JEE Main 2020 Physics Question Paper with Answer and Solution

(A) The root mean square $(rms)$ speed of a gas molecule is given by the formula: $V_{rms} = \sqrt{\frac{3RT}{M}}$.
Given that the $rms$ speed of $H_{2}$ is equal to the $rms$ speed of $N_{2}$:
$V_{rms(H_{2})} = V_{rms(N_{2})}$
$\sqrt{\frac{3RT_{H_{2}}}{M_{H_{2}}}} = \sqrt{\frac{3RT_{N_{2}}}{M_{N_{2}}}}$
Squaring both sides and canceling common terms $(3R)$:
$\frac{T_{H_{2}}}{M_{H_{2}}} = \frac{T_{N_{2}}}{M_{N_{2}}}$
Given $T_{N_{2}} = 300^{\circ}C = 300 + 273 = 573 \ K$,$M_{N_{2}} = 28 \ g/mol$,and $M_{H_{2}} = 2 \ g/mol$:
$\frac{T_{H_{2}}}{2} = \frac{573}{28}$
$T_{H_{2}} = \frac{573 \times 2}{28} = \frac{573}{14} \approx 40.928 \ K$.
Rounding to the nearest integer,we get $41 \ K$.

(C) Since the impulsive force acts at the pivot,the angular momentum about the pivot is conserved during the collision.
Initial angular momentum $L_i = m v L$.
Final angular momentum $L_f = I_{total} \omega$,where $I_{total} = I_{rod} + I_{particle} = \frac{M L^2}{3} + m L^2$.
Equating $L_i = L_f$:
$m v L = \left( \frac{M L^2}{3} + m L^2 \right) \omega$
Substituting the given values $(M = 0.9\, kg, m = 0.1\, kg, L = 1\, m, v = 80\, m/s)$:
$0.1 \times 80 \times 1 = \left( \frac{0.9 \times 1^2}{3} + 0.1 \times 1^2 \right) \omega$
$8 = (0.3 + 0.1) \omega$
$8 = 0.4 \omega$
$\omega = \frac{8}{0.4} = 20\, rad/s$.

(D) Given: Mass $m = 2\, kg$,Power $P = 1\, J/s$,Time $t = 9\, s$,Initial velocity $u = 0$.
Power is defined as $P = F \cdot v = (m \cdot a) \cdot v = m \cdot \frac{dv}{dt} \cdot v$.
Rearranging the terms: $m \cdot v \cdot dv = P \cdot dt$.
Integrating both sides from rest ($v=0$ at $t=0$): $\int_{0}^{v} m \cdot v \cdot dv = \int_{0}^{t} P \cdot dt$.
$m \cdot \frac{v^2}{2} = P \cdot t \Rightarrow v^2 = \frac{2Pt}{m} \Rightarrow v = \sqrt{\frac{2Pt}{m}}$.
Since $v = \frac{dx}{dt}$,we have $\frac{dx}{dt} = \sqrt{\frac{2P}{m}} \cdot t^{1/2}$.
Integrating to find distance $x$: $\int_{0}^{x} dx = \sqrt{\frac{2P}{m}} \int_{0}^{t} t^{1/2} dt$.
$x = \sqrt{\frac{2P}{m}} \cdot \frac{t^{3/2}}{3/2} = \sqrt{\frac{2P}{m}} \cdot \frac{2}{3} \cdot t^{3/2}$.
Substituting the values $m = 2$,$P = 1$,$t = 9$: $x = \sqrt{\frac{2 \cdot 1}{2}} \cdot \frac{2}{3} \cdot 9^{3/2} = 1 \cdot \frac{2}{3} \cdot (3^2)^{3/2} = \frac{2}{3} \cdot 27 = 18\, m$.

(A) From the first position at point $B$,the angle with the vertical is $45^{\circ}$. Thus,the angle with the ground is $90^{\circ} - 45^{\circ} = 45^{\circ}$.
$\tan 45^{\circ} = \frac{h_{1}}{d} \Rightarrow 1 = \frac{h_{1}}{d} \Rightarrow h_{1} = d$.
When the balloon is at height $h_{1} + h_{2}$,the girl is at point $C$. The distance from $A$ to $C$ is $d + 2.464d = 3.464d$. The angle with the vertical is $60^{\circ}$,so the angle with the ground is $90^{\circ} - 60^{\circ} = 30^{\circ}$.
$\tan 30^{\circ} = \frac{h_{1} + h_{2}}{3.464d}$.
Given $\tan 30^{\circ} = 0.5774$,we have $0.5774 = \frac{d + h_{2}}{3.464d}$.
$d + h_{2} = 0.5774 \times 3.464d \approx 2d$.
$h_{2} = 2d - d = d$.

(C) In a resonance tube,the distance between two consecutive resonance positions is equal to half the wavelength of the sound wave,i.e.,$\frac{\lambda}{2} = l_2 - l_1$.
Given $l_1 = 17.0 \, cm$ and $l_2 = 24.5 \, cm$.
$\frac{\lambda}{2} = 24.5 \, cm - 17.0 \, cm = 7.5 \, cm$.
Therefore,$\lambda = 2 \times 7.5 \, cm = 15.0 \, cm = 0.15 \, m$.
The relationship between velocity $(v)$,frequency $(f)$,and wavelength $(\lambda)$ is $v = f \lambda$.
Given $v = 330 \, m/s$,we have $330 = f \times 0.15$.
$f = \frac{330}{0.15} = \frac{33000}{15} = 2200 \, Hz$.

(C) $1$. First,find the velocity of the helicopter at height $h$. Since it starts from rest with acceleration $g$,$v^2 = u^2 + 2as \Rightarrow v^2 = 0 + 2gh \Rightarrow v = \sqrt{2gh}$.
$2$. When the packet is dropped,it has an initial upward velocity $u = \sqrt{2gh}$.
$3$. The packet moves upward,reaches a maximum height,and then falls to the ground. The maximum height reached by the packet from the point of release is $H_{max} = \frac{u^2}{2g} = \frac{2gh}{2g} = h$.
$4$. The total height of the packet from the ground is $H_{total} = h + h = 2h$.
$5$. Using the equation of motion $s = ut + \frac{1}{2}at^2$ for the downward journey (taking upward as positive,$s = -h$,$u = \sqrt{2gh}$,$a = -g$):
$-h = \sqrt{2gh} \cdot t - \frac{1}{2}gt^2$
$\frac{1}{2}gt^2 - \sqrt{2gh} \cdot t - h = 0$
Multiplying by $2/g$: $t^2 - 2\sqrt{\frac{2h}{g}} \cdot t - \frac{2h}{g} = 0$.
$6$. Solving for $t$ using the quadratic formula $t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$:
$t = \frac{2\sqrt{2h/g} + \sqrt{8h/g + 8h/g}}{2} = \sqrt{\frac{2h}{g}} + \sqrt{\frac{4h}{g}} = (\sqrt{2} + 2) \sqrt{\frac{h}{g}} \approx (1.414 + 2) \sqrt{\frac{h}{g}} = 3.414 \sqrt{\frac{h}{g}}$.

(B) For the shell to float just submerged,the weight of the shell must be equal to the weight of the water displaced.
Weight of shell = $V_{material} \times \rho_{material} \times g = \frac{4}{3} \pi (R^3 - r^3) \rho_{material} g$.
Weight of displaced water = $V_{total} \times \rho_{water} \times g = \frac{4}{3} \pi R^3 \rho_{water} g$.
Equating the two: $\frac{4}{3} \pi (R^3 - r^3) \rho_{material} g = \frac{4}{3} \pi R^3 \rho_{water} g$.
$(R^3 - r^3) \rho_{material} = R^3 \rho_{water}$.
Given specific gravity $\frac{\rho_{material}}{\rho_{water}} = \frac{27}{8}$.
So,$(R^3 - r^3) \frac{27}{8} = R^3$.
$R^3 - r^3 = \frac{8}{27} R^3$.
$r^3 = R^3 - \frac{8}{27} R^3 = \frac{19}{27} R^3$.
$r = R \left( \frac{19}{27} \right)^{1/3} = \frac{R}{3} (19)^{1/3}$.
Since $(19)^{1/3} \approx 2.668$,$r \approx \frac{2.668}{3} R \approx 0.889 R$.

(C) $1$. Internal energy of an ideal gas depends only on temperature: $\Delta U = nC_v \Delta T$. Since all processes start at point $A$ (temperature $T_2$) and end at different points $B, C, D$ which lie on different isotherms,we must check the final temperatures. Point $B$ lies on isotherm $T_1$. Points $C$ and $D$ lie on an isotherm higher than $T_1$. Therefore,the final temperatures are $T_B = T_1$ and $T_C = T_D > T_1$. Thus,$\Delta U_{AB} < \Delta U_{AC} = \Delta U_{AD}$.
$2$. Work done $W$ is the area under the $P-V$ curve. For $A \rightarrow B$,volume increases,so $W_{AB} > 0$. For $A \rightarrow C$,it is an isochoric process (vertical line),so $W_{AC} = 0$. For $A \rightarrow D$,volume decreases,so $W_{AD} < 0$.
$3$. Comparing these,we get $E_{AB} < E_{AC} = E_{AD}$ and $W_{AB} > 0, W_{AC} = 0, W_{AD} < 0$. Note: The provided options seem to have a typo in the question's intended logic. Based on the graph,$E_{AB} < E_{AC} = E_{AD}$ and $W_{AB} > 0, W_{AC} = 0, W_{AD} < 0$. Option $C$ is the closest match if we interpret the internal energy comparison correctly.

(B) The pressure amplitude $\Delta p$ in a sound wave is related to the displacement amplitude $s$ by the formula $\Delta p = B k s$,where $B$ is the bulk modulus and $k$ is the wave number.
Since $B = \rho v^2$ and $k = \frac{\omega}{v} = \frac{2 \pi f}{v}$,we have $\Delta p = (\rho v^2) \times (\frac{2 \pi f}{v}) \times s$.
Taking the multiplicative constant to be $1$ (ignoring $2 \pi$),the relation simplifies to $\Delta p = \rho v \omega s$.
Rearranging for $s$,we get $s = \frac{\Delta p}{\rho v \omega} = \frac{\Delta p}{\rho v (2 \pi f)}$.
Substituting the given values: $s \approx \frac{10}{1 \times 300 \times 2 \pi \times 1000} \, m$.
Ignoring the factor of $2 \pi$ as per the instruction to take the multiplicative constant as $1$: $s \approx \frac{10}{300 \times 1000} \, m = \frac{1}{30000} \, m$.
Converting to millimeters: $s \approx \frac{1}{30000} \times 1000 \, mm = \frac{1}{30} \, mm \approx 0.033 \, mm = \frac{3}{100} \, mm$.

(B) The initial kinetic energy of the bullet is $K = \frac{1}{2} m v^2$.
Given $m = 5\, g$ and $v = 210\, m/s = 21000\, cm/s$.
$K = \frac{1}{2} \times 5 \times (21000)^2 = 1.1025 \times 10^9\, ergs$.
One half of this energy is converted into heat in the bullet:
$Q = \frac{1}{2} K = \frac{1}{2} \times 1.1025 \times 10^9 = 5.5125 \times 10^8\, ergs$.
We know $Q = m s \Delta T$,where $s = 0.030\, cal/(g \cdot ^{\circ}C) = 0.030 \times 4.2 \times 10^7\, ergs/(g \cdot ^{\circ}C) = 1.26 \times 10^6\, ergs/(g \cdot ^{\circ}C)$.
Substituting the values:
$5.5125 \times 10^8 = 5 \times (1.26 \times 10^6) \times \Delta T$.
$\Delta T = \frac{5.5125 \times 10^8}{6.3 \times 10^6} = \frac{551.25}{6.3} = 87.5^{\circ}C$.

(C) The pressure $P$ is given by $P = h \rho g$,where $h = 2\, cm$,$\rho = 13.6\, g/cm^{3}$,and $g = 980\, cm/s^{2}$.
$P = 2 \times 13.6 \times 980 = 26656\, dyne/cm^{2}$.
The volume $V = 4\, cm^{3}$.
The mean kinetic energy of a monoatomic gas molecule is given by $E = \frac{3}{2} kT = 4 \times 10^{-14}\, erg$.
From the ideal gas law,$PV = NkT$,where $N$ is the number of molecules.
We can write $N = \frac{PV}{kT}$.
Since $kT = \frac{2}{3} E = \frac{2}{3} \times 4 \times 10^{-14} = \frac{8}{3} \times 10^{-14}\, erg$.
Substituting the values: $N = \frac{26656 \times 4}{\frac{8}{3} \times 10^{-14}} = \frac{106624 \times 3}{8 \times 10^{-14}} = 13328 \times 3 \times 10^{14} = 39984 \times 10^{14} \approx 4.0 \times 10^{18}$.

(B) The given relation is $z = \frac{a^2 b^{2/3}}{c^{1/2} d^3}$.
The relative error in $z$ is given by the formula: $\frac{\Delta z}{z} = 2 \frac{\Delta a}{a} + \frac{2}{3} \frac{\Delta b}{b} + \frac{1}{2} \frac{\Delta c}{c} + 3 \frac{\Delta d}{d}$.
Given percentage errors are: $\frac{\Delta a}{a} \times 100 = 2\%$,$\frac{\Delta b}{b} \times 100 = 1.5\%$,$\frac{\Delta c}{c} \times 100 = 4\%$,and $\frac{\Delta d}{d} \times 100 = 2.5\%$.
Substituting these values into the error formula:
$\frac{\Delta z}{z} \times 100 = 2(2\%) + \frac{2}{3}(1.5\%) + \frac{1}{2}(4\%) + 3(2.5\%)$.
$\frac{\Delta z}{z} \times 100 = 4\% + 1\% + 2\% + 7.5\% = 14.5\%$.
Thus,the percentage error in $z$ is $14.5\%$.

(C) According to the law of conservation of angular momentum:
$I\omega + 3I \times 0 = (I + 3I)\omega'$
$I\omega = 4I\omega'$
$\omega' = \frac{\omega}{4}$
Initial kinetic energy $(KE)_i = \frac{1}{2}I\omega^2$
Final kinetic energy $(KE)_f = \frac{1}{2}(I + 3I)(\omega')^2 = \frac{1}{2}(4I)\left(\frac{\omega}{4}\right)^2 = 2I \times \frac{\omega^2}{16} = \frac{I\omega^2}{8}$
Loss in kinetic energy $\Delta KE = (KE)_i - (KE)_f = \frac{1}{2}I\omega^2 - \frac{1}{8}I\omega^2 = \frac{3}{8}I\omega^2$
Fractional loss in kinetic energy = $\frac{\Delta KE}{(KE)_i} = \frac{\frac{3}{8}I\omega^2}{\frac{1}{2}I\omega^2} = \frac{3}{8} \times 2 = \frac{3}{4}$

(D) The acceleration due to gravity at a height $h$ is given by $g_{h} = \frac{GM}{(R+h)^{2}}$.
Given $h = \frac{R}{2}$,we have $g_{1} = \frac{GM}{(R + R/2)^{2}} = \frac{GM}{(3R/2)^{2}} = \frac{4GM}{9R^{2}} \ldots(1)$
The acceleration due to gravity at a depth $d$ is given by $g_{d} = g(1 - \frac{d}{R}) = \frac{GM}{R^{2}}(1 - \frac{d}{R}) = \frac{GM(R-d)}{R^{3}} \ldots(2)$
Since $g_{1} = g_{d}$,we equate equations $(1)$ and $(2)$:
$\frac{4GM}{9R^{2}} = \frac{GM(R-d)}{R^{3}}$
$\frac{4}{9} = \frac{R-d}{R}$
$\frac{4}{9} = 1 - \frac{d}{R}$
$\frac{d}{R} = 1 - \frac{4}{9} = \frac{5}{9}$

(B) The torque $\vec{\tau}$ about a point $\vec{r}_2$ is given by $\vec{\tau} = (\vec{r}_1 - \vec{r}_2) \times \vec{F}$.
First,calculate the position vector relative to the point of rotation:
$\vec{r} = \vec{r}_1 - \vec{r}_2 = (4\hat{i} + 3\hat{j} - \hat{k}) - (\hat{i} + 2\hat{j} + \hat{k}) = 3\hat{i} + \hat{j} - 2\hat{k}$.
Now,calculate the cross product $\vec{\tau} = \vec{r} \times \vec{F}$:
$\vec{\tau} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 3 & 1 & -2 \\ 1 & 2 & 3 \end{vmatrix}$
$= \hat{i}(1(3) - (-2)(2)) - \hat{j}(3(3) - (-2)(1)) + \hat{k}(3(2) - 1(1))$
$= \hat{i}(3 + 4) - \hat{j}(9 + 2) + \hat{k}(6 - 1) = 7\hat{i} - 11\hat{j} + 5\hat{k}$.
The magnitude of the torque is $|\vec{\tau}| = \sqrt{7^2 + (-11)^2 + 5^2} = \sqrt{49 + 121 + 25} = \sqrt{195}$.
Comparing this with $\sqrt{x}$,we get $x = 195$.

(B) To reach the smaller planet,the body must cross the point of zero gravitational field (neutral point) between the two planets.
Let the distance of the neutral point from the center of the smaller planet be $x$.
Equating the gravitational fields from both planets at this point:
$\frac{GM}{x^2} = \frac{G(16M)}{(10a - x)^2}$
Taking the square root on both sides:
$\frac{1}{x} = \frac{4}{10a - x} \implies 10a - x = 4x \implies x = 2a$.
Now,apply the Law of Conservation of Mechanical Energy between the surface of the larger planet and the neutral point.
The potential at the surface of the larger planet (radius $2a$,mass $16M$) is $V_L = -\frac{G(16M)}{2a} - \frac{GM}{8a} = -\frac{8GM}{a} - \frac{GM}{8a} = -\frac{65GM}{8a}$.
The potential at the neutral point (distance $8a$ from larger,$2a$ from smaller) is $V_P = -\frac{G(16M)}{8a} - \frac{GM}{2a} = -\frac{2GM}{a} - \frac{GM}{2a} = -\frac{5GM}{2a}$.
Using $KE_i + PE_i = KE_f + PE_f$,where $KE_f = 0$ at the neutral point:
$\frac{1}{2}mv^2 + m V_L = 0 + m V_P$
$\frac{1}{2}v^2 = V_P - V_L = -\frac{5GM}{2a} - (-\frac{65GM}{8a}) = \frac{-20GM + 65GM}{8a} = \frac{45GM}{8a}$.
$v^2 = \frac{45GM}{4a} \implies v = \sqrt{\frac{45GM}{4a}} = \frac{3}{2} \sqrt{\frac{5GM}{a}}$.

(A) The rods are identical,meaning they have the same length $(\ell)$ and area of cross-section $(A)$.
Since the rods are joined in series,the rate of heat flow (heat current) is the same for all rods in a steady state.
Let the heat current be $H = \frac{\Delta Q}{\Delta t}$.
$H = \frac{K_{1} A (100 - 70)}{\ell} = \frac{K_{2} A (70 - 20)}{\ell} = \frac{K_{3} A (20 - 0)}{\ell}$
Simplifying the equation:
$30 K_{1} = 50 K_{2} = 20 K_{3}$
Dividing by $10$:
$3 K_{1} = 5 K_{2} = 2 K_{3}$
From $3 K_{1} = 2 K_{3}$,we get $\frac{K_{1}}{K_{3}} = \frac{2}{3}$ or $K_{1}: K_{3} = 2: 3$.
From $5 K_{2} = 2 K_{3}$,we get $\frac{K_{2}}{K_{3}} = \frac{2}{5}$ or $K_{2}: K_{3} = 2: 5$.
Thus,the correct relationship is $K_{1}: K_{3} = 2: 3$ and $K_{2}: K_{3} = 2: 5$.

(C) Given force $\overrightarrow{F} = m\overrightarrow{a} = k(v_y \hat{i} + v_x \hat{j})$.
Thus,$a_x = \frac{dv_x}{dt} = \frac{k}{m} v_y$ and $a_y = \frac{dv_y}{dt} = \frac{k}{m} v_x$.
Dividing the two equations: $\frac{dv_y}{dv_x} = \frac{v_x}{v_y} \implies v_y dv_y = v_x dv_x$.
Integrating both sides: $\int v_y dv_y = \int v_x dv_x \implies v_y^2 = v_x^2 + C \implies v_y^2 - v_x^2 = \text{constant}$.
Now,calculate $\overrightarrow{v} \times \overrightarrow{a}$:
$\overrightarrow{v} \times \overrightarrow{a} = (v_x \hat{i} + v_y \hat{j}) \times \frac{k}{m}(v_y \hat{i} + v_x \hat{j})$
$= \frac{k}{m} [v_x^2 (\hat{i} \times \hat{j}) + v_y^2 (\hat{j} \times \hat{i})]$
$= \frac{k}{m} [v_x^2 \hat{k} - v_y^2 \hat{k}] = \frac{k}{m} (v_x^2 - v_y^2) \hat{k}$.
Since $v_y^2 - v_x^2$ is constant,$\overrightarrow{v} \times \overrightarrow{a}$ is constant in time.

(D) The moment of inertia $I$ is given by $I = \int r^{2} dm = \int_{0}^{L} x^{2} \lambda(x) dx$.
Substituting $\lambda(x) = \lambda_{0}(1 + \frac{x}{L})$:
$I = \int_{0}^{L} x^{2} \lambda_{0}(1 + \frac{x}{L}) dx = \lambda_{0} \int_{0}^{L} (x^{2} + \frac{x^{3}}{L}) dx$
$I = \lambda_{0} [\frac{x^{3}}{3} + \frac{x^{4}}{4L}]_{0}^{L} = \lambda_{0} (\frac{L^{3}}{3} + \frac{L^{3}}{4}) = \frac{7}{12} \lambda_{0} L^{3} \quad ...(i)$
Now,calculate the total mass $M$:
$M = \int_{0}^{L} \lambda(x) dx = \int_{0}^{L} \lambda_{0}(1 + \frac{x}{L}) dx = \lambda_{0} [x + \frac{x^{2}}{2L}]_{0}^{L} = \lambda_{0} (L + \frac{L}{2}) = \frac{3}{2} \lambda_{0} L$
Thus,$\lambda_{0} L = \frac{2}{3} M \quad ...(ii)$
Substituting $(ii)$ into $(i)$:
$I = \frac{7}{12} (\lambda_{0} L) L^{2} = \frac{7}{12} (\frac{2}{3} M) L^{2} = \frac{14}{36} M L^{2} = \frac{7}{18} M L^{2}$
$I \approx 0.388 M L^{2}$.

(C) $1$. The given readings are $5.50\, mm, 5.55\, mm, 5.45\, mm,$ and $5.65\, mm$. All these readings have $3$ significant figures.
$2$. The average value is $5.5375\, mm$. According to the rules of significant figures,the result should be rounded off to the same number of significant figures as the measured values,which is $3$. Thus,$5.5375$ is rounded to $5.54\, mm$.
$3$. The standard deviation is $0.07395\, mm$. The uncertainty (error) should generally be expressed to one or two significant figures. Rounding $0.07395$ to one significant figure gives $0.07\, mm$.
$4$. Therefore,the diameter should be recorded as $(5.54 \pm 0.07)\, mm$.

(B) The given equation of motion is $y = y_{0} \sin^{2} \omega t$.
Using the trigonometric identity $\sin^{2} \theta = \frac{1 - \cos 2\theta}{2}$,we can rewrite the equation as:
$y = \frac{y_{0}}{2} (1 - \cos 2\omega t)$
$y - \frac{y_{0}}{2} = -\frac{y_{0}}{2} \cos 2\omega t$
This represents simple harmonic motion about the equilibrium position $y_{eq} = \frac{y_{0}}{2}$ with an amplitude $A = \frac{y_{0}}{2}$.
For a vertical spring-mass system,the equilibrium position is at a distance $y_{eq} = \frac{mg}{k}$ from the unstretched position.
Therefore,$\frac{y_{0}}{2} = \frac{mg}{k}$,which implies $\frac{k}{m} = \frac{2g}{y_{0}}$.
The angular frequency of the oscillation is given by $\Omega = \sqrt{\frac{k}{m}}$.
From the equation $y = \frac{y_{0}}{2} - \frac{y_{0}}{2} \cos 2\omega t$,the angular frequency of the motion is $2\omega$.
Thus,$2\omega = \sqrt{\frac{k}{m}} = \sqrt{\frac{2g}{y_{0}}}$.
Solving for $\omega$,we get $\omega = \frac{1}{2} \sqrt{\frac{2g}{y_{0}}} = \sqrt{\frac{2g}{4y_{0}}} = \sqrt{\frac{g}{2y_{0}}}$.

(A) The mean free path $\lambda$ is given by $\lambda = \frac{k_B T}{\sqrt{2} \pi d^2 P}$,where $k_B$ is the Boltzmann constant and $d$ is the molecular diameter.
For a constant pressure $P$,$\lambda \propto T$.
The average speed of a molecule $v_{avg}$ is given by $v_{avg} = \sqrt{\frac{8 k_B T}{\pi m}}$,so $v_{avg} \propto \sqrt{T}$.
The mean time between successive collisions $t_0$ is defined as $t_0 = \frac{\lambda}{v_{avg}}$.
Substituting the proportionalities: $t_0 \propto \frac{T}{\sqrt{T}} = \sqrt{T}$.
Therefore,the mean time between successive collisions varies with $T$ as $\sqrt{T}$.

(B) For a horizontal pipe,the height $h$ is constant,so $h_1 = h_2$. According to Bernoulli's equation:
$P_1 + \frac{1}{2}\rho v_1^2 = P_2 + \frac{1}{2}\rho v_2^2$
Given $P_1 = P$,$v_1 = v$,$P_2 = \frac{P}{2}$,and $v_2 = V$.
Substituting these values:
$P + \frac{1}{2}\rho v^2 = \frac{P}{2} + \frac{1}{2}\rho V^2$
Subtract $\frac{P}{2}$ from both sides:
$\frac{P}{2} + \frac{1}{2}\rho v^2 = \frac{1}{2}\rho V^2$
Multiply the entire equation by $\frac{2}{\rho}$:
$\frac{P}{\rho} + v^2 = V^2$
Taking the square root on both sides:
$V = \sqrt{\frac{P}{\rho} + v^2}$

(D) Initial velocity of particle $A$: $\vec{u}_{1} = (\sqrt{3} \hat{i} + \hat{j}) \, m/s$.
Initial velocity of particle $B$: $\vec{u}_{2} = 0$.
Given $m_{1} = 2m_{2}$.
Final velocity of particle $A$: $\vec{V}_{1} = (\hat{i} + \sqrt{3} \hat{j}) \, m/s$.
By the law of conservation of linear momentum:
$m_{1} \vec{u}_{1} + m_{2} \vec{u}_{2} = m_{1} \vec{V}_{1} + m_{2} \vec{V}_{2}$
$2m_{2}(\sqrt{3} \hat{i} + \hat{j}) + 0 = 2m_{2}(\hat{i} + \sqrt{3} \hat{j}) + m_{2} \vec{V}_{2}$
Dividing by $m_{2}$:
$2(\sqrt{3} \hat{i} + \hat{j}) = 2(\hat{i} + \sqrt{3} \hat{j}) + \vec{V}_{2}$
$\vec{V}_{2} = 2(\sqrt{3} \hat{i} + \hat{j} - \hat{i} - \sqrt{3} \hat{j}) = 2[(\sqrt{3}-1) \hat{i} + (1-\sqrt{3}) \hat{j}] = 2(\sqrt{3}-1) (\hat{i} - \hat{j})$.
Now,the angle $\theta$ between $\vec{V}_{1}$ and $\vec{V}_{2}$ is given by $\cos \theta = \frac{\vec{V}_{1} \cdot \vec{V}_{2}}{|\vec{V}_{1}| |\vec{V}_{2}|}$.
$\vec{V}_{1} \cdot \vec{V}_{2} = (\hat{i} + \sqrt{3} \hat{j}) \cdot [2(\sqrt{3}-1) (\hat{i} - \hat{j})] = 2(\sqrt{3}-1) (1 - \sqrt{3}) = -2(\sqrt{3}-1)^{2}$.
$|\vec{V}_{1}| = \sqrt{1^{2} + (\sqrt{3})^{2}} = 2$.
$|\vec{V}_{2}| = 2(\sqrt{3}-1) \sqrt{1^{2} + (-1)^{2}} = 2\sqrt{2}(\sqrt{3}-1)$.
$\cos \theta = \frac{-2(\sqrt{3}-1)^{2}}{2 \cdot 2\sqrt{2}(\sqrt{3}-1)} = \frac{-(\sqrt{3}-1)}{2\sqrt{2}} = \frac{1-\sqrt{3}}{2\sqrt{2}}$.
Since $\cos 105^{\circ} = \cos(60^{\circ} + 45^{\circ}) = \cos 60^{\circ} \cos 45^{\circ} - \sin 60^{\circ} \sin 45^{\circ} = \frac{1}{2} \cdot \frac{1}{\sqrt{2}} - \frac{\sqrt{3}}{2} \cdot \frac{1}{\sqrt{2}} = \frac{1-\sqrt{3}}{2\sqrt{2}}$,therefore $\theta = 105^{\circ}$.

(D) Let the velocity of rain be $\vec{v}_r = -v_r \hat{j}$ and the velocity of the car be $\vec{v}_m = v_m \hat{i}$.
When the car moves with speed $v$,the relative velocity of rain with respect to the car is $\vec{v}_{r/m} = \vec{v}_r - \vec{v}_m = -v_r \hat{j} - v \hat{i}$.
The angle $\theta$ with the horizontal is given by $\tan \theta = \frac{|v_r|}{|v_m|}$.
Given $\theta = 60^{\circ}$ and $v_m = v$,we have $\tan 60^{\circ} = \frac{v_r}{v} = \sqrt{3}$,so $v_r = v\sqrt{3}$.
When the car speed increases to $(1+\beta)v$,the new angle is $45^{\circ}$.
Thus,$\tan 45^{\circ} = \frac{v_r}{(1+\beta)v} = 1$.
Substituting $v_r = v\sqrt{3}$,we get $\frac{v\sqrt{3}}{(1+\beta)v} = 1$.
$\sqrt{3} = 1 + \beta$.
$\beta = \sqrt{3} - 1 \approx 1.732 - 1 = 0.732$.
Therefore,the value of $\beta$ is close to $0.73$.

(A) The work done per cycle is the area enclosed by the $P-V$ diagram:
$W = \text{Area} = (2V_0 - V_0) \times (3P_0 - P_0) = V_0 \times 2P_0 = 2P_0V_0$.
Heat is absorbed during processes $AB$ and $BC$:
For process $AB$ (isochoric): $Q_{AB} = nC_V \Delta T = n \left( \frac{3R}{2} \right) \left( \frac{P_B V_A}{nR} - \frac{P_A V_A}{nR} \right) = \frac{3}{2} V_A (P_B - P_A) = \frac{3}{2} V_0 (3P_0 - P_0) = 3P_0V_0$.
For process $BC$ (isobaric): $Q_{BC} = nC_P \Delta T = n \left( \frac{5R}{2} \right) \left( \frac{P_B V_C}{nR} - \frac{P_B V_B}{nR} \right) = \frac{5}{2} P_B (V_C - V_B) = \frac{5}{2} (3P_0) (2V_0 - V_0) = \frac{15}{2} P_0V_0$.
Total heat absorbed $Q_{\text{in}} = Q_{AB} + Q_{BC} = 3P_0V_0 + 7.5P_0V_0 = 10.5P_0V_0 = \frac{21}{2} P_0V_0$.
Efficiency $\eta = \frac{W}{Q_{\text{in}}} \times 100 = \frac{2P_0V_0}{(21/2)P_0V_0} \times 100 = \frac{4}{21} \times 100 \approx 19.04\%$.
Thus,the efficiency is close to $19\%$.

(B) The formula for the distance of the centre of mass of a solid hemisphere from the centre of its flat base is given by $X = \frac{3R}{8}$,where $R$ is the radius of the hemisphere.
Given,$R = 8 \, cm$.
Substituting the value of $R$ in the formula:
$X = \frac{3 \times 8}{8} \, cm$
$X = 3 \, cm$.
Therefore,the value of $X$ is $3$.

(A) According to the law of conservation of angular momentum,the angular momentum of the satellite remains constant throughout its orbit.
At the closest point (periapsis) and farthest point (apoapsis),the velocity vector is perpendicular to the position vector.
Therefore,$L = m r_{\min} V_{\max} = m r_{\max} V_{\min}$.
This simplifies to $r_{\min} V_{\max} = r_{\max} V_{\min} \quad \dots(i)$.
Given that the velocity at the farthest point is $6$ times less than the velocity at the closest point,we have $V_{\min} = \frac{V_{\max}}{6}$,or $\frac{V_{\min}}{V_{\max}} = \frac{1}{6}$.
Substituting this into equation $(i)$,we get $\frac{r_{\min}}{r_{\max}} = \frac{V_{\min}}{V_{\max}} = \frac{1}{6}$.
Thus,the ratio of the distances is $1:6$.

(B) The least count $(LC)$ of a screw gauge is calculated as:
$LC = \frac{\text{Pitch}}{\text{Number of divisions on circular scale}}$
Given,$\text{Pitch} = 0.5\, mm$ and $\text{Number of divisions} = 50$.
$LC = \frac{0.5\, mm}{50} = 0.01\, mm = 10\, \mu m$.
Since the circular scale is $4$ units ahead of the pitch scale marking (the zero of the circular scale is above the reference line),the zero error is positive.
Therefore,the nature of zero error is positive and the least count is $10\, \mu m$.

(A) An elastic wire acts like a spring with a force constant $k$ given by the formula $k = \frac{YA}{L}$.
The frequency of oscillation $f$ for a mass-spring system is given by $f = \frac{1}{2 \pi} \sqrt{\frac{k}{m}}$.
Substituting the value of $k$ into the frequency formula:
$f = \frac{1}{2 \pi} \sqrt{\frac{YA/L}{m}}$
Therefore,the frequency is $f = \frac{1}{2 \pi} \sqrt{\frac{YA}{mL}}$.

(D) Let $\theta$ be the angle the radius makes with the vertical at the point where the insect starts slipping. At this point,the component of gravity along the tangent balances the maximum static friction.
$mg \sin \theta = f_{max} = \mu N$
Since the normal force $N = mg \cos \theta$,we have:
$mg \sin \theta = \mu mg \cos \theta$
$\tan \theta = \mu = 0.75 = \frac{3}{4}$
From the geometry of the hemisphere,the height $h$ from the bottom is given by:
$h = R - R \cos \theta = R(1 - \cos \theta)$
Given $\tan \theta = \frac{3}{4}$,we have $\sin \theta = \frac{3}{5}$ and $\cos \theta = \frac{4}{5}$.
Substituting the values:
$h = 1 \times (1 - \frac{4}{5}) = 1 \times \frac{1}{5} = 0.2\, m$.

(A) The length of the second's hand is the radius of the circular path,$R = 0.1\, m$.
The time period for the second's hand is $T = 60\, s$.
The angular velocity $\omega$ is given by $\omega = \frac{2\pi}{T} = \frac{2\pi}{60} \approx 0.105\, rad/s$.
The centripetal acceleration $a$ of the tip of the hand is given by $a = \omega^2 R$.
Substituting the values: $a = (0.105)^2 \times 0.1$.
$a \approx 0.011025 \times 0.1 = 0.0011025\, m/s^2$.
This can be written as $1.1 \times 10^{-3}\, m/s^2$.
Therefore,the average acceleration is of the order of $10^{-3}$.

(A) Consider a hollow cone of mass $M$,height $H$,and base radius $R$. The slant height is $L = \sqrt{R^2 + H^2}$.
Let $\theta$ be the semi-vertical angle,so $\tan \theta = R/H$.
Consider a thin circular ring element at a distance $y$ from the apex along the axis,with thickness $dy$ along the slant height.
The radius of this ring is $r = y \tan \theta$.
The slant length of the element is $dl = dy / \cos \theta$.
The surface area of the cone is $A = \pi R L = \pi R \sqrt{R^2 + H^2}$.
The mass per unit area is $\sigma = M / A = M / (\pi R \sqrt{R^2 + H^2})$.
The area of the ring element is $dA = 2 \pi r dl = 2 \pi (y \tan \theta) (dy / \cos \theta)$.
The mass of the element is $dm = \sigma dA = \frac{M}{\pi R \sqrt{R^2 + H^2}} \cdot 2 \pi (y \tan \theta) \frac{dy}{\cos \theta} = \frac{2M}{R \sqrt{R^2 + H^2}} \cdot \frac{\tan \theta}{\cos \theta} y dy$.
Since $\tan \theta = R/H$ and $\cos \theta = H/L = H/\sqrt{R^2 + H^2}$,we have $\frac{\tan \theta}{\cos \theta} = \frac{R/H}{H/L} = \frac{RL}{H^2} = \frac{R\sqrt{R^2+H^2}}{H^2}$.
Thus,$dm = \frac{2M}{R\sqrt{R^2+H^2}} \cdot \frac{R\sqrt{R^2+H^2}}{H^2} y dy = \frac{2M}{H^2} y dy$.
The moment of inertia of this ring about the axis is $dI = dm \cdot r^2 = (\frac{2M}{H^2} y dy) (y \tan \theta)^2 = \frac{2M}{H^2} \tan^2 \theta \cdot y^3 dy$.
Integrating from $y=0$ to $y=H$:
$I = \int_0^H \frac{2M}{H^2} (R/H)^2 y^3 dy = \frac{2MR^2}{H^4} \int_0^H y^3 dy = \frac{2MR^2}{H^4} [\frac{y^4}{4}]_0^H = \frac{2MR^2}{H^4} \cdot \frac{H^4}{4} = \frac{MR^2}{2}$.

(C) The potential energy is given by $U(r) = -\frac{A}{r^6} + \frac{B}{r^{12}}$.
At equilibrium,the force $F = -\frac{dU}{dr} = 0$.
Calculating the derivative: $\frac{dU}{dr} = -A(-6r^{-7}) + B(-12r^{-13}) = \frac{6A}{r^7} - \frac{12B}{r^{13}}$.
Setting $\frac{dU}{dr} = 0$: $\frac{6A}{r^7} = \frac{12B}{r^{13}} \Rightarrow r^6 = \frac{12B}{6A} = \frac{2B}{A}$.
Thus,the equilibrium separation is $r = \left(\frac{2B}{A}\right)^{1/6}$.
Substituting this value into the potential energy expression:
$U = -\frac{A}{(2B/A)} + \frac{B}{(2B/A)^2} = -\frac{A^2}{2B} + \frac{B \cdot A^2}{4B^2} = -\frac{A^2}{2B} + \frac{A^2}{4B} = -\frac{A^2}{4B}$.

(C) Let the corners of the square be at $(0,0)$,$(\ell, 0)$,$(\ell, \ell)$,and $(0, \ell)$. The axis passes through $(0,0)$ and is parallel to the diagonal connecting $(0, \ell)$ and $(\ell, 0)$. The equation of this diagonal is $x + y = \ell$. The perpendicular distance $r$ of a point $(x, y)$ from the line $x + y - \ell = 0$ is given by $r = \frac{|x + y - \ell|}{\sqrt{1^2 + 1^2}} = \frac{|x + y - \ell|}{\sqrt{2}}$.
For the four masses:
$1$. At $(0,0)$: $r_1 = \frac{|0 + 0 - \ell|}{\sqrt{2}} = \frac{\ell}{\sqrt{2}}$.
$2$. At $(\ell, 0)$: $r_2 = \frac{|\ell + 0 - \ell|}{\sqrt{2}} = 0$.
$3$. At $(0, \ell)$: $r_3 = \frac{|0 + \ell - \ell|}{\sqrt{2}} = 0$.
$4$. At $(\ell, \ell)$: $r_4 = \frac{|\ell + \ell - \ell|}{\sqrt{2}} = \frac{\ell}{\sqrt{2}}$.
The moment of inertia $I$ about the axis is $I = \sum mr_i^2 = m \left( \left(\frac{\ell}{\sqrt{2}}\right)^2 + 0^2 + 0^2 + \left(\frac{\ell}{\sqrt{2}}\right)^2 \right) = m \left( \frac{\ell^2}{2} + \frac{\ell^2}{2} \right) = m\ell^2$.
Angular momentum $L = I\omega = m\ell^2\omega$. Therefore,the missing value is $1$.

(D) The total number of degrees of freedom $(f)$ is the sum of translational and rotational degrees of freedom: $f = 3 + 2 = 5$.
The total internal energy $(U)$ for one mole of an ideal gas is given by the formula $U = \frac{f}{2} RT$. Substituting $f = 5$,we get $U = \frac{5}{2} RT$.
The adiabatic index $\gamma$ is defined as the ratio of molar heat capacities,$\gamma = \frac{C_P}{C_V} = 1 + \frac{2}{f}$.
Substituting $f = 5$ into the formula for $\gamma$,we get $\gamma = 1 + \frac{2}{5} = \frac{7}{5}$.

(D) The frequency $f$ heard by the observer is given by the Doppler effect formula: $f = f_{0} \left( \frac{v_{s}}{v_{s} - v \cos \theta} \right)$,where $v_{s}$ is the speed of sound,$v$ is the speed of the source,and $\theta$ is the angle between the velocity vector of the source and the line joining the source to the observer.
Before the source reaches the point of closest approach $(t < t_{0})$,the source is approaching the observer,so $\theta$ is acute,$\cos \theta > 0$,and the observed frequency $f > f_{0}$. As the source approaches the point of closest approach,$\theta$ increases towards $90^{\circ}$,so $\cos \theta$ decreases,and $f$ decreases towards $f_{0}$.
At the point of closest approach $(t = t_{0})$,$\theta = 90^{\circ}$,$\cos \theta = 0$,and $f = f_{0}$.
After the source passes the point of closest approach $(t > t_{0})$,the source is moving away from the observer,so $\theta$ is obtuse,$\cos \theta < 0$,and the observed frequency $f < f_{0}$. As the source moves further away,$\theta$ increases,$\cos \theta$ becomes more negative,and $f$ decreases further below $f_{0}$.
Thus,the frequency $f$ starts above $f_{0}$,decreases to $f_{0}$ at $t = t_{0}$,and continues to decrease below $f_{0}$ for $t > t_{0}$. This behavior is best represented by Graph $D$.

(B) Let the two bodies have mass $m$ and initial speed $v_0$. Let the angle between their initial velocity vectors be $2\theta$.
By symmetry,the combined mass $2m$ will move along the angle bisector of the initial velocity vectors with a final speed $v_f = v_0/2$.
Applying the law of conservation of linear momentum along the direction of the final velocity:
$m v_0 \cos \theta + m v_0 \cos \theta = (2m) v_f$
$2 m v_0 \cos \theta = 2 m (v_0/2)$
$2 m v_0 \cos \theta = m v_0$
$\cos \theta = 1/2$
$\theta = 60^\circ$
The angle between the initial velocities is $2\theta = 2 \times 60^\circ = 120^\circ$.

(D) The density $\rho$ of a sphere is given by $\rho = \frac{M}{V} = \frac{M}{\frac{4}{3} \pi (D/2)^3} = \frac{6M}{\pi D^3}$.
Taking the natural logarithm on both sides: $\ln \rho = \ln(6/\pi) + \ln M - 3 \ln D$.
Differentiating both sides to find the relative error: $\frac{d\rho}{\rho} = \frac{dM}{M} - 3 \frac{dD}{D}$.
For the maximum percentage error,we add the absolute values of the relative errors: $\left( \frac{d\rho}{\rho} \times 100 \right)_{\text{max}} = \left( \frac{dM}{M} \times 100 \right) + 3 \left( \frac{dD}{D} \times 100 \right)$.
Given $\frac{dM}{M} \times 100 = 6.0 \%$ and $\frac{dD}{D} \times 100 = 1.5 \%$.
Substituting these values: $\text{Max error} = 6.0 + 3(1.5) = 6.0 + 4.5 = 10.5 \%$.
We are given the error as $\left(\frac{x}{100}\right) \% = 10.5 \%$.
Therefore,$\frac{x}{100} = 10.5 \implies x = 1050$.

(A) Let the initial number of moles be $n$. The initial state is given by $P_{1}V_{1} = nRT_{1}$,where $T_{1} = 250\, K$.
When $25\%$ of the diatomic molecules $(X_{2})$ dissociate,the reaction is $X_{2} \rightarrow 2X$. If $25\%$ of $n$ moles dissociate,the number of moles of $X_{2}$ remaining is $0.75n$,and the number of moles of $X$ produced is $2 \times 0.25n = 0.5n$.
The total number of moles in the final state is $n' = 0.75n + 0.5n = 1.25n = \frac{5n}{4}$.
The final state is given by $P_{2}V_{2} = n'RT_{2}$,where $V_{2} = 2V_{1}$ and $T_{2} = 2000\, K$.
Substituting the values: $P_{2}(2V_{1}) = \left(\frac{5n}{4}\right)R(2000)$.
Dividing the two equations: $\frac{P_{2}(2V_{1})}{P_{1}V_{1}} = \frac{(5n/4)R(2000)}{n R(250)}$.
$2 \frac{P_{2}}{P_{1}} = \frac{5}{4} \times \frac{2000}{250} = \frac{5}{4} \times 8 = 10$.
Therefore,$\frac{P_{2}}{P_{1}} = \frac{10}{2} = 5$.

(C) Let the elongation of the spring be $\Delta x$. The total length of the spring becomes $l = l_{0} + \Delta x$.
The centripetal force required for the circular motion of the mass $m$ is provided by the spring force $F_{s} = k \Delta x$.
Equating the spring force to the centripetal force: $k \Delta x = m \omega^{2} (l_{0} + \Delta x)$.
Expanding the equation: $k \Delta x = m \omega^{2} l_{0} + m \omega^{2} \Delta x$.
Rearranging the terms to solve for $\Delta x$: $k \Delta x - m \omega^{2} \Delta x = m \omega^{2} l_{0}$.
$\Delta x (k - m \omega^{2}) = m \omega^{2} l_{0}$.
Therefore,the elongation is $\Delta x = \frac{m \omega^{2} l_{0}}{k - m \omega^{2}}$.

(D) Let $h$ be the distance above and below the surface of the earth where the weight of the body is the same. This implies the acceleration due to gravity at height $h$ $(g_h)$ must be equal to the acceleration due to gravity at depth $h$ $(g_d)$.
The formula for acceleration due to gravity at height $h$ is $g_h = g \left( 1 + \frac{h}{R} \right)^{-2} \approx g \left( 1 - \frac{2h}{R} \right)$ (for $h \ll R$). However,for a general solution,we use $g_h = \frac{g R^2}{(R+h)^2}$.
The formula for acceleration due to gravity at depth $h$ is $g_d = g \left( 1 - \frac{h}{R} \right)$.
Equating the two: $\frac{g R^2}{(R+h)^2} = g \left( 1 - \frac{h}{R} \right)$.
$\frac{1}{(1 + h/R)^2} = 1 - \frac{h}{R}$.
Let $x = \frac{h}{R}$. Then $\frac{1}{(1+x)^2} = 1 - x$.
$1 = (1-x)(1+x)^2 = (1-x)(1 + 2x + x^2) = 1 + 2x + x^2 - x - 2x^2 - x^3$.
$1 = 1 + x - x^2 - x^3$.
$x^3 + x^2 - x = 0$.
Since $x \neq 0$,we have $x^2 + x - 1 = 0$.
Using the quadratic formula $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$,we get $x = \frac{-1 \pm \sqrt{1^2 - 4(1)(-1)}}{2(1)} = \frac{-1 \pm \sqrt{5}}{2}$.
Since $x$ must be positive,$x = \frac{\sqrt{5} - 1}{2}$.
Therefore,$h = \frac{\sqrt{5} - 1}{2} R = \frac{\sqrt{5} R - R}{2}$.

(A) When the bar magnet enters the coil with a constant speed,the magnetic flux linked with the coil changes,and an $e.m.f.$ is induced,causing the galvanometer to deflect in one direction (e.g.,positive).
When the magnet is completely inside the coil,the magnetic flux linked with the coil remains constant,so the rate of change of flux is zero,and the reading of the galvanometer is zero.
When the bar magnet exits the coil,the magnetic flux linked with the coil changes again,and an $e.m.f.$ is induced in the opposite direction to that of the entry phase,causing the galvanometer to deflect in the opposite direction (e.g.,negative).

(D) Given:
Electromotive force $E = 3.0 \ V$
Terminal voltage $V = 2.5 \ V$
Power dissipated by external resistor $P_R = 0.5 \ W$
We know that the terminal voltage $V$ is given by $V = E - ir$,where $i$ is the current and $r$ is the internal resistance.
$2.5 = 3.0 - ir$
$ir = 0.5 \ V$
The power dissipated in the internal resistance is $P_r = i^2r = (ir) \cdot i$.
Since $V = iR = 2.5 \ V$,we have $i = \frac{2.5}{R}$.
Also,$P_R = i^2R = 0.5 \ W$.
Substituting $i = \frac{2.5}{R}$ into $P_R = i^2R$:
$0.5 = (\frac{2.5}{R})^2 \cdot R = \frac{6.25}{R}$
$R = \frac{6.25}{0.5} = 12.5 \ \Omega$.
Now,find the current $i$:
$i = \frac{V}{R} = \frac{2.5}{12.5} = 0.2 \ A$.
Finally,calculate the power dissipated in the internal resistance $P_r$:
$P_r = i^2r = i(ir) = 0.2 \ A \times 0.5 \ V = 0.10 \ W$.

(C) The potential energy of a charge $q$ at a point is given by $U = qV$,where $V$ is the electric potential due to the fixed charges.
Initial position: $x_i = 0$. The distances from $4q$ and $-q$ are $r_1 = d/2$ and $r_2 = d/2$.
Initial potential $V_i = \frac{1}{4\pi\varepsilon_0} \left( \frac{4q}{d/2} + \frac{-q}{d/2} \right) = \frac{1}{4\pi\varepsilon_0} \left( \frac{3q}{d/2} \right) = \frac{6q}{4\pi\varepsilon_0 d}$.
Final position: $x_f = d$. The distances from $4q$ and $-q$ are $r_1' = d + d/2 = 3d/2$ and $r_2' = d - d/2 = d/2$.
Final potential $V_f = \frac{1}{4\pi\varepsilon_0} \left( \frac{4q}{3d/2} + \frac{-q}{d/2} \right) = \frac{1}{4\pi\varepsilon_0} \left( \frac{8q}{3d} - \frac{2q}{d} \right) = \frac{1}{4\pi\varepsilon_0} \left( \frac{8q - 6q}{3d} \right) = \frac{2q}{12\pi\varepsilon_0 d} = \frac{q}{6\pi\varepsilon_0 d}$.
Change in potential energy $\Delta U = q(V_f - V_i) = q \left( \frac{q}{6\pi\varepsilon_0 d} - \frac{6q}{4\pi\varepsilon_0 d} \right) = \frac{q^2}{4\pi\varepsilon_0 d} \left( \frac{2}{3} - 6 \right) = \frac{q^2}{4\pi\varepsilon_0 d} \left( -\frac{16}{3} \right) = -\frac{4q^2}{3\pi\varepsilon_0 d}$.
The negative sign indicates a decrease in potential energy by $\frac{4q^2}{3\pi\varepsilon_0 d}$.

(A) Thin infinite uniformly charged planes produce a uniform electric field. Therefore,the electric field lines must be parallel straight lines,which rules out options $B$ and $C$ as they show curved field lines.
The electric field due to a positively charged sheet is directed away from it,and the field due to a negatively charged sheet is directed towards it. The magnitude of the electric field is given by $E = \frac{\sigma}{2\epsilon_0}$.
Since $\left|\sigma_{+}\right| > \left|\sigma_{-}\right|$,the magnitude of the electric field produced by the positively charged sheet is greater than that produced by the negatively charged sheet. This implies that the density of field lines originating from the positive sheet is higher than the density of field lines terminating on the negative sheet. Option $A$ correctly depicts the field lines originating from the positive sheet and terminating on the negative sheet with the appropriate relative densities,making it the correct representation.

(A) The circuit consists of two Zener diodes connected in series in reverse polarity.
When the input voltage $V_{in}$ is positive, the first Zener diode is in reverse bias and the second is in forward bias.
When $V_{in}$ exceeds the breakdown voltage of $6\, V$, the first Zener diode operates in the breakdown region, maintaining a constant output voltage of $6\, V$.
When $V_{in}$ is between $0\, V$ and $6\, V$, the output voltage follows the input voltage.
Similarly, when the input voltage $V_{in}$ is negative, the second Zener diode is in reverse bias and the first is in forward bias.
When $V_{in}$ becomes more negative than $-6\, V$, the second Zener diode operates in the breakdown region, maintaining a constant output voltage of $-6\, V$.
Thus, the output voltage is clipped at $+6\, V$ and $-6\, V$, resulting in a waveform that follows the input sine wave between $-6\, V$ and $+6\, V$ and remains constant at $\pm 6\, V$ outside this range.
Comparing this with the given options, the correct graph is represented by option $A$.

(A) Let $u_1$ be the object distance from the objective lens. Given $f_o = 1 \, cm$. The image distance $v_1$ from the objective lens is given by the lens formula: $\frac{1}{v_1} - \frac{1}{-u_1} = \frac{1}{f_o} \Rightarrow \frac{1}{v_1} = 1 - \frac{1}{u_1} = \frac{u_1 - 1}{u_1} \Rightarrow v_1 = \frac{u_1}{u_1 - 1}$.
The magnification of the objective lens is $m_o = \frac{v_1}{u_1} = \frac{1}{u_1 - 1}$.
The tube length $L = v_1 + |u_e| = 20 \, cm$,where $u_e$ is the object distance for the eye-piece. Thus,$|u_e| = 20 - v_1 = 20 - \frac{u_1}{u_1 - 1} = \frac{20u_1 - 20 - u_1}{u_1 - 1} = \frac{19u_1 - 20}{u_1 - 1}$.
The magnification of the eye-piece is $m_e = \frac{D}{|u_e|} = \frac{25}{|u_e|} = \frac{25(u_1 - 1)}{19u_1 - 20}$.
Total magnification $M = m_o \times m_e = 100 \Rightarrow \left(\frac{1}{u_1 - 1}\right) \times \left(\frac{25(u_1 - 1)}{19u_1 - 20}\right) = 100$.
$\frac{25}{19u_1 - 20} = 100 \Rightarrow 19u_1 - 20 = 0.25 \Rightarrow 19u_1 = 20.25 \Rightarrow u_1 = \frac{20.25}{19} \approx 1.0658 \, cm$.
Now,$|u_e| = 20 - v_1 = 20 - \frac{1.0658}{1.0658 - 1} = 20 - \frac{1.0658}{0.0658} \approx 20 - 16.2 = 3.8 \, cm$.
Using the lens formula for the eye-piece: $\frac{1}{v_e} - \frac{1}{u_e} = \frac{1}{f_e} \Rightarrow \frac{1}{-25} - \frac{1}{-3.8} = \frac{1}{f_e} \Rightarrow \frac{1}{f_e} = \frac{1}{3.8} - \frac{1}{25} \approx 0.263 - 0.04 = 0.223$.
$f_e \approx \frac{1}{0.223} \approx 4.48 \, cm$. The closest option is $4.5 \, cm$.

(A) The wavelength $\lambda$ is given by $\frac{1}{\lambda} = R \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right)$,where $R$ is the Rydberg constant. Let $C = \frac{1}{R}$. Then $\lambda = C \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right)^{-1}$.
For the Lyman series $(n_1 = 1)$:
Shortest wavelength $\lambda_{L,s} = C \left( \frac{1}{1^2} - \frac{1}{\infty^2} \right)^{-1} = C$.
Longest wavelength $\lambda_{L,l} = C \left( \frac{1}{1^2} - \frac{1}{2^2} \right)^{-1} = \frac{4C}{3}$.
Difference $\Delta \lambda_L = \frac{4C}{3} - C = \frac{C}{3} = 304\,\mathring{A} \implies C = 912\,\mathring{A}$.
For the Paschen series $(n_1 = 3)$:
Shortest wavelength $\lambda_{P,s} = C \left( \frac{1}{3^2} - \frac{1}{\infty^2} \right)^{-1} = 9C$.
Longest wavelength $\lambda_{P,l} = C \left( \frac{1}{3^2} - \frac{1}{4^2} \right)^{-1} = C \left( \frac{1}{9} - \frac{1}{16} \right)^{-1} = C \left( \frac{7}{144} \right)^{-1} = \frac{144C}{7}$.
Difference $\Delta \lambda_P = \frac{144C}{7} - 9C = \frac{144C - 63C}{7} = \frac{81C}{7}$.
Substituting $C = 912\,\mathring{A}$:
$\Delta \lambda_P = \frac{81 \times 912}{7} \approx 10553.14\,\mathring{A}$.

(A) Let the potential at point $A$ be $V_A = 0 \ V$.
Applying Kirchhoff's Current Law $(KCL)$ at junction $C$:
The current entering from the battery branch is $i_1 = 1 \ A$.
The current flowing through the $2 \ \Omega$ resistor is $i_3$ (downwards from $D$ to $C$).
The current leaving towards $F$ is $i_2 = 2 \ A$.
According to $KCL$ at junction $C$: $i_1 + i_3 = i_2$.
Substituting the values: $1 \ A + i_3 = 2 \ A$,which gives $i_3 = 1 \ A$.
Now,we find the potential at $D$ by traversing from $A$ to $D$ via $C$:
$V_D = V_A + 1 \ V - i_3 \times (2 \ \Omega) = 0 + 1 - (1 \times 2) = -1 \ V$.
Now,we find the potential at $B$ by traversing from $D$ to $B$:
$V_B = V_D + 2 \ V = -1 + 2 = 1 \ V$.
Thus,the potential at point $B$ with respect to point $A$ is $V_B - V_A = 1 - 0 = 1 \ V$.

(C) Initial capacitance $C_i = \frac{\varepsilon_0 A}{d} = \frac{\varepsilon_0 (lw)}{d}$.
Initial energy $U_i = \frac{1}{2} C_i V^2$.
After inserting the slab of length $x$,the capacitor acts as two capacitors in parallel: one with dielectric (length $x$) and one with air (length $l-x$).
$C_f = C_1 + C_2 = \frac{k \varepsilon_0 (xw)}{d} + \frac{\varepsilon_0 ((l-x)w)}{d}$.
Given $U_f = 2 U_i$,and since $V$ is constant,$C_f = 2 C_i$.
$\frac{\varepsilon_0 w}{d} [kx + l - x] = 2 \frac{\varepsilon_0 lw}{d}$.
$kx + l - x = 2l$.
Substitute $k = 4$: $4x + l - x = 2l$.
$3x = l \Rightarrow x = \frac{l}{3}$.

(C) The electric potential $V$ at the centre is the algebraic sum of potentials due to individual charges: $V = \sum \frac{kq_i}{R} = \frac{k}{R} \sum q_i$. Since there are five charges of $(+q)$ and five charges of $(-q)$, the net charge $\sum q_i = 5(+q) + 5(-q) = 0$. Thus, $V = 0$.
For the electric field $E$ at the centre, each charge $(+q)$ produces a field $\vec{E}_0$ directed away from it, and each charge $(-q)$ produces a field $\vec{E}_0$ directed towards it. Due to the symmetry of the ten charges arranged at equal angular intervals of $36^\circ$, every charge has an diametrically opposite charge of equal magnitude but opposite sign. For example, charge $1$ $(+q)$ and charge $6$ $(-q)$ are diametrically opposite. The field due to charge $1$ is $\vec{E}_1$ (away from $1$) and the field due to charge $6$ is $\vec{E}_6$ (towards $6$). Since $1$ and $6$ are opposite, $\vec{E}_1$ and $\vec{E}_6$ point in the same direction and add up to $2\vec{E}_0$. This results in five such vectors of magnitude $2E_0$ separated by $72^\circ$ each. The vector sum of these five equal vectors separated by equal angles is zero. Therefore, $E = 0$.

(D) The magnetic field inside the solenoid is $B_0 = \mu_0 n i$.
When an iron rod is placed inside,the magnetic field becomes $B = \mu_r B_0 = \mu_r \mu_0 n i$.
The magnetization $M_z$ is given by $M_z = \frac{B - \mu_0 H}{\mu_0} = (\mu_r - 1) H$.
Since $\mu_r = 1000 \gg 1$,we can approximate $M_z \approx \mu_r H = \mu_r n i$.
The magnetic moment $m$ is given by $m = M_z \times V$,where $V$ is the volume of the rod.
Given: $V = 10^{-3} \ m^3$,$\mu_r = 1000$,$n = 10 \ turns/cm = 1000 \ turns/m$,$i = 0.5 \ A$.
$m = (\mu_r n i) V = 1000 \times 1000 \times 0.5 \times 10^{-3}$.
$m = 10^6 \times 0.5 \times 10^{-3} = 0.5 \times 10^3 = 500 \ Am^2$.
$m = 5 \times 10^2 \ Am^2$.

(B) The magnetic field $B$ at a distance $r$ from an infinitely long straight wire carrying current $I$ is given by:
$B = \frac{\mu_{0} I}{2 \pi r}$
The motional electromotive force $(EMF)$ induced in a conductor of length $l$ moving with velocity $v$ perpendicular to a magnetic field $B$ is given by:
$e = Bvl$
Substituting the expression for $B$ into the formula for $e$:
$e = \left( \frac{\mu_{0} I}{2 \pi r} \right) vl = \frac{\mu_{0} Ivl}{2 \pi r}$
The induced current $i$ in the loop is given by Ohm's law,$i = \frac{e}{R}$:
$i = \frac{1}{R} \left( \frac{\mu_{0} Ivl}{2 \pi r} \right) = \frac{\mu_{0} Ivl}{2 \pi Rr}$

(A) When a radioactive nucleus decays by two different processes with decay constants $\lambda_1$ and $\lambda_2$,the total decay constant is $\lambda_{\text{eff}} = \lambda_1 + \lambda_2$.
Since the decay constant $\lambda = \frac{\ln 2}{T}$,we can write the relation for half-lives as:
$\frac{\ln 2}{T_{\text{eff}}} = \frac{\ln 2}{T_1} + \frac{\ln 2}{T_2}$
$\frac{1}{T_{\text{eff}}} = \frac{1}{T_1} + \frac{1}{T_2}$
Given $T_1 = 10 \ s$ and $T_2 = 100 \ s$,we have:
$\frac{1}{T_{\text{eff}}} = \frac{1}{10} + \frac{1}{100} = \frac{10 + 1}{100} = \frac{11}{100}$
$T_{\text{eff}} = \frac{100}{11} \approx 9.09 \ s$
Thus,the effective half-life is close to $9 \ s$.

(B) Let the potential of point $O$ be $V_{O} = 0 \, V$.
Let the potential of the upper junction be $x$.
The capacitors are connected to the batteries as shown. The potential at the left plate of the $2\, \mu F$ capacitor is $6\, V$ and the potential at the right plate of the $4\, \mu F$ capacitor is $6\, V$.
Using nodal analysis at the upper junction $x$:
Sum of charges on the plates connected to junction $x$ is zero: $q_{1} + q_{2} + q_{3} = 0$.
$2(x - 6) + 4(x - 6) + 5(x - 0) = 0$.
$2x - 12 + 4x - 24 + 5x = 0$.
$11x = 36$.
$x = \frac{36}{11} \, V$.
The charge on the $5\, \mu F$ capacitor is $q_{3} = C_{3} \cdot V_{3} = 5 \cdot (x - 0) = 5 \cdot \frac{36}{11} = \frac{180}{11} \, \mu C$.
$q_{3} \approx 16.36 \, \mu C$.

(B) $x = \frac{1}{\sqrt{\mu_{0} \varepsilon_{0}}}$ represents the speed of light in vacuum,so its dimension is $[x] = [L^{1}T^{-1}]$.
$y = \frac{E}{B}$ represents the speed of an electromagnetic wave,so its dimension is $[y] = [L^{1}T^{-1}]$.
$z = \frac{l}{CR}$ where $RC$ is the time constant $(\tau)$,so $z = \frac{l}{\tau}$. The dimension is $[z] = \frac{[L]}{[T]} = [L^{1}T^{-1}]$.
Since all three quantities have the same dimension $[L^{1}T^{-1}]$,the correct option is $B$.

(A) The figure of merit $(k)$ of a galvanometer is defined as the current required to produce a unit deflection in the galvanometer.
It is given by the formula: $k = \frac{I}{\theta}$
Given:
Current $I = 6 \, mA = 6 \times 10^{-3} \, A$
Deflection $\theta = 2^{\circ}$
Substituting the values:
$k = \frac{6 \times 10^{-3} \, A}{2^{\circ}} = 3 \times 10^{-3} \, A/\text{div}$
Therefore,the figure of merit is $3 \times 10^{-3} \, A/\text{div}$.

(A) $1$. For $V_{\text{in}} < 4\, V$: Neither Zener diode $A$ $(6\, V)$ nor $B$ $(4\, V)$ is in the breakdown region. Thus,the output voltage $V_{0}$ follows the input voltage $V_{\text{in}}$,i.e.,$V_{0} = V_{\text{in}}$.
$2$. For $4\, V < V_{\text{in}} < 6\, V$: Zener diode $B$ reaches its breakdown voltage of $4\, V$. Since it is in parallel with the load,the output voltage $V_{0}$ is clamped at $4\, V$. The excess voltage $(V_{\text{in}} - 4\, V)$ drops across the series resistance associated with diode $B$.
$3$. For $V_{\text{in}} > 6\, V$: Zener diode $A$ also reaches its breakdown voltage of $6\, V$. Since $A$ is in parallel with the load,the output voltage $V_{0}$ is now clamped at $6\, V$. The excess voltage drops across the series resistance of the source or the circuit.
$4$. Therefore,the graph of $V_{0}$ versus time will show a linear increase up to $4\, V$,a constant plateau at $4\, V$,and then a step increase to a constant plateau at $6\, V$.

(D) The wavelength ranges for the given electromagnetic radiations are as follows:
$(a)$ Microwave: The wavelength is approximately $10^{-3} \, m$ to $0.3 \, m$. Thus, $(a) \rightarrow (iv)$.
$(b)$ Gamma rays: These have the shortest wavelengths, typically less than $10^{-13} \, m$ (often cited as $10^{-15} \, m$ in this context). Thus, $(b) \rightarrow (ii)$.
$(c)$ $A.M.$ radio waves: These have very long wavelengths, typically in the range of $100 \, m$ to $1000 \, m$. Thus, $(c) \rightarrow (i)$.
$(d)$ $X$-rays: These have wavelengths in the range of $10^{-8} \, m$ to $10^{-12} \, m$ (often cited as $10^{-10} \, m$). Thus, $(d) \rightarrow (iii)$.
Therefore, the correct matching is $(a)-(iv), (b)-(ii), (c)-(i), (d)-(iii)$.

(A) According to Einstein's photoelectric equation,the maximum kinetic energy $K$ is given by $K = E - \phi$,where $E$ is the photon energy and $\phi$ is the work function.
For the first case: $K_{1} = E_{1} - \phi = 4 - \phi$.
For the second case: $K_{2} = E_{2} - \phi = 2.5 - \phi$.
Since $K = \frac{1}{2}mv_{max}^{2}$,the ratio of kinetic energies is $\frac{K_{1}}{K_{2}} = \left(\frac{v_{1}}{v_{2}}\right)^{2}$.
Given $\frac{v_{1}}{v_{2}} = 2$,we have $\frac{K_{1}}{K_{2}} = 2^{2} = 4$,so $K_{1} = 4K_{2}$.
Substituting the expressions for $K_{1}$ and $K_{2}$: $4 - \phi = 4(2.5 - \phi)$.
$4 - \phi = 10 - 4\phi$.
$3\phi = 6$.
$\phi = 2 \ eV$.

(A) For a thin prism,the angle of minimum deviation $\delta_{\min}$ is given by the formula:
$\delta_{\min} = (\mu - 1)A$
Given:
Prism angle $A = 1^{\circ}$
Refractive index $\mu = 1.5$
Substituting the values:
$\delta_{\min} = (1.5 - 1) \times 1^{\circ}$
$\delta_{\min} = 0.5^{\circ}$
We are given that $\delta_{\min} = N/10$.
So,$0.5 = N/10$
$N = 0.5 \times 10 = 5$
Thus,the value of $N$ is $5$.

(B) The activity of a radioactive substance is given by $R = R_{0} e^{-\lambda t}$.
Taking the natural logarithm on both sides,we get $\ln R = \ln R_{0} - \lambda t$.
This is the equation of a straight line $y = mx + c$,where the slope $m = -\lambda$.
From the graph,the slope $\lambda$ for each substance is:
For $A$: $\lambda_{A} = \frac{6 - 0}{10 - 0} = 0.6 = \frac{3}{5}$.
For $B$: $\lambda_{B} = \frac{4 - 0}{5 - 0} = 0.8 = \frac{4}{5}$.
For $C$: $\lambda_{C} = \frac{2 - 0}{5 - 0} = 0.4 = \frac{2}{5}$.
The half-life is given by $T_{\frac{1}{2}} = \frac{\ln 2}{\lambda}$.
Thus,$T_{\frac{1}{2}}(A) : T_{\frac{1}{2}}(B) : T_{\frac{1}{2}}(C) = \frac{1}{\lambda_{A}} : \frac{1}{\lambda_{B}} : \frac{1}{\lambda_{C}} = \frac{5}{3} : \frac{5}{4} : \frac{5}{2}$.
Multiplying by the $LCM$ of the denominators $(12)$,we get $20 : 15 : 30$,which simplifies to $4 : 3 : 6$.

(A) The total energy of the system is conserved. The initial potential energy of the small piece of charge $q$ at the surface of the sphere (distance $R$ from the center) is $U_i = \frac{kQq}{R} + mgy_0$ (taking the reference level at the bottom of the sphere).
When the piece has fallen by a height $y$,its distance from the center of the sphere is $(R+y)$.
The final energy is $U_f + K_f = \frac{kQq}{R+y} + mg(y_0 - y) + \frac{1}{2}mv^2$.
Equating initial and final energy:
$\frac{kQq}{R} + mgy_0 = \frac{kQq}{R+y} + mg(y_0 - y) + \frac{1}{2}mv^2$
$\frac{1}{2}mv^2 = \frac{kQq}{R} - \frac{kQq}{R+y} + mgy$
$\frac{1}{2}mv^2 = kQq \left[ \frac{R+y-R}{R(R+y)} \right] + mgy$
$\frac{1}{2}mv^2 = \frac{kQqy}{R(R+y)} + mgy$
$v^2 = 2 \left[ \frac{kQqy}{mR(R+y)} + gy \right] = 2y \left[ \frac{kQq}{mR(R+y)} + g \right]$
Substituting $k = \frac{1}{4\pi\epsilon_0}$:
$v^2 = 2y \left[ \frac{qQ}{4\pi\epsilon_0 mR(R+y)} + g \right]$.

(A) photodiode is operated in reverse bias mode. When light of energy greater than the bandgap energy falls on the photodiode,electron-hole pairs are generated in the depletion region.
As the reverse biasing voltage is increased,the electric field in the depletion region increases,which helps in sweeping the generated charge carriers across the junction more effectively,thereby increasing the photocurrent.
However,once all the generated charge carriers are collected,further increase in the reverse bias voltage does not increase the photocurrent significantly,and it reaches a saturation value. This is shown in the $I-V$ characteristic curve of a photodiode.

(D) The lens formula is given by $\frac{1}{v} - \frac{1}{u} = \frac{1}{f}$.
For a concave lens,the focal length $f$ is negative,so let $f = -|f|$.
Thus,$\frac{1}{v} - \frac{1}{u} = -\frac{1}{|f|}$,which simplifies to $\frac{1}{v} = \frac{1}{u} - \frac{1}{|f|} = \frac{|f|-u}{u|f|}$.
Therefore,$v = \frac{u|f|}{|f|-u}$.
As $u \to 0$,$v \to 0$.
As $u \to |f|$,$v \to \infty$.
As $u \to \infty$,$v \to -|f|$.
Since the question uses magnitudes for $u$ and $v$ in the graphs,we consider the virtual image formed by a concave lens,where $v$ is always between $0$ and $f$. The correct graphical representation for the magnitude of image distance $v$ versus object distance $u$ is shown in option $D$.

(B) The power delivered is $P_{out} = 1 \, kW = 1000 \, W$ at a voltage $V = 220 \, V$.
The current $I$ flowing through the line is given by $I = \frac{P_{out}}{V} = \frac{1000}{220} = \frac{50}{11} \, A$.
The power loss in the transmission line due to its resistance $R = 2 \, \Omega$ is $P_{loss} = I^2 R$.
$P_{loss} = \left( \frac{50}{11} \right)^2 \times 2 = \frac{2500}{121} \times 2 = \frac{5000}{121} \approx 41.32 \, W$.
The total power generated at the source is $P_{in} = P_{out} + P_{loss} = 1000 + 41.32 = 1041.32 \, W$.
The efficiency $\eta$ is given by $\eta = \left( \frac{P_{out}}{P_{in}} \right) \times 100$.
$\eta = \left( \frac{1000}{1041.32} \right) \times 100 \approx 96.03 \%$.
Therefore, the efficiency is approximately $96 \%$.

(A) The magnetic field produced by the long wire at the center of the loop is $B = \frac{\mu_{0} I}{2 \pi b}$.
The magnetic moment of the square loop is $M = I \times \text{Area} = I \times (2a)^2 = 4a^2 I$.
The torque on the loop is given by $\tau = M \times B \times \sin(\theta)$.
Since the magnetic moment vector is along the $y$-axis and the magnetic field is along the $x$-axis,the angle between them is $90^{\circ}$.
Thus,$\tau = (4a^2 I) \times (\frac{\mu_{0} I}{2 \pi b}) \times \sin(90^{\circ}) = \frac{2 \mu_{0} I^{2} a^{2}}{\pi b}$.

(B) Let $i_{g}$ be the full-scale deflection current of the galvanometer.
For the first case,the range is $0-1\, V$ with resistance $R_{1}$ in series:
$1 = i_{g}(G + R_{1}) \quad \dots(1)$
For the second case,the range is $0-2\, V$ with an additional resistance $R_{2}$ in series:
$2 = i_{g}(G + R_{1} + R_{2}) \quad \dots(2)$
Dividing equation $(2)$ by equation $(1)$:
$\frac{2}{1} = \frac{i_{g}(G + R_{1} + R_{2})}{i_{g}(G + R_{1})}$
$2 = \frac{G + R_{1} + R_{2}}{G + R_{1}}$
$2(G + R_{1}) = G + R_{1} + R_{2}$
$2G + 2R_{1} = G + R_{1} + R_{2}$
$R_{2} = G + R_{1}$
Thus,the additional resistance required is $G + R_{1}$.

(B) The electric field is given by $\overrightarrow{E} = E_0 \sin(kx - \omega t) \hat{j}$,where $E_0 = 30 \, V/m$,$\omega = 1.5 \times 10^7 \, rad/s$,and $k = 5 \times 10^{-2} \, rad/m$.
The amplitude of the magnetic field $B_0$ is related to the electric field amplitude $E_0$ by $B_0 = \frac{E_0}{c}$.
$B_0 = \frac{30}{3 \times 10^8} = 10^{-7} \, T$.
The electron moves along the $y$-axis with velocity $\overrightarrow{v} = 0.1 c \hat{j} = 0.1 \times 3 \times 10^8 \hat{j} = 3 \times 10^7 \hat{j} \, m/s$.
The magnetic force is given by $\overrightarrow{F}_{mag} = q(\overrightarrow{v} \times \overrightarrow{B})$. Since $\overrightarrow{v}$ is along the $y$-axis and $\overrightarrow{E}$ is along the $y$-axis,the wave propagates along the $x$-axis. The magnetic field $\overrightarrow{B}$ must be perpendicular to both the direction of propagation ($x$-axis) and the electric field ($y$-axis),so $\overrightarrow{B}$ is along the $z$-axis.
$F_{max} = q v B_0 = (1.6 \times 10^{-19} \, C) \times (3 \times 10^7 \, m/s) \times (10^{-7} \, T)$.
$F_{max} = 1.6 \times 3 \times 10^{-19} = 4.8 \times 10^{-19} \, N$.

(D) Initial charges on the capacitors are:
$Q_1 = C \times V = CV$
$Q_2 = 2C \times 2V = 4CV$
Since they are connected with opposite polarities (positive to negative),the net charge available is:
$Q_{net} = Q_2 - Q_1 = 4CV - CV = 3CV$
When connected in parallel,the equivalent capacitance is:
$C_{eq} = C + 2C = 3C$
The common potential $V_c$ is given by:
$V_c = \frac{Q_{net}}{C_{eq}} = \frac{3CV}{3C} = V$
The final energy $U_f$ stored in the configuration is:
$U_f = \frac{1}{2} C_{eq} V_c^2 = \frac{1}{2} \times (3C) \times V^2 = 1.5 CV^2$

(B) The magnetic field $B$ at the center of coil $C_{2}$ due to current $I$ is given by $B = \frac{\mu_{0} N_{2} I}{2 R_{2}}$.
The magnetic flux $\phi$ linked with coil $C_{1}$ is $\phi = B \cdot A_{1} \cdot N_{1} = \left( \frac{\mu_{0} N_{2} I}{2 R_{2}} \right) (\pi r_{1}^{2}) N_{1}$.
Substituting the values: $N_{1} = 500$,$r_{1} = 0.01\; m$,$N_{2} = 200$,$R_{2} = 0.2\; m$,$\mu_{0} = 4\pi \times 10^{-7}\; T\cdot m/A$.
$\phi = \frac{(4\pi \times 10^{-7}) \times 200 \times (5t^{2} - 2t + 3)}{2 \times 0.2} \times \pi \times (0.01)^{2} \times 500$.
$\phi = \frac{4\pi^{2} \times 10^{-7} \times 200 \times 500 \times 10^{-4}}{0.4} \times (5t^{2} - 2t + 3) = (10\pi^{2} \times 10^{-6}) \times (5t^{2} - 2t + 3)$.
The induced $emf$ is $\varepsilon = -\frac{d\phi}{dt} = -10\pi^{2} \times 10^{-6} \times (10t - 2)$.
At $t = 1\; s$,$\varepsilon = -10\pi^{2} \times 10^{-6} \times (10(1) - 2) = -80\pi^{2} \times 10^{-6}\; V$.
Taking magnitude,$|\varepsilon| = 80\pi^{2} \times 10^{-6}\; V = 80\pi^{2} \times 10^{-3}\; mV \approx 80 \times 9.87 \times 10^{-3} \approx 0.789\; mV$.
Given $|\varepsilon| = \frac{4}{x} = 0.5\; mV$ (assuming $\pi^{2} \approx 10$),then $x = 8$.

(C) According to Bragg's law for diffraction, $2d \sin \theta = n\lambda$. For the first maximum, $n = 1$, so $2d \sin \theta = \lambda$.
Given $d = 1 \, Å = 10^{-10} \, m$ and $\theta = 60^{\circ}$.
Thus, $\lambda = 2 \times 10^{-10} \times \sin(60^{\circ}) = 2 \times 10^{-10} \times \frac{\sqrt{3}}{2} = \sqrt{3} \times 10^{-10} \, m$.
Using the de Broglie wavelength formula, $\lambda = \frac{h}{\sqrt{2mE}}$, we have $\sqrt{2mE} = \frac{h}{\lambda}$.
Squaring both sides, $2mE = \frac{h^2}{\lambda^2}$, so $E = \frac{h^2}{2m\lambda^2}$.
Substituting the values: $E = \frac{(6.64 \times 10^{-34})^2}{2 \times 9.1 \times 10^{-31} \times (\sqrt{3} \times 10^{-10})^2} = \frac{44.0896 \times 10^{-68}}{2 \times 9.1 \times 10^{-31} \times 3 \times 10^{-20}} = \frac{44.0896 \times 10^{-68}}{54.6 \times 10^{-51}} \approx 0.8075 \times 10^{-17} \, J$.
To convert to $eV$, divide by $1.6 \times 10^{-19} \, J/eV$:
$E = \frac{0.8075 \times 10^{-17}}{1.6 \times 10^{-19}} \approx 50.47 \, eV$.
Rounding to the nearest integer, $E \approx 50 \, eV$.

(D) Let $m = 200 \, MeV/c^2$ be the mass of the particle and $M = 1000 \, MeV/c^2$ be the mass of the hydrogen atom.
Let $v$ be the initial velocity of the particle and $V$ be the recoil velocity of the atom.
By conservation of linear momentum: $mv = MV = p$,where $p$ is the momentum.
The energy required to excite the hydrogen atom to its first excited state is $\Delta E = 13.6 \, eV \times (1 - 1/4) = 10.2 \, eV$.
By conservation of energy: $K_{initial} = K_{final} + \Delta E$.
$K_{initial} = \frac{p^2}{2m}$ and $K_{final} = \frac{p^2}{2M}$.
Thus,$\frac{p^2}{2m} - \frac{p^2}{2M} = 10.2 \, eV$.
$\frac{p^2}{2m} (1 - \frac{m}{M}) = 10.2 \, eV$.
Given $m/M = 200/1000 = 0.2$,we have $K_{initial} (1 - 0.2) = 10.2 \, eV$.
$K_{initial} (0.8) = 10.2 \, eV$.
$K_{initial} = \frac{10.2}{0.8} = 12.75 \, eV$.
We are given $K_{initial} = \frac{N}{4}$,so $\frac{N}{4} = 12.75$.
$N = 12.75 \times 4 = 51$.

(A) For minimum strain on the eye,the final image is formed at infinity.
For the eyepiece,the object must be placed at its focal length,so $u_e = -f_e = -5 \, cm$.
The separation between the lenses is $L = 10 \, cm$. The image formed by the objective lens $(v_o)$ acts as the object for the eyepiece.
Thus,$v_o = L - |u_e| = 10 \, cm - 5 \, cm = 5 \, cm$.
Using the lens formula for the objective lens: $\frac{1}{v_o} - \frac{1}{u_o} = \frac{1}{f_o}$.
Substituting the values: $\frac{1}{5} - \frac{1}{u_o} = \frac{1}{1} \Rightarrow -\frac{1}{u_o} = 1 - \frac{1}{5} = \frac{4}{5}$.
Therefore,$u_o = -\frac{5}{4} \, cm$. The distance is $|u_o| = \frac{5}{4} \, cm = \frac{50}{40} \, cm$.
Comparing this with $\frac{n}{40} \, cm$,we get $n = 50$.

(B) For an electromagnetic wave,the magnitude of the electric field $E_0$ is related to the magnetic field $B_0$ by $E_0 = c B_0$.
Given $c = 3 \times 10^{8} \text{ m/s}$ and $B_0 = 1.2 \times 10^{-7} \text{ T}$,we have $E_0 = 3 \times 10^{8} \times 1.2 \times 10^{-7} = 36 \text{ V/m}$.
The wave propagates in the direction of $-x$ because the argument of the sine function is $(kx + \omega t)$.
The magnetic field is along the $\hat{k}$ direction ($z$-axis).
Since the direction of propagation is $\vec{E} \times \vec{B}$,and the propagation is along $-\hat{i}$,we have $\hat{j} \times \hat{k} = \hat{i}$. To get $-\hat{i}$,we need $(-\hat{j}) \times \hat{k} = -\hat{i}$.
Thus,the electric field is $\overrightarrow{ E }( x , t ) = [-36 \sin (0.5 \times 10^{3} x + 1.5 \times 10^{11} t) \hat{ j }] \text{ V/m}$.

(D) For a double convex lens, the radii of curvature are $R_1 = R$ and $R_2 = -R$. Using the lens maker's formula, the power $P$ is given by:
$P = (\mu - 1) \left( \frac{1}{R_1} - \frac{1}{R_2} \right) = (\mu - 1) \left( \frac{1}{R} - \frac{1}{-R} \right) = (\mu - 1) \left( \frac{2}{R} \right) \quad ...(i)$
For a plano-convex lens, the radii of curvature are $R_1' = R'$ and $R_2' = \infty$. The power $P'$ is given by:
$P' = (\mu - 1) \left( \frac{1}{R'} - \frac{1}{\infty} \right) = (\mu - 1) \left( \frac{1}{R'} \right) \quad ...(ii)$
Given that $P' = 1.5P = \frac{3}{2}P$, we substitute the expressions from $(i)$ and $(ii)$:
$(\mu - 1) \left( \frac{1}{R'} \right) = \frac{3}{2} \left[ (\mu - 1) \left( \frac{2}{R} \right) \right]$
$\frac{1}{R'} = \frac{3}{R}$
Therefore, $R' = \frac{R}{3} \approx 0.33R$.

(A) To verify $Ohm's$ law,we need to measure the current flowing through the resistor and the potential difference across it.
$1$. Ammeter: An ammeter is designed to measure the current flowing through a circuit. To ensure the same current passes through the ammeter as the circuit component,it must be connected in series with the resistor.
$2$. Voltmeter: $A$ voltmeter is designed to measure the potential difference (voltage) between two points. To measure the voltage across a specific resistor,it must be connected in parallel with that resistor.
Therefore,the correct configuration is that the ammeter is connected in series and the voltmeter is connected in parallel.

(A) According to Gauss's Law,for a uniformly charged spherical shell of radius $R$ and total charge $Q$:
$1$. For points outside the shell $(r > R)$,the shell behaves as if all its charge is concentrated at the centre. Thus,the electric field is $E = \frac{1}{4 \pi \varepsilon_{0}} \frac{Q}{r^{2}}$. The force on a charge $q$ is $F = qE = \frac{1}{4 \pi \varepsilon_{0}} \frac{Qq}{r^{2}}$.
$2$. For points inside the shell $(r < R)$,the electric field due to the shell is zero $(E = 0)$. Therefore,the force on a charge $q$ placed inside is $F = qE = 0$.
Comparing these results with the given options,statement $A$ is correct.

(A) For a process to be allowed by energy conservation,the mass of the reactants must be greater than or equal to the mass of the products (considering the energy equivalent of mass).
Let us evaluate $A$: $n + p \rightarrow d + \gamma$.
Mass of reactants: $m_{n} + m_{p} = 1.0087 + 1.0072 = 2.0159 \ u$.
Mass of product: $m_{d} = 2.0141 \ u$.
Since $2.0159 \ u > 2.0141 \ u$,the mass difference is released as energy (photon $\gamma$),satisfying conservation laws.
Evaluating $B$: $e^{+} + e^{-} \rightarrow \gamma$ violates momentum conservation (a single photon cannot conserve both energy and momentum in the center-of-mass frame).
Evaluating $D$: $p \rightarrow n + e^{+} + \bar{v}$ is not possible for a free proton because $m_{p} < m_{n} + m_{e}$.
Thus,the correct process is $A$.

(C) The potential energy of two dipoles $\overrightarrow{p}_{1}$ and $\overrightarrow{p}_{2}$ separated by distance $r$ along the $x$-axis is given by $U = \frac{1}{4\pi\varepsilon_{0}} \left[ \frac{\overrightarrow{p}_{1} \cdot \overrightarrow{p}_{2} - 3(\overrightarrow{p}_{1} \cdot \hat{r})(\overrightarrow{p}_{2} \cdot \hat{r})}{r^{3}} \right]$.
Here,$\overrightarrow{p}_{1} = p\hat{i}$,$\overrightarrow{p}_{2} = -p\hat{i}$,and $\hat{r} = \hat{i}$.
So,$U = \frac{1}{4\pi\varepsilon_{0}} \left[ \frac{(p\hat{i}) \cdot (-p\hat{i}) - 3(p\hat{i} \cdot \hat{i})(-p\hat{i} \cdot \hat{i})}{a^{3}} \right] = \frac{1}{4\pi\varepsilon_{0}} \left[ \frac{-p^{2} - 3(p)(-p)}{a^{3}} \right] = \frac{1}{4\pi\varepsilon_{0}} \left[ \frac{2p^{2}}{a^{3}} \right] = \frac{p^{2}}{2\pi\varepsilon_{0}a^{3}}$.
By conservation of energy,$KE_{i} + PE_{i} = KE_{f} + PE_{f}$.
Initially,$KE_{i} = 0$ and $PE_{i} = \frac{p^{2}}{2\pi\varepsilon_{0}a^{3}}$.
Finally,at infinite separation,$PE_{f} = 0$ and $KE_{f} = 2 \times (\frac{1}{2}mv^{2}) = mv^{2}$.
Thus,$0 + \frac{p^{2}}{2\pi\varepsilon_{0}a^{3}} = mv^{2} + 0$.
$v^{2} = \frac{p^{2}}{2\pi\varepsilon_{0}ma^{3}}$.
$v = \frac{p}{a} \sqrt{\frac{1}{2\pi\varepsilon_{0}ma}}$.

(C) To find the current in the $10 \, V$ battery,we can simplify the circuit using Thevenin's theorem or nodal analysis. Let the node between the $5 \, \Omega$ and $10 \, \Omega$ resistors be $A$ and the bottom node be $B$.
Applying Kirchhoff's Current Law at the node above the $10 \, \Omega$ resistor,let the potential be $V$.
$\frac{V - 20}{5 + 2} + \frac{V}{10} + \frac{V - 10}{4} = 0$
$\frac{V - 20}{7} + \frac{V}{10} + \frac{V - 10}{4} = 0$
Multiplying by $140$ ($LCM$ of $7, 10, 4$):
$20(V - 20) + 14V + 35(V - 10) = 0$
$20V - 400 + 14V + 35V - 350 = 0$
$69V = 750$
$V = \frac{750}{69} \approx 10.87 \, V$
The current $I$ through the $10 \, V$ battery is $I = \frac{V - 10}{4} = \frac{10.87 - 10}{4} = \frac{0.87}{4} = 0.2175 \, A$.
Since the potential $V$ is greater than $10 \, V$,the current flows from the positive terminal to the negative terminal.

(B) The $r.m.s.$ velocity of a gas molecule is given by $v_{rms} = \sqrt{\frac{3kT}{m}}$.
The de$-$Broglie wavelength is given by $\lambda = \frac{h}{mv}$.
Substituting $v_{rms}$ for $v$, we get $\lambda = \frac{h}{m \sqrt{\frac{3kT}{m}}} = \frac{h}{\sqrt{3kTm}}$.
Given values: $h = 6.63 \times 10^{-34} \ J \cdot s$, $k = 1.38 \times 10^{-23} \ J/K$, $T = 400 \ K$, and $m = 4.64 \times 10^{-26} \ kg$.
Calculating the denominator: $\sqrt{3 \times 1.38 \times 10^{-23} \times 400 \times 4.64 \times 10^{-26}} = \sqrt{7.68768 \times 10^{-45}} \approx 2.77 \times 10^{-22} \ kg \cdot m/s$.
Now, $\lambda = \frac{6.63 \times 10^{-34}}{2.77 \times 10^{-22}} \approx 2.39 \times 10^{-11} \ m$.
Converting to $\mathring{A}$s: $2.39 \times 10^{-11} \ m = 0.239 \ \mathring{A} \approx 0.24 \ \mathring{A}$.

(D) The magnetic moment $M$ of a current loop is given by $M = iA$,where $i$ is the current and $A$ is the area of the loop.
For a particle of charge $q$ moving in a circle of radius $r$ with speed $v$,the time period $T$ is $T = \frac{2 \pi r}{v}$.
The equivalent current is $i = \frac{q}{T} = \frac{qv}{2 \pi r}$.
The area of the loop is $A = \pi r^2$.
Thus,the magnitude of the magnetic moment is $M = iA = \left( \frac{qv}{2 \pi r} \right) \times (\pi r^2) = \frac{qvr}{2}$.
For a charged particle moving in a magnetic field,the radius of the circular path is $r = \frac{mv}{qB}$.
Substituting $r$ into the expression for $M$,we get $M = \frac{qv}{2} \times \left( \frac{mv}{qB} \right) = \frac{mv^2}{2B}$.
From the right-hand rule,the direction of the magnetic moment is opposite to the direction of the magnetic field for a positively charged particle moving in a circle. Therefore,in vector form,$\overrightarrow{M} = -\frac{mv^2}{2B} \hat{B}$.
Since $\hat{B} = \frac{\vec{B}}{B}$,we have $\overrightarrow{M} = -\frac{mv^2}{2B} \left( \frac{\vec{B}}{B} \right) = -\frac{mv^2 \vec{B}}{2B^2}$.

(A) The magnetic field $B$ produced by the long wire at a distance $r$ is given by $B = \frac{\mu_{0} I}{2 \pi r}$.
For the two sides of the loop parallel to the $z$-axis,the distance from the wire is $r = \sqrt{b^2 + a^2}$.
The force on each of these sides is $F = B I (2a) = \frac{\mu_{0} I}{2 \pi \sqrt{b^2 + a^2}} \cdot I \cdot 2a = \frac{\mu_{0} I^2 a}{\pi \sqrt{b^2 + a^2}}$.
The torque $\tau$ about the $z$-axis is produced by the vertical components of these forces. The component of force perpendicular to the lever arm is $F \cos \theta$,where $\cos \theta = \frac{b}{\sqrt{b^2 + a^2}}$.
The total torque is $\tau = 2 \cdot (F \cos \theta) \cdot a = 2 \cdot \left( \frac{\mu_{0} I^2 a}{\pi \sqrt{b^2 + a^2}} \right) \cdot \left( \frac{b}{\sqrt{b^2 + a^2}} \right) \cdot a$.
Simplifying this,we get $\tau = \frac{2 \mu_{0} I^2 a^2 b}{\pi (a^2 + b^2)}$.

(C) The $AC$ current gain $\beta_{ac}$ is defined as the ratio of the change in collector current $\Delta I_{C}$ to the change in base current $\Delta I_{B}$ at a constant collector-emitter voltage $V_{CE}$.
From the graph,at $V_{CE} = 10\, V$,the collector current $I_{C} = 4.0\, mA$ corresponds to a base current $I_{B} = 30\, \mu A$.
To find $\beta_{ac}$ around this operating point,we consider the adjacent curves for $I_{B} = 20\, \mu A$ and $I_{B} = 40\, \mu A$.
At $V_{CE} = 10\, V$:
For $I_{B} = 20\, \mu A$,$I_{C} = 3.0\, mA$.
For $I_{B} = 40\, \mu A$,$I_{C} = 6.0\, mA$.
Therefore,the change in base current is $\Delta I_{B} = 40\, \mu A - 20\, \mu A = 20\, \mu A = 20 \times 10^{-6}\, A$.
The change in collector current is $\Delta I_{C} = 6.0\, mA - 3.0\, mA = 3.0\, mA = 3.0 \times 10^{-3}\, A$.
The $AC$ current gain is $\beta_{ac} = \frac{\Delta I_{C}}{\Delta I_{B}} = \frac{3.0 \times 10^{-3}}{20 \times 10^{-6}} = \frac{3000}{20} = 150$.

(D) Given: $P = 400 \ W$,$V_{rms} = 250 \ V$,$f = 50 \ Hz$,$\cos \phi = 0.8$.
$1$. Calculate impedance $Z$:
$P = V_{rms} I_{rms} \cos \phi = \frac{V_{rms}^2}{Z} \cos \phi$
$400 = \frac{250^2}{Z} \times 0.8$
$Z = \frac{62500 \times 0.8}{400} = \frac{50000}{400} = 125 \ \Omega$.
$2$. Calculate resistance $R$ and inductive reactance $X_L$:
$R = Z \cos \phi = 125 \times 0.8 = 100 \ \Omega$.
$X_L = Z \sin \phi = 125 \times \sqrt{1 - 0.8^2} = 125 \times 0.6 = 75 \ \Omega$.
$3$. For unity power factor,the circuit must be in resonance,so $X_C = X_L = 75 \ \Omega$.
$X_C = \frac{1}{2 \pi f C} = 75$
$C = \frac{1}{2 \pi \times 50 \times 75} = \frac{1}{7500 \pi} \ F = \frac{10^6}{7500 \pi} \ \mu F = \frac{10000}{75 \pi} \ \mu F = \frac{400}{3 \pi} \ \mu F$.
Comparing with $C = (\frac{n}{3 \pi}) \ \mu F$,we get $n = 400$.

(A) The intensity at any point in a Young's double-slit experiment is given by $I = I_1 + I_2 + 2\sqrt{I_1 I_2} \cos \phi$,where $\phi$ is the phase difference.
Given that the intensity at a path difference of $\lambda$ is $K$,and since the path difference $\lambda$ corresponds to a phase difference of $2\pi$,the intensity is $I_{max} = 4I_0 = K$,where $I_0$ is the intensity of each slit.
Thus,$I_0 = K/4$.
For a path difference $\Delta x = \frac{\lambda}{6}$,the phase difference $\phi$ is $\frac{2\pi}{\lambda} \times \Delta x = \frac{2\pi}{\lambda} \times \frac{\lambda}{6} = \frac{\pi}{3}$.
The intensity at this point is $I = I_0 + I_0 + 2\sqrt{I_0 I_0} \cos(\frac{\pi}{3}) = 2I_0 + 2I_0 \cos(\frac{\pi}{3})$.
Substituting $I_0 = K/4$ and $\cos(\frac{\pi}{3}) = 1/2$,we get $I = 2(K/4) + 2(K/4)(1/2) = K/2 + K/4 = 3K/4$.
We are given $I = \frac{nK}{12}$,so $\frac{3K}{4} = \frac{nK}{12}$.
Solving for $n$,we get $n = \frac{3 \times 12}{4} = 9$.

(A) The time constant of the $RC$ circuit is $\tau = RC = (1 \times 10^3 \ \Omega) \times (10 \times 10^{-9} \ \text{F}) = 10 \times 10^{-6} \ \text{s} = 10 \ \mu\text{s}$.
During the first $5 \ \mu\text{s}$ $(0 \le t \le 5 \ \mu\text{s})$,the capacitor charges from $0 \ \text{V}$ towards $5 \ \text{V}$ according to $V_0(t) = 5(1 - e^{-t/\tau})$.
At $t = 5 \ \mu\text{s}$,$V_0(5 \ \mu\text{s}) = 5(1 - e^{-5/10}) = 5(1 - e^{-0.5}) \approx 5(1 - 0.6065) = 5(0.3935) \approx 1.9675 \ \text{V} \approx 2 \ \text{V}$.
During the next $5 \ \mu\text{s}$ $(5 \ \mu\text{s} \le t \le 10 \ \mu\text{s})$,the input is $0 \ \text{V}$,so the capacitor discharges from $2 \ \text{V}$ towards $0 \ \text{V}$ according to $V_0(t) = 2e^{-(t-5)/\tau}$.
At $t = 10 \ \mu\text{s}$,$V_0(10 \ \mu\text{s}) = 2e^{-(10-5)/10} = 2e^{-0.5} \approx 2(0.6065) \approx 1.213 \ \text{V}$.
During the next $5 \ \mu\text{s}$ $(10 \ \mu\text{s} \le t \le 15 \ \mu\text{s})$,the capacitor charges from $1.213 \ \text{V}$ towards $5 \ \text{V}$ according to $V_0(t) = 5 - (5 - 1.213)e^{-(t-10)/\tau} = 5 - 3.787e^{-(t-10)/\tau}$.
At $t = 15 \ \mu\text{s}$,$V_0(15 \ \mu\text{s}) = 5 - 3.787e^{-0.5} \approx 5 - 3.787(0.6065) \approx 5 - 2.297 = 2.703 \ \text{V} \approx 2.7 \ \text{V}$.
Comparing this behavior with the given options,Option $A$ correctly shows the charging and discharging cycles with the calculated values.

(B) Let $E_{1}$ be the electric field at $O$ due to charge $Q_{1}$ at $A$,and $E_{2}$ be the electric field at $O$ due to charge $Q_{2}$ at $B$.
$E_{1} = \frac{k|Q_{1}|}{x_{1}^{2}}$ along $OA$ (assuming $Q_{1}$ is positive).
$E_{2} = \frac{k|Q_{2}|}{x_{2}^{2}}$ along $OB$ (assuming $Q_{2}$ is positive).
Let $\angle OAB = \alpha$. Then $\angle OBA = 90^{\circ} - \alpha$.
The resultant electric field $E_{net}$ is perpendicular to the hypotenuse $AB$.
From the geometry of the triangle,the angle that the resultant field $E_{net}$ makes with $E_{1}$ is $\alpha$,and with $E_{2}$ is $90^{\circ} - \alpha$.
Using the vector addition property,$\tan(\alpha) = \frac{E_{2}}{E_{1}}$.
From the triangle $OAB$,$\tan(\alpha) = \frac{OB}{OA} = \frac{x_{2}}{x_{1}}$.
Equating the two expressions for $\tan(\alpha)$:
$\frac{E_{2}}{E_{1}} = \frac{x_{2}}{x_{1}}$
$\frac{k|Q_{2}| / x_{2}^{2}}{k|Q_{1}| / x_{1}^{2}} = \frac{x_{2}}{x_{1}}$
$\frac{|Q_{2}| x_{1}^{2}}{|Q_{1}| x_{2}^{2}} = \frac{x_{2}}{x_{1}}$
$\frac{|Q_{1}|}{|Q_{2}|} = \frac{x_{1}^{2}}{x_{2}^{2}} \cdot \frac{x_{1}}{x_{2}} = \frac{x_{1}^{3}}{x_{2}^{3}}$
Thus,$Q_{1} / Q_{2}$ is proportional to $\frac{x_{1}^{3}}{x_{2}^{3}}$.

(B) When a charged particle enters a uniform magnetic field perpendicular to its velocity,it follows a circular path.
The radius $R$ of this circular path is given by $R = \frac{mv}{qB_{0}}$.
For the particle not to hit the screen placed at a distance $d$ from its initial position,the radius of the circular path must be less than or equal to $d$.
Therefore,$R \leq d$.
Substituting the expression for $R$,we get $\frac{mv}{qB_{0}} \leq d$.
Solving for $v$,we get $v \leq \frac{q B_{0} d}{m}$.
However,the question asks for the minimum value of $v$ such that the particle does not hit the screen. If the particle's path is tangent to the screen,it will just graze it without hitting it. This occurs when $R = d$.
Thus,$v = \frac{q B_{0} d}{m}$ is the required value.

(D) The de-Broglie wavelength $\lambda$ is given by the formula: $\lambda = \frac{h}{p} = \frac{h}{\sqrt{2m(KE)}}$.
Since the kinetic energy $(KE)$ and Planck's constant $(h)$ are the same for all particles,we have $\lambda \propto \frac{1}{\sqrt{m}}$.
Comparing the masses of the particles: $m_{He^{++}} > m_{P} > m_{e}$.
Since $\lambda$ is inversely proportional to the square root of mass,the particle with the smallest mass will have the largest wavelength.
Therefore,the relation is: $\lambda_{e} > \lambda_{P} > \lambda_{He^{++}}$.