JEE Main 2022 Physics Question Paper with Answer and Solution

(B) The acceleration of a system of two masses $M_{1}$ and $M_{2}$ connected by a string over a frictionless pulley is given by the formula:
$a = \frac{|M_{2} - M_{1}|}{M_{1} + M_{2}} g$
Case $1$: Given $M_{2} = 2M_{1}$.
$a_{1} = \frac{|2M_{1} - M_{1}|}{M_{1} + 2M_{1}} g = \frac{M_{1}}{3M_{1}} g = \frac{g}{3}$
Case $2$: Given $M_{2} = 3M_{1}$.
$a_{2} = \frac{|3M_{1} - M_{1}|}{M_{1} + 3M_{1}} g = \frac{2M_{1}}{4M_{1}} g = \frac{g}{2}$
Now,calculating the ratio $\frac{a_{1}}{a_{2}}$:
$\frac{a_{1}}{a_{2}} = \frac{g/3}{g/2} = \frac{2}{3}$
Therefore,the correct option is $B$.

(A) The breaking stress of the material is constant,as it is an intrinsic property of the material.
Breaking stress = $\frac{\text{Maximum load}}{\text{Area of cross-section}}$.
Let $A_1 = 2.5 \times 10^{-4} \, m^2$ be the initial area for a load $L_1 = 10$ metric tons.
Let $A_2$ be the required area for a load $L_2 = 25$ metric tons.
Since the breaking stress is constant: $\frac{L_1}{A_1} = \frac{L_2}{A_2}$.
Substituting the values: $\frac{10}{2.5 \times 10^{-4}} = \frac{25}{A_2}$.
$A_2 = \frac{25 \times 2.5 \times 10^{-4}}{10}$.
$A_2 = 6.25 \times 10^{-4} \, m^2$.

(A) Given vectors are $\overrightarrow{A} = (2\hat{i} + 3\hat{j} - \hat{k}) \; m$ and $\overrightarrow{B} = (\hat{i} + 2\hat{j} + 2\hat{k}) \; m$.
The component of vector $\overrightarrow{A}$ along vector $\overrightarrow{B}$ is given by the projection formula: $\text{Component} = \overrightarrow{A} \cdot \hat{B} = \frac{\overrightarrow{A} \cdot \overrightarrow{B}}{|\overrightarrow{B}|}$.
First,calculate the dot product $\overrightarrow{A} \cdot \overrightarrow{B}$:
$\overrightarrow{A} \cdot \overrightarrow{B} = (2)(1) + (3)(2) + (-1)(2) = 2 + 6 - 2 = 6$.
Next,calculate the magnitude of vector $\overrightarrow{B}$:
$|\overrightarrow{B}| = \sqrt{1^2 + 2^2 + 2^2} = \sqrt{1 + 4 + 4} = \sqrt{9} = 3$.
Finally,calculate the component:
$\text{Component} = \frac{6}{3} = 2 \; m$.

(A) The moment of inertia $I$ of a uniform rod of mass $m$ and length $\ell$ about an axis passing through its center and perpendicular to its length is given by $I = \frac{m \ell^{2}}{12}$.
By definition,the moment of inertia is also expressed as $I = mk^{2}$,where $k$ is the radius of gyration.
Equating the two expressions: $mk^{2} = \frac{m \ell^{2}}{12}$.
Solving for $k$: $k^{2} = \frac{\ell^{2}}{12} \Rightarrow k = \frac{\ell}{\sqrt{12}} = \frac{\ell}{2 \sqrt{3}}$.
Given $\ell = 10 \sqrt{3} \ m$,we substitute this value:
$k = \frac{10 \sqrt{3}}{2 \sqrt{3}} = 5 \ m$.
Thus,the radius of gyration is $5 \ m$.

(B) The elongation $\Delta \ell$ of a rod of length $\ell$,mass $m$,and cross-sectional area $A$ due to its own weight is given by the formula:
$\Delta \ell = \frac{mg\ell}{2AY}$
Given values:
$m = 20\,kg$
$A = 0.4\,m^{2}$
$\ell = 20\,m$
$Y = 2 \times 10^{11}\,N/m^{2}$
$g = 10\,m/s^{2}$
Substituting these values into the formula:
$\Delta \ell = \frac{20 \times 10 \times 20}{2 \times 0.4 \times 2 \times 10^{11}}$
$\Delta \ell = \frac{4000}{1.6 \times 10^{11}}$
$\Delta \ell = 2500 \times 10^{-11}\,m$
$\Delta \ell = 25 \times 10^{-9}\,m$
Comparing this with $x \times 10^{-9}\,m$,we get $x = 25$.

(D) For figure $(a)$,the springs are in series. The equivalent spring constant $K_{eq}$ is given by:
$K_{eq} = \frac{K \times 2K}{K + 2K} = \frac{2K}{3}$
The period of oscillation $T$ is:
$T = 2\pi \sqrt{\frac{m}{K_{eq}}} = 2\pi \sqrt{\frac{m}{2K/3}} = 2\pi \sqrt{\frac{3m}{2K}} = 3 \text{ s}$
For figure $(b)$,the springs are in parallel. The equivalent spring constant $K'_{eq}$ is:
$K'_{eq} = K + 2K = 3K$
The period of oscillation $T'$ is:
$T' = 2\pi \sqrt{\frac{m}{3K}}$
Dividing $T'$ by $T$:
$\frac{T'}{T} = \frac{2\pi \sqrt{\frac{m}{3K}}}{2\pi \sqrt{\frac{3m}{2K}}} = \sqrt{\frac{m}{3K} \times \frac{2K}{3m}} = \sqrt{\frac{2}{9}} = \frac{\sqrt{2}}{3}$
Given $T = 3 \text{ s}$,we have:
$T' = 3 \times \frac{\sqrt{2}}{3} = \sqrt{2} \text{ s}$
Comparing $T' = \sqrt{x} \text{ s}$ with $T' = \sqrt{2} \text{ s}$,we get $x = 2$.

(C) Let the acceleration of the $100 \, kg$ block be $a_{1}$ towards the right.
For the $m_{2} = 20 \, kg$ block,it is moving upward with acceleration $a_{y} = 2 \, m/s^{2}$ relative to the $100 \, kg$ block. The tension $T$ in the string acts upward,and gravity $20g$ acts downward. The pseudo force $20a_{1}$ acts horizontally to the left.
Applying Newton's second law in the vertical direction for $m_{2}$:
$T - m_{2}g = m_{2}a_{y}$
$T - 20(10) = 20(2)$
$T - 200 = 40 \Rightarrow T = 240 \, N$.
Now,consider the $m_{1} = 10 \, kg$ block on top of the $100 \, kg$ block. It is connected to the $20 \, kg$ block via the string. The horizontal acceleration of $m_{1}$ relative to the $100 \, kg$ block is $a_{x} = 2 \, m/s^{2}$ (since the string is inextensible,the horizontal acceleration of $m_{1}$ must match the vertical acceleration of $m_{2}$).
Applying Newton's second law for $m_{1}$ in the horizontal direction:
$T = m_{1}a_{1} \Rightarrow 240 = 10a_{1} \Rightarrow a_{1} = 24 \, m/s^{2}$.
Finally,consider the whole system $(M + m_{1} + m_{2})$ moving with acceleration $a_{1}$:
$F = (M + m_{1} + m_{2})a_{1}$
$F = (100 + 10 + 20) \times 24$
$F = 130 \times 24 = 3120 \, N$.

(B) For a particle in $SHM$,its velocity $v$ depends on displacement $x$ as:
$v = \omega \sqrt{A^{2} - x^{2}}$
Where $\omega$ is the angular frequency and $A$ is the amplitude.
Squaring both sides,we get:
$v^{2} = \omega^{2} (A^{2} - x^{2})$
$v^{2} = \omega^{2} A^{2} - \omega^{2} x^{2}$
Rearranging the terms:
$v^{2} + \omega^{2} x^{2} = \omega^{2} A^{2}$
Dividing by $\omega^{2} A^{2}$:
$\frac{v^{2}}{(\omega A)^{2}} + \frac{x^{2}}{A^{2}} = 1$
This equation is of the form $\frac{y^{2}}{b^{2}} + \frac{x^{2}}{a^{2}} = 1$,which represents an ellipse.
Therefore,the graph between velocity $v$ and displacement $x$ is elliptical.

(B) For any ideal gas,the change in internal energy $\Delta U$ depends only on the change in temperature $\Delta T$ and is given by the formula:
$\Delta U = n C_V \Delta T$
For a monoatomic ideal gas,the molar heat capacity at constant volume is $C_V = \frac{3}{2} R$.
Given values are:
$n = 7 \text{ mol}$
$\Delta T = 40 K$
$R = 8.3 J K^{-1} mol^{-1}$
Substituting these values into the formula:
$\Delta U = 7 \times \left( \frac{3}{2} \times 8.3 \right) \times 40$
$\Delta U = 7 \times 3 \times 8.3 \times 20$
$\Delta U = 21 \times 166 = 3486 J$
Thus,the increase in internal energy is $3486 J$.

(C) process with constant entropy is an adiabatic process.
For an adiabatic process,the relation between pressure and volume is given by $PV^{\gamma} = \text{constant}$.
For a monoatomic gas,the adiabatic index $\gamma = \frac{5}{3}$.
Given,initial volume $V_1 = V$ and final volume $V_2 = \frac{V}{8}$.
Using the adiabatic equation: $P_1 V_1^{\gamma} = P_2 V_2^{\gamma}$.
Substituting the values: $P \cdot V^{5/3} = P_2 \cdot (\frac{V}{8})^{5/3}$.
$P_2 = P \cdot (\frac{V}{V/8})^{5/3} = P \cdot (8)^{5/3}$.
$P_2 = P \cdot (2^3)^{5/3} = P \cdot 2^5$.
$P_2 = 32P$.

(C) Initial radius $R = 1\,cm = 10^{-2}\,m$. Surface tension $T = 75\,dyne/cm = 75 \times 10^{-3}\,N/m$.
Initial surface area $A_i = 4\pi R^2 = 4\pi(10^{-2})^2 = 4\pi \times 10^{-4}\,m^2$.
Initial surface energy $U_i = T \times A_i = 75 \times 10^{-3} \times 4\pi \times 10^{-4} = 300\pi \times 10^{-7}\,J$.
Let $r$ be the radius of each small droplet. By volume conservation: $\frac{4}{3}\pi R^3 = 729 \times \frac{4}{3}\pi r^3$.
$r^3 = \frac{R^3}{729} \implies r = \frac{R}{9} = \frac{1}{9}\,cm = \frac{1}{9} \times 10^{-2}\,m$.
Final surface area $A_f = 729 \times 4\pi r^2 = 729 \times 4\pi \times (\frac{1}{9} \times 10^{-2})^2 = 729 \times 4\pi \times \frac{1}{81} \times 10^{-4} = 9 \times 4\pi \times 10^{-4} = 36\pi \times 10^{-4}\,m^2$.
Final surface energy $U_f = T \times A_f = 75 \times 10^{-3} \times 36\pi \times 10^{-4} = 2700\pi \times 10^{-7}\,J$.
Gain in surface energy $\Delta U = U_f - U_i = (2700\pi - 300\pi) \times 10^{-7} = 2400\pi \times 10^{-7}\,J$.
Using $\pi \approx 3.14$,$\Delta U = 2400 \times 3.14 \times 10^{-7} = 7536 \times 10^{-7} = 7.536 \times 10^{-4}\,J$.
Rounding to the first decimal place,the gain is $7.5 \times 10^{-4}\,J$.

(A) The acceleration due to gravity at a height $h$ above the surface of the Earth is given by $g' = g(1 - \frac{2h}{R})$,where $R$ is the radius of the Earth and $h << R$.
The change in acceleration due to gravity is $\Delta g = g - g' = g(\frac{2h}{R})$.
The fractional decrease in weight (which is proportional to the change in $g$) is given by $\frac{\Delta g}{g} = \frac{2h}{R}$.
To find the percentage decrease,we multiply by $100$:
$\text{Percentage decrease} = \frac{\Delta g}{g} \times 100 = \frac{2h}{R} \times 100$.
Substituting the given values $h = 32 \ km$ and $R = 6400 \ km$:
$\text{Percentage decrease} = 2 \times \frac{32}{6400} \times 100 = 2 \times \frac{1}{200} \times 100 = 1 \%$.

(B) Let the mass of each block be $m = 250\,g = 0.25\,kg$ and the spring constant be $k = 2\,N/m$.
At the moment of maximum elongation $x$,both blocks momentarily come to rest in the center-of-mass frame.
By the law of conservation of energy,the initial kinetic energy of the system is converted into the elastic potential energy of the spring at maximum elongation.
The total initial kinetic energy is $K_i = \frac{1}{2}mv^2 + \frac{1}{2}mv^2 = mv^2$.
The potential energy stored in the spring at maximum elongation $x$ is $U_f = \frac{1}{2}kx^2$.
Equating the two: $mv^2 = \frac{1}{2}kx^2$.
Solving for $x$: $x^2 = \frac{2mv^2}{k} \implies x = v \sqrt{\frac{2m}{k}}$.
Substituting the given values $m = 0.25\,kg$ and $k = 2\,N/m$:
$x = V \sqrt{\frac{2 \times 0.25}{2}} = V \sqrt{0.25} = 0.5V = \frac{V}{2}$.

(C) Free Body Diagram ($F$.$B$.$D$) of the monkey while moving downward:
Using Newton's second law: $mg - T = ma_1$
$500 - T = 50 \times 4 \Rightarrow T = 300\,N$.
Free Body Diagram ($F$.$B$.$D$) of the monkey while moving upward:
Using Newton's second law of motion: $T - mg = ma_2$
$T - 500 = 50 \times 5 \Rightarrow T = 750\,N$.
Breaking strength of the string is $350\,N$.
Since $750\,N > 350\,N$,the string will break while the monkey is moving upward.

(C) The time taken to reach the maximum height for a projectile is given by the formula:
$t = \frac{u \sin \theta}{g}$
Given that both projectiles reach the maximum height in the same time,we have:
$t_1 = t_2$
$\frac{u_1 \sin \theta_1}{g} = \frac{u_2 \sin \theta_2}{g}$
Substituting the given angles $\theta_1 = 30^{\circ}$ and $\theta_2 = 45^{\circ}$:
$u_1 \sin 30^{\circ} = u_2 \sin 45^{\circ}$
$u_1 \left( \frac{1}{2} \right) = u_2 \left( \frac{1}{\sqrt{2}} \right)$
Rearranging to find the ratio $\frac{u_1}{u_2}$:
$\frac{u_1}{u_2} = \frac{1/\sqrt{2}}{1/2} = \frac{2}{\sqrt{2}} = \sqrt{2}$
Therefore,the ratio of their initial velocities is $\sqrt{2}: 1$.

(B) The least count $(L.C.)$ of the screw gauge is given by $L.C. = \frac{\text{Pitch}}{\text{Number of divisions}} = \frac{0.5\,mm}{50} = 0.01\,mm = 0.001\,cm$.
The diameter $(D)$ of the wire is calculated as: $D = \text{Main scale reading} + (\text{Circular scale reading} \times L.C.)$.
$D = 1.5\,mm + (7 \times 0.01\,mm) = 1.5\,mm + 0.07\,mm = 1.57\,mm = 0.157\,cm$.
The length $(l)$ of the wire is $6.8\,cm$.
The curved surface area $(A)$ of the wire is given by $A = \pi D l$.
$A = 3.14 \times 0.157\,cm \times 6.8\,cm \approx 3.353\,cm^2$.
Rounding to two significant figures (as the length $6.8$ has two significant figures),we get $A = 3.4\,cm^2$.

(C) The initial horizontal velocity is given as $u_x = 1\,m/s$.
The equation of the trajectory is $y = 5x - 5x^2$.
Differentiating the equation with respect to time $t$,we get the vertical velocity component $v_y = \frac{dy}{dt} = \frac{dy}{dx} \cdot \frac{dx}{dt}$.
First,find $\frac{dy}{dx} = \frac{d}{dx}(5x - 5x^2) = 5 - 10x$.
Since $v_x = \frac{dx}{dt} = u_x = 1$ (assuming constant horizontal velocity),we have $v_y = (5 - 10x) \cdot 1$.
At the initial point,$x = 0$.
Therefore,the initial vertical velocity $u_y = (5 - 10(0)) = 5\,m/s$.
Thus,the $y$-component vector of the initial velocity is $5\,\hat{j}$.

(C) Let $m$ be the mass of the disc and the particle. The moment of inertia of the disc about the axis is $I_{\text{disc}} = \frac{1}{2} mR^2$. The moment of inertia of the particle at distance $R$ is $I_{\text{particle}} = mR^2$. The total moment of inertia is $I = I_{\text{disc}} + I_{\text{particle}} = \frac{1}{2} mR^2 + mR^2 = \frac{3}{2} mR^2$.
Using the law of conservation of mechanical energy,the loss in potential energy of the particle equals the gain in rotational kinetic energy of the system.
The particle moves from the highest point to the lowest point,a vertical distance of $2R$.
Loss in potential energy $= mg(2R) = 2mgR$.
Gain in rotational kinetic energy $= \frac{1}{2} I \omega^2 = \frac{1}{2} \left( \frac{3}{2} mR^2 \right) \omega^2 = \frac{3}{4} mR^2 \omega^2$.
Equating the two: $2mgR = \frac{3}{4} mR^2 \omega^2$.
Solving for $\omega^2$: $\omega^2 = \frac{8g}{3R}$.
Given $\omega = 4 \sqrt{\frac{x}{3R}}$,so $\omega^2 = 16 \left( \frac{x}{3R} \right) = \frac{16x}{3R}$.
Equating the expressions for $\omega^2$: $\frac{16x}{3R} = \frac{8g}{3R} \implies 16x = 8g$.
Substituting $g = 10 \text{ m s}^{-2}$: $16x = 8(10) = 80$.
$x = \frac{80}{16} = 5$.

(C) Young's modulus is given by $Y = \frac{FL}{A\Delta L} = \frac{mgL}{(\pi d^2/4)\Delta L} = \frac{4mgL}{\pi d^2 \Delta L}$.
Given: $L = 1\;m$,$m = 1\;kg$,$g = 10\;m/s^2$,$\Delta L = 0.4 \times 10^{-3}\;m$,$d = 0.4 \times 10^{-3}\;m$.
$Y = \frac{4 \times 1 \times 10 \times 1}{\pi \times (0.4 \times 10^{-3})^2 \times 0.4 \times 10^{-3}} = \frac{40}{\pi \times 0.064 \times 10^{-9}} \approx \frac{40}{3.14 \times 0.064 \times 10^{-9}} \approx 1.99 \times 10^{11}\;N/m^2$.
The relative error in $Y$ is given by $\frac{\Delta Y}{Y} = \frac{\Delta m}{m} + \frac{\Delta L}{L} + 2\frac{\Delta d}{d} + \frac{\Delta(\Delta L)}{\Delta L}$.
Assuming $\Delta m = 0$ and $\Delta L = 0$ (as length is exact),$\frac{\Delta Y}{Y} = 2\frac{\Delta d}{d} + \frac{\Delta(\Delta L)}{\Delta L} = 2 \times \frac{0.01}{0.4} + \frac{0.02}{0.4} = 0.05 + 0.05 = 0.1$.
$\Delta Y = 0.1 \times Y = 0.1 \times 1.99 \times 10^{11} = 1.99 \times 10^{10}\;N/m^2$.
Comparing with $x \times 10^{10}$,we get $x \approx 2$.

(D) Let $f_0$ be the actual frequency of the horn,$C$ be the speed of sound,and $V_s$ be the speed of the car.
When the car approaches the observer: $f_1 = f_0 \left( \frac{C}{C - V_s} \right) = 100 \, Hz$.
When the car moves away from the observer: $f_2 = f_0 \left( \frac{C}{C + V_s} \right) = 50 \, Hz$.
Dividing the two equations: $\frac{f_1}{f_2} = \frac{100}{50} = 2 = \frac{C + V_s}{C - V_s}$.
Solving for $V_s$: $2(C - V_s) = C + V_s \implies 2C - 2V_s = C + V_s \implies C = 3V_s \implies V_s = \frac{C}{3}$.
Substituting $V_s$ into the first equation: $100 = f_0 \left( \frac{C}{C - C/3} \right) = f_0 \left( \frac{C}{2C/3} \right) = f_0 \left( \frac{3}{2} \right)$.
Thus,$f_0 = 100 \times \frac{2}{3} = \frac{200}{3} \, Hz$.
If the observer moves with the car,the relative velocity between the source and the observer is zero,so the observed frequency is equal to the actual frequency $f_0$.
Therefore,$f = f_0 = \frac{200}{3} \, Hz$.
Comparing this with $\frac{x}{3} \, Hz$,we get $x = 200$.

(D) The argument of the $\sin$ function must be dimensionless,so $[\frac{\alpha x}{kt}] = [M^{0}L^{0}T^{0}]$.
Given $x = [L]$,$k = [ML^{2}T^{-2}\theta^{-1}]$,and $t = [\theta]$.
Thus,$[\alpha] = [\frac{kt}{x}] = \frac{[ML^{2}T^{-2}\theta^{-1}][\theta]}{[L]} = [MLT^{-2}]$.
Energy density $u$ is energy per unit volume,so $[u] = [ML^{-1}T^{-2}]$.
From the expression $u = \frac{\alpha}{\beta}$,we have $[\beta] = \frac{[\alpha]}{[u]}$.
$[\beta] = \frac{[MLT^{-2}]}{[ML^{-1}T^{-2}]} = [L^{2}]$.
Therefore,the dimensions of $\beta$ are $[M^{0}L^{2}T^{0}]$.

(D) The equation of trajectory is $y = x \tan \theta - \frac{gx^2}{2u^2 \cos^2 \theta}$.
Given $\theta = 45^{\circ}$,$x = 20$,$y = 10$,and $g = 10 \ m/s^2$.
$10 = 20(1) - \frac{10(20)^2}{2u^2(1/2)} \Rightarrow 10 = 20 - \frac{4000}{u^2} \Rightarrow \frac{4000}{u^2} = 10 \Rightarrow u^2 = 400 \Rightarrow u = 20 \ m/s$.
The time of flight $T = \frac{2u \sin \theta}{g} = \frac{2(20)(1/\sqrt{2})}{10} = 2\sqrt{2} \ s$.
At time $t = \frac{T}{\sqrt{2}} = \frac{2\sqrt{2}}{\sqrt{2}} = 2 \ s$.
The velocity components are $v_x = u \cos \theta = 20(1/\sqrt{2}) = 10\sqrt{2} \ m/s$ and $v_y = u \sin \theta - gt = 20(1/\sqrt{2}) - 10(2) = 10\sqrt{2} - 20 \ m/s$.
The momentum vector $\vec{p} = m\vec{v} = 10(10\sqrt{2} \hat{i} + (10\sqrt{2} - 20) \hat{j}) = 100\sqrt{2} \hat{i} + (100\sqrt{2} - 200) \hat{j} \ kg \cdot m/s$.

(A) Since the block slides down with constant velocity,the net force acting on it is zero.
The forces acting on the block are the gravitational force $(Mg)$,the normal force $(N)$,and the kinetic friction force $(f)$.
Resolving the gravitational force into components,we have:
$N = Mg \cos \theta$ (perpendicular to the plane)
$f = Mg \sin \theta$ (parallel to the plane)
The contact force $(R)$ is the resultant of the normal force $(N)$ and the friction force $(f)$:
$R = \sqrt{N^2 + f^2}$
Substituting the values of $N$ and $f$:
$R = \sqrt{(Mg \cos \theta)^2 + (Mg \sin \theta)^2}$
$R = \sqrt{M^2g^2(\cos^2 \theta + \sin^2 \theta)}$
Since $\sin^2 \theta + \cos^2 \theta = 1$,we get:
$R = \sqrt{M^2g^2(1)}$
$R = Mg$

(C) Since the lift is moving with uniform velocity,its acceleration is zero. Therefore,the effective acceleration due to gravity on the block remains '$g$'.
For a block sliding down a frictionless incline,the acceleration is $a = g \sin \theta$.
Using the equation of motion $s = ut + \frac{1}{2}at^2$,where $u = 0$ and $s = l$:
$l = \frac{1}{2} (g \sin 30^{\circ}) (2)^2 = \frac{1}{2} g (0.5) (4) = g$.
Now,for the incline of $45^{\circ}$,let the time taken be '$t$':
$l = \frac{1}{2} (g \sin 45^{\circ}) t^2$.
Since '$l$' is constant,we equate the two expressions:
$g = \frac{1}{2} g \left(\frac{1}{\sqrt{2}}\right) t^2$.
$1 = \frac{1}{2\sqrt{2}} t^2 \Rightarrow t^2 = 2\sqrt{2} \approx 2.828$.
$t = \sqrt{2.828} \approx 1.68\,s$.

(C) Let the initial velocity be $V$ and the constant retardation be $a$.
Using the equation of motion $v^2 = u^2 - 2as$:
For the first $4 \ cm$,the velocity becomes $V/3$:
$(V/3)^2 = V^2 - 2a(4) \Rightarrow V^2/9 = V^2 - 8a \Rightarrow 8a = 8V^2/9 \Rightarrow a = V^2/9$.
Now,for the bullet to stop completely,the final velocity is $0$ at a distance $s = 4+x$:
$0^2 = V^2 - 2a(4+x) \Rightarrow V^2 = 2(V^2/9)(4+x)$.
Dividing both sides by $V^2$:
$1 = (2/9)(4+x) \Rightarrow 9/2 = 4+x \Rightarrow 4.5 = 4+x \Rightarrow x = 0.5 \ cm$.

(B) According to the law of conservation of energy,the total energy at the surface of the earth is equal to the total energy at the maximum height $h$ from the center of the earth.
Total energy at surface = Total energy at height $h$
$-\frac{GMm}{R} + \frac{1}{2}m(\lambda v_{e})^{2} = -\frac{GMm}{h} + 0$
Since the escape velocity $v_{e} = \sqrt{\frac{2GM}{R}}$,we have $v_{e}^{2} = \frac{2GM}{R}$.
Substituting this into the equation:
$-\frac{GMm}{R} + \frac{1}{2}m\lambda^{2}(\frac{2GM}{R}) = -\frac{GMm}{h}$
$-\frac{GMm}{R} + \frac{\lambda^{2}GMm}{R} = -\frac{GMm}{h}$
Dividing both sides by $-GMm$:
$\frac{1}{R} - \frac{\lambda^{2}}{R} = \frac{1}{h}$
$\frac{1-\lambda^{2}}{R} = \frac{1}{h}$
$h = \frac{R}{1-\lambda^{2}}$

(D) The total elongation $\Delta \ell$ is the sum of the individual elongations of the steel wire $(\Delta \ell_{S})$ and the copper wire $(\Delta \ell_{C})$:
$\Delta \ell = \Delta \ell_{S} + \Delta \ell_{C}$
Using the formula for elongation $\Delta \ell = \frac{F \ell}{AY}$,we have:
$\Delta \ell = \frac{F \ell_{S}}{A Y_{S}} + \frac{F \ell_{C}}{A Y_{C}} = \frac{F}{A} \left( \frac{\ell_{S}}{Y_{S}} + \frac{\ell_{C}}{Y_{C}} \right)$
Given $r = 1.4 \, mm = 1.4 \times 10^{-3} \, m$,the cross-sectional area $A = \pi r^{2} = \frac{22}{7} \times (1.4 \times 10^{-3})^{2} = \frac{22}{7} \times 1.96 \times 10^{-6} = 6.16 \times 10^{-6} \, m^{2}$.
Substituting the values:
$1.4 \times 10^{-3} = \frac{F}{6.16 \times 10^{-6}} \left( \frac{3.2}{2.0 \times 10^{11}} + \frac{4.4}{1.1 \times 10^{11}} \right)$
$1.4 \times 10^{-3} = \frac{F}{6.16 \times 10^{-6}} \left( 1.6 \times 10^{-11} + 4.0 \times 10^{-11} \right)$
$1.4 \times 10^{-3} = \frac{F}{6.16 \times 10^{-6}} \times 5.6 \times 10^{-11}$
$F = \frac{1.4 \times 10^{-3} \times 6.16 \times 10^{-6}}{5.6 \times 10^{-11}} = \frac{8.624 \times 10^{-9}}{5.6 \times 10^{-11}} = 1.54 \times 10^{2} = 154 \, N$.

(A) First case: The efficiency of a Carnot engine is given by $\eta = 1 - \frac{T_L}{T_H}$.
For the first case,$\eta = 1 - \frac{100}{300} = 1 - \frac{1}{3} = \frac{2}{3}$.
Second case: Two engines are connected in series. The net efficiency $\eta_{\text{net}}$ of two engines in series is given by $\eta_{\text{net}} = \eta_1 + \eta_2 - \eta_1 \eta_2$.
For the first engine $(E_1)$ operating between $300 \, K$ and $200 \, K$,$\eta_1 = 1 - \frac{200}{300} = 1 - \frac{2}{3} = \frac{1}{3}$.
For the second engine $(E_2)$ operating between $200 \, K$ and $100 \, K$,$\eta_2 = 1 - \frac{100}{200} = 1 - \frac{1}{2} = \frac{1}{2}$.
Now,calculating the net efficiency: $\eta_{\text{net}} = \frac{1}{3} + \frac{1}{2} - (\frac{1}{3} \times \frac{1}{2}) = \frac{1}{3} + \frac{1}{2} - \frac{1}{6} = \frac{2+3-1}{6} = \frac{4}{6} = \frac{2}{3}$.
Thus,the efficiency in the $2^{\text{nd}}$ case is the same as in the $1^{\text{st}}$ case.

(B) Statement $A$ is incorrect because energy is stored in each degree of freedom independently,not as $n^2$.
Statement $B$ is correct according to the Law of Equipartition of Energy,which states that each degree of freedom contributes $\frac{1}{2} RT$ per mole to the internal energy.
Statement $C$ is incorrect because a monoatomic gas molecule has $0$ rotational degrees of freedom,while a diatomic molecule has $2$ rotational degrees of freedom.
Statement $D$ is correct. $CH_{4}$ is a non-linear polyatomic molecule. The total degrees of freedom $f$ for a non-linear molecule with $N$ atoms is $3N$. For $CH_{4}$,$N=5$,so $f = 3 \times 5 = 15$. However,if we consider rigid molecules at moderate temperatures,the degrees of freedom are $3$ (translational) $+ 3$ (rotational) $= 6$. Thus,$D$ is considered correct in the context of rigid body dynamics.
Therefore,$B$ and $D$ are the correct statements.

(B) The formula for Young's modulus $Y$ is given by $Y = \frac{F/A}{\Delta l/L} = \frac{F \cdot L}{A \cdot \Delta l}$.
Rearranging this,we get the ratio of extension to load as $\frac{\Delta l}{F} = \frac{L}{YA}$.
Here,the graph plots $\frac{\Delta l}{F}$ on the $y$-axis and $L$ on the $x$-axis.
The slope of this graph is $m = \frac{\Delta l/F}{L} = \frac{1}{YA}$.
From the graph,we can calculate the slope $m = \frac{0.25 \times 10^{-5}}{1} = 0.25 \times 10^{-5}\,m/N$.
Given cross-sectional area $A = 2\,mm^2 = 2 \times 10^{-6}\,m^2$.
Substituting these values into the slope equation: $0.25 \times 10^{-5} = \frac{1}{Y \times 2 \times 10^{-6}}$.
$Y = \frac{1}{0.25 \times 10^{-5} \times 2 \times 10^{-6}} = \frac{1}{0.5 \times 10^{-11}} = 2 \times 10^{11}\,N/m^2$.
Comparing this with $x \times 10^{11}\,N/m^2$,we get $x = 2$.

(D) The frequency of the $n^{\text{th}}$ harmonic is given by $f_n = \frac{nv}{2L} = 400\,Hz$.
The frequency of the $(n+1)^{\text{th}}$ harmonic is $f_{n+1} = \frac{(n+1)v}{2L} = 450\,Hz$.
The difference between consecutive harmonics is $\Delta f = f_{n+1} - f_n = \frac{v}{2L} = 450 - 400 = 50\,Hz$.
Given length $L = 0.3\,m$,we have $\frac{v}{2(0.3)} = 50$,which implies $v = 50 \times 0.6 = 30\,m/s$.
The wave speed $v$ is related to tension $T$ and linear mass density $\mu$ by $v = \sqrt{\frac{T}{\mu}}$.
Substituting the values: $30 = \sqrt{\frac{2700}{\mu}}$.
Squaring both sides: $900 = \frac{2700}{\mu}$.
Therefore,$\mu = \frac{2700}{900} = 3\,kg/m$.

(D) Let $P_0$ be the atmospheric pressure,$P_2$ be the pressure in the region between the two bubbles,and $P_1$ be the pressure inside the smaller bubble.
For the outer bubble of radius $R_2 = 6\,cm$,the excess pressure is $P_2 - P_0 = \frac{4T}{R_2} = \frac{4T}{6}$.
For the inner bubble of radius $R_1 = 3\,cm$,the excess pressure is $P_1 - P_2 = \frac{4T}{R_1} = \frac{4T}{3}$.
Adding these two equations,we get the total excess pressure inside the smaller bubble relative to the atmosphere:
$P_1 - P_0 = (P_1 - P_2) + (P_2 - P_0) = \frac{4T}{3} + \frac{4T}{6} = \frac{8T + 4T}{6} = \frac{12T}{6} = 2T$.
For a single soap bubble of radius $r$,the excess pressure is $P_{excess} = \frac{4T}{r}$.
Equating the two,we have $\frac{4T}{r} = 2T$,which gives $r = \frac{4}{2} = 2\,cm$.

(C) By the principle of conservation of energy,the loss in potential energy equals the gain in kinetic energy (translational + rotational).
$mgh = \frac{1}{2}mv^2 + \frac{1}{2}I\omega^2$
For a solid cylinder,the moment of inertia $I = \frac{1}{2}mR^2$. Since the string unwinds without slipping,the condition $v = R\omega$ holds,which implies $\omega = \frac{v}{R}$.
Substituting these into the energy equation:
$mgh = \frac{1}{2}mv^2 + \frac{1}{2}(\frac{1}{2}mR^2)(\frac{v}{R})^2$
$mgh = \frac{1}{2}mv^2 + \frac{1}{4}mv^2$
$mgh = \frac{3}{4}mv^2$
$gh = \frac{3}{4}v^2$
Given $g = 10\,ms^{-2}$ and $v = 4\,ms^{-1}$:
$10h = \frac{3}{4}(4)^2$
$10h = \frac{3}{4} \times 16$
$10h = 12$
$h = 1.2\,m = 120\,cm$.

(C) $1$. Motion from $A$ to $B$: The block is projected with velocity $u$ such that it just reaches $B$ $(v=0)$. By conservation of energy, $\frac{1}{2}mu^2 = mgh$. Thus, $u = \sqrt{2gh} = \sqrt{2 \times 10 \times 10} = 10\sqrt{2} \ m/s$.
$2$. The acceleration along the incline $AB$ is $a_1 = -g \sin 45^{\circ} = -\frac{10}{\sqrt{2}} \ m/s^2$.
$3$. Using $v = u + a_1t_1$, we get $0 = 10\sqrt{2} - \frac{10}{\sqrt{2}}t_1$, which gives $t_1 = 2 \ s$.
$4$. Motion from $B$ to $C$: The block starts from rest $(u=0)$ and slides down incline $BC$. The length of $BC$ is $L = \frac{h}{\sin 30^{\circ}} = \frac{10}{0.5} = 20 \ m$.
$5$. The acceleration along $BC$ is $a_2 = g \sin 30^{\circ} = 10 \times 0.5 = 5 \ m/s^2$.
$6$. Using $s = ut_2 + \frac{1}{2}a_2t_2^2$, we get $20 = 0 + \frac{1}{2}(5)t_2^2$, so $t_2^2 = 8$, which means $t_2 = 2\sqrt{2} \ s$.
$7$. Total time $T = t_1 + t_2 = 2 + 2\sqrt{2} = 2(\sqrt{2} + 1) \ s$.
$8$. Comparing with $t(\sqrt{2}+1)$, we find $t = 2$.

(B) The dimensional formula for torque $(\tau)$ is given by $[\tau] = [M^1 L^2 T^{-2}]$.
According to the theory of propagation of errors,if a physical quantity $X$ is given by $X = M^a L^b T^c$,the relative error is $\frac{\Delta X}{X} = a \frac{\Delta M}{M} + b \frac{\Delta L}{L} + c \frac{\Delta T}{T}$.
Given that the percentage error in mass,length,and time is $5 \%$ each,i.e.,$\frac{\Delta M}{M} \times 100 = 5 \%$,$\frac{\Delta L}{L} \times 100 = 5 \%$,and $\frac{\Delta T}{T} \times 100 = 5 \%$.
The percentage error in torque is calculated as:
$\frac{\Delta \tau}{\tau} \times 100 = (1 \times \% M) + (2 \times \% L) + (2 \times \% T)$
Substituting the given values:
$\frac{\Delta \tau}{\tau} \times 100 = 1(5 \%) + 2(5 \%) + 2(5 \%)$
$\frac{\Delta \tau}{\tau} \times 100 = 5 \% + 10 \% + 10 \% = 25 \%$.

(A) The bullet is moving downwards. Let the downward direction be negative. The initial velocity is $u = -100\,m/s$.
The acceleration due to gravity is $g = 10\,m/s^2$ downwards,so $a = -10\,m/s^2$.
The velocity at any time $t$ (for $0 \le t \le 10\,s$) is given by $v = u + at = -100 - 10t$.
At $t = 0\,s$,$v = -100\,m/s$.
At $t = 10\,s$,$v = -100 - 10(10) = -200\,m/s$.
After $t = 10\,s$,the bullet hits the ground and comes to rest,so $v = 0$ for $10\,s < t \le 20\,s$.
The graph starts at $-100\,m/s$,decreases linearly to $-200\,m/s$ at $t = 10\,s$,and then stays at $0$ from $t = 10\,s$ to $t = 20\,s$. This corresponds to the graph in option $A$.

(D) Given: Rate of mass flow $\frac{dm}{dt} = 0.5 \, kg s^{-1}$,Velocity of belt $v = 5 \, m s^{-1}$.
When sand falls on the belt,the belt must exert a force to accelerate the sand to the velocity of the belt.
The force required to maintain the constant velocity is equal to the thrust force: $F = \frac{dm}{dt} \times v$.
$F = 0.5 \, kg s^{-1} \times 5 \, m s^{-1} = 2.5 \, N$.
The power required to keep the belt moving at a constant velocity is $P = F \times v$.
$P = 2.5 \, N \times 5 \, m s^{-1} = 12.5 \, W$.

(B) When the bag is dropped on the belt,it experiences a kinetic frictional force $f_k = \mu N = \mu mg$.
This force provides an acceleration $a = f_k / m = \mu g$ to the bag in the direction of the belt's motion.
Given $\mu = 0.4$ and $g = 10\,m/s^2$,the acceleration is $a = 0.4 \times 10 = 4\,m/s^2$.
In the frame of the belt,the initial velocity of the bag is $u = 2\,m/s$ (relative to the belt) and the final velocity is $v = 0$ (when it stops slipping).
Using the equation of motion $v^2 = u^2 - 2as$ (where $s$ is the distance relative to the belt):
$0^2 = 2^2 - 2(4)s$
$0 = 4 - 8s$
$8s = 4$
$s = 0.5\,m$.

(B) The initial heights are $h_{1} = 1.0\,m$ and $h_{2} = 1.5\,m$. The cross-sectional area is $A = 16\,cm^{2} = 16 \times 10^{-4}\,m^{2}$.
When interconnected,the final height $h$ in both vessels will be the average of the initial heights: $h = \frac{h_{1} + h_{2}}{2} = \frac{1.0 + 1.5}{2} = 1.25\,m$.
The work done by gravity is equal to the decrease in potential energy: $W = U_{i} - U_{f}$.
Initial potential energy $U_{i} = (m_{1}g \frac{h_{1}}{2}) + (m_{2}g \frac{h_{2}}{2}) = \rho A h_{1} g \frac{h_{1}}{2} + \rho A h_{2} g \frac{h_{2}}{2} = \frac{\rho A g}{2} (h_{1}^{2} + h_{2}^{2})$.
Final potential energy $U_{f} = (m_{1}+m_{2})g \frac{h}{2} = (2 \rho A h) g \frac{h}{2} = \rho A g h^{2} = \rho A g (\frac{h_{1}+h_{2}}{2})^{2}$.
$W = \frac{\rho A g}{2} [h_{1}^{2} + h_{2}^{2} - 2(\frac{h_{1}+h_{2}}{2})^{2}] = \frac{\rho A g}{4} [2h_{1}^{2} + 2h_{2}^{2} - (h_{1}+h_{2})^{2}] = \frac{\rho A g}{4} (h_{1}-h_{2})^{2}$.
Substituting the values: $W = \frac{10^{3} \times 16 \times 10^{-4} \times 10}{4} (1.5 - 1.0)^{2} = \frac{16}{4} (0.5)^{2} = 4 \times 0.25 = 1\,J$.

(B) The total mechanical energy $E$ of a satellite of mass $m$ orbiting at a distance $r$ from the center of the Earth is given by $E = -\frac{GMm}{2r}$.
From this expression,we can see that $E \propto \frac{m}{r}$.
Given the mass ratio $\frac{m_A}{m_B} = \frac{4}{3}$ and the radius ratio $\frac{r_A}{r_B} = \frac{3r}{4r} = \frac{3}{4}$.
Therefore,the ratio of the total mechanical energy of $A$ to $B$ is $\frac{E_A}{E_B} = \frac{m_A}{m_B} \times \frac{r_B}{r_A}$.
Substituting the given values: $\frac{E_A}{E_B} = \frac{4}{3} \times \frac{4}{3} = \frac{16}{9}$.

(C) In the steady state,the rate of heat flow through the steel rod must be equal to the rate of heat flow through the copper rod.
$\frac{dQ}{dt} = \frac{K_{1} A_{1} (T_{1} - T)}{L_{1}} = \frac{K_{2} A_{2} (T - T_{2})}{L_{2}}$
Given: $T_{1} = 450^{\circ}C$,$T_{2} = 0^{\circ}C$,$\frac{K_{2}}{K_{1}} = 9$,$\frac{A_{1}}{A_{2}} = 2$,$\frac{L_{1}}{L_{2}} = 2$.
Substituting these values into the equation:
$\frac{450 - T}{T - 0} = \frac{K_{2}}{K_{1}} \times \frac{A_{2}}{A_{1}} \times \frac{L_{1}}{L_{2}}$
$\frac{450 - T}{T} = 9 \times \frac{1}{2} \times 2 = 9$
$450 - T = 9T$
$10T = 450$
$T = 45^{\circ}C$

(A) $A.$ According to Newton's law of cooling,the rate of loss of heat $\frac{dQ}{dt} \propto \Delta T$. If $\Delta T$ is doubled,the rate of loss of heat becomes twice. Thus,$A$ is correct.
$B.$ The power radiated by a body is $P = \sigma e A T^4$. The ratio of radiation emitted is $\frac{H_P}{H_Q} = \left(\frac{T_P}{T_Q}\right)^4$. Converting temperatures to Kelvin: $T_P = 10 + 273 = 283 K$ and $T_Q = 20 + 273 = 293 K$. Thus,$\frac{H_P}{H_Q} = \left(\frac{283}{293}\right)^4 \approx (0.9658)^4 \approx 0.87$. However,using the approximation $\frac{293}{283} \approx 1.035$,then $(1.035)^4 \approx 1.15$. Thus,$1:1.15$ is correct. $B$ is correct.
$C.$ Efficiency $\eta = 1 - \frac{T_L}{T_H} = 1 - \frac{100}{400} = 1 - 0.25 = 0.75$ or $75\%$. Thus,$C$ is correct.
$D.$ Since $\frac{dQ}{dt} \propto \Delta T$,if $\Delta T$ is quadrupled,the rate of loss of heat becomes four times,not twice. Thus,$D$ is incorrect.
Therefore,statements $A, B,$ and $C$ are correct.

(C) According to the Kinetic Theory of Gases $(KTG)$:
$1$. The $r.m.s.$ velocity of gas molecules is given by $V_{rms} = \sqrt{\frac{3RT}{M_m}}$. Since the temperature $T$ and the molar mass $M_m$ are the same for both vessels,the $r.m.s.$ velocity is the same. Thus,statement $A$ is correct.
$2$. From the ideal gas equation $PV = NkT$,where $N$ is the number of molecules and $k$ is the Boltzmann constant,we have $P = \frac{NkT}{V}$. Since $V$,$T$,and $k$ are constant,$P \propto N$. Therefore,the ratio of pressures is $\frac{P_1}{P_2} = \frac{N_1}{N_2} = 1:4$. Thus,statement $B$ is correct.
Since statements $A$ and $B$ are correct,the correct option is $C$.

(D) Given: $1\,MSD = 1\,mm = 0.1\,cm$.
$10\,VSD = 9\,MSD$,so $1\,VSD = 0.9\,MSD = 0.9\,mm = 0.09\,cm$.
Least Count $(L.C.) = 1\,MSD - 1\,VSD = 1\,mm - 0.9\,mm = 0.1\,mm = 0.01\,cm$.
Positive Zero Error $= + (4 \times L.C.) = + (4 \times 0.01\,cm) = +0.04\,cm$.
Observed Reading $= \text{Main Scale Reading} + (\text{Vernier Coincidence} \times L.C.) = 4.1\,cm + (6 \times 0.01\,cm) = 4.16\,cm$.
Correct Diameter $= \text{Observed Reading} - \text{Zero Error} = 4.16\,cm - 0.04\,cm = 4.12\,cm$.
$4.12\,cm = 412 \times 10^{-2\,cm}$.

(C) The total energy of the $SHM$ is given by $E = \frac{1}{2} k A^2$. At the mean position,the potential energy is zero,so the total energy is purely kinetic: $E = \frac{p^2}{2m}$,where $p$ is the momentum.
Since the mass is added at the mean position,the velocity $v$ changes,but the momentum $p$ is conserved during the instantaneous placement of the mass.
Initial mass $m_1 = 0.9 \, kg = 900 \, g$. Final mass $m_2 = 900 \, g + 124 \, g = 1024 \, g$.
Since $E = \frac{1}{2} k A^2 = \frac{p^2}{2m}$,we have $A \propto \frac{1}{\sqrt{m}}$ for a constant momentum $p$.
Therefore,$\frac{A_1}{A_2} = \sqrt{\frac{m_2}{m_1}} = \sqrt{\frac{1024}{900}} = \frac{32}{30} = \frac{16}{15}$.
Given $\frac{A_1}{A_2} = \frac{\alpha}{\alpha-1} = \frac{16}{16-1}$.
Comparing the terms,we get $\alpha = 16$.

(C) The shear modulus $\eta$ is given by the formula: $\eta = \frac{F/A}{x/L}$,where $F$ is the shearing force,$A$ is the area of the face on which the force is applied,$x$ is the displacement,and $L$ is the side length of the slab.
Rearranging for displacement $x$: $x = \frac{F \cdot L}{A \cdot \eta}$.
Given: $F = 18.0 \times 10^{4} \, N$,$L = 60 \, cm = 0.6 \, m$,thickness $t = 15 \, cm = 0.15 \, m$,and $\eta = 25 \times 10^{9} \, N m^{-2}$.
The area $A$ of the narrow face is $L \times t = 0.6 \, m \times 0.15 \, m = 0.09 \, m^2$.
Substituting the values: $x = \frac{18.0 \times 10^{4} \times 0.6}{0.09 \times 25 \times 10^{9}}$.
$x = \frac{10.8 \times 10^{4}}{2.25 \times 10^{9}} = 4.8 \times 10^{-5} \, m = 48 \times 10^{-6} \, m = 48 \, \mu m$.

(A) The torque applied is $\tau = F \cdot r = (12t - 3t^2) \cdot 1.5 = 18t - 4.5t^2$.
Using $\tau = I\alpha$,we have $18t - 4.5t^2 = 4.5\alpha$,which gives $\alpha = 4t - t^2$.
Since $\alpha = \frac{d\omega}{dt}$,we integrate to find angular velocity: $\omega = \int (4t - t^2) dt = 2t^2 - \frac{t^3}{3}$.
The motion reverses when $\omega = 0$,so $t^2(2 - \frac{t}{3}) = 0$,giving $t = 6\,s$.
The angular displacement $\theta$ is $\int_{0}^{6} (2t^2 - \frac{t^3}{3}) dt = [\frac{2t^3}{3} - \frac{t^4}{12}]_{0}^{6} = \frac{2(216)}{3} - \frac{1296}{12} = 144 - 108 = 36\,rad$.
The number of rotations is $N = \frac{\theta}{2\pi} = \frac{36}{2\pi} = \frac{18}{\pi}$.
Comparing with $\frac{K}{\pi}$,we get $K = 18$.

(A) Let the initial velocity of the first ball be $u_1$ and the second ball be $u_2$.
For the first ball thrown vertically upward,the time of flight is $T_1 = \frac{2u_1}{g}$.
For the second ball thrown at an angle $\theta$ with the vertical,the vertical component of velocity is $u_{2y} = u_2 \cos \theta$. The time of flight is $T_2 = \frac{2u_2 \cos \theta}{g}$.
Given $T_1 = T_2$,we have $u_1 = u_2 \cos \theta$.
The maximum height attained by the first ball is $H_1 = \frac{u_1^2}{2g}$.
The maximum height attained by the second ball is $H_2 = \frac{(u_2 \cos \theta)^2}{2g} = \frac{u_2^2 \cos^2 \theta}{2g}$.
Since $u_1 = u_2 \cos \theta$,it follows that $H_1 = H_2$.
The ratio of the heights is $\frac{H_1}{H_2} = 1 = \frac{1}{x}$.
Therefore,$x = 1$.

(D) Efficiency $\eta$ is a dimensionless quantity. Therefore,$[\eta] = [M^0 L^0 T^0]$.
In the expression $\eta = \frac{\alpha \beta}{\sin \theta} \log_{e} \frac{\beta x}{k T}$,the argument of the logarithm must be dimensionless. Thus,$[\frac{\beta x}{k T}] = [M^0 L^0 T^0]$.
Since $[k T] = \text{Energy} = [M L^2 T^{-2}]$ and $[x] = [L]$,we have $[\beta] = \frac{[M L^2 T^{-2}]}{[L]} = [M L T^{-2}]$,which is the dimension of force.
Now,since $\eta$ is dimensionless and $\sin \theta$ is dimensionless,the term $\frac{\alpha \beta}{\sin \theta}$ must be dimensionless. Thus,$[\alpha \beta] = [M^0 L^0 T^0]$.
Since $[\beta] = [M L T^{-2}]$,then $[\alpha] = [M^{-1} L^{-1} T^2]$.
Checking the options:
$A$: $[\beta] = [M L T^{-2}]$ (Force),which is correct.
$B$: $[\alpha^{-1} x] = [M L T^{-2} \cdot L] = [M L^2 T^{-2}]$ (Energy),which is correct.
$C$: $[\eta^{-1} \sin \theta] = [1] = [\alpha \beta]$,which is correct.
$D$: $[\alpha] = [M^{-1} L^{-1} T^2]$ and $[\beta] = [M L T^{-2}]$. These are not the same. Thus,this is the incorrect option.

(B) The position vector of the particle at any time $t$ is given by $\vec{r}(t) = x(t) \hat{i} + y(t) \hat{j} + z(t) \hat{k}$.
Given $x = 3t$,$y = 5t^3$,and $z = 7$ (constant).
So,$\vec{r}(t) = 3t \hat{i} + 5t^3 \hat{j} + 7 \hat{k}$.
The velocity vector $\vec{v}$ is the first derivative of position with respect to time: $\vec{v} = \frac{d\vec{r}}{dt} = \frac{d}{dt}(3t) \hat{i} + \frac{d}{dt}(5t^3) \hat{j} + \frac{d}{dt}(7) \hat{k} = 3 \hat{i} + 15t^2 \hat{j}$.
The acceleration vector $\vec{a}$ is the derivative of velocity with respect to time: $\vec{a} = \frac{d\vec{v}}{dt} = \frac{d}{dt}(3) \hat{i} + \frac{d}{dt}(15t^2) \hat{j} = 0 \hat{i} + 30t \hat{j}$.
At $t = 1 \, \text{s}$,the acceleration is $\vec{a} = 30(1) \hat{j} = 30 \hat{j} \, \text{cm/s}^2$.

(B) The magnetic field at the centre of a circular coil with $N$ turns and radius $R$ carrying current $i$ is given by $B = \frac{N \mu_{0} i}{2 R}$.
For the first coil: $B_{1} = \frac{N_{1} \mu_{0} i}{2 R_{1}}$,where $N_{1} = 2$.
When the wire is unwound and rewound,the total length of the wire $L = N_{1} (2 \pi R_{1}) = N_{2} (2 \pi R_{2})$ remains constant.
Thus,$R_{2} = R_{1} \frac{N_{1}}{N_{2}} = R_{1} \frac{2}{5}$.
For the second coil: $B_{2} = \frac{N_{2} \mu_{0} i}{2 R_{2}}$,where $N_{2} = 5$.
Taking the ratio: $\frac{B_{2}}{B_{1}} = \frac{N_{2}}{N_{1}} \times \frac{R_{1}}{R_{2}} = \frac{N_{2}}{N_{1}} \times \frac{N_{2}}{N_{1}} = \left( \frac{N_{2}}{N_{1}} \right)^{2}$.
Substituting the values: $\frac{B_{2}}{B_{1}} = \left( \frac{5}{2} \right)^{2} = \frac{25}{4}$.

(A) The capacitance of an isolated conducting sphere of radius $R_{1}$ is given by $C_{1} = 4 \pi \varepsilon_{0} R_{1}$.
When this sphere is enclosed by a concentric conducting sphere of radius $R_{2}$ which is connected to the earth,the system forms a spherical capacitor. The new capacitance is $C_{2} = \frac{4 \pi \varepsilon_{0} R_{1} R_{2}}{R_{2} - R_{1}}$.
According to the problem,the new capacitance is $n$ times the original capacitance: $C_{2} = n C_{1}$.
Substituting the expressions: $\frac{4 \pi \varepsilon_{0} R_{1} R_{2}}{R_{2} - R_{1}} = n (4 \pi \varepsilon_{0} R_{1})$.
Simplifying the equation: $\frac{R_{2}}{R_{2} - R_{1}} = n$.
Dividing the numerator and denominator of the left side by $R_{1}$: $\frac{R_{2}/R_{1}}{(R_{2}/R_{1}) - 1} = n$.
Let $x = \frac{R_{2}}{R_{1}}$. Then $\frac{x}{x - 1} = n$.
$x = n(x - 1) \Rightarrow x = nx - n \Rightarrow n = nx - x \Rightarrow n = x(n - 1)$.
Therefore,$x = \frac{R_{2}}{R_{1}} = \frac{n}{n - 1}$.

(D) The de Broglie wavelength $\lambda$ of a charged particle accelerated through a potential $V$ is given by $\lambda = \frac{h}{\sqrt{2mqV}}$.
Given the ratio of wavelengths $\frac{\lambda_{p}}{\lambda_{d}} = \frac{1}{\sqrt{2}}$.
Using the formula,we have $\frac{\lambda_{p}}{\lambda_{d}} = \sqrt{\frac{m_{d} q_{d} V_{d}}{m_{p} q_{p} V_{p}}}$.
For a proton,$m_{p} = m$ and $q_{p} = e$. For a deuteron,$m_{d} = 2m$ and $q_{d} = e$.
Substituting these values: $\frac{1}{\sqrt{2}} = \sqrt{\frac{(2m)(e)V_{d}}{(m)(e)V_{p}}} = \sqrt{\frac{2V_{d}}{V_{p}}}$.
Squaring both sides: $\frac{1}{2} = \frac{2V_{d}}{V_{p}}$.
Therefore,$\frac{V_{p}}{V_{d}} = 4$,which is $4 : 1$.

(B) $1$. Applying the lens formula for the initial state:
$\frac{1}{v} - \frac{1}{u} = \frac{1}{f}$
Given $v = 12\,cm = 0.12\,m$ and $u = -2.4\,m$:
$\frac{1}{0.12} - \frac{1}{-2.4} = \frac{1}{f} \Rightarrow \frac{100}{12} + \frac{1}{2.4} = \frac{1}{f} \Rightarrow \frac{25}{3} + \frac{1}{2.4} = \frac{20+1}{2.4} = \frac{21}{2.4} = \frac{1}{f} \Rightarrow f = \frac{2.4}{21} = \frac{0.8}{7}\,m$.
$2$. When a glass slab of thickness $t$ and refractive index $\mu$ is introduced,the image shifts by $\Delta x = t(1 - \frac{1}{\mu})$.
$\Delta x = 1\,cm \times (1 - \frac{1}{1.5}) = 1\,cm \times (1 - \frac{2}{3}) = \frac{1}{3}\,cm$.
$3$. The new image position $v'$ required to be on the screen is $v' = 12\,cm - \frac{1}{3}\,cm = \frac{35}{3}\,cm = \frac{35}{300}\,m = \frac{7}{60}\,m$.
$4$. Applying the lens formula again for the new object distance $u'$:
$\frac{1}{v'} - \frac{1}{u'} = \frac{1}{f}$
$\frac{60}{7} - \frac{1}{u'} = \frac{21}{2.4} = \frac{210}{24} = \frac{35}{4}$
$\frac{1}{u'} = \frac{60}{7} - \frac{35}{4} = \frac{240 - 245}{28} = -\frac{5}{28}$
$u' = -\frac{28}{5} = -5.6\,m$.
$5$. The shift in the object position is $|u' - u| = |-5.6 - (-2.4)| = |-3.2| = 3.2\,m$.

(A) The given equation of the electric field is $E_{y} = 540 \sin \pi \times 10^{4}(x - ct) \text{ Vm}^{-1}$.
Comparing this with the standard wave equation $E = E_{0} \sin(kx - \omega t)$,we identify the peak value of the electric field as $E_{0} = 540 \text{ Vm}^{-1}$.
The relationship between the peak electric field $E_{0}$ and the peak magnetic field $B_{0}$ is given by $B_{0} = \frac{E_{0}}{c}$.
Substituting the given values,$B_{0} = \frac{540}{3 \times 10^{8}} \text{ T}$.
$B_{0} = 180 \times 10^{-8} \text{ T} = 18 \times 10^{-7} \text{ T}$.
Thus,the peak value of the magnetic field is $18 \times 10^{-7} \text{ T}$.

(B) metal detector consists of an $LC$ circuit that is tuned to a specific resonant frequency. When a metal object enters the magnetic field produced by the coil,it changes the inductance of the circuit,thereby shifting the resonant frequency. This change is detected by the circuit,which triggers an alarm. Thus,the device operates on the principle of resonance in $AC$ circuits.

(C) The frequency of revolution $f$ of a charged particle moving in a magnetic field is given by the formula:
$f = \frac{eB}{2 \pi m}$
Given values:
Charge of electron $e = 1.6 \times 10^{-19} \, C$
Magnetic field $B = 1 \times 10^{-4} \, Wb/m^2$
Mass of electron $m = 9.0 \times 10^{-31} \, kg$
Substituting these values into the formula:
$f = \frac{1.6 \times 10^{-19} \times 10^{-4}}{2 \times 3.14159 \times 9.0 \times 10^{-31}}$
$f = \frac{1.6 \times 10^{-23}}{56.548 \times 10^{-31}}$
$f \approx 0.2829 \times 10^8 \, Hz = 2.8 \times 10^6 \, Hz$
Thus,the frequency of revolution is $2.8 \times 10^6 \, Hz$.

(D) The circuit consists of a $5\,k\Omega$ resistor in series with a parallel combination of a $10\,k\Omega$ resistor and a $5\,k\Omega$ resistor,followed by another $10\,k\Omega$ resistor in series.
First,calculate the equivalent resistance of the parallel part $(R_p)$:
$R_p = \frac{10\,k\Omega \times 5\,k\Omega}{10\,k\Omega + 5\,k\Omega} = \frac{50}{15}\,k\Omega = \frac{10}{3}\,k\Omega$.
The total current $I = 15\,mA$ flows through the $5\,k\Omega$ resistor,then splits into the parallel branches,and finally flows through the $10\,k\Omega$ resistor.
The potential difference $V_{AB}$ is the sum of the potential drops across each component in the path from $A$ to $B$:
$V_{AB} = I \times R_{5k} + I \times R_p + I \times R_{10k}$
$V_{AB} = (15\,mA \times 5\,k\Omega) + (15\,mA \times \frac{10}{3}\,k\Omega) + (15\,mA \times 10\,k\Omega)$
$V_{AB} = 75\,V + 50\,V + 150\,V = 275\,V$.

(B) The energy of an electron in the $n$-th orbit of a hydrogen atom is given by $E_n = -\frac{Rch}{n^2}$.
The energy of the emitted photon is the difference between the energy of the excited state $(E_n)$ and the ground state $(E_1)$:
$E_{\text{photon}} = E_n - E_1 = \frac{hc}{\lambda}$
Substituting the energy values:
$-\frac{Rch}{n^2} - (-\frac{Rch}{1^2}) = \frac{hc}{\lambda}$
Dividing both sides by $hc$:
$-\frac{R}{n^2} + R = \frac{1}{\lambda}$
Rearranging the terms:
$R - \frac{1}{\lambda} = \frac{R}{n^2}$
$\frac{\lambda R - 1}{\lambda} = \frac{R}{n^2}$
$n^2 = \frac{\lambda R}{\lambda R - 1}$
Therefore,$n = \sqrt{\frac{\lambda R}{\lambda R - 1}}$.

(B) The energy of a photon emitted during a transition from level $n_2$ to $n_1$ is given by $E = 13.6 \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right) \text{ eV}$.
For transition $(i)$ from $n=3$ to $n=2$: $E_1 = 13.6 \left( \frac{1}{2^2} - \frac{1}{3^2} \right) = 13.6 \left( \frac{1}{4} - \frac{1}{9} \right) = 13.6 \left( \frac{5}{36} \right)$.
For transition $(ii)$ from $n=\infty$ to $n=2$: $E_2 = 13.6 \left( \frac{1}{2^2} - \frac{1}{\infty^2} \right) = 13.6 \left( \frac{1}{4} - 0 \right) = 13.6 \left( \frac{1}{4} \right)$.
The ratio is $\frac{E_1}{E_2} = \frac{13.6(5/36)}{13.6(1/4)} = \frac{5/36}{1/4} = \frac{5}{9}$.
Given the ratio is $\frac{x}{x+4}$,we have $\frac{x}{x+4} = \frac{5}{9}$.
Cross-multiplying gives $9x = 5x + 20$,which simplifies to $4x = 20$,so $x = 5$.

(B) Let the potential gradient of the potentiometer wire be $k$.
For the first cell,the balance point is at $l_1 = 36\, cm$ for emf $E_1 = 1.20\, V$.
$E_1 = k \cdot l_1 \implies 1.20 = k \times 36 \implies k = \frac{1.20}{36} = \frac{1}{30}\, V/cm$.
For the second cell,the balance point is at $l_2$ for emf $E_2 = 1.80\, V$.
$E_2 = k \cdot l_2 \implies 1.80 = \frac{1}{30} \times l_2$.
$l_2 = 1.80 \times 30 = 54\, cm$.
The difference in balancing lengths is $\Delta l = l_2 - l_1 = 54 - 36 = 18\, cm$.

(D) When a potential difference is applied between $A$ and $B$ such that $A$ is at a higher potential $(+)$ and $B$ is at a lower potential $(-)$,the upper diode becomes forward-biased and acts as a short circuit (zero resistance).
The lower diode becomes reverse-biased and acts as an open circuit (infinite resistance).
In this state,the circuit simplifies to a series combination of the $20\,\Omega$ resistor and the $15\,\Omega$ resistor.
Therefore,the equivalent resistance $R_{eq} = 20\,\Omega + 15\,\Omega = 35\,\Omega$.

(C) $1$. Initial state: The capacitors $C$ and $3C$ are in parallel and connected to an $18V$ battery. The total charge $Q_{total}$ stored in the system is:
$Q_{total} = (C + 3C) \times 18V = 4C \times 18V = 72CV$.
$2$. After disconnecting the battery: The total charge $Q_{total} = 72CV$ remains constant because the system is isolated.
$3$. Dielectric insertion: $A$ dielectric of constant $K=9$ is inserted into the capacitor of capacity $C$. Its new capacitance becomes $C^{\prime} = K \times C = 9C$. The other capacitor remains $3C$.
$4$. Final state: The capacitors are still in parallel. Let the new common potential difference be $V^{\prime}$. The total capacitance of the combination is $C_{eq} = C^{\prime} + 3C = 9C + 3C = 12C$.
$5$. Conservation of charge: $Q_{total} = C_{eq} \times V^{\prime}$
$72CV = 12C \times V^{\prime}$
$V^{\prime} = \frac{72CV}{12C} = 6V$.

(D) For the lens,using the lens formula: $\frac{1}{v} - \frac{1}{u} = \frac{1}{f}$.
Given $u = -60\,cm$ and $f = +20\,cm$,we have $\frac{1}{v} - \frac{1}{-60} = \frac{1}{20}$.
$\frac{1}{v} + \frac{1}{60} = \frac{1}{20} \Rightarrow \frac{1}{v} = \frac{1}{20} - \frac{1}{60} = \frac{3-1}{60} = \frac{2}{60} = \frac{1}{30}$.
So,$v = 30\,cm$. This image is formed $30\,cm$ behind the lens.
For the final image to coincide with the object,the rays must strike the mirror normally (perpendicularly). This happens if the rays are directed towards the center of curvature $(C)$ of the mirror.
The distance of the image from the mirror is $30\,cm - 10\,cm = 20\,cm$. Thus,the radius of curvature $R = 20\,cm$.
The focal length of the mirror is $f_m = \frac{R}{2} = \frac{20}{2} = 10\,cm$.

(D) Given magnetic flux $\phi = 8t^2 - 9t + 5$.
According to Faraday's law of induction,the induced $emf$ is given by $\varepsilon = -\frac{d\phi}{dt}$.
$\varepsilon = -\frac{d}{dt}(8t^2 - 9t + 5) = -(16t - 9) = 9 - 16t$.
At $t = 0.25\,s$,the induced $emf$ is $\varepsilon = 9 - 16(0.25) = 9 - 4 = 5\,V$.
The magnitude of the induced current $I$ is given by $I = \frac{|\varepsilon|}{R}$.
Given resistance $R = 20\,\Omega$,so $I = \frac{5\,V}{20\,\Omega} = 0.25\,A$.
Converting to milliamperes $(mA)$,$I = 0.25 \times 1000\,mA = 250\,mA$.

(A) The electric displacement vector is defined as $\overrightarrow{D} = \epsilon_{0} \overrightarrow{E}$.
Dimensional analysis of electric displacement: $[D] = [\epsilon_{0}][E]$.
Since the electric field $E$ is related to surface charge density $\sigma$ by $E = \frac{\sigma}{\epsilon_{0}}$,we have $\epsilon_{0} E = \sigma$.
Therefore,the dimensions of electric displacement $[D]$ are equal to the dimensions of surface charge density $[\sigma]$.
Both have dimensions of $[Q L^{-2}]$,where $Q$ is charge and $L$ is length.

(A) The relationship between charge $(Q)$,capacitance $(C)$,and potential difference $(V)$ for a capacitor is given by the formula: $Q = CV$.
Rearranging this for potential difference,we get: $V = \frac{1}{C} Q$.
This equation represents a straight line passing through the origin,where the slope of the line is $\frac{1}{C}$.
Given $C = 2\,\mu F = 2 \times 10^{-6}\,F$ and the maximum charge $Q = 5\,C$,the maximum potential difference is:
$V = \frac{Q}{C} = \frac{5}{2 \times 10^{-6}} = 2.5 \times 10^{6}\,V$.
Thus,the graph should be a straight line starting from $(0,0)$ and ending at $(5, 2.5 \times 10^{6})$.
Comparing this with the given options,graph $A$ correctly represents this linear relationship.

(B) The radius $R$ of a circular path for a charged particle moving in a magnetic field $B$ is given by $R = \frac{mv}{qB}$.
Since kinetic energy $K = \frac{1}{2}mv^2$,we have $mv = \sqrt{2mK}$.
Substituting this into the radius formula,we get $R = \frac{\sqrt{2mK}}{qB}$.
Rearranging for charge $q$,we get $q = \frac{\sqrt{2mK}}{RB}$.
Given that $K$ and $B$ are the same for both particles,the ratio of charges is $\frac{q_1}{q_2} = \sqrt{\frac{m_1}{m_2}} \times \frac{R_2}{R_1}$.
Given $\frac{R_1}{R_2} = \frac{6}{5}$ and $\frac{m_1}{m_2} = \frac{9}{4}$,we have $\frac{q_1}{q_2} = \sqrt{\frac{9}{4}} \times \frac{5}{6} = \frac{3}{2} \times \frac{5}{6} = \frac{15}{12} = \frac{5}{4}$.

(C) The resonant frequency $f_r$ of a series $LCR$ circuit is given by the formula:
$f_r = \frac{1}{2 \pi \sqrt{LC}}$
To increase the resonant frequency $f_r$,the product $LC$ must decrease.
When a capacitor is added in series with the existing capacitor,the equivalent capacitance $C_{eq}$ is given by $\frac{1}{C_{eq}} = \frac{1}{C_1} + \frac{1}{C_2}$. This results in a decrease in the total capacitance $(C_{eq} < C_1)$.
Since $f_r \propto \frac{1}{\sqrt{C}}$,a decrease in total capacitance leads to an increase in the resonant frequency.

(C) Let a current $I$ flow through the outer square loop of side $L$. The magnetic field $B$ produced by this loop at its centre $O$ is given by the sum of the fields due to its four sides:
$B = 4 \times \left( \frac{\mu_{0} I}{4 \pi (L/2)} \times (\sin 45^{\circ} + \sin 45^{\circ}) \right) = 4 \times \left( \frac{\mu_{0} I}{2 \pi L} \times 2 \times \frac{1}{\sqrt{2}} \right) = \frac{2 \sqrt{2} \mu_{0} I}{\pi L}$.
Since $L \gg l$,we can assume this magnetic field is uniform over the area of the small inner loop.
The magnetic flux $\phi$ linked with the inner loop of side $l$ is:
$\phi = B \times \text{Area} = \left( \frac{2 \sqrt{2} \mu_{0} I}{\pi L} \right) \times l^{2}$.
The mutual inductance $M$ is defined as $M = \frac{\phi}{I}$.
Therefore,$M = \frac{2 \sqrt{2} \mu_{0} l^{2}}{\pi L}$.

(B) The current in a capacitor is given by the formula $I = \frac{V}{X_C}$,where $X_C = \frac{1}{\omega C}$ is the capacitive reactance.
Substituting $X_C$,we get $I = V \omega C$.
Rearranging for capacitance $C$,we have $C = \frac{I}{V \omega}$.
Given values are $I = 6.9 \times 10^{-6}\,A$,$V = 230\,V$,and $\omega = 600\,rad/s$.
Substituting these values: $C = \frac{6.9 \times 10^{-6}}{230 \times 600}$.
$C = \frac{6.9 \times 10^{-6}}{138000} = 0.05 \times 10^{-9}\,F$.
$C = 50 \times 10^{-12}\,F = 50\,pF$.

(A) In a primary rainbow,the red colour is at the top and the violet colour is at the bottom because the deviation for red is minimum and for violet is maximum.
In a primary rainbow,light undergoes one total internal reflection,whereas in a secondary rainbow,it undergoes two total internal reflections.
Due to the two internal reflections in a secondary rainbow,more light is lost,making the secondary rainbow less bright than the primary rainbow. Therefore,the primary rainbow is brighter than the secondary rainbow.

(A) The refractive index is given by $\mu = c/v$,where $c$ is the speed of light in vacuum and $v$ is the speed in the medium.
Given $\mu_{A}/\mu_{B} = 1/2$,we have $(c/v_{A}) / (c/v_{B}) = v_{B}/v_{A} = 1/2$,which implies $v_{A} = 2v_{B}$.
Let the thickness of both materials be $d$.
The time taken to travel through the material is $t = d/v$.
Given $t_{2} - t_{1} = 5 \times 10^{-10} \text{ s}$,where $t_{1} = d/v_{A}$ and $t_{2} = d/v_{B}$.
Substituting the values: $d/v_{B} - d/v_{A} = 5 \times 10^{-10}$.
$d(1/v_{B} - 1/v_{A}) = 5 \times 10^{-10}$.
$d((v_{A} - v_{B}) / (v_{A}v_{B})) = 5 \times 10^{-10}$.
Since $v_{A} = 2v_{B}$,we have $d((2v_{B} - v_{B}) / (2v_{B} \cdot v_{B})) = 5 \times 10^{-10}$.
$d(v_{B} / 2v_{B}^{2}) = 5 \times 10^{-10}$.
$d / (2v_{B}) = 5 \times 10^{-10}$.
$d = 10 \times 10^{-10} v_{B} = 10^{-9} v_{B} \text{ m}$.
Alternatively,using $v_{B} = v_{A}/2$,we get $d / (2(v_{A}/2)) = 5 \times 10^{-10} \Rightarrow d/v_{A} = 5 \times 10^{-10} \Rightarrow d = 5 \times 10^{-10} v_{A} \text{ m}$.

(C) According to Einstein's photoelectric equation,the maximum kinetic energy $K_{max}$ is given by $K_{max} = \frac{hc}{\lambda} - \phi$,where $\phi$ is the work function.
For $\lambda_1 = 800 \, nm$,$K_1 = \frac{1230}{800} - \phi = 1.5375 - \phi$ --- $(1)$
For $\lambda_2 = 500 \, nm$,$K_2 = \frac{1230}{500} - \phi = 2.46 - \phi$ --- $(2)$
Given that $K_2 = 2K_1$,we substitute the expressions:
$2.46 - \phi = 2(1.5375 - \phi)$
$2.46 - \phi = 3.075 - 2\phi$
$2\phi - \phi = 3.075 - 2.46$
$\phi = 0.615 \, eV$.

(A) According to Bohr's postulate,the angular momentum $L$ of an electron in the $n^{\text{th}}$ orbit is quantized as:
$L = mvr = \frac{nh}{2\pi}$
where $m$ is the mass of the electron,$v$ is its velocity,$r$ is the radius of the orbit,$h$ is Planck's constant,and $n$ is the principal quantum number.
To find the linear momentum $p = mv$,we rearrange the equation:
$mv = \frac{nh}{2\pi r}$
Therefore,the momentum of the electron is $\frac{nh}{2\pi r}$.

(B) The magnetic moment $\vec{\mu}$ of a current loop is given by $\vec{\mu} = I \vec{A}$.
For an electron of charge $-e$ revolving in an orbit of radius $r$ with speed $v$,the equivalent current is $I = \frac{-e}{T} = \frac{-ev}{2\pi r}$.
The area of the orbit is $A = \pi r^2$.
Thus,$\vec{\mu} = I \vec{A} = \left( \frac{-ev}{2\pi r} \right) (\pi r^2) = \frac{-evr}{2}$.
The orbital angular momentum is $\vec{L} = mvr$.
Substituting $vr = \frac{L}{m}$,we get $\vec{\mu} = -\frac{e}{2m} \vec{L}$.

(A) In the given circuit,if either input $A$ or $B$ is at $0 \, V$ (logical $0$),the corresponding diode ($D_1$ or $D_2$) becomes forward-biased. This pulls the output potential $Y$ down to approximately $0 \, V$ (logical $0$).
If both inputs $A$ and $B$ are at $5 \, V$ (logical $1$),both diodes are reverse-biased. No current flows through the diodes,and the output $Y$ is pulled up to $5 \, V$ (logical $1$) through the resistor $R$.
This behavior corresponds to an $AND$ gate,where the output is $1$ only if both inputs are $1$. The truth table is:

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

(C) The number of electric field lines per unit area is equivalent to the electric field intensity $E$ at that surface.
For a uniformly charged sphere of radius $R$ and volume charge density $\rho$,the electric field at the surface $(r = R)$ is given by Gauss's Law as $E = \frac{q}{4 \pi \epsilon_{0} R^{2}}$.
Since $q = \rho \times \text{Volume} = \rho \times \frac{4}{3} \pi R^{3}$,we substitute this into the expression for $E$:
$E = \frac{\rho \times \frac{4}{3} \pi R^{3}}{4 \pi \epsilon_{0} R^{2}} = \frac{\rho R}{3 \epsilon_{0}}$.
Given $\rho = 2 \, \mu C \, m^{-3} = 2 \times 10^{-6} \, C \, m^{-3}$,$R = 6 \, m$,and $\epsilon_{0} = 8.85 \times 10^{-12} \, C^{2} N^{-1} m^{-2}$.
$E = \frac{2 \times 10^{-6} \times 6}{3 \times 8.85 \times 10^{-12}} = \frac{12 \times 10^{-6}}{26.55 \times 10^{-12}} \approx 0.4519 \times 10^{6} \times 10^{6} = 0.4519 \times 10^{12} \, N C^{-1}$.
Rounding to the nearest integer as per the options,we get $45 \times 10^{10} \, N C^{-1}$.

(B) To find the value of $V_{0}$,we apply Kirchhoff's Current Law $(KCL)$ at the node where $V_{0}$ is defined.
Let the node voltage be $V_{0}$. The currents leaving the node through the three parallel branches must sum to zero.
Using nodal analysis:
$\frac{V_{0}-2}{1 \text{ k}\Omega} + \frac{V_{0}-4}{1 \text{ k}\Omega} + \frac{V_{0}-6}{1 \text{ k}\Omega} = 0$
Since the resistances are equal,we can multiply the entire equation by $1 \text{ k}\Omega$:
$(V_{0}-2) + (V_{0}-4) + (V_{0}-6) = 0$
$3V_{0} - 12 = 0$
$3V_{0} = 12$
$V_{0} = 4 \text{ V}$

(B) The resistance of a single wire of length $l$ and diameter $d$ is given by $r = \rho \frac{l}{A} = \rho \frac{l}{\pi (d/2)^2} = \rho \frac{4l}{\pi d^2}$.
When $8$ such wires are connected in parallel,the equivalent resistance $R$ is given by $R = \frac{r}{8} = \frac{1}{8} \left( \rho \frac{4l}{\pi d^2} \right) = \frac{\rho l}{2 \pi d^2}$.
For a single wire of length $2l$ and diameter $d_1$ to have the same resistance $R$,we have $R = \rho \frac{2l}{\pi (d_1/2)^2} = \rho \frac{8l}{\pi d_1^2}$.
Equating the two expressions for $R$: $\frac{\rho l}{2 \pi d^2} = \frac{8 \rho l}{\pi d_1^2}$.
Simplifying,we get $\frac{1}{2 d^2} = \frac{8}{d_1^2}$,which implies $d_1^2 = 16 d^2$.
Taking the square root,we find $d_1 = 4d$.

(A) The energy of a photon emitted by an $LED$ is given by the formula $E = \frac{hc}{\lambda}$.
Given,wavelength $\lambda = 4000 \, \mathring{A} = 400 \, \text{nm}$.
Using the relation $E \approx \frac{1240}{\lambda (\text{in nm})} \, \text{eV}$:
$E = \frac{1240}{400} \, \text{eV} = 3.1 \, \text{eV}$.
Rounding off to the nearest integer,we get $3 \, \text{eV}$.

(A) The coverage distance $d$ of a $TV$ tower of height $h$ is given by $d = \sqrt{2Rh}$.
Given the radius of the Earth $R = 6400 \, km$.
The area covered by the tower is $A = \pi d^2 = \pi (2Rh)$.
Given the population $P = 6.03 \times 10^5$ and population density $\rho = 100 \, \text{people/km}^2$.
The area $A$ is also given by $A = \frac{P}{\rho} = \frac{6.03 \times 10^5}{100} = 6030 \, \text{km}^2$.
Equating the two expressions for area: $6030 = \pi \times 2 \times 6400 \times h$ (where $h$ is in $km$).
$h = \frac{6030}{2 \times \pi \times 6400} \, km$.
$h = \frac{6030}{40212} \approx 0.150 \, km$.
Converting to meters: $h = 0.150 \times 1000 = 150 \, m$.

(B) When two conductors are connected by a conducting wire,charge flows until their potentials become equal,i.e.,$V_A = V_B$.
Since the potential of a spherical conductor is $V = \frac{KQ}{R}$,we have $\frac{KQ_A}{R_A} = \frac{KQ_B}{R_B}$.
This implies $\frac{Q_A}{Q_B} = \frac{R_A}{R_B} = \frac{5 \ mm}{10 \ mm} = \frac{1}{2}$.
The electric field at the surface of a sphere is given by $E = \frac{KQ}{R^2}$.
Therefore,the ratio of the electric fields is $\frac{E_A}{E_B} = \frac{KQ_A / R_A^2}{KQ_B / R_B^2} = \frac{Q_A}{Q_B} \times \left(\frac{R_B}{R_A}\right)^2$.
Substituting the ratio of charges: $\frac{E_A}{E_B} = \left(\frac{R_A}{R_B}\right) \times \left(\frac{R_B}{R_A}\right)^2 = \frac{R_B}{R_A}$.
Given $R_A = 5 \ mm$ and $R_B = 10 \ mm$,we get $\frac{E_A}{E_B} = \frac{10}{5} = \frac{2}{1}$.

(A) The given equation for the magnetic field is $B_{y} = B_{0} \sin(kx - \omega t)$.
Comparing this with the given equation $B_{y} = 5 \times 10^{-6} \sin(5000\pi x - 4 \times 10^{11}\pi t)$,we identify the amplitude of the magnetic field as $B_{0} = 5 \times 10^{-6} \text{ T}$.
The speed of the electromagnetic wave $c$ is related to the electric field amplitude $E_{0}$ and magnetic field amplitude $B_{0}$ by the relation $E_{0} = c B_{0}$.
Using the speed of light $c = 3 \times 10^{8} \text{ m/s}$,we calculate:
$E_{0} = (3 \times 10^{8} \text{ m/s}) \times (5 \times 10^{-6} \text{ T})$
$E_{0} = 15 \times 10^{2} \text{ V/m} = 1500 \text{ V/m}$.

(A) The speed of light in a medium is given by $V = \frac{c}{n}$,where $c$ is the speed of light in vacuum and $n$ is the refractive index.
Since $n = \frac{c}{V}$,the refractive index is inversely proportional to the speed of light.
For total internal reflection to occur,light must travel from a denser medium $(M_{1})$ to a rarer medium $(M_{2})$.
The condition for the critical angle $i_{c}$ is $n_{1} \sin i_{c} = n_{2} \sin 90^{\circ}$.
Thus,$\sin i_{c} = \frac{n_{2}}{n_{1}} = \frac{V_{1}}{V_{2}}$.
Given $V_{1} = 1.5 \times 10^{8} \text{ m/s}$ and $V_{2} = 2.0 \times 10^{8} \text{ m/s}$.
$\sin i_{c} = \frac{1.5 \times 10^{8}}{2.0 \times 10^{8}} = \frac{1.5}{2.0} = \frac{3}{4}$.
If $\sin i_{c} = \frac{3}{4}$,then the opposite side is $3$ and the hypotenuse is $4$.
The adjacent side is $\sqrt{4^{2} - 3^{2}} = \sqrt{16 - 9} = \sqrt{7}$.
Therefore,$\tan i_{c} = \frac{\text{opposite}}{\text{adjacent}} = \frac{3}{\sqrt{7}}$.
Hence,$i_{c} = \tan^{-1}\left(\frac{3}{\sqrt{7}}\right)$.

(B) Given:
$V_{\max} = 60\,V$
$V_{\min} = 20\,V$
The modulation index $\mu$ is given by the formula:
$\mu = \frac{V_{\max} - V_{\min}}{V_{\max} + V_{\min}}$
Substituting the values:
$\mu = \frac{60 - 20}{60 + 20} = \frac{40}{80} = 0.5$
The percentage modulation index is $\mu \times 100\% = 0.5 \times 100\% = 50\%$.

(C) According to the law of conservation of linear momentum,since the initial nucleus is at rest,the total initial momentum is zero.
Therefore,the magnitudes of the momenta of the two parts must be equal and opposite:
$|\vec{P}_1| = |\vec{P}_2| = P$
The de Broglie wavelength $\lambda$ is given by the relation:
$\lambda = \frac{h}{P}$
Since both parts have the same magnitude of momentum $P$,their de Broglie wavelengths will also be equal.
Therefore,the ratio of their de Broglie wavelengths is $\lambda_1 : \lambda_2 = 1 : 1$.

(A) The circuit contains a Wheatstone bridge structure. Let us check the ratio of the resistors in the arms: $\frac{3\,\Omega}{6\,\Omega} = \frac{1}{2}$ and $\frac{3\,\Omega}{6\,\Omega} = \frac{1}{2}$.
Since the ratios are equal,the bridge is balanced.
Therefore,no current flows through the central $5\,\Omega$ resistor,and it can be removed from the circuit.
Now,the upper branch has two $3\,\Omega$ resistors in series,giving $R_1 = 3 + 3 = 6\,\Omega$.
The lower branch has two $6\,\Omega$ resistors in series,giving $R_2 = 6 + 6 = 12\,\Omega$.
These two branches are in parallel,so the equivalent resistance of the bridge part is $R_p = \frac{R_1 \times R_2}{R_1 + R_2} = \frac{6 \times 12}{6 + 12} = \frac{72}{18} = 4\,\Omega$.
This equivalent resistance is in series with the $2\,\Omega$ resistor connected to the battery.
Total resistance $R_{eq} = 4 + 2 = 6\,\Omega$.
The current drawn from the battery is $I = \frac{V}{R_{eq}} = \frac{6\,V}{6\,\Omega} = 1\,A$.

(C) Case $I$: Switch $K$ is closed.
Both capacitors are in parallel with the source $V$.
Equivalent capacitance $C_{eq} = C + C = 2C$.
Total energy $E_{1} = \frac{1}{2} C_{eq} V^{2} = \frac{1}{2} (2C) V^{2} = CV^{2}$.
Case $II$: Switch $K$ is opened.
The left capacitor remains connected to the source $V$,so its potential remains $V$ and its new capacitance is $C' = K_{d}C = 5C$. Its energy is $E_{L} = \frac{1}{2} (5C) V^{2} = \frac{5}{2} CV^{2}$.
The right capacitor is disconnected,so its charge $Q = CV$ remains constant. Its new capacitance is $C' = K_{d}C = 5C$. Its energy is $E_{R} = \frac{Q^{2}}{2C'} = \frac{(CV)^{2}}{2(5C)} = \frac{CV^{2}}{10}$.
Total energy $E_{2} = E_{L} + E_{R} = \frac{5}{2} CV^{2} + \frac{1}{10} CV^{2} = \frac{25+1}{10} CV^{2} = \frac{26}{10} CV^{2} = \frac{13}{5} CV^{2}$.
Ratio $\frac{E_{1}}{E_{2}} = \frac{CV^{2}}{\frac{13}{5} CV^{2}} = \frac{5}{13}$.

(B) The magnetic moment of a current loop is given by $\vec{M} = I \vec{A}$,where $\vec{A}$ is the area vector directed according to the right-hand rule.
For the inner loop (radius $r_{1} = 0.3 \, m$),the current flows counter-clockwise,so $\vec{M}_{1} = I \pi r_{1}^{2} \hat{k} = 7 \times \pi \times (0.3)^{2} \hat{k} = 0.63 \pi \hat{k} \, A \cdot m^{2}$.
For the outer loop (radius $r_{2} = 0.5 \, m$),the current flows clockwise,so $\vec{M}_{2} = -I \pi r_{2}^{2} \hat{k} = -7 \times \pi \times (0.5)^{2} \hat{k} = -1.75 \pi \hat{k} \, A \cdot m^{2}$.
The net magnetic moment is $\vec{M}_{net} = \vec{M}_{1} + \vec{M}_{2} = (0.63 \pi - 1.75 \pi) \hat{k} = -1.12 \pi \hat{k} \, A \cdot m^{2}$.
Using $\pi \approx \frac{22}{7}$,we get $\vec{M}_{net} = -1.12 \times \frac{22}{7} \hat{k} = -0.16 \times 22 \hat{k} = -3.52 \hat{k} \, A \cdot m^{2}$.
This is approximately $-\frac{7}{2} \hat{k} \, A \cdot m^{2}$.

(A) For a particle to pass undeflected in a velocity selector,the electric force must balance the magnetic force: $qE = qvB$,which implies $E = vB$.
First,calculate the velocity $v$ of the electron using its kinetic energy $K = 728 \, eV$:
$K = \frac{1}{2} mv^2$
$728 \times 1.6 \times 10^{-19} \, J = \frac{1}{2} \times 9.1 \times 10^{-31} \, kg \times v^2$
$v^2 = \frac{2 \times 728 \times 1.6 \times 10^{-19}}{9.1 \times 10^{-31}}$
$v^2 = 256 \times 10^{12} \, m^2/s^2$
$v = 16 \times 10^6 \, m/s$
Now,calculate the electric field $E$:
$E = vB = (16 \times 10^6 \, m/s) \times (12 \times 10^{-3} \, T)$
$E = 192 \times 10^3 \, V/m = 192 \, kV/m$.

(C) The radius of a nucleus is given by $R = R_{0} A^{\frac{1}{3}}$,where $A$ is the mass number and $R_{0}$ is a constant.
The density of a nucleus $\rho$ is defined as the ratio of the mass of the nucleus to its volume.
$\rho = \frac{\text{Mass of nucleus}}{\text{Volume of nucleus}} = \frac{m \times A}{\frac{4}{3} \pi R^{3}}$,where $m$ is the average mass of a nucleon (proton or neutron).
Substituting the expression for $R$:
$\rho = \frac{m \times A}{\frac{4}{3} \pi (R_{0} A^{\frac{1}{3}})^{3}} = \frac{m \times A}{\frac{4}{3} \pi R_{0}^{3} A}$.
Simplifying the expression,the $A$ terms cancel out:
$\rho = \frac{m}{\frac{4}{3} \pi R_{0}^{3}}$.
Since $\rho$ is independent of the mass number $A$,the nuclear density is constant for all nuclei.
Therefore,the ratio of the nuclear densities of two nuclei is $1: 1$.

(A) According to Snell's law at the interface of the glass prism and the liquid at face $AC$:
$\mu_{glass} \sin(i) = n \sin(r)$
Given that the incident angle $i = 60^{\circ}$ and the refracted light grazes along the face $AC$,the angle of refraction $r = 90^{\circ}$.
Substituting the values:
$1.5 \times \sin(60^{\circ}) = n \times \sin(90^{\circ})$
$1.5 \times \frac{\sqrt{3}}{2} = n \times 1$
$n = \frac{1.5 \sqrt{3}}{2} = \frac{3 \sqrt{3}}{4}$
We are given $n = \frac{\sqrt{x}}{4}$.
Equating the two expressions for $n$:
$\frac{\sqrt{x}}{4} = \frac{3 \sqrt{3}}{4}$
$\sqrt{x} = 3 \sqrt{3} = \sqrt{9 \times 3} = \sqrt{27}$
Therefore,$x = 27$.

(B) The energy released in a nuclear reaction is given by the difference between the total binding energy of the products and the total binding energy of the reactants.
Total binding energy of the product $({ }_{2}^{4} Y)$: $4 \times 7.6 \, MeV = 30.4 \, MeV$.
Total binding energy of the reactants $(2 \times { }_{1}^{2} X)$: $2 \times (2 \times 1.1 \, MeV) = 4.4 \, MeV$.
Energy released = (Total binding energy of products) - (Total binding energy of reactants)
Energy released = $30.4 \, MeV - 4.4 \, MeV = 26 \, MeV$.

(B) The current gain $\beta$ in $CE$ configuration is given by $\beta = \frac{\Delta I_C}{\Delta I_B}$.
From the graph,we can choose two points to calculate the change in collector current $\Delta I_C$ and base current $\Delta I_B$.
Let us take the points $(I_B = 100\,\mu A, I_C = 5\,mA)$ and $(I_B = 200\,\mu A, I_C = 10\,mA)$.
$\Delta I_C = (10 - 5)\,mA = 5\,mA = 5 \times 10^{-3}\,A$.
$\Delta I_B = (200 - 100)\,\mu A = 100\,\mu A = 100 \times 10^{-6}\,A$.
Therefore,$\beta = \frac{5 \times 10^{-3}}{100 \times 10^{-6}} = \frac{5000}{100} = 50$.
The voltage gain $A_V$ is given by $A_V = \beta \times \frac{R_C}{R_{in}}$.
Given $R_C = 2\,k\Omega$ and $R_{in} = 0.50\,k\Omega$.
$A_V = 50 \times \frac{2\,k\Omega}{0.50\,k\Omega} = 50 \times 4 = 200$.

(C) Given charges: $q_A = 5 \, \mu C$,$q_B = 0.16 \, \mu C$,$q_C = 0.3 \, \mu C$.
Distances: $r_{AB} = 3 \, cm = 3 \times 10^{-2} \, m$,$r_{AC} = 3 \, cm = 3 \times 10^{-2} \, m$.
Coulomb's Law: $F = \frac{k q_1 q_2}{r^2}$,where $k = 9 \times 10^9 \, N \cdot m^2/C^2$.
Force on $A$ due to $C$ $(F_1)$: $F_1 = \frac{9 \times 10^9 \times (5 \times 10^{-6}) \times (0.3 \times 10^{-6})}{(3 \times 10^{-2})^2} = \frac{9 \times 10^9 \times 1.5 \times 10^{-12}}{9 \times 10^{-4}} = 1.5 \times 10 = 15 \, N$.
Force on $A$ due to $B$ $(F_2)$: $F_2 = \frac{9 \times 10^9 \times (5 \times 10^{-6}) \times (0.16 \times 10^{-6})}{(3 \times 10^{-2})^2} = \frac{9 \times 10^9 \times 0.8 \times 10^{-12}}{9 \times 10^{-4}} = 0.8 \times 10 = 8 \, N$.
Since $A$ is the right-angle corner,$F_1$ and $F_2$ are perpendicular.
Net force $F_{net} = \sqrt{F_1^2 + F_2^2} = \sqrt{15^2 + 8^2} = \sqrt{225 + 64} = \sqrt{289} = 17 \, N$.

(C) Given,$\phi = \frac{2}{3}(9 - t^2)$.
The flux becomes zero when $\phi = 0$,so $\frac{2}{3}(9 - t^2) = 0$,which gives $t^2 = 9$,or $t = 3 \, \text{s}$.
The induced electromotive force $(EMF)$ is given by Faraday's law: $e = -\frac{d\phi}{dt}$.
$e = -\frac{d}{dt} [\frac{2}{3}(9 - t^2)] = -\frac{2}{3}(0 - 2t) = \frac{4t}{3}$.
The heat produced $H$ in the coil is given by $H = \int_{0}^{t} \frac{e^2}{R} dt$.
Substituting the values $e = \frac{4t}{3}$ and $R = 8 \, \Omega$:
$H = \int_{0}^{3} \frac{(\frac{4t}{3})^2}{8} dt = \int_{0}^{3} \frac{16t^2}{9 \times 8} dt = \int_{0}^{3} \frac{2t^2}{9} dt$.
$H = \frac{2}{9} [\frac{t^3}{3}]_{0}^{3} = \frac{2}{9} \times \frac{27}{3} = \frac{2}{9} \times 9 = 2 \, \text{J}$.

(D) Let the resistance of the potentiometer wire be $R$.
The current $i$ flowing through the potentiometer wire is given by $i = \frac{4}{R + 780}$.
The potential difference across the entire potentiometer wire $AB$ is $V_{AB} = iR = \frac{4R}{R + 780}$.
The potential gradient along the wire is $k = \frac{V_{AB}}{L} = \frac{4R}{(R + 780) \times 300}$.
The null point is obtained at a length $l = 60\,cm$ for a cell of emf $E = 20\,mV = 20 \times 10^{-3}\,V$.
At the null point,the potential difference across the length $l$ is equal to the emf of the cell:
$E = k \times l$
$20 \times 10^{-3} = \left( \frac{4R}{(R + 780) \times 300} \right) \times 60$
Simplifying the equation:
$20 \times 10^{-3} = \frac{4R \times 60}{(R + 780) \times 300}$
$0.02 = \frac{4R}{5(R + 780)}$
$0.1(R + 780) = 4R$
$0.1R + 78 = 4R$
$3.9R = 78$
$R = \frac{78}{3.9} = 20\,\Omega$.

(B) Given: Frequency $f_{1} = 6\,MHz = 6 \times 10^{6}\,Hz$ and $f_{2} = 10\,MHz = 10 \times 10^{6}\,Hz$.
Speed of light $c = 3 \times 10^{8}\,m/s$.
The wavelength $\lambda$ is given by $\lambda = \frac{c}{f}$.
For $f_{1} = 6 \times 10^{6}\,Hz$,$\lambda_{1} = \frac{3 \times 10^{8}}{6 \times 10^{6}} = 50\,m$.
For $f_{2} = 10 \times 10^{6}\,Hz$,$\lambda_{2} = \frac{3 \times 10^{8}}{10 \times 10^{6}} = 30\,m$.
The wavelength bandwidth is the difference between the two wavelengths: $\Delta\lambda = \lambda_{1} - \lambda_{2} = 50\,m - 30\,m = 20\,m$.

(C) The disintegration rate $A$ follows the law $A = A_0 e^{-\lambda t}$.
At $t = 0$,$A_0 = 4250 \, \text{dpm}$.
At $t = 10 \, \text{min}$,$A = 2250 \, \text{dpm}$.
Substituting these values into the equation:
$2250 = 4250 e^{-\lambda (10)}$
$e^{-10\lambda} = \frac{2250}{4250} = \frac{45}{85} = \frac{9}{17}$
Taking the natural logarithm on both sides:
$-10\lambda = \ln\left(\frac{9}{17}\right)$
$10\lambda = \ln\left(\frac{17}{9}\right) \approx \ln(1.888)$
$10\lambda \approx 0.6356$
$\lambda \approx 0.06356 \, \min^{-1}$.
Thus,the approximate decay constant is $0.063 \, \min^{-1}$.