AIPMT 1994 Physics Question Paper with Answer and Solution

(A) The angle $\theta$ between two vectors $\vec A$ and $\vec B$ is given by the formula: $\cos \theta = \frac{\vec A \cdot \vec B}{|A||B|}$.
First,calculate the dot product $\vec A \cdot \vec B = (3)(3) + (4)(4) + (5)(-5) = 9 + 16 - 25 = 0$.
Next,calculate the magnitudes $|A| = \sqrt{3^2 + 4^2 + 5^2} = \sqrt{9 + 16 + 25} = \sqrt{50}$ and $|B| = \sqrt{3^2 + 4^2 + (-5)^2} = \sqrt{9 + 16 + 25} = \sqrt{50}$.
Substituting these values into the formula: $\cos \theta = \frac{0}{\sqrt{50} \cdot \sqrt{50}} = \frac{0}{50} = 0$.
Since $\cos \theta = 0$,the angle $\theta = 90^\circ$.

(B) The work done $W$ by a constant force $\vec{F}$ during a displacement $\vec{r}$ is given by the dot product: $W = \vec{F} \cdot \vec{r}$.
Given the force $\vec{F} = (-2\hat{i} + 15\hat{j} + 6\hat{k})\,N$.
The displacement is $10\,m$ along the $Y$-axis,so $\vec{r} = 10\hat{j}\,m$.
Substituting these into the formula:
$W = (-2\hat{i} + 15\hat{j} + 6\hat{k}) \cdot (10\hat{j})$
$W = (-2 \times 0) + (15 \times 10) + (6 \times 0)$
$W = 0 + 150 + 0 = 150\,J$.
Therefore,the work done is $150\,J$.

(B) Let $\overrightarrow{v_b}$ be the velocity of the boat relative to the water and $\overrightarrow{v_r}$ be the velocity of the river water relative to the ground.
The resultant velocity of the boat relative to the ground is given by $\overrightarrow{v_{bg}} = \overrightarrow{v_b} + \overrightarrow{v_r}$.
Since the boat crosses the river perpendicular to the flow,the vectors $\overrightarrow{v_b}$ and $\overrightarrow{v_r}$ are perpendicular to each other.
Therefore,the magnitude of the resultant velocity is $v_{bg} = \sqrt{v_b^2 + v_r^2}$.
Given $v_{bg} = 10 \, km/h$ and $v_b = 8 \, km/h$,we have:
$10 = \sqrt{8^2 + v_r^2}$
$100 = 64 + v_r^2$
$v_r^2 = 100 - 64 = 36$
$v_r = 6 \, km/h$.

(C) Pressure is defined as the force applied per unit area.
$P = \frac{F}{A}$
The dimensional formula for force $(F)$ is $[MLT^{-2}]$.
The dimensional formula for area $(A)$ is $[L^2]$.
Therefore,the dimensions of pressure are:
$[P] = \frac{[MLT^{-2}]}{[L^2]} = [ML^{-1}T^{-2}]$.

(D) The displacement is given by $s = t^3 - 6t^2 + 3t + 4$.
Velocity $v$ is the first derivative of displacement with respect to time: $v = \frac{ds}{dt} = 3t^2 - 12t + 3$.
Acceleration $a$ is the derivative of velocity with respect to time: $a = \frac{dv}{dt} = 6t - 12$.
To find the time when acceleration is zero,set $a = 0$:
$6t - 12 = 0 \implies t = 2 \ s$.
Now,substitute $t = 2$ into the velocity equation:
$v = 3(2)^2 - 12(2) + 3 = 3(4) - 24 + 3 = 12 - 24 + 3 = -9 \ m s^{-1}$.

(D) Let the car accelerate at a constant rate $\alpha$ for time $t_1$. The maximum velocity $v$ attained is given by $v = u + \alpha t_1$. Since it starts from rest,$u = 0$,so $v = \alpha t_1$.
Next,the car decelerates at a constant rate $\beta$ for the remaining time $(t - t_1)$ until it comes to rest. Using the equation $v = u + at$ for the deceleration phase,we have $0 = v - \beta(t - t_1)$.
Substituting $v = \alpha t_1$ into the equation,we get $0 = \alpha t_1 - \beta t + \beta t_1$.
Rearranging the terms,we get $(\alpha + \beta) t_1 = \beta t$,which gives $t_1 = \frac{\beta}{\alpha + \beta} t$.
Substituting $t_1$ back into the expression for $v$,we get $v = \alpha \left( \frac{\beta}{\alpha + \beta} t \right) = \frac{\alpha \beta}{\alpha + \beta} t$.

(D) The instantaneous velocity of a particle is given by the slope of the displacement-time graph at that point.
Mathematically,$v = \frac{ds}{dt}$.
- At point $C$,the slope is positive as the displacement is increasing.
- At point $D$,the slope is zero because it is at the peak of the curve.
- At point $E$,the slope is negative because the displacement is decreasing with time.
- At point $F$,the slope is positive as the displacement is increasing again.
Therefore,the instantaneous velocity is negative at point $E$.

(A) In uniform circular motion,the body moves with a constant speed along a circular path.
Since the centripetal force $F$ acts towards the center of the circle and the displacement $ds$ is always along the tangent to the circle,the angle $\theta$ between the force and the displacement is $90^{\circ}$.
The work done $W$ is given by $W = \int F \cdot ds = \int F \cos(90^{\circ}) ds = 0$.
Therefore,no work is done on the body by the centripetal force.

(D) The time period of a simple pendulum is given by $T = 2\pi \sqrt{\frac{l}{g_{eff}}}$,where $g_{eff}$ is the effective acceleration due to gravity.
For the period $T$ to decrease,the effective acceleration $g_{eff}$ must increase.
When a rocket moves up with a uniform acceleration $a$,the effective acceleration experienced by the pendulum is $g_{eff} = g + a$.
Since $g + a > g$,the effective gravity increases,which leads to a decrease in the time period $T$.
Therefore,the correct option is $D$.

(C) The force acting on the satellite is given by Newton's second law for a variable mass system: $F = \frac{d}{dt}(Mv)$.
Since the satellite is in force-free space,the net external force $F = 0$.
Expanding the derivative,we get: $F = M \frac{dv}{dt} + v \frac{dM}{dt} = 0$.
Given the rate of mass accumulation is $\frac{dM}{dt} = \alpha v$,we substitute this into the equation:
$M \frac{dv}{dt} + v(\alpha v) = 0$.
Rearranging to solve for acceleration $a = \frac{dv}{dt}$:
$M \frac{dv}{dt} = -\alpha v^2$.
Therefore,the deceleration (or acceleration) is $a = -\frac{\alpha v^2}{M}$.

(D) The work done by a variable force is given by the integral of the force with respect to displacement: $W = \int_{x_1}^{x_2} F(x) \, dx$.
Given $F(x) = 7 - 2x + 3x^2$,$x_1 = 0$,and $x_2 = 5$.
$W = \int_{0}^{5} (7 - 2x + 3x^2) \, dx$.
Integrating term by term:
$W = [7x - x^2 + x^3]_{0}^{5}$.
Substituting the limits:
$W = (7(5) - (5)^2 + (5)^3) - (7(0) - (0)^2 + (0)^3)$.
$W = (35 - 25 + 125) - 0$.
$W = 135 \, J$.

(B) According to the work-energy theorem,the work done by the net force on an object is equal to the change in its kinetic energy.
Since the object starts from rest,the initial kinetic energy is $0$.
Work done $W = F \times d$.
Therefore,the final kinetic energy $K.E. = F \times d$.
Since the force $F$ and the distance $d$ are constant,the kinetic energy acquired depends only on the product of the force and the distance.
It does not depend on the mass $m$ of the object.
Thus,the kinetic energy is independent of $m$.

(B) The gravitational force provides the necessary centripetal force for the particle to move in a circular orbit.
For a circular orbit,the centripetal force is given by $F_c = \frac{mv^2}{R}$.
According to the problem,the gravitational force is $F_g \propto \frac{1}{R}$,which can be written as $F_g = \frac{K}{R}$ for some constant $K$.
Equating the two forces: $\frac{mv^2}{R} = \frac{K}{R}$.
Canceling $R$ from both sides,we get $mv^2 = K$.
Since $m$ and $K$ are constants,$v^2$ is constant,which means $v$ is constant.
Therefore,$v \propto R^0$.

(A) According to Kepler's third law of planetary motion,the square of the time period $T$ is proportional to the cube of the semi-major axis $R$ of the orbit: $T^2 \propto R^3$.
Given:
$R_1 = 10^{13} \ m$ (Neptune)
$R_2 = 10^{12} \ m$ (Saturn)
The ratio of the time periods is given by:
$\frac{T_1}{T_2} = \left( \frac{R_1}{R_2} \right)^{3/2}$
Substituting the values:
$\frac{T_1}{T_2} = \left( \frac{10^{13}}{10^{12}} \right)^{3/2}$
$\frac{T_1}{T_2} = (10)^{3/2}$
$\frac{T_1}{T_2} = 10^1 \cdot 10^{1/2} = 10\sqrt{10}$

(A) Simple harmonic motion $(SHM)$ is a type of periodic motion where the restoring force is directly proportional to the displacement and acts in the direction opposite to that of displacement.
In a string fixed at both ends,the standing waves formed are a superposition of two traveling waves,and the particles of the string execute simple harmonic motion about their mean positions.
Therefore,the motion of particles in a string fixed at both ends is an example of simple harmonic motion.

(A) For a particle executing Simple Harmonic Motion $(S.H.M.)$,the restoring force is directly proportional to the negative of the displacement from the mean position.
Mathematically,this is expressed as $F = -kx$,where $k$ is the force constant.
Given that $A$ and $K$ are positive constants,the force equation for this specific motion is $F = -AKx$.
Therefore,the correct option is $A$.

(D) The general equation of a simple harmonic progressive wave is given by $y = A \sin (\omega t - kx + \phi)$ or $y = A \sin (kx - \omega t + \phi)$.
Option $D$,$Y = A \sin (at - bx + c)$,represents a progressive wave because it is a function of $(at - bx)$,which satisfies the wave equation $\frac{\partial^2 y}{\partial t^2} = v^2 \frac{\partial^2 y}{\partial x^2}$,where the wave speed $v = \frac{a}{b}$.
Options $A$,$B$,and $C$ do not represent a traveling wave as they lack the necessary dependence on both space $(x)$ and time $(t)$ in the required functional form.

(A) The given equation for the standing wave is $Y = A \sin(100t) \cos(0.01x)$.
Comparing this with the standard equation of a standing wave $Y = A \sin(\omega t) \cos(kx)$,we identify the angular frequency $\omega = 100 \ rad/s$ and the wave number $k = 0.01 \ rad/m$.
The velocity of the wave $v$ is given by the ratio of angular frequency to wave number:
$v = \frac{\omega}{k} = \frac{100}{0.01} = 10000 \ m/s$.

(B) At a fixed end,a node is always formed.
Given that a node is formed at a distance of $10 \, cm$ from the fixed end,this distance represents the distance between two consecutive nodes.
The distance between two consecutive nodes is given by $\frac{\lambda}{2} = 10 \, cm$.
Therefore,the wavelength $\lambda = 20 \, cm = 0.2 \, m$.
The speed of the wave $v$ is given by the formula $v = f \lambda$,where $f = 100 \, Hz$.
$v = 100 \times 0.2 = 20 \, m/s$.

(B) The beat frequency is the absolute difference between the two frequencies.
Given,for frequency $\nu$ and $200 \;Hz$,the beat frequency is $5 \;Hz$.
So,$\nu = 200 \pm 5$,which means $\nu = 195 \;Hz$ or $\nu = 205 \;Hz$ ... $(i)$
For the second harmonic $2\nu$ and $420 \;Hz$,the beat frequency is $10 \;Hz$.
So,$2\nu = 420 \pm 10$,which means $2\nu = 410 \;Hz$ or $2\nu = 430 \;Hz$.
Dividing by $2$,we get $\nu = 205 \;Hz$ or $\nu = 215 \;Hz$ ... $(ii)$
Comparing equations $(i)$ and $(ii)$,the common value is $\nu = 205 \;Hz$.

(D) In an elastic collision between two bodies of equal mass,the velocities of the bodies are interchanged after the collision.
Given: Initial velocity of ball $A$ $(u_A)$ = $0.5 \, m s^{-1}$ and initial velocity of ball $B$ $(u_B)$ = $-0.3 \, m s^{-1}$.
Since the masses are identical and the collision is elastic,the final velocity of ball $A$ $(v_A)$ will be equal to $u_B$,and the final velocity of ball $B$ $(v_B)$ will be equal to $u_A$.
Therefore,$v_A = -0.3 \, m s^{-1}$ and $v_B = 0.5 \, m s^{-1}$.
The question asks for the velocities of $B$ and $A$ respectively,which are $v_B$ and $v_A$.
Thus,the velocities are $0.5 \, m s^{-1}$ and $-0.3 \, m s^{-1}$.

(D) For an adiabatic process,the relationship between pressure $P$ and temperature $T$ is given by $T^\gamma P^{1-\gamma} = \text{constant}$.
Rearranging this,we get $P^{1-\gamma} \propto T^{-\gamma}$,which implies $P \propto T^{-\frac{\gamma}{1-\gamma}}$ or $P \propto T^{\frac{\gamma}{\gamma-1}}$.
Comparing this with the given relation $P \propto T^C$,we find $C = \frac{\gamma}{\gamma-1}$.
For a monoatomic gas,the adiabatic index $\gamma = 5/3$.
Substituting the value of $\gamma$:
$C = \frac{5/3}{5/3 - 1} = \frac{5/3}{2/3} = 5/2$.

(B) Given mass $(m) = 5\; kg$ and time period $(T) = 2\pi\; s$.
The formula for the time period of a spring-mass system is $T = 2\pi \sqrt{\frac{m}{k}}$.
Substituting the values: $2\pi = 2\pi \sqrt{\frac{5}{k}}$.
Squaring both sides: $1 = \frac{5}{k}$,which gives the spring constant $k = 5\; N/m$.
When the body is hanging,the spring is stretched by an extension $x$ due to the weight $mg$. According to Hooke's Law,$mg = kx$.
Therefore,the decrease in length $x = \frac{mg}{k}$.
Substituting $m = 5\; kg$ and $k = 5\; N/m$: $x = \frac{5g}{5} = g\; m$.

(C) The thrust force $F$ exerted on a rocket is given by the formula $F = v \frac{dm}{dt}$,where $v$ is the velocity of the ejected fuel relative to the rocket and $\frac{dm}{dt}$ is the rate of fuel consumption.
Given:
Rate of fuel consumption $\frac{dm}{dt} = 1\; kg/s$
Velocity of ejected fuel $v = 60\; km/s = 60 \times 10^3\; m/s$
Substituting these values into the formula:
$F = (60 \times 10^3\; m/s) \times (1\; kg/s)$
$F = 60000\; N$
Therefore,the force exerted on the rocket is $60000\; N$.

(A) Given: Initial velocity $u = 20 \; km/h$,final velocity $v = 60 \; km/h$,and time $t = 4 \; h$.
First,we calculate the acceleration $a$ using the equation $v = u + at$:
$60 = 20 + (a \times 4)$
$40 = 4a$
$a = 10 \; km/h^2$.
Now,we calculate the distance $d$ using the equation $d = ut + \frac{1}{2}at^2$:
$d = (20 \times 4) + \frac{1}{2} \times 10 \times (4)^2$
$d = 80 + 5 \times 16$
$d = 80 + 80 = 160 \; km$.

(A) For a particle executing $S.H.M.$ with time period $T$,the kinetic energy $(K)$ and potential energy $(U)$ are given by:
$K = \frac{1}{2} k A^2 \cos^2(\omega t)$
$U = \frac{1}{2} k A^2 \sin^2(\omega t)$
The difference between kinetic and potential energy is:
$K - U = \frac{1}{2} k A^2 (\cos^2(\omega t) - \sin^2(\omega t)) = \frac{1}{2} k A^2 \cos(2\omega t)$
Since the angular frequency of the difference is $2\omega$,the new time period $T'$ is given by:
$T' = \frac{2\pi}{2\omega} = \frac{T}{2}$
Given $T = 4\; sec$,we have $T' = \frac{4}{2} = 2\; sec$.

(A) Given: Radius of Earth $R_e = 6400 \; km$,Radius of Mars $R_m = 3200 \; km$.
Mass of Earth $M_e = 10 M_m$,Weight on Earth $W_e = 200 \; N$.
The weight of an object is given by $W = mg$,where $g = \frac{GM}{R^2}$.
Therefore,the ratio of weights is $\frac{W_m}{W_e} = \frac{m g_m}{m g_e} = \frac{M_m}{M_e} \times \left(\frac{R_e}{R_m}\right)^2$.
Substituting the values: $\frac{W_m}{200} = \frac{1}{10} \times \left(\frac{6400}{3200}\right)^2$.
$\frac{W_m}{200} = \frac{1}{10} \times (2)^2 = \frac{4}{10} = 0.4$.
$W_m = 200 \times 0.4 = 80 \; N$.

(D) The total kinetic energy $(K_{total})$ of a rolling sphere is the sum of its linear kinetic energy $(K_{linear})$ and rotational kinetic energy $(K_{rotational})$.
$K_{total} = K_{linear} + K_{rotational} = \frac{1}{2}mv^2 + \frac{1}{2}I\omega^2$
For a solid sphere,the moment of inertia is $I = \frac{2}{5}mr^2$ and the angular velocity is $\omega = \frac{v}{r}$.
Substituting these values into the rotational kinetic energy expression:
$K_{rotational} = \frac{1}{2} \left(\frac{2}{5}mr^2\right) \left(\frac{v}{r}\right)^2 = \frac{1}{5}mv^2$.
Now,calculating the total kinetic energy:
$K_{total} = \frac{1}{2}mv^2 + \frac{1}{5}mv^2 = \left(\frac{5+2}{10}\right)mv^2 = \frac{7}{10}mv^2$.
The ratio of rotational kinetic energy to total kinetic energy is:
$\frac{K_{rotational}}{K_{total}} = \frac{\frac{1}{5}mv^2}{\frac{7}{10}mv^2} = \frac{1}{5} \times \frac{10}{7} = \frac{2}{7}$.
Thus,the ratio is $2:7$.

(D) In a uniform electric field $E$,the potential energy $U$ of an electric dipole with dipole moment $p$ is given by the formula $U = -p \cdot E = -pE \cos \theta$,where $\theta$ is the angle between the dipole moment vector $p$ and the electric field vector $E$.
Initially,the dipole is in stable equilibrium,which means the angle between $p$ and $E$ is $0^\circ$.
When the dipole is rotated through an angle $\theta$ from this initial position,the new angle between the dipole moment and the electric field becomes $\theta$.
Therefore,the potential energy of the electric dipole in the final position is $U = -pE \cos \theta$.

(A) Given:
Length $l = 50\, cm = 0.5\, m = 50 \times 10^{-2}\, m$
Area of cross-section $A = 1\, mm^2 = 1 \times 10^{-6}\, m^2$
Current $I = 4\, A$
Voltage $V = 2\, V$
Using Ohm's Law,the resistance $R$ is:
$R = \frac{V}{I} = \frac{2}{4} = 0.5\, \Omega$
The formula for resistance in terms of resistivity $\rho$ is:
$R = \rho \frac{l}{A}$
Rearranging for $\rho$:
$\rho = \frac{R \cdot A}{l} = \frac{0.5 \times 10^{-6}}{0.5} = 1 \times 10^{-6}\, \Omega\cdot m$
Thus,the resistivity of the wire is $1 \times 10^{-6}\, \Omega\cdot m$.

(C) The power $P$ and voltage $V$ of an electric bulb are related to its resistance $R$ by the formula: $P = \frac{V^2}{R}$.
Rearranging for $R$,we get $R = \frac{V^2}{P}$.
Given $V = 220\,V$ and $P = 60\,W$.
Substituting the values: $R = \frac{220 \times 220}{60} = \frac{48400}{60} \approx 806.67\,\Omega$.
Rounding to the nearest whole number,we get $R = 807\,\Omega$.

(A) Given:
Area $A = 0.01\,m^2$,Current $I = 10\,A$,Magnetic field $B = 0.1\,T$.
The loop is held perpendicular to the magnetic field,which means the angle between the area vector $\vec{A}$ and the magnetic field vector $\vec{B}$ is $\theta = 0^{\circ}$.
Formula:
The torque $\vec{\tau}$ acting on a current-carrying loop in a magnetic field is given by $\vec{\tau} = \vec{M} \times \vec{B}$,where $\vec{M}$ is the magnetic moment.
The magnitude of the torque is $\tau = M B \sin \theta$,where $M = I A$.
Calculation:
$M = I \times A = 10\,A \times 0.01\,m^2 = 0.1\,A\cdot m^2$.
Since the loop is perpendicular to the field,the angle $\theta$ between the normal to the loop (area vector) and the magnetic field is $0^{\circ}$.
$\tau = M B \sin(0^{\circ}) = 0.1 \times 0.1 \times 0 = 0\,N\cdot m$.

(A) The time period of a bar magnet oscillating in a magnetic field is given by $T = 2\pi \sqrt{\frac{I}{MB}}$,where $I$ is the moment of inertia,$M$ is the magnetic moment,and $B$ is the magnetic field.
For a bar magnet of mass $m$ and length $l$,the moment of inertia is $I = \frac{ml^2}{12}$.
Substituting this into the formula,we get $T = 2\pi \sqrt{\frac{ml^2}{12MB}}$.
This shows that $T \propto \sqrt{m}$.
If the mass $m$ is quadrupled $(m' = 4m)$,the new time period $T'$ becomes $T' = 2\pi \sqrt{\frac{4ml^2}{12MB}} = 2 \times (2\pi \sqrt{\frac{ml^2}{12MB}}) = 2T$.
The motion remains $S.H.M.$ because the restoring torque $\tau = -MB \sin \theta \approx -MB \theta$ for small oscillations.

(D) The induced electromotive force $(emf)$ in an inductor is given by the formula $E = -L \frac{dI}{dt}$.
In the first half of the time interval,the current $I$ increases linearly with time,so the rate of change of current $\frac{dI}{dt}$ is a positive constant. Consequently,the induced $emf$ $E = -L \frac{dI}{dt}$ is a negative constant.
In the second half of the time interval,the current $I$ decreases linearly with time,so the rate of change of current $\frac{dI}{dt}$ is a negative constant. Consequently,the induced $emf$ $E = -L \frac{dI}{dt}$ is a positive constant.
Comparing this with the given options,the graph that shows a negative constant value followed by a positive constant value is represented by option $D$.

(A) At resonance,the inductive reactance $(X_L)$ is equal to the capacitive reactance $(X_C)$,i.e.,$X_L = X_C$.
In an $L-C-R$ series circuit,the net impedance $(Z)$ is given by $Z = \sqrt{R^2 + (X_L - X_C)^2}$.
At resonance,since $X_L - X_C = 0$,the impedance becomes $Z = R$.
This means the circuit behaves as a purely resistive circuit.
In a purely resistive circuit,the voltage and current are in the same phase.
Therefore,the phase difference $(\phi)$ between the applied voltage and the current is $0$.

(D) For a $NAND$ gate,the output $C$ is given by the Boolean expression $C = \overline{A \cdot B}$.
Checking the values:
$1$. For $A = 0, B = 0$: $C = \overline{0 \cdot 0} = \overline{0} = 1$.
$2$. For $A = 0, B = 1$: $C = \overline{0 \cdot 1} = \overline{0} = 1$.
$3$. For $A = 1, B = 0$: $C = \overline{1 \cdot 0} = \overline{0} = 1$.
$4$. For $A = 1, B = 1$: $C = \overline{1 \cdot 1} = \overline{1} = 0$.
Comparing these results with the given table,the output matches the $NAND$ gate logic.

(D) Ray optics is based on the assumption that light travels in straight lines. This approximation holds true when the characteristic dimensions of the obstacles or apertures (such as the size of a slit or an object) are much larger than the wavelength of light $(\lambda)$. If the dimensions are comparable to or smaller than $\lambda$, wave effects like diffraction become significant, and ray optics is no longer valid. Therefore, the correct condition is that the dimensions must be much larger than the wavelength of light.

(B) The light from the source will be cut off if the disc covers the area corresponding to the critical angle $\theta_c$.
For a point source at depth $h$,the radius $r$ of the disc is given by $r = h \tan \theta_c$.
Since $\sin \theta_c = \frac{1}{\mu}$,we have $\tan \theta_c = \frac{1}{\sqrt{\mu^2 - 1}}$.
Substituting the given values $h = 4 \; m$ and $\mu = 5/3$:
$r = \frac{4}{\sqrt{(5/3)^2 - 1}} = \frac{4}{\sqrt{25/9 - 1}} = \frac{4}{\sqrt{16/9}} = \frac{4}{4/3} = 3 \; m$.
The minimum diameter of the disc is $D = 2r = 2 \times 3 = 6 \; m$.

(A) According to Rayleigh's law of scattering, the intensity of scattered light $I$ is inversely proportional to the fourth power of its wavelength, given by $I \propto \frac{1}{\lambda^4}$.
Since the wavelength of blue light $(\lambda_{blue})$ is the shortest among the visible spectrum, it undergoes maximum scattering by the atmospheric particles.
Therefore, the sky appears blue to our eyes.

(C) In an electromagnetic wave,the oscillating electric field vector $\vec{E}$ and the magnetic field vector $\vec{B}$ are always mutually perpendicular to each other and also perpendicular to the direction of wave propagation.
Furthermore,these fields are in phase,which means they reach their maximum and minimum values at the same time and at the same spatial location.
Therefore,the correct option is $C$.

(A) The $Big$ $Bang$ theory is widely accepted as the most satisfactory scientific explanation for the origin of the universe.
It proposes that the universe began as a hot,dense point approximately $13.8$ billion years ago and has been expanding and cooling ever since.
This theory is supported by significant observational evidence,such as the cosmic microwave background radiation and the observed redshift of distant galaxies.

(A) The electromagnetic spectrum is ordered by frequency and wavelength. The order of increasing wavelength is: $\gamma-$ rays < $X-$ rays < $UV$ rays < Visible light < Infrared < Microwaves < Radio waves.
Since $\gamma-$ rays have the highest frequency,they possess the smallest wavelength among the given options.

(A) The magnification $M$ of an astronomical telescope is given by the formula $M = -\frac{f_o}{f_e}$,where $f_o$ is the focal length of the objective lens and $f_e$ is the focal length of the eyepiece.
To obtain the largest magnification,the ratio $\left| \frac{f_o}{f_e} \right|$ must be maximized.
This requires the objective lens to have the largest focal length $(f_o = 150\; cm)$ and the eyepiece to have the smallest focal length $(f_e = 15\; cm)$.
Thus,the focal length of the eyepiece should be $15\; cm$.

(D) The phenomenon known as the $Skin \text{ } Effect$ causes the majority of $A.C.$ current to flow near the surface of a conductor.
For $A.C.$ transmission, using a single thick wire is inefficient because the interior of the wire carries very little current.
By using multiple thin strands, the total surface area is increased, which reduces the effective resistance for $A.C.$ and improves efficiency.
For $D.C.$ transmission, the current is distributed uniformly throughout the cross-section of the conductor, so either a single thick wire or multiple thin strands will function effectively.
Therefore, multiple strands are preferred for $A.C.$, while either type is suitable for $D.C.$

(B) The root mean square $(I_{\text{rms}})$ value of an alternating current is defined as the square root of the mean of the squares of the instantaneous current over one complete cycle.
For a sinusoidal alternating current given by $I = I_{0} \sin(\omega t)$,the $I_{\text{rms}}$ is calculated as:
$I_{\text{rms}} = \sqrt{\frac{1}{T} \int_{0}^{T} I^{2} dt}$
Substituting $I = I_{0} \sin(\omega t)$:
$I_{\text{rms}} = \sqrt{\frac{1}{T} \int_{0}^{T} I_{0}^{2} \sin^{2}(\omega t) dt}$
$I_{\text{rms}} = I_{0} \sqrt{\frac{1}{T} \int_{0}^{T} \frac{1 - \cos(2\omega t)}{2} dt}$
$I_{\text{rms}} = I_{0} \sqrt{\frac{1}{2T} [t - \frac{\sin(2\omega t)}{2\omega}]_{0}^{T}}$
Since $\sin(2\omega T) = \sin(4\pi) = 0$,we get:
$I_{\text{rms}} = I_{0} \sqrt{\frac{T}{2T}} = \frac{I_{0}}{\sqrt{2}}$
Therefore,the correct relation is $I_{\text{rms}} = \frac{1}{\sqrt{2}} I_{0}$.

(A) Good conductors of electricity are materials that allow electric current to flow through them easily due to the presence of free electrons.
Metals like $Cu$ (Copper),$Ag$ (Silver),and $Au$ (Gold) have a large number of free electrons in their metallic lattice,making them excellent conductors.
$Si$ (Silicon) and $Ge$ (Germanium) are semiconductors.
$Diamond$ is an insulator.
$NaCl$ (Sodium Chloride) is an ionic solid that conducts electricity only in a molten state or aqueous solution,not in its solid state.
Therefore,the set containing $Cu, Ag$,and $Au$ consists entirely of good conductors.

(B) Electric field lines originate from a positive charge and terminate on a negative charge.
In the given figure,the electric field lines are directed towards both charges $q_1$ and $q_2$.
Since the field lines are terminating at both charges,both $q_1$ and $q_2$ must be negative charges.
Therefore,both $q_1$ and $q_2$ are negative.

(B) The electric potential $V$ at any point on a circular path of radius $r$ due to a charge $q_{2}$ at the center is given by $V = \frac{1}{4 \pi \varepsilon_{0}} \frac{q_{2}}{r}$.
Since $q_{2}$ and $r$ are constant,the potential $V$ is constant at all points on the circular path,making it an equipotential surface.
The work done $W$ in moving a charge $q_{1}$ between two points $A$ and $B$ is given by $W = q_{1}(V_{B} - V_{A})$.
Since the path is equipotential,the potential at the starting point and the ending point (after one full rotation) is the same,i.e.,$V_{A} = V_{B}$.
Therefore,the work done $W = q_{1}(V - V) = 0$.

(C) For a hollow metallic sphere,the electric field inside is zero $(E = 0)$.
Since the electric field is the negative gradient of the potential $(E = -dV/dr)$,if $E = 0$,then the potential $V$ must be constant throughout the interior.
Therefore,the potential at any point inside the sphere,including the centre,is equal to the potential at its surface.
Given that the potential at the surface is $80 \; V$,the potential at the centre is also $80 \; V$.

(B) The circuit can be simplified by observing the symmetry and series-parallel combinations.
$1$. The resistors along $AF$ and $FE$ are in series. Their equivalent resistance is $R_1 = 3 \; \Omega + 3 \; \Omega = 6 \; \Omega$.
$2$. This $R_1$ is in parallel with the $6 \; \Omega$ resistor connected along $AE$. The equivalent resistance $R_{AE}'$ is given by $\frac{1}{R_{AE}'} = \frac{1}{6} + \frac{1}{6} = \frac{2}{6} = \frac{1}{3} \Rightarrow R_{AE}' = 3 \; \Omega$.
$3$. Similarly, the resistors along $ED$ and $DC$ are in series with the $6 \; \Omega$ resistor along $AD$ in parallel, leading to an equivalent resistance of $3 \; \Omega$ between $A$ and $C$.
$4$. Finally, the circuit reduces to three resistors of $3 \; \Omega$ each connected between $A$ and $B$ (one directly, and two through path $C$).
$5$. The equivalent resistance $R_{eq}$ is the parallel combination of the direct $3 \; \Omega$ resistor and the series combination of the other two $3 \; \Omega$ resistors $(3+3=6 \; \Omega)$.
$6$. $R_{eq} = \frac{3 \times 6}{3 + 6} = \frac{18}{9} = 2 \; \Omega$.

(D) The number of electrons flowing per unit time is given by $\frac{n}{t} = 10^{7} \; s^{-1}$.
The electric current $I$ is defined as the rate of flow of charge,$I = \frac{q}{t}$.
Since $q = ne$,where $e$ is the elementary charge $(e = 1.6 \times 10^{-19} \; C)$,we have $I = \frac{ne}{t} = \left(\frac{n}{t}\right) \times e$.
Substituting the given values: $I = 10^{7} \times (1.6 \times 10^{-19} \; C) = 1.6 \times 10^{-12} \; A$.