AIIMS 2009 Physics Question Paper with Answer and Solution

(D) Let $h$ be the common height when the vessels are connected. By the conservation of mass,the total volume of liquid remains constant:
$A h_1 + A h_2 = A h + A h = 2Ah$
$h = \frac{h_1 + h_2}{2}$
The initial potential energy of the system is the sum of the potential energies of the liquid in each vessel. The center of gravity of the liquid in a cylindrical vessel is at half its height:
$U_i = (A h_1 \rho) g \frac{h_1}{2} + (A h_2 \rho) g \frac{h_2}{2} = \frac{1}{2} A \rho g (h_1^2 + h_2^2)$
After connecting the vessels,the final height in each vessel is $h = \frac{h_1 + h_2}{2}$. The final potential energy of the system is:
$U_f = 2 \times (A h \rho) g \frac{h}{2} = A \rho g h^2 = A \rho g \left( \frac{h_1 + h_2}{2} \right)^2 = \frac{1}{4} A \rho g (h_1 + h_2)^2$
The work done by gravity is the decrease in potential energy:
$W = U_i - U_f = \frac{1}{2} A \rho g (h_1^2 + h_2^2) - \frac{1}{4} A \rho g (h_1 + h_2)^2$
$W = \frac{1}{4} A \rho g [2(h_1^2 + h_2^2) - (h_1^2 + h_2^2 + 2h_1 h_2)]$
$W = \frac{1}{4} A \rho g (h_1^2 + h_2^2 - 2h_1 h_2) = \frac{1}{4} A \rho g (h_1 - h_2)^2$

(C) For a particle moving under gravity,the acceleration is always equal to the acceleration due to gravity $(g)$,which is approximately $9.8\,ms^{-2}$ directed downwards.
Since the acceleration is constant throughout the motion,the acceleration at $t = 1\,s$ is $a_1 = 9.8\,ms^{-2}$.
Similarly,the acceleration at $t = 2\,s$ is $a_2 = 9.8\,ms^{-2}$.
The ratio of the accelerations is $\frac{a_1}{a_2} = \frac{9.8}{9.8} = 1$.

(C) The time taken by the body to reach the highest point is given by the equation of motion $v = u - gt$.
At the highest point,the final velocity $v = 0$.
Given $u = 19.6 \, ms^{-1}$ and $g = 9.8 \, ms^{-2}$.
$0 = 19.6 - 9.8 \times t$
$t = \frac{19.6}{9.8} = 2 \, s$.
Since the time taken to go up is equal to the time taken to come down,the body will return to the starting point in $2 \, s + 2 \, s = 4 \, s$.

(C) Given the position function: $x(t) = 4t^3 - 3t^2 + 2$.
Velocity $v(t)$ is the first derivative of position with respect to time: $v(t) = \frac{dx}{dt} = \frac{d}{dt}(4t^3 - 3t^2 + 2) = 12t^2 - 6t$.
Acceleration $a(t)$ is the derivative of velocity with respect to time: $a(t) = \frac{dv}{dt} = \frac{d}{dt}(12t^2 - 6t) = 24t - 6$.
At $t = 2 \, s$:
Velocity $v(2) = 12(2)^2 - 6(2) = 12(4) - 12 = 48 - 12 = 36 \, ms^{-1}$.
Acceleration $a(2) = 24(2) - 6 = 48 - 6 = 42 \, ms^{-2}$.
Therefore,the acceleration is $42 \, ms^{-2}$ and the velocity is $36 \, ms^{-1}$.

(C) When the stone moves up,both gravity $(mg)$ and the water drag force $(F_{drag})$ act downwards. Therefore,the net retarding force is $F_{up} = mg + F_{drag}$,and the acceleration is $a_{up} = g + (F_{drag}/m)$.
When the stone moves down,gravity acts downwards while the water drag force acts upwards. Therefore,the net accelerating force is $F_{down} = mg - F_{drag}$,and the acceleration is $a_{down} = g - (F_{drag}/m)$.
Since $a_{up} > a_{down}$,the stone experiences a higher deceleration while moving up and a lower acceleration while moving down.
For the same displacement $h$,using $h = \frac{1}{2}at^2$,we have $t = \sqrt{\frac{2h}{a}}$.
Since $a_{up} > a_{down}$,it follows that $t_{up} < t_{down}$.

(C) Let the two forces be $\vec{F}_1$ (acting towards the East) and $\vec{F}_2$ (acting towards the North).
Since the magnitudes are equal,let $|\vec{F}_1| = |\vec{F}_2| = F$.
According to the parallelogram law of vector addition,the resultant force $\vec{R}$ is given by $\vec{R} = \vec{F}_1 + \vec{F}_2$.
Since the angle between the East and North directions is $90^{\circ}$,the resultant force acts along the diagonal of the square formed by these two vectors.
This diagonal points exactly in the North-East direction.
Therefore,the body will displace in the North-East direction.

(B) Given: Magnitudes of vectors $P = x$ and $Q = x$. The angle between them is $\theta = 45^\circ$. The resultant magnitude is $R = \sqrt{2 + \sqrt{2}}$.
Using the vector addition formula: $R = \sqrt{P^2 + Q^2 + 2PQ \cos \theta}$.
Substituting the values: $R = \sqrt{x^2 + x^2 + 2(x)(x) \cos 45^\circ}$.
Since $\cos 45^\circ = \frac{1}{\sqrt{2}}$,we have $R = \sqrt{2x^2 + 2x^2 \left(\frac{1}{\sqrt{2}}\right)}$.
Simplifying the expression: $R = \sqrt{2x^2 + \sqrt{2}x^2} = \sqrt{x^2(2 + \sqrt{2})} = x\sqrt{2 + \sqrt{2}}$.
Equating this to the given resultant: $x\sqrt{2 + \sqrt{2}} = \sqrt{2 + \sqrt{2}}$.
Therefore,$x = 1$.

(B) The horizontal range $R$ of a projectile is given by: $R = \frac{u^2 \sin 2\theta}{g} = \frac{2u^2 \sin \theta \cos \theta}{g}$.
The maximum height $H$ of a projectile is given by: $H = \frac{u^2 \sin^2 \theta}{2g}$.
Dividing $H$ by $R$:
$\frac{H}{R} = \frac{u^2 \sin^2 \theta}{2g} \times \frac{g}{2u^2 \sin \theta \cos \theta}$.
Simplifying the expression:
$\frac{H}{R} = \frac{\sin \theta}{4 \cos \theta} = \frac{1}{4} \tan \theta$.
Therefore,$\frac{R}{H} = \frac{4}{\tan \theta} = 4 \cot \theta$.

(A) In non-uniform circular motion,the total acceleration $a$ is the vector sum of the radial acceleration $a_r$ and the tangential acceleration $a_t$. Since these two components are perpendicular to each other,the magnitude of the resultant acceleration is given by $a = \sqrt{a_r^2 + a_t^2}$.
Given $a = 5\, ms^{-2}$.
Checking option $A$: $a_r = 3\, ms^{-2}$ and $a_t = 4\, ms^{-2}$.
$a = \sqrt{3^2 + 4^2} = \sqrt{9 + 16} = \sqrt{25} = 5\, ms^{-2}$.
This matches the given total acceleration. Thus,option $A$ is correct.

(C) An inertial frame of reference is defined as a frame in which Newton's laws of motion are valid.
Since the vehicle is moving with a constant speed on a straight road,its acceleration is $0$. Therefore,it is an inertial frame of reference.
Thus,the $Assertion$ is correct.
However,the $Reason$ states that a frame where Newton's laws are applicable is non-inertial,which is false. Newton's laws are applicable in inertial frames,not non-inertial frames.
Therefore,the $Assertion$ is correct but the $Reason$ is incorrect.

(C) Let the tennis ball bounce with an initial velocity $u$. It will rise to a height $h$ where its final velocity becomes $0$. Using the equation of motion $v^2 - u^2 = 2as$,we get:
$0^2 - u^2 = 2(-g')h$,where $g'$ is the acceleration due to gravity on the hill.
Therefore,$h = \frac{u^2}{2g'}$.
Since the acceleration due to gravity on the hill $(g')$ is less than that on the surface of the earth $(g)$ due to the increase in altitude,the height $h$ will be greater on the hill.
Thus,the $Assertion$ is correct,but the $Reason$ is incorrect because $g'$ is actually less than $g$.

(A) The total mass of the system is $M = 80\, kg + 5\, kg = 85\, kg$.
The forces acting on the system are the gravitational force $W = Mg$ acting downwards and the upward air resistance force $F_{up}$ acting upwards.
According to Newton's second law,the net force is $F_{net} = Mg - F_{up} = Ma$,where $a = 2.8\, m/s^2$ is the downward acceleration.
Rearranging the equation to solve for $F_{up}$:
$F_{up} = M(g - a)$.
Substituting the given values:
$F_{up} = 85\, kg \times (9.8\, m/s^2 - 2.8\, m/s^2)$.
$F_{up} = 85\, kg \times 7.0\, m/s^2$.
$F_{up} = 595\, N$.

(C) According to the law of conservation of linear momentum,the total momentum before the explosion must be equal to the total momentum after the explosion.
Initial momentum $P_i = mv$.
After the shell breaks,one part of mass $m_1 = m/3$ has velocity $v_1 = 0$.
The other part has mass $m_2 = m - m/3 = 2m/3$ and velocity $v_2 = v'$.
Final momentum $P_f = m_1v_1 + m_2v_2 = (m/3)(0) + (2m/3)v' = (2m/3)v'$.
Equating initial and final momentum:
$mv = (2m/3)v'$
$v' = \frac{mv \times 3}{2m} = \frac{3}{2}v$.

(B) The work done by an external agent in stretching a spring is equal to the change in the elastic potential energy of the spring.
Given,spring constant $k = 1\, N/m$ and displacement $x = 2\, m$.
The work done $W$ is given by the formula:
$W = \int_{0}^{x} kx\, dx = \frac{1}{2} k x^2$
Substituting the values:
$W = \frac{1}{2} \times 1\, N/m \times (2\, m)^2$
$W = \frac{1}{2} \times 1 \times 4 = 2\, J$.
Therefore,the work done by Saroj is $2\, J$.

(A) Since no external torque is applied,the angular momentum of the body remains conserved.
$L_1 = L_2$
$I_1 \omega_1 = I_2 \omega_2$
We know that the moment of inertia $I$ can be expressed in terms of the radius of gyration $K$ as $I = MK^2$,where $M$ is the mass of the body.
Substituting this into the conservation equation:
$M K_1^2 \omega_1 = M K_2^2 \omega_2$
$K_1^2 \omega_1 = K_2^2 \omega_2$
Rearranging the terms to find the ratio of radii of gyration $K_1/K_2$:
$\frac{K_1^2}{K_2^2} = \frac{\omega_2}{\omega_1}$
Taking the square root on both sides:
$\frac{K_1}{K_2} = \sqrt{\frac{\omega_2}{\omega_1}} = \sqrt{\omega_2} : \sqrt{\omega_1}$

(A) For a disc rolling without slipping,the translational kinetic energy is $K_t = \frac{1}{2}mv^2$.
The rotational kinetic energy is $K_r = \frac{1}{2}I\omega^2$. Since $I = \frac{1}{2}mR^2$ and $\omega = v/R$,we have $K_r = \frac{1}{2}(\frac{1}{2}mR^2)(v/R)^2 = \frac{1}{4}mv^2$.
The total kinetic energy is $K_{total} = K_t + K_r = \frac{1}{2}mv^2 + \frac{1}{4}mv^2 = \frac{3}{4}mv^2$.
The ratio of translational kinetic energy to total kinetic energy is $\frac{K_t}{K_{total}} = \frac{\frac{1}{2}mv^2}{\frac{3}{4}mv^2} = \frac{1/2}{3/4} = \frac{2}{3}$.

(C) The system consists of two particles $m_1$ and $m_2$ interacting via an internal force of attraction.
Since there is no external force acting on the system,the net external force $F_{ext} = 0$.
According to the property of the centre of mass,the acceleration of the centre of mass is given by $a_{CM} = F_{ext} / (m_1 + m_2) = 0$.
Since the particles are initially at rest,the initial velocity of the centre of mass $v_{CM} = 0$.
Because $a_{CM} = 0$ and $v_{CM} = 0$,the centre of mass remains at rest throughout the motion.

(C) The position of the centre of mass of a body depends on the shape,size,and distribution of mass of the body.
Therefore,the $Assertion$ is correct.
The centre of mass of a body does not necessarily lie at the geometric centre of the body. For example,in a non-uniform object,the centre of mass shifts towards the heavier side.
Furthermore,the centre of mass does not even need to lie within the material of the object,such as in the case of a ring or a horseshoe.
Therefore,the $Reason$ is incorrect.
Thus,the correct option is $C$.

(B) For path $iaf$:
$Q = 50 \, cal$,$W = 20 \, cal$.
According to the first law of thermodynamics,$\Delta U = Q - W$.
$\Delta U = 50 - 20 = 30 \, cal$.
Since internal energy is a state function,$\Delta U$ is the same for any path from $i$ to $f$.
$(i)$ For path $ibf$:
$Q = 36 \, cal$,$\Delta U = 30 \, cal$.
$W = Q - \Delta U = 36 - 30 = 6 \, cal$.
$(ii)$ For path $fi$ (reverse path):
$W = -13 \, cal$,$\Delta U_{fi} = -\Delta U_{if} = -30 \, cal$.
$Q = \Delta U + W = -30 + (-13) = -43 \, cal$.
$(iii)$ Given $E_{int,i} = 10 \, cal$:
$E_{int,f} = E_{int,i} + \Delta U = 10 + 30 = 40 \, cal$.

(A) If $f$ is the degree of freedom,then the ratio $\gamma = \frac{C_p}{C_v}$ is given by $\gamma = 1 + \frac{2}{f}$.
For a monoatomic gas,$f = 3$.
Therefore,$\gamma_{\text{mono}} = 1 + \frac{2}{3} = \frac{5}{3} \approx 1.67$.
For a diatomic gas,$f = 5$.
Therefore,$\gamma_{\text{dia}} = 1 + \frac{2}{5} = \frac{7}{5} = 1.4$.
Since $1.4 < 1.67$,the ratio for a diatomic gas is indeed less than that for a monoatomic gas.
The reason is correct because the degree of freedom $f$ is higher for diatomic molecules $(f=5)$ than for monoatomic atoms $(f=3)$,which leads to a smaller value of $\gamma$.

(A) The displacement of the bob in simple harmonic motion is given by $x = A \sin(\frac{2\pi}{T}t)$.
In one complete oscillation (time $T$),the bob travels from the mean position to the extreme position $A$,back to the mean position,to the other extreme position $-A$,and returns to the mean position.
The total distance covered in one time period $T$ is $A + A + A + A = 4A$.
The average speed is defined as the total distance divided by the total time.
Average speed $= \frac{\text{Total distance}}{\text{Total time}} = \frac{4A}{T}$.

(A) $SHM$ is defined as a to-and-fro motion about a mean position. The restoring force,and consequently the acceleration,is always directed towards the mean position to bring the body back. The acceleration in $SHM$ is given by $a = -\omega^2 x$,where $x$ is the displacement from the mean position. Since the negative sign indicates that the acceleration is always opposite to the direction of displacement,it is always directed towards the mean position. The reason provided correctly explains that the body must stop at the extreme position and return to the mean position due to this acceleration. Therefore,both statements are correct and the reason explains the assertion.

(C) The speed of a particle performing $SHM$ is given by $v = \omega \sqrt{A^2 - x^2}$,where $A$ is the amplitude and $x$ is the displacement from the mean position. As the particle moves away from the mean position,the magnitude of displacement $|x|$ increases,which causes the speed $v$ to decrease. Thus,the Assertion is correct.
The acceleration in $SHM$ is given by $a = -\omega^2 x$. The acceleration is directed towards the mean position. When the particle moves away from the mean position,the velocity is directed away from the mean position,so the acceleration and velocity are in opposite directions (deceleration). However,when the particle moves towards the mean position,the velocity is directed towards the mean position,so the acceleration and velocity are in the same direction (acceleration). Therefore,the Reason is incorrect because the acceleration is not always opposite to the velocity.

(B) The standard equation of a travelling wave is given by $y = A \sin (\omega t - kx)$.
Comparing the given equation $x = A \sin (2t - 0.1x)$ with the standard form,we identify the wave number $k$ as $k = 0.1$.
The wave number $k$ is related to the wavelength $\lambda$ by the formula $k = \frac{2\pi}{\lambda}$.
Substituting the value of $k$:
$0.1 = \frac{2\pi}{\lambda}$
$\lambda = \frac{2\pi}{0.1} = 20\pi$.
Therefore,the wavelength of the wave is $20\pi$.

(A) The wavelength $\lambda$ of the sound wave is given by $\lambda = \frac{v}{f} = \frac{340\, m/s}{340\, Hz} = 1\, m = 100\, cm$.
The resonant lengths $l$ for a tube closed at one end are given by $l = \frac{(2n-1)\lambda}{4}$ for $n = 1, 2, 3, \dots$.
For $n=1$,$l_1 = \frac{\lambda}{4} = \frac{100\, cm}{4} = 25\, cm$.
For $n=2$,$l_2 = \frac{3\lambda}{4} = \frac{3 \times 100\, cm}{4} = 75\, cm$.
For $n=3$,$l_3 = \frac{5\lambda}{4} = \frac{5 \times 100\, cm}{4} = 125\, cm$.
Since the tube height is $120\, cm$,only $l_1 = 25\, cm$ and $l_2 = 75\, cm$ are possible.
To find the minimum height of water,we need the maximum possible air column length for resonance,which is $l_2 = 75\, cm$.
Minimum height of water $= \text{Total height} - l_2 = 120\, cm - 75\, cm = 45\, cm$.

(A) According to the Laplace correction for the speed of sound in an ideal gas,the velocity $v$ is given by $v = \sqrt{\frac{\gamma P}{\rho}}$,where $\gamma$ is the adiabatic index,$P$ is the pressure,and $\rho$ is the density.
Given that the density is denoted by $d$,the formula becomes $v = \sqrt{\frac{\gamma P}{d}}$.
This can be derived from the ideal gas equation $PV = nRT$,where $n = \frac{m}{M}$.
Thus,$PV = \frac{m}{M}RT$,which implies $P = \frac{m}{V} \cdot \frac{RT}{M} = d \cdot \frac{RT}{M}$.
Therefore,$\frac{P}{d} = \frac{RT}{M}$.
Substituting this into the expression $v = \sqrt{\frac{\gamma RT}{M}}$,we get $v = \sqrt{\frac{\gamma P}{d}}$.

(A) $1$. For wind instruments (like a flute or organ pipe),the frequency is given by $f = \frac{v}{2L}$ (or similar depending on the pipe). As the temperature increases,the speed of sound $v$ in the air increases. Since $f \propto v$,the frequency (pitch) rises.
$2$. For string instruments (like a guitar or violin),the frequency is given by $f = \frac{1}{2L} \sqrt{\frac{T}{\mu}}$. As the temperature increases,the string expands due to thermal expansion,increasing its length $L$. Additionally,the tension $T$ in the string decreases due to the expansion of the instrument's body. Both factors lead to a decrease in frequency (pitch).
$3$. Thus,the Assertion is correct. The Reason correctly identifies that the speed of sound increases with temperature and explains the physical changes in string instruments that lead to a decrease in pitch. Therefore,the Reason is a correct explanation of the Assertion.

(D) Let the nodes be $C$ and $D$ at the bottom junctions. The circuit consists of two resistors $R$ in series with the parallel combination of the remaining three resistors.
Specifically,the two resistors $R$ at the bottom are in series,giving $R + R = 2R$.
This $2R$ is in parallel with the middle resistor $R$,giving an equivalent resistance $R_p = \frac{(2R \cdot R)}{(2R + R)} = \frac{2R^2}{3R} = \frac{2}{3}R$.
Adding the two outer resistors $R$ in series,the total equivalent resistance is $R_{eq} = R + \frac{2}{3}R + R = \frac{8}{3}R$.
Given $R = 3\,\Omega$,we have $R_{eq} = \frac{8}{3} \times 3 = 8\,\Omega$.
Since $8\,\Omega$ is not among the options,the correct choice is $D$.

(C) According to Gauss's Law,the electric flux through a closed Gaussian surface is given by $\oint \vec{E} \cdot d\vec{A} = \frac{Q_{\text{enclosed}}}{\epsilon_0}$.
For a point at a distance $r$ such that $r_1 < r < r_2$,we consider a spherical Gaussian surface of radius $r$ concentric with the shells.
The only charge enclosed by this Gaussian surface is the charge $Q_1$ on the inner shell.
Therefore,$Q_{\text{enclosed}} = Q_1$.
Applying Gauss's Law: $E(4\pi r^2) = \frac{Q_1}{\epsilon_0}$.
Thus,the electric field is $E = \frac{Q_1}{4\pi \epsilon_0 r^2}$.

(C) The electric potential $V$ at any point on the circumference of a circle of radius $R$ due to a charge $+q$ at its centre is given by $V = \frac{kq}{R}$.
Since the potential is the same at all points on the circumference,the circle represents an equipotential surface.
The work done $W$ in moving a charge $q_0$ between two points $B$ and $C$ on an equipotential surface is given by $W = q_0(V_C - V_B)$.
Since $V_B = V_C$,the work done $W = q_0(0) = 0$.

(B) Initial capacitance $C = \frac{\varepsilon_0 A}{d}$.
When the capacitor is half-filled with a dielectric of dielectric constant $K=5$ in series (as shown in the figure),the system acts as two capacitors in series,each with plate separation $d/2$.
Capacitance of the dielectric part: $C_1 = \frac{K \varepsilon_0 A}{d/2} = \frac{2 K \varepsilon_0 A}{d} = \frac{10 \varepsilon_0 A}{d}$.
Capacitance of the air part: $C_2 = \frac{\varepsilon_0 A}{d/2} = \frac{2 \varepsilon_0 A}{d}$.
The equivalent capacitance $C'$ is given by:
$\frac{1}{C'} = \frac{1}{C_1} + \frac{1}{C_2} = \frac{d}{10 \varepsilon_0 A} + \frac{d}{2 \varepsilon_0 A} = \frac{d}{\varepsilon_0 A} \left( \frac{1}{10} + \frac{1}{2} \right) = \frac{d}{\varepsilon_0 A} \left( \frac{1+5}{10} \right) = \frac{6d}{10 \varepsilon_0 A} = \frac{3d}{5 \varepsilon_0 A}$.
Therefore,$C' = \frac{5}{3} \frac{\varepsilon_0 A}{d} = \frac{5}{3} C$.
Percentage increase in capacitance = $\frac{C' - C}{C} \times 100 = \left( \frac{5/3 C - C}{C} \right) \times 100 = \left( \frac{5}{3} - 1 \right) \times 100 = \frac{2}{3} \times 100 = 66.67\% \approx 66.6\%$.

(C) The charge stored on the two plates of a capacitor are $+Q$ and $-Q$. The total charge is $Q + (-Q) = 0$. Thus,the Assertion is correct.
The electric field outside a parallel plate capacitor is zero because the fields produced by the two plates (each of magnitude $\frac{\sigma}{2\varepsilon_0}$) are equal and opposite. The field between the plates is $\frac{\sigma}{\varepsilon_0}$.
As shown in the figure,by drawing a Gaussian surface $ABCD$,we can apply Gauss's Law: $\oint \vec{E} \cdot d\vec{s} = \frac{q_{enclosed}}{\varepsilon_0}$. Since the net charge enclosed by the surface is $Q - Q = 0$,the net flux is zero,which implies the electric field outside the plates is zero. Therefore,the Reason is incorrect.

(C) For an isolated capacitor,the charge $Q$ remains constant.
The force $F$ between the plates is given by $F = \frac{Q^2}{2A\epsilon_0 K}$. Since $K > 1$ for a dielectric,the force $F$ decreases.
Thus,the Assertion is correct.
The electric field $E$ between the plates is given by $E = \frac{\sigma}{K\epsilon_0} = \frac{E_0}{K}$. Since $K > 1$,the electric field $E$ decreases.
Therefore,the Reason is incorrect.
Hence,the correct option is $C$.

(C) In a conductor,there are a large number of free electrons. When we close the circuit,an electric field is established instantly throughout the conductor at the speed of electromagnetic waves (approximately $3 \times 10^8 \ m/s$).
This electric field exerts a force on all free electrons simultaneously,causing them to drift.
Consequently,the current is established in the entire circuit instantly.
The current does not depend on the time taken for an individual electron to travel from one end of the conductor to the other.
The drift velocity of electrons is actually very small (typically in the order of $10^{-4} \ m/s$).
Therefore,the Assertion is correct,but the Reason is incorrect.

(D) The magnetic field $B$ due to a straight wire of length $L$ at a perpendicular distance $r$ is given by $B = \frac{\mu_0 I}{4 \pi r} (\sin \theta_1 + \sin \theta_2)$.
For a square loop of side $a$,the distance from the centre to each side is $r = a/2$.
The angles subtended by the corners at the centre are $\theta_1 = 45^{\circ}$ and $\theta_2 = 45^{\circ}$.
The magnetic field due to one side is $B_1 = \frac{\mu_0 I}{4 \pi (a/2)} (\sin 45^{\circ} + \sin 45^{\circ}) = \frac{\mu_0 I}{2 \pi a} (\frac{1}{\sqrt{2}} + \frac{1}{\sqrt{2}}) = \frac{\mu_0 I}{2 \pi a} (\frac{2}{\sqrt{2}}) = \frac{\mu_0 I}{\sqrt{2} \pi a}$.
Since there are $4$ identical sides,the total magnetic field is $B = 4 \times B_1 = 4 \times \frac{\mu_0 I}{\sqrt{2} \pi a} = 2\sqrt{2} \frac{\mu_0 I}{\pi a}$.
Wait,re-evaluating: $B = 4 \times \frac{\mu_0 I}{4 \pi (a/2)} (2 \sin 45^{\circ}) = \frac{\mu_0 I}{\pi a} \times 2 \times \frac{1}{\sqrt{2}} = \sqrt{2} \frac{\mu_0 I}{\pi a}$.

(B) The radius of a circular path for a charged particle in a magnetic field is given by the formula: $r = \frac{mv}{Bq}$.
Here,the mass $m$ and charge $q$ of the electron remain constant.
Therefore,the relationship between the variables is $r \propto \frac{v}{B}$.
This implies: $\frac{r_2}{r_1} = \frac{v_2}{v_1} \cdot \frac{B_1}{B_2}$.
According to the problem,the new speed $v_2 = 2v$ and the new magnetic field $B_2 = \frac{B}{2}$.
Substituting these values: $\frac{r_2}{r} = \frac{2v}{v} \cdot \frac{B}{B/2} = 2 \cdot 2 = 4$.
Thus,$r_2 = 4r$.
Therefore,the correct option is $B$.

(D) stationary charge produces only an electric field in the surrounding space.
When a charge is in motion,it constitutes an electric current,which produces both an electric field and a magnetic field in the surrounding space.
Therefore,the Assertion is incorrect because a stationary charge does not produce a magnetic field.
The Reason is also incorrect because moving charges produce both electric and magnetic fields,not just an electric field.

(C) The radius $r$ of a charged particle moving in a uniform magnetic field $B$ is given by $r = \frac{mv}{qB} = \frac{\sqrt{2mK}}{qB}$,where $K$ is the kinetic energy.
For a proton,$m_p = m$ and $q_p = e$. Thus,$r_p = \frac{\sqrt{2mK}}{eB}$.
For an alpha particle,$m_{\alpha} = 4m$ and $q_{\alpha} = 2e$. Thus,$r_{\alpha} = \frac{\sqrt{2(4m)K}}{(2e)B} = \frac{2\sqrt{2mK}}{2eB} = \frac{\sqrt{2mK}}{eB}$.
Since $r_p = r_{\alpha}$,the Assertion is correct.
The Reason states that any two charged particles with equal kinetic energies in a uniform magnetic field will have equal radii. However,the radius depends on the ratio $\frac{\sqrt{m}}{q}$. Since this ratio is not the same for all particles (e.g.,for an electron and a proton),the Reason is incorrect.

(A) The given $AC$ voltage is the $rms$ value,so $E_{rms} = 220\,V$.
The peak voltage $E_{0}$ is related to $E_{rms}$ by the formula $E_{rms} = \frac{E_{0}}{\sqrt{2}}$,which gives $E_{0} = E_{rms} \times \sqrt{2} = 220\sqrt{2}\,V$.
The average $emf$ over a positive half cycle is given by the formula $E_{avg} = \frac{2}{\pi} E_{0}$.
Substituting the value of $E_{0}$,we get $E_{avg} = \frac{2}{\pi} \times 220\sqrt{2}$.
Using $\pi \approx 3.14$ and $\sqrt{2} \approx 1.414$,we get $E_{avg} = \frac{2}{3.14} \times 220 \times 1.414 \approx 0.637 \times 311.13 \approx 198.18\,V$.
Rounding to the nearest integer,the average $emf$ is $198\,V$.

(D) The Assertion is incorrect because Ohm's law $(V = IR)$ can be applied to $a.c.$ circuits using the concept of impedance $(Z)$. For an $a.c.$ circuit, the relation is $V = IZ$, where $Z$ is the impedance.
The Reason is also incorrect because the resistance offered by a capacitor is called capacitive reactance $(X_C)$, not resistance. While it is true that $X_C = 1 / (2\pi fC)$ depends on frequency, calling it 'resistance' is technically inaccurate in the context of general circuit theory, and the statement as a whole does not justify the false assertion.

(B) $1$. The fish sees the direct image of the bulb through refraction at the water surface. The real distance of the bulb from the surface is $50 \, cm$. Due to refraction,the apparent height of the bulb from the surface is $h' = \mu_w \times 50 = (4/3) \times 50 = 200/3 \, cm$. The fish is at a depth of $30 \, cm$ from the surface. So,the apparent distance of the bulb from the fish is $d_1 = 30 + 200/3 = 290/3 \, cm$.
$2$. The fish also sees the image of the bulb formed by the reflecting bottom surface. The bulb is $50 \, cm$ above the water surface and the water is $60 \, cm$ deep. The total distance of the bulb from the bottom is $50 + 60 = 110 \, cm$. The mirror forms an image at a distance of $110 \, cm$ below the bottom surface. The total distance of this image from the water surface is $60 + 110 = 170 \, cm$. The apparent distance of this image from the surface is $h'' = \mu_w \times 170 = (4/3) \times 170 = 680/3 \, cm$. Since the fish is $30 \, cm$ below the surface,the apparent distance of this image from the fish is $d_2 = 30 + 680/3 = 770/3 \, cm$.
$3$. The distance between the two images seen by the fish is $d_2 - d_1 = 770/3 - 290/3 = 480/3 = 160 \, cm$. Wait,re-evaluating: The fish sees the bulb directly at $d_1 = 30 + (4/3) \times 50 = 290/3 \, cm$. The fish sees the reflected image at $d_2 = (60 - 30) + 60 + (4/3) \times 50 = 30 + 60 + 200/3 = 90 + 200/3 = 470/3 \, cm$. The distance between them is $470/3 - 290/3 = 180/3 = 60 \, cm$. Let's re-read: The image is formed by the bottom mirror. The object distance from the mirror is $50 + 60 = 110 \, cm$. The image is $110 \, cm$ below the bottom. Total depth from surface is $60 + 110 = 170 \, cm$. Apparent depth from surface is $(4/3) \times 170 = 680/3 \, cm$. Distance from fish = $680/3 + (60 - 30) = 680/3 + 30 = 770/3 \, cm$. The first image is at $30 + (4/3) \times 50 = 290/3 \, cm$. The difference is $770/3 - 290/3 = 480/3 = 160 \, cm$. Given the options,let's check the sum: $770/3 + 290/3 = 1060/3$. None match. Re-calculating: The fish sees the bulb at $30 + (4/3) \times 50 = 290/3$. The fish sees the image in the mirror. The mirror is at $60 \, cm$ depth. The bulb is $50 \, cm$ above. Total distance to mirror = $110 \, cm$. Image is $110 \, cm$ below mirror. Total distance from surface = $60 + 110 = 170 \, cm$. Apparent distance from surface = $(4/3) \times 170 = 680/3 \, cm$. Distance from fish = $680/3 + 30 = 770/3 \, cm$. The distance between the two images is $770/3 - 290/3 = 480/3 = 160 \, cm$. If the question asks for the sum of distances,it is $770/3 + 290/3 = 1060/3$. If the question implies the distance of the images from the fish,and the fish is between them,the distance is $290/3 + 470/3 = 760/3 \, cm$.

(A) virtual object is formed when incident rays are converging towards a point behind the mirror. In this case,the point $P$ behind the mirror acts as a virtual object.
When these rays strike the plane mirror,they are reflected and actually meet at a point $Q$ in front of the mirror.
Since the reflected rays actually meet at point $Q$,the image formed at $Q$ is a real image.
Therefore,the image of a virtual object due to a plane mirror is indeed real.
Both the Assertion and the Reason are correct,and the Reason correctly explains why the image is real.

(C) The mirror formula $\frac{1}{f} = \frac{1}{v} + \frac{1}{u}$ is derived using the paraxial approximation,which assumes that the rays are close to the principal axis and the mirror aperture is small compared to the radius of curvature. Thus,the Assertion is correct.
The laws of reflection (angle of incidence equals angle of reflection) are universal and hold true for any reflecting surface,whether plane or spherical,regardless of size. Therefore,the Reason is incorrect.

(A) The formula for fringe width in Young's double-slit experiment is given by $\beta = \frac{D \lambda}{d}$,where $D$ is the distance between the screen and the slits,$\lambda$ is the wavelength of light,and $d$ is the distance between the slits.
When the experiment is performed in a medium like glycerine with refractive index $\mu$,the wavelength of light changes to $\lambda^{\prime} = \frac{\lambda}{\mu}$.
Substituting this into the fringe width formula,the new fringe width $\beta^{\prime}$ becomes $\beta^{\prime} = \frac{D \lambda^{\prime}}{d} = \frac{D \lambda}{\mu d} = \frac{\beta}{\mu}$.
Since the refractive index of glycerine $\mu > 1$,it follows that $\beta^{\prime} < \beta$.
Therefore,the fringe width decreases or shrinks.

(B) Unpolarised light consists of electric field vectors vibrating in all possible directions perpendicular to the direction of propagation.
When unpolarised light of intensity $I_0$ passes through a polaroid,the polaroid allows only those components of the electric field to pass that are parallel to its transmission axis.
The average value of $\cos^2 \theta$ over all angles from $0$ to $2\pi$ is $\frac{1}{2}$.
Therefore,the intensity of the transmitted light $I$ is given by $I = \frac{I_0}{2}$.
The ratio of the intensity of the transmitted light to the initial intensity is $\frac{I}{I_0} = \frac{1}{2}$.

(B) The angular width of the central maximum in a single slit diffraction is given by $\theta = \frac{2\lambda}{d}$,where $\lambda$ is the wavelength and $d$ is the slit width.
Since $\theta \propto \frac{1}{d}$,if the slit width $d$ is doubled,the angular width $\theta$ becomes half,meaning the central maximum becomes narrower.
As the width of the central maximum decreases,the energy is concentrated in a smaller area,which increases the intensity of the central maximum,making it brighter.

(A) The fringe width $\beta$ in a Young's double-slit experiment is given by the formula $\beta = \frac{\lambda D}{d}$,where $\lambda$ is the wavelength of light,$D$ is the distance between the slits and the screen,and $d$ is the distance between the two coherent sources (slits).
From the formula,it is clear that $\beta \propto \frac{1}{d}$.
As the two coherent sources become infinitely close to each other,$d \to 0$.
Consequently,the fringe width $\beta \to \infty$.
When the fringe width becomes extremely large,the entire screen may be covered by a single bright or dark fringe,making it impossible to observe a distinct interference pattern.
Therefore,both the Assertion and the Reason are correct,and the Reason provides the correct explanation for the Assertion.

(A) According to Einstein's photoelectric equation,the stopping potential $V_s$ is given by $eV_s = h\nu - \phi$,where $\phi$ is the work function.
For frequency $\nu$,the stopping potential is $V_s = \frac{h\nu - \phi}{e}$.
For frequency $\frac{3\nu}{2}$,the stopping potential is $2V_s = \frac{h(\frac{3\nu}{2}) - \phi}{e}$.
Substituting the first equation into the second: $2(\frac{h\nu - \phi}{e}) = \frac{1.5h\nu - \phi}{e}$.
Multiplying by $e$: $2h\nu - 2\phi = 1.5h\nu - \phi$.
Rearranging the terms: $2h\nu - 1.5h\nu = 2\phi - \phi$.
Therefore,$\phi = 0.5h\nu = \frac{h\nu}{2}$.

(C) The momentum of each photon is $p = \frac{h}{\lambda} = \frac{6.63 \times 10^{-34}}{500 \times 10^{-9}} = 1.33 \times 10^{-27} \, kg \cdot m/s$.
The number of photons striking per unit area per unit time is $n = \frac{I}{hc/\lambda} = \frac{I \lambda}{hc}$.
Given $I = 0.5 \, W/cm^2 = 0.5 \times 10^4 \, W/m^2$,the number of photons per unit area per second is $n = \frac{0.5 \times 500 \times 10^{-9}}{6.63 \times 10^{-34} \times 3 \times 10^8} \approx 1.25 \times 10^{22} \, photons/(m^2 \cdot s)$.
For perfectly inelastic collisions,the force exerted by a photon is $F = \frac{dp}{dt} = p \cdot n \cdot A_{eff}$.
The effective area $A_{eff}$ projected perpendicular to the beam is the area of the circular base of the hemisphere,$A = \pi r^2 = \pi \times (0.01)^2 = \pi \times 10^{-4} \, m^2$.
Force $F = (1.33 \times 10^{-27}) \times (1.25 \times 10^{22}) \times (\pi \times 10^{-4}) \approx 5.22 \times 10^{-9} \, N$.

(B) According to Moseley's law,the energy of a photon emitted during an electronic transition in an atom is given by $\Delta E = Rch(Z-1)^2 \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right)$.
For the $K_{\alpha}$ line,the transition is from $n_2 = 2$ to $n_1 = 1$. Thus,$\Delta E_{\alpha} \propto \left( \frac{1}{1^2} - \frac{1}{2^2} \right) = \left( 1 - \frac{1}{4} \right) = \frac{3}{4} = 0.75$.
For the $K_{\beta}$ line,the transition is from $n_2 = 3$ to $n_1 = 1$. Thus,$\Delta E_{\beta} \propto \left( \frac{1}{1^2} - \frac{1}{3^2} \right) = \left( 1 - \frac{1}{9} \right) = \frac{8}{9} \approx 0.889$.
Since $\Delta E_{\alpha} < \Delta E_{\beta}$,the ratio $\frac{\Delta E_{\alpha}}{\Delta E_{\beta}}$ is less than $1$.

(C) According to Bohr's postulate,the angular momentum $L$ of an electron in an orbit is given by $L = \frac{nh}{2\pi}$,where $n$ is the principal quantum number.
For the ground state,$n = 1$.
For the first excited state,$n = 2$.
For the second excited state,$n = 3$.
Therefore,for the $2^{nd}$ excited state,the angular momentum is $L = \frac{3h}{2\pi}$.

(A) The binding energy per nucleon is a measure of the stability of a nucleus. $A$ higher binding energy per nucleon indicates that the nucleus is more tightly bound and therefore more stable. Stable nuclei do not decay readily and are more likely to persist over long periods,which is why they are found to be abundantly available in nature.

(A) The activity of a radioactive sample is given by $R = R_0 e^{-\lambda t}$.
Given that in $t = 3\, \text{days}$, the activity becomes $R = R_0/3$:
$\frac{1}{3} = e^{-\lambda \times 3} = e^{-3\lambda}$ .........$(1)$
We need to find the activity $R'$ after $t = 9\, \text{days}$:
$R' = R_0 e^{-\lambda \times 9} = R_0 (e^{-3\lambda})^3$
Substituting the value from equation $(1)$:
$R' = R_0 \times (1/3)^3$
$R' = R_0 \times (1/27)$
Therefore, the activity becomes $(1/27)$ of the original value.

(C) Let the equivalent resistance of the infinite network be $x$.
Since the network is infinite,adding one more section to the front does not change the total resistance. Thus,the network can be represented as two $1\,\Omega$ resistors in series with the parallel combination of a $1\,\Omega$ resistor and the equivalent resistance $x$.
The equivalent resistance $x$ is given by:
$x = 1 + \left( \frac{1 \times x}{1 + x} \right) + 1$
$x = 2 + \frac{x}{1 + x}$
$x - 2 = \frac{x}{1 + x}$
$(x - 2)(x + 1) = x$
$x^2 + x - 2x - 2 = x$
$x^2 - 2x - 2 = 0$
Using the quadratic formula $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$:
$x = \frac{2 \pm \sqrt{(-2)^2 - 4(1)(-2)}}{2(1)}$
$x = \frac{2 \pm \sqrt{4 + 8}}{2} = \frac{2 \pm \sqrt{12}}{2} = \frac{2 \pm 2\sqrt{3}}{2} = 1 \pm \sqrt{3}$
Since resistance cannot be negative,we take the positive value:
$x = (1 + \sqrt{3})\,\Omega$