MHT CET 2023 11th May Evening Shift (Physics & Chemistry) Question Paper with Solutions

$$\begin{aligned} \frac{1}{\alpha}-\frac{1}{\beta} & =\frac{1}{0.98}-\frac{1}{4.9} \\ & =1.204-0.204=1 \end{aligned}$$

Given: $\mathrm{v}=320 \mathrm{~m} / \mathrm{s}, \mathrm{f}=480 \mathrm{~Hz}, \mathrm{~N}=180$

$$\begin{aligned} \mathrm{v} & =\mathrm{f} \lambda \\ \therefore \quad \lambda & =\frac{\mathrm{v}}{\mathrm{f}} \end{aligned}$$

Substituting the values, we get

$\lambda=\frac{320}{480}=\frac{2}{3}$

$\therefore \quad$ The total distance covered after 180 vibrations is

$$\begin{aligned} & D=N \times \lambda \\ & D=180 \times \frac{2}{3} \\ & D=120 \mathrm{~m} \end{aligned}$$

The formula for excess pressure inside the bubble is given as $P_0=\frac{4 T}{R}$

Where, $\mathrm{T}$ is surface tension and $\mathrm{R}$ is the radius of the sphere

Rearranging the formula for $\mathrm{R}$ and substituting the values,

$$\begin{aligned} & R=\frac{4 T}{P_0} \\ & R=\frac{4 \times 3.2 \times 10^{-2}}{25.6} \\ & R=0.5 \times 10^{-2} \mathrm{~m}=0.5 \mathrm{~cm} \end{aligned}$$

The diameter then becomes, $2 \mathrm{R}=1 \mathrm{~cm}$

The equation for self-inductance is:

$\mathrm{L}=\frac{\mu_0 \mathrm{~N}^2 \mathrm{~A}}{l}$

$\therefore \quad$ The self-inductance $\mathrm{L}$ of a solenoid of length $l$ and area of cross section '$A$', with a fixed number of turns '$\mathrm{N}$' increases as $l$ decreases and A increases.

$$\begin{aligned} & \mathrm{y}_1=\mathrm{a}_1 \sin \left(\omega \mathrm{t}-\frac{2 \pi \mathrm{x}}{\lambda}\right) \\ & \mathrm{y}_2=\mathrm{a}_2 \cos \left(\omega \mathrm{t}-\frac{2 \pi \mathrm{x}}{\lambda}+\phi\right) \end{aligned}$$

$\mathrm{y}_2$ can also be written as

$$\begin{aligned} \mathrm{y}_2 & =\mathrm{a}_2 \sin \left[\frac{\pi}{2}+\left(\omega \mathrm{t}-\frac{2 \pi \mathrm{x}}{\lambda}+\phi\right)\right] \\ & =\mathrm{a}_2 \sin \left(\omega \mathrm{t}-\frac{2 \pi \mathrm{x}}{\lambda}+\phi+\frac{\pi}{2}\right) \end{aligned}$$

The phase difference between the two waves is

Path difference $=\frac{\lambda}{2 \pi} \times$ Phase difference

Path difference $=\frac{\lambda}{2 \pi} \times\left(\phi+\frac{\pi}{2}\right)$

Moment of inertia of disc is given by

$\mathrm{I}_{\text {disc }}=\mathrm{I}_{\mathrm{r}}+\mathrm{I}_{\text {hole }}$ $\ldots .\left\{I_r=\right.$ M.I. of remaining part $\}$

$\therefore \quad \mathrm{I}_{\mathrm{r}}=\mathrm{I}_{\text {disc }}-\mathrm{I}_{\text {hole }}$ ..... (i)

$\mathrm{I}_{\mathrm{disc}}=\frac{\mathrm{MR}^2}{2}$ ..... (ii)

By parallel axes theorem we get,

$\mathrm{I}_{\text {hole }}=\left[\frac{\frac{\mathrm{M}}{4}\left(\frac{\mathrm{R}}{2}\right)^2}{2}+\frac{\mathrm{M}}{4}\left(\frac{\mathrm{R}}{2}\right)^2\right]$

$$\ldots .\left\{\begin{array}{l}\because \mathrm{M}_{\text {hole }}=\frac{\mathrm{M}_{\text {dise }}}{4} \\ \because \text { the surface density is same }\end{array}\right\}$$

$\therefore \quad \mathrm{I}_{\text {hole }}=\left[\frac{\mathrm{MR}^2}{32}+\frac{\mathrm{MR}^2}{16}\right]$ ..... (iii)

Substituting eq (iii) and eq (ii) in eq (i) we get,

$$\begin{aligned} \mathrm{I}_{\mathrm{r}} & =\frac{\mathrm{MR}^2}{2}-\frac{\mathrm{MR}^2}{32}-\frac{\mathrm{MR}^2}{16} \\ & =\mathrm{MR}^2\left[\frac{1}{2}-\frac{1}{32}-\frac{1}{16}\right] \\ & =\frac{13}{32} \mathrm{MR}^2 \end{aligned}$$

$\therefore\quad$ Orbital magnetic moment:

$M_0=\frac{-e}{2 m_e} L$

$\therefore\quad$ The orbital magnetic moment is directly proportional to angular momentum L of the electron

According to Faraday's law of electromagnetic induction,

$$\begin{aligned} & \varepsilon=\frac{\Delta \phi}{\Delta t} \\ & \mathrm{IR}=\frac{\Delta \phi}{\Delta \mathrm{t}} \\ & \mathrm{I}=\frac{\Delta \phi}{\Delta \mathrm{t} \times \mathrm{R}} \\ & \mathrm{I} \times \Delta \mathrm{t}=\frac{\Delta \phi}{\mathrm{R}} \end{aligned}$$

$\therefore$ The total quantity of electric charge passing through the circuit is

$\mathrm{Q}=\frac{\Delta \phi}{\mathrm{R}}$

The angle of incidence is $\mathrm{i}=45^{\circ}$

The angle of deviation is $\delta=15^{\circ}$

The angle of refraction is

$$\begin{aligned} & \delta=i-r \\ & r=i-\delta \\ & r=45^{\circ}-15^{\circ} \\ & r=30^{\circ} \end{aligned}$$

The relation between the speed of the wave in the medium, the angle of incidence and the angle of refraction is given by the formula:

$$\begin{aligned} & \frac{v_2}{v_1}=\frac{\sin r}{\sin i} \\ & \frac{v_2}{v_1}=\frac{\sin 30^{\circ}}{\sin 45^{\circ}} \\ & \frac{v_2}{v_1}=\frac{1}{\sqrt{2}}\\ & \begin{aligned} & \mathrm{v}_2=\frac{\mathrm{v}_1}{\sqrt{2}} \\ & \mathrm{v}_2=\frac{3 \times 10^8}{\sqrt{2}} \\ & \mathrm{v}_2=2.1 \times 10^8 \mathrm{~m} / \mathrm{s} \end{aligned} \end{aligned}$$

The equation for critical velocity is given as: $v=k \frac{\eta}{\rho r}$

$\therefore$ The critical velocity is directly proportional to $\eta$.

The formula for frequency of a sonometer is $\mathrm{f}=\frac{1}{2 l} \sqrt{\frac{\mathrm{T}}{\pi \rho \mathrm{D}^2}}$

Here $l$ is length, T is tension, D is diameter and $\rho$ is density.

The frequency of both the wires is same.

The frequency of the wire $A$ is $f_A=\frac{1}{2 l} \sqrt{\frac{T}{\pi \rho_1 D^2}}$

The frequency of the wire $B$ is $f_B=\frac{1}{2 l} \sqrt{\frac{2 T}{\pi \rho_2(2 D)^2}}$

Equating both the frequencies

$$\begin{aligned} & \frac{1}{2 l} \sqrt{\frac{\mathrm{T}}{\pi \rho \mathrm{D}^2}}=\frac{1}{2 l} \sqrt{\frac{2 \mathrm{~T}}{\pi \rho_2(2 \mathrm{D})^2}} \\ & \sqrt{\frac{1}{\rho_1}}=\sqrt{\frac{1}{2 \rho_2}} \\ & \frac{1}{\rho_1}=\frac{1}{2 \rho_2} \\ \therefore \quad & \rho_2=\frac{\rho_1}{2} \end{aligned}$$

According to Bohr's theory of hydrogen atom, the equation for total energy of the electron in the $\mathrm{n}^{\text {th }}$ stationary orbit is,

$$\begin{aligned} & \mathrm{E}_{\mathrm{n}}=\frac{-m \mathrm{Z}^2 \mathrm{e}^4}{8 \varepsilon_0{ }^2 \mathrm{~h}^2 \mathrm{n}^2} \\ \therefore \quad & \mathrm{E}_{\mathrm{n}} \propto \frac{1}{\mathrm{n}^2} \end{aligned}$$

The equation for angular momentum is

$\mathrm{L}=\sqrt{2 \mathrm{~K}_{\text {Rot }} \times \mathrm{I}}$

So, $\mathrm{L} \propto \sqrt{\mathrm{I}}$

$\therefore \quad$ The ratio of angular momentum of the two bodies is

$\frac{\mathrm{L}_1}{\mathrm{~L}_2}=\sqrt{\frac{\mathrm{I}_1}{\mathrm{I}_2}}$

$\frac{\mathrm{L}_1}{\mathrm{~L}_2} =\sqrt{\frac{\mathrm{I}}{2 \mathrm{I}}} \quad .... (\text{given} ~\mathrm{I}_2=2 \mathrm{I})$

$\therefore \quad \frac{\mathrm{L}_1}{\mathrm{~L}_2} =\frac{1}{\sqrt{2}}$

For bifocal convex lens

$$\begin{aligned} \frac{1}{f} & =(\mu-1)\left(\frac{1}{R_1}+\frac{1}{R_2}\right) \\ & =\frac{(\mu-1) \times 2}{R} \quad \ldots .\left(R_1=R_2=R\right) \end{aligned}$$

For plane surface $R_2=\infty$

For half plane-convex lens

$$\begin{aligned} & \frac{1}{\mathrm{f}^{\prime}}=(\mu-1) \frac{1}{\mathrm{R}} \\ & \frac{1 / \mathrm{f}}{1 / \mathrm{f}^{\prime}}=\frac{(\mu-1)}{\mathrm{R}} \times 2 \times \frac{\mathrm{R}}{\mu-1}=2 \\ & \frac{\mathrm{f}}{\mathrm{f}}=2 \\ & \mathrm{f}^{\prime}=2 \mathrm{f} \end{aligned}$$

As focal length is $\mathrm{F}, \mathrm{f}^{\prime}=2 \mathrm{~F}$.

The formula for the time period of a spring mass system is $T=2 \pi \sqrt{\frac{m}{k}}$

For mass $\mathrm{m}+1, \mathrm{~T}^{\prime}=2 \pi \sqrt{\frac{\mathrm{m}+1}{\mathrm{k}}}$

Taking the ratio,

$$\begin{aligned} \frac{\mathrm{T}}{\mathrm{T}^{\prime}} & =\frac{2 \pi \sqrt{\frac{\mathrm{m}}{\mathrm{k}}}}{2 \pi \sqrt{\frac{\mathrm{m}+1}{\mathrm{k}}}} \\ \frac{\mathrm{T}}{\mathrm{T}^{\prime}} & =\sqrt{\frac{\mathrm{m}}{\mathrm{m}+1}} \\ \frac{3}{5} & =\sqrt{\frac{\mathrm{m}}{\mathrm{m}+1}} \\ \frac{9}{25} & =\frac{\mathrm{m}}{\mathrm{m}+1} \\ \therefore \quad 9 \mathrm{~m} & +9=25 \mathrm{~m} \\ \therefore \quad 16 \mathrm{~m} & =9 \\ \therefore \quad \mathrm{m} & =\frac{9}{16} \mathrm{~kg} \end{aligned}$$

The force per unit length of the conductors is given as:

$\mathrm{F}=\frac{\mu_0 \mathrm{I}_1 \mathrm{I}_2}{2 \pi \mathrm{d}}$

When the value and direction of current in the first conductor and the distance between the conductors are changed,

$$\begin{aligned} \therefore \quad \mathrm{F}_2 & =\frac{-\mu_0 2 \mathrm{I}_1 \mathrm{I}_2}{2 \pi \times 3 \mathrm{~d}} \\ \therefore \quad & \frac{\mathrm{F}_2}{\mathrm{~F}}=\frac{-2}{3} \\ \mathrm{~F}_2 & =-\frac{2 \mathrm{~F}}{3} \end{aligned}$$

Let total current be denoted as I. The given circuit is a Wheatstone bridge.

$$\begin{aligned} & \Rightarrow \mathrm{R}_1=6+3=9 \Omega \\ & \Rightarrow \mathrm{R}_2=8+4=12 \Omega \end{aligned}$$

According to $\mathrm{KCL}$, the current will get divided into two parts $I_1$ and $I_2$

$\therefore \quad \mathrm{I}_1=\frac{\mathrm{R}_2 \mathrm{I}}{\left(\mathrm{R}_1+\mathrm{R}_2\right)}$

Substituting the values,

$$\begin{aligned} & I_1=\frac{12}{9+12} \times 5 \\ & I_1=2.85 \mathrm{~A} \end{aligned}$$

Potential difference between $\mathrm{P}$ and $\mathrm{Q}$ is

$$\begin{aligned} & V=I_1 R \\ & V=2.85 \times 6 \\ & V=17 V \end{aligned}$$

$\therefore \quad$ e.m.f e $=200 \sin (50 \mathrm{t}) \mathrm{~V}$ .... (i)

$\therefore \quad$ Current is given by:

$\mathrm{I}_{\mathrm{o}}=\frac{\mathrm{e}_{\mathrm{o}}}{\mathrm{R}}=\frac{200}{50}=4 \mathrm{~A}$ .... (from (i))

$\therefore \quad$ R.M.S value of current:

$\mathrm{I}_{\mathrm{r} \mathrm{ms}}=\frac{\mathrm{I}_{\mathrm{o}}}{\sqrt{2}}=\frac{4}{\sqrt{2}}=2.828 \mathrm{~A}$

The magnetic field due to current carrying circular arc is $B=\frac{\mu_0 I}{2 r}\left(\frac{\theta}{2 \pi}\right)$

Here, $\theta=\frac{\pi}{16}$

$$\begin{aligned} \therefore \quad \mathrm{B} & =\frac{\mu_0 \mathrm{I}}{2 \mathrm{r}}\left(\frac{1}{2 \pi} \times \frac{\pi}{16}\right) \\ \mathrm{B} & =\frac{\mu_0 \mathrm{I}}{2 \mathrm{r}}\left(\frac{1}{32}\right) \\ \mathrm{B} & =\frac{\mu_0 \mathrm{I}}{64 \mathrm{r}} \end{aligned}$$

From Stefan - Boltzmann's law

$\frac{\mathrm{dQ}}{\mathrm{dt}}=\mathrm{e}\left(\sigma \mathrm{AT}^4\right)$

Given $\mathrm{A}$ and $\frac{\mathrm{dQ}}{\mathrm{dt}}$ are same for both the bodies

$$\begin{aligned} & \Rightarrow \mathrm{e}_1 \mathrm{~T}_1^4=\mathrm{e}_2 \mathrm{~T}_2^4 \\ \therefore \quad & \left(\frac{\mathrm{T}_1}{\mathrm{~T}_2}\right)^4=\frac{\mathrm{e}_2}{\mathrm{e}_1}=\frac{3}{1} \\ & \Rightarrow \frac{\mathrm{T}_1}{\mathrm{~T}_2}=\frac{\sqrt[4]{3}}{1}=\frac{3^{\frac{1}{4}}}{1} \end{aligned}$$

The emf across the $\mathrm{AC}$ source is given as:

$$\begin{aligned} \mathrm{e} & =\sqrt{\left(\mathrm{V}_{\mathrm{R}}\right)^2+\left(\mathrm{V}_{\mathrm{C}}-\mathrm{V}_{\mathrm{L}}\right)^2}=\sqrt{(40)^2+(80-40)^2} \\ \mathrm{e} & =\sqrt{3200} \\ \therefore \quad \mathrm{e} & =40 \sqrt{2} \mathrm{~V} \end{aligned}$$

The formula for induced emf is $\mathrm{e}=\frac{\Delta \phi}{\Delta \mathrm{t}}$, where $\phi=\mathrm{BA}$

Here, the area is constant and the magnetic field is changing.

$$\begin{aligned} & \therefore & \Delta \phi & =\Delta \mathrm{BA} \\ & \therefore & \Delta \phi & =\mathrm{A} \cdot \Delta \mathrm{B} \\ & \therefore & \Delta \mathrm{B} & =\mathrm{B}_1-\mathrm{B}_2 \end{aligned}$$

$\mathrm{B}_1=\mathrm{B} \text { and } \mathrm{B}_2=\frac{25}{200} \mathrm{~B}=\frac{1}{4} \mathrm{~B}$

$$\begin{aligned} & \therefore \quad \mathrm{B}=\mathrm{B}-\frac{1}{4} \mathrm{~B} \\ & \therefore \quad \mathrm{B}=\frac{3}{4} \mathrm{~B} \end{aligned}$$

Substituting the values,

$$\begin{aligned} \mathrm{e} & =\frac{\Delta \phi}{\Delta \mathrm{t}} \\ \mathrm{e} & =\frac{\mathrm{A} \times \frac{3}{4} \mathrm{~B}}{1} \\ \therefore \quad \mathrm{e} & =\frac{3}{4} \mathrm{AB} \end{aligned}$$

When a $𝑝$-$𝑛$ junction diode is forward biased, the external applied voltage opposes the built-in potential barrier across the junction.

This reduction in the effective barrier potential pushes majority carriers towards the junction, thereby decreasing the width of the depletion layer.

Hence, the correct option is (D).

The given equation is $y=A \sin 2(\omega t-k x)$

$\therefore \quad$ Velocity of the particle, $v=\frac{d y}{d t}$

$$\begin{array}{ll} & =2 \mathrm{~A} \omega \cos 2(\omega \mathrm{t}-\mathrm{kx}) \\ \therefore \quad & \text { Maximum velocity }=2 \mathrm{~A} \omega \\ & \text { Velocity of the wave }=\frac{\omega}{\mathrm{k}} \end{array}$$

$$\begin{aligned} & \text { Given } 2 \mathrm{~A} \omega=\frac{\omega}{\mathrm{k}} \\ & \therefore \quad \mathrm{A}=\frac{1}{2 \mathrm{k}}=\frac{\lambda}{(2 \pi)^2}=\frac{\lambda}{4 \pi} \end{aligned}$$

$$\begin{aligned} & \operatorname{mr} \omega^2=\frac{G M m}{r^2} \\ & \omega^2=\frac{G M}{r^3}=\frac{g R^2}{r^3} \quad \ldots .\left(\because g=\frac{G M}{R^2}\right) \end{aligned}$$

$\therefore \quad$ Radius of the orbit of the satellite is:

$r=\left(\frac{g R^2}{\omega^2}\right)^{\frac{1}{3}}$

The distance between the central bright fringe and the first dark fringe is given as:

$$\begin{aligned} y_{n_d} & =\frac{n \lambda D}{d} \\ y_{1_d} & =\frac{1 \times 600 \times 10^{-9} \times 2}{10^{-3}} \\ & =1.2 \times 10^{-3}=1.2 \mathrm{~mm} \end{aligned}$$

$\therefore \quad$ The distance between the first dark fringe on either side of the central bright fringe is $2 \mathrm{y}_{\mathrm{n}}=2 \times 1.2=2.4 \mathrm{~mm}$

The units of PV can be calculated as follows:

The unit of pressure is $\mathrm{Kg} \mathrm{m}^{-1} \mathrm{~s}^{-2}$

The unit of volume is $\mathrm{m}^3$

$\therefore\quad$ The unit of PV is

Unit $=\mathrm{Kgm}^{-1} \mathrm{~s}^{-2} \mathrm{~m}^3=\mathrm{Kgm}^2 \mathrm{~s}^{-2}$

This is a unit of energy.

For stopping potential $\mathrm{V}, \mathrm{eV}=\frac{\mathrm{hc}}{\lambda}-\frac{\mathrm{hc}}{\lambda_0}$

For stopping potential $\frac{\mathrm{V}}{4}, \frac{\mathrm{eV}}{4}=\frac{\mathrm{hc}}{2 \lambda}-\frac{\mathrm{hc}}{\lambda_0}$

Taking the ratio, we get

$$\begin{aligned} & 4=\frac{\left(\frac{1}{\lambda}-\frac{1}{\lambda_0}\right)}{\frac{1}{2 \lambda}-\frac{1}{\lambda_0}} \\ & 4\left(\frac{1}{2 \lambda}-\frac{1}{\lambda_0}\right)=\frac{1}{\lambda}-\frac{1}{\lambda_0} \\ & \left(\frac{2}{\lambda}-\frac{4}{\lambda_0}\right)=\frac{1}{\lambda}-\frac{1}{\lambda_0} \\ & \left(\frac{2}{\lambda}-\frac{1}{\lambda}\right)=-\frac{1}{\lambda_0}+\frac{4}{\lambda_0} \\ & \quad\left(\frac{1}{\lambda}\right)=\frac{3}{\lambda_0} \\ & \therefore \quad \lambda_0=3 \lambda \end{aligned}$$

We know

$\mathrm{V}_{\mathrm{rms}}=\sqrt{\frac{3 \mathrm{RT}}{\mathrm{m}}}$

The temperature of the same gas molecule is raised.

$\mathrm{V}_{\mathrm{ms}} \propto \sqrt{\mathrm{T}}$

$\therefore \quad$ The ratio of the velocities is

$$\begin{aligned} & \frac{\mathrm{V}_1}{\mathrm{~V}_2}=\frac{\sqrt{\mathrm{T}_1}}{\sqrt{\mathrm{T}_2}} \\ & \frac{\mathrm{V}_1}{\mathrm{~V}_2}=\frac{\sqrt{400}}{\sqrt{800}} \\ & \frac{\mathrm{V}_1}{\mathrm{~V}_2}=\frac{1}{\sqrt{2}} \end{aligned}$$

Convert temperatures given in centigrade to Kelvin before calculation.

Energy of incident light:

$$\begin{aligned} E & =\frac{h c}{\mathrm{e} \lambda} \\ & =\frac{1240 \times 10^{-9}}{310 \times 10^{-9}} \\ & =4 \mathrm{~ev} \end{aligned}$$

Stopping potential $\mathrm{V}_0=\frac{\mathrm{hc}}{\mathrm{e} \lambda}-\frac{\phi_0}{\mathrm{e}}$

$\mathrm{V}_0=4-1.13=2.87 \mathrm{~V}$

We know,

$$\begin{aligned} \mathrm{mvr} & =\frac{\mathrm{nh}}{2 \pi} \\ \therefore \quad \mathrm{vr} & =\frac{\mathrm{nh}}{2 \pi \mathrm{m}}\quad \text{.... (i)} \end{aligned}$$

Also,

$$\begin{aligned} & \mathrm{qvB}=\frac{\mathrm{mv}^2}{\mathrm{r}} \\ & \therefore \quad \mathrm{mv}=\mathrm{qBr} \quad \text{.... (ii)}\\ & \mathrm{mv}^2 \mathrm{r}=\mathrm{qBr} \times \frac{\mathrm{nh}}{2 \pi \mathrm{m}} \quad \text{....(Multiplying (i) with (ii))}\\ & \mathrm{E}=\frac{1}{2} \mathrm{mv}^2=\mathrm{n}\left[\frac{\mathrm{qBh}}{4 \pi \mathrm{m}}\right] \end{aligned}$$

We know,

$$\begin{aligned} & V_{\text {rms }}=\sqrt{\frac{3 R T}{M_0}} \\ & \Rightarrow T \propto V^2 \\ & \frac{T_2}{T_1}=\frac{V_2^2}{V_1^2}=\frac{\left(\frac{V_1}{3}\right)^2}{V_1^2}=\frac{1}{9} \quad \text{.... (i)}\\ & \text { Also, } T V^{\gamma-1}=\text { Constant } \\ & \Rightarrow \frac{V_2}{V_1}=\left(\frac{T_1}{T_2}\right)^{\frac{1}{\gamma-1}} \\ & \frac{V_2}{V_1}=(9)^2=81 \end{aligned}$$

The potential difference across the plates of the capacitor is maintained constant at $𝑉$ by the connected battery.