KCET 2020 Question Paper with Solutions

The value of acceleration due to gravity at height $(ℎ=10\mathrm{km})$ is $𝑥$.

$\begin{array}{cc}& \text{ i.e., }{𝑔}_{ℎ}=𝑥\\ & \Rightarrow \frac{𝑔}{{(1+\frac{ℎ}{{𝑅}_{𝑒}})}^2}=𝑥\\ & \Rightarrow 𝑔(1-\frac{2ℎ}{{𝑅}_{𝑒}})=𝑥\text{.... (i)}\end{array}$

The value of acceleration due to gravity at depth $𝑑$ below the surface of earth,

$\begin{array}{cc}{𝑔}_{𝑑}& =𝑥\\ \Rightarrow 𝑔(1-\frac{𝑑}{{𝑅}_{𝑒}})& =𝑥\text{.... (ii)}\end{array}$

From Eqs. (i) and (ii), we have

$\begin{array}{cc}\Rightarrow & 𝑔(1-\frac{2ℎ}{{𝑅}_{𝑒}})=𝑔(1-\frac{𝑑}{{𝑅}_{𝑒}})\\ \Rightarrow & 1-\frac{2ℎ}{{𝑅}_{𝑒}}=1-\frac{𝑑}{{𝑅}_{𝑒}}\Rightarrow \frac{2ℎ}{{𝑅}_{𝑒}}=\frac{𝑑}{{𝑅}_{𝑒}}\\ \Rightarrow & 𝑑=2ℎ=2\times 10=20\mathrm{km}\end{array}$

For a perfect rigid body, strain is always zero, for all values of stress.

As, Young's modulus, $𝑌=\frac{\text{ Normal stress }}{\text{ Longitudinal strain }}$

So, its Young's modulus would be infinity.

Initial angular velocity of wheel, ${𝜔}_0=0$

Final angular velocity, $𝜔=10\mathrm{rad}/s$

$𝑡=5s$

By first equation of rotational motion,

$\begin{array}{cc}& 𝜔={𝜔}_0+𝛼𝑡\\ \Rightarrow & 10=0+𝛼\times 5\\ \Rightarrow & 𝛼=\frac{10}{5}=2\mathrm{rad}/s^2\end{array}$

If $𝜃$ be the total angle through which wheel has turned, then from,

$\begin{array}{cc}𝜃& ={𝜔}_0𝑡+\frac{1}{2}𝛼{𝑡}^2\\ & =0\times 5+\frac{1}{2}\times 2\times 5^2=0+25\\ \Rightarrow 𝜃& =25\mathrm{rad}\end{array}$

Density of ice, ${𝜌}_{𝑖}=0.917g{\mathrm{cm}}^{-3}$

Fraction of the volume of iceberg submerged in water is given as

$\frac{\text{ Volume of iceberg submerged in water }}{\text{ Volume of iceberg }}$

$=\frac{\text{ density of ice }}{\text{ density of water }}=\frac{0.917g{\mathrm{cm}}^{-3}}{1g{\mathrm{cm}}^{-3}}=0.917$

The rate of heat transfer is directly proportional to the surface area of body. Since, circular plate has greatest surface area and sphere has least surface area amongst the given three bodies. Hence, when a sphere, a cube and a thin circular plate are allowed to cool down under similar conditions, then the plate will cool the fastest and sphere the slowest.

In an adiabatic expansion, internal energy decreases and hence temperature also decreases. Hence, from ideal gas equation,

$𝑝𝑉=𝑛𝑅𝑇$

Since, $𝑇$ decreases.

Hence, $𝑝𝑉$ also decreases.

Suppose Q amount of heat is supplied to monoatomic gas which expands at constant pressure. This heat is then converted into the internal energy and work. According to the first law of thermodynamics,

$\begin{array}{cc}𝑄& =𝑈+𝑊\\ \text{ or }𝑛{𝐶}_{𝑝}\Delta 𝑇& =𝑛{𝐶}_{𝑉}\Delta 𝑇+𝑊\end{array}$

where, ${𝐶}_{𝑝}$ and ${𝐶}_{𝑉}$ are the specific heat capacity at constant pressure and volume, respectively.

$\therefore$ The fraction of heat converted to work is given as

$\frac{𝑊}{𝑄}=\frac{𝑛{𝐶}_{𝑝}\Delta 𝑇-𝑛{𝐶}_{𝑉}\Delta 𝑇}{𝑛{𝐶}_{𝑝}\Delta 𝑇}$

For monoatomic gas, $𝑛=1$

$\begin{array}{cc}\Rightarrow \frac{𝑊}{𝑄}=\frac{{𝐶}_{𝑝}-{𝐶}_{𝑉}}{{𝐶}_{𝑝}}& =\frac{\frac{5}{2}𝑅-\frac{3}{2}𝑅}{\frac{5}{2}𝑅}\\ & =\frac{𝑅}{\frac{5}{2}𝑅}=\frac{2}{5}\end{array}$

Mass of tray, $𝑚=12\mathrm{kg}$

Time period, $𝑇=1.5s$

If $𝑘$ be the spring constant of each spring, then

${𝑘}_1={𝑘}_2=𝑘$

since, springs are connected in parallel, hence

${𝑘}_{\text{baA }}=𝑘+𝑘$

Time period, $𝑇=2𝜋\sqrt{\surd \frac{𝑚}{{𝑘}_{\text{net }}}}=2𝜋\sqrt{\surd \frac{𝑚}{{𝑘}_1+{𝑘}_2}}$

$\begin{array}{cc}\Rightarrow & 1.5=2𝜋\sqrt{\surd \frac{12}{𝑘+𝑘}}\Rightarrow 1.5=2𝜋\sqrt{\surd \frac{12}{2𝑘}}\\ \Rightarrow & 1.5=2𝜋\sqrt{\surd \frac{6}{𝑘}}\end{array}$

Squaring both side, we have

$\begin{array}{cc}2.25& =4{𝜋}^2\cdot \frac{6}{𝑘}\\ \Rightarrow 𝑘& =\frac{4{𝜋}^2\times 6}{2.25}\\ & =105.17{\mathrm{Nm}}^{-1}\approx 105{\mathrm{Nm}}^{-1}\end{array}$

Since, the observer is stationary, so when train is approaching towards the observer or moving towards the observer, apparent frequency is given as,

${𝑛}^{\prime }=𝑛\left(\frac{𝑣}{𝑣-{𝑣}_{𝑠}}\right)$

where, ${𝑣}_{𝑠}$ is the speed of sound.

It is clear from the above expression that ${𝑛}^{\prime }>𝑛$.

However, when the train is going away from the observer, then apparent frequency,

$\begin{array}{cc}& {𝑛}^{\prime }=𝑛\left(\frac{𝑣}{𝑣+{𝑣}_{𝑠}}\right)\\ \therefore & {𝑛}^{\prime }<𝑛\end{array}$

Hence, graph represented in option (d) is correct.

If charge is placed at the one corner of a cube, then it can be shared further by 8 cubes.

$\therefore$ Charge on one cube $=\frac{𝑞}{8}$

Now, in any given cube this charge will be touching 3 out of 9 of cube faces. So, the area vector of that side and the electric field vector will be perpendicular. So, flux through those 3 sides will be zero. However, equal amount of flux will flow from other 3 sides.

So, flux through one side $=\left(\frac{𝑞/8}{3}\right)\cdot \frac{1}{{𝜀}_0}=\frac{𝑞}{24{𝜀}_0}$

Hence, flux through one face $\left(𝐴𝐵𝐶𝐷\right)=\frac{𝑞}{24{𝜀}_0}$

According to given figure,

Electric field at point $𝐴,{𝐸}_{𝐴}=40{\mathrm{Vm}}^{-1}$

Since, electric field lines on the left have twice the separation on those on the right (at point $𝐵$), hence electric field at point $𝐵$

$\begin{array}{cc}& {𝐸}_{𝐵}=\frac{{𝐸}_{𝐴}}{2}=\frac{40}{2}\\ & {𝐸}_{𝐵}=20{\mathrm{Vm}}^{-1}\end{array}$

Force on charge $𝑞$ kept at $𝐵$ is

$\begin{array}{cc}& 𝐹=𝑞{𝐸}_{𝐵}\\ \text{ Given, }& 𝑞=20𝜇C=20\times {10}^{-6}C\\ \Rightarrow & 𝐹=20\times {10}^{-6}\times 20=4\times {10}^{-4}N\end{array}$

Given, charge density of uniformly long thin wire,

$\begin{array}{cc}𝜆& =\frac{1}{4}\times {10}^{-2}{\mathrm{cm}}^{-1}\\ & =\frac{1}{4}\times {10}^{-2}\times {10}^2{m}^{-1}=\frac{1}{4}{m}^{-1}\\ 𝑟& =20\mathrm{cm}=0.2m\end{array}$

Electric field at a distance $𝑟$ from the axis of wire is given as

$\begin{array}{cc}𝐸& =\frac{𝜆}{2𝜋{𝜀}_0𝑟}=\frac{2𝜆}{4𝜋{𝜀}_0𝑟}\\ & =\frac{9\times {10}^9\times 2\times \frac{1}{4}}{0.2}\\ & =2.25\times {10}^{10}{\mathrm{NC}}^{-1}\end{array}$

No given option is correct.

Torque on electric dipole placed in uniform electric field $𝐸$,

$𝜏=𝑝𝐸\sin 𝜃\text{.... (i)}$

where, $𝑝=$ electric dipole moment,

For small angle $𝜃,\sin 𝜃\approx 𝜃$

$\therefore$ From Eq. (i), we have

$𝜏=𝑝𝐸𝜃\text{.... (ii)}$

but $𝜏=𝐼\times 𝛼\text{.... (iii)}$

where $𝛼$ is angular acceleration.

From Eqs. (ii) and (iii), we have

$𝐼𝛼=𝑝𝐸𝜃$

$𝛼=\frac{𝑝𝐸}{𝐼}\cdot 𝜃\Rightarrow \frac{𝜃}{𝛼}=\frac{𝐼}{𝑝𝐸}\text{..... (iv)}$

$\therefore$ Time period for the oscillation of dipole,

$𝑇=2𝜋\sqrt{\surd \frac{𝜃}{𝛼}}=2𝜋\sqrt{\surd \frac{𝐼}{𝑝𝐸}}$

Given, ${𝐶}_1={𝐶}_2=𝐶$ (identical capacitors)

When both capacitors are connected in parallel, then equivalent capacitance

$\begin{array}{cc}& {𝐶}_{𝑝}={𝐶}_1+{𝐶}_2=𝐶+𝐶\\ & {𝐶}_{𝑝}=2𝐶\text{.... (i)}\end{array}$

When both capacitors are connected in series, then equivalent capacitance,

$\begin{array}{cc}\frac{1}{{𝐶}_{𝑠}}& =\frac{1}{{𝐶}_1}+\frac{1}{{𝐶}_2}=\frac{1}{𝐶}+\frac{1}{𝐶}\\ \Rightarrow & \frac{1}{{𝐶}_{𝑠}}=\frac{2}{𝐶}\Rightarrow {𝐶}_{𝑠}=\frac{𝐶}{2}\end{array}$

According to question,

$\begin{array}{cc}{𝐶}_{𝑝}-{𝐶}_{𝑠}& =6𝜇F\\ \Rightarrow 2𝐶-\frac{𝐶}{2}& =6𝜇F\Rightarrow \frac{3𝐶}{2}=6𝜇F\\ \Rightarrow 𝐶& =4𝜇F\end{array}$

According to given figure, line $𝐴𝐵$ is perpendicular to direction of electric field lines. Hence, surface passing through line $𝐴𝐵$ and perpendicular to electric field lines behaves like a equipotential surface, therefore

${𝑉}_{𝐴}={𝑉}_{𝐵}\text{..... (i)}$

Electric field and electric potential are related as

$𝐸=-\frac{𝑑𝑉}{𝑑𝑥}\Rightarrow 𝑉=-\int 𝐸𝑑𝑥$

Which indicates that electric potential decreases in the direction of electric field, i.e.

${𝑉}_{𝐵}>{𝑉}_{𝐶}\text{... (ii)}$

From Eqs. (i) and (ii), we have

${𝑉}_{𝐴}={𝑉}_{𝐵}>{𝑉}_{𝐶}$

When a soap bubble is charged, then the bubble expands because the charged particles uniformly distributed on it causes them to repel each other due to the electrostatic force. Hence, radius of soap bubble increases.

Given, anode potential, $𝑉=1200V$

Electron will accelerate with the effect of anode potential.

Hence, $\frac{1}{2}𝑚{𝑣}^2=\mathrm{eV}$

$\begin{array}{cc}𝑣& =\sqrt{\surd \frac{2\mathrm{eV}}{𝑚}}=\sqrt{\surd \frac{2\times 1.6\times {10}^{-19}\times 1200}{9.1\times {10}^{-31}}}\\ & =\sqrt{\surd 421.98\times {10}^{12}}=20.5\times {10}^6{\mathrm{ms}}^{-1}\\ & =2.05\times {10}^7{\mathrm{ms}}^{-1}\simeq 2.1\times {10}^7{\mathrm{ms}}^{-1}\end{array}$

Given, length, $𝑙=10\mathrm{cm}$

Resistance of metal rod.

$𝑅=𝜌\cdot \frac{𝑙}{𝐴}$

i.e., $𝑅\propto \frac{1}{𝐴}$

Since, among given faces in the options, area is minimum corresponding to face $1\mathrm{cm}\times \frac{1}{2}\mathrm{cm}$.

Hence, resistance will be maximum when the battery is connected across the face of dimension $1\mathrm{cm}\times \frac{1}{2}\mathrm{cm}$.

Given, emf of battery, $𝐸=12V$

Internal resistance, $𝑟=2\times {10}^{-2}\Omega$

Current, $𝐼=80A$

$\therefore$ Terminal voltage when starter is $\mathrm{ON}$ is given as

$\begin{array}{cc}𝑉& =𝐸-𝐼𝑟\\ & =12-80\times 2\times {10}^{-2}\\ & =12-1.6=10.4V\end{array}$

No given option is correct.

Given, length of potentiometer wire

$𝑙=5m$

emf of battery, $𝐸=10V$

Potential gradient across the potentiometer wire,

$𝐾=\frac{𝐸}{𝑙}=\frac{10}{5}=2{\mathrm{Vm}}^{-1}$

When secondary cell is connected to the circuit of potentiometer, then balancing length,

${𝑙}_1=200\mathrm{cm}=2m$

$\therefore$ emf of the secondary cell, ${𝐸}_{𝑠}=𝐾{𝑙}_1=2\times 2=4V$

Resistance of carbon resistor,

$\begin{array}{cc}𝑅& =0.2k\Omega \pm 10\%\\ & =200\Omega \pm 10\%\\ & =20\times {10}^1\Omega \pm 10\%\end{array}$

According to the colour code for carbon resistor, the colour assigned to number 2, 0 and 1 are red, black and brown. For colour used for tolerance of $10\%$ is silver.

$\therefore$ colour code are red, black, brown and silver. No given option is correct.

The given network of resistances can be shown as follows,

Let $6𝐼$ be the current through the cell, which is distributed as shown in the figure.

Applying Kirchhoff's law to the loop $𝐴𝐵𝐶{𝐶}^{\prime }𝐸𝐴$, we have

$\begin{array}{cc}& -2𝐼𝑟-𝐼𝑟-2𝐼𝑟+𝐸=0\\ & \Rightarrow 𝐸=5𝐼𝑟\\ & \text{ but }𝑟=1\Omega \text{ given, }\\ \therefore & 𝐸=5𝐼\end{array}$

If ${𝑅}^{\prime }$ be the equivalent resistance of the network, then

${𝑅}^{\prime }=\frac{\text{ total emf }}{\text{ total current }}=\frac{5𝐼}{6𝐼}=\frac{5}{6}\Omega$

$𝐼-𝑉$ characteristics of copper wire of length $𝐿$ and area of cross-section $𝐴$ is shown in the figure.

Since, $𝐼-𝑉$ characteristic is a straight line, hence it obeys Ohm's law. According to which resistance of copper wire,

$\begin{array}{cc}& 𝑅=\frac{𝑉}{𝐼}=\cot 𝜃\\ & \Rightarrow 𝑅=\cot 𝜃\\ & \Rightarrow 𝜌\cdot \frac{𝑙}{𝐴}=\cot 𝜃\end{array}$

or $\tan 𝜃\propto \left(\frac{𝐴}{𝜌𝑙}\right)$

Hence, slope $(\tan 𝜃)$ of the curve becomes

(i) less if the length of the wire is increased.

(ii) more if the area of the wire is increased.

(iii) less if a wire of steel of same dimensions is used as ${𝜌}_{\text{steel }}>{𝜌}_{\text{copper }}$.

(iv) less if experiment is performed at higher temperature.

According to the figure, magnetic field at point is given as

${𝐵}_{\text{net }}=$ magnetic field due to straight wire $𝐴𝐵+$ magnetic field due to straight wire $𝐶𝐷+$ magnetic field due circular wire $𝐵𝐶$

Since, point $𝑂$ is along the length of the wire $𝐵𝐴$. So, ${𝐵}_{𝐴𝐵}=0$

$\begin{array}{cc}{𝐵}_{\text{net }}& =0+\frac{{𝜇}_0}{4𝜋}\cdot \frac{𝐼}{𝑟}+\frac{270}{360}\left(\frac{{𝜇}_0𝐼}{2𝑟}\right)\\ & =\frac{{𝜇}_0𝐼}{4𝜋𝑟}+\frac{3}{4}\cdot \frac{{𝜇}_0𝐼}{2𝑟}\\ & =\frac{{𝜇}_0𝐼}{4𝜋𝑟}+\frac{3{𝜇}_0𝐼}{8𝑟}\text{ (downward) }\end{array}$

According to Biot-Savart's law, magnetic field at the origin due to a current element $𝑖𝑑l$ placed at a point with position vector $𝐫$ is given as.

$𝐵=\frac{{𝜇}_0𝑖}{4𝜋}\cdot \frac{𝑑𝐥\times 𝐫}{{𝑟}^3}$

Magnetic field due to long cylindrical wire of radius $𝑅$ inside the cylinder at a distance $r<r$< p="">

$\begin{array}{cc}& {𝐵}_{\text{in }}=\frac{{𝜇}_0}{2𝜋}\cdot \frac{𝐼𝑟}{{𝑅}^2}\text{ [where, }𝐼=\text{ current] }\\ \therefore & {𝐵}_{\text{in }}\propto 𝑟\dots \left(i\right)\end{array}$

Magnetic field outside the current carrying cylindrical wire of radius $𝑅$ at a distance $𝑟>𝑅$ is given as

${𝐵}_{\text{out }}=\frac{{𝜇}_0𝐼}{2𝜋𝑟}$

i.e. ${𝐵}_{\text{out }}\propto \frac{1}{𝑟}\text{... (ii)}$

From Eqs. (i) and (ii), graph between $𝐵$ and $𝑟$ drawn as

When a charged particle is accelerated by a potential difference $𝑉$ volts, then the radius of the trajectory of the path of the particle is given as,

$\begin{array}{cc}𝑟& =\frac{\sqrt{\surd 2𝐾𝑚}}{𝐵𝑞}=\frac{\sqrt{\surd 2𝑞𝑉𝑚}}{𝐵𝑞}\\ \Rightarrow 𝐾& =\frac{1}{2}\frac{{𝐵}^2{𝑞}^2{𝑟}^2}{𝑚}\text{ or }𝐾\propto \frac{{𝑞}^2}{𝑚}\text{.... (i)}\end{array}$

For protons $\left({}_1^{1}H\right)$, deuterons $\left({}_2^{1}H\right)$ and $𝛼$-particles $\left({}_4^2\mathrm{He}\right)$,

$\begin{array}{cc}{𝑞}_{𝑝}:{𝑞}_{𝑑}:{𝑞}_{𝛼}& =1:1:2\text{.... (ii)}\\ {𝑚}_{𝑝}:{𝑚}_{𝑑}:{𝑚}_{𝛼}& =1:2:4\text{.... (iii)}\end{array}$

From Eqs. (i), (ii) and (iii), we can write

$\Rightarrow {𝐾}_{𝑝}:{𝐾}_{𝑑}:{𝐾}_{𝛼}=1:\frac{1}{4}:1$

Thus, minimum kinetic energy is gained by deuterons.

Given, ${𝑀}_1=8{\mathrm{Am}}^{-1},{𝐵}_1=0.6T,{𝑇}_1=4K,{𝐵}_2=0.2T$ and ${𝑇}_2=16K$

According to Curie's law, for paramagnetic materials,

$\begin{array}{cc}& 𝑀=\frac{𝐶𝐵}{𝑇}\\ & \Rightarrow {𝑀}_1=\frac{𝐶{𝐵}_1}{{𝑇}_1}\\ & \text{ and }{𝑀}_2=\frac{𝐶{𝐵}_2}{{𝑇}_2}\\ & \therefore \frac{{𝑀}_2}{{𝑀}_1}\frac{\frac{𝐶{𝐵}_2}{{𝑇}_2}}{\frac{𝐶{𝐵}_1}{{𝑇}_1}}=\frac{{𝐵}_2}{{𝐵}_1}\times \frac{{𝑇}_1}{{𝑇}_2}\\ & =\frac{0.2}{0.6}\times \frac{4K}{16K}=\frac{1}{3}\times \frac{1}{4}\\ & \Rightarrow \frac{{𝑀}_2}{{𝑀}_1}=\frac{1}{12}\Rightarrow {𝑀}_2=\frac{{𝑀}_1}{12}=\frac{8}{12}\\ & \Rightarrow {𝑀}_2=\frac{2}{3}{\mathrm{Am}}^{-1}\end{array}$

Magnetic field at the centre of current carrying circular coil

$𝐵=\frac{{𝜇}_0𝐼}{2𝑟}$

where, $𝐼=$ current and $𝑟=$ radius.

Magnetic dipole moment,

$\begin{array}{cc}& 𝑚=𝐼𝐴=𝐼\cdot 𝜋{𝑟}^2\\ \text{ Given, }& \frac{𝐵}{𝑚}=𝑥\\ \Rightarrow & \frac{\frac{{𝜇}_0𝐼}{2𝑟}}{𝜋𝐼{𝑟}^2}=𝑥\\ \Rightarrow & \frac{{𝜇}_0}{2𝜋{𝑟}^3}=𝑥\text{.... (i)}\end{array}$

When current and radius, both are doubled, then

$\begin{array}{cc}\frac{{𝐵}^{\prime }}{{𝑚}^{\prime }}& =\frac{\frac{{𝜇}_{0}2𝐼}{2\cdot 2𝑟}}{𝜋\left(2𝐼\right)(2𝑟{)}^2}\\ & =\frac{\frac{{𝜇}_0𝐼}{2𝑟}}{8𝜋{𝑟}^2𝐼}=\frac{{𝜇}_0}{16𝜋{𝑟}^3}\\ & =\frac{1}{8}\cdot \frac{{𝜇}_0}{2𝜋{𝑟}^3}=\frac{𝑥}{8}\text{ [from Eq. (i)] }\end{array}$

At room temperature, all domains of permanent magnet are perfectly aligned like ferromagnetic material.

Given, $𝑙=2m,𝑣=5m/s,𝐵=0.04T$ and $𝑅=3\Omega$

The current induced in a conducting coil, when moved in a uniform magnetic field is

$𝐼=\frac{𝐵𝑙𝑣}{𝑅}=\frac{0.04\times 2\times 5}{3}=0.133A=133\mathrm{mA}$

Given, $𝐿=0.2H,{𝐼}_1=5A,{𝐼}_2=2A$ and $\Delta 𝑡=0.5s$

The magnitude of induced emf, $\left|𝑒\right|=𝐿\frac{𝑑𝐼}{𝑑𝑡}$

$=𝐿\frac{\Delta 𝐼}{\Delta 𝑡}=\frac{0.2\times (5-2)}{0.5}=1.2V$

Given, ${𝑉}_{𝐶}=30V,{𝑉}_{𝐿}=110V$ and ${𝑉}_{𝑅}=60V$

The peak voltage across series $𝐿-𝐶-𝑅$ circuit is given by

$\begin{array}{cc}{𝑉}_{𝑂}& =\sqrt{\surd {𝑉}_2^{𝑅}+{({𝑉}_{𝐿}-{𝑉}_{𝐶})}^2}\\ & =\sqrt{\surd (60{)}^2+(110-30{)}^2}\\ & =\sqrt{\surd (60{)}^2+(80{)}^2}=100V\end{array}$

$\therefore$ Rms value of voltage, ${𝑉}_{\mathrm{rms}}=\frac{{𝑉}_0}{\sqrt{\surd 2}}=\frac{100}{\sqrt{\surd 2}}=70.7V$

Given, power factor $=\frac{1}{\sqrt{\surd 3}}$

Inductive reactance, ${𝑋}_{𝐿}=2\Omega$

The power factor in a $𝑅$-$𝐿$ circuit is given by

power factor $=\frac{\text{ Resistance }}{\text{ Impedance }}=\frac{𝑅}{\sqrt{\surd {𝑅}^2+{𝑋}_2^{𝐿}}}$

$\Rightarrow \frac{1}{\sqrt{\surd 3}}=\frac{𝑅}{\sqrt{\surd {𝑅}^2+4}}$

Squaring both sides, we get

$\begin{array}{cc}\frac{1}{3}& =\frac{{𝑅}^2}{{𝑅}^2+4}\\ \Rightarrow {𝑅}^2+4& =3{𝑅}^2\Rightarrow 𝑅=\sqrt{\surd 2}\Omega \end{array}$

Given, $𝐿=0.5\mathrm{mH}=0.5\times {10}^{-3}H$

and $𝐶=20𝜇F=20\times {10}^{-6}F$

The resonant frequency of $𝐿-𝐶$ circuit is

$\begin{array}{cc}& 𝑓=\frac{1}{2𝜋\sqrt{\surd 𝐿𝐶}}=\frac{1}{2𝜋\sqrt{\surd 0.5\times {10}^{-3}\times 20\times {10}^{-6}}}\\ & =\frac{1}{2𝜋\times \sqrt{\surd 10\times {10}^{-9}}}=\frac{{10}^4}{2𝜋}=1592.3\simeq 1592\mathrm{Hz}\end{array}$

Given, intensity, $𝐼=20W/{\mathrm{cm}}^2$

Area of surface, $𝐴=(25\times 15){\mathrm{cm}}^2$

and time, $𝑡=1s$

The intensity of light of energy $𝐸$ is given by

$\begin{array}{cc}& 𝐼=\frac{𝐸}{𝐴𝑡}\Rightarrow 20=\frac{𝐸}{25\times 15\times 1}\\ \Rightarrow & =20\times 25\times 15J\end{array}$

The momentum imparted to a perfectly reflecting surface is given by

$𝑝=\frac{2𝐸}{𝑐}=\frac{2\times 20\times 25\times 15}{3\times {10}^8}=5\times {10}^{-5}\mathrm{kg}{\mathrm{ms}}^{-1}$

To understand the behavior of the image as an object approaches a converging lens, we begin with the lens formula:

$\frac{1}{𝑓}=\frac{1}{𝑣}-\frac{1}{𝑢}$

where $𝑓$ is the focal length, $𝑣$ is the image distance, and $𝑢$ is the object distance.

By differentiating this equation with respect to time, we obtain:

$0=-\frac{1}{{𝑣}^2}\frac{𝑑𝑣}{𝑑𝑡}+\frac{1}{{𝑢}^2}\frac{𝑑𝑢}{𝑑𝑡}$

This implies:

$\frac{𝑑𝑣}{𝑑𝑡}=\left(\frac{{𝑣}^2}{{𝑢}^2}\right)\frac{𝑑𝑢}{𝑑𝑡}$

Here, $\frac{𝑑𝑣}{𝑑𝑡}$ is the speed of the image (${𝑣}_{𝑖}$), and $\frac{𝑑𝑢}{𝑑𝑡}$ is the speed of the object (${𝑣}_0=5\text{m/s}$). Additionally, $𝑚$ represents the magnification. So, we derive:

${𝑣}_{𝑖}={𝑚}^2{𝑣}_0\text{.... (i)}$

For acceleration, we differentiate with respect to time again:

$\frac{{𝑑}^2{𝑣}_{𝑖}}{𝑑{𝑡}^2}={𝑣}_0\cdot 2𝑚\cdot \frac{𝑑𝑚}{𝑑𝑡}$

This gives:

${𝑎}_{𝑖}=2𝑚{𝑣}_0\frac{𝑑𝑚}{𝑑𝑡}\text{.... (ii)}$

where ${𝑎}_{𝑖}$ is the acceleration of the image.

When the object approaches the lens with a constant speed of $5\text{m/s}$ and stops at the focus, equations (i) and (ii) indicate that the image will move away from the lens with a non-uniform acceleration. The variability in acceleration arises because $\frac{𝑑𝑚}{𝑑𝑡}$ changes as the object moves, resulting in non-uniform motion of the image.

Given:

Angle of the prism $𝐴$

Refractive index of the prism material $𝜇=\cot \frac{𝐴}{2}$

We know that the refractive index $𝜇$ is related to the angle of minimum deviation ${𝛿}_{𝑚}$ as follows:

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

Substitute the given refractive index:

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

This simplifies to:

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

Therefore:

$\sin ({90}^{\circ }-\frac{𝐴}{2})=\sin \left(\frac{𝐴+{𝛿}_{𝑚}}{2}\right)$

Equating the angles, we get:

$\frac{{180}^{\circ }-𝐴}{2}=\frac{𝐴+{𝛿}_{𝑚}}{2}$

Solving for ${𝛿}_{𝑚}$ gives:

${𝛿}_{𝑚}={180}^{\circ }-2𝐴$

When the beam of light converges at point $𝑃$, it acts as a virtual object for the concave lens.

To determine the convergence distance $𝑥$ from the lens:

The object distance $𝑢$ is given as $12\text{cm}$.

The focal length $𝑓$ of the concave lens is $-16\text{cm}$ (negative because it's a concave lens).

Using the lens formula:

$\frac{1}{𝑣}-\frac{1}{𝑢}=\frac{1}{𝑓}$

Substitute the known values into the formula:

$\frac{1}{𝑣}=\frac{1}{𝑓}+\frac{1}{𝑢}=-\frac{1}{16}+\frac{1}{12}=\frac{-3+4}{48}=\frac{1}{48}$