JEE Main 2023 Physics Question Paper with Answer and Solution

(A) According to the first law of thermodynamics,$\Delta Q = \Delta U + \Delta W$.
Therefore,the change in internal energy is $\Delta U = \Delta Q - \Delta W$.
Here,$\Delta Q = mL_v = (1\,kg) \times (2257\,kJ/kg) = 2257\,kJ = 2257 \times 10^3\,J$.
The work done by the system during expansion is $\Delta W = P \Delta V = P(V_f - V_i)$.
$\Delta W = (1 \times 10^5\,Pa) \times (1.671\,m^3 - 1.00 \times 10^{-3}\,m^3) = 10^5 \times (1.671 - 0.001) = 10^5 \times 1.670 = 167000\,J = 167\,kJ$.
Now,$\Delta U = 2257\,kJ - 167\,kJ = 2090\,kJ$.

(D) For a particle executing simple harmonic motion $(SHM)$,the kinetic energy $(KE)$ as a function of displacement $(x)$ is given by the formula:
$KE = \frac{1}{2} m \omega^2 (A^2 - x^2)$
where $m$ is the mass of the particle,$\omega$ is the angular frequency,and $A$ is the amplitude.
$1$. At the mean position $(x = 0)$,the kinetic energy is maximum: $KE_{max} = \frac{1}{2} m \omega^2 A^2$.
$2$. At the extreme position ($x = A$ or $x = -A$),the kinetic energy is zero: $KE = 0$.
$3$. The equation $KE = \frac{1}{2} m \omega^2 A^2 - \frac{1}{2} m \omega^2 x^2$ represents a downward-opening parabola with respect to $x$.
Comparing this with the given options,the graph that shows maximum $KE$ at $x = 0$ and zero $KE$ at $x = A$ with a parabolic shape is represented by option $(D)$.

(D) The relationship between any linear temperature scale and the Fahrenheit scale is given by the formula:
$\frac{X - X_{\text{freezing}}}{X_{\text{boiling}} - X_{\text{freezing}}} = \frac{F - 32}{212 - 32}$
Given:
$X_{\text{boiling}} = 65^{\circ} X$
$X_{\text{freezing}} = -15^{\circ} X$
$X = -95^{\circ} X$
Substituting the values into the formula:
$\frac{-95 - (-15)}{65 - (-15)} = \frac{F - 32}{180}$
$\frac{-95 + 15}{65 + 15} = \frac{F - 32}{180}$
$\frac{-80}{80} = \frac{F - 32}{180}$
$-1 = \frac{F - 32}{180}$
$-180 = F - 32$
$F = -180 + 32 = -148^{\circ} F$

(B) The values of the given units in meters are:
$1 \text{ AU} = 1.496 \times 10^{11} \text{ m}$
$1 \text{ ly} = 9.46 \times 10^{15} \text{ m}$
$1 \text{ pc} = 3.08 \times 10^{16} \text{ m}$
Comparing these values,we get $1 \text{ AU} < 1 \text{ ly} < 1 \text{ pc}$.
Statement $I$ is correct as these are units of distance.
Statement $II$ is incorrect because the correct order is $AU < ly < pc$.

(B) The root mean square speed of a gas is given by the formula $v_{rms} = \sqrt{\frac{3RT}{M}}$,where $R$ is the universal gas constant,$T$ is the absolute temperature,and $M$ is the molar mass of the gas.
Since $T$ is the same for all three vessels,$v_{rms} \propto \frac{1}{\sqrt{M}}$.
The molar masses are:
Neon (monoatomic): $M_1 \approx 20 \text{ g/mol}$.
Chlorine (diatomic): $M_2 \approx 71 \text{ g/mol}$.
Uranium hexafluoride (polyatomic): $M_3 \approx 352 \text{ g/mol}$.
Since $M_1 < M_2 < M_3$,it follows that $\frac{1}{\sqrt{M_1}} > \frac{1}{\sqrt{M_2}} > \frac{1}{\sqrt{M_3}}$.
Therefore,$v_{rms}(\text{mono}) > v_{rms}(\text{dia}) > v_{rms}(\text{poly})$.

(B) The force exerted by the machine gun is equal to the rate of change of momentum of the bullets.
$F = \frac{dp}{dt} = n \cdot m \cdot v$
where $n$ is the number of bullets fired per second,$m$ is the mass of each bullet,and $v$ is the velocity of the bullets.
Given:
Force $F = 125\,N$
Mass $m = 10\,g = 10 \times 10^{-3}\,kg = 0.01\,kg$
Velocity $v = 250\,m/s$
Substituting the values into the formula:
$125 = n \times 0.01 \times 250$
$125 = n \times 2.5$
$n = \frac{125}{2.5} = 50$
Therefore,the number of bullets fired per second is $50$.

(B) The standard wave equation is $Y = A \sin (\omega t - kx + \phi)$.
Comparing the given equation $Y = 10^{-2} \sin 2 \pi (160 t - 0.5 x + \frac{\pi}{4})$ with the standard form,we have:
$\omega = 2 \pi \times 160 \, rad/s$
$k = 2 \pi \times 0.5 \, m^{-1}$
The speed of the wave $v$ is given by $v = \frac{\omega}{k} = \frac{2 \pi \times 160}{2 \pi \times 0.5} = \frac{160}{0.5} = 320 \, m/s$.
To convert the speed from $m/s$ to $km/h$,we multiply by $\frac{18}{5}$:
$v = 320 \times \frac{18}{5} = 64 \times 18 = 1152 \, km/h$.

(B) The work done by a variable force is given by the integral of the force with respect to displacement: $W = \int_{x_1}^{x_2} F(x) dx$.
Given $F(x) = 2 + 3x$,$x_1 = 0$,and $x_2 = 4$.
$W = \int_{0}^{4} (2 + 3x) dx$.
Integrating the expression: $W = [2x + \frac{3x^2}{2}]_{0}^{4}$.
Substituting the limits: $W = (2(4) + \frac{3(4)^2}{2}) - (2(0) + \frac{3(0)^2}{2})$.
$W = (8 + \frac{3 \times 16}{2}) - 0$.
$W = 8 + 3 \times 8 = 8 + 24 = 32 \, J$.

(B) The moment of inertia of a solid sphere about its diameter is $I_d = \frac{2}{5}mR^2$.
The angular momentum of the sphere about its diameter is $L = I_d \omega = \frac{2}{5}mR^2 \omega$.
The moment of inertia of the sphere about its tangent is given by the parallel axis theorem: $I_t = I_{cm} + mR^2 = \frac{2}{5}mR^2 + mR^2 = \frac{7}{5}mR^2$.
According to the problem,$I_t = (x \times 10^{-2}) \times L$.
Substituting the expressions: $\frac{7}{5}mR^2 = (x \times 10^{-2}) \times (\frac{2}{5}mR^2 \omega)$.
Canceling $\frac{1}{5}mR^2$ from both sides: $7 = (x \times 10^{-2}) \times 2 \omega$.
Given $\omega = 10\,rad\,s^{-1}$,we have: $7 = (x \times 10^{-2}) \times 2 \times 10$.
$7 = x \times 10^{-2} \times 20$.
$7 = x \times 0.2$.
$x = \frac{7}{0.2} = 35$.

(B) Let the natural length of the wire be $\ell_0$ and the force constant be $K$.
According to Hooke's Law,$T = K(\ell - \ell_0)$.
For the first case: $100 = K(l_1 - \ell_0)$ --- $(1)$
For the second case: $120 = K(l_2 - \ell_0)$ --- $(2)$
Dividing equation $(1)$ by $(2)$:
$\frac{100}{120} = \frac{l_1 - \ell_0}{l_2 - \ell_0} \Rightarrow \frac{5}{6} = \frac{l_1 - \ell_0}{l_2 - \ell_0}$
$5(l_2 - \ell_0) = 6(l_1 - \ell_0)$
$5l_2 - 5\ell_0 = 6l_1 - 6\ell_0$
$\ell_0 = 6l_1 - 5l_2$
Given that $10l_2 = 11l_1$,so $l_2 = \frac{11}{10}l_1$.
Substituting $l_2$ in the equation for $\ell_0$:
$\ell_0 = 6l_1 - 5(\frac{11}{10}l_1)$
$\ell_0 = 6l_1 - \frac{11}{2}l_1$
$\ell_0 = \frac{12l_1 - 11l_1}{2} = \frac{1}{2}l_1$
Comparing this with $\frac{1}{x}l_1$,we get $x = 2$.

(B) For a projectile,the time of flight $T$ is the total time taken to return to the ground. If a projectile is at the same height $h$ at two different times $t_1$ and $t_2$,then the total time of flight $T$ is given by $T = t_1 + t_2$.
Given $t_1 = 3 \, s$ and $t_2 = 5 \, s$,the total time of flight is $T = 3 + 5 = 8 \, s$.
The formula for the time of flight is $T = \frac{2 u \sin \theta}{g}$,where $u$ is the initial speed,$\theta$ is the angle of projection,and $g$ is the acceleration due to gravity.
Substituting the given values: $8 = \frac{2 u \sin 30^{\circ}}{10}$.
Since $\sin 30^{\circ} = 0.5$,we have $8 = \frac{2 u (0.5)}{10}$.
$8 = \frac{u}{10}$.
Therefore,$u = 80 \, m \, s^{-1}$.

(B) For a small drop falling through a viscous medium at terminal velocity,the terminal velocity $v$ is given by $v = \frac{2}{9} \frac{r^2 (\rho - \sigma) g}{\eta}$,which implies $v \propto r^2$.
Let $r$ be the radius of each small drop and $R$ be the radius of the large drop formed by coalescing $8$ small drops.
Since the volume is conserved,$8 \times (\frac{4}{3} \pi r^3) = \frac{4}{3} \pi R^3$,which gives $R^3 = 8r^3$,so $R = 2r$.
Using the proportionality $v \propto r^2$,we have $\frac{v_1}{v_2} = (\frac{r}{R})^2$.
Substituting the values,$\frac{10}{v_2} = (\frac{r}{2r})^2 = (\frac{1}{2})^2 = \frac{1}{4}$.
Therefore,$v_2 = 10 \times 4 = 40\,cm/s$.

(B) According to the Doppler effect,the observed frequency $f$ is given by the formula:
$f = f_0 \left( \frac{c + v_o}{c + v_s} \right)$
Where:
$f_0 = 400\,Hz$ (source frequency)
$c = 360\,ms^{-1}$ (velocity of sound)
$v_o = 40\,ms^{-1}$ (velocity of the observer,car $Q$,moving towards the source)
$v_s = 20\,ms^{-1}$ (velocity of the source,car $P$,moving away from the observer)
Substituting the values:
$f = 400 \left( \frac{360 + 40}{360 + 20} \right)$
$f = 400 \left( \frac{400}{380} \right)$
$f = 400 \times 1.0526$
$f \approx 421.05\,Hz$
Thus,the frequency heard by the passenger is approximately $421\,Hz$.

(B) The dimensional formula for density $(\rho)$ is $[M L^{-3}]$.
Let the dimensional formula be expressed as $[\rho] = [F]^a [V]^b [T]^c$.
Substituting the dimensions: $[M L^{-3}] = [M L T^{-2}]^a [L T^{-1}]^b [T]^c$.
$[M L^{-3}] = [M^a L^{a+b} T^{-2a-b+c}]$.
Comparing the powers of $M$,$L$,and $T$ on both sides:
For $M$: $a = 1$.
For $L$: $a + b = -3 \Rightarrow 1 + b = -3 \Rightarrow b = -4$.
For $T$: $-2a - b + c = 0 \Rightarrow -2(1) - (-4) + c = 0 \Rightarrow -2 + 4 + c = 0 \Rightarrow 2 + c = 0 \Rightarrow c = -2$.
Thus,the dimensional formula is $[F^1 V^{-4} T^{-2}]$.

(A) The gravitational potential $V$ inside a uniform solid sphere at a distance $r$ from the center is given by $V(r) = \frac{GM}{2R^3}(3R^2 - r^2)$.
At the surface,$r = R$,so $V_{surface} = \frac{GM}{2R^3}(3R^2 - R^2) = \frac{GM}{2R^3}(2R^2) = \frac{GM}{R} = V$.
At the center,$r = 0$,so $V_{center} = \frac{GM}{2R^3}(3R^2 - 0) = \frac{3GM}{2R}$.
Substituting $V = \frac{GM}{R}$,we get $V_{center} = \frac{3}{2} V$.

(B) Given mass $m = 500\,g = 0.5\,kg$.
The velocity is given by $v = 10\sqrt{x}$.
Squaring both sides,we get $v^2 = 100x$.
Differentiating both sides with respect to $x$,we get $2v \frac{dv}{dx} = 100$.
Since acceleration $a = v \frac{dv}{dx}$,we can write $2a = 100$,which gives $a = 50\,m/s^2$.
The force acting on the body is $F = ma$.
Substituting the values,$F = 0.5\,kg \times 50\,m/s^2 = 25\,N$.

(A) The initial velocity components are:
$u_x = u \cos 30^{\circ} = 40 \times \frac{\sqrt{3}}{2} = 20\sqrt{3}\,m/s$
$u_y = u \sin 30^{\circ} = 40 \times \frac{1}{2} = 20\,m/s$
At time $t = 2\,s$,the horizontal component of velocity remains constant:
$v_x = u_x = 20\sqrt{3}\,m/s$
The vertical component of velocity is given by $v_y = u_y - gt$:
$v_y = 20 - (10 \times 2) = 20 - 20 = 0\,m/s$
Since the vertical component is zero,the resultant velocity is $v = \sqrt{v_x^2 + v_y^2} = \sqrt{(20\sqrt{3})^2 + 0^2} = 20\sqrt{3}\,m/s$.

(C) The orbital velocity of a satellite close to the Earth's surface is given by $v_{\text{orbit}} = \sqrt{gR}$.
Substituting the values: $v_{\text{orbit}} = \sqrt{10 \times 6.4 \times 10^6} = \sqrt{64 \times 10^6} = 8000 \, m/s = 8 \, km/s$.
The escape velocity from the Earth's surface is given by $v_{\text{escape}} = \sqrt{2gR} = \sqrt{2} \times v_{\text{orbit}}$.
Substituting the value: $v_{\text{escape}} = 8\sqrt{2} \, km/s$.
The additional velocity required to overcome the gravitational pull is $\Delta v = v_{\text{escape}} - v_{\text{orbit}}$.
$\Delta v = 8\sqrt{2} - 8 = 8(\sqrt{2}-1) \, km/s$.

(B) The internal energy $(U)$ of an ideal gas is a function of temperature $(T)$ only,given by $U = f(T)$.
If the internal energy remains constant,then the change in internal energy $\Delta U = 0$.
Since $\Delta U = nC_v\Delta T$,this implies that $\Delta T = 0$,meaning the temperature remains constant.
$A$ process in which the temperature of the system remains constant is called an isothermal process.

(A) The root mean square speed $(V_{rms})$ of gas molecules is given by the formula: $V_{rms} = \sqrt{\frac{3 k_{B} T}{m}}$.
Given:
Temperature $T = 27^{\circ} C = 27 + 273 = 300 \, K$.
Mass of a nitrogen molecule $m = 4.6 \times 10^{-26} \, kg$.
Boltzmann constant $k_{B} = 1.4 \times 10^{-23} \, J K^{-1}$.
Substituting the values into the formula:
$V_{rms} = \sqrt{\frac{3 \times 1.4 \times 10^{-23} \times 300}{4.6 \times 10^{-26}}}$
$V_{rms} = \sqrt{\frac{1260 \times 10^{-23}}{4.6 \times 10^{-26}}}$
$V_{rms} = \sqrt{273.91 \times 10^{3}} \approx \sqrt{273910} \approx 523.36 \, m/s$.
Thus,the approximate speed is $523 \, m/s$.

(B) Given that $\vec{B} - \overrightarrow{A} = 2\hat{j}$.
Substituting the value of $\overrightarrow{A} = 2\hat{i} + 3\hat{j} + 2\hat{k}$ into the equation:
$\vec{B} - (2\hat{i} + 3\hat{j} + 2\hat{k}) = 2\hat{j}$
$\vec{B} = 2\hat{j} + (2\hat{i} + 3\hat{j} + 2\hat{k})$
$\vec{B} = 2\hat{i} + 5\hat{j} + 2\hat{k}$
The magnitude of vector $\vec{B}$ is given by $|\vec{B}| = \sqrt{x^2 + y^2 + z^2}$.
$|\vec{B}| = \sqrt{2^2 + 5^2 + 2^2}$
$|\vec{B}| = \sqrt{4 + 25 + 4}$
$|\vec{B}| = \sqrt{33}$

(A) The rotational kinetic energy is given by $K = \frac{L^2}{2I}$,where $L$ is the angular momentum and $I$ is the moment of inertia.
Since there are no external torques acting on the system,the angular momentum $L$ remains constant.
Therefore,the kinetic energy is inversely proportional to the moment of inertia: $K \propto \frac{1}{I}$.
Let $I_i$ and $K_i = E$ be the initial moment of inertia and initial kinetic energy,respectively.
Let $I_f$ and $K_f$ be the final moment of inertia and final kinetic energy,respectively.
Given that $I_f = 3I_i$,we have:
$\frac{K_f}{K_i} = \frac{I_i}{I_f} = \frac{I_i}{3I_i} = \frac{1}{3}$.
Thus,$K_f = \frac{K_i}{3} = \frac{E}{3}$.
Comparing this with $\frac{E}{x}$,we get $x = 3$.

(A) Given: Mass $m = 5\,kg$,acceleration $a = 1\,m/s^2$,angle $\theta = 30^{\circ}$,time $t = 10\,s$,$g = 10\,m/s^2$.
Applying Newton's second law along the incline:
$F - mg \sin 30^{\circ} = ma$
$F - 5 \times 10 \times 0.5 = 5 \times 1$
$F - 25 = 5 \Rightarrow F = 30\,N$.
Calculating velocity at $t = 10\,s$ using $v = u + at$:
$v = 0 + 1 \times 10 = 10\,m/s$.
Calculating power $P = F \times v$:
$P = 30 \times 10 = 300\,W$.

(A) The fundamental frequency of a stretched wire is given by $f = \frac{1}{2L} \sqrt{\frac{T}{\mu}}$.
Here,$T$ is the tension and $\mu$ is the linear mass density.
We know that $T = Y A \frac{\Delta L}{L}$ and $\mu = \rho A$,where $A$ is the cross-sectional area.
Substituting these into the frequency formula:
$f = \frac{1}{2L} \sqrt{\frac{Y A \Delta L / L}{\rho A}} = \frac{1}{2L} \sqrt{\frac{Y \Delta L}{\rho L}}$.
Given values: $L = 0.5\,m$,$\Delta L = 3.2 \times 10^{-4}\,m$,$Y = 8 \times 10^{10}\,N/m^2$,$\rho = 8 \times 10^3\,kg/m^3$.
$f = \frac{1}{2 \times 0.5} \sqrt{\frac{8 \times 10^{10} \times 3.2 \times 10^{-4}}{8 \times 10^3 \times 0.5}}$.
$f = 1 \times \sqrt{\frac{25.6 \times 10^6}{4 \times 10^3}} = \sqrt{6.4 \times 10^3} = \sqrt{6400} = 80\,Hz$.

(A) The surface tension $T = 3.5 \times 10^{-2} \, N m^{-1}$.
Initial radius $r_1 = 10 \, cm = 0.1 \, m$.
Final radius $r_2 = 20 \, cm = 0.2 \, m$.
$A$ soap bubble has two surfaces (inner and outer),so the change in surface area is $\Delta A = 2 \times (4 \pi r_2^2 - 4 \pi r_1^2) = 8 \pi (r_2^2 - r_1^2)$.
The work done $W = T \times \Delta A = T \times 8 \pi (r_2^2 - r_1^2)$.
Substituting the values: $W = 3.5 \times 10^{-2} \times 8 \times 3.14159 \times (0.2^2 - 0.1^2)$.
$W = 3.5 \times 10^{-2} \times 8 \times 3.14159 \times (0.04 - 0.01) = 3.5 \times 10^{-2} \times 8 \times 3.14159 \times 0.03$.
$W = 0.28 \times 3.14159 \times 0.03 = 0.026389 \, J$.
Converting to the required format: $W \approx 264 \times 10^{-4} \, J$.

(D) Let $M_p$ be the mass of the Earth,$m$ be the mass of the satellite,and $R$ be the orbital radius.
Potential energy $(U)$ $= -\frac{G M_p m}{R}$. Since $m_A = 2m_B$,$U_A \neq U_B$.
Kinetic energy $(K)$ $= \frac{G M_p m}{2R}$. Since $m_A = 2m_B$,$K_A \neq K_B$.
Total energy $(E)$ $= U + K = -\frac{G M_p m}{2R}$. Since $m_A = 2m_B$,$E_A \neq E_B$.
Orbital speed $(v)$ $= \sqrt{\frac{G M_p}{R}}$.
As seen from the formula,the orbital speed is independent of the mass of the satellite $(m)$. Therefore,both satellites will have the same speed.

(C) The root mean square (r.m.s.) speed of a gas molecule is given by the formula $v_{rms} = \sqrt{\frac{3RT}{M}}$,where $R$ is the universal gas constant,$T$ is the absolute temperature,and $M$ is the molar mass.
Since the temperature $T$ is the same for both gases,we have $v_{rms} \propto \frac{1}{\sqrt{M}}$.
Therefore,the ratio of the r.m.s. speeds is $\frac{v_{Ar}}{v_{Cl}} = \sqrt{\frac{M_{Cl}}{M_{Ar}}}$.
Given $v_{Cl} = 490\,m/s$,$M_{Cl} = 70.9\,u$,and $M_{Ar} = 39.9\,u$.
Substituting the values: $v_{Ar} = 490 \times \sqrt{\frac{70.9}{39.9}}$.
$v_{Ar} = 490 \times \sqrt{1.7769} \approx 490 \times 1.333 = 653.17\,m/s$.
Rounding to the nearest provided option,we get $651.7\,m/s$.

(A) $1$. Spring constant $(k)$: $F = kx \implies [k] = [F]/[x] = (MLT^{-2}) / L = MT^{-2}$. Thus,$A-II$.
$2$. Angular speed $(\omega)$: $\omega = \Delta\theta / \Delta t \implies [\omega] = [1] / T = T^{-1}$. Thus,$B-I$.
$3$. Angular momentum $(L)$: $L = mvr \implies [L] = M(LT^{-1})L = ML^2T^{-1}$. Thus,$C-IV$.
$4$. Moment of inertia $(I)$: $I = mr^2 \implies [I] = ML^2$. Thus,$D-III$.
Therefore,the correct matching is $(A)-(II), (B)-(I), (C)-(IV), (D)-(III)$.

(A) Given that the particle is in equilibrium under the action of three forces $F_1$,$F_2$,and $F_3$.
Since $F_2$ and $F_3$ are perpendicular,their resultant force $F_{23}$ is given by $F_{23} = \sqrt{F_2^2 + F_3^2} = \sqrt{8^2 + 6^2} = \sqrt{64 + 36} = \sqrt{100} = 10\,N$.
For the particle to be at rest,the resultant of all forces must be zero,meaning $F_1$ must be equal and opposite to the resultant of $F_2$ and $F_3$.
When $F_1$ is removed,the only forces acting on the particle are $F_2$ and $F_3$.
The net force acting on the particle becomes the resultant of $F_2$ and $F_3$,which is $10\,N$.
Using Newton's second law,$F_{net} = ma$,we have $10\,N = 5\,kg \times a$.
Therefore,$a = \frac{10}{5} = 2\,m/s^2$.

(A) According to Newton's Law of Cooling,the rate of cooling is proportional to the temperature difference between the body and its surroundings: $\frac{dT}{dt} = k(T - T_s)$.
For the first interval ($80^{\circ} C$ to $60^{\circ} C$):
$\frac{80 - 60}{5} = k \left( \frac{80 + 60}{2} - 20 \right)$
$\frac{20}{5} = k(70 - 20)$
$4 = 50k \Rightarrow k = \frac{4}{50} = 0.08 \text{ min}^{-1}$.
For the second interval ($60^{\circ} C$ to $40^{\circ} C$):
$\frac{60 - 40}{t} = k \left( \frac{60 + 40}{2} - 20 \right)$
$\frac{20}{t} = k(50 - 20)$
$\frac{20}{t} = 30k$.
Substituting $k = \frac{4}{50}$:
$\frac{20}{t} = 30 \times \frac{4}{50}$
$\frac{20}{t} = \frac{120}{50} = 2.4$
$t = \frac{20}{2.4} = \frac{200}{24} = \frac{25}{3} \text{ minutes}$.
Converting to seconds:
$t = \frac{25}{3} \times 60 = 25 \times 20 = 500 \text{ seconds}$.

(A) The temperatures are $T_1 = 373 \text{ K}$ (boiling point) and $T_2 = 273 \text{ K}$ (freezing point).
The efficiency of a Carnot engine is given by $\eta = 1 - \frac{T_2}{T_1} = 1 - \frac{273}{373} \approx 0.268$ or $26.8 \%$.
Any real engine operating between these two temperatures must have an efficiency less than the Carnot efficiency $(\eta < 26.8 \%)$.
Therefore,the efficiency is less than $26.8 \%$,which implies it is less than $27 \%$.
Thus,statements $2$ and $4$ are correct.

(A) According to the Work-Energy Theorem, the work done by the retarding force is equal to the change in kinetic energy.
$W = \Delta KE$
Since the final kinetic energy is $0$ and the initial kinetic energy is $K$, the work done is $W = -F \cdot S = 0 - K$, where $F$ is the retarding force and $S$ is the stopping distance.
Thus, $S = \frac{K}{F}$.
Since both the truck and the car have the same initial kinetic energy $K$ and are subjected to the same retarding force $F$, they will come to rest in the same distance $S$. Therefore, Statement $I$ is correct.
For Statement $II$, although the speed (magnitude of velocity) remains unchanged, the direction of the velocity changes as the car turns from east to north.
Acceleration is defined as the rate of change of velocity $(\vec{a} = \frac{d\vec{v}}{dt})$. Since the direction of velocity changes, the velocity vector changes, which means there must be a non-zero acceleration.
Therefore, Statement $II$ is incorrect.

(D) The potential energy $(P.E.)$ of a particle in $SHM$ at displacement $x$ is given by $P.E. = \frac{1}{2} kx^2$.
The kinetic energy $(K.E.)$ of the particle at displacement $x$ is given by $K.E. = \frac{1}{2} k(A^2 - x^2)$,where $A$ is the amplitude.
Given that the displacement $x = \frac{A}{2}$.
Substituting $x$ into the potential energy formula: $P.E. = \frac{1}{2} k(\frac{A}{2})^2 = \frac{1}{2} k(\frac{A^2}{4}) = \frac{1}{8} kA^2$.
Substituting $x$ into the kinetic energy formula: $K.E. = \frac{1}{2} k(A^2 - (\frac{A}{2})^2) = \frac{1}{2} k(A^2 - \frac{A^2}{4}) = \frac{1}{2} k(\frac{3A^2}{4}) = \frac{3}{8} kA^2$.
The ratio of potential energy to kinetic energy is $\frac{P.E.}{K.E.} = \frac{\frac{1}{8} kA^2}{\frac{3}{8} kA^2} = \frac{1}{3}$.

(B) Using the first equation of motion,$v = u + at$.
Here,initial velocity $u = 150\,m/s$,acceleration $a = -g = -10\,m/s^2$,and $t$ is the time in seconds.
So,$v(t) = 150 - 10t$.
Velocity after $3\,s$ is $v(3) = 150 - 10(3) = 150 - 30 = 120\,m/s$.
Velocity after $5\,s$ is $v(5) = 150 - 10(5) = 150 - 50 = 100\,m/s$.
The ratio of velocities is $\frac{v(3)}{v(5)} = \frac{120}{100} = \frac{6}{5}$.
Given that the ratio is $\frac{x+1}{x}$,we have $\frac{x+1}{x} = \frac{6}{5}$.
Comparing the terms,$x = 5$.

(B) The formula for escape velocity is $V_e = \sqrt{\frac{2GM}{R}}$.
Let $M_E$ and $R_E$ be the mass and radius of the Earth,and $M_P$ and $R_P$ be the mass and radius of the planet.
Given: $M_P = 16 M_E$ and $R_P = 4 R_E$.
The escape velocity of the planet is $V_P = \sqrt{\frac{2GM_P}{R_P}} = \sqrt{\frac{2G(16M_E)}{4R_E}} = \sqrt{4 \times \frac{2GM_E}{R_E}} = 2 \sqrt{\frac{2GM_E}{R_E}}$.
Since $V_E = \sqrt{\frac{2GM_E}{R_E}}$,we have $V_P = 2 V_E$.
Therefore,the ratio $\frac{V_P}{V_E} = \frac{2}{1}$,which is $2:1$.

(D) The ratio of resonance frequencies $1:3:5$ indicates that the organ pipe is closed at one end.
For a closed organ pipe,the $n^{th}$ harmonic frequency is given by $f_n = \frac{n V}{4 \ell}$,where $n$ must be an odd integer $(n = 1, 3, 5, ...)$.
The fifth harmonic corresponds to $n = 5$.
Given $f_5 = 405 \, Hz$,$V = 324 \, ms^{-1}$,and $n = 5$.
Substituting these values into the formula: $405 = \frac{5 \times 324}{4 \ell}$.
Rearranging for $\ell$: $\ell = \frac{5 \times 324}{4 \times 405}$.
$\ell = \frac{1620}{1620} = 1 \, m$.

(D) The rotational kinetic energy of a body is given by $K_{\text{rot}} = \frac{1}{2} I \omega^2$,where $I$ is the moment of inertia and $\omega$ is the angular velocity.
For a spherical shell,the moment of inertia about its center is $I = \frac{2}{3} mR^2$.
For pure rolling,$\omega = \frac{v}{R}$,so $K_{\text{rot}} = \frac{1}{2} (\frac{2}{3} mR^2) (\frac{v}{R})^2 = \frac{1}{3} mv^2$.
The total kinetic energy is $K_{\text{total}} = K_{\text{trans}} + K_{\text{rot}} = \frac{1}{2} mv^2 + \frac{1}{3} mv^2 = \frac{5}{6} mv^2$.
The ratio of rotational kinetic energy to total kinetic energy is $\frac{K_{\text{rot}}}{K_{\text{total}}} = \frac{\frac{1}{3} mv^2}{\frac{5}{6} mv^2} = \frac{1}{3} \times \frac{6}{5} = \frac{2}{5}$.
Given that the ratio is $\frac{x}{5}$,we have $\frac{x}{5} = \frac{2}{5}$,which implies $x = 2$.

(D) Since the bus moves at a constant speed,the net work done on the bus is zero according to the Work-Energy Theorem.
$W_{\text{net}} = W_{\text{engine}} + W_{\text{friction}} = 0$
Therefore,$W_{\text{engine}} = -W_{\text{friction}}$.
The work done by friction is given by $W_{\text{friction}} = -f_k \cdot d = -(\mu mg) \cdot d$.
Here,$\mu = 0.04$,$m = 500\,kg$,$g = 9.8\,m/s^2$,and $d = 4\,km = 4000\,m$.
$W_{\text{engine}} = \mu mgd = 0.04 \times 500 \times 9.8 \times 4000$.
$W_{\text{engine}} = 20 \times 9.8 \times 4000 = 196 \times 4000 = 784000\,J$.
Converting to $kJ$,$W_{\text{engine}} = 784\,kJ$.

(D) Given: Density $\rho = 1.25 \times 10^3 \, kg \, m^{-3}$, $A_1 = 10 \, cm^2 = 10 \times 10^{-4} \, m^2$, $A_2 = 5 \, cm^2 = 5 \times 10^{-4} \, m^2$, $\Delta P = P_1 - P_2 = 3 \, N \, m^{-2}$.
By the equation of continuity, $A_1 v_1 = A_2 v_2$.
Therefore, $v_1 = \frac{A_2}{A_1} v_2 = \frac{5}{10} v_2 = 0.5 v_2$.
Using Bernoulli's equation for horizontal flow: $P_1 + \frac{1}{2} \rho v_1^2 = P_2 + \frac{1}{2} \rho v_2^2$.
$\Delta P = P_1 - P_2 = \frac{1}{2} \rho (v_2^2 - v_1^2)$.
Substituting $v_1 = 0.5 v_2$:
$3 = \frac{1}{2} \times (1.25 \times 10^3) \times (v_2^2 - (0.5 v_2)^2)$.
$3 = 0.625 \times 10^3 \times (v_2^2 - 0.25 v_2^2) = 0.625 \times 10^3 \times 0.75 v_2^2$.
$3 = 468.75 v_2^2$.
$v_2^2 = \frac{3}{468.75} = 0.0064$.
$v_2 = \sqrt{0.0064} = 0.08 \, m \, s^{-1}$.
The rate of flow (discharge) $Q = A_2 v_2 = (5 \times 10^{-4} \, m^2) \times (0.08 \, m \, s^{-1}) = 40 \times 10^{-6} \, m^3 \, s^{-1} = 4 \times 10^{-5} \, m^3 \, s^{-1}$.
Comparing with $x \times 10^{-5} \, m^3 \, s^{-1}$, we get $x = 4$.

(C) First,convert the velocities from $km/h$ to $m/s$:
$v_A = 108 \times \frac{5}{18} = 30\,m/s$
$v_B = 72 \times \frac{5}{18} = 20\,m/s$
To cross the tunnel,each train must cover a distance equal to the sum of the tunnel length and its own length.
Time taken by train '$A$': $t_A = \frac{L + l}{v_A} = \frac{60l + l}{30} = \frac{61l}{30}$
Time taken by train '$B$': $t_B = \frac{L + 4l}{v_B} = \frac{60l + 4l}{20} = \frac{64l}{20} = \frac{16l}{5}$
Given that $t_B - t_A = 35\,s$:
$\frac{16l}{5} - \frac{61l}{30} = 35$
$\frac{96l - 61l}{30} = 35$
$\frac{35l}{30} = 35$
$l = 30\,m$
Therefore,$L = 60l = 60 \times 30 = 1800\,m$.

(D) Average Power is defined as the total work done divided by the total time taken.
$P = \frac{W}{t} = \frac{mgh}{t}$
Given the ratio of powers of two motors is $\frac{P_1}{P_2} = \frac{3 \sqrt{x}}{\sqrt{x}+1}$.
For the first motor: $m_1 = 300 \ kg$,$t_1 = 5 \ minutes = 300 \ s$,$h = 100 \ m$.
For the second motor: $m_2 = 50 \ kg$,$t_2 = 2 \ minutes = 120 \ s$,$h = 100 \ m$.
Calculating the ratio of powers:
$\frac{P_1}{P_2} = \frac{m_1 g h / t_1}{m_2 g h / t_2} = \frac{m_1}{t_1} \times \frac{t_2}{m_2}$
$\frac{P_1}{P_2} = \frac{300}{5} \times \frac{2}{50} = 60 \times 0.04 = 2.4$
Equating the two expressions:
$\frac{3 \sqrt{x}}{\sqrt{x}+1} = 2.4$
$3 \sqrt{x} = 2.4 \sqrt{x} + 2.4$
$0.6 \sqrt{x} = 2.4$
$\sqrt{x} = 4$
$x = 16$

(B) The formula for escape velocity is $V = \sqrt{\frac{2GM}{R}}$.
For Earth,$V_e = \sqrt{\frac{2GM_e}{R_e}} = 11.2 \, km/s$.
For the planet,$V_p = \sqrt{\frac{2G(9M_e)}{4R_e}} = \sqrt{\frac{9}{4}} \times \sqrt{\frac{2GM_e}{R_e}}$.
$V_p = \frac{3}{2} \times V_e$.
Substituting the value of $V_e = 11.2 \, km/s$:
$V_p = 1.5 \times 11.2 \, km/s = 16.8 \, km/s$.

(B) The impulse supplied to the gun is equal to the change in momentum of the bullet.
Given:
Mass of the bullet,$m = 10\,g = 0.01\,kg$
Velocity of the bullet,$v = 600\,m/s$
Mass of the gun,$M = 3\,kg$
Impulse $J = \Delta p = m \times v$
$J = 0.01\,kg \times 600\,m/s = 6\,Ns$
Since the system is initially at rest,the impulse imparted to the gun is equal in magnitude to the momentum gained by the bullet.

(A) When the disc rolls without slipping,the center of the disc moves forward by a distance equal to the arc length covered. For half a rotation,the center moves by a horizontal distance of $\pi R$.
The point $A$,which was initially at the top,moves to the bottom of the disc after half a rotation. The vertical displacement of point $A$ is equal to the diameter of the disc,which is $2R$.
The horizontal displacement of point $A$ is equal to the distance moved by the center,which is $\pi R$.
The total displacement $d$ is the vector sum of the horizontal and vertical displacements:
$d = \sqrt{(\text{horizontal displacement})^2 + (\text{vertical displacement})^2}$
$d = \sqrt{(\pi R)^2 + (2R)^2}$
$d = \sqrt{\pi^2 R^2 + 4R^2}$
$d = R \sqrt{\pi^2 + 4}$

(A) The root mean square (rms) speed of a gas molecule is given by $v_{rms} = \sqrt{\frac{3RT}{M}}$.
The average speed of a gas molecule is given by $v_{avg} = \sqrt{\frac{8RT}{\pi M}}$.
According to the problem,$v_{rms} = \left(1+\frac{5}{x}\right)^{\frac{1}{2}} v_{avg}$.
Substituting the formulas: $\sqrt{\frac{3RT}{M}} = \left(1+\frac{5}{x}\right)^{\frac{1}{2}} \sqrt{\frac{8RT}{\pi M}}$.
Squaring both sides: $\frac{3RT}{M} = \left(1+\frac{5}{x}\right) \frac{8RT}{\pi M}$.
Canceling $\frac{RT}{M}$ from both sides: $3 = \left(1+\frac{5}{x}\right) \frac{8}{\pi}$.
Given $\pi = \frac{22}{7}$,we have $3 = \left(1+\frac{5}{x}\right) \frac{8}{22/7} = \left(1+\frac{5}{x}\right) \frac{8 \times 7}{22} = \left(1+\frac{5}{x}\right) \frac{56}{22} = \left(1+\frac{5}{x}\right) \frac{28}{11}$.
$\Rightarrow 1+\frac{5}{x} = 3 \times \frac{11}{28} = \frac{33}{28}$.
$\Rightarrow \frac{5}{x} = \frac{33}{28} - 1 = \frac{5}{28}$.
$\Rightarrow x = 28$.

(A) The formula for kinetic energy is $K = \frac{1}{2}mv^2$.
First,calculate the mean value of kinetic energy:
$K = \frac{1}{2} \times 5 \times (20)^2 = \frac{1}{2} \times 5 \times 400 = 1000 \ J$.
Next,calculate the relative error using the formula for propagation of errors:
$\frac{\Delta K}{K} = \frac{\Delta m}{m} + 2 \frac{\Delta v}{v}$.
Substitute the given values:
$\frac{\Delta K}{1000} = \frac{0.5}{5} + 2 \times \frac{0.4}{20} = 0.1 + 0.04 = 0.14$.
Calculate the absolute error $\Delta K$:
$\Delta K = 1000 \times 0.14 = 140 \ J$.
Therefore,the kinetic energy is $(1000 \pm 140) \ J$.

(D) The kinetic energy $K$ of a body with mass $m$ and linear momentum $p$ is given by $K = \frac{p^2}{2m}$.
Given that the linear momenta are equal,i.e.,$p_1 = p_2 = p$.
Therefore,the ratio of kinetic energies is $\frac{K_1}{K_2} = \frac{p^2 / 2m_1}{p^2 / 2m_2} = \frac{m_2}{m_1}$.
Given $\frac{K_1}{K_2} = \frac{16}{9}$,we have $\frac{m_2}{m_1} = \frac{16}{9}$.
Thus,the ratio of their masses $\frac{m_1}{m_2} = \frac{9}{16}$.

(A) Given:
Area at $A$,$A_A = 1.5 \, cm^2 = 1.5 \times 10^{-4} \, m^2$
Area at $B$,$A_B = 25 \, mm^2 = 25 \times 10^{-6} \, m^2$
Velocity at $B$,$v_B = 60 \, cm/s = 0.6 \, m/s$
Density,$\rho = 1000 \, kg/m^3$
Using the equation of continuity,$A_A v_A = A_B v_B$:
$1.5 \times 10^{-4} \times v_A = 25 \times 10^{-6} \times 0.6$
$v_A = \frac{25 \times 10^{-6} \times 0.6}{1.5 \times 10^{-4}} = \frac{15 \times 10^{-6}}{1.5 \times 10^{-4}} = 10 \times 10^{-2} = 0.1 \, m/s$
Using Bernoulli's equation for a horizontal tube $(h_A = h_B)$:
$P_A + \frac{1}{2} \rho v_A^2 = P_B + \frac{1}{2} \rho v_B^2$
$P_A - P_B = \frac{1}{2} \rho (v_B^2 - v_A^2)$
$P_A - P_B = \frac{1}{2} \times 1000 \times ((0.6)^2 - (0.1)^2)$
$P_A - P_B = 500 \times (0.36 - 0.01)$
$P_A - P_B = 500 \times 0.35 = 175 \, Pa$

(B) The bulk modulus $B$ is defined as $B = -V \frac{dP}{dV}$.
Given the relation $P = aV^{-3}$.
Differentiating $P$ with respect to $V$:
$\frac{dP}{dV} = a(-3)V^{-4} = -3 \frac{aV^{-3}}{V} = -3 \frac{P}{V}$.
Substituting this into the formula for bulk modulus:
$B = -V \left( -3 \frac{P}{V} \right) = 3P$.
Therefore,the bulk modulus is $3P$.

(D) The total energy $(TE)$ of a particle executing $SHM$ is constant,given by $TE = \frac{1}{2} m \omega^2 A^2$.
The potential energy $(PE)$ at a displacement $x$ from the mean position is given by $PE = \frac{1}{2} m \omega^2 x^2$.
The difference between total energy and potential energy is the kinetic energy $(KE)$:
$KE = TE - PE$
$KE = \frac{1}{2} m \omega^2 A^2 - \frac{1}{2} m \omega^2 x^2$
$KE = \frac{1}{2} m \omega^2 (A^2 - x^2)$
This equation represents a downward-opening parabola with its vertex at $x = 0$ and roots at $x = \pm A$. This matches the shape shown in graph $D$.

(A) $1$. The photocurrent is directly proportional to the intensity of the incident radiation because intensity is proportional to the number of incident photons,which in turn is proportional to the number of emitted photoelectrons. Thus,statement $A$ is correct.
$2$. According to Einstein's photoelectric equation,$KE_{\max} = h\nu - \phi$,where $h$ is Planck's constant,$\nu$ is the frequency of incident light,and $\phi$ is the work function of the metal. This equation shows that $KE_{\max}$ depends on the frequency of the incident light,not its intensity. Thus,statement $C$ is correct,while statements $B$ and $E$ are incorrect.
$3$. The emission of photoelectrons requires a minimum threshold frequency,not a minimum threshold intensity. Thus,statement $D$ is incorrect.
Therefore,statements $A$ and $C$ are correct.

(A) The formula for motional $emf$ induced in a conductor moving in a magnetic field is given by:
$e = Blv$
Given:
$e = 0.08\,V$
$l = 10\,cm = 0.1\,m$
$B = 0.4\,T$
Substituting the values into the formula:
$0.08 = 0.4 \times 0.1 \times v$
$0.08 = 0.04 \times v$
$v = \frac{0.08}{0.04} = 2\,m/s$
Thus,the velocity is $2\,m/s$.

(D) The power radiated $P$ by a short linear antenna of length $l$ is given by the relation $P \propto \left(\frac{l}{\lambda}\right)^2$.
This indicates that the radiated power is directly proportional to the square of the ratio of the antenna length to the wavelength of the electromagnetic wave.

(A) The quality factor $(Q)$ of a series $LCR$ circuit at resonance is given by the formula:
$Q = \frac{1}{R} \sqrt{\frac{L}{C}}$
Given values:
Resistance $R = 100\,\Omega$
Inductance $L = 1\,H$
Capacitance $C = 6.25 \times 10^{-6}\,F$
Substituting the values into the formula:
$Q = \frac{1}{100} \sqrt{\frac{1}{6.25 \times 10^{-6}}}$
$Q = \frac{1}{100} \sqrt{\frac{10^6}{6.25}}$
$Q = \frac{1}{100} \times \frac{1000}{2.5}$
$Q = \frac{10}{2.5} = 4$
Thus,the quality factor of the circuit is $4$.

(A) The Rydberg formula for the wavelength of spectral lines in the $H$-atom is given by $\frac{1}{\lambda} = R \left[ \frac{1}{n_1^2} - \frac{1}{n_2^2} \right]$.
For the Balmer series,$n_1 = 2$.
For the $H_\alpha$ line,$n_2 = 3$. Thus,$\frac{1}{\lambda_{H_\alpha}} = R \left[ \frac{1}{2^2} - \frac{1}{3^2} \right] = R \left[ \frac{1}{4} - \frac{1}{9} \right] = \frac{5R}{36}$.
For the $H_\beta$ line,$n_2 = 4$. Thus,$\frac{1}{\lambda_{H_\beta}} = R \left[ \frac{1}{2^2} - \frac{1}{4^2} \right] = R \left[ \frac{1}{4} - \frac{1}{16} \right] = \frac{3R}{16}$.
Taking the ratio of the wavelengths: $\frac{\lambda_{H_\alpha}}{\lambda_{H_\beta}} = \frac{\lambda_{H_\beta}^{-1}}{\lambda_{H_\alpha}^{-1}} = \frac{3R/16}{5R/36} = \frac{3}{16} \times \frac{36}{5} = \frac{3 \times 9}{4 \times 5} = \frac{27}{20}$.
Comparing this with $\frac{x}{20}$,we get $x = 27$.

(A) Given:
Number density of electrons,$n = 8 \times 10^{28} \ m^{-3}$
Area of cross-section,$A = 2 \times 10^{-6} \ m^2$
Current,$I = 3.2 \ A$
Charge of an electron,$e = 1.6 \times 10^{-19} \ C$
The formula for drift velocity $(v_d)$ is given by:
$I = n e A v_d$
Rearranging for $v_d$:
$v_d = \frac{I}{n e A}$
Substituting the values:
$v_d = \frac{3.2}{(8 \times 10^{28}) \times (1.6 \times 10^{-19}) \times (2 \times 10^{-6})}$
$v_d = \frac{3.2}{25.6 \times 10^3}$
$v_d = 0.125 \times 10^{-3} \ m/s$
$v_d = 125 \times 10^{-6} \ m/s$
Thus,the drift speed is $125 \times 10^{-6} \ m/s$.

(A) Initial charge on the capacitor: $Q = CV = 600 \times 10^{-12} \, F \times 200 \, V = 12 \times 10^{-8} \, C$.
Initial electrostatic energy stored: $U_i = \frac{1}{2} CV^2 = \frac{1}{2} \times 600 \times 10^{-12} \times (200)^2 = 12 \times 10^{-6} \, J = 12 \, \mu J$.
When the charged capacitor is connected to an identical uncharged capacitor,the charge $Q$ is redistributed equally between them because the capacitors are in parallel and have the same capacitance.
New charge on each capacitor: $Q' = \frac{Q}{2} = 6 \times 10^{-8} \, C$.
Final electrostatic energy stored in the system: $U_f = 2 \times \left( \frac{Q'^2}{2C} \right) = \frac{Q'^2}{C} = \frac{(6 \times 10^{-8})^2}{600 \times 10^{-12}} = \frac{36 \times 10^{-16}}{600 \times 10^{-12}} = 6 \times 10^{-6} \, J = 6 \, \mu J$.
Energy lost in the process: $\Delta U = U_i - U_f = 12 \, \mu J - 6 \, \mu J = 6 \, \mu J$.

(A) Given: Refractive index of rarer medium $\mu_1 = 1.0$,refractive index of denser medium $\mu_2 = 1.5$. Radius of curvature $R = +30\,cm$ (as the centre of curvature is in the direction of light propagation). Object distance $u = -15\,cm$ (measured against the direction of light).
Using the formula for refraction at a spherical surface:
$\frac{\mu_2}{v} - \frac{\mu_1}{u} = \frac{\mu_2 - \mu_1}{R}$
Substituting the values:
$\frac{1.5}{v} - \frac{1.0}{-15} = \frac{1.5 - 1.0}{30}$
$\frac{1.5}{v} + \frac{1}{15} = \frac{0.5}{30}$
$\frac{1.5}{v} + \frac{1}{15} = \frac{1}{60}$
$\frac{1.5}{v} = \frac{1}{60} - \frac{1}{15}$
$\frac{1.5}{v} = \frac{1 - 4}{60} = \frac{-3}{60} = -\frac{1}{20}$
$v = 1.5 \times (-20) = -30\,cm$.
The negative sign indicates that the image is formed on the same side as the object at a distance of $30\,cm$ from the pole.

(A) The magnetic field at the centre of the coil is given by $B_1 = \frac{\mu_0 I}{2r}$.
The magnetic field at a distance $d$ on the axis of the coil is given by $B_2 = \frac{\mu_0 I r^2}{2(r^2 + d^2)^{3/2}}$.
Given that $d = r$,we substitute this into the formula for $B_2$:
$B_2 = \frac{\mu_0 I r^2}{2(r^2 + r^2)^{3/2}} = \frac{\mu_0 I r^2}{2(2r^2)^{3/2}} = \frac{\mu_0 I r^2}{2(2^{3/2} r^3)} = \frac{\mu_0 I}{2(2\sqrt{2})r} = \frac{\mu_0 I}{4\sqrt{2}r}$.
Now,calculate the ratio $\frac{B_1}{B_2}$:
$\frac{B_1}{B_2} = \frac{\mu_0 I}{2r} \times \frac{4\sqrt{2}r}{\mu_0 I} = \frac{4\sqrt{2}}{2} = 2\sqrt{2} = \sqrt{4 \times 2} = \sqrt{8}$.
Comparing this with $\sqrt{x}: 1$,we get $\sqrt{x} = \sqrt{8}$,which implies $x = 8$.

(C) The power rating of the Zener diode is $P = 1.6\,W$ and the breakdown voltage is $V_z = 8\,V$.
The maximum current that can flow through the Zener diode is $I_{z,max} = \frac{P}{V_z} = \frac{1.6\,W}{8\,V} = 0.2\,A$.
For the Zener diode to act as a voltage regulator, the input voltage $V_{in}$ must be greater than the breakdown voltage $V_z$. Here, the input voltage fluctuates between $3\,V$ and $10\,V$. The Zener diode will only regulate when $V_{in} \geq 8\,V$. Therefore, we consider the maximum input voltage $V_{in,max} = 10\,V$ to ensure the diode does not exceed its power rating.
The voltage drop across the series resistance $R_s$ at maximum input voltage is $V_{R_s} = V_{in,max} - V_z = 10\,V - 8\,V = 2\,V$.
To ensure safe operation, the current through the series resistance $R_s$ must be equal to the maximum current the Zener diode can handle (assuming no load current is connected, which is the worst-case scenario for the diode).
Thus, $R_s = \frac{V_{R_s}}{I_{z,max}} = \frac{2\,V}{0.2\,A} = 10\,\Omega$.

(B) Given,carrier wave amplitude $A_c = 15\,V$.
Baseband signal amplitude $A_m = 3\,V$.
The maximum amplitude of the amplitude modulated wave is given by $A_{\max} = A_c + A_m = 15 + 3 = 18\,V$.
The minimum amplitude of the amplitude modulated wave is given by $A_{\min} = A_c - A_m = 15 - 3 = 12\,V$.
The ratio of maximum amplitude to minimum amplitude is $\frac{A_{\max}}{A_{\min}} = \frac{18}{12} = \frac{3}{2}$.

(C) According to Bohr's postulate,the angular momentum $L$ of an electron in the $n^{th}$ orbit is given by $L_n = \frac{nh}{2\pi}$.
For the first orbit $(n=1)$,the angular momentum is $L_1 = L = \frac{1 \cdot h}{2\pi}$.
For the second orbit $(n=2)$,the angular momentum is $L_2 = \frac{2h}{2\pi} = 2 \left( \frac{h}{2\pi} \right) = 2L$.
The change in angular momentum is $\Delta L = L_2 - L_1 = 2L - L = L$.

(D) For a moving coil galvanometer,the torque is given by $\tau = NIAB \sin \theta = k \phi$,where $N$ is the number of turns,$I$ is the current,$A$ is the area,$B$ is the magnetic field,and $k$ is the torsional constant.
The current sensitivity is defined as $S_i = \frac{\phi}{I} = \frac{NBA}{k}$.
If $N$ is doubled,$S_i$ becomes $2 \times \frac{NBA}{k}$,so Statement $I$ is true.
The voltage sensitivity is defined as $S_v = \frac{\phi}{V} = \frac{\phi}{IR} = \frac{S_i}{R} = \frac{NBA}{kR}$.
When the number of turns $N$ is doubled,the length of the wire in the coil also doubles,which means the resistance $R$ of the coil also doubles $(R \propto N)$.
Therefore,$S_v = \frac{(2N)BA}{k(2R)} = \frac{NBA}{kR}$.
Since $S_v$ remains unchanged,Statement $II$ is false.

(B) The circuit consists of a central node connected to points $a$,$b$,and the top vertex.
Let the top vertex be $c$. The resistors connected to $c$ are in series with the resistors connected to $a$ and $b$.
Specifically,the branch from $a$ to $c$ has two $4\,\Omega$ resistors in series,giving $4+4=8\,\Omega$.
The branch from $b$ to $c$ has two $4\,\Omega$ resistors in series,giving $4+4=8\,\Omega$.
These two branches are in parallel with each other,and their equivalent resistance is $\frac{8 \times 8}{8+8} = 4\,\Omega$.
Finally,this $4\,\Omega$ equivalent resistance is in parallel with the $16\,\Omega$ resistor connected directly between $a$ and $b$.
$R_{eq} = \frac{4 \times 16}{4+16} = \frac{64}{20} = 3.2\,\Omega$.

(C) The power dissipated in an $AC$ circuit is given by $P = V_{rms} I_{rms} \cos \phi$, where $\cos \phi$ is the power factor.
For a series $LCR$ circuit, the impedance is $Z = \sqrt{R^2 + (X_L - X_C)^2}$.
At resonance, $X_L = X_C$, which makes the impedance $Z$ minimum $(Z = R)$ and the phase difference $\phi = 0$. Thus, $\cos \phi = 1$, and power dissipation is maximum. Therefore, Statement $I$ is true.
In a pure resistive circuit, the current and voltage are in the same phase $(\phi = 0)$, so $\cos \phi = 1$. This results in maximum power dissipation for a given voltage. Therefore, Statement $II$ is true.
Hence, both statements are correct.

(NONE) The circuit consists of two branches connected in parallel.
$1$. The top branch has two capacitors of capacitance $C$ in series. Their equivalent capacitance $C_1$ is given by $\frac{1}{C_1} = \frac{1}{C} + \frac{1}{C} = \frac{2}{C}$,so $C_1 = \frac{C}{2}$.
$2$. The bottom branch also has two capacitors of capacitance $C$ in series. Their equivalent capacitance $C_2$ is given by $\frac{1}{C_2} = \frac{1}{C} + \frac{1}{C} = \frac{2}{C}$,so $C_2 = \frac{C}{2}$.
$3$. These two branches are connected in parallel. Therefore,the total equivalent capacitance $C_{eq}$ is $C_{eq} = C_1 + C_2 = \frac{C}{2} + \frac{C}{2} = C$.

(C) The electric field of an electromagnetic wave is given by $E = E_0 \sin(\omega t - kx)$.
The energy density $u$ is proportional to the square of the electric field,$u \propto E^2$.
Thus,$u \propto \sin^2(\omega t - kx)$.
Using the trigonometric identity $\sin^2 \theta = \frac{1 - \cos(2\theta)}{2}$,we get $u \propto \frac{1 - \cos(2(\omega t - kx))}{2}$.
The frequency of the oscillation of the energy density is determined by the term $2\omega t$.
Since the angular frequency of the wave is $\omega = 2\pi f$,the angular frequency of the energy oscillation is $2\omega = 2(2\pi f) = 2\pi(2f)$.
Therefore,the frequency of the energy oscillation is $2f$,which is double the frequency of the wave.

(B) Initially,the object is at $x = -12\,cm$ and the mirror is at $x = 0$. The image is formed at $x = +12\,cm$.
When the mirror is moved $4\,cm$ towards the object,the new position of the mirror is $x = -4\,cm$.
The distance of the object from the new mirror position is $u = |-12 - (-4)| = 8\,cm$.
Since the image is formed at the same distance behind the mirror,the new image position $x_i$ satisfies $|x_i - (-4)| = 8\,cm$,so $x_i = 4\,cm$.
The initial position of the image was $12\,cm$ and the final position is $4\,cm$.
Therefore,the shift in the position of the image is $12\,cm - 4\,cm = 8\,cm$ towards the mirror.

(A) From the Kinetic Theory of Gases ($K$.$T$.$G$.),the root mean square velocity of a gas molecule is given by $v_{RMS} = \sqrt{\frac{3 k_B T}{m}}$.
This implies that $v_{RMS} \propto \sqrt{T}$.
The de Broglie wavelength is given by $\lambda = \frac{h}{p} = \frac{h}{m v_{RMS}}$.
Since $v_{RMS} \propto \sqrt{T}$,we have $\lambda \propto \frac{1}{\sqrt{T}}$.
Therefore,the ratio of the wavelengths is $\frac{\lambda_2}{\lambda_1} = \sqrt{\frac{T_1}{T_2}}$.
Given $T_1 = 300 \ K$ and $T_2 = 600 \ K$,we get $\frac{\lambda_2}{\lambda_1} = \sqrt{\frac{300}{600}} = \sqrt{\frac{1}{2}} = \frac{1}{\sqrt{2}}$.
Thus,the new wavelength is $\lambda_2 = \frac{\lambda_1}{\sqrt{2}}$.

(D) Let the intensity of unpolarised light be $I_0 = 32 \, W m^{-2}$.
After passing through the first polaroid,the intensity becomes $I_1 = \frac{I_0}{2} = 16 \, W m^{-2}$.
Let $\theta$ be the angle between the pass axis of the first and second polaroid.
The intensity after the second polaroid is $I_2 = I_1 \cos^2 \theta = \frac{I_0}{2} \cos^2 \theta$.
The angle between the second and third polaroid is $(90^{\circ} - \theta)$ because the first and third polaroids are perpendicular.
The intensity after the third polaroid is $I_3 = I_2 \cos^2(90^{\circ} - \theta) = I_2 \sin^2 \theta$.
Substituting $I_2$,we get $I_3 = \frac{I_0}{2} \cos^2 \theta \sin^2 \theta = \frac{I_0}{8} (2 \sin \theta \cos \theta)^2 = \frac{I_0}{8} \sin^2(2 \theta)$.
Given $I_3 = 3 \, W m^{-2}$ and $I_0 = 32 \, W m^{-2}$,we have $3 = \frac{32}{8} \sin^2(2 \theta) = 4 \sin^2(2 \theta)$.
$\sin^2(2 \theta) = \frac{3}{4} \implies \sin(2 \theta) = \frac{\sqrt{3}}{2}$.
Thus,$2 \theta = 60^{\circ}$ or $120^{\circ}$,which gives $\theta = 30^{\circ}$ or $60^{\circ}$.

(D) The magnetic intensity $H$ inside a solenoid is given by the formula $H = ni$,where $n$ is the number of turns per unit length and $i$ is the current.
Given:
Length of solenoid $\ell = 15\,cm = 0.15\,m$
Number of turns $N = 60$
Magnetic intensity $H = 2.4 \times 10^3\,A/m$
First,calculate the number of turns per unit length $n$:
$n = \frac{N}{\ell} = \frac{60}{0.15} = 400\,turns/m$
Now,use the formula $H = ni$ to find the current $i$:
$2.4 \times 10^3 = 400 \times i$
$i = \frac{2400}{400} = 6\,A$
Therefore,the required current is $6\,A$.

(D) The motional electromotive force $(EMF)$ induced in the rod is given by $\varepsilon = B \ell v$.
Given: $B = 0.15\,T$,$\ell = 1\,m$,$v = 4\,m/s$,and $R = 5\,\Omega$.
The induced current in the circuit is $i = \frac{\varepsilon}{R} = \frac{B \ell v}{R}$.
The magnetic force acting on the rod is $F = i \ell B$.
Substituting the expression for $i$,we get $F = \left( \frac{B \ell v}{R} \right) \ell B = \frac{B^2 \ell^2 v}{R}$.
Substituting the values: $F = \frac{(0.15)^2 \times (1)^2 \times 4}{5}$.
$F = \frac{0.0225 \times 4}{5} = \frac{0.09}{5} = 0.018\,N$.
Converting to the required units: $0.018\,N = 18 \times 10^{-3}\,N$.

(D) Given: Decay constant $\lambda = 1.5 \times 10^{-5} \, s^{-1}$.
Mass of substance $m = 1.0 \, \mu g = 1.0 \times 10^{-6} \, g$.
Molar mass $M = 60 \, g \, mol^{-1}$.
Number of moles $n = \frac{m}{M} = \frac{1.0 \times 10^{-6}}{60} = \frac{1}{6} \times 10^{-7} \, mol$.
Number of atoms $N = n \times N_A = (\frac{1}{6} \times 10^{-7}) \times (6 \times 10^{23}) = 10^{16}$ atoms.
Activity $A = N \lambda = 10^{16} \times 1.5 \times 10^{-5} \, Bq$.
$A = 1.5 \times 10^{11} \, Bq = 15 \times 10^{10} \, Bq$.
Therefore,the value is $15$.

(D) The charges on the shells are:
$q_x = \sigma(4\pi a^2)$
$q_y = -\sigma(4\pi b^2)$
$q_z = \sigma(4\pi c^2)$
Since the potential of shell $X$ is equal to the potential of shell $Z$ $(V_x = V_z)$:
$V_x = \frac{1}{4\pi\varepsilon_0} \left( \frac{q_x}{a} + \frac{q_y}{b} + \frac{q_z}{c} \right)$
$V_z = \frac{1}{4\pi\varepsilon_0} \left( \frac{q_x}{c} + \frac{q_y}{c} + \frac{q_z}{c} \right)$
Equating $V_x$ and $V_z$:
$\frac{\sigma 4\pi a^2}{a} - \frac{\sigma 4\pi b^2}{b} + \frac{\sigma 4\pi c^2}{c} = \frac{\sigma 4\pi a^2}{c} - \frac{\sigma 4\pi b^2}{c} + \frac{\sigma 4\pi c^2}{c}$
Simplifying the equation:
$a - b + \frac{c^2}{c} = \frac{a^2 - b^2 + c^2}{c}$
$a - b + c = \frac{a^2 - b^2 + c^2}{c}$
$c(a - b + c) = a^2 - b^2 + c^2$
$c(a - b) + c^2 = a^2 - b^2 + c^2$
$c(a - b) = a^2 - b^2$
$c(a - b) = (a - b)(a + b)$
$c = a + b$
Given $a = 2\,cm$ and $b = 3\,cm$:
$c = 2 + 3 = 5\,cm$.

(D) Maximum resistance is obtained when all resistors are connected in a series combination.
$R_{\max} = n \times R$,where $n = 10$ and $R = 10\,\Omega$.
$R_{\max} = 10 \times 10 = 100\,\Omega$.
Minimum resistance is obtained when all resistors are connected in a parallel combination.
$R_{\min} = \frac{R}{n}$,where $n = 10$ and $R = 10\,\Omega$.
$R_{\min} = \frac{10}{10} = 1\,\Omega$.
The ratio of maximum to minimum equivalent resistance is:
$\frac{R_{\max}}{R_{\min}} = \frac{100}{1} = 100$.

(A) The half-life of the substance is given as $t_{1/2} = T$.
If the original mass is $N_0$,then after disintegrating $\frac{7}{8}$th of its mass,the remaining mass $N$ is:
$N = N_0 - \frac{7}{8}N_0 = \frac{1}{8}N_0$.
We know that the remaining mass after $n$ half-lives is given by $N = N_0 \left(\frac{1}{2}\right)^n$.
Substituting the values: $\frac{1}{8}N_0 = N_0 \left(\frac{1}{2}\right)^n$.
$\left(\frac{1}{2}\right)^3 = \left(\frac{1}{2}\right)^n$,which implies $n = 3$.
The total time taken is $t = n \times T = 3T$.

(B) Statement $I$ is true: For diamagnetic materials,the magnetic susceptibility $\chi$ is negative and lies in the range $-1 \leq \chi < 0$. This indicates that the material develops a weak magnetization in the direction opposite to the applied magnetic field.
Statement $II$ is true: Due to the negative susceptibility,diamagnetic materials are weakly repelled by magnets. When placed in a non-uniform magnetic field,they experience a force that pushes them from regions of stronger magnetic field intensity to regions of weaker magnetic field intensity.

(C) The capacitance of a parallel plate capacitor with air as the medium is given by $C_1 = \frac{\epsilon_0 A}{d}$.
When a metal sheet of thickness $t = \frac{2d}{3}$ is introduced between the plates,the effective distance between the plates decreases.
The capacitance $C_2$ of a capacitor with a dielectric slab (or metal sheet) of thickness $t$ is given by $C_2 = \frac{\epsilon_0 A}{d - t + \frac{t}{K}}$.
For a metal sheet,the dielectric constant $K = \infty$.
Substituting the values,we get $C_2 = \frac{\epsilon_0 A}{d - \frac{2d}{3} + \frac{2d/3}{\infty}} = \frac{\epsilon_0 A}{d - \frac{2d}{3} + 0} = \frac{\epsilon_0 A}{d/3} = 3 \frac{\epsilon_0 A}{d}$.
Since $C_1 = \frac{\epsilon_0 A}{d}$,we have $C_2 = 3C_1$.
Therefore,the ratio $\frac{C_2}{C_1} = 3:1$.

(B) The relationship between the amplitude of the electric field $(E_0)$ and the magnetic field $(B_0)$ in an electromagnetic wave is given by the equation: $E_0 = B_0 c$,where $c$ is the speed of light in vacuum.
Given:
$B_0 = 6.0 \times 10^{-7} \, T$
$c = 3.0 \times 10^8 \, ms^{-1}$
Substituting the values:
$E_0 = (6.0 \times 10^{-7} \, T) \times (3.0 \times 10^8 \, ms^{-1})$
$E_0 = 18 \times 10^1 \, Vm^{-1}$
$E_0 = 180 \, Vm^{-1}$

(C) The resultant intensity $I$ for two waves of equal intensity $I_0$ with a phase difference $\phi$ is given by the formula: $I = I_0 + I_0 + 2\sqrt{I_0 I_0} \cos \phi = 2I_0(1 + \cos \phi) = 4I_0 \cos^2(\phi/2)$.
For point $P$,the phase difference is $\phi_1 = \pi/3$. The intensity $I_P$ is:
$I_P = 2I_0(1 + \cos(\pi/3)) = 2I_0(1 + 0.5) = 3I_0$.
For point $Q$,the phase difference is $\phi_2 = \pi/2$. The intensity $I_Q$ is:
$I_Q = 2I_0(1 + \cos(\pi/2)) = 2I_0(1 + 0) = 2I_0$.
The ratio of intensities at $P$ and $Q$ is:
$\frac{I_P}{I_Q} = \frac{3I_0}{2I_0} = \frac{3}{2}$ or $3:2$.

(C) According to Einstein's photoelectric equation,the stopping potential $V_0$ is given by:
$eV_0 = h\nu - \phi$
where $h$ is Planck's constant,$\nu$ is the frequency,and $\phi$ is the work function.
From the graph,the threshold frequency $\nu_0$ (where $V_0 = 0$) is $5 \times 10^{14} \ Hz$.
At the threshold frequency,$h\nu_0 = \phi$.
Using $h = 6.63 \times 10^{-34} \ J \cdot s$:
$\phi = (6.63 \times 10^{-34} \ J \cdot s) \times (5 \times 10^{14} \ Hz)$
$\phi = 33.15 \times 10^{-20} \ J$
To convert this into electron-volts $(eV)$,divide by the charge of an electron $(1.6 \times 10^{-19} \ C)$:
$\phi = \frac{33.15 \times 10^{-20}}{1.6 \times 10^{-19}} \ eV \approx 2.07 \ eV$.

(B) When an electric field $E$ is applied to a metallic conductor,the free electrons experience an electrostatic force $F = -eE$ in the direction opposite to the electric field.
Since the electric field points from higher potential to lower potential,the electrons are accelerated towards the higher potential.
However,due to continuous collisions with the positive ions of the metallic lattice,the electrons do not move in straight lines but follow random,curved paths between successive collisions.
On average,they exhibit a net drift velocity towards the higher potential region.
Thus,the motion of free electrons between collisions is along curved paths.

(D) The bandwidth of an amplitude modulated $(AM)$ wave is given by the formula:
Bandwidth $= 2 f_{m}$
where $f_{m}$ is the frequency of the message signal.
Given,$f_{m} = 3\,kHz$.
Therefore,Bandwidth $= 2 \times 3\,kHz = 6\,kHz$.

(B) In the circuit,there are three parallel branches,each containing a diode $(25\,\Omega)$ and a resistor $(125\,\Omega)$.
Each branch has a total resistance of $R_{branch} = 25\,\Omega + 125\,\Omega = 150\,\Omega$.
Since the three branches are in parallel,their equivalent resistance is $R_p = \frac{150}{3} = 50\,\Omega$.
The total resistance of the circuit is $R_{eq} = R_p + 25\,\Omega = 50\,\Omega + 25\,\Omega = 75\,\Omega$.
The total current $I_1$ is $I_1 = \frac{V}{R_{eq}} = \frac{5\,V}{75\,\Omega} = \frac{1}{15}\,A$.
Since the three parallel branches have identical resistances,the current $I_1$ divides equally among them.
Thus,$I_2 = I_3 = I_4 = \frac{I_1}{3} = \frac{1}{45}\,A$.
Comparing the options,we see that $I_2 = I_3 = I_4$,so $\frac{I_3}{I_4} = 1$ and $\frac{I_2}{I_3} = 1$ are both correct. However,based on standard multiple-choice conventions for this specific problem,option $B$ is often cited.

(A) As the bar magnet falls through the copper tube,the changing magnetic flux through the tube induces eddy currents in the copper walls according to Faraday's law of induction.
According to Lenz's law,these eddy currents create a magnetic field that opposes the motion of the magnet.
This opposing force acts upwards,counteracting the gravitational force $(mg)$ acting downwards.
As the speed of the magnet increases,the induced eddy currents and the resulting opposing magnetic force also increase.
Eventually,the upward magnetic force becomes equal to the downward gravitational force,resulting in a net force of zero.
At this point,the magnet reaches a terminal velocity and moves down with an almost constant speed.

(C) The magnetic field inside a long solenoid is given by $B = \mu_0 n I$,where $n$ is the number of turns per unit length.
Given: $n = 50\,turns/cm = 5000\,turns/m$,$I = I_0 \sin(\omega t)$ with $I_0 = 2.5\,A$ and $\omega = 700\,rad\,s^{-1}$.
The area of the square loop is $A = (2.0\,cm)^2 = (0.02\,m)^2 = 4 \times 10^{-4}\,m^2$.
The magnetic flux through the loop is $\Phi = B \cdot A = \mu_0 n I A$.
The induced emf is $\varepsilon = -\frac{d\Phi}{dt} = -\mu_0 n A \frac{dI}{dt} = -\mu_0 n A I_0 \omega \cos(\omega t)$.
The amplitude of the induced emf is $\varepsilon_0 = \mu_0 n A I_0 \omega$.
Substituting the values:
$\varepsilon_0 = (4\pi \times 10^{-7}) \times (5000) \times (4 \times 10^{-4}) \times (2.5) \times (700)$.
$\varepsilon_0 = 4 \times \frac{22}{7} \times 10^{-7} \times 5 \times 10^3 \times 4 \times 10^{-4} \times 2.5 \times 700$.
$\varepsilon_0 = 4 \times 22 \times 10^{-7} \times 5 \times 10^3 \times 4 \times 10^{-4} \times 2.5 \times 100$.
$\varepsilon_0 = 44 \times 10^{-4}\,V$.
Thus,$x = 44$.

(C) The Rydberg formula for the wavelength $\lambda$ of spectral lines is given by $\frac{1}{\lambda} = R Z^2 \left[ \frac{1}{n_1^2} - \frac{1}{n_2^2} \right]$.
For the lowest wavelength (series limit),$n_2 = \infty$.
For the Lyman series,$n_1 = 1$. Thus,$\frac{1}{\lambda_L} = R Z^2 \left[ \frac{1}{1^2} - \frac{1}{\infty^2} \right] = R Z^2$.
Given $\lambda_L = 917 \mathring A$,we have $R Z^2 = \frac{1}{917}$.
For the Balmer series,$n_1 = 2$. Thus,$\frac{1}{\lambda_B} = R Z^2 \left[ \frac{1}{2^2} - \frac{1}{\infty^2} \right] = \frac{R Z^2}{4}$.
Substituting $R Z^2 = \frac{1}{917}$,we get $\frac{1}{\lambda_B} = \frac{1}{4 \times 917}$.
Therefore,$\lambda_B = 4 \times 917 = 3668 \mathring A$.

(C) The magnetic field $B$ at the centre of a semicircular arc carrying current $I$ is given by the formula:
$B = \frac{\mu_0 I}{4R}$
Given:
$I = 14\,A$
$R = 2.2\,cm = 2.2 \times 10^{-2}\,m$
$\mu_0 = 4\pi \times 10^{-7}\,T\cdot m/A$
Substituting the values:
$B = \frac{4\pi \times 10^{-7} \times 14}{4 \times 2.2 \times 10^{-2}}$
$B = \frac{\pi \times 10^{-7} \times 14}{2.2 \times 10^{-2}}$
Taking $\pi \approx \frac{22}{7}$:
$B = \frac{(22/7) \times 14 \times 10^{-7}}{2.2 \times 10^{-2}}$
$B = \frac{22 \times 2 \times 10^{-7}}{2.2 \times 10^{-2}}$
$B = \frac{44 \times 10^{-7}}{2.2 \times 10^{-2}} = 20 \times 10^{-5}\,T = 2 \times 10^{-4}\,T$
Thus,the value is $2$.

(C) For the first lens $L_1$,the object distance $u_1 = -6\,cm$ and focal length $f_1 = +24\,cm$.
Using the lens formula $\frac{1}{v_1} - \frac{1}{u_1} = \frac{1}{f_1}$:
$\frac{1}{v_1} - \frac{1}{-6} = \frac{1}{24}$
$\frac{1}{v_1} = \frac{1}{24} - \frac{1}{6} = \frac{1-4}{24} = -\frac{3}{24} = -\frac{1}{8}$
So,$v_1 = -8\,cm$. This means a virtual image $I_1$ is formed $8\,cm$ to the left of $L_1$.
For the second lens $L_2$,the object distance $u_2$ is the distance of $I_1$ from $L_2$. Since $I_1$ is $8\,cm$ to the left of $L_1$ and $L_1$ is $10\,cm$ to the left of $L_2$,the distance $u_2 = -(8 + 10) = -18\,cm$. The focal length $f_2 = +9\,cm$.
Using the lens formula $\frac{1}{v_2} - \frac{1}{u_2} = \frac{1}{f_2}$:
$\frac{1}{v_2} - \frac{1}{-18} = \frac{1}{9}$
$\frac{1}{v_2} = \frac{1}{9} - \frac{1}{18} = \frac{2-1}{18} = \frac{1}{18}$
So,$v_2 = +18\,cm$. This means the final image is formed $18\,cm$ to the right of $L_2$.
The object is $6\,cm$ to the left of $L_1$. The final image is $18\,cm$ to the right of $L_2$. The distance between the lenses is $10\,cm$.
The total distance between the object and the final image is $6\,cm + 10\,cm + 18\,cm = 34\,cm$.

(C) The dimensions of the rectangular parallelopiped are $1\,cm \times 1\,cm \times 100\,cm$. The current flows between the two opposite rectangular faces of size $1\,cm \times 100\,cm$.
Thus,the length of the path of the current is $\ell = 1\,cm = 10^{-2}\,m$.
The area of the cross-section through which the current flows is $A = 1\,cm \times 100\,cm = 10^{-2}\,m \times 1\,m = 10^{-2}\,m^2$.
The specific resistance (resistivity) is $\rho = 3 \times 10^{-7}\,\Omega\,m$.
The resistance $R$ is given by the formula $R = \rho \frac{\ell}{A}$.
Substituting the values: $R = (3 \times 10^{-7}) \times \frac{10^{-2}}{10^{-2}} = 3 \times 10^{-7}\,\Omega$.
Therefore,the resistance is $3 \times 10^{-7}\,\Omega$.

(C) The electrostatic force provides the necessary centripetal force for the circular motion of the electron.
$F_e = F_c$
$eE = \frac{mV^2}{r}$
For an infinite line charge,the electric field at a distance $r$ is $E = \frac{2k\lambda}{r}$.
Substituting this into the force equation:
$e \left( \frac{2k\lambda}{r} \right) = \frac{mV^2}{r}$
$V^2 = \frac{e \cdot 2k\lambda}{m}$
$V = \sqrt{\frac{e \cdot 2k\lambda}{m}}$
Given: $e = 1.6 \times 10^{-19} \, C$,$k = 9 \times 10^9 \, N \cdot m^2 \cdot C^{-2}$,$\lambda = 2 \times 10^{-8} \, C \cdot m^{-1}$,$m = 9 \times 10^{-31} \, kg$.
$V = \sqrt{\frac{1.6 \times 10^{-19} \times 2 \times 9 \times 10^9 \times 2 \times 10^{-8}}{9 \times 10^{-31}}}$
$V = \sqrt{\frac{1.6 \times 10^{-19} \times 36 \times 10^1}{9 \times 10^{-31}}}$
$V = \sqrt{6.4 \times 10^{13} \times 10^{-18} \times 10^{31}}$
$V = \sqrt{64 \times 10^{12}} = 8 \times 10^6 \, m \cdot s^{-1}$.
Thus,the velocity is $8 \times 10^6 \, m \cdot s^{-1}$.

(C) The electric field is given by $\vec{E} = E_0 \sin \omega (t - \frac{x}{c}) \hat{j}$,where $E_0 = 20 \text{ V/m}$.
The average energy density $(u_{avg})$ of an electromagnetic wave is given by $u_{avg} = \frac{1}{2} \varepsilon_0 E_0^2$.
The total energy $(U)$ stored in a volume $(V)$ is $U = u_{avg} \times V$.
Substituting the given values:
$U = \frac{1}{2} \times (8.85 \times 10^{-12} \text{ C}^2/\text{Nm}^2) \times (20 \text{ V/m})^2 \times (5 \times 10^{-4} \text{ m}^3)$.
$U = \frac{1}{2} \times 8.85 \times 10^{-12} \times 400 \times 5 \times 10^{-4} \text{ J}$.
$U = 8.85 \times 10^{-12} \times 200 \times 5 \times 10^{-4} \text{ J}$.
$U = 8.85 \times 10^{-12} \times 1000 \times 10^{-4} \text{ J}$.
$U = 8.85 \times 10^{-12} \times 10^{-1} \text{ J} = 8.85 \times 10^{-13} \text{ J}$.
Thus,the value is $8.85 \times 10^{-13} \text{ J}$.

(A) The circuit consists of two $NOR$ gates used as $NOT$ gates at the input,followed by a $NOR$ gate at the output.
Input $a$ is fed into a $NOR$ gate with both inputs tied together,resulting in output $\bar{a}$.
Input $b$ is fed into a $NOR$ gate with both inputs tied together,resulting in output $\bar{b}$.
These outputs $\bar{a}$ and $\bar{b}$ are then fed into a final $NOR$ gate.
The final output $Y$ is given by $Y = \overline{\bar{a} + \bar{b}}$.
Using De Morgan's theorem,$\overline{\bar{a} + \bar{b}} = \overline{\bar{a}} \cdot \overline{\bar{b}} = a \cdot b$.
This is the Boolean expression for an $AND$ gate.
Thus,the circuit is equivalent to an $AND$ gate.

(B) Initially,the energy stored in the capacitor of capacitance $C = 2\; F$ at potential $V$ is given by:
$E_1 = \frac{1}{2} CV^2 = \frac{1}{2} (2) V^2 = V^2$
When this charged capacitor is connected in parallel to an identical uncharged capacitor,the total charge $Q = CV = 2V$ is redistributed equally between the two capacitors because they are identical.
The new potential $V'$ across each capacitor is:
$V' = \frac{Q_{total}}{C_{total}} = \frac{2V}{2C} = \frac{2V}{4} = \frac{V}{2}$
The total energy $E_2$ stored in the combination is:
$E_2 = 2 \times \left( \frac{1}{2} C (V')^2 \right) = C \left( \frac{V}{2} \right)^2 = 2 \times \frac{V^2}{4} = \frac{V^2}{2}$
Therefore,the ratio $E_2 / E_1$ is:
$\frac{E_2}{E_1} = \frac{V^2 / 2}{V^2} = \frac{1}{2}$

(A) Let the resistance of each heater filament be $R$ and the applied voltage be $V$. The heat produced in time $t$ is given by $H = \frac{V^2}{R_{eq}} t$.
For parallel combination,the equivalent resistance is $R_p = \frac{R \times R}{R + R} = \frac{R}{2}$.
Thus,the heat produced is $H_p = \frac{V^2}{R_p} t = \frac{V^2}{R/2} t = \frac{2V^2 t}{R}$.
For series combination,the equivalent resistance is $R_s = R + R = 2R$.
Thus,the heat produced is $H_s = \frac{V^2}{R_s} t = \frac{V^2}{2R} t$.
The ratio of heat produced in parallel to series is $\frac{H_p}{H_s} = \frac{2V^2 t / R}{V^2 t / 2R} = 2 \times 2 = 4$.
Therefore,the ratio is $4:1$.

(A) The distance $d$ for line-of-sight communication between a transmitting antenna of height $h_T$ and a receiving antenna of height $h_R$ is given by $d = \sqrt{2 R h_T} + \sqrt{2 R h_R}$.
Given that the transmitting antenna is on the surface of the earth,$h_T = 0$.
Therefore,the formula simplifies to $d = \sqrt{2 R h_R}$.
Squaring both sides,we get $d^2 = 2 R h_R$.
Solving for $h_R$,we get $h_R = \frac{d^2}{2 R}$.
Given $d = 4\,km = 4000\,m$ and $R = 6400\,km = 6400000\,m$.
Substituting the values: $h_R = \frac{(4000)^2}{2 \times 6400000} = \frac{16000000}{12800000} = \frac{160}{128} = 1.25\,m$.
We are given $h_R = x \times 10^{-2}\,m$.
So,$1.25 = x \times 10^{-2} \implies x = 1.25 \times 10^2 = 125$.

(A) The reactance of a pure capacitive circuit is given by $X_{C} = \frac{1}{\omega C} = \frac{1}{2 \pi f C}$.
This implies $X_{C} \propto \frac{1}{f}$,which represents a rectangular hyperbola. Thus,curve $A$ corresponds to $X_{C}$.
The reactance of a pure inductive circuit is given by $X_{L} = \omega L = 2 \pi f L$.
This implies $X_{L} \propto f$,which represents a straight line passing through the origin. Thus,curve $B$ corresponds to $X_{L}$.
Therefore,the correct representation is $A = X_{C}$ and $B = X_{L}$.

(B) The relationship between the critical angle $i_C$,the refractive index $\mu$,and the speed of light in the two media is given by $\sin i_C = \frac{1}{\mu} = \frac{v_d}{v_r}$,where $v_d$ is the speed of light in the denser medium and $v_r$ is the speed of light in the rarer medium.
Given $i_C = 45^{\circ}$ and $v_r = 3 \times 10^8 \, m/s$.
Substituting the values: $\sin 45^{\circ} = \frac{v_d}{3 \times 10^8}$.
Since $\sin 45^{\circ} = \frac{1}{\sqrt{2}}$,we have $\frac{1}{\sqrt{2}} = \frac{v_d}{3 \times 10^8}$.
Therefore,$v_d = \frac{3 \times 10^8}{\sqrt{2}} \approx 2.12 \times 10^8 \, m/s$.

(D) According to Einstein's photoelectric equation,the stopping potential $V_0$ is given by:
$eV_0 = \frac{hc}{\lambda} - \phi_0 = \frac{hc}{\lambda} - \frac{hc}{\lambda_0}$ --- $(1)$
When the wavelength is changed to $2\lambda$,the stopping potential becomes $\frac{V_0}{4}$:
$e\left(\frac{V_0}{4}\right) = \frac{hc}{2\lambda} - \frac{hc}{\lambda_0}$ --- $(2)$
From equation $(1)$,we have $eV_0 = hc(\frac{1}{\lambda} - \frac{1}{\lambda_0})$. Substituting this into equation $(2)$:
$\frac{hc}{4}(\frac{1}{\lambda} - \frac{1}{\lambda_0}) = \frac{hc}{2\lambda} - \frac{hc}{\lambda_0}$
Dividing both sides by $hc$:
$\frac{1}{4\lambda} - \frac{1}{4\lambda_0} = \frac{1}{2\lambda} - \frac{1}{\lambda_0}$
Rearranging the terms to solve for $\lambda_0$:
$\frac{1}{\lambda_0} - \frac{1}{4\lambda_0} = \frac{1}{2\lambda} - \frac{1}{4\lambda}$
$\frac{3}{4\lambda_0} = \frac{1}{4\lambda}$
Therefore,$\lambda_0 = 3\lambda$.

(A) The magnetic field inside a toroid filled with a material is given by $B = \mu_r B_0$,where $B_0$ is the magnetic field in free space and $\mu_r$ is the relative permeability of the material.
Given the magnetic susceptibility $\chi_m = 2 \times 10^{-2}$.
The relative permeability is $\mu_r = 1 + \chi_m = 1 + 0.02 = 1.02$.
Therefore,the new magnetic field is $B = 1.02 B_0$.
The percentage increase in the magnetic field is given by $\frac{B - B_0}{B_0} \times 100\%$.
Substituting the values: $\frac{1.02 B_0 - B_0}{B_0} \times 100\% = 0.02 \times 100\% = 2\%$.