MHT CET 2025 26th April Evening Shift (Physics & Chemistry) Question Paper with Solutions

Step 1. Recall expression for orbital magnetic moment

An electron in the $𝑛$th Bohr orbit has current loop behavior. The orbital magnetic moment is

$𝜇=\frac{𝑒}{2𝑚}𝐿,$

where $𝐿$ is the orbital angular momentum.

Step 2. Angular momentum in Bohr model

From Bohr’s quantization condition:

${𝐿}_{𝑛}=𝑛\hslash .$

So,

${𝜇}_{𝑛}=\frac{𝑒}{2𝑚}\left(𝑛\hslash \right).$

Step 3. Relation to Bohr magneton

${𝜇}_{𝐵}=\frac{𝑒\hslash }{2𝑚}.$

So,

${𝜇}_{𝑛}=𝑛{𝜇}_{𝐵}.$

Answer:

The orbital magnetic moment is proportional to $𝑛$.

✅ Correct option: A. $𝑛$

Moment of inertia of disc $t_d=\frac{{\mathrm{MR}}^2}{2}$

Moment of inertia of ring $I_r={\mathrm{MR}}^2$

$\therefore {𝐼}_{𝑟}>{𝐼}_{𝑑}$

Angular acceleration $𝛼=\frac{𝜏}{1}$

As torque is same, $𝛼\propto \frac{1}{1}$

$\because {𝛼}_{𝑑}>{𝛼}_{𝑟}$

$\therefore$ Disc will rotate with greater angular frequency

Step 1: Formula for charge stored

The charge stored in a capacitor is given by the formula: $𝑄=𝐶𝑉$ where $𝑄$ is the charge, $𝐶$ is the capacitance, and $𝑉$ is the voltage.

Step 2: Charge after voltage decreases

When the voltage is reduced by ${𝑉}_1$, the new voltage becomes $𝑉-{𝑉}_1$. The new charge stored is: ${𝑄}_1=𝐶(𝑉-{𝑉}_1)$

Step 3: Isolate $𝐶$ (capacitance) from the second equation

We can find $𝐶$ from the second equation: $𝐶=\frac{{𝑄}_1}{𝑉-{𝑉}_1}$

Step 4: Substitute $𝐶$ back into the first equation

Now, substitute $𝐶$ into the first equation: $𝑄=\left(\frac{{𝑄}_1}{𝑉-{𝑉}_1}\right)𝑉$

Step 5: Rearranging to solve for $𝑉$

Multiply both sides by $(𝑉-{𝑉}_1)$: $𝑄(𝑉-{𝑉}_1)={𝑄}_1𝑉$

Expand and bring terms involving $𝑉$ to one side: $𝑄𝑉-𝑄{𝑉}_1={𝑄}_1𝑉$

Now, get all $𝑉$ terms together: $𝑄𝑉-{𝑄}_1𝑉=𝑄{𝑉}_1$

Factor $𝑉$ on the left: $𝑉(𝑄-{𝑄}_1)=𝑄{𝑉}_1$

Now, solve for $𝑉$: $𝑉=\frac{𝑄{𝑉}_1}{𝑄-{𝑄}_1}$

When a p-n junction diode is forward biased, the positive terminal of the battery is connected to the p-side, and the negative terminal is connected to the n-side.

• This reduces the potential barrier of the junction.

• As a result, the depletion layer width decreases, allowing charge carriers (electrons & holes) to cross the junction more easily, leading to current conduction.

✅ Correct Answer:

Option B: width of depletion layer decreases.

$\frac{{𝑄}_1}{{𝑄}_2}={\left(\frac{{𝑅}_1}{{𝑅}_2}\right)}^2\times {\left(\frac{{𝑇}_1}{{𝑇}_2}\right)}^4$

$\begin{array}{cc}& \therefore \text{ for same power, }\\ & \begin{array}{cc}& {\left(\frac{R_1}{R_2}\right)}^2={\left(\frac{T_2}{T_1}\right)}^4\\ \therefore & \left(\frac{R_1}{R_2}\right)={\left(\frac{T_2}{T_1}\right)}^2\\ & \left(\frac{R_2}{R_1}\right)={\left(\frac{T_1}{T_2}\right)}^2\end{array}\end{array}$

The Lorentz force due to a magnetic field only is:

$\overrightarrow{𝐹}=𝑞\overrightarrow{𝑉}\times \overrightarrow{𝐵}.$

Now, the work done by a force in moving a particle is:

$\frac{𝑑𝑊}{𝑑𝑡}=\overrightarrow{𝐹}\cdot \overrightarrow{𝑉}.$

Substitute:

$\overrightarrow{𝐹}\cdot \overrightarrow{𝑉}=(𝑞\overrightarrow{𝑉}\times \overrightarrow{𝐵})\cdot \overrightarrow{𝑉}.$

But $\overrightarrow{𝑉}\times \overrightarrow{𝐵}$ is perpendicular to $\overrightarrow{𝑉}$, so the dot product is zero.

Hence,

$\frac{𝑑𝑊}{𝑑𝑡}=0⟹𝑊=0.$

✅ Correct Answer: Option A (zero).

The formula for fringe width in Young's double slit experiment is:

$𝑋=\frac{𝜆𝐷}{𝑑}$

Here, $𝑋$ is the fringe width, $𝜆$ is the wavelength, $𝐷$ is the distance from the slits to the screen, and $𝑑$ is the distance between the two slits.

The graph shows how $𝑋$ (fringe width) changes with $𝐷$ (distance to the screen). This means the slope of the graph is $\frac{𝑋}{𝐷}$.

From the formula above, if we write it as $\frac{𝑋}{𝐷}=\frac{𝜆}{𝑑}$, we see that the slope of the graph equals $\frac{𝜆}{𝑑}$.

So, you can calculate $𝜆$ (wavelength) using:

$𝜆=\text{slope}\times 𝑑$

Here, ‘slope’ means the value you get from the graph ($\frac{𝑋}{𝐷}$), and $𝑑$ is the distance between the slits.

Electric field at ' $𝑂$ ' due to charge ' $𝑄$ ' is given by,

$E=\frac{\mathrm{kQ}}{r^2}$

Electric field at ' O ' due to charges q at A and $𝐷$ are equal and opposite and hence they will cancel each other. Similarly electric field at ' O ' due to charges -q at F and C will also cancel each other.

Electric field at ' O ' due to charge -q at B is given by,

${𝐸}^{\prime }=\frac{𝑘𝑞}{{𝑟}^2}$

Now, $E^{\prime }=2E$

$\therefore \frac{\mathrm{kq}}{r^2}=\frac{2\mathrm{kQ}}{r^2}\Rightarrow Q=\frac{q}{2}$

The inductors in the circuit are connected in parallel.

When we have two inductors, ${𝐿}_1$ and ${𝐿}_2$, in parallel, the total inductance (${𝐿}_{𝑇}$) is found using the formula: $\frac{1}{{𝐿}_{𝑇}}=\frac{1}{{𝐿}_1}+\frac{1}{{𝐿}_2}$

In this circuit, both inductors are the same (${𝐿}_1={𝐿}_2=𝐿$). So, $\frac{1}{{𝐿}_{𝑇}}=\frac{1}{𝐿}+\frac{1}{𝐿}=\frac{2}{𝐿}$

This means ${𝐿}_{𝑇}=\frac{𝐿}{2}$

The capacitors are also connected in parallel.

For two capacitors, ${𝐶}_1$ and ${𝐶}_2$, in parallel, the total capacitance (${𝐶}_{𝑇}$) is: ${𝐶}_{𝑇}={𝐶}_1+{𝐶}_2$

In this circuit, both capacitors are the same (${𝐶}_1={𝐶}_2=𝐶$), so ${𝐶}_{𝑇}=𝐶+𝐶=2𝐶$

But the explanation used the formula for capacitors in series, so this may be an important point to check when solving problems. If the capacitors were in series, the formula would be $\frac{1}{{𝐶}_{𝑇}}=\frac{1}{{𝐶}_1}+\frac{1}{{𝐶}_2}$ and ${𝐶}_{𝑇}=\frac{𝐶}{2}$

Now, the resonant frequency ($𝑓$) for an LC circuit is: $𝑓=\frac{1}{2𝜋\sqrt{\surd {𝐿}_{𝑇}{𝐶}_{𝑇}}}$

Substituting ${𝐿}_{𝑇}=\frac{𝐿}{2}$ and ${𝐶}_{𝑇}=\frac{𝐶}{2}$, we get: $𝑓=\frac{1}{2𝜋\sqrt{\surd \frac{𝐿}{2}\cdot \frac{𝐶}{2}}}$

This simplifies to: $𝑓=\frac{1}{𝜋\sqrt{\surd 𝐿𝐶}}$

Given the radius of drop is R .

Let $𝑟$ be the radius of smaller droplets.

As the total volume remains the same,

$\begin{array}{cc}& \frac{4}{3}𝜋{𝑅}^3=64\times \frac{4}{3}𝜋{𝑟}^3\\ \therefore & 𝑟=\frac{𝑅}{(64{)}^{\frac{1}{3}}}=\frac{𝑅}{4}\end{array}$

Initial surface energy ${𝐸}_1=4𝜋{𝑅}^2𝑇$

Final surface energy

${𝐸}_2=64\times 4𝜋\times {\left(\frac{𝑅}{4}\right)}^2\times 𝑇=16𝜋{𝑅}^2𝑇$

$\therefore$ Work done, $𝑊={𝐸}_2-{𝐸}_1$

$W=12𝜋{\mathrm{TR}}^2$

$\begin{array}{cc}& \text{ Given data: }{𝜇}_g=1.5,R_1=R_2=40\mathrm{cm},{𝜇}_{𝑙}=1.25\\ & P_1=(\frac{{𝜇}_g}{{𝜇}_a}-1)(\frac{1}{R_1}-\frac{1}{R_2})=(1.5-1)(\frac{1}{R_1}-\frac{1}{R_2})\\ & \therefore P_2=(\frac{{𝜇}_g}{{𝜇}_{𝑙}}-1)(\frac{1}{R_1}-\frac{1}{R_2})=(\frac{1.5}{1.25}-1)(\frac{1}{R_1}-\frac{1}{R_2})\\ & =(1.2-1)(\frac{1}{R_1}-\frac{1}{R_2})\end{array}$

$\begin{array}{cc}& \text{ Taking ratio, }\\ & \frac{P_2}{P_1}=\frac{1.2-1}{1.5-1}=\frac{2}{5}\end{array}$

Step 1: Energy conservation

At the top, the sphere has potential energy:

${𝐸}_{\text{top}}=𝑚𝑔ℎ$

At the bottom, the energy is entirely kinetic (translational + rotational):

${𝐸}_{\text{bottom}}=\frac{1}{2}𝑚{𝑣}^2+\frac{1}{2}𝐼{𝜔}^2$

Step 2: Moment of inertia of a solid sphere

For a solid sphere about its diameter:

$𝐼=\frac{2}{5}𝑚{𝑅}^2$

Also, rolling without slipping condition:

$𝑣=𝜔𝑅\Rightarrow 𝜔=\frac{𝑣}{𝑅}$

Step 3: Substitute

${𝐸}_{\text{bottom}}=\frac{1}{2}𝑚{𝑣}^2+\frac{1}{2}\left(\frac{2}{5}𝑚{𝑅}^2\right){\left(\frac{𝑣}{𝑅}\right)}^2$

$=\frac{1}{2}𝑚{𝑣}^2+\frac{1}{5}𝑚{𝑣}^2=\frac{7}{10}𝑚{𝑣}^2$

Step 4: Equating to initial energy

$𝑚𝑔ℎ=\frac{7}{10}𝑚{𝑣}^2$

Cancel $𝑚$:

$𝑔ℎ=\frac{7}{10}{𝑣}^2$

${𝑣}^2=\frac{10}{7}𝑔ℎ$

$𝑣=\sqrt{\surd \frac{10𝑔ℎ}{7}}$

✅ Final Answer:

The speed of the sphere at the bottom is:

$\overline{){\left(\frac{10𝑔ℎ}{7}\right)}^{1/2}}$

So the correct option is B.

For a simple pendulum, time period is

$𝑇=2𝜋\sqrt{\surd \frac{𝑙}{𝑔}}$

When the lift accelerates upward with acceleration $𝑎$, the effective acceleration due to gravity becomes

${𝑔}^{\prime }=𝑔+𝑎$

So the new time period becomes

${𝑇}^{\prime }=2𝜋\sqrt{\surd \frac{𝑙}{𝑔+𝑎}}$

Given that the new period is reduced to half:

${𝑇}^{\prime }=\frac{𝑇}{2}$

Now substitute:

$2𝜋\sqrt{\surd \frac{𝑙}{𝑔+𝑎}}=\frac{1}{2}\left(2𝜋\sqrt{\surd \frac{𝑙}{𝑔}}\right)$

Cancel $2𝜋$:

$\sqrt{\surd \frac{𝑙}{𝑔+𝑎}}=\frac{1}{2}\sqrt{\surd \frac{𝑙}{𝑔}}$

Squaring both sides:

$\frac{𝑙}{𝑔+𝑎}=\frac{1}{4}\cdot \frac{𝑙}{𝑔}$

Cancel $𝑙$:

$\frac{1}{𝑔+𝑎}=\frac{1}{4𝑔}$

So,

$𝑔+𝑎=4𝑔$

$𝑎=3𝑔$

Hence, the correct answer is:

$\overline{)3𝑔}$

Option B

$\text{ Let the capacitors be }{C}_1,C_2,C_3,C_4\text{ and }{C}_5$

Between P and Q- ${𝐶}_2,{𝐶}_3,{𝐶}_4$ and ${𝐶}_5$ are connected in series and $C_1$ is connected in parallel with them.

For the series combination

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

Equivalent capacitance,

${𝐶}_{𝑃𝑄}={𝐶}_{\text{series }}+𝐶=\frac{𝐶}{4}+𝐶=\frac{5𝐶}{4}...\left(𝑖\right)$

Between points P and $R-C_2,C_3$ and $C_4$ are connected in series and ${𝐶}_1$ and ${𝐶}_5$ connected in series are connected in parallel with them

Between points P and R , capacitors $C_2,C_3$, and $C_4$ are connected in series, and this series combination is in parallel with another series combination of capacitors $C_1$ and $C_5$

$\begin{array}{cc}& \frac{1}{{𝐶}_{\text{series }}}=\frac{1}{𝐶}+\frac{1}{𝐶}+\frac{1}{𝐶}=\frac{3}{𝐶}\Rightarrow {𝐶}_{\text{series }1}=\frac{𝐶}{3}\\ & \frac{1}{{𝐶}_{\text{series }2}}=\frac{1}{𝐶}+\frac{1}{𝐶}=\frac{2}{𝐶}\Rightarrow {𝐶}_{\text{series }2}=\frac{𝐶}{2}\end{array}$

$\begin{array}{cc}& {𝐶}_{𝑃𝑅}={𝐶}_{\text{series }1}+{𝐶}_{\text{series }2}=\frac{𝐶}{3}+\frac{𝐶}{2}=\frac{5𝐶}{6}...\left(𝑖𝑖\right)\\ & \text{ From (i) and (ii), }\\ & \frac{{𝐶}_{𝑃𝑅}}{{𝐶}_{𝑃𝑄}}=\frac{5𝐶/6}{5𝐶/4}=\frac{2}{3}\end{array}$

Step 1: Write the energy of the photon

The energy of a photon is given by: ${𝐸}_{𝑝}=ℎ𝜈=\frac{ℎ𝑐}{{𝜆}_{𝑝}}$

We know from the question that the photon’s de-Broglie wavelength (${𝜆}_{𝑝}$) is twice the electron’s de-Broglie wavelength (${𝜆}_{𝑒}$): ${𝜆}_{𝑝}=2{𝜆}_{𝑒}$

So, we substitute ${𝜆}_{𝑝}$ into the energy formula: ${𝐸}_{𝑝}=\frac{ℎ𝑐}{2{𝜆}_{𝑒}}$

Step 2: Write the kinetic energy of the electron

The kinetic energy of the electron is: ${𝐸}_{𝑐}=\frac{{𝑝}^2}{2𝑚}$ where $𝑝$ is momentum and $𝑚$ is the mass of the electron.

The momentum $𝑝$ can also be written as: $𝑝=\frac{ℎ}{{𝜆}_{𝑒}}$

Substitute this into the kinetic energy equation: ${𝐸}_{𝑐}=\frac{1}{2𝑚}{\left(\frac{ℎ}{{𝜆}_{𝑒}}\right)}^2=\frac{{ℎ}^2}{2𝑚{𝜆}_2^{𝑒}}$

Step 3: Find the ratio $\frac{{𝐸}_{𝑝}}{{𝐸}_{𝑐}}$

Now substitute the formulas from Step 1 and Step 2:

$\frac{{𝐸}_{𝑝}}{{𝐸}_{𝑐}}=\frac{\frac{ℎ𝑐}{2{𝜆}_{𝑒}}}{\frac{{ℎ}^2}{2𝑚{𝜆}_2^{𝑒}}}$

The denominators are complex, but if we multiply by the reciprocal, it looks like this:

$\frac{{𝐸}_{𝑝}}{{𝐸}_{𝑐}}=\frac{ℎ𝑐}{2{𝜆}_{𝑒}}\times \frac{2𝑚{𝜆}_2^{𝑒}}{{ℎ}^2}$

Simplify the terms:

$\frac{{𝐸}_{𝑝}}{{𝐸}_{𝑐}}=\frac{𝑚𝑐{𝜆}_{𝑒}}{ℎ}$

Step 4: Replace ${𝜆}_{𝑒}$ using speed of electron

The de-Broglie wavelength of electron is: ${𝜆}_{𝑒}=\frac{ℎ}{𝑝}$ where $𝑝$ is the electron’s momentum.

Since the electron’s speed $𝑣=\frac{𝑐}{100}$, then $𝑝=𝑚\times 𝑣=𝑚\times \frac{𝑐}{100}$.

So, ${𝜆}_{𝑒}=\frac{ℎ}{𝑚\times \frac{𝑐}{100}}=\frac{100ℎ}{𝑚𝑐}$

Substitute this back into the ratio:

$\frac{{𝐸}_{𝑝}}{{𝐸}_{𝑐}}=\frac{𝑚𝑐}{ℎ}\times \frac{100ℎ}{𝑚𝑐}$

The $𝑚$, $𝑐$, and $ℎ$ cancel out, so:

$\frac{{𝐸}_{𝑝}}{{𝐸}_{𝑐}}=100$

For a closed pipe, fundamental frequency

$n_1=\frac{v}{4L}=150\mathrm{Hz}$

For an open pipe, fundamental frequency

$n_{\prime }^1=\frac{v}{2L}=2n_1=300\mathrm{Hz}$

In an open pipe all multiples of the fundamental are produced. Hence, frequencies produced can be $300\mathrm{Hz},600\mathrm{Hz}$ and so on.

To find the apparent frequency when the observer is moving, we use the formula:

$F^{\prime }=\left[\frac{V\pm V_0}{V\mp V_{𝑠}}\right]F$

Here, $V$ is the speed of sound, $V_0$ is the speed of the observer, and $V_{𝑠}$ is the speed of the source.

In the problem, the source does not move, so ${𝑉}_{𝑠}=0$. The observer moves with speed ${𝑉}_1$, so ${𝑉}_0={𝑉}_1$.

When the observer moves towards the source, the formula becomes: ${𝐹}_1=\left[\frac{𝑉+{𝑉}_1}{𝑉}\right]𝐹$

When the observer moves away from the source, the formula becomes: ${𝐹}_2=\left[\frac{𝑉-{𝑉}_1}{𝑉}\right]𝐹$

Now, divide ${𝐹}_1$ by ${𝐹}_2$: $\frac{{𝐹}_1}{{𝐹}_2}=\frac{𝑉+{𝑉}_1}{𝑉-{𝑉}_1}$

The problem gives $\frac{{𝐹}_1}{{𝐹}_2}=2$. So: $2=\frac{𝑉+{𝑉}_1}{𝑉-{𝑉}_1}$

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

Multiply both sides by $𝑉-{𝑉}_1$: $2(𝑉-{𝑉}_1)=𝑉+{𝑉}_1$

Expand: $2𝑉-2{𝑉}_1=𝑉+{𝑉}_1$

Move all terms with $𝑉$ to one side and all with ${𝑉}_1$ to the other side: $2𝑉-𝑉=2{𝑉}_1+{𝑉}_1$ $𝑉=3{𝑉}_1$

Divide both sides by ${𝑉}_1$: $\frac{𝑉}{{𝑉}_1}=3$

Process ABCA is cyclic,

$\Delta U=0$

Using first law of thermodynamics,

$\Delta 𝑄=\Delta 𝑊=5𝐽...\left(𝑖\right)$

From the graph,

$\begin{array}{cc}& {𝑊}_{𝐴𝐵}=𝑃\Delta 𝑉=10(2-1)=10...\left(𝑖𝑖\right)\\ & {𝑊}_{𝐵𝐶}=𝑃\Delta 𝑉=𝑃\left(0\right)=0...\left(𝑖𝑖𝑖\right)\end{array}$

Total work done,

$\begin{array}{cc}& 𝑊={𝑊}_{𝐴𝐵}+{𝑊}_{𝐵𝐶}+{𝑊}_{𝐶𝐴}\\ & 5𝐽=10+0+{𝑊}_{𝐶𝐴}\dots \left[\text{ From (i), (ii) and (iii) }\right]\end{array}$

${𝑊}_{𝐶𝐴}=-5J$

Step 1: Path difference at a point on screen

Suppose the slits are ${𝑆}_1$ and ${𝑆}_2$, separated by distance $𝑑$. Consider the point $𝑃$ on the screen which is exactly in front of slit ${𝑆}_2$. Then:

• Path from ${𝑆}_2$ to $𝑃$ = $𝐷$.

• Path from ${𝑆}_1$ to $𝑃$: slightly longer because it is displaced by $𝑑$.

Using geometry, the path difference is approximately

$\Delta =\frac{𝑑\cdot 𝑥}{𝐷}$

where $𝑥$ is the distance of point $𝑃$ from the central axis (the midpoint between slits projected onto screen).

But here point $𝑃$ is directly in front of a slit (say slit ${𝑆}_2$). The distance of that point on screen from the central axis is simply $𝑥=\frac{𝑑}{2}$.

(Reason: midpoint between slits is at $𝑥=0$; each slit is $\pm \frac{𝑑}{2}$ from that axis; projecting forward keeps that offset).

Thus,

$\Delta =\frac{𝑑\cdot (𝑑/2)}{𝐷}=\frac{{𝑑}^2}{2𝐷}.$

Step 2: Condition for minimum

Nth minimum condition in Young’s setup:

$\Delta =(𝑚+\frac{1}{2})𝜆,𝑚=0,1,2,...$

The second minimum means $𝑚=1$ (since $𝑚=0$ is first minimum).

So:

$\Delta =\frac{3}{2}𝜆.$

Step 3: Equating path difference

$\frac{{𝑑}^2}{2𝐷}=\frac{3}{2}𝜆$

$𝜆=\frac{{𝑑}^2}{3𝐷}.$

✅ Final Answer:

$\overline{)\frac{{𝑑}^2}{3𝐷}}$

That corresponds to Option B.

Step 1: Recall key concepts

• True power (P) = $𝑉𝐼\cos 𝜙$

• Apparent power (S) = $𝑉𝐼$

• Power factor (p.f.) = $\frac{𝑃}{𝑆}=\cos 𝜙$

So, the ratio of true power to apparent power is $\cos 𝜙$.

Step 2: Express $\cos 𝜙$ in terms of $𝑅$ and $𝑍$

For a series $𝑅𝐿𝐶$ circuit:

• Impedance magnitude: $𝑍=\sqrt{\surd {𝑅}^2+({𝑋}_{𝐿}-{𝑋}_{𝐶}{)}^2}$

• Phase angle: $\tan 𝜙=\frac{{𝑋}_{𝐿}-{𝑋}_{𝐶}}{𝑅}$

• Therefore, $\cos 𝜙=\frac{𝑅}{𝑍}$

Step 3: Compare with answer choices

Option A. $\frac{𝑍}{𝑅}$ → reciprocal, not correct.

Option B. $\tan 𝜙$ → not correct.

Option C. $\cot 𝜙$ → not correct.

Option D. $\frac{𝑅}{𝑍}$ → matches $\cos 𝜙$. ✅

✅ Final Answer:

$\text{The ratio of true power to apparent power is }\frac{𝑅}{𝑍}.$

Correct option: D

The formula for torque is:

$𝜏=𝐼𝐴𝐵\sin 𝜃$ where $𝜏$ is torque, $𝐼$ is current, $𝐴$ is the area of the loop, $𝐵$ is the magnetic field, and $𝜃$ is the angle between the field and the normal to the loop.

Since the current, magnetic field, and angle are the same for both shapes, the torque mainly depends on the area $𝐴$ of each loop.

For a given length of wire (same perimeter), a circle always has a bigger area than a square.

So, the circular loop will experience a greater torque than the square loop.

Let the water equivalent of the calorimeter be $𝑥$ grams. This means the calorimeter acts like it contains $𝑥$ grams of water for heat calculations.

Formula: The way the temperature of water falls is given by: $𝑚𝑐\frac{𝑑𝑇}{𝑑𝑡}=-𝑘[\frac{{𝑇}_1+{𝑇}_2}{2}-{𝑇}_{𝑜}]$ where $𝑚$ is the total mass (water + calorimeter equivalent), $𝑐$ is specific heat, ${𝑇}_1$ and ${𝑇}_2$ are starting and ending temperatures, and ${𝑇}_{𝑜}$ is the air temperature outside.

First case (10 g water):

$(10+𝑥)𝑐\frac{(20-15)}{10}=-𝑘[\frac{20+15}{2}-{𝑇}_{𝑜}]$

Second case (20 g water):

$(20+𝑥)𝑐\frac{(20-15)}{15}=-𝑘[\frac{20+15}{2}-{𝑇}_{𝑜}]$

The bracketed factor on the right is the same in both cases. The left side is different because of the different amounts of water and time.

Setting the left sides as equal ratios:

$\frac{10+𝑥}{10}=\frac{20+𝑥}{15}$

Cross-multiplied, this gives: $15(10+𝑥)=10(20+𝑥)$

Expand: $150+15𝑥=200+10𝑥$

Move $10𝑥$ to the left and $150$ to the right: $15𝑥-10𝑥=200-150$

$5𝑥=50$

So, $𝑥=10$ grams.

The water equivalent of the calorimeter is 10 grams.

Number of moles is conserved, so

${𝑃}_1{𝑉}_1+{𝑃}_2{𝑉}_2={𝑃}_3𝑉$

But, $𝑃=\frac{4𝑇}{𝑟}$ where, $𝑟$ is the radius of the bubble

$\begin{array}{cc}\therefore & \frac{4𝑇}{{𝑟}_1}\left(\frac{4}{3}𝜋{𝑟}_3^1\right)+\frac{4𝑇}{{𝑟}_2}\left(\frac{4}{3}𝜋{𝑟}_3^2\right)=\frac{4𝑇}{𝑟}\left(\frac{4}{3}𝜋{𝑟}^3\right)\\ & {𝑟}_2^1+{𝑟}_2^2={𝑟}^2\\ & 𝑟={({𝑟}_2^1+{𝑟}_2^2)}^{\frac{1}{2}}\end{array}$

$𝑣=𝑅{𝑍}^2(\frac{1}{{𝑛}^2}-\frac{1}{{𝑚}^2})$

Series limit of Balmer: $n=2,m=\infty$

$\therefore v_1=\frac{{\mathrm{RZ}}^2}{4}$

Series limit of Paschen: $n=3,m=\infty$

$\therefore {𝑣}_2=\frac{𝑅{𝑍}^2}{9}$

$1^{\text{st }}$ line of Balmer series: $n_1=2,n_2=3$

$\therefore v_3={\mathrm{RZ}}^2(\frac{1}{2^2}-\frac{1}{3^2})=\frac{{\mathrm{RZ}}^2}{4}-\frac{{\mathrm{RZ}}^2}{9}=v_1-v_2$

Step 1: Recall specific heats for diatomic ideal gas

For a diatomic gas (ignoring vibrational d.o.f. at moderate temperatures):

${𝐶}_{𝑣}=\frac{5}{2}𝑅,{𝐶}_{𝑝}={𝐶}_{𝑣}+𝑅=\frac{7}{2}𝑅$

$𝛾=\frac{{𝐶}_{𝑝}}{{𝐶}_{𝑣}}=\frac{7/2}{5/2}=\frac{7}{5}$

Step 2: Heat supplied in each case

• For cylinder A (isobaric):

$𝑄=𝑛{𝐶}_{𝑝}𝑑{𝑇}_{𝐴}$

• For cylinder B (isochoric) with temperature rise $𝑑{𝑇}_{𝐵}$:

$𝑄=𝑛{𝐶}_{𝑣}𝑑{𝑇}_{𝐵}$

Since $𝑄$ is the same in both cases:

$𝑛{𝐶}_{𝑝}𝑑{𝑇}_{𝐴}=𝑛{𝐶}_{𝑣}𝑑{𝑇}_{𝐵}$

Step 3: Relating temperature rises

$𝑑{𝑇}_{𝐵}=\frac{{𝐶}_{𝑝}}{{𝐶}_{𝑣}}𝑑{𝑇}_{𝐴}=𝛾𝑑{𝑇}_{𝐴}$

Final Answer:

The rise in temperature of the gas in B is

$\overline{)𝛾𝑑{𝑇}_{𝐴}}$

Correct option: C

NAND gate

$\begin{array}{ccc}\text{ A }& \text{ B }& \text{ Y }\\ 0& 0& 1\\ 0& 1& 1\\ 1& 0& 1\\ 1& 1& 0\end{array}$

The sum of the two forces,

$\overrightarrow{{𝐹}_1}=\overrightarrow{𝐴}+\overrightarrow{𝐵}...\left(𝑖\right)$

The difference of the two forces,

${\overrightarrow{𝐹}}_2=\overrightarrow{𝐴}-\overrightarrow{𝐵}...\left(𝑖𝑖\right)$

Since sum of the two forces is perpendicular to their difference,

$\begin{array}{cc}& {\overrightarrow{𝐹}}_1\cdot {\overrightarrow{𝐹}}_2=0\\ \Rightarrow & (\overrightarrow{𝐴}+\overrightarrow{𝐵})\cdot (\overrightarrow{𝐴}-\overrightarrow{𝐵})=0\\ \Rightarrow & {𝐴}^2-\overrightarrow{𝐴}\cdot \overrightarrow{𝐵}+\overrightarrow{𝐵}\cdot \overrightarrow{𝐴}-{𝐵}^2=0\\ \therefore & {𝐴}^2={𝐵}^2\Rightarrow \left|𝐴\right|=\left|𝐵\right|\end{array}$

Thus, the forces are equal to each other in magnitude.

The total voltage across A and B is 16 V . Since the bridge is balanced, the circuit can be simplified by considering the parallel combination of the two paths:

Path ACB: $R_{\mathrm{AC}}+R_{\mathrm{CB}}=4\Omega +4\Omega =8\Omega$

Path ADB: $R_{\mathrm{AD}}+R_{\mathrm{DB}}=4\Omega +4\Omega =8\Omega$

The equivalent resistance between A and B is:

$R_{\mathrm{eq}}=\frac{8\times 8}{8+8}=\frac{64}{16}=4\Omega$

The total current from the 16 V battery is:

${𝐼}_{\text{total }}=\frac{𝑉}{{𝑅}_{𝑒𝑞}}=\frac{16}{4}=4𝐴$

This total current of 4A splits equally between the two parallel paths ACB and ADB because the resistances are equal:

${𝐼}_{𝐴𝐶𝐵}=\frac{{𝐼}_{\text{total }}}{2}=\frac{4}{2}=2A$

Step 1: Recall the SHM relations

For a particle in SHM,

$𝑥\left(𝑡\right)=𝐴\cos (𝜔𝑡+𝜙)$

where:

• $𝑥\left(𝑡\right)$ = displacement

• $𝐴$ = amplitude

• $𝜔$ = angular frequency

• $𝜙$ = phase constant

The acceleration is related to displacement by:

$𝑎\left(𝑡\right)=\ddot{𝑥}\left(𝑡\right)=-{𝜔}^2𝑥\left(𝑡\right)$

Step 2: Relationship between $𝑥\left(𝑡\right)$ and $𝑎\left(𝑡\right)$

From the expression:

$𝑎\left(𝑡\right)=-{𝜔}^2𝑥\left(𝑡\right)$

This shows that acceleration is directly proportional to displacement, but opposite in direction.

Step 3: Phase difference

• Both $𝑎\left(𝑡\right)$ and $𝑥\left(𝑡\right)$ vary sinusoidally with the same frequency.

• However, $𝑎\left(𝑡\right)$ has a negative sign compared to displacement.

• This corresponds to a phase difference of $𝜋$ radians (180°).

✅ Final Answer:

$\text{Phase difference}=𝜋\text{rad}$

Correct option: A. $𝜋\text{rad}$

Flux, $𝜙=\mathrm{LI}$

Comparing the above equation with $y=\mathrm{mx}$, $L=𝜙/I$, which is the slope of the graph. Since slope of ${𝐿}_1$ is greater, the value of selfinductance for ${𝐿}_1$ is greater than that of ${𝐿}_2$.

The energy of a vibrating object (oscillation) is given by the formula:

$𝐸=\frac{1}{2}𝑚{𝜔}^2{𝐴}^2$ Here, $𝐸$ is energy, $𝑚$ is mass, $𝜔$ is angular frequency, and $𝐴$ is amplitude.

Since $𝜔=2𝜋𝑛$ (where $𝑛$ is the frequency), the formula becomes:

$𝐸\propto {𝑛}^2{𝐴}^2$ This means energy is directly related to the square of frequency and the square of amplitude.

The problem says both $𝑃$ and $𝑄$ have the same energy, so:

${𝑛}_2^{𝑃}{𝐴}_2^{𝑃}={𝑛}_2^{𝑄}{𝐴}_2^{𝑄}$

We know ${𝑛}_{𝑄}=\frac{{𝑛}_{𝑃}}{4}$. Substitute this value:

${𝑛}_2^{𝑃}{𝐴}_2^{𝑃}={\left(\frac{{𝑛}_{𝑃}}{4}\right)}^2{𝐴}_2^{𝑄}$

This gives:

${𝑛}_2^{𝑃}{𝐴}_2^{𝑃}=\frac{{𝑛}_2^{𝑃}}{16}{𝐴}_2^{𝑄}$