Thus, the time period returns to its original value. Therefore, the period of oscillation first increases and then decreases to its original value.

(B) When an observer moves towards a stationary source,the apparent frequency $n'$ is given by the Doppler effect formula:
$n' = \left( \frac{v + v_O}{v} \right) n$
Given that the velocity of the observer $v_O = \frac{v}{5}$,where $v$ is the velocity of sound.
Substituting the value of $v_O$ into the formula:
$n' = \left( \frac{v + v/5}{v} \right) n = \left( \frac{6v/5}{v} \right) n = 1.2n$
The increase in frequency is $\Delta n = n' - n = 1.2n - n = 0.2n$.
The percentage increase in frequency is given by:
$\text{Percentage increase} = \left( \frac{\Delta n}{n} \right) \times 100 = \left( \frac{0.2n}{n} \right) \times 100 = 20\%$.

(C) For a complete circular disc of mass $M'$ and radius $r$,the moment of inertia about an axis perpendicular to its plane passing through the center is $I = \frac{1}{2} M' r^2$.
$A$ semicircular disc is exactly half of a complete circular disc.
If the mass of the semicircular disc is $M$,then the mass of the corresponding full circular disc would be $2M$.
Substituting $M' = 2M$ into the formula for the full disc:
$I = \frac{1}{2} (2M) r^2 = M r^2$.
Therefore,the moment of inertia of the semicircular disc about the axis perpendicular to its plane passing through the center is $M r^2$.

(D) The motion of the centre of mass of a system depends only on the external forces acting on it.
In this case,the only external force acting on the body $A$ (and subsequently on the system of parts $B$ and $C$) is the gravitational force,which results in an acceleration equal to $g$ downwards.
Since the external force remains unchanged during the breaking process,the acceleration of the centre of mass remains $g$.
Therefore,the trajectory of the centre of mass of the system consisting of parts $B$ and $C$ follows the same path as the original body $A$.
Thus,the centre of mass does not shift relative to the path of the original body $A$.

(A) For an object to undergo pure translational motion,the applied force must pass through its center of mass $(CM)$.
Let the mass of the horizontal rod $AB$ be $m$. Since the vertical rod $CD$ has length $2l$ and the horizontal rod $AB$ has length $l$,assuming uniform density,the mass of rod $CD$ is $2m$.
The center of mass of rod $AB$ is at its midpoint $D$,which is at a distance $2l$ from $C$. Let this be $y_1 = 2l$.
The center of mass of rod $CD$ is at its midpoint,which is at a distance $l$ from $C$. Let this be $y_2 = l$.
Taking $C$ as the origin,the position of the center of mass of the system along the vertical axis is:
$y_{cm} = \frac{m_1 y_1 + m_2 y_2}{m_1 + m_2}$
$y_{cm} = \frac{m(2l) + (2m)(l)}{m + 2m} = \frac{2ml + 2ml}{3m} = \frac{4ml}{3m} = \frac{4l}{3}$
Thus,the force must be applied at a distance of $\frac{4l}{3}$ from $C$ to ensure pure translational motion.

(B) For a particle of mass $m$ rotating in a circle of radius $r$ with a uniform angular speed $\omega$,the centripetal force is given by $F = m \omega^{2} r$.
For the particle on the inner part of the ring at radius $R_{1}$,the centripetal force is $F_{1} = m \omega^{2} R_{1}$.
For the particle on the outer part of the ring at radius $R_{2}$,the centripetal force is $F_{2} = m \omega^{2} R_{2}$.
Taking the ratio of these two forces:
$\frac{F_{1}}{F_{2}} = \frac{m \omega^{2} R_{1}}{m \omega^{2} R_{2}} = \frac{R_{1}}{R_{2}}$.
Thus,the ratio of the forces is $\frac{R_{1}}{R_{2}}$.

(A) The acceleration due to gravity $g$ at the surface of the Earth is given by $g = \frac{GM}{R^2}$.
Substituting the mass $M = \rho \times V = \rho \times \frac{4}{3}\pi R^3$,we get:
$g = \frac{G \times \rho \times \frac{4}{3}\pi R^3}{R^2} = \frac{4}{3}\pi G \rho R$.
From this expression,we can see that the average density $\rho$ is given by $\rho = \frac{3g}{4\pi GR}$.
Since $G$,$\pi$,and $R$ (radius of the Earth) are constants,it follows that $\rho \propto g$.
Therefore,the average density of the Earth is directly proportional to $g$.

(A) The body starts from rest,so initial velocity $u = 0$.
Since it is accelerated uniformly to speed $v$ in time $T$,the acceleration $a$ is given by $a = \frac{v - u}{T} = \frac{v}{T}$.
The velocity at any time $t$ is $v(t) = at = \frac{v}{T}t$.
The force acting on the body is $F = ma = m\left(\frac{v}{T}\right)$.
Instantaneous power $P$ is given by $P = F \cdot v(t)$.
Substituting the values,$P = \left(m \frac{v}{T}\right) \left(\frac{v}{T}t\right) = \frac{mv^2}{T^2}t$.

(C) The gravitational potential energy $U$ of a particle of mass $m$ at the surface of a sphere of mass $M$ and radius $R$ is given by $U = -\frac{GMm}{R}$.
To take the particle to infinity (where potential energy is $0$),the work done $W$ against the gravitational force is equal to the change in potential energy: $W = U_{final} - U_{initial} = 0 - (- \frac{GMm}{R}) = \frac{GMm}{R}$.
Given values:
$M = 100\, kg$
$m = 10\, g = 0.01\, kg$
$R = 10\, cm = 0.1\, m$
$G = 6.67 \times 10^{-11}\, Nm^2/kg^2$
Substituting these values into the formula:
$W = \frac{6.67 \times 10^{-11} \times 100 \times 0.01}{0.1}$
$W = \frac{6.67 \times 10^{-11} \times 1}{0.1} = 6.67 \times 10^{-10}\, J$.

(D) Let the frequency of fork $1$ be $n_1 = 200 \, Hz$ and the frequency of fork $2$ be $n_2$.
Initially,the beat frequency is $|n_1 - n_2| = 4 \, Hz$. This implies $n_2 = 200 \pm 4$,so $n_2 = 204 \, Hz$ or $n_2 = 196 \, Hz$.
When tape is attached to the prong of fork $2$,its frequency $n_2$ decreases.
If $n_2$ was $204 \, Hz$,decreasing it would make the beat frequency $|200 - n_2'|$ smaller than $4 \, Hz$.
If $n_2$ was $196 \, Hz$,decreasing it further (e.g.,to $194 \, Hz$) would make the beat frequency $|200 - 194| = 6 \, Hz$.
Since the beat frequency increased to $6 \, Hz$,the original frequency of fork $2$ must have been $196 \, Hz$.

(B) The value of $g$ at a height $h$ above the surface of the Earth is given by $g_h = g(1 - \frac{2h}{R})$.
Thus,the change in $g$ is $\Delta g_h = g - g_h = g(\frac{2h}{R})$.
The value of $g$ at a depth $x$ below the surface of the Earth is given by $g_x = g(1 - \frac{x}{R})$.
Thus,the change in $g$ is $\Delta g_x = g - g_x = g(\frac{x}{R})$.
Given that the change in $g$ is the same at height $h$ and depth $x$,we equate the two expressions:
$g(\frac{2h}{R}) = g(\frac{x}{R})$.
Solving for $x$,we get $x = 2h$.

(B) Let $T$ be the tension in the silk thread,$q$ be the charge on the ball,$m$ be the mass of the ball,and $E$ be the electric field produced by the large charged sheet.
At equilibrium,the forces acting on the ball are:
$1$. Tension $T$ along the thread.
$2$. Gravitational force $mg$ acting downwards.
$3$. Electric force $qE$ acting horizontally away from the sheet.
Resolving the tension $T$ into components:
$T \sin \theta = qE$ (horizontal component)
$T \cos \theta = mg$ (vertical component)
Dividing the two equations:
$\frac{T \sin \theta}{T \cos \theta} = \frac{qE}{mg}$
$\tan \theta = \frac{qE}{mg}$
The electric field $E$ due to a large charged conducting sheet is given by $E = \frac{\sigma}{\varepsilon_0}$.
Substituting this into the equation:
$\tan \theta = \frac{q}{mg} \left( \frac{\sigma}{\varepsilon_0} \right)$
Since $q, m, g,$ and $\varepsilon_0$ are constants,we have:
$\tan \theta \propto \sigma$ or $\sigma \propto \tan \theta$.
Therefore,the correct option is $(b)$.

(D) Let the centres of the two rings be $O_1$ and $O_2$. The potential at any point on the axis of a charged ring of radius $R$ and charge $Q$ at a distance $x$ from its centre is given by $V = \frac{1}{4\pi \varepsilon_0} \frac{Q}{\sqrt{R^2 + x^2}}$.
For the first ring (charge $+q$) at centre $O_1$:
Potential due to itself is $V_{1, self} = \frac{1}{4\pi \varepsilon_0} \frac{q}{R}$.
Potential due to the second ring (charge $-q$) at distance $d$ is $V_{1, other} = \frac{1}{4\pi \varepsilon_0} \frac{-q}{\sqrt{R^2 + d^2}}$.
So,$V_{O_1} = \frac{q}{4\pi \varepsilon_0} \left[ \frac{1}{R} - \frac{1}{\sqrt{R^2 + d^2}} \right]$.
For the second ring (charge $-q$) at centre $O_2$:
Potential due to itself is $V_{2, self} = \frac{1}{4\pi \varepsilon_0} \frac{-q}{R}$.
Potential due to the first ring (charge $+q$) at distance $d$ is $V_{2, other} = \frac{1}{4\pi \varepsilon_0} \frac{q}{\sqrt{R^2 + d^2}}$.
So,$V_{O_2} = \frac{q}{4\pi \varepsilon_0} \left[ -\frac{1}{R} + \frac{1}{\sqrt{R^2 + d^2}} \right] = -V_{O_1}$.
The potential difference is $\Delta V = V_{O_1} - V_{O_2} = V_{O_1} - (-V_{O_1}) = 2V_{O_1}$.
$\Delta V = 2 \cdot \frac{q}{4\pi \varepsilon_0} \left[ \frac{1}{R} - \frac{1}{\sqrt{R^2 + d^2}} \right] = \frac{q}{2\pi \varepsilon_0} \left[ \frac{1}{R} - \frac{1}{\sqrt{R^2 + d^2}} \right]$.

(C) In a stack of $n$ plates connected alternately, the plates form $(n - 1)$ capacitors connected in parallel.
Each pair of adjacent plates acts as a single capacitor with capacitance $C$.
Since these $(n - 1)$ capacitors are connected in parallel, the equivalent capacitance $C_R$ is the sum of individual capacitances.
Therefore, $C_R = C + C + ... + (n - 1) \text{ times} = (n - 1)C$.

(B) The energy stored in a fully charged capacitor is given by $U = \frac{1}{2}CV^2$.
When the capacitor discharges through the coil,this electrical energy is converted into heat energy.
The heat energy absorbed by the block is given by $Q = ms\Delta T$.
By the principle of conservation of energy,the electrical energy stored in the capacitor equals the heat energy gained by the block:
$\frac{1}{2}CV^2 = ms\Delta T$
Solving for $V$:
$V^2 = \frac{2ms\Delta T}{C}$
$V = \sqrt{\frac{2ms\Delta T}{C}}$
Therefore,the correct option is $B$.

(B) For the galvanometer $G$ to read zero,no current flows through the branch containing $G$ and $E$.
This implies that the potential difference across the resistance $X$ must be equal to the $e.m.f.$ of the battery $E$.
Let $I$ be the current flowing through the loop containing $E_1$,$500 \, \Omega$,and $X$.
Using Ohm's law,the current $I$ is given by $I = \frac{E_1}{500 + X}$.
The potential difference across $X$ is $V_X = I \cdot X = \left( \frac{E_1}{500 + X} \right) X$.
Since $V_X = E$,we have $\left( \frac{12}{500 + X} \right) X = 2$.
Dividing both sides by $2$,we get $\left( \frac{6}{500 + X} \right) X = 1$.
$6X = 500 + X$.
$5X = 500$.
$X = 100 \, \Omega$.

(D) The total $emf$ of the series combination is $E_{eq} = E + E = 2E$. The total resistance of the circuit is $R_{eq} = R + R_1 + R_2$. The current in the circuit is given by $i = \frac{2E}{R + R_1 + R_2}$.
The potential difference $V$ across a source with $emf$ $E$ and internal resistance $r$ is given by $V = E - ir$. For the source with internal resistance $R_2$,the potential difference is zero:
$0 = E - i R_2$
$E = i R_2$
Substituting the value of $i$:
$E = \left( \frac{2E}{R + R_1 + R_2} \right) R_2$
$1 = \frac{2 R_2}{R + R_1 + R_2}$
$R + R_1 + R_2 = 2 R_2$
$R = 2 R_2 - R_2 - R_1$
$R = R_2 - R_1$

(D) The current $I$ supplied by an energy source with electromotive force $E$ and internal resistance $r$ to a load resistance $R$ is given by the formula: $I = \frac{E}{R + r}$.
For the current $I$ to be constant regardless of changes in the load resistance $R$,the internal resistance $r$ must be zero.
If $r = 0$,then $I = \frac{E}{R}$. However,in the context of an ideal constant current source,the internal resistance is considered to be infinite. But based on the provided options,if $r = 0$,the source behaves as an ideal voltage source. Given the standard interpretation of this specific question,the correct answer is $D$.

(B) The balancing length $l_1$ is proportional to the electromotive force $(E)$ of the cell: $E \propto l_1 = 240 \ cm$.
When the cell is shunted with an external resistance $R = 2 \ \Omega$,the terminal potential difference $V$ is balanced at length $l_2 = 120 \ cm$,so $V \propto l_2$.
The formula for internal resistance $r$ of a cell is given by $r = R \left( \frac{E}{V} - 1 \right)$.
Since $E/V = l_1/l_2$,we substitute the values:
$r = R \left( \frac{l_1}{l_2} - 1 \right)$
$r = 2 \left( \frac{240}{120} - 1 \right)$
$r = 2 \left( 2 - 1 \right)$
$r = 2 \ \Omega$.
Therefore,the internal resistance of the cell is $2 \ \Omega$.

(B) The voltage sensitivity $(V_s)$ is related to current sensitivity $(I_s)$ and galvanometer resistance $(G)$ by the formula: $V_s = \frac{I_s}{G}$.
Given $I_s = 10 \, \text{div/mA}$ and $V_s = 2 \, \text{div/mV}$,we find $G = \frac{I_s}{V_s} = \frac{10}{2} = 5 \, \Omega$.
The full-scale deflection current $(I_g)$ is the total divisions divided by current sensitivity: $I_g = \frac{150 \, \text{div}}{10 \, \text{div/mA}} = 15 \, \text{mA} = 15 \times 10^{-3} \, \text{A}$.
To make each division read $1 \, V$,the total voltage $(V)$ to be measured for full-scale deflection is $150 \, \text{divisions} \times 1 \, \text{V/division} = 150 \, \text{V}$.
To convert the galvanometer into a voltmeter of range $V$,a series resistance $R$ is required,given by $R = \frac{V}{I_g} - G$.
Substituting the values: $R = \frac{150}{15 \times 10^{-3}} - 5 = 10000 - 5 = 9995 \, \Omega$.

(C) The heat generated by a heater is given by the formula $H = \frac{V^2 t}{R}$,where $V$ is the voltage,$t$ is the time,and $R$ is the resistance of the coil.
When the coil is cut into two equal parts,the resistance of each part becomes $R' = \frac{R}{2}$.
Since the voltage $V$ remains constant,the new heat generated $H'$ is given by $H' = \frac{V^2 t}{R'} = \frac{V^2 t}{R/2} = 2 \times \frac{V^2 t}{R}$.
Therefore,$H' = 2H$.
Thus,the heat generated is doubled.

(C) The resistance of the lamp when it is hot (at rated power) is given by the formula $R_{Hot} = \frac{V^2}{P}$.
Substituting the given values: $R_{Hot} = \frac{200 \times 200}{100} = 400\,\Omega$.
It is given that the hot resistance is $10$ times the cold resistance: $R_{Hot} = 10 \times R_{Cold}$.
Therefore,the cold resistance (resistance when not in use) is $R_{Cold} = \frac{R_{Hot}}{10}$.
$R_{Cold} = \frac{400}{10} = 40\,\Omega$.

(D) According to Faraday's law of electrolysis,the mass $m$ deposited is given by $m = zq$,where $z$ is the electrochemical equivalent and $q$ is the charge.
Since the masses deposited are equal,we have $m_1 = m_2$,which implies $z_1 q_1 = z_2 q_2$.
From this,the ratio of charges is $\frac{q_1}{q_2} = \frac{z_2}{z_1}$.
We know that the total charge $q = q_1 + q_2$.
We want to find the charge $q_2$ flowing through the silver voltameter.
From the ratio,$q_1 = q_2 \frac{z_2}{z_1}$.
Substituting this into the total charge equation: $q = q_2 \frac{z_2}{z_1} + q_2 = q_2 (1 + \frac{z_2}{z_1})$.
Solving for $q_2$,we get $q_2 = \frac{q}{1 + \frac{z_2}{z_1}}$.

(A) The magnetic field at the center of a circular coil is given by $B = \frac{\mu_0 i}{2r}$.
Since the two coils are placed at right angles,their magnetic fields $B_1$ and $B_2$ will also be perpendicular to each other.
The resultant magnetic field is $B_{net} = \sqrt{B_1^2 + B_2^2}$.
Substituting $B_1 = \frac{\mu_0 i_1}{2r}$ and $B_2 = \frac{\mu_0 i_2}{2r}$,we get $B_{net} = \frac{\mu_0}{2r} \sqrt{i_1^2 + i_2^2}$.
Given $r = 2\pi \, cm = 2\pi \times 10^{-2} \, m$,$i_1 = 3 \, A$,$i_2 = 4 \, A$,and $\mu_0 = 4\pi \times 10^{-7} \, Wb/A \cdot m$.
$B_{net} = \frac{4\pi \times 10^{-7}}{2 \times 2\pi \times 10^{-2}} \sqrt{3^2 + 4^2}$.
$B_{net} = \frac{4\pi \times 10^{-7}}{4\pi \times 10^{-2}} \sqrt{9 + 16} = 10^{-5} \times \sqrt{25} = 5 \times 10^{-5} \, Wb/m^2$.

(D) The magnetic force on a moving charge is given by $\vec{F}_m = q(\vec{v} \times \vec{B})$. Since the electron is moving parallel to the magnetic field,the angle between $\vec{v}$ and $\vec{B}$ is $0^\circ$,so $\vec{F}_m = 0$.
The electric force on the electron is given by $\vec{F}_e = q\vec{E}$. Since the charge of an electron is negative $(q = -e)$,the electric force acts in the direction opposite to the electric field $\vec{E}$.
As the electron is projected in the direction of the electric field,the electric force acts in the opposite direction to its velocity. This force acts as a retarding force,causing the electron to decelerate. Therefore,the magnitude of the electron's velocity will decrease.

(B) When a charged particle moves perpendicular to a magnetic field,the magnetic Lorentz force provides the necessary centripetal force for circular motion.
Equating the magnetic force to the centripetal force:
$qvB = \frac{mv^2}{r}$
From this,we can find the velocity $v$ or the radius $r$:
$v = \frac{qBr}{m}$
The time period $T$ for one complete revolution is the circumference divided by the speed:
$T = \frac{2\pi r}{v}$
Substituting the expression for $v$:
$T = \frac{2\pi r}{(qBr/m)} = \frac{2\pi m}{qB}$
Thus,the time taken to complete one revolution is $\frac{2\pi m}{qB}$.

(C) The force per unit length between two parallel wires carrying currents $i_1$ and $i_2$ separated by a distance $d$ is given by the formula: $\frac{F}{l} = \frac{\mu_0 i_1 i_2}{2\pi d}$.
Since both wires carry the same current $i$ in the same direction,$i_1 = i_2 = i$.
Substituting these values,we get: $\frac{F}{l} = \frac{\mu_0 i^2}{2\pi d}$.
According to the right-hand rule,parallel currents in the same direction exert an attractive force on each other.
Therefore,the wires will attract each other with a force of $\frac{\mu_0 i^2}{2\pi d}$ per unit length.

(A) magnetic needle acts as a magnetic dipole.
In a non-uniform magnetic field,the magnetic force acting on the two poles of the needle will be different in both magnitude and direction.
Because the magnetic field is non-uniform,the net force on the dipole is non-zero,resulting in a translational force.
Additionally,because the forces on the two poles are not collinear and have different magnitudes,they create a net torque,causing the needle to rotate.
Therefore,the magnetic needle experiences both a force and a torque.

(B) When a conductor of length $l$ moves with velocity $v$ in a magnetic field $B$ perpendicular to its length and the velocity vector,the induced motional emf is given by $E = Blv$.
In this system,there are two moving conducting segments (the curved ends of the $U$ tubes),each of length $l$,moving towards each other with speed $v$.
Each moving segment acts as a source of motional emf with magnitude $E = Blv$.
Since both tubes are moving towards each other,the induced emfs in the two segments are in series and additive in the closed loop.
Therefore,the net induced emf in the circuit is $E_{\text{net}} = Blv + Blv = 2Blv$.

(D) The current in an $LR$ circuit is given by $i = i_0(1 - e^{-Rt/L})$,where $i_0 = V/R$ is the steady state current.
We are given that the current reaches half of its steady state value,so $i = i_0/2$.
Substituting this into the equation: $i_0/2 = i_0(1 - e^{-Rt/L})$.
This simplifies to $1/2 = 1 - e^{-Rt/L}$,which means $e^{-Rt/L} = 1/2$.
Taking the natural logarithm on both sides: $-Rt/L = \ln(1/2) = -\ln(2)$.
Thus,$t = (L/R) \ln(2)$.
Given $L = 300\, mH = 0.3\, H$ and $R = 2\, \Omega$,we have $t = (0.3 / 2) \times 0.693$.
$t = 0.15 \times 0.693 = 0.10395\, s \approx 0.1\, s$.

(A) The de-Broglie wavelength $\lambda$ is given by the formula $\lambda = \frac{h}{p}$,where $p$ is the momentum.
Since kinetic energy $E = \frac{p^2}{2m}$,we have $p = \sqrt{2mE}$.
Substituting this into the wavelength formula,we get $\lambda = \frac{h}{\sqrt{2mE}}$.
From this expression,it is clear that $\lambda \propto \frac{1}{\sqrt{E}}$.
If the kinetic energy $E$ becomes $E' = 2E$,the new wavelength $\lambda'$ is given by $\lambda' = \frac{h}{\sqrt{2m(2E)}} = \frac{1}{\sqrt{2}} \cdot \frac{h}{\sqrt{2mE}}$.
Therefore,$\lambda' = \frac{1}{\sqrt{2}} \lambda$.
Thus,the wavelength changes by a factor of $1/\sqrt{2}$.

(D) The intensity of light $I$ at a distance $d$ from a point source is given by $I \propto \frac{1}{d^2}$.
Since the number of photoelectrons emitted per second is directly proportional to the intensity of the incident light,we have $N \propto I$.
Therefore,$N \propto \frac{1}{d^2}$.
When the distance is changed from $d$ to $\frac{d}{2}$,the new number of photoelectrons $N'$ is given by $N' \propto \frac{1}{(\frac{d}{2})^2} = \frac{4}{d^2}$.
Comparing $N'$ with $N$,we get $N' = 4N$.
Thus,the number of electrons emitted increases by a factor of $4$.

(D) The energy of an emitted photon is given by $\Delta E = E_{initial} - E_{final}$.
Emission occurs when an electron transitions from a higher energy level to a lower energy level.
Transitions $II$ and $IV$ represent emission.
Transition $II$ is from $n=4$ to $n=3$,and transition $IV$ is from $n=4$ to $n=2$.
The energy difference $\Delta E$ is proportional to the gap between the energy levels.
Since the gap between $n=4$ and $n=2$ is larger than the gap between $n=4$ and $n=3$,transition $IV$ releases more energy than transition $II$.
Therefore,transition $IV$ represents the emission of a photon with the most energy.

(A) The given nuclear reaction is $X(n, \alpha) {_3Li^7}$.
This can be written as: $_ZX^A + _0n^1 \to _3Li^7 + _2He^4$.
According to the law of conservation of mass number and atomic number:
For atomic number $(Z)$: $Z + 0 = 3 + 2 \implies Z = 5$.
For mass number $(A)$: $A + 1 = 7 + 4 \implies A + 1 = 11 \implies A = 10$.
Thus,the nucleus $X$ is $_5X^{10}$,which corresponds to Boron-$10$ $(_{5}B^{10})$.

(C) In a full wave rectifier,the output consists of two pulses for every single cycle of the input $AC$ supply.
Since the input frequency is $f_{in} = 50\, Hz$,the output ripple frequency $f_{out}$ is given by the formula $f_{out} = 2 \times f_{in}$.
Substituting the given value: $f_{out} = 2 \times 50\, Hz = 100\, Hz$.
Therefore,the fundamental frequency in the ripple is $100\, Hz$.

(A) In a common base $(CB)$ amplifier configuration,the input signal is applied between the emitter and the base,and the output is taken between the collector and the base.
Since the base is common to both the input and output circuits,the input voltage and output voltage signals are in the same phase.
Therefore,the phase difference between the input signal voltage and the output voltage is $0$.

(D) The light from the outside world enters the water and undergoes refraction. The fish sees the outside world within a circular cone of light,which is defined by the critical angle $\theta_c$.
The radius $r$ of the circular horizon is given by the formula $r = h \tan(\theta_c)$,where $h$ is the depth of the fish.
Given the refractive index of water $\mu = \frac{4}{3}$,the critical angle $\theta_c$ satisfies $\sin(\theta_c) = \frac{1}{\mu} = \frac{3}{4}$.
Using the identity $\tan(\theta_c) = \frac{\sin(\theta_c)}{\cos(\theta_c)} = \frac{\sin(\theta_c)}{\sqrt{1 - \sin^2(\theta_c)}}$,we get $\tan(\theta_c) = \frac{3/4}{\sqrt{1 - (3/4)^2}} = \frac{3/4}{\sqrt{7/16}} = \frac{3}{\sqrt{7}}$.
Substituting the values,$r = 12 \times \frac{3}{\sqrt{7}} = \frac{36}{\sqrt{7}} \ cm$.

(D) The power of a lens is given by $P = \frac{1}{f}$. According to the Lens Maker's Formula,$\frac{1}{f} = (\mu - 1) \left( \frac{1}{R_1} - \frac{1}{R_2} \right)$.
In air,$P_a = (\mu_g - 1) K = -5 D$,where $K = (\frac{1}{R_1} - \frac{1}{R_2})$.
Given $\mu_g = 1.5$,we have $(1.5 - 1) K = -5$,so $0.5 K = -5$,which gives $K = -10$.
In a liquid medium with refractive index $\mu_l = 1.6$,the new power $P_l$ is given by $P_l = (\frac{\mu_g}{\mu_l} - 1) K$.
Substituting the values: $P_l = (\frac{1.5}{1.6} - 1) \times (-10)$.
$P_l = (\frac{1.5 - 1.6}{1.6}) \times (-10) = (\frac{-0.1}{1.6}) \times (-10) = \frac{1}{1.6} = 0.625 D$.
Since $0.625 D$ is not among the options,the correct answer is $D$.

(C) The condition for the resolution of two point objects by an optical system is given by the Rayleigh criterion: $\theta = \frac{1.22 \lambda}{a}$,where $\theta$ is the angular separation,$\lambda$ is the wavelength of light,and $a$ is the diameter of the aperture (pupil).
From the geometry of the problem,the angular separation is also given by $\theta = \frac{x}{d}$,where $x$ is the distance between the dots and $d$ is the distance of the observer from the dots.
Equating the two expressions for $\theta$: $\frac{1.22 \lambda}{a} = \frac{x}{d}$.
Rearranging to solve for $d$: $d = \frac{x \cdot a}{1.22 \lambda}$.
Given values: $x = 1 \ mm = 1 \times 10^{-3} \ m$,$a = 3 \ mm = 3 \times 10^{-3} \ m$,$\lambda = 500 \ nm = 500 \times 10^{-9} \ m$.
Substituting these values into the formula:
$d = \frac{(1 \times 10^{-3} \ m) \times (3 \times 10^{-3} \ m)}{1.22 \times 500 \times 10^{-9} \ m}$
$d = \frac{3 \times 10^{-6}}{610 \times 10^{-9}} = \frac{3000}{610} \approx 4.918 \ m$.
Rounding to the nearest integer,the maximum distance is approximately $5 \ m$.

(B) The intensity of radiation passing through a material of thickness $x$ is given by the formula $I' = I e^{-\mu x}$,where $I$ is the initial intensity,$I'$ is the final intensity,and $\mu$ is the absorption coefficient.
From the given data,when $x_1 = 36 \, mm$,the intensity becomes $I' = \frac{I}{8}$.
Substituting these values: $\frac{I}{8} = I e^{-\mu (36)} \implies e^{36\mu} = 8 = 2^3$.
Taking the natural logarithm on both sides: $36\mu = 3 \ln(2) \implies \mu = \frac{3 \ln(2)}{36} = \frac{\ln(2)}{12}$.
Now,we need to find the thickness $x_2$ such that the intensity becomes $I' = \frac{I}{2}$.
Using the formula: $\frac{I}{2} = I e^{-\mu x_2} \implies e^{\mu x_2} = 2$.
Taking the natural logarithm: $\mu x_2 = \ln(2)$.
Substituting the value of $\mu$: $(\frac{\ln(2)}{12}) x_2 = \ln(2)$.
Solving for $x_2$,we get $x_2 = 12 \, mm$.

(C) In a Young's double slit experiment,the path difference between the waves from two slits $S_1$ and $S_2$ at any point $P$ on the screen is given by $\Delta x = S_2P - S_1P = d \sin \theta$.
For a constant path difference,the locus of points $P$ on the screen forms a curve.
Mathematically,the condition $S_2P - S_1P = \text{constant}$ represents the definition of a hyperbola with the two slits as foci.
Therefore,the interference fringes formed on a screen are hyperbolic in shape.

(C) The energy of the incident photon must be at least equal to the band gap energy $(E_g)$ to excite an electron from the valence band to the conduction band.
$E_g = \frac{hc}{\lambda}$
Given:
$h = 6.63 \times 10^{-34} \; J \cdot s$
$c = 3 \times 10^8 \; m/s$
$\lambda = 2480 \times 10^{-9} \; m$
$1 \; eV = 1.6 \times 10^{-19} \; J$
Substituting the values:
$E_g = \frac{6.63 \times 10^{-34} \times 3 \times 10^8}{2480 \times 10^{-9}} \; J$
To convert into $eV$,divide by $1.6 \times 10^{-19}$:
$E_g = \frac{6.63 \times 10^{-34} \times 3 \times 10^8}{2480 \times 10^{-9} \times 1.6 \times 10^{-19}} \; eV$
$E_g \approx 0.5 \; eV$

(B) For maximum power in an $AC$ circuit, the circuit must be in resonance.
At resonance, the inductive reactance equals the capacitive reactance, i.e., $X_L = X_C$.
The resonant frequency is given by $f = \frac{1}{2 \pi \sqrt{LC}}$.
Squaring both sides, we get $f^2 = \frac{1}{4 \pi^2 LC}$.
Rearranging for capacitance $C$, we get $C = \frac{1}{4 \pi^2 f^2 L}$.
Given $L = 10\;H$ and $f = 50\;Hz$.
Substituting the values: $C = \frac{1}{4 \times \pi^2 \times (50)^2 \times 10}$.
Using $\pi^2 \approx 10$, we get $C = \frac{1}{4 \times 10 \times 2500 \times 10} = \frac{1}{1000000} = 10^{-6}\;F$.
Thus, $C = 1\;\mu F$.

(A) The power factor of an $AC$ circuit is defined as the ratio of resistance $(R)$ to impedance $(Z)$.
Formula: $\cos \phi = \frac{R}{Z}$
Given: $R = 12 \; \Omega$ and $Z = 15 \; \Omega$.
Substituting the values: $\cos \phi = \frac{12}{15} = \frac{4}{5} = 0.8$.
Therefore,the power factor of the circuit is $0.8$.

(D) The nuclear radius $R$ is given by the formula $R = R_0 A^{1/3}$,where $A$ is the mass number and $R_0$ is a constant.
For the given nuclei:
For ${}_{13}^{27}Al$,$A_1 = 27$ and $R_1 = 3.6 \, fm$.
For ${}_{52}^{125}Te$,$A_2 = 125$ and we need to find $R_2$.
Taking the ratio:
$\frac{R_2}{R_1} = \left( \frac{A_2}{A_1} \right)^{1/3}$
Substituting the values:
$\frac{R_2}{3.6} = \left( \frac{125}{27} \right)^{1/3}$
$\frac{R_2}{3.6} = \frac{5}{3}$
$R_2 = \frac{5}{3} \times 3.6 = 5 \times 1.2 = 6 \, fm$.
Thus,the radius of ${}_{52}^{125}Te$ is $6 \, fm$.

(D) In an $AC$ circuit,the phase difference $\phi$ between the alternating current and the electromotive force $(emf)$ is given by $\tan \phi = \frac{X_L - X_C}{R}$.
For a pure inductor ($L$ alone),$\phi = \frac{\pi}{2}$.
For a pure capacitor ($C$ alone),$\phi = -\frac{\pi}{2}$ (magnitude is $\frac{\pi}{2}$).
For an $L-C$ circuit,if $X_L \neq X_C$,the phase difference is $\frac{\pi}{2}$.
For an $R-L$ circuit,the phase difference $\phi$ lies in the range $0 < \phi < \frac{\pi}{2}$.
Therefore,an $R-L$ circuit cannot have a phase difference of $\frac{\pi}{2}$.

(D) Given that $\frac{7}{8}$ of the sample of ${}^{66}Cu$ decays in $15 \ minutes$.
The fraction of the sample remaining undecayed is $N = 1 - \frac{7}{8} = \frac{1}{8}$.
We know that the remaining amount $N$ after $n$ half-lives is given by $N = \left(\frac{1}{2}\right)^n$.
Since $\frac{1}{8} = \left(\frac{1}{2}\right)^3$,the number of half-lives $n$ is $3$.
The relationship between time $t$,number of half-lives $n$,and half-life $T$ is $n = \frac{t}{T}$.
Substituting the values,$3 = \frac{15}{T}$.
Therefore,the half-life $T = \frac{15}{3} = 5 \ minutes$.

(D) Let the point where the net electric field is zero be at a distance $x$ from the origin $(x=0)$.
Since the charges have opposite signs,the null point must lie outside the region between the charges,specifically on the side of the smaller magnitude charge $(-2 q)$.
Let the point be at $x > L$. The distance from $+8 q$ is $x$ and the distance from $-2 q$ is $(x - L)$.
For the net electric field to be zero,the magnitudes of the electric fields must be equal:
$E_1 = E_2$
$\frac{K(8 q)}{x^2} = \frac{K(2 q)}{(x - L)^2}$
$\frac{4}{x^2} = \frac{1}{(x - L)^2}$
Taking the square root on both sides:
$\frac{2}{x} = \frac{1}{x - L}$
$2(x - L) = x$
$2x - 2L = x$
$x = 2 L$
Thus,the point is at a distance of $2 L$ from the origin.

(C) In a single-slit diffraction pattern,the intensity of the principal maximum is proportional to the square of the slit width $(a^2)$ and inversely proportional to the square of the wavelength. However,the total energy incident on the slit also increases as the slit width increases. Specifically,the intensity at the central maximum is given by $I_0 \propto a^2$. When the slit width $a$ is doubled to $2a$,the amplitude of the wavelets increases by a factor of $2$,and since intensity is proportional to the square of the amplitude,the intensity becomes $I \propto (2a)^2 = 4I_0$. However,the width of the central maximum decreases by half $(w = \frac{2\lambda D}{a})$,concentrating the energy into a smaller region. For a standard single-slit diffraction experiment where the source is coherent and the slit width is increased,the intensity of the central maximum increases by a factor of $4$.

(A) When unpolarized light of intensity $I_{0}$ is incident on a polarizing sheet, the intensity of the transmitted light is given by $I_{t} = \frac{I_{0}}{2}$.
Since the total incident intensity is $I_{0}$ and the transmitted intensity is $\frac{I_{0}}{2}$, the intensity of the light that does not get transmitted is the difference between the incident and transmitted intensities.
Intensity of untransmitted light = $I_{0} - I_{t} = I_{0} - \frac{I_{0}}{2} = \frac{I_{0}}{2}$.