MHT CET 2021 22nd September Morning Shift (Physics & Chemistry) Question Paper with Solutions

Here's a detailed explanation of the solution:

The photoelectric effect is governed by Einstein's photoelectric equation:

$h\nu = \phi + \frac{1}{2}mv^2$

where:

• $h$ is Planck's constant
• $ν$ is the frequency of the incident light
• $\phi$ is the work function of the metal (the minimum energy required to eject an electron)
• $m$ is the mass of the electron
• $v$ is the maximum speed of the emitted photoelectron

We can rewrite the equation in terms of wavelength using the relationship $c = νλ$ (where $c$ is the speed of light):

$hc/\lambda = \phi + \frac{1}{2}mv^2$

Let's analyze the scenario:

Initially, with wavelength $\lambda$, the maximum speed is $v$. So we have:

$hc/\lambda = \phi + \frac{1}{2}mv^2$

Now, the wavelength is changed to $\frac{2 \lambda}{3}$. Let's call the new maximum speed $v'$. The equation becomes:

$hc/(\frac{2 \lambda}{3}) = \phi + \frac{1}{2}mv'^2$

Simplifying the second equation:

$\frac{3hc}{2 \lambda} = \phi + \frac{1}{2}mv'^2$

Notice that the work function ($\phi$) remains the same for the same metal. Subtracting the first equation from the second equation, we get:

$\frac{3hc}{2 \lambda} - \frac{hc}{\lambda} = \frac{1}{2}mv'^2 - \frac{1}{2}mv^2$

Simplifying further:

$\frac{hc}{2 \lambda} = \frac{1}{2}m(v'^2 - v^2)$

We can now see that the left-hand side of the equation has increased by a factor of 1.5. Therefore, the right-hand side must also increase by a factor of 1.5. This means:

$v'^2 - v^2 = 1.5v^2$

Solving for $v'$:

$v'^2 = 2.5v^2$

$v' = \sqrt{2.5}v = \sqrt{1.5^2}v = 1.5v$

Therefore, the maximum speed of emitted photoelectrons will be greater than $v(1.5)^{\frac{1}{2}}$.

The correct answer is Option C: greater than $v(1.5)^{\frac{1}{2}}$

Kirchhoff's laws are fundamental principles in electrical circuit theory and are based on the conservation principles of physics.

Kirchhoff's Current Law (KCL): This law states that the total current entering a junction in a circuit is equal to the total current leaving the junction. Mathematically, it can be expressed as:

$\sum I_{in} = \sum I_{out}$

This is based on the principle of conservation of charge. Since charge cannot accumulate at a junction, the amount of charge (and therefore current) flowing into the junction must equal the amount flowing out.

Kirchhoff's Voltage Law (KVL): This law states that the sum of all electrical potential differences (voltages) around any closed loop in a circuit is zero. Mathematically, it can be expressed as:

$\sum V = 0$

This is based on the principle of conservation of energy. Because energy cannot be created or destroyed, the energy gained by the charges in a loop must equal the energy lost.

Considering these explanations, the correct option is:

Option A: charge, energy

Let the reactance of the capacitor at the original frequency '$\omega$' be $X_C$.

The impedance of the circuit at the original frequency is given by:

$Z = \sqrt{R^2 + X_C^2}$

The current in the circuit at the original frequency is given by:

$I = \frac{V}{Z} = \frac{V}{\sqrt{R^2 + X_C^2}}$

When the frequency is changed to $\frac{\omega}{3}$, the reactance of the capacitor becomes $3X_C$ (since reactance is inversely proportional to frequency). The impedance of the circuit at this new frequency is given by:

$Z' = \sqrt{R^2 + (3X_C)^2} = \sqrt{R^2 + 9X_C^2}$

The current in the circuit at this new frequency is given by:

$I' = \frac{V}{Z'} = \frac{V}{\sqrt{R^2 + 9X_C^2}}$

We are given that the current in the circuit is halved when the frequency is changed to $\frac{\omega}{3}$. Therefore, we can write:

$I' = \frac{I}{2}$

$\frac{V}{\sqrt{R^2 + 9X_C^2}} = \frac{1}{2} \cdot \frac{V}{\sqrt{R^2 + X_C^2}}$

Squaring both sides, we get:

$\frac{1}{R^2 + 9X_C^2} = \frac{1}{4(R^2 + X_C^2)}$

$4R^2 + 4X_C^2 = R^2 + 9X_C^2$

$3R^2 = 5X_C^2$

Therefore, the ratio of reactance to resistance at the original frequency '$\omega$' is:

$\frac{X_C}{R} = \sqrt{\frac{3}{5}}$

So the correct answer is Option D: $\sqrt{\frac{3}{5}}$

The correct answer is Option B: 4 $\pi^2$n$^2$R.

Here's why:

Understanding Centripetal Acceleration

Centripetal acceleration is the acceleration that keeps an object moving in a circular path. It always points towards the center of the circle. The magnitude of centripetal acceleration (a_c) is given by:

$a_c = \frac{v^2}{R}$

Where:

• v is the object's speed


• R is the radius of the circular path

Relating Frequency and Speed

Frequency (n) represents the number of revolutions the object completes per second. We can relate frequency to speed (v) as follows:

In one revolution, the object covers a distance equal to the circumference of the circle (2πR). So, the speed is:

$v = 2\pi Rn$

Putting it Together

Substituting the expression for v in the centripetal acceleration formula, we get:

$a_c = \frac{(2\pi Rn)^2}{R}$

$a_c = \frac{4\pi^2 R^2 n^2}{R}$

$a_c = \boxed{4\pi^2 n^2 R}$

Therefore, the centripetal acceleration of a body performing uniform circular motion is 4π²n²R.

When a nucleus breaks into two parts, conservation of momentum applies. Let's denote the masses of the two nuclear parts as $m_1$ and $m_2$, and their velocities as $v_1$ and $v_2$ respectively. According to the conservation of momentum:

$m_1 v_1 = m_2 v_2$

Given that the velocity ratio is $v_1 : v_2 = 2 : 1$, we can write:

$\frac{v_1}{v_2} = 2 \implies v_1 = 2v_2$

Substituting this into the momentum equation:

$m_1 (2v_2) = m_2 v_2 \implies 2m_1 = m_2 \implies \frac{m_1}{m_2} = \frac{1}{2}$

Now, the nuclear radius $R$ is related to the mass number $A$ (which corresponds to the mass) by the relation:

$R \propto A^{1/3}$

Thus, the ratio of the radii $R_1$ and $R_2$ of the two nuclear parts will be:

$\frac{R_1}{R_2} = \left( \frac{A_1}{A_2} \right)^{1/3} = \left( \frac{m_1}{m_2} \right)^{1/3} = \left( \frac{1}{2} \right)^{1/3} = \frac{1}{2^{1/3}}$

Therefore, the correct answer is:

Option C

$\frac{1}{2^{1 / 3}}$

The magnetic field at the centre of a circular coil is given by

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

If $\mathrm{T}$ is the periodic time of revolving electron, then $\mathrm{I}=\frac{\mathrm{e}}{\mathrm{T}}$

Also, $T=\frac{2 \pi r}{V}$

$$\begin{array}{ll} \therefore & I=\frac{e V}{2 \pi r} \\ \therefore & B=\frac{\mu_0 e V}{4 \pi r^2} \quad \text{....(1)} \end{array}$$

For electron in ground state,

$V=\frac{\mathrm{e}^2}{2 \epsilon_0 h} \text { and } r=\frac{h^2 \epsilon_0}{\pi m e^2}$

Putting these values in eq. (1) and simplifying we get

$\therefore \mathrm{B}=\frac{\mu_0 \mathrm{e}^7 \pi \mathrm{m}^2}{8 \epsilon_0 \mathrm{~h}^5}$

Here's a step-by-step explanation to determine the relationship between the angular momenta of the two rotating bodies:

Understanding the Concepts

• Rotational Kinetic Energy: The kinetic energy of a rotating body is given by:

$K.E. = \frac{1}{2} I \omega^2$

where:

• $I$ is the moment of inertia


• $\omega$ is the angular velocity


• Angular Momentum: The angular momentum of a rotating body is given by:

$L = I \omega$

Analyzing the Problem

1. Equal Kinetic Energy: The problem states that the kinetic energies of the two bodies are equal:

$\frac{1}{2} I_P \omega_P^2 = \frac{1}{2} I_Q \omega_Q^2$

1. Moment of Inertia Relationship: We are given that $I_Q > I_P$.


2. Solving for Angular Velocities: From the equal kinetic energy equation, we can see that:

$\omega_Q^2 = \frac{I_P}{I_Q} \omega_P^2$

Since $I_Q > I_P$, this implies $\omega_Q^2 < \omega_P^2$, and therefore $\omega_Q < \omega_P$.

1. Comparing Angular Momenta: Now let's compare the angular momenta:

• $L_P = I_P \omega_P$


• $L_Q = I_Q \omega_Q$

Since $I_Q > I_P$ and $\omega_Q < \omega_P$, the overall effect is that $L_Q > L_P$.

Conclusion

Therefore, the correct answer is Option D: $L_Q > L_P$.

To find the distance between the two points in the path of a wave, we need to use the relationship between the phase difference, wavelength, and distance. The phase difference $\Delta \phi$ is given by:

$\Delta \phi = \frac{2 \pi d}{\lambda}$

where:

• $\Delta \phi$ is the phase difference
• $d$ is the distance between the two points
• $\lambda$ is the wavelength

We are given:

• Phase difference $\Delta \phi = \frac{\pi}{3}$
• Velocity of sound $v = 330 \, \mathrm{m/s}$
• Frequency $f = 50 \, \mathrm{Hz}$

First, we find the wavelength $\lambda$ using the relation:

$\lambda = \frac{v}{f}$

Substituting the given values:

$\lambda = \frac{330 \, \mathrm{m/s}}{50 \, \mathrm{Hz}} = 6.6 \, \mathrm{m}$

Now, substitute the values of $\Delta \phi$ and $\lambda$ into the phase difference formula to find $d$:

$\frac{\pi}{3} = \frac{2 \pi d}{6.6}$

Solving for $d$:

$d = \frac{(\frac{\pi}{3}) \cdot 6.6}{2 \pi} = \frac{6.6}{6} = 1.1 \, \mathrm{m}$

Thus, the distance between those two points is 1.1 m. Therefore, the correct answer is:

Option A: 1.1 m

To determine the correct relationship between the gas constant $\mathrm{R}$ and the molar specific heat at constant volume $\mathrm{C}_{\mathrm{v}}$ for a gas molecule with 6 degrees of freedom, we can use the concepts from the kinetic theory of gases. For a gas molecule, the degrees of freedom (f) directly influence the molar specific heat capacities.

Firstly, let's recall the general relationship between the degrees of freedom (f), the molar specific heat at constant volume ($\mathrm{C}_{\mathrm{v}}$), and the molar specific heat at constant pressure ($\mathrm{C}_{\mathrm{p}}$):

$\mathrm{C}_{\mathrm{v}} = \frac{f}{2} \mathrm{R}$

$\mathrm{C}_{\mathrm{p}} = \mathrm{C}_{\mathrm{v}} + \mathrm{R}$

For a gas molecule with 6 degrees of freedom:

$\mathrm{C}_{\mathrm{v}} = \frac{6}{2} \mathrm{R} = 3\mathrm{R}$

Now substituting $\mathrm{C}_{\mathrm{v}}$ in the first equation:

$3 \mathrm{R} = \mathrm{C}_{\mathrm{v}}$

Therefore, solving for $\mathrm{R}$ we get:

$\mathrm{R} = \frac{\mathrm{C}_{\mathrm{v}}}{3}$

Given this derivation, the correct relation is Option A:

$\mathrm{R}=\frac{\mathrm{C}_{\mathrm{v}}}{3}$

To solve this problem, we'll first calculate the current entering the emitter and then determine the current amplification factor. Given that the charge of an electron is 1.6 $\times$ 10$^{-19}$ C, and 200 electrons enter the emitter in 10$^{-8}$ seconds, we can calculate the emitter current.

The current entering the emitter ($I_E$) is given by:

$I_E = \frac{n \cdot e}{t}$

where

$n$ is the number of electrons,

$e$ is the charge of an electron, and

$t$ is the time.

Substituting the given values:

$I_E = \frac{200 \times 1.6 \times 10^{-19}}{10^{-8}}$

Calculating this gives:

$I_E = \frac{320 \times 10^{-19}}{10^{-8}} = 320 \times 10^{-11} = 3.2 \times 10^{-9} \text{ A}$

Next, we know that 1% of the electrons are lost in the base. This means 99% of the electrons reach the collector. The collector current ($I_C$) can be calculated as:

$I_C = 0.99 \cdot I_E$

So,

$I_C = 0.99 \cdot 3.2 \times 10^{-9} \text{ A}$

Calculating this, we get:

$I_C = 3.168 \times 10^{-9} \text{ A}$

The current amplification factor ($\beta$) is given by the ratio of the collector current to the base current. First, let's find the base current ($I_B$). The base current can be found using:

$I_B = I_E - I_C$

Substituting the known values:

$I_B = 3.2 \times 10^{-9} \text{ A} - 3.168 \times 10^{-9} \text{ A} = 0.032 \times 10^{-9} \text{ A} = 3.2 \times 10^{-11} \text{ A}$

The current amplification factor ($\beta$) is:

$\beta = \frac{I_C}{I_B}$

Substituting the known values:

$\beta = \frac{3.168 \times 10^{-9}}{3.2 \times 10^{-11}}$

Calculating this, we get:

$\beta = 99$

Therefore, the current that enters the emitter is 3.2 $\times$ 10$^{-9}$ A and the current amplification factor is 99. The correct option is:

Option B: 3.2 $\times$ 10$^{-9}$ A and 99

In an LC oscillator circuit, the frequency of the output signal is determined by the inductance $L$ and the capacitance $C$. The expression for the frequency $F$ of an LC oscillator is given by:

$F = \frac{1}{2\pi\sqrt{LC}}$

Given that the initial frequency is $\mathrm{F}$ with a capacitance of $0.1 \mu \mathrm{F}$, we can write:

$\mathrm{F} = \frac{1}{2\pi\sqrt{L \cdot 0.1 \times 10^{-6}}}$

Now, if the capacitance is increased to $0.2 \mu \mathrm{F}$, the new frequency $F_{\text{new}}$ can be expressed as:

$F_{\text{new}} = \frac{1}{2\pi\sqrt{L \cdot 0.2 \times 10^{-6}}}$

To find the ratio of the new frequency to the original frequency, we can divide the expression for $F_{\text{new}}$ by the expression for $F$:

$\frac{F_{\text{new}}}{F} = \frac{\frac{1}{2\pi\sqrt{L \cdot 0.2 \times 10^{-6}}}}{\frac{1}{2\pi\sqrt{L \cdot 0.1 \times 10^{-6}}}}$

This simplifies to:

$\frac{F_{\text{new}}}{F} = \frac{\sqrt{L \cdot 0.1 \times 10^{-6}}}{\sqrt{L \cdot 0.2 \times 10^{-6}}}$

Since the inductance $L$ and the constants are the same, we get:

$\frac{F_{\text{new}}}{F} = \sqrt{\frac{0.1}{0.2}} = \sqrt{\frac{1}{2}} = \frac{1}{\sqrt{2}}$

So, the new frequency $F_{\text{new}}$ is:

$F_{\text{new}} = \frac{F}{\sqrt{2}} \mathrm{~Hz}$

Therefore, the correct option is:

Option A: $\frac{\mathrm{F}}{\sqrt{2}} \mathrm{~Hz}$

The given angular displacement of the body performing circular motion is expressed as:

$\theta = 5 \sin \frac{\pi t}{6}$

To determine the angular velocity, we need to differentiate the angular displacement $\theta$ with respect to time $t$. The angular velocity $\omega$ is given by:

$\omega = \frac{d\theta}{dt}$

Let's find the derivative of $\theta = 5 \sin \frac{\pi t}{6}$ with respect to $t$.

First, apply the chain rule to differentiate the sine function:

$\frac{d\theta}{dt} = 5 \cdot \frac{d}{dt} \left( \sin \frac{\pi t}{6} \right)$

Next, differentiate $\sin \frac{\pi t}{6}$ with respect to $t$. Using the chain rule:

$\frac{d}{dt} \left( \sin \frac{\pi t}{6} \right) = \cos \frac{\pi t}{6} \cdot \frac{d}{dt} \left( \frac{\pi t}{6} \right)$

$= \cos \frac{\pi t}{6} \cdot \frac{\pi}{6}$

Now, substituting this back into the expression for $\frac{d\theta}{dt}$:

$\frac{d\theta}{dt} = 5 \cdot \cos \frac{\pi t}{6} \cdot \frac{\pi}{6}$

$\frac{d\theta}{dt} = \frac{5\pi}{6} \cos \frac{\pi t}{6}$

We need to evaluate the angular velocity at $t = 3$ seconds:

$\omega(t=3) = \frac{5\pi}{6} \cos \left( \frac{\pi \cdot 3}{6} \right)$

$\omega(t=3) = \frac{5\pi}{6} \cos \left( \frac{\pi}{2} \right)$

Given that $\cos \frac{\pi}{2} = 0$:

$\omega(t=3) = \frac{5\pi}{6} \cdot 0$

$\omega(t=3) = 0$

Therefore, the angular velocity of the body at $t = 3$ seconds is:

Option D: zero $\frac{\mathrm{rad}}{\mathrm{s}}$

To determine the power delivered by the engine of a car for doubling its velocity, we need to analyze the work done and the time over which this work is performed.

Let's start with the given information:

• Initial velocity of the car: $u$
• Final velocity of the car: $2u$ (since the velocity is doubled)
• Time taken to double the velocity: $t$
• Mass of the car: $m$

First, we need to calculate the initial and final kinetic energy of the car:

The initial kinetic energy $(KE_i)$ is given by:

$KE_i = \frac{1}{2} m u^2$

The final kinetic energy $(KE_f)$ is given by:

$KE_f = \frac{1}{2} m (2u)^2 = \frac{1}{2} m \cdot 4u^2 = 2 m u^2$

The work done by the engine to double the velocity is equal to the change in kinetic energy:

$\text{Work Done (W)} = KE_f - KE_i = 2 m u^2 - \frac{1}{2} m u^2 = \frac{3}{2} m u^2$

Power is defined as the work done per unit time:

$\text{Power (P)} = \frac{\text{Work Done}}{\text{Time}} = \frac{\frac{3}{2} m u^2}{t} = \frac{3 m u^2}{2 t}$

Thus, the correct option is:

Option A: $\frac{3 m u^2}{2 t}$

The phenomenon of capillary rise is driven by surface tension, and the height to which the liquid rises in a capillary tube is given by the Jurin's Law. The formula to determine the height of water rise in a capillary tube is:

$\mathrm{h} = \frac{2 \gamma \cos \theta}{\rho g r}$

where:

• $\gamma$ is the surface tension of water.
• $\theta$ is the contact angle.
• $\rho$ is the density of water.
• $g$ is the acceleration due to gravity.
• $r$ is the radius of the capillary tube.

From the above formula, it is evident that the height '$h$' is inversely proportional to the radius '$r$'.

If the radius of the capillary tube is reduced to $\frac{r}{3}$, the new height '$h'$' of water rise will be:

$h' = \frac{2 \gamma \cos \theta}{\rho g \left(\frac{r}{3}\right)} = 3 \times \frac{2 \gamma \cos \theta}{\rho g r} = 3h$

The mass of water in a capillary tube of radius '$r$' up to height '$h$' is:

$\mathrm{m} = \rho \times (\text{Volume of water}) = \rho \times (\pi r^2 h)$

Substituting height $h'$ and radius $\frac{r}{3}$ into the new volume, the new volume $V'$ will be:

$V' = \pi \left(\frac{r}{3}\right)^2 \times 3h = \pi \frac{r^2}{9} \times 3h = \pi \frac{r^2 h}{3}$

Thus, the new mass $m'$ will be:

$\mathrm{m'} = \rho \times V' = \rho \times \pi \frac{r^2 h}{3} = \frac{\mathrm{m}}{3}$

Therefore, the mass of water that will rise in a capillary tube of radius $\frac{r}{3}$ is:

Option B: $\frac{\mathrm{m}}{3}$

$W_1=8 \pi r^2 T$

where $\mathrm{T}=$ surface tension

$\mathrm{W}_2=8 \pi(2 r)^2 \mathrm{~T}^{\prime}=32 \mathrm{r}^2 \mathrm{~T}^{\prime}$

Since soap solution is heated, its surface tension decreases.

$$\begin{array}{ll} \therefore \mathrm{T}^{\prime}<\mathrm{T} \\ \therefore \mathrm{W}_2<4 \mathrm{~W}_1 \text { or } 4 \mathrm{~W}_1>\mathrm{W}_2 \end{array}$$

Let's start by recalling the formula for the energy stored in a capacitor. The energy $U$ stored in a capacitor with capacitance $C$ and charge $Q$ is given by:

$U = \frac{Q^2}{2C}$

Since we do not know the capacitance $C$, we will manipulate the formula using given information. If the charge on the capacitor is increased by 2 C, the new charge $Q'$ becomes $Q + 2$. Thus, the new energy $U'$ becomes:

$U' = \frac{(Q + 2)^2}{2C}$

We are given that this new energy is 21% greater than the original energy. Therefore, we can write:

$U' = 1.21U$

Substituting the expressions for $U$ and $U'$, we get:

$\frac{(Q + 2)^2}{2C} = 1.21 \cdot \frac{Q^2}{2C}$

Since $C$ is a common factor in the numerator and denominator, it cancels out. Hence, we can simplify to:

$(Q + 2)^2 = 1.21Q^2$

Expanding and rearranging the equation, we obtain:

$Q^2 + 4Q + 4 = 1.21Q^2$

$0.21Q^2 - 4Q - 4 = 0$

Solving this quadratic equation using the quadratic formula, where $a = 0.21$, $b = -4$, and $c = -4$, we get:

$Q = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$

Substitute the values of $a$, $b$, and $c$:

$Q = \frac{4 \pm \sqrt{(-4)^2 - 4 \cdot 0.21 \cdot (-4)}}{2 \cdot 0.21}$

$Q = \frac{4 \pm \sqrt{16 + 3.36}}{0.42}$

$Q = \frac{4 \pm \sqrt{19.36}}{0.42}$

$Q = \frac{4 \pm 4.4}{0.42}$

Therefore, we have two possible solutions for $Q$:

$Q = \frac{4 + 4.4}{0.42} = 20$

$Q = \frac{4 - 4.4}{0.42} = -0.95 \ (not \ practical)$

Since a negative charge is not practical in this context, the original charge on the capacitor is:

Option C: 20 C

In Young's double slit experiment, the position of the $n^{\text{th}}$ maximum for a given wavelength can be calculated using the formula:

$Y = \frac{n \lambda D}{d}$

where:

• $Y$ is the distance from the central maximum,
• $n$ is the order of the maximum,
• $\lambda$ is the wavelength of the light,
• $D$ is the distance between the slits and the screen,
• $d$ is the distance between the two slits.

According to the given problem:

1. The $10^{\text{th}}$ maximum for wavelength $\lambda_1$ is at a distance $Y_1$ from the central maximum:

   $Y_1 = \frac{10 \lambda_1 D}{d}$

2. The $5^{\text{th}}$ maximum for wavelength $\lambda_2$ is at a distance $Y_2$ from the central maximum:

   $Y_2 = \frac{5 \lambda_2 D}{d}$

We need to find the ratio $\frac{Y_1}{Y_2}$:

$\frac{Y_1}{Y_2} = \frac{\frac{10 \lambda_1 D}{d}}{\frac{5 \lambda_2 D}{d}} = \frac{10 \lambda_1}{5 \lambda_2} = \frac{2 \lambda_1}{\lambda_2}$

Therefore, the correct answer is:

Option A

$\frac{2 \lambda_1}{\lambda_2}$

$$\begin{aligned} & \text { Distance of nth maximum } \mathrm{y}_{\mathrm{n}}=\left(\mathrm{n}+\frac{1}{2}\right) \frac{\lambda \mathrm{D}}{\mathrm{a}} \\ & \therefore \quad \mathrm{y}_2=\left(2+\frac{1}{2}\right) \frac{\lambda_1 \mathrm{D}}{\mathrm{a}} \text { and } \mathrm{y}_3=\left(3+\frac{1}{2}\right) \frac{\lambda_2 \mathrm{D}}{\mathrm{a}} \\ & \therefore \quad 2.5 \lambda_1=3.5 \lambda_2 \\ & \therefore \quad 2.5 \times 6000=3.5 \lambda_2 \\ & \therefore \quad \lambda_2=\frac{2.5 \times 6000}{3.5} \approx 4300 \mathop A\limits^o\end{aligned}$$

The width of central maximum is given by

$\mathrm{W}=\frac{2 \lambda \mathrm{D}}{\mathrm{a}}$

W depends on wavelength of light $\lambda$, which is related to the frequency. Hence question is not proper.

To determine the amplitude of the oscillation for the given simple harmonic motion (SHM), we first use some basic principles and formulas related to SHM.

Given:

• Mass, $m = 0.4 \mathrm{~kg}$
• Frequency, $f = \frac{16}{\pi} \mathrm{Hz}$
• Kinetic energy, $KE = 2 \mathrm{~J}$ at a certain displacement
• Potential energy, $PE = 1.2 \mathrm{~J}$ at the same displacement

First, the total energy in the SHM system is the sum of kinetic energy and potential energy:

$E_{\text{total}} = KE + PE = 2 \mathrm{~J} + 1.2 \mathrm{~J} = 3.2 \mathrm{~J}$

In SHM, the total energy is also given by the formula:

$E_{\text{total}} = \frac{1}{2} m \omega^2 A^2$

where:

• $m$ is the mass.
• $\omega$ is the angular frequency.
• $A$ is the amplitude of oscillation.

First, we need to find the angular frequency $\omega$ from the given frequency $f$:

$\omega = 2\pi f$

Substituting the given frequency, we get:

$\omega = 2\pi \left(\frac{16}{\pi}\right) = 32 \mathrm{~rad/s}$

Now, we use the total energy formula to find the amplitude $A$:

$3.2 = \frac{1}{2} \times 0.4 \times 32^2 \times A^2$

Simplifying, we get:

$3.2 = 0.2 \times 1024 \times A^2$

$3.2 = 204.8 A^2$

$A^2 = \frac{3.2}{204.8} = 0.015625$

$A = \sqrt{0.015625} = 0.125 \mathrm{~m}$

Hence, the amplitude of oscillation is $0.125 \mathrm{~m}$.

Therefore, the correct option is:

Option B

$0.125 \mathrm{~m}$

To find the escape velocity from the planet, we need to use the formula for escape velocity. The escape velocity $V$ from a celestial body is given by:

$V = \sqrt{\frac{2GM}{R}}$

where:

• $G$ is the gravitational constant,
• $M$ is the mass of the celestial body, and
• $R$ is the radius of the celestial body.

For Earth, the escape velocity is:

$V_e = \sqrt{\frac{2GM_e}{R_e}}$

where $M_e$ is the mass of the Earth and $R_e$ is the radius of the Earth.

Now, for the planet in question, the mass of the planet $M_p$ is 6 times that of the Earth, i.e.,

$M_p = 6M_e$

The radius of the planet $R_p$ is twice that of the Earth, i.e.,

$R_p = 2R_e$

The escape velocity $V_p$ from the planet is then given by:

$V_p = \sqrt{\frac{2GM_p}{R_p}}$

Substituting the values of $M_p$ and $R_p$ in the escape velocity formula, we get:

$V_p = \sqrt{\frac{2G(6M_e)}{2R_e}}$

Simplifying the expression inside the square root:

$V_p = \sqrt{\frac{12GM_e}{2R_e}}$

$V_p = \sqrt{6 \frac{GM_e}{R_e}}$

Since we know that:

$V_e = \sqrt{\frac{2GM_e}{R_e}}$

we can substitute $V_e^2$ in place of $\frac{2GM_e}{R_e}$:

$V_e^2 = \frac{2GM_e}{R_e}$

Therefore:

$V_p = \sqrt{3 \cdot V_e^2}$

$V_p = \sqrt{3} \cdot V_e$

Thus, the escape velocity from the planet is $\sqrt{3} \cdot V_e$.

The correct answer is Option A: $\sqrt{3} \mathrm{~V}_{\mathrm{e}}$.