MHT CET 2023 13th May Morning Shift (Physics & Chemistry) Question Paper with Solutions

We know

$$\begin{aligned}& \mathrm{T}_{\mathrm{g}}=\frac{\mathrm{T}_1-\mathrm{T}_2}{\mathrm{x}} \\ & \Rightarrow \frac{\mathrm{T}_1-10}{0.75}=40 \end{aligned}$$

$\therefore \quad$ Temperature of the hotter end is:

$$\begin{aligned} & \mathrm{T}_1=(40 \times 0.75)+10 \\ & \mathrm{~T}_1=40^{\circ} \mathrm{C} \end{aligned}$$

Time period of simple pendulum $\mathrm{T}=2 \pi \sqrt{\frac{l}{\mathrm{~g}}}$

$\Rightarrow \mathrm{T} \propto \frac{1}{\sqrt{\mathrm{g}}}$

Net downward force acting on the bob inside the liquid $=$ Weight of bob $-$ Upthrust

$\Rightarrow \mathrm{Vpg}-\mathrm{V} \frac{\rho}{\rho} \mathrm{g}=\frac{7}{8} \mathrm{~V} \rho \mathrm{g}$

The value of $g$ inside the liquid will be $\frac{g}{8}$

$\therefore \quad$ Time period in liquid $\mathrm{T}_1=\frac{1}{2 \pi} \sqrt{\frac{\frac{l}{7}}{8}} =\sqrt{\frac{8}{7}} \mathrm{~T}$

Given: $\mathrm{u}=-\mathrm{x}, \mathrm{v}=+\mathrm{x}$

We know,

$$\begin{aligned} & \frac{\mathrm{n}_2}{\mathrm{v}}-\frac{\mathrm{n}_1}{\mathrm{u}}=\frac{\mathrm{n}_2-\mathrm{n}_1}{\mathrm{R}} \\ \Rightarrow & \frac{1.5}{\mathrm{x}}+\frac{1}{\mathrm{x}}=\frac{0.5}{\mathrm{R}} \\ & \frac{2.5}{\mathrm{x}}=\frac{0.5}{\mathrm{R}} \\ \therefore \quad & \mathrm{x}=5 \mathrm{R} \end{aligned}$$

If threshold frequency is $\nu_0$, then light frequency becomes $1.5 ~\nu_0$.

If we make it half it becomes $0.75 ~\nu_0$, which is smaller than threshold frequency, therefore photoelectric current is zero.

Given $\phi_1=\frac{\pi}{6}$ and $\phi_2=\frac{\pi}{3}$

$\therefore \quad \Delta \phi=\frac{\pi}{3}-\left(\frac{-\pi}{6}\right)=\frac{\pi}{2}$

As the bridge is balanced, no current will flow through the galvanometer. Hence, the current through the $40 \Omega$ resistor will be $\mathrm{I}_2$.

From Stefan $-$ Boltzmann's Law

$$\begin{aligned} & E=\sigma T^4 \\ & \Rightarrow E \propto T^4 \end{aligned}$$

$$\begin{aligned} \therefore \quad \frac{\mathrm{E}_1}{\mathrm{E}_2} & =\left(\frac{\mathrm{T}_1}{\mathrm{~T}_2}\right)^4=\left(\frac{400}{1200}\right)^4 \\ \frac{\mathrm{E}_1}{\mathrm{E}_2} & =\frac{1}{81} \\ \mathrm{E}_2 & =81 \times \mathrm{E}_1 \\ & =81 \times 5 \\ & =405 \mathrm{~cal} / \mathrm{cm}^2 \mathrm{~s} \end{aligned}$$

$\text { Given } \mathrm{AB}=\mathrm{AC}=2 \mathrm{a}$

$\mathrm{V}_{\mathrm{D}}=\mathrm{V}_{\mathrm{E}}$ $(\because \mathrm{D}$ and $\mathrm{E}$ are mid-points)

$\therefore \quad$ The work done in taking a charge $\mathrm{q}$ from $\mathrm{D}$ to $\mathrm{E}$ is

$\mathrm{W}=\mathrm{q} \Delta \mathrm{V}=\mathrm{q}\left(\mathrm{V}_{\mathrm{D}}-\mathrm{V}_{\mathrm{E}}\right)$

$\therefore \quad \mathrm{W}=0$ (Equipotential surfaces)

Now, using the impulse-momentum theorem to find the final velocity

$$\begin{aligned} & 12=2 \times(-6) \\ & v-6=\frac{12}{2}=6 \\ & v=12 \mathrm{~m} / \mathrm{s} \end{aligned}$$

So, with the given force-time data $(0,6)$ and $(4,0)$, the final velocity of the particle is $12 \mathrm{~m} / \mathrm{s}$

According to Charles's Law, the volume of an ideal gas is directly proportional to its absolute temperature when pressure is held constant. i.e., $\mathrm{V} \propto \mathrm{T}$

This is depicted by graph $\mathrm{Q}$.

Magnetic field along the axis of the coil is:

$\mathrm{B}_{\mathrm{a}}=\frac{\mu_0}{4 \pi}\left[\frac{2 \pi n \mathrm{Ir}^2}{\left(\mathrm{r}^2+\mathrm{h}^2\right)^{\frac{3}{2}}}\right]$ .... (i)

Magnetic field at the centre of the coil is:

$$\begin{aligned} \mathrm{B}_{\mathrm{C}} & =\frac{\mu_0}{4 \pi}\left(\frac{2 \pi n \mathrm{I}}{\mathrm{r}}\right) \quad \text{.... (ii)}\\ \frac{\mathrm{B}_{\mathrm{C}}}{\mathrm{B}_{\mathrm{a}}} & =\frac{\left(\mathrm{r}^2+\mathrm{h}^2\right)^{\frac{3}{2}}}{\mathrm{r}^3} \\ \therefore \quad \mathrm{B}_{\mathrm{C}} & =\mathrm{B}_{\mathrm{a}}\left[1+\frac{\mathrm{h}^2}{\mathrm{r}^2}\right]^{\frac{3}{2}} \end{aligned}$$

Given: $\mathrm{e}=0$

$\therefore \quad \mathrm{r}_2=0, \mathrm{~A}=\mathrm{r}_1$

Since 'i' is small, Snell's law of refraction can be modified to,

$$\begin{aligned} \mu & =\frac{i}{r_1} \\ \therefore \quad i & =\mu r_1=\mu A \end{aligned}$$

Work function $\phi_0=\frac{\mathrm{hc}}{\lambda_0}$

$$\begin{aligned} \therefore \quad \phi_0 & \propto \frac{1}{\lambda_0} \\ & \frac{\phi_{01}}{\phi_{0_2}}=\frac{\lambda_{0_2}}{\lambda_{0_1}}=\frac{600}{300}=\frac{2}{1} \end{aligned}$$

The situation described involves the Doppler effect, but with a key distinction: both the source of the sound (the train's engine) and the observer (the passenger) are moving at the same speed in the same direction since they are both on the train.

In the Doppler effect, the frequency of a wave changes due to the relative motion between the source and the observer. However, if the source and the observer are moving together at the same velocity (as is the case here with the passenger and the train's engine), there is no relative motion between them. Consequently, the frequency of the sound heard by the observer (passenger) will be the same as the actual frequency of the sound emitted by the source (train's engine).

Therefore, the correct answer is :

Option A : $\mathrm{f} = \mathrm{n}$

Density $\propto \mathrm{P} / \mathrm{T}$

So, $\frac{\mathrm{d}_2}{\mathrm{~d}_1}=\frac{\mathrm{P}_2}{\mathrm{~T}_2} \times \frac{\mathrm{T}_1}{\mathrm{P}_1}$

$\therefore \quad$ The new density is:

$$\begin{aligned} & \mathrm{d}_2=\rho_0 \times \frac{3 \mathrm{P}_0}{2 \mathrm{~T}_0} \times \frac{\mathrm{T}_0}{\mathrm{P}_0} \\ & \mathrm{~d}_2=\frac{3}{2} \rho_0 \end{aligned}$$

$$\begin{aligned} & \text { Total phase difference }=\phi_1+\phi_2 \\ & \begin{aligned} \phi_1 & =54 \times \frac{\pi}{180}=0.3 \pi \\ \phi_2 & =\frac{2 \pi}{\lambda} \times(\mathrm{PB}-\mathrm{PA}) \\ & =\frac{2 \pi}{\lambda} \times 2.5 \lambda=5 \pi \\ \therefore \quad \phi_1 & +\phi_2=5 \pi+0.3 \pi=5.3 \pi \end{aligned} \end{aligned}$$

Efficiency of a carnot engine is $\eta=1-\frac{T_C}{T_H}$

For case (i),

$\mathrm{T}_{\mathrm{C}}=100 \mathrm{~K}$ and $\mathrm{T}_{\mathrm{H}}=600 \mathrm{~K}$

$\therefore \quad \eta_1=1-\frac{100}{600}=\frac{5}{6}$

For case (ii),

$\mathrm{T}_{\mathrm{C}}=\mathrm{T} \mathrm{K} \text { and } \mathrm{T}_{\mathrm{H}}=960 \mathrm{~K}$

$\therefore \quad \eta_2=1-\frac{\mathrm{T}}{960}$

Given $\eta_1=\eta_2$

$$\begin{aligned} \therefore \quad \frac{5}{6} & =1-\frac{\mathrm{T}}{960} \\ \frac{5}{6} & =\frac{960-\mathrm{T}}{960} \end{aligned}$$

Solving for $\mathrm{T}$, we get $\mathrm{T}=160 \mathrm{~K}$

First resonance will occur at $l_1+\mathrm{x}=\frac{\lambda}{4}\quad$ .... (i)

Second resonance will occur at $l_2+\mathrm{x}=\frac{3 \lambda}{4}\quad$ ..... (ii)

$$\begin{aligned} & l_2+\mathrm{x}=3\left(l_1+\mathrm{x}\right) \\ & l_2+\mathrm{x}=3 l_1+3 \mathrm{x} \\ & 2 \mathrm{x}=l_2-3 l_1 \\ & \therefore \quad \mathrm{x}=\frac{l_2-3 l_1}{2} \\ & \quad=\frac{70.2-68.1}{2}=1.05 \mathrm{~cm} \end{aligned}$$

$\therefore \quad$ Third resonance occurs at $l_3+\mathrm{x}=\frac{5 \lambda}{4}$

$$\begin{aligned} \therefore \quad l_3 & =5\left(l_1+\mathrm{x}\right)-\mathrm{x} \\ & =5 l_1+4 \mathrm{x} \\ & =113.5+4.2 \\ & =117.7 \mathrm{~cm} \end{aligned}$$

Diamagnetic substances have a small negative value for magnetic susceptibility and is independent of temperature.

When a particle of mass $m$ is taken from the Earth's surface to a height $h=n R$, then the change in P.E. can be calculated as,

$$\begin{array}{ll} & \Delta \mathrm{U}=\mathrm{mgR}\left(\frac{\mathrm{n}}{\mathrm{n}+1}\right) \\ \therefore & \frac{\mathrm{mgR}}{5}=\operatorname{mgR}\left(\frac{\mathrm{n}}{\mathrm{n}+1}\right) \\ \therefore & \mathrm{n}+1=5 \mathrm{n} \\ \therefore & \mathrm{n}=\frac{1}{4} \\ \therefore & \mathrm{h}=\frac{\mathrm{R}}{4} \end{array}$$

Magnetic field at the centre of a circular loop

$\mathrm{B}=\frac{\mu_0 \mathrm{NI}}{2 \mathrm{R}}$

Net Magnetic field $B_{\text {net }}=0$ (given)

$$\begin{aligned} \Rightarrow & B_{\text {net }}=\frac{\mu_0 \mathrm{~N}_1 \times 0.5}{2 \times(0 \cdot 2)}-\frac{\mu_0 \mathrm{~N}_1 \times \mathrm{x}}{2 \times(0 \cdot 1)} \\ \Rightarrow & \frac{\mu_0 \mathrm{~N}_1 \mathrm{x}}{0.2}=\frac{\mu_0 \mathrm{~N}_1 0.5}{0.4} \\ \therefore \quad \mathrm{x} & =\frac{0.5 \times 0.2}{0.4} \\ & =0.25 \mathrm{~A} \text { in clockwise direction } \end{aligned}$$

The combinations $\mathrm{P}$ and $\mathrm{S}$ are series combinations. Hence, the effective inductance cannot be $1 \mathrm{H}$.

This leaves combinations $\mathrm{R}$ and $\mathrm{Q}$.

For combination $\mathrm{R}$,

$\mathrm{L}_1+\mathrm{L}_2=5 \mathrm{H}$

$\frac{1}{\mathrm{~L}_1+\mathrm{L}_2}+\frac{\mathrm{L}}{\mathrm{L}_3}=\frac{1}{5}+\frac{1}{6}=\frac{6+5}{30}=\frac{11}{30}$

$\therefore \quad \mathrm{L}_{\text {eff }}=\frac{30}{11}=2.72 \mathrm{H}$

$\therefore \quad$ The correct combination would be $\mathrm{Q}$.

$$\begin{aligned} & \quad \frac{1}{\mathrm{~L}_1}+\frac{1}{\mathrm{~L}_2}+\frac{1}{\mathrm{~L}_3}=\frac{1}{2}+\frac{1}{3}+\frac{1}{6}=\frac{3}{6}+\frac{2}{6}+\frac{1}{6} \\ & \therefore \quad \mathrm{L}_{\mathrm{eff}}=1 \mathrm{H} \end{aligned}$$

Let's analyze each statement to determine which one is wrong :

Option A : A body can have zero velocity and still be accelerated.

• This statement is true. For example, at the highest point of its trajectory, a thrown ball has zero velocity, but it is still accelerating due to gravity.

Option B : A body can have a constant velocity and still have a varying speed.

• This statement is false. Velocity is a vector quantity that includes both speed and direction. If the velocity is constant, it means both the speed and the direction of the body are constant. Therefore, a body with constant velocity cannot have a varying speed.

Option C : A body can have a constant speed and still have a varying velocity.

• This statement is true. A body moving in a circular path at a constant speed has a constantly changing velocity because the direction of the velocity vector changes even though its magnitude (speed) remains constant.

Option D : The direction of the velocity of a body can change when its acceleration is constant.

• This statement is true. A common example is an object in free fall under gravity. The direction of its velocity changes as it goes up, stops, and comes back down, even though the acceleration (due to gravity) is constant.

Therefore, the wrong statement is Option B : A body can have a constant velocity and still have a varying speed.

Angular momentum,

$$\begin{aligned} \mathrm{L} & =\sqrt{2 \mathrm{KI}} \\ \mathrm{K} & =4 \mathrm{~K} \quad \text{(given)}\\ \therefore \quad \mathrm{L} & =\sqrt{2 \times 4 \mathrm{KI}}=2 \sqrt{2 \mathrm{KI}}=2 \mathrm{~L} \end{aligned}$$

$$\begin{array}{ll} & \text { Fringe width } Z=\frac{\lambda D}{d} \\ \therefore \quad & \frac{Z}{D}=\frac{\lambda}{d}=\text { constant, as } \lambda \text { and } d \text { are constant } \\ \therefore \quad & \frac{Z_1}{D_1}=\frac{Z_2}{D_2}=\frac{Z_3}{D_3}=\frac{Z_4}{D_4} \end{array}$$

Electromagnetic waves can propagate through vacuum. This can be clearly understood from Maxwell's equations for EM waves. Self-sustaining electric and magnetic fields enable waves to travel through empty space.

${ }_{\mathrm{Z}} \mathrm{X}^{\mathrm{A}} \stackrel{\alpha}{\longrightarrow}{ }_{\mathrm{Z}-2} \mathrm{Y}^{\mathrm{A}-4} \stackrel{2 \mathrm{\beta}^{-}}{\longrightarrow}{ }_{\mathrm{Z}} \mathrm{X}^{\mathrm{A}-4}$

$$\begin{aligned} F & =\frac{G \times m \times m}{(2 R)^2}=\frac{G \times\left(\frac{4}{3} \pi R^3 \rho\right)^2}{4 R^2} \\ & =\frac{4}{9} \pi^2 \rho^2 R^4 \\ \therefore \quad F & \propto R^4 \end{aligned}$$

Under isochoric (constant volume) conditions, the heat energy (dQ) supplied to an ideal gas contributes solely to the change in its internal energy (dU).

Mathematically, this can be expressed as dQ = dU + dW

$\mathrm{\Rightarrow d Q=d U}\quad$ ($\because\mathrm{dW=0}$)

When the reverse voltage increases beyond the breakdown voltage, the current suddenly increases.

Translational kinetic energy of a ring is:

$\mathrm{KE}_{\text {Trans }}=\frac{1}{2} \mathrm{mv}^2$

Rotational kinetic energy of the ring is:

$$\begin{aligned} \mathrm{KE}_{\text {Rolling }} & =\frac{1}{2} \mathrm{I}^2 \\ & =\frac{1}{2} \mathrm{mR}^2 \omega^2 \\ \mathrm{KE}_{\text {Rolling }} & =\frac{1}{2} \mathrm{~mv}^2 \quad (\because \mathrm{v=\omega R}) \end{aligned}$$

We know,

$\frac{V_s}{V_p}=\frac{N_s}{N_p}$

$\therefore \quad$ The maximum value of secondary output voltage is:

$$\begin{aligned} V_s & =\frac{N_s}{N_p} \times V_p=\frac{100}{20} \times 600 \\ V_s & =3000 V \end{aligned}$$

$$\begin{aligned} & E_{\text {axial }}=E_{\text {equatorial }} \Rightarrow k \frac{2 p}{r_1^3}=\frac{k p}{r_2^3} \\ & \Rightarrow \frac{r_1}{r_2}=\frac{2^{1 / 3}}{1}=\sqrt[3]{2}: 1 \end{aligned}$$

Radius of each droplet = r

Radius of the drop = R

As volume remains constant,

$$\begin{aligned} & \mathrm{n} \times \frac{4}{3} \pi \mathrm{r}^3=\frac{4}{3} \pi \mathrm{R}^3 \\ & \therefore \quad \mathrm{n}=\frac{\mathrm{R}^3}{\mathrm{r}^3} \end{aligned}$$

Decrease in surface area $=4 \pi r^2 n-4 \pi R^2 n$

$$\begin{aligned} \Delta \mathrm{A} & =4 \pi\left[\mathrm{nr}^2-\mathrm{R}^2\right] \\ & =4 \pi\left[\frac{\mathrm{R}^3}{\mathrm{r}^3} \mathrm{r}^2-\mathrm{R}^2\right] \\ & =4 \pi \mathrm{R}^3\left[\frac{1}{\mathrm{r}}-\frac{1}{\mathrm{R}}\right] \end{aligned}$$

Energy released $\mathrm{W}=\mathrm{T} \times \Delta \mathrm{T}$

$\quad=\mathrm{T} \times 4 \pi \mathrm{R}^3\left[\frac{1}{\mathrm{r}}-\frac{1}{\mathrm{R}}\right]$ ..... (i)

Heat produced $\mathrm{Q}=\frac{\mathrm{W}}{\mathrm{J}}$ ..... (ii)

$\mathrm{Q}=\mathrm{m} \cdot \mathrm{s} \cdot \Delta \theta$

Put (i) and (iii) into (ii)

$$\begin{aligned} & \mathrm{m} \cdot \mathrm{S} \Delta \theta=\frac{4 \pi \mathrm{R}^3 \mathrm{~T}}{\mathrm{~J}}\left[\frac{1}{\mathrm{r}}-\frac{1}{\mathrm{R}}\right] \\ & \frac{4}{3} \pi \mathrm{R} \rho_{\text {water }} \mathrm{S}_{\text {water }} \Delta \theta=\frac{4 \pi \mathrm{R}^3 \mathrm{~T}}{\mathrm{~J}}\left[\frac{1}{\mathrm{r}}-\frac{1}{\mathrm{R}}\right] \\ & \Delta \mathrm{Q}=\frac{3 \mathrm{~T}}{\mathrm{~J}}\left[\frac{1}{\mathrm{r}}-\frac{1}{\mathrm{R}}\right] \end{aligned}$$

Maximum energy stored in a capacitor, $\mathrm{E}_1=\frac{\mathrm{Q}^2}{2 \mathrm{C}}$

When energy is stored equally between the electric and magnetic fields, then energy in the capacitor is $\mathrm{E}_2=\frac{1}{2} \mathrm{E}_1$

If $\mathrm{Q}^{\prime}$ is the charge on the capacitor in this case, then $\mathrm{E}_2=\frac{\mathrm{Q}^{\prime 2}}{2 \mathrm{C}}$.

$$\begin{aligned} \therefore \quad & \frac{\mathrm{Q}^{\prime 2}}{2 \mathrm{C}}=\frac{1}{2} \frac{\mathrm{Q}^2}{2 \mathrm{C}} \\ \mathrm{Q}^{\prime} & =\frac{\mathrm{Q}}{\sqrt{2}}. \end{aligned}$$

We know

$\mathrm{X}_{\mathrm{L}}=\mathrm{L} \omega \text { and } \mathrm{X}_{\mathrm{C}}=\frac{1}{\mathrm{C} \omega}$

$\Rightarrow$ When the frequency increases, $X_L$ increases and $X_C$ decreases.

$\therefore \quad$ The impedance of an LCR series circuit decreases at first, becomes minimum and then increases.

$\mathrm{m}=\frac{50}{10}=5 \mathrm{~kg}\quad$ $(\because \mathrm{W}=\mathrm{mg})$

Tension in the mid-point of the wire is:

$\mathrm{T}=\frac{\mathrm{m}}{2} \mathrm{~g}=\frac{5}{2} \times 10=25 \mathrm{~N}$

$\therefore \quad$ Speed of the wave at mid-point of the wire is:

$$\begin{aligned} \quad \mathrm{v} & =\sqrt{\frac{\mathrm{T}}{\mu}}=\sqrt{\frac{25}{\left(\frac{5}{20}\right)}} \quad\left(\because \mu=\frac{\mathrm{m}}{\mathrm{L}}\right) \\ \therefore \quad \mathrm{v} & =10 \mathrm{~m} / \mathrm{s} \end{aligned}$$

The equation for the induced emf is:

$\mathrm{e}=\mathrm{B} / \mathrm{V}$

Relative velocity between the coil and the magnet is:

$\mathrm{v}_{\mathrm{r}}=2 \mathrm{v}$

$\therefore \quad$ The new induced emf in the coil is:

$\mathrm{e}_{\text {new }}=\mathrm{B} l \cdot 2 \mathrm{~V}=2 \mathrm{e}$

From

$$\begin{aligned} \mathrm{W} =4 \pi \mathrm{r}^2 \mathrm{~T}\left(\mathrm{n}-\mathrm{n}^{2 / 3}\right) \\ \mathrm{W} =4 \pi \times\left(0.1 \times 10^{-3}\right)^2 \times 0.072\left[27-(3)^{2 / 3}\right] \\ =1.627 \times 10^{-7} \quad\left(\mathrm{~T}_{\text {water }}=0.075 \mathrm{~N} / \mathrm{m}\right) \end{aligned}$$

With increasing voltage, the LED's current rises, leading to light emission due to charge carrier recombination.

A Venturi meter measures liquid flow in a pipe using a narrow section and works on the basis of Bernoulli's principle.

Given:

Total time, $\mathrm{T}=10$ years

Half life, $\mathrm{T}_{1 / 2}=5$ years

$\therefore \quad$ No. of half lives, $\mathrm{n}=\frac{10}{5}=2$

From $\frac{\mathrm{N}}{\mathrm{N}_0}=\left(\frac{1}{2}\right)^{\mathrm{n}}$

$\therefore \quad \frac{\mathrm{N}}{\mathrm{N}_0}=\left(\frac{1}{2}\right)^2=\frac{1}{4}$