Class 12th

Kindly go through the solution

Concept based.

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

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

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

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

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

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

$\mathrm{n}=12$

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

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

$6f_e=60f_e=10$

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)

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