MHT CET 2023 9th May Evening Shift (Physics & Chemistry) Question Paper with Solutions

Using Biot-Savart's Law,

$$\begin{aligned} \mathrm{B} & =\frac{\mu_0 \mathrm{I}}{2 \pi \mathrm{R}} \\ \therefore \quad \mathrm{R} & =\frac{\mu_0 \mathrm{I}}{2 \pi \mathrm{B}} \\ & =\frac{4 \pi \times 10^{-7} \times 12}{2 \pi \times 3 \times 10^{-5}} \\ & =8 \times 10^{-2} \mathrm{~m} \\ & =80 \mathrm{~mm} \end{aligned}$$

For least energetic photon emitted in Lyman series, $E=E_2-E_1=10.2 \mathrm{~eV}$

$$\begin{aligned} \lambda=\frac{\mathrm{hc}}{\mathrm{E}} & =\frac{1240}{10.2} \mathrm{~nm} \\ & =121.57 \mathrm{~nm} \approx 122 \mathrm{~nm} \end{aligned}$$

We know, $\mathrm{Q}=\mathrm{C} . \mathrm{V}$

$$\begin{array}{ll} \therefore \quad Q_1=10 \mathrm{C} \text { and } \mathrm{Q}_2=5 \mathrm{C} & \\ \text { Energy stored, } \mathrm{E}=\frac{1}{2} \mathrm{CV}^2 \\ \therefore \quad \mathrm{E}_1=\frac{1}{2} \mathrm{C}_1 \mathrm{~V}^2=\frac{1}{2} \times 2 \mathrm{C} \times 25 =25 \mathrm{~J} \\ \text { Similarly, } \mathrm{E}_2=\frac{1}{2} \mathrm{C}_2 \mathrm{~V}^2=\frac{1}{2} \times \mathrm{C} \times 25 =12.5~\mathrm{J}\\ \therefore \quad \frac{\mathrm{E}_1-\mathrm{E}_2}{\mathrm{Q}_1-\mathrm{Q}_2}=\frac{12.5}{5}=\frac{5}{2} \end{array}$$

$$\begin{aligned} & \mathrm{eV}_0=\mathrm{h} v-\mathrm{h} v_0 \quad \text{..... (i)}\\ & \frac{\mathrm{eV}_0}{4}=\frac{\mathrm{h} v}{2}-\mathrm{hv} v_0 \quad \text{..... (ii)} \end{aligned}$$

Dividing equation (i) by equation (ii),

$$\begin{array}{ll} & 4=\frac{v-v_0}{\frac{v}{2}-v_0} \\ \therefore \quad & 2 v-4 v_0=v-v_0 \\ \therefore \quad & 3 v_0=v \\ \therefore \quad & v_0=\frac{v}{3} \end{array}$$

In diffraction of light by single slit, the width of central maximum is given as

$\mathrm{W}_{\mathrm{c}}=\frac{2 \lambda \mathrm{D}}{\mathrm{d}}$

Given: $\mathrm{W}_{\mathrm{c}}=\mathrm{d}$

$$\begin{aligned} \therefore \quad d & =\frac{2 \lambda D}{d} \\ \Rightarrow D & =\frac{d^2}{2 \lambda} \end{aligned}$$

The Reynolds number (R_e) is given by the formula :

R_e $= \frac{\rho v d}{\eta}$

where $\rho$ is the density of the fluid, $v$ is the velocity of the fluid, $d$ is the characteristic linear dimension (diameter of the pipe in this case), and $\eta$ is the dynamic viscosity of the fluid.

Based on this formula, the correct answer would be Option B : $\frac{d \rho v}{\eta}$.

$\mathrm{T}_1=2 \pi \sqrt{\frac{\mathrm{m}}{\mathrm{k}}} .$

Also, spring constant $(\mathrm{k}) \propto \frac{1}{\text { Length }(l)}$

When the spring is half in length, then $k$ becomes twice.

$$\begin{array}{ll} \therefore & \mathrm{T}_2=2 \pi \sqrt{\frac{\mathrm{m}}{2 \mathrm{k}}} \\ \therefore & \frac{\mathrm{T}_1}{\mathrm{~T}_2}=\frac{1}{\frac{1}{\sqrt{2}}}=\sqrt{2} \end{array}$$

$$\begin{aligned} & \text { We know } v_{\text {rms }}=\sqrt{\frac{3 R T}{M}} \\ & \Rightarrow \mathrm{v}_{\text {rms }} \propto \sqrt{\mathrm{T}} \\ & \therefore \quad \frac{\mathrm{v}_{\mathrm{mm}}}{\mathrm{v}_{\mathrm{mm}_2}}=\sqrt{\frac{\mathrm{T}_1}{\mathrm{~T}_2}}=\sqrt{\frac{300}{1200}}=\sqrt{\frac{1}{4}} \\ & \frac{\mathrm{v}_{\mathrm{ms}}}{\mathrm{v}_{\mathrm{ms}_2}}-\frac{1}{2} \\ & \therefore \quad \mathrm{v}_{\mathrm{mss}_2}=2 \cdot \mathrm{v}_{\mathrm{rms} 1} \\ & \end{aligned}$$

$$\begin{aligned} & \mathrm{R}=\frac{\mathrm{V}}{\mathrm{I}_{\mathrm{g}}}-\mathrm{G} \\ & \text { Here, } \mathrm{I}_{\mathrm{g}}=\frac{50}{5}=10 \mathrm{~mA} \\ \therefore \quad \mathrm{R} & =\frac{25}{10 \times 10^{-3}}-20 \\ & =2480 \Omega \text { in series } \end{aligned}$$

As the change is sudden, the process is adiabatic

$$\begin{array}{ll} \therefore & \mathrm{P}_1 \mathrm{~V}_1^\gamma=\mathrm{P}_2 \mathrm{~V}_2^\gamma \\ \therefore & \frac{\mathrm{P}_2}{\mathrm{P}_1}=\left[\frac{\mathrm{V}_1}{\mathrm{~V}_2}\right]^\gamma=\left[\frac{4}{1}\right]^{3 / 2}=\frac{8}{1} \end{array}$$

For an open pipe, $\mathrm{e}=0.6 \mathrm{~d}$

$$\begin{array}{ll} \therefore & \mathrm{d}=\frac{\mathrm{e}}{0.6} \\ \therefore & 2 \mathrm{r}=\frac{\mathrm{e}}{0.6} \\ \therefore & \mathrm{r}=\frac{0.8}{1.2}=\frac{2}{3} \mathrm{~cm} \end{array}$$

Coefficient of coupling,

$\mathrm{K}=\frac{\mathrm{M}}{\sqrt{\mathrm{L}_1 \mathrm{~L}_2}}=\frac{3}{\sqrt{36}}=0.5$

Magnetic moment of $\mathrm{rod}=\mathrm{M}$

Let $r$ be the radius after bending it into a semicircular arc.

$\therefore \quad$ The separation between the two ends is $2 \mathrm{r}$. Here length $=$ circumference of the semicircle i.e., $\mathrm{L}=\pi \mathrm{r}$

$\therefore \quad r=\frac{L}{\pi}$

Also $M=m \times L$ and $m=\frac{M}{L}$

$$\begin{aligned} \therefore \quad M_{\text {new }}=m(2 r) & =\frac{M}{L} \times \frac{2 L}{\pi} \\ & =\frac{2 M}{\pi} \end{aligned}$$

From charge configuration, at the centre electric field is non-zero.

Potential at the centre due to $2 \mathrm{q}$ charge $\mathrm{V}_{2 \mathrm{q}}=\frac{2 \mathrm{q}}{\mathrm{r}}$ and potential due to $-\mathrm{q}$ charge

$\mathrm{V}_{-\mathrm{q}}=-\frac{\mathrm{q}}{\mathrm{r}} \quad(\mathrm{r}=$ distance of centre point)

$\therefore \quad$ total potential $\mathrm{V}=\mathrm{V}_{2 \mathrm{q}}+\mathrm{V}_{-\mathrm{q}}+\mathrm{V}_{-\mathrm{q}}=0$

Given: $\mathrm{y}_1=0.1 \sin \left(100 \pi \mathrm{t}+\frac{\pi}{3}\right)$ and

$$\begin{aligned} \mathrm{y}_2=0.1 \cos (100 \pi \mathrm{t}) & \\ \therefore \quad \mathrm{v}_1=\frac{\mathrm{dy_{1 }}}{\mathrm{dt}} & =0.1 \times 100 \pi \cos \left(100 \pi \mathrm{t}+\frac{\pi}{3}\right) \\ \mathrm{v}_2=\frac{\mathrm{dy_{2 }}}{\mathrm{dt}} & =-0.1 \times 100 \pi \sin (100 \pi \mathrm{t}) \\ & =(0.1 \times 100 \pi) \cos \left(100 \pi \mathrm{t}+\frac{\pi}{2}\right) \end{aligned}$$

Phase difference of velocity of first particle with respect to the velocity of second particle at $t=0$ is

$\therefore \quad \Delta \phi=\phi_1-\phi_2=\frac{\pi}{3}-\frac{\pi}{2}=\frac{-\pi}{6}$

According to the kinetic theory of gases, the pressure $𝑃$ exerted by an ideal gas is given by

$𝑃=\frac{1}{3}𝜌\overline{{𝑣}^2}$

where $𝜌$ is the density of the gas and $\overline{{𝑣}^2}$ is the mean square speed of the gas molecules.

For a given density,

$𝑃\propto \overline{{𝑣}^2}$

Therefore, the pressure is directly proportional to the mean of the square of the speed of the gas molecules.

Hence, the correct option is (B).

In a parallel $\mathrm{LC}$ circuit at $\omega^2=\frac{1}{\mathrm{LC}}$, the current is minimum.

As current and voltage are out of phase by $90^{\circ}$, the voltage would be maximum.

General equation of a simple harmonic progressive wave,

$y=A \sin 2 \pi\left[\frac{t}{T}-\frac{x}{\lambda}\right]$

Given: $\mathrm{y}=\mathrm{a} \sin 2 \pi(b t-c x)$

$\Rightarrow \mathrm{A}=\mathrm{a}, \frac{1}{\mathrm{~T}}=\mathrm{b} \text { and } \frac{1}{\lambda}=\mathrm{cx}$

Also, $\left(\mathrm{v}_{\mathrm{p}}\right)_{\max }=\mathrm{a} \omega=\mathrm{a}(2 \pi \mathrm{n})=\frac{\mathrm{a} 2 \pi}{\mathrm{T}}$

$=\frac{\mathrm{A} 2 \pi}{\mathrm{T}}=\mathrm{a} 2 \pi \mathrm{b}$

From $\mathrm{v}=\frac{\lambda}{\mathrm{T}}=\frac{1 / \mathrm{c}}{1 / \mathrm{b}}=\frac{\mathrm{b}}{\mathrm{c}}$

Given: $\left(\mathrm{v}_{\mathrm{p}}\right)_{\max }=\frac{1}{2} \mathrm{v}$

$$\begin{aligned} & \Rightarrow 2 \pi a b=\frac{1}{2} \times \frac{b}{c} \\ \therefore \quad & c=\frac{1}{4 \pi a} \end{aligned}$$

$\mathrm{F}=\frac{\mu_0}{4 \pi} \frac{2 \mathrm{I}_1 \mathrm{I}_2}{\mathrm{r}} l=10^{-3} \mathrm{~N}$

When current in both the wires is doubled, then

$\mathrm{F}^{\prime}=\frac{\mu_0}{4 \pi} \frac{2\left(2 \mathrm{I}_1 \times 2 \mathrm{I}_2\right)}{\mathrm{r}} l=4 \times 10^{-3} \mathrm{~N}$

Given : $l=2$ m, $\theta=60^\circ$, m = 200 g = 2g

From the figure,

$\mathrm{T}=\mathrm{mr} \omega^2$ ..... (i)

also $\mathrm{T} \cos \theta=\mathrm{mg}$ ..... (ii)

putting (i) into (ii)

$m r \omega^2-\cos \theta=m g$ ..... (iii)

putting the given values into equation (iii)

$$\begin{aligned} & 2 \times 2 \times \omega^2 \frac{1}{2}=2 \times 10 \quad \ldots\left(\because \cos 60=\frac{1}{2}\right) \\ & \omega^2=10 \\ & \Rightarrow \omega=\sqrt{10} \quad \ldots .(\because \sqrt{10}=\sqrt{2 \times 2 \times 2.5}) \\ & =2 \sqrt{2.5} ~\mathrm{rad} / \mathrm{s} \end{aligned}$$

$$\begin{aligned} & \text { Using } 3^{\text {rd }} \text { equation of motion, } \\ & v^2=u^2+2 a s \\ \therefore \quad v^2 & =2 a_t s \quad \text{.... (For u = 0)}\\ \therefore \quad & a_t=\frac{v^2}{2 s} \end{aligned}$$

By the end of $3^{\text {rd }}$ revolution, distance covered, $\mathrm{s}=3(2 \pi \mathrm{r})=6 \pi \mathrm{r}$

$\therefore \quad a_{\mathrm{t}}=\frac{\mathrm{v}^2}{2 \times 6 \pi \mathrm{r}}$ .... (i)

Also, $\frac{1}{2} \mathrm{mv}^2=\mathrm{E}$

$\therefore \quad v^2=\frac{2 \mathrm{E}}{\mathrm{m}}$ ..... (ii)

Substituting equation (ii) in equation (i),

$a_{\mathrm{t}}=\frac{2 \mathrm{E}}{2 \times 6 \pi \mathrm{r} \times \mathrm{m}}=\frac{\mathrm{E}}{6 \pi \mathrm{rm}}$

According to Snell's Law,

$\frac{\sin i}{\sin r}=\frac{1}{n}$

But $\mathrm{i}=\mathrm{r}$,

$\therefore \quad \frac{\sin \mathrm{r}}{\sin \mathrm{r}^{\prime}}=\frac{1}{\mathrm{n}}$

From figure $r^{\prime}+r+90=180^{\circ}$

$\therefore \quad r^{\prime}=180^{\circ}-90^{\circ}-r=90^{\circ}-r$

$\therefore \quad \Rightarrow \frac{\sin \mathrm{r}}{\sin \left(90^{\circ}-r\right)}=\frac{1}{\mathrm{n}}$

$\frac{\sin r}{\cos r}=\frac{1}{n}$

$$\begin{array}{ll} \therefore \quad & \tan \mathrm{r}=\frac{\mathrm{l}}{\mathrm{n}} \\ & \text { Critical angle is given by } \sin \mathrm{i}_{\mathrm{c}}=\frac{1}{\mathrm{n}} \\ \therefore \quad & \tan \mathrm{r}=\sin \mathrm{i}_{\mathrm{c}} \\ \therefore \quad & \mathrm{i}_{\mathrm{c}}=\sin ^{-1}(\tan \mathrm{r}) \end{array}$$

Electric force, $F=q E$ and potential difference, $\mathrm{V}=\mathrm{Ed}$

$\therefore \quad \mathrm{V}=\frac{\mathrm{Fd}}{\mathrm{q}}=\frac{3000 \times 10^{-2}}{3}=10 \mathrm{~V}$

[Note: Framing of question is modified to arrive at the correct answer.]

$$\begin{aligned} \mathrm{T}=2 \pi \sqrt{\frac{(\mathrm{R}+\mathrm{h})^3}{\mathrm{gR}^2}} & =2 \pi \sqrt{\frac{(2 \mathrm{R})^3}{\mathrm{gR}^2}} \quad \cdots .(\because \mathrm{h}=\mathrm{R}) \\ & =4 \pi \sqrt{\frac{2 \mathrm{R}}{\mathrm{g}}} \end{aligned}$$

In volumetric expansion for a cube,

Change in volume $\Delta \mathrm{V}=\mathrm{V} 3 \alpha \Delta \mathrm{T}$

$$\begin{aligned} & \Delta \mathrm{V}=3 \times 18 \times 10^{-6} \times 100 \\ & \Delta \mathrm{V}=54 \times 10^{-4} \mathrm{~m}^3 \end{aligned}$$

$$\begin{aligned} \text { K.E } & =\frac{1}{2} m \omega^2\left(A^2-x^2\right) \text { and } \\ \text { P.E } & =\frac{1}{2} m \omega^2 x^2 \\ \therefore \quad \text { T.E } & =\text { K.E }+ \text { P.E } \\ & =\frac{1}{2} m \omega^2\left(A^2-x^2\right)+\frac{1}{2} m \omega^2 x^2 \\ & =\frac{1}{2} m \omega^2 A^2 \\ & \text { Given: K.E }=\text { P.E } \\ \therefore \quad A^2 & -x^2=x^2 \\ \therefore \quad & A^2=2 x^2 \\ \therefore \quad x & =\sqrt{\frac{A^2}{2}}=\frac{A}{\sqrt{2}} \\ \therefore \quad x & =\frac{4}{\sqrt{2}}=2 \sqrt{2} ~\mathrm{c m} \end{aligned}$$

$$\begin{aligned} X_C & =\frac{1}{2 \pi f c} \\ \therefore \quad C & =\frac{1}{2 \pi f(5)} \quad \ldots .\left(\because X_C=5 \Omega\right) \end{aligned}$$

New reactance,

$X_C^{\prime}=\frac{1}{2 \pi f^{\prime} C}=\frac{1}{2 \pi(2 f) C}=\frac{X_C}{2}=\frac{1}{2} \times 5=2.5 \Omega$

The total electromagnetic force (Lorentz force) acting on a charge $𝑞$ moving with velocity $\overrightarrow{𝑣}$ in the presence of both an electric field $\overrightarrow{𝐸}$ and a magnetic field $\overrightarrow{𝐵}$ is the vector sum of the electric force and the magnetic force:

$\overrightarrow{𝐹}={\overrightarrow{𝐹}}_{𝑒}+{\overrightarrow{𝐹}}_{𝑚}$

Here, ${\overrightarrow{𝐹}}_{𝑒}=𝑞\overrightarrow{𝐸}$ and ${\overrightarrow{𝐹}}_{𝑚}=𝑞(\overrightarrow{𝑣}\times \overrightarrow{𝐵})$.

Therefore,

$\overrightarrow{𝐹}=𝑞\overrightarrow{𝐸}+𝑞(\overrightarrow{𝑣}\times \overrightarrow{𝐵})$

Hence, the correct option is (D).

Given: Fringe width $\mathrm{W}=2 \mathrm{~mm}$

The distance of the $\mathrm{n}^{\text {th }}$ bright fringe from centre of the screen $y_n=\frac{n \lambda D}{d}$ ...... (i)

The distance of the $\mathrm{n}^{\text {th }}$ dark fringe from centre of the screen $y_n^{\prime}=(2 n-1) \frac{\lambda D}{2 d}$ ..... (ii)

Substituting, $\mathrm{n}=13$ in (i) and $\mathrm{n}=4$ in (ii) we get $\mathrm{y}_{13}=\frac{13 \lambda \mathrm{D}}{\mathrm{d}}$ and $\mathrm{y}_4^{\prime}=\frac{7}{2} \frac{\lambda \mathrm{D}}{\mathrm{d}}$

$\therefore \quad$ The separation between the $13^{\text {th }}$ bright fringe and the $7^{\text {th }}$ dark fringe is

$$\begin{aligned} & \mathrm{y}_{13}-\mathrm{y}_4^{\prime} \\ & =\frac{13 \lambda \mathrm{D}}{\mathrm{d}}-\frac{7}{2} \frac{\lambda \mathrm{D}}{\mathrm{d}} \\ & =\left(13-\frac{7}{2}\right) \frac{\lambda \mathrm{D}}{\mathrm{d}} \\ & =\frac{19}{2} \frac{\lambda \mathrm{D}}{\mathrm{d}}=\frac{19}{2} \mathrm{~W} \\ & \text { but } \mathrm{W}=2 \mathrm{~mm} \\ \therefore \quad & \mathrm{y}_{13}-\mathrm{y}_4^{\prime}=\frac{19}{2} \times 2=19 \mathrm{~mm} \end{aligned}$$

M.I. of disc, $I=\frac{1}{2} ~\mathrm{MR}_{\mathrm{d}}^2$ ..... (i)

M.I. of sphere, $\mathrm{I}_{\text {sphere }}=\frac{2}{5}~ \mathrm{MR}_{\mathrm{s}}^2$ .... (ii)

$\because \quad$ volume of disc = volume of sphere

$$\begin{array}{ll} \therefore & \pi \mathrm{R}_{\mathrm{d}}^2\left(\frac{\mathrm{R}_{\mathrm{d}}}{6}\right)=\frac{4}{3} \pi \mathrm{R}_{\mathrm{s}}^3 \\ \therefore & \mathrm{R}_{\mathrm{d}}^3=8 \mathrm{R}_{\mathrm{s}}^3 \\ \therefore & \mathrm{R}_{\mathrm{S}}=\frac{\mathrm{R}_{\mathrm{d}}}{2} \quad \text{.... (iii)} \end{array}$$

Substitute equation (iii) in equation (ii)

$$\begin{aligned} \therefore \quad I_{\text {sphere }} & =\frac{2}{5} \mathrm{M}\left(\frac{\mathrm{R}_{\mathrm{d}}}{2}\right)^2=\frac{2}{5} \times \frac{1}{4} \mathrm{MR}_{\mathrm{d}}^2 \\ & =\frac{1}{5}\left(\frac{1}{2} \mathrm{MR}_{\mathrm{d}}^2\right)=\frac{1}{5} \quad \text{.... [from (i)]} \end{aligned}$$

In a p-n junction diode :

• Reverse Biasing : This occurs when the p-type material is connected to the negative terminal of the battery and the n-type material to the positive terminal. In reverse bias, the built-in potential barrier increases, which tends to widen the depletion region.

Given this, let's evaluate the options :

Option A : The width of the depletion layer decreases.

• This is incorrect. In reverse bias, the width of the depletion layer increases.

Option B : The width of the depletion layer increases.

• This is correct. In reverse bias, the depletion layer widens as the potential barrier across the junction increases.

Option C : The number of minority charge carriers increase.

• This is not necessarily true. While the flow of minority carriers constitutes the reverse current, reverse bias does not increase their number.

Option D : The number of majority charge carriers increase.

• This is incorrect. Reverse biasing does not increase the number of majority charge carriers.

The correct answer is Option B : the width of the depletion layer increases.

For LR circuit,

$\mathrm{Z}=\sqrt{\mathrm{R}^2+(\omega \mathrm{L})^2}$

Comparing equation $E=4 \cos (1000 t)$ with standard equation, $\mathrm{E}=\mathrm{E}_0 \cos \omega \mathrm{t}$, we get $\omega=1000$ units and $E_0=4 \mathrm{~V}$

$$\begin{aligned} \therefore \quad & Z=\sqrt{16+\left(1000 \times 3^2 \times 10^{-6}\right)} \\ & Z=\sqrt{16+9}=5 \Omega \\ \therefore \quad & I=\frac{E_0}{Z}=\frac{4}{5}=0.8 A \end{aligned}$$

Using Kirchhoff's current Law,

Current flowing in = Current flowing out $10+2.5+5=6+x$

$\therefore \quad x=17.5-6$

$$\begin{aligned} \therefore \quad \mathrm{x} & =17.5-6 \\ & =11.5 \mathrm{~A} \end{aligned}$$

$\therefore \quad$ Magnitude and direction of the current in 'OP' should be $11.5 \mathrm{~A}$ from $\mathrm{O}$ to $\mathrm{P}$.

$\mathrm{PV}=\mathrm{Nk}_{\mathrm{B}} \mathrm{T}$

and, K.E./molecule $=\frac{3}{2} \mathrm{k}_{\mathrm{B}} \mathrm{T}=\frac{3}{2} \frac{\mathrm{PV}}{\mathrm{N}}$

$$\begin{aligned} \therefore \quad \mathrm{N} & =\frac{3}{2} \times \frac{\mathrm{PV}}{(\text { K.E./molecule })} \\ & =\frac{3}{2} \times \frac{2 \times 10^5 \times 500 \times 10^{-6}}{6 \times 10^{-21}}=2.5 \times 10^{22} \end{aligned}$$