JEE Advanced 2020 Paper 2 Question Paper with Solutions

Weight = Upthrust

$mg = {F_u} \Rightarrow 480 \times 10 = \rho Vg$

$480 \times 10 = {\rho _0}{e^{ - {h \over {{h_0}}}}} \Rightarrow 480 \times 10 = {\rho _0}{e^{ - {{100} \over {6000}}}}Vg$ .... (i)

$(480 - N \times 1)10 = \rho 'Vg$

$(480 - N)10 = {\rho _0}{e^{ - {{150} \over {6000}}}}Vg$ .... (ii)

Dividing Eq. (i) by Eq. (ii), we get

${{480} \over {480 - N}} = {e^{\left( {{{150 - 100} \over {6000}}} \right)}}$

${{480} \over {480 - N}} = {e^{{{50} \over {6000}}}} \Rightarrow {{480} \over {480 - N}} = {e^{{1 \over {120}}}}$

N = 4

Consider ray of light from object $\mathrm{O}$ moves along $\mathrm{OB}$ at an angle of incidence $i$ and refracted through air parallel to line of sight, that is, $\angle r=45^{\circ}$. Another ray slight above $\mathrm{OA}$ is refracted along $\mathrm{AC}$ and image appears to be formed at $I_1$ having height $h$.


At point B; from Snell's law

$$\begin{aligned} & \frac{\sin i}{\sin 45}=\frac{1}{4 / 3} \Rightarrow \sin i=\frac{3}{4} \sin 45^{\circ} \\ & \Rightarrow \sin =\frac{3}{4} \times \frac{1}{\sqrt{2}}=\frac{3}{4 \times 1.4142}=0.5303 \end{aligned}$$

So, $i=32.03^{\circ}$.

In $\triangle \mathrm{OAB}, \quad \angle B=\left(90^{\circ}+i\right)=\left(90^{\circ}+32.03^{\circ}\right) \Rightarrow 122.03^{\circ}$.

$$\begin{aligned} & \angle A=45^{\circ} \\ & \angle O=\left(180-45^{\circ}-122.03^{\circ}\right) \cong 13^{\circ} \end{aligned}$$

From sine law, we have

$$\begin{aligned} & \frac{12}{\sin 122^{\circ}}=\frac{\mathrm{AB}}{\sin 13^{\circ}} \Rightarrow \mathrm{AB}=\frac{12}{\sin 122^{\circ}} \times \sin 13^{\circ} \\ & \mathrm{AB}=\frac{12}{0.8480} \times 0.2249=3.1833 \mathrm{~cm} \end{aligned}$$

In $\triangle \mathrm{BAE}$,

$$\begin{aligned} & \frac{\mathrm{BE}}{\mathrm{AB}}=\sin 45^{\circ} \Rightarrow \mathrm{BE}=\mathrm{AB} \sin 45^{\circ} \\ & \mathrm{BE}=31833 \times \frac{1}{\sqrt{2}}=\frac{3.1833}{1.4142}=2.25 \mathrm{~cm} \end{aligned}$$

Since, $\mathrm{BE}=\mathrm{FG}$ is position of image $I_1$ at upper half of square, $h=2.25 \mathrm{~cm}$.

Similarly, in lower half of square image $I_2$ will be at a distance of $2.25 \mathrm{~cm}$.

$I_1 I_2=2 \times 2.25=4.50 \mathrm{~cm}=4.00 \mathrm{~cm}$

According to conservation of angular momentum about suspension point,

$mvx = \left( {{{m{L^2}} \over 3} + m{x^2}} \right)\omega$

$vx = \left( {{{{L^2}} \over 3} + {x^2}} \right)\omega \Rightarrow \omega = {{3vx} \over {{L^2} + 3{x^2}}}$

For maximum,

${{d\omega } \over {dx}} = 0 \Rightarrow x = {L \over {\sqrt 3 }} \Rightarrow \omega = {{\sqrt 3 v} \over {2L}}$

Given that, $C_1=(2000 \pm 10) \mathrm{pF} ; \quad C_2=(3000 \pm 15) \mathrm{pF}$ and $V=(5.00 \pm 8.02) \mathrm{V}$.

Combine capacitance of series combination

$$\begin{aligned} \frac{1}{C} & =\frac{1}{C_1}+\frac{1}{C_2} ~~~~....(1)\\ \Rightarrow & \frac{1}{C}=\frac{1}{2000}+\frac{1}{3000}=\frac{(6+4)}{12000} \\ \frac{1}{C} & =\frac{10}{12000} \Rightarrow C=1200 \mathrm{pF} \end{aligned}$$

Differentiating Eq. (1), we get

$$\begin{aligned} -\frac{1}{C^2} \cdot d C & =-\frac{1}{C_1^2} d C_1+\left(-\frac{1}{C_2^2}\right) \cdot d C_2 \\ \Rightarrow \frac{d C}{C^2} & =\left(\frac{d C_1}{C_1^2}+\frac{d C_2}{C_2^2}\right) \\ \Rightarrow \frac{d C}{C} & =\left[\frac{d C_1}{C_1^2}+\frac{d C_2}{C_2^2}\right] \cdot C \\ \Rightarrow \frac{d C}{C} & =\left[\frac{10}{(2000)^2}+\frac{15}{(3000)^2}\right] \times 1200 \\ \Rightarrow \frac{d C}{C} & =\left[\frac{10}{4} \times 10^{-6}+\frac{15}{9} \times 10^{-6}\right] \times 1200 \\ & =\left[\frac{10}{4}+\frac{15}{9}\right] \times 12 \times 10^{-4} \end{aligned}$$

$\Rightarrow \frac{d C}{C}=[2.5+1.66] \times 12 \times 10^{-4}=[4.17 \times 12] \times 10^{-4}$

And $\frac{d V}{V}=\frac{0.02}{5}$

Energy stored in the capacitor is given by

$U=\frac{1}{2} C V^2$    .....(2)

Differentiating Eq. (2), we get

$$\begin{aligned} \frac{d U}{U} \times 100 & =\frac{d C}{C} \times 100+2 \frac{d V}{V} \times 100 \\ & =\left[50.04 \times 10^{-4} \times 100+\frac{2 \times 0.02}{5} \times 100\right] \\ \left(\frac{d U}{U}\right)^{100} & =[0.5004+0.8] \% \Rightarrow 1.30 \% \end{aligned}$$

U_i = 0

${U_f} = {{kqp} \over {{{\left( {2l\sin {\alpha \over 2}} \right)}^2}}}$ + mgh .... (i)

Now, from $\Delta$OAB,

$\alpha$ + 90 $-$ $\theta$ + 90 $-$ $\theta$ = 180

$\Rightarrow$ $\alpha$ = 2$\theta$

From $\Delta$ABC, h = 2l sin$\left( {{\alpha \over 2}} \right)$ sin$\theta$

h = 2l sin$\left( {{\alpha \over 2}} \right)$ sin$\left( {{\alpha \over 2}} \right)$ $\Rightarrow$ h = 2l sin²$\left( {{\alpha \over 2}} \right)$

Now, charge is in equilibrium at point B.

So, using sine rule,

${{mg} \over {\sin \left( {90 + {\alpha \over 2}} \right)}} = {{qE} \over {\sin 1(80 - 2\theta )}}$

$\Rightarrow {{mg} \over {\cos {\alpha \over 2}}} = {{qE} \over {\sin 2\theta }}$

$\Rightarrow {{mg} \over {\cos {\alpha \over 2}}} = {{qE} \over {\sin \alpha }} \Rightarrow {{mg} \over {\cos {\alpha \over 2}}} = {{qE} \over {2\sin {\alpha \over 2}\cos {\alpha \over 2}}}$

$\Rightarrow qE = mg2\sin \left( {{\alpha \over 2}} \right)$

$\Rightarrow {{q2kp} \over {{{\left( {2l\sin {\alpha \over 2}} \right)}^3}}} = mg2\sin \left( {{\alpha \over 2}} \right)$

$\Rightarrow {{kpq} \over {{{\left( {2l\sin {\alpha \over 2}} \right)}^2}}} = mg\sin \left( {{\alpha \over 2}} \right) \times \left( {2l\sin {\alpha \over 2}} \right)$

$\Rightarrow {{kpq} \over {{{\left( {2l\sin {\alpha \over 2}} \right)}^2}}} = mgh$

Substituting this in Eq. (i), we get

${U_f} = mgh + {{kpq} \over {{{\left( {2l\sin {\alpha \over 2}} \right)}^2}}} \Rightarrow {U_f} = 2mgh$

W = $\Delta$U = 2mgh = Nmgh

$\therefore$ N = 2

Applying Bernoulli's equation,

${p_0} + {1 \over 2}\rho v_1^2 = p + {1 \over 2}\rho {v^2}$

${p_0} - p = {1 \over 2}\rho ({v^2} - v_1^2)$ .... (i)

From equation of continuity,

$4{S_t}{v_t} = v \times 3{S_t}$

$\Rightarrow v = {4 \over 3}{v_t}$ ..... (ii)

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

${p_0} - p = {1 \over 2}\rho \left( {{{16} \over 9}v_t^2 - v_t^2} \right) = {1 \over 2}\rho {{7v_t^2} \over 9}$

$\therefore$ N = 9

${{dV} \over V} = - {{dp} \over B}$ (where, B = bulk modulus)

$V = {l^3} \Rightarrow {{\Delta V} \over V} = 3{{\Delta l} \over l}$

$3{{\Delta l} \over l} = \left| { - {{\Delta p} \over B}} \right| = {{\rho gh} \over B} \Rightarrow \Delta l = {{\rho ghl} \over {3B}}$

Substituting the given values, we get $\Delta$l = 0.24 mm

To start with, both the compartments have same volume. Under the action of weight, moving down by distance $x$, length of upper compartment will be $(4+x)$ and that of lower compartment $(4-x)$.


Volumes of two compartments are

$V_1=(4+x) A \text { and } V_2=(4-x) A$

At equilibrium,

$F_2=F_1+m g$

$$\begin{aligned} & P_2 A=P_1 A+m g \\ & P_2=P_1+\frac{m g}{A} \end{aligned}$$

From $P V=n R T$, we get $P=\frac{n R T}{V}$

$$\begin{aligned} & \frac{n R T}{V_2}=\frac{n R T}{V_1}+\frac{m g}{A} \\ \Rightarrow & n R T\left[\frac{1}{V_2}-\frac{1}{V_1}\right]=\frac{m g}{A} \\ \Rightarrow & n R T\left[\frac{1}{A(4-x)}-\frac{1}{4(4+x)}\right]=\frac{m g}{A} \\ \Rightarrow & \frac{n R T}{A}\left[\frac{(4+x)-(4-x)}{(4-x)(4+x)}\right]=\frac{m g}{A} \\ \Rightarrow & n R T\left[\frac{2 x}{\left(16-x^2\right)}\right]=m g \\ \Rightarrow & 0.1 \times 8.3 \times 300\left[\frac{2 x}{\left(16-x^2\right)}\right]=8.3 \times 10 \\ \Rightarrow & \frac{6 x}{16-x^2}=1 \Rightarrow 16-x^2=6 x \\ \Rightarrow & x^2+6 x-16=0 \\ \Rightarrow & x=\frac{-6 \pm \sqrt{36+64}}{2} \\ & x=\frac{-6 \pm \sqrt{100}}{2}=\frac{-6 \pm 10}{2} \\ & x=\frac{10-6}{2} \text { or } \frac{-10-6}{2}=-8 \text { or } 2 \end{aligned}$$

Neglecting negative $\operatorname{sign} x=2$

Partition from top $=4+2=6 \mathrm{~m}$

$qE = Mg$

$\Rightarrow q = {{mg} \over E} = \left( {{{mg} \over {V/d}}} \right)$

$= {{900 \times {4 \over 5}\pi (8 \times {{10}^{ - 7}}) \times 10} \over {{{200} \over {0.01}}}}$

$N = {q \over e} = {{900 \times 4\pi \times {8^3} \times {{10}^{ - 21}} \times 10 \times 0.01} \over {200 \times 1.6 \times {{10}^{ - 19}}}}$

N = 6

${(R - h)^2} + {r^2} = {R^2}$

${R^2} + {h^2} - 2Rh + {r^2} = {R^2}$

${h^2} + {r^2} = 2Rh \Rightarrow {{{h^2} + {r^2}} \over {2h}} = R$ .....(i)

When r > > > h, then

$R = {{{r^2}} \over {2h}} \Rightarrow h = {{{\omega ^2}{r^2}} \over {2g}}$

$R = {{{r^2}2g} \over {2{\omega ^2}{r^2}}} \Rightarrow R = {g \over {{\omega ^2}}}$ ..... (ii)

${{{\mu _1}} \over v} - {{{\mu _2}} \over u} = {{{\mu _1} - {\mu _2}} \over R}$

$\Rightarrow {1 \over v} + {4 \over {3u}} = {{1 - {4 \over 3}} \over R}$

${1 \over v} = - \left( {{1 \over {3R}} + {4 \over {3(H - h)}}} \right)$

${1 \over v} = - \left( {{1 \over {3R}} + {4 \over {3H}}} \right)$

Putting the value of Eq. (ii) h < < H

${1 \over v} = - \left[ {{4 \over {3H}}\left( {1 + {{3H} \over 4}{{{\omega ^2}} \over {39}}} \right)} \right]$

$v = - \left[ {{{3H} \over 4}{{\left( {1 + {{{\omega ^2}H} \over 4}} \right)}^{ - 1}}} \right]$

${{dQ} \over {dt}} = e\sigma A({T^4} - T_0^4)$

For small temperature change,

${{dQ} \over {dt}} = e\sigma A{T^3}\Delta T$ .... (i)

${{mCdT} \over {dt}} = e\sigma A{T^3}\Delta T$

$\Rightarrow {{dT} \over {dt}} = {{e\sigma A{T^3}} \over {mC}}\Delta T$

${{e\sigma A{T^3}} \over {mC}}$ $\to$ constant for Newton law of cooling

$\therefore$ ${{e\sigma A{T^3}} \over {mC}} = 0.001$

$\Rightarrow e\sigma A{T^3} = mC \times 0.001 = 1 \times 4200 \times 0.001$

$e\sigma A{T^3} = 4.2$ .... (ii)

${{dQ} \over {dt}} = 700 \times 0.05 = 35$ W ..... (iii)

Putting the value of Eqs. (ii) and (iii) in Eq. (i), we get

$35 = 4.2\Delta T \Rightarrow {{35} \over {4.2}} = \Delta T \Rightarrow \Delta T = 8.33$

As change in the kinetic energy of the surface of the bubble is zero.

$\therefore$ W_external + W_water + W_gas = 0 ....(i)

where W_external = work done by external agent

W_water = work done by water

W_gas = work done by gas

For adiabatic process, ${P_1}V_1^\gamma = {P_2}V_2^\gamma$

Here, ${P_0}{\left[ {{4 \over 3}\pi {R^3}} \right]^{{{41} \over {30}}}} = P{\left[ {{4 \over 3}\pi {{\left( {R - a} \right)}^3}} \right]^{{{41} \over {30}}}}$

$\Rightarrow$ P = ${P_0}{\left[ {{R \over {R - a}}} \right]^{{{41} \over {10}}}}$

Now, W_gas = ${{{P_1}{V_1} - {P_2}{V_2}} \over {\gamma - 1}}$

= ${{{P_0} \times {4 \over 3}\pi {R^3} - P \times {4 \over 3}\pi {{\left( {R - a} \right)}^3}} \over {{{41} \over {30}} - 1}}$

= ${{{P_0} \times {4 \over 3}\pi {R^3} - {P_0}{{\left[ {{R \over {R - a}}} \right]}^{{{41} \over {10}}}} \times {4 \over 3}\pi {{\left( {R - a} \right)}^3}} \over {{{41} \over {30}} - 1}}$

= ${{{P_0} \times {4 \over 3}\pi {R^3}\left[ {1 - {{\left( {{R \over {R - a}}} \right)}^{{{41} \over {10}}}}{{\left( {{{R - a} \over R}} \right)}^3}} \right]} \over {{{11} \over {30}}}}$

= ${{{40} \over {11}}{P_0}\pi {R^3}\left[ {1 - {{\left( {{{R - a} \over R}} \right)}^{ - {{41} \over {10}}}}{{\left( {{{R - a} \over R}} \right)}^3}} \right]}$

= ${{40} \over {11}}{P_0}\pi {R^3}\left[ {1 - {{\left( {1 - {a \over R}} \right)}^{ - {{11} \over {10}}}}} \right]$

= ${{40} \over {11}}{P_0}\pi {R^3}\left[ {1 - 1 - {{11a} \over {10R}} - {{\left( { - {{11} \over {10}}} \right)\left( { - {{11} \over {10}} - 1} \right)} \over 2}{{{a^2}} \over {{R^2}}}} \right]$

= -4P₀$\pi$R²$a$ - ${{40 \times 21} \over {100 \times 2}}{P_0}\pi R{a^2}$

= -4P₀$\pi$R²$a$ - $4.2{P_0}\pi R{a^2}$

W_water = P₀dV

= P₀${\left[ {{4 \over 3}\pi {R^3} - {4 \over 3}\pi {{\left( {R - a} \right)}^3}} \right]}$

= ${{{4{P_0}} \over 3}\pi \left[ {{R^3} - {{\left( {R - a} \right)}^3}} \right]}$

= ${{{4{P_0}} \over 3}\pi \left[ {\left[ {R - \left( {R - a} \right)} \right]\left[ {{R^2} + R\left( {R - a} \right) + {{\left( {R - a} \right)}^2}} \right]} \right]}$

= ${{{4{P_0}} \over 3}\pi \left[ {\left[ a \right]\left[ {3{R^2} - 3Ra + {a^2}} \right]} \right]}$

= ${{{4{P_0}} \over 3}\pi \left[ {\left( {3{R^2}a - 3R{a^2} + {a^3}} \right)} \right]}$

= ${4{P_0}\pi \left[ {{R^2}a - R{a^2}} \right]}$ [ignore ${{a^3}}$ term as no ${{a^3}}$ term in the question]

$\therefore$ W_gas + W_water = -$4{P_0}\pi R{a^2}$ - $4.2{P_0}\pi R{a^2}$

= $- 4{P_0}\pi R{a^2}\left[ {1 + 1.05} \right]$

= $- 4{P_0}\pi R{a^2}\left[ {2.05} \right]$

From (i), W_external = - (W_gas + W_water)

= $4{P_0}\pi R{a^2}\left[ {2.05} \right]$

$\therefore$ X = 2.05

Mutual inductance is producing flux in same direction as self-inductance.

$\therefore$ $U = {1 \over 2}{L_1}I_1^2 + {1 \over 2}{L_2}I_2^2 + M{I_1}{I_2}$

$\Rightarrow U = {1 \over 2} \times (10 \times {10^{ - 3}}){1^2} + {1 \over 2} \times (20 \times {10^{ - 3}}) \times {2^2} + (5 \times {10^{ - 3}}) \times 1 \times 2$

$= 55$ mJ

$R{'_3} = 300(1 + \alpha \Delta T) = 312\Omega$

Now,



${I_1} = {{50} \over {372}}$ and ${I_2} = {{50} \over {600}}$

${V_S} - {V_T} = 312{I_1} - 500{I_2}$

$= 41.94 - 41.67 = 0.27$ V

$\lambda \propto {1 \over V}$

So, when V double, then cut-off wavelength become half. Characteristic does not depend on voltage. When I is halved, then intensity will decrease.

Apply energy conservation,

${1 \over 2}mv_0^2 = mgh + {1 \over 2}m{v^2}$

${1 \over 2}m{v^2} = {1 \over 2}mv_0^2 - mgh$ ..... (i)

At top point y, the centripetal force provided by the component of mg,

mg sin$\theta$ = centripetal force = ${{m{v^2}} \over R}$

${{m{v^2}} \over R} = mg\sin \theta$

Putting this value in Eq. (i), we get

${{mgR\sin \theta } \over 2} = {1 \over 2}mv_0^2 - mgh$

${{mgR} \over 2}{1 \over 2} = {1 \over 2}mv_0^2 - mgh$

${{gR} \over 2} = v_0^2 - 2gh$

At point x and a of circular path, the points are at same height but less than h. So the velocity is more than point y.

So required centripetal force = $\frac{mv^{2}}{r}$ is more.

Given, equation $y = {{{x^2}} \over 2}$

${{dy} \over {dt}} = {{2x} \over 2}{{dx} \over {dt}} \Rightarrow {v_y} = x{v_x}$

${{d{v_y}} \over {dt}} = x{{d{v_x}} \over {dt}} + {{dx} \over {dt}}{v_x} \Rightarrow {a_y} = x{a_x} + v_x^2$

For option A :

when particle is at origin then x = 0.

$\therefore$ ${a_y} = 0 \times{a_x} + v_x^2$ = $v_x^2$

Given that at origin, v_x = 1 m/s

So, ${a_y}$ = (1)² = 1 ms^-2

At origin, $0 \times{a_x}$ = 0 means ${a_y}$ does not depend on the value of ${a_x}$.

For option B :

When ${a_x}$ = 0 $\Rightarrow$ ${v_x}$ = constant

So ${v_x}$ will stay 1 ms^-1.

when a_x = 0,

${a_y} = x \times0 + v_x^2$ = $v_x^2$ = 1 ms^-2

For option C :

Velocity is always tangential to the path of motion of the particle. So at origin path is along x axis as tangent at origin along x axis.

For option D :

When ${a_x}$ = 0 $\Rightarrow$ ${v_x}$ = constant

So ${v_x}$ will stay 1 ms^-1.

At t = 1, x = 1 m

$\therefore$ ${v_y} = x{v_x}$ = 1 $\times$ 1 = 1

Now, slope of velocity at t = 1 is,

tan$\theta$ = ${{{v_y}} \over {{v_x}}}$ = ${1 \over 1}$ = 1

$\Rightarrow$ $\theta$ = 45^o

In air,



T cos$\theta$ = mg ..... (i)

T sin$\theta$ = qE ..... (ii)

qE = mg tan$\theta$ ..... (iii)

In liquid,



T' cos$\theta$ = mg $-$ U

T' cos$\theta$ = mg - $\rho$vg ..... (iv)

T' sin$\theta$ = ${{qE} \over K}$ ...... (v)

Dividing Eq. (v) by Eq. (ii), we get

${{T'} \over T} = {1 \over K} \Rightarrow T' = {T \over K} = {T \over {21}}$ ..... (vi)

Dividing Eq. (iv) by Eq. (v), we get

${{qE} \over K} =$ (mg - $\rho$vg)tan $\theta$

$qE =$ K(mg - $\rho$vg)tan $\theta$ ..... (vii)

From Eqs. (vii) and (iii), we get

mg tan$\theta$ = K(mg - $\rho$vg)tan $\theta$

$\Rightarrow$ m = K(m - $\rho$v)

$\Rightarrow$ dv = K(dv - $\rho$v)

$\Rightarrow$ d = Kd - K$\rho$

$\Rightarrow$ d = 21d - 21$\rho$ [ $\because$ K = 21]

$\Rightarrow$ 20d = 21$\rho$

$\Rightarrow$ 20d = 21 $\times$ 800

$\Rightarrow$ d = 840 kg/m³

Aniline (P) is treated with NaNO₂ and HCl in cold condition to form benzene diazonium chloride. The process of conversion of primary aromatic amines into its diazonium salt is called diazotisation.



$\beta$-naphthol couples with phenyldiazonium electrophile to produce an intense orange-red dye (Q) as major product.



Given that, volume of aniline (P) = 9.3 mL (density of P = 1.00 g/mL)

So, mass of aniline = 9.3 g

Molecular mass of aniline (C₆H₇N) = 93 g/mol

Therefore, moles of aniline = 9.3 / 93 = 0.1 mol of P.

Molecular mass of 2 napthol aniline orange dye (Q) = 248 g/mol

$\Rightarrow$ 0.1 mol of aniline (P) will produce 0.1 mol of compound (Q).

But, according to the question the major product Q from P is 75%.

Therefore, mass of 'Q' produced

= (0.1 $\times$ 248 $\times$ 0.75)g = 18.60 g

$ZnS(s)\buildrel {} \over \longrightarrow \mathop {Z{n^{2 + }}(aq)}\limits_{(0.05\,M)} + {S^{2 - }}(aq)$

${K_{sp}}(ZnS) = [Z{n^{2 + }}][{S^{2 - }}] = 1.25 \times {10^{ - 22}}$

$0.05 \times [{S^{2 - }}] = 1.25 \times {10^{ - 22}}$

$\Rightarrow [{S^{2 - }}] = {{1.25 \times {{10}^{ - 22}}} \over {0.05}} \Rightarrow 25 \times {10^{ - 22}}M$

For H₂S, $\mathop {{H_2}S}\limits_{(0.1M)} \buildrel {} \over \longrightarrow 2{H^ + } + \mathop {{S^{ - 2}}}\limits_{(25 \times {{10}^{ - 22}}M)}$

${K_{net}} = 1 \times {10^{ - 21}} = {{{{[{H^ + }]}^2}[{S^{2 - }}]} \over {[{H_2}S]}}$

$1 \times {10^{ - 21}} = {{{{[{H^ + }]}^2} \times 25 \times {{10}^{ - 22}}} \over {[0.1]}}$

${[{H^ + }]^2} = {1 \over {25}}$

$[{H^ + }] = {1 \over 5} = 0.2M$

Sn²⁺ of stannous chloride dihydrate (SnCl₂ . 2H₂O) tends to convert into Sn⁴⁺.

Hence, statement (a) is correct.

(b) SnO₂ reacts with KOH and gives K₂SnO₃ . 3H₂O or K₂[Sn(OH)₆] because it is amphoteric in nature.

$Sn{O_2} + KOH\buildrel {} \over \longrightarrow {K_2}Sn{O_3} + {H_2}O$

or ${K_2}[Sn{(OH)_6}]$

Hence, statement (b) is correct.

(c) In conc. HCl, PbCl₂ exists as chloroplumbous acid, H₂[PbCl₄]

$\mathop {PbC{l_2}}\limits^{II} + 2HCl\buildrel {} \over \longrightarrow \mathop {{H_2}[PbC{l_4}]}\limits^{II}$

Hence, statement (c) is incorrect.

(d) Pb₃O₄ is a mixture of $(2\mathop {PbO}\limits^{ + 2} + {\mathop {PbO}\limits^{ + 4} _2})$

$P{b_3}{O_4} + 4HN{O_3}\buildrel {} \over \longrightarrow \mathop {2Pb{{(N{O_3})}_2}}\limits^{ + 2} + \mathop {Pb{O_2}}\limits^{ + 4} + 2{H_2}O$

It is not a redox reaction. Thus, the statement (d) is incorrect.

Paramagnetism is a form of magnetism whereby some materials are attracted by an externally applied magnetic field, and form internal, induced magnetic fields in the direction of the applied magnetic field. So, magnetic balance shows downward deflection. While diamagnetic substance shows repulsion in magnetic field and magnetic balance shows upward deflection.

(a) X = H₂O(l)
(Water has no unpaired electrons and is thus diamagnetic). Hence, statement (a) is correct.

(b) X = K₄[Fe(CN)₆](s)
(CN is a strong field ligand which forces the d-orbital electrons to pair up ($t_{2g}^6e_g^0$) and making it diamagnetic. Hence, statement (b) is correct.

(c) X = O₂ (g)
Here, O₂(g) is paramagnetic due to two-unpaired electrons present in $\pi$* (antibonding orbitals). Hence, statement (c) is correct.

(d) X = C₆H₆(l)
(Here, C₆H₆ is diamagnetic due to presence of 0 unpaired electrons. Hence, statement (d) is incorrect.

Using Raoult's law equation for a mixture of volatile liquids.

${P_T} = p_A^o{\chi _A} + p_B^o{\chi _B}$

$0.3 = 0.25\chi p_A^o + 0.75\chi p_B^o$ ....(i)

$0.4 = 0.5\chi p_A^o + 0.5\chi p_B^o$ ....(ii)

By solving equation (i) and (ii)

$p_A^o$ = 0.6 bar and $p_B^o$ = 0.2 bar

Thus, the vapour pressure of pure liquid B in bar is 0.2.

At pH = 2,

There are two $-$NH₂ group, and + 1 charge on each group because all amino groups exist in the form of $-$NH$_3^ \oplus$.

Therefore, |Z₁| = 2.

At pH = 6,

NH₂ of lysine (+ 1) (pH = 9.47) and COOH ($-$1) of glutamic (pH = 3.08) acid, so because of dipolar ion exists, therefore |Z₂| = 0.

At pH = 11,

COOH of glutamic acid has ($-$1), COOH of lysine ($-$1) and OH of phenol ($-$1).

Therefore, |Z₃| = |$-$3| = 3 (All COOH and OH exist in the form of $-$COO^$-$ and $-$O^$-$).

$\therefore$ |Z₁| + |Z₂| + |Z₃| = 2 + 0 + 3 = 5

When a solution of K₂CrO₄ is treated with amyl alcohol and acidified H₂O₂, the layer of amyl alcohol turns blue because acidified H₂O₂ converts K₂CrO₄ to CrO₅ to given the blue colouration,



The structure of CrO₅ is :


Number of oxygen atom bonded with chromium with single bond is = 4.

By observing the values of different ionisation energies, I₁, I₂ and I₃ for
atomic number (n + 2), it is observed that there is very large difference between the second ionisation energy and first ionisation energy (I₂ >> I₁).

This indicates that number of valence shell electrons is 1 and atomic number (n + 2) should be an alkali metal.

Also for atomic number (n + 3), I₃ >> I₂. This indicates that it will be an alkaline earth metal which suggests that atomic number (n + 1) should be a noble gas and atomic number (n) should belong to halogen family. Since, n < 10; hence, n = 9 (F atom)