AIIMS 2010 Physics Question Paper with Answer and Solution

(C) The viscous force $F$ acting on a fluid layer is given by Newton's law of viscosity: $F = -\eta A \frac{dv}{dx}$.
Here,$F$ is the force,$\eta$ is the coefficient of viscosity,$A$ is the area,and $\frac{dv}{dx}$ is the velocity gradient.
Rearranging for $\eta$: $\eta = \frac{F}{A (dv/dx)}$.
Substituting the dimensions: $[F] = [M L T^{-2}]$,$[A] = [L^2]$,$[dv] = [L T^{-1}]$,and $[dx] = [L]$.
Thus,$[\eta] = \frac{[M L T^{-2}]}{[L^2] [L T^{-1} / L]} = \frac{[M L T^{-2}]}{[L^2] [T^{-1}]} = [M L^{-1} T^{-1}]$.
Therefore,the correct option is $C$.

(C) Let the student catch the bus after time $t \, s$. The distance covered by the student is $s_s = ut$.
The distance covered by the bus starting from rest is $s_b = \frac{1}{2}at^2 = \frac{1}{2}(1)t^2 = \frac{t^2}{2}$.
For the student to catch the bus,the total distance covered by the student must equal the initial distance plus the distance covered by the bus:
$ut = 50 + \frac{t^2}{2}$.
Rearranging for $u$:
$u = \frac{50}{t} + \frac{t}{2}$.
To find the minimum velocity $u$,we differentiate $u$ with respect to $t$ and set it to zero:
$\frac{du}{dt} = -\frac{50}{t^2} + \frac{1}{2} = 0$.
Solving for $t$:
$\frac{50}{t^2} = \frac{1}{2} \implies t^2 = 100 \implies t = 10 \, s$.
Substituting $t = 10 \, s$ back into the equation for $u$:
$u = \frac{50}{10} + \frac{10}{2} = 5 + 5 = 10 \, m/s$.

(A) The gravitational force exerted by the Earth on the satellite always acts towards the centre of the Earth. According to Newton's second law,$F = ma$,the acceleration of the satellite is always directed towards the centre of the Earth.
Since the gravitational force is a central force,the torque acting on the satellite about the centre of the Earth is zero. Therefore,the angular momentum $L$ of the satellite remains constant in both magnitude and direction.
According to the law of conservation of energy,in the absence of non-conservative forces,the total mechanical energy of the satellite remains constant throughout its orbit.
Since the distance $r$ of the satellite from the Earth varies in an elliptical orbit,the orbital velocity $v$ must change to conserve angular momentum $(L = mvr \sin \theta)$. Consequently,the linear momentum $p = mv$ is not constant.

(D) For an ideal gas,the speeds are defined as:
${v_{rms}} = \sqrt{\frac{3kT}{m}}$,${v_p} = \sqrt{\frac{2kT}{m}}$,and $\bar{v} = \sqrt{\frac{8kT}{\pi m}}$.
Comparing these values:
${v_p} = \sqrt{2} \sqrt{\frac{kT}{m}} \approx 1.414 \sqrt{\frac{kT}{m}}$
$\bar{v} = \sqrt{\frac{8}{3.14}} \sqrt{\frac{kT}{m}} \approx 1.596 \sqrt{\frac{kT}{m}}$
${v_{rms}} = \sqrt{3} \sqrt{\frac{kT}{m}} \approx 1.732 \sqrt{\frac{kT}{m}}$
Thus,${v_p} < \bar{v} < {v_{rms}}$,which confirms option $(c)$ is correct.
For the average kinetic energy:
${E_{av}} = \frac{1}{2} m v_{rms}^2 = \frac{1}{2} m \left( \frac{3kT}{m} \right) = \frac{3}{2} kT$.
Since ${v_p}^2 = \frac{2kT}{m}$,we have $kT = \frac{1}{2} m v_p^2$.
Substituting this into the energy equation:
${E_{av}} = \frac{3}{2} \left( \frac{1}{2} m v_p^2 \right) = \frac{3}{4} m v_p^2$,which confirms option $(b)$ is correct.
Therefore,both $(b)$ and $(c)$ are correct.

(C) The process is isothermal,therefore $T = \text{constant}$.
Since $PV = \mu RT$,we have $P = \frac{\mu RT}{V}$.
For chamber $A$,the change in pressure is $\Delta P = P_i - P_f = \frac{\mu_A RT}{V} - \frac{\mu_A RT}{2V} = \frac{\mu_A RT}{2V} \dots (i)$.
For chamber $B$,the change in pressure is $1.5 \Delta P = P_i - P_f = \frac{\mu_B RT}{V} - \frac{\mu_B RT}{2V} = \frac{\mu_B RT}{2V} \dots (ii)$.
Dividing equation $(i)$ by $(ii)$,we get $\frac{\Delta P}{1.5 \Delta P} = \frac{\mu_A}{\mu_B} \implies \frac{1}{1.5} = \frac{\mu_A}{\mu_B} \implies \frac{\mu_A}{\mu_B} = \frac{2}{3}$.
Since $\mu = \frac{m}{M}$,where $M$ is the molar mass,we have $\frac{m_A/M}{m_B/M} = \frac{2}{3} \implies \frac{m_A}{m_B} = \frac{2}{3}$.
Therefore,$3m_A = 2m_B$.

(B) According to Wien's displacement law,the product of the wavelength of maximum emission $(\lambda_m)$ and the absolute temperature $(T)$ of the black body is a constant $(b)$:
$\lambda_m T = b$
Given:
$\lambda_m = 2.93 \times 10^{-10} \ m$
$b = 2.93 \times 10^{-3} \ m \cdot K$
Substituting the values into the formula:
$T = \frac{b}{\lambda_m} = \frac{2.93 \times 10^{-3}}{2.93 \times 10^{-10}}$
$T = 10^{-3 - (-10)} \ K = 10^7 \ K$
Thus,the maximum temperature attained is of the order of $10^7 \ K$.

(A) The kinetic energy $E$ is given by the formula $E = \frac{1}{2}mv^2$.
Taking the relative error,we use the formula for propagation of errors:
$\frac{\Delta E}{E} = \frac{\Delta m}{m} + 2\frac{\Delta v}{v}$.
Given that the percentage error in mass $\frac{\Delta m}{m} \times 100 = 1\%$ and the percentage error in velocity $\frac{\Delta v}{v} \times 100 = 2\%$.
Substituting these values into the error equation:
$\frac{\Delta E}{E} \times 100 = (\frac{\Delta m}{m} \times 100) + 2 \times (\frac{\Delta v}{v} \times 100)$
$\frac{\Delta E}{E} \times 100 = 1\% + 2 \times 2\% = 1\% + 4\% = 5\%$.
Since both the Assertion and the Reason are correct and the Reason correctly explains the propagation of error formula used in the Assertion,the correct option is $A$.

(A) The formula for the angle of banking in a horizontal circular turn is given by $\tan \theta = \frac{v^2}{rg}$.
Given values are speed $v = 150\, m/s$,angle $\theta = 12^\circ$,$g = 10\, m/s^2$,and $\tan 12^\circ = 0.2125$.
Substituting these values into the formula:
$0.2125 = \frac{(150)^2}{r \times 10}$
$0.2125 = \frac{22500}{10r}$
$0.2125 = \frac{2250}{r}$
$r = \frac{2250}{0.2125} \approx 10588.2\, m$.
Converting the radius into kilometers:
$r \approx 10.588\, km \approx 10.6\, km$.

(B) An inertial frame is a frame of reference that is either at rest or moving with a constant velocity (zero acceleration).
In option $A$,the child is in circular motion,which involves centripetal acceleration.
In option $B$,the car is moving with a constant speed on a straight road,meaning its velocity is constant and its acceleration is zero. Therefore,it acts as an inertial frame.
In option $C$,the aeroplane is accelerating during takeoff.
In option $D$,the cyclist is changing direction,which involves acceleration.
Thus,the driver in the sports car is the correct observer in an inertial frame.

(A) In uniform circular motion,the speed of the particle remains constant,but the direction of motion changes continuously. Since velocity is a vector quantity,a change in direction implies a change in velocity.
Similarly,the centripetal acceleration is directed towards the center of the circle. As the particle moves,the direction of the center relative to the particle changes,so the direction of acceleration also changes.
The centripetal acceleration is given by the formula $a_c = \omega^2 r$,where $\omega$ is the angular velocity and $r$ is the radius of the circle. Thus,the $Reason$ is also correct and explains why the acceleration vector changes (as $\omega$ or the position changes).
Therefore,both $Assertion$ and $Reason$ are correct,and the $Reason$ explains the $Assertion$.

(A) According to Newton's third law of motion,when a horse pushes the ground backward with its hooves,the ground exerts an equal and opposite reaction force on the horse in the forward direction. This reaction force from the ground is what causes the horse to move forward.

(A) Let $u$ be the initial downward velocity and $h = 12.4\, m$ be the height. The velocity $v$ of the ball just before hitting the ground is given by $v^2 = u^2 + 2gh$.
Kinetic energy just before impact is $K_1 = \frac{1}{2}mv^2 = \frac{1}{2}m(u^2 + 2gh)$.
The ball loses $15\%$ of its kinetic energy,so the kinetic energy after impact is $K_2 = 0.85 K_1$.
Let $v_2$ be the velocity just after impact. Then $\frac{1}{2}mv_2^2 = 0.85 \times \frac{1}{2}m(u^2 + 2gh)$,which simplifies to $v_2^2 = 0.85(u^2 + 2gh)$.
For the ball to reach the same height $h$,the upward velocity $v_2$ must satisfy $v_2^2 = 2gh$.
Equating the two expressions for $v_2^2$: $0.85(u^2 + 2gh) = 2gh$.
Substituting $g = 9.8\, m/s^2$ and $h = 12.4\, m$:
$0.85(u^2 + 2 \times 9.8 \times 12.4) = 2 \times 9.8 \times 12.4$.
$0.85(u^2 + 243.04) = 243.04$.
$u^2 + 243.04 = \frac{243.04}{0.85} \approx 285.93$.
$u^2 = 285.93 - 243.04 = 42.89$.
$u = \sqrt{42.89} \approx 6.55\, m/s$.

(D) According to the law of conservation of linear momentum,the total momentum before the collision is equal to the total momentum after the collision.
Let $m_1 = 0.1\, kg$ and $m_2 = 0.4\, kg$ be the masses.
Let $v_1 = 1\, m/s$ and $v_2 = -0.1\, m/s$ be their respective velocities (taking the direction of the first body as positive).
After the collision,they stick together,so they move with a common velocity $v$.
$m_1 v_1 + m_2 v_2 = (m_1 + m_2)v$
$(0.1)(1) + (0.4)(-0.1) = (0.1 + 0.4)v$
$0.1 - 0.04 = 0.5v$
$0.06 = 0.5v$
$v = \frac{0.06}{0.5} = 0.12\, m/s$
The distance covered in $t = 10\, s$ is given by $d = v \times t$.
$d = 0.12\, m/s \times 10\, s = 1.2\, m$.

(A) In an elastic collision,kinetic energy is conserved,therefore the ball rebounds with the same velocity $u$ in the opposite direction.
According to Newton's second law,the force $F$ exerted by the surface is equal to the rate of change of momentum.
The change in momentum for one ball is $\Delta p = m(u - (-u)) = 2mu$.
Since $n$ balls collide per second,the total change in momentum per second is $n \times 2mu = 2mnu$.
Therefore,the force experienced by the surface is $F = 2mnu$.
Both the $Assertion$ and $Reason$ are correct,and the $Reason$ explains why the change in momentum is $2mu$ per ball.

(D) The $Assertion$ is incorrect because many helicopters operate with a single main rotor and a smaller tail rotor,not necessarily two main propellers.
The $Reason$ is also incorrect because the purpose of the second rotor (tail rotor) is to balance the torque produced by the main rotor to prevent the helicopter from spinning in the opposite direction,which is based on the conservation of angular momentum,not linear momentum.

(C) The length of the wire is $l$ and its mass is $m$.
When the wire is bent into a circular ring of radius $r$,the circumference of the ring is equal to the length of the wire.
$2\pi r = l$
Therefore,the radius of the ring is $r = \frac{l}{2\pi}$.
The moment of inertia $I$ of a ring of mass $m$ and radius $r$ about its central axis is given by the formula:
$I = mr^2$
Substituting the value of $r$ into the formula:
$I = m \left( \frac{l}{2\pi} \right)^2$
$I = m \left( \frac{l^2}{4\pi^2} \right)$
$I = \left( \frac{1}{4\pi^2} \right) ml^2$

(D) When polar ice melts, the water flows from the poles towards the equator.
This redistribution of mass increases the moment of inertia $(I)$ of the Earth because mass is moved further from the axis of rotation.
According to the principle of conservation of angular momentum $(L = I\omega)$, if the angular momentum $L$ remains constant and the moment of inertia $I$ increases, the angular velocity $(\omega)$ must decrease.
Since the angular velocity $\omega = 2\pi / T$ decreases, the time period $(T)$ of the Earth's rotation increases.
Therefore, the length of the day becomes longer, not shorter.
Thus, the $Assertion$ is incorrect and the $Reason$ is also incorrect.

(B) The effective acceleration due to gravity $g'$ at a latitude $\lambda$ is given by $g' = g - R\omega^2 \cos^2 \lambda$,where $g$ is the acceleration due to gravity if the earth were stationary,$R$ is the radius of the earth,and $\omega$ is the angular velocity of the earth.
The increase in the value of $g$ when the earth stops rotating is $\Delta g = g - g' = R\omega^2 \cos^2 \lambda$.
Given:
$R = 6400 \times 10^3 \ m = 6.4 \times 10^6 \ m$
$\lambda = 45^{\circ} \implies \cos^2 45^{\circ} = (1/\sqrt{2})^2 = 0.5$
$\omega = \frac{2\pi}{T} = \frac{2 \times 3.14}{24 \times 3600} \approx 7.27 \times 10^{-5} \ rad/sec$
Substituting the values:
$\Delta g = (6.4 \times 10^6) \times (7.27 \times 10^{-5})^2 \times 0.5$
$\Delta g = (6.4 \times 10^6) \times (52.85 \times 10^{-10}) \times 0.5$
$\Delta g \approx 0.0169 \ m/sec^2$
Converting to $C.G.S.$ units $(cm/sec^2)$:
$1 \ m/sec^2 = 100 \ cm/sec^2$
$\Delta g = 0.0169 \times 100 = 1.69 \ cm/sec^2 \approx 1.68 \ cm/sec^2$.

(D) According to the principle of flotation,for the boy and the log to float,the total weight of the system must be equal to the upthrust (buoyant force) exerted by the water.
Let $V$ be the volume of the wooden log.
Mass of the boy = $60\, kg$.
Density of wood $\rho_{wood} = 0.6 \times 1000 = 600\, kg/m^3$.
Density of water $\rho_{water} = 1000\, kg/m^3$.
Total weight of the system = Weight of boy + Weight of log
$= 60g + (V \times 600)g$
Upthrust (Buoyant force) = Weight of water displaced by the log
$= V \times 1000 \times g$
Equating the two:
$60g + 600Vg = 1000Vg$
$60 = 1000V - 600V$
$60 = 400V$
$V = \frac{60}{400} = \frac{3}{20}\, m^3$.

(C) The $Assertion$ is correct. When a thin film of water is placed between two glass plates,a large force is required to separate them.
This is primarily due to the surface tension of water and the adhesive forces between the water molecules and the glass surface.
The pressure inside the thin water film becomes less than the atmospheric pressure due to the concave meniscus formed at the edges,creating a pressure difference that holds the plates together.
The $Reason$ is incorrect because water does not act as a 'glue' in the chemical sense; rather,the phenomenon is explained by the physics of surface tension and capillary action,not by the adhesive properties of glue.

(A) When two thin blankets are placed together,a layer of air is trapped between them.
Since air is a very poor conductor of heat (a good insulator),this trapped air layer prevents the heat from the body from escaping to the surroundings.
Therefore,the combination of two thin blankets provides better insulation and warmth compared to a single thick blanket of the same total thickness,as the air layer acts as an additional thermal barrier.

(D) For an isothermal process,the temperature $T$ remains constant.
For $n$ moles of an ideal gas,the heat exchanged $\Delta Q$ during an isothermal expansion from $V_1$ to $V_2$ is equal to the work done $W$,because the change in internal energy $\Delta U = 0$.
$W = \int_{V_1}^{V_2} P \, dV = \int_{V_1}^{V_2} \frac{nRT}{V} \, dV = nRT \ln \frac{V_2}{V_1}$.
Given $n = 1$ mole,the heat exchanged is $\Delta Q = RT \ln \frac{V_2}{V_1}$.
The change in entropy $\Delta S$ is defined as $\Delta S = \frac{\Delta Q}{T}$.
Substituting the value of $\Delta Q$,we get $\Delta S = \frac{RT \ln \frac{V_2}{V_1}}{T} = R \ln \frac{V_2}{V_1}$.

(D) The time period of a simple pendulum is given by the formula $T = 2 \pi \sqrt{\frac{l}{g_{eff}}}$,where $g_{eff}$ is the effective acceleration due to gravity.
When an elevator accelerates downwards with an acceleration $a$,the effective acceleration becomes $g_{eff} = (g - a)$.
Since $T \propto \frac{1}{\sqrt{g_{eff}}}$,as $g_{eff}$ decreases,the time period $T$ increases.
Among the given options,when the elevator is accelerating downwards,$g_{eff}$ is minimized (assuming $a < g$),resulting in the greatest time period.

(D) According to the definition of Young's modulus $(E)$:
$E = \frac{F L}{A \Delta L}$
where $F$ is the force,$L$ is the original length,$A$ is the cross-sectional area,and $\Delta L$ is the extension.
Rearranging for force $(F)$:
$F = \left( \frac{EA}{L} \right) \Delta L$ --- $(1)$
According to Hooke's law,the restoring force is given by:
$F = k \Delta L$ --- $(2)$
Comparing equations $(1)$ and $(2)$,we get the force constant $(k)$:
$k = \frac{EA}{L}$
The time period $(T)$ of a spring-mass system is given by:
$T = 2\pi \sqrt{\frac{M}{k}}$
Substituting the value of $k$:
$T = 2\pi \sqrt{\frac{M}{EA/L}} = 2\pi \sqrt{\frac{ML}{EA}}$
Thus,the time period is $2\pi \sqrt{\frac{ML}{EA}}$ and the force constant is $\frac{EA}{L}$.

(A) The time period of a simple pendulum is given by the formula $T = 2\pi \sqrt{\frac{L}{g}}$.
Here,$L$ is the effective length of the pendulum and $g$ is the acceleration due to gravity.
As observed from the formula,the time period $T$ is independent of the mass of the bob.
Since the length of the pendulum and the acceleration due to gravity remain unchanged,the time period will remain the same regardless of the change in mass.
Therefore,the new time period is $2\, s$.

(A) The equation of motion for a particle in $SHM$ starting from the mean position is given by $y = A \sin(\omega t)$,where $A = 25\, cm$ and $\omega = \frac{2\pi}{T} = \frac{2\pi}{3}$.
To find the time taken to reach $y = 12.5\, cm$ from the mean position $(y = 0)$:
$12.5 = 25 \sin(\frac{2\pi}{3} t)$
$\frac{1}{2} = \sin(\frac{2\pi}{3} t)$
Since $\sin(\frac{\pi}{6}) = \frac{1}{2}$,we have $\frac{\pi}{6} = \frac{2\pi}{3} t$.
Solving for $t$,we get $t = \frac{1}{4} = 0.25\, s$.
The particle moves from $-12.5\, cm$ to $+12.5\, cm$ passing through the mean position. The total time taken is $2t = 2 \times 0.25 = 0.5\, s$.

(C) The $Assertion$ is correct. Resonance occurs when the driving frequency $(\omega)$ equals the natural frequency $(\omega_{0})$ of the system,resulting in maximum amplitude.
The $Reason$ is incorrect. The amplitude of forced vibrations does not simply increase with the frequency of the external force. Instead,the amplitude follows a resonance curve; it increases as the driving frequency approaches the natural frequency and decreases as it moves away from it.
The amplitude of oscillation for a forced,damped oscillator is given by:
$A = \frac{F_{0} / m}{\sqrt{(\omega^{2} - \omega_{0}^{2})^{2} + (b \omega / m)^{2}}}$
where $b$ is the damping constant and $\omega_{0} = \sqrt{k / m}$ is the natural frequency. As $\omega$ approaches $\omega_{0}$,the denominator decreases,causing the amplitude $A$ to reach its maximum value.

(B) For the formation of stationary waves,it is necessary that the medium should be bounded. When a wave propagates in a bounded medium,it reflects at the boundary,producing a wave of the same frequency and amplitude travelling in the opposite direction. The superposition of these two waves results in a stationary wave. Thus,the $Assertion$ is correct.
In stationary waves,there are specific points in the medium where the displacement is always zero. These points are called nodes,and the particles at these positions remain permanently at rest. Thus,the $Reason$ is also correct.
However,the fact that some particles remain at rest (nodes) is a property of stationary waves,not the cause of why the medium must be bounded. Therefore,the $Reason$ is not the correct explanation of the $Assertion$.

(B) The core of a transformer is laminated to minimize the energy losses due to eddy currents. When a time-varying magnetic flux passes through the metallic core,it induces eddy currents in the core,which lead to heating and energy dissipation. By using thin,insulated laminated sheets,the path for these eddy currents is restricted,significantly reducing the magnitude of the currents and thus minimizing energy loss.

(D) The electric potential $V$ due to an electric dipole at a point $(r, \theta)$ is given by the formula:
$V = \frac{1}{4 \pi \varepsilon_{0}} \frac{p \cos \theta}{r^{2}}$
Given values:
$V = 1.8 \times 10^{5} \, V$
$\theta = 60^{\circ}$
$r = 50 \, cm = 0.5 \, m$
$\frac{1}{4 \pi \varepsilon_{0}} = 9 \times 10^{9} \, N \cdot m^2/C^2$
Substituting these values into the formula:
$1.8 \times 10^{5} = (9 \times 10^{9}) \times \frac{p \cos 60^{\circ}}{(0.5)^{2}}$
Since $\cos 60^{\circ} = 0.5$:
$1.8 \times 10^{5} = 9 \times 10^{9} \times \frac{p \times 0.5}{0.25}$
$1.8 \times 10^{5} = 9 \times 10^{9} \times p \times 2$
$p = \frac{1.8 \times 10^{5}}{18 \times 10^{9}}$
$p = 0.1 \times 10^{-4} = 10^{-5} \, C-m$

(B) When a capacitor is charged and then disconnected from the battery,the charge $Q$ on the plates remains constant.
When a dielectric slab of dielectric constant $K$ is introduced between the plates,the capacitance increases to $C' = KC$.
Since $Q$ is constant,the new potential difference $V' = \frac{Q}{C'} = \frac{Q}{KC} = \frac{V}{K}$,which means the potential difference decreases.
The stored energy $U = \frac{Q^2}{2C}$ becomes $U' = \frac{Q^2}{2C'} = \frac{Q^2}{2KC} = \frac{U}{K}$,which means the stored energy decreases.
Therefore,the potential difference and stored energy decrease,while the charge remains unchanged.

(D) The electric field $E$ at the surface of a conducting sphere of radius $r$ is given by $E = \frac{V}{r}$,where $V$ is the potential of the sphere.
Given,the maximum electric field $E_{max} = 10^7\,V/m$ and the radius $r = 0.10\,m$.
To find the maximum potential $V_{max}$,we use the relation $V_{max} = E_{max} \times r$.
Substituting the values: $V_{max} = 10^7\,V/m \times 0.10\,m$.
$V_{max} = 10^7 \times 10^{-1} = 10^6\,V$.
Thus,the maximum potential to which the sphere can be charged is $10^6\,V$.

(A) The equivalent capacitance of the circuit is given by:
$\frac{1}{C_{eq}} = \frac{1}{C_3} + \frac{1}{C_1 + C_2}$
(since $C_1$ and $C_2$ are in parallel,and this combination is in series with $C_3$).
Thus,$\frac{1}{C_{eq}} = \frac{C_1 + C_2 + C_3}{C_3(C_1 + C_2)}$
$\therefore C_{eq} = \frac{C_3(C_1 + C_2)}{C_1 + C_2 + C_3}$
Since $V$ is the voltage of the battery,the initial total charge is $q = C_{eq} V = \frac{C_3(C_1 + C_2) V}{C_1 + C_2 + C_3}$.
If the capacitor $C_3$ breaks down (acts as a short circuit),the effective capacitance becomes $C_{eq}' = C_1 + C_2$.
The new charge is $q' = C_{eq}' V = (C_1 + C_2) V$.
The change in total charge is $\Delta q = q' - q = (C_1 + C_2) V - \frac{C_3(C_1 + C_2) V}{C_1 + C_2 + C_3}$.
Factoring out $(C_1 + C_2) V$,we get:
$\Delta q = (C_1 + C_2) V \left[ 1 - \frac{C_3}{C_1 + C_2 + C_3} \right]$.

(A) Let the inner shell $B$ have radius $R_1$ and charge $Q_1$,and the outer shell $A$ have radius $R_2$ and charge $Q_2$.
The potential at the surface of the outer shell $A$ is $V_A = \frac{1}{4 \pi \varepsilon_0} \frac{Q_1 + Q_2}{R_2}$.
The potential at the surface of the inner shell $B$ is $V_B = \frac{1}{4 \pi \varepsilon_0} \left( \frac{Q_1}{R_1} + \frac{Q_2}{R_2} \right)$.
The potential difference between the shells is $V_B - V_A = \frac{1}{4 \pi \varepsilon_0} \left( \frac{Q_1}{R_1} + \frac{Q_2}{R_2} - \frac{Q_1 + Q_2}{R_2} \right) = \frac{1}{4 \pi \varepsilon_0} Q_1 \left( \frac{1}{R_1} - \frac{1}{R_2} \right)$.
This shows that the potential difference depends only on the charge of the inner shell $Q_1$. Thus,the Assertion is correct.
The potential due to the charge of the outer shell $Q_2$ at any point inside it (including the surface of the inner shell) is constant and equal to $\frac{1}{4 \pi \varepsilon_0} \frac{Q_2}{R_2}$. Thus,the Reason is also correct and explains why the $Q_2$ term cancels out in the potential difference calculation.

(A) Let $V_{AB}$ be the potential difference between $A$ and $B$ and $V_{BC}$ be the potential difference between $B$ and $C$. Let $k$ be the potential gradient of the potentiometer wire.
When connected between $A$ and $B$,the balance length $l_1 = 203.6 \, cm$. Thus,$V_{AB} = k \cdot l_1 = k \cdot 203.6 \, cm$.
When the end is shifted to $C$,the total potential difference is $V_{AC} = V_{AB} + V_{BC}$. The balance length is $l_2 = 24.6 \, cm$. However,since $V_{AC} > V_{AB}$,the balance length should increase. The problem implies the polarity is reversed or the connection is $V_{AB} - V_{BC}$. Given the balance point is $24.6 \, cm$,we have $V_{AB} - V_{BC} = k \cdot 24.6 \, cm$.
Subtracting the two equations:
$V_{AB} = k \cdot 203.6 \, cm$
$V_{AB} - V_{BC} = k \cdot 24.6 \, cm$
$V_{BC} = k \cdot (203.6 - 24.6) = k \cdot 179.0 \, cm$.
Thus,the balance point for $B$ and $C$ is $179.0 \, cm$.

(D) The resistance of a wire is given by $R = \frac{\rho L}{A}$. Since the diameter is the same for all wires,the cross-sectional area $A$ is constant.
Calculating resistance for each wire:
$R_1 = \frac{\rho L}{A}$
$R_2 = \frac{(1.2\rho)(1.2L)}{A} = 1.44 \frac{\rho L}{A}$
$R_3 = \frac{(0.9\rho)(0.9L)}{A} = 0.81 \frac{\rho L}{A}$
$R_4 = \frac{\rho(1.5L)}{A} = 1.5 \frac{\rho L}{A}$
Comparing the resistances: $R_3 < R_1 < R_2 < R_4$.
The rate of energy dissipation (power) for a constant potential difference $V$ is given by $P = \frac{V^2}{R}$.
Since $P \propto \frac{1}{R}$,the wire with the smallest resistance will have the greatest power dissipation.
Therefore,the order of power dissipation from greatest to least is: $P_3 > P_1 > P_2 > P_4$.

(D) Given: Galvanometer resistance $G = 50\,\Omega$,full-scale deflection current $I_G = 100\,\mu A = 100 \times 10^{-6}\,A$,and the desired range $I = 10\,A$.
To convert a galvanometer into an ammeter,a shunt resistance $S$ must be connected in parallel with the galvanometer.
The formula for shunt resistance is $S = \left( \frac{I_G}{I - I_G} \right) G$.
Substituting the values: $S = \left( \frac{100 \times 10^{-6}}{10 - 100 \times 10^{-6}} \right) \times 50$.
Since $100 \times 10^{-6} = 10^{-4}$ is very small compared to $10$,the denominator $10 - 10^{-4} \approx 10$.
Thus,$S \approx \left( \frac{10^{-4}}{10} \right) \times 50 = 10^{-5} \times 50 = 5 \times 10^{-4}\,\Omega$.

(D) Given:
Magnetic field $B = 1.5 \times 10^{-2} \, T$
Speed $v = 6 \times 10^7 \, m/s$
Specific charge $\frac{e}{m} = 1.7 \times 10^{11} \, C/kg$
Angle $\theta = 90^{\circ}$ (since it moves at right angles),so $\sin \theta = 1$.
The radius $r$ of the circular path of a charged particle in a magnetic field is given by the formula:
$r = \frac{mv}{qB}$
Since the specific charge is $\frac{q}{m} = 1.7 \times 10^{11} \, C/kg$,we can write the radius as:
$r = \frac{v}{B(q/m)}$
Substituting the values:
$r = \frac{6 \times 10^7}{(1.5 \times 10^{-2}) \times (1.7 \times 10^{11})}$
$r = \frac{6 \times 10^7}{2.55 \times 10^9}$
$r \approx 2.35 \times 10^{-2} \, m$
Converting to centimeters:
$r = 2.35 \, cm$.

(C) The magnetic field $B$ produced by a solenoid is given by $B = \mu_0 n I$. When the current $I$ is reversed,the direction of the magnetic field $B$ is reversed,but its magnitude remains the same.
The magnetic field energy density $u$ is given by the formula $u = \frac{B^2}{2\mu_0}$.
Since the energy density depends on the square of the magnetic field magnitude $(B^2)$,the sign of $B$ does not affect the energy density.
Therefore,the total magnetic field energy stored in the solenoid,$U = u \times V$ (where $V$ is the volume),remains unchanged.
However,the Reason states that the energy density is proportional to the magnetic field,which is incorrect because it is proportional to the square of the magnetic field $(B^2)$.
Thus,the Assertion is correct,but the Reason is incorrect.

(B) The equation $\oint \vec{B} \cdot d\vec{A} = 0$ represents Gauss's Law for magnetism.
It states that the net magnetic flux through any closed surface is always zero.
This implies that there are no isolated magnetic charges (magnetic monopoles) in nature.
Magnetic field lines always form closed loops,meaning every north pole must be accompanied by a south pole.

(B) The induced $emf$ across a rotating spoke is given by the formula $e = \frac{1}{2} B \omega \ell^2$.
Given:
$B = 0.40\,G = 0.40 \times 10^{-4}\,T$
$l = 0.50\,m$
$f = 120\,rev/min = \frac{120}{60}\,rev/s = 2\,Hz$
Angular velocity $\omega = 2 \pi f = 2 \times \pi \times 2 = 4\pi\,rad/s$.
Substituting the values:
$e = \frac{1}{2} \times (0.40 \times 10^{-4}) \times (4\pi) \times (0.50)^2$
$e = \frac{1}{2} \times 0.40 \times 10^{-4} \times 4 \times 3.14 \times 0.25$
$e = 0.20 \times 10^{-4} \times 4 \times 3.14 \times 0.25$
$e = 0.628 \times 10^{-4}\,V$.

(B) Given: $L = 0.7 \, H$,$R = 220 \, \Omega$,$V_{rms} = 220 \, V$,$f = 50 \, Hz$.
First,calculate the inductive reactance $X_L$:
$X_L = 2 \pi f L = 2 \times \frac{22}{7} \times 50 \times 0.7 = 220 \, \Omega$.
Phase angle $\phi$ is given by:
$\tan \phi = \frac{X_L}{R} = \frac{220}{220} = 1 \implies \phi = 45^o$.
The total impedance $Z$ is:
$Z = \sqrt{R^2 + X_L^2} = \sqrt{220^2 + 220^2} = 220\sqrt{2} \, \Omega$.
The $rms$ current $I_{rms}$ is:
$I_{rms} = \frac{V_{rms}}{Z} = \frac{220}{220\sqrt{2}} = \frac{1}{\sqrt{2}} \, A$.
The wattless component of current is $I_{rms} \sin \phi$:
$I_{wattless} = I_{rms} \sin 45^o = \frac{1}{\sqrt{2}} \times \frac{1}{\sqrt{2}} = 0.5 \, A$.
Thus,the phase angle is $45^o$ and the wattless component is $0.5 \, A$.

(D) In a pure inductor,the current lags behind the potential difference across the inductor by a phase angle of $\pi/2$ radians (or $90^{\circ}$).
This phase relationship is independent of the frequency of the $A.C.$ source.
Therefore,the phase angle between the current and the potential difference across the inductor $L$ remains constant at $90^{\circ}$ regardless of the change in frequency.
However,if the question refers to the phase angle $\phi$ of the entire $LR$ series circuit,it is given by $\tan \phi = \frac{X_L}{R} = \frac{\omega L}{R} = \frac{2\pi f L}{R}$.
As the frequency $f$ increases,$\tan \phi$ increases,and thus the phase angle $\phi$ between the source voltage and current increases.
Given the specific phrasing 'phase angle between current and the potential difference across $L$',the answer is that it remains constant. Since this is not an option,the question likely intends to ask for the phase angle of the circuit,which increases as frequency increases.

(A) The given equations are $V = V_0 \sin(\omega t + \phi_1)$ and $i = i_0 \sin(\omega t + \phi_2)$.
Here,$V_0 = 200 \text{ V}$,$i_0 = 50 \text{ mA} = 50 \times 10^{-3} \text{ A} = 0.05 \text{ A}$.
The phase difference $\phi = \phi_1 - \phi_2 = -\frac{\pi}{6} - \frac{\pi}{6} = -\frac{\pi}{3}$.
The average power dissipated is given by $P = V_{rms} I_{rms} \cos \phi = \frac{V_0}{\sqrt{2}} \frac{i_0}{\sqrt{2}} \cos \phi = \frac{V_0 i_0}{2} \cos \phi$.
Substituting the values: $P = \frac{200 \times 0.05}{2} \cos(-\frac{\pi}{3}) = \frac{10}{2} \times \frac{1}{2} = 5 \times 0.5 = 2.5 \text{ W}$.

(D) An inductor in a $d.c.$ circuit behaves as an open circuit at $t = 0$ (infinite resistance) and as a short circuit at steady state ($t \to \infty$,zero resistance).
Since the resistance changes from infinity to zero,the assertion that it is 'always constant' is incorrect.
In a steady state,an ideal inductor has zero resistance,so the statement that it is 'non-zero' is also incorrect.
Therefore,both the Assertion and the Reason are incorrect.

(C) The magnifying power $M$ of a telescope in normal adjustment is given by $M = -\frac{f_o}{f_e}$.
However,for maximum magnifying power,the image is formed at the least distance of distinct vision $(d = 25 \, cm)$.
The formula for maximum magnifying power is $M = -\frac{f_o}{f_e} \left(1 + \frac{f_e}{d}\right)$.
Given $f_o = 200 \, cm$,$f_e = 5 \, cm$,and $d = 25 \, cm$.
Substituting the values: $M = -\frac{200}{5} \left(1 + \frac{5}{25}\right)$.
$M = -40 \left(1 + 0.2\right) = -40 \times 1.2$.
$M = -48$.

(A) The lens formula is given by $\frac{1}{f} = \frac{1}{v} - \frac{1}{u}$.
As the object distance $u$ approaches infinity $(\infty)$,the term $\frac{1}{u}$ approaches $0$.
Substituting this into the lens formula,we get $\frac{1}{v} = \frac{1}{f}$,which implies $v = f$.
Thus,the image position approaches the focus when the object is at infinity.
This is a fundamental property of lenses,and the reason provided correctly explains that rays parallel to the principal axis (which originate from an object at infinity) converge at the focus after refraction.
Therefore,both the Assertion and the Reason are correct,and the Reason is the correct explanation of the Assertion.

(B) The energy of a photon is given by the formula $E = \frac{hc}{\lambda}$.
For the first photon:
$E_1 = 2.5\,\text{eV}$ and $\lambda_1 = 5000\,\mathring{A}$.
So,$2.5 = \frac{hc}{5000\,\mathring{A}} \implies hc = 2.5 \times 5000\,\text{eV} \cdot \mathring{A}$.
For the $X$-ray photon:
$E_2 = \frac{hc}{\lambda_2}$ where $\lambda_2 = 1\,\mathring{A}$.
Substituting the value of $hc$:
$E_2 = \frac{2.5 \times 5000}{1} = 12500\,\text{eV}$.
Thus,the energy is $2.5 \times 5000\,\text{eV}$.

(D) The Assertion is incorrect because while soft and hard $X-$ rays differ in frequency,they both travel with the same velocity,which is the speed of light $(c \approx 3 \times 10^8 \ m/s)$ in a vacuum.
The Reason is correct because hard $X-$ rays have higher energy and frequency,which gives them greater penetrating power compared to soft $X-$ rays.
Therefore,the Assertion is incorrect and the Reason is correct.

(A) photoemissive cell consists of two electrodes enclosed in a glass bulb,which may be evacuated or filled with an inert gas at low pressure.
When an inert gas is present,the photoelectrons emitted from the cathode collide with the gas atoms,causing ionization. This process increases the number of charge carriers,resulting in a greater current compared to an evacuated cell.
Therefore,both the Assertion and the Reason are correct,and the Reason correctly explains why an inert gas is used in such cells.

(D) The decay rate (activity) is given by the formula: $\frac{dN}{dt} = \lambda N$.
First, calculate the decay constant $\lambda$ using the half-life $T_{1/2} = 1.2 \times 10^7 \, s$:
$\lambda = \frac{0.693}{T_{1/2}} = \frac{0.693}{1.2 \times 10^7} \, s^{-1}$.
Now, substitute the values of $\lambda$ and $N = 4.0 \times 10^{15}$ atoms into the decay rate formula:
$\frac{dN}{dt} = \left( \frac{0.693}{1.2 \times 10^7} \right) \times (4.0 \times 10^{15})$
$\frac{dN}{dt} = \frac{0.693 \times 4.0}{1.2} \times 10^{15-7}$
$\frac{dN}{dt} = \frac{2.772}{1.2} \times 10^8$
$\frac{dN}{dt} = 2.31 \times 10^8 \, \text{atoms/s}$.
Rounding to the appropriate significant figures, the decay rate is $2.3 \times 10^8 \, \text{atoms/s}$.

(B) The conductivity $\sigma$ of an intrinsic semiconductor is given by the formula:
$\sigma = e(n_e \mu_e + n_h \mu_h)$
Given:
$e = 1.6 \times 10^{-19} \, C$
$n_e = n_h = 2.5 \times 10^{19} \, /m^3$
$\mu_e = 0.35 \, m^2/V-s$
$\mu_h = 0.18 \, m^2/V-s$
Substituting the values:
$\sigma = 1.6 \times 10^{-19} \times (2.5 \times 10^{19} \times 0.35 + 2.5 \times 10^{19} \times 0.18)$
$\sigma = 1.6 \times 10^{-19} \times 2.5 \times 10^{19} \times (0.35 + 0.18)$
$\sigma = 1.6 \times 2.5 \times 0.53$
$\sigma = 4 \times 0.53 = 2.12 \, S/m$

(A) The wavelength $\lambda$ of the emitted light is given by the relation:
$\lambda = \frac{hc}{E_g}$
where $h = 6.63 \times 10^{-34} \, J \cdot s$ is Planck's constant,$c = 3 \times 10^8 \, m/s$ is the speed of light,and $E_g$ is the energy gap.
Given $E_g = 1.9 \, eV = 1.9 \times 1.6 \times 10^{-19} \, J$.
Substituting the values:
$\lambda = \frac{6.63 \times 10^{-34} \times 3 \times 10^8}{1.9 \times 1.6 \times 10^{-19}} \, m$
$\lambda \approx \frac{19.89 \times 10^{-26}}{3.04 \times 10^{-19}} \, m$
$\lambda \approx 6.54 \times 10^{-7} \, m$
$\lambda \approx 654 \times 10^{-9} \, m = 654 \, nm$.
Rounding to the nearest provided option,the wavelength is $650 \, nm$.

(D) Given that the collector current $I_{c} = 10 \, mA$.
Since $90 \%$ of the electrons emitted from the emitter reach the collector,we have the relation $I_{c} = 0.90 \times I_{e}$.
To find the emitter current $I_{e}$,we rearrange the formula:
$I_{e} = \frac{I_{c}}{0.90} = \frac{10 \, mA}{0.9} = \frac{100}{9} \, mA$.
Calculating this value,$I_{e} \approx 11.11 \, mA$.
Therefore,the emitter current is nearly $11 \, mA$.

(D) Ground wave propagation is a method of radio wave propagation that uses the area between the surface of the Earth and the ionosphere for waveguide.
Ground waves are typically used for low-frequency and medium-frequency communication.
However,the specific range mentioned in the options refers to the wavelength range used for ground-based communication systems,which is typically between $10^{-3}\, m$ and $0.1\, m$ for specific microwave applications or line-of-sight communication.