JEE MAIN 2020 - 03 SEP S2 2020

${Be}< {Mg}$ (atomic radius) ${Be}> {\text{Al}}$ (I.E1 greater than ${Al}$ because of fully filled valence shell of ${Be}$ ) Charge /radius ratio of ${Be}$ is less than that of ${\text{Al}}$. Both Be and ${\text{Al}}$ form mainly covalent compound.

Volume strength $=11.2 \\times$ molarity $\Rightarrow$ molarity $=\ \frac{5.6}{11.2}=0.5$ Assuming 1 litre solution; mass of solution $=1000 \ {ml} \\times 1 \ {g} / \ {ml}=1000 \ {g}$ mass of solute $=$ moles $\\times$ molar mass $=0.5 \ {mol} \\times 34 \ {g} / {mol}$ $=17 {gm}$ \Rightarrow ${\text{mass}} \%= \frac{17}{1000} \\times 100=1.7 \%$

$l=0$ to $( {n}+1)$ $n=1$ $( {n}+l) \Rightarrow \frac{1 {s}}{1} \frac{1 {p}}{2} \frac{1{d}}{3}$ $\ {n}=2$ $l=0,1,2,3$ $\frac{2 {s}}{2} \frac{2 {p}}{3} \frac{2 {d}}{4} \frac{2 {f}}{5}$ $\ {n}=3$ $l=0,1,2,3,4$ $\frac{3 {s}}{3} \frac{3 {p}}{4} \ \frac{3 {d}}{5} \frac{3 {f}}{6} \frac{3 {g}}{7}$ Now, in order to write electronic configuration, we need to apply $(\ {n}+l)$ rule Energy order $: 1 \ {s} < 1 {p} < 2 {s} < 1 {d} < 2 {p}$ $< 3 {s} < 2 {d} .....$ Option1) $13: 1 {s}^{2} 1 {p}^{6} 2 {s}^{2} 1 {d}^{3}$ is not half filled Option 2 ) $9: 1 {s}^{2} 1 {p}^{6} 2 {s}^{1}$ is the first alkali metal because after losing one electron, it will achieve first noble gas configuration Option 3 ) $8: 1 {s}^{2} 1 {p}^{6}$ is the first noble gas because after $1 {p}^{6} {e}^{-}$ willenter 2 s hence new period Option 4) $6: 1 {s}^{2} 1 {p}^{4}$ has $1 {p}$ valencesubshell.

Steep rise in $\ {pH}$ around the equivalence point for titration of strong acid with strong base.

(a) Acid rain is harmful for trees and animal life. It can also cause respiratory ailments in animals.

Let suppose element X⇒ $X_{(g)}\ { \to }^{ IE_1}_{800}X {(g)^{+}}\ { \to }^{IE_2 }_{2427}X {(g)^{+2}} \ { \to }^{IE_3}_{3658}$ $X {(g)^{+3}}\ { \to }^{IE_4}_{25024} X {(g)^{+4}}\ { \to }^{IE_5}_{32824} X {(g)^{+5}}$ ${X}^{+3}$ has stable inert gas configuration as there is high jump after $\ {IE}_{3}$ So valence electrons are 3

According to Dalton's law of partial pressure$\ {p}_{ {i}}=\ {x}_{ {i}}\times {P}_{ {T}}$ $\ {p}_{ {i}}=$ partial pressure of the ${i}^{ {\text{th}}}$ component ${x}_{ {i}}=$ mole fraction of the ${i}^{ {\text{th}}}$ component ${p}_{ {T}}=$ total pressure of mixture $\Rightarrow 2 {\text{atm}}=({{n}_{ {H}_{2}}}/{ {n}_{ {H}_{2}}+ {n}_{ {H}_{ {e}}}+ {n}_{ {O}_{2}}} )$ $\times {p}_{ {T}}$ $\Rightarrow {p}_{ {T}}=2 {\text{atm}} \\times \frac{3}{1}=6 \ {\text{atm}}$

$\text{table} [{Ru}( {en})_{3} ] {Cl}_{2}, {R}_{ {u}} \Rightarrow 4 {d} \text{series} ;,en \Rightarrow \text{chelating} \text{ligand} ; {CN}=6{,} \text{octahedral} \text{splitting}, \text{hence} \text{laye} \text{splitting} \text{of} \text{d}-\text{subshell}$ [{Fe}{{H}_{2} {O} )_{6} ] {Cl}_{2}, \Rightarrow $H_2O\Rightarrow$ Weak filled ligand ${Fe}^{+2} \Rightarrow [ {Ar}] 3 {d}^{6} 4 {s}^{0}$less plitting $\ {CN}=6$ octahedral splitting

Correct Ans. is (2)

Reactivity order of various carbonylcompounds $\to$ Aldehydes $>$ Ketones

Option 1 ) Manganate $\Rightarrow {MnO}_{4}^{2-}$ , Permanganate $\Rightarrow {MnO}_{4}^{-}$ After excitation $2 \\times 2 {p}_{\pi }-3 {d}_{\pi \sigma }$ $2 \\times 2 \ {P}_{\pi }-3 \ {d}_{\pi }$ $1 \\times 2 \ {P}_{\pi }-4 \ {P}_{\pi }$ (2) $\ {MnO}_{4}^{2-} \Rightarrow$ green $\ {MnO}_{ {4}}^{-} \Rightarrow$ purple/violet (3) Manganate contains 1 unpaired electron hence it is paramagnetic where as permanganetic contains no unpaired electrons hence it is diamagnetic. (4) Both have $\ {d}^{3}$ s hybridisation hence both have tetrahedral geometry.

(a) B will be more crystalline due to more inter molecular interactions hence more efficient packing. (b) B will have higher boiling point due to higher intermolecular interactions. (c) B will be more soluble in water than A as B will have more extent of H-bonding in water So all three statements are correct {Solubility date ⇒ O-salicylic acid = 2g/L P-salicylic acid = 5g/L}

due to pressure of $\ {b}$

Raniticline→ Antacid Nard1→ Antidepressant Chloramphenicol→ Antibiotic Dimetane→ Antihistamine

For $a A+b B \to c C$ $\frac{-1}{a} \frac{d[A]}{d t}= \frac{-1}{b} \frac{d[B]}{d t}$ $= \frac{1}{c} \frac{d[C]}{d t}$ $\therefore \frac{-1}{2} \frac{{d}[ {A}]}{ {dt}}=\ \frac{-1}{3} \frac{{d}[ {B}]}{ {dt}}$ $= \frac{-2}{3}\frac{{d}[ {C}]}{ {dt}}= \frac{1}{3} \frac{{d}[ {p}]}{ {dt}}$

% mass of water $= \frac{x \\times 18}{(12+6 \\times 16+35 \\times 3+52)}$ $\times 100=13.5$ $\Rightarrow \ x= \frac{265 \\times 13.5}{18 \\times 100} \simeq 2$

Moles of ${e}^{\ominus }=(\frac{8 \\times 60 \\times 2}{96000} )$ Using stoichiometry; theoritically ${{n}_{ {e}^{ {⊖}}} \text{used }}/{6}={{n}_{ {cr}^{+3}} \text{produced }}/{2}$ $\Rightarrow {n}_{ {cr}^{+3}}$ produced $= \frac{2}{6}\times \frac{8 \\times 60 \\times 2}{96000}$ $= \frac{0.02}{6}$ $\Rightarrow {wt}_{ {cr}^{+3}}$ theoritically produced $=(\frac{0.02}{6} \\times 52 ) {g}$ $\Rightarrow \%$ efficiency $= \frac{0.104 \ {g}}{(\frac{0.02 \\times 52}{6}){g}} \\times 100$ $= 60$

moles $= \frac{\text{number of molecules}}{6 \\times 10^{23}}$ $= \frac{\text{ given mass }}{\text{ molar mass }}$ $\Rightarrow$ molar mas $= \frac{10 \\times 6.023 \\times 10^{23}}{6.023 \\times 10^{22}}$ $=100 \ {g} / {mol}$ $\Rightarrow$ molarity $= {\text{ moles of solute }}/{\text{volume of } {\text{sol}}^{{n}}(l)}=\frac{(5 / 100)}{2}$ $= 0.025$

$H _3PO_2 + NaOH \to NaH_2PO_2$ $+ H _2 O$ ${{n}_{ {H}_{3} {PO}_{2}} \text{ reacted }}/{1}={{n}_{ {NaOH}} \text{ reacted}}/{1}$ $\Rightarrow \frac{0.1 \\times 10}{1}=0.1 \\times {V}_{ {NaOH}}$ $\Rightarrow {V}_{ {NaOH}}=10 {ml}$

Let molar mass of protein A = x g/mol Let molar mass of protein B = y g/mol $\pi _{ {A}}=$ osmotic pressure of protein ${A}={( \frac{0.73}{ {x}} )}/{0.25} {RT}$ $\pi _{ {B}}=$ osmotic pressure of protein ${B}={(\frac{1.65}{ {y}} )}/{1} {RT}$ $\pi _{ {A}}=\pi _{ {B}}$ $\Rightarrow (\frac{0.73}{x \\times 0.25} ) {RT}=(\frac{1.65}{ {y}} ) {RT}$ $\Rightarrow (\frac{{x}}{ {y}} ) = \frac{0.73}{0.25 \\times 1.65}$ $=1.769 \cong 1.77$

$\frac{{d}}{ {dt}} (6 {a}^{2} )=3.6 \Rightarrow12 {a} \frac{{da}}{ {dt}}=3.6$ ${a} \frac{{da}}{ {dt}}=0.3$ $\frac{{dv}}{ {dt}}=\frac{{d}}{ {dt}}({a}^{3} )=3 {a}({a} \frac{{da}}{ {dt}} )$ $=3 \\times 10 \\times 0.3=9$

$\int _{0} ^{1 / 2}\frac{((x^{2}-1 )+1 )}{ (1-x^{2} )^{3 / 2}} d x$ $\int _{0} ^{1 / 2}\frac{d x}{ (1-x^{2} )^{3 / 2}}-\int _{0} ^{1 / 2} {d x}/{\\sqrt{1-x^{2}}}$ $\int _{0} ^{1 / 2}\frac{x^{-3}}{ (x^{-2}-1 )^{3 / 2}} d x- (\sin ^{-1} x )_{0}^{1 / 2}$ Let $x^{-2}-1=t^{2} \Rightarrowx^{-3} d x=-t d t$ $∫_{\infty }^{\\sqrt{3}} \frac{-t d t}{t^{3}}- \frac{\pi }{6}= ∫^{\infty }_{\\sqrt{3}} \frac{d t}{t^{2}}- \frac{\pi }{6}$ $= \frac{1}{\\sqrt{3}}- \frac{\pi }{6}= \frac{k}{6}$ $k=2 \\sqrt{3}-\pi$

Let $a^{2}+b^{2} \in Q\ \&\ b^{2}+c^{2} \in Q$ eg. $a=2+\\sqrt{3} \ \&\ b=2-\\sqrt{3}$ $a^{2}+b^{2}=14 \in Q$ Let $\ c=(1+2 \\sqrt{3})$ ${b}^{2}+ {c}^{2}=20 \in {Q}$ But $a^{2}+c^{2}=(2+\sqrt{3})^{2}$ $+(1+2 \\sqrt{3})^{2} \in Q$ for $R_{2}$ Let $a^{2}=1, b^{2}=\\sqrt{3}\ \&\ c^{2}=2$ ${a}^{2}+ {b}^{2}∉{Q} \ \&\ {b}^{2}+ {c}^{2}∉{Q}$ But $a^{2}+c^{2} \in Q$

Length of latus rectum = 4${DB}=2$ ${C}_{1} {B}=\sqrt{({C}_{1} {D} )^{2}-( {DB})^{2}}=4$ ${C}_{1} {C}_{2}=8$

Put $x=\tan ^{2} \theta \Rightarrow d x$ $=2 \tan \theta \sec ^{2} \theta d \theta$ \int \theta . (2 \tan \theta \.\sec ^{2} \theta ) {d} \theta $=\theta . \tan ^{2} \theta -\int \tan ^{2} \theta {d} \theta$ $=\theta . \tan ^{2} \theta -∫(\sec ^{2} \theta -1 ) {d} \theta$ $=\theta (1+\tan ^{2} \theta )-\tan \theta + {C}$ $=\tan ^{-1}(\sqrt{{x}})(1+ {x})-\sqrt{{x}}+ {C}$

First Case: Choose two non-zero digits ${ }^{9} {C}_{2}$ Now, number of 5 -digit numbers containing both digits $=2^{5}-2$ Second Case: Choose one non-zero & one zero as digit ${ }^{9} {C}_{1}$. Number of 5 -digit numbers containg one non zero and one zero both $= (2^{4}-1 )$ Required prob. $=\frac{({ }^{9} {C}_{2} \times (2^{5}-2 )+{ }^{9} {C}_{1}\times (2^{4}-1))}{9 \\times 10^{4}}$ $= \frac{36 \\times (32-2)+9 \\times (16-1)}{9 \\times 10^{4}}$ $= \frac{4 \\times 30+15}{10^{4}}= \frac{135}{10^{4}}$

Let orthocentre is ${H}({x}_{0}, {y}_{0} )$ ${m}_{ {AH}} .{m}_{ {BC}}=-1$ $\Rightarrow \ (\frac{y_{0}-7}{x_{0}+1})(\frac{1+5}{-7-5} )=-1$ ⇒ $2 {x}_{0}- {y}_{0}+9=0$ ........ (1) and ${m}_{ {BH}} .{m}_{ {AC}}=-1$ $\Rightarrow \ (\frac{{y}_{0}-1}{ {x}_{0}+7})(\frac{7-(-5)}{-1-5} )=-1$ $\Rightarrow \ x_{0}-2 y_{0}+9=0$ ........ (2) Solving equation (1) and (2) we get $({x}_{0}, {y}_{0} ) \equiv (-3,3)$

${\text{Re}}( {z})=| {z}-1|$ $\Rightarrow \ {x}=\\sqrt{( {x}-1)^{2}+( {y}-0)^{2}} \;\;( {x}>0)$ $\Rightarrow \ {y}^{2}=2 {x}-1$ $=4 .\frac{1}{2}({x}- \frac{1}{2} )$ $\Rightarrow$ a parabola with focus (1,0) $\ \&$ directrix as imaginary axis. \therefore \ Vertex $=(\frac{1}{2}, 0 )$ ${A}({z}_{1} ) \ \& \ {B}({z}_{2} )$ are two points on it such that slope of $\ {AB}=\tan \frac{\pi }{6}= \frac{1}{\\sqrt{3}}$ $(\arg (z_{1}-z_{2} )= \frac{\pi }{6} )$ for $\;\;\;y^{2}=4 a x$ Let $\;\;A (a t_{1}^{2}, 2 a t_{1} )\ \&\ B (a t_{2}^{2}, 2 a t_{2})$ ${m}_{ {AB}}= {2}{ {t}_{1}+ {t}_{2}}$ $= \frac{4 {a}}{ {y}_{1}+ {y}_{2}}= \frac{1}{\\sqrt{3}}$ $($ Here ${a}= \frac{1}{2} )$ $\Rightarrow \ {y}_{1}+ {y}_{2}$ $=4 {a} \\sqrt{3}=2 \\sqrt{3}$

PA = PB ⇒ $PA^2 = PB^2$ $\Rightarrow (\alpha -4)^2 + (\beta + 2)^2 + (\gamma -3)^2$ $=(\alpha -2)^{2}+(\beta -4)^{2}+(\gamma +1)^{2}$ $\Rightarrow -4 \alpha +12 \beta -8 \gamma =-8$ $\Rightarrow 2 x-6 y+4 z=4$

Required limit ${L}=\lim _{ {h} \to 0} \frac{({a}+2( {a}+ {h}))^{1 / 3}-(3( {a}+ {h}))^{1 / 3}}{(3 {a}+ {a}+ {h})^{1 / 3}-(4( {a}+ {h}))^{1 / 3}}$ $=\lim _{ {h} \to 0}\frac{(3 a)^{1 / 3} (1+ \frac{2 h}{3 a} )^{1 / 3}-(3 a)^{1 / 3} (1+ \frac{h}{a} )^{1 / 3}}{(4 a)^{1 / 3} (1+ \frac{h}{4 a} )^{1 / 3}-(4 a)^{1 / 3} (1+ \frac{h}{a} )^{1 / 3}}$ $=\lim _{ {h} \to 0}( \frac{3^{1 / 3}}{4^{1 / 3}})[\frac{ (1+ \frac{2 h}{9 a} )- (1+ \frac{h}{3 a} )}{ (1+ \frac{h}{12 a} )- (1+ \frac{h}{3 a} )} ]$ $=(\frac{3}{4} )^{1 / 3} \frac{(\frac{2}{9}- \frac{1}{3} )}{\frac{1}{12}- \frac{1}{3} )}$ $=(\frac{3}{4} )^{1 / 3}(\frac{8-12}{3-12} )$ $=(\frac{3}{4} )^{1 / 3}(\frac{-4}{-9} )= {4^{1- \frac{1}{3}}}/{3^{2- \frac{1}{3}}}$ $= \frac{4^{2 / 3}}{3^{5 / 3}}$ $=\ \frac{(8 \\times 2)^{1 / 3}}{(27 \\times 9)^{1 / 3}}$ $=\ \frac{2}{3}(\frac{2}{9} )^{1 / 3}$

$C={\text{adj}} \ {A}=\begin{bmatrix}2 & -1 & 1 \\ -1 & 0 & 2 \\ 1 & -2 & -1\end{bmatrix}$ $|C|=| {\text{adj}} A|$ $=+2(0+4)+1 .(1-2)+1 .(2,4)$ $=+8-1+2$ $| {\text{adj}} {A}|=| {A}|^{2}=9=9$ $\lambda=| {A}|=\±3$ $|\lambda|=3$ $\ {B}= {\text{adj}} {C}$ $| {B}|=| {\text{adj}} {C}|$ $=| {C}|^{2}=81$ $|({B}^{-1})^{ {T}} |=| {B}|^{-1}= \frac{1}{81}$ $(|\lambda|, \mu)= (3, \frac{1}{81} )$