MHT CET 2024 11th May Morning Shift (Physics & Chemistry) Question Paper with Solutions

Answer: Option C (is conserved)

Explanation:

For a particle moving in a vertical circle under the influence of gravity, assuming no non-conservative forces like air resistance or friction, the total mechanical energy (sum of kinetic energy and gravitational potential energy) remains constant throughout the motion.

At the lowest point, the particle has maximum kinetic energy and minimum potential energy.

At the highest point, it has minimum kinetic energy and maximum potential energy.

At intermediate points, the energy distribution between kinetic and potential energy changes, but the total remains constant.

Thus, the total energy is conserved.

The self-inductance of a coil is given by the formula:

$𝐿=𝜇\frac{{𝑁}^2𝐴}{𝑙}$

where:

$𝐿$ is the self-inductance,

$𝜇$ is the permeability of the core material,

$𝑁$ is the number of turns in the coil,

$𝐴$ is the cross-sectional area of the coil, and

$𝑙$ is the length of the coil.

When the number of turns $𝑁$ is doubled, the expression for self-inductance becomes:

${𝐿}^{\prime }=𝜇\frac{(2𝑁{)}^2𝐴}{𝑙}=𝜇\frac{4{𝑁}^2𝐴}{𝑙}=4\left(𝜇\frac{{𝑁}^2𝐴}{𝑙}\right)=4𝐿$

Thus, the self-inductance becomes 4 times the original value. Therefore, the correct answer is:

Option A: becomes 4 times.

First, find the electric field $𝐸$ between the plates using the formula:

$𝐸=\frac{𝑉}{𝑑},$

where

$𝑉=980\text{V}$ is the potential difference,

$𝑑=8\text{cm}=0.08\text{m}$ is the separation between the plates.

Substituting the values:

$𝐸=\frac{980}{0.08}=12250\text{V/m}.$

The charged particle experiences an electric force given by:

${𝐹}_{𝑒}=𝑞𝐸,$

where

$𝑞=1.6\times {10}^{-19}\text{C}$ is the charge of the particle.

Therefore,

${𝐹}_{𝑒}=(1.6\times {10}^{-19})\times 12250=1.96\times {10}^{-15}\text{N}.$

Since the particle is suspended in equilibrium, the upward electric force balances the downward gravitational force. The gravitational force is:

${𝐹}_{𝑔}=𝑚𝑔,$

where

$𝑔=9.8{\text{m/s}}^2$ is the acceleration due to gravity,

$𝑚$ is the mass of the particle.

Setting the forces equal:

$𝑞𝐸=𝑚𝑔.$

Solve for the mass $𝑚$:

$𝑚=\frac{𝑞𝐸}{𝑔}=\frac{1.96\times {10}^{-15}}{9.8}\approx 2\times {10}^{-16}\text{kg}.$

Thus, the correct option is Option A:

$2\times {10}^{-16}\text{kg}.$

Given that two spheres, ${𝑆}_1$ and ${𝑆}_2$, have the same radius but different temperatures ${𝑇}_1$ and ${𝑇}_2$, their emissive powers are the same with emissivities in the ratio $1:4$. We need to determine the ratio ${𝑇}_1:{𝑇}_2$.

First, recall that emissive power is given by the formula:

$\frac{{𝑄}_1}{{𝐴}_1𝑡}=\frac{{𝑄}_2}{{𝐴}_2𝑡}$

Since the radii of the spheres are the same, it follows that:

${𝐴}_1={𝐴}_2$

Thus, we have:

${𝑄}_1={𝑄}_2$

Using the Stefan-Boltzmann law, which states that emissive power $𝐸$ is given by:

$𝐸=𝑒𝜎𝐴{𝑇}^4$

We can write:

${𝑒}_1𝜎𝐴{𝑇}_4^1={𝑒}_2𝜎𝐴{𝑇}_4^2$

Given the emissivity ratio $\frac{{𝑒}_2}{{𝑒}_1}=\frac{4}{1}$, we have:

$\frac{{𝑇}_4^1}{{𝑇}_4^2}=\frac{{𝑒}_2}{{𝑒}_1}=\frac{4}{1}$

Taking the fourth root on both sides, we find:

$\frac{{𝑇}_1}{{𝑇}_2}={\left(\frac{4}{1}\right)}^{\frac{1}{4}}=\frac{\sqrt{\surd 2}}{1}$

Hence, the ratio ${𝑇}_1:{𝑇}_2$ is $\sqrt{\surd 2}:1$.

$g=\frac{\mathrm{GM}}{R^2}=\frac{4}{3}𝜋R𝜌G$

As, $𝑔$ is same on earth and the planet, $R\propto \frac{1}{𝜌}$

$\begin{array}{cc}\therefore & \frac{R_2}{R}=\frac{𝜌}{3𝜌}\\ \therefore & R_2=\frac{R}{3}\end{array}$

Let $𝑅$ be the radius of bigger drop and $𝑟$ be the radius of single small water drop.

Volume of big drop $=n$ (Volume of small drop)

$\begin{array}{cc}\therefore & \frac{4}{3}𝜋{𝑅}^3=𝑛\times \frac{4}{3}𝜋{𝑟}^3\\ & \Rightarrow {𝑅}^3=𝑛{𝑟}^3\\ & 𝑅={𝑛}^{\frac{1}{3}}𝑟\end{array}$

Surface energy of $𝑛$ drops $\left({𝐸}_{𝑛}\right)=𝑛\times 4𝜋{𝑟}^2\times 𝑇$

Surface energy of big drop $\left(𝐸\right)=4𝜋{𝑅}^2𝑇$

$\therefore \frac{E_n}{E}=\frac{{\mathrm{nr}}^2}{R^2}=\frac{{\mathrm{nr}}^2}{{\left(n^{\frac{1}{3}}\right)}^2}=\frac{{\mathrm{nr}}^2}{n^{\frac{2}{3}}{r}^2}=n^{\frac{1}{3}}=\text{exception 25:}:1$

$\tan 𝜙=\frac{X_L-X_C}{R}=\left(\frac{𝜔L-\frac{1}{𝜔C}}{R}\right)$

$\begin{array}{cc}& =\frac{(2\times 𝜋\times 50)\times (\frac{300}{𝜋}\times {10}^{-3})-\frac{1}{(2\times 𝜋\times 50)\times (\frac{1}{𝜋}\times {10}^{-3})}}{20}\\ & \text{ i.e. }\tan 𝜙=1\\ & 𝜙={\tan }^{-1}\left(1\right)\end{array}$

$\begin{array}{cc}& \text{ Potential energy of particle at extreme position is}\\ & \begin{array}{cc}& P.E.=\frac{1}{2}𝑀{𝜔}^2{𝐴}^2\\ & =\frac{1}{2}𝑀\times \frac{𝑔}{𝐿}\times {𝐴}^2\dots ..(\because 𝜔=\sqrt{\surd \frac{𝑔}{𝑙}})\end{array}\end{array}$

In a Fraunhofer diffraction pattern, the slit width is 0.5 mm, and the screen is placed 2 meters away from the lens. The light used has a wavelength of $5500\text{Å}$. To find the distance between the first minima on either side of the central maximum, we use the following formulas (assuming $𝜃$ is small and measured in radians):

The distance of the first minimum from the central maximum is given by:

${𝑦}_{1𝑑}=\frac{𝜆𝐷}{𝑎}$

Where:

$𝜆=5500\times {10}^{-10}\text{m}$ (wavelength converted from angstroms to meters)

$𝐷=2\text{m}$ (distance to the screen)

$𝑎=0.5\times {10}^{-3}\text{m}$ (slit width converted from millimeters to meters)

The distance between the first minima on either side of the central maximum is:

$2{𝑦}_{1𝑑}=\frac{2𝜆𝐷}{𝑎}=\frac{2\times 5500\times {10}^{-10}\times 2}{0.5\times {10}^{-3}}=4.4\text{mm}$

Therefore, the distance between the first minima on either side of the central maximum is 4.4 mm.

When the velocity of the bullet changes from V to $\frac{V}{2}$ the distance travelled by the bullet is 30 cm . Using $3^{\text{rd }}$ kinematic equation,

$\begin{array}{cc}& {𝑣}^2={𝑢}^2+2𝑎𝑠\\ & {\left(\frac{𝑉}{2}\right)}^2={𝑉}^2+2𝑎\left(30\right)\\ & \frac{{𝑉}^2}{4}={𝑉}^2+60𝑎\\ & \frac{-3{𝑉}^2}{4}=60𝑎\\ & 𝑎=\frac{-{𝑉}^2}{80}\end{array}$

Further, when a bullet penetrates it comes to rest. So, the final velocity of the bullet becomes zero.

Using the relation,

$\begin{array}{cc}& v^2=u^2+2\mathrm{as}\\ & 0={\left(\frac{v}{2}\right)}^2+2(-\frac{V^2}{80})s\\ & \frac{V^2}{4}=\left(\frac{V^2}{40}\right)s\\ & s=\frac{40}{4}=10\mathrm{cm}\end{array}$

$\therefore$ Given combination of gates is equivalent to an X-OR gate.

We can find the induced e.m.f. using the formula for motional e.m.f.:

$𝜖=𝐵𝐿𝑣\sin 𝜃$

where

$𝐵=3.6\times {10}^{-5}\text{ T}$ is the Earth's magnetic field.

$𝐿=2\text{ m}$ is the length of the rod.

$𝑣=2.00\text{ m/s}$ is the speed of the boat.

$𝜃$ is the angle between the velocity of the rod and the magnetic field.

In this problem:

The boat is moving due east.

The Earth's magnetic field is due north.

Because the velocity (east) and the magnetic field (north) are perpendicular, the angle is $𝜃={90}^{\circ }$ and $\sin {90}^{\circ }=1$.

Plugging in the values:

$𝜖=(3.6\times {10}^{-5}\text{ T})\left(2\text{ m}\right)\left(2.00\text{ m/s}\right)\left(1\right)$

$𝜖=(3.6\times {10}^{-5})\left(4\right)\text{ V}$

$𝜖=1.44\times {10}^{-4}\text{ V}$

To convert this into millivolts (mV):

$1.44\times {10}^{-4}\text{ V}=0.144\text{ mV}$

Thus, the magnitude of the induced e.m.f. in the rod is 0.144 mV, which corresponds to Option D.

$\begin{array}{cc}& \text{ Using KCL, }\\ & 2+4=1+2+𝐼\\ \therefore & 𝐼=3A\end{array}$

$\begin{array}{cc}& 𝑎=\frac{𝑔\sin 𝜃}{(1+\frac{{𝐾}^2}{R^2})}=\frac{𝑔\sin {30}^{\circ }}{(1+\frac{2}{5})}\\ & \therefore 𝑎=\frac{5𝑔}{7}\times \left(\frac{1}{2}\right)=\frac{5𝑔}{14}\end{array}$

When light passes through a medium, it bends due to refraction. In this case, when a glass slab is placed over an ink mark, the ink mark appears to be closer than it actually is. This apparent shift can be described through the concept of apparent and real depth, where the apparent depth $𝑥$ is related to the real depth $𝑡$ by the formula:

$𝑥=\frac{𝑡}{𝜇}$

where $𝑥$ is the apparent shift, $𝑡$ is the real thickness of the glass slab, and $𝜇$ is the refractive index of the glass slab. Rearranging this equation to solve for the actual thickness $𝑡$ gives:

$𝑡=𝜇𝑥$

However, the apparent shift $𝑥$ is given as the distance the ink mark appears to be raised. To find the relationship where $𝑡$ is expressed in terms of $𝑥$, we consider:

$𝑥=𝑡-\frac{𝑡}{𝜇}$

Simplifying:

$𝑥=𝑡(1-\frac{1}{𝜇})$

$𝑥=𝑡\frac{𝜇-1}{𝜇}$

Solving for $𝑡$:

$𝑡=\frac{𝜇𝑥}{𝜇-1}$

Thus, the actual thickness of the glass slab $𝑡$ in terms of $𝑥$, $𝜇$, and the shift is given by:

Option B:

$\frac{𝜇𝑥}{𝜇-1}$

Let $E_1$ be the energy of the first energy level,

then ${𝐸}_2=\frac{{𝐸}_1}{4}$

For highest level $E_{\infty }=0$

$\begin{array}{cc}\therefore & {𝐸}_2-{𝐸}_1=\frac{{𝐸}_1}{4}-{𝐸}_1=-\frac{3}{4}{𝐸}_1\\ & {𝐸}_{\infty }-{𝐸}_2=0-\frac{{𝐸}_1}{4}=-\frac{{𝐸}_1}{4}\\ \therefore & \frac{{𝐸}_2-{𝐸}_1}{{𝐸}_{\infty }-{𝐸}_2}=3:1\end{array}$

The magnetic induction inside an air-core solenoid is given by the formula:

$𝐵=\frac{{𝜇}_0𝑁𝐼}{𝐿}$

where:

${𝜇}_0$ is the permeability of free space ($4𝜋\times {10}^{-7}$ SI unit),

$𝑁$ is the number of turns,

$𝐼$ is the current,

$𝐿$ is the length of the solenoid.

The magnetic flux ($𝜙$) within the solenoid is expressed as:

$𝜙=𝐵𝐴=\frac{{𝜇}_0𝑁𝐼𝐴}{𝐿}$

where $𝐴$ is the cross-sectional area of the solenoid.

We are tasked with finding the magnetic moment, which can be derived from:

$\text{Magnetic moment}=𝑁𝐼𝐴=\frac{𝜙𝐿}{{𝜇}_0}$

Substituting the given values, we solve for the magnetic moment:

$\text{Magnetic moment}=\frac{(1.57\times {10}^{-6}\text{Wb})\times 0.8\text{m}}{4𝜋\times {10}^{-7}\text{SI unit}}$

$\text{Magnetic moment}=1{\text{Am}}^2$

Thus, the magnetic moment of the solenoid is $1{\text{Am}}^2$.

To determine the intensity of the resultant wave formed by two identical light waves with a phase difference $𝜙$ propagating in the same direction, we start with the formula for the amplitude $𝐴$ of the resultant wave:

${𝐴}^2={𝑎}_2^1+{𝑎}_2^2+2{𝑎}_1{𝑎}_2\cos 𝜙$

Given that the amplitudes of the individual waves are equal, ${𝑎}_1={𝑎}_2=𝑎$, we can substitute to find:

${𝐴}^2=2{𝑎}^2(1+\cos 𝜙)$

Using the trigonometric identity, $\cos 𝜙=2{\cos }^2\left(\frac{𝜙}{2}\right)-1$, the equation becomes:

${𝐴}^2=2{𝑎}^2(1+2{\cos }^2\frac{𝜙}{2}-1)$

Simplifying this, we obtain:

${𝐴}^2=2{𝑎}^2\cdot 2{\cos }^2\frac{𝜙}{2}=4{𝑎}^2{\cos }^2\frac{𝜙}{2}$

Thus,

${𝐴}^2\propto {\cos }^2\frac{𝜙}{2}$

Since intensity $𝐼$ is proportional to the square of the amplitude ($𝐼\propto {𝐴}^2$), we conclude:

$𝐼\propto {\cos }^2\frac{𝜙}{2}$

This shows that the intensity of the resultant wave is proportional to ${\cos }^2\frac{𝜙}{2}$.

For isothermal process (Gas A):

In an isothermal process, the product of pressure and volume remains constant:

${𝑃}_1{𝑉}_1={𝑃}_2{𝑉}_2$

Given:

${𝑃}_0\cdot \left(8{𝑉}_0\right)={𝑃}_2\cdot {𝑉}_0$

Solving this relation gives:

${𝑃}_2=8{𝑃}_0$

For adiabatic process (Gas B):

In an adiabatic process, $𝑃{𝑉}^{𝛾}=\text{constant}$.

Thus:

$\frac{{𝑃}_2}{{𝑃}_1}=(8{)}^{𝛾}$

Therefore:

${𝑃}_2=(8{)}^{𝛾}{𝑃}_0$

The ratio of the final pressures of gas B to gas A is:

$\frac{({𝑃}_2{)}_{𝐵}}{({𝑃}_1{)}_{𝐴}}=\frac{(8{)}^{𝛾}{𝑃}_0}{8{𝑃}_0}=(8{)}^{𝛾-1}$

Given $𝛾=\frac{5}{3}$:

$\frac{({𝑃}_2{)}_{𝐵}}{({𝑃}_1{)}_{𝐴}}=8^{2/3}=\text{exception 25:}=4$

Thus, the inverse ratio is:

$\frac{({𝑃}_1{)}_{𝐴}}{({𝑃}_2{)}_{𝐵}}=\frac{1}{4}$

Phosphorus is a pentavalent impurity.

Hence, $n_e\gg n_h$.

To determine the frequency of tuning fork C, let's denote its frequency as $𝑓$.

Tuning fork A has a frequency that is 1.4% more than tuning fork C. Therefore, the frequency of A, ${𝑓}_{𝐴}$, is given by:

${𝑓}_{𝐴}=𝑓+0.014𝑓=1.014𝑓$

Similarly, tuning fork B has a frequency that is 2.6% less than that of tuning fork C. Therefore, the frequency of B, ${𝑓}_{𝐵}$, is given by:

${𝑓}_{𝐵}=𝑓-0.026𝑓=0.974𝑓$

When tuning forks A and B are sounded together, they produce 10 beats per second, which implies that the absolute difference in their frequencies is 10 Hz:

$|{𝑓}_{𝐴}-{𝑓}_{𝐵}|=10$

Substituting the expressions for ${𝑓}_{𝐴}$ and ${𝑓}_{𝐵}$:

$|1.014𝑓-0.974𝑓|=10$

This simplifies to:

$\left|0.04𝑓\right|=10$

Solving for $𝑓$:

$0.04𝑓=10\Rightarrow 𝑓=\frac{10}{0.04}=250$

Thus, the frequency of tuning fork C is 250 Hz.

Option A: 250 Hz

The phase difference between the current and the voltage is given by $\tan 𝜙=\frac{𝜔L-\frac{1}{𝜔C}}{R}$

$\begin{array}{cc}\therefore & 𝜔L-\frac{1}{𝜔C}=R\dots (\because \tan 𝜙=\tan {45}^{\circ }=1)\\ \therefore & 𝜔L=R+\frac{1}{𝜔C}\\ \therefore & L=\frac{R}{𝜔}+\frac{1}{{𝜔}^{2}C}=\frac{R𝜔C+1}{{𝜔}^{2}C}\\ \therefore & L=\frac{1+2𝜋\mathrm{fCR}}{4{𝜋}^2{f}^{2}C}\dots .(\because 𝜔=2𝜋f)\end{array}$

$\begin{array}{cc}& {𝐶}_{𝑒𝑞}=\frac{3𝐶\times 𝐶}{3𝐶+𝐶}=\frac{3{𝐶}^2}{4𝐶}\\ & {𝐶}_{𝑒𝑞}=\frac{3𝐶}{4}\\ \therefore & 𝐶=\frac{4{𝐶}_{𝑒𝑞}}{3}=\frac{4\times 4.5}{3}=6𝜇F\end{array}$

For an ideal gas, $\mathrm{PV}=\mathrm{RT}$ and $M=𝜌V$

$\begin{array}{cc}\therefore & \frac{𝑃𝑀}{𝜌}=𝑅𝑇\\ \therefore & \frac{{𝑃}_{𝐴}{𝑀}_{𝐴}}{{𝜌}_{𝐴}}=\frac{{𝑃}_{𝐵}{𝑀}_{𝐵}}{{𝜌}_{𝐵}}\\ & \frac{{𝑀}_{𝐴}}{{𝑀}_{𝐵}}=\frac{{𝜌}_{𝐴}}{{𝜌}_{𝐵}}\times \frac{{𝑃}_{𝐵}}{{𝑃}_{𝐴}}=\frac{3}{2}\times \frac{1}{2}=\frac{3}{4}\end{array}$

$\begin{array}{cc}& {𝑔}_{𝑑}=𝑔(1-\frac{𝑑}{𝑅})\\ & 𝑔\left(\frac{1}{𝑛-1}\right)=𝑔(1-\frac{𝑑}{𝑅})\dots (\text{ Given: }{𝑔}_{𝑑}=𝑔\left(\frac{1}{𝑛-1}\right)\\ \therefore & \frac{𝑑}{𝑅}=1-\left(\frac{1}{𝑛-1}\right)\\ \therefore & 𝑑=𝑅\left(\frac{𝑛-2}{𝑛-1}\right)\end{array}$

$\begin{array}{cc}& {\mathrm{eV}}_s=h𝜈-W_0\\ \therefore & \text{ If }𝜈\text{ increases, }{V}_s\text{ will increase. }\end{array}$

When a resonance tube closed at one end is used, it vibrates in certain patterns or modes. For the tube to resonate in its first mode, the length of the air column above the water should be a quarter of the wavelength of the sound produced by the tuning fork.

Given:

Frequency of tuning fork, $𝑛=340\text{Hz}$

Speed of sound in air, $𝑉=340\text{m/s}$

Maximum height of the tube, $1.5\text{m}$

In the first mode, the length of the air column, ${𝐿}_1$, should be:

${𝐿}_1=\frac{𝑉}{4𝑛}=\frac{340}{4\times 340}=0.25\text{m}=25\text{cm}$

This means the wavelength $𝜆$ for the first mode is:

$𝜆=4\times {𝐿}_1=4\times 25\text{cm}=100\text{cm}$

Resonance occurs not only at $\frac{𝜆}{4}$ but also at higher harmonics where the length is $\frac{3𝜆}{4},\frac{5𝜆}{4},\frac{7𝜆}{4}$, etc.

Calculating for $\frac{5𝜆}{4}$:

$𝐿=\frac{5\times 100}{4}=125\text{cm}$

Since the height of the tube is 150 cm (1.5 m), resonance is achieved when:

The air column length is 125 cm.

Thus, the height of the water column needed for this setup is:

$\text{Height of water}=\text{Total tube height}-\text{Air column length}=150\text{cm}-125\text{cm}=25\text{cm}$

Therefore, the smallest height of water column for resonance is 25 cm.

Rise of water in capillary tube is given by

$h=\frac{2T\cos 𝜃}{\mathrm{rpg}}$

For water, $\cos 𝜃=1$

Also, the radius of capillary tube becomes (r/3).

$\therefore h^{\prime }=\frac{3\cdot 2T}{\mathrm{rgg}}\Rightarrow h^{\prime }=3h$

Now, $𝑚=𝜋{𝑟}^{2}h\times 𝜌$

$\therefore m^{\prime }=𝜋(r/3{)}^2(3h)\times 𝜌=\frac{𝜋r^{2}h𝜌}{3}=\frac{m}{3}$

Force ${𝐹}_1$ due to charge $3𝑄$ on $𝑄$ will be,

$F_1=\frac{1}{4𝜋{𝜀}_0}\frac{3Q^2}{{𝑙}^2}$

Force ${𝐹}_2$ due to charge $𝑞$ on charge $𝑄$ will be,

$F_2=\frac{1}{4𝜋{𝜀}_0}\frac{\mathrm{Qq}}{{\left(\frac{2𝑙}{3}\right)}^2}$

$\therefore F_2=\frac{1}{4𝜋{𝜀}_0}\frac{9\mathrm{Qq}}{4{𝑙}^2}$

For the force on charge Q to be zero,

$F_1+F_2=0$ or $F_1=-F_2$

$\begin{array}{cc}\therefore & \frac{3Q^2}{{𝑙}^2}=-\frac{9\mathrm{Qq}}{4{𝑙}^2}\\ \therefore & Q=-\frac{3q}{4}\\ \therefore & q=-\frac{4}{3}Q\end{array}$

$\begin{array}{cc}& \text{ Intensity of magnetization }\\ & \begin{array}{cc}& =\frac{{𝑀}_{\text{nat }}}{\text{ Volume }}\\ & =\frac{{𝑀}_{\text{net }}}{\text{ length }\times \text{ area of cross-section }}\\ & =\frac{6}{(4\times {10}^{-2})\times (2\times {10}^{-1})}=7.5\times {10}^{5}A/m\end{array}\end{array}$

To understand how a wire vibrates in unison with a tuning fork at different locations on Earth, we need to consider the fundamental principles affecting its frequency. The frequency $𝑛$ for a stretched string is given by:

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

where:

$𝑙$ is the length of the wire,

$𝑇$ is the tension in the wire, and

$𝑚$ is the mass per unit length of the wire.

If we assume the tension $𝑇$ is due to a weight $𝑀𝑔$ (where $𝑀$ is the mass and $𝑔$ is the acceleration due to gravity), then the frequency becomes:

$𝑛=\frac{1}{2𝑙}\sqrt{\surd \frac{𝑀𝑔}{𝑚}}$

Considering the poles and the equator, let:

${𝑙}_1$ and ${𝑔}_1$ be the length and acceleration due to gravity at the poles,

${𝑙}_2$ and ${𝑔}_2$ be the length and acceleration due to gravity at the equator.

For the frequency to remain the same at both the poles and the equator:

$𝑛=\frac{1}{2{𝑙}_1}\sqrt{\surd \frac{𝑀{𝑔}_1}{𝑚}}=\frac{1}{2{𝑙}_2}\sqrt{\surd \frac{𝑀{𝑔}_2}{𝑚}}$

From this, we obtain:

$\frac{\sqrt{\surd {𝑔}_1}}{{𝑙}_1}=\frac{\sqrt{\surd {𝑔}_2}}{{𝑙}_2}\text{or}\frac{{𝑙}_2}{{𝑙}_1}=\sqrt{\surd \frac{{𝑔}_2}{{𝑔}_1}}$

Since the acceleration due to gravity at the equator (${𝑔}_2$) is less than at the poles (${𝑔}_1$), it follows that:

${𝑔}_2<{𝑔}_1$

Therefore, the ratio $\frac{{𝑙}_2}{{𝑙}_1}=\sqrt{\surd \frac{{𝑔}_2}{{𝑔}_1}}<1$, indicating that:

${𝑙}_2<{𝑙}_1$

Thus, the vibrating length of the wire should be decreased when moving from the poles to the equator to maintain the same resonance with the tuning fork.

Displacement of the particle, $x=A\sin 𝜔t$

Velocity of the particle,

$𝑣=\frac{𝑑𝑥}{𝑑𝑡}=𝐴𝜔\cos 𝜔𝑡\text{.... (i)}$

Given that,

$\begin{array}{cc}& 𝑣=𝜋m/s,T=12s,\\ \therefore & 𝜔=\frac{2𝜋}{T}=\frac{𝜋}{6}\mathrm{rad}/s\end{array}$

Substituting in equation (i), we get,

$\begin{array}{cc}& 𝜋=A\times \frac{𝜋}{6}\times \cos (\frac{𝜋}{6}\times 2)\\ \therefore & 1=\frac{𝐴}{6}\cos \left(\frac{𝜋}{3}\right)=\frac{𝐴}{6}\times \frac{1}{2}\\ \therefore & 𝐴=12m\end{array}$

Total internal reflection (TIR) takes place only when:

Light travels from a denser medium to a rarer medium (i.e., ${𝑛}_{\text{denser}}>{𝑛}_{\text{rarer}}$).

The angle of incidence $𝑖$ is greater than the critical angle ${𝑖}_c$.

Thus, among the options :

$\overline{)\text{(C) light travels from denser to rarer medium and }𝑖>{𝑖}_c.}$

We know

$X_L=L𝜔\text{ and }{X}_C=\frac{1}{C𝜔}$

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

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

Let's break down the problem step by step.

Two identical air capacitors, each of capacitance $𝐶$, are connected in series. The effective capacitance in series is given by

$\frac{1}{{𝐶}_{\text{series}}}=\frac{1}{𝐶}+\frac{1}{𝐶}=\frac{2}{𝐶},$

so

${𝐶}_{\text{series}}=\frac{𝐶}{2}.$

Now, one capacitor is completely filled with a dielectric of constant $𝐾$. When a capacitor is fully filled with a dielectric, its capacitance increases by a factor of $𝐾$, so its new capacitance becomes $𝐾𝐶$. The other capacitor remains at $𝐶$.

For two capacitors in series (one with capacitance $𝐾𝐶$ and the other with $𝐶$), the effective capacitance becomes:

${𝐶}^{\prime }=\frac{\left(𝐾𝐶\right)\cdot 𝐶}{𝐾𝐶+𝐶}=\frac{𝐾{𝐶}^2}{𝐶(𝐾+1)}=\frac{𝐾𝐶}{𝐾+1}.$

The change in effective capacitance is the difference between the new effective capacitance and the original:

$\Delta 𝐶={𝐶}^{\prime }-\frac{𝐶}{2}=\frac{𝐾𝐶}{𝐾+1}-\frac{𝐶}{2}.$

To combine the terms, find a common denominator:

$\Delta 𝐶=\frac{2𝐾𝐶-𝐶(𝐾+1)}{2(𝐾+1)}=\frac{2𝐾𝐶-𝐶𝐾-𝐶}{2(𝐾+1)}=\frac{𝐶(𝐾-1)}{2(𝐾+1)}.$

Thus, the change in the effective capacitance is

$\frac{𝐶}{2}\left(\frac{𝐾-1}{𝐾+1}\right),$

which corresponds to Option C.

The angular momentum of the Earth revolving around the Sun can be derived using the formula for angular momentum in circular motion:

$𝐿=𝑚𝑣𝑟$

where:

$𝐿$ is the angular momentum,

$𝑚$ is the mass of the Earth,

$𝑣$ is the tangential velocity,

$𝑟$ is the radius of the circular orbit, given as $𝑅$.

For circular orbits under the gravitational influence of the Sun, the centripetal force is provided by gravitational force:

$\frac{𝑚{𝑣}^2}{𝑟}=\frac{𝐺𝑀𝑚}{{𝑟}^2}$

where:

$𝑀$ is the mass of the Sun,

$𝐺$ is the gravitational constant.

From this, solving for the velocity $𝑣$, we have:

${𝑣}^2=\frac{𝐺𝑀}{𝑟}$

Taking the square root:

$𝑣=\sqrt{\surd \frac{𝐺𝑀}{𝑟}}$

Substituting this back into the angular momentum expression:

$𝐿=𝑚\times \sqrt{\surd \frac{𝐺𝑀}{𝑟}}\times 𝑟$