MHT CET 2025 25th April Evening Shift (Physics & Chemistry) Question Paper with Solutions

For a pipe open at both ends, the fundamental frequency is given by $n_1=\frac{V}{2L}$

For a pipe closed at one end, the fundamental frequency is given by $n_2=\frac{v}{4L}$

$\therefore n_1=2n_2$

A square plate of side $𝐿$ is subjected to force $𝐹$. The pressure is

$𝑃=\frac{𝐹}{{𝐿}^2}.$

Step 1: Error propagation formula

For a quantity of the form

$𝑄=\frac{{𝐹}^{𝑎}}{{𝐿}^{𝑏}},$

the percentage error is

$\frac{\Delta 𝑄}{𝑄}\times 100=𝑎\cdot \frac{\Delta 𝐹}{𝐹}\times 100+𝑏\cdot \frac{\Delta 𝐿}{𝐿}\times 100.$

Step 2: Apply to our case

Here

$𝑃=\frac{𝐹}{{𝐿}^2},𝑎=1,𝑏=2.$

So

$\text{percentage error in }𝑃=\left(1\right)\left(\%\text{ error in }𝐹\right)+\left(2\right)\left(\%\text{ error in }𝐿\right).$

Step 3: Substitute values

$=1\cdot 3\%+2\cdot 2\%=3\%+4\%=7\%.$

✅ Answer: Option A, 7%

$\begin{array}{cc}& \text{ According to law of conservation of energy, }\\ & \begin{array}{cc}& \text{ P.E }=\text{ K.E }\\ \therefore & \mathrm{mgh}=\frac{1}{2}{\mathrm{mv}}^2+\frac{1}{2}I{𝜔}^2\\ \therefore \mathrm{mgh}=& \frac{1}{2}m(𝜔𝑟{)}^2+\frac{1}{2}\left(\frac{{\mathrm{mr}}^2}{2}\right){𝜔}^2\\ & \dots .(\because v=r𝜔,I=\frac{{\mathrm{mr}}^2}{2})\end{array}\end{array}$

$\begin{array}{cc}\therefore & \mathrm{mgh}=\frac{3}{4}{m}^2{r}^2\\ \therefore & \mathrm{gh}=\frac{3}{4}{𝜔}^2{r}^2\\ \therefore & 𝜔=\sqrt{\surd \frac{4\mathrm{gh}}{3r^2}}=\frac{2}{r}\sqrt{\surd \frac{\mathrm{gh}}{3}}\end{array}$

When there is a cut in the ring, e.m.f. will be induced in it but there is no induced current in the ring. Hence there is no opposition to falling magnet. Therefore, acceleration is equal to ' $𝑔$ '.

If $I_0$ is the intensity of each wave and $𝜙$ is the phase difference, then the resultant intensity is given by

$𝐼=4{𝐼}_0{\cos }^2\frac{𝜙}{2}$

When path difference $=0$, phase difference $𝜙=0$, $I_1=4I_0{\cos }^{2}0=4I_0$

When path difference $=\frac{𝜆}{4}$, phase difference

$\begin{array}{cc}& 𝜙=\frac{𝜋}{2}\\ \therefore & {𝐼}_2=4{𝐼}_0{\cos }^2\frac{𝜋}{4}=2{𝐼}_0\\ \therefore & \frac{{𝐼}_1}{{𝐼}_2}=\frac{4{𝐼}_0}{2{𝐼}_0}=\frac{2}{1}\end{array}$

$\begin{array}{ccccc}𝐀& 𝐁& 𝐂& 𝐁+𝐂& 𝐘=(𝐁+𝐂)\cdot 𝐀\\ 0& 0& 0& 0& 0\\ 0& 1& 0& 1& 0\\ 1& 0& 0& 0& 0\\ 1& 0& 1& 1& 1\end{array}$

Initial temperature $T_1={27}^{\circ }C=27+273=300K$

Final temperature $T_2={327}^{\circ }C=327+273=600K$

Average kinetic energy of an ideal gas molecule is directly propartional to the abosulate temperature T .

$\begin{array}{cc}& \mathrm{KE}\propto T\\ \therefore & \frac{{\mathrm{KE}}_2}{{\mathrm{KE}}_1}=\frac{T_2}{T_1}\Rightarrow \frac{{\mathrm{KE}}_2}{{\mathrm{KE}}_1}=\frac{600}{300}\\ \therefore & \frac{{\mathrm{KE}}_2}{{\mathrm{KE}}_1}=2\\ \therefore & \frac{{\mathrm{KE}}_2}{E}=2\\ \therefore & {\mathrm{KE}}_2=2E\end{array}$

In a semiconductor, increasing temperature increases the number of charge carriers, which enhances conductivity. Thus, the current in the circuit will increase.

As required, equivalent capacitance should be,

$C_{\mathrm{eq}}=\frac{10}{11}𝜇F$

For combination of the five capacitors in parallel and two in series,

$\begin{array}{cc}& \therefore \frac{1}{{𝐶}_{\text{eq }}}=\frac{1}{5𝐶}+[\frac{1}{𝐶}+\frac{1}{𝐶}]=\frac{𝐶+10𝐶}{5{𝐶}^2}\\ & \therefore {𝐶}_{\text{eq }}=\frac{5𝐶}{11}\\ & \\ & =\frac{5\times 2}{11}\\ & =\frac{10}{11}𝜇\mathrm{Fs}\end{array}$

$\therefore$ The required combination is 5 capacitors in parallel and 2 in series

Step 1: Write down given data

• $𝐿=\frac{0.3}{𝜋}\text{H}$

• $𝑅=40\Omega$

• Supply voltage $𝑉=230\text{V (rms)}$

• Frequency $𝑓=50\text{Hz}$

Step 2: Reactance of inductor

${𝑋}_{𝐿}=2𝜋𝑓𝐿$

${𝑋}_{𝐿}=2𝜋\left(50\right)\left(\frac{0.3}{𝜋}\right)=2\times 50\times 0.3=30\Omega$

Step 3: Impedance of circuit

$𝑍=\sqrt{\surd {𝑅}^2+{𝑋}_2^{𝐿}}$

$𝑍=\sqrt{\surd {40}^2+{30}^2}=\sqrt{\surd 1600+900}=\sqrt{\surd 2500}=50\Omega$

Step 4: Current

$𝐼=\frac{𝑉}{𝑍}=\frac{230}{50}=4.6\text{A}$

✅ Final Answer:

• Impedance: $50\Omega$

• Current: $4.6\text{A}$

Correct option: D

Step 1: Gravity inside the Earth

For a uniform Earth of radius $𝑅$,

$𝑔\left(ℎ\right)=𝑔(1-\frac{ℎ}{𝑅})$

where:

• $𝑔$ is surface gravity,

• $ℎ$ is depth below the surface.

Equivalently, at radial distance $𝑟=𝑅-ℎ$:

$𝑔\left(𝑟\right)=𝑔\cdot \frac{𝑟}{𝑅}$

Step 2: Set condition

We want:

$𝑔\left(ℎ\right)=\frac{1}{𝑛}𝑔$

So:

$𝑔(1-\frac{ℎ}{𝑅})=\frac{1}{𝑛}𝑔$

Cancel $𝑔$:

$1-\frac{ℎ}{𝑅}=\frac{1}{𝑛}$

Step 3: Solve for depth

$\frac{ℎ}{𝑅}=1-\frac{1}{𝑛}=\frac{𝑛-1}{𝑛}$

$ℎ=\frac{𝑅(𝑛-1)}{𝑛}$

✅ Final Answer:

$\overline{)\frac{𝑅(𝑛-1)}{𝑛}}$

Correct Option: A

A horizontal pipeline with flowing water. At one section:

• Cross-sectional area ${𝐴}_1=10{\text{cm}}^2=10\times {10}^{-4}{\text{m}}^2=1.0\times {10}^{-3}{\text{m}}^2$

• Velocity ${𝑣}_1=1.0\text{m/s}$

• Pressure ${𝑃}_1=2000\text{Pa}$

Density of water $𝜌=1000{\text{kg/m}}^3$.

At another section: ${𝐴}_2=5{\text{cm}}^2=5\times {10}^{-4}{\text{m}}^2$. Pressure ${𝑃}_2$?

Step 1: Continuity equation

${𝐴}_1{𝑣}_1={𝐴}_2{𝑣}_2$

$(1.0\times {10}^{-3})\left(1\right)=(5\times {10}^{-4})\left({𝑣}_2\right)$

${𝑣}_2=\frac{1.0\times {10}^{-3}}{5\times {10}^{-4}}=2.0\text{m/s}$

Step 2: Apply Bernoulli’s theorem

Since pipe is horizontal (no height change):

${𝑃}_1+\frac{1}{2}𝜌{𝑣}_2^1={𝑃}_2+\frac{1}{2}𝜌{𝑣}_2^2$

So:

${𝑃}_2={𝑃}_1+\frac{1}{2}𝜌({𝑣}_2^1-{𝑣}_2^2)$

$=2000+\frac{1}{2}\left(1000\right)(1^2-2^2)$

$=2000+500(1-4)$

$=2000-1500=500\text{Pa}$

✅ Final Answer:

The pressure at the second point is

Option C: 500 Pa

Total energy of a simple pendulum,

$𝐸=\frac{1}{2}𝑀{𝜔}^2{𝐴}^2$

For a simple pendulum,

$𝜔=\sqrt{\surd 𝑔/𝐿}$

For pendulum A:

${𝐸}_{𝐴}=\frac{1}{2}{𝑀}_1\left(\frac{𝑔}{𝐿}\right){𝐴}_2^1$

For pendulum B :

${𝐸}_{𝐵}=\frac{1}{2}{𝑀}_2\left(\frac{𝑔}{𝐿}\right){𝐴}_2^2$

Given:

$\begin{array}{cc}& {𝐸}_{𝐴}={𝐸}_{𝐵}\\ & \frac{1}{2}{𝑀}_1\left(\frac{𝑔}{{𝐿}_1}\right){𝐴}_2^1=\frac{1}{2}{𝑀}_2\left(\frac{𝑔}{{𝐿}_2}\right){𝐴}_2^2\\ \therefore & \frac{{𝐴}_2^1}{{𝐴}_2^2}=\frac{{𝐿}_1}{{𝐿}_2}\dots \dots .({𝑀}_1={𝑀}_2)\end{array}$

$\therefore A_2^1=2A_2^2\dots \dots (L_1=2L_2)$

$\begin{array}{cc}& \therefore A_1>A_2\\ & \therefore \text{ Amplitude of }𝐵\text{ is smaller than amplitude of }𝐴\end{array}$

By photoelectric effect,

${\mathrm{eV}}_0=h𝑣-𝜙$

$\therefore V_0=\frac{h𝑣-𝜙}{e}\dots (V_0=$ initial stopping potential)

Initial frequency $\left(𝑣\right)$ is doubled.

$\begin{array}{cc}& \begin{array}{cc}\therefore & {𝑉}_{\prime }^0=\frac{ℎ\left(2𝑣\right)-{𝜙}_0}{𝑒}\dots ({𝑉}_{\prime }^0=\text{ New stopping potential })\\ & {𝑉}_{\prime }^0=\frac{2ℎ𝑣-{𝜙}_0}{𝑒}\\ & {𝑉}_{\prime }^0=2\cdot \frac{ℎ𝑣-𝜙}{𝑒}+\frac{𝜙}{𝑒}=2{𝑉}_0+\frac{𝜙}{𝑒}\\ \therefore & {𝑉}_{\prime }^0>2{𝑉}_0\end{array}\\ \therefore & \text{ Stopping potential becomes more than double. }\end{array}$

     Work done $=\overrightarrow{F}\cdot \overrightarrow{r}$

$\therefore$  Work done $=(3\widehat{𝑖}-2\widehat{𝑗}-\widehat{𝑘})\cdot (2\widehat{𝑖}-3\widehat{𝑗}-3\widehat{𝑘})$

$\therefore$  Work done $=(6+6+3)$

$\therefore$  Work done $=15$ joule

Power $=\frac{\text{ work done }}{\text{ time }}$

$\therefore$  Work done $=\frac{15}{2}$

$\therefore$   Work done $=7.5$ watt.

In a common emitter amplifier, the voltage gain $A_v$ :

$\begin{array}{cc}& {𝐴}_{𝑣}=𝛽\cdot \frac{{𝑅}_{𝑐}}{{𝑅}_{\text{in }}}\\ \therefore & {𝐴}_{𝑣}=62\cdot \frac{5000}{500}=62.10\\ \therefore & {𝐴}_{𝑣}=620\end{array}$

Now, output voltage $V_{\text{out }}$ is:

$\begin{array}{cc}& {𝑉}_{\text{out }}={𝐴}_{𝑉}{𝑉}_{\text{in }}\\ \therefore & {𝑉}_{\text{out }}=620\times \left(0.01\right)\\ \therefore & {𝑉}_{\text{out }}=6.2V\end{array}$

Step 1: Concept

• The rod swings like a physical pendulum about one end.

• When its angular speed is maximum, it must be passing through the lowest vertical position (since at the lowest position, kinetic energy is max and potential energy min).

• At the highest point, angular speed becomes zero (purely potential energy).

Thus, energy conservation applies:

$\text{Maximum rotational KE}=\text{Gain in gravitational PE of CM}.$

Step 2: Energy expressions

• For a rod of mass $𝑀$, length $𝐿$, pivot at one end:

Moment of inertia about one end is

$𝐼=\frac{1}{3}𝑀{𝐿}^2.$

• Rotational KE when angular speed = $𝜔$:

$𝐾=\frac{1}{2}𝐼{𝜔}^2=\frac{1}{2}\cdot \frac{1}{3}𝑀{𝐿}^2{𝜔}^2=\frac{1}{6}𝑀{𝐿}^2{𝜔}^2.$

• Rise in potential energy of centre of mass:

• CM is initially below pivot by distance $𝐿/2$.

• If it rises a vertical height $ℎ$, gravitational potential energy increases by:

$\Delta 𝑈=𝑀𝑔ℎ.$

Step 3: Apply conservation of energy

$\frac{1}{6}𝑀{𝐿}^2{𝜔}^2=𝑀𝑔ℎ.$

$ℎ=\frac{{𝐿}^2{𝜔}^2}{6𝑔}.$

Step 4: Final Answer

$\overline{)\frac{{𝐿}^2{𝜔}^2}{6𝑔}}$

That corresponds to Option D. ✅

$\begin{array}{cc}& \begin{array}{cc}& W=\frac{𝜆D}{d}\\ \therefore & \frac{W_1}{W_2}=\frac{{𝜆}_1}{{𝜆}_2}\\ \therefore & \frac{W_2-W_1}{W_1}=\frac{{𝜆}_2-{𝜆}_1}{{𝜆}_1}=\frac{6400-5000}{5000}=\frac{1400}{5000}\\ & =\frac{7}{25}=0.28=28\%\end{array}\\ & \text{ Therefore, the fringe width will increase by }28\%\text{. }\end{array}$

Let distance covered by girl running from point $𝐵$ be ' $𝑥$ ',

$\therefore x=V_{1}t...\left(𝑖\right)$

Let distance covered by girl running from point A in time ' $𝑡$ ' be ' $𝑦$ ',

$\therefore 𝑦={𝑉}_2𝑡...\left(𝑖𝑖\right)$

Given that, B follows a perpendicular path and A follows the shortest path, i.e. $𝑦=\sqrt{\surd {𝑏}^2+{𝑥}^2}$

$\therefore y^2=b^2+x^2$

Substituting from equation (i) \& (ii),

$\begin{array}{cc}& {𝑉}_2^2{𝑡}^2={𝑏}^2+{𝑉}_2^1{𝑡}^2\\ \therefore & 𝑡=\frac{𝑏}{\sqrt{\surd {𝑉}_2^2-{𝑉}_2^1}}\end{array}$

Step 1: Recall the dependence of self-inductance

For a given geometry (same core, radius, length, etc.), the self-inductance scales as

$𝐿\propto {𝑁}^2$

where $𝑁$ is the number of turns.

Step 2: Set up ratio

Given:

• ${𝑁}_1=600$, ${𝐿}_1=108\text{mH}$

• ${𝑁}_2=500$, ${𝐿}_2=?$

So,

$\frac{{𝐿}_2}{{𝐿}_1}={\left(\frac{{𝑁}_2}{{𝑁}_1}\right)}^2$

${𝐿}_2={𝐿}_1{\left(\frac{{𝑁}_2}{{𝑁}_1}\right)}^2$

Step 3: Substitute values

${𝐿}_2=108\times {\left(\frac{500}{600}\right)}^2$

$\frac{500}{600}=\frac{5}{6}$

${\left(\frac{5}{6}\right)}^2=\frac{25}{36}$

${𝐿}_2=108\times \frac{25}{36}$

$=3\times 25=75\text{mH}$

✅ Final Answer:

The self-inductance is

Option B: 75 mH

For constant pressure,

Using first law of thermodynamics,

$\begin{array}{cc}& W=Q-\Delta U\\ & W={\mathrm{mC}}_p\Delta T-{\mathrm{mC}}_v\Delta T\end{array}$

For monoatomic gas $𝛾=\frac{C_p}{C_v}=\frac{5}{3}$

$\begin{array}{cc}\Rightarrow & {𝐶}_{𝑣}=\frac{3}{5}{𝐶}_{𝑝}\\ \therefore & 𝑊=𝑚{𝐶}_{𝑝}\Delta 𝑇-𝑚\left(\frac{3}{5}{𝐶}_{𝑝}\right)\Delta 𝑇\\ \therefore & 𝑊=\frac{2}{5}𝑚{𝐶}_{𝑝}\Delta 𝑇=\frac{2}{5}\times 𝑄\Rightarrow 𝑊=\frac{2}{5}\times 100\\ \therefore & 𝑊=40𝐽\end{array}$

Step 1: Recall the quantisation condition

Bohr's quantisation condition for angular momentum:

$𝐿=𝑛\frac{ℎ}{2𝜋},𝑛=1,2,3,\dots$

Step 2: Express the rotational energy

Rotational kinetic energy is:

$𝐸=\frac{{𝐿}^2}{2𝐼}$

Substitute $𝐿=𝑛\frac{ℎ}{2𝜋}$:

${𝐸}_{𝑛}=\frac{1}{2𝐼}{\left(𝑛\frac{ℎ}{2𝜋}\right)}^2$

${𝐸}_{𝑛}=\frac{{𝑛}^2{ℎ}^2}{8{𝜋}^2𝐼}$

Step 3: For the second orbit (n=2)

${𝐸}_2=\frac{(2{)}^2{ℎ}^2}{8{𝜋}^2𝐼}=\frac{4{ℎ}^2}{8{𝜋}^2𝐼}=\frac{{ℎ}^2}{2𝐼{𝜋}^2}$

✅ Final Answer:

$\overline{)\text{Option B: }\frac{{ℎ}^2}{2𝐼{𝜋}^2}}$

Potential at surface of a sphere is,

$𝑉=𝐾\frac{𝑞}{𝑟}=20V$

where, $𝑞$ and $𝑟$ are the charge and radius of the small drop respectively.

As the volume of 27 small drops equals volume of the large drop,

$27\left(\frac{4}{3}𝜋{𝑟}^3\right)=\frac{4}{3}𝜋{𝑅}^3$

where, $𝑅$ is radius of large drop.

$\therefore 𝑅=3𝑟$

The total charge on large drop is $Q=27q$ Hence, potential at surface of this drop is

$\begin{array}{cc}& {𝑉}^{\prime }=\frac{𝐾𝑄}{𝑅}=𝐾\frac{\left(27𝑞\right)}{\left(3𝑟\right)}=9\frac{𝐾𝑞}{𝑟}=9\times 20\\ \therefore & {𝑉}^{\prime }=180𝑉\end{array}$

Step 1: Write momentum conservation

Collision is assumed to be in 1D and no external force acts.

Initial momentum:

${𝑝}_{𝑖}=𝑚𝑢+𝑀\cdot 0=𝑚𝑢$

Final momentum (since mass $𝑚$ stops after collision, its velocity = 0; velocity of $𝑀=𝑉$):

${𝑝}_{𝑓}=𝑚\cdot 0+𝑀𝑉=𝑀𝑉$

By conservation of momentum:

$𝑚𝑢=𝑀𝑉\Rightarrow 𝑉=\frac{𝑚}{𝑀}𝑢$

Step 2: Coefficient of restitution formula

Coefficient of restitution $𝑒$ is defined as:

$𝑒=\frac{\text{relative velocity after}}{\text{relative velocity before}}$

Relative velocity before collision:

${𝑢}_{\text{rel,before}}=𝑢-0=𝑢$

Relative velocity after collision:

Object $𝑚$ has velocity = 0, object $𝑀$ has velocity $𝑉=\frac{𝑚}{𝑀}𝑢$.

So relative velocity after collision:

${𝑢}_{\text{rel,after}}=𝑉-0=\frac{𝑚}{𝑀}𝑢$

Thus,

$𝑒=\frac{\frac{𝑚}{𝑀}𝑢}{𝑢}=\frac{𝑚}{𝑀}$

Final Answer:

The coefficient of restitution is

$\overline{)\frac{𝑚}{𝑀}}$

Correct option: C

Magnetic fields due to different portions 1, 2 and 3 are respectively,

$B_1=0$,

$B_2=\frac{{𝜇}_0}{4𝜋}\cdot \frac{𝜋I}{r}$ (directed outside the paper)

$B_3=\frac{{𝜇}_0}{4𝜋}\cdot \frac{I}{r}$ (directed outside the paper)

$\therefore {𝐵}_{𝑛𝑒𝑡}={𝐵}_2+{𝐵}_3=\frac{{𝜇}_0𝐼}{4𝜋𝑟}+\frac{{𝜇}_0𝐼}{4𝑟}$

Initial temperture $\left(T_1\right)={27}^{\circ }C=300K$

Final temperature $\left(T_2\right)=-3^{\circ }C=270K$

From combined Gas Law,

$\frac{{𝑃}_1{𝑉}_1}{{𝑇}_1}=\frac{{𝑃}_2{𝑉}_2}{{𝑇}_2}$

Substituting values:

$\begin{array}{cc}\therefore & \frac{1\times 500}{300}=\frac{0.5\times V_2}{270}\\ \therefore & V_2=\frac{5\times 270}{3\times 0.5}\\ \therefore & V_2=900m^3\end{array}$

Case 1: With both springs ${𝑆}_1\&{𝑆}_2$

Effective spring constant: ${𝐾}_{\text{eff }}:𝐾+𝐾=2𝐾$

Oscillation frequency: $𝑓=\frac{1}{2𝜋}\sqrt{\surd \frac{{𝐾}_{\text{eff }}}{𝑚}}=\frac{1}{2𝜋}\sqrt{\surd \frac{2𝐾}{𝑚}}$

Case II: After removing ${𝑆}_2$

Now only one spring with spring constant K

New frequency: $f^{\prime }=\frac{1}{2𝜋}\sqrt{\surd \frac{K}{m}}$

Divide both cases: $\frac{{𝐹}^{\prime }}{𝐹}=\frac{\sqrt{\surd \frac{𝐾}{𝑚}}}{\sqrt{\surd \frac{2𝐾}{𝑚}}}$

$\begin{array}{cc}\Rightarrow & \frac{{𝐹}^{\prime }}{𝐹}=\frac{1}{\sqrt{\surd 2}}\\ \therefore & {𝐹}^{\prime }=\frac{𝐹}{\sqrt{\surd 2}}\end{array}$

Step 1: Recall relation between volumetric and linear expansion.

• Volumetric expansion coefficient: $𝛽=3𝛼$

  where $𝛼$ = coefficient of linear expansion.

Relative change in volume:

$\frac{\Delta 𝑉}{𝑉}=𝛽\Delta 𝑇=3𝛼\Delta 𝑇$

Step 2: Express given data.

• Percentage change in volume = $0.33\%=0.0033$ (in decimal).

So,

$\frac{\Delta 𝑉}{𝑉}=0.0033$

• Temperature rise = $\Delta 𝑇={50}^{\circ }C$.

Step 3: Use formula.

$0.0033=3𝛼\left(50\right)$

$0.0033=150𝛼$

$𝛼=\frac{0.0033}{150}$

$𝛼=2.2\times {10}^{-5}{/}^{\circ }C$

✅ Final Answer:

$\overline{)2.2\times {10}^{-5}{/}^{\circ }C}$

Correct option: A

Step 1: Concept

The work done to increase the size of bubble goes into increasing surface energy.

• Surface energy $𝐸=\text{surface area}\times \text{surface tension (T)}$.

• For a soap bubble, there are two surfaces (inner and outer), hence we need to account for 2 × area.

So,

$𝐸=2\times \left(4𝜋{𝑟}^2\right)\times 𝑇=8𝜋{𝑟}^2𝑇$

Step 2: Initial and Final Energies

• Initial radius ${𝑟}_{𝑖}=\frac{𝑑}{2}$.

• Final radius ${𝑟}_{𝑓}=\frac{𝐷}{2}$.

• Initial energy:

${𝐸}_{𝑖}=8𝜋{\left(\frac{𝑑}{2}\right)}^2𝑇=8𝜋\frac{{𝑑}^2}{4}𝑇=2𝜋{𝑑}^2𝑇$

• Final energy:

${𝐸}_{𝑓}=8𝜋{\left(\frac{𝐷}{2}\right)}^2𝑇=8𝜋\frac{{𝐷}^2}{4}𝑇=2𝜋{𝐷}^2𝑇$

Step 3: Work Done

$𝑊={𝐸}_{𝑓}-{𝐸}_{𝑖}=2𝜋𝑇({𝐷}^2-{𝑑}^2)$

Final Answer:

$\overline{)2𝜋({𝐷}^2-{𝑑}^2)𝑇}$

✅ That corresponds to Option B.

Step 1: Relation between phase angle and reactance

For a series $𝑅𝐿$ circuit, the phase angle $𝜙$ between voltage and current is:

$\tan 𝜙=\frac{{𝑋}_{𝐿}}{𝑅}$

where ${𝑋}_{𝐿}=𝜔𝐿$ = inductive reactance.

Step 2: Apply the given data

$𝜙={45}^{\circ },\tan {45}^{\circ }=1$

So,

$\tan 𝜙=\frac{{𝑋}_{𝐿}}{𝑅}=1$

$⟹{𝑋}_{𝐿}=𝑅$

✅ Final Answer:

The value of inductive reactance is $𝑅$.

Correct Option: A (R) ✅

Step 1: Behavior of open organ pipe

• In an open organ pipe, both ends are open.

• At open ends, there are pressure nodes (displacement antinodes).

• Therefore, the fundamental mode (first harmonic) has two open ends, meaning:

• The pipe length $𝐿=\frac{𝜆}{2}$.

Step 2: Fundamental frequency

From wave relation:

$𝑓=\frac{𝑣}{𝜆}$

For fundamental,

$𝐿=\frac{𝜆}{2}\Rightarrow 𝜆=2𝐿$

So,

${𝑓}_1=\frac{𝑣}{𝜆}=\frac{𝑣}{2𝐿}.$

Step 3: Harmonics present

• In an open-open pipe, all harmonics are present:

  Fundamental, 2nd, 3rd, etc.

✅ Final Answer

$\overline{)\frac{𝑉}{2𝐿}\text{ and all harmonics are present.}}$

That corresponds to Option A.

Applying Kirchhoff's voltage law to the Given loop QPQ ,

$\begin{array}{cc}& -2I+8-4-(1\times I)-9I=0\Rightarrow I=\frac{1}{3}A\\ & \therefore \text{ Potential difference across }\mathrm{PQ}=\frac{1}{3}\times 9=3V\end{array}$

Step 1. Recall Doppler effect formula

When the source is moving (with constant velocity) and the observer is stationary, the apparent frequency ${𝑓}^{\prime }$ is given by:

${𝑓}^{\prime }=𝑓\cdot \frac{𝑣}{𝑣+{𝑣}_{𝑠}}$

where

• $𝑓$ = actual frequency,

• $𝑣$ = speed of sound in air,

• ${𝑣}_{𝑠}$ = speed of source (positive if source is moving away from observer).

Step 2. Condition given

We want the apparent frequency to be half the original:

${𝑓}^{\prime }=\frac{𝑓}{2}$

Substitute in the formula:

$\frac{𝑓}{2}=𝑓\cdot \frac{𝑣}{𝑣+{𝑣}_{𝑠}}$

Cancel $𝑓$:

$\frac{1}{2}=\frac{𝑣}{𝑣+{𝑣}_{𝑠}}$

Step 3. Solve for ${𝑣}_{𝑠}$

Cross multiply:

$𝑣+{𝑣}_{𝑠}=2𝑣$

${𝑣}_{𝑠}=𝑣$

Step 4. Final Answer

The source should move with speed equal to the speed of sound $𝑉$.

Correct Option: D (V). ✅

$\begin{array}{cc}& \text{ For SHM, }\\ & \begin{array}{cc}& \text{ P.E. }=\frac{1}{2}𝑚{𝜔}^2{𝑥}^2...\left(𝑖\right)\\ & \text{ and K.E. }=\frac{1}{2}𝑚{𝜔}^2({𝐴}^2-{𝑥}^2)...\left(𝑖𝑖\right)\\ & \text{ Given }𝑡=\frac{𝑇}{6}\end{array}\end{array}$