JEE Main 2025 22nd January Evening Shift Question Paper with Solutions

$$\begin{aligned} & B=\frac{\mu_0(5)}{2 \pi \times .01}-\frac{\mu_0 4}{2 \pi \times 0.04} \\ & =-\frac{100 \mu_0}{4 \pi} \\ & =-100 \times 10^{-7} \\ & =-1 \times 10^{-5} \mathrm{~T} \end{aligned}$$

$$\begin{aligned} & \mathrm{mg}-\mathrm{F}_{\mathrm{B}}-\mathrm{f}=0 \\ & \Rightarrow \mathrm{mg}-\frac{\mathrm{mg}}{2}-\mathrm{f}=0 \\ & \therefore \mathrm{f}=\frac{\mathrm{mg}}{2} \end{aligned}$$

$$\begin{aligned} & \mathrm{R}_{\mathrm{eq}}=2 \Omega \\ & \mathrm{I}=\frac{2}{2}=1 \mathrm{~A} \end{aligned}$$

$$\begin{aligned} & \mathrm{F}=2 \mathrm{M} \omega^2 \frac{\ell}{2}=\mathrm{Mw}^2 \ell \\ & \omega=\sqrt{\frac{\mathrm{F}}{\mathrm{M} \ell}} \end{aligned}$$

$$\begin{aligned} & \mathrm{J}_{\mathrm{d}}=\frac{\mathrm{I}}{\mathrm{~A}}=\frac{6}{16} \\ & \therefore \text { I through small area }=\mathrm{J}_{\mathrm{d}} \times A^{\prime}=\frac{6}{16} \times 3.2= \\ & 1.2 \mathrm{~A}=1200 \mathrm{~mA} \end{aligned}$$

For transmitted green light to be maxima, reflected green should be minima.

$$\begin{aligned} & \Delta \mathrm{P}=2 \mu_0 \mathrm{t}=\mathrm{n} \lambda \\ & \Rightarrow \mathrm{t}=\frac{\mathrm{n} \lambda}{2 \mu_0} \therefore \mathrm{t}_{\min }=\frac{\lambda}{2 \mu_0}=\frac{550}{2 \times 2}=137.5 \end{aligned}$$

$$\begin{aligned} & \mathrm{A}_1 \mathrm{v}_1=\mathrm{A}_2 \mathrm{~V}_2 \Rightarrow 2 \pi(2 \mathrm{R})^2=\mathrm{V}_2 \pi \mathrm{R}^2 \\ & \therefore \mathrm{~V}_2=8 \mathrm{~m} / \mathrm{s} \end{aligned}$$

$$\begin{aligned} & \mathrm{k}_{\mathrm{i}}=\frac{1}{2} \mathrm{mv}^2 \\ & \mathrm{k}_{\mathrm{f}}=\frac{1}{2} \mathrm{~m}\left(\mathrm{v} \cos 60^{\circ}\right)^2=\frac{1}{8} \mathrm{mv}^2 \\ & \Delta \mathrm{k}=\mathrm{k}_{\mathrm{i}}-\mathrm{k}_{\mathrm{f}}=\frac{3}{8} \mathrm{mv}^2=\frac{3}{8} \times 0.1 \times 400=15 \mathrm{~J} \end{aligned}$$

Let's break down the problem step by step.

The photoelectric equation is given by:

$K_{\text{max}} = \frac{hc}{\lambda} - \phi$

where:

$K_{\text{max}}$ is the maximum kinetic energy of the electrons.

$\frac{hc}{\lambda}$ is the energy of the incident photon.

$\phi$ is the work function of the metal.

For the initial light source of wavelength $\lambda$, we know:

$2 \text{ eV} = \frac{hc}{\lambda} - 1 \text{ eV}$

Solving for $\frac{hc}{\lambda}$ gives:

$\frac{hc}{\lambda} = 2 \text{ eV} + 1 \text{ eV} = 3 \text{ eV}$

Now, if the wavelength is halved to $\frac{\lambda}{2}$, the photon energy becomes:

$\frac{hc}{\lambda/2} = \frac{2hc}{\lambda} = 2 \times 3 \text{ eV} = 6 \text{ eV}$

The maximum kinetic energy for the new wavelength is then:

$K'_{\text{max}} = 6 \text{ eV} - 1 \text{ eV} = 5 \text{ eV}$

Therefore, the maximum kinetic energy of the ejected electrons when the surface is illuminated by a light source of wavelength $\frac{\lambda}{2}$ is 5 eV.

The correct answer is Option A.

$C=\frac{q}{V}=\frac{q \cdot q}{V \cdot q}=\frac{q^2}{W D}=\frac{C^2}{M L L^2 T^{-2}}=C^2 M^{-1} L^{-2} T^2$

Let's analyze the problem step by step:

For a rigid diatomic molecule (without vibrational modes):

Degrees of freedom:

Translation: 3

Rotation: 2

Total: 5

The molar specific heat at constant volume is given by:

$C_v = \frac{5}{2}R$

At constant pressure, it is:

$C_p = C_v + R = \frac{5}{2}R + R = \frac{7}{2}R$

Therefore, the ratio is:

$\gamma_1 = \frac{C_p}{C_v} = \frac{\frac{7}{2}R}{\frac{5}{2}R} = \frac{7}{5} \approx 1.4$

For the diatomic molecule that also has vibrational modes (assuming the vibrational mode is fully excited):

Additional vibrational mode contributions:

Each vibrational mode contributes 2 degrees of freedom (one kinetic and one potential).

Therefore, the degrees of freedom become:

$3\ (\text{translation}) + 2\ (\text{rotation}) + 2\ (\text{vibration}) = 7$

The molar specific heat at constant volume now is:

$C_v = \frac{7}{2}R$

And at constant pressure:

$C_p = C_v + R = \frac{7}{2}R + R = \frac{9}{2}R$

Thus, the ratio is:

$\gamma_2 = \frac{C_p}{C_v} = \frac{\frac{9}{2}R}{\frac{7}{2}R} = \frac{9}{7} \approx 1.2857$

Comparison:

For the rigid molecule, $\gamma_1 \approx 1.4$.

For the molecule with vibrational modes, $\gamma_2 \approx 1.2857$.

Clearly, $\gamma_2 < \gamma_1$.

Therefore, the correct option is:

Option A: $\gamma_2 < \gamma_1$

$$\begin{aligned} & \frac{1}{2} \mathrm{~m} \times 100+0=\frac{1}{2} \mathrm{mV}_{\mathrm{B}}^2+\mathrm{mg}\left(\mathrm{R}-\frac{\mathrm{R} \sqrt{3}}{2}\right) \\ & 100=\mathrm{V}_{\mathrm{B}}^2+2 \mathrm{gR}\left(1-\frac{\sqrt{3}}{2}\right] \\ & \mathrm{V}_{\mathrm{B}}^2=100-20(2-\sqrt{3}) \\ & \left.\mathrm{V}_{\mathrm{B}}^2=60+20 \sqrt{3}\right) \\ & \mathrm{K}_{\mathrm{E}} \mathrm{E}_{\mathrm{B}}=\frac{1}{2} \mathrm{mV}_{\mathrm{B}}^2=\frac{\mathrm{m}}{2}(60+20 \sqrt{3}) \\ & \frac{1}{2} \mathrm{~m}(100)=\frac{1}{2} \mathrm{mV}_{\mathrm{C}}^2+\mathrm{mg}\left(\frac{3 \mathrm{R}}{2}\right) \end{aligned}$$

$$\begin{aligned} & 100=V_{\mathrm{C}}^2=60 \\ & \mathrm{~V}_{\mathrm{C}}^2=40 \\ & \mathrm{~K} \cdot \mathrm{E}_{\mathrm{C}}=\frac{1}{2} m V_{\mathrm{C}}^2=\frac{1}{2} \mathrm{~m}(40) \\ & \mathrm{K} \cdot \mathrm{E}_{\mathrm{B}}=\frac{60+20 \sqrt{3}}{40}=\frac{3}{2}+\frac{\sqrt{3}}{2}=\frac{3+\sqrt{3}}{2} \end{aligned}$$

$$\begin{aligned} & E_A=\frac{2 k P}{r^3}=E_0 \& V_A=\frac{k P}{r^2}=v_0 \\ & E_B=\frac{k P}{(2 r)^3}=\frac{E_0}{16} \& V_B=\frac{k \vec{p} \cdot \hat{r}}{r^2}=0 \end{aligned}$$

$$\begin{aligned} &\begin{aligned} & \mathrm{r}=\frac{\mathrm{mv}}{\mathrm{eB}} \& \mathrm{mvr}=\frac{\mathrm{nh}}{2 \pi} \Rightarrow(\mathrm{eBr}) \mathrm{r}=\frac{\mathrm{nh}}{2 \pi} \\ & \Rightarrow \mathrm{r}=\sqrt{\frac{\mathrm{nh}}{2 \pi \mathrm{eB}}} \end{aligned}\\ &\text { first excited state } \mathrm{n}=2 \therefore \mathrm{r}=\sqrt{\frac{\mathrm{h}}{\pi \mathrm{eB}}} \end{aligned}$$

To determine the maximum percentage error in the density of the wire, follow these steps:

The density of the wire is given by the formula for a cylinder:

$\rho = \frac{m}{\pi r^2 l}$

When calculating the maximum percentage (relative) error, you add the relative errors from each variable, taking into account the power to which each variable is raised. Thus, for density:

$\frac{\Delta \rho}{\rho} = \frac{\Delta m}{m} + 2\frac{\Delta r}{r} + \frac{\Delta l}{l}$

Here:

$\frac{\Delta m}{m}$ is the relative error in the mass.

$2\frac{\Delta r}{r}$ is due to the radius being squared.

$\frac{\Delta l}{l}$ is the relative error in the length.

Now plug in the given values:

Mass: $m = 0.60 \, \text{g}$ with an error $\Delta m = 0.003 \, \text{g}$

$\frac{\Delta m}{m} = \frac{0.003}{0.60} = 0.005$ or 0.5%.

Radius: $r = 0.50 \, \text{cm}$ with an error $\Delta r = 0.01 \, \text{cm}$

$\frac{\Delta r}{r} = \frac{0.01}{0.50} = 0.02$ or 2%. Since the radius is squared in the formula, multiply by 2:

$2\frac{\Delta r}{r} = 2 \times 0.02 = 0.04$ or 4%.

Length: $l = 10.00 \, \text{cm}$ with an error $\Delta l = 0.05 \, \text{cm}$

$\frac{\Delta l}{l} = \frac{0.05}{10.00} = 0.005$ or 0.5%.

Sum these contributions to find the overall maximum relative error:

$\frac{\Delta \rho}{\rho} = 0.005 + 0.04 + 0.005 = 0.05$

Converting this relative error into a percentage gives:

$0.05 \times 100\% = 5\%$

Thus, the maximum percentage error in the density of the wire is 5%, which corresponds to Option C.

Motional emf : $\varepsilon=\mathrm{B} \ell \mathrm{v}=\mathrm{constant}$

Extensive variables depends on size or mass of system ex : internal energy, volume, mass

For NOR gate : $\overline{\mathrm{A}} \overline{\mathrm{B}}=\overrightarrow{\mathrm{A}}+\mathrm{B}$

$\therefore$ Truth table $$\begin{array}{ccc} \mathrm{A} & \mathrm{~B} & \mathrm{Y} \\ 0 & 0 & 1 \\ 0 & 1 & 1 \\ 1 & 0 & 0 \\ 1 & 1 & 1 \end{array}$$

$\therefore$ Bonus

$$\begin{aligned} & g=\frac{G M}{R^2} \\ & g^{\prime}=\frac{G(4 M)}{(2 R)^2}=g \end{aligned}$$

A is correct, $R$ is correct ; but since $T=2 \pi \sqrt{\frac{\ell}{g}}$ doesn't depend on mass ; R doesn't explain A .

Let's break down the problem step by step:

When the proton moves undeflected in crossed electric and magnetic fields, the electric and magnetic forces balance each other. That is,

$qE = qvB,$

which simplifies to

$E = vB.$

After the electric field is switched off, the proton moves in a circular path under the action of the magnetic force. The magnetic force provides the required centripetal force:

$qvB = \frac{mv^2}{r}.$

Solving for the magnetic field $B$, we get:

$B = \frac{mv}{qr}.$

Now substitute this expression for $B$ back into the equilibrium condition:

$E = vB = v\left(\frac{mv}{qr}\right) = \frac{mv^2}{qr}.$

Plug in the given values:

Proton mass, $m = 1.6 \times 10^{-27} \, \text{kg}$

Speed, $v = 2 \times 10^5 \, \text{m/s}$

Radius, $r = 2 \, \text{cm} = 0.02 \, \text{m}$

Proton charge, $q = 1.6 \times 10^{-19} \, \text{C}$

Thus,

$E = \frac{(1.6 \times 10^{-27} \, \text{kg})(2 \times 10^5 \, \text{m/s})^2}{(1.6 \times 10^{-19} \, \text{C})(0.02 \, \text{m})}.$

Calculate the numerator:

$(2 \times 10^5)^2 = 4 \times 10^{10},$

So, $(1.6 \times 10^{-27}) \times (4 \times 10^{10}) = 6.4 \times 10^{-17}.$

Calculate the denominator:

$(1.6 \times 10^{-19}) \times (0.02) = 3.2 \times 10^{-21}.$

Now compute the electric field:

$E = \frac{6.4 \times 10^{-17}}{3.2 \times 10^{-21}} = 2 \times 10^4 \, \text{N/C}.$

The problem states that the magnitude of the electric field is $x \times 10^4 \, \text{N/C}.$ Since we found

$E = 2 \times 10^4 \, \text{N/C},$

it follows that

$x = 2.$

$$\begin{aligned} & \beta=\frac{\lambda \mathrm{D}}{\mathrm{~d}} \& \lambda_{\mathrm{R}}>\lambda_{\mathrm{b}} \\ & \therefore \beta_{\mathrm{R}}>\beta_{\mathrm{b}} \end{aligned}$$

Initially, $\mathrm{I}_0=\frac{\varepsilon_{\mathrm{m}}}{\mathrm{R}}$

Finally, $I_0^1=\frac{\varepsilon_m}{2 R}=\frac{I_0}{2}$

$\vec{\tau}=\overrightarrow{\mathrm{r}} \times \overrightarrow{\mathrm{F}}=\left|\begin{array}{ccc}\hat{\mathrm{i}} & \hat{\mathrm{j}} & \hat{\mathrm{k}} \\ 1 & 1 & 1 \\ 2 & 1 & 2\end{array}\right|=\hat{\mathrm{i}}-0 \hat{\mathrm{j}}-\hat{\mathrm{k}}$

To determine the value of $b$ such that the work done on the particle is zero, follow these steps:

The work done by a force on a displacement is given by the dot product:

$\text{Work} = \overrightarrow{F} \cdot \overrightarrow{d}.$

Given the force

$\overrightarrow{F} = 2\hat{i} + b\hat{j} + \hat{k},$

and the displacement

$\overrightarrow{d} = \hat{i} - 2\hat{j} - \hat{k},$

compute the dot product:

$\overrightarrow{F} \cdot \overrightarrow{d} = (2)(1) + (b)(-2) + (1)(-1).$

Simplify the expression:

$\overrightarrow{F} \cdot \overrightarrow{d} = 2 - 2b - 1 = 1 - 2b.$

Since the work done is zero:

$1 - 2b = 0.$

Solve for $b$:

$2b = 1,$

$b = \frac{1}{2}.$

Thus, the value of $b$ is $\frac{1}{2}$, which corresponds to Option B.

$\mathrm{Ni}^{2+}$ with $(\mathrm{dmg})$ forms $\left[\mathrm{Ni}(\mathrm{dmg})_2\right]^{2+}$ having 2 H -Bonds as shown :

The no. of H atoms $=14$

First, let us re-state the two statements clearly:

Statement (I): An element in the extreme left of the periodic table forms acidic oxides.

Statement (II): Acid is formed during the reaction between water and oxide of a reactive element present in the extreme right of the periodic table.

Analyzing Statement (I)

The extreme left of the periodic table corresponds to the alkali metals (Group 1) and alkaline earth metals (Group 2).

These metals typically form basic (or occasionally amphoteric, in the case of some Group 2 elements) oxides, not acidic oxides.

For example, Na₂O, K₂O, MgO, CaO, etc., all form basic solutions (e.g., Na₂O + H₂O → 2 NaOH).

Hence, Statement (I)—that an element in the extreme left forms acidic oxides—is false.

Analyzing Statement (II)

The extreme right of the periodic table corresponds to the nonmetals in Groups 15, 16, 17 (and noble gases in Group 18).

Nonmetal oxides (such as those of sulfur, phosphorus, chlorine) are generally acidic.

When these oxides dissolve in water, they typically form acids.

Example: SO₃ + H₂O → H₂SO₄ (sulfuric acid)

Example: P₂O₅ + 3 H₂O → 2 H₃PO₄ (phosphoric acid)

Example: Cl₂O₇ + H₂O → 2 HClO₄ (perchloric acid)

Thus, Statement (II)—that acid is formed when water reacts with an oxide of a reactive element in the extreme right—is true.

Conclusion

Statement (I) is false.

Statement (II) is true.

Therefore, the correct choice is:

(D) Statement (I) is false but Statement (II) is true.

$\textbf{Oxidation States:}$

In $\mathrm{ClO_4^-}$ (perchlorate), chlorine is in the +7 oxidation state.

In $\mathrm{ClO_3^-}$ (chlorate), chlorine is in the +5 oxidation state.

In $\mathrm{ClO_2^-}$ (chlorite), chlorine is in the +3 oxidation state.

In $\mathrm{ClO^-}$ (hypochlorite), chlorine is in the +1 oxidation state.

$\textbf{Disproportionation Reaction:}$

A disproportionation reaction is one in which a species simultaneously undergoes oxidation and reduction. For this to occur, the element must be in an intermediate oxidation state such that it can be oxidized to a higher state and reduced to a lower one.

In $\mathrm{ClO^-}$ and $\mathrm{ClO_2^-}$, chlorine is in lower oxidation states (+1 and +3, respectively), making them susceptible to disproportionation. For example, hypochlorite can disproportionate in basic solution as follows:

$3\,\mathrm{ClO^-} \rightarrow 2\,\mathrm{Cl^-} + \mathrm{ClO_3^-}$

In $\mathrm{ClO_3^-}$, chlorine is in an intermediate oxidation state (+5) that, under certain conditions, can undergo disproportionation.

However, in $\mathrm{ClO_4^-}$, chlorine is in its highest possible oxidation state (+7) and cannot be oxidized further. Since disproportionation requires one part of the species to be oxidized and the other reduced, $\mathrm{ClO_4^-}$ is thermodynamically stable and does not undergo disproportionation.

$\boxed{\mathrm{ClO_4^-} \text{ (perchlorate ion) does not undergo disproportionation.}}$

$$\begin{aligned} & \mathrm{C}_2 \mathrm{H}_6(\mathrm{~g})+\frac{7}{2} \mathrm{O}_2(\mathrm{~g}) \rightarrow 2 \mathrm{CO}_2(\mathrm{~g})+3 \mathrm{H}_2 \mathrm{O}(\mathrm{I}) \Delta \mathrm{H}_1^{\circ} =-1550 \ldots \text { (i) } \\ & \mathrm{C} \text { (graphite) }+\mathrm{O}_2(\mathrm{~g}) \rightarrow \mathrm{CO}_2(\mathrm{~g}) \Delta \mathrm{H}_2^{\circ}=-393.5 \ldots \text { (ii) } \\ & \mathrm{H}_2(\mathrm{~g})+\frac{1}{2} \mathrm{O}_2(\mathrm{~g}) \rightarrow \mathrm{H}_2 \mathrm{O}(\mathrm{I}) \Delta \mathrm{H}_3^{\circ}=-286 .......(iii) \end{aligned}$$

To determine the final concentration of the NaOH solution, we use the formula for mixing solutions :

$M_F = \frac{M_1 \times V_1 + M_2 \times V_2}{V_1 + V_2}$

Where :

$M_1 = 2 \, \text{M}$ and $V_1 = 20 \, \text{mL}$: Concentration and volume of the first solution.

$M_2 = 0.5 \, \text{M}$ and $V_2 = 400 \, \text{mL}$: Concentration and volume of the second solution.

Substitute the values into the equation:

$M_F = \frac{2 \times 20 + 0.5 \times 400}{420}$

Calculating each term:

$2 \times 20 = 40$

$0.5 \times 400 = 200$

Add these results:

$M_F = \frac{40 + 200}{420} = \frac{240}{420} \approx 0.571 \, \text{M}$

Convert this to scientific notation as the problem specifies the answer should be in $\times 10^{-2}$ form:

$M_F = 57.1 \times 10^{-2} \, \text{M}$

Rounded to the nearest integer, the final concentration is:

$57$

We need to determine the number of electrons each element has in its 4d orbitals and then add them together.

1. Niobium $\mathrm{(Nb, Z = 41)}$

The ground‐state electronic configuration of niobium is:

$[\mathrm{Kr}]\,4d^4\,5s^1.$

Hence, Niobium has 4 electrons in its 4d orbitals.

2. Ruthenium $\mathrm{(Ru, Z = 44)}$

The ground‐state electronic configuration of ruthenium is:

$[\mathrm{Kr}]\,4d^7\,5s^1.$

Hence, Ruthenium has 7 electrons in its 4d orbitals.

3. Sum of 4d Electrons

$x \;=\; 4 \quad (\text{for Nb}), \quad y \;=\; 7 \quad (\text{for Ru}).$

Therefore,

$x + y \;=\; 4 \;+\; 7 \;=\; 11.$

Answer: 11

4-Ethyl-3,4-dimethyloctane

We need to compare the experimental (or well‐established) dipole moments of the given molecules: $\mathrm{HBr}, \mathrm{H_2S}, \mathrm{NF_3},$ and $\mathrm{CHCl_3}$.

1. $\mathbf{NF_3}$

Structure: Trigonal pyramidal (like $\mathrm{NH_3}$), but each bond is more polar toward fluorine.

Net dipole moment: Quite small, about $0.23\text{–}0.24\,D$ (the bond dipoles partly oppose the lone‐pair contribution).

2. $\mathbf{HBr}$

Structure: Simple diatomic.

Dipole moment: About $0.78\,D$.

3. $\mathbf{H_2S}$

Structure: Bent (like $\mathrm{H_2O}$) but S is less electronegative than O, and the S–H bond angle is wider than the O–H angle in water.

Dipole moment: About $0.95\,D$.

4. $\mathbf{CHCl_3}$ (chloroform)

Structure: Tetrahedral around carbon with three Cl and one H.

Dipole moment: About $1.01\,D$.

Ordering from smallest to largest dipole moment

$\boxed{\mathrm{NF_3} \;<\; \mathrm{HBr} \;<\; \mathrm{H_2S} \;<\; \mathrm{CHCl_3}.}$

Hence, the correct choice is:

$\boxed{\text{Option (C)}\quad NF_3 < HBr < H_2S < CHCl_3.}$