COMEDK 2024 Afternoon Shift Question Paper with Solutions

To solve this problem, we can use the principles of an ideal transformer along with Ohm’s law. Here’s the step-by-step process:

Determine the Voltage Ratio:

An ideal transformer follows the relationship:

$\frac{{𝑉}_{𝑠}}{{𝑉}_{𝑝}}=\frac{{𝑁}_{𝑠}}{{𝑁}_{𝑝}}$

Given:

Number of primary turns, ${𝑁}_{𝑝}=400$

Number of secondary turns, ${𝑁}_{𝑠}=800$

Primary voltage, ${𝑉}_{𝑝}=20\text{V}$

Compute the secondary voltage:

${𝑉}_{𝑠}={𝑉}_{𝑝}\times \frac{{𝑁}_{𝑠}}{{𝑁}_{𝑝}}=20\text{V}\times \frac{800}{400}=20\text{V}\times 2=40\text{V}$

Calculate the Secondary Current:

With the load resistance on the secondary being 4 ohms, use Ohm's law:

${𝐼}_{𝑠}=\frac{{𝑉}_{𝑠}}{𝑅}=\frac{40\text{V}}{4\Omega }=10\text{A}$

Find the Primary Current Using Power Conservation:

For an ideal transformer, power is conserved (ignoring losses):

${𝑉}_{𝑝}\times {𝐼}_{𝑝}={𝑉}_{𝑠}\times {𝐼}_{𝑠}$

Solve for the primary current (${𝐼}_{𝑝}$):

${𝐼}_{𝑝}=\frac{{𝑉}_{𝑠}\times {𝐼}_{𝑠}}{{𝑉}_{𝑝}}=\frac{40\text{V}\times 10\text{A}}{20\text{V}}=\frac{400}{20}=20\text{A}$

Thus, the primary current is 20 A, which corresponds to Option B.

Let's analyze the problem step by step.

• In the first situation, the object is at a distance

${𝑢}_1=12\text{ cm}.$

Since a virtual image is formed by a convex lens, the image distance is negative (using the sign convention where the object distance is positive and real image distances are positive). Let the image distance be

${𝑣}_1=-\left|{𝑣}_1\right|.$

The lens formula is given by

$\frac{1}{𝑓}=\frac{1}{𝑣}+\frac{1}{𝑢}.$

Thus, for the first case, we have

$\frac{1}{𝑓}=\frac{1}{{𝑣}_1}+\frac{1}{12}=-\frac{1}{\left|{𝑣}_1\right|}+\frac{1}{12}.$

• In the second situation, the object is moved 4 cm further from the lens. Thus,

${𝑢}_2=12+4=16\text{ cm},$

and a real image is formed (so its image distance is positive), say

${𝑣}_2.$

The lens formula now gives

$\frac{1}{𝑓}=\frac{1}{{𝑣}_2}+\frac{1}{16}.$

• We are told that the images have the same size. The magnification is given by

$𝑚=-\frac{𝑣}{𝑢}.$

For the magnitudes of the images to be the same, we require that

$\mid \frac{-{𝑣}_1}{12}\mid =\mid \frac{-{𝑣}_2}{16}\mid .$

Since for the first case we have

$\left|{𝑣}_1\right|=-{𝑣}_1,$

this condition becomes

$\frac{\left|{𝑣}_1\right|}{12}=\frac{{𝑣}_2}{16}.$

Hence,

${𝑣}_2=\frac{16}{12}\left|{𝑣}_1\right|=\frac{4}{3}\left|{𝑣}_1\right|.$

• Now, equate the two expressions for $\frac{1}{𝑓}$ from the lens formulas:

From the first case:

$\frac{1}{𝑓}=\frac{1}{12}-\frac{1}{\left|{𝑣}_1\right|},$

and from the second case:

$\frac{1}{𝑓}=\frac{1}{{𝑣}_2}+\frac{1}{16}.$

Substitute ${𝑣}_2=\frac{4}{3}\left|{𝑣}_1\right|$ into the second equation:

$\frac{1}{𝑓}=\frac{1}{\frac{4}{3}\left|{𝑣}_1\right|}+\frac{1}{16}=\frac{3}{4\left|{𝑣}_1\right|}+\frac{1}{16}.$

Thus, we set:

$\frac{1}{12}-\frac{1}{\left|{𝑣}_1\right|}=\frac{3}{4\left|{𝑣}_1\right|}+\frac{1}{16}.$

• Rearranging the equation, subtract $\frac{1}{16}$ from both sides:

$\frac{1}{12}-\frac{1}{16}-\frac{1}{\left|{𝑣}_1\right|}=\frac{3}{4\left|{𝑣}_1\right|}.$

Compute $\frac{1}{12}-\frac{1}{16}$ by finding a common denominator:

$\frac{1}{12}-\frac{1}{16}=\frac{4}{48}-\frac{3}{48}=\frac{1}{48}.$

So,

$\frac{1}{48}-\frac{1}{\left|{𝑣}_1\right|}=\frac{3}{4\left|{𝑣}_1\right|}.$

• Add $\frac{1}{\left|{𝑣}_1\right|}$ to both sides:

$\frac{1}{48}=\frac{1}{\left|{𝑣}_1\right|}+\frac{3}{4\left|{𝑣}_1\right|}=\frac{1+\frac{3}{4}}{\left|{𝑣}_1\right|}=\frac{\frac{7}{4}}{\left|{𝑣}_1\right|}.$

Thus,

$\frac{1}{48}=\frac{7}{4\left|{𝑣}_1\right|}.$

Solving for $\left|{𝑣}_1\right|$:

$\left|{𝑣}_1\right|=\frac{7\times 48}{4}=7\times 12=84\text{ cm}.$

So, the virtual image distance is

${𝑣}_1=-84\text{ cm}.$

• Now, substitute the value of $\left|{𝑣}_1\right|$ back into the first lens equation to find the focal length:

$\frac{1}{𝑓}=\frac{1}{12}-\frac{1}{84}.$

Find a common denominator:

$\frac{1}{12}-\frac{1}{84}=\frac{7}{84}-\frac{1}{84}=\frac{6}{84}=\frac{1}{14}.$

Therefore,

$𝑓=14\text{ cm}.$

Thus, the focal length of the lens is $\overline{)14\text{ cm}}.$

This corresponds to Option B.

The speed of the object is given by the slope of the distance-time curve:

$𝑣=\frac{𝑑𝑠}{𝑑𝑡}$

Since $𝑣\ge 0$ throughout, the distance $𝑠$ increases monotonically with time.

At both the beginning and the end of the motion, $𝑣=0$, so the slope of the $𝑠\text{--}𝑡$ curve must be zero at these points.

While the speed increases to its maximum value, $\frac{{𝑑}^2𝑠}{𝑑{𝑡}^2}>0$, making the curve concave upward. As the speed decreases back to zero, $\frac{{𝑑}^2𝑠}{𝑑{𝑡}^2}<0$, making the curve concave downward.

Hence, the correct option is (D).

Given, $𝑚=1\text{kg}$, $𝐼=1{\text{kg m}}^2$, $ℎ=4\text{m}$, and $𝑡=2\text{s}$. Taking $𝑔=10{\text{m s}}^{-2}$.

Using the equation of motion for the falling block,

$ℎ=𝑢𝑡+\frac{1}{2}𝑎{𝑡}^2$

$4=0+\frac{1}{2}𝑎(2{)}^2⟹𝑎=2{\text{m s}}^{-2}$

For the vertical motion of the mass,

$𝑚𝑔-𝑇=𝑚𝑎⟹𝑇=𝑚(𝑔-𝑎)$

$𝑇=1\times (10-2)=8\text{N}$

For the rotation of the wheel,

$𝜏=𝑇𝑅=𝐼𝛼=𝐼\left(\frac{𝑎}{𝑅}\right)$

$𝑇{𝑅}^2=𝐼𝑎$

$8{𝑅}^2=1\times 2⟹{𝑅}^2=\frac{2}{8}=0.25{\text{m}}^2$

$𝑅=0.5\text{m}$

Hence, the correct option is (D).

Given, for Earth of mass $𝑀$ and radius $𝑅$, the gravitational potential at the surface is $𝑉=-\frac{𝐺𝑀}{𝑅}$ and the magnitude of the gravitational field strength is $𝑔=\frac{𝐺𝑀}{{𝑅}^2}$.

For the planet, mass ${𝑀}^{\prime }=2𝑀$ and radius ${𝑅}^{\prime }=2𝑅$.

The gravitational potential at the surface of the planet is

${𝑉}^{\prime }=-\frac{𝐺{𝑀}^{\prime }}{{𝑅}^{\prime }}=-\frac{𝐺\left(2𝑀\right)}{2𝑅}=-\frac{𝐺𝑀}{𝑅}=𝑉$

The magnitude of the gravitational field strength at the surface of the planet is

${𝑔}^{\prime }=\frac{𝐺{𝑀}^{\prime }}{({𝑅}^{\prime }{)}^2}=\frac{𝐺\left(2𝑀\right)}{(2𝑅{)}^2}=\frac{2𝐺𝑀}{4{𝑅}^2}=\frac{𝑔}{2}$

This corresponds to row C.

Hence, the correct option is (A).

We have two tuning forks:

A known fork of frequency ${𝑛}_2=310\text{Hz}.$

An unknown fork of frequency ${𝑛}_1$.

When they are sounded together (initially), they produce $4$ beats per second.

Therefore,

$|{𝑛}_1-310|=4⟹{𝑛}_1=310\pm 4.$

So ${𝑛}_1$ could be either $314$ Hz or $306$ Hz initially.

The fork of unknown frequency is then filed, and it is noted that the beat frequency with the $310$-Hz fork remains $4$.

Important physical fact about “filing a tuning fork”:

Typically, if you file (remove mass from) the tips of the fork’s prongs, you make the prongs effectively shorter, and this raises the frequency of the fork.

If one wanted to lower the fork’s frequency by filing, that filing would have to be done near the “root” (base) of the prongs, which is less common in many standard problems (unless explicitly stated).

In most textbook treatments, “filing the fork” is taken to mean removing a small amount of mass at or near the free ends, thereby increasing its frequency.

Figuring Out Which Value ${𝑛}_1$ Was Initially

We want to see which initial frequency ($306$ or $314$) is consistent with the beat frequency remaining $4$ after the fork is filed.

If the fork’s frequency increases on filing:

Case A: If ${𝑛}_1$ was $314$ Hz initially, any further increase would lead to, say, $315$, $316$, … etc. Then the difference $|{𝑛}_1-310|$ would become $5,6,\dots$, which would not stay at $4$.

Case B: If ${𝑛}_1$ was $306$ Hz initially, increasing it could take it up to $307,308,\dots ,314,\dots$. If finally it lands at $314$, then the difference with $310$ is again $4$. Hence the beat frequency would still be $4$.

Therefore, the scenario consistent with “filing raises the frequency and the beats remain $4$” is that the fork started at $306$ Hz. After filing, its frequency rose (possibly through some intermediate values) until it reached $314$ Hz—still exactly $4$ Hz away from $310$ Hz.

Final Answer

$\overline{)\text{The initial frequency of the unknown fork is }306\text{Hz}.}$

We start by noting that the induced emf (by Faraday’s law) is given by

$𝜀=-\frac{\Delta \Phi }{\Delta 𝑡},$

where the flux change

$\Delta \Phi ={\Phi }_{\text{final}}-{\Phi }_{\text{initial}}.$

In our problem the magnetic field is perpendicular to the loop (so we don’t have to worry about any angle factors) and its value doubles when the loop is changed in 0.1 sec. Let’s set up the steps:

1. Define l as the length of the wire (in cm) and a as the side of the square (in cm) into which the wire is reformed. (In many problems the wire length l is used; here the given value “1 cm” is simply a particular case, so we write our answer in terms of l and a.)

2. When the wire is bent into a circle, its circumference is l. That means the radius r of the circle is

$𝑟=\frac{𝑙}{2𝜋}.$

Therefore, the area of the circular loop is

${𝐴}_{\text{circle}}=𝜋{𝑟}^2=𝜋{\left(\frac{𝑙}{2𝜋}\right)}^2=\frac{{𝑙}^2}{4𝜋}.$

3. When the loop is changed into a square of side a, its area is simply

${𝐴}_{\text{square}}={𝑎}^2.$

(Note: In a typical wire‐reshaping problem the new shape’s perimeter would equal l so that a would be l/4. Here, however, the side a is given explicitly, so we use it as an independent parameter.)

4. The initial magnetic flux is that through the circular loop in a field B:

${\Phi }_{\text{initial}}=𝐵{𝐴}_{\text{circle}}=𝐵\frac{{𝑙}^2}{4𝜋}.$

After 0.1 sec the loop is made into a square and, at the same time, the magnetic flux density is doubled to 2B. Therefore the final flux is

${\Phi }_{\text{final}}=\left(2𝐵\right){𝐴}_{\text{square}}=2𝐵{𝑎}^2.$

5. Thus the change in flux is

$\Delta \Phi ={\Phi }_{\text{final}}-{\Phi }_{\text{initial}}=2𝐵{𝑎}^2-𝐵\frac{{𝑙}^2}{4𝜋}.$

6. Plug this into Faraday’s law with the time interval $\Delta 𝑡=0.1\text{sec}$:

$𝜀=-\frac{\Delta \Phi }{\Delta 𝑡}=-\frac{2𝐵{𝑎}^2-\frac{𝐵{𝑙}^2}{4𝜋}}{0.1}.$

Multiplying numerator and denominator by 10 (since $1/0.1=10$) gives

$𝜀=-10𝐵(2{𝑎}^2-\frac{{𝑙}^2}{4𝜋}).$

7. To combine the terms inside the parentheses, rewrite $2{𝑎}^2$ with a common denominator:

$2{𝑎}^2=\frac{8𝜋{𝑎}^2}{4𝜋}.$

Thus we have

$𝜀=-10𝐵\left(\frac{8𝜋{𝑎}^2-{𝑙}^2}{4𝜋}\right)=-\frac{10𝐵}{4𝜋}(8𝜋{𝑎}^2-{𝑙}^2).$

Simplify the factor $\frac{10}{4}=\frac{5}{2}$, so

$𝜀=-\frac{5𝐵}{2𝜋}(8𝜋{𝑎}^2-{𝑙}^2).$

This is exactly the expression given in option C:

$\text{Option C:}𝜀=\frac{-5𝐵(8𝜋{𝑎}^2-{𝑙}^2)}{2𝜋}.$

Thus, the induced emf is

$𝜀=-\frac{5𝐵(8𝜋{𝑎}^2-{𝑙}^2)}{2𝜋},$

and the correct answer is Option C.

Let's solve the problem step by step.

The iron piece is heated in a furnace to a temperature $𝑇$ (in °C) and then transferred to a calorimeter. After some time, the system equilibrates at ${60}^{\circ }𝐶$.

When the iron (of mass $200𝑔=0.200𝑘𝑔$) is placed in the calorimeter, it cools from $𝑇$ to ${60}^{\circ }𝐶$. The energy lost by the iron is given by:

${𝑄}_{\text{iron}}={𝑚}_{\text{iron}}{𝑐}_{\text{iron}}(𝑇-60)$

where

${𝑚}_{\text{iron}}=0.200𝑘𝑔$,

${𝑐}_{\text{iron}}=470𝐽𝑘{𝑔}^{-1}{𝐶}^{-1}$.

The calorimeter contains $230𝑔$ of water and has a water equivalent of $20𝑔$. The total effective mass of water is:

$0.230𝑘𝑔+0.020𝑘𝑔=0.250𝑘𝑔$.

The water heats up from ${20}^{\circ }𝐶$ to ${60}^{\circ }𝐶$, a temperature rise of ${40}^{\circ }𝐶$. The energy gained by the water and calorimeter is:

${𝑄}_{\text{water}}={𝑚}_{\text{eff}}{𝑐}_{\text{water}}(60-20)$

where we use

${𝑚}_{\text{eff}}=0.250𝑘𝑔$,

${𝑐}_{\text{water}}\approx 4200𝐽𝑘{𝑔}^{-1}{𝐶}^{-1}$.

According to the conservation of energy:

${𝑄}_{\text{iron}}={𝑄}_{\text{water}}$

Substituting the expressions:

$0.200\times 470\times (𝑇-60)=0.250\times 4200\times 40$

First, calculate the constants:

For iron: $0.200\times 470=94𝐽{/}^{\circ }𝐶$,

For water: $0.250\times 4200\times 40=42000𝐽$.

So, the equation becomes:

$94(𝑇-60)=42000$

Solving for $𝑇$:

$𝑇-60=\frac{42000}{94}\approx 446.81$

$𝑇\approx 60+446.81={506.81}^{\circ }𝐶$

Thus, the temperature of the furnace is approximately ${506.8}^{\circ }𝐶$.

This corresponds to Option C.

To find the angular position of the first minimum in single-slit diffraction, we use the condition:

$𝑎\sin 𝜃=𝑚𝜆$

where:

$𝑎$ is the slit width,

$𝜆$ is the wavelength of the light, and

$𝑚$ is the order of the minimum (for the first minimum, $𝑚=1$).

Given:

$𝑎=10\times {10}^{-7}\text{ m}={10}^{-6}\text{ m}$

$𝜆=500\text{ nm}=500\times {10}^{-9}\text{ m}=5\times {10}^{-7}\text{ m}$

Follow these steps:

Substitute the values into the formula for the first minimum ($𝑚=1$):

${10}^{-6}\sin 𝜃=5\times {10}^{-7}$

Solve for $\sin 𝜃$:

$\sin 𝜃=\frac{5\times {10}^{-7}}{{10}^{-6}}=0.5$

Find the corresponding angle $𝜃$:

$𝜃={\sin }^{-1}\left(0.5\right)={30}^{\circ }$

Thus, the angular position of the first minimum is ${30}^{\circ }$. This corresponds to Option B.

From the given $𝑃\text{--}𝑉$ diagram, the process $B\to C$ is an isobaric expansion at a constant pressure $𝑃=15{\text{N m}}^{-2}$.

The work done by the gas during the process $B\to C$ is

${𝑊}_{\mathrm{BC}}=𝑃({𝑉}_C-{𝑉}_B)=15\times (9-3)=90\text{J}$

Given, ${𝑈}_C=20\text{J}$ and ${𝑈}_B=10\text{J}$, the change in internal energy is

$\Delta {𝑈}_{\mathrm{BC}}={𝑈}_C-{𝑈}_B=20-10=10\text{J}$

Using the first law of thermodynamics,

${𝑄}_{\mathrm{BC}}=\Delta {𝑈}_{\mathrm{BC}}+{𝑊}_{\mathrm{BC}}=10+90=100\text{J}$

Hence, the correct option is (C).

The kinetic energy of an ideal gas is related to its temperature. According to the kinetic theory of gases, the average kinetic energy of gas molecules is given by:

$𝐾{𝐸}_{\text{avg}}=\frac{3}{2}𝑘𝑇$

where:

$𝐾{𝐸}_{\text{avg}}$ is the average kinetic energy of the gas molecules,

$𝑘$ is the Boltzmann constant,

$𝑇$ is the absolute temperature.

Since the temperature is kept constant in this scenario, the kinetic energy, which depends solely on the temperature for an ideal gas, will not change.

Therefore, the correct answer is:

Option C: Not change

In Young's double slit experiment:

• Bright fringes occur when the path difference is an integer multiple of the wavelength. Mathematically, the condition is:

$𝑑\sin 𝜃=𝑚𝜆(𝑚=0,1,2,\dots )$

• Dark fringes occur when the path difference is an odd multiple of half the wavelength, i.e.,

$𝑑\sin 𝜃=(𝑚+\frac{1}{2})𝜆(𝑚=0,1,2,\dots )$

Now, let’s determine the phase differences for the fringes in question.

For any fringe, the phase difference is related to the path difference by:

$𝜙=\frac{2𝜋}{𝜆}\times \left(\text{path difference}\right)$

For the third bright fringe (taking $𝑚=3$):

The path difference is:

${\Delta }_{\text{bright}}=3𝜆$

Therefore, the phase difference is:

${𝜙}_{\text{bright}}=\frac{2𝜋}{𝜆}\times 3𝜆=6𝜋$

For the third dark fringe (with $𝑚=2$; since the dark fringes occur at $𝑚+\frac{1}{2}$, the third dark fringe corresponds to $𝑚=2$):

The path difference is:

${\Delta }_{\text{dark}}=(2+\frac{1}{2})𝜆=\frac{5𝜆}{2}$

Thus, the phase difference is:

${𝜙}_{\text{dark}}=\frac{2𝜋}{𝜆}\times \frac{5𝜆}{2}=5𝜋$

Now, the ratio of the phase differences between the third bright fringe and the third dark fringe is:

$\frac{{𝜙}_{\text{bright}}}{{𝜙}_{\text{dark}}}=\frac{6𝜋}{5𝜋}=\frac{6}{5}$

This corresponds to Option C: $\frac{6}{5}.$

When an electromagnetic wave passes from one medium to another, the frequency remains unchanged. Here's why:

In vacuum, the speed of light is $𝑐$ and the wavelength is given by the formula:

${𝜆}_0=\frac{𝑐}{𝑓}$

When the wave enters the dielectric medium, its speed becomes:

$𝑣=\frac{𝑐}{\sqrt{\surd {𝜖}_{𝑟}{𝜇}_{𝑟}}}$

Given that ${𝜇}_{𝑟}=1$ and ${𝜖}_{𝑟}=4.0$, we have:

$𝑣=\frac{𝑐}{\sqrt{\surd 4.0}}=\frac{𝑐}{2}$

The frequency $𝑓$ remains the same as in vacuum, so the new wavelength ${𝜆}^{\prime }$ in the medium is:

${𝜆}^{\prime }=\frac{𝑣}{𝑓}=\frac{𝑐/2}{𝑓}=\frac{1}{2}\left(\frac{𝑐}{𝑓}\right)=\frac{{𝜆}_0}{2}$

Thus, the wavelength is halved, and the frequency remains unchanged.

The correct option is:

Option D

Wavelength is halved and frequency remains unchanged.

The devices consume a real power of

$𝑃=500\text{W}.$

The AC voltage is given by

$𝐸\left(𝑡\right)=200\sin (3.14𝑡+\frac{𝜋}{4}).$

Here, the peak voltage is

${𝑉}_{𝑚}=200\text{V},$

so its rms value is

${𝑉}_{𝑟𝑚𝑠}=\frac{{𝑉}_{𝑚}}{\sqrt{\surd 2}}=\frac{200}{\sqrt{\surd 2}}\text{V}.$

In AC circuits the real power is given by the formula

$𝑃=\frac{1}{2}{𝑉}_{𝑚}{𝐼}_{𝑚}\cos 𝜙,$

where

${𝐼}_{𝑚}$ is the peak current, and

$𝜙$ is the phase difference between the voltage and the current.

In this problem the voltage is written with a phase angle of $𝜋/4$. This phase angle is usually associated with the load’s impedance and hence the phase difference between voltage and current. Assuming that the current has zero phase (i.e. taken as the reference), then

$𝜙=\frac{𝜋}{4}.$

And note that

$\cos \left(\frac{𝜋}{4}\right)=\frac{1}{\sqrt{\surd 2}}.$

Substitute the given values into the power equation:

$500=\frac{1}{2}\times 200\times {𝐼}_{𝑚}\times \frac{1}{\sqrt{\surd 2}}.$

Simplify this step-by-step:

First, compute the constant:

$\frac{1}{2}\times 200=100.$

Then the equation becomes:

$500=100\times {𝐼}_{𝑚}\times \frac{1}{\sqrt{\surd 2}}.$

Solve for ${𝐼}_{𝑚}$:

${𝐼}_{𝑚}=\frac{500\sqrt{\surd 2}}{100}=5\sqrt{\surd 2}\text{A}.$

Now, convert the peak current to rms current using

${𝐼}_{𝑟𝑚𝑠}=\frac{{𝐼}_{𝑚}}{\sqrt{\surd 2}}.$

Substitute ${𝐼}_{𝑚}=5\sqrt{\surd 2}\text{A}:$

${𝐼}_{𝑟𝑚𝑠}=\frac{5\sqrt{\surd 2}}{\sqrt{\surd 2}}=5\text{A}.$

Thus, the rms value of the circuit current is 5 A, which corresponds to Option D.

In summary:

Given the voltage, the power is calculated using the peak values and the cosine of the phase difference.

After finding the peak current, it is converted to its rms value.

The final answer is $\overline{)5\text{A}}.$

The electric field ($𝐸$) across the depletion layer of a p-n junction can be determined using the relationship between the electric field, potential difference ($𝑉$), and the thickness of the depletion layer ($𝑑$). The formula is:

$𝐸=\frac{𝑉}{𝑑}$

Where:

$𝑉=0.05\text{V}$ (potential difference across the depletion layer)

$𝑑=1𝜇\text{m}=1\times {10}^{-6}\text{m}$ (thickness of the depletion layer)

Substitute the given values into the formula:

$𝐸=\frac{0.05}{1\times {10}^{-6}}=5\times {10}^4\text{N/C}$

Therefore, the electric field across the depletion layer is $5\times {10}^4\text{N/C}$.

The correct answer is Option D, which is:

$5\times {10}^4$

For wire $Q$ to remain suspended directly above wire $P$, the magnetic force exerted by wire $P$ on wire $Q$ must be repulsive and directed vertically upwards to balance the downward gravitational force.

Since parallel currents in the same direction attract and currents in opposite directions repel, the current in wire $Q$ must flow in the direction opposite to that in $P$, which is from right to left.

Equating the upward magnetic force per unit length to the downward gravitational force per unit length:

$\frac{{𝜇}_0{𝐼}_P𝐼}{2𝜋𝑑}=𝑚𝑔$

Substituting ${𝐼}_P=50\text{ A}$:

$\frac{{𝜇}_0\left(50\right)𝐼}{2𝜋𝑑}=𝑚𝑔⟹𝑑=\frac{25𝐼{𝜇}_0}{𝜋𝑚𝑔}$

Hence, the correct option is (C).

Let's break down the solution step by step:

Convert the initial angular velocity from revolutions per minute (rpm) to radians per second (rad/s):

Since $1\text{ revolution}=2𝜋\text{ rad}$ and $1\text{ minute}=60\text{ s}$,

${𝜔}_0=45\times \frac{2𝜋}{60}=\frac{3𝜋}{2}\text{ rad/s}$

The record player decelerates at a constant rate. Since it "slows down," the angular acceleration is negative:

$𝛼=-0.8{\text{ rad/s}}^2$

We use the kinematic equation for rotational motion:

${𝜔}_2^{𝑓}={𝜔}_2^0+2𝛼𝜃$

When the record player stops, ${𝜔}_{𝑓}=0$. Thus,

$0={𝜔}_2^0+2𝛼𝜃$

Solving for the angular displacement ($𝜃$):

$𝜃=-\frac{{𝜔}_2^0}{2𝛼}$

Substituting the values:

$𝜃=-\frac{{\left(\frac{3𝜋}{2}\right)}^2}{2(-0.8)}=\frac{{\left(\frac{3𝜋}{2}\right)}^2}{1.6}$

Simplify the expression:

First, compute the square:

${\left(\frac{3𝜋}{2}\right)}^2=\frac{9{𝜋}^2}{4}$

Then, substitute back in:

$𝜃=\frac{\frac{9{𝜋}^2}{4}}{1.6}=\frac{9{𝜋}^2}{4\times 1.6}=\frac{9{𝜋}^2}{6.4}$

For a numerical approximation:

Using ${𝜋}^2\approx 9.87$,

$𝜃\approx \frac{9\times 9.87}{6.4}\approx \frac{88.83}{6.4}\approx 13.88\text{ rad}$

Therefore, the angular displacement is approximately $13.88\text{ rad}$.

The correct option is:

Option A: 13.88 rad

We can solve the problem using the concept of temperature dependence of resistance, which is given by the relation:

$𝑅={𝑅}_0(1+𝛼𝑇)$

Here,

• $𝑅$ is the resistance at temperature $𝑇$ (measured from 0°C),

• ${𝑅}_0$ is the resistance at 0°C, and

• $𝛼$ is the temperature coefficient.

However, the wire’s resistance is given at room temperature (20°C) as 10 Ω. So first, we find the resistance at 0°C:

Write the relation at 20°C:

$𝑅\left(20\right)={𝑅}_0(1+𝛼\times 20)$

Substitute the known values:

$10={𝑅}_0(1+0.002\times 20)={𝑅}_0(1+0.04)=1.04{𝑅}_0$

Solve for ${𝑅}_0$:

${𝑅}_0=\frac{10}{1.04}\approx 9.615\Omega$

Now, when the resistance increases by 10%, the new resistance becomes:

${𝑅}^{\prime }=10\Omega \times 1.10=11\Omega$

This new resistance at an unknown temperature $𝑇$ (measured from 0°C) is given by:

$11={𝑅}_0(1+𝛼𝑇)$

Substitute ${𝑅}_0\approx 9.615\Omega$ and $𝛼=0.002$:

$11=9.615(1+0.002𝑇)$

Now, solve for $𝑇$:

Divide both sides by 9.615:

$1+0.002𝑇=\frac{11}{9.615}\approx 1.144$

Subtract 1 from both sides:

$0.002𝑇=1.144-1=0.144$

Solve for $𝑇$:

$𝑇=\frac{0.144}{0.002}=72{}^{\circ }\text{C}$

Thus, the new temperature of the wire is ${72}^{\circ }\text{C}$, which corresponds to Option D.

The two cells of EMF $14\text{V}$ and $28\text{V}$ are connected in opposition in the closed loop.

The net EMF in the circuit is

${𝐸}_{\text{net}}=28\text{V}-14\text{V}=14\text{V}$

The capacitors of capacitance $4𝜇\text{F}$ and $8𝜇\text{F}$ are in series. The equivalent capacitance is

${𝐶}_{\text{eq}}=\frac{4\times 8}{4+8}𝜇\text{F}=\frac{8}{3}𝜇\text{F}$

The magnitude of charge on each capacitor in steady state is

$𝑞={𝐶}_{\text{eq}}{𝐸}_{\text{net}}=\frac{8}{3}\times 14=\frac{112}{3}𝜇\text{C}$

The potential difference across the $4𝜇\text{F}$ capacitor is

$𝑉=\frac{𝑞}{{𝐶}_1}=\frac{112}{3\times 4}=\frac{28}{3}\text{V}\approx 9.33\text{V}$

Taking into account the polarity and potential drop across this section,

${𝑉}_P-{𝑉}_Q\approx -9.3\text{V}$

Hence, the correct option is (A).

To determine the dimensions of time when using mass (M), length (L), and the gravitational constant (G) as fundamental units, follow these steps:

Recognize that the gravitational constant has the dimensions:

$\left[𝐺\right]=\frac{{𝐿}^3}{𝑀{𝑇}^2}$

Rearrange this equation to solve for time (T). Start by writing:

${𝑇}^2=\frac{{𝐿}^3}{𝑀𝐺}$

Take the square root of both sides to isolate T:

$𝑇=\sqrt{\surd \frac{{𝐿}^3}{𝑀𝐺}}$

Express the result using exponents:

$𝑇={𝐿}^{3/2}{𝑀}^{-1/2}{𝐺}^{-1/2}$

Comparing with the provided options, this matches:

Option C:

$M^{-1/2}{L}^{3/2}{G}^{-1/2}$

Thus, the correct answer is Option C.

To calculate the power of the electric motor, we need to determine the work done in lifting the mass and then divide it by the time taken.

Step 1: Calculate the Work Done

The work done $𝑊$ in lifting the mass is given by:

$𝑊=𝑚\times 𝑔\times ℎ$

Where:

$𝑚=1.5\text{kg}$ (mass)

$𝑔=9.81{\text{m/s}}^2$ (acceleration due to gravity)

$ℎ=1.128\text{m}$ (height)

Calculating:

$𝑊=1.5\text{kg}\times 9.81{\text{m/s}}^2\times 1.128\text{m}=16.59852\text{J}$

Step 2: Calculate the Power

Power $𝑃$ is the work done divided by the time $𝑡$:

$𝑃=\frac{𝑊}{𝑡}$

Given $𝑡=4.79\text{s}$:

$𝑃=\frac{16.59852\text{J}}{4.79\text{s}}\approx 3.465\text{W}$

Step 3: Determine Significant Figures

We need to express the power to the appropriate number of significant figures. Let's look at the significant figures in the given data:

Mass $𝑚=1.5\text{kg}$ has 2 significant figures.

Height $ℎ=1.128\text{m}$ has 4 significant figures.

Time $𝑡=4.79\text{s}$ has 3 significant figures.

Acceleration $𝑔=9.81{\text{m/s}}^2$ has 3 significant figures.

The calculated work $𝑊$ and power $𝑃$ should be reported with the least number of significant figures from the given data, which is 2 significant figures (from the mass).

Therefore, we round the power to 2 significant figures:

$𝑃\approx 3.5\text{W}$

Answer: Option D: 3.5 W

An electron of mass $𝑚$ and charge $𝑒$ starts from rest ($𝑢=0$) and is accelerated through a potential difference $𝑉$.

By the work-energy theorem,

$\frac{1}{2}𝑚{𝑣}^2=𝑒𝑉$

$𝑣=\sqrt{\surd \frac{2𝑒}{𝑚}}{𝑉}^{1/2}$

$\therefore {𝑣}^2\propto 𝑉$

This represents the equation of a parabola opening along the positive horizontal axis ($𝑉$-axis) with its vertex at the origin $(0,0)$.

Differentiating $𝑣$ with respect to $𝑉$,

$\frac{d𝑣}{d𝑉}=\frac{1}{2}\sqrt{\surd \frac{2𝑒}{𝑚}}{𝑉}^{-1/2}>0$

$\frac{d^2𝑣}{d{𝑉}^2}=-\frac{1}{4}\sqrt{\surd \frac{2𝑒}{𝑚}}{𝑉}^{-3/2}<0$

The curve passes through the origin, has an initial steep slope, and is concave downwards as the slope decreases with increasing $𝑉$. This corresponds to graph (C).

Hence, the correct option is (C).

To determine when the emf is zero, follow these steps:

Recall that the induced emf in an inductor is given by:

$𝜀=-𝐿\frac{𝑑𝐼}{𝑑𝑡}.$

Since the self-inductance $𝐿$ is a nonzero constant, the emf will be zero when the derivative $\frac{𝑑𝐼}{𝑑𝑡}$ is zero.

Differentiate the given current equation with respect to time:

$𝐼\left(𝑡\right)=3{𝑡}^2-3𝑡+8.$

Calculating the derivative:

$\frac{𝑑𝐼}{𝑑𝑡}=6𝑡-3.$

Set the derivative equal to zero and solve for $𝑡$:

$6𝑡-3=0⟹6𝑡=3⟹𝑡=\frac{3}{6}=0.5\text{sec}.$

Thus, the emf becomes zero at 0.5 sec, which corresponds to Option C.

Let's analyze the forces acting on the ball when it is in equilibrium:

The ball of mass $𝑚$ experiences a gravitational force downward: $𝑚𝑔$.

The ball experiences a horizontal electric force due to the uniform electric field, which is given by: $𝑞𝐸$.

The string exerts a tension $𝑇$ along its length, making an angle of ${45}^{\circ }$ with the vertical.

Since the ball is in equilibrium, the forces in both the vertical and horizontal directions must balance:

Vertical component:

The vertical component of the tension is

$𝑇\cos {45}^{\circ }$

This balances the weight:

$𝑇\cos {45}^{\circ }=𝑚𝑔.$

Horizontal component:

The horizontal component of the tension is

$𝑇\sin {45}^{\circ }$

This balances the electric force:

$𝑇\sin {45}^{\circ }=𝑞𝐸.$

Now, divide the horizontal equation by the vertical equation to eliminate $𝑇$:

$\frac{𝑇\sin {45}^{\circ }}{𝑇\cos {45}^{\circ }}=\frac{𝑞𝐸}{𝑚𝑔}.$

Since

$\frac{\sin {45}^{\circ }}{\cos {45}^{\circ }}=\tan {45}^{\circ }=1,$

the equation simplifies to:

$1=\frac{𝑞𝐸}{𝑚𝑔}.$

Solving for $𝐸$ gives:

$𝐸=\frac{𝑚𝑔}{𝑞}.$

Thus, the magnitude of the uniform electric field is

$𝐸=\frac{𝑚𝑔}{𝑞}{\text{NC}}^{-1}.$

This corresponds to Option A.

When light travels from one medium to another (e.g., from air to water), it refracts, or bends, due to a change in speed as it passes through media with different refractive indices. When a fish views a fisherman standing on the bank of a lake, the light rays from the fisherman bend at the air-water interface due to refraction.

Using Snell's Law:

${𝑛}_1\sin {𝜃}_1={𝑛}_2\sin {𝜃}_2$

Where:

${𝑛}_1$ is the refractive index of air (approximately 1.0),

${𝑛}_2$ is the refractive index of water (approximately 1.33),

${𝜃}_1$ is the angle of incidence in air,

${𝜃}_2$ is the angle of refraction in water.

Since the refractive index of water is higher than that of air, light bends towards the normal when it enters the water. As a result, the fisherman appears to be at a shallower angle than he actually is. This optical phenomenon causes the fisherman to appear taller to the fish.

Thus, the correct option is:

Option C: Taller than what he actually is

To solve this problem, we need to compare the time taken for the block to slide down the inclined plane with and without friction.

Without Friction:

When there is no friction, the only force acting along the inclined plane is the component of the gravitational force. The acceleration ${𝑎}_{\text{nf}}$ (where "nf" stands for no friction) can be given by:

${𝑎}_{\text{nf}}=𝑔\sin \left(𝜃\right)$

Where:

$𝑔$ is the acceleration due to gravity.

$𝜃={45}^{\circ }$, so $\sin \left({45}^{\circ }\right)=\frac{\sqrt{\surd 2}}{2}$.

Thus:

${𝑎}_{\text{nf}}=𝑔\frac{\sqrt{\surd 2}}{2}$

With Friction:

When friction is present, the net force along the plane is reduced by the frictional force. The acceleration ${𝑎}_{\text{f}}$ (where "f" stands for friction) is:

${𝑎}_{\text{f}}=𝑔\sin \left(𝜃\right)-𝜇𝑔\cos \left(𝜃\right)$

Where:

$𝜇$ is the coefficient of friction.

$\cos \left({45}^{\circ }\right)=\frac{\sqrt{\surd 2}}{2}$.

Thus:

${𝑎}_{\text{f}}=𝑔\frac{\sqrt{\surd 2}}{2}-𝜇𝑔\frac{\sqrt{\surd 2}}{2}$

${𝑎}_{\text{f}}=𝑔\frac{\sqrt{\surd 2}}{2}(1-𝜇)$

Relation between Times:

From the problem, the time with friction ${𝑡}_{\text{f}}$ is three times the time without friction ${𝑡}_{\text{nf}}$. Thus:

${𝑡}_{\text{f}}=3{𝑡}_{\text{nf}}$

Since time $𝑡$ under constant acceleration is inversely proportional to the square root of the acceleration:

$𝑡\propto \frac{1}{\sqrt{\surd 𝑎}}$

Therefore:

$\frac{1}{\sqrt{\surd {𝑎}_{\text{f}}}}=3\cdot \frac{1}{\sqrt{\surd {𝑎}_{\text{nf}}}}$

Squaring both sides gives:

$\frac{1}{{𝑎}_{\text{f}}}=9\cdot \frac{1}{{𝑎}_{\text{nf}}}$

Hence:

${𝑎}_{\text{f}}=\frac{{𝑎}_{\text{nf}}}{9}$

Substitute for ${𝑎}_{\text{f}}$ and ${𝑎}_{\text{nf}}$:

$𝑔\frac{\sqrt{\surd 2}}{2}(1-𝜇)=\frac{𝑔\frac{\sqrt{\surd 2}}{2}}{9}$

Canceling terms common on both sides gives:

$1-𝜇=\frac{1}{9}$

Solving for $𝜇$:

$𝜇=1-\frac{1}{9}$

$𝜇=\frac{8}{9}$

Therefore, the coefficient of friction is approximately 0.88. So, the correct option is:

Option C: 0.88

Let's break it down step by step:

Magnetic susceptibility, $𝜒$, is defined as:

$𝜒=\frac{𝑀}{𝐻},$

where $𝑀$ is the magnetization (magnetic moment per unit volume) and $𝐻$ is the applied magnetic field.

In diamagnetic materials, the induced magnetization is opposite to the applied magnetic field, which means the susceptibility is negative.

An ideal diamagnetic substance is one in which the induced magnetization exactly cancels the applied magnetic field inside the material. This situation corresponds to having a relative permeability ${𝜇}_{𝑟}$ given by:

${𝜇}_{𝑟}=1+𝜒=0.$

Solving for $𝜒$:

$𝜒=-1.$

Therefore, the magnetic susceptibility of an ideal diamagnetic substance is $-1$.

The correct answer is Option A: $-1$.

To solve the problem, we use the principle of continuity, which tells us that the flow rate is constant. This means that

${𝐴}_1{𝑣}_1={𝐴}_2{𝑣}_2,$

where

• ${𝐴}_1={10}^{-3}{m}^2$ is the area at the tap,

• ${𝑣}_1=0.5m/s$ is the initial speed,

• ${𝐴}_2$ is the area 0.19 m below the tap, and

• ${𝑣}_2$ is the speed 0.19 m below the tap.

Since we assume the pressure is constant and neglect viscosity, the water accelerates under gravity. We can find the speed ${𝑣}_2$ using energy conservation:

${𝑣}_2=\sqrt{\surd {𝑣}_2^1+2𝑔ℎ},$

where

• $𝑔\approx 9.8{m/s}^2$ is the acceleration due to gravity, and

• $ℎ=0.19m$ is the vertical distance.

Now, calculate ${𝑣}_2$:

Compute the increase in speed squared:

$2𝑔ℎ=2\times 9.8\times 0.19\approx 3.724m^2/s^2.$

Add the initial speed squared:

${𝑣}_2^1=(0.5{)}^2=0.25m^2/s^2,$