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

Let us find the ratio step by step.

Step 1: Formula for Wavelength in Hydrogen Spectrum

The wavelength of a spectral line in the hydrogen spectrum is given by:

$\frac{1}{𝜆}=𝑅{𝑍}^2(\frac{1}{{𝑛}_2^1}-\frac{1}{{𝑛}_2^2})$

For hydrogen, $𝑍=1$, and $𝑅$ is the Rydberg constant.

Step 2: Lyman Series

• Lyman series: ${𝑛}_1=1$

The last line corresponds to the transition from ${𝑛}_2=\infty$ to ${𝑛}_1=1$.

So,

$\frac{1}{{𝜆}_{𝐿}}=𝑅(\frac{1}{1^2}-\frac{1}{{\infty }^2})=𝑅(1-0)=𝑅$

Therefore,

${𝜆}_{𝐿}=\frac{1}{𝑅}$

Step 3: Balmer Series

• Balmer series: ${𝑛}_1=2$

The last line corresponds to the transition from ${𝑛}_2=\infty$ to ${𝑛}_1=2$.

So,

$\frac{1}{{𝜆}_{𝐵}}=𝑅(\frac{1}{2^2}-\frac{1}{{\infty }^2})=𝑅(\frac{1}{4}-0)=\frac{𝑅}{4}$

Therefore,

${𝜆}_{𝐵}=\frac{1}{𝑅/4}=\frac{4}{𝑅}$

Step 4: Ratio Calculation

The required ratio:

$\text{Ratio}=\frac{{𝜆}_{𝐿}}{{𝜆}_{𝐵}}=\frac{\frac{1}{𝑅}}{\frac{4}{𝑅}}=\frac{1}{4}=0.25$

Final Answer:

Option C: 0.25

For a perfectly black body:

• By definition, a black body absorbs all the radiation incident on it.

• According to Kirchhoff’s law of radiation, for a body in thermal equilibrium, the ratio of emissive power to absorptive power is the same for all bodies at a given temperature and wavelength.

• Since a black body has absorptivity = 1, it also has emissivity = 1.

Therefore,

Correct option: B. unity ✅

$\text{ Given circuit can also be drawn as, }$

$\begin{array}{cc}& \text{ By Kirchhoff’s voltage law, }\\ & \begin{array}{cc}& {𝑉}_{𝐴}-(2\times 2)-\left(3\right)-(2\times 1)-{𝑉}_{𝐵}=0\\ \therefore & {𝑉}_{𝐴}-4-3-2-{𝑉}_{𝐵}=0\\ \therefore & {𝑉}_{𝐴}-{𝑉}_{𝐵}=+9𝑉\end{array}\end{array}$

An inertial frame of reference is a frame that is either at rest or moving with a constant velocity, having zero acceleration.

A decelerating train, a revolving merry-go-round, and an aeroplane during takeoff all possess non-zero acceleration, making them non-inertial frames of reference.

A bus moving with a constant velocity has an acceleration of zero, so a reference frame attached to the driver is an inertial frame of reference.

Hence, the correct option is (C).

Let the thickness of both discs be $𝑡$, and their material has density $𝜌$.

Let disc A have radius $𝑅$.

Let disc B have radius $3𝑅$.

Step 1: Find Mass of Each Disc

Volume of a disc = Area × Thickness

For disc of radius $𝑟$ and thickness $𝑡$:

$\text{Volume}=𝜋{𝑟}^2𝑡$

So, mass $𝑀$ of a disc:

$𝑀=\left(\text{Volume}\right)\times \left(\text{Density}\right)=𝜋{𝑟}^2𝑡𝜌$

For disc A:

${𝑀}_{𝐴}=𝜋{𝑅}^2𝑡𝜌$

For disc B (radius $3𝑅$):

${𝑀}_{𝐵}=𝜋(3𝑅{)}^2𝑡𝜌=𝜋\cdot 9{𝑅}^2𝑡𝜌=9𝜋{𝑅}^2𝑡𝜌$

Step 2: Moment of Inertia of a Uniform Disc about Its Axis

Formula:

$𝐼=\frac{1}{2}𝑀{𝑟}^2$

For A:

${𝐼}_{𝐴}=\frac{1}{2}{𝑀}_{𝐴}{𝑅}^2=\frac{1}{2}\left(𝜋{𝑅}^2𝑡𝜌\right){𝑅}^2=\frac{1}{2}𝜋𝑡𝜌{𝑅}^4$

For B:

${𝐼}_{𝐵}=\frac{1}{2}{𝑀}_{𝐵}(3𝑅{)}^2=\frac{1}{2}(9𝜋{𝑅}^2𝑡𝜌)\times (9{𝑅}^2)$

${𝐼}_{𝐵}=\frac{1}{2}\cdot 9𝜋{𝑅}^2𝑡𝜌\cdot 9{𝑅}^2=\frac{1}{2}\cdot 81𝜋𝑡𝜌{𝑅}^4=\frac{81}{2}𝜋𝑡𝜌{𝑅}^4$

Step 3: Ratio of Moments of Inertia

$\frac{{𝐼}_{𝐴}}{{𝐼}_{𝐵}}=\frac{\frac{1}{2}𝜋𝑡𝜌{𝑅}^4}{\frac{81}{2}𝜋𝑡𝜌{𝑅}^4}=\frac{1}{81}$

So,

${𝐼}_{𝐴}:{𝐼}_{𝐵}=1:81$

Answer:

Option C $(1:81)$ is correct.

Let:

• Actual frequency of the source = $𝑓$

• Velocity of sound = $𝑣$

• Speed of observer = ${𝑉}_1$

**Case 1: Observer moves *towards* stationary source**

The apparent frequency (${𝐹}_1$) is:

${𝐹}_1=𝑓\left(\frac{𝑣+{𝑉}_1}{𝑣}\right)$

**Case 2: Observer moves *away from* stationary source**

The apparent frequency (${𝐹}_2$) is:

${𝐹}_2=𝑓\left(\frac{𝑣-{𝑉}_1}{𝑣}\right)$

Given:

$\frac{{𝐹}_1}{{𝐹}_2}=2$

Now, substitute the values of ${𝐹}_1$ and ${𝐹}_2$:

$\frac{\frac{𝑣+{𝑉}_1}{𝑣}𝑓}{\frac{𝑣-{𝑉}_1}{𝑣}𝑓}=2$

$\frac{𝑣+{𝑉}_1}{𝑣-{𝑉}_1}=2$

Now, solve for $\frac{𝑣}{{𝑉}_1}$:

Cross-multiply:

$𝑣+{𝑉}_1=2(𝑣-{𝑉}_1)$

$𝑣+{𝑉}_1=2𝑣-2{𝑉}_1$

$𝑣+{𝑉}_1-2𝑣+2{𝑉}_1=0$

$-𝑣+3{𝑉}_1=0$

$𝑣=3{𝑉}_1$

So,

$\frac{𝑣}{{𝑉}_1}=3$

Final Answer:

Option C: 3

Given reaction is:

${}_6{C}^{11}⟶{}_5{B}^{11}+𝛽+𝑋$

Let's break it down step-by-step:

1. Identify $𝛽$ particle:

A $𝛽$ particle in nuclear reactions usually refers to a beta-minus (${𝛽}^{-}$) particle, which is an electron (${𝑒}^{-}$).

2. Write nuclear equations:

A general ${𝛽}^{-}$ decay reaction is:

${}_{𝐴}^{𝑍}𝑋\to {}_{𝐴}^{𝑍+1}𝑌+{𝛽}^{-}+\overline{𝜈}$

Where:

• $𝑋$ is the parent nucleus,

• $𝑌$ is the daughter nucleus,

• ${𝛽}^{-}$ is the emitted electron,

• $\overline{𝜈}$ is the anti-neutrino.

1. Apply to given reaction:

In this reaction:

• The atomic number decreases: from $6$ (Carbon) to $5$ (Boron)

• This fits beta-plus (${𝛽}^{+}$) decay, not beta-minus

But the question writes $𝛽$ (generic). However, with decrease in atomic number, it should be ${𝛽}^{+}$ (positron), not ${𝛽}^{-}$. But let's carry forward with the usual nuclear conventions, where

${}_{𝐴}^{𝑍}𝑋\to {}_{𝐴}^{𝑍-1}𝑌+{𝛽}^{+}+𝜈$

Here, the neutrino ($𝜈$) is emitted along with the ${𝛽}^{+}$.

4. Determine particle 'X'

Comparing with the general reaction:

• For beta-positive (${𝛽}^{+}$) decay,

  ${}_{11}^{6}C\to {}_{11}^{5}B+{𝛽}^{+}+𝜈$

Thus, $𝑋=𝜈=$ neutrino.

Final Answer:

Option B: neutrino

Given:

• Time taken for sound to reach from open end to closed end = $𝑡$ seconds.

• The pipe is closed at one end and open at the other (closed pipe).

• Neglect end correction.

Step 1: Express length of pipe in terms of $𝑡$ and speed of sound

Let the length of pipe be $𝐿$.

Let speed of sound in air = $𝑣$.

Time to travel from open to closed end:

$𝐿=𝑣𝑡$

Step 2: Write the fundamental frequency for closed pipe

For a closed pipe, the fundamental (first harmonic) forms a node at closed end and antinode at open end:

Length of air column ($𝐿$) in fundamental mode:

$𝐿=\frac{𝜆}{4}$

where $𝜆$ = wavelength.

So,

$𝜆=4𝐿$

Step 3: Express frequency in terms of $𝑡$

Frequency,

$𝑓=\frac{𝑣}{𝜆}$

Substitute $𝜆$ and $𝐿$:

$𝑓=\frac{𝑣}{4𝐿}$

Since $𝐿=𝑣𝑡$, we substitute:

$𝑓=\frac{𝑣}{4\cdot 𝑣𝑡}$

$𝑓=\frac{1}{4𝑡}$

So, the correct option is:

$\overline{)(4𝑡{)}^{-1}}$

That is, Option A.

Let's analyse each wire section's contribution to magnetic field at centre $𝑂$:

1. Straight Segment AB

For a straight wire passing through the centre, and with infinite length, field at the centre is zero by symmetry because the contributions from both sides cancel. The segment AB is straight and passes directly through $𝑂$, so:

${𝐵}_{𝐴𝐵}=0$

2. Semicircular Section BC

Magnetic field at centre, due to a current $𝐼$ in a semicircular arc of radius $𝑅$ is:

${𝐵}_{𝐵𝐶}=\frac{{𝜇}_0𝐼}{4𝑅}$

Direction: Using right hand rule, direction is into the page (or use NCERT convention, if you imagine current flowing from B to C).

3. Straight Segment CD

Again, this is a straight segment passing through $𝑂$, so its field at $𝑂$ is:

${𝐵}_{𝐶𝐷}=0$

4. Total Magnetic Field at $𝑂$

Adding all three contributions:

${𝐵}_{\text{total}}={𝐵}_{𝐴𝐵}+{𝐵}_{𝐵𝐶}+{𝐵}_{𝐶𝐷}=0+\frac{{𝜇}_0𝐼}{4𝑅}+0=\overline{)\frac{{𝜇}_0𝐼}{4𝑅}}$

Correct Option:

$\overline{)\text{Option A: }\frac{{𝜇}_0𝐼}{4𝑅}}$

Given times for 20 oscillations: $30\text{s},32\text{s},35\text{s},35\text{s}$

Minimum division of clock: $1\text{s}$

Step 1: Calculate the mean time

Mean time, $\overline{𝑡}$:

$\overline{𝑡}=\frac{30+32+35+35}{4}=\frac{132}{4}=33\text{s}$

Step 2: Find the absolute errors

List the deviations from the mean:

• $|30-33|=3$

• $|32-33|=1$

• $|35-33|=2$

• $|35-33|=2$

Sum of absolute errors: $3+1+2+2=8$

Step 3: Calculate mean absolute error

Mean absolute error, $\Delta 𝑡$:

$\Delta 𝑡=\frac{8}{4}=2\text{s}$

Step 4: Instrument Least Count

The least count is $1\text{s}$. Mean absolute error $\left(2\text{s}\right)$ is greater than least count, so error is $2\text{s}$.

Step 5: Express final result

Final result:

$(33\pm 2)\text{s}$

Correct answer: Option A

$(33\pm 2)$

The photoelectric effect occurs only when the energy of the incident photon is greater than or equal to the work function of the metal.

The energy of the incident photon of wavelength $𝜆$ is:

$𝐸=\frac{ℎ𝑐}{𝜆}$

The work function ($𝜙$) is given as:

$𝜙=\frac{ℎ𝑐}{{𝜆}_0}$

For photoelectric emission:

$𝐸\ge 𝜙$

So,

$\frac{ℎ𝑐}{𝜆}\ge \frac{ℎ𝑐}{{𝜆}_0}$

Divide both sides by $ℎ𝑐$:

$\frac{1}{𝜆}\ge \frac{1}{{𝜆}_0}$

Taking reciprocal (and reversing the inequality since reciprocals of positive numbers reverse the inequality sign):

$𝜆\le {𝜆}_0$

Correct option:

Option C: $𝜆\le {𝜆}_0$

We first recall some basic facts from Uniform Circular Motion (UCM):

• The particle is moving with constant angular speed.

• Its velocity is always tangent to the circle.

• The acceleration is called centripetal acceleration, and it always points towards the centre of the circle.

• The acceleration is perpendicular to the velocity.

Now let us check each option:

Option A:

“The velocity vector is tangent to the circle.”

→ This is true.

Option B:

“The acceleration vector is tangent to the circle.”

→ This is false, because acceleration is directed towards the centre, not tangent.

Option C:

“The velocity and acceleration vectors are perpendicular to each other.”

→ This is true.

Option D:

“The acceleration vector points to the centre of the circle.”

→ This is true.

Correct Answer: Option B

We are given a glass plate with thickness $𝑡=0.04\mathrm{mm}$.

When this glass plate is placed on one slit, the fringe pattern shifts. The amount the pattern shifts (lateral displacement) is found using:

${𝑑}^{\prime }=\frac{𝑡(𝜇-1)𝐷}{𝑑}$

Here's what the variables mean:

• $𝑡$ = Thickness of glass plate ($0.04\mathrm{mm}$)
• $𝜇$ = Refractive index of glass ($1.5$)
• $𝐷$ = Distance from slits to screen ($1m$)
• $𝑑$ = Distance between the slits ($2\mathrm{mm}$)

Let's put in the numbers. First, change all measurements to meters:

• $𝑡=0.04\mathrm{mm}=0.04\times {10}^{-3}m$
• $𝑑=2\mathrm{mm}=2\times {10}^{-3}m$

Now substitute the values into the formula:

${𝑑}^{\prime }=\frac{0.04\times {10}^{-3}\times (1.5-1)\times 1}{2\times {10}^{-3}}$

Calculate $1.5-1=0.5$.

So, ${𝑑}^{\prime }=\frac{0.04\times {10}^{-3}\times 0.5\times 1}{2\times {10}^{-3}}$

Multiply the top: $0.04\times 0.5=0.02$. So the top becomes $0.02\times {10}^{-3}$.

Now divide: ${𝑑}^{\prime }=\frac{0.02\times {10}^{-3}}{2\times {10}^{-3}}$

Divide $0.02$ by $2$ to get $0.01$.

So, ${𝑑}^{\prime }=0.01m$

$0.01m=1\mathrm{cm}$

Therefore, the fringe pattern shifts by $1\mathrm{cm}$.

For an ideal diode:

• In forward bias, the diode conducts perfectly. Its resistance is taken as zero.

• In reverse bias, the diode does not conduct at all. Its resistance is taken as infinite.

So, the correct option is:

Option D: zero, infinite. ✅

Let the distance between the slits and the screen be $𝐷$ and slit separation be $𝑑$. The width of the screen that is illuminated and on which fringes are observed is $𝐿$.

The fringe width (distance between two consecutive bright or dark fringes) is given by:

$𝛽=\frac{𝜆𝐷}{𝑑}$

The total number of fringes ($𝑛$) that can be seen on the screen (of length $𝐿$) is:

$𝑛=\frac{𝐿}{𝛽}$

So,

$𝑛=\frac{𝐿}{\frac{𝜆𝐷}{𝑑}}=\frac{𝐿𝑑}{𝜆𝐷}$

This shows that $𝑛\propto \frac{1}{𝜆}$ if $𝐿$, $𝑑$, and $𝐷$ are unchanged.

Let ${𝑛}_1$ be the initial number of fringes for wavelength ${𝜆}_1$ and ${𝑛}_2$ the number for wavelength ${𝜆}_2$:

$\frac{{𝑛}_1}{{𝑛}_2}=\frac{{𝜆}_2}{{𝜆}_1}$

Given:

• ${𝑛}_1=18$ for ${𝜆}_1=600\mathrm{nm}$

• ${𝜆}_2=400\mathrm{nm}$

• ${𝑛}_2=?$

Substitute the values:

$\frac{18}{{𝑛}_2}=\frac{400}{600}$

$\frac{18}{{𝑛}_2}=\frac{2}{3}$

${𝑛}_2=\frac{18\times 3}{2}=27$

Correct answer:

$\overline{)27}$

So, the correct option is Option B: 27.

$\begin{array}{cc}\overrightarrow{𝑃}& =\overrightarrow{𝐴}+\overrightarrow{𝐵}+\overrightarrow{𝐶}\\ & =\widehat{𝑖}+\widehat{𝑗}+3\widehat{𝑘}-\widehat{𝑖}+\widehat{𝑗}+4\widehat{𝑘}+2\widehat{𝑖}-2\widehat{𝑗}-8\widehat{𝑘}\end{array}$

$\begin{array}{cc}& \begin{array}{cc}\overrightarrow{P}& =2\hat{i}-\hat{k}\\ \overrightarrow{Q}& =\overrightarrow{A}\times \overrightarrow{B}=\mid \begin{array}{ccc}\hat{i}& \hat{j}& \hat{k}\\ 1& 1& 3\\ -1& 1& 4\end{array}\mid \\ & =\hat{i}(4-3)-\hat{j}(4+3)+\hat{k}(1+1)\\ \overrightarrow{Q}& =\hat{i}-7\hat{j}+2\hat{k}\end{array}\\ & \text{ Angle between }\overrightarrow{𝑃}\text{ and }\overrightarrow{𝑄}\text{ is given by }\\ & \begin{array}{cc}\cos 𝜃& =\frac{\overrightarrow{p}\cdot \overrightarrow{Q}}{\left|\overrightarrow{P}\right|\left|\overrightarrow{Q}\right|}\\ & =\frac{(2\hat{i}-\hat{k})\cdot (\hat{i}-7\hat{j}+2\hat{k})}{\sqrt{\surd (2{)}^2+(0{)}^2+(1{)}^2}\cdot \sqrt{\surd (1{)}^2+(7{)}^2+(2{)}^2}}\\ \therefore \cos 𝜃& =0S\\ \therefore 𝜃=90& \end{array}\end{array}$

Let us analyze each option step by step as per NCERT methods:

The coil (inductor) has inductive reactance ${𝑋}_{𝐿}=2𝜋𝑓𝐿$.

The total impedance of the circuit is $𝑍=\sqrt{\surd {𝑅}^2+{𝑋}_2^{𝐿}}$, where $𝑅$ is the resistance of the bulb.

The brightness of the bulb depends on the current in the circuit:

$𝐼=\frac{𝑉}{𝑍}$

More impedance $𝑍$ $⟹$ less current $⟹$ less brightness.

Option A: An iron rod is inserted in the coil.

• Inserting iron increases the core's permeability, which increases inductance $𝐿$.

• ${𝑋}_{𝐿}=2𝜋𝑓𝐿$ also increases.

• Higher ${𝑋}_{𝐿}$ $⟹$ higher $𝑍$ $⟹$ current decreases.

• Brightness decreases. ✅

Option B: Frequency of a.c. source is decreased.

• ${𝑋}_{𝐿}=2𝜋𝑓𝐿$, so if frequency $𝑓$ decreases, ${𝑋}_{𝐿}$ decreases.

• Lower ${𝑋}_{𝐿}$ $⟹$ lower $𝑍$ $⟹$ current increases.

• Brightness increases. ❌

Option C: Number of turns in the coil is reduced.

• Inductance $𝐿$ depends on the number of turns. If turns decrease, $𝐿$ decreases.

• ${𝑋}_{𝐿}$ decreases $⟹$ $𝑍$ decreases $⟹$ current increases.

• Brightness increases. ❌

Option D: A capacitance of reactance $({𝑋}_{𝐶}-{𝑋}_{𝐿})$ is included in the same circuit.

• If a capacitor is added in series, the total impedance becomes

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

• If ${𝑋}_{𝐶}={𝑋}_{𝐿}⟹𝑍=𝑅$ (current maximum, brightness maximum).

• If ${𝑋}_{𝐶}<{𝑋}_{𝐿}$, $𝑍$ is less compared to only ${𝑋}_{𝐿}$.

• Including a capacitor generally decreases overall impedance (except at resonance).

• Brightness increases. ❌

Final answer:

The brightness of the bulb decreases when an iron rod is inserted in the coil.

Correct option: Option A

Given:

• Period of SHM, $𝑇=16\text{seconds}$

• Times: ${𝑡}_1=2\text{s}$ and ${𝑡}_2=4\text{s}$

Step 1: Find the angular frequency $𝜔$

The formula is:

$𝜔=\frac{2𝜋}{𝑇}$

Substituting $𝑇=16$ s:

$𝜔=\frac{2𝜋}{16}=\frac{𝜋}{8}\text{rad/s}$

Step 2: Calculate the phase at ${𝑡}_1$ and ${𝑡}_2$

The general phase at time $𝑡$ (ignoring initial phase, as only difference is required):

$\text{Phase at}𝑡=𝜔𝑡$

So,

• At ${𝑡}_1=2$ s, phase = $𝜔\times 2=\frac{𝜋}{8}\times 2=\frac{𝜋}{4}$

• At ${𝑡}_2=4$ s, phase = $𝜔\times 4=\frac{𝜋}{8}\times 4=\frac{𝜋}{2}$

Step 3: Find the phase difference

$\text{Phase difference}=\mid \frac{𝜋}{2}-\frac{𝜋}{4}\mid =\frac{𝜋}{4}$

Final Answer:

Option C: $\overline{)\frac{𝜋}{4}}$

According to Newton's law of cooling,

$\frac{{𝜃}_1-{𝜃}_2}{t}=K[\frac{{𝜃}_1+{𝜃}_2}{2}-{𝜃}_0]$

where, ${𝜃}_0=$ temperature of surrounding

$\begin{array}{cc}\therefore & \frac{60-40}{6}=𝐾[\frac{60+40}{2}-10]\\ & \frac{20}{6}=𝐾\times 40...\left(𝑖\right)\\ & 𝐾=\frac{1}{12}\end{array}$

After another t min , let the temperature be x .

$\begin{array}{cc}\therefore & \frac{40-𝑥}{6}=\frac{1}{12}[\frac{40+𝑥}{2}-10]\text{ ...[using (i)] }\\ \therefore & 40-𝑥=\frac{20+𝑥}{4}\\ \therefore & 160-4𝑥=20+𝑥\\ \therefore & 5𝑥=140\\ \therefore & 𝑥=\frac{140}{5}={28}^{\circ }C\end{array}$

Given:

• Power of incident light, $𝑃=90$ W

• Surface absorbs $50\%$ of the light and reflects $50\%$

• Speed of light, $𝑐=3\times {10}^8$ m/s

Let's follow the NCERT method:

Step 1: Find the force due to absorption

When light is absorbed, the force exerted on the surface is:

${𝐹}_{\text{abs}}=\frac{{𝑃}_{\text{absorbed}}}{𝑐}$

Since $50\%$ is absorbed:

${𝑃}_{\text{absorbed}}=0.5\times 90=45\text{W}$

So,

${𝐹}_{\text{abs}}=\frac{45}{3\times {10}^8}=1.5\times {10}^{-7}\text{N}$

Step 2: Find the force due to reflection

When light is reflected normally (for normal incidence and perfect reflection), the change in momentum is twice that of absorption.

Force due to reflection:

${𝐹}_{\text{ref}}=\frac{2{𝑃}_{\text{reflected}}}{𝑐}$

Since $50\%$ is reflected:

${𝑃}_{\text{reflected}}=0.5\times 90=45\text{W}$

Thus,

${𝐹}_{\text{ref}}=\frac{2\times 45}{3\times {10}^8}=\frac{90}{3\times {10}^8}=3\times {10}^{-7}\text{N}$

Step 3: Total force exerted

${𝐹}_{\text{total}}={𝐹}_{\text{abs}}+{𝐹}_{\text{ref}}$

${𝐹}_{\text{total}}=1.5\times {10}^{-7}\text{N}+3\times {10}^{-7}\text{N}$

${𝐹}_{\text{total}}=4.5\times {10}^{-7}\text{N}$

Correct answer:

$\overline{)4.5\times {10}^{-7}\text{N}}$ (Option A)

We use Gay-Lussac’s Law (Pressure Law) for this question. According to this law, for a fixed amount of gas at constant volume:

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

where,

• ${𝑃}_1=270\text{kPa}$

• ${𝑇}_1={27}^{\circ }\text{C}=27+273=300\text{K}$

• ${𝑇}_2={37}^{\circ }\text{C}=37+273=310\text{K}$

• ${𝑃}_2=?$ (final pressure)

Let's substitute the values:

$\frac{270}{300}=\frac{{𝑃}_2}{310}$

Now solve for ${𝑃}_2$:

${𝑃}_2=\frac{270\times 310}{300}$

${𝑃}_2=\frac{83700}{300}=279\text{kPa}$

Final Answer:

Option D: 279 kPa

At resonance, ${𝑋}_{𝐿}={𝑋}_{𝐶}$

$\begin{array}{cc}& \Rightarrow 𝜔𝐿=\frac{1}{𝜔𝐶}\\ & \Rightarrow {𝜔}^2=\frac{1}{\sqrt{\surd 𝐿𝐶}}\\ & \Rightarrow 𝑓=\frac{1}{2𝜋\sqrt{\surd 𝐿𝐶}}...\left(𝑖\right)\end{array}$

From (i), we can see that the resonant frequency is independent of R .(resistance).

So, if the resistance is made ${\left(\frac{1}{3}\right)}^{\text{rd }}$, the resonant frequency remain unchanged.

The time period $𝑇$ of a mass-spring system is given by:

$𝑇=2𝜋\sqrt{\surd \frac{𝑚}{𝑘}}$

where

$𝑚=$ mass suspended from the spring

$𝑘=$ spring constant

Given:

For $𝑚$, ${𝑇}_1=2$ s

Now, for mass $4𝑚$, the new period ${𝑇}_2$ will be:

${𝑇}_2=2𝜋\sqrt{\surd \frac{4𝑚}{𝑘}}$

$=2𝜋\sqrt{\surd 4}\sqrt{\surd \frac{𝑚}{𝑘}}$

$=2𝜋\times 2\sqrt{\surd \frac{𝑚}{𝑘}}$

$=2\times \left(2𝜋\sqrt{\surd \frac{𝑚}{𝑘}}\right)$

But $2𝜋\sqrt{\surd \frac{𝑚}{𝑘}}={𝑇}_1=2$ s

So,

${𝑇}_2=2\times 2=4\text{s}$

Answer: Option D (4 s)

We know that the current through the inductor is increasing at a constant rate. This means that $\frac{\mathrm{dI}}{\mathrm{dt}}$ is constant.

The formula for the induced emf (e) in an inductor is: $e=-L\frac{\mathrm{dI}}{\mathrm{dt}}$.

Since $\frac{\mathrm{dI}}{\mathrm{dt}}$ is constant, the induced emf $e$ will also be constant (but negative because of the minus sign).

This means that if we plot $e$ versus $\frac{\mathrm{dI}}{\mathrm{dt}}$, the graph will be a straight line. The line will have a slope of $-L$ and will pass through the origin.

This is shown in graph A, which matches option (B) in the choices.

Let the three charges $+𝑄$, $+2𝑄$, and $𝑞$ be placed at the vertices of an equilateral triangle of side $𝑎$.

Step 1: Potential Energy for a System of 3 Charges

Total electrostatic potential energy, $𝑈$, for three charges ${𝑞}_1,{𝑞}_2,{𝑞}_3$ at the corners of an equilateral triangle (each side = $𝑎$) is:

$𝑈=\frac{1}{4𝜋{𝜖}_0}[\frac{{𝑞}_1{𝑞}_2}{𝑎}+\frac{{𝑞}_2{𝑞}_3}{𝑎}+\frac{{𝑞}_3{𝑞}_1}{𝑎}]$

Step 2: Substitute Given Values

Given: ${𝑞}_1=𝑄$, ${𝑞}_2=2𝑄$, ${𝑞}_3=𝑞$.

So,

$𝑈=\frac{1}{4𝜋{𝜖}_0}[\frac{𝑄\cdot 2𝑄}{𝑎}+\frac{2𝑄\cdot 𝑞}{𝑎}+\frac{𝑞\cdot 𝑄}{𝑎}]$

$𝑈=\frac{1}{4𝜋{𝜖}_0}[\frac{2{𝑄}^2}{𝑎}+\frac{2𝑄𝑞}{𝑎}+\frac{𝑄𝑞}{𝑎}]$

$𝑈=\frac{1}{4𝜋{𝜖}_0}\left[\frac{2{𝑄}^2+3𝑄𝑞}{𝑎}\right]$

Step 3: Set Potential Energy to Zero

For $𝑈=0$:

$2{𝑄}^2+3𝑄𝑞=0$

$3𝑄𝑞=-2{𝑄}^2$

$𝑞=-\frac{2{𝑄}^2}{3𝑄}$

$𝑞=-\frac{2}{3}𝑄$

Step 4: Select the Correct Option

The correct answer is:

$\overline{)\text{Option C: }𝑞=-\frac{2}{3}𝑄}$

Let $𝑅$ be radius of bigger drop and $𝑟$ be radius of smaller droplet.

As mass of liquid is conserved before and after spreading of drop,

$\therefore$ mass of big drop $=𝑛\times$ mass of droplet

$\begin{array}{cc}& 𝜌\times \frac{4}{3}𝜋{𝑅}^3=𝜌\times \frac{4}{3}𝜋{𝑟}^3\times 1000\\ \therefore & {𝑅}^3=1000{𝑟}^3\\ \therefore & 𝑅=10𝑟\end{array}$

Surface energy of bigger drop $=V=\mathrm{AT}$

$=\left(4𝜋{𝑅}^2\right)T$

Total surface energy of small droplets

$\begin{array}{cc}{𝑉}^{\prime }& =𝑛{𝐴}^{\prime }𝑇=1000\times 4𝜋{𝑟}^2\times 𝑇\\ & =1000\times 4𝜋\times {\left(\frac{𝑅}{10}\right)}^2\times 𝑇=\frac{1000}{100}\times 4𝜋{𝑅}^2𝑇\\ \therefore {𝑉}^{\prime }& =10\times 𝑉\end{array}$

Let's analyse each option based on NCERT concepts of transistors in Common Emitter (CE) mode:

Option A: The base-emitter junction is NOT reverse biased.

• In CE mode, base-emitter is forward biased, and collector-base is reverse biased.

Option B: The input impedance is high and the output impedance is low.

• In CE mode, input impedance is low (a few kΩ) and output impedance is high (tens of kΩ).

Option C: Input signal is applied between the base and emitter and the output is obtained between collector and emitter.

• This is correct as per textbook diagram and working:

• Input: between base and emitter

• Output: between collector and emitter.

Option D: The input voltage and output voltage are NOT in phase; they are 180° out of phase in CE mode.

Final answer:

Option C is correct.

Input signal is applied between the base and emitter and the output is obtained between collector and emitter.

Let us solve the problem step by step.

1. Pressure of an Ideal Gas:

   According to the kinetic theory of gases, the pressure ($𝑃$) exerted by an ideal gas is given by:

   $𝑃=\frac{1}{3}\frac{𝑁𝑚⟨{𝑣}^2⟩}{𝑉}$

   where

   $𝑁$ = number of molecules,

   $𝑚$ = mass of one molecule,

   $⟨{𝑣}^2⟩$ = mean square speed,

   $𝑉$ = volume.

2. Kinetic Energy and Temperature:

   The average kinetic energy of one molecule is

   $\frac{1}{2}𝑚⟨{𝑣}^2⟩=\frac{3}{2}{𝑘}_{𝐵}𝑇$

   Thus,

   $𝑚⟨{𝑣}^2⟩=3{𝑘}_{𝐵}𝑇$

3. Expressing Pressure in Terms of Temperature:

   Substitute the value from step 2 into the pressure equation:

   $𝑃=\frac{1}{3}\cdot \frac{𝑁}{𝑉}\cdot \left(3{𝑘}_{𝐵}𝑇\right)$

   $𝑃=\frac{𝑁}{𝑉}{𝑘}_{𝐵}𝑇$

   For a fixed number of molecules and volume, $𝑁$ and $𝑉$ are constant.

4. Force on the Walls:

   Pressure ($𝑃$) is force ($𝐹$) per unit area ($𝐴$). So,

   $𝑃=\frac{𝐹}{𝐴}⟹𝐹=𝑃\cdot 𝐴$

   For a given container (fixed area), $𝐹\propto 𝑃$.

5. Dependence on Temperature:

   From above, $𝑃\propto 𝑇$

   So, $𝐹\propto 𝑇$

6. Value of $𝑥$:

   Force on walls $\propto {𝑇}^{𝑥}$,

   So, $𝑥=1$.

Final Answer:

Option A

$\overline{)1}$

As per Gauss' law the net flux through closed surface is,

${𝜙}_A+{𝜙}_B+{𝜙}_C=\frac{q}{{𝜀}_0}$

Due to symmetry, the same flux passes through plane surfaces A and C , i.e., ${𝜙}_A={𝜙}_C$

$</mo><msub><mi>𝜙</mi><mrow><mrow><mi>B</mi></mrow></mrow></msub><mo>=</mo><mi>𝜙</mi><mo>$

For an adiabatic process:

$𝑇{𝑉}^{𝛾-1}=\text{constant}$

Let the initial state be $({𝑇}_{𝑖},{𝑉}_0)$ and the final state be $({𝑇}_{𝑓},{𝑉}_{𝑓})$.

Given:

• $𝛾=\frac{7}{5}$

• ${𝑉}_{𝑓}=\frac{{𝑉}_0}{32}$

• ${𝑇}_{𝑓}=𝑥{𝑇}_{𝑖}$

We can write:

${𝑇}_{𝑖}{𝑉}_{𝛾-1}^0={𝑇}_{𝑓}{𝑉}_{𝛾-1}^{𝑓}$

Substitute the values:

${𝑇}_{𝑖}{𝑉}_{𝛾-1}^0=𝑥{𝑇}_{𝑖}{\left(\frac{{𝑉}_0}{32}\right)}^{𝛾-1}$

Divide both sides by ${𝑇}_{𝑖}$:

${𝑉}_{𝛾-1}^0=𝑥{\left(\frac{{𝑉}_0}{32}\right)}^{𝛾-1}$

Rewrite the right side:

$=𝑥\cdot {𝑉}_{𝛾-1}^0\cdot {32}^{-(𝛾-1)}$

Now, cancel ${𝑉}_{𝛾-1}^0$ from both sides:

$1=𝑥\cdot {32}^{-(𝛾-1)}$

Bring ${32}^{𝛾-1}$ to the left:

$𝑥={32}^{𝛾-1}$

Now, calculate $𝛾-1$:

$𝛾-1=\frac{7}{5}-1=\frac{2}{5}$

Now:

$𝑥={32}^{2/5}$

Recall, $32=2^5$:

${32}^{2/5}=(2^5{)}^{2/5}=2^2=4$

So,

$𝑥=4$

Correct Option: B (4)

Given:

• Coefficient of mutual induction, $𝑀=2\text{H}$

• Induced emf, $𝐸=2\text{kV}=2000\text{V}$

• Change in current in primary, $\Delta 𝐼=6\text{A}-3\text{A}=3\text{A}$ (it is reduced, so $\Delta 𝐼=-3\text{A}$ but magnitude is enough)

• Required: Time taken, $\Delta 𝑡$

According to NCERT, the induced emf due to mutual induction is:

$𝐸=𝑀\frac{\Delta 𝐼}{\Delta 𝑡}$

Rearrange to solve for $\Delta 𝑡$:

$\Delta 𝑡=𝑀\frac{\Delta 𝐼}{𝐸}$

Substitute the values:

$\Delta 𝑡=\frac{2\times 3}{2000}$

Calculate:

$\Delta 𝑡=\frac{6}{2000}=0.003\text{s}$

or

$\Delta 𝑡=3\times {10}^{-3}\text{s}$

Correct option: Option D ($3\times {10}^{-3}s$)