AIPMT 2014 Physics Question Paper with Answer and Solution

(D) The work done in a cyclic process on a $P-V$ diagram is equal to the net area enclosed by the cycle.
In the given figure,the cycle consists of two triangles: $\triangle AOD$ and $\triangle BOC$.
For the process $A \rightarrow O \rightarrow D$,the volume increases,so the work done is positive. The area of $\triangle AOD = \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times (2V_0 - V_0) \times (2P_0 - P_0) = \frac{1}{2} \times V_0 \times P_0 = \frac{P_0V_0}{2}$.
For the process $B \rightarrow O \rightarrow C$,the volume decreases,so the work done is negative. The area of $\triangle BOC = \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times (2V_0 - V_0) \times (3P_0 - 2P_0) = \frac{1}{2} \times V_0 \times P_0 = \frac{P_0V_0}{2}$.
The total work done $W = W_{AOD} + W_{BOC} = \frac{P_0V_0}{2} + (-\frac{P_0V_0}{2}) = 0$.

(D) Let mass $m \propto F^a V^b T^c$.
Then $m = k F^a V^b T^c$,where $k$ is a dimensionless constant.
Writing dimensions on both sides:
$[M L^0 T^0] = [M L T^{-2}]^a [L T^{-1}]^b [T]^c$
$[M L^0 T^0] = [M^a L^{a+b} T^{-2a-b+c}]$
Comparing the powers of $M, L,$ and $T$ on both sides:
For $M$: $a = 1$
For $L$: $a + b = 0 \implies 1 + b = 0 \implies b = -1$
For $T$: $-2a - b + c = 0 \implies -2(1) - (-1) + c = 0 \implies -2 + 1 + c = 0 \implies c = 1$
Substituting the values of $a, b,$ and $c$ in the equation,we get:
$[m] = [F^1 V^{-1} T^1] = [F V^{-1} T]$.

(A) The equation of the trajectory of a projectile is given by: $y = x \tan \theta - \frac{g x^2}{2 u^2 \cos^2 \theta}$.
For two trajectories to be identical with the same angle of projection $\theta$,the term $\frac{g}{u^2}$ must be constant.
Therefore,$\frac{g_{earth}}{u_{earth}^2} = \frac{g_{planet}}{u_{planet}^2}$.
Given $g_{earth} = 9.8 \, m s^{-2}$,$u_{earth} = 5 \, m s^{-1}$,and $u_{planet} = 3 \, m s^{-1}$.
Substituting the values: $\frac{9.8}{5^2} = \frac{g_{planet}}{3^2}$.
$\frac{9.8}{25} = \frac{g_{planet}}{9}$.
$g_{planet} = \frac{9.8 \times 9}{25} = \frac{88.2}{25} = 3.528 \, m s^{-2}$.
Rounding to the nearest provided option,the value is $3.5 \, m s^{-2}$.

(D) At time $t_1 = 0 \, s$,the position vector of the particle is $\vec{r}_1 = 2\hat{i} + 3\hat{j}$.
At time $t_2 = 5 \, s$,the position vector of the particle is $\vec{r}_2 = 13\hat{i} + 14\hat{j}$.
The displacement $\Delta \vec{r}$ from $t = 0$ to $t = 5 \, s$ is given by $\Delta \vec{r} = \vec{r}_2 - \vec{r}_1$.
$\Delta \vec{r} = (13\hat{i} + 14\hat{j}) - (2\hat{i} + 3\hat{j}) = 11\hat{i} + 11\hat{j}$.
The average velocity $\vec{v}_{av}$ is defined as the total displacement divided by the total time interval $\Delta t = t_2 - t_1 = 5 - 0 = 5 \, s$.
$\vec{v}_{av} = \frac{\Delta \vec{r}}{\Delta t} = \frac{11\hat{i} + 11\hat{j}}{5} = \frac{11}{5}(\hat{i} + \hat{j}) \, m/s$.

(C) The driving force for the system is the weight of mass $m_1$,which is $m_1g$.
The opposing forces are the frictional forces acting on masses $m_2$ and $m_3$.
The frictional force on $m_2$ is $f_2 = \mu m_2g$.
The frictional force on $m_3$ is $f_3 = \mu m_3g$.
Using Newton's second law for the system,the net force is $F_{net} = m_1g - \mu m_2g - \mu m_3g$.
The total mass of the system is $M = m_1 + m_2 + m_3$.
The acceleration $a$ is given by $a = \frac{F_{net}}{M} = \frac{m_1g - \mu m_2g - \mu m_3g}{m_1 + m_2 + m_3}$.
Given $m_1 = m_2 = m_3 = m$,we substitute these values:
$a = \frac{mg - \mu mg - \mu mg}{m + m + m} = \frac{mg - 2\mu mg}{3m} = \frac{g(1 - 2\mu)}{3}$.

(C) The change in momentum of a particle is equal to the impulse,which is the area under the force-time $(F-t)$ graph.
Change in momentum = Area under $F-t$ graph
$= \text{Area of triangle } ABC + \text{Area of rectangle } CDEF + \text{Area of rectangle } FGHI$
$= (\frac{1}{2} \times \text{base} \times \text{height}) + (\text{width} \times \text{height}) + (\text{width} \times \text{height})$
$= (\frac{1}{2} \times 2 \times 6) + (2 \times -3) + (4 \times 3)$
$= 6 - 6 + 12$
$= 12 \, N-s$.

(A) Let $F$ be the upthrust of the air. As the balloon is descending down with an acceleration $a$:
$mg - F = ma$ ... $(i)$
Let mass $m_0$ be removed from the balloon so that it starts moving up with an acceleration $a$. Then:
$F - (m - m_0)g = (m - m_0)a$
$F - mg + m_0g = ma - m_0a$ ... $(ii)$
Adding equation $(i)$ and equation $(ii)$,we get:
$m_0g = 2ma - m_0a$
$m_0(g + a) = 2ma$
$m_0 = \frac{2ma}{g + a}$

(B) Let $\vec{v}'$ be the velocity of the third piece of mass $2m$.
Initial momentum,$\vec{P}_i = 0$ (as the body is at rest).
Final momentum,$\vec{P}_f = m v \hat{i} + m v \hat{j} + 2m \vec{v}'$.
According to the law of conservation of momentum,$\vec{P}_i = \vec{P}_f$.
$0 = m v \hat{i} + m v \hat{j} + 2m \vec{v}'$.
$\vec{v}' = -\frac{v}{2} \hat{i} - \frac{v}{2} \hat{j}$.
The magnitude of $v'$ is $v' = \sqrt{(-\frac{v}{2})^2 + (-\frac{v}{2})^2} = \frac{v}{\sqrt{2}}$.
Total kinetic energy generated due to the explosion is the sum of the kinetic energies of the three pieces:
$K.E. = \frac{1}{2} m v^2 + \frac{1}{2} m v^2 + \frac{1}{2} (2m) (v')^2$.
$K.E. = m v^2 + m (\frac{v}{\sqrt{2}})^2$.
$K.E. = m v^2 + m (\frac{v^2}{2}) = \frac{3}{2} m v^2 = 1.5 m v^2$.

(A) Let $T_s$ be the temperature of the surroundings.
According to $Newton's$ law of cooling,the rate of cooling is given by:
$\frac{T_1 - T_2}{t} = K \left( \frac{T_1 + T_2}{2} - T_s \right)$
For the first $5$ minutes:
$T_1 = 70^\circ C, T_2 = 60^\circ C, t = 5$ minutes.
$\frac{70 - 60}{5} = K \left( \frac{70 + 60}{2} - T_s \right)$
$2 = K(65 - T_s)$ --- $(i)$
For the next $5$ minutes:
$T_1 = 60^\circ C, T_2 = 54^\circ C, t = 5$ minutes.
$\frac{60 - 54}{5} = K \left( \frac{60 + 54}{2} - T_s \right)$
$1.2 = K(57 - T_s)$ --- $(ii)$
Dividing equation $(i)$ by equation $(ii)$:
$\frac{2}{1.2} = \frac{65 - T_s}{57 - T_s}$
$\frac{5}{3} = \frac{65 - T_s}{57 - T_s}$
$5(57 - T_s) = 3(65 - T_s)$
$285 - 5T_s = 195 - 3T_s$
$2T_s = 90$
$T_s = 45^\circ C$

(B) The mean free path $\lambda$ of a gas molecule is given by the formula:
$\lambda = \frac{1}{\sqrt{2} n \pi d^2}$
where $n$ is the number density of molecules and $d$ is the diameter of the molecule.
Since the radius of the molecule is $r$,the diameter is $d = 2r$.
Substituting this into the formula:
$\lambda = \frac{1}{\sqrt{2} n \pi (2r)^2} = \frac{1}{4\sqrt{2} n \pi r^2}$
From this expression,it is clear that $\lambda \propto \frac{1}{r^2}$.
Therefore,the mean free path is inversely proportional to $r^2$.

(D) Given: Mass of the cylinder, $M = 50\, kg$. Radius of the cylinder, $R = 0.5\, m$. Angular acceleration, $\alpha = 2\, \text{rev } s^{-2}$.
Converting angular acceleration to $SI$ units: $\alpha = 2 \times 2\pi\, \text{rad } s^{-2} = 4\pi\, \text{rad } s^{-2}$.
The torque $\tau$ produced by the tension $T$ in the string is $\tau = T \times R$.
The moment of inertia $I$ of a solid cylinder about its central axis is $I = \frac{1}{2}MR^2$.
Using the relation $\tau = I\alpha$, we have $TR = (\frac{1}{2}MR^2)\alpha$.
Solving for tension $T$: $T = \frac{1}{2}MR\alpha$.
Substituting the values: $T = \frac{1}{2} \times 50 \times 0.5 \times 4\pi = 50\pi$.
Using $\pi \approx 3.14$, $T = 50 \times 3.14 = 157\, N$.

(A) Acceleration of the solid sphere slipping down the incline without rolling is
${a_{slipping}} = g \sin \theta \,\,\,\,\,\,\,(i)$
Acceleration of the solid sphere rolling down the incline without slipping is
${a_{rolling}} = \frac{g \sin \theta}{1 + \frac{k^2}{R^2}} = \frac{g \sin \theta}{1 + \frac{2}{5}}$
(For a solid sphere,$\frac{k^2}{R^2} = \frac{2}{5}$)
$= \frac{5}{7} g \sin \theta \,\,\,\,\,(ii)$
Dividing equation $(ii)$ by equation $(i)$,we get
$\frac{a_{rolling}}{a_{slipping}} = \frac{5}{7}$

(C) For an object to become a black hole,its escape velocity must be equal to the speed of light $(c)$.
The formula for escape velocity is $v_{esc} = \sqrt{\frac{2GM}{R}}$.
Setting $v_{esc} = c$,we get $c = \sqrt{\frac{2GM}{R}}$,which implies $R = \frac{2GM}{c^2}$.
Given $G = 6.67 \times 10^{-11} \, N \cdot m^2/kg^2$,$M = 5.98 \times 10^{24} \, kg$,and $c = 3 \times 10^8 \, m/s$:
$R = \frac{2 \times 6.67 \times 10^{-11} \times 5.98 \times 10^{24}}{(3 \times 10^8)^2}$
$R = \frac{79.77 \times 10^{13}}{9 \times 10^{16}}$
$R \approx 8.86 \times 10^{-3} \, m \approx 10^{-2} \, m$.

(A) For a point inside the Earth,i.e.,$r < R$:
$E = -\frac{GM}{R^3}r$
Where $M$ and $R$ are the mass and radius of the Earth,respectively. The negative sign indicates that the field is directed towards the center.
At the center,$r = 0$,therefore $E = 0$.
For a point outside the Earth,i.e.,$r > R$:
$E = -\frac{GM}{r^2}$
On the surface of the Earth,i.e.,$r = R$:
$E = -\frac{GM}{R^2}$
The magnitude of the gravitational field increases linearly with distance $r$ inside the Earth and decreases inversely with the square of the distance $r$ outside the Earth. The graph representing this variation is shown in the provided solution image.

(B) Given that the volume $V$ of the copper wire is constant,we have $V = A \cdot l$,where $A$ is the cross-sectional area and $l$ is the length of the wire.
From this,we can express the area as $A = \frac{V}{l}$.
According to Hooke's Law,Young's modulus $Y$ is defined as $Y = \frac{\text{Stress}}{\text{Strain}} = \frac{F/A}{\Delta l/l} = \frac{F \cdot l}{A \cdot \Delta l}$.
Rearranging this formula to solve for the extension $\Delta l$,we get $\Delta l = \frac{F \cdot l}{Y \cdot A}$.
Substituting $A = \frac{V}{l}$ into the equation,we get $\Delta l = \frac{F \cdot l}{Y \cdot (V/l)} = \frac{F \cdot l^2}{Y \cdot V}$.
Since $F$,$Y$,and $V$ are constants,we have $\Delta l \propto l^2$.
Therefore,the graph of $\Delta l$ versus $l^2$ will be a straight line passing through the origin.

(C) Let $n$ droplets each of radius $r$ coalesce to form a big drop of radius $R$.
Volume of $n$ droplets = Volume of big drop
$n \times \frac{4}{3}\pi r^3 = \frac{4}{3}\pi R^3 \Rightarrow n = \frac{R^3}{r^3}$ $(i)$
Volume of big drop,$V = \frac{4}{3}\pi R^3$ $(ii)$
Initial surface area of $n$ droplets,$A_i = n \times 4\pi r^2 = \frac{R^3}{r^3} \times 4\pi r^2 = \frac{4\pi R^3}{r} = \left( \frac{4}{3}\pi R^3 \right) \frac{3}{r} = \frac{3V}{r}$ $(Using (i) \text{ and } (ii))$
Final surface area of big drop,$A_f = 4\pi R^2 = \left( \frac{4}{3}\pi R^3 \right) \frac{3}{R} = \frac{3V}{R}$ $(Using (ii))$
Decrease in surface area,$\Delta A = A_i - A_f = \frac{3V}{r} - \frac{3V}{R} = 3V \left( \frac{1}{r} - \frac{1}{R} \right)$
Energy released = Surface tension $\times$ Decrease in surface area
Energy released $= T \times \Delta A = 3VT \left( \frac{1}{r} - \frac{1}{R} \right)$.

(C) Step $1$: Isothermal expansion.
For an isothermal process, $PV = \text{constant}$.
Initial state: $(P, V)$.
Final state after isothermal expansion: $(P', 2V)$.
$PV = P'(2V) \implies P' = \frac{P}{2}$.
Step $2$: Adiabatic expansion.
For an adiabatic process, $PV^\gamma = \text{constant}$.
Initial state for this process: $(P', 2V) = (P/2, 2V)$.
Final state: $(P_f, 16V)$.
$P'(2V)^\gamma = P_f(16V)^\gamma$.
Substituting $P' = P/2$ and $\gamma = 5/3$:
$\frac{P}{2} (2V)^{5/3} = P_f (16V)^{5/3}$.
$P_f = \frac{P}{2} \left( \frac{2V}{16V} \right)^{5/3} = \frac{P}{2} \left( \frac{1}{8} \right)^{5/3}$.
Since $8 = 2^3$, we have $\left( \frac{1}{2^3} \right)^{5/3} = \frac{1}{2^5} = \frac{1}{32}$.
$P_f = \frac{P}{2} \times \frac{1}{32} = \frac{P}{64}$.

(C) Given the displacement equation: $x = A \cos \omega t$.
The velocity $v$ is the first derivative of displacement with respect to time:
$v = \frac{dx}{dt} = \frac{d}{dt}(A \cos \omega t) = -A \omega \sin \omega t$.
The acceleration $a$ is the derivative of velocity with respect to time:
$a = \frac{dv}{dt} = \frac{d}{dt}(-A \omega \sin \omega t) = -A \omega^2 \cos \omega t$.
Comparing this with the original displacement equation $x = A \cos \omega t$,we see that $a = -\omega^2 x$. At $t = 0$,$x = A$,so $a = -A \omega^2$. This means the acceleration starts at a negative maximum value and follows a negative cosine curve. Graph $C$ represents this variation correctly.

(C) The fundamental frequency of a stretched string is given by $n = \frac{1}{2l} \sqrt{\frac{T}{\mu}}$.
Since the tension $T$ and mass per unit length $\mu$ are constant for the string,we have $n \propto \frac{1}{l}$,which implies $l = \frac{k}{n}$,where $k$ is a constant.
When the string is divided into three segments of lengths $l_{1}, l_{2}, l_{3}$ with fundamental frequencies $n_{1}, n_{2}, n_{3}$ respectively,we have $l_{1} = \frac{k}{n_{1}}$,$l_{2} = \frac{k}{n_{2}}$,and $l_{3} = \frac{k}{n_{3}}$.
The total length of the string is $l = l_{1} + l_{2} + l_{3}$.
Substituting the expressions for length in terms of frequency,we get $\frac{k}{n} = \frac{k}{n_{1}} + \frac{k}{n_{2}} + \frac{k}{n_{3}}$.
Dividing by $k$,we obtain the relation $\frac{1}{n} = \frac{1}{n_{1}} + \frac{1}{n_{2}} + \frac{1}{n_{3}}$.

(C) The speed of the motorcyclist (observer) is $v_o = 36\, km\, h^{-1} = 36 \times \frac{5}{18} = 10\, m s^{-1}$.
Since the motorcyclist is moving towards the source (car),the observer's velocity is taken as positive.
The speed of the car (source) is $v_s = 18\, km\, h^{-1} = 18 \times \frac{5}{18} = 5\, m s^{-1}$.
Since the source is moving away from the observer,the source's velocity is taken as positive.
The speed of sound is $v = 343\, m s^{-1}$.
The frequency of the source is $f_0 = 1392\, Hz$.
Using the Doppler effect formula for frequency heard by the observer,$f' = f_0 \left( \frac{v + v_o}{v + v_s} \right)$,
$f' = 1392 \left( \frac{343 + 10}{343 + 5} \right)$,
$f' = 1392 \left( \frac{353}{348} \right)$,
$f' = 4 \times 353 = 1412\, Hz$.

(C) The fundamental frequency $(f_1)$ of a pipe closed at one end is given by $f_1 = \frac{v}{4L}$.
Given: $v = 340 \, m s^{-1}$ and $L = 85 \, cm = 0.85 \, m$.
Substituting the values: $f_1 = \frac{340}{4 \times 0.85} = \frac{340}{3.4} = 100 \, Hz$.
The natural frequencies of a closed pipe are odd harmonics of the fundamental frequency: $f_n = (2n - 1)f_1$,where $n = 1, 2, 3, \dots$.
The frequencies are: $100 \, Hz, 300 \, Hz, 500 \, Hz, 700 \, Hz, 900 \, Hz, 1100 \, Hz, 1300 \, Hz, \dots$.
We need to find the number of frequencies below $1250 \, Hz$.
The frequencies are $100, 300, 500, 700, 900, 1100$.
Counting these,we get $6$ possible natural oscillations.

(D) Let $m$ be the mass of steam condensed into water.
Heat lost by steam $=$ Heat gained by water.
Heat lost by steam consists of two parts: latent heat of condensation and cooling of condensed water from $100\,^{\circ}C$ to $80\,^{\circ}C$.
Heat lost $= m \times L_v + m \times s_w \times \Delta T_1 = m \times 540 + m \times 1 \times (100 - 80) = 540m + 20m = 560m$.
Heat gained by $20\,g$ of water to reach $80\,^{\circ}C$ from $10\,^{\circ}C$ is:
Heat gained $= m_w \times s_w \times \Delta T_2 = 20 \times 1 \times (80 - 10) = 20 \times 70 = 1400\,cal$.
Equating heat lost and gained: $560m = 1400$.
$m = \frac{1400}{560} = 2.5\,g$.
The total mass of water present is the initial mass plus the condensed steam mass:
Total mass $= 20\,g + 2.5\,g = 22.5\,g$.

(A) For the light ray to retrace its path,it must strike the silvered surface normally (at an angle of $90^{\circ}$).
In the triangle formed inside the prism,the angle at the silvered surface is $90^{\circ}$,and the angle at the apex is $A$. Thus,the angle of refraction $r$ at the first surface must be $r = 90^{\circ} - A$.
According to Snell's Law at the first surface: $\mu = \frac{\sin i}{\sin r}$.
Given $i = 2A$ and $r = 90^{\circ} - A$,we have:
$\mu = \frac{\sin(2A)}{\sin(90^{\circ} - A)}$
$\mu = \frac{2 \sin A \cos A}{\cos A}$
$\mu = 2 \sin A$.

(D) The electric field $\vec{E}$ is given by the negative gradient of the potential $V$: $\vec{E} = -\nabla V = -\left( \frac{\partial V}{\partial x} \hat{i} + \frac{\partial V}{\partial y} \hat{j} + \frac{\partial V}{\partial z} \hat{k} \right)$.
Given $V(x, y, z) = 6x - 8xy - 8y + 6yz$.
Calculating partial derivatives:
$E_x = -\frac{\partial V}{\partial x} = -(6 - 8y) = -6 + 8y$.
$E_y = -\frac{\partial V}{\partial y} = -(-8x - 8 + 6z) = 8x + 8 - 6z$.
$E_z = -\frac{\partial V}{\partial z} = -(6y) = -6y$.
At point $(1, 1, 1)$:
$E_x = -6 + 8(1) = 2 \ V/m$.
$E_y = 8(1) + 8 - 6(1) = 10 \ V/m$.
$E_z = -6(1) = -6 \ V/m$.
Thus,$\vec{E} = 2\hat{i} + 10\hat{j} - 6\hat{k} \ V/m$.
The magnitude of the electric field is $|\vec{E}| = \sqrt{2^2 + 10^2 + (-6)^2} = \sqrt{4 + 100 + 36} = \sqrt{140} = 2\sqrt{35} \ V/m$.
The electric force $\vec{F} = q\vec{E}$. Given $q = 2 \ C$,the magnitude of the force is $F = q|\vec{E}| = 2 \times 2\sqrt{35} = 4\sqrt{35} \ N$.

(B) For a conducting sphere,the charge $Q$ resides entirely on its outer surface.
The electric field $E$ inside a conducting sphere is zero because the charges are distributed on the surface such that they cancel each other's field at every point inside.
The electric potential $V$ inside a conducting sphere is constant and equal to the potential at its surface.
Therefore,the potential at the centre is $V = \frac{1}{4\pi \varepsilon_0} \frac{Q}{R}$.
The electric field at the centre is $E = 0$.

(D) The electric field $E$ between the plates of a parallel plate capacitor is given by $E = \frac{\sigma}{\varepsilon_0 K}$,where $\sigma$ is the surface charge density and $K$ is the dielectric constant of the medium.
In the regions where the dielectric is absent (air/vacuum),$K = 1$,so the electric field is $E_0 = \frac{\sigma}{\varepsilon_0}$.
In the regions where the dielectric slabs are present,the electric field is $E_1 = \frac{\sigma}{\varepsilon_0 K_1}$ and $E_2 = \frac{\sigma}{\varepsilon_0 K_2}$.
Since $K_1 < K_2$,it follows that $\frac{1}{K_1} > \frac{1}{K_2}$,which implies $E_1 > E_2$.
Also,both $E_1$ and $E_2$ are less than the electric field $E_0$ in the air gap ($E_1 < E_0$ and $E_2 < E_0$).
Therefore,the electric field is highest in the air gaps,and among the dielectric regions,the field is higher in the slab with the smaller dielectric constant $(K_1)$.
Comparing this with the given options,the graph that shows the highest field in the air gaps and a higher field in the $K_1$ region compared to the $K_2$ region is represented by option $(D)$.

(B) The distance between the two cities is $d = 150\, km$.
The total resistance of the copper wire is $R = (0.5\, \Omega/km) \times (150\, km) = 75\, \Omega$.
The total voltage drop across the wire is $V = (8\, V/km) \times (150\, km) = 1200\, V$.
The power loss in the wire is given by the formula $P = \frac{V^2}{R}$.
Substituting the values,$P = \frac{(1200\, V)^2}{75\, \Omega} = \frac{1440000}{75}\, W = 19200\, W$.
Converting to kilowatts,$P = 19.2\, kW$.

(A) The internal resistance $r$ of a cell using a potentiometer is given by the formula:
$r = \left( \frac{l_1}{l_2} - 1 \right) R$
where $l_1$ is the balancing length when the cell is in an open circuit (i.e.,$R = \infty$) and $l_2$ is the balancing length when an external resistance $R$ is connected across the cell.
Given:
$l_1 = 3\,m$
$l_2 = 2.85\,m$
$R = 9.5\,\Omega$
Substituting the values into the formula:
$r = \left( \frac{3}{2.85} - 1 \right) \times 9.5\,\Omega$
$r = \left( \frac{3 - 2.85}{2.85} \right) \times 9.5\,\Omega$
$r = \left( \frac{0.15}{2.85} \right) \times 9.5\,\Omega$
$r = \frac{15}{285} \times 9.5\,\Omega$
$r = \frac{1}{19} \times 9.5\,\Omega = 0.5\,\Omega$
Thus,the internal resistance of the cell is $0.5\,\Omega$.

(B) In the first case,at the balance point:
$\frac{5}{R} = \frac{l_1}{100 - l_1}$ $....(i)$
In the second case,the resistance $R$ is shunted with an equal resistance $R$,so the equivalent resistance becomes $R' = \frac{R \times R}{R + R} = \frac{R}{2}$.
The new balance point is at $1.6\,l_1$. Thus,at the balance point:
$\frac{5}{R/2} = \frac{1.6\,l_1}{100 - 1.6\,l_1}$ $....(ii)$
Dividing equation $(i)$ by equation $(ii)$:
$\frac{5/R}{5/(R/2)} = \frac{l_1 / (100 - l_1)}{1.6\,l_1 / (100 - 1.6\,l_1)}$
$\frac{1}{2} = \frac{l_1}{100 - l_1} \times \frac{100 - 1.6\,l_1}{1.6\,l_1}$
$\frac{1}{2} = \frac{100 - 1.6\,l_1}{1.6(100 - l_1)}$
$0.8(100 - l_1) = 100 - 1.6\,l_1$
$80 - 0.8\,l_1 = 100 - 1.6\,l_1$
$0.8\,l_1 = 20$
$l_1 = 25\,cm$
Substituting $l_1 = 25\,cm$ into equation $(i)$:
$\frac{5}{R} = \frac{25}{100 - 25} = \frac{25}{75} = \frac{1}{3}$
$R = 15\,\Omega$.

(C) Let the total current be $I$ and the resistance of the galvanometer be $G$.
The current through the galvanometer is $I_G = 0.2\% \text{ of } I = \frac{0.2}{100} I = \frac{1}{500} I$.
The current through the shunt resistance $S$ is $I_S = I - I_G = I - \frac{1}{500} I = \frac{499}{500} I$.
Since the galvanometer and shunt are in parallel, the potential difference across them is the same:
$I_G G = I_S S$
$\left( \frac{1}{500} I \right) G = \left( \frac{499}{500} I \right) S$
$S = \frac{G}{499}$.
The resistance of the ammeter $R_A$ is the equivalent resistance of the parallel combination of $G$ and $S$:
$\frac{1}{R_A} = \frac{1}{G} + \frac{1}{S} = \frac{1}{G} + \frac{499}{G} = \frac{500}{G}$.
Therefore, $R_A = \frac{G}{500}$.

(C) The direction of the magnetic dipole moment is from the south pole to the north pole of the magnet.
In configuration $(1)$,the angle between the two magnetic moments is $90^{\circ}$. The net magnetic dipole moment is $m_{\text{net}} = \sqrt{m^2 + m^2 + 2m^2 \cos 90^{\circ}} = \sqrt{2m^2} = m\sqrt{2} \approx 1.414m$.
In configuration $(2)$,the magnetic moments are in opposite directions,so the angle is $180^{\circ}$. The net magnetic dipole moment is $m_{\text{net}} = m - m = 0$.
In configuration $(3)$,the angle between the two magnetic moments is $30^{\circ}$. The net magnetic dipole moment is $m_{\text{net}} = \sqrt{m^2 + m^2 + 2m^2 \cos 30^{\circ}} = \sqrt{2m^2 + 2m^2(\frac{\sqrt{3}}{2})} = m\sqrt{2 + \sqrt{3}} \approx m\sqrt{3.732} \approx 1.932m$.
In configuration $(4)$,the angle between the two magnetic moments is $60^{\circ}$. The net magnetic dipole moment is $m_{\text{net}} = \sqrt{m^2 + m^2 + 2m^2 \cos 60^{\circ}} = \sqrt{2m^2 + 2m^2(\frac{1}{2})} = m\sqrt{3} \approx 1.732m$.
Comparing the values,configuration $(3)$ has the highest net magnetic dipole moment.

(D) The magnifying power of a compound microscope is given by $m = \left(\frac{L}{f_o}\right) \left(\frac{D}{f_e}\right)$,where $f_o$ is the focal length of the objective lens,$f_e$ is the focal length of the eyepiece,$L$ is the tube length,and $D$ is the least distance of distinct vision.
From this formula,it is clear that $m \propto \frac{1}{f_o}$. Therefore,if $f_o$ increases,the magnifying power of the microscope will decrease.
The magnifying power of an astronomical telescope is given by $m = \frac{f_o}{f_e}$,where $f_o$ is the focal length of the objective lens and $f_e$ is the focal length of the eyepiece.
From this formula,it is clear that $m \propto f_o$. Therefore,if $f_o$ increases,the magnifying power of the telescope will increase.

(B) According to Einstein's photoelectric equation,the kinetic energy $K$ of emitted photoelectrons is given by $K = E - \phi_0$,where $E$ is the energy of incident radiation and $\phi_0$ is the work function of the metal.
Initially,the energy of incident radiation is $E$. Given $K_1 = 0.5\, eV$,we have:
$0.5 = E - \phi_0$ ..... $(i)$
When the energy of incident radiation is increased by $20\%$,the new energy becomes $E' = E + 0.2E = 1.2E$. The new kinetic energy is $K_2 = 0.8\, eV$. Thus:
$0.8 = 1.2E - \phi_0$ ..... $(ii)$
From equation $(i)$,we get $E = 0.5 + \phi_0$. Substituting this into equation $(ii)$:
$0.8 = 1.2(0.5 + \phi_0) - \phi_0$
$0.8 = 0.6 + 1.2\phi_0 - \phi_0$
$0.8 - 0.6 = 0.2\phi_0$
$0.2 = 0.2\phi_0$
$\phi_0 = 1.0\, eV$.

(D) The de Broglie wavelength is given by the formula:
$\lambda = \frac{h}{\sqrt{2mK}}$ ...... $(i)$
where $m$ is the mass and $K$ is the kinetic energy of the particle.
When the kinetic energy of the particle is increased to $16$ times its initial value,the new kinetic energy $K' = 16K$.
The new de Broglie wavelength $\lambda'$ becomes:
$\lambda' = \frac{h}{\sqrt{2m(16K)}} = \frac{h}{4\sqrt{2mK}} = \frac{\lambda}{4}$ (Using equation $(i)$).
The percentage change in the de Broglie wavelength is calculated as:
$\text{Percentage change} = \frac{\lambda - \lambda'}{\lambda} \times 100$
$= \left(1 - \frac{\lambda'}{\lambda}\right) \times 100$
$= \left(1 - \frac{1}{4}\right) \times 100 = \frac{3}{4} \times 100 = 75\%$.

(B) The motional $emf$ induced in a curved conductor moving in a magnetic field is equal to the $emf$ induced in a straight conductor connecting the two ends of the curve.
The effective length of the semicircular ring $PQR$ is the straight distance between its ends $P$ and $R,$ which is the diameter $l = 2r.$
The induced $emf$ is given by $\varepsilon = Bvl = Bv(2r) = 2rBv.$
According to Fleming's Right-Hand Rule,for a downward velocity $v$ and a magnetic field $B$ directed into the plane,the force on positive charges is directed towards $R.$ Thus,$R$ is at a higher potential than $P.$

(B) The energy flux (intensity) is given by $I = 25 \times 10^4 \, W/m^2$.
The surface area is $A = 15 \, cm^2 = 15 \times 10^{-4} \, m^2$.
The speed of light is $c = 3 \times 10^8 \, m/s$.
For a perfectly reflecting surface,the momentum transferred per unit time (force) is given by the formula $F = \frac{2IA}{c}$.
Substituting the values:
$F = \frac{2 \times (25 \times 10^4) \times (15 \times 10^{-4})}{3 \times 10^8}$
$F = \frac{2 \times 25 \times 15 \times 10^0}{3 \times 10^8}$
$F = \frac{750}{3} \times 10^{-8} \, N$
$F = 250 \times 10^{-8} \, N = 2.50 \times 10^{-6} \, N$.

(B) The intensity at any point on the screen is given by $I = 4I_0 \cos^2(\phi/2)$,where $I_0$ is the intensity of each individual wave and $\phi$ is the phase difference.
The phase difference $\phi$ is related to the path difference $\Delta x$ by the formula $\phi = (2\pi / \lambda) \times \Delta x$.
Case $1$: When path difference $\Delta x = \lambda$,the phase difference is $\phi = (2\pi / \lambda) \times \lambda = 2\pi$.
Substituting this into the intensity formula: $I = 4I_0 \cos^2(2\pi / 2) = 4I_0 \cos^2(\pi) = 4I_0(1)^2 = 4I_0$. Given this intensity is $K$,we have $K = 4I_0$.
Case $2$: When path difference $\Delta x = \lambda / 4$,the phase difference is $\phi = (2\pi / \lambda) \times (\lambda / 4) = \pi / 2$.
Substituting this into the intensity formula: $I' = 4I_0 \cos^2((\pi / 2) / 2) = 4I_0 \cos^2(\pi / 4) = 4I_0 (1 / \sqrt{2})^2 = 4I_0 (1 / 2) = 2I_0$.
Since $K = 4I_0$,then $2I_0 = K / 2$. Therefore,the intensity is $K / 2$.

(D) Given: $\lambda = 600 \, nm = 600 \times 10^{-9} \, m$,slit width $a = 1 \, mm = 10^{-3} \, m$,and distance $D = 2 \, m$.
The distance between the first dark fringes on either side of the central bright fringe is equal to the width of the central maximum.
The formula for the width of the central maximum is given by $w = \frac{2 \lambda D}{a}$.
Substituting the values:
$w = \frac{2 \times (600 \times 10^{-9} \, m) \times 2 \, m}{10^{-3} \, m}$
$w = \frac{2400 \times 10^{-9}}{10^{-3}} \, m$
$w = 2400 \times 10^{-6} \, m = 2.4 \times 10^{-3} \, m = 2.4 \, mm$.

(C) The energy of the incident photon is given by $E = \frac{hc}{\lambda}$.
Using $hc \approx 12400 \; \text{eV} \cdot \mathring{A}$,we get $E = \frac{12400}{975} \approx 12.75 \; \text{eV}$.
The energy levels of a hydrogen atom are given by $E_n = -\frac{13.6}{n^2} \; \text{eV}$.
For the ground state $(n=1)$,$E_1 = -13.6 \; \text{eV}$.
After absorbing the photon,the new energy level is $E_n = E_1 + E = -13.6 + 12.75 = -0.85 \; \text{eV}$.
Since $E_n = -\frac{13.6}{n^2} = -0.85 \; \text{eV}$,we have $n^2 = \frac{13.6}{0.85} = 16$,so $n = 4$.
The electron is excited to the $n = 4$ state.
The number of spectral lines emitted when the electron transitions from $n$ to the ground state is given by $\frac{n(n-1)}{2}$.
For $n = 4$,the number of spectral lines is $\frac{4(4-1)}{2} = \frac{4 \times 3}{2} = 6$.

(C) Given the ratio of the amount of radioisotope $X$ to stable isotope $Y$ is $\frac{X}{Y} = \frac{1}{7}$.
The total amount of the initial sample is $X + Y$. The fraction of the remaining radioisotope $X$ is given by $\frac{X}{X+Y} = \frac{1}{1+7} = \frac{1}{8}$.
We know that the remaining fraction after $n$ half-lives is $\left(\frac{1}{2}\right)^{n}$.
Equating the two,$\left(\frac{1}{2}\right)^{n} = \frac{1}{8} = \left(\frac{1}{2}\right)^{3}$.
Thus,the number of half-lives elapsed is $n = 3$.
The age of the rock $t$ is given by $t = n \times T$,where $T$ is the half-life.
$t = 3 \times 1.4 \times 10^{9} \text{ years} = 4.2 \times 10^{9} \text{ years}$.

(D) The binding energy of a nucleus is given by $BE = (\text{number of nucleons}) \times (\text{binding energy per nucleon})$.
For ${}_3^7Li$: $BE_1 = 7 \times 5.60\,MeV = 39.2\,MeV$.
For ${}_1^1H$: The binding energy is $0\,MeV$ as it is a single proton.
For $2{}_2^4He$: $BE_2 = 2 \times (4 \times 7.06\,MeV) = 2 \times 28.24\,MeV = 56.48\,MeV$.
The energy released $Q$ is the difference between the total binding energy of the products and the reactants:
$Q = BE_{\text{products}} - BE_{\text{reactants}}$
$Q = 56.48\,MeV - (39.2\,MeV + 0\,MeV)$
$Q = 17.28\,MeV \approx 17.3\,MeV$.

(D) The barrier potential $(V_b)$ of a $p-n$ junction is determined by the following factors:
$1$. Type of semiconductor material: The band gap energy differs for different materials. For $Si$,$V_b \approx 0.7 \ V$,and for $Ge$,$V_b \approx 0.3 \ V$.
$2$. Amount of doping: The concentration of impurity atoms affects the width of the depletion region and the potential difference across it.
$3$. Temperature: The barrier potential decreases with an increase in temperature because the intrinsic carrier concentration increases,which affects the Fermi level and the built-in potential.
Therefore,all three factors $(1, 2, 3)$ influence the barrier potential.

(A) The $V-I$ characteristic curve shown in the graph is characteristic of a solar cell.
In this graph,the $V$-axis represents the voltage and the $I$-axis represents the current.
Point $A$ lies on the $V$-axis where the current $I = 0$,which corresponds to the open circuit voltage $(V_{OC})$.
Point $B$ lies on the $I$-axis where the voltage $V = 0$,which corresponds to the short circuit current $(I_{SC})$.
Therefore,the graph represents the $V-I$ characteristic of a solar cell where $A$ is the open circuit voltage and $B$ is the short circuit current.

(C) The magnetic field produced by a long straight current-carrying wire at a distance $d$ is given by $B = \frac{\mu_0 I}{2 \pi d}$.
Since the wires $AOB$ and $COD$ are placed at right angles to each other,the magnetic fields $B_1$ and $B_2$ produced by them at a point $P$ (at distance $d$ from $O$ along the normal to the plane) will also be perpendicular to each other.
Thus,the magnitude of the resultant magnetic field $B$ is given by $B = \sqrt{B_1^2 + B_2^2}$.
Substituting the values,$B_1 = \frac{\mu_0 I_1}{2 \pi d}$ and $B_2 = \frac{\mu_0 I_2}{2 \pi d}$.
Therefore,$B = \sqrt{\left( \frac{\mu_0 I_1}{2 \pi d} \right)^2 + \left( \frac{\mu_0 I_2}{2 \pi d} \right)^2}$.
$B = \frac{\mu_0}{2 \pi d} \sqrt{I_1^2 + I_2^2} = \frac{\mu_0}{2 \pi d} (I_1^2 + I_2^2)^{1/2}$.

(D) Given: Efficiency $\eta = 90 \% = 0.9$,Primary voltage $V_P = 200 \ V$,Primary power $P_P = 3 \ kW = 3000 \ W$,Secondary current $I_S = 6 \ A$.
First,calculate the primary current $I_P$ using $P_P = V_P \times I_P$:
$I_P = \frac{P_P}{V_P} = \frac{3000 \ W}{200 \ V} = 15 \ A$.
Next,calculate the output power (secondary power) $P_S$ using efficiency: $P_S = \eta \times P_P = 0.9 \times 3000 \ W = 2700 \ W$.
Finally,calculate the secondary voltage $V_S$ using $P_S = V_S \times I_S$:
$V_S = \frac{P_S}{I_S} = \frac{2700 \ W}{6 \ A} = 450 \ V$.
Thus,the secondary voltage is $450 \ V$ and the primary current is $15 \ A$.