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

The kinetic energy of a body is:

$\text{ K.E. }=\frac{1}{2}𝑚{𝑣}^2$

The percentage error in K.E. is,

$\frac{\Delta K.E.}{K.E.}=\frac{\Delta m}{m}+2\frac{\Delta v}{v}$

Given: $\frac{\Delta 𝑚}{𝑚}\times 100=3\%$ and $\frac{\Delta 𝑣}{𝑣}\times 100=4\%$ :

$\therefore \frac{\Delta \text{ K.E. }}{\text{ K.E. }}\times 100=3\%+2\left(4\%\right)=3\%+8\%=11\%$

$\begin{array}{cc}& {𝑣}_{max}=\sqrt{\surd 𝑟𝑔\left(\frac{\tan 𝜃+{𝜇}_{𝑠}}{1-{𝜇}_{𝑠}\tan 𝜃}\right)}\\ & {𝑣}_{max}\propto \sqrt{\surd 𝑟}\\ \therefore & \frac{{𝑣}_2}{{𝑣}_1}=\sqrt{\surd \frac{{𝑟}_2}{{𝑟}_1}}\\ \therefore & {\left(\frac{{𝑣}_2}{{𝑣}_1}\right)}^2=\frac{{𝑟}_2}{{𝑟}_1}\\ \therefore & {𝑟}_2={\left(\frac{{𝑣}_2}{{𝑣}_1}\right)}^2{𝑟}_1.\end{array}$

When safety speed is increased by $20\%$,

${𝑣}_2=1.2{𝑣}_1$

Also given that, $r_1=20m$

$\therefore {𝑟}_2=(1.2{)}^2\times 20=1.44\times 20=28.8m$

$\therefore$ Increase in radius $=28.8-20=8.8m$

$\text{ From the figure, }$

$\begin{array}{cc}& \text{ Restoring force }F=-\mathrm{mg}\sin 𝜃\\ & \begin{array}{cc}& \sin 𝜃=𝜃=\frac{𝑥}{𝑅-𝑥}\\ \therefore & 𝐹=-𝑚𝑔\left(\frac{𝑥}{𝑅-𝑟}\right)\Rightarrow 𝐹=-\left(\frac{𝑚𝑔}{𝑅-𝑟}\right)𝑥\\ \therefore & 𝐾=\frac{𝑚𝑔}{𝑅-𝑟}\\ \therefore & 𝑚{𝜔}^2=\frac{𝑚𝑔}{𝑅-𝑟}\\ \therefore & 𝜔=\sqrt{\surd \frac{𝑔}{𝑅-𝑟}}\\ \therefore & 𝑇=\frac{2𝜋}{𝜔}=2𝜋\sqrt{\surd \frac{𝑅-𝑟}{𝑔}}\end{array}\end{array}$

Given: $𝑙=6\mathrm{cm}=6\times {10}^{-2}m,q=3𝜇C$

Potential energy of any 2 given charges will be

$U=\frac{1}{4𝜋{𝜀}_0}\frac{q_1{q}_2}{r}=\frac{1}{4𝜋{𝜀}_0}\frac{q^2}{r}$

$\therefore$ Net potential energy of the system will be

$\begin{array}{cc}{U}_{\text{net }}=3\times \frac{1}{4𝜋{𝜀}_0}\frac{{(3\times {10}^{-6})}^2}{6\times {10}^{-2}}& =3\times 9\times {10}^9\times \frac{9\times {10}^{-12}}{6\times {10}^{-2}}\\ & =4.05J\approx 4.1J\end{array}$

Step 1: Recall what hysteresis means

• Magnetic hysteresis refers to the lag between magnetization (M) and the applied magnetic field (H).

• When a ferromagnetic material is magnetized and then the field is removed, it retains residual magnetization (remanence). To demagnetize, a reverse field (coercivity) must be applied.

• This hysteresis loop is a characteristic property of ferromagnetic materials.

Step 2: Analyze the options

• Paramagnetic: Do not show hysteresis. Magnetization disappears instantly when external field is removed.

• Diamagnetic: Weak and opposite response, no hysteresis.

• Ferromagnetic: Show hysteresis because of domain alignment and tendency to retain magnetization.

• Both para and ferromagnetic: Incorrect, since only ferromagnetic materials show hysteresis.

✅ Correct Answer: Option C – only ferromagnetic

$\begin{array}{cc}& \overrightarrow{𝐴}=(2\widehat{𝑖}-3\widehat{𝑗}+\widehat{𝑘})\\ & \overrightarrow{𝐵}=(3\widehat{𝑖}+\widehat{𝑗}-2\widehat{𝑘})\\ & \begin{array}{cc}& \overrightarrow{𝐶}=(3\widehat{𝑖}+2\widehat{𝑗}+\widehat{𝑘})\\ & \overrightarrow{𝐴}+\overrightarrow{𝐵}=(2+3)\widehat{𝑖}+(-3+1)\widehat{𝑗}+(1-2)\widehat{𝑘}\\ & =5\widehat{𝑖}-2\widehat{𝑗}-\widehat{𝑘}\\ & \begin{array}{cc}(\overrightarrow{𝐴}+\overrightarrow{𝐵})\cdot \overrightarrow{𝐶}& =(5\widehat{𝑖}-2\widehat{𝑗}-\widehat{𝑘})\cdot (3\widehat{𝑖}+2\widehat{𝑗}+\widehat{𝑘})\\ & =15-4-1\\ & =10\end{array}\end{array}\end{array}$

We’re asked to compute something involving vectors:

$\overrightarrow{𝐴}=(2\widehat{𝑖}-3\widehat{𝑗}+\widehat{𝑘}),\overrightarrow{𝐵}=(3\widehat{𝑖}+\widehat{𝑗}-2\widehat{𝑘}),\overrightarrow{𝐶}=(3\widehat{𝑖}+2\widehat{𝑗}+\widehat{𝑘}).$

Expression: $(\overrightarrow{𝐴}+\overrightarrow{𝐵})\cdot \overrightarrow{𝐶}$.

Step 1. Add A and B:

$\overrightarrow{𝐴}+\overrightarrow{𝐵}=(2+3)\widehat{𝑖}+(-3+1)\widehat{𝑗}+(1-2)\widehat{𝑘}=5\widehat{𝑖}-2\widehat{𝑗}-\widehat{𝑘}.$

So $\overrightarrow{𝐴}+\overrightarrow{𝐵}=(5,-2,-1).$

$\overrightarrow{𝐶}=(3,2,1).$

Step 2. Dot product:

$(\overrightarrow{𝐴}+\overrightarrow{𝐵})\cdot \overrightarrow{𝐶}=\left(5\right)\left(3\right)+(-2)\left(2\right)+(-1)\left(1\right)=15-4-1=10.$

✅ Final Answer:

$(\overrightarrow{𝐴}+\overrightarrow{𝐵})\cdot \overrightarrow{𝐶}=10$

So the correct option is A. 10.

$\text{ Truth table for the following logic circuit, }$

$\overline{)\begin{array}{ccccc}A& B& C& \overline{B+C}& Y\\ 0& 0& 0& 1& 0\\ 0& 0& 1& 0& 0\\ 0& 1& 0& 0& 0\\ 0& 1& 1& 0& 0\\ 1& 0& 0& 1& 1\\ 1& 0& 1& 0& 0\\ 1& 1& 0& 0& 0\\ 1& 1& 1& 0& 0\end{array}}$

$𝑥$-component of velocity

$v_x=\frac{\mathrm{dx}}{\mathrm{dt}}=\frac{d}{\mathrm{dt}}\left({\mathrm{at}}^2\right)=2\mathrm{at}$

$𝑦$ - component of velocity

${𝑣}_{𝑦}=\frac{𝑑𝑦}{𝑑𝑡}=\frac{𝑑}{𝑑𝑡}\left(𝑏{𝑡}^2\right)=2𝑏𝑡$

Speed of the particle is the magnitude of the velocity $v=\sqrt{\surd v_2^x+v_2^y}$

$\begin{array}{cc}\therefore 𝑣& =\sqrt{\surd (2𝑎𝑡{)}^2+(2𝑏𝑡{)}^2}\\ & =\sqrt{\surd 4{𝑎}^2{𝑡}^2+4{𝑏}^2{𝑡}^2}\\ & =2𝑡\sqrt{\surd {𝑎}^2+{𝑏}^2}\end{array}$

Speed of the particle is $2𝑡\sqrt{\surd {𝑎}^2+{𝑏}^2}$

$A\to B+{}_2{\mathrm{He}}^4;B\to C+2e^{-}$

Let $𝐴$ be : ${}_{𝑧}{𝐴}^{𝐴}$

In $A\to B$ decay, ${}_2{\mathrm{He}}^4$ is released

$\therefore A\to B$ is alpha decay reaction

$\therefore {}_{𝑧}{A}^{\wedge }\to {}_{𝑧-2}{B}^{\wedge -4}+{}_2{\mathrm{He}}^4$

In $B\to C$ decay, $2e^{-}$is released

$\therefore B\to C$ is ${𝛽}^{-}$decay x action

$z-2B^{N-4}\to {}_z{C}^{\Lambda -4}$          ( $\because$ As each ${𝛽}^{-}$decay increases Z by 1 )

Now A and C have the same atomic number but different mass numbers.

$\therefore$ A and C are isotope.

Both the semi-circular arc of a ring and the complete ring have the same moment of inertia because the mass in both cases is distributed at the same radial distance R from the central axis. Thus, the moment of inertia is determined solely by the distance R and the total mass M , not by the shape of the distribution along the ring.

M.I. of ring about diameter $I=\frac{{\mathrm{MR}}^2}{2}\dots$. (i)

$\because L=𝜋R\Rightarrow R=L/𝜋$

$\therefore$ From equation (i), $I=\frac{{\mathrm{ML}}^2}{2{𝜋}^2}$

Speed at displacement x

$𝑢=𝜔\sqrt{\surd {𝑎}^2-{𝑥}^2}$

Speed of particle at a certain instant is half that of maximum speed $V_{\text{max }}$

$\begin{array}{cc}\therefore & u=\frac{V_{max}}{2}=\frac{𝜔a}{2}\\ \therefore & \frac{𝜔a}{2}=𝜔\sqrt{\surd a^2-x^2}\\ & a^2-x^2=\frac{a^2}{4}\\ & x^2=\frac{3}{4}{a}^2\\ & x=\frac{\sqrt{\surd 3}}{2}a\end{array}$

Energy will remain the same due to conservation of energy

In case 1 , each capacitor is charged to p.d. V So, when they are connected in series, the p.d. will be nV

Step 1. Recall Bohr’s quantization condition

Bohr’s model says:

$𝑚𝑣𝑟=𝑛\hslash \left(1\right)$

And also, de Broglie’s condition:

$2𝜋𝑟=𝑛𝜆\left(2\right)$

Thus, the circumference of the orbit equals an integer multiple of the de Broglie wavelength.

Step 2. Bohr’s orbit radius

For a hydrogen atom,

${𝑟}_{𝑛}={𝑎}_0{𝑛}^2,$

where ${𝑎}_0$ is the Bohr radius.

Step 3. Relation between λ and n

From equation (2):

$𝜆=\frac{2𝜋{𝑟}_{𝑛}}{𝑛}.$

Substitute ${𝑟}_{𝑛}={𝑎}_0{𝑛}^2$:

$𝜆=\frac{2𝜋\left({𝑎}_0{𝑛}^2\right)}{𝑛}=2𝜋{𝑎}_0𝑛.$

Step 4. Conclusion

So, the de Broglie wavelength is directly proportional to $𝑛$.

$\overline{)𝜆\propto 𝑛}$

Correct Option: D ✅

From Newton's second Law

$F=\mathrm{ma}\Rightarrow a=\frac{F}{m}$

Acceleration $A_1$ in mass $m_1$,

${𝐴}_1=\frac{𝐹}{{𝑚}_1}\Rightarrow {𝑚}_1=\frac{𝐹}{{𝐴}_1}$

Acceleration ${𝐴}_2$ in mass ${𝑚}_2$,

${𝐴}_2=\frac{𝐹}{{𝑚}_2}\Rightarrow {𝑚}_2=\frac{𝐹}{{𝐴}_2}$

Acceleration ' a ' of combined Mass.

$\therefore 𝑎=\frac{𝐹}{{𝑚}_1+{𝑚}_2}=\frac{𝐹}{\frac{𝐹}{{𝐴}_1}+\frac{𝐹}{{𝐴}_2}}=\frac{1}{\frac{1}{{𝐴}_1}+\frac{1}{{𝐴}_2}}=\frac{{𝐴}_1{𝐴}_2}{{𝐴}_1+{𝐴}_2}$

Step 1: Potentiometer principle

In potentiometer experiments,

$\text{E.M.F.}\propto \text{balancing length}.$

So if a cell of emf $𝐸$ alone is connected, balancing length $𝑙\propto 𝐸$.

Thus,

${𝐸}_1\propto {𝑙}_1.$

Step 2: When cells oppose

If another cell ${𝐸}_2$ is connected in opposition, the effective emf across the potentiometer = ${𝐸}_1-{𝐸}_2$.

Its balancing length should be proportional:

${𝐸}_1-{𝐸}_2\propto {𝑙}_2.$

Step 3: Ratio relations

From step 1:

$\frac{{𝐸}_1}{{𝐸}_1-{𝐸}_2}=\frac{{𝑙}_1}{{𝑙}_2}$.

⇒ ${𝐸}_1{𝑙}_2=({𝐸}_1-{𝐸}_2){𝑙}_1$.

⇒ ${𝐸}_1{𝑙}_2={𝐸}_1{𝑙}_1-{𝐸}_2{𝑙}_1$.

⇒ ${𝐸}_2{𝑙}_1={𝐸}_1({𝑙}_1-{𝑙}_2)$.

⇒ $\frac{{𝐸}_1}{{𝐸}_2}=\frac{{𝑙}_1}{{𝑙}_1-{𝑙}_2}$.

Step 4: Final Answer

So the ratio is

$\overline{)\frac{{𝐸}_1}{{𝐸}_2}=\frac{{𝑙}_1}{{𝑙}_1-{𝑙}_2}}$

Correct Option: (B) $\frac{{𝑙}_1}{{𝑙}_1-{𝑙}_2}$. ✅

Flux, $𝜙=\mathrm{LI}$

Comparing the above equation with $y=\mathrm{mx}$, L is the slope of the graph.

Since line P has highest slope, it indicates the largest value of self-inductance.

Threshold frequency is given as $𝑣$

Work function ${𝜙}_0=h𝜈$

Now, $E=$ K.E. $+{𝜙}_0$

$\begin{array}{cc}& 4ℎ𝑣=\frac{1}{2}𝑚{𝑣}^2+ℎ𝑣\\ \therefore & \frac{1}{2}𝑚{𝑣}^2=4ℎ𝑣-ℎ𝑣\\ \therefore & \frac{1}{2}𝑚{𝑣}^2=3ℎ𝑣\\ \therefore & 𝑣=\sqrt{\surd \frac{6ℎ𝑣}{𝑚}}\end{array}$

Step 1: Recall the moments of inertia

• Solid cylinder about its symmetry axis:

${𝐼}_{\text{cyl}}=\frac{1}{2}𝑀{𝑅}^2$

• Solid sphere about its diameter:

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

Step 2: Rotational KE formula

$\text{Rotational KE}=\frac{1}{2}𝐼{𝜔}^2$

Step 3: Compute KEs

• Cylinder (angular speed = $𝜔$):

${𝐾}_{\text{cyl}}=\frac{1}{2}\cdot \frac{1}{2}𝑀{𝑅}^2\cdot {𝜔}^2=\frac{1}{4}𝑀{𝑅}^2{𝜔}^2$

• Sphere (angular speed = $\frac{𝜔}{2}$):

${𝐾}_{\text{sphere}}=\frac{1}{2}\cdot \frac{2}{5}𝑀{𝑅}^2\cdot (\frac{𝜔}{2}{)}^2$

$=\frac{1}{2}\cdot \frac{2}{5}𝑀{𝑅}^2\cdot \frac{{𝜔}^2}{4}=\frac{1}{20}𝑀{𝑅}^2{𝜔}^2$

Step 4: Ratio

$\frac{{𝐾}_{\text{sphere}}}{{𝐾}_{\text{cyl}}}=\frac{\frac{1}{20}𝑀{𝑅}^2{𝜔}^2}{\frac{1}{4}𝑀{𝑅}^2{𝜔}^2}=\frac{1}{20}\cdot \frac{4}{1}=\frac{1}{5}$

So the ratio = $1:5$.

✅ Final Answer:

Option C: $1:5$

$\begin{array}{cc}& {𝑉}_{𝑒𝑞}=200V-10V=190V\\ & {𝑉}_{𝑒𝑞}=𝐼𝑅\\ & 𝐼=\frac{{𝑉}_{𝑒𝑞}}{𝑅}=\frac{190}{38}=5A\end{array}$

Step 1: Recall expressions

• Gravitational potential energy of a body of mass $𝑚$ at distance $𝑟$ from Earth's centre:

$𝑈=-\frac{𝐺𝑀𝑚}{𝑟}.$

So, the magnitude is:

$\left|𝑈\right|=\frac{𝐺𝑀𝑚}{𝑟}.$

• Weight (force of gravitation) at distance $𝑟$:

$𝐹=\frac{𝐺𝑀𝑚}{{𝑟}^2}.$

Step 2: Relating to the given $𝐸$

At $𝑟=𝑅$, the magnitude of potential energy:

$𝐸=\frac{𝐺𝑀𝑚}{𝑅}.$

So:

$𝐺𝑀𝑚=𝐸𝑅.$

Step 3: Force at $𝑟=1.5𝑅$

At $1.5𝑅$:

$𝐹=\frac{𝐺𝑀𝑚}{(1.5𝑅{)}^2}=\frac{𝐸𝑅}{(1.5𝑅{)}^2}=\frac{𝐸}{\left({1.5}^2𝑅\right)}.$

Since ${1.5}^2=2.25=\frac{9}{4}$,

$𝐹=\frac{𝐸}{\frac{9}{4}𝑅}=\frac{4𝐸}{9𝑅}.$

Step 4: Match with options

That corresponds to Option C:

$\overline{)\frac{4𝐸}{9𝑅}}$

Step 1: Recall de Broglie wavelength and thermal relation

The de Broglie wavelength:

$𝜆=\frac{ℎ}{𝑝}$

For a particle in thermal equilibrium, its average kinetic energy is proportional to temperature:

$\frac{1}{2}𝑚{𝑣}^2\sim {𝑘}_{𝐵}𝑇$

Thus,

$𝑣\propto \sqrt{\surd 𝑇},𝑝=𝑚𝑣\propto \sqrt{\surd 𝑇}.$

So:

$𝜆\propto \frac{1}{\sqrt{\surd 𝑇}}.$

Step 2: Convert temperatures to Kelvin

• At ${27}^{\circ }C$: ${𝑇}_1=27+273=300\text{K}$.

• At ${927}^{\circ }C$: ${𝑇}_2=927+273=1200\text{K}$.

Step 3: Ratio of wavelengths

$\frac{{𝜆}_2}{{𝜆}_1}=\sqrt{\surd \frac{{𝑇}_1}{{𝑇}_2}}=\sqrt{\surd \frac{300}{1200}}=\sqrt{\surd \frac{1}{4}}=\frac{1}{2}.$

So:

${𝜆}_2=\frac{{𝜆}_0}{2}.$

✅ Final Answer:

Option C: $\frac{{𝜆}_0}{2}$

Step 1: Capacitance of a conducting isolated sphere

For a conducting sphere of radius $𝑅$,

$𝐶=4𝜋{𝜀}_0𝑅$

Step 2: Express $𝑅$ in terms of $𝑉$ (volume) and $𝐴$ (surface area)

• Volume of sphere:

$𝑉=\frac{4}{3}𝜋{𝑅}^3$

• Surface area of sphere:

$𝐴=4𝜋{𝑅}^2$

We need to eliminate $𝑅$.

Step 3: Relationship between $𝑉$ and $𝐴$

From $𝐴=4𝜋{𝑅}^2⟹{𝑅}^2=\frac{𝐴}{4𝜋}$.

From $𝑉=\frac{4}{3}𝜋{𝑅}^3⟹{𝑅}^3=\frac{3𝑉}{4𝜋}$.

Now compute $𝑅$ in terms of $𝑉$ and $𝐴$.

$𝑅=\frac{3𝑉}{𝐴}$

(Reason: $𝑉=\frac{4}{3}𝜋{𝑅}^3$ and $𝐴=4𝜋{𝑅}^2$. Divide:

$\frac{𝑉}{𝐴}=\frac{\frac{4}{3}𝜋{𝑅}^3}{4𝜋{𝑅}^2}=\frac{𝑅}{3}⟹𝑅=\frac{3𝑉}{𝐴}.$)

Step 4: Substitute into capacitance formula

$𝐶=4𝜋{𝜀}_0𝑅=4𝜋{𝜀}_0\cdot \frac{3𝑉}{𝐴}$

$𝐶=\frac{12𝜋{𝜀}_0𝑉}{𝐴}$

✅ Correct Answer:

Option C: $\frac{12𝜋{𝜀}_0𝑉}{𝐴}$

Step 1: Understand the motion

• Time period: $𝑇=8\text{s}$.

• Therefore, angular frequency:

$𝜔=\frac{2𝜋}{𝑇}=\frac{2𝜋}{8}=\frac{𝜋}{4}\text{rad/s}.$

• At $𝑡=0$, the particle is at the mean position.

So displacement function:

$𝑥\left(𝑡\right)=𝐴\sin \left(𝜔𝑡\right).$

• Velocity: $𝑣\left(𝑡\right)=\frac{𝑑𝑥}{𝑑𝑡}=𝐴𝜔\cos \left(𝜔𝑡\right).$

Step 2: Distance travelled in 1st second (0 to 1s)

Distance travelled = integral of $\left|𝑣\right|$.

For $𝑡=0$ to small values, $\cos \left(𝜔𝑡\right)>0$ (since $𝜔=𝜋/4$, in 1 second $𝜔𝑡=𝜋/4\approx {45}^{\circ }$, still positive). So velocity doesn’t change sign in this interval → monotonic increase of $𝑥$.

Hence distance travelled = displacement magnitude:

${𝑠}_1=𝑥\left(1\right)-𝑥\left(0\right)=𝐴\sin (\frac{𝜋}{4}\cdot 1)-0=𝐴\sin \frac{𝜋}{4}=\frac{𝐴}{\sqrt{\surd 2}}.$

Step 3: Distance travelled in 2nd second (1 to 2s)

Check velocity sign between $𝑡=1$s and $𝑡=2$s.

At $𝑡=1$, $𝜔𝑡=𝜋/4$, cos positive. At $𝑡=2$, $𝜔𝑡=𝜋/2$, cos zero. So velocity gradually reduces to zero without changing sign. So again monotonic increase in displacement.

Thus distance = displacement change:

${𝑠}_2=𝑥\left(2\right)-𝑥\left(1\right).$

$𝑥\left(2\right)=𝐴\sin (\frac{𝜋}{4}\cdot 2)=𝐴\sin \frac{𝜋}{2}=𝐴.$

So

${𝑠}_2=𝐴-\frac{𝐴}{\sqrt{\surd 2}}=𝐴(1-\frac{1}{\sqrt{\surd 2}}).$

Step 4: Ratio ${𝑠}_1:{𝑠}_2$

$\frac{{𝑠}_1}{{𝑠}_2}=\frac{\frac{𝐴}{\sqrt{\surd 2}}}{𝐴(1-\frac{1}{\sqrt{\surd 2}})}=\frac{\frac{1}{\sqrt{\surd 2}}}{1-\frac{1}{\sqrt{\surd 2}}}.$

Simplify denominator: $1-\frac{1}{\sqrt{\surd 2}}=\frac{\sqrt{\surd 2}-1}{\sqrt{\surd 2}}$.

So:

$\frac{{𝑠}_1}{{𝑠}_2}=\frac{\frac{1}{\sqrt{\surd 2}}}{\frac{\sqrt{\surd 2}-1}{\sqrt{\surd 2}}}=\frac{1}{\sqrt{\surd 2}-1}.$

Multiply numerator and denominator:

$\frac{1}{\sqrt{\surd 2}-1}=\sqrt{\surd 2}+1.$

Thus ratio:

${𝑠}_1:{𝑠}_2=(\sqrt{\surd 2}+1):1.$

Step 5: Match with options

They ask ratio of distances in 1st and 2nd second = $1:(\sqrt{\surd 2}-1)$ or $2:1$ etc.

We got $(\sqrt{\surd 2}+1):1$.

So equivalently $1:(\sqrt{\surd 2}-1)$? Let’s check:

$(\sqrt{\surd 2}+1):1$ = multiply numerator and denominator by $(\sqrt{\surd 2}-1)$:

$(\sqrt{\surd 2}+1):1=1:(\sqrt{\surd 2}-1).$

Yes, same.

✅ Final Answer:

$\overline{)1:(\sqrt{\surd 2}-1)}$

Option A

Step 1: Volume conservation

A single drop of radius $𝑅$ is broken into $𝑛$ drops of radius $𝑟$.

So volumes must be equal:

$\frac{4}{3}𝜋{𝑅}^3=𝑛\cdot \frac{4}{3}𝜋{𝑟}^3$

${𝑅}^3=𝑛{𝑟}^3\Rightarrow 𝑛={\left(\frac{𝑅}{𝑟}\right)}^3$

Step 2: Surface energy of a drop

Surface energy = Surface area $\times$ surface tension.

• Initial energy:

${𝐸}_{𝑖}=4𝜋{𝑅}^2𝑇$

• Final energy:

${𝐸}_{𝑓}=𝑛\cdot 4𝜋{𝑟}^2𝑇=4𝜋𝑇𝑛{𝑟}^2$

Step 3: Energy required

$\Delta 𝐸={𝐸}_{𝑓}-{𝐸}_{𝑖}=4𝜋𝑇(𝑛{𝑟}^2-{𝑅}^2)$

Substituting $𝑛=\frac{{𝑅}^3}{{𝑟}^3}$:

$\Delta 𝐸=4𝜋𝑇(\frac{{𝑅}^3}{{𝑟}^3}\cdot {𝑟}^2-{𝑅}^2)=4𝜋𝑇(\frac{{𝑅}^3}{𝑟}-{𝑅}^2)$

Factorize:

$\Delta 𝐸=4𝜋𝑇{𝑅}^2(\frac{𝑅}{𝑟}-1)$

✅ Final Answer:

$\overline{)4𝜋𝑇{𝑅}^2(\frac{𝑅}{𝑟}-1)}$

Correct Option: B

We are dealing with a pipe closed at one end (closed organ pipe).

Key points:

• In a closed organ pipe, the closed end is always a node, and the open end is always an antinode.

• Only odd harmonics are possible:

• 1st harmonic (fundamental: $𝑛=1$)

• 2nd harmonic (first overtone, $𝑛=3$)

• 3rd harmonic (second overtone, $𝑛=5$)

• For the $𝑛=5$ mode (second overtone), the standing wave inside the pipe will have multiple nodes and antinodes.

Determining the pattern:

For the $𝑛=5$ mode in a closed pipe, the pattern corresponds to $\frac{5}{4}𝜆$.

That means the pipe length contains 5 quarter wavelengths.

• Each quarter wavelength segment contains either a node or an antinode alternately.

• Starting from the closed end = node → A–N–A… until open end = antinode.

So, for 5 quarter wavelengths, we get:

• Node (at closed end)

• Antinode

• Node

• Antinode

• Node

• Antinode (at open end)

Thus:

• Nodes = 3

• Antinodes = 3

✅ Correct Answer:

Option A: three nodes and three antinodes.

We are asked about heat conduction through a rod.

The formula for the rate of heat conduction is Fourier’s law:

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

Where

• $𝐾$ = thermal conductivity

• $𝐴$ = cross-sectional area

• $𝑥$ = length of the rod

• $({𝑇}_1-{𝑇}_2)$ = temperature difference

• $\frac{𝑄}{𝑡}$ = rate of heat flow

Now we rearrange for $𝐾$:

$𝐾=\frac{\frac{𝑄}{𝑡}\cdot 𝑥}{𝐴({𝑇}_1-{𝑇}_2)}$

$𝐾=\frac{𝑥𝑄}{𝑡𝐴({𝑇}_1-{𝑇}_2)}$

✅ Correct option is:

Option B:

$\frac{𝑥𝑄}{𝑡𝐴({𝑇}_1-{𝑇}_2)}$

$\begin{array}{cc}& \text{ Magnetic force formula }\\ & \begin{array}{cc}& \overrightarrow{𝐹}=𝑞(\overrightarrow{𝑉}\times \overrightarrow{𝐵})\\ \therefore & \overrightarrow{𝑉}\times \overrightarrow{𝐵}=𝑎\widehat{𝑖}\times (𝑏\widehat{𝑗}+𝑐\widehat{𝑘})\end{array}\end{array}$

$\begin{array}{cc}& =𝑎𝑏(\widehat{𝑖}\times \widehat{𝑗})+𝑎𝑐(\widehat{𝑖}\times \widehat{𝑘})\\ & =𝑎𝑏\widehat{𝑘}-𝑎𝑐\widehat{𝑗}\end{array}$

Magnitude of Force

$\begin{array}{cc}\left|\overrightarrow{F}\right|& =𝑞|\stackrel{\to \leftarrow }{V}\times \overrightarrow{B}|\\ & =𝑞\sqrt{\surd (𝑎𝑏{)}^2+(𝑎𝑐{)}^2}\\ & =𝑞𝑎\sqrt{\surd {𝑏}^2+{𝑐}^2}\end{array}$

$\therefore$ Magnitude of Force $\left|\overrightarrow{F}\right|=\mathrm{qa}\sqrt{\surd b^2+c^2}$

A Zener diode is operated in the reverse breakdown region for voltage regulation, where the voltage across the diode remains almost constant despite large variations in the reverse current.

In the given $𝐼\text{-}𝑉$ characteristics curve, the segment $𝑑𝑒$ represents this reverse breakdown region.

Hence, the correct option is (D).

Step 1: Recall the formula for excess pressure inside a soap bubble.

For a soap bubble of radius $𝑟$:

$\Delta 𝑃=\frac{4𝑇}{𝑟}$

where $𝑇$ is the surface tension.

Step 2: Compare the pressures.

Let the first bubble radius = ${𝑟}_1$,

the second bubble radius = ${𝑟}_2$.

Given:

$\Delta {𝑃}_1=1.5\Delta {𝑃}_2$

So:

$\frac{4𝑇}{{𝑟}_1}=1.5\cdot \frac{4𝑇}{{𝑟}_2}$

Canceling $4𝑇$:

$\frac{1}{{𝑟}_1}=\frac{1.5}{{𝑟}_2}$

${𝑟}_2=1.5{𝑟}_1$

Step 3: Relating volumes.

Volume of a bubble:

$𝑉=\frac{4}{3}𝜋{𝑟}^3$

So:

$\frac{{𝑉}_2}{{𝑉}_1}=\frac{{𝑟}_3^2}{{𝑟}_3^1}=(1.5{)}^3=\frac{27}{8}$

Thus:

$𝑥=\frac{27}{8}$

Final Answer:

$\overline{)\frac{27}{8}}\text{(Option D)}$

From the law of length for a vibrating string

$\begin{array}{cc}& n_2{𝑙}_2=n_1{𝑙}_1\\ \therefore & n_2=\frac{{𝑙}_1}{{𝑙}_2}\times n_1=\frac{{𝑙}_1}{1.25{𝑙}_1}{n}_1=0.8n_1\\ \therefore & n_1-n_2=n_1(1-0.8)=0.2n_1\\ & \frac{n_1-n_2}{n_1}=0.2=20\%\end{array}$