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

Given,

$𝑇=2𝜋\sqrt{\surd \frac{𝐿}{𝑔}}$

We need percentage error in $𝑔$.

First write $𝑔$ in terms of $𝐿$ and $𝑇$:

${𝑇}^2=4{𝜋}^2\frac{𝐿}{𝑔}$

So,

$𝑔=\frac{4{𝜋}^2𝐿}{{𝑇}^2}$

For percentage error, constants like $4{𝜋}^2$ have no error.

Since

$𝑔\propto \frac{𝐿}{{𝑇}^2}$

the fractional error in $𝑔$ is

$\frac{\Delta 𝑔}{𝑔}=\frac{\Delta 𝐿}{𝐿}+2\frac{\Delta 𝑇}{𝑇}$

Now,

$𝐿=10\text{ cm}$

Accuracy in length measurement is $1\text{ mm}=0.1\text{ cm}$.

So,

$\frac{\Delta 𝐿}{𝐿}\times 100=\frac{0.1}{10}\times 100=1\%$

Time for $100$ oscillations is $50\text{ s}$.

The watch has $1\text{ s}$ resolution, so error in time measurement is

$\Delta 𝑡=1\text{ s}$

Percentage error in time is

$\frac{\Delta 𝑇}{𝑇}\times 100=\frac{1}{50}\times 100=2\%$

Since $𝑔$ depends on ${𝑇}^2$, this error becomes double:

$2\times 2\%=4\%$

So total percentage error in $𝑔$ is

$1\%+4\%=5\%$

Hence, the percentage error in measurement of $𝑔$ is

$\overline{)5\%}$

Correct option: C.

Given that the vector sum $\overrightarrow{𝐴}+\overrightarrow{𝐵}$ is perpendicular to the vector difference $\overrightarrow{𝐴}-\overrightarrow{𝐵}$.

When two vectors are perpendicular, their dot product is zero.

So,

$(\overrightarrow{𝐴}+\overrightarrow{𝐵})\cdot (\overrightarrow{𝐴}-\overrightarrow{𝐵})=0$

Using distributive property of dot product,

$\overrightarrow{𝐴}\cdot \overrightarrow{𝐴}-\overrightarrow{𝐴}\cdot \overrightarrow{𝐵}+\overrightarrow{𝐵}\cdot \overrightarrow{𝐴}-\overrightarrow{𝐵}\cdot \overrightarrow{𝐵}=0$

Now, $\overrightarrow{𝐴}\cdot \overrightarrow{𝐵}=\overrightarrow{𝐵}\cdot \overrightarrow{𝐴}$, so the middle terms cancel:

${𝐴}^2-{𝐵}^2=0$

So,

${𝐴}^2={𝐵}^2$

Hence,

$𝐴=𝐵$

So the two forces have equal magnitudes.

Answer: Option D — equal in magnitude.

According to Boyle’s law, $𝑃𝑉$ remains constant.

The SI unit of pressure is

$\text{Pa}={\text{N m}}^{-2}$

The SI unit of volume is

${\text{m}}^3$

So the unit of $𝑃𝑉$ is

$𝑃𝑉=\left({\text{N m}}^{-2}\right)\left({\text{m}}^3\right)$

$𝑃𝑉=\text{N m}$

Now, $\text{N m}$ is the SI unit of work or energy, which is called joule $\left(\text{J}\right)$.

So, the SI unit of $𝑃𝑉$ is the same as that of energy.

Answer: Option B — energy

The mass moves in a horizontal circle, so the tension in the string provides the required centripetal force through its horizontal component.

If the string makes an angle $𝜃$ with the vertical, then the radius of the circular path is

$𝑟=𝐿\sin 𝜃$

The horizontal component of tension is

$𝑇\sin 𝜃$

This provides centripetal force:

$𝑇\sin 𝜃=𝑚{𝜔}^2𝑟$

Substitute $𝑟=𝐿\sin 𝜃$:

$𝑇\sin 𝜃=𝑚{𝜔}^2𝐿\sin 𝜃$

Cancelling $\sin 𝜃$,

$𝑇=𝑚{𝜔}^2𝐿$

The frequency is given as

$𝑛=\frac{3}{𝜋}\text{r.p.s.}$

Angular velocity is

$𝜔=2𝜋𝑛$

So,

$𝜔=2𝜋\times \frac{3}{𝜋}=6{\text{rad s}}^{-1}$

Now,

$𝑇=𝑚(6{)}^2𝐿$

$𝑇=36𝑚𝐿$

So the correct answer is

$\overline{)36𝑚𝐿}$

Option B.

Since the body is initially stationary, its initial momentum is zero.

After explosion, the two parts move in opposite directions. So by conservation of momentum,

${𝑀}_1{𝑉}_1={𝑀}_2{𝑉}_2$

This gives

$\frac{{𝑉}_1}{{𝑉}_2}=\frac{{𝑀}_2}{{𝑀}_1}$

Now, kinetic energy of a body is

$𝐾=\frac{1}{2}𝑚{𝑣}^2$

So the ratio of kinetic energies is

$\frac{{𝐾}_1}{{𝐾}_2}=\frac{\frac{1}{2}{𝑀}_1{𝑉}_2^1}{\frac{1}{2}{𝑀}_2{𝑉}_2^2}$

$\frac{{𝐾}_1}{{𝐾}_2}=\frac{{𝑀}_1}{{𝑀}_2}\times \frac{{𝑉}_2^1}{{𝑉}_2^2}$

Using

$\frac{{𝑉}_1}{{𝑉}_2}=\frac{{𝑀}_2}{{𝑀}_1}$

we get

$\frac{{𝐾}_1}{{𝐾}_2}=\frac{{𝑀}_1}{{𝑀}_2}\times {\left(\frac{{𝑀}_2}{{𝑀}_1}\right)}^2$

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

So the correct option is

$\overline{)\text{Option B }\frac{{𝑀}_2}{{𝑀}_1}}$

For the body sliding down the smooth inclined plane, there is no rotational motion.

If the vertical height of the inclined plane is $ℎ$, then by conservation of mechanical energy,

$𝑚𝑔ℎ=\frac{1}{2}𝑚{𝑉}^2$

So,

${𝑉}^2=2𝑔ℎ$

Now, when the body is a ring and it rolls down the inclined plane, its energy at the bottom is partly translational and partly rotational.

For a ring,

$𝐼=𝑚{𝑅}^2$

Also, for rolling without slipping,

$𝑣=𝑅𝜔$

Using conservation of energy,

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

Substitute $𝐼=𝑚{𝑅}^2$ and $𝜔=\frac{𝑣}{𝑅}$:

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

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

$𝑚𝑔ℎ=𝑚{𝑣}^2$

So,

${𝑣}^2=𝑔ℎ$

But from the sliding case,

${𝑉}^2=2𝑔ℎ$

Hence,

${𝑣}^2=\frac{{𝑉}^2}{2}$

$𝑣=\frac{𝑉}{\sqrt{\surd 2}}$

So the correct option is

$\overline{)\frac{𝑉}{\sqrt{\surd 2}}}$

Option A.

The minute hand completes one full rotation in $60$ minutes.

So its angular speed is

${𝜔}_{𝑚}=\frac{2𝜋}{60}\text{rad/min}$

The hour hand completes one full rotation in $12$ hours $=720$ minutes.

So its angular speed is

${𝜔}_{ℎ}=\frac{2𝜋}{720}\text{rad/min}$

The relative angular speed is the difference of their angular speeds:

${𝜔}_{\text{relative}}={𝜔}_{𝑚}-{𝜔}_{ℎ}$

$=\frac{2𝜋}{60}-\frac{2𝜋}{720}$

Taking LCM $720$,

$=\frac{24𝜋}{720}-\frac{2𝜋}{720}$

$=\frac{22𝜋}{720}$

$=\frac{11𝜋}{360}\text{rad/min}$

So, if the unit is rad/min, the correct option is

$\overline{)\frac{11𝜋}{360}}$

Hence, the answer is Option B.

If the answer is required in rad/s, then

$\frac{11𝜋}{360}÷60=\frac{11𝜋}{21600}\text{rad/s}$

which is Option C.

For a disc rotating about an axis through its centre and perpendicular to its plane, the moment of inertia is

$𝐼=\frac{1}{2}𝑀{𝑅}^2$

Given,

$𝑀=1\text{kg},𝑅=0.4\text{m},𝛼=10{\text{rad s}}^{-2}$

Torque required is

$𝜏=𝐼𝛼$

Also, when a tangential force $𝐹$ is applied at the rim,

$𝜏=𝐹𝑅$

So,

$𝐹𝑅=𝐼𝛼$

$𝐹=\frac{𝐼𝛼}{𝑅}$

Substitute $𝐼=\frac{1}{2}𝑀{𝑅}^2$:

$𝐹=\frac{\frac{1}{2}𝑀{𝑅}^2𝛼}{𝑅}$

$𝐹=\frac{1}{2}𝑀𝑅𝛼$

Now substitute the values:

$𝐹=\frac{1}{2}\times 1\times 0.4\times 10$

$𝐹=2\text{N}$

So, the tangential force applied to the rim is

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

Correct option: Option B.

The velocity of projection is half of escape velocity.

Escape velocity from the earth’s surface is

${𝑣}_{𝑒}=\sqrt{\surd \frac{2𝐺𝑀}{𝑅}}$

So the given velocity is

$𝑢=\frac{{𝑣}_{𝑒}}{2}$

Squaring,

${𝑢}^2=\frac{{𝑣}_2^{𝑒}}{4}=\frac{1}{4}\cdot \frac{2𝐺𝑀}{𝑅}=\frac{𝐺𝑀}{2𝑅}$

Now use conservation of mechanical energy.

At the earth’s surface,

${𝐸}_{𝑖}=\frac{1}{2}𝑚{𝑢}^2-\frac{𝐺𝑀𝑚}{𝑅}$

Substitute ${𝑢}^2=\frac{𝐺𝑀}{2𝑅}$:

${𝐸}_{𝑖}=\frac{1}{2}𝑚\left(\frac{𝐺𝑀}{2𝑅}\right)-\frac{𝐺𝑀𝑚}{𝑅}$

${𝐸}_{𝑖}=\frac{𝐺𝑀𝑚}{4𝑅}-\frac{𝐺𝑀𝑚}{𝑅}=-\frac{3𝐺𝑀𝑚}{4𝑅}$

At the maximum height, the velocity becomes zero. If maximum height is $ℎ$, then distance from the earth’s centre is $𝑅+ℎ$.

So energy at the highest point is

${𝐸}_{𝑓}=-\frac{𝐺𝑀𝑚}{𝑅+ℎ}$

Using conservation of energy,

$-\frac{3𝐺𝑀𝑚}{4𝑅}=-\frac{𝐺𝑀𝑚}{𝑅+ℎ}$

Cancel common terms:

$\frac{3}{4𝑅}=\frac{1}{𝑅+ℎ}$

So,

$𝑅+ℎ=\frac{4𝑅}{3}$

$ℎ=\frac{4𝑅}{3}-𝑅=\frac{𝑅}{3}$

Hence, the maximum height reached is

$\overline{)\frac{𝑅}{3}}$

Correct option: A.

For a liquid drop, surface energy is

$\text{Surface energy}=𝑇\times \text{surface area}$

where $𝑇$ is the surface tension.

Suppose each small drop has radius $𝑟$.

When two drops coalesce, volume remains conserved.

Volume before coalescence:

$2\times \frac{4}{3}𝜋{𝑟}^3$

Let the radius of the big drop be $𝑅$. Its volume is

$\frac{4}{3}𝜋{𝑅}^3$

So,

$\frac{4}{3}𝜋{𝑅}^3=2\times \frac{4}{3}𝜋{𝑟}^3$

${𝑅}^3=2{𝑟}^3$

$𝑅=2^{1/3}𝑟$

Surface energy before:

${𝐸}_{\text{before}}=𝑇\times 2\left(4𝜋{𝑟}^2\right)$

${𝐸}_{\text{before}}=8𝜋{𝑟}^2𝑇$

Surface energy after:

${𝐸}_{\text{after}}=𝑇\times 4𝜋{𝑅}^2$

Substitute $𝑅=2^{1/3}𝑟$:

${𝐸}_{\text{after}}=4𝜋𝑇(2^{1/3}𝑟{)}^2$

${𝐸}_{\text{after}}=4𝜋𝑇\cdot 2^{2/3}{𝑟}^2$

Now the required ratio is

$\frac{{𝐸}_{\text{after}}}{{𝐸}_{\text{before}}}=\frac{4𝜋𝑇\cdot 2^{2/3}{𝑟}^2}{8𝜋{𝑟}^2𝑇}$

$=\frac{2^{2/3}}{2}$

$=2^{-1/3}$

So the ratio of total surface energies after and before is

$2^{-1/3}:1$

Correct option: C.

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.

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

Also, the rise of liquid is same, so $ℎ$ is same.

Thus,

$\frac{\cos 𝜃}{𝜌}=\text{constant}$

or

$𝜌\cos 𝜃=\text{constant}$

Since

${𝜌}_1>{𝜌}_2>{𝜌}_3$

the value of $\cos 𝜃$ must be smaller for the liquid having greater density.

So,

$\cos {𝜃}_1<\cos {𝜃}_2<\cos {𝜃}_3$

For capillary rise, $𝜃$ is less than ${90}^{\circ }$, and in this range, as $\cos 𝜃$ decreases, $𝜃$ increases.

Hence,

${𝜃}_1>{𝜃}_2>{𝜃}_3$

So the correct option is

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

The work done in splitting the drop is equal to the increase in surface energy.

For a liquid drop, surface energy is

$\text{Surface energy}=𝑇\times \text{surface area}$

Initial radius of the water drop is $𝑅$.

Initial surface area:

${𝐴}_1=4𝜋{𝑅}^2$

Now it is split into $64$ identical droplets. Let the radius of each small droplet be $𝑟$.

Since volume remains the same,

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

Canceling common terms,

${𝑅}^3=64{𝑟}^3$

${𝑟}^3=\frac{{𝑅}^3}{64}$

$𝑟=\frac{𝑅}{4}$

Total final surface area of $64$ droplets:

${𝐴}_2=64\times 4𝜋{𝑟}^2$

Substitute $𝑟=\frac{𝑅}{4}$:

${𝐴}_2=64\times 4𝜋{\left(\frac{𝑅}{4}\right)}^2$

${𝐴}_2=64\times 4𝜋\frac{{𝑅}^2}{16}$

${𝐴}_2=16𝜋{𝑅}^2$

Increase in surface area:

$\Delta 𝐴={𝐴}_2-{𝐴}_1$

$\Delta 𝐴=16𝜋{𝑅}^2-4𝜋{𝑅}^2$

$\Delta 𝐴=12𝜋{𝑅}^2$

So, work done is

$𝑊=𝑇\Delta 𝐴$

$𝑊=12𝜋{𝑅}^2𝑇$

So the correct option is:

$\overline{)12𝜋{𝑅}^2𝑇}$

Option B.

In boiling water, the initial temperature is

${100}^{\circ }\text{C}={212}^{\circ }\text{F}$

When the Fahrenheit thermometer shows ${140}^{\circ }\text{F}$, convert it to Celsius using

$\frac{𝐶}{5}=\frac{𝐹-32}{9}$

Put $𝐹=140$:

$\frac{𝐶}{5}=\frac{140-32}{9}$

$\frac{𝐶}{5}=\frac{108}{9}=12$

$𝐶={60}^{\circ }\text{C}$

So the temperature has fallen from ${100}^{\circ }\text{C}$ to ${60}^{\circ }\text{C}$.

$\text{Fall in temperature}=100-60={40}^{\circ }\text{C}$

So the correct answer is:

$\overline{){40}^{\circ }\text{C}}$

Option B.

For a black body, the rate of radiation is given by Stefan’s law:

$𝑅=𝜎𝐴{𝑇}^4$

Here, area is $𝐴$ and temperature is $𝑇$.

For the second body, emissivity is

$𝑒=0.2$

and temperature is

$3𝑇$

So its rate of radiation will be

${𝑅}^{\prime }=𝑒𝜎𝐴(3𝑇{)}^4$

${𝑅}^{\prime }=0.2\times 𝜎𝐴\times 81{𝑇}^4$

Since

$𝜎𝐴{𝑇}^4=𝑅$

we get

${𝑅}^{\prime }=0.2\times 81𝑅$

${𝑅}^{\prime }=16.2𝑅$

So the correct option is

$\overline{)16.2𝑅}$

Option C.

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

$𝐸\propto 𝐴{𝑇}^4$

where $𝐴$ is the surface area and $𝑇$ is the absolute temperature in kelvin.

Initial temperature:

${27}^{\circ }𝐶=27+273=300𝐾$

Final temperature:

${327}^{\circ }𝐶=327+273=600𝐾$

So the temperature becomes double:

$\frac{{𝑇}_2}{{𝑇}_1}=\frac{600}{300}=2$

Now, length and breadth are each reduced to $\frac{1}{4}$ of their initial values.

So the new area becomes:

${𝐴}_2=\frac{1}{4}\times \frac{1}{4}𝐴=\frac{𝐴}{16}$

Using Stefan’s law:

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

$\frac{{𝐸}_2}{𝐸}=\frac{1}{16}\times (2{)}^4$

$\frac{{𝐸}_2}{𝐸}=\frac{1}{16}\times 16=1$

So,

${𝐸}_2=𝐸$

Hence, the correct option is:

$\overline{)𝐸}$

Option A.

At constant pressure, the gas expands from volume $𝑉$ to $2𝑉$.

For an ideal gas,

$𝑃𝑉=𝑛𝑅𝑇$

Since pressure is constant, temperature is directly proportional to volume.

Initial state:

$𝑃𝑉=𝑛𝑅{𝑇}_1$

Final state:

$𝑃\left(2𝑉\right)=𝑛𝑅{𝑇}_2$

So,

${𝑇}_2-{𝑇}_1=\frac{2𝑃𝑉}{𝑛𝑅}-\frac{𝑃𝑉}{𝑛𝑅}$

$\Delta 𝑇=\frac{𝑃𝑉}{𝑛𝑅}$

Change in internal energy of an ideal gas is

$\Delta 𝑈=𝑛{𝐶}_{𝑣}\Delta 𝑇$

Substitute $\Delta 𝑇$:

$\Delta 𝑈=𝑛{𝐶}_{𝑣}\times \frac{𝑃𝑉}{𝑛𝑅}$

$\Delta 𝑈=\frac{{𝐶}_{𝑣}}{𝑅}𝑃𝑉$

Now,

$𝛾=\frac{{𝐶}_{𝑝}}{{𝐶}_{𝑣}}$

and

${𝐶}_{𝑝}-{𝐶}_{𝑣}=𝑅$

So,

$𝑅={𝐶}_{𝑣}(𝛾-1)$

Hence,

$\frac{{𝐶}_{𝑣}}{𝑅}=\frac{1}{𝛾-1}$

So,

$\Delta 𝑈=\frac{𝑃𝑉}{𝛾-1}$

The correct option is:

$\overline{)\text{Option A }\frac{𝑃𝑉}{𝛾-1}}$

For an ideal gas, the r.m.s. velocity is related to temperature as

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

So,

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

The r.m.s. velocity 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,

$𝑇{𝑉}^{𝛾-1}=\text{constant}$

Given,

$𝛾=1.5$

So,

$𝛾-1=0.5=\frac{1}{2}$

Thus,

$𝑇{𝑉}^{1/2}=\text{constant}$

So,

${𝑇}_1{𝑉}_{1/2}^1={𝑇}_2{𝑉}_{1/2}^2$

Using ${𝑇}_2=\frac{{𝑇}_1}{9}$,

${𝑇}_1{𝑉}_{1/2}^1=\frac{{𝑇}_1}{9}{𝑉}_{1/2}^2$

Cancel ${𝑇}_1$:

${𝑉}_{1/2}^1=\frac{1}{9}{𝑉}_{1/2}^2$

So,

${𝑉}_{1/2}^2=9{𝑉}_{1/2}^1$

Squaring both sides,

${𝑉}_2=81{𝑉}_1$

Hence, the gas has to be expanded $81$ times.

Correct option: B, $81$ times.

For a simple pendulum, the tension in the string at an angular displacement $𝜙$ from the vertical is

$𝑇=𝑚𝑔\cos 𝜙+\frac{𝑚{𝑣}^2}{𝑙}$

At the extreme position, $𝜙=𝜃$ and the speed is zero.

So the tension is minimum there:

${𝑇}_{min}=𝑚𝑔\cos 𝜃$

At the mean position, $𝜙=0$ and the speed is maximum.

Using conservation of energy,

$\frac{1}{2}𝑚{𝑣}^2=𝑚𝑔𝑙(1-\cos 𝜃)$

So,

${𝑣}^2=2𝑔𝑙(1-\cos 𝜃)$

Now tension at the mean position is

${𝑇}_{max}=𝑚𝑔+\frac{𝑚{𝑣}^2}{𝑙}$

Substitute ${𝑣}^2$:

${𝑇}_{max}=𝑚𝑔+\frac{𝑚}{𝑙}\left[2𝑔𝑙\right(1-\cos 𝜃\left)\right]$

${𝑇}_{max}=𝑚𝑔+2𝑚𝑔(1-\cos 𝜃)$

${𝑇}_{max}=𝑚𝑔(3-2\cos 𝜃)$

Given that maximum tension is twice the minimum tension:

${𝑇}_{max}=2{𝑇}_{min}$

$𝑚𝑔(3-2\cos 𝜃)=2𝑚𝑔\cos 𝜃$

Cancel $𝑚𝑔$:

$3-2\cos 𝜃=2\cos 𝜃$

$3=4\cos 𝜃$

$\cos 𝜃=\frac{3}{4}$

So,

$𝜃={\cos }^{-1}\left(0.75\right)$

Hence, the correct option is

$\overline{)\text{Option A }{\cos }^{-1}\left(0.75\right)}$

For a mass-spring system in S.H.M., the angular frequency is

$𝜔=\sqrt{\surd \frac{𝑘}{𝑚}}$

Given,

$𝑘=10\text{N/m},𝑚=10\text{kg}$

So,

$𝜔=\sqrt{\surd \frac{10}{10}}=1{\text{rad s}}^{-1}$

The velocity of a particle in S.H.M. at displacement $𝑥$ is

$𝑣=𝜔\sqrt{\surd {𝐴}^2-{𝑥}^2}$

Given velocity:

$𝑣=40\text{cm/s}=0.4\text{m/s}$

Amplitude:

$𝐴=0.5\text{m}$

Now substitute the values:

$0.4=1\sqrt{\surd (0.5{)}^2-{𝑥}^2}$

Squaring both sides,

$0.16=0.25-{𝑥}^2$

So,

${𝑥}^2=0.25-0.16=0.09$

$𝑥=0.3\text{m}$

So the displacement at that instant is

$\overline{)0.3\text{m}}$

Correct option: B.

The distance between two consecutive crests is the wavelength.

So,

$𝜆=40\text{ m}$

The velocity of the wave is

$𝑣=10\text{ m/s}$

For waves, we use the relation

$𝑣=\frac{𝜆}{𝑇}$

where $𝑇$ is the time period, that is, the time between two consecutive crests reaching the boat.

So,

$𝑇=\frac{𝜆}{𝑣}$

$𝑇=\frac{40}{10}$

$𝑇=4\text{ s}$

So, the consecutive waves strike the boat after

$\overline{)4\text{ s}}$

Correct option: C.

For a sound wave, the energy carried by the wave is proportional to the square of its amplitude and the square of its frequency.

So,

$𝐸\propto {𝐴}^2{𝑛}^2$

For instrument P:

${𝐸}_{𝑃}\propto {𝐴}^2{𝑛}^2$

For instrument Q, frequency is $\frac{𝑛}{4}$. Let its amplitude be ${𝐴}_{𝑄}$.

${𝐸}_{𝑄}\propto {𝐴}_2^{𝑄}{\left(\frac{𝑛}{4}\right)}^2$

Given that both waves have equal energies,

${𝐴}^2{𝑛}^2={𝐴}_2^{𝑄}{\left(\frac{𝑛}{4}\right)}^2$

${𝐴}^2{𝑛}^2={𝐴}_2^{𝑄}\frac{{𝑛}^2}{16}$

Cancel ${𝑛}^2$ from both sides:

${𝐴}^2=\frac{{𝐴}_2^{𝑄}}{16}$

${𝐴}_2^{𝑄}=16{𝐴}^2$

${𝐴}_{𝑄}=4𝐴$

So, the amplitude of waves produced by Q will be

$\overline{)4𝐴}$

Correct option: C.

In a stationary wave, the particles of the medium do not travel forward with the wave. Each particle vibrates about its mean position.

For a stationary wave, displacement can be written as

$𝑦=2𝑎\sin 𝑘𝑥\cos 𝜔𝑡$

Here, for a fixed particle at position $𝑥$,

$2𝑎\sin 𝑘𝑥$

is its amplitude, and

$\cos 𝜔𝑡$

shows that all vibrating particles have the same angular frequency $𝜔$, so they have the same time period.

But the amplitude depends on the position $𝑥$.

At nodes,

$\sin 𝑘𝑥=0$

so the amplitude is zero. Hence particles at nodes do not vibrate.

At other points, particles vibrate in S.H.M. with the same period but with different amplitudes.

So the correct option is:

Option A: except at node vibrate in S.H.M. of same period but of different amplitudes.

In a pipe closed at one end, the fundamental mode has a node at the closed end and an antinode at the open end.

So, the length of the air column is one-fourth of the wavelength:

$𝐿=\frac{𝜆}{4}$

The sound wave takes $𝑡$ seconds to travel this length $𝐿$ up to the closed end.

So the speed of sound is

$𝑣=\frac{𝐿}{𝑡}$

Substitute $𝐿=\frac{𝜆}{4}$:

$𝑣=\frac{𝜆}{4𝑡}$

Now frequency is

$𝑓=\frac{𝑣}{𝜆}$

So,

$𝑓=\frac{\frac{𝜆}{4𝑡}}{𝜆}$

$𝑓=\frac{1}{4𝑡}$

$𝑓=\frac{0.25}{𝑡}$

So the correct option is

$\overline{)\text{Option B }\frac{0.25}{𝑡}}$

For a sonometer wire, both ends are fixed. So the two ends are always nodes.

The fundamental mode is called the first harmonic.

So,

• First overtone = second harmonic

• Second overtone = third harmonic

• Third overtone = fourth harmonic

For the fourth harmonic, the wire is divided into $4$ equal loops.

Each loop has one antinode, so number of antinodes is

$4$

For a string vibrating in the ${𝑛}^{\text{th}}$ harmonic:

$\text{Number of nodes}=𝑛+1$

Here $𝑛=4$, so

$\text{Number of nodes}=4+1=5$

So, in the third overtone, the wire has

$5\text{ nodes and }4\text{ antinodes}$

Correct option: Option A.

Using Gauss’s law, the total electric flux through the closed surface of the cube is

${\Phi }_{\text{total}}=\frac{𝑄}{{𝜖}_0}$

Since the charge $𝑄$ is placed at the centre of the cube, the flux is distributed equally through all $6$ faces of the cube.

So, flux through one face is

${\Phi }_{\text{one face}}=\frac{𝑄}{6{𝜖}_0}$

The question asks for the electric flux through two opposite faces, so

${\Phi }_{\text{two faces}}=2\times \frac{𝑄}{6{𝜖}_0}$

${\Phi }_{\text{two faces}}=\frac{𝑄}{3{𝜖}_0}$

So the correct option is

$\overline{)\frac{𝑄}{3{𝜖}_0}}$

Option B.

For a parallel plate capacitor, when a dielectric slab of thickness $𝑡$ is introduced between the plates, the effective capacitance becomes

$𝐶=\frac{{𝜀}_0𝐴}{𝑑-𝑡+\frac{𝑡}{𝐾}}$

where $𝐾$ is the dielectric constant.

The original capacitance is

${𝐶}_0=\frac{{𝜀}_0𝐴}{𝑑}$

Given,

$𝐶=\frac{7}{6}{𝐶}_0$

So,

$\frac{{𝜀}_0𝐴}{𝑑-𝑡+\frac{𝑡}{𝐾}}=\frac{7}{6}\cdot \frac{{𝜀}_0𝐴}{𝑑}$

Cancel ${𝜀}_0𝐴$:

$\frac{1}{𝑑-𝑡+\frac{𝑡}{𝐾}}=\frac{7}{6𝑑}$

So,

$𝑑-𝑡+\frac{𝑡}{𝐾}=\frac{6𝑑}{7}$

Now,

$𝑡=\frac{2𝑑}{3}$

Hence,

$𝑑-𝑡=𝑑-\frac{2𝑑}{3}=\frac{𝑑}{3}$

Substitute this:

$\frac{𝑑}{3}+\frac{1}{𝐾}\cdot \frac{2𝑑}{3}=\frac{6𝑑}{7}$

Cancel $𝑑$:

$\frac{1}{3}+\frac{2}{3𝐾}=\frac{6}{7}$

Multiply by $3$:

$1+\frac{2}{𝐾}=\frac{18}{7}$

$\frac{2}{𝐾}=\frac{18}{7}-1$

$\frac{2}{𝐾}=\frac{11}{7}$

$𝐾=\frac{14}{11}$

So, the dielectric constant is

$\overline{)\frac{14}{11}}$

Correct option: B.

Let the capacitance of each identical air capacitor be $𝐶$.

When one capacitor is inserted in a liquid of dielectric constant $𝐾$, its capacitance becomes

$𝐾𝐶$

The other capacitor remains an air capacitor, so its capacitance is still

$𝐶$

Now the two capacitors are in series across a battery of emf $𝑉$.

For capacitors in series, charge on both capacitors is the same. The equivalent capacitance is

${𝐶}_{\text{eq}}=\frac{𝐾𝐶\times 𝐶}{𝐾𝐶+𝐶}$

${𝐶}_{\text{eq}}=\frac{𝐾{𝐶}^2}{𝐶(𝐾+1)}$

${𝐶}_{\text{eq}}=\frac{𝐾𝐶}{𝐾+1}$

So the charge supplied by the battery is

$𝑄={𝐶}_{\text{eq}}𝑉$

$𝑄=\frac{𝐾𝐶}{𝐾+1}𝑉$

The potential difference across the other capacitor, whose capacitance is $𝐶$, is

${𝑉}_{\text{other}}=\frac{𝑄}{𝐶}$

${𝑉}_{\text{other}}=\frac{\frac{𝐾𝐶}{𝐾+1}𝑉}{𝐶}$

${𝑉}_{\text{other}}=\frac{𝐾𝑉}{𝐾+1}$

So the correct option is

$\overline{)\frac{𝐾𝑉}{𝐾+1}}$

Option A.

Initially, the capacitor of capacitance ${𝐶}_1$ is charged to potential ${𝑉}_1$.

So, charge on it is

$𝑄={𝐶}_1{𝑉}_1$

Now it is disconnected from the battery, so this charge remains conserved.

When the uncharged capacitor of capacitance ${𝐶}_2$ is connected in parallel with it, both capacitors come to the same common potential ${𝑉}_2$.

The equivalent capacitance in parallel is

${𝐶}_1+{𝐶}_2$

So the total charge after connection is

$𝑄=({𝐶}_1+{𝐶}_2){𝑉}_2$

Using conservation of charge,

${𝐶}_1{𝑉}_1=({𝐶}_1+{𝐶}_2){𝑉}_2$

Hence,

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

So the correct option is

$\overline{)\text{Option B }\frac{{𝐶}_1{𝑉}_1}{{𝐶}_1+{𝐶}_2}}$

The potentiometer wire is $4\text{m}$ long, that is

$4\text{m}=400\text{cm}$

A potential difference of $3\text{V}$ is maintained across the whole wire.

So, potential drop per cm of the wire is

$\frac{3}{400}\text{V/cm}$

The cell balances against a length of $100\text{cm}$.

So the e.m.f. of the cell is equal to the potential drop across $100\text{cm}$ of the wire:

$𝐸=\frac{3}{400}\times 100$

$𝐸=\frac{300}{400}$

$𝐸=0.75\text{V}$

So, the correct option is

$\overline{)0.75\text{V}}$

Option C.

In a metre bridge, at the null point,

$\frac{{𝑅}_{𝐴}}{{𝑅}_{𝐵}}=\frac{𝑙}{100-𝑙}$

Here wire A is in the left gap and wire B is in the right gap.

The null point is at $40\text{ cm}$ from the left end, so

$𝑙=40\text{ cm}$

and

$100-𝑙=60\text{ cm}$

So,

$\frac{{𝑅}_{𝐴}}{{𝑅}_{𝐵}}=\frac{40}{60}=\frac{2}{3}$

Now resistance of a wire is

$𝑅=𝜌\frac{𝐿}{𝐴}$

So,

$𝜌=𝑅\frac{𝐴}{𝐿}$

The two wires have equal lengths, so ${𝐿}_{𝐴}={𝐿}_{𝐵}$.

Hence,

$\frac{{𝜌}_{𝐴}}{{𝜌}_{𝐵}}=\frac{{𝑅}_{𝐴}}{{𝑅}_{𝐵}}\times \frac{{𝐴}_{𝐴}}{{𝐴}_{𝐵}}$

The diameters are in the ratio

${𝑑}_{𝐴}:{𝑑}_{𝐵}=3:1$

Area of cross-section is proportional to ${𝑑}^2$, so

${𝐴}_{𝐴}:{𝐴}_{𝐵}=3^2:1^2=9:1$

Thus,

$\frac{{𝜌}_{𝐴}}{{𝜌}_{𝐵}}=\frac{2}{3}\times 9=6$

So the ratio of specific resistance of A to that of B is

$\overline{)6:1}$

Correct option: C.

Magnetic flux through the rod is given by

$𝜙=𝐵𝐴$

So the magnetic flux density $𝐵$ is

$𝐵=\frac{𝜙}{𝐴}$

Given,

$𝜙=6\times {10}^{-4}\text{Wb}$

and cross-sectional area,

$𝐴=3{\text{cm}}^2$

Since,

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

so,

$𝐴=3\times {10}^{-4}{\text{m}}^2$

Now,

$𝐵=\frac{6\times {10}^{-4}}{3\times {10}^{-4}}$

$𝐵=2\text{T}$

Magnetic permeability is related to magnetic field intensity by

$𝐵=𝜇𝐻$

So,

$𝜇=\frac{𝐵}{𝐻}$

Given,

$𝐻=2000\text{A/m}$

Thus,

$𝜇=\frac{2}{2000}$

$𝜇={10}^{-3}\text{Wb/A-m}$

So the correct answer is

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

Option C.

For a long air-core solenoid, the magnetic field inside it near the axis is

$𝐵={𝜇}_0𝑛𝐼$

where $𝑛=\frac{𝑁}{𝑙}$ is the number of turns per unit length.

So,

$𝐵={𝜇}_0\frac{𝑁}{𝑙}𝐼$

Magnetic flux through the cross-section of the solenoid is

$𝜙=𝐵𝐴$

Substituting the value of $𝐵$,

$𝜙={𝜇}_0\frac{𝑁}{𝑙}𝐼𝐴$

Now, magnetic moment of a solenoid is

$𝑀=𝑁𝐼𝐴$

So,

$𝜙=\frac{{𝜇}_0𝑀}{𝑙}$

Hence,

$𝑀=\frac{𝜙𝑙}{{𝜇}_0}$

Given,

$𝜙=\frac{𝜋}{2}\times {10}^{-6}\text{Wb}$

$𝑙=60\text{cm}=0.60\text{m}$

${𝜇}_0=4𝜋\times {10}^{-7}$

So,

$𝑀=\frac{(\frac{𝜋}{2}\times {10}^{-6})\left(0.60\right)}{4𝜋\times {10}^{-7}}$

$𝑀=\frac{0.30𝜋\times {10}^{-6}}{4𝜋\times {10}^{-7}}$

$𝑀=0.75{\text{A m}}^2$

So the magnetic moment is

$\overline{)0.75{\text{A m}}^2}$