JEE Advanced 2018 Paper 2 Question Paper with Solutions

For a point charge Q at the origin, we have

$E = {1 \over {4\pi {\varepsilon _0}}}{Q \over {{d^2}}}$

$\Rightarrow E \propto {1 \over {{d^2}}}$

Thus, the correct mapping is $R \to 1$.

For a small dipole with point charges Q at (0, 0, l) and $-$Q at (0, 0, $-$l), 2l << d, we have

$E = {1 \over {4\pi {\varepsilon _0}}}{{2Q \times 2l} \over {{d^3}}}$

$\Rightarrow E \propto {1 \over {{d^3}}}$

Thus, the correct mapping is $S \to 2$.

For an infinite line charge coincident with x-axis with uniform linear charge density $\lambda$, we have

$E = {\lambda \over {2\pi {\varepsilon _0}d}}$

$\Rightarrow E \propto {1 \over d}$

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

For two infinite wires carrying uniform linear charge density parallel to the x-axis, we have

$E = {\lambda \over {2\pi {\varepsilon _0}(d - l)}} - {\lambda \over {2\pi {\varepsilon _0}(d + l)}} = {\lambda \over {2\pi {\varepsilon _0}}}\left( {{1 \over {d - l}} - {1 \over {d + l}}} \right)$

$\Rightarrow E = {\lambda \over {2\pi {\varepsilon _0}}}\left( {{{d + l - d + l} \over {(d - l)(d + l)}}} \right) = {\lambda \over {2\pi {\varepsilon _0}}}{{(2l)} \over {({d^2} - {l^2})}}$

Since $2l < < d$

$\Rightarrow E = {\lambda \over {2\pi {\varepsilon _0}}}{{2l} \over {{d^2}}}$

$\Rightarrow E \propto {1 \over {{d^2}}}$

Thus, the correct mapping is $R \to 4$.

For an infinite plane charge coincident with the x-y plane with uniform surface charge density, we have

$E = {\sigma \over {2\pi {\varepsilon _0}}}$

Therefore, E is independent of d.

Thus, the correct mapping is $P \to 5$.

Therefore, option (B) is correct.

We know that $\Delta L = {W \over {(YA/L)}}$

where W is weight or load = mg = 1.2 $\times$ 10 = 12 kg m s^$-$2, Y is Young's modulus = 2 $\times$ 10¹¹ N m^$-$2, L is length of wire with load = 1.0 m, A is area of steel wire = $= \pi {r^2} = {\pi \over 4}{d^2} = {\pi \over 4} \times {(0.5 \times {10^{ - 3}})^2}$

Therefore,

$\Delta L = {{1.2 \times 10} \over {2 \times {{10}^{11}} \times {\pi \over 4}{{(0.5 \times {{10}^{ - 3}})}^2} \times {1 \over {1.0\,m}}}}$

$= {{1.2 \times 10 \times 4} \over {2 \times {{10}^{11}} \times \pi \times {{(0.5)}^2} \times {{10}^{ - 6}}}}$

$\Rightarrow \Delta L = 0.3 \times {10^{ - 3}}$ m = 0.3 mm

Now, least count of vernier scale $= \left( {1 - {9 \over {10}}} \right)$ mm = 0.1 mm

Therefore, Vernier reading $= {{\Delta L} \over {least\,count}}$

Vernier reading $= {{0.3\,mm} \over {0.1\,mm}} = 3$

Therefore, 3^rd vernier scale division coincides with the main scale division.

Given, mass of particle = 10^$-$3 kg, charge on particle = 1.0 C, electric field $\overrightarrow E (t) = {E_0}\sin \omega t\,\widehat i$, E₀ = 1.0 N C^$-$1, $\omega$ = 10³ rad s^$-$1

Force on particle is given by

$\Rightarrow \overrightarrow F = q{E_0}\sin \omega t\,\widehat i = 1.0 \times 1.0 \times \sin ({10^3}t)\widehat i$

$\Rightarrow \overrightarrow F = \sin ({10^3}t)\widehat i$

We know that $\overrightarrow F = m\overrightarrow a \Rightarrow \overrightarrow a = {{\overrightarrow F } \over m}$

$\Rightarrow a = {{\sin ({{10}^3}t)} \over {{{10}^{ - 3}}}} \Rightarrow a = {10^3}\sin ({10^3}t)$

We know $a = {{dv} \over {dt}}$

$\Rightarrow {{dv} \over {dt}} = {10^3}\sin ({10^3}t)$

$\Rightarrow dv = {10^3}\sin ({10^3}t)dt$

Now, integrating it, we get

$\int\limits_0^v {dv = \int\limits_0^t {{{10}^3}\sin ({{10}^3}t)dt} }$

$\Rightarrow \left. v \right|_0^v = {10^3}\left( {{{ - \left. {\cos ({{10}^3}t)} \right|_0^t} \over {{{10}^3}}}} \right) \Rightarrow \left. v \right|_0^v = \left. { - \cos ({{10}^3}t)} \right|_0^t$

$\Rightarrow v = ( - \cos {10^3}t + \cos 0)$

$\Rightarrow v = (1 - \cos ({10^3}t))$

We know that cos$\theta$ can take values between $-$1 and 1. Therefore, maximum speed attained by the particle when cos$\theta$ = $-$1. Thus,

${v_{\max }} = 1 - ( - 1) = 2$ m s^$-$1

Given,

${}_{90}^{232}Th\buildrel {} \over \longrightarrow {}_{82}^{212}pb$

We know that $\alpha$-decay is given as

${}_Z^AX\buildrel {} \over \longrightarrow {}_{Z - 2}^{A - 4}Y + {\alpha ^ - }$ particle

and $\beta$-decay is given as

${}_Z^AX\buildrel {} \over \longrightarrow {}_{Z + 1}^AX + {\beta ^ - }$ particle

In above decay process change in mass number A is 20 and change in atomic number Z is 8.

Therefore, change in mass number A $\Rightarrow$ number of $\alpha$-particles emitted are

${N_\alpha } = {{20} \over 4} = 5$

Now, emission of 5 $\alpha$ particles implies atomic number reduces by 10 but change in atomic number is 8, therefore, number of $\beta$-particles emitted is 2.

Therefore, N_$\alpha$ = 5 and N_$\beta$ = 2

So, option (A) and option (C) are correct.

The block was initially at rest and its velocity just after the application of impulse is $v(0) = {v_0}{e^{ - 0/\tau }} = {v_0}$. The applied impulse is equal to the change in linear momentum of the block i.e., J = mv₀, which gives

v₀ = J/m = 1/0.4 = 2.5 m/s.

The velocity of the particle is given as

$v(t) = {v_0}{e^{ - t/\tau }}$.

Integrate to get the displacement

$x(t) = \int_0^t {{v_0}{e^{ - t/\tau }}dt = {v_0}\tau (} 1 - {e^{ - t/\tau }})$.

Substitute t = $\tau$ = 4 s and v₀ = 2.5 m/s to get x($\tau$) = (2.5) (4) (1 $-$ e^$-$1) = 6.3 m.

The degrees of freedom of a monatomic gas is f = 3. Its molar specific heats are ${C_v} = {f \over 2}RT$ and ${C_p} = {C_v} + R = {{(f + 2)} \over 2}RT$. The ratio of specific heats is given by

$\gamma = {C_p}/{C_v} = (f + 2)/f = 5/3$.

In an adiabatic process, $TV_{}^{\gamma - 1}$ = constant. Hence, ${T_1}V_1^{\gamma - 1} = {T_2}V_2^{\gamma - 1}$, which gives

${T_2} = {T_1}{({V_1}/{V_2})^{\gamma - 1}} = 100{(1/8)^{5/3 - 1}} = 25\,K$,

where we used T₁ = 100 K and V₂ = 8V₁. The change in internal energy of one mole of the ideal gas is

$\Delta U = {U_2} - {U_1} = {f \over 2}R{T_2} - {f \over 2}R{T_1} = {f \over 2}R({T_2} - {T_1})$

$= (3/2)(8)(25 - 100) = - 900\,J$.

Thus, the decrease in internal energy of the gas is 900 J.

Frequency of tuning fork = 500 Hz

Length of n₁ harmonic = 50.7 cm

Length of n₂ harmonic = 83.9 cm

We know that

n₂ harmonic $-$ n₁ harmonic = ${\lambda \over 2}$

$\Rightarrow (83.9 - 50.7) = {\lambda \over 2} \Rightarrow {\lambda \over 2} = 33.2$ cm

$\Rightarrow$ $\lambda$ = 66.4 cm

Therefore, wavelength of sound wave is 66.4 cm, so option (C) is correct.

Now, speed of sound is given as

v = f $\lambda$ = 500 $\times$ 66.4 $\times$ 10^$-$2

$\Rightarrow$ v = 332 m s^$-$1

Thus, speed of sound determined from this experiment is 332 n s^$-$1, so option (A) is correct.

Now, $(2n - 1){\lambda \over 4} = 50.7 + e$

For n = 2, $(4 - 1){\lambda \over 4} = 50.7 + e \Rightarrow {{3\lambda } \over 4} = 50.7 + e$

$e = {{3\lambda } \over 4} - 50.7 \Rightarrow e = {3 \over 4} \times 66.4 - 50.7 = 49.8 - 50.7$

$\Rightarrow e = - 0.9$ cm

Thus, the end correction in this experiment is 0.9 cm. Therefore, option (B) is correct.

Force of viscosity can be written as follows:

$f = \eta A{{dv} \over {dy}}$

Here $\eta$ is coefficient of viscosity, A is face area of plate and dv / dy is velocity gradient. We can assume that velocity of fluid at the bottom (h depth below) is zero; hence ${{dv} \over {dy}} = {{{u_0}} \over h}$; hence force of viscocity can be written as follows:

$f = \eta A{{{u_0}} \over h}$

We can see that force of viscosity is inversely proportional to h; hence option (a) is correct.

We can see that force of viscosity is proportional to area of plate; hence option (b) is wrong.

Tangential stress can be written as ${f \over A} = \eta {{{u_0}} \over h}$.

We can see that options (c) and (d) are also correct.

We know that,

$\Delta {E_{i - j}} = 13.6\left[ {{1 \over {n_j^2}} - {1 \over {n_i^2}}} \right]{Z^2}$

Therefore,

$\Delta {E_{2 - 1}} = 13.6\left[ {{1 \over {{1^2}}} - {1 \over {{2^2}}}} \right]{Z^2} = 13.6\left[ {{1 \over 1} - {1 \over 4}} \right]{Z^2}$

$\Delta {E_{2 - 1}} = 13.6 \times {3 \over 4} \times {Z^2}$

Similarly,

$\Delta {E_{3 - 2}} = 13.6\left[ {{1 \over {{2^2}}} - {1 \over {{3^2}}}} \right]{Z^2} = 13.6\left[ {{1 \over 4} - {1 \over 9}} \right]{Z^2}$

$\Delta {E_{3 - 2}} = 13.6 \times {5 \over {36}} \times {Z^2}$

Given, $\Delta {E_{2 - 1}} = \Delta {E_{3 - 2}} + 74.8$

$\Rightarrow 13.6 \times {3 \over 4} \times {Z^2} = 13.6 \times {5 \over {36}} \times {Z^2} + 74.8$

$\Rightarrow 13.6 \times {3 \over 4} \times {Z^2} - 13.6 \times {5 \over {36}} \times {Z^2} = 74.8$

$\Rightarrow 13.6 \times {Z^2}\left[ {{3 \over 4} - {5 \over {36}}} \right] = 74.8$

$\Rightarrow 13.6 \times {Z^2}\left[ {{{27 - 5} \over {36}}} \right] = 74.8$

$\Rightarrow 13.6 \times {Z^2} \times {{22} \over {36}} = 74.8$

$\Rightarrow {Z^2} = {{74.8 \times 36} \over {13.6 \times 22}} = 9$

$\Rightarrow Z = 3$

Let Gaussian surface be same as the non-conducting spherical shell. The charge enclosed by this Gaussian surface is the charge on the line segment PQ i.e.,

${q_{enc}} = PQ\lambda = \sqrt 3 R\lambda$,

where we used law of cosines in triangle OPQ to get $PQ = \sqrt {{R^2} + {R^2} - 2(R)(R)\cos 120^\circ } = \sqrt 3 R$. The electric flux through the shell (Gaussian surface) is given by Gauss's law

$\phi = {q_{enc}}/{ \in _0} = \sqrt 3 R\lambda /{ \in _0}$.

The electric field at a point due to the infinitely long line charge is radial i.e., perpendicular to the line PQ. The non-conducting spherical shell does not affect or alter the electric field. Thus, electric field at a point lying on the surface of the shell is perpendicular to the z-axis i.e., its z component is zero. The figure shows electric field on the surface of the shell for a positive line charge. Note that the magnitude of electric field is inversely proportional to the distance of the point from the line charge.

Process I is adiabatic, therefore, $\Delta$Q = 0

$\Rightarrow$ $\Delta$Q = $\Delta$U + W $\Rightarrow$ W = $-$$\Delta$U

Thus, W < 0 since volume of gas is decreasing $\Rightarrow$ $\Delta$U > 0 Therefore, temperature of gas increases and no heat is exchanged between gas and surroundings.

Thus, the correct mapping is P $\to$ 3.

Process II is isobaric, therefore, pressure remains constant

$\Rightarrow$ W = P$\Delta$V = 3P₀[3V₀ $-$ V₀] = 6P₀V₀

Therefore, work done by gas in process II is 6P₀V₀.

Thus, the correct mapping is Q $\to$ 4.

Process III is isochoric, therefore, volume remains constant

$\Rightarrow$ $\Delta$Q = $\Delta$U + W $\Rightarrow$ $\Delta$Q = $\Delta$U

and work done by gas is zero.

Thus, the correct mapping is R $\to$ 1.

Process IV is isothermal process, therefore temperature remains constant

$\Rightarrow$ $\Delta$Q = $\Delta$U + W $\Rightarrow$ $\Delta$Q = W and $\Delta$ = 0

Thus, the correct mapping is S $\to$ 2.

Therefore, option (C) is correct.

Given : ${{{m_1}} \over {{m_2}}} = 2$ and ${{{R_1}} \over {{R_2}}} = {1 \over 4}$

Now,

${{GMm} \over {{R^2}}} = {{m{v^2}} \over R}$

$\Rightarrow {{GM} \over R} = {v^2} \Rightarrow v = \sqrt {{{GM} \over R}}$

$\Rightarrow {{{v_1}} \over {{v_2}}} = {{\sqrt {{{GM} \over {{R_1}}}} } \over {\sqrt {{{GM} \over {{R_2}}}} }} \Rightarrow {{{v_1}} \over {{v_2}}} = \sqrt {{{{R_2}} \over {{R_1}}}}$

$\Rightarrow {{{v_1}} \over {{v_2}}} = \sqrt {{4 \over 1}} \Rightarrow {{{v_1}} \over {{v_2}}} = 2$

Thus, the correct mapping is P $\to$ 3.

Now, L = mvR

$\Rightarrow {{{L_1}} \over {{L_2}}} = {{{m_1}{v_1}{R_1}} \over {{m_2}{v_2}{R_2}}} = {{{m_1}} \over {{m_2}}} \times {{{v_1}} \over {{v_2}}} \times {{{R_1}} \over {{R_2}}}$

$\Rightarrow {{{L_1}} \over {{L_2}}} = 2 \times 2 \times {1 \over 4} = 1$

Thus, the correct mapping is Q $\to$ 2.

Now,

$K = {1 \over 2}m{v^2}$

$\Rightarrow {{{K_1}} \over {{K_2}}} = {{{1 \over 2}{m_1}v_1^2} \over {{1 \over 2}{m_2}v_2^2}} = {{{m_1}} \over {{m_2}}} \times {\left( {{{{v_1}} \over {{v_2}}}} \right)^2}$

$\Rightarrow {{{K_1}} \over {{K_2}}} = 2 \times {(2)^2} = 8$

Thus, the correct mapping is R $\to$ R.

Now,

$T = {{2\pi R} \over v}$

$\Rightarrow {{{T_1}} \over {{T_2}}} = {{{{2\pi {R_1}} \over {{v_1}}}} \over {{{2\pi {R_2}} \over {{v_2}}}}} \Rightarrow {{{T_1}} \over {{T_2}}} = {{{R_1}} \over {{R_2}}} \times {{{v_2}} \over {{v_1}}}$

$\Rightarrow {{{T_1}} \over {{T_2}}} = {1 \over 4} \times {1 \over 2} \Rightarrow {{{T_1}} \over {{T_2}}} = {1 \over 8}$

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

Therefore, option (B) is correct.

The linear momentum $\overrightarrow p$ of a particle is conserved if force on it is zero i.e., $\overrightarrow F = m\overrightarrow a = \overrightarrow 0$. If $\overrightarrow p$ is conserved then kinetic energy $K = {p^2}/2m$ is also conserved. The angular momentum $\overrightarrow L$ of a particle is conserved if torque on it is zero i.e., $\overrightarrow \tau = \overrightarrow r \times \overrightarrow F = \overrightarrow 0$. The total energy E is conserved for conservative forces.

For P : $\overrightarrow r (t) = \alpha t\widehat i + \beta t\widehat j$

$\overrightarrow v = {{d\overrightarrow r } \over {dt}} = \alpha \widehat i + \beta \widehat j$ = constant

$\Rightarrow \overrightarrow p$ is constant (conserved)

Now, $\left| {\overrightarrow v } \right| = \sqrt {{\alpha ^2} + {\beta ^2}}$ = constant

Therefore, $K = {1 \over 2}m{v^2}$ is constant (conserved)

Now, $\overrightarrow a = {{d\overrightarrow v } \over {dt}} = 0 \Rightarrow \overrightarrow F = m\overrightarrow a = 0$

Thus, $F = \nabla U \Rightarrow U$ is constant (conserved)

Now, $E = U + K \Rightarrow E$ is constant, since U and K are constant

Now, $\overrightarrow L = m(\overrightarrow r \times \overrightarrow v ) = 0 \Rightarrow \overrightarrow L$ is constant (conserved)

Thus, the correct mapping is $P \to 1,2,3,4,5$.

For Q : $\overrightarrow r (t) = \alpha \cos \omega t\,\widehat i + \beta \sin \omega t\,\widehat j$

$\Rightarrow v = {{d\overrightarrow r } \over {dt}} = - \alpha \omega \sin \omega t\,\widehat i + \beta \omega \cos \omega t\,\widehat j$

So v is not a constant $\Rightarrow \overrightarrow p$ is not conserved for the path.

Now, $\left| {\overrightarrow v } \right| = \sqrt {{\alpha ^2}{\omega ^2}{{\sin }^2}\omega t + {\beta ^2}{\omega ^2}\sin \omega t}$ which is not constant. Therefore,

$K = {1 \over 2}m{v^2}$ is not constant (not conserved)

Now, $\overrightarrow a = {{d\overrightarrow v } \over {dt}} = - \alpha {\omega ^2}\cos \omega t\,\widehat i + \beta {\omega ^2}\sin \omega t\,\widehat j$

$\Rightarrow \overrightarrow a$ is not constant

$\Rightarrow \overrightarrow F$ is not constant

$\Rightarrow \overrightarrow U$ is not constant (not conserved)

E = U + K is constant (conserved)

Now, $\overrightarrow L = m(\overrightarrow r \times \overrightarrow v ) = m\omega \alpha \beta \widehat k$

$\Rightarrow \overrightarrow L$ is constant (conserved)

Thus, the correct mapping is $Q \to 2,5$.

For R : $\overrightarrow r (t) = \alpha (\cos \omega t\,\widehat i + \sin \omega t\,\widehat j)$

$\overrightarrow v = {{d\overrightarrow r } \over {dt}} = \alpha \omega ( - \sin \omega t\,\widehat i + \cos \omega t\,\widehat j)$

$\Rightarrow \overrightarrow v$ is not a constant

$\Rightarrow \overrightarrow p$ is not a constant (not conserved)

Now, $\left| {\overrightarrow v } \right| = \sqrt {{\alpha ^2}{\omega ^2}({{\sin }^2}\omega t + {{\cos }^2}\omega t)} = \alpha \omega$

$\Rightarrow \left| {\overrightarrow v } \right|$ is constant

$\Rightarrow K = {1 \over 2}m{v^2}$ is a constant (conserved)

Now, $\overrightarrow a = {{dv} \over {dt}} = \alpha {\omega ^2}( - \sin \omega t\,\widehat i - \cos \omega t\,\widehat j)$

$\Rightarrow \overrightarrow a$ is not constant

Since $E = U + K$ is constant (conserved)

$\Rightarrow U$ is constant (conserved)

Now, $\overrightarrow L = m(\overrightarrow r \times \overrightarrow v ) = m\omega {\alpha ^2}\widehat k$

$\Rightarrow \overrightarrow L$ is constant (conserved)

Thus, the correct mapping is $R \to 2,3,4,5.$

For S : $\overrightarrow r (t) = \alpha t\,\widehat i + {\beta \over 2}{t^2}\,\widehat j$

$\overrightarrow v = {{d\overrightarrow r } \over {dt}} = \alpha \widehat i + \beta t\,\widehat j$

$\Rightarrow \overrightarrow v$ is not conserved

$\Rightarrow \overrightarrow p$ is not conserved

Now, $\left| {\overrightarrow v } \right| = \sqrt {{\alpha ^2} + {\beta ^2}\,{t^2}}$

$\Rightarrow \left| {\overrightarrow v } \right|$ is not conserved

$\Rightarrow K = {1 \over 2}m{v^2}$ is not conserved

$\overrightarrow a = {{d\overrightarrow v } \over {dt}} = \beta \widehat j$

$\Rightarrow E = U + K$ is conserved

But, U is not conserved.

Now, $\overrightarrow L = m(\overrightarrow r \times \overrightarrow v ) = {1 \over 2}\alpha \beta {t^2}\widehat k$

$\Rightarrow \overrightarrow L$ is not conserved.

Thus, the correct mapping is $S \to 5$.

Therefore, option (A) is correct.

The frequency of incident light is just above the threshold frequency. Hence, the energy of each photon is equal to the work function (E_p = $\phi$ = 6.25 eV) and the kinetic energy of emitted photo-electron is zero (K_e0 = 0). The energy incident per second on the cathode is incident power P = 200 W. Thus, number of photons incident per second is

N_p = P/E_p = P/$\phi$.

The photo-electron emission efficiency is 100%. Thus, number of photo-electron emitted per second is equal to the number of photons incident per second i.e., N_e = N_p. These photo-electrons are accelerated by a potential difference V = 500 V. Thus, gain in potential energy of each photo-electron is $\Delta$U = eV = 500 eV. By conservation of energy, kinetic energy of the photo-electron when it reaches the anode is

K_e = K_e0 + $\Delta$U = 500 eV.

The linear momentum of photo-electron of mass m_e and kinetic energy K_e is given by

${p_e} = \sqrt {2{m_e}{K_e}}$.

The photo-electron transfer its entire linear momentum to the anode (absorbed by the anode). Thus, gain in linear momentum of the anode by absorbing one photo-electron is

$\Delta$p_a = p_e.

The force on the anode is equal to the increase in its linear momentum per second. Since N_e photo-electrons strikes the anode per second, the force acting on the anode is given by

${F_a} = {N_e}\Delta {p_a} = {N_e}{p_e}$ ($\because$ $\Delta {p_a} = {p_e}$)

$= {N_p}{p_e} = (P/\phi ){p_e}$ ($\because$ ${N_e} = {N_p} = P/\phi$)

$= (P/\phi )\sqrt {2{m_e}{K_e}}$ ($\because$ ${p_e} = \sqrt {2{m_e}{K_e}}$)

$= (P/\phi )\sqrt {2{m_e}eV}$ ($\because$ ${K_e} = eV$

$= 2.4 \times {10^{ - 4}}N$.

Consider the point F, the focus of the concave mirror. All rays emanating from F will be reflected back parallel to the principal axis PF. Thus, virtual image F' of the point F is formed at $\infty$.

Consider the point L at an object distance u = $-$ | f | / 2. For convenience, we use symbol $\widehat f$ for | f | i.e., $\widehat f$ = | f |. Use mirror formula, 1/v + 1/u = 1/f, with f = $-$ $\widehat f$ to get the image distance v = $\widehat f$. Thus, the virtual image L' of the point L is formed on the left side of the mirror at a distance $\widehat f$.

The image of a point on LF will lie on the line L'F'. Now, consider the point M. Its object distance is u = $-$ $\widehat f$/2. Thus, the virtual image M' of the point M is also formed at a distance $\widehat f$ from the mirror. The distance of M' above the principal axis is given by h_M' = $-$ (v/u) (LM) = $-$ ($-$2) ($\widehat f$/2) = $\widehat f$ (magnification by a spherical mirror).

Now consider a point N on the line segment FM. Let N₁ be the projection of N on the principal axis PF such that FN₁ = x. The object distance is u = $-$ ($\widehat f$ $-$ x). The mirror formula gives the image distance v = $\widehat f$ ($\widehat f$ $-$ x) / x. The distance of the image N' above the principal axis is given by h_N' = $-$ (v/u) (N₁N) = ($\widehat f$/x) (x) = $\widehat f$ (it is independent of x). Since, N is an arbitrary point on FM, the image of all points on FM will be formed at a distance $\widehat f$ above the optic axis.

Given, a particle at rest is subjected to force. Motion of particle is along x-axis.

Change in kinetic energy is given as

$K = {{dK} \over {dt}} = \gamma t$ ..... (1)

where $\gamma$ is positive constant.

We know that kinetic energy is given by

$K = {1 \over 2}m{v^2}$...... (2)

Now, differentiate Eq. (2), we get

${{dK} \over {dt}} = {2 \over 2}mv{{dv} \over {dt}} = mv{{dv} \over {dt}}$

${{dK} \over {dt}} = mv{{dv} \over {dt}}$ ....... (3)

From Eqs. (1) and (3), we get

$mv{{dv} \over {dt}} = \gamma t$

$\Rightarrow mv\,dv = \gamma t\,dt$ ...... (4)

Integrating Eq. (4), we get

$\int {mv\,dv = \int {\gamma t\,dt \Rightarrow m\int {v\,dv = \gamma \int {t\,dt} } } }$

${{m{v^2}} \over 2} = {{\gamma {t^2}} \over 2} \Rightarrow m{v^2} = \gamma {t^2}$

$\Rightarrow {v^2} = {\gamma \over m}{t^2} \Rightarrow v = \sqrt {{\gamma \over m}} t$

$\Rightarrow v \propto t$

Since, $\gamma$ and m are constant so the speed of particle is proportional to time therefore option (B) is true.

Also, $a = {{dv} \over {dt}} \Rightarrow a = \sqrt {{\gamma \over m}}$ = constant

F = ma = constant

So, option (A) is true.

Also, $v = {{dr} \over {dt}}$

$\Rightarrow {{dr} \over {dt}} = \sqrt {{\gamma \over m}} t$

$\Rightarrow dr = \sqrt {{\gamma \over m}} t\,dt$ ...... (5)

Integrating Eq. (5), we get

$\int {dr = \int {\sqrt {{\gamma \over m}} t\,dt \Rightarrow r = \sqrt {{\gamma \over m}} {{{t^2}} \over 2}} }$

$\Rightarrow r \propto {t^2}$

So, the distance of the particle from the origin increases at t² so option (C) is false.

A force is conservative if work done by the force in any closed loop is zero (i.e., work done is independent of the path). If force is constant (both magnitude and direction) then work done by the force in a closed loop is given by

$W = \oint {\overrightarrow F \,.\,d\overrightarrow r = \overrightarrow F \,.\,\oint {d\overrightarrow r = \overrightarrow F \,.\,\overrightarrow 0 } = 0}$.

Note that we are able to take $\overrightarrow F$ out of integral sign because $\overrightarrow F$ is constant and $\oint {d\overrightarrow r }$ is zero because integration is carried out over a closed loop. The argument 'since force is constant, hence it is conservative' should be understood properly. The kinetic friction force and viscous drag force (when body moves with a terminal speed) are both constant (in magnitude) but are non-conservative. These forces cannot start the particle motion from the rest. The situation given in this problem is similar to a particle dropped from rest under the gravitational force. This force is constant, can initiate the motion given in the problem and is conservative.