MHT CET 2025 25th April Morning Shift (Physics & Chemistry) Question Paper with Solutions

We know,

$𝑔=\frac{𝐺𝑀}{{𝑅}^2}$

If the density is $𝜌$, then

$\begin{array}{c}\Rightarrow M=\frac{4}{3}𝜋R^3𝜌\dots (\because M=V\times d)\\ \therefore g=\frac{4}{3}𝜋\mathrm{GR}𝜌\end{array}$

From given,

$\begin{array}{cc}{g}_A& =\frac{4}{3}𝜋{\mathrm{Gr}}_1{𝜌}_1\\ g_B& =\frac{4}{3}𝜋{\mathrm{Gr}}_2{𝜌}_2\end{array}$

Therefore,

$\frac{{𝑔}_{𝐴}}{{𝑔}_{𝐵}}=\frac{{𝑟}_1}{{𝑟}_2}\times \frac{{𝜌}_1}{{𝜌}_2}\Rightarrow \frac{{𝑔}_{𝐴}}{{𝑔}_{𝐵}}=\frac{{𝑟}_1{𝜌}_1}{{𝑟}_2{𝜌}_2}$

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

$𝑅=𝜎𝐴{𝑇}^4$

where

• $𝑅$ = rate of radiation

• $𝜎$ = Stefan-Boltzmann constant

• $𝐴$ = surface area

• $𝑇$ = absolute temperature in kelvin

Initially,

${𝑅}_1=𝐸$

Initial temperature:

${127}^{\circ }\text{C}=127+273=400\text{ K}$

Let initial area be $𝐴$.

So,

$𝐸=𝜎𝐴(400{)}^4$

Now temperature is increased by ${400}^{\circ }\text{C}$:

${127}^{\circ }\text{C}+{400}^{\circ }\text{C}={527}^{\circ }\text{C}$

In kelvin:

$527+273=800\text{ K}$

New area is half of the initial area:

${𝐴}_2=\frac{𝐴}{2}$

So new rate of radiation:

${𝑅}_2=𝜎\left(\frac{𝐴}{2}\right)(800{)}^4$

Now compare with initial rate:

$\frac{{𝑅}_2}{𝐸}=\frac{𝜎\left(\frac{𝐴}{2}\right)(800{)}^4}{𝜎𝐴(400{)}^4}$

$\frac{{𝑅}_2}{𝐸}=\frac{1}{2}{\left(\frac{800}{400}\right)}^4$

$\frac{{𝑅}_2}{𝐸}=\frac{1}{2}(2{)}^4$

$\frac{{𝑅}_2}{𝐸}=\frac{16}{2}=8$

Hence,

${𝑅}_2=8𝐸$

So the correct option is:

$\overline{)8𝐸}$

Option D

If amplitudes are different, then in the region of destructive interference, the two waves do not cancel each other completely. Hence, there is some intensity of light.

Given, $X_C=X\Omega$

$\Rightarrow \frac{1}{2𝜋\mathrm{fC}}=X\Omega$

New Capacitance $C=2C$ and new frequency

$f^{\prime }=2f$

$\therefore$ New capacitive reactance $X_{\prime }^C=\frac{1}{2𝜋\left(2f\right)\left(2C\right)}$

$\begin{array}{cc}& X_{\prime }^C=\frac{1}{\left(2𝜋\right)\left(4\mathrm{fC}\right)}\\ & X_{\prime }^C=\frac{1}{4}{X}_C\\ & X_{\prime }^C=\frac{X_C}{4}\Omega \end{array}$

Step 1: Conservation principle

When the two discs are suddenly coupled (say by clamping them together along the same axis),

• Angular momentum is conserved (because external torque about the axis is zero).

• Kinetic energy is not necessarily conserved (some energy may be lost as heat/friction during coupling).

Step 2: Total angular momentum before coupling

For disc 1:

${𝐿}_1={𝐼}_1{𝜔}_1$

For disc 2:

${𝐿}_2={𝐼}_2{𝜔}_2$

So, total angular momentum:

$𝐿={𝐼}_1{𝜔}_1+{𝐼}_2{𝜔}_2$

Step 3: Common angular velocity after coupling

When they rotate together: Let the final angular velocity = $𝜔$.

Moment of inertia = ${𝐼}_1+{𝐼}_2$.

From conservation of angular momentum:

$({𝐼}_1+{𝐼}_2)𝜔={𝐼}_1{𝜔}_1+{𝐼}_2{𝜔}_2$

$𝜔=\frac{{𝐼}_1{𝜔}_1+{𝐼}_2{𝜔}_2}{{𝐼}_1+{𝐼}_2}$

Step 4: Final rotational kinetic energy

$𝐾=\frac{1}{2}({𝐼}_1+{𝐼}_2){𝜔}^2$

Substitute $𝜔$ from above:

$𝐾=\frac{1}{2}({𝐼}_1+{𝐼}_2){\left(\frac{{𝐼}_1{𝜔}_1+{𝐼}_2{𝜔}_2}{{𝐼}_1+{𝐼}_2}\right)}^2$

$𝐾=\frac{({𝐼}_1{𝜔}_1+{𝐼}_2{𝜔}_2{)}^2}{2({𝐼}_1+{𝐼}_2)}$

Final Answer:

$\overline{)\frac{({𝐼}_1{𝜔}_1+{𝐼}_2{𝜔}_2{)}^2}{2({𝐼}_1+{𝐼}_2)}}$

That corresponds to Option D.

Here, $\frac{\Delta m}{m}=\frac{.003}{0.3},\frac{\Delta r}{r}=\frac{0.005}{0.5},\frac{\Delta L}{L}=\frac{0.06}{6}$

As $𝜌=\frac{m}{\left(𝜋r^2\right)L}$,

$\begin{array}{cc}\therefore \left(\frac{\Delta 𝜌}{𝜌}\right)\times 100& =(\frac{\Delta 𝑚}{𝑚}+\frac{\Delta 𝑟}{𝑟}+\frac{\Delta 𝐿}{𝐿})\times 100\\ & =(\frac{0.003}{0.3}+\frac{2\times 0.005}{0.5}+\frac{0.06}{6})\times 100\\ & =1+2+1=4\%\end{array}$

$\begin{array}{cc}& \text{ Phase difference, }\\ & \begin{array}{cc}& \Delta 𝜙=(\frac{2𝜋}{T_1}-\frac{2𝜋}{T_2})t=(\frac{2𝜋}{T}-\frac{2𝜋}{\frac{3}{2}T})t\\ & \text{ At }t=2T\\ & \Delta 𝜙=(2𝜋-\frac{4𝜋}{3})\frac{2T}{T}=\frac{4𝜋}{3}\mathrm{rad}\\ & \Delta 𝜙=\frac{4𝜋}{3}\mathrm{rad}\end{array}\end{array}$

$\text{ The given circuit can be drawn as: }$

From the figure, we can see that this is a balanced Wheatstone bridge.

$\begin{array}{cc}\therefore & \frac{1}{R}=\frac{1}{15}+\frac{1}{30}=\frac{3}{30}=\frac{1}{10}\\ \therefore & R=10\Omega \\ \therefore & I=\frac{5}{10}\\ & \\ & =0.5A\end{array}$

Finding the energy difference when the electron jumps:

The electron moves from the 3rd orbit ($𝑛=3$) to the ground state ($𝑛=1$).

The energy in the ${𝑛}^{\text{th}}$ orbit is ${𝐸}_{𝑛}=\frac{-13.6}{{𝑛}^2}$ eV.

So, ${𝐸}_1=\frac{-13.6}{1^2}=-13.6$ eV and ${𝐸}_3=\frac{-13.6}{3^2}=\frac{-13.6}{9}=-1.51$ eV.

The energy difference is: ${𝐸}_{\text{photon}}={𝐸}_1-{𝐸}_3=(-13.6)-(-1.51)=-13.6+1.51=-12.09\text{ eV}$ But since energy emitted is positive, use $12.09$ eV.

Linking energy to the photon:

This energy is carried by the photon that is released: $ℎ𝜈=12.09\text{ eV}$

Using the photoelectric equation to find stopping potential:

The photon causes electrons to be emitted from a material that has a work function (${𝜙}_0$) of $4.1$ eV.

The equation is: $ℎ𝜈={𝜙}_0+𝑒{𝑉}_{𝑠}$ where ${𝑉}_{𝑠}$ is the stopping potential.

Plug in the known values: $12.09=4.1+{𝑉}_{𝑠}$

Solve for ${𝑉}_{𝑠}$: ${𝑉}_{𝑠}=12.09-4.1=8.0\text{ V}$

So, the stopping potential is about $8\text{ V}$.

We have

$\left|\overrightarrow{𝐴}\right|=5\sqrt{\surd 3},\left|\overrightarrow{𝐵}\right|=10,𝜃={30}^{\circ }.$

The magnitude of the cross product is

$|\overrightarrow{𝐴}\times \overrightarrow{𝐵}|=\left|\overrightarrow{𝐴}\right|\left|\overrightarrow{𝐵}\right|\sin 𝜃.$

Substitute the values:

$|\overrightarrow{𝐴}\times \overrightarrow{𝐵}|=\left(5\sqrt{\surd 3}\right)\left(10\right)(\sin {30}^{\circ }).$

$=50\sqrt{\surd 3}\cdot \frac{1}{2}$

$=25\sqrt{\surd 3}.$

✅ Correct Answer: Option C: $25\sqrt{\surd 3}$ units

Step 1: Recall the relation

Magnetic field intensity $𝐻$ (measured in A/m) is given by

$𝐻=\frac{𝑁𝐼}{𝑙}$

where $𝑁$ is number of turns, $𝑙$ is length of solenoid, and $𝐼$ is current.

For $𝑛$ turns per unit length,

$\frac{𝑁}{𝑙}=𝑛⟹𝐻=𝑛𝐼.$

Step 2: Magnetic flux density $𝐵$

The magnetic induction inside solenoid is

$𝐵={𝜇}_0𝐻={𝜇}_0𝑛𝐼.$

But the question specifically asks for magnetic field intensity $𝐻$ (not $𝐵$).

Final Answer:

$𝐻=𝑛𝐼$

Correct Option = C. ✅

Step 1: Recall definition of radius of gyration

Radius of gyration is defined by:

$𝑘=\sqrt{\surd \frac{𝐼}{{𝑀}_{\text{total}}}}$

where $𝐼$ = moment of inertia of the system about the given axis, and ${𝑀}_{\text{total}}$ is the total mass.

Step 2: Distance of each mass from axis

The axis is perpendicular to the square and passes through its centre.

So, the distance of each mass (corner of square) from centre = half of the diagonal =

$𝑟=\frac{\sqrt{\surd 2}}{2}𝐿=\frac{𝐿}{\sqrt{\surd 2}}$

Step 3: Moment of inertia

Each mass contributes $𝑀{𝑟}^2$ where $𝑟=𝐿/\sqrt{\surd 2}$.

So,

$𝐼=4\cdot 𝑀{\left(\frac{𝐿}{\sqrt{\surd 2}}\right)}^2=4𝑀\cdot \frac{{𝐿}^2}{2}=2𝑀{𝐿}^2$

Step 4: Total mass

${𝑀}_{\text{total}}=4𝑀$

Step 5: Radius of gyration

$𝑘=\sqrt{\surd \frac{𝐼}{{𝑀}_{\text{total}}}}=\sqrt{\surd \frac{2𝑀{𝐿}^2}{4𝑀}}=\sqrt{\surd \frac{{𝐿}^2}{2}}=\frac{𝐿}{\sqrt{\surd 2}}$

✅ Final Answer:

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

That corresponds to Option B.

Distance of $1^{\text{st }}$ minima from central maxima

${𝑦}_{1𝑑}=\frac{𝜆𝐷}{𝑎}$

Distance between two minima on either side of the central maxima is

$2y_{1d}=\frac{2𝜆D}{a}=\frac{2\times 4500\times {10}^{-10}\times 1.5}{0.3\times {10}^{-3}}=4.5\mathrm{mm}$

Distance covered by boy running from point A in time ' t ':

$d_1=\mathrm{vt}$

Distance covered by boy running from point B in time ' t ':

$\begin{array}{cc}& {𝑑}_2={𝑣}_1𝑡\\ & \text{ By Pythagoras theorem, }\\ & \begin{array}{cc}& d_2^1=x^2+d_2^2\\ \therefore & (\mathrm{vt}{)}^2=x^2+{\left(v_{1}t\right)}^2\\ \therefore & (v^2-v_2^1)t^2=x^2\\ \therefore & t={\left(\frac{x^2}{v^2-v_2^1}\right)}^{1/2}\end{array}\end{array}$

Step 1: Formula for r.m.s. speed of a gas

${𝑣}_{𝑟𝑚𝑠}=\sqrt{\surd \frac{3𝑘𝑇}{𝑚}}$

where:

• $𝑘$ is Boltzmann constant,

• $𝑇$ is absolute temperature,

• $𝑚$ is mass of one molecule.

Step 2: Relation with temperature

${𝑣}_{𝑟𝑚𝑠}\propto \sqrt{\surd 𝑇}$

Step 3: Ratio of r.m.s. velocities

Let the r.m.s. velocity at ${𝑇}_1=100\text{K}$ be $𝑥$. At that point:

${𝑣}_1=𝑥$

At ${𝑇}_2=400\text{K}$:

$\frac{{𝑣}_2}{{𝑣}_1}=\sqrt{\surd \frac{{𝑇}_2}{{𝑇}_1}}=\sqrt{\surd \frac{400}{100}}=\sqrt{\surd 4}=2$

So,

${𝑣}_2=2𝑥$

✅ Final Answer:

Option B: $2𝑥$

Step 1: Relation for torque

The maximum torque on a current-carrying coil in a magnetic field is:

${𝜏}_{\text{max}}=𝑛𝐼𝐴𝐵$

where:

• $𝑛$ = number of turns,

• $𝐼$ = current,

• $𝐴$ = area of coil,

• $𝐵$ = magnetic field strength.

Here, number of turns $𝑛=1$.

Step 2: Find radius of the coil

The wire has length $𝐿$. It is bent into a circle.

$𝐿=2𝜋𝑟\Rightarrow 𝑟=\frac{𝐿}{2𝜋}$

Step 3: Area of coil

$𝐴=𝜋{𝑟}^2=𝜋{\left(\frac{𝐿}{2𝜋}\right)}^2=\frac{{𝐿}^2}{4𝜋}$

Step 4: Torque

${𝜏}_{\text{max}}=𝐼\cdot 𝐴\cdot 𝐵=𝐼\cdot \frac{{𝐿}^2}{4𝜋}\cdot 𝐵$

${𝜏}_{\text{max}}=\frac{{𝐿}^2𝐼𝐵}{4𝜋}$

✅ Final Answer:

Option B: $\frac{{𝐿}^2𝐼𝐵}{4𝜋}$

All the inductors are connected in parallel with each other. Hence, the equivalent inductance between A and B is: $\frac{1}{L}=\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{6}$

$\therefore 𝐿=\frac{4}{5}𝐻$

${𝐶}_1,{𝐶}_2\&{𝐶}_3$ are in parallel, whereas ${𝐶}_4$ and ${𝐶}_5$ are in series.

$\begin{array}{cc}\therefore & C_{\text{eq1 }}=C_1+C_2+C_3=3C\\ \therefore & C_{\text{eq2 }}=\frac{1}{C_4}+\frac{1}{C_5}=\frac{C}{2}\end{array}$

$C_{\text{eq }}$ is in parallel with $C_{\text{eq }2}$

$\begin{array}{cc}\therefore C_{\mathrm{eq}}& =C_{\mathrm{eq}1}+C_{\mathrm{eq}2}\\ C_{\mathrm{eq}}& =\frac{7}{2}C\end{array}$

Now we know,

$\begin{array}{cc}& C_{\mathrm{eq}}=14𝜇F\\ \therefore & \frac{7}{2}C=14𝜇F\\ \therefore & C=4𝜇F\end{array}$

1. Given data:

• Mass of bullet, $𝑚=35\text{g}=0.035\text{kg}$

• Speed of bullet, $𝑣=600\text{m/s}$

• Maximum force person can apply (i.e. average resisting force), $𝐹=147\text{N}$

We want: number of bullets per second, i.e. rate of fire $𝑛$.

2. Momentum per bullet:

Impulse per bullet = change in momentum of one bullet =

$\Delta 𝑝=𝑚𝑣=0.035\times 600=21\text{kg·m/s}.$

3. Time rate of momentum (Force):

When $𝑛$ bullets are fired per second, the rate of momentum imparted per second is $𝑛\times 21$.

This equals the average force:

$𝐹=𝑛\Delta 𝑝$

So,

$147=𝑛\times 21$

$𝑛=\frac{147}{21}=7.$

Answer:

$\overline{)7}$

So the correct option is C (7).

According to the kinetic theory of gases, molecular collisions are assumed to be perfectly elastic.

• In an elastic collision, both kinetic energy and momentum are conserved.

• This assumption is crucial for deriving the Maxwell-Boltzmann distribution and gas laws.

✅ Correct Answer: Option A

Both kinetic energy and momentum are conserved.

1. For a pure inductor:

   The relation between voltage and current is:

   $𝑣\left(𝑡\right)=𝐿\frac{𝑑𝑖\left(𝑡\right)}{𝑑𝑡}$

2. If applied voltage is sinusoidal:

   $𝑣\left(𝑡\right)={𝑉}_{𝑚}\sin \left(𝜔𝑡\right)$

3. Current:

   $𝑖\left(𝑡\right)=\frac{1}{𝐿}\int 𝑣\left(𝑡\right)𝑑𝑡=\frac{1}{𝐿}\int {𝑉}_{𝑚}\sin \left(𝜔𝑡\right)𝑑𝑡$

   $𝑖\left(𝑡\right)=-\frac{{𝑉}_{𝑚}}{𝜔𝐿}\cos \left(𝜔𝑡\right)=\frac{{𝑉}_{𝑚}}{𝜔𝐿}\sin (𝜔𝑡-\frac{𝜋}{2})$

4. Phase relation:

• Current lags voltage by $\frac{𝜋}{2}$.

✅ Correct Answer:

Option C: lags behind the voltage by ${\left(\frac{𝜋}{2}\right)}^{𝑐}$

Step 1: Recall the First Law of Thermodynamics

$\Delta 𝑈=𝑄-𝑊$

• $\Delta 𝑈$ = change in internal energy

• $𝑄$ = heat added to the system

• $𝑊$ = work done by the system

Step 2: Isothermal Expansion of an Ideal Gas

For an ideal gas:

$\Delta 𝑈=0(\text{since }\Delta 𝑈\propto \Delta 𝑇,\text{and }\Delta 𝑇=0)$

So:

$0=𝑄-𝑊\Rightarrow 𝑄=𝑊$

That is, the heat supplied equals the work done by the gas.

Step 3: Interpreting the Signs

The problem states:

The gas does $(-150\text{J})$ of work.

This implies using the given sign convention:

• If $𝑊>0$, work is done by the gas.

• If $𝑊<0$, work is done on the gas.

So here $𝑊=-150\text{J}$, meaning the surroundings do 150 J of work on the gas.

From:

$𝑄=𝑊=-150\text{J}$

So, $𝑄<0$. That means 150 J of heat is removed from the gas.

✅ Answer:

Option B. 150 J of heat has been removed from the gas.

Step 1. Condition for secondary maxima

For a single slit of width $𝑎$, the diffraction pattern has intensity minima at:

$𝑎\sin 𝜃=𝑚𝜆,𝑚=\pm 1,\pm 2,\pm 3,\dots$

Between two minima, there is a secondary maximum. The approximate positions of secondary maxima are given by:

$𝑎\sin 𝜃\approx (𝑚+\frac{1}{2})𝜆,𝑚=1,2,3,\dots$

Thus:

• 1st secondary max: $𝑎\sin 𝜃\approx 1.5𝜆$

• 2nd secondary max: $𝑎\sin 𝜃\approx 2.5𝜆$

• 3rd secondary max: $𝑎\sin 𝜃\approx 3.5𝜆$

etc.

Step 2. Apply the given condition

For wavelength ${𝜆}_1=6384\text{Å}$:

• 2nd secondary maximum → $𝑎\sin 𝜃=2.5{𝜆}_1$.

For wavelength ${𝜆}_0$:

• 3rd secondary maximum → $𝑎\sin 𝜃=3.5{𝜆}_0$.

These coincide at the same angular position $𝜃$, so:

$2.5{𝜆}_1=3.5{𝜆}_0$

Step 3. Solve for ${𝜆}_0$

${𝜆}_0=\frac{2.5}{3.5}{𝜆}_1=\frac{5}{7}{𝜆}_1$

${𝜆}_0=\frac{5}{7}\times 6384\approx 4560\text{Å}$

✅ Final Answer:

${𝜆}_0=4560\text{Å}$

Correct option: B.

Given that: $V_{{\text{max }}_1}=V$ and $V_{{\text{max }}_2}=2V$ From Einstein's photoelectric equation,

$E=\frac{\mathrm{hc}}{{𝜆}_1}-{𝜙}_0\dots (\text{ where }𝐸=\frac{1}{2}𝑚{𝑉}_2^{max})$

From the given data, we get

$\begin{array}{cc}& \frac{1}{2}𝑚{𝑉}^2=\frac{ℎ𝑐}{{𝜆}_1}-{𝜙}_0...\left(𝑖\right)\\ & \frac{1}{2}𝑚(2𝑉{)}^2=\frac{4ℎ𝑐}{{𝜆}_2}-4{𝜙}_0...(𝑖𝑖)\end{array}$

Subtracting (i) from (ii),

$\begin{array}{cc}\therefore & \frac{\mathrm{hc}}{{𝜆}_1}-\frac{4\mathrm{hc}}{{𝜆}_2}=-3{𝜙}_0\\ & {𝜙}_0=\frac{\mathrm{hc}}{3}[\frac{4}{{𝜆}_2}-\frac{1}{{𝜆}_1}]=\frac{\mathrm{hc}}{3}\left[\frac{4{𝜆}_1-{𝜆}_2}{{𝜆}_1{𝜆}_2}\right]\end{array}$

$\begin{array}{cc}& \text{ By Conservation of energy, }\\ & \begin{array}{cc}& {\mathrm{KE}}_A+{\mathrm{PE}}_A={\mathrm{KE}}_C+{\mathrm{PE}}_C\\ \therefore & 2\mathrm{mgl}=\frac{1}{2}{\mathrm{mv}}_2^c+\mathrm{mgl}\dots (\because \mathrm{At}A,v_A=0)\end{array}\end{array}$

$\therefore v_2^C=2\mathrm{gl}$

We know,

Centripetal acceleration $=\frac{v^2}{r}$

$\dots (r=$ radius of circular motion $)$

$\therefore$ At point A ,

${𝑎}_{𝐴}=\frac{{𝑣}_2^{𝐴}}{𝑙}\Rightarrow {𝑎}_{𝐴}=0\dots (\because {𝑣}_{𝐴}=0)$

$\therefore$ At point C ,

$a_C=\frac{v_2^C}{l}\Rightarrow a_C=2g$

$\therefore$ Increase in the centripetal acceleration of the point mass from point A to point C is 2 g

According to Newton's law of cooling,

$\frac{{𝜃}_1-{𝜃}_2}{𝑡}=𝐾(\frac{{𝜃}_1+{𝜃}_2}{2}-{𝜃}_0)$

where ${𝜃}_0=$ temperature of the surroundings.

Situation I: $\frac{80-50}{5}=𝐾(\frac{80+50}{2}-20)$

$\frac{30}{5}=𝐾\left(45\right)...\left(𝑖\right)$

Situation II: $\frac{50-30}{t}=K(\frac{50+30}{2}-20)$

$\frac{20}{𝑡}=𝐾\left(20\right)...\left(𝑖𝑖\right)$

Dividing equation (i) by equation (ii), we get

$\begin{array}{cc}& \frac{30}{5}\times \frac{𝑡}{20}=\frac{45}{20}\\ & 𝑡=\frac{45\times 5}{30}\\ \therefore & 𝑡=7.5\mathrm{mins}\end{array}$

$\therefore$ Total time taken from ${80}^{\circ }C$ to ${30}^{\circ }C$

$=7.5+5=12.5\mathrm{mins}$

$\begin{array}{cc}& \text{ Magnetic flux }\left(𝜙\right)=\overrightarrow{𝐵}\cdot \overrightarrow{𝐴}\\ & \text{ Here, }\overrightarrow{𝐵}={𝐵}_0(2\widehat{𝑖}+3\widehat{𝑗}+4\widehat{𝑘})\\ & \overrightarrow{𝐴}={𝐿}^2\widehat{𝑘}\\ \therefore & 𝜙=\left({𝐵}_0\right(2\widehat{𝑖}+3\widehat{𝑗}+4\widehat{𝑘}\left)\right)\cdot \left({𝐿}^2\widehat{𝑘}\right)\\ \therefore & 𝜙=(2{𝐵}_0\widehat{𝑖}+3{𝐵}_0\widehat{𝑗}+4{𝐵}_0\widehat{𝑘})\cdot \left({𝐿}^2\widehat{𝑘}\right)\\ \therefore & 𝜙=4{𝐵}_0{𝐿}^2\end{array}$

$\begin{array}{cc}& \because \text{ Total potential energy }\\ & \begin{array}{cc}& U_{\text{Toal }}=U_{12}+U_{13}+U_{23}=0\text{ ...(given) }\\ \therefore & U_{\text{Total }}=-\frac{2{\mathrm{kQq}}^2}{𝑙}-\frac{2{\mathrm{kQq}}^2}{𝑙}+\frac{4{\mathrm{kq}}^2}{𝑙\sqrt{\surd 2}}=0\\ & \frac{4{\mathrm{kq}}^2}{𝑙\sqrt{\surd 2}}=\frac{4{\mathrm{kQq}}^2}{𝑙}\\ & Q=\frac{4{\mathrm{kq}}^2}{𝑙\sqrt{\surd 2}}\cdot \frac{𝑙}{4\mathrm{kq}}=\frac{q}{\sqrt{\surd 2}}\end{array}\end{array}$

If ${𝑟}_1$ and ${𝑟}_2$ are the radii, then with the given data,

$\begin{array}{cc}& a=𝜋{𝑟}_2^1\text{ and }4a=𝜋{𝑟}_2^2\\ \therefore & \frac{r_2^1}{r_2^2}=\frac{a}{4a}=\frac{1}{4}\\ \therefore & \frac{r_1}{r_2}=\frac{1}{2}\\ & h\propto \frac{1}{r}\\ \therefore & \frac{h_2}{h_1}=\frac{r_1}{r_2}=\frac{1}{2}\\ \therefore & h_2=\frac{h_1}{2}=\frac{h}{2}\end{array}$

Step 1: Recall the situation

A small sphere (or particle) oscillates in a spherical surface of large radius $𝑅$. If displaced slightly from equilibrium, the restoring force comes from component of gravity along the surface.

For small oscillations, the motion approximates simple harmonic motion with period:

$𝑇=2𝜋\sqrt{\surd \frac{𝑅}{𝑔}}$

where $𝑅$ is the radius of curvature of the spherical surface.

Step 2: Substitute values

Here $𝑅=1.6\text{m}$, $𝑔=10{\text{m/s}}^2$.

$𝑇=2𝜋\sqrt{\surd \frac{1.6}{10}}$

$𝑇=2𝜋\sqrt{\surd 0.16}$

$𝑇=2𝜋\left(0.4\right)=0.8𝜋\text{s}.$

Step 3: Final Answer

So the correct option is:

Option D: $0.8𝜋s$ ✅

The logic gate for which the output goes 'HIGH' ('1') only when an odd number of 'HIGH' or '1' are at its input is:

• OR → Output is 1 if any input is 1 (not necessarily odd).

• NAND → Output is 0 only when all inputs are 1.

• Ex-OR (XOR) → Output is 1 when the number of 1s at the input is odd. ✅

• NOR → Output is 1 only if all inputs are 0.

👉 Therefore, the correct answer is:

Option C: Ex-OR gate ✅

When the point is inside the wire,

By Ampere's Law:

B. $2𝜋𝑥={𝜇}_0{𝐼}_{\text{enclosed }}$

Since current is uniformly distributed, the current enclosed at radius x is :

$\begin{array}{cc}& {𝐼}_{\text{enclosed }}=𝐼\left(\frac{𝜋{𝑥}^2}{𝜋{𝑟}^2}\right)=𝐼\left(\frac{{𝑥}^2}{{𝑟}^2}\right)\\ \therefore & 𝐵\cdot 2𝜋𝑥={𝜇}_0{𝐼}_{\text{enclosed }}\\ \therefore & 𝐵=\frac{{𝜇}_0𝐼𝑥}{2𝜋{𝑟}^2}\end{array}$

For outside the wire, entire I is enclosed so: ${𝐵}_{1}2𝜋𝑥.3r={𝜇}_{0}I$

$\begin{array}{cc}& \therefore {𝐵}_1=\frac{\frac{{𝜇}_0𝐼𝑥}{2𝜋{𝑟}^2}}{\frac{{𝜇}_0𝐼}{6𝜋𝑟}}\\ & \therefore \frac{𝐵}{{𝐵}_1}=\frac{6𝑥}{2𝑟}\\ & \therefore \frac{𝐵}{{𝐵}_1}=\frac{3}{2}\dots (\because 𝑥=\frac{𝑟}{2})\end{array}$

Step 1. Formula for Doppler effect when source approaches stationary observer:

${𝑓}^{\prime }=𝑓\cdot \frac{𝑣}{𝑣-{𝑣}_{𝑠}}$

where

• ${𝑓}^{\prime }$ = apparent frequency

• $𝑓$ = source frequency (real)

• $𝑣$ = speed of sound

• ${𝑣}_{𝑠}$ = speed of source (toward observer)

Step 2. Given condition:

${𝑣}_{𝑠}=\frac{1}{10}𝑣$

So,