JEE Main 2024 29th January Morning Shift Question Paper with Solutions

$\mathrm{R}=\frac{\mathrm{V}}{1}$

According to error analysis

$$\begin{aligned} & \frac{\mathrm{dR}}{\mathrm{R}}=\frac{\mathrm{dV}}{\mathrm{V}}+\frac{\mathrm{dI}}{\mathrm{I}} \\ & \frac{\mathrm{dR}}{\mathrm{R}}=\frac{5}{200}+\frac{0.2}{20} \\ & \frac{\mathrm{dR}}{\mathrm{R}}=\frac{7}{200} \\ & \% \text { error } \frac{\mathrm{dR}}{\mathrm{R}} \times 100=\frac{7}{200} \times 100=3.5 \% \end{aligned}$$

We know that the de-Broglie wavelength $\lambda$ of a particle is given by:

$\lambda = \frac{h}{p}$

where:

• $h$ is Planck's constant,
• $p$ is the momentum of the particle.

For a photon (which has zero rest mass), its energy $E$ and momentum $p$ are related by the equation:

$E = cp$

and its de-Broglie wavelength $\lambda$ is given by:

$\lambda = \frac{h}{E} \times \frac{1}{c}$

Now, since the photon and electron are said to have the same de-Broglie wavelength:

$\lambda_{electron} = \lambda_{photon}$

$\frac{h}{p_{electron}} = \frac{h}{E_{photon}} \times \frac{1}{c}$

For the electron, its momentum $p_{electron}$ is given by:

$p_{electron} = m_{e}v_{electron}$

where:

• $m_{e}$ is the electron's rest mass,
• $v_{electron}$ is the electron's velocity.

The kinetic energy $K.E.$ of the electron is:

$K.E._{electron} = \frac{1}{2}m_{e}v_{electron}^2$

The question states that $v_{electron}$ is $25\%$ ($0.25c$) of the speed of light $c$. So we write:

$v_{electron} = 0.25c$

Plugging this into the kinetic energy formula, we get:

$K.E._{electron} = \frac{1}{2}m_{e}(0.25c)^2 = \frac{1}{2}m_{e} \times \frac{1}{16}c^2$

$K.E._{electron} = \frac{1}{32}m_{e}c^2$

For a photon, $p_{photon} = \frac{E_{photon}}{c}$ and hence its kinetic energy (which, for a photon, is simply its energy) is:

$K.E._{photon} = E_{photon} = cp_{photon}$

Now, comparing the kinetic energies:

$\frac{K.E._{electron}}{K.E._{photon}} = \frac{\frac{1}{32}m_{e}c^2}{cp_{photon}}$

Since $p_{electron} = p_{photon}$ (from the de-Broglie relation), we can replace $p_{photon}$ with $p_{electron}$ which is $m_{e}v_{electron}$:

$\frac{K.E._{electron}}{K.E._{photon}} = \frac{\frac{1}{32}m_{e}c^2}{m_{e}v_{electron}c}$

$\frac{K.E._{electron}}{K.E._{photon}} = \frac{\frac{1}{32}c}{0.25c}$

$\frac{K.E._{electron}}{K.E._{photon}} = \frac{1}{32} \times \frac{1}{0.25}$

$\frac{K.E._{electron}}{K.E._{photon}} = \frac{1}{32} \times 4$

$\frac{K.E._{electron}}{K.E._{photon}} = \frac{1}{8}$

So the correct answer is Option C $\frac{1}{8}$.

$$\begin{aligned} & g_p=\frac{g R^2}{(R+h)^2} \\ & g_q=g\left(1-\frac{h}{R}\right) \\ & g_p=g_q \\ & \frac{g}{\left(1+\frac{h}{R}\right)^2}=g\left(1-\frac{h}{R}\right) \\ & \left(1-\frac{h^2}{R^2}\right)\left(1+\frac{h}{R}\right)=1 \end{aligned}$$

Take $\frac{\mathrm{h}}{\mathrm{R}}=\mathrm{x}$

So

$$\begin{aligned} & \mathrm{x}^3-\mathrm{x}+\mathrm{x}^2=0 \\ & \mathrm{x}=\frac{\sqrt{5}-1}{2} \\ & \mathrm{~h}=\frac{\mathrm{R}}{2}(\sqrt{5}-1) \end{aligned}$$

$$\begin{aligned} & I=\frac{V}{R_{\text {eq }}} I=\frac{V}{R_{\text {eq }}}=\frac{9}{1+\frac{12 \times 4}{12+4}}=\frac{9}{4} \\ & I_1=\frac{9}{4} \times \frac{4}{16}=\frac{9}{16} \\ & V_A-V_B=I_1 \times 8=\frac{9}{16} \times 8=\frac{9}{2} V \\ & \therefore U=\frac{1}{2} \times 4 \times \frac{81}{4} \mu J \\ & \therefore U=\frac{81}{2} \mu J \\ & \therefore x=81 \end{aligned}$$

Let x = current/division

After applying shunt

Now $5 \mathrm{x} \times \mathrm{G}=20 \mathrm{x} \times 24$

$$\begin{aligned} & \mathrm{G}=4 \times 24 \\ & \mathrm{G}=96 \Omega \end{aligned}$$

$$\begin{aligned} & \frac{\mathrm{P}_{\mathrm{A}} \mathrm{V}_{\mathrm{A}}}{\mathrm{P}_{\mathrm{B}} \mathrm{V}_{\mathrm{B}}}=\frac{\mathrm{n}_{\mathrm{A}} R T_{\mathrm{A}}}{\mathrm{n}_{\mathrm{B}} \mathrm{RT}_{\mathrm{B}}} \\ & \text { Given } \mathrm{V}_{\mathrm{A}}=\mathrm{V}_{\mathrm{B}} \\ & \text { And } \mathrm{T}_{\mathrm{A}}=\mathrm{T}_{\mathrm{B}} \\ & \frac{\mathrm{P}_{\mathrm{A}}}{\mathrm{P}_{\mathrm{B}}}=\frac{\mathrm{n}_{\mathrm{A}}}{\mathrm{n}_{\mathrm{B}}} \\ & \frac{\mathrm{P}_{\mathrm{A}}}{\mathrm{P}_{\mathrm{B}}}=\frac{1 / 2}{1 / 32}=16 \end{aligned}$$

To calculate the amount of electric charge $Q$ that crosses through a section of the wire over a period of $20$ seconds, given the current varies with time as $I = I_0 + \beta t$, where $I_0 = 20$ A (initial current) and $\beta = 3$ A/s (rate of change of current with time), we use the concept of integration from calculus because the current is not constant but changes linearly with time.

The electric charge $Q$ is the integral of current $I$ over the time interval from $0$ to $20$ seconds. The formula for $Q$ is given by:

$Q = \int_{0}^{T} I(t) \, dt$

Where $I(t) = I_0 + \beta t$ and $T = 20$ s. Substituting the given values:

$Q = \int_{0}^{20} (20 + 3t) \, dt$

This integral can be solved in two parts:

1. The integral of the constant term $20$:

$\int 20 \, dt = 20t$

2. The integral of the linear term $3t$:

$\int 3t \, dt = \frac{3}{2}t^2$

Thus, combining these and evaluating from $0$ to $20$ seconds:

$Q = \left[20t + \frac{3}{2}t^2\right]_{0}^{20}$

$Q = \left[20(20) + \frac{3}{2}(20)^2\right] - \left[20(0) + \frac{3}{2}(0)^2\right]$

$Q = 400 + 600 = 1000$ C

Therefore, the amount of electric charge that crosses through a section of the wire in $20$ seconds is $1000$ C.

Given $\mathrm{G}=10 \Omega$

$$\begin{aligned} & \mathrm{I}_{\mathrm{g}}=3 \mathrm{~mA} \\ & \mathrm{I}=8 \mathrm{~A} \end{aligned}$$

In case of conversion of galvanometer into ammeter.

Let's denote the constant acceleration with which the body moves as $a$. We know that the displacement $S$ covered by a body starting from rest under a constant acceleration $a$ in time $t$ is given by the equation of motion: $S = \frac{1}{2} a t^2$.

Considering the displacement $\mathrm{S}_1$ in first $(p-1)$ seconds, we apply the equation of motion:

Similarly, for the displacement $\mathrm{S}_2$ in first $p$ seconds:

To find out the total time it will take to cover the displacement $\mathrm{S}_1+\mathrm{S}_2$, we first find the sum of these two displacements:

Let's simplify this:

If we consider the total displacement $\mathrm{S}_1+\mathrm{S}_2$ is to be covered in a time $t$ seconds from rest, we should set this equal to the equation of motion:

Equating the two equations:

Since $a \neq 0$, we can simplify by dividing both sides by $\frac{1}{2} a$:

To find $t$, we take the square root of both sides:

Therefore, the time taken to cover the displacement $\mathrm{S}_1+\mathrm{S}_2$ will be:

Hence, the correct option would be:

$$\begin{aligned} & \text { Let total energy }=\mathrm{E}=\frac{1}{2} \mathrm{KA}^2 \\ & \mathrm{U}=\frac{1}{2} \mathrm{~K}\left(\frac{\mathrm{A}}{3}\right)^2=\frac{\mathrm{KA}^2}{2 \times 9}=\frac{\mathrm{E}}{9} \\ & \mathrm{KE}=\mathrm{E}-\frac{\mathrm{E}}{9}=\frac{8 \mathrm{E}}{9} \\ & \text { Ratio } \frac{\text { Total }}{\mathrm{KE}}=\frac{\mathrm{E}}{\frac{8 \mathrm{E}}{9}}=\frac{9}{8} \\ & \mathrm{x}=9 \end{aligned}$$

The ball needs to just cross 4 steps to just hit $5^{\text {th }}$ step

Therefore, horizontal range $(R)=0.4 \mathrm{~m}$

$\mathrm{R}=\text { u.t }$

Similarly, in vertical direction

$$\begin{aligned} & \mathrm{h}=\frac{1}{2} \mathrm{gt}^2 \\ & 0.4=\frac{1}{2} \mathrm{gt}^2 \\ & 0.4=\frac{1}{2} \mathrm{~g}\left(\frac{0.4}{\mathrm{u}}\right)^2 \\ & \mathrm{u}^2=2 \\ & \mathrm{u}=\sqrt{2} \mathrm{~m} / \mathrm{s} \end{aligned}$$

Therefore, $\mathrm{x}=2$

To determine the acceleration of a cylinder rolling down an inclined plane without slipping, we can use Newton's second law and the concept of rolling motion. For an inclined plane at an angle $\theta$, the component of gravitational acceleration along the plane is $g \sin \theta$. However, because the cylinder is rolling and not sliding, not all of this component accelerates the center of mass; some of it goes into causing rotational acceleration about the center of mass.

For a rolling cylinder, the moment of inertia $I$ is $I = \frac{1}{2} m r^2$, where $m$ is the mass of the cylinder and $r$ is the radius. The condition for rolling without slipping is that the linear acceleration $a$ of the center of mass is equal to the radius $r$ times the angular acceleration $\alpha$, i.e., $a = r \alpha$.

To find the linear acceleration $a$, we use the torque $\tau$ about the center of mass caused by the gravitational force down the incline. The torque due to gravity is $\tau = mg \sin \theta \cdot r$, and from Newton's second law for rotation, the angular acceleration is given by

Now using $a = r \alpha$:

But we must account for the fact that only a portion of the gravitational acceleration goes into translating the cylinder down the plane due to the rolling condition. This is where we apply the concept of the ``rolling factor'' for a cylinder, which is $\frac{2}{3}$ for a solid cylinder, meaning $\frac{2}{3}$ of the gravitational component is used for translation.

The acceleration of the center of mass for the cylinder is therefore:

Now we plug in the values of $\theta = 60^{\circ}$ (which has $\sin 60^{\circ} = \frac{\sqrt{3}}{2}$) and $g = 10 \; m/s^2$:

To match the given expression $\frac{x}{\sqrt{3}} m / s^2$, let's manipulate our expression for $a$:

Hence, the value of $x$ is $10$.

Given $\mathrm{R}=30 \mathrm{~cm}$

$\mathrm{f}=\mathrm{R} / 2=+15 \mathrm{~cm}$

Magnification $(\mathrm{m})= \pm \frac{1}{2}$

For convex mirror, virtual image is formed for real object.

Therefore, $\mathrm{m}$ is +ve

$$\begin{aligned} & \frac{1}{2}=\frac{f}{f-u} \\ & u=-15 \mathrm{~cm} \end{aligned}$$

$$\begin{aligned} & \mathrm{u}=1 \mathrm{~m} / \mathrm{s} ; \mathrm{a}=-\frac{\sigma \mathrm{e}}{2 \varepsilon_0 \mathrm{~m}} \\ & \mathrm{t}=1 \mathrm{~s} \\ & \mathrm{~S}=-1 \mathrm{~m} \\ & \text { Using } \mathrm{S}=\mathrm{ut}+\frac{1}{2} \mathrm{at}^2 \\ & -1=1 \times 1-\frac{1}{2} \times \frac{\sigma \mathrm{e}}{2 \varepsilon_0 \mathrm{~m}} \times(1)^2 \\ & \therefore \sigma=8 \frac{\varepsilon_0 \mathrm{~m}}{\mathrm{e}} \\ & \therefore \alpha=8 \end{aligned}$$

Ampere-Maxwell law

$\rightarrow \oint \overrightarrow{\mathrm{B}} \cdot \overrightarrow{\mathrm{dl}}=\mu_0 \mathrm{i}_{\mathrm{c}}+\mu_0 \varepsilon_0 \frac{\mathrm{d} \phi_{\mathrm{E}}}{\mathrm{dt}}$

Faraday law $\rightarrow \oint \vec{E} \cdot \overrightarrow{d l}=\frac{d \phi_{\mathrm{B}}}{d t}$

Gauss' law for electricity $\rightarrow \oint \overrightarrow{\mathrm{E}} \cdot \overrightarrow{\mathrm{dA}}=\frac{\mathrm{Q}}{\varepsilon_0}$

Gauss ' law for magnetism $\rightarrow \oint \vec{B} \cdot \overrightarrow{d A}=0$

Path difference for minima at $\mathrm{P}$

$$\begin{aligned} & 2 \sqrt{\mathrm{D}^2+\mathrm{d}^2}-2 \mathrm{D}=\frac{\lambda}{2} \\ & \therefore \sqrt{\mathrm{D}^2+\mathrm{d}^2}-\mathrm{D}=\frac{\lambda}{4} \\ & \therefore \sqrt{\mathrm{D}^2+\mathrm{d}^2}=\frac{\lambda}{4}+\mathrm{D} \\ & \Rightarrow \mathrm{D}^2+\mathrm{d}^2=\mathrm{D}^2+\frac{\lambda^2}{16}+\frac{\mathrm{D} \lambda}{2} \\ & \Rightarrow \mathrm{d}^2=\frac{\mathrm{D} \lambda}{2}+\frac{\lambda^2}{16} \\ & \Rightarrow \mathrm{d}^2=\frac{0.2 \times 400 \times 10^{-9}}{2}+\frac{4 \times 10^{-14}}{4} \\ & \Rightarrow \mathrm{d}^2 \approx 400 \times 10^{-10} \\ & \therefore \mathrm{d}=20 \times 10^{-5} \\ & \Rightarrow \mathrm{d}=0.20 \mathrm{~mm} \end{aligned}$$

$\mathrm{V}_{\mathrm{z}}=3 \mathrm{~V}$

Let potential at $\mathrm{B}=0 \mathrm{~V}$

Potential at $E\left(V_E\right)=10 V$

$$\begin{aligned} & V_C=V_A=3 V \\ & I_z+I_1=I \\ & I=\frac{10-3}{1000}=\frac{7}{1000} A \\ & I_1=\frac{3}{2000} A \end{aligned}$$

Therefore $I_z=\frac{7-1.5}{1000}=5.5 \mathrm{~mA}$

$$\begin{aligned} & \text { Given } U=2 x^2+3 y^3+2 z \\ & F_x=-\frac{\partial U}{\partial x}=-4 x \end{aligned}$$

At $x=1$ magnitude of $F_x$ is $4 N$

$$\begin{aligned} & \overrightarrow{\mathrm{A}}=(0.1)^2 \hat{\mathrm{j}} \\ & \overrightarrow{\mathrm{B}}=\frac{0.2}{\sqrt{2}} \hat{\mathrm{i}}+\frac{0.2}{\sqrt{2}} \hat{\mathrm{j}} \end{aligned}$$

Magnitude of induced emf

$\mathrm{e}=\frac{\Delta \phi}{\Delta \mathrm{t}}=\frac{\overrightarrow{\mathrm{B}} \cdot \overrightarrow{\mathrm{A}}-0}{1}=\sqrt{2} \times 10^{-3} \mathrm{~V}$

Work done $\mathrm{AB}=\frac{1}{2}(8000+4000)$ Dyne$/\mathrm{cm}^2 \times 4 \mathrm{~m}^3=\left(6000\right.$ Dyne$\left./\mathrm{cm}^2\right) \times 4 \mathrm{~m}^3$

Work done $\mathrm{BC}=-\left(4000\right.$ Dyne$\left./\mathrm{cm}^2\right) \times 4 \mathrm{~m}^3$

Total work done $=2000$ Dyne$/\mathrm{cm}^2 \times 4 \mathrm{~m}^3$

$$\begin{aligned} =2 \times 10^3 & \times \frac{1}{10^5} \frac{\mathrm{N}}{\mathrm{cm}^2} \times 4 \mathrm{~m}^3 \\ & =2 \times 10^{-2} \times \frac{\mathrm{N}}{10^{-4} \mathrm{~m}^2} \times 4 \mathrm{~m}^3 \\ & =2 \times 10^2 \times 4 \mathrm{Nm}=800 \mathrm{~J} \end{aligned}$$

By energy conservation

$$\begin{aligned} & \frac{1}{2} \mathrm{CV}^2=\frac{1}{2} \mathrm{LI}_{\text {max }}^2 \\ & \mathrm{I}_{\max }=\sqrt{\frac{\mathrm{C}}{\mathrm{L}}} \mathrm{V} \\ & =\sqrt{\frac{100 \times 10^{-6}}{6.4 \times 10^{-3}}} \times 12 \\ & =\frac{12}{8}=\frac{3}{2}=1.5 \mathrm{~A} \end{aligned}$$

$$\begin{aligned} & \text { Given } \mathrm{m}=100 \mathrm{~kg} \\ & \mathrm{~s}=10 \mathrm{~m} \\ & \mu=0.4 \\ & \text { As } \mathrm{f}=\mu \mathrm{mg}=0.4 \times 100 \times 10=400 \mathrm{~N} \\ & \text { Now } \mathrm{W}=\mathrm{f} . \mathrm{s}=400 \times 10=4000 \mathrm{~J} \end{aligned}$$

\begin{aligned} & { }_3 \mathrm{Li}^6+{ }_0 \mathrm{n}^1 \rightarrow{ }_2 \mathrm{He}^4+{ }_1 \mathrm{H}^3 \\ & { }_1 \mathrm{H}^2+{ }_1 \mathrm{H}^3 \rightarrow{ }_2 \mathrm{He}^4+{ }_0 \mathrm{n}^1 \\ & \hline{ }_3 \mathrm{Li}^6+{ }_1 \mathrm{H}^2 \rightarrow 2\left({ }_2 \mathrm{He}^4\right) \\ & \hline \end{aligned}

Energy released in process

$$\begin{aligned} & \mathrm{Q}=\Delta \mathrm{mc}^2 \\ & \mathrm{Q}=\left[\mathrm{M}(\mathrm{Li})+\mathrm{M}\left(\mathrm{H}^2\right)-2 \times \mathrm{M}\left({ }_2 \mathrm{He}^4\right)\right] \times 931.5 \mathrm{~MeV} \\ & \mathrm{Q}=[6.01690+2.01471-2 \times 4.00388] \times 931.5 \mathrm{~MeV} \\ & \mathrm{Q}=22.216 \mathrm{~MeV} \\ & \mathrm{Q}=22.22 \mathrm{~MeV} \end{aligned}$$

$$\begin{aligned} & \mathrm{V}=\frac{\mu_0}{4 \pi} \frac{\mathrm{M}}{\mathrm{r}^2} \\ & \Rightarrow 1.5 \times 10^{-5}=10^{-7} \times \frac{\mathrm{M}}{\left(20 \times 10^{-2}\right)^2} \\ & \Rightarrow \mathrm{M}=\frac{1.5 \times 10^{-5} \times 20 \times 20 \times 10^{-4}}{10^{-7}} \\ & \mathrm{M}=1.5 \times 4=6 \end{aligned}$$

Surface tension will be less as temperature increases

$\mathrm{h}=\frac{2 \mathrm{~T} \cos \theta}{\rho \mathrm{gr}}$

Height of capillary rise will be smaller in hot water and larger in cold water.

$$\begin{aligned} & \mathrm{F}=\frac{1}{2} \rho\left(\mathrm{v}_1^2-\mathrm{v}_2^2\right) \mathrm{A} \\ & \mathrm{F}=\frac{1}{2} \times 1.2 \times\left(70^2-65^2\right) \times 2 \\ & =810 \mathrm{~N} \end{aligned}$$

$$\begin{aligned} & \Delta \mathbf{E}=\mathbf{1 0 . 2} \mathrm{eV} \\ & \text { Recoil speed }(\mathrm{v})=\frac{\Delta \mathrm{E}}{\mathrm{mc}} \\ & =\frac{10.2 \mathrm{eV}}{1.6 \times 10^{-27} \times 3 \times 10^8} \\ & =\frac{10.2 \times 1.6 \times 10^{-19}}{1.6 \times 10^{-27} \times 3 \times 10^8} \\ & \mathrm{v}=3.4 \mathrm{~m} / \mathrm{s}=\frac{17}{5} \mathrm{~m} / \mathrm{s} \end{aligned}$$

Therefore, $x=17$