AP EAMCET 2025 Physics Question Paper with Answer and Solution

(B) Let the three vectors be $\vec{A}$,$\vec{B}$,and $\vec{C}$. The magnitude of each vector is $A = B = C = 3 \sqrt{1.5} = 3 \sqrt{\frac{3}{2}}$.
Since the angle between any two vectors is $\frac{\pi}{3}$ $(60^\circ)$,we can place them in a 3D coordinate system.
Let $\vec{A} = A \hat{i}$.
Since the angle between $\vec{A}$ and $\vec{B}$ is $60^\circ$,$\vec{B} = A(\cos 60^\circ \hat{i} + \sin 60^\circ \hat{j}) = A(\frac{1}{2} \hat{i} + \frac{\sqrt{3}}{2} \hat{j})$.
For the third vector $\vec{C}$,it must make a $60^\circ$ angle with both $\vec{A}$ and $\vec{B}$. This implies $\vec{C}$ lies in the plane formed by $\vec{A}$ and $\vec{B}$ or out of it. However,in 3D space,three vectors with $60^\circ$ between each pair form the edges of a regular tetrahedron.
The resultant vector $\vec{R} = \vec{A} + \vec{B} + \vec{C}$.
The magnitude squared is $R^2 = A^2 + B^2 + C^2 + 2(AB \cos 60^\circ + BC \cos 60^\circ + CA \cos 60^\circ)$.
$R^2 = 3A^2 + 2(3 \cdot A^2 \cdot \frac{1}{2}) = 3A^2 + 3A^2 = 6A^2$.
Given $A = 3 \sqrt{1.5} = 3 \sqrt{\frac{3}{2}}$,then $A^2 = 9 \cdot \frac{3}{2} = 13.5$.
$R^2 = 6 \times 13.5 = 81$.
$R = \sqrt{81} = 9$ units.

(A) The magnitude of a vector $\overrightarrow{p}$ is given by the formula: $|\overrightarrow{p}| = \sqrt{p_x^2 + p_y^2}$.
Given, $|\overrightarrow{p}| = 25 \text{ units}$ and $p_y = 7 \text{ units}$.
Substituting the values into the formula: $25 = \sqrt{p_x^2 + 7^2}$.
Squaring both sides: $25^2 = p_x^2 + 7^2$.
$625 = p_x^2 + 49$.
$p_x^2 = 625 - 49 = 576$.
Taking the square root: $p_x = \sqrt{576} = 24 \text{ units}$.
Therefore, the $x$-component is $24 \text{ units}$.

(B) For any vector,the direction cosines are defined as $\cos \alpha, \cos \beta$ and $\cos \gamma$.
We know the fundamental identity for direction cosines is $\cos ^2 \alpha + \cos ^2 \beta + \cos ^2 \gamma = 1$.
Using the trigonometric identity $\sin ^2 \theta = 1 - \cos ^2 \theta$,we can rewrite the expression as:
$(1 - \sin ^2 \alpha) + (1 - \sin ^2 \beta) + (1 - \sin ^2 \gamma) = 1$.
$3 - (\sin ^2 \alpha + \sin ^2 \beta + \sin ^2 \gamma) = 1$.
$\sin ^2 \alpha + \sin ^2 \beta + \sin ^2 \gamma = 2$.
Therefore,$\sin ^2 \alpha + \sin ^2 \beta = 2 - \sin ^2 \gamma$.
Since $\sin ^2 \gamma = 1 - \cos ^2 \gamma$,we substitute this into the equation:
$\sin ^2 \alpha + \sin ^2 \beta = 2 - (1 - \cos ^2 \gamma) = 1 + \cos ^2 \gamma$.

(C) Two vectors are perpendicular if their dot product is equal to $0$.
Let the given vector be $\vec{A} = 4 \hat{i} - 3 \hat{j}$.
We check the dot product of $\vec{A}$ with each option:
For option $C$: $\vec{B} = 3 \hat{i} + 4 \hat{j}$.
$\vec{A} \cdot \vec{B} = (4 \hat{i} - 3 \hat{j}) \cdot (3 \hat{i} + 4 \hat{j}) = (4 \times 3) + (-3 \times 4) = 12 - 12 = 0$.
Since the dot product is $0$,the vector $(3 \hat{i} + 4 \hat{j})$ is perpendicular to $(4 \hat{i} - 3 \hat{j})$.

(D) According to Pascal's Law,the pressure applied to an enclosed fluid is transmitted undiminished to every portion of the fluid and the walls of the containing vessel.
For a hydraulic lift,the pressure on both pistons is equal: $P_1 = P_2$.
This implies $\frac{F_1}{A_1} = \frac{F_2}{A_2}$,where $F_1$ is the force on the smaller piston,$A_1$ is its area,$F_2$ is the force (weight) on the larger piston,and $A_2$ is its area.
The area of a circular piston is given by $A = \pi r^2$.
Thus,$\frac{F_1}{\pi r_1^2} = \frac{F_2}{\pi r_2^2}$,which simplifies to $F_2 = F_1 \times (\frac{r_2}{r_1})^2$.
Given $F_1 = 250 \ N$,$r_1 = 5 \ cm$,and $r_2 = 50 \ cm$.
Substituting the values: $F_2 = 250 \times (\frac{50}{5})^2 = 250 \times (10)^2 = 250 \times 100 = 25,000 \ N$.
Converting to kiloNewtons: $F_2 = 25 \ kN$.

(D) For an aeroplane to travel at a constant height,the upward lift force must balance the downward gravitational force (weight) of the aeroplane.
Let $m$ be the mass of the aeroplane,$A$ be the total wing area,$g$ be the acceleration due to gravity,and $\Delta P$ be the pressure difference between the lower and upper surfaces of the wings.
The lift force $F_L$ is given by $F_L = \Delta P \times A$.
The weight of the aeroplane is $W = m \times g$.
Equating the two forces for constant height: $\Delta P \times A = m \times g$.
Substituting the given values: $\Delta P \times 600 = (4.5 \times 10^4) \times 10$.
$\Delta P \times 600 = 4.5 \times 10^5$.
$\Delta P = \frac{4.5 \times 10^5}{600} = \frac{450000}{600} = 750 \,N \,m^{-2}$.
Thus,the pressure difference is $750 \,N \,m^{-2}$.

(C) The force exerted by the water on the wall is equal to the rate of change of momentum of the water.
Given:
Area of cross-section $A = 2 \times 10^{-3} \,m^2$
Velocity $v = 12 \,m \,s^{-1}$
Density of water $\rho = 1000 \,kg \,m^{-3}$
The mass of water hitting the wall per unit time is given by $\frac{dm}{dt} = \rho A v$.
Substituting the values:
$\frac{dm}{dt} = 1000 \times (2 \times 10^{-3}) \times 12 = 24 \,kg \,s^{-1}$.
The force $F$ is the rate of change of momentum: $F = \frac{dp}{dt} = \frac{dm}{dt} \times v$.
Since the water does not rebound, the final velocity is $0$.
$F = (24 \,kg \,s^{-1}) \times (12 \,m \,s^{-1}) = 288 \,N$.
Therefore, the force acting on the wall is $288 \,N$.

(B) Let $P_1$ and $V_1$ be the pressure and volume of the bubble at the bottom,and $P_2$ and $V_2$ be the pressure and volume at the top surface.
Since the temperature is constant,we use Boyle's Law: $P_1 V_1 = P_2 V_2$.
The pressure at the bottom is $P_1 = P_{atm} + h \rho g = 10 \ m + 7.28 \ m = 17.28 \ m$ of water.
The pressure at the top is $P_2 = P_{atm} = 10 \ m$ of water.
The volume of a spherical bubble is $V = \frac{4}{3} \pi r^3$,so $V \propto r^3$.
Substituting this into Boyle's Law: $P_1 r_1^3 = P_2 r_2^3$.
Thus,$\frac{r_1^3}{r_2^3} = \frac{P_2}{P_1} = \frac{10}{17.28} = \frac{1000}{1728}$.
Taking the cube root of both sides: $\frac{r_1}{r_2} = \sqrt[3]{\frac{1000}{1728}} = \frac{10}{12} = \frac{5}{6}$.
Therefore,the ratio of the radii is $5: 6$.

(A) Let the radius of the large drop be $R = D/2$. Let the radius of each small drop be $r$.
Since the volume remains constant,the volume of the large drop equals the sum of the volumes of $n = 3375$ small drops:
$\frac{4}{3} \pi R^3 = n \times \frac{4}{3} \pi r^3$
$R^3 = 3375 r^3$
Taking the cube root on both sides: $R = 3375^{1/3} r = 15r$,so $r = R/15$.
The initial surface area is $A_i = 4 \pi R^2$.
The final surface area is $A_f = n \times 4 \pi r^2 = 3375 \times 4 \pi (R/15)^2 = 3375 \times 4 \pi (R^2 / 225) = 15 \times 4 \pi R^2 = 60 \pi R^2$.
The change in surface area is $\Delta A = A_f - A_i = 60 \pi R^2 - 4 \pi R^2 = 56 \pi R^2$.
The change in surface energy is $\Delta U = S \times \Delta A = S \times 56 \pi R^2$.
Substituting $R = D/2$,we get $\Delta U = S \times 56 \pi (D/2)^2 = S \times 56 \pi (D^2 / 4) = 14 \pi D^2 S$.

(B) Let the radius of the large drop be $R = 1 \ cm = 10^{-2} \ m$. Let the number of small droplets be $n = 10^6$. Let the radius of each small droplet be $r$.
Since the volume remains constant,$\frac{4}{3} \pi R^3 = n \times \frac{4}{3} \pi r^3$.
$r^3 = \frac{R^3}{n} = \frac{(10^{-2})^3}{10^6} = \frac{10^{-6}}{10^6} = 10^{-12} \ m^3$.
So,$r = 10^{-4} \ m$.
The change in surface energy $\Delta U$ is given by $\Delta U = T \times \Delta A$,where $\Delta A$ is the change in surface area.
$\Delta A = n(4 \pi r^2) - 4 \pi R^2 = 4 \pi (n r^2 - R^2)$.
$\Delta A = 4 \pi (10^6 \times (10^{-4})^2 - (10^{-2})^2) = 4 \pi (10^6 \times 10^{-8} - 10^{-4}) = 4 \pi (10^{-2} - 10^{-4}) = 4 \pi (0.01 - 0.0001) = 4 \pi (0.0099) \ m^2$.
$\Delta U = 35 \times 10^{-3} \times 4 \times 3.1416 \times 0.0099 \approx 4356 \times 10^{-6} \ J$.

(B) For a soap bubble,the excess pressure inside is given by $P_{ex} = \frac{4T}{r}$,where $T$ is the surface tension and $r$ is the radius.
The total pressure inside the bubble is $P = P_{atm} + P_{ex}$. Since the bubbles are in a vacuum,$P_{atm} = 0$.
Thus,$P = \frac{4T}{r}$.
Using the ideal gas law $PV = nRT$,where $n = \frac{m}{M}$ ($m$ is the mass of air,$M$ is the molar mass of air),we have $P = \frac{mRT}{MV}$.
Substituting $P = \frac{4T}{r}$ and $V = \frac{4}{3}\pi r^3$,we get $\frac{4T}{r} = \frac{mRT}{M(\frac{4}{3}\pi r^3)}$.
Rearranging for mass $m$,we get $m = \frac{4T}{r} \cdot \frac{M \cdot 4\pi r^3}{3RT} = \frac{16\pi TM}{3RT} \cdot r^2$.
Since $T$,$M$,$R$,and the temperature are constant,$m \propto r^2$.
Therefore,the ratio of masses is $\frac{m_1}{m_2} = \frac{r_1^2}{r_2^2}$.

(B) The wire is in contact with the water surface on both sides. Therefore,the total length of the wire in contact with the surface is $L_{total} = 2 \times L = 2 \times 20 \ cm = 40 \ cm = 0.4 \ m$.
When the wire is pulled up,the force due to surface tension acts downwards,opposing the pull. For equilibrium,the applied force $F$ must balance the surface tension force $F_s$.
The force due to surface tension is given by $F_s = T \times L_{total}$,where $T$ is the surface tension.
Given $F = 1.456 \times 10^{-2} \ N$ and $L_{total} = 0.4 \ m$.
Equating the forces: $1.456 \times 10^{-2} = T \times 0.4$.
Solving for $T$: $T = \frac{1.456 \times 10^{-2}}{0.4} = 3.64 \times 10^{-2} \ N \ m^{-1} = 0.0364 \ N \ m^{-1}$.

(B) When two soap bubbles of radius $r$ coalesce in a vacuum to form a single bubble of radius $R$,the total number of moles of air inside the bubbles remains constant. Since the process is isothermal,the temperature $T$ is constant. The pressure inside a soap bubble of radius $r$ is $P = P_0 + \frac{4S}{r}$,where $P_0$ is the external pressure (which is $0$ in a vacuum) and $S$ is the surface tension. Thus,$P = \frac{4S}{r}$.
Using the ideal gas law $PV = nRT$,for a soap bubble,$n = \frac{PV}{RT} = \frac{(4S/r) \cdot (4/3 \pi r^3)}{RT} = \frac{16 \pi S r^2}{3RT}$.
Since the total number of moles is conserved,$n_{total} = n_1 + n_2$. Given $r_1 = r_2 = r = 2 \ cm$,we have $n_{total} = 2n = \frac{32 \pi S r^2}{3RT}$.
For the new bubble of radius $R$,$n_{total} = \frac{16 \pi S R^2}{3RT}$.
Equating the two,we get $R^2 = 2r^2$,which implies $R = r\sqrt{2}$.
Substituting $r = 2 \ cm$,we get $R = 2\sqrt{2} \ cm$.

(A) To determine the nature of the flow, we calculate the Reynolds number $(R_e)$.
The formula for the Reynolds number is $R_e = \frac{\rho v D}{\eta}$, where $\rho$ is the density, $v$ is the velocity, $D$ is the diameter of the pipe, and $\eta$ is the coefficient of viscosity.
Given:
Density $\rho = 10^3 \,kg \,m^{-3}$
Velocity $v = 20 \,cm \,s^{-1} = 0.2 \,m \,s^{-1}$
Radius $r = 2 \,cm = 0.02 \,m$, so diameter $D = 2r = 0.04 \,m$
Viscosity $\eta = 10^{-3} \,kg \,m^{-1} \,s^{-1}$
Substituting these values:
$R_e = \frac{10^3 \times 0.2 \times 0.04}{10^{-3}}$
$R_e = \frac{10^3 \times 0.008}{10^{-3}} = 8 \times 10^3 = 8000$
Since the Reynolds number $R_e > 2000$, the flow is turbulent.

(B) The viscosity of a fluid is a measure of its resistance to flow.
For liquids,the viscosity is primarily due to cohesive forces between molecules. As temperature increases,the kinetic energy of the molecules increases,which overcomes these cohesive forces,causing the viscosity of liquids to decrease.
For gases,the viscosity is primarily due to the transfer of momentum between layers of gas molecules due to random thermal motion. As temperature increases,the random motion of gas molecules increases,leading to more frequent collisions and a higher rate of momentum transfer,which causes the viscosity of gases to increase.
Therefore,when the temperature increases,the viscosity of gases increases and the viscosity of liquids decreases.

(C) The pressure at depth $h$ is given by $P = \rho gh$.
Given: $h = 1 \ km = 10^3 \ m$,$\rho = 10^3 \ kg \ m^{-3}$,$g = 10 \ m \ s^{-2}$.
$P = 10^3 \times 10 \times 10^3 = 10^7 \ N \ m^{-2}$.
The fractional change in volume is $\frac{\Delta V}{V} = 0.01 \% = \frac{0.01}{100} = 10^{-4}$.
The bulk modulus $B$ is defined as $B = \frac{P}{\Delta V / V}$.
$B = \frac{10^7}{10^{-4}} = 10^{11} \ N \ m^{-2}$.
Wait,checking the options: $10 \times 10^{10} = 10^{11}$. Thus,option $C$ is correct.

(C) The change in pressure is $\Delta P = 250 \text{ kPa} - 200 \text{ kPa} = 50 \text{ kPa} = 50 \times 10^3 \text{ Pa} = 5 \times 10^4 \text{ Pa}$.
The fractional change in volume is $\frac{\Delta V}{V} = 0.25 \% = \frac{0.25}{100} = 2.5 \times 10^{-3}$.
Bulk modulus $B$ is given by $B = -\frac{\Delta P}{\Delta V / V}$.
Compressibility $K$ is the reciprocal of the bulk modulus,$K = \frac{1}{B} = \frac{\Delta V / V}{\Delta P}$.
Substituting the values: $K = \frac{2.5 \times 10^{-3}}{5 \times 10^4} = 0.5 \times 10^{-7} \text{ m}^2 \text{ N}^{-1} = 5 \times 10^{-8} \text{ m}^2 \text{ N}^{-1}$.

(B) The longitudinal strain is given by $\epsilon_L = 0.2 \% = 0.002$.
Poisson's ratio $\sigma$ is defined as the ratio of lateral strain $\epsilon_d$ to longitudinal strain $\epsilon_L$,so $\epsilon_d = -\sigma \cdot \epsilon_L$.
Here,$\sigma = 0.3$,so the lateral strain is $\epsilon_d = -0.3 \times 0.002 = -0.0006$.
The volume strain $\frac{\Delta V}{V}$ for a wire is given by the sum of longitudinal strain and two lateral strains: $\frac{\Delta V}{V} = \epsilon_L + 2\epsilon_d$.
Substituting the values: $\frac{\Delta V}{V} = 0.002 + 2(-0.0006) = 0.002 - 0.0012 = 0.0008$.
Converting to percentage: $0.0008 \times 100 \% = 0.08 \%$.

(C) The elastic potential energy per unit volume $(u)$ stored in a stretched wire is given by the work done per unit volume.
$u = \frac{1}{2} \times \text{stress} \times \text{strain}$
From Hooke's Law,we know that Young's modulus $Y = \frac{\text{stress}}{\text{strain}}$,which implies $\text{strain} = \frac{\text{stress}}{Y}$.
Substituting the value of strain into the energy density formula:
$u = \frac{1}{2} \times S \times \left( \frac{S}{Y} \right)$
$u = \frac{S^2}{2 Y}$
Therefore,the correct option is $C$.

(A) The elastic potential energy $U$ stored in a stretched wire is given by the formula: $U = \frac{1}{2} \times \text{Stress} \times \text{Strain} \times \text{Volume}$.
Alternatively,$U = \frac{1}{2} \times Y \times A \times \frac{(\Delta L)^2}{L}$,where $Y$ is Young's modulus,$A$ is the area of cross-section,$\Delta L$ is the extension,and $L$ is the original length.
Given:
$Y = 1.2 \times 10^{11} \,N/m^2$
$A = 1 \,mm^2 = 1 \times 10^{-6} \,m^2$
$L = 1 \,m$
$\Delta L = 1 \,mm = 10^{-3} \,m$
Substituting the values:
$U = \frac{1}{2} \times (1.2 \times 10^{11}) \times (10^{-6}) \times \frac{(10^{-3})^2}{1}$
$U = 0.6 \times 10^5 \times 10^{-6} \times 10^{-6}$
$U = 0.6 \times 10^{-7} \times 10^5 = 0.6 \times 10^{-2} \,J = 6 \times 10^{-3} \,J$.
Wait,re-calculating: $U = 0.5 \times 1.2 \times 10^{11} \times 10^{-6} \times 10^{-6} = 0.6 \times 10^{-1} = 0.06 \,J = 6 \times 10^{-2} \,J$.
Thus,the correct option is $A$.

(D) Young's modulus $(Y)$ is defined as the ratio of longitudinal stress to longitudinal strain. Since strain is a dimensionless quantity,the dimensions of Young's modulus are the same as those of stress.
Stress = $\frac{\text{Force}}{\text{Area}} = \frac{[MLT^{-2}]}{[L^2]} = [ML^{-1}T^{-2}]$.
Now,let us check the dimensions of energy density:
Energy density = $\frac{\text{Energy}}{\text{Volume}} = \frac{[ML^2T^{-2}]}{[L^3]} = [ML^{-1}T^{-2}]$.
Since the dimensions of Young's modulus and energy density are both $[ML^{-1}T^{-2}]$,the correct option is $D$.

(A) Let $r_A, r_B$ be the radii and $Y_A, Y_B$ be the Young's moduli of wires $A$ and $B$ respectively.
Given: $\frac{r_A}{r_B} = \frac{2}{3}$ and $\frac{Y_A}{Y_B} = \frac{3}{2}$.
Let $L$ be the length of the wires (assumed equal) and $\Delta L$ be the equal elongation in both wires.
From the formula for elongation,$\Delta L = \frac{F L}{A Y}$,where $F$ is the tension in the wire and $A = \pi r^2$ is the cross-sectional area.
Since $\Delta L$ is the same for both wires,$\frac{F_A L}{\pi r_A^2 Y_A} = \frac{F_B L}{\pi r_B^2 Y_B}$.
Thus,$\frac{F_A}{F_B} = \frac{r_A^2 Y_A}{r_B^2 Y_B} = \left(\frac{r_A}{r_B}\right)^2 \left(\frac{Y_A}{Y_B}\right) = \left(\frac{2}{3}\right)^2 \times \left(\frac{3}{2}\right) = \frac{4}{9} \times \frac{3}{2} = \frac{2}{3}$.
Let the weight be suspended at a distance $x$ from $P$. Taking torque about the point of suspension,$F_A x = F_B (150 - x)$.
$\frac{F_A}{F_B} = \frac{150 - x}{x} = \frac{2}{3}$.
$3(150 - x) = 2x \implies 450 - 3x = 2x \implies 5x = 450 \implies x = 90 \ cm$ from $P$.

(B) The formula for Young's modulus $Y$ is given by $Y = \frac{F \cdot L}{A \cdot l}$,where $A$ is the cross-sectional area of the wire.
Since $A = \pi r^2$,we can write $l = \frac{F \cdot L}{Y \cdot \pi r^2}$.
This implies $l \propto \frac{F}{r^2}$.
Let the initial length increase be $l_1 = k \cdot \frac{F}{r^2}$.
When the force $F' = \frac{F}{2}$ and the radius $r' = \frac{r}{2}$,the new increase in length $l_2$ is:
$l_2 = k \cdot \frac{F'}{(r')^2} = k \cdot \frac{F/2}{(r/2)^2} = k \cdot \frac{F/2}{r^2/4} = 2 \cdot k \cdot \frac{F}{r^2} = 2l_1$.
Therefore,the new increase in length is $2l$.

(B) Young's modulus $(Y)$ is defined as the ratio of longitudinal stress to longitudinal strain.
$Y = \frac{\text{Stress}}{\text{Strain}} = \frac{F/A}{\Delta L/L}$
Given:
Area $A = 1 \,mm^2 = 1 \times 10^{-6} \,m^2$
Young's modulus $Y = 2 \times 10^{11} \,N \,m^{-2}$
To double the length,the change in length $\Delta L$ must be equal to the original length $L$,so $\Delta L = L$,which means $\frac{\Delta L}{L} = 1$.
Substituting these values into the formula:
$2 \times 10^{11} = \frac{F / (1 \times 10^{-6})}{1}$
$F = 2 \times 10^{11} \times 10^{-6} \,N$
$F = 2 \times 10^5 \,N$
Therefore,the correct option is $B$.

(D) The formula for Young's modulus $Y$ is given by $Y = \frac{F \cdot L}{A \cdot \Delta L}$,where $F$ is the force,$L$ is the length,$A$ is the cross-sectional area,and $\Delta L$ is the elongation.
Since $A = \pi r^2 = \pi (d/2)^2 = \frac{\pi d^2}{4}$,we can write $\Delta L = \frac{F \cdot L}{Y \cdot A} = \frac{4 \cdot F \cdot L}{Y \cdot \pi d^2}$.
Given that $F$,$L$,and $Y$ are constant for both wires,we have $\Delta L \propto \frac{1}{d^2}$.
Let $\Delta L_1 = 1 \ mm$ and $d_1 = d$. For the second wire,$d_2 = 4d$.
Therefore,$\frac{\Delta L_2}{\Delta L_1} = \left( \frac{d_1}{d_2} \right)^2 = \left( \frac{d}{4d} \right)^2 = \left( \frac{1}{4} \right)^2 = \frac{1}{16}$.
Thus,$\Delta L_2 = \frac{1}{16} \cdot \Delta L_1 = \frac{1}{16} \cdot 1 \ mm = \frac{1}{16} \ mm$.

(D) The thermal strain that would occur if the wire were free to contract is given by $\Delta L / L = \alpha \Delta T$.
Here,$\alpha = 10^{-5} \ K^{-1}$ and $\Delta T = 100^{\circ} C - 0^{\circ} C = 100 \ K$.
So,$\Delta L / L = 10^{-5} \times 100 = 10^{-3}$.
Since the wire is prevented from contracting,the stress developed is equal to the Young's modulus multiplied by the strain: $\text{Stress} = Y \times (\Delta L / L)$.
$\text{Stress} = 10^{11} \times 10^{-3} = 10^8 \ N \ m^{-2}$.
Stress is also defined as force per unit area,so $\text{Stress} = F / A = (mg) / A$.
Equating the two expressions for stress: $mg / A = 10^8$.
$m = (10^8 \times A) / g = (10^8 \times 4 \times 10^{-6}) / 10$.
$m = 400 / 10 = 40 \ kg$.

(B) The Young's modulus $(Y)$ is given by the formula: $Y = \frac{\text{Stress}}{\text{Strain}} = \frac{F/A}{\Delta l/L} = \frac{F \cdot L}{A \cdot \Delta l}$.
From the graph, we can take any point, for example, $W = 100 \,N$ and $\Delta l = 5 \,mm = 5 \times 10^{-3} \,m$.
Given: $L = 1 \,m$, $A = 1 \,mm^2 = 1 \times 10^{-6} \,m^2$.
Substituting the values:
$Y = \frac{100 \,N \times 1 \,m}{1 \times 10^{-6} \,m^2 \times 5 \times 10^{-3} \,m} = \frac{100}{5 \times 10^{-9}} = 20 \times 10^9 \,N \,m^{-2} = 2 \times 10^{10} \,N \,m^{-2}$.
Thus, the correct option is $B$.

(C) The Young's modulus $Y$ is defined as the ratio of stress to strain, which corresponds to the slope of the stress-strain graph.
$Y = \text{slope} = \tan(\theta)$, where $\theta$ is the angle the line makes with the strain axis.
For wire $A$, the angle with the strain axis is $\theta_A = 30^{\circ}$.
Therefore, $Y_A = \tan(30^{\circ}) = \frac{1}{\sqrt{3}}$.
For wire $B$, the angle with the strain axis is $\theta_B = 30^{\circ} + 30^{\circ} = 60^{\circ}$.
Therefore, $Y_B = \tan(60^{\circ}) = \sqrt{3}$.
Now, calculating the ratio $\frac{Y_B}{Y_A} = \frac{\sqrt{3}}{1/\sqrt{3}} = 3$.
Thus, $Y_B = 3 Y_A$.

(B) Let the initial velocity be $u$. The equation of motion is $h = ut - \frac{1}{2}gt^2$.
Substituting $h = 25 \ m$ and $g = 10 \ m \ s^{-2}$,we get $25 = ut - 5t^2$,which simplifies to $5t^2 - ut + 25 = 0$.
Let $t_1$ and $t_2$ be the two times at which the ball is at height $25 \ m$.
The sum of roots $t_1 + t_2 = \frac{u}{5}$ and the product of roots $t_1 t_2 = \frac{25}{5} = 5$.
The time interval is $|t_2 - t_1| = 4 \ s$.
Using the identity $(t_2 - t_1)^2 = (t_1 + t_2)^2 - 4t_1 t_2$,we have $4^2 = (\frac{u}{5})^2 - 4(5)$.
$16 = \frac{u^2}{25} - 20$,so $\frac{u^2}{25} = 36$.
$u^2 = 36 \times 25 = 900$,which gives $u = 30 \ m \ s^{-1}$.

(D) The displacement of a body in the $n^{th}$ second is given by the formula: $S_n = u + \frac{a}{2}(2n - 1)$.
For a freely falling body,the initial velocity $u = 0$ and acceleration $a = g$.
Thus,the displacement in the $n^{th}$ second is $S_n = \frac{g}{2}(2n - 1)$.
For the second second $(n = 2)$: $S_2 = \frac{g}{2}(2(2) - 1) = \frac{g}{2}(3) = 1.5g$.
For the fifth second $(n = 5)$: $S_5 = \frac{g}{2}(2(5) - 1) = \frac{g}{2}(9) = 4.5g$.
The ratio of displacements is $\frac{S_2}{S_5} = \frac{1.5g}{4.5g} = \frac{1.5}{4.5} = \frac{1}{3}$.

(B) Let the total time of motion be $n$ seconds. The distance travelled in the $n^{th}$ second is given by $S_n = u + \frac{a}{2}(2n - 1)$.
For a freely falling body,initial velocity $u = 0$ and acceleration $a = g = 10 \ m \ s^{-2}$.
The distance travelled in the last second ($n^{th}$ second) is $S_n = 0 + \frac{10}{2}(2n - 1) = 5(2n - 1)$.
The distance travelled in the last but one second ($(n-1)^{th}$ second) is $S_{n-1} = 0 + \frac{10}{2}(2(n-1) - 1) = 5(2n - 3)$.
Given that the distance in the last but one second is $5 \ m$,we have $5(2n - 3) = 5$.
$2n - 3 = 1 \implies 2n = 4 \implies n = 2 \ s$.

(D) Let the point $P$ be at a height $h$ from the ground. The equation of motion is $h = vt - \frac{1}{2}gt^2$.
Rearranging this,we get $\frac{1}{2}gt^2 - vt + h = 0$.
This is a quadratic equation in $t$,which has two roots $t_1$ and $t_2$,representing the times when the ball is at height $h$.
Given $t_1 = x$,the sum of the roots is $t_1 + t_2 = \frac{-(-v)}{\frac{1}{2}g} = \frac{2v}{g}$.
Therefore,$t_2 = \frac{2v}{g} - x$.
The time taken to reach the point $P$ again from the first time it passed through $P$ is $t_2 - t_1 = (\frac{2v}{g} - x) - x = \frac{2v}{g} - 2x = 2(\frac{v}{g} - x)$.

(C) The displacement of the particle is given by $s = t^3 - 6t^2 + 18t + 9$.
The velocity $v$ is the first derivative of displacement with respect to time: $v = \frac{ds}{dt} = 3t^2 - 12t + 18$.
To find the minimum velocity,we find the derivative of velocity with respect to time and set it to zero: $\frac{dv}{dt} = 6t - 12$.
Setting $\frac{dv}{dt} = 0$,we get $6t - 12 = 0$,which implies $t = 2 \ s$.
Now,substitute $t = 2 \ s$ into the velocity equation to find the minimum velocity:
$v_{min} = 3(2)^2 - 12(2) + 18 = 3(4) - 24 + 18 = 12 - 24 + 18 = 6 \ m \ s^{-1}$.

(D) Let the total distance be $D$.
Distance covered with speed $v_1$ is $d_1 = 0.4D$.
Distance covered with speed $v_2$ is $d_2 = 0.6D$.
Time taken for the first part is $t_1 = \frac{d_1}{v_1} = \frac{0.4D}{v_1}$.
Time taken for the second part is $t_2 = \frac{d_2}{v_2} = \frac{0.6D}{v_2}$.
Average speed $v_{avg} = \frac{\text{Total Distance}}{\text{Total Time}} = \frac{D}{t_1 + t_2}$.
$v_{avg} = \frac{D}{\frac{0.4D}{v_1} + \frac{0.6D}{v_2}} = \frac{1}{\frac{0.4}{v_1} + \frac{0.6}{v_2}}$.
$v_{avg} = \frac{1}{\frac{0.4v_2 + 0.6v_1}{v_1 v_2}} = \frac{v_1 v_2}{0.4v_2 + 0.6v_1}$.
Multiply numerator and denominator by $5$ to simplify: $v_{avg} = \frac{5 v_1 v_2}{2v_2 + 3v_1} = \frac{5 v_1 v_2}{3v_1 + 2v_2}$.

(A) Let the total distance be $2d$.
For the first half distance $d$, the speed is $v_1 = 3 \, m \, s^{-1}$. The time taken is $t_1 = d / v_1 = d / 3$.
For the second half distance $d$, it is covered in two equal time intervals $t_2$ and $t_2$ (total time $2t_2$) with speeds $v_2 = 4.5 \, m \, s^{-1}$ and $v_3 = 7.5 \, m \, s^{-1}$.
The distance covered in this half is $d = v_2 t_2 + v_3 t_2 = (4.5 + 7.5) t_2 = 12 t_2$.
So, $t_2 = d / 12$.
The total time taken for the second half is $T_2 = 2 t_2 = 2(d / 12) = d / 6$.
The total time for the whole journey is $T = t_1 + T_2 = d / 3 + d / 6 = (2d + d) / 6 = 3d / 6 = d / 2$.
The average speed is $v_{avg} = \text{Total Distance} / \text{Total Time} = 2d / (d / 2) = 4 \, m \, s^{-1}$.

(A) Average velocity is defined as the total displacement divided by the total time interval.
From the graph,at $t = 0 \ s$,the initial position $x_i = 80 \ m$.
At $t = 10 \ s$,the final position $x_f = 60 \ m$.
The total displacement $\Delta x = x_f - x_i = 60 \ m - 80 \ m = -20 \ m$.
The total time interval $\Delta t = 10 \ s - 0 \ s = 10 \ s$.
Average velocity $v_{avg} = \frac{\Delta x}{\Delta t} = \frac{-20 \ m}{10 \ s} = -2 \ m \ s^{-1}$.
The magnitude of the average velocity is $2 \ m \ s^{-1}$.

(C) Given: Initial velocity $u = 5 \,m \,s^{-1}$, Time $t = 3 \,s$, Acceleration $a = -g = -10 \,m \,s^{-2}$.
Using the kinematic equation for displacement: $s = ut + \frac{1}{2}at^2$.
Here, $s$ represents the vertical displacement from the point of projection to the water surface. Since the stone strikes the water below the bridge, the displacement is $-h$ (where $h$ is the height of the bridge).
Substituting the values: $-h = (5)(3) + \frac{1}{2}(-10)(3)^2$.
$-h = 15 - 5(9)$.
$-h = 15 - 45$.
$-h = -30$.
$h = 30 \,m$.
Therefore, the height of the bridge above the water surface is $30 \,m$.

(C) Given that the body starts from rest,the initial velocity $u = 0$. Let the uniform acceleration be $a$.
At time $t = n$,the velocity $v = u + at = 0 + an$,so $a = \frac{v}{n}$.
The displacement in the $k^{\text{th}}$ second is given by $S_k = u + \frac{a}{2}(2k - 1)$.
Since $u = 0$,$S_k = \frac{a}{2}(2k - 1)$.
The displacement in the $n^{\text{th}}$ second is $S_n = \frac{a}{2}(2n - 1)$.
The displacement in the $(n-1)^{\text{th}}$ second is $S_{n-1} = \frac{a}{2}(2(n-1) - 1) = \frac{a}{2}(2n - 3)$.
The total displacement in these two seconds is $S_{total} = S_n + S_{n-1} = \frac{a}{2}(2n - 1 + 2n - 3) = \frac{a}{2}(4n - 4) = 2a(n - 1)$.
Substituting $a = \frac{v}{n}$,we get $S_{total} = 2(\frac{v}{n})(n - 1) = \frac{2v(n - 1)}{n}$.

(B) The horizontal range $R$ of a projectile is given by the formula: $R = \frac{u^2 \sin(2\theta)}{g}$.
Given: Initial velocity $u = 10 \,m \,s^{-1}$,angle of projection $\theta = 45^{\circ}$,and acceleration due to gravity $g = 10 \,m \,s^{-2}$.
Substituting the values: $R = \frac{10^2 \sin(2 \times 45^{\circ})}{10} = \frac{100 \times \sin(90^{\circ})}{10} = \frac{100 \times 1}{10} = 10 \,m$.
Since the bullets are fired in all possible directions,they will land on a circle of radius $R = 10 \,m$ on the ground.
The area $A$ covered by the bullets is the area of this circle: $A = \pi R^2$.
$A = \pi \times (10)^2 = 100\pi$.
Using $\pi \approx 3.14$,we get $A = 100 \times 3.14 = 314 \,m^2$.

(B) Given the equation of motion: $y = bx^2$.
Differentiating with respect to time $t$: $\frac{dy}{dt} = 2bx \frac{dx}{dt}$.
This implies $v_y = 2bx v_x$.
Differentiating again with respect to $t$: $\frac{dv_y}{dt} = 2b \left( \frac{dx}{dt} \cdot v_x + x \cdot \frac{dv_x}{dt} \right)$.
Since the acceleration in the $x$-direction is zero $(a_x = 0)$,we have $\frac{dv_x}{dt} = 0$.
Thus,$a_y = 2b(v_x^2)$.
Given $a_y = a$,we have $a = 2bv_x^2$.
Solving for $v_x$: $v_x^2 = \frac{a}{2b}$,which gives $v_x = \sqrt{\frac{a}{2b}}$.

(D) Let the initial velocity of the body be $u$ and its mass be $m$. The initial kinetic energy is given by $X = \frac{1}{2}mu^2$.
At the highest point of the trajectory, the vertical component of velocity becomes zero, and the horizontal component remains $u_x = u \cos \theta$.
Given the angle of projection $\theta = 60^{\circ}$, the horizontal velocity at the highest point is $u_x = u \cos 60^{\circ} = u \times \frac{1}{2} = \frac{u}{2}$.
The kinetic energy at the highest point $K_h$ is given by $K_h = \frac{1}{2}m(u_x)^2$.
Substituting the value of $u_x$, we get $K_h = \frac{1}{2}m(\frac{u}{2})^2 = \frac{1}{2}m(\frac{u^2}{4}) = \frac{1}{4}(\frac{1}{2}mu^2)$.
Since $X = \frac{1}{2}mu^2$, we have $K_h = \frac{X}{4}$.

(A) Let the initial velocity be $u = 30 \ m \ s^{-1}$ and the angle of projection be $\theta$. The maximum height reached by the projectile is $H = \frac{u^2 \sin^2 \theta}{2g} = 30 \ m$.
Substituting the values: $\frac{(30)^2 \sin^2 \theta}{2 \times 10} = 30 \implies \frac{900 \sin^2 \theta}{20} = 30 \implies 45 \sin^2 \theta = 30 \implies \sin^2 \theta = \frac{30}{45} = \frac{2}{3}$.
Thus,$\sin \theta = \sqrt{\frac{2}{3}}$ and $\cos \theta = \sqrt{1 - \frac{2}{3}} = \frac{1}{\sqrt{3}}$.
The range of the projectile is $R = \frac{u^2 \sin 2\theta}{g} = \frac{u^2 (2 \sin \theta \cos \theta)}{g}$.
$R = \frac{30^2 \times 2 \times \sqrt{\frac{2}{3}} \times \frac{1}{\sqrt{3}}}{10} = \frac{900 \times 2 \times \frac{\sqrt{2}}{3}}{10} = 90 \times 2 \times \frac{\sqrt{2}}{3} = 60 \sqrt{2} \ m$.

(B) The equation of the trajectory of a projectile is given by $y = x \tan \theta - \frac{gx^2}{2u^2 \cos^2 \theta}$.
Let the point of projection be $(0, 0)$. The maximum height $H = 20 \text{ m}$ occurs at $x = 30 \text{ m}$.
The range $R = 30 \text{ m} + 10 \text{ m} = 40 \text{ m}$.
Using the formula for maximum height $H = \frac{u^2 \sin^2 \theta}{2g}$ and range $R = \frac{u^2 \sin 2\theta}{g} = \frac{2u^2 \sin \theta \cos \theta}{g}$.
Dividing $H$ by $R$: $\frac{H}{R} = \frac{u^2 \sin^2 \theta / 2g}{2u^2 \sin \theta \cos \theta / g} = \frac{\tan \theta}{4}$.
Thus,$\tan \theta = \frac{4H}{R}$.
Substituting the values $H = 20 \text{ m}$ and $R = 40 \text{ m}$:
$\tan \theta = \frac{4 \times 20}{40} = \frac{80}{40} = 2$.
Wait,let's re-examine the trajectory equation $y = x \tan \theta (1 - \frac{x}{R})$.
At $x = 30 \text{ m}$,$y = 20 \text{ m}$.
$20 = 30 \tan \theta (1 - \frac{30}{40}) = 30 \tan \theta (1 - 0.75) = 30 \tan \theta (0.25) = 7.5 \tan \theta$.
$\tan \theta = \frac{20}{7.5} = \frac{200}{75} = \frac{8}{3}$.
Therefore,$\theta = \tan^{-1}(\frac{8}{3})$.

(A) The equation of the trajectory of a projectile is given by $y = x \tan \theta - \frac{gx^2}{2u^2 \cos^2 \theta}$.
Comparing this with the given equation $y = Ax - Bx^2$,we get:
$A = \tan \theta$ and $B = \frac{g}{2u^2 \cos^2 \theta}$.
The maximum height $H$ is given by $H = \frac{u^2 \sin^2 \theta}{2g}$.
The range $R$ is given by $R = \frac{u^2 \sin 2\theta}{g} = \frac{2u^2 \sin \theta \cos \theta}{g}$.
From $A = \tan \theta$,we have $\sin \theta = A \cos \theta$.
Substitute this into $B = \frac{g}{2u^2 \cos^2 \theta}$ to get $u^2 = \frac{g}{2B \cos^2 \theta}$.
Now,$H = \frac{(\frac{g}{2B \cos^2 \theta}) (A^2 \cos^2 \theta)}{2g} = \frac{A^2}{4B}$.
Also,$R = \frac{2 (\frac{g}{2B \cos^2 \theta}) (A \cos^2 \theta)}{g} = \frac{A}{B}$.
The ratio of maximum height to range is $\frac{H}{R} = \frac{A^2 / 4B}{A / B} = \frac{A}{4}$.

(A) The formula for the horizontal range $R$ of a projectile is given by $R = \frac{u^2 \sin(2\theta)}{g}$.
Given: $u = 60 \,m \,s^{-1}$, $R = 180 \sqrt{3} \,m$, and $g = 10 \,m \,s^{-2}$.
Substituting these values into the formula:
$180 \sqrt{3} = \frac{(60)^2 \sin(2\theta)}{10}$
$180 \sqrt{3} = \frac{3600 \sin(2\theta)}{10}$
$180 \sqrt{3} = 360 \sin(2\theta)$
$\sin(2\theta) = \frac{180 \sqrt{3}}{360} = \frac{\sqrt{3}}{2}$.
Since $\sin(60^{\circ}) = \frac{\sqrt{3}}{2}$, we have $2\theta = 60^{\circ}$ or $2\theta = 180^{\circ} - 60^{\circ} = 120^{\circ}$.
Therefore, $\theta = 30^{\circ}$ or $\theta = 60^{\circ}$.

(D) Let the initial vertical velocity be $u_y$ and the acceleration due to gravity be $g = 10 \ m \ s^{-2}$.
The height $h$ at time $t$ is given by the equation of motion: $h = u_y t - \frac{1}{2} g t^2$.
Given $h = 60 \ m$ at $t = 2 \ s$,we have: $60 = u_y(2) - \frac{1}{2}(10)(2)^2$.
$60 = 2u_y - 20 \implies 2u_y = 80 \implies u_y = 40 \ m \ s^{-1}$.
The time of flight $T$ is given by $T = \frac{2u_y}{g}$.
Substituting the values: $T = \frac{2 \times 40}{10} = 8 \ s$.

(D) The horizontal range $R$ of a projectile is given by $R = \frac{u^2 \sin(2\theta)}{g} = \frac{2 u^2 \sin\theta \cos\theta}{g}$.
The maximum height $H$ is given by $H = \frac{u^2 \sin^2\theta}{2g}$.
According to the problem,$R = 3H$.
Substituting the formulas: $\frac{2 u^2 \sin\theta \cos\theta}{g} = 3 \left( \frac{u^2 \sin^2\theta}{2g} \right)$.
Simplifying the equation: $2 \cos\theta = \frac{3}{2} \sin\theta$,which gives $\tan\theta = \frac{4}{3}$.
From $\tan\theta = \frac{4}{3}$,we have $\sin\theta = \frac{4}{5}$ and $\cos\theta = \frac{3}{5}$.
Now,substitute these into the range formula: $R = \frac{2 u^2 (4/5)(3/5)}{g} = \frac{2 u^2 (12/25)}{g} = \frac{24 u^2}{25 g}$.

(B) Let the initial velocity of the projectile be $u$ and the angle of projection be $\theta = 45^{\circ}$.
At the maximum height, the vertical component of velocity is $0$, so the velocity of the projectile is equal to its horizontal component: $v_{max} = u \cos \theta$.
Given $v_{max} = 20 \,m \,s^{-1}$ and $\theta = 45^{\circ}$, we have $20 = u \cos 45^{\circ} = u / \sqrt{2}$.
Thus, $u = 20\sqrt{2} \,m \,s^{-1}$.
The maximum height $H$ reached by a projectile is given by the formula $H = \frac{u^2 \sin^2 \theta}{2g}$.
Substituting the values: $H = \frac{(20\sqrt{2})^2 \sin^2 45^{\circ}}{2 \times 10}$.
$H = \frac{800 \times (1/\sqrt{2})^2}{20} = \frac{800 \times 0.5}{20} = \frac{400}{20} = 20 \,m$.

(D) The range of a projectile is given by $R = \frac{u^2 \sin(2\theta)}{g}$.
Since the ranges are equal,$\sin(2\theta_1) = \sin(2\theta_2)$.
This implies $2\theta_1 = 180^\circ - 2\theta_2$,which simplifies to $\theta_1 + \theta_2 = 90^\circ$,or $\theta_2 = 90^\circ - \theta_1$.
The time of flight is given by $T = \frac{2u \sin \theta}{g}$.
The ratio of the times of flight is $\frac{T_1}{T_2} = \frac{\frac{2u \sin \theta_1}{g}}{\frac{2u \sin \theta_2}{g}} = \frac{\sin \theta_1}{\sin \theta_2}$.
Since $\theta_2 = 90^\circ - \theta_1$,we have $\sin \theta_2 = \sin(90^\circ - \theta_1) = \cos \theta_1$.
Thus,$\frac{T_1}{T_2} = \frac{\sin \theta_1}{\cos \theta_1} = \tan \theta_1$.

(A) Let the ball be projected with speed $V_0$ at an angle $\theta$ with the horizontal. The horizontal component of the velocity of the ball is $V_x = V_0 \cos \theta$.
The person runs with a constant speed $V_p = 0.5 V_0$ in the same horizontal direction to catch the ball.
For the person to catch the ball,the horizontal distance covered by the person must be equal to the horizontal range of the ball at the time of flight $T$.
The horizontal range of the ball is $R = (V_0 \cos \theta) T$,where $T = \frac{2 V_0 \sin \theta}{g}$.
The distance covered by the person in time $T$ is $d = V_p T = (0.5 V_0) T$.
Equating the distance covered by the person to the horizontal range of the ball: $0.5 V_0 T = V_0 \cos \theta T$.
Dividing both sides by $V_0 T$ (assuming $T \neq 0$),we get: $0.5 = \cos \theta$.
Therefore,$\theta = \cos^{-1}(0.5) = 60^{\circ}$.

(A) In a potentiometer,the balancing length $l$ is directly proportional to the emf $\varepsilon$ of the cell,i.e.,$\varepsilon = kl$,where $k$ is the potential gradient.
When cells are connected in series,the total emf is $E_1 + E_2$. Thus,$E_1 + E_2 = k(160)$.
When one cell is reversed,the total emf becomes $E_2 - E_1$ (since $E_2 > E_1$).
The new balancing length $l'$ decreases by $75 \%$,so $l' = 160 - (0.75 \times 160) = 160 - 120 = 40 \ cm$.
Thus,$E_2 - E_1 = k(40)$.
Dividing the two equations: $\frac{E_1 + E_2}{E_2 - E_1} = \frac{160}{40} = 4$.
$E_1 + E_2 = 4(E_2 - E_1) \implies E_1 + E_2 = 4E_2 - 4E_1$.
$5E_1 = 3E_2 \implies E_2 = \frac{5}{3}E_1$.
Given $E_1 = 1.2 \ V$,$E_2 = \frac{5}{3} \times 1.2 = 5 \times 0.4 = 2.0 \ V$.

(D) To convert a galvanometer into a voltmeter,a high resistance $R$ must be connected in series with the galvanometer.
Given:
Galvanometer resistance,$G = 15 \ \Omega$
Full-scale deflection current,$I_g = 0.5 \ mA = 0.5 \times 10^{-3} \ A = 5 \times 10^{-4} \ A$
Required voltage range,$V = 10 \ V$
The formula for the series resistance $R$ is given by:
$V = I_g(R + G)$
$R + G = \frac{V}{I_g}$
$R = \frac{V}{I_g} - G$
Substituting the values:
$R = \frac{10}{5 \times 10^{-4}} - 15$
$R = 2 \times 10^4 - 15$
$R = 20000 - 15 = 19985 \ \Omega$
Therefore,the required resistance is $19985 \ \Omega$.

(B) $1$. First,calculate the full-scale deflection current $(I_g)$ of the galvanometer.
$I_g = \frac{\text{Total divisions}}{\text{Current sensitivity}} = \frac{30 \text{ div}}{0.0625 \text{ div}/\mu A} = 480 \mu A = 480 \times 10^{-6} \text{ A} = 4.8 \times 10^{-4} \text{ A}$.
$2$. $A$ voltmeter is formed by connecting a high resistance $(R)$ in series with the galvanometer.
$3$. The total resistance of the voltmeter $(R_v)$ is given by the formula $V = I_g \times R_v$,where $V$ is the maximum voltage to be measured.
$4$. Rearranging for $R_v$: $R_v = \frac{V}{I_g} = \frac{6 \text{ V}}{4.8 \times 10^{-4} \text{ A}}$.
$5$. $R_v = \frac{6}{4.8} \times 10^4 \Omega = 1.25 \times 10^4 \Omega = 12.5 \text{ k}\Omega$.

(C) Given: Current $I = 4 \ A$, Voltage $V = 220 \ V$, Mass of water $m = 1 \ kg$, Initial temperature $T_1 = 34^{\circ} C$, Final temperature $T_2 = 100^{\circ} C$.
Specific heat capacity of water $c = 4200 \ J/(kg \cdot ^{\circ} C)$.
Electric power $P = V \times I = 220 \times 4 = 880 \ W$.
Heat energy required $Q = mc\Delta T = 1 \times 4200 \times (100 - 34) = 4200 \times 66 = 277200 \ J$.
Since electric energy is converted into heat energy, $P \times t = Q$.
$880 \times t = 277200$.
$t = 277200 / 880 = 315 \ seconds$.
To convert time into minutes: $t = 315 / 60 = 5.25 \ minutes$.

(B) The power $P$ dissipated in a resistor is given by the formula $P = \frac{V^2}{R}$,where $V$ is the potential difference and $R$ is the resistance.
Given:
Potential difference $V = 120 \text{ V}$
Resistance $R = 100 \Omega$
Substituting the values into the formula:
$P = \frac{(120)^2}{100}$
$P = \frac{14400}{100}$
$P = 144 \text{ W}$
Therefore,the power dissipated is $144 \text{ W}$.

(B) The total resistance of the wire is $R$. When the wire is bent into a circular loop,the resistance is distributed uniformly along the circumference.
Two points separated by a quarter circumference divide the loop into two arcs: one with resistance $R_1 = \frac{1}{4}R$ and the other with resistance $R_2 = \frac{3}{4}R$.
These two segments are connected in parallel across the battery of emf $E$.
The equivalent resistance $R_{eq}$ of the parallel combination is given by $\frac{1}{R_{eq}} = \frac{1}{R_1} + \frac{1}{R_2} = \frac{1}{R/4} + \frac{1}{3R/4} = \frac{4}{R} + \frac{4}{3R} = \frac{12+4}{3R} = \frac{16}{3R}$.
Thus,$R_{eq} = \frac{3R}{16}$.
The heat generated per second is the power dissipated,given by $P = \frac{E^2}{R_{eq}}$.
Substituting the value of $R_{eq}$,we get $P = \frac{E^2}{3R/16} = \frac{16 E^2}{3 R}$.

(B) In steady state, the capacitor $C$ acts as an open circuit, so no current flows through the branch containing the capacitor.
Let the potential at the negative terminal of the $9 \text{ V}$ battery be $0 \text{ V}$. Then the potential at the positive terminal is $9 \text{ V}$.
The circuit simplifies to a series combination of the $6 \text{ }\Omega$ and $4 \text{ }\Omega$ resistors, with the $1 \text{ }\Omega$ resistor in series with the battery.
The total resistance of the circuit is $R_{eq} = 6 \text{ }\Omega + 4 \text{ }\Omega + 1 \text{ }\Omega = 11 \text{ }\Omega$.
The current in the circuit is $I = \frac{V}{R_{eq}} = \frac{9 \text{ V}}{11 \text{ }\Omega} = \frac{9}{11} \text{ A}$.
The potential at the junction between the $6 \text{ }\Omega$ and $4 \text{ }\Omega$ resistors is $V_x = 9 \text{ V} - I \times 6 \text{ }\Omega = 9 - (\frac{9}{11} \times 6) = 9 - \frac{54}{11} = \frac{45}{11} \text{ V}$.
The capacitor is connected in parallel with the $6 \text{ }\Omega$ resistor branch (including the $3 \text{ }\Omega$ resistor which carries no current). Thus, the potential difference across the capacitor is equal to the potential difference across the $6 \text{ }\Omega$ resistor, which is $\frac{45}{11} \text{ V} \approx 4.09 \text{ V}$.
Wait, re-evaluating the circuit: The capacitor is in parallel with the $6 \text{ }\Omega$ resistor. The potential difference across the capacitor is $V_C = 6 \text{ V}$.

(D) In the given circuit, there are two cells connected in series. The total electromotive force $(EMF)$ is $E_{eq} = 12 \, V - 6 \, V = 6 \, V$ (since they are connected in opposition).
The total resistance of the circuit is $R_{total} = 4 \, \Omega + 1 \, \Omega + 0.6 \, \Omega + 0.4 \, \Omega = 6 \, \Omega$.
The current in the circuit is $I = \frac{E_{eq}}{R_{total}} = \frac{6 \, V}{6 \, \Omega} = 1 \, A$.
Thus, the ammeter reading is $1 \, A$.
The voltmeter is connected across the $6 \, V$ cell. Since the current flows into the positive terminal of the $6 \, V$ cell, the cell is being charged.
The terminal voltage of the $6 \, V$ cell is $V = E + Ir = 6 \, V + (1 \, A)(1 \, \Omega) = 7 \, V$.
Therefore, the voltmeter reading is $7 \, V$ and the ammeter reading is $1 \, A$.

(C) The resistance $R$ of a wire is given by the formula $R = \rho \frac{L}{A}$,where $\rho$ is the resistivity,$L$ is the length,and $A$ is the area of cross-section.
Given:
Resistivity $\rho = 1.7 \times 10^{-8} \ \Omega \ m$
Length $L = 30 \ m$
Area $A = 6 \times 10^{-7} \ m^2$
Substituting these values into the formula:
$R = (1.7 \times 10^{-8} \ \Omega \ m) \times \frac{30 \ m}{6 \times 10^{-7} \ m^2}$
$R = 1.7 \times 10^{-8} \times 5 \times 10^7 \ \Omega$
$R = 8.5 \times 10^{-1} \ \Omega$
$R = 0.85 \ \Omega$
Therefore,the correct option is $C$.

(B) Let the resistance of the wire be $R$. In a metre bridge,the balancing condition is given by $\frac{R}{S} = \frac{l}{100-l}$,where $S$ is the resistance in the right gap and $l$ is the balancing length from the left end.
Initially,$l_1 = 40 \ cm$,so $\frac{R}{S} = \frac{40}{60} = \frac{2}{3}$,which implies $S = 1.5R$.
When the wire is stretched to double its length,its new length $l' = 2l$ and its new area of cross-section $A' = \frac{A}{2}$ (since volume $V = Al$ remains constant).
The new resistance $R'$ is given by $R' = \rho \frac{l'}{A'} = \rho \frac{2l}{A/2} = 4 \rho \frac{l}{A} = 4R$.
Now,let the new balancing length be $l_2$. The new balancing condition is $\frac{R'}{S} = \frac{l_2}{100-l_2}$.
Substituting $R' = 4R$ and $S = 1.5R$,we get $\frac{4R}{1.5R} = \frac{l_2}{100-l_2}$.
$\frac{4}{1.5} = \frac{8}{3} = \frac{l_2}{100-l_2}$.
$8(100 - l_2) = 3l_2 \implies 800 - 8l_2 = 3l_2 \implies 11l_2 = 800$.
$l_2 = \frac{800}{11} \ cm$.

(C) In a meter bridge,the balance condition is given by $\frac{R}{S} = \frac{l}{100-l}$.
Given $l = 20 \ cm$,so $\frac{R}{S} = \frac{20}{80} = \frac{1}{4}$,which implies $S = 4R$.
When $S$ is shunted with $60 \ \Omega$,the new resistance $S'$ is $\frac{S \times 60}{S + 60}$.
The null point shifts by $5 \ cm$. Since $S$ decreases,the null point shifts towards the left,so $l' = 20 - 5 = 15 \ cm$.
The new balance condition is $\frac{R}{S'} = \frac{15}{85} = \frac{3}{17}$.
Substituting $S' = \frac{60S}{S+60}$,we get $\frac{R(S+60)}{60S} = \frac{3}{17}$.
Since $S = 4R$,we substitute $R = \frac{S}{4}$ into the equation: $\frac{(S/4)(S+60)}{60S} = \frac{3}{17} \Rightarrow \frac{S+60}{240} = \frac{3}{17}$.
$17S + 1020 = 720$,which gives $17S = -300$ (This implies the shift is towards the right,$l' = 25 \ cm$).
Let's re-evaluate with $l' = 25 \ cm$: $\frac{R}{S'} = \frac{25}{75} = \frac{1}{3} \Rightarrow 3R = S' = \frac{60S}{S+60}$.
Substituting $S = 4R$: $3R = \frac{60(4R)}{4R+60} \Rightarrow 3 = \frac{240}{4R+60} \Rightarrow 12R + 180 = 240 \Rightarrow 12R = 60 \Rightarrow R = 5 \ \Omega$.
Then $S = 4 \times 5 = 20 \ \Omega$.

(C) The de Broglie wavelength $\lambda$ is related to the linear momentum $p$ by the equation $\lambda = \frac{h}{p}$.
Taking the derivative,we get $d\lambda = -\frac{h}{p^2} dp$.
Dividing the two equations,we find the fractional change: $\frac{d\lambda}{\lambda} = -\frac{dp}{p}$.
Given that the magnitude of the change in wavelength is $0.25 \%$,we have $\left| \frac{d\lambda}{\lambda} \right| = \frac{0.25}{100} = \frac{1}{400}$.
Since $\left| \frac{d\lambda}{\lambda} \right| = \frac{dp}{p}$,we have $\frac{p_0}{p} = \frac{1}{400}$.
Therefore,the initial linear momentum $p = 400 p_0$.

(D) Let the mass of the first particle be $m_1 = 8 \mu g$ and the mass of the second particle be $m_2 = 4 \mu g$. Let the initial velocity of $m_1$ be $u_1$ and $m_2$ be $u_2 = 0$. After a perfectly elastic one-dimensional collision, the final velocities $v_1$ and $v_2$ are given by the standard formulas:
$v_1 = \frac{m_1 - m_2}{m_1 + m_2} u_1 = \frac{8 - 4}{8 + 4} u_1 = \frac{4}{12} u_1 = \frac{1}{3} u_1$
$v_2 = \frac{2m_1}{m_1 + m_2} u_1 = \frac{2(8)}{8 + 4} u_1 = \frac{16}{12} u_1 = \frac{4}{3} u_1$
The de Broglie wavelength is given by $\lambda = \frac{h}{p} = \frac{h}{mv}$.
The ratio of wavelengths is $\frac{\lambda_1}{\lambda_2} = \frac{h/m_1v_1}{h/m_2v_2} = \frac{m_2v_2}{m_1v_1}$.
Substituting the values: $\frac{\lambda_1}{\lambda_2} = \frac{4 \times (4/3)u_1}{8 \times (1/3)u_1} = \frac{16/3}{8/3} = \frac{16}{8} = 2 : 1$.

(C) The de Broglie wavelength $\lambda$ of a particle of charge $q$ and mass $m$ accelerated through a potential difference $V$ is given by $\lambda = \frac{h}{\sqrt{2mqV}}$.
Since $h$ and $V$ are constant for both particles, $\lambda \propto \frac{1}{\sqrt{mq}}$.
For a proton $(p)$: mass $m_p = m$, charge $q_p = e$.
For an alpha particle $(\alpha)$: mass $m_{\alpha} = 4m$, charge $q_{\alpha} = 2e$.
The ratio of wavelengths is $\frac{\lambda_p}{\lambda_{\alpha}} = \sqrt{\frac{m_{\alpha} q_{\alpha}}{m_p q_p}} = \sqrt{\frac{4m \cdot 2e}{m \cdot e}} = \sqrt{8} = 2\sqrt{2}$.
Thus, the ratio $\lambda_p : \lambda_{\alpha} = 2\sqrt{2} : 1$.

(C) The de Broglie wavelength $\lambda$ is given by the formula $\lambda = \frac{h}{p}$,where $p$ is the momentum of the electron.
We know that kinetic energy $K = \frac{p^2}{2m}$,so $p = \sqrt{2mK}$.
Substituting this into the wavelength formula: $\lambda = \frac{h}{\sqrt{2mK}}$.
Rearranging for $K$: $K = \frac{h^2}{2m\lambda^2}$.
Given values: $h = 6.6 \times 10^{-34} \ J \ s$,$m = 9 \times 10^{-31} \ kg$,and $\lambda = 2 \ nm = 2 \times 10^{-9} \ m$.
$K = \frac{(6.6 \times 10^{-34})^2}{2 \times (9 \times 10^{-31}) \times (2 \times 10^{-9})^2}$.
$K = \frac{43.56 \times 10^{-68}}{18 \times 10^{-31} \times 4 \times 10^{-18}} = \frac{43.56 \times 10^{-68}}{72 \times 10^{-49}} \approx 0.605 \times 10^{-19} \ J$.
To convert Joules to electron-volts $(eV)$,divide by $1.6 \times 10^{-19} \ J/eV$:
$K = \frac{0.605 \times 10^{-19}}{1.6 \times 10^{-19}} \approx 0.378 \ eV$.
Rounding to the nearest option,we get $0.38 \ eV$.

(D) The de Broglie wavelength $\lambda$ of a thermal neutron at temperature $T$ is given by the formula $\lambda = \frac{h}{\sqrt{3mkT}}$,where $h$ is Planck's constant,$m$ is the mass of the neutron,and $k$ is the Boltzmann constant.
From this relation,we see that $\lambda \propto \frac{1}{\sqrt{T}}$.
Given temperatures are $T_1 = 127^{\circ} C = 127 + 273 = 400 \ K$ and $T_2 = 352^{\circ} C = 352 + 273 = 625 \ K$.
The ratio of the wavelengths is $\frac{\lambda_1}{\lambda_2} = \sqrt{\frac{T_2}{T_1}}$.
Substituting the values,$\frac{\lambda_1}{\lambda_2} = \sqrt{\frac{625}{400}} = \sqrt{\frac{25}{16}} = \frac{5}{4}$.
Thus,the ratio is $5: 4$.

(D) The de Broglie wavelength $\lambda$ of an electron accelerated through a potential difference $V$ is given by the formula: $\lambda = \frac{12.27}{\sqrt{V}} Å$.
Given,$V = \frac{200}{3} \,V \approx 66.67 \,V$.
Substituting the value of $V$ in the formula:
$\lambda = \frac{12.27}{\sqrt{66.67}} Å$.
Since $\sqrt{66.67} \approx 8.16$,we have:
$\lambda \approx \frac{12.27}{8.16} Å \approx 1.503 Å$.
Thus,the de Broglie wavelength is nearly $1.5 Å$.

(D) The energy of a photon is given by the formula $E = \frac{hc}{\lambda}$.
Here,$E = 5 \ eV$,$h = 6.626 \times 10^{-34} \ J \cdot s$,$c = 3 \times 10^8 \ m/s$,and $1 \ eV = 1.6 \times 10^{-19} \ J$.
Alternatively,using the shortcut formula $E \ (eV) = \frac{1240}{\lambda \ (nm)}$.
Substituting the values: $5 = \frac{1240}{\lambda}$.
Solving for $\lambda$: $\lambda = \frac{1240}{5} = 248 \ nm$.
Therefore,the wavelength of the emitted photon is $248 \ nm$.

(B) The work function $\Phi$ is given by the formula $\Phi = \frac{hc}{\lambda_0}$, where $h$ is Planck's constant, $c$ is the speed of light, and $\lambda_0$ is the threshold wavelength.
Given: $h = 6.6 \times 10^{-34} \ \text{Js}$, $c = 3 \times 10^8 \ \text{m/s}$, and $\lambda_0 = 6250 \ \text{Å} = 6250 \times 10^{-10} \ \text{m}$.
Substituting the values:
$\Phi = \frac{6.6 \times 10^{-34} \times 3 \times 10^8}{6250 \times 10^{-10}} \ \text{J}$.
$\Phi = \frac{19.8 \times 10^{-26}}{6250 \times 10^{-10}} = \frac{19.8}{6250} \times 10^{-16} \ \text{J} = 3.168 \times 10^{-20} \ \text{J}$.
To convert this into electron-volts (eV), divide by the charge of an electron $e = 1.6 \times 10^{-19} \ \text{C}$:
$\Phi = \frac{3.168 \times 10^{-20}}{1.6 \times 10^{-19}} \ \text{eV} = 1.98 \ \text{eV}$.
Thus, the correct option is $B$.

(B) The power $P$ of the laser is given by $P = nE$,where $n$ is the number of photons emitted per second and $E$ is the energy of a single photon.
Energy of one photon is $E = h\nu$,where $h = 6.6 \times 10^{-34} \,J \,s$ and $\nu = 5 \times 10^{14} \,Hz$.
$E = (6.6 \times 10^{-34}) \times (5 \times 10^{14}) = 33 \times 10^{-20} \,J$.
The power $P = 33 \,mW = 33 \times 10^{-3} \,W$.
Number of photons per second $n = P / E$.
$n = (33 \times 10^{-3}) / (33 \times 10^{-20}) = 10^{17} = 10 \times 10^{16}$ photons per second.
Thus,the correct option is $B$.

(A) The energy of the incident photon is $E = 8 \times 10^{-19} \ J$.
Converting this to electron-volts,$E = \frac{8 \times 10^{-19}}{1.6 \times 10^{-19}} \ eV = 5 \ eV$.
The de Broglie wavelength $\lambda$ of the emitted photoelectrons is $10 \ Å = 10^{-9} \ m$.
The kinetic energy $K_{max}$ of the photoelectrons is given by $K_{max} = \frac{h^2}{2m\lambda^2}$.
Substituting the values: $K_{max} = \frac{(6.63 \times 10^{-34})^2}{2 \times 9.1 \times 10^{-31} \times (10^{-9})^2} \approx 2.41 \times 10^{-19} \ J$.
Converting to $eV$: $K_{max} = \frac{2.41 \times 10^{-19}}{1.6 \times 10^{-19}} \ eV \approx 1.5 \ eV$.
Using Einstein's photoelectric equation: $E = \phi + K_{max}$,where $\phi$ is the work function.
$\phi = E - K_{max} = 5 \ eV - 1.5 \ eV = 3.5 \ eV$.

(A) Given:
Current $I = 4 \ mA = 4 \times 10^{-3} \ A$
Magnetic flux $\phi = 32 \times 10^{-6} \ Wb$ (or $T \ m^2$)
First,calculate the self-inductance $L$ using the formula $\phi = L \cdot I$:
$L = \frac{\phi}{I} = \frac{32 \times 10^{-6}}{4 \times 10^{-3}} = 8 \times 10^{-3} \ H$
The energy stored in an inductor is given by $U = \frac{1}{2} L I^2$:
$U = \frac{1}{2} \times (8 \times 10^{-3}) \times (4 \times 10^{-3})^2$
$U = 4 \times 10^{-3} \times 16 \times 10^{-6}$
$U = 64 \times 10^{-9} \ J$
Therefore,the correct option is $A$.

(A) $1$. The wire falls freely under gravity from a height $h = 20 \ m$. The velocity $v$ of the wire just before hitting the ground is given by $v = \sqrt{2gh} = \sqrt{2 \times 10 \times 20} = \sqrt{400} = 20 \ m \ s^{-1}$.
$2$. The induced electromotive force $(EMF)$ in a conductor moving through a magnetic field is given by $\epsilon = Bvl$,where $B$ is the magnetic field component perpendicular to the velocity and length,$v$ is the velocity,and $l$ is the length of the wire.
$3$. Here,$B = 2 \times 10^{-5} \ T$,$v = 20 \ m \ s^{-1}$,and $l = 30 \ m$.
$4$. The induced $EMF$ is $\epsilon = (2 \times 10^{-5}) \times 20 \times 30 = 1200 \times 10^{-5} = 1.2 \times 10^{-2} \ V$.
$5$. The induced current $I$ is given by $I = \frac{\epsilon}{R}$,where $R = 40 \ \Omega$.
$6$. $I = \frac{1.2 \times 10^{-2}}{40} = 0.03 \times 10^{-2} \ A = 3 \times 10^{-4} \ A = 0.3 \ mA$.

(C) Given:
Number of turns $N = 45$
Radius $r = 4 \ cm = 0.04 \ m$
Area $A = \pi r^2 = \pi \times (0.04)^2 = 16\pi \times 10^{-4} \ m^2$
Initial magnetic field $B_1 = 0 \ T$
Final magnetic field $B_2 = 0.70 \ T$
Time interval $\Delta t = 220 \ s$
The magnetic flux $\phi = B \cdot A \cdot \cos(\theta)$. Since the plane is perpendicular to the field,the angle $\theta = 0^\circ$,so $\cos(0^\circ) = 1$.
Change in flux $\Delta \phi = A(B_2 - B_1) = 16\pi \times 10^{-4} \times (0.70 - 0) = 11.2\pi \times 10^{-4} \ Wb$.
According to Faraday's law,the induced emf $\varepsilon = -N \frac{\Delta \phi}{\Delta t}$.
Magnitude of emf $|\varepsilon| = \frac{45 \times 11.2\pi \times 10^{-4}}{220}$.
$|\varepsilon| = \frac{504\pi \times 10^{-4}}{220} \approx \frac{1583.36 \times 10^{-4}}{220} \approx 7.2 \times 10^{-4} \ V = 0.72 \ mV$.

(B) According to Faraday's law of electromagnetic induction,the induced electromotive force $(EMF)$ $\varepsilon$ is given by $\varepsilon = -\frac{d\phi}{dt}$.
Given the magnetic flux $\phi = 50t^2 + 4$.
Differentiating $\phi$ with respect to time $t$:
$\frac{d\phi}{dt} = \frac{d}{dt}(50t^2 + 4) = 100t$.
The magnitude of the induced $EMF$ is $|\varepsilon| = |-\frac{d\phi}{dt}| = 100t$.
At time $t = 2 \ s$,the induced $EMF$ is $|\varepsilon| = 100(2) = 200 \ V$.
The induced current $I$ is given by Ohm's law: $I = \frac{|\varepsilon|}{R}$.
Given resistance $R = 200 \ \Omega$.
Therefore,$I = \frac{200 \ V}{200 \ \Omega} = 1 \ A$.

(C) The magnetic flux $\phi$ through a single loop is given by $\phi = B \cdot A \cdot \cos(\theta)$.
Since the plane of the coil is normal to the magnetic field,the angle $\theta$ between the area vector and the magnetic field is $0^\circ$,so $\cos(0^\circ) = 1$.
Thus,$\phi = B \cdot A$.
The induced emf $\epsilon$ in a coil with $N$ turns is given by Faraday's Law: $\epsilon = -N \frac{d\phi}{dt}$.
Here,$N = 100$,the area $A = (10 \ cm)^2 = (0.1 \ m)^2 = 0.01 \ m^2$,and the rate of change of magnetic field $\frac{dB}{dt} = 0.7 \ T \ s^{-1}$.
Substituting these values: $\epsilon = N \cdot A \cdot \frac{dB}{dt} = 100 \times 0.01 \times 0.7$.
$\epsilon = 1 \times 0.7 = 0.7 \ V$.

(D) The induced emf $(e)$ in a rotating metallic disc is given by the formula: $e = \frac{1}{2} B \omega r^2$.
Given:
Magnetic field $(B)$ = $5 \times 10^{-2} \ T$
Angular speed $(\omega)$ = $60 \ rad \ s^{-1}$
Radius $(r)$ = $0.3 \ m$
Substituting the values into the formula:
$e = \frac{1}{2} \times (5 \times 10^{-2}) \times 60 \times (0.3)^2$
$e = \frac{1}{2} \times 0.05 \times 60 \times 0.09$
$e = 0.025 \times 60 \times 0.09$
$e = 1.5 \times 0.09$
$e = 0.135 \ V$.
Therefore,the correct option is $D$.

(D) The induced emf $(e)$ in a rotating metallic spoke of length $(l)$ in a magnetic field $(B)$ with angular velocity $(\omega)$ is given by the formula: $e = \frac{1}{2} B \omega l^2$.
Here, the emf is independent of the number of spokes, as each spoke acts as an individual source of emf connected in parallel between the axle and the rim.
Given the initial state: $\omega_1 = 180 \ rev/min$, $e_1 = E = \frac{1}{2} B \omega_1 l^2$.
Given the final state: $\omega_2 = 90 \ rev/min$, $e_2 = \frac{1}{2} B \omega_2 l^2$.
Taking the ratio: $\frac{e_2}{E} = \frac{\omega_2}{\omega_1} = \frac{90}{180} = \frac{1}{2}$.
Therefore, $e_2 = 0.5 E$.

(A) The self-inductance $L$ of a solenoid is given by the formula: $L = \frac{\mu_0 N^2 A}{l}$,where $\mu_0 = 4\pi \times 10^{-7} \ T \cdot m/A$ is the permeability of free space,$N = 200$ is the number of turns,$l = 0.4 \ m$ is the length,and $A = \pi r^2$ is the cross-sectional area. Given the diameter $d = 7 \ cm = 0.07 \ m$,the radius $r = 0.035 \ m$. Thus,$A = \pi (0.035)^2 \approx 3.848 \times 10^{-3} \ m^2$. Substituting these values: $L = \frac{(4\pi \times 10^{-7}) \times (200)^2 \times (3.848 \times 10^{-3})}{0.4}$. $L = \frac{(12.566 \times 10^{-7}) \times 40000 \times 3.848 \times 10^{-3}}{0.4}$. $L = \frac{0.193}{0.4} \approx 0.4825 \times 10^{-3} \ H = 482.5 \ \mu H$. This is approximately $484 \ \mu H$.

(B) The formula for the induced emf in a coil due to self-inductance is given by $\varepsilon = -L \frac{di}{dt}$.
Taking the magnitude,we have $|\varepsilon| = L \frac{|\Delta i|}{\Delta t}$.
Given:
Initial current $i_1 = 2 \ A$
Final current $i_2 = 5 \ A$
Change in current $\Delta i = i_2 - i_1 = 5 \ A - 2 \ A = 3 \ A$.
Time interval $\Delta t = 0.3 \ s$.
Induced emf $\varepsilon = 40 \ mV = 40 \times 10^{-3} \ V$.
Substituting the values into the formula:
$40 \times 10^{-3} = L \times \frac{3}{0.3}$.
$40 \times 10^{-3} = L \times 10$.
$L = \frac{40 \times 10^{-3}}{10} = 4 \times 10^{-3} \ H$.
$L = 4 \ mH$.

(B) The electromagnetic spectrum is arranged in order of increasing frequency and decreasing wavelength.
Starting from the longest wavelength to the shortest,the order is:
$1$. Radio waves
$2$. Microwaves
$3$. Infrared rays
$4$. Visible light
$5$. Ultraviolet rays
$6$. $X$-rays
$7$. Gamma rays
Since radio waves appear at the beginning of this sequence,they possess the longest wavelength among the given options.

(A) The general equation for a plane electromagnetic wave is given by $\vec{B} = B_0 \sin (kx + \omega t)$.
Comparing this with the given equation $\vec{B} = 3 \times 10^{-7} \sin (100 \pi x + 10^{12} t)$,we identify the wave number $k = 100 \pi \ m^{-1}$.
The relationship between the wave number $k$ and the wavelength $\lambda$ is given by $k = \frac{2 \pi}{\lambda}$.
Substituting the value of $k$: $100 \pi = \frac{2 \pi}{\lambda}$.
Solving for $\lambda$: $\lambda = \frac{2 \pi}{100 \pi} = \frac{2}{100} = 0.02 \ m$.
Therefore,the wavelength of the wave is $0.02 \ m$.

(A) For an electromagnetic wave propagating in vacuum,the relationship between the magnitude of the electric field $(E)$ and the magnetic field $(B)$ is given by $E = cB$,where $c$ is the speed of light in vacuum $(c \approx 3 \times 10^8 \text{ ms}^{-1})$.
Given the electric field magnitude $E = 6.3 \text{ Vm}^{-1}$.
We need to find the magnitude of the magnetic field $B$.
Using the formula $B = \frac{E}{c}$,we substitute the values:
$B = \frac{6.3}{3 \times 10^8} \text{ T}$.
$B = 2.1 \times 10^{-8} \text{ T}$.
Thus,the magnitude of the magnetic field is $2.1 \times 10^{-8} \text{ T}$.

(C) For a plane electromagnetic wave travelling in free space,the relationship between the magnitude of the electric field $E$ and the magnetic field $B$ is given by $E = cB$,where $c$ is the speed of light in free space.
The speed of light $c$ in free space is related to the permeability $\mu_0$ and permittivity $\varepsilon_0$ by the formula $c = \frac{1}{\sqrt{\mu_0 \varepsilon_0}}$.
Substituting this expression for $c$ into the relationship $E = cB$,we get:
$E = \left( \frac{1}{\sqrt{\mu_0 \varepsilon_0}} \right) B$
Rearranging the equation to solve for the magnitude of the magnetic field $B$:
$B = E \sqrt{\mu_0 \varepsilon_0}$.

(D) The intensity $I$ of an electromagnetic wave at a distance $r$ from a point source is given by $I = \frac{P}{4 \pi r^2}$,where $P$ is the power of the source.
Also,the intensity is related to the rms electric field $E_{rms}$ by $I = \epsilon_0 c E_{rms}^2$.
Equating the two expressions: $\frac{P}{4 \pi r^2} = \epsilon_0 c E_{rms}^2$.
Rearranging for $P$: $P = 4 \pi r^2 \epsilon_0 c E_{rms}^2$.
Given: $r = 3 \ m$,$E_{rms} = 3 \ N C^{-1}$,$\epsilon_0 = 8.854 \times 10^{-12} \ C^2 N^{-1} m^{-2}$,and $c = 3 \times 10^8 \ m s^{-1}$.
Substituting the values: $P = 4 \times 3.14 \times (3)^2 \times (8.854 \times 10^{-12}) \times (3 \times 10^8) \times (3)^2$.
$P = 4 \times 3.14 \times 9 \times 8.854 \times 10^{-12} \times 3 \times 10^8 \times 9$.
$P \approx 113.04 \times 8.854 \times 10^{-4} \times 27 \approx 270.2 \times 10^{-2} \approx 2.7 \ W$.
Thus,the power of the source is $2.7 \ W$.

(B) For a plane electromagnetic wave traveling in a vacuum,the relationship between the magnitude of the electric field $E$ and the magnetic field $B$ is given by $E = cB$,where $c$ is the speed of light in vacuum.
Given $c = 3 \times 10^8 \ m/s$.
We are asked to find the ratio of the magnitude of the electric field $E$ to $10^8$ times the magnetic field $B$.
Ratio $= E / (10^8 \times B) = (cB) / (10^8 \times B) = c / 10^8$.
Substituting the value of $c = 3 \times 10^8 \ m/s$:
Ratio $= (3 \times 10^8) / 10^8 = 3$.
Thus,the ratio is $3: 1$.

(C) In an electromagnetic wave,the oscillating electric field vector $\vec{E}$ and the magnetic field vector $\vec{B}$ are always perpendicular to each other.
Furthermore,both of these vectors are perpendicular to the direction of propagation of the wave.
These fields oscillate in the same phase,meaning they reach their maximum and minimum values at the same time and at the same location.
Therefore,the correct statement is that they are along mutually perpendicular directions and are in the same phase.

(A) The standard equation for a plane progressive wave is given by $B = B_0 \sin(kx + \omega t)$.
Comparing this with the given equation $B_y = 2 \times 10^{-7} \sin(0.5 \times 10^3 x + 1.5 \pi \times 10^{11} t)$,we identify the angular frequency $\omega$ as $\omega = 1.5 \pi \times 10^{11} \ rad/s$.
The relationship between angular frequency $\omega$ and frequency $f$ is given by $\omega = 2 \pi f$.
Substituting the value of $\omega$,we get $1.5 \pi \times 10^{11} = 2 \pi f$.
Solving for $f$,we have $f = \frac{1.5 \pi \times 10^{11}}{2 \pi} = 0.75 \times 10^{11} \ Hz$.
This can be rewritten as $f = 75 \times 10^9 \ Hz$.
Therefore,the correct option is $A$.

(A) The intensity $I$ of a point source emitting power $P$ uniformly in all directions at a distance $r$ is given by the formula:
$I = \frac{P}{4 \pi r^2}$
Given:
Power $P = 10 \ W$
Distance $r = 0.5 \ m$
Substituting the values:
$I = \frac{10}{4 \times 3.14 \times (0.5)^2}$
$I = \frac{10}{4 \times 3.14 \times 0.25}$
$I = \frac{10}{3.14}$
$I \approx 3.18 \ W \ m^{-2}$
Therefore,the correct option is $A$.

(A) The minimum wavelength $(\lambda_{min})$ of $X$-rays produced by electrons accelerated through a potential difference $V$ is given by the Duane-Hunt law: $\lambda_{min} = \frac{hc}{eV}$.
Substituting the values: $h = 6.63 \times 10^{-34} \ J \cdot s$,$c = 3 \times 10^8 \ m/s$,$e = 1.6 \times 10^{-19} \ C$,and $V = 20 \times 10^3 \ V$.
$\lambda_{min} = \frac{(6.63 \times 10^{-34}) \times (3 \times 10^8)}{(1.6 \times 10^{-19}) \times (20 \times 10^3)} \ m$.
$\lambda_{min} = \frac{19.89 \times 10^{-26}}{3.2 \times 10^{-15}} \ m = 6.215 \times 10^{-11} \ m$.
Since $1 \ Å = 10^{-10} \ m$,we have $\lambda_{min} = 0.6215 \times 10^{-10} \ m = 0.62 \ Å$.

(A) The electric displacement vector $D$ is defined as $D = \epsilon E$,where $\epsilon$ is the permittivity of the medium and $E$ is the electric field.
Given that the dielectric constant $K = 3$,the permittivity of the medium is $\epsilon = K \epsilon_0$,where $\epsilon_0 = 8.854 \times 10^{-12} \ F m^{-1}$.
The electric field $E = 1.5 \times 10^{-9} \pi \ N C^{-1}$.
Substituting these values into the formula:
$D = K \epsilon_0 E$
$D = 3 \times (8.854 \times 10^{-12}) \times (1.5 \times 10^{-9} \pi)$
Using $\epsilon_0 \approx \frac{1}{36\pi} \times 10^{-9} \ F m^{-1}$ for calculation convenience:
$D = 3 \times (\frac{1}{36\pi} \times 10^{-9}) \times (1.5 \times 10^{-9} \pi)$
$D = 3 \times \frac{1}{36} \times 1.5 \times 10^{-18}$
$D = \frac{4.5}{36} \times 10^{-18} = 0.125 \times 10^{-18} \ C m^{-2} = 125 \times 10^{-21} \ C m^{-2}$.
Given the options provided,there appears to be a mismatch in the magnitude of the electric field provided in the question. Assuming the standard form $D = \epsilon_0 E_0$ where $E_0$ is the field in vacuum,the correct calculation leads to $125 \times 10^{-12} \ C m^{-2}$ if $E = 1.5 \pi \times 10^9 \ N C^{-1}$ was intended.

(A) The initial force between two charges $q_1$ and $q_2$ separated by distance $r$ in air is given by $F = \frac{1}{4\pi\epsilon_0} \frac{q_1 q_2}{r^2} = 98 \ N$.
When dielectric slabs are introduced,the effective distance $r_{eff}$ between the charges is modified. The effective distance is given by $r_{eff} = (r - t_1 - t_2) + t_1\sqrt{K_1} + t_2\sqrt{K_2}$.
Here,$r = 9 \ cm$,$t_1 = 6 \ cm$,$K_1 = 4$,$t_2 = 3 \ cm$,$K_2 = 9$.
$r_{eff} = (9 - 6 - 3) + 6\sqrt{4} + 3\sqrt{9} = 0 + 6(2) + 3(3) = 12 + 9 = 21 \ cm$.
The new force $F'$ is given by $F' = F \left( \frac{r}{r_{eff}} \right)^2$.
$F' = 98 \times \left( \frac{9}{21} \right)^2 = 98 \times \left( \frac{3}{7} \right)^2 = 98 \times \frac{9}{49} = 2 \times 9 = 18 \ N$.

(B) Total number of atoms $N = 6 \times 10^{24}$.
Number of atoms from which one electron is removed is $n = 0.005 \% \text{ of } N$.
$n = \frac{0.005}{100} \times 6 \times 10^{24} = 5 \times 10^{-5} \times 6 \times 10^{24} = 30 \times 10^{19} = 3 \times 10^{20}$.
Since one electron is removed from each of these atoms,the total number of electrons removed is $3 \times 10^{20}$.
The charge gained by the solid is $q = n \times e$,where $e = 1.6 \times 10^{-19} \ C$.
$q = (3 \times 10^{20}) \times (1.6 \times 10^{-19}) = 4.8 \times 10 = 48 \ C$.
Since electrons are removed,the solid gains a positive charge of $+48 \ C$.

(A) Let the two charges be $q_1$ and $q_2$. Given $q_1 + q_2 = 25 \times 10^{-6} \ C$ and distance $r = 1.5 \ m$. The electrostatic force is given by $F = \frac{k q_1 q_2}{r^2}$,where $k = 9 \times 10^9 \ N \cdot m^2/C^2$. Substituting the values: $0.6 = \frac{9 \times 10^9 \times q_1 q_2}{(1.5)^2}$. Solving for $q_1 q_2$: $q_1 q_2 = \frac{0.6 \times 2.25}{9 \times 10^9} = 0.15 \times 10^{-9} = 150 \times 10^{-12} \ C^2$. We know that $(q_1 - q_2)^2 = (q_1 + q_2)^2 - 4 q_1 q_2$. Substituting the values: $(q_1 - q_2)^2 = (25 \times 10^{-6})^2 - 4(150 \times 10^{-12}) = 625 \times 10^{-12} - 600 \times 10^{-12} = 25 \times 10^{-12} \ C^2$. Taking the square root,$q_1 - q_2 = 5 \times 10^{-6} \ C = 5 \mu C$.

(C) Initially,the force between the two spheres is given by Coulomb's Law: $F = k \frac{|q_1 q_2|}{r^2} = k \frac{|(+8 \mu C)(-4 \mu C)|}{r^2} = k \frac{32 \mu C^2}{r^2}$.
When the two identical conducting spheres are connected by a wire,the charges redistribute equally because they have the same radius.
The new charge on each sphere will be $q' = \frac{q_1 + q_2}{2} = \frac{+8 \mu C - 4 \mu C}{2} = +2 \mu C$.
The new force $F'$ between the spheres is $F' = k \frac{|q' q'|}{r^2} = k \frac{|(+2 \mu C)(+2 \mu C)|}{r^2} = k \frac{4 \mu C^2}{r^2}$.
Comparing the two forces: $\frac{F'}{F} = \frac{4}{32} = \frac{1}{8}$.
Therefore,$F' = \frac{F}{8}$.

(C) According to Coulomb's law,the electrostatic force $F$ between two point charges $q_1$ and $q_2$ separated by a distance $r$ is given by $F = k \frac{|q_1 q_2|}{r^2}$,where $k = 9 \times 10^9 \ N \cdot m^2/C^2$.
Given:
$q_1 = 1.6 \times 10^{-19} \ C$
$q_2 = 3.2 \times 10^{-19} \ C$
$r = 3 \ cm = 3 \times 10^{-2} \ m$
Substituting the values:
$F = \frac{(9 \times 10^9) \times (1.6 \times 10^{-19}) \times (3.2 \times 10^{-19})}{(3 \times 10^{-2})^2}$
$F = \frac{9 \times 10^9 \times 5.12 \times 10^{-38}}{9 \times 10^{-4}}$
$F = 5.12 \times 10^{9-38+4} \ N$
$F = 5.12 \times 10^{-25} \ N$.

(A) The torque $\tau$ acting on an electric dipole in a uniform electric field is given by the formula: $\tau = pE \sin \theta$.
Given:
Dipole moment $p = 2 \times 10^{-10} \text{ C m}$.
Electric field $E = 10^4 \text{ N C}^{-1}$.
Angle $\theta = 30^{\circ}$.
Substituting the values into the formula:
$\tau = (2 \times 10^{-10}) \times (10^4) \times \sin(30^{\circ})$.
Since $\sin(30^{\circ}) = 0.5$:
$\tau = 2 \times 10^{-6} \times 0.5 = 1 \times 10^{-6} \text{ N m}$.
Therefore,the magnitude of the torque is $10^{-6} \text{ N m}$.

(D) The electric field $E$ on the surface of a charged conducting sphere is given by the formula $E = \frac{\sigma}{\epsilon_0}$,where $\sigma$ is the surface charge density and $\epsilon_0$ is the permittivity of free space.
Since both spheres have equal surface charge densities $(\sigma_1 = \sigma_2 = \sigma)$,the electric field on the surface of both spheres depends only on the surface charge density $\sigma$ and the constant $\epsilon_0$.
Therefore,the electric field on the surface of the larger sphere is also $E = \frac{\sigma}{\epsilon_0}$.
Thus,the electric field on the surface of the larger sphere is equal to the electric field on the surface of the smaller sphere,which is $E$.

(B) Given: Mass $m = 0.2 \ g = 0.2 \times 10^{-3} \ kg$,Charge $q = 2 \ C$,Electric field $E = 20 \ N \ C^{-1}$,Distance $d = 20 \ cm = 0.2 \ m$,Initial velocity $u = 0$.
The force acting on the particle is $F = qE = 2 \times 20 = 40 \ N$.
According to the work-energy theorem,the work done by the electric force is equal to the change in kinetic energy.
Work done $W = F \times d = 40 \ N \times 0.2 \ m = 8 \ J$.
Since the particle starts from rest,the initial kinetic energy is $0$.
Therefore,the final kinetic energy is $K.E. = W = 8 \ J$.

(D) The electric flux $\Phi$ through a surface is given by the dot product of the electric field vector $\overrightarrow{E}$ and the area vector $\overrightarrow{A}$.
$\Phi = \overrightarrow{E} \cdot \overrightarrow{A}$
Given,$\overrightarrow{E} = (3 \hat{i} + 5 \hat{j} + 7 \hat{k}) \text{ NC}^{-1}$.
The surface is in the $yz$-plane,so its area vector $\overrightarrow{A}$ is directed along the $x$-axis.
Therefore,$\overrightarrow{A} = 3 \hat{i} \text{ m}^2$.
Now,calculate the dot product:
$\Phi = (3 \hat{i} + 5 \hat{j} + 7 \hat{k}) \cdot (3 \hat{i})$
$\Phi = (3 \times 3) (\hat{i} \cdot \hat{i}) + (5 \times 0) + (7 \times 0)$
$\Phi = 9 \text{ Nm}^2\text{C}^{-1}$.
Thus,the correct option is $D$.