MHT CET 2026 15th April Evening Shift (Physics & Chemistry) Question Paper with Solutions

The correct equation is

$\overrightarrow{𝐴}+\overrightarrow{𝐵}=\overrightarrow{𝐵}+\overrightarrow{𝐴}$

So, the correct option is Option C.

Vector addition is commutative. This means that changing the order of adding two vectors does not change the resultant vector.

For example, if we add $\overrightarrow{𝐴}$ first and then $\overrightarrow{𝐵}$, or add $\overrightarrow{𝐵}$ first and then $\overrightarrow{𝐴}$, the final resultant remains the same:

$\overrightarrow{𝐴}+\overrightarrow{𝐵}=\overrightarrow{𝐵}+\overrightarrow{𝐴}$

Now check the other options:

Option A is not correct because dot product is commutative:

$\overrightarrow{𝐴}\cdot \overrightarrow{𝐵}=\overrightarrow{𝐵}\cdot \overrightarrow{𝐴}$

not negative.

Option B is not correct because cross product is not commutative. Instead,

$\overrightarrow{𝐴}\times \overrightarrow{𝐵}=-(\overrightarrow{𝐵}\times \overrightarrow{𝐴})$

Option D is not correct because vector subtraction is not commutative:

$\overrightarrow{𝐴}-\overrightarrow{𝐵}\ne \overrightarrow{𝐵}-\overrightarrow{𝐴}$

Hence, the correct answer is

$\overline{)\overrightarrow{𝐴}+\overrightarrow{𝐵}=\overrightarrow{𝐵}+\overrightarrow{𝐴}}$

For a wire, density is

$𝜌=\frac{\text{mass}}{\text{volume}}$

The volume of a cylindrical wire is

$𝑉=𝜋{𝑟}^2𝑙$

So,

$𝜌=\frac{𝑚}{𝜋{𝑟}^2𝑙}$

For maximum percentage error, when quantities are multiplied or divided, their percentage errors are added. Also, since $𝑟$ has power $2$, its percentage error is taken twice.

So,

$\frac{\Delta 𝜌}{𝜌}\times 100=\frac{\Delta 𝑚}{𝑚}\times 100+2\frac{\Delta 𝑟}{𝑟}\times 100+\frac{\Delta 𝑙}{𝑙}\times 100$

Now calculate each percentage error:

$\frac{\Delta 𝑚}{𝑚}\times 100=\frac{0.003}{0.3}\times 100=1\%$

$\frac{\Delta 𝑟}{𝑟}\times 100=\frac{0.005}{0.5}\times 100=1\%$

$\frac{\Delta 𝑙}{𝑙}\times 100=\frac{0.06}{6}\times 100=1\%$

Hence,

$\frac{\Delta 𝜌}{𝜌}\times 100=1\%+2\left(1\%\right)+1\%$

$=4\%$

So, the maximum error in the measurement of density is

$\overline{)4\%}$

Correct option: A.

The ball is hit with speed $𝑣$ making an angle of ${60}^{\circ }$ with the vertical.

So, its angle with the horizontal is

${90}^{\circ }-{60}^{\circ }={30}^{\circ }$

The horizontal component of velocity remains constant throughout the projectile motion because there is no acceleration in the horizontal direction.

Initial horizontal component is

${𝑢}_{𝑥}=𝑣\cos {30}^{\circ }$

Using

$\cos {30}^{\circ }=\frac{\sqrt{\surd 3}}{2}$

we get

${𝑢}_{𝑥}=\frac{\sqrt{\surd 3}}{2}𝑣$

After some time, the velocity makes an angle of ${60}^{\circ }$ with the horizontal. If the speed at that instant is $𝑉$, then its horizontal component is

$𝑉\cos {60}^{\circ }$

Since horizontal velocity remains constant,

$𝑉\cos {60}^{\circ }=\frac{\sqrt{\surd 3}}{2}𝑣$

Using

$\cos {60}^{\circ }=\frac{1}{2}$

we get

$𝑉\times \frac{1}{2}=\frac{\sqrt{\surd 3}}{2}𝑣$

So,

$𝑉=\sqrt{\surd 3}𝑣$

Hence, the speed of the ball at that instant is

$\overline{)\sqrt{\surd 3}𝑣}$

Option B.

For circular motion,

centripetal acceleration is

${𝑎}_{𝑐}=\frac{{𝑉}^2}{𝑟}$

and tangential acceleration is

${𝑎}_{𝑡}=𝑟𝛼$

Given that centripetal acceleration is one-third of tangential acceleration,

${𝑎}_{𝑐}=\frac{1}{3}{𝑎}_{𝑡}$

So,

$\frac{{𝑉}^2}{𝑟}=\frac{1}{3}\left(𝑟𝛼\right)$

Multiplying both sides by $𝑟$,

${𝑉}^2=\frac{{𝑟}^2𝛼}{3}$

Taking square root,

$𝑉=𝑟\sqrt{\surd \frac{𝛼}{3}}$

So the correct option is

$\overline{)𝑉=𝑟\sqrt{\surd \frac{𝛼}{3}}}$

Option C is correct.

At the highest point of a convex bridge, the car needs centripetal force towards the centre of curvature.

For a convex bridge, the centre of curvature is below the car at the highest point. So the required centripetal force is downward.

Forces on the car:

• Weight $𝑚𝑔$ acts downward.

• Normal reaction $𝑁$ from the bridge acts upward.

So the net downward force is:

$𝑚𝑔-𝑁$

This provides the centripetal force:

$𝑚𝑔-𝑁=\frac{𝑚{𝑣}^2}{𝑅}$

So,

$𝑁=𝑚𝑔-\frac{𝑚{𝑣}^2}{𝑅}$

The thrust exerted by the car on the bridge is equal in magnitude to the normal reaction.

Hence, the thrust is:

$\overline{)𝑚𝑔-\frac{𝑚{𝑣}^2}{𝑅}}$

Correct option: C.

For water not to fall down, the most critical point is the highest point of the vertical circle.

At the highest point, the water needs centripetal force towards the centre, that is downward. Its weight $𝑚𝑔$ also acts downward.

For just preventing the water from falling,

$\frac{𝑚{𝑣}^2}{𝑟}=𝑚𝑔$

Cancel $𝑚$:

$\frac{{𝑣}^2}{𝑟}=𝑔$

So,

$𝑣=\sqrt{\surd 𝑟𝑔}$

Now, the period of revolution is

$𝑇=\frac{\text{circumference}}{\text{speed}}$

$𝑇=\frac{2𝜋𝑟}{\sqrt{\surd 𝑟𝑔}}$

$𝑇=2𝜋\sqrt{\surd \frac{𝑟}{𝑔}}$

So the required period is

$\overline{)2𝜋\sqrt{\surd \frac{𝑟}{𝑔}}}$

Correct option: B.

In a perfectly elastic collision, both momentum and kinetic energy are conserved.

So, the total kinetic energy of the bodies before collision is equal to the total kinetic energy of the bodies after collision.

Mathematically,

$\text{Total K.E. before collision}=\text{Total K.E. after collision}$

Also, for a perfectly elastic collision, the coefficient of restitution is

$𝑒=1$

So option D is not correct because $𝑒=0$ is for a perfectly inelastic collision.

The correct answer is:

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

For a planet revolving around the Sun, the gravitational force provides the required centripetal force.

So,

$\frac{𝐺𝑀𝑚}{{𝑟}^2}=\frac{𝑚{𝑣}^2}{𝑟}$

Cancelling $𝑚$ and simplifying,

${𝑣}^2=\frac{𝐺𝑀}{𝑟}$

So,

$𝑣=\sqrt{\surd \frac{𝐺𝑀}{𝑟}}$

This shows that the orbital speed varies inversely as the square root of the distance from the Sun:

$𝑣\propto \frac{1}{\sqrt{\surd 𝑟}}$

For planets A and B,

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

Given,

${𝑟}_{𝐵}=100{𝑟}_{𝐴}$

So,

$\frac{{𝑣}_{𝐴}}{{𝑣}_{𝐵}}=\sqrt{\surd \frac{100{𝑟}_{𝐴}}{{𝑟}_{𝐴}}}$

$\frac{{𝑣}_{𝐴}}{{𝑣}_{𝐵}}=\sqrt{\surd 100}=10$

So, the ratio of the speed of planet A to that of planet B is

$\overline{)10}$

Correct option: C.

For a soap film, there are two free surfaces: one on the front side and one on the back side.

Surface tension force acts along the boundary of the film.

For one surface, force due to surface tension along the square frame is

$𝐹=𝑇\times \text{perimeter}$

The perimeter of the square frame is

$4𝐿$

So for one surface,

$𝐹=𝑇\times 4𝐿=4𝑇𝐿$

But a soap film has two surfaces, so the total force is

$𝐹=2\times 4𝑇𝐿$

$𝐹=8𝑇𝐿$

So the correct option is

$\overline{)8𝑇𝐿}$

Option B.

For capillary rise, the height of rise is given by

$ℎ=\frac{2𝑇\cos 𝜃}{𝑟𝜌𝑔}$

Here, the three capillaries are identical, so $𝑟$ is same.

Also, the liquids have the same surface tension, so $𝑇$ is same.

The rise of liquid is also same, so $ℎ$ is same.

Thus,

$ℎ=\frac{2𝑇\cos 𝜃}{𝑟𝜌𝑔}$

Since $ℎ,𝑇,𝑟,$ and $𝑔$ are constant,

$\frac{\cos 𝜃}{𝜌}=\text{constant}$

So,

$\cos 𝜃\propto 𝜌$

Given,

${𝜌}_1<{𝜌}_2<{𝜌}_3$

So,

$\cos {𝜃}_1<\cos {𝜃}_2<\cos {𝜃}_3$

For capillary rise, the angle of contact is acute, and in this range, as $\cos 𝜃$ increases, $𝜃$ decreases.

Hence,

${𝜃}_1>{𝜃}_2>{𝜃}_3$

So the correct option is

$\overline{)\text{Option B: }{𝜃}_1>{𝜃}_2>{𝜃}_3}$

For a soap bubble, the excess pressure inside it is

$\Delta 𝑃=\frac{4𝑇}{𝑅}$

where $𝑇$ is the surface tension and $𝑅$ is the radius of the bubble.

For the first bubble,

$\Delta {𝑃}_1=\frac{4𝑇}{{𝑅}_1}$

For the second bubble,

$\Delta {𝑃}_2=\frac{4𝑇}{{𝑅}_2}$

It is given that the excess pressure inside the first bubble is three times that inside the second bubble:

$\Delta {𝑃}_1=3\Delta {𝑃}_2$

So,

$\frac{4𝑇}{{𝑅}_1}=3\times \frac{4𝑇}{{𝑅}_2}$

Cancel $4𝑇$ from both sides:

$\frac{1}{{𝑅}_1}=\frac{3}{{𝑅}_2}$

So,

${𝑅}_2=3{𝑅}_1$

Now, volume of a bubble is proportional to the cube of its radius:

$𝑉\propto {𝑅}^3$

Hence,

$\frac{{𝑉}_1}{{𝑉}_2}=\frac{{𝑅}_3^1}{{𝑅}_3^2}$

Since ${𝑅}_2=3{𝑅}_1$,

$\frac{{𝑉}_1}{{𝑉}_2}=\frac{{𝑅}_3^1}{(3{𝑅}_1{)}^3}$

$\frac{{𝑉}_1}{{𝑉}_2}=\frac{1}{27}$

So, the ratio of volumes is

${𝑉}_1:{𝑉}_2=1:27$

Correct option: A, $1:27$

Thermal resistance of a rod is given by

$𝑅=\frac{𝐿}{𝐾𝐴}$

where $𝐿$ is length, $𝐾$ is thermal conductivity, and $𝐴$ is area of cross-section.

Here, both rods have the same area of cross-section and the same thermal resistance.

So,

$\frac{{𝐿}_1}{{𝐾}_1𝐴}=\frac{{𝐿}_2}{{𝐾}_2𝐴}$

Since $𝐴$ is same, it cancels out:

$\frac{{𝐿}_1}{{𝐾}_1}=\frac{{𝐿}_2}{{𝐾}_2}$

So,

$\frac{{𝐿}_1}{{𝐿}_2}=\frac{{𝐾}_1}{{𝐾}_2}$

Given,

${𝐾}_1:{𝐾}_2=6:5$

Hence,

${𝐿}_1:{𝐿}_2=6:5$

So the correct option is Option B: $6:5$.

For a black surface, the energy emitted per second is given by Stefan’s law:

$𝐸\propto 𝐴{𝑇}^4$

Initial temperature:

${127}^{\circ }𝐶=127+273=400𝐾$

Final temperature:

${527}^{\circ }𝐶=527+273=800𝐾$

The length and breadth are both reduced to half, so the new area becomes:

${𝐴}^{\prime }=\frac{1}{2}\times \frac{1}{2}𝐴=\frac{𝐴}{4}$

Now compare the emitted energies:

$\frac{{𝐸}^{\prime }}{𝐸}=\frac{{𝐴}^{\prime }}{𝐴}{\left(\frac{{𝑇}^{\prime }}{𝑇}\right)}^4$

$\frac{{𝐸}^{\prime }}{𝐸}=\frac{1}{4}{\left(\frac{800}{400}\right)}^4$

$\frac{{𝐸}^{\prime }}{𝐸}=\frac{1}{4}(2{)}^4$

$\frac{{𝐸}^{\prime }}{𝐸}=\frac{16}{4}=4$

So,

${𝐸}^{\prime }=4𝐸$

The correct option is Option C: $4𝐸$.

For a black body, the wavelength at which energy is maximum is related to temperature by Wien’s displacement law:

${𝜆}_{𝑚}𝑇=\text{constant}$

Initially,

${𝜆}_{𝑚}=𝜆,𝑇={𝑇}_1$

After changing the temperature,

${𝜆}_{𝑚}=\frac{2𝜆}{3},𝑇={𝑇}_2$

Using Wien’s law:

$𝜆{𝑇}_1=\frac{2𝜆}{3}{𝑇}_2$

Cancel $𝜆$:

${𝑇}_1=\frac{2}{3}{𝑇}_2$

So,

$\frac{{𝑇}_2}{{𝑇}_1}=\frac{3}{2}$

Now, according to Stefan’s law, the power radiated by a black body is proportional to the fourth power of temperature:

$𝑃\propto {𝑇}^4$

So the factor by which power increases is

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

$\frac{{𝑃}_2}{{𝑃}_1}={\left(\frac{3}{2}\right)}^4$

$\frac{{𝑃}_2}{{𝑃}_1}=\frac{81}{16}$

So, the power radiated increases by a factor of

$\overline{)\frac{81}{16}}$

Correct option: Option B.

In a cyclic process, the system returns to its initial state.

So, the change in internal energy is zero:

$\Delta 𝑈=0$

According to the first law of thermodynamics,

$\Delta 𝑄=\Delta 𝑈+\Delta 𝑊$

Here, $\Delta 𝑊$ is the work done by the system. Since $\Delta 𝑈=0$ for a cyclic process,

$\Delta 𝑄=\Delta 𝑊$

So, the work done by the system is equal to the heat given to the system.

Correct option: B. equal to the heat given to system.

The average translational kinetic energy of one gas molecule is

${𝐸}_1=\frac{3}{2}𝑘𝑇$

Here $𝑘$ is Boltzmann constant. Since

$𝑘=\frac{𝑅}{𝑁}$

where $𝑁$ is the number of molecules in one mole, we get

${𝐸}_1=\frac{3}{2}\frac{𝑅}{𝑁}𝑇$

The kinetic energy gained by an electron accelerated from rest through a potential difference $𝑉$ is

${𝐸}_2=𝑒𝑉$

Given that

${𝐸}_1={𝐸}_2$

So,

$\frac{3}{2}\frac{𝑅}{𝑁}𝑇=𝑒𝑉$

Solving for $𝑇$,

$𝑇=\frac{2𝑒𝑉𝑁}{3𝑅}$

So the correct option is

$\overline{)\frac{2𝑒𝑉𝑁}{3𝑅}}$

Option B.

For an ideal gas, the r.m.s. speed of molecules is

${𝑣}_{\text{rms}}=\sqrt{\surd \frac{3𝑅𝑇}{𝑀}}$

So,

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

The r.m.s. speed is reduced $3$ times, so

$\frac{{𝑣}_2}{{𝑣}_1}=\frac{1}{3}$

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

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

So,

$\frac{1}{3}=\sqrt{\surd \frac{{𝑇}_2}{{𝑇}_1}}$

Squaring both sides,

$\frac{{𝑇}_2}{{𝑇}_1}=\frac{1}{9}$

For an adiabatic process,

$𝑇{𝑉}^{𝛾-1}=\text{constant}$

So,

${𝑇}_1{𝑉}_{𝛾-1}^1={𝑇}_2{𝑉}_{𝛾-1}^2$

Given $𝛾=1.5$,

$𝛾-1=0.5=\frac{1}{2}$

Thus,

$\frac{{𝑇}_2}{{𝑇}_1}={\left(\frac{{𝑉}_1}{{𝑉}_2}\right)}^{𝛾-1}$

$\frac{1}{9}={\left(\frac{{𝑉}_1}{{𝑉}_2}\right)}^{1/2}$

Squaring,

$\frac{1}{81}=\frac{{𝑉}_1}{{𝑉}_2}$

So,

$\frac{{𝑉}_2}{{𝑉}_1}=81$

The gas has to be expanded $81$ times.

Correct option: D $81$ times

For simple harmonic motion, the restoring force is given by

$𝐹=𝑘𝑥$

where $𝑘$ is the force constant and $𝑥$ is the displacement from the mean position.

Here,

$𝐹=0.4\text{N}$

and

$𝑥=4\text{cm}=0.04\text{m}$

So,

$𝑘=\frac{𝐹}{𝑥}$

$𝑘=\frac{0.4}{0.04}$

$𝑘=10\text{N/m}$

Now, the magnitude of acceleration is found using Newton’s second law:

$𝐹=𝑚𝑎$

So,

$𝑎=\frac{𝐹}{𝑚}$

Given,

$𝑚=0.4\text{kg}$

$𝑎=\frac{0.4}{0.4}$

$𝑎=1{\text{m/s}}^2$

So, the force constant is $10\text{N/m}$ and the magnitude of acceleration is $1{\text{m/s}}^2$.

Hence, the correct option is

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

For a simple harmonic motion, we try to write the displacement in the form

$𝑥=𝐴\sin (𝜔𝑡+𝜙)$

where $𝜙$ is the phase constant.

Given,

${𝑥}_1=\frac{1}{\sqrt{\surd 2}}\sin 𝜔𝑡+\frac{1}{\sqrt{\surd 2}}\cos 𝜔𝑡$

Using

$\sin (𝜔𝑡+𝜙)=\sin 𝜔𝑡\cos 𝜙+\cos 𝜔𝑡\sin 𝜙$

and

$\sin \frac{𝜋}{4}=\cos \frac{𝜋}{4}=\frac{1}{\sqrt{\surd 2}}$

we get

${𝑥}_1=\sin (𝜔𝑡+\frac{𝜋}{4})$

Now,

${𝑥}_2=\sin 𝜔𝑡+\cos 𝜔𝑡$

Factor out $\sqrt{\surd 2}$:

${𝑥}_2=\sqrt{\surd 2}(\frac{1}{\sqrt{\surd 2}}\sin 𝜔𝑡+\frac{1}{\sqrt{\surd 2}}\cos 𝜔𝑡)$

So,

${𝑥}_2=\sqrt{\surd 2}\sin (𝜔𝑡+\frac{𝜋}{4})$

Both motions have the same phase angle:

$𝜔𝑡+\frac{𝜋}{4}$

Only their amplitudes are different.

So, the phase difference is

$0$

Hence, the correct option is

$\overline{)\text{Option A: zero}}$

A pendulum clock running slow means it is taking more time for each oscillation than it should.

For a simple pendulum, the time period is

$𝑇=2𝜋\sqrt{\surd \frac{𝑙}{𝑔}}$

Here, $𝑇$ is the time period and $𝑙$ is the length of the pendulum.

So, if the clock is slow, $𝑇$ is too large. To make the clock run faster, we must reduce $𝑇$.

From the formula, $𝑇$ decreases when the length $𝑙$ is reduced.

The mass of the bob does not affect the time period. Also, for small oscillations, amplitude does not affect the time period in the NCERT treatment.

So, the correct option is:

$\overline{)\text{Option C: reduce the length of pendulum.}}$

For the wave

$𝑦=𝐴\sin (120𝜋𝑡+3𝑥)$

the phase of the wave is

$𝜙=120𝜋𝑡+3𝑥$

For two points on the wave at the same instant, the time $𝑡$ is the same. So the phase difference depends only on the difference in position.

If the distance between the two points is $\Delta 𝑥$, then

$\Delta 𝜙=3\Delta 𝑥$

Given phase difference is

$\Delta 𝜙=\frac{𝜋}{3}$

So,

$3\Delta 𝑥=\frac{𝜋}{3}$

$\Delta 𝑥=\frac{𝜋}{9}$

Hence, the distance between the two points is

$\overline{)\frac{𝜋}{9}\text{m}}$

Correct option: Option B.

For the pipe open at both ends, suppose its full length is $𝐿$.

Its fundamental frequency is

$𝑓=\frac{𝑣}{2𝐿}$

where $𝑣$ is the speed of sound in air.

Now the pipe is dipped in water so that $\frac{2}{3}$ of its length is in water. So only the remaining length contains air:

${𝐿}_{\text{air}}=𝐿-\frac{2𝐿}{3}=\frac{𝐿}{3}$

The water surface acts like a closed end, and the top of the pipe is open. So the air column becomes a pipe closed at one end and open at the other.

For a closed pipe, the fundamental frequency is

${𝑓}^{\prime }=\frac{𝑣}{4{𝐿}_{\text{air}}}$

Substitute ${𝐿}_{\text{air}}=\frac{𝐿}{3}$:

${𝑓}^{\prime }=\frac{𝑣}{4\left(\frac{𝐿}{3}\right)}$

${𝑓}^{\prime }=\frac{3𝑣}{4𝐿}$

Now compare it with the original frequency:

$𝑓=\frac{𝑣}{2𝐿}$

So,

$\frac{{𝑓}^{\prime }}{𝑓}=\frac{\frac{3𝑣}{4𝐿}}{\frac{𝑣}{2𝐿}}$

$\frac{{𝑓}^{\prime }}{𝑓}=\frac{3}{2}$

Hence,

${𝑓}^{\prime }=\frac{3}{2}𝑓$

So the correct option is

$\overline{)\text{Option B }\frac{3}{2}𝑓}$

For a stretched string, the frequency of transverse oscillations is

$𝑓=\frac{1}{2𝑙}\sqrt{\surd \frac{𝑇}{𝜇}}$

Here, length $𝑙$ and linear mass density $𝜇$ remain constant, so

$𝑓\propto \sqrt{\surd 𝑇}$

The frequency is increased by $40\%$, so the new frequency is

${𝑓}_2=1.40{𝑓}_1$

Since

$\frac{{𝑓}_2}{{𝑓}_1}=\sqrt{\surd \frac{{𝑇}_2}{{𝑇}_1}}$

we get

$1.40=\sqrt{\surd \frac{{𝑇}_2}{{𝑇}_1}}$

Squaring both sides,

$\frac{{𝑇}_2}{{𝑇}_1}=(1.40{)}^2=1.96$

So the new tension is $196\%$ of the original tension.

Increase in tension:

$196\%-100\%=96\%$

Hence, the tension must be increased by

$\overline{)96\%}$

Correct option: Option C.

For the same pipe, let the length be $𝐿$ and speed of sound be $𝑣$.

For an open pipe, the fundamental frequency is

$𝑓=\frac{𝑣}{2𝐿}$

So, the first overtone of an open pipe is the second harmonic:

${𝑓}_{\text{open first overtone}}=2𝑓$

Now the same pipe is closed from one end. Its length is still $𝐿$.

For a closed pipe, only odd harmonics are present:

$𝑓,3𝑓,5𝑓,\dots$

But the fundamental frequency of the closed pipe is

$\frac{𝑣}{4𝐿}$

Since

$\frac{𝑣}{4𝐿}=\frac{1}{2}\cdot \frac{𝑣}{2𝐿}=\frac{𝑓}{2}$

So the second overtone of the closed pipe is the fifth harmonic:

${𝑓}_{\text{closed second overtone}}=5\left(\frac{𝑓}{2}\right)=\frac{5𝑓}{2}$

Given that this is higher than the first overtone of the open pipe by $100$ Hz:

$\frac{5𝑓}{2}-2𝑓=100$

$\frac{𝑓}{2}=100$

$𝑓=200\text{Hz}$

So, the fundamental frequency of the open pipe is

$\overline{)200\text{Hz}}$

Correct option: A.

For a uniformly charged infinite plane sheet, the electric field is

$𝐸=\frac{𝜎}{2{𝜖}_0}$

Here, $𝜎$ is the surface charge density and ${𝜖}_0$ is the permittivity of free space.

In this formula, there is no term for distance $𝑑$. This means the electric field $𝐸$ does not depend on how far the point is from the sheet.

So, as the distance $𝑑$ changes, the electric field remains constant.

A constant electric field is shown by a straight horizontal line parallel to the $𝑑$-axis.

Hence, graph (C) is correct.

Suppose the two charges $+𝑞$ and $+𝑞$ are separated by a distance $2𝑎$.

So the charge $𝑄$ is at the midpoint, at a distance $𝑎$ from each $+𝑞$.

For the whole system to be in equilibrium, the net force on each charge must be zero.

The force on the middle charge $𝑄$ due to the two equal charges $+𝑞$ is already zero, because both forces are equal and opposite.

Now consider one end charge $+𝑞$.

It is repelled by the other end charge $+𝑞$.

Distance between the two $+𝑞$ charges is $2𝑎$, so this repulsive force is

${𝐹}_1=\frac{𝑘{𝑞}^2}{(2𝑎{)}^2}$

${𝐹}_1=\frac{𝑘{𝑞}^2}{4{𝑎}^2}$

To balance this force, the middle charge $𝑄$ must attract the end charge $+𝑞$.

So $𝑄$ must be negative.

Force between $𝑄$ and $+𝑞$ is

${𝐹}_2=\frac{𝑘𝑞\left|𝑄\right|}{{𝑎}^2}$

For equilibrium,

${𝐹}_1={𝐹}_2$

$\frac{𝑘{𝑞}^2}{4{𝑎}^2}=\frac{𝑘𝑞\left|𝑄\right|}{{𝑎}^2}$

Cancel common terms:

$\left|𝑄\right|=\frac{𝑞}{4}$

Since $𝑄$ must be negative,

$𝑄=-\frac{𝑞}{4}$

So the correct option is

$\overline{)\text{Option D }\frac{-𝑞}{4}}$

For capacitors connected in series, the charge on each capacitor is the same.

The equivalent capacitance of ${𝐶}_1$ and ${𝐶}_2$ in series is

$𝐶=\frac{{𝐶}_1{𝐶}_2}{{𝐶}_1+{𝐶}_2}$

If the combination is connected to a supply of $𝑉$ volt, then the charge on the combination is

$𝑄=𝐶𝑉$

So,

$𝑄=\frac{{𝐶}_1{𝐶}_2}{{𝐶}_1+{𝐶}_2}𝑉$

The potential difference across capacitor ${𝐶}_2$ is

${𝑉}_2=\frac{𝑄}{{𝐶}_2}$

Substituting the value of $𝑄$,

${𝑉}_2=\frac{1}{{𝐶}_2}\times \frac{{𝐶}_1{𝐶}_2}{{𝐶}_1+{𝐶}_2}𝑉$

${𝑉}_2=\frac{{𝐶}_1𝑉}{{𝐶}_1+{𝐶}_2}$

So, the potential difference across capacitor ${𝐶}_2$ is

$\overline{)\frac{{𝐶}_1𝑉}{{𝐶}_1+{𝐶}_2}}$

Correct option: Option B.

For a parallel plate air capacitor, capacitance is

$𝐶=\frac{{𝜖}_0𝐴}{𝑑}$

Here, the two capacitors are connected in parallel, so their capacitances will add:

${𝐶}_{\text{eq}}={𝐶}_1+{𝐶}_2$

For the first capacitor:

Plate area $=\frac{𝐴}{2}$, separation $=𝑑$

${𝐶}_1=\frac{{𝜖}_0\left(\frac{𝐴}{2}\right)}{𝑑}=\frac{𝐴{𝜖}_0}{2𝑑}$

For the second capacitor:

Plate area $=\frac{𝐴}{2}$, separation $=2𝑑$

${𝐶}_2=\frac{{𝜖}_0\left(\frac{𝐴}{2}\right)}{2𝑑}=\frac{𝐴{𝜖}_0}{4𝑑}$

Now add them:

${𝐶}_{\text{eq}}=\frac{𝐴{𝜖}_0}{2𝑑}+\frac{𝐴{𝜖}_0}{4𝑑}$

${𝐶}_{\text{eq}}=\frac{2𝐴{𝜖}_0}{4𝑑}+\frac{𝐴{𝜖}_0}{4𝑑}$

${𝐶}_{\text{eq}}=\frac{3𝐴{𝜖}_0}{4𝑑}$

So, the equivalent capacitance is

$\overline{)\frac{3𝐴{𝜖}_0}{4𝑑}}$

Correct option: B.

In a metre bridge, at balance condition,

$\frac{𝑅}{𝑆}=\frac{𝑙}{100-𝑙}$

Here, resistance in the left gap is

$𝑅=15\Omega$

and resistance in the right gap is

$𝑆=45\Omega$

So,

$\frac{15}{45}=\frac{𝑙}{100-𝑙}$

$\frac{1}{3}=\frac{𝑙}{100-𝑙}$

On cross-multiplying,

$100-𝑙=3𝑙$

$100=4𝑙$

$𝑙=25\text{ cm}$

So, the null point is $25$ cm from the left end of the wire.

The centre of the metre bridge wire is at $50$ cm.

Hence, distance of null point from the centre is

$50-25=25\text{ cm}$

So the correct option is

$\overline{)25\text{ cm}}$

Option D.

The potential difference across the full potentiometer wire is $2\text{ V}$ and its length is $4\text{ m}$.

So, the potential gradient is

$𝑘=\frac{𝑉}{𝐿}=\frac{2}{4}=0.5{\text{ V m}}^{-1}$

The balancing length is $80\text{ cm}$.

Convert it into metres:

$80\text{ cm}=0.80\text{ m}$

The e.m.f. of the cell is equal to the potential drop along the balancing length:

$𝐸=𝑘𝑙$

$𝐸=0.5\times 0.80$

$𝐸=0.40\text{ V}$

So, the e.m.f. of the cell is

$\overline{)0.40\text{ V}}$

Correct option: C.

The rod is placed parallel to the magnetic field, so the magnetic flux density $𝐵$ through the rod is

$𝐵=\frac{𝜙}{𝐴}$

Given,

$𝜙=3\times {10}^{-4}\text{Wb}$

and

$𝐴=1.5{\text{cm}}^2$

Convert area into square metre:

$1{\text{cm}}^2={10}^{-4}{\text{m}}^2$

So,

$𝐴=1.5\times {10}^{-4}{\text{m}}^2$

Now,

$𝐵=\frac{3\times {10}^{-4}}{1.5\times {10}^{-4}}$

$𝐵=2{\text{Wb/m}}^2$

We know that

$𝐵=𝜇𝐻$

So,

$𝜇=\frac{𝐵}{𝐻}$

Given,

$𝐻=1000\text{A/m}$

Thus,

$𝜇=\frac{2}{1000}$

$𝜇=2\times {10}^{-3}\text{Wb/A-m}$

So, the correct option is

$\overline{)2\times {10}^{-3}}$

Option B.

For a charged particle moving perpendicular to a uniform magnetic field, the magnetic force provides the centripetal force.

$𝑞𝑣𝐵=\frac{𝑚{𝑣}^2}{𝑟}$

So,

$𝑟=\frac{𝑚𝑣}{𝑞𝐵}$

The time period of revolution is

$𝑇=\frac{2𝜋𝑟}{𝑣}$

Substitute the value of $𝑟$:

$𝑇=\frac{2𝜋}{𝑣}\times \frac{𝑚𝑣}{𝑞𝐵}$

$𝑇=\frac{2𝜋𝑚}{𝑞𝐵}$

So the time period depends on the mass $𝑚$, charge $𝑞$, and magnetic field $𝐵$.

It does not depend on the velocity $𝑣$ of the particle.

So the correct option is:

Option B: velocity of the particle.

For a circular coil, the magnetic field at the centre is

$𝐵=\frac{{𝜇}_0𝑁𝐼}{2𝑟}$

where $𝑁$ is the number of turns and $𝑟$ is the radius.

Suppose the length of the wire is $𝐿$.

When the wire is bent into one circular turn of radius $𝑅$,

$𝐿=2𝜋𝑅$

So the magnetic field is

${𝐵}_1=\frac{{𝜇}_0𝐼}{2𝑅}$

Now the same wire is bent into a coil of $𝑛$ turns with smaller radius $𝑟$.

The total length of wire is still the same, so

$𝐿=𝑛\left(2𝜋𝑟\right)$

Since $𝐿=2𝜋𝑅$,

$2𝜋𝑅=2𝜋𝑛𝑟$

$𝑅=𝑛𝑟$

So,

$𝑟=\frac{𝑅}{𝑛}$

For the $𝑛$-turn coil,

${𝐵}_2=\frac{{𝜇}_0𝑛𝐼}{2𝑟}$

Putting $𝑟=\frac{𝑅}{𝑛}$,

${𝐵}_2=\frac{{𝜇}_0𝑛𝐼}{2\left(\frac{𝑅}{𝑛}\right)}$

${𝐵}_2=\frac{{𝜇}_0{𝑛}^2𝐼}{2𝑅}$

So,

$\frac{{𝐵}_1}{{𝐵}_2}=\frac{\frac{{𝜇}_0𝐼}{2𝑅}}{\frac{{𝜇}_0{𝑛}^2𝐼}{2𝑅}}=\frac{1}{{𝑛}^2}$

Hence, the ratio is

${𝐵}_1:{𝐵}_2=1:{𝑛}^2$

Correct option: Option C, $1:{𝑛}^2$

A uniformly charged ring is rotating, so the charge is moving in a circle. Moving charge is equivalent to an electric current.

The ring makes $𝑓$ revolutions per second.

In one complete revolution, the total charge passing through any point of the ring is $𝑞$.

So, current is

$𝐼=\frac{\text{charge}}{\text{time}}=𝑞𝑓$

For a circular ring of radius $𝑅$, the magnetic induction at the centre is

$𝐵=\frac{{𝜇}_0𝐼}{2𝑅}$

Substituting $𝐼=𝑞𝑓$,

$𝐵=\frac{{𝜇}_0𝑞𝑓}{2𝑅}$

So the correct option is

$\overline{)\frac{{𝜇}_0𝑞𝑓}{2𝑅}}$

Option C.

Since ${𝑟}_2\ll {𝑟}_1$, the smaller coil lies very close to the centre of the larger coil. So the magnetic field due to the larger coil can be taken as nearly uniform over the whole area of the smaller coil.

Magnetic field at the centre of a circular coil of radius ${𝑟}_1$ carrying current $𝐼$ is

$𝐵=\frac{{𝜇}_0𝐼}{2{𝑟}_1}$

Area of the smaller coil is

$𝐴=𝜋{𝑟}_2^2$

So magnetic flux linked with the smaller coil is

$𝜙=𝐵𝐴$

$𝜙=\frac{{𝜇}_0𝐼}{2{𝑟}_1}\times 𝜋{𝑟}_2^2$

Mutual inductance is

$𝑀=\frac{𝜙}{𝐼}$

$𝑀=\frac{{𝜇}_0𝜋{𝑟}_2^2}{2{𝑟}_1}$