Physics

$\mathrm{J}=\sigma \mathrm{E}\Rightarrow \frac{\mathrm{E}}{\rho }=\frac{\mathrm{E}\mathcal{l}}{\mathrm{R}\mathrm{A}}=\frac{10*10*\pi }{10*{10}^{-4}}$

$\Rightarrow {10}^5\mathrm{A}/{\mathrm{m}}^2$

$\mathrm{R}={\mathrm{R}}_0(1+\alpha \mathrm{\Delta }\mathrm{T})$

$\begin{array}{rr}& \Rightarrow 6.8=2[1+\alpha *(80-0\left)\right]\\ & \Rightarrow \alpha =\frac{3.4-1}{80}\\ & =0.03\\ & =3*{10}^{-2}{}^{?}{\mathrm{C}}^{-1}\end{array}$

Because angle of incidence is Brewster's angle so that angle between reflected and refracted ray is ${90}^{?}$

$\mathrm{t}\mathrm{a}\mathrm{n}?i_p=\mu =\sqrt[]{3}$

${\mathrm{i}}_{\mathrm{p}}={60}^{?}=\mathrm{i}$

${\mathrm{V}}_{\mathrm{P}}={\mathrm{V}}_{\mathrm{q}}+{\mathrm{V}}_{-\mathrm{q}}$

$\begin{array}{rr}& =\left(\frac{\mathrm{K}\mathrm{q}}{5-3}+\frac{\mathrm{K}(-\mathrm{q})}{5+3}\right)*{10}^2\\ & =\left(\frac{\mathrm{K}\mathrm{q}}{2}-\frac{\mathrm{K}\mathrm{q}}{8}\right)*{10}^2\\ & =\left(\frac{3\mathrm{K}\mathrm{q}}{8}\right)*{10}^2\end{array}$

First excited state $\Rightarrow n=2$ $$

${\mathrm{T}}_1=13.6\frac{{\mathrm{z}}^2}{{\mathrm{n}}^2}=-\frac{13.6}{4}\mathrm{e}\mathrm{V}$

Second excited state $\Rightarrow \mathrm{n}=3$ $$

${\mathrm{T}}_2=13.6\frac{{\mathrm{z}}^2}{{\mathrm{n}}^2}=-\frac{13.6}{9}\mathrm{e}\mathrm{V}$

${\mathrm{T}}_1:{\mathrm{T}}_2=\frac{1}{4}:\frac{1}{9}=9:4$

From $x-t$ graph,

$\begin{array}{rr}& A=1,T=8\\ & \Rightarrow \omega =\frac{2\pi }{T}\\ & \Rightarrow \omega =\frac{\pi }{4}\\ & \text{at}t=2,x=1\\ & a=-{\omega }^{2}x\\ & \Rightarrow \mathrm{a}=\frac{-{\pi }^2}{16}*1\\ & \Rightarrow \mathrm{a}=\frac{-{\pi }^2}{16}\mathrm{m}/{\mathrm{s}}^2\end{array}$

Time period of satellite

$\mathrm{T}=2\pi \sqrt[]{\frac{{\mathrm{R}}^3}{\mathrm{G}\mathrm{M}}}$

$=2\pi \sqrt[]{\frac{{\mathrm{R}}^3}{\mathrm{G}\mathrm{d}\frac{4}{3}\pi {\mathrm{R}}^3}}$

$\Rightarrow \mathrm{T}=\sqrt[]{\frac{3\pi }{\mathrm{G}\mathrm{d}}}$ $\Rightarrow \frac{3\pi }{\mathrm{G}\mathrm{d}}={\mathrm{T}}^2$

$\frac{1}{2}\mathrm{m}{\left(\frac{\mathrm{u}}{3}\right)}^2-\frac{1}{2}{\mathrm{m}\mathrm{u}}^2=-{\mathrm{F}}_{\mathrm{R}}*24$

$0-\frac{1}{2}{\mathrm{m}\mathrm{u}}^2=-{\mathrm{F}}_{\mathrm{R}}*\mathrm{d}$

$\frac{\frac{1}{2}{\mathrm{m}\mathrm{u}}^2}{\frac{1}{2}m{u}^2*\frac{8}{9}}=\frac{\mathrm{d}}{24}$

$\mathrm{d}=24*\frac{9}{8}=27\mathrm{c}\mathrm{m}$

$\begin{array}{rr}& {}_{11}^{22}\mathrm{N}\mathrm{a}\to \mathrm{X}+{\mathrm{e}}^{+}+\mathrm{v}\\ & {}_{11}^{22}\mathrm{N}\mathrm{a}\to {}_{10}^{22}\mathrm{N}\mathrm{e}+{\mathrm{e}}^{+}+\mathrm{v}\end{array}$