AIIMS 2000 Physics Question Paper with Answer and Solution

(B) Given that $|\overrightarrow A \times \overrightarrow B | = |\overrightarrow A \cdot \overrightarrow B |$.
Using the definitions of cross product and dot product:
$|\overrightarrow A | |\overrightarrow B | \sin \theta = |\overrightarrow A | |\overrightarrow B | \cos \theta$.
Dividing both sides by $|\overrightarrow A | |\overrightarrow B | \cos \theta$ (assuming $\overrightarrow A, \overrightarrow B \neq 0$):
$\frac{\sin \theta}{\cos \theta} = 1$.
$\tan \theta = 1$.
Since $\tan 45^\circ = 1$,the angle $\theta = 45^\circ$.

(B) For a body to be in equilibrium,the vector sum of all forces acting on it must be zero,i.e.,$\sum \vec{F} = 0$.
If there are only two forces,they must be equal in magnitude and opposite in direction to cancel each other out,which means they must be collinear.
Since the problem states that the forces are non-collinear,two forces cannot satisfy the equilibrium condition.
Therefore,at least three non-collinear forces are required to form a closed triangle of vectors,such that their resultant is zero.
An example of this is three equal forces acting at angles of $120^{\circ}$ to each other.
Thus,the minimum number of forces is $3$.

(B) According to Newton's law of universal gravitation,the force $F$ between two masses $m_1$ and $m_2$ separated by a distance $d$ is given by $F = \frac{G m_1 m_2}{d^2}$.
Rearranging the formula to solve for the gravitational constant $G$,we get $G = \frac{F d^2}{m_1 m_2}$.
The dimensional formula for force $F$ is $[MLT^{-2}]$,for distance $d$ is $[L]$,and for mass $m$ is $[M]$.
Substituting these into the expression for $G$: $[G] = \frac{[MLT^{-2}][L^2]}{[M][M]} = \frac{[ML^3T^{-2}]}{[M^2]} = [M^{-1}L^3T^{-2}]$.

(B) The distance covered in the $n^{th}$ second is given by the formula: $S_n = u + \frac{g}{2}(2n - 1)$.
Given: Initial velocity $u = 10\; m/s$,acceleration $g = 10\; m/s^2$.
For the $3^{rd}$ second $(n=3)$:
$S_3 = 10 + \frac{10}{2}(2 \times 3 - 1) = 10 + 5(5) = 10 + 25 = 35\; m$.
For the $2^{nd}$ second $(n=2)$:
$S_2 = 10 + \frac{10}{2}(2 \times 2 - 1) = 10 + 5(3) = 10 + 15 = 25\; m$.
The ratio of the distances is $\frac{S_3}{S_2} = \frac{35}{25} = \frac{7}{5}$.

(A) When a man wants to jump off a platform,he must exert an additional downward force on the platform to gain upward momentum according to Newton's $3^{rd}$ law of motion.
This additional force causes the spring balance reading to increase momentarily.
As the man leaves the platform,the contact force becomes zero,and consequently,the reading of the spring balance drops to zero.

(B) The momentum $p$ of a body is defined as the product of its mass $m$ and velocity $v$,given by $p = mv$.
If the momentum $p$ is constant and the mass $m$ of the body is assumed to be constant,then the velocity $v = p/m$ must also be constant.
Since velocity is constant,the acceleration $a = dv/dt$ must be zero.
According to Newton's second law,the net force $F = ma$ acting on the body must also be zero.
Therefore,if momentum is constant,the velocity must be constant.

(A) The work done against friction is given by the formula $W = f_k \times S$,where $f_k$ is the kinetic frictional force and $S$ is the displacement.
Since the surface is horizontal,the normal force $N = mg$.
The frictional force is $f_k = \mu N = \mu mg$.
Given: mass $m = 50\, kg$,distance $S = 1\, m$,coefficient of friction $\mu = 0.2$,and acceleration due to gravity $g = 9.8\, m/s^2$.
Substituting the values: $W = 0.2 \times 50 \times 9.8 \times 1$.
$W = 10 \times 9.8 = 98\, J$.

(D) In a one-dimensional elastic collision between two bodies of equal mass,the velocities of the bodies are interchanged after the collision.
Given: Initial velocity of mass $m_1$ is $u_1 = +3 \, m/s$ and initial velocity of mass $m_2$ is $u_2 = -5 \, m/s$.
Since $m_1 = m_2$,after the elastic collision,the final velocity of mass $m_1$ becomes $v_1 = u_2 = -5 \, m/s$ and the final velocity of mass $m_2$ becomes $v_2 = u_1 = +3 \, m/s$.

(A) Let the initial height be $h_1 = 10\,m$ and the rebound height be $h_2$.
Since the body loses $20\%$ of its energy during the impact,the remaining energy is $80\%$ of the initial potential energy.
Thus,$mgh_2 = 0.80 \times mgh_1$.
This simplifies to $\frac{h_2}{h_1} = 0.8$.
The coefficient of restitution $e$ for a body rebounding from a floor is given by $e = \sqrt{\frac{h_2}{h_1}}$.
Substituting the value,$e = \sqrt{0.8} \approx 0.894$.
Therefore,the coefficient of restitution is approximately $0.89$.

(A) The weight of a body at the surface of the earth is given by $W = mg = 72 \ N$.
At a height $h$ above the surface of the earth,the acceleration due to gravity $g'$ is given by the formula $g' = g \left( \frac{R}{R + h} \right)^2$,where $R$ is the radius of the earth.
Given $h = \frac{R}{2}$,we substitute this into the formula:
$g' = g \left( \frac{R}{R + \frac{R}{2}} \right)^2 = g \left( \frac{R}{\frac{3R}{2}} \right)^2 = g \left( \frac{2}{3} \right)^2 = \frac{4}{9}g$.
The weight of the body at height $h$ is $W' = mg' = m \left( \frac{4}{9}g \right) = \frac{4}{9} W$.
Substituting the value of $W = 72 \ N$:
$W' = \frac{4}{9} \times 72 = 4 \times 8 = 32 \ N$.

(D) The gravitational potential energy $U$ at a distance $r$ from the center of the earth is given by $U = -\frac{GMm}{r}$.
At the earth's surface,$r = R_e$,so $U_1 = -\frac{GMm}{R_e} = -mgR_e$ (since $g = \frac{GM}{R_e^2}$).
At a height $h = R_e$ from the surface,the distance from the center is $r = R_e + h = R_e + R_e = 2R_e$.
The potential energy at this height is $U_2 = -\frac{GMm}{2R_e}$.
Substituting $GM = gR_e^2$,we get $U_2 = -\frac{(gR_e^2)m}{2R_e} = -\frac{1}{2}mgR_e$.

(C) The formula for escape velocity is given by $v_e = \sqrt{\frac{2GM}{R}}$.
For the Earth,$v_e = \sqrt{\frac{2GM_e}{R_e}} = 11.2 \, km/s$.
For the Moon,the mass $M_m = \frac{M_e}{81}$ and the radius $R_m = \frac{R_e}{4}$.
The escape velocity on the Moon is $v_m = \sqrt{\frac{2GM_m}{R_m}} = \sqrt{\frac{2G(M_e/81)}{(R_e/4)}} = \sqrt{\frac{2GM_e}{R_e} \times \frac{4}{81}}$.
Substituting the values,$v_m = \sqrt{\frac{4}{81}} \times \sqrt{\frac{2GM_e}{R_e}} = \frac{2}{9} \times 11.2 \, km/s$.
$v_m = 0.222 \times 11.2 \approx 2.488 \, km/s$,which is approximately $2.5 \, km/s$.

(A) The potential energy $U$ of a satellite of mass $m$ at a distance $r$ from the center of the Earth is given by $U = -\frac{GMm}{r}$.
Here,$r = R_e + h$,where $R_e$ is the radius of the Earth and $h$ is the height of the satellite.
Given $h = 6.4 \times 10^6 \ m$ and $R_e \approx 6.4 \times 10^6 \ m$,we have $r = R_e + R_e = 2R_e$.
Using the relation $g = \frac{GM}{R_e^2}$,we can write $GM = gR_e^2$.
Substituting these into the potential energy formula:
$U = -\frac{(gR_e^2)m}{2R_e} = -\frac{1}{2}mgR_e = -0.5 \, mgR_e$.

(C) According to Kepler's third law of planetary motion,the square of the time period $T$ is proportional to the cube of the orbital radius $R$,i.e.,$T^2 \propto R^3$.
Given for the first satellite: $T_1 = T$ and $R_1 = R$.
For the second satellite: $R_2 = 4R$ and we need to find $T_2$.
Using the ratio formula:
$\frac{T_2}{T_1} = \left( \frac{R_2}{R_1} \right)^{3/2}$
Substituting the values:
$\frac{T_2}{T} = \left( \frac{4R}{R} \right)^{3/2}$
$\frac{T_2}{T} = (4)^{3/2}$
$\frac{T_2}{T} = (2^2)^{3/2} = 2^3 = 8$
Therefore,$T_2 = 8T$.

(C) The isothermal bulk modulus $(K_i)$ of an ideal gas is defined as the pressure $(P)$ of the gas.
For a gas at atmospheric pressure,the pressure is $P = 1\,atm$.
We know that $1\,atm = 1.013 \times 10^5\,N/m^2$.
Therefore,the isothermal bulk modulus is $K_i = 1.013 \times 10^5\,N/m^2$.

(B) When a rubber band is stretched,work is done against the internal elastic forces of the material.
This work done is stored within the rubber band in the form of elastic potential energy.
Therefore,a stretched rubber band possesses increased potential energy compared to its unstretched state.

(B) The work done $(W)$ in increasing the surface area of a soap film is given by $W = T \times \Delta A_{total}$.
Since a soap film has two surfaces,the total change in area is $\Delta A_{total} = 2 \times (A_{final} - A_{initial})$.
Initial area $A_i = 10 \, cm \times 6 \, cm = 60 \, cm^2 = 60 \times 10^{-4} \, m^2$.
Final area $A_f = 10 \, cm \times 11 \, cm = 110 \, cm^2 = 110 \times 10^{-4} \, m^2$.
Change in area $\Delta A = A_f - A_i = (110 - 60) \times 10^{-4} \, m^2 = 50 \times 10^{-4} \, m^2$.
Total change in area $\Delta A_{total} = 2 \times 50 \times 10^{-4} \, m^2 = 100 \times 10^{-4} \, m^2 = 10^{-2} \, m^2$.
Given $W = 3 \times 10^{-4} \, J$.
Using $W = T \times \Delta A_{total}$,we get $T = \frac{W}{\Delta A_{total}} = \frac{3 \times 10^{-4}}{10^{-2}} = 3 \times 10^{-2} \, N/m$.

(A) The work done $W$ in increasing the area of a soap film is given by $W = T \times \Delta A \times 2$,where $T$ is the surface tension and the factor $2$ accounts for the two surfaces of the soap film.
Initial area $A_1 = 10\;cm \times 6\;cm = 60\;cm^2 = 60 \times 10^{-4}\;m^2$.
Final area $A_2 = 10\;cm \times 11\;cm = 110\;cm^2 = 110 \times 10^{-4}\;m^2$.
Change in area $\Delta A = A_2 - A_1 = (110 - 60) \times 10^{-4}\;m^2 = 50 \times 10^{-4}\;m^2$.
Given $W = 2 \times 10^{-4}\;J$.
Using the formula $W = 2T \Delta A$,we get $T = \frac{W}{2 \Delta A}$.
$T = \frac{2 \times 10^{-4}}{2 \times (50 \times 10^{-4})} = \frac{1}{50} = 0.02\;N/m = 2 \times 10^{-2}\;N/m$.

(B) The excess pressure $\Delta P$ inside a soap bubble of radius $r$ is given by the formula $\Delta P = \frac{4T}{r}$,where $T$ is the surface tension of the soap solution.
From this relation,we can see that $\Delta P \propto \frac{1}{r}$.
Let the radius of the first bubble be $r_1 = r$ and the radius of the second bubble be $r_2 = 4r$.
The ratio of their excess pressures is $\frac{\Delta P_1}{\Delta P_2} = \frac{r_2}{r_1} = \frac{4r}{r} = \frac{4}{1}$.
Wait,re-evaluating: If the first bubble has radius $r$ and the second has $4r$,then $\Delta P_1 = \frac{4T}{r}$ and $\Delta P_2 = \frac{4T}{4r} = \frac{T}{r}$.
Thus,the ratio $\Delta P_1 : \Delta P_2 = 4:1$.

(C) The excess pressure $\Delta P$ inside a spherical drop is given by the formula $\Delta P = \frac{2T}{R}$.
Given:
Surface tension $T = 70 \times 10^{-3}\, N/m$
Radius $R = 1\, mm = 1 \times 10^{-3}\, m$
Substituting the values:
$\Delta P = \frac{2 \times 70 \times 10^{-3}}{1 \times 10^{-3}}$
$\Delta P = 2 \times 70 = 140\, N/m^2$.
Therefore,the correct option is $C$.

(B) The apparent expansion of a liquid is the difference between its real expansion and the expansion of the container.
The volume expansion coefficient of the glass vessel $(\gamma_{vessel})$ is related to the linear expansion coefficient $(\alpha)$ by the formula $\gamma_{vessel} = 3\alpha$.
$\gamma_{vessel} = 3 \times 0.000009 = 0.000027 \text{ /}^{\circ}\text{C}$.
The apparent volume expansion coefficient $(\gamma_{app})$ is given by $\gamma_{app} = \gamma_{real} - \gamma_{vessel}$.
$\gamma_{app} = 0.000597 - 0.000027 = 0.00057 \text{ /}^{\circ}\text{C}$.

(A) When air in a cylinder is suddenly compressed,the process is adiabatic.
Due to the sudden compression,the work done on the gas increases its internal energy,causing the temperature of the system to rise significantly.
After the compression,the piston is held at a fixed position,meaning the volume remains constant.
Since the system is at a higher temperature than the surroundings,heat flows from the system to the surroundings until thermal equilibrium is reached.
As the temperature of the gas decreases while the volume remains constant,according to the ideal gas law $(PV = nRT)$,the pressure of the gas must decrease.

(C) The primary mode of heat transfer in fluids (like air) is convection. When air near a fire is heated,it becomes less dense and rises due to buoyancy. This rising current of hot air carries a significant amount of heat upwards. Therefore,the temperature is much higher directly above the fire compared to the sides,where the air is not being actively replaced by the rising hot currents. Thus,option $C$ is correct.

(B) According to Wien's displacement law,the product of the wavelength corresponding to maximum intensity $(\lambda_{\max})$ and the absolute temperature $(T)$ of a black body is constant.
$\lambda_{\max} T = b$ (constant)
Therefore,$T \propto \frac{1}{\lambda_{\max}}$.
Let $T_S$ and $\lambda_S$ be the temperature and wavelength of maximum intensity for the sun,and $T_N$ and $\lambda_N$ be those for the north star.
Given: $\lambda_S = 510\;nm$ and $\lambda_N = 350\;nm$.
The ratio of the surface temperatures is given by:
$\frac{T_S}{T_N} = \frac{\lambda_N}{\lambda_S} = \frac{350}{510} \approx 0.686$.
Rounding to two decimal places,we get $0.69$.

(C) According to Stefan-Boltzmann law,the total energy $E$ radiated per unit surface area of a black body per unit time is directly proportional to the fourth power of its absolute temperature $T$.
Mathematically,this is expressed as $E = \sigma T^4$,where $\sigma$ is the Stefan-Boltzmann constant.
Therefore,the amount of radiation emitted is proportional to the fourth power of the temperature on the ideal gas scale.

(D) According to Stefan-Boltzmann Law,the rate of heat radiation $H$ is proportional to the fourth power of the absolute temperature $T$ $(H \propto T^4)$.
First,convert the temperatures from Celsius to Kelvin:
$T_A = 727 + 273 = 1000 \ K$
$T_B = 327 + 273 = 600 \ K$
The ratio of the rates of heat radiated is given by:
$\frac{H_A}{H_B} = \left( \frac{T_A}{T_B} \right)^4$
$\frac{H_A}{H_B} = \left( \frac{1000}{600} \right)^4 = \left( \frac{10}{6} \right)^4 = \left( \frac{5}{3} \right)^4$
$\frac{H_A}{H_B} = \frac{5^4}{3^4} = \frac{625}{81}$
Therefore,the ratio is $625:81$.

(A) The time period of a mass-spring system is given by $T = 2\pi \sqrt{\frac{m}{k}}$.
For the initial state,$T_1 = 2 \, s$ and mass is $m_1 = m$.
So,$2 = 2\pi \sqrt{\frac{m}{k}} \implies 1 = \pi \sqrt{\frac{m}{k}}$.
For the final state,the mass is increased by $2 \, kg$,so $m_2 = m + 2$. The time period increases by $1 \, s$,so $T_2 = 2 + 1 = 3 \, s$.
So,$3 = 2\pi \sqrt{\frac{m+2}{k}}$.
Dividing the two equations:
$\frac{T_2}{T_1} = \frac{3}{2} = \sqrt{\frac{m+2}{m}}$.
Squaring both sides:
$\frac{9}{4} = \frac{m+2}{m}$.
Cross-multiplying:
$9m = 4(m + 2) \implies 9m = 4m + 8$.
$5m = 8 \implies m = \frac{8}{5} \, kg = 1.6 \, kg$.

(A) The given equation is $y = 10\sin(0.01\pi x - 2\pi t)$.
Comparing this with the standard wave equation $y = A\sin(kx - \omega t)$,we get the amplitude $A = 10 \ cm$ and the angular frequency $\omega = 2\pi \ rad/s$.
The maximum transverse speed of a particle in the rope is given by the formula $v_{\max} = A\omega$.
Substituting the values,we get $v_{\max} = 10 \times 2\pi$.
$v_{\max} = 20 \times 3.14159 = 62.83 \ cm/s$.
Rounding to the nearest integer,we get $v_{\max} \approx 63 \ cm/s$.

(D) The standard equation of a progressive wave is given by $y = a \sin(kx - \omega t)$.
Comparing the given equation $y = a \sin(0.01x - 2t)$ with the standard equation,we get:
Wave number $k = 0.01 \, cm^{-1}$
Angular frequency $\omega = 2 \, rad/s$
The velocity of propagation of the wave is given by $v = \frac{\omega}{k}$.
Substituting the values,we get $v = \frac{2}{0.01} = 200 \, cm/s$.

(A) The distance between two consecutive nodes is $\frac{\lambda}{2}$.
In a standing wave with $3$ nodes and $2$ antinodes,there are $2$ loops.
The total length $L$ between the two extreme nodes is given by $L = 2 \times \frac{\lambda}{2} = \lambda$.
Given that the distance between the two atoms (which act as the extreme nodes) is $1.21 \; \mathring{A}$,we have $L = 1.21 \; \mathring{A}$.
Therefore,the wavelength $\lambda = 1.21 \; \mathring{A}$.

(B) The Doppler effect occurs when there is a relative velocity between the source and the observer along the line joining them.
In this problem,the vehicle is moving in a direction perpendicular to the line joining the observer and the vehicle.
Therefore,the component of the velocity of the source along the line joining the observer and the source is $v_s \cos(90^{\circ}) = 0$.
Since there is no relative motion along the line of sight,the frequency perceived by the observer remains the same as the source frequency.
Thus,the observed frequency is $n' = n$.
Given that the observed frequency is $n + n_1$,we have $n + n_1 = n$,which implies $n_1 = 0$.

(B) According to Wien's displacement law,the wavelength corresponding to maximum emission intensity is inversely proportional to the absolute temperature of the black body: $\lambda_{max} \propto \frac{1}{T}$.
Let $T_S$ and $\lambda_S$ be the temperature and maximum wavelength of the Sun,and $T_X$ and $\lambda_X$ be the temperature and maximum wavelength of star $X$.
Given: $\lambda_S = 510 \, nm$ and $\lambda_X = 350 \, nm$.
The ratio of the temperatures is given by: $\frac{T_S}{T_X} = \frac{\lambda_X}{\lambda_S}$.
Substituting the values: $\frac{T_S}{T_X} = \frac{350}{510} \approx 0.686$.
Rounding to two decimal places,the ratio is $0.68$.

(B) According to the work-energy theorem,the work done is given by $W = \frac{1}{2} I \omega^2$.
Since the angular speed $\omega$ is constant,the work done is minimum when the moment of inertia $I$ is minimum.
Let the axis of rotation pass through a point at a distance $x$ from the $0.3 \ kg$ mass. Then the distance from the $0.7 \ kg$ mass is $(1.4 - x)$.
The moment of inertia $I$ is given by $I = 0.3x^2 + 0.7(1.4 - x)^2$.
To find the minimum value of $I$,we differentiate with respect to $x$ and set it to zero:
$\frac{dI}{dx} = 0.3(2x) + 0.7(2)(1.4 - x)(-1) = 0$
$0.6x - 1.4(1.4 - x) = 0$
$0.6x - 1.96 + 1.4x = 0$
$2.0x = 1.96$
$x = 0.98 \ m$.
Thus,the axis should pass at a distance of $0.98 \ m$ from the $0.3 \ kg$ mass.

(A) For the upward motion,the final velocity at the highest point is $v = 0$.
Using the equation of motion $v = u + at$,for the last second of motion,the initial velocity $u'$ at the start of that last second is $u' = v - at = 0 - (-g)(1) = g$.
The distance covered in the last second is $s = u't + \frac{1}{2}at^2 = g(1) + \frac{1}{2}(-g)(1)^2 = g - \frac{g}{2} = \frac{g}{2}$.
Taking $g \approx 10 \ m/s^2$,we get $s = \frac{10}{2} = 5 \ m$.
This distance is independent of the initial velocity of projection.
The reason is also correct because the motion is symmetric; the distance covered in the last second of upward motion is identical to the distance covered in the first second of free fall (downward motion) starting from rest $(u=0)$.
Thus,both Assertion and Reason are correct,and the Reason explains the Assertion.

(A) On Earth,a balloon filled with hydrogen rises because the buoyant force (upthrust) is greater than the weight of the balloon.
On the moon,there is no atmosphere,which means there is no air to provide a buoyant force.
Therefore,the balloon will undergo free fall under the influence of the moon's gravity alone.
The acceleration due to gravity on the moon is $\frac{g}{6}$,where $g$ is the acceleration due to gravity on Earth.
Since the balloon is in free fall,its acceleration will be equal to the acceleration due to gravity on the moon,which is $\frac{g}{6}$.
Thus,both the $Assertion$ and $Reason$ are correct,and the $Reason$ correctly explains why the balloon falls with that specific acceleration.

(B) The velocity of efflux is given by Torricelli's law: $v = \sqrt{2gh}$.
The total pressure at the bottom of the tank is the sum of atmospheric pressure $(P_{atm})$ and gauge pressure due to the water column $(h\rho g)$.
Given,$P_{total} = P_{atm} + h\rho g = 3 \, atm$.
Since $P_{atm} = 1 \, atm$,the gauge pressure is $h\rho g = 3 \, atm - 1 \, atm = 2 \, atm$.
Substituting the values: $h\rho g = 2 \times 10^5 \, N/m^2$.
$gh = \frac{2 \times 10^5}{\rho} = \frac{2 \times 10^5}{10^3} = 200 \, m^2/s^2$.
Now,substituting $gh$ into the velocity formula:
$v = \sqrt{2 \times (gh)} = \sqrt{2 \times 200} = \sqrt{400} \, m/s$.

(B) monoatomic gas atom has $3$ degrees of freedom because it can only undergo translational motion along the $X-$,$Y-$,and $Z-$ axes. Rotational and vibrational degrees of freedom are not present for a single point-like atom.
Therefore,the Assertion is correct.
The expression $\frac{C_P}{C_V} = \gamma$ is a standard thermodynamic relation for the ratio of molar specific heats,which is also correct.
However,the value of the degrees of freedom is determined by the structure of the molecule (monoatomic,diatomic,etc.),not by the ratio of specific heats $\gamma$. Thus,the Reason is not the correct explanation of the Assertion.

(B) The speed of sound in a medium is given by the formula $v = \sqrt{\frac{E}{\rho}}$,where $E$ is the modulus of elasticity and $\rho$ is the density of the medium.
Sound travels faster in solids than in gases because the elasticity $(E)$ of solids is significantly higher than that of gases.
While it is true that solids generally possess greater density than gases,density is in the denominator of the velocity formula $(v \propto \frac{1}{\sqrt{\rho}})$. Therefore,higher density actually tends to decrease the speed of sound.
The reason sound travels faster in solids is primarily due to the much higher elasticity of solids,which outweighs the effect of their higher density. Thus,the Reason is a true statement,but it is not the correct explanation for the Assertion.

(A) The electric potential $V$ at any point on a circle of radius $r$ due to a charge $q$ at its center is given by $V = \frac{kq}{r}$.
Since the radius $r$ is constant for all points on the circle, the electric potential is the same at every point on the circumference.
Therefore, the circle acts as an equipotential surface.
The work done $W$ in moving a charge $q_0$ between two points with potential difference $\Delta V$ is given by $W = q_0 \Delta V$.
Since the charge is moved around the circle once, the initial and final points are the same, so $\Delta V = 0$.
Thus, the work done $W = 1 \times 0 = 0\,units$.

(C) When capacitors are connected in series,the equivalent capacitance $C_{eq}$ is given by the formula: $\frac{1}{C_{eq}} = \frac{1}{C_1} + \frac{1}{C_2} + \frac{1}{C_3}$.
Substituting the given values: $\frac{1}{C_{eq}} = \frac{1}{3} + \frac{1}{10} + \frac{1}{15}$.
Finding a common denominator $(30)$: $\frac{1}{C_{eq}} = \frac{10 + 3 + 2}{30} = \frac{15}{30} = \frac{1}{2}$.
Therefore,$C_{eq} = 2\,\mu F$.
The total charge $Q$ supplied by the source is $Q = C_{eq} \times V = 2\,\mu F \times 100\,V = 200\,\mu C$.
In a series combination,the charge on each capacitor is the same and is equal to the total charge supplied by the source.
Thus,the charge on the $15\,\mu F$ capacitor is $200\,\mu C$.

(B) Let the given capacitor have capacitance $C = 8\,\mu F$ and voltage rating $V = 250\,V$. The required composite capacitor has capacitance $C' = 16\,\mu F$ and voltage rating $V' = 1000\,V$.
Suppose $m$ rows of capacitors are connected in parallel,and each row contains $n$ capacitors connected in series.
The potential difference across each capacitor in a series row is $V = \frac{V'}{n}$.
Substituting the values: $250 = \frac{1000}{n} \implies n = 4$.
The equivalent capacitance of the network is $C' = \frac{mC}{n}$.
Substituting the values: $16 = \frac{m \times 8}{4} \implies 16 = 2m \implies m = 8$.
The total number of capacitors required is $N = n \times m = 4 \times 8 = 32$.
Alternatively,using the formula: $N = \frac{C'}{C} \times \left( \frac{V'}{V} \right)^2 = \frac{16}{8} \times \left( \frac{1000}{250} \right)^2 = 2 \times (4)^2 = 2 \times 16 = 32$.

(A) Kirchhoff's first law,also known as the Kirchhoff's Current Law $(KCL)$,states that the algebraic sum of currents meeting at a junction in an electric circuit is zero,i.e.,$\Sigma i = 0$.
This law implies that the total charge entering a junction must equal the total charge leaving the junction in the same time interval.
Since electric charge cannot be created or destroyed at a junction,this law is a direct consequence of the law of conservation of charge.

(A) To find the current $i$ in the circuit,we apply Kirchhoff's voltage law $(KVL)$ to the loop.
Starting from point $A$ and moving clockwise through the upper branch:
$-10i + 5 - 20i - 2 = 0$
Combining the terms:
$-30i + 3 = 0$
$30i = 3$
$i = \frac{3}{30} = 0.1 \, A$
Therefore,the current in the circuit is $0.1 \, A$.

(D) Let the resistance of each heater wire be $R$.
When connected in series,the equivalent resistance is $R_s = R + R = 2R$.
When connected in parallel,the equivalent resistance is $R_p = \frac{R \cdot R}{R + R} = \frac{R}{2}$.
For a constant potential difference $V$,the heat produced $H$ is given by $H = \frac{V^2}{R_{eq}} \cdot t$.
Therefore,$H \propto \frac{1}{R_{eq}}$.
The ratio of heat produced in series $(H_s)$ to parallel $(H_p)$ is $\frac{H_s}{H_p} = \frac{R_p}{R_s} = \frac{R/2}{2R} = \frac{1}{4}$.
Thus,the ratio is $1:4$.

(B) The time period of oscillation of a magnet in a magnetic field is given by $T = 2\pi \sqrt{\frac{I}{M B_H}}$,where $I$ is the moment of inertia and $M$ is the magnetic moment.
This implies $T \propto \frac{1}{\sqrt{B_H}}$,or $B_H \propto \frac{1}{T^2}$.
At the first place,the frequency is $40 \text{ oscillations/min}$,so the time period $T_1 = \frac{60}{40} = 1.5 \,s$. The magnetic field is $(B_H)_1 = 0.1 \times 10^{-5} \,T = 10^{-6} \,T$.
At the second place,the time period $T_2 = 2.5 \,s$.
Using the ratio $\frac{(B_H)_2}{(B_H)_1} = \left( \frac{T_1}{T_2} \right)^2$,we get:
$(B_H)_2 = (B_H)_1 \times \left( \frac{1.5}{2.5} \right)^2$
$(B_H)_2 = 10^{-6} \times \left( \frac{3}{5} \right)^2 = 10^{-6} \times \frac{9}{25} = 10^{-6} \times 0.36 = 0.36 \times 10^{-6} \,T$.

(B) In a tangent galvanometer,the magnetic field $B$ produced by the coil is related to the horizontal component of the Earth's magnetic field $B_H$ and the deflection angle $\theta$ by the formula:
$B = B_H \tan \theta$
Given:
$B_H = 0.34 \times 10^{-4} \, T$
$\theta = 30^\circ$
Substituting the values:
$B = (0.34 \times 10^{-4}) \times \tan 30^\circ$
Since $\tan 30^\circ = \frac{1}{\sqrt{3}} \approx 0.577$:
$B = 0.34 \times 10^{-4} \times 0.577$
$B \approx 0.196 \times 10^{-4} \, T$
$B = 1.96 \times 10^{-5} \, T$

(C) Magnetic permeability $(\mu)$ of a material is defined as its ability to support the formation of a magnetic field within itself when placed in an external magnetic field.
For diamagnetic substances, $\mu < \mu_0$.
For paramagnetic substances, $\mu > \mu_0$.
For ferromagnetic substances, $\mu \gg \mu_0$.
Therefore, magnetic permeability is maximum for ferromagnetic substances.

(C) Diamagnetic materials are weakly repelled by magnetic fields. When placed in a non-uniform magnetic field,they experience a force that pushes them towards the region of lower magnetic field intensity. Therefore,a diamagnetic material moves from the stronger to the weaker parts of the field.

(B) The induced $e.m.f.$ $(e)$ in a dynamo armature is given by the formula $e = N B A \omega \sin(\omega t)$, where $N$ is the number of turns, $B$ is the magnetic field, $A$ is the area of the coil, and $\omega$ is the angular (rotational) velocity.
From this relation, we can see that the induced $e.m.f.$ is directly proportional to the rotational velocity, i.e., $e \propto \omega$.
If the rotational velocity $\omega$ is doubled $(2\omega)$, the induced $e.m.f.$ will also become two times the original value.

(D) Transformer works on the principle of mutual induction, which requires a changing magnetic flux to induce an electromotive force $(e.m.f.)$ in the secondary coil.
Since a Leclanche cell provides a constant direct current $(dc)$, the current flowing through the primary coil is constant.
A constant current produces a constant magnetic flux, which does not change with time.
According to Faraday's law of electromagnetic induction, an induced $e.m.f.$ is produced only when there is a change in magnetic flux $(\frac{d\phi}{dt} \neq 0)$.
Therefore, for a $dc$ input, the induced $e.m.f.$ in the secondary coil is $0 \, V$.

(B) In an $LR$ series circuit,the impedance $Z$ is given by the formula $Z = \sqrt{R^2 + X_L^2}$.
Here,$X_L$ is the inductive reactance,which is defined as $X_L = \omega L$.
Since the angular frequency $\omega = 2\pi f$,we can substitute this into the expression for $X_L$ to get $X_L = 2\pi fL$.
Substituting $X_L$ back into the impedance formula,we get $Z = \sqrt{R^2 + (2\pi fL)^2}$.
Therefore,$Z = \sqrt{R^2 + 4\pi^2 f^2 L^2}$.

(D) The particle nature of light is demonstrated by the photoelectric effect,where light behaves as a stream of photons. The wave nature of electrons is demonstrated by electron microscopy,where an electron beam behaves as a wave with a specific wavelength (de Broglie wavelength) to achieve high resolution. Thus,both natures are exhibited by these phenomena.

(C) neutrino is an elementary subatomic particle that interacts only via the weak subatomic force and gravity. It has no electric charge (it is chargeless) and possesses an intrinsic angular momentum (spin) of $1/2$. Therefore,the correct description is that it is chargeless and has spin.

(B) The decay constant $\lambda$ is given by $\lambda = \frac{0.693}{T_{1/2}} = \frac{0.693}{20} = 0.03465 \ min^{-1}$.
The time $t$ required for a substance to decay is given by $t = \frac{2.303}{\lambda} \log_{10} \left( \frac{N_0}{N} \right)$.
For $33\%$ disintegration,the remaining amount $N = N_0 - 0.33N_0 = 0.67N_0$. Thus,$t_1 = \frac{2.303}{0.03465} \log_{10} \left( \frac{100}{67} \right) \approx 66.46 \times 0.1739 \approx 11.56 \ min$.
For $67\%$ disintegration,the remaining amount $N = N_0 - 0.67N_0 = 0.33N_0$. Thus,$t_2 = \frac{2.303}{0.03465} \log_{10} \left( \frac{100}{33} \right) \approx 66.46 \times 0.4815 \approx 32.00 \ min$.
The time difference is $\Delta t = t_2 - t_1 = 32.00 - 11.56 = 20.44 \ min \approx 20 \ min$.

(B) To create an $N$-type semiconductor,we need to add a pentavalent impurity (an element with $5$ valence electrons) to an intrinsic semiconductor like germanium $(Ge)$.
When a pentavalent atom (such as phosphorus,arsenic,or antimony) is added,$4$ of its valence electrons form covalent bonds with the neighboring germanium atoms,while the $5$th electron remains free to conduct electricity.
Therefore,the valence of the impurity atom must be $5$.

(C) In semiconductors,the forbidden energy gap (the energy separation between the conduction band and the valence band) is typically of the order of $1 \, eV$. For example,Silicon has a band gap of approximately $1.1 \, eV$ and Germanium has a band gap of approximately $0.7 \, eV$.

(D) To obtain an $N-$ type semiconductor,an intrinsic semiconductor like germanium (a group $14$ element) must be doped with a pentavalent impurity (a group $15$ element).
Pentavalent impurities have $5$ valence electrons.
When added to the germanium crystal lattice,$4$ electrons form covalent bonds with neighboring germanium atoms,and the $5$th electron becomes a free charge carrier (electron).
Both Phosphorus $(P)$ and Arsenic $(As)$ are pentavalent elements.
Therefore,doping germanium with either Phosphorus or Arsenic results in an $N-$ type semiconductor.
Thus,the correct option is $(d)$.

(B) In a forward-biased $P-N$ junction,the potential barrier is reduced,which allows majority charge carriers to cross the junction easily. This process is known as diffusion,which becomes the dominant mechanism for current flow.
In a reverse-biased $P-N$ junction,the potential barrier increases,which prevents majority charge carriers from crossing the junction. However,minority charge carriers can still cross the junction due to the electric field present in the depletion region. This process is known as drift,which becomes the dominant mechanism for the small reverse saturation current.

(C) Given the ratio of intensities of two waves is $\frac{I_1}{I_2} = \frac{9}{1}$.
Let the amplitudes of the two waves be $A_1$ and $A_2$. Since intensity $I \propto A^2$,the ratio of amplitudes is $\frac{A_1}{A_2} = \sqrt{\frac{I_1}{I_2}} = \sqrt{\frac{9}{1}} = \frac{3}{1}$.
The ratio of maximum to minimum intensity is given by the formula:
$\frac{I_{\max}}{I_{\min}} = \left( \frac{A_1 + A_2}{A_1 - A_2} \right)^2$.
Substituting the values:
$\frac{I_{\max}}{I_{\min}} = \left( \frac{3 + 1}{3 - 1} \right)^2 = \left( \frac{4}{2} \right)^2 = (2)^2 = \frac{4}{1}$.
Thus,the ratio is $4:1$.

(C) cyclotron operates on the principle that the frequency of the oscillating electric field matches the cyclotron frequency of the particle,which is given by $f = \frac{qB}{2\pi m}$.
Because the mass of an electron $(m)$ is very small,it gains speed very rapidly when accelerated.
According to the theory of relativity,as the speed of the electron approaches the speed of light,its relativistic mass increases significantly $(m = \frac{m_0}{\sqrt{1 - v^2/c^2}})$.
This increase in mass causes the cyclotron frequency to change,leading to a mismatch between the frequency of the $a.c.$ source and the particle's rotation frequency in the Dees.
Consequently,the electron falls out of phase with the electric field and is no longer accelerated effectively. Thus,both the assertion and the reason are correct,and the reason is the correct explanation for the assertion.

(C) Given: Refractive index of diamond with respect to air,$\mu_d = \sqrt{6}$. Refractive index of liquid with respect to air,$\mu_l = \sqrt{3}$.
The refractive index of diamond with respect to the liquid is given by $\mu = \frac{\mu_d}{\mu_l} = \frac{\sqrt{6}}{\sqrt{3}} = \sqrt{2}$.
For total internal reflection to occur,the angle of incidence $i$ must be greater than the critical angle $C$. The critical angle $C$ is given by $\sin C = \frac{1}{\mu} = \frac{1}{\sqrt{2}}$.
Thus,$C = 45^{\circ}$.
Since the given angle of incidence $i = 30^{\circ}$ is less than the critical angle $C = 45^{\circ}$,total internal reflection will not occur. Therefore,the Assertion is incorrect.
The Reason states $\mu = \frac{1}{\sin C}$,which is the correct formula for the critical angle where $\mu$ is the refractive index of the denser medium with respect to the rarer medium. Thus,the Reason is correct.

(C) The assertion is correct: The setting sun appears red because, at sunset, the sun's light travels a longer distance through the atmosphere. During this journey, most of the shorter wavelengths (blue and violet) are scattered away by atmospheric particles, leaving mostly the longer wavelengths (red) to reach our eyes.
The reason is incorrect: According to Rayleigh's law of scattering, the intensity of scattered light $(I)$ is inversely proportional to the fourth power of its wavelength $(\lambda)$, i.e., $I \propto 1/\lambda^4$. Therefore, scattering is inversely proportional to the wavelength, not directly proportional.

(D) According to the theory of relativity,the mass of a particle increases with its speed $v$ as $m = \frac{m_0}{\sqrt{1 - \frac{v^2}{c^2}}}$.
However,the electric charge of a particle is an invariant quantity and does not change with its speed.
Therefore,the Assertion is incorrect because the charge remains constant,while the Reason is correct as it accurately describes the relativistic mass variation.

(B) The momentum of a photon is given by $p = \frac{h}{\lambda}$. Since $p = mc$ (where $m$ is the relativistic mass of the photon and $c$ is the speed of light), we have $mc = \frac{h}{\lambda}$, which implies $m = \frac{h}{c\lambda}$. Thus, the mass $m$ of a moving photon is inversely proportional to its wavelength $\lambda$. The Assertion is correct.
The Reason states $E = mc^2$. While this is Einstein's mass-energy equivalence relation, it describes the energy of a particle at rest or the energy equivalent of a mass. It does not explain the relationship between the mass of a photon and its wavelength. Therefore, the Reason is not the correct explanation for the Assertion.