Physics

Given:   ${\mathrm{C}}_1+{\mathrm{C}}_2=10\mu \mathrm{F}$

$\begin{array}{rr}& 4\left(\frac{1}{2}{C}_1{V}^2\right)=\frac{1}{2}{C}_2{V}^2\\ & \Rightarrow \mathrm{}4C_1=C_2\end{array}$

From equation (i) and (ii)

$\begin{array}{rr}& C_1=2\mu F\\ & C_2=8\mu F\end{array}$ If they are in series

$C_{\text{eq}}=\frac{C_1{C}_2}{C_1+C_2}=1.6\mu F$

${90}^{?}$

Sol. $\mathrm{}|\stackrel{?}{\mathrm{P}}+\stackrel{?}{\mathrm{Q}}|=\left|\stackrel{?}{\mathrm{P}}\right|$

${\mathrm{P}}^2+{\mathrm{Q}}^2+2\mathrm{P}\mathrm{Q}\mathrm{c}\mathrm{o}\mathrm{s}?\theta ={\mathrm{P}}^2$

$\Rightarrow \mathrm{}Q+2P\mathrm{c}\mathrm{o}\mathrm{s}?\theta =0$

$\Rightarrow \mathrm{}\mathrm{c}\mathrm{o}\mathrm{s}?\theta =-\frac{\mathrm{Q}}{2\mathrm{P}}$

$\mathrm{t}\mathrm{a}\mathrm{n}?\alpha =\frac{2\mathrm{P}\mathrm{s}\mathrm{i}\mathrm{n}?\theta }{2\mathrm{P}\mathrm{c}\mathrm{o}\mathrm{s}?\theta +\mathrm{Q}}=\mathrm{\infty }$

$?\mathrm{}[2\mathrm{P}\mathrm{c}\mathrm{o}\mathrm{s}?\theta +\mathrm{Q}=0]$

$\alpha ={90}^{?}$

$\rho ={\rho }_0\left(1-\frac{r^2}{R^2}\right)\mathrm{}0<r\le R$

$\begin{array}{rr}& \mathrm{m}\mathrm{g}=\mathrm{B}\\ & \int \rho \left(4\pi r^{2}dr\right)={\rho }_{\mathrm{L}}\frac{4}{3}\pi {\mathrm{R}}^3;\mathrm{}{\int }_0^{\mathrm{R}}??{\rho }_0\left(1-\frac{{\mathrm{r}}^2}{{\mathrm{R}}^2}\right)4\pi r^{2}dr={\rho }_{\mathrm{L}}\frac{4}{3}\pi {\mathrm{R}}^3\\ & {\int }_0^{\mathrm{R}}??{\rho }_{0}4\pi \left({\mathrm{r}}^2-\frac{{\mathrm{r}}^4}{{\mathrm{R}}^2}\right)\mathrm{d}\mathrm{r}={\rho }_{0}4\pi {\left(\frac{{\mathrm{r}}^3}{3}-\frac{{\mathrm{r}}^5}{5{\mathrm{R}}^2}\right)}_0^{\mathrm{R}}={\rho }_{\mathrm{L}}\frac{4}{3}\pi {\mathrm{R}}^3;\mathrm{}\frac{2}{5}{\rho }_0={\rho }_{\mathrm{L}}\end{array}$

$\mathrm{}{r}_{\text{in}}=\left(\frac{{\mathcal{l}}_1-{\mathcal{l}}_2}{{\mathcal{l}}_2}\right)R_{\text{ext}}=\frac{60}{500}*10$

$r=\frac{6}{5}=1.2\mathrm{\Omega }$

$\mathrm{n}=12$

$m=\frac{f_o}{f_e}\mathrm{}5=\frac{f_o}{f_e}$

${\mathrm{f}}_{\mathrm{o}}=5{\mathrm{f}}_{\mathrm{e}}\mathrm{}{\mathrm{f}}_{\mathrm{o}}+{\mathrm{f}}_{\mathrm{e}}=60$

$6f_e=60f_e=10$

$\stackrel{?}{E}=\left(9\mathrm{s}\mathrm{i}\mathrm{n}?\left(1.6*{10}^{3}x+48*{10}^{10}\mathrm{t}\right)\stackrel{\mathrm{ˆ}}{\mathrm{k}}?/\mathrm{m}\right)$

$4000*\mathrm{V}+\mathrm{m}\mathrm{g}*\mathrm{V}=\mathrm{P}$

$\begin{array}{rr}& \frac{60*746}{2000*4000}=V\\ & V=1.86\mathrm{m}/\mathrm{s}\approx 1.9\mathrm{m}/\mathrm{s}\end{array}$

6

Sol. Common potential after connection.

${\mathrm{V}}_{\text{common}}=\frac{{\mathrm{C}}_1{\mathrm{V}}_1+{\mathrm{C}}_2{\mathrm{V}}_2}{{\mathrm{C}}_1+{\mathrm{C}}_2}=\frac{60*20+0}{120}=10\mathrm{V}\mathrm{o}\mathrm{l}\mathrm{t}$

Loss of energy $=\frac{1}{2}{\mathrm{C}\mathrm{V}}^2-\frac{1}{2}\left(2\mathrm{C}\right)*{\mathrm{V}}_{\text{Common}}^2$

$\begin{array}{rr}& =\frac{1}{2}*60*{10}^{-12}*(20{)}^2-60*{10}^{-12}*(10{)}^2\\ & =60*{10}^{-12}(200-100)\\ & =6000*{10}^{-12}\\ & =6\mathrm{n}\mathrm{J}\end{array}$

A logic GATE is reversible if we can recover input data fro the output eg. NOT gate.

$\mathrm{m}=\mathrm{L},\mathrm{b}=\mathrm{R},\mathrm{k}=\frac{1}{\mathrm{c}}$ (speed of light in vacuum)

$\mathrm{}\frac{{\mathrm{E}}_0}{{\mathrm{B}}_0}=\mathrm{C}$

So, ${\mathrm{E}}_0={\mathrm{B}}_0\mathrm{C}=3*{10}^{-8}*3*{10}^8=9\mathrm{N}/\mathrm{C}$