Class 12th

20

Sol. Angular momentum conservation:

$\begin{array}{rr}& \Rightarrow \mathrm{}{\mathrm{I}}_1{\omega }_1+{\mathrm{I}}_2{\omega }_2=\left({\mathrm{I}}_1+{\mathrm{I}}_2\right){\omega }_{\mathrm{f}}\\ & \Rightarrow \mathrm{}\frac{{\mathrm{M}\mathrm{R}}^2}{2}{\omega }_{\mathrm{o}}=\left(\frac{{\mathrm{M}\mathrm{R}}^2}{2}+\frac{{\mathrm{M}\mathrm{R}}^2}{8}\right){\omega }_{\mathrm{f}}\\ & \Rightarrow \mathrm{}{\omega }_{\mathrm{f}}=\frac{4}{5}{\omega }_{\mathrm{o}}\\ & \Rightarrow \mathrm{}{\mathrm{K}\mathrm{E}}_{\text{final}}=\frac{1}{2}\left({\mathrm{I}}_1+{\mathrm{I}}_2\right){\omega }_{\mathrm{f}}^2=\frac{{\mathrm{M}\mathrm{R}}^2{\omega }^2}{5}\\ & \Rightarrow \mathrm{}\mathrm{K}{E}_{\text{initial}}=\frac{1}{2}\mathrm{I},{\omega }_0^2=\frac{{\mathrm{M}\mathrm{R}}^2{\omega }^2}{4}\\ & \Rightarrow \mathrm{}\mathrm{\%}\text{loss}\Rightarrow 20\mathrm{\%}.\end{array}$

10553.33

Sol. $\frac{1}{{\lambda }_{\text{min}}}=R\left[\frac{1}{1}-\frac{1}{\mathrm{\infty }}\right]=R\mathrm{}[n=\mathrm{\infty }\to n=1]$

$\frac{1}{{\lambda }_{\mathrm{m}\mathrm{a}\mathrm{x}}}=\mathrm{R}\left[\frac{1}{1}-\frac{1}{4}\right]=\frac{3\mathrm{R}}{4}\mathrm{}[\mathrm{n}=2\to \mathrm{n}=1]$

$\Rightarrow \mathrm{\Delta }\lambda \Rightarrow \frac{4}{3R}-\frac{1}{R}\mathrm{}\Rightarrow \mathrm{}\frac{1}{3R}=340$

For Paschan $\Rightarrow \frac{1}{{\lambda }_{\text{min}}}=R\left[\frac{1}{9}\right]\mathrm{}[\mathrm{n}=\mathrm{\infty }\to \mathrm{h}=3]$

$\Rightarrow \frac{1}{{\lambda }_{\mathrm{m}\mathrm{a}\mathrm{x}}}=\mathrm{R}\left[\frac{1}{9}-\frac{1}{16}\right]=\frac{7\mathrm{R}}{144}$

$\Rightarrow \mathrm{\Delta }\lambda =\frac{81}{7\mathrm{R}}$

E = 440 sin 100 t $\omega =100\pi$

$L=\frac{\sqrt[]{2}}{\pi }H.$ $X_L=\omega L=100\pi .\frac{\sqrt[]{2}}{\pi }=100\sqrt[]{2}\Omega .$

$=\frac{220}{100}=2.2A$

R = R1 + R2

$\Rightarrow \frac{2\mathcal{l}}{\sigma A}=\frac{\mathcal{l}}{{\sigma }_{1}A}+\frac{\mathcal{l}}{{\sigma }_{2}A}\Rightarrow \frac{2}{\sigma }=\frac{1}{{\sigma }_1}+\frac{1}{{\sigma }_2}\Rightarrow \sigma =\frac{2{\sigma }_1{\sigma }_2}{{\sigma }_1+{\sigma }_2}$

Sol.

$\begin{array}{ll}\mathrm{A}\mathrm{H}\to \frac{\mathrm{H}}{2}& \mathrm{H}=\frac{\sqrt[]{3}\mathrm{a}}{2}\\ \mathrm{A}\mathrm{G}\to \frac{\mathrm{H}}{\sqrt[]{3}}& \mathrm{A}\mathrm{l}=\frac{\mathrm{H}}{2\sqrt[]{3}}\\ {\mathrm{M}}_{\mathrm{A}\mathrm{B}\mathrm{C}}=\mathrm{M}& {\mathrm{M}}_{\mathrm{A}\mathrm{D}\mathrm{E}}=\frac{\mathrm{M}}{4}\end{array}$

$\Rightarrow {\mathrm{I}}_{\mathrm{G}}=\frac{M{a}^2}{12}-\left[\frac{\mathrm{M}}{4}\frac{{\left[\frac{\mathrm{a}}{2}\right]}^2}{12}+\frac{\mathrm{M}}{4}{\left[\frac{\mathrm{a}}{2\sqrt[]{3}}\right]}^2\right]=\left(\frac{11}{16}\right)\frac{{\mathrm{M}\mathrm{a}}^2}{12}$

$S=2+\frac{6}{7}+\frac{12}{7^2}+\frac{20}{7^3}+\frac{30}{7^4}+........(i)$  

$\frac{1}{7}S=\frac{2}{7}+\frac{6}{7^2}+\frac{12}{7^3}+\frac{20}{7^4}+........(ii)$               

(i) - (ii)

$\frac{6}{7}S=2+\frac{4}{7}=\frac{6}{7^2}+\frac{8}{7^3}+...........(iii)$               

$\frac{6}{7^2}S=\frac{2}{7}+\frac{4}{7^2}+\frac{6}{7^3}+........(iv)$              

(iii) – (iv)

$=2+\left[\frac{1}{1-\frac{1}{7}}\right]=2\left(\frac{7}{6}\right)$

$\therefore 4S=8{\left(\frac{7}{6}\right)}^3={\left(\frac{7}{3}\right)}^3$      

Lindlar's catalyst reduces alkynes to cis-alkene.

Lindlar's catalyst reduces alkynes to cis-alkene.