JEE Main 2024 31st January Evening Shift Question Paper with Solutions

$$\begin{aligned} & \mathrm{P}=\mathrm{i}^2 \mathrm{R} \\ & \mathrm{P}_{\text {int }}=\mathrm{I}_{\text {int }}^2 \mathrm{R} \\ & \mathrm{P}_{\text {final }}=\left(0.8 \mathrm{I}_{\text {int }}\right)^2 \mathrm{R} \end{aligned}$$

% change in power $=$

$\frac{P_{\text {final }}-P_{\text {int }}}{P_{\text {int }}} \times 100=(0.64-1) \times 100=-36 \%$

$$\begin{aligned} & \mathrm{L}=\mathrm{mu} \cos \theta \frac{\mathrm{u}^2 \sin ^2 \theta}{2 \mathrm{~g}} \\ & =\mathrm{mu}^3 \frac{1}{4 \sqrt{2} \mathrm{~g}} \Rightarrow \mathrm{x}=8 \end{aligned}$$

$$\begin{aligned} & \mathrm{T}=2 \pi \sqrt{\frac{\ell}{\mathrm{g}}} \\ & \mathrm{g}=\frac{4 \pi^2 \ell}{\mathrm{T}^2} \\ & \frac{\Delta \mathrm{g}}{\mathrm{g}}=\frac{\Delta \ell}{\ell}+\frac{2 \Delta \mathrm{T}}{\mathrm{T}} \\ & =\frac{0.2}{20}+2\left(\frac{1}{40}\right) \\ & =\frac{1.2}{20} \end{aligned}$$

Percentage change $=\frac{1.2}{20} \times 100=6 \%$

\begin{aligned} & \vec{F}=\left(6 t \hat{i}+6 t^2 \hat{j}\right) N \\ & \vec{F}=m \vec{a}=\left(6 t \hat{i}+6 t^2 \hat{j}\right) \\ & \vec{a}=\frac{\vec{F}}{m}=\left(3 t \hat{i}+3 t^2 \hat{j}\right) \\ & \vec{v}=\int_\limits0^t \vec{a} d t=\frac{3 t^2}{2} \hat{i}+t^3 \hat{j} \\ & P=\vec{F} \cdot \vec{v}=\left(9 t^3+6 t^5\right) W \end{aligned}

In the problem, you're dealing with refraction at a convex surface. The light is coming from a point source located in air (with a refractive index, $\mu_1 = 1.0$) and entering a medium with a refractive index of $\mu_2 = 1.5$. The convex surface has a radius of curvature $R = 20 \, \text{cm}$, and the source is placed $100 \, \text{cm}$ from the convex surface.

The formula used to find the image distance $v$ due to refraction at a spherical surface is:

$\frac{\mu_2}{v} - \frac{\mu_1}{u} = \frac{\mu_2 - \mu_1}{R}$

Substituting the given values:

$\frac{1.5}{v} - \frac{1}{-100} = \frac{1.5 - 1}{20}$

This equation allows us to solve for $v$, the distance from the convex surface to the image. Upon solving, we found that $v = 100 \, \text{cm}$, which means the image forms $100 \, \text{cm}$ on the other side of the convex surface, away from the point of refraction.

The distance from the object to the image isn't just $v$, the distance from the surface to where the image forms. Since the object is $100 \, \text{cm}$ from the convex surface and the image also forms $100 \, \text{cm}$ from the convex surface but on the opposite side, the total distance between the object and the image is the sum of these distances:

• Distance from the object to the convex surface: $100 \, \text{cm}$
• Distance from the convex surface to the image: $100 \, \text{cm}$

Therefore, the total distance between the object and the image is $100 \, \text{cm} + 100 \, \text{cm} = 200 \, \text{cm}$.

This calculation accounts for the physical layout where the object and the image are on opposite sides of the convex surface, and to determine the distance between them, you sum the distances from each to the surface. This results in the image being formed $200 \, \text{cm}$ from the object, which means if you were to measure directly from the object to its image, the total distance covered would be $200 \, \text{cm}$.

$$\begin{aligned} & \mathrm{B}_{\mathrm{P}}=\frac{\mu_0 \mathrm{Ni}_1}{2 \mathrm{r}}=\frac{\mu_0 \times 1 \times 100}{2 \pi}=2 \times 10^{-3} \mathrm{~T} \\ & \mathrm{~B}_{\mathrm{Q}}=\frac{\mu_0 \mathrm{Ni}_2}{2 \mathrm{r}}=\frac{\mu_0 \times 2 \times 100}{2 \pi}=4 \times 10^{-3} \mathrm{~T} \\ & \mathrm{~B}_{\text {net }}=\sqrt{\mathrm{B}_{\mathrm{P}}^2+\mathrm{B}_{\mathrm{Q}}^2} \\ & =\sqrt{20} \mathrm{mT} \\ & \mathrm{x}=20 \end{aligned}$$

The speed of sound in a gas at standard temperature and pressure (STP) can be calculated using the following formula derived from the ideal gas law and the speed of sound relation in a gas:

$v = \sqrt{\gamma \frac{R T}{M}}$

Where:

• $v$ = speed of sound in the gas
• $\gamma$ = adiabatic index (ratio of specific heats, $C_p/C_v$)
• $R$ = universal gas constant
• $T$ = temperature in Kelvin
• $M$ = molar mass of the gas in kilograms per mole (kg/mol)

Given:

• $R = 8.3 \, J \cdot mol^{-1} \cdot K^{-1}$


• $\gamma = 1.4$ (For diatomic gases such as oxygen)


• $T = 273.15 \, K$ (Standard temperature, 0°C in Kelvin)


• Molar mass of Oxygen ($O_2$) $M = 32 \times 10^{-3} \, kg/mol$ (32 g/mol converted to kg/mol)

Plugging these values into the formula:

$v = \sqrt{1.4 \times \frac{8.3 \times 273.15}{32 \times 10^{-3}}}$

Calculating the values inside the square root:

$v = \sqrt{1.4 \times \frac{2268.745}{0.032}}$

$v = \sqrt{1.4 \times 70896.40625}$

$v = \sqrt{99304.96875}$

$v \approx 315 \, m/s$

$$\begin{aligned} & <\mathrm{P}>=\mathrm{IV} \cos \phi \\ & =\frac{20}{\sqrt{2}} \times \frac{10}{\sqrt{2}} \times \cos 60^{\circ} \\ & =50 \mathrm{~W} \end{aligned}$$

$$\begin{aligned} & \frac{25}{\mathrm{r} \ell_1}=\frac{\mathrm{X}}{\mathrm{r} \ell_2} \quad \text{.... (i)}\\ & \frac{25}{2 \mathrm{r} \ell_1^{\prime}}=\frac{\mathrm{X}}{2 \mathrm{r} \ell^{\prime}{ }_2} \quad \text{.... (ii)} \end{aligned}$$

From (i) and (ii)

$\ell_2^{\prime}=\ell_2=40 \mathrm{~cm}$

$$\begin{aligned} & a=\frac{\left(m_1-m_2\right) g}{\left(m_1+m_2\right)}=\frac{g}{8} \\ & 8 m_1-8 m_2=m_1+m_2 \\ & 7 m_1=9 m_2 \\ & \frac{m_1}{m_2}=\frac{9}{7} \end{aligned}$$

For a nucleus

Volume: $\mathrm{V}=\frac{4}{3} \pi \mathrm{R}^3$

$$\begin{aligned} & \mathrm{R}=\mathrm{R}_0(\mathrm{A})^{1 / 3} \\ & \mathrm{~V}=\frac{4}{3} \pi \mathrm{R}_0^3 \mathrm{A} \\ & \Rightarrow \frac{\mathrm{V}_2}{\mathrm{~V}_1}=\frac{\mathrm{A}_2}{\mathrm{~A}_1}=4 \end{aligned}$$

$$\begin{aligned} & \varepsilon=-\left(\frac{\mathrm{d} \phi}{\mathrm{dt}}\right)=10 \mathrm{t}-36 \\ & \text { at } \mathrm{t}=2, \varepsilon=16 \mathrm{~V} \\ & \mathrm{i}=\frac{\varepsilon}{\mathrm{R}}=\frac{16}{8}=2 \mathrm{~A} \end{aligned}$$

$$\begin{aligned} \text { If } \mathrm{A} & =0 ; \overline{\mathrm{A}}=1 \\ \mathrm{~A} & =1 ; \overline{\mathrm{A}}=0 \\ \mathrm{~B} & =0 ; \overline{\mathrm{B}}=1 \\ \mathrm{~B} & =1 ; \overline{\mathrm{B}}=0 \\ \mathrm{Y} & =\overline{(\mathrm{A}+\overline{\mathrm{B}})+(\overline{\mathrm{A}}+\mathrm{B})}=\overline{(1+1)}=0 \end{aligned}$$

$$\begin{aligned} & \frac{1}{2} \varepsilon_0 \mathrm{E}^2=\frac{\mathrm{B}^2}{2 \mu_0} \\ & \because \mathrm{E}=\mathrm{CB} \text { and } \mathrm{C}=\frac{1}{\mu_0 \varepsilon_0} \end{aligned}$$

Since $\frac{\mathrm{f}}{2}<\mathrm{f}_0$

i.e. the incident frequency is less than threshold frequency. Hence there will be no emission of photoelectrons.

$\Rightarrow \text { current }=0$

$$\begin{aligned} & \mathrm{R}_1=\frac{\mathrm{R}_2}{2} \\ & \mathrm{R}_0\left(\mathrm{~A}_1\right)^{1 / 3}=\frac{\mathrm{R}_0}{2}\left(\mathrm{~A}_2\right)^{1 / 3} \\ & \mathrm{~A}_1=\frac{1}{8} \mathrm{~A}_2 \\ & \mathrm{~A}_1=\frac{192}{8}=24 \end{aligned}$$

$$\begin{aligned} \mathrm{f}_{\mathrm{K}} & =\mu \mathrm{mg} \cos \theta \\ & =0.1 \times \frac{50 \times \sqrt{3}}{2} \\ & =2.5 \sqrt{3} \mathrm{~N} \end{aligned}$$

$$\begin{aligned} \mathrm{F}_1= & \mathrm{mg} \sin \theta+\mathrm{f}_{\mathrm{K}} \\ = & 25+2.5 \sqrt{3} \end{aligned}$$

$$\begin{aligned} \mathrm{F}_2= & \mathrm{mg} \sin \theta-\mathrm{f}_{\mathrm{K}} \\ & =25-2.5 \sqrt{3} \\ \therefore \mathrm{F}_1 & -\mathrm{F}_2=5 \sqrt{3} \mathrm{~N} \end{aligned}$$

By Brewster's law

At complete reflection refracted ray and reflected ray are perpendicular.

In air $F=\frac{1}{4 \pi \epsilon_0} \frac{\mathrm{q}_1 \mathrm{q}_2}{\mathrm{r}_2}$

In medium $\mathrm{F}^{\prime}=\frac{1}{4 \pi\left(\mathrm{K} \epsilon_0\right)} \frac{\mathrm{q}_1 \mathrm{q}_2}{\left(\mathrm{r}^{\prime}\right)^2}=\frac{25}{4 \pi\left(5 \epsilon_0\right)} \frac{\mathrm{q}_1 \mathrm{q}_2}{(\mathrm{r})^2}=5 \mathrm{~F}$

$$\begin{aligned} & \mathrm{I}=\left(\frac{2}{5} \mathrm{mR}^2+\mathrm{md}^2\right) \times 2 \\ & \mathrm{I}=2\left(\frac{2}{5} \times 2 \times\left(\frac{1}{2}\right)^2+2 \times\left(\frac{3}{4}\right)^2\right)=\frac{53}{20} \mathrm{~kg}-\mathrm{m}^2 \\ & \mathrm{X}=53 \end{aligned}$$

$$\begin{aligned} & \mathrm{T}=\left(\frac{2 \mathrm{~m}_1 \mathrm{~m}_2}{\mathrm{~m}_1+\mathrm{m}_2}\right) \mathrm{g}=\frac{80}{3} \mathrm{~N} \\ & \mathrm{~A}=\pi \mathrm{r}^2=16 \pi \times 10^{-10} \mathrm{~m}^2 \\ & \text { Strain }=\frac{\Delta \ell}{\ell}=\frac{\mathrm{F}}{\mathrm{AY}}=\frac{\mathrm{T}}{\mathrm{AY}} \\ & =\frac{80 / 3}{16 \pi \times 10^{-10} \times 2 \times 10^{11}}=\frac{1}{12 \pi} \\ & \alpha=12 \end{aligned}$$

$$\begin{aligned} & \mathrm{R}_{\text {eq }}=\frac{4}{3} \Omega \\ & \therefore \mathrm{P}=\frac{\mathrm{V}^2}{\mathrm{R}_{\text {eq }}}=\frac{4}{4 / 3}=3 \mathrm{~W} \end{aligned}$$

$\mathrm{k}_{\mathrm{eq}}=\frac{2 \mathrm{k} \cdot \mathrm{k}}{3 \mathrm{k}}+\mathrm{k}=\frac{5 \mathrm{k}}{3}$

Angular frequency of oscillation $(\omega)=\sqrt{\frac{\mathrm{k}_{\mathrm{eq}}}{\mathrm{m}}}$

$(\omega)=\sqrt{\frac{5 \mathrm{k}}{3 \mathrm{~m}}}$

Period of oscillation $(\tau)=\frac{2 \pi}{\omega}=2 \pi \sqrt{\frac{3 \mathrm{~m}}{5 \mathrm{k}}}$

$=\pi \sqrt{\frac{12 \mathrm{~m}}{5 \mathrm{k}}}$

$$\begin{aligned} & {[\mathrm{B}]=\mathrm{L}^2} \\ & \mathrm{~A}=\frac{\mathrm{x}^2}{\mathrm{tE}}=\frac{\mathrm{L}^2}{\mathrm{TML}^2 \mathrm{~T}^{-2}}=\frac{1}{\mathrm{MT}^{-1}} \\ & {[\mathrm{~A}]=\mathrm{M}^{-1 \mathrm{~T}}} \\ & {[\mathrm{AB}]=\left[\mathrm{L}^2 \mathrm{M}^{-1} \mathrm{~T}^1\right]} \end{aligned}$$

$$\begin{aligned} & \mathrm{V}=\frac{\mathrm{K} \overrightarrow{\mathrm{p}} \cdot \overrightarrow{\mathrm{r}}}{\mathrm{r}^3}=\frac{9 \times 10^9(6 \mathrm{q} \ell)}{\mathrm{r}^2} \cos \left(120^{\circ}\right) \\ & =-(27)\left(\frac{\mathrm{q} \ell}{\mathrm{r}^2}\right) \times 10^9 \mathrm{Nm}^2 \mathrm{c}^{-2} \\ & \Rightarrow \alpha=27 \end{aligned}$$

$$\begin{aligned} & \overrightarrow{\mathrm{M}}=\mathrm{i} \overrightarrow{\mathrm{A}} \\ & =5 \times(0.2) \times(0.1)(-\hat{\mathrm{i}}) \\ & =0.1(-\hat{\mathrm{i}}) \\ & \vec{\tau}=\overrightarrow{\mathrm{M}} \times \overrightarrow{\mathrm{B}}=0.1(-\hat{\mathrm{i}}) \times\left(2 \times 10^{-3}\right)(\hat{\mathrm{j}}) \\ & =2 \times 10^{-4}(-\hat{\mathrm{k}}) \mathrm{N}-\mathrm{m} \end{aligned}$$

To determine the total internal energy of the gas mixture, we adhere to the equipartition theorem, which dictates that each degree of freedom contributes $\frac{1}{2} RT$ to the internal energy per mole, where $R$ is the gas constant and $T$ is the temperature.

An argon atom, being a noble gas, is monoatomic, with 3 translational degrees of freedom. Since we neglect all vibrational modes, and monoatomic gases have no rotational or vibrational degrees of freedom that contribute to energy at our specified temperature, each mole of argon has $3 \times \frac{1}{2} RT = \frac{3}{2} RT$ of energy.

Oxygen, on the other hand, is a diatomic molecule. This implies that under normal conditions, it has 3 translational and 2 rotational degrees of freedom. So, each mole of oxygen has $5 \times \frac{1}{2} RT = \frac{5}{2} RT$ of energy as we are neglecting vibrational modes which typically become relevant only at higher temperatures.

Hence, the total internal energy ($U$) of the gas mixture is:

$U = (\text{Energy per mole of Ar} \times \text{Number of moles of Ar}) + (\text{Energy per mole of O}_{2} \times \text{Number of moles of O}_{2})$

$U = \left(\frac{3}{2} RT \times 8\right) + \left(\frac{5}{2} RT \times 6\right)$

$U = \left(12 RT + 15 RT\right)$

$U = 27 RT$

Thus, the total internal energy of the system is $27 RT$.

The magnitude of resultant vector

$R^{\prime}=\sqrt{a^2+b^2+2 a b \cos \theta}$

Here $a=b=R$

Then $R^{\prime}=\sqrt{R^2+R^2+2 R^2 \cos \theta}$

$$\begin{aligned} & =R \sqrt{2} \sqrt{1+\cos \theta} \\ & =\sqrt{2} R \sqrt{2 \cos ^2 \frac{\theta}{2}} \\ & =2 R \cos \frac{\theta}{2} \end{aligned}$$

$$\begin{aligned} & \mathrm{V}_{\text {escape }}=\sqrt{\frac{2 \mathrm{GM}}{\mathrm{R}}} \\ & \mathrm{V}_{\text {planet }}=\sqrt{\frac{2 \mathrm{GM}}{\mathrm{R}}}=\mathrm{V} \\ & \mathrm{V}_{\text {Moon }}=\sqrt{\frac{2 \mathrm{GM} \times 16}{144 \mathrm{R}}}=\frac{1}{3} \sqrt{\frac{2 \mathrm{GM}}{\mathrm{R}}} \\ & \mathrm{V}_{\text {Moon }}=\frac{\mathrm{V}_{\text {Planet }}}{3}=\frac{\mathrm{V}}{3} \end{aligned}$$

Since density is negligible hence Buoyancy force will be negligible

At terminal velocity.

$\mathrm{Mg} =6 \pi \eta \mathrm{rv}$

$\mathrm{V} \propto \frac{1}{\mathrm{r}} \quad$ (as mass is constant)

Now, $\frac{\mathrm{v}}{\mathrm{v}^{\prime}}=\frac{\mathrm{r}^{\prime}}{\mathrm{r}}$

$r^{\prime}=2 \mathrm{r}$

So, $v^{\prime}=\frac{v}{2}$

$$\begin{aligned} & \mathrm{CaCO}_3(\mathrm{s}) \xrightarrow{\Delta} \mathrm{CaO}(\mathrm{s})+\mathrm{CO}_2(\mathrm{g}) \\ & \mathrm{MgCO}_3(\mathrm{s}) \xrightarrow{\Delta} \mathrm{MgO}(\mathrm{s})+\mathrm{CO}_2(\mathrm{g}) \end{aligned}$$

Let the weight of $\mathrm{CaCO}_3$ be $\mathrm{x}$ gm

$\therefore$ weight of $\mathrm{MgCO}_3=(2.21-\mathrm{x}) \mathrm{~gm}$

Moles of $\mathrm{CaCO}_3$ decomposed $=$ moles of $\mathrm{CaO}$ formed

$\frac{\mathrm{x}}{100}=$ moles of $\mathrm{CaO}$ formed

$\therefore$ weight of $\mathrm{CaO}$ formed $=\frac{\mathrm{x}}{100} \times 56$

Moles of $\mathrm{MgCO}_3$ decomposed $=$ moles of $\mathrm{MgO}$ formed

$\frac{(2.21-x)}{84}=$ moles of $\mathrm{MgO}$ formed

$\therefore$ weight of $\mathrm{MgO}$ formed $=\frac{2.21-\mathrm{x}}{84} \times 40$

$\Rightarrow \frac{2.21-\mathrm{x}}{84} \times 40+\frac{\mathrm{x}}{100} \times 56=1.152$

$\therefore \mathrm{x}=1.1886 \mathrm{~g}=$ weight of $\mathrm{CaCO}_3$

& weight of $\mathrm{MgCO}_3=1.0214 \mathrm{~g}$

Vitamins A, D, E, K and B$_{12}$ are stored in liver and adipose tissue.

(A) $\mathrm{Mn}_2 \mathrm{O}_7$ is green oil at room temperature.

(B) $\mathrm{V}_2 \mathrm{O}_4$ dissolve in acids to give $\mathrm{VO}^{2+}$ salts.

(C) $\mathrm{CrO}$ is basic oxide

(D) $\mathrm{V}_2 \mathrm{O}_5$ is amphoteric it reacts with acid as well as base.

It is isothermal reversible expansion, so work done negative

$$\begin{aligned} & \mathrm{W}=-2.303 \mathrm{nRT} \log \left(\frac{\mathrm{V}_2}{\mathrm{~V}_1}\right) \\ & =-2.303 \times 5 \times 8.314 \times 300 \log \left(\frac{100}{10}\right) \\ & =-28720.713 \mathrm{~J} \\ & \equiv-28721 \mathrm{~J} \end{aligned}$$

$$\begin{aligned} & {\left[\mathrm{Cr}\left(\mathrm{H}_2 \mathrm{O}\right)_6\right]^{3+} \text { Contains } \mathrm{Cr}^{3+}:[\mathrm{Ar}] 3 \mathrm{~d}^3: \mathrm{t}_{2 \mathrm{~g}}^3 \mathrm{e}_{\mathrm{g}}^{\mathrm{o}}} \\ & {\left[\mathrm{Fe}\left(\mathrm{H}_2 \mathrm{O}\right)_6\right]^{3+} \text { Contains } \mathrm{Fe}^{3+}:[\mathrm{Ar}] 3 \mathrm{~d}^5: \mathrm{t}_{2 \mathrm{~g}}^3 \mathrm{e}_{\mathrm{g}}^2} \\ & {\left[\mathrm{Ni}\left(\mathrm{H}_2 \mathrm{O}\right)_6\right]^{2+} \text { Contains } \mathrm{Ni}^{2+}:[\mathrm{Ar}] 3 \mathrm{~d}^8: \mathrm{t}_{2 \mathrm{~g}}^6 \mathrm{e}_{\mathrm{g}}^2} \\ & {\left[\mathrm{~V}\left(\mathrm{H}_2 \mathrm{O}\right)_6\right]^{3+} \text { Contains } \mathrm{V}^{3+}:[\mathrm{Ar}] 3 \mathrm{~d}^2: \mathrm{t}_{2 \mathrm{~g}}^2 \mathrm{eg}^{\mathrm{o}}} \end{aligned}$$