JEE Main 2022 Physics Question Paper with Answer and Solution

(C) The horizontal range $R$ is given by $R = \frac{u^2 \sin 2\theta}{g}$.
For maximum range,$\theta = 45^{\circ}$,so $R_{\max} = \frac{u^2}{g}$.
For the second object,the range is $R' = \frac{R_{\max}}{2} = \frac{u^2}{2g}$.
Equating the range formula for the second object: $\frac{u^2 \sin 2\theta'}{g} = \frac{u^2}{2g}$.
This simplifies to $\sin 2\theta' = \frac{1}{2}$.
Therefore,$2\theta' = 30^{\circ}$ or $2\theta' = 150^{\circ}$.
Solving for $\theta'$,we get $\theta' = 15^{\circ}$ or $\theta' = 75^{\circ}$.

(C) The acceleration due to gravity at a height $h$ above the Earth's surface is given by $g_h = g(1 - \frac{2h}{R_e})$ for $h \ll R_e$.
The acceleration due to gravity at a depth $d$ below the Earth's surface is given by $g_d = g(1 - \frac{d}{R_e})$.
According to the problem,$g_h = g_d$,so:
$g(1 - \frac{2h}{R_e}) = g(1 - \frac{d}{R_e})$
Canceling $g$ from both sides and simplifying:
$1 - \frac{2h}{R_e} = 1 - \frac{d}{R_e}$
$\frac{2h}{R_e} = \frac{d}{R_e}$
$d = 2h$
Given that $d = \alpha h$,we compare the two expressions:
$\alpha h = 2h$
$\alpha = 2$.

(C) For an adiabatic process,the relation between pressure $P$ and density $d$ is given by $P \propto d^{\gamma}$.
Thus,$\frac{P_{2}}{P_{1}} = \left(\frac{d_{2}}{d_{1}}\right)^{\gamma}$.
Given $\frac{d_{2}}{d_{1}} = 32$ and $\gamma = \frac{7}{5}$,we have $\frac{P_{2}}{P_{1}} = (32)^{7/5} = (2^5)^{7/5} = 2^7 = 128$.
Using the ideal gas law $P = \frac{dRT}{M}$,we have $T \propto \frac{P}{d}$.
Therefore,$\frac{T_{2}}{T_{1}} = \frac{P_{2}}{P_{1}} \times \frac{d_{1}}{d_{2}}$.
Substituting the values,$\frac{T_{2}}{T_{1}} = 128 \times \frac{1}{32} = 4$.
So,the temperature becomes $4$ times its initial temperature.

(D) The molar specific heat at constant volume for a mixture is given by the formula: $C_{V_{mix}} = \frac{n_1 C_{V_1} + n_2 C_{V_2}}{n_1 + n_2}$.
For a monoatomic gas,$C_{V_1} = \frac{3}{2} R$ and $n_1 = 1$.
For a diatomic gas without vibrational modes,$C_{V_2} = \frac{5}{2} R$ and $n_2 = 3$.
Substituting these values into the formula:
$C_{V_{mix}} = \frac{1 \cdot (\frac{3}{2} R) + 3 \cdot (\frac{5}{2} R)}{1 + 3} = \frac{\frac{3}{2} R + \frac{15}{2} R}{4} = \frac{\frac{18}{2} R}{4} = \frac{9 R}{4}$.
Given that $C_{V_{mix}} = \frac{\alpha^2}{4} R$,we equate the two expressions:
$\frac{9 R}{4} = \frac{\alpha^2}{4} R$.
This implies $\alpha^2 = 9$,so $\alpha = 3$ (taking the positive value).

(D) For complete absorption,the force $F$ exerted by light on a surface is given by $F = \frac{I \cdot A}{c}$,where $I$ is the intensity (energy flux),$A$ is the area,and $c$ is the speed of light $(3 \times 10^{8} \, m/s)$.
Given: $A = 36 \, cm^{2} = 36 \times 10^{-4} \, m^{2}$,$F = 7.2 \times 10^{-9} \, N$.
Rearranging for $I$: $I = \frac{F \cdot c}{A}$.
Substituting the values: $I = \frac{7.2 \times 10^{-9} \times 3 \times 10^{8}}{36 \times 10^{-4}}$.
$I = \frac{21.6 \times 10^{-1}}{36 \times 10^{-4}} = 0.6 \times 10^{3} \, W/m^{2} = 600 \, W/m^{2}$.
To convert to $W/cm^{2}$: $I = \frac{600 \, W}{10^{4} \, cm^{2}} = 0.06 \, W/cm^{2}$.

(B) The lens maker's formula for power $P$ in a medium with refractive index $\mu_2$ is given by:
$P = \frac{1}{f} = \left( \frac{\mu_1}{\mu_2} - 1 \right) \left( \frac{1}{R_1} - \frac{1}{R_2} \right)$
Here,$\mu_1 = 1.5$ (refractive index of lens),$R_1 = 0.2\,m$,$R_2 = -0.4\,m$ (for a biconvex lens),and $P = 1.25\,m^{-1}$.
Substituting the values:
$1.25 = \left( \frac{1.5}{\mu_2} - 1 \right) \left( \frac{1}{0.2} - \frac{1}{-0.4} \right)$
$1.25 = \left( \frac{1.5 - \mu_2}{\mu_2} \right) \left( 5 + 2.5 \right)$
$1.25 = \left( \frac{1.5 - \mu_2}{\mu_2} \right) (7.5)$
$\frac{1.25}{7.5} = \frac{1.5 - \mu_2}{\mu_2}$
$\frac{1}{6} = \frac{1.5 - \mu_2}{\mu_2}$
$\mu_2 = 9 - 6\mu_2$
$7\mu_2 = 9$
$\mu_2 = \frac{9}{7}$

(C) According to Einstein's photoelectric equation,the maximum kinetic energy $K_{max}$ is given by $K_{max} = E - \phi$,where $E$ is the energy of the incident photon and $\phi$ is the work function.
For the first case,$E_1 = 5\phi$. Therefore,$K_{max,1} = 5\phi - \phi = 4\phi$.
Since $K_{max,1} = \frac{1}{2}mv_1^2$,we have $\frac{1}{2}mv_1^2 = 4\phi$.
For the second case,$E_2 = 10\phi$. Therefore,$K_{max,2} = 10\phi - \phi = 9\phi$.
Since $K_{max,2} = \frac{1}{2}mv_2^2$,we have $\frac{1}{2}mv_2^2 = 9\phi$.
Taking the ratio of the two equations:
$\frac{\frac{1}{2}mv_1^2}{\frac{1}{2}mv_2^2} = \frac{4\phi}{9\phi}$
$\frac{v_1^2}{v_2^2} = \frac{4}{9}$
Taking the square root on both sides,we get $\frac{v_1}{v_2} = \frac{2}{3}$.
Thus,the ratio of the maximum velocities is $2:3$.

(A) Let the initial quantity be $N_0$.
Given that the sample decays by $\frac{7}{8}$ of its original quantity,the remaining quantity $N$ is:
$N = N_0 - \frac{7}{8}N_0 = \frac{1}{8}N_0$.
We know the radioactive decay law is $N = N_0 (\frac{1}{2})^n$,where $n$ is the number of half-lives.
$\frac{1}{8}N_0 = N_0 (\frac{1}{2})^n$
$(\frac{1}{2})^3 = (\frac{1}{2})^n$
Thus,$n = 3$.
Since $n = \frac{t}{T_{1/2}}$,where $t = 15$ minutes and $T_{1/2}$ is the half-life:
$3 = \frac{15}{T_{1/2}}$
$T_{1/2} = \frac{15}{3} = 5$ minutes.

(B) The voltage gain $A_v$ of a common emitter amplifier is given by the formula:
$A_v = \frac{v_{\text{out}}}{v_{\text{in}}} = \beta \frac{R_{\text{out}}}{R_{\text{in}}}$
Given:
$\beta = 100$
$R_{\text{in}} = 1 \text{ k}\Omega = 10^3 \, \Omega$
$R_{\text{out}} = 10 \text{ k}\Omega = 10^4 \, \Omega$
$v_{\text{in}} = 1 \text{ mV} = 10^{-3} \, \text{V}$
Substituting the values:
$v_{\text{out}} = v_{\text{in}} \times \beta \times \frac{R_{\text{out}}}{R_{\text{in}}}$
$v_{\text{out}} = 10^{-3} \times 100 \times \frac{10 \times 10^3}{1 \times 10^3}$
$v_{\text{out}} = 10^{-3} \times 100 \times 10$
$v_{\text{out}} = 1 \, \text{V}$

(D) Given:
Modulating frequency $f_m = 20\,kHz$.
Deviation ratio $\beta = 10$.
The deviation ratio is defined as $\beta = \frac{\Delta f}{f_m}$,where $\Delta f$ is the frequency deviation.
Therefore,$\Delta f = \beta \times f_m = 10 \times 20\,kHz = 200\,kHz$.
According to Carson's rule,the bandwidth $BW$ required for an $FM$ signal is given by:
$BW = 2(\Delta f + f_m)$
$BW = 2(200\,kHz + 20\,kHz)$
$BW = 2(220\,kHz) = 440\,kHz$.

(A) First,calculate the resistance of each bulb using the formula $R = \frac{V^2}{P}$.
For the $100\,W$ bulb: $R_1 = \frac{220^2}{100} = 484\,\Omega$.
For the $60\,W$ bulb: $R_2 = \frac{220^2}{60} = \frac{4840}{6} = 806.67\,\Omega$.
When connected in series,the total resistance $R_{eq} = R_1 + R_2 = 484 + 806.67 = 1290.67\,\Omega$.
The current $I$ flowing through the series combination is $I = \frac{V_{total}}{R_{eq}} = \frac{220}{1290.67} \approx 0.17045\,A$.
The power consumed by the $100\,W$ bulb is $P_1 = I^2 R_1 = (0.17045)^2 \times 484 \approx 0.02905 \times 484 \approx 14.06\,W$.
Thus,the power consumed is approximately $14\,W$.

(D) Just after closing the switch $S$ $(t = 0)$,the inductor opposes any change in current,so it behaves as an open circuit (infinite resistance).
Therefore,the branch containing the inductor carries no current.
The circuit simplifies to the $6\,V$ battery in series with the $2\,\Omega$ resistor and the $4\,\Omega$ resistor.
The total resistance of the circuit is $R_{eq} = 2\,\Omega + 4\,\Omega = 6\,\Omega$.
Using Ohm's law,the current through the battery is $I = \frac{V}{R_{eq}} = \frac{6\,V}{6\,\Omega} = 1\,A$.

(D) Given: Initial distance of object $u_0 = -100\,cm$. Radius of curvature $R = -200\,cm$. Focal length $f = R/2 = -100\,cm$.
Speed of object $v_{obj} = 2\,cm/s$ towards the mirror.
After time $t = 10\,s$,the distance moved by the object is $d = v_{obj} \times t = 2\,cm/s \times 10\,s = 20\,cm$.
The new position of the object from the mirror is $u = u_0 + d = -100\,cm + 20\,cm = -80\,cm$.
Using the mirror formula: $\frac{1}{v} + \frac{1}{u} = \frac{1}{f}$.
Substituting the values: $\frac{1}{v} + \frac{1}{-80} = \frac{1}{-100}$.
$\frac{1}{v} = \frac{1}{80} - \frac{1}{100} = \frac{5 - 4}{400} = \frac{1}{400}$.
Therefore,$v = 400\,cm$.

(B) According to Newton's lens formula,when distances are measured from the focus,the relationship between the image distance $(v')$ and the object distance $(u')$ is given by $v' u' = f^2$,where $f$ is the focal length of the lens.
Given the equation of the curve is $v' u' = 225$.
Comparing this with $v' u' = f^2$,we get $f^2 = 225$.
Taking the square root of both sides,$f = \sqrt{225} = 15 \, cm$.
Therefore,the magnitude of the focal length of the lens is $15 \, cm$.

(C) The energy density $u$ associated with a magnetic field $B$ is given by the formula $u = \frac{B^{2}}{2\mu_{0}}$.
Here,$u$ represents the energy per unit volume.
The dimensions of energy are $[ML^{2}T^{-2}]$ and the dimensions of volume are $[L^{3}]$.
Therefore,the dimensions of energy density $u$ are $\frac{[ML^{2}T^{-2}]}{[L^{3}]} = [ML^{-1}T^{-2}]$.
Since $\frac{B^{2}}{\mu_{0}} = 2u$,the dimensions of $\left(\frac{B^{2}}{\mu_{0}}\right)$ are the same as the dimensions of $u$,which is $[ML^{-1}T^{-2}]$.

(A) Let the initial capacitance of each capacitor be $C = 40\,\mu F$.
When a dielectric of constant $K$ is inserted into one capacitor,its new capacitance becomes $C' = KC$.
The two capacitors are connected in series,so the equivalent capacitance $C_{eq}$ is given by:
$C_{eq} = \frac{C \cdot C'}{C + C'} = \frac{C \cdot (KC)}{C + KC} = \frac{KC}{K + 1}$
Given $C_{eq} = 24\,\mu F$ and $C = 40\,\mu F$,we have:
$24 = \frac{K \cdot 40}{K + 1}$
$24(K + 1) = 40K$
$24K + 24 = 40K$
$16K = 24$
$K = \frac{24}{16} = 1.5$

(A) Let the original length be $L_{1}$ and the original area be $A_{1}$. The original resistance is $R_{1} = \rho \frac{L_{1}}{A_{1}}$.
When the length is increased by twice its original length,the new length $L_{2} = L_{1} + 2L_{1} = 3L_{1}$.
Since the volume of the wire remains constant,$A_{1}L_{1} = A_{2}L_{2}$.
Thus,$A_{2} = A_{1} \frac{L_{1}}{L_{2}} = A_{1} \frac{L_{1}}{3L_{1}} = \frac{A_{1}}{3}$.
The new resistance is $R_{2} = \rho \frac{L_{2}}{A_{2}} = \rho \frac{3L_{1}}{A_{1}/3} = 9 \rho \frac{L_{1}}{A_{1}}$.
Therefore,the ratio $\frac{R_{2}}{R_{1}} = \frac{9 \rho (L_{1}/A_{1})}{\rho (L_{1}/A_{1})} = 9:1$.

(D) The current sensitivity $(I_s)$ of a galvanometer is defined as the deflection produced per unit current,given by the formula:
$I_s = \frac{\theta}{i} = \frac{NAB}{K}$
where:
$N$ = number of turns in the coil
$A$ = area of the coil
$B$ = magnetic field strength
$K$ = torsional constant of the spring
From the formula,$I_s \propto N$,$I_s \propto A$,$I_s \propto B$,and $I_s \propto \frac{1}{K}$.
To increase the current sensitivity,we must:
$1$. Increase the number of turns $(N)$.
$2$. Increase the area of the coil $(A)$.
$3$. Increase the magnetic field $(B)$.
$4$. Decrease the torsional constant of the spring $(K)$.
Comparing these requirements with the given options,increasing the magnetic field $(B)$ and decreasing the torsional constant $(D)$ will increase the current sensitivity. Therefore,the correct option is $(B)$ and $(D)$ only.

(A) The forces acting on the rod are its weight $(mg)$ acting vertically downwards, the normal force $(N)$ perpendicular to the inclined plane, and the magnetic force $(F_m = ILB)$ acting horizontally (since the magnetic field $B$ is vertical and the current $I$ is horizontal).
To keep the rod stationary, the component of the gravitational force along the inclined plane must be balanced by the component of the magnetic force along the inclined plane.
The component of the gravitational force along the incline is $mg \sin 45^{\circ}$.
The magnetic force $F_m = ILB$ acts horizontally. Its component along the inclined plane is $F_m \cos 45^{\circ} = ILB \cos 45^{\circ}$.
Equating these two components for equilibrium:
$mg \sin 45^{\circ} = ILB \cos 45^{\circ}$
Since $\sin 45^{\circ} = \cos 45^{\circ}$, we can simplify the equation:
$mg = ILB$
Given the linear density $\lambda = \frac{m}{L} = 0.45\,kg/m$, $g = 10\,m/s^2$, and $B = 0.15\,T$:
$I = \frac{mg}{LB} = \left(\frac{m}{L}\right) \frac{g}{B}$
Substituting the values:
$I = \frac{0.45 \times 10}{0.15} = \frac{4.5}{0.15} = 30\,A$
Therefore, the minimum current required is $30\,A$.

(D) The given current equation is $i = i_0 \sin (\omega t - 30^{\circ})$,where $i_0 = 5 \, A$ and $\omega = 49 \pi \, rad/s$.
In a purely inductive circuit,the inductive reactance is $X_L = \omega L$.
Given $L = 30 \, mH = 30 \times 10^{-3} \, H$.
$X_L = (49 \times \frac{22}{7}) \times 30 \times 10^{-3} = (7 \times 22) \times 30 \times 10^{-3} = 154 \times 30 \times 10^{-3} = 4.62 \, \Omega$.
The peak voltage is $v_0 = i_0 X_L = 5 \times 4.62 = 23.1 \, V$.
In a purely inductive circuit,the voltage leads the current by $90^{\circ}$.
Therefore,the phase of the voltage is $(-30^{\circ} + 90^{\circ}) = +60^{\circ}$.
The voltage equation is $v = v_0 \sin (\omega t + 60^{\circ}) = 23.1 \sin (49 \pi t + 60^{\circ})$.

(A) The refractive index of a medium is given by $n = \sqrt{\mu_{r} \varepsilon_{r}}$.
For medium $2$,the refractive index is $n_{2} = \sqrt{\mu_{r2} \varepsilon_{r2}} = \sqrt{1 \times 9} = 3$.
The speed of light in a medium is related to the speed of light in vacuum $c$ by the formula $v = \frac{c}{n}$.
Therefore,the speed of light in medium $2$ is $v_{2} = \frac{c}{n_{2}} = \frac{3 \times 10^{8} \, ms^{-1}}{3} = 1 \times 10^{8} \, ms^{-1}$.
Thus,the value is $1.0 \times 10^{8} \, ms^{-1}$.

(A) In normal adjustment,the length of the telescope $L$ is given by $L = f_o + f_e$,where $f_o$ is the focal length of the objective and $f_e$ is the focal length of the eyepiece.
Given $L = 30\,cm$,so $f_o + f_e = 30$ (Equation $1$).
The angular magnification $m$ for a telescope in normal adjustment is given by $m = \frac{f_o}{f_e}$.
Given $m = 2$,so $\frac{f_o}{f_e} = 2$,which implies $f_o = 2f_e$ (Equation $2$).
Substituting Equation $2$ into Equation $1$:
$2f_e + f_e = 30$
$3f_e = 30$
$f_e = 10\,cm$.
Now,substituting $f_e$ back into Equation $2$:
$f_o = 2 \times 10 = 20\,cm$.
Thus,the focal length of the objective is $20\,cm$.

(D) The de-Broglie wavelength is given by $\lambda = \frac{h}{p}$.
Since kinetic energy $K = \frac{p^2}{2m}$,we have $p = \sqrt{2mK}$.
For an electron accelerated through a potential $V$,the kinetic energy $K = eV$,where $e$ is the charge of the electron.
Thus,$\lambda = \frac{h}{\sqrt{2meV}}$.
Substituting the values of Planck's constant $h = 6.626 \times 10^{-34} \text{ J s}$,mass of electron $m = 9.11 \times 10^{-31} \text{ kg}$,and charge $e = 1.602 \times 10^{-19} \text{ C}$:
$\lambda = \frac{6.626 \times 10^{-34}}{\sqrt{2 \times 9.11 \times 10^{-31} \times 1.602 \times 10^{-19} \times V}} \text{ m}$.
Calculating this gives $\lambda \approx \frac{1.227 \times 10^{-9}}{\sqrt{V}} \text{ m} = \frac{1.227}{\sqrt{V}} \text{ nm}$.
Comparing this with $\lambda = \frac{1.227}{x} \text{ nm}$,we find $x = \sqrt{V}$.

(C) Given that the half-life $T_{1/2} = 60 \ days$.
If $\frac{7}{8}$ of the original mass disintegrates,the remaining mass $N$ is $N = N_0 - \frac{7}{8}N_0 = \frac{1}{8}N_0$.
The relation between remaining mass and initial mass is given by $N = N_0 \left(\frac{1}{2}\right)^n$,where $n$ is the number of half-lives.
Substituting the values: $\frac{1}{8}N_0 = N_0 \left(\frac{1}{2}\right)^n$.
This simplifies to $\left(\frac{1}{2}\right)^3 = \left(\frac{1}{2}\right)^n$,which gives $n = 3$.
The total time taken $t = n \times T_{1/2} = 3 \times 60 \ days = 180 \ days$.

(B) solar cell is a $p-n$ junction diode that converts light energy into electrical energy. It operates in the fourth quadrant of the $I-V$ characteristic curve. In this quadrant,the voltage is positive (forward bias) while the current is negative (as the device acts as a source,delivering power to an external circuit). Therefore,the characteristic curve of a solar cell is represented by the graph in the fourth quadrant.

(A) The modulation index is defined as $\mu = \frac{A_m}{A_c}$,where $A_m$ is the amplitude of the message signal and $A_c$ is the amplitude of the carrier wave.
To avoid distortion in the amplitude modulated wave,the modulation index must satisfy the condition $\mu \leq 1$.
If $\mu > 1$,over-modulation occurs,which leads to the distortion of the signal and interference between the carrier frequency and the message frequency.

(C) The radioactive decay law is given by $A = A_0 \times (1/2)^{t/T}$,where $A$ is the activity at time $t$,$A_0$ is the initial activity,and $T$ is the half-life.
Given,$A_0 = 64 \times A_{safe}$ and we want to find the time $t$ when $A = A_{safe}$.
Substituting these values: $A_{safe} = 64 \times A_{safe} \times (1/2)^{t/T}$.
This simplifies to $1/64 = (1/2)^{t/T}$.
Since $64 = 2^6$,we have $(1/2)^6 = (1/2)^{t/T}$.
Therefore,$t/T = 6$,which means $t = 6 \times T$.
Given the half-life $T = 2$ hours $30$ minutes = $2.5$ hours.
Thus,$t = 6 \times 2.5 = 15$ hours.

(C) The fringe width $\beta$ in a Young's double slit experiment is given by $\beta = \frac{\lambda D}{d}$.
Since $D$ and $d$ are constant,we have $\beta \propto \lambda$.
Therefore,$\frac{\beta_2}{\beta_1} = \frac{\lambda_2}{\lambda_1}$.
Given $\beta_1 = 7.2\,mm$,$\lambda_1 = 560\,nm$,and $\beta_2 = 8.1\,mm$.
Substituting the values: $\frac{8.1}{7.2} = \frac{\lambda_2}{560}$.
$\lambda_2 = \frac{8.1}{7.2} \times 560 = \frac{9}{8} \times 560$.
$\lambda_2 = 9 \times 70 = 630\,nm$.

(A) In an $LCR$ series circuit,the frequencies at which the current amplitude becomes $\frac{1}{\sqrt{2}}$ times its maximum value are known as half-power frequencies,denoted as $\omega_1$ and $\omega_2$.
The bandwidth of the circuit is given by $\Delta\omega = \omega_2 - \omega_1$.
Given $\omega_1 = 212\,rad\,s^{-1}$ and $\omega_2 = 232\,rad\,s^{-1}$,the bandwidth is $\Delta\omega = 232 - 212 = 20\,rad\,s^{-1}$.
The formula for bandwidth in an $LCR$ series circuit is $\Delta\omega = \frac{R}{L}$.
Substituting the given values: $20 = \frac{5}{L}$.
Solving for $L$: $L = \frac{5}{20} = 0.25\,H$.
Converting to millihenries $(mH)$: $L = 0.25 \times 1000 = 250\,mH$.

(A) The potential difference across the potentiometer wire $AB$ is given by $V_{AB} = I \times R_{AB}$,where $I$ is the current in the primary circuit.
The current $I$ is given by $I = \frac{E}{R + R_{AB}}$,where $E = 4\,V$ and $R_{AB} = 20\,\Omega$.
So,$V_{AB} = \left( \frac{4}{R + 20} \right) \times 20 = \frac{80}{R + 20}$.
The potential gradient along the wire is $k = \frac{V_{AB}}{L}$,where $L = 300\,cm$.
The emf of the secondary cell is $E' = 20\,mV = 20 \times 10^{-3}\,V$,and the null point length is $l = 60\,cm$.
At the null point,$E' = k \times l = \left( \frac{V_{AB}}{L} \right) \times l$.
Substituting the values: $20 \times 10^{-3} = \left( \frac{80}{R + 20} \right) \times \left( \frac{60}{300} \right)$.
$20 \times 10^{-3} = \left( \frac{80}{R + 20} \right) \times \left( \frac{1}{5} \right)$.
$20 \times 10^{-3} = \frac{16}{R + 20}$.
$R + 20 = \frac{16}{20 \times 10^{-3}} = \frac{16}{0.02} = 800$.
$R = 800 - 20 = 780\,\Omega$.

(A) The maximum torque experienced by an electric dipole in a uniform electric field is given by the formula $|\tau|_{\max} = PE$,where $P$ is the dipole moment and $E$ is the electric field strength.
For the first dipole: $P_1 = 1.2 \times 10^{-30} \, C \cdot m$ and $E_1 = 5 \times 10^{4} \, N \cdot C^{-1}$.
For the second dipole: $P_2 = 2.4 \times 10^{-30} \, C \cdot m$ and $E_2 = 15 \times 10^{4} \, N \cdot C^{-1}$.
The ratio of the maximum torques is given by:
$\frac{\tau_1}{\tau_2} = \frac{P_1 E_1}{P_2 E_2} = \frac{(1.2 \times 10^{-30}) \times (5 \times 10^{4})}{(2.4 \times 10^{-30}) \times (15 \times 10^{4})}$
Simplifying the expression:
$\frac{\tau_1}{\tau_2} = \left(\frac{1.2}{2.4}\right) \times \left(\frac{5}{15}\right) = \left(\frac{1}{2}\right) \times \left(\frac{1}{3}\right) = \frac{1}{6}$
Comparing this to $\frac{1}{x}$,we find that $x = 6$.

(B) Let the initial charges on spheres $A$ and $B$ be $q_A = q_B = q$. The initial force is $F = \frac{Kq^2}{r^2}$.
When sphere $C$ (initially uncharged) is placed in contact with $A$,the charge redistributes equally: $q_A' = q_C' = \frac{q}{2}$.
Now,sphere $C$ is placed in contact with $B$. The total charge is $q + \frac{q}{2} = \frac{3q}{2}$. Upon separation,each gets $q_B' = q_C'' = \frac{3q/2}{2} = \frac{3q}{4}$.
Now,$A$ has charge $\frac{q}{2}$ and $B$ has charge $\frac{3q}{4}$. Sphere $C$ is placed at the midpoint (distance $r/2$ from both).
The force on $C$ due to $A$ is $F_1 = \frac{K(q/2)(3q/4)}{(r/2)^2} = \frac{3Kq^2/8}{r^2/4} = \frac{3Kq^2}{2r^2} = \frac{3F}{2}$ (towards $A$).
The force on $C$ due to $B$ is $F_2 = \frac{K(3q/4)(3q/4)}{(r/2)^2} = \frac{9Kq^2/16}{r^2/4} = \frac{9Kq^2}{4r^2} = \frac{9F}{4}$ (towards $A$).
The net force on $C$ is $F_{net} = F_2 - F_1 = \frac{9F}{4} - \frac{3F}{2} = \frac{9F - 6F}{4} = \frac{3F}{4}$.

(C) When two large conducting plates with charges $q_{1}$ and $q_{2}$ are placed parallel to each other,the charge on the inner surfaces is given by $q_{inner} = \frac{q_{1}-q_{2}}{2}$.
The electric field $E$ between the plates is due to these inner charges: $E = \frac{\sigma}{\varepsilon_{0}} = \frac{q_{inner}}{A\varepsilon_{0}} = \frac{q_{1}-q_{2}}{2A\varepsilon_{0}}$.
The potential difference $V$ between the plates is given by $V = E \cdot d$,where $d$ is the separation between the plates.
Since the capacitance of a parallel plate capacitor is $C = \frac{A\varepsilon_{0}}{d}$,we can write $V = \frac{q_{1}-q_{2}}{2A\varepsilon_{0}} \cdot d = \frac{q_{1}-q_{2}}{2(A\varepsilon_{0}/d)}$.
Substituting $C$,we get $V = \frac{q_{1}-q_{2}}{2C}$.

$(A)$ Standard resistance coils require a resistance value that remains stable despite changes in ambient temperature.
The temperature coefficient of resistance $(\alpha)$ determines how much the resistance changes with temperature according to the formula $R_t = R_0(1 + \alpha \Delta T)$.
Alloys like constantan and manganin are specifically chosen because they possess a very low temperature coefficient of resistance.
This property ensures that the resistance of the coils remains nearly constant even if the temperature fluctuates.
Therefore, Assertion $A$ is true, Reason $R$ is true, and $R$ is the correct explanation of $A$.

(B) Let the total length of the wire be $L = 1\,m$. Let the length of part $X$ be $\ell_X$ and the length of part $Y$ be $\ell_Y$. Thus,$\ell_X + \ell_Y = 1$.
The resistance of a wire is given by $R = \rho \frac{\ell}{A}$.
For part $X$,$R_X = \rho \frac{\ell_X}{A_X}$. For part $Y$,$R_Y = \rho \frac{\ell_Y}{A_Y}$.
When wire $X$ is stretched to length $\ell_W = 2\ell_X$,its volume remains constant $(A_X \ell_X = A_W \ell_W)$.
Since $\ell_W = 2\ell_X$,we have $A_W = \frac{A_X}{2}$.
The resistance of wire $W$ is $R_W = \rho \frac{\ell_W}{A_W} = \rho \frac{2\ell_X}{A_X/2} = 4 \left( \rho \frac{\ell_X}{A_X} \right) = 4R_X$.
Given that $R_W = 2R_Y$,we substitute $R_W = 4R_X$ to get $4R_X = 2R_Y$,which implies $R_Y = 2R_X$.
Since both parts $X$ and $Y$ were cut from the same original wire,they have the same cross-sectional area $A$ and resistivity $\rho$. Thus,$R \propto \ell$.
Therefore,$\frac{R_X}{R_Y} = \frac{\ell_X}{\ell_Y}$.
Substituting $R_Y = 2R_X$,we get $\frac{R_X}{2R_X} = \frac{\ell_X}{\ell_Y} = \frac{1}{2}$.

(A) The force per unit length between two parallel wires carrying currents $I_1$ and $I_2$ separated by a distance $r$ is given by $f = \frac{\mu_0 I_1 I_2}{2 \pi r}$.
Here,$I_1 = 2 \; A$,$I_2 = 3 \; A$,$r = 5 \; cm = 0.05 \; m$,and the length of wire $X$ is $\ell = 50 \; cm = 0.5 \; m$.
The total force on wire $X$ is $F = f \times \ell = \frac{\mu_0 I_1 I_2 \ell}{2 \pi r}$.
Substituting the values:
$F = \frac{(4 \pi \times 10^{-7} \; T \cdot m/A) \times (2 \; A) \times (3 \; A) \times (0.5 \; m)}{2 \pi \times (0.05 \; m)}$
$F = \frac{2 \times 10^{-7} \times 6 \times 0.5}{0.05} = \frac{6 \times 10^{-7}}{0.05} = 120 \times 10^{-7} = 1.2 \times 10^{-5} \; N$.
Since the currents are in the same direction,the force is attractive,meaning wire $X$ is pulled towards wire $Y$.

(D) For element $X$,the current is in phase with the voltage,so $X$ is a resistor.
The resistance $R = \frac{V_0}{I_0} = \frac{100}{5} = 20\,\Omega$.
For element $Y$,the current lags behind the voltage by $\frac{\pi}{2}$,so $Y$ is an inductor.
The inductive reactance $X_L = \frac{V_0}{I_0} = \frac{100}{5} = 20\,\Omega$.
When $X$ and $Y$ are connected in series,the total impedance $Z = \sqrt{R^2 + X_L^2} = \sqrt{20^2 + 20^2} = 20\sqrt{2}\,\Omega$.
The peak current in the series circuit is $I_0 = \frac{V_0}{Z} = \frac{100}{20\sqrt{2}} = \frac{5}{\sqrt{2}}\,A$.
The rms current is $I_{rms} = \frac{I_0}{\sqrt{2}} = \frac{5/\sqrt{2}}{\sqrt{2}} = \frac{5}{2}\,A$.

(C) When unpolarised light of intensity $I_{in}$ passes through a polaroid, the intensity of the emergent polarised light is $I_{1} = \frac{1}{2} I_{in}$.
Given $I_{in} = 2 I_{0}$, the intensity after passing through polaroid $P$ is $I_{1} = \frac{1}{2} (2 I_{0}) = I_{0}$.
According to Malus' Law, when polarised light of intensity $I_{1}$ passes through a second polaroid whose transmission axis makes an angle $\theta$ with the incident light's polarisation direction, the emergent intensity $I_{2}$ is given by $I_{2} = I_{1} \cos^{2} \theta$.
Here, $\theta = 30^{\circ}$, so $I_{2} = I_{0} \cos^{2} 30^{\circ}$.
Since $\cos 30^{\circ} = \frac{\sqrt{3}}{2}$, we have $\cos^{2} 30^{\circ} = \frac{3}{4}$.
Therefore, $I_{2} = I_{0} \cdot \frac{3}{4} = \frac{3 I_{0}}{4}$.

(B) The kinetic energy $K$ acquired by a particle of charge $q$ accelerated through a potential difference $V$ is given by $K = qV$.
The linear momentum $p$ is related to kinetic energy by the formula $p = \sqrt{2mK} = \sqrt{2mqV}$.
For an $\alpha$ particle,the mass $m_{\alpha} = 4m_p$ and the charge $q_{\alpha} = 2e$,where $m_p$ is the mass of a proton and $e$ is the elementary charge.
For a proton,the mass is $m_p$ and the charge is $q_p = e$.
The ratio of their momenta is $\frac{p_{\alpha}}{p_p} = \sqrt{\frac{m_{\alpha} q_{\alpha}}{m_p q_p}} = \sqrt{\frac{4m_p \cdot 2e}{m_p \cdot e}} = \sqrt{4 \cdot 2} = \sqrt{8} = 2\sqrt{2}$.
Thus,the ratio is $2\sqrt{2} : 1$.

(B) The radius of a nucleus is given by $R = R_0 A^{1/3}$,where $A$ is the mass number.
The volume $V$ of the nucleus is given by $V = \frac{4}{3} \pi R^3 = \frac{4}{3} \pi (R_0 A^{1/3})^3 = \frac{4}{3} \pi R_0^3 A$.
Since $\frac{4}{3} \pi R_0^3$ is a constant,the volume $V$ is directly proportional to the mass number $A$. Thus,statement $(A)$ is correct.
The density $\rho$ of the nucleus is given by $\rho = \frac{\text{Mass}}{\text{Volume}} = \frac{m A}{V}$,where $m$ is the average mass of a nucleon.
Substituting $V \propto A$,we get $\rho = \frac{m A}{k A} = \frac{m}{k}$,where $k$ is a constant.
Since $\rho$ is a constant,the density of the nucleus is independent of the mass number $A$. Thus,statement $(E)$ is correct.
Therefore,statements $(A)$ and $(E)$ are correct.

(D) The vertical component of the earth's magnetic field $(B_V)$ is related to the resultant magnetic field $(B)$ and the angle of dip $(\delta)$ by the formula:
$B_V = B \sin \delta$
Given that $B_V = 6 \times 10^{-5} \text{ T}$ and $\delta = 37^{\circ}$.
Since $\tan 37^{\circ} = \frac{3}{4}$,we can form a right-angled triangle with opposite side $3$ and adjacent side $4$. The hypotenuse is $\sqrt{3^2 + 4^2} = 5$.
Therefore,$\sin 37^{\circ} = \frac{3}{5}$.
Substituting the values into the formula:
$6 \times 10^{-5} = B \times \frac{3}{5}$
$B = \frac{6 \times 10^{-5} \times 5}{3}$
$B = 2 \times 10^{-5} \times 5$
$B = 10 \times 10^{-5} \text{ T} = 10^{-4} \text{ T}$

(C) Given modulating signal frequency $\omega_{m} = 6.28 \times 10^{6} \text{ rad/s}$.
Frequency $f_{m} = \frac{\omega_{m}}{2\pi} = \frac{6.28 \times 10^{6}}{2 \times 3.14} = 1 \times 10^{6} \text{ Hz} = 1 \text{ MHz}$.
Carrier signal frequency $\omega_{c} = 12.56 \times 10^{9} \text{ rad/s}$.
Frequency $f_{c} = \frac{\omega_{c}}{2\pi} = \frac{12.56 \times 10^{9}}{2 \times 3.14} = 2 \times 10^{9} \text{ Hz} = 2000 \text{ MHz}$.
The square law device produces frequencies: $2f_{c}, f_{c}+f_{m}, f_{c}, f_{c}-f_{m}, 2f_{m}, f_{m}$.
The band pass filter centered at $f_{c}$ allows the frequencies $f_{c}-f_{m}, f_{c}, f_{c}+f_{m}$ to pass.
The bandwidth of the output signal is $(f_{c}+f_{m}) - (f_{c}-f_{m}) = 2f_{m}$.
Bandwidth $= 2 \times 1 \text{ MHz} = 2 \text{ MHz}$.

(A) The voltage across the series resistor $R$ is given by the difference between the supply voltage $V_{S}$ and the Zener breakdown voltage $V_{Z}$.
$V_{R} = V_{S} - V_{Z} = 20\,V - 8\,V = 12\,V$.
To find the minimum value of the series resistance $R$,we must use the maximum Zener current $I_{Z,max} = 25\,mA = 25 \times 10^{-3}\,A$.
Using Ohm's law,$V_{R} = I_{Z,max} \times R$.
$12\,V = (25 \times 10^{-3}\,A) \times R$.
$R = \frac{12}{25 \times 10^{-3}}\,\Omega = \frac{12000}{25}\,\Omega = 480\,\Omega$.

(A) The number of nuclei remaining at time $t$ is given by $N(t) = N_0 e^{-\lambda t}$.
Given that initially both have the same number of nuclei,$N_{0A} = N_{0B} = N_0$.
The number of nuclei of $A$ at time $t$ is $N_A = N_0 e^{-25 \lambda t}$.
The number of nuclei of $B$ at time $t$ is $N_B = N_0 e^{-16 \lambda t}$.
We are given that the ratio $\frac{N_B}{N_A} = e$ at time $t = \frac{1}{a \lambda}$.
$\frac{N_B}{N_A} = \frac{N_0 e^{-16 \lambda t}}{N_0 e^{-25 \lambda t}} = e^{(-16 \lambda + 25 \lambda) t} = e^{9 \lambda t}$.
Setting this equal to $e^1$,we get $9 \lambda t = 1$,which implies $t = \frac{1}{9 \lambda}$.
Comparing this with $t = \frac{1}{a \lambda}$,we find $a = 9$.

(A) The energy stored in the capacitor is given by $U = \frac{1}{2} CV^2$.
Substituting the given values,$U = \frac{1}{2} \times (500 \times 10^{-6} \, F) \times (100 \, V)^2 = \frac{1}{2} \times 500 \times 10^{-6} \times 10^4 = 2.5 \, J$.
In an $LC$ circuit,the total energy is conserved. The maximum current $I_{max}$ occurs when the entire energy of the capacitor is transferred to the inductor.
Thus,$\frac{1}{2} LI_{max}^2 = U_{total}$.
$\frac{1}{2} \times (50 \times 10^{-3} \, H) \times I_{max}^2 = 2.5 \, J$.
$I_{max}^2 = \frac{2.5 \times 2}{50 \times 10^{-3}} = \frac{5}{0.05} = 100$.
$I_{max} = \sqrt{100} = 10 \, A$.

(C) According to Gauss's Law,the electric flux through a Gaussian surface is given by $\oint \vec{E} \cdot d\vec{s} = \frac{Q_{\text{in}}}{\varepsilon_{0}}$.
For a spherical Gaussian surface of radius $r$ $(r < R)$,the electric field $E$ is uniform and directed radially outward,so $\oint \vec{E} \cdot d\vec{s} = E(4\pi r^2)$.
The charge enclosed $Q_{\text{in}}$ is calculated by integrating the charge density $\rho(r)$ over the volume of the sphere of radius $r$:
$Q_{\text{in}} = \int_{0}^{r} \rho(r') 4\pi r'^2 dr' = \int_{0}^{r} \rho_{0} \left(\frac{3}{4} - \frac{r'}{R}\right) 4\pi r'^2 dr'$
$Q_{\text{in}} = 4\pi \rho_{0} \int_{0}^{r} \left(\frac{3}{4}r'^2 - \frac{r'^3}{R}\right) dr' = 4\pi \rho_{0} \left[ \frac{3}{4} \cdot \frac{r^3}{3} - \frac{r^4}{4R} \right] = 4\pi \rho_{0} \left( \frac{r^3}{4} - \frac{r^4}{4R} \right) = \pi \rho_{0} r^3 \left( 1 - \frac{r}{R} \right)$.
Applying Gauss's Law:
$E(4\pi r^2) = \frac{\pi \rho_{0} r^3}{\varepsilon_{0}} \left( 1 - \frac{r}{R} \right)$
$E = \frac{\rho_{0} r}{4\varepsilon_{0}} \left( 1 - \frac{r}{R} \right)$.

(A) Statement $I$ is true: In electrostatic equilibrium,the interior of a conductor has no net electric field $(E = 0)$. Since $E = -dV/dr$,if $E = 0$,then the potential $V$ must be constant throughout the volume and on the surface of the conductor.
Statement $II$ is true: Since the conductor is an equipotential surface,any electric field component tangential to the surface would cause charges to move along the surface. To maintain equilibrium,the electric field must have no tangential component,meaning it must be strictly perpendicular to the surface at every point.

(C) Given: $E = 440 \sin(100 \pi t)$ and $L = \frac{\sqrt{2}}{\pi} \text{ H}$.
Comparing with $E = E_0 \sin(\omega t)$,we get $E_0 = 440 \text{ V}$ and $\omega = 100 \pi \text{ rad/s}$.
The inductive reactance $X_L$ is given by $X_L = \omega L = (100 \pi) \times \left( \frac{\sqrt{2}}{\pi} \right) = 100 \sqrt{2} \, \Omega$.
The peak current $I_0$ is $I_0 = \frac{E_0}{X_L} = \frac{440}{100 \sqrt{2}} = \frac{4.4}{\sqrt{2}} \text{ A}$.
An $AC$ ammeter measures the $RMS$ value of the current.
$I_{rms} = \frac{I_0}{\sqrt{2}} = \frac{4.4 / \sqrt{2}}{\sqrt{2}} = \frac{4.4}{2} = 2.2 \text{ A}$.
Thus,the reading of the ammeter is $2.2 \text{ A}$.

(C) The current in an $LR$ circuit is given by $i = \frac{E}{R}(1 - e^{-t/\tau})$,where $\tau = \frac{L}{R} = \frac{1}{100} = 0.01\,s = 10\,ms$.
$(a)$ For the current to reach half its steady-state value $(i = \frac{E}{2R})$:
$\frac{E}{2R} = \frac{E}{R}(1 - e^{-t/\tau})$
$0.5 = 1 - e^{-t/\tau} \implies e^{-t/\tau} = 0.5$
$t = \tau \ln 2 = 10\,ms \times 0.693 = 6.93\,ms \approx 7\,ms$.
$(b)$ At $t = 15\,ms$,$t/\tau = 15/10 = 1.5$:
$i = \frac{6}{100}(1 - e^{-1.5}) = 0.06(1 - 0.25) = 0.06 \times 0.75 = 0.045\,A$.
The energy stored is $U = \frac{1}{2}Li^2 = \frac{1}{2} \times 1 \times (0.045)^2 = 0.5 \times 0.002025 \approx 0.001\,J = 1\,mJ$.

(A) $UV$ rays are used for water purification as they kill bacteria.
$X-$rays are used as a diagnostic tool in medicine to detect fractures.
Microwaves are used for communication and radar systems.
Infrared waves have longer wavelengths and show less scattering,therefore they are used for improving visibility in foggy conditions.

(B) According to Einstein's photoelectric equation,the kinetic energy $E$ is given by:
$E = \frac{hc}{\lambda} - \phi$ --- $(i)$
When the kinetic energy is doubled to $2E$,let the new wavelength be $\lambda^{\prime}$:
$2E = \frac{hc}{\lambda^{\prime}} - \phi$ --- (ii)
Subtracting equation $(i)$ from equation (ii):
$(2E - E) = (\frac{hc}{\lambda^{\prime}} - \phi) - (\frac{hc}{\lambda} - \phi)$
$E = \frac{hc}{\lambda^{\prime}} - \frac{hc}{\lambda}$
Rearranging the terms to solve for $\lambda^{\prime}$:
$E + \frac{hc}{\lambda} = \frac{hc}{\lambda^{\prime}}$
$\frac{E\lambda + hc}{\lambda} = \frac{hc}{\lambda^{\prime}}$
$\lambda^{\prime} = \frac{hc\lambda}{E\lambda + hc}$

(A) The energy of an electron in the $n^{th}$ orbit of a hydrogen atom is given by $E_n = -\frac{13.6}{n^2} \text{ eV}$.
For the transition from the second energy level $(n=2)$ to the first energy level $(n=1)$,the energy of the emitted photon is:
$\Delta E_1 = E_2 - E_1 = -\frac{13.6}{2^2} - (-\frac{13.6}{1^2}) = 13.6(1 - \frac{1}{4}) = 13.6 \times \frac{3}{4} \text{ eV}$.
For the transition from the highest permitted energy level $(n=\infty)$ to the first energy level $(n=1)$,the energy of the emitted photon is:
$\Delta E_2 = E_{\infty} - E_1 = 0 - (-\frac{13.6}{1^2}) = 13.6 \text{ eV}$.
The ratio of the energies is:
$\frac{\Delta E_1}{\Delta E_2} = \frac{13.6 \times \frac{3}{4}}{13.6} = \frac{3}{4}$.

(A) The modulation index $m$ is defined as $m = \frac{A_m}{A_c}$,where $A_m$ is the amplitude of the modulating signal and $A_c$ is the amplitude of the carrier wave.
Given the variation in amplitude is $2A_m = 8\,V$,we find $A_m = 4\,V$.
The maximum amplitude of the $AM$ wave is given by $A_{max} = A_c + A_m = 9\,V$.
Substituting $A_m = 4\,V$,we get $A_c + 4 = 9$,which implies $A_c = 5\,V$.
Therefore,the modulation index is $m = \frac{4}{5} = 0.8$.

(D) By analyzing the circuit,we can see that all three resistors of $9\,\Omega$ are connected in parallel across the $6\,V$ battery.
Let the equivalent resistance be $R_{eq}$.
Since the resistors are in parallel,$\frac{1}{R_{eq}} = \frac{1}{9} + \frac{1}{9} + \frac{1}{9} = \frac{3}{9} = \frac{1}{3}$.
Therefore,$R_{eq} = 3\,\Omega$.
Using Ohm's law,$I = \frac{V}{R_{eq}} = \frac{6\,V}{3\,\Omega} = 2\,A$.
Thus,the current flowing through the circuit is $2\,A$.