JEE MAIN 2020 - 08 JAN S2 2020

In blast furnace (metallugy of iron) involved reactions are (a) $CaO + SiO_2 \to CaSiO_3$(b) $3Fe_2O_3 + CO \to 2Fe_3O_4+CO_2$

Alcohol has more boiling point than ether (due to hydrogen bonding).So, In (A),-OH group is present, so inter-molecular Hbonding is possible while in (B), due to methoxy group there is no possibility of Inter-molecular H-bonding. So A is having higher boiling point than $B$.

$\log K=\frac{-Ea}{2.303RT}+\log A$ Acrroding to Arrhenius equation plot of 'log K' Vs $1/T$ is linear with. $\text{Slope} =\frac{-Ea}{2.303R}$ From plot we conclude : ${|\text{slope}|}_{ \text{magnitude}} : c > a > d > b$ $\therefore E_c> E_a > E_d > E_b$ → More is the slope more is the activation energy but slope should be highly negative.

If the given elements are arranged according to their position in periodic tableAtomic radius $\text{table} C ,>, O, >, F;,,,,∧;,,,,Cl;,,,,∧;,,,,Br$B r > Cl > C > O > Fc < b < a < d < e

Kjeldahl's method for estimation of nitrogen is not applicable for nitrobenzene $C_6H_5NO_2$. because reaction with $H_2SO_4$, nitrobenzene can not give ammonia.

Since AgBr has intermediate radius ratio∴ it shows both schottky & Frenkel defectsZnS → Frenkel defectsKBr, CsCl → Schottky defects

$H_2O(l) ⇌{H^{+}}_(aq)+{{OH}^{-}}_{(aq)}$For ionization of $H_2O : \DeltaH > O$ \Rightarrow ENDOTHERMICOn temperature increase reaction shifts forward⇒ both $[H^{+}]$ and $[OH^{-}]$ increase⇒ pH & pOH decreases.

Bond length order in carbon halogen bonds are in the order of C - F < C - C l < C - B r < C - IHence, Bond energy order$C-F>C-Cl>C-Br>C-I$

$P_4 + 3NaOH + 3H_2O$ → ${3NaH_2P O_2}_{(X)} +PH_3$ $NaH_2PO_2 + HCl$ $\to NaCl + {H_3PO_2}_{(Y)}$

$r_n={n^2\times a_0}/Z$For 2nd Bohr orbit of $Li^{+2 }$n = 2z = 3$\Rightarrow r_n={2^2\times a_0}/3={4a_0}/3$

$[Pt(NH_3)_3Cl]^{+}$ & $[Pt(NH_3)Cl_5]^{-}$ does not show geometrical isomerism

Two monomers in maltose are α-D-glucose & α-D-glucose.

Hydrogen has three isotopes $\text{table} \text{Isotopes} ,\text{Number} \text{of} \text{neutrons};\text{Protium} (^1_1H), 0;\text{Deutrium} (^2_1H), 1;\text{Tritium} (^3_1H), 2$Hence the sum o f neutrons are 3

Bakelite formation is example of electrophilic substitution and dehydration.

Group 7-9 elements of the periodic table show variable valencies so they have maximum activity because of the increase in adsorption rate.

$\text{table} [Ni(CO)_4] ,\mu _m= 0 B.M.;[Ni(H_2O)_6]Cl_2,\mu _m = 2.8 B.M.;Na_2[Ni(CN)_4],\mu _m = 0 B.M.;[PdCl_2(PPh_3)_2] ,\mu _m =0 B.M$ (A) ≈ (C) ≈ (D) < (B)

Cell reaction is : $Sn(s) + Pb^{+2}(aq)$ $\to Sn^{+2}(aq) + Pb(s)$ Apply Nemst equation : E_\text{cell}=E°_\text{cell}-0.06/2\log\[\text{Sn}^{+2}]$/[Pb^{+2}]$...(1)$E°-\text{cell} = -0.13 + 0.14 = 0.01 V$At equilibrium :$E_\text{cell} = 0$Substituting in (1)$0=0.01-0.06/2\log$[\text{Sn}^{+2}]$/[Pb^{+2}]$$\Rightarrow 1/3= \log$[\text{Sn}^{+2}]$/[Pb^{+2}]$$\Rightarrow $[\text{Sn}^{+2}]$/$[Pb^{+2}]$=2.15$

For ideal gas :$\DeltaU=nC_V[T_2-T_1]$ $\Rightarrow 5000 = 4 \times C_V[500 - 300]$ $\Rightarrow C_V=6.25\text{J}\text{mole}^{-1}\text{K}^{-1}$

First, we use the Ideal Gas Law, $PV=nRT$, to determine the number of moles of $O_2$ required.
$n_{O_2} = \frac{PV}{RT} = \frac{(1 \text{ atm})(492 \text{ L})}{(0.082 \text{ L atm mol}^{-1} \text{K}^{-1})(300 \text{ K})} = 20 \text{ mol}$
The provided reaction is $NaClO_3(s) + Fe(s) \to O_2(g) + NaCl(s) + FeO(s)$. This equation is balanced, showing that 1 mole of $NaClO_3$ produces 1 mole of $O_2$.
Thus, to produce 20 moles of $O_2$, we need 20 moles of $NaClO_3$.
The molar mass of $NaClO_3$ is $23 + 35.5 + 3(16) = 106.5 \text{ g/mol}$.
Finally, we calculate the required mass of $NaClO_3$:
$\text{Mass} = \text{moles} \times \text{Molar Mass} = 20 \text{ mol} \times 106.5 \text{ g/mol} = 2130 \text{ g}$

Total 13 atom are present in same plane (7 carbon & 6 hydrogen atoms. )

The $(NiL_5)$ complex, with an ideal square pyramidal geometry, has 4 basal ligands in a square plane and 1 apical ligand. This results in 8 L-M-L angles of 90° (4 basal-basal, 4 apical-basal) and 2 angles of 180° (opposite basal ligands).

The $(FeL_5)$ complex, with a trigonal bipyramidal geometry, has 3 equatorial and 2 axial ligands. This gives 6 angles of 90° (axial-equatorial), 3 angles of 120° (equatorial-equatorial), and 1 angle of 180° (axial-axial).

We sum the counts for each specified angle type across both complexes:
Total number of 90° angles = $8 (\text{from } NiL_5) + 6 (\text{from } FeL_5) = 14$.
Total number of 120° angles = $0 (\text{from } NiL_5) + 3 (\text{from } FeL_5) = 3$.
Total number of 180° angles = $2 (\text{from } NiL_5) + 1 (\text{from } FeL_5) = 3$.

The total sum is the sum of these counts: $14 + 3 + 3 = 20$.

$\vec{b}\times \vec{c}-\vec{b}\times \vec{a}=\vec{0}$ $\vec{b}(\vec{c}-\vec{a})=\vec{0}$ $\vec{b}=\lambda (\vec{c}-\vec{a})$...(i)$\vec{a}.\vec{b}=\lambda (\vec{a}.\vec{c}-\vec{a}^2)$from (i) $\vec{b}=-2/3(\vec{c}-\vec{a})$ $\vec{c}=-3/2\vec{b}+\vec{a}$ $=-1/2( \hat{i}+\hat{j}+\hat{k})$ $\vec{b}.\vec{c}=-1/2$

Area $=∫^{1}_{-3}(3-2x-x^2)dx=32/3$

$x^2 + 2xy - 3y^2 = 0$ $m_N$ = slope o f normal drawn to curve at (2,2) is -1 L : x + y = 4.perpendicular distance of L from (0,0) $={|0+0-4|}/\sqrt2=2\sqrt2$

$f(x)= {dx}/\sqrt{2x^{3-}9x^{2+}12x+4}$ $f'(x)=\frac{-6(x-1)(x-2)}{2(2x^{3-}9x^{2+}12x+4)^{3/2}}$ $\therefore f(x)$ is decreasing in (1,2)$f(1)=1/3,f(2)=1/\sqrt8$ $1/3 < I < 1/\sqrt8$ $\Rightarrow I^2 \in (1/9,1/8)$

Slope of tangent to $x^2 + y^2 = 1$ at $P(1/\sqrt2,1/\sqrt2)$ $2x + 2yy' = 0$ $\Rightarrow m_T|_P=-1$y = mx + c is tangent to $(x - 3)^2 + y^2 = 1$y = x + c is tangent to $(x - 3)^2 + y^2 = 1$ $|{ {c+3}/\sqrt2}|=1$ $\Rightarrow c^{2+}6c+7=0$

(BONUS) option (1), (2), (3) are incorrect for f(x) = constant and option (4) is incorrect $\frac{f(1)-f(c)}{1-c}=f'(c)$ where c < a < 1 (use LMVT)Also for $f(x) = x^2$ option (4) is incorrect

$~(p \vee ~q)\to p\vee q$ $(~p \wedge q) \to p \vee q$ $~\{(~P \wedge q) \wedge (~p \wedge ~q)\}$ $~(~P\wedge q)$

$T_10=1/20=a+9d$ $T_20=1/10=a+19d$ $a=1/200=d$Hence, $S_200=200/2[2/200+199/200]=201/2$

$f(x)=\{\text{table} x/{x^{2+}1}{,},x \in (1,2);\frac{2x}{x^{2+}1}{,},x \in [2,3)$f(x) is decreasing function$\therefore f(x)\in (2/5,1/2)⋃(3/5,4/5]$

$D=\begin{vmatrix}\lambda & 3 & 2 \\ 2\lambda & 3 & 5 \\ 4 & \lambda & 6\end{vmatrix}=(\lambda +8)(2-\lambda )$for $\lambda = 2 ; D_1\ne 0$Hence, no solution for $\lambda =2$

$2[^6C_0 .x^6 + ^6C _2x^4(x^2 -1 )$ $+ ^6C_4x^2(x^2 - 1)^2 + ^6C_6(x^2 -10^3]$ $\alpha = -96$ & $\beta = 36$ $\therefore \alpha -\beta = -132$

Using L.H. Rule $\lim_{x\to 0}{x\sin(10x)}/1=0$

$A=(\text{table} 2,2;9,4)$; $|A|=8-18=-10$ $A^{-1}=\frac{\text{adj} A}{|A|}= (\text{table} 4,-2;-9,2)/{-10}$ $10A^{-1} = (\text{table} 4,-2;-9,2)=A-6I$

${ ∑x_i}/20=10$ $\Rightarrow ∑x_i=200$...(i)${∑x_i^2}/20-100=4$ $\Rightarrow ∑x_i^2=2080$...(ii)Actual mean $={200-9+11}/20=202/20$Variance $={2080-81+121}/20-(202/20)^2$ $=3.99$

$x^2/36-y^2/b^2=1$...(i)P(10,16) lies on (i) get $b^2 = 144$ $x^2/36-y^2/144=1$Equation of normal is${a^2x}/x_1+{b^2y}/y_1=a^2e^2$ $2x + 5y = 100$

$P(A) + P(B) - 2P(A ∩ B) =2/5$ $P(A) + P(B) - P(A ∩ B) =1/2$ $P(A ∩ B)=1/10$

Point on plane $R(-2/3,1/3,4/3)$ Normal vector of plane is $10/3\hat{i}+10/3\hat{j}+10/3\hat{k}$Equation of require plane is $x + y + z = 1$Hence (1, -1, 1) lies on plane

Let $3 ^x = t ; t > 0$ $t(t - 1) + 2 = |t - 1| + |t - 2|$ $t^2 - t + 2 = |t - 1| + |t - 2|$Case-I : $t < 1$ $t^{2-} t + 2= 1- t + 2 - t$ $t^2 + 2 = 3 - t$ $t^2 + t - 1 = 0$ $t={-1±\sqrt5}/2$ $t={\sqrt5-1}/2$ is only acceptableCase-II :$1 < t < 2$ $t^{2-} t + 2 = t - 1 + 2 - t$ $t^2 - t + 1 = 0$ $D < 0$ no real solutionCase-Ill :$t > 2$ $t^{2-} t + 2 = t - 1+ t - 2$ $t^2 - 3t 5 = 0$ $\Rightarrow D < 0$ no real solution

$\alpha =\omega$ $a=(1+\omega )(1+\omega +\omega ^{2+}\omega ^{4+}...+\omega ^200)$ $a=(1+\omega )\frac{(1-(\omega ^2)^101)}{1-\omega ^2}=1$ $b = 1 + \omega ^3 + \omega ^6 + ..... +\omega ^300 = 101$ $x^2 - 102x + 101 = 0$

$2x = 4by' \Rightarrow y'=\frac{2x}{4b}$Required D.E. is $x^2={2x}/y'y+(x/y')^2$ $x(y')^2 = 2yy' + x$

${\sqrt2\sin\alpha }/ {\sqrt2\cos\alpha }=1/7$ $\Rightarrow \tan\alpha =1/7$ $\sin\beta =1/\sqrt10\Rightarrow \tan\beta =1/3$ $\Rightarrow \tan2\beta =3/4$ $\tan (\alpha +\beta )=\frac{\tan\alpha +\tan2\beta }{1-\tan\alpha \tan2\beta }=1$

$f"(x) = \lambda (x - 1)$ $f'(x)={\lambda x^2}/2-\lambda x+C$ $\Rightarrow f'(-1)=0\Rightarrow c={-3\lambda }/2$ $f(x)={\lambda x^3}/6-{\lambda x^2}/2-{3\lambda }/2x+d$ $f(1)=-6$ $\Rightarrow -11\lambda +6d=-36$...(i)$f(-1)=10\Rightarrow 5\lambda +6d=60$...(ii)from (i) & (ii) \lambda = 6 \;\;&\;\; d = 5 $f(x) = x^3 - 3x^2 - 9x + 5$Which has minima at x = 3

ΔOPQ = 4$1/2{|{\begin{vmatrix}0 & 0 & 1 \\ t^2 & t & 1 \\ -t^2 & 0 & 1\end{vmatrix}}|}=4$ $t=2 ( \because t > 0)$ $\therefore m=1/2$

$1/4(∑ ^{7}_{n=1}2n^{3+}∑ ^{7}_{n=1}3n^{2+}∑ ^{7}_{n=1}n)$ $=504$

N → 2, A → 2, I→ 2, E, X, M, T, O → 1 Category Selection Arrangement 2 alike of one kind and 2 alike of other kind $^3C_2=3$ $3\times \frac{4!}{2!2!}=18$ 2 alike and 2 different $^3C_1\times ^7C_2$ $^3C_1\times ^7C_2\times 4!/2!=756$ All 4 different $^8C_4$ $^8C_4\times 4!=1680$Total = 2454