MHT CET 2024 2nd May Morning Shift (Physics & Chemistry) Question Paper with Solutions

At room temperature, semiconductors exhibit thermal excitation of electrons from the valence band to the conduction band due to the relatively small energy gap between these bands (typically around 1 eV). This means that:

The valence band is not completely filled because some electrons have moved to the conduction band, leaving behind holes.

The conduction band is not completely empty because it now contains some thermally excited electrons.

Both the conduction band and the valence band are partially filled and partially empty, respectively.

Therefore, the correct statement is:

Option C: The conduction band is partially filled and the valence band is partially empty.

Answer: Option C

Given that, $𝛼=4\mathrm{rad}/s^2$

Centripetal (radial) acceleration, $a_r=r{𝜔}^2$

Tangential acceleration, ${𝑎}_{𝑡}=𝑟𝛼$

If ${𝑎}_{𝑟}={𝑎}_{𝑡}$, then $𝑟{𝜔}^2=𝑟𝛼$

$\begin{array}{cc}\therefore & {𝜔}^2=𝛼=4\\ \therefore & 𝜔=\sqrt{\surd 4}=2\mathrm{rad}/s\\ & \text{ But, }𝜔={𝜔}_0+𝛼t=0+𝛼t=𝛼t\\ \therefore & 2=4t\\ \therefore & t=\frac{1}{2}s\end{array}$

The output of the NAND gate will be $\overline{A\cdot B}$.

The output of the NOR gate will be

$\overline{\stackrel{–}{A\cdot B}+\stackrel{–}{A\cdot B}}=A\cdot B$

The output of the NOT gate will be $\overline{A.B}$

Thus, the given network is equivalent to a NAND gate.

Two coherent sources of intensities $I_1$ and $I_2$ produce,

maximum intensity, $I_{\text{max }}={(\sqrt{\surd {𝐼}_1}+\sqrt{\surd {𝐼}_2})}^2$ and

minimum intensity, ${𝐼}_{\text{min }}={(\sqrt{\surd {𝐼}_1}-\sqrt{\surd {𝐼}_2})}^2$ in an interference pattern.

$\begin{array}{cc}& \frac{{𝐼}_{max}}{{𝐼}_{min}}=\frac{{(\sqrt{\surd {𝐼}_1}+\sqrt{\surd {𝐼}_2})}^2}{{(\sqrt{\surd {𝐼}_1}-\sqrt{\surd {𝐼}_2})}^2}=\frac{{𝐼}_1+{𝐼}_2+2\sqrt{\surd {𝐼}_{{𝐼}_2}}}{{𝐼}_1+{𝐼}_2-2\sqrt{\surd {𝐼}_1{𝐼}_2}}\end{array}$

Given $I_1=2I_2$

$\begin{array}{cc}\therefore & \frac{I_{\text{max }}}{I_{min}}=\frac{3I_2+2\sqrt{\surd 2}{I}_2}{3I_2-2\sqrt{\surd 2}{I}_2}=\frac{3+2\sqrt{\surd 2}}{3-2\sqrt{\surd 2}}=34\end{array}$

The ideal gas law is given by:

$𝑃𝑉=𝑛𝑅𝑇$

Where:

$𝑃$ is the pressure,

$𝑉$ is the volume,

$𝑛$ is the number of moles,

$𝑅$ is the ideal gas constant,

$𝑇$ is the absolute temperature.

For the two gases, since the mass, volume, and pressure are the same for both oxygen ($O_2$) and hydrogen ($H_2$), we can set up the following expressions for each gas:

For oxygen:

$𝑃𝑉={𝑛}_{O_2}𝑅{𝑇}_{O_2}$

For hydrogen:

$𝑃𝑉={𝑛}_{H_2}𝑅{𝑇}_{H_2}$

Since $𝑃𝑉$ is the same for both gases, we have:

${𝑛}_{O_2}𝑅{𝑇}_{O_2}={𝑛}_{H_2}𝑅{𝑇}_{H_2}$

This simplifies to:

${𝑛}_{O_2}{𝑇}_{O_2}={𝑛}_{H_2}{𝑇}_{H_2}$

The number of moles ($𝑛$) is given by:

$𝑛=\frac{\text{mass}}{\text{molar mass}}$

Let $𝑚$ be the mass of each gas. Then:

For oxygen:

${𝑛}_{O_2}=\frac{𝑚}{32}$

For hydrogen:

${𝑛}_{H_2}=\frac{𝑚}{2}$

Substitute into the equation:

$\frac{𝑚}{32}{𝑇}_{O_2}=\frac{𝑚}{2}{𝑇}_{H_2}$

Cancel $𝑚$ from both sides:

$\frac{1}{32}{𝑇}_{O_2}=\frac{1}{2}{𝑇}_{H_2}$

Cross-multiply to find the ratio of temperatures:

$2{𝑇}_{O_2}=32{𝑇}_{H_2}$

Simplify:

${𝑇}_{O_2}=16{𝑇}_{H_2}$

Thus, the ratio of the absolute temperatures is:

${𝑇}_{O_2}:{𝑇}_{H_2}=16:1$

The correct option is Option C: $16:1$.

To find the height above the Earth's surface where the acceleration due to gravity (${𝑔}^{\prime }$) becomes $\frac{1}{𝑛}$ times the value at the surface ($𝑔$), we can use the formula for gravitational acceleration at a distance:

${𝑔}^{\prime }=\frac{𝐺𝑀}{(𝑅+ℎ{)}^2}$

where:

$𝐺$ is the gravitational constant,

$𝑀$ is the mass of the Earth,

$𝑅$ is the radius of the Earth,

$ℎ$ is the height above the Earth's surface,

$𝑔=\frac{𝐺𝑀}{{𝑅}^2}$ is the acceleration due to gravity at the surface of the Earth.

Given that ${𝑔}^{\prime }=\frac{1}{𝑛}\cdot 𝑔$, we have:

$\frac{𝐺𝑀}{(𝑅+ℎ{)}^2}=\frac{1}{𝑛}\cdot \frac{𝐺𝑀}{{𝑅}^2}$

The gravitational constant $𝐺$ and the Earth's mass $𝑀$ cancel out, leading to:

$\frac{1}{(𝑅+ℎ{)}^2}=\frac{1}{𝑛\cdot {𝑅}^2}$

Taking the reciprocal of both sides and then taking the square root:

$(𝑅+ℎ{)}^2=𝑛\cdot {𝑅}^2$

$𝑅+ℎ=\sqrt{\surd 𝑛}\cdot 𝑅$

Isolating $ℎ$, we find:

$ℎ=\sqrt{\surd 𝑛}\cdot 𝑅-𝑅$

Therefore, the height $ℎ$ is:

$ℎ=(\sqrt{\surd 𝑛}-1)\cdot 𝑅$

The correct answer is Option D: $(\sqrt{\surd 𝑛}-1)\cdot 𝑅$.

Using Bernoulli's equation,

$\begin{array}{cc}& {𝑃}_1+\frac{1}{2}𝜌{𝑣}_2^1+𝜌{\mathrm{gh}}_1={𝑃}_2+\frac{1}{2}𝜌{𝑣}_2^2+𝜌𝑔{ℎ}_2\\ & {𝑃}_1+\frac{1}{2}𝜌{𝑣}_2^1={𝑃}_2+\frac{1}{2}𝜌{𝑣}_2^2\text{...(given horizontal pipe)}\end{array}$

Substituting the given values,

$\begin{array}{cc}& 𝑃+\frac{1}{2}𝜌{𝑣}_2^1={𝑃}_2+\frac{1}{2}𝜌(3𝑣{)}^2\\ & {𝑃}_2=𝑃+(\frac{1}{2}𝜌{𝑣}^2-\frac{9}{2}𝜌{𝑣}^2)=𝑃-4𝜌{𝑣}^2\end{array}$

The given current in the a.c. circuit is expressed as :

$𝐼=100\sin \left(200𝜋𝑡\right).$

To determine the time required for the current to achieve its peak value, we need to analyze the sinusoidal function $\sin \left(200𝜋𝑡\right)$. The peak value of a sine function occurs when the argument of the sine function is an odd multiple of $\frac{𝜋}{2}$ (i.e., $\frac{𝜋}{2}$, $\frac{3𝜋}{2}$, etc.).

Since we are interested in the first peak, we consider :

$200𝜋𝑡=\frac{𝜋}{2}.$

Solving for $𝑡$ gives :

Divide both sides by $200𝜋$ :

$𝑡=\frac{𝜋/2}{200𝜋}=\frac{1}{400}.$

Therefore, the time required for the current to achieve its first peak value is :

$\frac{1}{400}\text{seconds}.$

The correct option is D : $\frac{1}{400}s$.

The work done to move a charge in an electric field is related to the change in electric potential by the equation:

$𝑊=𝑞\Delta 𝑉$

where:

$𝑊=90\text{J}$ (work done),

$𝑞=3\text{C}$ (charge),

$\Delta 𝑉={𝑉}_1-(-10\text{V})={𝑉}_1+10\text{V}$ (change in electric potential).

We can solve for ${𝑉}_1$ using the formula:

$90=3({𝑉}_1+10)$

Divide both sides by 3:

$30={𝑉}_1+10$

Subtract 10 from both sides:

${𝑉}_1=20\text{V}$

Thus, the value of ${𝑉}_1$ is 20 V, corresponding to Option B.

$\begin{array}{cc}& {𝜔}^2=\frac{k}{m},r=\frac{b}{2m}\\ & \text{ Angular frequency, }\\ & \begin{array}{cc}{𝜔}^{\prime }& =\sqrt{\surd ({𝜔}^2-{𝑟}^2)}\\ & =\sqrt{\surd \frac{𝑘}{𝑚}-{\left(\frac{𝑏}{2𝑚}\right)}^2}\end{array}\end{array}$

The magnetic energy stored in an inductor can be calculated using the formula:

$𝐸=\frac{1}{2}𝐿{𝐼}^2$

where:

$𝐸$ is the energy stored in joules,

$𝐿$ is the inductance in henries,

$𝐼$ is the current in amperes.

Given:

$𝐿=5𝜇H=5\times {10}^{-6}H$,

$𝐼=2A$.

Substitute these values into the formula:

$𝐸=\frac{1}{2}\times 5\times {10}^{-6}\times (2{)}^2$

Calculate the energy:

$𝐸=\frac{1}{2}\times 5\times {10}^{-6}\times 4$

$𝐸=10\times {10}^{-6}J=10𝜇J$

Thus, the magnetic energy stored in the inductor is $10𝜇J$.

The correct option is Option C $10𝜇J$.

$\begin{array}{cc}& \begin{array}{cc}& (M+m)g-T=(M+m)a\\ & T-\mathrm{Mg}=\mathrm{Ma}\end{array}\\ & \text{ Adding the above equations, }\\ & \begin{array}{cc}& (𝑀+𝑚)𝑔-𝑀𝑔=(2𝑀+𝑚)𝑎\\ \therefore & 𝑎=\frac{𝑚𝑔}{(2𝑀+𝑚)}\end{array}\end{array}$

The radius of the orbit in a hydrogen atom is determined by the formula for the Bohr model of the hydrogen atom. The radius of the $𝑛$-th orbit is given by:

${𝑟}_{𝑛}={𝑛}^2\times {𝑟}_1$

where ${𝑟}_1$ is the radius of the innermost orbit (ground state, $𝑛=1$) and $𝑛$ is the principal quantum number.

Given:

The radius of the innermost orbit (ground state) is ${𝑟}_1=5.3\times {10}^{-11}\text{m}$.

For the fourth orbit ($𝑛=4$), the radius is calculated as follows:

${𝑟}_4=4^2\times {𝑟}_1=16\times 5.3\times {10}^{-11}\text{m}$

Calculate ${𝑟}_4$:

${𝑟}_4=16\times 5.3\times {10}^{-11}=84.8\times {10}^{-11}\text{m}$

Converting meters to ångströms (where $1\text{Å}={10}^{-10}\text{m}$):

${𝑟}_4=8.48\text{Å}$

Therefore, the radius of the fourth allowed orbit of the hydrogen atom is:

Option A

8.48 $\stackrel{𝑜}{𝐴}$

$\begin{array}{cc}& {𝑃}_1+{𝑃}_2=9𝐷\text{.... (i)}\\ & {𝑃}_1+{𝑃}_2-𝑑{𝑃}_1{𝑃}_2=\frac{27}{5}\text{.... (ii)}\end{array}$

where $𝑑=20\mathrm{cm}=0.2m$

Substituting value of $(P_1+P_2)$ in equation (ii) we get

$\begin{array}{cc}& 9-0.2{𝑃}_1{𝑃}_2=\frac{27}{5}\\ \therefore & 0.2{𝑃}_1{𝑃}_2=9-\frac{27}{5}=\frac{18}{5}\\ \therefore & {𝑃}_1{𝑃}_2=\frac{18}{0.2\times 5}=18\end{array}$

Since ${𝑃}_1+{𝑃}_2=9$ and ${𝑃}_1{𝑃}_2=18$

$\therefore {𝑃}_1=3$ and ${𝑃}_2=6$

$\begin{array}{cc}{𝑦}_1& ={𝑎}_1\sin (\frac{2𝜋𝑥}{𝜆}-𝜔𝑡)\text{.... (i)}\\ {𝑦}_2& ={𝑎}_2\cos (\frac{2𝜋𝑥}{𝜆}-𝜔𝑡+𝜙)\\ & ={𝑎}_2\sin (\frac{2𝜋𝑥}{𝜆}-𝜔𝑡+𝜙+\frac{𝜋}{2})\text{.... (ii)}\end{array}$

From (ii) and (i)

$\therefore$ Phase difference $=𝜙+\frac{𝜋}{2}$

Outside pressure $=1\mathrm{atm}$

Pressure inside soap bubble $A=1.01\mathrm{atm}$

Pressure inside soap bubble $𝐵=1.02\mathrm{atm}$

$\therefore$ Excess pressures will be

$\begin{array}{cc}& \Delta P_A=1.01-1=0.01\mathrm{atm}\text{ and }\\ & \Delta P_B=1.02-1=0.02\mathrm{atm}\end{array}$

Now, $\Delta P\propto \frac{1}{r}\Rightarrow r\propto \frac{1}{\Delta P}$

$\begin{array}{cc}\therefore & \frac{r_A}{r_B}=\frac{\Delta P_B}{\Delta P_A}=\frac{0.02}{0.01}=\frac{2}{1}\\ & \text{ Now, }V=\frac{4}{3}𝜋r^3\\ & \Rightarrow V\propto r^3\\ \therefore & \frac{V_A}{V_B}={\left(\frac{r_A}{r_B}\right)}^3={\left(\frac{2}{1}\right)}^3=\frac{8}{1}\end{array}$

As the angular velocity is constant, the linear velocity changes from 0 to $R𝜔$ as we move from the axle to the rim.

$\begin{array}{cc}& \therefore \text{ Average velocity }{v}_{\text{avg }}=\frac{R𝜔}{2}\\ & \text{ We know motional e.m.f, }{E}_{\mathrm{mot}}=B/v\\ & \therefore E_{\mathrm{mot}}=\frac{B\times R\times R𝜔}{2}=\frac{{\mathrm{BR}}^2\times 2𝜋F}{2}\\ & ={B𝜋R^2}^2\end{array}$

For a simple pendulum,

$T=2𝜋\sqrt{\surd \frac{𝑙}{g}}\Rightarrow T\propto \frac{1}{g}$

When the point of suspension is moved vertically upwards with acceleration a, $𝑔$ becomes $𝑔+𝑎$. Hence, T decreases.

$\begin{array}{cc}& \text{ From Einstein’s equation, }\\ & \mathrm{hv}=e{V}_0+h{𝑣}_0\\ & \therefore \frac{\mathrm{hc}}{𝜆}-\frac{\mathrm{hc}}{{𝜆}_0}=𝑒V_0\\ & \text{ case (i) }𝜆=𝜆;V_0=V\\ & \frac{\mathrm{hc}}{𝜆}-\frac{\mathrm{hc}}{{𝜆}_0}=\mathrm{eV}\text{... (i)}\\ & \text{ case (ii) }𝜆=3𝜆;V_0=\frac{V}{6}\\ & \frac{\mathrm{hc}}{3𝜆}-\frac{\mathrm{hc}}{{𝜆}_0}=\frac{\mathrm{eV}}{6}\text{... (ii)}\end{array}$

$\begin{array}{cc}\therefore & \frac{(\frac{\mathrm{hc}}{𝜆}-\frac{\mathrm{hc}}{{𝜆}_0})}{(\frac{\mathrm{hc}}{3𝜆}-\frac{\mathrm{hc}}{{𝜆}_0})}=6\\ \therefore & \frac{1}{𝜆}-\frac{1}{{𝜆}_0}=6(\frac{1}{3𝜆}-\frac{1}{{𝜆}_0})\end{array}$

$\begin{array}{cc}\therefore & \frac{1}{𝜆}-\frac{1}{{𝜆}_0}=\frac{2}{𝜆}-\frac{6}{{𝜆}_0}\\ \therefore & \frac{-1}{{𝜆}_0}+\frac{6}{{𝜆}_0}=\frac{2}{𝜆}-\frac{1}{𝜆}\\ \therefore & \frac{5}{{𝜆}_0}=\frac{1}{𝜆}\Rightarrow {𝜆}_0=5𝜆\end{array}$

$\begin{array}{cc}& A=𝜋r^2\Rightarrow 𝑟=\sqrt{\surd \frac{A}{𝜋}}\\ & \therefore B=\frac{{𝜇}_{0}I}{2r}=\frac{{𝜇}_{0}I}{2\sqrt{\surd \frac{A}{𝜋}}}\\ & \therefore I=\frac{2B}{{𝜇}_0}\sqrt{\surd \frac{A}{𝜋}}\\ & \\ & \text{ Magnetic moment }m=\mathrm{IA}\\ & \\ & \mathrm{IA}=\frac{2B}{{𝜇}_0}\sqrt{\surd \frac{A}{𝜋}}\times A=\frac{2{\mathrm{BA}}^{3/2}}{{𝜇}_0{𝜋}^{1/2}}\end{array}$

The adiabatic curve is steeper (has a greater slope).This steepness reflects the rapid change in pressure with volume during an adiabatic process, due to the absence of heat transfer which allows the gas to expand or contract more quickly in response to changes in volume. In the given $P-V$ graph, adiabatic process is shown by region $𝐴𝐷$ and $𝐵𝐶$ only as their slope is greater.

$\begin{array}{cc}& \text{ Now, }𝐺=\left(\frac{𝐺𝑆}{𝑆+𝐺}\right)+{𝑆}^{\prime }\\ \therefore & 𝐺-\frac{𝐺𝑆}{𝑆+𝐺}={𝑆}^{\prime }\\ \therefore & {𝑆}^{\prime }=\frac{{𝐺}^2}{𝑆+𝐺}\end{array}$

$\begin{array}{cc}& 𝑉=100\sin \left(100t\right)V\\ & I=100\sin (100t+\frac{𝜋}{3})\mathrm{mA}\end{array}$

Comparing these equations with the standard forms,

$V=V_0\sin 𝜔t$ and $I=I_0\sin 𝜔t$

we get,

$V_0=100V,I_0=100\times {10}^{-3}A\text{ and }𝜙=\frac{𝜋}{3}$

$\begin{array}{cc}\therefore P=V_{\mathrm{rms}}{I}_{\mathrm{rms}}\cos 𝜙& =\frac{V_0}{\sqrt{\surd 2}}\times \frac{I_0}{\sqrt{\surd 2}}\cos \frac{𝜋}{3}\\ & =\frac{100}{\sqrt{\surd 2}}\times \frac{100\times {10}^{-3}}{\sqrt{\surd 2}}\times \frac{1}{2}=2.5W\end{array}$

$\begin{array}{cc}& \text{ The moment of inertia of the annular ring is }\\ & \begin{array}{cc}𝐼& =\frac{1}{2}𝑀({𝑅}_2^1+{𝑅}_2^2)\\ & =\frac{10(100+25)}{2}=625{\mathrm{kgm}}^2\end{array}\end{array}$

The increase in length of a segment of railway track due to thermal expansion can be calculated using the formula for linear expansion:

$\Delta 𝐿={𝐿}_0\cdot 𝛼\cdot \Delta 𝑇$

where:

$\Delta 𝐿$ is the change in length of the track,

${𝐿}_0=10\text{m}$ is the original length of the segment,

$𝛼=1.2\times {10}^{-5}/{}^{\circ }C$ is the coefficient of linear expansion for steel,

$\Delta 𝑇={45}^{\circ }C-{17}^{\circ }C={28}^{\circ }C$ is the change in temperature.

Plugging the values into the formula gives:

$\Delta 𝐿=10\text{m}\times 1.2\times {10}^{-5}/{}^{\circ }C\times {28}^{\circ }C$

Calculating further:

$\Delta 𝐿=10\times 1.2\times {10}^{-5}\times 28=3.36\times {10}^{-3}\text{m}$

The increase in length in terms of $𝑥\times {10}^{-5}\text{m}$ is:

$\Delta 𝐿=336\times {10}^{-5}\text{m}$

Thus, the value of $𝑥$ is 336.

The correct option is Option C: 336.

Let n be frequency of tuning fork.

Let $n_1,n_2$ be frequency of wire at tension $T_1,T_2$ respectively.

$\begin{array}{cc}n& \propto \sqrt{\surd T}\text{.... (i)}\\ n_1& =n-6\text{.... (ii)}\\ n_2& =n+6\text{.... (iii)}\end{array}$

$\begin{array}{cc}\therefore & \frac{n_1}{n_2}=\sqrt{\surd \frac{T_1}{T_2}}=\sqrt{\surd \frac{225}{256}}=\frac{15}{16}\\ \therefore & \frac{n-6}{n+6}=\frac{15}{16}\text{... from (i), (ii), (iii)}\\ \therefore & 16n-96=15n+90\\ \therefore & 𝑛=186\mathrm{Hz}\end{array}$

$\begin{array}{cc}& 𝐸\propto \frac{{𝑍}^2}{{𝑛}^2}\\ & \therefore \frac{E_H}{E_{\mathrm{Hc}}}=\frac{{\left(Z^2\right)}_H}{(n{)}_2^H}\times \frac{(n{)}_2^{\mathrm{He}}}{{\left(Z^2\right)}_{\mathrm{He}}}\\ & =\frac{1}{3^2}\times \frac{5^2}{2^2}=\frac{25}{36}\\ & \therefore E_{\mathrm{He}}=\frac{36}{25}{E}_H=\frac{36}{25}E\end{array}$

The mean free path of gas molecules at standard temperature and pressure (S.T.P.) is given by:

${𝜆}_0=1500𝑑$

where ${𝜆}_0$ is the mean free path at S.T.P., and $𝑑$ is the diameter of a gas molecule.

At constant volume, the mean free path $𝜆$ is affected by temperature according to the relationship:

$𝜆\propto \frac{𝑇}{𝑃}$

Since the pressure remains constant (as volume is constant), the relationship simplifies to:

$𝜆\propto 𝑇$

Thus, to find the mean free path at a new temperature ${𝑇}_1=373\text{K}$, we use the ratio of the temperatures:

${𝜆}_1=\left(\frac{{𝑇}_1}{{𝑇}_0}\right){𝜆}_0$

Substituting the known temperatures:

${𝑇}_0=273\text{K}$ (S.T.P.)

${𝑇}_1=373\text{K}$

The mean free path at 373 K becomes:

${𝜆}_1=\left(\frac{373}{273}\right)\times 1500𝑑$

Thus, the correct answer is Option B:

$\frac{373}{273}\times 1500d$

Potential energy $𝑈=\frac{1}{4𝜋{𝜀}_0}\frac{{𝑞}_1{𝑞}_2}{𝑎}$

$\therefore$ Potential energy at $𝐴$ due to $𝐵=\frac{q^2}{4𝜋{𝜀}_{0}a}$

Potential energy at $𝐵$ due to $𝐶=\frac{𝑄𝑞}{4𝜋{𝜀}_0𝑎}$

Potential energy at C due to $A=\frac{\mathrm{Qq}}{4𝜋{𝜀}_{0}a}$

$\begin{array}{cc}\therefore & \text{ Total P.E }=\frac{𝑞}{4𝜋{𝜀}_0𝑎}(𝑞+𝑄+𝑄)=0\\ \therefore & 2𝑄+𝑞=0\Rightarrow 𝑄=-\frac{𝑞}{2}\end{array}$

For an adiabatic process,

$W=\frac{\mathrm{nR}(T_i-T_f)}{𝛾-1}$

$\begin{array}{ccc}& \therefore 6𝑅& =\frac{𝑅(𝑇-{𝑇}_{𝑓})}{(\frac{5}{3}-1)}\text{... (Given : n = 1)}\\ & \therefore {𝑇}_{𝑓}& =(𝑇-4)𝐾\end{array}$

In a Young's double slit experiment where one slit is covered by a red filter and the other slit by a blue filter, the interference pattern will not form. This is because the two slits are illuminated with light of different wavelengths (red and blue), and for a meaningful interference pattern, coherent sources of the same wavelength (or nearly the same) are necessary.

Thus, the correct option is:

Option C: there shall be no interference fringes.

Interference occurs when waves overlap and superimpose to create a pattern; however, this requires the waves to have a constant phase relationship. This condition cannot be met when using light of different colors, or wavelengths, because they will not have the necessary coherent phase relationship across the slits. Therefore, the result is no visible interference pattern of fringes on the observing screen.

Using Bernoulli's equation,

$\begin{array}{cc}& \text{ Using Bernoulli’s equatuon, }\\ & {𝑃}_1+\frac{1}{2}𝜌{𝑣}_2^1+𝜌{\mathrm{gh}}_1={𝑃}_2+\frac{1}{2}𝜌{𝑣}_2^2+𝜌{\mathrm{gh}}_2\\ & {𝑃}_1+\frac{1}{2}𝜌{𝑣}_2^1={𝑃}_2+\frac{1}{2}𝜌{𝑣}_2^2\\ & {𝑃}_1-{𝑃}_2=\frac{1}{2}(𝜌{𝑣}_2^2-𝜌{𝑣}_2^1)\\ & {𝑣}_2^2-{𝑣}_2^1=\frac{2\times 3}{1.25\times {10}^3}=4.8\times {10}^{-3}\text{.... (i)}\end{array}$

From equation of continuity,

$\begin{array}{cc}& A_1{v}_1=A_2{V}_2\\ & \frac{v_1}{V_2}=\frac{A_2}{A_1}=\frac{5}{10}=0.5\\ & v_1=0.5v_2\text{.... (ii)}\end{array}$

Substituting (ii) into (i),

$\begin{array}{cc}0.75v_2^2=4.8\times {10}^{-3}\Rightarrow v_2& =0.08m/s\\ \text{ Rate of flow of glycerine }& =A_2{v}_2\\ & =5\times {10}^{-4}\times 0.08\\ & =4\times {10}^{-5}{m}^3/s\end{array}$

In an NPN transistor, the current amplification factors are defined as:

Alpha ($𝛼$): It is the ratio of the collector current (${𝐼}_{𝐶}$) to the emitter current (${𝐼}_{𝐸}$) and is given by $𝛼=\frac{{𝐼}_{𝐶}}{{𝐼}_{𝐸}}$

Beta ($𝛽$): It is the ratio of the collector current (${𝐼}_{𝐶}$) to the base current (${𝐼}_{𝐵}$) and is defined as $𝛽=\frac{{𝐼}_{𝐶}}{{𝐼}_{𝐵}}$

To find these values, first, calculate the emitter current. The number of electrons entering the emitter is ${10}^{10}$ in ${10}^{-6}$ seconds:

The charge of one electron is approximately $1.6\times {10}^{-19}$ coulombs. Therefore, the emitter current ${𝐼}_{𝐸}$ is:

${𝐼}_{𝐸}=\frac{\text{Number of electrons}\times \text{Charge of one electron}}{\text{Time}}=\frac{{10}^{10}\times 1.6\times {10}^{-19}}{{10}^{-6}}=1.6\times {10}^{-3}\text{A}$

Given that $2\%$ of electrons recombine in the base, the remaining $98\%$ reach the collector. Thus, the collector current ${𝐼}_{𝐶}$ is:

${𝐼}_{𝐶}=0.98\times {𝐼}_{𝐸}=0.98\times 1.6\times {10}^{-3}=1.568\times {10}^{-3}\text{A}$

The base current ${𝐼}_{𝐵}$ can be found as:

${𝐼}_{𝐵}={𝐼}_{𝐸}-{𝐼}_{𝐶}=1.6\times {10}^{-3}-1.568\times {10}^{-3}=0.032\times {10}^{-3}\text{A}$

Now compute $𝛼$ and $𝛽$:

Alpha ($𝛼$):

$𝛼=\frac{{𝐼}_{𝐶}}{{𝐼}_{𝐸}}=\frac{1.568\times {10}^{-3}}{1.6\times {10}^{-3}}=0.98$

Beta ($𝛽$):

$𝛽=\frac{{𝐼}_{𝐶}}{{𝐼}_{𝐵}}=\frac{1.568\times {10}^{-3}}{0.032\times {10}^{-3}}=49$

Therefore, the current gain ratios are $𝛼=0.98$ and $𝛽=49$, which corresponds to Option A: $0.98,49$.