MHT CET 2020 19th October Evening Shift (Physics & Chemistry) Question Paper with Solutions

To determine the initial frequency of tuning fork $𝐴$, denoted as $𝑓$, we start by considering the beats produced. A beat frequency of 5 beats per second is observed with another tuning fork of 480 Hz. This implies:

$𝑓=480\pm 5$

This results in two possible initial frequencies for fork $𝐴$: 485 Hz or 475 Hz.

Next, when a small amount of wax is added to a prong of fork $𝐴$, its frequency decreases. Subsequently, the beat frequency changes to 2 beats per second. This reduction in beat frequency is only feasible when the frequency of fork $𝐴$ initially was the higher of the two possibilities, which is 485 Hz.

Therefore, the original frequency of tuning fork $𝐴$ before adding the wax was 485 Hz.

The terminal velocity of rain drop is

$\begin{array}{cc}& {𝑣}_{𝑇}=\frac{2(𝜎-𝜌){𝑟}^2𝑔}{9𝜂}\\ \Rightarrow & {𝑣}_{𝑇}\propto {𝑟}^2\text{.... (i)}\end{array}$

Also, surface area of rain drop, $𝐴=4𝜋{𝑟}^2$

$\Rightarrow 𝐴\propto {𝑟}^2\text{.... (ii)}$

$\begin{array}{cc}& \text{ From Eqs. (i) and (ii), we get }\\ & \frac{{𝐴}_1}{{𝐴}_2}=\frac{{𝑉}_{{𝑇}_1}}{{𝑉}_{{𝑇}_2}}=\frac{16}{9}\text{ or }16:9\end{array}$

Let ${𝑥}_1$ and ${𝑥}_2$ be the position of masses ${𝑚}_1$ and ${𝑚}_2$, respectively.

The position of centre of mass is

${𝑥}_{\mathrm{CM}}=\frac{{𝑥}_1{𝑚}_1+{𝑥}_2{𝑚}_2}{{𝑚}_1+{𝑚}_2}$

If $\Delta {𝑥}_1$ and $\Delta {𝑥}_2$ be the changes in positions, then change in the position of centre of mass,

$\Delta {𝑥}_{\mathrm{CM}}=\frac{\Delta {𝑥}_1{𝑚}_1+\Delta {𝑥}_2{𝑚}_2}{{𝑚}_1+{𝑚}_2}$

Given that, the centre of mass remains unchanged e., $\Delta {𝑥}_{𝐶𝑀}=0$ and $\Delta {𝑥}_1=𝑑$

$\Rightarrow 0=\frac{𝑑{𝑚}_1+{𝑚}_2\Delta {𝑥}_2}{{𝑚}_1+{𝑚}_2}$

or $\Delta {𝑥}_2=-\frac{{𝑚}_1}{{𝑚}_2}𝑑$

Here, negative sign shows that the second particle should be moved towards the centre of mass.

The fringe width is given by

$𝑍=\frac{𝐷𝜆}{𝑑}$

As, $𝜆$ and $𝑑$ are constants.

$\therefore 𝑍\propto 𝐷$

So, the graph between $𝑍$ and $𝐷$ will be a straight line inclined to D-axis, as shown in graph (C).

The work-function of a surface is

$𝜙=ℎ{𝜈}_0$

where, $ℎ=$ Planck's constant

and ${𝜈}_0=$ threshold frequency.

From graph it is clear that,

$\begin{array}{c}{\left({𝜈}_0\right)}_{𝑝}<{\left({𝜈}_0\right)}_0\\ \therefore {𝜙}_{𝑝}<{𝜙}_{𝑄}\end{array}$

The magnetic field at the centre of a circular orbit of a moving electron is given by

$\begin{array}{cc}& 𝐵=\frac{{𝜇}_0}{4𝜋}\times \frac{𝑒}{{𝑟}^3}(𝐯\times 𝐫)\\ & 𝐵=\frac{{𝜇}_0}{4𝜋}\times \frac{𝑒𝑣\sin 𝜃}{{𝑟}^2}\end{array}$

As, in a circular path, the angle between radius vector and velocity vector is ${90}^{\circ }$.

$\begin{array}{cc}\Rightarrow & 𝐵=\frac{{𝜇}_0}{4𝜋}\times \frac{𝜃𝑉}{{𝑟}^2}\\ \therefore & 𝑟={\left(\frac{{𝜇}_0𝑒𝑣}{4𝜋𝐵}\right)}^{1/2}\end{array}$

The maximum kinetic energy of emitted photoelectron is

${𝐾}_{\text{max }}=𝑒{𝑉}_{𝑠}$

where, ${𝑉}_{𝑠}=$ stopping potential.

$\Rightarrow {𝑉}_{𝑠}=\frac{{𝐾}_{max}}{𝜃}=\frac{3.2\times {10}^{-19}}{1.6\times {10}^{-19}}=2V$

Also, the work-function of a metal is

$\begin{array}{cc}𝜙& =\frac{1242}{{𝜆}_0\left(\mathrm{nm}\right)}\mathrm{eV}\\ {𝜆}_0& =\frac{1242\times 1.6\times {10}^{-19}}{𝜙}\mathrm{nm}\\ \begin{array}{}\end{array}\end{array}$

$\begin{array}{cc}\Rightarrow & =\frac{1242\times 1.6\times {10}^{-19}}{6.63\times {10}^{-19}}\mathrm{nm}\\ & \approx 3000\stackrel{𝑜}{𝐴}\end{array}$

The root mean square (rms) velocity of gas molecules is given by the formula:

${𝑣}_{\mathrm{rms}}=\sqrt{\surd \frac{3𝑅𝑇}{𝑀}}$

This shows that:

${𝑣}_{\mathrm{rms}}\propto \sqrt{\surd \frac{𝑇}{𝑀}}$

If we have two scenarios with different conditions and we are comparing their rms velocities, the relationship can be expressed as:

$\frac{({𝑣}_{\mathrm{rms}}{)}_1}{({𝑣}_{\mathrm{rms}}{)}_2}=\sqrt{\surd \frac{{𝑇}_1}{{𝑀}_1}}\times \sqrt{\surd \frac{{𝑀}_2}{{𝑇}_2}}$

Here, when the molecular weight is doubled, ${𝑀}_2=2{𝑀}_1$, and when the absolute temperature is halved, ${𝑇}_2=\frac{{𝑇}_1}{2}$.

Substituting these values into the equation, we get:

$\frac{({𝑣}_{\mathrm{rms}}{)}_1}{({𝑣}_{\mathrm{rms}}{)}_2}=\sqrt{\surd \frac{{𝑇}_1}{{𝑀}_1}}\times \sqrt{\surd \frac{4{𝑀}_1}{{𝑇}_1}}=2$

This implies that:

$({𝑣}_{\mathrm{rms}}{)}_2=\frac{({𝑣}_{\mathrm{rms}}{)}_1}{2}=\frac{200}{2}=100\text{m/s}$

The given situation can be drawn as

The gravitational potential at $𝑃$ is

$\begin{array}{cc}{𝑉}_{𝑝}& =-(\frac{𝐺{𝑀}_1}{\frac{𝑟}{3}}+\frac{𝐺{𝑀}_2}{\frac{2𝑟}{3}})\\ & =\frac{-3𝐺(2{𝑀}_1+{𝑀}_2)}{2𝑟}\end{array}$

The work done to escape the mass $𝑀$ to infinity is

$𝑊=𝑀({𝑉}_{\infty }-{𝑉}_{𝑝})=\frac{3𝐺𝑀(2{𝑀}_1+{𝑀}_2)}{2𝑟}$

As, work done is equal to kinetic energy of mass $𝑀$.

$\begin{array}{cc}\Rightarrow & \frac{1}{2}𝑀{𝑣}_2^{𝑒}=\frac{3𝐺𝑀(2{𝑀}_1+{𝑀}_2)}{2𝑟}\\ & {𝑣}_{𝑒}={\left[\frac{3𝐺}{𝑟}(2{𝑀}_1+{𝑀}_2)\right]}^{1/2}\\ \text{ or }& {𝑣}_{𝑒}={\left[\frac{6𝐺}{𝑟}({𝑀}_1+\frac{{𝑀}_2}{2})\right]}^{1/2}\end{array}$

The height of liquid column in a capillary tube is

$ℎ=\frac{2𝑇\cos 𝜃}{𝜌𝑟𝑔}\Rightarrow ℎ\propto \frac{1}{𝑔}$

$\therefore$ When the setup is kept in a lift going down with acceleration $𝑎<𝑔$, then the height of water in the tube increases as net downward acceleration becomes $(𝑔-𝑎)$.

The Van de Graaff generator is specifically designed to produce a very high voltage while only providing a very low current. Here’s a brief explanation of why:

The generator uses a moving belt to transport electrical charge to a large, hollow metal sphere.

This accumulation of charge creates a high electric potential (voltage) on the sphere.

However, due to its design, the amount of current (the flow of charge per unit time) is very limited.

So, the correct answer is:

$\text{Option B: high voltage and low current}.$

Modulation Index

$\left(\mathrm{MI}\right)=\frac{\text{ Peak voltage of modulating signal }\left({𝑉}_{𝑚}\right)}{\text{ Peak voltage of carrier signal }\left({𝑉}_{𝑐}\right)}$

Given, $\mathrm{MI}=75\%$ and ${𝑉}_{𝑐}=16V$

$\therefore {𝑉}_{𝑚}=\mathrm{MI}\times {𝑉}_{𝑐}=\frac{75}{100}\times 16=12V$

As, we know that the difference of energy between two level is equal to the energy of photon.

$\begin{array}{cc}& \Delta 𝐸=ℎ𝑣=\frac{ℎ𝑐}{𝜆}\Rightarrow 𝜆=\frac{ℎ𝑐}{\Delta 𝐸}\\ \text{or}& 𝜆\propto \frac{1}{\Delta 𝐸}\end{array}$

The value of $\Delta 𝐸$ is minimum for $𝐴$, so radiation corresponding to $𝐴$ have maximum wavelength and value of $\Delta 𝐸$ is maximum for $𝐷$, so radiation corresponding to $𝐷$ have minimum wavelength.

To convert a milliammeter into a voltmeter with a range of $0-15\text{V}$, the following information is given:

Maximum Voltage, $𝑉=15\text{V}$

Current through the galvanometer, ${𝐼}_{𝑔}=30\text{mA}=30\times {10}^{-3}\text{A}$

Internal resistance of the galvanometer, $𝐺=40\Omega$

We need to determine the resistance $𝑅$ that should be added in series. The relationship for the voltmeter is given by:

$𝑉={𝐼}_{𝑔}(𝑅+𝐺)$

Substituting the known values:

$15=30\times {10}^{-3}(𝑅+40)$

Solving for $𝑅$:

$\begin{array}{cc}15& =30\times {10}^{-3}(𝑅+40)\\ 15& =0.03(𝑅+40)\\ \frac{15}{0.03}& =𝑅+40\\ 500& =𝑅+40\\ 𝑅& =500-40=460\Omega \end{array}$

Therefore, the resistance that should be used in series is $460\Omega$.

When child moves in a circular track, he is acted upon by two force as shown below,

Here, ${𝑓}_{𝑐}=𝑓\sin 𝜃$ and ${𝑓}_{𝑡}=𝑓\cos 𝜃$.

As, ${𝑓}_{𝑐}$ is the centripetal force and ${𝑓}_{𝑡}$ is the tangential force. So,

$\begin{array}{cc}& {𝑓}_{𝑐}=\frac{𝑚{𝑣}^2}{𝑟}=𝑓\sin 𝜃\\ \text{and }& {𝑓}_{𝑡}=𝑚𝑎=𝑓\cos 𝜃\end{array}$

$\therefore$ Resultant force, ${𝑓}_{𝑅}=\sqrt{\surd (𝑓\sin 𝜃{)}^2+(𝑓\cos 𝜃{)}^2}$

$=\sqrt{\surd {\left(\frac{𝑚{𝑣}^2}{𝑟}\right)}^2+(𝑚𝑎{)}^2}$

Also, when the shoes starts slipping, the friction becomes equal to resultant force.

$\begin{array}{cc}\therefore & 𝑓={𝑓}_{𝑅}\\ \Rightarrow & 𝜇𝑚𝑔=\sqrt{\surd {\left(\frac{𝑚{𝑣}^2}{𝑟}\right)}^2+(𝑚𝑎{)}^2}\\ \Rightarrow & {𝜇}^2{𝑚}^2{𝑔}^2=\frac{{𝑚}^2{𝑣}^4}{{𝑟}^2}+{𝑚}^2{𝑎}^2\\ \Rightarrow & {𝜇}^2{𝑔}^2=\frac{(𝑎𝑡{)}^4+{𝑎}^2{𝑟}^2}{{𝑟}^2}(\because 𝑎=\frac{𝑣}{𝑡})\\ \text{ or }& 𝜇=\frac{{({𝑎}^4{𝑡}^4+{𝑎}^2{𝑟}^2)}^{\frac{1}{2}}}{𝑔𝑟}\end{array}$

The work done to elongate a metal rod can be calculated using the formula:

$𝑊=\frac{1}{2}\frac{𝐴𝑌}{𝐿}(\Delta 𝐿{)}^2$

Where:

$𝑊$ is the work done,

$𝐴$ is the cross-sectional area,

$𝑌$ is Young's modulus,

$𝐿$ is the original length of the rod,

$\Delta 𝐿$ is the change in length, or elongation.

Given that $\Delta 𝐿=\ell$, we can rewrite the expression for work done as:

$𝑊\propto {\ell }^2$

This shows that the work done is proportional to the square of the elongation, $\ell$.

The emf of a cell in a potentiometer is

$𝐸=\frac{𝑉}{𝐿}𝐼$

where, $𝐼=$ length of wire at null point,

$𝑉=$ voltage of source

and $𝐿=$ total length of potentiometer wire.

$\therefore 𝐸\propto 𝐼$

When two cell of emf ${𝐸}_1$ and ${𝐸}_2({𝐸}_1>{𝐸}_2)$ are connected to assist each other, then

${𝐸}_1+{𝐸}_2=490\text{... (i)}$

When the cell of emf ${𝐸}_2$ is connected, so as to oppose cell ${𝐸}_1$, then

${𝐸}_1-{𝐸}_2=90\text{.... (ii)}$

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

$\begin{array}{cc}& {𝐸}_1=290V\text{ and }{𝐸}_2=200V\\ \therefore & \frac{{𝐸}_1}{{𝐸}_2}=\frac{290}{200}=\frac{29}{20}=1.45\end{array}$

The given situation is as shown below

We have to find the moment of inertia about axis $𝑋{𝑋}^{\prime }$.

The distance of line $𝑋{𝑋}^{\prime }$ from base of triangle is

$𝑑=\sqrt{\surd {𝐼}^2-\frac{{𝑙}^2}{4}}=\frac{\sqrt{\surd 3}𝐼}{2}$

$\therefore$ Moment of inertia, due to each mass,

$\begin{array}{cc}𝐼& =𝑚\times 0+𝑚{\left(\frac{\sqrt{\surd 3}}{2}𝐼\right)}^2+𝑚{\left(\frac{\sqrt{\surd 3}}{2}𝐼\right)}^2\\ & =\frac{3𝑚{𝑙}^2}{4}+\frac{3𝑚{𝑙}^2}{4}\\ & =\frac{6𝑚{𝑙}^2}{4}=\frac{3𝑚{𝑙}^2}{2}\end{array}$

The orbital period of revolution of electron in $𝑛$th orbit is

${𝑇}_{𝑛}=\frac{2𝜋{𝑟}_{𝑛}}{{𝑣}_{𝑛}}$

As, we know, ${𝑟}_{𝑛}=\left(\frac{{ℎ}^2{𝜀}_0}{𝜋𝑚{𝑒}^2}\right)\frac{{𝑛}^2}{𝑍}$

and ${𝑣}_{𝑛}=\left(\frac{{𝑒}^2}{2ℎ{𝜀}_0}\right)\frac{𝑧}{𝑛}$

$\therefore {𝑇}_{𝑛}=2𝜋\frac{{ℎ}^2{𝜀}_0{𝑛}^2}{𝜋𝑚{𝑒}^2𝑍}\times \frac{2ℎ{𝜀}_0𝑛}{{𝑒}^2𝑍}$

$=\frac{4{𝜀}_2^0{𝑛}^3{ℎ}^3}{𝑚{𝑒}^4{𝑍}^2}$

For hydrogen atom, $𝑍=1$

$\therefore {𝑇}_{𝑛}=\frac{4{𝜀}_2^0{𝑛}^3{ℎ}^3}{𝑚{𝑒}^4}$

Dimensions of Stefan's constant,

$\begin{array}{cc}\left[𝜎\right]& =\frac{\left[𝑢\right]}{\left[𝐴\right][𝑇{]}^4}=\frac{\left[{\mathrm{ML}}^2{T}^{-2}\right]/\left[𝑇\right]}{\left[{𝐿}^2{K}^4\right]}\\ & =\left[{\mathrm{MT}}^{-3}{K}^{-4}\right]\end{array}$

Dimensions of Wien's constant,

$\begin{array}{cc}& \left[𝑏\right]=\left[𝜆\right]\left[𝑇\right]=\left[\mathrm{LK}\right]\\ & \therefore \text{ Dimensions of }\left[𝜎𝑏\right]=\left[{\mathrm{MT}}^{-3}{K}^{-4}\right]\left[\mathrm{LK}\right]\\ & =\left[{𝐿}^1{𝑀}^{\top }{𝑇}^{-3}{𝐾}^{-3}\right]\end{array}$

The fundamental frequency of a vibrating string is given by

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

where

• $𝐿$ is the vibrating length of the wire,

• $𝑇$ is the tension in the wire, and

• $𝜇$ is the mass per unit length.

When the string is stretched by a hanging mass, the tension $𝑇$ is due to the weight of the mass, i.e.,

$𝑇=𝑚𝑔,$

where $𝑔$ is the acceleration due to gravity.

At the poles, the acceleration due to gravity is slightly higher (${𝑔}_{\text{poles}}$) compared to the equator (${𝑔}_{\text{eq}}$), where ${𝑔}_{\text{eq}}<{𝑔}_{\text{poles}}$. Therefore:

At the poles, the tension is ${𝑇}_{\text{poles}}=𝑚{𝑔}_{\text{poles}}$,

At the equator, the tension is ${𝑇}_{\text{eq}}=𝑚{𝑔}_{\text{eq}}$.

Since the frequency is proportional to $\sqrt{\surd 𝑇}$, using the same wire length at the equator would yield a lower frequency:

${𝑓}_{\text{eq}}=\frac{1}{2𝐿}\sqrt{\surd \frac{𝑚{𝑔}_{\text{eq}}}{𝜇}}<{𝑓}_{\text{poles}}=\frac{1}{2𝐿}\sqrt{\surd \frac{𝑚{𝑔}_{\text{poles}}}{𝜇}}.$

To have resonance with the same tuning fork (which has the frequency ${𝑓}_{\text{poles}}$), we need to increase the fundamental frequency of the wire at the equator. One way to do this is by decreasing the vibrating length $𝐿$ because frequency is inversely proportional to length.

Let the new required vibrating length be ${𝐿}^{\prime }$. For resonance, we equate the frequency at the equator with that at the poles:

$\frac{1}{2{𝐿}^{\prime }}\sqrt{\surd \frac{𝑚{𝑔}_{\text{eq}}}{𝜇}}=\frac{1}{2𝐿}\sqrt{\surd \frac{𝑚{𝑔}_{\text{poles}}}{𝜇}}.$

Canceling common factors, we get:

$\frac{\sqrt{\surd {𝑔}_{\text{eq}}}}{{𝐿}^{\prime }}=\frac{\sqrt{\surd {𝑔}_{\text{poles}}}}{𝐿},$

which leads to

${𝐿}^{\prime }=𝐿\sqrt{\surd \frac{{𝑔}_{\text{eq}}}{{𝑔}_{\text{poles}}}}.$

Since ${𝑔}_{\text{eq}}<{𝑔}_{\text{poles}}$, it follows that

${𝐿}^{\prime }<𝐿.$

Thus, to resonate with the same tuning fork at the equator, the vibrating length of the wire must be decreased.

Answer: Option C – the vibrating length of the wire should be decreased.

The length of wire, which is bent in form of semicircular arc of radius $𝑟$ is

$\begin{array}{cc}& 𝐼=𝜋𝑟\\ & 𝑟=\frac{𝐼}{𝜋}\end{array}$

The diameter of semicircular wire,

$𝑑=2𝑟=\frac{2}{𝜋}$

$\therefore$ Magnetic moment of semicircular wire,

${𝑀}^{\prime }=𝑚\times 𝑑=𝑚\times \frac{2}{𝜋}\text{... (i)}$

The magnetic moment of straight wire,

$𝑀=𝑚𝑙$

Substituting in Eq. (i), we get

${𝑀}^{\prime }=\frac{2𝑀}{𝜋}$

Given, $𝐀=4\widehat{𝐢}+3\widehat{𝐣}+12\widehat{𝐤}$

$\begin{array}{cc}& \text{ Also, }𝐀={𝐀}_{𝑥}+{𝐀}_{𝑦}+{𝐀}_{𝑧}\\ & \Rightarrow {𝐴}_{𝑥}=4\end{array}$

The angle subtended by vector $𝐀$ with $𝑋$-axis is

$\begin{array}{cc}\cos 𝜙& =\frac{{𝐴}_{𝑥}}{\left|𝐴\right|}=\frac{4}{\sqrt{\surd 4^2+3^2+{12}^2}}\\ \cos 𝜙& =\frac{4}{13}\\ 𝜙& ={\cos }^{-1}\left(\frac{4}{13}\right)\end{array}$

The change in momentum of one ball,

$\Delta 𝑝=𝑚𝑢-(-𝑚𝑢)=2𝑚𝑢$

The force exerted on the wall by $𝑁$ balls,

$\begin{array}{cc}𝐹& =\frac{\text{ Change in momentum for }𝑁\text{ balls }}{\text{ Total time }\left(𝑡\right)}\\ & =\frac{2𝑚\mathrm{Nu}}{1}(\because 𝑡=1𝑠)\\ & =2𝑚\mathrm{Nu}\end{array}$

According to right hand thumb rule, the direction of magnetic field in ring $𝑃$, due to current in wire will be perpendicularly outward from plane of paper.

Using Lenz's law, when the current in the wire is increasing, the direction of induced current in ring $𝑃$ will be such that, it opposes the increasing current in wire. For this, the upper face of the ring acts like South pole and lower face acts like North pole. Thus, the direction of induced current in ring $𝑃$ is clockwise.

Similarly, the direction of induced current in ring $𝑄$ is anti-clockwise.

The equivalent resistance of the circuit,

${𝑅}_{\mathrm{eq}}=20+10=30\Omega$

The current flowing through the wire,

$𝑖=\frac{𝑉}{𝑅}=\frac{3V}{30\Omega }=0.1A$

The potential gradient along the wire,

$\begin{array}{cc}& =\frac{\text{ Potential difference across }20\Omega }{\text{ Length of }20\Omega \text{ wire }}\\ & =\frac{0.1\times 20\Omega }{10m}=0.20V/m\end{array}$

Given, number of atoms, $𝑁=8\times {10}^{10}$ atoms and dipole moment, $𝑀=9\times {10}^{-24}{\mathrm{Am}}^2$

The maximum dipole moment is given by

$\begin{array}{cc}{𝑀}_{max}& =𝑁𝑀\\ & =8\times {10}^{10}\times 9\times {10}^{-24}\\ & =7.2\times {10}^{-13}{\mathrm{Am}}^2\end{array}$

The volume of cubic domain,

$\begin{array}{cc}𝑉& ={𝐼}^3={\left({10}^{-6}\right)}^3[\because 𝑙=1𝜇m={10}^{-6}m]\\ & ={10}^{-18}{m}^3\end{array}$

The magnetisation of domain

$=\frac{{𝑀}_{max}}{𝑉}=\frac{7.2\times {10}^{-13}}{{10}^{-18}}=7.2\times {10}^5{\mathrm{Am}}^{-1}$

Given, surface tension, $𝑇=3\times {10}^{-2}N/m$ and surface area of soap film, $𝐴=20\mathrm{cm}\times 5\mathrm{cm}$

$\begin{array}{cc}& =100{\mathrm{cm}}^2\\ & =100\times {10}^{-4}{m}^2\end{array}$

The work done in forming a soap film, $𝑊=2\left(𝑇𝐴\right)$

$\begin{array}{cc}& =2(3\times {10}^{-2}\times 100\times {10}^{-4})\\ & =6\times {10}^{-4}J\end{array}$

The compressibility of a fluid, such as water, can be quantified using the formula:

$𝐾=\frac{-\Delta 𝑉}{\Delta 𝑝𝑉}$

Where:

$𝐾$ is the compressibility factor

$\Delta 𝑉$ is the change in volume

$\Delta 𝑝$ is the change in pressure

$𝑉$ is the original volume

Given:

Compressibility of water, $𝐾=5\times {10}^{-10}{\text{m}}^2/\text{N}$

Pressure change, $\Delta 𝑝=15\times {10}^6\text{Pa}$

Initial volume, $𝑉=100\text{mL}$

To find the change in volume ($\Delta 𝑉$) when the pressure is applied, we rearrange the formula as follows:

$\Delta 𝑉=-𝐾\Delta 𝑝𝑉$

Substituting the given values into the equation:

$\Delta 𝑉=-(5\times {10}^{-10}\times 15\times {10}^6\times 100)$

Calculating this yields:

$\Delta 𝑉=-0.75\text{mL}$

Thus, the volume of water decreases by 0.75 mL when the specified pressure is applied.

The maximum velocity of body in SHM,

$𝑣=𝐴𝜔$

where, $𝐴$ is amplitude of body and $𝜔$ is the angulat frequency.

It is given that, the maximum velocities of bodies are equal.

$\begin{array}{cc}{𝑣}_{𝐴}& ={𝑣}_{𝐵}\\ {𝐴}_{𝐴}{𝜔}_{𝐴}& ={𝐴}_{𝐵}{𝜔}_{𝐵}\\ {𝐴}_{𝐴}\sqrt{\surd \frac{{𝑘}_1}{𝑚}}& ={𝐴}_{𝐵}\sqrt{\surd \frac{{𝑘}_2}{𝑚}}(\because 𝜔=\sqrt{\surd \frac{𝑘}{𝑚}})\\ \frac{{𝐴}_{𝐴}}{{𝐴}_{𝐵}}& =\sqrt{\surd \frac{{𝑘}_2}{{𝑘}_1}}\end{array}$

The circuit diagram of given situation is as shown below

The equivalent capacitance of $2𝜇F$ and $4𝜇F$ capacitors connected in parallel is

${𝐶}_{\text{eq }}=2+4=6𝜇F$

The circuit now becomes

As, both the capacitors are of same capacitance, so potential of 12 V is equally divided in themi.e.

${𝑉}_1={𝑉}_2=6V$

In parallel combination, potential remains same.

$\therefore$ Charge on $2𝜇F,𝑄=2\times {𝑉}_1=2\times 6=12𝜇C$

The escape velocity of body from earth's surface is

${𝑣}_{𝑒}=\sqrt{\surd \frac{2𝐺𝑀}{𝑅}}$

where, $𝐺$ is universal gravitational constant, ${𝑀}^{\text{is }}$ mass of earth and $𝑅$ is the radius of earth.

For the planet, ${𝑀}_1=6𝑀$ and ${𝑅}_1=2𝑅$

$\therefore$ The escape velocity from the planet will be

${𝑣}_{𝑝}=\sqrt{\surd \frac{2𝐺\left(6𝑀\right)}{\left(2𝑅\right)}}=\sqrt{\surd 3}{𝑣}_{𝑒}$

Consider the figure shown below

In $△𝑃𝑄𝑅$,

$\begin{array}{cc}& 𝛼={180}^{\circ }-{90}^{\circ }-𝐴\\ & 𝛼={90}^{\circ }-𝐴\end{array}$

Also, $𝑟+𝛼={90}^{\circ }$

$\begin{array}{cc}\Rightarrow & 𝑟={90}^{\circ }-𝛼={90}^{\circ }-{90}^{\circ }+𝐴\\ \text{ or }& 𝑟=𝐴\end{array}$

Using Snell's law,

${𝜇}_1\sin 𝑖={𝜇}_2\sin 𝑟$

(1) $\sin 𝑖=𝑛\sin 𝐴$ (given, ${𝜇}_2=𝑛$ for glass prism)

$\Rightarrow \sin 𝑖=𝑛\sin 𝐴$

For small angle,

$\begin{array}{cc}& \sin 𝑖\simeq 𝑖\text{ and }\sin 𝐴\simeq 𝐴\\ \therefore & 𝑖=𝑛𝐴\end{array}$

The vector $(𝐴+𝐵)$ will be in plane containing vector $𝐴$ and vector $𝐵$.

Vector $(𝐀\times 𝐁)$ will be perpendicular to the plane containing vector $𝐴$ and $𝐵$.

Thus, the angle between $(𝐴+𝐵)$ and $(𝐴\times 𝐵)$ is ${90}^{\circ }$ or $\frac{𝜋}{2}\mathrm{rad}$.

Alternate solution

$\begin{array}{cc}& (𝐴+𝐵)\cdot (𝐴\times 𝐵)=|𝐴+𝐵||𝐴\times 𝐵|\cos 𝛼\\ & 𝐴\cdot (𝐴\times 𝐵)+𝐵\cdot (𝐴\times 𝐵)=|𝐴+𝐵||𝐴\times 𝐵|\cos 𝛼\\ & 0+0=|𝐴+𝐵||𝐴\times 𝐵|\cos 𝛼\\ & \Rightarrow \cos 𝛼=0\\ & 𝛼={90}^{\circ }\text{ or }\frac{𝜋}{2}\mathrm{rad}\end{array}$

Consider the free body diagram shown below,

The torque about centre of cylinder,

$𝜏=𝐹\times 𝑟=25\times 0.4=10\mathrm{Nm}$

The relation between angular acceleration ( $𝛼$ ) and torque $\left(𝜏\right)$ is given as

$\begin{array}{cc}𝜏& =𝑙𝛼\\ & =\frac{𝑀{𝑟}^2}{2}𝛼(\text{ for solid cylinder, }𝐼=\frac{𝑀{𝑅}^2}{2})\\ 10& =\frac{1\times (0.4{)}^2}{2}𝛼\Rightarrow 𝛼=125\mathrm{rad}/s^2\end{array}$

To determine the change in collector current ($\Delta {𝐼}_{𝐶}$) in a transistor connected in common emitter configuration, we can use the relation between the base current (${𝐼}_{𝐵}$), collector current (${𝐼}_{𝐶}$), and the current gain ($𝛽$).

Firstly, the relationship between $𝛼$ and $𝛽$ is given by: