JEE Main 2023 Physics Question Paper with Answer and Solution

(C) The radius of an orbit in Bohr's model is given by the formula $r_n = a_0 \frac{n^2}{Z}$,where $n$ is the principal quantum number and $Z$ is the atomic number.
For $He^{+}$ $(Z=2)$,the radius of the $2^{nd}$ orbit $(n=2)$ is $r_1 = a_0 \frac{2^2}{2} = 2a_0$.
For $Be^{3+}$ $(Z=4)$,the radius of the $4^{th}$ orbit $(n=4)$ is $r_2 = a_0 \frac{4^2}{4} = 4a_0$.
The ratio $\frac{r_2}{r_1} = \frac{4a_0}{2a_0} = 2$.
Since the ratio is $x : 1$,we have $x = 2$.

(C) The electric field due to an infinite line charge is $E_L = \frac{\lambda}{2 \pi \varepsilon_0 r}$ and due to an infinite sheet charge is $E_S = \frac{\sigma}{2 \varepsilon_0}$.
Since the points $P$ and $Q$ are between the line and the sheet,the fields are in opposite directions.
At point $P$ $(r_P = \frac{3}{\pi} \ m)$: $E_P = \frac{\lambda}{2 \pi \varepsilon_0 (3/\pi)} - \frac{\sigma}{2 \varepsilon_0} = \frac{1}{2 \varepsilon_0} (\frac{\lambda}{3} - \sigma)$.
At point $Q$ $(r_Q = \frac{4}{\pi} \ m)$: $E_Q = \frac{\lambda}{2 \pi \varepsilon_0 (4/\pi)} - \frac{\sigma}{2 \varepsilon_0} = \frac{1}{2 \varepsilon_0} (\frac{\lambda}{4} - \sigma)$.
Given $2|\sigma| = |\lambda|$,we take $\lambda = 2\sigma$.
$E_P = \frac{1}{2 \varepsilon_0} (\frac{2\sigma}{3} - \sigma) = \frac{1}{2 \varepsilon_0} (-\frac{\sigma}{3})$. Magnitude $|E_P| = \frac{\sigma}{6 \varepsilon_0}$.
$E_Q = \frac{1}{2 \varepsilon_0} (\frac{2\sigma}{4} - \sigma) = \frac{1}{2 \varepsilon_0} (-\frac{\sigma}{2})$. Magnitude $|E_Q| = \frac{\sigma}{4 \varepsilon_0}$.
$\frac{E_P}{E_Q} = \frac{\sigma / 6 \varepsilon_0}{\sigma / 4 \varepsilon_0} = \frac{4}{6}$.
Comparing with $\frac{4}{a}$,we get $a = 6$.

(C) The energy stored in an inductor is given by $U = \frac{1}{2} L I^2$.
The rate of energy storage is $P = \frac{dU}{dt} = L I \frac{dI}{dt}$.
For an $R-L$ circuit,the current at time $t$ is $I = \frac{E}{R} (1 - e^{-tR/L})$.
The rate of change of current is $\frac{dI}{dt} = \frac{E}{L} e^{-tR/L}$.
Substituting these into the power equation:
$P = L \left[ \frac{E}{R} (1 - e^{-tR/L}) \right] \left[ \frac{E}{L} e^{-tR/L} \right] = \frac{E^2}{R} (e^{-tR/L} - e^{-2tR/L})$.
To find the maximum rate,we differentiate $P$ with respect to $t$ and set it to zero:
$\frac{dP}{dt} = \frac{E^2}{R} \left( -\frac{R}{L} e^{-tR/L} + \frac{2R}{L} e^{-2tR/L} \right) = 0$.
This implies $e^{-tR/L} = 2 e^{-2tR/L}$,so $e^{tR/L} = 2$,or $t = \frac{L}{R} \ln 2$.
At this time,$e^{-tR/L} = \frac{1}{2}$.
Substituting this back into the power equation:
$P_{max} = \frac{E^2}{R} \left( \frac{1}{2} - (\frac{1}{2})^2 \right) = \frac{E^2}{R} (\frac{1}{4}) = \frac{E^2}{4R}$.
Given $R = 25 \, \Omega$,we have $P_{max} = \frac{E^2}{4 \times 25} = \frac{E^2}{100}$.
Comparing this with $\frac{E^a}{2b}$,we get $a = 2$ and $b = 50$.
Therefore,$\frac{b}{a} = \frac{50}{2} = 25$.

(C) Let the velocity of the fish be $v_f = 8\,ms^{-1}$ (upward) and the velocity of the bird be $v_b$ (downward).
The apparent velocity of the bird as seen by the fish is given by $v_{b/f} = 12\,ms^{-1}$ (downward).
According to the formula for apparent velocity due to refraction at a plane surface,the apparent velocity of an object in a rarer medium as seen from a denser medium is $v_{app} = \mu \cdot v_{actual}$.
Here,the bird is in air $(\mu_1 = 1)$ and the fish is in water $(\mu_2 = 4/3)$.
The velocity of the bird relative to the water surface is $v_b$. The velocity of the fish relative to the water surface is $v_f = 8\,ms^{-1}$.
The apparent velocity of the bird relative to the fish is $v_{b/f} = v_{b,app} - v_f$.
Since the bird is in air,its apparent velocity as seen from water is $v_{b,app} = \mu \cdot v_b = \frac{4}{3} v_b$.
Given $v_{b/f} = 12\,ms^{-1}$ (downward),we have:
$12 = \frac{4}{3} v_b + 8$
$12 - 8 = \frac{4}{3} v_b$
$4 = \frac{4}{3} v_b$
$v_b = 3\,ms^{-1}$.

(C) The current gain $\beta$ is defined as the slope of the $I_C$ versus $I_B$ graph: $\beta = \frac{\Delta I_C}{\Delta I_B}$.
From the graph,taking two points $(100\,\mu A, 10\,mA)$ and $(200\,\mu A, 20\,mA)$:
$\beta = \frac{(20 - 10) \times 10^{-3} \, A}{(200 - 100) \times 10^{-6} \, A} = \frac{10 \times 10^{-3}}{100 \times 10^{-6}} = 100$.
The power gain $A_P$ is given by $A_P = \beta^2 \times \frac{R_C}{R_B}$.
Substituting the values: $A_P = (100)^2 \times \frac{1 \times 10^3 \, \Omega}{10 \times 10^3 \, \Omega} = 10000 \times 0.1 = 1000 = 10^3$.
Comparing this with $10^x$,we get $x = 3$.

(D) The binding energy per nucleon curve shows that for nuclei with mass numbers between $30$ and $170$, the binding energy per nucleon is nearly constant (approximately $8 \text{ MeV}$ per nucleon).
This happens because the nuclear force is short-ranged, meaning a nucleon only interacts with its nearest neighbors.
As the mass number increases, the number of neighbors remains effectively constant for a given nucleon, making the binding energy per nucleon independent of the total number of nucleons.
Therefore, both the Assertion and the Reason are true, and the short-range nature of the nuclear force is the reason for the saturation of binding energy per nucleon.

(D) $NAND$ gate produces a low output $(0)$ only when both inputs are high $(1)$. Otherwise, it produces a high output $(1)$. The truth table for a $NAND$ gate is:
| $A$ | $B$ | $Y = \overline{A \cdot B}$ |
|---|---|---|
| $0$ | $0$ | $1$ |
| $0$ | $1$ | $1$ |
| $1$ | $0$ | $1$ |
| $1$ | $1$ | $0$ |
By analyzing the input waveforms for $A$ and $B$ at different time intervals:
$1$. For the first interval, $A=1, B=1$, so $Y=0$.
$2$. For the second interval, $A=0, B=0$, so $Y=1$.
$3$. For the third interval, $A=0, B=1$, so $Y=1$.
$4$. For the fourth interval, $A=1, B=0$, so $Y=1$.
$5$. For the fifth interval, $A=1, B=1$, so $Y=0$.
$6$. For the sixth interval, $A=0, B=0$, so $Y=1$.
$7$. For the seventh interval, $A=0, B=1$, so $Y=1$.
Comparing this sequence $(0, 1, 1, 1, 0, 1, 1)$ with the given options, the waveform in image $D$ matches this output.

(A) In the steady state,the capacitor acts as an open circuit,meaning no current flows through the branch containing the capacitor.
The current $I$ flows through the outer loop containing the $3 \text{ V}$ battery and the two $6 \,\Omega$ and $4 \,\Omega$ resistors in series.
$I = \frac{V}{R_{eq}} = \frac{3 \text{ V}}{6 \,\Omega + 4 \,\Omega} = \frac{3}{10} \text{ A} = 0.3 \text{ A}$.
The potential difference across the $6 \,\Omega$ resistor (which is in parallel with the capacitor branch) is:
$V_{cap} = I \times R = 0.3 \text{ A} \times 6 \,\Omega = 1.8 \text{ V}$.
Since the capacitor is connected in parallel to this $6 \,\Omega$ resistor,the potential difference across the capacitor is $1.8 \text{ V}$.
The charge $q$ accumulated in the capacitor is given by:
$q = C \times V_{cap} = 4 \,\mu\text{F} \times 1.8 \text{ V} = 7.2 \,\mu\text{C}$.

(B) The direction of propagation of an electromagnetic wave is given by the direction of the Poynting vector,which is $\overrightarrow{S} = \overrightarrow{E} \times \overrightarrow{B}$.
Given,the electric field $\overrightarrow{E} = -E_0 \hat{k}$ (along negative $z$-axis).
The magnetic field $\overrightarrow{B} = B_0 \hat{i}$ (along positive $x$-axis).
The direction of propagation is $\hat{k}_{prop} = \hat{E} \times \hat{B} = (-\hat{k}) \times (\hat{i})$.
Using the cross product rules for unit vectors ($\hat{i} \times \hat{j} = \hat{k}$,$\hat{j} \times \hat{k} = \hat{i}$,$\hat{k} \times \hat{i} = \hat{j}$),we get:
$(-\hat{k}) \times \hat{i} = -(\hat{k} \times \hat{i}) = -\hat{j}$.
Therefore,the direction of propagation is along the negative $y$-axis.

(A) The magnetic force on a moving charge is given by the Lorentz force formula: $\overrightarrow{F} = q(\overrightarrow{v} \times \overrightarrow{B})$.
For an electron,$q = -e$. The velocity vector is $\overrightarrow{v} = v\hat{i}$ and the magnetic field is $\overrightarrow{B} = -B\hat{k}$.
Substituting these values: $\overrightarrow{F} = -e(v\hat{i} \times -B\hat{k}) = evB(\hat{i} \times \hat{k}) = evB(-\hat{j})$.
Thus,the force acts along the negative $y$-axis ($B$ is correct).
Since the magnetic force is always perpendicular to the velocity,it acts as a centripetal force,causing the electron to move in a circular path ($E$ is correct).
Therefore,the correct options are $B$ and $E$.

(A) Statement $I$ is false because electrical resonance in an $AC$ circuit requires the presence of both an inductor $(L)$ and a capacitor $(C)$ so that the inductive reactance $(X_L = \omega L)$ and capacitive reactance $(X_C = 1/\omega C)$ can cancel each other out,resulting in a phase angle $\phi = 0$.
Statement $II$ is false because for a pure inductor or a pure capacitor,the phase difference between voltage and current is $\pi/2$. The power factor is $\cos(\pi/2) = 0$. Therefore,the average power consumed by a pure inductor or a pure capacitor is $P = V_{rms} I_{rms} \cos(\phi) = 0$. They do not consume high power; they consume zero power.
Since both statements are incorrect,the correct option is $A$.

(B) For an antenna to radiate electromagnetic signals with high efficiency,its length must be comparable to the wavelength of the signal. Specifically,the minimum length required for an antenna to act as an effective radiator is $\frac{\lambda}{4}$,which is known as a quarter-wave antenna.

(A) Statement $I$: The energy of a photon is given by $E = hf$. Ultraviolet $(UV)$ rays have a higher frequency compared to infrared rays and microwaves. Since $E = hf$,$UV$ rays carry more energy per photon,making them more effective at overcoming the work function of a metal to emit electrons. Thus,Statement $I$ is true.
Statement $II$: According to Einstein's photoelectric equation,$KE_{\max} = hf - \phi$,where $hf$ is the energy of the incident photon and $\phi = hf_0$ is the work function. This equation shows that $KE_{\max}$ increases linearly with the frequency $f$ of the incident light,not inversely. Thus,Statement $II$ is false.

(B) Let the total charge $q = 10\,\mu C$ be divided into two parts $x$ and $(q - x)$.
The electrostatic repulsive force between them at a distance $r$ is given by Coulomb's Law:
$F = \frac{K x(q - x)}{r^2}$
To find the condition for maximum force,we differentiate $F$ with respect to $x$ and set it to zero:
$\frac{dF}{dx} = \frac{K}{r^2} \frac{d}{dx} (qx - x^2) = 0$
$\frac{K}{r^2} (q - 2x) = 0$
Since $\frac{K}{r^2} \neq 0$,we have $q - 2x = 0$,which implies $x = \frac{q}{2}$.
Given $q = 10\,\mu C$,the two parts are:
$x = \frac{10\,\mu C}{2} = 5\,\mu C$
$(q - x) = 10\,\mu C - 5\,\mu C = 5\,\mu C$.
Thus,the charges are $5\,\mu C$ and $5\,\mu C$.

(A) Given the ratio of amplitudes of light from the two slits is $\frac{A_1}{A_2} = \frac{2}{1}$.
The intensity $I$ is proportional to the square of the amplitude,$I \propto A^2$.
The formula for the ratio of maximum intensity to minimum intensity is given by:
$\frac{I_{\max}}{I_{\min}} = \left( \frac{A_1 + A_2}{A_1 - A_2} \right)^2$.
Substituting the given values:
$\frac{I_{\max}}{I_{\min}} = \left( \frac{2 + 1}{2 - 1} \right)^2 = \left( \frac{3}{1} \right)^2 = \frac{9}{1}$.
Therefore,the ratio of maximum to minimum intensity is $9:1$.

(B) The power of a lens is given by $P = \frac{1}{f(m)}$.
For the original lens,$f = 10\,cm = 0.1\,m$,so the power $P = \frac{1}{0.1} = 10\,D$.
When a lens is cut perpendicular to the principal axis,the focal length of each part becomes double the original focal length $(f' = 2f = 20\,cm = 0.2\,m)$.
The power of each new lens is $P' = \frac{1}{f'} = \frac{1}{0.2} = 5\,D$.

(B) The net energy absorbed by the atom is given by the difference between the energy of the absorbed photon and the emitted photon.
$E_{\text{net}} = E_{\text{absorbed}} - E_{\text{emitted}} = \frac{hc}{\lambda_1} - \frac{hc}{\lambda_2} = hc \left( \frac{1}{\lambda_1} - \frac{1}{\lambda_2} \right)$
Substituting the given values:
$E_{\text{net}} = (6.6 \times 10^{-34}) \times (3 \times 10^8) \left( \frac{1}{500 \times 10^{-9}} - \frac{1}{600 \times 10^{-9}} \right)$
$E_{\text{net}} = 19.8 \times 10^{-26} \left( \frac{600 - 500}{300000 \times 10^{-18}} \right) = 19.8 \times 10^{-26} \left( \frac{100}{3 \times 10^{-13}} \right)$
$E_{\text{net}} = 19.8 \times 10^{-26} \times 33.33 \times 10^{13} = 6.6 \times 10^{-20}\,J$
To convert this energy into $eV$,divide by $1.6 \times 10^{-19}\,J/eV$:
$E_{\text{net}} = \frac{6.6 \times 10^{-20}}{1.6 \times 10^{-19}} = 4.125 \times 10^{-1}\,eV$
$E_{\text{net}} = 4125 \times 10^{-4}\,eV$
Comparing this with $n \times 10^{-4}\,eV$,we get $n = 4125$.

(B) Let the charges be $q_1 = q$ at $x=0$,$q_2 = -2q$ at $x=\frac{3}{4}R$,and $q_3 = 2q$ at $x=R$.
The force on $q_2$ due to $q_1$ is $F_{21} = \frac{k |q_1 q_2|}{r_{21}^2} = \frac{k |q(-2q)|}{(\frac{3}{4}R)^2} = \frac{2kq^2}{\frac{9}{16}R^2} = \frac{32kq^2}{9R^2}$ (directed towards the origin,i.e.,in $-x$ direction).
The force on $q_2$ due to $q_3$ is $F_{23} = \frac{k |q_2 q_3|}{r_{23}^2} = \frac{k |(-2q)(2q)|}{(R - \frac{3}{4}R)^2} = \frac{4kq^2}{(\frac{1}{4}R)^2} = \frac{4kq^2}{\frac{1}{16}R^2} = \frac{64kq^2}{R^2}$ (directed towards $x=R$,i.e.,in $+x$ direction).
The net force on $-2q$ is $F_{net} = F_{23} - F_{21} = \frac{64kq^2}{R^2} - \frac{32kq^2}{9R^2} = \frac{576kq^2 - 32kq^2}{9R^2} = \frac{544kq^2}{9R^2}$.
Substituting values $k = 9 \times 10^9 \, N \cdot m^2/C^2$,$q = 2 \times 10^{-6} \, C$,and $R = 0.02 \, m$:
$F_{net} = \frac{544 \times 9 \times 10^9 \times (2 \times 10^{-6})^2}{9 \times (0.02)^2} = \frac{544 \times 10^9 \times 4 \times 10^{-12}}{4 \times 10^{-4}} = 544 \times 10^1 = 5440 \, N$.

(B) The circuit consists of two parallel branches connected across a $12\,V$ source.
Branch $1$ has resistors of $3\,\Omega$ and $9\,\Omega$ in series. The current in this branch is $I_1 = \frac{12}{3+9} = 1\,A$.
The potential at point $C$ relative to point $A$ is $V_A - V_C = I_1 \times 3 = 1 \times 3 = 3\,V$.
Branch $2$ has resistors of $4\,\Omega$ and $2\,\Omega$ in series. The current in this branch is $I_2 = \frac{12}{4+2} = 2\,A$.
The potential at point $D$ relative to point $A$ is $V_A - V_D = I_2 \times 4 = 2 \times 4 = 8\,V$.
The potential difference across the capacitor is $V_{CD} = |(V_A - V_D) - (V_A - V_C)| = |8 - 3| = 5\,V$.
The energy stored in the capacitor is $U = \frac{1}{2} CV^2 = \frac{1}{2} \times 6\,\mu F \times (5\,V)^2 = 3 \times 25 = 75\,\mu J$.
Thus,$n = 75$.

(B) The change in magnetic flux $\Delta \phi$ is given by $\Delta \phi = N A (B_2 - B_1)$.
Given: $N = 100$,$A = 24\,cm^2 = 24 \times 10^{-4}\,m^2$,$R = 12\,\Omega$,$B_1 = 1.5\,T$,and $B_2 = -1.5\,T$.
Thus,$\Delta \phi = 100 \times 24 \times 10^{-4} \times (-1.5 - 1.5) = 0.24 \times (-3) = -0.72\,Wb$.
The induced charge $Q$ is given by $Q = \frac{|\Delta \phi|}{R}$.
$Q = \frac{0.72}{12} = 0.06\,C$.
Converting to $mC$: $0.06\,C = 60\,mC$.

(B) Given:
Mass of the wire,$m = 40\,g = 40 \times 10^{-3}\,kg$
Length of the wire,$\ell = 50\,cm = 50 \times 10^{-2}\,m = 0.5\,m$
Magnetic field,$B = 0.40\,T$
Acceleration due to gravity,$g = 10\,ms^{-2}$
To remove the tension in the supporting leads,the upward magnetic force on the wire must balance the downward gravitational force (weight) of the wire.
Magnetic force,$F_m = I\ell B$
Weight of the wire,$W = mg$
For equilibrium,$F_m = W$
$I\ell B = mg$
Substituting the values:
$I \times 0.5 \times 0.4 = 40 \times 10^{-3} \times 10$
$I \times 0.2 = 0.4$
$I = \frac{0.4}{0.2} = 2\,A$
Thus,the magnitude of the current required is $2\,A$.

(C) The electric field $E$ due to a short electric dipole at a distance $r$ on its equatorial plane is given by the formula:
$E = \frac{kp}{r^3}$
where $k$ is the Coulomb constant and $p$ is the dipole moment.
Since $k$ and $p$ are constants,the electric field $E$ is proportional to $\frac{1}{r^3}$.
Therefore,the electric field varies with distance as $\frac{1}{r^3}$.

(D) The maximum line-of-sight distance $d_{\max}$ between a transmitting antenna of height $h_t$ and a receiving antenna of height $h_r$ is given by the formula:
$d_{\max} = \sqrt{2Rh_t} + \sqrt{2Rh_r}$
Given:
$R = 6400\,km = 64 \times 10^5\,m$
$h_t = 180\,m$
$h_r = 245\,m$
Substituting the values:
$d_{\max} = \sqrt{2 \times 64 \times 10^5 \times 180} + \sqrt{2 \times 64 \times 10^5 \times 245}$
$d_{\max} = \sqrt{128 \times 180 \times 10^5} + \sqrt{128 \times 245 \times 10^5}$
$d_{\max} = \sqrt{23040 \times 10^5} + \sqrt{31360 \times 10^5}$
$d_{\max} = \sqrt{2304 \times 10^6} + \sqrt{3136 \times 10^6}$
$d_{\max} = (48 \times 10^3) + (56 \times 10^3)\,m$
$d_{\max} = 48\,km + 56\,km = 104\,km$.

(C) The half-life $T_{1/2} = 5$ years.
Total time $t = 15$ years.
The number of half-lives $n = \frac{t}{T_{1/2}} = \frac{15}{5} = 3$.
The fraction of nuclei remaining after $n$ half-lives is given by $\frac{N}{N_0} = (\frac{1}{2})^n = (\frac{1}{2})^3 = \frac{1}{8}$.
The fraction of the sample that decays is $1 - \frac{N}{N_0} = 1 - \frac{1}{8} = \frac{7}{8}$.

(C) The de Broglie wavelength $\lambda$ of an electron with kinetic energy $E$ is given by the formula: $\lambda = \frac{h}{\sqrt{2mE}}$.
When the kinetic energy becomes $E' = \frac{E}{4}$,the new de Broglie wavelength $\lambda'$ is:
$\lambda' = \frac{h}{\sqrt{2mE'}} = \frac{h}{\sqrt{2m(\frac{E}{4})}}$.
Simplifying the expression:
$\lambda' = \frac{h}{\sqrt{\frac{2mE}{4}}} = \frac{h}{\frac{1}{2}\sqrt{2mE}} = 2 \left( \frac{h}{\sqrt{2mE}} \right)$.
Substituting $\lambda$ into the equation:
$\lambda' = 2\lambda$.

(C) For the voltmeter:
The resistance $R$ to be connected in series is given by $R = \frac{V}{I_g} - G$.
Given $V = 50\,V$,$I_g = 1\,mA = 10^{-3}\,A$,and $G = 54\,\Omega$.
$R = \frac{50}{10^{-3}} - 54 = 50000 - 54 = 49946\,\Omega \approx 50\,k\Omega$.
Thus,statement $(A)$ is correct.
For the ammeter:
The shunt resistance $r$ to be connected in parallel is given by $r = \frac{I_g G}{I - I_g}$.
Given $I = 10\,mA = 10^{-2}\,A$,$I_g = 1\,mA = 10^{-3}\,A$,and $G = 54\,\Omega$.
$r = \frac{10^{-3} \times 54}{10 \times 10^{-3} - 1 \times 10^{-3}} = \frac{54 \times 10^{-3}}{9 \times 10^{-3}} = 6\,\Omega$.
Thus,statement $(C)$ is correct.
Therefore,the correct option is $(A)$ and $(C)$.

(B) For Statement $I$: In a series combination,the equivalent resistance is given by $R_{eq} = R_1 + R_2 + ... + R_n$. Since all resistances are positive,$R_{eq}$ is always greater than any individual resistance in the combination. Therefore,Statement $I$ is false.
For Statement $II$: The resistivity of a material is temperature-dependent. For metals,resistivity increases with an increase in temperature according to the relation $\rho_T = \rho_0 [1 + \alpha(T - T_0)]$. Therefore,Statement $II$ is false.
Conclusion: Both Statement $I$ and Statement $II$ are false.

(D) The initial current $I$ in the circuit is given by Ohm's law: $I = \frac{V}{R} = \frac{12\,V}{6\,\Omega} = 2\,A$.
When the switch is opened,the current drops from $2\,A$ to $0\,A$ in a time interval $\Delta t = 1\,ms = 10^{-3}\,s$.
The magnitude of the induced emf is given by the formula: $|\varepsilon| = L \left| \frac{\Delta I}{\Delta t} \right|$.
Substituting the given values: $20 = L \times \frac{2 - 0}{10^{-3}}$.
$20 = L \times \frac{2}{10^{-3}}$.
$20 = L \times 2000$.
$L = \frac{20}{2000} = 0.01\,H$.
Converting to millihenry $(mH)$: $L = 0.01 \times 1000\,mH = 10\,mH$.

(B) The wavelength ranges for the given electromagnetic waves are as follows:
$(A)$ Microwave: The wavelength range is $0.1\,m$ to $1\,mm$ (matches $(IV)$).
$(B)$ Ultraviolet: The wavelength range is $400\,nm$ to $1\,nm$ (matches $(I)$).
$(C)$ $X$-Ray: The wavelength range is $1\,nm$ to $10^{-3}\,nm$ (matches $(II)$).
$(D)$ Infra-red: The wavelength range is $1\,mm$ to $700\,nm$ (matches $(III)$).
Therefore,the correct matching is $(A)-(IV), (B)-(I), (C)-(II), (D)-(III)$.

(B) For the first minimum in a single-slit diffraction pattern, the condition is given by the formula:
$a \sin \theta = n \lambda$
For the first minimum, $n = 1$, so the equation becomes:
$a \sin \theta = \lambda$
Given:
$\lambda = 600 \, nm = 600 \times 10^{-9} \, m$
$\theta = 30^{\circ}$
Substituting the values into the equation:
$a \sin 30^{\circ} = 600 \times 10^{-9} \, m$
Since $\sin 30^{\circ} = 0.5$:
$a \times 0.5 = 600 \times 10^{-9} \, m$
$a = 1200 \times 10^{-9} \, m$
$a = 1.2 \times 10^{-6} \, m$
$a = 1.2 \, \mu m$

(A) In the given circuit,diodes $D_1$ and $D_3$ are connected in forward bias,while diode $D_2$ is connected in reverse bias.
Therefore,the branch containing $D_2$ acts as an open circuit.
The circuit simplifies to two parallel branches connected to the $10\,V$ battery.
The first branch consists of a $10\,\Omega$ resistor in series with $D_1$ and another $10\,\Omega$ resistor,giving a total resistance of $R_1 = 10\,\Omega + 10\,\Omega = 20\,\Omega$.
The second branch consists of a $10\,\Omega$ resistor in series with $D_3$,giving a resistance of $R_2 = 10\,\Omega$.
The equivalent resistance $(R_{eq})$ of these two parallel branches is given by:
$\frac{1}{R_{eq}} = \frac{1}{R_1} + \frac{1}{R_2} = \frac{1}{20} + \frac{1}{10} = \frac{1+2}{20} = \frac{3}{20}$
$R_{eq} = \frac{20}{3}\,\Omega$
Using Ohm's law,the current $(I)$ through the battery is:
$I = \frac{V}{R_{eq}} = \frac{10}{20/3} = \frac{30}{20} = 1.5\,A$.

(A) Given:
Length of the rod,$\ell = 20\,cm = 0.2\,m$
Angular velocity,$\omega = 210\,rpm = 210 \times \frac{2\pi}{60}\,rad/s = 7\pi\,rad/s$
Magnetic field,$B = 0.2\,T$
The motional emf induced in a rod rotating in a uniform magnetic field is given by the formula:
$\varepsilon = \frac{1}{2} B \omega \ell^2$
Substituting the given values:
$\varepsilon = \frac{1}{2} \times 0.2 \times (7\pi) \times (0.2)^2$
$\varepsilon = 0.1 \times 7 \times \frac{22}{7} \times 0.04$
$\varepsilon = 0.1 \times 22 \times 0.04$
$\varepsilon = 0.088\,V$
Converting to millivolts $(mV)$:
$\varepsilon = 0.088 \times 1000\,mV = 88\,mV$

(A) The circuit consists of two parallel branches connected across the $9\,V$ battery.
Branch $1$ (left side) has a total resistance of $2\,\Omega + 4\,\Omega = 6\,\Omega$. The current in this branch is $I_1 = \frac{9\,V}{6\,\Omega} = 1.5\,A$.
The potential at point $A$ relative to the left junction (let's call it $C$) is $V_C - V_A = I_1 \times 2\,\Omega = 1.5 \times 2 = 3\,V$.
Branch $2$ (right side) has a total resistance of $4\,\Omega + 2\,\Omega = 6\,\Omega$. The current in this branch is $I_2 = \frac{9\,V}{6\,\Omega} = 1.5\,A$.
The potential at point $B$ relative to the left junction $C$ is $V_C - V_B = I_1 \times 4\,\Omega = 1.5 \times 4 = 6\,V$.
Now,the potential difference between $A$ and $B$ is $|V_A - V_B| = |(V_C - 3) - (V_C - 6)| = |6 - 3| = 3\,V$.

(A) The energy levels for a hydrogen atom are given by $n=1$ (ground state),$n=2$ (first excited state),$n=3$ (second excited state),and $n=4$ (third excited state).
From the figure,$A$ corresponds to $n=2$,$B$ corresponds to $n=3$,and $C$ corresponds to $n=4$.
The wavelength $\lambda$ for a transition from $n_2$ to $n_1$ is given by the Rydberg formula: $\frac{1}{\lambda} = R Z^2 \left[ \frac{1}{n_1^2} - \frac{1}{n_2^2} \right]$.
For transition $\lambda_1$ (from $n=3$ to $n=2$): $\frac{1}{\lambda_1} = R(1)^2 \left[ \frac{1}{2^2} - \frac{1}{3^2} \right] = R \left[ \frac{1}{4} - \frac{1}{9} \right] = R \left( \frac{5}{36} \right)$.
For transition $\lambda_2$ (from $n=4$ to $n=3$): $\frac{1}{\lambda_2} = R(1)^2 \left[ \frac{1}{3^2} - \frac{1}{4^2} \right] = R \left[ \frac{1}{9} - \frac{1}{16} \right] = R \left( \frac{7}{144} \right)$.
Now,calculate the ratio $\frac{\lambda_1}{\lambda_2}$:
$\frac{\lambda_1}{\lambda_2} = \frac{1/\lambda_2}{1/\lambda_1} = \frac{R(7/144)}{R(5/36)} = \frac{7}{144} \times \frac{36}{5} = \frac{7}{4 \times 5} = \frac{7}{20}$.
Given the ratio is $\frac{7}{4n}$,we have $\frac{7}{20} = \frac{7}{4n}$,which implies $4n = 20$,so $n = 5$.

(A) The formula for the refractive index $\mu$ of a prism is given by $\mu = \frac{\sin((D_{\min} + A)/2)}{\sin(A/2)}$.
Given that the prism is equilateral,the angle of the prism $A = 60^{\circ}$.
The refractive index $\mu = \sqrt{2}$.
Substituting these values into the formula:
$\sqrt{2} = \frac{\sin((D_{\min} + 60^{\circ})/2)}{\sin(60^{\circ}/2)}$
$\sqrt{2} = \frac{\sin((D_{\min} + 60^{\circ})/2)}{\sin(30^{\circ})}$
Since $\sin(30^{\circ}) = 1/2$,we have:
$\sqrt{2} = \frac{\sin((D_{\min} + 60^{\circ})/2)}{1/2}$
$\frac{\sqrt{2}}{2} = \sin((D_{\min} + 60^{\circ})/2)$
$\frac{1}{\sqrt{2}} = \sin((D_{\min} + 60^{\circ})/2)$
Since $\sin(45^{\circ}) = 1/\sqrt{2}$,we get:
$(D_{\min} + 60^{\circ})/2 = 45^{\circ}$
$D_{\min} + 60^{\circ} = 90^{\circ}$
$D_{\min} = 30^{\circ}$.

(D) The magnetic field $B$ at the center of a circular loop carrying current $I$ is given by $B = \frac{\mu_0 I}{2r}$.
The current $I$ produced by an electron revolving with speed $v$ in an orbit of radius $r$ is $I = \frac{ev}{2\pi r}$.
Substituting $I$ into the magnetic field formula: $B = \frac{\mu_0}{2r} \left( \frac{ev}{2\pi r} \right) = \frac{\mu_0 ev}{4\pi r^2}$.
Given values: $e = 1.6 \times 10^{-19} \, C$,$v = 6.76 \times 10^6 \, m/s$,$r = 0.52 \times 10^{-10} \, m$,and $\frac{\mu_0}{4\pi} = 10^{-7} \, T \cdot m/A$.
$B = 10^{-7} \times \frac{1.6 \times 10^{-19} \times 6.76 \times 10^6}{(0.52 \times 10^{-10})^2}$.
$B = 10^{-7} \times \frac{10.816 \times 10^{-13}}{0.2704 \times 10^{-20}} = 10^{-7} \times 40 \times 10^7 = 40 \, T$.

(A) The capacitors are connected in parallel,so the potential difference across each capacitor is the same,$V = 10\,V$.
The charge on each capacitor is given by $Q = CV$.
Charge on $C_1$ is $Q_1 = C_1 V = 2\,\mu F \times 10\,V = 20\,\mu C$.
Charge on $C_2$ is $Q_2 = C_2 V = x\,\mu F \times 10\,V = 10x\,\mu C$.
Charge on $C_3$ is $Q_3 = C_3 V = 3\,\mu F \times 10\,V = 30\,\mu C$.
The total charge stored in the combination is $Q_{total} = Q_1 + Q_2 + Q_3$.
Given $Q_{total} = 100\,\mu C$,we have:
$20 + 10x + 30 = 100$
$50 + 10x = 100$
$10x = 50$
$x = 5$.