MHT CET 2026 17th April Morning Shift (Physics & Chemistry) Question Paper with Solutions

Let the angle between $\overrightarrow{𝑃}$ and $\overrightarrow{𝑄}$ be $𝜃$.

The magnitude of their resultant is $𝐴$:

${𝐴}^2={𝑃}^2+{𝑄}^2+2𝑃𝑄\cos 𝜃$

When $\overrightarrow{𝑄}$ is reversed, the resultant becomes $\overrightarrow{𝑃}-\overrightarrow{𝑄}$, with magnitude $𝐵$:

${𝐵}^2={𝑃}^2+{𝑄}^2-2𝑃𝑄\cos 𝜃$

Adding these equations,

${𝐴}^2+{𝐵}^2=2{𝑃}^2+2{𝑄}^2$

Hence,

$\overline{){𝐴}^2+{𝐵}^2=2({𝑃}^2+{𝑄}^2)}$

So, the correct answer is Option C.

Density is given by

$𝜌=\frac{𝑚}{𝑉}$

Substituting the measured values,

$𝜌=\frac{6.00}{2.00}=3.00{\text{kg/m}}^3$

For division, the maximum fractional errors are added:

$\frac{\Delta 𝜌}{𝜌}=\frac{\Delta 𝑚}{𝑚}+\frac{\Delta 𝑉}{𝑉}$

$\frac{\Delta 𝜌}{𝜌}=\frac{0.03}{6.00}+\frac{0.02}{2.00}=0.005+0.010=0.015$

Hence, the absolute error in density is

$\Delta 𝜌=3.00\times 0.015=0.045{\text{kg/m}}^3$

Thus,

$\overline{)𝜌=(3.00\pm 0.045){\text{kg/m}}^3}$

So, Option D is correct.

For a projectile, the maximum height is

$𝐻=\frac{{𝑢}^2{\sin }^2𝜃}{2𝑔}$

Since both balls reach the same maximum height,

${𝑢}_2^{𝐴}{\sin }^2{45}^{\circ }={𝑢}_2^{𝐵}{\sin }^2{60}^{\circ }$

Taking the positive square root,

${𝑢}_{𝐴}\sin {45}^{\circ }={𝑢}_{𝐵}\sin {60}^{\circ }$

Thus,

$\frac{{𝑢}_{𝐴}}{{𝑢}_{𝐵}}=\frac{\sin {60}^{\circ }}{\sin {45}^{\circ }}=\frac{\frac{\sqrt{\surd 3}}{2}}{\frac{1}{\sqrt{\surd 2}}}=\frac{\sqrt{\surd 3}}{\sqrt{\surd 2}}$

Hence,

$\overline{){𝑢}_{𝐴}:{𝑢}_{𝐵}=\sqrt{\surd 3}:\sqrt{\surd 2}}$

So, Option D is correct.

Let the mass of each sphere be $𝑚$.

Take the center of sphere A as the origin. So, the position of A on the straight line is ${𝑥}_{𝐴}=0$.

The centers of the three spheres lie on the same straight line. So their positions along this line are:

${𝑥}_{𝐴}=0$

${𝑥}_{𝐵}=𝐴𝐵$

${𝑥}_{𝐶}=𝐴𝐶$

For point masses lying on a straight line, the position of the center of mass is:

${𝑥}_{𝑐𝑚}=\frac{𝑚{𝑥}_{𝐴}+𝑚{𝑥}_{𝐵}+𝑚{𝑥}_{𝐶}}{𝑚+𝑚+𝑚}$

Now substitute the positions of A, B, and C:

${𝑥}_{𝑐𝑚}=\frac{𝑚\left(0\right)+𝑚\left(𝐴𝐵\right)+𝑚\left(𝐴𝐶\right)}{3𝑚}$

Since $𝑚$ is common in the numerator and denominator, it cancels out:

${𝑥}_{𝑐𝑚}=\frac{𝐴𝐵+𝐴𝐶}{3}$

So, the distance of the center of mass of the system from A is $\frac{𝐴𝐵+𝐴𝐶}{3}$.

For a point mass, the moment of inertia about an axis is

$𝐼=𝑚{𝑟}^2$

where $𝑟$ is its perpendicular distance from the axis through $𝐷$.

In the square:

• At $𝐴$: $𝑟=𝐷𝐴=𝑏$

• At $𝐵$: $𝑟=𝐷𝐵=\sqrt{\surd 2}𝑏$ (diagonal)

• At $𝐶$: $𝑟=𝐷𝐶=𝑏$

• At $𝐷$: $𝑟=0$

Hence,

$\begin{array}{cc}𝐼& =𝑚(𝑏{)}^2+2𝑚(\sqrt{\surd 2}𝑏{)}^2+3𝑚(𝑏{)}^2+4𝑚(0{)}^2\\ & =𝑚{𝑏}^2+4𝑚{𝑏}^2+3𝑚{𝑏}^2\\ & =8𝑚{𝑏}^2\end{array}$

Thus, the correct answer is

$\overline{)8𝑚{𝑏}^2}$

Option C.

The system has $5$ solid spheres. Each sphere has mass $𝑚$ and radius $𝑟$.

From the figure, the axis AA' passes through the centres of $3$ spheres. For a solid sphere, the moment of inertia about an axis through its centre is

${𝐼}_{𝑐𝑚}=\frac{2}{5}𝑚{𝑟}^2$

So, the total moment of inertia of these $3$ spheres is

${𝐼}_1=3\times \frac{2}{5}𝑚{𝑟}^2=\frac{6}{5}𝑚{𝑟}^2$

The other $2$ spheres do not have the axis passing through their centres. The axis AA' passes through their point of contact. So, the centre of each of these spheres is at a distance $𝑟$ from the axis.

Using the parallel axis theorem, the moment of inertia of one such sphere about AA' is

$𝐼={𝐼}_{𝑐𝑚}+𝑚{𝑑}^2$

Here, $𝑑=𝑟$. So,

$𝐼=\frac{2}{5}𝑚{𝑟}^2+𝑚{𝑟}^2=\frac{7}{5}𝑚{𝑟}^2$

For $2$ such spheres, the total moment of inertia is

${𝐼}_2=2\times \frac{7}{5}𝑚{𝑟}^2=\frac{14}{5}𝑚{𝑟}^2$

Now add the moment of inertia of all the spheres:

$𝐼={𝐼}_1+{𝐼}_2$

$𝐼=\frac{6}{5}𝑚{𝑟}^2+\frac{14}{5}𝑚{𝑟}^2$

$𝐼=\frac{20}{5}𝑚{𝑟}^2=4𝑚{𝑟}^2$

The rotational kinetic energy in terms of angular momentum is

$𝐾=\frac{{𝐿}^2}{2𝐼}$

Since both bodies have the same rotational kinetic energy,

$\frac{{𝐿}_2^{𝐴}}{2{𝐼}_{𝐴}}=\frac{{𝐿}_2^{𝐵}}{2{𝐼}_{𝐵}}$

Cancelling $2$ and rearranging,

$\frac{{𝐿}_2^{𝐴}}{{𝐿}_2^{𝐵}}=\frac{{𝐼}_{𝐴}}{{𝐼}_{𝐵}}$

Taking the square root,

$\overline{)\frac{{𝐿}_{𝐴}}{{𝐿}_{𝐵}}=\sqrt{\surd \frac{{𝐼}_{𝐴}}{{𝐼}_{𝐵}}}}$

Hence, Option C is correct.

For a person standing at the equator, the centrifugal force due to Earth’s rotation acts opposite to gravity.

If Earth rotates with angular speed $𝜔$, the apparent weight is

${𝑊}^{\prime }=𝑚𝑔-𝑚{𝜔}^2𝑅$

The required apparent weight is $\frac{3}{5}$ of the present weight:

$𝑚𝑔-𝑚{𝜔}^2𝑅=\frac{3}{5}𝑚𝑔$

Cancelling $𝑚$ and rearranging,

${𝜔}^2𝑅=𝑔-\frac{3𝑔}{5}=\frac{2𝑔}{5}$

Thus,

$𝜔=\sqrt{\surd \frac{2𝑔}{5𝑅}}$

Hence, the correct answer is Option B.

For a spherical water drop, the excess pressure is

$\Delta 𝑃=\frac{2𝑇}{𝑟}$

where $𝑇$ is the surface tension and $𝑟$ is the radius.

Given that the excess pressure inside drop A is four times that inside drop B,

$\frac{\Delta {𝑃}_{𝐴}}{\Delta {𝑃}_{𝐵}}=4$

Since both drops are of water, their surface tension is the same. Thus,

$\frac{\Delta {𝑃}_{𝐴}}{\Delta {𝑃}_{𝐵}}=\frac{{𝑟}_{𝐵}}{{𝑟}_{𝐴}}=4$

So,

$\frac{{𝑟}_{𝐴}}{{𝑟}_{𝐵}}=\frac{1}{4}$

The mass of a water drop is proportional to its volume:

$𝑚\propto {𝑟}^3$

Hence,

$\frac{{𝑚}_{𝐴}}{{𝑚}_{𝐵}}={\left(\frac{{𝑟}_{𝐴}}{{𝑟}_{𝐵}}\right)}^3={\left(\frac{1}{4}\right)}^3=\frac{1}{64}$

Thus, the required ratio is

$\overline{){𝑚}_{𝐴}:{𝑚}_{𝐵}=1:64}$

Option D.

For capillary rise,

$ℎ=\frac{2𝑇\cos 𝜃}{𝜌𝑔𝑟}$

For the same liquid and tube material, $𝑇$, $𝜃$, $𝜌$, and $𝑔$ remain constant. Hence,

$ℎ\propto \frac{1}{𝑟}$

The cross-sectional area is

$𝑎=𝜋{𝑟}^2$

If the area becomes $4𝑎$,

$𝜋{𝑅}^2=4𝜋{𝑟}^2$

so,

$𝑅=2𝑟$

Since capillary rise is inversely proportional to radius,

$𝐻=ℎ\frac{𝑟}{𝑅}=ℎ\frac{𝑟}{2𝑟}=\frac{ℎ}{2}$

Thus, water will rise to a height

$\overline{)\frac{ℎ}{2}}$

Option B.

A soap bubble has two surfaces, one inner and one outer. Hence, the increase in surface energy is

$𝑊=2𝑇\Delta 𝐴$

For a spherical surface,

$\Delta 𝐴=4𝜋({𝑟}_2^2-{𝑟}_2^1)$

So,

$𝑊=2𝑇\cdot 4𝜋({𝑟}_2^2-{𝑟}_2^1)$

Given

$𝑇=25\text{dyne/cm},{𝑟}_1=\frac{1}{\sqrt{\surd 𝜋}}\text{cm},{𝑟}_2=\frac{3}{\sqrt{\surd 𝜋}}\text{cm}$

Thus,

${𝑟}_2^2-{𝑟}_2^1=\frac{9}{𝜋}-\frac{1}{𝜋}=\frac{8}{𝜋}$

Therefore,

$𝑊=8𝜋\times 25\times \frac{8}{𝜋}=1600\text{dyne-cm}$

Since $1\text{dyne-cm}=1\text{erg}$,

$\overline{)𝑊=1600\text{erg}}$

Hence, Option C is correct.

According to Stefan’s law of radiation, the cooling rate of a sphere is proportional to

${𝑇}_4^{𝑠}-{𝑇}_4^0$

where ${𝑇}_{𝑠}$ is the temperature of the sphere and ${𝑇}_0$ is the temperature of the surroundings.

Initially, the sphere is at $3𝑇$ and the surroundings are at $𝑇$:

${𝑅}_1\propto (3𝑇{)}^4-{𝑇}^4=81{𝑇}^4-{𝑇}^4=80{𝑇}^4$

When the sphere cools to $2𝑇$:

${𝑅}_2\propto (2𝑇{)}^4-{𝑇}^4=16{𝑇}^4-{𝑇}^4=15{𝑇}^4$

Hence,

$\frac{{𝑅}_2}{{𝑅}_1}=\frac{15}{80}=\frac{3}{16}$

So,

$\overline{){𝑅}_2=\frac{3}{16}{𝑅}_1}$

Option D.

The incident heat is $𝑥$ joule.

The reflected and transmitted heat together is $𝑦$ joule. Hence, the heat absorbed by the body is

${𝑄}_{\text{absorbed}}=𝑥-𝑦$

The absorption coefficient is defined as

$𝑎=\frac{\text{heat absorbed}}{\text{heat incident}}$

So,

$𝑎=\frac{𝑥-𝑦}{𝑥}$

Hence, the correct answer is Option B, $\overline{)\frac{𝑥-𝑦}{𝑥}}$.

The rate of heat conduction through a rod is given by Fourier’s law:

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

Both rods have the same length $𝐿$ and the same temperature difference $({𝑇}_1-{𝑇}_2)$. Since their rates of heat loss are equal,

$\frac{{𝐾}_1{𝐴}_1({𝑇}_1-{𝑇}_2)}{𝐿}=\frac{{𝐾}_2{𝐴}_2({𝑇}_1-{𝑇}_2)}{𝐿}$

Cancelling the common terms,

${𝐴}_1{𝐾}_1={𝐴}_2{𝐾}_2$

Specific heat does not affect the steady rate of heat conduction.

Hence, the correct answer is Option B.

For a Carnot engine, efficiency is

$𝜂=1-\frac{{𝑇}_2}{{𝑇}_1}$

Initially, $𝜂=0.40$ and ${𝑇}_1=600\text{K}$.

$0.40=1-\frac{{𝑇}_2}{600}$

${𝑇}_2=360\text{K}$

For the same exhaust temperature and new efficiency $𝜂=0.60$:

$0.60=1-\frac{360}{{𝑇}_{\prime }^1}$

$\frac{360}{{𝑇}_{\prime }^1}=0.40$

${𝑇}_{\prime }^1=\frac{360}{0.40}=900\text{K}$

Hence, the required intake temperature is

$\overline{)900\text{K}}$

Option B

For nitrogen gas at room temperature, ${\text{N}}_2$ behaves as a diatomic gas. Its molar heat capacity at constant pressure is

${𝐶}_{𝑃}=\frac{7}{2}𝑅$

Number of moles of nitrogen:

$𝑛=\frac{14}{28}=\frac{1}{2}\text{ mol}$

The temperature change is

$\Delta 𝑇={48}^{\circ }\text{C}=48\text{ K}$

At constant pressure, the required heat is

$𝑄=𝑛{𝐶}_{𝑃}\Delta 𝑇$

$𝑄=\frac{1}{2}\times \frac{7}{2}𝑅\times 48=84𝑅$

Hence, the correct answer is

$\overline{)84𝑅}$

Option B.

The average speed is

${𝑣}_{\text{avg}}=\frac{1+3+5+7+2+4+6}{7}=\frac{28}{7}=4\text{km/s}$

The r.m.s. speed is

${𝑣}_{\text{rms}}=\sqrt{\surd \frac{1^2+3^2+5^2+7^2+2^2+4^2+6^2}{7}}$

$=\sqrt{\surd \frac{140}{7}}=\sqrt{\surd 20}=2\sqrt{\surd 5}\text{km/s}$

Hence,

${𝑣}_{\text{rms}}:{𝑣}_{\text{avg}}=2\sqrt{\surd 5}:4=\sqrt{\surd 5}:2$

So, the correct answer is Option D: $\sqrt{\surd 5}:2$.

For a simple pendulum, the time period is

$𝑇=2𝜋\sqrt{\surd \frac{𝐿}{𝑔}}$

Squaring,

${𝑇}^2=\frac{4{𝜋}^2𝐿}{𝑔}$

Thus, for the two pendulums,

${𝑇}_2^1=\frac{4{𝜋}^2{𝐿}_1}{𝑔},{𝑇}_2^2=\frac{4{𝜋}^2{𝐿}_2}{𝑔}$

Subtracting,

${𝑇}_2^1-{𝑇}_2^2=\frac{4{𝜋}^2({𝐿}_1-{𝐿}_2)}{𝑔}$

If $𝑇$ is the time period of the pendulum of length $({𝐿}_1-{𝐿}_2)$, then

${𝑇}^2=\frac{4{𝜋}^2({𝐿}_1-{𝐿}_2)}{𝑔}={𝑇}_2^1-{𝑇}_2^2$

Hence,

$\overline{)𝑇=\sqrt{\surd {𝑇}_2^1-{𝑇}_2^2}}$

So, the correct answer is Option B.

For a particle in S.H.M., potential energy at displacement $𝑥$ is

$𝑈=\frac{1}{2}𝑘{𝑥}^2$

Its kinetic energy is

$𝐾=\frac{1}{2}𝑘({𝐴}^2-{𝑥}^2)$

Given that the potential energy is $8$ times the kinetic energy,

$\frac{1}{2}𝑘{𝑥}^2=8\left[\frac{1}{2}𝑘\right({𝐴}^2-{𝑥}^2\left)\right]$

Cancelling the common terms,

${𝑥}^2=8({𝐴}^2-{𝑥}^2)$

$9{𝑥}^2=8{𝐴}^2$

$\left|𝑥\right|=\frac{2\sqrt{\surd 2}}{3}𝐴$

Thus, the particle can be at either side of the mean position:

$𝑥=\pm \frac{2\sqrt{\surd 2}}{3}𝐴$

The required magnitude is

$\overline{)\frac{2\sqrt{\surd 2}}{3}𝐴}$

Hence, Option C is correct.

For a particle in simple harmonic motion, speed at displacement $𝑥$ is

$𝑣=𝜔\sqrt{\surd {𝐴}^2-{𝑥}^2}$

The maximum speed occurs at the mean position:

${𝑣}_{max}=𝜔𝐴$

Given that the speed is one-fourth of the maximum speed,

$𝜔\sqrt{\surd {𝐴}^2-{𝑥}^2}=\frac{𝜔𝐴}{4}$

Cancelling $𝜔$ and squaring,

${𝐴}^2-{𝑥}^2=\frac{{𝐴}^2}{16}$

${𝑥}^2=\frac{15{𝐴}^2}{16}$

Hence, the magnitude of displacement is

$\left|𝑥\right|=\frac{𝐴\sqrt{\surd 15}}{4}$

So, the correct answer is Option D:

$\overline{)\frac{𝐴\sqrt{\surd 15}}{4}}$

The third harmonic of wire A is

${𝑓}_{𝐴}=3\times 256=768\text{Hz}$

The second harmonic of wire B is

${𝑓}_{𝐵}=2\times 382=764\text{Hz}$

The beat frequency is the difference between the two frequencies:

${𝑓}_{\text{beat}}=|768-764|=4\text{Hz}$

So, the number of beats heard in $2$ seconds is

$𝑁=4\times 2=8$

Hence, the correct answer is Option A: $8$.

The correct option is D.

For a stretched string fixed at both ends, the allowed frequencies are

${𝑓}_{𝑛}=𝑛\frac{𝑣}{2𝐿},𝑛=1,2,3,\dots$

The fundamental frequency is the first harmonic:

${𝑓}_1=\frac{𝑣}{2𝐿}$

The first overtone is the next allowed frequency, which is the second harmonic:

${𝑓}_2=2{𝑓}_1$

So, the frequency of the first overtone is the same as that of the second harmonic.

The other options are incorrect because:

• Consecutive antinodes are separated by $\frac{𝜆}{2}$, not $𝜆$.

• A pipe open at both ends contains both odd and even harmonics.

• A pipe closed at one end contains only odd harmonics.

Hence, Option D is correct.

The correct answer is Option B: Saxophone.

A saxophone is a wind instrument because sound is produced by blowing air through it. Drum, daphali, and xylophone are percussion instruments.

For a gas, the speed of sound is

$𝑣=\sqrt{\surd \frac{𝛾𝑃}{𝜌}}$

Here, the box is closed and has rigid walls, so the volume and mass of air remain constant. Hence, the density $𝜌$ remains constant. Also, $𝛾$ remains unchanged.

Thus,

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

Given that sound travels $1.4$ times faster,

$1.4=\sqrt{\surd \frac{{𝑃}_2}{{𝑃}_1}}$

Squaring both sides,

$\frac{{𝑃}_2}{{𝑃}_1}=(1.4{)}^2=1.96$

Hence, the correct answer is $\overline{)1.96}$, which is Option B.

The wrong statement is Option C: Polar molecules have symmetrical shape.

Polar molecules have an unequal distribution of positive and negative charges. Hence, the centres of positive and negative charges do not coincide, producing a permanent dipole moment. They behave like tiny electric dipoles.

A symmetrical molecule generally has its bond dipole moments cancelling each other, so it is usually non-polar.

Thus, the correct choice is:

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

For an electric dipole in a uniform electric field, the torque is

$𝜏=𝑝𝐸\sin 𝜃$

Here, dipole moment $𝑝=𝑞𝑙$, where $𝑙$ is the separation between the charges.

Given:

$𝑙=3\text{ cm}=3\times {10}^{-2}\text{ m}$

$𝐸=5\times {10}^4\text{ N/C},𝜃={60}^{\circ }$

$𝜏=9\sqrt{\surd 3}\text{ Nm}$

Substituting in $𝜏=𝑞𝑙𝐸\sin 𝜃$:

$9\sqrt{\surd 3}=𝑞(3\times {10}^{-2})(5\times {10}^4)\left(\frac{\sqrt{\surd 3}}{2}\right)$

Cancelling $\sqrt{\surd 3}$:

$9=𝑞\times \frac{3\times 5\times {10}^2}{2}$

$𝑞=\frac{18}{1500}=1.2\times {10}^{-2}\text{ C}$

Hence, the correct answer is

$\overline{)1.2\times {10}^{-2}\text{ C}}$

Option B

From the circuit diagram, ${𝐶}_1$ and ${𝐶}_2$ are connected in parallel because they are connected between the same two points.

For capacitors in parallel, capacitances add directly. So their equivalent capacitance is

${𝐶}_{12}={𝐶}_1+{𝐶}_2$

Now this equivalent capacitor ${𝐶}_{12}$ is connected in series with ${𝐶}_3$.

For capacitors in series, the charge on each capacitor is the same. The equivalent capacitance of ${𝐶}_{12}$ and ${𝐶}_3$ is

${𝐶}_{𝑒𝑞}=\frac{{𝐶}_{12}{𝐶}_3}{{𝐶}_{12}+{𝐶}_3}=\frac{({𝐶}_1+{𝐶}_2){𝐶}_3}{{𝐶}_1+{𝐶}_2+{𝐶}_3}$

The charge supplied by the voltage source is

$𝑄={𝐶}_{𝑒𝑞}𝑉=\frac{({𝐶}_1+{𝐶}_2){𝐶}_3𝑉}{{𝐶}_1+{𝐶}_2+{𝐶}_3}$

The same charge appears on ${𝐶}_3$ because it is in series with the equivalent capacitor ${𝐶}_{12}$.

So the voltage across ${𝐶}_3$ is

${𝑉}_3=\frac{𝑄}{{𝐶}_3}$

Putting the value of $𝑄$, we get

${𝑉}_3=\frac{1}{{𝐶}_3}\times \frac{({𝐶}_1+{𝐶}_2){𝐶}_3𝑉}{{𝐶}_1+{𝐶}_2+{𝐶}_3}$

${𝑉}_3=\frac{({𝐶}_1+{𝐶}_2)𝑉}{{𝐶}_1+{𝐶}_2+{𝐶}_3}$

Let

${𝐶}_1=𝐶,{𝐶}_2=2𝐶$

For the series combination, equivalent capacitance is

${𝐶}_{𝑠}=\frac{{𝐶}_1{𝐶}_2}{{𝐶}_1+{𝐶}_2}=\frac{𝐶\left(2𝐶\right)}{𝐶+2𝐶}=\frac{2𝐶}{3}$

For the parallel combination,

${𝐶}_{𝑝}={𝐶}_1+{𝐶}_2=3𝐶$

Energy stored in a capacitor combination is

$𝑈=\frac{1}{2}{𝐶}_{\text{eq}}{𝑉}^2$

Since the energies in both cases are equal,

$\frac{1}{2}{𝐶}_{𝑠}{𝑉}_2^{𝑠}=\frac{1}{2}{𝐶}_{𝑝}{𝑉}_2^{𝑝}$

Substituting the equivalent capacitances,

$\frac{2𝐶}{3}{𝑉}_2^{𝑠}=3𝐶{𝑉}_2^{𝑝}$

$\frac{{𝑉}_2^{𝑠}}{{𝑉}_2^{𝑝}}=\frac{9}{2}$

Hence,

$\frac{{𝑉}_{𝑠}}{{𝑉}_{𝑝}}=\frac{3}{\sqrt{\surd 2}}$

Thus,

$\overline{){𝑉}_{𝑠}:{𝑉}_{𝑝}=3:\sqrt{\surd 2}}$

Option B.

Let the three junctions from left to right be ${𝐽}_1$, ${𝐽}_2$, and ${𝐽}_3$.

We will use Kirchhoff's Current Law (KCL). It says that at any junction, total current entering is equal to total current leaving.

At junction ${𝐽}_1$, currents $4.5$ A and $1.7$ A are entering. Currents $2.3$ A and ${𝐼}_1$ are leaving.

$4.5+1.7=2.3+{𝐼}_1$

${𝐼}_1=6.2-2.3=3.9\text{ A}$

So, the current flowing from ${𝐽}_1$ to ${𝐽}_2$ is $3.9$ A.

At junction ${𝐽}_2$, currents ${𝐼}_1$, $3.2$ A, and $4.9$ A are entering. Currents $2.6$ A and ${𝐼}_2$ are leaving.

${𝐼}_1+3.2+4.9=2.6+{𝐼}_2$

$3.9+3.2+4.9=2.6+{𝐼}_2$

$12.0=2.6+{𝐼}_2$

${𝐼}_2=9.4\text{ A}$

So, the current flowing from ${𝐽}_2$ to ${𝐽}_3$ is $9.4$ A.

At junction ${𝐽}_3$, current ${𝐼}_2$ is entering. Currents $1.9$ A and $𝐼$ are leaving.

${𝐼}_2=1.9+𝐼$

$9.4=1.9+𝐼$

$𝐼=7.5\text{ A}$

So, the required current is $7.5$ A.

For a galvanometer, the full-scale deflection current remains the same when its voltage range is increased.

For the range $0$–$15\text{V}$, the total resistance is

${𝑅}_1=𝑋+100$

To double the range to $0$–$30\text{V}$, an additional resistance of $1500\Omega$ is connected in series. The new total resistance is

${𝑅}_2=𝑋+100+1500=𝑋+1600$

Since the voltage range is doubled while the current remains unchanged, the total resistance must also double:

$𝑋+1600=2(𝑋+100)$

$𝑋+1600=2𝑋+200$

$𝑋=1400\Omega$

Hence, the correct answer is Option C: $1400\Omega$.

For a circular coil of $𝑁$ turns, the magnetic field at its centre is

$𝐵=\frac{{𝜇}_0𝑁𝐼}{2𝑅}$

Initially, the wire forms one turn of radius $𝑅$. Its length is

$𝐿=2𝜋𝑅$

When the same wire forms two turns, let the radius of each turn be $𝑟$. Then

$𝐿=2\left(2𝜋𝑟\right)=4𝜋𝑟$

Since the wire length remains the same,

$2𝜋𝑅=4𝜋𝑟$

Hence,

$𝑟=\frac{𝑅}{2}$

The new magnetic field is

${𝐵}^{\prime }=\frac{{𝜇}_0\left(2\right)𝐼}{2(𝑅/2)}=\frac{2{𝜇}_0𝐼}{𝑅}$

The original field was

$𝐵=\frac{{𝜇}_0𝐼}{2𝑅}$

Thus,

$\frac{{𝐵}^{\prime }}{𝐵}=4$

So,

$\overline{){𝐵}^{\prime }=4𝐵}$

Hence, Option A: $4𝐵$.

For a long straight wire carrying current $𝐼$, the magnetic field at a perpendicular distance $𝑟$ is

$𝐵=\frac{{𝜇}_0𝐼}{2𝜋𝑟}$

Thus, the magnetic field is inversely proportional to the distance:

$𝐵\propto \frac{1}{𝑟}$

At distance $2𝑟$,

${𝐵}^{\prime }=\frac{{𝜇}_0𝐼}{2𝜋\left(2𝑟\right)}=\frac{1}{2}\left(\frac{{𝜇}_0𝐼}{2𝜋𝑟}\right)=\frac{𝐵}{2}$

Hence, the correct answer is Option B: $\frac{𝐵}{2}$.

Given inductances are ${𝐿}_1=2$ H, ${𝐿}_2=3$ H, and ${𝐿}_3=6$ H.

We need the circuit whose equivalent inductance between A and B is $\frac{18}{11}$ H.

In circuit (P), all three inductors are connected in series. For inductors in series, we add their inductances:

${𝐿}_{𝑒𝑞}={𝐿}_1+{𝐿}_2+{𝐿}_3=2+3+6=11$ H.

So circuit (P) does not give $\frac{18}{11}$ H.

In circuit (Q), all three inductors are connected in parallel. For inductors in parallel:

$\frac{1}{{𝐿}_{𝑒𝑞}}=\frac{1}{{𝐿}_1}+\frac{1}{{𝐿}_2}+\frac{1}{{𝐿}_3}$

$\frac{1}{{𝐿}_{𝑒𝑞}}=\frac{1}{2}+\frac{1}{3}+\frac{1}{6}=\frac{3+2+1}{6}=1$

So, ${𝐿}_{𝑒𝑞}=1$ H.

Thus, circuit (Q) also does not give $\frac{18}{11}$ H.

In circuit (R), ${𝐿}_1$ and ${𝐿}_2$ are in series. Their combination is in parallel with ${𝐿}_3$.

First, the series combination is:

${𝐿}_1+{𝐿}_2=2+3=5$ H.

Now this $5$ H combination is in parallel with $6$ H:

${𝐿}_{𝑒𝑞}=\frac{({𝐿}_1+{𝐿}_2){𝐿}_3}{({𝐿}_1+{𝐿}_2)+{𝐿}_3}$

${𝐿}_{𝑒𝑞}=\frac{(2+3)\times 6}{(2+3)+6}=\frac{30}{11}$ H.

So circuit (R) does not give $\frac{18}{11}$ H.

In circuit (S), ${𝐿}_2$ and ${𝐿}_3$ are in series. Their combination is in parallel with ${𝐿}_1$.

First, the series combination is:

${𝐿}_2+{𝐿}_3=3+6=9$ H.

Now this $9$ H combination is in parallel with ${𝐿}_1=2$ H:

${𝐿}_{𝑒𝑞}=\frac{{𝐿}_1({𝐿}_2+{𝐿}_3)}{{𝐿}_1+({𝐿}_2+{𝐿}_3)}$

${𝐿}_{𝑒𝑞}=\frac{2\times (3+6)}{2+(3+6)}=\frac{18}{11}$ H.

Circuit (S) gives the required effective inductance.

For an inductor, the instantaneous power supplied is

$𝑃=𝐿𝑖\frac{𝑑𝑖}{𝑑𝑡}$

Since the power is the same for both coils and their currents increase at the same constant rate,

${𝐿}_1{𝑖}_1\frac{𝑑𝑖}{𝑑𝑡}={𝐿}_2{𝑖}_2\frac{𝑑𝑖}{𝑑𝑡}$

Hence,

${𝐿}_1{𝑖}_1={𝐿}_2{𝑖}_2$

or

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

The energy stored in an inductor is

$𝑈=\frac{1}{2}𝐿{𝑖}^2$

Thus,

$\frac{{𝑈}_1}{{𝑈}_2}=\frac{{𝐿}_1{𝑖}_2^1}{{𝐿}_2{𝑖}_2^2}=\frac{{𝐿}_1}{{𝐿}_2}{\left(\frac{{𝐿}_2}{{𝐿}_1}\right)}^2=\frac{{𝐿}_2}{{𝐿}_1}$

So, the correct answer is

$\overline{)\frac{{𝐿}_2}{{𝐿}_1}}$

Option D.