KCET 2026 Question Paper with Solutions

For each quantity, we write its dimensional formula.

Torque is given by

$\text{Torque}=\text{Force}\times \text{distance}$

Force has dimensions

$\left[𝑀𝐿{𝑇}^{-2}\right]$

So,

$\left[\text{Torque}\right]=\left[𝑀𝐿{𝑇}^{-2}\right]\left[𝐿\right]=\left[𝑀{𝐿}^2{𝑇}^{-2}\right]$

So, torque matches (iv).

For gravitational constant $𝐺$, from Newton’s law of gravitation,

$𝐹=\frac{𝐺{𝑚}_1{𝑚}_2}{{𝑟}^2}$

So,

$𝐺=\frac{𝐹{𝑟}^2}{{𝑚}_1{𝑚}_2}$

Now,

$\left[𝐺\right]=\frac{\left[𝑀𝐿{𝑇}^{-2}\right]\left[{𝐿}^2\right]}{\left[{𝑀}^2\right]}$

$\left[𝐺\right]=\left[{𝑀}^{-1}{𝐿}^3{𝑇}^{-2}\right]$

So, gravitational constant matches (iii).

Capacitance is

$𝐶=\frac{𝑄}{𝑉}$

Charge has dimensions

$\left[𝑄\right]=\left[𝐴𝑇\right]$

Potential difference is

$𝑉=\frac{\text{Work}}{\text{Charge}}$

So,

$\left[𝑉\right]=\frac{\left[𝑀{𝐿}^2{𝑇}^{-2}\right]}{\left[𝐴𝑇\right]}=\left[𝑀{𝐿}^2{𝑇}^{-3}{𝐴}^{-1}\right]$

Hence,

$\left[𝐶\right]=\frac{\left[𝐴𝑇\right]}{\left[𝑀{𝐿}^2{𝑇}^{-3}{𝐴}^{-1}\right]}$

$\left[𝐶\right]=\left[{𝑀}^{-1}{𝐿}^{-2}{𝑇}^4{𝐴}^2\right]$

So, capacitance matches (i).

Planck’s constant $ℎ$ is related to energy by

$𝐸=ℎ𝜈$

So,

$ℎ=\frac{𝐸}{𝜈}$

Energy has dimensions

$\left[𝑀{𝐿}^2{𝑇}^{-2}\right]$

Frequency has dimensions

$\left[{𝑇}^{-1}\right]$

Therefore,

$\left[ℎ\right]=\frac{\left[𝑀{𝐿}^2{𝑇}^{-2}\right]}{\left[{𝑇}^{-1}\right]}=\left[𝑀{𝐿}^2{𝑇}^{-1}\right]$

So, Planck’s constant matches (ii).

Thus the correct matching is:

$𝑎\to 𝑖𝑣,𝑏\to 𝑖𝑖𝑖,𝑐\to 𝑖,𝑑\to 𝑖𝑖$

Correct option: Option B.

The velocity is given as a function of position:

$𝑣={𝑥}^2-5𝑥+4$

We need acceleration when the velocity is zero.

First find the positions where $𝑣=0$:

${𝑥}^2-5𝑥+4=0$

Factorising,

$(𝑥-1)(𝑥-4)=0$

So,

$𝑥=1\text{ m}\text{or}𝑥=4\text{ m}$

Now, since velocity is given as a function of $𝑥$, acceleration is not simply $\frac{𝑑𝑣}{𝑑𝑥}$.

We use

$𝑎=\frac{𝑑𝑣}{𝑑𝑡}$

Using chain rule,

$𝑎=\frac{𝑑𝑣}{𝑑𝑥}\cdot \frac{𝑑𝑥}{𝑑𝑡}$

But

$\frac{𝑑𝑥}{𝑑𝑡}=𝑣$

So,

$𝑎=𝑣\frac{𝑑𝑣}{𝑑𝑥}$

Here, when the velocity is zero,

$𝑣=0$

Hence,

$𝑎=0\cdot \frac{𝑑𝑣}{𝑑𝑥}=0$

So the magnitude of acceleration is

$0{\text{ m/s}}^2$

Correct option: C, Zero.

For average speed, we use

$\text{Average speed}=\frac{\text{Total distance}}{\text{Total time}}$

Here, the car covers equal distances at two different speeds.

Let the total distance be $2𝑑$ km.

So, first half distance $=𝑑$ km and second half distance $=𝑑$ km.

Time taken for first half:

$\frac{𝑑}{40}$

Time taken for second half:

$\frac{𝑑}{50}$

Total time:

$\frac{𝑑}{40}+\frac{𝑑}{50}$

Now average speed:

$\text{Average speed}=\frac{2𝑑}{\frac{𝑑}{40}+\frac{𝑑}{50}}$

Taking $𝑑$ common in the denominator:

$\text{Average speed}=\frac{2𝑑}{𝑑(\frac{1}{40}+\frac{1}{50})}$

Cancel $𝑑$:

$\text{Average speed}=\frac{2}{\frac{1}{40}+\frac{1}{50}}$

Now,

$\frac{1}{40}+\frac{1}{50}=\frac{50+40}{2000}=\frac{90}{2000}$

So,

$\text{Average speed}=\frac{2}{\frac{90}{2000}}=\frac{4000}{90}=44.44\text{ km/h}$

So, the correct option is

$\overline{)\text{Option B }44.44\text{ km/h}}$

For a projectile, the maximum height is given by

$𝐻=\frac{{𝑢}^2{\sin }^2𝜃}{2𝑔}$

Here both bodies are projected with the same velocity, so $𝑢$ is the same for both. Also, $𝑔$ is the same.

So maximum height depends only on ${\sin }^2𝜃$.

For the first body,

$𝜃={30}^{\circ }$

${𝐻}_1=\frac{{𝑢}^2{\sin }^2{30}^{\circ }}{2𝑔}$

For the second body,

$𝜃={45}^{\circ }$

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

So,

$\frac{{𝐻}_1}{{𝐻}_2}=\frac{{\sin }^2{30}^{\circ }}{{\sin }^2{45}^{\circ }}$

We know,

$\sin {30}^{\circ }=\frac{1}{2}$

and

$\sin {45}^{\circ }=\frac{1}{\sqrt{\surd 2}}$

So,

$\frac{{𝐻}_1}{{𝐻}_2}=\frac{{\left(\frac{1}{2}\right)}^2}{{\left(\frac{1}{\sqrt{\surd 2}}\right)}^2}$

$=\frac{\frac{1}{4}}{\frac{1}{2}}=\frac{1}{2}$

Hence,

${𝐻}_1:{𝐻}_2=1:2$

The correct option is Option B, $1:2$.

The body is on a frictionless table, so the applied force produces acceleration without any loss due to friction.

Given,

$𝐹=5\text{N}$

$𝑚=5\text{kg}$

Using Newton’s second law,

$𝐹=𝑚𝑎$

So,

$𝑎=\frac{𝐹}{𝑚}=\frac{5}{5}=1{\text{m s}}^{-2}$

The body starts from rest, so

$𝑢=0$

Time,

$𝑡=10\text{s}$

Using

$𝑣=𝑢+𝑎𝑡$

we get

$𝑣=0+1\times 10=10{\text{m s}}^{-1}$

Initial kinetic energy is zero because the body is initially at rest.

Final kinetic energy is

$𝐾=\frac{1}{2}𝑚{𝑣}^2$

$𝐾=\frac{1}{2}\times 5\times (10{)}^2$

$𝐾=\frac{1}{2}\times 5\times 100=250\text{J}$

So, the change in kinetic energy of the body in $10$ s is

$250\text{J}$

Correct option: Option D.

Let ${𝑇}_1$ be the tension in the horizontal string, ${𝑇}_2$ be the tension in the slanted string, and ${𝑇}_3$ be the tension in the vertical string.

The mass $𝑀$ is at rest, so it is in equilibrium. The vertical string supports the weight of the mass. Hence,

${𝑇}_3=𝑀𝑔$

Now consider the knot $𝑂$. Three tensions act on it: ${𝑇}_1$ horizontally, ${𝑇}_2$ along the slanted string, and ${𝑇}_3$ vertically downward.

Since the knot is also in equilibrium, the forces in the vertical direction must balance.

${𝑇}_2\sin {45}^{\circ }={𝑇}_3=𝑀𝑔$

The forces in the horizontal direction must also balance.

${𝑇}_1={𝑇}_2\cos {45}^{\circ }$

Now divide the vertical balance equation by the horizontal balance equation:

$\frac{{𝑇}_2\sin {45}^{\circ }}{{𝑇}_2\cos {45}^{\circ }}=\frac{𝑀𝑔}{{𝑇}_1}$

$\tan {45}^{\circ }=\frac{𝑀𝑔}{{𝑇}_1}$

Since $\tan {45}^{\circ }=1$,

$1=\frac{𝑀𝑔}{{𝑇}_1}$

So,

${𝑇}_1=𝑀𝑔$

Thus, the tension in the horizontal string is $𝑀𝑔$.

When the lift moves upwards with acceleration, the man feels heavier.

The weighing scale shows the normal reaction $𝑅$ exerted by the scale on the man.

For upward acceleration,

$𝑅=𝑚(𝑔+𝑎)$

Here,

$𝑚=80\text{ kg},𝑔=10{\text{ m/s}}^2,𝑎=6{\text{ m/s}}^2$

So,

$𝑅=80(10+6)$

$𝑅=80\times 16=1280\text{ N}$

The scale is marked in kg, so it divides the force by $𝑔=10{\text{ m/s}}^2$:

$\text{Scale reading}=\frac{1280}{10}=128\text{ kg}$

So the man’s apparent weight would be

$\overline{)128\text{ kg}}$

Correct option: Option D.

For a body having momentum $𝑝$ and mass $𝑚$, kinetic energy is

$𝐾=\frac{{𝑝}^2}{2𝑚}$

Here, both bodies have equal linear momentum, so $𝑝$ is the same for both.

So kinetic energy is inversely proportional to mass:

$𝐾\propto \frac{1}{𝑚}$

Given,

${𝐾}_1:{𝐾}_2=3:1$

So,

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

$\frac{3}{1}=\frac{{𝑚}_2}{{𝑚}_1}$

Hence,

${𝑚}_1:{𝑚}_2=1:3$

The ratio of their masses is

$\overline{)1:3}$

So the correct option is Option B.

When the earth contracts suddenly, no external torque is acting on it. So its angular momentum remains conserved.

Angular momentum is

$𝐿=𝐼𝜔$

For a spherical body,

$𝐼\propto {𝑅}^2$

If the radius becomes half,

${𝑅}^{\prime }=\frac{𝑅}{2}$

So the new moment of inertia becomes

${𝐼}^{\prime }=𝐼{\left(\frac{1}{2}\right)}^2=\frac{𝐼}{4}$

Using conservation of angular momentum,

$𝐼𝜔={𝐼}^{\prime }{𝜔}^{\prime }$

$𝐼𝜔=\frac{𝐼}{4}{𝜔}^{\prime }$

${𝜔}^{\prime }=4𝜔$

So the earth will rotate $4$ times faster.

Time period is inversely proportional to angular velocity, so

${𝑇}^{\prime }=\frac{𝑇}{4}$

Present duration of a day is

$𝑇=24\text{ h}$

Hence,

${𝑇}^{\prime }=\frac{24}{4}=6\text{ h}$

So the duration of the day would be

$\overline{)6\text{ h}}$

Correct option: Option A.

The angular momentum of a moving body remains constant if net external torque is not applied.

So the correct option is:

Option D: net external torque is not applied

According to the principle of conservation of angular momentum, if the net external torque acting on a body or system is zero, then its angular momentum remains constant.

Mathematically,

${𝜏}_{\text{ext}}=\frac{𝑑𝐿}{𝑑𝑡}$

If no net external torque acts,

${𝜏}_{\text{ext}}=0$

So,

$\frac{𝑑𝐿}{𝑑𝑡}=0$

This means angular momentum $𝐿$ does not change with time.

Hence, the angular momentum remains constant when net external torque is not applied.

Let the mass of the Earth be ${𝑀}_{𝑒}$, the mass of the Moon be ${𝑀}_{𝑚}$, and the distance between their centres be $𝑑$.

Suppose the spaceship is at a distance $𝑟$ from the centre of the Earth. Then its distance from the centre of the Moon is $(𝑑-𝑟)$.

The Earth pulls the passenger towards the Earth, and the Moon pulls the passenger towards the Moon. These two pulls are in opposite directions, so the net gravitational force on the passenger of mass $𝑀$ is:

${𝐹}_{𝑛𝑒𝑡}=\frac{𝐺{𝑀}_{𝑒}𝑀}{{𝑟}^2}-\frac{𝐺{𝑀}_{𝑚}𝑀}{(𝑑-𝑟{)}^2}$

The spaceship moves with constant velocity. So the distance $𝑟$ from the Earth increases uniformly with time $𝑡$.

At the surface of the Earth, that is at $𝑡=0$, the pull due to the Moon is very small compared to the pull due to the Earth. So the net force is almost equal to the weight on Earth:

${𝐹}_{𝑛𝑒𝑡}\approx \frac{𝐺{𝑀}_{𝑒}𝑀}{{𝑅}_2^{𝑒}}=𝑀𝑔$

As the spaceship moves towards the Moon, the gravitational pull of the Earth decreases because the spaceship is getting farther from the Earth.

At the same time, the gravitational pull of the Moon increases because the spaceship is getting closer to the Moon.

At one point between the Earth and the Moon, both gravitational pulls become equal in magnitude and opposite in direction. So the net gravitational force becomes zero. This point is called the neutral point.

Since the Earth is much more massive than the Moon, ${𝑀}_{𝑒}\approx 81{𝑀}_{𝑚}$. So the neutral point is much closer to the Moon than to the Earth.

This means the force becomes zero at a time closer to ${𝑡}_0$, not near the beginning of the journey.

After crossing the neutral point, the Moon’s gravitational pull becomes greater than the Earth’s pull.

At the surface of the Moon, that is at $𝑡={𝑡}_0$, the net force is almost equal to the weight on the Moon:

$\left|{𝐹}_{𝑛𝑒𝑡}\right|\approx \frac{𝐺{𝑀}_{𝑚}𝑀}{{𝑅}_2^{𝑚}}=𝑀{𝑔}^{\prime }$

The weight here means the magnitude of the net gravitational force, so it cannot be negative.

So the graph must start from $𝑀𝑔$, decrease to zero near the Moon, and then increase to $𝑀{𝑔}^{\prime }$.

Curve C shows this behaviour correctly.

For any planet, acceleration due to gravity on its surface is

$𝑔=\frac{𝐺𝑀}{{𝑅}^2}$

where $𝑀$ is the mass of the planet and $𝑅$ is its radius.

The new planet has the same density as the earth. Also, it is two times bigger in size, so we take its radius as

${𝑅}^{\prime }=2𝑅$

Now, mass of a spherical planet is

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

Since volume of a sphere is proportional to ${𝑅}^3$, for the same density,

$𝑀\propto {𝑅}^3$

So, if the radius becomes $2𝑅$, the mass becomes

${𝑀}^{\prime }=2^3𝑀=8𝑀$

Now use the formula for gravity on the new planet:

${𝑔}^{\prime }=\frac{𝐺{𝑀}^{\prime }}{{𝑅}^{\prime 2}}$

Substitute ${𝑀}^{\prime }=8𝑀$ and ${𝑅}^{\prime }=2𝑅$:

${𝑔}^{\prime }=\frac{𝐺\left(8𝑀\right)}{(2𝑅{)}^2}$

${𝑔}^{\prime }=\frac{8𝐺𝑀}{4{𝑅}^2}$

${𝑔}^{\prime }=2\frac{𝐺𝑀}{{𝑅}^2}$

Since

$𝑔=\frac{𝐺𝑀}{{𝑅}^2}$

we get

${𝑔}^{\prime }=2𝑔$

So the correct option is:

$\overline{)\text{Option C }{𝑔}^{\prime }=2𝑔}$

For a wire, the extension produced by a load is given by

$\Delta 𝐿=\frac{𝐹𝐿}{𝐴𝑌}$

Here,

• $𝐹$ is the load applied,

• $𝐿$ is the length of the wire,

• $𝐴$ is the area of cross-section,

• $𝑌$ is Young’s modulus.

Both wires have the same material, so $𝑌$ is the same.

Both have the same length, and the same load is applied.

So,

$\Delta 𝐿\propto \frac{1}{𝐴}$

Now, area of cross-section of a wire is

$𝐴=\frac{𝜋{𝑑}^2}{4}$

So area is proportional to the square of diameter:

$𝐴\propto {𝑑}^2$

Given,

${𝑑}_2=2{𝑑}_1$

So,

${𝐴}_2=4{𝐴}_1$

Since extension is inversely proportional to area,

$\frac{\Delta {𝐿}_1}{\Delta {𝐿}_2}=\frac{{𝐴}_2}{{𝐴}_1}=4$

Hence, the ratio of extensions is

$\Delta {𝐿}_1:\Delta {𝐿}_2=4:1$

So the correct option is Option D: $4:1$.

In normal conditions, water rises in a capillary tube because the upward force due to surface tension balances the weight of the water column.

The capillary rise is

$ℎ=\frac{2𝑇\cos 𝜃}{𝜌𝑔𝑟}$

Here, $𝑔$ is the acceleration due to gravity.

In a freely falling elevator, the effective acceleration due to gravity becomes zero:

${𝑔}_{\text{eff}}=0$

So the water column has no effective weight. Since there is no downward weight to balance the upward pull due to surface tension, the water will rise as much as possible in the tube.

But the tube is only $30$ cm long. So the water can fill the whole tube.

Hence, the length of the water column becomes

$30\text{ cm}$

Correct option: C, $30$ cm.

The correct option is:

Option B: Bernoulli's Principle

A carburetor mixes air and fuel in the correct ratio for combustion in an automobile engine.

Inside the carburetor, air passes through a narrow region called a venturi. When air flows through this narrow part, its speed increases. According to Bernoulli's Principle, when the speed of a fluid increases, its pressure decreases.

So, the pressure in the narrow region becomes less than the pressure in the fuel chamber. Due to this pressure difference, fuel is drawn into the moving air stream. This forms the air-fuel mixture needed for combustion.

So, the instrument works on Bernoulli's Principle.

All the given processes are irreversible in nature.

Free expansion of a gas is irreversible because the gas spreads into the available space on its own. It cannot return to its original volume without some external work being done.

Combustion of petrol and air is also irreversible. Once petrol burns, it forms new substances like carbon dioxide, water vapour, etc. These products cannot change back into petrol and air by simply reversing the process.

Leaking of gas from a kitchen cylinder is irreversible because the gas moves from high pressure inside the cylinder to lower pressure outside. It will not collect back into the cylinder on its own.

Transfer of heat from a hotter part of a liquid to a colder part is irreversible because heat naturally flows from higher temperature to lower temperature. The reverse does not happen by itself.

So, all four processes are irreversible.

$\overline{)\text{Option D: I, II, III and IV}}$

The statement that is not true is:

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

In an isobaric process, the pressure remains constant, not the volume.

The word “isobaric” means constant pressure. So for an isobaric process,

$𝑃=\text{constant}$

If the volume remains constant, that process is called an isochoric process.

The other statements are correct:

In an isothermal process,

$𝑇=\text{constant}$

In an adiabatic process, no heat is exchanged with the surroundings, so the system is insulated:

$𝑄=0$

Also, for a reversible adiabatic process of an ideal gas,

$𝑃{𝑉}^{𝛾}=\text{constant}$

So the incorrect statement is Option B.

According to the ideal gas equation, $𝑃𝑉=𝑛𝑅𝑇$.

Here, the gas is $1$ mole, so $𝑛=1$. The temperature is constant, so $𝑇$ is also constant.

Since $𝑅$ is already a constant, the whole quantity $𝑛𝑅𝑇$ becomes constant.

So, we get $𝑃𝑉=\text{constant}$.

This means that when volume $𝑉$ increases, pressure $𝑃$ decreases, and when volume $𝑉$ decreases, pressure $𝑃$ increases.

So, pressure is inversely proportional to volume:

$𝑃\propto \frac{1}{𝑉}$

The graph of $𝑃$ versus $𝑉$ for this relation is a rectangular hyperbola.

Among the given graphs, Graph-IV shows this rectangular hyperbola.

For SHM,

$𝑥=𝐴\cos (𝜔𝑡+𝜙)$

Given displacement is

$𝑥=6.0\cos (100𝑡+𝜋/4)\text{ cm}$

So the amplitude is

$𝐴=6.0\text{ cm}=0.06\text{ m}$

and angular frequency is

$𝜔=100{\text{ rad s}}^{-1}$

In SHM, maximum kinetic energy is equal to the total mechanical energy:

${𝐾}_{max}=\frac{1}{2}𝑚{𝜔}^2{𝐴}^2$

Substitute the values:

${𝐾}_{max}=\frac{1}{2}\left(1\right)\left(100{)}^2\right(0.06{)}^2$

${𝐾}_{max}=\frac{1}{2}\times 10000\times 0.0036$

${𝐾}_{max}=18\text{ J}$

So the maximum kinetic energy is

$\overline{)18\text{ J}}$

Correct option: Option D.

For beats, the number of beats per second is equal to the difference in the two frequencies.

Given that the source of frequency $𝜈$ gives $6$ beats/second with $200\text{ Hz}$:

$|𝜈-200|=6$

So,

$𝜈=200+6=206\text{ Hz}$

or

$𝜈=200-6=194\text{ Hz}$

Now the second harmonic of the source has frequency $2𝜈$.

It gives $8$ beats/second with $420\text{ Hz}$, so

$|2𝜈-420|=8$

This gives

$2𝜈=420+8=428$

or

$2𝜈=420-8=412$

So,

$𝜈=214\text{ Hz}$

or

$𝜈=206\text{ Hz}$

The common value from both conditions is

$𝜈=206\text{ Hz}$

So the correct option is Option B: $206$ Hz.

Work done in moving a charge from one point to another is related to potential difference by

$𝑊=𝑞({𝑉}_{𝐵}-{𝑉}_{𝐴})$

Here,

$𝑊=200\text{ J}$

$𝑞=5\text{ C}$

${𝑉}_{𝐴}=-20\text{ V}$

So,

$200=5[{𝑉}_{𝐵}-(-20\left)\right]$

$200=5({𝑉}_{𝐵}+20)$

Divide both sides by $5$:

$40={𝑉}_{𝐵}+20$

${𝑉}_{𝐵}=20\text{ V}$

So, the value of potential at point B is

$\overline{)20\text{ V}}$

Correct option: Option B.

Statement-I is correct.

For an electric dipole placed in a uniform electric field, the torque acting on it is

$\overrightarrow{𝜏}=\overrightarrow{𝑃}\times \overrightarrow{𝐸}$

This torque tends to rotate the dipole so that $\overrightarrow{𝑃}$ becomes aligned with $\overrightarrow{𝐸}$.

Statement-II is also correct.

The potential energy of an electric dipole in a uniform electric field is

$𝑈=-\overrightarrow{𝑃}\cdot \overrightarrow{𝐸}$

or

$𝑈=-𝑃𝐸\cos 𝜃$

where $𝜃$ is the angle between $\overrightarrow{𝑃}$ and $\overrightarrow{𝐸}$.

Statement-III is incorrect.

In a uniform electric field, the positive and negative charges of the dipole experience equal and opposite forces. So the net force on the dipole is zero.

${\overrightarrow{𝐹}}_{\text{net}}=0$

Only Statement-I and Statement-II are correct.

So the correct option is:

$\overline{)\text{Option B: I and II only}}$

According to Gauss's law, the net electric flux through a closed surface depends only on the total charge inside that surface.

It is given by $\Phi =\frac{{𝑄}_{\text{enclosed}}}{{𝜀}_0}$.

For surface ${𝑆}_1$, the charges inside it are $-2𝑄$ and $𝑄$.

So, the net charge enclosed by ${𝑆}_1$ is $-2𝑄+𝑄=-𝑄$.

Hence, the net flux through ${𝑆}_1$ is $\frac{-𝑄}{{𝜀}_0}$.

So, (a) matches with (ii).

For surface ${𝑆}_2$, the charges inside it are $𝑄$ and $-𝑄$.

So, the net charge enclosed by ${𝑆}_2$ is $𝑄+(-𝑄)=0$.

Hence, the net flux through ${𝑆}_2$ is zero.

So, (b) matches with (iii).

For surface ${𝑆}_3$, all three charges $-2𝑄$, $𝑄$, and $-𝑄$ are inside it.

So, the net charge enclosed by ${𝑆}_3$ is $-2𝑄+𝑄+(-𝑄)=-2𝑄$.

Hence, the net flux through ${𝑆}_3$ is $\frac{-2𝑄}{{𝜀}_0}$.

So, (c) matches with (i).

The correct matching is a - ii, b - iii, c - i.

According to Gauss’s law, the total electric flux through a closed surface is

$\Phi =\frac{{𝑞}_{\text{enclosed}}}{{𝜀}_0}$

Here, the charge $𝑄$ is placed at $𝐵$, which is the midpoint of an edge of the cube. So the charge is not fully inside this one cube.

To handle this situation, imagine placing three more identical cubes around the same edge. Then all four cubes together surround the charge symmetrically.

For this larger closed surface made of 4 cubes, the enclosed charge is $𝑄$. So the total flux through all four cubes together is

$\frac{𝑄}{{𝜀}_0}$

By symmetry, each cube gets an equal share of this flux. Since there are 4 identical cubes, the flux through the given cube is

${\Phi }_{\text{cube}}=\frac{1}{4}\times \frac{𝑄}{{𝜀}_0}$

${\Phi }_{\text{cube}}=\frac{𝑄}{4{𝜀}_0}$

The electron starts from rest, so its initial velocity is

$𝑢=0$

The electric force on the electron is

$𝐹=𝑒𝐸$

Using Newton’s second law,

$𝐹=𝑚𝑎$

so the acceleration is

$𝑎=\frac{𝑒𝐸}{{𝑚}_{𝑒}}$

Substitute the given values:

$𝑎=\frac{(1.6\times {10}^{-19})(2.0\times {10}^4)}{9.11\times {10}^{-31}}$

$𝑎=\frac{3.2\times {10}^{-15}}{9.11\times {10}^{-31}}$

$𝑎\approx 3.51\times {10}^{15}{\text{m/s}}^2$

The distance covered is

$𝑠=1.5\text{cm}=1.5\times {10}^{-2}\text{m}$

Since the electron starts from rest,

$𝑠=\frac{1}{2}𝑎{𝑡}^2$

So,

$𝑡=\sqrt{\surd \frac{2𝑠}{𝑎}}$

$𝑡=\sqrt{\surd \frac{2(1.5\times {10}^{-2})}{3.51\times {10}^{15}}}$

$𝑡=\sqrt{\surd 8.55\times {10}^{-18}}$

$𝑡\approx 2.9\times {10}^{-9}\text{s}$

So the correct option is

$\overline{)2.9\times {10}^{-9}\text{s}}$

Option A.

The correct option is:

$\overline{)\text{Option B: B observes both electric and magnetic fields}}$

The point charge is placed in the moving train.

For passenger A sitting inside the train, the charge is at rest. A stationary charge produces only an electric field around it. So A observes only electric field, not magnetic field.

For person B standing on the ground, the train is moving. So the charge is also moving with the train. A moving charge is equivalent to an electric current, and current produces a magnetic field. Since the charge itself also has electric field, B observes both electric and magnetic fields.

So, the correct statement is:

$\overline{)\text{B observes both electric and magnetic fields}}$

The $9𝜇\text{F}$ and $3𝜇\text{F}$ capacitors are connected in parallel. In parallel combination, capacitances are added directly.

${𝐶}_{𝑝}=9+3=12𝜇\text{F}$

Now this $12𝜇\text{F}$ equivalent capacitor is in series with the $4𝜇\text{F}$ capacitor. For two capacitors in series, the equivalent capacitance is:

${𝐶}_{𝑒𝑞}=\frac{4\times 12}{4+12}=\frac{48}{16}=3𝜇\text{F}$

The battery voltage is $12\text{V}$. So the total charge supplied by the battery is:

$𝑄={𝐶}_{𝑒𝑞}\times 𝑉=3𝜇\text{F}\times 12\text{V}=36𝜇\text{C}$

In a series combination, each capacitor has the same charge. So the charge on the $4𝜇\text{F}$ capacitor is also $36𝜇\text{C}$.

The potential difference across the $4𝜇\text{F}$ capacitor is:

${𝑉}_4=\frac{𝑄}{{𝐶}_4}=\frac{36𝜇\text{C}}{4𝜇\text{F}}=9\text{V}$

For a parallel plate capacitor, the energy stored is

$𝑈=\frac{1}{2}𝐶{𝑉}^2$

For a parallel plate capacitor in air or vacuum,

$𝐶=\frac{{𝜀}_0𝐴}{𝑑}$

Also, the electric field between the plates is uniform, so

$𝐸=\frac{𝑉}{𝑑}$

Hence,

$𝑉=𝐸𝑑$

Substituting these in the energy formula,

$𝑈=\frac{1}{2}\left(\frac{{𝜀}_0𝐴}{𝑑}\right)(𝐸𝑑{)}^2$

$𝑈=\frac{1}{2}\frac{{𝜀}_0𝐴}{𝑑}{𝐸}^2{𝑑}^2$

$𝑈=\frac{1}{2}{𝜀}_0{𝐸}^2𝐴𝑑$

So, the energy stored in the capacitor is

$\overline{)\frac{1}{2}{𝜀}_0{𝐸}^2𝐴𝑑}$

Correct option: C.

At any junction, Kirchhoff's Current Law says that the total current entering the junction is equal to the total current leaving it.

For the first junction, $5\text{ A}$ enters. The currents leaving are $1\text{ A}$ and ${𝐼}_1$.

So,

$5=1+{𝐼}_1$

$\Rightarrow {𝐼}_1=5-1=4\text{ A}$

For the second junction, ${𝐼}_1$ enters. The currents leaving are $1.5\text{ A}$ and ${𝐼}_2$.

So,

${𝐼}_1=1.5+{𝐼}_2$

Putting ${𝐼}_1=4\text{ A}$,

$4=1.5+{𝐼}_2$

$\Rightarrow {𝐼}_2=4-1.5=2.5\text{ A}$

For the third junction, ${𝐼}_2$ enters. The currents leaving are $1.5\text{ A}$ and ${𝐼}_3$.

So,

${𝐼}_2=1.5+{𝐼}_3$

Putting ${𝐼}_2=2.5\text{ A}$,

$2.5=1.5+{𝐼}_3$

$\Rightarrow {𝐼}_3=2.5-1.5=1\text{ A}$

The values of ${𝐼}_1,{𝐼}_2$ and ${𝐼}_3$ are $4\text{ A},2.5\text{ A}$ and $1\text{ A}$ respectively.

To verify Ohm's law, we have to measure two quantities for a resistor: the current passing through it and the potential difference across it.

An ammeter measures current. So it must be connected in series with the resistor, because the same current should pass through both the resistor and the ammeter.

A voltmeter measures potential difference. So it must be connected in parallel across the resistor, because it has to measure the potential difference between the two ends of the resistor.

The terminals of the instruments must also be connected correctly. The positive terminal of the ammeter and the positive terminal of the voltmeter should be connected towards the positive terminal of the battery, which is at higher potential.

In the given circuit diagrams, the longer line of the battery shows the positive terminal. It is on the left side. So current flows from the positive terminal of the battery through the circuit.

So, the correct circuit should have the ammeter in series with the resistor and the voltmeter in parallel across the resistor, with correct polarities.

Option (B) shows this arrangement correctly.

Electric current is taken in the direction of motion of positive charge.

For negative charge, current is opposite to its direction of motion.

Given:

$𝑒=1.6\times {10}^{-19}\text{C}$

Number of particles moving per second:

${10}^{19}$

Electrons have charge $-𝑒$ and move to the left.

So current due to electrons is towards the right:

${𝐼}_{𝑒}={10}^{19}\times 1.6\times {10}^{-19}=1.6\text{A}$

Protons have charge $+𝑒$ and move to the left.

So current due to protons is towards the left:

${𝐼}_{𝑝}=1.6\text{A}$

Alpha-particles have charge $+2𝑒$ and move to the right.

So current due to alpha-particles is towards the right:

${𝐼}_{𝛼}={10}^{19}\times 2\times 1.6\times {10}^{-19}$

${𝐼}_{𝛼}=3.2\text{A}$

Now take right direction as positive.

${𝐼}_{\text{net}}=1.6+3.2-1.6$

${𝐼}_{\text{net}}=3.2\text{A}$

So the resulting electric current is

$\overline{)3.2\text{A towards right}}$

Correct option: B.

The total charge flowing through the wire is given by

$𝑄=\int 𝐼𝑑𝑡$

This means charge is equal to the area under the current-time graph.

Here, the current decreases linearly from $10$ A to $0$ A in $4$ s. So the graph forms a triangle.

Area of triangle:

$𝑄=\frac{1}{2}\times \text{base}\times \text{height}$

Here,

$\text{base}=4\text{ s}$

and

$\text{height}=10\text{ A}$

So,

$𝑄=\frac{1}{2}\times 4\times 10$

$𝑄=20\text{ C}$

So, the total charge flowing through the wire is

$\overline{)20\text{ C}}$

Correct option: B.

Statement-I is false.

For a conductor, especially a metal, resistivity depends on temperature. When temperature increases, the ions in the metal vibrate more strongly. This makes it harder for free electrons to move, so the resistivity increases.

So, resistivity of a conductor is not independent of temperature.

Statement-II is true.

For a semiconductor, when temperature increases, more electrons get enough energy to move to the conduction band. This increases the number of charge carriers, so conductivity increases. Hence, resistivity decreases with increase in temperature.

So the correct option is:

$\overline{)\text{Option D: Statement I is false but Statement II is true}}$

Power of the lamp is

$𝑃=60\text{ W}$

Voltage is

$𝑉=120\text{ V}$

Using the relation

$𝑃=𝑉𝐼$

we get the current through the filament:

$𝐼=\frac{𝑃}{𝑉}$

$𝐼=\frac{60}{120}$

$𝐼=0.5\text{ A}$

Now, current means charge flowing per second.

So charge flowing per second is

$0.5\text{ C}$

Charge of one electron is

$𝑒=1.6\times {10}^{-19}\text{ C}$

Number of electrons flowing per second is

$𝑛=\frac{𝐼}{𝑒}$

$𝑛=\frac{0.5}{1.6\times {10}^{-19}}$

$𝑛=3.125\times {10}^{18}$

Nearly,

$𝑛\approx 3.1\times {10}^{18}$

So the correct option is

$\overline{)3.1\times {10}^{18}}$

Option C.

The wrong statements are II and III.

For magnetic substances, remember these NCERT-based facts:

• Diamagnetic substances have small negative susceptibility:

  $-1\le 𝜒<0$

  So statement I is correct.

• Paramagnetic substances have small positive susceptibility:

  $0<𝜒\ll 1$

  They do not have $𝜒\gg 1$.

  So statement II is wrong.

• Ferromagnetic substances have very large positive susceptibility:

  $𝜒\gg 1$

  They do not have $𝜒\ll 1$.

  So statement III is wrong.

• For ferromagnetic substances, relative permeability is also very large:

  ${𝜇}_{𝑟}\gg 1$

  So statement IV is correct.

Hence, the wrong statements are:

$\overline{)\text{II and III}}$

Correct option: Option C.

A paramagnetic substance is weakly attracted by a magnetic field.

So, if a paramagnetic bar is brought near a bar magnet, it will be attracted by whichever pole is near it. This happens because paramagnetic materials get weakly magnetised in the same direction as the applied magnetic field.

So it is attracted by both the North pole and the South pole of the bar magnet.

Correct option: Option A

Attracted by both the poles of the bar magnet.

The magnetic field at a point on the axis of a circular current loop is

$𝐵=\frac{{𝜇}_0𝐼{𝑅}^2}{2({𝑅}^2+{𝑥}^2{)}^{3/2}}$

Here, $𝑅$ is the radius of the loop and $𝑥$ is the distance of point P from the centre of the loop along its axis.

For both identical loops, point P is at the same distance from the centre. From the figure, this distance is $𝑥=\sqrt{\surd 3}𝑅$.

So, the magnetic field due to each loop has the same magnitude:

${𝐵}_1={𝐵}_2=\frac{{𝜇}_0𝐼{𝑅}^2}{2({𝑅}^2+(\sqrt{\surd 3}𝑅{)}^2{)}^{3/2}}$

${𝐵}_1={𝐵}_2=\frac{{𝜇}_0𝐼{𝑅}^2}{2({𝑅}^2+3{𝑅}^2{)}^{3/2}}$

${𝐵}_1={𝐵}_2=\frac{{𝜇}_0𝐼{𝑅}^2}{2(4{𝑅}^2{)}^{3/2}}$

Since $(4{𝑅}^2{)}^{3/2}=8{𝑅}^3$, we get

${𝐵}_1={𝐵}_2=\frac{{𝜇}_0𝐼{𝑅}^2}{2\left(8{𝑅}^3\right)}=\frac{{𝜇}_0𝐼}{16𝑅}$

The magnetic field due to a circular loop at an axial point is along the axis of the loop. So, if the axis of the first loop is along the x-axis, then

${\overrightarrow{𝐵}}_1=\frac{{𝜇}_0𝐼}{16𝑅}\widehat{𝑖}$

The axis of the second loop makes an angle of ${45}^{\circ }$ with the first loop's axis. Hence, its magnetic field also makes an angle of ${45}^{\circ }$ with the x-axis.

Resolving ${\overrightarrow{𝐵}}_2$ into x and y components:

${\overrightarrow{𝐵}}_2=\frac{{𝜇}_0𝐼}{16𝑅}(\cos {45}^{\circ }\widehat{𝑖}+\sin {45}^{\circ }\widehat{𝑗})$

Since $\cos {45}^{\circ }=\sin {45}^{\circ }=\frac{1}{\sqrt{\surd 2}}$,

${\overrightarrow{𝐵}}_2=\frac{{𝜇}_0𝐼}{16𝑅}(\frac{1}{\sqrt{\surd 2}}\widehat{𝑖}+\frac{1}{\sqrt{\surd 2}}\widehat{𝑗})$

The resultant magnetic field at P is the vector sum of the two fields:

${\overrightarrow{𝐵}}_{𝑛𝑒𝑡}={\overrightarrow{𝐵}}_1+{\overrightarrow{𝐵}}_2$

${\overrightarrow{𝐵}}_{𝑛𝑒𝑡}=\frac{{𝜇}_0𝐼}{16𝑅}\widehat{𝑖}+\frac{{𝜇}_0𝐼}{16𝑅}(\frac{1}{\sqrt{\surd 2}}\widehat{𝑖}+\frac{1}{\sqrt{\surd 2}}\widehat{𝑗})$

${\overrightarrow{𝐵}}_{𝑛𝑒𝑡}=\frac{{𝜇}_0𝐼}{16𝑅}((1+\frac{1}{\sqrt{\surd 2}})\widehat{𝑖}+\frac{1}{\sqrt{\surd 2}}\widehat{𝑗})$

${\overrightarrow{𝐵}}_{𝑛𝑒𝑡}=\frac{{𝜇}_0𝐼}{16𝑅}(\frac{\sqrt{\surd 2}+1}{\sqrt{\surd 2}}\widehat{𝑖}+\frac{1}{\sqrt{\surd 2}}\widehat{𝑗})$

${\overrightarrow{𝐵}}_{𝑛𝑒𝑡}=\frac{{𝜇}_0𝐼}{16\sqrt{\surd 2}𝑅}\left[\right(\sqrt{\surd 2}+1)\widehat{𝑖}+\widehat{𝑗}]$