Class 12th

$\frac{x-1}{2}=\frac{y-2}{1}=\frac{z-1}{3}.......\left(i\right)$

$\frac{x-2}{1}=\frac{y-\lambda }{2}=\frac{z-3}{4}......\left(ii\right)$

$\overrightarrow{OA}=\left(i+2j+k\right)\overrightarrow{OB}=\left(2\widehat{i}+\lambda \widehat{j}+3\widehat{k}\right)$

$\overrightarrow{a}=2i+j+3k\&\overrightarrow{b}=i+2j+4k$

$\therefore$ shortest distance = $\left|\frac{\left(\overrightarrow{OB}-\overrightarrow{OA}\right).\left(\overrightarrow{a}*\overrightarrow{b}\right)}{\overrightarrow{a}*\overrightarrow{b}}\right|=\frac{1}{\sqrt[]{38}}$

$\Rightarrow \left|14-5\lambda \right|=1$           

$\Rightarrow \lambda =3\&\frac{13}{5}$          

Integer value of $\lambda$ = 3

$\left(\frac{A}{B}\right)*\left(\frac{B}{C}\right)*\left(\frac{C}{D}\right)=\frac{A}{D}$

$\Rightarrow \frac{5}{6}*\frac{13}{14}*\frac{9}{4}=\frac{195}{112}$

The new time is $\frac{1}{3}$  of the old time. If time becomes $\frac{1}{3}$ , then new speed should have been   $\frac{3}{1}$ of the initial speed. Therefore

$\frac{3S-S}{S}*100=200\mathrm{\%}$

Using ; Nernst equation

$E_{cell}=E_{cell}^o-\frac{0.06}{n}lo{g}_{10}\frac{\left[C{u}^{2+}\right]}{{\left[A{g}^{+}\right]}^2}$              

$E_{cell}=2.97-\frac{0.06}{2}lo{g}_{10}\frac{0.25}{10^{-6}}V$              

$=2.97-0.03\left(log2.5+5\right)V$              

$E_{cell}=2.8V$              

Ans. is 3 (the nearest integer)

Using the successive percentage change rule, we have

$-25+x-\frac{25x}{100}=20$

$\frac{3x}{4}=45$

x =60%

We have to find last non zero digit by 4 so we will eliminate 0 in question so now it becomes 104925 = 4 as 4odd = 4

Density, $\rho =7.62gc{m}^{-3}$  

Cell edge, a = 0.4518nm = 0.458 * 10^-7 cm

Effective number of atoms, Z = 4 for CCP.

Using ; $\rho =\frac{Z*A}{N_A*a^3}$  

$\rho =\frac{4*A}{6.022*10^{23}*{\left(0.458*10^{-7}\right)}^3}$              

$7.62=\frac{4*A}{6.022*{\left(0.4518\right)}^3*10^2}$              

A = 105.79 g/mol

Ans. is 106 (the nearest integers)