JEE Advanced 2015 Paper 2 Question Paper with Solutions

${}_{92}^{236}U \to {}_{54}^{140}Xe + {}_{38}^{94}Sr + x + y$

K_x = 2 MeV, K_y = 2 MeV, K_Xe = ?, K_Sr = ?

By conservation of charge number and mass number, x $\equiv$ y $\equiv$ n

B.E. per nucleon of ${}_{92}^{236}U = 7.5$ MeV

B.E. per nucleon of ${}_{54}^{140}Xe$ or ${}_{38}^{94}Sr = 8.5$ MeV

Q value of reaction,

Q = Net kinetic energy gained in the process

$= {K_{Xe}} + {K_{Sr}} + 2 + 2 - 0 = {K_{Xe}} + {K_{Sr}} + 4$ ..... (1)

As number of nucleons is conserved in a reaction, so Q = Difference of binding energies of the nuclei

$= 140 \times 8.5 + 94 \times 8.5 - 236 \times 7.5 = 219$ MeV ...... (2)

From eqns. (i) and (ii)

${K_{Xe}} + {K_{Sr}} = 219 - 4 = 215$ MeV ....... (3)

The linear momentum of a particle of mass m and kinetic energy K is given by $p = \sqrt {2mK}$. Since the masses and kinetic energies of x and y are very small in comparison to that of ${}_{54}^{140}Xe$ and ${}_{38}^{94}Sr$, we can neglect the linear momentum of these particles. Initially, the linear momentum of ${}_{92}^{236}U$ is zero (at rest). Finally, the products ${}_{54}^{140}Xe$ and ${}_{38}^{94}Sr$ will move in opposite direction with equal linear momentum (by conservation of linear momentum). Thus,

$\sqrt {2{M_{Xe}}{K_{Xe}}} = \sqrt {2{M_{Sr}}{K_{Sr}}}$, i.e.,

$140{K_{Xe}} = 94{K_{Sr}}$. ...... (4)

Solve equations (3) and (4) to get K_Xe = 86 MeV and K_Sr = 129 MeV.

Law of radioactivity : $N = {N_0}{e^{ - \lambda t}}$ where $\lambda$ = decay constant

Activity $\left| A \right| = \left| {{{ - dN} \over {dt}}} \right| = {N_0}\lambda {e^{ - \lambda t}}$

Rate of activity $R = {{d|A|} \over {dt}} = {N_0}{\lambda ^2}{e^{ - \lambda t}}$

At t = 0, A₁ = A₂. Therefore,

${N_{OP}}{\lambda _P} = {N_{OQ}}{\lambda _Q}$

At $t = 2\tau ,\,{{{R_P}} \over {{R_Q}}} = {\left( {{{{\lambda _P}} \over {{\lambda _Q}}}} \right)^2}\left( {{{{N_{OP}}} \over {{N_{OQ}}}}} \right){{{e^{ - \lambda P(25)}}} \over {{e^{ - \lambda Q(25)}}}} = {{{\lambda _P}} \over {{\lambda _Q}}}{e^{({\lambda _Q} - {\lambda _P})25}}$

Since mean life is given by $\tau = {1 \over \lambda }$.

Therefore,

${{{R_P}} \over {{R_Q}}} = {{{\lambda _P}} \over {{\lambda _Q}}}{e^{{{\left( {{1 \over {25}} - {1 \over 5}} \right)}^{25}}}} = {{{\lambda _P}} \over {{\lambda _Q}}}{e^{ - 1}}$

${{{R_P}} \over {{R_Q}}} = {{{\lambda _\{ }} \over {{\lambda _Q}}}{1 \over e} = {n \over e}$

$n = {{{\lambda _P}} \over {{\lambda _Q}}} = {{2\tau } \over \tau } = 2$

The acceleration due to gravity at a radial distance r (r < R) from the centre of a sphere of constant mass density $\rho$ is given by

$g = {{G\left( {{4 \over 3}\pi {r^3}} \right)\rho } \over {{r^2}}} = {{4\pi \rho G} \over 3}r$.

Consider a spherical shell of radius r and thickness dr.

Let P and P + dP be the fluid pressures inside and outside the shell. Consider a small element of area A and mass $dm = \rho Adr$ on the shell. The forces on this element are gravitational force (dm g) direted inwards, the force due to pressure of the fluid outside the shell $((P + dP)A)$ directed inwards, and the force due to pressure of the fluid inside the shell (PA) directed outwards. In equilibrium, net force on the element is zero, i.e.,

$(\rho Adr){4 \over 3}\pi \rho Gr + (P + dP)A = PA$, or

$dP = - {4 \over 3}\pi G{\rho ^2}r\,dr$.

Note that the pressure decreases with increase in r. The pressure is zero at the surface of the sphere i.e., P = 0 at r = R. Integrate to get

$\int_0^{P(r)} {dP = P(r) = - {4 \over 3}\pi G{\rho ^2}\int_R^r {rdr} }$

$= {2 \over 3}\pi G{\rho ^2}({R^2} - {r^2})$.

Now, $P(r = 0) = {{2\pi } \over 3}{\rho ^2}G{R^2} \ne 0$

${{P(r = 3R/4)} \over {P(r = 2R/3)}} = {{{R^2} - {{(3R/4)}^2}} \over {{R^2} - {{(2R/3)}^2}}} = {7 \over {16}} \times {9 \over 5} = {{63} \over {80}}$

${{P(r = 3R/5)} \over {P(r = 2R/5)}} = {{{R^2} - {{(3R/5)}^2}} \over {{R^2} - {{(2R/5)}^2}}} = {{16} \over {25}} \times {{25} \over {21}} = {{16} \over {21}}$

${{P(r = R/2)} \over {P(r = R/3)}} = {{{R^2} - {{(R/2)}^2}} \over {{R^2} - {{(R/3)}^2}}} = {3 \over 4} \times {9 \over 8} = {{27} \over {32}} \ne {{20} \over {27}}$

$E(t) = {A^2}{e^{ - \alpha t}}$ ...... (i)

$\alpha$ = 0.2 s^$-$1

$\left( {{{dA} \over A}} \right) \times 100 = 1.25\%$

$\left( {{{dt} \over t}} \right) \times 100 = 1.50$

$\Rightarrow (dt \times 100) = 1.5t = 1.5 \times 5 = 7.5$

Differentiating on both sides of equation (i), we get

$d \mathrm{E}=(2 \mathrm{~A} d \mathrm{~A}) e^{-\alpha t}+\mathrm{A}^2 e^{-\alpha t}(-\alpha d t)$

Dividing throughout by $\mathrm{E}=\mathrm{A}^2 e^{-\alpha t}$

$\frac{d \mathrm{E}}{\mathrm{E}}=\frac{2}{\mathrm{~A}} d \mathrm{~A}+\alpha d t$

(Considering worst possible case)

$\therefore$ $\left( {{{dE} \over E}} \right) \times 100 = 2\left( {{{dA} \over A}} \right) \times 100 + \alpha (dt \times 100)$

$= 2(1.25) + 0.2(7.5)$

$= 2.5 + 1.5$

$= 4\%$

Let the whole structure be placed in the medium of refractive index n₀. From geometry, if the angle of incidence at Q is $\theta$ then the angle of refraction at P is 90$^\circ$ $-$ $\theta$.

The ray will undergo total internal reflection at Q if the angle of incidence at Q is greater than or equal to the critical angle i.e.,

$\theta$ $\ge$ $\theta$_c = sin^$-$1(n₂ / n₁). ........ (1)

Apply Snell's law for refraction at P to get

n₀ sin i = n₁ sin(90$^\circ$ $-$ $\theta$) = n₁ cos$\theta$

= ${n_1}\sqrt {1 - {{\sin }^2}\theta }$. ...... (2)

From equation (2), the angle of incidence is maximum (i = i_m) when $\theta$ is minimum i.e., when $\theta$ = $\theta$_c (from equation (1)). Thus, the numerical aperture is given by

$NA = \sin {i_m} = {{{n_1}\sqrt {1 - {{\sin }^2}{\theta _c}} } \over {{n_0}}} = {{\sqrt {n_1^2 - n_2^2} } \over {{n_0}}}$ ........ (3)

Substitute ${n_1} = \sqrt {45} /4$ and ${n_2} = 3/2$ in equation (3) to get the numerical aperture for the structure S₁ is

$N{A_1} = {{\sqrt {45/16 - 9/4} } \over {{n_0}}} = {3 \over {4{n_0}}}$ ...... (4)

Similarly, substitute n₁ = 8/4 and n₂ = 7/5 in equation (3) to get the numerical aperture for the structure S₂ as

$N{A_2} = {{\sqrt {64/25 - 49/25} } \over {{n_0}}} = {{\sqrt {15} } \over {5{n_0}}}$ ...... (5)

In case (A), substitute n₀ = 4/3 in equation (4) to get NA₁ = 9/16 and substitute n₀ = 16/3$\sqrt15$ in equation (5) to get NA₂ = 9/16.

In case (B), substitute n₀ = 6/$\sqrt15$ in equation (4) to get NA₁ = $\sqrt15$/8 and substitute n₀ = 4/3 in equation (5) to get NA₂ = 3$\sqrt15$/20.

In case (C), substitute n₀ = 1 in equation (4) to get NA₁ = 3/4 and substitute n₀ = 4/$\sqrt15$ equation (5) to get NA₂ = 3/4.

In case (D), substitute n₀ = 1 in equation (4) to get NA₁ = 3/4 and substitute n₀ = 4/3 equation (5) to get NA₂ = 3$\sqrt15$/20.

We can think of this configuration to be made up of three parts:

(a) Capacitor of plate area s/2, plate separation d, and filled with a dielectric of relative permittivity $\in$₁ = 2 (lower half). The capacitance of this part is ${C_a} = { \in _1}{ \in _0}{{s/2} \over d} = { \in _0}{s \over d} = {C_1}$.

(b) Capacitor of plate area s/2, plate separation d/2, and filled with a dielectric of relative permittivity $\in$₁ = 2 (left upper half). The capacitance of this part is ${C_b} = { \in _1}{ \in _0}{{s/2} \over {d/2}} = 2{ \in _0}{s \over d} = 2{C_1}$.

(c) Capacitor of plate area s/2, plate separation d/2, and filled with a dielectric of relative permittivity $\in$₂ = 4 (right upper half). The capacitance of this part is ${C_c} = { \in _2}{ \in _0}{{s/2} \over {d/2}} = 4{ \in _0}{s \over d} = 4{C_1}$.

The capacitors C_b and C_c are connected in series with equivalent capacitance

${C_{bc}} = {{{C_b}{C_c}} \over {{C_b} + {C_c}}} = {{(2{C_1})(4{C_1})} \over {(2{C_1}) + (4{C_1})}} = {4 \over 3}{C_1}$.

The capacitor C_bc is connected in parallel with C_a. The equivalent capacitance is

$\left. {{C_2} = {C_{abc}} = {C_a}} \right\|{C_{bc}} = {C_1} + {4 \over 3}{C_1} = {7 \over 3}{C_1}$.

Angular momentum $mvr = {{nh} \over {2\pi }}$ where $r = 3{a_0}$ where $n = 3$, that is, electron in $L{i^{2 + }}$ is in second excited state

$\lambda = {h \over {mv}} = p\pi {a_0}$

$\Rightarrow n = p\pi (mv{a_0}) = p\pi \left( {{{mvr} \over 3}} \right) = {{p\pi } \over 3}\left( {{{3h} \over {2\pi }}} \right) = {{ph} \over 2}$

Therefore, $p = 2$.

Let $\rho$ be the charge density of the spherical charge distribution of radius r₁ centred at the origin O. A spherical cavity of radius r₂ centred at P with distance OP = a = r₁ $-$ r₂ is made in the spherical charge distribution.

The sphere with cavity is equivalent to a sphere of uniform charge density $-$$\rho$ and radius r₂ centred at P embedded in the original sphere. Thus, the electric field at a point Q in the cavity is superposition of (i) electric field at Q due to the sphere of charge density $\rho$ and radius r₁ centred at O (say $\overrightarrow E$₁), and (ii) electric field at Q due to the sphere of charge density $-$$\rho$ and radius r₂ centred at P (say $\overrightarrow E$₂). Let $\overrightarrow a$, $\overrightarrow r$, and $\overrightarrow r$ $-$ $\overrightarrow a$ be the vectors as shown in the figure. The electric fields $\overrightarrow E$₁, $\overrightarrow E$₂, and their superposition $\overrightarrow E$₁₂ are given by

${\overrightarrow E _1} = {1 \over {4\pi { \in _0}}}{{{4 \over 3}\pi |\overrightarrow r {|^3}\rho } \over {|\overrightarrow r {|^2}}}\widehat r = {\rho \over {3{ \in _0}}}\overrightarrow r$,

${\overrightarrow E _2} = - {\rho \over {3{ \in _0}}}(\overrightarrow r - \overrightarrow a )$,

${\overrightarrow E _{12}} = {\overrightarrow E _1} + {\overrightarrow E _2} = {\rho \over {3{ \in _0}}}\overrightarrow a$.

Thus, the electric field at a point within the cavity is uniform and its magnitude and direction both depend on $\overrightarrow a$.

$\sin i_m=n_1 \sin \left(90-\theta_c\right)$

$\Rightarrow \sin i_m=n_1 \cos \theta_c$

$$\begin{aligned} \Rightarrow N A & =n_1 \sqrt{1-\sin ^2 \theta_c} \\ & =n_1 \sqrt{1-\frac{n_2^2}{n_1^2}}=\sqrt{n_1^2-n_2^2} \end{aligned}$$

Substituting the values we get,

$N A_1=\frac{3}{4}$

$\text { and } N A_2=\frac{\sqrt{15}}{5}=\sqrt{\frac{3}{4}}$

and $N A_2 < N A_1$

Therefore, the numerical aperture of combined structure is equal to the lesser of the two numerical aperture, which is $N A_2$.

At equilibrium,

$$\begin{aligned} & \mathrm{F}_{\mathrm{M}} =\mathrm{F}_{\mathrm{E}} \\ & q \mathrm{v} \mathrm{B} =q \mathrm{E} \\ & \mathrm{E} =\mathrm{v} \cdot \mathrm{B} \\ & i =n e \mathrm{~A} v_d \end{aligned}$$

where $v_d=$ drift speed of electrons and $n=$ electron density

$$\begin{aligned} \mathrm{E} & =\frac{i}{n e \mathrm{~A}} \mathrm{~B} \\ & =\frac{i}{n e(\omega d)} \end{aligned}$$

Area of cross section is always taken normal to the direction of flow of current. Potential difference,

$\mathrm{V}=\mathrm{E} \omega=\frac{i \mathrm{~B}} { ned }$

For the two strips

$\mathrm{V}_1=\frac{i \mathrm{~B}}{n e}\left(\frac{1}{d_1}\right)$

$\mathrm{V}_2=\frac{i B}{n e}\left(\frac{1}{d_2}\right)$

For option (A)

$\frac{\mathrm{V}_1}{\mathrm{~V}_2}=\frac{d_2}{d_1}=\frac{1}{2}$

For option (B)

$\frac{\mathrm{V}_1}{\mathrm{~V}_2}=1, \frac{d_2}{d_1}=\frac{1}{2}$

For option (C)

$\frac{\mathrm{V}_1}{\mathrm{~V}_2}=\frac{1}{2}, \frac{d_2}{d_1}=1$

For option (D)

$\frac{\mathrm{V}_1}{\mathrm{~V}_2}=\frac{d_2}{d_1}=1$

The intensity of a wave is proportional to the square of its amplitude i.e., I₀ = cA², where c is a constant. The amplitudes of four harmonic waves are equal as their intensities are equal. Let these waves be travelling along the x direction with wave vector k and angular frequency $\omega$. The resultant displacement of these waves is given by

$y = {y_1} + {y_2} + {y_3} + {y_4}$

$= A\sin (\omega t - kx + 0) + A\sin (\omega t - kx + \pi /3) + Asin(\omega t - kx + 2\pi /3) + A\sin (\omega t - kx + \pi )$

$= A\sin (\omega t - kx + \pi /3) + Asin(\omega t - kx + 2\pi /3)$

$= 2A\sin (\omega t - kx + \pi /2)\cos (\pi /6)$

$= \sqrt 3 A\cos (\omega t - kx)$.

The amplitude of the resultant wave is ${A_r} = \sqrt 3 A$ and its intensity is ${I_r} = cA_r^2 = 3c{A^2} = 3{I_0}$.

Using Snell's law:

$\sin 60^\circ = n\sin {r_1}$ ..... (1)

$\sin {r_1} = {{\sqrt 3 } \over {2 \times \sqrt 3 }} = {1 \over 2}$

${r_1} = 30^\circ$

Also,

$n\sin {r_2} = 1\sin \theta$

Also

${r_1} + {r_2} = A = 60^\circ$

Therefore,

$n\sin (60^\circ - {r_1}) = 1\sin \theta$ ....... (2)

Differentiating on both sides, we get

$\sin (60^\circ - {r_1}) - n\cos (60^\circ - {r_1}){{d{r_1}} \over {dn}} = \cos \theta {{d\theta } \over {dn}}$

Differentiating Eq. (1) on both sides, we get

$0 = \sin {r_1} + n\cos {r_1}{{d{r_1}} \over {dn}}$

$0 = {1 \over 2} + \sqrt 3 .{{\sqrt 3 } \over 2}{{d{r_1}} \over {dn}}$

Therefore,

${{d{r_1}} \over {dn}} = {{ - 1} \over 3}$

Hence, substituting ${r_1} = 30^\circ$, we get

${{d{r_1}} \over {dn}} = {{ - 1} \over 3}$

Now, $\sin 30^\circ - \sqrt 3 \cos 30^\circ \left( { - {1 \over 3}} \right) = \cos 60^\circ {{d\theta } \over {dn}}$

${1 \over 2} + {3 \over {2 \times 3}} = {1 \over 2}{{d\theta } \over {dn}}$

${{d\theta } \over {dn}} = 2$

Consider the following figure:

ACGA constitutes a Wheatstone bridge; hence, 8 $\Omega$ is redundant and hence can be removed. Therefore,

${R_{AG}} = {{3 \times 6} \over 9} = \,2\,\Omega$

AGDFA again constitutes a Wheatstone bridge 10 $\Omega$ which is redundant and hence can be removed.

${R_{AB}} = {{6 \times 18} \over {24}} = 4.5\,\,\Omega$

$I = {{6.5} \over {6.5}} = 1\,A$

Let $V=\frac{4}{3} \pi r^3$ be the volume of the spheres $P$ and $Q$ of equal radii $r$. The forces acting on the sphere $P$ are its weight $\rho_1 V g$, tension from the string $T$, and the buoyancy force $\sigma_1 V g$.



Similarly, forces on the sphere $Q$ are $\rho_2 V g, T$, and $\sigma_1 V g$. In equilibrium, the net force on the spheres $P$ and $Q$ are separately zero i.e.,

$$\begin{aligned} & T+\rho_1 V g=\sigma_1 V g .........(1) \\ & T+\sigma_2 V g=\rho_2 V g ..........(2) \end{aligned}$$

The tension $T > 0$ because the string is taut. Thus, equation (1) gives $\rho_1<\sigma_1$ and equation (2) gives $\rho_2>$ $\sigma_2$. Eliminate $T$ from equations (1) and (2) to get

$\sigma_1-\rho_1=\rho_2-\sigma_2$ ..........(3)



Now, consider the situation when the sphere $P$ moves in liquid $L_2$ and the sphere $Q$ moves in liquid $L_1$. These spheres will attain the terminal velocities $\vec{v}_{\mathrm{P}}$ and $\vec{v}_{\mathrm{Q}}$ after some time. The direction of the velocity (upwards or downwards) will depend on the density of the sphere in comparison to the density of the liquid. Let us consider the case when $\rho_1>\sigma_2$. In this case, the velocity of the sphere $P$ is downwards. From equation (3), if $\rho_1>\sigma_2$ then $\rho_2<\sigma_1$. If the density of a sphere is less than the density of the liquid in which it is immersed, it will move up. Thus, the velocity of the sphere $Q$ is upwards i.e., the directions of $\vec{v}_{\mathrm{P}}$ and $\vec{v}_{\mathrm{Q}}$ are opposite.

Hence, $\vec{v}_{\mathrm{P}} \cdot \vec{v}_{\mathrm{Q}}<0$.

The forces on the sphere $P$ are its weight $\rho_1 V g$, buoyancy force $\sigma_2 V g$, and viscous drag $6 \pi \eta_2 r v_{\mathrm{P}}$ (see figure). Similarly, the forces on the sphere $Q$ are $\rho_2 V g, \sigma_1 V g$ and $6 \pi \eta_1 r v_{\mathrm{Q}}$. Net forces on the spheres are zero when they move with terminal velocities i.e.,

$$\begin{aligned} & 6 \pi \eta_2 r v_{\mathrm{P}}+\rho_1 V g=\sigma_2 V g ...........(4) \\ & 6 \pi \eta_1 r v_{\mathrm{Q}}+\sigma_2 V g=\rho_1 V g ..........(5) \end{aligned}$$

Solving equations (4) and (5), we get

$$\begin{aligned} & v_{\mathrm{P}}=\frac{2 r^2\left(\rho_1-\sigma_2\right)}{9 \eta_2} .........(6) \\ & v_{\mathrm{Q}}=\frac{2 r^2\left(\sigma_1-\rho_2\right)}{9 \eta_1} .........(7) \end{aligned}$$

By dividing equation (6) by (7), we get

$\frac{\left|\vec{v}_{\mathrm{P}}\right|}{\left|\vec{v}_{\mathrm{Q}}\right|}=\frac{\eta_1}{\eta_2} \frac{\rho_1-\sigma_2}{\sigma_2-\rho_2}=\frac{\eta_1}{\eta_2} .$

The speed of light in vacuum is given by, $c = {1 \over {\sqrt {{\mu _0}{\varepsilon _0}} }}$ and The impedance (resistance) of free space is defined a $R = \sqrt {{{{\mu _0}} \over {{\varepsilon _0}}}}$. Using this check the dimensional correctness of equations.

(a) ${\mu _0}{I^2} = {\varepsilon _0}{V^2}$

${{{\mu _0}} \over {{\varepsilon _0}}} = {{{V^2}} \over {{I^2}}} = {R^2}$

$\therefore$ ${R^2} = {R^2}$ which is dimensionally correct.

(b) ${\varepsilon _0}I = {\mu _0}V \Rightarrow {{{\varepsilon _0}} \over {{\mu _0}}} = {V \over I} \Rightarrow {1 \over {{R^2}}} = R$

which is dimensionally incorrect.

(c) $I = {\varepsilon _0}cV \Rightarrow {I \over V} = {\varepsilon _0}C = {{{\varepsilon _0}} \over {\sqrt {{\varepsilon _0}{\mu _0}} }}$ = $\sqrt {{{{\varepsilon _0}} \over {{\mu _0}}}}$ = ${1 \over R}$

${1 \over R} = \sqrt {{{{\varepsilon _0}} \over {{\mu _0}}}} \Rightarrow {1 \over R} = {1 \over R}$

which is dimensionally correct.

(d) ${\mu _0}cI = {\varepsilon _0}V \Rightarrow {{{\mu _0}c} \over {{\varepsilon _0}}} = {V \over I}$

${{{\mu _0}} \over {{\varepsilon _0}}}{1 \over {\sqrt {{\varepsilon _0}{\mu _0}} }} = R \Rightarrow {1 \over {{\varepsilon _0}}}\sqrt {{{{\mu _0}} \over {{\varepsilon _0}}}} = R$

${R \over {{\varepsilon _0}}} = R$, which is dimensionally incorrect.

Consider a spherical shell of radius r and small thickness dr.

The volume of the shell is $dV = 4\pi {r^2}dr$ and its mass is

$dm = \rho dV = 4\pi \rho {r^2}dr$.

The moment of inertia of the spherical shell of mass dm and radius r about an axis passing through its centre O is given by $dI = {2 \over 3}dm\,{r^2}$. Substitute the expressions for dm and $\rho$ and then integrate to get the moment of inertia of the two spheres.

${I_A} = \int_0^R {{2 \over 3}(4\pi {\rho _A}{r^4})dr = {{8\pi k} \over {3R}}\int_0^R {{r^5}dr = {{8\pi k{R^5}} \over {18}}} }$,

${I_B} = \int_0^R {{2 \over 3}(4\pi {\rho _B}{r^4})dr = {{8\pi k} \over {3{R^5}}}\int_0^R {{r^9}dr = {{8\pi k{R^5}} \over {30}}} }$.

Divide to get ${I_B}/{I_A} = 6/10$.

Initially both the compartments has same pressure as they are in equilibrium.

Suppose spring is compressed by x on heating the gas.

Let A be the area of cross-section of piston. As gas is ideal monoatomic, so

${{{P_1}{V_1}} \over {{T_1}}} = {{{P_2}{V_2}} \over {{T_2}}}$ ...... (i)

Force on spring by gas = kx

$\therefore$ ${P_2} = {P_1} + {{kx} \over A}$ ...... (ii)

Case I : When ${V_2} = 2{V_1}$, ${T_2} = 3{T_1}$

From Eqn. (i)

${{{P_1}{V_1}} \over {{T_1}}} = {{{P_2}(2{V_1})} \over {3{T_1}}} \Rightarrow {P_2} = {3 \over 2}{P_1}$

Putting this value in eqn. (ii) we get

${3 \over 2}{P_1} = {P_1} + {{kx} \over A} \Rightarrow kx = {{{P_1}A} \over 2}$

$x = {{{V_2} - {V_1}} \over A} = {{2{V_1} - {V_1}} \over A} = {{{V_1}} \over A}$

Energy stored in the spring

$= {1 \over 2}k{x^2} = {1 \over 2}(kx)(x) = {{{P_1}{V_1}} \over 4}$

So, option (a) is correct.

Change in internal energy,

$\Delta U = {f \over 2}({P_2}{V_2} - {P_1}{V_1}) = {3 \over 2}\left( {{3 \over 2}{P_1} \times 2{V_1} - {P_1}{V_1}} \right) = 3{P_1}{V_1}$

So, option (b) is correct.

Case II : When ${V_2} = 3{V_1}$ and ${T_2} = 4{T_1}$

From Eqn. (i),

${{{P_1}{V_1}} \over {{T_1}}} = {{{P_2}(3{V_1})} \over {4{T_1}}} \Rightarrow {P_2} = {4 \over 3}{P_1}$

$x = {{{V_2} - {V_1}} \over A} = {{2{V_1}} \over A}$

From eqn. (ii),

${4 \over 3}{P_1} = {P_1} + {{kx} \over A} \Rightarrow kx = {{{P_1}A} \over 3}$

Gas is heated very slowly so pressure on the other compartment remains same.

Work done by gas = Work done by gas on atmosphere + Energy stored in spring.

${W_g} = {P_1}Ax + {1 \over 2}k{x^2} = {P_1}(2{V_1}) + {1 \over 2}\left( {{{{P_1}A} \over 3}} \right)\left( {{{2{V_1}} \over A}} \right)$

$= 2{P_1}{V_1} + {1 \over 3}{P_1}{V_1} = {7 \over 3}{P_1}{V_1}$

So, option (c) is correct.

Heat supplied to the gas,

$\Delta Q = {W_g} + \Delta U$

$= {7 \over 3}{P_1}{V_1} + {3 \over 2}({P_2}{V_2} - {P_1}{V_1})$

$= {7 \over 3}{P_1}{V_1} + {3 \over 2}\left( {{4 \over 3}{P_1} \times 3{V_1} - {P_1}{V_1}} \right)$

$= {7 \over 3}{P_1}{V_1} + {9 \over 2}{P_1}{V_1} = {{41} \over 6}{P_1}{V_1}$

So, option (d) is incorrect.

We know, $Y = {{Stress} \over {Strain}}$

According to graph,

Slope of curve $= {{Change\,in\,strain} \over {Change\,in\,stress}} = {1 \over Y}$

(Slope)_P > (Slope)_Q $\therefore$ Y_P < Y_Q

P has more tensile strength than Q as it sustains more stress after elastic limit.

There is large deformation between the elastic limit and the fracture point for material P as compared to material Q. Hence, P is more ductile than Q.

After the elastic limit, Q breaks soon as compared to P. So, Q is more brittle than P.

$$\text { and } \quad \begin{aligned} & \mathrm{V}_1 =\frac{i \mathrm{~B}_1}{n_1 e d} \\ & \mathrm{~V}_2 =\frac{i \mathrm{~B}_2}{n_2 e d} \\ & \frac{\mathrm{V}_1}{\mathrm{~V}_2} =\frac{\mathrm{B}_1}{\mathrm{~B}_2} \frac{n_2}{n_1} \end{aligned}$$

For option (A) :

$\frac{\mathrm{V}_1}{\mathrm{~V}_2}=\frac{1}{2}$

and $\frac{\mathrm{B}_1}{\mathrm{~B}_2} \frac{n_2}{n_1}=1 \times \frac{1}{2}=\frac{1}{2}$

For option (B) :

$$\begin{aligned} \frac{\mathrm{V}_1}{\mathrm{~V}_2} & =1 \\ \text { and } \frac{\mathrm{B}_1}{\mathrm{~B}_2} \frac{n_2}{n_1} & =1 \times \frac{1}{2}=\frac{1}{2} \end{aligned}$$

For option (C) :

$$\begin{aligned} \frac{\mathrm{V}_1}{\mathrm{~V}_2} & =2 \\ \text { and } \frac{\mathrm{B}_1}{\mathrm{~B}_2} \frac{n_2}{n_1} & =2 \times 1=2 \end{aligned}$$

For option (D) :

$$\begin{aligned} \frac{\mathrm{V}_1}{\mathrm{~V}_2} & =1 \\ \text { and } \frac{\mathrm{B}_1}{\mathrm{~B}_2} \frac{n_2}{n_1} & =2 \times 1=2 \end{aligned}$$

The acceleration $\vec{a}$ of the point masses are equal because they are connected by a massless rigid rod.


Consider the situation when tension in the rod is zero. The gravitational forces on the two point masses are shown in the figure. The forces $f_1=\frac{G M m}{r^2}$ and $f_3=\frac{G M m}{(r+l)^2}$ are due to the attraction by the larger mass $M$. The force $f_2=\frac{G m m}{l^2}$ is due to mutual attraction between the two point masses. Apply Newton's second law on the two point masses to get

$\frac{G M m}{r^2}-\frac{G m m}{l^2}=m a$ ....... (1)

$\frac{G M m}{(r+l)^2}+\frac{G m m}{l^2}=m a$ ......... (2)

From eqn. (1) and (2), we get

$$\begin{aligned} & \frac{G M}{9 l^2}-\frac{G m}{l^2}=\frac{G M}{16 l^2}+\frac{G m}{l^2} \\ & \frac{M}{9}-\frac{M}{16}=m+m \Rightarrow \frac{7 M}{144}=2 m \end{aligned}$$

$m=\frac{7 M}{288}=k\left(\frac{M}{288}\right)$

$\therefore$ k = 7

In complex, $\mathop {{{[Fe{{({C_2}{O_4})}_2}{{({H_2}O)}_2}]}^{2 - }}}\limits_{Diaquodioxalatoferrate\,(II)}$,

Fe is in +2 oxidation state.

In acidic medium, $KMn{O_4}$ oxidises $F{e^{2 + }}$ to $F{e^{3 + }}$,

$2MnO_4^ - + 16{H^ + } + 10F{e^{2 + }} \to 2M{n^{2 + }} + 8{H_2}O + 10F{e^{3 + }}$

or $MnO_4^ - + 8{H^ + } + 5F{e^{2 + }} \to M{n^{2 + }} + 4{H_2}O + 5F{e^{3 + }}$

${{Rate\,of\,change\,of\,[{H^ + }]} \over {Rate\,of\,change\,of\,[MnO_4^ - ]}} = {8 \over 1} = 8$