MHT CET 2025 19th April Morning Shift (Physics & Chemistry) Question Paper with Solutions

Given:

$P=\frac{{𝑥}^3𝑦}{{𝑧}^2}$

Percentage errors:

Error in $𝑥=0.6\%$

Error in $𝑦=3\%$

Error in $𝑧=1.3\%$

Step 1: Formula for percentage error propagation

For a formula $𝑄=\frac{{𝑎}^{𝑝}{𝑏}^{𝑞}}{{𝑐}^{𝑟}}$,

the total percentage error in $𝑄$ is:

$\text{Percentage error in }𝑄=\left|𝑝\right|\left(\text{Error in }𝑎\right)+\left|𝑞\right|\left(\text{Error in }𝑏\right)+\left|𝑟\right|\left(\text{Error in }𝑐\right)$

Step 2: Apply the values for exponents and errors

Here,

• Power of $𝑥$ is $3$.

• Power of $𝑦$ is $1$.

• Power of $𝑧$ is $2$ (in the denominator, so power is $-2$, but we take its absolute value).

So,

$\text{Percentage error in }𝑃=3\times \left(\text{Error in }𝑥\right)+1\times \left(\text{Error in }𝑦\right)+2\times \left(\text{Error in }𝑧\right)$

Step 3: Substitute the given values

$=3\times 0.6\%+1\times 3\%+2\times 1.3\%$

$=1.8\%+3\%+2.6\%$

$=(1.8+3+2.6)\%$

$=7.4\%$

Step 4: Write the final answer

The percentage error in the measurement of $𝑃$ is $\overline{)7.4\%}$.

Correct option: D

Given:

• Work done: $𝑊$ joule

• Charge: $𝑞$ coulombs

• Potential at $𝐴$: ${𝑉}_{𝐴}=-5$ volt

• Potential at $𝐵$: ${𝑉}_{𝐵}=𝑉$ volt

Step 1:

Work done in moving charge $𝑞$ from $𝐴$ to $𝐵$ is given by:

$𝑊=𝑞({𝑉}_{𝐵}-{𝑉}_{𝐴})$

Step 2:

Substitute the values:

• ${𝑉}_{𝐵}=𝑉$

• ${𝑉}_{𝐴}=-5$

So,

$𝑊=𝑞(𝑉-(-5\left)\right)$

$𝑊=𝑞(𝑉+5)$

Step 3:

Solve for $𝑉$:

$𝑊=𝑞(𝑉+5)$

Divide both sides by $𝑞$:

$\frac{𝑊}{𝑞}=𝑉+5$

Subtract $5$ from both sides:

$𝑉=\frac{𝑊}{𝑞}-5$

Answer:

The correct value of $𝑉$ is:

$\overline{)\frac{𝑊}{𝑞}-5}$

So, correct option: D.

Step 1: Tension at Highest and Lowest Points

At the highest point,

Let tension = ${𝑇}_{𝐻}$, speed = ${𝑣}_{𝐻}$.

For a body moving in a vertical circle:

${𝑇}_{𝐻}+𝑚𝑔=\frac{𝑚{𝑣}_2^{𝐻}}{𝐿}$

So,

${𝑇}_{𝐻}=\frac{𝑚{𝑣}_2^{𝐻}}{𝐿}-𝑚𝑔$

At the lowest point,

Let tension = ${𝑇}_{𝐿}$, speed = ${𝑣}_{𝐿}$.

${𝑇}_{𝐿}-𝑚𝑔=\frac{𝑚{𝑣}_2^{𝐿}}{𝐿}$

So,

${𝑇}_{𝐿}=\frac{𝑚{𝑣}_2^{𝐿}}{𝐿}+𝑚𝑔$

Step 2: Ratio of Maximum to Minimum Tension

Given:

$\frac{{𝑇}_{𝐿}}{{𝑇}_{𝐻}}=3$

Step 3: Energy Conservation Between Highest and Lowest Point

Mechanical energy is conserved:

$\frac{1}{2}𝑚{𝑣}_2^{𝐿}=\frac{1}{2}𝑚{𝑣}_2^{𝐻}+𝑚𝑔\left(2𝐿\right)$

${𝑣}_2^{𝐿}={𝑣}_2^{𝐻}+4𝑔𝐿$

Step 4: Substitute Tensions

Plug in $𝑚=1$ (from question):

• ${𝑇}_{𝐻}=\frac{{𝑣}_2^{𝐻}}{𝐿}-𝑔$

• ${𝑇}_{𝐿}=\frac{{𝑣}_2^{𝐿}}{𝐿}+𝑔$

Given ratio:

$\frac{{𝑇}_{𝐿}}{{𝑇}_{𝐻}}=\frac{\frac{{𝑣}_2^{𝐿}}{𝐿}+𝑔}{\frac{{𝑣}_2^{𝐻}}{𝐿}-𝑔}=3$

Step 5: Substitute ${𝑣}_2^{𝐿}={𝑣}_2^{𝐻}+4𝑔𝐿$

$\frac{\frac{{𝑣}_2^{𝐻}+4𝑔𝐿}{𝐿}+𝑔}{\frac{{𝑣}_2^{𝐻}}{𝐿}-𝑔}=3$

Simplify numerator:

$\frac{\frac{{𝑣}_2^{𝐻}}{𝐿}+4𝑔+𝑔}{\frac{{𝑣}_2^{𝐻}}{𝐿}-𝑔}=3$

$\frac{\frac{{𝑣}_2^{𝐻}}{𝐿}+5𝑔}{\frac{{𝑣}_2^{𝐻}}{𝐿}-𝑔}=3$

Step 6: Cross Multiply and Solve

$\frac{{𝑣}_2^{𝐻}}{𝐿}+5𝑔=3(\frac{{𝑣}_2^{𝐻}}{𝐿}-𝑔)$

$\frac{{𝑣}_2^{𝐻}}{𝐿}+5𝑔=3\frac{{𝑣}_2^{𝐻}}{𝐿}-3𝑔$

Bring terms together:

$5𝑔+3𝑔=3\frac{{𝑣}_2^{𝐻}}{𝐿}-\frac{{𝑣}_2^{𝐻}}{𝐿}$

$8𝑔=2\frac{{𝑣}_2^{𝐻}}{𝐿}$

$\frac{{𝑣}_2^{𝐻}}{𝐿}=4𝑔$

${𝑣}_2^{𝐻}=4𝑔𝐿$

${𝑣}_{𝐻}=\sqrt{\surd 4𝑔𝐿}$

Step 7: Put the Value of $𝐿=\frac{5}{3}$

${𝑣}_{𝐻}=\sqrt{\surd 4𝑔𝐿}=2\sqrt{\surd 𝑔𝐿}$

Final Answer:

$\overline{)\sqrt{\surd 4𝑔𝐿}}$

This matches Option C.

Let the total current be $𝐼$.

Only $5\%$ of $𝐼$ should pass through the galvanometer.

So, current through galvanometer,

${𝐼}_{𝑔}=\frac{5}{100}𝐼=\frac{𝐼}{20}$

Current through shunt,

${𝐼}_{𝑠ℎ}=𝐼-{𝐼}_{𝑔}=𝐼-\frac{𝐼}{20}=\frac{19𝐼}{20}$

If $𝐺$ is the resistance of galvanometer and $𝑆$ is the resistance of the shunt, both are in parallel across the same voltage.

By Ohm's Law:

$\text{Voltage across }𝐺=\text{Voltage across }𝑆$

${𝐼}_{𝑔}\cdot 𝐺={𝐼}_{𝑠ℎ}\cdot 𝑆$

Substitute the values:

$\frac{𝐼}{20}\cdot 𝐺=\frac{19𝐼}{20}\cdot 𝑆$

Divide both sides by $𝐼$:

$\frac{1}{20}𝐺=\frac{19}{20}𝑆$

Divide both sides by $\frac{19}{20}$:

$𝑆=\frac{1}{20}𝐺\times \frac{1}{\frac{19}{20}}$

$𝑆=\frac{𝐺}{19}$

Correct Option:

$\overline{)\frac{𝐺}{19}}$

So, the right answer is Option C.

Given:

$\overrightarrow{𝐹}=5\widehat{𝑖}-10\widehat{𝑗}$

$\overrightarrow{𝑟}=4\widehat{𝑖}-3\widehat{𝑗}$

Step 1:

Torque ($\overrightarrow{𝜏}$) is given by the cross product:

$\overrightarrow{𝜏}=\overrightarrow{𝑟}\times \overrightarrow{𝐹}$

Step 2:

Write vectors in component form:

$\overrightarrow{𝑟}=4\widehat{𝑖}+(-3)\widehat{𝑗}+0\widehat{𝑘}$

$\overrightarrow{𝐹}=5\widehat{𝑖}+(-10)\widehat{𝑗}+0\widehat{𝑘}$

Step 3:

Use the determinant method for cross product:

$\overrightarrow{𝜏}=\mid \begin{array}{ccc}\widehat{𝑖}& \widehat{𝑗}& \widehat{𝑘}\\ 4& -3& 0\\ 5& -10& 0\end{array}\mid$

Step 4:

Expand the determinant:

$\overrightarrow{𝜏}=\widehat{𝑖}\mid \begin{array}{cc}-3& 0\\ -10& 0\end{array}\mid -\widehat{𝑗}\mid \begin{array}{cc}4& 0\\ 5& 0\end{array}\mid +\widehat{𝑘}\mid \begin{array}{cc}4& -3\\ 5& -10\end{array}\mid$

Calculate each minor:

• $\mid \begin{array}{cc}-3& 0\\ -10& 0\end{array}\mid =(-3)\left(0\right)-\left(0\right)(-10)=0$

• $\mid \begin{array}{cc}4& 0\\ 5& 0\end{array}\mid =\left(4\right)\left(0\right)-\left(0\right)\left(5\right)=0$

• $\mid \begin{array}{cc}4& -3\\ 5& -10\end{array}\mid =\left(4\right)(-10)-(-3)\left(5\right)=-40+15=-25$

So,

$\overrightarrow{𝜏}=0\widehat{𝑖}-0\widehat{𝑗}-25\widehat{𝑘}$

$\overrightarrow{𝜏}=-25\widehat{𝑘}$

Final Answer:

$\overline{)-25\widehat{𝑘}}$

So, the correct option is Option D.

Magnetic field produced by two wires

$B_1=\frac{{𝜇}_0{I}_1}{2𝜋X}\text{ and }{B}_2=\frac{{𝜇}_0{I}_2}{2𝜋X}$

From Figure,

$B_{\mathrm{net}}=\sqrt{\surd B_2^1+B_2^2}=\frac{{𝜇}_0}{2𝜋X}\sqrt{\surd I_2^1+I_2^2}$

Also, using Pythagoras theorem, $2X^2=5\times 5$

$\begin{array}{cc}\therefore & 𝑋=\frac{5}{\sqrt{\surd 2}}\mathrm{cm}\\ & {𝐵}_{\text{net }}=\frac{4𝜋\times {10}^{-7}}{2𝜋\times \frac{5}{\sqrt{\surd 2}}\times {10}^{-2}}\sqrt{\surd 4^2+3^2}=2\sqrt{\surd 2}\times {10}^{-5}T\end{array}$

Let the initial radius be $𝑅$ and initial temperature be $𝑇$.

Step 1: Write the formula for power radiated by a black body

According to Stefan-Boltzmann law,

$𝑃=𝜎\cdot 𝐴\cdot {𝑇}^4$

where $𝜎$ is the Stefan-Boltzmann constant, $𝐴$ is the surface area, and $𝑇$ is the temperature.

For a sphere, $𝐴=4𝜋{𝑅}^2$.

So,

$𝑃=𝜎\cdot 4𝜋{𝑅}^2{𝑇}^4$

Step 2: Write the expression for the new power

When temperature is doubled $\left(2𝑇\right)$ and radius is also doubled $\left(2𝑅\right)$:

Let the new power be ${𝑃}^{\prime }$,

${𝑃}^{\prime }=𝜎\cdot 4𝜋\left(2𝑅{)}^2\right(2𝑇{)}^4$

Step 3: Expand $(2𝑅{)}^2$ and $(2𝑇{)}^4$

$(2𝑅{)}^2=4{𝑅}^2$

$(2𝑇{)}^4=16{𝑇}^4$

So,

${𝑃}^{\prime }=𝜎\cdot 4𝜋\cdot 4{𝑅}^2\cdot 16{𝑇}^4$

Step 4: Multiply and simplify

${𝑃}^{\prime }=𝜎\cdot 4𝜋\cdot 4\cdot 16\cdot {𝑅}^2{𝑇}^4$

${𝑃}^{\prime }=𝜎\cdot 4𝜋\cdot 64\cdot {𝑅}^2{𝑇}^4$

But the original power $𝑃=𝜎\cdot 4𝜋{𝑅}^2{𝑇}^4$

So,

${𝑃}^{\prime }=64\cdot 𝑃$

Final Answer:

$\overline{)64𝑃}$

So, Option D is correct.

Given:

• Source voltage, $𝑉=230$ V

• Frequency, $𝑓=50$ Hz

• Resistance, $𝑅=80\Omega$

• Inductive reactance, ${𝑋}_{𝐿}=70\Omega$

• Capacitive reactance, ${𝑋}_{𝐶}=130\Omega$

Let us find the power factor step-by-step:

Step 1: Calculate Net Reactance

$𝑋={𝑋}_{𝐿}-{𝑋}_{𝐶}=70-130=-60\Omega$

Step 2: Calculate Impedance

$𝑍=\sqrt{\surd {𝑅}^2+{𝑋}^2}=\sqrt{\surd (80{)}^2+(-60{)}^2}=\sqrt{\surd 6400+3600}=\sqrt{\surd 10000}=100\Omega$

Step 3: Power Factor Formula

Power factor, $𝑥=\cos 𝜙=\frac{𝑅}{𝑍}$

$𝑥=\frac{80}{100}=0.8$

Step 4: Correct Option

Option B: 0.8

Given:

• Time period in stationary lift, ${𝑇}_1=\sqrt{\surd 3}$ s

• Acceleration due to gravity = $𝑔$

• Lift moves upwards with acceleration $𝑎=\frac{𝑔}{3}$

Step 1: Time period formula for a simple pendulum

In general,

$𝑇=2𝜋\sqrt{\surd \frac{𝑙}{{𝑔}_{\text{eff}}}}$

where ${𝑔}_{\text{eff}}$ is the effective acceleration due to gravity.

Step 2: Effective gravity in the stationary case

When the lift is stationary, ${𝑔}_{\text{eff}}=𝑔$

So,

${𝑇}_1=2𝜋\sqrt{\surd \frac{𝑙}{𝑔}}$

Step 3: Effective gravity when lift moves upward

When lift accelerates upwards with $𝑎$,

${𝑔}_{\text{eff}}=𝑔+𝑎=𝑔+\frac{𝑔}{3}=\frac{4𝑔}{3}$

Step 4: New time period

The new time period is

${𝑇}_2=2𝜋\sqrt{\surd \frac{𝑙}{{𝑔}_{\text{eff}}}}=2𝜋\sqrt{\surd \frac{𝑙}{\frac{4𝑔}{3}}}$

${𝑇}_2=2𝜋\sqrt{\surd \frac{3𝑙}{4𝑔}}$

Express $𝑙$ in terms of ${𝑇}_1$:

${𝑇}_1=2𝜋\sqrt{\surd \frac{𝑙}{𝑔}}$

Square both sides:

${𝑇}_2^1=4{𝜋}^2\frac{𝑙}{𝑔}⟹𝑙=\frac{𝑔{𝑇}_2^1}{4{𝜋}^2}$

Now, substitute $𝑙$ into ${𝑇}_2$:

${𝑇}_2=2𝜋\sqrt{\surd \frac{3}{4𝑔}\cdot \frac{𝑔{𝑇}_2^1}{4{𝜋}^2}}$

${𝑇}_2=2𝜋\sqrt{\surd \frac{3{𝑇}_2^1}{16{𝜋}^2}}$

${𝑇}_2=2𝜋\cdot \frac{{𝑇}_1}{4𝜋}\sqrt{\surd 3}$

${𝑇}_2=\frac{{𝑇}_1}{2}\sqrt{\surd 3}$

Step 5: Substitute ${𝑇}_1=\sqrt{\surd 3}$ s

${𝑇}_2=\frac{\sqrt{\surd 3}}{2}\times \sqrt{\surd 3}=\frac{3}{2}=1.5\text{seconds}$

Final Answer:

Option A) 1.5 s

Given:

• Three inductors, each of inductance $𝐿$, connected in series.

• Current passing through the circuit = $𝐼$

Step 1: Equivalent inductance in series

When inductors are connected in series, their inductances add up:

${𝐿}_{\text{eq}}={𝐿}_1+{𝐿}_2+{𝐿}_3=𝐿+𝐿+𝐿=3𝐿$

Step 2: Formula for energy stored in an inductor

The energy stored in an inductor of inductance ${𝐿}_{\text{eq}}$ carrying current $𝐼$ is:

$𝐸=\frac{1}{2}{𝐿}_{\text{eq}}{𝐼}^2$

Step 3: Substitute the values

Substitute ${𝐿}_{\text{eq}}=3𝐿$:

$𝐸=\frac{1}{2}\times 3𝐿\times {𝐼}^2=\frac{3}{2}𝐿{𝐼}^2$

Final Answer:

Option B

$\overline{)\frac{3}{2}𝐿{𝐼}^2}$

First, let us identify the transitions for the first members of the Balmer and Paschen series in the hydrogen atom:

• Balmer series (visible): transitions to ${𝑛}_2=2$

• First member: ${𝑛}_1=3$ to ${𝑛}_2=2$

• Paschen series (infrared): transitions to ${𝑛}_2=3$

• First member: ${𝑛}_1=4$ to ${𝑛}_2=3$

Let wavelengths be:

• ${𝜆}_1$ = Balmer, $3\to 2$

• ${𝜆}_2$ = Paschen, $4\to 3$

The wavelength for a transition in the hydrogen atom is given by the Rydberg formula:

$\frac{1}{𝜆}={𝑅}_{𝐻}(\frac{1}{{𝑛}_2^2}-\frac{1}{{𝑛}_2^1})$

For the first Balmer line $(3\to 2)$:

$\frac{1}{{𝜆}_1}={𝑅}_{𝐻}(\frac{1}{2^2}-\frac{1}{3^2})={𝑅}_{𝐻}(\frac{1}{4}-\frac{1}{9})$

Calculate:

$\frac{1}{4}-\frac{1}{9}=\frac{9-4}{36}=\frac{5}{36}$

So,

$\frac{1}{{𝜆}_1}={𝑅}_{𝐻}\cdot \frac{5}{36}$

For the first Paschen line $(4\to 3)$:

$\frac{1}{{𝜆}_2}={𝑅}_{𝐻}(\frac{1}{3^2}-\frac{1}{4^2})={𝑅}_{𝐻}(\frac{1}{9}-\frac{1}{16})$

Calculate:

$\frac{1}{9}-\frac{1}{16}=\frac{16-9}{144}=\frac{7}{144}$

So,

$\frac{1}{{𝜆}_2}={𝑅}_{𝐻}\cdot \frac{7}{144}$

The frequency is given by $𝑓=\frac{𝑐}{𝜆}$

Therefore,

${𝑓}_1=\frac{𝑐}{{𝜆}_1},{𝑓}_2=\frac{𝑐}{{𝜆}_2}$

$\frac{{𝑓}_1}{{𝑓}_2}=\frac{{𝜆}_2}{{𝜆}_1}$

From above:

$\frac{1}{{𝜆}_1}={𝑅}_{𝐻}\cdot \frac{5}{36}⟹{𝜆}_1=\frac{36}{5{𝑅}_{𝐻}}$

$\frac{1}{{𝜆}_2}={𝑅}_{𝐻}\cdot \frac{7}{144}⟹{𝜆}_2=\frac{144}{7{𝑅}_{𝐻}}$

Therefore,

$\frac{{𝑓}_1}{{𝑓}_2}=\frac{{𝜆}_2}{{𝜆}_1}=\frac{\frac{144}{7{𝑅}_{𝐻}}}{\frac{36}{5{𝑅}_{𝐻}}}=\frac{144}{7}\times \frac{5}{36}$

$=\frac{144\times 5}{7\times 36}$

$=\frac{720}{252}$

$=\frac{720÷36}{252÷36}=\frac{20}{7}$

Correct answer:

$\overline{)20:7}$

So, the answer is Option A.

1. Angular Momentum ($𝐿$):

   For an electron revolving in a circular orbit,

   $𝐿=𝑚𝑣𝑟$

   where $𝑚$ is the mass, $𝑣$ is the velocity, and $𝑟$ is the radius.

2. Orbital Magnetic Dipole Moment (${𝑚}_{\text{orb}}$):

   Current due to electron motion is $𝐼=\frac{𝑒}{𝑇}=\frac{𝑒𝑣}{2𝜋𝑟}$

   Area of orbit $𝐴=𝜋{𝑟}^2$

   So,

   ${𝑚}_{\text{orb}}=𝐼\times 𝐴=\left(\frac{𝑒𝑣}{2𝜋𝑟}\right)\times \left(𝜋{𝑟}^2\right)=\frac{𝑒𝑣𝑟}{2}$

3. Finding the ratio $\frac{𝐿}{{𝑚}_{\text{orb}}}$:

   $\frac{𝐿}{{𝑚}_{\text{orb}}}=\frac{𝑚𝑣𝑟}{\left(\frac{𝑒𝑣𝑟}{2}\right)}=\frac{𝑚𝑣𝑟}{\frac{𝑒𝑣𝑟}{2}}=\frac{𝑚𝑣𝑟\times 2}{𝑒𝑣𝑟}=\frac{2𝑚}{𝑒}$

So, the correct answer is:

$\overline{)\frac{2𝑚}{𝑒}}$

This matches Option B.

Given:

• Three samples of a gas ($𝑋,𝑌,𝑍$) have equal initial volumes ($𝑉$) and temperatures.

• Volume is doubled for each sample ($𝑉\to 2𝑉$) by different processes:

• X: Isothermal

• Y: Adiabatic ($𝛾=\frac{3}{2}$)

• Z: Isobaric

Let the initial pressures be ${𝑃}_{𝑋},{𝑃}_{𝑌},{𝑃}_{𝑍}$.

Let the final pressures (after the process) for all be $𝑃$ (equal for all).

We have to find ${𝑃}_{𝑋}:{𝑃}_{𝑌}:{𝑃}_{𝑍}$.

Step 1: Isothermal Process (Sample X)

For isothermal process:

${𝑃}_{𝑋}𝑉=𝑃\cdot 2𝑉$

Solve for ${𝑃}_{𝑋}$:

${𝑃}_{𝑋}=2𝑃$

Step 2: Adiabatic Process (Sample Y)

For adiabatic:

${𝑃}_{𝑌}{𝑉}^{𝛾}=𝑃(2𝑉{)}^{𝛾}$

Plug in $𝛾=\frac{3}{2}$:

${𝑃}_{𝑌}{𝑉}^{3/2}=𝑃\cdot (2𝑉{)}^{3/2}$

${𝑃}_{𝑌}=𝑃\cdot \frac{(2{)}^{3/2}}{1}$

${𝑃}_{𝑌}=𝑃\cdot 2^{3/2}$

$2^{3/2}=(\sqrt{\surd 2}{)}^3=2\sqrt{\surd 2}$

${𝑃}_{𝑌}=2\sqrt{\surd 2}𝑃$

Step 3: Isobaric Process (Sample Z)

For isobaric, pressure is constant during the process, so:

${𝑃}_{𝑍}=𝑃$

Step 4: Ratio

${𝑃}_{𝑋}:{𝑃}_{𝑌}:{𝑃}_{𝑍}=2𝑃:2\sqrt{\surd 2}𝑃:𝑃$

$=2:2\sqrt{\surd 2}:1$

Answer:

Option B $\overline{)2:2\sqrt{\surd 2}:1}$

Self-inductance ($𝐿$) of a circuit is defined as the ratio of the magnetic flux linked with the circuit to the current producing it:

$𝐿=\frac{\Phi }{𝐼}$

The work done ($𝑊$) in establishing a current $𝐼$ against the induced emf is given by:

$𝑊=\frac{1}{2}𝐿{𝐼}^2$

Now, if the current is unity ($𝐼=1$):

$𝑊=\frac{1}{2}𝐿(1{)}^2=\frac{1}{2}𝐿$

Therefore,

$𝐿=2\times \text{(work done in establishing unit current)}$

Thus, the correct answer is:

Option B

twice the work done in establishing the magnetic flux associated with unit current in the circuit.

When a source of sound moves towards a stationary observer, the Doppler effect takes place.

According to the NCERT formula:

If the source moves towards a stationary observer, the apparent frequency ${𝑓}^{\prime }$ heard by the observer is:

${𝑓}^{\prime }=\frac{𝑣}{𝑣-{𝑣}_{𝑠}}𝑓$

where,

$𝑓$ = actual frequency of the source

$𝑣$ = speed of sound in air

${𝑣}_{𝑠}$ = speed of the source towards the observer

Step-by-step reasoning:

1. Since the denominator $𝑣-{𝑣}_{𝑠}$ is less than $𝑣$, ${𝑓}^{\prime }>𝑓$.

   $⟹$ The apparent frequency increases.

2. The wavelength ${𝜆}^{\prime }$ of the sound received by the observer is

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

   Since ${𝑣}_{𝑠}$ is positive (source moves towards observer), $𝑣-{𝑣}_{𝑠}<𝑣$, hence ${𝜆}^{\prime }<𝜆$ (where $𝜆=\frac{𝑣}{𝑓}$).

   $⟹$ Wavelength decreases.

Correct option:

Option B: increases while wavelength decreases.

Let the rise of liquid in capillary $𝑄$ be ${ℎ}_{𝑄}$.

Let:

• Height in capillary $𝑃={ℎ}_{𝑃}=2.4\text{cm}$

• Diameter of capillary $𝑃={𝑑}_{𝑃}$

• Diameter of capillary $𝑄={𝑑}_{𝑄}=0.8{𝑑}_{𝑃}$

Step 1: Formula for capillary rise

The capillary rise $ℎ$ is given by:

$ℎ=\frac{2𝑇\cos 𝜃}{𝜌𝑔𝑟}$

where

• $𝑇$ = surface tension,

• $𝜃$ = angle of contact,

• $𝜌$ = density of liquid,

• $𝑔$ = acceleration due to gravity,

• $𝑟$ = radius of the capillary.

Step 2: Relation between height and radius

Since $ℎ\propto \frac{1}{𝑟}$, if the radius decreases, the height increases.

Diameter of $𝑄$ is $80\%$ of $𝑃$:

• ${𝑑}_{𝑄}=0.8{𝑑}_{𝑃}$

• ${𝑟}_{𝑄}=0.8{𝑟}_{𝑃}$

So,

${ℎ}_{𝑄}=\frac{{ℎ}_{𝑃}{𝑟}_{𝑃}}{{𝑟}_{𝑄}}$

But since ${𝑟}_{𝑄}=0.8{𝑟}_{𝑃}$,

${ℎ}_{𝑄}=\frac{{ℎ}_{𝑃}{𝑟}_{𝑃}}{0.8{𝑟}_{𝑃}}=\frac{{ℎ}_{𝑃}}{0.8}$

Step 3: Substitute the values

Given ${ℎ}_{𝑃}=2.4\text{cm}$,

${ℎ}_{𝑄}=\frac{2.4}{0.8}=3\text{cm}$

Final Answer:

Option C

$3\text{cm}$

Let the length of the closed pipe be ${𝐿}_1$ and the length of the open pipe be ${𝐿}_2$.

Step 1: Frequencies in closed and open pipes

• For a closed pipe, only odd harmonics are present.

• $𝑛$-th overtone $=$ $(2𝑛+1)$-th harmonic.

  So, Fourth overtone $(𝑛=4)$ is:

• Harmonic number $=2\times 4+1=9$

• Frequency:

  ${𝑓}_{𝑐𝑙𝑜𝑠𝑒𝑑}=\frac{9𝑣}{4{𝐿}_1}$

  (where $𝑣$ is speed of sound)

• For an open pipe, all harmonics are present.

• $𝑛$-th overtone $=$ $(𝑛+1)$-th harmonic.

  So, Fifth overtone $(𝑛=5)$ is:

• Harmonic number $=5+1=6$

• Frequency:

  ${𝑓}_{𝑜𝑝𝑒𝑛}=\frac{6𝑣}{2{𝐿}_2}=\frac{3𝑣}{{𝐿}_2}$

Step 2: Set frequencies equal (in unison)

$\frac{9𝑣}{4{𝐿}_1}=\frac{3𝑣}{{𝐿}_2}$

Cancel $𝑣$ from both sides:

$\frac{9}{4{𝐿}_1}=\frac{3}{{𝐿}_2}$

Cross-multiplied,

$9{𝐿}_2=12{𝐿}_1$

$\frac{{𝐿}_1}{{𝐿}_2}=\frac{9}{12}=\frac{3}{4}$

Step 3: Final Answer

The ratio of length of closed pipe to open pipe is

$\overline{)3:4}$

So, the correct option is Option B.

Step 1: Calculate the initial capacitance of the parallel plate capacitor.

When the medium between the plates is air (or vacuum), the capacitance (${𝐶}_0$) is given by:

${𝐶}_0=\frac{{𝜖}_0𝐴}{𝑑}$

where:

• ${𝜖}_0$ is the permittivity of free space.

• $𝐴$ is the area of each plate.

• $𝑑$ is the separation between the plates.

Step 2: Calculate the capacitance after introducing the dielectric slab.

When a dielectric slab of thickness $𝑡$ and dielectric constant $𝐾$ is introduced between the plates of a parallel plate capacitor, the new capacitance (${𝐶}_{𝑓}$) is given by the formula:

${𝐶}_{𝑓}=\frac{{𝜖}_0𝐴}{𝑑-𝑡+\frac{𝑡}{𝐾}}$

In this problem, the dielectric constant $𝐾=3$. So, substituting this value:

${𝐶}_{𝑓}=\frac{{𝜖}_0𝐴}{𝑑-𝑡+\frac{𝑡}{3}}$

To simplify the denominator:

${𝐶}_{𝑓}=\frac{{𝜖}_0𝐴}{𝑑-\frac{3𝑡}{3}+\frac{𝑡}{3}}$

${𝐶}_{𝑓}=\frac{{𝜖}_0𝐴}{𝑑-\frac{2𝑡}{3}}$

Step 3: Use the given condition about the increase in capacity.

The problem states that the capacity increases by $50\%$. This means the final capacitance ${𝐶}_{𝑓}$ is $150\%$ of the initial capacitance ${𝐶}_0$.

${𝐶}_{𝑓}={𝐶}_0+50\%\text{ of }{𝐶}_0$

${𝐶}_{𝑓}={𝐶}_0+0.5{𝐶}_0$

${𝐶}_{𝑓}=1.5{𝐶}_0$

Step 4: Equate the expressions for ${𝐶}_{𝑓}$ and solve for the thickness $𝑡$.

Now, substitute the expressions for ${𝐶}_{𝑓}$ and ${𝐶}_0$ into the equation from Step 3:

$\frac{{𝜖}_0𝐴}{𝑑-\frac{2𝑡}{3}}=1.5\left(\frac{{𝜖}_0𝐴}{𝑑}\right)$

We can cancel out ${𝜖}_0𝐴$ from both sides of the equation:

$\frac{1}{𝑑-\frac{2𝑡}{3}}=\frac{1.5}{𝑑}$

Now, cross-multiply to solve for $𝑡$:

$𝑑=1.5(𝑑-\frac{2𝑡}{3})$

$𝑑=1.5𝑑-1.5\times \frac{2𝑡}{3}$

$𝑑=1.5𝑑-\frac{3}{2}\times \frac{2𝑡}{3}$

$𝑑=1.5𝑑-𝑡$

Rearrange the equation to isolate $𝑡$:

$𝑡=1.5𝑑-𝑑$

$𝑡=0.5𝑑$

$𝑡=\frac{𝑑}{2}$

Thus, the thickness of the dielectric slab is $\frac{𝑑}{2}$.

The final answer is $\overline{)\frac{d}{2}}$.

Let the two particles, each of mass $𝑚$, move in a circle of radius $𝑟$ due to their mutual gravitational attraction.

Step 1: Gravitational Force

The gravitational force between the particles separated by distance $2𝑟$ is:

$𝐹=\frac{𝐺{𝑚}^2}{(2𝑟{)}^2}=\frac{𝐺{𝑚}^2}{4{𝑟}^2}$

Step 2: Required Centripetal Force

Each particle moves in a circle of radius $𝑟$ about the center of mass. The centripetal force for each is:

${𝐹}_{\text{centripetal}}=\frac{𝑚{𝑣}^2}{𝑟}$

Step 3: Equate Gravitational and Centripetal Force

The only force providing this centripetal acceleration is the gravitational force:

$\frac{𝑚{𝑣}^2}{𝑟}=\frac{𝐺{𝑚}^2}{4{𝑟}^2}$

Step 4: Solve for $𝑣$

Divide both sides by $𝑚$:

$\frac{{𝑣}^2}{𝑟}=\frac{𝐺𝑚}{4{𝑟}^2}$

Multiply both sides by $𝑟$:

${𝑣}^2=\frac{𝐺𝑚}{4𝑟}$

Take the square root:

$𝑣=\sqrt{\surd \frac{𝐺𝑚}{4𝑟}}$

So, the correct answer is:

$\overline{)\sqrt{\surd \frac{\mathrm{Gm}}{4r}}}$

Option A is correct.

Let the four particles each of mass $𝑀$ be placed at the corners of a square of side $𝐿$.

Step 1: Calculate the distance of each particle from the centre

• The centre of the square is equidistant from all corners.

• The distance from the centre to a corner (let’s call it $𝑟$):

$𝑟=\text{half the diagonal of the square}=\frac{𝐿}{\sqrt{\surd 2}}$

Step 2: Calculate the moment of inertia about the given axis

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

• Moment of inertia for each mass about the axis:

${𝐼}_{\text{each}}=𝑀{𝑟}^2=𝑀{\left(\frac{𝐿}{\sqrt{\surd 2}}\right)}^2=𝑀\cdot \frac{{𝐿}^2}{2}$

• For 4 masses:

${𝐼}_{\text{total}}=4\cdot \frac{𝑀{𝐿}^2}{2}=2𝑀{𝐿}^2$

Step 3: Total mass of the system

${𝑀}_{\text{total}}=4𝑀$

Step 4: Formula for radius of gyration $𝐾$

The moment of inertia is also given by: $𝐼={𝑀}_{\text{total}}{𝐾}^2$

So,

${𝐼}_{\text{total}}={𝑀}_{\text{total}}{𝐾}^2⟹2𝑀{𝐿}^2=4𝑀{𝐾}^2$

Step 5: Solve for $𝐾$

$4𝑀{𝐾}^2=2𝑀{𝐿}^2{𝐾}^2=\frac{2𝑀{𝐿}^2}{4𝑀}{𝐾}^2=\frac{{𝐿}^2}{2}𝐾=\frac{𝐿}{\sqrt{\surd 2}}$

Correct option:

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

So, the answer is Option D.

The correct answer is:

Option C: electric field is maximum.

Explanation:

• In the depletion layer of a reverse biased p-n junction, immobile ions create a strong electric field.

• This electric field opposes the flow of majority charge carriers.

• The electric field is strongest (maximum) across the depletion region because this is where the immobile charges are present.

Therefore, the correct answer is Option C.

Let the original lengths of the rods be ${𝑙}_1$ and ${𝑙}_2$.

Let the coefficients of linear expansion be ${𝛼}_1$ and ${𝛼}_2$.

Step 1 (Write the expansion formula):

When temperature increases by $\Delta 𝑇$, the new lengths are:

• First rod: ${𝑙}_{\prime }^1={𝑙}_1(1+{𝛼}_1\Delta 𝑇)$

• Second rod: ${𝑙}_{\prime }^2={𝑙}_2(1+{𝛼}_2\Delta 𝑇)$

Step 2 (Find the difference in lengths):

Difference after expansion:

$\text{Difference}={𝑙}_{\prime }^1\left(𝑇\right)-{𝑙}_{\prime }^2\left(𝑇\right)={𝑙}_1(1+{𝛼}_1\Delta 𝑇)-{𝑙}_2(1+{𝛼}_2\Delta 𝑇)$

Simplify:

$=({𝑙}_1-{𝑙}_2)+({𝑙}_1{𝛼}_1-{𝑙}_2{𝛼}_2)\Delta 𝑇$

Step 3 (Condition given in the question):

The difference should be independent of temperature.

So coefficient of $\Delta 𝑇$ must be zero:

${𝑙}_1{𝛼}_1-{𝑙}_2{𝛼}_2=0$

${𝑙}_1{𝛼}_1={𝑙}_2{𝛼}_2$

$\frac{{𝑙}_1}{{𝑙}_2}=\frac{{𝛼}_2}{{𝛼}_1}$

Step 4 (Choose the correct option):

This matches Option B.

Final Answer:

$\overline{)\frac{{𝑙}_1}{{𝑙}_2}=\frac{{𝛼}_2}{{𝛼}_1}}$

So, the correct answer is Option B.

Let us analyze the situation step by step:

1. Understanding Mutual Inductance ($𝑀$):

• Mutual inductance between two coils is given by

  $𝑀=\frac{{\Phi }_{21}}{{𝐼}_1}$

  where ${\Phi }_{21}$ is the magnetic flux linked with the second coil due to the current ${𝐼}_1$ in the first coil.

1. Loops Arrangement:

• Both loops are concentric and coplanar, with ${𝑅}_1>{𝑅}_2$.

1. Field due to Large Loop (${𝑅}_1$) at its Centre:

• Magnetic field at the centre of a loop of radius ${𝑅}_1$ carrying current ${𝐼}_1$:

  $𝐵=\frac{{𝜇}_0{𝐼}_1}{2{𝑅}_1}$

1. Flux through the Smaller Loop (${𝑅}_2$):

• The flux through the smaller loop is:

  ${\Phi }_{21}=𝐵\cdot \text{Area of smaller loop}$

  ${\Phi }_{21}=\left(\frac{{𝜇}_0{𝐼}_1}{2{𝑅}_1}\right)𝜋{𝑅}_2^2$

1. Calculating $𝑀$:

   $𝑀=\frac{{\Phi }_{21}}{{𝐼}_1}=\frac{{𝜇}_0}{2{𝑅}_1}𝜋{𝑅}_2^2$

• So, $𝑀\propto \frac{{𝑅}_2^2}{{𝑅}_1}$

Final Answer:

$\overline{)\frac{{𝑅}_2^2}{{𝑅}_1}}$

So, the correct option is Option D.

Given:

• Time for one complete swing (period) $𝑇=0.1$ seconds

• The spring is not stretched when the mass is at its highest point (this is the farthest point from the middle position).

• Gravitational acceleration, $𝑔=10{m/s}^2$

Let's find the answer step by step.

Step 1: Formula for Maximum Speed in S.H.M.

In simple harmonic motion, the highest speed is given by:

${𝑣}_{𝑚𝑎𝑥}=𝐴𝜔$

Where:

• $𝐴$ is the amplitude (the maximum distance from the center position)

• $𝜔$ is the angular frequency (tells us how fast the object goes in a circle or swings back and forth)

Step 2: Find Angular Frequency $𝜔$

The formula connecting period and angular frequency is:

$𝜔=\frac{2𝜋}{𝑇}$

Plug in $𝑇=0.1$:

$𝜔=\frac{2𝜋}{0.1}=20𝜋$

Step 3: Find Amplitude $𝐴$

At the highest position, the spring is just relaxed (not stretched). This point is $𝐴$ distance from the mean (middle) position.

• At the middle (mean position), gravity balances the pull of the spring: $𝑚𝑔=𝑘{𝑥}_0$, where ${𝑥}_0$ is the stretch from natural length.

• At highest point: extension is $0$ (spring just relaxed). This distance from the middle to the top is our amplitude $𝐴$.

• So, $𝐴={𝑥}_0$.

Thus, to find $𝐴$ we find ${𝑥}_0$.

From $𝑚𝑔=𝑘{𝑥}_0$:

${𝑥}_0=\frac{𝑚𝑔}{𝑘}$

Step 4: Connect $𝑘$ and $𝑚$ to $𝜔$

For a spring and mass, angular frequency is:

$𝜔=\sqrt{\surd \frac{𝑘}{𝑚}}$

This means $\frac{𝑘}{𝑚}={𝜔}^2$.

So, substitute into earlier expression for ${𝑥}_0$:

${𝑥}_0=\frac{𝑚𝑔}{𝑘}=\frac{𝑔}{𝑘/𝑚}=\frac{𝑔}{{𝜔}^2}$

Step 5: Plug Amplitude into the Speed Formula

Recall:

${𝑣}_{𝑚𝑎𝑥}=𝐴𝜔$

Since $𝐴={𝑥}_0=\frac{𝑔}{{𝜔}^2}$, then:

${𝑣}_{𝑚𝑎𝑥}=\frac{𝑔}{{𝜔}^2}\cdot 𝜔=\frac{𝑔}{𝜔}$

Step 6: Substitute the Values

• $𝑔=10$

• $𝜔=20𝜋$

Now calculate maximum speed:

${𝑣}_{𝑚𝑎𝑥}=\frac{10}{20𝜋}=\frac{1}{2𝜋}m/s$

Step 7: Final Answer

The correct answer is:

$\overline{)\frac{1}{2𝜋}m/s}$