MHT CET 2025 22nd April Morning Shift (Physics & Chemistry) Question Paper with Solutions

$\text{ Horizontal road }$

$\begin{array}{cc}& N=\mathrm{mg}...\left(𝑖\right)\\ & \text{ Concave road }\end{array}$

$\begin{array}{cc}& N=\frac{{\mathrm{mv}}^2}{r}+\mathrm{mg}\cos 𝜃\\ \therefore & N_{max}=\frac{{\mathrm{mv}}^2}{r}+\mathrm{mg}\dots \text{ (ii) }\end{array}$

                                                …. (When $\cos 𝜃=1$ )

Convex road

$\begin{array}{cc}& 𝑁+\frac{𝑚{𝑣}^2}{𝑟}=𝑚𝑔\cos 𝜃\\ \therefore & 𝑁=𝑚𝑔\cos 𝜃-\frac{𝑚{𝑣}^2}{𝑟}\\ \therefore & {𝑁}_{max}=𝑚𝑔-\frac{𝑚{𝑣}^2}{𝑟}\dots \text{ (iii) }\\ & \dots (\text{ When }\cos 𝜃=1)\end{array}$

From equations (i), (ii) and (iii), we get: Normal reaction on the vehicle is maximum for a concave shaped road.

Ra-226 (${}_{226}^{88}𝑅𝑎$) decays to Pb-206 (${}_{206}^{82}𝑃𝑏$) by emission of $𝛼$ and $𝛽$ particles. We need to find how many $𝛼$'s and $𝛽$'s are emitted.

Step 1: Write data

• Initial: $𝐴=226,𝑍=88$.

• Final: $𝐴=206,𝑍=82$.

Step 2: Effect of alpha and beta emissions

• An $𝛼$-particle emission: decreases mass number by 4, atomic number by 2. ($𝐴\to 𝐴-4$, $𝑍\to 𝑍-2$)

• A ${𝛽}^{-}$-particle emission: mass number unchanged, atomic number increases by 1 ($𝑍\to 𝑍+1$).

Step 3: Compare initial and final mass numbers

Mass number difference: $226-206=20$.

• Each $𝛼$ reduces mass number by 4.

• To reduce by 20, we need $\frac{20}{4}=5$ $𝛼$ emissions.

So number of $𝛼=5$.

Step 4: Effect on atomic number

After 5 $𝛼$ decays, atomic number decreases by $5\times 2=10$.

Initial $𝑍=88$. After 5 $𝛼$, $𝑍=88-10=78$.

But final $𝑍$ must be 82.

So we must increase atomic number from 78 to 82 → increase by 4.

Each ${𝛽}^{-}$ increases $𝑍$ by 1. Thus, 4 ${𝛽}^{-}$ emissions.

Step 5: Answer

• Number of $𝛼$: 5

• Number of ${𝛽}^{-}$: 4

So the correct answer = Option A: 5,4 ✅

Step 1: Recall the Young’s double-slit fringe position formula

In Young’s double slit experiment, the position of the ${𝑚}^{\text{th}}$ order maximum is given by:

${𝑦}_{𝑚}=𝑚\frac{𝜆𝐷}{𝑑}$

where

• $𝑚$ = order of the fringe,

• $𝜆$ = wavelength,

• $𝐷$ = distance between slits and screen,

• $𝑑$ = slit separation.

Thus, the fringe position is proportional to $𝑚𝜆$ for fixed $𝐷/𝑑$.

Step 2: Use the given information

We are told that:

• First order maximum ($𝑚=1$) at $𝜆=6000\text{Å}$ lies at ${𝑦}_1=14.50\text{mm}$.

• Tenth order maximum ($𝑚=10$) at same wavelength lies at ${𝑦}_{10}=16.75\text{mm}$.

So, fringe positions are not measured from the central fringe (zero-order maxima), but from some reference point.

Step 3: Find the difference

For wavelength $6000\text{Å}$:

${𝑦}_{10}-{𝑦}_1=16.75-14.50=2.25\text{mm}.$

This distance corresponds to 9 fringes (from $𝑚=1$ to $𝑚=10$).

So, fringe width is:

$\Delta 𝑦=\frac{2.25}{9}=0.25\text{mm}.$

Step 4: Locate zero-order maximum with original wavelength

Since for $𝜆=6000\text{Å}$, each fringe spacing is 0.25 mm:

• First order maximum is at 14.50 mm.

• Therefore, zero-order maximum is 1 fringe earlier:

${𝑦}_0=14.50-0.25=14.25\text{mm}.$

So the zero-order maximum lies at 14.25 mm.

Step 5: Check tenth order

From calculated zero position:

${𝑦}_{10}={𝑦}_0+10\Delta 𝑦=14.25+(10\times 0.25)=14.25+2.50=16.75\text{mm}.$

✔ Matches the given.

So reference point is consistent.

Step 6: For new wavelength $𝜆=5500\text{Å}$

Fringe width (direct proportionality to wavelength):

$\Delta {𝑦}^{\prime }=\Delta 𝑦\times \frac{5500}{6000}=0.25\times \frac{5500}{6000}.$

$\Delta {𝑦}^{\prime }=0.2292\text{mm (approx)}.$

Step 7: Zero-order position

The zero order maximum position does not depend on wavelength (it’s central bright fringe), so it remains at:

${𝑦}_{\prime }^0=14.25\text{mm}.$

Step 8: Tenth-order position with new wavelength

${𝑦}_{\prime }^{10}={𝑦}_{\prime }^0+10\times \Delta {𝑦}^{\prime }=14.25+10\times 0.2292.$

${𝑦}_{\prime }^{10}=14.25+2.292\approx 16.55\text{mm}.$

🔑 Final Answer:

Zero-order and tenth-order maxima positions are:

$\overline{)14.25\text{mm},16.55\text{mm}}$

That corresponds to Option A. ✅

Step 1. Identify given data:

• Intrinsic carrier concentration at 301 K (intrinsic = electrons = holes):

${𝑛}_{𝑖}={𝑝}_{𝑖}={10}^{16}{\text{m}}^{-3}.$

• Doping: ${𝑁}_{𝐷}={10}^{21}{\text{m}}^{-3}$ of phosphorus (donors).

• We want new hole concentration.

Step 2. Law of mass action:

In thermal equilibrium for silicon:

$𝑛\cdot 𝑝={𝑛}_2^{𝑖},$

where:

• $𝑛$ = electron concentration,

• $𝑝$ = hole concentration,

• ${𝑛}_{𝑖}$ = intrinsic carrier concentration.

Step 3. Doping effect:

For an n-type semiconductor (phosphorus donor):

• The electron concentration in conduction band is approximately:

$𝑛\approx {𝑁}_{𝐷}(\text{since }{𝑁}_{𝐷}\gg {𝑛}_{𝑖}).$

Step 4. Calculate hole concentration:

$𝑝=\frac{{𝑛}_2^{𝑖}}{𝑛}\approx \frac{{𝑛}_2^{𝑖}}{{𝑁}_{𝐷}}.$

• Compute: ${𝑛}_2^{𝑖}=({10}^{16}{)}^2={10}^{32}.$

• Divide by ${𝑁}_{𝐷}={10}^{21}$:

$𝑝=\frac{{10}^{32}}{{10}^{21}}={10}^{11}.$

Answer:

The new hole concentration is

$\overline{){10}^{11}{\text{m}}^{-3}}$

Option B.

Step 1: Acceleration of the vehicle

The vehicle moves in a circular horizontal track with

• Speed: $𝑣=10\text{m/s}$

• Radius: $𝑅=20\text{m}$

Centripetal acceleration of the vehicle (and therefore of the bob inside it):

${𝑎}_{𝑐}=\frac{{𝑣}^2}{𝑅}=\frac{{10}^2}{20}=\frac{100}{20}=5{\text{m/s}}^2$

Step 2: Forces on the bob

• Weight: $𝑚𝑔$ downward.

• String tension has vertical and horizontal components:

• Vertical component balances gravity

• Horizontal component provides centripetal acceleration (in the non-inertial frame of the vehicle, this is due to pseudo force).

So, if the string makes angle $𝜃$ with the vertical, then

$\tan 𝜃=\frac{\text{horizontal acceleration}}{\text{vertical acceleration}}=\frac{{𝑎}_{𝑐}}{𝑔}$

Step 3: Substitute values

$\tan 𝜃=\frac{5}{10}=0.5$

$𝜃={\tan }^{-1}\left(0.5\right)$

Final Answer:

$\overline{){\tan }^{-1}\left(0.5\right)}\text{(Option A)}$

$\begin{array}{cc}\text{ R.P. }& =\frac{1.22𝜆}{2n\sin 𝜃}=\frac{1.22\times (500\times {10}^{-9}m)}{2\times 1\times \left(\frac{0.25}{25}\right)}\\ & =3.05\times {10}^{-5}m\approx 30𝜇m\end{array}$

By Gauss’s law, the net outward electric flux through a closed Gaussian surface is

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

Total charge enclosed is

${𝑞}_{\text{enclosed}}=2𝜇𝐶+(-3𝜇𝐶)+4𝜇𝐶+(-4𝜇𝐶)+(-1𝜇𝐶)$

${𝑞}_{\text{enclosed}}=(2-3+4-4-1)𝜇𝐶$

${𝑞}_{\text{enclosed}}=-2𝜇𝐶$

So,

$\Phi =\frac{-2𝜇𝐶}{{𝜀}_0}$

Since the charge is negative, the net flux is inward, so the outward flux is negative:

$\Phi =-\frac{2}{{𝜀}_0}𝜇𝑉\text{-}𝑚$

But among the given options, only the magnitude is present:

$\left|\Phi \right|=\frac{2}{{𝜀}_0}$

Therefore, the correct option is:

$\overline{)\text{Option A }\frac{2}{{𝜀}_0}}$

${𝐼}_{\text{sphere }}={𝐼}_s=\frac{2}{5}𝑚{𝑅}^2$

Let ${𝜔}_{𝑠}$ be angular speed of sphere,

$\begin{array}{cc}\therefore {𝐸}_{\text{sphere }}& =\frac{1}{2}{𝐼}_{𝑠}{𝜔}_2^{𝑠}\\ & =\frac{1}{2}\left(\frac{2}{5}𝑚{𝑅}^2\right){𝜔}_2^{𝑠}...\left(𝑖\right)\end{array}$

Similarly,

$I_{\text{cylinder }}=I_c=\frac{1}{2}{\mathrm{mR}}^2$

Let ${𝜔}_{𝑐}$ be the angular speed of cylinder, Then it is given

$\begin{array}{cc}{𝜔}_{𝑐}=2{𝜔}_{𝑠}& \\ \therefore E_{\text{cylinder }}& =\frac{1}{2}{I}_{𝑐}{𝜔}_2^{𝑐}\\ & =\frac{1}{2}\left(\frac{1}{2}{\mathrm{mR}}^2\right){\left(2{𝜔}_{𝑠}\right)}^2\dots \text{ (ii) }\\ \therefore \frac{E_{\text{sphere }}}{E_{\text{cylinder }}}& =\frac{\frac{1}{2}\left(\frac{2}{5}{\mathrm{mR}}^2\right){𝜔}_2^{𝑠}}{\frac{1}{2}\left(\frac{1}{2}{\mathrm{mR}}^2\right)\left(4{𝜔}_2^{𝑠}\right)}\dots \left[\text{ From (i) and (ii) }\right]\\ & =\frac{1}{5}\end{array}$

A person stands between two parallel cliffs, fires a gun, and hears two echoes:

• First echo after 1 second

• Second echo after 3 seconds

Speed of sound $𝑣=340m/s$.

We want the distance between the two cliffs.

Step 1: Relation for echo

For an echo, sound travels from the person to the cliff and back.

So if distance from the person to a cliff is $𝑑$, the round-trip distance is $2𝑑$.

Echo time is

$𝑡=\frac{2𝑑}{𝑣}.$

So,

$𝑑=\frac{𝑣𝑡}{2}.$

Step 2: Find distances to the two cliffs

• For the first echo ($𝑡=1s$):

${𝑑}_1=\frac{340\times 1}{2}=170m.$

• For the second echo ($𝑡=3s$):

${𝑑}_2=\frac{340\times 3}{2}=510m.$

Step 3: Distance between cliffs

The person stands between cliffs, so one cliff is $170m$ away, the other is $510m$ away.

Thus, the distance between the two cliffs is

${𝑑}_1+{𝑑}_2=170+510=680m.$

✅ Final Answer:

Option B: 680 m

Case 1: Series combination of $10$ capacitors, each ${𝐶}_1$.

• Equivalent capacitance of capacitors in series:

$\frac{1}{{𝐶}_{\text{eq1}}}=\frac{1}{{𝐶}_1}+\frac{1}{{𝐶}_1}+\dots \left(10\text{ terms}\right)=\frac{10}{{𝐶}_1}$

${𝐶}_{\text{eq1}}=\frac{{𝐶}_1}{10}.$

• Energy stored when connected across $4\text{V}$:

${𝑈}_1=\frac{1}{2}{𝐶}_{\text{eq1}}{𝑉}^2=\frac{1}{2}\cdot \frac{{𝐶}_1}{10}\cdot (4{)}^2=\frac{1}{2}\cdot \frac{{𝐶}_1}{10}\cdot 16=\frac{8}{10}{𝐶}_1=0.8{𝐶}_1.$

Case 2: Parallel combination of $8$ capacitors, each of capacitance ${𝐶}_2$.

• Equivalent capacitance of capacitors in parallel:

${𝐶}_{\text{eq2}}=8{𝐶}_2.$

• Energy stored when connected across $𝑉$:

${𝑈}_2=\frac{1}{2}{𝐶}_{\text{eq2}}{𝑉}^2=\frac{1}{2}\left(8{𝐶}_2\right){𝑉}^2=4{𝐶}_2{𝑉}^2.$

Condition: Energy equal

${𝑈}_1={𝑈}_2$

$0.8{𝐶}_1=4{𝐶}_2{𝑉}^2$

${𝐶}_2=\frac{0.8}{4{𝑉}^2}{𝐶}_1=\frac{0.2}{{𝑉}^2}{𝐶}_1=\frac{{𝐶}_1}{5{𝑉}^2}.$

Check the problem statement

The second system ("parallel combination… charged by a source of potential difference $𝑉$") uses the same total energy as the first.

So the relation indeed gives:

${𝐶}_2=\frac{{𝐶}_1}{5{𝑉}^2}.$

Now, look at the given options:

• A. $\frac{{𝐶}_1}{5}$

• B. $\frac{8}{5}{𝐶}_1$

• C. $\frac{64}{5}{𝐶}_1$

• D. $\frac{{𝐶}_1}{40}$

This suggests in the problem, $𝑉$ may actually equal 1 V (i.e., the wording "charged by source potential difference 'V'" intends $𝑉=1$ volt).

If $𝑉=1$:

${𝐶}_2=\frac{{𝐶}_1}{5}.$

✅ Final Answer:

$\overline{)\frac{{𝐶}_1}{5}}$

So the correct option is A.

$\begin{array}{cc}\therefore & I_s=\frac{2}{5}{\mathrm{MR}}^2,I_h=\frac{2}{3}{\mathrm{MR}}^2\\ \therefore & \frac{I_s}{I_h}=\frac{\left(\frac{2}{5}{\mathrm{MR}}^2\right)}{\left(\frac{2}{3}{\mathrm{MR}}^2\right)}=\frac{3}{5}<1\\ \therefore & \frac{I_s}{I_h}<1\Rightarrow I_h>I_s\end{array}$

• Faraday’s law:

  Induced emf $E=-\frac{𝑑\Phi }{𝑑𝑡}$, where $\Phi$ is the magnetic flux through the loop.

• The induced current is $𝐼=\frac{E}{𝑅}$, where $𝑅$ is the resistance of the loop.

• The total charge that flows through the circuit during the entire process:

  $𝑄=\int 𝐼𝑑𝑡=\frac{1}{𝑅}\int E𝑑𝑡.$

Step 2: Relation to flux.

$𝑄=\frac{1}{𝑅}\int -\frac{𝑑\Phi }{𝑑𝑡}𝑑𝑡=\frac{1}{𝑅}({\Phi }_{\text{initial}}-{\Phi }_{\text{final}}).$

Thus, the total charge depends on the net change in magnetic flux, not on the rate at which it changes.

Answer:

$\overline{)\text{Option C: the total change in magnetic flux.}}$

$𝑊=\text{Surface tension}\times \text{increase in surface area}$

But since a film has two surfaces,

$𝑊=2𝑇\Delta 𝐴$

Step 2: Initial and final areas

${𝐴}_{𝑖}=5\times 4=20{\text{cm}}^2$

${𝐴}_{𝑓}=7\times 8=56{\text{cm}}^2$

So,

$\Delta 𝐴={𝐴}_{𝑓}-{𝐴}_{𝑖}=36{\text{cm}}^2$

Convert to m²:

$1{\text{cm}}^2={10}^{-4}{\text{m}}^2$

$\Delta 𝐴=36\times {10}^{-4}=3.6\times {10}^{-3}{\text{m}}^2$

Step 3: Apply formula

$𝑊=2𝑇\Delta 𝐴$

$3\times {10}^{-4}=2𝑇(3.6\times {10}^{-3})$

$3\times {10}^{-4}=7.2\times {10}^{-3}𝑇$

$𝑇=\frac{3\times {10}^{-4}}{7.2\times {10}^{-3}}\approx 0.0417\text{N/m}$

Answer:

$\overline{)0.04\text{N/m}}$

Correct option: B

The temperature of a body on Kelvin scale = $𝑥$ K

On Fahrenheit scale = $𝑥$ ${}^{\circ }𝐹$

Find $𝑥$.

Step 1: Relation between Kelvin and Celsius

$𝑇{(}^{\circ }𝐶)=𝑇(𝐾)-273$

So if $𝑇=𝑥$ K, then $𝑇=𝑥-273$ ${}^{\circ }𝐶$.

Step 2: Relation between Celsius and Fahrenheit

$𝑇{(}^{\circ }𝐹)=\frac{9}{5}𝑇{(}^{\circ }𝐶)+32$

So,

$𝑥=\frac{9}{5}(𝑥-273)+32$

Step 3: Simplify

$𝑥=\frac{9}{5}𝑥-\frac{9}{5}\cdot 273+32$

$𝑥-\frac{9}{5}𝑥=-\frac{9}{5}\cdot 273+32$

$(1-\frac{9}{5})𝑥=-\frac{9}{5}\cdot 273+32$

$(-\frac{4}{5})𝑥=-\frac{2457}{5}+32$

$(-\frac{4}{5})𝑥=-491.4+32$

$(-\frac{4}{5})𝑥=-459.4$

Step 4: Solve for $𝑥$

$𝑥=\frac{-459.4\cdot 5}{-4}$

$𝑥=\frac{2297}{4}\approx 574.25$

Final Answer:

$𝑥\approx 574$

✔ Correct Option: C (574)

Concept

The formula for capillary rise $ℎ$ is:

$ℎ=\frac{2𝑇\cos 𝜃}{𝜌𝑔𝑟}$

where

• $𝑇$: surface tension of the liquid,

• $𝜃$: angle of contact,

• $𝜌$: density of liquid,

• $𝑔$: acceleration due to gravity,

• $𝑟$: radius of capillary.

So:

$ℎ\propto \frac{1}{𝑔}.$

On Earth

Let ${ℎ}_{𝐸}=\frac{2𝑇\cos 𝜃}{𝜌{𝑔}_{𝐸}𝑟}$.

On Moon

Since ${𝑔}_{𝐸}=6{𝑔}_{𝑀}$:

${ℎ}_{𝑀}=\frac{2𝑇\cos 𝜃}{𝜌{𝑔}_{𝑀}𝑟}.$

$\frac{{ℎ}_{𝑀}}{{ℎ}_{𝐸}}=\frac{{𝑔}_{𝐸}}{{𝑔}_{𝑀}}=\frac{6{𝑔}_{𝑀}}{{𝑔}_{𝑀}}=6.$

Final Answer

On the Moon, the liquid column will rise six times higher than on Earth.

Correct Option: B) six times that on the earth's surface. ✅

Two thin lenses are in contact, so the resultant focal length will be:

$\begin{array}{cc}& \frac{1}{𝑓}=\frac{1}{{𝑓}_1}+\frac{1}{{𝑓}_2}\\ & 𝑓=\frac{{𝑓}_1{𝑓}_2}{{𝑓}_1+{𝑓}_2}\end{array}$

$\therefore$ Power of lens: $𝑃=\frac{1}{𝑓}=\frac{{𝑓}_1+{𝑓}_2}{{𝑓}_1{𝑓}_2}$

Since $2𝑓=𝑅$, we have:

Power of the lens, $𝑃=\frac{\frac{{𝑅}_1}{2}+\frac{{𝑅}_2}{2}}{\frac{{𝑅}_1}{2}\frac{{𝑅}_2}{2}}=2\frac{{𝑅}_1+{𝑅}_2}{{𝑅}_1{𝑅}_2}$

According to the kinetic theory of gases, the average kinetic energy (K.E.) of gas molecules is given as:

$\text{ K.E. }=\frac{3}{2}\mathrm{kT}$

This means that at a given temperature, the average kinetic energy of all gas molecules will be the same.

Step 1: Recall basic property

In a common emitter (CE) transistor amplifier:

• The voltage gain is negative.

• This means that the output signal is inverted relative to the input signal.

Step 2: Phase relationship

• A signal inversion corresponds to a phase shift of ${180}^{\circ }$ (or $𝜋$ radians) between input and output.

Step 3: Match with options

• Option A: $0^{\circ }$ → No shift. ❌

• Option B: $𝜋/2$ → 90°. ❌

• Option C: $3𝜋/4$ → 135°. ❌

• Option D: $𝜋$ → 180°. ✅

✅ Correct Answer:

Option D: $𝜋$ (i.e., 180° phase difference)

• Mass: $𝑚=100\text{g}=0.1\text{kg}$

• Spring constant: $𝑘=8\text{N/m}$

• Angular speed: $𝜔=8\text{rad/s}$

• Motion: Mass is moving in circle on smooth surface, spring attached to center.

We are asked: ratio of extension ($𝑥$) to natural length ($𝑙$).

Step 1: Force Balance

Centripetal force is provided by spring force:

$𝑘𝑥=𝑚{𝜔}^2𝑟$

Here $𝑟$ is radius of circular motion = natural length $𝑙$ + extension $𝑥$:

$𝑟=𝑙+𝑥$

So:

$𝑘𝑥=𝑚{𝜔}^2(𝑙+𝑥)$

Step 2: Plug Known Values

Substitute $𝑚=0.1$, $𝜔=8$, $𝑘=8$:

$8𝑥=\left(0.1\right)\left(64\right)(𝑙+𝑥)$

$8𝑥=6.4(𝑙+𝑥)$

Step 3: Rearrange

$8𝑥-6.4𝑥=6.4𝑙$

$1.6𝑥=6.4𝑙$

$\frac{𝑥}{𝑙}=\frac{6.4}{1.6}=4$

Final Answer:

$\frac{\text{extension}}{\text{natural length}}=4:1$

✅ Correct Option: D (4:1)

Step 1: Understand

• By default, voltmeter is rated 10 V → At 10 V, the voltmeter draws current

  $𝐼=\frac{𝑉}{𝑅}=\frac{10}{50}=0.2\text{ A}$

So the full-scale deflection current = 0.2 A.

Step 2: New Range

We want voltmeter to read up to 15 V (instead of 10 V).

For 15 V applied, same current (0.2 A) must flow through the meter.

More voltage → we must drop extra 5 V somewhere with some series resistance.

Step 3: Series Resistance Calculation.

At full-scale deflection, current through meter = 0.2 A (constant).

Additional resistance:

${𝑅}_{𝑠𝑒𝑟𝑖𝑒𝑠}=\frac{{𝑉}_{𝑒𝑥𝑡𝑟𝑎}}{𝐼}=\frac{5}{0.2}=25\Omega$

Step 4: Check Options

• Option A: 125 Ω parallel → modifies current division, not correct.

• Option B: 125 Ω series → too large.

• Option C: 25 Ω parallel → wrong effect.

• Option D: 25 Ω series → ✅ correct.

Final Answer:

$\overline{)25\Omega \text{ resistance in series (Option D)}}$

We are given:

• Thermal energy incident = ${𝑄}_{\text{incident}}=120\text{J}$

• Heat transmitted = ${𝑄}_{\text{transmitted}}=12\text{J}$

• Coefficient of absorption = $𝑎=0.6$

Step 1: Absorbed heat

Coefficient of absorption is

$𝑎=\frac{{𝑄}_{\text{absorbed}}}{{𝑄}_{\text{incident}}}$

So,

${𝑄}_{\text{absorbed}}=𝑎\cdot {𝑄}_{\text{incident}}=0.6\times 120=72\text{J}$

Step 2: Energy conservation

Total incident energy is partitioned as:

${𝑄}_{\text{incident}}={𝑄}_{\text{absorbed}}+{𝑄}_{\text{reflected}}+{𝑄}_{\text{transmitted}}$

So,

$120=72+{𝑄}_{\text{reflected}}+12$

${𝑄}_{\text{reflected}}=120-84=36\text{J}$

✅ Answer:

The amount of heat reflected is

36 J

Correct Option: C (36 J)

For a circular coil of radius $𝑅$, carrying a current $𝐼$, with $𝑛$ turns:

$𝐵=\frac{{𝜇}_0𝑛𝐼}{2𝑅}$

Step 2: Substitute known values

• ${𝜇}_0=12.56\times {10}^{-7}$ H/m

• Radius: $𝑅=8\text{ cm}=0.08\text{ m}$

• Current: $𝐼=0.4\text{ A}$

• Magnetic field: $𝐵=3.14\times {10}^{-4}\text{ T}$

Step 3: Rearrange for $𝑛$

$𝑛=\frac{2𝑅𝐵}{{𝜇}_0𝐼}$

Step 4: Plug in numbers

$𝑛=\frac{2\times 0.08\times 3.14\times {10}^{-4}}{(12.56\times {10}^{-7})\left(0.4\right)}$

First, numerator:

$2\times 0.08=0.16,0.16\times 3.14\times {10}^{-4}=0.5024\times {10}^{-4}=5.024\times {10}^{-5}$

Denominator:

$12.56\times {10}^{-7}\times 0.4=5.024\times {10}^{-7}$

So,

$𝑛=\frac{5.024\times {10}^{-5}}{5.024\times {10}^{-7}}=100$

✅ Final Answer:

The number of turns is

Option C: 100

There are two identical small holes on the opposite sides of a tank, hence velocities of liquid flowing out of both the holes are

$v_1=\sqrt{\surd 2g(h+x)}\text{ and }{v}_2=\sqrt{\surd 2\mathrm{gx}}$

Now, we know:

Volume of the liquid discharged per second at a hole: $V=\mathrm{Av}$.

Thus, mass of the liquid discharged per second:

$𝑀=𝜌𝑉=𝜌𝐴𝑣$

$\therefore$ Momentum of the liquid discharged per second

$𝑝=𝜌𝐴{𝑣}^2$

Now, Net force on the tank = Change in momentum

$\begin{array}{cc}F& =A𝜌v_2^1-A𝜌v_2^2=A𝜌(v_2^1-v_2^2)\\ & =A𝜌\left[2g\right(h+x)-2\mathrm{gx}]\\ & =2A𝜌\mathrm{gh}\end{array}$

$\Rightarrow 𝐹\propto ℎ$

In the graph, $V_A$ and $V_B$ are the voltages across parallel plate capacitors A and B corresponding to charge Q on each of the capacitors.

$V_A=Q/C_A$ and $V_B=Q/C_B$

or $\frac{{𝑉}_{𝐵}}{{𝑉}_{𝐴}}=\frac{\frac{𝑄}{{𝐶}_{𝐵}}}{\frac{𝑄}{{𝐶}_{𝐴}}}=\frac{{𝐶}_{𝐴}}{{𝐶}_{𝐵}}$

Since, ${𝑉}_{𝐵}>{𝑉}_{𝐴},\frac{{𝑉}_{𝐵}}{{𝑉}_{𝐴}}=\frac{{𝐶}_{𝐴}}{{𝐶}_{𝐵}}>1$

$\therefore {𝐶}_{𝐴}>{𝐶}_{𝐵}$, i.e. capacitor $𝐴$ has the higher capacitance.

$\begin{array}{cc}\overrightarrow{𝜏}=\overrightarrow{𝑟}\times \overrightarrow{𝐹}& =\mid \begin{array}{ccc}\widehat{𝑖}& \widehat{𝑗}& \widehat{𝑘}\\ 7& 3& 1\\ -3& 1& 5\end{array}\mid \\ & =\widehat{𝑖}(15-1)-\widehat{𝑗}(35+3)+\widehat{𝑘}(7+9)\\ & =(14\widehat{𝑖}-38\widehat{𝑗}+16\widehat{𝑘})N-m\end{array}$

For a paramagnetic substance, the magnetic susceptibility (χ):

• Is positive (since paramagnetic materials are weakly attracted to a magnetic field).

• But it is relatively small in magnitude compared to ferromagnetic substances.

So, the correct answer is:

Option D — positive and small. ✅

Step 1: Recall relation between Cv, Cp, and γ

$𝛾=\frac{{𝐶}_{𝑝}}{{𝐶}_{𝑣}},{𝐶}_{𝑝}-{𝐶}_{𝑣}=𝑅$

Given:

$𝛾=\frac{5}{3}$

Step 2: Express ${𝐶}_{𝑣}$ and ${𝐶}_{𝑝}$ in terms of R

$\frac{{𝐶}_{𝑝}}{{𝐶}_{𝑣}}=\frac{5}{3}$

${𝐶}_{𝑝}=\frac{5}{3}{𝐶}_{𝑣}$

And:

${𝐶}_{𝑝}-{𝐶}_{𝑣}=𝑅⟹\frac{5}{3}{𝐶}_{𝑣}-{𝐶}_{𝑣}=𝑅⟹\frac{2}{3}{𝐶}_{𝑣}=𝑅$

So:

${𝐶}_{𝑣}=\frac{3}{2}𝑅,{𝐶}_{𝑝}=\frac{5}{2}𝑅$

Step 3: Fraction of heat used for work in isobaric process

At constant pressure, if temperature rises by $\Delta 𝑇$:

• Heat supplied:

$𝑄=𝑛{𝐶}_{𝑝}\Delta 𝑇$

• Work done:

$𝑊=𝑛𝑅\Delta 𝑇$

• Fraction used for work:

$\frac{𝑊}{𝑄}=\frac{𝑛𝑅\Delta 𝑇}{𝑛{𝐶}_{𝑝}\Delta 𝑇}=\frac{𝑅}{{𝐶}_{𝑝}}$

Step 4: Substitute values

$\frac{𝑊}{𝑄}=\frac{𝑅}{\frac{5}{2}𝑅}=\frac{2}{5}=0.4$

So:

$40\%$

✅ Final Answer:

Option D: $40\%$

The potential difference across resistance $𝑅$ is 6 V . Hence, current, $i=\frac{10-6}{4}=\frac{4}{4}=1A$ Thus, $1\times R=6$

$R=6\Omega$

From $\frac{\mathrm{hc}}{𝜆}-𝜙=\mathrm{eV}$,

We can write,

$\frac{\mathrm{hc}}{{𝜆}_1}={\mathrm{eV}}_1+𝜙...\left(𝑖\right)$

and

$\frac{\mathrm{hc}}{3{𝜆}_1}=\frac{{\mathrm{eV}}_1}{6}+𝜙...\left(𝑖𝑖\right)$

Equation (ii) can be rewritten as,

$\frac{2\mathrm{hc}}{{𝜆}_1}={\mathrm{eV}}_1+6𝜙...\left(𝑖𝑖𝑖\right)$

On subtracting equation (i) from (iii), we have $\frac{\mathrm{hc}}{{𝜆}_1}=5𝜙$ and $𝜙=\frac{\mathrm{hc}}{{𝜆}_0}$

$\therefore {𝜆}_0=5{𝜆}_1$

Step 1: Recall the formula

For an ideal gas, the mean kinetic energy per molecule is proportional to absolute temperature:

$⟨𝐾𝐸⟩=\frac{3}{2}{𝑘}_{𝐵}𝑇$

So:

$⟨𝐾𝐸⟩\propto 𝑇\left(\text{in Kelvin}\right)$

Step 2: Initial conditions

Given that at ${𝑇}_1={399}^{\circ }\text{C}$:

${𝑇}_1=399+273=672\text{K}$

The mean kinetic energy $𝐸\propto {𝑇}_1$.

Step 3: Required condition

We want $𝐸/2$. Since KE is proportional to $𝑇$:

$\frac{𝐸}{2}\propto \frac{{𝑇}_1}{2}$

So the required temperature:

${𝑇}_2=\frac{{𝑇}_1}{2}=\frac{672}{2}=336\text{K}$

Step 4: Convert back to Celsius

${𝑇}_2\left({}^{\circ }\text{C}\right)=336-273={63}^{\circ }\text{C}$

✅ Final Answer:

$\overline{){63}^{\circ }\text{C}}$

Correct option: D.

Step 1: Equivalent capacitance (for series)

For capacitors in series:

$\frac{1}{{𝐶}_{\text{eq}}}=\frac{1}{𝐶}+\frac{1}{𝐶}+\frac{1}{𝐶}=\frac{3}{𝐶}$

So,

${𝐶}_{\text{eq}}=\frac{𝐶}{3}.$

Step 2: Breakdown voltage

Each capacitor can withstand a maximum voltage of $𝑉$ before breakdown.

When three are in series, the total applied voltage is divided (equally, since they have the same capacitance). Thus, each capacitor gets $\frac{{𝑉}_{\text{total}}}{3}$.

For safe operation, none should exceed $𝑉$.

So the maximum total breakdown voltage is:

${𝑉}_{\text{eq}}=3𝑉$

Final Answer:

${𝐶}_{\text{eq}}=\frac{𝐶}{3},{𝑉}_{\text{eq}}=3𝑉$

✅ Correct option: D