JEE Main 2024 Physics Question Paper with Answer and Solution

(B) Given position: $x = t^3 - 6t^2 + 20t + 15 \ m$.
Velocity $v$ is the first derivative of position with respect to time: $v = \frac{dx}{dt} = 3t^2 - 12t + 20 \ m/s$.
Acceleration $a$ is the derivative of velocity with respect to time: $a = \frac{dv}{dt} = 6t - 12 \ m/s^2$.
Set acceleration to zero to find the time: $6t - 12 = 0 \implies t = 2 \ s$.
Substitute $t = 2 \ s$ into the velocity equation: $v = 3(2)^2 - 12(2) + 20 = 12 - 24 + 20 = 8 \ m/s$.

(C) The equivalent degree of freedom $f_{eq}$ for a mixture of gases is given by the formula:
$f_{eq} = \frac{n_1 f_1 + n_2 f_2}{n_1 + n_2}$
For a diatomic gas,the degree of freedom is $f_{eq} = 5$.
Given:
$n_1 = N$,$f_1 = 6$ (polyatomic gas)
$n_2 = 2$,$f_2 = 3$ (monoatomic gas)
Substituting these values into the formula:
$5 = \frac{(N)(6) + (2)(3)}{N + 2}$
Multiplying both sides by $(N + 2)$:
$5(N + 2) = 6N + 6$
$5N + 10 = 6N + 6$
Rearranging the terms to solve for $N$:
$10 - 6 = 6N - 5N$
$N = 4$

(A) The Young's modulus $Y$ is given by $Y = \frac{\text{Stress}}{\text{Strain}} = \frac{F/A}{\Delta l/l} = \frac{Fl}{A\Delta l}$.
Rearranging for extension,we get $\Delta l = \frac{Fl}{AY}$.
Since volume $V = A \times l$,we can write $l = \frac{V}{A}$.
Substituting $l$ into the extension formula: $\Delta l = \frac{F(V/A)}{AY} = \frac{FV}{A^2Y}$.
Since $Y$ and $V$ are the same for both wires,$\Delta l \propto \frac{F}{A^2}$.
Given $\Delta l_1 = \Delta l_2$,we have $\frac{F_1}{A_1^2} = \frac{F_2}{A_2^2}$.
Therefore,$\frac{F_1}{F_2} = \frac{A_1^2}{A_2^2} = \left(\frac{A_1}{A_2}\right)^2$.
Given the ratio of areas $\frac{A_1}{A_2} = \frac{4}{1}$,we get $\frac{F_1}{F_2} = (4)^2 = 16$.

(A) The equation for displacement starting from the mean position is $x(t) = A \sin(\omega t)$.
Given $T = 6 \pi \text{ s}$,the angular frequency is $\omega = \frac{2 \pi}{T} = \frac{2 \pi}{6 \pi} = \frac{1}{3} \text{ rad/s}$.
At $x = A$,the particle is at the extreme position. The time taken to reach the mean position $(x=0)$ from $x=A$ is $T/4 = (6 \pi)/4 = 1.5 \pi \text{ s}$.
However,the question asks for the time to travel from $x=A$ to $x=\frac{\sqrt{3}}{2} A$.
Using the phasor diagram,the position $x = A \sin(\theta)$ corresponds to the projection on the vertical axis.
At $x = A$,the phase angle $\theta_1 = \frac{\pi}{2}$.
At $x = \frac{\sqrt{3}}{2} A$,$\sin(\theta_2) = \frac{\sqrt{3}}{2}$,so $\theta_2 = \frac{\pi}{3}$.
The change in phase is $\Delta \theta = \theta_1 - \theta_2 = \frac{\pi}{2} - \frac{\pi}{3} = \frac{\pi}{6}$.
Since $\Delta \theta = \omega \Delta t$,we have $\frac{\pi}{6} = \frac{1}{3} \Delta t$.
Therefore,$\Delta t = \frac{3 \pi}{6} = \frac{\pi}{2} \text{ s}$.
Comparing this with $\frac{\pi}{x} \text{ s}$,we get $x = 2$.

(A) Given that the normal acceleration $a_c = \frac{v^2}{r}$ and tangential acceleration $a_t = \frac{dv}{dt}$ are equal:
$\frac{v^2}{r} = \frac{dv}{dt}$
Integrating with respect to time from $t=0$ $(v=4 \ m/s)$ to $t$:
$\int_{4}^{v} \frac{dv}{v^2} = \int_{0}^{t} \frac{dt}{r}$
$\left[ -\frac{1}{v} \right]_{4}^{v} = \frac{t}{r}$
$-\frac{1}{v} + \frac{1}{4} = \frac{t}{0.5} = 2t$
$\frac{1}{v} = \frac{1}{4} - 2t = \frac{1-8t}{4} \implies v = \frac{4}{1-8t}$
Since $v = \frac{ds}{dt}$,we integrate to find the distance $s$ for one revolution $(s = 2\pi r = 2\pi(0.5) = \pi \ m)$:
$\int_{0}^{\pi} ds = \int_{0}^{t} \frac{4}{1-8t} dt$
$\pi = 4 \left[ \frac{\ln(1-8t)}{-8} \right]_{0}^{t}$
$\pi = -\frac{1}{2} \ln(1-8t)$
$-2\pi = \ln(1-8t)$
$e^{-2\pi} = 1-8t$
$8t = 1 - e^{-2\pi}$
$t = \frac{1}{8} [1 - e^{-2\pi}] \ s$
Comparing with the given form,$\alpha = 8$.

(B) The equation of the line is $x - y + 4 = 0$.
The angular momentum $L$ of a particle about the origin is given by $L = mvd$,where $m$ is the mass,$v$ is the speed,and $d$ is the perpendicular distance from the origin to the line of motion.
The perpendicular distance $d$ from the origin $(0,0)$ to the line $Ax + By + C = 0$ is $d = \frac{|C|}{\sqrt{A^2 + B^2}}$.
Here,$A = 1$,$B = -1$,and $C = 4$. So,$d = \frac{|4|}{\sqrt{1^2 + (-1)^2}} = \frac{4}{\sqrt{2}} = 2\sqrt{2} \,m$.
Given mass $m = 5 \,kg$ and speed $v = 3\sqrt{2} \,m/s$.
Therefore,$L = 5 \times (3\sqrt{2}) \times (2\sqrt{2}) = 5 \times 3 \times 2 \times 2 = 60 \,kg \,m^2/s$.

(C) $1$. Coefficient of Viscosity $(\eta)$: From $F = \eta A \frac{dv}{dy}$,we have $\eta = \frac{F}{A(dv/dy)}$. Dimensions: $\frac{[M L T^{-2}]}{[L^2][T^{-1}]} = [M L^{-1} T^{-1}]$. (Matches $III$)
$2$. Surface Tension $(S)$: $S = \frac{F}{l}$. Dimensions: $\frac{[M L T^{-2}]}{[L]} = [M L^0 T^{-2}]$. (Matches $IV$)
$3$. Angular momentum $(L)$: $L = mvr$. Dimensions: $[M][L T^{-1}][L] = [M L^2 T^{-1}]$. (Matches $II$)
$4$. Rotational Kinetic energy $(K)$: $K = \frac{1}{2} I \omega^2$. Dimensions: $[M L^2][T^{-1}]^2 = [M L^2 T^{-2}]$. (Matches $I$)
Thus,the correct matching is $A-III, B-IV, C-II, D-I$.

(B) Let the acceleration of the $4 \,kg$ block be $a$ downwards. Due to the constraint, the acceleration of the $2 \,kg$ block up the incline will be $2a$.
For the $4 \,kg$ block: $4g - 2T = 4a \Rightarrow 2g - T = 2a$ (Equation $1$)
For the $2 \,kg$ block: $T - m_2g \sin(30^{\circ}) = m_2(2a) \Rightarrow T - 2g(0.5) = 2(2a) \Rightarrow T - g = 4a$ (Equation $2$)
Adding Equation $1$ and Equation $2$: $(2g - T) + (T - g) = 2a + 4a \Rightarrow g = 6a \Rightarrow a = \frac{g}{6}$.
The acceleration of the $2 \,kg$ block is $2a = 2(\frac{g}{6}) = \frac{g}{3}$.

(C) Young's modulus is an intrinsic property of the material of the wire.
It depends only on the nature of the material and the temperature,not on the dimensions of the wire such as its length $L$ or cross-sectional area $A$.
Therefore,even if the length is doubled and the cross-sectional area is halved,the Young's modulus $Y$ remains unchanged.

(B) Since the track is frictionless,the total mechanical energy of the particle is conserved.
Applying the Law of Conservation of Mechanical Energy between points $A$ and $B$:
$KE_A + PE_A = KE_B + PE_B$
Here,$KE_A = 0$ (as it is gently pushed,initial velocity is negligible),$PE_A = mgh_A$,$KE_B = \frac{1}{2}mv^2$,and $PE_B = mgh_B$.
Given: $h_A = 1 \ m$,$h_B = 0.5 \ m$,and $g = 10 \ m/s^2$.
Substituting the values:
$0 + mg(1) = \frac{1}{2}mv^2 + mg(0.5)$
$mg(1 - 0.5) = \frac{1}{2}mv^2$
$mg(0.5) = \frac{1}{2}mv^2$
$g = v^2$
$v = \sqrt{g} = \sqrt{10} \ m/s$.

(A) Let $R_E$ be the radius of the Earth and $h$ be the height of the point above the surface. The distance from the center of the Earth is $r = R_E + h$.
Gravitational potential $V = -\frac{G M_E}{r} = -5.12 \times 10^7 \,J/kg$ ... $(i)$
Acceleration due to gravity $g' = \frac{G M_E}{r^2} = 6.4 \,m/s^2$ ... $(ii)$
Dividing equation $(i)$ by equation $(ii)$:
$\frac{V}{g'} = \frac{-G M_E / r}{G M_E / r^2} = -r$
$r = -\frac{V}{g'} = -\frac{-5.12 \times 10^7}{6.4} = 0.8 \times 10^7 \,m = 8000 \,km$
Since $r = R_E + h$, we have $h = r - R_E = 8000 \,km - 6400 \,km = 1600 \,km$.

(C) The equation of a polytropic process is $PV^x = \text{constant}$,which can be written as $\log P + x \log V = \text{constant}$,or $\log P = -x \log V + \text{constant}$.
Comparing this with the equation of a line $y = mx + c$,the slope $m = -x$.
For process $A$,the slope is $\tan(\theta_A) = \gamma$. Thus,$-x_A = \gamma$,so $x_A = -\gamma$.
The molar heat capacity for a polytropic process is $C = C_V + \frac{R}{1-x}$.
For process $A$,$C_A = C_V + \frac{R}{1 - (-\gamma)} = C_V + \frac{R}{1+\gamma}$. Since $\gamma > 1$,$C_A$ is a finite positive value.
For process $B$,the slope is $\tan(45^\circ) = 1$. Thus,$-x_B = 1$,so $x_B = -1$.
For process $B$,$C_B = C_V + \frac{R}{1 - (-1)} = C_V + \frac{R}{2}$.
Comparing the values:
$C_P = C_V + R$
$C_B = C_V + 0.5R$
$C_A = C_V + \frac{R}{1+\gamma}$ (where $1 < \gamma < 1.67$,so $0.37R < \frac{R}{1+\gamma} < 0.5R$)
Thus,$C_P > C_B > C_A > C_V$.

(D) The impulse $\vec{I}$ is equal to the change in momentum $\Delta \vec{P} = \vec{P}_f - \vec{P}_i$.
Mass $m = 100 \,g = 0.1 \,kg$.
Velocity just before hitting the ground: $v_i = -\sqrt{2gh_1} = -\sqrt{2 \times 9.8 \times 10} = -\sqrt{196} = -14 \,m/s$.
Velocity just after rebounding: $v_f = \sqrt{2gh_2} = \sqrt{2 \times 9.8 \times 5} = \sqrt{98} = 7\sqrt{2} \approx 9.9 \,m/s$.
Impulse $I = m(v_f - v_i) = 0.1 \times (9.9 - (-14)) = 0.1 \times (23.9) = 2.39 \,kg \,m/s$.

(A) The angular momentum $L$ of a particle about the point of projection is given by $L = \vec{r} \times \vec{p} = m(\vec{r} \times \vec{v})$.
At the maximum height, the vertical component of velocity is zero, so the velocity is purely horizontal: $v_x = u \cos \theta$.
The horizontal distance (range) at maximum height is $x = \frac{R}{2} = \frac{u^2 \sin \theta \cos \theta}{g}$.
The maximum height is $h = \frac{u^2 \sin^2 \theta}{2g}$.
The angular momentum is $L = m v_x h = m (u \cos \theta) \left( \frac{u^2 \sin^2 \theta}{2g} \right)$.
Substituting $\theta = 30^{\circ}$, $\cos 30^{\circ} = \frac{\sqrt{3}}{2}$, and $\sin 30^{\circ} = \frac{1}{2}$:
$L = m u \left( \frac{\sqrt{3}}{2} \right) \frac{u^2 (1/2)^2}{2g} = m u \left( \frac{\sqrt{3}}{2} \right) \frac{u^2}{8g} = \frac{\sqrt{3} m u^3}{16g}$.

(D) The root mean square (r.m.s.) velocity of a gas molecule is given by the formula $v_{rms} = \sqrt{\frac{3RT}{M}}$, where $R$ is the universal gas constant, $T$ is the absolute temperature in Kelvin, and $M$ is the molar mass of the gas.
Given, the r.m.s. velocity of hydrogen $(H_2)$ at temperature $T_H$ is equal to the r.m.s. velocity of oxygen $(O_2)$ at temperature $T_O = 47^{\circ} C = 47 + 273 = 320 \,K$.
The molar mass of hydrogen $M_H = 2 \,g/mol$ and the molar mass of oxygen $M_O = 32 \,g/mol$.
Equating the velocities: $\sqrt{\frac{3RT_H}{M_H}} = \sqrt{\frac{3RT_O}{M_O}}$.
Squaring both sides: $\frac{T_H}{M_H} = \frac{T_O}{M_O}$.
Substituting the values: $\frac{T_H}{2} = \frac{320}{32}$.
$T_H = 2 \times 10 = 20 \,K$.

(D) Let the mass of each block be $m = 3 \,kg$. The total mass of the system is $3m = 9 \,kg$.
The driving force is the weight of block $R$,which is $F = mg = 3 \times 10 = 30 \,N$.
The acceleration of the system is $a = \frac{F}{\text{total mass}} = \frac{30}{3m} = \frac{30}{9} = \frac{10}{3} \,m/s^2$.
Wire $B$ connects block $R$ to block $Q$. The tension $T_1$ in wire $B$ can be found by considering the motion of block $R$:
$mg - T_1 = ma$
$30 - T_1 = 3 \times \frac{10}{3} = 10$
$T_1 = 30 - 10 = 20 \,N$.
The stress in wire $B$ is $\sigma = \frac{T_1}{A}$,where $A = 0.005 \,cm^2 = 0.005 \times 10^{-4} \,m^2 = 5 \times 10^{-7} \,m^2$.
$\sigma = \frac{20}{5 \times 10^{-7}} = 4 \times 10^7 \,N/m^2$.
The longitudinal strain is $\epsilon = \frac{\sigma}{Y} = \frac{4 \times 10^7}{2 \times 10^{11}} = 2 \times 10^{-4}$.
Thus,the strain is $2 \times 10^{-4}$.

(B) Let the initial velocity at time $t$ be $u$ and acceleration be $a$.
Given that the increase in velocity in $1 \ s$ is $50 \ m/s$,we have $v = u + a(1) = u + 50$.
Thus,$a = 50 \ m/s^2$.
The displacement in the interval $t$ to $t+1$ is given by $s = ut + \frac{1}{2}at^2$. For the interval of $1 \ s$ starting at $t$,the displacement is $s = u(1) + \frac{1}{2}a(1)^2 = 125$.
Substituting $a = 50$,we get $u + 25 = 125$,which implies $u = 100 \ m/s$.
The distance travelled in the $(t+2)^{th}$ second is given by $S_n = u + \frac{a}{2}(2n - 1)$,where $n = t+2$.
However,the question asks for the distance in the $(t+2)^{th}$ second relative to the start of the motion. Since $u$ is the velocity at time $t$,the distance in the next second (the $(t+1)^{th}$ second) is $125 \ m$. The distance in the $(t+2)^{th}$ second is $S = (u+a) + \frac{a}{2} = 100 + 50 + 25 = 175 \ m$.

(D) The moment of inertia of the disc is $I = \frac{1}{2} M R^2 = \frac{1}{2} \times 5 \times (2)^2 = 10 \,kg \cdot m^2$.
The initial angular momentum is $L_i = I \omega_i = 10 \times 10 = 100 \,kg \cdot m^2/s$.
The initial kinetic energy is $E_i = \frac{1}{2} I \omega_i^2 = \frac{1}{2} \times 10 \times (10)^2 = 500 \,J$.
When an identical disc is placed on top, the final moment of inertia becomes $I_f = I + I = 2I = 20 \,kg \cdot m^2$.
By the conservation of angular momentum, $L_i = L_f$, so $100 = I_f \omega_f = 20 \omega_f$.
Thus, the final angular velocity is $\omega_f = \frac{100}{20} = 5 \,rad/s$.
The final kinetic energy is $E_f = \frac{1}{2} I_f \omega_f^2 = \frac{1}{2} \times 20 \times (5)^2 = 10 \times 25 = 250 \,J$.
The energy dissipated is $\Delta E = E_i - E_f = 500 \,J - 250 \,J = 250 \,J$.

(A) For a closed organ pipe,the fundamental frequency is given by $f = \frac{v}{4L}$.
Initially,$f_1 = 30 \,Hz$ and $v = 330 \,m/s$,so $L_1 = \frac{330}{4 \times 30} = \frac{330}{120} = 2.75 \,m$.
After pouring water,the new length of the air column is $L_2$. The new frequency is $f_2 = 110 \,Hz$.
$L_2 = \frac{330}{4 \times 110} = \frac{330}{440} = 0.75 \,m$.
The change in the length of the air column is $\Delta L = L_1 - L_2 = 2.75 - 0.75 = 2.0 \,m = 200 \,cm$.
The volume of water poured is $V = A \times \Delta L = 2 \,cm^2 \times 200 \,cm = 400 \,cm^3$.
Since the density of water is $1 \,g/cm^3$,the mass of water is $400 \,g$.

(D) The Vernier constant $(VC)$ is defined as the difference between one main scale division $(MSD)$ and one Vernier scale division $(VSD)$.
Given: $50 \,VSD = 49 \,MSD$.
Therefore, $1 \,VSD = \frac{49}{50} \,MSD$.
The smallest reading of the main scale $(MSD)$ is $0.5 \,mm$.
$VC = 1 \,MSD - 1 \,VSD = 1 \,MSD - \frac{49}{50} \,MSD = \frac{1}{50} \,MSD$.
Substituting the value of $MSD = 0.5 \,mm$:
$VC = \frac{0.5 \,mm}{50} = \frac{0.5}{50} \,mm = 0.01 \,mm$.

(B) The normal force $N$ acting on the block is given by $N = mg \cos \theta$.
Given: $m = 1 \ kg$,$g = 10 \ m/s^2$,$\theta = 60^{\circ}$,and $\mu_s = 0.1$.
$N = 1 \times 10 \times \cos(60^{\circ}) = 10 \times 0.5 = 5 \ N$.
The frictional force $f$ is given by $f = \mu_s N$.
$f = 0.1 \times 5 = 0.5 \ N$.
The work done against the frictional force over a distance $d = 10 \ m$ is:
$W = f \times d = 0.5 \times 10 = 5 \ J$.

(C) The process of heating ice at $-10^{\circ} C$ to steam at $100^{\circ} C$ involves several stages:
$1$. Heating ice from $-10^{\circ} C$ to $0^{\circ} C$: Temperature increases linearly with heat supplied.
$2$. Melting ice at $0^{\circ} C$ to water at $0^{\circ} C$: Temperature remains constant (phase change,latent heat of fusion).
$3$. Heating water from $0^{\circ} C$ to $100^{\circ} C$: Temperature increases linearly with heat supplied.
$4$. Boiling water at $100^{\circ} C$ to steam at $100^{\circ} C$: Temperature remains constant (phase change,latent heat of vaporization).
Thus,the heating curve should show two distinct horizontal segments representing the phase changes at $0^{\circ} C$ and $100^{\circ} C$,separated by a rising segment for water heating,and followed by a rising segment for steam heating. Graph $C$ correctly depicts these two phase changes (horizontal lines) and the intermediate heating stages.

(A) In a $P-V$ diagram,the slope of an adiabatic process is $\gamma$ times the slope of an isothermal process. Therefore,the steeper curve represents an adiabatic process.
$1$. Curve $B$ is steeper than curve $A$,so process $B$ is adiabatic. The equation for an adiabatic process is $PV^\gamma = k$.
$2$. Curve $A$ is less steep,so process $A$ is isothermal. The equation for an isothermal process is $PV = k$.
Comparing this with the given options,option $A$ correctly identifies process $B$ as adiabatic $(PV^\gamma = k)$ and process $A$ as isothermal $(PV = k)$.

(A) The condition for a block to remain at rest on an inclined surface without slipping is that the angle of inclination $\theta$ must be less than or equal to the angle of repose $\phi$,where $\tan \phi = \mu$.
Given the surface equation $y = x^2 / 4$,the slope at any point $x$ is given by the derivative $\frac{dy}{dx} = \tan \theta$.
$\frac{dy}{dx} = \frac{d}{dx} (x^2 / 4) = \frac{2x}{4} = \frac{x}{2}$.
For the block to be at the maximum height without slipping,the slope must equal the coefficient of friction $\mu = 0.5$.
So,$\frac{x}{2} = 0.5$,which gives $x = 1$.
Substituting $x = 1$ into the equation of the surface $y = x^2 / 4$,we get the maximum height $y = (1)^2 / 4 = 1/4 \ m$.

(D) The formula for escape velocity is $V_e = \sqrt{\frac{2GM}{R}}$.
Given for Earth: $V_E = 11.2 \,km/s$, $M_E$, and $R_E$.
For the planet: $M_P = \frac{M_E}{6}$ and $R_P = \frac{R_E}{3}$.
The escape velocity for the planet is $V_P = \sqrt{\frac{2GM_P}{R_P}}$.
Substituting the values: $V_P = \sqrt{\frac{2G(M_E/6)}{(R_E/3)}} = \sqrt{\frac{2GM_E}{R_E} \times \frac{3}{6}} = \sqrt{\frac{2GM_E}{R_E} \times \frac{1}{2}}$.
Since $V_E = \sqrt{\frac{2GM_E}{R_E}} = 11.2 \,km/s$, we have $V_P = \frac{V_E}{\sqrt{2}}$.
$V_P = \frac{11.2}{1.414} \approx 7.9 \,km/s$.

(A) Given the formula for mass: $m = k c^{P} G^{-1/2} h^{1/2}$.
We use the dimensional formulas for each constant:
$c$ (speed of light) $= [L T^{-1}]$
$G$ (gravitational constant) $= [M^{-1} L^3 T^{-2}]$
$h$ (Planck's constant) $= [M L^2 T^{-1}]$
Substituting these dimensions into the equation:
$[M^1 L^0 T^0] = [L T^{-1}]^{P} [M^{-1} L^3 T^{-2}]^{-1/2} [M L^2 T^{-1}]^{1/2}$
$[M^1 L^0 T^0] = [L^P T^{-P}] [M^{1/2} L^{-3/2} T^1] [M^{1/2} L^1 T^{-1/2}]$
Combining the powers of $M$,$L$,and $T$ on the right side:
$M: 1/2 + 1/2 = 1$
$L: P - 3/2 + 1 = P - 1/2$
$T: -P + 1 - 1/2 = -P + 1/2$
Comparing the powers of $L$ on both sides:
$P - 1/2 = 0 \implies P = 1/2$.
Thus,the value of $P$ is $1/2$.

(C) For a monoatomic gas,the degrees of freedom $f_1 = 3$. For a diatomic gas,the degrees of freedom $f_2 = 5$.
Given $n_1 = 3$ moles and $n_2 = 2$ moles.
The degrees of freedom for the mixture is given by $f_{\text{mix}} = \frac{n_1 f_1 + n_2 f_2}{n_1 + n_2}$.
Substituting the values: $f_{\text{mix}} = \frac{3(3) + 2(5)}{3 + 2} = \frac{9 + 10}{5} = \frac{19}{5} = 3.8$.
The adiabatic exponent $\gamma$ is related to the degrees of freedom $f$ by the formula $\gamma = 1 + \frac{2}{f}$.
Therefore,$\gamma_{\text{mix}} = 1 + \frac{2}{f_{\text{mix}}} = 1 + \frac{2}{3.8} = 1 + \frac{20}{38} = 1 + \frac{10}{19} = \frac{29}{19} \approx 1.526$.
Rounding to two decimal places,we get $\gamma_{\text{mix}} = 1.52$.

(A) The total mass of the system is $M = m_A + m_B + m_C = 5 \text{ kg} + 3 \text{ kg} + 2 \text{ kg} = 10 \text{ kg}$.
Since the surface is smooth,the acceleration $a$ of the system is given by $a = \frac{F}{M} = \frac{80 \text{ N}}{10 \text{ kg}} = 8 \text{ m/s}^2$.
For block $A$ $(5 \text{ kg})$,the only horizontal force acting is the tension $T_1$. Therefore,$T_1 = m_A \times a = 5 \text{ kg} \times 8 \text{ m/s}^2 = 40 \text{ N}$.
For block $B$ $(3 \text{ kg})$,the forces acting are $T_2$ (pulling forward) and $T_1$ (pulling backward). Therefore,$T_2 - T_1 = m_B \times a$.
Substituting the values,$T_2 - 40 \text{ N} = 3 \text{ kg} \times 8 \text{ m/s}^2 = 24 \text{ N}$.
Thus,$T_2 = 40 \text{ N} + 24 \text{ N} = 64 \text{ N}$.
Therefore,the tensions $T_1$ and $T_2$ are $40 \text{ N}$ and $64 \text{ N}$ respectively.

(D) Let the angles with the horizontal be $\theta_A$ and $\theta_B$. Given angles with vertical are $45^{\circ}$ and $60^{\circ}$,so $\theta_A = 90^{\circ} - 45^{\circ} = 45^{\circ}$ and $\theta_B = 90^{\circ} - 60^{\circ} = 30^{\circ}$.
For a projectile thrown from height $h$,the time of flight $T$ is given by $h = -u \sin \theta T + \frac{1}{2} g T^2$. Since $T$ is the same for both,the vertical components of initial velocity must be equal: $v_A \sin \theta_A = v_B \sin \theta_B$.
$v_A \sin 45^{\circ} = v_B \sin 30^{\circ} \implies v_A (\frac{1}{\sqrt{2}}) = v_B (\frac{1}{2}) \implies \frac{v_A}{v_B} = \frac{\sqrt{2}}{2} = \frac{1}{\sqrt{2}}$.
Also,the range $R = (v \cos \theta) T$. Since $R$ and $T$ are same,$v_A \cos \theta_A = v_B \cos \theta_B$.
$v_A \cos 45^{\circ} = v_B \cos 30^{\circ} \implies v_A (\frac{1}{\sqrt{2}}) = v_B (\frac{\sqrt{3}}{2}) \implies \frac{v_A}{v_B} = \frac{\sqrt{6}}{2} = \sqrt{\frac{3}{2}}$.
Since the two conditions lead to different ratios,the problem statement is physically inconsistent. However,if we assume the vertical components are equal to satisfy the time of flight condition,the ratio is $1 : \sqrt{2}$.

(A) Let $r$ be the radius of each small drop and $R$ be the radius of the big drop.
Since the volume remains conserved during the coalescence of $1000$ small drops into one big drop:
$1000 \times (\frac{4}{3} \pi r^3) = \frac{4}{3} \pi R^3$
$R^3 = 1000 r^3 \implies R = 10r$.
Surface energy $E$ is given by $E = S \times A$,where $S$ is surface tension and $A$ is the surface area.
Total surface energy of $1000$ small drops: $E_1 = 1000 \times (4 \pi r^2 \times S) = 4000 \pi r^2 S$.
Surface energy of the big drop: $E_2 = 4 \pi R^2 S = 4 \pi (10r)^2 S = 400 \pi r^2 S$.
The ratio $E_1 : E_2 = \frac{4000 \pi r^2 S}{400 \pi r^2 S} = \frac{10}{1}$.
Therefore,$x = 10$.

(C) According to the Law of Conservation of Angular Momentum ($C$.$O$.$A$.$M$.),the total angular momentum before contact equals the total angular momentum after contact.
$I_1 \omega_1 + I_2 \omega_2 = (I_1 + I_2) \omega_0$
Substituting the given values: $(4 \times 10) + (2 \times 4) = (4 + 2) \omega_0$
$40 + 8 = 6 \omega_0 \implies 48 = 6 \omega_0 \implies \omega_0 = 8 \ rad/s$.
Initial kinetic energy $E_1 = \frac{1}{2} I_1 \omega_1^2 + \frac{1}{2} I_2 \omega_2^2 = \frac{1}{2}(4)(10)^2 + \frac{1}{2}(2)(4)^2 = 200 + 16 = 216 \ J$.
Final kinetic energy $E_2 = \frac{1}{2}(I_1 + I_2) \omega_0^2 = \frac{1}{2}(4 + 2)(8)^2 = \frac{1}{2}(6)(64) = 3 \times 64 = 192 \ J$.
The loss in kinetic energy $\Delta E = E_1 - E_2 = 216 \ J - 192 \ J = 24 \ J$.

(A) First,calculate the magnitude of vector $\overrightarrow{A} = 3 \hat{i} + 4 \hat{j}$.
$|\overrightarrow{A}| = \sqrt{3^2 + 4^2} = \sqrt{9 + 16} = \sqrt{25} = 5$.
Next,find the unit vector in the direction of $\overrightarrow{B} = 4 \hat{i} + 3 \hat{j}$.
$|\overrightarrow{B}| = \sqrt{4^2 + 3^2} = \sqrt{16 + 9} = 5$.
$\hat{B} = \frac{\overrightarrow{B}}{|\overrightarrow{B}|} = \frac{4 \hat{i} + 3 \hat{j}}{5}$.
The required vector $\overrightarrow{V}$ has magnitude $5$ and is parallel to $\overrightarrow{B}$,so $\overrightarrow{V} = |\overrightarrow{A}| \hat{B} = 5 \times \frac{4 \hat{i} + 3 \hat{j}}{5} = 4 \hat{i} + 3 \hat{j}$.
Comparing this with the given components $X \hat{i} + 3 \hat{j}$,we find $X = 4$.

(D) The acceleration due to gravity at a height $h$ above the earth's surface is given by $g' = g \left( \frac{R}{R+h} \right)^2$.
Given $h = R$,the acceleration due to gravity becomes $g' = g \left( \frac{R}{R+R} \right)^2 = g \left( \frac{1}{2} \right)^2 = \frac{g}{4}$.
The time period of a simple pendulum is given by $T = 2\pi \sqrt{\frac{\ell}{g'}}$.
Substituting $\ell = 4 \ m$ and $g' = \frac{g}{4}$,we get $T = 2\pi \sqrt{\frac{4}{g/4}} = 2\pi \sqrt{\frac{16}{g}}$.
Given $g = \pi^2 \ ms^{-2}$,we substitute this into the equation:
$T = 2\pi \sqrt{\frac{16}{\pi^2}} = 2\pi \left( \frac{4}{\pi} \right) = 8 \ s$.

(A) For a point source,the intensity $I$ at a distance $r$ is given by the inverse square law: $I \propto \frac{1}{r^2}$.
Let $I_0$ be the intensity at $r_0 = 1 \text{ m}$,where $I_0 = 16 \times 10^{-8} \text{ W m}^{-2}$.
The intensity at distance $r$ is $I(r) = \frac{I_0 \cdot r_0^2}{r^2} = \frac{16 \times 10^{-8} \times (1)^2}{r^2} = \frac{16 \times 10^{-8}}{r^2} \text{ W m}^{-2}$.
At $r_1 = 2 \text{ m}$,the intensity is $I_1 = \frac{16 \times 10^{-8}}{2^2} = \frac{16 \times 10^{-8}}{4} = 4 \times 10^{-8} \text{ W m}^{-2}$.
At $r_2 = 4 \text{ m}$,the intensity is $I_2 = \frac{16 \times 10^{-8}}{4^2} = \frac{16 \times 10^{-8}}{16} = 1 \times 10^{-8} \text{ W m}^{-2}$.
The difference in intensity is $|I_1 - I_2| = |4 \times 10^{-8} - 1 \times 10^{-8}| = 3 \times 10^{-8} \text{ W m}^{-2}$.
Thus,the value is $3$.

(A) According to the Kinetic Theory of Gases,the average translational kinetic energy of a gas molecule is given by the formula: $KE_{avg} = \frac{3}{2} kT$,where $k$ is the Boltzmann constant and $T$ is the absolute temperature.
Since $k$ is a universal constant and $T$ is given as constant for all gases,the average kinetic energy depends only on the temperature.
Therefore,at a given temperature,the average kinetic energy of molecules of all gases remains the same,regardless of the nature of the gas.

(A) Given the relation: $t = \alpha x^2 + \beta x$.
Differentiating with respect to $x$:
$\frac{dt}{dx} = 2 \alpha x + \beta$.
Since $v = \frac{dx}{dt}$,we have $\frac{dt}{dx} = \frac{1}{v}$.
Therefore,$\frac{1}{v} = 2 \alpha x + \beta$.
Differentiating both sides with respect to $t$:
$-\frac{1}{v^2} \frac{dv}{dt} = 2 \alpha \frac{dx}{dt}$.
Since $a = \frac{dv}{dt}$ and $v = \frac{dx}{dt}$,we substitute these:
$-\frac{1}{v^2} a = 2 \alpha v$.
Rearranging for $a$:
$a = -2 \alpha v^3$.

(B) Let the side length of the square be $a$. Consider one of the masses at a corner. It experiences gravitational forces from the other three masses.
Two masses are at a distance $a$ (adjacent corners),and one mass is at a distance $\sqrt{2}a$ (opposite corner).
The forces from the two adjacent masses are $F = \frac{Gm^2}{a^2}$ each,acting at an angle of $90^\circ$. Their resultant is $\sqrt{F^2 + F^2} = \sqrt{2}F = \sqrt{2} \frac{Gm^2}{a^2}$.
The force from the opposite mass is $F' = \frac{Gm^2}{(\sqrt{2}a)^2} = \frac{Gm^2}{2a^2}$.
Both these forces act along the diagonal of the square,so the net force is:
$F_{\text{net}} = \sqrt{2} \frac{Gm^2}{a^2} + \frac{Gm^2}{2a^2} = \frac{Gm^2}{a^2} \left( \sqrt{2} + \frac{1}{2} \right) = \frac{Gm^2}{a^2} \left( \frac{2\sqrt{2} + 1}{2} \right)$.
Given that $F_{\text{net}} = \left( \frac{2\sqrt{2} + 1}{32} \right) \frac{Gm^2}{L^2}$,we equate the two expressions:
$\frac{Gm^2}{a^2} \left( \frac{2\sqrt{2} + 1}{2} \right) = \left( \frac{2\sqrt{2} + 1}{32} \right) \frac{Gm^2}{L^2}$.
$\frac{1}{2a^2} = \frac{1}{32L^2} \implies a^2 = 16L^2 \implies a = 4L$.

(D) For an ideal gas,the equation of state is $PV = nRT$.
Rearranging for $V$ as a function of $T$,we get $V = \left(\frac{nR}{P}\right)T$.
Comparing this with the equation of a straight line $y = mx$,where $y = V$ and $x = T$,the slope of the $V-T$ graph is given by $\text{Slope} = \frac{nR}{P}$.
Since $n$ and $R$ are constants,the slope is inversely proportional to the pressure,i.e.,$\text{Slope} \propto \frac{1}{P}$.
From the given figure,the slope of the line corresponding to $P_2$ is greater than the slope of the line corresponding to $P_1$,i.e.,$(\text{Slope})_2 > (\text{Slope})_1$.
Since $\text{Slope} \propto \frac{1}{P}$,a higher slope implies a lower pressure. Therefore,$P_2 < P_1$ or $P_1 > P_2$.

(B) The resistance $R$ of a wire is given by the formula $R = \rho \frac{L}{A}$,where $A = \pi \frac{d^2}{4}$.
Substituting $A$,we get $R = \frac{4 \rho L}{\pi d^2}$.
Taking the relative error,we have $\frac{\Delta R}{R} = \frac{\Delta L}{L} + 2 \frac{\Delta d}{d}$.
Given that $\frac{\Delta L}{L} = 0.1 \%$ and $\frac{\Delta d}{d} = 0.1 \%$.
Substituting these values,$\frac{\Delta R}{R} = 0.1 \% + 2(0.1 \%) = 0.1 \% + 0.2 \% = 0.3 \%$.

(A) According to the principle of homogeneity of dimensions,each term in the equation must have the same dimensions as force $F$.
$1$. For the term $ax^2$:
$[ax^2] = [F] = [MLT^{-2}]$
$[a] = [F] / [x^2] = [MLT^{-2}] / [L^2] = [ML^{-1}T^{-2}]$
$2$. For the term $bt^{1/2}$:
$[bt^{1/2}] = [F] = [MLT^{-2}]$
$[b] = [F] / [t^{1/2}] = [MLT^{-2}] / [T^{1/2}] = [MLT^{-5/2}]$
$3$. Calculating the dimensions of $b^2/a$:
$[b^2/a] = [b]^2 / [a] = ([MLT^{-5/2}])^2 / [ML^{-1}T^{-2}]$
$[b^2/a] = [M^2 L^2 T^{-5}] / [ML^{-1}T^{-2}] = [M^{2-1} L^{2-(-1)} T^{-5-(-2)}] = [ML^3 T^{-3}]$

(B) For block $M$ $(M=10 \text{ kg})$:
$M g \sin 53^{\circ} - \mu M g \cos 53^{\circ} - T = M a$
$10 \times 10 \times 0.8 - 0.25 \times 10 \times 10 \times 0.6 - T = 10 \times 2$
$80 - 15 - T = 20$
$T = 80 - 15 - 20 = 45 \text{ N}$
For block $m$:
$T - m g \sin 37^{\circ} - \mu m g \cos 37^{\circ} = m a$
$45 - m \times 10 \times 0.6 - 0.25 \times m \times 10 \times 0.8 = m \times 2$
$45 - 6m - 2m = 2m$
$45 = 10m$
$m = 4.5 \text{ kg}$

(D) Let $L_1$ be the length of the closed organ pipe and $L_2$ be the length of the open organ pipe.
The fundamental frequency of a closed organ pipe is given by $f_c = \frac{v}{4L_1}$.
The frequencies of an open organ pipe are given by $f_n = \frac{nv}{2L_2}$, where $n = 1, 2, 3, ...$.
The first overtone of an open organ pipe corresponds to $n = 2$, so $f_{o1} = \frac{2v}{2L_2} = \frac{v}{L_2}$.
According to the problem, the fundamental frequency of the closed pipe is equal to the first overtone frequency of the open pipe:
$f_c = f_{o1}$
$\frac{v}{4L_1} = \frac{v}{L_2}$
$L_2 = 4L_1$
Given $L_2 = 60 \,cm$, we have:
$60 \,cm = 4L_1$
$L_1 = \frac{60}{4} \,cm = 15 \,cm$.
Therefore, the length of the closed pipe is $15 \,cm$.

(B) When a steel ball is dropped in a viscous liquid like glycerine,it experiences three forces: gravitational force $(mg)$ downwards,buoyant force $(F_B)$ upwards,and viscous drag force $(F_v = 6 \pi \eta r v)$ upwards.
The equation of motion is: $mg - F_B - F_v = ma$
Substituting the expressions for forces: $(\rho \frac{4}{3} \pi r^3) g - (\rho_L \frac{4}{3} \pi r^3) g - 6 \pi \eta r v = m \frac{dv}{dt}$
Let $K_1 = \frac{4}{3} \pi r^3 g (\rho - \rho_L) / m$ and $K_2 = \frac{6 \pi \eta r}{m}$. Then the equation becomes: $\frac{dv}{dt} = K_1 - K_2 v$
Integrating this differential equation with initial conditions ($v=0$ at $t=0$): $\int_0^v \frac{dv}{K_1 - K_2 v} = \int_0^t dt$
This yields: $v = \frac{K_1}{K_2} (1 - e^{-K_2 t})$
As $t \to \infty$,the velocity approaches a constant terminal velocity $v_T = \frac{K_1}{K_2}$. The graph of $v$ versus $t$ is an exponential curve that starts from the origin and asymptotically approaches the terminal velocity,which corresponds to Graph $B$.

(C) For the coin to remain on the disc without slipping,the centripetal force required for circular motion must be provided by the static frictional force.
The normal force acting on the coin is $N = mg$.
The maximum static frictional force available is $f_{max} = \mu N = \mu mg$.
The required centripetal force for the coin moving in a circle of radius $r$ with angular velocity $\omega$ is $F_c = m \omega^2 r$.
For the coin not to slip,the centripetal force must be less than or equal to the maximum static friction:
$m \omega^2 r \leq \mu mg$
Solving for $\omega$:
$\omega^2 \leq \frac{\mu g}{r}$
Therefore,the maximum angular velocity is $\omega_{max} = \sqrt{\frac{\mu g}{r}}$.

(B) According to the law of conservation of linear momentum,the magnitude of the momentum of the artillery piece $(p_1)$ and the shell $(p_2)$ must be equal immediately after firing,as the initial momentum of the system was zero.
$|p_1| = |p_2| = p$
Kinetic energy $(KE)$ is related to momentum $(p)$ and mass $(m)$ by the formula $KE = \frac{p^2}{2m}$.
Since the momentum $p$ is the same for both,$KE \propto \frac{1}{m}$.
Therefore,the ratio of the kinetic energy of the artillery $(KE_1)$ to that of the shell $(KE_2)$ is:
$\frac{KE_1}{KE_2} = \frac{p^2 / 2M_1}{p^2 / 2M_2} = \frac{M_2}{M_1}$.

(D) According to the work-energy theorem, the work done $W$ is equal to the change in kinetic energy $\Delta KE$.
Since the disc is rolling, its total kinetic energy $KE$ is the sum of translational and rotational kinetic energy: $KE = \frac{1}{2}mv^2 + \frac{1}{2}I\omega^2$.
For a solid disc, the moment of inertia $I = \frac{1}{2}mR^2$ and for rolling without slipping, $\omega = \frac{v}{R}$.
Substituting these, $KE = \frac{1}{2}mv^2 + \frac{1}{2}(\frac{1}{2}mR^2)(\frac{v}{R})^2 = \frac{1}{2}mv^2 + \frac{1}{4}mv^2 = \frac{3}{4}mv^2$.
Given $m = 50 \,kg$ and $v = 0.4 \,m/s$, the initial kinetic energy is $KE_i = \frac{3}{4} \times 50 \times (0.4)^2 = \frac{3}{4} \times 50 \times 0.16 = 0.75 \times 8 = 6 \,J$.
To stop the disc, the final kinetic energy $KE_f = 0$.
Thus, $W = KE_f - KE_i = 0 - 6 = -6 \,J$.
The absolute value of the work done is $|W| = 6 \,J$.

(A) Let the body fall from height $H$. It hits the inclined plane at a height $h$ from the ground,meaning it has fallen a distance of $(H-h)$.
The time taken to fall the distance $(H-h)$ is $t_1 = \sqrt{\frac{2(H-h)}{g}}$.
At this point,the impact is perfectly elastic and the velocity becomes horizontal. The body then falls from height $h$ with an initial vertical velocity of $0$. The time taken to fall this remaining distance $h$ is $t_2 = \sqrt{\frac{2h}{g}}$.
The total time of flight $T$ is $T = t_1 + t_2 = \sqrt{\frac{2(H-h)}{g}} + \sqrt{\frac{2h}{g}}$.
To find the value of $h$ for which $T$ is maximum,we differentiate $T$ with respect to $h$ and set it to $0$:
$\frac{dT}{dh} = \sqrt{\frac{2}{g}} \left( \frac{1}{2\sqrt{H-h}} \cdot (-1) + \frac{1}{2\sqrt{h}} \right) = 0$.
This implies $\frac{1}{\sqrt{h}} = \frac{1}{\sqrt{H-h}}$,which gives $h = H - h$,or $2h = H$.
Therefore,$\frac{H}{h} = 2$.

(D) The bulk modulus $\beta$ is defined as $\beta = -\frac{\Delta P}{\Delta V / V}$.
Here,the change in pressure $\Delta P$ at a depth $h$ is given by $\Delta P = \rho gh$.
The fractional change in volume is $\frac{\Delta V}{V} = -0.02 \% = -\frac{0.02}{100} = -2 \times 10^{-4}$.
Substituting the values into the formula:
$\rho gh = -\beta \left( \frac{\Delta V}{V} \right)$
$10^3 \times 10 \times h = -(9 \times 10^8) \times (-2 \times 10^{-4})$
$10^4 \times h = 18 \times 10^4$
$h = 18 \ m$.

(A) The velocity of a particle in simple harmonic motion is given by $v = \omega \sqrt{A^2 - x^2}$.
At displacement $x = \frac{2A}{3}$,the velocity is $v = \omega \sqrt{A^2 - (\frac{2A}{3})^2} = \omega \sqrt{A^2 - \frac{4A^2}{9}} = \omega \sqrt{\frac{5A^2}{9}} = \frac{\sqrt{5}A\omega}{3}$.
When the speed is increased to three times,the new velocity $v' = 3v = 3 \times \frac{\sqrt{5}A\omega}{3} = \sqrt{5}A\omega$.
Let the new amplitude be $A'$. The new velocity at the same position $x = \frac{2A}{3}$ is $v' = \omega \sqrt{(A')^2 - x^2}$.
Equating the expressions: $\sqrt{5}A\omega = \omega \sqrt{(A')^2 - (\frac{2A}{3})^2}$.
Squaring both sides: $5A^2 = (A')^2 - \frac{4A^2}{9}$.
$(A')^2 = 5A^2 + \frac{4A^2}{9} = \frac{45A^2 + 4A^2}{9} = \frac{49A^2}{9}$.
Taking the square root: $A' = \frac{7A}{3}$.
Comparing this with $\frac{nA}{3}$,we get $n = 7$.

(A) For a system of two masses $m_1$ and $m_2$ connected by a light string over a smooth fixed pulley,the magnitude of acceleration $a$ is given by the formula:
$a = \frac{|m_1 - m_2| g}{m_1 + m_2}$
Given that the acceleration $a = g / 8$,we have:
$\frac{|m_1 - m_2| g}{m_1 + m_2} = \frac{g}{8}$
Assuming $m_1 > m_2$,we get:
$\frac{m_1 - m_2}{m_1 + m_2} = \frac{1}{8}$
Cross-multiplying gives:
$8(m_1 - m_2) = m_1 + m_2$
$8m_1 - 8m_2 = m_1 + m_2$
$7m_1 = 9m_2$
Therefore,the ratio of the masses is:
$\frac{m_1}{m_2} = \frac{9}{7}$

(B) The given circuit consists of two $AND$ gates and one $OR$ gate. The inputs to the first $AND$ gate are $A$ and $B$,so its output is $A \cdot B$.
The input $A$ passes through a $NOT$ gate,becoming $\overline{A}$. This $\overline{A}$ and input $B$ are fed into the second $AND$ gate,so its output is $\overline{A} \cdot B$.
These two outputs are then fed into an $OR$ gate,giving the final output $Y = (A \cdot B) + (\overline{A} \cdot B)$.
Using Boolean algebra: $Y = (A + \overline{A}) \cdot B$.
Since $A + \overline{A} = 1$,we get $Y = 1 \cdot B = B$.
Thus,the output $Y$ is equal to input $B$. Checking the truth table for $Y = B$:
If $A=0, B=0$,then $Y=0$.
If $A=0, B=1$,then $Y=1$.
If $A=1, B=0$,then $Y=0$.
If $A=1, B=1$,then $Y=1$.
This matches option $B$.

(C) The average power dissipated in an $a.c.$ circuit is given by the formula: $P_{\text{avg}} = V_{\text{rms}} I_{\text{rms}} \cos(\phi)$.
Given $V = 100 \sin(100t) \ V$,the peak voltage $V_0 = 100 \ V$.
Given $I = 100 \sin(100t + \frac{\pi}{3}) \ mA$,the peak current $I_0 = 100 \ mA = 100 \times 10^{-3} \ A = 0.1 \ A$.
The phase difference $\phi = \frac{\pi}{3}$.
$V_{\text{rms}} = \frac{V_0}{\sqrt{2}} = \frac{100}{\sqrt{2}} \ V$.
$I_{\text{rms}} = \frac{I_0}{\sqrt{2}} = \frac{0.1}{\sqrt{2}} \ A$.
$P_{\text{avg}} = \left(\frac{100}{\sqrt{2}}\right) \times \left(\frac{0.1}{\sqrt{2}}\right) \times \cos\left(\frac{\pi}{3}\right)$.
$P_{\text{avg}} = \frac{100 \times 0.1}{2} \times \frac{1}{2} = \frac{10}{4} = 2.5 \ W$.

(C) For a virtual image, magnification $m = +2$.
Using the magnification formula for a spherical mirror, $m = -v/u$.
Since the image is virtual, it is formed behind the mirror, so $v$ is positive. Let the object distance be $u$ (where $u$ is negative by sign convention).
The distance between the object and the image is $|v - u| = 15 \,cm$.
Since the object is in front of the mirror $(u < 0)$ and the virtual image is behind the mirror $(v > 0)$, the distance is $v - u = 15$.
Substituting $v = -mu = -2u$:
$-2u - u = 15
-3u = 15
u = -5 \,cm$.
Then, $v = -2(-5) = 10 \,cm$.
Using the mirror formula, $\frac{1}{f} = \frac{1}{v} + \frac{1}{u}$:
$\frac{1}{f} = \frac{1}{10} + \frac{1}{-5} = \frac{1 - 2}{10} = -\frac{1}{10}$.
Therefore, $f = -10 \,cm$.

(B) The radius of a circular path for a charged particle in a magnetic field is given by $R = \frac{mv}{qB}$.
Since the particle is accelerated through a potential difference $V$,its kinetic energy is $KE = qV = \frac{1}{2}mv^2$,which implies $v = \sqrt{\frac{2qV}{m}}$.
Substituting $v$ into the radius formula: $R = \frac{m}{qB} \sqrt{\frac{2qV}{m}} = \frac{\sqrt{2mqV}}{qB} = \frac{1}{B} \sqrt{\frac{2mV}{q}}$.
Since $q$,$V$,and $B$ are constant for both particles,we have $R \propto \sqrt{m}$,which implies $m \propto R^2$.
Therefore,the ratio of the masses is $\frac{m_X}{m_Y} = \left(\frac{R_1}{R_2}\right)^2$.

(A) The path difference is given by $\Delta x = \frac{7 \lambda}{4}$.
The phase difference $\phi$ is related to the path difference by the formula $\phi = \frac{2 \pi}{\lambda} \Delta x$.
Substituting the value of $\Delta x$: $\phi = \frac{2 \pi}{\lambda} \times \frac{7 \lambda}{4} = \frac{7 \pi}{2}$.
The intensity $I$ at any point is given by $I = I_{\max} \cos^2\left(\frac{\phi}{2}\right)$.
Therefore,the ratio $\frac{I}{I_{\max}} = \cos^2\left(\frac{7 \pi}{2 \times 2}\right) = \cos^2\left(\frac{7 \pi}{4}\right)$.
Using the property $\cos(2 \pi - \theta) = \cos(\theta)$,we get $\cos^2\left(2 \pi - \frac{\pi}{4}\right) = \cos^2\left(\frac{\pi}{4}\right)$.
Since $\cos\left(\frac{\pi}{4}\right) = \frac{1}{\sqrt{2}}$,we have $\left(\frac{1}{\sqrt{2}}\right)^2 = \frac{1}{2}$.

(A) The relationship between the electric field $E$ and the magnetic field $B$ in an electromagnetic wave is given by $E/B = c$,where $c$ is the speed of light in free space $(c = 3 \times 10^8 \ m/s)$.
First,calculate the magnitude of the magnetic field:
$B = E/c = 9.6 / (3 \times 10^8) = 3.2 \times 10^{-8} \ T$.
Next,determine the direction of the magnetic field using the property that the direction of propagation $\hat{v}$ is given by $\hat{v} = \hat{E} \times \hat{B}$.
Rearranging this,the direction of the magnetic field is $\hat{B} = \hat{v} \times \hat{E}$.
Given the wave travels in the $X$-direction $(\hat{v} = \hat{i})$ and the electric field is in the $Y$-direction $(\hat{E} = \hat{j})$:
$\hat{B} = \hat{i} \times \hat{j} = \hat{k}$.
Therefore,the magnetic field vector is $\overrightarrow{B} = 3.2 \times 10^{-8} \ \hat{k} \ T$.

(A) First, calculate the equivalent resistance of the circuit. The resistors $R_2$ and $R_3$ are in parallel, so their equivalent resistance $R_p$ is given by:
$R_p = \frac{R_2 \times R_3}{R_2 + R_3} = \frac{4 \times 4}{4 + 4} = \frac{16}{8} = 2 \,\Omega$
Now, the total equivalent resistance $R_{eq}$ of the circuit is the sum of $R_1$, $R_p$, and $R_4$ in series:
$R_{eq} = R_1 + R_p + R_4 = 2 \,\Omega + 2 \,\Omega + 1 \,\Omega = 5 \,\Omega$
The total current $i$ flowing through the circuit is:
$i = \frac{V}{R_{eq}} = \frac{10 \,V}{5 \,\Omega} = 2 \,A$
This total current $i = 2 \,A$ flows through $R_1$, then splits into two parallel branches $R_2$ and $R_3$. Since $R_2 = R_3 = 4 \,\Omega$, the current divides equally between them:
$i_{R_3} = i \times \left( \frac{R_2}{R_2 + R_3} \right) = 2 \,A \times \left( \frac{4}{4 + 4} \right) = 2 \times \frac{4}{8} = 1 \,A$

(C) According to Rutherford's atomic model,most of the mass of the atom and all its positive charge are concentrated in a tiny nucleus,and the electrons revolve around it. This confirms that $Statement$ $I$ is true.
According to Thomson's atomic model (often called the plum pudding model),an atom is a spherical cloud of positive charge with electrons embedded in it. This is not a special case of Rutherford's model; rather,it is a completely different model that was superseded by Rutherford's model. Therefore,$Statement$ $II$ is false.
Hence,$Statement$ $I$ is true but $Statement$ $II$ is false.

(C) The electric flux $\phi$ through a surface is given by the dot product of the electric field vector $\vec{E}$ and the area vector $\vec{A}$.
Given:
$\vec{E} = (6 \hat{i} + 5 \hat{j} + 3 \hat{k}) \ N/C$
$\vec{A} = 30 \hat{i} \ m^2$
Using the formula $\phi = \vec{E} \cdot \vec{A}$:
$\phi = (6 \hat{i} + 5 \hat{j} + 3 \hat{k}) \cdot (30 \hat{i})$
Since $\hat{i} \cdot \hat{i} = 1$,$\hat{i} \cdot \hat{j} = 0$,and $\hat{i} \cdot \hat{k} = 0$:
$\phi = 6 \times 30 = 180 \ N \cdot m^2/C$
Thus,the electric flux is $180 \ N \cdot m^2/C$.

(B) Given: Length of the wire $l = 5 \ m$.
Horizontal component of earth's magnetic field $B_H = 0.60 \times 10^{-4} \ Wb \ m^{-2}$.
Velocity of the wire $v = 10 \ m \ s^{-1}$.
The induced emf $(e)$ in a conductor moving in a magnetic field is given by the formula $e = B_H v l$.
Substituting the values:
$e = (0.60 \times 10^{-4}) \times 10 \times 5$
$e = 0.60 \times 10^{-3} \times 5$
$e = 3.0 \times 10^{-3} \ V$.
Thus,the instantaneous value of the induced emf is $3 \times 10^{-3} \ V$.

(C) To observe any line of the Balmer series, the hydrogen atom must be excited to at least the $n = 3$ energy level, as the first line of the Balmer series corresponds to the transition from $n = 3$ to $n = 2$.
The energy required to excite an electron from the ground state $(n = 1)$ to the $n = 3$ state is given by:
$\Delta E = E_3 - E_1 = 13.6 \left( 1 - \frac{1}{3^2} \right) \,eV$
$\Delta E = 13.6 \left( 1 - \frac{1}{9} \right) \,eV = 13.6 \times \frac{8}{9} \,eV$
$\Delta E = 12.088... \,eV \approx 12.09 \,eV$
Since the energy of the bombarding electron is $eV$, the potential difference $V$ required is $12.09 \,V$.
Given that the potential difference is $\frac{\alpha}{10} \,V$, we have:
$\frac{\alpha}{10} = 12.1$
$\alpha = 121$.

(D) Given: Charge $q = 4.0 \mu C = 4.0 \times 10^{-6} \ C$.
Velocity $\vec{v} = 4.0 \times 10^6 \hat{j} \ m/s$.
Magnetic field $\vec{B} = 2 \hat{k} \ T$.
The magnetic force $\vec{F}$ on a moving charge is given by the Lorentz force formula: $\vec{F} = q(\vec{v} \times \vec{B})$.
Substituting the values: $\vec{F} = (4.0 \times 10^{-6} \ C) \times (4.0 \times 10^6 \hat{j} \ m/s \times 2 \hat{k} \ T)$.
Using the cross product rule $\hat{j} \times \hat{k} = \hat{i}$,we get: $\vec{F} = (4.0 \times 10^{-6}) \times (8.0 \times 10^6) \hat{i} \ N$.
$\vec{F} = 32 \hat{i} \ N$.
Comparing this with $\vec{F} = x \hat{i} \ N$,we find $x = 32$.

(B) At steady state,capacitors behave as open circuits. When points $A$ and $B$ are connected by a wire,the capacitors are bypassed in the $DC$ circuit path.
The circuit effectively consists of the $6 \ \Omega$ resistor and the $3 \ \Omega$ resistor in series connected to the $9 \ V$ source.
The equivalent resistance is $R_{\text{eq}} = 6 \ \Omega + 3 \ \Omega = 9 \ \Omega$.
The current flowing through the circuit is $i = \frac{V}{R_{\text{eq}}} = \frac{9 \ V}{9 \ \Omega} = 1 \ A$.
Since points $A$ and $B$ are connected,they are at the same potential. The potential at $A$ (and $B$) relative to the ground is the potential drop across the $3 \ \Omega$ resistor: $V_B = i \times 3 \ \Omega = 1 \ A \times 3 \ \Omega = 3 \ V$.
The potential at the top terminal is $9 \ V$. Thus,the potential difference across the $6 \ \mu F$ capacitor (connected between the $9 \ V$ terminal and point $B$) is $\Delta V = 9 \ V - 3 \ V = 6 \ V$.
The charge stored in the $6 \ \mu F$ capacitor is $Q = C \Delta V = 6 \ \mu F \times 6 \ V = 36 \ \mu C$.

(C) For the $n^{\text{th}}$ minima in a single slit diffraction pattern,the condition is given by:
$b \sin \theta = n \lambda$
Since $\lambda$ is very small,$\sin \theta \approx \tan \theta = \frac{y_n}{D}$.
Thus,the position of the $n^{\text{th}}$ minima is $y_n = \frac{n \lambda D}{b}$.
The position of the first minima $(n=1)$ is $y_1 = \frac{\lambda D}{b}$.
The position of the third minima $(n=3)$ is $y_3 = \frac{3 \lambda D}{b}$.
The distance between the first and third minima is $\Delta y = y_3 - y_1 = \frac{2 \lambda D}{b}$.
Given $\Delta y = 3 \text{ mm} = 3 \times 10^{-3} \text{ m}$,$\lambda = 6000 \mathring A = 6000 \times 10^{-10} \text{ m}$,and $D = 50 \text{ cm} = 0.5 \text{ m}$.
Substituting the values:
$3 \times 10^{-3} = \frac{2 \times 6000 \times 10^{-10} \times 0.5}{b}$
$b = \frac{2 \times 6000 \times 10^{-10} \times 0.5}{3 \times 10^{-3}}$
$b = \frac{6000 \times 10^{-10}}{3 \times 10^{-3}} = 2000 \times 10^{-7} = 2 \times 10^{-4} \text{ m}$.
Therefore,the width of the slit is $2 \times 10^{-4} \text{ m}$. The correct option is $C$.

(D) Given,the current through the $20 \ \Omega$ resistor is $i_1 = 0.3 \ A$. The total current measured by the ammeter is $I = i_1 + i_2 + i_3 = 0.9 \ A$.
Since the resistors are connected in parallel,the potential difference $V_{AB}$ across them is the same.
$V_{AB} = i_1 \times 20 \ \Omega = 0.3 \ A \times 20 \ \Omega = 6 \ V$.
Now,calculate the current $i_2$ through the $15 \ \Omega$ resistor:
$i_2 = \frac{V_{AB}}{15 \ \Omega} = \frac{6 \ V}{15 \ \Omega} = 0.4 \ A$.
Using the total current equation:
$i_1 + i_2 + i_3 = 0.9 \ A$
$0.3 \ A + 0.4 \ A + i_3 = 0.9 \ A$
$0.7 \ A + i_3 = 0.9 \ A$
$i_3 = 0.2 \ A$.
Finally,calculate $R_1$ using the potential difference $V_{AB}$ and current $i_3$:
$R_1 = \frac{V_{AB}}{i_3} = \frac{6 \ V}{0.2 \ A} = 30 \ \Omega$.

(C) The circuit consists of several resistors connected in series across a $4 \ V$ $DC$ source.
First,calculate the equivalent resistance $R_{eq}$ of the entire circuit:
$R_{eq} = 3.3 \ k\Omega + 100 \ \Omega + 100 \ \Omega + 100 \ \Omega + 100 \ \Omega + 100 \ \Omega + 100 \ \Omega + 100 \ \Omega$
$R_{eq} = 3300 \ \Omega + 700 \ \Omega = 4000 \ \Omega$
Next,calculate the total current $i$ flowing through the circuit using Ohm's law:
$i = \frac{V}{R_{eq}} = \frac{4 \ V}{4000 \ \Omega} = 10^{-3} \ A = 1 \ mA$
The output voltage $V_0$ is measured across the last five $100 \ \Omega$ resistors (as shown in the figure,the arrow points to the junction after the first three resistors,and $V_0$ is across the remaining five resistors).
Resistance across which $V_0$ is measured: $R_{out} = 5 \times 100 \ \Omega = 500 \ \Omega$
Therefore,$V_0 = i \times R_{out} = 1 \times 10^{-3} \ A \times 500 \ \Omega = 0.5 \ V$.

(B) For photoelectric emission $(P.E.E.)$,the condition is $\lambda \leq \frac{hc}{W_0}$.
Given the work function $W_0 = 3.0 \ eV$.
Using the relation $\lambda_{\max} = \frac{hc}{W_0}$,where $hc \approx 1240 \ eV \cdot nm$.
$\lambda_{\max} = \frac{1240 \ eV \cdot nm}{3.0 \ eV}$.
$\lambda_{\max} \approx 413.33 \ nm$.
Therefore,the longest wavelength is approximately $414 \ nm$.

(C) For an electron in a hydrogen atom,the kinetic energy $(KE)$ and potential energy $(PE)$ are related by the virial theorem.
$KE = -\frac{1}{2} PE$
Taking the magnitude of the potential energy,we have $|PE| = 2 KE$.
Therefore,the ratio of the magnitude of the kinetic energy to the potential energy is given by $\frac{KE}{|PE|} = \frac{1}{2}$.
This relationship holds true for any orbit $n$,including the $5^{\text{th}}$ excited state $(n = 6)$.

(B) The electric field is given by $\vec{E} = E_0 \cos(\omega t - kz) \hat{i}$.
In an electromagnetic wave, the relationship between the electric field amplitude $E_0$ and the magnetic field amplitude $B_0$ is given by $B_0 = \frac{E_0}{C}$, where $C$ is the speed of light.
The direction of propagation of the wave is given by the direction of the vector $\vec{E} \times \vec{B}$.
Here, the wave propagates in the $+z$ direction $(\hat{k})$.
Given $\vec{E}$ is in the $\hat{i}$ direction, we have $\hat{i} \times \hat{B} = \hat{k}$.
This implies $\hat{B} = \hat{j}$.
Thus, the magnetic field vector is $\vec{B} = \frac{E_0}{C} \cos(\omega t - kz) \hat{j}$.

(C) The magnetic field at the centre of a circular loop of radius '$a$' carrying current '$I$' is given by $B = \frac{\mu_0 I}{2 a}$.
Since the two loops $A$ and $B$ are perpendicular to each other,the magnetic fields produced by them at the centre $O$ will also be perpendicular.
Let $B_A$ be the magnetic field due to loop $A$ and $B_B$ be the magnetic field due to loop $B$. Then $B_A = B_B = \frac{\mu_0 I}{2 a}$.
The net magnetic field $B_{net}$ at the centre is the vector sum of $B_A$ and $B_B$:
$B_{net} = \sqrt{B_A^2 + B_B^2} = \sqrt{\left(\frac{\mu_0 I}{2 a}\right)^2 + \left(\frac{\mu_0 I}{2 a}\right)^2}$
$B_{net} = \sqrt{2 \left(\frac{\mu_0 I}{2 a}\right)^2} = \sqrt{2} \cdot \frac{\mu_0 I}{2 a} = \frac{\mu_0 I}{\sqrt{2} a}$.

(A) The width of the $n^{\text{th}}$ secondary maxima in a single-slit diffraction pattern is given by $\beta = \frac{\lambda D}{a}$,where $\lambda$ is the wavelength,$D$ is the distance of the screen (focal length of the lens),and $a$ is the slit width.
Given:
$\lambda = 400 \ nm = 400 \times 10^{-9} \ m$
$a = 0.2 \ mm = 0.2 \times 10^{-3} \ m$
$D = 100 \ cm = 1 \ m$
Substituting the values:
$\text{Width} = \frac{400 \times 10^{-9} \ m \times 1 \ m}{0.2 \times 10^{-3} \ m}$
$= \frac{400}{0.2} \times 10^{-6} \ m$
$= 2000 \times 10^{-6} \ m$
$= 2 \times 10^{-3} \ m = 2 \ mm$.

(A) For an ideal transformer, the ratio of voltages is equal to the ratio of the number of turns:
$\frac{V_p}{V_s} = \frac{N_p}{N_s}$
Given $V_p = 220 \,V$, $N_p = 100$, and $N_s = 10$, we have:
$\frac{220}{V_s} = \frac{100}{10} = 10$
$V_s = \frac{220}{10} = 22 \,V$
This secondary voltage $V_s = 22 \,V$ is applied across two resistors in series: $R_1 = 15 \,k\Omega$ and $R_2 = 7 \,k\Omega$.
The output voltage $V_0$ is the voltage across the $7 \,k\Omega$ resistor, which can be calculated using the voltage divider rule:
$V_0 = V_s \times \left( \frac{R_2}{R_1 + R_2} \right)$
$V_0 = 22 \times \left( \frac{7 \,k\Omega}{15 \,k\Omega + 7 \,k\Omega} \right)$
$V_0 = 22 \times \left( \frac{7}{22} \right) = 7 \,V$

(C) Given: Resistance at room temperature $R_0 = 60 \ \Omega$ at $T_0 = 27^{\circ} C$.
Voltage $V = 220 \ V$ and current $I = 2.75 \ A$.
The resistance at the final temperature $T$ is given by $R_T = \frac{V}{I} = \frac{220}{2.75} = 80 \ \Omega$.
The temperature dependence of resistance is given by $R_T = R_0(1 + \alpha \Delta T)$,where $\Delta T = T - T_0$.
Substituting the values: $80 = 60[1 + 2 \times 10^{-4}(T - 27)]$.
$80/60 = 1 + 2 \times 10^{-4}(T - 27)$.
$1.333 - 1 = 2 \times 10^{-4}(T - 27)$.
$0.333 = 2 \times 10^{-4}(T - 27)$.
$T - 27 = \frac{0.333}{2 \times 10^{-4}} = 1665$.
$T = 1665 + 27 = 1692^{\circ} C$.
Rounding to the nearest option,the temperature is approximately $1694^{\circ} C$.

(C) The Zener diode is in the breakdown region, so the voltage across it is constant at $10 \, V$.
The current through the $500 \, \Omega$ load resistor $(I_L)$ is:
$I_L = \frac{V_Z}{R_L} = \frac{10 \, V}{500 \, \Omega} = 0.02 \, A = 20 \, mA$
The current through the series resistor $(I_S)$ is:
$I_S = \frac{V_{in} - V_Z}{R_S} = \frac{20 \, V - 10 \, V}{200 \, \Omega} = \frac{10 \, V}{200 \, \Omega} = 0.05 \, A = 50 \, mA$
Applying Kirchhoff's Current Law at the junction, the current through the Zener diode $(I_Z)$ is:
$I_Z = I_S - I_L = 50 \, mA - 20 \, mA = 30 \, mA$

(B) The electrostatic potential $V$ at a point due to an electric dipole is given by the formula:
$V = \frac{1}{4\pi\epsilon_0} \frac{p \cos \theta}{r^2}$
Where $p$ is the dipole moment,$\theta$ is the angle between the dipole moment vector and the position vector,and $r$ is the distance from the center of the dipole.
From this expression,it is clear that the potential $V$ is inversely proportional to the square of the distance $r$.
Therefore,$V \propto \frac{1}{r^2}$.

(C) For a series $L, R$ circuit,the power factor is given by $\cos \phi = \frac{R}{Z} = \frac{R}{\sqrt{R^2 + X_L^2}}$,where $X_L = \omega L$.
Given the initial angular frequency $\omega_1 = 1000 \ rad/s$ and power factor $\cos \phi_1 = \frac{1}{\sqrt{2}}$.
Since $\cos \phi_1 = \frac{1}{\sqrt{2}}$,the phase angle $\phi_1 = 45^{\circ}$.
Thus,$\tan \phi_1 = \frac{X_{L1}}{R} = \frac{\omega_1 L}{R} = \tan 45^{\circ} = 1$.
This implies $R = \omega_1 L = 1000 L$.
When the source is changed to $E = (20 \sin 2000 t) \ V$,the new angular frequency is $\omega_2 = 2000 \ rad/s$.
The new inductive reactance is $X_{L2} = \omega_2 L = 2000 L = 2(\omega_1 L) = 2R$.
The new power factor is $\cos \phi_2 = \frac{R}{\sqrt{R^2 + X_{L2}^2}} = \frac{R}{\sqrt{R^2 + (2R)^2}} = \frac{R}{\sqrt{R^2 + 4R^2}} = \frac{R}{\sqrt{5R^2}} = \frac{1}{\sqrt{5}}$.

(A) The horizontal component of the Earth's magnetic field is $B_H = 3.5 \times 10^{-5} \,T$.
The current in the conductor is $i = \sqrt{2} \,A$.
The conductor is placed from South-East to North-West,which makes an angle of $\theta = 45^\circ$ with the horizontal magnetic field (North-South direction).
The force per unit length on a current-carrying conductor is given by $\frac{F}{\ell} = i B_H \sin \theta$.
Substituting the values: $\frac{F}{\ell} = \sqrt{2} \times (3.5 \times 10^{-5}) \times \sin(45^\circ)$.
Since $\sin(45^\circ) = \frac{1}{\sqrt{2}}$,we get $\frac{F}{\ell} = \sqrt{2} \times 3.5 \times 10^{-5} \times \frac{1}{\sqrt{2}} = 3.5 \times 10^{-5} \,N/m$.
Converting to the required format: $3.5 \times 10^{-5} = 35 \times 10^{-6} \,N/m$.
Therefore,the value is $35$.

(B) The total electromotive force (emf) of the circuit is $E_{eq} = E_1 - E_2 = 8 \ V - 2 \ V = 6 \ V$.
The total resistance of the circuit is $R_{eq} = r_1 + r_2 = 2 \ \Omega + 4 \ \Omega = 6 \ \Omega$.
The current flowing in the circuit is $I = \frac{E_{eq}}{R_{eq}} = \frac{6 \ V}{6 \ \Omega} = 1 \ A$.
Since the cell $E_2$ is being charged (current enters its positive terminal),its terminal potential difference $V$ is given by $V = E_2 + Ir_2$.
Substituting the values,we get $V = 2 \ V + (1 \ A \times 4 \ \Omega) = 2 \ V + 4 \ V = 6 \ V$.

(C) The energy of an electron in the $n^{th}$ orbit of a hydrogen atom is given by $E_n = -\frac{13.6}{n^2} \ eV$.
Given $E_n = -0.85 \ eV$,we have:
$-\frac{13.6}{n^2} = -0.85$
$n^2 = \frac{13.6}{0.85} = 16$
$n = 4$.
The number of possible transitions from an excited state $n$ to lower energy levels is given by the formula $\frac{n(n-1)}{2}$.
Substituting $n = 4$:
Number of transitions $= \frac{4(4-1)}{2} = \frac{4 \times 3}{2} = 6$.

(A) For a real image,magnification $m = -2$.
Since $m = \frac{v}{u}$,we have $\frac{v}{u} = -2$,which implies $v = -2u$.
Given the distance between the object and the image is $45 \,cm$,and for a real image formed by a convex lens,the object and image are on opposite sides,the distance is $|v| + |u| = 45$.
Substituting $v = -2u$ (where $u$ is negative),we get $|-2u| + |u| = 45$,so $3|u| = 45$,which gives $|u| = 15 \,cm$.
Thus,$u = -15 \,cm$ and $v = +30 \,cm$.
Using the lens formula $\frac{1}{f} = \frac{1}{v} - \frac{1}{u}$:
$\frac{1}{f} = \frac{1}{30} - \frac{1}{-15} = \frac{1}{30} + \frac{2}{30} = \frac{3}{30} = \frac{1}{10}$.
Therefore,$f = 10 \,cm$.

(C) Initial energy of the first capacitor: $E_1 = \frac{1}{2} C V^2 = E$.
Initial energy of the second capacitor: $E_2 = \frac{1}{2} (2C) (2V)^2 = \frac{1}{2} (2C) (4V^2) = 4 C V^2 = 8E$.
Total initial energy: $E_i = E_1 + E_2 = E + 8E = 9E$.
Total charge: $Q_{total} = Q_1 + Q_2 = CV + (2C)(2V) = CV + 4CV = 5CV$.
Total capacitance: $C_{eq} = C + 2C = 3C$.
Common potential: $V_{common} = \frac{Q_{total}}{C_{eq}} = \frac{5CV}{3C} = \frac{5}{3} V$.
Final energy: $E_f = \frac{1}{2} C_{eq} V_{common}^2 = \frac{1}{2} (3C) (\frac{5}{3} V)^2 = \frac{1}{2} (3C) (\frac{25}{9} V^2) = \frac{25}{6} C V^2 = \frac{25}{3} E$.
Loss of energy: $\Delta E = E_i - E_f = 9E - \frac{25}{3} E = \frac{27E - 25E}{3} = \frac{2}{3} E$.
Comparing with $\frac{x}{3} E$,we get $x = 2$.

(B) The vertical component of the Earth's magnetic field is given by $B_V = B \sin(\delta)$,where $B = 0.5 \ G = 0.5 \times 10^{-4} \ T$ and $\delta = 30^{\circ}$.
$B_V = 0.5 \times 10^{-4} \times \sin(30^{\circ}) = 0.5 \times 10^{-4} \times 0.5 = 0.25 \times 10^{-4} \ T = 0.25 \times 10^{-4} \ T = \frac{1}{4} \times 10^{-4} \ T$.
The angular velocity $\omega$ is given by $\omega = \frac{2 \pi n}{60}$,where $n = 1200 \ rpm$.
$\omega = \frac{2 \pi \times 1200}{60} = 40 \pi \ rad/s$.
The induced $EMF$ across a rotating rod is $\varepsilon = \frac{1}{2} B_V \omega \ell^2$,where $\ell = 80 \ cm = 0.8 \ m$.
$\varepsilon = \frac{1}{2} \times (0.25 \times 10^{-4}) \times (40 \pi) \times (0.8)^2$.
$\varepsilon = 0.5 \times 0.25 \times 10^{-4} \times 40 \pi \times 0.64$.
$\varepsilon = 0.125 \times 10^{-4} \times 40 \pi \times 0.64 = 5 \pi \times 10^{-4} \times 0.64 = 3.2 \pi \times 10^{-4} = 32 \pi \times 10^{-5} \ V$.
Comparing this with $N \pi \times 10^{-5} \ V$,we get $N = 32$.

(D) According to Einstein's photoelectric equation,the maximum kinetic energy $(E_{k})$ of the emitted photoelectrons is given by:
$E_{k} = h\nu - \phi$
where $h$ is Planck's constant,$\nu$ is the frequency of incident photons,and $\phi$ is the work function of the metal.
Comparing this equation with the equation of a straight line $y = mx + c$,where $y = E_{k}$,$x = \nu$,$m$ is the slope,and $c$ is the intercept:
The slope of the graph is $m = \tan \theta = h$.
Therefore,the slope of the graph gives Planck's constant.

(C) The initial mass of the parent nucleus is $M$. It splits into three daughter nuclei,each of mass $m = M/3$.
The energy released in the fission reaction is given by Einstein's mass-energy equivalence: $E = \Delta M c^2$.
This energy is converted into the kinetic energy of the three daughter nuclei. Let $v$ be the speed of each daughter nucleus.
The total kinetic energy is $K.E. = 3 \times (\frac{1}{2} m v^2) = 3 \times (\frac{1}{2} \times \frac{M}{3} \times v^2) = \frac{1}{2} M v^2$.
Equating the energy released to the kinetic energy: $\Delta M c^2 = \frac{1}{2} M v^2$.
Solving for $v$: $v^2 = \frac{2 \Delta M c^2}{M}$.
Therefore,$v = c \sqrt{\frac{2 \Delta M}{M}}$.

(B) The magnetic moment $\mu$ of an electron revolving in an orbit is given by $\mu = iA$,where $i$ is the current and $A$ is the area of the orbit.
$i = \frac{e}{T} = \frac{ev}{2\pi r}$,where $v$ is the velocity and $r$ is the radius.
$A = \pi r^2$.
Therefore,$\mu = \left(\frac{ev}{2\pi r}\right) \pi r^2 = \frac{evr}{2}$.
According to Bohr's theory,$r \propto n^2$ and $v \propto \frac{1}{n}$.
Substituting these relations: $\mu \propto \left(\frac{1}{n}\right) \cdot n^2 = n^1$.
Comparing $\mu \propto n^1$ with $\mu \propto n^x$,we get $x = 1$.

(B) The given voltage is $V(t) = 220 \sin(100 \pi t)$. Since the load is purely resistive,the current $I(t)$ is in phase with the voltage: $I(t) = I_0 \sin(100 \pi t)$,where $I_0 = V_0 / R = 220 / 50 = 4.4 \ A$.
We need the time taken for the current to rise from $I_0 / 2$ to $I_0$.
At $t_1$,$I(t_1) = I_0 \sin(100 \pi t_1) = I_0 / 2 \implies 100 \pi t_1 = \pi / 6 \implies t_1 = 1 / 600 \ s$.
At $t_2$,$I(t_2) = I_0 \sin(100 \pi t_2) = I_0 \implies 100 \pi t_2 = \pi / 2 \implies t_2 = 1 / 200 \ s$.
The time taken is $\Delta t = t_2 - t_1 = 1/200 - 1/600 = (3 - 1) / 600 = 2 / 600 = 1 / 300 \ s$.
$\Delta t = 0.00333 \ s = 3.33 \ ms$.

(B) For complete absorption of radiation, the momentum $p$ delivered to a surface is related to the energy $E$ transferred by the relation $p = \frac{E}{c}$, where $c$ is the speed of light in vacuum $(c = 3 \times 10^8 \,m/s)$.
Given energy $E = 6.48 \times 10^5 \,J$.
Substituting the values:
$p = \frac{6.48 \times 10^5}{3 \times 10^8} \,kg \cdot m/s$
$p = 2.16 \times 10^{-3} \,kg \cdot m/s$.
Thus, the correct option is $B$.

(A) $1$. When unpolarised light of intensity $I_0$ passes through the first polaroid $A$,the intensity of the transmitted light becomes $I_1 = I_0 / 2$.
$2$. According to Malus' Law,when this polarised light passes through a second polaroid $B$ whose transmission axis is at an angle $\theta = 45^{\circ}$ with respect to the first polaroid,the final intensity $I_2$ is given by $I_2 = I_1 \cos^2 \theta$.
$3$. Substituting the values: $I_2 = (I_0 / 2) \cos^2(45^{\circ})$.
$4$. Since $\cos(45^{\circ}) = 1 / \sqrt{2}$,then $\cos^2(45^{\circ}) = 1/2$.
$5$. Therefore,$I_2 = (I_0 / 2) \times (1 / 2) = I_0 / 4$.

(B) The electric field $E$ at a distance $r$ from an infinitely long line charge is given by $E = \frac{2 k \lambda}{r}$.
The electrostatic force providing the centripetal force is $F = qE = \frac{2 k \lambda q}{r}$.
Equating this to the centripetal force $m \omega^2 r$,we get $\frac{2 k \lambda q}{r} = m \omega^2 r$.
Solving for angular velocity $\omega$,we have $\omega^2 = \frac{2 k \lambda q}{m r^2}$,which implies $\omega = \frac{1}{r} \sqrt{\frac{2 k \lambda q}{m}}$.
Since the time period $T = \frac{2 \pi}{\omega}$,we substitute $\omega$ to get $T = 2 \pi r \sqrt{\frac{m}{2 k \lambda q}}$.

(A) The circuit contains a $15 \, V$ $DC$ source, a Germanium $(Ge)$ diode, a Silicon $(Si)$ diode, a $1.5 \, k\Omega$ resistor, and a load resistor $R_L = 2.5 \, k\Omega$ in series.
The barrier potential for a $Ge$ diode is $V_{Ge} = 0.3 \, V$ and for a $Si$ diode is $V_{Si} = 0.7 \, V$.
Applying Kirchhoff's Voltage Law $(KVL)$ in the loop:
$15 - V_{Ge} - V_{Si} - i(1.5 \, k\Omega) - i(2.5 \, k\Omega) = 0$
$15 - 0.3 - 0.7 = i(1.5 + 2.5) \, k\Omega$
$14 = i(4 \, k\Omega)$
$i = \frac{14}{4} \, mA = 3.5 \, mA$
The voltage across the load resistance $R_L$ is:
$V_L = i \times R_L = 3.5 \, mA \times 2.5 \, k\Omega = 8.75 \, V$

(D) The initial power dissipated is given by $W = \frac{V^2}{R} \quad ...(i)$
When the wire is cut into two equal halves,the resistance of each half becomes $R' = \frac{R}{2}$.
When these two halves are connected in parallel,the equivalent resistance $R_{eq}$ is given by:
$\frac{1}{R_{eq}} = \frac{1}{R'} + \frac{1}{R'} = \frac{2}{R} + \frac{2}{R} = \frac{4}{R}$
So,$R_{eq} = \frac{R}{4}$.
The new power dissipation $W'$ with the same potential difference $V$ is:
$W' = \frac{V^2}{R_{eq}} = \frac{V^2}{R/4} = 4 \left( \frac{V^2}{R} \right)$
Substituting equation $(i)$ into this expression,we get:
$W' = 4W$.

(D) The Maxwell's equations in integral form are as follows:
$A$. Gauss's law of magnetostatics states that the net magnetic flux through any closed surface is zero: $\oint \vec{B} \cdot d\vec{a} = 0$ $(A-II)$.
$B$. Faraday's law of electromagnetic induction states that the induced electromotive force in a closed loop is equal to the negative rate of change of magnetic flux through the loop: $\oint \vec{E} \cdot d\vec{l} = -\frac{d}{dt} \int \vec{B} \cdot d\vec{a}$ $(B-III)$.
$C$. Ampere's law (in its original form) relates the line integral of the magnetic field around a closed loop to the current passing through the surface enclosed by the loop: $\oint \vec{B} \cdot d\vec{l} = \mu_0 I$ $(C-IV)$.
$D$. Gauss's law of electrostatics states that the total electric flux through a closed surface is equal to the total charge enclosed divided by the permittivity of free space: $\oint \vec{E} \cdot d\vec{a} = \frac{1}{\epsilon_0} \int \rho dV$ $(D-I)$.
Thus,the correct matching is $A-II, B-III, C-IV, D-I$.

(A) The input power $P_{\text{in}}$ is given by the product of primary voltage $V_p$ and primary current $I_p$.
$P_{\text{in}} = V_p \times I_p = 2300 \text{ V} \times 5 \text{ A} = 11500 \text{ W}$.
Efficiency $\eta$ is defined as the ratio of output power $P_{\text{out}}$ to input power $P_{\text{in}}$.
Given $\eta = 90\% = 0.9$,so $P_{\text{out}} = \eta \times P_{\text{in}} = 0.9 \times 11500 \text{ W} = 10350 \text{ W}$.
The output power is also given by $P_{\text{out}} = V_s \times I_s$,where $V_s = 230 \text{ V}$ is the output voltage.
$10350 \text{ W} = 230 \text{ V} \times I_s$.
$I_s = \frac{10350}{230} \text{ A} = 45 \text{ A}$.

(D) According to Newton's form of the lens equation,when distances $x$ and $y$ are measured from the focal points,the relation is given by $xy = f^2$.
From the provided graph,we can observe a point on the curve where $x = 20 \ cm$ and $y = 20 \ cm$.
Substituting these values into the equation:
$20 \times 20 = f^2$
$f^2 = 400$
$f = 20 \ cm$.
Alternatively,the graph shows that at $x = 10 \ cm$,$y = 40 \ cm$ (from the intercept on the y-axis),and at $x = 40 \ cm$,$y = 10 \ cm$ (from the intercept on the x-axis).
Using $xy = f^2$,we get $10 \times 40 = f^2$,which implies $f^2 = 400$,so $f = 20 \ cm$.

(B) The magnetic field due to a straight wire of length $L$ at a perpendicular distance $d$ is given by $B = \frac{\mu_0 i}{4 \pi d} (\sin \theta_1 + \sin \theta_2)$.
For a square loop of side $a = 1 \ m$,the distance from the center to any side is $d = a/2 = 0.5 \ m$.
The angles at the center subtended by the corners are $\theta_1 = \theta_2 = 45^\circ$,so $\sin 45^\circ = 1/\sqrt{2}$.
The magnetic field due to one side is $B_1 = \frac{\mu_0 i}{4 \pi (a/2)} (\frac{1}{\sqrt{2}} + \frac{1}{\sqrt{2}}) = \frac{\mu_0 i}{2 \pi a} (\frac{2}{\sqrt{2}}) = \frac{\mu_0 i \sqrt{2}}{2 \pi a}$.
For a square loop,the total magnetic field $B = 4 \times B_1 = 4 \times \frac{\mu_0 i \sqrt{2}}{2 \pi a} = \frac{2 \mu_0 i \sqrt{2}}{\pi a}$.
Given $\mu_0 = 4 \pi \times 10^{-7} \ T \cdot m/A$,$i = 5 \ A$,and $a = 1 \ m$:
$B = \frac{2 \times (4 \pi \times 10^{-7}) \times 5 \times \sqrt{2}}{\pi \times 1} = 40 \sqrt{2} \times 10^{-7} \ T$.
Comparing this with $X \sqrt{2} \times 10^{-7} \ T$,we get $X = 40$.

(C) Let $\theta$ be the angle each string makes with the vertical. Since the total angle between the strings is $37^{\circ}$,$\theta = 37^{\circ}/2 = 18.5^{\circ}$.
In air,the equilibrium condition is $\tan \theta = \frac{F_e}{mg} = \frac{F_e}{\rho_B V g}$,where $\rho_B$ is the density of the sphere and $V$ is its volume.
In the liquid,the effective weight is $mg' = V(\rho_B - \rho_L)g$ and the electrostatic force becomes $F_e' = \frac{F_e}{K}$,where $K$ is the dielectric constant.
The equilibrium condition in the liquid is $\tan \theta = \frac{F_e'}{mg'} = \frac{F_e}{K V(\rho_B - \rho_L)g}$.
Since $\theta$ remains the same,we equate the two expressions for $\tan \theta$:
$\frac{F_e}{\rho_B V g} = \frac{F_e}{K V(\rho_B - \rho_L)g}$
$\rho_B = K(\rho_B - \rho_L)$
$1.4 = K(1.4 - 0.7)$
$1.4 = K(0.7)$
$K = \frac{1.4}{0.7} = 2$.

(B) Let the resistance of the voltmeter be $R_V$. The voltmeter is connected in parallel with the $100 \ \Omega$ resistor. The equivalent resistance of this parallel combination is $R_p = \frac{100 \cdot R_V}{100 + R_V}$.
The circuit now consists of $R_p$ in series with the $200 \ \Omega$ resistor,connected to a $4 \ V$ battery.
According to the voltage divider rule,the voltage across the parallel combination $(R_p)$ is given by $V_p = V \cdot \frac{R_p}{R_p + 200}$.
Given $V_p = 1 \ V$ and $V = 4 \ V$,we have $1 = 4 \cdot \frac{R_p}{R_p + 200}$.
$R_p + 200 = 4 R_p \implies 3 R_p = 200 \implies R_p = \frac{200}{3} \ \Omega$.
Equating the two expressions for $R_p$: $\frac{100 R_V}{100 + R_V} = \frac{200}{3}$.
$300 R_V = 200(100 + R_V) \implies 300 R_V = 20000 + 200 R_V$.
$100 R_V = 20000 \implies R_V = 200 \ \Omega$.

(C) The circuit consists of two $NOT$ gates connected to the inputs $A$ and $B$,followed by a $NAND$ gate.
$1$. The inputs to the $NAND$ gate are $\overline{A}$ and $\overline{B}$.
$2$. The output $Y$ of the $NAND$ gate is given by $Y = \overline{\overline{A} \cdot \overline{B}}$.
$3$. According to De Morgan's law,$\overline{X \cdot Y} = \overline{X} + \overline{Y}$.
$4$. Applying this to our expression: $Y = \overline{\overline{A}} + \overline{\overline{B}} = A + B$.
$5$. The expression $Y = A + B$ represents the $OR$ logic operation.

(D) The formula for the refractive index $\mu$ of a prism in terms of the apex angle $A$ and the angle of minimum deviation $\delta_m$ is given by:
$\mu = \frac{\sin((A + \delta_m)/2)}{\sin(A/2)}$
Given $\mu = \cot(A/2) = \frac{\cos(A/2)}{\sin(A/2)}$,we equate the two expressions:
$\frac{\cos(A/2)}{\sin(A/2)} = \frac{\sin((A + \delta_m)/2)}{\sin(A/2)}$
Canceling $\sin(A/2)$ from both sides,we get:
$\cos(A/2) = \sin((A + \delta_m)/2)$
Using the trigonometric identity $\cos(\theta) = \sin(90^{\circ} - \theta)$,we write:
$\sin(90^{\circ} - A/2) = \sin((A + \delta_m)/2)$
Equating the angles:
$90^{\circ} - A/2 = (A + \delta_m)/2$
$180^{\circ} - A = A + \delta_m$
$\delta_m = 180^{\circ} - 2A$

(C) The magnetic force on a current-carrying wire is given by $\vec{F} = i (\vec{L}_{eff} \times \vec{B})$.
Here,the effective length vector $\vec{L}_{eff}$ is the displacement vector from the starting point to the ending point of the wire within the magnetic field.
From the figure,the current enters at the bottom left and exits at the bottom right of the semicircular loop. The displacement vector $\vec{L}_{eff}$ for the semicircular part is $2R \hat{i}$.
The magnetic field is $\vec{B} = B_0 \hat{k}$.
Therefore,$\vec{F} = i (2R \hat{i} \times B_0 \hat{k})$.
Since $\hat{i} \times \hat{k} = -\hat{j}$,we get $\vec{F} = i (2R B_0) (-\hat{j}) = -2 i B_0 R \hat{j}$.