JEE Main 2019 Physics Question Paper with Answer and Solution

(A) When unpolarized light of intensity $I_0$ passes through the first polarizer $P_1$,the transmitted intensity is $I_1 = I_0/2$.
The pass axis of $P_3$ is crossed with respect to $P_1$,meaning the angle between their axes is $90^{\circ}$.
The pass axis of $P_2$ is inclined at $60^{\circ}$ to the pass axis of $P_3$. Therefore,the angle between the pass axis of $P_2$ and $P_1$ is $90^{\circ} - 60^{\circ} = 30^{\circ}$.
Using Malus' Law,the intensity transmitted through $P_2$ is $I_2 = I_1 \cos^2(30^{\circ}) = (I_0/2) \times (3/4) = 3I_0/8$.
The angle between the pass axis of $P_3$ and $P_2$ is $60^{\circ}$.
Using Malus' Law again,the final intensity transmitted through $P_3$ is $I = I_2 \cos^2(60^{\circ}) = (3I_0/8) \times (1/4) = 3I_0/32$.
Thus,the ratio $(I_0/I) = 32/3 \approx 10.67$.

(C) First,we find the constant $k$ in terms of $Q$ and $R$ by integrating the charge density over the volume of the sphere:
$2Q = \int_{0}^{R} \rho(r) 4\pi r^2 dr = \int_{0}^{R} (kr) 4\pi r^2 dr = 4\pi k \int_{0}^{R} r^3 dr = 4\pi k \frac{R^4}{4} = \pi k R^4$.
Thus,$k = \frac{2Q}{\pi R^4}$.
Next,we find the electric field $E$ at distance $a$ from the centre using Gauss's Law:
$E(4\pi a^2) = \frac{q_{enc}}{\varepsilon_0} = \frac{1}{\varepsilon_0} \int_{0}^{a} (kr) 4\pi r^2 dr = \frac{4\pi k}{\varepsilon_0} \frac{a^4}{4} = \frac{\pi k a^4}{\varepsilon_0}$.
$E = \frac{k a^2}{4\varepsilon_0} = \frac{(2Q/\pi R^4) a^2}{4\varepsilon_0} = \frac{Q a^2}{2\pi \varepsilon_0 R^4}$.
For charge $A$ (or $B$) to experience no force,the force due to the sphere must be balanced by the force due to the other charge:
$|qE| = \frac{1}{4\pi \varepsilon_0} \frac{|Q||-Q|}{(2a)^2} \implies Q \left( \frac{Q a^2}{2\pi \varepsilon_0 R^4} \right) = \frac{Q^2}{16\pi \varepsilon_0 a^2}$.
Simplifying: $\frac{a^2}{2 R^4} = \frac{1}{16 a^2} \implies a^4 = \frac{R^4}{8} \implies a = \frac{R}{8^{1/4}} = 8^{-1/4}R$.

(C) The radius of the circular path of the electron in the magnetic field is given by $R = \frac{mv}{qB} = \frac{\sqrt{2m(KE)}}{qB}$.
Substituting the given values: $R = \frac{\sqrt{2 \times 9.1 \times 10^{-31} \times 100 \times 1.6 \times 10^{-19}}}{1.6 \times 10^{-19} \times 1.5 \times 10^{-3}} \approx 0.02248 \, m = 2.248 \, cm$.
From the geometry,$\sin \theta = \frac{x}{R} = \frac{2}{2.248} \approx 0.8896$,so $\theta \approx 62.8^\circ$.
The vertical displacement $y$ at the exit point $T$ is $y = R(1 - \cos \theta) = 2.248(1 - \cos(62.8^\circ)) \approx 2.248(1 - 0.457) \approx 1.22 \, cm$.
The electron travels in a straight line after leaving the magnetic field. The distance from the exit point $T$ to the screen is $8 \, cm - 2 \, cm = 6 \, cm$.
The additional vertical displacement is $\Delta y = (6 \, cm) \tan \theta = 6 \times \tan(62.8^\circ) \approx 6 \times 1.945 \approx 11.67 \, cm$.
The total distance $d = y + \Delta y = 1.22 + 11.67 = 12.89 \, cm$. The closest option is $12.87 \, cm$.

(A) Let the initial number of nuclei be $N_0$ for both $A$ and $B$.
For nucleus $A$,half-life $T_{1/2, A} = 10 \, min$. After $t = 60 \, min$,the number of half-lives $n_A = 60/10 = 6$.
The number of undecayed nuclei $N_A = N_0 / 2^6 = N_0 / 64$.
The number of decayed nuclei $D_A = N_0 - N_A = N_0(1 - 1/64) = 63N_0 / 64$.
For nucleus $B$,half-life $T_{1/2, B} = 20 \, min$. After $t = 60 \, min$,the number of half-lives $n_B = 60/20 = 3$.
The number of undecayed nuclei $N_B = N_0 / 2^3 = N_0 / 8$.
The number of decayed nuclei $D_B = N_0 - N_B = N_0(1 - 1/8) = 7N_0 / 8$.
The ratio of decayed nuclei $D_A / D_B = (63N_0 / 64) / (7N_0 / 8) = (63/64) \times (8/7) = 9/8$.

(A) The modulation index $\mu$ is defined as the ratio of the amplitude of the modulating signal $(A_m)$ to the amplitude of the carrier wave $(A_c)$:
$\mu = \frac{A_m}{A_c} = \frac{2}{4} = 0.5$.
The carrier frequency $f_c$ is obtained from $2 \pi f_c = 20000 \pi$,which gives $f_c = 10000 \text{ Hz} = 10 \text{ kHz}$.
The modulating frequency $f_m$ is obtained from $2 \pi f_m = 2000 \pi$,which gives $f_m = 1000 \text{ Hz} = 1 \text{ kHz}$.
The lower sideband frequency $(LSB)$ is given by $f_c - f_m = 10 \text{ kHz} - 1 \text{ kHz} = 9 \text{ kHz}$.
Therefore,the modulation index is $0.5$ and the lower sideband frequency is $9 \text{ kHz}$.

(B) The current in an $LR$ circuit as a function of time $t$ after closing the switch is given by
$i(t) = \frac{E}{R}\left(1 - e^{-\frac{Rt}{L}}\right)$.
The charge $q$ that passes through the battery in time interval $t = 0$ to $t = \frac{L}{R}$ is given by the integral of current with respect to time:
$q = \int_{0}^{\frac{L}{R}} i(t)\, dt = \int_{0}^{\frac{L}{R}} \frac{E}{R}\left(1 - e^{-\frac{Rt}{L}}\right) dt$
$q = \frac{E}{R} \int_{0}^{\frac{L}{R}} \left(1 - e^{-\frac{Rt}{L}}\right) dt$
$q = \frac{E}{R} \left[ t + \frac{L}{R} e^{-\frac{Rt}{L}} \right]_{0}^{\frac{L}{R}}$
$q = \frac{E}{R} \left[ \left(\frac{L}{R} + \frac{L}{R} e^{-1}\right) - \left(0 + \frac{L}{R} e^{0}\right) \right]$
$q = \frac{E}{R} \left[ \frac{L}{R} + \frac{L}{Re} - \frac{L}{R} \right]$
$q = \frac{E}{R} \cdot \frac{L}{Re} = \frac{EL}{eR^2}$
Since $e \approx 2.718$, we have:
$q = \frac{EL}{2.7R^2}$

(C) The electromagnetic wave propagates in the $+z$ direction,so the wave equation must be of the form $B = B_0 \sin(kz - \omega t)$.
The amplitude of the magnetic field is $B_0 = \frac{E_0}{c} = \frac{60}{3 \times 10^8} = 2 \times 10^{-7} \, T$.
The angular frequency is $\omega = 2\pi n = 2 \times 3.14 \times 23.9 \times 10^9 \approx 1.5 \times 10^{11} \, rad/s$.
The wave number is $k = \frac{\omega}{c} = \frac{1.5 \times 10^{11}}{3 \times 10^8} = 0.5 \times 10^3 \, m^{-1}$.
Since the wave propagates in the $+z$ direction and the electric field is typically in the $x$ or $y$ direction,the magnetic field must be perpendicular to both the direction of propagation $(z)$ and the electric field. Thus,for an electric field in the $y$ direction,the magnetic field is in the $x$ direction. The correct form is $\vec B = 2 \times 10^{-7} \sin(0.5 \times 10^3 z - 1.5 \times 10^{11} t) \hat i$.

(D) The magnetic field $B$ at a distance $d$ from a finite straight wire carrying current $I$ is given by $B = \frac{\mu_0 I}{4\pi d}(\sin \theta_1 + \sin \theta_2)$.
Given,length of wire $AB = 6\, cm$,so the distance from the center to each end is $3\, cm$.
The perpendicular distance $d$ from point $P$ to the wire is $4\, cm$ (from the $3-4-5$ triangle geometry).
Here,$\theta_1 = \theta_2 = \theta$,where $\sin \theta = \frac{\text{opposite}}{\text{hypotenuse}} = \frac{3}{5} = 0.6$.
Thus,$B = \frac{\mu_0 I}{4\pi d}(2 \sin \theta)$.
Substituting the values: $B = \frac{(10^{-7} \times 4\pi) \times 5}{4\pi \times (4 \times 10^{-2})} \times 2 \times \frac{3}{5}$.
$B = \frac{10^{-7} \times 5}{4 \times 10^{-2}} \times \frac{6}{5} = \frac{10^{-5} \times 5}{4} \times 1.2 = 1.25 \times 1.2 \times 10^{-5} = 1.5 \times 10^{-5}\, T$.

(A) The Rydberg formula for the wavelength of emitted photons is given by $\frac{1}{\lambda} = R \left( \frac{1}{n_f^2} - \frac{1}{n_i^2} \right)$.
For the first jump from the third excited state $(n_i = 4)$ to the second excited state $(n_f = 3)$:
$\frac{1}{\lambda_1} = R \left( \frac{1}{3^2} - \frac{1}{4^2} \right) = R \left( \frac{1}{9} - \frac{1}{16} \right) = R \left( \frac{16 - 9}{144} \right) = R \left( \frac{7}{144} \right)$.
For the second jump from the second excited state $(n_i = 3)$ to the first excited state $(n_f = 2)$:
$\frac{1}{\lambda_2} = R \left( \frac{1}{2^2} - \frac{1}{3^2} \right) = R \left( \frac{1}{4} - \frac{1}{9} \right) = R \left( \frac{9 - 4}{36} \right) = R \left( \frac{5}{36} \right)$.
Now,calculating the ratio $\frac{\lambda_1}{\lambda_2} = \frac{1/\lambda_2}{1/\lambda_1} = \frac{R(5/36)}{R(7/144)} = \frac{5}{36} \times \frac{144}{7} = \frac{5 \times 4}{7} = \frac{20}{7}$.

(D) Let $C$ be the critical angle for the interface between the cube $(\mu_2)$ and the liquid $(\mu_1)$.
For total internal reflection $(TIR)$ at point $E$,the angle of incidence $i'$ at face $BC$ must be greater than the critical angle $C$.
Thus,$i' > C$.
From the geometry,the angle of refraction $r$ at face $AB$ is related to $i'$ by $r = 90^\circ - i'$.
Since $i' > C$,we have $90^\circ - r > C$,or $r < 90^\circ - C$.
Applying Snell's Law at face $AB$: $\mu_1 \sin \theta = \mu_2 \sin r$.
Since $\sin$ is an increasing function,$\sin r < \sin(90^\circ - C) = \cos C$.
Therefore,$\mu_1 \sin \theta < \mu_2 \cos C$.
We know $\sin C = \frac{\mu_1}{\mu_2}$,so $\cos C = \sqrt{1 - \sin^2 C} = \sqrt{1 - \frac{\mu_1^2}{\mu_2^2}} = \frac{\sqrt{\mu_2^2 - \mu_1^2}}{\mu_2}$.
Substituting this into the inequality: $\mu_1 \sin \theta < \mu_2 \left( \frac{\sqrt{\mu_2^2 - \mu_1^2}}{\mu_2} \right) = \sqrt{\mu_2^2 - \mu_1^2}$.
$\sin \theta < \frac{\sqrt{\mu_2^2 - \mu_1^2}}{\mu_1} = \sqrt{\frac{\mu_2^2}{\mu_1^2} - 1}$.
Thus,$\theta < \sin^{-1} \sqrt{\frac{\mu_2^2}{\mu_1^2} - 1}$.

(B) For an ammeter,a shunt resistance $R_A$ is connected in parallel with the galvanometer.
$I_g G = (I_0 - I_g) R_A$
$R_A = \left( \frac{I_g}{I_0 - I_g} \right) G$
For a voltmeter,a series resistance $R_V$ is connected in series with the galvanometer.
$I_g (G + R_V) = V = G I_0$
$G + R_V = \frac{G I_0}{I_g}$
$R_V = \frac{G I_0}{I_g} - G = G \left( \frac{I_0 - I_g}{I_g} \right)$
Now,calculating the product:
$R_A R_V = \left( \frac{I_g}{I_0 - I_g} G \right) \times \left( \frac{I_0 - I_g}{I_g} G \right) = G^2$
Calculating the ratio:
$\frac{R_A}{R_V} = \frac{\left( \frac{I_g}{I_0 - I_g} \right) G}{G \left( \frac{I_0 - I_g}{I_g} \right)} = \left( \frac{I_g}{I_0 - I_g} \right)^2$