WBJEE 2025 Paper 2 (Physics & Chemistry) Question Paper with Solutions

$\text{ Equation of line using }𝑣-𝑡\text{ graph }$

$\begin{array}{cc}𝑣& =-\frac{50}{10}𝑡+50\\ 𝑜𝑟𝑣& =-5𝑡+50\\ 𝑎𝑡𝑡& =0s\\ {𝑣}_{𝑖}& =50m/s\\ 𝑎𝑡𝑡& =2s\\ {𝑣}_{𝑓}& =40m/s\end{array}$

$\begin{array}{cc}& \text{ Using work-energy theorem, }\\ & \begin{array}{cccc}& & 𝑊& =\Delta 𝐾𝐸={\mathrm{KE}}_{𝑓}-{\mathrm{KE}}_{𝑖}\\ \Rightarrow & & 𝑊& =\frac{1}{2}𝑚{𝑣}_2^{𝑓}-\frac{1}{2}𝑚{𝑣}_2^{𝑖}\\ \Rightarrow & & 𝑊& =\frac{1}{2}\times 10\left[\right(40{)}^2-\left(50{)}^2\right]\\ & & & =-5\times 900\\ \Rightarrow & & 𝑊& =-4500J\end{array}\end{array}$

The frequency of the AC supply is 50 Hz. This means the angular frequency, $𝜔$, is calculated as:

$𝜔=2𝜋𝑓=2𝜋\times 50=100𝜋\text{ radians per second.}$

The given 220 V is the Root Mean Square (RMS) value of the voltage. The peak (maximum) voltage, ${𝑉}_{\text{max}}$, is related to the RMS value by:

${𝑉}_{\text{max}}=220\sqrt{\surd 2}.$

The general expression for an AC voltage is:

$𝑉\left(𝑡\right)={𝑉}_{\text{max}}\cos \left(𝜔𝑡\right).$

Substituting the values we calculated:

$𝑉\left(𝑡\right)=220\sqrt{\surd 2}\cos \left(100𝜋𝑡\right).$

Comparing with the options, we see that Option A:

$V\left(𝑡\right)=220\sqrt{\surd 2}\cos \left(100𝜋𝑡\right)$

is the correct answer.

Let's analyze the decay chain step-by-step. We start with a nucleus $𝑋$ having atomic number $𝑍$ and mass number $𝐴$. The decay chain is as follows:

$𝑋\stackrel{𝛼}{\to }{𝑋}_1$

In alpha decay, an alpha particle (helium nucleus) is emitted, which has a mass number of 4 and an atomic number of 2.

Therefore:

${𝑋}_1:\text{atomic number }=𝑍-2,\text{mass number }=𝐴-4$

${𝑋}_1\stackrel{𝛽}{\to }{𝑋}_2$

In beta decay, a neutron is converted into a proton, increasing the atomic number by 1 while the mass number remains unchanged.

Therefore:

${𝑋}_2:\text{atomic number }=(𝑍-2)+1=𝑍-1,\text{mass number }=𝐴-4$

${𝑋}_2\stackrel{𝛼}{\to }{𝑋}_3$

Again, alpha decay reduces the atomic number by 2 and the mass number by 4.

Therefore:

${𝑋}_3:\text{atomic number }=(𝑍-1)-2=𝑍-3,\text{mass number }=(𝐴-4)-4=𝐴-8$

${𝑋}_3\stackrel{𝛾}{\to }{𝑋}_4$

Gamma decay does not change either the atomic number or the mass number.

Therefore:

${𝑋}_4:\text{atomic number }=𝑍-3,\text{mass number }=𝐴-8$

We are given that ${𝑋}_4$ has a mass number of 172 and an atomic number of 69. This gives us two equations:

$\begin{array}{cc}𝑍-3& =69,\\ 𝐴-8& =172.\end{array}$

Now, solve these equations:

From the first equation:

$𝑍=69+3=72.$

From the second equation:

$𝐴=172+8=180.$

Thus, the original nucleus $𝑋$ has an atomic number of 72 and a mass number of 180.

Answer: Option A (72, 180).

Fringe positions are given by the single‐slit minima condition

$𝑎\sin 𝜃=𝑚𝜆(𝑚=\pm 1,\pm 2,\dots )$

so the angular fringe spacing scales as $\Delta 𝜃\sim 𝜆/𝑎$. Blue light has a smaller wavelength than red, hence its diffraction fringes are closer together (narrower angular width).

Answer: Option B.

$\begin{array}{cc}\because & \frac{𝑑𝑚}{𝑑𝑣}=𝜌\\ \Rightarrow & 𝑑𝑚=𝛼𝑑𝑥𝜌[\because 𝑑𝑣=𝑎𝑑𝑡]\\ \Rightarrow & 𝑑𝑚=\frac{𝛼{𝜌}_0{𝑥}^2}{{𝐿}^2}𝑑𝑥\end{array}$

$\begin{array}{cc}& \text{ Now, }\\ & \begin{array}{cc}& {𝑋}_{\mathrm{CM}}=\frac{\int 𝑥𝑑𝑚}{\int 𝑑𝑚}=\frac{\frac{𝛼{𝜌}_0}{{𝐿}^2}{\int }_{𝐿}^0{𝑥}^3𝑑𝑥}{\frac{𝛼{𝜌}_0}{{𝐿}^2}{\int }_{𝐿}^0{𝑥}^2𝑑𝑥}\\ & {𝑋}_{\mathrm{CM}}=\frac{{\left[\frac{{𝑥}^4}{4}\right]}_{𝐿}^0}{{\left[\frac{{𝑥}^3}{3}\right]}_{𝐿}^0}=\frac{3}{4}𝐿\end{array}\end{array}$

At height $ℎ=𝑅$ above the Earth’s surface, the distance from Earth’s centre is $2𝑅$. Gravitational acceleration there is

${𝑔}^{\prime }=𝑔\frac{{𝑅}^2}{(2𝑅{)}^2}=\frac{𝑔}{4}.$

The time period of a simple pendulum of length $𝐿$ in gravitational field ${𝑔}^{\prime }$ is

$𝑇=2𝜋\sqrt{\surd \frac{𝐿}{{𝑔}^{\prime }}}=2𝜋\sqrt{\surd \frac{𝐿}{𝑔/4}}=2𝜋\sqrt{\surd \frac{4𝐿}{𝑔}}=4𝜋\sqrt{\surd \frac{𝐿}{𝑔}}.$

Substituting $𝐿=4.0$ m and $𝑔={𝜋}^2$ m/s${}^2$:

$\sqrt{\surd \frac{𝐿}{𝑔}}=\sqrt{\surd \frac{4}{{𝜋}^2}}=\frac{2}{𝜋},𝑇=4𝜋\times \frac{2}{𝜋}=8\text{ s}.$

Answer: 8 s (Option C).

The equation of motion is

$𝑚\ddot{𝑥}=𝑞{𝐸}_0\sin 𝜔𝑡,𝑚=0.001kg,q=1C,E_0=2N/C,𝜔=1000s^{-1}.$

Integrate once (with $𝑣\left(0\right)=0$) to get

$𝑣\left(𝑡\right)={\int }_{𝑡}^0\frac{𝑞{𝐸}_0}{𝑚}\sin \left(𝜔{𝑡}^{\prime }\right)𝑑{𝑡}^{\prime }=\frac{𝑞{𝐸}_0}{𝑚𝜔}[1-\cos \left(𝜔𝑡\right)]=\frac{2}{0.001\cdot 1000}(1-\cos 𝜔𝑡)=2(1-\cos 𝜔𝑡)(m/s).$

Since $\cos 𝜔𝑡$ ranges from $-1$ to $1$, the maximum speed is

${𝑣}_{max}=2(1-(-1))=4m/s.$

Answer: 4 m/s.

Given:

• Coefficient of friction = $𝜇$

• Angle of incline = $𝜃$, where $\tan 𝜃=2𝜇$

Let the weight of the body be $𝑚𝑔$.

Step 1: Forces Acting on the Body

• Along the plane:

• Component of weight down the incline = $𝑚𝑔\sin 𝜃$

• Normal reaction = $𝑚𝑔\cos 𝜃$

• Frictional force = $𝜇𝑚𝑔\cos 𝜃$

• Let $𝐹$ be the applied force parallel to the incline (as is usual in such problems).

A. Pushing the Body Up (Minimum ${𝐹}_1$)

To just start moving up, applied force ${𝐹}_1$ must overcome:

• gravity component $𝑚𝑔\sin 𝜃$

• plus friction (acts downward) $𝜇𝑚𝑔\cos 𝜃$

So,

${𝐹}_1=𝑚𝑔\sin 𝜃+𝜇𝑚𝑔\cos 𝜃$

B. Preventing from Sliding Down (Minimum ${𝐹}_2$)

To just prevent sliding down, applied force ${𝐹}_2$ must balance:

• gravity component $𝑚𝑔\sin 𝜃$

• minus friction (acts upward if object tends to slide down) $𝜇𝑚𝑔\cos 𝜃$

So,

${𝐹}_2=𝑚𝑔\sin 𝜃-𝜇𝑚𝑔\cos 𝜃$

C. Ratio $\frac{{𝐹}_1}{{𝐹}_2}$

$\frac{{𝐹}_1}{{𝐹}_2}=\frac{𝑚𝑔\sin 𝜃+𝜇𝑚𝑔\cos 𝜃}{𝑚𝑔\sin 𝜃-𝜇𝑚𝑔\cos 𝜃}=\frac{\sin 𝜃+𝜇\cos 𝜃}{\sin 𝜃-𝜇\cos 𝜃}$

D. Express in terms of $\tan 𝜃$

Divide numerator and denominator by $\cos 𝜃$:

$=\frac{\tan 𝜃+𝜇}{\tan 𝜃-𝜇}$

Given: $\tan 𝜃=2𝜇$

So,

$\frac{{𝐹}_1}{{𝐹}_2}=\frac{2𝜇+𝜇}{2𝜇-𝜇}=\frac{3𝜇}{𝜇}=3$

Final Answer:

$\overline{)3}$

Option D is correct.

$\begin{array}{cc}& \\ 𝐴𝑠,& \frac{𝑑𝑣}{𝑑𝑡}=\tan 𝜃=𝑎\end{array}$

From the equation of motion

$\begin{array}{cc}& \begin{array}{cc}𝑇-𝑚𝑔& =𝑚𝑎\\ & \\ \Rightarrow & =𝑚𝑔+𝑚𝑎=𝑚(𝑔+𝑎)\\ & =10(10+2)=120N\end{array}\\ & \therefore \text{ Young’s modulus, }\end{array}$

$\begin{array}{c}𝑌=\frac{𝑇/𝐴}{\frac{\Delta 𝑙}{𝑙}}\\ \Rightarrow \frac{\Delta 𝑙}{𝑙}=\frac{𝑇}{𝐴𝑌}=\frac{120}{2\times {10}^{-4}\times 2\times {10}^{11}}=3\times {10}^{-6}\end{array}$

Let the escalator have “length’’ $𝐿$ (in steps).

Walking alone (escalator off):

${𝑣}_1{𝑡}_1=𝐿$.

Standing alone (escalator on):

${𝑣}_2{𝑡}_2=𝐿$.

Both walking and moving: net speed ${𝑣}_1+{𝑣}_2$, so time

$𝑡=\frac{𝐿}{{𝑣}_1+{𝑣}_2}=\frac{{𝑣}_1{𝑡}_1}{{𝑣}_1+{𝑣}_2}=\frac{\frac{𝐿}{{𝑡}_1}{𝑡}_1}{\frac{𝐿}{{𝑡}_1}+\frac{𝐿}{{𝑡}_2}}=\frac{𝐿}{𝐿(\frac{1}{{𝑡}_1}+\frac{1}{{𝑡}_2})}=\frac{1}{\frac{1}{{𝑡}_1}+\frac{1}{{𝑡}_2}}=\frac{{𝑡}_1{𝑡}_2}{{𝑡}_1+{𝑡}_2}.$

Answer: $\frac{{𝑡}_1{𝑡}_2}{{𝑡}_1+{𝑡}_2}$ (Option B).

Given:

$𝑋={𝜀}_0𝐿\frac{\Delta 𝑉}{\Delta 𝑡}$

where

• ${𝜀}_0$: permittivity of free space

• $𝐿$: length

• $\Delta 𝑉$: potential difference (voltage)

• $\Delta 𝑡$: time interval

We need to find the dimensions of $𝑋$ and compare with the options.

Step 1: Dimension of ${𝜀}_0$

Permittivity of free space:

$\left[{𝜀}_0\right]=\left[{𝑀}^{-1}{𝐿}^{-3}{𝑇}^4{𝐴}^2\right]$

(from standard tables or derivation)

Step 2: Dimension of $𝐿$

Length:

$\left[𝐿\right]=\left[𝐿\right]$

Step 3: Dimension of $\Delta 𝑉$

Potential difference:

$\left[𝑉\right]=\left[𝑀{𝐿}^2{𝑇}^{-3}{𝐴}^{-1}\right]$

Step 4: Dimension of $\Delta 𝑡$

Time:

$\left[\Delta 𝑡\right]=\left[𝑇\right]$

Step 5: Combine all

Recall:

$𝑋={𝜀}_0\times 𝐿\times \frac{\Delta 𝑉}{\Delta 𝑡}$

So

$\begin{array}{cc}\left[𝑋\right]& =\left[{𝜀}_0\right]\left[𝐿\right]\frac{\left[𝑉\right]}{\left[𝑇\right]}\\ & =\left[{𝑀}^{-1}{𝐿}^{-3}{𝑇}^4{𝐴}^2\right]\left[𝐿\right]\frac{\left[𝑀{𝐿}^2{𝑇}^{-3}{𝐴}^{-1}\right]}{\left[𝑇\right]}\\ & =\left[{𝑀}^{-1}{𝐿}^{-3}{𝑇}^4{𝐴}^2\right]\left[𝐿\right]\left[𝑀{𝐿}^2{𝑇}^{-3}{𝐴}^{-1}\right]\left[{𝑇}^{-1}\right]\end{array}$

Now, combine:

• $𝑀$: ${𝑀}^{-1}\times {𝑀}^1={𝑀}^0$

• $𝐿$: ${𝐿}^{-3}\times {𝐿}^1\times {𝐿}^2={𝐿}^0$

• $𝑇$: ${𝑇}^4\times {𝑇}^{-3}\times {𝑇}^{-1}={𝑇}^{4-3-1}={𝑇}^0$

• $𝐴$: ${𝐴}^2\times {𝐴}^{-1}={𝐴}^1$

So,

$\left[𝑋\right]=\left[{𝐴}^1\right]$

Step 6: Compare with Options

• Resistance: $\left(𝑀{𝐿}^2{𝑇}^{-3}{𝐴}^{-2}\right)$

• Charge: $(𝐴\cdot 𝑇)$

• Voltage: $\left(𝑀{𝐿}^2{𝑇}^{-3}{𝐴}^{-1}\right)$

• Current: $\left(𝐴\right)$

Our answer $\left[𝐴\right]$ matches the dimension of current.

Final Answer:

Option D: Current

According to the given options, in reverse bias only resistance is considered and it will follow Ohm's law, $𝑉\propto 𝐼$

In forward biasing initially current flows through resistance, then after knee voltage, it flows through $𝑝-𝑛$ diode.

• Lyman series corresponds to electron transitions from higher energy levels (${𝑛}_2$) to the first energy level (${𝑛}_1=1$) in a hydrogen atom.

Step 1: Wavelength formula for Lyman series

The wavelength $𝜆$ is given by:

$\frac{1}{𝜆}=𝑅(1-\frac{1}{{𝑛}_2^2})$

where $𝑅$ is the Rydberg constant, ${𝑛}_1=1$, and ${𝑛}_2=2,3,4,...$

Step 2: Minimum wavelength ($𝑃$) of Lyman series

• Minimum wavelength occurs when the denominator is the largest, that is, when ${𝑛}_2\to \infty$:

$\frac{1}{𝑃}=𝑅(1-\frac{1}{{\infty }^2})=𝑅(1-0)=𝑅⟹𝑃=\frac{1}{𝑅}$

Step 3: Maximum wavelength of Lyman series

• Maximum wavelength occurs when ${𝑛}_2$ is the smallest possible (${𝑛}_2=2$):

$\frac{1}{{𝜆}_{𝑚𝑎𝑥}}=𝑅(1-\frac{1}{2^2})=𝑅(1-\frac{1}{4})=𝑅\left(\frac{3}{4}\right)$

${𝜆}_{𝑚𝑎𝑥}=\frac{1}{𝑅}\cdot \frac{4}{3}=\frac{4}{3}\cdot \frac{1}{𝑅}$

But from Step 2, $𝑃=\frac{1}{𝑅}$. So:

${𝜆}_{𝑚𝑎𝑥}=\frac{4}{3}𝑃$

Step 4: Answer

Correct option is:

$\overline{)\frac{4𝑃}{3}}$

Which is Option A.

Step 1: Calculate the energy difference when electron jumps from $𝑛=2$ to $𝑛=1$.

Given,

Energy in $𝑛$th orbit: ${𝐸}_{𝑛}=-\frac{13.6}{{𝑛}^2}$ eV

For $𝑛=2$:

${𝐸}_2=-\frac{13.6}{2^2}=-\frac{13.6}{4}=-3.4\text{ eV}$

For $𝑛=1$:

${𝐸}_1=-\frac{13.6}{1^2}=-13.6\text{ eV}$

Energy difference $\left(\Delta 𝐸\right)$:

$\Delta 𝐸={𝐸}_1-{𝐸}_2=(-13.6)-(-3.4)=-13.6+3.4=-10.2\text{ eV}$

But, when electron drops from higher to lower orbit, energy emitted is $+10.2$ eV (just use magnitude).

Step 2: This energy is the energy of emitted photon:

${𝐸}_{\text{photon}}=10.2\text{ eV}$

Step 3: Apply Photoelectric Equation

Einstein’s equation:

${𝐸}_{\text{photon}}=𝜙+𝑒{𝑉}_0$

Where:

• $𝜙$ = work function = 4.2 eV

• ${𝑉}_0$ = stopping potential

So,

$10.2=4.2+{𝑉}_0$

${𝑉}_0=10.2-4.2=6.0\text{ V}$

Final Answer:

Option C: $6\text{V}$

When you drop the ball from height $ℎ$, it hits the floor and rebounds to a height

${ℎ}_1={𝑒}^2ℎ.$

Each subsequent bounce reaches ${𝑒}^2$ times the previous height, so the up-and-down distances form a geometric series.

Total distance $𝐷$ is

$𝐷=ℎ+2({ℎ}_1+{ℎ}_2+{ℎ}_3+\cdots )=ℎ+2ℎ({𝑒}^2+{𝑒}^4+{𝑒}^6+\cdots ).$

The infinite sum ${𝑒}^2+{𝑒}^4+{𝑒}^6+\cdots$ equals $\frac{{𝑒}^2}{1-{𝑒}^2}$. Hence

$𝐷=ℎ+2ℎ\frac{{𝑒}^2}{1-{𝑒}^2}=ℎ\frac{1-{𝑒}^2+2{𝑒}^2}{1-{𝑒}^2}=ℎ\frac{1+{𝑒}^2}{1-{𝑒}^2}.$

That matches option B:

$\overline{)\frac{1+{𝑒}^2}{1-{𝑒}^2}ℎ}.$

We have,

$\begin{array}{c}𝑃={𝑉}_{𝑧}{𝐼}_{𝑧}\Rightarrow \frac{1}{4}=5.6\times {𝐼}_{𝑧}\\ {𝐼}_{𝑧}=\frac{1}{4\times 5.6}\end{array}$

Also,  ${𝐼}_{𝑧}={𝐼}_{𝑠}$

Now, ${𝑉}_{𝑠}={𝐼}_{𝑠}{𝑅}_{𝑠}\Rightarrow {𝑅}_{𝑠}=\frac{{𝑉}_{𝑠}}{{𝐼}_{𝑠}}$

$\Rightarrow {𝑅}_{𝑠}=4.4\times 4\times 5.6=98.56\Omega$

Using Newton’s second law,

$\overrightarrow{𝐹}=𝑚\overrightarrow{𝑎}⟹\overrightarrow{𝑎}=\frac{\overrightarrow{𝐹}}{𝑚}=(\frac{𝑎}{𝑚},\frac{𝑏}{𝑚},\frac{𝑐}{𝑚}).$

Starting from rest at the origin, the displacement after time $𝑡$ is

$\overrightarrow{𝑟}\left(𝑡\right)=\frac{1}{2}\overrightarrow{𝑎}{𝑡}^2=\frac{1}{2}(\frac{𝑎}{𝑚},\frac{𝑏}{𝑚},\frac{𝑐}{𝑚}){𝑡}^2=(\frac{𝑎{𝑡}^2}{2𝑚},\frac{𝑏{𝑡}^2}{2𝑚},\frac{𝑐{𝑡}^2}{2𝑚}).$

Hence the correct choice is Option A.

We know that,

$\Rightarrow \begin{array}{cc}& 𝛿=𝑖+𝑒-𝐴\\ & 𝐴=𝑖+𝑒-𝛿\\ & =60+15-30={45}^{\circ }\end{array}$

Since, electrostatic force is conservative in nature, we use work energy theorem,

$\begin{array}{cc}& {𝑊}_{𝐶𝑆𝐷}=\text{ change in potential energy },\\ & {𝑊}_{𝐶𝑆𝐷}=𝑄({𝑉}_{𝐷}-{𝑉}_{𝐶})[\because {𝑉}_{𝐶}=0]\\ & {𝑊}_{𝐶𝑆𝐷}=𝑄[\frac{𝑞}{4𝜋{𝜀}_0\cdot 3𝐿}-\frac{𝑞}{4𝜋{𝜀}_0\cdot 𝐿}]\\ & {𝑊}_{𝐶𝑆𝐷}=\frac{𝑄𝑞}{4𝜋{𝜀}_0𝐿}[\frac{1}{3}-1]=\frac{-2𝑄𝑞}{12𝜋{𝜀}_0𝐿}\\ & {𝑊}_{𝐶𝑆𝐷}=-\frac{𝑄𝑞}{6𝜋{𝜀}_0𝐿}\end{array}$

And also, ${𝑊}_{𝐶𝑆𝐷}={𝑊}_{𝐶𝐵𝐷}$

                           ${𝑊}_1={𝑊}_2$

(Independent of path)

$\begin{array}{cc}& \text{ Output of logic gate is, }\\ & \begin{array}{cc}𝑌& =\overline{\overline{𝐴}+\overline{𝐵}}\\ & =\overline{𝐴}\cdot \overline{𝐵}=𝐴𝐵\\ & =\text{ Output of AND gate. }\end{array}\end{array}$

We have,

$𝑁={𝑁}_0{𝑒}^{-𝜆𝑡}$

Differentiate with respect to $𝑡$,

$\begin{array}{cc}& \frac{𝑑𝑁}{𝑑𝑡}={𝑁}_0{𝑒}^{-𝜆𝑡}(-𝜆)\\ & 𝐴=-\frac{𝑑𝑁}{𝑑𝑡}=𝜆𝑁\end{array}$

$\Rightarrow 𝐴=𝜆𝑁$ (straight line passing through origin. )

Given, $𝑝-𝑇$ diagram for cyclic process $𝐴𝐵𝐶𝐴$ Use ideal gas law.

$𝑉\propto \frac{𝑇}{𝑝}$

Process analysis $𝐴𝐵:𝑝$ constant, $𝑇<text>\downarrow </text><text>\Rightarrow </text>𝑉<text>\downarrow </text><text>\to </text>$ isobaric compression $𝐵𝐶:𝑇$ constant, $𝑝<text>\downarrow </text><text>\Rightarrow </text>𝑉<text>\uparrow </text><text>\to </text>$ isothermal expansion $𝐶𝐴:𝑝\uparrow ,𝑇<text>\uparrow </text><text>\Rightarrow </text>𝑉$ constant.

From the options, only option (c) shows $𝐴𝐵$ as horizontal line (constant $𝑝$ ), $𝐵𝐶$ as smooth curve (isothermal), $𝐶𝐴$ as vertical line (constant $𝑉$ )

The fractional (percentage) error in a quotient $𝑅=𝑉/𝐼$ is the sum of the fractional errors of $𝑉$ and $𝐼$:

$\frac{\Delta 𝑅}{𝑅}=\frac{\Delta 𝑉}{𝑉}+\frac{\Delta 𝐼}{𝐼}.$

Here

$\frac{\Delta 𝑉}{𝑉}=\frac{0.4}{25}=0.016=1.6\%$,

$\frac{\Delta 𝐼}{𝐼}=\frac{3}{200}=0.015=1.5\%$.

So

$\frac{\Delta 𝑅}{𝑅}=0.016+0.015=0.031=3.1\%.$

Answer: 3.1 % (Option C).

The magnetic moment is

$𝜇=𝐼𝐴=\frac{𝑞𝜔}{2𝜋}\times \left(𝜋{𝑟}^2\right)=\frac{𝑞𝜔{𝑟}^2}{2},$

and the angular momentum is

$𝐿=𝑚{𝑟}^2𝜔.$

Thus

$\frac{𝜇}{𝐿}=\frac{\frac{𝑞𝜔{𝑟}^2}{2}}{𝑚{𝑟}^2𝜔}=\frac{𝑞}{2𝑚},$

which depends only on $𝑞$ and $𝑚$. (Option C)

Let the initial mass of the bus be ${𝑚}_1$.

Initial speed $𝑣$ ⇒

${𝐾}_1=\frac{1}{2}{𝑚}_1{𝑣}^2,{𝜆}_1=\frac{ℎ}{{𝑚}_1𝑣}\equiv 𝜆.$

After some passengers leave, the mass becomes ${𝑚}_2$.

New speed $2𝑣$ ⇒

${𝐾}_2=\frac{1}{2}{𝑚}_2(2𝑣{)}^2=2{𝑚}_2{𝑣}^2.$

Given ${𝐾}_2=2{𝐾}_1=2(\frac{1}{2}{𝑚}_1{𝑣}^2)={𝑚}_1{𝑣}^2$, so

$2{𝑚}_2{𝑣}^2={𝑚}_1{𝑣}^2⟹{𝑚}_2=\frac{{𝑚}_1}{2}.$

New de Broglie wavelength

${𝜆}_2=\frac{ℎ}{{𝑚}_2\left(2𝑣\right)}=\frac{ℎ}{(\frac{{𝑚}_1}{2})2𝑣}=\frac{ℎ}{{𝑚}_1𝑣}=𝜆.$

Answer: $𝜆$.

Displacement:

$𝑦\left(𝑡\right)=2\sin (\frac{𝜋𝑡}{2}+𝜙)cm$

so the angular frequency is

$𝜔=\frac{𝜋}{2}{s}^{-1}.$

Acceleration amplitude is

${𝑎}_{max}={𝜔}^2𝐴=(\frac{𝜋}{2}{)}^2\cdot 2=\frac{{𝜋}^2}{2}cm/s^2.$

Answer: $\frac{{𝜋}^2}{2}cm/s^2$.

$\text{ After steady state of current, }$

$\begin{array}{cc}& 𝐼=\frac{6}{3\times {10}^3}=2\times {10}^{-3}A\\ & \\ & \begin{array}{cc}𝑁𝑜𝑤,{𝑉}_{𝐴𝐵}& =𝐼𝑅=2\times {10}^{-3}\times 2\times {10}^3=4V\\ {𝑞}_1& ={𝐶}_1𝑉=1\times 4=4𝜇𝐶\\ {𝑞}_2& ={𝐶}_2𝑉=2\times 4=8𝜇𝐶\end{array}\end{array}$

Pressure on left side,

${𝑝}_1={𝑝}_{\mathrm{atm}}+{𝜌}_1ℎ𝑔+{𝜌}_3\frac{ℎ}{2}𝑔$

Pressure on right side,

${𝑝}_2={𝑝}_{\mathrm{atm}}+{𝜌}_2ℎ𝑔$

Equating the pressures,

$\begin{array}{cc}{𝑝}_1& ={𝑝}_2\\ {𝑝}_{\mathrm{atm}}+{𝜌}_1ℎ𝑔+{𝜌}_3\frac{ℎ}{2}𝑔& ={𝑝}_{\mathrm{atm}}+{𝜌}_2ℎ𝑔\\ \Rightarrow {𝜌}_3& =2({𝜌}_2-{𝜌}_1)\end{array}$

$=$ Sum of Buoyant forces from water and mercury.

$F=F_w+F_m$

$$\begin{aligned}Weight\,\, of \,\,granite,\,\, W & =\rho g V \\ F_w & =\rho_1 g V_w \\and\,\, F_m & =\rho_2 g V_m \end{aligned}$$

Since, the granite is floating,

$\rho g V=\rho_1 g V_W+\rho_2 g V_m \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(i)$

and also,

$$\begin{aligned} V & =V_m+V_W \\ V_W & =V-V_m \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(ii)\end{aligned}$$

Using Eq. (i) and (ii)

$V_m=\frac{\left(\rho-\rho_1\right) V}{\left(\rho_2-\rho_1\right)}, V_w=\frac{\left(\rho_2-\rho\right) V}{\left(\rho_2-\rho_1\right)}$

Now,

$\frac{V_w}{V_m}=\frac{\left(\rho_2-\rho\right)}{\left(\rho-\rho_1\right)}$