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

To determine the ratio of the increase in the volume of rod $A$ to that of rod $B$, we need to consider the formula for the volumetric expansion due to heating. The change in volume ($\Delta V$) of a rod due to a change in temperature can be given by:

$\Delta V = \beta V \Delta T$

where:

$\beta$ = coefficient of volumetric expansion (which is the same for both rods since they are made of brass),

$V$ = initial volume of the rod,

$\Delta T$ = change in temperature (which is the same for both rods).

First, let's calculate the volume of each rod. The volume $V$ of a cylindrical rod is:

$V = \pi r^2 l$

For rod $A$:

Length = $l$,

Radius = $2r$.

Therefore, the initial volume of rod $A$ ($V_A$) is:

$V_A = \pi (2r)^2 l = \pi \cdot 4r^2 \cdot l = 4 \pi r^2 l$

For rod $B$:

Length = $2l$,

Radius = $r$.

Therefore, the initial volume of rod $B$ ($V_B$) is:

$V_B = \pi r^2 (2l) = 2 \pi r^2 l$

Now, the increase in volume ($\Delta V$) of each rod due to heating is given by:

For rod $A$:

$\Delta V_A = \beta V_A \Delta T = \beta (4 \pi r^2 l) \Delta T = 4 \beta \pi r^2 l \Delta T$

For rod $B$:

$\Delta V_B = \beta V_B \Delta T = \beta (2 \pi r^2 l) \Delta T = 2 \beta \pi r^2 l \Delta T$

To find the ratio of the increase in the volume of rod $A$ to that of rod $B$, we take:

$\frac{\Delta V_A}{\Delta V_B} = \frac{4 \beta \pi r^2 l \Delta T}{2 \beta \pi r^2 l \Delta T} = \frac{4}{2} = 2$

Thus, the ratio of the increase in volume of rod $A$ to that of rod $B$ is: 2:1.

Therefore, the correct option is Option C: 2:1.

From $T=2 \pi \sqrt{\frac{L}{g}}$

The effective length increases due to this flow of water. Therefore, $\mathrm{T}$ increases. As the water flows out, the length decreases and becomes equal to the original length. Hence, the time period becomes equal to the value at the beginning.

The fundamental frequency of a string is given by

Given: $l=l_1+l_2+l_3$ ..... (i)

$\mathrm{n}=\frac{1}{2 l} \sqrt{\frac{\mathrm{T}}{\mathrm{m}}}$

$\Rightarrow \mathrm{n} \propto \frac{1}{l}$ or $\mathrm{n} l=\mathrm{k}$

$\therefore \quad l_1=\frac{\mathrm{k}}{\mathrm{n}_1}, l_2=\frac{\mathrm{k}}{\mathrm{n}_2} \text { and } l_3=\frac{\mathrm{k}}{\mathrm{n} 3}$ .... (ii)

$\therefore \quad$ Original length $l=\frac{\mathrm{k}}{\mathrm{n}}$ .... (iii)

Putting eq (ii) and (iii) into eq (i)

$$\begin{aligned} & \frac{\mathrm{k}}{\mathrm{n}}=\frac{\mathrm{k}}{\mathrm{n}_1}+\frac{\mathrm{k}}{\mathrm{n}_2}+\frac{\mathrm{k}}{\mathrm{n}_3} \\ \therefore \quad \frac{1}{\mathrm{n}} & =\frac{1}{\mathrm{n}_1}+\frac{1}{\mathrm{n}_2}+\frac{1}{\mathrm{n}_3} \end{aligned}$$

In outer space, g = 0. Therefore, T = $\infty$.

Given $\mathrm{T}^{\prime}=\frac{1}{3} \mathrm{~T}$ and $\mathrm{A}^{\prime}=2 \mathrm{~A}$

$\therefore \quad \omega^{\prime}=\frac{2 \pi}{T^{\prime}}=\frac{2 \pi}{\left(\frac{1}{3} T\right)}=\frac{6 \pi}{T}=3 \omega$

$$\begin{aligned} & \therefore \quad \text { The new maximum velocity } \\ & \qquad \begin{aligned} \mathrm{V}^{\prime} & =\mathrm{A}^{\prime} \omega^{\prime} \\ & =(2 \mathrm{~A}) \times(3 \omega) \\ & =6 \mathrm{~A} \omega \\ & =6 \mathrm{~V} \end{aligned} \end{aligned}$$

Acceleration due to gravity at depth d,

$$\begin{aligned} g_d & =g\left(1-\frac{d}{R}\right) \\ & =g\left(1-\frac{1}{2}\right) \quad \because d=\frac{R}{2} \\ & =\frac{g}{2} \end{aligned}$$

$\therefore \quad$ Weight of the body at depth $\mathrm{d}_1$

$\mathrm{W}_{\mathrm{d}}=\mathrm{mg}_{\mathrm{d}}=300 \times \frac{1}{2}=150 \mathrm{~N}$

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

For jar $\mathrm{P}$, we have

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

For jar Q, we have,

(2P) $\left(\frac{V}{4}\right)=\mathrm{N}_2 \mathrm{k}_{\mathrm{B}}(2 \mathrm{~T})$

$\Rightarrow \mathrm{PV}=4 \mathrm{~N}_2 \mathrm{k}_{\mathrm{B}} \mathrm{T}$ ..... (ii)

From equations (i) and (ii)

$$\begin{aligned} & N_1=4 N_2 \quad \Rightarrow \frac{N_1}{N_2}=4 \\ \therefore \quad & N_1: N_2=4: 1 \end{aligned}$$

We know the electric field inside a shell is zero. So, we only consider the electric field due to the solid sphere.

$$\begin{aligned} & \therefore \quad \text { Using } E=\frac{1}{4 \pi \varepsilon_0} \frac{q}{r^2} \text {, we get } \\ & \quad E=\frac{1}{4 \pi \varepsilon_0} \frac{(3 Q)}{R^2} \end{aligned}$$

Given $\mathrm{R}=10 ~\Omega, \phi=6 \mathrm{t}^2+7 \mathrm{t}+1 ~\mathrm{mWb}, \mathrm{t}=1 \mathrm{~s}$

From $\mathrm{e}=\frac{\mathrm{d} \phi}{\mathrm{dt}}$,

$e=\frac{d}{d t}\left(6 t^2+7 t+1\right)=12 t+7$

Put $\mathrm{t}=1$ in the above equation,

$\therefore \quad \mathrm{e}=19 ~\mathrm{mV}$

The correct option for materials suitable for making electromagnets is Option B: low retentivity and low coercivity.

To understand why, let's dive into the terms retentivity and coercivity in the context of magnetic materials:

• Retentivity (or remanence) is the ability of a magnetic material to retain a certain amount of residual magnetism after an external magnetizing field is removed. High retentivity means the material retains a significant amount of magnetization, while low retentivity means it loses most of the magnetization when the external field is removed.


• Coercivity is the measure of the coercive force required to reduce the magnetization of a magnetic material to zero after the magnetization of the sample has been driven to saturation. Essentially, it indicates how difficult it is to demagnetize the material. High coercivity means it is difficult to demagnetize, while low coercivity indicates it is easy to demagnetize.

Electromagnets are designed to be turned on and off (magnetized and demagnetized) easily with the application and removal of an electric current. Therefore, materials used for electromagnets should have low retentivity so they can easily lose their magnetism when the current is turned off and low coercivity so they can be magnetized and demagnetized easily with small changes in current. This ensures the electromagnet is effective in applications that require rapid switching between the magnetized and demagnetized states.

Materials with high retentivity and high coercivity are more suitable for permanent magnets, which are intended to maintain their magnetic field without a continuous external power source.

Using equation of continuity,

$A_1 v_1=A_2 v_2$

In streamlined flow, as the product of $\mathrm{Av}$ is a constant, the velocity at the narrowest part will be maximum.

By using Bernoulli's principle,

$\mathrm{P}+\frac{1}{2} \rho \mathrm{v}^2+\rho \mathrm{gh}=\text { Constant }$

We see that $\rho g h$ is constant because the pipe is horizontal.

As velocity increases, for the equation to remain constant, $\mathrm{P}$ has to decrease.

$\therefore \quad$ Velocity will be maximum and pressure is minimum.

From de Broglie's hypothesis,

$\lambda=\frac{\mathrm{h}}{\mathrm{p}_{\mathrm{n}}}=\frac{\mathrm{h}}{\mathrm{mv}} \quad \text{.... (i)}$

and from Bohr's atomic model,

$\mathrm{L}=\frac{\mathrm{nh}}{2 \pi}$

Also,

$\mathrm{L}=\mathrm{mvr}_{\mathrm{n}}$

Due to quantization of angular momentum, we can write,

$$\begin{aligned} & \frac{\mathrm{nh}}{2 \pi}=\mathrm{mvr}_{\mathrm{n}} \\ & \therefore \quad \mathrm{v}=\frac{\mathrm{nh}}{2 \pi \mathrm{mr}} \quad \text{.... (ii)} \end{aligned}$$

putting (ii) into (i), we get,

$\therefore \quad \lambda=\frac{\mathrm{h}}{\mathrm{m}(\mathrm{nh} / 2 \pi \mathrm{mr})}=\frac{2 \pi \mathrm{r}}{\mathrm{n}}$

Given $\mathrm{I}_1=\mathrm{I}_2=25 \mathrm{~A}, l=1 \mathrm{~m}$,

$$\begin{aligned} & \mathrm{B}=\frac{\mu_0 \mathrm{I}}{2 \pi \mathrm{h}} \\ & \mathrm{F}=\mathrm{BI} l \sin \theta=\mathrm{BI} l \end{aligned}$$

Force applied on $\mathrm{PQ}=$ Weight of the smaller current carrying wire

$$\begin{array}{ll} \text { i.e, } \mathrm{mg}=\frac{\mu_0 \mathrm{I}^2 l}{2 \pi \mathrm{h}} \\ \therefore \quad \mathrm{h}=\frac{4 \pi \times 10^{-7} \times 250 \times 25 \times 1}{2 \pi \times 2.5 \times 10^{-3} \times 9.8}=5 \mathrm{~mm} \end{array}$$

Under isothermal condition, $\mathrm{T}$ is constant.

This means the surface energy of the bubbles before combining will be equal to the surface energy after combining.

$$\begin{array}{ll} & \text { i.e. } 4 \pi r_1^2 T+4 \pi r_2^2 T=4 \pi R^2 T \\ & \Rightarrow r_1^2+r_2^2=R^2 \\ \therefore \quad & R=\sqrt{r_1^2+r_2^2} \end{array}$$

Radius of circular path in a cyclotron is given by

$\mathrm{R}=\frac{\mathrm{mV}}{\mathrm{qB}}$

Here $\mathrm{q}=\mathrm{e}$,

$\therefore \quad \mathrm{R}=\frac{\mathrm{mV}}{\mathrm{eB}}$ ..... (i)

On entering the field,

$$\begin{aligned} & \mathrm{KE}=\mathrm{eV}=\frac{1}{2} \mathrm{mv}^2 \\ & \therefore \quad \mathrm{v}=\sqrt{\frac{2 \mathrm{eV}}{\mathrm{m}}} \quad \text{.... (ii)}\\ & \text { Putting (ii) into (i), } \\ & \mathrm{R}=\frac{\mathrm{m} \sqrt{\frac{2 \mathrm{eV}}{\mathrm{m}}}}{\mathrm{eB}}=\sqrt{\frac{2 \mathrm{Vm}}{\mathrm{eB}^2}} \end{aligned}$$

An adiabatic process is a thermodynamic process in which no heat is exchanged between the system and its surroundings, so that $\Delta 𝑄=0$.

In contrast, temperature remains constant in an isothermal process, pressure remains constant in an isobaric process, and volume remains constant in an isochoric process.

Hence, the correct option is (D).

Truth table for Y, with the possible values of C is,

For gate G

$\mathrm{G}$ is not a NOT gate as NOT gate takes only one input. (II) indicates $\mathrm{G}$ is not a OR gate as OR gate would give high output for the inputs in (II). Also, (II) indicates it is not a XOR gate as XOR would also give high output for inputs in (II). Hence, the given truth table is satisfied only by AND gate.

To find the ratio of the wavelengths for the transition of electrons from the $2^{\text{nd}}$ orbit to the $1^{\text{st}}$ orbit for Helium ion $\text{He}^{++}$ and Lithium ion $\mathrm{Li}^{+}$, we can use the formula derived from the Rydberg equation for wavelengths of emitted photons during electron transitions in hydrogen-like atoms.

The Rydberg formula for the wavelength $\lambda$ is given by:

$\frac{1}{\lambda} = RZ^2 \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right)$

where,

• $R$ is the Rydberg constant for hydrogen,


• $Z$ is the atomic number of the nucleus,


• $n_1$ is the lower energy level,


• $n_2$ is the higher energy level.

For a transition from the $2^{\text{nd}}$ orbit to the $1^{\text{st}}$ orbit ($n_2 = 2$ and $n_1 = 1$), the formula simplifies to:

$\frac{1}{\lambda} = RZ^2 \left( \frac{1}{1^2} - \frac{1}{2^2} \right) = RZ^2 \left( \frac{3}{4} \right)$

Therefore, $\lambda = \frac{4}{3RZ^2}$.

For Helium ion $\text{He}^{++}$ ($Z = 2$):

$\lambda_{\text{He}} = \frac{4}{3R \cdot 2^2} = \frac{4}{12R} = \frac{1}{3R}$

For Lithium ion $\mathrm{Li}^{+}$ ($Z = 3$):

$\lambda_{\text{Li}} = \frac{4}{3R \cdot 3^2} = \frac{4}{27R}$

To find the ratio of the wavelengths ($\lambda_{\text{He}} : \lambda_{\text{Li}}$), we divide the expressions obtained for $\lambda_{\text{He}}$ and $\lambda_{\text{Li}}$:

$\frac{\lambda_{\text{He}}}{\lambda_{\text{Li}}} = \frac{\frac{1}{3R}}{\frac{4}{27R}} = \frac{27}{12} = \frac{9}{4}$

Thus, the ratio of the wavelengths for the transition of electrons from the $2^{\text{nd}}$ orbit to the $1^{\text{st}}$ orbit of Helium ion ($\text{He}^{++}$) to Lithium ion ($\mathrm{Li}^{+}$) is $9 : 4$.

Therefore, the correct answer is Option A: $9: 4$.

Consider two diametrically and equal small and equal mass segments $\mathrm{dM}_1$ and $\mathrm{dM}_2$

$\therefore \quad$ Force at the centre due to $\mathrm{dM}_1$ is

$\mathrm{F}_1=\frac{\mathrm{GmdM}_1}{\mathrm{r}^2}$

Similarly,

Force at the centre due to $\mathrm{dM}_2$ is

$F_2=\frac{G m d M_2}{r^2}$

But $\mathrm{F}_1=-\mathrm{F}_2$

$\Rightarrow \mathrm{F}_1+\mathrm{F}_2=0 \quad (\because$ the forces cancel each other out as they are equal and opposite)

If this process is done for all such $\mathrm{dM}$ segments, we find the net force at the centre of the circle to be zero.

The ratio of moments of inertia of two discs of the same mass and same thickness but of different densities is given by $\frac{I_1}{I_2}=\frac{R_1^2}{R_2^2}=\frac{d_2}{d_1}$

$$\begin{aligned} \frac{I_g}{I} & =\frac{S}{S+G}=\frac{0.1 G}{0.1 G+G}=\frac{1}{11} \\ \therefore \quad I_g & =\frac{1}{11} I \end{aligned}$$

Given: Angle of Prism $=\mathrm{A}, \mathrm{I}=2 \mathrm{~A}$

As DE is falling normally on the silvered side AC,

$$\begin{aligned} & \mathrm{r}=90-(90-\mathrm{A}) \\ & \mathrm{r}=\mathrm{A} \end{aligned}$$

Using Snell's law,

$\frac{\sin \mathrm{i}}{\sin \mathrm{r}}=\frac{\sin 2 \mathrm{~A}}{\sin \mathrm{A}}=\frac{2 \sin \mathrm{A} \cos \mathrm{A}}{\sin \mathrm{A}}=2 \cos \mathrm{A}$

$$\begin{aligned} & \mathrm{P}=\mathrm{I}_{\mathrm{rms}} \mathrm{V}_{\mathrm{rms}} \cos \phi \\ & \phi=90^{\circ} \\ \therefore & \quad \mathrm{P}=0 \quad \left(\because \cos 90^{\circ}=0\right) \\ \end{aligned}$$

$\gamma$ and degrees of freedom is related by

$\gamma=\frac{\mathrm{f}+2}{\mathrm{f}}$

Where $f$ is the number of degrees of freedoms.

Given $\mathrm{f}=\mathrm{X}$,

$\therefore \quad \gamma=\frac{X+2}{X}=1+\frac{2}{X}$

We know, $v_0=\frac{Q^2}{2 C}$

On inserting the slab of dielectric constant $\mathrm{k}$, the new capacitance $\mathrm{C}^{\prime}=\mathrm{KC}$

$\therefore \quad$ New potential energy $v_0^{\prime}=\frac{Q^2}{2 C^{\prime}}$

$v_0^1=\frac{\mathrm{Q}^2}{2 \mathrm{KC}}=\frac{v_0}{\mathrm{~K}}$

$$\begin{array}{ll} & \frac{1}{\lambda}=\mathrm{R}\left[\frac{1}{\mathrm{n}_1^2}-\frac{1}{\mathrm{n}_2^2}\right] \\ \therefore \quad & \frac{1}{\lambda}=\left[\frac{1}{2^2}-\frac{1}{3^2}\right]=\frac{5}{36} \mathrm{R} \\ \therefore \quad & \mathrm{f}=\frac{\mathrm{c}}{\lambda}=\frac{5}{36} \mathrm{Rc} \end{array}$$

In Adiabatic compression,

$\mathrm{PV}^\gamma=$ constant

Given $V_{\text {new }}=\frac{1}{4} V$ and $\gamma=\frac{3}{2}$

$$\begin{array}{ll} \therefore & \frac{\mathrm{P}_{\text {new }}}{\mathrm{P}}=\left(\frac{\mathrm{V}}{\mathrm{V}_{\text {new }}}\right)^\gamma=\left(\frac{\mathrm{V}}{\frac{1}{4} \mathrm{~V}}\right)^{3 / 2} \\ \therefore & \mathrm{P}_{\text {new }}=8 \mathrm{P} \end{array}$$

For single slit diffraction,

Fringe width $\mathrm{W=\frac{\lambda D}{d}}$

Also, $\mathrm{I \propto D^2}$

$\mathrm{I}\propto \mathrm{D}^2$ This means I will be 4 times the original value of $\mathrm{I}$. As only 1 option mentions this, the correct answer is D.

Using Dopper's effect formula for approaching frequency when both source and listener are approaching each other, the observed frequency of sound is given by,

$\mathrm{n}=\mathrm{n}_0\left[\frac{\mathrm{V}+\mathrm{V}_{\mathrm{L}}}{\mathrm{V}-\mathrm{V}_{\mathrm{s}}}\right]$

The self-inductance $𝐿$ of a solenoid having length $𝑙$, cross-sectional area $𝐴$, and total number of turns $𝑁$ is given by

$𝐿=\frac{{𝜇}_0{𝑁}^2𝐴}{𝑙}$

Therefore, the self-inductance is directly proportional to the area of cross-section of the solenoid ($𝐿\propto 𝐴$) and is independent of the current flowing through it.

Hence, the correct option is (C).

$$\begin{aligned} & \Delta \mathrm{P} \propto \frac{1}{\mathrm{r}} \Rightarrow \frac{\mathrm{r}_1}{\mathrm{r}_2}=\frac{\Delta \mathrm{P}_2}{\Delta \mathrm{P}_1}=\frac{1}{3} \\ \therefore \quad & \frac{\mathrm{V}_1}{\mathrm{~V}_2}=\left(\frac{\mathrm{r}_1}{\mathrm{r}_2}\right)^3=\frac{1}{27} \end{aligned}$$

$$\begin{aligned} & N=100, A=2 \mathrm{~m}^2, \omega=30 \mathrm{rad} / \mathrm{s} \\ & B=2 \times 10^{-2} \mathrm{~T}, \mathrm{R}=600 \Omega \end{aligned}$$

Maximum power dissipated in the circuit

$$\begin{aligned} P_{\max }=E_{r m s} \times I_{r m s} & =\frac{E_0}{\sqrt{2}} \times \frac{I_0}{\sqrt{2}} \\ & =\frac{E_0 I_0}{2} \quad \text{..... (i)} \end{aligned}$$

Putting (2) into (1) we get,

$\mathrm{I}_0=\frac{\mathrm{E}_0{ }^2}{2 \mathrm{R}}$

But $\mathrm{E}_0=\mathrm{NAB} \omega$

$$\begin{aligned} \mathrm{E}_0 & =100 \times 2 \times 2 \times 10^{-2} \times 30 \\ & =120 \mathrm{~V} \\ \therefore \quad \mathrm{P}_{\max }= & \frac{120 \times 120}{2 \times 600}=12 \mathrm{~W} \end{aligned}$$

$y=\frac{1}{\sqrt{a}} \sin \omega t \pm \frac{1}{\sqrt{b}} \sin \left(\omega t+\frac{\pi}{2}\right)$

Here phase difference $=\frac{\pi}{2}$

The resultant amplitude

$=\sqrt{\left(\frac{1}{\sqrt{a}}\right)^2+\left(\frac{1}{\sqrt{b}}\right)^2}=\sqrt{\frac{1}{a}+\frac{1}{b}}=\sqrt{\frac{a+b}{a b}}$

P.E. of the stone projected vertically is,

P.E. $=\mathrm{mgh}$

But $h=\frac{v^2}{2 g}$

$$\begin{aligned} \therefore \quad P . E_1 & =m g\left(\frac{v^2}{2 g}\right) \\ & =\frac{\mathrm{mv}^2}{2} \quad \text{...... (i)} \end{aligned}$$

For the second stone thrown at an angle $\theta_{\text {to the }}$ horizontal,

$$\begin{aligned} & h=\frac{v^2 \sin ^2 \theta}{2 g}=\frac{v^2 \sin ^2 30^{\circ}}{2 g}=\frac{v^2}{8 g} \\ \therefore \quad & \text { P.E. } 2=m g\left(\frac{v^2}{8 g}\right)=\frac{m v^2}{8} \quad \text{...... (ii)} \end{aligned}$$

Dividing equation (i) by equation (ii)

$\frac{P . E_1}{P . E_2}=\frac{\left(\frac{m v^2}{2}\right)}{\left(\frac{m^2}{8}\right)}=4: 1$