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

Series :

$$\begin{aligned} & \mathrm{R}_{\mathrm{eq}}=\mathrm{R}_1+\mathrm{R}_2 \\ & 2 \mathrm{R}\left(1+\alpha_{\mathrm{eq}} \Delta \theta\right)=\mathrm{R}\left(1+\alpha_1 \Delta \theta\right)+\mathrm{R}\left(1+\alpha_2 \Delta \theta\right) \\ & 2 \mathrm{R}\left(1+\alpha_{\mathrm{eq}} \Delta \theta\right)=2 \mathrm{R}+\left(\alpha_1+\alpha_2\right) \mathrm{R} \Delta \theta \\ & \alpha_{\mathrm{eq}}=\frac{\alpha_1+\alpha_2}{2} \end{aligned}$$

Parallel :

$$\begin{aligned} & \frac{1}{R_{\text {eq }}}=\frac{1}{R_1}+\frac{1}{R_2} \\ & \frac{1}{\frac{R}{2}\left(1+\alpha_{\text {eq }} \Delta \theta\right)}=\frac{1}{R\left(1+\alpha_1 \Delta \theta\right)}+\frac{1}{\mathrm{R}\left(1+\alpha_2 \Delta \theta\right)} \end{aligned}$$

$$\begin{aligned} & \frac{2}{1+\alpha_{\mathrm{eq}} \Delta \theta}=\frac{1}{1+\alpha_1 \Delta \theta}+\frac{1}{1+\alpha_2 \Delta \theta} \\ & \frac{2}{1+\alpha_{\mathrm{eq}} \Delta \theta}=\frac{1+\alpha_2 \Delta \theta+1+\alpha_1 \Delta \theta}{\left(1+\alpha_1 \Delta \theta\right)\left(1+\alpha_2 \Delta \theta\right)} \\ & 2\left[\left(1+\alpha_1 \Delta \theta\right)\left(1+\alpha_2 \Delta \theta\right)\right] \\ & =\left[2+\left(\alpha_1+\alpha_2\right) \Delta \theta\right]\left[1+\alpha_{\mathrm{eq}} \Delta \theta\right] \\ & 2\left[1+\alpha_1 \Delta \theta+\alpha_2 \Delta \theta+\alpha_1 \alpha_2 \Delta \theta\right] \end{aligned}$$

$$\begin{aligned} & = \\ & 2+2 \alpha_{\mathrm{eq}} \Delta \theta+\left(\alpha_1+\alpha_2\right) \Delta \theta+\alpha_{\mathrm{eq}}\left(\alpha_1+\alpha_2\right) \Delta \theta^2 \end{aligned}$$

Neglecting small terms

$$\begin{aligned} & 2+2\left(\alpha_1+\alpha_2\right) \Delta \theta=2+2 \alpha_{\mathrm{eq}} \Delta \theta+\left(\alpha_1+\alpha_2\right) \Delta \theta \\ & \left(\alpha_1+\alpha_2\right) \Delta \theta=2 \alpha_{\mathrm{eq}} \Delta \theta \\ & \alpha_{\mathrm{eq}}=\frac{\alpha_1+\alpha_2}{2} \end{aligned}$$

The average energy density of an electromagnetic wave is given by the formula:

$U = \frac{1}{2} \varepsilon_0 E^2 + \frac{1}{2} \frac{B^2}{\mu_0}$

For a plane electromagnetic wave, the electric field (E) and the magnetic field (B) contribute equally to the energy density. Therefore, we can focus on just one part to find the total energy density. The formula for the energy density due to the electric field is:

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

Given that the electric field amplitude ($E$) is $50 \, \mathrm{Vm}^{-1}$ and the permittivity of free space ($\varepsilon_0$) is $8.85 \times 10^{-12} \, \mathrm{C}^2/\mathrm{Nm}^2$, we can calculate the energy density due to the electric field as follows:

$U_E = \frac{1}{2} \times 8.85 \times 10^{-12} \times (50)^2 = 1.10625 \times 10^{-8} \, \mathrm{J/m^3}$

Since the energy density contributions from the electric and magnetic fields are equal, the total average energy density of the electromagnetic wave is just this value.

$$\begin{aligned} & \mu=\frac{\sin \left(\frac{A+\delta m}{2}\right)}{\sin \frac{A}{2}} \\ & \frac{\cos \frac{A}{2}}{\sin \frac{A}{2}}=\frac{\sin \left(\frac{A+\delta m}{2}\right)}{\sin \frac{A}{2}} \\ & \sin \left(\frac{\pi}{2}-\frac{A}{2}\right)=\sin \left(\frac{A+\delta_m}{2}\right) \\ & \frac{\pi}{2}-\frac{A}{2}=\frac{A}{2}+\frac{\delta m}{2} \\ & \delta_m=\pi-2 A \end{aligned}$$

$6\Omega$ is short circuit

$\mathrm{R}_{\mathrm{eq}}=3 \times \frac{1}{3}=1 \Omega$

$\mathrm{KE}=\frac{\mathrm{f}}{2} \mathrm{kT}$

$$\begin{aligned} & \beta=\frac{-\Delta \mathrm{P}}{\frac{\Delta \mathrm{V}}{\mathrm{V}}} \\ & \Delta \mathrm{P}=-\beta \frac{\Delta \mathrm{V}}{\mathrm{V}} \\ & \rho \mathrm{gh}=-\beta \frac{\Delta \mathrm{V}}{\mathrm{V}} \\ & 10^3 \times 10 \times \mathrm{h}=-9 \times 10^8 \times\left(-\frac{0.02}{100}\right) \\ & \Rightarrow \mathrm{h}=18 \mathrm{~m} \end{aligned}$$

When the coin is on the disc and the disc starts rotating, centrifugal force acts on the coin, trying to push it away from the center. The friction between the coin and the disc opposes this motion. For the coin to not slip, the frictional force must be equal to the centrifugal force acting on the coin.

The normal force (N) acting on the coin is equal to the weight of the coin, which can be represented as:

$N = mg$

where $m$ is the mass of the coin, and $g$ is the acceleration due to gravity.

The centrifugal force ($f$) that acts on the coin due to the rotation of the disc is given by:

$f = m\omega^2r$

where $\omega$ is the angular velocity, and $r$ is the distance of the coin from the center of the disc.

Friction force (f) is also given by the formula:

$f = \mu N$

where $\mu$ is the coefficient of static friction between the coin and the disc.

For the coin to not slip, the centrifugal force must be equal to the frictional force, hence:

$\mu mg = m\omega^2r$

Dividing both sides by $mr$, we get:

$\omega = \sqrt{\frac{\mu g}{r}}$

This equation shows that the maximum angular velocity ($\omega$) that can be given to the disc to prevent the coin from slipping off depends on the coefficient of friction ($\mu$), the acceleration due to gravity ($g$), and the distance of the coin from the center ($r$).

$$\begin{aligned} & \mathrm{f}_1=\frac{\mathrm{v}}{4 \mathrm{~L}_1} \\ & \mathrm{f}_1=\mathrm{f}_2 \\ & \frac{\mathrm{v}}{4 \mathrm{~L}_1}=\frac{\mathrm{v}}{\mathrm{L}_2} \\ & \Rightarrow \mathrm{L}_2=4 \mathrm{~L}_1 \\ & 60=4 \times \mathrm{L}_1 \\ & \mathrm{~L}_1=15 \mathrm{~cm} \end{aligned}$$

$\mathrm{Y}=\overline{\overline{\mathrm{A}} \cdot \overline{\mathrm{B}}}=\overline{\overline{\mathrm{A}}}+\overline{\overline{\mathrm{B}}}=\mathrm{A}+\mathrm{B}$

(De-Morgan's law)

Using work energy theorem

$$\begin{aligned} & \mathrm{W}=\Delta \mathrm{KE}=0-\left(\frac{1}{2} \mathrm{mv}^2+\frac{1}{2} \mathrm{I} \omega^2\right) \\ & \mathrm{W}=0-\frac{1}{2} \mathrm{mv}^2\left(1+\frac{\mathrm{K}^2}{\mathrm{R}^2}\right) \\ & =-\frac{1}{2} \times 50 \times 0.4^2\left(1+\frac{1}{2}\right)=-6 \mathrm{~J} \end{aligned}$$

Absolute work $=+6 \mathrm{~J}$

$W=-6 J \quad|W|=6 J$

Note : Direction of magnetic field is in $+\hat{k}$

$$\begin{aligned} & \overrightarrow{\mathrm{F}}=\mathrm{i} \vec{\ell} \times \overrightarrow{\mathrm{B}} \\ & \ell=2 \mathrm{R} \\ & \overrightarrow{\mathrm{F}}=-2 \mathrm{iRB} \hat{\mathrm{j}} \end{aligned}$$

$$\begin{aligned} & \mathrm{t}=\alpha \mathrm{x}^2+\beta \mathrm{x} \text { (differentiating wrt time) } \\ & \frac{\mathrm{dt}}{\mathrm{dx}}=2 \alpha \mathrm{x}+\beta \\ & \frac{1}{\mathrm{v}}=2 \alpha \mathrm{x}+\beta \\ & (\text { differentiating wrt time) } \\ & -\frac{1}{\mathrm{v}^2} \frac{\mathrm{dv}}{\mathrm{dt}}=2 \alpha \frac{\mathrm{dx}}{\mathrm{dt}} \\ & \frac{\mathrm{dv}}{\mathrm{dt}}=-2 \alpha \mathrm{v}^3 \end{aligned}$$

Flux linkage for inner loop.

$$\begin{aligned} & \phi=\mathrm{B}_{\text {center }} \cdot \ell^2 \\ & =4 \times \frac{\mu_0 \mathrm{i}}{4 \pi \frac{\mathrm{L}}{2}}(\sin 45+\sin 45) \ell^2 \\ & \phi=2 \sqrt{2} \frac{\mu_0 \mathrm{i}}{\pi \mathrm{L}} \ell^2 \\ & \mathrm{M}=\frac{\phi}{\mathrm{i}}=\frac{2 \sqrt{2} \mu_0 \ell^2}{\pi \mathrm{L}}=2 \sqrt{2} \frac{\mu_0}{\pi} \\ & =2 \sqrt{2} \frac{4 \pi}{\pi} \times 10^{-7} \\ & =8 \sqrt{2} \times 10^{-7} \mathrm{H} \\ & =\sqrt{128} \times 10^{-7} \mathrm{H} \\ & \mathrm{x}=128 \end{aligned}$$

To find the new amplitude of the motion when the speed is increased to three times at a given displacement, we use the concepts of simple harmonic motion (SHM) and its formulas.

In SHM, the velocity $v$ of a particle at a displacement $x$ from the mean position can be given by the formula:

$v = \omega \sqrt{A^2 - x^2}$

where:

• $\omega$ is the angular frequency of the motion,
• $A$ is the amplitude, and
• $x$ is the displacement at that instance.

Given:

• Displacement at the instance, $x = \frac{2A}{3}$,
• Initial velocity is increased to three times at this displacement.

Thus, let's find the initial velocity $v$ at $x = \frac{2A}{3}$:

$v = \omega \sqrt{A^2 - \left(\frac{2A}{3}\right)^2} = \omega \sqrt{A^2 - \frac{4A^2}{9}} = \omega \sqrt{\frac{5A^2}{9}} = \frac{\sqrt{5}A\omega}{3}$

With the velocity increased to three times, the new velocity $v'$ becomes:

$v' = 3v = 3 \times \frac{\sqrt{5}A\omega}{3} = \sqrt{5}A\omega$

For the new amplitude $A'$, the velocity $v'$ at the same displacement $x$ is:

$v' = \omega \sqrt{{A'}^2 - \left(\frac{2A}{3}\right)^2}$

Setting the expressions for $v'$ equal gives:

$\sqrt{5}A\omega = \omega \sqrt{{A'}^2 - \frac{4A^2}{9}}$

$\sqrt{5}A = \sqrt{{A'}^2 - \frac{4A^2}{9}}$

Solving for $A'$ gives:

${A'}^2 = 5A^2 + \frac{4A^2}{9} = \frac{45A^2 + 4A^2}{9} = \frac{49A^2}{9}$

$A' = \sqrt{\frac{49A^2}{9}} = \frac{7A}{3}$

Therefore, the new amplitude of the motion is $\frac{7A}{3}$, which means the value of $n$ is 7.

$$\begin{aligned} & \mathrm{mg}-\mathrm{F}_{\mathrm{B}}-\mathrm{F}_{\mathrm{v}}=\mathrm{ma} \\ & \left(\rho \frac{4}{3} \pi \mathrm{r}^3\right) g-\left(\rho_{\mathrm{L}} \frac{4}{3} \pi \mathrm{r}^3\right) g-6 \pi \eta r v=m \frac{d v}{d t} \end{aligned}$$

Let $\frac{4}{3 m} \pi R^3 g\left(\rho-\rho_L\right)=K_1$ and $\frac{6 \pi \eta r}{m}=K_2$

$\frac{\mathrm{dv}}{\mathrm{dt}}=\mathrm{K}_1-\mathrm{K}_2 \mathrm{~V}$

\begin{aligned} & \int_\limits0^v \frac{d v}{K_1-K_2 v}=\int_\limits0^t d t \\ & -\frac{1}{K_2} \ln \left[K_1-K_2 v\right]_0^v=t \\ & \ell n\left(\frac{K_1-K_2 v}{K_1}\right)=-K_2 t \\ & K_1-K_2 v=K_1 e^{-K_2 t} \\ & v=\frac{K_1}{K_2}\left[1-e^{-K_2 t}\right] \end{aligned}

To determine the dimensions of $\frac{b^2}{a}$, let's start by identifying the dimensions of each term in the equation $F=a x^2+b t^{\frac{1}{2}}$, where $F$ represents force, $x$ represents distance, and $t$ represents time.

The dimension of force ($F$) is given by [MLT^-2], where $M$ is mass, $L$ is length, and $T$ is time.

The term $ax^2$ has the same dimension as force, so:

$[a] = \frac{[F]}{[x]^2} = \frac{MLT^{-2}}{L^2} = M L^{-1} T^{-2}$

The term $bt^{\frac{1}{2}}$ also has the same dimension as force, which gives:

$[b] = \frac{[F]}{[t]^{\frac{1}{2}}} = \frac{MLT^{-2}}{T^{\frac{1}{2}}} = M L T^{-\frac{5}{2}}$

Now, to find $\frac{b^2}{a}$, we substitute the dimensions of $b$ and $a$:

$\left[\frac{b^2}{a}\right] = \frac{\left(M^2 L^2 T^{-5}\right)}{\left(M L^{-1} T^{-2}\right)} = [M^1 L^3 T^{-3}]$

Therefore, the dimensions of $\frac{b^2}{a}$ are $\left[\mathrm{ML}^3 \mathrm{~T}^{-3}\right]$, which corresponds to mass times length cubed per time cubed.

Total time of flight $=\mathrm{T}$

$T=\sqrt{\frac{2 h}{g}}+\sqrt{\frac{2(H-h)}{g}}$

For max. time $=\frac{\mathrm{dT}}{\mathrm{dh}}=0$

$$\begin{aligned} & \sqrt{\frac{2}{\mathrm{~g}}}\left(\frac{-1}{2 \sqrt{\mathrm{H}-\mathrm{h}}}+\frac{1}{2 \sqrt{\mathrm{h}}}\right)=0 \\ & \sqrt{\mathrm{H}-\mathrm{h}}=\sqrt{\mathrm{h}} \\ & \mathrm{h}=\frac{\mathrm{H}}{2} \Rightarrow \frac{\mathrm{H}}{\mathrm{h}}=2 \end{aligned}$$

$$\begin{aligned} & \left(\vec{E}_{\text {net }}\right)_P=0 \\ & \frac{\mathrm{kq}}{\mathrm{x}^2}=\frac{\mathrm{k} \cdot 3 \mathrm{q}}{(\mathrm{r}-\mathrm{x})^2} \\ & (\mathrm{r}-\mathrm{x})^2=3 \mathrm{x}^2 \\ & \mathrm{r}-\mathrm{x}=\sqrt{3} \mathrm{x} \\ & \mathrm{x}=\frac{\mathrm{r}}{\sqrt{3}+1} \end{aligned}$$

Without dielectric

$\mathrm{Q}=\frac{\mathrm{A} \in_0}{\mathrm{~d}} \mathrm{~V}$

with dielectric

$Q=\frac{A \in_0 V}{d-t+\frac{t}{K}}$

given

$$\begin{aligned} & \frac{\mathrm{A} \in_0 \mathrm{~V}}{\mathrm{~d}-\mathrm{t}+\frac{\mathrm{t}}{\mathrm{K}}}=(1.25) \frac{\mathrm{A} \in_0 \mathrm{~V}}{\mathrm{~d}} \\ & \Rightarrow 1.25\left(3+\frac{2}{\mathrm{~K}}\right)=5 \\ & \Rightarrow \mathrm{K}=2 \end{aligned}$$

$$\begin{aligned} & \left|\overrightarrow{\mathrm{p}_1}\right|=\left|\overrightarrow{\mathrm{p}_2}\right| \\ & \mathrm{KE}=\frac{\mathrm{p}^2}{2 \mathrm{M}} ; \mathrm{p} \text { same } \\ & \mathrm{KE} \propto \frac{1}{\mathrm{~m}} \\ & \frac{\mathrm{KE}_1}{\mathrm{KE}_2}=\frac{\mathrm{p}^2 / 2 \mathrm{M}_1}{\mathrm{p}^2 / 2 \mathrm{M}_2}=\frac{\mathrm{M}_2}{\mathrm{M}_1} \end{aligned}$$

For M block

$$\begin{aligned} & 10 \mathrm{~g} \sin 53^{\circ}-\mu(10 \mathrm{~g}) \cos 53^{\circ}-\mathrm{T}=10 \times 2 \\ & \mathrm{~T}=80-15-20 \\ & \mathrm{~T}=45 \mathrm{~N} \end{aligned}$$

For $\mathrm{m}$ block

$$\begin{aligned} & \mathrm{T}-\mathrm{mg} \sin 37^{\circ}-\mu \mathrm{mg} \cos 37^{\circ}=\mathrm{m} \times 2 \\ & 45=10 \mathrm{~m} \\ & \mathrm{~m}=4.5 \mathrm{~kg} \end{aligned}$$

For incoherent wave $\mathrm{I}_1=\mathrm{I}_{\mathrm{A}}+\mathrm{I}_{\mathrm{B}} \Rightarrow \mathrm{I}_1=\mathrm{I}_0+9 \mathrm{I}_0$

$\mathrm{I}_1=10 \mathrm{I}_0$

For coherent wave $\mathrm{I_2=I_A+I_B+2 \sqrt{I_A I_B} \cos 60^{\circ}}$

$$\begin{aligned} & \mathrm{I}_2=\mathrm{I}_0+9 \mathrm{I}_0+2 \sqrt{9 \mathrm{I}_0^2} \cdot \frac{1}{2}=13 \mathrm{I}_0 \\ & \frac{\mathrm{I}_1}{\mathrm{I}_2}=\frac{10}{13} \end{aligned}$$

$$\begin{aligned} & \overrightarrow{\mathrm{F}}=\mathrm{q}(\overrightarrow{\mathrm{v}} \times \overrightarrow{\mathrm{B}}) \\ & 5 \mathrm{e} \hat{\mathrm{k}}=\mathrm{e}(3 \hat{\mathrm{i}}+5 \hat{\mathrm{j}}) \times\left(\mathrm{B}_0 \hat{\mathrm{i}}+2 \mathrm{~B}_0 \hat{\mathrm{j}}\right) \\ & 5 \mathrm{e} \hat{\mathrm{k}}=\mathrm{e}\left(6 \mathrm{~B}_0 \hat{\mathrm{k}}-5 \mathrm{~B}_0 \hat{\mathrm{k}}\right) \\ & \Rightarrow \mathrm{B}_0=5 \mathrm{~T} \end{aligned}$$

\begin{aligned} & \mathrm{E}=\phi+\mathrm{K}_{\max } \\ & \phi=\frac{\mathrm{hc}}{\lambda_0} \\ & \mathrm{~K}_{\max }=\mathrm{eV}_0 \\ & 8 \mathrm{e}=\frac{\mathrm{hc}}{\lambda}-\frac{\mathrm{hc}}{\lambda_0} \ldots . . \text { (i) } \\ & 2 \mathrm{e}=\frac{\mathrm{hc}}{3 \lambda}-\frac{\mathrm{hc}}{\lambda_0} \ldots . . . \text { (ii) } \\ & \text { on solving (i) & (ii) } \\ & \lambda_0=9 \lambda \end{aligned}

$$\begin{aligned} \varepsilon & =\mathrm{N}\left(\frac{\Delta \phi}{\mathrm{t}}\right) \\ \Delta \phi & =(\Delta \mathrm{B}) \mathrm{A} \\ \mathrm{B}_{\mathrm{i}} & =5000 \mathrm{~T}, \\ \mathrm{~B}_{\mathrm{f}} & =3000 \mathrm{~T} \\ \mathrm{~d} & =0.02 \mathrm{~m} \\ \mathrm{r} & =0.01 \mathrm{~m} \\ \Delta \phi & =(\Delta \mathrm{B}) \mathrm{A} \\ & =(2000) \pi(0.01)^2=0.2 \pi \\ \varepsilon & =\mathrm{N}\left(\frac{\Delta \phi}{\mathrm{t}}\right) \Rightarrow 22=\mathrm{N}\left(\frac{0.2 \pi}{2}\right) \\ \mathrm{N} & =70 \end{aligned}$$

The two isobaric processes depicted in the figure show changes in the volume of an ideal gas with temperature under constant pressure conditions.

Isobaric processes follow the equation $PV = nRT$, where P is pressure, V is volume, n is the number of moles of the gas, R is the ideal gas constant, and T is temperature.

When we rearrange the equation in terms of V (Volume), we get $V = \left(\frac{nR}{P}\right)T$. The slope of the volume-temperature graph in such a scenario is given by $\frac{nR}{P}$. This implies that the slope is inversely proportional to the pressure (i.e., slope $\propto \frac{1}{P}$).

From this relationship, if one process has a higher slope compared to another, its corresponding pressure must be lower. Observing the given figure and applying this understanding, if Process 2 has a greater slope than Process 1, it indicates that $P_2 < P_1$.

$$\begin{aligned} & F_{\text {net }}=\sqrt{2} F+F^{\prime} \\ & F=\frac{G m^2}{a^2} \text { and } F^{\prime}=\frac{G^2}{(\sqrt{2} \mathrm{a})^2} \\ & F_{\text {net }}=\sqrt{2} \frac{\mathrm{Gm}^2}{\mathrm{a}^2}+\frac{\mathrm{Gm}^2}{2 \mathrm{a}^2} \\ & \left(\frac{2 \sqrt{2}+1}{32}\right) \frac{\mathrm{Gm}^2}{\mathrm{~L}^2}=\frac{\mathrm{Gm}^2}{\mathrm{a}^2}\left(\frac{2 \sqrt{2}+1}{2}\right) \\ & \mathrm{a}=4 \mathrm{~L} \end{aligned}$$

$$\begin{aligned} & \frac{1}{\lambda}=\frac{13.6 \mathrm{z}^2}{\mathrm{hc}}\left[\frac{1}{1^2}-\frac{1}{2^2}\right] ...... \text{(i)}\\ & \frac{1}{\lambda^{\prime}}=\frac{13.6 \mathrm{z}^2}{\mathrm{hc}}\left[\frac{1}{1^2}-\frac{1}{3^2}\right] ...... \text{(ii)} \end{aligned}$$

On dividing (i) & (ii)

$\lambda^{\prime}=\frac{27}{32} \lambda$

$$\begin{aligned} & \mathrm{R}=\frac{\rho \mathrm{L}}{\pi \frac{\mathrm{d}^2}{4}} \\ & \frac{\Delta \mathrm{R}}{\mathrm{R}}=\frac{\Delta \mathrm{L}}{\mathrm{L}}+\frac{2 \Delta \mathrm{d}}{\mathrm{d}} \\ & \frac{\Delta \mathrm{L}}{\mathrm{L}}=0.1 \% \text { and } \frac{\Delta \mathrm{d}}{\mathrm{d}}=0.1 \% \\ & \frac{\Delta \mathrm{R}}{\mathrm{R}}=0.3 \% \end{aligned}$$

$$\begin{aligned} & \mathrm{E}=\Delta \mathrm{mc}^2 \\ & =0.4 \times 10^{-3} \times\left(3 \times 10^8\right)^2 \\ & =3600 \times 10^7 \mathrm{kWs} \\ & =\frac{3600 \times 10^7}{3600} \mathrm{kWh}=1 \times 10^7 \mathrm{kWh} \end{aligned}$$

$\mathrm{NH}_3$ act as WFL with $\mathrm{Ni}^{2+}$

$\mathrm{Ni}^{2+}=3 \mathrm{~d}^8$

No. of unpaired electron $=2$

$$\begin{aligned} \mu=\sqrt{\mathrm{n}(\mathrm{n}+2)} & =\sqrt{8}=2.82 \mathrm{~BM} \\ & =28.2 \times 10^{-1} \mathrm{~BM} \\ \mathrm{x} & =28 \end{aligned}$$

Product benzene has zero hydroxyl group