JEE Main 2025 24th January Evening Shift Question Paper with Solutions

$$\begin{aligned} & \frac{\mathrm{T}_2-\mathrm{T}_1}{\mathrm{t}}=\mathrm{K}\left[\mathrm{~T}_{\text {avg }}-\mathrm{T}_{\mathrm{s}}\right] \\ & \mathrm{T}_1=24^{\circ} \mathrm{C} ; \mathrm{T}_2=40^{\circ} \mathrm{C}, \mathrm{t}=4, \mathrm{~T}_{\mathrm{s}}=16^{\circ} \mathrm{C} \\ & \frac{40-24}{4}=\mathrm{K}[32-16] \\ & \mathrm{K}=\frac{4}{16}=\frac{1}{4} \\ & \text { Now } \frac{24-\mathrm{T}}{4}=\mathrm{K}\left[\frac{\mathrm{~T}+24}{2}-16\right] \\ & 24-\mathrm{T}=\frac{\mathrm{T}-16}{2}+16 \\ & \frac{3 \mathrm{~T}}{2}=28 \\ & \mathrm{~T}=\frac{56}{3} \mathrm{C} \end{aligned}$$

$$\begin{aligned} & \mathrm{B}_{\text {in }}=\frac{\mu_0(\overline{\mathrm{~J}} \times \overline{\mathrm{r}})}{2} \\ & \mathrm{~B}_{\text {in }} \propto \mathrm{r} \\ & \mathrm{~B}_{\text {out }}=\frac{\mu_0 \mathrm{I}}{2 \pi \mathrm{r}} \\ & \mathrm{~B}_{\text {net }} \propto \frac{1}{\mathrm{r}} \end{aligned}$$

(A) $\mathrm{T}_1 \mathrm{~V}_1^{\gamma-1}=\mathrm{T}_2 \mathrm{~V}_2^{\gamma-1}$

Temp increases

R) Free expansion is assumed fast, so Adiabatic

$$\begin{aligned} & I_A=\frac{N q}{\frac{2 \pi}{\omega}} \\ & I_A=\frac{N q \omega}{2 \pi}, I_B=0 \\ & I_A-I_B=\frac{N q \omega}{2 \pi} \end{aligned}$$

$$\begin{aligned} &\text { Since bulk modulus is given as }\\ &\begin{aligned} & \mathrm{B}=\frac{-\Delta \mathrm{P}}{\left(\frac{\Delta \mathrm{~V}}{\mathrm{~V}}\right)} \\ & 2.15 \times 10^9=\frac{-\Delta \mathrm{P}}{-\left(\frac{0.2}{100}\right)} \\ & \Delta \mathrm{P}=2.15 \times 10^9 \times 2 \times 10^{-3} \\ & =4.3 \times 10^6=43 \times 10^5 \mathrm{~N} / \mathrm{m}^2 \end{aligned} \end{aligned}$$

$$\begin{aligned} & \overline{\mathrm{F}}=\mathrm{q}(\overline{\mathrm{v}} \times \overline{\mathrm{B}}) \\ & \overline{\mathrm{F}}=0 \\ & \overline{\mathrm{~V}} \| \overline{\mathrm{B}} \\ & \theta=0 \text { or } 180 \end{aligned}$$

$$\begin{aligned} & \frac{\text { Linear KE }}{\text { Rotational K.E }}=\frac{\frac{1}{2} \mathrm{mv}_{\mathrm{cm}}^2}{\frac{1}{2} \mathrm{I} \omega^2} \\ & \frac{\mathrm{mv}_{\mathrm{cm}}^2}{\frac{2}{5} \mathrm{mR}^2 \omega^2}=\frac{5}{2} \quad(\mathrm{~V}=\omega \mathrm{R}) \end{aligned}$$

$$\begin{aligned} & \mathrm{t}=\sqrt{\frac{2 \ell}{\mathrm{a}_{\mathrm{cm}}}} \\ & \mathrm{a}_{\mathrm{cm}}=\frac{\mathrm{g} \sin \theta}{1+\frac{\mathrm{I}_{\mathrm{cm}}}{\mathrm{MR}^2}} \\ & \mathrm{a}_1=\mathrm{a}_{\mathrm{cm}_1}=\frac{5 \mathrm{~g} \sin \theta}{7} \ldots . . \text { Solid } \\ & \mathrm{a}_2=\mathrm{a}_{\mathrm{cm}_2}=\frac{3 \mathrm{~g} \sin \theta}{5} \ldots . \text { Hollow } \\ & \mathrm{a}_1>\mathrm{a}_2 \\ & \mathrm{t}_1<\mathrm{t}_2 \end{aligned}$$

after passing through third poleriser, Intensity of both the waves must be $\frac{I_0}{4}$

now, at a point where path diff is $\frac{\lambda}{3}$, phase difference

$$\begin{aligned} & \Delta \phi=2 \mathrm{~K}\left(\frac{\Delta \mathrm{x}}{\lambda}\right)=\frac{2 \pi}{3} \\ & \therefore \mathrm{I}_{\text {res }}=\sqrt{\left(\frac{\mathrm{I}_0}{4}\right)^2+\left(\frac{\mathrm{I}_0}{4}\right)^2+2\left(\frac{\mathrm{I}_0}{4}\right)^2} \cos \frac{2 \pi}{3} \\ & =\frac{\mathrm{I}_0}{4} \end{aligned}$$

$\frac{C}{5}=\frac{F-32}{9} \Rightarrow C=\frac{5 F}{9}-\frac{160}{9}$

$$\begin{aligned} &\text { Since power emitting by a source is given as }\\ &\begin{aligned} & =\frac{\text { Total energy emitted }}{\text { time }} \\ & =\frac{\left(E_1 \text { photon }\right) \times \text { Number of photons }(N)}{t} \\ & P_1=\left(E_1\right) n \end{aligned} \end{aligned}$$

$$\begin{aligned} &\begin{aligned} & \frac{\mathrm{P}_1}{\mathrm{P}_2}=\frac{\left(\mathrm{E}_1\right) \mathrm{n}_1}{\left(\mathrm{E}_2\right) \mathrm{n}_2}=\frac{\left(\frac{\mathrm{hC}}{\lambda_1}\right) \mathrm{n}_1}{\left(\frac{\mathrm{hC}}{\lambda_2}\right) \mathrm{n}_2} \\ & \frac{\mathrm{P}_1}{\mathrm{P}_2}=\left(\frac{\lambda_2}{\lambda_1}\right) \frac{\mathrm{n}_1}{\mathrm{n}_2} \end{aligned}\\ &\text { Substituting the given values }\\ &\begin{aligned} & 2=\left(\frac{300}{600}\right) \times \frac{2 \times 10^{15}}{\mathrm{n}_2} \\ & \mathrm{n}_2=\frac{1}{2} \times 10^{15}=5 \times 10^{14} \text { Photon } / \mathrm{sec} \end{aligned} \end{aligned}$$

$$\begin{aligned} & x=x_0 \sin ^2 \frac{t}{2}=x_0\left(\frac{1-\cos t}{2}\right) \\ & x-\frac{x_0}{2}=\frac{-\cos t}{2} \\ & \text { where } x_0=1 \\ & x-\frac{1}{2}=\frac{-\cos t}{2} \end{aligned}$$

Particle is oscillating between $\mathrm{x}=0$ to $\mathrm{x}=1$

$$\begin{aligned} & \mathrm{hv}=\phi+\mathrm{KE}_{\max } \\ & \mathrm{KE}_{\max }=\mathrm{eV}_0 \\ & \mathrm{~V}_0=\frac{\mathrm{hv}-\phi}{\mathrm{e}} \end{aligned}$$

(A) $\mathrm{V}_0 \mathrm{v} / \mathrm{s} \mathrm{V}$ is linear correct

(B) Slope

$\mathrm{v}_0=\left(\frac{\mathrm{h}}{\mathrm{e}}\right) \mathrm{v}-\frac{\phi}{\mathrm{e}} \text { Wrong }$

Slope $\frac{\mathrm{h}}{\mathrm{e}}$

(C) Correct

(D) Incorrect

(E) Correct

$$\begin{aligned} & \mathrm{F}=\frac{\mathrm{k}\left(\frac{\theta}{2}\right)\left(\frac{\theta}{2}\right)}{\mathrm{r}^2} \\ & 9 \times 10^{-3}=\frac{9 \times 10^9 \times\left(4 \times 10^{-8}\right) \times 4 \times 10^{-8}}{4 \times \mathrm{r}^2} \\ & \mathrm{r}^2=\frac{9 \times 10^9 \times 16 \times 10^{-16}}{4 \times 9 \times 10^{-3}}=4 \times 10^{-4} \\ & \mathrm{r}=2 \times 10^{-2} \mathrm{~m} \Rightarrow 2 \mathrm{~cm} \end{aligned}$$

$$\begin{aligned} & {[E]=M^1 L^2 T^{-2}} \\ & {[P c]=M^1 L^1 T^{-1} \cdot L^1 T^{-1}=M^1 L^2 T^{-2}} \\ & {\left[\mathrm{mc}^2\right]=\mathrm{M}^1 L^2 T^{-2}} \\ & E^2=\mathrm{M}^1 \mathrm{~L}^2 \mathrm{~T}^{-2} \\ & E^2=\mathrm{P}^2 \mathrm{c}^2+\mathrm{m}^2 \mathrm{c}^4 \end{aligned}$$

$$\begin{aligned} & \overrightarrow{\mathrm{r}}=5 \mathrm{t}^2 \hat{\mathrm{i}}-5 \hat{\mathrm{t}} \\ & \overrightarrow{\mathrm{v}}=10 \mathrm{t} \hat{\mathrm{i}}-5 \hat{\mathrm{j}} \\ & \overrightarrow{\mathrm{v}}=20 \hat{\mathrm{i}}-5 \hat{\mathrm{j}} \quad \text { at } \mathrm{t}=2 \mathrm{sec} \end{aligned}$$

$$\begin{aligned} & \tan \theta=\frac{20}{5}=4 \\ & \theta=\tan ^{-1} 4 \\ & \text { From-veY-axis } \end{aligned}$$

$$\begin{aligned} & \mathrm{W}=\frac{1}{2} \pi \mathrm{R}^2 \\ & =\frac{1}{2} \times \pi \times\left(\frac{200}{2} \times 10^3\right) \times \frac{200}{2} \times 10^{-6} \\ & =\frac{10 \pi}{2}=5 \pi \mathrm{~J} \end{aligned}$$

Since time period of a revolving charge is $\frac{2 \pi \mathrm{~m}}{\mathrm{qB}}$

Where $\mathrm{B}=$ magnetic field

due to a solenoid $=\mu_0 \mathrm{nI}$

$$\begin{aligned} & \therefore \mathrm{T}=\frac{2 \pi \mathrm{~m}}{\mathrm{q}\left(\mu_0 \mathrm{nI}\right)} \\ & 75 \times 10^{-9}=\frac{(2 \pi)\left(9 \times 10^{-31}\right)}{1.6 \times 10^{-19} \times 4 \pi \times 10^{-7} \times \mathrm{n} \times 1.5} \\ & \mathrm{~N}=250 \end{aligned}$$

$$\begin{aligned} & \mathrm{U}_1=\frac{4 \mathrm{Kq}_0^2}{\mathrm{a}}+\frac{2 \mathrm{Kq}_0^2}{\sqrt{2} \mathrm{a}}=\frac{\mathrm{Kq}_0^2}{\mathrm{a}}(4+\sqrt{2}) \\ & \mathrm{U}_2=\frac{\mathrm{Kq}_0^2}{\left(\frac{\mathrm{a}}{\sqrt{2}}\right)}(4+\sqrt{2})=\frac{\mathrm{Kq}_0^2}{\mathrm{a}}(4 \sqrt{2}+2) \\ & \mathrm{U}_2-\mathrm{U}_1=\frac{\mathrm{Kq}_0^2}{\mathrm{a}}(3 \sqrt{2}-2) \end{aligned}$$

$$\begin{aligned} & \mathrm{T} \cos \theta=\mathrm{mg} \\ & \mathrm{~T} \sin \theta=\mathrm{M} \omega^2 \mathrm{R} \\ & \text { Using equation (2) } \\ & \mathrm{T} \sin \theta=\mathrm{M} \omega^2(\mathrm{~L} \sin \theta) \\ & \mathrm{T}=\mathrm{M} \omega^2 \mathrm{~L}=\mathrm{M}\left(\frac{3}{\pi} \times 2 \pi\right)^2 \mathrm{~L} \\ & \mathrm{~T}=36 \mathrm{ML} \end{aligned}$$

$$\begin{aligned} &\mathrm{H}=18 \mathrm{~km}\\ &\text { Size of camera film }=2 \mathrm{~cm} \times 2 \mathrm{~cm}\\ &\begin{aligned} & \mathrm{A}_{\text {image }}=400 \mathrm{~km}^2 \\ & \mathrm{x}=20 \times 10^3 \mathrm{~m}=2 \times 10^4 \mathrm{~m} \\ & \mathrm{y}=2 \times 10^{-2} \mathrm{~m} \\ & \frac{\mathrm{x}}{\mathrm{y}}=10^6=\frac{18 \mathrm{Km}}{\mathrm{f}} \\ & \mathrm{f}=18 \times 10^{-3} \mathrm{~m}=18 \mathrm{~mm} \\ & \mathrm{f}=1.8 \mathrm{~cm} \end{aligned} \end{aligned}$$

$\because$ acceleration due to gravity on surface is given by

$\mathrm{g}=\frac{\mathrm{GM}}{\mathrm{R}_{\mathrm{e}}^2}$

Now since diameter is reduced to $1 / 3^{\text {rd }}$, radius also reduces to $1 / 3^{\text {rd }}$, keeping mass constant

New value of acceleration due to gravity on Earth's surface is

$g^{\prime}=\frac{\mathrm{GM}}{\left(\frac{\mathrm{R}_{\mathrm{e}}}{3}\right)^2}=9 \frac{\mathrm{GMe}}{\mathrm{R}_{\mathrm{e}}^2}=9 \mathrm{~g}$

In Young's double-slit experiment, the fringe width ($\beta$) can be calculated using the formula:

$\beta = \frac{\lambda' \cdot D}{d}$

where:

$\lambda'$ is the wavelength of light in the medium,

$D$ is the distance between the slits and the screen,

$d$ is the separation between the slits.

The wavelength in the medium ($\lambda'$) is given by:

$\lambda' = \frac{\lambda_0}{n}$

where:

$\lambda_0 = 690 \, \text{nm}$ is the wavelength of light in air,

$n = 1.44$ is the refractive index of the liquid.

Substituting the values,

$\lambda' = \frac{690 \, \text{nm}}{1.44} \approx 479.17 \, \text{nm}$

Now, convert $\lambda'$ to meters:

$\lambda' = 479.17 \, \text{nm} = 479.17 \times 10^{-9} \, \text{m}$

Next, use the values for $D$ and $d$:

$D = 0.72 \, \text{m}$

$d = 1.5 \, \text{mm} = 1.5 \times 10^{-3} \, \text{m}$

Calculating the fringe width:

$\beta = \frac{479.17 \times 10^{-9} \, \text{m} \cdot 0.72 \, \text{m}}{1.5 \times 10^{-3} \, \text{m}}$

$\beta \approx \frac{344.2 \times 10^{-9} \, \text{m}}{1.5 \times 10^{-3} \, \text{m}}$

$\beta \approx 0.229 \times 10^{-3} \, \text{m} = 0.229 \, \text{mm}$

The closest answer choice is:

Option D: 0.23 mm

$$\begin{array}{lllll} \mathrm{C} & : & \mathrm{H} & : & \mathrm{O} \\ \frac{54.2}{12} & : & 9.2 & : & \frac{36.6}{16} \\ 4.516 & : & 9.2 & : & 2.287 \\ \frac{4.516}{2.287} & : & \frac{9.2}{2.287} & : & \frac{2.287}{2.287} \\ 1.97 & : & 4.02 & : & 1 \end{array}$$

$\mathrm{C}_2 \mathrm{H}_4 \mathrm{O} \Rightarrow$ Empirical formula

E.F. mass $=24+4+16=44$

and molar mass $=132$

$$\begin{aligned} \text { Hence molecular formula } & =\left(\mathrm{C}_2 \mathrm{H}_4 \mathrm{O}\right)_3 \\ & =\mathrm{C}_6 \mathrm{H}_{12} \mathrm{O}_3 \end{aligned}$$

Correct Option (3)

$\mathrm{H}_2+\mathrm{I}_2 \rightleftharpoons 2 \mathrm{HI}$

Concentration of $\mathrm{H}_2$ and $\mathrm{I}_2$ decreases untill equilibrium condition and concentration of HI increases till equilibrium condition and after equilibrium concentration of all the reactant and products remain constant. Correct option (2)

Number of sp and $\mathrm{sp}^2$ hybridised carbon atom are 3 and 5.

$\underset{(\mathrm{g})}{\mathrm{SO}_2}+\frac{1}{2} \underset{(\mathrm{~g})}{\mathrm{O}_2} \longrightarrow \underset{(\mathrm{~g})}{\mathrm{SO}_3} \quad \Delta \mathrm{H}=-\mathrm{y}$

$$\begin{aligned} &\begin{aligned} & \Delta \mathrm{H}_{\mathrm{r}}=\left(\Delta \mathrm{H}_{\mathrm{f}}\right)_{\mathrm{SO}_3}-\left(\Delta \mathrm{H}_{\mathrm{f}}\right)_{\mathrm{SO}_2} \\ & -\mathrm{y}=-2 \mathrm{x}-\left(\Delta \mathrm{H}_{\mathrm{f}}\right)_{\mathrm{SO}_2} \\ & \left(\Delta \mathrm{H}_{\mathrm{f}}\right)_{\mathrm{SO}_2}=\mathrm{y}-2 \mathrm{x} \end{aligned}\\ &\text { Option (2) } \end{aligned}$$

List-I

$\mathrm{CoS} \xrightarrow{\text { Aquaregia }} \mathrm{CoCl}_2 \xrightarrow[2 \cdot \mathrm{AcOH}]{\mathrm{1.KNO}_2 / \mathrm{NH}_4 \mathrm{OH}} \mathrm{K}_3\left[\mathrm{Co}\left(\mathrm{NO}_2\right)_6\right] \downarrow \text { yellow }$

Due to resonance bond length is identical in ozone. Therefore statement I is true and statement II is false.

Higher the number of milli moles of acid or base reacted higher will be temperature rise.

Option (4) $n_{\text {acid }}$ or $n_{\text {base }}$ reacted $=30 \mathrm{~m} \mathrm{~mol}$

Option (2) $n_{\text {acid }}$ or $n_{\text {base }}$ reacted $=30 \mathrm{~m} \mathrm{~mol}$

but less energy will be released by neutralisation

reaction of weak acid hence option (2) can not be correct.

Option (3) $\Rightarrow 20 \mathrm{~m} \mathrm{~mol}$

Option (4) $\Rightarrow 25 \mathrm{~m} \mathrm{~mol}$

Hence Correct Option (1)

(B) $-$CN is meta directing But -OH is ortho / pera directing.

(E) Halides are deactivating groups.