MHT CET 2025 20th April Evening Shift (Physics & Chemistry) Question Paper with Solutions

The terminal velocity is given by,

$\begin{array}{cc}& V_T=\frac{2}{9}\frac{(𝜌-𝜎)r^{2}g}{𝜂}\\ \therefore & \frac{v_{T_1}}{v_{T_2}}=\frac{r_2^1}{r_2^2}\\ \therefore & \frac{r_1}{r_2}=\frac{3}{2}\end{array}\Rightarrow \frac{r_2^1}{r_2^2}=\frac{9}{4}$

Now, Volume of the drop will be:

$𝑉=\frac{4}{3}𝜋{𝑟}^3$

$\therefore$ Ratio of Volumes:

$\frac{{𝑉}_1}{{𝑉}_2}={\left(\frac{{𝑟}_1}{{𝑟}_2}\right)}^3={\left(\frac{3}{2}\right)}^3=\frac{27}{8}=27:8$

$\begin{array}{cc}& \text{ Since, }{V}_L=V_C\text{ it is a resistive circuit }\\ & \therefore 𝑉=300V\text{ and }𝐼=\frac{𝑉}{𝑅}=\frac{300}{50}=6A\end{array}$

Let the initial angular velocity be ${𝜔}_0=0$ (since the wheel is initially at rest).

Let the angular acceleration be $𝛼$.

Angular displacement in time $𝑡$ is:

$𝜃={𝜔}_0𝑡+\frac{1}{2}𝛼{𝑡}^2$

First 2 seconds ($𝑡=2$ s)

${𝜃}_1=0\cdot 2+\frac{1}{2}𝛼(2{)}^2=\frac{1}{2}𝛼\cdot 4=2𝛼$

Next 2 seconds ($𝑡=4$ s total, $𝑡=2$ s extra)

Total angle covered in 4 seconds:

${𝜃}_{\text{total}}=0\cdot 4+\frac{1}{2}𝛼(4{)}^2=\frac{1}{2}𝛼\cdot 16=8𝛼$

So, angle covered in the next 2 seconds (${𝜃}_2$):

${𝜃}_2={𝜃}_{\text{total}}-{𝜃}_1=8𝛼-2𝛼=6𝛼$

Ratio ${𝜃}_1:{𝜃}_2$

${𝜃}_1:{𝜃}_2=2𝛼:6𝛼=1:3$

Correct option: D

Let mass of the cube be $𝑚$ and length of the side be $𝑎$.

Density, $𝜌$, is given by:

$𝜌=\frac{𝑚}{{𝑎}^3}$

Let the percentage error in mass be $\Delta 𝑚\%=5\%$ and in length be $\Delta 𝑎\%=6\%$.

According to error propagation rules (in products and powers):

$\frac{\Delta 𝜌}{𝜌}\times 100=(\frac{\Delta 𝑚}{𝑚}+3\frac{\Delta 𝑎}{𝑎})\times 100$

But since we are given the percentage errors directly, sum them (with powers as multipliers):

$\text{Total percentage error in }𝜌=\text{percentage error in }𝑚+3\times \text{percentage error in }𝑎$

Substitute the values:

$\text{Total percentage error}=5\%+3\times 6\%$

$\text{Total percentage error}=5\%+18\%=23\%$

Correct option:

Option B: $23\%$

As the equation of displacement is given in terms of cosine, the particle starts from the extreme position.

Comparing the given equation with,

$\begin{array}{cc}& 𝑥=𝐴\cos \left(𝜔𝑡\right),\\ & 𝜔=2𝜋\\ \therefore & \frac{2𝜋}{𝑇}=2𝜋\\ \therefore & 𝑇=1𝑠.\end{array}$

Now, velocity of the particle performing S.H.M. is maximum at the mean position. Thus, starting from extreme position, the time taken to reach the mean position for the first time is $\frac{T}{4}$.

i.e., $\frac{1}{4}$ seconds.

If a gas is compressed isothermally, its temperature remains constant.

The root mean square (r.m.s.) velocity of gas molecules is given by:

${𝑣}_{𝑟𝑚𝑠}=\sqrt{\surd \frac{3𝑘𝑇}{𝑚}}$

where:

• $𝑘$ = Boltzmann constant,

• $𝑇$ = absolute temperature,

• $𝑚$ = mass of one molecule.

Since the r.m.s. velocity depends only on temperature $𝑇$, and isothermal means temperature does not change, the r.m.s. velocity remains the same.

Correct answer: Option C - remains the same.

$\begin{array}{cc}\overrightarrow{𝑟}& =\overrightarrow{𝑎}+\overrightarrow{𝑏}+\overrightarrow{𝑐}=4\widehat{𝑖}-\widehat{𝑗}-3\widehat{𝑖}+2\widehat{𝑗}-\widehat{𝑘}\\ & =\widehat{𝑖}+\widehat{𝑗}-\widehat{𝑘}\\ \widehat{𝑟}& =\frac{\overrightarrow{𝑟}}{\left|𝑟\right|}=\frac{\widehat{𝑖}+\widehat{𝑗}-\widehat{𝑘}}{\sqrt{\surd 1^2+1^2+(-1{)}^2}}=\frac{\widehat{𝑖}+\widehat{𝑗}-\widehat{𝑘}}{\sqrt{\surd 3}}\end{array}$

According to Malus' law,

$𝐼={𝐼}_0{\cos }^2𝜃$

When beam passed through first polaroid,

$I=\frac{I_0}{2}$

When beam passed through second polaroid,

$\begin{array}{cc}& I_2=\frac{I_0}{2}{\cos }^2{30}^{\circ }\\ \therefore & I_2=\frac{3I_0}{8}\end{array}$

When beam passed through third polaroid,

${𝐼}_3=\frac{3{𝐼}_0}{8}{\cos }^2{30}^{\circ }=\frac{3{𝐼}_0}{8}\times {\left(\frac{\sqrt{\surd 3}}{2}\right)}^2=\frac{9{𝐼}_0}{32}$

When beam passed through fourth polaroid,

${𝐼}_4=\frac{9{𝐼}_0}{32}{\cos }^2{30}^{\circ }=\frac{9{𝐼}_0}{32}\times {\left(\frac{\sqrt{\surd 3}}{2}\right)}^2=\frac{27{𝐼}_0}{128}$

For body B moving with acceleration a, initial velocity is zero and final velocity is $𝑢$.

$\therefore {𝑢}^2=2𝑎𝑠\Rightarrow 𝑠=\frac{{𝑢}^2}{2𝑎}$

If the time taken to attain this velocity is $𝑡$, then $u=\mathrm{at}\Rightarrow t=\frac{u}{a}$

For body A, distance travelled is given by,

${𝑠}^{\prime }=𝑢𝑡=𝑢\times \frac{𝑢}{𝑎}=\frac{{𝑢}^2}{𝑎}$

Hence distance between $𝐴$ and $𝐵$ is

$s^{\prime }-s=\frac{u^2}{a}-\frac{u^2}{2a}=\frac{u^2}{2a}$

For hydrogen atom, according to Bohr’s model:

Step 1: Angular momentum in the ${𝑛}^{\text{th}}$ orbit

$𝐿=𝑚{𝑣}_{𝑛}{𝑟}_{𝑛}=𝑛\frac{ℎ}{2𝜋}$

Step 2: Velocity of electron in the ${𝑛}^{\text{th}}$ orbit

The velocity in the ${𝑛}^{\text{th}}$ orbit is:

${𝑣}_{𝑛}=\frac{2𝜋𝑘{𝑒}^2}{𝑛ℎ}$

where $𝑘$ is Coulomb's constant and $𝑒$ is the electron charge.

Step 3: Ratio of angular momentum to velocity

We want:

$\frac{𝐿}{{𝑣}_{𝑛}}$

From above,

• $𝐿=𝑛\frac{ℎ}{2𝜋}$

• ${𝑣}_{𝑛}=\frac{2𝜋𝑘{𝑒}^2}{𝑛ℎ}$

So,

$\frac{𝐿}{{𝑣}_{𝑛}}=\frac{𝑛\frac{ℎ}{2𝜋}}{\frac{2𝜋𝑘{𝑒}^2}{𝑛ℎ}}=𝑛\frac{ℎ}{2𝜋}\cdot \frac{𝑛ℎ}{2𝜋𝑘{𝑒}^2}=\frac{{𝑛}^2{ℎ}^2}{4{𝜋}^2𝑘{𝑒}^2}$

This ratio is proportional to ${𝑛}^2$.

Final Answer:

Option A: $\overline{){𝑛}^2}$

Given:

• Element $\stackrel{\to \leftarrow }{\Delta 𝑙}=\Delta 𝑥\widehat{𝑖}$ at the origin,

• Current, $𝐼=10\text{A}$

• Position for the magnetic field: $(0,0.5,0)$ (i.e., on Y-axis, $𝑟=0.5\text{m}$)

• $\Delta 𝑥=1\text{cm}=0.01\text{m}$

• $\frac{{𝜇}_0}{4𝜋}={10}^{-7}\text{SI units}$

• $\sin {90}^{\circ }=1$

Step 1: Biot-Savart Law for a Small Element

The magnetic field at point $𝑃$ due to a small current element is:

$𝑑\overrightarrow{𝐵}=\frac{{𝜇}_0}{4𝜋}\frac{𝐼𝑑\overrightarrow{𝑙}\times \widehat{𝑟}}{{𝑟}^2}$

Step 2: Set Up the Cross Product

• $𝑑\overrightarrow{𝑙}=\Delta 𝑥\widehat{𝑖}$

• $\overrightarrow{𝑟}=0\widehat{𝑖}+0.5\widehat{𝑗}+0\widehat{𝑘}$

  So,

• $\widehat{𝑟}=\frac{\overrightarrow{𝑟}}{\left|\overrightarrow{𝑟}\right|}=\widehat{𝑗}$

• Angle between $𝑑\overrightarrow{𝑙}$ and $\overrightarrow{𝑟}$ is ${90}^{\circ }$ (since $𝑑\overrightarrow{𝑙}$ along $𝑥$, $\overrightarrow{𝑟}$ along $𝑦$).

So,

$𝑑\overrightarrow{𝑙}\times \widehat{𝑟}=\Delta 𝑥\widehat{𝑖}\times \widehat{𝑗}=\Delta 𝑥\widehat{𝑘}$

Step 3: Calculate the Magnitude of $𝑑𝐵$

$𝑑𝐵=\frac{{𝜇}_0}{4𝜋}\frac{𝐼\Delta 𝑥\sin {90}^{\circ }}{{𝑟}^2}$

$𝑑𝐵=\frac{{𝜇}_0}{4𝜋}\frac{𝐼\Delta 𝑥}{{𝑟}^2}$

Substitute values:

• $𝐼=10\text{A}$

• $\Delta 𝑥=0.01\text{m}$

• $𝑟=0.5\text{m}$

• $\frac{{𝜇}_0}{4𝜋}={10}^{-7}$

$𝑑𝐵={10}^{-7}\times \frac{10\times 0.01}{(0.5{)}^2}$

$𝑑𝐵={10}^{-7}\times \frac{0.1}{0.25}$

$𝑑𝐵={10}^{-7}\times 0.4$

$𝑑𝐵=4\times {10}^{-8}T$

Correct option:

$\overline{)4\times {10}^{-8}T}$

So the answer is Option C.

Given:

• Collector current, ${𝐼}_{𝐶}=9\text{mA}$

• $90\%$ of the electrons emitted reach the collector (i.e., $𝛼=0.9$)

Let's use the NCERT terminology and steps.

Step 1: Meaning of $𝛼$ and $𝛽$

• $𝛼$ is the current gain in common base configuration:

  $𝛼=\frac{{𝐼}_{𝐶}}{{𝐼}_{𝐸}}$

• $𝛽$ is the current gain in common emitter configuration:

  $𝛽=\frac{{𝐼}_{𝐶}}{{𝐼}_{𝐵}}$

  or

  $𝛽=\frac{𝛼}{1-𝛼}$

Step 2: Using the data

Since $90\%$ electrons emitted from the emitter reach the collector,

$𝛼=0.9$

Step 3: Calculate $𝛽$

The relationship is:

$𝛽=\frac{𝛼}{1-𝛼}$

Put $𝛼=0.9$:

$𝛽=\frac{0.9}{1-0.9}=\frac{0.9}{0.1}=9$

Step 4: Check options

Only option C matches:

$\overline{)𝛼=0.9,𝛽=9}$

Final Answer:

Option C: $𝛼=0.9,𝛽=9$

For open pipe first overtone, ${𝑛}_1=\frac{𝑣}{𝐿}$

For closed pipe first overtone, $n_{\prime }^1=\frac{3v}{4L}$

$\therefore$ Number of beats produced are,

$\begin{array}{cc}& n_1-n_{\prime }^1=\frac{v}{L}-\frac{3v}{4L}=3\\ \therefore & \frac{v}{4L}=3\\ \therefore & \frac{v}{L}=12\dots .\left(i\right)\end{array}$

When length of open pipe is made $\frac{L}{3}$, the fundamental frequency becomes,

$n=\frac{v}{2\left(\frac{L}{3}\right)}=\frac{3v}{2L}$

When length of closed pipe is made 3 times, the fundamental frequency becomes,

$n^{\prime }=\frac{v}{4\left(3L\right)}=\frac{v}{12L}$

$\begin{array}{cc}\therefore & \text{ Beats produced }=𝑛-{𝑛}^{\prime }\\ & =\frac{3𝑣}{2𝐿}-\frac{𝑣}{12𝐿}\\ & =\frac{17}{12}\times \frac{𝑣}{𝐿}=\frac{17}{12}\times 12\text{ ....[From (i)] }\\ & =17\end{array}$

The spectral emissive power ${𝐸}_{𝜆}$ represents the energy radiated per unit area per unit time in the wavelength interval between $𝜆$ and $𝜆+𝑑𝜆$.

The total area under the ${𝐸}_{𝜆}$ versus $𝜆$ curve is:

$\text{Area}=\underset{0}{\overset{\infty }{\int }}{𝐸}_{𝜆}𝑑𝜆$

By the Stefan–Boltzmann law, this integral gives the total emissive power, which is the total energy emitted per unit time per unit surface area of the black body over all wavelengths.

Hence, the correct option is (D).

Maximum current occurs at Resonance

$\begin{array}{cc}\therefore & {𝑋}_{𝐿}={𝑋}_{𝐶}\Rightarrow 𝜔𝐿=\frac{1}{𝜔𝐶}\\ & {𝜔}^2=\frac{1}{𝐿𝐶}\Rightarrow 𝐶=\frac{1}{{𝜔}^2𝐿}\\ & 𝜔=100𝜋\Rightarrow {𝜔}^2=(100𝜋{)}^2={10}^4.10={10}^5\\ \therefore & 𝐶=\frac{1}{{10}^5\times 2.5}=\frac{4}{{10}^6}=4\times {10}^{-6}F=4𝜇F\end{array}$

Maximum current at resonance

$I_{max}=\frac{V_0}{R}=\frac{230}{11.5}=20A$

$\therefore C=4𝜇F$ and maximum current $I_{max}=20A$

Let initial separation between Earth and satellite be $𝑟$ and its time period is $𝑇=24$ hours.

If the separation is decreased to one fourth, new separation ${𝑟}^{\prime }=\frac{𝑟}{4}$.

From NCERT, the time period of a satellite in orbit at distance $𝑟$ from the center of Earth is

$𝑇=2𝜋\sqrt{\surd \frac{{𝑟}^3}{𝐺𝑀}}$

where $𝐺$ is gravitational constant and $𝑀$ is mass of the Earth.

Let new time period be ${𝑇}^{\prime }$:

${𝑇}^{\prime }=2𝜋\sqrt{\surd \frac{({𝑟}^{\prime }{)}^3}{𝐺𝑀}}$

Substitute ${𝑟}^{\prime }=\frac{𝑟}{4}$:

${𝑇}^{\prime }=2𝜋\sqrt{\surd \frac{{\left(\frac{𝑟}{4}\right)}^3}{𝐺𝑀}}=2𝜋\sqrt{\surd \frac{{𝑟}^3}{𝐺𝑀}\times \frac{1}{4^3}}$

$=2𝜋\sqrt{\surd \frac{{𝑟}^3}{𝐺𝑀}\times \frac{1}{64}}$

$=2𝜋\sqrt{\surd \frac{{𝑟}^3}{𝐺𝑀}}\times \frac{1}{8}$

But $2𝜋\sqrt{\surd \frac{{𝑟}^3}{𝐺𝑀}}=𝑇=24$ hours

So,

${𝑇}^{\prime }=𝑇\times \frac{1}{8}=24\times \frac{1}{8}=3\text{ hours}$

Correct option: A) 3 hours

At the extreme position, the bob of the pendulum at a height ' $ℎ$ ' from the mean position.

$ℎ=𝐿-𝐿\cos 𝜃=𝐿(1-\cos 𝜃)=2𝐿{\sin }^2\left(\frac{𝜃}{2}\right)$

When the bob reaches the mean position, it loses potential energy and gains equal kinetic energy.

$\begin{array}{cc}\therefore & \frac{1}{2}{\mathrm{mv}}^2=\mathrm{mgh}=\mathrm{mg}\times 2{\mathrm{Lsin}}^2\left(\frac{𝜃}{2}\right)\\ \therefore & v=2\sin \left(\frac{𝜃}{2}\right)\sqrt{\surd \mathrm{gL}}\end{array}$

Induced emf (using Instantaneous max velocity) At lowest point, entire wire of length $𝐿$ is moving with speed v .

So, maximum emf is:

$\begin{array}{cc}& {𝜀}_{max}=𝐵\cdot 𝐿\cdot 𝑣=𝐵\cdot 𝐿\cdot 2\sqrt{\surd 𝑔𝐿}\sin \left(\frac{𝜃}{2}\right)\\ \therefore & {𝜀}_{max}=2𝐵𝐿\sqrt{\surd 𝑔𝐿}\sin \left(\frac{𝜃}{2}\right)\end{array}$

In a cyclic process, the system returns to its initial state. According to the first law of thermodynamics,

$\Delta 𝑈=𝑄-𝑊$

where $\Delta 𝑈$ is the change in internal energy, $𝑄$ is the heat given to the system, and $𝑊$ is the work done by the system.

For a cyclic process:

• The system returns to its original state, so $\Delta 𝑈=0$.

So,

$0=𝑄-𝑊⟹𝑄=𝑊$

Correct answer:

Option B: equal to the heat given to the system.

According to Newton's law of cooling,

$\mid \frac{\mathrm{dT}}{\mathrm{dt}}\mid =k(T-{𝜃}_0)$

At $T={80}^{\circ },1.5=k[80-{𝜃}_0]$

At $T={40}^{\circ },0.3=k[40-{𝜃}_0]$

Dividing (i) and (ii)

$\begin{array}{cc}& \frac{1.5}{0.3}=\frac{80-{𝜃}_0}{40-{𝜃}_0}\\ \Rightarrow & 200-5{𝜃}_0=80-{𝜃}_0\\ \Rightarrow & 4{𝜃}_0=120\\ \Rightarrow & {𝜃}_0={30}^{\circ }C\end{array}$

From the circuit diagram,

$Y=\overline{(\stackrel{–}{A)}\cdot \stackrel{–}{\left(B\right)}}...\left(𝑖\right)$

De Morgan's Law is given as,

$\bar{A}+\bar{B}=\overline{\mathrm{AB}}...\left(𝑖𝑖\right)$

$\begin{array}{cc}& 𝑌=\overline{\stackrel{–}{𝐴+𝐵}}\dots \left[𝐹𝑟𝑜𝑚\right(𝑖\left)𝑎𝑛𝑑\right(𝑖𝑖\left)\right]\\ & 𝑌=𝐴+𝐵\end{array}$

This is the equation for OR gate.

Alternate method:

A and B are given to two NOT gates. Let $A^{\prime }$ and ${𝐵}^{\prime }$ be the outputs of the two NOT gates that are given to a NAND gate. We can write the truth table as shown below.

$\begin{array}{ccccc}A& B& A^{\prime }& B^{\prime }& Y\\ 0& 0& 1& 1& 0\\ 0& 1& 1& 0& 1\\ 1& 0& 0& 1& 1\\ 1& 1& 0& 1& 1\end{array}$

$\text{ This is truth table of OR gate. }$

$\begin{array}{cc}& \begin{array}{cc}& \text{ Mass }𝑚=5𝑘𝑔\\ & \overrightarrow{𝐹}=(4\widehat{𝑖}-2\widehat{𝑗}+3\widehat{𝑘})𝑁\end{array}\\ & \text{ Using Newton’s second law }\overrightarrow{F}=𝑚\overrightarrow{a}\\ & \begin{array}{cc}& \overrightarrow{𝑎}=\frac{\overrightarrow{𝐹}}{𝑚}=\frac{4\widehat{𝑖}-2\widehat{𝑗}+3\widehat{𝑘}}{5}\\ & \overrightarrow{𝑎}=(\frac{4}{5}\widehat{𝑖}-\frac{2}{5}\widehat{𝑗}+\frac{3}{5}\widehat{𝑘})m/s^2\\ & \overrightarrow{𝑣}={\overrightarrow{𝑣}}_0+\overrightarrow{𝑎}𝑡\end{array}\end{array}$

$\begin{array}{cc}& \begin{array}{cc}& {\overrightarrow{v}}_0=0\because \text{ body was initially at rest }\\ & \overrightarrow{v}=(\frac{4}{5}\hat{i}-\frac{2}{5}\hat{j}+\frac{3}{5}\hat{k})\cdot 10\\ & \overrightarrow{v}=8\hat{i}-4\hat{j}+6\hat{k}\end{array}\\ & \text{ Magnitude of final velocity }\\ & \begin{array}{cc}\left|\overrightarrow{𝑣}\right|& =\sqrt{\surd {𝑣}_2^{𝑥}+{𝑣}_2^{𝑦}+{𝑣}_2^{𝑥}}\\ & =\sqrt{\surd 8^2+(-4{)}^2+6^2}\\ & =\sqrt{\surd 64+16+36}\\ & =\sqrt{\surd 116}\\ \left|\overrightarrow{𝑣}\right|& =2\sqrt{\surd 29}m/s\end{array}\end{array}$

For a closed pipe

The frequency of $n^{\text{th }}$ overtone is given by

$𝑓=(2𝑛+1){𝑓}_1=(2𝑛+1)\frac{𝑉}{4𝐿}$

For an open pipe

The frequency of $n^{\text{th }}$ overtone is given by

$\begin{array}{cc}& {𝑓}^{\prime }=(𝑛+1){𝑓}_1=(𝑛+1)\frac{𝑉}{2𝐿}\\ \therefore & \frac{𝑓}{{𝑓}^{\prime }}=\frac{(2𝑛+1)}{2(𝑛+1)}\end{array}$

Net impedance $𝑍=\sqrt{\surd {𝑅}^2+{({𝑋}_{𝐿}-{𝑋}_{𝐶})}^2}$

$\begin{array}{cc}𝑍& =\sqrt{\surd {𝑅}^2+(3𝑅-𝑅{)}^2}=\sqrt{\surd {𝑅}^2+4{𝑅}^2}\\ & =\sqrt{\surd 5}𝑅\\ {𝐼}_{\text{rms }}& =\frac{{𝐸}_{\text{ms }}}{𝑍}=\frac{{𝐸}_0/\sqrt{\surd 2}}{\sqrt{\surd 5}𝑅}=\frac{{𝐸}_0}{\sqrt{\surd 10}𝑅}\end{array}$

Average power dissipated,

$𝑃={𝐼}_2^{\text{rim }}\cdot 𝑅={\left(\frac{{𝐸}_0}{\sqrt{\surd 10}𝑅}\right)}^2𝑅=\frac{{𝐸}_2^0𝑅}{10{𝑅}^2}=\frac{{𝐸}_2^0}{10𝑅}$

When the diameter doubles, the volume increases by $2^3=8$ times

Using Boyle's law

$\begin{array}{cc}& {𝑃}_1{𝑉}_1={𝑃}_2{𝑉}_2\\ & {𝑉}_2=8{𝑉}_1\Rightarrow {𝑃}_{\text{bottom }}=8{𝑃}_{\mathrm{atm}}\end{array}$

Pressure in terms of mercury column

$P_{\mathrm{atm}}={𝜌}_{\mathrm{Hg}}\mathrm{gh}$

$\begin{array}{cc}& \therefore {𝑃}_{\text{bottom }}={𝑃}_{\mathrm{atm}}+{𝜌}_{\text{water }}𝑔𝑑\\ & 8𝜌\cdot {𝜌}_{\text{water }}𝑔ℎ=𝜌\cdot {𝜌}_{\text{water }}𝑔ℎ+{𝜌}_{\text{water }}𝑔𝑑(\because {𝜌}_{𝐻𝑔}=𝜌\cdot {𝜌}_{\text{water }})\\ & 8𝜌ℎ=𝜌ℎ+𝑑\\ & 𝑑=7ℎ𝜌\end{array}$

When a star moves away from the Earth, the observed wavelength $\left({𝜆}^{\prime }\right)$ becomes longer than the actual wavelength $\left(𝜆\right)$. This phenomenon is known as the redshift due to the Doppler effect.

According to the Doppler effect for light (for velocities much less than $𝑐$):

$\frac{\Delta 𝜆}{𝜆}=\frac{𝑣}{𝑐}$

where

$\Delta 𝜆={𝜆}^{\prime }-𝜆$ is the change in wavelength,

$𝑣$ is the velocity of the star away from the Earth,

$𝑐$ is the speed of light.

Given:

The apparent wavelength is $0.02\%$ more than the actual wavelength.

So,

$\frac{\Delta 𝜆}{𝜆}=0.02\%=\frac{0.02}{100}=0.0002$

Substitute in the Doppler formula:

$0.0002=\frac{𝑣}{3\times {10}^8}$

Solving for $𝑣$:

$𝑣=0.0002\times 3\times {10}^8$

$𝑣=6\times {10}^{4}m/s$

$𝑣=60\mathrm{km}/s$

Correct option:

Option B $60\mathrm{km}/s$

When dielectrics are placed in series, the equivalent dielectric constant is:

$\begin{array}{cc}& \frac{1}{K_{\mathrm{eq}}}=\frac{1}{2K_1}+\frac{1}{2K_2}\\ & \frac{1}{K_{\mathrm{eq}}}=\left(\frac{K_1+K_2}{2K_1{K}_2}\right)\\ & K_{\mathrm{eq}}=\left(\frac{2K_1{K}_2}{K_1+K_2}\right)\end{array}$

So, new capacitance:

$\begin{array}{cc}& {𝐶}_{𝐾}={𝐾}_{𝑒𝑞}\cdot {𝐶}_{𝑃}=\left(\frac{2{𝐾}_1{𝐾}_2}{{𝐾}_1+{𝐾}_2}\right){𝐶}_{𝑃}\\ & \frac{{𝐶}_{𝑃}}{{𝐶}_{𝐾}}=\frac{{𝐾}_1+{𝐾}_2}{2{𝐾}_1{𝐾}_2}\end{array}$

Change in internal energy is given by,

$\Delta 𝑈=𝑛{𝐶}_{𝑣}\Delta 𝑇...\left(𝑖\right)$

Given, $\frac{C_P}{C_v}=𝛾$ and $C_P-C_v=R$

$\begin{array}{cc}& 1+\frac{𝑅}{{𝐶}_{𝑉}}=𝛾\\ & {𝐶}_{𝑉}=\frac{𝑅}{(𝛾-1)}\\ \therefore & \Delta 𝑈=𝑛\left(\frac{𝑅}{𝛾-1}\right)\Delta 𝑇\dots \left[𝐹𝑟𝑜𝑚\right(𝑖\left)\right]\end{array}$

$\begin{array}{cc}& \text{ Using }𝑃\Delta 𝑉=𝑛𝑅\Delta 𝑇\text{ under constant }𝑃\text{, }\\ & \Delta 𝑈=\frac{𝑃\Delta 𝑉}{𝛾-1}=\frac{𝑃(2𝑉-𝑉)}{𝛾-1}=\frac{𝑃𝑉}{(𝛾-1)}\end{array}$

If the length of the potentiometer wire is increased while the potential difference (V) across it remains constant, let's see what happens:

The potential gradient ($𝑘$) along the potentiometer wire is:

$𝑘=\frac{\text{Potential difference across the wire (V)}}{\text{Length of the wire (L)}}$

So,

$𝑘=\frac{𝑉}{𝐿}$

If $𝐿$ increases and $𝑉$ is constant, then $𝑘$ decreases.

To obtain the null point, the potential difference across a certain length $𝑙$ of the potentiometer wire must be equal to the emf of the cell connected to it:

$\text{emf}=𝑘\cdot 𝑙$

Since $𝑘$ has decreased, to balance the same emf, $𝑙$ (the length to the null point) must increase.

Correct option:

Option C

the null point is obtained at larger distance.

Sphere's Kinetic Energy = Translational energy + Rotational energy

$\begin{array}{cc}& =\frac{1}{2}𝑚{𝑣}^2+\frac{1}{5}𝑚{𝑣}^2\\ & =\frac{7}{10}𝑚{𝑣}^2\end{array}$

At maximum height h ,

Total K.E. = Potential energy

$\begin{array}{cc}& \frac{7}{10}{\mathrm{mv}}^2=\mathrm{mgh}\\ & h=\frac{7v^2}{10g}=\frac{7\times 2^2}{10\times 10}=0.28m\end{array}$

Distance along the incline

$𝑑=\frac{ℎ}{\sin 𝜃}=\frac{0.28}{\sin \left({30}^{\circ }\right)}=0.56m=56\mathrm{cm}$

In Young's double slit experiment, the fringe width $\left(𝛽\right)$ is given by:

$𝛽=\frac{𝜆𝐷}{𝑑}$

where

• $𝜆=$ wavelength of light = $600\mathrm{nm}=600\times {10}^{-9}m$

• $𝐷=$ distance between slits and screen

• $𝑑=$ separation between the slits = ${10}^{-3}m$

Let the initial fringe width be ${𝛽}_1$ at distance ${𝐷}_1$, and the new fringe width be ${𝛽}_2$ at distance ${𝐷}_2$.

Given,

${𝛽}_2-{𝛽}_1=3\times {10}^{-5}m$

So,

${𝛽}_2=\frac{𝜆{𝐷}_2}{𝑑}$

${𝛽}_1=\frac{𝜆{𝐷}_1}{𝑑}$

Therefore,

${𝛽}_2-{𝛽}_1=\frac{𝜆}{𝑑}({𝐷}_2-{𝐷}_1)=3\times {10}^{-5}m$