MHT CET 2023 10th May Evening Shift (Physics & Chemistry) Question Paper with Solutions

$$\begin{aligned} S_2 P & =\left(D^2+d^2\right)^{1 / 2} \\ & =D\left[1+\frac{d^2}{D^2}\right]^{1 / 2} \end{aligned}$$

Using binomial equation,

$\mathrm{S}_2 \mathrm{P}=\mathrm{D}\left[1+\frac{1}{2} \frac{\mathrm{d}^2}{\mathrm{D}^2}\right]^{1 / 2}=\mathrm{D}+\frac{\mathrm{d}^2}{2 \mathrm{D}}$

$\Rightarrow$ Path difference $=\frac{d^2}{2 D}$

For dark fringe, $\frac{d^2}{2 D}=\frac{\lambda}{2}$

$\therefore \quad \lambda=\frac{d^2}{D}$

From Stefan-Boltzmann's law,

$\mathrm{P}=\frac{\mathrm{dQ}}{\mathrm{dt}}=\mathrm{A \sigma T}^4$

Also, from Wien's displacement law,

$$\begin{aligned} & \lambda_{\max }=\frac{\mathrm{b}}{\mathrm{T}}(\mathrm{b} \rightarrow \text { Wien's constant }) \\ & \Rightarrow \mathrm{T}=\frac{\mathrm{b}}{\lambda} \end{aligned}$$

$$\begin{aligned} & \therefore \quad \mathrm{P}=\mathrm{A} \cdot \sigma\left(\frac{\mathrm{b}}{\lambda}\right)^4 \\ & \Rightarrow \mathrm{P} \propto \frac{1}{(\lambda)^4} \end{aligned}$$

$\therefore \quad$ Ratio of power dissipated is $\frac{\mathrm{P}_2}{\mathrm{P}_1}=\left(\frac{\lambda_1}{\lambda_2}\right)^4$ Given $\lambda_1=\lambda$ and $\lambda_2=\frac{2 \lambda}{3}$

$\therefore \quad \frac{\mathrm{P}_2}{\mathrm{P}_1}=\frac{(\lambda)^4}{\left(\frac{2 \lambda}{3}\right)^4}=\frac{81}{16}$

As the currents in an inductor and capacitor are out of phase by $180^{\circ}$, we can write

Current through the capacitor $\mathrm{I}_{\mathrm{C}}=0.9 \mathrm{~A}$

Current through the inductor $\mathrm{I}_{\mathrm{L}}=-(0.6 \mathrm{~A})$

$$\begin{aligned} \therefore \quad \text { Total current drawn from the source } & =\mathrm{I}_{\mathrm{C}}+\mathrm{I}_{\mathrm{L}} \\ & =0.9-0.6 \\ & =0.3 \mathrm{~A} \end{aligned}$$

From the question, it can be understood that the total distance the ball falls is $(\mathrm{S}+\mathrm{h})$

The spring is compressed through a length $h$

$\therefore \quad$ Loss of P.E. by the ball $=\mathrm{mg}(\mathrm{S}+\mathrm{h})$

Work done on the spring $=\frac{1}{2} \mathrm{Kh}^2$

Using law of conservation of energy,

$$\begin{aligned} & \frac{1}{2} \mathrm{Kh}^2=\mathrm{mg}(\mathrm{S}+\mathrm{h}) \\ \therefore \quad & \mathrm{K}=\frac{2 \mathrm{mg}(\mathrm{S}+\mathrm{h})}{\mathrm{h}^2} \end{aligned}$$

Equation of the given harmonic progressive wave $y=A \sin (100 \pi t+3)$ ...... (i)

General equation of a harmonic wave

$\mathrm{y}=\mathrm{A} \sin (\mathrm{wt}+\mathrm{kx})$ ..... (ii)

From equations (i) and (ii), $\omega=100 \pi, k=3$

But, $\mathrm{k}=\frac{2 \pi}{\lambda} \Rightarrow 3=\frac{2 \pi}{\lambda}$

$\therefore \quad \lambda=\frac{2 \pi}{3}$

We also know,

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

$$\begin{aligned} \therefore \quad \Delta x & =\frac{2 \pi}{3 \times 2 \pi} \times \frac{\pi}{3} \quad \ldots .\left(\text { Given } \Delta \phi=\frac{\pi}{3}\right) \\ & =\frac{\pi}{9} \mathrm{~m} \end{aligned}$$

Kinetic energy loss of the gas is

$\Delta \mathrm{E}=\frac{1}{2} M V^2 \cdot \mathrm{n}$ ..... (i) ....(where $\mathrm{n}$ is the no. of moles of the gas)

Heat gained by the gas due to temperature change $\Delta \mathrm{T}$ is $\Delta Q={ }_n C_V \Delta T$

But, $\mathrm{C}_{\mathrm{V}}=\frac{3}{2} \mathrm{R}$ ....(gas is monoatomic)

$\therefore \quad \Delta \mathrm{Q}=\frac{3}{2} \mathrm{R} \cdot \mathrm{n} \Delta \mathrm{T}$ ..... (ii)

Equating (i) and (ii),

$$\begin{aligned} & \Delta \mathrm{E}=\Delta \mathrm{Q} \\ & \frac{1}{2} \mathrm{MV}^2 \cdot \mathrm{n}=\frac{3}{2} \mathrm{R} \cdot \mathrm{n} \Delta \mathrm{T} \\ & \Delta \mathrm{T}=\frac{\mathrm{MV}}{3 \mathrm{R}} \end{aligned}$$

Fundamental frequency of the first wire is

$\mathrm{n}=\frac{1}{2 l_1} \sqrt{\frac{\mathrm{T}}{\mathrm{m}}}=\frac{1}{2 l_1} \sqrt{\frac{\mathrm{T}}{\pi \mathrm{r}_1^2 \rho}}=\frac{1}{2 l_1 \mathrm{r}_1} \sqrt{\frac{\mathrm{T}}{\pi \rho}}$

The first overtone $n_1=2 n=\frac{1}{l_1 r_1} \sqrt{\frac{T}{\pi \rho}}$

Similarly, the second overtone of the second wire will be,

$\mathrm{n}_2=\frac{3}{2 l_2 \mathrm{r}_2} \sqrt{\frac{\mathrm{T}}{\pi \rho}}$

Given that $\mathrm{n}_1=\mathrm{n}_2$

$$\begin{aligned} & \therefore \quad \frac{1}{l_1 \mathrm{r}_1} \sqrt{\frac{\mathrm{T}}{\pi \rho}}=\frac{3}{2 l_2 \mathrm{r}_2} \sqrt{\frac{\mathrm{T}}{\pi \rho}} \\ & \therefore \quad 3 l_1 \mathrm{r}_1=2 l_2 \mathrm{r}_2 \\ & \frac{l_1}{l_2}=\frac{2 r_2}{3 r_1} \\ & =\frac{2 r_2}{3\left(2 r_2\right)} \quad\ldots\left(\because r_1=2 r_2\right) \\ & =\frac{1}{3} \\ \end{aligned}$$

Truth table of a NOR gate

The output matches with that of a NOR gate.

$\therefore$ The combination represents a NOR gate.

Isochoric processes are those in which the volume of the system remains constant. In a Pressure-Volume (P-V) graph, an isochoric process is represented by a vertical line because the volume does not change, while the pressure can vary.

Looking at the provided graph :

• Graph A shows a diagonal line, indicating changes in both pressure and volume.


• Graph B shows a horizontal line, indicating constant pressure, not constant volume. This represents an isobaric process.


• Graph C shows a vertical line, which is characteristic of an isochoric process, where the volume remains constant while the pressure changes.


• Graph D shows a curve, indicating that both pressure and volume are changing.

Therefore, the correct option that represents an isochoric process is :

Option C : C

Moment of Inertia $\mathrm{I}=\mathrm{MR}^2$

Radius of the $\mathrm{n}^{\text {th }}$ Bohr orbit is,

$\mathrm{r}_{\mathrm{n}}=\frac{\varepsilon_0 \mathrm{~h}^2 \mathrm{n}^2}{\pi \mathrm{me}^2}$

For $\mathrm{n}=2$,

$\mathrm{r}_2=\frac{4 \varepsilon_0 \mathrm{~h}^2}{\pi \mathrm{me}^2}$

$\therefore \quad$ Moment of inertia of the electron in the $2^{\text {nd }}$ orbit is

$$\begin{aligned} \text { M.I } & =m \times\left[\frac{4 \varepsilon_0 h^2}{\pi m e^2}\right]^2 \\ & =m \times \frac{16 \varepsilon_0^2 h^4}{\pi^2 m^2 e^4} \\ & =\frac{16 \varepsilon_0^2 h^4}{\pi^2 m e^4} \end{aligned}$$

Energy stored in an inductor $\mathrm{E}=\frac{1}{2} \mathrm{LI}^2$

$$\begin{aligned} & =\frac{1}{2} \times\left(100 \times 10^{-3}\right) \times 1 \\ & =0.05 \mathrm{~J} \end{aligned}$$

As a bar magnet falls through the pipe, an emf is induced in the body of the pipe due to change in magnetic flux. However, the free fall of the magnet will be opposed due to the development of eddy currents in accordance with Lenz's law. This process continues till the bar magnet attains a constant terminal velocity.

Width of central maximum $\mathrm{W}_{\mathrm{c}}=2\left[\frac{\lambda \mathrm{D}}{\mathrm{a}}\right]$

$\Rightarrow \mathrm{W}_{\mathrm{c}} \propto \frac{1}{\mathrm{a}}$

$\therefore \quad$ Angular width of principal maximum decreases with increase in width of the slit.

For a Metre bridge,

$\frac{\mathrm{X}}{\mathrm{R}}=\frac{l_{\mathrm{X}}}{l_{\mathrm{R}}}$

After doubling and interchanging the left and right gap resistances, the ratio of the resistances do not change.

This means the null point does not change.

$\therefore \quad$ The new position will remain at $(100-l)$.

Given: Density of Si atoms $=4 \times 10^{28}$ atoms $/ \mathrm{m}^3$

After doping with $1 \mathrm{~ppm}$ of Sb,

No. of Sb atoms $=\frac{4 \times 10^{28}}{10^6}$

$=4 \times 10^{22}$

The above number of $\mathrm{Sb}$ atoms donates 1 electron each.

$\therefore \quad$ The total number of free electrons will be $4 \times 10^{22} \mathrm{~m}^{-3}$

From the fig., we can see that the forces due to masses at $\mathrm{Q}$ and $\mathrm{R}$ cancel each other as they are equal and opposite. The force at $\mathrm{P}$ is only due to $\mathrm{m}_1$.

In $\triangle \mathrm{PQS}$,

$\mathrm{h}=\frac{\mathrm{L}}{3} \cos 30^{\circ}=\frac{\mathrm{L} \sqrt{3}}{6}$

$\therefore \quad$ Force on $\mathrm{m}_2$ due to $\mathrm{m}_1$ at $\mathrm{P}$ is

$$\begin{aligned} F & =\frac{G m_1 m_2}{\left(\frac{L \sqrt{3}}{6}\right)^2} \\ & =\frac{G m_1 m_2 \cdot 12}{L^2} \\ & =\frac{12 \mathrm{Gm}_1 \mathrm{~m}_2}{\mathrm{~L}^2} \end{aligned}$$

Given: Resistance of the galvanometer = G

Range of the galvanometer $=\mathrm{V}_{\mathrm{g}}$

The series resistance value to be used for converting the galvanometer into a voltmeter of range 0 to $\mathrm{V}_{\mathrm{g}^{\prime}}$ is,

$\mathrm{R}=\frac{\mathrm{V}_8}{\mathrm{I}_8}-\mathrm{G}$

Also,

$I_g=\frac{V_8}{G}$

To increase the measuring range to $\mathrm{V}$, the new resistance value

$$\begin{aligned} \mathrm{R}^{\prime} & =\frac{\mathrm{V}}{\left(\frac{\mathrm{V}_{\mathrm{g}}}{\mathrm{G}}\right)}-\mathrm{G} \\ & =\frac{\mathrm{VG}}{\mathrm{V}_{\mathrm{g}}}-\mathrm{G}=\mathrm{G}\left(\frac{\mathrm{V}}{\mathrm{V}_{\mathrm{g}}}-1\right) \end{aligned}$$

We know,

Coefficient of Linear expansion $\alpha=\frac{\mathrm{L}_2-\mathrm{L}_1}{\mathrm{~L}_1 \Delta \mathrm{T}}$ ...... (i)

Given: $\Delta \mathrm{T}=60-0=60^{\circ} \mathrm{C}$

$\mathrm{L}_1=2 \mathrm{~m}$ and $\mathrm{L}_2=2.0016$

Substituting the given values into (i),

$\alpha=\frac{.0016}{120}=1.33 \times 10^{-5} /{ }^{\circ} \mathrm{C}$

Given, $\mathrm{X}_{\mathrm{C}}=\mathrm{X} \Omega$

$\Rightarrow \frac{1}{2 \pi \mathrm{fC}}=\mathrm{X} \Omega$

New Capacitance $\mathrm{C}=2 \mathrm{C}$ and new frequency $\mathrm{f}^{\prime}=2 \mathrm{f}$

$\therefore \quad$ New capacitive reactance

$\mathrm{X}_{\mathrm{C}}^{\prime}=\frac{1}{2 \pi(2 \mathrm{f})(2 \mathrm{C})}$

$$\begin{aligned} & =\frac{1}{(2 \pi)(4 \mathrm{fC})} \\ & =\frac{1}{4} X_C \\ & =\frac{X}{4} \Omega \end{aligned}$$

At the critical temperature of a liquid, the surface tension becomes zero. This happens because at the critical temperature, the properties of the liquid and gas phases become indistinguishable, leading to the disappearance of the liquid-gas interface. Surface tension is a property that arises due to the difference in intermolecular forces between the liquid and gas phases. When these phases become indistinguishable at the critical point, there's no longer a distinct interface, and thus no surface tension.

Therefore, the correct answer is :

Option A : zero

$\frac{4}{3} \pi R^3=n \times \frac{4}{3} \pi r^3\quad$ ..... (given)

$\therefore \quad R^3 =n r^3$

Energy released,

$\Delta U =T \times 4 \pi r^2 \times n-T \times 4 \pi R^2$

$=T \times 4 \pi \frac{R^3}{r}-T \times 4 \pi R^2$

This energy is converted into K.E

$$\begin{array}{ll} \therefore \quad & \frac{1}{2} m v^2=T \times 4 \pi R^3\left[\frac{1}{r}-\frac{1}{R}\right] \\ & \Rightarrow \frac{1}{2} \rho \times \frac{4}{3} \pi R^3 \times v^2=T \times 4 \pi R^3\left[\frac{1}{r}-\frac{1}{R}\right] \\ & v^2=\frac{6 T}{\rho}\left[\frac{1}{r}-\frac{1}{R}\right] \\ \therefore \quad & v=\sqrt{\frac{6 T}{\rho}\left[\frac{1}{r}-\frac{1}{R}\right]} \end{array}$$

From law of conservation of linear momentum,

$$\begin{aligned} & M v \hat{i}+2 M(3 v \hat{j})=3 M \vec{v} \\ & v \hat{i}+6 v \hat{j}=3 \vec{v} \\ & \vec{v}=\frac{v}{3} \hat{i}+2 v \hat{j} \end{aligned}$$

Relative velocity of one train w. r. t other $=5+0 =15 \mathrm{~m} / \mathrm{s}$

Total length to cross $(\mathrm{L})=30+30=60 \mathrm{~m}$

$\therefore \quad t=\frac{L}{V}=\frac{60}{15}=4 s$

Fundamental frequency of a closed pipe, $\mathrm{n}_1=\frac{\mathrm{v}}{4 \mathrm{~L}_1}$

Frequency of the second overtone, $n_2=\frac{3 v}{2 L_2}$

Given $\mathrm{n}_1=\mathrm{n}_2$

$\Rightarrow \frac{\mathrm{v}}{4 \mathrm{~L}_1}=\frac{3 \mathrm{v}}{2 \mathrm{~L}_2}$

$\therefore \quad \frac{\mathrm{L}_1}{\mathrm{~L}_2}=\frac{1}{6}$

$P=\frac{1}{f}$

For combination of lenses,

$\mathrm{P}=\mathrm{P}_1+\mathrm{P}_2$

Given: $P_1=-15 D$ and $P_2=+5 F$

$$\begin{aligned} & \therefore \quad P=-10 D \\ & \Rightarrow f=-\frac{1}{10} m \\ & \therefore \quad f=-10 \mathrm{~cm} \end{aligned}$$

$$\begin{aligned} & \text { From } \lambda=\frac{\mathrm{h}}{\sqrt{2 \mathrm{mqV}}}, \\ & \lambda \propto \frac{1}{\sqrt{\mathrm{V}}} \end{aligned}$$

If potential difference is doubled, $\lambda \propto \frac{1}{\sqrt{2 \mathrm{~V}}}$

$\therefore \quad \lambda$ is decreased by $\frac{1}{\sqrt{2}}$ times.

In the active region of operation, the emitter-base junction is forward biased in both $n\text{-}p\text{-}n$ and $p\text{-}n\text{-}p$ transistors, causing the emitter to inject its majority charge carriers into the base region.

In an $n\text{-}p\text{-}n$ transistor, the emitter is of $n$-type material and injects electrons into the $p$-type base. In a $p\text{-}n\text{-}p$ transistor, the emitter is of $p$-type material and injects holes into the $n$-type base.

Hence, the correct option is (C).

Given: $v=\frac{v_e}{2}$

If body is projected with velocity $\mathrm{v}\left(\mathrm{v} < \mathrm{v}_{\mathrm{e}}\right)$ then height up to which it will rise, $h=\frac{R}{\left(\frac{v_c^2}{v^2}-1\right)}$

$\therefore \quad \mathrm{h}=\frac{\mathrm{R}}{\left(\frac{\mathrm{v}_{\mathrm{e}}}{\mathrm{v}_{\mathrm{e}} / 2}\right)^2-1}=\frac{\mathrm{R}}{4-1}=\frac{\mathrm{R}}{3}$

We know,

Potential Energy $\mathrm{E}_{\mathrm{P}}=\frac{1}{2} \mathrm{Kx}^2$

$\therefore \quad \mathrm{E}_1=\frac{1}{2} \mathrm{Kx}^2 \Rightarrow \mathrm{x}=\sqrt{\frac{2 \mathrm{E}_1}{\mathrm{~K}}}$ .... (i)

and

$E_2=\frac{1}{2} K y^2 \Rightarrow y=\sqrt{\frac{2 E_2}{K}}$ ..... (ii)

Given, total displacement $=(x+y)$

$\therefore \quad$ Potential energy at displacement $(x+y)$,

$\mathrm{E}$ is $\frac{1}{2} \mathrm{~K}(\mathrm{x}+\mathrm{y})^2$

$$\begin{aligned} & =\frac{1}{2} K\left(\sqrt{\frac{2 E_1}{K}}+\sqrt{\frac{2 E_2}{K}}\right)^2 \\ & =\frac{1}{2} K\left[\frac{2 E_1}{K}+\frac{2 E_2}{K}+2\left(\sqrt{\frac{E_1}{K}}\right)\left(\sqrt{\frac{2 E_2}{K}}\right)\right] \\ & =\left(\sqrt{E_1}+\sqrt{E_2}\right)^2=\left(E_1+E_2+2 \sqrt{E_1 E_2}\right) \end{aligned}$$

According to the gas equation, $\mathrm{PV}=\mathrm{Nk}_{\mathrm{B}} \mathrm{T}$

For the first gas, we have,

$\mathrm{PV}=\mathrm{N}_1 \mathrm{k}_{\mathrm{B}} \mathrm{T}$ ..... (i)

For the second gas, we have,

$$\begin{aligned} & (2 \mathrm{P})\left(\frac{\mathrm{V}}{4}\right)=\mathrm{N}_2 \mathrm{k}_{\mathrm{B}}(2 \mathrm{~T}) \\ & \mathrm{PV}=4 \mathrm{~N}_2 \mathrm{k}_{\mathrm{B}} \mathrm{T} \quad \text{.... (ii)} \end{aligned}$$

From equations (i) and (ii)

$\mathrm{N}_1=4 \mathrm{~N}_2 \Rightarrow \frac{\mathrm{N}_1}{\mathrm{~N}_2}=4$

Let velocity of pulse at lower end be $v_1$ and at top be $v_2$

$\therefore \quad \frac{\lambda_2}{\lambda_1}=\frac{\mathrm{v}_2}{\mathrm{v}_1} \quad\left(\because \lambda=\frac{\mathrm{v}}{\mathrm{n}} \text { and } \mathrm{n}=\text { constant }\right)$

Velocity of transverse wave on a string is

$v=\sqrt{\frac{T}{m}}$

where, $\mathrm{m}$ is linear density.

In this case, $\mathrm{v} \propto \sqrt{\mathrm{T}}$

$\therefore \quad \frac{\lambda_2}{\lambda_1}=\frac{\mathrm{v}_2}{\mathrm{v}_1}=\sqrt{\frac{\mathrm{T}_2}{\mathrm{~T}_1}}=\sqrt{\frac{\left(\mathrm{m}_2+\mathrm{m}_1\right)}{\mathrm{m}_2}}$

Where, $T_2$ is tension at upper end of rope and $T_1$ is tension at lower end of rope.

$\Rightarrow \frac{\lambda_1}{\lambda_2}=\sqrt{\frac{m_2}{m_2+m_1}}$

Given: $\mathrm{n}=500$ turns$/ \mathrm{m}, \mathrm{I}=3 \mathrm{~A}$

$$\begin{aligned} & \mu_{\mathrm{r}}=5001 \\ & \therefore \quad \mu=n I=500 \times 3 \mathrm{~A} \\ & =1500 \mathrm{~A} / \mathrm{m} \\ & \text { But, } \chi_{\mathrm{m}}=\mu_{\mathrm{r}}-1 \\ & =5001-1 \\ & =5000 \\ & \therefore \quad \text { Magnetization } \mathrm{M}=\chi_{\mathrm{m}} \mathrm{H} \\ & =5000 \times 1500 \\ & =7.5 \times 10^6 \mathrm{Am}^{-1} \\ \end{aligned}$$

\begin{aligned} & d V=\vec{E} \cdot d x \\ & \int_\limits{V_0}^{V_A} d V=-\int_\limits0^2 30 x^2 d x \\ & V_A-V_0=-\left[10 x^3\right]_0^2=-80 J \end{aligned}

From parallel axis theorem,

$\mathrm{I}_{\mathrm{o}}=\mathrm{I}_{\mathrm{c}}+\mathrm{Mh}^2$

Let the moment of inertia of sphere 1 be

$\mathrm{I}_1=\frac{2}{5} \mathrm{M}\left(\frac{\mathrm{R}}{2}\right)^2+\mathrm{M}(2 \mathrm{R})^2$

and,

Let the moment of inertia of sphere 2 be

$\mathrm{I}_2=\frac{2}{5} \mathrm{M}\left(\frac{\mathrm{R}}{2}\right)^2$

Moment of inertia of the $\operatorname{rod} \mathrm{I}_3=0$

$\therefore \quad$ Moment of inertia of the system,

$$\begin{aligned} \mathrm{I} & =\mathrm{I}_1+\mathrm{I}_2+\mathrm{I}_3 \\ \mathrm{I} & =\frac{2}{5} \mathrm{M}\left(\frac{\mathrm{R}}{2}\right)^2+\mathrm{M}(2 \mathrm{R})^2+\frac{2}{5} \mathrm{M}\left(\frac{\mathrm{R}}{2}\right)^2 \\ & =\frac{4}{5} \mathrm{M}\left(\frac{\mathrm{R}}{2}\right)^2+4 \mathrm{MR}^2 \\ & =\frac{1}{5} \mathrm{MR}^2+4 \mathrm{MR}^2 \\ & =\frac{21}{5} \mathrm{MR}^2 \end{aligned}$$