JEE Advanced 2020 Paper 1 Question Paper with Solutions

We have, h₁ + h₂ = 0.29 $\times$ 2 + 0.1

h₁ + h₂ = 0.68 .......(i)

$\Rightarrow$ p₀ + $\rho$_kg(0.1) +$\rho$_wg(h₁ $-$ 0.1) $-$ $\rho$_wgh₂ = p₀

where, $\rho$_k = density of kerosene and $\rho$_w = density of water.

$\Rightarrow$ $\rho$_kg(0.1) + $\rho$_wgh₁ $-$ $\rho$_wg $\times$ (0.1) = $\rho$_wgh₂

$\Rightarrow$ 800 $\times$ 10 $\times$ 0.1 + 1000 $\times$ 10 $\times$ h₁ $-$ 1000 $\times$ 10 $\times$ 0.1 = 1000 $\times$ 10 $\times$ h₂

$\Rightarrow$ 10000(h₁ $-$ h₂) = 200

$\Rightarrow$ h₁ $-$ h₂ = 0.01 .......(ii)

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

h₁ = 0.35

and h₂ = 0.33

So, ${{{h_1}} \over {{h_2}}} = {{35} \over {33}}$

$f \propto {1 \over {{l_1}}} \Rightarrow f = {k \over {{l_1}}}$ ....(i)

(where, l₁ $\Rightarrow$ initial length of pipe)

$\left( {{v \over {v - {v_T}}}} \right)f = {k \over {{l_1}}}$ ....(ii)

(where, v_T = speed of tuning fork, l₂ = new length of pipe)

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

${{v - {v_T}} \over c} = {{{l_2}} \over {{l_1}}} \Rightarrow {{{l_2}} \over {{l_1}}} - 1 = {{v - {v_T}} \over v} - 1$

${{{l_2} - {l_1}} \over {{l_1}}} = {{ - {v_T}} \over v}$

${{{l_2} - {l_1}} \over {{l_1}}} \times 100 = {{ - 2} \over {320}} \times 100 = - 0.625$

Therefore, smallest value of percentage change required in the length of pipe is 0.625.

${n_1} > > ({n_1} - {n_2}) = \Delta n$

$\because$ ${p_1} = {{{n_1}RT} \over {{N_A}}}$ and ${p_2} = {{{n_2}RT} \over {{N_A}}}$

$F = ({n_1} - {n_2}){k_B}TS = \Delta n{k_B}TS$

$V = {{\Delta n{k_B}TS} \over \beta }$

$\Delta n{k_B}TS = {\ln _1}S\beta v$

$\Rightarrow {n_1}\beta vl = \Delta n{k_B}T$

Total number of molecules/second

$= {{({n_1}vdt)S} \over {dt}}$

$= {n_1}vS = {{\Delta n{k_B}TvS} \over {\beta vl}} = \left( {{{\Delta n} \over l}} \right)\left( {{{{k_B}T} \over \beta }} \right)S$

As $\Delta$n will decrease with time, therefore rate of molecules getting transfer decreases with time.

Here, A = 64 mm², T = 2500K (where, A = surface area of filament, T = temperature of filament, d is distance of bulb from observer and R_e = radius of pupil of eye)

Point source, d = 100 m, R_e = 3 mm

(a) P = $\sigma$AeT^$-$4

= 5.67 $\times$ 10^$-$8 $\times$ 64 $\times$ 10^$-$6 $\times$ 1 $\times$ (2500)⁴

($\because$ e = 1 for black body)

= 141.75 W

(b) Power reaching to the eye

= ${P \over {4\pi {d^2}}} \times (\pi R_e^2)$

$= {{141.75} \over {4\pi \times {{(100)}^2}}} \times \pi \times {(3 \times {10^{ - 3}})^2}$

$= 3.189375 \times {10^{ - 8}}$ W

(c) ${\lambda _m}T = b$

${\lambda _m} \times 2500 = 2.9 \times {10^{ - 3}}$

$\Rightarrow {\lambda _m} = 1.16 \times {10^{ - 6}}$ = 1160 nm

(d) Power received by one eye of observer = $\left( {{{hc} \over \lambda }} \right) \times {{\Delta N} \over {\Delta t}}$

where, ${{\Delta N} \over {\Delta t}}$ = number of photons.

entering into eye per second.

$= 3.189375 \times {10^{ - 8}}$

$= {{6.63 \times {{10}^{ - 34}} \times 3 \times {{10}^8}} \over {1740 \times {{10}^{ - 9}}}} \times {{\Delta N} \over {\Delta t}}$

$\Rightarrow {{\Delta N} \over {\Delta t}} = 2.79 \times {10^{11}}$

Electrostatic force = Spring force

At l, $Fe = {F_{sp}}kl = {{2\alpha pq} \over {{l^3}}}$ (Here, $\alpha = {1 \over {4\pi {\varepsilon _0}}}$)



Now, the mass m is displaced by $\Delta$l = x from the mean position.

${F_{net}} = {F_{sp}} - Fe = k(l + x) - {{q(2\alpha p)} \over {{{(l + x)}^3}}}$

$= k(x + 1) - {{q(2\alpha p)} \over {{l^3}{{(l + x/l)}^3}}}$

$= kx + kl - q\left( {{{2\alpha p} \over {{l^3}}}} \right)\left( {1 - {{3x} \over l}} \right)$

$= kx + kl - q\left( {{{2\alpha p} \over {{l^3}}}} \right) + {{2\alpha p} \over {{l^3}}}.{{3x} \over l}$

Substituting ${{2\alpha pq} \over {{l^3}}} = kl$, we get

${F_{net}} = kx + kl\left( {{{3x} \over l}} \right) = 4kx$

This is restoring in nature.

Hence, ${k_{eq}} = 4k$

or $T = 2\pi \sqrt {{m \over {4k}}} = \pi \sqrt {{m \over k}}$

$\therefore$ $f = {1 \over \pi }\sqrt {{k \over m}}$

So, $\delta = \pi = 3.14$

$|Force| = \left| { - {{dV} \over {dr}}} \right| \Rightarrow F = {{m{v^2}} \over r}$ .....(i)

$mvr = {{nh} \over {2\pi }}$ .....(ii)

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

$\therefore$ $F = {m \over r}{\left( {{{nh} \over {2\pi mr}}} \right)^2}$ $\because$ $\left( {v = {{nh} \over {2\pi mr}}} \right)$

or $F = {{m{n^2}{h^2}} \over {4{\pi ^2}{m^2}{r^3}}}$

or $F = {{{n^2}{h^2}} \over {4{\pi ^2}m{r^3}}}$

${r^3} \propto \,{n^2}$ (as F = constant)

$r \propto \,{n^{2/3}}$

$v = {{nh} \over {2\pi mr}}$

$v \propto {n \over {{n^{2/3}}}}$

or $v \propto {n^{1/3}}$

$\therefore$ $KE = {1 \over 2}m{v^2} = {{Fr} \over 2}$

$\because$ $TE = PE + KE = Fr + {{Fr} \over 2}$

$\Rightarrow {{3Fr} \over 2} = TE$

or $TE = E = {{3F} \over 2} \times {\left( {{{{n^2}{h^2}} \over {4{\pi ^2}rnF}}} \right)^{1/3}}$

$= {3 \over 2}{\left( {{{{n^2}{h^2}{F^2}} \over {4{\pi ^2}m}}} \right)^{1/3}}$

$a = {{qE} \over m}{10^{10}} \times 400\sqrt 3 \,m/{s^2}$

or $a = 4\sqrt 3 \times {10^{12}}\,m/{s^2}$

$\therefore$ $R = {{{u^2}\sin 2\theta } \over a}$

$\Rightarrow \sin 2\theta = {{Ra} \over {{u^2}}}$

or $\sin 2\theta = {{5 \times 4\sqrt 3 \times {{10}^{12}}} \over {4 \times 10 \times {{10}^{12}}}}$

$\Rightarrow \sin 2\theta = {{\sqrt 3 } \over 2}$

$\Rightarrow 2\theta = 60^\circ ,\,120^\circ$

$\Rightarrow \theta = 30^\circ$ or $\,60^\circ$ for same range

$\therefore$ $T = {{2u\sin \theta } \over a}$

$= {{2 \times 2\sqrt {10} \times {{10}^6}} \over {2 \times 4\sqrt 3 \times {{10}^{12}}}}$

$\Rightarrow T = \sqrt {{{10} \over {12}}} \times {10^{ - 6}} \Rightarrow T = \sqrt {{5 \over 6}} \times {10^{ - 6}}$

Here, $\mu$_s = 0.40, $\mu$_k = 0.32

$\because$ $\mu$_kN₃ = $\mu$_sN₄

and x₁N₃ = 40 $\times$ N₄

So, ${{{\mu _k}} \over {{x_1}}} = {{{\mu _s}} \over {40}} \Rightarrow {{0.32} \over {{x_1}}} = {{0.4} \over {40}}$

$\Rightarrow {{0.32} \over {{x_1}}} = {1 \over {100}} \Rightarrow {x_1} = 32$ cm



${{{\mu _s}{N_5}} \over {{x_1}{N_5}}} = {{{\mu _s}{N_6}} \over {{x_g}{N_6}}} \Rightarrow {{0.40} \over {10 \times 32}} \times {{0.32} \over {{x_g}}}$

$\Rightarrow {x_g} = 25.6$

For no slipping at ground,

v_c = $\omega$R

(where, r is the radius of roller)

$\therefore$ Velocity of scale = (v_c + $\omega$r) (where, r is the radius of axle)

Given, v_c . t = 50 cm

$\therefore$ Distance moved by scale = (v_c + $\omega$r)t = $\left( {{v_c} + {{{v_c}r} \over r}} \right)t = {{3{v_c}} \over 2}.t = 75cm$

Therefore, relative displacement (with respect to centre of roller) is (75 $-$ 50) cm = 25 cm.

Let us consider a ring element of radius r and thickness dr.

Surface charge density of disc of radius R,

$\sigma \left( r \right) = {\sigma _0}\left( {1 - {r \over R}} \right)$

Charge of disc element, dq = ${\sigma _0}\left( {1 - {r \over R}} \right)2\pi rdr$

${\phi _0} = {{\int {dq} } \over {{\varepsilon _0}}} = {{\int\limits_0^R {{\sigma _0}\left( {1 - {r \over R}} \right)2\pi r\,dr} } \over {{\varepsilon _0}}}$

${\phi _0} = {{\int {dq} } \over {{\varepsilon _0}}} = {{\int\limits_0^{R/4} {{\sigma _0}\left( {1 - {r \over R}} \right)2\pi r\,dr} } \over {{\varepsilon _0}}}$

$\therefore$ ${{{\phi _0}} \over \phi } = {{{\sigma _0}2\pi \int\limits_0^R {\left( {r - {{{r^2}} \over R}} \right)} \,dr} \over {{\sigma _0}2\pi \int\limits_0^{R/4} {\left( {r - {{{r^2}} \over R}} \right)} \,dr}}$

$= {{{{{R^2}} \over 2} - {{{R^2}} \over 3}} \over {{{{R^2}} \over {32}} - {{{R^2}} \over {3 \times 64}}}} = {{32} \over 5} = 6.40$

When the magnet is moved, it creates a state, where the plate moves through the magnetic flux, due to which an electromotive force is generated in the plate and eddy currents are induced. These currents are such that it opposes the relative motion $\Rightarrow$ disc will rotate in the direction of rotation of magnet.

$[x] = {x^\alpha } \Rightarrow {p \over v} = time$

$[v] = {x^\beta }$

$[a] = {x^p} \Rightarrow {v \over a} = time$

$[p] = {x^q}$

$[F] = {x^r}$

$\Rightarrow \left( {{{position} \over {speed}}} \right) = \left( {{{speed} \over {acceleration}}} \right) = \left( {{p \over F}} \right)$

$\Rightarrow {{{x^\alpha }} \over {{x^\beta }}} = {{{x^\beta }} \over {{x^\alpha }}} = {{{x^q}} \over {{x^r}}}$

$\Rightarrow {x^{\alpha - \beta }} = {x^{\beta - p}} = {x^{q - r}}$

$\therefore$ $\alpha - \beta = \beta - p = q - r$

$\alpha + p = 2\beta$

$q + p = \beta + r$

$p + q - r = \beta$

Clearly middle part of glass is diverging and upper and lower part are conversing so correct shape of the emergent wavefront is as shown in the figure.

All the elements are in parallel.

$\therefore$ $\int {{l \over {dr}} = \int\limits_{{R_1}}^{{R_2}} {{{t\,dx} \over {\rho \,\pi x}}} }$

${l \over r} = {t \over {\pi \rho }}\ln \left( {{{{R_2}} \over {{R_1}}}} \right)$

Resistance = ${{\pi \rho } \over {t\ln \left( {{{{R_2}} \over {{R_1}}}} \right)}}$

$\therefore$ $I = {V \over {{\mathop{\rm Re}\nolimits} sis\tan ce}}$

$= {{{V_t}} \over {\pi \rho }}\ln \left( {{{{R_2}} \over {{R_1}}}} \right)$



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

$I = neA{V_d}$

${{{V_0}tdr} \over {\rho \pi r}} = ne(drt)V$

$\Rightarrow V = \sqrt {{{{V_0}} \over {\rho \pi ner}}}$

$\therefore$ $E = {m \over {er}}{{V_0^2{t^2}} \over {{\rho ^2}{\pi ^2}{n^2}{e^2}{r^2}}}$

$\Rightarrow E = {{mV_0^2} \over {{e^3}{r^2}{\rho ^2}{\pi ^2}{n^2}}}$

$dV = - Edr \Rightarrow dV = - {K \over {{r^3}}}dr$

$V = K\int\limits_{{R^1}}^{{R_2}} {{r^{ - 3}}dr}$

$\Delta V = {{mV_0^2} \over {{e^3}{\rho ^2}{\pi ^2}{n^2}}}\left( {{1 \over {r_1^2}} - {1 \over {r_2^2}}} \right)$

$\Delta V \propto V_0^2$

$\Rightarrow$ $\Delta$V $\propto$ I² ( As $I \propto {V_0}$ )

For $\theta$_max, the football is about to roll, then N₂ = 0 and all the forces (mg and N₁) must pass through contact point.

$\therefore$ $\cos (90^\circ - {\theta _{\max }}) = {r \over R}$

$\Rightarrow \sin {\theta _{\max }} = {r \over R}$

${{{p_1}} \over 4}{(4{V_1})^{5/3}} = {p_2}{(32{V_1})^{5/3}}$

${p_2} = {{{p_1}} \over 4}{\left( {{1 \over 8}} \right)^{5/3}} = {{{p_1}} \over {128}}$

${W_{adi}} = {{{p_1}{V_1} - {p_2}{V_2}} \over {\gamma - 1}}$

$= {{{p_1}{V_1} - {{{p_1}} \over {128}}(32{V_1})} \over {{5 \over 3} - 1}}$

$= {{{p_1}{V_1}(3/4)} \over {2/3}} = {9 \over 8}{p_1}{V_1}$

${W_{iso}} = {p_1}{V_1}\ln \left( {{{4{V_1}} \over {{V_1}}}} \right) = 2{p_1}{V_1}\ln 2$

$\therefore$ ${{{W_{iso}}} \over {{W_{adi}}}} = {{16} \over 9}\ln 2$

$\Rightarrow f = {{16} \over 9} = 1.7778 \approx 1.78$

Torque experienced by circular loop = M $\times$ B

where, M is magnetic moment and B is magnetic field.

$\therefore$ $\tau$ = i$\pi$R²NB₀

(at the instant shown $\theta$ = $\pi$/2)

$\therefore$ $\tau$dt = dL = i$\pi$R²NB₀dt

= Q$\pi$R²NB₀ ($\because$ idt = Q)

Pressure at the bottom of disc = Pressure due to surface tension

$\rho gh = T\left( {{1 \over {{R_1}}} + {1 \over {{R_2}}}} \right)$

${R_1} > > > {R_2}$

So, ${1 \over {{R_1}}} < < < {1 \over {{R_2}}}$ and ${R_2} = h/2$

$\therefore$ $\rho gh = T\left( {{1 \over {{R_1}}} + {1 \over {{R_2}}}} \right) = T\left( {0 + {1 \over {h/2}}} \right)$

${h^2} = {{2T} \over {\rho g}} \Rightarrow h = \sqrt {{{2T} \over {\rho g}}}$

$\Rightarrow \, = \sqrt {{{2 \times 0.07} \over {{{10}^3} \times 10}}} = \sqrt {{{14 \times 100} \over {{{10}^4} \times 100}}}$

$h = \sqrt {14}$ mm = 3.741

(a) $P{b_3}{O_4} + {H_2}O\buildrel {} \over \longrightarrow$ No reaction Pb₃O₄ is insoluble in water or do not react with water.

(b) $K{O_2} + 2{H_2}O\buildrel {} \over \longrightarrow KOH + {H_2}{O_2} + 1/2{O_2}$

Potassium superoxide is a strong oxidant, able to convert oxides into peroxides or molecular oxygen. Hydrolysis gives oxygen gas, hydrogen peroxide and potassium hydroxide.

(c) $N{a_2}{O_2} + 2{H_2}O\buildrel {} \over \longrightarrow 2NaOH + {H_2}{O_2}$

When sodium peroxide dissolves in water, it is hydrolysed and forms sodium hydroxide and hydrogen peroxide. The reaction is highly exothermic.

(d) $L{i_2}{O_2} + 2{H_2}O\buildrel {} \over \longrightarrow 2LiOH + {H_2}{O_2}$

The reactivity of Li₂O₂ toward water differs from LiO₂, in Li₂O₂ results in H₂O₂ as a product.

Hence, the correct option is (b).

(a) The empirical formula of the compound :

Contribution of M and X :

${M_{\left( {2 \times {1 \over 2}} \right)}}{X_{\left( {4 \times {1 \over 4}} \right)}} \Rightarrow MX$

(b) Coordination number of both M and X is 8.

(c) Distance between M and X

$= \sqrt {{{{a^2}} \over 4} + {{{a^2}} \over 2}} = \sqrt {{3 \over 4}} a = {{\sqrt 3 } \over 2}a$

$\Rightarrow$ 0.866 a

(d) ${r_M}\,:\,{r_X} = (\sqrt 3 - 1)\,:\,1 \Rightarrow 0.732:1$, thus statement (d) is incorrect.

IUPAC name of P, Q, R and S are :



IUPAC name of compounds Q and R are same, thus the correct option is (c).

Compound (P) undergoes Friedel-Craft acylation reaction in presence of succinic anhydride and AlCl₃ to gives an acid (Q).

Compound (Q) on treatment with Zn-Hg/HCl undergoes Clemmensen reduction, where ketone group change into alkane and on further reaction with dehydrating agent H₃PO₄ gives compound (R).

Compound (R) on reaction with Grignard reagent and heated in presence of acid gives dehydrating product (S).



Thus, the option (b) and (d) are correct.

X : Ag Y : Pb P : AgCl Q : PbCl₂



Hence, the correct option is (a).

Vessel is insulated, thus q = 0

For the given reaction : \eqalign{ & {H_2}(g) + 1/2{O_2}(g)\buildrel {} \over \longrightarrow {H_2}O(l); \cr & {E^o} = 1.23 - 0.00 = 1.23V \cr}

$\Delta$G^o = $-$nFE^o

= $-$2 $\times$ 96500 $\times$ 1.23 J/mol

Therefore, work derived from this fuel cell using 70% efficiency and on consumption of 1.0 $\times$ 10^$-$3 mol of H₂(g)

= 2 $\times$ 96500 $\times$ 1.23 $\times$ 0.7 $\times$ 1 $\times$ 10^$-$3

= 166.17 J

This work done = change in internal energy (for monoatomic gas,

${C_{V'm}} = 3R/2$),

$166.17 = n{C_V}{,_m}\Delta T$

$\Rightarrow \Delta T = {{166.17 \times 2} \over {1 \times 3 \times 8.314}}$

$\Rightarrow 13.32\,K$

Given, $w = - \int {{p_{ext}}dV}$

For 1 mole van der Walls' gas

$p = \left( {{{RT} \over {V - b}} - {a \over {{V^2}}}} \right)$

For reversible process, p_ext = p_gas

$w = - \int {\left( {{{RT} \over {V - b}} - {a \over {{V^2}}}} \right)dV}$

But, it is not applicable for irreversible process which are carried out very fast. So, work done is calculated assuming final pressure remains constant throughout the process. Thus, statement (a), (b) and (c) correct while statement (d) is incorrect.

Triphenylmethane (I) is acidic because its conjugate base is stabilised by resonance.

${(Ph)_3}CH\buildrel {} \over \longrightarrow \mathop {{{(Ph)}_3}C}\limits_{Triphenylmethyl\,carbanion} + {H^ + }$

Cyclopentadiene (IV) is acidic because its conjugate base is aromatic.



Nitrobenzene is more acidic than benzene because nitro group is electron withdrawing. It will stabilise the conjugate base of benzene by $-$R and $-$I effect.

The acidic strength order on the basis of pK_a data is

IV > V > I > II > III.

Hence, the correct options are (a), (b) and (c) only.

(a) In [FeCl₄]^-, oxidation number of Fe atom = +3

Electronic configuration of Fe in ground state = 3d⁶4s²

Electronic configuration of

Fe³⁺ = 3d⁵4s⁰4p⁰



Thus, [FeCl₄]^- has tetrahedral geometry.

(b) [Co(en)(NH₃)₂Cl₂]⁺ have three geometrical isomers. Thus, statement (b) is incorrect.



(c) Fe³⁺ in [FeCl₄]^- is sp³-hybridised with 5 unpaired electrons. (higher spin-only magnetic moment

= $\sqrt {n(n + 2)}$ = 5.92 BM). While Co³⁺ in [Co(en)(NH₃)₂Cl₂]⁺ is d²sp³-hybridised with zero unpaired electrons (low spin-only magnetic moment = $\sqrt {n(n + 2)}$ = 0 BM).

Thus, the statement (c) is correct.

(d) Co³⁺[Co(en)(NH₃)₂Cl₂]⁺



Co³⁺ in [Co(en)(NH₃)₂Cl₂]⁺ is d²sp³-hybridised and has octahedral geometry with 0 unpaired electron. Thus, statement (d) is incorrect.

Fraction of molecules vs velocity graph is Maxwell distribution curve.

This curve is slightly unsymmetrical as shown below.

The ratio of most probable, the average and the root mean square speeds for this graph is 1 : 1.128 : 1.224

But the graph in question is completely symmetrical. Therefore, the most probable and the average speed will be same here but root mean square speed will be greater than average speed. So, the correct ratio is 1 : 1 : 1.224.

Aluminium reacts with sulphuric acid to form aluminium sulphate and hydrogen.

$\mathop {2Al}\limits_{\left( {{{5.4} \over {27}} = 0.2\,mol} \right)} + \mathop {3{H_2}S{O_4}}\limits_{\left( {{{50 \times 5} \over {1000}} = 0.25\,mol} \right)} \buildrel {} \over \longrightarrow A{l_2}{(S{O_4})_3} + 3{H_2}$

H₂SO₄ is limiting reagent and moles of H₂(g) produced = 0.25 mol

Using ideal gas equation,

pV = nRT

$\Rightarrow$ $V = {{0.25 \times 0.082 \times 300} \over {1\,atm}} = 6.15\,L$

First of all, we need to convert wedge-dash formula into Fischer projection formula.



Now we can clearly see, that compound P is same as given compound, thus compound P is identical to D-erythrose.

Compound Q and D-erythrose are not mirror images of one another and are non-superimposable on one another. Thus, they are diastereomers.

Compound R and D-erythrose are also not mirror images of one another and are non-superimposable on one another. Thus, they are diastereomers.

Compound S and D-erythrose are chiral molecules. They are mirror images of one another. Furthermore, the molecules are non-superimposable on one another. This means that the molecules cannot be placed on top of one another and give the same molecule. Chiral molecules with one or more stereocenters can be enantiomers.

Hence, the correct option is (c).

$_{92}^{238}U\buildrel {} \over \longrightarrow _{82}^{206}Pb + 8_2^4He + 6_{ - 1}^0\beta$

Number of moles of $_{92}^{238}U$ present initially

= ${{68 \times {{10}^{ - 6}}} \over {238}}$

After three half-lifes, moles of $_{92}^{238}U$ decayed

$= {{68 \times {{10}^{ - 6}}} \over {238}} \times \left( {1 - {1 \over {{2^3}}}} \right)$

$= {{68 \times {{10}^{ - 6}}} \over {238}} \times {7 \over 8}$

Therefore, number of $\alpha$-particles emitted

$= {{68 \times {{10}^{ - 6}}} \over {238}} \times {7 \over 8} \times 8 \times 6.023 \times {10^{23}}$

$= 1.204 \times {10^{18}}$

$\approx 1.2 \times {10^{18}}$

Thus, the correct answer is 1.2.

Given : Image

Using $\Delta G = \Delta {G^o} + RT\ln {K_p}$

At equilibrium : $\Delta {G^o} = - RT\ln {K_p}$

$\Delta G_1^0 = - R{T_1}\ln {K_{p1}}$ ... (i)

$\Delta G_2^0 = - R{T_2}\ln {K_{p2}}$ ...(ii)

From Eqs. (i) and (ii),

${{\Delta G_1^o} \over {\Delta G_2^o}} = {{{T_1}} \over {{T_2}}} \times {{\ln K{p_1}} \over {\ln K{p_2}}}$

$= {{1000} \over {2000}} \times {{\ln (10)} \over {\ln (100)}} = {1 \over 4} = 0.25$

According to the IUPAC rules, first of all select the longest carbon chain (6-carbon atoms) then numbering should be start from most prior functional group, which is carboxylic acid in the given compound, thus



3-ethynyl-2hydroxy-4-methylhex-3-en-5-ynoic acid

(a) Order of acid strength different oxyacids of chlorine are :

$\mathop {HClO}\limits_{(Hypochlorous\,acid)}^{( + 1)} < \mathop {HCl{O_3}}\limits_{(Chloric\,acid)}^{( + 5)} < \mathop {HCl{O_4}}\limits_{(Perchloric\,acid)}^{( + 7)}$

Weak acid have strong conjugate base thus hypochlorite ion has strongest conjugate base. Therefore, statement (a) is correct.

(b) Hypochlorite ion is linear and perchlorate ion is tetrahedral and there is no effect of lone pair on hypochlorite ion. Thus, statement (b) is correct.



(c) In the disproportionation reaction, chlorate ion Cl(+5) is oxidised to perchlorate, Cl(+7) and reduce to chloride, Cl($-$1).

$4ClO_3^ - \mathrel{\mathop{\kern0pt\longrightarrow} \limits_{}} 3ClO_4^ - + C{l^ - }$

While in hypochlorite ion, chlorite ion Cl(+1) is oxidised to chlorate, Cl(+5) and reduced to chloride, Cl($-$1) ion.

$3ClO_{}^ - \mathrel{\mathop{\kern0pt\longrightarrow} \limits_{}} ClO_3^ - + 2C{l^ - }$

Thus, statement (c) is incorrect.

(d) The hypochlorite ion oxidises the sulphite ion to sulphate ion, because HOCl is the strongest oxidising Cloxyacids,

$Cl{O^ - } + SO_3^{2 - }\buildrel {} \over \longrightarrow SO_4^{2 - } + C{l^ - }$

Thus, statement (d) is correct.