AIPMT 2006 Physics Question Paper with Answer and Solution

(A) Consider a small element of the liquid of length $dx$ at a distance $x$ from the axis of rotation. The mass of this element is $dM = (M/L)dx$.
The centripetal force required for this element to rotate in a circle of radius $x$ is $dF = (dM)\omega^2 x$.
Substituting $dM$,we get $dF = (M/L)dx \cdot \omega^2 x = (M\omega^2/L)x dx$.
The total force $F$ exerted by the liquid at the outer end is the integral of these centripetal forces from $x = 0$ to $x = L$:
$F = \int_0^L \frac{M\omega^2}{L} x dx = \frac{M\omega^2}{L} \left[ \frac{x^2}{2} \right]_0^L = \frac{M\omega^2}{L} \cdot \frac{L^2}{2} = \frac{1}{2}ML\omega^2$.

(D) The time of reverberation is defined as the time during which the intensity of sound in an auditorium becomes one millionth of its initial intensity. Sabine's formula for reverberation time is given by:
$T = \frac{0.16 V}{\sum a s}$
Where $V$ is the volume of the hall in $m^3$ and $\sum a s$ is the total absorption of the hall.
Since the dimensions of the room are doubled,the volume $V$ becomes $V' = (2L)(2W)(2H) = 8V$.
The total surface area $S$ (which is proportional to $\sum a s$) becomes $S' = (2L)(2W) + (2W)(2H) + (2H)(2L) = 4S$.
Using the ratio of the reverberation times:
$\frac{T'}{T} = \frac{V'}{S'} \times \frac{S}{V} = \frac{8V}{4S} \times \frac{S}{V} = \frac{8}{4} = 2$
Given $T = 1 \; s$,we find $T' = 2 \times 1 = 2 \; s$.

(A) The position of the particle is given by $x = 40 + 12t - t^3$.
Velocity $v$ is the rate of change of position with respect to time: $v = \frac{dx}{dt} = \frac{d}{dt}(40 + 12t - t^3) = 12 - 3t^2$.
For the particle to come to rest,its velocity must be zero: $v = 0$.
$12 - 3t^2 = 0 \implies 3t^2 = 12 \implies t^2 = 4 \implies t = 2 \text{ s}$ (since time cannot be negative).
The distance traveled by the particle from $t = 0$ to $t = 2$ is given by the change in position:
$x(0) = 40 + 12(0) - (0)^3 = 40 \text{ m}$.
$x(2) = 40 + 12(2) - (2)^3 = 40 + 24 - 8 = 56 \text{ m}$.
Distance traveled = $x(2) - x(0) = 56 - 40 = 16 \text{ m}$.

(C) The distance traveled in one complete circular lap is equal to the circumference of the track,which is $2\pi r$.
Given $r = 100\,m$ and time $t = 62.8\,s$.
Average speed is defined as the total distance divided by the total time:
$\text{Average speed} = \frac{2\pi r}{t} = \frac{2 \times 3.14 \times 100}{62.8} = \frac{628}{62.8} = 10\,m/s$.
Since the car completes one full lap,the final position is the same as the initial position.
Therefore,the net displacement is $0\,m$.
Average velocity is defined as the net displacement divided by the total time:
$\text{Average velocity} = \frac{\text{net displacement}}{t} = \frac{0}{62.8} = 0\,m/s$.
Thus,the average velocity is $0\,m/s$ and the average speed is $10\,m/s$.

(A) The time $t$ taken by a body to reach the ground when dropped from a height $h$ is given by the formula $t = \sqrt{\frac{2h}{g}}$.
Since both bodies are dropped,their initial velocity is $0$ and they are under the influence of the same acceleration due to gravity $g$.
The ratio of the time taken by body $A$ $(t_A)$ to the time taken by body $B$ $(t_B)$ is:
$\frac{t_A}{t_B} = \frac{\sqrt{\frac{2h_A}{g}}}{\sqrt{\frac{2h_B}{g}}} = \sqrt{\frac{h_A}{h_B}}$
Given $h_A = 16\,m$ and $h_B = 25\,m$:
$\frac{t_A}{t_B} = \sqrt{\frac{16}{25}} = \frac{4}{5} = 0.8$
Thus,the ratio of the time taken is $0.8$.

(B) The horizontal range $R$ of a projectile is given by the formula: $R = \frac{u^2 \sin(2\alpha)}{g}$,where $u$ is the initial velocity,$\alpha$ is the angle of projection,and $g$ is the acceleration due to gravity.
For the first angle of projection $\alpha_1 = (45^\circ - \theta)$:
$R_1 = \frac{u^2 \sin[2(45^\circ - \theta)]}{g} = \frac{u^2 \sin(90^\circ - 2\theta)}{g} = \frac{u^2 \cos(2\theta)}{g}$.
For the second angle of projection $\alpha_2 = (45^\circ + \theta)$:
$R_2 = \frac{u^2 \sin[2(45^\circ + \theta)]}{g} = \frac{u^2 \sin(90^\circ + 2\theta)}{g} = \frac{u^2 \cos(2\theta)}{g}$.
Comparing the two ranges:
$\frac{R_1}{R_2} = \frac{\frac{u^2 \cos(2\theta)}{g}}{\frac{u^2 \cos(2\theta)}{g}} = \frac{1}{1}$.
Thus,the ratio of the horizontal ranges is $1:1$.

(C) Given:
Mass of the ball,$m = 0.5\, kg$
Speed of the ball,$v = 12\, m/s$
Angle with the wall,$\theta = 30^\circ$
Time of contact,$\Delta t = 0.25\, s$
The component of momentum parallel to the wall remains unchanged. The component of momentum perpendicular to the wall changes direction.
Initial momentum perpendicular to the wall: $p_i = -mv \sin \theta$ (taking direction towards the wall as negative).
Final momentum perpendicular to the wall: $p_f = mv \sin \theta$ (taking direction away from the wall as positive).
Change in momentum,$\Delta p = p_f - p_i = mv \sin \theta - (-mv \sin \theta) = 2mv \sin \theta$.
Average force,$F = \frac{\Delta p}{\Delta t} = \frac{2mv \sin \theta}{\Delta t}$
$F = \frac{2 \times 0.5 \times 12 \times \sin 30^\circ}{0.25}$
$F = \frac{2 \times 0.5 \times 12 \times 0.5}{0.25}$
$F = \frac{6}{0.25} = 24\, N$.

(D) Given: Mass $m = 3\, kg$,displacement $s = \frac{1}{3}t^2$.
First,find the velocity $v$ by differentiating $s$ with respect to $t$:
$v = \frac{ds}{dt} = \frac{d}{dt}(\frac{1}{3}t^2) = \frac{2}{3}t$.
Next,find the acceleration $a$ by differentiating $v$ with respect to $t$:
$a = \frac{dv}{dt} = \frac{d}{dt}(\frac{2}{3}t) = \frac{2}{3}\, m/s^2$.
The force $F$ acting on the body is $F = m \times a = 3 \times \frac{2}{3} = 2\, N$.
The work done $W$ is given by $W = \int F \, ds$. Since $ds = v \, dt$,we have $W = \int_0^2 F \cdot v \, dt$.
$W = \int_0^2 2 \cdot (\frac{2}{3}t) \, dt = \frac{4}{3} \int_0^2 t \, dt$.
$W = \frac{4}{3} [\frac{t^2}{2}]_0^2 = \frac{4}{3} \times \frac{4}{2} = \frac{8}{3}\, J$.

(C) The total work done on the block is $W_{total} = 300\, J$.
This work is used to increase the potential energy of the block and to overcome friction.
The increase in potential energy is given by $\Delta U = mgh$.
Substituting the given values: $\Delta U = 2\, kg \times 10\, m/s^2 \times 10\, m = 200\, J$.
The work done against friction $(W_f)$ is the difference between the total work done and the work done against gravity:
$W_f = W_{total} - \Delta U$
$W_f = 300\, J - 200\, J = 100\, J$.

(B) According to $Wien's$ displacement law,the product of the wavelength of maximum intensity $(\lambda_{\max})$ and the absolute temperature $(T)$ is constant:
$\lambda_{\max} T = b$ (constant)
Given:
Initial temperature $T_1 = 1227^{\circ}C = 1227 + 273 = 1500\;K$
Initial wavelength $\lambda_{\max 1} = 5000\;\mathring{A}$
Final temperature $T_2 = 1227^{\circ}C + 1000^{\circ}C = 2227^{\circ}C = 2227 + 273 = 2500\;K$
Using the relation $\lambda_{\max 1} T_1 = \lambda_{\max 2} T_2$:
$\lambda_{\max 2} = \frac{\lambda_{\max 1} T_1}{T_2}$
$\lambda_{\max 2} = \frac{5000 \times 1500}{2500}$
$\lambda_{\max 2} = 5000 \times 0.6 = 3000\;\mathring{A}$
Therefore,the maximum intensity will be observed at $3000\;\mathring{A}$.

(C) The torque $\tau$ about point $A$ is due to the gravitational force acting at the center of mass of the rod,which is at a distance $l/2$ from $A.$
$\tau = mg \times \frac{l}{2} = \frac{mgl}{2}$
Using the relation between torque and angular acceleration,$\tau = I\alpha$,where $I$ is the moment of inertia about $A$ and $\alpha$ is the angular acceleration.
Given $I = \frac{ml^2}{3}$,we have:
$\alpha = \frac{\tau}{I} = \frac{mgl/2}{ml^2/3} = \frac{mgl}{2} \times \frac{3}{ml^2} = \frac{3g}{2l}$

(C) The escape velocity from the surface of the Earth is given by $v = \sqrt{\frac{2GM}{R}} = \sqrt{2gR}$.
For a body at a height $h = R$ from the surface,the distance from the center of the Earth is $r = R + h = 2R$.
The total energy required to escape from this distance must be zero:
$-\frac{GMm}{2R} + \frac{1}{2}m(fv)^2 = 0$.
Solving for the escape velocity at height $R$ $(v')$:
$\frac{1}{2}m(v')^2 = \frac{GMm}{2R} \Rightarrow v' = \sqrt{\frac{GM}{R}}$.
Since $v = \sqrt{\frac{2GM}{R}}$,we can write $v' = \frac{1}{\sqrt{2}} \sqrt{\frac{2GM}{R}} = \frac{v}{\sqrt{2}}$.
Given that the escape velocity from the platform is $fv$,we have $fv = \frac{v}{\sqrt{2}}$.
Therefore,$f = \frac{1}{\sqrt{2}}$.

(B) The molar specific heat at constant pressure is given as ${C_p} = \frac{7}{2}R$.
Using Mayer's relation,${C_p} - {C_V} = R$,we can find the molar specific heat at constant volume ${C_V}$.
${C_V} = {C_p} - R = \frac{7}{2}R - R = \frac{5}{2}R$.
The ratio of specific heat at constant pressure to that at constant volume is denoted by $\gamma = \frac{{C_p}}{{C_V}}$.
Substituting the values,$\gamma = \frac{(7/2)R}{(5/2)R} = \frac{7}{5}$.

(D) The efficiency of a Carnot engine is given by $\eta = 1 - \frac{T_2}{T_1}$,where $T_2$ is the sink temperature and $T_1$ is the source temperature.
Given $\eta = 40\% = 0.4$ and $T_2 = 300\, K$.
$0.4 = 1 - \frac{300}{T_1} \implies \frac{300}{T_1} = 0.6 \implies T_1 = \frac{300}{0.6} = 500\, K$.
The efficiency is to be increased by $50\%$ of its original efficiency.
New efficiency $\eta' = \eta + 0.5 \times \eta = 0.4 + 0.5(0.4) = 0.4 + 0.2 = 0.6 = 60\%$.
Let the new source temperature be $T_1'$.
$0.6 = 1 - \frac{300}{T_1'} \implies \frac{300}{T_1'} = 0.4 \implies T_1' = \frac{300}{0.4} = 750\, K$.
The increase in source temperature is $\Delta T = T_1' - T_1 = 750 - 500 = 250\, K$.

(C) Let $l$ be the length of the block immersed in the liquid at equilibrium. When the block is floating,the weight of the block is balanced by the buoyant force:
$mg = A l \rho g$
If the block is given a small vertical displacement $y$ downwards,the additional buoyant force acting upwards is:
$F_{restoring} = -(A(l+y)\rho g - mg) = -(Al\rho g + Ay\rho g - mg)$
Since $mg = Al\rho g$,we get:
$F_{restoring} = -A\rho g y$
This is the equation of simple harmonic motion $F = -ky$,where the spring constant $k = A\rho g$.
The time period $T$ of the oscillation is given by:
$T = 2\pi \sqrt{\frac{m}{k}} = 2\pi \sqrt{\frac{m}{A\rho g}}$
From this expression,we can see that $T \propto \frac{1}{\sqrt{A}}$.

(A) The frequency $f$ of a sound wave is given by the formula $f = \frac{v}{\lambda}$,where $v$ is the velocity and $\lambda$ is the wavelength.
For the first wave: $f_1 = \frac{330}{5.0} = 66\, Hz$.
For the second wave: $f_2 = \frac{330}{5.5} = 60\, Hz$.
The number of beats per second is equal to the absolute difference between the two frequencies: $|f_1 - f_2|$.
Beats per second $= |66 - 60| = 6\, Hz$.

(A) The given wave equation is $y(x, t) = 8.0 \sin \left( 0.5 \pi x - 4 \pi t - \frac{\pi}{4} \right)$.
Comparing this with the standard wave equation $y(x, t) = A \sin (kx - \omega t + \phi)$:
Here,the wave number $k = 0.5 \pi$ and the angular frequency $\omega = 4 \pi$.
The speed of the wave $v$ is given by the ratio of angular frequency to wave number:
$v = \frac{\omega}{k}$
Substituting the values:
$v = \frac{4 \pi}{0.5 \pi} = \frac{4}{0.5} = 8 \ m/s$.
Therefore,the speed of the wave is $8 \ m/s$.

(C) According to the principle of dimensional homogeneity,the dimensions of each term in an equation must be the same.
$1$. For the term $c$: Since $c$ is added to $t$ (time),the dimension of $c$ must be the same as the dimension of $t$. Therefore,$[c] = [T]$.
$2$. For the term $at$: The dimension of $at$ must be equal to the dimension of velocity $v$. Thus,$[a][T] = [LT^{-1}]$,which gives $[a] = [LT^{-2}]$.
$3$. For the term $\frac{b}{t+c}$: The dimension of the entire term must be equal to the dimension of velocity $v$. Since $[t+c] = [T]$,we have $\frac{[b]}{[T]} = [LT^{-1}]$. Therefore,$[b] = [L]$.
Thus,the dimensions are $a = LT^{-2}, b = L, c = T$.

(B) Sound waves are mechanical waves that require a material medium for propagation. In air,sound waves propagate as longitudinal waves,where the particles of the medium oscillate parallel to the direction of wave propagation.
Light waves are electromagnetic waves that do not require a material medium for propagation. They are transverse waves,where the electric and magnetic field vectors oscillate perpendicular to the direction of wave propagation.
Therefore,the correct statement is that sound waves in air are longitudinal while light waves are transverse.

(A) The moment of inertia of a uniform circular disc about an axis passing through its centre and perpendicular to its plane is $I_{G} = \frac{1}{2}MR^2$.
By the theorem of parallel axes,the moment of inertia about an axis passing through the edge and normal to the disc is given by $I = I_{G} + Md^2$,where $d = R$ is the distance between the parallel axes.
Substituting the values,we get $I = \frac{1}{2}MR^2 + M(R)^2 = \frac{3}{2}MR^2$.

(D) The magnitude of the sum of two vectors is given by $|\vec{A}+\vec{B}|^2 = |\vec{A}|^2 + |\vec{B}|^2 + 2|\vec{A}||\vec{B}| \cos \theta$.
Similarly,the magnitude of the difference of two vectors is given by $|\vec{A}-\vec{B}|^2 = |\vec{A}|^2 + |\vec{B}|^2 - 2|\vec{A}||\vec{B}| \cos \theta$.
Given that $|\vec{A}+\vec{B}| = |\vec{A}-\vec{B}|$,we square both sides to get $|\vec{A}+\vec{B}|^2 = |\vec{A}-\vec{B}|^2$.
Substituting the expressions,we have $|\vec{A}|^2 + |\vec{B}|^2 + 2|\vec{A}||\vec{B}| \cos \theta = |\vec{A}|^2 + |\vec{B}|^2 - 2|\vec{A}||\vec{B}| \cos \theta$.
Canceling the common terms $|\vec{A}|^2$ and $|\vec{B}|^2$ from both sides,we get $2|\vec{A}||\vec{B}| \cos \theta = -2|\vec{A}||\vec{B}| \cos \theta$.
Rearranging the terms,we get $4|\vec{A}||\vec{B}| \cos \theta = 0$.
Since the vectors are non-zero,$|\vec{A}| \neq 0$ and $|\vec{B}| \neq 0$,therefore $\cos \theta = 0$.
This implies $\theta = 90^{\circ}$.

(A) The potential energy $U$ stored in a spring stretched by a distance $x$ is given by the formula $U = \frac{1}{2} k x^2$,where $k$ is the spring constant.
For the initial stretch $x_1 = 2\,cm$,the potential energy is $U = \frac{1}{2} k (2)^2 = 2k$.
For the final stretch $x_2 = 8\,cm$,the new potential energy $U'$ is $U' = \frac{1}{2} k (8)^2 = 32k$.
Dividing the two equations:
$\frac{U'}{U} = \frac{\frac{1}{2} k (8)^2}{\frac{1}{2} k (2)^2} = \left(\frac{8}{2}\right)^2 = (4)^2 = 16$.
Therefore,$U' = 16\,U$.

(B) The Curie temperature $(T_C)$ is a characteristic property of ferromagnetic materials.
At temperatures below $T_C$,the material exhibits ferromagnetic properties due to the alignment of magnetic domains.
As the temperature increases and reaches $T_C$,the thermal agitation becomes strong enough to overcome the exchange coupling that maintains the alignment of these domains.
Consequently,above the Curie temperature,the material loses its spontaneous magnetization and transitions into a paramagnetic state.
Therefore,the correct option is $(B)$.

(C) When magnetic flux linked with a coil changes,an induced $emf$ is produced in it,and the induced current flows through the material. These induced currents are set up in the conductor in the form of closed loops. These currents resemble eddies or whirlpools and are known as $Eddy$ currents. These currents oppose the cause of their origin; therefore,due to $Eddy$ currents,a significant amount of energy is wasted in the form of heat energy. If the core of the transformer is laminated,the path for these currents is restricted,and their effect is minimized.

(A) The work done $W$ in rotating an electric dipole of dipole moment $\vec{p}$ in a uniform electric field $\vec{E}$ from an initial angle $\theta_1$ to a final angle $\theta_2$ is given by the formula:
$W = \int_{\theta_1}^{\theta_2} pE \sin \theta \, d\theta = pE (\cos \theta_1 - \cos \theta_2)$
Given that the dipole is initially lying along the electric field,the initial angle $\theta_1 = 0^{\circ}$.
The final angle is $\theta_2 = 90^{\circ}$.
Substituting these values into the formula:
$W = pE (\cos 0^{\circ} - \cos 90^{\circ})$
Since $\cos 0^{\circ} = 1$ and $\cos 90^{\circ} = 0$,we get:
$W = pE (1 - 0) = pE$.

(D) The electric flux $\phi$ through a surface is defined as the dot product of the electric field vector $\vec{E}$ and the area vector $\vec{S}$.
$\phi = \vec{E} \cdot \vec{S} = ES \cos \theta$
Here,the electric field $\vec{E}$ lies in the plane of the paper.
The area vector $\vec{S}$ for a surface in the plane of the paper is perpendicular to the plane of the paper.
Therefore,the angle $\theta$ between the electric field vector $\vec{E}$ and the area vector $\vec{S}$ is $90^{\circ}$.
Since $\cos 90^{\circ} = 0$,the electric flux $\phi$ associated with the surface is:
$\phi = ES \cos 90^{\circ} = ES(0) = 0$.

(B) The capacitance of a parallel plate capacitor is given by $C = \frac{\varepsilon_{0} A}{d}$,where $A$ is the area of the plates and $d$ is the distance between them.
When the distance $d$ is increased,the capacitance $C$ decreases.
Since the battery is disconnected,the charge $Q$ on the plates remains constant.
The relationship between charge,capacitance,and potential difference is $V = \frac{Q}{C}$.
Since $Q$ is constant and $C$ decreases,the potential difference $V$ must increase.

(C) The circuit consists of two parallel branches. The upper branch has resistors $1 \,\Omega$ and $3 \,\Omega$ in series,so its total resistance is $R_1 = 1 + 3 = 4 \,\Omega.$ The lower branch has a resistor $R_2 = 8 \,\Omega.$
Since the branches are in parallel,the potential difference $V$ across both branches is the same.
The power dissipated in a resistor is given by $P = \frac{V^2}{R}.$
For the $8 \,\Omega$ resistor,$P_2 = \frac{V^2}{8} = 2 \,\text{W},$ which implies $V^2 = 16 \,\text{V}^2.$
The current $i_1$ flowing through the upper branch is $i_1 = \frac{V}{R_1} = \frac{V}{4}.$
The power dissipated in the $3 \,\Omega$ resistor is $P_{3\Omega} = i_1^2 \times 3 = \left(\frac{V}{4}\right)^2 \times 3 = \frac{V^2}{16} \times 3.$
Substituting $V^2 = 16,$ we get $P_{3\Omega} = \frac{16}{16} \times 3 = 3 \,\text{W}.$

(C) The total $e.m.f.$ 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 $I$ flowing through the circuit is given by $I = \frac{2E}{R + r_1 + r_2}$.
The potential difference $V_1$ across the first cell (with internal resistance $r_1$) is given by $V_1 = E - I r_1$.
Given that $V_1 = 0$,we have $E - I r_1 = 0$,which implies $E = I r_1$.
Substituting the expression for $I$:
$E = \left( \frac{2E}{R + r_1 + r_2} \right) r_1$.
Dividing both sides by $E$ (assuming $E \neq 0$):
$1 = \frac{2r_1}{R + r_1 + r_2}$.
$R + r_1 + r_2 = 2r_1$.
$R = 2r_1 - r_1 - r_2 = r_1 - r_2$.
Thus,the correct option is $C$.

(A) Kirchhoff's first law,also known as the Junction Rule,states that the algebraic sum of currents meeting at a junction is zero. This is a direct consequence of the law of conservation of charge,as charge cannot accumulate at a junction.
Kirchhoff's second law,also known as the Loop Rule,states that the algebraic sum of potential differences in any closed loop is zero. This is a direct consequence of the law of conservation of energy,as the work done in moving a unit charge around a closed loop in an electrostatic field is zero.

(B) Let the potential at point $C$ be $V_C$ and at point $D$ be $V_D$. Let $V_C - V_D = V$.
The circuit consists of two parallel branches: $CAD$ and $CBD$.
In branch $CAD$,the total resistance is $4 \, \Omega + 4 \, \Omega = 8 \, \Omega$. The current is $I = \frac{V}{8} \, A$.
The potential at $A$ relative to $C$ is $V_C - V_A = I \times 4 \, \Omega = \frac{V}{8} \times 4 = \frac{V}{2}$. Thus,$V_A = V_C - \frac{V}{2}$.
In branch $CBD$,the total resistance is $1 \, \Omega + 3 \, \Omega = 4 \, \Omega$. The current is $I' = \frac{V}{4} \, A$.
The potential at $B$ relative to $C$ is $V_C - V_B = I' \times 1 \, \Omega = \frac{V}{4} \times 1 = \frac{V}{4}$. Thus,$V_B = V_C - \frac{V}{4}$.
Comparing the potentials,$V_B = V_C - 0.25V$ and $V_A = V_C - 0.5V$. Since $V_B > V_A$,the current will flow from $B$ to $A$ when a wire is connected between them.

(D) The force acting on a charged particle moving with velocity $\vec{v}$ in a magnetic field $\vec{B}$ is given by the Lorentz force formula:
$\vec{F} = q(\vec{v} \times \vec{B})$
The magnitude of this force is given by:
$F = qvB \sin \theta$
where $\theta$ is the angle between the velocity vector $\vec{v}$ and the magnetic field vector $\vec{B}$.
For the force to be non-zero $(F \neq 0)$,we must have $\sin \theta \neq 0$.
Since $\sin \theta = 0$ at $\theta = 0^{\circ}$ and $\theta = 180^{\circ}$,the force is zero in these cases.
Therefore,for the force to be non-zero,the angle $\theta$ can have any value other than $0^{\circ}$ and $180^{\circ}$.

(A) The focal length of the convex lens is $f_{1} = +25\, cm$.
The focal length of the concave lens is $f_{2} = -25\, cm$.
The power $P$ of a lens in diopters is given by $P = \frac{100}{f(cm)}$.
The power of the convex lens is $P_{1} = \frac{100}{25} = +4\, D$.
The power of the concave lens is $P_{2} = \frac{100}{-25} = -4\, D$.
When lenses are in contact,the total power of the combination is $P = P_{1} + P_{2}$.
Therefore,$P = 4\, D + (-4\, D) = 0\, D$.

(C) The apparent depth of the mark when viewed through the glass slab is given by the formula: $\text{Apparent depth} = \frac{\text{Real depth}}{\mu} = \frac{3\, cm}{1.5} = 2\, cm$.
The shift in the position of the mark is given by: $\text{Shift} = \text{Real depth} - \text{Apparent depth} = 3\, cm - 2\, cm = 1\, cm$.
Since the mark appears to be raised by $1\, cm$ towards the microscope,the microscope must be moved upwards by $1\, cm$ to bring the mark back into focus.

(A) The energy of the photon is given as $E = 1 \, MeV = 1 \times 10^6 \, eV$.
Converting energy into Joules: $E = 10^6 \times 1.6 \times 10^{-19} \, J = 1.6 \times 10^{-13} \, J$.
The momentum $p$ of a photon is related to its energy $E$ by the formula $p = E/c$,where $c$ is the speed of light $(c \approx 3 \times 10^8 \, m/s)$.
Substituting the values: $p = \frac{1.6 \times 10^{-13} \, J}{3 \times 10^8 \, m/s}$.
$p \approx 0.533 \times 10^{-21} \, kg \, m/s$.
$p \approx 5.33 \times 10^{-22} \, kg \, m/s$.
Rounding to the nearest given option,the correct value is $5 \times 10^{-22} \, kg \, m/s$.

(A) Let $K$ and $K^{\prime}$ be the maximum kinetic energy of photoelectrons for incident light of frequency $f$ and $2f$ respectively.
According to Einstein's photoelectric equation:
$K = hf - E_0$ ....... $(i)$
For the doubled frequency $2f$,the new maximum kinetic energy $K^{\prime}$ is:
$K^{\prime} = h(2f) - E_0$ ...... $(ii)$
We can rewrite equation $(ii)$ as:
$K^{\prime} = 2hf - E_0$
$K^{\prime} = hf + hf - E_0$
Substituting $hf = K + E_0$ from equation $(i)$ into the expression:
$K^{\prime} = (K + E_0) + hf - E_0$
$K^{\prime} = K + hf$

(C) The photoelectric current is directly proportional to the intensity of the incident light (illumination).
According to the photoelectric effect,when light of a suitable frequency falls on a metal surface,electrons are emitted.
The number of photoelectrons emitted per second is directly proportional to the intensity of the incident radiation.
Therefore,a change in the intensity of illumination results in a corresponding change in the photoelectric current.

(D) The mutual inductance $M$ between two coils is given by the formula $M = K \sqrt{L_1 L_2}$,where $K$ is the coefficient of coupling.
Since the flux in one coil is completely linked with the other,the coils are perfectly coupled,meaning $K = 1$.
Given $L_1 = 2 \, mH$ and $L_2 = 8 \, mH$.
Substituting these values into the formula:
$M = 1 \times \sqrt{2 \, mH \times 8 \, mH}$
$M = \sqrt{16 \, mH^2}$
$M = 4 \, mH$.

(D) Given:
Inductive reactance $X_{L} = 31\,\Omega$
Resistance $R = 8\,\Omega$
Capacitive reactance $X_{C} = 25\,\Omega$
The impedance $Z$ of a series $LCR$ circuit is given by the formula:
$Z = \sqrt{R^{2} + (X_{L} - X_{C})^{2}}$
Substituting the given values:
$Z = \sqrt{8^{2} + (31 - 25)^{2}}$
$Z = \sqrt{64 + 6^{2}}$
$Z = \sqrt{64 + 36}$
$Z = \sqrt{100} = 10\,\Omega$
The power factor of the circuit is defined as:
$\cos \phi = \frac{R}{Z}$
Substituting the values of $R$ and $Z$:
$\cos \phi = \frac{8}{10} = 0.8$

(D) The frequency of an $LC$ oscillator is given by the formula $f = \frac{1}{2\pi\sqrt{LC}}$.
Let the initial frequency be $f_1 = f$ with inductance $L_1 = L$ and capacitance $C_1 = C$.
Let the new frequency be $f_2$ with new inductance $L_2 = 2L$ and new capacitance $C_2 = 4C$.
Using the ratio of frequencies:
$\frac{f_2}{f_1} = \frac{\frac{1}{2\pi\sqrt{L_2 C_2}}}{\frac{1}{2\pi\sqrt{L_1 C_1}}} = \sqrt{\frac{L_1 C_1}{L_2 C_2}}$
Substituting the values:
$\frac{f_2}{f} = \sqrt{\frac{L \times C}{2L \times 4C}} = \sqrt{\frac{1}{8}} = \frac{1}{2\sqrt{2}}$
Therefore,the new frequency is $f_2 = \frac{f}{2\sqrt{2}}$.

(C) The ionization potential of a hydrogen atom is $13.6 \ eV$. The energy of the $n^{th}$ orbit is given by $E_n = -\frac{13.6}{n^2} \ eV$.
When the hydrogen atom in the ground state $(n=1)$ absorbs a photon of energy $12.1 \ eV$,it is excited to a higher energy level $n$. The energy difference is given by:
$E = E_n - E_1$
$12.1 = -\frac{13.6}{n^2} - (-13.6)$
$12.1 = 13.6 - \frac{13.6}{n^2}$
$\frac{13.6}{n^2} = 13.6 - 12.1 = 1.5$
$n^2 = \frac{13.6}{1.5} \approx 9.06 \approx 9$
$n = 3$
The number of spectral lines emitted when the electron transitions from the $n^{th}$ state back to the ground state is given by the formula $\frac{n(n-1)}{2}$.
For $n=3$,the number of spectral lines = $\frac{3(3-1)}{2} = \frac{3 \times 2}{2} = 3$.

(C) The nuclear fusion reaction is given by: $_{1}^{2}H + _{1}^{2}H \rightarrow _{2}^{4}He + Q$.
The energy released $(Q)$ in a nuclear reaction is equal to the difference between the total binding energy of the products and the total binding energy of the reactants.
$Q = B.E.(_{2}^{4}He) - 2 \times B.E.(_{1}^{2}H)$.
Given:
$B.E.(_{2}^{4}He) = 28 \, MeV$
$B.E.(_{1}^{2}H) = 2.2 \, MeV$
Substituting the values:
$Q = 28 - 2(2.2) \, MeV$
$Q = 28 - 4.4 \, MeV$
$Q = 23.6 \, MeV$.
Therefore,the energy released is $23.6 \, MeV$.

(B) According to the law of radioactive decay,the activity $R$ at any time $t$ is given by $R = R_0 e^{-\lambda t}$.
At time $t_1$,the activity is $R_1 = R_0 e^{-\lambda t_1}$.
At time $t_2$,the activity is $R_2 = R_0 e^{-\lambda t_2}$.
Dividing the two equations,we get:
$\frac{R_1}{R_2} = \frac{R_0 e^{-\lambda t_1}}{R_0 e^{-\lambda t_2}} = e^{-\lambda t_1} \cdot e^{\lambda t_2} = e^{-\lambda(t_1 - t_2)}$.
Therefore,$R_1 = R_2 e^{-\lambda(t_1 - t_2)}$.

(B) By observing the given voltage waveforms,we can construct the truth table for the inputs $A$ and $B$ and the output $C$:

$A$	$B$	$C$
$1$	$1$	$1$
$1$	$0$	$0$
$0$	$1$	$0$
$0$	$0$	$0$

Comparing this truth table with the standard truth tables of logic gates,we find that the output $C$ is $1$ only when both inputs $A$ and $B$ are $1$. This is the characteristic behavior of an $AND$ gate. Therefore,the logic circuit gate is an $AND$ gate.

(D) The change in base current is $\Delta I_B = 150 \, \mu A - 100 \, \mu A = 50 \, \mu A = 50 \times 10^{-6} \, A$.
The change in collector current is $\Delta I_C = 10 \, mA - 5 \, mA = 5 \, mA = 5 \times 10^{-3} \, A$.
The current gain $\beta$ in common emitter configuration is defined as the ratio of the change in collector current to the change in base current:
$\beta = \frac{\Delta I_C}{\Delta I_B} = \frac{5 \times 10^{-3} \, A}{50 \times 10^{-6} \, A} = \frac{5000 \times 10^{-6}}{50 \times 10^{-6}} = 100$.
Therefore,the current gain $\beta$ is $100$.

(D) $p-n$ junction diode is said to be forward-biased if the potential at the $p$-side $(V_p)$ is higher than the potential at the $n$-side $(V_n)$,i.e.,$V_p > V_n$.
Let us analyze each option:
$A$: $V_p = -4 \text{ V}$,$V_n = -3 \text{ V}$. Here,$-4 < -3$,so $V_p < V_n$ (Reverse biased).
$B$: $V_p = -2 \text{ V}$,$V_n = +2 \text{ V}$. Here,$-2 < +2$,so $V_p < V_n$ (Reverse biased).
$C$: $V_p = 3 \text{ V}$,$V_n = 5 \text{ V}$. Here,$3 < 5$,so $V_p < V_n$ (Reverse biased).
$D$: $V_p = 0 \text{ V}$,$V_n = -2 \text{ V}$. Here,$0 > -2$,so $V_p > V_n$ (Forward biased).
Therefore,option $D$ represents a forward-biased diode.

(B) The magnetic field at the centre of a circular coil is given by $B = \frac{\mu_0 I}{2r}$.
Given that the magnetic field at the centre is the same for both coils,we have:
$\frac{\mu_0 I_1}{2(2r)} = \frac{\mu_0 I_2}{2(r)} \Rightarrow \frac{I_1}{I_2} = 2 \quad ...(i)$
Since the coils are made from the same wire,their resistance $R$ is proportional to their length $l$ $(R = \rho \frac{l}{A})$.
The lengths are $l_1 = 2\pi(2r) = 4\pi r$ and $l_2 = 2\pi(r) = 2\pi r$.
Thus,the ratio of resistances is $\frac{R_1}{R_2} = \frac{l_1}{l_2} = \frac{4\pi r}{2\pi r} = 2 \quad ...(ii)$
Using Ohm's law,$V = IR$,we have $I = V/R$.
Substituting this into equation $(i)$:
$\frac{V_1/R_1}{V_2/R_2} = 2 \Rightarrow \frac{V_1}{V_2} = 2 \times \frac{R_1}{R_2}$
Substituting the ratio from equation $(ii)$:
$\frac{V_1}{V_2} = 2 \times 2 = 4$.

(A) The radius of a nucleus is given by the relation $R = R_0 A^{1/3}$,where $A$ is the number of nucleons.
Therefore,the ratio of the radii of two nuclei is given by $\frac{R_1}{R_2} = \left(\frac{A_1}{A_2}\right)^{1/3}$.
Given that $R_{Ge} = 2 R_{Be}$ and $A_{Be} = 9$.
Substituting these values into the formula:
$2 = \left(\frac{A_{Ge}}{9}\right)^{1/3}$.
Cubing both sides,we get $2^3 = \frac{A_{Ge}}{9}$.
$8 = \frac{A_{Ge}}{9}$.
$A_{Ge} = 8 \times 9 = 72$.
Thus,the number of nucleons in germanium is $72$.