MHT CET 2025 23rd April Evening Shift (Physics & Chemistry) Question Paper with Solutions

$\begin{array}{cc}\overrightarrow{𝑠}& =\overrightarrow{𝑄}-\overrightarrow{𝑃}\\ & =(2\widehat{𝑖}+3\widehat{𝑗}+4\widehat{𝑘})𝑚-(3\widehat{𝑖}+4\widehat{𝑗}+5\widehat{𝑘})𝑚\\ & =-1\widehat{𝑖}-1\widehat{𝑗}-1\widehat{𝑘}\\ 𝑊& =\overrightarrow{𝐹}\cdot \overrightarrow{𝑠}\\ & =(3\widehat{𝑖}+4\widehat{𝑗}+5\widehat{𝑘})(-1\widehat{𝑖}-1\widehat{𝑗}\_1\widehat{𝑘})\\ & =-3-4-5=-12𝐽\end{array}$

Step 1: Current due to revolving electron

An electron going around in a circular orbit of radius $𝑟$ constitutes a current loop.

• Time period of revolution:

$𝑇=\frac{\text{circumference}}{\text{speed}}=\frac{2𝜋𝑟}{𝑣}.$

• Current:

$𝐼=\frac{\text{charge}}{\text{time period}}=\frac{𝑒}{𝑇}=\frac{𝑒}{(2𝜋𝑟/𝑣)}=\frac{𝑒𝑣}{2𝜋𝑟}.$

Step 2: Magnetic moment formula

Magnetic moment of a current loop:

$𝜇=𝐼\cdot 𝐴,$

where $𝐴$ is area of loop: $𝐴=𝜋{𝑟}^2$.

So:

$𝜇=\left(\frac{𝑒𝑣}{2𝜋𝑟}\right)\cdot 𝜋{𝑟}^2=\frac{𝑒𝑣𝑟}{2}.$

Step 3: Match with options

Thus, the associated magnetic moment is

$\overline{)\frac{𝑒𝑣𝑟}{2}}$

That corresponds to Option B.

Step 1: Formula

Pressure:

$𝑃=\frac{𝐹}{𝐴}$

For a square plate of side $𝐿$:

$𝐴={𝐿}^2,𝑃=\frac{𝐹}{{𝐿}^2}$

Step 2: Error propagation

For quantities with powers, percentage (relative) errors add with exponents in absolute value:

$\frac{\Delta 𝑃}{𝑃}=\frac{\Delta 𝐹}{𝐹}+2\frac{\Delta 𝐿}{𝐿}$

Step 3: Substitute given errors

• Error in $𝐹$: $4\%$

• Error in $𝐿$: $2\%$

So:

$\%\text{error in }𝑃=4\%+2\times 2\%=4\%+4\%=8\%$

✅ Final Answer:

Option D: $8\%$

• Three charges on a straight line of length $𝑙$.

• At $0$: charge $+3𝑞$.

• At $𝑙/2$: charge $𝑄$.

• At $𝑙$: charge $+𝑞$.

• Condition: Net force on the charge $+𝑞$ (at $𝑙$) is zero.

• We want: $𝑄=𝑥𝑞$. Find $𝑥$.

Step 1: Forces on the $+𝑞$ (at $𝑥=𝑙$).

• Due to $+3𝑞$ at $0$:

  Repulsive force, directed to the right.

  Magnitude:

  ${𝐹}_{3𝑞\to 𝑞}=𝑘\cdot \frac{\left(3𝑞\right)\left(𝑞\right)}{(𝑙-0{)}^2}=\frac{3𝑘{𝑞}^2}{{𝑙}^2}.$

• Due to $𝑄$ at $𝑙/2$:

  Distance: $𝑙-𝑙/2=𝑙/2$.

  Force magnitude:

  ${𝐹}_{𝑄\to 𝑞}=𝑘\cdot \frac{\left|𝑄\right|𝑞}{(𝑙/2{)}^2}=\frac{4𝑘\left|𝑄\right|𝑞}{{𝑙}^2}.$

  Direction depends on sign of $𝑄$:

• If $𝑄$ is positive, then repulsion → force on $+𝑞$ is to the right.

• If $𝑄$ is negative, then attraction → force on $+𝑞$ is to the left.

Step 2: Net force condition.

We want net force on $+𝑞$ to be zero.

Equation in vector form:

${𝐹}_{3𝑞\to 𝑞}\left(\text{to the right}\right)+{𝐹}_{𝑄\to 𝑞}\left(\text{depends on sign}\right)=0.$

So:

$\frac{3𝑘{𝑞}^2}{{𝑙}^2}+\frac{4𝑘𝑄𝑞}{{𝑙}^2}=0.$

(Note: The formula automatically accounts for sign of $𝑄$.)

Step 3: Solve for $𝑄$.

$3{𝑞}^2+4𝑄𝑞=0$

$4𝑄𝑞=-3{𝑞}^2$

$𝑄=-\frac{3}{4}𝑞.$

Thus,

$𝑥=-\frac{3}{4}.$

Final Answer:

$\overline{)\frac{-3}{4}}$

Correct option: B

Checking Translational K.E.:

The formula for translational kinetic energy is $\text{Translational K.E.}=\frac{3}{2}\mathrm{PV}.$

This means that options (A) and (D) are wrong, because they do not match this formula.

Finding r.m.s. Speed:

The r.m.s. speed ($v_{\mathrm{rms}}$) is given by $v_{\mathrm{rms}}=\sqrt{\surd \frac{3𝑅𝑇}{M}}.$

This shows that r.m.s. speed increases as the square root of the temperature: $v_{\mathrm{rms}}\propto \sqrt{\surd T}.$

Checking Option (C):

For option (C): $T_1={100}^{\circ }C$, $T_2={627}^{\circ }C$

First, convert to Kelvin:
$T_1=100+273=373K$
$T_2=627+273=900K$

Calculate the ratio of speeds:

$\frac{v_1}{v_2}=\sqrt{\surd \frac{T_1}{T_2}}=\sqrt{\surd \frac{373}{900}}=\frac{\sqrt{\surd 373}}{30}$

This means:

$v_2=\frac{30}{\sqrt{\surd 373}}{v}_1$

So, option (C) is not correct.

Checking Option (B):

For option (B): $T_1={73}^{\circ }C$, $T_2={527}^{\circ }C$

Convert to Kelvin:
$T_1=73+273=346K$
$T_2=527+273=800K$

Calculate the ratio of speeds:

$\frac{v_1}{v_2}=\sqrt{\surd \frac{T_1}{T_2}}=\sqrt{\surd \frac{346}{800}}=\frac{1}{2}$

This means:

$v_2=2v_1$

So, option (B) is the only correct answer.

Using Lens maker's formula,

$\frac{1}{f}=P=(𝜇-1)(\frac{1}{R_1}-\frac{1}{R_2})$

When placed in liquid the power is given by

$P^{\prime }=({𝜇}^{\prime }-1)(\frac{1}{R_1}-\frac{1}{R_2})$

Where refractive index of lens in the liquid,

$\begin{array}{cc}& {𝜇}^{\prime }=\frac{𝜇}{{𝜇}_{𝑙}}=\frac{1.5}{2}=\frac{3}{4}\\ \therefore & \frac{P^{\prime }}{P}=\frac{{𝜇}^{\prime }-1}{𝜇-1}=\frac{\frac{3}{4}-1}{1.5-1}=\frac{-\frac{1}{4}}{0.5}=-\frac{1}{2}\\ & P^{\prime }=-\frac{1}{2}\times P=-\frac{1}{2}\times 3=-1.5D\end{array}$

Step 1: Faraday’s law relation

The induced emf (total across coil):

$E=-𝑛\frac{𝑑\Phi }{𝑑𝑡}$

where $\Phi =𝐵𝐴$ is the flux through one turn (if field is uniform and perpendicular).

Step 2: Relation between emf and current

$E=𝐼𝑅$

Step 3: Charge flow

Total charge that flows over the discharge interval:

$𝑞=\int 𝐼𝑑𝑡=\frac{1}{𝑅}\int E𝑑𝑡$

Step 4: Evaluate the integral

$\int E𝑑𝑡=-𝑛\int 𝑑\Phi =𝑛\Delta \Phi$

Here, change in flux per turn is initial $𝐵𝐴$ to zero, so $\Delta \Phi =𝐵𝐴$.

Thus:

$𝑞=\frac{1}{𝑅}\left(𝑛𝐵𝐴\right)$

Step 5: Solve for $𝐵$

$𝐵=\frac{𝑞𝑅}{𝑛𝐴}$

Final Answer:

$\overline{)\frac{𝑞𝑅}{𝑛𝐴}}$

This corresponds to Option B.

Photodiode is operated in reverse bias

Reverse saturation current $\propto$ Intensity of illumination

$\therefore I_4$ is maximum intensity.

Step 1: Relation for maximum height

For a projectile, the vertical component of velocity:

${𝑢}_{𝑦}=𝑢\sin 𝜃$

Maximum height:

$𝐻=\frac{{𝑢}_2^{𝑦}}{2𝑔}=\frac{(𝑢\sin 𝜃{)}^2}{2𝑔}$

$\therefore 𝑢\sin 𝜃=\sqrt{\surd 2𝑔𝐻}$

Step 2: Total time of flight

Time of flight:

$𝑇=\frac{2𝑢\sin 𝜃}{𝑔}$

Substitute $𝑢\sin 𝜃=\sqrt{\surd 2𝑔𝐻}$:

$𝑇=\frac{2\sqrt{\surd 2𝑔𝐻}}{𝑔}$

$𝑇=2\sqrt{\surd \frac{2𝐻}{𝑔}}$

✅ Final Answer:

$\overline{)2\sqrt{\surd \frac{2𝐻}{𝑔}}}$

Correct option: C

For Doppler effect in sound, the apparent frequency is

${𝑓}^{\prime }=𝑓\cdot \frac{𝑣\pm {𝑣}_{𝑜}}{𝑣\mp {𝑣}_{𝑠}}$

where:

• $𝑣$ = speed of sound

• ${𝑣}_{𝑜}$ = speed of observer

• ${𝑣}_{𝑠}$ = speed of source

Now apply the sign convention carefully:

• Observer is moving away from the source, so frequency decreases.

  Hence numerator becomes:

$𝑣-{𝑣}_{𝑜}$

• Source is moving towards the observer, so frequency increases.

  Hence denominator becomes:

$𝑣-{𝑣}_{𝑠}$

Given:

${𝑣}_{𝑜}=\frac{𝑉}{4},{𝑣}_{𝑠}=\frac{𝑉}{3}$

So,

${𝑓}^{\prime }=𝐹\cdot \frac{𝑉-\frac{𝑉}{4}}{𝑉-\frac{𝑉}{3}}$

Now simplify:

${𝑓}^{\prime }=𝐹\cdot \frac{\frac{3𝑉}{4}}{\frac{2𝑉}{3}}$

${𝑓}^{\prime }=𝐹\cdot \frac{3𝑉}{4}\cdot \frac{3}{2𝑉}$

${𝑓}^{\prime }=𝐹\cdot \frac{9}{8}$

Therefore, the apparent frequency is

$\overline{)\frac{9}{8}𝐹}$

So, the correct option is:

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

Step 1: Recall hydrogen energy level formula

For hydrogen:

${𝐸}_{𝑛}=-\frac{13.6\text{eV}}{{𝑛}^2}$

The energy of the photon emitted in a transition ${𝑛}_{𝑖}\to {𝑛}_{𝑓}$ is:

$\Delta 𝐸=13.6(\frac{1}{{𝑛}_2^{𝑓}}-\frac{1}{{𝑛}_2^{𝑖}})\text{eV}$

Step 2: Ultraviolet vs Visible vs Infrared

• Large energy difference → higher frequency → shorter wavelength → ultraviolet.

• Moderate energy difference → visible.

• Small energy difference → longer wavelength → infrared.

Step 3: Analyze options

• (A) $𝑛=3\to 𝑛=1$:

$\Delta 𝐸=13.6(1-\frac{1}{9})=13.6\times \frac{8}{9}\approx 12.1\text{eV}$.

This corresponds to wavelength $𝜆\approx 1240/12.1\approx 102\text{nm}$.

That is deep UV. Not IR.

• (B) $𝑛=4\to 𝑛=2$:

$\Delta 𝐸=13.6(\frac{1}{4}-\frac{1}{16})=13.6\times \frac{3}{16}=2.55\text{eV}$.

$𝜆\approx 1240/2.55\approx 486\text{nm}$ → visible (blue-green). Not IR.

• (C) $𝑛=6\to 𝑛=2$:

$\Delta 𝐸=13.6(\frac{1}{4}-\frac{1}{36})=13.6\times \frac{8}{36}\approx 3.02\text{eV}$.

$𝜆\approx 1240/3.02\approx 410\text{nm}$. Visible (violet). Not IR.

• (D) $𝑛=5\to 𝑛=3$:

$\Delta 𝐸=13.6(\frac{1}{9}-\frac{1}{25})=13.6\times \frac{16}{225}=0.967\text{eV}$.

$𝜆\approx 1240/0.967\approx 1280\text{nm}$. That's Infrared.

✅ Correct Answer: Option D ($𝑛=5\to 𝑛=3$)

Weight of the cylinder $=$ Upthrust due to liquid $1+$ Upthrust due liquid 2

$\begin{array}{cc}& =\left({𝜌}_1𝑔{𝑉}_1\right)+\left({𝜌}_2𝑔{𝑉}_2\right)\\ & =\left(𝜌𝑔{𝐿}_1{𝐴}_1\right)+\left(3𝜌𝑔{𝐿}_2{𝐴}_2\right)\\ & =𝜌𝑔\frac{31}{4}𝐴+3𝜌𝑔\frac{1}{4}𝐴=\frac{3}{2}𝜌𝑔𝐴𝑙\\ & =\frac{3}{2}𝜌\end{array}$

Step 1: Find the total charge before connecting.

Total charge before connecting = charge on ${𝐶}_1$ + charge on ${𝐶}_2$: ${𝑄}_{\text{total}}={𝐶}_1{𝑉}_1+{𝐶}_2{𝑉}_2$

Step 2: After connecting, both capacitors reach a common voltage $𝑉$.

Total charge = $({𝐶}_1+{𝐶}_2)𝑉$

Step 3: Set the total charges equal (since charge is conserved):

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

Step 4: Plug in the values.

${𝐶}_1=100\times {10}^{-6}F$, ${𝑉}_1=20V$ ${𝐶}_2=50\times {10}^{-6}F$, ${𝑉}_2=40V$

$𝑉=\frac{100\times {10}^{-6}\times 20+50\times {10}^{-6}\times 40}{100\times {10}^{-6}+50\times {10}^{-6}}$

Step 5: Calculate:

$100\times {10}^{-6}\times 20=2\times {10}^{-3}$ $50\times {10}^{-6}\times 40=2\times {10}^{-3}$ Total numerator $=4\times {10}^{-3}$

$100\times {10}^{-6}+50\times {10}^{-6}=150\times {10}^{-6}=1.5\times {10}^{-4}$

$𝑉=\frac{4\times {10}^{-3}}{1.5\times {10}^{-4}}=\frac{400}{15}=\frac{80}{3}$

A Light Emitting Diode (LED) operates in forward bias to emit radiation through the spontaneous radiative recombination of electron-hole pairs at the junction.

In the forward-bias condition, the current remains negligibly small until the applied potential difference exceeds the threshold (knee) voltage. Beyond this knee voltage, the current increases rapidly with an increase in voltage. Consequently, the $𝐼\text{--}𝑉$ characteristic curve of an LED is represented in the first quadrant, which corresponds to graph (b).

Hence, the correct option is (B).

$\begin{array}{cc}\overrightarrow{𝑣}& =\overrightarrow{𝜔}\times \overrightarrow{𝑟}=\mid \begin{array}{ccc}\widehat{𝑖}& \widehat{𝑗}& \widehat{𝑘}\\ 3& -4& 1\\ 5& -6& 6\end{array}\mid \\ & =\widehat{𝑖}(-24+6)-\widehat{𝑗}(18-5)+\widehat{𝑘}(-18+20)\\ \overrightarrow{𝑣}& =-18\widehat{𝑖}-13\widehat{𝑗}+2\widehat{𝑘}\end{array}$

As the Since ta $𝑛{𝜃}_{𝐵}=𝜇$ and refractive index becomes $1/𝜇$ when going from glass to air tan ${𝜃}^{\prime }{}_B=1/𝜇$.

Given that ${𝜃}_B=𝜃$ and $𝜇=\tan 𝜃$, we get; $\tan {𝜃}^{\prime }{}_B=1/\tan 𝜃=\cot 𝜃\therefore {𝜃}^{\prime }{}_B=𝜋/2-𝜃$

As the temperature increases, the conductors are usually more resistant to the flow of electricity, resulting in an increase in their specific resistance (resistivity). On the other hand, semiconductors tend to become less resistant to electrical flow, decreasing their specific resistance as the temperature increases.

Step 1: Isothermal Expansion

For isothermal process:

$𝑃𝑉={𝑃}_1{𝑉}_1$

Initial state: $𝑃,𝑉$.

After isothermal expansion to $3𝑉$:

${𝑃}_{𝑖}{𝑉}_{𝑖}={𝑃}_1{𝑉}_1⟹𝑃\cdot 𝑉={𝑃}_2\cdot \left(3𝑉\right)$

${𝑃}_2=\frac{𝑃}{3}$

So, after first process:

$({𝑃}_2,{𝑉}_2)=(\frac{𝑃}{3},3𝑉).$

Step 2: Adiabatic Expansion

Now expand adiabatically from ${𝑉}_2=3𝑉$ to ${𝑉}_3=24𝑉$.

Adiabatic relation:

$𝑃{𝑉}^{𝛾}=\text{constant}.$

So:

${𝑃}_2\left({𝑉}_{𝛾}^2\right)={𝑃}_3\left({𝑉}_{𝛾}^3\right).$

${𝑃}_3={𝑃}_2{\left(\frac{{𝑉}_2}{{𝑉}_3}\right)}^{𝛾}.$

Substitute:

${𝑃}_3=\frac{𝑃}{3}{\left(\frac{3𝑉}{24𝑉}\right)}^{𝛾}=\frac{𝑃}{3}{\left(\frac{1}{8}\right)}^{𝛾}.$

Step 3: Value of γ for a Polyatomic Gas

For a polyatomic gas, "moderate temperatures" → assume non-linear polyatomic molecule, having $𝑓=6$ degrees of freedom (3 translational + 3 rotational, neglecting vibrations).

${𝐶}_{𝑣}=\frac{𝑓}{2}𝑅=3𝑅,{𝐶}_{𝑝}={𝐶}_{𝑣}+𝑅=4𝑅$

$𝛾=\frac{{𝐶}_{𝑝}}{{𝐶}_{𝑣}}=\frac{4𝑅}{3𝑅}=\frac{4}{3}.$

Step 4: Compute Final Pressure

${𝑃}_3=\frac{𝑃}{3}{\left(\frac{1}{8}\right)}^{4/3}.$

Now simplify.

$(1/8{)}^{4/3}={\left(8^{-1}\right)}^{4/3}=8^{-4/3}.$

Since $8=2^3$,

$8^{4/3}=(2^3{)}^{4/3}=2^4=16.$

So:

$(1/8{)}^{4/3}=\frac{1}{16}.$

Therefore:

${𝑃}_3=\frac{𝑃}{3}\cdot \frac{1}{16}=\frac{𝑃}{48}.$

✅ Final Answer:

$\overline{)\frac{𝑃}{48}}$

Correct option: D.

$\begin{array}{cc}& 𝑊=𝑘\frac{{𝑞}_1{𝑞}_2}{{𝑟}_2}-𝑘\frac{{𝑞}_1{𝑞}_2}{{𝑟}_1}\\ & =9\times {10}^9\times (4\times {10}^{-6})\times (10\times {10}^{-6})(\frac{1}{0.08}-\frac{1}{0.1})\\ & =360\times 2.5\times {10}^{-3}\\ & =900\times {10}^{-3}=0.9J\end{array}$

$\begin{array}{cc}& \text{ Given that, }f=\frac{1}{𝜋}\mathrm{rps}\text{, }\\ & \therefore 𝜔=2𝜋f=2𝜋\times \frac{1}{𝜋}=2\mathrm{rad}/s\end{array}$

The centripetal force is provided by the horizontal component of the tension i.e., ( $T\sin 𝜃$ ).

$\begin{array}{cc}\therefore & \text{ Tsin }𝜃=𝑚𝑟{𝜔}^2\\ & \text{ But }𝑟=𝑙\sin 𝜃\text{ and }\\ & {𝜔}^2=4\\ \therefore & 𝑇\sin 𝜃=𝑚\times (𝑙\sin 𝜃)\times 4\\ \therefore & 𝑇=4𝑚𝑙\end{array}$

Step 1. Induced emf (Faraday’s law):

Induced emf:

$E=𝑛\cdot \frac{\Delta 𝜙}{\Delta 𝑡}=𝑛\cdot \frac{{𝜙}_1-{𝜙}_2}{𝑡}$

(sign is not important here, they ask for current magnitude).

Step 2. Total resistance:

Coil resistance = $𝑅$.

Extra resistance = $\frac{𝑅}{4}$.

Total resistance ${𝑅}_{\text{eq}}=𝑅+\frac{𝑅}{4}=\frac{5𝑅}{4}$.

Step 3. Current:

$𝐼=\frac{E}{{𝑅}_{\text{eq}}}=\frac{𝑛({𝜙}_1-{𝜙}_2)/𝑡}{(5𝑅/4)}=\frac{4𝑛({𝜙}_1-{𝜙}_2)}{5𝑅𝑡}$

Final Answer:

$\overline{)\frac{4𝑛({𝜙}_1-{𝜙}_2)}{5𝑅𝑡}}$

which corresponds to Option B.

Gravitational pull depends upon the acceleration due to gravity on that planet.

$\begin{array}{cccc}& & {𝑀}_{𝑚}& =\frac{1}{6}{𝑀}_{𝑒},{𝑔}_{𝑚}=\frac{1}{8}{𝑔}_{𝑒}\\ & \therefore & \frac{{𝑅}_{𝑒}}{{𝑅}_{𝑚}}& ={(\frac{{𝑀}_{𝑒}}{{𝑀}_{𝑚}}\times \frac{{𝑔}_{𝑚}}{{𝑔}_{𝑒}})}^{1/2}={(6\times \frac{1}{8})}^{1/2}\\ & \therefore & {𝑅}_{𝑚}& =\sqrt{\surd \frac{4}{3}}{𝑅}_{𝑒}\end{array}$

$𝑒{𝑉}_{𝑜}=ℎ𝜈-𝜙$

From graph

At C,

${\mathrm{eV}}_1=h{𝑣}_1-𝜙...\left(𝑖\right)$

At D,

$𝑒{𝑉}_2=ℎ{𝑣}_2-𝜙...\left(𝑖𝑖\right)$

Subtracting eq (ii) from (i),

$\begin{array}{cc}& 𝑒{𝑉}_1-𝑒{𝑉}_2=ℎ{𝑣}_1-ℎ{𝑣}_2\\ & ℎ=\frac{𝑒({𝑉}_1-{𝑉}_2)}{({𝑣}_1-{𝑣}_2)}\end{array}$

Step 1: Recall the concept

• A soap bubble has two surfaces (inner and outer).

• Work is done against surface tension when increasing surface area.

• Work done = Increase in surface energy = $\Delta 𝑈=𝑇\times \Delta 𝐴$.

• Since there are two surfaces,

$\Delta 𝑈=2𝑇\times \left(\Delta {𝐴}_{\text{sphere}}\right).$

Step 2: Surface area of a sphere

For diameter $𝑑$, radius $𝑟=\frac{𝑑}{2}$.

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

Step 3: Increase in total area

$\Delta 𝐴={𝐴}_2-{𝐴}_1=𝜋{𝑑}_2^2-𝜋{𝑑}_2^1=𝜋({𝑑}_2^2-{𝑑}_2^1).$

Since there are two surfaces:

$\Delta {𝐴}_{\text{total}}=2𝜋({𝑑}_2^2-{𝑑}_2^1).$

Step 4: Work done

$𝑊=𝑇\times \Delta {𝐴}_{\text{total}}=2𝜋({𝑑}_2^2-{𝑑}_2^1)𝑇.$

✅ Final Answer:

$\overline{)2𝜋({𝑑}_2^2-{𝑑}_2^1)𝑇}$

That corresponds to Option D.

Step 1: Calculate potential energy (P.E.)

The object has potential energy because it is 2000 m above the ground. We use the formula:

$\text{P.E.}=𝑚𝑔ℎ$

Here, $𝑚=3.5\mathrm{kg}$ (mass of object), $𝑔=10{m/s}^2$ (acceleration due to gravity), $ℎ=2000m$ (height).

Step 2: Understand energy conversion

When the object falls, its potential energy turns into heat energy, which is used to melt some of the ice. All the energy goes into melting the ice because the object immediately comes to rest.

We set the energy from falling equal to the heat needed to melt $𝑀$ kg of ice:

$\text{Latent heat needed}=𝑀𝐿$

$\text{Potential energy lost}=𝑚𝑔ℎ$

So, $𝑀𝐿=𝑚𝑔ℎ$

Step 3: Solve for the mass of melted ice ($𝑀$)

Given latent heat of ice, $𝐿=3.5\times {10}^{5}J/\mathrm{kg}$.

Plug in the values:

$𝑀\times 3.5\times {10}^5=3.5\times 10\times 2000$

$𝑀\times 3.5\times {10}^5=70,000$

$𝑀=\frac{70,000}{3.5\times {10}^5}$

$𝑀=0.2\mathrm{kg}$

So, $0.2\mathrm{kg}$ or $200$ grams of ice melts.

$\begin{array}{cc}& \text{ From figure, }\\ & \begin{array}{cc}& 𝑁\sin 𝜃=𝑚𝑔\\ & 𝑁\cos 𝜃=\frac{𝑚{𝑉}^2}{𝑅}\\ \therefore & \tan 𝜃=\frac{𝑅𝑔}{{𝑉}^2}\\ \therefore & \frac{𝑅}{ℎ}=\frac{𝑅𝑔}{{𝑉}^2}\\ \therefore & ℎ=\frac{{𝑉}^2}{𝑔}\\ \therefore & 𝑉=\sqrt{\surd ℎ𝑔}=\sqrt{\surd 0.1\times 10}=1m/s\end{array}\end{array}$

We have two stones:

• Stone A (lighter): mass = $𝑚$, radius of circular motion = $𝑟$, tangential speed = ${𝑣}_{𝐴}$.

• Stone B (heavier): mass = $3𝑚$, radius of circular motion = $\frac{𝑟}{3}$, tangential speed = ${𝑣}_{𝐵}$.

It is given:

${𝑣}_{𝐴}=𝑛{𝑣}_{𝐵}$

We need: condition when centripetal forces are equal in magnitude.

Step 1. Write centripetal force formula

$𝐹=\frac{𝑚{𝑣}^2}{𝑅}$

Step 2. Forces for two stones

• For Stone A:

${𝐹}_{𝐴}=\frac{𝑚{𝑣}_2^{𝐴}}{𝑟}$

• For Stone B:

${𝐹}_{𝐵}=\frac{\left(3𝑚\right){𝑣}_2^{𝐵}}{\frac{𝑟}{3}}=\frac{3𝑚{𝑣}_2^{𝐵}}{(𝑟/3)}=\frac{9𝑚{𝑣}_2^{𝐵}}{𝑟}$

Step 3. Equating forces

${𝐹}_{𝐴}={𝐹}_{𝐵}$

$\frac{𝑚{𝑣}_2^{𝐴}}{𝑟}=\frac{9𝑚{𝑣}_2^{𝐵}}{𝑟}$

Cancel $𝑚$ and $𝑟$:

${𝑣}_2^{𝐴}=9{𝑣}_2^{𝐵}$

$\frac{{𝑣}_{𝐴}}{{𝑣}_{𝐵}}=3$

So:

${𝑣}_{𝐴}=3{𝑣}_{𝐵}$

That means $𝑛=3$.

✅ Answer: 3 (Option B)

We have an LC oscillation setup:

• Capacitance $𝐶=4𝜇𝐹=4\times {10}^{-6}𝐹$

• Initial capacitor voltage $𝑉=10\text{V}$

• Inductance $𝐿=10\text{mH}=10\times {10}^{-3}𝐻=0.01𝐻$

Step 1: Initial energy stored in the capacitor

${𝐸}_{𝐶}=\frac{1}{2}𝐶{𝑉}^2$

${𝐸}_{𝐶}=\frac{1}{2}\cdot 4\times {10}^{-6}\cdot \left({10}^2\right)$

${𝐸}_{𝐶}=2\times {10}^{-6}\cdot 100=2\times {10}^{-4}\text{J}$

Step 2: Maximum energy in the inductor

At peak current, all energy resides in the inductor:

${𝐸}_{𝐿}=\frac{1}{2}𝐿{𝐼}_2^{\text{max}}$

Equating:

$\frac{1}{2}𝐿{𝐼}_2^{\text{max}}={𝐸}_{𝐶}$

$\frac{1}{2}\cdot 0.01\cdot {𝐼}_2^{\text{max}}=2\times {10}^{-4}$

$0.005{𝐼}_2^{\text{max}}=2\times {10}^{-4}$

${𝐼}_2^{\text{max}}=\frac{2\times {10}^{-4}}{0.005}=\frac{2\times {10}^{-4}}{5\times {10}^{-3}}=0.04$

${𝐼}_{\text{max}}=0.2\text{A}$

✅ Final Answer:

${𝐼}_{\text{max}}=0.2\text{A}$

Correct Option: A (0.2 A)

$\begin{array}{cc}& \begin{array}{cc}& \text{ Total energy }=𝐾.𝐸.+𝑃.𝐸.=𝐾+𝑃=\frac{1}{2}𝑚{𝜔}^2{𝐴}^2\\ & {𝐴}^2=\frac{2(𝐾+𝑃)}{𝑚{𝜔}^2}\\ & 𝐴=\frac{1}{𝜔}\sqrt{\surd \frac{2(𝐾+𝑃)}{𝑚}}=\frac{1}{2𝜋𝑛}\sqrt{\surd \frac{2(𝐾+𝑃)}{𝑚}}=\frac{1}{𝜋𝑛}\sqrt{\surd \frac{𝐾+𝑃}{2𝑚}}\end{array}\\ & \text{ Substituting the values, }\\ & 𝐴=\frac{1}{𝜋}\times \frac{𝜋}{25}\sqrt{\surd \frac{1+0.6}{2\times 0.2}}=0.08m\end{array}$

$\begin{array}{cc}𝐵& =\frac{{𝜇}_0𝑁}{2𝑅}\sqrt{\surd {𝐼}_2^1+{𝐼}_2^2}\\ \therefore 𝐵& =\frac{{𝜇}_0𝑁𝐼}{2𝑅}\sqrt{\surd 1^2+1^2}\\ \therefore 𝐵& =\frac{{𝜇}_0𝑁𝐼}{\sqrt{\surd 2}𝑅}\end{array}$