NEET 2020 Physics Question Paper with Answer and Solution

(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 torque $\vec{\tau}$ about the origin is given by the cross product of the position vector $\vec{r}$ and the force vector $\vec{F}$.
Given: $\vec{F} = 3 \hat{j} \text{ N}$ and $\vec{r} = 2 \hat{k} \text{ m}$.
$\vec{\tau} = \vec{r} \times \vec{F} = (2 \hat{k}) \times (3 \hat{j})$.
Using the properties of unit vector cross products: $\hat{k} \times \hat{j} = -\hat{i}$.
Therefore,$\vec{\tau} = 6 (\hat{k} \times \hat{j}) = 6(-\hat{i}) = -6 \hat{i} \text{ Nm}$.

(C) The ideal gas equation is given by $PV = nRT$.
Since $n = \frac{m}{M_w}$,we have $PV = \frac{m}{M_w} RT$.
Rearranging for density $\rho = \frac{m}{V}$,we get $\rho = \frac{PM_w}{RT}$.
Given: Pressure $P = 249\; kPa = 249 \times 10^{3}\; Pa$,Temperature $T = 27^{\circ} C = 300\; K$,Gas constant $R = 8.3\; J\; mol^{-1} K^{-1}$,and Molar mass of hydrogen gas $(H_2)$ $M_w = 2 \times 10^{-3}\; kg/mol$.
Substituting the values:
$\rho = \frac{249 \times 10^{3} \times 2 \times 10^{-3}}{8.3 \times 300}$
$\rho = \frac{498}{2490} = 0.2\; kg/m^{3}$.

(C) The process described is known as free expansion,where a gas expands into a vacuum.
Since the system is thermally insulated,there is no heat exchange with the surroundings,meaning $Q = 0$.
Because the gas expands into a vacuum,it does no work against any external pressure,meaning $W = 0$.
According to the first law of thermodynamics,$\Delta U = Q - W$. Since $Q = 0$ and $W = 0$,the change in internal energy $\Delta U = 0$.
For an ideal gas,internal energy depends only on temperature. Therefore,if $\Delta U = 0$,the temperature remains constant.
However,the defining characteristic of a process where $Q = 0$ is that it is an adiabatic process.

(D) Let $m_1 = 5\, kg$ and $m_2 = 10\, kg$ be the masses of the two particles.
Let $r = 1\, m = 100\, cm$ be the length of the rod.
Let $r_1$ be the distance of the centre of mass from the $5\, kg$ particle.
The formula for the distance of the centre of mass from mass $m_1$ is given by $r_1 = \frac{m_2 r}{m_1 + m_2}$.
Substituting the values: $r_1 = \frac{10\, kg \times 100\, cm}{5\, kg + 10\, kg} = \frac{1000}{15}\, cm$.
$r_1 = 66.67\, cm \approx 67\, cm$.

(A) Given:
Initial velocity,$u = 20\; m/s$
Final velocity,$v = 80\; m/s$
Acceleration due to gravity,$g = 10\; m/s^2$
Using the third equation of motion:
$v^2 = u^2 + 2gh$
Substituting the values:
$80^2 = 20^2 + 2 \times 10 \times h$
$6400 = 400 + 20h$
$6000 = 20h$
$h = \frac{6000}{20} = 300\; m$
Therefore,the height of the tower is $300\; m$.

(C) According to the law of equipartition of energy,the average energy associated with each degree of freedom is $\frac{1}{2} k_{B} T$.
For a mono-atomic gas,the number of degrees of freedom $(f)$ is $3$ (all translational).
Therefore,the average thermal energy $E = f \times \frac{1}{2} k_{B} T = 3 \times \frac{1}{2} k_{B} T = \frac{3}{2} k_{B} T$.

(D) The height of water rise in a capillary tube is given by $h = \frac{2T \cos \theta}{r \rho g}$.
The mass of water in the capillary tube is $m = V \rho = (\pi r^2 h) \rho$.
Substituting the expression for $h$:
$m = \pi r^2 \left( \frac{2T \cos \theta}{r \rho g} \right) \rho = \frac{2 \pi r T \cos \theta}{g}$.
Since $T$,$\theta$,and $g$ are constants,we find that $m \propto r$.
For the first tube,$m_1 = 5 \, g$ and $r_1 = r$.
For the second tube,$r_2 = 2r$.
Using the proportionality $m \propto r$:
$\frac{m_2}{m_1} = \frac{r_2}{r_1} = \frac{2r}{r} = 2$.
Therefore,$m_2 = 2 \times m_1 = 2 \times 5 \, g = 10 \, g$.

(A) Young's modulus $(Y)$ is defined as the ratio of longitudinal stress to longitudinal strain.
Stress = $\frac{F}{A} = \frac{Mg}{A}$
Strain = $\frac{\Delta L}{L} = \frac{L_{1}-L}{L}$
Therefore,$Y = \frac{\text{Stress}}{\text{Strain}} = \frac{Mg/A}{(L_{1}-L)/L} = \frac{MgL}{A(L_{1}-L)}$.

(D) To convert energy from Joules $(J)$ to electron-volts $(eV)$,we divide the energy value by the charge of an electron,which is $1.6 \times 10^{-19} \ C$.
Given energy $E = 10^{-20} \ J$.
$E \text{ (in } eV) = \frac{10^{-20} \ J}{1.6 \times 10^{-19} \ J/eV}$
$E = \frac{1}{1.6} \times 10^{-1} \ eV$
$E = 0.625 \times 0.1 \ eV$
$E = 0.0625 \ eV$.

(C) The mean free path $(\lambda)$ is the average distance traveled by a molecule between two successive collisions.
According to the kinetic theory of gases,the mean free path is given by the formula:
$\lambda = \frac{1}{\sqrt{2} \pi d^2 n}$
where:
$d$ = molecular diameter
$n$ = number density (number of molecules per unit volume)
Thus,the correct expression is $\frac{1}{\sqrt{2} n \pi d^2}$.

(C) The frequency of a stretched string is given by $f = \frac{1}{2L} \sqrt{\frac{T}{\mu}}$. Thus,$f \propto \sqrt{T}$.
When the tension $T$ in string $B$ is decreased,its frequency $f_B$ decreases.
The initial beat frequency is $|f_A - f_B| = 6 \, Hz$. Given $f_A = 530 \, Hz$,the possible values for $f_B$ are $530 + 6 = 536 \, Hz$ or $530 - 6 = 524 \, Hz$.
Case $1$: If $f_B = 536 \, Hz$,decreasing the tension decreases $f_B$. As $f_B$ moves towards $530 \, Hz$,the beat frequency $|530 - f_B|$ decreases (e.g.,$536 \to 535 \implies$ beats $6 \to 5$). This contradicts the problem statement.
Case $2$: If $f_B = 524 \, Hz$,decreasing the tension decreases $f_B$. As $f_B$ moves away from $530 \, Hz$,the beat frequency $|530 - f_B|$ increases (e.g.,$524 \to 523 \implies$ beats $6 \to 7$). This matches the problem statement.
Therefore,the original frequency of $B$ is $524 \, Hz$.

(B) The displacement $(x)$ of a particle in $SHM$ is given by $x = A \sin(\omega t + \phi)$.
The velocity $(v)$ is the first derivative of displacement: $v = \frac{dx}{dt} = A \omega \cos(\omega t + \phi)$.
The acceleration $(a)$ is the second derivative of displacement: $a = \frac{d^2x}{dt^2} = -A \omega^2 \sin(\omega t + \phi)$.
Using the trigonometric identity $-\sin(\theta) = \sin(\theta + \pi)$,we can rewrite the acceleration as $a = A \omega^2 \sin(\omega t + \phi + \pi)$.
Comparing the phase of displacement $(\omega t + \phi)$ and acceleration $(\omega t + \phi + \pi)$,the phase difference is $\pi \; rad$.

(C) In subtraction,the number of decimal places in the result should be equal to the number of decimal places of the term in the operation that contains the fewest decimal places.
Given values:
$9.99\, m$ (has $2$ decimal places)
$0.0099\, m$ (has $4$ decimal places)
Performing the subtraction:
$9.99 - 0.0099 = 9.9801$
According to the rule of significant figures for subtraction,the result must be rounded to the same number of decimal places as the term with the fewest decimal places,which is $2$ decimal places.
Rounding $9.9801$ to $2$ decimal places gives $9.98$.
Thus,the final result is $9.98\, m$.

(A) Stress is defined as the restoring force per unit area.
$\text{Stress} = \frac{\text{Force}}{\text{Area}}$
Substituting the dimensional formulas for force $[M L T^{-2}]$ and area $[L^2]$:
$\text{Stress} = \frac{[M L T^{-2}]}{[L^2]}$
$\text{Stress} = [M L^{1-2} T^{-2}]$
$\text{Stress} = [M L^{-1} T^{-2}]$

(D) For a system of two masses $m_1$ and $m_2$ connected by a string over a frictionless pulley,the acceleration $a$ is given by the formula:
$a = \frac{(m_2 - m_1)g}{m_1 + m_2}$
Here,$m_1 = 4 \, kg$ and $m_2 = 6 \, kg$.
Substituting the values into the formula:
$a = \frac{(6 - 4)g}{6 + 4}$
$a = \frac{2g}{10}$
$a = \frac{g}{5}$
Thus,the acceleration of the system is $\frac{g}{5}$.

(D) The least count $(L.C.)$ of a screw gauge is defined by the formula:
$L.C. = \frac{\text{Pitch}}{\text{Number of divisions on circular scale}}$
Given:
$L.C. = 0.01\, mm$
Number of divisions = $50$
Substituting the values into the formula:
$0.01\, mm = \frac{\text{Pitch}}{50}$
Therefore,the pitch is:
$\text{Pitch} = 0.01\, mm \times 50 = 0.5\, mm$.

(B) The heat required to change the temperature of a body is given by the formula $\Delta Q = M s \Delta T$,where $M$ is the mass,$s$ is the specific heat capacity,and $\Delta T$ is the change in temperature.
Since both spheres are made of copper,the specific heat capacity $s$ is the same for both.
Given that the temperature change $\Delta T = 1 \ K$ is the same for both,we have $\Delta Q \propto M$.
The mass $M$ of a sphere is given by $M = V \rho = \frac{4}{3} \pi r^3 \rho$,where $V$ is the volume,$r$ is the radius,and $\rho$ is the density.
Since $\rho$ is the same for both copper spheres,$M \propto r^3$.
Therefore,the ratio of heat required is $\frac{\Delta Q_1}{\Delta Q_2} = \frac{M_1}{M_2} = \left( \frac{r_1}{r_2} \right)^3$.
Given $r_1 = 1.5 r_2$,we have $\frac{r_1}{r_2} = 1.5 = \frac{3}{2}$.
Thus,$\frac{\Delta Q_1}{\Delta Q_2} = \left( \frac{3}{2} \right)^3 = \frac{27}{8}$.

(B) The mean free path $\ell$ of a gas molecule is defined as the average distance traveled by a molecule between two successive collisions.
The formula for the mean free path is given by:
$\ell = \frac{1}{\sqrt{2} n \pi d^{2}}$
where $n$ is the number density of molecules and $d$ is the diameter of the molecule.
From the formula,it is clear that $\ell$ is inversely proportional to the square of the diameter $d$.
Therefore,$\ell \propto \frac{1}{d^{2}}$.

(A) The arithmetic mean of the given values is taken as the true value.
$t_{\text{mean}} = \frac{1.25 + 1.24 + 1.27 + 1.21 + 1.28}{5} = \frac{6.25}{5} = 1.25 \; s$.
The absolute errors in each measurement are:
$|\Delta t_1| = |1.25 - 1.25| = 0 \; s$
$|\Delta t_2| = |1.24 - 1.25| = 0.01 \; s$
$|\Delta t_3| = |1.27 - 1.25| = 0.02 \; s$
$|\Delta t_4| = |1.21 - 1.25| = 0.04 \; s$
$|\Delta t_5| = |1.28 - 1.25| = 0.03 \; s$
The mean absolute error is:
$\Delta t_{\text{mean}} = \frac{0 + 0.01 + 0.02 + 0.04 + 0.03}{5} = \frac{0.10}{5} = 0.02 \; s$.
The percentage relative error is:
$\text{Percentage error} = \frac{\Delta t_{\text{mean}}}{t_{\text{mean}}} \times 100 = \frac{0.02}{1.25} \times 100 = 1.6 \%$.

(D) The ideal gas equation is given by $PV = \mu RT$.
Here,$\mu$ is the number of moles,which is defined as $\mu = \frac{M}{M_{0}}$,where $M$ is the total mass of the gas and $M_{0}$ is the molar mass.
Substituting this into the ideal gas equation,we get $PV = \left(\frac{M}{M_{0}}\right) RT$.
Rearranging the terms,we get $P = \left(\frac{M}{V}\right) \frac{RT}{M_{0}}$.
Since mass density $\rho$ is defined as $\rho = \frac{M}{V}$,we can substitute it into the equation to get $P = \frac{\rho RT}{M_{0}}$.
Thus,$\rho$ represents the mass density and $M_{0}$ represents the molar mass.

(C) The fundamental frequency of a stretched string is given by $n = \frac{1}{2L} \sqrt{\frac{T}{\mu}}$.
Since the tension $T$ and mass per unit length $\mu$ are constant,the frequency is inversely proportional to the length of the string: $n \propto \frac{1}{L}$.
Therefore,$n_1 L_1 = n_2 L_2$.
Given $n_1 = 120 \;Hz$,$L_1 = 90 \;cm$,and $n_2 = 180 \;Hz$.
Substituting the values: $120 \times 90 = 180 \times L_2$.
$L_2 = \frac{120 \times 90}{180} = \frac{10800}{180} = 60 \;cm$.
Thus,the string should be pressed at $60 \;cm$ from one end to produce the required frequency.

(C) The pressure exerted by a liquid column is given by the formula $P = h \rho g$,where $h$ is the height,$\rho$ is the density,and $g$ is the acceleration due to gravity.
For the mercury barometer,the pressure is $P = h_{Hg} \rho_{Hg} g = 0.76 \; m \times 13600 \; kg/m^3 \times g$.
For the liquid barometer,the pressure is $P = h' \rho' g = h' \times 760 \; kg/m^3 \times g$.
Since the atmospheric pressure is the same in both cases,we equate the two expressions:
$h' \times 760 = 0.76 \times 13600$
Solving for $h'$:
$h' = \frac{0.76 \times 13600}{760}$
$h' = \frac{10336}{760} = 13.6 \; m$.

(D) In a $P-V$ diagram,the $y$-axis represents pressure $(P)$ and the $x$-axis represents volume $(V)$.
From the given figure,the line representing the process is horizontal,which means the pressure $(P)$ remains constant as the volume $(V)$ changes from the initial state to the final state.
$A$ thermodynamic process in which the pressure of the system remains constant is known as an isobaric process.
Therefore,the correct option is $D$.

(C) The efficiency $\eta$ of a Carnot engine is given by the formula:
$\eta = 1 - \frac{T_{2}}{T_{1}}$
where $T_{1}$ is the absolute temperature of the source and $T_{2}$ is the absolute temperature of the sink.
Therefore,the efficiency of a Carnot engine depends on both the temperature of the source and the temperature of the sink.

(D) The angle of contact $\theta$ determines the wetting behavior of a liquid on a solid surface.
If $\theta < 90^{\circ},$ the liquid wets the surface (e.g.,water on glass).
If $\theta > 90^{\circ},$ the cohesive forces between liquid molecules are stronger than the adhesive forces between the liquid and the solid surface.
Consequently,the liquid does not wet the solid surface (e.g.,mercury on glass).
Therefore,the correct condition for a liquid not to wet a solid surface is that the angle of contact is greater than $90^{\circ}.$

(A) According to Wien's Displacement Law,the product of the wavelength of maximum emission $\lambda_{m}$ and the absolute temperature $T$ is constant,i.e.,$\lambda_{m} T = b$.
This implies $T \propto \frac{1}{\lambda_{m}}$.
The wavelength of light corresponding to colors follows the order: $\lambda_{\text{blue}} < \lambda_{\text{yellow}} < \lambda_{\text{red}}$.
Given that star $A$ is bluish,star $C$ is yellowish,and star $B$ is reddish,their wavelengths are $\lambda_{A} < \lambda_{C} < \lambda_{B}$.
Since temperature is inversely proportional to wavelength,we have $T_{A} > T_{C} > T_{B}$.

(D) Let the three spheres be located at coordinates $(0, 0)$,$(2, 0)$,and $(0, 2)$ in the $xy$-plane.
Since all spheres have the same mass $M$,the coordinates of the center of mass $(x_{cm}, y_{cm})$ are given by:
$x_{cm} = \frac{M(0) + M(2) + M(0)}{M + M + M} = \frac{2M}{3M} = \frac{2}{3} \; m$
$y_{cm} = \frac{M(0) + M(0) + M(2)}{M + M + M} = \frac{2M}{3M} = \frac{2}{3} \; m$
Therefore,the position vector of the center of mass is $\vec{r}_{cm} = x_{cm}\hat{i} + y_{cm}\hat{j} = \frac{2}{3}(\hat{i} + \hat{j}) \; m$.

(B) We know that $1^{\circ} = 60^{\prime}$ (minutes of arc).
Therefore,$1^{\prime} = (1/60)^{\circ}$.
To convert degrees to radians,we multiply by $\frac{\pi}{180}$.
So,$1^{\prime} = \left(\frac{1}{60}\right) \times \left(\frac{\pi}{180}\right) \text{ rad}$.
Using $\pi \approx 3.14159$,we get:
$1^{\prime} = \frac{3.14159}{10800} \text{ rad} \approx 2.9088 \times 10^{-4} \text{ rad}$.
Rounding to three significant figures,we get $2.91 \times 10^{-4} \text{ rad}$.

(C) The value of acceleration due to gravity at a depth $d$ below the surface of the Earth is given by the formula:
$g' = g \left(1 - \frac{d}{R}\right)$
where $g$ is the acceleration due to gravity at the surface and $R$ is the radius of the Earth.
According to the problem,the value of acceleration at depth $d$ is $\frac{1}{n}$ times the value at the surface:
$g' = \frac{g}{n}$
Substituting this into the formula:
$\frac{g}{n} = g \left(1 - \frac{d}{R}\right)$
Dividing both sides by $g$:
$\frac{1}{n} = 1 - \frac{d}{R}$
Rearranging the terms to solve for $d$:
$\frac{d}{R} = 1 - \frac{1}{n}$
$\frac{d}{R} = \frac{n-1}{n}$
$d = R \left(\frac{n-1}{n}\right)$

(B) Initial angular speed $\omega_0 = 360 \; rpm = 360 \times \frac{2\pi}{60} \; rad/s = 12\pi \; rad/s$.
Final angular speed $\omega = 1200 \; rpm = 1200 \times \frac{2\pi}{60} \; rad/s = 40\pi \; rad/s$.
Time taken $t = 14 \; s$.
Using the equation of rotational motion $\omega = \omega_0 + \alpha t$,where $\alpha$ is the angular acceleration:
$40\pi = 12\pi + \alpha(14)$.
$28\pi = 14\alpha$.
$\alpha = \frac{28\pi}{14} = 2\pi \; rad/s^2$.

(B) Free Body Diagram ($F$.$B$.$D$.) of the trolley:
$T - f = m_{T} a$
Where $f = \mu m_{T} g = 0.05 \times 10 \times 10 = 5\; N$.
So,$T - 5 = 10a$ --- $(i)$
Free Body Diagram ($F$.$B$.$D$.) of the block:
$m_{b} g - T = m_{b} a$
$2 \times 10 - T = 2a$
$20 - T = 2a$ --- (ii)
Adding equations $(i)$ and (ii):
$(T - 5) + (20 - T) = 10a + 2a$
$15 = 12a$
$a = \frac{15}{12} = 1.25\; m/s^{2}$.

(D) At the lowest point of a vertical circle,the forces acting on the mass '$m$' are the tension '$T$' acting upwards and the weight '$mg$' acting downwards.
The net centripetal force required for circular motion is provided by the difference between tension and weight:
$T - mg = \frac{mv^2}{r}$
Given that the velocity at the lowest point is $v = \sqrt{7gr}$,we substitute this into the equation:
$T - mg = \frac{m(\sqrt{7gr})^2}{r}$
$T - mg = \frac{m(7gr)}{r}$
$T - mg = 7mg$
$T = 7mg + mg = 8mg$
Therefore,the tension in the string at the lowest point is $8mg$.

(C) Spherometer is a precision instrument designed to measure the sagitta (height) of a spherical surface. By knowing the distance between the legs of the Spherometer and measuring the sagitta,we can calculate the radius of curvature $(R)$ of the curved surface using the formula $R = \frac{a^2}{6h} + \frac{h}{2}$,where $a$ is the distance between the legs and $h$ is the sagitta. Therefore,a Spherometer is used to measure the radius of curvature of the curved surface.

(D) function $f(t)$ is periodic if it repeats its value after a fixed interval of time $T$,such that $f(t) = f(t + T)$.
$1$. The functions $e^{-\omega t}$,$e^{\omega t}$,and $\log_{e}(\omega t)$ are non-periodic functions because they do not repeat their values at regular intervals.
$2$. For the function $f(t) = \sin \omega t + \cos \omega t$,we can check for periodicity by replacing $t$ with $t + T$:
$f(t + T) = \sin(\omega(t + T)) + \cos(\omega(t + T)) = \sin(\omega t + \omega T) + \cos(\omega t + \omega T)$.
$3$. If we choose $T = \frac{2\pi}{\omega}$,then $\omega T = 2\pi$. Substituting this:
$f(t + T) = \sin(\omega t + 2\pi) + \cos(\omega t + 2\pi) = \sin \omega t + \cos \omega t = f(t)$.
Since the function repeats itself after a time period $T = \frac{2\pi}{\omega}$,it represents a periodic motion.

(C) Let $u$ be the velocity of the ball at the top of the window.
Using the second equation of motion for the motion of the ball across the window:
$S = ut + \frac{1}{2}at^2$
Here,$S = 1.5 \; m$,$t = 0.1 \; s$,and $a = g = 10 \; m/s^2$.
Substituting the values:
$1.5 = u(0.1) + \frac{1}{2} \times 10 \times (0.1)^2$
$1.5 = 0.1u + 5 \times 0.01$
$1.5 = 0.1u + 0.05$
$0.1u = 1.5 - 0.05$
$0.1u = 1.45$
$u = \frac{1.45}{0.1} = 14.5 \; m/s$.

(A) The color code for the resistor is Yellow,Violet,Brown,and Gold.
According to the standard color code table:
$1$. The first color is Yellow,which corresponds to the digit $4$.
$2$. The second color is Violet,which corresponds to the digit $7$.
$3$. The third color is Brown,which acts as the multiplier $10^{1}$.
$4$. The fourth color is Gold,which represents the tolerance of $\pm 5 \%$.
Thus,the resistance value is $R = 47 \times 10^{1} \; \Omega = 470 \; \Omega$.
The tolerance is $5 \%$.
Therefore,the resistance is $470 \; \Omega, 5 \%$.

(D) For a prism,the angle of the prism is given by $A = r_1 + r_2$.
Since the ray emerges normally from the opposite surface,the angle of emergence $e = 0$,which implies the angle of refraction at the second surface $r_2 = 0$.
Substituting $r_2 = 0$ into the prism equation,we get $r_1 = A$.
Applying Snell's law at the first surface: $\sin i = \mu \sin r_1$.
For small angles,$\sin i \approx i$ and $\sin r_1 \approx r_1$.
Therefore,$i = \mu r_1$.
Substituting $r_1 = A$,we get $i = \mu A$.

(B) The relative permeability $\mu_{r}$ is related to magnetic susceptibility $\chi_{m}$ by the formula: $\mu_{r} = 1 + \chi_{m}$.
Given $\chi_{m} = 599$,we have $\mu_{r} = 1 + 599 = 600$.
The absolute permeability $\mu$ of the material is given by $\mu = \mu_{0} \mu_{r}$.
Substituting the values,$\mu = (4 \pi \times 10^{-7} \, T m A^{-1}) \times 600$.
$\mu = 2400 \pi \times 10^{-7} \, T m A^{-1} = 2.4 \pi \times 10^{-4} \, T m A^{-1}$.

(A) For effective transistor action,the base region must be very thin and lightly doped to ensure that most of the charge carriers injected from the emitter reach the collector without recombining in the base.

(D) The intensity (flux) $I$ is defined as the energy $E$ incident per unit area $A$ per unit time $t$,given by $I = \frac{E}{At}$.
To find the total energy $E$,we rearrange the formula: $E = I \times A \times t$.
Given values:
Intensity $I = 20 \, W/cm^2$
Area $A = 20 \, cm^2$
Time $t = 1 \, minute = 60 \, seconds$
Substituting the values into the equation:
$E = 20 \, W/cm^2 \times 20 \, cm^2 \times 60 \, s$
$E = 400 \times 60 \, J$
$E = 24000 \, J = 24 \times 10^3 \, J$.

(C) The electric potential $V$ due to a short electric dipole at a point $(r, \theta)$ is given by the formula:
$V = \frac{1}{4 \pi \epsilon_{0}} \frac{p \cos \theta}{r^{2}}$
Given values:
Dipole moment $p = 16 \times 10^{-9} \, Cm$
Distance $r = 0.6 \, m$
Angle $\theta = 60^{\circ}$
Constant $k = \frac{1}{4 \pi \epsilon_{0}} = 9 \times 10^{9} \, Nm^{2}/C^{2}$
Substituting the values into the formula:
$V = \frac{9 \times 10^{9} \times 16 \times 10^{-9} \times \cos(60^{\circ})}{(0.6)^{2}}$
Since $\cos(60^{\circ}) = 0.5$:
$V = \frac{9 \times 16 \times 0.5}{0.36}$
$V = \frac{72}{0.36}$
$V = 200 \, V$

(C) In a meter bridge,the balancing condition is given by $\frac{R}{S} = \frac{\ell_1}{\ell_2}$,where $R$ is the resistance in the left gap and $S$ is the resistance in the right gap.
Given $S = 10\, \Omega$ and the ratio $\frac{\ell_1}{\ell_2} = \frac{3}{2}$.
Substituting these values: $\frac{R}{10} = \frac{3}{2} \implies R = 15\, \Omega$.
The total length of the resistance wire $R$ is $1.5\, m$.
Therefore,the length per unit resistance is $\frac{1.5\, m}{15\, \Omega} = 0.1\, m/\Omega$.
Converting this to the required form: $0.1\, m = 10 \times 10^{-2}\, m$.
Thus,the value is $10$.

(B) The nuclear fission reaction is given by the conservation of mass number and atomic number.
Let the unknown nucleus be ${ }_{Z}^{A} X$.
The reaction is: ${ }_{92}^{235} U + { }_{0}^{1} n \rightarrow { }_{36}^{89} Kr + { }_{Z}^{A} X + 3({ }_{0}^{1} n)$.
Conservation of mass number $(A)$: $235 + 1 = 89 + A + 3(1) \Rightarrow 236 = 92 + A \Rightarrow A = 144$.
Conservation of atomic number $(Z)$: $92 + 0 = 36 + Z + 3(0) \Rightarrow 92 = 36 + Z \Rightarrow Z = 56$.
The element with atomic number $56$ is Barium $(Ba)$.
Thus,the missing nucleus is ${ }_{56}^{144} Ba$.

(B) The magnetic field $B$ at the centre of a long solenoid is given by the formula: $B = \mu_{0} n I$,where $n = \frac{N}{\ell}$.
Given:
Length $\ell = 50\, cm = 0.5\, m$
Number of turns $N = 100$
Current $I = 2.5\, A$
Permeability $\mu_{0} = 4\pi \times 10^{-7}\, T\, m\, A^{-1}$
Substituting the values:
$B = (4\pi \times 10^{-7}) \times \left(\frac{100}{0.5}\right) \times 2.5$
$B = (4 \times 3.14159 \times 10^{-7}) \times 200 \times 2.5$
$B = 12.566 \times 10^{-7} \times 500$
$B = 6283.18 \times 10^{-7} = 6.283 \times 10^{-4}\, T$
To express this in the form $...... \times 10^{-5}\, T$:
$B = 62.83 \times 10^{-5}\, T$
Rounding to the nearest option,the value is $62.8$.

(C) In an electromagnetic wave $(EMW)$,the energy density associated with the electric field is $u_E = \frac{1}{2} \epsilon_0 E^2$ and the energy density associated with the magnetic field is $u_B = \frac{1}{2} \frac{B^2}{\mu_0}$.
Since $E = cB$ and $c = \frac{1}{\sqrt{\mu_0 \epsilon_0}}$,we have $u_E = u_B$.
Because the intensity of the wave is proportional to the energy density,the contributions of the electric field and magnetic field components to the total intensity are equal.
Therefore,the ratio is $1:1$.

(B) The limit of resolution $(\Delta \theta)$ of a telescope is given by the formula: $\Delta \theta = \frac{1.22 \lambda}{D}$, where $\lambda$ is the wavelength of light and $D$ is the diameter of the objective lens.
Given: $\lambda = 600\, nm = 600 \times 10^{-9}\, m = 6 \times 10^{-7}\, m$ and $D = 2\, m$.
Substituting the values into the formula:
$\Delta \theta = \frac{1.22 \times 6 \times 10^{-7}}{2}$
$\Delta \theta = 1.22 \times 3 \times 10^{-7}$
$\Delta \theta = 3.66 \times 10^{-7}\, rad$.
Thus, the value is $3.66$.

(B) The relationship between electric field $\vec{E}$ and electric potential $V$ is given by the gradient formula: $\vec{E} = -\nabla V$.
Since the electric potential $V$ is constant $(5 \ V)$ throughout the given region,its spatial derivative (gradient) is zero.
Therefore,$\vec{E} = -\frac{dV}{dr} = 0$.
Thus,the magnitude of the electric field in this region is $0 \ N/C$.

(A) The de Broglie wavelength $\lambda$ of an electron accelerated through a potential difference $V$ is given by the formula: $\lambda = \frac{12.27}{\sqrt{V}} \, \mathring{A}$.
Given $\lambda = 1.227 \times 10^{-2} \, nm$.
Since $1 \, nm = 10 \, \mathring{A}$,we have $\lambda = 1.227 \times 10^{-2} \times 10 \, \mathring{A} = 0.1227 \, \mathring{A}$.
Substituting this into the formula: $0.1227 = \frac{12.27}{\sqrt{V}}$.
Rearranging for $\sqrt{V}$: $\sqrt{V} = \frac{12.27}{0.1227} = 100 = 10^{2}$.
Squaring both sides: $V = (10^{2})^{2} = 10^{4} \, V$.

(D) Given: Capacitance $C = 40 \, \mu F = 40 \times 10^{-6} \, F$, Voltage $V_{rms} = 200 \, V$, Frequency $f = 50 \, Hz$.
The capacitive reactance is given by $X_C = \frac{1}{\omega C} = \frac{1}{2 \pi f C}$.
The rms current is given by $I_{rms} = \frac{V_{rms}}{X_C} = V_{rms} \times 2 \pi f C$.
Substituting the values:
$I_{rms} = 200 \times 2 \times 3.1416 \times 50 \times 40 \times 10^{-6}$.
$I_{rms} = 200 \times 314.16 \times 40 \times 10^{-6} = 2.513 \, A$.
Rounding to the nearest value, we get $I_{rms} \approx 2.5 \, A$.

(C) In a $p-n$ junction diode,when a reverse bias is applied,the positive terminal of the external battery is connected to the $n$-region and the negative terminal to the $p$-region.
This causes the majority charge carriers (electrons in the $n$-region and holes in the $p$-region) to be pulled away from the junction.
As a result,the number of uncovered immobile ions near the junction increases,which leads to an increase in the width of the depletion region.

(D) According to Brewster's law,$\tan i_b = \frac{\mu_2}{\mu_1}$.
Assuming the first medium is air,$\mu_1 = 1$.
Thus,$\tan i_b = \mu_2$.
Since the refractive index of any denser medium $\mu_2$ is greater than $1$ (i.e.,$\mu_2 > 1$),we have $\tan i_b > 1$.
Since $\tan 45^{\circ} = 1$,for $\tan i_b > 1$,the angle $i_b$ must be greater than $45^{\circ}$.
Also,the angle of incidence must be less than $90^{\circ}$ for the interface.
Therefore,$45^{\circ} < i_b < 90^{\circ}$.

(C) For a point outside a spherical conductor,the conductor acts as a point charge located at its centre.
Given: Charge $Q = 3.2 \times 10^{-7} \, C$,distance $r = 15 \, cm = 0.15 \, m$,and $k = 9 \times 10^{9} \, Nm^{2}/C^{2}$.
The formula for the electric field is $E = \frac{kQ}{r^{2}}$.
Substituting the values: $E = \frac{9 \times 10^{9} \times 3.2 \times 10^{-7}}{(0.15)^{2}}$.
$E = \frac{28.8 \times 10^{2}}{0.0225} = \frac{2880}{0.0225} = 1.28 \times 10^{5} \, N/C$.

(D) The capacitance of a parallel plate capacitor with air is given by $C_{0} = 6\, \mu F$.
When a dielectric medium is introduced,the new capacitance is $C_{m} = 30\, \mu F$.
The dielectric constant $\epsilon_{r}$ is defined as the ratio of the capacitance with the medium to the capacitance with air:
$\epsilon_{r} = \frac{C_{m}}{C_{0}} = \frac{30}{6} = 5$.
The permittivity of the medium $\epsilon$ is given by $\epsilon = \epsilon_{0} \cdot \epsilon_{r}$.
Substituting the given values:
$\epsilon = 8.85 \times 10^{-12} \times 5 = 44.25 \times 10^{-12} = 0.4425 \times 10^{-10} \approx 0.44 \times 10^{-10} C^{2} N^{-1} m^{-2}$.

(D) In a series $LCR$ circuit,the phase difference $\phi$ is given by $\tan \phi = \frac{|X_L - X_C|}{R}$.
When $L$ is removed,the circuit becomes an $RC$ circuit. The phase difference is $\tan \phi = \frac{X_C}{R} = \tan(\frac{\pi}{3}) = \sqrt{3}$.
When $C$ is removed,the circuit becomes an $RL$ circuit. The phase difference is $\tan \phi = \frac{X_L}{R} = \tan(\frac{\pi}{3}) = \sqrt{3}$.
Equating the two,we get $\frac{X_C}{R} = \frac{X_L}{R}$,which implies $X_L = X_C$.
Since $X_L = X_C$,the circuit is at resonance.
At resonance,the impedance $Z = R$.
The power factor is $\cos \phi = \frac{R}{Z} = \frac{R}{R} = 1.0$.

(A) The threshold frequency is denoted by $\nu_0$ and the work function is $\phi_0 = h\nu_0$.
In the first case,the incident frequency is $\nu_1 = 1.5\nu_0$. Since $\nu_1 > \nu_0$,photoelectric emission occurs.
In the second case,the frequency is halved,so the new frequency is $\nu_2 = \frac{1.5\nu_0}{2} = 0.75\nu_0$.
Since the new incident frequency $\nu_2$ is less than the threshold frequency $\nu_0$ $(\nu_2 < \nu_0)$,no photoelectric emission can take place regardless of the intensity of the light.
Therefore,the photoelectric current will be $0$.

(A) The temperature coefficient of resistance $(\alpha)$ is defined by the relation $R_t = R_0(1 + \alpha \Delta T)$.
For metals, resistance increases with temperature, so $\alpha$ is positive.
For insulators and semiconductors, the number of charge carriers increases significantly with an increase in temperature, which leads to a decrease in resistance.
Therefore, insulators and semiconductors exhibit a negative temperature coefficient of resistance.

(C) The mobility $\mu$ of a charged particle is defined as the ratio of its drift velocity $v_d$ to the applied electric field $E$.
Formula: $\mu = \frac{v_d}{E}$
Given:
$v_d = 7.5 \times 10^{-4} \, m s^{-1}$
$E = 3 \times 10^{-10} \, V m^{-1}$
Calculation:
$\mu = \frac{7.5 \times 10^{-4}}{3 \times 10^{-10}}$
$\mu = 2.5 \times 10^{-4 - (-10)}$
$\mu = 2.5 \times 10^{6} \, m^{2} V^{-1} s^{-1}$

(D) For metals like copper,the resistivity ( $\rho$ ) increases with an increase in temperature $(T)$.
According to the relation $\rho_T = \rho_0 [1 + \alpha(T - T_0)]$,the resistivity increases linearly with temperature for a wide range of temperatures.
At very low temperatures,the curve becomes non-linear and approaches a finite value as $T$ approaches $0 \ K$.
Among the given options,graph $D$ correctly represents this characteristic behavior of resistivity versus temperature for a metal like copper.

(D) The formula for fringe width in Young's double slit experiment is given by $\beta = \frac{\lambda D}{d}$,where $\lambda$ is the wavelength,$D$ is the distance of the screen from the sources,and $d$ is the separation between the sources.
Let the initial fringe width be $\beta = \frac{\lambda D}{d}$.
According to the problem,the new distance $D^{\prime} = 2D$ and the new separation $d^{\prime} = \frac{d}{2}$.
The new fringe width $\beta^{\prime}$ is given by $\beta^{\prime} = \frac{\lambda D^{\prime}}{d^{\prime}}$.
Substituting the new values: $\beta^{\prime} = \frac{\lambda (2D)}{d/2} = 4 \times \frac{\lambda D}{d}$.
Therefore,$\beta^{\prime} = 4\beta$.
The fringe width becomes $4$ times the original value.

(B) The given circuit consists of two $NOT$ gates followed by a $NOR$ gate.
Let the inputs be $A$ and $B$.
The outputs of the two $NOT$ gates are $\bar{A}$ and $\bar{B}$.
These are the inputs to the $NOR$ gate.
The output $Y$ of the $NOR$ gate is given by the Boolean expression:
$Y = \overline{\bar{A} + \bar{B}}$
Using De Morgan's theorem,$\overline{\bar{A} + \bar{B}} = \overline{\bar{A}} \cdot \overline{\bar{B}} = A \cdot B$.
Thus,the circuit acts as an $AND$ gate.
The truth table for an $AND$ gate is:

$A$	$B$	$Y$
$0$	$0$	$0$
$0$	$1$	$0$
$1$	$0$	$0$
$1$	$1$	$1$

Therefore,the correct option is $B$.

(C) According to Einstein's mass-energy equivalence principle,the energy $E$ is given by the formula $E = mc^2$.
Given mass $m = 0.5\, g = 0.5 \times 10^{-3}\, kg$.
The speed of light $c = 3 \times 10^8\, m/s$.
Substituting these values into the equation:
$E = (0.5 \times 10^{-3}\, kg) \times (3 \times 10^8\, m/s)^2$
$E = 0.5 \times 10^{-3} \times 9 \times 10^{16}\, J$
$E = 4.5 \times 10^{13}\, J$.

(A) The Bohr model is strictly applicable only to hydrogen-like species,which are atoms or ions containing exactly one electron.
$1$. Hydrogen atom $(H)$ has $1$ electron.
$2$. Singly ionized helium atom $(He^+)$ has $2 - 1 = 1$ electron.
$3$. Deuteron atom (an isotope of hydrogen) has $1$ electron.
$4$. Singly ionized neon atom $(Ne^+)$ has $10 - 1 = 9$ electrons.
Since $Ne^+$ has more than one electron,the Bohr model is not valid for it.

(B) An intrinsic semiconductor (such as $Si$ or $Ge$) is converted into an $N$-type extrinsic semiconductor by doping it with a pentavalent impurity element (Group $15$ elements).
Examples of pentavalent dopants include Phosphorus $(P)$,Arsenic $(As)$,and Antimony $(Sb)$.
When a pentavalent atom is added to the crystal lattice,four of its valence electrons form covalent bonds with the neighboring semiconductor atoms,while the fifth electron remains loosely bound and becomes available for conduction.
Since the dopant atom donates an extra electron,it is called a donor impurity,resulting in an $N$-type semiconductor.

(B) The activity $(A)$ of a radioactive sample is defined as the rate of decay,given by the formula: $A = \lambda N$.
Here,$\lambda$ is the decay constant,which is related to the half-life $(T_{1/2})$ by the expression: $\lambda = \frac{0.693}{T_{1/2}}$.
Given:
Number of nuclei $(N)$ = $2.0 \times 10^{21}$
Half-life $(T_{1/2})$ = $1.4 \times 10^{17} \; s$
Substituting these values into the activity formula:
$A = N \times \frac{0.693}{T_{1/2}}$
$A = (2.0 \times 10^{21}) \times \frac{0.693}{1.4 \times 10^{17}}$
$A = \frac{1.386}{1.4} \times 10^{4}$
$A \approx 0.99 \times 10^{4} \; Bq$
$A \approx 10^{4} \; Bq$.

(D) The electromagnetic spectrum is ordered by wavelength.
Gamma rays have the shortest wavelength,typically ranging from $10^{-10} \ m$ to $10^{-14} \ m$.
Comparing the given options:
$1$. Microwaves: $10^{-3} \ m$ to $10^{-1} \ m$
$2$. Ultraviolet rays: $10^{-7} \ m$ to $10^{-8} \ m$
$3$. $X$-rays: $10^{-8} \ m$ to $10^{-10} \ m$
$4$. Gamma-rays: $< 10^{-10} \ m$
Therefore,Gamma-rays have the shortest wavelength.

(C) Applying Kirchhoff's Voltage Law $(KVL)$ to the closed loop:
Starting from point $B$ and moving clockwise:
$-I(4) - 2 - I(1) - 4 - I(1) = 0$
$-6I - 6 = 0$
$-6I = 6$
$I = -1 \text{ A}$
The magnitude of the current is $1 \text{ A}$. The negative sign indicates that the current flows in the opposite direction to the assumed path.

(C) From the given figure,we can observe that the two resistors of $4 \ \Omega$ and $8 \ \Omega$ in the upper branch are in series. Their equivalent resistance is $R_1 = 4 \ \Omega + 8 \ \Omega = 12 \ \Omega$.
This $12 \ \Omega$ resistor is in parallel with the $6 \ \Omega$ resistor connected between the same two nodes. The equivalent resistance $R_p$ of this parallel combination is given by $\frac{1}{R_p} = \frac{1}{12} + \frac{1}{6} = \frac{1+2}{12} = \frac{3}{12} = \frac{1}{4}$,so $R_p = 4 \ \Omega$.
Now,the circuit simplifies to a series combination of the lower $4 \ \Omega$ resistor,the equivalent $4 \ \Omega$ resistor,and the lower $8 \ \Omega$ resistor.
Therefore,the total equivalent resistance between $A$ and $B$ is $R_{AB} = 4 \ \Omega + 4 \ \Omega + 8 \ \Omega = 16 \ \Omega$.

(B) Consider one spoke $(OP)$ as shown in the diagram.
The induced emf across one spoke $(OP)$ is given by:
$e = \frac{B \omega l^2}{2}$
Given:
$B = 0.4 \,G = 0.4 \times 10^{-4} \,T$
$l = 1 \,m$
$f = 120 \,rpm = \frac{120}{60} \,rps = 2 \,rps$
Angular velocity $\omega = 2 \pi f = 2 \pi \times 2 = 4 \pi \,rad/s$
Substituting the values:
$e = \frac{1}{2} \times (0.4 \times 10^{-4}) \times (4 \pi) \times (1)^2$
$e = 0.2 \times 10^{-4} \times 4 \times 3.14159$
$e = 0.8 \times 3.14159 \times 10^{-4} \approx 2.51 \times 10^{-4} \,V$
Since all spokes are connected in parallel between the axle and the rim,the net induced emf is equal to the emf induced in a single spoke:
$e_{\text{Net}} = e = 2.51 \times 10^{-4} \,V$

(A) $PN$ junction diode is forward biased when the potential at the $P$-side $(V_P)$ is greater than the potential at the $N$-side $(V_N)$,i.e.,$V_P > V_N$.
For option $(A)$: $V_P = 0 \text{ V}$,$V_N = -3 \text{ V}$. Since $0 > -3$,$V_P > V_N$. This is forward biased.
For option $(B)$: $V_P = -4 \text{ V}$,$V_N = -2 \text{ V}$. Since $-4 < -2$,$V_P < V_N$. This is reverse biased.
For option $(C)$: $V_P = 2 \text{ V}$,$V_N = 5 \text{ V}$. Since $2 < 5$,$V_P < V_N$. This is reverse biased.
For option $(D)$: $V_P = -2 \text{ V}$,$V_N = 2 \text{ V}$. Since $-2 < 2$,$V_P < V_N$. This is reverse biased.
Therefore,option $(A)$ is the correct answer.

(D) The voltage drop across the load resistance $R_C$ is given by Ohm's law as:
$V_L = I_C \times R_C$
Given that the voltage drop $V_L = 0.8 \; V$ and the load resistance $R_C = 800 \; \Omega$.
Substituting these values into the equation:
$0.8 \; V = I_C \times 800 \; \Omega$
$I_C = \frac{0.8}{800} \; A$
$I_C = 0.001 \; A$
Since $1 \; A = 1000 \; mA$,we have:
$I_C = 0.001 \times 1000 \; mA = 1 \; mA$.
Therefore,the collector current is $1 \; mA$.

(C) The resistance $R$ of a conductor is given by the formula:
$R = \rho \frac{l}{A}$
where $\rho$ is the resistivity,$l$ is the length,and $A$ is the cross-sectional area.
Given that both conductors are made of the same material,their resistivities are equal: $\rho_{1} = \rho_{2} = \rho$.
Given that they have the same length: $l_{1} = l_{2} = l$.
Given that they have the same resistance: $R_{1} = R_{2} = R$.
From the formula,we can express the area as $A = \frac{\rho l}{R}$.
Since $\rho$,$l$,and $R$ are identical for both conductors,the cross-sectional areas must also be equal:
$A_{1} = \frac{\rho l}{R}$ and $A_{2} = \frac{\rho l}{R}$
Therefore,$\frac{A_{1}}{A_{2}} = 1$.

(B) In an interference pattern,the condition for constructive interference is that the path difference $\Delta x$ must be an integral multiple of the wavelength $\lambda$,i.e.,$\Delta x = n\lambda$ where $n = 0, 1, 2, \dots$.
The central maximum occurs at the point where the path difference between the two waves is zero $(\Delta x = 0)$.
The relationship between phase difference $\phi$ and path difference $\Delta x$ is given by $\phi = \frac{2\pi}{\lambda} \Delta x$.
Substituting $\Delta x = 0$ into the equation,we get $\phi = \frac{2\pi}{\lambda} \times 0 = 0$.
Therefore,for the central maximum,the phase difference is $0$.

(C) To apply Kirchhoff's Voltage Law $(KVL)$ to the loop $BCDEB$,we traverse the loop starting from point $B$ in the clockwise direction.
$1$. Moving from $B$ to $C$ through resistor $R_2$ in the direction of current $i_2$,the potential drop is $-i_2 R_2$.
$2$. Moving from $C$ to $D$ through battery $E_2$,we go from the positive terminal to the negative terminal,so the potential change is $-E_2$.
$3$. Moving from $D$ to $E$ through battery $E_3$,we go from the negative terminal to the positive terminal,so the potential change is $+E_3$.
$4$. Moving from $E$ to $B$ through resistor $R_1$ against the direction of current $i_3$,the potential change is $+i_3 R_1$.
Summing these changes to zero:
$-i_2 R_2 - E_2 + E_3 + i_3 R_1 = 0$
Multiplying by $-1$ gives:
$i_2 R_2 + E_2 - E_3 - i_3 R_1 = 0$

(C) The magnitude of the induced emf $(\varepsilon)$ is given by Faraday's law of induction: $|\varepsilon| = |\frac{d\phi}{dt}|$.
Given the magnetic flux $\phi = 5t^2 + 3t + 16$.
Differentiating $\phi$ with respect to time $t$:
$\frac{d\phi}{dt} = \frac{d}{dt}(5t^2 + 3t + 16) = 10t + 3$.
To find the magnitude of the induced emf at the fourth second $(t = 4 \, s)$:
$|\varepsilon| = 10(4) + 3 = 40 + 3 = 43 \, V$.

(A) The general equation for a plane electromagnetic wave is given by $B = B_{0} \sin (kx + \omega t)$.
Comparing this with the given equation $B_{y} = 2 \times 10^{-7} \sin (\pi \times 10^{3} x + 3 \pi \times 10^{11} t) \; T$, we identify the wave number $k$ as:
$k = \pi \times 10^{3} \; \text{rad/m}$.
The relationship between the wave number $k$ and the wavelength $\lambda$ is given by $k = \frac{2\pi}{\lambda}$.
Substituting the value of $k$:
$\pi \times 10^{3} = \frac{2\pi}{\lambda}$.
Solving for $\lambda$:
$\lambda = \frac{2\pi}{\pi \times 10^{3}} = 2 \times 10^{-3} \; m$.

(A) For an electric dipole with dipole moment $\overrightarrow{p}$,the electric field at a point on the equatorial plane at a large distance $r$ ($r >> a$,where $2a$ is the separation between charges) is given by the formula:
$\overrightarrow{E} = -\frac{1}{4 \pi \epsilon_{0}} \frac{\overrightarrow{p}}{r^{3}}$
The negative sign indicates that the direction of the electric field at the equatorial point is opposite to the direction of the dipole moment vector $\overrightarrow{p}$ (which is from $-q$ to $+q$).

(B) When an iron rod is inserted into the inductor,its self-inductance $L$ increases because the permeability of the core increases.
The inductive reactance is given by $X_L = \omega L$. As $L$ increases,$X_L$ also increases.
The impedance of the circuit is $Z = \sqrt{R^2 + X_L^2}$. Since $X_L$ increases,the total impedance $Z$ of the circuit increases.
The current in the circuit is given by $I = \frac{V}{Z}$. As $Z$ increases,the current $I$ flowing through the circuit decreases.
The power dissipated in the bulb is given by $P = I^2 R$. Since the current $I$ decreases,the power dissipated in the bulb decreases,and therefore,the glow of the light bulb decreases.

(B) The de Broglie wavelength $\lambda$ of an electron is given by the formula:
$\lambda = \frac{12.27}{\sqrt{K}} \; \mathring{A}$,where $K$ is the kinetic energy in $eV$.
Given $K = 144 \; eV$.
Substituting the value of $K$ in the formula:
$\lambda = \frac{12.27}{\sqrt{144}} \; \mathring{A}$
$\lambda = \frac{12.27}{12} \; \mathring{A} = 1.0225 \; \mathring{A}$.
Since $1 \; \mathring{A} = 0.1 \; nm$,we have:
$\lambda = 1.0225 \times 0.1 \; nm = 0.10225 \; nm$.
Converting this to the form $102 \times 10^{-x} \; nm$:
$0.10225 \; nm = 102.25 \times 10^{-3} \; nm$.
Rounding to the nearest value,we get $102 \times 10^{-3} \; nm$.

(C) According to the Bohr model of the hydrogen atom,the total energy of an electron in the $n^{th}$ stationary orbit is given by the formula:
$E_{n} = -\frac{13.6}{n^{2}} \; eV$
where $n$ is the principal quantum number representing the orbit number $(n = 1, 2, 3, \dots)$.

(A) The length of the wire is $L$,which forms the circumference of the circular loop.
Therefore,$L = 2 \pi R$,where $R$ is the radius of the circle.
From this,the radius is $R = \frac{L}{2 \pi}$.
The area $A$ of the circular loop is given by $A = \pi R^{2}$.
Substituting the value of $R$,we get $A = \pi \left( \frac{L}{2 \pi} \right)^{2} = \pi \left( \frac{L^{2}}{4 \pi^{2}} \right) = \frac{L^{2}}{4 \pi}$.
The magnetic moment $M$ of a current-carrying loop is given by $M = I A$.
Substituting the value of $A$,we get $M = I \left( \frac{L^{2}}{4 \pi} \right) = \frac{I L^{2}}{4 \pi} \; A \cdot m^{2}$.

(B) Using the mirror formula: $\frac{1}{v} + \frac{1}{u} = \frac{1}{f}$.
For a concave mirror,the focal length $f$ is taken as negative,so $f_{mirror} = -f$.
The object distance $u$ is also negative,so $u = -1.5 f = -\frac{3 f}{2}$.
Substituting these values into the mirror formula:
$\frac{1}{v} + \frac{1}{-1.5 f} = \frac{1}{-f}$
$\frac{1}{v} - \frac{2}{3 f} = -\frac{1}{f}$
$\frac{1}{v} = -\frac{1}{f} + \frac{2}{3 f}$
$\frac{1}{v} = \frac{-3 + 2}{3 f}$
$\frac{1}{v} = -\frac{1}{3 f}$
Therefore,$v = -3 f$. The image is formed at a distance of $3 f$ in front of the mirror.

(D) The electrostatic force between an electron and a proton is given by Coulomb's Law: $F = \frac{1}{4 \pi \varepsilon_{0}} \frac{e^{2}}{r^{2}}$.
According to Newton's second law,the acceleration of the electron is $a_{e} = \frac{F}{m_{e}}$.
Substituting the expression for force: $a_{e} = \frac{1}{4 \pi \varepsilon_{0}} \frac{e^{2}}{m_{e} r^{2}}$.
Given $r = 1.6 \; \mathring{A} = 1.6 \times 10^{-10} \; m$,$e = 1.6 \times 10^{-19} \; C$,$m_{e} = 9 \times 10^{-31} \; kg$,and $\frac{1}{4 \pi \varepsilon_{0}} = 9 \times 10^{9} \; Nm^{2} C^{-2}$.
$a_{e} = \frac{9 \times 10^{9} \times (1.6 \times 10^{-19})^{2}}{9 \times 10^{-31} \times (1.6 \times 10^{-10})^{2}}$.
$a_{e} = \frac{9 \times 10^{9} \times 2.56 \times 10^{-38}}{9 \times 10^{-31} \times 2.56 \times 10^{-20}}$.
$a_{e} = \frac{10^{9} \times 10^{-38}}{10^{-31} \times 10^{-20}} = \frac{10^{-29}}{10^{-51}} = 10^{22} \; m/s^{2}$.

(C) When an excited nucleus emits $\gamma$-radiation,it transitions from a higher energy state to a lower energy state.
The process is represented as: $_{Z}X^{A} \xrightarrow{\gamma \text{ decay}} _{Z}X^{A} + \gamma$.
Since $\gamma$-rays are high-energy electromagnetic photons and carry no charge or mass,the emission of $\gamma$-radiation does not change the number of protons or neutrons in the nucleus.
Therefore,both the mass number $(A)$ and the atomic number $(Z)$ remain unchanged.

(C) The relationship between the critical angle $C$ and the refractive index $\mu$ is given by $\sin C = \frac{1}{\mu}$.
Given $C = 45^{\circ}$,we have $\sin 45^{\circ} = \frac{1}{\mu}$.
Since $\sin 45^{\circ} = \frac{1}{\sqrt{2}}$,it follows that $\frac{1}{\sqrt{2}} = \frac{1}{\mu}$,which implies $\mu = \sqrt{2}$.
The velocity of light in a medium $v$ is related to the speed of light in vacuum $c$ by $v = \frac{c}{\mu}$.
Substituting the values,$v = \frac{3 \times 10^{8}}{\sqrt{2}} \; m/s$.

(C) The original capacitance of the parallel plate capacitor with air is given by $C_{0} = \frac{\varepsilon_{0} A}{d}$.
When a dielectric slab of thickness $t$ and dielectric constant $K$ is inserted between the plates,the new capacitance $C$ is given by the formula:
$C = \frac{\varepsilon_{0} A}{d - t + \frac{t}{K}}$
Given $t = \frac{d}{2}$ and $K = 4$,we substitute these values into the formula:
$C = \frac{\varepsilon_{0} A}{d - \frac{d}{2} + \frac{d/2}{4}}$
$C = \frac{\varepsilon_{0} A}{\frac{d}{2} + \frac{d}{8}}$
$C = \frac{\varepsilon_{0} A}{\frac{4d + d}{8}} = \frac{\varepsilon_{0} A}{\frac{5d}{8}}$
$C = \frac{8}{5} \frac{\varepsilon_{0} A}{d} = \frac{8}{5} C_{0}$
Therefore,the ratio of the new capacitance to the original capacitance is $\frac{C}{C_{0}} = \frac{8}{5}$.