JEE Main 2024 Physics Question Paper with Answer and Solution

(A) According to the law of conservation of linear momentum,since the initial nucleus is at rest,the initial momentum is $0$.
Let the masses of the two nuclei be $m_1$ and $m_2$ and their velocities be $v_1$ and $v_2$ respectively.
$p_i = p_f$
$0 = m_1 v_1 + m_2 v_2$
$m_1 v_1 = -m_2 v_2$
Taking the magnitude of the velocities,we have $m_1 |v_1| = m_2 |v_2|$,which implies $\frac{|v_1|}{|v_2|} = \frac{m_2}{m_1}$.
Given the ratio of masses is $\frac{m_1}{m_2} = \frac{2}{1}$,we substitute this into the ratio of speeds:
$\frac{|v_1|}{|v_2|} = \frac{1}{2}$.
The negative sign indicates that the nuclei move in opposite directions.
Therefore,they move in opposite directions with speeds in the ratio of $1:2$.

(C) For the rays to emerge parallel to the principal axis after passing through two convex lenses,the intermediate image formed by the first lens must coincide with the focal point of the second lens.
Since the incident rays are parallel to the principal axis,they converge at the focal point of the first lens $L_1$ at a distance $f_1 = 10 \,cm$ from it.
For these rays to emerge parallel from the second lens $L_2$,this focal point must also be the focal point of the second lens $L_2$. Therefore,the distance between the two lenses must be $d = f_1 + f_2$.
Given $f_1 = 10 \,cm$ and $f_2 = 15 \,cm$,the distance $d = 10 \,cm + 15 \,cm = 25 \,cm$.

(D) The energy of the ground state of a hydrogen atom is $E_1 = -13.6 \ eV$.
When an energy of $10.2 \ eV$ is provided,the new energy level $E_n$ is given by $E_n = E_1 + 10.2 \ eV = -13.6 \ eV + 10.2 \ eV = -3.4 \ eV$.
Since $E_n = -13.6/n^2 \ eV$,we have $-3.4 = -13.6/n^2$,which gives $n^2 = 4$,so $n = 2$.
The electron is excited to the first excited state $(n = 2)$.
The number of spectral lines emitted when an electron transitions from state $n$ to the ground state is given by the formula $N = n(n-1)/2$.
For $n = 2$,$N = 2(2-1)/2 = 1$.
Therefore,only $1$ spectral line will be emitted.

(D) The relationship between the amplitude of the electric field $E_0$ and the magnetic field $B_0$ is given by $E_0 = B_0 c$,where $c$ is the speed of light in vacuum $(c \approx 3 \times 10^8 \ ms^{-1})$.
Given $B_0 = 3.5 \times 10^{-7} \ T$,we calculate $E_0 = (3.5 \times 10^{-7}) \times (3 \times 10^8) = 105 \ Vm^{-1}$.
The wave propagates in the negative $x$-direction (indicated by the $+kx$ term). The magnetic field is in the $y$-direction $(B_y)$. Since the electric field,magnetic field,and direction of propagation are mutually perpendicular,the electric field must be in the $z$-direction $(E_z)$.
Thus,the electric field is $E_z = 105 \sin (1.5 \times 10^3 x + 0.5 \times 10^{11} t) \ Vm^{-1}$.

$(C)$ The side length of the square loop is $L = 15 \ cm = 0.15 \ m$.
The speed of the loop is $v = 2 \ cm/s = 0.02 \ m/s$.
The magnetic field width is $W = 50 \ cm = 0.5 \ m$.
At $t=0$, the front edge enters the magnetic field.
The time taken for the entire loop to enter the magnetic field is $t_{in} = \frac{L}{v} = \frac{15 \ cm}{2 \ cm/s} = 7.5 \ s$.
At $t = 7.5 \ s$, the entire loop is inside the magnetic field.
The loop will remain completely inside the magnetic field until the back edge reaches the boundary of the magnetic field.
The time taken for the back edge to reach the magnetic field is $t_{out} = \frac{W}{v} = \frac{50 \ cm}{2 \ cm/s} = 25 \ s$.
Since $t = 10 \ s$ lies between $7.5 \ s$ and $25 \ s$, the entire loop is inside the magnetic field region.
When the entire loop is inside a uniform magnetic field, the magnetic flux $\phi$ linked with the loop remains constant.
According to Faraday's law of electromagnetic induction, the induced emf $e = -\frac{d\phi}{dt}$.
Since $\phi$ is constant, $\frac{d\phi}{dt} = 0$, therefore $e = 0$.

(C) The given $I-V$ characteristic curve shows a sharp breakdown in the reverse bias region at a specific voltage.
This behavior is characteristic of a Zener diode.
$A$ Zener diode is specifically designed to operate in the reverse breakdown region without being damaged.
Due to this property,it is widely used as a voltage regulator to maintain a constant output voltage across a load despite variations in input voltage or load current.

(B) Due to the symmetry of the circuit about the horizontal axis, the potential at the top and bottom nodes is the same. Thus, no current flows through the vertical resistors, and they can be removed.
After removing the vertical resistors, the circuit simplifies to three parallel branches between the two central nodes, each having a resistance of $R + R = 2R$.
The equivalent resistance of these three parallel branches is given by $\frac{1}{R_{eq, parallel}} = \frac{1}{2R} + \frac{1}{2R} + \frac{1}{2R} = \frac{3}{2R}$, which gives $R_{eq, parallel} = \frac{2R}{3}$.
Finally, this equivalent resistance is in series with the two resistors connected to terminals $A$ and $B$. Therefore, the total effective resistance is $R_{total} = R + \frac{2R}{3} + R = 2R + \frac{2R}{3} = \frac{8R}{3}$.

(B) According to Gauss's law,the electric flux $\phi$ through a closed surface is given by $\phi = \frac{q_{\text{in}}}{\epsilon_0}$,where $q_{\text{in}}$ is the net charge enclosed by the surface.
From the figure,the charges enclosed by the surface $S$ are $+q, +5q,$ and $-2q$.
The charges $+3q$ and $-4q$ are outside the surface $S$,so they do not contribute to the electric flux through the surface.
Therefore,the net enclosed charge is $q_{\text{in}} = (+q) + (+5q) + (-2q) = 4q$.
Substituting this into Gauss's law,we get $\phi = \frac{4q}{\epsilon_0}$.

(C) In a uniform magnetic field,the radius $R$ of the circular path of a charged particle is given by:
$R = \frac{mv}{qB} = \frac{\sqrt{2m(K.E.)}}{qB}$
Since both particles have the same kinetic energy $(K.E.)$ and move in the same magnetic field $(B)$,we have:
$R \propto \frac{\sqrt{m}}{q}$
For a proton,$m_p = m$ and $q_p = e$. For a deuteron,$m_d = 2m$ and $q_d = e$.
The ratio of the radii is:
$\frac{r_d}{r_p} = \frac{\sqrt{m_d}/q_d}{\sqrt{m_p}/q_p} = \sqrt{\frac{m_d}{m_p}} \times \frac{q_p}{q_d}$
Substituting the values:
$\frac{r_d}{r_p} = \sqrt{\frac{2m}{m}} \times \frac{e}{e} = \sqrt{2} \times 1 = \sqrt{2}$
Therefore,the ratio $r_d : r_p$ is $\sqrt{2} : 1$.

(B) According to Einstein's photoelectric equation,the energy of the incident photon is equal to the sum of the work function of the metal and the maximum kinetic energy of the ejected photoelectrons.
$E_{\text{photon}} = \Phi + K.E_{\max}$
Given:
Energy of incident photon $(E_{\text{photon}})$ = $4.13 eV$
Work function $(\Phi)$ = $3.13 eV$
Substituting these values into the equation:
$4.13 eV = 3.13 eV + K.E_{\max}$
$K.E_{\max} = 4.13 eV - 3.13 eV$
$K.E_{\max} = 1 eV$
Therefore,the maximum kinetic energy of the ejected photoelectrons is $1 eV$.

(C) In the fusion reaction,$4$ hydrogen nuclei release $26.7 \ MeV$ of energy.
Energy released per hydrogen nucleus $= \frac{26.7}{4} \ MeV$.
Number of hydrogen nuclei in $2 \ kg = \frac{2000 \ g}{1 \ g/mol} \times N_{A} = 2000 \ N_{A}$.
Total energy $E_{H} = 2000 \ N_{A} \times \frac{26.7}{4} \ MeV = 500 \times 26.7 \ N_{A} \ MeV = 13350 \ N_{A} \ MeV$.
In the fission reaction,$1$ nucleus of ${ }^{235} U$ releases $200 \ MeV$ of energy.
Number of uranium nuclei in $2 \ kg = \frac{2000 \ g}{235 \ g/mol} \times N_{A} = \frac{2000}{235} \ N_{A}$.
Total energy $E_{U} = \frac{2000}{235} \ N_{A} \times 200 \ MeV = \frac{400000}{235} \ N_{A} \ MeV \approx 1702.13 \ N_{A} \ MeV$.
The ratio $\frac{E_{H}}{E_{U}} = \frac{13350 \ N_{A}}{1702.13 \ N_{A}} \approx 7.84$.
Given the options,the closest value is $7.62$.

(A) The given circuit consists of an $AND$ gate,two $NOT$ gates,another $AND$ gate,and a $NOR$ gate.
Let the inputs be $A$ and $B$. The output of the first $AND$ gate is $A \cdot B$.
The inputs to the second $AND$ gate are $\bar{A}$ and $\bar{B}$,so its output is $\bar{A} \cdot \bar{B}$.
These two outputs are fed into a $NOR$ gate,so the final output $E$ is $\overline{(A \cdot B) + (\bar{A} \cdot \bar{B})}$.
For $X$: $A = 0, B = 1$.
$E = \overline{(0 \cdot 1) + (\bar{0} \cdot \bar{1})} = \overline{0 + (1 \cdot 0)} = \overline{0 + 0} = \overline{0} = 1$. Thus,$X = 1$.
For $Y$: $A = 1, B = 0$.
$E = \overline{(1 \cdot 0) + (\bar{1} \cdot \bar{0})} = \overline{0 + (0 \cdot 1)} = \overline{0 + 0} = \overline{0} = 1$. Thus,$Y = 1$.
Therefore,the values of $X$ and $Y$ are $1, 1$.

(A) Let the pole strength of the magnetic strip be $m$ and its length be $\ell$. The initial magnetic moment is $M_1 = m \ell = 44 \text{ Am}^2$.
When the strip is bent into a semicircle of radius $R$,the length of the strip becomes the arc length of the semicircle,so $\ell = \pi R$,which means $R = \frac{\ell}{\pi}$.
The new magnetic moment $M_2$ is the product of the pole strength and the straight-line distance between the poles (the diameter of the semicircle),which is $2R$.
$M_2 = m \times (2R) = m \times \left( \frac{2\ell}{\pi} \right) = \frac{2}{\pi} (m \ell)$.
Substituting the given values: $M_2 = \frac{2}{\pi} \times 44 = \frac{2}{(22/7)} \times 44 = \frac{2 \times 7}{22} \times 44 = \frac{14}{22} \times 44 = 14 \times 2 = 28 \text{ Am}^2$.

(C) Given:
Reactance of capacitor $X_C = 4 \sqrt{3} \Omega$
Resistance $R = 4 \Omega$
Peak voltage $V_0 = 8 \sqrt{2} \text{ V}$
Step $1$: Calculate the impedance $Z$ of the $RC$ circuit.
$Z = \sqrt{R^2 + X_C^2}$
$Z = \sqrt{4^2 + (4 \sqrt{3})^2} = \sqrt{16 + 16 \times 3} = \sqrt{16 + 48} = \sqrt{64} = 8 \Omega$
Step $2$: Calculate the $RMS$ voltage $V_{\text{rms}}$.
$V_{\text{rms}} = \frac{V_0}{\sqrt{2}} = \frac{8 \sqrt{2}}{\sqrt{2}} = 8 \text{ V}$
Step $3$: Calculate the $RMS$ current $I_{\text{rms}}$.
$I_{\text{rms}} = \frac{V_{\text{rms}}}{Z} = \frac{8}{8} = 1 \text{ A}$
Step $4$: Calculate the power dissipation $P$.
Power is dissipated only in the resistor.
$P = I_{\text{rms}}^2 \times R = (1)^2 \times 4 = 4 \text{ W}$

(D) Given,electric field $\overrightarrow{E} = 2x\hat{i} \text{ NC}^{-1}$.
The cube is placed between $x = 2 \text{ m}$ and $x = 4 \text{ m}$. The side length of the cube is $a = 2 \text{ m}$,so the area of each face is $A = a^2 = (2)^2 = 4 \text{ m}^2$.
Electric flux is given by $\phi = \int \overrightarrow{E} \cdot d\overrightarrow{A}$.
For the left face at $x = 2 \text{ m}$,the area vector is $\overrightarrow{A}_1 = -4\hat{i} \text{ m}^2$ and $\overrightarrow{E}_1 = 2(2)\hat{i} = 4\hat{i} \text{ NC}^{-1}$.
$\phi_{\text{in}} = \overrightarrow{E}_1 \cdot \overrightarrow{A}_1 = (4\hat{i}) \cdot (-4\hat{i}) = -16 \text{ Nm}^2 \text{C}^{-1}$.
For the right face at $x = 4 \text{ m}$,the area vector is $\overrightarrow{A}_2 = 4\hat{i} \text{ m}^2$ and $\overrightarrow{E}_2 = 2(4)\hat{i} = 8\hat{i} \text{ NC}^{-1}$.
$\phi_{\text{out}} = \overrightarrow{E}_2 \cdot \overrightarrow{A}_2 = (8\hat{i}) \cdot (4\hat{i}) = 32 \text{ Nm}^2 \text{C}^{-1}$.
The net electric flux through the cube is $\phi_{\text{net}} = \phi_{\text{in}} + \phi_{\text{out}} = -16 + 32 = 16 \text{ Nm}^2 \text{C}^{-1}$.

(B) From the circuit diagram,the voltmeter is connected in parallel with the $10 \text{ k}\Omega$ resistor and the resistor $R$. However,the voltmeter measures the voltage across the parallel combination of $R$ and $10 \text{ k}\Omega$. Let $R_p$ be the equivalent resistance of $R$ and $10 \text{ k}\Omega$ in parallel.
$R_p = \frac{R \times 10^4}{R + 10^4}$
From the $V-I$ graph,we can take any point,for example,$V = 4 \text{ V}$ and $I = 2 \text{ mA} = 2 \times 10^{-3} \text{ A}$.
Using Ohm's law,$V = I R_p$,so $R_p = \frac{V}{I} = \frac{4}{2 \times 10^{-3}} = 2000 \text{ } \Omega$.
Now,equate the two expressions for $R_p$:
$2000 = \frac{R \times 10^4}{R + 10^4}$
$2000(R + 10000) = 10000R$
$2000R + 2 \times 10^7 = 10000R$
$8000R = 2 \times 10^7$
$R = \frac{20000000}{8000} = 2500 \text{ } \Omega$.

(D) The shift in the central maximum due to the introduction of a glass plate of thickness $t$ and refractive index $\mu$ is given by the formula: $\Delta x = \frac{(\mu - 1) t D}{d}$.
Given that the central maximum shifts to the position of the $4^{th}$ bright fringe, the shift is equal to the path difference of the $4^{th}$ bright fringe, which is $4\lambda$.
Equating the path difference introduced by the glass plate to the shift: $(\mu - 1) t = n \lambda$.
Substituting the given values: $(1.5 - 1) t = 4 \times 500 \, nm$.
$0.5 \times t = 2000 \, nm$.
$t = 4000 \, nm$.
Since $1 \, \mu m = 1000 \, nm$, we have $t = 4 \, \mu m$.

(B) The resistance of a conductor at a temperature $T$ is given by the formula: $R = R_0(1 + \alpha \Delta T)$.
Here,$R_0 = 50 \Omega$ at $T_0 = 27^{\circ} C$,$R = 62 \Omega$,and $\alpha = 2.4 \times 10^{-4} { }^{\circ} C^{-1}$.
Substituting the values into the equation:
$62 = 50(1 + 2.4 \times 10^{-4} \Delta T)$
$1.24 = 1 + 2.4 \times 10^{-4} \Delta T$
$0.24 = 2.4 \times 10^{-4} \Delta T$
$\Delta T = \frac{0.24}{2.4 \times 10^{-4}} = 1000^{\circ} C$.
Since $\Delta T = T - T_0$,we have $T = T_0 + \Delta T = 27^{\circ} C + 1000^{\circ} C = 1027^{\circ} C$.