MHT CET 2020 16th October Evening Shift (Physics & Chemistry) Question Paper with Solutions

The linear velocity of the body, $𝑣=\sqrt{\surd 2𝑔ℎ}$

The velocity of the sphere about its centre,

${𝑣}_{\mathrm{CM}}=\sqrt{\surd \frac{2𝑔ℎ}{1+\frac{{𝐾}^2}{{𝑅}^2}}}=\frac{𝑣}{\sqrt{\surd 1+\frac{{𝐾}^2}{{𝑅}^2}}}\text{... (i)}$

For uniform solid sphere,

$\frac{{𝐾}^2}{{𝑅}^2}=\frac{2}{5}$

$\begin{array}{cc}& \text{ Substituting value in Eq. (i), we get }\\ & {𝑣}_{\mathrm{CM}}=\frac{𝑣}{\sqrt{\surd 1+\left(\frac{2}{5}\right)}}=\sqrt{\surd \frac{5}{7}}𝑣\end{array}$

The linear width of the principal maxima in a diffraction pattern is given by:

$𝛽=\frac{2𝐷𝜆}{𝑑}$

It is stated that the linear width of the principal maxima is equal to the width of the slit. Therefore, we have:

$𝑑=\frac{2𝐷𝜆}{𝑑}$

Solving for $𝐷$, we can rearrange the equation to find:

$𝐷=\frac{{𝑑}^2}{2𝜆}$

Given, damping force $\propto$ velocity

$\begin{array}{cc}& 𝐹\propto 𝑣\\ & 𝐹=𝑘𝑣\end{array}$

$\begin{array}{cc}& \text{ where, }𝑘\text{ is the proportionality constant. }\\ & \Rightarrow 𝑘=\frac{𝐹}{𝑣}=\frac{\mathrm{kg}-m/s^2}{m/s}=\frac{\mathrm{kg}}{s}=\mathrm{kg}{s}^{-1}\end{array}$

For paramagnetic substance, magnetic susceptibility is given by

$\begin{array}{cc}{𝜒}_{𝑚}𝑇& =\text{ constant }\\ \Rightarrow {\left({𝜒}_{𝑚}\right)}_1{𝑇}_1& ={\left({𝜒}_{𝑚}\right)}_2{𝑇}_2\text{... (i)}\end{array}$

$\begin{array}{cc}\text{ Here, }{\left({𝜒}_{𝑚}\right)}_1& =0.0075,\\ {𝑇}_1& =-{73}^{\circ }C=273-73=200K\\ \text{ and }{𝑇}_2& =-{173}^{\circ }C=273-173=100K\end{array}$

Substituting the given values in Eq. (i), we get

$\begin{array}{cc}0.0075\times 200& ={\left({𝜒}_{𝑚}\right)}_2\times 100\\ {\left({𝜒}_{𝑚}\right)}_2& =0.0150\end{array}$

The displacement of particle in SHM at any instant,

$\begin{array}{cc}𝑥& =𝐴\sin 𝜔𝑡\\ \frac{𝐴}{2}& =𝐴\sin 𝜔𝑡(\text{ given, }𝑥=\frac{𝐴}{2})\\ \sin 𝜔𝑡& =\frac{1}{2}=\sin \frac{𝜋}{6}\Rightarrow 𝜔𝑡=\frac{𝜋}{6}\end{array}$

$\begin{array}{ccc}\Rightarrow & \frac{2𝜋}{𝑇}\cdot 𝑡=\frac{𝜋}{6}& (\because 𝜔=\frac{2𝜋}{𝑇})\\ \Rightarrow & \frac{2𝜋}{3}𝑡=\frac{𝜋}{6}& (\because 𝑇=3s)\\ \Rightarrow & 𝑡=\frac{1}{4}s& \end{array}$

The fundamental frequency of a sonometer wire is given by:

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

Here, $𝐿$ (the length) and $𝜇$ (the linear mass density) are constants, so the frequency $𝑛$ is dependent on the tension $𝑇$.

We have:

$𝑛\propto \sqrt{\surd 𝑇}$

When considering the wire's vibration after submerging the load in water, the change in frequency can be determined by comparing the tension in both scenarios (before and after immersion):

$\frac{{𝑛}_2}{{𝑛}_1}=\sqrt{\surd \frac{{𝑇}_2}{{𝑇}_1}}$

Where:

$(\text{Weight}{)}_{𝑠}=\text{Weight of the object in air}=𝑉\cdot {𝜌}_{𝑠}\cdot 𝑔$

$(\text{Weight}{)}_{𝑤}=\text{Weight of the object in water}=𝑉\cdot {𝜌}_{𝑤}\cdot 𝑔$

Thus:

$\frac{{𝑛}_2}{{𝑛}_1}=\sqrt{\surd \frac{𝑉\cdot {𝜌}_{𝑠}\cdot 𝑔-𝑉\cdot {𝜌}_{𝑤}\cdot 𝑔}{𝑉\cdot {𝜌}_{𝑠}\cdot 𝑔}}=\sqrt{\surd \frac{\frac{{𝜌}_{𝑠}}{{𝜌}_{𝑤}}-1}{\frac{{𝜌}_{𝑠}}{{𝜌}_{𝑤}}}}=\sqrt{\surd \frac{𝜌-1}{𝜌}}$

Given that the specific gravity of the sonometer wire is:

$\text{Specific gravity, }𝜌=\frac{{𝜌}_{𝑠}}{{𝜌}_{𝑤}}$

The frequency ${𝑛}_2$ after submerging the load in water is:

${𝑛}_2=𝑛\sqrt{\surd \frac{𝜌-1}{𝜌}}$

Let us assume, a particle $𝑋$ having atomic number $𝑍$ and mass number $𝐴$.

When an $𝛼$-particle is emitted by a nucleus, then its atomic number decreases by 2 and mass number decreases by 4. So, for given case,

${𝑧}^{{𝑋}^{𝐴}}\stackrel{4𝛼\text{-particle }}{\to \leftarrow }𝑧-8^{{𝑌}^{𝐴-16}}$

When a $𝛽$-particle is emitted by a nucleus its atomic number increases by one and mass number remains unchanged. So, for given case,

$𝑍-8^{{𝑌}^{𝐴-16}}\stackrel{7𝛽\text{-particle }}{\to \leftarrow }𝑍-1^{{𝑍}^{𝐴-16}}$

$\begin{array}{cc}\therefore \frac{\text{ Number of neutrons }}{\text{ Number of protons }}& =\frac{(𝐴-16)-(𝑍-1)}{(𝑍-1)}\\ & =\frac{𝐴-𝑍-15}{(𝑍-1)}\end{array}$

Consider the figure shown below

A point at a distance $\frac{𝐿}{3}$ from one end is at a distance of $𝑑=\frac{𝐿}{2}-\frac{𝐿}{3}=\frac{𝐿}{6}$ from the centre of rod.

The moment of inertia of rod about an axis passing through centre and perpendicular to it is

${𝐼}_{𝑐}=\frac{𝑀{𝐿}^2}{12}$

Using parallel axis theorem,

$\begin{array}{cc}{𝐼}_{𝑑}& ={𝐼}_{𝑐}+𝑀{𝑑}^2\\ & =\frac{𝑀{𝐿}^2}{12}+𝑀{\left(\frac{𝐿}{6}\right)}^2=\frac{𝑀{𝐿}^2}{9}\end{array}$

When an obstacle moves towards a sound source, the sound waves reflect off the obstacle. To find the wavelength of the reflected wave, denoted as ${𝜆}_{𝑟}$, we start by considering the changes in frequency due to the movement of the obstacle.

Step-by-Step Explanation

Frequency of the Reflected Sound Wave:

The frequency ${𝑓}_{𝑟}$ of the reflected sound wave when the obstacle is moving towards the source can be determined by the formula:

${𝑓}_{𝑟}=𝑓\left(\frac{𝑐+𝑣}{𝑐-𝑣}\right)$

Here, $𝑐$ is the speed of sound, $𝑣$ is the velocity of the obstacle, and $𝑓$ is the original frequency of the sound wave.

No Change in Velocity:

The velocity of the sound wave remains unchanged, even after it is reflected. This means the speed of sound $𝑐$ is constant before and after reflection.

Wavelength of the Reflected Wave:

Since the velocity is unchanged and frequency changes, it affects the wavelength. Using the relationship between speed, frequency, and wavelength:

$\frac{𝑐}{{𝜆}_{𝑇}}=\frac{𝑐}{𝜆}\left(\frac{𝑐+𝑣}{𝑐-𝑣}\right)$

Simplifying this equation gives the wavelength of the reflected wave, ${𝜆}_{𝑇}$, as:

${𝜆}_{𝑇}=\left(\frac{𝑐-𝑣}{𝑐+𝑣}\right)𝜆$

This equation describes how the wavelength of the sound wave changes upon reflection from a moving obstacle.

Thus, the wavelength of the reflected wave ${𝜆}_{𝑟}$ is given by:

${𝜆}_{𝑇}=\left(\frac{𝑐-𝑣}{𝑐+𝑣}\right)𝜆$

The horizontal component of earth's magnetic field,

$𝐻=𝐵\cos 𝛿$

At place $𝐴$,

$\begin{array}{cc}{𝐻}_{𝐴}& =𝐵\cos {𝛿}_{𝐴}\\ & =𝐵\cos {30}^{\circ }=\frac{\sqrt{\surd 3}𝐵}{2}\end{array}$

At place $𝐵$,

$\begin{array}{cc}{𝐻}_{𝐵}& =𝐵\cos {𝛿}_{𝐵}\\ & =𝐵\cos {45}^{\circ }\\ & =\frac{𝐵}{\sqrt{\surd 2}}\end{array}$

The ratio of ${𝐻}_{𝐴}:{𝐻}_{𝐵}$ becomes,

$\frac{{𝐻}_{𝐴}}{{𝐻}_{𝐵}}=\frac{\sqrt{\surd 3}𝐵}{2}\times \frac{\sqrt{\surd 2}}{𝐵}=\frac{\sqrt{\surd 3}}{\sqrt{\surd 2}}$

The energy required to raise a satellite of mass m to a height h,

${𝐸}_1=\frac{𝑚𝑔ℎ}{(1+\frac{ℎ}{𝑅})}\text{... (i)}$

$\begin{array}{cc}& \text{ The energy required to put it into orbit, }\\ & {𝐸}_2=\frac{1}{2}𝑚{𝑣}_2^0=\frac{1}{2}𝑚\left(\frac{𝐺𝑀}{𝑟}\right)\end{array}$

$(\because \text{ orbital velocity, }{𝑣}_0=\frac{𝐺𝑀}{𝑟})$

$\begin{array}{cc}& =\frac{1}{2}𝑚\left(\frac{𝐺𝑀}{𝑅+ℎ}\right)=\frac{1}{2}\frac{𝑚𝑔{𝑅}^2}{𝑅(1+\frac{ℎ}{𝑅})}(\because 𝐺𝑀=𝑔{𝑅}^2)\\ & =\frac{𝑚𝑔𝑅}{2(1+\frac{ℎ}{𝑅})}\end{array}$

$\begin{array}{cc}& \text{ The ratio of }{𝐸}_1:{𝐸}_2\text{ becomes, }\\ & \frac{{𝐸}_1}{{𝐸}_2}=\frac{𝑚𝑔ℎ}{(1+\frac{ℎ}{𝑅})}\times \frac{2(1+\frac{ℎ}{𝑅})}{𝑚𝑔𝑅}=\frac{2ℎ}{𝑅}\end{array}$

Consider the figure shown below,

The magnetic field due to the circular coil at its centre is given by

${𝐵}_1=\frac{{𝜇}_0{𝐼}_1}{2𝑅}\text{ (outward) }$

$\begin{array}{cc}& \text{ The magnetic field due to the wire at }𝑂\text{ is given by }\\ & {𝐵}_2=\frac{{𝜇}_0{𝐼}_2}{2𝜋𝑑}\end{array}$

$\begin{array}{cc}& \text{ The net magnetic field at the centre of coil is zero. }\\ & \begin{array}{cc}\therefore {𝐵}_1-{𝐵}_2& =0\\ \frac{{𝜇}_0{𝐼}_1}{2𝑅}-\frac{{𝜇}_0{𝐼}_2}{2𝜋𝑑}& =0\\ 𝑑& =\left(\frac{{𝐼}_2}{{𝐼}_1}\right)\frac{𝑅}{𝜋}\end{array}\end{array}$

First, calculate the total resistance between points $𝑃$ and $𝑄$ :

$\begin{array}{cc}{𝑅}_{𝑃𝑄}& =\left(\frac{30\Omega \times 10\Omega }{30\Omega +10\Omega }\right)+0.5\Omega \\ \\ & =\frac{300\Omega }{40\Omega }+0.5\Omega \\ \\ & =7.5\Omega +0.5\Omega \\ \\ & =8\Omega \end{array}$

Next, determine the current flowing through the circuit using Ohm's law : $𝑉=𝐼𝑅$

$𝐼=\frac{𝑉}{{𝑅}_{𝑃𝑄}}$

Where $𝑉$ is the voltage of the source. Therefore,

$𝐼=\frac{16\text{ V}}{8\Omega }=2\text{ A}$

The current $𝐼$ flowing through the circuit is thus 2 A.

The radius of circular path of charged particlein uniform magnetic field,

$𝑅=\frac{𝑚𝑣}{𝑞𝐵}\text{.... (i)}$

The kinetic energy of charged particle,

$\begin{array}{cc}& 𝐾=\frac{1}{2}𝑚{𝑣}^2\\ \Rightarrow & 𝑣=\sqrt{\surd \frac{2𝐾}{𝑚}}\text{.... (ii)}\end{array}$

Substituting Eq. (ii) in Eq. (i), we get

$𝑅=\frac{𝑚\sqrt{\surd \frac{2𝐾}{𝑚}}}{𝑞𝐵}=\frac{\sqrt{\surd 2𝑚𝐾}}{𝑞𝐵}$

For proton,

${𝑅}_{𝑝}=\frac{\sqrt{\surd 2{𝑚}_{𝑝}{𝐾}_{𝑝}}}{{𝑞}_{𝑝}𝐵}\text{.... (iii)}$

where, mass of proton is ${𝑚}_{𝑝}$, kinetic energy of proton is ${𝐾}_{𝑝}$ and charge on proton is ${𝑞}_{𝑝}$.

Similarly, for $𝛼$-particle,

${𝑅}_{𝛼}=\frac{\sqrt{\surd 2{𝑚}_{𝛼}{𝐾}_{𝛼}}}{{𝑞}_{𝛼}𝐵}\text{.... (iv)}$

where, mass of $𝛼$-particle is ${𝑚}_{𝛼}$, charge on $𝛼$-particle is ${𝑞}_{𝛼}$ and kinetic energy of $𝛼$-particle is ${𝐾}_{𝛼}$.

Here, ${𝑚}_{𝑎}=4{𝑚}_{𝑝}$ and ${𝑞}_{𝑎}=2{𝑞}_{𝑝}$ Substituting value in Eq. (iv), we get

${𝑅}_{𝑎}=\frac{\sqrt{\surd 2⟨4{𝑚}_{𝑝}){𝐾}_{𝑎}}}{𝐵\left(2{𝑞}_{𝑝}\right)}=\frac{\sqrt{\surd 2{𝑚}_{𝑝}{𝐾}_{𝑎}}}{𝐵{𝑞}_{𝑝}}$

$\begin{array}{c}\therefore \frac{{𝑅}_{𝑝}}{{𝑅}_{𝛼}}=\sqrt{\surd \frac{{𝐾}_{𝑝}}{{𝐾}_{𝛼}}}\\ \text{ As, }{𝑅}_{𝑝}={𝑅}_{𝛼}\text{(given)}\\ \therefore {𝐾}_{𝛼}={𝐾}_{𝑝}=10\mathrm{eV}\end{array}$

A repeater is a combination of at least one transmitter, one receiver and one carrier operated relay (COR) configured in a manner where a signal is received and transmitter rebroadcasts the received signal on another frequency.

Consider the figure shown below

$\text{ Stress }=\frac{\text{ Force }}{\text{ Area }}$

As, force on both wire is same and their areas are also same.

Hence, the stress on both wire will be same.

The strain is given by

$\text{ Strain }=\frac{\text{ Stress }}{𝑌}$

where, $𝑌$ is the Young's modulus.

For different material the value of Young's modulus is different.

Hence, strain for both wires are different.

The resolving power of optical instruments is inversely proportional to the wavelength of light used, as represented by the formula:

$\text{Resolving Power}\propto \frac{1}{𝜆}$

Given that the wavelengths for optical instruments A and B are ${𝜆}_{𝐴}=4500$ Å and ${𝜆}_{𝐵}=6000$ Å, the ratio of their resolving powers can be calculated as follows:

$\frac{(\text{Resolving Power}{)}_{𝐴}}{(\text{Resolving Power}{)}_{𝐵}}=\frac{{𝜆}_{𝐵}}{{𝜆}_{𝐴}}=\frac{6000}{4500}=\frac{4}{3}$

Therefore, the ratio of resolving power for instrument A to B is $4:3$.

photoelectric effect, the maximum kinetic energy possessed by the particle,

$\begin{array}{cc}{\mathrm{KE}}_{max}& =ℎ𝑣-𝜙\\ \frac{1}{2}𝑚{𝑣}^2& =\frac{ℎ𝑐}{𝜆}-𝜙(\because 𝑣=\frac{𝑐}{𝜆})\\ \Rightarrow {𝑣}^2& =\frac{2(ℎ𝑐-𝜙𝜆)}{𝜆𝑚}\\ 𝑣& ={\left[\frac{2(ℎ𝑐-𝜆𝜙)}{𝑚𝜆}\right]}^{1/2}\end{array}$

When a thin, lightweight rod made of a diamagnetic material such as silver is suspended in a uniform external magnetic field, it will align itself with its length perpendicular to the magnetic field.

Diamagnetic substances are materials that become weakly magnetized in the direction opposite to the applied external magnetic field. As a result, in a uniform magnetic field, a freely suspended diamagnetic rod will position itself perpendicular to the magnetic field.

The impulse is equals to change in linear momentum.

$\begin{array}{cc}\text{ Change in momentum }& =𝑚(𝑣-𝑢)\\ & =0.2(-6-6)=-2.4\end{array}$

The negative sign shows the direction of impulse is from batsman to bowler.

Let’s break down the problem to find which lenses form a telescope with a length of 10 cm and a magnifying power of 4.

Given Lenses and Focal Lengths:

${𝐿}_1=2\text{cm}$

${𝐿}_2=4\text{cm}$

${𝐿}_3=6\text{cm}$

${𝐿}_4=8\text{cm}$

Telescope Characteristics:

Length ($𝐿$) = 10 cm

Magnifying Power ($𝑀$) = 4

Telescope Equations:

Magnifying power, $𝑀=\frac{{𝑓}_{𝑜}}{{𝑓}_{𝑒}}$, where ${𝑓}_{𝑜}$ is the focal length of the objective lens and ${𝑓}_{𝑒}$ is the focal length of the eyepiece lens. Given $𝑀=4$, so:

${𝑓}_{𝑜}=4{𝑓}_{𝑒}\text{(i)}$

Total length of the telescope is the sum of the focal lengths of the objective and the eyepiece:

$𝐿={𝑓}_{𝑜}+{𝑓}_{𝑒}$

Substituting the known values:

$10={𝑓}_{𝑜}+{𝑓}_{𝑒}$

$10=4{𝑓}_{𝑒}+{𝑓}_{𝑒}\text{(from equation i)}$

$5{𝑓}_{𝑒}=10$

${𝑓}_{𝑒}=2\text{cm}$

Then, ${𝑓}_{𝑜}=4\times 2=8\text{cm}$

Conclusion:

The eyepiece has a focal length of 2 cm (${𝑓}_{𝑒}=2\text{cm}$) and corresponds to lens ${𝐿}_1$.

The objective lens has a focal length of 8 cm (${𝑓}_{𝑜}=8\text{cm}$) and corresponds to lens ${𝐿}_4$.

Therefore, the correct setup for the telescope uses ${𝐿}_1$ as the eyepiece and ${𝐿}_4$ as the objective lens.

$\begin{array}{cc}& \text{ The radius of gyration is given by }\\ & 𝐾=\sqrt{\surd \frac{𝐼}{𝑀}}\Rightarrow 𝐾\propto \sqrt{\surd 1}\end{array}$

Moment of inertia of ring about tangential axis is calculated by parallel axis theorem, ${𝐼}_{\text{ring }}={𝐼}_{𝑐}+𝑀{𝑅}^2=𝑀{𝑅}^2+𝑀{𝑅}^2=2𝑀{𝑅}^2$

Similarly,

$\begin{array}{cc}& {𝐼}_{\text{disc }}={𝐼}_{𝑐}+𝑀{𝑅}^2=\frac{1}{2}𝑀{𝑅}^2+𝑀{𝑅}^2=\frac{3}{2}𝑀{𝑅}^2\\ & \therefore \frac{{𝐾}_{\text{ring }}}{{𝐾}_{\text{disc }}}=\sqrt{\surd \frac{{𝐼}_{\text{ring }}}{{𝐼}_{\text{disc }}}}=\sqrt{\surd \frac{2𝑀{𝑅}^2}{\frac{3}{2}𝑀{𝑅}^2}}=\frac{2}{\sqrt{\surd 3}}\end{array}$

The frequency higher than the fundamental frequency of sound is known as an overtone.

For an open organ pipe, the first overtone is given by:

${𝑣}_{𝑜}=\frac{2𝑣}{2{𝐿}_{𝑜}}=\frac{𝑣}{{𝐿}_{𝑜}}$

For a closed organ pipe, the first overtone is:

${𝑣}_{𝑐}=\frac{3𝑣}{4{𝐿}_{𝑐}}$

It is provided that the first overtones of both pipes are identical:

${𝑣}_{𝑜}={𝑣}_{𝑐}$

Substituting the expressions for the overtones, we have:

$\frac{𝑣}{{𝐿}_{𝑜}}=\frac{3𝑣}{4{𝐿}_{𝑐}}$

By simplifying this equation, we find the ratio of the lengths of the open pipe to the closed pipe:

$\frac{{𝐿}_{𝑜}}{{𝐿}_{𝑐}}=\frac{4}{3}$

To find the magnitude of the average velocity of a particle moving along a circular path with angular acceleration, we first consider the kinematics of rotational motion.

The angular displacement, $𝜃$, is given by the rotational kinematics equation for a particle starting from rest:

$𝜃=\frac{1}{2}𝛼{𝑡}^2$

Solving for the time $𝑡$:

$𝑡=\sqrt{\surd \frac{2𝜃}{𝛼}}$

The tangential velocity $𝑣$ at any time $𝑡$ for a particle traveling along the circumference of a circle with angular velocity $𝜔$ is related by:

$𝑣=𝑟𝜔$

Since the angular velocity $𝜔$ is related to angular acceleration $𝛼$ by:

$𝜔=𝛼𝑡$

Substituting $𝑡=\sqrt{\surd \frac{2𝜃}{𝛼}}$:

$𝜔=𝛼\sqrt{\surd \frac{2𝜃}{𝛼}}=\sqrt{\surd 2𝛼𝜃}$

Therefore, the tangential velocity $𝑣$ becomes:

$𝑣=𝑟\sqrt{\surd 2𝛼𝜃}$

Average velocity $\overline{𝑣}$ over the time interval during which the particle covers the angular displacement $𝜃$ is computed as the total arc length traveled divided by the time taken. The arc length $𝑠$ is given by:

$𝑠=𝑟𝜃$

Thus, the average velocity is:

$\overline{𝑣}=\frac{𝑠}{𝑡}=\frac{𝑟𝜃}{\sqrt{\surd \frac{2𝜃}{𝛼}}}$

Simplifying:

$\overline{𝑣}=𝑟𝜃\cdot \sqrt{\surd \frac{𝛼}{2𝜃}}=𝑟\sqrt{\surd \frac{𝛼𝜃}{2}}$

Therefore, the correct choice is:

Option D: $𝑟{\left(\frac{𝛼𝜃}{2}\right)}^{\frac{1}{2}}$

The power radiated by a black body can be described by the formula:

$𝐸=𝜀𝜎𝐴{𝑇}^4$

Where:

$𝜀$ is the emissivity of the body,

$𝜎$ is the Stefan-Boltzmann constant,

$𝐴$ is the surface area,

$𝑇$ is the surface temperature.

For a spherical black body, the area $𝐴$ is given by $4𝜋{𝑟}^2$.

Given that two spherical bodies radiate the same power, we have:

${𝐸}_1={𝐸}_2$

So:

$𝜀𝜎\left(4𝜋{𝑟}_2^1\right){𝑇}_4^1=𝜀𝜎\left(4𝜋{𝑟}_2^2\right){𝑇}_4^2$

Since emissivity, the Stefan-Boltzmann constant, and $4𝜋$ are common, they cancel out:

${𝑇}_4^1{𝑟}_2^1={𝑇}_4^2{𝑟}_2^2$

Rearranging gives:

$\frac{{𝑟}_1}{{𝑟}_2}={\left(\frac{{𝑇}_2}{{𝑇}_1}\right)}^2$

The potentiometer is used for measuring voltage or potential difference by comparison of an unknown voltage with the known voltage, hence it is not a direct reading instrument.

Using kinematic relation,

${𝑣}^2={𝑢}^2-2𝑎𝑠\text{.... (i)}$

The momentum $𝑝$ is given by

$\begin{array}{cc}𝑝& =𝑀𝑢\\ 𝑢& =\frac{𝑝}{𝑀}\end{array}$

and acceleration is given as $𝑎=𝜇𝑔$

Substituting values in Eq. (i), we get

$0={𝑢}^2-2\text{ as }(\because \text{ final velocity },𝑣=0)$

$0={\left(\frac{𝑝}{𝑀}\right)}^2-2𝜇𝑔𝑠$

${\left(\frac{𝑝}{𝑀}\right)}^2=2𝜇𝑔𝑠$

$\Rightarrow 𝑠=\frac{{𝑝}^2}{2{𝑀}^2𝜇𝑔}$

$\begin{array}{cc}& \text{ Using principle of continuity, }\\ & {𝐴}_1{𝑣}_1={𝐴}_2{𝑣}_2\text{... (i)}\end{array}$

where, ${𝐴}_1$ is the area of square hole, ${𝐴}_1={𝑎}^2$,

${𝐴}_2$ is the area of circular hole, ${𝐴}_2=𝜋{𝑟}^2$,

${𝑣}_1$ is the velocity at depth $𝑦,{𝑣}_1=\sqrt{\surd 2𝑔𝑦}$

and ${𝑣}_2$ is the velocity at depth $16y,{𝑣}_2=\sqrt{\surd 2𝑔\left(16𝑦\right)}$.

Substituting values in Eq. (i), we get

$\begin{array}{cc}{𝑎}^2\left(\sqrt{\surd 2𝑔𝑦}\right)& =𝜋{𝑟}^2\left(\sqrt{\surd 2𝑔\left(16𝑦\right)}\right)\\ {𝑎}^2\left(\sqrt{\surd 2𝑔𝑦}\right)& =4𝜋{𝑟}^2\left(\sqrt{\surd 2𝑔𝑦}\right)\\ {𝑟}^2& =\frac{{𝑎}^2}{4𝜋}\end{array}$

$\therefore 𝑟=\frac{𝑎}{2\sqrt{\surd 𝜋}}$

The refractive index of a medium, denoted as $𝜇$, describes how light propagates through that medium. It is interconnected with the wavelength $𝜆$ of the light. To explore this relationship, consider the following:

Definition:

$𝜇=\frac{𝑐}{𝑣}$

where $𝑐$ is the speed of light in a vacuum, and $𝑣$ is the speed of light in the medium.

Relation with Wavelength:

The speed of light in a medium ($𝑣$) can also be expressed using the equation:

$𝑣=𝜆𝑓$

where $𝜆$ is the wavelength of light, and $𝑓$ is its frequency.

Considering Constant Frequency:

When light transitions from one medium to another, its frequency $𝑓$ remains unchanged. With this in mind, the equation for $𝜇$ becomes:

$𝜇=\frac{𝑐}{𝑣}=\frac{𝑐}{𝜆𝑓}$

Since both $𝑐$ and $𝑓$ are constants, $𝜇$ is inversely proportional to $𝜆$:

$𝜇\propto \frac{1}{𝜆}$

This relationship indicates that as the wavelength increases, the refractive index decreases, assuming the frequency remains constant.

The voltage gain of oscillator circuit is given by

${𝐴}_{𝑓}=\frac{𝐴}{1-𝐴𝛽}$

where, $𝐴$ is the voltage gain without feedback and $𝛽$ is called feedback factor.

For sustained oscillation, from Barkhausen criterion

$𝐴𝛽=1$

A spherical rubber balloon carries a charge uniformly distributed over its surface. As the balloon is inflated and its size increases, the total electric flux emanating from the surface does not change.

The electric field intensity due to a point charge placed at the center of a sphere with radius $𝑟$ is given by:

$𝐸=\frac{1}{4𝜋{𝜀}_0}\frac{𝑞}{{𝑟}^2}$

As the balloon expands, the charge enclosed by the Gaussian surface doesn't change. Therefore, the electric field intensity $𝐸$ remains the same. The electric flux $𝜙$ is defined by the relationship:

$𝜙=𝐸\cdot 𝑑𝑆$

Since the electric field $𝐸$ remains constant, the electric flux $𝜙$ remains unchanged.

For a diatomic gas undergoing an adiabatic change, the relationship between pressure ($𝑝$) and temperature ($𝑇$) is defined by:

$\frac{{𝑇}^{𝛾}}{{𝑝}^{𝛾-1}}=\text{constant}$

This equation can be rearranged to:

${𝑇}^{𝛾}=\text{constant}\times {𝑝}^{𝛾-1}$

By simplifying further, we find:

$𝑝\propto {𝑇}^{\frac{𝛾}{𝛾-1}}$

Given that $𝑝\propto {𝑇}^{𝑥}$, we equate the exponents:

$𝑥=\frac{𝛾}{𝛾-1}$

For a diatomic gas, the adiabatic index $𝛾$ is $\frac{7}{5}$. Substituting this value, we calculate:

$𝑥=\frac{\frac{7}{5}}{\frac{7}{5}-1}$

This simplifies to:

$𝑥=\frac{\frac{7}{5}}{\frac{2}{5}}=\frac{7}{2}=3.5$

Therefore, for a diatomic gas undergoing an adiabatic process, the value of $𝑥$ is 3.5.

The energy of photons is given by

$𝐸=𝑅ℎ𝑐(\frac{1}{{𝑛}_2^1}-\frac{1}{{𝑛}_2^2})$

where, $𝑅$ is Rydberg constant, $ℎ$ is Planck's constant and $𝑐$ is the speed of light.

(i) Energy of photon produced from second to first energy level,

$\begin{array}{cc}& {𝐸}_1=𝑅ℎ𝑐(\frac{1}{1^2}-\frac{1}{2^2})\\ & {𝐸}_1=\frac{3}{4}𝑅ℎ𝑐\end{array}$

(ii) Energy of photon produced from highest energy level (i.e., $\infty$) to second level,

$\begin{array}{cc}& {𝐸}_2=𝑅ℎ𝑐(\frac{1}{2^2}-\frac{1}{\infty })=\frac{1}{4}𝑅ℎ𝑐\\ \therefore & \frac{{𝐸}_1}{{𝐸}_2}=\frac{\frac{3}{4}𝑅ℎ𝑐}{\frac{1}{4}𝑅ℎ𝑐}=\frac{3}{1}\text{ or }3:1\end{array}$

The acceleration due to gravity increases as we go from equator of earth to pole of earth because the equatorial radius is more than the polar radius of earth and $𝑔\propto \frac{1}{𝑅}$.

Stress $=\frac{\text{ Force }}{\text{ Area }}$

The stress on wire $𝐴$,

$(\text{Stress}{)}_{𝐴}=\frac{𝐹}{𝜋{𝑟}_2^{𝐴}}=\frac{𝐹}{𝜋{\left(2{𝑟}_{𝐵}\right)}^2}\text{ (given, }{𝑟}_{𝐴}=2{𝑟}_{𝐵}\text{)}$

$\begin{array}{cc}& \text{ The stress on wire }𝐵\text{, }\\ & \begin{array}{cc}& (\text{ Stress }{)}_{𝐵}=\frac{𝐹}{𝜋{𝑟}_2^{𝐵}}\\ \therefore & \frac{(\text{ Stress }{)}_{𝐵}}{(\text{ Stress }{)}_{𝐴}}=\frac{\frac{𝐹}{𝜋{𝑟}_2^{𝐵}}}{\frac{𝐹}{𝜋{\left(2{𝑟}_{𝐵}\right)}^2}}=4\\ \Rightarrow & (\text{ Stress }{)}_{𝐵}=4(\text{Stress}{)}_{𝐴}\end{array}\end{array}$

When a tension $𝑇$ is applied to a sonometer wire of length $𝐿$, it vibrates with a fundamental frequency $𝑛$. If the tension is increased by 8 N, the fundamental frequency becomes three times the initial frequency. We need to find the initial tension applied to the wire.

The formula for the fundamental frequency of a vibrating wire is given by:

$𝑛=\frac{1}{2𝐿}\sqrt{\surd \frac{𝑇}{𝜇}}\text{.... (i)}$

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

For the increased tension, the fundamental frequency becomes:

$3𝑛=\frac{1}{2𝐿}\sqrt{\surd \frac{𝑇+8}{𝜇}}\text{... (ii)}$

To find the relationship between the initial and final tensions, we divide equation (ii) by equation (i):

$\frac{3𝑛}{𝑛}=\frac{\frac{1}{2𝐿}\sqrt{\surd \frac{𝑇+8}{𝜇}}}{\frac{1}{2𝐿}\sqrt{\surd \frac{𝑇}{𝜇}}}$

This simplifies to:

$3=\sqrt{\surd \frac{𝑇+8}{𝑇}}$

Squaring both sides gives:

$9𝑇=𝑇+8$

Solving for $𝑇$:

$8𝑇=8\Rightarrow 𝑇=1\text{ N}$

Thus, the initial tension applied to the wire was $1\text{ N}$.

The pure silicon has negative temperature coefficient of resistivity.

At absolute zero temperature, its resistance becomes infinite and acts like an insulator.

Given that the resultant capacitance ${𝐶}_{\text{eq}}=2𝜇\text{F}$:

The capacitors are connected in series. Therefore, the formula for the equivalent capacitance of capacitors in series is:

$\frac{1}{{𝐶}_{\text{eq}}}=\frac{1}{𝐶}+\frac{1}{𝐶}+\frac{1}{𝐶}+\frac{1}{𝐶}+\frac{1}{𝐶}$

This simplifies to:

$\frac{1}{{𝐶}_{\text{eq}}}=\frac{5}{𝐶}$

Substitute the given equivalent capacitance:

$\frac{1}{2𝜇\text{F}}=\frac{5}{𝐶}$

Solving for $𝐶$:

$𝐶=10𝜇\text{F}$

Therefore, each capacitor has a capacitance of $10𝜇\text{F}$.

The work done in blowing a soap bubble,

$𝑊=𝑇\times 𝐴$

where, $𝑇$ is the surface tension and $𝐴$ is the area of soap bubble.

$\begin{array}{cc}& 𝑊=𝑇\times 4𝜋{𝑅}^2\\ & 𝑊\propto {𝑅}^2\end{array}$

So, $\frac{{𝑊}_1}{{𝑊}_2}=\frac{{𝑅}_2^1}{{𝑅}_2^2}$

$\Rightarrow {𝑊}_2={𝑊}_1{\left(\frac{2𝑅}{𝑅}\right)}^2=4𝑊$