MHT CET 2024 10th May Morning Shift (Physics & Chemistry) Question Paper with Solutions

In simple harmonic motion (S.H.M.), the maximum velocity (${𝑣}_{\text{max}}$) of a particle can be given by the formula:

${𝑣}_{\text{max}}=𝐴𝜔$

where $𝐴$ is the amplitude and $𝜔$ is the angular frequency. The angular frequency $𝜔$ can be related to the period $𝑇$ by:

$𝜔=\frac{2𝜋}{𝑇}$

Initially, the particle has an amplitude $𝐴$ and period $𝑇$, so the initial maximum velocity is:

$𝑉=𝐴\cdot \frac{2𝜋}{𝑇}$

Now, the amplitude is doubled to $2𝐴$ and the period becomes $\left(\frac{1}{3}\right)𝑇$. The new angular frequency ${𝜔}^{\prime }$ is:

${𝜔}^{\prime }=\frac{2𝜋}{(1/3)𝑇}=\frac{6𝜋}{𝑇}$

Thus, the new maximum velocity ${𝑣}_{\prime }^{\text{max}}$ is:

${𝑣}_{\prime }^{\text{max}}=\left(2𝐴\right)\cdot \frac{6𝜋}{𝑇}=12𝐴\cdot \frac{𝜋}{𝑇}$

Since the original maximum velocity $𝑉$ was:

$𝑉=𝐴\cdot \frac{2𝜋}{𝑇}$

Relating the new maximum velocity to the original, we get:

${𝑣}_{\prime }^{\text{max}}=6\cdot (𝐴\cdot \frac{2𝜋}{𝑇})=6𝑉$

Therefore, the maximum velocity, when the amplitude is doubled and the period is reduced to one-third, is:

Option A: 6V

The magnetic induction at the center of a circular arc can be determined using the formula:

$𝐵=\frac{{𝜇}_0𝐼}{4𝜋𝑟}\times 𝜃$

Given that the angle $𝜃$ subtended by the arc is $\frac{𝜋}{8}$, we substitute $𝜃$ in the formula:

$𝐵=\frac{{𝜇}_0𝐼}{4𝜋𝑟}\times \frac{𝜋}{8}$

Simplifying this expression gives:

$𝐵=\frac{{𝜇}_0𝐼}{32𝑟}$

Moment of Inertia $\left(I\right)=I_{\mathrm{CM}}+{\mathrm{Md}}^2$

where $𝑑$ is distance of axis from COM

The $I_{\mathrm{CM}}$ and M is same for all the points and only ' d ' is varying

As axis B is having more distance from COM and thus will have more moment of inertia.

When an inductor coil of inductance $𝐿$ is divided into two equal parts, each part will have an inductance of $\frac{𝐿}{2}$. When both parts are connected in parallel, the combined or net inductance ${𝐿}_{\text{net}}$ is found using the formula for inductors in parallel:

$\frac{1}{{𝐿}_{\text{net}}}=\frac{1}{{𝐿}_1}+\frac{1}{{𝐿}_2}$

In this case, since both inductors have the same inductance of $\frac{𝐿}{2}$:

$\frac{1}{{𝐿}_{\text{net}}}=\frac{1}{\frac{𝐿}{2}}+\frac{1}{\frac{𝐿}{2}}$

This simplifies to:

$\frac{1}{{𝐿}_{\text{net}}}=\frac{2}{𝐿}+\frac{2}{𝐿}=\frac{4}{𝐿}$

Thus, the net inductance is:

${𝐿}_{\text{net}}=\frac{𝐿}{4}$

Therefore, the correct choice is Option D: $\frac{𝐿}{4}$.

K.E. $=\frac{{𝑝}^2}{2𝑚}$

When momentum increases by $20\%$, new momentum is $p^{\prime }$

$\begin{array}{cc}{p}^{\prime }& =p+20\%p\\ & =p+0.2p=1.2p\end{array}$

New kinetic energy,

$\begin{array}{cc}K.E^{\prime }& =\frac{p^{\prime 2}}{2m}\\ & =\frac{(1.2p{)}^2}{2m}=\frac{1.44p^2}{2m}\\ \therefore K.E^{\prime }& =1.44K.E.\end{array}$

Increase in kinetic energy is calculated by change in kinetic energy.

$\begin{array}{cc}\Delta \mathrm{KE}& =(K\cdot E^{\prime }-K\cdot E)\\ & =(1.44K.E-K\cdot E)\\ & =0.44K.E.\\ & \Rightarrow 44\%\end{array}$

To achieve an effective capacitance of $\frac{6}{13}𝜇F$ using seven capacitors, each with a capacitance of $2𝜇F$, you should arrange them with 3 capacitors in parallel and 4 capacitors in series.

Here's how the calculation works:

Parallel Connection: When capacitors are connected in parallel, their capacitances add up. So for 3 capacitors in parallel:

${𝐶}_{\text{parallel}}=3\times 𝐶=3\times 2𝜇F=6𝜇F$

Series Connection: When capacitors are connected in series, the reciprocal of the total capacitance is the sum of the reciprocals of individual capacitances. For 4 capacitors, each of capacitance $𝐶$, in series, the total capacitance is:

$\frac{1}{{𝐶}_{\text{series}}}=\frac{1}{𝐶}+\frac{1}{𝐶}+\frac{1}{𝐶}+\frac{1}{𝐶}=\frac{4}{𝐶}$

Therefore:

${𝐶}_{\text{series}}=\frac{𝐶}{4}=\frac{2𝜇F}{4}=0.5𝜇F$

Combining Series and Parallel: For combining the parallel and series connections, where ${𝐶}_{\text{total}}$ is the effective capacitance:

$\frac{1}{{𝐶}_{\text{total}}}=\frac{1}{{𝐶}_{\text{parallel}}}+\frac{1}{{𝐶}_{\text{series}}}=\frac{1}{6}+\frac{1}{0.5}=\frac{1+12}{6}=\frac{13}{6}$

Consequently:

${𝐶}_{\text{total}}=\frac{6}{13}𝜇F$

Therefore, the required configuration is 3 capacitors in parallel followed by 4 capacitors in series to yield an overall capacitance of $\frac{6}{13}𝜇F$.

$\begin{array}{cc}& 𝑃={𝐸}_{𝑟𝑚𝑠}{𝐼}_{𝑟𝑚𝑠}\cos 𝜙\\ & \cos 𝜙=\frac{𝑅}{𝑍}\\ & \text{ Also, }{𝐼}_{𝑚𝑠}=\frac{{𝐸}_{𝑚𝑠}}{𝑍}=\frac{{𝐸}_0}{𝑍\sqrt{\surd 2}}\\ \therefore & 𝑃=\frac{{𝐸}_0}{\sqrt{\surd 2}}\times \frac{{𝐸}_0}{𝑍\sqrt{\surd 2}}\times \frac{𝑅}{𝑍}\\ & =\frac{{𝐸}_2^0𝑅}{2{𝑍}^2}\text{.... (i)}\\ & \text{ Given }{𝑋}_{𝐿}=𝑅\\ \therefore & 𝑍=\sqrt{\surd {𝑅}^2+{𝑅}^2}=\sqrt{\surd 2}𝑅\\ \therefore & 𝑃=\frac{{𝐸}_2^0}{4𝑅}\text{... [From (i)]}\end{array}$

$\begin{array}{cc}& {𝑣}_{𝑐}=\sqrt{\surd \frac{2𝐺𝑀}{𝑅}}\\ & {𝑣}_{𝑐}=𝑅\sqrt{\surd \frac{8}{3}𝜋𝐺𝜌}\dots .(\because 𝑀=\frac{4}{3}𝜋{𝑅}^3𝜌)\\ & \text{ Also, }{𝜌}_{𝑝}=4{𝜌}_{𝐸}\text{ and }{𝑅}_{𝑝}=4{𝑅}_{𝐸}\\ & \frac{{𝑣}_{𝑐}}{{𝑣}_{𝑝}}=\frac{{𝑅}_{𝐸}\sqrt{\surd {𝜌}_{𝐸}}}{4{𝑅}_{𝐸}\sqrt{\surd 4{𝜌}_{𝐸}}}=\frac{1}{8}=1:8\end{array}$

To determine the change in the rate of radiation when the radius and temperature of a black sphere are altered, consider these steps:

Initial Formula for Radiation:

The rate of radiation $𝐸$ is given by the formula:

$𝐸=𝑒𝐴𝜎{𝑇}^4$

where $𝑒$ is the emissivity, $𝐴$ is the surface area, $𝜎$ is the Stefan-Boltzmann constant, and $𝑇$ is the temperature.

Surface Area of a Sphere:

For a sphere, the surface area $𝐴$ is:

$𝐴=4𝜋{𝑅}^2$

Thus, the initial radiation rate is:

$𝐸=𝑒\left(4𝜋{𝑅}^2\right)𝜎{𝑇}^4$

Revised Conditions:

The sphere's radius is reduced to $\frac{𝑅}{2}$, and the temperature is increased to $4𝑇$.

New Radiation Formula:

With the new radius and temperature, let the new rate of radiation be ${𝐸}_2$. Substitute the new values into the radiation formula:

${𝐸}_2=𝑒\left(4𝜋{\left(\frac{𝑅}{2}\right)}^2\right)𝜎(4𝑇{)}^4$

Comparing Both Rates:

To find the ratio of the initial and new radiation rates:

$\frac{𝐸}{{𝐸}_2}=\frac{{𝑅}^2{𝑇}^4}{{\left(\frac{𝑅}{2}\right)}^2(4𝑇{)}^4}$

Calculation:

Simplify the ratio:

$\frac{𝐸}{{𝐸}_2}=\frac{{𝑅}^2{𝑇}^4}{\frac{{𝑅}^2}{4}\cdot 256{𝑇}^4}=\frac{1}{\frac{1}{4}\cdot 256}=\frac{1}{64}$

Therefore, ${𝐸}_2=64𝐸$.

From these calculations, when the radius is halved and the temperature is increased by four times, the rate of radiation increases to $64$ times the original rate.

Minimum energy required to excite from ground state,

$E_{min}=13.6[\frac{1}{1^2}-\frac{1}{2^2}]=10.2\mathrm{eV}$

$\begin{array}{cc}& \text{ According to Bernoulli’s equation, }\\ & \begin{array}{cc}& {𝑃}_1+\frac{1}{2}𝜌{𝑣}_2^1={𝑃}_2+\frac{1}{2}𝜌{𝑣}_2^2\\ \therefore & {𝑣}_2^2=\frac{2({𝑃}_1-{𝑃}_2)}{𝜌}[\because {𝑣}_1=0]\\ & {𝑣}_2=\sqrt{\surd \frac{2({𝑃}_1-{𝑃}_2)}{𝜌}}\end{array}\end{array}$

In the context of a single-slit diffraction pattern, the first minimum occurs when the light of wavelength $5000\text{Å}$ is diffracted at an angle of ${30}^{\circ }$. This scenario arises from the relationship for minima in single-slit diffraction as described by:

$𝑎\sin {𝜃}_{𝑛}=𝑛𝜆$

For the first minimum ($𝑛=1$), we can express it as:

$𝑎\sin {30}^{\circ }=𝜆\text{(Equation i)}$

Given the value of $\sin {30}^{\circ }=\frac{1}{2}$, we substitute to get:

$𝑎\cdot \frac{1}{2}=𝜆$

Next, to find the angle for the first secondary maximum, we consider:

$𝑎\sin {𝜃}_{𝑛}=(2𝑛+1)\frac{𝜆}{2}$

For the second secondary maximum ($𝑛=1$), the equation becomes:

$𝑎\sin {𝜃}_{𝑛}=(2\cdot 1+1)\frac{𝜆}{2}=\frac{3𝜆}{2}\text{(Equation ii)}$

By dividing Equation i by Equation ii, we get:

$\frac{\frac{1}{2}}{\sin {𝜃}_{𝑛}}=\frac{2}{3}$

Solving for $\sin {𝜃}_{𝑛}$, we find:

$\sin {𝜃}_{𝑛}=\frac{3}{4}$

Thus, the angle ${𝜃}_{𝑛}$ for the first secondary maximum is:

${𝜃}_{𝑛}={\sin }^{-1}\left(\frac{3}{4}\right)$

To understand the relationship between the angle of incidence, the refractive index, and the angle of deviation for a ray of light passing through a medium, we start with the following equations:

$𝛿=𝑖+𝑟$: The angle of deviation ($𝛿$) is the sum of the angle of incidence ($𝑖$) and the angle of refraction ($𝑟$).

$𝜇=\frac{\sin 𝑖}{\sin 𝑟}$: Snell's Law, where $𝜇$ is the refractive index of the medium.

To express $𝜇$ in terms of the angle of deviation, we derive that:

$𝜇=\frac{\sin 𝑖}{\sin (𝑖-𝛿)}$

We then rearrange to find:

$\frac{1}{𝜇}=\frac{\sin (𝑖-𝛿)}{\sin 𝑖}$

Using the trigonometric identity for $\sin (𝑖-𝛿)$, which is $\sin 𝑖\cos 𝛿-\cos 𝑖\sin 𝛿$, we can write:

$\frac{1}{𝜇}=\frac{\sin 𝑖\cos 𝛿-\cos 𝑖\sin 𝛿}{\sin 𝑖}$

Simplifying further, we get:

$\frac{1}{𝜇}=\cos 𝛿-\frac{\sin 𝛿}{\tan 𝑖}$

This equation reveals how the inverse of the refractive index is related to the angles involved through cosine and sine functions, as well as the tangent of the angle of incidence.

For an air column vibrating in the fundamental mode:

For a pipe open at both ends, the fundamental wavelength is given by:

${𝜆}_1=2𝐿$

Therefore, the frequency is:

${𝑛}_1=\frac{𝑣}{{𝜆}_1}=\frac{𝑣}{2𝐿}$

For a pipe closed at one end (and open at the other), the fundamental mode has:

${𝜆}_2=4𝐿$

Thus, the fundamental frequency is:

${𝑛}_2=\frac{𝑣}{{𝜆}_2}=\frac{𝑣}{4𝐿}$

Comparing the two frequencies:

$\frac{{𝑛}_1}{{𝑛}_2}=\frac{\frac{𝑣}{2𝐿}}{\frac{𝑣}{4𝐿}}=2$

This gives:

${𝑛}_1=2{𝑛}_2$

Thus, the correct answer is Option A.

In given circuit, the diode $D_2$ is connected in reverse bias. Hence, no current flows through $40\Omega$ resistor.

$\therefore I=\frac{V}{R}=\frac{10}{40}=0.25A$

In the biprism experiment, if the 5th bright band with wavelength ${𝜆}_1$ coincides with the 6th dark band with wavelength ${𝜆}_2$, we can find the ratio $\frac{{𝜆}_1}{{𝜆}_2}$ using the following equations:

Given that the position of the 5th bright band for ${𝜆}_1$ is the same as the position of the 6th dark band for ${𝜆}_2$:

${𝑦}_5={𝑦}_{\prime }^6$

The position of the bright band is given by:

${\left(\frac{𝑛𝜆𝐷}{𝑑}\right)}_5={\left[\frac{(2𝑛-1)𝜆𝐷}{2𝑑}\right]}_6$

Substituting the values for the 5th bright band and 6th dark band:

$\frac{5{𝜆}_1𝐷}{𝑑}=\frac{11{𝜆}_2𝐷}{2𝑑}$

By solving this equation, we find:

$\frac{{𝜆}_1}{{𝜆}_2}=\frac{11}{10}$

To solve the problem, we start by noting the given information:

The difference in the wavelengths at which bodies P and Q radiate maximum energy is $3𝜇\text{m}$. This can be expressed as:

${𝜆}_Q-{𝜆}_P=3𝜇\text{m}\text{.... (i)}$

According to Wien's law, the relationship between the wavelength of maximum emission ($𝜆$) and temperature ($𝑇$) for a black body is given by:

${𝜆}_{𝑃}{𝑇}_{𝑃}={𝜆}_{𝑄}{𝑇}_{𝑄}$

We are informed that the temperature ${𝑇}_{𝑃}$ of body P is four times that of body Q (${𝑇}_{𝑄}$), therefore:

${𝜆}_{𝑃}\cdot 4{𝑇}_{𝑄}={𝜆}_{𝑄}\cdot {𝑇}_{𝑄}$

Simplifying the equation above gives us:

${𝜆}_{𝑄}=4{𝜆}_{𝑃}\text{.... (ii)}$

Using equation (ii), we substitute back into equation (i):

Substitute ${𝜆}_{𝑄}=4{𝜆}_{𝑃}$ into ${𝜆}_{𝑄}-{𝜆}_{𝑃}=3𝜇\text{m}$:

$4{𝜆}_{𝑃}-{𝜆}_{𝑃}=3𝜇\text{m}$

Solving for ${𝜆}_{𝑃}$:

$3{𝜆}_{𝑃}=3𝜇\text{m}⟹{𝜆}_{𝑃}=1𝜇\text{m}$

Finally, using ${𝜆}_{𝑃}=1𝜇\text{m}$ in the expression ${𝜆}_{𝑄}=4{𝜆}_{𝑃}$:

Solve for ${𝜆}_{𝑄}$:

${𝜆}_{𝑄}=4\times 1𝜇\text{m}=4𝜇\text{m}$

Thus, the wavelength at which body Q radiates maximum energy is $4𝜇\text{m}$.

The kinetic energy of the bullet before it strikes the wall can be expressed as:

${𝐸}_{𝑘}=\frac{1}{2}𝑚{𝑉}^2$

where $𝑚$ is the mass of the bullet and $𝑉$ is the velocity.

Since $75\%$ of the kinetic energy is converted into heat, the heat energy produced is:

$𝑄=0.75\times \frac{1}{2}𝑚{𝑉}^2=\frac{3}{8}𝑚{𝑉}^2$

According to the mechanical equivalent of heat, the relationship between heat energy and temperature change is given by:

$𝑄=𝑚𝑠\Delta 𝑇$

where $𝑠$ is the specific heat capacity of lead, and $\Delta 𝑇$ is the change in temperature.

Substitute the expression for $𝑄$:

$\frac{3}{8}𝑚{𝑉}^2=𝑚𝑠\Delta 𝑇$

Divide both sides by $𝑚𝑠$:

$\Delta 𝑇=\frac{3{𝑉}^2}{8𝑠}$

Since the conversion from mechanical energy to heat energy involves the mechanical equivalent of heat $𝐽$, the formula should be adjusted to:

$\Delta 𝑇=\frac{3{𝑉}^2}{8𝐽𝑠}$

This matches with the answer option A:

$\overline{)\frac{3V^2}{8\mathrm{Js}}}$

Photocurrent is independent of the frequency of photon. However when intensity of incident photon increases number of electrons emitted from the surface also increases, thereby increasing photocurrent.

The average force applied on the wall of a closed container can be related to the temperature of an ideal gas using its dependency on temperature $T^x$. To find the value of $𝑥$, we can derive it step by step:

The pressure $𝑃$ is defined as force $𝐹$ per unit area $𝐴$:

$𝑃=\frac{𝐹}{𝐴}$

Therefore, pressure is proportional to force:

$𝑃\propto 𝐹\text{.... (i)}$

According to the ideal gas law, the pressure $𝑃$ is also given by:

$𝑃=\frac{𝑛𝑅𝑇}{𝑉}$

where $𝑛$ is the number of moles, $𝑅$ is the ideal gas constant, $𝑇$ is the temperature, and $𝑉$ is the volume. If the volume $𝑉$ is constant:

$𝑃\propto 𝑇\text{.... (ii)}$

Combining equations (i) and (ii), we find the relationship between force and temperature:

$𝐹\propto {𝑇}^{𝑥}$

From the proportionality derived:

$𝑥=1$

${\mathrm{DP}}^2=(2L{)}^2+L^2=4L^2+L^2=5L^2$

$\begin{array}{cc}& \begin{array}{cc}\therefore \mathrm{DP}& =\sqrt{\surd 5}L=\mathrm{CP}\\ \mathrm{AP}& =\mathrm{BP}=L\end{array}\\ & \therefore \text{ Potential at point }P\text{ due to the four charges, }\end{array}$

$\begin{array}{cc}𝑉& =\frac{1}{4𝜋{𝜀}_0}(\frac{𝑞}{𝐿}+\frac{𝑞}{𝐿}-\frac{𝑞}{\sqrt{\surd 5}𝐿}-\frac{𝑞}{\sqrt{\surd 5}𝐿})\\ & =\frac{1}{4𝜋{𝜀}_0}(\frac{2𝑞}{𝐿}-\frac{2𝑞}{\sqrt{\surd 5}𝐿})\\ & =\frac{1}{4𝜋{𝜀}_0}\cdot \frac{2𝑞}{𝐿}(1-\frac{1}{\sqrt{\surd 5}})\end{array}$

When an electron with mass $𝑚$ and charge $𝑞$ travels in a circular path of radius $𝑅$ at a right angle to a uniform magnetic field $𝐵$ with speed $𝑣$, the radius of the path is determined by the balance between the magnetic force and the centripetal force. This relationship is given by:

$𝑞𝑣𝐵\sin {90}^{\circ }=\frac{𝑚{𝑣}^2}{𝑅}$

Thus, the radius $𝑅$ can be expressed as:

$𝑅=\frac{𝑚𝑣}{𝑞𝐵}\text{(Equation 1)}$

Now, suppose the speed of the electron is halved, i.e., ${𝑣}^{\prime }=\frac{𝑣}{2}$, and the magnetic field is doubled, i.e., ${𝐵}^{\prime }=2𝐵$.

Using these new values, the new radius ${𝑅}^{\prime }$ can be calculated as follows:

${𝑅}^{\prime }=\frac{𝑚{𝑣}^{\prime }}{𝑞{𝐵}^{\prime }}=\frac{𝑚\left(\frac{𝑣}{2}\right)}{𝑞\left(2𝐵\right)}=\frac{1}{4}\frac{𝑚𝑣}{𝑞𝐵}$

From Equation 1, we know that $𝑅=\frac{𝑚𝑣}{𝑞𝐵}$, so substituting this back, we get:

${𝑅}^{\prime }=\frac{1}{4}𝑅$

Thus, the resulting path has a radius $\frac{1}{4}$ of the original radius $𝑅$.

$\begin{array}{cc}& \left|{𝑒}_{𝑠}\right|=𝑀\frac{𝑑{𝐼}_{𝑝}}{𝑑𝑡}\\ & \left|{𝑒}_{𝑠}\right|=𝑀\frac{𝑑}{𝑑𝑡}{𝐼}_0\sin 𝜔𝑡\\ & \left|{𝑒}_{𝑠}\right|=𝑀{𝐼}_0𝜔\cos 𝜔𝑡\\ \therefore & {\left|{𝑒}_{𝑠}\right|}_{max}=𝑀{𝐼}_0𝜔=0.003\times 8\times 100𝜋\times 1\\ \therefore & {\left|{𝑒}_{𝑠}\right|}_{max}=\left(2.4𝜋\right)\text{ volt }\end{array}$

To find the beat frequency and maximum intensity produced by the interference of the two sound waves, use the following concepts:

Beat Frequency: The beat frequency is given by the absolute difference between the frequencies of the two interfering waves.

Maximum Intensity: This is determined by the sum of the amplitudes of the two waves when they interfere constructively.

Solution

Identify Frequencies:

The standard form of a wave is $𝑥=𝐴\sin \left(2𝜋𝑓𝑡\right)$, where $𝑓$ is the frequency in Hz.

For the wave ${𝑥}_1=2\sin \left(1000𝜋𝑡\right)$:

The angular frequency is $1000𝜋$.

Frequency ${𝑓}_1=\frac{1000𝜋}{2𝜋}=500\text{Hz}$.

For the wave ${𝑥}_2=3\sin \left(1006𝜋𝑡\right)$:

The angular frequency is $1006𝜋$.

Frequency ${𝑓}_2=\frac{1006𝜋}{2𝜋}=503\text{Hz}$.

Calculate Beat Frequency:

The beat frequency ${𝑓}_{\text{beat}}$ is the difference between the two frequencies:

${𝑓}_{\text{beat}}=|{𝑓}_2-{𝑓}_1|=|503-500|=3\text{Hz}$

Calculate Maximum Intensity:

When two waves interfere constructively, the maximum amplitude is the sum of the amplitudes.

Amplitude of ${𝑥}_1=2$ and amplitude of ${𝑥}_2=3$.

Maximum amplitude = $2+3=5$.

The intensity (which is proportional to the square of the amplitude) is given by:

$\text{Maximum Intensity}=(2+3{)}^2=5^2=25\text{units}$

Conclusion

Thus, the waves produce a beat frequency of $3\text{Hz}$ with a maximum intensity of $25\text{units}$.

Correct Option: C

3 beats/s with maximum intensity 25 units

To determine the velocity of water flowing out of the hole at the bottom of the tank, we use Torricelli’s theorem, which is derived from Bernoulli’s principle. According to Torricelli's theorem, the speed $𝑣$ of efflux of a fluid under the force of gravity through a small hole at depth $ℎ$ is given by:

$𝑣=\sqrt{\surd 2𝑔ℎ}$

Here, $𝑔$ is the acceleration due to gravity and $ℎ$ is the height of the fluid column above the hole.

Given:

The total pressure at the bottom of the tank is $4𝐻$.

The atmospheric pressure is $𝐻$.

The pressure exerted by the water column is the difference between the total pressure at the bottom and the atmospheric pressure:

${𝑃}_{\text{water}}=4𝐻-𝐻=3𝐻$

Now, equate the pressure due to the water column to $𝜌𝑔ℎ$:

$𝜌𝑔ℎ=3𝐻$

Solving for $𝑔ℎ$, we find:

$𝑔ℎ=\frac{3𝐻}{𝜌}$

Substituting this into Torricelli's theorem gives:

$𝑣=\sqrt{\surd 2𝑔ℎ}=\sqrt{\surd 2\times \frac{3𝐻}{𝜌}}=\sqrt{\surd \frac{6𝐻}{𝜌}}$

Thus, the velocity of water flowing out of the hole is:

Option C

$\sqrt{\surd \frac{6𝐻}{𝜌}}$

In a series resonant circuit, the resonant frequency is given by the formula:

$𝑓=\frac{1}{2𝜋\sqrt{\surd 𝐿𝐶}}$

When the inductor $𝐿$ is changed to $3𝐿$, and the capacitor $𝐶$ is changed to $6𝐶$, the new resonant frequency ${𝑓}^{\prime }$ becomes:

${𝑓}^{\prime }=\frac{1}{2𝜋\sqrt{\surd \left(3𝐿\right)\left(6𝐶\right)}}$

Simplifying the expression inside the square root:

${𝑓}^{\prime }=\frac{1}{2𝜋\sqrt{\surd 18𝐿𝐶}}$

This can be further simplified to:

${𝑓}^{\prime }=\frac{1}{2𝜋\sqrt{\surd (9\times 2)𝐿𝐶}}=\frac{1}{2𝜋\cdot 3\cdot \sqrt{\surd 2𝐿𝐶}}$

This simplifies to:

${𝑓}^{\prime }=\frac{1}{3\sqrt{\surd 2}}\cdot \frac{1}{2𝜋\sqrt{\surd 𝐿𝐶}}$

Given that the original resonant frequency $𝑓$ is:

$𝑓=\frac{1}{2𝜋\sqrt{\surd 𝐿𝐶}}$

The new frequency ${𝑓}^{\prime }$ can be expressed in terms of $𝑓$ as:

${𝑓}^{\prime }=\frac{𝑓}{3\sqrt{\surd 2}}$

Thus, the resonant frequency will become:

Option D: $\frac{f}{3\sqrt{\surd 2}}$

$\begin{array}{cc}& T=2𝜋\sqrt{\surd \frac{𝑙}{𝑔}}\\ & \text{ As the periodic time increases by }20\%\text{, }\\ & \begin{array}{cc}& T_2=T_1+\frac{20}{100}=\frac{120}{100}{T}_1\\ & \frac{T_2}{T_1}=\sqrt{\surd \frac{{𝑙}_2}{{𝑙}_1}}\\ & {\left(\frac{T_2}{T_1}\right)}^2=\frac{{𝑙}_2}{{𝑙}_1}\\ & {\left(\frac{120T_1}{100\times T_1}\right)}^2=\frac{{𝑙}_2}{{𝑙}_1}\\ & \frac{36}{25}=\frac{{𝑙}_2}{{𝑙}_1}\\ & \frac{{𝑙}_2}{{𝑙}_1}=1.44\end{array}\end{array}$

Let's start by writing down the Doppler effect formulas for both cases:

When the observer moves toward the stationary source with speed ${𝑉}_1$, the apparent frequency is:

${𝐹}_1=𝑓\times \frac{𝑉+{𝑉}_1}{𝑉},$

where $𝑓$ is the source frequency and $𝑉$ is the speed of sound.

When the observer moves away from the source with speed ${𝑉}_1$, the frequency becomes:

${𝐹}_2=𝑓\times \frac{𝑉-{𝑉}_1}{𝑉}.$

The problem tells us that:

$\frac{{𝐹}_1}{{𝐹}_2}=2.$

Substitute the expressions for ${𝐹}_1$ and ${𝐹}_2$:

$\frac{{𝐹}_1}{{𝐹}_2}=\frac{𝑓\times \frac{𝑉+{𝑉}_1}{𝑉}}{𝑓\times \frac{𝑉-{𝑉}_1}{𝑉}}=\frac{𝑉+{𝑉}_1}{𝑉-{𝑉}_1}=2.$

Now, solve for $𝑉$ in terms of ${𝑉}_1$:

Write the equation:

$\frac{𝑉+{𝑉}_1}{𝑉-{𝑉}_1}=2.$

Cross-multiply:

$𝑉+{𝑉}_1=2(𝑉-{𝑉}_1).$

Expand the right-hand side:

$𝑉+{𝑉}_1=2𝑉-2{𝑉}_1.$

Rearrange the equation to isolate terms with $𝑉$ and ${𝑉}_1$:

$𝑉+{𝑉}_1-2𝑉+2{𝑉}_1=0,$

$-𝑉+3{𝑉}_1=0.$

Solve for $𝑉$:

$𝑉=3{𝑉}_1.$

Thus, the ratio is:

$\frac{𝑉}{{𝑉}_1}=3.$

So, the correct answer is Option B.

In an isochoric process, the volume of the system remains constant, that is, $\Delta 𝑉=0$.

On a pressure-volume ($𝑃\text{--}𝑉$) diagram where volume is plotted along the horizontal axis, a process at constant volume is represented by a straight line parallel to the pressure axis (a vertical line).

Among the given plots, graph (B) represents a constant-volume process.

Hence, the correct option is (D).

Total $(K.E{)}_{\text{ring }}={\mathrm{Mv}}^2\text{.... (i)}$

Total (K.E) ${)}_{\text{disc }}=\frac{3}{4}{\mathrm{Mv}}^2\text{.... (ii)}$

$\begin{array}{cc}& \text{ Dividing equation (ii) by equation (i) }\\ & \begin{array}{cc}& \frac{(K.E{)}_{\text{disc }}}{(K.E{)}_{\text{ring }}}=\frac{\frac{3}{4}{\mathrm{Mv}}^2}{{\mathrm{Mv}}^2}\\ \therefore & (K.E{)}_{\text{disc }}=(K.E{)}_{\text{ring }}\times \frac{3}{4}=6\times \frac{3}{4}=\frac{9}{2}J\end{array}\end{array}$

In a potentiometer experiment, when cells with electromotive forces (emfs) ${𝐸}_1$ and ${𝐸}_2$ are connected in series $({𝐸}_1>{𝐸}_2)$, the balancing length of the wire is 80 cm. If the polarity of ${𝐸}_2$ is reversed, the balancing length changes to 20 cm. To find the ratio $\frac{{𝐸}_1}{{𝐸}_2}$, we use the following formula for potentiometer readings:

$\frac{{𝐸}_1}{{𝐸}_2}=\frac{{𝑙}_1+{𝑙}_2}{{𝑙}_1-{𝑙}_2}$

Plugging in the given values:

$\frac{{𝐸}_1}{{𝐸}_2}=\frac{80+20}{80-20}=\frac{100}{60}=\frac{5}{3}$

Thus, the ratio $\frac{{𝐸}_1}{{𝐸}_2}$ is $5:3$.

M.I. of the solid sphere about a diameter

$I=\frac{2}{5}{\mathrm{MR}}^2\text{.... (i)}$

M.I. of the disc about an axis through its edge and perpendicular to its plane is

$I=\frac{3}{2}{\mathrm{Mr}}^2\text{... (ii)}$

$\begin{array}{cc}\therefore & \frac{2}{5}𝑀{𝑅}^2=\frac{3}{2}𝑀{𝑟}^2\text{.... [From (i) and (ii)]}\\ \therefore & 𝑟=\frac{2}{\sqrt{\surd 15}}R\end{array}$

In Young's double-slit experiment, the position of the nth maximum for a wavelength ${𝜆}_1$ is given by:

${𝑦}_1=\frac{𝑛{𝜆}_1𝐷}{𝑑}\text{(Equation 1)}$

where $𝐷$ is the distance from the slits to the screen, and $𝑑$ is the separation between the slits.

When the wavelength is changed to ${𝜆}_2$, the position of the ${\left(\frac{𝑛}{3}\right)}^{\text{th}}$ maximum is given by:

${𝑦}_2=\frac{\frac{𝑛}{3}{𝜆}_2𝐷}{𝑑}\text{(Equation 2)}$

To find the ratio $\frac{{𝑦}_1}{{𝑦}_2}$, substitute the expressions for ${𝑦}_1$ and ${𝑦}_2$ from Equations 1 and 2:

$\frac{{𝑦}_1}{{𝑦}_2}=\frac{\frac{𝑛{𝜆}_1𝐷}{𝑑}}{\frac{\frac{𝑛}{3}{𝜆}_2𝐷}{𝑑}}$

Simplifying this expression:

$\frac{{𝑦}_1}{{𝑦}_2}=\frac{3{𝜆}_1}{{𝜆}_2}$

Thus, the ratio $\frac{{𝑦}_1}{{𝑦}_2}$ is $\frac{3{𝜆}_1}{{𝜆}_2}$.

Answer: Option B

Explanation:

In uniform circular motion, the speed (magnitude of the linear velocity) remains constant, but the direction of the velocity continuously changes. The linear velocity at any point along the circular path is always directed tangentially to the circle. Since the speed is constant, its magnitude does not change.

Key points to note:

Linear velocity:

For a particle moving in a circle, the instantaneous linear velocity vector is directed along the tangent to the circle at the particle’s position. The magnitude of this velocity remains constant in uniform circular motion.

Linear acceleration (centripetal acceleration):

Although not asked directly in the correct option, the acceleration in uniform circular motion is directed radially inward (towards the center of the circle), not tangentially or along any axis perpendicular to the plane of the circle.

Thus, the only correct statement given the options is that the linear velocity is always tangential to the circular path and its magnitude remains unchanged in uniform circular motion.