Class 12th

 Al equilibrium

${\mathrm{K}}_{\mathrm{f}}\left. [{\mathrm{H}}_2\right][\mathrm{N}\mathrm{O}{]}^2=\frac{{\mathrm{K}}_{\mathrm{b}}\left. [{\mathrm{N}}_2\right]{\left. [{\mathrm{H}}_2\mathrm{O}\right]}^{+}}{\left. [{\mathrm{H}}_2\right]}$

$\Rightarrow I_{g}G=\left(l-l_g\right)8\Rightarrow l_g*72=\left(l-l_g\right)8\Rightarrow 9l_g=l-l_g$

$10l_g=l{l}_g=\frac{1}{10}l$

% of I which passes through galvanometer

$$ $l_g=10\mathrm{\%}$  of I

$6\mathrm{N}\mathrm{a}\mathrm{O}\mathrm{H}+3{\mathrm{C}\mathrm{l}}_2?5\mathrm{N}\mathrm{a}\mathrm{C}\mathrm{l}+{\mathrm{N}\mathrm{a}\mathrm{C}\mathrm{l}\mathrm{O}}_3+3{\mathrm{H}}_2\mathrm{O}$

$2\mathrm{C}\mathrm{a}(\mathrm{O}\mathrm{H}{)}_2+{\mathrm{C}\mathrm{l}}_2?\mathrm{C}\mathrm{a}(\mathrm{O}\mathrm{C}\mathrm{l}{)}_2+{\mathrm{C}\mathrm{a}\mathrm{C}\mathrm{l}}_2+{\mathrm{H}}_2\mathrm{O}$

In forward bias diode act as a short circuit wire. Hence, the equivalent circuit is now.

So,  ${}_{}\frac{}{}$ $V_{ab}=\frac{30}{5+10}*5=10\mathrm{V}$

$\mathrm{}50.00$

Electric field at $\mathrm{t}=0$ & $(\mathrm{x},\mathrm{y},\mathrm{z})=\left(\mathrm{0,0},\frac{\pi }{\mathrm{k}}\right)$  is

$\begin{array}{rr}& \stackrel{?}{E}=-\left(\frac{\hat{i}+\hat{j}}{\sqrt[]{2}}\right)E_0\text{and}\stackrel{?}{F}=\stackrel{?}{E}q\\ & \therefore \mathrm{}\stackrel{?}{F}=-\left(\frac{\hat{i}+\hat{j}}{\sqrt[]{2}}\right)q\end{array}$

Which is antiparallel to $\left(\frac{\stackrel{\mathrm{ˆ}}{\mathrm{i}}+\stackrel{\mathrm{ˆ}}{\mathrm{j}}}{\sqrt[]{2}}\right)$

$A=A_0{e}^{-\lambda t}$

$\begin{array}{rr}& 500=700{\mathrm{e}}^{-\lambda \mathrm{t}}\\ & \lambda \mathrm{t}=\mathrm{l}\mathrm{n}?\frac{7}{5}\\ & \frac{\mathrm{l}\mathrm{n}?2}{{\mathrm{t}}_{1/2}}*30=\mathrm{l}\mathrm{n}?\frac{7}{5}\\ & ?\mathrm{}{\mathrm{t}}_{1/2}=\frac{\mathrm{l}\mathrm{n}?2*30}{\mathrm{l}\mathrm{n}?\frac{7}{5}}\end{array}$

${\mathrm{t}}_{1/2}=61.8\mathrm{m}\mathrm{i}\mathrm{n}\approx 62\mathrm{m}\mathrm{i}\mathrm{n}$

$2Mn{O}_4^{-}+6H^{+}+5H_2{O}_2\to 2M{n}^{2+}+8H_{2}O+5O_2$

$FeO+\underset{\left(Acidicflux\right)}{Si{O}_2}\to \underset{\left(Slag\right)}{FeSi{O}_3}$