JEE Main 2024 30th January Evening Shift Question Paper with Solutions

$$\begin{aligned} & \mathrm{i}=\frac{14}{4}=3.5 \mathrm{~mA} \\ & \mathrm{~V}_{\mathrm{L}}=\mathrm{iR}_{\mathrm{L}}=3.5 \times 2.5 \text { volt } \\ & \quad=8.75 \text { volt } \end{aligned}$$

$$\begin{aligned} & \frac{2 \mathrm{k} \lambda \mathrm{q}}{\mathrm{r}}=\mathrm{m} \omega^2 \mathrm{r} \\ & \omega^2=\frac{2 \mathrm{k} \lambda \mathrm{q}}{\mathrm{mr}^2} \\ & \left(\frac{2 \pi}{\mathrm{T}}\right)^2=\frac{2 \mathrm{k} \lambda \mathrm{q}}{\mathrm{mr}^2} \\ & \mathrm{~T}=2 \pi \mathrm{r} \sqrt{\frac{\mathrm{m}}{2 \mathrm{k} \lambda \mathrm{q}}} \end{aligned}$$

$$\begin{array}{ll} \mathrm{f}_1=3, & \mathrm{f}_2=5 \\ \mathrm{n}_1=3, & \mathrm{n}_2=2 \end{array}$$

$$\begin{aligned} & \mathrm{f}_{\text {mixture }}=\frac{\mathrm{n}_1 \mathrm{f}_1+\mathrm{n}_2 \mathrm{f}_2}{\mathrm{n}_1+\mathrm{n}_2}=\frac{9+10}{\mathrm{f}}=\frac{19}{5} \\ & \gamma_{\text {mixture }}=1+\frac{2 \times 5}{19}=\frac{29}{19}=1.52 \end{aligned}$$

$$\begin{aligned} & \mathrm{R}_{\mathrm{P}}=\frac{\mathrm{R}_{\mathrm{E}}}{3}, \mathrm{M}_{\mathrm{P}}=\frac{\mathrm{M}_{\mathrm{E}}}{6} \\ & \mathrm{~V}_{\mathrm{c}}=\sqrt{\frac{2 \mathrm{GM}_{\mathrm{e}}}{\mathrm{R}_{\mathrm{e}}}} \quad \text{.... (i)}\\ & \mathrm{V}_{\mathrm{P}}=\sqrt{\frac{2 \mathrm{GM}_{\mathrm{P}}}{\mathrm{R}_{\mathrm{P}}}} \quad \text{.... (ii)}\\ & \frac{\mathrm{V}_{\mathrm{e}}}{\mathrm{V}_{\mathrm{p}}}=\sqrt{2} \\ & \mathrm{~V}_{\mathrm{P}}=\frac{\mathrm{V}_{\mathrm{e}}}{\sqrt{2}}=\frac{11.2}{\sqrt{2}}=7.9 \mathrm{~km} / \mathrm{sec} \end{aligned}$$

Acceleration due to gravity g' $=\frac{g}{4}$

$$\begin{aligned} & \mathrm{T}=2 \pi \sqrt{\frac{4 \ell}{\mathrm{g}}} \\ & \mathrm{T}=2 \pi \sqrt{\frac{4 \times 4}{\mathrm{~g}}} \\ & \mathrm{~T}=2 \pi \frac{4}{\pi}=8 \mathrm{~s} \end{aligned}$$

$$\begin{aligned} & \rho\left({ }_3^4 \pi r^3\right) 1000={ }_3^4 \pi R^3 \rho \\ & R=10 r \\ & E_1=1000 \times 4 \pi r^2 \times S \\ & E_2=4 \pi(10 r)^2 S \\ & \frac{E_1}{E_2}=\frac{10}{1}, x=10 \end{aligned}$$

To solve this problem, we need to understand the relationship between the intensity of sound waves and the distance from the source. The intensity $I$ of sound waves from a point source decreases with the square of the distance $r$ from the source, according to the inverse square law, which can be expressed as:

$I \propto \frac{1}{r^2}$

This means that if the distance is doubled, the intensity becomes one-fourth of its initial value because $(2r)^2 = 4r^2$.

The initial intensity given at the origin (source) is:

$I_0 = 16 \times 10^{-8} \mathrm{~Wm}^{-2}$

Let's call $I_1$ the intensity at $r = 2 \text{ m}$ and $I_2$ the intensity at $r = 4 \text{ m}$. Using the inverse square law, we can write:

$I_1 = \frac{I_0}{(2)^2} = \frac{I_0}{4}$

and

$I_2 = \frac{I_0}{(4)^2} = \frac{I_0}{16}$

Now substitute the given value for $I_0$ to find $I_1$ and $I_2$:

$I_1 = \frac{16 \times 10^{-8}}{4} = 4 \times 10^{-8} \mathrm{~Wm}^{-2}$

$I_2 = \frac{16 \times 10^{-8}}{16} = 1 \times 10^{-8} \mathrm{~Wm}^{-2}$

Now to find the difference in intensity (magnitude only) between the two points, we subtract $I_2$ from $I_1$:

$\Delta I = | I_1 - I_2 |$

$\Delta I = | 4 \times 10^{-8} - 1 \times 10^{-8} |$

$\Delta I = 3 \times 10^{-8} \mathrm{~Wm}^{-2}$

So the difference in intensity (magnitude only) at the two points is:

$3 \times 10^{-8} \mathrm{~Wm}^{-2}$

To find the loss in kinetic energy when two spinning discs are brought together, we use the principle of conservation of angular momentum and the formula for kinetic energy. Here's how:

First, because angular momentum before and after they touch must be the same, we have:

$I_1 \omega_1 + I_2 \omega_2 = (I_1 + I_2) \omega_0$

where:

• $I_1$ and $I_2$ are the moments of inertia for the two discs.
• $\omega_1$ and $\omega_2$ are their angular speeds before contact.
• $\omega_0$ is their common angular speed after contact.

Plugging in the given values, we find that:

$\omega_0 = 8 \mathrm{rad/s}$

Next, to calculate the loss in kinetic energy, we first find the total kinetic energy before and after they touch:

Before: $E_1 = \frac{1}{2} I_1 \omega_1^2 + \frac{1}{2} I_2 \omega_2^2 = 216 \,\mathrm{J}$

After: $E_2 = \frac{1}{2}(I_1 + I_2) \omega_0^2 = 192 \,\mathrm{J}$

The loss in kinetic energy ($\Delta E$) is the difference:

$\Delta E = E_1 - E_2 = 24 \,\mathrm{J}$

So, when the two discs are brought together, the system loses 24 J of kinetic energy.

$$\begin{aligned} & \frac{\mathrm{v}^2}{\mathrm{R}}=\mathrm{W} \qquad \text{.... (i)}\\ & \frac{\mathrm{v}^2}{\frac{1}{2}\left(\frac{\mathrm{R}}{2}\right)}=\mathrm{W}^{\prime} \quad \text{.... (ii)} \end{aligned}$$

From (i) & (ii), we get

$W^{\prime}=4 W$

$$\begin{aligned} & T \cos \theta=m g \\ & T \sin \theta=F_e \\ & \tan \theta=\frac{F_e}{m g} \end{aligned}$$

$\tan \theta=\frac{F_c}{\rho_B V g}$ ..... (i)

$\tan \theta=\frac{F_e}{\frac{k}{\left(\rho_B-\rho_L\right) V g}}$ ..... (ii)

From Eq. (i) & (ii)

$$\begin{aligned} & \rho_{\mathrm{B}} \mathrm{Vg}=\left(\rho_{\mathrm{B}}-\rho_{\mathrm{L}}\right) \mathrm{kVg} \\ & 1.4=0.7 \mathrm{k} \\ & \mathrm{k}=2 \end{aligned}$$

$a_A=a_B=a_c=\frac{F}{5+3+2}=\frac{80}{10}=8 \mathrm{~m} / \mathrm{s}^2$

$\mathrm{T_1=5\times8=40}$

$\mathrm{T_2-T_1=3\times8\Rightarrow T_2=64}$

Let's identify each law listed in List I and match it with the corresponding mathematical expression listed in List II.

$\oint \vec{B} \cdot \vec{da} = 0$

So, (A) matches with (II).

$\oint \vec{E} \cdot \vec{dl} = -\frac{d}{dt} \int \vec{B} \cdot \vec{da}$

Hence, (B) matches with (III).

$\oint \vec{B} \cdot \vec{dl} = \mu_0 I$

Consequently, (C) matches with (IV).

Lastly, Gauss's law of electrostatics states that the total electric flux out of a closed surface is proportional to the charge enclosed within the surface:

$\oint \vec{E} \cdot \vec{da} = \frac{1}{\varepsilon_0} \int \rho dV$

Therefore, (D) matches with (I).

Based on these matches, the correct answer must link A-II, B-III, C-IV, and D-I:

The correct option is:

Option C

A-II, B-III, C-IV, D-I

For $\mathrm{u}_{\mathrm{A}}$ & $\mathrm{u}_{\mathrm{B}}$ time of flight and range can not be same. So above options are incorrect.

$$\begin{aligned} & \frac{1}{f+20}-\frac{1}{-(f+20)}=\frac{1}{f} \\ & \frac{2}{f+20}=\frac{1}{f} \quad f=20 \mathrm{~cm} \end{aligned}$$

Or $\mathrm{x}_1 \mathrm{x}_2=\mathrm{f}^2$ gives $\mathrm{f}=20 \mathrm{~cm}$

$\mathrm{p=\frac{E}{C}=\frac{6.48 \times 10^5}{3 \times 10^8}=2.16 \times 10^{-3}}$

$$\begin{aligned} & \mathrm{m}=\mathrm{kc}^{\mathrm{P}} \mathrm{G}^{-1 / 2} \mathrm{~h}^{1 / 2} \\ & \mathrm{M}^1 \mathrm{~L}^0 \mathrm{~T}^0=\left[\mathrm{LT}^{-1}\right]^{\mathrm{P}}\left[\mathrm{M}^{-1} \mathrm{~L}^3 \mathrm{~T}^{-2}\right]^{-1 / 2}\left[\mathrm{ML}^2 \mathrm{~T}^{-1}\right]^{1 / 2} \end{aligned}$$

By comparing $\mathrm{P}=1 / 2$

Work done again frictional force

$$\begin{aligned} & =\mu \mathrm{N} \times 10 \\ & =0.1 \times 5 \times 10=5 \mathrm{~J} \end{aligned}$$

$$\begin{aligned} & P_i=2300 \times 5 \text { watt } \\ & P_0=2300 \times 5 \times 0.9=230 \times I_2 \\ & I_2=45 A \end{aligned}$$

$$\begin{aligned} & \begin{aligned} & (\mathrm{X}) \rightarrow(\mathrm{Y})+(\mathrm{Z})+(\mathrm{P}) \\ & \begin{array}{llll} \mathrm{M} & \mathrm{M} / 3 & \mathrm{M} / 3 & \mathrm{M} / 3 \end{array} \\ \end{aligned} \\ & \Delta \mathrm{Mc}^2=\frac{1}{2} \frac{\mathrm{M}}{3} \mathrm{~V}^2+\frac{1}{2} \frac{\mathrm{M}}{3} \mathrm{~V}^2+\frac{1}{2} \frac{\mathrm{M}}{3} \mathrm{~V}^2 \\ & \mathrm{~V}=\mathrm{c} \sqrt{\frac{2 \Delta \mathrm{M}}{\mathrm{M}}} \end{aligned}$$

Steeper curve (B) is adiabatic

Adiabatic $\Rightarrow \mathrm{PV}^v=$ const.

Or $\mathrm{P}\left(\frac{\mathrm{T}}{\mathrm{P}}\right)^v=$ const.

$\frac{\mathrm{T}^v}{\mathrm{P}^{v-1}}=\text { const. }$

Curve (A) is isothermal

$\mathrm{T}=$ const.

$\mathrm{PV}=$ const.

$$\begin{aligned} & 50 \mathrm{~V}+\mathrm{S}=49 \mathrm{~S}+\mathrm{S} \\ & \mathrm{S}=50(\mathrm{~S}-\mathrm{V}) \\ & 5=50(\mathrm{~S}-\mathrm{V}) \\ & \mathrm{S}-\mathrm{V}=\frac{0.5}{50}=\frac{1}{100}=0.01 \mathrm{~mm} \end{aligned}$$

$$\begin{aligned} & \mathrm{B}=4 \times \frac{\mu_0 \mathrm{i}}{4 \pi(1 / 2)}\left(\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}}\right) \\ & =4 \times 10^{-7} \times 5 \times 2 \times \sqrt{2} \\ & -40 \sqrt{2} \times 10^{-7} \mathrm{~T} \end{aligned}$$

$$\begin{aligned} & \text { Magnetic moment }=\mathrm{i} \pi \mathrm{r}^2 \\ & \mu=\frac{\text { evr }}{2} \\ & \mu \propto \left(\frac{1}{\mathrm{n}}\right) \mathrm{n}^2 \\ & \mu \propto \mathrm{n} \\ & \mathrm{x}=1 \end{aligned}$$

$$\begin{aligned} & \frac{\mathrm{dy}}{\mathrm{dx}}=\tan \theta=\frac{\mathrm{x}}{2}=\mu=\frac{1}{2} \\ & \mathrm{x}=1, \mathrm{y}=1 / 4 \end{aligned}$$

$$\begin{aligned} & \frac{R_v 100}{R_v+100}=\frac{200}{3} \\ & 3 R_v=2 R_v+200 \\ & R_v=200 \end{aligned}$$

Intensity of emergent light

$=\frac{\mathrm{I}_0}{2} \cos ^2 45^{\circ}=\frac{\mathrm{I}_0}{4}$

Rising half to peak

$$\begin{aligned} & \mathrm{t}=\mathrm{T} / 6 \\ & \mathrm{t}=\frac{2 \pi}{6 \omega}=\frac{\pi}{3 \omega}=\frac{\pi}{300 \pi}=\frac{1}{300}=3.33 \mathrm{~ms} \end{aligned}$$

To find a vector that has the same magnitude as vector $\vec{A}$ and is parallel to vector $\vec{B}$, we use the formula:

$\vec{N} = |\vec{A}| \hat{B}$

First, let's find the magnitude of $\vec{A}$, which is $|\vec{A}|$.

$|\vec{A}| = \sqrt{3^2 + 4^2}=5$

We're given that $\vec{B} = 4 \hat{i}+3 \hat{j}$, so to make $\vec{N}$ parallel to $\vec{B}$ and have it have the same magnitude as $\vec{A}$, they should be the same when $\vec{N}$'s magnitude is divided by 5, since $|\vec{A}|=5$.

So, $\vec{N} = \frac{5(4\hat{i}+3\hat{j})}{5} = 4\hat{i}+3\hat{j}$.

This means the $x$ component of the vector in the first quadrant is 4, and the $y$ component is given as 3. So, $x=4$.

$$\begin{aligned} & \text { K.E. }=\mathrm{hf}-\phi \\ & \tan \theta=\mathrm{h} \end{aligned}$$

$$\begin{array}{ccc} & \text { 1M Benzoic acid } & +1 \mathrm{M} \text { Sodium Benzoate } \\ & \left(\mathrm{V}_{\mathrm{a}} \mathrm{ml}\right) & \left(\mathrm{V}_{\mathrm{s}} \mathrm{ml}\right) \\ \text { Millimole } & \mathrm{V}_{\mathrm{a}} \times 1 & \mathrm{~V}_{\mathrm{s}} \times 1 \end{array}$$

$$\begin{aligned} & \mathrm{pH}=4.5 \\ & \mathrm{pH}=\mathrm{pka}+\log \frac{[\text { salt }]}{[\text { acid }]} \\ & 4.5=4.2+\log \left(\frac{\mathrm{V}_{\mathrm{s}}}{\mathrm{V}_{\mathrm{a}}}\right) \\ & \frac{\mathrm{V}_{\mathrm{s}}}{\mathrm{V}_{\mathrm{a}}}=2 \quad \text{..... (1)}\\ & \mathrm{~V}_{\mathrm{s}}+\mathrm{V}_{\mathrm{a}}=300 \quad \text{..... (2)}\\ & \mathrm{~V}_{\mathrm{a}}=100 \mathrm{~ml} \end{aligned}$$

$\Delta \mathrm{T}_{\mathrm{f}}$ is maximum when $\mathrm{i} \times \mathrm{m}$ is maximum.

1) $\mathrm{m}_1=\frac{180}{60}=3, \mathrm{i}=1+\alpha$

Hence

$\Delta \mathrm{T}_{\mathrm{f}}=(1+\alpha) \cdot \mathrm{k}_{\mathrm{f}}=3 \times 1.86=5.58^{\circ} \mathrm{C}(\alpha<<1)$

2) $\mathrm{m}_2=\frac{180}{60}=3, \mathrm{i}=0.5, \Delta \mathrm{T}_{\mathrm{f}}=\frac{3}{2} \times \mathrm{k}_{\mathrm{f}}{ }^{\prime}=7.68^{\circ} \mathrm{C}$

3) $\mathrm{m}_3=\frac{180}{122}=1.48, \mathrm{i}=0.5, \Delta \mathrm{T}_{\mathrm{f}}=\frac{1.48}{2} \times \mathrm{k}_{\mathrm{f}}{ }^{\prime}=3.8^{\circ} \mathrm{C}$

4) $\mathrm{m}_4=\frac{180}{180}=1, \mathrm{i}=1, \Delta \mathrm{T}_{\mathrm{f}}=1 \cdot \mathrm{k}_{\mathrm{f}}{ }^{\prime}=1.86^{\circ} \mathrm{C}$

As per NCERT, $\mathrm{k}_{\mathrm{f}}{ }^{\prime}\left(\mathrm{H}_2 \mathrm{O}\right)=1.86 \mathrm{~k} \cdot \mathrm{~kg} \mathrm{~mol}^{-1}$

$\mathrm{k}_{\mathrm{f}}{ }^{\prime}(\text { Benzene })=5.12 \mathrm{~k} \cdot \mathrm{~kg} \mathrm{~mol}^{-1}$

3 stereocenteres, symmetrical

Total Geometrical isomers $\rightarrow 4$. EE, ZZ, EZ (two isomers)

The purification method among the ones listed that is based on the principle of "Solubility" in two different solvents is Option D, Differential Extraction.

Differential extraction, also known as liquid-liquid extraction or solvent extraction, is a technique to separate compounds based on their relative solubilities in two different immiscible liquids, usually water and an organic solvent. It exploits the fact that the compound of interest has different solubility properties in two solvents which are not miscible with each other. A solution containing the desired compound and impurities is mixed with a second solvent in which the desired compound is more soluble. The two solvents form layers due to their immiscibility, and the compound partitions into the layer where it is more soluble. By separating the layers and repeating the process if necessary, a higher degree of purification can be achieved for the target compound.

Let's quickly review why the other options do not primarily use solubility in two different solvents:

• Distillation (Option A): Distillation is a process that separates the components of a mixture based on differences in their boiling points. It does not rely on differential solubility in two solvents.


• Sublimation (Option B): Sublimation is the phase transition of a substance from the solid to the gas phase without passing through the intermediate liquid phase. This process is based on the volatility of the substance, not on its solubility in solvents.


• Column Chromatography (Option C): Column chromatography separates compounds based on their differing affinities to a stationary phase (often a solid) and a mobile phase (often a liquid). The separation is achieved by partitioning between the phases, but it does not primarily involve the solubility of the compounds in two different immiscible solvents.

Intermediate oxidation state of element can undergo disproportionation.

$\mathrm{H}_2 \mathrm{O}_2, \mathrm{ClO}_3^{-}, \mathrm{P}_4, \mathrm{Cl}_2, \mathrm{Cu}^{+1}, \mathrm{NO}_2$

$\mathrm{Cr{O_2} + 4NaOH \to \mathop {N{a_2}Cr{O_4}}\limits_{(A)} + 2NaCl + 2{H_2}O}$

$\mathrm{N{a_2}Cr{O_4} + 2{H_2}{O_2} + 2HCl \to \mathop {Cr{O_5}}\limits_{(B)} + \mathop {2NaCl}\limits_{(Mis\sin g\,from\,balanced\,equation)} + 3{H_2}O}$

A. RNA is regarded as the reserve of genetic information. (False)

B. DNA molecule self-duplicates during cell division. (True)

C. DNA synthesizes proteins in the cell. (False)

D. The message for the synthesis of particular proteins is present in DNA. (True)

E. Identical DNA strands are transferred to daughter cells. (True)

Chemical reactivity of elements decreases along the period therefore statement - I is false.

Group - 1 elements from basic nature oxides while group - 17 elements form acidic oxides therefore statement - II is true.

All the following metal ions will respond to flame test.

$\mathrm{Sr}^{2+}, \mathrm{Ba}^{2+}, \mathrm{Ca}^{2+}, \mathrm{Cu}^{2+}$

$\mathrm{BrF_5}$

Higher the value of $\oplus$ve SRP (Std. reduction potential) more is tendency to undergo reduction, so better is oxidising power of reactant.

Hence, ox. Power:- $\mathrm{BrO}_4^{-}>\mathrm{IO}_4^{-}>\mathrm{ClO}_4^{-}$