JEE Main 2025 28th January Evening Shift Question Paper with Solutions

To find the volume contraction of the copper cube, we need to use the formula for volumetric strain under pressure, which involves the bulk modulus (K). The volumetric strain is given by the formula:

$\text{Volumetric strain} = \frac{\Delta V}{V} = -\frac{P}{K}$

Where:

$\Delta V$ is the change in volume or volume contraction,

$V$ is the original volume of the cube,

$P$ is the hydraulic pressure applied,

$K$ is the bulk modulus of the material.

Given:

Edge length of the cube, $a = 10 \, \text{cm} = 0.1 \, \text{m}$

Hydraulic pressure, $P = 7 \times 10^6 \, \text{Pa}$

Bulk modulus of copper, $K = 1.4 \times 10^{11} \, \text{Pa}$

Calculate the original volume ($V$) of the cube:

$V = a^3 = (0.1 \, \text{m})^3 = 0.001 \, \text{m}^3$

Calculate the volumetric strain:

$\text{Volumetric strain} = -\frac{P}{K} = -\frac{7 \times 10^6}{1.4 \times 10^{11}} = -5 \times 10^{-5}$

Calculate the volume contraction ($\Delta V$):

$\Delta V = V \times \text{Volumetric strain} = 0.001 \, \text{m}^3 \times (-5 \times 10^{-5}) = -5 \times 10^{-8} \, \text{m}^3$

Convert $\Delta V$ from cubic meters to cubic millimeters:

1 cubic meter = $10^9$ cubic millimeters. Therefore,

$\Delta V = -5 \times 10^{-8} \, \text{m}^3 \times 10^9 \, \text{mm}^3/\text{m}^3 = -50 \, \text{mm}^3$

Thus, the volume contraction of the copper cube, when subjected to the given hydraulic pressure, is $50 \, \text{mm}^3$.

$\mathrm{V}=\mathrm{V}_0 \sin \omega \mathrm{t}$

$\begin{aligned} & \text { Distance }=\text { Area under } \mathrm{v} \mathrm{v} / \mathrm{s} \text { t graph } \\ & \text { Distance }=\frac{1}{2} \times 2 \times 10+2 \times 10=30 \mathrm{~m}\end{aligned}$

$$\begin{aligned} & \mathrm{m}=-3=-\frac{\mathrm{v}}{\mathrm{u}} \text { and } \mathrm{v}-\mathrm{u}=20 \mathrm{~cm} \\ & \mathrm{f}=\frac{\mathrm{vu}}{\mathrm{v}+\mathrm{u}}=\frac{(-30)(-10)}{-30-10} \\ & \therefore \mathrm{R}=+15 \end{aligned}$$

Matching the quantities from List - I with their correct dimensions from List - II is crucial to understanding their physical significance. Here is the correct matching:

(A) Angular Impulse: The dimensions of angular impulse are equivalent to those of angular momentum, represented as $M L^2 T^{-1}$. Hence, (A) corresponds to (III).

(B) Latent Heat: Latent heat has dimensions of energy per unit mass, which simplifies to $M^0 L^2 T^{-2}$. Hence, (B) corresponds to (I).

(C) Electrical resistivity: Electrical resistivity has the dimensions of resistance multiplied by length, which can be represented as $M L^3 T^{-3} A^{-2}$. Hence, (C) corresponds to (IV).

(D) Electromotive force: Electromotive force (emf) is similar to electric potential difference and has dimensions $M L^2 T^{-3} A^{-1}$. Hence, (D) corresponds to (II).

Therefore, the correct answer is Option C:

(A)-(III)

(B)-(I)

(C)-(IV)

(D)-(II)

$$\begin{aligned} & \mathrm{Mg}=\mathrm{F}_{\mathrm{B}} \Rightarrow\left(400 \times 10^{-3}\right)=10^3 \times \mathrm{V}_{\mathrm{d}} \\ & \mathrm{~V}_{\mathrm{d}}=400 \times 10^{-6} \mathrm{~m}^3 \\ & (\text { Vol. })_{\text {outside }}=\left(10 \times 10^{-2}\right)^3-400 \times 10^{-6} \\ & =600 \times 10^{-6} \mathrm{~m}^2=600 \mathrm{~cm}^3 \end{aligned}$$

$$\begin{aligned} & \tau_{\text {Net }}=0 \Rightarrow(400 \mathrm{~g} \times 30)=(250 \mathrm{~g} \times 10)(\mathrm{mg} \times 50) \\ & \mathrm{m}=\frac{12000-2500}{50}=\frac{9500}{50} \\ & \mathrm{M}=190 \mathrm{~g} \end{aligned}$$

Without considering uncentainity in $\ell$.

$B=\frac{\mu_0}{4 \pi} \frac{m}{r^3}$

$\mathrm{B} \propto \frac{1}{\mathrm{r}^3}$

$\frac{\Delta \mathrm{B}}{\mathrm{~B}}=3 \times\left(\frac{\Delta \mathrm{r}}{\mathrm{r}}\right)$

$\%$ uncertainity in $\mathrm{B}=3 \%$

Method-2 :

With considering uncentainity in $\ell$.

$$\begin{aligned} & \mathrm{B} \propto \frac{1}{\mathrm{r}^3} \\ & \frac{\Delta \mathrm{~B}}{\mathrm{~B}}=\frac{\Delta \ell}{\ell}+3 \times\left(\frac{\Delta \mathrm{r}}{\mathrm{r}}\right)=1+3 \times 1=4 \% \end{aligned}$$

$\%$ uncertainity in $B=4 \%$

Maxima condition

$2 \mu \mathrm{t}=\mathrm{n} \lambda \Rightarrow \mathrm{t}=\frac{\mathrm{n} \lambda}{2 \mu} \Rightarrow \mathrm{t}=\frac{\lambda}{2 \mu}, \frac{2 \lambda}{2 \mu}, \ldots \ldots$

Minima condition $2 \mu \mathrm{t}=(2 \mathrm{n}-1) \lambda / 2$

$$\begin{aligned} & \Rightarrow \mathrm{t}=\frac{(2 \mathrm{n}-1) \lambda}{4 \mu} \Rightarrow \mathrm{t}=\frac{\lambda}{4 \mu}, \frac{3 \lambda}{4 \mu}, \ldots . . \\ & \Delta \mathrm{t}=\frac{2 \lambda}{4 \mu} \end{aligned}$$

Rate of evaporation $=\frac{\mathrm{A}(\Delta \mathrm{t})}{\text { time }}=54 \times 10^{-13} \mathrm{~m}^3 / \mathrm{s}$

The escape velocity ($v_{\text{esc}}$) for a planet with mass $M$ and radius $R$ is given by the formula:

$v_{\text{esc}} = \sqrt{\frac{2GM}{R}}$

where $G$ is the gravitational constant.

For Earth, let's denote its mass as $M_E$ and its radius as $R_E$. The escape velocity is given as 11.2 km/s. Therefore:

$v_{\text{esc, Earth}} = \sqrt{\frac{2G M_E}{R_E}} = 11.2 \text{ km/s}$

For the planet in question, its mass $M_P$ is $M_E/8$ and its radius $R_P$ is $R_E/2$. The escape velocity from the planet is:

$v_{\text{esc, Planet}} = \sqrt{\frac{2G M_P}{R_P}} = \sqrt{\frac{2G (M_E/8)}{R_E/2}}$

Simplifying the expression:

$v_{\text{esc, Planet}} = \sqrt{\frac{2G (M_E/8)}{R_E/2}} = \sqrt{\frac{2G M_E \cdot 2}{8R_E}} = \sqrt{\frac{G M_E}{2R_E}}$

Now, relate it to the escape velocity from Earth:

$v_{\text{esc, Planet}} = \frac{1}{\sqrt{4}} \cdot v_{\text{esc, Earth}}$

Calculating further:

$v_{\text{esc, Planet}} = \frac{1}{2} \cdot 11.2 \text{ km/s} = 5.6 \text{ km/s}$

Thus, the escape velocity from the planet is 5.6 km/s.

Option C: 5.6 km/s is the correct answer.

$$\begin{aligned} & \frac{\mathrm{n}_2}{\mathrm{v}}-\frac{\mathrm{n}_1}{\mathrm{u}} \frac{\mathrm{n}_2-\mathrm{n}_1}{\mathrm{R}} \\ & \frac{1.4}{\mathrm{v}}-\frac{1.3}{-13}=\frac{0.1}{\mathrm{R}} \\ & \frac{1.4}{\mathrm{v}}=\frac{1-\mathrm{R}}{10 \mathrm{R}} \\ & \frac{1.4}{\mathrm{v}}=\frac{1-\mathrm{R}}{10 \mathrm{R}} \\ & \mathrm{~m}=\frac{\mathrm{v} / \mathrm{n}_2}{\mathrm{u} / \mathrm{n}_1} \\ & -2 \times \frac{(-13)}{1.3}=\frac{10 \mathrm{R}}{1-\mathrm{R}} \\ & \mathrm{R}=\frac{2}{3} \mathrm{~cm} \end{aligned}$$

$$\begin{aligned} & \mathrm{B}=0.4 \mathrm{~T} \\ & \mathrm{r}=20 \mathrm{~cm} \\ & \omega=10 \pi \mathrm{rad} / \mathrm{s} \\ & \mathrm{E}=\frac{1}{2} \mathrm{~B} \omega \mathrm{R}^2 \\ & =0.2512 \mathrm{~V} \end{aligned}$$

$$\begin{aligned} & \mathrm{x}=\mathrm{A} \sin (\omega \mathrm{t}+\phi) \\ & \mathrm{x}_0=\mathrm{A} \sin \phi \quad\text{.... (1)}\\ & \mathrm{p}=\mathrm{mA} \omega \cos (\omega \mathrm{t}+\phi) \\ & \mathrm{p}_0=\mathrm{mA} \omega \cos \phi \quad\text{.... (2)} \end{aligned}$$

$(2) /(1) \Rightarrow \tan \phi=\left(\frac{x_0}{p_0}\right) m \omega$

$\sin \phi=\frac{\mathrm{x}_0 \mathrm{~m} \omega}{\sqrt{\left(\mathrm{~m} \omega \mathrm{x}_0\right)^2+\mathrm{p}_0^2}}$

From (1), $A=\frac{x_0}{\sin \phi}=\frac{\sqrt{\left(m \omega x_0\right)^2+p_0^2}}{m \omega}$

This means we can explain assertion with the given reason.

$$\begin{aligned} & (\mathrm{KE})_{\text {Transsational }}=\left[\frac{3}{2} \mathrm{KT}\right] \times \text { no. of molecule } \\ & \text { No. of molecule }=\left[\frac{50}{44} \times 6.023 \times 10^{23}\right] \\ & (\mathrm{KE})_{\text {Transsational }}=4108.644 \mathrm{~J} \end{aligned}$$

Solution Explanation

Among the four listed phenomena:

Reflection of light

Wave theory (Huygens’s principle, for example) adequately explains reflection.

Refraction of light

Wave theory also explains refraction via the change of wave speed in different media.

Diffraction of light

Wave theory (again, Huygens–Fresnel principle) explains the bending of waves around obstacles.

Compton effect

This involves inelastic scattering of photons by electrons (where the photon loses energy and is scattered with a longer wavelength).

Wave theory alone cannot explain the discrete energy exchange and wavelength shift; it requires the particle (photon) concept from quantum theory.

Hence, the phenomenon not explained by the (classical) wave theory is the Compton effect.

Answer: (D) Compton effect.

Frequency of revolution $\propto \frac{1}{\mathrm{n}^3}$

$$\begin{aligned} & \mathrm{V}_{\mathrm{A}}=\mathrm{V}_{\mathrm{B}} \Rightarrow \text { the bridge is balanced } \\ & \Rightarrow \frac{10}{\mathrm{R}}=\frac{20}{40} \\ & \mathrm{R}=20 \Omega \\ & \mathrm{I}=\frac{40}{20}=2 \mathrm{~A} \end{aligned}$$

$$\begin{aligned} &\begin{aligned} & F-m g=m a \\ & F=m a+m g \\ & F-(m-x) g=(m-x) 3 a \end{aligned}\\ &\text { Put F }\\ &\begin{aligned} & \mathrm{Ma}+\mathrm{mg}-\mathrm{mg}+\mathrm{xg}=3 \mathrm{ma}-3 \mathrm{xa} \\ & \mathrm{x}=\frac{2 \mathrm{ma}}{\mathrm{~g}+3 \mathrm{a}} \end{aligned} \end{aligned}$$

$$\begin{aligned} & \mathrm{B}_1=\frac{\mu_0 \mathrm{i}}{4 \pi \mathrm{a}} \otimes \\ & \mathrm{~B}_2=\frac{\mu_0}{4 \pi} \frac{\mathrm{i}}{\mathrm{a}}\left(\frac{3 \pi}{2}\right) \otimes \\ & \mathrm{B}_3=0 \\ & \mathrm{~B}=\frac{\mu_0}{4 \pi} \frac{\mathrm{i}}{\mathrm{a}}\left(1 \frac{3 \pi}{2}\right) \otimes \end{aligned}$$

The energy density ($u$) between the plates of a capacitor is given by the formula:

$u = \frac{1}{2} \varepsilon_0 E^2$

where $\varepsilon_0$ is the permittivity of free space ($8.85 \times 10^{-12} \, \text{F/m}$) and $E$ is the electric field between the plates. The electric field $E$ is related to the potential difference $V$ and the separation $d$ between the plates by:

$E = \frac{V}{d}$

Given:

Capacitance ($C$) = 1 µF = $1 \times 10^{-6} \, \text{F}$

Potential Difference ($V$) = 20 V

Distance ($d$) = 1 µm = $1 \times 10^{-6} \, \text{m}$

First, calculate the electric field $E$:

$E = \frac{V}{d} = \frac{20 \, \text{V}}{1 \times 10^{-6} \, \text{m}} = 2 \times 10^7 \, \text{V/m}$

Now plug this into the formula for energy density:

$u = \frac{1}{2} \times 8.85 \times 10^{-12} \, \text{F/m} \times (2 \times 10^7 \, \text{V/m})^2$

Calculate $u$:

$u = \frac{1}{2} \times 8.85 \times 10^{-12} \times 4 \times 10^{14}$

$u = 2 \times 8.85 \times 10^2$

$u = 1770 \, \text{J/m}^3$

This value is approximately $1.8 \times 10^3 \, \text{J/m}^3$. Therefore, the correct option is:

Option A: $1.8 \times 10^3 \, \text{J/m}^3$

$$\begin{aligned} &\frac{\rho_1}{\rho_2}=\frac{4}{25}\\ &\text { Ratio of rms velocities }=\sqrt{\frac{\rho_2}{\rho_1}}=\frac{5}{2} \end{aligned}$$

$$\begin{aligned} & \langle\mathrm{p}\rangle=\frac{(2 \hat{\mathrm{i}}+3 \hat{\mathrm{j}}) \cdot(3 \hat{\mathrm{i}}+6 \hat{\mathrm{j}})}{4}=6 \\ & \overrightarrow{\mathrm{a}}=\left(\frac{\overrightarrow{\mathrm{F}}}{\mathrm{~m}}=\frac{1}{2} \hat{\mathrm{i}}+\frac{3}{4} \hat{\mathrm{j}}\right) \\ & \overrightarrow{\mathrm{v}} \text { at } \mathrm{t}=4 \mathrm{sec}=\left(\frac{1}{2} \hat{\mathrm{i}}+\frac{3}{4} \hat{\mathrm{j}}\right) \times 4=(2 \hat{\mathrm{i}}+3 \hat{\mathrm{j}}) \\ & P_{\text {ins }}=(2 \hat{\mathrm{i}}+3)(2 \hat{\mathrm{i}}+3 \hat{\mathrm{j}})=13 \\ & \frac{\langle\mathrm{P}\rangle}{P_{\text {ins }}}=\frac{6}{13} \end{aligned}$$

Note: Given data is not matching.

$$\begin{aligned} & S=u t+\frac{1}{2} a t^2 \\ & S=0+\frac{1}{2} \frac{(2 \hat{i}+3 \hat{j})}{4}(4)^2=4 \hat{i}+6 \hat{\mathrm{j}} \end{aligned}$$

If $\overrightarrow{\mathrm{r}}_{\mathrm{i}}=3 \hat{\mathrm{i}}+4 \hat{\mathrm{j}}$ then $\overrightarrow{\mathrm{r}}_{\mathrm{f}}=7 \hat{\mathrm{i}}+10 \hat{\mathrm{j}}$

But Final position given in the question is $(6,10)$.

$$\begin{aligned} & \mathrm{p}=6 \times 10^{-6} \mathrm{Cm} \\ & \mathrm{E}=10^6 \mathrm{v} / \mathrm{m} \\ & \mathrm{~W}=\Delta \mathrm{U}=-\mathrm{pE}\left(\cos \theta_{\mathrm{f}}-\cos \theta_{\mathrm{i}}\right) \\ & \mathrm{W}=2 \mathrm{pE}=12 \mathrm{~J} \end{aligned}$$

$\mathrm{E}=\ell \mathrm{vB}$

$E=\frac{2 x}{\sqrt{3}} \times v B$ and $x=v t$

$\mathrm{E}=\frac{2}{\sqrt{3}} \mathrm{v}^2 \mathrm{Bt}$

$E \propto t^1$

$$\begin{aligned} & \vec{B}=\left(\frac{\sqrt{3}}{2} \hat{i}+\frac{1}{2} \hat{j}\right) 30 \sin \left[\omega\left(t-\frac{z}{c}\right)\right] \\ & \vec{E}=\vec{B} \times \vec{c} \text { and } E=B_0 c \\ & \text { Here } \vec{E}\left(\frac{\sqrt{3}}{2}(-\hat{j})+\frac{1}{2} \hat{i}\right) \\ & E_0=30 c \\ & \vec{E}=\left(\frac{1}{2} \hat{i}-\frac{\sqrt{3}}{2} \hat{j}\right) 30 c \sin \left[\omega\left(t-\frac{z}{c}\right)\right] \end{aligned}$$

(A) Starch $\xrightarrow{\mathrm{H}^{+} / \mathrm{H}_2 \mathrm{O}}$ Glucose

(B) $\mathrm{\mathop {Cane\,sugar}\limits_{(Sucrose)} \buildrel {{H^ + }/{H_2}O} \over \longrightarrow \mathop {glu\cos e}\limits_{50\% } \, + \mathop {fructose}\limits_{50\% }}$

(C) $\mathrm{\mathop {Milk\,sugar}\limits_{(Lactose)} \buildrel {{H^ + }/{H_2}O} \over \longrightarrow glu\cos e\, + \,galactose}$

(D) Amylopectin $\xrightarrow{\mathrm{H}^{+} / \mathrm{H}_2 \mathrm{O}}$ Glucose

(E) Amylose $\xrightarrow{\mathrm{H}^{+} / \mathrm{H}_2 \mathrm{O}}$ Glucose

So, correct options are B, C and E only

$$\begin{aligned} & \mathrm{R}_1=\mathrm{K}[\mathrm{~A}]^1[\mathrm{~B}]^1 \\ & \mathrm{R}_1=\mathrm{K}\left[\frac{\mathrm{n}_{\mathrm{A}}}{\mathrm{~V}}\right]^1\left[\frac{\mathrm{n}_{\mathrm{B}}}{\mathrm{~V}}\right]^1 \\ & \mathrm{R}_2=\mathrm{K}\left[\frac{3 \mathrm{n}_{\mathrm{A}}}{\mathrm{~V}}\right]^1\left[\frac{3 \mathrm{n}_{\mathrm{B}}}{\mathrm{~V}}\right]^1 \\ & \mathrm{R}_2=9 \mathrm{R}_1 \end{aligned}$$

Because no. of bacteria initial $=\mathrm{N}_0$ and No. of bacteria at any time $\mathrm{t}=\mathrm{N}$

Since bacterial growth is given as

$\mathrm{N}=\mathrm{N}_0 \mathrm{e}^{\mathrm{Kt}}$

Where K = growth constant for bacterial growth

$$\begin{aligned} &\mathrm{V}_2 \mathrm{O}_5+\text { alkali } \rightarrow \mathrm{VO}_4^{3-}\\ &\text { In } \mathrm{VO}_4^{3-} \text { ion, vanadium is in }+5 \text { oxidation state } \end{aligned}$$

Based on the Ksp values and salt analysis cation identification, we can say that order of Ksp value is:

$\mathrm{HgS}<\mathrm{PbS}<\mathrm{AgBr}<\mathrm{Ca}(\mathrm{OH})_2$

Ksp values

$$\begin{aligned} & \mathrm{HgS} \rightarrow 4 \times 10^{-53} \\ & \mathrm{PbS} \rightarrow 8 \times 10^{-28} \\ & \mathrm{AgBr} \rightarrow 5 \times 10^{-13} \\ & \mathrm{Ca}(\mathrm{OH})_2 \rightarrow 5.5 \times 10^{-6} \end{aligned}$$

Living cell $=0.9 \mathrm{gm}$ in 100 gm of solution $\% \mathrm{w} / \mathrm{w}=0.9$

Solution is have equal moles of glucose and water $=0.5$

Weight of solution $=0.5 \times 180+0.5 \times 18=99 \mathrm{gm}$ $\% \mathrm{w} / \mathrm{w} \simeq 90 \%$

Concentrated solution $=$ Cell will shrink

Law of octaves was arranged in the increasing order of their atomic weight.

Lothar Meyer plotted the physical properties such as atomic volume, melting point and boiling point against atomic weight.

$\% \mathrm{w} / \mathrm{w}$ of $\mathrm{HNO}_3=75 \%$

means 100 gm of solution containing 75 g of $\mathrm{HNO}_3$

$\&\left(\frac{\mathrm{gm}}{\mathrm{m}_1}\right)_{\text {solution }}=1.25=\frac{100 \mathrm{gm}}{\mathrm{V}}$

$\mathrm{V}_{\mathrm{ml}}$ of 100 gm solution $=\frac{100}{1.25} \mathrm{ml}$

$\because 75 \mathrm{gm}$ of $\mathrm{HNO}_3$ present in $\frac{100}{1.25} \mathrm{ml}$ solution

$\therefore 30 \mathrm{gm}$ of $\mathrm{HNO}_3$ present in

$\frac{100}{1.25 \times 75} \times 30=32 \mathrm{ml} \text { solution }$

As internal energy ' $U$ ' is a state function, its cyclic integral must be zero in a cyclic process

$\therefore \Delta U \text { case }(I)=\Delta U \text { case }(I I)=\Delta U \text { case }(\text { III })$

(A) $\mathrm{R}-\mathrm{NH}_2 \xrightarrow[\mathrm{HCl}]{\mathrm{NaNO}_2} \mathrm{R}-\mathrm{N}_2^{\oplus} \mathrm{Cl}^{\ominus}$

(B)

(C) Only primary amine gives carbyl amine test

(D)

(E) Tertiary amine do not react with $\mathrm{Ph}-\mathrm{SO}_2 \mathrm{Cl}$

So correct options are (B) and (D) only

$\mathrm{O}_2 \rightarrow 2$ unpaired electrons according to MOT

$\mathrm{O}_2{ }^{+} \rightarrow 1$ unpaired electrons according to MOT

$\mathrm{O}_2^{-} \rightarrow 1$ unpaired electrons according to MOT

$\mathrm{NO} \rightarrow$ odd electron species

$\mathrm{NO}_2 \rightarrow$ odd electron species

$\mathrm{K}_2\left[\mathrm{NiCl}_4\right] \rightarrow \mathrm{Ni}^{2+} \Rightarrow 3 \mathrm{~d}^8$ weak Ligand, C.N. $=4$

$\Rightarrow$ Tetrahedral, Paramagnetic with 2 unpaired electrons

(A) Sucrose $\rightarrow \alpha_1-\beta_2$ Glycosidic linkage

(B) Maltose $\rightarrow \alpha$ 1-4 Glycosidic linkage

(C) Lactose $\rightarrow \beta$ 1-4 Glycosidic linkage

(D) Amylopectin $\rightarrow \alpha$ 1-4 and $\alpha$ 1-6 Glycosidic linkage

$\mathrm{A}-\mathrm{III}, \mathrm{~B}-\mathrm{I}, \mathrm{C}-\mathrm{IV}, \mathrm{D}-\mathrm{II}$