JEE Advanced 2019 Paper 1 Question Paper with Solutions

Given, V at surface of the sphere

${V_0} = {{KQ} \over R}$

Here, $K = {1 \over {4\pi {\varepsilon _0}}}$ = constant

V at point C,

${V_C} = {{KQ} \over R} - {{K\alpha Q} \over R} = {V_0}(1 - \alpha )$

V at point B,

${V_B} = {{KQ} \over R} - {{K(\alpha Q)} \over {R/2}} = {V_0}(1 - 2\alpha )$

$\therefore$ ${{{V_C}} \over {{V_B}}} = {{1 - \alpha } \over {1 - 2\alpha }}$

$[M] = [Mass] = [{M^0}{L^0}{T^0}]$

[J] = [Angular momentum] = [ML²T^-1]

[L] = [Length]

Now, $[M{L^2}{T^{ - 1}}] = [{M^0}{L^0}{T^0}]$

$\therefore$ [L²] = [T]

Power [P] = [MLT².LT^-1] = [ML²T^-3]

= [L²L^-6]

[P] = [L^-4\

Energy/Work [W] = [MLT^-2.L]

= [L²L^-4] = [L^-2]

Force [F] = [MLT^-2] = [L.L^-4] = [L^-3]

Linear momentum [p] = [MLT^-1] = [L.L^-2]

[p] = [L^-1]

Just after closing of switch charge on any capacitor is zero.

$\therefore$ Replace all capacitors by conducting wires.



Current flow in the circuit,

$i = {5 \over {70 + 100 + 30}} = {5 \over {200}} = 25$ mA

Now S₁ is kept closed for long time circuit is in steady state.



${q \over {10}} + {q \over {80}} + {q \over {80}} - 5 = 0 \Rightarrow {{10q} \over {80}} = 5$

$\therefore$ $q = 40\mu C$

$\therefore$ V across ${C_1} = {{40} \over {10}} = 4V$

Now just after closing S₂ charge on each capacitor remains same.



Applying KVL,

$- 10 + x \times 30 + {{40} \over {10}} + y \times 70 = 0$

30x + 70y = 6 .....(1)

$- {{40} \over {80}} + 5 + (x - y)30 - {{40} \over {80}} + (x + y) \times 100 - 10 + x \times 30 = 0$

$160x - 130y - 6 = 0$ .... (2)

$y = {{96} \over {1510}}$

x = 0.05 A

${{{T_s}} \over 2} = {T_c}{{\sqrt 3 } \over 2}$

${T_s} = \sqrt 3 {T_c}$

$\Delta l = {{Tl} \over {Ay}}$

$\therefore$ ${{\Delta {l_c}} \over {\Delta {l_s}}} = \left( {{{{T_c}} \over {{T_s}}}} \right)\left( {{{{l_c}} \over {{l_s}}}} \right)\left( {{{{Y_s}} \over {{Y_c}}}} \right)$

$= \left( {{1 \over {\sqrt 3 }}} \right)\left( {{{\sqrt 3 } \over 1}} \right)\left( {{{2 \times {{10}^{11}}} \over {1 \times {{10}^{11}}}}} \right) = 2.00$

(a) h > 2R and r > R



$\phi = {Q \over {{\varepsilon _0}}}$, clearly from Gauss' Law

(b) suppose $h = {{8R} \over 5}$ and $r = {{3R} \over 5}$



$\therefore$ $\phi$ = 0

so, for $h < {{8R} \over 5}$ then $\phi$ = 0.

(c) For h = 3R and r = ${{4R} \over 5}$



Shaded charge = $2\pi (1 - \cos \,53^\circ ) \times {Q \over {4\pi }} = {Q \over 5}$

$\therefore$ ${q_{enclosed}} = {{2Q} \over 5}$

$\therefore$ $\phi = {{2Q} \over {5{\varepsilon _0}}}$

$\therefore$ for h > 2R and $r = {{4R} \over 5}$

$\therefore$ $\phi = {{2Q} \over {5{\varepsilon _0}}}$

(d) like option c for h = 2R and $r = {{3R} \over 5}$

${q_{enclosed}} = 2 \times 2\pi (1 - \cos \,37^\circ ){Q \over {4\pi }} = {Q \over 5}$

$\therefore$ Electric flux, $\phi = {Q \over {5{\varepsilon _0}}}$

Heat capacity, ${{dQ} \over {dt}} = H{{dT} \over {dt}}$

Power of the rod, $P = H.{T_0}.\beta .{1 \over 4}.{t^{ - 3/4}}$

${{4P} \over {{T_0}.\beta }} = {t^{ - 3/4}}.H$

$\Rightarrow$ $H = \left( {{{4P} \over {{T_0}\beta }}} \right){t^{3/4}}$ .... (i)

Now, $T - {T_0} = {T_0}\beta {t^{1/4}}$

So, ${t^{3/4}} = {\left( {{{T - {T_0}} \over {{T_0}\beta }}} \right)^3}$

Substituting this value of t^3/4 in equation (i) we get,

$H = {{4P{{(T - {T_0})}^3}} \over {T_0^4{\beta ^4}}}$

Case - I 5C $\times$ 50 + 5L = C₂ $\times$ 30 ....(i)

Case - II 80C[50$-$30] = C₂ [80$-$50] ....(ii)

By Eq. (i) and (ii)

1600C = 250 + 5L

$\therefore$ ${L \over C} = {{1350} \over 5} = 270^\circ C$

d = F.dr

d = $\alpha$ydx + 2$\alpha$xdy

A$\to$B, y = 1, dy = 0

then ${W_{A \to B}} = \int {\alpha ydx}$$= \alpha 1\int_0^1 {dx} = \alpha$

B$\to$C, x = 1, dx = 0

then ${W_{B \to C}} = 2\alpha 1\int_1^{0.5} {dy} = - 2\alpha (0.5) = - \alpha$

C$\to$D, y = 0.5, dy = 0

then ${W_{C \to D}} = \int_1^{0.5} {\alpha ydx} = \alpha .{1 \over 2}\int_1^{0.5} {dx} = - {\alpha \over 4}$

D$\to$E, x = 0.5, dx = 0

then ${W_{D \to E}} = 2\alpha \int {xdy} = 2\alpha .{1 \over 2}\int\limits_{0.5}^0 {dy} = - {\alpha \over 2}$

E$\to$F, y = 0, dy = 0 then W_EF = 0

F$\to$A, x = 0, dx = 0 then W_F$\to$A = 0

$\therefore$ $W = \alpha - \alpha - {\alpha \over 4} - {\alpha \over 2} = - {{3\alpha } \over 4}$

Given, $\alpha = - 1$

$\Rightarrow W = {3 \over 4}J = 0.75J$

Given, thin convex lens made of two materials.

Radius of curvature of left spherical surface $=$ radius of curvature of right spherical surface

When $n_1=n_2=n$, focal length $=f$

When $n_1=n$ and $n_2=n+\Delta n$, focal length $=f+\Delta f$

Here, $\Delta n \ll(n-1)$ and $1< n < 2$

Now, when $n_1=n_2=n$, from lens maker's formula

$$\begin{aligned} \frac{1}{f} & =(n-1)\left(\frac{1}{R_1}-\frac{1}{R_2}\right)=(n-1)\left(\frac{1}{R}-\frac{1}{(-R)}\right) \\ \Rightarrow \frac{1}{f} & =(n-1)\left(\frac{2}{R}\right) ........(1) \end{aligned}$$

When $n_1=n$ and $n_2=n+\Delta n$

$\frac{1}{f+\Delta f}=(n-1) \frac{1}{R_1}-(n+\Delta n-1) \frac{1}{+R_2}$

$$\begin{aligned} & \quad=(n-1) \frac{1}{R}-(n+\Delta n-1) \frac{1}{(-R)} \\ & \Rightarrow \frac{1}{f+\Delta f}=\frac{1}{R}(n-1+n+\Delta n-1)=\frac{1}{R}(2 n+\Delta n-2) \\ & \Rightarrow \frac{1}{f+\Delta f}=\frac{1}{R}(2 n+\Delta n-2) ........(2) \end{aligned}$$

Subtracting Eq. (2) from Eq. (1), we get

$$\begin{aligned} & \frac{1}{f}-\frac{1}{f+\Delta f}=(n-1)\left(\frac{2}{R}\right)-\frac{1}{R}(2 n+\Delta n-2) \\ & =\frac{1}{R}(2 n-2-2 n-\Delta n+2)=-\frac{\Delta n}{R} \\ & \Rightarrow \frac{1}{f}-\frac{1}{f+\Delta f}=-\frac{\Delta n}{R} \Rightarrow \frac{(f+\Delta f)-f}{f(f+\Delta f)}=-\frac{\Delta n}{R} \\ & \Rightarrow \frac{\Delta f}{f(f+\Delta f)}=-\frac{\Delta n}{R} \sim \frac{\Delta f}{f}=-\frac{\Delta n}{R} \\ & \Rightarrow \frac{\Delta f}{f}=-\frac{\Delta n}{R} \times f \end{aligned}$$

Using Eq. (1), we get $f=\frac{R}{2(n-1)}$

$$\begin{aligned} & \Rightarrow \frac{\Delta f}{f}=-\frac{\Delta n}{R} \times \frac{R}{2(n-1)} \Rightarrow \frac{\Delta f}{f}=-\frac{\Delta n}{2(n-1)} \\ & \Rightarrow\left|\frac{\Delta f}{f}\right|=\frac{1}{2}\left|\frac{\Delta n}{n-1}\right| \\ & \Rightarrow\left|\frac{\Delta f}{f}\right|>\left|\frac{\Delta n}{n}\right| \end{aligned}$$

Hence, option (C) is incorrect.

Now, for $n=1.5, \Delta n=10^{-3}$ and $f=20 \mathrm{~cm}$. Using relation $\frac{\Delta f}{f(f+\Delta f)}=-\frac{\Delta n}{R}$ and Eq. (1), we get

$$\begin{aligned} & \frac{\Delta f}{f(f+\Delta f)}=-\frac{\Delta n}{2(n-1) f} \Rightarrow \frac{\Delta f}{f+\Delta f}=-\frac{\Delta n}{2 n-2} \\ & \Rightarrow \Delta f(2 n-2)=-\Delta n(f+\Delta f) \\ & \Rightarrow \Delta f(2 n-2)=-\Delta n f-\Delta n \Delta f \\ & \Rightarrow \Delta f(2 n-2+\Delta n)=-\Delta n f \\ & \Rightarrow \Delta f=\frac{-\Delta n \times f}{2 n-2+\Delta n} \\ & \Rightarrow|\Delta f|=\frac{\Delta n+f}{2 n-2+\Delta n} \end{aligned}$$

Substituting all values, we get

$|\Delta f|=\frac{10^{-3} \times 20}{2 \times 1.5-2+10^{-3}}=0.02 \mathrm{~cm}$

Hence, option (B) is correct.

Since $\frac{\Delta f}{f}=-\frac{\Delta n}{2(n-1)}$, so if $\frac{\Delta n}{n}<0$ then $\frac{\Delta f}{f}>0$.

Hence, option $(\mathrm{A})$ is correct.

Since the relation between $\frac{\Delta f}{f}$ and $\frac{\Delta n}{n}$ is independent of radius of curvature so the relation remains unchanged only the sign of focal length changes.

Hence, option (D) is correct.

According to total internal reflection (TIR),

$1.5\sin {\theta _c} = 1.44\sin 90^\circ$

$\sin {\theta _c} = {{1.44} \over {1.50}} = {{24} \over {25}}$

$\therefore$ $\sin {\theta _c} = {x \over d} = {{24} \over {25}} \Rightarrow d = {{25x} \over {24}}$

$\therefore$ Total length travelled by light,

$\therefore$ $t = {S \over {\left( {{c \over {{n_1}}}} \right)}} = {{10} \over {{{3 \times {{10}^8}} \over {1.5}}}} = {1 \over 2} \times {10^{ - 7}} = 5 \times {10^{ - 8}}$

t = 50 ns $\Rightarrow$ t = 50 $\times$ 10^-9

V_sound = 330 m/s



${f_1} = 120\left[ {{{330 + 10\,\cos \,53^\circ } \over {330 - 30\,\cos 37^\circ }}} \right]Hz$

${f_2} = 120\left[ {{{330 + 10} \over {330}}} \right]Hz$

$\Delta f = ({f_2} - {f_1}) = 120 \times \left[ {{{336} \over {306}} - {{34} \over {33}}} \right] = 8.13Hz$

Gravitational force = Centripetal force of the earth

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

($\because$ M = total mass from 0 to r)

= ${2 \over r}\left( {{1 \over 2}m{v^2}} \right)$

$\Rightarrow {{GMm} \over {{r^2}}} = {{2K} \over r} \Rightarrow M = {{2Kr} \over {Gm}}$

$\because$ $\left( {{1 \over 2}m{v^2} = K} \right)$

Differentiate on both sides, we get

$\Rightarrow dM = {{2K} \over {Gm}}dr$

$\Rightarrow 4\pi {r^2}dr\rho = {{2K} \over {Gm}}dr$

($\because$ volume = mass $\times$ density)

$\therefore$ $\rho = {K \over {2\pi Gm{r^2}}}$

$\therefore$ ${\rho \over m} = {K \over {2\pi {r^2}{m^2}G}}$ $\because$ $\left( {{\rho \over m} = volume} \right)$

$V = 100 \times {10^{ - 3}} = {10^{ - 1}}V$

$V = {{\mathop{\rm l}\nolimits} _g}({R_g} + {R_v})$

${{{{10}^{ - 1}}} \over {2 \times {{10}^{ - 6}}}}={R_g} + {R_v}$

$5 \times {10^4}\Omega \approx {R_v}$ ($\because$${R_v} < {10^5}\Omega$)



I_gR_g = (I - I_g)S


$S = {{2 \times {{10}^{ - 6}} \times 10} \over {{{10}^{ - 3}} - 2 \times {{10}^{ - 6}}}}$

$S = 2 \times {10^{ - 5}} \times {10^3} = 2 \times {10^{ - 2}} = 20m\Omega$

${R_A} = {{S{R_g}} \over {S + {R_g}}} = {{20 \times {{10}^{ - 3}} \times 10} \over {10 + 20 \times {{10}^{ - 3}}}}$

$\approx 20 \times {10^{ - 3}}\Omega$

$i = {\varepsilon \over {\left( {{{1000 \times 50 \times {{10}^3}} \over {51 \times {{10}^3}}}} \right)}} = {{51\varepsilon } \over {5 \times {{10}^4}}}$

($\because$ R_A $\to$ 0)

$i' = i\left( {{{{R_v}} \over {51 \times {{10}^3}}}} \right) = {\varepsilon \over {1000}}$

Measured resistance,

$\therefore$ ${R_m} = {{i' \times 1000} \over i} = {\varepsilon \over {51\varepsilon }} \times 5 \times 10 = 980.4\Omega$

If the voltmeter shows full scale deflection then

${\varepsilon \over {980}} \times \left( {{{1000} \over {51 \times {{10}^3}}}} \right) \times 5 \times {10^4} = {10^{ - 1}}$

$\varepsilon = 999.6$ mV

Since, i_A = 1 mA so maximum reading of R can be

${{999.6\,mV} \over {1\,mA}} = 999.6\Omega$

at t = 0 dissipated energy,

${{dN} \over {dt}} = - ({\lambda _1} + {\lambda _2}) \times N$

$\Rightarrow$ ${{dN} \over N} = - ({\lambda _1} + {\lambda _2})dt$

Integration on both sides, we get

${\log _e}\left( {{N \over {{N_0}}}} \right) = - ({\lambda _1} + {\lambda _2})t$

$2.3 \times {\log _{10}}\left( {{{{N_0}} \over {{N_0}/100}}} \right) = (5 \times {10^{ - 10}})t$

${{2.303 \times 2} \over {5 \times {{10}^{ - 10}}}} = t$

$2.303 \times 0.4 \times {10^{10}} = t$

$\Rightarrow t = 9.2 \times {10^9}Yr$

Heights if only single material tubes are used of sufficient length,

${h_1} = {{2R\cos \theta } \over {\rho rg}}$

$= {{2 \times 0.075 \times \cos 0^\circ } \over {1000 \times 2 \times {{10}^{ - 4}} \times 10}} = 7.5\,cm$

${h_2} = {{2T\cos \theta '} \over {\rho rg}}$

$= {{2 \times 0.075 \times \cos 60^\circ } \over {1000 \times 2 \times {{10}^{ - 4}} \times 10}} = 3.75\,cm$

Case - I

Case - II

2. Liquid will rise only upto height of 5 cm and meniscus will adjust by changing is radius of curvature. If the liquid goes up in tube 2 then it will not be able to support the weight of the liquid.

3. Weight of water in meniscus will be different in two cases because angle of contact is different.

Motional emf across the length dy is,

$d\varepsilon = B{V_0}\,dy = {B_0}\left[ {1 + {{\left( {{y \over L}} \right)}^\beta }} \right]{V_0}dy$

$\therefore$ $\varepsilon = \int\limits_0^L {{B_0}} \left( {1 + {{\left( {{y \over L}} \right)}^\beta }} \right){V_0}dy$

$= {B_0}{V_0}\left[ {1 + {1 \over {\beta + 1}}} \right]$

emf in loop is proportional to L for given value of $\beta$,

$\beta = 0,\,\varepsilon = 2{B_0}{V_0}L$

$\beta = 2,\,\varepsilon = {B_0}{V_0}L\left[ {1 + {1 \over 3}} \right] = {4 \over 3}{B_0}{V_0}L$

The length of projection of the wire Y = X of length $\sqrt 2 L$ on the y-axis is thus, the answer remain unchanged.

Therefore, correct options are (a), (b) and (d)

P-V graph of the given V-T graph is shown below.



(d) $\left| {{{\Delta {Q_{1 \to 2}}} \over {\Delta {Q_{3 \to 4}}}}} \right| = \left| {{{N{C_p}\Delta {T_{1 \to 2}}} \over {N{C_p}\Delta {T_{3 \to 4}}}}} \right| = {{{T_0}} \over {{T_0}/2}} = 2$

(b) $\left| {{{\Delta {Q_{1 \to 2}}} \over {\Delta {Q_{2 \to 3}}}}} \right| = \left| {{{N{C_p}\Delta {T_{1 \to 2}}} \over {N{C_p}\Delta {T_{3 \to 4}}}}} \right| = {{{C_p}} \over {{C_V}}} = {5 \over 3}$

(a) ${W_{cycle}} = {p_0}{V_0} = nR\left[ {{{{T_0}} \over 2}} \right]$

Note For ideal gas equation,

pV = nRT

$\therefore$ (pV)₄ = (nRT)₄

or p₀V₀ = $nR{{{T_0}} \over 2}$ = $R{{{T_0}} \over 2}$

as n = 1

(c) Wrong as no adiabatic process is involved.

Parallel plate capacitor,

${x \over m} = {d \over N}$

$d\left( {{1 \over C}} \right) = {{dx} \over {{K_m}{\varepsilon _0}A}} = {{dx} \over {K{\varepsilon _0}A\left( {1 + {m \over N}} \right)}} = {{dx} \over {K{\varepsilon _0}A\left( {1 + {x \over d}} \right)}}$

Integration on both sides, we get

${1 \over {{C_{eq}}}} = \int {d\left( {{1 \over C}} \right) = \int_0^D {{{d\,dx} \over {K{\varepsilon _0}A(d + x)}}} }$

${1 \over {{C_{eq}}}} = \int {{d \over {K{\varepsilon _0}A}}\ln 2} \Rightarrow {C_{eq}} = {{K{\varepsilon _0}A} \over {d\,\ln 2}}$

Therefore, $\alpha$ = 1.

At normal or low concentration, sodium stearate (CH₃(CH₂)₁₆COO^-Na⁺] behaves as strong electrolyte and for strong electrolyte, molar conductance ($\Lambda$_m) decreases with increase in concentration. Above particular concentration, sodium stearate forms aggregates known as micelles. The concentration is called as CMC. Since, number of ions decreases and hence $\Lambda$_m also decreases.

Hence, option (c) is correct.

Borax bead test is performed only for coloured salt. Borax (sodium pyroborate), Na₂B₄O₇.10H₂O on heating gets fused and lose water of crystallisation. It swells up into fluffy white porous mass which melts into a colourless liquid which later form a clear transparent glassy bead consisting of boric anhydride and sodium metaborate.



Boric anhydride is non-volatile. When it react with Cr(III) salt then deep green complex is formed.



Hence, option (c) is correct.

ZnCO₃ - Calamine (zinc ore)

CuCO₃.Cu(OH)₂ - Malachite (copper ore)

Fe₃O₄ - Magnetite (iron ore)

Na₃AlF₆ - Cryolite (aluminum ore)

Thus, option (d) is correct.

Note : ( Remember all those ores names. Any one of those can be asked in the exam.)

Oxides Ores :

(1) ZnO - Zincite

(2) Fe₂O₃ - Haematite

(3) Fe₃O₄ - Magnetite (FeO + Fe₂O₃ mixture)

(4) Fe₂O₃ . 3H₂O - Limonite

(5) MnO₂ - Pyrolusite

(6) Cu₂O - Cuprite or Ruby Copper

(7) TiO₂ - Rutile

(8) FeCr₂O₄ - Chromite (FeO + Cr₂O₃)

(9) FeTiO₃ - Illmenite (FeO + TiO₂)

(10) Na₂B₄O₇ . 10H₂O - Borax or Tincal

(11) U₃O₈ - Pitch Blende

(12) SnO₂ - Tin Stone or Cassiterite

(13) Ca₂B₆O₁₁ . 5H₂O - Colemanite (2 Cao + 3 B₂O₃)

(14) Al₂O₃ . 2H₂O - Bauxite

(15) Al₂O₃ . H₂O - Diaspore

(16) Al₂O₃ - Corundum

Sulphides Ores :

(1) ZnS - Zinc Blende or Sphalerite

(2) PbS - Galena

(3) Ag₂S - Argentite or Silver Glance

(4) HgS - Cinnabar

(5) Cu₂S - Chalcocite or Copper glance

(6) CuFeS₂ - Copper pyrites or Chalco pyrites (Cu₂S + Fe₂S₃ mixture)

(7) FeS₂ - Iron pyrites or Fool's Gold

(8) 3Ag₂S . Sb₂S₂ - Pyrargyrite or ruby silver

Halides Ores :

(1) NaCl - Rock Salt

(2) KCl - Sylvine

(3) Na₃AlF₆ - Cryolite [3NaF + AlF₆]

(4) CaF₂ - Fluorspar

(5) KCl . MgCl₂ . 6H₂O - Carnalite

(6) AgCl - Horn Silver

Carbonates Ores :

(1) CaCO₃ - Limestone

(2) MgCO₃ - Magnesite

(3) CaCO₃ . MgCO₃ - Dolomite

(4) ZnCO₃ - Calamine

(5) PbCO₃ - Cerrusite

(6) FeCO₃ - Siderite

(7) CuCO₃ . Cu(OH)₂ or Cu₂CO₃(OH)₂ - Malachite or Basic Copper Carbonates

(8) 2 CuCO₃ . Cu(OH)₂ - Azurite

Sulphates Ores :

(1) CuSO₄ . 2H₂O - Gypsum

(2) MgSO₄ . 7H₂O - Epson Salt

(3) Na₂SO₄ . 10 H₂O - Glauber's Salt

(4) PbSO₄ - Anglesite

(5) ZnSO₄ . 7H₂O - White Vitriol

(6) FeSO₄ . 7H₂O - Green Vitriol

(7) CuSO₄ . 5H₂O - Blue Vitriol or Chalcanthite

Nitrate Ores :

(1) KNO₃ - Indian Saltpetre

(2) NaNO₃ - Chile Saltpetre

Arsenides Ores :

(1) NiAs - Kupfernickel

(2) NiAsS - Nickel glance

Acidic nature depends upon nature of electron withdrawing group and electronegativity. Electronegativity further depends on % of s character. Higher the s-character, greater will be the electronegativity and hence tendency to loose H increases thus acidic character also increases.



sp- hybridisation (50% s character) (pK_a = 1.86)



sp² - hybridisation (33-33% s character) (pK_a = 4.3)



sp² - hybridisation (Resonance effect) (pK_a = 4.5)



sp³ - hybridisation (25% s-character) (pK_a = 4.8)

Hence, acidic order I > II > III > IV.

II is more acidic than III since electron donating group ($- OC{H_3}$) is attached to benzene ring in III which decreases the acidic character.

On the other hand, pK_a value also determined acidic nature, lower pK_a value gives maximum acidic character.

Hence, option (b) is correct.

Sn can exist in +2 or +4 oxidation state. So, Q in the given reactions can be SnCl₂ or SnCl₄. Since X is monoanion having trigonal pyramidal geometry (sp³ with one lone pair as per VSEPR), so Q is SnCl₂, and the first reaction is

The second reaction is Stephen’s reaction, where nitrile is converted to aldehyde via formation of iminium salt.


There is a coordinate bond in the Cl₂Sn.N(CH₃)₃ in between nitrogen and Sn metal.

Z is oxidised product and oxidation state of Sn is +4 in Z compound. Structure of SnCl₄ (Z) is

$MnO_4^{2 - }$ ion has one unpaired electrons, therefore it gives d-d transition to form green colour. Y complex has paramagnetic nature due to presence of one unpaired electron.

In aqueous solution,



$MnO_4^{2 - }$ ions gives charge transfer spectrum in which a fraction of electronic charge is transferred between the molecular entities.

$\because$ $\mathop {MnO_4^{2 - }}\limits_{(Y)} \mathrel{\mathop{\kern0pt\longrightarrow} \limits_{oxidation}^{Electrolytic}} \mathop {MnO_4^ - }\limits_{(Z)} + {e^ - }$

In acidic medium, Y undergoes disproportionation reaction.



$\mathop {MnO_4^{2 - }}\limits_{(Y)}$ and $\mathop {MnO_4^ - }\limits_{(Z)}$ both ions form $\pi$-bonding between p-orbitals of oxygen and d-orbitals of manganese.

Thus, options (a, c, d) are correct.

The explanation of given statements are as follows :

(a) U_rms is inversely proportional to the square root of its molecular mass.

${U_{rms}} = \sqrt {{{3RT} \over M}}$

Hence, option (a) is correct.

(b) When temperature is increased four times then U_rms become doubled.

${U_{rms}} = \sqrt {{{3R} \over M} \times 4T}$

${U_{rms}} = 2 \times \sqrt {{{3RT} \over M}}$

Hence, option (b) is correct.

(c) and (d) E_av is directly proportional to temperature but does not depends on its molecular mass at a given temperature as ${E_{av}} = {3 \over 2}KT$. If temperature raised four times than E_av becomes four time multiple.

Thus, option (c) is incorrect and option (d) is correct.

The molecules which gives permanent dipole moment are polar in nature.





Thus, options (a, c) are correct.

N₂O, N₂O₄, H₂S₂O₃ and H₂S₂O₈ molecules are containing covalent bond between two atoms.


B₂H₆ and B₃N₃H₆ have polar bond, but do not have same kind of atom.

When 0.5 g of non-volatile solute dissolve into 39 gm of benzene then relative lowering of vapour pressure occurs. Hence, vapour pressure decreases from 650 mmHg to 640 mmHg.

Given, vapour pressure of solvent (p$^\circ$) = 650 mmHg

Vapour pressure of solution (p_s) = 640 mmHg

Weight of non-volatile solute = 0.5 g

Weight of solvent (benzene) = 39 g

From relative lowering of vapour pressure,

${{p^\circ - {p_s}} \over {p^\circ }} = {x_{Solute}} = {{{n_{solute}}} \over {{n_{solute}} + {n_{solvent}}}}$

${{650 - 640} \over {650}} = {{{{0.5} \over {molar\,mass}}} \over {{{0.5} \over {molar\,mass}} + {{39} \over {78}}}}$

${{10} \over {650}} = {{{{0.5} \over {molar\,mass}}} \over {{{0.5} \over {molar\,mass}} + 0.5}}$

$0.5 + 0.5 \times molar\,mass = 65 \times 0.5$

$\therefore$ Molar mass of solute = 64 g

From molal depression of freezing point,

$\Delta {T_f} = {K_f} \times molality$

$= {{{K_f} \times {w_{solute}}} \over {{{(MW)}_{solute}} \times {w_{solvent}}}}$

$\Rightarrow$ $\Delta {T_f} = 5.12 \times {{0.5 \times 1000} \over {64 \times 39}}$

$\Rightarrow \Delta {T_f} = 1.02K$

Given, equilibrium constant (K_c) at 298 K = 1.6 $\times$ 10¹⁷


[Here, K_c >>10³, thus limiting reagent will be consumed almost completely, 0.03 $-$ X = 0 $\therefore$ X = 0.03]

From equilibrium constant,

${K_c} = {{[FeS]} \over {[F{e^{2 + }}][{S^{2 - }}]}}$

${K_c} = {1 \over {X \times 0.07}}$

[For $\mathop {Fes(s)}\limits_{(Pure\,solid)} = 1\,mol\,{L^{ - 1}}]$]

$1.6 \times {10^{17}} = {1 \over {X \times 0.07}}$

$X = {1 \over {1.6 \times {{10}^{17}} \times 0.07}}$

$= 8.9 \times {10^{ - 17}}$

Given, X = Y $\times$ 10^-17 = 8.9 $\times$ 10^-17

$\therefore$ Y = 8.9