JEE Advanced 2017 Paper 1 Question Paper with Solutions

The velocity of the transverse wave is given by

$v = \sqrt {{T \over \mu }}$ ..... (i)

Where T is the tension in the rope and is given by

T = Mg + $\mu$yg

Where y = distance of point P on the rope from its bottom

L = total length of the rope

Put the value of T in eqn. (i), we get

$v = \sqrt {{{Mg + \mu yg} \over \mu }}$ ...... (ii)

Also, speed of pulse 1 and pulse 2 are same at each point because speed is a characteristic of the medium. Hence, speed of the two pulses are same at the mid point (x = l/2). However, since pulse 1 is moving in +x direction and pulse 2 is moving in −x direction, their velocities (which is a vector quantity) at the mid point are not same.

So, option (b) is incorrect.

Since both the pulse travel in the same medium and velocity is medium dependent. The velocity of any pulse along the rope depends on tension in the rope and linear mass density so it is independent of its frequency(characteristic of the source of pulse generation) and wavelength.

So, option (d) is correct.

Also, v = $\nu$$\lambda$ or $\lambda$ $\propto$ v

($\because$ frequency of a wave depends on the source)

For pulse 1, the velocity decreases at point A as tension decreases. So the wavelength of pulse 1 becomes shorter when it reaches point A.

So, option (c) is incorrect.

From eqn. (ii), the velocity is given as

$v = \sqrt {{{Mg + \mu yg} \over \mu }}$ or ${{dy} \over {dt}} = \sqrt {{{Mg + \mu yg} \over \mu }} \Rightarrow dt = {{dy} \over {\sqrt {{{Mg + \mu yg} \over \mu }} }}$

Integrating both sides,

$\int\limits_0^T {dt = \int\limits_0^L {{{dy} \over {\sqrt {{{Mg} \over \mu } + yg} }}} }$; $\therefore$ $T = {2 \over g}\left( {\sqrt {{{Mg} \over \mu } + Lg} - \sqrt {{{Mg} \over \mu }} } \right)$

Where T is time taken by pulse to travel distance L.

Since T depends upon g, M, L and $\mu$ which are constant.

$\therefore$ Time taken by pulse 1 to reach point A and time taken by pulse 2 to reach point O is same.

$\therefore$ T_AO = T_OA; So option (a) is correct.

The particle will describe a helical path with axis along +z direction if (i) component of net force in x-y plane provides the centripetal acceleration and (ii) the particle has a non-zero velocity and/or non-zero force along +z direction. A proton with initial velocity $\overrightarrow v = 2{{{E_0}} \over {{B_0}}}\widehat x$ will move in a helical path with axis along +z direction when placed in an electric field $\overrightarrow E = {E_0}\widehat z$ and magnetic field $\overrightarrow B = {B_0}\widehat z$.

Applying Snell's law, we have nsin$\theta$ = (n $-$ m$\Delta$n)sin90$^\circ$ it is given that n = 1.6, $\theta$ = 30$^\circ$, $\Delta$n = 0.1. The above equation becomes

1.6sin30 = [1.6 $-$ m(0.1)]sin90

1.6 $\times$ ${1 \over 2}$ = (1.6 $-$ 0.1m) $\times$ 1

0.8 = (1.6 $-$ 0.1m)

0.1m = 1.6 $-$ 0.8

0.1m = 0.8

m = 8

Given V = V₀ sin $\omega$t

At resonant frequency, current and voltage will be in same phase i.e., $\omega L = {1 \over {\omega C}}$

$\therefore$ Resonant frequency $\omega = {1 \over {\sqrt {LC} }}$ ...... (i)

Here, L = 1 $\mu$H = 10^$-$6 H, C = 10^$-$6 F

$\therefore$ $\omega = {1 \over {\sqrt {{{10}^{ - 12}}} }} = {10^6}$ rad s^$-$1

So, option (a) is incorrect.

Since, ${X_C} = {1 \over {\omega C}},\,{X_L} = \omega L$ ....... (ii)

For large $\omega$, X_C $\sim$ 0, so circuit will behave like a inductive circuit.

So, option (d) is incorrect.

From Eqn. (i)

$\omega = {1 \over {\sqrt {LC} }}$

At resonant frequency or the frequency at which the current will be in phase with the voltage is independent of R.

$\therefore$ option (b) is correct.

From Eqn. (ii), at $\omega$ $\sim$ 0, X_C $\sim$ $\infty$ and X_L = 0

$\therefore$ The current flowing through the circuit becomes nearly zero.

So, option (c) is correct.

Adiabatic process is used as a correction in the determination of the speed of sound in an ideal gas. The p-V diagram of this process is shown in (Q). The work done on the gas in an adiabatic process is given by ${W_{1 \to 2}} = {{{p_2}{V_2} - {p_1}{V_1}} \over {\gamma - 1}}$.

We discuss the options as follows:

$\bullet$ For option (A) : For the angle of incidence i₁ = A, we have minimum deviation and the ray inside the prism is parallel to the base of the prism.

Hence, option (A) is correct.

$\bullet$ For option (C) : The angle of deviation is given as

$\delta$ = i₁ + i₂ $-$ A

At minimum deviation, i₁ = i₂ and r₁ = r₂. Therefore,

i₁ = A ....... (1)

Also, we know that

r₁ + r₂ = A

As r₁ = r₂, we get

2r₁ = A $\Rightarrow$ r₁ = ${A \over 2}$

From Eq. (1), we get

${r_1} = {{{i_1}} \over 2}$

Hence, option (C) is correct.

According to Snell's law, we have

${{\sin {i_2}} \over {\sin {r_2}}} = \mu$

For emergent ray to be tangential to the surface, we have

i₂ = 90$^\circ$

$\Rightarrow$ sin90$^\circ$ = $\mu$sinr₂ $\Rightarrow$ $\mu$sinr₂ = 1

sinr₂ = ${1 \over \mu }$ $\Rightarrow$ r₂ = sin^$-$1$\left( {{1 \over \mu }} \right)$ ....... (2)

From the relation r₁ + r₂ = A, we get

r₁ = A $-$ r₂

$\Rightarrow$ sinr₁ = sin(A $-$ r₂)

Using sin(a $-$ b) = sinacosb $-$ sinbcosa, we have

$\Rightarrow$ sinr₁ = sinAcosr₂ $-$ sinr₂cosA ......... (3)

Using Eq. (2), we have

sinr₂ = ${{1 \over \mu }}$

Squaring it, we get

sin²r₂ = ${{1 \over {{\mu ^2}}}}$

Using the relation sin²x + cos²x = 1, we get

sin²x = 1 $-$ cos²x

$\Rightarrow$ 1 $-$ cos² r₂ = ${{1 \over {{\mu ^2}}}}$

$\Rightarrow$ cos² r₂ = 1 $-$ ${{1 \over {{\mu ^2}}}}$

That is,

cosr₂ = $\sqrt {1 - {1 \over {{\mu ^2}}}}$

Substituting the values of sinr₂ and cosr₂ in Eq. (3), we get

sinr₁ = sin A$\sqrt {1 - {1 \over {{\mu ^2}}}}$ $-$ ${{1 \over \mu }}$cos A

According to Snell's law, we have

${{\sin {i_1}} \over {\sin {r_1}}} = \mu$

That is,

$\sin {i_1} = \mu \sin {r_1} = \mu \left( {\sin A\sqrt {1 - {1 \over {{\mu ^2}}}} - {1 \over \mu }\cos A} \right)$

$\sin {i_1} = \mu \left( {\sin A{{\sqrt {{\mu ^2} - 1} } \over \mu } - {{\cos A} \over \mu }} \right)$

$\sin {i_1} = \sin A\sqrt {{\mu ^2} - 1} - \cos A$ ....... (4)

Now, at minimum deviation, we have A = i₁ and r₁ = ${A \over 2}$. Thus, from Snell's law, we get

$\mu = {{\sin {i_1}} \over {\sin {r_1}}} = {{\sin A} \over {\sin (A/2)}}$

Using $\sin x = 2\sin {x \over 2}\cos {x \over 2}$, we get

$\mu = {{2\sin A/2\cos A/2} \over {\sin A/2}} \Rightarrow \mu = 2\cos {A \over 2}$ ....... (5)

Now, Eq. (4) becomes

$\sin {i_1} = \sin A\sqrt {{{\left( {2\cos {A \over 2}} \right)}^2} - 1 - \cos A}$

$= \sin A\sqrt {4{{\cos }^2}{A \over 2} - 1} - \cos A$

${i_1} = {\sin ^{ - 1}}\left[ {\sin A\sqrt {4{{\cos }^2}{A \over 2} - 1} - \cos A} \right]$

Hence, option (D) is correct.

$\bullet$ For option (B) : From Eq. (5), we have

$\mu = 2\cos {A \over 2}$

$\cos {A \over 2} = {\mu \over 2}$

$\Rightarrow {A \over 2} = {\cos ^{ - 1}}\left( {{\mu \over 2}} \right)$

$\Rightarrow A = 2{\cos ^{ - 1}}\left( {{\mu \over 2}} \right)$

Hence, option (B) is incorrect.

$\because$ dW = PdV

and there is no change in volume in isochoric process.

$\therefore$ dW = 0

We discuss the options as follows:

$\bullet$ For option (B) : From law of conservation of momentum, we have

mv = MV ...... (1)

where v is the velocity of point mass m and V is velocity of block of mass M.

Now, by using law of conservation of energy, after the point mass is released, we have the following:

Loss in P.E. of mass m = Gain in K.E. of both mass M and the mass m.

$\Rightarrow mgh = {1 \over 2}m{v^2} + {1 \over 2}M{V^2}$

Substituting $V = {{mu} \over M}$ [from Eq. (1)], we get

$mgh = {1 \over 2}m{v^2} + {1 \over 2}M{\left( {{{mv} \over M}} \right)^2}$

$mgh = {1 \over 2}m{v^2} + {1 \over 2}M{{{m^2}{v^2}} \over {{M^2}}}$

$mgR = {1 \over 2}m{v^2} + {1 \over 2}{{{m^2}{v^2}} \over M}$

$mgR = {1 \over 2}m{v^2}\left( {1 + {m \over M}} \right)$

$gR = {1 \over 2}{v^2}\left( {1 + {m \over M}} \right)$

Rearranging this equation, we get

$2gR = {v^2}\left( {1 + {m \over M}} \right)$

${v^2} = {{2gR} \over {\left( {1 + {m \over M}} \right)}}$

$\Rightarrow v = \sqrt {{{2gR} \over {\left( {1 + {m \over M}} \right)}}}$

Hence, option (B) is correct.

$\bullet$ For option (C) : Now, from Eq. (1), we have

$V = {m \over M}v = {m \over M}\sqrt {{{2gR} \over {\left( {1 + {m \over M}} \right)}}}$

$V = \sqrt {{{{m^2}2gR} \over {{M^2}\left( {1 + {m \over M}} \right)}}} = \sqrt {{{{m^2}2gR} \over {{M^2}\left( {{{M + m} \over M}} \right)}}}$

$V = \sqrt {{{{m^2}2gR} \over {M(m + M)}}} \Rightarrow V = m\sqrt {{{2gR} \over {M(m + M)}}}$

Since, there is no external force acting on the system, the centre of mass does not change.

Now, if the change in position of mass m is $\Delta$x and the change in position of mass M is $\Delta$X, then

m$\Delta$x + M$\Delta$x = 0 (since centre of mass will not change)

Now, we know that the change in position of point mass m is

$\Delta$x = R $-$ x

and the change in position of block of mass M is

$\Delta$X = $-$x

Therefore,

m(R $-$ x) $-$ Mx = 0

m(R $-$ x) = Mx

mR $-$ mx = Mx

mR = Mx + mx

mR = (M + m)x

$\Rightarrow$ x = ${{mR} \over {M + m}}$

The change in position of block of mass M is

$\Delta$x = $-$x = ${{ - mR} \over {M + m}}$

Thus, x-component of the displacement of the centre of mass of the block is

$M = {{ - mR} \over {M + m}}$

Hence, option (C) is also correct.

Consider the coordinate axes shown in the figure. The magnetic field is into the paper, i.e., $\overrightarrow B = B\widehat k$. Let loop 1 be the loop of area A and loop 2 be the loop of area 2A. At t = 0, both loops are in the plane of the paper. They start rotating about the x-axis with a constant angular velocity $\overrightarrow \omega = \omega \widehat i$. Let the area vectors of loop 1 and 2 be ${\overrightarrow A _1}$ and ${\overrightarrow A _2}$, respectively. At t = 0, ${\overrightarrow A _1}$ = $A\widehat k$ and ${\overrightarrow A _2} = 2A\widehat k$ (we chose area into the plane of the paper as positive). In time t, the loops rotate by an angle $\theta$ = $\omega$t. Their area vectors make an angle $\theta$ = $\omega$t with z-axis (direction of the magnetic field). Thus, the magnetic fluxes passing through the loop 1 and the loop 2 at time t are given by

$\phi$₁ = $\overrightarrow B .{\overrightarrow A _1}$ = B A₁ cos$\theta$ = B A cos $\omega$t,

$\phi$₂ = $\overrightarrow B .{\overrightarrow A _2}$ = B A₂ cos$\theta$ = 2B A cos $\omega$t.

By Faraday's law of electromagnetic induction, induced emfs in the loop 1 and in the loop 2 are given by

${\varepsilon _1} = - {{d{\phi _1}} \over {dt}} = B\,A\omega \sin \omega t$,

${\varepsilon _2} = - {{d{\phi _2}} \over {dt}} = 2B\,A\omega \sin \omega t$.

The rate of change of flux is maximum when $\omega$t = $\pi$/2 i.e., when the plane of the loops is perpendicular to the plane to the paper.

Let T = 2$\pi$/$\omega$ be the time period to complete one revolution. The fluxes $\phi$₁ and $\phi$₂ decrease with time if t $\in$ (0, T/4). By Lenz's law, the induced current should oppose reductions in $\phi$₁ and $\phi$₂. Thus, induced current in both the loops should be clockwise. Hence, the polarity of 'equivalent batteries' are as shown in the figure. These batteries are connected with reverse polarity. Thus, net emf induced due to both the loop is

$\varepsilon = {\varepsilon _2} - {\varepsilon _1} = B\,A\omega \sin \omega t$.

Note that the net emf $\varepsilon$ is proportional to the difference in areas of two loops and $\left| \varepsilon \right| = \left| {{\varepsilon _1}} \right|$.

Lorentz force on a charged particle having a charge q, moving with a velocity $\overrightarrow v$ in a region of uniform electric field $\overrightarrow E$ and uniform magnetic field $\overrightarrow B$, is given by

${\overrightarrow F _{net}} = {\overrightarrow F _E} + {\overrightarrow F _B} = q\overrightarrow E + q\overrightarrow v \times \overrightarrow B$. ...... (1)

The particle will move along $- \widehat y$ axis if (i) initial velocity $\overrightarrow v$ is along $-\widehat y$ and net force is either zero or directed along $-\widehat y$ (ii) initial velocity $\overrightarrow v$ is zero and net force is directed along $-\widehat y$. A proton with initial velocity $\overrightarrow v = \overrightarrow 0$ placed in an electric field $\overrightarrow E = - {E_0}\widehat y$ and magnetic field $\overrightarrow B = - {B_0}\widehat y$ will experience a net force ${\overrightarrow F _{net}} = - q{E_0}\widehat y$. It moves in a straight line along $- \widehat y$.

It is given that the source emits sound of frequency f₀ = 492 Hz. The car is approaching the source and the speed of car is v = 2 m/s.

Also, the speed of sound in air, v_s = 330 m/s.

The frequency of sound received by car is given as

${f_1} = \left( {{{{V_s} + V} \over {{V_s}}}} \right){f_0} = \left( {{{330 + 2} \over {330}}} \right)492$

Here, f₁ = 494.98 Hz, which is the frequency reflected by the car towards the source.

Therefore, now, the car acts as the source. The frequency of sound received by the source is

${f_2} = \left( {{{{V_s}} \over {{V_s} - v}}} \right){f_1} = \left( {{{330} \over {330 - 2}}} \right)494.98$

Here, f₂ = 498 Hz. Therefore, the beat frequency of the resulting signal is

$\left| {{f_0} - {f_2}} \right| = \left| {492 - 498} \right| = 6$ Hz

Compare the given relation, $\Delta$U = $\Delta$Q $-$ p$\Delta$V, with the first law of thermodynamics, $\Delta$Q = $\Delta$U + $\Delta$W, to get work done by the gas as $\Delta$W = p$\Delta$V. This is the work done in an isobaric process. The p-V diagram of this process is shown in (P). The work done in this process is ${W_{1 \to 2}} = \int {pdV = p{V_1} - p{V_2}}$.

The decay rate is related to half life by $\lambda = {{\ln 2} \over {{t_{1/2}}}}$. A sample having N₀ radioactive nuclei at time t = 0 will have $N = {N_0}{e^{ - \lambda t}}$ radioactive nuclei at time t. Thus, activity (A = $\lambda$N) of a sample reduces from its initial value A₀ to a value $A = {A_0}{e^{ - \lambda t}}$ at time t.

Let V be the total volume of blood and N be the total number of radioactive nuclei at time t. The nuclei are distributed uniformly in blood. A sample of volume v will have n = (v/V)N radioactive nuclei in it. Thus, activity of this sample is

$a = \lambda n = \lambda \left( {{v \over V}N} \right) = {v \over V}(\lambda N) = {v \over V}A = {v \over V}{A_0}{e^{ - \lambda t}}$.

Substitute a = 115 Bq, v = 2.5 ml, A₀ = 2.4 $\times$ 10⁵ Bq, $\lambda$ = ln 2/t_1/2 $\approx$ 0.7/(8 $\times$ 24) hr^$-$1 and t = 11.5 hr to get

$V = {{(2.5)(2.4 \times {{10}^5})} \over {115}}{e^{ - {{0.7 \times 11.5} \over {8 \times 24}}}}$

$\approx {{(2.5)(2.4 \times {{10}^5})} \over {115}}\left( {1 - {{0.7 \times 11.5} \over {8 \times 24}}} \right)$

= 4998 ml $\approx$ 5 litre.

The energy of a hydrogen atom with an electron in n^th orbit is given by

${E_n} = - {{13.6} \over {{n^2}}}eV$.

The potential energy in n^th orbit is related to E_n by V_n = E_n/2. Thus, potential energies in orbits with quantum numbers n_i and n_f are given by

${V_i} = - {{13.6} \over {2n_i^2}}eV$, and ${V_f} = - {{13.6} \over {2n_f^2}}eV$, which gives

${{{V_i}} \over {{V_f}}} = {{n_f^2} \over {n_i^2}} = 6.25$ (given).

Take square root to get n_f = 2.5n_i. Since n_i and n_f are positive integers, smallest possible integral value of n_f is 5 for n_i = 2.

Let a spherical drop of liquid of radius R divides itself into K spherical drops, each of radius r. The density of liquid does not change in this process. Hence, conservation of mass gives

${4 \over 3}\pi {R^3} = {4 \over 3}\pi {r^3}K$, i.e., $r = R{K^{ - 1/3}}$. ...... (1)

The surface energy of a drop of radius R is ${U_i} = 4\pi {R^2}S$ and total surface energy of K drops of radius r is ${U_f} = 4\pi {r^2}KS$. Note that the surface energy increases when a bigger drop is divided into multiple smaller drops. Total change in surface energy is

$\Delta U = {U_f} - {U_i} = 4\pi S(K{r^2} - {R^2})$ ...... (2)

Eliminate r from equations (1) and (2) to get

$K = {\left( {{{\Delta U} \over {4\pi S{R^2}}} + 1} \right)^3}$.

Substitute $\Delta$U = 10^$-$3 J, S = ${{0.1} \over {4\pi }}$ N/m and R = 10^$-$2 m to get $K = {(100 + 1)^3} \approx {10^6}$.

The rate of radiation energy emitted by a body of surface area A, emissivity e, and kept at an absolute temperature T is given by Stefan's law

dQ_e/dt = $\sigma$eAT⁴,

where $\sigma$ is Stefan's constant. If this body is kept at a surrounding temperature T₀ then rate of radiation energy absorbed by the body is

dQ_a/dt = $\sigma$eAT$_0^4$.

Thus, the net rate of energy loss by the body is

$d{Q_i}/dt = d{Q_e}/dt - d{Q_a}/dt = \sigma eA({T^4} - T_0^4)$.

If surrounding temperature is reduced by $\Delta$T₀ (<< T₀) then net rate of energy lost by the body becomes

${{d{Q_f}} \over {dt}} = \sigma eA\left[ {{T^4} - {{({T_0} - \Delta {T_0})}^4}} \right]$

$= \sigma eA\left[ {{T^4} - T_0^4{{\left( {1 - {{\Delta {T_0}} \over {{T_0}}}} \right)}^4}} \right]$

$\approx \sigma eA\left[ {{T^4} - T_0^4\left( {1 - 4{{\Delta {T_0}} \over {{T_0}}}} \right)} \right]$

$= {{d{Q_i}} \over {dt}} + 4\sigma eAT_0^3\Delta {T_0}$.

Thus, decrease in the surrounding temperature increases the rate of energy loss by $4\sigma eAT_0^3\Delta {T_0}$. To maintain the body temperature, the human being needs to increase the rate of energy generation (through chemical reactions in the body) by $4\sigma eAT_0^3\Delta {T_0}$ to balance the increase in rate of energy loss.

The human beings can also maintain the body temperature (without increasing the rate of energy generation) by reducing the surface area (e.g., by curling up) of their body. If surrounding temperature is reduced by $\Delta$T₀ and surface area of the body is reduced by $\Delta$A (<< A) then net rate of energy loss by the body becomes

${{d{Q_f}} \over {dt}} = 4\sigma e(A - \Delta A)\left[ {{T^4} - {{({T_0} - \Delta {T_0})}^4}} \right]$

$\approx {{d{Q_i}} \over {dt}} + 4\sigma eAT_0^3\Delta {T_0} - \sigma e\Delta A({T^4} - \Delta T_0^4)$.

Thus, the rate of energy loss by radiation remains same $(d{Q_f}/dt = d{Q_i}/dt)$ if reduction in surface area is

$\Delta A = {{4AT_0^3\Delta {T_0}} \over {{T^4} - T_0^4}} \approx A{{\Delta {T_0}} \over {\Delta T}}$,

where, the body temperature is $T = {T_0} + \Delta T$.

We assume body to be black i.e., e = 1. The amount of energy emitted by the body in one second is

$d{Q_e}/dt = \sigma eA{T^4} = \sigma eA{({T_0} + \Delta T)^4}$

$\approx \sigma eAT_0^4(1 + 4\Delta T/{T_0})$

$= (1)(1)(460)(1 + 4(10)/300) = 521$ J.

The amount of energy lost by the body in one second is

$d{Q_i}/dt = \sigma eA({T^4} - T_0^4) = \sigma eA{({T_0} + \Delta T)^4} - T_0^4)$

$\approx eA\sigma T_0^4(4\Delta T/{T_0}) = 61$ J.

It is given that the average speed of gas molecules is u; the speed of the plate is v; the force on the plate if F.

$\bullet$ Just before the collision : The gas molecules are approaching the leading and trailing faces of the plate with speed u.

$\bullet$ Just after the collision : The gas molecules bounce back with speed u₁ and u₂.

$\bullet$ At trailing face

The speed before collision is u $-$ v.

The speed after collision is u₂ + v.

From law of conservation of linear momentum, we have

u₂ + v = u $-$ v

That is,

u₂ = u $-$ 2v ....... (1)

and

$\Delta$u₂ = 2u $-$ 2v ....... (2)

$\bullet$ At leading face

The speed before collision is u + v.

The speed after collision is u₁ $-$ v.

From law of conservation of linear momentum, we have

u₁ $-$ v = u + v

That is,

u₁ = u + 2v ..... (3)

and

$\Delta$u₁ = 2u + 2v ...... (4)

Now, from Newton's second law of motion, we have

$F = {{dp} \over {dt}}$

$\bullet$ For leading face : ${F_1} = {{d{p_1}} \over {dt}}$.

The volume swept by the plate is A(u + v). Therefore,

${F_1} = {{d{p_1}} \over {dt}} = \rho A(u + v)\Delta {u_1}$

${F_1} = \rho A(u + v)(2u + 2v)$ ....... (5)

$\bullet$ For trailing face : ${F_2} = {{d{p_2}} \over {dt}}$

The volume swept by the plate is A(u $-$ v). Therefore,

${F_2} = {{d{p_2}} \over {dt}} = \rho A(u - v)\Delta {u_2}$

${F_2} = \rho A(u - v)(2u - 2v)$ ....... (6)

The net force on the plate is

${F_{net}} = {F_1} - {F_2}$

$= \rho A(u + v)(2u + 2v) - \rho A(u - v)(2u - 2v)$

$= 2\rho A(u + v)(u + v) - 2\rho A(u - v)(u - v)$

$= 2\rho A{(u + v)^2} - 2\rho A{(u - v)^2}$

$= 2\rho A[{(u + v)^2} - {(u - v)^2}]$

$= 2\rho A[{u^2} + {v^2} + 2uv - ({u^2} + {v^2} - 2uv)]$

$= 2\rho A({u^2} + {v^2} + 2uv - {u^2} - {v^2} + 2uv)$

$= 2\rho A(4uv)$

$= 8\rho Auv$

Therefore, the pressure difference is

${{{F_{net}}} \over A} = {{8\rho Auv} \over A} = 8\rho uv$

Thus, the pressure difference between the leading and trailing faces of the plate is proportional to uv. Hence, option (A) is correct.

Now,

${F_{net}} = 8\rho Auv$

The net resistive force is

$F - (8\rho Au)v = {{mdv} \over {dt}}$

where F is the constant force applied by the gas molecules on the plate.

The resistive force experienced by the plate is proportional to v.

Hence, option (B) is also correct.

At a later time, the velocity v becomes sufficient and the external force $F( = 8\rho Auv)$ balances the resistive force.

Hence, option (D) is also correct.