JEE MAIN 2020 - 02 SEP S1 2020

2,5 -dimethyl-6-oxo-hex-3-enoic acid

Photoelectric effect (option 2), atomic spectrum (option 3 ) and Black body radiations (option 4 $)$ may be explained by quantum theory. As on increasing temperature, all the values of internal energy becomes possible, it is not directly explained from quantum theory.

Slow or fast process is kinetic parameter but extent less or more is thermodynamic parameter.

(1) If ${AB}_{4}$ molecule is a square pyramidal then it has one lone pair and their structure should beand it should be polar because dipole moment of lone pair of 'A' never be cancelled by others. (2) If $\ {AB}_{4}$ molecule is a tetrahedral then it has no lone pair and their structure should be and it should be non polar due to perfect symmetry. (3) If $\ {AB}_{4}$ molecule is a square planar thenit should be non polar because vector sum of dipole moment is zero. (4) If $\ {AB}_{4}$ molecule is a rectangular planar thenit should be non polar because vector sum of dipole moment is zero.

Basic gas (C) must be ammonia (NH $_{3}$ ). It means (B) gas should be $\ {N}_{2}$ which is formed by heating of compound (A). (1) $({NH}_{4} )_{2} {Cr}_{2} {O}_{7} \ { \to }^{\Delta}$ ${N}_{2} ↑+\ {Cr}_{2} {O}_{3}+4 {H}_{2} {O} \uparrow$ (2) ${Pb} ({NO}_{3} )_{2} \ { \to }^{\Delta}$ ${PbO}+2 {NO}_{2} ↑+ \frac{1}{2} {O}_{2} \uparrow$ (3) $2 {NaN}_{3}\ { \to }^{\Delta}$ ${Na}+3 {N}_{2} ↑$ (4) ${NH}_{4} {NO}_{2}\ { \to }^{\Delta}$ ${N}_{2} \uparrow +2 {H}_{2} {O} \uparrow$ So, (A) should not be ${Pb}({NO}_{3} )_{2}$

In general across a period atomic radius decreases while ionisation enthalpy, electron gain enthalpy and electronegativity increases because effective nuclear charge $({Z}_{ {\text{eff}}} )$ increases.

In the stratosphere, CFCs release chlorine freeradical (\text{C}\text{l}^{ •}$)${CF}{2} {Cl}{2}( {g})\ \xrightarrow{UV} \text{C}\text{l}^{ •}( {g})$$+{{C^{ •}F}}{2} {Cl}( {g})$which react with${O}{3}$to give chlorine oxide$( {ClO^{ •}})$radical not chlorine dioxide$({ClO^{ •}}{2} )$radical.${Cl^{ •}}( {g})+ {O}{3}( {g})→{ClO^{ •}}( {g})+ {O}_{2}( {g})$

Cs used in photoelectric cell as it has least ionisation energy.

(I) Under weak field ligand, octahedral Mn(II) and tetrahedral ${\text{Ni}}( {\text{II}})$ both the complexes are high spin complex. (II) Tetrahedral Ni(II) complex can very rarely be low spin because square planar (under strong ligand) complexes of ${\text{Ni}}( {\text{II}})$ are low spin complexes. (III) With strong field ligands Mn (II) complexes can be low spin because they have less number of unpaired electron (unpaired electron $=1$ ) While with weak field ligands Mn(II) complexes can be high spin because they have more number of unpaired electron (unpaired electron $=5$ ) (IV) Aqueous solution of Mn(II) ions is pink in colour.

If spin only magnetic moment of the complex is $4.90 \ {BM},$ it means number of unpaired electrons should be 4 . (A) In octahedral complex: $[{M}({H}_{2} {O} )_{6} ]^{2+}$ ${ C.F.S.E. } = (-0.4 \Delta_{0} ) \times 4$ $+ (+0.6 \Delta_{0} ) \\times 2+0 \times {P}$ $=-0.4 \Delta_{0}$ (B) In tetrahedral complex: $[{M}({H}_{2} {O} )_{4} ]^{2+}$ ${ C.F.S.E. } = (-0.6 \Delta_{t} ) \\times 3$ $+ (+0.4 \Delta_{t} ) \\times 3+0 \\times \ {P}$ $=-0.6 \Delta_{ {t}}$

In Carius method mass of organic compound $=0.172 \ {gm}$ mass of Bromine $=0.08 \ {gm}$ Hence $\%$ of Bromine $= \frac{0.08}{0.172} \\times 100$ $=46.51 \%$ $[\% \ {Br}=\ \frac{80}{172} \\times 100 ]=46.51 \%$ (2) ${CH}_{3} {CH}_{2} {Br}\ \; {C}_{2} { H}_{5} {Br} \;\;\;$ $$ (3)(4) $CH_3Br$

With addition of solute in solvent, surface area for vapourisation decreases causes lowering invapour pressure

For ideal gas $PM = dRT$ $d=[\;\frac{P M}{R}] \;\frac{1}{T}$ So graph between $d Vs T$ is not straight line.

As language given, we have to go with option (1) as stereochemistry of chiral centre is not distortet.

Increasing order of reactivity towards HCN addition Greater the electrophilicity on - C^{∥ }^{\text{O}}- group greater the reactivity in nucleophilic addition. (iii) < (i) < (iv) < (ii)

Lab manual

(i) Glucose + dry HCl$\ {────── ▸}^{\text{ROH} }$ Acetal $\ {────── ▸}^{x \text{Eq}. }_{(\text{CH} _3\text{CO})_2\text{O}}$ acetyl derivative (i) Glucose $\ {────── ▸}^{\text{Ni}/\text{H}_2}$ A $\ {────── ▸}^{y \text{Eq} }_{(\text{CH} _3\text{CO})_2\text{O}}$ acetyl derivative (i) Glucose + $\ {\to }^{z \text{Eq} }_{(\text{CH} _3\text{CO})_2\text{O}}$ acetyl derivative due to presence of -OH group in Glucose the reaction is

Bredig's Arc method is used to from metal colloids.

$\Delta{G}=\Delta{G}°+ {RT} \ln [{ {Sn}^{+2}}{ {Cu}^{+2}} ]$ $=-2\times 96500[(-0.16)-0.34]$ $+ {RT} \ln (\frac{1}{1} )$ $=96500 \ {J}$

$\frac{{x}}{ {m}}= {k} {p}^{ {x}}$ ....(i) Given $c=\log {k}=0.4771$ or $\ {k}=3$ slope $x=2$ put in eq.(1) $\ \frac{x}{m}=3 \\times (4)^{2} \Rightarrow 48$

No. of chiral centres

(A) ${Na}_{4}[{Fe} ^{x}({CN})_{5} ( {NOS})]$ $(+1) 4+x+(-1) 5+(-1) 1=0$ $[x=+2]$ (B) ${Na}_{4}[{{Fe}^{y}O}_{4} ]$ $(+1) 4+y+(-2) 4=0$ $[y=+4]$ (C) $[ {Fe}^{z}_{2}( {CO})_{9} ]$ $2 {z}+0 \\times 9=0$ $[{z}=0 ]$ so $( {x}+ {y}+ {z}) =+2+4+0$ $=6$

$H _2O( l)⇌ H _2O(g)$ 90 gm of $H_2 O$ $\DeltaH = \DeltaU + \Delta n_gRT$ \Rightarrow 5 moles of $H_2 O$ 5 \\times 41000 \ {J}=\Delta{U}+1 \ $\times 8.314 \\times 373 \\times 5$ $\Delta{U}=189494.39$ Joule

$| {x}| < 1,| {y}| < 1, {x}\ne {y}$ $(x+y)+ (x^{2}+x y+y^{2} )+ (x^{3}$ $+x^{2} y+x y^{2}+y^{3} )$ $+......$ By multiplying and dividing $x-y:$ $\frac{({x}^{2}- {y}^{2})+({x}^{3}- {y}^{3} )+({x}^{4}- {y}^{4} )+....}{ {x}- {y}}$ $=\frac{({x}^{2}+ {x}^{3}+\ {x}^{4}+....)-({y}^{2}+\ {y}^{3}+ {y}^{4}+..... .)}{ {x}- {y}}$ $={\frac{x^{2}}{1-x}-\ \frac{y^{2}}{1-y}}/{x-y}$ $=\frac{(x^{2}-y^{2} )-x y(x-y)}{(1-x)(1-y)(x-y)}$ $= [\frac{x+y-x y}{(1-x)(1-y)}]$

Let $t_{r+1}$ denotes ${r}+1^{ {\text{th} }}$ term of $(\alpha {x}^{ \frac{1}{9}}+\beta {x}^{- \frac{1}{6}} )^{10}$ ${t}_{ {r}+1}={ }^{10} {C}_{ {r}} \; \alpha ^{10- {r}}( {x})^{ \frac{10- {r}}{9}}.\;\beta ^{ {r}} \; {x}^{- \frac{{r}}{6}}$ $={ }^{10} {C}_{ {r}} \; \alpha ^{10- {r}} \; \beta ^{ {r}}( {x})^{ \frac{10- {r}}{9}-\frac{{r}}{6}}$ If $t_{r+1}$ is independent of $x$ $\frac{10- {r}}{9}-\frac{{r}}{6}=0 \Rightarrow {r}=4$ maximum value of ${t}_{5}$ is $10 {K}$ (given) $\Rightarrow{ }^{10} \ {C}_{4} \alpha ^{6} \beta ^{4}$ is maximum By AM $\ge$ GM (for positive numbers) ${ \frac{\alpha ^{3}}{2}+ \frac{\alpha ^{3}}{2}+ \frac{\beta ^{2}}{2}+ \frac{\beta ^{2}}{2}}/4 \ge ( \frac{\alpha ^{6} \beta ^{4}}{16} )^{ \frac{1}{4}}$ $\Rightarrow \alpha ^{6} \beta ^{4} \le 16$ So, $10 \ {K}={ }^{10} {C}_{4} 16$ $\Rightarrow {K}=336$

{f}( {x})=\begin{cases}{ae}^{ {x}}+ {be}^{- {x}} & -1 \le {x} <1 \\ {cx}^{2} & 1\le {x} \le 3 \\ {ax}^{2}+2 {cx} & 3< {x} \le 4 For continuity at $x=1$ $\lim _{x\to 1^{-}} {f}( {x})=\lim _{x \to 1^{+}} {f}( {x})$ $\Rightarrowa e+b e^{-1}=c$ $\Rightarrow b=c e-a e^{2}$\end{cases}..(1) For continuity at x = 3 $\lim _{x\to 3^{-}} {f}( {x})=\lim _{x \to 3^{+}} {f}( {x})$ $\Rightarrow 3c=9 a+6 c$ $\Rightarrowc=3 a$ ....(2) $f'(0) + f'(2) = e$ $(ae^x - be^x )_{x = 0 }+ (2cx)_{x = 2} = e$ \Rightarrow $a-b+4 c=e$ ....(3) From (1), (2) & (3) $a - 3ae + ae^2 + 12a = e$ \Rightarrow $a(e^2 + 13 - 3e) = e$ \Rightarrow $a=\frac{{e}}{ {e}^{2}-3 {e}+13}$

Let ${B}_{1}$ be the event where Box-I is selected. $\& \ {B}_{2} \to$ where box-II selected ${P}({B}_{1} )= {P}({B}_{2} )= \frac{1}{2}$ Let E be the event where selected card is nonprime. For $B_1$ : Prime numbers : {2, 3, 5, 7, 11, 13, 17, 19, 23, 29} For $B_2$ : Prime numbers : {31, 37, 41, 43, 47} ${P}( {E})= {P}({B}_{1} ) \times {P}(\frac{{E}}{ {B}_{1}} )$ $+ {P}({B}_{2} ) {P}(\frac{{E}}{ {B}_{2}} )$ $= \frac{1}{2} \times \frac{20}{30}+ \frac{1}{2} \times \frac{15}{20}$ Required probability : $P(\frac{B_{1}}{E} )= \frac{\frac{1}{2}\times \frac{20}{30}}{ \frac{1}{2}\times \frac{20}{30}+ \frac{1}{2} \times \frac{15}{20}}$ $=\frac{\frac{2}{3}}{ \frac{2}{3}+ \frac{3}{4}}= \frac{8}{17}$

$\frac{|x|}{2}+ \frac{|y|}{3}=1$ $\frac{x^{2}}{4}+ \frac{y^{2}}{9}=1$Area of Ellipse $=\pi a b=6 \pi$ Required area,$=\pi \times 2 \\times 3-($ Area of quadrilateral $)$ $=6 \pi - \frac{1}{2} 6 \\times 4$ $=6 \pi -12$ $=6(\pi -2)$

$2 x-y+2 z=2$ $x-2 y+\lambda z=-4$ $x+\lambday+z=4$For no solution : $D=\begin{vmatrix}2 & -1 & 2 \\ 1 & -2 & \lambda \\ 1 & \lambda & 1\end{vmatrix}$ $\Rightarrow 2 (-2-\lambda^{2} )+1(1-\lambda)$ $+2(\lambda+2)=0$ $\Rightarrow-2 \lambda^{2}+\lambda+1=0$ $\Rightarrow \lambda=1,- \frac{1}{2}$ $D_x=\begin{vmatrix}2 & -1 & 2 \\ -4 & 2 & \lambda \\ 4 & \lambda & 1\end{vmatrix}$ $=2\begin{vmatrix}1 & -1 & 2 \\ -2 & -2 & \lambda \\ \lambda & \lambda & 1\end{vmatrix}$ $=2(1+\lambda)$ whichis not equal to zero for$\lambda=1,-\frac{1}{2}$

$| {A}| \ne 0$ For $( {P}): {A} \ne {I}_{2}$ So, $A=\begin{bmatrix}0 & 1 \\ 1 & 0\end{bmatrix}$ or $\begin{bmatrix}1 & 1 \\ 1 & 0\end{bmatrix}$ or $\begin{bmatrix}0 & 1 \\ 1 & 1\end{bmatrix}$ or $\begin{bmatrix}1 & 1 \\ 0 & 1\end{bmatrix}$ or $\begin{bmatrix}1 & 0 \\ 1 & 1\end{bmatrix}$ $|A|$ can be -1 or 1 So (P) is false. For $( {Q}) ; | {A}|=1$ $A=\begin{bmatrix}1 & 0 \\ 0 & 1\end{bmatrix}$ or $\begin{bmatrix}1 & 1 \\ 0 & 1\end{bmatrix}$ or $\begin{bmatrix}1 & 0 \\ 1 & 1\end{bmatrix}$ \Rightarrow ${\text{tr}}( {A})=2$ ⇒ Q is true

Let p denotes statement p : I reach the station in time. q : I will catch the train. Contrapositive of p→ q is $~q \to ~p$ $~q\to ~p$ : I will not catch the train, then I do not reach the station in time.

$\frac{2+\sin x}{y+1} \frac{d y}{d x}$ $=-\cos x, y>0$ $\Rightarrow \frac{d y}{y+1}= \frac{-\cos x}{2+\sin x} d x$ By integrating both sides : $\text{l}{n}| {y}+1 |=-\text{l}{n}| 2+\sin {x}|$ $+\text{l}{nK}$ $\Rightarrow {y}+1=\frac{{K}}{2+\sin {x}} \;\;$ $( {y}+1>0)$ $\Rightarrow {y}( {x})=\frac{{K}}{2+\sin {x}}-1$ Given ${y}(0)=1 \Rightarrow{K}=4$ So, $y(x)= \frac{4}{2+\sin x}-1$ ${a}= {y}(\pi )=1$ ${b}={\frac{{dy}}{ {dx}} ]}_{ {x}=\pi }$ $={ \frac{-\cos {x}}{2+\sin{x}}( {y}( {x})+1) ]}_{ {x}=\pi }=1$ ${\text{So}},( {a}, {b})=(1,1)$

$\sigma ^{2}=$ variance $\mu=$ mean $\sigma ^{2}= \frac{\sum _{i=1} ^{n} (x_{i}-\mu)^{2}}{{n}}$ $\mu=17$ $\Rightarrow {\sum _{x=1}^{ {17}}(a x+b)}/{17}=17$ $\Rightarrow 9 a+b=17$ ......(1) $\sigma ^{2}=216$ $\Rightarrow {\sum _{x=1}^{ {17}}(a x+b-17)^{2}}/{17}=216$ $\Rightarrow {\sum _{x=1}^{ {17}} a^{2}(x-9)^{2}}/{17}=216$ $\Rightarrow a^{2} 81-18 \\times 9 a^{2}$ $+a^{2} 3 \times (35)=216$ $\Rightarrow a^{2}= \frac{216}{24}=9$ $\Rightarrow a=3(a>0)$ $\Rightarrow$ From $(1), b=-10$ So, $a+b=-7$

Slope of tangent to the curve $y=x+\sin y$ at $( {a}, {b})$ is $\frac{2- \frac{3}{2}}{ \frac{1}{2}-0}=1$ $\Rightarrow {\frac{ {dy}}{ {dx}} ]}_{ {x}= {a}}=1$ $\frac{{dy}}{ {dx}}= 1+\cos {y} .\frac{ {dy}}{ {dx}}$ (from equation of curve) ${\frac{{dy}}{ {dx}} ]}_{ {x}= {a}}$ $={1+\cos {b}.\frac{{dy}}{ {dx}} ]}_{ {x}= {a}}$ $\Rightarrow \cos {b}=0$ $\Rightarrow \sin \ {b}=\±1$ Now, from curve ${y}= {x}+\sin {y}$ $b=a+\sin b$ $\Rightarrow|b-a|=|\sin b|=1$

Let $L$ be the common normal to parabola $y=x^{2}+7 x+2$ and line $y=3 x-3$ $\Rightarrow$ slope of tangent of ${y}= {x}^{2}+7 {x}+2$ at ${P}=3$ $\Rightarrow {\frac{{dy}}{ {dx}} ]}_{ {\text{For}} {P}}=3$ $\Rightarrow2 x+7=3 \Rightarrowx=-2 \Rightarrowy=-8$ So ${P}(-2,-8)$ Normal at ${P}: {x}+3 {y}+ {C}=0$ $\Rightarrow {C}=26$ (P satisfies the line) Normal $\ : x+3 y+26=0$

Two points on the line (L say) $\frac{x}{3}= \frac{y}{2}, z=1$ are(0,0,1)\ \ &(3,2,1) So dr's of the line is $< 3,2,0 >$ Line passing through $(1,2,1),$ parallel to $\ {L}$ and coplanar with given plane is ${\vec{r}}=\hat{i}+2 \hat{j}+\hat{k}+ {t}(3 \hat{i}+2 {j}),$ ${t}\in {R}(-2,0,1)$ satisfies the line (for $\ {t}=-1$ ) $\Rightarrow(-2,0,1)$ lies on given plane. Answer of the question is (2) We can check other options by finding eqution of plane Equation plane : $\begin{vmatrix}x-1 & y-2 & z-1 \\ 1+2 & 2-0 & 1-1 \\ 2+2 & 1-0 & 2-1\end{vmatrix}=0$ $\Rightarrow2(x-1)-3(y-2)$ $-5(z-1)=0$ $\Rightarrow2 x-3 y-5 z+9=0$