MHT CET 2019 3rd May Morning Shift (Physics & Chemistry) Question Paper with Solutions

Modulation index, $𝜇=\frac{{𝐴}_{𝑚}}{{𝐴}_{𝑐}}\le 1$ is carried by the carrier wave, i.e. there is no signal distortion.

The problem involves comparing the frequencies of the ${𝑃}^{\text{th}}$ overtone of a pipe open at both ends with one closed at one end, given they have the same length.

Calculating Frequencies

Open Pipe:

The frequency of the ${𝑃}^{\text{th}}$ overtone for a pipe open at both ends is given by:

${𝑓}_{\text{open}}=(𝑝+1)\frac{𝑣}{2𝐿}$

Here $𝑣$ is the speed of sound and $𝐿$ is the length of the pipe.

Closed Pipe:

The frequency of the ${𝑃}^{\text{th}}$ overtone for a pipe closed at one end is given by:

${𝑓}_{\text{closed}}=(2𝑝+1)\frac{𝑣}{4𝐿}$

Ratio of Frequencies

To find the ratio of the ${𝑃}^{\text{th}}$ overtone frequency of the open pipe to the closed pipe, we use the expressions derived:

$\text{Ratio}=\frac{{𝑓}_{\text{open}}}{{𝑓}_{\text{closed}}}=\frac{(𝑝+1)\frac{𝑣}{2𝐿}}{(2𝑝+1)\frac{𝑣}{4𝐿}}$

Simplifying this expression yields:

$\text{Ratio}=\frac{2(𝑝+1)}{(2𝑝+1)}$

Therefore, the ratio of frequencies for the ${𝑃}^{\text{th}}$ overtone of the open pipe to the closed pipe is $\frac{2(𝑝+1)}{2𝑝+1}$.

The refractive index of glass with respect to air is denoted as $𝜇$, and the critical angle for light transitioning from glass to air is $𝜃$.

We have:

$𝜇=\frac{1}{\sin 𝜃}\text{...(i)}$

When a ray of light originates from air and strikes the glass at an angle of incidence $𝜃$, we need to determine the angle of refraction $𝑟$. According to Snell's Law, the relationship is given by:

$\begin{array}{cc}𝜇& =\frac{\sin 𝜃}{\sin 𝑟}\\ \Rightarrow \sin 𝑟& =\frac{\sin 𝜃}{𝜇}=\frac{1/𝜇}{𝜇}\text{[using Eq. (i)]}\\ \Rightarrow \sin 𝑟& =\frac{1}{{𝜇}^2}\\ 𝑟& ={\sin }^{-1}\left(\frac{1}{{𝜇}^2}\right)\end{array}$

Thus, the corresponding angle of refraction is ${\sin }^{-1}\left(\frac{1}{{𝜇}^2}\right)$.

Maximum kinetic energy gained by charged particle in a cyclotion is given by

${𝐸}_{𝐾}=\frac{{𝑞}^2{𝐵}^2{𝑅}^2}{2𝑚}$

where, $𝑞=$ charge of the cyclotron,

$𝐵=$ intensity of magnetic field,

$𝑅=$ radius of orbit

and $𝑚=$ mass of the particle,

Hence, ${𝐸}_{𝐾}$ is independent of frequency of revolution.

A molecule on the surface experiences a net force (Cohesive force) in downward direction due to net number of molecules pulling it in the downward direction (molecule $𝐴$ as shown in figure underneath).

In contrast, a molecule well inside is equally pulled in all directions by molecules around at (molecule $𝐵$ as shown in figure).

For final image at infinity, magnifying power of a telescope is given by

$𝑚=\frac{{𝑓}_{𝑜}}{{𝑓}_{𝑒}}=9$

where, $𝑚=$ magnification,

${𝑓}_{𝑜}=$ focal length of objective

and ${𝑓}_{𝜃}=$ focal length of eyepiece

$\Rightarrow {𝑓}_{𝑜}=9{𝑓}_{𝑒}\text{.... (i)}$

Also, distance between objective and eyepiece

$\begin{array}{cc}& ={𝑓}_{𝑜}+{𝑓}_{𝑒}=20\text{ (given) }\\ & \Rightarrow 9{𝑓}_{𝑒}+{𝑓}_{𝑒}=20\Rightarrow {𝑓}_{𝑒}=2\mathrm{cm}\\ & {𝑓}_{𝑜}=9{𝑓}_{𝑒}=18\mathrm{cm}\end{array}$

According to Stefan's law, energy emitted by a body per unit area per second is proportional to fourth power of the absolute temperature.

$𝐸=𝜎{𝑇}^4$

where, $𝐸=$ energy emitted/area/second, $𝑇=$ absolute temperature in kelvin and $𝜎=$ Stefan's constant. $\Rightarrow 𝜎=\frac{𝐸}{{𝑇}^4}$

Unit of Stefan's constant

$\begin{array}{cc}& \text{ Unit of Stefan’s }\\ & \Rightarrow \frac{J/m^{2}s}{K^4}=J/m^{2}s{K}^4\\ & \text{ nimonsions of Stefan’s constant, }\end{array}$

Dimensions of Stefan's constant,

$\begin{array}{cc}& 𝜎=\frac{\left[\text{ Energy }\right]}{\left[\text{ Area }\right]\times \left[\text{ Time }\right]\times [\text{ Temperature }{]}^4}\\ & =\frac{\left[{\mathrm{ML}}^2{T}^{-2}\right]}{[L^{2}T\times K^4]}=\left[M^1{L}^0{T}^{-3}{K}^{-4}\right]\end{array}$

Magnetic moment of revolving electron,

${𝜇}_e=𝑖\times 𝐀$

where, $𝑖=$ current and $𝐴=$ area,

$\therefore {𝜇}_{𝑒}=\frac{𝑒}{𝑇}\times 𝜋{𝑟}^2$ (As current, $𝑖=\frac{\text{ charge }}{\text{ time }}=\frac{𝑒}{𝑇})$

where, T = time period of revolution and r = radius of orbit.

$\begin{array}{cc}{𝜇}_{𝑒}=& \frac{𝑒}{2𝜋𝑟}\times 𝑣\times 𝜋{𝑟}^2\\ & (\because \text{ Time period }(𝑇)=\frac{\text{ distance }}{\text{ speed }})\end{array}$

where, $𝐯=$ velocity of revolving electron

$\Rightarrow {𝜇}_{𝑒}=\frac{𝑒𝑣𝑟}{2}$

Hence, the magnetic moment of revolving electron is $\frac{𝑒𝑣𝑟}{2}$.

Since, an LED is forward biased diode, so its characteristics are those of forward bias, i.e., Here, option (a) is characteristics of a diode in forward and reverse bias. Option (c) $\to$ characteristics of solar cell.

Option (d) $\to$ reverse bias characteristics 1 drawn in first quadrant can be photodiode or Zener diode.

The stopping potential of photoelectrons is potential needed to stop the electrons from reaching the collector depends only on the frequency of incident radiation. It is directly proportional to it.

This is because, more the frequency of incident light, more will be the energy of the photons incident photons. $(𝐸=ℎ𝜈)$ Thus, more will be the energy (kinetic) acquired the electrons. More the kinetic energy of emitted electrons, higher is the potential needed to stop them (stopping potential).

We know, refractive index of crystal w.r.t. oil,

${𝜇}_{𝑐𝑜}=\frac{{𝜇}_{𝑐}}{{𝜇}_{𝑜}}=\frac{\text{ speed of light in oil }}{\text{ speed of light in crystal }}$

Where, ${𝜇}_{𝑐}=$ absolute refractive index of crystal w.r.t. vacuum.

and ${𝜇}_0=$ absolute refractive index of oil w.r.t. vacuum.

${𝜇}_{𝑐𝑜}=\frac{1.68}{1.2}=\frac{7}{5}=\frac{\text{ speed in oil }}{\text{ speed in crystal }}$

Fractional change in velocity, when light travel from oil to crystal

$=\frac{\text{ speed in crystal }}{\text{ speed in oil }}=\frac{5}{7}$

When a body suspended by the string situated at position $𝑃$ as shown in the figure, where latitude is $𝜆$, then body is also rotated with angular frequency $\left(𝜔\right)$ of earth, hence tension on the string is given by

$\begin{array}{cc}& 𝑇=𝑚𝑔-𝑚𝑟{𝜔}^2\cos 𝜆\\ 𝑇& =𝑚\cdot \frac{𝐺𝑀}{{𝑅}^2}-𝑚𝑟{𝜔}^2\cos 𝜆[\therefore 𝑔=\frac{𝐺𝑀}{{𝑅}^2}]\\ 𝑇=& \frac{𝐺𝑀𝑚}{{𝑅}^2}-𝑚𝑟{𝜔}^2\cos 𝜆\text{.... (i)}\end{array}$

When body is suspended at equator, then

$𝜆=0\text{ and }𝑟=𝑅$

$\therefore$ From Eq. (i), we have,

$\begin{array}{cc}& 𝑇=\frac{𝐺𝑀𝑚}{{𝑅}^2}-𝑚𝑅{𝜔}^2\cos 0^{\circ }\\ & 𝑇=\frac{𝐺𝑀𝑚}{{𝑅}^2}-𝑚𝑅{𝜔}^2\end{array}$

Net current due to all wires,

$\begin{array}{c}{𝑖}_{\text{net }}={𝑖}_1+{𝑖}_2+{𝑖}_3+{𝑖}_4+{𝑖}_5+{𝑖}_6\\ {𝑖}_{\text{net }}=10-13+10+7-12+18=20A\end{array}$

We know, magnetic field due to an infinitely long straight conductor at a perpendicular distance $𝑟$ from it is given by

$𝐵=\frac{{𝜇}_0𝑖}{2𝜋𝑟}=\frac{{𝜇}_0{𝑖}_{\text{net }}}{2𝜋𝑟}$

where, $𝑖=$ current in wire

and $𝑟=$ perpendicular distance.

$\begin{array}{cc}& =\frac{4𝜋\times {10}^{-7}\times 20}{2𝜋\times 10\times {10}^{-2}}=4\times {10}^{-5}\\ 𝐵& =40𝜇T\end{array}$

The fundamental frequency of a sonometer wire increases by 9 Hz when its tension is increased by 69%, with the length remaining unchanged. We need to determine the original frequency of the wire.

The frequency of vibration of a stretched string is determined by the formula:

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

Where:

$𝜈$ is the frequency,

$𝑇$ is the tension in the string,

$𝑚$ is the mass per unit length,

$𝐼$ is the length of the string (though not included in this formula directly, it’s relevant for the context of length being constant).

When the tension is increased by 69%, the new frequency becomes $𝜈+9$.

Let's calculate the new frequency using the increased tension:

${𝜈}^{\prime }=𝜈+9=\frac{1}{2}\sqrt{\surd \frac{𝑇+(69/100)𝑇}{𝑚}}$

On simplifying, we find the ratio of the old and new frequencies:

$\frac{𝜈}{{𝜈}^{\prime }}=\sqrt{\surd \frac{𝑇}{\frac{169}{100}𝑇}}\Rightarrow \frac{𝜈}{𝜈+9}=\sqrt{\surd \frac{100}{169}}=\frac{10}{13}$

Solving the equation for the original frequency $𝜈$:

$13𝜈=10𝜈+90$

This simplifies to:

$𝜈=30\text{ Hz}$

Therefore, the frequency of the wire is 30 Hz.

To determine the dimensions of the electric field and the electric dipole moment concerning the power of mass, let's analyze each one:

Electric Field:

The expression for the electric field $𝐸$ is given by:

$𝐸=\frac{𝐹}{𝑞}$

Where $𝐹$ is the force and $𝑞$ is the electric charge.

The dimensions of force $𝐹$ are:

$\left[𝐹\right]=\left[{\text{MLT}}^{-2}\right]$

The dimensions of charge $𝑞$ are:

$\left[𝑞\right]=\left[𝐴𝑇\right]$

Therefore, the dimensions of the electric field $𝐸$ are:

$\left[𝐸\right]=\frac{\left[{\text{MLT}}^{-2}\right]}{\left[𝐴𝑇\right]}=\left[{\text{MLT}}^{-3}{\text{A}}^{-1}\right]$

This shows that the power of mass in the dimensions of the electric field is $1$.

Electric Dipole Moment:

The electric dipole moment $𝑚$ is defined as $𝑚=1\times 𝐴$, where $𝐼$ is the current and $𝐴$ is the area.

The dimensions are given by:

$\left[𝑚\right]=\left[𝐴{𝐿}^2\right]=\left[{𝑀}^0{𝐿}^2{𝑇}^0𝐴\right]$

This indicates that the power of mass in the dimensions of the electric dipole moment is $0$.

In summary, for the electric field, the power of mass is $1$, and for the electric dipole moment, it is $0$.

For cell of emf $𝐸$ balancing length $=\frac{𝐿}{3}=\frac{1}{3}\times 𝐿($ length of wire $)$

Increased length of wire $=𝐿+\frac{𝐿}{2}=\frac{3𝐿}{2}$

Since, the cell is the same, therefore the new balancing length $=\frac{1}{3}$ (new increased length)

$\begin{array}{cc}& =\frac{1}{3}\times \frac{3}{2}𝐿\\ & =\frac{𝐿}{2}\end{array}$

Therefore, for the same cell the balance point will be as length $\frac{𝐿}{2}$ from positive end.

The Poisson's ratio for homogeneous isotropic material must be between -1.0 to +0.5 because of the requirement. For Young's modulus, the shear modulus and bulk modulus to have positive values.

Hence, 0.8 cannot be the value of Poisson's ratio for homogeneous isotropic material.

Angular velocity $=𝜔$

Linear velocity $=𝑣$

Length of string = Radius $=𝑟$

If string is halved, ${𝑟}^{\prime }=\frac{𝑟}{2}$

Angular momentum, $𝐿=𝐫\times 𝐩$

where, $𝐫=$ radius vector and $𝐩=$ linear momentum.

$\begin{array}{cc}=𝑟\times 𝑚𝑣& [\because 𝜃={90}^{\circ }]\\ =𝑚𝑣𝑟& \end{array}$

As $𝐿$ remain constant,

$𝑚𝑣𝑟=\text{ constant }$

$\Rightarrow$ If ${𝑟}^{\prime }\to \frac{𝑟}{2},{𝑣}^{\prime }=2𝑣$ such that

$\begin{array}{cc}𝑚{𝑣}^{\prime }{𝑟}^{\prime }& =𝑚2𝑣\times \frac{𝑟}{2}\\ & =𝑚𝑣𝑟\end{array}$

Hence, linear velocity will be $2𝑣$.

Let / be the length of metal wire.

When wire is bent into a circular coil of radius $𝑟$, then

$\begin{array}{cc}& 𝑟=\frac{1}{2𝜋}\\ \therefore \text{Area},& 𝐴=𝜋{\left(\frac{1}{2𝜋}\right)}^2=𝜋\frac{{𝑙}^2}{4{𝜋}^2}\end{array}$

$\therefore$ Magnetic dipole moment associated with circular coil,

$\begin{array}{cc}& {𝜇}_{𝑐}=𝑖𝐴=𝑖𝜋\left(\frac{{𝑙}^2}{4{𝜋}^2}\right)\\ & {𝜇}_{𝑐}=\frac{𝑖{𝐼}^2}{4𝜋}\text{.... (i)}\end{array}$

When metal wire is bent into a square coil then side of square

$𝑎=\frac{𝐼}{4}$

$\therefore$ Area, $𝐴={𝑎}^2=\frac{1^2}{16}$

$\therefore$ Magnetic dipole moment associated with square coil,

$\begin{array}{cc}{𝜇}_{𝑠}=𝑖𝐴& =𝑖\frac{{𝑙}^2}{16}\\ \therefore \frac{{𝜇}_{𝑐}}{{𝜇}_{𝑠}}& =\frac{\frac{𝑖{𝑙}^2}{4𝜋}}{\frac{1{𝑙}^2}{16}}=\frac{16}{4𝜋}=\frac{4}{𝜋}\end{array}$

Hence, the ratio of magnetic dipole moment of circular coil and square coil is $4:𝜋$.

In Young's double slit experiment, the position of the $𝑛$th dark fringe is determined by the formula:

${𝑥}_{𝑛}=\frac{𝐷}{𝑑}(2𝑛-1)\frac{𝜆}{2}$

From this, the wavelength $𝜆$ can be expressed as:

$𝜆=\frac{2{𝑥}_{𝑛}𝑑}{𝐷(2𝑛-1)}\text{.... (i)}$

Given that the fifth dark fringe is formed directly across one of the slits, we have:

$2{𝑥}_5=𝑑$

Substituting $𝑛=5$ into Eq. (i), we find:

$𝜆=\frac{2{𝑥}_5𝑑}{𝐷(2\times 5-1)}$

Since $2{𝑥}_5=𝑑$, it follows that:

$𝜆=\frac{𝑑\cdot 𝑑}{𝐷(10-1)}$

Simplifying gives:

$𝜆=\frac{{𝑑}^2}{9𝐷}$

Magnetic induction at the centre of circular current carrying loop.

${𝐵}_1=\frac{{𝜇}_0𝑖}{2𝑟}$ (downward direction)

Magnetic induction at the centre $𝑂$ due to straight wire,

${𝐵}_2=\frac{{𝜇}_0}{2𝜋}\cdot \frac{𝑖}{𝑟}$ (upward direction)

Net magnetic induction at $𝑂$,

$\begin{array}{cc}𝐵& ={𝐵}_1-{𝐵}_2=\frac{{𝜇}_0𝑖}{2𝑟}-\frac{{𝜇}_0}{2𝜋}\cdot \frac{𝑖}{𝑟}\\ & =\frac{{𝜇}_0𝑖}{2𝑟}[1-\frac{1}{𝜋}]\text{ [downward direction] }\end{array}$

Hence, there are no option correct.

Given, input voltage supplied to transformer,

${𝑉}_1=220V$

Current at output circuit of transformer,

$\begin{array}{c}{𝑖}_2=2A\\ {𝑉}_2=440V\\ \frac{\text{ Output power }\left({𝑃}_2\right)}{\text{ Input power }\left({𝑃}_1\right)}=0.8\Rightarrow \frac{{𝑉}_2{𝑖}_2}{{𝑉}_1{𝑖}_1}=0.8\\ \frac{440\times 2}{220\times {𝑖}_1}=0.8\\ \frac{4}{{𝑖}_1}=0.8\Rightarrow {𝑖}_1=\frac{4}{0.8}=5A\end{array}$

Hence, current drawn by the primary windings is 5 A .

When sonometer is stretched by weight $𝑤$ (tension), then frequency of vibration in the string

$\begin{array}{cc}& 𝜈=\frac{1}{2}\sqrt{\surd \frac{𝑇}{𝑚}}\\ & 𝜈=\frac{1}{2{𝐿}_1}\sqrt{\surd \frac{𝑤}{𝑚}}\text{.... (i)}\end{array}$

Where, $𝐿$ is the resonating length and $𝑚$ is mass per unit length of the string.

Similarly, when weight reduces to $\frac{𝑊}{4}$ and resonating length is ${𝐿}_2$, then

$𝜈=\frac{1}{2{𝐿}_2}\sqrt{\surd \frac{\frac{𝑤}{4}}{𝑚}}\Rightarrow 𝑣=\frac{1}{4{𝐿}_2}\sqrt{\surd \frac{𝑤}{𝑚}}\text{.... (ii)}$

$\begin{array}{cc}& \text{ From Eqs. (i) and (ii), we have }\\ & \begin{array}{cc}& \frac{1}{2{𝐿}_1}\sqrt{\surd \frac{𝑤}{𝑚}}=\frac{1}{4{𝐿}_2}\sqrt{\surd \frac{𝑤}{𝑚}}\Rightarrow {𝐿}_1=2{𝐿}_2\\ \therefore & \frac{{𝐿}_1}{{𝐿}_2}=2\\ \therefore & {𝐿}_1:{𝐿}_2=2:1\end{array}\end{array}$

Speed of electron of hydrogen atom in ground state,

$\begin{array}{cc}& {𝑣}_1=22\times {10}^{6}m/s\\ & \text{ Since, }𝑣\propto \frac{1}{𝑛}\\ & \therefore \frac{{𝑣}_1}{{𝑣}_2}=\frac{{𝑛}_2}{{𝑛}_1}=\frac{2.2\times {10}^6}{{𝑣}_2}\\ & \text{ [for third excited state, }{𝑛}_2=4\text{ ] }\\ & \Rightarrow {𝑣}_2=\frac{2.2\times {10}^6}{4}=0.55\times {10}^{6}m/s\\ & =5.5\times {10}^{5}m/s\end{array}$

Length of rod = $𝑙$

Moment of couple,

$\begin{array}{cc}& 𝜏=𝐹𝑙\sin 𝜃\\ \therefore & 𝐹=\frac{𝜏}{𝑙\sin 𝜃}=\frac{𝜏}{𝑙\sin {30}^{\circ }}=\frac{𝜏}{𝑙\times \frac{1}{2}}\end{array}$

Force, $𝐹=\frac{2𝜏}{𝑙}$

Vectors $(𝐀+𝐁)$ and $(𝐀-𝐁)$ are at right angle to each other, therefore

$\begin{array}{cc}& (𝐴+𝐵)\cdot (𝐴-𝐵)=0\\ & 𝐴\cdot 𝐴+𝐵\cdot 𝐴-𝐴\cdot 𝐵-𝐵\cdot 𝐵=0\\ & \left|𝐴\right|+𝐵𝐴\cos 𝜃-𝐴𝐵\cos 𝜃-|𝐵{|}^2=0\\ & |𝐴{|}^2=|𝐵{|}^2\\ \text{Hence,}& \left|𝐴\right|=\left|𝐵\right|\end{array}$

Given, inductance of a coil, $𝐿=2H$

Reactance of coil, when it is connected to $𝐴𝐶$ source,

$\begin{array}{cc}& {\left({𝑋}_{𝐿}\right)}_{𝐴𝐶}=\frac{1}{𝜔𝐿}\text{ where, }𝜔=\text{ angular frequency }\\ & {\left({𝑋}_{𝐿}\right)}_{𝜔}=\frac{1}{2𝜇𝜔}\end{array}$

For DC source, inductor coil behaves as pure conductor, hence ${\left({𝑋}_{𝐿}\right)}_{\mathrm{DC}}=0$

$\therefore \frac{{\left({𝑋}_{𝐿}\right)}_{𝐴𝐶}}{{\left({𝑋}_{𝐿}\right)}_{𝐷𝐶}}=\frac{\frac{1}{2𝜔}}{0}=\infty \text{(at infinity)}$

Time period of simple pendulum inside a stationary lift,

$𝑇=2𝜋\sqrt{\surd \frac{𝑙}{𝑔}}\text{.... (i)}$

where, $𝐼=$ length of string

and $𝑔=$ gravitational acceleration

Acceleration of lift in upward direction,

$𝑎=\frac{𝑔}{3}$

$\therefore$ Time period of pendulum,

$\begin{array}{c}\text{ d of pendulum, }\\ \begin{array}{cc}{𝑇}^{\prime }& =2𝜋\sqrt{\surd \frac{𝑙}{𝑔+𝑎}}=2𝜋\sqrt{\surd \frac{𝑙}{𝑔+\frac{𝑔}{3}}}\\ & =2𝜋\sqrt{\surd \frac{3𝑙}{4𝑔}}=2𝜋\cdot \frac{\sqrt{\surd 3}}{2}\cdot \sqrt{\surd \frac{𝑙}{𝑔}}\\ & =\frac{\sqrt{\surd 3}}{2}\cdot 2𝜋\sqrt{\surd \frac{𝑙}{𝑔}}=\frac{\sqrt{\surd 3}}{2}𝑇\text{ [from Eq. (i)] }\end{array}\end{array}$

Given that a charged conductor produces an electric field intensity of ${10}^{3}V/m$ just outside its surface in a vacuum, we need to determine the electric field $𝐸$ when the conductor is placed in a medium with a dielectric constant of 4.

The electric field in a medium with a dielectric constant $𝐾$ is given by:

$𝐸=\frac{{𝐸}_0}{𝐾}$

Substituting the given values:

$𝐸=\frac{{10}^3}{4}=250V/m$

Thus, the electric field intensity just outside the surface of the conductor when placed in the medium is $250V/m$.

In damped SHM, damping force is proportional to the velocity of the oscillator.

i.e., ${𝐹}_{𝑑}\propto 𝑣\Rightarrow {𝐹}_{𝑑}=𝑏𝑣$

where, $𝑏$ is a damping constant.

$\therefore$ Damping constant, $𝑏=\frac{{𝐹}_{𝑑}}{𝑣}$

SI unit of damping constant

$\begin{array}{cc}𝑏& =\frac{\text{ SI unit of force }}{\text{ SI unit of velocity }}=\frac{\mathrm{kg}\cdot m\cdot s^{-2}}{m\cdot s^{-1}}\\ & =\mathrm{kg}\cdot s^{-1}=\frac{\mathrm{kg}}{s}\end{array}$

Hence, the SI unit of damping constant is $\mathrm{kg}/s$.

Let’s analyze the given situation with the two wires made from the same material:

Initial Wire:

Length, $𝐿$

Radius, $𝑟$

Force applied, $𝐹$

Increase in length (elongation), $𝐼$

Using Young’s modulus ($𝑌$), we know:

$𝑌=\frac{\frac{𝐹}{𝜋{𝑟}^2}}{\frac{𝐼}{𝐿}}=\frac{𝐹𝐿}{𝜋{𝑟}^2𝐼}\text{(Equation 1)}$

Second Wire:

Length, ${𝐿}^{\prime }=2𝐿$

Radius, ${𝑟}^{\prime }=2𝑟$

Force applied, ${𝐹}^{\prime }=2𝐹$

The elongation for the second wire is denoted as ${𝐼}^{\prime }$.

The Young’s modulus for the second wire is:

$𝑌=\frac{\frac{{𝐹}^{\prime }}{𝜋({𝑟}^{\prime }{)}^2}}{\frac{{𝐼}^{\prime }}{{𝐿}^{\prime }}}=\frac{{𝐹}^{\prime }{𝐿}^{\prime }}{𝜋({𝑟}^{\prime }{)}^2{𝐼}^{\prime }}$

Substituting the given dimensions and force:

$𝑌=\frac{2𝐹\cdot 2𝐿}{𝜋(2𝑟{)}^2\cdot {𝐼}^{\prime }}=\frac{4𝐹𝐿}{4𝜋{𝑟}^2\cdot {𝐼}^{\prime }}$

Simplifying gives:

$𝑌=\frac{𝐹𝐿}{𝜋{𝑟}^2{𝐼}^{\prime }}\text{(Equation 2)}$

Comparing Equation 1 and Equation 2:

Both equations are equal since the material's Young's modulus $𝑌$ remains constant, giving us:

$\frac{𝐹𝐿}{𝜋{𝑟}^2𝐼}=\frac{𝐹𝐿}{𝜋{𝑟}^2{𝐼}^{\prime }}$

Hence, $𝐼={𝐼}^{\prime }$.

Therefore, the increase in length of the second wire is the same as the first, making the final elongation ${𝐼}^{\prime }=𝐼$.

Given for a metal, wavelength of light used $=𝜆$

Stopping potential $=𝑉$

If ${𝜆}_0$ be the threshold wavelength, then maximum kinetic energy of emitted electrons

${𝐾}_{max}=ℎ𝑐(\frac{1}{𝜆}-\frac{1}{{𝜆}_0})\text{.... (i)}$

Again, wavelength of used light

${𝜆}^{\prime }=2𝜆$

Stopping potential, ${𝑉}^{\prime }=\frac{𝑉}{3}$

then ${𝐾}_{max}=ℎ𝑐(\frac{1}{{𝜆}^{\prime }}-\frac{1}{{𝜆}_0})$

$\begin{array}{cc}\Rightarrow & 𝑒{𝑉}^{\prime }=ℎ𝑐(\frac{1}{2𝜆}-\frac{1}{{𝜆}_0})\\ \Rightarrow & \frac{𝑒𝑉}{3}=ℎ𝑐(\frac{1}{2𝜆}-\frac{1}{{𝜆}_0})\text{.... (ii)}\end{array}$

From Eqs. (i) and (ii), we have

$\begin{array}{c}\frac{𝑒𝑉}{\frac{𝑒𝑉}{3}}=\frac{ℎ𝑐(\frac{1}{𝜆}-\frac{1}{{𝜆}_0})}{ℎ𝑐(\frac{1}{2𝜆}-\frac{1}{{𝜆}_0})}\\ 3(\frac{1}{2𝜆}-\frac{1}{{𝜆}_0})=\frac{1}{𝜆}-\frac{1}{{𝜆}_0}\\ \Rightarrow {𝜆}_0=4𝜆\end{array}$

So, threshold wavelength is 4 times of wavelengths of light.

In a transistor, emitter region is highly doped because it injects charge carriers toward base and collector region.

Collector region is moderately doped because in recombination of charge carriers takes place in this region.

Base region are lightly doped, so that recombination of charge carriers is small.

Given, distance of bottom of hole from the surface of earth $\left(𝑑\right)=$ half the radius of earth $=\frac{{𝑅}_{𝑒}}{2}$

If $𝑔$ be the value of gravitational acceleration on the surface of earth, then weight of body

$𝑚𝑔=300N$

If ${𝑔}^{\prime }$ be the gravitational acceleration at the bottom of hole, then

${𝑔}^{\prime }=𝑔(1-\frac{𝑑}{{𝑅}_{𝜃}})=𝑔(1-\frac{\frac{{𝑅}_{𝜃}}{2}}{{𝑅}_{𝑒}})\Rightarrow {𝑔}^{\prime }=\frac{𝑔}{2}$

$\therefore$ Weight of the body on the bottom of hole,

$𝑚{𝑔}^{\prime }=\frac{𝑚𝑔}{2}=\frac{300}{2}=150N$

Given, terminal velocity of each small drop

${𝑣}_1=10\mathrm{cm}/s$

When 8 small drops are combined to form bigger drop, then volume will be same.

If ${𝑟}_1$ be radius of small drop and ${𝑟}_2$ be radius of bigger drop, then ${𝑣}_2={𝑣}_1$

$\frac{4}{3}𝜋{𝑟}_3^2=8\times \frac{4}{3}𝜋{𝑟}_3^1\Rightarrow {𝑟}_2=2{𝑟}_1$

$\therefore$ Terminal velocity of the drop

$\begin{array}{cc}& 𝑣=\frac{2}{9}\frac{{𝑟}^2(𝜌-𝜎)}{9𝜂}\\ \text{i.e.,}& 𝑣\propto {𝑟}^2\\ \therefore & \frac{{𝑣}_1}{{𝑣}_2}=\frac{{𝑟}_2^1}{{𝑟}_2^2}=\frac{{𝑟}_2^1}{{\left(2{𝑟}_1\right)}^2}\Rightarrow \frac{{𝑣}_1}{{𝑣}_2}=\frac{1}{4}\\ & {𝑣}_2=4{𝑣}_1=4\times 10=40\mathrm{cm}/s\end{array}$

Given, height of inclined plane, $ℎ=7m$

$𝑔=10m/s^2$

By conservation of energy,

Potential energy lost by the solid sphere in rolling down the inclined plane $=$ Kinetic energy gained by the sphere.

$\begin{array}{cc}𝑚𝑔ℎ& =\frac{1}{2}𝑚{𝑣}^2+\frac{1}{2}𝑙{𝜔}^2\\ & =\frac{1}{2}𝑚{𝑣}^2+\frac{1}{2}\cdot \frac{2}{5}𝑚{𝑅}^2\cdot {\left(\frac{𝑣}{𝑅}\right)}^2\\ & =\frac{1}{2}𝑚{𝑣}^2+\frac{1}{5}𝑚{𝑣}^2\\ 𝑚𝑔ℎ& =\frac{7}{10}𝑚{𝑣}^2\\ \therefore {𝑣}^2& =\frac{10}{7}𝑔ℎ\\ \Rightarrow 𝑣& =\sqrt{\surd \frac{10}{7}𝑔ℎ}=\sqrt{\surd \frac{10}{7}\times 10\times 7}=\sqrt{\surd 100}m/s\end{array}$

When a body perform uniform circular motion, then centripetal force acts always in perpendicular direction to its velocity, hence work done by centripetal force is always zero. Hence, work done by centripetal force is zero, when it through $\left(\frac{2}{3}\right)$ rd of the circular path.

Electric polarisation $\left(𝑃\right)$ is given by

$𝑃={𝜀}_0𝜒𝐸$

where, $𝐸$ is electric field and $𝜒$ is electric susceptibility and it is dimensionless quantity,

Dimensional formula of $𝑃=$ Dimensional formula of $\left({𝜀}_0\right)\times$ Dimensional formula of $𝐸$

$=\left[{𝑀}^{-1}{𝐿}^{-3}{𝑇}^4{𝐼}^2\right]\left[𝑀𝐿{𝑇}^{-3}{𝐼}^{-1}\right]=\left[{𝑀}^0{𝐿}^{-2}{𝑇}^1{𝐼}^1\right]$