MHT CET 2026 16th April Morning Shift (Physics & Chemistry) Question Paper with Solutions

Density of the wire is

$𝜌=\frac{𝑚}{𝑉}$

Since the wire is cylindrical,

$𝑉=𝜋{𝑟}^2𝑙$

So,

$𝜌=\frac{𝑚}{𝜋{𝑟}^2𝑙}$

For maximum percentage error, powers are added as multipliers:

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

Now,

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

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

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

So,

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

$=4\%$

The maximum percentage error in density is

$\overline{)4\%}$

Correct option: C.

Given,

$\overrightarrow{𝑃}=\widehat{𝑖}+\widehat{𝑗}+\widehat{𝑘}$

and

$\overrightarrow{𝑄}=-(\widehat{𝑖}+\widehat{𝑗}+\widehat{𝑘})$

So,

$\overrightarrow{𝑄}=-\overrightarrow{𝑃}$

Now,

$\overrightarrow{𝑃}-\overrightarrow{𝑄}=\overrightarrow{𝑃}-(-\overrightarrow{𝑃})$

$\overrightarrow{𝑃}-\overrightarrow{𝑄}=2\overrightarrow{𝑃}$

This means $\overrightarrow{𝑃}-\overrightarrow{𝑄}$ is in the same direction as $\overrightarrow{𝑃}$.

The angle between two vectors in the same direction is

$0^{\circ }$

So the correct option is

$\overline{)\text{Option C }{0}^{\circ }}$

When the ball is released from height $ℎ$, it falls freely under gravity.

For free fall from rest,

$ℎ=\frac{1}{2}𝑔{𝑡}^2$

So the time taken to reach the ground is

$𝑡=\sqrt{\surd \frac{2ℎ}{𝑔}}$

Since the collision with the ground is perfectly elastic, the ball rebounds with the same speed with which it hit the ground. So it rises back to the same height $ℎ$.

The time taken to rise back to height $ℎ$ is also

$\sqrt{\surd \frac{2ℎ}{𝑔}}$

So one complete period of the motion is

$𝑇=\sqrt{\surd \frac{2ℎ}{𝑔}}+\sqrt{\surd \frac{2ℎ}{𝑔}}$

$𝑇=2\sqrt{\surd \frac{2ℎ}{𝑔}}$

Frequency is the reciprocal of time period:

$𝑓=\frac{1}{𝑇}$

$𝑓=\frac{1}{2\sqrt{\surd \frac{2ℎ}{𝑔}}}$

$𝑓=\frac{1}{2}\sqrt{\surd \frac{𝑔}{2ℎ}}$

So the correct option is

$\overline{)\text{Option A }\frac{1}{2}\sqrt{\surd \frac{𝑔}{2ℎ}}}$

Take point $𝐴$ as the origin and line $𝐴𝐵$ along the horizontal direction.

So initially,

$𝐴𝐵=𝑎$

The boy at $𝐵$ runs perpendicular to $𝐴𝐵$ with speed ${𝑉}_1$.

In time $𝑡$, he covers distance

${𝑉}_1𝑡$

So at the time of catching, his position is like this: he is still horizontally at distance $𝑎$ from $𝐴$, and vertically at distance ${𝑉}_1𝑡$ from the line $𝐴𝐵$.

The boy from $𝐴$ catches him in time $𝑡$ with speed $𝑉$.

So distance covered by boy at $𝐴$ is

$𝑉𝑡$

This distance is the straight-line distance from $𝐴$ to the catching point.

Using Pythagoras theorem,

$(𝑉𝑡{)}^2={𝑎}^2+({𝑉}_1𝑡{)}^2$

${𝑉}^2{𝑡}^2={𝑎}^2+{𝑉}_2^1{𝑡}^2$

Bring the ${𝑡}^2$ terms together:

${𝑉}^2{𝑡}^2-{𝑉}_2^1{𝑡}^2={𝑎}^2$

${𝑡}^2({𝑉}^2-{𝑉}_2^1)={𝑎}^2$

${𝑡}^2=\frac{{𝑎}^2}{{𝑉}^2-{𝑉}_2^1}$

So,

$𝑡={\left[\frac{{𝑎}^2}{{𝑉}^2-{𝑉}_2^1}\right]}^{\frac{1}{2}}$

Hence, the correct option is

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

This is possible only when $𝑉>{𝑉}_1$.

Tangential acceleration is

${𝑎}_{𝑡}=𝑟𝛼$

and centripetal acceleration is

${𝑎}_{𝑐}=𝑟{𝜔}^2$

Given angular acceleration,

$𝛼=4{\text{rad/s}}^2$

For both accelerations to have equal magnitude,

${𝑎}_{𝑡}={𝑎}_{𝑐}$

So,

$𝑟𝛼=𝑟{𝜔}^2$

Cancel $𝑟$ from both sides:

$𝛼={𝜔}^2$

$4={𝜔}^2$

$𝜔=2\text{rad/s}$

Since the particle starts from rest, we use

$𝜔=𝛼𝑡$

$2=4𝑡$

$𝑡=\frac{2}{4}$

$𝑡=0.5\text{s}$

So the correct option is:

$\overline{)0.5\text{s}}$

Option C.

The forces on the particle are:

• Weight $𝑚𝑔$ vertically downward

• Normal reaction $𝑁$ from the smooth surface

Since the particle moves in a horizontal circle, the horizontal component of $𝑁$ provides the centripetal force, and the vertical component balances the weight.

Suppose the particle is at height $ℎ$ from the vertex and the radius of the circle is $𝑟$.

From the geometry of the cone, the ratio of horizontal component of $𝑁$ to vertical component of $𝑁$ is

$\frac{{𝑁}_{\text{horizontal}}}{{𝑁}_{\text{vertical}}}=\frac{ℎ}{𝑟}$

Now,

${𝑁}_{\text{horizontal}}=\frac{𝑚{𝑣}^2}{𝑟}$

and

${𝑁}_{\text{vertical}}=𝑚𝑔$

So,

$\frac{\frac{𝑚{𝑣}^2}{𝑟}}{𝑚𝑔}=\frac{ℎ}{𝑟}$

Cancel $𝑚$ and $𝑟$:

$\frac{{𝑣}^2}{𝑔}=ℎ$

Given,

$𝑣=0.5\text{m/s},𝑔=10{\text{m/s}}^2$

So,

$ℎ=\frac{(0.5{)}^2}{10}$

$ℎ=\frac{0.25}{10}=0.025\text{m}$

Converting into centimetres:

$0.025\text{m}=2.5\text{cm}$

So, the height of the plane of the circle from the vertex is

$\overline{)2.5\text{cm}}$

Correct option: B.

When the block compresses the spring by a maximum distance $𝑥$, its velocity becomes zero for an instant. At that instant, all the kinetic energy of the block is stored as elastic potential energy of the spring.

So,

$\text{Kinetic energy of block}=\text{Potential energy stored in spring}$

$\frac{{𝑝}^2}{2𝑚}=\frac{1}{2}𝐾{𝑥}^2$

where $𝑝$ is the momentum of the block.

Cancelling $\frac{1}{2}$,

$\frac{{𝑝}^2}{𝑚}=𝐾{𝑥}^2$

${𝑝}^2=𝑚𝐾{𝑥}^2$

$𝑝=𝑥\sqrt{\surd 𝑚𝐾}$

So, the maximum momentum of the block is

$\overline{)𝑥\sqrt{\surd 𝑚𝐾}}$

Hence, the correct option is Option B: $\sqrt{\surd 𝐾𝑚}𝑥$.

For a solid sphere, the moment of inertia about an axis passing through its centre is

${𝐼}_{\text{cm}}=\frac{2}{5}𝑀{𝑅}^2$

The given axis is parallel to this central axis and passes through a point at a distance

$𝑑=\frac{𝑅}{3}$

from the centre.

Using the parallel axis theorem,

$𝐼={𝐼}_{\text{cm}}+𝑀{𝑑}^2$

Substitute the values:

$𝐼=\frac{2}{5}𝑀{𝑅}^2+𝑀{\left(\frac{𝑅}{3}\right)}^2$

$𝐼=\frac{2}{5}𝑀{𝑅}^2+\frac{1}{9}𝑀{𝑅}^2$

Taking LCM $45$,

$𝐼=(\frac{18}{45}+\frac{5}{45})𝑀{𝑅}^2$

$𝐼=\frac{23}{45}𝑀{𝑅}^2$

So the correct option is

$\overline{)\frac{23𝑀{𝑅}^2}{45}}$

Option C.

At the point of projection, the body is at a distance

${𝑟}_1=\frac{2𝑑}{3}$

from the centre of mass ${𝑀}_1$.

Since the centres of ${𝑀}_1$ and ${𝑀}_2$ are separated by distance $𝑑$, the distance of the body from ${𝑀}_2$ is

${𝑟}_2=𝑑-\frac{2𝑑}{3}=\frac{𝑑}{3}$

For minimum speed to escape to infinity, the total mechanical energy should just become zero.

At infinity, gravitational potential energy is zero and speed becomes zero for minimum escape speed.

So,

$\frac{1}{2}𝑚{𝑣}^2-\frac{𝐺{𝑀}_1𝑚}{{𝑟}_1}-\frac{𝐺{𝑀}_2𝑚}{{𝑟}_2}=0$

Substitute ${𝑟}_1=\frac{2𝑑}{3}$ and ${𝑟}_2=\frac{𝑑}{3}$:

$\frac{1}{2}𝑚{𝑣}^2=\frac{𝐺{𝑀}_1𝑚}{2𝑑/3}+\frac{𝐺{𝑀}_2𝑚}{𝑑/3}$

$\frac{1}{2}{𝑣}^2=\frac{3𝐺{𝑀}_1}{2𝑑}+\frac{3𝐺{𝑀}_2}{𝑑}$

Multiplying by $2$,

${𝑣}^2=\frac{3𝐺{𝑀}_1}{𝑑}+\frac{6𝐺{𝑀}_2}{𝑑}$

${𝑣}^2=\frac{3𝐺}{𝑑}({𝑀}_1+2{𝑀}_2)$

Hence,

$𝑣=\sqrt{\surd \frac{3𝐺}{𝑑}({𝑀}_1+2{𝑀}_2)}$

So the correct option is:

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

When the rectangular block is floating in equilibrium, its weight is balanced by the buoyant force.

Now suppose the block is pushed down slightly by a distance $𝑥$.

Because of this extra downward displacement, an extra volume of liquid displaced is

$𝐴𝑥$

So the extra buoyant force acting upward is

$𝜚𝑔𝐴𝑥$

This extra buoyant force acts opposite to the displacement, so it is a restoring force.

Hence,

$𝐹=-𝜚𝑔𝐴𝑥$

Using Newton’s second law,

$𝑚\frac{{𝑑}^2𝑥}{𝑑{𝑡}^2}=-𝜚𝑔𝐴𝑥$

So,

$\frac{{𝑑}^2𝑥}{𝑑{𝑡}^2}+\frac{𝜚𝑔𝐴}{𝑚}𝑥=0$

This is the equation of simple harmonic motion, where

${𝜔}^2=\frac{𝜚𝑔𝐴}{𝑚}$

So,

$𝜔=\sqrt{\surd \frac{𝜚𝑔𝐴}{𝑚}}$

Frequency is

$𝑓=\frac{𝜔}{2𝜋}$

Therefore,

$𝑓=\frac{1}{2𝜋}\sqrt{\surd \frac{𝐴𝜚𝑔}{𝑚}}$

So the correct option is

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

When the pipe is closed, the liquid is at rest, so the gauge shows pressure ${𝑃}_1$.

When the valve is opened, water starts flowing out. Due to the motion of water, some pressure energy changes into kinetic energy. So the pressure inside the pipe decreases to ${𝑃}_2$.

Using Bernoulli’s principle at the same height:

${𝑃}_1={𝑃}_2+\frac{1}{2}𝜚{𝑣}^2$

So,

${𝑃}_1-{𝑃}_2=\frac{1}{2}𝜚{𝑣}^2$

Now solving for $𝑣$:

${𝑣}^2=\frac{2({𝑃}_1-{𝑃}_2)}{𝜚}$

$𝑣={\left[\frac{2({𝑃}_1-{𝑃}_2)}{𝜚}\right]}^{\frac{1}{2}}$

So the correct option is:

$\overline{)\text{Option B }{\left[\frac{2({𝑃}_1-{𝑃}_2)}{𝜚}\right]}^{\frac{1}{2}}}$

For flowing water, we use the equation of continuity.

Since water is incompressible, the volume of water passing per second through the pipe and through the nozzle must be the same.

If the pipe diameter is $𝑑$ and the nozzle diameter is $𝐷$, then

${𝐴}_1𝑉={𝐴}_2{𝑉}_1$

Area of cross-section of pipe:

${𝐴}_1=\frac{𝜋{𝑑}^2}{4}$

Area of cross-section of nozzle:

${𝐴}_2=\frac{𝜋{𝐷}^2}{4}$

So,

$\frac{𝜋{𝑑}^2}{4}𝑉=\frac{𝜋{𝐷}^2}{4}{𝑉}_1$

Cancel $\frac{𝜋}{4}$ from both sides:

${𝑑}^2𝑉={𝐷}^2{𝑉}_1$

So,

${𝐷}^2=\frac{{𝑑}^2𝑉}{{𝑉}_1}$

Taking square root,

$𝐷=𝑑\sqrt{\surd \frac{𝑉}{{𝑉}_1}}$

So, the diameter of the nozzle is

$\overline{)𝑑\sqrt{\surd \frac{𝑉}{{𝑉}_1}}}$

Correct option: Option A.

The correct option is Option A: attracted by other molecules in the sphere of influence.

In a liquid, each molecule attracts nearby molecules due to intermolecular forces. But this attraction is effective only up to a certain small distance. This region around a molecule is called its sphere of influence.

So, a molecule at the centre of its sphere of influence is attracted by the other liquid molecules present inside that sphere.

Since the liquid is at rest and the molecule is well inside the liquid, the attractions from all sides balance each other. So the resultant force may be zero, but the molecule is still being attracted by the surrounding molecules.

Molecules outside the sphere of influence do not have any appreciable effect.

So the answer is:

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

For heat conduction through a cylindrical rod, the rate of flow of heat is

$𝐻=\frac{𝐾𝐴({𝑇}_1-{𝑇}_2)}{𝐿}$

where

$𝐾$ is thermal conductivity,

$𝐴$ is area of cross-section,

$𝐿$ is length of the rod,

and $({𝑇}_1-{𝑇}_2)$ is the temperature difference.

Initially,

${𝐻}_1=\frac{𝐾𝐴({𝑇}_1-{𝑇}_2)}{𝐿}$

Now all dimensions of the rod become double.

So the length becomes

${𝐿}^{\prime }=2𝐿$

Since the rod is cylindrical, if its radius becomes double, then the area of cross-section becomes

${𝐴}^{\prime }=𝜋(2𝑟{)}^2=4𝜋{𝑟}^2=4𝐴$

The temperature difference remains the same.

So the new rate of flow of heat is

${𝐻}_2=\frac{𝐾\left(4𝐴\right)({𝑇}_1-{𝑇}_2)}{2𝐿}$

${𝐻}_2=2\frac{𝐾𝐴({𝑇}_1-{𝑇}_2)}{𝐿}$

${𝐻}_2=2{𝐻}_1$

So the correct option is

$\overline{)2{𝐻}_1}$

Option C.

For a black body, Wien’s displacement law says

${𝜆}_{𝑚}𝑇=\text{constant}$

Initially, the wavelength at which energy is maximum is $𝜆$, and temperature is $𝑇$.

After changing the temperature, the wavelength of maximum energy becomes

${𝜆}^{\prime }=\frac{2𝜆}{3}$

Using Wien’s law,

$𝜆𝑇={𝜆}^{\prime }{𝑇}^{\prime }$

Substitute ${𝜆}^{\prime }=\frac{2𝜆}{3}$:

$𝜆𝑇=\frac{2𝜆}{3}{𝑇}^{\prime }$

Cancel $𝜆$:

$𝑇=\frac{2}{3}{𝑇}^{\prime }$

So,

${𝑇}^{\prime }=\frac{3𝑇}{2}$

Now, according to Stefan-Boltzmann 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

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

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

$\frac{{𝑃}^{\prime }}{𝑃}=\frac{81}{16}$

So, the power radiated increases by a factor

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

Correct option: B.

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

$𝐸\propto 𝐴{𝑇}^4$

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

Initial temperature:

${27}^{\circ }𝐶=27+273=300𝐾$

Final temperature:

${327}^{\circ }𝐶=327+273=600𝐾$

Now, length and breadth are both reduced to half.

So the new area becomes:

${𝐴}^{\prime }=\frac{𝑙}{2}\times \frac{𝑏}{2}=\frac{𝑙𝑏}{4}=\frac{𝐴}{4}$

Using Stefan’s law,

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

Substitute the values:

$\frac{{𝐸}^{\prime }}{𝐸}=\frac{1}{4}{\left(\frac{600}{300}\right)}^4$

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

$\frac{{𝐸}^{\prime }}{𝐸}=\frac{1}{4}\times 16=4$

So,

${𝐸}^{\prime }=4𝐸$

The correct answer is Option B: $4𝐸$.

For the first expansion, the gas expands isothermally from volume $𝑉$ to $3𝑉$.

In an isothermal process,

$𝑃𝑉=\text{constant}$

So,

$𝑃\times 𝑉={𝑃}_1\times 3𝑉$

${𝑃}_1=\frac{𝑃}{3}$

Now the gas expands adiabatically from volume $3𝑉$ to $81𝑉$.

For an adiabatic process,

$𝑃{𝑉}^{𝛾}=\text{constant}$

So,

${𝑃}_1(3𝑉{)}^{𝛾}={𝑃}_{𝑓}(81𝑉{)}^{𝛾}$

Substitute ${𝑃}_1=\frac{𝑃}{3}$ and $𝛾=\frac{5}{3}$:

${𝑃}_{𝑓}=\frac{𝑃}{3}{\left(\frac{3𝑉}{81𝑉}\right)}^{𝛾}$

${𝑃}_{𝑓}=\frac{𝑃}{3}{\left(\frac{1}{27}\right)}^{5/3}$

Since $27=3^3$,

${\left(\frac{1}{27}\right)}^{5/3}={\left(\frac{1}{3^3}\right)}^{5/3}=\frac{1}{3^5}$

So,

${𝑃}_{𝑓}=\frac{𝑃}{3}\times \frac{1}{3^5}$

${𝑃}_{𝑓}=\frac{𝑃}{3^6}$

So the final pressure is

$\overline{)\frac{𝑃}{(3{)}^6}}$

Correct option: D.

For an adiabatic process, we use the relation

$𝑇{𝑉}^{𝛾-1}=\text{constant}$

So,

${𝑇}_1{𝑉}_{𝛾-1}^1={𝑇}_2{𝑉}_{𝛾-1}^2$

This gives

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

Given,

${𝑇}_1={27}^{\circ }\text{C}=300\text{ K}$

and

${𝑉}_2=\frac{8}{27}{𝑉}_1$

So,

$\frac{{𝑉}_1}{{𝑉}_2}=\frac{27}{8}$

Also,

$𝛾=\frac{5}{3}$

So,

$𝛾-1=\frac{5}{3}-1=\frac{2}{3}$

Now,

$\frac{{𝑇}_2}{300}={\left(\frac{27}{8}\right)}^{2/3}$

Since,

$\frac{27}{8}={\left(\frac{3}{2}\right)}^3$

we get

${\left(\frac{27}{8}\right)}^{2/3}={\left(\frac{3}{2}\right)}^2=\frac{9}{4}$

So,

${𝑇}_2=300\times \frac{9}{4}=675\text{ K}$

The rise in temperature is

$\Delta 𝑇={𝑇}_2-{𝑇}_1$

$\Delta 𝑇=675-300=375\text{ K}$

So, the rise in temperature is

$\overline{)375\text{ K}}$

Correct option: Option C.

The translational kinetic energy of gas molecules depends directly on the absolute temperature.

For one molecule, the average translational kinetic energy is

$𝐾=\frac{3}{2}𝑘𝑇$

where $𝑘$ is Boltzmann constant and $𝑇$ is the absolute temperature.

Since $𝐾\propto 𝑇$, to double the translational kinetic energy, the absolute temperature must also be doubled.

So the temperature should be increased from $𝑇$ to $2𝑇$.

Correct option: Option D, increasing $𝑇$ to $2𝑇$.

For a particle performing S.H.M., the restoring force is always opposite to the displacement.

So, if displacement is $𝑥$, then restoring force is

$𝐹=-𝑘𝑥$

where $𝑘$ is the force constant.

The potential energy in S.H.M. at displacement $𝑥$ is

$𝐸=\frac{1}{2}𝑘{𝑥}^2$

Now,

$2𝐸=𝑘{𝑥}^2$

Using $𝐹=-𝑘𝑥$,

$\frac{2𝐸}{𝐹}=\frac{𝑘{𝑥}^2}{-𝑘𝑥}$

$\frac{2𝐸}{𝐹}=-𝑥$

So,

$\frac{2𝐸}{𝐹}+𝑥=0$

Hence, the correct option is

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

For a progressive wave, the phase change at a fixed point in a time interval $\Delta 𝑡$ is

$\Delta 𝜙=𝜔\Delta 𝑡$

where

$𝜔=2𝜋𝑓$

Given,

$𝑓=50\text{Hz}$

and

$\Delta 𝑡=0.01\text{s}$

So,

$\Delta 𝜙=2𝜋𝑓\Delta 𝑡$

$\Delta 𝜙=2𝜋\times 50\times 0.01$

$\Delta 𝜙=2𝜋\times 0.5$

$\Delta 𝜙=𝜋\text{rad}$

So, the change in phase is

$\overline{)𝜋\text{rad}}$

Correct option: Option C.

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

In a stretched string fixed at both ends:

• Fundamental mode is the first harmonic.

• First overtone is the second harmonic.

• Second overtone is the third harmonic.

• Third overtone is the fourth harmonic.

So, in the third overtone, the wire vibrates in the fourth harmonic.

For the $𝑛$th harmonic:

$\text{Number of antinodes}=𝑛$

$\text{Number of nodes}=𝑛+1$

Here, $𝑛=4$.

So,

$\text{Number of antinodes}=4$

$\text{Number of nodes}=5$

Hence, the correct option is:

$\overline{)\text{Option A }4,5}$

For a stationary source and a moving observer, the apparent frequency is given by Doppler effect.

When the observer moves towards the source,

${𝐹}_1=𝐹\left(\frac{𝑉+{𝑉}_1}{𝑉}\right)$

When the observer moves away from the source,

${𝐹}_2=𝐹\left(\frac{𝑉-{𝑉}_1}{𝑉}\right)$

Given,

$\frac{{𝐹}_1}{{𝐹}_2}=2$

So,

$\frac{𝐹\left(\frac{𝑉+{𝑉}_1}{𝑉}\right)}{𝐹\left(\frac{𝑉-{𝑉}_1}{𝑉}\right)}=2$

Canceling common terms,

$\frac{𝑉+{𝑉}_1}{𝑉-{𝑉}_1}=2$

Now solve:

$𝑉+{𝑉}_1=2(𝑉-{𝑉}_1)$

$𝑉+{𝑉}_1=2𝑉-2{𝑉}_1$

$3{𝑉}_1=𝑉$

So,

$\frac{𝑉}{{𝑉}_1}=3$

Hence the correct option is

$\overline{)3}$

Option B.

The given wave is

$𝑌=6\sin (12𝜋𝑡-0.02𝜋𝑥+\frac{𝜋}{3})$

Compare it with the standard equation of a progressive wave:

$𝑌=𝐴\sin (𝜔𝑡-𝑘𝑥+𝜙)$

Here,

$𝜔=12𝜋\text{ rad/s}$

and

$𝑘=0.02𝜋\text{ rad/m}$

The velocity of a wave is

$𝑣=\frac{𝜔}{𝑘}$

So,

$𝑣=\frac{12𝜋}{0.02𝜋}$

$𝑣=\frac{12}{0.02}$

$𝑣=600\text{ m/s}$

So, the velocity of the wave is

$\overline{)600\text{ m/s}}$

Correct option: D.

For a square area parallel to the $𝑌\text{-}𝑍$ plane, its area vector is perpendicular to the plane.

Since the plane is parallel to $𝑌\text{-}𝑍$, the perpendicular direction is along the $𝑋$-axis. So,

$\overrightarrow{𝐴}={𝑙}^2\widehat{𝑖}$

The electric field is

$\overrightarrow{𝐸}=𝑎\widehat{𝑖}+𝑏\widehat{𝑗}$

Electric flux is given by

$𝜙=\overrightarrow{𝐸}\cdot \overrightarrow{𝐴}$

So,

$𝜙=(𝑎\widehat{𝑖}+𝑏\widehat{𝑗})\cdot \left({𝑙}^2\widehat{𝑖}\right)$

$𝜙=𝑎{𝑙}^2$

The $𝑏\widehat{𝑗}$ component does not contribute because it is parallel to the square surface, so it does not pass through the area.

Hence, the net electric flux is

$\overline{)𝑎{𝑙}^2}$

Correct option: D

The potential energy of two point charges separated by a distance $𝑟$ is

$𝑈=\frac{1}{4𝜋{𝜖}_0}\frac{{𝑞}_1{𝑞}_2}{𝑟}$

Here,

${𝑞}_1=6𝜇𝐶=6\times {10}^{-6}𝐶$

${𝑞}_2=4𝜇𝐶=4\times {10}^{-6}𝐶$

Initial distance:

${𝑟}_1=10\text{cm}=0.10\text{m}$

Since ${𝑞}_2$ is moved towards ${𝑞}_1$ by $2\text{cm}$, the final distance becomes

${𝑟}_2=10-2=8\text{cm}=0.08\text{m}$

Increase in potential energy is

$\Delta 𝑈={𝑈}_2-{𝑈}_1$

$\Delta 𝑈=\frac{1}{4𝜋{𝜖}_0}{𝑞}_1{𝑞}_2(\frac{1}{{𝑟}_2}-\frac{1}{{𝑟}_1})$

Substituting the values,

$\Delta 𝑈=9\times {10}^9\times 6\times {10}^{-6}\times 4\times {10}^{-6}(\frac{1}{0.08}-\frac{1}{0.10})$

Now,

$9\times {10}^9\times 24\times {10}^{-12}=0.216$

and

$\frac{1}{0.08}-\frac{1}{0.10}=12.5-10=2.5$

So,

$\Delta 𝑈=0.216\times 2.5=0.54\text{J}$

The increase in potential energy is

$\overline{)0.54\text{J}}$

So the correct option is Option C.

The force on each plate is due to the electric field produced by the other plate only.

For a parallel plate capacitor, the total electric field between the plates is

$𝐸$

This field is due to both plates together. Since the plates have equal and opposite charges, each plate contributes half of the total field between them.

So, the electric field due to one plate is

$\frac{𝐸}{2}$

Now, force on a plate of charge $𝑄$ is

$𝐹=𝑄\times \text{electric field due to the other plate}$

So,

$𝐹=𝑄\times \frac{𝐸}{2}$

$𝐹=\frac{𝑄𝐸}{2}$

Hence, the correct option is

$\overline{)\text{Option C }\frac{𝑄𝐸}{2}}$

Initially, the two condensers are joined in parallel and charged to potential $𝑉$.

Their capacitances are $2𝐶$ and $𝐶$.

For parallel combination, total capacitance is

${𝐶}_{\text{initial}}=2𝐶+𝐶=3𝐶$

So the total charge stored initially is

$𝑄={𝐶}_{\text{initial}}𝑉=3𝐶𝑉$

Now the battery is disconnected. So no charge can enter or leave the system. Hence, the total charge remains the same:

$𝑄=3𝐶𝑉$

When the condenser of capacity $𝐶$ is completely filled with a dielectric of dielectric constant $𝐾$, its new capacitance becomes

$𝐾𝐶$

The other condenser still has capacitance $2𝐶$.

So the new equivalent capacitance in parallel is

${𝐶}_{\text{final}}=2𝐶+𝐾𝐶=(𝐾+2)𝐶$

Let the new common potential difference be ${𝑉}^{\prime }$. Then

$𝑄={𝐶}_{\text{final}}{𝑉}^{\prime }$

$3𝐶𝑉=(𝐾+2)𝐶{𝑉}^{\prime }$

Cancel $𝐶$:

${𝑉}^{\prime }=\frac{3𝑉}{𝐾+2}$

So the potential difference across the capacitors in the second case is

$\overline{)\frac{3𝑉}{𝐾+2}}$

Correct option: A.

The potential difference across the full potentiometer wire is $3\text{ V}$.

Length of the wire is

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

So, potential drop per cm, called the potential gradient, is

$𝑘=\frac{3}{400}\text{ V/cm}$

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

So its e.m.f. is

$𝐸=𝑘\times 𝑙$

$𝐸=\frac{3}{400}\times 100$

$𝐸=\frac{3}{4}=0.75\text{ V}$

So the correct answer is:

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

Option A.

For a balanced metre bridge,

$\frac{𝑅}{𝑆}=\frac{𝑙}{100-𝑙}$

Here, resistance in the left gap is

$𝑅=20\Omega$

and resistance in the right gap is

$𝑆=60\Omega$

So,

$\frac{20}{60}=\frac{𝑙}{100-𝑙}$

$\frac{1}{3}=\frac{𝑙}{100-𝑙}$

Cross-multiplying,

$100-𝑙=3𝑙$

$100=4𝑙$

$𝑙=25\text{cm}$

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

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

Hence, distance of the null point from the centre is

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

So, the correct option is

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

Option C.

Magnetization is magnetic moment per unit volume.

Given magnetic moment,

$𝑚=8\times {10}^{-7}{\text{A m}}^2$

Mass of sample,

$𝑀=2\text{g}$

Density,

$𝜌=4{\text{g/cm}}^3$

Volume of the sample is

$𝑉=\frac{\text{mass}}{\text{density}}$

$𝑉=\frac{2}{4}=0.5{\text{cm}}^3$

Now convert this into ${\text{m}}^3$:

$1{\text{cm}}^3={10}^{-6}{\text{m}}^3$

So,

$𝑉=0.5\times {10}^{-6}{\text{m}}^3$

$𝑉=5\times {10}^{-7}{\text{m}}^3$

Magnetization is

$𝐼=\frac{𝑚}{𝑉}$

$𝐼=\frac{8\times {10}^{-7}}{5\times {10}^{-7}}$

$𝐼=1.6\text{A/m}$

So, the magnetization of the sample is

$\overline{)1.6\text{A/m}}$

Correct option: Option A.

The magnetic force on a current-carrying straight wire is given by

$\overrightarrow{𝐹}=𝐼(\overrightarrow{𝐿}\times \overrightarrow{𝐵})$

Here, the wire is along the $𝑥$-axis, so the length vector is

$\overrightarrow{𝐿}=𝐿\widehat{𝑖}$

The magnetic field is

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

So,

$\overrightarrow{𝐹}=𝐼[𝐿\widehat{𝑖}\times 𝐵(\widehat{𝑖}+2\widehat{𝑗}-2\widehat{𝑘}\left)\right]$

$\overrightarrow{𝐹}=𝐼𝐿𝐵[\widehat{𝑖}\times \widehat{𝑖}+2(\widehat{𝑖}\times \widehat{𝑗})-2(\widehat{𝑖}\times \widehat{𝑘}\left)\right]$

Now,

$\widehat{𝑖}\times \widehat{𝑖}=0$

$\widehat{𝑖}\times \widehat{𝑗}=\widehat{𝑘}$

$\widehat{𝑖}\times \widehat{𝑘}=-\widehat{𝑗}$

So,

$\overrightarrow{𝐹}=𝐼𝐿𝐵[0+2\widehat{𝑘}-2(-\widehat{𝑗}\left)\right]$

$\overrightarrow{𝐹}=𝐼𝐿𝐵(2\widehat{𝑗}+2\widehat{𝑘})$

Now the magnitude is

$\left|\overrightarrow{𝐹}\right|=𝐼𝐿𝐵\sqrt{\surd 2^2+2^2}$

$\left|\overrightarrow{𝐹}\right|=𝐼𝐿𝐵\sqrt{\surd 8}$

So the correct option is

$\overline{)\sqrt{\surd 8}𝐼𝐿𝐵}$

Option A.

For a charged particle moving perpendicular to a uniform magnetic field, the radius of circular path is

$𝑟=\frac{𝑚𝑣}{𝑞𝐵}$

Since $𝑚$, $𝑞$, and $𝐵$ are constant,

$𝑟\propto 𝑣$

The kinetic energy of the particle is

$𝐾=\frac{1}{2}𝑚{𝑣}^2$