Class 12th

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

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

$6f_e=60f_e=10$

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

In damped oscillation

$\therefore \mathrm{}{v}_2=\frac{1}{8}*\frac{1}{\mathrm{T}}=\frac{1}{8}*\frac{{10}^{16}}{1.6}=7.8*{10}^{14}$

$ma+bv+kx=0$

In the circuit

$m\frac{d^{2}x}{{d{t}^2}^2}+b\frac{dx}{dt}+kx=0$

$-\mathrm{i}\mathrm{R}-\mathrm{L}\frac{\mathrm{d}\mathrm{i}}{\mathrm{d}\mathrm{t}}-\frac{\mathrm{q}}{\mathrm{C}}=0$

Comparing equation (i) and (ii)

$L\frac{d^{2}q}{d{t}^2}+R\frac{dq}{dt}+\frac{1}{c}q=0$

$=5\mathrm{M}\left({\mathrm{u}}^2-\frac{119{\mathrm{G}\mathrm{M}}_{\mathrm{e}}}{200\mathrm{R}}\right)$

$T=\frac{2\pi }{\omega }=2\pi \frac{r}{v}\propto \frac{n^2}{\frac{1}{n}}\propto n^3$

$\therefore \mathrm{}v=\frac{1}{\mathrm{T}}\propto \frac{1}{{\mathrm{n}}^3}$

$\frac{v_2}{v_1}=\frac{1^3}{2^3}=\frac{1}{8}$

Potential gradient $=\frac{5}{1000}=\frac{V_P}{1200}$

$V_P=6\mathrm{V}$ $\frac{}{}\frac{{}_{}}{}$

and  ${\mathrm{R}}_{\mathrm{P}}=\frac{{\mathrm{V}}_{\mathrm{P}}}{\mathrm{l}}=\frac{6}{60*{10}^{-3}}=100\mathrm{\Omega }$ ${}_{}\frac{{}_{}}{}\frac{}{{}^{}}$

$\mathrm{}N\propto \frac{1}{{\mathrm{s}\mathrm{i}\mathrm{n}}^4?\left(\frac{\theta }{2}\right)}$

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

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