Class 12th

If S3 gets the most, S5 ranks the lowest and occupies seventh place. Since S7 always ranks above S1 and S2, . so at the maximum, S1 and S2 will occupy fifth and sixth places. Hence, S7 will not rank lower than fourth.

$\Delta H={\left(E_{act}\right)}_f-{\left(E_{act}\right)}_b$

= x – (x + y) = -y

              = -45 KJ/mol

              Ans. = 45

  $\overrightarrow{b}*\overrightarrow{c}=\left|\begin{array}{ccc}\hfill \widehat{i}\hfill & \hfill \widehat{j}\hfill & \hfill \widehat{k}\hfill \\ \hfill 1\hfill & \hfill 3\hfill & \hfill \beta \hfill \\ \hfill -1\hfill & \hfill 2\hfill & \hfill -3\hfill \end{array}\right|=\widehat{i}\left(-9-2\beta \right)-\widehat{j}\left(-3+\beta \right)+\widehat{k}\left(5\right)$             

$\left|\overrightarrow{b}*\overrightarrow{c}\right|=5\sqrt[]{3}gives\sqrt[]{{\left(9-2\beta \right)}^2+{\left(\beta -3\right)}^2+25}=5\sqrt[]{3}$         

$81+4{\beta }^2+36\beta +{\beta }^2+9-6\beta +25=75$

$\beta =-2,-4$    

also $\overrightarrow{a}\perp \overrightarrow{b}so\overrightarrow{a}.\overrightarrow{b}=0i.e.1+15+\alpha \beta =0$

So, ${\left|\overrightarrow{a}\right|}^2=1+25+{\alpha }^2=90$

According to the question, S7 > S1; S1 > S2 S7 > S1 > S2

If S3 scores the highest, we have S3 > … > S5

If S4 scores the highest, we have S4 > … > S2 or S4 > … > S6

If S7 ranks fifth, then S1 and S2 coming after S7 will occupy sixth and seventh places respectively i.e., S2 ranks the lest.

So, S4 will score the highest.

HgS, PbS, CuS, Sb₂S₃, As₂S₃, & CdS are given sulphides

              CdS, PbS, As₂S₃ & CuS are soluble in 50% HNO₃ but Sb₂S₃ & HgS are not soluble.

              Ans. = 4

Let f (x) = 3x⁴ + 4x³ – 12x² + 4

So, f' (x) = 12x³ + 12x² – 24x = 12x (x² + x – 2)

=12x (x + 2) (x – 1)

So, number of distinct real roots = 4

Change in oxidation number = 2

$\Delta G°=-n{E}_{cell}^{o}F$

$\Delta G°=-n{E}_{cell}^{o}F$

=-2 * 4.315 * 96487 Jmol^-1

^{$\therefore \Delta G°=\Delta H°-T\Delta S°$}

^{$\Delta S°=\frac{\Delta H°-\Delta G°}{T}$}

$=\frac{-825.2*1000-\left(-2*4.315*96487\right)}{298.15}J{K}^{-1}$

$=\frac{7482.81}{298.15}=25.09J{K}^{-1}$

$\approx$ 25.1 JK^-1

Ans. = 25

$l={\displaystyle \underset{0}{\overset{1}{\int }}\frac{\sqrt[]{x}dx}{\left(1+x\right)\left(1+3x\right)\left(3+x\right)}}$

Put x = t² then dx = 2tdt

$l={\displaystyle \underset{0}{\overset{1}{\int }}\frac{dt}{\left(1+t^2\right)\left(3+t^2\right)}-{\displaystyle \underset{0}{\overset{1}{\int }}\frac{dt}{\left(1+3t^2\right)\left(3+t^2\right)}}}$

$=\frac{1}{2}{\left(ta{n}^{-1}t\right)}_0^1-\frac{3}{8\sqrt[]{3}}{\left(ta{n}^{-1}\frac{t}{\sqrt[]{3}}\right)}_0^1-\frac{8}{8\sqrt[]{3}}{\left(ta{n}^{-1}\sqrt[]{3}t\right)}_0^1$

$=\frac{\pi }{8}\left(1-\frac{\sqrt[]{3}}{2}\right)$      

Given $2\mathcal{l}+2m-n=0........\left(i\right)$

$mn+n\mathcal{l}+\mathcal{l}m=0...........\left(ii\right)$

$\&wehave{\mathcal{l}}^2+m^2+n^2=1..........\left(iii\right)$

$\left(i\right)\Rightarrow 2\left(\mathcal{l}+m\right)=n$  

  $\left(ii\right)\Rightarrow \mathcal{l}m+n\left(\mathcal{l}+m\right)=0$

$2{\mathcal{l}}^2+2m^2+5\mathcal{l}m=0$             

(a) $\frac{\mathcal{l}}{m}=-2$  

(i) $\Rightarrow \frac{2\mathcal{l}}{m}+2-\frac{n}{m}=0$  

$\frac{n}{m}=-2$  

$So,\left(\mathcal{l},m,n\right)=\left(-2m,m,-2m\right)$  

= (-2, 1, -2)

(b)   $\frac{\mathcal{l}}{m}=\frac{-1}{2}givesn=-2\mathcal{l}$

$\left(\mathcal{l},m,n\right)=\left(\mathcal{l},-2\mathcal{l},-2\mathcal{l}\right)=\left(1,-2,-2\right)$

$Now,cos\theta =\frac{-2-2+4}{3*3}=0$

$\Rightarrow \theta =\frac{\pi }{2}$  

P1 is living in white house and earning low salary.