MHT CET 2025 19th April Evening Shift (Physics & Chemistry) Question Paper with Solutions

$\begin{array}{cc}& \overrightarrow{𝐴}+\overrightarrow{𝐵}=\overrightarrow{𝐶}\overrightarrow{𝐴}\\ & {𝐶}^2={𝐴}^2+{𝐵}^2+2𝐴𝐵\cos 𝜃...\left(𝑖\right)\end{array}$

and $\tan 𝛼=\frac{B\sin 𝜃}{A+B\cos 𝜃}$

When $𝐵$ is doubled, resultant is perpendicular to $\overrightarrow{𝐴}$

$\therefore {𝐶}_2^1={𝐴}^2+4{𝐵}^2+4𝐴𝐵\cos 𝜃...\left(𝑖𝑖\right)$

From right angled triangle PSR

$\begin{array}{cc}& (2𝐵{)}^2={𝐶}_2^1+{𝐴}^2\\ & {𝐶}_2^1=4{𝐵}^2-{𝐴}^2\end{array}$

Substituting in (ii) and solving,

$A^2+2\mathrm{AB}\cos 𝜃=0...\left(𝑖𝑖𝑖\right)$

Substituting (iii) in (i), $C=B$

Given:

Focal length of convex lens, $𝑓=\frac{1}{3}\text{m}$

Magnification, $𝑚=-2$ (negative because the image is real and inverted)

Step 1: Magnification formula for lens:

$𝑚=\frac{{ℎ}^{\prime }}{ℎ}=\frac{𝑣}{𝑢}$

So,

$𝑚=\frac{𝑣}{𝑢}$

Given $𝑚=-2$, so

$\frac{𝑣}{𝑢}=-2⟹𝑣=-2𝑢$

Step 2: Lens formula:

$\frac{1}{𝑓}=\frac{1}{𝑣}-\frac{1}{𝑢}$

Put $𝑣=-2𝑢$ and $𝑓=\frac{1}{3}\text{m}$

$\frac{1}{1/3}=\frac{1}{-2𝑢}-\frac{1}{𝑢}$

$3=-\frac{1}{2𝑢}-\frac{1}{𝑢}$

Get common denominator:

$3=-\frac{1+2}{2𝑢}$

$3=-\frac{3}{2𝑢}$

Step 3: Solve for $𝑢$:

$3=-\frac{3}{2𝑢}$

Multiply both sides by $2𝑢$:

$6𝑢=-3$

$𝑢=-\frac{1}{2}\text{m}$

Step 4: Take the magnitude (distance is always positive):

Distance of the object from the lens is $\overline{)0.5\text{m}}$

Correct option: A

Tuning forks resonate when their frequencies are equal or integral multiples of each other (harmonics).

Let us check the options:

• Option A: $256\text{Hz}$ (same frequency)

• Option B: $512\text{Hz}$ ($256\times 2$)

• Option C: $754\text{Hz}$ (not a multiple of 256)

• Option D: $768\text{Hz}$ ($256\times 3$)

Conclusion:

The tuning fork of frequency $754\text{Hz}$ will not resonate with the $256\text{Hz}$ tuning fork.

Answer:

Option C: $754$ Hz

Given:

• Charge on the particle, $𝑞=1000\times 𝑒=1000\times 1.6\times {10}^{-19}C$

• Frequency of rotation, $𝑓=1\mathrm{Hz}$

• Radius of path, $𝑟$ (to be found)

• Magnetic field at centre, $𝐵=𝑥{𝜇}_0$ where $𝑥=2\times {10}^{-16}$

• ${𝜇}_0$ is the permeability of vacuum

Step 1: Formula for Magnetic Field at the Centre

When a charge $𝑞$ moves in a circle of radius $𝑟$ with frequency $𝑓$, it makes current $𝐼$:

$𝐼=𝑞\times 𝑓$

Magnetic field at centre of the circle:

$𝐵=\frac{{𝜇}_0𝐼}{2𝑟}$

Substitute $𝐼$:

$𝐵=\frac{{𝜇}_0𝑞𝑓}{2𝑟}$

Step 2: Substitute Values

Given $𝐵=𝑥{𝜇}_0$:

$𝑥{𝜇}_0=\frac{{𝜇}_0𝑞𝑓}{2𝑟}$

${𝜇}_0$ cancels out:

$𝑥=\frac{𝑞𝑓}{2𝑟}$

Step 3: Rearrangement for $𝑟$

$𝑟=\frac{𝑞𝑓}{2𝑥}$

Step 4: Substitute $𝑞$, $𝑓$, and $𝑥$

$𝑞=1000\times 1.6\times {10}^{-19}=1.6\times {10}^{-16}\text{ C}$

$𝑓=1\text{ Hz}$

$𝑥=2\times {10}^{-16}$

Substituting,

$𝑟=\frac{1.6\times {10}^{-16}\times 1}{2\times 2\times {10}^{-16}}$

$𝑟=\frac{1.6\times {10}^{-16}}{4\times {10}^{-16}}$

$𝑟=\frac{1.6}{4}$

$𝑟=0.4\text{m}$

Answer:

Option D: $0.4$

Step 1: Write the formula for tangential velocity.

The tangential velocity $𝑣$ for a particle in uniform circular motion is given by:

$𝑣=\frac{\text{distance travelled}}{\text{time}}$

Step 2: Find the distance travelled in $𝑥$ revolutions.

• In 1 revolution, distance = circumference of circle $=2𝜋𝑟$

• For $𝑥$ revolutions, distance $=𝑥\times 2𝜋𝑟$

The radius is given as $𝑟=\frac{𝜋}{2}\text{m}$.

So,

$\text{distance}=𝑥\times 2𝜋\left(\frac{𝜋}{2}\right)$

$\text{distance}=𝑥\times 𝜋\times 𝜋=𝑥{𝜋}^2$

Step 3: Divide total distance by total time.

Total time taken $=𝑡$

$𝑣=\frac{𝑥{𝜋}^2}{𝑡}$

Step 4: Match with the given options.

Option C

$\overline{)\frac{{𝜋}^2𝑥}{𝑡}}$

This is the correct answer.

According to Bohr's atomic model, the radius of the electron's orbit in the $𝑛$th level is:

${𝑟}_{𝑛}=\frac{{𝜀}_0{𝑛}^2{ℎ}^2}{𝜋{𝑚}_{𝑒}𝑍{𝑒}^2}$

The speed of the electron in the $𝑛$th orbit is:

${𝑣}_{𝑛}=\frac{𝑍{𝑒}^2}{2{𝜀}_0𝑛ℎ}$

Finding angular frequency:

The angular frequency ($𝜔$) is the speed divided by the radius:

$𝜔=\frac{𝑣}{𝑟}=\frac{\left(\frac{𝑍{𝑒}^2}{2{𝜀}_0𝑛ℎ}\right)}{\left(\frac{{𝜀}_0{𝑛}^2{ℎ}^2}{𝜋{𝑚}_{𝑒}𝑍{𝑒}^2}\right)}=\frac{𝜋{𝑚}_{𝑒}{𝑒}^4}{2{𝜀}_2^0{ℎ}^3{𝑛}^3}$

Finding frequency of revolution:

The frequency ($𝑓$) is $𝜔$ divided by $2𝜋$:

$𝑓=\frac{𝜔}{2𝜋}=\frac{\left(\frac{𝜋{𝑚}_{𝑒}{𝑒}^4}{2{𝜀}_2^0{ℎ}^3{𝑛}^3}\right)}{2𝜋}=\frac{{𝑚}_{𝑒}{𝑒}^4}{4{𝜀}_2^0{ℎ}^3{𝑛}^3}$

Relation to $𝑛$:

$\therefore 𝑓\propto \frac{1}{{𝑛}^3}$

Given:

• Initial temperature, ${𝑇}_1=(38.6\pm 0.2{)}^{\circ }C$

• Final temperature, ${𝑇}_2=(82.3\pm 0.3{)}^{\circ }C$

Step 1: Calculate the rise in temperature

$\Delta 𝑇={𝑇}_2-{𝑇}_1=82.3-38.6={43.7}^{\circ }C$

Step 2: Calculate the error in the rise

When we subtract (or add) two measured quantities, the absolute errors get added:

$\Delta \left(\Delta 𝑇\right)=\Delta {𝑇}_1+\Delta {𝑇}_2=0.2+0.3={0.5}^{\circ }C$

Step 3: Write the result with proper error limits

$\Delta 𝑇=(43.7\pm 0.5{)}^{\circ }C$

Final Answer:

Option D

$\overline{)(43.7\pm 0.5{)}^{\circ }C}$

Let us solve the problem step by step:

Step 1: Terminal velocity of a water drop

The terminal velocity (${𝑉}_{𝑡}$) of a sphere of radius $𝑟$, falling through a fluid with viscosity $𝜂$, according to Stokes’ Law, is:

$𝑉=\frac{2}{9}\frac{{𝑟}^2(𝜌-𝜎)𝑔}{𝜂}$

where

$𝜌$ = density of the sphere (water drop)

$𝜎$ = density of air

$𝑔$ = acceleration due to gravity

Step 2: Relation between radius of small and big drop

When $𝑛$ small drops of radius $𝑟$ combine to form one large drop of radius $𝑅$, the total volume is conserved:

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

$𝑛{𝑟}^3={𝑅}^3$

$𝑅={𝑛}^{1/3}𝑟$

Step 3: Terminal velocity of the big drop

Let the terminal velocity of the big drop be ${𝑉}^{\prime }$. By Stokes' Law:

${𝑉}^{\prime }=\frac{2}{9}\frac{{𝑅}^2(𝜌-𝜎)𝑔}{𝜂}$

Substitute $𝑅={𝑛}^{1/3}𝑟$:

${𝑉}^{\prime }=\frac{2}{9}\frac{\left({𝑛}^{1/3}𝑟{)}^2\right(𝜌-𝜎)𝑔}{𝜂}=\frac{2}{9}\frac{{𝑛}^{2/3}{𝑟}^2(𝜌-𝜎)𝑔}{𝜂}$

Step 4: Find the ratio of ${𝑉}^{\prime }$ and $𝑉$

Terminal velocity of a single small drop:

$𝑉=\frac{2}{9}\frac{{𝑟}^2(𝜌-𝜎)𝑔}{𝜂}$

Therefore,

$\frac{{𝑉}^{\prime }}{𝑉}=\frac{{𝑛}^{2/3}{𝑟}^2}{{𝑟}^2}={𝑛}^{2/3}$

But we are to answer in terms of $𝑉$, $𝑟$, and $𝑅$ only.

Recall ${𝑅}^3=𝑛{𝑟}^3$, so $𝑛=\frac{{𝑅}^3}{{𝑟}^3}$

So,

${𝑉}^{\prime }={𝑛}^{2/3}𝑉={\left(\frac{{𝑅}^3}{{𝑟}^3}\right)}^{2/3}𝑉={\left(\frac{𝑅}{𝑟}\right)}^2𝑉$

So,

${𝑉}^{\prime }=𝑉\frac{{𝑅}^2}{{𝑟}^2}$

Step 5: Choose the correct option

The answer is:

Option A: $\overline{)\frac{{\mathrm{VR}}^2}{r^2}}$

Given:

• Period of oscillation, $𝑇=6$ s

• Mass suspended, $𝑚$

• Acceleration due to gravity, $𝑔={𝜋}^2=10{\text{m/s}}^2$

Let the extension in the spring at equilibrium (rest position) be $𝑥$.

Step 1: Relation between period and spring constant

The period for a spring-mass system:

$𝑇=2𝜋\sqrt{\surd \frac{𝑚}{𝑘}}$

where $𝑘$ is the spring constant.

Step 2: Express $𝑘$ in terms of $𝑚$ and $𝑇$

Squaring both sides and rearranging for $𝑘$,

$𝑇=2𝜋\sqrt{\surd \frac{𝑚}{𝑘}}$

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

$\Rightarrow {\left(\frac{𝑇}{2𝜋}\right)}^2=\frac{𝑚}{𝑘}$

$\Rightarrow 𝑘=\frac{𝑚}{{\left(\frac{𝑇}{2𝜋}\right)}^2}$

Step 3: Extension at equilibrium

At rest,

$𝑚𝑔=𝑘𝑥$

$\Rightarrow 𝑥=\frac{𝑚𝑔}{𝑘}$

Step 4: Substitute value of $𝑘$

From above,

$𝑥=\frac{𝑚𝑔}{𝑘}$

$=𝑚𝑔\times \frac{{\left(\frac{𝑇}{2𝜋}\right)}^2}{𝑚}$

$=𝑔{\left(\frac{𝑇}{2𝜋}\right)}^2$

Step 5: Plug the values

Given $𝑇=6$ s and $𝑔={𝜋}^2=10$,

$𝑥=10{\left(\frac{6}{2𝜋}\right)}^2$

$=10{\left(\frac{3}{𝜋}\right)}^2$

$=10\times \frac{9}{{𝜋}^2}$

But ${𝜋}^2=10$, so:

$𝑥=10\times \frac{9}{10}=9\text{m}$

Final Answer:

Option D: 9 m

Given:

Electric potential,

$𝑉=4{𝑥}^2+8𝑥-3$

The relation between electric field $𝐸$ and potential $𝑉$ (along $𝑥$-axis) is:

$𝐸=-\frac{𝑑𝑉}{𝑑𝑥}$

Step 1: Differentiate $𝑉$ with respect to $𝑥$

$\frac{𝑑𝑉}{𝑑𝑥}=\frac{𝑑}{𝑑𝑥}(4{𝑥}^2+8𝑥-3)$

Use rules of differentiation:

• $\frac{𝑑}{𝑑𝑥}\left({𝑥}^2\right)=2𝑥$

• $\frac{𝑑}{𝑑𝑥}\left(𝑥\right)=1$

• $\frac{𝑑}{𝑑𝑥}\left(\text{constant}\right)=0$

So,

$\frac{𝑑𝑉}{𝑑𝑥}=4\times 2𝑥+8\times 1+0$

$\frac{𝑑𝑉}{𝑑𝑥}=8𝑥+8$

Step 2: Write expression for $𝐸$

$𝐸=-\frac{𝑑𝑉}{𝑑𝑥}=-(8𝑥+8)$

$𝐸=-8𝑥-8$

Step 3: Substitute $𝑥=0.5$ m

$𝐸=-8\left(0.5\right)-8$

$𝐸=-4-8$

$𝐸=-12\text{V/m}$

Final Answer:

$\overline{)-12\text{V/m}}$

Option B is correct.

Lenz's law in induction obeys the law of conservation of energy.

Explanation:

• Lenz's law states that the induced current in a coil due to a change in magnetic flux always flows in such a direction that it opposes the change in flux.

• This opposition ensures that energy is not created or destroyed, which is in accordance with the law of conservation of energy.

Correct Option:

Option B: Lenz's law in induction.

$\begin{array}{cccccc}𝐀& 𝐗=\overline{𝐀}& 𝐁& 𝐂& 𝐙=\overline{𝐁\cdot 𝐂}& 𝐘=𝐗+𝐙\\ 0& 1& 0& 0& 1& 1\\ 1& 0& 1& 1& 0& 0\end{array}$

The average kinetic energy of a gas molecule is given by:

$\text{Average kinetic energy}=\frac{3}{2}𝑘𝑇$

where $𝑘$ is the Boltzmann constant and $𝑇$ is the absolute temperature.

Let ${𝑇}_{𝐴}=350\text{K}$ (for gas A) and ${𝑇}_{𝐵}=420\text{K}$ (for gas B).

The ratio of average kinetic energies:

$\frac{\text{Average KE of B}}{\text{Average KE of A}}=\frac{\frac{3}{2}𝑘{𝑇}_{𝐵}}{\frac{3}{2}𝑘{𝑇}_{𝐴}}=\frac{{𝑇}_{𝐵}}{{𝑇}_{𝐴}}$

Substitute the values:

$\frac{{𝑇}_{𝐵}}{{𝑇}_{𝐴}}=\frac{420}{350}=\frac{42}{35}=\frac{6}{5}$

Therefore, the ratio is $6:5$.

Correct answer: Option A

Displacement and force (ma) are out of phase ( $\Delta 𝛼=𝜋$ ) in S.H.M. Therefore, the correct graph will be graph (c).

Given:

• Resistance, $𝑅=450\Omega$

• Self-inductance, $𝐿=1.5\text{H}$

• Frequency, $𝑓=\frac{150}{𝜋}\text{Hz}$

Step 1: Calculate angular frequency $𝜔$:

$𝜔=2𝜋𝑓=2𝜋\times \frac{150}{𝜋}=2\times 150=300\text{rad/s}$

Step 2: Inductive reactance ${𝑋}_{𝐿}$:

${𝑋}_{𝐿}=𝜔𝐿=300\times 1.5=450\Omega$

Step 3: Phase difference formula:

The phase difference $𝜙$ between voltage and current in a series $𝐿𝑅$ circuit is given by:

$\tan 𝜙=\frac{{𝑋}_{𝐿}}{𝑅}$

Substituting the values:

$\tan 𝜙=\frac{450}{450}=1$

Step 4: Hence,

$𝜙={\tan }^{-1}\left(1\right)$

Correct Option: Option B

$\overline{){\tan }^{-1}\left(1\right)}$

Given:

• Density of sphere = $𝜌$

• Density of liquid = $𝜎$ (where $𝜎>𝜌$)

• Height dropped = $ℎ$

• We are to neglect damping forces (ignore viscosity and friction)

• We have to find the maximum depth the sphere sinks after falling from height $ℎ$ into the liquid.

Let us proceed step by step:

Step 1: Find the velocity of the sphere just before entering the liquid

When the sphere falls from height $ℎ$ under gravity:

• Initial velocity, $𝑢=0$

• Acceleration, $𝑎=𝑔$

• Distance, $𝑠=ℎ$

Using the equation of motion:

${𝑣}^2={𝑢}^2+2𝑎𝑠$

So,

${𝑣}^2=0+2𝑔ℎ⟹𝑣=\sqrt{\surd 2𝑔ℎ}$

Step 2: Forces acting on the sphere inside the liquid

• Weight of sphere (downward): $𝑊=𝑚𝑔=𝑉𝜌𝑔$

• Buoyant force (upward): $𝐵=𝑉𝜎𝑔$

Net force acting on the sphere upward inside the liquid:

${𝐹}_{\text{net}}=𝐵-𝑊=𝑉𝜎𝑔-𝑉𝜌𝑔=𝑉𝑔(𝜎-𝜌)$

(Since $𝜎>𝜌$, the net force is upward.)

Step 3: Acceleration of the sphere inside the liquid

By Newton's second law:

$𝑚𝑎=𝑉𝑔(𝜎-𝜌)$

But $𝑚=𝑉𝜌$, so:

$𝑉𝜌𝑎=𝑉𝑔(𝜎-𝜌)$

$𝑎=\frac{𝑔(𝜎-𝜌)}{𝜌}$

This acceleration is upward.

Step 4: Maximum depth calculation

Inside the liquid, the sphere enters with velocity $𝑣=\sqrt{\surd 2𝑔ℎ}$ (downward).

It will go down, but since the net force is upward, it will slow down, stop momentarily, and then start rising.

Let the sphere sink to depth $𝑥$ before coming to rest.

We use the kinematic equation (taking downward as positive):

Final velocity, ${𝑣}_{𝑓}=0$ (momentarily stops)

Initial velocity, ${𝑣}_0=\sqrt{\surd 2𝑔ℎ}$ (downward)

Acceleration, $𝑎=-\frac{𝑔(𝜎-𝜌)}{𝜌}$ (upward, so take as negative compared to downward direction)

Distance travelled in liquid = $𝑥$

Use:

${𝑣}_2^{𝑓}={𝑣}_2^0+2𝑎𝑥$

So,

$0=\left(2𝑔ℎ\right)+2(-\frac{𝑔(𝜎-𝜌)}{𝜌})𝑥$

Simplify:

$0=2𝑔ℎ-2𝑔\frac{(𝜎-𝜌)}{𝜌}𝑥$

$2𝑔\frac{(𝜎-𝜌)}{𝜌}𝑥=2𝑔ℎ$

$𝑥=\frac{2𝑔ℎ\cdot \frac{𝜌}{2𝑔(𝜎-𝜌)}}{1}$

$𝑥=\frac{ℎ𝜌}{𝜎-𝜌}$

Step 5: Final Answer

The maximum depth up to which the sphere sinks is:

$\overline{)\frac{ℎ𝜌}{𝜎-𝜌}}$

Correct option: Option D

Let the two slabs have:

• Slab 1: Thermal conductivity $𝐾$, thickness $𝑥$

• Slab 2: Thermal conductivity $2𝐾$, thickness $4𝑥$

Let the temperature at the interface be $𝑇$.

Step 1: Write heat transfer equations for each slab

For steady-state, the rate of heat transfer ($𝑄$) through each layer is same:

For Slab 1:

$𝑄=\frac{𝐾𝐴({𝑇}_2-𝑇)}{𝑥}$

For Slab 2:

$𝑄=\frac{2𝐾𝐴(𝑇-{𝑇}_1)}{4𝑥}$

Step 2: Equate the rates and solve for $𝑇$

Set the two $𝑄$ values equal:

$\frac{𝐾𝐴({𝑇}_2-𝑇)}{𝑥}=\frac{2𝐾𝐴(𝑇-{𝑇}_1)}{4𝑥}$

Simplify:

• $𝐾$ and $𝐴$ cancel out.

• $2/\left(4𝑥\right)=1/\left(2𝑥\right)$:

$\frac{{𝑇}_2-𝑇}{𝑥}=\frac{𝑇-{𝑇}_1}{2𝑥}$

Cross-multiplied:

$2({𝑇}_2-𝑇)=𝑇-{𝑇}_1$

$2{𝑇}_2-2𝑇=𝑇-{𝑇}_1$

$2{𝑇}_2+{𝑇}_1=3𝑇$

$𝑇=\frac{2{𝑇}_2+{𝑇}_1}{3}$

Step 3: Substitute $𝑇$ back to find $𝑄$ in terms of ${𝑇}_2$ and ${𝑇}_1$

Recall:

$𝑄=\frac{𝐾𝐴({𝑇}_2-𝑇)}{𝑥}$

Use $𝑇=\frac{2{𝑇}_2+{𝑇}_1}{3}$:

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

So,

$𝑄=\frac{𝐾𝐴}{𝑥}\cdot \frac{{𝑇}_2-{𝑇}_1}{3}$

Or,

$𝑄=\left[\frac{𝐴({𝑇}_2-{𝑇}_1)𝐾}{𝑥}\right]\cdot \frac{1}{3}$

Step 4: Compare with the question's expression

Here,

$𝑓=\frac{1}{3}$

Final Answer:

Option D: $𝑓=\frac{1}{3}$

Step 1: Understanding the overtones in a pipe closed at one end

• In an organ pipe closed at one end, only odd harmonics are present.

• Fundamental frequency (first harmonic): $𝑓$

• First overtone (third harmonic): $3𝑓$

• Second overtone (fifth harmonic): $5𝑓$

• Third overtone (seventh harmonic): $7𝑓$

But the question asks for the first three overtones, i.e.,

• First overtone = $3𝑓$

• Second overtone = $5𝑓$

• Third overtone = $7𝑓$

Step 2: Writing the sum of first three overtones

Sum of frequencies:

$3𝑓+5𝑓+7𝑓=15𝑓$

Given that:

$15𝑓=3930\text{Hz}$

Step 3: Solving for fundamental frequency $𝑓$

$𝑓=\frac{3930}{15}$

$𝑓=262\text{Hz}$

Step 4: Final answer

The frequency of the fundamental mode of the organ pipe is 262 Hz.

Correct option:

Option B (262 Hz)

Given:

• Diameter of sphere, $𝑑=3.5\text{cm}=3.5\times {10}^{-2}\text{m}$

• Surface charge density, $𝜎=20𝜇\text{C}/{\text{m}}^2=20\times {10}^{-6}\text{C}/{\text{m}}^2$

• Permittivity of free space, ${𝜀}_0=8.85\times {10}^{-12}{\text{C}}^2/{\text{Nm}}^2$

Step 1: Formula for Electric Flux (Gauss’s Law)

Total electric flux, $𝜙$, leaving the surface is:

$𝜙=\frac{𝑞}{{𝜀}_0}$

where $𝑞$ is the total charge on the sphere.

Step 2: Calculate Area of the Sphere

Radius, $𝑟=\frac{𝑑}{2}=\frac{3.5}{2}\text{cm}=1.75\times {10}^{-2}\text{m}$

Area, $𝐴$:

$𝐴=4𝜋{𝑟}^2=4𝜋(1.75\times {10}^{-2}{)}^2$

Calculate:

$𝐴=4\times 3.14\times (1.75\times {10}^{-2}{)}^2=12.56\times (3.0625\times {10}^{-4})=12.56\times 3.0625\times {10}^{-4}\approx 38.48\times {10}^{-4}=3.848\times {10}^{-3}{\text{m}}^2$

Step 3: Find Total Charge

Total charge, $𝑞=𝜎\times 𝐴$:

$𝑞=20\times {10}^{-6}{\text{C/m}}^2\times 3.848\times {10}^{-3}{\text{m}}^2=76.96\times {10}^{-9}\text{C}=7.696\times {10}^{-8}\text{C}$

Step 4: Calculate the Total Electric Flux

$𝜙=\frac{𝑞}{{𝜀}_0}=\frac{7.696\times {10}^{-8}}{8.85\times {10}^{-12}}$

Calculate denominator:

$𝜙=\frac{7.696}{8.85}\times {10}^{-8+12}=0.870\times {10}^4=8.7\times {10}^3\text{Wb}$

Step 5: Rounding off

Among the given options, the closest is:

$8.7\times {10}^3\text{Wb}\approx 87\times {10}^2\text{Wb}$

Final Answer:

Option C: $87\times {10}^2\text{Wb}$

Given:

• Initial kinetic energy of the stone = $𝐸$

• Angle of projection with horizontal = $𝜃$

Step 1: Find the initial velocity using kinetic energy

Initial kinetic energy $𝐸$ is given by:

$𝐸=\frac{1}{2}𝑚{𝑢}^2$

where $𝑢$ is the initial speed and $𝑚$ is the mass.

So,

$𝑢=\sqrt{\surd \frac{2𝐸}{𝑚}}$

Step 2: Break the velocity into horizontal and vertical components

• Horizontal component: ${𝑢}_{𝑥}=𝑢\cos 𝜃$

• Vertical component: ${𝑢}_{𝑦}=𝑢\sin 𝜃$

Step 3: At the highest point

At the highest point, the vertical component of velocity becomes zero (${𝑣}_{𝑦}=0$), but the horizontal component remains unchanged (${𝑢}_{𝑥}$).

So, velocity at the highest point is:

${𝑣}_{\text{highest}}=𝑢\cos 𝜃$

Step 4: Kinetic energy at the highest point

Only the horizontal component remains, so

$\text{Kinetic energy at highest point}=\frac{1}{2}𝑚(𝑢\cos 𝜃{)}^2=\frac{1}{2}𝑚{𝑢}^2{\cos }^2𝜃$

Recall from Step 1:

$\frac{1}{2}𝑚{𝑢}^2=𝐸$

So, kinetic energy at highest point:

$=𝐸{\cos }^2𝜃$

Final Answer:

The correct option is Option C: $𝐸{\cos }^2𝜃$

Given:

• Intensity at path difference $\frac{𝜆}{4}$ is $\frac{𝐾}{2}$.

• We are to find the intensity at path difference $𝜆$.

Step 1: Write the general formula for intensity

In Young's Double Slit Experiment, the intensity at a point on the screen is:

$𝐼={𝐼}_0[1+\cos 𝜙]$

Here,

• ${𝐼}_0$: intensity due to each slit (if both slits have equal amplitude).

• $𝜙$: phase difference at that point.

The phase difference $𝜙$ is related to path difference $𝑥$ as:

$𝜙=\frac{2𝜋}{𝜆}\times \text{path difference}$

Step 2: Use the condition at path difference $=\frac{𝜆}{4}$

At $\text{path difference}=\frac{𝜆}{4}$:

• ${𝜙}_1=\frac{2𝜋}{𝜆}\cdot \frac{𝜆}{4}=\frac{𝜋}{2}$

The intensity at this point is:

${𝐼}_1={𝐼}_0[1+\cos \left(\frac{𝜋}{2}\right)]$

Since $\cos \left(\frac{𝜋}{2}\right)=0$,

${𝐼}_1={𝐼}_0[1+0]={𝐼}_0$

But it is given that this intensity is $\frac{𝐾}{2}$.

So,

${𝐼}_0=\frac{𝐾}{2}$

Step 3: Find intensity at path difference $=𝜆$

Now,

• ${𝜙}_2=\frac{2𝜋}{𝜆}\times 𝜆=2𝜋$

So, the intensity at this point:

${𝐼}_2={𝐼}_0[1+\cos (2𝜋\left)\right]$

Since $\cos \left(2𝜋\right)=1$,

${𝐼}_2={𝐼}_0(1+1)=2{𝐼}_0$

Substitute ${𝐼}_0=\frac{𝐾}{2}$:

${𝐼}_2=2\times \frac{𝐾}{2}=𝐾$

Step 4: Check answer options

The correct answer is

Option C: $𝐾$

The correct NCERT relation between $𝐵$, $𝐻$, and $𝑀$ is:

$𝐵={𝜇}_0(𝐻+𝑀)$

Let's break it down:

1. Magnetic field inside the material ($𝐵$):

• $𝐵$ is the net magnetic field inside a magnetised material.

1. External magnetic field ($𝐻$):

• $𝐻$ is the magnetic field intensity produced by external currents (without the material).

1. Magnetisation ($𝑀$):

• $𝑀$ is the magnetic moment per unit volume, representing how much the material is magnetised.

1. Relation:

• According to NCERT, the relation is:

  $𝐵={𝜇}_0(𝐻+𝑀)$

• Here, ${𝜇}_0$ is the permeability of free space.

Correct option:

$\overline{)𝐵={𝜇}_0(𝐻+𝑀)}$

So, the answer is Option D.

Change in momentum is given by the area under the F - t graph.

$\begin{array}{cc}& \text{ Area }=8\times 0.5+4\times 0.5\\ & =4+2=6N-s\\ \therefore & \mathrm{mv}=6N-s\\ \therefore & v=\frac{6}{3}=2m/s\end{array}$

Given:

An electron accelerated by potential difference $𝑉$ has de-Broglie wavelength $𝜆$.

We have to find: The de-Broglie wavelength when it is accelerated by $9𝑉$.

Step 1: Formula for de-Broglie wavelength of an electron

The de-Broglie wavelength $𝜆$ of an electron accelerated through a potential $𝑉$ is:

$𝜆=\frac{ℎ}{𝑝}$

When an electron is accelerated by $𝑉$, its kinetic energy $𝐾=𝑒𝑉$.

Momentum:

$𝐾=\frac{1}{2}𝑚{𝑣}^2=𝑒𝑉⟹𝑣=\sqrt{\surd \frac{2𝑒𝑉}{𝑚}}$

$𝑝=𝑚𝑣=𝑚\sqrt{\surd \frac{2𝑒𝑉}{𝑚}}=\sqrt{\surd 2𝑚𝑒𝑉}$

So,

$𝜆=\frac{ℎ}{\sqrt{\surd 2𝑚𝑒𝑉}}$

Step 2: Relationship between wavelength and potential

From the formula:

$𝜆\propto \frac{1}{\sqrt{\surd 𝑉}}$

Step 3: Compare for $𝑉$ and $9𝑉$

Let ${𝜆}_1$ be for $𝑉$ and ${𝜆}_2$ for $9𝑉$.

$\frac{{𝜆}_2}{{𝜆}_1}=\sqrt{\surd \frac{𝑉}{9𝑉}}=\sqrt{\surd \frac{1}{9}}=\frac{1}{3}$

${𝜆}_2=\frac{{𝜆}_1}{3}$

Step 4: Conclusion

The answer is:

Option B: $\frac{𝜆}{3}$

Given:

• Maximum tension in thread, ${𝑇}_{\text{max}}=3.7\text{kg wt}=3.7\times 10\text{N}=37\text{N}$

• Mass of stone, $𝑚=500\text{g}=0.5\text{kg}$

• Radius of circle, $𝑟=4\text{m}$

• Acceleration due to gravity, $𝑔=10{\text{m/s}}^2$

Step 1: Identify the location of maximum tension

• In the vertical circular motion, the tension will be maximum at the lowest point.

Step 2: Write the equation for tension at the lowest point

At the lowest point,

${𝑇}_{\text{max}}=𝑚{𝜔}^2𝑟+𝑚𝑔$

Where,

• $𝑚{𝜔}^2𝑟$ is the centripetal force,

• $𝑚𝑔$ is the weight (acting downwards).

Step 3: Substitute the given values

$37=0.5{𝜔}^2\times 4+0.5\times 10$

$37=2{𝜔}^2+5$

Step 4: Solve for $𝜔$

Subtract $5$ from both sides:

$37-5=2{𝜔}^2$

$32=2{𝜔}^2$

${𝜔}^2=\frac{32}{2}=16$

$𝜔=\sqrt{\surd 16}=4\text{rad/s}$

Answer:

Option B ($4\text{rad/s}$) is correct.

First, according to Poiseuille's law for flow between two horizontal layers, the velocity of liquid increases linearly with the height from the stationary surface (assuming laminar flow):

Let,

• $𝑦$ = height from the stationary bottom

• $𝑣$ = velocity of water at height $𝑦$

For flow between layers, velocity at height $𝑦$ is proportional to $𝑦$:

$𝑣\propto 𝑦$

Let ${𝑣}_{𝐴}$ and ${𝑣}_{𝐵}$ be the velocities at heights ${𝑦}_{𝐴}$ and ${𝑦}_{𝐵}$ respectively.

Given:

• ${𝑦}_{𝐴}=40$ cm, ${𝑣}_{𝐴}=12$ cm/s

• ${𝑦}_{𝐵}=90$ cm, ${𝑣}_{𝐵}=?$

Using the proportionality,

$\frac{{𝑣}_{𝐴}}{{𝑣}_{𝐵}}=\frac{{𝑦}_{𝐴}}{{𝑦}_{𝐵}}$

Substitute the values:

$\frac{12}{{𝑣}_{𝐵}}=\frac{40}{90}$

Cross-multiply:

$12\times 90={𝑣}_{𝐵}\times 40$

${𝑣}_{𝐵}=\frac{12\times 90}{40}$

${𝑣}_{𝐵}=\frac{1080}{40}$

${𝑣}_{𝐵}=27\text{cm/s}$

Correct Option:

Option C

$27\mathrm{cm}/s$

Given:

• Frequency, $𝑓=400\text{Hz}$

• Velocity, $𝑣=336\text{m/s}$

• Phase difference, $\Delta 𝜙={60}^{\circ }$

Step 1: Calculate the wavelength $\left(𝜆\right)$.

We know,

$𝑣=𝑓𝜆$

So,

$𝜆=\frac{𝑣}{𝑓}=\frac{336}{400}=0.84\text{m}$

Step 2: Relate the phase difference to distance $\left(𝑥\right)$.

A phase difference ${360}^{\circ }$ (or $2𝜋$ radians) corresponds to one full wavelength:

$\frac{\Delta 𝜙}{{360}^{\circ }}=\frac{𝑥}{𝜆}$

Substitute the values:

• $\Delta 𝜙={60}^{\circ }$

• $𝜆=0.84\text{m}$

So,

$\frac{{60}^{\circ }}{{360}^{\circ }}=\frac{𝑥}{0.84}$

$\frac{1}{6}=\frac{𝑥}{0.84}$

$𝑥=\frac{0.84}{6}=0.14\text{m}$

Final Answer:

$\overline{)0.14\text{m}}$

Correct option: Option B

Given:

• Initial flux, ${𝜙}_1=4\times {10}^{-4}\text{Wb}$

• Final flux, ${𝜙}_2=30\%$ of ${𝜙}_1=0.3\times 4\times {10}^{-4}=1.2\times {10}^{-4}\text{Wb}$

• Induced emf, $𝑒=0.56\text{mV}=0.56\times {10}^{-3}\text{V}$

• Time interval, $𝑡=?$

Step 1: Formula for average induced emf

$𝑒=\frac{|{𝜙}_2-{𝜙}_1|}{𝑡}$

Step 2: Substituting the values

$𝑒=\frac{\mid 1.2\times {10}^{-4}-4\times {10}^{-4}\mid }{𝑡}$

$𝑒=\frac{\mid -2.8\times {10}^{-4}\mid }{𝑡}$

$𝑒=\frac{2.8\times {10}^{-4}}{𝑡}$

Given $𝑒=0.56\times {10}^{-3}\text{V}=5.6\times {10}^{-4}\text{V}$ (But $0.56\text{mV}$ is $0.56\times {10}^{-3}$, or $5.6\times {10}^{-4}$ — check carefully.)

But $0.56\text{mV}=0.56\times {10}^{-3}=5.6\times {10}^{-4}\text{V}$