KCET 2024 Question Paper with Solutions

The ratio of molar specific heats $({𝐶}_{𝑝}$ and ${𝐶}_{𝑉})$ is

$𝛾=1+\frac{2}{𝑓}$

$\begin{array}{cc}& \text{ For }{O}_2\text{, }\\ & \begin{array}{cc}& 𝑓=5\\ \therefore & 𝛾=1+\frac{2}{5}=1+0.4=1.4\end{array}\end{array}$

We know that velocity $𝑣$ and acceleration of a particle in SHM.

$𝑣=𝜔\sqrt{\surd {𝐴}^2-{𝑦}^2}\text{ and }𝑎=-{𝜔}^2𝑦$

$\therefore$ At mean position, $𝑦=0$

$\begin{array}{cc}\therefore & 𝑣=𝜔\sqrt{\surd {𝐴}^2-0}\\ \Rightarrow & {𝑣}_{max}=𝜔𝐴\text{ and }𝑎=0\end{array}$

To find the number of beats heard by the motor-cyclist:

Convert the speed of the motor-cyclist:

The speed given is $18\text{ km/h}$.

Convert km/h to m/s:

${𝑣}_{𝑠}=18\times \frac{5}{18}\text{m/s}=5\text{m/s}$

Determine the properties given:

Frequency of the source, $𝑓=325\text{ Hz}$.

Speed of sound in air, $𝑣=330\text{ m/s}$.

Calculate the number of beats heard:

The formula for the frequency change due to the Doppler effect, when the observer and the source are moving towards each other, is:

$\Delta 𝑓=𝑓\left(\frac{{𝑣}_{𝑠}}{𝑣-{𝑣}_{𝑠}}\right)$

Substitute the known values:

$\Delta 𝑓=325\left(\frac{5}{330-5}\right)=5$

Conclusion:

The number of beats heard by the motor-cyclist is 5.

Charge on body, $𝑞=-3.2𝜇C=-3.2\times {10}^{-6}C$

$\therefore$ Number of excess electrons,

$\begin{array}{cc}𝑛& =\frac{𝑞}{𝑒}=\frac{-3.2\times {10}^{-6}}{-1.6\times {10}^{-19}}\\ & =2\times {10}^{13}\end{array}$

Electrostatic force on each charges placed at certain distance is equal in magnitude but opposite in direction.

$\begin{array}{cc}& \text{ Given, }𝐄=3\times {10}^5\widehat{𝐣}{\mathrm{NC}}^{-1}\\ & \begin{array}{cc}& 𝐴=10\mathrm{cm}\times 30\mathrm{cm}=3\times {10}^{-2}{m}^2\\ & \therefore 𝐀=3\times {10}^{-2}\widehat{𝐣}{m}^2\text{ (For }\mathrm{ZX}\text{-plane) }\\ & \begin{array}{cc}\therefore \text{ Electric flux, }𝜙& =𝐄\cdot 𝐀\\ & =3\times {10}^5\widehat{𝐣}\times 3\times {10}^{-2}\widehat{𝐣}\\ & =9\times {10}^{3}V-m\end{array}\end{array}.\begin{array}{c}\end{array}\end{array}$

The total electric flux through a closed surface is given by Gauss's Law, which states that:

$\Phi =\frac{{𝑄}_{\text{enc}}}{{𝜀}_0}$

Where $\Phi$ is the electric flux, ${𝑄}_{\text{enc}}$ is the total charge enclosed by the surface, and ${𝜀}_0$ is the permittivity of free space.

In the case of an electric dipole, the dipole consists of two equal and opposite charges separated by some distance. Therefore, the net charge ${𝑄}_{\text{enc}}$ enclosed by the spherical surface is zero, because the positive and negative charges cancel each other out.

Thus, the total electric flux through the closed spherical surface enclosing an electric dipole is:

$\Phi =\frac{0}{{𝜀}_0}=0$

Therefore, the correct option is:

Option A: zero

Since, there is no electric field inside the conductor, hence according to Coulomb's law, the mutual repulsion of like charges demands that the charges be as for apart as possible. Hence, on the surface of the conductor, the net electric charge resides entirely on its surface. Electric field on the surface of a charged conductor is $𝜎/{𝜀}_0$. Electric potential inside a charged conductor is constant (non-zero). The net electric field lines is perpendicular to the surface of the conductor.

Induced dipole moment, $𝑝=0.2\times {10}^{-5}C$

Number of molecules, $𝑛=100$ molecule/ ${\mathrm{cm}}^3$

$\begin{array}{cc}& =100\times {10}^6\text{ molecule }/m^3\\ 𝐸& =4N/C\end{array}$

$\therefore$ Electric susceptibility,

$\begin{array}{cc}{𝜒}_{𝑒}& =\frac{𝑝}{𝐸}\times 𝑛=\frac{0.2\times {10}^{-6}\times 100\times {10}^6}{4}\\ & =5C^2{N}^{-1}{m}^{-2}\end{array}$

Given, battery emf, $𝑒=10V$

$\begin{array}{cc}𝐶& =5𝜇F\end{array}=5\times {10}^{-6}F.$

We know that stored energy,

$\begin{array}{cc}𝑈& =\frac{1}{2}𝐶{𝑉}^2\\ \therefore \frac{𝑑𝑈}{𝑑𝑡}& =\frac{1}{2}𝐶\cdot 2𝑉\frac{𝑑𝑉}{𝑑𝑡}=𝐶𝑉\frac{𝑑𝑉}{𝑑𝑡}\\ & =5\times {10}^{-6}\times 4\times 0.6\\ & =12\times {10}^{-6}W=12𝜇W\end{array}$

$\begin{array}{cc}& \text{ We know that, }𝑉=𝐼𝑅\text{ (Ohm’s law) }\\ & \begin{array}{cc}& 𝑉=𝐼\cdot 𝜌\cdot \frac{𝑙}{𝐴}\Rightarrow \frac{𝑉}{𝑙}=\frac{𝐼}{𝐴}\cdot 𝜌\\ \Rightarrow & 𝐸=𝐽𝜌\Rightarrow 𝐸=𝜌𝐽\end{array}\end{array}$

1st and 2nd networks form balanced Wheatstone bridge.

Hence, circuit diagram can be redrawn as

$\begin{array}{cc}\therefore {𝑅}_{𝐴𝐵}& =2𝑅\Vert 2𝑅+4𝑅\Vert 4𝑅\\ & =\frac{2𝑅\times 2𝑅}{2𝑅+2𝑅}+\frac{4𝑅\times 4𝑅}{4𝑅+4𝑅}=𝑅+2𝑅=3𝑅\\ \therefore 𝐼& =\frac{{𝑉}_0}{{𝑅}_{𝐴𝐵}}=\frac{{𝑉}_0}{3𝑅}\end{array}$

Total charge flowing the cross-section of the conductor is equal to area under $𝐼-𝑡$ graph.

$\begin{array}{cc}\therefore 𝑄& =\frac{(100+300)}{2}10\times {10}^{-3}+(20-10)\times 300\times {10}^{-3}\\ & =1+3=4C\end{array}$

Since, $𝑄=𝑛𝑒$

$\therefore 𝑛=\frac{𝑄}{𝑒}=\frac{4}{1.6\times {10}^{-19}}=3.125\times {10}^{19}$

From the I-V graph,

Resistance of first conductor,

${𝑅}_1=\frac{1}{\tan {45}^{\circ }}=1\Omega$

Resistance of second conductor,

${𝑅}_2=\frac{1}{\tan {30}^{\circ }}=\sqrt{\surd 3}\Omega$

$\therefore$ Temperature coefficient of resistance of the conductor.

$\begin{array}{cc}𝛼& =\frac{{𝑅}_2-{𝑅}_1}{{𝑅}_1{𝑡}_2-{𝑅}_2{𝑡}_1}=\frac{\sqrt{\surd 3}-1}{1\times 400-\sqrt{\surd 3}\times 100}\\ & =3.22\times {10}^{-3}{/}^{\circ }C\simeq 3\times {10}^{-3}{/}^{\circ }C\end{array}$

Resistance of the bulb,

$\begin{array}{cc}& 𝑅=\frac{{𝑉}^2}{𝑃}=\frac{120\times 120}{60}=240\Omega \\ \therefore & 𝐼=\frac{𝑃}{𝑉}=\frac{60}{120}=0.5A\end{array}$

Since, bulb is connected with 220 V supply.

$\therefore$ Resistance required,

${𝑅}^{\prime }=\frac{𝑉}{𝐼}=\frac{220}{0.5}=440\Omega$

Thus, for the bulb to glow properly required resistance in series

$\begin{array}{cc}& ={𝑅}^{\prime }-𝑅\\ & =440-240=200\Omega \end{array}$

Standard resistance, $𝑆=4\Omega$

$\begin{array}{cc}& \text{ At }{𝑡}_1=0^{\circ }C,{𝑙}_1=50\mathrm{cm}\\ & \therefore {𝑅}_1=\frac{{𝑙}_1}{100-𝑙}\times 𝑆=\frac{50}{100-50}\times 4=4\Omega \end{array}$

At ${𝑡}_2={100}^{\circ },{𝑙}_2=60\mathrm{cm}$

$\therefore {𝑅}_2=\frac{{𝑙}_2}{100-{𝑙}_2}\times 𝑆=\frac{60}{100-60}\times 4=6\Omega$

$\therefore$ Temperature coefficient of the coil,

$\begin{array}{cc}𝛼& =\frac{{𝑅}_2-{𝑅}_1}{{𝑅}_1{𝑡}_2-{𝑅}_2{𝑡}_1}=\frac{6-4}{4\times 100-6\times 0}=\frac{1}{200}\\ & ={0.005}^{\circ }{C}^{-1}\end{array}$

A moving electron produces electric field and magnetic field both, while static charge produces only electric field.

Case - I

$\begin{array}{cc}& \text{ Number of turns, }{𝑁}_1=9\\ & \therefore \\ & \therefore {𝐵}_1=\frac{{𝜇}_0{𝑁}_1𝐼}{2𝑅}=\frac{9{𝜇}_0𝐼}{2𝑅}\end{array}$

Case-II ${𝑁}_2=3$

If radius of turns be ${𝑅}^{\prime }$, then

$9\times 2𝜋𝑅=3\times 2𝜋{𝑅}^{\prime }$

$\begin{array}{cc}\Rightarrow & {𝑅}^{\prime }=3𝑅\\ \therefore & {𝐵}_2=\frac{{𝜇}_0{𝑁}_2𝐼}{2{𝑅}^{\prime }}=\frac{{𝜇}_0\times 3\times 𝐼}{2\times 3𝑅}=\frac{{𝜇}_0𝐼}{2𝑅}\end{array}$

$\therefore \frac{{𝐵}_2}{{𝐵}_1}=\frac{\frac{{𝜇}_0𝐼}{2𝑅}}{\frac{9{𝜇}_0𝐸}{2𝑅}}=\frac{1}{9}\Rightarrow {𝐵}_2=\frac{{𝐵}_1}{9}$

We know that time period, $𝑇=\frac{2𝜋𝑚}{𝑞𝐵}=\frac{2𝜋}{\frac{𝑞}{𝑚}𝐵}1s$

Since, the particle will be at point $𝑃$ after time,

$𝑡=\frac{1}{12}s=\frac{𝑇}{12}s$

It will be deviated by an angle, $𝜃=2𝜋/12={30}^{\circ }$

$\begin{array}{cc}& \text{ Hence, velocity at point }𝑃\text{, }\\ & \begin{array}{cc}𝐯& =10(\cos {30}^{\circ }\widehat{𝐢}+\sin {30}^{\circ }\widehat{𝐣})\\ & =10(\frac{\sqrt{\surd 3}}{2}\widehat{𝐢}+\frac{1}{2}\widehat{𝐣})=5(\sqrt{\surd 3\widehat{𝑖}}+\widehat{𝐣})\end{array}\end{array}$

The magnetic field at the center of a circular coil with radius $𝑅$ carrying a current $𝐼$ is given by:

${𝐵}_{\text{center}}=\frac{{𝜇}_0𝐼}{2𝑅}$

where ${𝜇}_0$ is the permeability of free space.

The magnetic field at a distance $𝑥$ on the axis from the center of the coil is given by:

${𝐵}_{𝑥}=\frac{{𝜇}_0𝐼{𝑅}^2}{2({𝑅}^2+{𝑥}^2{)}^{3/2}}$

According to the problem, the field at the center is 64 times the field at distance $𝑥$:

${𝐵}_{\text{center}}=64{𝐵}_{𝑥}$

Substituting the expressions for ${𝐵}_{\text{center}}$ and ${𝐵}_{𝑥}$, we get:

$\frac{{𝜇}_0𝐼}{2𝑅}=64\cdot \frac{{𝜇}_0𝐼{𝑅}^2}{2({𝑅}^2+{𝑥}^2{)}^{3/2}}$

Simplifying, and cancelling out common terms, we have:

$\frac{1}{𝑅}=64\cdot \frac{{𝑅}^2}{({𝑅}^2+{𝑥}^2{)}^{3/2}}$

Rearranging gives:

$({𝑅}^2+{𝑥}^2{)}^{3/2}=64{𝑅}^3$

Taking the cube roots on both sides:

${𝑅}^2+{𝑥}^2=(64{𝑅}^3{)}^{2/3}$

Calculating the right-hand side:

${𝑅}^2+{𝑥}^2={64}^{2/3}{𝑅}^2$

Since $64=4^3$, we can simplify ${64}^{2/3}$ as:

${64}^{2/3}=4^2=16$

So the equation becomes:

${𝑅}^2+{𝑥}^2=16{𝑅}^2$

Therefore:

${𝑥}^2=15{𝑅}^2$

Thus, the value of $𝑥$ is:

$𝑥=𝑅\sqrt{\surd 15}$

The correct answer is Option D: $𝑅\sqrt{\surd 15}$.

Magnetic hysteresis is exhibited by ferromagnetic materials.

In ferromagnetism, materials like iron and cobalt show hysteresis when subjected to an external magnetic field. The process involves the alignment of magnetic domains in response to the applied field and retains a certain level of magnetization even when the field is removed. This characteristic behavior, which results from the energy loss in the system, manifests as a lag between changes in magnetization and the external magnetic field, forming a hysteresis loop.

Therefore, the correct choice is:

Option C: Only ferro

$\begin{array}{cc}& \text{ We know that magnetic susceptibility. }\\ & \begin{array}{cc}𝜒& \propto \frac{1}{𝑇}\\ \Rightarrow \frac{{𝜒}_2}{{𝜒}_1}& =\frac{{𝑇}_1}{{𝑇}_2}\Rightarrow {𝜒}_2=\frac{{𝑇}_1}{{𝑇}_2}\times {𝜒}_1\\ & =1.2\times {10}^{-5}\times \frac{300}{200}=1.8\times {10}^{-5}\end{array}\end{array}$

$\begin{array}{cc}& \text{ Given, }\\ & \begin{array}{cc}𝐵& =2\mathrm{mT}\\ & =2\times {10}^{-3}T,𝑅=\frac{10}{\sqrt{\surd 𝜋}}\mathrm{cm}\end{array}\end{array}$

$\begin{array}{cc}& \text{ Magnetic flux passing through surface }{𝑆}_1\text{, }\\ & \begin{array}{cc}{𝜙}_{{𝑆}_1}& =𝐵{𝑆}_1\cos {180}^{\circ }\\ & =2\times {10}^{-3}\times 𝜋{𝑅}^2(-1)\\ & =-2\times {10}^{-3}\times 𝜋\times \frac{100}{𝜋}\times {10}^{-4}\\ & =-20\times {10}^{-6}\mathrm{Wb}=-20𝜇\mathrm{Wb}\end{array}\end{array}$

Since, Total entering magnetic flux = Total leaving magnetic flux

$\therefore {𝜙}_{{𝑆}_2}=-{𝜙}_{{𝑆}_1}=20𝜇\mathrm{Wb}$

Magnetic field around the current carrying straight long conductor.

$𝐵=\frac{{𝜇}_0𝐼}{2𝜋𝑟}$

where, $𝑟$ is distance from the conductor.

Hence, magnetic field increases on moving towards the conductor.

Since, metal ring is moving towards the conducting wire, hence flux linked with ring, increases continuously in the out of plane direction. Hence, the induced current produces a magnetic field into the plane of paper. Thus, induced current will produce in the clockwise direction.

$\begin{array}{c}\text{ Given, }𝐼=2A,𝑅=10\Omega \\ 𝑡=1\mathrm{ms}={10}^{-3}s\end{array}$

$\begin{array}{cc}& \text{ We know that, }\\ & \begin{array}{cc}\left|𝜀\right|& =\frac{\Delta 𝜙}{\Delta 𝑡}\Rightarrow 𝐼𝑅=\frac{\Delta 𝜙}{\Delta 𝑡}\\ \Rightarrow \Delta 𝜙& =𝐼𝑅\Delta 𝑡=2\times 10\times {10}^{-3}\\ & =0.02\mathrm{Wb}=2\times {10}^{-2}\mathrm{Wb}\end{array}\end{array}$

As the mutual inductance is independent with time, hence graph shown in option (c) is correct.

For better tunning of an $𝐿-𝐶-𝑅$ circuit used in for communication, quality factor $𝑄$ must be high.

$\therefore 𝑄=\frac{1}{𝑅}\sqrt{\surd 𝑓𝑟𝑎𝑐𝐿𝐶}$

In option (d), $𝑅$ and $𝐶$ is minimum and $𝐿$ is maximum. Thus, the value of $𝑅,𝐿$ and $𝐶$ given in option (d) suitable for better communication.

Resonance frequency in $𝐿-𝐶-𝑅$ series circuit,

${𝑓}_0=\frac{1}{2𝜋\sqrt{\surd 𝐿𝐶}}\Rightarrow {𝑓}_0\propto \frac{1}{\sqrt{\surd 𝐶}}$

i.e. ${𝑓}_0$ decreases with $𝐶$, hence graph shown in option (c) is correct.

For a capacitive circuit, ${𝑋}_{𝐶}>{𝑋}_{𝐿}$

For the given diagram in question, the value ${𝑋}_{𝐶}$ corresponding to point $𝐴$ is greater than the value of ${𝑋}_{𝐿}$.

The force exerted by electromagnetic waves on a perfectly reflecting surface can be derived using the concept of radiation pressure. When electromagnetic waves are incident on a reflecting surface, they exert a pressure due to the change in momentum as they reflect.

For a perfectly reflecting surface, the radiation pressure $𝑃$ is given by:

$𝑃=\frac{2𝐼}{𝑐}$

where:

$𝐼$ is the intensity of the incident radiation,

$𝑐$ is the speed of light.

The force $𝐹$ exerted on a surface due to radiation pressure is given by multiplying the radiation pressure by the surface area $𝐴$:

$𝐹=𝑃\times 𝐴=\frac{2𝐼}{𝑐}\times 𝐴=\frac{2𝐼𝐴}{𝑐}$

Therefore, the correct expression for the force exerted by the electromagnetic wave on the reflecting surface is:

Option A

$\frac{2𝐼𝐴}{𝑐}$

The final image formed by an astronomical telescope is:

Option D

virtual, inverted and magnified

In an astronomical telescope, the objective lens (or mirror) creates a real, inverted image of a distant object at its focal plane. The eyepiece, which functions as a magnifying glass, then takes this real image and forms a virtual image that is also inverted but magnified for the viewer's eye. This is why when you look through an astronomical telescope, the image appears larger and upside down.

Given that the angle of minimum deviation ${𝛿}_{𝑚}$ is equal to the angle of the prism $𝐴$ for an equilateral prism, we have:

$𝐴={60}^{\circ },{𝛿}_{𝑚}={60}^{\circ }$

The refractive index $𝜇$ of the prism can be determined using the formula:

$𝜇=\frac{\sin \left(\frac{𝐴+{𝛿}_{𝑚}}{2}\right)}{\sin \frac{𝐴}{2}}$

Substitute the values:

$𝜇=\frac{\sin \left(\frac{{60}^{\circ }+{60}^{\circ }}{2}\right)}{\sin \frac{{60}^{\circ }}{2}}=\frac{\sin {60}^{\circ }}{\sin {30}^{\circ }}=\frac{\sqrt{\surd 3}/2}{1/2}=\sqrt{\surd 3}$

Therefore, if $𝑣$ is the speed of light inside the prism, the relationship with $𝜇$ is:

$𝜇=\frac{𝑐}{𝑣}$

Rearrange to find $𝑣$:

$𝑣=\frac{𝑐}{𝜇}=\frac{3\times {10}^8\text{m/s}}{\sqrt{\surd 3}}=\sqrt{\surd 3}\times {10}^8\text{m/s}$

$\begin{array}{cc}& \text{ From the given diagram in question, }\\ & \begin{array}{c}{𝑛}_1=\frac{4}{3}{𝑛}_2=\frac{3}{2}\\ 𝑢=-2𝑅,\end{array}\end{array}$

$\therefore$ Using refraction formula through convex spherical surface.

$\frac{{𝑛}_2}{𝑣}-\frac{{𝑛}_1}{𝜇}=\frac{{𝑛}_2-{𝑛}_1}{𝑅}$

$\Rightarrow \frac{\frac{3}{2}}{𝑣}-\frac{4}{3(-2𝑅)}=\frac{\frac{3}{2}-\frac{4}{3}}{𝑅}\Rightarrow 𝑣=-3𝑅$

$\begin{array}{cc}& =-3\times 10[\because 𝑅=10\mathrm{cm}]\\ & =-30\mathrm{cm}\end{array}$

Since, $𝑣$ is negative, thus image is obtained al a distance from $𝑃$ equal to 30 cm in the rarer medium.

$\begin{array}{cc}& \text{ From the diagram, }\\ & \begin{array}{cc}& \frac{1}{{𝑓}_1}=({𝜇}_1-1)(\frac{1}{{𝑅}_1}-\frac{1}{{𝑅}_2})=({𝜇}_1-1)(\frac{1}{𝑅}-\frac{1}{\infty })\\ & \frac{1}{{𝑓}_1}=\frac{{𝜇}_1-1}{𝑅}\text{ and }\frac{1}{{𝑓}_2}=({𝜇}_2-1)(\frac{1}{\infty }-\frac{1}{-𝑅})\\ & \frac{1}{{𝑓}_2}=\frac{{𝜇}_2-1}{𝑅}\end{array}\end{array}$

$\begin{array}{cc}& \therefore \text{ Combined focal length, }\\ & \begin{array}{cc}& \begin{array}{cc}\frac{1}{𝑓}& =\frac{1}{{𝑓}_1}+\frac{1}{{𝑓}_2}=\frac{{𝜇}_1-1}{𝑅}+\frac{{𝜇}_2-1}{𝑅}\\ & =\frac{{𝜇}_1+{𝜇}_2-2}{𝑅}=\frac{1.5+1.2-2}{14}\\ \Rightarrow \frac{1}{𝑓}& =\frac{0.7}{14}\\ \Rightarrow 𝑓& =20\mathrm{cm}\end{array}\end{array}\end{array}$

Again, $𝑣=-40\mathrm{cm},𝑣=$ ?

$\therefore$ Using lens formula,

$\begin{array}{cc}& \frac{1}{𝑓}=\frac{1}{𝑣}-\frac{1}{𝑢}\\ & \therefore \frac{1}{20}=\frac{1}{𝑣}-\frac{1}{(-40)}\Rightarrow \frac{1}{𝑣}=\frac{1}{20}-\frac{1}{40}=\frac{1}{40}\\ & \Rightarrow 𝑣=40\mathrm{cm}\end{array}$

The Doppler effect describes the change in frequency or wavelength of a wave in relation to an observer moving relative to the source of the waves. For light, this effect is observed in the form of redshift or blueshift, depending on whether the object is moving away from or towards the observer, respectively.

The formula to calculate the redshift $𝑧$ can be written as:

$𝑧=\frac{{𝜆}_{\text{observed}}-{𝜆}_{\text{original}}}{{𝜆}_{\text{original}}}$

where:

${𝜆}_{\text{observed}}=601\text{ nm}$ (the wavelength observed),

${𝜆}_{\text{original}}=600\text{ nm}$ (the original wavelength).

Substitute the given values:

$𝑧=\frac{601\text{nm}-600\text{nm}}{600\text{nm}}=\frac{1}{600}$

The redshift is also related to the velocity $𝑣$ of the galaxy and the speed of light $𝑐$ by the formula:

$𝑧=\frac{𝑣}{𝑐}$

Using this relationship, we can solve for $𝑣$, the speed of the galaxy:

$𝑣=𝑧\cdot 𝑐=\left(\frac{1}{600}\right)\cdot (3\times {10}^5\text{km/s})$

Calculating this gives:

$𝑣=\frac{3\times {10}^5\text{km/s}}{600}=500\text{km/s}$

Thus, the speed of the galaxy with respect to the Earth is $500\text{km/s}$.

Correct Option: A $500\mathrm{km}{s}^{-1}$

For unpolarised light falling on the first polaroid, the transmitted intensity becomes half.

So after sheet $1$,

${𝐼}_1=\frac{{𝐼}_0}{2}$

Now the pass axis of sheet $2$ makes an angle ${30}^{\circ }$ with sheet $1$.

Using Malus’ law,

${𝐼}_2={𝐼}_1{\cos }^2{30}^{\circ }$

Since

$\cos {30}^{\circ }=\frac{\sqrt{\surd 3}}{2}\Rightarrow {\cos }^2{30}^{\circ }=\frac{3}{4}$

we get

${𝐼}_2=\frac{{𝐼}_0}{2}\cdot \frac{3}{4}=\frac{3{𝐼}_0}{8}$

Now sheet $3$ makes ${90}^{\circ }$ with sheet $1$, while sheet $2$ makes ${30}^{\circ }$ with sheet $1$.

So the angle between sheets $2$ and $3$ is

${90}^{\circ }-{30}^{\circ }={60}^{\circ }$

Again using Malus’ law,

${𝐼}_3={𝐼}_2{\cos }^2{60}^{\circ }$

and

$\cos {60}^{\circ }=\frac{1}{2}\Rightarrow {\cos }^2{60}^{\circ }=\frac{1}{4}$

Hence

${𝐼}_3=\frac{3{𝐼}_0}{8}\cdot \frac{1}{4}=\frac{3{𝐼}_0}{32}$

So the intensity of emergent light through sheet $3$ is

$\frac{3{𝐼}_0}{32}$

Option B is correct.

In Young's double slit experiment, the fringe width $𝛽$ is determined by the formula:

$𝛽=\frac{𝜆𝐷}{𝑑}$

where $𝜆$ is the wavelength of the particles involved, $𝐷$ is the distance from the slits to the screen, and $𝑑$ is the separation between the two slits.

For particles like electrons and protons, the de Broglie wavelength $𝜆$ is given by:

$𝜆=\frac{ℎ}{𝑚𝑣}$

where $ℎ$ is Planck's constant, $𝑚$ is the mass of the particle, and $𝑣$ is the velocity (or speed) of the particle.

Both electrons and protons are used at the same speed $𝑣$ in this scenario, and the experimental setup (slit separation $𝑑$ and distance to screen $𝐷$) remains unchanged. Therefore, the ratio of their fringe widths, ${𝛽}_1$ (for electrons) and ${𝛽}_2$ (for protons), is determined by their respective de Broglie wavelengths:

Calculate the wavelength of the electrons:

${𝜆}_{\text{electron}}=\frac{ℎ}{{𝑚}_{\text{electron}}\cdot 𝑣}$

Calculate the wavelength of the protons:

${𝜆}_{\text{proton}}=\frac{ℎ}{{𝑚}_{\text{proton}}\cdot 𝑣}$

Since ${𝑚}_{\text{proton}}>{𝑚}_{\text{electron}}$, it follows that ${𝜆}_{\text{proton}}<{𝜆}_{\text{electron}}$ when both have the same speed. The narrower wavelength of protons results in narrower fringes. Therefore:

${𝛽}_1=\frac{{𝜆}_{\text{electron}}𝐷}{𝑑}>\frac{{𝜆}_{\text{proton}}𝐷}{𝑑}={𝛽}_2$

Thus, the correct relation between ${𝛽}_1$ and ${𝛽}_2$ is:

Option D: ${𝛽}_1>{𝛽}_2$

Given, work function, ${𝜙}_0=𝐸/3$

According to Einstein's photoelectric equation,

$\begin{array}{cc}{𝐾}_{max}& =𝐸-{𝜙}_0=𝐸-\frac{𝐸}{3}\\ \Rightarrow {𝐾}_{max}& =\frac{2𝐸}{3}\end{array}$

$\therefore$ Kinetic energy lies between 0 to $2𝐸/3$