MHT CET 2021 23rd September Morning Shift (Physics & Chemistry) Question Paper with Solutions

To determine the angular frequency of a molecule consisting of two atoms, we need to use the concept of rotational kinetic energy. The average rotational kinetic energy of a diatomic molecule at room temperature is given by:

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

Here, $E$ is the average rotational kinetic energy, $I$ is the moment of inertia, and $\omega$ is the angular frequency.

The moment of inertia for the system can be calculated using the formula:

$I = \mu d^2$

where $\mu$ is the reduced mass of the two-atom system and $d$ is the separation distance. For two atoms of mass $m$, the reduced mass $\mu$ is given by:

$\mu = \frac{m \cdot m}{m + m} = \frac{m^2}{2m} = \frac{m}{2}$

Thus, the moment of inertia becomes:

$I = \frac{m}{2} d^2$

Substituting $I$ back into the rotational kinetic energy equation, we get:

$E = \frac{1}{2} \left( \frac{m}{2} d^2 \right) \omega^2$

Simplifying this, we obtain:

$E = \frac{m d^2 \omega^2}{4}$

To solve for the angular frequency $\omega$, we rearrange the equation:

$\omega^2 = \frac{4E}{m d^2}$

Taking the square root of both sides, we find:

$\omega = \sqrt{\frac{4E}{m d^2}}$

Which simplifies to:

$\omega = \frac{2}{d} \sqrt{\frac{E}{m}}$

Thus, the correct answer is:

Option A: $\frac{2}{\mathrm{~d}} \sqrt{\frac{\mathrm{E}}{\mathrm{m}}}$

Eddy currents are induced currents that are generated within conductors when they are exposed to a changing magnetic field. According to Faraday's Law of Electromagnetic Induction, a change in magnetic flux through a conductor induces an electromotive force (emf) in the conductor. If the conductor is a solid piece, like a thick metal plate, circulating currents (eddy currents) are produced within it.

Given the options, let us analyze each to determine when eddy currents are produced:

Option A: A thick metal plate is kept in a steady magnetic field.

A steady magnetic field means that there is no change in magnetic flux. Without a change in magnetic flux, no emf is induced, and hence no eddy currents are produced.

Option B: A circular coil is placed in a steady magnetic field.

Similar to Option A, a steady magnetic field will not cause changes in magnetic flux linkage through the circular coil, so no eddy currents are produced.

Option C: A steady current is passed through a coil.

A steady current in a coil generates a steady magnetic field, but since the field is not changing, it does not induce emf in nearby conductors. Therefore, no eddy currents are produced.

Option D: A thick metal plate is kept in a varying magnetic field.

A varying magnetic field causes changes in magnetic flux through the metal plate, inducing emf within it. This induced current forms closed loops within the conductor, known as eddy currents.

Therefore, the correct answer is:

Option D: a thick metal plate is kept in a varying magnetic field

The inductive reactance ($X_L$) of an inductor in an AC circuit is given by the formula:

$X_L = 2 \pi f L$

where $f$ is the frequency of the AC supply and $L$ is the inductance.

From this formula, we see that the inductive reactance ($X_L$) is directly proportional to the frequency ($f$). This means that as the frequency increases, the inductive reactance increases as well. Therefore, the correct option is:

Option B: increases as it directly proportional to frequency

Here's how to solve this problem:

For an adiabatic process, we have the following relationship:

$p_1V_1^\gamma = p_2V_2^\gamma$

where:

* $p_1$ and $V_1$ are the initial pressure and volume.

* $p_2$ and $V_2$ are the final pressure and volume.

* $\gamma$ is the adiabatic index (ratio of specific heats).

We're given that the density doubles. Density is mass per unit volume, so if the density doubles, the volume is halved. Therefore, $V_2 = \frac{1}{2}V_1$.

Substituting these values into the adiabatic equation:

$p_1V_1^\gamma = p_2 \left(\frac{1}{2}V_1\right)^\gamma$

Simplifying:

$p_1 = p_2 \left(\frac{1}{2}\right)^\gamma$

Solving for $p_2$:

$p_2 = p_1 \left(\frac{1}{2}\right)^{-\gamma} = p_1 \times 2^\gamma$

Substituting the given value of $\gamma = \frac{7}{5}$:

$p_2 = p_1 \times 2^{7/5} \approx p_1 \times 2.63$

Therefore, the final pressure of the gas is approximately 2.63p. The correct answer is Option D.

Moment of inertia of the upper sphere $=\frac{2}{5} \mathrm{MR}^2$

For each lower sphere M.I. $=\frac{2}{5} \mathrm{MR}^2+\mathrm{MR}^2=\frac{7}{5} \mathrm{MR}^2$

$\therefore$ Total moment of inertia $\mathrm{I}=\frac{2}{5} \mathrm{MR}^2+2 \times \frac{7}{5} \mathrm{MR}^2=\frac{16}{5} \mathrm{MR}^2$

Rate of radiation $\mathrm{R} \propto \mathrm{T}^4$

$$\begin{aligned} & \therefore \frac{R_2}{R_1}=\left(\frac{T_2}{T_1}\right)^4=(1.5)^4 \approx 5 \\ & \therefore R_2=5 R_1 \\ & R_2-R_1=4 R_1 \\ & \frac{\mathrm{R}_2-\mathrm{R}_1}{\mathrm{R}_1}=4 \\ \end{aligned}$$

$\text { Percentage increase }=\left(\frac{R_2-R_1}{R_1}\right) \times 100=400 \%$

To determine the intensity of magnetization (I) of the bar magnet, we can use the formula:

$I = \frac{M}{V}$

where $M$ is the magnetic moment and $V$ is the volume of the bar magnet. The volume can be obtained using the length ($l$) and the cross-sectional area ($A$) of the bar magnet:

$V = A \cdot l$

Given that the length of the bar magnet $l$ is $3 \, \mathrm{cm} = 3 \times 10^{-2} \, \mathrm{m}$ and the cross-sectional area $A$ is $2 \, \mathrm{cm}^2 = 2 \times 10^{-4} \, \mathrm{m}^2$, we can calculate the volume as:

$V = 2 \times 10^{-4} \, \mathrm{m}^2 \times 3 \times 10^{-2} \, \mathrm{m} = 6 \times 10^{-6} \, \mathrm{m}^3$

Now, substituting the given magnetic moment $M = 3 \, \mathrm{Am}^2$ and the calculated volume into the intensity of magnetization formula, we get:

$I = \frac{M}{V} = \frac{3 \, \mathrm{Am}^2}{6 \times 10^{-6} \, \mathrm{m}^3} = \frac{3}{6 \times 10^{-6}} \, \mathrm{A/m} = \frac{1}{2 \times 10^{-6}} \, \mathrm{A/m} = 0.5 \times 10^6 \, \mathrm{A/m} = 5 \times 10^5 \, \mathrm{A/m}$

Therefore, the correct answer is:

Option D

$5 \times 10^5 \, \mathrm{A/m}$

To find the magnitude of the induced electromotive force (e.m.f.) in the coil, we'll use Faraday's Law of Electromagnetic Induction. Faraday's Law states that the induced e.m.f. in a coil is equal to the negative rate of change of magnetic flux through the coil. Mathematically, this can be expressed as:

$\text{e.m.f.} = -\frac{d\phi}{dt}$

Given that the magnetic flux $\phi$ varies with time $t$ as:

$\phi = 3t^2 + 4t + 7$

We need to find the derivative of $\phi$ with respect to $t$ in order to determine the rate of change of the magnetic flux. Differentiating $\phi$ with respect to $t$, we get:

$\frac{d\phi}{dt} = \frac{d}{dt}(3t^2 + 4t + 7)$

Applying the differentiation rule, we get:

$\frac{d\phi}{dt} = 6t + 4$

Now, we need to find the induced e.m.f. at $t = 2 \, \text{s}$. Substitute $t = 2$ into the derivative:

$\frac{d\phi}{dt} \bigg|_{t=2} = 6(2) + 4 = 12 + 4 = 16$

According to Faraday's Law, the induced e.m.f. is the negative of this value. However, since the question asks for the magnitude of the induced e.m.f., we will consider the positive value:

$\text{Magnitude of induced e.m.f.} = 16 \, \text{V}$

Hence, the correct answer is:

Option B: 16 V

To determine the work done in lifting a body of mass 5 kg through a distance of 2 m on the spherical planet with the given conditions, we’ll proceed step-by-step.

First, let's establish the gravitational force acting on the body on the planet, specifically the relationship between the gravitational forces on Earth and the planet.

Given:

• The mass of the planet is 4 times the mass of Earth.


• The radius (R) of the planet is the same as that of Earth.


• The gravitational acceleration on Earth, $g = 10 \, \text{m/s}^2$.

The gravitational force at the surface of a planet is given by Newton's law of gravitation:

$F = \frac{G M m}{R^2}$

where:

• $G$ is the gravitational constant,


• $M$ is the mass of the planet,


• $m$ is the mass of the object, and


• $R$ is the radius of the planet.

Since the planet has 4 times the mass of Earth and the same radius, the gravitational acceleration $g'$ on the planet can be written as:

$g' = \frac{G \cdot (4 M_{\text{Earth}})}{R^2}$

Comparing with Earth's gravity:

$g_{\text{Earth}} = \frac{G M_{\text{Earth}}}{R^2}$

Thus,

$g' = 4 g_{\text{Earth}}$

Given $g_{\text{Earth}} = 10 \, \text{m/s}^2$, we have:

$g' = 4 \times 10 \, \text{m/s}^2 = 40 \, \text{m/s}^2$

Now, the work done in lifting a body through a vertical distance $h$ is given by:

$W = m g' h$

Substituting the given values:

• $m = 5 \, \text{kg}$


• $g' = 40 \, \text{m/s}^2$


• $h = 2 \, \text{m}$

$W = 5 \, \text{kg} \times 40 \, \text{m/s}^2 \times 2 \, \text{m} = 400 \, \text{J}$

Therefore, the work done is:

400 J

The correct answer is Option A: 400 J.

$$\begin{aligned} & \mathrm{h}=\frac{2 \mathrm{~T} \cos \theta}{\mathrm{r} \rho \mathrm{g}} \\ & \therefore \mathrm{h} \propto \frac{1}{\mathrm{r}} \\ & \therefore \frac{\mathrm{h}_2}{\mathrm{~h}_1}=\frac{\mathrm{r}_1}{\mathrm{r}_2} \\ & \text { Area } A=\pi r^2 \quad \therefore A \propto r^2 \\ & \therefore \frac{A_1}{A_2}=\frac{r_1^2}{r_2^2} \text { or } \frac{r_1}{r_2}=\sqrt{\frac{A_1}{A_2}}=\sqrt{16} \\ & \frac{r_1}{r_2}=4 \\ & \therefore \frac{\mathrm{h}_2}{\mathrm{~h}_1}=4 \text { or } \quad \mathrm{h}_2=4 \mathrm{~h}_1=4 \times 2=8 \mathrm{~cm} \\ & \end{aligned}$$

To determine the number of beats produced per second when two waves propagate in the same direction, we must first understand the concept of beat frequency. The beat frequency is given by the absolute difference between the frequencies of the two waves.

Given the two waves:

$\mathrm{Y}_1 = 0.25 \sin(316 \mathrm{t})$

$\mathrm{Y}_2 = 0.25 \sin(310 \mathrm{t})$

The general form of a sine wave is:

$\mathrm{Y} = A \sin(2 \pi f t)$

Here, $\mathrm{Y}_1$ and $\mathrm{Y}_2$ follow the form $\mathrm{Y} = A \sin(\omega t)$, where $\omega$ is the angular frequency.

For wave $\mathrm{Y}_1$, the angular frequency $\omega_1$ is 316, thus the frequency $f_1$ is given by:

$\omega_1 = 2 \pi f_1 \implies f_1 = \frac{316}{2 \pi}$

For wave $\mathrm{Y}_2$, the angular frequency $\omega_2$ is 310, thus the frequency $f_2$ is given by:

$\omega_2 = 2 \pi f_2 \implies f_2 = \frac{310}{2 \pi}$

To find the beat frequency $f_{\text{beat}}$, we take the absolute difference between the frequencies $f_1$ and $f_2$:

$f_{\text{beat}} = |f_1 - f_2|$

Substituting the respective frequencies:

$f_{\text{beat}} = \left| \frac{316}{2 \pi} - \frac{310}{2 \pi} \right|$

Simplifying this:

$f_{\text{beat}} = \left| \frac{316 - 310}{2 \pi} \right| = \frac{|6|}{2 \pi} = \frac{6}{2 \pi} = \frac{3}{\pi}$

Therefore, the number of beats produced per second is:

$\boxed{\frac{3}{\pi}}$

The correct answer is Option A: $\frac{3}{\pi}$.

In the absence of the dielectric the electric field between the plates is given by

$\mathrm{E}=\frac{\mathrm{q}}{\mathrm{A} \varepsilon_0}$

when dielectric is placed, the field gets reduced by the factor $k$.

The potential difference between the plates is given by

$$\begin{aligned} V & =E_1 t_1+E_2 t_2+E_2 t_3 \\ & =\frac{E}{k_1} \cdot t_1+\frac{E}{k_2} \cdot t_2+\frac{E}{k_3} \cdot t_3 \\ & =E\left(\frac{t_1}{k_1}+\frac{t_2}{k_2}+\frac{t_3}{k_3}\right) \\ & =\frac{q}{A \varepsilon_0}\left(\frac{t_1}{k_1}+\frac{t_2}{k_2}+\frac{t_3}{k_3}\right) \\ C & =\frac{q}{V}=\frac{A \varepsilon_0}{\left(\frac{t_1}{k_1}+\frac{t_2}{k_2}+\frac{t_3}{k_3}\right)} \end{aligned}$$

An isobaric process is one in which the pressure remains constant as the system undergoes a change. Let's analyze the given options:

Option A: The pressure of the system remains constant

This statement is correct because, by definition, an isobaric process maintains a constant pressure.

Option B: There is change in volume, when work is done

This statement is also correct. In an isobaric process, the work done by the gas is the product of the pressure and the change in volume. This can be represented as:

$W = P \Delta V$

Option C: Temperature of the system remains constant

This statement is incorrect. In an isothermal process, the temperature remains constant, but in an isobaric process, it is not necessarily the case. The temperature can change even though the pressure remains constant. Therefore, this option is wrong for an isobaric process.

Option D: Energy exchanged is used to do work to change internal energy

This statement is correct. According to the first law of thermodynamics, the energy exchanged (heat energy) in an isobaric process is used to do work and change the internal energy of the system. The energy balance can be expressed as:

$Q = \Delta U + W$

where $Q$ is the heat added to the system, $\Delta U$ is the change in internal energy, and $W$ is the work done by the system.

In summary, Option C is the wrong statement for an isobaric process.

To determine the phase difference between the two waves represented by the equations:

$\mathrm{y}_1 = \mathrm{A} \sin (\omega \mathrm{t} + \mathrm{kx} + 0.57) \mathrm{m}$

and

$\mathrm{y}_2 = \mathrm{A} \cos (\omega \mathrm{t} + \mathrm{kx}) \mathrm{m}$

we need to express both waves in a similar trigonometric form to compare their phases directly. The cosine function can be rewritten in terms of the sine function using the identity:

$\cos(\theta) = \sin\left(\theta + \frac{\pi}{2}\right)$

Hence, we can rewrite $\mathrm{y}_2$ as follows:

$\mathrm{y}_2 = \mathrm{A} \cos (\omega \mathrm{t} + \mathrm{kx}) = \mathrm{A} \sin \left( (\omega \mathrm{t} + \mathrm{kx}) + \frac{\pi}{2} \right)$

Now, compare $\mathrm{y}_1$ and the rewritten form of $\mathrm{y}_2$:

$\mathrm{y}_1 = \mathrm{A} \sin (\omega \mathrm{t} + \mathrm{kx} + 0.57) \mathrm{m}$

$\mathrm{y}_2 = \mathrm{A} \sin \left( (\omega \mathrm{t} + \mathrm{kx}) + \frac{\pi}{2} \right) \mathrm{m}$

The wave $\mathrm{y}_1$ has a phase term of $(\omega \mathrm{t} + \mathrm{kx} + 0.57)$, and the rewritten wave $\mathrm{y}_2$ has a phase term of $(\omega \mathrm{t} + \mathrm{kx} + \frac{\pi}{2})$.

Thus, the phase difference between the two waves, $\phi$, is the difference between these two phase terms:

$\phi = \left( (\omega \mathrm{t} + \mathrm{kx} + \frac{\pi}{2}) \right) - \left( \omega \mathrm{t} + \mathrm{kx} + 0.57 \right)$

$\phi = \frac{\pi}{2} - 0.57$

Now, we need to evaluate $\frac{\pi}{2}$ in radians. Recall that $\pi \approx 3.14159265$. Therefore,

$\frac{\pi}{2} \approx \frac{3.14159265}{2} \approx 1.57 \, \text{radians}$

Substituting this value, we get:

$\phi = 1.57 - 0.57 = 1.0 \, \text{radians}$

Therefore, the phase difference between the two waves is:

Option B

1.0 radian

To solve this problem, we need to understand the relationship between the coefficients of linear expansion, initial lengths, and the condition given that ($\mathrm{L}_2 - \mathrm{L}_1$) is the same at all temperatures.

Linear expansion can be described by the formula for the change in length, $\Delta \mathrm{L}$, which is proportional to the initial length $\mathrm{L}$ and the change in temperature $\Delta \mathrm{T}$. For a given material, this can be written as:

$\Delta \mathrm{L}_1 = \mathrm{L}_1 \alpha_1 \Delta \mathrm{T}$

$\Delta \mathrm{L}_2 = \mathrm{L}_2 \alpha_2 \Delta \mathrm{T}$

Given that ($\mathrm{L}_2 - \mathrm{L}_1$) is the same at all temperatures, the difference in the change in lengths due to temperature changes should also be equal at all temperatures. This implies:

$\Delta \mathrm{L}_2 - \Delta \mathrm{L}_1 = 0$

Substitute the expressions for $\Delta \mathrm{L}_1$ and $\Delta \mathrm{L}_2$:

$\mathrm{L}_2 \alpha_2 \Delta \mathrm{T} - \mathrm{L}_1 \alpha_1 \Delta \mathrm{T} = 0$

Since $\Delta \mathrm{T}$ is common and non-zero, we can divide both sides by $\Delta \mathrm{T}$:

$\mathrm{L}_2 \alpha_2 - \mathrm{L}_1 \alpha_1 = 0$

Therefore, the correct relation is:

$\mathrm{L}_1 \alpha_1 = \mathrm{L}_2 \alpha_2$

So the correct option is:

Option D

According to Bohr's theory, the energy levels of an electron in a hydrogen atom are given by:

$E_n = -\frac{13.6 \, \text{eV}}{n^2}$

Given that the energy of the electron in the excited state is $-3.4 \, \text{eV}$, we can set up the following equation to find the principal quantum number $n$:

$-3.4 = -\frac{13.6}{n^2}$

Solving for $n$, we get:

$n^2 = \frac{13.6}{3.4} = 4$

So, $n = 2$. Now, according to Bohr's theory, the angular momentum $L$ of the electron in the $n$th excited state is given by:

$L = n \frac{h}{2 \pi}$

By substituting the value of $n = 2$, we get:

$L = 2 \frac{h}{2 \pi} = \frac{h}{\pi}$

Therefore, the angular momentum of the electron in that excited state is:

Option C

$\frac{\mathrm{h}}{\pi}$

The magnifying power of a refracting astronomical telescope is given by the formula:

$\mathrm{m}=\frac{f_o}{f_e}$

Here, $f_o$ is the focal length of the objective lens and $f_e$ is the focal length of the eyepiece.

Given that the initial magnifying power is $\mathrm{m}$, we can write:

$\mathrm{m}=\frac{f_o}{f_e}$

If the focal length of the eyepiece is doubled, the new focal length of the eyepiece becomes $2 f_e$. The new magnifying power $\mathrm{m'}$ will be:

$\mathrm{m'}=\frac{f_o}{2 f_e}$

Now, we can substitute $\frac{f_o}{f_e}$ with $\mathrm{m}$ to find the new magnifying power:

$\mathrm{m'}=\frac{\mathrm{m}}{2}$

Therefore, if the focal length of the eyepiece is doubled, the magnifying power will become $\frac{m}{2}$.

So, the correct option is:

Option C: $\frac{m}{2}$

Let $\mathrm{K}$ represent kinetic energy

$$\begin{aligned} \therefore \quad \mathrm{K}_1 & =\frac{\mathrm{hc}}{\lambda_1}-\mathrm{W}_0 \\ \text { and } & \mathrm{K}_2=\frac{\mathrm{hc}}{\lambda_2}-\mathrm{W}_0 \\ \mathrm{~K}_2 & =3 \mathrm{~K}_1 \end{aligned}$$

$$\begin{aligned} & \therefore \quad \frac{h c}{\lambda_2}-W_0=3 \frac{h c}{\lambda_1}-3 W_0 \\ & \therefore \quad 2 W_0=\frac{3 h c}{\lambda_1}-\frac{h c}{\lambda_2} \\ & \therefore \quad W_0=\frac{h c}{2}\left(\frac{3 \lambda_2-\lambda_1}{\lambda_1 \lambda_2}\right) \end{aligned}$$

To solve this problem, we need to compute the magnetic field (magnetic induction) at the center of the rotating ring due to the current created by the moving charges. Here’s a step-by-step explanation:

Step 1: Determine the current ($I$) created by the rotating charges.

Since the ring has a charge $q$ uniformly spread on it and it rotates with a frequency $f$ revolutions per second, each charge completes one full revolution every $1/f$ seconds. The ring's rotation creates a circular current $I$.

The formula for the current due to the rotation of charges is:

$I = q \cdot f$

Step 2: Use the Biot-Savart law to find the magnetic field at the center of the ring.

For a circular loop, the magnetic field at the center is given by:

$B = \frac{\mu_0 I}{2R}$

Substituting the current $I$ from step 1 into this equation, we have:

$B = \frac{\mu_0 (q \cdot f)}{2 R}$

Thus, the magnetic induction at the center of the ring is:

$B = \frac{\mu_0 q f}{2 R}$

Conclusion:

The correct answer is:

Option C

$\frac{\mu_0 q f}{2 R}$

Wire has a resitance $1 \Omega \mathrm{cm}^{-1}$.

$\therefore$ Resistance of $40 \mathrm{~cm}$ of wire $=40 \Omega$ and resistance of $60 \mathrm{~cm}$ of wire $=60 \Omega$

No current flows through the galvanometer since the bridge in balanced. Hence the branch containing the galvanometer can be removed.

For balanced bridge $\frac{4}{40}=\frac{Y}{60}$

$\therefore Y=\frac{4 \times 60}{40}=6 \Omega$

$4 \Omega$ and $6 \Omega$ are in series. Their equivalent resistance is $10 \Omega$.

$40 \Omega$ and $60 \Omega$ are in series. Their equivalent resistance is $100 \Omega$.

$10 \Omega$ and $100 \Omega$ are in parallel

Their equivalent resistance is $\frac{1000}{110} \Omega$.

$\therefore$ The current $\mathrm{I}=\frac{\mathrm{V}}{\mathrm{R}}=\frac{6 \times 110}{1000}=0.66 \mathrm{~A}$

To solve this problem, let's first understand the fundamental frequencies and harmonics in the two types of pipes described.

Pipe 'A' is closed at one end. For such a pipe, the fundamental frequency (first harmonic) is given by:

$f_A = \frac{v}{4L_A}$

where $v$ is the speed of sound and $L_A$ is the length of the pipe. The harmonics in a closed pipe are odd multiples of the fundamental frequency. So, the frequencies are:

$f_A, 3f_A, 5f_A, \ldots$

Pipe 'B' is open at both ends. For such a pipe, the fundamental frequency (first harmonic) is given by:

$f_B = \frac{v}{2L_B}$

where $L_B$ is the length of the pipe. The harmonics in an open pipe are whole number multiples of the fundamental frequency. So, the frequencies are:

$f_B, 2f_B, 3f_B, \ldots$

According to the problem, the fundamental frequency of pipe 'A' coincides with the second overtone of pipe 'B'. The second overtone of pipe 'B' is the third harmonic, which is:

$3f_B = 3 \left(\frac{v}{2L_B}\right) = \frac{3v}{2L_B}$

We set this equal to the fundamental frequency of pipe 'A':

$\frac{v}{4L_A} = \frac{3v}{2L_B}$

By canceling the common terms and rearranging, we get:

$\frac{1}{4L_A} = \frac{3}{2L_B}$

Cross-multiplying gives:

$2L_B = 12L_A$

Simplifying, we find:

$L_B = 6L_A$

Thus, the ratio of the length of pipe 'A' to that of pipe 'B' is:

$\frac{L_A}{L_B} = \frac{L_A}{6L_A} = \frac{1}{6}$

Therefore, the ratio is:

Option C 1 : 6

Young's double-slit experiment relies on the interference of light waves from two slits to produce fringes—a pattern of dark and bright bands. The basic principle behind this phenomenon is the constructive and destructive interference of coherent light waves emanating from the two slits. When these waves superimpose, they create alternating bright (constructive interference) and dark (destructive interference) fringes on a screen placed behind the slits.

If one of the slits is covered with black opaque paper, there won't be any interference because only one light source will be contributing to the pattern. The process of interference requires two coherent sources of light. Without the second slit, there is no second wave to interfere with the first wave, and therefore, the interference pattern of fringes cannot be formed.

Hence, the fringes will:

Option D: not be observed

To determine at what rate a single conductor should cut the magnetic flux to produce a current of $1.5 \mathrm{~mA}$ when a resistance of $5 \Omega$ is connected across its ends, we can use Faraday's law of electromagnetic induction and Ohm's law.

First, let's start with Ohm's law, which states:

$V = IR$

Where:

• $V$ is the voltage (emf) induced in the conductor
• $I$ is the current
• $R$ is the resistance

Given that the current $I = 1.5 \mathrm{~mA} = 1.5 \times 10^{-3} \mathrm{~A}$ and the resistance $R = 5 \Omega$, we can substitute these values into Ohm's law to find the induced voltage $V$:

$V = (1.5 \times 10^{-3} \mathrm{~A}) \times (5 \Omega)$

$V = 7.5 \times 10^{-3} \mathrm{~V}$

Next, we use Faraday's law of electromagnetic induction, which states that the induced emf (voltage) in a conductor moving through a magnetic field is equal to the rate of change of magnetic flux linkage:

$V = \frac{d\Phi}{dt}$

Where:

• $V$ is the induced emf
• $\frac{d\Phi}{dt}$ is the rate of change of magnetic flux

Given that we have already found $V = 7.5 \times 10^{-3} \mathrm{~V}$, we can substitute this value in to find the rate of change of magnetic flux:

$\frac{d\Phi}{dt} = 7.5 \times 10^{-3} \mathrm{~wb/s}$

Therefore, the conductor should cut the magnetic flux at a rate of $7.5 \times 10^{-3} \frac{\mathrm{wb}}{\mathrm{s}}$.

The correct option is:

Option D $7.5 \times 10^{-3} \frac{\mathrm{wb}}{\mathrm{s}}$

To determine the maximum height reached by the balls, we need to first understand the motion of the balls being thrown vertically upwards. When a ball is thrown upwards, it decelerates under the influence of gravity until it comes to a stop at its highest point. The time to reach the maximum height can be determined by using the kinematic equation:

$v = u + at$

where:

$v$ is the final velocity (0 m/s at the maximum height),

$u$ is the initial velocity,

$a$ is the acceleration (which is $-g$ due to gravity), and

$t$ is the time.

At the maximum height:

$0 = u - gt$

Solving for $t$ gives us:

$t = \frac{u}{g}$

The given problem states that the time interval between the throws is 3 seconds. Therefore, this time must equal the time taken to reach the maximum height. Thus:

$t = 3 \text{ seconds}$

Using the above formula:

$t = \frac{u}{g} = 3$

Solving for $u$, we get:

$u = 3g = 3 \times 10 = 30 \text{ m/s}$

Now, to find the maximum height, we use another kinematic equation:

$h = ut - \frac{1}{2}gt^2$

Substituting the values, we get:

$h = 30 \times 3 - \frac{1}{2} \times 10 \times 3^2$

$h = 90 - \frac{1}{2} \times 10 \times 9$

$h = 90 - 45$

$h = 45 \text{ meters}$

Therefore, the maximum height of the balls is 45 meters, corresponding to Option A.

Answer: Option A (45 m)

$\frac{\mathrm{E}_1}{\mathrm{E}_2}=\frac{\ell_1+\ell_2}{\ell_1+\ell_2}=\frac{64+32}{64-32}=\frac{96}{32}=3$

Truth table :

This is truth table of AND gate.

$\therefore \mathrm{Y=A . B}$

To determine the coefficient of restitution, let's analyze the given problem using the principles of momentum conservation and the definition of the coefficient of restitution.

First, we note the initial conditions:

• Mass of the moving block: $\mathrm{m}$
• Velocity of the moving block: $\mathrm{V}$
• Mass of the block at rest: $4 \mathrm{~m}$
• Velocity of the block at rest: $0$

After the collision:

• The block of mass $\mathrm{m}$ comes to rest, so its final velocity is $0$.
• Let $V_2$ be the final velocity of the block with mass $4 \mathrm{~m}$ after the collision.

By the law of conservation of linear momentum, the total momentum before and after the collision must be equal:

$\mathrm{m} \cdot \mathrm{V} + 4 \mathrm{~m} \cdot 0 = \mathrm{m} \cdot 0 + 4 \mathrm{~m} \cdot V_2$

This simplifies to:

$\mathrm{V} = 4 V_2 \implies V_2 = \frac{\mathrm{V}}{4}$

Next, we use the coefficient of restitution (e), which is defined as the relative speed of separation divided by the relative speed of approach. Mathematically, for a one-dimensional collision:

$e = \frac{\text{Relative speed after collision}}{\text{Relative speed before collision}}$

Before the collision, the relative speed of approach is:

$\mathrm{V} - 0 = \mathrm{V}$

After the collision, the relative speed of separation is:

$0 - \left( \frac{\mathrm{V}}{4} \right) = - \frac{\mathrm{V}}{4}$