MHT CET 2021 20th September Evening Shift (Physics & Chemistry) Question Paper with Solutions

If the length of potentiometer wire increases, the potential gradient will decrease. Hence to balance the same p.d., the length will increase.

$R=500 \Omega, \phi=60^{\circ}$

When capacitance is removed

$$\begin{aligned} & \tan \phi=\tan 60^{\circ}=\frac{\mathrm{X}_{\mathrm{L}}}{\mathrm{R}} \\ \therefore & \frac{\mathrm{X}_{\mathrm{L}}}{\mathrm{R}}=\sqrt{3} \quad \text { OR } \quad \mathrm{X}_{\mathrm{L}}=\mathrm{R} \sqrt{3} \end{aligned}$$

When inductance is removed,

$$\begin{aligned} & \frac{X_{\mathrm{C}}}{R}=\tan 60^{\circ}=\sqrt{3} \\ \therefore \quad X_C & =R \sqrt{3} \\ \therefore \quad X_C & =X_L \end{aligned}$$

Impedance $\mathrm{Z}=\sqrt{\mathrm{R}^2+\left(\mathrm{X}_{\mathrm{L}}-\mathrm{X}_{\mathrm{C}}\right)^2}=\mathrm{R}=500 \Omega$

$\mathrm{hv}=4.2 \mathrm{~eV}, \omega_0=2.4 \mathrm{~eV},(\mathrm{KE})_{\max }=4.2-2.4=1.8 \mathrm{~eV}$

The electrons will not be able to escape from the surface when its potential becomes $1.8 \mathrm{~V}$.

$$\begin{aligned} \mathrm{V} & =\frac{1}{4 \pi \epsilon_0} \frac{\mathrm{q}}{\mathrm{r}} \\ \therefore 1.8 & =9 \times 10^9 \times \frac{\mathrm{q}}{0.1} \\ \therefore \mathrm{q} & =2 \times 10^{-8} \mathrm{C} \\ \therefore \mathrm{ne} & =\mathrm{q} \\ \mathrm{n} & =\frac{\mathrm{q}}{\mathrm{e}}=\frac{2 \times 10^{-11}}{1.8 \times 10^{-14}}=1.25 \times 10^8 \end{aligned}$$

$$\begin{aligned} & \mathrm{T}=\frac{2 \pi}{\sqrt{3}} \mathrm{~s}, \mathrm{ZA}=4 \mathrm{~cm} \\ & \therefore \mathrm{A}=2 \mathrm{~cm}=2 \times 10^{-2} \mathrm{~m} \\ & \text { Acceleration, } a=\omega^2 \mathrm{x} \\ & \text { Velocity, } v=\omega \sqrt{A^2-x^2} \\ & \therefore \quad \omega^2 \mathrm{x}=\omega \sqrt{A^2-\mathrm{x}^2} \\ & \omega x=\sqrt{A^2-x^2} \\ & \omega^2 x^2=A^2-x^2 \\ & \therefore \quad 3 \mathrm{x}^2=\mathrm{A}^2-\mathrm{x}^2 \quad\left(\therefore \omega=\frac{2 \pi}{\mathrm{T}}=\sqrt{3}\right) \\ & \therefore \quad 4 \mathrm{x}^2=\mathrm{A}^2 \\ & \mathrm{c}=\frac{\mathrm{A}}{2}=\frac{2}{2}=1 \mathrm{~cm} \\ & \end{aligned}$$

$$\begin{aligned} g^{\prime} & =g\left(1-\frac{d}{R}\right) \\ \therefore \frac{g}{n} & =g\left(1-\frac{d}{R}\right)\end{aligned}$$

$$\begin{aligned} & \therefore \frac{1}{n}=\left(1-\frac{d}{R}\right) \\ & \therefore \frac{d}{R}=1-\frac{1}{n}=\frac{n-1}{n} \\ & \therefore d=\frac{R(n-1)}{n}\end{aligned}$$

A rectifier is an electrical device composed of one or more diodes that converts alternating current (a.c.) into direct current (d.c.) by allowing current to flow through it in one direction only.

Hence, the correct option is (A).

$$\begin{aligned} & \mathrm{F}_1=\mathrm{K}_1 \mathrm{x} \quad \mathrm{F}_2=\mathrm{K}_2 \mathrm{x} \quad \mathrm{F}=\left(\mathrm{K}_1+\mathrm{K}_2\right) \mathrm{x} \\ & \therefore T_1=2 \pi \sqrt{\frac{m}{K}} \quad T_2=2 \pi \sqrt{\frac{\mathrm{m}}{\mathrm{K}}} \\ & \therefore \mathrm{T}_1^2=4 \pi^2 \frac{\mathrm{m}}{\mathrm{K}_1} \quad \mathrm{~T}_2^2=4 \pi^2 \frac{\mathrm{m}}{\mathrm{K}_2} \\ & T^2=4 \pi^2 \frac{m}{K_1+K_2} \\ & \therefore \frac{1}{\mathrm{~T}^2}=\frac{\mathrm{K}_1+\mathrm{K}_2}{4 \pi^2 \mathrm{~m}} \quad=\frac{\mathrm{K}_1}{4 \pi^2 \mathrm{~m}}+\frac{\mathrm{K}_2}{4 \pi^2 \mathrm{~m}} \\ & =\frac{1}{\mathrm{~T}_1^2}+\frac{1}{\mathrm{~T}_2^2}=\frac{\mathrm{T}_1^2+\mathrm{T}_2^2}{\mathrm{~T}_1^2 \mathrm{~T}_2^2} \\ & \therefore \mathrm{T}^2=\frac{\mathrm{T}_1^2 \mathrm{~T}_2^2}{\mathrm{~T}_1^2+\mathrm{T}_2^2} \\ & \therefore \mathrm{T}=\frac{\mathrm{T}_1 \mathrm{~T}_2}{\sqrt{\mathrm{T}_1^2+\mathrm{T}_2^2}} \\ & \end{aligned}$$

$\mathrm{M=\frac{enh}{4\pi m_e}}$

$\mathrm{n}=264 \mathrm{~Hz}, \mathrm{~V}=300 \mathrm{~m} / \mathrm{s}$

For fundamental frequency, $\mathrm{n}=\frac{\mathrm{V}}{4 l}$

$l=\frac{\mathrm{V}}{4 \mathrm{n}} \quad=\frac{330}{4 \times 264}=0.3125 \mathrm{n} \quad=31.25 \mathrm{~cm}$

For fundamental mode, $l=\frac{\lambda}{4}$

other possible lengths are $\frac{3 \lambda}{4}, \frac{5 \lambda}{4}, \ldots \ldots$

$\text { i.e. } 93.75 \mathrm{~cm}, 156.25 \mathrm{~cm}, \ldots \ldots$

Hence, $62.50 \mathrm{~m}$ is not possible.

$$\begin{aligned} & \mathrm{y}_1=\mathrm{a} \sin 2000 \pi \mathrm{t} \\ & \therefore 2 \pi \mathrm{n}_1=2000 \pi \quad \therefore \mathrm{n}_1=1000 \mathrm{~Hz}\end{aligned}$$

$$\begin{array}{ll}\mathrm{y}_2=\mathrm{a} \sin 2008 \pi \mathrm{t} & \\ \therefore 2 \pi \mathrm{n}_2=2008 \pi & \therefore \mathrm{n}_2=1004 \mathrm{~Hz} \\ \therefore \text { Beat frequency } & =\mathrm{n}_2-\mathrm{n}_1 \\ & =4 \mathrm{~Hz}\end{array}$$

According to Wien's law, $\lambda$T = constant

$\therefore \mathrm{T \propto \frac{1}{\lambda}}$

Wavelength of blue light is less than that of yellow light. Hence, temperature of Q is greater than temperature of P.

The velocity of the body when it reaches the surface of the lake is given by

$\mathrm{v}^2=2 \mathrm{gh}$ ...... (1)

When the body is in air the gravitational force acting on it is

$\mathrm{F=mg=V \rho g}$ ...... (2)

Where $\mathrm{V}$ is the volume of the body.

When the body enters the lake, there will be an upthrust acting on the body. The upthrust is given by

$\mathrm{U}=\mathrm{V} \delta \mathrm{g}$

Since $\delta > \rho$, upthrust will be greater then the gravitational force. The net force acting on the body will be

$\mathrm{F}^{\prime}=\mathrm{V} \delta \mathrm{g}-\mathrm{V} \rho \mathrm{g} \quad \mathrm{Vg}=(\delta-\rho)$ ....... (3)

By eqs. (2) and (3)

$$\begin{aligned} & \frac{\mathrm{F}^{\prime}}{\mathrm{F}}=\frac{\delta-\rho}{\rho} \\ & \frac{\mathrm{a}}{\mathrm{g}}=\frac{\delta-\rho}{\rho} \\ & \therefore a=\left(\frac{\delta-\rho}{\rho}\right) \mathrm{g} \quad \therefore \frac{\mathrm{g}}{\mathrm{a}}=\frac{\rho}{\delta-\rho} \\ & \text { If } \mathrm{a} \text { is the accelertion } \end{aligned}$$

If $\mathrm{a}$ is the acceleration retardation in the liquid then

v$^2$ = 2ad ...... (4)

$\begin{gathered} \text { BY eqs (1) and (4) } \\ 2 a d=2 \mathrm{gh} \end{gathered}$

$\therefore \mathrm{d}=\frac{\mathrm{g}}{\mathrm{a}} \mathrm{h} \quad \therefore \mathrm{d}=\frac{\rho}{\delta-\rho} \mathrm{h}$

The Boolean expression for a two-input OR gate is given by

$𝑌=𝐴+𝐵$

The output $𝑌$ is '1' (HIGH) whenever at least one of the inputs $𝐴$ or $𝐵$ is '1', which includes the condition where both inputs are '1'. The output is '0' only when both inputs are '0'.

Hence, the correct option is (D).

Relation between $\mathrm{P}$ and $\mathrm{V}$ is given as

$\mathrm{PV}^{3 / 2}=\text { constant }$

For adiabatic process, $\mathrm{PV}^\gamma=$ constant

$\therefore \gamma=\frac{3}{2}$

The relation between $\mathrm{T}$ and $\mathrm{V}$ is given by

$$\begin{aligned} & \mathrm{TV}^{\gamma-1}=\text { constant } \\ \therefore & \mathrm{T}_1 \mathrm{~V}_1^{\gamma-1}=\mathrm{T}_0 \mathrm{~V}_0^{\gamma-1} \\ \therefore & \mathrm{T}_1 \mathrm{~V}_1^{1 / 2}=\mathrm{T}_0 \mathrm{~V}_0^{1 / 2} \\ \therefore & \frac{\mathrm{T}_2}{\mathrm{~T}_1}=\left(\frac{\mathrm{V}_1}{\mathrm{~V}_2}\right)^{1 / 2}=2^{1 / 2}=\sqrt{2} \\ \therefore & \mathrm{T}_2=\sqrt{2} \mathrm{~T}_1=\sqrt{2} \mathrm{~T} \end{aligned}$$

Sound waves in air propagate via alternating compressions and rarefactions, in which the particles of the medium oscillate parallel to the direction of wave propagation, making them longitudinal waves.

Light waves are electromagnetic waves in which electric and magnetic fields oscillate perpendicular to the direction of wave propagation, making them transverse waves.

Hence, the correct option is (A).

Gyromagnetic ratio = $\mathrm{\frac{e}{2m}}$

It is independent of the orbit of the electron.

At the highest point the particle will come to rest momentarily, hence it is at extreme position and has maximum force and acceleration. Since the spring is unstretched, the restoring forceto provided by the weight of the particle

$\therefore \mathrm{mA} \omega^2=\mathrm{mg}$

or $A \omega^2=\mathrm{g}$

$$\begin{aligned} \therefore A & =\frac{g}{\omega^2} \\ \omega & =2 \pi f=2 \pi \times 5=10 \pi \\ \therefore A & =\frac{10}{100 \pi^2}=\frac{1}{10 \pi^2} \\ V_{\max } & =A \omega=\frac{1}{10 \pi^2} \times 10 \pi=\frac{1}{\pi}\end{aligned}$$

In the region $0 \leq x \leq 1$, the potential energy of the particle is $E$.

Total energy is $\mathrm{nE}$.

Hence, kinetic energy, $K=n E-E=(n-1) E$

Its momentum, $\mathrm{p}_1=\sqrt{2 \mathrm{mK}}=\sqrt{2 \mathrm{~m}(\mathrm{n}-1) \mathrm{E}}$

$\therefore \lambda_1=\frac{\mathrm{h}}{\mathrm{p}_1}=\frac{\mathrm{h}}{\sqrt{2 \mathrm{~m}(\mathrm{n}-1) \mathrm{E}}}$

In the region $x > 1, \mathrm{PE}$ is zero.

Hence, total energy is kinetic energy.

$$\begin{aligned} & \therefore \mathrm{K}=\mathrm{nE} \\ & \therefore \lambda_2=\frac{\mathrm{h}}{\mathrm{p}_2}=\frac{\mathrm{h}}{\sqrt{2 \mathrm{mK}}} \\ & \therefore \frac{\lambda_1}{\lambda_2}=\sqrt{\frac{\mathrm{n}}{\mathrm{n}-1}} \end{aligned}$$

$$\begin{aligned} s_1 & =\frac{I_g G}{3 I-I_g}, s_2=\frac{I_g G}{4 I-I_g} \\ \therefore \quad \frac{s_2}{s_1} & =\frac{3 I-I_g}{4 I-I_g}\end{aligned}$$

The energy emitted by a black body per unit area per unit time (or power per unit area) is given by Stefan-Boltzmann Law :

$\sigma T^4$

Where :

• $\sigma$ is the Stefan-Boltzmann constant.
• $T$ is the absolute temperature of the body.

Given :

Initial energy emitted $E$ for area $A$ at temperature $27^\circ C = 273 + 27 = 300 \text{K}$

$E = A \sigma (300^4)$

Now, the area is reduced to $(1/3)^2$ of its original value :

$A' = \left(\frac{1}{3}\right)^2 A = \frac{A}{9}$

The temperature is raised to $327^\circ C = 273 + 327 = 600 \text{K}$

The new energy emitted $E'$ will be :

$E' = A' \sigma (600^4)$

Dividing the two equations :

$\frac{E'}{E} = \frac{A' \sigma (600^4)}{A \sigma (300^4)}$

Given $A' = \frac{A}{9}$ and using $(600^4)/(300^4) = 2^4 = 16$ :

$\frac{E'}{E} = \frac{16}{9}$

So, $E' = \frac{16E}{9}$

The correct answer is :

Option C

$\frac{16E}{9}$

$$\begin{array}{rlrl}\mathrm{EMF}, \mathrm{e} & =\pi \mathrm{r}^2 \mathrm{Bf} \quad=\pi \mathrm{r}^2 \mathrm{~B} \frac{\omega}{2 \pi} \quad=\frac{1}{2} \mathrm{Br}^2 \omega \\ & =\frac{1}{2} \times 0.2 \times 10^{-4} \times(1)^2 \times 5 & \\ & =0.5 \times 10^{-4} \mathrm{~V} & =50 \mu \mathrm{V}\end{array}$$

$\lambda_1$ = 1 nm, $\lambda_2$ = 0.5 nm

$\therefore \lambda_2=\frac{\lambda_1}{2}$

$\lambda=\frac{h}{p}\quad \therefore P_2=2P_1$

Kinetic energy, $\mathrm{E}=\frac{\mathrm{P}^2}{2 \mathrm{~m}}$

$$\begin{aligned} & \therefore \frac{E_2}{E_1}=\left(\frac{P_2}{P_1}\right)^2=(2)^2=4 \\ & \therefore E_2=4 E_1 \\ & \therefore E_2-E_1=3 E_1 \end{aligned}$$

Since it is compressed isothermally, the temperature remains constant. The rms speed is given by,

$c=\sqrt{\frac{3RT}{M}}$

Since temperature T remains constant, the rms speed remains unchanged.

The electric field at the centre due to a small element ds of the ring is given by

$\mathrm{dE}=\frac{1}{4 \pi \epsilon_0} \frac{\lambda \mathrm{ds}}{\mathrm{r}^2}$

The field can be resolved along $\mathrm{X}$ and $\mathrm{Y}$ directions as shown in the figure. Similarly field due to symmetrically situated another element can also be resolved along $\mathrm{X}$ and $\mathrm{Y}$ directions. The X-components are in the same direction and get added while the Y-components are in opposite directions and cancel each other. Hence, the total field due to the range is

$$\begin{array}{rlrl} \mathrm{E} & =\int \mathrm{dE} \cos \theta \\ \therefore \mathrm{E} & =\int \frac{1}{4 \pi \epsilon_0} \frac{\lambda d s}{r^2} \cos \theta \\ & =\int \frac{1}{4 \pi \epsilon_0} \frac{\lambda d \theta \cos \theta}{\mathrm{r}^2} \quad=\frac{1}{4 \pi \epsilon_0} \frac{\lambda}{\mathrm{r}} \int_{-\pi / 2}^{\pi / 2} \cos \theta d \theta \\ & =\frac{1}{4 \pi \epsilon_0} \frac{\lambda}{\mathrm{r}}[\sin \theta]_{0 \pi / 2}^{\pi / 2} =\frac{1}{4 \pi \epsilon_0} \frac{\lambda}{\mathrm{r}}[1-(-1)] \\ & =\frac{1}{4 \pi \epsilon_0} \frac{\lambda}{r} 2 \quad=\frac{\lambda}{2 \pi \epsilon_0 r} \end{array}$$

Magnification, $\mathrm{m}=-\frac{1}{4}$

(Since the image is real and inverted, it is taken as negative)

$$\begin{aligned} & \therefore \mathrm{m}=\frac{\mathrm{v}}{\mathrm{u}}=-\frac{1}{4} \\ & \therefore \mathrm{v}=-\frac{\mathrm{u}}{4} \end{aligned}$$

We have, $\frac{1}{\mathrm{v}}-\frac{1}{\mathrm{u}}=\frac{1}{\mathrm{f}}$

$$\begin{aligned} & -\frac{4}{\mathrm{u}}-\frac{1}{\mathrm{u}}=\frac{1}{\mathrm{f}} \\ & -\frac{5}{\mathrm{u}}=\frac{1}{\mathrm{f}} \\ \therefore \quad & \mathrm{u}=-5 \mathrm{f} \end{aligned}$$

When two bodies of equal masses collide elastically, they exchange their velocities. Since the two masses are exchanging their velocities, their masses must be equal.

Hence, $\frac{\mathrm{m}_{\mathrm{a}}}{\mathrm{m}_{\mathrm{b}}}=1$

Excess pressure in a water drop $=\frac{2 T}{R}$

$$\begin{aligned} \mathrm{T} & =7.2 \times 10^{-2} \mathrm{Nm}^{-2} \\ \therefore 72 & =\frac{2 \times 7.2 \times 10^{-2}}{\mathrm{R}} \\ \therefore \mathrm{R} & =\frac{2 \times 7.2 \times 10^{-2}}{72}=2 \times 10^{-3} \mathrm{~m}=2 \mathrm{~mm} \end{aligned}$$

Focal length in air is given by

$\frac{1}{f}=\left(n_g-1\right)\left(\frac{1}{R_1}-\frac{1}{R_2}\right)$ ..... (1)

Focal length in liquid is given by

$\frac{1}{f^{\prime}}=\left(n^{\prime}-1\right)\left(\frac{1}{R_1}-\frac{1}{R_2}\right)$ ..... (2)

where $\mathrm{n}^{\prime}=\frac{\mathrm{n}_{\mathrm{g}}}{\mathrm{n}_{\ell}}=\frac{3}{2} \times \frac{5}{9}=\frac{5}{6}$

Dividing eq. (1) and (2),

$$\begin{aligned} \frac{\mathrm{f}^{\prime}}{\mathrm{f}} & =\frac{\mathrm{n}_{\mathrm{g}}-1}{\mathrm{n}^{\prime}-1}=\frac{\frac{3}{2}-1}{\frac{5}{6}-1}=\frac{\frac{1}{2}}{-\frac{1}{6}}=-3 \\ \therefore \mathrm{f}^{\prime} & =-3 \mathrm{f} \quad=-3 \times 15=-45 \mathrm{~cm}\end{aligned}$$

$\frac{\mathrm{V}_2}{\mathrm{~V}_1}=\frac{1}{8}, \gamma=\frac{5}{3}$

For adiabatic process,

$$\begin{aligned} & P_1 V_1^\gamma=P_2 V_2^\gamma \\ \therefore & \frac{P_2}{P_1}=\left(\frac{V_1}{V_2}\right)^\gamma=(8)^{5 / 3}=(2)^5=32 \end{aligned}$$

$$\begin{aligned} V & =\frac{1}{4 \pi \epsilon_0} \frac{q}{R}-\frac{1}{4 \pi \epsilon_0} \frac{q}{5 R} \\ & =\frac{1}{4 \pi \epsilon_0} \frac{q}{R}\left(1-\frac{1}{5}\right) \\ & =\frac{1}{4 \pi \epsilon_0} \frac{q}{R} \frac{4}{5} \\ E & =\frac{1}{4 \pi \varepsilon_0} \cdot \frac{q}{25 R^2} \\ \therefore \frac{E}{V} & =\frac{1}{20 R} \text { or } E=\frac{V}{20 R}\end{aligned}$$

Current of $5 \mathrm{~A}$ and $2 \mathrm{~A}$ are enclosed in the loop.

The currents are in opposite direction.

The net current enclosed by the loop is $3 \mathrm{~A}$.

According to Ampere law,

$\oint \overrightarrow{\mathrm{B}} \mathrm{d} \overrightarrow{\mathrm{i}}=\mu_0 \mathrm{I}=3 \mu_0$

C$_1$ = C$_2$ = C$_3$ = C$_4$ = 6 $\mu$F

C$_2$ and C$_3$ are in parallel.

Hence their equivalent capacitance, C$_5$ = 2 $\times$ 6 = 12 $\mu$F

C$_5$ and C$_1$ in series.

Hence their equivalent capacitance C$_6$ is given by

${1 \over {{C_6}}} = {1 \over {12}} + {1 \over 6} = {3 \over {12}} = {1 \over 4}$

$\therefore$ C$_6$ = 4 $\mu$F

C$_6$ and C$_4$ are in parallel.

Hence their equivalent capacitance is C = 4 + 6 = 10 $\mu$F

$\mathrm{I}=\mathrm{MK}^2$

For a ring, $\quad \mathrm{MR}^2=\mathrm{MK}_{\mathrm{r}}^2 \quad \therefore \mathrm{K}_{\mathrm{r}}=\mathrm{R}$

For a disc, $\quad \frac{\mathrm{MR}^2}{2}=\mathrm{MK}_{\mathrm{d}}^2 \quad \therefore \mathrm{K}_{\mathrm{d}}=\frac{\mathrm{R}}{\sqrt{2}}$

$\therefore \frac{\mathrm{K}_{\mathrm{r}}}{\mathrm{K}_{\mathrm{d}}}=\sqrt{2}$

For an ideal transformer,

$\frac{V_s}{V_p}=\frac{I_p}{I_s}$

Hence, if $V_s < V_p$ then $I_p < I_s$

For maxima, path difference = n$\lambda$

For first maximum, n = 1

$\therefore$ Path difference, $\lambda$ = 6300 $\mathop A\limits^o$