MHT CET 2025 21st April Evening Shift (Physics & Chemistry) Question Paper with Solutions

We are asked:

Error in measurement of radius of a sphere is $2\%$. The error in the calculated value of its volume is?

Step 1: Recall formula of volume of a sphere

$𝑉=\frac{4}{3}𝜋{𝑟}^3$

Step 2: Propagation of error rule

If a quantity depends as ${𝑥}^{𝑛}$, the percentage error in the quantity is:

$\frac{\Delta 𝑄}{𝑄}\times 100\%=𝑛\cdot \frac{\Delta 𝑥}{𝑥}\times 100\%$

Step 3: Apply it here

$𝑉\propto {𝑟}^3$

So percentage error in $𝑉$ is

$\Delta 𝑉\%=3\cdot \Delta 𝑟\%$

Step 4: Substitute value

$\Delta 𝑟=2\%\Rightarrow \Delta 𝑉=3\times 2\%=6\%.$

✅ Correct Option: C) 6%

In the circuit shown, there is no resistor connected between point B \& D .

As a result. point 'B' \& point 'D' become equipotential

$\Rightarrow {𝑉}_{𝐵}={𝑉}_{𝐷}$

No current will flow through the wire even if the switch is closed.

$\therefore$ Resistor R \& Galvanometer will be in series.

$\therefore I_R=I_G$

We are asked:

A unit vector in the direction of the resultant vector of

$\overrightarrow{𝐴}=-2\widehat{𝑖}+3\widehat{𝑗}+\widehat{𝑘},\overrightarrow{𝐵}=\widehat{𝑖}+2\widehat{𝑗}-4\widehat{𝑘}.$

Step 1: Find the resultant

$\overrightarrow{𝑅}=\overrightarrow{𝐴}+\overrightarrow{𝐵}=(-2+1)\widehat{𝑖}+(3+2)\widehat{𝑗}+(1-4)\widehat{𝑘}=(-1)\widehat{𝑖}+5\widehat{𝑗}-3\widehat{𝑘}.$

So,

$\overrightarrow{𝑅}=-\widehat{𝑖}+5\widehat{𝑗}-3\widehat{𝑘}.$

Step 2: Magnitude of $\overrightarrow{𝑅}$

$\left|\overrightarrow{𝑅}\right|=\sqrt{\surd (-1{)}^2+5^2+(-3{)}^2}=\sqrt{\surd 1+25+9}=\sqrt{\surd 35}.$

Step 3: Unit vector in the direction of $\overrightarrow{𝑅}$

$\widehat{𝑅}=\frac{\overrightarrow{𝑅}}{\left|\overrightarrow{𝑅}\right|}=\frac{-\widehat{𝑖}+5\widehat{𝑗}-3\widehat{𝑘}}{\sqrt{\surd 35}}.$

✅ Final Answer:

Option D

$\frac{-\widehat{𝑖}+5\widehat{𝑗}-3\widehat{𝑘}}{\sqrt{\surd 35}}$

Step 1: Initial kinetic energy

Only first ring is rotating:

$𝐾{𝐸}_{\text{initial}}=\frac{1}{2}𝐼{𝜔}^2$

Step 2: Conservation of angular momentum

System is isolated (no external torque about axis).

Let final angular velocity = ${𝜔}_{𝑓}$.

Initial angular momentum:

${𝐿}_{\text{initial}}=𝐼𝜔$

Final total moment of inertia = $𝐼+𝐼=2𝐼$.

So,

${𝐿}_{\text{final}}=\left(2𝐼\right){𝜔}_{𝑓}$

By conservation:

$𝐼𝜔=2𝐼{𝜔}_{𝑓}\Rightarrow {𝜔}_{𝑓}=\frac{𝜔}{2}$

Step 3: Final kinetic energy

$𝐾{𝐸}_{\text{final}}=\frac{1}{2}\left(2𝐼\right){\left(\frac{𝜔}{2}\right)}^2=𝐼\cdot \frac{{𝜔}^2}{4}=\frac{𝐼{𝜔}^2}{4}$

Step 4: Loss in kinetic energy

$\Delta 𝐾𝐸=𝐾{𝐸}_{\text{initial}}-𝐾{𝐸}_{\text{final}}=\frac{1}{2}𝐼{𝜔}^2-\frac{1}{4}𝐼{𝜔}^2=\frac{1}{4}𝐼{𝜔}^2$

✅ Final Answer:

$\overline{)\frac{𝐼{𝜔}^2}{4}}$

That corresponds to Option C.

We’re asked:

Two coherent sources interfere. Intensity ratio = $𝑏$. Then compute

$\frac{{𝐼}_{max}+{𝐼}_{min}}{{𝐼}_{max}-{𝐼}_{min}}.$

Step 1. Express intensities of sources

Let the two sources have intensities ${𝐼}_1$ and ${𝐼}_2$.

Their ratio:

$\frac{{𝐼}_1}{{𝐼}_2}=𝑏.$

Assume ${𝐼}_1=𝑏{𝐼}_2$.

Let ${𝐼}_2={𝐼}_0$. Then ${𝐼}_1=𝑏{𝐼}_0$.

Step 2. Amplitudes

Amplitude is proportional to $\sqrt{\surd 𝐼}$.

So,

${𝐴}_1=\sqrt{\surd {𝐼}_1}=\sqrt{\surd 𝑏{𝐼}_0}=\sqrt{\surd 𝑏}\sqrt{\surd {𝐼}_0},$

${𝐴}_2=\sqrt{\surd {𝐼}_2}=\sqrt{\surd {𝐼}_0}.$

Step 3. Constructive and destructive interference

• Maximum amplitude: ${𝐴}_{max}={𝐴}_1+{𝐴}_2=(\sqrt{\surd 𝑏}+1)\sqrt{\surd {𝐼}_0}$.

So maximum intensity:

${𝐼}_{max}=({𝐴}_{max}{)}^2=(\sqrt{\surd 𝑏}+1{)}^2{𝐼}_0.$

• Minimum amplitude: ${𝐴}_{min}=|{𝐴}_1-{𝐴}_2|=|\sqrt{\surd 𝑏}-1|\sqrt{\surd {𝐼}_0}$.

So minimum intensity:

${𝐼}_{min}=({𝐴}_{min}{)}^2=(\sqrt{\surd 𝑏}-1{)}^2{𝐼}_0.$

Step 4. Compute the required ratio

$\frac{{𝐼}_{max}+{𝐼}_{min}}{{𝐼}_{max}-{𝐼}_{min}}=\frac{(\sqrt{\surd 𝑏}+1{)}^2{𝐼}_0+(\sqrt{\surd 𝑏}-1{)}^2{𝐼}_0}{(\sqrt{\surd 𝑏}+1{)}^2{𝐼}_0-(\sqrt{\surd 𝑏}-1{)}^2{𝐼}_0}.$

Cancel ${𝐼}_0$:

$=\frac{(\sqrt{\surd 𝑏}+1{)}^2+(\sqrt{\surd 𝑏}-1{)}^2}{(\sqrt{\surd 𝑏}+1{)}^2-(\sqrt{\surd 𝑏}-1{)}^2}.$

Expand:

• Numerator: $(𝑏+1+2\sqrt{\surd 𝑏})+(𝑏+1-2\sqrt{\surd 𝑏})=2𝑏+2.$

• Denominator: $(𝑏+1+2\sqrt{\surd 𝑏})-(𝑏+1-2\sqrt{\surd 𝑏})=4\sqrt{\surd 𝑏}.$

So ratio:

$=\frac{2(𝑏+1)}{4\sqrt{\surd 𝑏}}=\frac{1+𝑏}{2\sqrt{\surd 𝑏}}.$

✅ Final Answer:

Option B:

$\overline{)\frac{1+𝑏}{2\sqrt{\surd 𝑏}}}$

We have two satellites $𝐴$ and $𝐵$ orbiting Earth in circular orbits.

• Radius of orbit of $𝐴$: ${𝑟}_{𝐴}=4𝑅$

• Radius of orbit of $𝐵$: ${𝑟}_{𝐵}=𝑅$

• Speed of $𝐵$: ${𝑣}_{𝐵}=6𝑉$

We need the speed of $𝐴$.

Step 1: Orbital velocity formula

For a satellite in circular orbit,

$𝑣=\sqrt{\surd \frac{𝐺𝑀}{𝑟}}$

where $𝑀$ is mass of Earth and $𝑟$ is the orbital radius.

So,

$\frac{{𝑣}_{𝐴}}{{𝑣}_{𝐵}}=\sqrt{\surd \frac{{𝑟}_{𝐵}}{{𝑟}_{𝐴}}}$

Step 2: Plug in given values

$\frac{{𝑣}_{𝐴}}{{𝑣}_{𝐵}}=\sqrt{\surd \frac{𝑅}{4𝑅}}=\sqrt{\surd \frac{1}{4}}=\frac{1}{2}$

Thus,

${𝑣}_{𝐴}=\frac{1}{2}{𝑣}_{𝐵}$

Step 3: Evaluate

${𝑣}_{𝐴}=\frac{1}{2}\times 6𝑉=3𝑉$

✅ Final Answer:

$\overline{)3𝑉}$

Option A: 3V

Step 1. Recall magnetic field in a toroid (without material)

For a toroid with $𝑁$ turns and current $𝐼$, inside magnetic field (without core) is:

${𝐵}_0={𝜇}_0\cdot \frac{𝑁𝐼}{𝑙}$

where $𝑙=2𝜋𝑟$ (mean circumference).

Step 2. With a material (relative permeability)

If the medium inside has magnetic susceptibility $𝜒$, then:

${𝜇}_{𝑟}=1+𝜒$

So the permeability is:

$𝜇={𝜇}_0(1+𝜒)$

Thus the new field inside:

$𝐵=𝜇\frac{𝑁𝐼}{𝑙}=(1+𝜒){𝐵}_0$

Step 3. Percentage increase

The increase:

$\Delta 𝐵=𝐵-{𝐵}_0=𝜒{𝐵}_0$

$\%\text{increase}=\frac{\Delta 𝐵}{{𝐵}_0}\times 100=𝜒\times 100$

✅ Final Answer:

$\overline{)𝜒\times 100}$

That corresponds to Option D.

We are tasked to find the rotational kinetic energy in the third orbit for a diatomic molecule using Bohr's quantization condition.

Step 1: Bohr’s quantization condition for angular momentum

$𝐿=𝑛\frac{ℎ}{2𝜋},𝑛=1,2,3,\dots$

Step 2: Rotational kinetic energy in terms of angular momentum

For a rotating rigid body (diatomic molecule modeled as a rigid rotor), the rotational kinetic energy is

$𝐸=\frac{{𝐿}^2}{2𝐼}$

Step 3: Substitute quantized angular momentum

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

${𝐸}_{𝑛}=\frac{{𝑛}^2{ℎ}^2}{8{𝜋}^2𝐼}$

Step 4: Energy in the third orbit $(𝑛=3)$

${𝐸}_3=\frac{3^2{ℎ}^2}{8{𝜋}^2𝐼}=\frac{9{ℎ}^2}{8{𝜋}^2𝐼}$

✅ Final Answer:

Option A:

$\overline{)\frac{9{ℎ}^2}{8{𝜋}^2𝐼}}$

We are asked about rotational kinetic energy $𝑥$, angular momentum $𝑦$, and the moment of inertia $𝐼$.

Step 1: Recall formulas

• Rotational kinetic energy:

$𝐾=\frac{1}{2}𝐼{𝜔}^2$

• Angular momentum:

$𝐿=𝐼𝜔$

Here:

$𝐾=𝑥,𝐿=𝑦$

Step 2: Eliminate $𝜔$

From angular momentum:

$𝜔=\frac{𝐿}{𝐼}=\frac{𝑦}{𝐼}$

Substitute into the kinetic energy:

$𝑥=\frac{1}{2}𝐼{\left(\frac{𝑦}{𝐼}\right)}^2$

$𝑥=\frac{1}{2}𝐼\cdot \frac{{𝑦}^2}{{𝐼}^2}$

$𝑥=\frac{1}{2}\frac{{𝑦}^2}{𝐼}$

Step 3: Solve for $𝐼$

$𝐼=\frac{{𝑦}^2}{2𝑥}$

Final Answer:

The moment of inertia is

$\overline{)\frac{{𝑦}^2}{2𝑥}}$

✔ This corresponds to Option B.

Step 1. Relation between velocity and refractive index

Velocity in medium:

$𝑣=\frac{𝑐}{𝜇}$

where $𝜇$ is the refractive index.

In rarer medium (say air): ${𝑣}_1=𝑐$.

In glass slab: ${𝑣}_2=0.8𝑐$ (since reduced by 20%).

So

$𝜇=\frac{{𝑣}_1}{{𝑣}_2}=\frac{𝑐}{0.8𝑐}=\frac{1}{0.8}=1.25$

Step 2. Use Snell’s Law

Snell’s law:

${𝑛}_1\sin 𝑖={𝑛}_2\sin 𝑟$

For air, ${𝑛}_1=1$, ${𝑛}_2=𝜇=1.25$.

So

$\sin 𝑟=\frac{\sin 𝑖}{1.25}$

For small angles (since given $𝑖$ is small),

$\sin 𝜃\approx 𝜃$ (in radians).

Therefore

$𝑟\approx \frac{𝑖}{1.25}=0.8𝑖$

Step 3. Find angle of deviation

Deviation $𝛿=𝑖-𝑟$.

$𝛿=𝑖-0.8𝑖=0.2𝑖=\frac{𝑖}{5}$

✅ Final Answer

$\overline{)\frac{𝑖}{5}}$

So, the correct option is A. $\frac{𝑖}{5}$.

We’re asked:

A body starts from rest and moves with uniform acceleration $𝑎$.

We want the ratio of the distance covered in the ${𝑛}^{\text{th}}$ second to the total distance covered in $𝑛$ seconds.

Step 1: Recall standard kinematic results

• Initial velocity $𝑢=0$.

• Total distance in $𝑛$ seconds:

${𝑆}_{𝑛}=\frac{1}{2}𝑎{𝑛}^2$

• Distance covered in the ${𝑛}^{\text{th}}$ second:

${𝑠}_{𝑛}=𝑢+\frac{1}{2}𝑎(2𝑛-1)=\frac{1}{2}𝑎(2𝑛-1)\text{(since }𝑢=0\text{)}$

Step 2: Ratio

$\text{Ratio}=\frac{{𝑠}_{𝑛}}{{𝑆}_{𝑛}}=\frac{\frac{1}{2}𝑎(2𝑛-1)}{\frac{1}{2}𝑎{𝑛}^2}=\frac{2𝑛-1}{{𝑛}^2}$

Step 3: Simplify

$\frac{2𝑛-1}{{𝑛}^2}=\frac{2𝑛}{{𝑛}^2}-\frac{1}{{𝑛}^2}=\frac{2}{𝑛}-\frac{1}{{𝑛}^2}$

Answer:

$\overline{)\frac{2}{𝑛}-\frac{1}{{𝑛}^2}}$

That corresponds to Option B. ✅

The self-inductance of a coil (approximated as a solenoid wound on a core) is given by the formula:

$𝐿=𝜇{𝑛}^2𝐴𝑙$

Where:

• $𝐿$ is the self-inductance.

• $𝜇$ is the permeability of the core material.

• $𝑛$ is the number of turns per unit length.

• $𝐴$ is the cross-sectional area of the core.

• $𝑙$ is the length of the coil (which is the length of the core along the axis of the coil).

Let's analyze how each parameter changes based on the problem statement:

1. Linear dimensions of the core: The problem states that "all the linear dimensions of core are increased by a factor 3".

   A core with a rectangular cross-section has three main linear dimensions: width ($𝑤$), height ($ℎ$), and length ($𝑙$).

   So, the new dimensions will be:

   • New width ${𝑤}^{\prime }=3𝑤$

   • New height ${ℎ}^{\prime }=3ℎ$

   • New length ${𝑙}^{\prime }=3𝑙$

2. Cross-sectional area ($𝐴$):

   The original cross-sectional area was $𝐴=𝑤\times ℎ$.

   The new cross-sectional area ${𝐴}^{\prime }$ will be ${𝐴}^{\prime }={𝑤}^{\prime }\times {ℎ}^{\prime }=\left(3𝑤\right)\times \left(3ℎ\right)=9𝑤ℎ=9𝐴$.

   So, ${𝐴}^{\prime }$ increases by a factor of 9.

3. Length of the coil ($𝑙$):

   As part of the linear dimensions of the core, the length of the coil (the length over which the turns are wound) also increases by a factor of 3.

   So, ${𝑙}^{\prime }=3𝑙$.

4. Number of turns per unit length ($𝑛$):

   The problem explicitly states that "number of turns per unit length of coil remains same".

   So, ${𝑛}^{\prime }=𝑛$.

5. Permeability ($𝜇$):

   The core material is not changed, so its permeability $𝜇$ remains constant.

Now, let's calculate the new self-inductance (${𝐿}^{\prime }$):

The original self-inductance was $𝐿=𝜇{𝑛}^2𝐴𝑙$.

The new self-inductance is ${𝐿}^{\prime }=𝜇({𝑛}^{\prime }{)}^2{𝐴}^{\prime }{𝑙}^{\prime }$.

Substitute the changed parameters:

${𝐿}^{\prime }=𝜇\left(𝑛{)}^2\right(9𝐴\left)\right(3𝑙)$

${𝐿}^{\prime }=𝜇{𝑛}^2𝐴𝑙\times (9\times 3)$

${𝐿}^{\prime }=𝜇{𝑛}^2𝐴𝑙\times 27$

Since $𝐿=𝜇{𝑛}^2𝐴𝑙$, we can write:

${𝐿}^{\prime }=27𝐿$

Therefore, the self-inductance increases by a factor of 27.

The final answer is $\overline{)27}$.

The question is about a charged soap bubble with radius $𝑅$ that breaks into 64 smaller soap bubbles, each with radius $𝑟$.

Electrostatic Pressure:

Because the bubble is charged, there is a force called electrostatic pressure, which is given by $\frac{{𝜎}^2}{2{𝜀}_0}$. When the bubble is stable, the extra pressure inside it matches this electrostatic pressure.

Total Volume is Constant:

When the big bubble breaks into 64 small ones, the total amount of soap (volume) stays the same.

$\frac{4}{3}𝜋{𝑅}^3=64\times \frac{4}{3}𝜋{𝑟}^3$

This means: ${𝑅}^3=64{𝑟}^3$ so $𝑟=\frac{𝑅}{4}$

Mechanical Pressure in Bubbles:

The extra pressure inside a soap bubble is given by $\frac{4𝑇}{𝑅}$ for the big bubble and $\frac{4𝑇}{𝑟}$ for a small bubble, where $𝑇$ is the surface tension.

So, ${𝑃}_{𝐵}=\frac{4𝑇}{𝑅}$ for the big bubble, and ${𝑃}_{𝑠}=\frac{4𝑇}{𝑟}$ for the small bubble.

Ratio of Mechanical Pressures:

Now, to find the ratio of the mechanical pressure of the big bubble to that of a small bubble: $\frac{{𝑃}_{𝐵}}{{𝑃}_{𝑠}}=\frac{4𝑇/𝑅}{4𝑇/𝑟}=\frac{𝑟}{𝑅}$

We already found that $𝑟=\frac{𝑅}{4}$, so: $\frac{{𝑃}_{𝐵}}{{𝑃}_{𝑠}}=\frac{𝑅/4}{𝑅}=\frac{1}{4}$

Step 1: Recall the photoelectric equation

${𝐾}_{\text{max}}={𝐸}_{\text{photon}}-𝜙$

where $𝜙$ is the work function.

Also,

${𝐾}_{\text{max}}=\frac{1}{2}𝑚{𝑣}^2.$

Step 2: Case 1

Photon energy ${𝐸}_1=2𝜙$.

${𝐾}_1={𝐸}_1-𝜙=2𝜙-𝜙=𝜙.$

So

$\frac{1}{2}𝑚{𝑉}_2^1=𝜙\Rightarrow {𝑉}_1=\sqrt{\surd \frac{2𝜙}{𝑚}}$

Step 3: Case 2

Photon energy ${𝐸}_2=3𝜙$.

${𝐾}_2=3𝜙-𝜙=2𝜙.$

So

$\frac{1}{2}𝑚{𝑉}_2^2=2𝜙\Rightarrow {𝑉}_2=\sqrt{\surd \frac{4𝜙}{𝑚}}=2\sqrt{\surd \frac{𝜙}{𝑚}}.$

Step 4: Ratio

$\frac{{𝑉}_1}{{𝑉}_2}=\frac{\sqrt{\surd 2𝜙/𝑚}}{2\sqrt{\surd 𝜙/𝑚}}=\frac{\sqrt{\surd 2}}{2}=\frac{1}{\sqrt{\surd 2}}.$

So

${𝑉}_1:{𝑉}_2=1:\sqrt{\surd 2}.$

✅ Final Answer:

Option D — $1:\sqrt{\surd 2}$

We are asked:

According to Huygen's wave theory of light, which one of the following statements is not correct?

Option A:

Different colours of light are due to different wavelengths of waves.

✅ Correct according to wave theory.

Option B:

Different colours of light are due to different sizes of the corpuscles.

❌ This belongs to Newton’s corpuscular theory, not Huygen’s wave theory.

Option C:

Speed of light in denser medium is less than that in rarer medium.

✅ Correct prediction of Huygens’ wave theory.

Option D:

It can explain laws of reflection and refraction.

✅ Correct.

✅ Correct Answer: Option B

"Different colours of light are due to different sizes of the corpuscles."

We need to find the force needed to keep the conveyor belt moving at the same speed when sand is dropped on it.

Newton's second law says that force is the change in momentum per second:

$𝐹=\frac{𝑑\overrightarrow{𝑝}}{𝑑𝑡}$

Here, momentum ($\overrightarrow{𝑝}$) is mass ($𝑚$) times velocity ($\overrightarrow{𝑣}$):

$𝐹=\frac{𝑑\left(𝑚\overrightarrow{𝑣}\right)}{𝑑𝑡}$

The velocity of the belt is constant ($𝑉$), and sand is added at a rate of $𝑀$ kg/s, so:

$𝐹=\frac{𝑑𝑚}{𝑑𝑡}\times 𝑉$

Since $\frac{𝑑𝑚}{𝑑𝑡}=𝑀$, we get:

$𝐹=𝑀𝑉$

We are asked about the magnetic field at the center of a circular current-carrying coil when the number of turns is changed.

Step 1: Formula for field

For a circular coil of radius $𝑅$, carrying current $𝐼$, the magnetic field at the center due to one turn is

$𝐵=\frac{{𝜇}_0𝐼}{2𝑅}.$

For $𝑁$ turns (closely wound):

$𝐵=\frac{{𝜇}_0𝑁𝐼}{2𝑅}.$

So $𝐵\propto \frac{𝑁}{𝑅}$.

Step 2: Initial case

Initially:

• ${𝑁}_1=9$,

• Suppose the length of wire used = $𝐿$.

Circumference of one turn = $2𝜋{𝑅}_1$.

So $𝐿={𝑁}_1\cdot 2𝜋{𝑅}_1=9\cdot 2𝜋{𝑅}_1$.

Thus

${𝑅}_1=\frac{𝐿}{18𝜋}.$

Magnetic field

${𝐵}_1=\frac{{𝜇}_0{𝑁}_1𝐼}{2{𝑅}_1}.$

Step 3: Second case

Rewound into 3 turns, same wire length $𝐿$.

Now circumference per turn = $2𝜋{𝑅}_2$.

So $𝐿={𝑁}_2\cdot 2𝜋{𝑅}_2=3\cdot 2𝜋{𝑅}_2$.

Thus

${𝑅}_2=\frac{𝐿}{6𝜋}.$

Field

${𝐵}_2=\frac{{𝜇}_0{𝑁}_2𝐼}{2{𝑅}_2}.$

Step 4: Ratio ${𝐵}_2/{𝐵}_1$

$\frac{{𝐵}_2}{{𝐵}_1}=\frac{\frac{{𝜇}_0{𝑁}_2𝐼}{2{𝑅}_2}}{\frac{{𝜇}_0{𝑁}_1𝐼}{2{𝑅}_1}}=\frac{{𝑁}_2/{𝑅}_2}{{𝑁}_1/{𝑅}_1}=\frac{{𝑁}_2{𝑅}_1}{{𝑁}_1{𝑅}_2}.$

Now plug in numbers:

${𝑅}_1=\frac{𝐿}{18𝜋},{𝑅}_2=\frac{𝐿}{6𝜋}.$

So

$\frac{{𝑅}_1}{{𝑅}_2}=\frac{𝐿/\left(18𝜋\right)}{𝐿/\left(6𝜋\right)}=\frac{1}{3}.$

Thus

$\frac{{𝐵}_2}{{𝐵}_1}=\frac{{𝑁}_2}{{𝑁}_1}\cdot \frac{{𝑅}_1}{{𝑅}_2}=\frac{3}{9}\cdot \frac{1}{3}=\frac{1}{9}.$

So

${𝐵}_2=\frac{{𝐵}_1}{9}.$

✅ Final Answer:

Option A: $\frac{{𝐵}_1}{9}$

We are solving:

$\frac{{ℎ}_1}{{ℎ}_2}=?$

Step 1: Formula for capillary rise

The height of liquid rise in a capillary tube is given by

$ℎ=\frac{2𝑇\cos 𝜃}{𝜌𝑔𝑟}$

where

• $𝑇$ = surface tension

• $𝜃$ = angle of contact

• $𝜌$ = density of liquid

• $𝑟$ = radius of the tube

Since both tubes have the same radius and same angle of contact, and $𝑔$ is constant,

$ℎ\propto \frac{𝑇}{𝜌}$

Step 2: Ratio of heights

$\frac{{ℎ}_1}{{ℎ}_2}=\frac{\frac{{𝑇}_1}{{𝜌}_1}}{\frac{{𝑇}_2}{{𝜌}_2}}=\frac{{𝑇}_1}{{𝑇}_2}\cdot \frac{{𝜌}_2}{{𝜌}_1}$

Step 3: Substitute given ratios

• Densities ratio: ${𝜌}_1:{𝜌}_2=4:3$

• Surface tensions ratio: ${𝑇}_1:{𝑇}_2=6:5$

So,

$\frac{{ℎ}_1}{{ℎ}_2}=\frac{6}{5}\cdot \frac{3}{4}=\frac{18}{20}=\frac{9}{10}$

✅ Final Answer:

$\frac{{ℎ}_1}{{ℎ}_2}=\frac{9}{10}$

Correct Option: B

Given:

• Inductor: $𝐿=\frac{1}{𝜋}\text{H}$

• Series resistor: $𝑅=300\Omega$

• AC source: $𝑉=20\text{V},𝑓=200\text{Hz}$

Step 1: Compute Inductive Reactance

${𝑋}_{𝐿}=2𝜋𝑓𝐿$

$=2𝜋\cdot 200\cdot \frac{1}{𝜋}$

$=400\Omega$

Step 2: Phase Angle

For an RL circuit in series, the phase angle between voltage and current is

$𝜙={\tan }^{-1}\left(\frac{{𝑋}_{𝐿}}{𝑅}\right)$

$={\tan }^{-1}\left(\frac{400}{300}\right)$

$={\tan }^{-1}\left(\frac{4}{3}\right)$

Final Answer:

The phase difference is:

Option B: ${\tan }^{-1}\left(\frac{4}{3}\right)$ ✅

Newton's law of cooling tells us how things cool down. It says:

$\frac{{𝜃}_1-{𝜃}_2}{t}=K(\frac{{𝜃}_1+{𝜃}_2}{2}-{𝜃}_0)$

Here, ${𝜃}_1$ is the starting temperature, ${𝜃}_2$ is the ending temperature, $𝑡$ is the time taken, $𝐾$ is a constant, and ${𝜃}_0$ is the room temperature.

Step 1: Use the formula for cooling from ${80}^{\circ }$C to ${60}^{\circ }$C

Plug in the values: $\frac{80-60}{60}=𝐾(\frac{80+60}{2}-30)$

This becomes: $\frac{20}{60}=𝐾\left(40\right)$

Simplify: $\frac{1}{3}=𝐾\left(40\right)...\left(𝑖\right)$

Step 2: Use the formula for cooling from ${60}^{\circ }$C to ${50}^{\circ }$C

Plug in the values: $\frac{60-50}{𝑡}=𝐾(\frac{60+50}{2}-30)$

This becomes: $\frac{10}{𝑡}=𝐾\left(25\right)...\left(𝑖𝑖\right)$

Step 3: Divide equation (i) by equation (ii) to find $𝑡$

Set up the equations:

$\frac{\frac{1}{3}}{𝐾\left(40\right)}÷\frac{\frac{10}{𝑡}}{𝐾\left(25\right)}$

This simplifies to:

$\frac{1/3}{10/𝑡}=\frac{40}{25}$

$\frac{1}{3}\times \frac{𝑡}{10}=\frac{40}{25}$

Now, solve for $𝑡$:

$\frac{𝑡}{30}=\frac{40}{25}$

$𝑡=\frac{40}{25}\times 30=\frac{1200}{25}=48$ seconds.

We are given:

• Efficiency of Carnot engine:

$𝜂=1-\frac{{𝑇}_2}{{𝑇}_1}$

where ${𝑇}_1=$ source temperature, ${𝑇}_2=$ sink temperature (both in Kelvin).

Step 1: Use first efficiency condition

$𝜂=\frac{1}{6}=1-\frac{{𝑇}_2}{{𝑇}_1}$

$\frac{{𝑇}_2}{{𝑇}_1}=1-\frac{1}{6}=\frac{5}{6}$

${𝑇}_2=\frac{5}{6}{𝑇}_1$

Step 2: Use second efficiency condition

New efficiency ${𝜂}^{\prime }=\frac{1}{3}$, and new sink temp is ${𝑇}_{\prime }^2={𝑇}_2-57$.

$\frac{1}{3}=1-\frac{{𝑇}_2-57}{{𝑇}_1}$

$\frac{{𝑇}_2-57}{{𝑇}_1}=1-\frac{1}{3}=\frac{2}{3}$

${𝑇}_2-57=\frac{2}{3}{𝑇}_1$

Step 3: Solve system

From Eqn 1: ${𝑇}_2=\frac{5}{6}{𝑇}_1$.

Substitute into Eqn 2:

$\frac{5}{6}{𝑇}_1-57=\frac{2}{3}{𝑇}_1$

$\frac{5}{6}{𝑇}_1-\frac{2}{3}{𝑇}_1=57$

$(\frac{5}{6}-\frac{4}{6}){𝑇}_1=57$

$\frac{1}{6}{𝑇}_1=57$

${𝑇}_1=342\text{K}$

✅ Answer:

The source temperature is

342 K

Correct Option: C (342 K)

We are given an LCR series circuit with applied alternating emf:

$𝑉={𝑉}_0\sin 𝜔𝑡.$

Condition:

$|{𝑋}_{𝐿}-{𝑋}_{𝐶}|=𝑅$

where ${𝑋}_{𝐿}=𝜔𝐿$ and ${𝑋}_{𝐶}=\frac{1}{𝜔𝐶}$.

We want the r.m.s. value of potential difference across capacitor.

Step 1: Circuit impedance

In series circuit:

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

Given condition: $|{𝑋}_{𝐿}-{𝑋}_{𝐶}|=𝑅$.

So:

$𝑍=\sqrt{\surd {𝑅}^2+{𝑅}^2}=\sqrt{\surd 2}𝑅.$

Step 2: Current in circuit

Peak current:

${𝐼}_0=\frac{{𝑉}_0}{𝑍}=\frac{{𝑉}_0}{\sqrt{\surd 2}𝑅}.$

r.m.s. current:

${𝐼}_{\text{rms}}=\frac{{𝐼}_0}{\sqrt{\surd 2}}=\frac{{𝑉}_0}{2𝑅}.$

Step 3: Potential difference across capacitor

Voltage across capacitor (rms) is:

${𝑉}_{𝐶,\text{rms}}={𝐼}_{\text{rms}}\cdot {𝑋}_{𝐶}.$

So:

${𝑉}_{𝐶,\text{rms}}=\frac{{𝑉}_0}{2𝑅}\cdot {𝑋}_{𝐶}.$

Now ${𝑋}_{𝐶}=\frac{1}{𝜔𝐶}$.

Thus:

${𝑉}_{𝐶,\text{rms}}=\frac{{𝑉}_0}{2𝑅}\cdot \frac{1}{𝜔𝐶}=\frac{{𝑉}_0}{2𝑅𝜔𝐶}.$

✅ Final Answer:

The correct option is:

$\overline{)\text{Option C: }\frac{{𝑉}_0}{2𝑅𝜔𝐶}}$

In simple pendulum,

$T=\mathrm{mg}\cos 𝜙+\frac{{\mathrm{mv}}^2}{𝑙}$

At lowest point (Tension is maximum):

$T_{max}=\mathrm{mg}+\frac{{\mathrm{mv}}^2}{𝑙}\dots .(𝜙=0^{\circ })$

At highest point (Tension is minimum):

$T_{min}=\mathrm{mg}\cos 𝜃$

$\dots .($ angular amplitude $=𝜃$ and $v=0)$

By energy conservation,

$\begin{array}{cc}& \frac{1}{2}{\mathrm{mv}}^2=\mathrm{mgl}(1-\cos 𝜃)\\ \therefore & \frac{{\mathrm{mv}}^2}{𝑙}=2\mathrm{mg}(1-\cos 𝜃)...\left(𝑖\right)\\ & \text{ But, }{T}_{max}=4T_{min}\\ \therefore & \mathrm{mg}+\frac{{\mathrm{mv}}^2}{𝑙}=4\mathrm{mg}\cos 𝜃\\ \therefore & \mathrm{mg}+2\mathrm{mg}(1-\cos 𝜃)=4\mathrm{mg}\cos 𝜃\\ \therefore & 1+2-2\cos 𝜃=4\cos 𝜃\\ \therefore & 3=6\cos 𝜃\\ \Rightarrow & \cos 𝜃=\frac{1}{2}\\ \therefore & 𝜃={\cos }^{-1}\left(0.5\right)\end{array}$

Step 1: Magnetic field at centre due to revolving electron

The electron in orbit acts like a current loop.

• Current:

$𝐼=\frac{𝑒}{𝑇}=\frac{𝑒𝑣}{2𝜋𝑟}$

• Magnetic field at centre of a current loop of radius $𝑟$:

$𝐵=\frac{{𝜇}_0𝐼}{2𝑟}=\frac{{𝜇}_0}{2𝑟}\cdot \frac{𝑒𝑣}{2𝜋𝑟}=\frac{{𝜇}_0𝑒𝑣}{4𝜋{𝑟}^2}$

So:

$𝐵\propto \frac{𝑣}{{𝑟}^2}.$

Step 2: Relation between $𝑣,𝑟,𝑛$ in Bohr model

• Bohr radius relation:

${𝑟}_{𝑛}={𝑛}^2{𝑎}_0\Rightarrow 𝑟\propto {𝑛}^2$

• Velocity:

${𝑣}_{𝑛}=\frac{{𝑣}_1}{𝑛}\Rightarrow 𝑣\propto \frac{1}{𝑛}.$

Step 3: Substitute

$𝐵\propto \frac{𝑣}{{𝑟}^2}\propto \frac{1/𝑛}{({𝑛}^2{)}^2}=\frac{1}{{𝑛}^5}.$

✅ Final Answer:

$\overline{){𝑛}^{-5}}$

Correct Option: D.

We need to use the formula for the root mean square (r.m.s.) speed:

${𝑣}_{\text{rms}}=\sqrt{\surd \frac{3𝑘𝑇}{𝑚}}$

So,

${𝑣}_{\text{rms}}\propto \sqrt{\surd 𝑇}$

We are asked to compare ${𝑣}_{\text{rms}}$ at $𝑇=800𝐾$ with that at $𝑇=200𝐾$.

$\frac{{𝑣}_{800}}{{𝑣}_{200}}=\sqrt{\surd \frac{800}{200}}=\sqrt{\surd 4}=2$

Thus, the r.m.s. speed at 800 K is twice the value at 200 K.

✅ Correct Answer:

Option B: twice the value at 200 K

Step 1. Recall capacitor energy formula

The energy in a capacitor in terms of charge $𝑄$ and capacitance $𝐶$ is:

$𝑈=\frac{{𝑄}^2}{2𝐶}.$

Step 2. Let original charge = $𝑄$

Original energy:

$𝑈=\frac{{𝑄}^2}{2𝐶}.$

New charge = $𝑄+3$.

New energy:

${𝑈}^{\prime }=\frac{(𝑄+3{)}^2}{2𝐶}.$

Step 3. Relating energies

We are told:

${𝑈}^{\prime }=𝑈(1+0.21)=1.21𝑈.$

So:

$\frac{(𝑄+3{)}^2}{2𝐶}=1.21\cdot \frac{{𝑄}^2}{2𝐶}.$

Cancel $\frac{1}{2𝐶}$:

$(𝑄+3{)}^2=1.21{𝑄}^2.$

Step 4. Expand

${𝑄}^2+6𝑄+9=1.21{𝑄}^2.$

$6𝑄+9=1.21{𝑄}^2-{𝑄}^2=0.21{𝑄}^2.$

$0.21{𝑄}^2-6𝑄-9=0.$

Step 5. Solve quadratic

Multiply through by 100 to eliminate decimals:

$21{𝑄}^2-600𝑄-900=0.$

Divide by 3:

$7{𝑄}^2-200𝑄-300=0.$

Quadratic:

$𝑄=\frac{200\pm \sqrt{\surd {200}^2+4\cdot 7\cdot 300}}{14}.$

$=\frac{200\pm \sqrt{\surd 40000+8400}}{14}=\frac{200\pm \sqrt{\surd 48400}}{14}.$

$\sqrt{\surd 48400}=220$.

So:

$𝑄=\frac{200\pm 220}{14}.$

Case 1: $𝑄=\frac{200+220}{14}=\frac{420}{14}=30.$

Case 2: $𝑄=\frac{200-220}{14}=\frac{-20}{14}=-\frac{10}{7}$ (not possible).

Final Answer:

$\overline{)30\text{C}}$

So the correct option is C (30 C).

The magnetic field produced by the currentcarrying straight wire will be directed outside the plane of the paper, within the conducting ring. As the ring falls toward the conductor, the magnetic flux through the ring will continuously increase. According to Lenz's Law, the induced current will oppose the change in magnetic flux. Therefore, the induced current in the ring will flow in a direction (clockwise) that opposes the increasing magnetic flux. Thus, the current in the conducting ring will be induced in a clockwise direction.