MHT CET 2021 21st September Morning Shift (Physics & Chemistry) Question Paper with Solutions

Magnetic field inside a solenoid is given by

$\mathrm{B}=\mu_0 \mathrm{nI}$

If $\mathrm{n}$ is doubled and $\mathrm{I}$ is made $\left(\frac{1}{3}\right)^{\mathrm{rd}}$, the value of $\mathrm{B}$ will become $\frac{2}{3} \mathrm{~B}$.

Magnetic potential energy $\mathrm{U}=\frac{1}{2} \mathrm{LI}^2$

$$\begin{aligned} \therefore \mathrm{L} & =\frac{2 \mathrm{U}}{\mathrm{I}^2}=\frac{2 \times 25 \times 10^{-3}}{\left(50 \times 10^{-3}\right)^2} \\ & =20 \mathrm{H} \end{aligned}$$

Initial current $\mathrm{I}=\frac{\mathrm{V}}{\mathrm{z}}$, Final current $\mathrm{I}^{\prime}=\frac{\mathrm{V}}{\mathrm{z}}$

$$\begin{aligned} & \frac{1}{2}=\frac{\mathrm{I}^{\prime}}{\mathrm{I}}=\frac{\mathrm{z}}{\mathrm{z}^{\prime}} \quad \therefore \frac{\mathrm{z}}{\mathrm{z}^{\prime}}=\frac{1}{2} \\ & \therefore \frac{\mathrm{z}^2}{\mathrm{z}^{\prime 2}}=\frac{1}{4} \quad \therefore \frac{\mathrm{R}^2+\mathrm{X}_{\mathrm{c}}^2}{\mathrm{R}^2+\mathrm{X}_{\mathrm{c}}^{\prime 2}}=\frac{1}{4} \\ & \therefore 4 \mathrm{R}^2+4 \mathrm{X}_{\mathrm{c}}^2=\mathrm{R}^2+\mathrm{X}_{\mathrm{c}}^{\prime 2} \\ & \therefore 3 \mathrm{R}^2=\mathrm{X}_{\mathrm{c}}^{\prime 2}-4 \mathrm{X}_{\mathrm{c}}^2 \quad \text{......(1)} \end{aligned}$$

$\mathrm{X}_{\mathrm{c}}=\frac{1}{\omega C}, \mathrm{X}^{\prime}=\frac{3}{\omega C} \quad \therefore \mathrm{X}^{\prime}=3 \mathrm{X}_{\mathrm{c}}$

Putting this value of $X^{\prime}$ in (1) we get

$3 R^2=9 X_c^2-4 X_c^2=5 X_c^2$

$$\begin{aligned} & \therefore \frac{\mathrm{X}_{\mathrm{c}}^2}{\mathrm{R}^2}=\frac{3}{5}=0.6 \\ & \therefore \frac{\mathrm{X}_{\mathrm{c}}}{\mathrm{R}}=\sqrt{0.6} \end{aligned}$$

The kinetic energy of the photo electrons and hence the stopping potential does not depend on the intensity of light.

$$\begin{aligned} & 21 \times \frac{\lambda_1 D}{d}=N \times \frac{\lambda_2 D}{d} \\ & \therefore N=21 \frac{\lambda_1}{\lambda_2}=\frac{21 \times 4800}{5600}=18\end{aligned}$$

Let $\mathrm{v}$ be the velocity of mass $\mathrm{m}$ and $\mathrm{v}^{\prime}$ be the velocity of mass $\mathrm{M}$ after collision. By law of conservation of momentum

$$\begin{aligned} \mathrm{mv} & =\mathrm{Mv}^{\prime} \\ \therefore \frac{\mathrm{v}^{\prime}}{\mathrm{v}}=\frac{\mathrm{m}}{\mathrm{M}} & \\ \text { Coefficient of restitutions } & =\frac{\text { Relative velocity after collision }}{\text { Re lative velocity before collision }} \\ & =\frac{\mathrm{v}^{\prime}}{\mathrm{v}} \end{aligned}$$

$\mathrm{n}=50 \mathrm{~Hz}, \mathrm{t}=0.01 \mathrm{~s}$

If $\mathrm{T}$ is the period and phase difference is $\phi$

$$\begin{aligned} & \text { then, } \frac{\mathrm{t}}{\mathrm{T}}=\frac{\phi}{2 \pi} \\ & \therefore \phi=2 \pi \frac{\mathrm{t}}{\mathrm{T}}=2 \pi \mathrm{nt} \\ & \therefore \phi=2 \pi \times 50 \times 0.01=\pi \mathrm{~rad} \end{aligned}$$

Given, mass of the disc $𝑀=1\text{kg}$, radius $𝑅=2\text{m}$, and moment of inertia about the axis through the centre of mass ${𝐼}_{\text{cm}}=2{\text{kg m}}^2$.

The required axis is parallel to the axis $\text{XY}$ and passes through the edge of the disc, so the distance between the axes is $𝑑=𝑅=2\text{m}$.

By the parallel axes theorem,

$𝐼={𝐼}_{\text{cm}}+𝑀{𝑑}^2$

$𝐼=2+\left(1\right)(2{)}^2=6{\text{kg m}}^2$

Hence, the correct option is (A).

Force due to electric field $\vec{F}_e=q \vec{E}$

Force due to magnetic field $\overrightarrow{\mathrm{F}}_{\mathrm{m}}=\mathrm{q}(\overrightarrow{\mathrm{v}} \times \overrightarrow{\mathrm{B}})$

$\therefore$ Total force $\overrightarrow{\mathrm{F}}=\mathrm{q} \overrightarrow{\mathrm{E}}+\mathrm{q}(\overrightarrow{\mathrm{v}} \times \overrightarrow{\mathrm{B}})$

Magnetic field at the centre $B=\frac{\mu_0 I}{R}$

$\phi=$ Magnetic flux passing through the smaller coil $=\pi \mathrm{r}^2 \mathrm{~B}$

$\therefore \phi=\pi r^2 \times \frac{\mu_0 I}{R}$

$\therefore M=\frac{\phi}{I}=\frac{\mu \pi r^2}{R} \quad \therefore M \propto \frac{r^2}{R}$

$$\begin{aligned} & \frac{5}{R}=\frac{\ell_1}{100 \ell_1} \quad \text{.......(1)}\\ & \frac{5 \times 2}{R}=\frac{1.6 \ell_1}{100-1.6 \ell} \\ & \therefore \quad \frac{5}{R}=\frac{0.8 \ell_1}{100-1.6 \ell_1} \quad \text{.......(2)} \end{aligned}$$

Equating (1) and (2) and solving we get $\ell_1=25 \mathrm{~cm}$ Putting this in Eq.(1), we get $\mathrm{R}=15 \Omega$

Surface tension is caused by cohesive intermolecular forces between liquid molecules. As the temperature of the liquid rises, the kinetic energy of the molecules increases, which weakens the intermolecular cohesive forces. Consequently, the surface tension of most liquids decreases with a rise in temperature.

Hence, the correct option is (C).

$$\begin{aligned} & \mathrm{C}=\frac{\mathrm{KA} \varepsilon_0}{\mathrm{~d}} \\ & \therefore \frac{\mathrm{C}_2}{\mathrm{C}_1}=\frac{\mathrm{k}_2}{\mathrm{k}_1} \cdot \frac{\mathrm{d}_1}{\mathrm{~d}_2}\end{aligned}$$

$$\begin{aligned} & \therefore \frac{6}{2}=\frac{\mathrm{k}_2}{1} \cdot \frac{1}{2} \\ & \therefore \mathrm{k}_2=6\end{aligned}$$

From the figure $A=r$

$\mathrm{n}=\frac{\sin \mathrm{r}}{\sin \mathrm{r}}=\frac{\sin \mathrm{i}}{\sin \mathrm{A}}=\frac{\mathrm{i}}{\mathrm{A}}$

(Angles are small)

$\therefore \mathrm{i}=\mathrm{An}$

Maximum particle velocity $=Y_0 \omega=2 \pi n Y_0$

Wave velocity $=n \lambda$

$$\begin{aligned} & \therefore \frac{2 \pi n Y_0}{8}=n \lambda \\ & \therefore \lambda=\frac{\pi Y_0}{4} \end{aligned}$$

$$\begin{aligned} & \mathrm{g}^{\prime}=\mathrm{g} \frac{\mathrm{R}^2}{(\mathrm{R}+\mathrm{h})^2} \quad \mathrm{~g}^{\prime}=\frac{\mathrm{g}}{3} \\ & \therefore \frac{1}{3}=\frac{\mathrm{R}^2}{(\mathrm{R}+\mathrm{h})^2} \quad \therefore \frac{1}{\sqrt{3}}=\frac{\mathrm{R}}{\mathrm{R}+\mathrm{h}} \\ & \therefore \sqrt{3} \mathrm{R}=\mathrm{R}+\mathrm{h} \\ & \therefore \mathrm{h}=(\sqrt{3}-1) \mathrm{R}\end{aligned}$$

$\mathrm{E}=\mathrm{e} \sigma \cdot \mathrm{A}\left(\mathrm{T}^4-\mathrm{T}_0^4\right) \text { and } \mathrm{A}=\ell \mathrm{b}$

When $\ell$ and $b$ change to $\frac{\ell}{3}$ and $\frac{b}{3}$

$$\begin{aligned} & A^{\prime}=\frac{A}{9} \\ & \frac{E^{\prime}}{E}=\frac{A^{\prime}}{A} \frac{(327+273)^4}{(27+273)^4} ; \quad \frac{E^{\prime}}{E}=\frac{1}{9}\left(\frac{600}{300}\right)^4 \\ & \therefore E^{\prime}=\frac{1}{9} \times(2)^4 \times E \quad \Rightarrow E^{\prime}=\frac{16 E}{9} \end{aligned}$$

By Wien's displacement law $\lambda T=$ constant

$$\begin{aligned} & \therefore \lambda \propto \frac{1}{\mathrm{~T}} \\ & \therefore \frac{\lambda_2}{\lambda_1}=\frac{\mathrm{T}_1}{\mathrm{~T}_1}=\frac{2000}{3000}=\frac{2}{3} \\ & \therefore \lambda_2=\frac{2}{3} \lambda_1 \end{aligned}$$

$$\begin{aligned} & \frac{R_A}{R_B}=\frac{40}{60}=\frac{2}{3} \\ & R=\rho \cdot \frac{\ell}{A}=\rho \cdot \frac{\ell}{\pi r^2} \end{aligned}$$

Length of the wires are same.

$$\begin{aligned} & \therefore \frac{\mathrm{R}_{\mathrm{A}}}{\mathrm{R}_{\mathrm{B}}}=\frac{\rho_{\mathrm{A}}}{\rho_{\mathrm{B}}}\left(\frac{\mathrm{r}_{\mathrm{B}}}{\mathrm{r}_{\mathrm{A}}}\right)^2 \\ & \therefore \frac{2}{3}=\frac{\rho_{\mathrm{A}}}{\rho_{\mathrm{B}}}\left(\frac{1}{3}\right)^2=\frac{\rho_{\mathrm{A}}}{\rho_{\mathrm{B}}} \cdot \frac{1}{9} \\ & \therefore \frac{\rho_{\mathrm{A}}}{\rho_{\mathrm{B}}}=6 \end{aligned}$$

Gravitational potential energy gained by the ball = Elastic potential energy in the spring

$$\begin{aligned} \mathrm{mgh} & =\frac{1}{2} \mathrm{Fx} \\ \therefore \quad \mathrm{h} & =\frac{\mathrm{Fx}}{2 \mathrm{mg}} \\ & =\frac{5 \times 0.2}{2 \times 25 \times 10^{-3} \times 10} \\ & =2 \mathrm{~m} \end{aligned}$$

Let $x$ be the real depth when the apparent depth is $10 \mathrm{~cm}$.

$\therefore \mu=\frac{x}{10}$

Let $\mathrm{y}$ be the real depth when the apparent depth is $6 \mathrm{~cm}$

$$\begin{aligned} & \therefore \mu=\frac{y}{6} \\ & \therefore \frac{x}{10}=\frac{y}{6} \quad \therefore x=\frac{5}{3} y \end{aligned}$$

Also $x+y=24$

$\therefore \frac{5}{3} y+y=24 \quad \therefore \frac{8}{3} y=24$

$$\begin{aligned} \therefore y & =9 \mathrm{~cm} \\ \mu & =\frac{y}{6}=\frac{9}{6}=1.5 \end{aligned}$$

Angular width of central maximum $\theta=\frac{2\lambda}{a}$

Where a is slit width

$\mathrm{f}=8 \mathrm{~cm}$, when the particle is at mean position, $\mathrm{u}=-14 \mathrm{~cm}$

$$\begin{aligned} & \frac{1}{\mathrm{v}}=\frac{1}{\mathrm{f}}+\frac{1}{\mathrm{u}}=\frac{1}{8}-\frac{1}{14}=\frac{3}{56} \\ & \therefore \quad v=\frac{56}{3} \quad \approx 19 \mathrm{~cm} \\ \end{aligned}$$

When the particle is at one of the extreme positions its distance from the lens is $14+1=15 \mathrm{~cm}$

$$\begin{aligned} & \therefore \mathrm{u}=-15 \mathrm{~cm} \\ & \text { Again, } \frac{1}{\mathrm{v}}=\frac{1}{\mathrm{f}}+\frac{1}{\mathrm{u}}=\frac{1}{8}-\frac{1}{15}=\frac{7}{120} \\ & \therefore \quad \mathrm{v}=\frac{120}{7} \quad \approx 17 \mathrm{~cm} \end{aligned}$$

Amplitude of the image $=19-17=2 \mathrm{~cm}$

$$\begin{aligned} & \gamma=\frac{C_p}{C_v}, \text { Also } C_p-C_v=R \\ & C_p=\gamma C_v \\ \therefore & \gamma C_v-C_v=R \\ \therefore & C_v(\gamma-1)=R \\ \therefore & C_v=\frac{R}{\gamma-1}\end{aligned}$$

$$\begin{aligned} & \sin 2 \theta=2 \sin \theta \cdot \cos \theta \\ & \begin{aligned} & \therefore \quad \sin \theta \cos \theta=\frac{\sin 2 \theta}{2} \\ & \mathrm{~V}=80 \sin (100 \pi \mathrm{t}) \cos (100 \pi \mathrm{t}) \\ &=\frac{80 \sin 200 \pi \mathrm{t}}{2}=40 \sin 200 \pi \mathrm{t} \end{aligned} \end{aligned}$$

Amplitude or the peak value of the voltage $=40 \mathrm{~V}$

Total surface energy before coalesce

$\mathrm{E}_1=2\left(4 \pi \mathrm{R}^2\right) \mathrm{T}$

But $\frac{4}{3} \pi R^2 \times 2=\frac{4}{3} \pi R^{\prime 3}$

Total surface energy after coalesce

$$\begin{aligned} & E_2=4 \pi R^{\prime 2} T=4 \pi 2^{2 / 3} R^2 T \\ & \therefore \frac{E_1}{E_2}=\frac{2\left(4 \pi R^2\right) T}{4 \pi 2^{2 / 3} R^2 T}=2^{1-\frac{2}{3}}=2^{1 / 3} \end{aligned}$$

De-Broglie wavelength $\lambda \propto \frac{1}{p}$ where $\mathrm{p}$ is momentum

For an electron velocity $\mathrm{v} \propto \frac{1}{\mathrm{n}}$,

hence $\mathrm{p} \propto \frac{1}{\mathrm{n}}$

$\therefore \frac{\lambda_1}{\lambda_4}=\frac{1}{4}$

$\therefore \lambda_1=\frac{\lambda_4}{4}$

For isothermal process: $\mathrm{P}_1 \mathrm{~V}_1=\mathrm{P}_2 \mathrm{~V}_2$

$\therefore \quad \mathrm{P}_2=\frac{\mathrm{P}_1 \mathrm{~V}_1}{\mathrm{~V}_2}=\frac{\mathrm{P}_1}{2} \quad\left(\because \frac{\mathrm{v}_1}{\mathrm{v}_2}=\frac{1}{2}\right)$

For adiabatic process : $\quad \mathrm{P}_2 \mathrm{~V}_2^\gamma=\mathrm{P}_3 \mathrm{~V}_3^\gamma$

$$\begin{aligned} \therefore \quad P_3 & =P_2\left(\frac{V_2}{V_3}\right)^\gamma=P_2\left(\frac{2 V_1}{V_3}\right)^\gamma=\frac{P_1}{2}\left(\frac{1}{8}\right)^{5 / 3} \quad\left[\because \frac{V_1}{V_3}=\frac{1}{16}\right] \\ & =\frac{P_1}{2} \cdot \frac{1}{32}=\frac{P_1}{64} \end{aligned}$$

Frequency of a simple pendulum is given by

$$\begin{aligned} \mathrm{n} & =\frac{1}{2 \pi} \sqrt{\frac{\mathrm{g}}{\ell}} \\ \therefore \frac{\mathrm{n}^{\prime}}{\mathrm{n}} & =\sqrt{\frac{\mathrm{g}^{\prime}}{\mathrm{g}}} \\ \mathrm{g}^{\prime} & =\mathrm{g}\left(1-\frac{\mathrm{d}}{\mathrm{R}}\right)=\mathrm{g}\left(1-\frac{1}{2}\right)=\frac{\mathrm{g}}{2} \\ \therefore \quad & \mathrm{n}^{\prime}=\frac{\mathrm{n}}{\sqrt{2}} \end{aligned}$$

Series limit of Lyman series is given by

$\frac{1}{\lambda_1}=\mathrm{R}\left(\frac{1}{1}-\frac{1}{\infty}\right)=\mathrm{R}$

Series limit of Balmer series is given by

$\frac{1}{\lambda_3}=\mathrm{R}\left(\frac{1}{4}-\frac{1}{\infty}\right)=\frac{\mathrm{R}}{4}$

First line of Lyman series is given by

$$\begin{aligned} & \frac{1}{\lambda_2}=\mathrm{R}\left(\frac{1}{1}-\frac{1}{4}\right)=\mathrm{R}-\frac{\mathrm{R}}{4}=\frac{1}{\lambda_1}-\frac{1}{\lambda_3} \\ \therefore & \frac{1}{\lambda_1}-\frac{1}{\lambda_2}=\frac{1}{\lambda_3} \end{aligned}$$

The plane is flying horizontally. Hence initial vertical component of the velocity is zero. If it reaches the ground in time $t$, then

$$\begin{aligned} \mathrm{h} & =\frac{1}{2} \mathrm{gt}^2 \quad \therefore \mathrm{t}^2=\frac{2 \mathrm{~h}}{\mathrm{~g}}=\frac{2 \times 980}{9.8}=200 \\ \therefore \mathrm{t} & =10 \sqrt{2} \mathrm{~s} \end{aligned}$$

The horizontal component of the velocity is

$\mathrm{V}=200 \mathrm{~km} / \mathrm{hr}=200 \times \frac{5}{18}=\frac{1000}{18} \mathrm{~m} / \mathrm{s}$

The horizontal distance to be covered is

$d=V t=\frac{1000}{18} \times 10 \sqrt{2}=\frac{10^4}{9 \sqrt{2}} \mathrm{~m}$

The diode is forward biased.

Potential difference $=5-3=2$ V

$\mathrm{I=\frac{V}{R}=\frac{2}{200}=10^{-2}~A}$

Initial frequency $v=2 v_0$

Final frequency $v^{\prime}=\frac{2 v_0}{3}$

$v^{\prime}$ is less than the threshold frequency $v_0$

Hence no photoelectrons will be emitted and photoelectric current will be zero.

In fundamental mode the length of the closed tube is $\frac{\lambda}{4}$

Time required to travel a distance $\frac{\lambda}{4}$ is t.

Hence time required to travel a distance $\lambda$ will be $4 \mathrm{t}$.

$\therefore$ Time period $\mathrm{T}=4 \mathrm{t}$;

Frequency $\mathrm{n}=\frac{1}{\mathrm{~T}}=\frac{1}{4 \mathrm{t}}=\frac{0.25}{\mathrm{t}}$

$$\begin{aligned} & \frac{\mathrm{Q}}{\mathrm{tA}}=\frac{\mathrm{k} \Delta \theta}{\mathrm{d}} \\ & \therefore \mathrm{k}=\frac{\mathrm{Q}}{\mathrm{tA}} \cdot \frac{\mathrm{d}}{\Delta \theta} \\ & \frac{\mathrm{Q}}{\mathrm{tA}}=900 \mathrm{~kcal} \text { per minute per } \mathrm{m}^2=\frac{900}{60}=15 \mathrm{~kcal} / \mathrm{sm}^2 \\ & \mathrm{~d}=3 \mathrm{~cm}=3 \times 10^{-2} \mathrm{~m}, \Delta \theta=15^{\circ} \mathrm{C} \\ & \therefore \mathrm{k}=\frac{15 \times 3 \times 10^{-2}}{15}=3 \times 10^{-2} \mathrm{kcal} / \mathrm{ms}^{\circ} \mathrm{C}\end{aligned}$$

$$\begin{aligned} & \mathrm{mr} \omega^2=\frac{\mathrm{k}}{\mathrm{r}} \\ & \therefore \omega^2=\frac{\mathrm{k}}{\mathrm{mr}^2} \quad \therefore \omega^2 \propto \frac{1}{\mathrm{r}^2} \\ & \therefore \quad \omega \propto \frac{1}{\mathrm{r}}, \mathrm{T}=\frac{2 \pi}{\omega} \quad \therefore \mathrm{T} \propto \mathrm{r}\end{aligned}$$

Relative permeability $\mu_r=1+\chi$

$\therefore \chi=\mu_r-1=0.85-1=-0.15$

$$\begin{aligned} & \mathrm{F}=\frac{1}{4 \pi \varepsilon_0} \cdot \frac{\mathrm{q}^2}{\mathrm{r}^2} \\ & \therefore \quad \frac{\mathrm{F}_2}{\mathrm{~F}_1}=\left(\frac{\mathrm{r}_1}{\mathrm{r}_2}\right)^2=(2)^2=4 \\ & \therefore \quad \mathrm{F}_2=4 \mathrm{~F}_1\end{aligned}$$