MHT CET 2026 20th April Evening Shift (Physics & Chemistry) Question Paper with Solutions

Given,

$𝑥=\sqrt{\surd 𝑎}{𝑏}^2{𝑐}^3{𝑑}^{-4}$

We can write this as

$𝑥={𝑎}^{1/2}{𝑏}^2{𝑐}^3{𝑑}^{-4}$

For multiplication and powers, the relative error in the result is found by multiplying each relative error by the modulus of its power and then adding them.

So,

$\frac{\Delta 𝑥}{𝑥}=\frac{1}{2}\frac{\Delta 𝑎}{𝑎}+2\frac{\Delta 𝑏}{𝑏}+3\frac{\Delta 𝑐}{𝑐}+4\frac{\Delta 𝑑}{𝑑}$

Here the power of $𝑑$ is $-4$, but while finding maximum relative error, we take the magnitude of the power, so it becomes $4$.

Given relative errors:

$\frac{\Delta 𝑎}{𝑎}=2\%,\frac{\Delta 𝑏}{𝑏}=1\%,\frac{\Delta 𝑐}{𝑐}=3\%,\frac{\Delta 𝑑}{𝑑}=4\%$

Substitute:

$\frac{\Delta 𝑥}{𝑥}=\frac{1}{2}\left(2\%\right)+2\left(1\%\right)+3\left(3\%\right)+4\left(4\%\right)$

$=1\%+2\%+9\%+16\%$

$=28\%$

So, the relative error in $𝑥$ is

$\overline{)28\%}$

Correct option: B.

For two vectors to be mutually perpendicular, their dot product must be zero.

Given,

$\overrightarrow{𝐴}=4\widehat{𝑖}+𝑛\widehat{𝑗}+2\widehat{𝑘}$

and

$\overrightarrow{𝐵}=2\widehat{𝑖}+2\widehat{𝑗}-\widehat{𝑘}$

Now,

$\overrightarrow{𝐴}\cdot \overrightarrow{𝐵}=4\left(2\right)+𝑛\left(2\right)+2(-1)$

Since the vectors are perpendicular,

$4\left(2\right)+𝑛\left(2\right)+2(-1)=0$

$8+2𝑛-2=0$

$6+2𝑛=0$

$2𝑛=-6$

$𝑛=-3$

So, the value of $𝑛$ is

$\overline{)-3}$

Correct option: Option C.

For one complete revolution, the angle covered is

$2𝜋\text{ rad}$

The particle completes $2$ revolutions, so the angular displacement is

$2\times 2𝜋=4𝜋\text{ rad}$

The radius $3\text{ cm}$ is not needed here because angular displacement depends only on the angle covered, not on the radius.

So, the correct option is

$\overline{)4𝜋}$

Option D.

The car is moving on a circular path, so it has two accelerations:

Tangential acceleration, because its speed is increasing:

${𝑎}_{𝑡}=2{\text{m/s}}^2$

Centripetal acceleration, because its direction of motion is changing:

${𝑎}_{𝑐}=\frac{{𝑣}^2}{𝑟}$

Given,

$𝑣=40\text{m/s},𝑟=40\text{m}$

So,

${𝑎}_{𝑐}=\frac{{40}^2}{40}$

${𝑎}_{𝑐}=\frac{1600}{40}=40{\text{m/s}}^2$

Tangential acceleration and centripetal acceleration are perpendicular to each other. So the net acceleration is

$𝑎=\sqrt{\surd {𝑎}_2^{𝑐}+{𝑎}_2^{𝑡}}$

$𝑎=\sqrt{\surd {40}^2+2^2}$

$𝑎=\sqrt{\surd 1600+4}$

$𝑎=\sqrt{\surd 1604}\approx 40{\text{m/s}}^2$

So the net acceleration is nearly

$\overline{)40{\text{m/s}}^2}$

Correct option: D.

For a projectile, the maximum height reached is

$𝐻=\frac{{𝑢}^2{\sin }^2𝜃}{2𝑔}$

Here, the maximum height is same for both spheres.

Let the initial velocity of the sphere projected at ${30}^{\circ }$ be ${𝑢}_1$.

Let the initial velocity of the sphere projected at ${45}^{\circ }$ be ${𝑢}_2$.

So,

$\frac{{𝑢}_2^1{\sin }^2{30}^{\circ }}{2𝑔}=\frac{{𝑢}_2^2{\sin }^2{45}^{\circ }}{2𝑔}$

Cancel $2𝑔$ from both sides:

${𝑢}_2^1{\sin }^2{30}^{\circ }={𝑢}_2^2{\sin }^2{45}^{\circ }$

Taking square root:

${𝑢}_1\sin {30}^{\circ }={𝑢}_2\sin {45}^{\circ }$

So,

$\frac{{𝑢}_1}{{𝑢}_2}=\frac{\sin {45}^{\circ }}{\sin {30}^{\circ }}$

Using the given values,

$\frac{{𝑢}_1}{{𝑢}_2}=\frac{\frac{1}{\sqrt{\surd 2}}}{0.5}$

Since $0.5=\frac{1}{2}$,

$\frac{{𝑢}_1}{{𝑢}_2}=\frac{\frac{1}{\sqrt{\surd 2}}}{\frac{1}{2}}=\frac{2}{\sqrt{\surd 2}}=\sqrt{\surd 2}$

So, the ratio of their initial velocities is

$\overline{)\sqrt{\surd 2}:1}$

Hence, the correct option is Option B.

Let the common acceleration of all three blocks be $𝑎$.

The surface is frictionless and the string is inextensible, so all the blocks move together with the same acceleration.

The total mass of the system is $2\text{ kg}+8\text{ kg}+12\text{ kg}=22\text{ kg}$.

So the acceleration of the whole system is

$𝑎=\frac{𝐹}{22}$

Now look at tension ${𝑇}_1$. This tension pulls the two blocks on its left, whose total mass is $2+8=10\text{ kg}$.

For these two blocks, using Newton’s second law,

${𝑇}_1=10𝑎$

Now look at tension ${𝑇}_2$. This tension pulls only the $2\text{ kg}$ block.

So for the $2\text{ kg}$ block,

${𝑇}_2=2𝑎$

Now take the ratio:

$\frac{{𝑇}_1}{{𝑇}_2}=\frac{10𝑎}{2𝑎}=\frac{5}{1}$

So, ${𝑇}_1:{𝑇}_2=5:1$.

For the solid sphere, moment of inertia about its diameter is

${𝐼}_{\text{diameter}}=\frac{2}{5}𝑀{𝑅}^2$

A tangent to the sphere is parallel to a diameter and at a distance $𝑅$ from the centre.

So, using the parallel axis theorem,

${𝐼}_{\text{tangent}}={𝐼}_{\text{diameter}}+𝑀{𝑅}^2$

${𝐼}_{\text{sphere}}=\frac{2}{5}𝑀{𝑅}^2+𝑀{𝑅}^2$

${𝐼}_{\text{sphere}}=\frac{7}{5}𝑀{𝑅}^2$

Given $𝑀=5\text{ kg}$,

${𝐼}_{\text{sphere}}=\frac{7}{5}\times 5{𝑅}^2=7{𝑅}^2$

For the disc, moment of inertia about a diameter in its plane is

${𝐼}_{\text{diameter}}=\frac{1}{4}𝑀{𝑅}^2$

The tangent is in the plane of the disc, so it is parallel to a diameter and at a distance $𝑅$ from the centre.

Using the parallel axis theorem,

${𝐼}_{\text{disc}}=\frac{1}{4}𝑀{𝑅}^2+𝑀{𝑅}^2$

${𝐼}_{\text{disc}}=\frac{5}{4}𝑀{𝑅}^2$

Given $𝑀=4\text{ kg}$,

${𝐼}_{\text{disc}}=\frac{5}{4}\times 4{𝑅}^2=5{𝑅}^2$

So the required ratio is

${𝐼}_{\text{sphere}}:{𝐼}_{\text{disc}}=7{𝑅}^2:5{𝑅}^2=7:5$

Hence,

$𝑥=7,𝑦=5$

Correct option: C.

Acceleration due to gravity is maximum at the earth’s surface, if we ignore small effects like rotation and shape of the earth.

When we go above the earth’s surface, the distance from the centre of the earth increases. Since

$g=\frac{GM}{r^2}$

as $r$ increases, $g$ decreases.

At a height $h$ above the earth’s surface,

$g_h=\frac{g}{\left(1+\frac{h}{R}\right)^2}$

So $g_h<g$.

When we go below the earth’s surface, only the mass of the earth inside that depth contributes to gravity. So acceleration due to gravity also decreases with depth.

At depth $d$,

$g_d=g\left(1-\frac{d}{R}\right)$

So $g_d<g$.

Now checking the other options:

Option B is not correct because for the same value $h=d$,

$g_h \approx g\left(1-\frac{2h}{R}\right)$

but

$g_d=g\left(1-\frac{d}{R}\right)$

So they are not equal.

Option C is not correct. Here,

$h=640\text{ km}, \quad R=6400\text{ km}$

So,

$\frac{h}{R}=\frac{640}{6400}=\frac{1}{10}$

Using the approximate formula,

$g_h \approx g\left(1-\frac{2h}{R}\right)$

$g_h \approx g\left(1-\frac{2}{10}\right)=0.8g$

So the decrease is about $20\%$, not $10\%$.

Option D is not correct because at the poles, the points lie on the axis of rotation. So there is no reduction in $g$ due to the rotation of the earth at the poles.

Hence, the correct statement is:

$\boxed{\text{Option A}}$

For capillary rise, the height $ℎ$ to which a liquid rises in a capillary tube of radius $𝑟$ is given by

$ℎ=\frac{2𝑇\cos 𝜃}{𝜌𝑔𝑟}$

Here,

$𝑇$ = surface tension of water,

$𝜃$ = angle of contact,

$𝜌$ = density of water,

$𝑔$ = acceleration due to gravity.

Since all tubes are dipped in the same liquid, water, the values of $𝑇$, $𝜃$, $𝜌$, and $𝑔$ remain constant.

So,

$ℎ\propto \frac{1}{𝑟}$

This means the height of water rise is inversely proportional to the radius of the capillary tube.

Hence,

$ℎ𝑟=\text{constant}$

So for different tubes,

${ℎ}_{𝑛}{𝑟}_{𝑛}=\text{constant}$

The correct option is:

$\overline{)\text{Option D}}$

When a small metal sphere falls through a viscous liquid, three forces act on it. Its weight $𝑊$ acts downward. The buoyant force ${𝐹}_{𝐵}$ acts upward. The viscous drag force ${𝐹}_{𝑣}$ also acts upward.

The equation of motion is:

$𝑚\frac{𝑑𝑣}{𝑑𝑡}=𝑊-{𝐹}_{𝐵}-{𝐹}_{𝑣}$

$𝑚\frac{𝑑𝑣}{𝑑𝑡}=𝑚𝑔-\frac{4}{3}𝜋{𝑟}^3𝜎𝑔-6𝜋𝜂𝑟𝑣$

At the start, when $𝑡=0$, the velocity is $𝑣=0$. So the viscous drag force is also zero. That is why the net downward force is maximum, and the acceleration is maximum at the beginning.

As the sphere speeds up, the viscous drag force $6𝜋𝜂𝑟𝑣$ increases because it depends on $𝑣$. This reduces the net force acting downward. So the acceleration keeps decreasing with time.

After some time, the upward forces become equal to the weight. Then the net force becomes zero, so acceleration becomes zero. After this, the sphere moves with a constant speed called terminal velocity ${𝑣}_{𝑇}$.

The velocity changes with time as:

$𝑣={𝑣}_{𝑇}(1-{𝑒}^{-𝑘𝑡/𝑚})$

This shows that velocity starts from zero, increases quickly at first, and then increases more and more slowly. It finally approaches a constant value and becomes nearly horizontal on the graph.

So the correct graph is the one that starts from the origin and gradually levels off at terminal velocity. This is graph (d).

Suppose the depths of the two holes below the free surface of water are ${ℎ}_1$ and ${ℎ}_2$.

Since the vessel is wide, the velocity of water coming out of a hole at depth $ℎ$ is given by Torricelli’s theorem:

$𝑣=\sqrt{\surd 2𝑔ℎ}$

The reaction force due to water coming out of a hole is equal to the rate of change of momentum of the escaping water.

For one hole,

$𝐹=𝜌𝐴{𝑣}^2$

Using ${𝑣}^2=2𝑔ℎ$,

$𝐹=𝜌𝐴\left(2𝑔ℎ\right)=2𝜌𝐴𝑔ℎ$

So for the two holes,

${𝐹}_1=2𝜌𝐴𝑔{ℎ}_1$

and

${𝐹}_2=2𝜌𝐴𝑔{ℎ}_2$

The holes are on opposite sides of the vessel, so the reaction forces are in opposite directions. Hence the resultant force is the difference of the two forces:

${𝐹}_{\text{resultant}}={𝐹}_2-{𝐹}_1$

${𝐹}_{\text{resultant}}=2𝜌𝐴𝑔({ℎ}_2-{ℎ}_1)$

The height difference between the two holes is $𝑥$, so

${ℎ}_2-{ℎ}_1=𝑥$

Thus,

${𝐹}_{\text{resultant}}=2𝜌𝐴𝑔𝑥$

So the correct option is:

$\overline{)2𝜌𝐴𝑔𝑥}$

Option B.

Using Wien’s displacement law,

${𝜆}_{max}𝑇=\text{constant}$

So, the temperature of a star is inversely proportional to the wavelength at which it radiates maximum energy:

$𝑇\propto \frac{1}{{𝜆}_{max}}$

For stars A and B,

$\frac{{𝑇}_{𝐴}}{{𝑇}_{𝐵}}=\frac{{𝜆}_{𝐵}}{{𝜆}_{𝐴}}$

Given,

${𝜆}_{𝐴}=3.9\times {10}^{-7}\text{m}$

${𝜆}_{𝐵}=5.2\times {10}^{-7}\text{m}$

So,

$\frac{{𝑇}_{𝐴}}{{𝑇}_{𝐵}}=\frac{5.2\times {10}^{-7}}{3.9\times {10}^{-7}}$

$\frac{{𝑇}_{𝐴}}{{𝑇}_{𝐵}}=\frac{5.2}{3.9}=\frac{52}{39}=\frac{4}{3}$

Hence, the ratio of the temperature of star A to that of star B is

$\overline{)4:3}$

Correct option: D.

Heat lost by the hot metal = heat gained by water.

The water has mass $1\text{kg}=1000\text{g}$.

Its temperature rises from $300\text{K}$ to $350\text{K}$, so

$\Delta {𝑇}_{\text{water}}=350-300=50\text{K}$

A temperature difference of $50\text{K}$ is the same as ${50}^{\circ }\text{C}$.

Heat gained by water:

$𝑄=𝑚𝑐\Delta 𝑇$

$𝑄=1000\times 1\times 50=50000\text{cal}$

Now convert calories into joules:

$1\text{cal}=4.2\text{J}$

$𝑄=50000\times 4.2=210000\text{J}$

So, the metal loses $210000\text{J}$ of heat.

The metal cools from $850\text{K}$ to $350\text{K}$, so

$\Delta {𝑇}_{\text{metal}}=850-350=500\text{K}$

If the heat capacity of the metal is $𝐶$, then

$𝑄=𝐶\Delta 𝑇$

$210000=𝐶\times 500$

$𝐶=\frac{210000}{500}$

$𝐶=420\text{J/K}$

So the correct option is:

$\overline{)420}$

Option A.

For cooling due to radiation, the rate of cooling is proportional to

${𝑇}^4-{𝑇}_4^0$

where $𝑇$ is the temperature of the sphere and ${𝑇}_0$ is the temperature of the surroundings.

Here, surrounding temperature is

${𝑇}_0=300\text{ K}$

At $600\text{ K}$, rate of cooling is $𝑅$.

So,

$𝑅\propto {600}^4-{300}^4$

At $400\text{ K}$, let the rate of cooling be ${𝑅}^{\prime }$.

${𝑅}^{\prime }\propto {400}^4-{300}^4$

Now take the ratio:

$\frac{{𝑅}^{\prime }}{𝑅}=\frac{{400}^4-{300}^4}{{600}^4-{300}^4}$

Cancel ${100}^4$ from all terms:

$\frac{{𝑅}^{\prime }}{𝑅}=\frac{4^4-3^4}{6^4-3^4}$

$\frac{{𝑅}^{\prime }}{𝑅}=\frac{256-81}{1296-81}$

$\frac{{𝑅}^{\prime }}{𝑅}=\frac{175}{1215}$

$\frac{{𝑅}^{\prime }}{𝑅}=\frac{35}{243}$

So,

${𝑅}^{\prime }=\frac{35𝑅}{243}$

Given that the rate at $400\text{ K}$ is $\frac{𝑥}{243}$,

$𝑥=35𝑅$

So the correct option is

$\overline{)35𝑅}$

Option D.

Using Wien’s displacement law,

${𝜆}_{𝑚}𝑇=𝑏$

Given,

$𝑇=5780\text{K}$

and

$𝑏=2.89\times {10}^6\text{nmK}$

So,

${𝜆}_{𝑚}=\frac{𝑏}{𝑇}$

${𝜆}_{𝑚}=\frac{2.89\times {10}^6}{5780}$

${𝜆}_{𝑚}=500\text{nm}$

This means that at temperature $5780\text{K}$, the black body emits maximum energy at wavelength $500\text{nm}$.

So the energy at $500\text{nm}$, which is ${𝑈}_2$, is maximum.

Hence,

${𝑈}_2>{𝑈}_1$

and

${𝑈}_2>{𝑈}_3$

So the correct relation is

${𝑈}_1<{𝑈}_2>{𝑈}_3$

Correct option: C.

For a Carnot engine, efficiency is given by

$𝜂=1-\frac{{𝑇}_2}{{𝑇}_1}$

Here,

${𝑇}_1=500\text{K},{𝑇}_2=300\text{K}$

So,

$𝜂=1-\frac{300}{500}$

$𝜂=1-0.6=0.4$

To express it as a percentage,

$𝜂=0.4\times 100=40\%$

So, the efficiency of the Carnot engine is

$\overline{)40\%}$

Correct option: Option D.

For the isothermal expansion, temperature remains constant.

For an ideal gas in an isothermal process,

$𝑃𝑉=\text{constant}$

Initial state:

${𝑃}_1=𝑃,{𝑉}_1=𝑉$

Final volume:

${𝑉}_2=3𝑉$

So,

$𝑃𝑉={𝑃}_{𝐼}\left(3𝑉\right)$

${𝑃}_{𝐼}=\frac{𝑃}{3}$

Now for the adiabatic expansion,

$𝑃{𝑉}^{𝛾}=\text{constant}$

So,

$𝑃{𝑉}^{𝛾}={𝑃}_{𝐴}(3𝑉{)}^{𝛾}$

${𝑃}_{𝐴}=\frac{𝑃{𝑉}^{𝛾}}{3^{𝛾}{𝑉}^{𝛾}}$

${𝑃}_{𝐴}=\frac{𝑃}{3^{𝛾}}$

Now the required ratio is

$\frac{{𝑃}_{𝐴}}{{𝑃}_{𝐼}}=\frac{\frac{𝑃}{3^{𝛾}}}{\frac{𝑃}{3}}$

$\frac{{𝑃}_{𝐴}}{{𝑃}_{𝐼}}=\frac{3}{3^{𝛾}}=3^{1-𝛾}$

So the correct option is

$\overline{)3^{(1-𝛾)}}$

Option C.

The given displacement is

$𝑥=𝑎\sin 𝜔𝑡+𝑏\cos 𝜔𝑡$

This is a combination of sine and cosine functions having the same angular frequency $𝜔$.

To check whether the motion is simple harmonic, differentiate twice with respect to time.

Velocity:

$𝑣=\frac{𝑑𝑥}{𝑑𝑡}=𝑎𝜔\cos 𝜔𝑡-𝑏𝜔\sin 𝜔𝑡$

Acceleration:

$\frac{{𝑑}^2𝑥}{𝑑{𝑡}^2}=-𝑎{𝜔}^2\sin 𝜔𝑡-𝑏{𝜔}^2\cos 𝜔𝑡$

Taking $-{𝜔}^2$ common,

$\frac{{𝑑}^2𝑥}{𝑑{𝑡}^2}=-{𝜔}^2(𝑎\sin 𝜔𝑡+𝑏\cos 𝜔𝑡)$

But

$𝑥=𝑎\sin 𝜔𝑡+𝑏\cos 𝜔𝑡$

So,

$\frac{{𝑑}^2𝑥}{𝑑{𝑡}^2}=-{𝜔}^2𝑥$

This is the standard condition for simple harmonic motion.

Now we find the amplitude. The expression

$𝑎\sin 𝜔𝑡+𝑏\cos 𝜔𝑡$

can be written as a single sine function:

$𝑥=𝑅\sin (𝜔𝑡+𝜙)$

where the amplitude is

$𝑅=\sqrt{\surd {𝑎}^2+{𝑏}^2}$

So the motion is simple harmonic motion with amplitude

$\sqrt{\surd {𝑎}^2+{𝑏}^2}$

Hence, the correct option is:

$\overline{)\text{Option B}}$

For a particle in S.H.M., at displacement $𝑥$ from the mean position,

$\text{Potential energy}=\frac{1}{2}𝑘{𝑥}^2$

and

$\text{Kinetic energy}=\frac{1}{2}𝑘({𝐴}^2-{𝑥}^2)$

Here,

$𝑥=\frac{2\sqrt{\surd 2}}{3}𝐴$

So,

${𝑥}^2={\left(\frac{2\sqrt{\surd 2}}{3}𝐴\right)}^2$

${𝑥}^2=\frac{8}{9}{𝐴}^2$

Now,

$\frac{\text{Potential energy}}{\text{Kinetic energy}}=\frac{\frac{1}{2}𝑘{𝑥}^2}{\frac{1}{2}𝑘({𝐴}^2-{𝑥}^2)}$

$=\frac{{𝑥}^2}{{𝐴}^2-{𝑥}^2}$

Substitute ${𝑥}^2=\frac{8}{9}{𝐴}^2$:

$\frac{\text{P.E.}}{\text{K.E.}}=\frac{\frac{8}{9}{𝐴}^2}{{𝐴}^2-\frac{8}{9}{𝐴}^2}$

$=\frac{\frac{8}{9}{𝐴}^2}{\frac{1}{9}{𝐴}^2}$

$=8$

So the potential energy is $8$ times the kinetic energy.

The value of $𝑛$ is

$\overline{)8}$

Option D is correct.

For a particle performing S.H.M., the speed at a displacement $𝑥$ from the mean position is

$𝑣=𝜔\sqrt{\surd {𝑎}^2-{𝑥}^2}$

Here, amplitude $=𝑎$ and time period $=𝑇$, so angular frequency is

$𝜔=\frac{2𝜋}{𝑇}$

The particle is at

$𝑥=\frac{𝑎}{3}$

So,

$𝑣=\frac{2𝜋}{𝑇}\sqrt{\surd {𝑎}^2-{\left(\frac{𝑎}{3}\right)}^2}$

$𝑣=\frac{2𝜋}{𝑇}\sqrt{\surd {𝑎}^2-\frac{{𝑎}^2}{9}}$

$𝑣=\frac{2𝜋}{𝑇}\sqrt{\surd \frac{8{𝑎}^2}{9}}$

$𝑣=\frac{2𝜋}{𝑇}\cdot \frac{2\sqrt{\surd 2}𝑎}{3}$

$𝑣=\frac{4\sqrt{\surd 2}𝜋𝑎}{3𝑇}$

So, the correct option is

$\overline{)\text{Option D }\frac{4\sqrt{\surd 2}𝜋𝑎}{3𝑇}}$

For two waves of the same amplitude $𝐴$, the resultant amplitude is

$𝑅=\sqrt{\surd {𝐴}^2+{𝐴}^2+2𝐴\cdot 𝐴\cos 𝜙}$

where $𝜙$ is the phase difference.

The path difference is

$\Delta 𝑥=\frac{𝜆}{4}$

So the phase difference is

$𝜙=\frac{2𝜋}{𝜆}\times \frac{𝜆}{4}$

$𝜙=\frac{𝜋}{2}$

Now,

$𝑅=\sqrt{\surd {𝐴}^2+{𝐴}^2+2{𝐴}^2\cos \frac{𝜋}{2}}$

Given,

$\cos \frac{𝜋}{2}=0$

So,

$𝑅=\sqrt{\surd 2{𝐴}^2}$

$𝑅=\sqrt{\surd 2}𝐴$

So the correct option is

$\overline{)\sqrt{\surd 2}𝐴}$

Option B.

For an ideal gas, the speed of sound is

$𝑣=\sqrt{\surd \frac{𝛾𝑅𝑇}{𝑀}}$

where $𝛾$ is the ratio of specific heats, $𝑅$ is the gas constant, $𝑇$ is the temperature, and $𝑀$ is the molar mass.

Here both gases are monoatomic, so for both gases,

$𝛾=\frac{5}{3}$

Also, both are kept at the same temperature. So $𝛾$, $𝑅$, and $𝑇$ are the same for both gases.

Thus, speed of sound depends only on the molecular mass as

$𝑣\propto \frac{1}{\sqrt{\surd 𝑚}}$

So,

$\frac{{𝑣}_1}{{𝑣}_2}=\sqrt{\surd \frac{{𝑚}_2}{{𝑚}_1}}$

Hence, the correct option is

$\overline{)\sqrt{\surd \frac{{𝑚}_2}{{𝑚}_1}}}$

Option B.

Given,

${𝑦}_1={𝑎}_1\sin (𝜔𝑡-\frac{2𝜋𝑥}{𝜆})$

and

${𝑦}_2={𝑎}_2\cos (𝜔𝑡-\frac{2𝜋𝑥}{𝜆}+\frac{𝜋}{6})$

To compare the phase of the two waves, both should be written in the same trigonometric form.

We know,

$\cos 𝜃=\sin (𝜃+\frac{𝜋}{2})$

So,

${𝑦}_2={𝑎}_2\sin (𝜔𝑡-\frac{2𝜋𝑥}{𝜆}+\frac{𝜋}{6}+\frac{𝜋}{2})$

${𝑦}_2={𝑎}_2\sin (𝜔𝑡-\frac{2𝜋𝑥}{𝜆}+\frac{2𝜋}{3})$

Now compare it with

${𝑦}_1={𝑎}_1\sin (𝜔𝑡-\frac{2𝜋𝑥}{𝜆})$

The phase difference is

$\Delta 𝜙=\frac{2𝜋}{3}$

Relation between phase difference and path difference is

$\Delta 𝜙=\frac{2𝜋}{𝜆}\Delta 𝑥$

So,

$\frac{2𝜋}{3}=\frac{2𝜋}{𝜆}\Delta 𝑥$

$\Delta 𝑥=\frac{𝜆}{3}$

So the path difference between the two waves is

$\overline{)\frac{𝜆}{3}}$

Correct option: C.

The given wave equation is

$𝑌=0.01\sin \left[2𝜋(\frac{𝑡}{0.02}-\frac{𝑥}{0.50})\right]$

Compare it with the standard equation of a travelling wave,

$𝑌=𝐴\sin \left[2𝜋(\frac{𝑡}{𝑇}-\frac{𝑥}{𝜆})\right]$

So,

$𝑇=0.02\text{s}$

and

$𝜆=0.50\text{m}$

The wave speed is

$𝑣=\frac{𝜆}{𝑇}$

$𝑣=\frac{0.50}{0.02}$

$𝑣=25{\text{m s}}^{-1}$

For a wave on a stretched string,

$𝑣=\sqrt{\surd \frac{𝑇}{𝜇}}$

So,

$𝑇=𝜇{𝑣}^2$

Given linear mass density,

$𝜇=0.02{\text{kg m}}^{-1}$

Thus,

$𝑇=0.02\times (25{)}^2$

$𝑇=0.02\times 625$

$𝑇=12.50\text{N}$

So, the tension in the string is

$\overline{)12.50\text{N}}$

Correct option: Option A.

On an equipotential surface, the electric potential is the same at every point.

If a charge $+𝑞$ is taken from one point to another point on the same equipotential surface, then the potential difference is

$\Delta 𝑉=0$

Work done in moving a charge is

$𝑊=𝑞\Delta 𝑉$

So,

$𝑊=𝑞\times 0=0$

Hence, no work is done in moving the charge over an equipotential surface.

So the correct option is:

$\overline{)\text{Option D: no work is done.}}$

The centre of the cube is at the same distance from all its corners.

For a cube of side $𝑟$, the body diagonal is

$𝑟\sqrt{\surd 3}$

So, the distance from the centre of the cube to any corner is half of the body diagonal:

$𝑑=\frac{𝑟\sqrt{\surd 3}}{2}$

The electric potential due to one charge $𝑄$ at distance $𝑑$ is

$𝑉=\frac{1}{4𝜋{𝜖}_0}\frac{𝑄}{𝑑}$

There are $8$ corners in a cube, and potential is a scalar quantity, so the total potential at the centre is the simple sum of potentials due to all $8$ charges:

${𝑉}_{\text{total}}=8\times \frac{1}{4𝜋{𝜖}_0}\frac{𝑄}{𝑑}$

Substitute

$𝑑=\frac{𝑟\sqrt{\surd 3}}{2}$

${𝑉}_{\text{total}}=8\times \frac{1}{4𝜋{𝜖}_0}\frac{𝑄}{\frac{𝑟\sqrt{\surd 3}}{2}}$

${𝑉}_{\text{total}}=8\times \frac{1}{4𝜋{𝜖}_0}\times \frac{2𝑄}{𝑟\sqrt{\surd 3}}$

${𝑉}_{\text{total}}=\frac{4𝑄}{𝜋{𝜖}_0𝑟\sqrt{\surd 3}}$

So the correct option is

$\overline{)\text{Option A }\frac{4𝑄}{𝜋{𝜖}_0𝑟\sqrt{\surd 3}}}$

The wrong statement is:

$\overline{)\text{Option D: They have symmetrical shape.}}$

A polar molecule is formed when the centre of positive charge and the centre of negative charge do not coincide. Because of this separation of charges, the molecule has a permanent dipole moment and behaves like a tiny electric dipole.

So, Options A, B, and C are correct.

Polar molecules usually have an unsymmetrical shape, because the bond dipoles do not cancel each other.

For example, water $\left({\text{H}}_2\text{O}\right)$ is polar because it has a bent shape, so its dipoles do not cancel.

Symmetrical molecules generally have zero net dipole moment, so they are usually non-polar.

Hence, the wrong statement is:

$\overline{)\text{D}}$

In air, the force between two point charges separated by distance $𝑅$ is

$𝐹=\frac{1}{4𝜋{𝜀}_0}\frac{{𝑞}_1{𝑞}_2}{{𝑅}^2}$

When the space between the charges is filled with a dielectric of dielectric constant $𝐾$, the force becomes $𝐾$ times weaker for the same distance.

So, if the distance is now ${𝑅}_1$, the force in the dielectric is

$𝐹=\frac{1}{4𝜋{𝜀}_0𝐾}\frac{{𝑞}_1{𝑞}_2}{{𝑅}_2^1}$

According to the question, this force is equal to the original force $𝐹$ in air.

So,

$\frac{1}{4𝜋{𝜀}_0}\frac{{𝑞}_1{𝑞}_2}{{𝑅}^2}=\frac{1}{4𝜋{𝜀}_0𝐾}\frac{{𝑞}_1{𝑞}_2}{{𝑅}_2^1}$

Cancel the common terms:

$\frac{1}{{𝑅}^2}=\frac{1}{𝐾{𝑅}_2^1}$

So,

$𝐾{𝑅}_2^1={𝑅}^2$

${𝑅}_2^1=\frac{{𝑅}^2}{𝐾}$

${𝑅}_1=\frac{𝑅}{\sqrt{\surd 𝐾}}$

Hence, the correct option is

$\overline{){𝑅}_1=\frac{𝑅}{\sqrt{\surd 𝐾}}}$

Option D.

Current in the primary circuit is:

$𝐼=\frac{𝑉}{𝑅+{𝑅}_{𝐴𝐵}}$

Here, the battery is $5\text{ V}$ and the total resistance in the primary circuit is $45\Omega +5\Omega$.

$𝐼=\frac{5}{45+5}=\frac{5}{50}=0.1\text{ A}$

The potential difference across the potentiometer wire $𝐴𝐵$ is:

${𝑉}_{𝐴𝐵}=𝐼\times {𝑅}_{𝐴𝐵}=0.1\times 5=0.5\text{ V}$

Now find the potential gradient, which means potential drop per metre of the wire.

$𝑘=\frac{{𝑉}_{𝐴𝐵}}{𝐿}=\frac{0.5}{10}=0.05\text{ V/m}$

For balance, the e.m.f. is equal to the potential drop across the balancing length $𝑙$.

$𝐸=𝑘\times 𝑙$

Substitute $𝐸=0.4\text{ V}$ and $𝑘=0.05\text{ V/m}$:

$0.4=0.05\times 𝑙$

$𝑙=\frac{0.4}{0.05}=8\text{ m}$

First, use Kirchhoff's Current Law, which says that at any junction, total current entering is equal to total current leaving.

At the first junction, the incoming currents are $2\text{ A}$ and $3\text{ A}$.

So the current leaving this junction is: $2\text{ A}+3\text{ A}=5\text{ A}$

Now look at the second junction. The incoming current is $5\text{ A}$.

One branch carries away $1\text{ A}$.

So the current going towards the third junction is: $5\text{ A}-1\text{ A}=4\text{ A}$

At the third junction, the incoming current is $4\text{ A}$.

The outgoing currents are $0.6\text{ A}$, $1.3\text{ A}$, and $𝑖$.

Again using Kirchhoff's Current Law: $4\text{ A}=0.6\text{ A}+1.3\text{ A}+𝑖$

Add the known outgoing currents: $0.6+1.3=1.9$

So, $4=1.9+𝑖$

Now subtract $1.9$ from both sides: $𝑖=4-1.9=2.1\text{ A}$

Answer: $2.1\text{ A}$

The de-Broglie wavelength of a particle is

$𝜆=\frac{ℎ}{𝑝}$

where $𝑝$ is the momentum of the particle.

For an electron moving in a magnetic field, the magnetic force is

$\overrightarrow{𝐹}=𝑞(\overrightarrow{𝑣}\times \overrightarrow{𝐵})$

Here,

$\overrightarrow{𝑣}={𝑉}_0\widehat{𝑗},\overrightarrow{𝐵}={𝐵}_0\widehat{𝑖}$

So the velocity is perpendicular to the magnetic field.

A magnetic force is always perpendicular to the velocity of the charged particle. Because of this, it does no work on the particle.

Since no work is done, the kinetic energy remains constant. So the speed also remains constant.

Momentum magnitude is

$𝑝=𝑚𝑣$

Since $𝑣$ remains constant, $𝑝$ remains constant.

So,

$𝜆=\frac{ℎ}{𝑝}$

also remains constant.

Thus, the de-Broglie wavelength remains constant.

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

For a current-carrying loop, the magnetic moment is

$𝑀=𝐼\times 𝐴$

where $𝐼$ is the current and $𝐴$ is the area enclosed by the loop.

The same wire is first in the shape of a circle and then changed into a square. So, the length of the wire remains the same.

Let the length of the wire be $𝐿$.

For the circular loop,

$𝐿=2𝜋𝑟$

So,

$𝑟=\frac{𝐿}{2𝜋}$

Area of the circle is

${𝐴}_{𝑐}=𝜋{𝑟}^2$

${𝐴}_{𝑐}=𝜋{\left(\frac{𝐿}{2𝜋}\right)}^2$

${𝐴}_{𝑐}=\frac{{𝐿}^2}{4𝜋}$

So the magnetic moment of the circular loop is

$𝑀=𝐼\times \frac{{𝐿}^2}{4𝜋}$

For the square loop, if its side is $𝑎$, then

$4𝑎=𝐿$

$𝑎=\frac{𝐿}{4}$

Area of the square is

${𝐴}_{𝑠}={𝑎}^2$

${𝐴}_{𝑠}={\left(\frac{𝐿}{4}\right)}^2$

${𝐴}_{𝑠}=\frac{{𝐿}^2}{16}$

So the magnetic moment of the square loop is

${𝑀}_1=𝐼\times \frac{{𝐿}^2}{16}$

Now,

$\frac{𝑀}{{𝑀}_1}=\frac{𝐼\times \frac{{𝐿}^2}{4𝜋}}{𝐼\times \frac{{𝐿}^2}{16}}$

Cancel common terms:

$\frac{𝑀}{{𝑀}_1}=\frac{16}{4𝜋}$