MHT CET 2026 13th April Evening Shift (Physics & Chemistry) Question Paper with Solutions

Volume of a cylinder is

$𝑉=𝜋{𝑟}^2𝑙$

For percentage error in a product with powers, we add the relative errors. So,

$\frac{\Delta 𝑉}{𝑉}=\frac{2\Delta 𝑟}{𝑟}+\frac{\Delta 𝑙}{𝑙}$

Here, the length is measured by a meter rod of least count $0.1\text{ cm}$.

So,

$\Delta 𝑙=0.1\text{ cm}$

Given,

$𝑙=8.0\text{ cm}$

Hence,

$\frac{\Delta 𝑙}{𝑙}=\frac{0.1}{8.0}$

The diameter is measured by vernier calipers of least count $0.01\text{ cm}$.

Since radius is half of diameter,

$𝑟=4.0\text{ cm}$

Diameter is

$𝑑=8.0\text{ cm}$

The relative error in radius is the same as the relative error in diameter, so

$\frac{\Delta 𝑟}{𝑟}=\frac{\Delta 𝑑}{𝑑}=\frac{0.01}{8.0}$

Now,

$\frac{\Delta 𝑉}{𝑉}=2\left(\frac{0.01}{8.0}\right)+\frac{0.1}{8.0}$

$\frac{\Delta 𝑉}{𝑉}=\frac{0.02}{8.0}+\frac{0.1}{8.0}$

$\frac{\Delta 𝑉}{𝑉}=\frac{0.12}{8.0}=0.015$

Percentage error:

$0.015\times 100=1.5\%$

So the correct option is

$\overline{)1.5\%}$

Option C.

Since $\widehat{𝑛}=𝑎\widehat{𝑖}+𝑏\widehat{𝑗}$ is perpendicular to $\widehat{𝑖}+\widehat{𝑗}$, their dot product must be zero.

So,

$(𝑎\widehat{𝑖}+𝑏\widehat{𝑗})\cdot (\widehat{𝑖}+\widehat{𝑗})=0$

Using $\widehat{𝑖}\cdot \widehat{𝑖}=1$, $\widehat{𝑗}\cdot \widehat{𝑗}=1$, and $\widehat{𝑖}\cdot \widehat{𝑗}=0$,

$𝑎+𝑏=0$

So,

$𝑏=-𝑎$

Also, $\widehat{𝑛}$ is a unit vector, so its magnitude is $1$.

$\left|\widehat{𝑛}\right|=1$

$\sqrt{\surd {𝑎}^2+{𝑏}^2}=1$

Squaring both sides,

${𝑎}^2+{𝑏}^2=1$

Putting $𝑏=-𝑎$,

${𝑎}^2+(-𝑎{)}^2=1$

$2{𝑎}^2=1$

${𝑎}^2=\frac{1}{2}$

$𝑎=\frac{1}{\sqrt{\surd 2}}$

Then,

$𝑏=-\frac{1}{\sqrt{\surd 2}}$

So,

$𝑎=\frac{1}{\sqrt{\surd 2}},𝑏=-\frac{1}{\sqrt{\surd 2}}$

Correct option: D.

In a velocity-time graph, the area under the graph gives displacement.

If the graph is above the time axis, the area is positive. If the graph is below the time axis, the area is negative.

Distance travelled is found by adding the magnitudes of all areas, so negative signs are ignored for distance.

Area from $0$ to $2$ s: ${𝐴}_1=8\times 2=16$ m

Area from $2$ to $4$ s: ${𝐴}_2=-4\times 2=-8$ m

Area from $4$ to $8$ s: ${𝐴}_3=4\times 4=16$ m

Area from $8$ to $10$ s: ${𝐴}_4=-4\times 2=-8$ m

Total displacement $={𝐴}_1+{𝐴}_2+{𝐴}_3+{𝐴}_4$

$=16-8+16-8=16$ m

Total distance $=\left|{𝐴}_1\right|+\left|{𝐴}_2\right|+\left|{𝐴}_3\right|+\left|{𝐴}_4\right|$

$=16+8+16+8=48$ m

Ratio of displacement to distance $=\frac{16}{48}=\frac{1}{3}$

Thus, the required ratio is $1:3$.

The motion is only under gravity because air friction is negligible.

For a body dropped from rest from height $ℎ$, we use

$𝑠=𝑢𝑡+\frac{1}{2}𝑔{𝑡}^2$

Here,

$𝑠=ℎ,𝑢=0$

So,

$ℎ=\frac{1}{2}𝑔{𝑡}^2$

$𝑡=\sqrt{\surd \frac{2ℎ}{𝑔}}$

This time depends only on the height $ℎ$ and acceleration due to gravity $𝑔$. It does not depend on mass, size, or material of the body.

So the insulating ball and the metallic ball will reach the earth’s surface at the same time.

Correct statements are:

Option A: Time taken will be independent of the properties of their materials.

Option C: Both will reach the earth’s surface simultaneously.

If only one option has to be chosen as the direct result, choose Option C.

For one ball, mass is $𝑚$ and speed is $𝑣$.

Since it strikes the wall normally and comes back with the same speed, its direction reverses.

Take the direction towards the wall as positive.

Initial momentum of one ball:

${𝑝}_{𝑖}=𝑚𝑣$

Final momentum of one ball after reflection:

${𝑝}_{𝑓}=-𝑚𝑣$

Change in momentum of one ball:

$\Delta 𝑝={𝑝}_{𝑓}-{𝑝}_{𝑖}$

$\Delta 𝑝=-𝑚𝑣-𝑚𝑣=-2𝑚𝑣$

So, the magnitude of change in momentum of one ball is

$2𝑚𝑣$

For $100$ balls, total change in momentum is

$100\times 2𝑚𝑣=200𝑚𝑣$

Force is the rate of change of momentum:

$𝐹=\frac{\Delta 𝑝}{𝑡}$

So,

$𝐹=\frac{200𝑚𝑣}{𝑡}$

Hence, the total magnitude of force exerted by the balls on the wall is

$\overline{)\frac{200𝑚𝑣}{𝑡}}$

Correct option: B.

For a solid sphere rolling without slipping, total kinetic energy is the sum of translational and rotational kinetic energy:

$𝐾=\frac{1}{2}𝑚{𝑣}^2+\frac{1}{2}𝐼{𝜔}^2$

For a solid sphere,

$𝐼=\frac{2}{5}𝑚{𝑅}^2$

Also, for rolling without slipping,

$𝑣=𝑅𝜔$

So,

$𝐾=\frac{1}{2}𝑚{𝑣}^2+\frac{1}{2}\left(\frac{2}{5}𝑚{𝑅}^2\right){\left(\frac{𝑣}{𝑅}\right)}^2$

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

$𝐾=\frac{7}{10}𝑚{𝑣}^2$

Given,

$𝐾=7\times {10}^{-3}\text{ J},𝑚=1\text{ kg}$

So,

$7\times {10}^{-3}=\frac{7}{10}\times 1\times {𝑣}^2$

${𝑣}^2={10}^{-2}$

$𝑣=0.1{\text{ m s}}^{-1}$

Now convert into ${\text{cm s}}^{-1}$:

$0.1{\text{ m s}}^{-1}=10{\text{ cm s}}^{-1}$

So, the speed of the center of mass is

$\overline{)10{\text{ cm s}}^{-1}}$

Correct option: B.

For a disc, the moment of inertia about any diameter is given as

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

The axis normal to the disc and passing through its centre is perpendicular to two mutually perpendicular diameters of the disc.

So, by the perpendicular axis theorem,

${𝐼}_{\text{centre, normal}}={𝐼}_{𝑥}+{𝐼}_{𝑦}$

Since both diameters are identical axes,

${𝐼}_{\text{centre, normal}}=\frac{𝑀{𝑅}^2}{4}+\frac{𝑀{𝑅}^2}{4}$

${𝐼}_{\text{centre, normal}}=\frac{𝑀{𝑅}^2}{2}$

Now we need the moment of inertia about an axis normal to the disc and passing through a point on its edge.

This axis is parallel to the central normal axis, and the distance between the two axes is $𝑅$.

Using the parallel axis theorem,

$𝐼={𝐼}_{\text{centre, normal}}+𝑀{𝑅}^2$

$𝐼=\frac{𝑀{𝑅}^2}{2}+𝑀{𝑅}^2$

$𝐼=\frac{3𝑀{𝑅}^2}{2}$

Given that

$𝐼=\frac{𝑥}{2}𝑀{𝑅}^2$

So,

$\frac{𝑥}{2}𝑀{𝑅}^2=\frac{3}{2}𝑀{𝑅}^2$

Hence,

$𝑥=3$

The correct option is B.

For a thin uniform rod, the rotational kinetic energy is

$𝐸=\frac{1}{2}𝐼{𝜔}^2$

The rod has length

$𝐿=2\text{ m}$

Its mass is

$𝑀=\text{volume}\times \text{density}$

Since volume $=𝐴𝐿$,

$𝑀=𝐴\left(2\right)𝑑=2𝐴𝑑$

For a uniform rod rotating about an axis through its centre and perpendicular to its length, the moment of inertia is

$𝐼=\frac{𝑀{𝐿}^2}{12}$

Substituting $𝑀=2𝐴𝑑$ and $𝐿=2$,

$𝐼=\frac{\left(2𝐴𝑑\right)(2{)}^2}{12}$

$𝐼=\frac{8𝐴𝑑}{12}$

$𝐼=\frac{2𝐴𝑑}{3}$

Now using rotational kinetic energy,

$𝐸=\frac{1}{2}\left(\frac{2𝐴𝑑}{3}\right){𝜔}^2$

$𝐸=\frac{𝐴𝑑}{3}{𝜔}^2$

So,

${𝜔}^2=\frac{3𝐸}{𝐴𝑑}$

$𝜔={\left(\frac{3𝐸}{𝐴𝑑}\right)}^{1/2}$

Comparing with

$𝜔={\left(\frac{𝛼𝐸}{𝐴𝑑}\right)}^{1/2}$

we get

$𝛼=3$

So the correct option is Option B: $3$.

Minimum velocity required for a rocket to pull out of the gravitational force of a planet is its escape velocity.

Escape velocity is given by

$𝑣=\sqrt{\surd \frac{2𝐺𝑀}{𝑅}}$

For earth,

${𝑣}_{𝑒}=\sqrt{\surd \frac{2𝐺{𝑀}_{𝑒}}{{𝑅}_{𝑒}}}$

Given that earth has mass nine times that of planet $𝑃$ and radius twice that of planet $𝑃$,

${𝑀}_{𝑒}=9{𝑀}_{𝑃}$

and

${𝑅}_{𝑒}=2{𝑅}_{𝑃}$

So,

${𝑣}_{𝑒}=\sqrt{\surd \frac{2𝐺\left(9{𝑀}_{𝑃}\right)}{2{𝑅}_{𝑃}}}$

${𝑣}_{𝑒}=\sqrt{\surd \frac{9𝐺{𝑀}_{𝑃}}{{𝑅}_{𝑃}}}$

For planet $𝑃$, escape velocity is

${𝑣}_{𝑃}=\sqrt{\surd \frac{2𝐺{𝑀}_{𝑃}}{{𝑅}_{𝑃}}}$

Now compare both:

$\frac{{𝑣}_{𝑒}}{{𝑣}_{𝑃}}=\sqrt{\surd \frac{9𝐺{𝑀}_{𝑃}/{𝑅}_{𝑃}}{2𝐺{𝑀}_{𝑃}/{𝑅}_{𝑃}}}$

$\frac{{𝑣}_{𝑒}}{{𝑣}_{𝑃}}=\sqrt{\surd \frac{9}{2}}$

${𝑣}_{𝑒}=\frac{3}{\sqrt{\surd 2}}{𝑣}_{𝑃}$

So,

${𝑣}_{𝑃}=\frac{{𝑣}_{𝑒}\sqrt{\surd 2}}{3}$

This matches the form

$\frac{{𝑣}_{𝑒}}{3}\sqrt{\surd 𝑥}$

Hence,

$𝑥=2$

So the correct option is A, $2$.

For a liquid drop, surface energy is proportional to its surface area.

Surface energy,

$𝐸=𝑇\times 𝐴$

where $𝑇$ is surface tension and $𝐴$ is surface area.

Since the liquid is the same, $𝑇$ is constant. So,

$𝐸\propto 𝐴$

Let the radius of one small drop be $𝑟$, and the radius of the large drop be $𝑅$.

Since $1000$ small drops combine to form one large drop, volume remains conserved.

$1000\times \frac{4}{3}𝜋{𝑟}^3=\frac{4}{3}𝜋{𝑅}^3$

Cancel common terms:

$1000{𝑟}^3={𝑅}^3$

$𝑅=10𝑟$

Now, surface area of a spherical drop is:

$𝐴=4𝜋{𝑟}^2$

So the ratio of surface energy of one large drop to one small drop is:

$\frac{{𝐸}_{\text{large}}}{{𝐸}_{\text{small}}}=\frac{4𝜋{𝑅}^2}{4𝜋{𝑟}^2}$

$=\frac{{𝑅}^2}{{𝑟}^2}=\frac{(10𝑟{)}^2}{{𝑟}^2}=100$

So the ratio is:

$100:1$

Correct option: A, $100:1$

For a liquid drop, surface energy is

$𝑢=𝑇\times 𝐴$

where $𝑇$ is surface tension and $𝐴$ is surface area.

Let the radius of the original drop be $𝑅$, and the radius of each smaller drop be $𝑟$.

Since the liquid only splits and evaporation is ignored, total volume remains the same.

$\frac{4}{3}𝜋{𝑅}^3=1000\times \frac{4}{3}𝜋{𝑟}^3$

Cancel common terms:

${𝑅}^3=1000{𝑟}^3$

$𝑅=10𝑟$

Now surface energy is proportional to surface area.

Initial surface energy:

${𝑢}_{𝑖}=𝑇\times 4𝜋{𝑅}^2$

Final total surface energy:

${𝑢}_{𝑓}=1000\times 𝑇\times 4𝜋{𝑟}^2$

So,

$\frac{{𝑢}_{𝑓}}{{𝑢}_{𝑖}}=\frac{1000\times 𝑇\times 4𝜋{𝑟}^2}{𝑇\times 4𝜋{𝑅}^2}$

$\frac{{𝑢}_{𝑓}}{{𝑢}_{𝑖}}=1000\times \frac{{𝑟}^2}{{𝑅}^2}$

Since $𝑅=10𝑟$,

$\frac{{𝑢}_{𝑓}}{{𝑢}_{𝑖}}=1000\times \frac{{𝑟}^2}{(10𝑟{)}^2}$

$\frac{{𝑢}_{𝑓}}{{𝑢}_{𝑖}}=1000\times \frac{1}{100}=10$

Given,

$\frac{{𝑢}_{𝑓}}{{𝑢}_{𝑖}}=\frac{10}{𝑥}$

So,

$\frac{10}{𝑥}=10$

$𝑥=1$

So the correct option is Option A: $1$.

The water coming out of the tap reduces the water level in the tank.

Using continuity of volume flow,

$\text{Volume flowing out per second}=\text{Decrease in volume of water in tank per second}$

Area of tank:

$𝐴=750{\text{ cm}}^2$

Area of tap:

$𝑎=500{\text{ mm}}^2$

Since

$1{\text{ cm}}^2=100{\text{ mm}}^2$

we get

$𝑎=\frac{500}{100}=5{\text{ cm}}^2$

Speed of water from tap:

$𝑣=30\text{ cm/s}$

So volume of water coming out per second is

$𝑎𝑣=5\times 30=150{\text{ cm}}^3/\text{s}$

If the water level falls at the rate $\frac{𝑑ℎ}{𝑑𝑡}$, then

$𝐴\mid \frac{𝑑ℎ}{𝑑𝑡}\mid =𝑎𝑣$

So,

$750\mid \frac{𝑑ℎ}{𝑑𝑡}\mid =150$

$\mid \frac{𝑑ℎ}{𝑑𝑡}\mid =\frac{150}{750}=0.2\text{ cm/s}$

Convert into m/s:

$0.2\text{ cm/s}=0.2\times {10}^{-2}\text{ m/s}$

$=2\times {10}^{-3}\text{ m/s}$

Since the water level is decreasing,

$\frac{𝑑ℎ}{𝑑𝑡}=-2\times {10}^{-3}\text{ m/s}$

The options use the magnitude, so

$𝑦=2$

Correct option: A.

For the same metal sphere, the rate of radiation is given by Stefan’s law:

$𝑅\propto {𝑇}^4$

Here temperature must be taken in kelvin, not in degree Celsius.

Given,

${𝑇}_1={527}^{\circ }𝐶=527+273=800𝐾$

${𝑇}_2={127}^{\circ }𝐶=127+273=400𝐾$

So,

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

$\frac{{𝑅}_1}{{𝑅}_2}={\left(\frac{800}{400}\right)}^4$

$\frac{{𝑅}_1}{{𝑅}_2}=2^4=16$

So the ratio of rates of radiation is

$16:1$

Correct option: B) $16:1$

The correct option is Option C: material of the rod.

The coefficient of thermal conductivity, usually written as $𝐾$, tells us how easily heat can pass through a substance.

For heat conduction through a rod,

$\frac{𝑄}{𝑡}=\frac{𝐾𝐴({𝑇}_1-{𝑇}_2)}{𝑙}$

Here,

• $𝐴$ is the area of cross section

• $𝑙$ is the length of the rod

• ${𝑇}_1-{𝑇}_2$ is the temperature difference

• $𝐾$ is the coefficient of thermal conductivity

The value of $𝐾$ depends on the nature of the material. For example, copper has a higher thermal conductivity than iron, so copper conducts heat better.

It does not depend on the rod’s area, length, or mass. These quantities affect the rate of heat flow, but not the coefficient of thermal conductivity itself.

So, the answer is:

$\overline{)\text{Option C: material of the rod}}$

For an adiabatic expansion, the relation between temperature and volume is

$𝑇{𝑉}^{𝛾-1}=\text{constant}$

Here,

$𝛾=\frac{3}{2}$

So,

$𝛾-1=\frac{3}{2}-1=\frac{1}{2}$

The gas expands to double its volume, so

${𝑉}_2=2{𝑉}_1$

Using the adiabatic relation,

${𝑇}_1{𝑉}_{𝛾-1}^1={𝑇}_2{𝑉}_{𝛾-1}^2$

Given ${𝑇}_1=𝑇$,

$𝑇{𝑉}_{1/2}^1={𝑇}_2(2{𝑉}_1{)}^{1/2}$

$𝑇\sqrt{\surd {𝑉}_1}={𝑇}_2\sqrt{\surd 2{𝑉}_1}$

${𝑇}_2=\frac{𝑇}{\sqrt{\surd 2}}$

Now, work done by the gas in an adiabatic process is

$𝑊=𝑛{𝐶}_{𝑉}({𝑇}_1-{𝑇}_2)$

For one mole,

${𝐶}_{𝑉}=\frac{𝑅}{𝛾-1}$

Since $𝛾-1=\frac{1}{2}$,

${𝐶}_{𝑉}=2𝑅$

So,

$𝑊=2𝑅(𝑇-\frac{𝑇}{\sqrt{\surd 2}})$

$𝑊=2𝑅𝑇(1-\frac{1}{\sqrt{\surd 2}})$

$𝑊=𝑅𝑇(2-\sqrt{\surd 2})$

So the correct option is

$\overline{)\text{Option D }𝑊=𝑅𝑇[2-\sqrt{\surd 2}]}$

For an ideal gas, according to kinetic theory,

$𝑃=\frac{2}{3}\times \frac{\text{kinetic energy}}{\text{volume}}$

Here, energy of gas per litre is given as

$600\text{J/litre}$

Since,

$1\text{litre}={10}^{-3}{\text{m}}^3$

So energy per unit volume is

$\frac{600}{{10}^{-3}}=600\times {10}^3=6\times {10}^5{\text{J/m}}^3$

Now,

$𝑃=\frac{2}{3}\times 6\times {10}^5$

$𝑃=4\times {10}^5{\text{N/m}}^2$

So the pressure of the gas is

$\overline{)4\times {10}^5{\text{N/m}}^2}$

Correct option: A.

For a non-rigid diatomic gas, the molecule has vibrational motion also.

According to the law of equipartition of energy, each degree of freedom contributes

$\frac{1}{2}𝑅$

to the molar specific heat at constant volume.

For a diatomic gas:

• Translational degrees of freedom $=3$

• Rotational degrees of freedom $=2$

• Vibrational degrees of freedom $=2$

So total degrees of freedom are

$𝑓=3+2+2=7$

Hence,

${𝐶}_{𝑉}=\frac{𝑓}{2}𝑅$

${𝐶}_{𝑉}=\frac{7}{2}𝑅$

So the correct option is:

$\overline{)\text{Option D }\frac{7}{2}𝑅}$

For an ideal gas, the ratio of molar specific heats is

$𝛾=\frac{{𝐶}_{𝑝}}{{𝐶}_{𝑣}}$

According to the law of equipartition of energy, if a gas molecule has $𝑓$ degrees of freedom, then

${𝐶}_{𝑣}=\frac{𝑓}{2}𝑅$

and

${𝐶}_{𝑝}={𝐶}_{𝑣}+𝑅=\frac{𝑓+2}{2}𝑅$

So,

$𝛾=\frac{{𝐶}_{𝑝}}{{𝐶}_{𝑣}}=\frac{𝑓+2}{𝑓}$

For a mono-atomic gas, the molecule has only $3$ translational degrees of freedom.

So,

$𝑓=3$

Hence,

${𝛾}_1=\frac{3+2}{3}=\frac{5}{3}$

For a diatomic gas considered as a rigid rotator, it has $3$ translational and $2$ rotational degrees of freedom.

So,

$𝑓=5$

Hence,

${𝛾}_2=\frac{5+2}{5}=\frac{7}{5}$

Now,

$\frac{{𝛾}_1}{{𝛾}_2}=\frac{\frac{5}{3}}{\frac{7}{5}}=\frac{5}{3}\times \frac{5}{7}=\frac{25}{21}$

So, the correct option is

$\overline{)\frac{25}{21}}$

Option C.

Since the surface is frictionless, the mechanical energy of the spring-block system remains conserved.

The block is pulled to $10$ cm and released from rest, so this is the maximum displacement, or amplitude.

$𝐴=10\text{ cm}=0.10\text{ m}$

At any position $𝑥$, the total energy is

$𝐸=\frac{1}{2}𝑘{𝐴}^2$

and the potential energy of the spring at position $𝑥$ is

$𝑈=\frac{1}{2}𝑘{𝑥}^2$

So the kinetic energy at position $𝑥$ is

$𝐾=𝐸-𝑈$

$𝐾=\frac{1}{2}𝑘{𝐴}^2-\frac{1}{2}𝑘{𝑥}^2$

$𝐾=\frac{1}{2}𝑘({𝐴}^2-{𝑥}^2)$

Given,

$𝐾=0.25\text{ J}$

$𝑥=5\text{ cm}=0.05\text{ m}$

Substitute the values:

$0.25=\frac{1}{2}𝑘\left[\right(0.10{)}^2-\left(0.05{)}^2\right]$

$0.25=\frac{1}{2}𝑘(0.01-0.0025)$

$0.25=\frac{1}{2}𝑘\left(0.0075\right)$

$0.25=0.00375𝑘$

$𝑘=\frac{0.25}{0.00375}$

$𝑘\approx 66.7{\text{ Nm}}^{-1}$

So the spring constant is nearly

$\overline{)67{\text{ Nm}}^{-1}}$

Correct option: C.

For a particle in SHM,

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

and

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

Here amplitude is

$𝐴=3\text{ cm}$

Kinetic energy is $25\%$ more than potential energy, so

$𝐾.𝐸.=125\%\text{ of }𝑃.𝐸.$

$𝐾.𝐸.=\frac{5}{4}𝑃.𝐸.$

Now substitute the expressions:

$\frac{1}{2}𝑘({𝐴}^2-{𝑥}^2)=\frac{5}{4}\times \frac{1}{2}𝑘{𝑥}^2$

Cancel $\frac{1}{2}𝑘$ from both sides:

${𝐴}^2-{𝑥}^2=\frac{5}{4}{𝑥}^2$

${𝐴}^2=\frac{9}{4}{𝑥}^2$

${𝑥}^2=\frac{4{𝐴}^2}{9}$

$𝑥=\frac{2𝐴}{3}$

Since $𝐴=3$ cm,

$𝑥=\frac{2\times 3}{3}=2\text{ cm}$

So the displacement is

$\overline{)2\text{ cm}}$

Correct option: B.

For SHM, the relation between velocity and displacement is

${𝑣}^2={𝜔}^2({𝐴}^2-{𝑦}^2)$

where $𝑦$ is the displacement, $𝐴$ is the amplitude, and $𝜔$ is the angular frequency.

Given,

$4{𝑣}^2=50-{𝑦}^2$

So,

${𝑣}^2=\frac{50-{𝑦}^2}{4}$

${𝑣}^2=\frac{50}{4}-\frac{{𝑦}^2}{4}$

Compare this with

${𝑣}^2={𝜔}^2{𝐴}^2-{𝜔}^2{𝑦}^2$

The coefficient of ${𝑦}^2$ gives

${𝜔}^2=\frac{1}{4}$

So,

$𝜔=\frac{1}{2}$

Time period of SHM is

$𝑇=\frac{2𝜋}{𝜔}$

$𝑇=\frac{2𝜋}{1/2}=4𝜋$

Using $𝜋=\frac{22}{7}$,

$𝑇=4\times \frac{22}{7}$

$𝑇=\frac{88}{7}\text{ s}$

Given time period is $\frac{𝑥}{7}$ s, so

$\frac{𝑥}{7}=\frac{88}{7}$

$𝑥=88$

So, the correct option is C, $88$.

Given,

${𝑦}_1=𝑎\sin (𝜔𝑡+𝑘𝑥+0.57)$

and

${𝑦}_2=𝑎\cos (𝜔𝑡+𝑘𝑥)$

To compare the phases, both equations should be written in the same trigonometric form.

We know that

$\cos 𝜃=\sin (𝜃+\frac{𝜋}{2})$

So,

${𝑦}_2=𝑎\sin (𝜔𝑡+𝑘𝑥+\frac{𝜋}{2})$

Using $𝜋=3.14$,

$\frac{𝜋}{2}=\frac{3.14}{2}=1.57$

So the phase of the first wave is

$𝜔𝑡+𝑘𝑥+0.57$

and the phase of the second wave is

$𝜔𝑡+𝑘𝑥+1.57$

The phase difference is

$1.57-0.57=1.00\text{ radian}$

So the correct answer is:

$\overline{)1.0\text{ radian}}$

Option D.

The frequency of a sound wave does not change when it passes from one medium to another. So the frequency remains

$𝑓=700\text{ Hz}$

For a wave,

$𝑣=𝑓𝜆$

So,

$𝜆=\frac{𝑣}{𝑓}$

In warm air:

${𝜆}_{\text{air}}=\frac{350}{700}=0.5\text{ m}$

In brass:

${𝜆}_{\text{brass}}=\frac{3500}{700}=5\text{ m}$

Now compare the wavelengths:

$\frac{{𝜆}_{\text{brass}}}{{𝜆}_{\text{air}}}=\frac{5}{0.5}=10$

So the wavelength increases by a factor of $10$ when the sound wave enters brass from warm air.

The correct option is:

$\overline{)\text{Option C: It increases by a factor }10}$

For a wave, the phase difference between two points separated by distance $𝑥$ is

$\Delta 𝜙=\frac{2𝜋}{𝜆}\times 𝑥$

Here the phase difference is ${60}^{\circ }$.

Since

${60}^{\circ }=\frac{𝜋}{3}$

So,

$\frac{𝜋}{3}=\frac{2𝜋}{𝜆}\times 𝑥$

Cancel $𝜋$ from both sides:

$\frac{1}{3}=\frac{2𝑥}{𝜆}$

So,

$𝜆=6𝑥$

The velocity of a wave is

$𝑣=𝑛𝜆$

where $𝑛$ is the frequency.

Thus,

$𝑣=𝑛\left(6𝑥\right)$

$𝑣=6𝑛𝑥$

So the correct option is

$\overline{)6𝑛𝑥}$

Option D.

The given wave is

$𝑦(𝑥,𝑡)=0.05\sin (8𝑥-4𝑡)$

A travelling wave moving along positive $𝑥$-direction is generally written as

$𝑦=𝐴\sin (𝑘𝑥-𝜔𝑡)$

where $𝑘$ is the wave number and $𝜔$ is the angular frequency.

Comparing,

$𝑘=8{\text{m}}^{-1}$

and

$𝜔=4{\text{s}}^{-1}$

The wave velocity is

$𝑣=\frac{𝜔}{𝑘}$

So,

$𝑣=\frac{4}{8}=0.5{\text{m s}}^{-1}$

Hence, the velocity of the wave is

$\overline{)0.5{\text{m s}}^{-1}}$

Correct option: C.

At the center of a charged circular arc, every small charge element is at the same distance $𝑅$ from the center. So the electric potential due to the whole arc is:

$𝑉=\frac{1}{4𝜋{𝜖}_0}\frac{𝑄}{𝑅}$

Here, each ring is a half ring. The length of a half ring of radius $𝑅$ is $𝜋𝑅$.

Since the linear charge density is $𝜆$, the charge on one half ring is:

$𝑄=𝜆\left(𝜋𝑅\right)$

Substitute this value of $𝑄$ in the formula for potential:

$𝑉=\frac{1}{4𝜋{𝜖}_0}\frac{𝜆𝜋𝑅}{𝑅}=\frac{𝜆}{4{𝜖}_0}$

So, the potential at the center due to one half ring is $\frac{𝜆}{4{𝜖}_0}$. Notice that $𝑅$ gets cancelled, so the potential does not depend on the radius of the half ring.

There are two concentric half rings, and both have the same linear charge density $𝜆$. Electric potential is a scalar quantity, so we add the potentials directly:

${𝑉}_{𝑛𝑒𝑡}={𝑉}_1+{𝑉}_2$

${𝑉}_{𝑛𝑒𝑡}=\frac{𝜆}{4{𝜖}_0}+\frac{𝜆}{4{𝜖}_0}=\frac{𝜆}{2{𝜖}_0}$

In a medium of dielectric constant $𝐾$, the electrostatic force between two charges is

${𝐹}_{\text{medium}}=\frac{1}{4𝜋{𝜀}_0𝐾}\frac{{𝑞}_1{𝑞}_2}{{𝑑}^2}$

In air, suppose the equivalent distance is $𝑥$. Then the force in air is

${𝐹}_{\text{air}}=\frac{1}{4𝜋{𝜀}_0}\frac{{𝑞}_1{𝑞}_2}{{𝑥}^2}$

For the same electrostatic force,

${𝐹}_{\text{air}}={𝐹}_{\text{medium}}$

So,

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

Cancelling common terms,

$\frac{1}{{𝑥}^2}=\frac{1}{𝐾{𝑑}^2}$

So,

${𝑥}^2=𝐾{𝑑}^2$

Taking square root,

$𝑥=𝑑\sqrt{\surd 𝐾}$

So the equivalent distance in air is

$\overline{)𝑑\sqrt{\surd 𝐾}}$

Correct option: Option A.

For a conducting sphere, charge density is

$𝜎=\frac{𝑄}{4𝜋{𝑟}^2}$

Initially both spheres have the same charge density $𝜎$.

For the smaller sphere of radius $𝑅$,

${𝑄}_1=𝜎\cdot 4𝜋{𝑅}^2$

For the larger sphere of radius $2𝑅$,

${𝑄}_2=𝜎\cdot 4𝜋(2𝑅{)}^2$

${𝑄}_2=16𝜋{𝑅}^2𝜎$

So total charge is

$𝑄={𝑄}_1+{𝑄}_2$

$𝑄=4𝜋{𝑅}^2𝜎+16𝜋{𝑅}^2𝜎$

$𝑄=20𝜋{𝑅}^2𝜎$

When the two spheres are connected by a conducting wire, charge flows until their potentials become equal.

Potential of a conducting sphere is

$𝑉=\frac{𝑘𝑄}{𝑟}$

So after connection,

$\frac{𝑘{𝑄}_{\prime }^1}{𝑅}=\frac{𝑘{𝑄}_{\prime }^2}{2𝑅}$

Cancel $𝑘$ and $𝑅$:

${𝑄}_{\prime }^2=2{𝑄}_{\prime }^1$

Total charge remains conserved:

${𝑄}_{\prime }^1+{𝑄}_{\prime }^2=20𝜋{𝑅}^2𝜎$

Using ${𝑄}_{\prime }^2=2{𝑄}_{\prime }^1$,

$3{𝑄}_{\prime }^1=20𝜋{𝑅}^2𝜎$

So,

${𝑄}_{\prime }^2=\frac{40}{3}𝜋{𝑅}^2𝜎$

Now the new charge density of the larger sphere is

${𝜎}_1=\frac{{𝑄}_{\prime }^2}{4𝜋(2𝑅{)}^2}$

${𝜎}_1=\frac{\frac{40}{3}𝜋{𝑅}^2𝜎}{16𝜋{𝑅}^2}$

${𝜎}_1=\frac{5}{6}𝜎$

Hence,

$\frac{{𝜎}_1}{𝜎}=\frac{5}{6}$

So the correct option is

$\overline{)5:6}$

Option D.

For a parallel plate capacitor,

$𝐶=\frac{𝐾{𝜀}_0𝐴}{𝑑}$

Initially, air is between the plates, so $𝐾=1$.

Given,

$𝐶=15\text{pF}$

Now the separation becomes twice, so new distance is $2𝑑$, and the dielectric constant is $𝐾=3.5$.

So the new capacitance is

${𝐶}^{\prime }=\frac{3.5{𝜀}_0𝐴}{2𝑑}$

Since

$\frac{{𝜀}_0𝐴}{𝑑}=15\text{pF}$

we get

${𝐶}^{\prime }=\frac{3.5}{2}\times 15$

${𝐶}^{\prime }=26.25\text{pF}$

Given that the new capacitance is $\frac{𝑥}{4}$ pF,

$\frac{𝑥}{4}=26.25$

$𝑥=26.25\times 4=105$

So the value of $𝑥$ is

$\overline{)105}$

Correct option: A.

The given circuit is a Wheatstone bridge.

In the balanced condition of a Wheatstone bridge, the ratio of resistances in one pair of arms is equal to the ratio of resistances in the other pair of arms.

Here,

$\frac{{𝑅}_{𝐴𝐵}}{{𝑅}_{𝐵𝐶}}=\frac{10}{10}=1$

and

$\frac{{𝑅}_{𝐴𝐷}}{{𝑅}_{𝐷𝐶}}=\frac{30}{30}=1$

Since $\frac{{𝑅}_{𝐴𝐵}}{{𝑅}_{𝐵𝐶}}=\frac{{𝑅}_{𝐴𝐷}}{{𝑅}_{𝐷𝐶}}$, the bridge is balanced. So, no current flows through the middle branch BD.

Now the circuit becomes two branches connected in parallel between A and C.

The resistance of the upper branch ABC is:

${𝑅}_1=10+10=20\Omega$

The resistance of the lower branch ADC is:

${𝑅}_2=30+30=60\Omega$

For these two parallel branches, the equivalent resistance ${𝑅}_{𝑒𝑞}$ is:

$\frac{1}{{𝑅}_{𝑒𝑞}}=\frac{1}{{𝑅}_1}+\frac{1}{{𝑅}_2}$

$\frac{1}{{𝑅}_{𝑒𝑞}}=\frac{1}{20}+\frac{1}{60}$

$\frac{1}{{𝑅}_{𝑒𝑞}}=\frac{3+1}{60}=\frac{4}{60}=\frac{1}{15}{\Omega }^{-1}$

So,

${𝑅}_{𝑒𝑞}=15\Omega$

The battery voltage is $3\text{ V}$. Using Ohm’s law,

$𝐼=\frac{𝑉}{{𝑅}_{𝑒𝑞}}$

$𝐼=\frac{3}{15}=0.2\text{ A}$

For a potentiometer, when the same wire and same current are used, the potential gradient remains constant.

So, the emf of a cell is directly proportional to the balancing length:

$\frac{{𝐸}_1}{{𝐸}_2}=\frac{{𝑙}_1}{{𝑙}_2}$

For the first cell,

${𝐸}_1=1.5\text{ V},{𝑙}_1=60\text{ cm}$

The null point length increases by $40$ cm for the second cell, so

${𝑙}_2=60+40=100\text{ cm}$

Now,

$\frac{1.5}{𝐸}=\frac{60}{100}$

So,

$𝐸=1.5\times \frac{100}{60}$

$𝐸=2.5\text{ V}$

Given,

$𝐸=\frac{𝑥}{10}\text{ V}$

So,

$\frac{𝑥}{10}=2.5$

$𝑥=25$

The value of $𝑥$ is

$\overline{)25}$

Correct option: B.