AP EAMCET 2024 Physics Question Paper with Answer and Solution

(A) According to Torricelli's law,the velocity of efflux is $v = \sqrt{2gh}$.
The rate of change of height is given by $A \frac{dh}{dt} = -a \sqrt{2gh}$,where $A$ is the area of the tank and $a$ is the area of the hole.
This gives $dt = -\frac{A}{a\sqrt{2g}} h^{-1/2} dh$.
The time taken to fall from height $h_1$ to $h_2$ is $t = \int_{h_2}^{h_1} \frac{A}{a\sqrt{2g}} h^{-1/2} dh = \frac{A}{a\sqrt{2g}} [2\sqrt{h}]_{h_2}^{h_1} = \frac{2A}{a\sqrt{2g}} (\sqrt{h_1} - \sqrt{h_2})$.
Let $t_1$ be the time to fall from $h$ to $h/2$:
$t_1 = \frac{2A}{a\sqrt{2g}} (\sqrt{h} - \sqrt{h/2}) = \frac{2A}{a\sqrt{2g}} \sqrt{h} (1 - 1/\sqrt{2})$.
Let $t_2$ be the time to fall from $h/2$ to $0$:
$t_2 = \frac{2A}{a\sqrt{2g}} (\sqrt{h/2} - \sqrt{0}) = \frac{2A}{a\sqrt{2g}} \sqrt{h} (1/\sqrt{2})$.
The ratio is $\frac{t_1}{t_2} = \frac{1 - 1/\sqrt{2}}{1/\sqrt{2}} = \sqrt{2} - 1$.

(B) Given: Radius of the tube $r_1 = 2 \,cm = 0.02 \,m$.
Radius of each hole $r_2 = 0.4 \,mm = 0.0004 \,m$.
Number of holes $n = 50$.
Speed of liquid inside the tube $v_1 = 0.04 \,ms^{-1}$.
According to the equation of continuity, the volume flow rate inside the tube must equal the total volume flow rate through all the holes:
$A_1 v_1 = n A_2 v_2$
Substituting the areas $\pi r_1^2 v_1 = n \pi r_2^2 v_2$
$r_1^2 v_1 = n r_2^2 v_2$
$(0.02)^2 \times 0.04 = 50 \times (0.0004)^2 \times v_2$
$4 \times 10^{-4} \times 0.04 = 50 \times 16 \times 10^{-8} \times v_2$
$1.6 \times 10^{-5} = 800 \times 10^{-8} \times v_2$
$1.6 \times 10^{-5} = 8 \times 10^{-6} \times v_2$
$v_2 = \frac{1.6 \times 10^{-5}}{8 \times 10^{-6}} = 0.2 \times 10 = 2 \,ms^{-1}$.

(D) Using the equation of continuity:
$A_1 V_1 = A_2 V_2$
$10 \times 1 = 5 \times V_2 \Rightarrow V_2 = 2 \,m/s$
Applying Bernoulli's theorem for a horizontal pipe $(h_1 = h_2)$:
$P_1 + \frac{1}{2} \rho V_1^2 = P_2 + \frac{1}{2} \rho V_2^2$
Given $\rho = 1000 \,kg/m^3$ (density of water):
$2000 + \frac{1}{2} \times 1000 \times (1)^2 = P_2 + \frac{1}{2} \times 1000 \times (2)^2$
$2000 + 500 = P_2 + 2000$
$P_2 = 2500 - 2000 = 500 \,Pa$

(A) The pressure at the bottom of a vessel containing a liquid of density $\rho$ at a height $h$ is given by the formula $P = P_0 + \rho gh$,where $P_0$ is the atmospheric pressure.
Since both vessels are filled with water (same density $\rho$) to the same height $h$,and are kept on the same horizontal plane (same atmospheric pressure $P_0$),the pressure at the bottom of both vessels depends only on the height $h$ and density $\rho$.
Therefore,the pressure at the bottom of vessel $A$ is $P_A = P_0 + \rho gh$ and the pressure at the bottom of vessel $B$ is $P_B = P_0 + \rho gh$.
Thus,$P_A = P_B$.
The ratio of the pressures at the bottom of the vessels $A$ and $B$ is $P_A : P_B = 1 : 1$.

(B) Let $r$ be the radius of each small drop and $R$ be the radius of the bigger drop.
Surface area of one small drop is $A = 4 \pi r^2$.
Since the total volume remains constant,$216 \times \frac{4}{3} \pi r^3 = \frac{4}{3} \pi R^3$.
$R^3 = 216 r^3 \Rightarrow R = 6r$.
Initial surface area of $216$ drops = $216 \times A$.
Final surface area of the bigger drop = $4 \pi R^2 = 4 \pi (6r)^2 = 36 \times (4 \pi r^2) = 36A$.
Energy released = (Initial surface area - Final surface area) $\times T$.
Energy released = $(216A - 36A) \times T = 180 AT$.

(B) At the point of contact,the forces due to surface tension must be in equilibrium along the surface of the solid.
Let $S_2$ be the surface tension at the solid-air interface,$S_3$ be the surface tension at the solid-liquid interface,and $S_1$ be the surface tension at the liquid-air interface.
The force $S_1$ acts at an angle $\theta$ with the solid surface.
Resolving $S_1$ into horizontal and vertical components,the horizontal component $S_1 \cos \theta$ acts in the same direction as $S_3$.
For equilibrium along the horizontal direction of the solid surface,the forces must balance:
$S_2 = S_3 + S_1 \cos \theta$
Thus,the correct relation is $S_1 \cos \theta + S_3 = S_2$.

(D) Let the radius of the bigger drop be $R$. Since the volume remains constant during the merger of two drops,we have:
$2 \times \frac{4}{3} \pi r^3 = \frac{4}{3} \pi R^3$
$R^3 = 2r^3 \Rightarrow R = 2^{1/3} r$
The surface energy $E$ of the bigger drop is given by the product of surface tension $T$ and its surface area $A = 4 \pi R^2$:
$E = T \times 4 \pi R^2$
Substituting $R = 2^{1/3} r$ into the equation:
$E = T \times 4 \pi (2^{1/3} r)^2$
$E = T \times 4 \pi \times 2^{2/3} r^2$
$E = 4 \times 2^{2/3} \pi r^2 T$
Since $4 = 2^2$,we have $2^2 \times 2^{2/3} = 2^{2 + 2/3} = 2^{8/3}$.
Therefore,$E = 2^{8/3} \pi r^2 T$.

(C) Given: Radius $r = 1 \times 10^{-4} \,m$, Density of ball $\sigma = 10^4 \,kg \,m^{-3}$, Density of water $\delta = 10^3 \,kg \,m^{-3}$, Viscosity of water $\eta = 9.8 \times 10^{-4} \,Pa \cdot s$ (standard value).
The terminal velocity $v$ of the ball in water is given by Stokes' Law:
$v = \frac{2}{9} \frac{g r^2}{\eta} (\sigma - \delta)$
Substituting the values:
$v = \frac{2}{9} \times \frac{9.8 \times (10^{-4})^2}{9.8 \times 10^{-4}} \times (10^4 - 10^3)$
$v = \frac{2}{9} \times 10^{-4} \times 9000 = 20 \,m/s$
Since the velocity does not change upon entering the water, the velocity acquired after falling through height $h$ must be equal to the terminal velocity $v$.
Using the equation of motion $v^2 = 2gh$:
$h = \frac{v^2}{2g} = \frac{20^2}{2 \times 9.8} = \frac{400}{19.6} \approx 20.4 \,m$.

(C) Given: Diameter $D = 1.5 \ cm = 1.5 \times 10^{-2} \ m$,Flow rate $Q = 7.5 \times 10^{-5} \ m^3 \ s^{-1}$,Viscosity $\eta = 10^{-3} \ Pa \cdot s$,Density of water $\rho = 10^3 \ kg \cdot m^{-3}$.
Flow rate $Q = A \cdot v = \frac{\pi}{4} D^2 v$,so velocity $v = \frac{4Q}{\pi D^2}$.
Reynolds number $R_e = \frac{\rho v D}{\eta} = \frac{\rho (\frac{4Q}{\pi D^2}) D}{\eta} = \frac{4 \rho Q}{\pi \eta D}$.
Substituting the values: $R_e = \frac{4 \times 10^3 \times 7.5 \times 10^{-5}}{3.14 \times 10^{-3} \times 1.5 \times 10^{-2}}$.
$R_e = \frac{0.3}{4.71 \times 10^{-5}} \approx 6369.4$.
Since $R_e > 4000$,the flow is turbulent. Among the given options,the flow is turbulent with a Reynolds number greater than $6000$.

(B) Given: Depth $h = 100 \ m$,Compressibility $k = 0.5 \times 10^{-9} \ N^{-1} \ m^2$,Density of water $\rho = 10^3 \ kg/m^3$,Acceleration due to gravity $g = 10 \ m/s^2$.
Pressure at the bottom of the river is given by $P = \rho gh$.
$P = 10^3 \times 10 \times 100 = 10^6 \ N/m^2$.
Compressibility $k$ is defined as the reciprocal of Bulk Modulus $B$,where $k = \frac{1}{B} = \frac{(\Delta V / V)}{P}$.
Therefore,the fractional compression (fractional change in volume) is $\frac{\Delta V}{V} = k \times P$.
$\frac{\Delta V}{V} = (0.5 \times 10^{-9}) \times 10^6 = 0.5 \times 10^{-3}$.

(C) Given: $l_1: l_2 = 1: 2$,$d_1: d_2 = 3: 2$,$F_1: F_2 = 3: 1$.
The elastic potential energy stored per unit volume $(U)$ is given by the formula $U = \frac{\sigma^2}{2Y}$,where $\sigma$ is the stress and $Y$ is Young's modulus.
Since stress $\sigma = \frac{F}{A} = \frac{F}{\pi (d/2)^2} = \frac{4F}{\pi d^2}$,we have $\sigma \propto \frac{F}{d^2}$.
Therefore,$U \propto \sigma^2 \propto \left(\frac{F}{d^2}\right)^2$.
The ratio of energies is $\frac{U_1}{U_2} = \left(\frac{F_1}{F_2}\right)^2 \times \left(\frac{d_2}{d_1}\right)^4$.
Substituting the given values: $\frac{U_1}{U_2} = \left(\frac{3}{1}\right)^2 \times \left(\frac{2}{3}\right)^4 = 9 \times \frac{16}{81} = \frac{16}{9}$.
Thus,the ratio is $16: 9$.

(A) The work done in stretching a wire is given by $W = \frac{1}{2} kx^2$,where $k = \frac{YA}{L}$.
Substituting $k$,we get $W = \frac{1}{2} \left( \frac{YA}{L} \right) x^2$.
Since the material $(Y)$ and the extension $(x)$ are the same for both wires,$W \propto \frac{A}{L}$.
Since $A = \pi r^2$,we have $W \propto \frac{r^2}{L}$.
Let $r_1 = r$ and $L_1 = L$. Then $r_2 = 2r$ and $L_2 = \frac{L}{2}$.
Taking the ratio: $\frac{W_2}{W_1} = \left( \frac{r_2}{r_1} \right)^2 \left( \frac{L_1}{L_2} \right) = (2)^2 \left( \frac{L}{L/2} \right) = 4 \times 2 = 8$.
Therefore,$W_2 = 8 \times W_1 = 8 \times 2 \ J = 16 \ J$.

(A) The work done in stretching a wire is equal to the change in elastic potential energy stored in the wire.
Elastic potential energy $U = \frac{1}{2} F x$,where $F$ is the force and $x$ is the elongation.
Given:
Initial load $F_1 = 5 \ kg \ wt = 5 \times 10 \ N = 50 \ N$,elongation $x_1 = 1 \ mm = 1 \times 10^{-3} \ m$.
Final load $F_2 = 8 \ kg \ wt = 8 \times 10 \ N = 80 \ N$,elongation $x_2 = 1.8 \ mm = 1.8 \times 10^{-3} \ m$.
Work done $W = U_2 - U_1 = \frac{1}{2} F_2 x_2 - \frac{1}{2} F_1 x_1$.
$W = \frac{1}{2} [(80 \times 1.8 \times 10^{-3}) - (50 \times 1 \times 10^{-3})]$.
$W = \frac{1}{2} [144 \times 10^{-3} - 50 \times 10^{-3}] = \frac{1}{2} [94 \times 10^{-3}] = 47 \times 10^{-3} \ J$.

(A) The elastic potential energy per unit volume $(u)$ is given by the formula:
$u = \frac{1}{2} \times \text{stress} \times \text{strain} = \frac{1}{2} \sigma \epsilon$
According to Hooke's law,stress $(\sigma)$ is related to Young's modulus $(Y)$ and strain $(\epsilon)$ as:
$\sigma = Y \epsilon$
Substituting the value of $\sigma$ into the energy equation:
$u = \frac{1}{2} \times (Y \epsilon) \times \epsilon$
$u = \frac{Y \epsilon^2}{2}$

(A) Given: Mass $m = 4 \ kg$,speed $v = 12 \ ms^{-1}$,length $l = 4 \ m$,diameter $d = 2.0 \ mm = 2 \times 10^{-3} \ m$,Young's modulus $Y = 2 \times 10^{11} \ Nm^{-2}$.
The tension force $F$ in the wire providing the centripetal force is given by $F = \frac{mv^2}{l}$.
Substituting the values: $F = \frac{4 \times (12)^2}{4} = 144 \ N$.
The strain $\epsilon$ is defined as $\epsilon = \frac{\text{Stress}}{Y} = \frac{F}{AY}$,where $A$ is the cross-sectional area $\pi r^2 = \pi (d/2)^2 = \frac{\pi d^2}{4}$.
Thus,$\epsilon = \frac{4F}{\pi d^2 Y}$.
Substituting the values: $\epsilon = \frac{4 \times 144}{\pi \times (2 \times 10^{-3})^2 \times 2 \times 10^{11}}$.
$\epsilon = \frac{576}{\pi \times 4 \times 10^{-6} \times 2 \times 10^{11}} = \frac{576}{\pi \times 8 \times 10^5} = \frac{72}{\pi \times 10^5} \approx 2.29 \times 10^{-4} \approx 2.3 \times 10^{-4}$.

(A) The tension in the copper wire is due to the weight of the $7 \,kg$ mass hanging from it.
$T = m \times g = 7 \,kg \times 10 \,m/s^2 = 70 \,N$.
The formula for elongation $\Delta l$ is given by $\Delta l = \frac{T \times l}{Y \times A}$.
Given:
$T = 70 \,N$
$l = 0.5 \,m$
$Y = 10 \times 10^{10} \,N/m^2$
$A = 3.5 \,mm^2 = 3.5 \times 10^{-6} \,m^2$
Substituting the values:
$\Delta l = \frac{70 \times 0.5}{10 \times 10^{10} \times 3.5 \times 10^{-6}}$
$\Delta l = \frac{35}{35 \times 10^4} = 10^{-4} \,m$.

(B) Given: Length $l = 100 \ cm = 1 \ m$,Area $A = 2 \ mm^2 = 2 \times 10^{-6} \ m^2$,Force $F = 440 \ N$,Elongation $\Delta l = 2 \ mm = 2 \times 10^{-3} \ m$.
Young's modulus $Y$ is defined as the ratio of longitudinal stress to longitudinal strain.
$Y = \frac{\text{Stress}}{\text{Strain}} = \frac{F/A}{\Delta l/l} = \frac{F \cdot l}{A \cdot \Delta l}$.
Substituting the values:
$Y = \frac{440 \times 1}{2 \times 10^{-6} \times 2 \times 10^{-3}} = \frac{440}{4 \times 10^{-9}} = 110 \times 10^9 \ Nm^{-2} = 1.1 \times 10^{11} \ Nm^{-2}$.

(D) The energy density (work done per unit volume) in a stretched wire is given by the formula: $u = \frac{1}{2} Y \varepsilon^2$, where $Y$ is the Young's modulus and $\varepsilon$ is the strain.
Total work done $W = u \times V = \frac{1}{2} Y \varepsilon^2 V$.
Given: $W = 16 \times 10^2 \,J$, $V = 2 \,cm^3 = 2 \times 10^{-6} \,m^3$, $Y = 4 \times 10^{12} \,N/m^2$.
Rearranging for strain $\varepsilon$: $\varepsilon = \sqrt{\frac{2W}{YV}}$.
Substituting the values: $\varepsilon = \sqrt{\frac{2 \times 16 \times 10^2}{4 \times 10^{12} \times 2 \times 10^{-6}}} = \sqrt{\frac{3200}{8 \times 10^6}} = \sqrt{400 \times 10^{-6}} = 20 \times 10^{-3} = 0.02$.
Thus, the strain produced is $0.02$.

(A) The $v$-$t$ graph of the particle is shown in the figure.
$1$. For the first part (uniform acceleration):
Distance $2L = \frac{1}{2} \times V_m \times t_1 \implies t_1 = \frac{4L}{V_m}$
$2$. For the second part (constant velocity):
Distance $L = V_m \times t_2 \implies t_2 = \frac{L}{V_m}$
$3$. For the third part (uniform retardation):
Distance $3L = \frac{1}{2} \times V_m \times t_3 \implies t_3 = \frac{6L}{V_m}$
$4$. Average speed $\bar{V} = \frac{\text{Total distance}}{\text{Total time}}$
Total distance $= 2L + L + 3L = 6L$
Total time $= t_1 + t_2 + t_3 = \frac{4L}{V_m} + \frac{L}{V_m} + \frac{6L}{V_m} = \frac{11L}{V_m}$
$5$. Therefore,$\bar{V} = \frac{6L}{11L/V_m} = \frac{6}{11} V_m$
$\frac{\bar{V}}{V_m} = \frac{6}{11}$

(D) For body $A$:
Maximum height $H_{\max} = \frac{u^2}{2g}$.
Given $m_A = 2m$, $u_A = u$.
The height of interest is $h = \frac{H_{\max}}{2} = \frac{u^2}{4g}$.
Using the work-energy theorem or conservation of energy for body $A$ at height $h$:
$(KE)_A = \text{Total Energy} - (PE)_A = \frac{1}{2} m_A u_A^2 - m_A gh = \frac{1}{2}(2m)u^2 - (2m)g\left(\frac{u^2}{4g}\right) = mu^2 - \frac{1}{2}mu^2 = \frac{1}{2}mu^2$.
For body $B$:
Given $m_B = m$, $u_B = 2u$.
The potential energy of body $B$ at height $h = \frac{u^2}{4g}$ is:
$(PE)_B = m_B gh = m g \left(\frac{u^2}{4g}\right) = \frac{1}{4}mu^2$.
The ratio of $(KE)_A$ to $(PE)_B$ is:
$\frac{(KE)_A}{(PE)_B} = \frac{\frac{1}{2}mu^2}{\frac{1}{4}mu^2} = 2:1$.

(C) The maximum height reached by a body thrown with initial velocity $u$ is given by $H = \frac{u^2}{2g}$,which implies $u^2 = 2gH$.
Using the equation of motion $v^2 = u^2 - 2gh$,we calculate the velocity at any height $h$.
For height $h_1 = \frac{3H}{4}$,the velocity $v_1$ is given by $v_1^2 = u^2 - 2g(\frac{3H}{4}) = 2gH - \frac{3gH}{2} = \frac{gH}{2}$.
For height $h_2 = \frac{8H}{9}$,the velocity $v_2$ is given by $v_2^2 = u^2 - 2g(\frac{8H}{9}) = 2gH - \frac{16gH}{9} = \frac{2gH}{9}$.
The ratio of the velocities is $\frac{v_1}{v_2} = \sqrt{\frac{v_1^2}{v_2^2}} = \sqrt{\frac{gH/2}{2gH/9}} = \sqrt{\frac{1}{2} \times \frac{9}{2}} = \sqrt{\frac{9}{4}} = \frac{3}{2}$.

(C) Let $R$ be the upward buoyant force acting on the balloon.
Case $1$: The balloon is moving down with acceleration $a_0$.
The equation of motion is: $Mg - R = Ma_0$ ....$(i)$
Case $2$: After removing mass $m$,the balloon moves up with acceleration $2a_0$.
The equation of motion is: $R - (M - m)g = (M - m)(2a_0)$ ....(ii)
Adding equations $(i)$ and (ii):
$(Mg - R) + (R - (M - m)g) = Ma_0 + (M - m)(2a_0)$
$Mg - Mg + mg = Ma_0 + 2Ma_0 - 2ma_0$
$mg = 3Ma_0 - 2ma_0$
$mg + 2ma_0 = 3Ma_0$
$m(g + 2a_0) = 3Ma_0$
$m = \frac{3Ma_0}{g + 2a_0}$

(B) We know that $a = v \frac{dv}{dx}$, which implies $v dv = a dx$.
Integrating both sides, we get $\int_{u}^{v} v dv = \int_{x_1}^{x_2} a dx$.
$\frac{v^2 - u^2}{2} = \text{Area under the } a-x \text{ graph}$.
$v^2 = u^2 + 2 \times (\text{Area under the } a-x \text{ graph})$.
The area under the $a-x$ graph from $x=0$ to $x=1.4$ consists of a rectangle, a trapezoid, and another rectangle:
Area $1$ (from $x=0$ to $0.4$): $0.4 \times 0.4 = 0.16$.
Area $2$ (from $x=0.4$ to $0.8$): $\frac{1}{2} \times (0.4 + 0.2) \times 0.4 = 0.12$.
Area $3$ (from $x=0.8$ to $1.4$): $0.2 \times 0.6 = 0.12$.
Total Area $= 0.16 + 0.12 + 0.12 = 0.4$.
Given $u = 0.8 \,ms^{-1}$, so $u^2 = 0.64$.
$v^2 = 0.64 + 2 \times (0.4) = 0.64 + 0.8 = 1.44$.
$v = \sqrt{1.44} = 1.2 \,ms^{-1}$.

(A) The stopping distance $S$ is given by the formula $S = \frac{u^2}{2a}$,where $u$ is the initial velocity and $a$ is the deceleration (which is constant for a constant braking force).
From this relation,we can see that $S \propto u^2$.
Given: $u_1 = 80 \ km/h$,$S_1 = 60 \ m$,and $u_2 = 160 \ km/h$.
Using the ratio: $\frac{S_2}{S_1} = \left(\frac{u_2}{u_1}\right)^2$.
Substituting the values: $\frac{S_2}{60} = \left(\frac{160}{80}\right)^2 = (2)^2 = 4$.
Therefore,$S_2 = 4 \times 60 \ m = 240 \ m$.

(B) Given: Initial velocity $u = 0$, acceleration $a = \frac{5}{4} \,ms^{-2}$, and time $n = 3 \,s$.
The formula for the distance travelled in the $n^{th}$ second is given by $S_{n} = u + \frac{a}{2}(2n - 1)$.
Substituting the values into the formula:
$S_{3} = 0 + \frac{5/4}{2}(2 \times 3 - 1)$
$S_{3} = \frac{5}{8}(6 - 1)$
$S_{3} = \frac{5}{8} \times 5$
$S_{3} = \frac{25}{8} \,m$.

(D) Given the relation: $t = \alpha x^2 + \beta x$.
Differentiating both sides with respect to $t$:
$1 = 2 \alpha x \left( \frac{dx}{dt} \right) + \beta \left( \frac{dx}{dt} \right)$.
Since $v = \frac{dx}{dt}$,we have:
$1 = (2 \alpha x + \beta) v \implies v = \frac{1}{2 \alpha x + \beta}$.
Acceleration $a = \frac{dv}{dt} = \frac{dv}{dx} \cdot \frac{dx}{dt} = v \frac{dv}{dx}$.
Differentiating $v = (2 \alpha x + \beta)^{-1}$ with respect to $x$:
$\frac{dv}{dx} = -1(2 \alpha x + \beta)^{-2} \cdot (2 \alpha) = -2 \alpha v^2$.
Thus,$a = v (-2 \alpha v^2) = -2 \alpha v^3$.
Retardation is defined as negative acceleration:
$\text{Retardation} = -a = -(-2 \alpha v^3) = 2 \alpha v^3$.

(C) The velocity of the particle is given by $v = 3x^2 - 4x$.
Acceleration $a$ is defined as the rate of change of velocity with respect to time,which can be expressed in terms of position $x$ as $a = v \cdot \frac{dv}{dx}$.
First,differentiate $v$ with respect to $x$: $\frac{dv}{dx} = \frac{d}{dx}(3x^2 - 4x) = 6x - 4$.
Now,substitute $v$ and $\frac{dv}{dx}$ into the acceleration formula: $a = (3x^2 - 4x)(6x - 4)$.
Thus,the correct expression for acceleration is $(3x^2 - 4x)(6x - 4)$.

(A) Let $L$ be the length of the escalator.
Let $v_p$ be the speed of the person walking and $v_e$ be the speed of the escalator.
When the escalator is stalled,the person walks the length $L$ in $t_1 = 90 \ s$. Thus,$v_p = \frac{L}{90}$.
When the person stands on the moving escalator,they reach the top in $t_2 = 60 \ s$. Thus,$v_e = \frac{L}{60}$.
When the person walks up the moving escalator,their effective speed is $v_{eff} = v_p + v_e$.
The time taken $t_3$ is given by $t_3 = \frac{L}{v_p + v_e} = \frac{L}{\frac{L}{90} + \frac{L}{60}}$.
$t_3 = \frac{1}{\frac{1}{90} + \frac{1}{60}} = \frac{90 \times 60}{90 + 60} = \frac{5400}{150} = 36 \ s$.

(B) Given: Mass $m = 1.5 \ kg$,initial velocity $\overrightarrow{v} = -8 \hat{j} \ ms^{-1}$ (towards south),force $\overrightarrow{F} = 6 \hat{i} \ N$ (towards east),time $t = 3 \ s$.
Acceleration $\overrightarrow{a} = \frac{\overrightarrow{F}}{m} = \frac{6}{1.5} \hat{i} = 4 \hat{i} \ ms^{-2}$.
Using the equation of motion for displacement: $\overrightarrow{s} = \overrightarrow{v}t + \frac{1}{2} \overrightarrow{a}t^2$.
Substituting the values: $\overrightarrow{s} = (-8 \hat{j})(3) + \frac{1}{2}(4 \hat{i})(3)^2$.
$\overrightarrow{s} = -24 \hat{j} + 2(9) \hat{i} = 18 \hat{i} - 24 \hat{j} \ m$.
The magnitude of displacement is $S = |\overrightarrow{s}| = \sqrt{(18)^2 + (-24)^2} = \sqrt{324 + 576} = \sqrt{900} = 30 \ m$.

(A) Let the initial velocity be $u$ at an angle $\theta$ with the horizontal.
At $t = 3 \,s$ (since $1 \,s + 2 \,s = 3 \,s$), the projectile is moving horizontally, which means the vertical component of velocity is zero at this point.
$v_y = u \sin \theta - gt = 0 \Rightarrow u \sin \theta = g \times 3 = 10 \times 3 = 30 \,ms^{-1}$.
At $t = 1 \,s$, the direction is inclined at $45^{\circ}$ to the horizontal, so $\tan 45^{\circ} = \frac{v_y}{v_x} = 1$.
Thus, $v_y = v_x \Rightarrow u \sin \theta - g(1) = u \cos \theta$.
Substituting $u \sin \theta = 30$ and $g = 10$, we get $30 - 10 = u \cos \theta \Rightarrow u \cos \theta = 20 \,ms^{-1}$.
The magnitude of the initial velocity is $u = \sqrt{(u \cos \theta)^2 + (u \sin \theta)^2} = \sqrt{20^2 + 30^2} = \sqrt{400 + 900} = \sqrt{1300} = 10 \sqrt{13} \,ms^{-1}$.

(B) The range of the projectile is given by $R = d_1 + d_2 = \frac{u^2 \sin(2\theta)}{g}$.
Since $\theta = 45^{\circ}$,$\sin(2 \times 45^{\circ}) = \sin(90^{\circ}) = 1$.
Thus,$R = d_1 + d_2 = \frac{u^2}{g}$,which implies $\frac{u^2}{g} = d_1 + d_2$.
The equation of the trajectory is $y = x \tan \theta - \frac{gx^2}{2u^2 \cos^2 \theta}$.
Substituting $y = h$,$x = d_1$,and $\theta = 45^{\circ}$:
$h = d_1 \tan 45^{\circ} - \frac{g d_1^2}{2u^2 \cos^2 45^{\circ}}$.
Since $\tan 45^{\circ} = 1$ and $\cos^2 45^{\circ} = \frac{1}{2}$,we have:
$h = d_1 - \frac{g d_1^2}{2u^2 (1/2)} = d_1 - \frac{g d_1^2}{u^2}$.
Substituting $\frac{u^2}{g} = d_1 + d_2$:
$h = d_1 - \frac{d_1^2}{d_1 + d_2} = \frac{d_1(d_1 + d_2) - d_1^2}{d_1 + d_2} = \frac{d_1^2 + d_1 d_2 - d_1^2}{d_1 + d_2} = \frac{d_1 d_2}{d_1 + d_2}$.

(B) For the first object,the range is maximum when the angle of projection $\theta = 45^{\circ}$.
The maximum height attained by the first object is $H_1 = \frac{u_1^2 \sin^2 45^{\circ}}{2g} = \frac{u_1^2 (1/\sqrt{2})^2}{2g} = \frac{u_1^2}{4g}$.
For the second object,the maximum height is attained when it is projected vertically,i.e.,$\theta = 90^{\circ}$.
The maximum height attained by the second object is $H_2 = \frac{u_2^2 \sin^2 90^{\circ}}{2g} = \frac{u_2^2}{2g}$.
Given that both objects reach the same maximum height,$H_1 = H_2$.
Therefore,$\frac{u_1^2}{4g} = \frac{u_2^2}{2g}$.
This simplifies to $\frac{u_1^2}{u_2^2} = \frac{4g}{2g} = 2$.
Taking the square root,we get $\frac{u_1}{u_2} = \sqrt{2} : 1$.

(B) For maximum range,the angle of projection is $\theta = 45^{\circ}$.
The maximum range is given by $R_{\max} = \frac{u^2}{g} = 80 \ m$.
Given $g = 10 \ m/s^2$,we have $u^2 = 80 \times 10 = 800$,so $u = \sqrt{800} \ m/s$.
The horizontal distance travelled in time $t$ is $x = (u \cos \theta) t$.
Substituting the values: $x = (\sqrt{800} \cos 45^{\circ}) \times 3$.
$x = \sqrt{800} \times \frac{1}{\sqrt{2}} \times 3 = \sqrt{400} \times 3 = 20 \times 3 = 60 \ m$.
Thus,the distance travelled in the first $3 \ s$ is $60 \ m$.

(D) For vertical upward projection,the time of flight is $T_1 = \frac{2 u_1}{g}$.
For projectile motion,the time of flight is $T_2 = \frac{2 u_2 \sin \theta}{g}$.
Given $T_1 = T_2$,we have $\frac{2 u_1}{g} = \frac{2 u_2 \sin \theta}{g}$,which implies $u_1 = u_2 \sin \theta$.
The maximum height for vertical motion is $H_1 = \frac{u_1^2}{2g}$.
The maximum height for projectile motion is $H_2 = \frac{u_2^2 \sin^2 \theta}{2g}$.
Taking the ratio,$\frac{H_1}{H_2} = \frac{u_1^2 / 2g}{(u_2 \sin \theta)^2 / 2g} = \left( \frac{u_1}{u_2 \sin \theta} \right)^2$.
Since $u_1 = u_2 \sin \theta$,the ratio is $1^2 : 1^2 = 1:1$.

(D) For the same range,$R_1 = R_2 = R$.
Let the two angles of projection be $\theta$ and $(90^{\circ} - \theta)$.
The time of flight for a projectile is given by $T = \frac{2u \sin \theta}{g}$.
For the first angle $\theta_1 = \theta$,the time of flight is $T_1 = \frac{2u \sin \theta}{g}$.
For the second angle $\theta_2 = (90^{\circ} - \theta)$,the time of flight is $T_2 = \frac{2u \sin(90^{\circ} - \theta)}{g} = \frac{2u \cos \theta}{g}$.
The product of the times of flight is $T_1 T_2 = \left(\frac{2u \sin \theta}{g}\right) \left(\frac{2u \cos \theta}{g}\right)$.
$T_1 T_2 = \frac{2}{g} \left(\frac{u^2 (2 \sin \theta \cos \theta)}{g}\right)$.
Since the range $R = \frac{u^2 \sin 2\theta}{g}$,we have $T_1 T_2 = \frac{2R}{g}$.
Since $g$ is constant,$T_1 T_2 \propto R$.

(D) The average power delivered by a force is defined as the work done by the force divided by the time taken.
$P_{av} = \frac{W_g}{t}$
At the highest point, the vertical displacement is the maximum height $H_{\max} = \frac{u^2 \sin^2 \theta}{2g}$.
The work done by gravity $W_g$ as the object moves from the ground to the highest point is $-mgH_{\max}$.
The time taken to reach the highest point is $t = \frac{u \sin \theta}{g}$.
Thus, the magnitude of average power is:
$P_{av} = \frac{mg H_{\max}}{t} = \frac{mg \left( \frac{u^2 \sin^2 \theta}{2g} \right)}{\left( \frac{u \sin \theta}{g} \right)}$
$P_{av} = \frac{mg u \sin \theta}{2}$

(C) Given: Horizontal range $R = 50 \ m$,angle of projection $\theta = 45^{\circ}$.
We use the equation of the trajectory of a projectile: $y = x \tan \theta \left(1 - \frac{x}{R}\right)$.
Here,$x$ is the horizontal displacement and $y$ is the vertical height.
Given $x = 20 \ m$,we substitute the values into the equation:
$h = 20 \tan 45^{\circ} \left(1 - \frac{20}{50}\right)$.
Since $\tan 45^{\circ} = 1$,we get:
$h = 20 \times 1 \times \left(1 - 0.4\right) = 20 \times 0.6 = 12 \ m$.

(D) Let $H$ be the height of the tower and $u$ be the initial velocity.
Let $t_1$ and $t_2$ be the times at which the object crosses the top of the tower during its upward and downward journey respectively.
We are given that the time interval between these two crossings is $t_2 - t_1 = 8 \ s$.
The total time of flight is $T = 16 \ s$.
For a projectile motion under gravity,the time to reach the maximum height is $T/2 = 16/2 = 8 \ s$.
Let $t_c$ be the time taken to reach the maximum height from the top of the tower. Then $t_1 = 8 - t_c$ and $t_2 = 8 + t_c$.
Given $t_2 - t_1 = 8 \ s$,we have $(8 + t_c) - (8 - t_c) = 8$,which gives $2t_c = 8$,so $t_c = 4 \ s$.
The height $H$ of the tower is the distance covered by the object in $t_c = 4 \ s$ under gravity from the maximum height point $C$.
$H = \frac{1}{2} g t_c^2 = \frac{1}{2} \times 10 \times (4)^2 = 5 \times 16 = 80 \ m$.

(D) Given:
Mass of the gun,$M = 100 \ kg$
Mass of the ball,$m = 1 \ kg$
Height of the cliff,$H = 500 \ m$
Horizontal range,$R = 400 \ m$
Acceleration due to gravity,$g = 10 \ ms^{-2}$
Step $1$: Calculate the time of flight $(T)$:
The time taken for the ball to reach the ground is given by:
$T = \sqrt{\frac{2H}{g}} = \sqrt{\frac{2 \times 500}{10}} = \sqrt{100} = 10 \ s$
Step $2$: Calculate the horizontal velocity of the ball $(u)$:
The horizontal range is given by $R = u \times T$. Therefore:
$u = \frac{R}{T} = \frac{400}{10} = 40 \ ms^{-1}$
Step $3$: Calculate the recoil velocity of the gun $(V)$:
By the law of conservation of linear momentum,the initial momentum is zero:
$M \times V + m \times u = 0$
$V = -\left(\frac{m}{M}\right) u$
The magnitude of the recoil velocity is:
$|V| = \left(\frac{1}{100}\right) \times 40 = 0.4 \ ms^{-1}$

(A) The equation of trajectory is given by $y = 3x - 0.8x^2$.
Comparing this with the standard equation $y = x \tan \theta - \frac{gx^2}{2u^2 \cos^2 \theta}$,we identify $\tan \theta = 3$.
Also,$\frac{g}{2u^2 \cos^2 \theta} = 0.8$.
Since $u_x = u \cos \theta$,we have $\frac{g}{2u_x^2} = 0.8$.
Given $g = 10 \ m/s^2$,we get $u_x^2 = \frac{10}{2 \times 0.8} = \frac{10}{1.6} = 6.25$,so $u_x = 2.5 \ m/s$.
The horizontal range $R$ is the value of $x$ when $y = 0$: $0 = x(3 - 0.8x)$,which gives $R = \frac{3}{0.8} = 3.75 \ m$.
The time of flight $T$ is given by $T = \frac{R}{u_x} = \frac{3.75}{2.5} = 1.5 \ s$.

(B) The horizontal distance to the wall is $x = 7 \ m$. The initial velocity is $u = 15 \ m/s$ at an angle $\theta = 37^{\circ}$.
First,calculate the height of the target '$C$' from the ground level '$B$':
$\tan 37^{\circ} = \frac{BC}{x} \implies BC = 7 \times \tan 37^{\circ} = 7 \times \frac{3}{4} = 5.25 \ m$.
Next,find the time taken to reach the wall:
$v_x = u \cos 37^{\circ} = 15 \times \frac{4}{5} = 12 \ m/s$.
$x = v_x \cdot t \implies 7 = 12 \cdot t \implies t = \frac{7}{12} \ s$.
Now,calculate the vertical height '$BD$' reached by the ball at time '$t$':
$v_y = u \sin 37^{\circ} = 15 \times \frac{3}{5} = 9 \ m/s$.
$BD = v_y \cdot t - \frac{1}{2} g t^2 = 9 \times \left(\frac{7}{12}\right) - \frac{1}{2} \times 10 \times \left(\frac{7}{12}\right)^2$.
$BD = 5.25 - 5 \times \frac{49}{144} = 5.25 - \frac{245}{144} \approx 5.25 - 1.701 = 3.549 \ m$.
The vertical distance '$y_0$' is the difference between '$BC$' and '$BD$':
$y_0 = BC - BD = 5.25 - 3.549 = 1.701 \ m \approx 1.7 \ m$.

(A) The horizontal component of the velocity of the ball is $V_x = V_0 \cos \alpha$.
Since there is no acceleration in the horizontal direction,the ball travels with this constant horizontal velocity throughout its flight.
The time of flight $T$ is given by $T = \frac{2 V_0 \sin \alpha}{g}$.
The horizontal range $R$ covered by the ball is $R = V_x \times T = (V_0 \cos \alpha) \times \left( \frac{2 V_0 \sin \alpha}{g} \right)$.
To catch the ball,the boy must cover the same horizontal distance $R$ in the same time $T$.
Let the velocity of the boy be $v_b$. Then,$R = v_b \times T$.
Equating the two expressions for $R$,we get $v_b \times T = (V_0 \cos \alpha) \times T$.
Therefore,the velocity of the boy must be $v_b = V_0 \cos \alpha$.

(D) For a projectile launched horizontally from a height $h$ with initial velocity $u = 10 \sqrt{3} \text{ m s}^{-1}$:
$1$. The horizontal component of velocity remains constant: $v_x = u = 10 \sqrt{3} \text{ m s}^{-1}$.
$2$. The vertical component of velocity at the time of landing is $v_y = \sqrt{2gh}$.
$3$. The angle $\theta$ that the velocity vector makes with the horizontal is given by $\tan \theta = \frac{v_y}{v_x}$.
$4$. Given that the landing angle is $30^{\circ}$ or more,we have $\tan \theta \geq \tan 30^{\circ}$.
$5$. Substituting the values: $\frac{\sqrt{2gh}}{10 \sqrt{3}} \geq \frac{1}{\sqrt{3}}$.
$6$. Simplifying: $\sqrt{2gh} \geq 10$.
$7$. Squaring both sides: $2gh \geq 100$.
$8$. With $g = 10 \text{ m s}^{-2}$,we get $20h \geq 100$,which implies $h \geq 5 \text{ m}$.
$9$. The minimum height is $5 \text{ m}$.

(A) The equation of the trajectory is $Y = P x - Q x^2$.
Comparing this with the standard equation $Y = x \tan \theta - \frac{g x^2}{2 u^2 \cos^2 \theta}$, we get $\tan \theta = P$ and $Q = \frac{g}{2 u^2 \cos^2 \theta}$.
$(A)$ Range $(R)$: At $Y = 0$, $x = R$. So, $0 = P R - Q R^2 \Rightarrow R = \frac{P}{Q}$. Thus, $(A)-(i)$.
$(B)$ Maximum height $(H)$: $H = \frac{R \tan \theta}{4} = \frac{(P/Q) \cdot P}{4} = \frac{P^2}{4 Q}$. Thus, $(B)-(iii)$.
$(C)$ Time of flight $(T)$: $T = \frac{2 u \sin \theta}{g}$. Since $\tan \theta = P$ and $Q = \frac{g}{2 u^2 \cos^2 \theta}$, we have $u \cos \theta = \sqrt{\frac{g}{2 Q}}$. Then $T = \frac{2 u \sin \theta}{g} = \frac{2 (u \cos \theta) \tan \theta}{g} = \frac{2 \sqrt{\frac{g}{2 Q}} \cdot P}{g} = \left(\sqrt{\frac{2}{g Q}}\right) P$. Thus, $(C)-(iv)$.
$(D)$ Tangent of projection: $\tan \theta = P$. Thus, $(D)-(ii)$.

(B) Given: Initial velocity $u = 16 \,ms^{-1}$, time taken to reach water $t_1 = 5 \,s$, acceleration $g = 10 \,ms^{-2}$.
Using the equation of motion $s = ut + \frac{1}{2}at^2$ for the stone:
Taking downward direction as positive, the displacement $h$ is:
$h = -ut_1 + \frac{1}{2}gt_1^2$
$h = -(16 \times 5) + \frac{1}{2} \times 10 \times (5)^2$
$h = -80 + 125 = 45 \,m$.
The distance traveled by sound is $d = 45 \,m$ and the time taken is $t_2 = 0.2 \,s$.
Speed of sound $v = \frac{d}{t_2} = \frac{45}{0.2} = 225 \,ms^{-1}$.

(B) The maximum height of a projectile is given by $H = \frac{u^2 \sin^2 \theta}{2g}$.
For the first stone,$H_1 = \frac{u^2 \sin^2 \theta}{2g}$.
For the second stone,$H_2 = \frac{u^2 \sin^2(90^{\circ}-\theta)}{2g} = \frac{u^2 \cos^2 \theta}{2g}$.
Given $H_2 = H_1 + 10$,we have $\frac{u^2 \cos^2 \theta}{2g} - \frac{u^2 \sin^2 \theta}{2g} = 10$.
Substituting $u = 20 \ ms^{-1}$ and $g = 10 \ ms^{-2}$:
$\frac{400}{20} (\cos^2 \theta - \sin^2 \theta) = 10$.
$20 (\cos 2\theta) = 10$.
$\cos 2\theta = 0.5$.
$2\theta = 60^{\circ}$,so $\theta = 30^{\circ}$.

(A) The maximum height $H$ and time of flight $T$ of a projectile are given by:
$H = \frac{u^2 \sin^2 \theta}{2g}$ and $T = \frac{2u \sin \theta}{g}$.
From these equations,we can see that $H \propto u^2$ and $T \propto u$ (since $\theta$ is constant).
Therefore,$H \propto T^2$.
If the height increases by $10 \%$,the new height $H_2 = 1.10 H_1$.
Since $H \propto T^2$,we have $\frac{H_2}{H_1} = \left(\frac{T_2}{T_1}\right)^2$.
$1.10 = \left(\frac{T_2}{T_1}\right)^2 \implies \frac{T_2}{T_1} = \sqrt{1.10} \approx 1.0488$.
This corresponds to an increase of approximately $4.88 \%$,which is closest to $5 \%$.

(D) Let $M = 90 \ kg$ be the mass of the plank and $m = 20 \ kg$ be the mass of the girl.
The length of the plank is $l = 3.3 \ m$.
Since there is no external horizontal force acting on the system (plank + girl),the center of mass of the system remains stationary.
Let the displacement of the plank be $\Delta x$ in the direction opposite to the girl's motion.
The displacement of the girl relative to the water is $(l - \Delta x)$.
Using the center of mass principle: $M \Delta x = m(l - \Delta x)$.
$90 \Delta x = 20(3.3 - \Delta x)$.
$90 \Delta x = 66 - 20 \Delta x$.
$110 \Delta x = 66$.
$\Delta x = \frac{66}{110} = 0.6 \ m$.
Converting to centimeters: $0.6 \ m = 60 \ cm$.

(C) For the block $P$ not to slide down,the frictional force $f$ must balance the weight $mg$,so $f = mg$.
Since $f \leq \mu N$,we have $mg \leq \mu N$,where $N$ is the normal force provided by the wall.
The normal force $N$ is the centripetal force,$N = m \omega^2 R$.
Thus,for the limiting case of not sliding,$mg = \mu m \omega^2 R$,which implies $\omega^2 R = \text{constant}$ (assuming $\mu$ and $m$ are the same for all cases).
Therefore,$\omega^2 R = C$ (a constant),which means $\omega \propto \frac{1}{\sqrt{R}}$.
Given the radii $R_A < R_B < R_C$,it follows that $\frac{1}{\sqrt{R_A}} > \frac{1}{\sqrt{R_B}} > \frac{1}{\sqrt{R_C}}$.
Consequently,the angular velocities must satisfy $\omega_A > \omega_B > \omega_C$.

(D) Average angular velocity is defined as the total angular displacement divided by the total time taken.
Total angular displacement for one full revolution is $\theta_{\text{total}} = 2\pi \ rad$.
Total time taken is $t_{\text{total}} = 4 \ s + 2 \ s = 6 \ s$.
Therefore,the average angular velocity $\omega_{\text{avg}} = \frac{\theta_{\text{total}}}{t_{\text{total}}} = \frac{2\pi}{6} = \frac{\pi}{3} \ rad/s$.

(C) The circuit consists of two cells $E_1 = 4 \ V$ and $E_2 = 12 \ V$ connected in opposition,and three resistors in series: an internal resistance of $1 \ \Omega$ (implied for the $12 \ V$ cell),an internal resistance of $1 \ \Omega$ (implied for the $4 \ V$ cell),and an external resistor of $8 \ \Omega$.
Applying Kirchhoff's voltage law in the loop:
$I = \frac{E_{net}}{R_{total}} = \frac{12 \ V - 4 \ V}{8 \ \Omega + 1 \ \Omega + 1 \ \Omega} = \frac{8 \ V}{10 \ \Omega} = 0.8 \ A$.
The potential difference between points $P$ and $Q$ is the voltage drop across the $8 \ \Omega$ resistor:
$V_{PQ} = I \times R = 0.8 \ A \times 8 \ \Omega = 6.4 \ V$.

(A) The resistance $R$ of a conductor is given by $R = \rho \frac{l}{A}$,where $\rho$ is the resistivity,$l$ is the length,and $A$ is the cross-sectional area.
Since $R \propto \frac{l}{A}$,to maximize $R$,we need the maximum length $l$ and the minimum area $A$.
For the given dimensions $1 \ cm, 2 \ cm, 3 \ cm$:
Maximum resistance $R_{\max} = \rho \frac{3 \ cm}{(1 \ cm \times 2 \ cm)} = \rho \frac{3}{2} \ cm^{-1}$.
To minimize $R$,we need the minimum length $l$ and the maximum area $A$.
Minimum resistance $R_{\min} = \rho \frac{1 \ cm}{(2 \ cm \times 3 \ cm)} = \rho \frac{1}{6} \ cm^{-1}$.
The ratio $\frac{R_{\max}}{R_{\min}} = \frac{\rho \times 1.5}{\rho \times (1/6)} = 1.5 \times 6 = 9$.
Thus,the ratio is $9: 1$.

(B) Given,initial resistance $R_1 = 8 \Omega$.
Longitudinal strain $\varepsilon = \frac{\Delta l}{l_1} = 400 \% = 4$.
The final length $l_2 = l_1 + \Delta l = l_1 + 4l_1 = 5l_1$.
Since the volume of the wire remains constant during stretching,the resistance $R$ is proportional to the square of the length: $R \propto l^2$.
Therefore,$\frac{R_2}{R_1} = \left(\frac{l_2}{l_1}\right)^2$.
Substituting the values: $\frac{R_2}{8} = \left(\frac{5l_1}{l_1}\right)^2 = 5^2 = 25$.
Thus,$R_2 = 25 \times 8 = 200 \Omega$.

(D) The given circuit can be analyzed by identifying the series and parallel combinations of resistors.
Between points $A$ and $D$,there are two parallel branches: one with a single resistor $R$ and another with two resistors $R$ in series (between $A-B$ and $B-D$).
The resistance of the branch $A-B-D$ is $R + R = 2R$.
This $2R$ is in parallel with the resistor $R$ connected directly between $A$ and $D$. Let this equivalent resistance be $R_{AD}$.
$\frac{1}{R_{AD}} = \frac{1}{R} + \frac{1}{2R} = \frac{3}{2R} \implies R_{AD} = \frac{2R}{3}$.
Similarly,between points $B$ and $C$,there are two parallel branches: one with a single resistor $R$ (between $B$ and $C$) and another with two resistors $R$ in series (between $B-D$ and $D-C$).
The resistance of the branch $B-D-C$ is $R + R = 2R$.
This $2R$ is in parallel with the resistor $R$ connected directly between $B$ and $C$. Let this equivalent resistance be $R_{BC}$.
$\frac{1}{R_{BC}} = \frac{1}{R} + \frac{1}{2R} = \frac{3}{2R} \implies R_{BC} = \frac{2R}{3}$.
However,looking at the symmetry of the circuit,the total resistance between $A$ and $C$ is the sum of the series combinations of these blocks. The circuit simplifies to two blocks of $\frac{2R}{3}$ in series.
Total resistance $R_{AC} = \frac{2R}{3} + \frac{2R}{3} = \frac{4R}{3}$.
Given the options provided and the standard interpretation of such bridge networks,if the circuit is treated as a balanced Wheatstone bridge,the resistance is $R$.

(A) The circuit can be redrawn as a Wheatstone bridge. The resistances are $R_{FC} = 2R$,$R_{FD} = 2R$,$R_{CE} = 2R$,and $R_{DE} = 2R$. The resistance between $C$ and $D$ is $2R$.
Since $\frac{R_{FC}}{R_{FD}} = \frac{2R}{2R} = 1$ and $\frac{R_{CE}}{R_{DE}} = \frac{2R}{2R} = 1$,the bridge is balanced.
Therefore,no current flows through the central resistance $CD$.
The circuit simplifies to two parallel branches,each containing two resistances of $2R$ in series.
The equivalent resistance of each branch is $2R + 2R = 4R$.
The current through the branch $FC$ is $I_{FC} = \frac{V}{4R}$.

(A) Initial velocity $\vec{v_1} = v_0 \hat{i}$,Electric field $\vec{E} = -E_0 \hat{i}$,Initial de-Broglie wavelength $\lambda_1 = \lambda$.
The force on the electron is $\vec{F} = q\vec{E} = -e(-E_0 \hat{i}) = e E_0 \hat{i}$.
Acceleration of the electron is $\vec{a} = \frac{\vec{F}}{m} = \left(\frac{e E_0}{m}\right) \hat{i}$.
Velocity after time $t$ is $\vec{v_2} = \vec{v_1} + \vec{a} t = v_0 \hat{i} + \left(\frac{e E_0 t}{m}\right) \hat{i} = \left(v_0 + \frac{e E_0 t}{m}\right) \hat{i}$.
The de-Broglie wavelength is given by $\lambda = \frac{h}{mv}$,which implies $\lambda \propto \frac{1}{v}$.
Therefore,$\frac{\lambda_2}{\lambda_1} = \frac{v_1}{v_2} = \frac{v_0}{v_0 + \frac{e E_0 t}{m}} = \frac{v_0}{v_0(1 + \frac{e E_0 t}{m v_0})} = \frac{1}{1 + \frac{e E_0 t}{m v_0}}$.
Thus,$\lambda_2 = \frac{\lambda}{1 + \frac{e E_0 t}{m v_0}}$.

(B) Let the initial kinetic energy be $K_1$ and the final kinetic energy be $K_2$. Given that the kinetic energy is decreased by $36 \%$,we have:
$K_2 = K_1 - 0.36 K_1 = 0.64 K_1$.
The de Broglie wavelength $\lambda$ is related to kinetic energy $K$ by the formula:
$\lambda = \frac{h}{\sqrt{2mK}} \Rightarrow \lambda \propto \frac{1}{\sqrt{K}}$.
Therefore,the ratio of the final wavelength $\lambda_2$ to the initial wavelength $\lambda_1$ is:
$\frac{\lambda_2}{\lambda_1} = \sqrt{\frac{K_1}{K_2}} = \sqrt{\frac{K_1}{0.64 K_1}} = \sqrt{\frac{1}{0.64}} = \frac{1}{0.8} = 1.25$.
This implies $\lambda_2 = 1.25 \lambda_1$.
The percentage increase in the de Broglie wavelength is:
$\frac{\Delta \lambda}{\lambda_1} \times 100 = \frac{\lambda_2 - \lambda_1}{\lambda_1} \times 100 = (1.25 - 1) \times 100 = 25 \%$.

(B) The initial momentum of the radiation incident on the surface is $P_1 = \frac{E}{c}$.
Since the surface is perfectly reflecting,the radiation is reflected back with the same energy $E$.
Therefore,the final momentum of the radiation is $P_2 = -\frac{E}{c}$ (taking the direction of incident radiation as positive).
The momentum transferred to the surface is the change in momentum of the radiation,which is given by $\Delta P = P_1 - P_2$.
Substituting the values,we get $\Delta P = \frac{E}{c} - \left(-\frac{E}{c}\right) = \frac{E}{c} + \frac{E}{c} = \frac{2E}{c}$.

(A) Given: Work function $\phi_0 = 4.2 \text{ eV}$,Wavelength $\lambda = 2000 \text{ Å}$.
Energy of incident photon $E = \frac{12400}{\lambda(\text{in Å})} \text{ eV} = \frac{12400}{2000} = 6.2 \text{ eV}$.
According to Einstein's photoelectric equation,the maximum kinetic energy $K_{\max} = E - \phi_0$.
$K_{\max} = 6.2 \text{ eV} - 4.2 \text{ eV} = 2 \text{ eV}$.
Converting kinetic energy to Joules: $K_{\max} = 2 \times 1.6 \times 10^{-19} \text{ J} = 3.2 \times 10^{-19} \text{ J}$.
Using $K_{\max} = \frac{1}{2} m v_{\max}^2$,where mass of electron $m = 9.1 \times 10^{-31} \text{ kg}$.
$v_{\max}^2 = \frac{2 \times K_{\max}}{m} = \frac{2 \times 3.2 \times 10^{-19}}{9.1 \times 10^{-31}} \approx 0.703 \times 10^{12} \text{ m}^2/\text{s}^2$.
$v_{\max} = \sqrt{0.703 \times 10^{12}} \approx 8.38 \times 10^5 \text{ m/s} \approx 8.4 \times 10^5 \text{ m/s}$.

(A) The work function $\phi_0$ is given as $9 \ eV$.
The threshold wavelength $\lambda_0$ is the longest wavelength capable of initiating the photoelectric effect.
The relationship between work function and threshold wavelength is given by $\phi_0 = \frac{hc}{\lambda_0}$.
Using the constant $hc \approx 1240 \ eV \cdot nm$,we have:
$\lambda_0 = \frac{1240 \ eV \cdot nm}{9 \ eV} \approx 137.77 \ nm$.
Converting nanometers to meters:
$\lambda_0 \approx 137.77 \times 10^{-9} \ m = 1.3777 \times 10^{-7} \ m$.
Rounding to the nearest option,we get $1.37 \times 10^{-7} \ m$.

(A) Given: Wavelength $\lambda = 4500 \ \text{Å} = 4500 \times 10^{-10} \ \text{m}$,Power $P = 150 \ \text{W}$,Efficiency $\eta = 8 \% = 0.08$.
The power of the light emitted by the lamp is $P_{\text{out}} = P \times \eta = 150 \times 0.08 = 12 \ \text{W}$.
The energy of a single photon is $E = \frac{hc}{\lambda}$.
The number of photons emitted per second $n$ is given by $n = \frac{P_{\text{out}}}{E} = \frac{P_{\text{out}} \times \lambda}{hc}$.
Using $hc \approx 1240 \ \text{eV} \cdot \text{nm} = 12400 \ \text{eV} \cdot \text{Å}$ and $1 \ \text{eV} = 1.6 \times 10^{-19} \ \text{J}$,we have $hc = 12400 \times 1.6 \times 10^{-19} \ \text{J} \cdot \text{Å} = 1.984 \times 10^{-15} \ \text{J} \cdot \text{m}$.
Substituting the values: $n = \frac{12 \times 4500 \times 10^{-10}}{6.626 \times 10^{-34} \times 3 \times 10^8} \approx 2.717 \times 10^{19} \text{ (or using } hc \approx 1.987 \times 10^{-16} \text{ J} \cdot \text{m, } n \approx 2.717 \times 10^{19} \text{)}$.
Re-evaluating with standard constants: $n = \frac{12 \times 4500 \times 10^{-10}}{6.63 \times 10^{-34} \times 3 \times 10^8} = 2.715 \times 10^{19}$. Given the options,the calculation $27.17 \times 10^{18}$ is the intended answer.

(B) The power $P$ of the transmitter is given by $P = \frac{n E_{photon}}{t}$,where $n$ is the number of photons emitted in time $t$ and $E_{photon} = \frac{hc}{\lambda}$.
Thus,the number of photons emitted per second is $\frac{n}{t} = \frac{P \lambda}{hc}$.
Given: $P = 10 \text{ kW} = 10^4 \text{ W}$,$\lambda = 500 \text{ m}$,$h = 6.63 \times 10^{-34} \text{ J s}$,and $c = 3 \times 10^8 \text{ m/s}$.
Substituting the values:
$\frac{n}{t} = \frac{10^4 \times 500}{6.63 \times 10^{-34} \times 3 \times 10^8}$
$\frac{n}{t} = \frac{5 \times 10^6}{19.89 \times 10^{-26}}$
$\frac{n}{t} \approx 0.251 \times 10^{32} = 2.51 \times 10^{31}$.
Therefore,the order of magnitude is $10^{31}$.

(C) According to Einstein's photoelectric equation,the maximum kinetic energy $K_{\max}$ is given by:
$K_{\max} = \frac{hc}{\lambda} - \phi_0 = \frac{hc}{\lambda} - \frac{hc}{\lambda_0}$
Since $K_{\max} = \frac{1}{2}mv_{\max}^2$,we have:
$\frac{1}{2}mv_{\max}^2 = hc \left( \frac{1}{\lambda} - \frac{1}{\lambda_0} \right)$
$v_{\max} = \sqrt{\frac{2hc}{m} \left( \frac{1}{\lambda} - \frac{1}{\lambda_0} \right)}$
For the first beam with $\lambda_1 = \frac{\lambda_0}{3}$:
$v_1 = \sqrt{\frac{2hc}{m} \left( \frac{3}{\lambda_0} - \frac{1}{\lambda_0} \right)} = \sqrt{\frac{2hc}{m} \cdot \frac{2}{\lambda_0}}$
For the second beam with $\lambda_2 = \frac{\lambda_0}{9}$:
$v_2 = \sqrt{\frac{2hc}{m} \left( \frac{9}{\lambda_0} - \frac{1}{\lambda_0} \right)} = \sqrt{\frac{2hc}{m} \cdot \frac{8}{\lambda_0}}$
The ratio of the maximum velocities is:
$\frac{v_1}{v_2} = \sqrt{\frac{2/\lambda_0}{8/\lambda_0}} = \sqrt{\frac{2}{8}} = \sqrt{\frac{1}{4}} = \frac{1}{2}$

(D) Given: $\lambda_1 = 300 \ nm$,$\lambda_2 = 500 \ nm$,and the ratio of maximum velocities $\frac{v_1}{v_2} = 3$.
Using Einstein's photoelectric equation: $E = \phi_0 + K_{max}$,where $K_{max} = \frac{1}{2}mv^2$.
Thus,$\frac{1}{2}mv^2 = \frac{hc}{\lambda} - \phi_0$.
Taking the ratio for the two cases:
$\left(\frac{v_1}{v_2}\right)^2 = \frac{E_1 - \phi_0}{E_2 - \phi_0} = \frac{\frac{1240}{\lambda_1} - \phi_0}{\frac{1240}{\lambda_2} - \phi_0}$.
Substituting the values: $3^2 = \frac{\frac{1240}{300} - \phi_0}{\frac{1240}{500} - \phi_0}$.
$9 = \frac{4.133 - \phi_0}{2.48 - \phi_0}$.
$9(2.48 - \phi_0) = 4.133 - \phi_0$.
$22.32 - 9\phi_0 = 4.133 - \phi_0$.
$8\phi_0 = 18.187$.
$\phi_0 \approx 2.273 \ eV \approx 2.28 \ eV$.

(B) The work function $\phi_0$ of a metal is defined by the threshold wavelength $\lambda_0$ as $\phi_0 = \frac{hc}{\lambda_0}$.
Given the threshold wavelength $\lambda_0 = 5420 Å$.
Using the relation $\phi_0 (\text{in eV}) = \frac{12400}{\lambda_0 (\text{in Å})}$.
Substituting the value: $\phi_0 = \frac{12400}{5420} eV$.
$\phi_0 \approx 2.29 eV$.
Thus,the work function of sodium is $2.29 eV$.

(B) Given:
Work function,$\phi_0 = 2 \text{ eV}$
Maximum kinetic energy,$K_{\max} = 2 \text{ eV}$
According to Einstein's photoelectric equation:
$E = \phi_0 + K_{\max}$
$E = 2 \text{ eV} + 2 \text{ eV} = 4 \text{ eV}$
The energy of a photon is given by $E = \frac{hc}{\lambda}$.
Using the relation $\lambda = \frac{12400 \text{ eV Å}}{E \text{ (in eV)}}$:
$\lambda = \frac{12400}{4} \text{ Å} = 3100 \text{ Å}$
Thus,the wavelength of the incident photon is $3100 \text{ Å}$.

(A) The magnetic energy $U$ stored in an inductor is given by the formula:
$U = \frac{1}{2} LI^2$
This implies that $U \propto I^2$.
Let the initial current be $I_1 = 2 \ A$ and the final current be $I_2 = 3 \ A$.
The ratio of the final energy $U_2$ to the initial energy $U_1$ is:
$\frac{U_2}{U_1} = \left( \frac{I_2}{I_1} \right)^2 = \left( \frac{3}{2} \right)^2 = \frac{9}{4} = 2.25$
Thus,$U_2 = 2.25 \ U_1$.
The percentage increase in the stored energy is given by:
$\Delta U \% = \left( \frac{U_2 - U_1}{U_1} \times 100 \right) \%$
$\Delta U \% = \left( \frac{2.25 \ U_1 - U_1}{U_1} \times 100 \right) \% = (1.25 \times 100) \% = 125 \%$

(B) The inductance of a solenoid is given by $L = \frac{\mu_0 N^2 A}{l}$,which implies $L \propto \frac{N^2}{l}$.
When a coil is divided into $n = 6$ equal parts,the number of turns in each part becomes $N' = \frac{N}{6}$ and the length of each part becomes $l' = \frac{l}{6}$.
The inductance of each part $L'$ is given by $L' = L \left( \frac{N'}{N} \right)^2 \left( \frac{l}{l'} \right) = L \left( \frac{1}{6} \right)^2 \left( \frac{l}{l/6} \right) = L \left( \frac{1}{36} \right) (6) = \frac{L}{6}$.
When these $6$ parts are connected in parallel,the equivalent inductance $L_e$ is given by $\frac{1}{L_e} = \sum \frac{1}{L'} = \frac{1}{L'} + \frac{1}{L'} + \dots + \frac{1}{L'} = \frac{6}{L'}$.
Substituting $L' = \frac{L}{6}$,we get $\frac{1}{L_e} = \frac{6}{L/6} = \frac{36}{L}$.
Therefore,$L_e = \frac{L}{36}$.

(B) When two coaxial coils carrying current in the same direction are brought closer to each other,the magnetic flux linked with each coil increases.
According to Lenz's law,the induced electromotive force $(EMF)$ will oppose this change in magnetic flux.
To oppose the increase in flux,the induced current flows in a direction opposite to the original current in each coil.
Consequently,the net current in both coils decreases.

(D) Given: Number of turns $N = 120$,Inductance $L = 40 \text{ mH} = 40 \times 10^{-3} \text{ H}$,Current $I = 30 \text{ mA} = 30 \times 10^{-3} \text{ A}$.
We know that the total flux linkage is given by $N\phi = LI$.
Therefore,the magnetic flux $\phi$ linked with the coil is given by $\phi = \frac{LI}{N}$.
Substituting the values: $\phi = \frac{40 \times 10^{-3} \times 30 \times 10^{-3}}{120}$.
$\phi = \frac{1200 \times 10^{-6}}{120} = 10 \times 10^{-6} \text{ Wb}$.

(B) Given: Side length $l = 0.1 \text{ m}$, Resistance of loop $R_{loop} = 1 \Omega$, Magnetic field $B = 2 \text{ Wb m}^{-2}$, Current $I = 1 \text{ mA} = 10^{-3} \text{ A}$.
First, calculate the equivalent resistance of the resistor network connected to the loop. The network consists of four $3 \Omega$ resistors in a bridge-like configuration. The equivalent resistance of this network is $R_{net} = 3 \Omega$.
The total resistance of the circuit is $R_{total} = R_{loop} + R_{net} = 1 \Omega + 3 \Omega = 4 \Omega$.
The motional electromotive force (emf) induced in the loop is given by $E = Bvl$.
Using Ohm's law, $I = \frac{E}{R_{total}} = \frac{Bvl}{R_{total}}$.
Substituting the values: $10^{-3} = \frac{2 \times v \times 0.1}{4}$.
$10^{-3} = \frac{0.2v}{4} = 0.05v$.
$v = \frac{10^{-3}}{0.05} = 0.02 \text{ m s}^{-1} = 2 \text{ cm s}^{-1}$.

(C) The self-inductance $L$ of a solenoid is given by the formula $L = \mu_0 n^2 A l$,where $n$ is the number of turns per unit length,$A$ is the cross-sectional area,and $l$ is the length of the solenoid.
Given for solenoids $A$ and $B$:
Ratio of turns per unit length: $\frac{n_A}{n_B} = \frac{1}{3}$
Ratio of lengths: $\frac{l_A}{l_B} = \frac{1}{2}$
Cross-sectional areas are equal: $A_A = A_B = A$
Since $L \propto n^2 l$,the ratio of self-inductances is:
$\frac{L_A}{L_B} = \left(\frac{n_A}{n_B}\right)^2 \times \left(\frac{l_A}{l_B}\right)$
Substituting the given values:
$\frac{L_A}{L_B} = \left(\frac{1}{3}\right)^2 \times \left(\frac{1}{2}\right) = \frac{1}{9} \times \frac{1}{2} = \frac{1}{18}$
Therefore,the ratio is $1: 18$.

(B) Given: Mutual inductance $M = 3 \text{ mH} = 3 \times 10^{-3} \text{ H}$,Resistance of circuit $Y$ $R_2 = 4 \text{ } \Omega$,Induced current in $Y$ $I_2 = 60 \times 10^{-4} \text{ A}$,Time interval $\Delta t = 0.02 \text{ s}$.
The induced electromotive force $(e_2)$ in circuit $Y$ due to change in current in circuit $X$ is given by $e_2 = M \frac{\Delta I_1}{\Delta t}$.
The induced current in circuit $Y$ is $I_2 = \frac{e_2}{R_2}$.
Substituting the expression for $e_2$,we get $I_2 = \frac{M \cdot \Delta I_1}{R_2 \cdot \Delta t}$.
Rearranging to solve for the change in current $\Delta I_1$ in circuit $X$:
$\Delta I_1 = \frac{I_2 \cdot R_2 \cdot \Delta t}{M}$.
Substituting the values:
$\Delta I_1 = \frac{60 \times 10^{-4} \times 4 \times 0.02}{3 \times 10^{-3}}$.
$\Delta I_1 = \frac{60 \times 10^{-4} \times 0.08}{3 \times 10^{-3}} = \frac{4.8 \times 10^{-4}}{3 \times 10^{-3}} = 1.6 \times 10^{-1} = 0.16 \text{ A}$.

(C) Given: Change in current $\Delta I = 14 \ A - 4 \ A = 10 \ A$.
Time interval $\Delta t = 0.2 \ ms = 0.2 \times 10^{-3} \ s$.
Induced emf $e = 150 \ V$.
The formula for induced emf in an inductor is $e = L \cdot \frac{\Delta I}{\Delta t}$.
Substituting the values: $150 = L \cdot \frac{10}{0.2 \times 10^{-3}}$.
$150 = L \cdot \frac{10}{2 \times 10^{-4}} = L \cdot 5 \times 10^4$.
$L = \frac{150}{5 \times 10^4} = 30 \times 10^{-4} \ H = 3 \times 10^{-3} \ H = 3 \ mH$.

(C) Solar radiation consists of electromagnetic waves emitted by the Sun. According to the properties of electromagnetic waves,they are transverse in nature,meaning the oscillating electric and magnetic fields are perpendicular to the direction of wave propagation. Therefore,solar radiation is a transverse electromagnetic $(EM)$ wave.

(C) The speed of an electromagnetic wave in a medium is given by $v = \frac{1}{\sqrt{\mu \varepsilon}}$.
We know that $\mu = \mu_0 \mu_r$ and $\varepsilon = \varepsilon_0 \varepsilon_r$.
Thus,$v = \frac{1}{\sqrt{\mu_0 \mu_r \varepsilon_0 \varepsilon_r}} = \frac{1}{\sqrt{\mu_0 \varepsilon_0}} \cdot \frac{1}{\sqrt{\mu_r \varepsilon_r}} = \frac{c}{\sqrt{\mu_r \varepsilon_r}}$,where $c = 3 \times 10^8 \ m/s$ is the speed of light in vacuum.
Given $v = 2 \times 10^8 \ m/s$ and $\mu_r = 1$.
Substituting these values: $2 \times 10^8 = \frac{3 \times 10^8}{\sqrt{1 \times \varepsilon_r}}$.
Squaring both sides: $4 = \frac{9}{\varepsilon_r}$.
Therefore,$\varepsilon_r = \frac{9}{4} = 2.25$.

(D) The magnetic field is given by $\vec{B} = B_0 r t \hat{k}$.
According to Faraday's law of induction in integral form,$\oint \vec{E} \cdot d\vec{l} = -\frac{d\phi_B}{dt}$.
For a circular path of radius $r$ centered on the $z$-axis,the line integral of the electric field is $E(2\pi r)$.
The magnetic flux $\phi_B$ through the circular area of radius $r$ is $\int_0^r B(r') (2\pi r') dr'$.
Substituting $B(r') = B_0 r' t$,we get $\phi_B = \int_0^r (B_0 r' t) (2\pi r') dr' = 2\pi B_0 t \int_0^r r'^2 dr' = 2\pi B_0 t \frac{r^3}{3}$.
Now,differentiating with respect to time $t$: $\frac{d\phi_B}{dt} = \frac{d}{dt} \left( \frac{2\pi B_0 t r^3}{3} \right) = \frac{2\pi B_0 r^3}{3}$.
Equating the magnitudes: $E(2\pi r) = \frac{2\pi B_0 r^3}{3}$.
Solving for $E$,we get $E = \frac{B_0 r^2}{3}$.

(D) The instantaneous electric energy density in an electromagnetic wave is given by $u_E = 1/2 \epsilon_0 E^2$.
Since $E = E_0 \sin(\omega t - kx)$,we have $u_E = 1/2 \epsilon_0 E_0^2 \sin^2(\omega t - kx)$.
The average value of $\sin^2(\theta)$ over a complete cycle is $1/2$.
Therefore,the average electric energy density is $\langle u_E \rangle = 1/2 \epsilon_0 E_0^2 \times \langle \sin^2(\omega t - kx) \rangle$.
$\langle u_E \rangle = 1/2 \epsilon_0 E_0^2 \times 1/2 = 1/4 \epsilon_0 E_0^2$.

(C) According to the principles of electromagnetism,a charge at rest produces only an electric field. $A$ charge moving with uniform velocity produces both an electric field and a magnetic field,but it does not radiate energy. An accelerating charge produces a time-varying electric field,which in turn produces a time-varying magnetic field. This continuous process results in the propagation of electromagnetic waves through space.

(D) The average energy density $U_{av}$ of an electromagnetic wave is given by the formula $U_{av} = \frac{1}{2} \varepsilon_0 E_0^2$,where $E_0$ is the peak electric field.
Given $E_{rms} = 660 \ NC^{-1}$.
We know that $E_0 = \sqrt{2} E_{rms} = \sqrt{2} \times 660 \ NC^{-1}$.
Substituting this into the energy density formula:
$U_{av} = \frac{1}{2} \varepsilon_0 (\sqrt{2} E_{rms})^2 = \frac{1}{2} \varepsilon_0 (2 E_{rms}^2) = \varepsilon_0 E_{rms}^2$.
Using $\varepsilon_0 = 8.85 \times 10^{-12} \ C^2 N^{-1} m^{-2}$:
$U_{av} = 8.85 \times 10^{-12} \times (660)^2$.
$U_{av} = 8.85 \times 10^{-12} \times 435600$.
$U_{av} \approx 3.85 \times 10^{-6} \ J \ m^{-3}$.

(A) For plane electromagnetic waves,the electric field $\vec{E}$ and magnetic field $\vec{B}$ are mutually perpendicular,i.e.,$\vec{E} \cdot \vec{B} = 0$. Also,the direction of propagation is given by $\vec{E} \times \vec{B}$.
Given the propagation is in the $+Z$-direction $(\hat{k})$,we must have $\vec{E} \times \vec{B} \propto \hat{k}$.
Checking option $A$: $\vec{E} = (-2 \hat{i} - 3 \hat{j})$ and $\vec{B} = (3 \hat{i} - 2 \hat{j})$.
Dot product: $\vec{E} \cdot \vec{B} = (-2)(3) + (-3)(-2) = -6 + 6 = 0$. This satisfies the perpendicularity condition.
Cross product: $\vec{E} \times \vec{B} = (-2 \hat{i} - 3 \hat{j}) \times (3 \hat{i} - 2 \hat{j}) = 4(\hat{i} \times \hat{j}) - 9(\hat{j} \times \hat{i}) = 4\hat{k} - 9(-\hat{k}) = 13\hat{k}$.
Since the result is in the $+Z$-direction,option $A$ is correct.

(C) The structure of solids is investigated using $X$-ray diffraction. Since the wavelength of $X$-rays is of the order of the interatomic spacing in crystals (approximately $1 \ \text{\AA}$), they are diffracted by the atomic planes in a solid. This phenomenon is used to determine the crystal structure of materials.

(A) According to Coulomb's law,the force between two charges is given by $F = \frac{k q_1 q_2}{r^2}$.
Given $q_1 = 6 \mu C$,$q_2 = 10 \mu C$,and $F = 30 \ N$.
So,$30 = \frac{k(6)(10)}{r^2} \Rightarrow \frac{k}{r^2} = \frac{30}{60} = 0.5$.
Now,if an additional charge of $-8 \mu C$ is added to each,the new charges are:
$q_1' = 6 \mu C - 8 \mu C = -2 \mu C$
$q_2' = 10 \mu C - 8 \mu C = 2 \mu C$
The new force $F'$ is $F' = \frac{k q_1' q_2'}{r^2} = \frac{k}{r^2} \times (-2) \times (2)$.
Substituting $\frac{k}{r^2} = 0.5$,we get $F' = 0.5 \times (-4) = -2 \ N$.
The negative sign indicates an attractive force. Thus,they attract with a force of $2 \ N$.

(A) For the particles to experience no net force,the electrostatic repulsive force must be balanced by the gravitational attractive force.
$F_e = F_g$
$\frac{1}{4 \pi \epsilon_0} \cdot \frac{q^2}{r^2} = \frac{G m^2}{r^2}$
Canceling $r^2$ from both sides:
$\frac{1}{4 \pi \epsilon_0} q^2 = G m^2$
Rearranging the terms to find $\frac{q}{m}$:
$\frac{q^2}{m^2} = 4 \pi \epsilon_0 G$
Taking the square root on both sides:
$\frac{q}{m} = \sqrt{4 \pi \epsilon_0 G}$

(B) Initial charges are $Q_1 = 80 \mu\text{C}$ and $Q_2 = 40 \mu\text{C}$.
Radii are $r_1 = 4 \text{ cm}$ and $r_2 = 6 \text{ cm}$.
When connected by a wire,charge flows until both spheres reach the same potential.
The new charge $Q_1^{\prime}$ on sphere $A$ is given by:
$Q_1^{\prime} = \left( \frac{r_1}{r_1 + r_2} \right) (Q_1 + Q_2) = \left( \frac{4}{4 + 6} \right) (80 + 40) = \left( \frac{4}{10} \right) (120) = 48 \mu\text{C}$.
The charge that flows from sphere $A$ to sphere $B$ is:
$\Delta Q = Q_1 - Q_1^{\prime} = 80 \mu\text{C} - 48 \mu\text{C} = 32 \mu\text{C}$.
Since the result is positive,the charge flows from $A$ to $B$.

(D) Let the distance between $+q$ and $+2q$ be $r$,and the distance between $+2q$ and $+4q$ be $r$. Thus,the distance between $+q$ and $+4q$ is $2r$.
The net electrostatic force $F_1$ on charge $+q$ is the sum of forces due to $+2q$ and $+4q$:
$F_1 = \frac{k(q)(2q)}{r^2} + \frac{k(q)(4q)}{(2r)^2} = \frac{2kq^2}{r^2} + \frac{4kq^2}{4r^2} = \frac{2kq^2}{r^2} + \frac{kq^2}{r^2} = \frac{3kq^2}{r^2}$.
The net electrostatic force $F_2$ on charge $+4q$ is the sum of forces due to $+2q$ and $+q$:
$F_2 = \frac{k(4q)(2q)}{r^2} + \frac{k(4q)(q)}{(2r)^2} = \frac{8kq^2}{r^2} + \frac{4kq^2}{4r^2} = \frac{8kq^2}{r^2} + \frac{kq^2}{r^2} = \frac{9kq^2}{r^2}$.
The ratio of the net electrostatic force on charges $+q$ and $+4q$ is:
$\frac{F_1}{F_2} = \frac{3kq^2/r^2}{9kq^2/r^2} = \frac{3}{9} = 1:3$.

(C) The electric field $\overrightarrow{E}$ at a point on the equatorial line of an electric dipole with dipole moment $\overrightarrow{P}$ is given by the formula: $\overrightarrow{E} = -\frac{1}{4\pi\epsilon_0} \frac{\overrightarrow{P}}{r^3}$.
This expression shows that the direction of the electric field $\overrightarrow{E}$ is opposite to the direction of the electric dipole moment $\overrightarrow{P}$.
Since the vectors are anti-parallel,the angle between them is $180^{\circ}$.

(B) The electric dipole moment is given by $p = q \cdot l$,where $q$ is the magnitude of the total positive or negative charge and $l$ is the separation distance between the centres of charge.
For a neutral $NH_3$ molecule,the total number of electrons (or protons) is $7 + (3 \times 1) = 10$.
Thus,the total charge $q = 10e = 10 \times 1.6 \times 10^{-19} \ C = 1.6 \times 10^{-18} \ C$.
Given $p = 5 \times 10^{-30} \ C \cdot m$.
Using the formula $l = \frac{p}{q}$:
$l = \frac{5 \times 10^{-30}}{1.6 \times 10^{-18}}$
$l = 3.125 \times 10^{-12} \ m$.

(A) The electric field intensity $(E)$ at a distance $r$ from a long,uniformly charged straight wire is given by the formula:
$E = \frac{\lambda}{2 \pi \epsilon_0 r} = \frac{2 k \lambda}{r}$
Given:
Linear charge density $\lambda = 0.2 \ \mu Cm^{-1} = 0.2 \times 10^{-6} \ Cm^{-1}$
Distance $r = 3 \ m$
Coulomb constant $k = 9 \times 10^9 \ Nm^2C^{-2}$
Substituting the values:
$E = \frac{2 \times (9 \times 10^9) \times (0.2 \times 10^{-6})}{3}$
$E = \frac{18 \times 10^9 \times 0.2 \times 10^{-6}}{3}$
$E = 6 \times 0.2 \times 10^3$
$E = 1.2 \times 10^3 \ Vm^{-1}$

(A) Given: Mass of deuteron $m = 3.2 \times 10^{-27} \ kg$,Charge of deuteron $q = e = 1.6 \times 10^{-19} \ C$,Acceleration due to gravity $g = 9.8 \ m/s^2$.
For the deuteron to be suspended freely in air,the upward electric force must balance the downward gravitational force.
Equating the forces: $qE = mg$.
Rearranging for the electric field: $E = \frac{mg}{q}$.
Substituting the values: $E = \frac{3.2 \times 10^{-27} \times 9.8}{1.6 \times 10^{-19}}$.
$E = 2 \times 9.8 \times 10^{-27+19} \ NC^{-1}$.
$E = 19.6 \times 10^{-8} \ NC^{-1}$.

(D) Given:
Mass $m = 0.5 \ g = 0.5 \times 10^{-3} \ kg$
Charge $q = 10 \ \mu C = 10 \times 10^{-6} \ C$
Electric field $E = 8 \ NC^{-1}$
Initial velocity $u = 0 \ ms^{-1}$
Time $t = 5 \ s$
The force acting on the particle is $F = qE$.
According to Newton's second law,$F = ma$,so the acceleration $a = \frac{qE}{m}$.
Substituting the values:
$a = \frac{10 \times 10^{-6} \times 8}{0.5 \times 10^{-3}} = \frac{80 \times 10^{-6}}{0.5 \times 10^{-3}} = 160 \times 10^{-3} = 0.16 \ ms^{-2}$.
Using the first equation of motion $v = u + at$:
$v = 0 + (0.16 \times 5) = 0.8 \ ms^{-1}$.

(D) According to Gauss's theorem,the total electric flux $\phi_{\text{total}}$ linked with a closed surface is given by $\phi_{\text{total}} = \frac{Q_{\text{enclosed}}}{\epsilon_0}$.
For a cube with a charge '$q$' at its centre,the total flux through all six faces is $\phi_{\text{total}} = \frac{q}{\epsilon_0}$.
Since the cube is symmetric,the electric flux linked with each of the six faces is equal.
Therefore,the flux through each face is $\phi_{\text{face}} = \frac{\phi_{\text{total}}}{6} = \frac{q}{6 \epsilon_0}$.

(D) Given: $\vec{E} = (30 \hat{i} + 40 \hat{j}) \text{ NC}^{-1}$ and $V(0,0) = 0 \text{ V}$.
We know that the relation between electric field and electric potential is given by $dV = -\vec{E} \cdot d\vec{r}$.
Here,$d\vec{r} = dx \hat{i} + dy \hat{j} + dz \hat{k}$.
Integrating from the origin $(0,0)$ to the point $(1,2)$:
$\int_{V(0,0)}^{V(1,2)} dV = -\int_{(0,0)}^{(1,2)} (30 \hat{i} + 40 \hat{j}) \cdot (dx \hat{i} + dy \hat{j})$
$V(1,2) - V(0,0) = -\left[ \int_{0}^{1} 30 dx + \int_{0}^{2} 40 dy \right]$
$V(1,2) - 0 = -[30(1) + 40(2)]$
$V(1,2) = -(30 + 80) = -110 \text{ V}$.

(A) Given: Mass $m = 2 \ g = 2 \times 10^{-3} \ kg$,Charge $q = 6 \ \mu C = 6 \times 10^{-6} \ C$,Potential difference $V = 60 \ V$.
According to the work-energy theorem,the kinetic energy gained by the particle is equal to the work done by the electric field:
$K.E. = qV$
$\frac{1}{2}mv^2 = qV$
$v^2 = \frac{2qV}{m}$
$v = \sqrt{\frac{2qV}{m}}$
Substituting the values:
$v = \sqrt{\frac{2 \times (6 \times 10^{-6} \ C) \times (60 \ V)}{2 \times 10^{-3} \ kg}}$
$v = \sqrt{\frac{720 \times 10^{-6}}{2 \times 10^{-3}}}$
$v = \sqrt{360 \times 10^{-3}} = \sqrt{0.36} = 0.6 \ ms^{-1}$.

(A) The electrostatic potential energy $U$ of a system of point charges is given by the sum of the potential energies of all unique pairs of charges.
For three charges $q_1, q_2, q_3$ at distances $r_{12}, r_{23}, r_{31}$,the potential energy is $U = \frac{1}{4 \pi \epsilon_0} \left( \frac{q_1 q_2}{r_{12}} + \frac{q_2 q_3}{r_{23}} + \frac{q_3 q_1}{r_{31}} \right)$.
In this case,$q_1 = q_2 = q_3 = q$ and $r_{12} = r_{23} = r_{31} = L$.
Substituting these values,we get:
$U = \frac{1}{4 \pi \epsilon_0} \left( \frac{q^2}{L} + \frac{q^2}{L} + \frac{q^2}{L} \right)$
$U = \frac{1}{4 \pi \epsilon_0} \cdot \frac{3 q^2}{L}$

(C) Given charges at the corners of a square of side $a = 1 \text{ m}$ are:
$q_1 = +1 \times 10^{-8} \text{ C}$
$q_2 = -2 \times 10^{-8} \text{ C}$
$q_3 = +3 \times 10^{-8} \text{ C}$
$q_4 = +2 \times 10^{-8} \text{ C}$
The distance $r$ from each corner to the centre of the square is half the diagonal length:
$r = \frac{\sqrt{2}a}{2} = \frac{a}{\sqrt{2}} = \frac{1}{\sqrt{2}} \text{ m}$
The electric potential $V$ at the centre is the algebraic sum of the potentials due to each charge:
$V = \frac{k q_1}{r} + \frac{k q_2}{r} + \frac{k q_3}{r} + \frac{k q_4}{r} = \frac{k}{r} (q_1 + q_2 + q_3 + q_4)$
Substituting the values:
$V = \frac{9 \times 10^9}{1/\sqrt{2}} (1 - 2 + 3 + 2) \times 10^{-8}$
$V = 9 \sqrt{2} \times 10^1 \times (4)$
$V = 36 \sqrt{2} \times 10 \approx 36 \times 1.414 \times 10 = 509.04 \text{ V}$
Rounding to the nearest given option, $V \approx 510 \text{ V}$.

(B) Given:
$q_1 = 5 \text{ nC} = 5 \times 10^{-9} \text{ C}$
$q_2 = -2 \text{ nC} = -2 \times 10^{-9} \text{ C}$
The distance between the charges is $r = (23 - 5) \text{ cm} = 18 \text{ cm} = 18 \times 10^{-2} \text{ m}$.
The electrostatic potential energy $U$ of a system of two point charges is given by:
$U = \frac{k q_1 q_2}{r}$
Substituting the values:
$U = \frac{(9 \times 10^9 \text{ N m}^2/\text{C}^2) \times (5 \times 10^{-9} \text{ C}) \times (-2 \times 10^{-9} \text{ C})}{18 \times 10^{-2} \text{ m}}$
$U = \frac{9 \times 5 \times (-2) \times 10^{9-9-9}}{18 \times 10^{-2}} \text{ J}$
$U = \frac{-90 \times 10^{-9}}{18 \times 10^{-2}} \text{ J}$
$U = -5 \times 10^{-7} \text{ J}$

(C) Given:
Electrostatic potential energy $U = 1.8 \mu \text{ J} = 1.8 \times 10^{-6} \text{ J}$
Separation $r = 10 \text{ cm} = 0.1 \text{ m}$
Charge $Q_1 = 4 \text{ nC} = 4 \times 10^{-9} \text{ C}$
Coulomb's constant $k = 9 \times 10^9 \text{ N m}^2/\text{C}^2$
The formula for electrostatic potential energy of a two-charge system is:
$U = \frac{k Q_1 Q}{r}$
Substituting the values:
$1.8 \times 10^{-6} = \frac{9 \times 10^9 \times 4 \times 10^{-9} \times Q}{0.1}$
$1.8 \times 10^{-6} = \frac{36 \times Q}{0.1}$
$1.8 \times 10^{-6} = 360 \times Q$
$Q = \frac{1.8 \times 10^{-6}}{360}$
$Q = 0.005 \times 10^{-6} \text{ C}$
$Q = 5 \times 10^{-9} \text{ C} = 5 \text{ nC}$

(C) Let the radius of each small sphere be $r$ and the charge on each be $q$. The potential on each small sphere is given by $V_s = \frac{kq}{r} = 60 \text{ mV} = 0.06 \text{ V}$.
When $125$ small spheres coalesce to form a big sphere of radius $R$, the volume remains conserved:
$125 \times (\frac{4}{3} \pi r^3) = \frac{4}{3} \pi R^3 \Rightarrow R^3 = 125 r^3 \Rightarrow R = 5r$.
The total charge on the big sphere is $Q = 125q$.
The potential on the big sphere is $V_B = \frac{kQ}{R} = \frac{k(125q)}{5r} = 25 \times (\frac{kq}{r})$.
Substituting the value of $V_s$: $V_B = 25 \times 0.06 \text{ V} = 1.5 \text{ V}$.

(D) The torque $\tau$ experienced by a current-carrying coil in a magnetic field is given by $\tau = MB \sin \theta$,where $\theta$ is the angle between the magnetic field and the normal to the plane of the coil.
Maximum torque $\tau_{\max} = MB$ (when $\theta = 90^{\circ}$).
Given that $\tau = 80 \%$ of $\tau_{\max} = 0.8 \tau_{\max} = \frac{4}{5} MB$.
Equating the two expressions: $MB \sin \theta = \frac{4}{5} MB$.
This simplifies to $\sin \theta = \frac{4}{5}$.
Using the trigonometric identity $\tan \theta = \frac{\sin \theta}{\cos \theta} = \frac{\sin \theta}{\sqrt{1 - \sin^2 \theta}}$,we get $\tan \theta = \frac{4/5}{\sqrt{1 - (4/5)^2}} = \frac{4/5}{\sqrt{9/25}} = \frac{4/5}{3/5} = \frac{4}{3}$.
Therefore,$\theta = \tan ^{-1}\left(\frac{4}{3}\right)$.