JEE Advanced 2013 Paper 2 Question Paper with Solutions

In alpha decay, atomic number decreases by 2 and mass number decreases by 4.

P $\to$ 2

In $\beta$⁺ decay, atomic number decreases by 1 and mass number remains unchanged.

Q $\to$ 1

In fission, heavy nucleus breaks into two light nuclei.

R $\to$ 4

In proton emission, both atomic number and mass number decreases by 1.

S $\to$ 3

Let's break down the dimensional analysis of each physical quantity:

P. Boltzmann Constant (${𝑘}_{𝐵}$)

The Boltzmann constant relates energy to temperature. Its dimensional formula is derived from the relationship: $𝐸={𝑘}_{𝐵}𝑇$.

Energy ($𝐸$) has dimensions of [ML²T^-2]. Temperature ($𝑇$) has dimensions of [K]. Therefore:

$\left[{𝑘}_{𝐵}\right]=\frac{\left[𝐸\right]}{\left[𝑇\right]}=\frac{\left[𝑀{𝐿}^2{𝑇}^{-2}\right]}{\left[𝐾\right]}=\left[𝑀{𝐿}^2{𝑇}^{-2}{𝐾}^{-1}\right]$

Q. Coefficient of Viscosity ($𝜂$)

Viscosity is a measure of a fluid's resistance to flow. It is defined as the ratio of shear stress to shear rate. Shear stress has dimensions of [ML^-1T^-2] (force per unit area), and shear rate has dimensions of [T^-1] (velocity gradient). Therefore:

$\left[𝜂\right]=\frac{\left[𝑀{𝐿}^{-1}{𝑇}^{-2}\right]}{\left[{𝑇}^{-1}\right]}=\left[𝑀{𝐿}^{-1}{𝑇}^{-1}\right]$

R. Plank Constant ($ℎ$)

Planck's constant relates energy to frequency. Its dimensional formula is derived from the relationship: $𝐸=ℎ𝜈$.

Energy ($𝐸$) has dimensions of [ML²T^-2]. Frequency ($𝜈$) has dimensions of [T^-1]. Therefore:

$\left[ℎ\right]=\frac{\left[𝐸\right]}{\left[𝜈\right]}=\frac{\left[𝑀{𝐿}^2{𝑇}^{-2}\right]}{\left[{𝑇}^{-1}\right]}=\left[𝑀{𝐿}^2{𝑇}^{-1}\right]$

S. Thermal Conductivity ($𝑘$)

Thermal conductivity is a measure of a material's ability to conduct heat. It is defined as the rate of heat transfer per unit area per unit temperature gradient. Heat transfer rate has dimensions of [ML²T^-3]. Area has dimensions of [L²], and temperature gradient has dimensions of [K/L]. Therefore:

$\left[𝑘\right]=\frac{\left[𝑀{𝐿}^2{𝑇}^{-3}\right]}{\left[{𝐿}^2\right][𝐾/𝐿]}=\left[𝑀𝐿{𝑇}^{-3}{𝐾}^{-1}\right]$

Matching the Dimensions:

Based on the dimensional analysis, we can match the quantities in List I with the dimensions in List II as follows:

Therefore, the correct answer is Option C.

Here we will use the concept of vectors and concept of electric field.

Let's take a point P in the overlapping region.

From $\Delta 𝑂𝑃𝑄$,

By triangle law of vector addition,

$\overrightarrow{{𝑟}_1}-\overrightarrow{{𝑟}_2}=\overrightarrow{𝑑}\text{... (1)}$

Net electrostatic field at point P;

${\overrightarrow{𝐸}}_{𝑛𝑒𝑡}=\frac{𝐾{𝑞}_1}{{𝑟}_3^1}\overrightarrow{{𝑟}_1}=\frac{𝐾{𝑞}_2}{{𝑟}_3^2}\overrightarrow{{𝑟}_2}$

$\Rightarrow {\overrightarrow{𝐸}}_{𝑛𝑒𝑡}=\frac{𝐾\left(𝜌\frac{4}{3}𝜋{𝑟}_3^1\right)\overrightarrow{{𝑟}_1}}{{𝑟}_3^1}-\frac{𝐾\left(𝜌\frac{4}{3}𝜋{𝑟}_3^2\right)\overrightarrow{{𝑟}_2}}{{𝑟}_3^2}$

$\Rightarrow {\overrightarrow{𝐸}}_{𝑛𝑒𝑡}=\frac{4}{3}𝜌𝜋\times \frac{1}{4𝜋{𝜀}_0}\left(\overrightarrow{{𝑟}_1}-\overrightarrow{{𝑟}_2}\right)$

$\Rightarrow {\overrightarrow{𝐸}}_{𝑛𝑒𝑡}=\frac{𝜌}{3{𝜀}_0}\overrightarrow{𝑑}$

Hence, the electrostatic field is constant in magnitude and has same direction.

Therefore, options (C) and (D) are correct.

P = Vi

$\therefore$ $i = {P \over V} = {{600 \times {{10}^3}} \over {4000}} = 150$ A

Total resistance of cables,

R = 0.4 $\times$ 20 = 8$\Omega$

$\therefore$ Power loss in cables = i²R

= (150)² (8)

= 180000 W = 180 kW

This loss is 30% of 600 kW.

From the given figure, it can be concluded that process FG is isothermal and process FH is adiabatic.

${\gamma _{mono}} = 1 + {2 \over f} = {5 \over 3}$

${P_0}V_G^{5/3} = 32{P_0}V_0^{5/3}$

${V_G} = {(32)^{3/5}}{V_0} = 8{V_0}$

$\Delta {W_{GE}} = {P_0}({V_0} - 32{V_0}) = - 31{P_0}{V_0}$

Thus, the correct mapping is (P) $\to$ (4).

$\Delta {W_{GH}} = {P_0}(8{V_0} - 32{V_0}) = - 24{P_0}{V_0}$

Thus, the correct mapping is (Q) $\to$ (3).

$\Delta {W_{FH}} = {{{P_0}(8{V_0}) - 32{P_0}{V_0}} \over {1 - (5/3)}} = {{ - 24{P_0}{V_0}} \over { - (2/3)}} = 36{P_0}{V_0}$

Thus the correct mapping is (R) $\to$ (2).

$\Delta {W_{FG}} = 32R{T_0}\ln 32 = 32R{T_0}\ln {2^5} = 160R{T_0}\ln 2 = 160{P_0}{V_0}\ln 2$

Thus, the correct mapping is (S) $\to$ (1).

The alpha decay of $_{84}^{210}Po$ is given by

$_{84}^{210}Po \to _{82}^{206}Pb + _2^4He$

The energy released during this process is

$Q = ({M_{Po}} - {M_{Pb}} - {M_{He}}){c^2}$

$= (209.982876 - 205.974455 - 4.002603)\,u \times {c^2}$

$= (0.005818\,u){c^2} = (0.005818\,u) \times 932\,MeV$

$= 5.422\,MeV = 5422\,keV$

Kinetic energy of a particle, ${K_\alpha } = {{(A - 4)Q} \over A}$

${K_\alpha } = {{(210 - 4)} \over {210}} \times 5422\,keV = {{206} \over {210}} \times 5422\,keV = 5319\,keV$

For e $\to$ f

$\mu$₂ > $\mu$₁, as ray bends towards the normal.

$\mu$₂ > $\mu$₃, as ray bends away from the normal.

P $\to$ 2

For e $\to$ g

$\mu$₁ = $\mu$₂ as there is no deviation.

Q $\to$ 3

For e $\to$ h

$\mu$₂ < $\mu$₁, as ray bends away from the normal.

$\mu$₂ > $\mu$₃, as ray bends away from the normal.

Also, $\mu$₁ < $\sqrt2$$\mu$₂ (No total internal reflection)

R $\to$ 4

For e $\to$ i

Total internal reflection takes place

$\therefore$ sin45$^\circ$ > sinC

But $\sin C = {{{\mu _2}} \over {{\mu _1}}}$

$\therefore$ ${1 \over {\sqrt 2 }} > {{{\mu _2}} \over {{\mu _1}}} \Rightarrow {\mu _1} > \sqrt 2 {\mu _2}$

$S \to 1$

We have

$m(_1^2H) + m(_2^4He) = 2.014102 + 4.002603 = 6.016705\,u$

$m(_3^6Li) = 6.015123\,u$

${m_1} + {m_2} > M$

Thus, option (A) is wrong.

$m(_1^1H) + m(_{83}^{209}Bi) = 1.007825\,u + 208.980388\,u = 209.988213\,u$

$m(_{84}^{210}Po) = 209.982876\,u$

${m_1} + {m_2} > M$

Thus, option (B) is wrong.

$m(_1^2H) + m(_2^4He) = 2.014102\,u + 4.002603\,u = 6.016705\,u$

$_3^6Li = 6.015123\,u$

$({m_3} + {m_4}) > M'$

Thus, option (C) is correct. Therefore, deuteron and alpha particle can go complete fusion.

$m(_{30}^{70}Zn) + _{34}^{82}Se = 69.925325\,u + 81.916709\,u = 151.842034\,u$

$_{64}^{152}Gd = 151.919803\,u$

${m_3} + {m_4} < M'$

Thus, option (D) is wrong.

At point Q, the forces acting on the particle are normal reaction ${𝑁}_{𝑄}$, gravitational force $𝑚𝑔$ and frictional force $𝑓$. Resolve these in the directions parallel and perpendicular to the path. The net force towards the centre $\left(𝑃\right)$ provides the centripetal acceleration. Apply Newton's second law in a direction normal to the path to get

${𝑁}_{𝑄}-𝑚𝑔\sin {30}^{\circ }=\frac{𝑚{𝑣}^2}{𝑅}$

$\Rightarrow$ ${𝑁}_{𝑄}=𝑚𝑔\sin {30}^{\circ }+\frac{𝑚{𝑣}^2}{𝑅}$

$=1\times 10\times \frac{1}{2}+\frac{100}{40}$

$=5+\frac{5}{2}=\frac{15}{2}$

$\Rightarrow {𝑁}_{𝑄}=7.5$ N

Hence, option (A) is correct.

Here we will use the law of conservation of energy as there are no non-conservative force involved so the total energy of mass m remains constant.

For minimum initial velocity to escape the gravitational field, the velocity at infinity will be zero.

And we know PE at infinity is zero.

By energy conservation,

$𝑃{𝐸}_{𝑖}+𝐾{𝐸}_{𝑖}=𝑃{𝐸}_{𝑓}+𝐾{𝐸}_{𝑓}$

$\Rightarrow \frac{-𝐺𝑀𝑚}{𝐿}-\frac{𝐺𝑀𝑚}{𝐿}+\frac{1}{2}𝑚{𝑣}^2=0+\frac{1}{2}𝑚(0{)}^2$

$\Rightarrow -\frac{2𝐺𝑚}{𝐿}𝑀+\frac{1}{2}𝑚{𝑣}^2=0$

$\Rightarrow \frac{1}{2}{𝑉}^2=\frac{2𝐺𝑚}{𝐿}$

$\Rightarrow 𝑉=2\sqrt{\surd \frac{𝐺𝑀}{𝐿}}$

Hence, options (B) and (D) are correct.

To determine how the error in $d$ behaves as $\theta$ increases, let’s examine the given expression:

$2d\sin \theta = \lambda$

We can rearrange this to solve for $d$:

$d = \frac{\lambda}{2 \sin \theta}$

Given that the wavelength $\lambda$ is exactly known, any error in the measurement of $\theta$ (denoted as $\Delta \theta$) will propagate through to the error in $d$. To analyze this, we need to consider how the error propagates in the expression for $d$.

Assuming the error propagation formula, the differential of $d$ with respect to $\theta$ is:

$d = \frac{\lambda}{2 \sin \theta}$

Differentiating with respect to $\theta$:

$\frac{d(d)}{d(\theta)} = \frac{d}{d(\theta)} \left( \frac{\lambda}{2 \sin \theta} \right)$

Using the chain rule, we get the partial derivative:

$\frac{d(d)}{d(\theta)} = -\frac{\lambda \cos \theta}{2 (\sin \theta)^2}$

The absolute error in $d$, denoted as $\Delta d$, is given by:

$\Delta d = \left| \frac{d(d)}{d(\theta)} \Delta \theta \right| = \left| -\frac{\lambda \cos \theta}{2 (\sin \theta)^2} \right| \Delta \theta = \frac{\lambda \cos \theta}{2 (\sin \theta)^2} \Delta \theta$

Here, $\Delta \theta$ is the constant error in the measurement of $\theta$. From the equation above, we observe that $\Delta d$ depends on $\cos \theta / (\sin \theta)^2$ which is a function of $\theta$.

As $\theta$ increases from $0^\circ$ to $90^\circ$, the behavior is as follows:

• When $\theta$ is close to $0^\circ$, $\cos \theta \approx 1$ and $\sin \theta \approx 0$, thus $\left( \frac{\cos \theta}{(\sin \theta)^2} \right)$ becomes very large, indicating that $\Delta d$ is very large.
• As $\theta$ approaches $90^\circ$, $\cos \theta \approx 0$, and $\sin \theta \approx 1$, which makes $\left( \frac{\cos \theta}{(\sin \theta)^2} \right)$ approach zero, thus making $\Delta d$ very small.

Therefore, as $\theta$ increases from $0^\circ$ to $90^\circ$, the absolute error in $d$ decreases.

To analyze the fractional error in $d$ (denoted as $\frac{\Delta d}{d}$), we note:

$d = \frac{\lambda}{2 \sin \theta}$

So the fractional error is:

$\frac{\Delta d}{d} = \frac{\Delta d}{\frac{\lambda}{2 \sin \theta}} = \frac{\frac{\lambda \cos \theta}{2 (\sin \theta)^2} \Delta \theta}{\frac{\lambda}{2 \sin \theta}} = \frac{\cos \theta \Delta \theta}{\sin \theta}$

As $\theta$ increases from $0^\circ$ to $90^\circ$:

• Initially (at $\theta \approx 0^\circ$), $\cos \theta \approx 1$ and $\sin \theta \approx 0$, making the fractional error large.
• As $\theta$ continues to increase, $\cos \theta$ decreases and $\sin \theta$ increases, thus the fractional error $\frac{\cos \theta \Delta \theta}{\sin \theta}$ decreases.

Hence, the fractional error in $d$ also decreases as $\theta$ increases.

So the correct answer is:

${M \over L} = {Q \over {2m}}$

$\therefore$ $M = \left( {{Q \over {2m}}} \right)L$

$\Rightarrow M \propto L$, where $\gamma = {Q \over {2m}}$

$\left( {{Q \over {2m}}} \right)(I\omega )$

$= \left( {{Q \over {2m}}} \right)(m{R^2}\omega ) = {{Q\omega {R^2}} \over 2}$

Induced electric field is opposite. Therefore,

$\omega ' = \omega - \alpha t$

$\alpha = {\tau \over I} = {{(QE)R} \over {m{R^2}}} = {{(Q)\left( {{{BR} \over 2}} \right)R} \over {m{R^2}}} = {{QB} \over {2m}}$

$\therefore$ $\omega ' = \omega - {{QB} \over {2m}}.1 = \omega - {{QB} \over {2m}}$

${M_f} = {{Q\omega '{R^2}} \over 2} = Q\left( {\omega - {{QB} \over {2m}}} \right){{{R^2}} \over 2}$

$\therefore$ $\Delta M = {M_f} - {M_i} = - {{{Q^2}B{R^2}} \over {4m}}$

$M = - \gamma {{QB{R^2}} \over 2}$ (as $\gamma = {Q \over {2m}}$)

To solve this problem, we need to understand the working principle of step-up and step-down transformers and the relationships between the voltages and the number of turns of the windings. The key formula we use is the transformer equation:

$\frac{V_p}{V_s} = \frac{N_p}{N_s}$

where:

• $V_p$ is the primary voltage
• $V_s$ is the secondary voltage
• $N_p$ is the number of turns in the primary winding
• $N_s$ is the number of turns in the secondary winding

Let's break down the problem:

1. The power plant produces an electric power of 600 kW at 4000 V. 2. A step-up transformer is used at the plant side with a turns ratio of 1 : 10, i.e., $N_p : N_s = 1 : 10$.

This tells us that:

$\frac{V_p}{V_s} = \frac{1}{10}$

Since the primary voltage $V_p$ is 4000 V, we can find the secondary voltage $V_s$ as follows:

$V_s = 10 \times V_p$

$V_s = 10 \times 4000$

$V_s = 40000$ V

So, at the output of the step-up transformer, the voltage is 40000 V. This high voltage is then transported over the cables to the consumers' location 20 km away. At the consumers' end, a step-down transformer is used to reduce the voltage to 200 V to supply power to the consumers.

Now, we are given that the power supplied at the consumers' end must be 200 V. We need to find the ratio of the number of turns in the primary to the number of turns in the secondary for the step-down transformer so that it outputs 200 V from an input of 40000 V.

Using the transformer equation again, we can write:

$\frac{V_p}{V_s} = \frac{N_p}{N_s}$

where $V_p = 40000$ V and $V_s = 200$ V. Plugging these values in, we get:

$\frac{40000}{200} = \frac{N_p}{N_s}$

This simplifies to:

$\frac{40000}{200} = 200$

So, the turns ratio $\frac{N_p}{N_s} = 200 : 1$.

Thus, the correct answer is:

Option A: 200 : 1

Here we will apply the concept of energy conservation and concept of SHM.

We know that for elastic collision, coefficient of restitution, e = 1

Hence, $𝑒=\frac{𝑣}{0.5{𝑢}_0}=1\Rightarrow 𝑣=0.5{𝑢}_0$

Therefore, particle rebounds with the same speed after collision.

Now, by energy conservation of the spring-mass system, we can say that the speed of the particle when it returns to its equilibrium position is ${𝑢}_0$.

By SHM,

$𝑣={𝑣}_{max}\cos 𝑤𝑡$

$\Rightarrow 0.5{𝑢}_0={𝑢}_0\cos \left(\sqrt{\surd \frac{𝐾}{𝑚}}𝑡\right)$

$\Rightarrow \frac{1}{2}=\cos \left(\sqrt{\surd \frac{𝐾}{𝑚}}𝑡\right)$

$\Rightarrow \frac{𝜋}{3}=\sqrt{\surd \frac{𝐾}{𝑚}}𝑡$

$\Rightarrow 𝑡=\frac{𝜋}{3}\sqrt{\surd \frac{𝑚}{𝐾}}$

Hence, time taken by the particle to reach the wall is $\frac{𝜋}{3}\sqrt{\surd \frac{𝑚}{𝐾}}$

So the time at which the particle passes through the equilibrium position for the first time is

$2𝑡=\frac{2𝜋}{3}\sqrt{\surd \frac{𝑚}{𝐾}}$

The time at which the maximum compression of the spring occurs is

${𝑡}^{\prime }=2𝑡+\frac{𝑇}{4}=\frac{2𝜋}{3}\sqrt{\surd \frac{𝑚}{𝐾}}+\frac{2𝜋}{4}\sqrt{\surd \frac{𝑚}{𝐾}}$

$=\left(\frac{2}{3}+\frac{1}{2}\right)𝜋\sqrt{\surd \frac{𝑚}{𝐾}}$

${𝑡}^{\prime }=\frac{7𝜋}{6}\sqrt{\surd \frac{𝑚}{𝐾}}$

The time at which the particle passes through the equilibrium position for the second time is

$={𝑡}^{\prime }+𝑇$

$=\frac{7𝜋}{6}\sqrt{\surd \frac{𝑚}{𝐾}}+\frac{𝜋}{2}\sqrt{\surd \frac{𝑚}{𝐾}}$

$=\left(\frac{7𝜋}{6}+\frac{𝜋}{2}\right)\sqrt{\surd \frac{𝑚}{𝐾}}=\frac{10𝜋}{6}\sqrt{\surd \frac{𝑚}{𝐾}}$

$=\frac{5𝜋}{3}\sqrt{\surd \frac{𝑚}{𝐾}}$

Hence, options (A) and (D) are correct.

Here we will use the concept of work-energy theorem. By work-energy theorem,

Total work done = Change in KE

$\Rightarrow$ Workdone by gravity + workdone by friction = $\Delta 𝐾𝐸$

$\Rightarrow 𝑚𝑔𝑅\sin {30}^{\circ }-150=\frac{1}{2}𝑚{𝑣}^2-0$

$\Rightarrow 1\times 10\times 40\times \frac{1}{2}-150=\frac{1}{2}{𝑣}^2$

$\Rightarrow 200-150=\frac{1}{2}{𝑣}^2$

$\Rightarrow {𝑣}^2=100$

$\Rightarrow 𝑣=10$ m/s

Hence, option (B) is correct.

Doppler's effect gives

${f_2} = \left[ {{{v + {u_0}} \over {v - {u_s}}}} \right]{f_1}$ ..... (1)

Let the wind speed be w and wind moves towards the source in case (i) and towards the observer in case (ii) (see figure).

In the case (i), u₀, the speed of observer w.r.t. medium considered positive when it moves towards the source is +(u $-$ w) and u_s, the speed of source w.r.t. medium considered positive when it moves towards observer is +(u + w). Substitute these values in equation (1) to get

${f_2} = \left[ {{{v + (u - w)} \over {v - (u + w)}}} \right]{f_1}$,

which is greater than f₁ (assuming u > w). In the case (ii), u₀ is +(u + w) and u_s is +(u $-$ w). Substitute these values in equation (1) to get

${f_2} = \left[ {{{v + (u + w)} \over {v - (u - w)}}} \right]f1$,

which is again greater than f₁.

We know that

(i) For a cylinder, B = 0 for 0 < r < R and $B = {{{\mu _0}} \over {4\pi }} \times {{2I} \over r}$ for r > R.

(ii) For a solenoid, B = $\mu$₀nI for 0 < r < R and B = 0 for r > R.

$\bullet$ Assuming that the magnetic induction to be B_C for cylinder and B_S for solenoid.

In the region R > r > 0, B_S = $\mu$₀nI while B_C = 0. Therefore, the total field is non-zero.

Hence, option (A) is correct.

$\bullet$ In the region 2R < r < R, the magnetic field is not along the axis of cylinder which is given as

$B = \sqrt {B_S^2 + B_C^2}$

Hence, option (B) is wrong.

$\bullet$ For checking option (C), in the region R < r < 2R, the magnetic field is tangential to the circle of radius r, centred on the axis. So this statement is wrong because magnetic field is not in the plane of circle.

Hence, option (C) is wrong.

$\bullet$ In the region r > 2, B_S = 0; while ${B_C} = {{{\mu _0}} \over {4\pi }} \times {{2I} \over r}$. Therefore total field is non-zero.

Hence, option (D) is correct.

The orbital angular momentum is

$L = {{nh} \over {2\pi }}$

${{3h} \over {2\pi }} = {{nh} \over {2\pi }} \Rightarrow n = 3$

The radius of the orbit is

$4.5({a_0}) = {a_0}\left( {{{{n^2}} \over Z}} \right)$

$Z = {{{n^2}} \over {4.5}} = {{{3^2}} \over {4.5}} = {9 \over {4.5}} = 2$

Thus, the possible transitions are 3$\to$2, 3$\to$1 and 2$\to$1. For the transition 3$\to$2, the wavelength is

${1 \over \lambda } = R{(2)^2}\left[ {{1 \over 4} - {1 \over 9}} \right] = 4R\left[ {{{9 - 4} \over {36}}} \right] = {{5R} \over 9}$

$\Rightarrow \lambda = {9 \over {5R}}$

For the transition 3$\to$1, the wavelength is

${1 \over \lambda } = R{(2)^2}\left[ {1 - {1 \over 9}} \right] = 4R\left( {{8 \over 9}} \right) = {{32R} \over 9}$

$\Rightarrow \lambda = {9 \over {32R}}$

For the transition 2$\to$1, the wavelength is

${1 \over \lambda } = R{(2)^2}\left[ {1 - {1 \over 4}} \right] = 4R\left( {{3 \over 4}} \right) = 3R$

$\Rightarrow \lambda = {1 \over {3R}}$

According to definition of specific heat capacity,

$C = {{\Delta Q} \over {m\Delta T}}$

$\Rightarrow \Delta Q = mC\Delta T$ $\therefore$ ${{\Delta Q} \over {\Delta t}} = mC{{\Delta T} \over {\Delta t}}$

Rate of heat absorbed $R = {{\Delta Q} \over {\Delta t}}$

$\Rightarrow {{\Delta Q} \over {\Delta t}} \propto C$

(a) In 0-100 K,

C increases with T but not linearly. So R increases but not linearly.

(b) As $\Delta Q = mC\Delta T$

$Q = m\int {C\Delta T}$

= m area under C-T curve

From the graph it is clear that area under C-T is more in 400-500 K than in 0-100 K.

Therefore, heat absorbed in 0-100 K is less than in 400-500 K.

(c) In 400-500 K,

C remains constant so there is no change in R.

(d) In 200-300 K,

C increases so R increases.

Let the point charge Q is moving in a circle of constant radius R in anticlockwise direction as shown in the figure.

The magnetic flux through the circular loop at time t is given by $\phi = B(\overrightarrow t )\,.\,(\pi {R^2}\widehat z)\, = \pi B(t){R^2}$. Faraday's law gives the induced emf e as

$e = - {{d\phi } \over {dt}} = - \pi {R^2}{{dB(t)} \over {dt}} = - \pi {R^2}B$ ..... (1)

Lenz's law gives the direction of induced current and hence electric field $(\overrightarrow E )$ as clockwise. The induced emf is related to $\overrightarrow E$ by

$e = \oint {\overrightarrow E .\,d\overrightarrow l = - E(2\pi R)}$ ..... (2)

Eliminate e from equations (1) and (2) to get E = BR/2.

From the graph, as the term pV increases, the temperature increases; and when pV decreases the temperature decreases.

The term PV increases K$\to$L, so this involves heating and is an isobaric process.

It increases from N$\to$K, so this involves heating and is an isochoric process.

It decreases from M$\to$N, so this involves cooling and is an isobaric process.

It decreases from L$\to$M, so this involves cooling and is an isochoric process.

The reactions involved are as follows:

(P) $Pb{O_2} + {H_2}S{O_4}\buildrel {Warm} \over \longrightarrow PbS{O_4} + {H_2}O + {1 \over 2}{O_2}$

(Q) $2N{a_2}{S_2}{O_3} + 2{H_2}O + C{l_2} \to 2NaCl + 2NaHS{O_4} + 2S$

(R) ${N_2}{H_4} + 2{I_2} \to {N_2} + 4HI$

(S) $Xe{F_2} + 2NO \to Xe + 2NOF$

The nuclear transmutation of beryllium (Be) is given as:

${}_9^4\mathrm{Be}+𝑋{\to }_8^4\mathrm{Be}+𝑌$

We examine the possible reactions to determine the values of X and Y:

(A) If $𝑋{=}_0^0𝛾$ and $𝑌{=}_1^0𝑛$

The reaction is:

${}_9^4\mathrm{Be}{+}_0^0𝛾{⟶}_8^4\mathrm{Be}{+}_1^0𝑛$

Using the conditions of:

(i) Mass balance:

Mass of reactants $=9+0=9$

Mass of products $=8+1=9$

(ii) Charge balance:

Charge of reactants $=4+0=4$

Charge of products $=4+0=4$

Both mass and charge are balanced. Therefore, Option (A) is correct.

(B) If $𝑋=𝑝{=}_1^1𝐻$ and $𝑌=𝐷{=}_2^1𝐻$

The reaction is:

${}_9^4\mathrm{Be}{+}_1^1𝐻{⟶}_8^4\mathrm{Be}{+}_2^1𝐻$

Using the conditions of:

(i) Mass balance:

Mass of reactants: $9+1=10$

Mass of products $=8+2=10$

(ii) Charge balance:

Charge of reactants $=4+1=5$

Charge on products $=4+1=5$

Both mass and charge are balanced. Therefore, Option (B) is correct.

(C) If $𝑋{=}_1^0𝑛$ and $𝑌=𝐷{=}_2^1𝐻$

The reaction is:

${}_9^4\mathrm{Be}{+}_1^0𝑛{⟶}_8^4\mathrm{Be}{+}_2^1𝐻$

Using the conditions of:

(i) Mass balance:

Mass of reactants $=9+1=10$

Mass of products $=8+2=10$

(ii) Charge balance:

Charge on reactants $=4+0=4$

Charge on products $=4+1=5$

Charge on reactants $\ne$ Charge on products. Since, charge is not balanced, Option (C) is incorrect.

(D) If $𝑋{=}_0^0𝛾$ and $𝑌=𝑝{=}_1^1𝐻$

The reaction is:

${}_9^4\mathrm{Be}{+}_0^0𝛾{⟶}_8^4\mathrm{Be}{+}_1^1𝐻$

Using the conditions of:

(i) Mass balance:

Mass of reactants: $9+0=9$

Mass of products: $8+1=9$

(ii) Charge balance:

Charge on products: $4+1=5$

Charge on reactants: $4+0=4$

Charge on reactants $\ne$ Charge on products. Since the charge is not balanced, Option (D) is incorrect.

Extraction of various metals from their respective concentrated ores is carried out by different methods:

(i) Carbon Reduction of Hematite:

The process for the extraction of iron from hematite (Fe₂O₃) involves carbon reduction:

This reaction occurs at temperatures between 500-800 K.

Carbon monoxide needed for the reduction is produced as follows:

The net reaction can be written as:

Thus, carbon and its oxides are utilized for the reduction of iron ores such as Fe₂O₃.

(ii) Carbon Reduction of Cassiterite (SnO₂):

The reduction process involves carbon:

Therefore, carbon is used for the reduction of cassiterite ore.

(iii) Electrolytic Refining of Aluminium (Al₂O₃):

Aluminium is extracted from bauxite using electrolysis. The reduction occurs at the cathode:

The reduction of Al³⁺ takes place at the cathode.

(iv) Dow's Process for Extraction of Magnesium from Dolomite (MgCO₃.CaCO₃):

Dolomite is treated with hydrochloric acid (HCl) to produce magnesium chloride (MgCl₂). Electrolysis of MgCl₂ reduces Mg²⁺ to magnesium metal at the cathode:

Magnesium metal is obtained by the reduction of Mg²⁺.

The reactions involved are

First, we will calculate the potential values needed for each of the queries in List – I based on the equations and the given standard reduction potential data:

For P. $E^o (\text{Fe}^{3+}, \text{Fe})$: This reaction involves a combination of two reactions:

$\text{Fe}^{3+} + e^- \to \text{Fe}^{2+} \quad \text{(E}^o = +0.77 \text{ V)}$

$\text{Fe}^{2+} + 2e^- \to \text{Fe} \quad \text{(E}^o = -0.44 \text{ V)}$