JEE Main 2024 27th January Morning Shift Question Paper with Solutions

$$\begin{aligned} & g=\frac{G M}{R^2} \Rightarrow g \propto \frac{1}{R^2} \\ & \frac{g_2}{g_1}=\frac{R_1^2}{R_2^2} \\ & g_2=4 g_1\left(R_2=\frac{R_1}{2}\right) \end{aligned}$$

Spherometer can be used to measure curvature of surface.

$\overrightarrow{\mathrm{v}}=\frac{\mathrm{ds}}{\mathrm{dt}}=4 \mathrm{t} \hat{\mathrm{j}}$

At $\mathrm{t}=1 \sec \overrightarrow{\mathrm{v}}=4 \hat{\mathrm{j}}$

Gases have less viscosity.

Due to insoluble impurities like detergent surface tension decreases

$$\begin{aligned} & \mathrm{V}_{\text {at mean position }}=\mathrm{A} \omega \Rightarrow 10=4 \omega \\ & \quad \omega=\frac{5}{2} \\ & \mathrm{~V}=\omega \sqrt{\mathrm{A}^2-\mathrm{x}^2} \\ & 5=\frac{5}{2} \sqrt{4^2-\mathrm{x}^2} \Rightarrow \mathrm{x}^2=16-4 \\ & \mathrm{x}=\sqrt{12} \mathrm{~cm} \end{aligned}$$

To find the change in the internal energy of air when it is heated at constant volume, we use the formula for heat transfer at constant volume, which is given by:

$\Delta U = m c_v \Delta T$

Where:

• $\Delta U$ is the change in internal energy,
• $m$ is the mass of the substance (in this case, air),
• $c_v$ is the specific heat at constant volume,
• $\Delta T$ is the change in temperature.

Given that:

• The mass of air, $m = 0.08 \, \text{kg},$
• The specific heat of air at constant volume, $c_v = 0.17 \, \text{kcal/kg}^{\circ}\text{C},$
• The change in temperature, $\Delta T = 5^{\circ} \text{C},$
• And the conversion factor from calories to Joules, $1 \, \text{cal} = 4.18 \, \text{J}.$

First, convert the specific heat from kcal to Joules:

$c_v = 0.17 \, \text{kcal/kg}^{\circ}\text{C} \times 1000 \, \text{cal/kcal} \times 4.18 \, \text{J/cal} = 710.6 \, \text{J/kg}^{\circ}\text{C}$

Now, substitute the values into the formula:

$\Delta U = 0.08 \times 710.6 \times 5$

$\Delta U = 284 \, \text{J}$

Thus, the change in internal energy is approximately 284 Joules.

Momentum will remain conserve

$$\begin{aligned} & 1000 \times 6=1200 \times v \\ & v=5 \mathrm{~m} / \mathrm{s} \end{aligned}$$

$\mathrm{P}$ end should be at higher potential for forward biasing.

$h_{\text {app }}=\frac{h_1}{\mu_1}+\frac{h_2}{\mu_2}=\frac{6}{3 / 2}+\frac{6}{8 / 5}=4+\frac{15}{4}=\frac{31}{4} \mathrm{~cm}$

$\tan \theta=\frac{v^2}{R g}=\frac{12 \times 12}{10 \times 400}$

$$\begin{aligned} & \tan \theta=\frac{\mathrm{h}}{1.5} \\ & \Rightarrow \frac{\mathrm{h}}{1.5}=\frac{144}{4000} \\ & \mathrm{~h}=5.4 \mathrm{~cm} \end{aligned}$$

For null point,

$$\begin{aligned} & \frac{4.5}{60}=\frac{R}{40} \\ & \text { Also, } R=\frac{\rho \ell}{A}=\frac{\rho \ell}{\pi r^2} \\ & 4.5 \times 40=\rho \times \frac{0.1}{\pi \times 7 \times 10^{-8}} \times 60 \\ & \rho=66 \times 10^{-7} \Omega \times \mathrm{m} \end{aligned}$$

To find the temperature at which the average kinetic energy of a monatomic molecule is $0.414 \, \text{eV}$, we use the equation for the average kinetic energy of a molecule in terms of temperature:

$K_{\text{avg}} = \frac{3}{2} k_B T$

Where:

• $K_{\text{avg}}$ is the average kinetic energy
• $k_B$ is the Boltzmann constant, given as $1.38 \times 10^{-23} \, \text{J/K}$
• $T$ is the temperature in Kelvin

First, we need to convert the average kinetic energy from eV to Joules since the Boltzmann constant is in Joules. The conversion factor is $1 \, \text{eV} = 1.6 \times 10^{-19} \, \text{J}$, so:

$K_{\text{avg}} = 0.414 \, \text{eV} = 0.414 \times 1.6 \times 10^{-19} \, \text{J}$

Substituting $K_{\text{avg}}$ and $k_B$ into the equation:

$\frac{3}{2} k_B T = 0.414 \times 1.6 \times 10^{-19} \, \text{J}$

Solving for $T$:

$T = \frac{0.414 \times 1.6 \times 10^{-19} \, \text{J}}{\frac{3}{2} \times 1.38 \times 10^{-23} \, \text{J/K}}$

After performing the calculations:

$T \approx 3200 \, \text{K}$

Therefore, the temperature at which the average kinetic energy of a monatomic molecule is $0.414 \, \text{eV}$ is approximately 3200 K.

According to Faraday's Law of Electromagnetic Induction, the induced emf in a closed circuit is equal to the negative rate of change of magnetic flux through the circuit. Mathematically, it is expressed as:

$\varepsilon = -\frac{d\Phi}{dt}$

Where $\varepsilon$ is the induced emf, and $\Phi$ is the magnetic flux.

To find the magnetic flux $\Phi$ through the rectangular loop, we use the formula:

$\Phi = B \cdot A \cdot \cos(\theta)$

Where:

• $B$ is the magnetic field strength,
• $A$ is the area of the loop, and
• $\theta$ is the angle between the magnetic field lines and the normal (perpendicular) to the plane of the loop.

The area $A$ of the rectangular loop is:

$A = \text{length} \times \text{width} = 2.5 \mathrm{~m} \times 2 \mathrm{~m} = 5 \mathrm{~m}^2$

Given the angle $\theta = 60^\circ$, we can calculate the initial magnetic flux $\Phi_{initial}$:

$\Phi_{initial} = B \cdot A \cdot \cos(60^\circ)$

$\Phi_{initial} = 4 \mathrm{~T} \cdot 5 \mathrm{~m}^2 \cdot \cos(60^\circ)$

$\Phi_{initial} = 4 \cdot 5 \cdot \frac{1}{2} = 10 \mathrm{~Wb}$

(Since $\cos(60^\circ) = \frac{1}{2}$)

When the loop is removed from the magnetic field, the final magnetic flux $\Phi_{final}$ is zero, because the loop is no longer within the magnetic field. Thus, the change in magnetic flux $\Delta\Phi$ is:

$\Delta\Phi = \Phi_{final} - \Phi_{initial} = 0 - 10 \mathrm{~Wb} = -10 \mathrm{~Wb}$

The loop is removed from the field in $t = 10$ seconds, so the rate of change of magnetic flux is:

$\frac{d\Phi}{dt} = \frac{\Delta\Phi}{\Delta t} = \frac{-10 \mathrm{~Wb}}{10 \mathrm{~s}} = -1 \mathrm{~Wb/s}$

Now we can find the average induced emf $\varepsilon$:

$\varepsilon = -\frac{d\Phi}{dt}$

$\varepsilon = -(-1 \mathrm{~Wb/s})$

$\varepsilon = +1 \mathrm{~V}$

Therefore, the average induced emf in the loop during this time is $+1 \mathrm{~V}$. The correct answer is:

Option C

$+1 \mathrm{~V}$

As amount of energy incident on cell is same so current will remain same.

Net force on particle must be zero i.e. $\mathrm{q} \overrightarrow{\mathrm{E}}+\mathrm{q} \overrightarrow{\mathrm{V}} \times \overrightarrow{\mathrm{B}}=0$

Possible cases are

(i) $\overrightarrow{\mathrm{E}} \& \overrightarrow{\mathrm{B}}=0$

(ii) $\overrightarrow{\mathrm{V}} \times \overrightarrow{\mathrm{B}}=0, \overrightarrow{\mathrm{E}}=0$

(iii) $q \overrightarrow{\mathrm{E}}=-\mathrm{q} \overrightarrow{\mathrm{V}} \times \overrightarrow{\mathrm{B}}$

$\overrightarrow{\mathrm{E}} \neq 0 \& \overrightarrow{\mathrm{B}} \neq 0$

$$\begin{aligned} & \mathrm{I}=\frac{1}{2} \varepsilon_0 \mathrm{E}_0^2 \times \mathrm{c} \\ & \mathrm{I}=\frac{1}{2} \times 8.85 \times 10^{-12} \times 4 \times 10^4 \times 3 \times 10^8 \\ & \mathrm{I}=53.1 \mathrm{~W} / \mathrm{m}^2 \end{aligned}$$

$$\begin{aligned} & \mathrm{B}=-\frac{\Delta \mathrm{P}}{\left(\frac{\Delta \mathrm{V}}{\mathrm{V}}\right)} \\ & -\left(\frac{\Delta \mathrm{V}}{\mathrm{V}}\right)=\frac{\rho \mathrm{gh}}{\mathrm{B}}=\frac{1000 \times 10 \times 4000}{2 \times 10^9} \\ & =2 \times 10^{-2}[-\mathrm{ve} \text { sign represent compression }] \end{aligned}$$

$$\begin{aligned} & 2 \mathrm{~T} \sin \frac{\mathrm{d} \theta}{2}=\frac{\mathrm{kq}_0}{\mathrm{R}^2} \cdot \lambda \mathrm{Rd} \theta \\ & {\left[\lambda=\frac{\mathrm{Q}}{2 \pi \mathrm{R}}\right]} \end{aligned}$$

$$\begin{aligned} & \Rightarrow \mathrm{T}=\frac{\mathrm{Kq}_0 \mathrm{Q}}{\left(\mathrm{R}^2\right) \times 2 \pi} \\ & =\frac{\left(9 \times 10^9\right)\left(2 \pi \times 30 \times 10^{-12}\right)}{(0.30)^2 \times 2 \pi} \\ & =\frac{9 \times 10^{-3} \times 30}{9 \times 10^{-2}}=3 \mathrm{~N} \end{aligned}$$

$$\begin{aligned} & \mathrm{V}_{\mathrm{A}}+\frac{10}{3}(1)-6(1)=\mathrm{V}_{\mathrm{B}} \\ & \mathrm{V}_{\mathrm{A}}-\mathrm{V}_{\mathrm{B}}=6-\frac{10}{3}=\frac{8}{3} \mathrm{volt} \\ & \mathrm{Q}=\mathrm{C}\left(\mathrm{V}_{\mathrm{A}}-\mathrm{V}_{\mathrm{B}}\right) \\ & =150 \times \frac{8}{3}=400 \mu \mathrm{C} \end{aligned}$$

$$\begin{aligned} & \cot \frac{\mathrm{A}}{2}=\frac{\sin \left(\frac{\mathrm{A}+\delta_{\min }}{2}\right)}{\sin \frac{\mathrm{A}}{2}} \\ & \Rightarrow \cos \frac{\mathrm{A}}{2}=\sin \left(\frac{\mathrm{A}+\delta_{\min }}{2}\right) \\ & \frac{\mathrm{A}+\delta_{\min }}{2}=\frac{\pi}{2}-\frac{\mathrm{A}}{2} \\ & \delta_{\min }=\pi-2 \mathrm{~A} \end{aligned}$$

Potential difference $=\frac{K Q}{r_1}-\frac{K Q}{r_2}$

$$\begin{aligned} & r_1=\sqrt{(\sqrt{3})^2+(\sqrt{3})^2} \\ & r_2=\sqrt{(\sqrt{6})^2+0} \end{aligned}$$

As $r_1=r_2=\sqrt{6} \mathrm{~m}$

So potential difference $=0$

$$\begin{aligned} & \mathrm{B}=\left(\frac{\mu_0 \mathrm{i}}{2 \pi \mathrm{a}}\right) \times 2=\frac{4 \pi \times 10^{-7} \times 10}{\pi \times\left(\frac{5}{2} \times 10^{-2}\right)} \\ & =16 \times 10^{-5}=160 \mu \mathrm{T} \end{aligned}$$

Resistance of each part $=\frac{R}{5}$

Total resistance $=\frac{1}{5} \times \frac{\mathrm{R}}{5}=\frac{\mathrm{R}}{25}$

To determine the energy released in a nuclear fission process, we use the difference in binding energy (BE) before and after the fission. The formula for energy released ($Q$ value) in the process is given by:

$Q = (\text{Total BE of products}) - (\text{Total BE of reactants})$

In this case, the reactant is a high mass nuclide with atomic mass $A \approx 236$ and a binding energy of $7.6 \mathrm{MeV}/\mathrm{nucleon}$. Each of the two middle mass nuclides formed as products has atomic mass $A \approx 118$ and a binding energy of $8.6 \mathrm{MeV}/\mathrm{nucleon}$.

Therefore, we calculate the total binding energy of reactant and products as follows:

For reactant:

$\text{BE}_{\text{reactant}} = 236 \times 7.6 \mathrm{MeV}$

For products (since there are two identical products):

$\text{BE}_{\text{products}} = 2 \times (118 \times 8.6) \mathrm{MeV}$

Thus, the energy released ($Q$ value) is:

$Q = \text{BE}_{\text{products}} - \text{BE}_{\text{reactant}}$

$Q = 2(118 \times 8.6) - (236 \times 7.6)$

$Q = 236 \times (8.6 - 7.6)$

$Q = 236 \times 1$

$Q = 236 \mathrm{MeV}$

This calculation demonstrates how the difference in binding energy per nucleon before and after fission leads to the release of energy, consistent with the mass-energy equivalence principle.

$$\begin{aligned} & \mathrm{I}=\mathrm{ma}^2+\mathrm{ma}^2+\mathrm{m}(\sqrt{2 \mathrm{a}})^2 \\ & =4 \mathrm{ma}^2 \\ & =4 \times 1 \times(2)^2=16 \end{aligned}$$

$$\begin{aligned} & {[\mathrm{h}]=\mathrm{ML}^2 \mathrm{~T}^{-1}} \\ & {[\mathrm{~L}]=\mathrm{ML}^2 \mathrm{~T}^{-1}} \\ & {[\mathrm{P}]=\mathrm{MLT}^{-1}} \\ & {[\tau]=\mathrm{ML}^2 \mathrm{~T}^{-2}} \end{aligned}$$

(Here $\mathrm{h}$ is Planck's constant, $\mathrm{L}$ is angular momentum, $\mathrm{P}$ is linear momentum and $\tau$ is moment of force)

To solve for the value of $\alpha$, which represents the square of the speed of the particle at the time its $x$-coordinate is $84 \mathrm{m}$, we need to first determine the time at which the particle reaches this $x$-coordinate, and then use this time to calculate its final velocity in both the $x$ and $y$ directions.

The motion of the particle in the $x$-direction can be described by the kinematic equation for uniformly accelerated motion:

$x = x_0 + v_{0x}t + \frac{1}{2}a_xt^2$

Given:

$x_0 = 0 \mathrm{~m}$

$v_{0x} = 5 \mathrm{~m/s}$

$a_x = 3 \mathrm{~m/s}^2$

$x = 84 \mathrm{~m}$ (at which we need to find the speed)

Substituting these values into the kinematic equation:

$84 \mathrm{~m} = 0 \mathrm{~m} + (5 \mathrm{~m/s})t + \frac{1}{2}(3 \mathrm{~m/s}^2)t^2$

Simplifying this equation, we get:

$0 = \frac{3}{2}t^2 + 5t - 84$

Using the quadratic formula to solve for $t$:

$t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$

where $a = \frac{3}{2}, b = 5,$ and $c = -84$.

$t = \frac{-5 \pm \sqrt{(5)^2 - 4(\frac{3}{2})(-84)}}{2(\frac{3}{2})}$

$t = \frac{-5 \pm \sqrt{25 + 504}}{3}$

$t = \frac{-5 \pm \sqrt{529}}{3}$

$t = \frac{-5 \pm 23}{3}$

In this scenario, since we're looking for a time when the particle reaches $84 \mathrm{m}$, we only consider the positive root because time cannot be negative.

$t = \frac{18}{3}$

$t = 6 \mathrm{s}$

Now we have the time at which the particle's $x$-coordinate is $84 \mathrm{m}$. Next, we find final velocities in $x$ and $y$ directions at $t = 6 \mathrm{s}$.

The final velocity in the $x$-direction can be found using the formula for velocity with constant acceleration:

$v_x = v_{0x} + a_xt$

$v_x = 5 \mathrm{~m/s} + (3 \mathrm{~m/s}^2)(6 \mathrm{s})$

$v_x = 5 \mathrm{~m/s} + 18 \mathrm{~m/s}$

$v_x = 23 \mathrm{~m/s}$

Similarly, for the $y$-direction:

$v_y = v_{0y} + a_yt$

Since the particle starts from the origin and is only subject to a force after $t=0$, its initial velocity in the $y$-direction is $0$.

$v_y = 0 + (2 \mathrm{~m/s}^2)(6 \mathrm{s})$

$v_y = 12 \mathrm{~m/s}$

Now we can compute the speed of the particle, which is the magnitude of the velocity vector:

$v = \sqrt{v_x^2 + v_y^2}$

$v = \sqrt{(23 \mathrm{~m/s})^2 + (12 \mathrm{~m/s})^2}$

$v = \sqrt{529 + 144}$

$v = \sqrt{673}$

Therefore, the speed of the particle at the time its $x$-coordinate is $84 \mathrm{m}$ is $\sqrt{673} \mathrm{m/s}$.

So, $\alpha = 673$.

The radius of the nth stationary orbit in the Bohr model of an atom is directly proportional to the square of its principal quantum number n and inversely proportional to the atomic number Z. This relationship is represented by the formula $r_n = \frac{n^2h^2}{4\pi^2kme^2Z},$ where :

• $n$ is the principal quantum number,

• $h$ is Planck's constant,

• $k$ is the Coulomb constant,

• $m$ is the mass of the electron,

• $e$ is the charge of the electron, and

• $Z$ is the atomic number of the nucleus (for hydrogen, Z=1).

To find the radius of the fourth orbit ($r_4$), in comparison to the third orbit ($r_3$), we apply the formula with $n=4$ for the fourth orbit and $n=3$ for the third orbit, and simplify as follows:

$$\begin{aligned} & \frac{r_4}{r_3} = \frac{(4^2)h^2}{4\pi^2kme^2Z} \div \frac{(3^2)h^2}{4\pi^2kme^2Z} = \frac{(4^2)}{(3^2)} \\ & = \frac{16}{9} \end{aligned}$$

Thus, the radius of the fourth stationary orbit is $\frac{16}{9}$ times the radius of the third stationary orbit, $R$. Therefore, $r_4 = \frac{16}{9} R$.

$$\begin{aligned} & \phi=\mathrm{Mi}=\mathrm{Mi}_0 \sin \omega \mathrm{t} \\ & \mathrm{EMF}=-\mathrm{M} \frac{\mathrm{di}}{\mathrm{dt}}=-0.002\left(\mathrm{i}_0 \omega \cos \omega \mathrm{t}\right) \\ & \mathrm{EMF}_{\max }=\mathrm{i}_0 \omega(0.002)=(5)(50 \pi)(0.002) \\ & \mathrm{EMF}_{\max }=\frac{\pi}{2} \mathrm{~V} \end{aligned}$$

$$\begin{aligned} & \frac{\mathrm{P}_1^2}{2 \mathrm{~m}_1}=\frac{\mathrm{P}_2^2}{2 \mathrm{~m}_2} \\ & \frac{\mathrm{P}_1}{\mathrm{P}_2}=\sqrt{\frac{\mathrm{m}_1}{\mathrm{~m}_2}}=\frac{2}{5} \end{aligned}$$

Fluorine is the element among the options that does not exhibit a variable oxidation state, unlike its fellow halogens. This is because fluorine is the most electronegative element, and it always forms compounds in the oxidation state of -1 as it has a strong tendency to gain one electron to achieve a stable electronic configuration.

To determine the number of compounds with a +4 oxidation state of sulfur, we need to examine the oxidation states of sulfur in each of the listed compounds. The +4 oxidation state means that sulfur has lost 4 electrons compared to its elemental state.

Let's go through each compound:

• $\mathrm{SO}_3$: Sulfur trioxide. In this compound, sulfur exhibits a +6 oxidation state because each oxygen contributes -2, for a total of -6 (3 oxygens), which must be balanced by sulfur to maintain a neutral charge. So, this compound does not have sulfur in the +4 oxidation state.


• $\mathrm{H}_2 \mathrm{SO}_3$: Sulfurous acid. Here again, with two hydrogens (each contributing +1 = total +2) and three oxygens (each contributing -2 = total -6), sulfur has an oxidation state of +4 to balance out the -4 from the oxygens and +2 from the hydrogens. This compound does have sulfur in the +4 oxidation state.


• $\mathrm{SOCl}_2$: Thionyl chloride. Sulfur in this compound is connected to two chlorine atoms and one oxygen atom. Chlorine generally has an oxidation state of -1 (total -2 for both Cl atoms) and oxygen -2. To balance the -4 charge (from one oxygen and two chlorines), sulfur must have an oxidation state of +4. This compound has sulfur in the +4 oxidation state.


• $\mathrm{SF}_4$: Sulfur tetrafluoride. Fluorine is almost always in the -1 oxidation state, and there are four fluorine atoms for a total of -4. To balance this, sulfur must have a +4 oxidation state, making this compound one with sulfur in the +4 oxidation state.


• $\mathrm{BaSO}_4$: Barium sulfate. Barium has a +2 oxidation state, and sulfate (SO4) as a whole must have a -2 oxidation state to balance the barium. In sulfate, sulfur has an oxidation state of +6 (since each oxygen is -2 for a total of -8, and +6 from sulfur balances it to -2 overall). Hence, sulfur does not have a +4 oxidation state in this compound.


• $\mathrm{H}_2 \mathrm{S}_2 \mathrm{O}_7$: Pyrosulfuric acid or oleum. Each hydrogen is +1 (total +2), and the seven oxygens are -2 each (total -14). To balance -12 (total of oxygens and hydrogens), each sulfur must have an oxidation state of +6. Therefore, sulfur is not in the +4 oxidation state in this molecule.

So, from the given list, only three compounds have sulfur in the +4 oxidation state: $\mathrm{H}_2 \mathrm{SO}_3$, $\mathrm{SOCl}_2$, and $\mathrm{SF}_4$. Therefore, the number of compounds with a +4 oxidation state of sulfur is three.

Let, $\mathrm{R}=\mathrm{k}[\mathrm{HI}]^{\mathrm{n}}$

using any two of given data,

$$\begin{aligned} & \frac{3 \times 10^{-3}}{7.5 \times 10^{-4}}=\left(\frac{0.01}{0.005}\right)^{\mathrm{n}} \\ & \mathrm{n}=2 \end{aligned}$$

Here's how to identify the meta-directing functional groups and determine the total :

• Meta-directing groups are electron-withdrawing. They destabilize the intermediate carbocation that forms during electrophilic aromatic substitution at the ortho and para positions more than at the meta position.


• Common meta-directing groups include:


• -NO₂ (nitro)


• -CN (cyano)


• -COR (carbonyl groups, like ketones and aldehydes)


• -COOH (carboxylic acid)


• -SO₃H (sulfonic acid)

• -OCH₃: Ortho/para-directing (electron-donating)


• -NO₂: Meta-directing


• -CN: Meta-directing


• -CH₃: Ortho/para-directing (electron-donating)


• -NHCOCH₃: Ortho/para-directing (electron-donating)


• -COR: Meta-directing


• -OH: Ortho/para-directing (electron-donating)


• -COOH: Meta-directing


• -Cl: Ortho/para-directing (weakly deactivating)

There are four meta-directing groups in the list: -NO₂, -CN, -COR, -COOH

The total number of meta-directing functional groups is 4.

Lead chromate ( $\mathrm{PbCrO}_4$ ) dissolves in hot NaOH solution to form products due to a chemical reaction. The reaction involves the formation of a compound where lead ( $\mathrm{Pb}$ ) is coordinated by hydroxide ions ( $\mathrm{OH}^-$ ).