MHT CET 2024 3rd May Morning Shift (Physics & Chemistry) Question Paper with Solutions

Maximum height attained by ball $A,h_1=\frac{u^2}{2g}$

Maximum height attained by ball B,

$\begin{array}{cc}{h}_2& =\frac{u^2{\sin }^2\left({60}^{\circ }\right)}{2g}\dots (\because 𝜃={90}^{\circ }-{30}^{\circ }={60}^{\circ })\\ & =\frac{3u^2}{8g}\\ \therefore \frac{h_1}{h_2}& =\frac{u^2}{2g}\times \frac{8g}{3u^2}=\frac{8}{6}=\frac{4}{3}\end{array}$

$\begin{array}{cc}T& =2𝜋\sqrt{\surd \frac{𝑙}{g}}\\ & =2𝜋\sqrt{\surd \frac{1.6}{10}}\dots (\because R=𝑙)\\ & =0.8𝜋\end{array}$

The path difference between two waves given by ${𝑌}_1={𝑎}_1\sin (𝜔𝑡-\frac{2𝜋𝑥}{𝜆})$ and ${𝑌}_2={𝑎}_2\cos (𝜔𝑡-\frac{2𝜋𝑥}{𝜆}+𝜙)$ can be calculated as follows:

First, express the second wave in terms of sine:

${𝑦}_2={𝑎}_2\sin (𝜔𝑡-\frac{2𝜋𝑥}{𝜆}+𝜙+\frac{𝜋}{2})$

This shows that the phase of ${𝑌}_2$ is $𝜔𝑡-\frac{2𝜋𝑥}{𝜆}+𝜙+\frac{𝜋}{2}$.

The phase difference between the two waves is thus given by:

$𝛿=𝜙+\frac{𝜋}{2}$

Using the relation for path difference $\Delta 𝑥=\frac{𝜆}{2𝜋}\cdot 𝛿$, we substitute the phase difference:

$\Delta 𝑥=\frac{𝜆}{2𝜋}(𝜙+\frac{𝜋}{2})$

When two equal point charges, each of magnitude $𝑞$, exert a force $𝐹$ on each other when placed distance $𝑥$ apart in air, the force is given by:

${𝐹}_1=\frac{1}{4𝜋{𝜀}_0}\frac{{𝑞}^2}{{𝑥}^2}$

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

When these charges are placed distance $𝑦$ apart in a medium with a dielectric constant $𝑘$, they exert the same force $𝐹$:

${𝐹}_2=\frac{1}{4𝜋{𝜀}_0𝑘}\frac{{𝑞}^2}{{𝑦}^2}$

Since the forces ${𝐹}_1$ and ${𝐹}_2$ are equal, we have:

$\frac{1}{4𝜋{𝜀}_0}\frac{{𝑞}^2}{{𝑥}^2}=\frac{1}{4𝜋{𝜀}_0𝑘}\frac{{𝑞}^2}{{𝑦}^2}$

Simplifying the equation:

$𝑘\frac{{𝑦}^2}{{𝑥}^2}=1$

From this, we determine the ratio of distance $𝑦$ to $𝑥$:

$\frac{𝑦}{𝑥}=\frac{1}{\sqrt{\surd 𝑘}}$

Thus, the ratio of distance $𝑦$ to $𝑥$ in the medium is $\frac{1}{\sqrt{\surd 𝑘}}$.

Alternating voltage: $𝑒=200\sqrt{\surd 2}\sin \left(100𝑡\right)$ volt

Comparing with $e=e_0\sin 𝜔𝑡$

$𝜔=100\mathrm{rad}/s,{𝑒}_0=200\sqrt{\surd 2}$

Capacitive reactance,

$\begin{array}{cc}& X_C=\frac{1}{𝜔C}=\frac{1}{100\times {10}^{-6}}\Omega ={10}^4\Omega \\ & I_0=\frac{e_0}{X_c}\\ & I_0=\frac{200\sqrt{\surd 2}}{{10}^4}\\ & I_0=2\sqrt{\surd 2}\times {10}^{-2}A\\ & I_{\mathrm{rms}}=\frac{I_0}{\sqrt{\surd 2}}=\frac{2\sqrt{\surd 2}\times {10}^{-2}}{\sqrt{\surd 2}}=2\times {10}^{-2}A=20\mathrm{mA}\end{array}$

From charge configuration, at the centre electric field is non-zero. Potential at the centre due to $2𝑞$ charge ${𝑉}_{2𝑞}=\frac{2𝑞}{𝑟}$

and potential due to $-$q charge

$V_{-q}=-\frac{q}{r}(r=$ distance of centre point $)$

$\therefore$ total potential $V=V_{2q}+V_{-q}+V_{-q}=0$

To determine the speed of water leaving the sprinkler, we can apply the principle of conservation of mass, also known as the continuity equation for incompressible fluids. According to this principle, the volumetric flow rate must be constant throughout the system. The flow rate can be expressed as the product of the cross-sectional area and the speed of the fluid.

Given:

Internal radius of the pipe: $𝑅$

Speed of water in the pipe: $𝑣$

Number of holes in the sprinkler: $𝑛$

Radius of each hole: $𝑟$

The cross-sectional area of the pipe is:

${𝐴}_{\text{pipe}}=𝜋{𝑅}^2$

The flow rate through the pipe is:

${𝑄}_{\text{pipe}}={𝐴}_{\text{pipe}}\cdot 𝑣=𝜋{𝑅}^2𝑣$

The total cross-sectional area of the holes in the sprinkler is the sum of the areas of the $𝑛$ holes:

${𝐴}_{\text{holes}}=𝑛\cdot 𝜋{𝑟}^2$

Let ${𝑣}_{𝑠}$ be the speed of water leaving the sprinkler. The flow rate out of the sprinkler is:

${𝑄}_{\text{sprinkler}}={𝐴}_{\text{holes}}\cdot {𝑣}_{𝑠}=𝑛\cdot 𝜋{𝑟}^2\cdot {𝑣}_{𝑠}$

By the continuity equation, the flow rate through the pipe is equal to the flow rate out of the sprinkler:

$𝜋{𝑅}^2𝑣=𝑛\cdot 𝜋{𝑟}^2\cdot {𝑣}_{𝑠}$

Solving for ${𝑣}_{𝑠}$, we have:

${𝑣}_{𝑠}=\frac{{𝑅}^2𝑣}{𝑛{𝑟}^2}$

Therefore, the speed of water leaving the sprinkler is given by option B:

$\frac{R^{2}v}{{\mathrm{nr}}^2}$

The time period for the oscillation of a magnetic needle performing simple harmonic motion in a magnetic field is given by the formula:

$𝑇=2𝜋\sqrt{\surd \frac{𝐼}{𝑚𝐵}}$

where:

$𝑇$ is the time period of one oscillation.

$𝐼=9.6\times {10}^{-5}\mathrm{kg}{m}^2$ is the moment of inertia.

$𝑚=6\times {10}^{-2}{\mathrm{Am}}^2$ is the magnetic moment.

$𝐵=0.01T$ is the magnetic field.

First, calculate the time period $𝑇$:

$𝑇=2𝜋\sqrt{\surd \frac{9.6\times {10}^{-5}}{6\times {10}^{-2}\times 0.01}}$

$=2𝜋\sqrt{\surd \frac{9.6\times {10}^{-5}}{6\times {10}^{-4}}}$

$=2𝜋\sqrt{\surd 0.16}$

$=2𝜋\left(0.4\right)$

Substituting the value of $𝜋=3.14$,

$𝑇=2\times 3.14\times 0.4=2.512s$

Since this $𝑇$ represents the time for one oscillation, the time taken for 10 oscillations is:

$10𝑇=10\times 2.512s=25.12s$

The correct answer is thus:

Option C: 25.12 s

The correct option is Option C.

Let us check each relation one by one.

1. Linear velocity

For rotational motion,

$\overrightarrow{𝑣}=\overrightarrow{𝜔}\times \overrightarrow{𝑟}$

So this is correct in B, C, D.

Option A is wrong because it writes

$\overrightarrow{𝑣}=\overrightarrow{𝑟}\times \overrightarrow{𝜔}$

But cross product is not commutative:

$\overrightarrow{𝑟}\times \overrightarrow{𝜔}=-(\overrightarrow{𝜔}\times \overrightarrow{𝑟})$

2. Relation between angular acceleration and linear acceleration

In the given options, the intended vector relation is

$\overrightarrow{𝛼}=\overrightarrow{𝑎}\times \overrightarrow{𝑟}$

So this matches B, C, D.

Option A has

$\overrightarrow{𝛼}=\overrightarrow{𝑟}\times \overrightarrow{𝑎}$

which has wrong order.

3. Angular momentum

Angular momentum is given by

$\overrightarrow{𝐿}=\overrightarrow{𝑟}\times \overrightarrow{𝑝}$

So only A and C have this correct.

Option B is wrong because it writes

$\overrightarrow{𝐿}=\overrightarrow{𝑝}\times \overrightarrow{𝑟}$

which is opposite in direction.

Option D is also wrong because it gives dot product:

$\overrightarrow{𝐿}=\overrightarrow{𝑝}\cdot \overrightarrow{𝑟}$

But angular momentum is a vector, so it must be a cross product, not a dot product.

4. Torque

Torque is

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

So this is correct in B, C, D.

Option A is wrong because it writes

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

which is in reverse order.

Final conclusion

Only Option C has all correct expressions.

$\overline{)\text{Option C}}$

To find the object distance when a convex lens creates a real, inverted image twice the size of the object, follow these steps:

Focal Length: Let the focal length be $𝑓$.

Magnification: The magnification ($𝑚$) of the lens is given by the ratio of the image height (${ℎ}^{\prime }$) to the object height ($ℎ$), and also by the ratio of the image distance ($𝑣$) to the object distance ($𝑢$):

$𝑚=\frac{𝑣}{𝑢}=\frac{{ℎ}^{\prime }}{ℎ}=-2$

Since the image is real and inverted, $𝑚=-2$. Therefore, the image distance is:

$𝑣=-2𝑢$

Lens Formula: The lens formula relates the image distance $𝑣$, object distance $𝑢$, and focal length $𝑓$:

$\frac{1}{𝑣}-\frac{1}{𝑢}=\frac{1}{𝑓}$

Substitute the value of $𝑣$:

$\frac{1}{-2𝑢}-\frac{1}{𝑢}=\frac{1}{𝑓}$

Simplify the equation:

$\frac{𝑢+2𝑢}{2{𝑢}^2}=\frac{1}{𝑓}$

$\frac{3𝑢}{2{𝑢}^2}=\frac{1}{𝑓}$

Solve for $𝑢$:

$\frac{1.5}{𝑢}=\frac{1}{𝑓}$

$𝑢=1.5𝑓$

Thus, the object distance from the lens is $1.5𝑓$.

To determine the ratio of the wavelength of a photon to that of an electron, both having the same energy $𝐸$, we'll use the following expressions:

Wavelength of a photon:

The energy $𝐸$ of a photon can be expressed using the equation:

$𝐸=\frac{ℎ𝑐}{{𝜆}_{\text{photon}}}$

where $ℎ$ is Planck's constant, $𝑐$ is the speed of light, and ${𝜆}_{\text{photon}}$ is the wavelength of the photon. Solving for the wavelength:

${𝜆}_{\text{photon}}=\frac{ℎ𝑐}{𝐸}$

Wavelength of an electron:

Using de Broglie's wavelength formula, the wavelength ${𝜆}_{\text{electron}}$ of an electron is:

${𝜆}_{\text{electron}}=\frac{ℎ}{𝑝}$

The energy $𝐸$ of the electron is given by:

$𝐸=\frac{{𝑝}^2}{2𝑚}$

Solving for $𝑝$, we get:

$𝑝=\sqrt{\surd 2𝑚𝐸}$

Substituting $𝑝$ into the expression for the wavelength:

${𝜆}_{\text{electron}}=\frac{ℎ}{\sqrt{\surd 2𝑚𝐸}}$

Ratio of the wavelengths:

The ratio of the wavelength of the photon to that of the electron is:

$\frac{{𝜆}_{\text{photon}}}{{𝜆}_{\text{electron}}}=\frac{\frac{ℎ𝑐}{𝐸}}{\frac{ℎ}{\sqrt{\surd 2𝑚𝐸}}}$

Simplifying this expression:

$\frac{{𝜆}_{\text{photon}}}{{𝜆}_{\text{electron}}}=\frac{ℎ𝑐}{𝐸}\times \frac{\sqrt{\surd 2𝑚𝐸}}{ℎ}$

Cancel out $ℎ$:

$=\frac{𝑐}{𝐸}\times \sqrt{\surd 2𝑚𝐸}$

Simplify further:

$=𝑐\sqrt{\surd \frac{2𝑚}{𝐸}}$

Hence, the correct option is Option D: $𝑐\sqrt{\surd \frac{2𝑚}{𝐸}}$.

To determine the final temperature and the root mean square (r.m.s.) velocity of the hydrogen gas when its volume becomes three times at constant pressure, the following relationships and concepts are used:

Ideal Gas Law at Constant Pressure:

At constant pressure, the volume ($𝑉$) of an ideal gas varies directly with its absolute temperature ($𝑇$):

$\frac{{𝑉}_1}{{𝑇}_1}=\frac{{𝑉}_2}{{𝑇}_2}$

Given that the volume triples (${𝑉}_2=3{𝑉}_1$), and initially, the gas is at Standard Temperature and Pressure (STP) where ${𝑇}_1=273\text{ K}$, we can solve for the final temperature ${𝑇}_2$:

$\frac{{𝑉}_1}{273}=\frac{3{𝑉}_1}{{𝑇}_2}$

Simplifying gives:

${𝑇}_2=819\text{K}$

RMS Velocity Calculation:

The r.m.s. velocity of a gas is given by:

$𝑢=\sqrt{\surd \frac{3𝑘𝑇}{𝑚}}$

where $𝑘$ is the Boltzmann constant and $𝑚$ is the molecular mass. The r.m.s. velocity is proportional to the square root of the temperature ($𝑢\propto \sqrt{\surd 𝑇}$).

Initially, the velocity ${𝑢}_1=𝑢$ at ${𝑇}_1=273\text{ K}$. For the final velocity ${𝑢}_2$ at ${𝑇}_2=819\text{ K}$:

${𝑢}_2=𝑢\sqrt{\surd \frac{{𝑇}_2}{{𝑇}_1}}=𝑢\sqrt{\surd \frac{819}{273}}=𝑢\sqrt{\surd 3}$

Thus, the final temperature of the gas is $819\text{K}$ and the final r.m.s. speed is $𝑢\sqrt{\surd 3}\text{m/s}$.

The correct option is Option A: $819\text{K},𝑢\sqrt{\surd 3}\text{m/s}$.

The orbital speed of a planet is determined by the formula for circular orbital velocity:

$𝑣=\sqrt{\surd \frac{𝐺𝑀}{𝑟}}$

where:

$𝑣$ is the orbital speed,

$𝐺$ is the gravitational constant,

$𝑀$ is the mass of the central object (in this case, the Sun),

$𝑟$ is the distance from the center of the planet to the center of the central object (orbital radius).

For two planets $𝐴$ and $𝐵$ with radii ${𝑟}_{𝐴}$ and ${𝑟}_{𝐵}$ (where ${𝑟}_{𝐵}=100{𝑟}_{𝐴}$), the orbital speed for each planet can be set as follows:

For planet A:

${𝑣}_{𝐴}=\sqrt{\surd \frac{𝐺𝑀}{{𝑟}_{𝐴}}}$

For planet B:

${𝑣}_{𝐵}=\sqrt{\surd \frac{𝐺𝑀}{{𝑟}_{𝐵}}}$

Since ${𝑟}_{𝐵}=100{𝑟}_{𝐴}$, substitute into the equation for ${𝑣}_{𝐵}$:

${𝑣}_{𝐵}=\sqrt{\surd \frac{𝐺𝑀}{100{𝑟}_{𝐴}}}$

This can be further simplified to:

${𝑣}_{𝐵}=\frac{1}{\sqrt{\surd 100}}\sqrt{\surd \frac{𝐺𝑀}{{𝑟}_{𝐴}}}$

${𝑣}_{𝐵}=\frac{1}{10}\sqrt{\surd \frac{𝐺𝑀}{{𝑟}_{𝐴}}}$

Thus, since ${𝑣}_{𝐴}=\sqrt{\surd \frac{𝐺𝑀}{{𝑟}_{𝐴}}}$, it follows that:

${𝑣}_{𝐵}=\frac{1}{10}{𝑣}_{𝐴}$

Therefore, the orbital speed of planet B is $\frac{𝑣}{10}$.

So, the correct answer is Option A:

$\frac{𝑣}{10}$

For a depression of y cm on one side, the level of liquid will be 2 y cm higher on the other side.

$\therefore$ Weight of extra liquid on the P -side $=2$ Aydg The above weight acts as the restoring force acting on mass M .

$\therefore$ Restoring acceleration $=-\frac{2\mathrm{Aydg}}{M}\dots (-\mathrm{ve}$ sign indicates the acceleration and force are opposite to each other) ... (i)

We know for simple harmonic motion, $𝑎=-\left(\frac{𝑘}{𝑚}\right)𝑥\text{.... (ii)}$

Comparing (i) and (ii), we see the motion is simple harmonic.

$\begin{array}{cc}\therefore T& =2𝜋\sqrt{\surd \frac{\text{ Displacement }}{\text{ Acceleration }}}=2𝜋\sqrt{\surd \frac{y}{\frac{2\mathrm{Aydg}}{M}}}\\ & =2𝜋\sqrt{\surd \frac{M}{2\mathrm{Adg}}}\end{array}$

An $\text{n-p-n}$ transistor consists of a p-type base between an n-type emitter and an n-type collector.

In a semiconductor diode, the symbol points from the p-type region (anode) to the n-type region (cathode). In the two-diode representation, the base-emitter junction corresponds to a diode directed from the base to the emitter, and the base-collector junction corresponds to a diode directed from the base to the collector.

Thus, the anodes of both diodes are connected at the common base terminal $\text{B}$, while their cathodes connect to the emitter $\text{E}$ and collector $\text{C}$ respectively. This configuration is shown in figure (b).

Hence, the correct option is (B).

The fundamental frequency of an air column in a pipe open at both ends is determined by the equation:

${𝑓}_1=\frac{𝑣}{2𝐿}$

where $𝑣$ is the speed of sound in air and $𝐿$ is the length of the air column.

When 80% of the pipe is immersed in water, only 20% of the length remains filled with air and acts as an open pipe, since the other end of the air column is now effectively 'closed' by the water surface. Therefore, the effective length ${𝐿}^{\prime }$ of the air column is:

${𝐿}^{\prime }=0.2𝐿$

For an open-closed column (since the top is open and the bottom is closed by water), the fundamental frequency is given by:

${𝑓}_2=\frac{𝑣}{4{𝐿}^{\prime }}$

Substituting ${𝐿}^{\prime }=0.2𝐿$ into the equation for ${𝑓}_2$:

${𝑓}_2=\frac{𝑣}{4\left(0.2𝐿\right)}=\frac{𝑣}{0.8𝐿}=\frac{5𝑣}{4𝐿}$

Now, to find the ratio ${𝑓}_1:{𝑓}_2$:

$\frac{{𝑓}_1}{{𝑓}_2}=\frac{\frac{𝑣}{2𝐿}}{\frac{5𝑣}{4𝐿}}=\frac{𝑣}{2𝐿}\times \frac{4𝐿}{5𝑣}=\frac{4}{10}=\frac{2}{5}$

The ratio ${𝑓}_1:{𝑓}_2$ is therefore $2:5$.

Option D: $2:5$ is the correct answer.

The binding energy of the body is given by

$\begin{array}{cc}\text{ B.E. }& =\frac{𝐺{𝑀}_1𝑀}{\frac{𝑟}{3}}+\frac{{\mathrm{GM}}_{2}M}{\frac{2𝑟}{3}}=\frac{3𝐺{𝑀}_1𝑀}{𝑟}+\frac{3{\mathrm{GM}}_2𝑀}{2r}\\ & =\frac{3𝐺𝑀}{𝑟}[{𝑀}_1+\frac{{𝑀}_2}{2}]\end{array}$

If v is the velocity given to the body, then

$\begin{array}{cc}& \frac{1}{2}𝑀{𝑣}^2=\frac{3𝐺𝑀}{𝑟}[{𝑀}_1+\frac{{𝑀}_2}{2}]\\ \therefore & 𝑣={\left[\frac{6𝐺}{𝑟}({𝑀}_1+\frac{{𝑀}_2}{2})\right]}^{\frac{1}{2}}\end{array}$

In an interference experiment, the condition for a bright fringe due to a light of wavelength ${𝜆}_1$ is given by the equation:

$𝑑\sin 𝜃=𝑛{𝜆}_1,$

where $𝑑$ is the slit separation, $𝜃$ is the angle of diffraction, and $𝑛$ is the order of the bright fringe ($𝑛=0,1,2,\dots$).

For a dark fringe due to a light of wavelength ${𝜆}_2$, the condition is:

$𝑑\sin 𝜃=(𝑚-\frac{1}{2}){𝜆}_2,$

where $𝑚$ is the order of the dark fringe ($𝑚=1,2,3,\dots$).

If both these fringes coincide, their path differences must be equal. Therefore:

$𝑛{𝜆}_1=(𝑚-\frac{1}{2}){𝜆}_2.$

Solving for the ratio $\frac{{𝜆}_1}{{𝜆}_2}$, we have:

$\frac{{𝜆}_1}{{𝜆}_2}=\frac{𝑚-\frac{1}{2}}{𝑛}.$

Multiplying the numerator and the denominator by 2 to clear the fraction, we obtain:

$\frac{{𝜆}_1}{{𝜆}_2}=\frac{2𝑚-1}{2𝑛}.$

So, the correct option is Option C: $\frac{2𝑚-1}{2𝑛}$.

Equivalent inductance or inductances connected in parallel is $L_p=\frac{1}{L_1}+\frac{1}{L_2}$

$\therefore L_p=\frac{1}{(\frac{2}{L}+\frac{2}{L})}=\frac{L}{4}$

Equivalent inductance in series

$\begin{array}{cc}{L}_s& =L_p+L_3\\ & =\frac{L}{4}+\frac{3}{4}L=L\end{array}$

For an isothermal process.

${𝑃}_1{𝑉}_1={𝑃}_2{𝑉}_2$

Given that $V_2=\frac{V_1}{2}$. It can be clearly understood that for $P_2{V}_2$ to remain constant,

$P_2=2P_1\text{.... (i)}$.

For an adiabatic process, ${\mathrm{PV}}^{𝛾}=$ constant.

$\begin{array}{cc}& P_1{V}_{𝛾}^1=P_2{V}_{𝛾}^2\\ & P_2=P_1{\left(\frac{V_1}{V_2}\right)}^{𝛾}=2P_{𝛾}^1\text{.... (ii)}\end{array}$

$𝛾$ is always greater than 1. From equations (i) and (ii), it can be seen that the pressure change in an adiabatic process would be greater than that in an isothermal process for the same change in volume.

In a single slit diffraction pattern, the position of secondary maxima can be determined using the angular position equation:

$𝑎\sin 𝜃=(𝑚+\frac{1}{2})𝜆$

where:

$𝑎$ is the width of the slit,

$𝜃$ is the angle of diffraction,

$𝑚$ is the order number of the secondary maximum,

$𝜆$ is the wavelength of light.

For the second secondary maximum (m = 2) of the light with wavelength $𝜆=6195\stackrel{𝑜}{𝐴}$:

$𝑎\sin 𝜃=(2+\frac{1}{2})𝜆=\frac{5}{2}\times 6195$

For the third secondary maximum (m = 3) with the unknown wavelength ${𝜆}_0$:

$𝑎\sin 𝜃=(3+\frac{1}{2}){𝜆}_0=\frac{7}{2}{𝜆}_0$

Since both maxima coincide, their angular positions are equal:

$\frac{5}{2}\times 6195=\frac{7}{2}{𝜆}_0$

Solving for ${𝜆}_0$:

${𝜆}_0=\frac{\frac{5}{2}\times 6195}{\frac{7}{2}}=\frac{5\times 6195}{7}$

${𝜆}_0=\frac{30975}{7}=4425\stackrel{𝑜}{𝐴}$

Therefore, the value of ${𝜆}_0$ is $4425\stackrel{𝑜}{𝐴}$.

Correct Option:

Option B: $4425\stackrel{𝑜}{𝐴}$

When the two rods, one made of aluminum and the other made of steel, are subjected to the same temperature increase, the length of each rod increases by the same amount. This implies:

For aluminum: $\Delta {𝐿}_{\text{aluminum}}={𝐿}_1{𝛼}_1𝑡$

For steel: $\Delta {𝐿}_{\text{steel}}={𝐿}_2{𝛼}_2𝑡$

Given that these increases in lengths are equal, so:

${𝐿}_1{𝛼}_1𝑡={𝐿}_2{𝛼}_2𝑡$

Cancelling the common factor $𝑡$ from both sides gives:

${𝐿}_1{𝛼}_1={𝐿}_2{𝛼}_2$

We need to find the ratio $\frac{{𝐿}_1}{{𝐿}_1+{𝐿}_2}$.

From ${𝐿}_1{𝛼}_1={𝐿}_2{𝛼}_2$, we can express ${𝐿}_2$ in terms of ${𝐿}_1$:

${𝐿}_2=\frac{{𝐿}_1{𝛼}_1}{{𝛼}_2}$

Substituting this into ${𝐿}_1+{𝐿}_2$:

${𝐿}_1+{𝐿}_2={𝐿}_1+\frac{{𝐿}_1{𝛼}_1}{{𝛼}_2}={𝐿}_1(1+\frac{{𝛼}_1}{{𝛼}_2})$

Therefore, the desired ratio is:

$\frac{{𝐿}_1}{{𝐿}_1+{𝐿}_2}=\frac{{𝐿}_1}{{𝐿}_1(1+\frac{{𝛼}_1}{{𝛼}_2})}$

Simplifying:

$\frac{{𝐿}_1}{{𝐿}_1(1+\frac{{𝛼}_1}{{𝛼}_2})}=\frac{1}{1+\frac{{𝛼}_1}{{𝛼}_2}}=\frac{{𝛼}_2}{{𝛼}_1+{𝛼}_2}$

So the correct option is:

Option C: $\frac{{𝛼}_2}{{𝛼}_1+{𝛼}_2}$.

When the plane of the coil in an alternating current (a.c.) generator is perpendicular to the magnetic field, the magnetic flux through the coil is maximum, because the area vector of the coil is aligned with the magnetic field lines.

At this position, the rate of change of magnetic flux, which is responsible for inducing the electromotive force (e.m.f.), is zero. Therefore, the induced e.m.f. at this point is zero.

Thus, the correct option is:

Option B: magnetic flux is maximum and induced e.m.f. is zero.

The capacitance of a parallel plate capacitor is given by the formula:

$𝐶=\frac{{𝜀}_0𝐴}{𝑑}$

where ${𝜀}_0$ is the permittivity of free space, $𝐴$ is the area of the plates, and $𝑑$ is the distance between the plates.

Initial Capacitance ${𝐶}_1$:

The area of each plate, ${𝐴}_1$, is:

${𝐴}_1=𝜋{𝑟}^2$

So, the initial capacitance ${𝐶}_1$ is:

${𝐶}_1=\frac{{𝜀}_0𝜋{𝑟}^2}{𝑑}$

Changed Capacitance ${𝐶}_2$:

The radius of each plate is increased to $\sqrt{\surd 2}\times 𝑟$, so the new area ${𝐴}_2$ is:

${𝐴}_2=𝜋(\sqrt{\surd 2}𝑟{)}^2=2𝜋{𝑟}^2$

The distance between the plates is decreased to $\frac{1}{2}𝑑$.

The new capacitance ${𝐶}_2$ is:

${𝐶}_2=\frac{{𝜀}_0\cdot 2𝜋{𝑟}^2}{\frac{1}{2}𝑑}=\frac{2\cdot {𝜀}_0𝜋{𝑟}^2}{\frac{1}{2}𝑑}=\frac{4{𝜀}_0𝜋{𝑟}^2}{𝑑}$

Ratio ${𝐶}_1:{𝐶}_2$:

$\frac{{𝐶}_1}{{𝐶}_2}=\frac{\frac{{𝜀}_0𝜋{𝑟}^2}{𝑑}}{\frac{4{𝜀}_0𝜋{𝑟}^2}{𝑑}}=\frac{1}{4}$

Therefore, the ratio of the capacitances ${𝐶}_1:{𝐶}_2$ is:

Option C: $1:4$

The radius of gyration ($𝑘$) is related to the moment of inertia ($𝐼$) and the mass ($𝑀$) of an object by the formula:

$𝑘=\sqrt{\surd \frac{𝐼}{𝑀}}$

For a circular disc:

Moment of inertia (${𝐼}_{\text{disc}}$) about its central axis is given by:

${𝐼}_{\text{disc}}=\frac{1}{2}𝑀{𝑅}^2$

Radius of gyration (${𝑘}_{\text{disc}}$) is:

${𝑘}_{\text{disc}}=\sqrt{\surd \frac{{𝐼}_{\text{disc}}}{𝑀}}=\sqrt{\surd \frac{\frac{1}{2}𝑀{𝑅}^2}{𝑀}}=\sqrt{\surd \frac{1}{2}}𝑅$

For a circular ring:

Moment of inertia (${𝐼}_{\text{ring}}$) about its central axis is given by:

${𝐼}_{\text{ring}}=𝑀{𝑅}^2$

Radius of gyration (${𝑘}_{\text{ring}}$) is:

${𝑘}_{\text{ring}}=\sqrt{\surd \frac{{𝐼}_{\text{ring}}}{𝑀}}=\sqrt{\surd \frac{𝑀{𝑅}^2}{𝑀}}=𝑅$

Ratio of the radii of gyration:

The ratio ${𝑘}_{\text{disc}}:{𝑘}_{\text{ring}}$ is:

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

Therefore, the ratio of the radius of gyration of a circular disc to that of a circular ring, each with the same mass and radius, is:

Option D: $1:\sqrt{\surd 2}$

The power dissipated in an AC circuit can be calculated using the formula for average power:

${𝑃}_{\text{avg}}=𝑉𝐼\cos 𝜙$

where $𝑉$ and $𝐼$ are the RMS (root mean square) values of voltage and current, and $𝜙$ is the phase difference between them.

Given:

Voltage $𝐸\left(𝑡\right)=200\sin \left(50𝑡\right)$ volts

Current $𝐼\left(𝑡\right)=100\sin (50𝑡+\frac{𝜋}{3})$ mA

Convert the peak values to RMS values:

For a sinusoidal function, the RMS value is given by:

${𝑉}_{\text{RMS}}=\frac{{𝑉}_{\text{peak}}}{\sqrt{\surd 2}}$

${𝐼}_{\text{RMS}}=\frac{{𝐼}_{\text{peak}}}{\sqrt{\surd 2}}$

Therefore:

${𝑉}_{\text{RMS}}=\frac{200}{\sqrt{\surd 2}}$

${𝐼}_{\text{RMS}}=\frac{100}{\sqrt{\surd 2}}$

The phase difference $𝜙=\frac{𝜋}{3}$.

Calculate the power factor:

The power factor is given by $\cos 𝜙$.

Given $𝜙=\frac{𝜋}{3}$, we use:

$\cos \frac{𝜋}{3}=\frac{1}{2}$

Calculate the average power:

Substituting the values into the power formula:

${𝑃}_{\text{avg}}=\left(\frac{200}{\sqrt{\surd 2}}\right)\left(\frac{100}{\sqrt{\surd 2}}\right)\left(\frac{1}{2}\right)$

${𝑃}_{\text{avg}}=\left(\frac{200\times 100}{2}\right)\times \frac{1}{2}$

${𝑃}_{\text{avg}}=10000\times \frac{1}{2}$

${𝑃}_{\text{avg}}=5000\text{mW}=5\text{W}$

Thus, the power dissipated in the circuit is 5 watts.

Option D: 5 watt

Kirchhoff's Current Law (KCL): According to KCL, the total current (I) through the circuit can be calculated by adding the individual currents:

$𝐼=1+1+0.5=2.5\text{A}$

Calculate Total Resistance (R_total): The total resistance in the section AB includes the unknown resistance $𝑋$ and a known resistance of $2\Omega$.

${𝑅}_{\text{total}}=𝑋+2\Omega$

Power Calculation: The power absorbed in the section AB is given as $50\text{W}$. Use the power formula $𝑃={𝐼}^2𝑅$:

$50=\left(2.5{)}^2\right(𝑋+2)$

Solve for X: Simplify the equation to find the value of $𝑋$.

$50=6.25(𝑋+2)$

$\Rightarrow 𝑋+2=8$

$\Rightarrow 𝑋=6\Omega$

Therefore, the value of the resistance $𝑋$ is $6\Omega$.

To solve this problem, we apply Snell's Law and the prism formula. Given:

Angle of incidence, $𝑖={60}^{\circ }$

Angle of the prism, $𝐴={30}^{\circ }$

Angle between emergent and incident ray, $\Delta ={30}^{\circ }$

The angle of deviation $𝛿$ is:

$𝛿=𝑖+𝑒-𝐴$

where $𝑒$ is the angle of emergence. Given $\Delta ={30}^{\circ }$, it follows that:

$𝛿=\Delta ={30}^{\circ }$

Thus:

$𝛿=𝑖+𝑒-𝐴={30}^{\circ }$

Substitute the known values:

${60}^{\circ }+𝑒-{30}^{\circ }={30}^{\circ }$

$𝑒=0^{\circ }$

This angle $𝑒$ cannot be $0{\cdot }^{\circ }$. Therefore, revising calculations leads us to Snell's law application at points where the ray enters and leaves the prism.

Using the prism formula for deviation $𝛿=(𝑖+𝑒)-𝐴$ and Snell’s law:

For the first interface:

$𝑛\cdot \sin {30}^{\circ }=\sin {60}^{\circ }$

The condition is:

$𝑛\times 0.5=\frac{\sqrt{\surd 3}}{2}$

Solve for $𝑛$:

$𝑛=\frac{\sqrt{\surd 3}}{2\times 0.5}$

$𝑛=\sqrt{\surd 3}$

Calculating:

$𝑛=1.732$

Thus, the refractive index of the prism is

Option A: 1.732

The absorption coefficient of a body, denoted by $𝑎$, is a measure of the amount of incident heat that is absorbed by the body. The absorption coefficient can be calculated using the formula:

$𝑎=\frac{\text{absorbed heat}}{\text{incident heat}}$

Given that the incident heat is $𝑥$ joules and the heat reflected and transmitted is $𝑦$ joules, the heat absorbed by the body can be expressed as:

$\text{absorbed heat}=𝑥-𝑦$

This is because the total heat accounted for includes reflected, transmitted, and absorbed heat, and it must equal the incident heat. Therefore, the absorption coefficient is:

$𝑎=\frac{𝑥-𝑦}{𝑥}$

Thus, the correct option is:

Option C

$\frac{𝑥-𝑦}{𝑥}$