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

m = 5 g = 5 $\times$ 10$^{-3}$ kg, A = 0.3 m, T = $\frac{\pi}{5}$s

$\mathrm{\omega=\frac{2\pi}{T}=10}$ rad/s

Maximum force

F = m$\omega^2$A = 5 $\times$ 10$^{-3} \times(10)^2\times0.3=0.15$ N

$$\begin{aligned} & \mathrm{E}_1=\frac{1}{2} \mathrm{I}_1 \omega_1^2 \\ & \mathrm{E}_2=\frac{1}{2} \mathrm{I}_2 \omega_2^2=\frac{1}{2} \cdot 3 \mathrm{I}_1 \cdot \omega_1^2 \quad\left[\because \mathrm{I}_2=3 \mathrm{I}_1\right] \\ & =\frac{3}{2} \mathrm{I}_1 \omega_2^2 \\ & \mathrm{E}_1=27 \mathrm{E}_2=27 \times \frac{3}{2} \times \mathrm{I}_1 \omega_2^2 \\ \end{aligned}$$

$$\begin{aligned} & \therefore \frac{1}{2} \mathrm{I}_1 \omega_2^2=\frac{81}{2} \mathrm{I}_1 \omega_2^2 \\ & \therefore \quad \omega_1=9 \omega_2 \\ & \frac{L_1}{L_2}=\frac{\mathrm{I}_1 \omega_1}{\mathrm{I}_2 \omega_2}=\frac{\mathrm{I}_1 \times 9 \omega_2}{3 \mathrm{I}_1 \times \omega_2}=3 \\ & \end{aligned}$$

Force due to surface tension

F = 2$\pi$rT

$=2\times\frac{22}{7}\times2\times10^{-3}\times70\times10^{-3}$

$=8.8\times10^{-3}$ N

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

$$\begin{aligned} & \mathrm{y}=10 \sin \left(\frac{2 \pi t}{30}+\alpha\right) \\ & \text { At } \mathrm{t}=0, \mathrm{y}=5 \mathrm{~cm} \quad \therefore 5=10 \sin \alpha \\ & \therefore \sin \alpha=\frac{1}{2} \quad \therefore \quad \alpha=\frac{\pi^{\mathrm{c}}}{6} \\ & \text { At } t=7.5 \text {, the total phase }\\& \phi=\frac{2 \pi \times 7.5}{30}+\alpha \\ & =\frac{\pi}{2}+\alpha \\ & =\frac{\pi}{2}+\frac{\pi}{6}=\frac{2 \pi}{3} \mathrm{rad} \\ & \end{aligned}$$

Power $\mathrm{P=\frac{Energy}{t}=\frac{nhc}{\lambda t}}$

$\therefore \mathrm{n=\frac{P\lambda t}{hC}}$

Excess pressure in a soap bubble is given by

$\mathrm{P=\frac{4T}{R}}$.

Hence if radius is increased, the pressure will decrease.

Distance between the first minima $= {{2\lambda D} \over a} = {{2 \times 5 \times {{10}^{ - 7}} \times 2} \over {0.2 \times {{10}^{ - 3}}}} = {10^{ - 3}}$ m

Inductive reactance when 100 V, 50 Hz source is connected is given by

$\mathrm{X_L=\frac{V}{I}=\frac{100}{8}=12.5\Omega}$

Resistance $\mathrm{R=\frac{V}{I}=\frac{100}{10}=10\Omega}$

When 100V, 40 Hz supply is connected the inductive reactance will get reduced

Since it is proportional to the frequency

$$\begin{aligned} & \therefore \quad \frac{\mathrm{X}_{\mathrm{L}}^{\prime}}{}=\frac{40}{50}=\frac{4}{5} \\ & \therefore \quad \mathrm{X}_{\mathrm{L}}^{\prime}=\frac{4}{5} \mathrm{X}_{\mathrm{L}}=\frac{4}{5} \times 12.5=10 \Omega\end{aligned}$$

The resistance will remain same. When they are connected in series the impedance will be given by

$$\begin{aligned} & Z=\sqrt{R^2+X_L^{\prime 2}}=\sqrt{(10)^2+(10)^2}=10 \sqrt{2} A \\ & I=\frac{V}{Z}=\frac{100}{10 \sqrt{2}}=5 \sqrt{2} \mathrm{~A} \end{aligned}$$

Since the clockwise current is increasing, the induced current in the loop will be such as to oppose the change in magnetic flux and hence it will flow in anticlockwise direction.

$$\begin{aligned} & \text { } \\ & T_1=2 \pi \sqrt{\frac{m_1}{k}}, T_2=2 \pi \sqrt{\frac{m_1+m_2}{k}} \\ & \therefore \frac{T_2}{T_1}=\sqrt{\frac{m_1+m_2}{m_1}} \\ & \therefore \frac{T_2^2}{T_1^2}=\frac{m_1+m_2}{m_1} \\ & \therefore \frac{T_2^2-T_1^2}{T_1^2}=\frac{m_2}{m_1} \end{aligned}$$

Gyromagnetic ratio $=\mathrm{\frac{e}{2m_e}=\frac{A}{2}}$

Due to charge $-$q at the centre of the shell, a charge q will be induced on the inner surface and $-$q on the outer surface. The charge on outer surface will become Q $-$ q. Hence, surface change densities will be $\frac{q}{4 \pi r_1^2}$ and $\frac{\mathrm{Q}-\mathrm{q}}{4 \pi \mathrm{r}_2^2}$

Magnetic field at the centre of a circular coil is given by

$B=\frac{\mu_0 I}{2 r}$

In this case the current $I=\frac{e}{T}$ where T is the period of revolution

$$\begin{aligned} \mathrm{T} & =\frac{2 \pi \mathrm{r}}{\mathrm{v}} \\ \therefore \mathrm{I} & =\frac{\mathrm{eV}}{2 \pi \mathrm{r}} \quad \therefore \quad \mathrm{B}=\frac{\mu_0 \mathrm{eV}}{4 \pi \mathrm{r}^2}\end{aligned}$$

If dielectric of dielectric constant k is placed between the plates of a capacitor its capacitance increase k times. If it is removed the capacitance will become $\mathrm{\frac{1}{k}}$ times.

The resistance of the galvanometer and the shunt are in parallel. Their equivalent resistance is given by

$$\begin{array}{rlrl} & \frac{1}{R}=\frac{1}{G}+\frac{1}{S} \\ \therefore \frac{1}{S} & =\frac{1}{R}-\frac{1}{G} \\ \therefore \frac{1}{S} & =\frac{1}{2.5}-\frac{1}{50}=\frac{19}{50} \quad \therefore S=\frac{50}{19} \Omega\end{array}$$

Heat supplied at constant pressure $\mathrm{Q_1}=\mathrm{nC}_{\mathrm{p}} \mathrm{dT}$

Heat supplied at constant volume $\mathrm{Q_2}=\mathrm{nC}_{\mathrm{v}} \mathrm{dT}$

Work done

$$\begin{aligned} & \mathrm{W}=\mathrm{Q}_1-\mathrm{Q}_2=\mathrm{n}\left(\mathrm{C}_{\mathrm{p}}-\mathrm{C}_{\mathrm{v}}\right) \mathrm{dT} \\ & \therefore \frac{\mathrm{W}}{\mathrm{Q}_2}=\frac{\mathrm{C}_{\mathrm{p}}-\mathrm{C}_{\mathrm{v}}}{\mathrm{C}_{\mathrm{v}}}=\frac{\mathrm{C}_{\mathrm{p}}}{\mathrm{C}_{\mathrm{v}}}-1=\frac{5}{3}-1=\frac{2}{3} \\ & \therefore \mathrm{Q}_2=\frac{3 \mathrm{~W}}{2} \end{aligned}$$

Force, F = Charge $\times$ Electric field = 2eE

Acceleration $=\mathrm{\frac{Force}{Mass}=\frac{2eE}{4m}=\frac{eE}{2m}}$

Resistivity of conductors increases with temperature and that of semiconductor decreases with temperature.

When temperature is decreased, the resistivity of copper will decrease while that of silicon was increase.

The orbital velocity near the surface of the earth

$\mathrm{V}=\sqrt{\frac{\mathrm{GM}}{\mathrm{R}}}$

At an altitude $\frac{R}{2}$, the orbital velocity

$\mathrm{V}^{\prime}=\sqrt{\frac{\mathrm{GM}}{\mathrm{R}+\frac{\mathrm{R}}{2}}}=\sqrt{\frac{2 \mathrm{GM}}{3 \mathrm{R}}} \quad \therefore \frac{\mathrm{V}^{\prime}}{\mathrm{V}}=\sqrt{\frac{2}{3}} \quad \text { or } \quad \mathrm{V}^{\prime}=\sqrt{\frac{2}{3}} \mathrm{~V}$

Initially the electron is in the third excited state i.e. n = 4. Hence we can have transitions :

$4\to3,4\to2,4\to1$

$3\to2,3\to1$

$2\to1$

Hence, six transitions are possible.

W$_1$ = 620 N, W$_2$ = 340 N

W$_1$ = m(g + a) ..... (i)

W$_2$ = m(g $-$ a) ..... (ii)

$\therefore \mathrm{\frac{W_1}{W_2}=\frac{620}{340}=\frac{g+a}{g-a}}$

$\therefore \frac{31}{17}=\mathrm{\frac{g+a}{g-a}}$

Solving : $\mathrm{a=\frac{7}{24}g}$

Putting this value in Eq. (i) and solving we get mg = 480 N

The induced emf is given by

e = B$l$v = $1.2\times0.6\times10=7.2$ V

Width of central maximum $=\frac{2\lambda D}{a}$

$\therefore a=\frac{2\lambda D}{a}$

$\therefore D=\frac{a^2}{2\lambda}$

When a d.c. voltage is applied only the resistance of the coil comes into play, its inductive reactance is zero.

$\mathrm{R=\frac{V}{I}=\frac{200}{1}=200~\Omega}$

When a.c. voltage is applied, the resistance and inductive reactance come into play and the coil has an impedance Z.

The impedance

$\mathrm{Z=\frac{V}{I}=\frac{200}{0.5}=400~\Omega}$

$\mathrm{Z}^2=\mathrm{R}^2+\mathrm{X}_{\mathrm{L}}^2$

$\therefore (400)^2=(200)^2+\mathrm{X}_{\mathrm{L}}^2$

$\therefore \mathrm{X}_{\mathrm{L}}^2=(400)^2-(200)^2=12 \times 10^4$

$\therefore \mathrm{X}_{\mathrm{L}}=2 \sqrt{3} \times 10^2=200 \sqrt{3} \Omega$ ..... (1)

Also $\mathrm{X}_{\mathrm{L}}=2 \pi \mathrm{fL}=2 \pi \mathrm{f} \times \frac{2 \sqrt{3}}{\pi}=4 \sqrt{3} \mathrm{f}$ ..... (2)

By (1) and (2) : $\quad 4 \sqrt{3} \mathrm{f}=200 \sqrt{3}$

$\therefore \quad \mathrm{f}=50 \mathrm{~Hz}$

Since reflected and refracted rays are perpendicular to each other,

$$\begin{aligned} & \mathrm{r}+\mathrm{r}_1=90^{\circ} \\ \therefore & \mathrm{r}_1=90^{\circ}-\mathrm{r} \end{aligned}$$

Refractive index of the denser medium

$\mu=\frac{\sin r_1}{\sin r}=\frac{\sin \left(90^{\circ}-r\right)}{\sin r}=\frac{\cos r}{\sin r}=\cot r$

Also, $\mu=\frac{1}{\sin C} \quad \therefore \quad \frac{1}{\sin C}=\cot r$

$$\begin{array}{ll}\text { or } & \sin C=\frac{1}{\cot r}=\tan r=\tan i \\ \therefore & i=\tan ^{-1}(\sin C) \end{array}$$

${{{P_2}} \over {{P_1}}} = {\left( {{{{T_2}} \over {{T_1}}}} \right)^4}$

$\therefore$ $16 = {\left( {{{{T_2}} \over {{T_1}}}} \right)^4}$

$\therefore$ ${{{T_2}} \over {{T_1}}} = 2$

If $\mathrm{q}$ is the charge on the ring, then the potential at the centre is given by

$\mathrm{V}=\frac{1}{4 \pi \varepsilon_0} \cdot \frac{\mathrm{q}}{\mathrm{R}}$

$$\begin{aligned} & \text { But } q=\sigma \times \pi R \\ & \therefore V=\frac{1}{4 \pi \varepsilon_0} \cdot \frac{\rho \pi R}{R}=\frac{\sigma}{4 \varepsilon_0} \end{aligned}$$

A dipole has two equal and opposite charges. In a uniform electric field, the two charges will experience equal and opposite force, hence net force will be zero.

The potential energy of an electric dipole is given by

$U=-p E \cos \theta$

where $\theta$ is the angle between the dipole moment and the electric field. The potential energy will be minimum when $\cos \theta$ is maximum.

For $\theta=0^{\circ}, \cos \theta$ has maximum value equal to 1.

$\therefore \mathrm{U}=-\mathrm{pE} \text { when } \theta=0^{\circ} \text {. }$

Frequency of vibration n will be same for both the segments. If p$_1$ and p$_2$ are the number of loops for the wires of radius r and 2r then we have

$n = {{{p_1}} \over {2lr}}\sqrt {{T \over {\pi p}}} = {{{p_2}} \over {4lr}}\sqrt {{T \over {\pi p}}}$

$\therefore$ ${{{p_1}} \over {{p_2}}} = {1 \over 2}$

The magnetic field

$\mathrm{B=\mu_0H}$

For a solenoid $\mathrm{B=\mu_0nI}$

$\therefore$ H = nI

Since the coils are connected to a dc voltage. Inductive reactance will be zero, only resistance has to be considered.

When the coil is broken into two identical points, each part will have resistance $\mathrm{\frac{R}{2}}$. When these are connected in parallel, their equivalent resistance will be $\mathrm{\frac{R}{4}}$. Hence the current I is given by $\mathrm{I=\frac{E}{R/4}=\frac{4E}{R}}$

For uniform velocity, the distance travelled is given by

$\mathrm{x}=\mathrm{vt}$

For uniform acceleration $x=\frac{1}{2}$ at

$$\begin{aligned} & \therefore \quad \frac{1}{2} \mathrm{at}^2=\mathrm{Vt} \\ & \therefore \mathrm{t}=\frac{2 \mathrm{~V}}{\mathrm{a}} \end{aligned}$$

If $p$ is the number of loops (or antinodes) then we have,

$\mathrm{Tp}^2=$ constant where $\mathrm{T}$ is the tension

$$\begin{aligned} & \therefore \mathrm{T}_1 \mathrm{P}_1^2=\mathrm{T}_2^2 \mathrm{P}_2^2 \\ & \therefore \frac{\mathrm{T}_2}{\mathrm{~T}_1}=\frac{\mathrm{P}_1^2}{\mathrm{P}_2^2}=\left(\frac{4}{2}\right)^2=4 \quad \therefore \mathrm{T}_2=4 \mathrm{~T}_1=4 \times 1=4 \mathrm{~kg}-\mathrm{wt} \end{aligned}$$

The volume of the bigger drop will be equal to the sum of the volumes of the smaller drops

$$\begin{aligned} & \therefore \quad \frac{4}{3} \pi \mathrm{R}^3=\frac{4}{3} \pi \mathrm{R}_1^3+\frac{4}{3} \pi \mathrm{R}_2^3 \\ & \therefore \quad \mathrm{R}=\left(\mathrm{R}_1^3+\mathrm{R}_2^3\right)^{1 / 3}\end{aligned}$$

Moment of inertia of the solid sphere I = $\frac{2}{5}$MR$^2$

Moment of inertia of the disc about an axis passing through its edge and perpendicular to the plane is given by

$$\begin{aligned} & \mathrm{I}^{\prime}=\frac{\mathrm{MR}^{\prime 2}}{2}+\mathrm{MR}^{\prime 2}=\frac{3}{2} \mathrm{MR}^{\prime 2} \\ & \mathrm{I}^{\prime}=\mathrm{I} \quad \therefore \frac{3}{2} \mathrm{MR}^{\prime 2}=\frac{2}{5} \mathrm{MR}^2 \\ & \therefore \quad \mathrm{R}^{\prime}=\frac{4}{15} \mathrm{R}^2 \quad \therefore \mathrm{R}^{\prime}=\frac{2}{\sqrt{15}} \mathrm{R} \\ & \end{aligned}$$

$x=x_1+x_2=A \sin (\omega t+\alpha)$

Maximum value of $x_1+x_2=A=\sqrt{2} a$

$$\begin{aligned} & A^2=a^2+a^2+2 a^2 \cos \alpha=2 a^2(1+\cos \alpha) \\ & \therefore 2 a^2=2 a^2(1+\cos \alpha) \quad \therefore 1=1+\cos \alpha \\ & \therefore \quad \cos \alpha=0 \quad \therefore \alpha=\frac{\pi}{2} \end{aligned}$$

In Celsius scale the freezing and boiling points are 0$^\circ$C and 100$^\circ$C. In the given imaginary scale the freezing and boiling points are 39$^\circ$ W and 239$^\circ$ W. Hnece we can write

$\mathrm{\frac{C}{100}=\frac{W-39}{200}}$

For C = 39$^\circ$, $\frac{39}{100}=\mathrm{\frac{W-39}{200}}$

Solving, W = 117$^\circ$C

$$\begin{aligned} & Y_1=\frac{\mathrm{n} \lambda_1 D}{\mathrm{~d}}, \quad Y_2=\frac{\mathrm{n} \lambda_2 \mathrm{D}}{\mathrm{d}} \\ & \therefore \frac{\mathrm{Y}_1}{\mathrm{Y}_2}=\frac{2 \lambda_1}{\lambda_2}\end{aligned}$$

In AND gate, output is 1 only if both the inputs are '1'. Hence output will be '0' for (0, 1) and (1, 0)

In NOR gate, output is '1' only if both the inputs are '0'. Hence, output will be '0' for (0, 1) and (1, 0)

The image is real and hence inverted.

$\therefore \mathrm{\frac{v}{u}=-n}$ or $\mathrm{u=-\frac{v}{n}}$

By lens equation,

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

$\therefore$ $\mathrm{\frac{1}{v}+\frac{n}{v}=\frac{1}{f}}$

$\therefore \mathrm{\frac{1+n}{v}=\frac{1}{f}}$

$\therefore$ v = f(1 + n)

$\mathrm{g}^{\prime}=\mathrm{g}\left(1-\frac{\mathrm{d}}{\mathrm{R}}\right)$

If $g^{\prime}=\frac{g}{n}$ then $\frac{g}{n}=g\left(1-\frac{d}{R}\right)$

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

Applying Kirchoff's voltage law to loop ABEFA we get

$$\begin{aligned} & -28 \mathrm{I}_1-6-8=0 \\ & -28 \mathrm{I}_1-14=0 \\ \therefore & -28 \mathrm{I}_1=14 \\ \therefore \quad & \mathrm{I}_1=\frac{-14}{28}=-\frac{1}{2} \mathrm{~A} \end{aligned}$$