MHT CET 2024 9th May Morning Shift (Physics & Chemistry) Question Paper with Solutions

The gyromagnetic ratio is given by the expression:

$\frac{𝑒}{2𝑚}$

This ratio describes the magnetic moment of a system per unit angular momentum. It is especially prominent in the context of electron spin and nuclear magnetic resonance.

The Bohr magneton, which represents the natural unit of the magnetic moment in atomic physics, is given by:

$\frac{𝑒\hslash }{2𝑚}$

where $\hslash$ is the reduced Planck's constant, defined as $\hslash =\frac{ℎ}{2𝜋}$.

For the given options, the correct expressions describing the gyromagnetic ratio and the Bohr magneton are found in Option A:

Gyromagnetic Ratio: $\frac{𝑒}{2𝑚}$

Bohr Magneton: $\frac{𝑒ℎ}{4𝜋𝑚}$ (since $\hslash =\frac{ℎ}{2𝜋}$, this simplifies to the Bohr magneton from the definition with $\hslash$)

Thus, Option A is the correct choice which matches these definitions.

$\begin{array}{cc}\frac{{𝐸}_{𝐴}}{{𝐸}_{𝐵}}& =\frac{{𝑀}_{𝐴}}{{𝑀}_{𝐵}}\times \frac{{𝑟}_{𝐵}}{{𝑟}_{𝐴}}\dots (𝐸\propto \frac{𝑀}{𝑟})\\ & =\frac{3}{1}\times \frac{4𝑟}{𝑟}\\ \frac{{𝐸}_{𝐴}}{{𝐸}_{𝐵}}& =\frac{12}{1}\end{array}$

$\begin{array}{cc}& \text{ Given: }𝛽=50,R_i=1k\Omega ={10}^3\Omega ,\\ & V_i=0.01V\\ & 𝛽=\frac{I_C}{I_B}=50\Rightarrow I_C=50\times I_B\\ & V_i=I_B\times R_i\\ \therefore & I_B=\frac{V_i}{R_i}=\frac{0.01}{{10}^3}={10}^{-5}\\ \therefore & I_C=50\times {10}^{-5}=500\times {10}^{-6}A=500𝜇A\end{array}$

For an ideal inductor and an ideal capacitor, the average power over a complete cycle of alternating current (AC) is zero for both components.

Explanation

Inductor:

An ideal inductor has no resistance and therefore no real power loss. The energy is stored temporarily in the magnetic field during one part of the AC cycle and then completely returned to the circuit in another part of the cycle. Therefore, the net energy transfer over a full cycle is zero, resulting in:

Average Power for Inductor = 0

Capacitor:

Similarly, an ideal capacitor temporarily stores energy in its electric field. During one half of the AC cycle, energy is stored, and during the other half, it is returned to the circuit. Since the energy is not dissipated, the net energy transfer over a full cycle is zero:

Average Power for Capacitor = 0

Thus, the average power over a complete cycle for both an ideal inductor and ideal capacitor is:

Option C: zero, zero

A conducting rod of length ' $𝑙$ ' whose one end is fixed, is rotated about the axis passing through its fixed end and perpendicular to its length with constant angular velocity $𝜔$.

$\begin{array}{cc}e& =B𝜋{𝑙}^{2}n\\ \therefore n& =\frac{e}{𝜋B{𝑙}^2}\end{array}$

$\begin{array}{cc}& 𝑌=𝐴\cos {30}^{\circ }=40\times \frac{\sqrt{\surd 3}}{2}=20\sqrt{\surd 3}\mathrm{cm}\\ & \text{ K.E. }=\frac{1}{2}𝑘({𝐴}^2-{𝑌}^2)\\ & 200=\frac{1}{2}𝑘(\frac{1600}{{10}^4}-\frac{1200}{{10}^4})\end{array}$

$\begin{array}{cc}& 200=\frac{1}{2}𝑘\left(\frac{400}{{10}^4}\right)\\ & 𝑘=10000=1\times {10}^4\\ & \therefore 𝑥=4\end{array}$

Let's consider the two waves given by

${𝑦}_1=𝐴\sin \left(𝜔𝑡\right)$

and

${𝑦}_2=𝐴\sin (𝜔𝑡+\frac{𝜋}{2}).$

When adding two sine waves of equal amplitude $𝐴$ with a phase difference $𝛿$, we can use the identity:

$\sin 𝜃+\sin (𝜃+𝛿)=2\sin (𝜃+\frac{𝛿}{2})\cos \left(\frac{𝛿}{2}\right).$

Here, $𝜃=𝜔𝑡$ and $𝛿=\frac{𝜋}{2}$. Plugging in the phase difference gives:

${𝑦}_1+{𝑦}_2=2𝐴\sin (𝜔𝑡+\frac{𝜋}{4})\cos \left(\frac{𝜋}{4}\right).$

Since

$\cos \left(\frac{𝜋}{4}\right)=\frac{\sqrt{\surd 2}}{2},$

the expression simplifies to:

${𝑦}_1+{𝑦}_2=2𝐴\cdot \frac{\sqrt{\surd 2}}{2}\sin (𝜔𝑡+\frac{𝜋}{4})=𝐴\sqrt{\surd 2}\sin (𝜔𝑡+\frac{𝜋}{4}).$

This shows that the maximum amplitude of the resultant wave is:

$𝐴\sqrt{\surd 2}.$

Thus, the correct answer is:

Option C: $\sqrt{\surd 2}𝐴$.

The correct answer is:

Option D: Clarinet

Here's why:

The clarinet is a woodwind instrument, meaning it produces sound when air is blown through a mouthpiece with a single reed.

The other instruments listed—Daphali, Sambal, and Cymbals—are percussion instruments, which produce sound when struck or shaken.

Thus, the clarinet is not a percussion instrument.

The correct answer is Option A.

Here's why:

When a dielectric is inserted between the plates of a capacitor, the dielectric material becomes polarized. This polarization creates an electric field that opposes the field due to the charges on the plates.

As a result, the effective electric field inside the capacitor is reduced. Since the potential difference (voltage) across the capacitor is related to the electric field by:

$𝑉=𝐸\cdot 𝑑$

(where $𝑑$ is the separation between the plates), reducing the electric field $𝐸$ also reduces the effective potential difference $𝑉$ between the plates.

With a lower effective potential difference for the same amount of charge, the capacitance, defined by

$𝐶=\frac{𝑄}{𝑉},$

increases. This explains why the dielectric increases the capacitance rather than decreasing it.

So, Option A, "to reduce the effective potential on plates," correctly describes the role of a dielectric in a capacitor.

The cooling of each sphere is mainly due to radiation and is described by the Stefan–Boltzmann law:

$𝑃=𝜖𝜎𝐴({𝑇}^4-{𝑇}_4^{\text{ambient}})$

where:

$𝜖$ is the emissivity of the surface,

$𝜎$ is the Stefan–Boltzmann constant,

$𝐴$ is the surface area,

$𝑇$ is the temperature of the sphere,

${𝑇}_{\text{ambient}}$ is the ambient temperature.

Since all spheres have the same surface area and are at the same temperature, the rate of cooling is directly proportional to their emissivities. Typically:

Black surfaces have high emissivity (close to 1),

Red surfaces generally have a moderate emissivity,

White surfaces tend to have low emissivity because they are more reflective.

Thus, the sphere with black color will radiate energy (and cool) the fastest, followed by the red sphere, while the white sphere cools the slowest.

Therefore, the order from maximum rate of cooling to minimum rate is:

Black

Red

White

This corresponds to Option A.

Shunt $S=\frac{I_{g}G}{I-I_g}$

On substituting the given values, we get,

$\begin{array}{cc}\therefore 5& =\frac{(\frac{4}{100}𝐼\times 𝐺)}{𝐼-\frac{4}{100}𝐼}=\frac{4𝐺}{96}\\ & \Rightarrow 𝐺=\frac{96\times 5}{4}=120\Omega \end{array}$

$\begin{array}{cc}& \text{ At resonance, }f=\frac{1}{2𝜋\sqrt{\surd \mathrm{LC}}}\Rightarrow f\propto \frac{1}{\sqrt{\surd \mathrm{LC}}}\\ \therefore & \frac{f^{\prime }}{f}=\sqrt{\surd \left(\frac{L_1{C}_1}{L_2{C}_2}\right)}={\left(\frac{L\times C}{3L\times 4C}\right)}^{1/2}=\frac{1}{(12{)}^{1/2}}\\ \therefore & \frac{f^{\prime }}{f}=\frac{1}{2\sqrt{\surd 3}}\Rightarrow f^{\prime }=\frac{f}{2\sqrt{\surd 3}}\end{array}$

A wavefront is defined as the locus of points that have the same phase of vibration. In an isotropic medium (where light travels uniformly in all directions), these wavefronts are surfaces perpendicular to the direction of propagation of light rays.

Hence, the correct choice is:

(A) perpendicular to the direction of propagation of light.

For a string under tension, the frequency of the fundamental mode is given by the formula:

$𝑓=\frac{1}{2𝐿}\sqrt{\surd \frac{𝑇}{𝜇}}$

where:

$𝑓$ is the frequency,

$𝐿$ is the length of the string,

$𝑇$ is the tension in the string,

$𝜇$ is the linear mass density.

Initially, let the tension be $𝑇$ and the initial frequency be ${𝑓}_1$. When the tension is increased by $3\mathrm{kg}𝜔t$, the new tension becomes $𝑇+3\mathrm{kg}𝜔t$, and the new frequency is ${𝑓}_2$.

The ratio of the frequencies is given as $\frac{{𝑓}_1}{{𝑓}_2}=\frac{2}{3}$.

Given this relationship:

$\frac{{𝑓}_2}{{𝑓}_1}=\frac{3}{2}$

Using the formula for frequency and the fact that frequency is proportional to the square root of tension:

$\frac{{𝑓}_2}{{𝑓}_1}=\sqrt{\surd \frac{𝑇+3}{𝑇}}$

Equating the ratios:

$\frac{3}{2}=\sqrt{\surd \frac{𝑇+3}{𝑇}}$

Squaring both sides:

${\left(\frac{3}{2}\right)}^2=\frac{𝑇+3}{𝑇}$

$\frac{9}{4}=\frac{𝑇+3}{𝑇}$

Cross-multiplying gives:

$9𝑇=4(𝑇+3)$

Expanding and simplifying:

$9𝑇=4𝑇+12$

$5𝑇=12$

Thus, the initial tension $𝑇$ is:

$𝑇=\frac{12}{5}=2.4\mathrm{kg}𝜔t$

Therefore, the initial tension in the string is $2.4\mathrm{kg}𝜔t$. The correct option is C.

$\begin{array}{cc}& {𝑚}_1{𝑥}_1={𝑚}_2{𝑥}_2\text{.... (i)}\\ & {𝑚}_1({𝑥}_1-𝑑)={𝑚}_2({𝑥}_2-{𝑑}^{\prime })\text{.... (ii)}\\ \therefore & {𝑚}_1{𝑥}_1-{𝑚}_1𝑑={𝑚}_2{𝑥}_2-{𝑚}_2{𝑑}^{\prime }\\ & {𝑚}_1𝑑={𝑚}_2{𝑑}^{\prime }\text{.... [From (i)]}\\ \therefore & {𝑑}^{\prime }=\frac{{𝑚}_1}{{𝑚}_2}𝑑\end{array}$

The potential energy stored in a spring when it is stretched or compressed is given by the formula:

$𝑈=\frac{1}{2}𝑘{𝑥}^2$

where:

$𝑈$ is the potential energy,

$𝑘$ is the spring constant,

$𝑥$ is the displacement from the equilibrium position.

Given that the potential energy is $𝑈$ when the spring is stretched by 3 cm, we have:

$𝑈=\frac{1}{2}𝑘(3{)}^2$

Now, if the spring is stretched by 9 cm, the potential energy ${𝑈}^{\prime }$ becomes:

${𝑈}^{\prime }=\frac{1}{2}𝑘(9{)}^2$

We can express ${𝑈}^{\prime }$ in terms of $𝑈$ as follows:

Calculate ${𝑈}^{\prime }$ based on the new displacement:

${𝑈}^{\prime }=\frac{1}{2}𝑘(9{)}^2=\frac{1}{2}𝑘\times 81$

Substitute $\frac{1}{2}𝑘=\frac{𝑈}{3^2}=\frac{𝑈}{9}$:

${𝑈}^{\prime }=\frac{𝑈}{9}\times 81=9𝑈$

Hence, when the spring is stretched by 9 cm, the potential energy stored in it is $9𝑈$. Therefore, the correct option is:

Option D: 9U

To determine the maximum height to which the center of mass of the swinging rod rises, we need to apply the principle of conservation of energy.

Initially, the entire energy of the system is kinetic because the rod is swinging with its maximum angular speed $𝜔$ around its center. The moment of inertia $𝐼$ of a thin uniform rod of length $𝐿$ about an axis through its center is given by:

$𝐼=\frac{1}{12}𝑀{𝐿}^2$

The initial kinetic energy (${𝐾}_{𝑖}$) of the rod is thus:

${𝐾}_{𝑖}=\frac{1}{2}𝐼{𝜔}^2=\frac{1}{2}\left(\frac{1}{12}𝑀{𝐿}^2\right){𝜔}^2=\frac{1}{24}𝑀{𝐿}^2{𝜔}^2$

At the highest point of the swing, all this kinetic energy is converted to potential energy (${𝑈}_{𝑓}$), assuming no energy is lost:

${𝑈}_{𝑓}=𝑀𝑔ℎ$

where $ℎ$ is the maximum height attained by the center of mass and $𝑔$ is the gravitational acceleration. By conservation of energy:

${𝐾}_{𝑖}={𝑈}_{𝑓}$

$\frac{1}{24}𝑀{𝐿}^2{𝜔}^2=𝑀𝑔ℎ$

Solving for $ℎ$, we get:

$ℎ=\frac{1}{24}\frac{{𝐿}^2{𝜔}^2}{𝑔}$

Thus, the maximum height of the center of mass is:

Option D:

$\frac{{𝜔}^2{𝐿}^2}{24g}$

To find the apparent depth of a vessel containing three immiscible transparent liquids with refractive indices of $\frac{3}{2}$, $\frac{4}{3}$, and $\frac{6}{5}$, and depths of 3 cm, 4 cm, and 6 cm respectively, we can use the formula for apparent depth.

The formula relating real depth (R) to apparent depth with refractive index ($𝜇$) is given by:

$𝜇=\frac{\text{Real depth (R)}}{\text{Apparent depth}}$

Thus, the apparent depth for each liquid can be calculated as:

$\text{Apparent depth}=\frac{𝑅}{𝜇}$

For the entire vessel, the total apparent depth is the sum of the apparent depths of each layer of liquid:

$\text{Apparent depth}=\frac{{𝑅}_1}{{𝜇}_1}+\frac{{𝑅}_2}{{𝜇}_2}+\frac{{𝑅}_3}{{𝜇}_3}$

Plugging in the values:

$\text{Apparent depth}=\frac{3}{\frac{3}{2}}+\frac{4}{\frac{4}{3}}+\frac{6}{\frac{6}{5}}$

Calculating each term:

For the first liquid:

$\frac{3}{\frac{3}{2}}=2$

For the second liquid:

$\frac{4}{\frac{4}{3}}=3$

For the third liquid:

$\frac{6}{\frac{6}{5}}=5$

Adding these gives:

$\text{Total apparent depth}=2+3+5=10\text{ cm}$

Therefore, the apparent depth of the vessel is 10 cm.

In the (simplified) Bohr model of the hydrogen atom:

The total energy ${𝐸}_{𝑛}$ in the ${𝑛}^{\text{th}}$ orbit is

${𝐸}_{𝑛}=-\frac{13.6\text{eV}}{{𝑛}^2}.$

The kinetic energy of the electron is

${𝐾}_{𝑛}=-{𝐸}_{𝑛}=\frac{13.6\text{eV}}{{𝑛}^2}.$

The potential energy of the electron is

${𝑈}_{𝑛}=2{𝐸}_{𝑛}=-\frac{27.2\text{eV}}{{𝑛}^2}.$

When the atom is excited from $𝑛=1$ (ground state) to $𝑛>1$ (an excited state):

${𝐸}_{𝑛}$ becomes less negative (it increases toward 0).

${𝐾}_{𝑛}$ decreases, because ${𝐾}_{𝑛}=13.6\text{eV}/{𝑛}^2$ gets smaller for larger $𝑛$.

${𝑈}_{𝑛}$ becomes less negative, which means ${𝑈}_{𝑛}$ is effectively increasing (going from a larger negative value to a smaller negative value).

Therefore, when a hydrogen atom is raised from the ground state to an excited state,

the potential energy increases (less negative),

the kinetic energy decreases.

Hence the correct option is:

$\overline{)\text{(A) potential energy increases and K.E. decreases.}}$

The electromotive force (e.m.f.) induced in the second coil due to a changing current in the first coil is given by Faraday's law of electromagnetic induction and the concept of mutual inductance. The e.m.f. $E$ in the second coil is related to the rate of change of current in the first coil as follows:

$E=-𝑀\frac{𝑑{𝐼}_1}{𝑑𝑡}$

where $𝑀$ is the mutual inductance and ${𝐼}_1$ is the current in the first coil.

Given:

$𝑀=5\times {10}^{-3}H$

${𝐼}_1={𝐼}_0\sin 𝜔𝑡$

${𝐼}_0=10A$

$𝜔=100𝜋\mathrm{rad}/s$

To find the maximum e.m.f., we first differentiate the current ${𝐼}_1$ with respect to time $𝑡$:

$\frac{𝑑{𝐼}_1}{𝑑𝑡}=\frac{𝑑}{𝑑𝑡}({𝐼}_0\sin 𝜔𝑡)={𝐼}_0𝜔\cos 𝜔𝑡$

Substituting the values, we have:

$\frac{𝑑{𝐼}_1}{𝑑𝑡}=10\times 100𝜋\cos \left(100𝜋𝑡\right)=1000𝜋\cos \left(100𝜋𝑡\right)$

The maximum value of $\cos \left(100𝜋𝑡\right)$ is 1, therefore the maximum rate of change of current is:

${\left(\frac{𝑑{𝐼}_1}{𝑑𝑡}\right)}_{\text{max}}=1000𝜋$

Now, substituting this into the equation for the induced e.m.f., we calculate the maximum e.m.f.:

$E_{\text{max}}=-𝑀\times 1000𝜋=-(5\times {10}^{-3})\times 1000𝜋$

Simplifying, we find:

$E_{\text{max}}=-5\times 𝜋=-5𝜋V$

Since e.m.f. is generally discussed in magnitude, the answer is:

Option D: $5𝜋V$

$\begin{array}{cc}& \begin{array}{cc}& {𝑃}_1{𝑉}_1={𝑃}_2{𝑉}_2\\ & \Rightarrow ({𝑃}_0+ℎ𝑝𝑔)\times \frac{4}{3}𝜋{𝑟}^3={𝑃}_0\times \frac{4}{3}𝜋(2𝑟{)}^3\end{array}\\ & \text{ Where, }h=\text{ depth of lake }\\ & \Rightarrow h𝜌g=7P_0\Rightarrow h=7\times \frac{\mathrm{Hpg}}{𝜌g}=7H\end{array}$

Kepler's Third Law of Planetary Motion states that the square of the period of orbit ($𝑇$) of a planet is directly proportional to the cube of the semi-major axis of its orbit ($𝑟$). Mathematically, this is given by:

${𝑇}^2\propto {𝑟}^3$

For the Earth, we have:

${𝑇}_2^{\text{earth}}=𝑘\cdot {𝑟}_3^{\text{earth}}$

For the planet with a period 8 times that of the Earth, we have:

$(8{𝑇}_{\text{earth}}{)}^2=𝑘\cdot {𝑟}_3^{\text{planet}}$

Simplifying this expression:

$64{𝑇}_2^{\text{earth}}=𝑘\cdot {𝑟}_3^{\text{planet}}$

Using the equation for Earth, ${𝑇}_2^{\text{earth}}=𝑘\cdot {𝑟}_3^{\text{earth}}$, we substitute:

$64(𝑘\cdot {𝑟}_3^{\text{earth}})=𝑘\cdot {𝑟}_3^{\text{planet}}$

The equation simplifies to:

$64{𝑟}_3^{\text{earth}}={𝑟}_3^{\text{planet}}$

Taking the cube root on both sides:

$4{𝑟}_{\text{earth}}={𝑟}_{\text{planet}}$

Therefore, the ratio of the radius of the planet's orbit to the radius of the Earth's orbit is 4.

Option A: 4

For an ideal solenoid (long, tightly wound), the magnetic field inside is given (in SI units) by

$𝐵={𝜇}_0𝑛𝐼={𝜇}_0(\frac{𝑁}{𝐿})𝐼,$

where

${𝜇}_0$ is the permeability of free space,

$𝑛=\frac{𝑁}{𝐿}$ is the number of turns per unit length,

$𝑁$ is the total number of turns,

$𝐿$ is the length of the solenoid, and

$𝐼$ is the current in the winding.

From this formula, we see that $𝐵$ does depend on:

the total number of turns $𝑁$ (through $𝑛$),

the length $𝐿$ (also through $𝑛$),

the current $𝐼$.

However, in an ideal solenoid model, the radius (thickness) of the wire that makes up the solenoid winding does not appear in the expression for $𝐵$. Thus, among the given options, the magnetic induction (field) inside an ideal solenoid is independent of:

(C) the radius of the wire.

Axis of ${𝐼}_1$ and ${𝐼}_2$ and that of ${𝐼}_3$ and ${𝐼}_4$ are pependicular to each other.

By theorem of perpendicular axis,

$I=I_1+I_2\text{ or }I=I_3+I_4\text{.... (i)}$

As it is a square,

$I_1=I_2\text{ and }{I}_3=I_4$

From (i),

$\begin{array}{cc}& I_1=I_2=\frac{I}{2}\\ & I_3=I_4=\frac{I}{2}\\ \therefore & I_3=I_1\\ \therefore & \text{ Momentof inertia of the plate }=I=I_1+I_3\end{array}$

The shortest wavelength in Balmer series is given by

$\frac{1}{{𝜆}_1}=R(\frac{1}{2^2}-\frac{1}{\infty })=\frac{R}{4}\text{.... (i)}$

The shortest wavelength in the Paschen series given by

$\frac{1}{{𝜆}_2}=R(\frac{1}{3^2}-\frac{1}{\infty })=\frac{𝑅}{9}\text{.... (ii)}$

Dividing equation (ii) by equation (i),

$\frac{{𝜆}_1}{{𝜆}_2}=\frac{R}{9}\times \frac{4}{R}=\frac{4}{9}$

To find the ratio ${𝐿}_{𝐴}:{𝐿}_{𝐵}$ such that the difference in lengths of rods A and B remains constant at all temperatures, we start by considering their lengths at a temperature $𝑡$.

The length of rod A at temperature $𝑡$ is given by:

${𝐿}_{{𝐴}_{𝑡}}={𝐿}_{𝐴}+{𝐿}_{𝐴}{𝛼}_{𝐴}\Delta 𝑡$

Similarly, the length of rod B at temperature $𝑡$ is:

${𝐿}_{{𝐵}_{𝑡}}={𝐿}_{𝐵}+{𝐿}_{𝐵}{𝛼}_{𝐵}\Delta 𝑡$

The difference in their lengths at temperature $𝑡$ is then:

$\Delta 𝐿={𝐿}_{{𝐴}_{𝑡}}-{𝐿}_{{𝐵}_{𝑡}}=({𝐿}_{𝐴}-{𝐿}_{𝐵})+({𝐿}_{𝐴}{𝛼}_{𝐴}-{𝐿}_{𝐵}{𝛼}_{𝐵})\Delta 𝑡$

For this difference to remain constant regardless of changes in temperature, the coefficient of $\Delta 𝑡$ must be zero:

${𝐿}_{𝐴}{𝛼}_{𝐴}-{𝐿}_{𝐵}{𝛼}_{𝐵}=0$

Thus, we have:

${𝐿}_{𝐴}{𝛼}_{𝐴}={𝐿}_{𝐵}{𝛼}_{𝐵}$

Rearranging to find the length ratio:

$\frac{{𝐿}_{𝐴}}{{𝐿}_{𝐵}}=\frac{{𝛼}_{𝐵}}{{𝛼}_{𝐴}}$

This relationship tells us that the ratio of lengths ${𝐿}_{𝐴}:{𝐿}_{𝐵}$ is $\frac{{𝛼}_{𝐵}}{{𝛼}_{𝐴}}$.

Given the problem, let's derive the needed information step-by-step:

Velocity in Mediums:

The velocity of light in diamond (${𝑣}_{𝑑}$) is $\left(\frac{5}{12}\right)$ times the velocity of light in air ($𝑐$).

The velocity of light in water (${𝑣}_{𝑤}$) is $\left(\frac{3}{4}\right)$ times the velocity of light in air ($𝑐$).

Refractive Indices:

Refractive index of diamond ($\frac{n_d}{n_w}$) relative to water can be calculated using the relation between velocities:

$\frac{n_d}{n_w}=\frac{\frac{𝑐}{{𝑣}_{𝑑}}}{\frac{𝑐}{{𝑣}_{𝑤}}}=\frac{\frac{12}{5}}{\frac{4}{3}}=\frac{12}{5}\times \frac{3}{4}=\frac{9}{5}$

Snell's Law:

Using Snell's Law:

$n_w{n}_d=\frac{\sin i}{\sin r}$

Plug in the values (with angle of refraction $𝑟={30}^{\circ }$ where $\sin {30}^{\circ }=\frac{1}{2}$):

$\sin i=n_w{n}_d\times \sin r=\frac{9}{5}\times \frac{1}{2}=\frac{9}{10}$

Angle of Incidence:

Therefore, the angle of incidence ($𝑖$) is:

$i={\sin }^{-1}\left(\frac{9}{10}\right)$

This calculation demonstrates how to determine the angle of incidence when light travels from water to diamond, based on the given velocities and angle of refraction.

Impedance for L-R circuit,

$Z_1=\sqrt{\surd R^2+X_2^L}$

Impedance for $L-C-R$ circuit,

$\begin{array}{cc}{Z}_2& =\sqrt{\surd R^2+{(X_L-X_c)}^2}\\ \therefore Z_2& <Z_1\end{array}$

As impedance decreases, the current increases.

The internal energy of a gas will increase when it is subjected to an adiabatic process in which work is done on the gas. This is because, in an adiabatic process, no heat is exchanged with the surroundings, so any work done on the gas results in a change in its internal energy.

Explanation

Adiabatic Process: A process in which no heat is transferred to or from the system. The change in internal energy is solely due to the work done on or by the system.

Compression vs. Expansion:

In adiabatic compression, work is done on the gas, which increases its internal energy. This results in a rise in temperature.

In adiabatic expansion, the gas does work on the surroundings, causing a decrease in its internal energy, thus leading to a drop in temperature.

Isothermal Process: A process that occurs at constant temperature. For an ideal gas, the internal energy depends only on temperature, hence there is no change in internal energy.

In isothermal compression or isothermal expansion, although work is done, the heat exchanged is such that the internal energy remains constant.

Therefore, the correct option is:

Option B: The internal energy of a gas will increase when it is compressed adiabatically.

The fundamental frequency of a wire under tension is given by the equation:

$𝑛=\frac{1}{2𝐿}\sqrt{\surd \frac{𝑇}{𝜇}}$

where:

$𝑛$ is the frequency,

$𝐿$ is the length of the wire,

$𝑇$ is the tension in the wire,

$𝜇$ is the linear mass density of the wire.

When the bob is not immersed in water, the tension $𝑇$ is due to the weight of the bob, $𝑀𝑔$. Therefore, the initial frequency ${𝑛}_1$ is:

${𝑛}_1=\frac{1}{2𝐿}\sqrt{\surd \frac{𝑀𝑔}{𝜇}}$

When the bob is immersed in water, the apparent weight of the bob changes due to the buoyant force. The new tension ${𝑇}^{\prime }$ is given by:

${𝑇}^{\prime }=𝑀𝑔-\text{buoyant force}=𝑀𝑔-{𝜌}_{\text{water}}𝑉𝑔$

where ${𝜌}_{\text{water}}$ is the density of water and $𝑉$ is the volume of the bob.

Therefore, the new frequency ${𝑛}_2$ is:

${𝑛}_2=\frac{1}{2𝐿}\sqrt{\surd \frac{{𝑇}^{\prime }}{𝜇}}$

Substituting the new tension:

${𝑛}_2=\frac{1}{2𝐿}\sqrt{\surd \frac{𝑀𝑔-{𝜌}_{\text{water}}𝑉𝑔}{𝜇}}$

The frequency relates to the relative density $𝑑$ (which is the density of the bob divided by the density of water) as:

${𝑛}_2=\frac{1}{2𝐿}\sqrt{\surd \frac{𝑀𝑔(1-\frac{1}{𝑑})}{𝜇}}$

Given the relationship between ${𝑛}_1$ and ${𝑛}_2$, the ratio of the squares of the frequencies allows determining:

$\frac{{𝑛}_2^1}{{𝑛}_2^2}=\frac{𝑀𝑔}{𝑀𝑔-{𝜌}_{\text{water}}𝑉𝑔}$

From the equation:

$\frac{{𝑛}_2^1}{{𝑛}_2^2}=\frac{𝑑}{𝑑-1}$

Solving for $𝑑$:

$𝑑=\frac{{𝑛}_2^1}{{𝑛}_2^1-{𝑛}_2^2}$

Thus, the relative density of the metal of the bob corresponds to Option C:

$\frac{n_2^1}{n_2^1-n_2^2}$

The magnetic field intensity inside a solenoid can be given by the formula:

$𝐻=\frac{𝑛\cdot 𝐼}{𝐿}$

where:

$𝐻$ is the magnetic field intensity,

$𝑛$ is the number of turns,

$𝐼$ is the current,

$𝐿$ is the length of the solenoid in meters.

Given values:

$𝐻=2.4\times {10}^{3}A/m$

$𝑛=60$

$𝐿=15\mathrm{cm}=0.15m$

To find the current $𝐼$, rearrange the formula to solve for $𝐼$:

$𝐼=\frac{𝐻\cdot 𝐿}{𝑛}$

Substitute in the given values:

$𝐼=\frac{2.4\times {10}^{3}A/m\times 0.15m}{60}$

Calculate:

$𝐼=\frac{360}{60}A$

$𝐼=6A$

Therefore, the current flowing in the solenoid is 6 A.

Correct Option:

Option B - 6 A

Gauss's law tells us that the net electric flux through a closed surface is related to the charge enclosed by the surface. Specifically, Gauss’s law is:

${\oint }_{𝑆}𝐄\cdot 𝑑𝐀=\frac{{𝑄}_{\text{enc}}}{{𝜀}_0}$

Here’s how to apply this to the problem:

The net electric flux through the surface is the algebraic sum of the flux leaving the surface and the flux entering the surface. If you take ${𝜙}_2$ as the flux exiting the surface and ${𝜙}_1$ as the flux entering, then the net flux is:

${\Phi }_{\text{net}}={𝜙}_2-{𝜙}_1$

By substituting this into Gauss’s law, we have:

${𝜙}_2-{𝜙}_1=\frac{{𝑄}_{\text{enc}}}{{𝜀}_0}$

Solving for the enclosed charge ${𝑄}_{\text{enc}}$, we multiply both sides by ${𝜀}_0$:

${𝑄}_{\text{enc}}={𝜀}_0({𝜙}_2-{𝜙}_1)$

Thus, the correct answer is:

Option D: ${𝜀}_0({𝜙}_2-{𝜙}_1)$

To find the increase in the temperature of the gas, we start with the initial conditions and apply the principles of ideal gas law. The initial pressure is given as ${𝑃}_1$, and the pressure increase is 4%, so the new pressure is:

${𝑃}_2={𝑃}_1+0.04{𝑃}_1=1.04{𝑃}_1$

Using the ideal gas equation $𝑃𝑉=𝑛𝑅𝑇$, we relate the initial and final states as follows, assuming the volume and the amount of gas remain constant:

$\frac{{𝑇}_1}{{𝑇}_2}=\frac{{𝑃}_1}{{𝑃}_2}$

The initial temperature ${𝑇}_1$ is given as ${100}^{\circ }C$, which converts to Kelvin as $373K$.

Supposing the final temperature is ${𝑇}_2={𝑇}_1+\Delta 𝑇$:

$\frac{373}{373+\Delta 𝑇}=\frac{1}{1.04}$

By cross-multiplying, we solve for $\Delta 𝑇$:

$373\left(1.04\right)=373+\Delta 𝑇$

$387.92=373+\Delta 𝑇$

$\Delta 𝑇=387.92-373=14.92K$

Since temperature changes in Celsius and Kelvin are equivalent:

$\Delta 𝑇={14.92}^{\circ }C$

Thus, the increase in the temperature of the gas is ${14.92}^{\circ }C$.

$\begin{array}{cc}& \frac{1}{𝜆}=R(\frac{1}{n_2^1}-\frac{1}{n_2^2})\\ & \text{ For Lyman series, }{n}_1=1\\ & \begin{array}{cc}& \frac{1}{{𝜆}_{max}}=𝑅(\frac{1}{1^2}-\frac{1}{2^2})=\frac{3}{4}𝑅\\ & \frac{1}{{𝜆}_{min}}=𝑅(\frac{1}{1^2}-\frac{1}{{\infty }^2})=𝑅\end{array}\\ & \therefore \frac{{𝜆}_{max}}{{𝜆}_{min}}=\frac{4}{3}\end{array}$

In the steady state of the given circuit, the current through the capacitor is zero. This implies that there is no current flowing through the 4 Ω resistor.

To find the effective resistance (R_eq) for the parallel combination of the other resistors (1 Ω, 2 Ω, and 3 Ω), we use:

$\frac{1}{{𝑅}_{\text{eq}}}=\frac{1}{1}+\frac{1}{2}+\frac{1}{3}=\frac{6+3+2}{6}=\frac{11}{6}$

The total current (I) flowing through the circuit is given by:

$𝐼=\frac{𝑉}{{𝑅}_{\text{eq}}}=\frac{6\text{ V}\times 11}{6}=11\text{ A}$

For the charge on the capacitor, we use the formula $𝑄=𝐶𝑉$, where C is the capacitance and V is the voltage across the capacitor:

${𝑄}_{𝐶}=0.5\text{ μF}\times 6\text{ V}=3\text{ μC}$