Class 12th

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

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

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

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

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

Case-I:

If final image is at least distance of clear vision

M.P. $=\frac{\mathrm{L}}{{\mathrm{f}}_0}\left(1+\frac{\mathrm{D}}{{\mathrm{f}}_{\mathrm{e}}}\right);375=\frac{150}{5}\left[1+\frac{25}{{\mathrm{f}}_{\mathrm{e}}}\right]$ $\frac{}{{}_{}}\left(\frac{}{{}_{}}\right)\frac{}{}\left[\frac{}{{}_{}}\right]$

$\frac{375}{30}=1+\frac{25}{{\mathrm{f}}_{\mathrm{e}}}\mathrm{};\mathrm{}\frac{345}{30}=\frac{25}{{\mathrm{f}}_{\mathrm{e}}}$

${\mathrm{f}}_{\mathrm{e}}=\frac{750}{345}=2.17\mathrm{c}\mathrm{m};{\mathrm{f}}_{\mathrm{e}}\approx 22\mathrm{m}\mathrm{m}$

Case-II:

In final image is at infinity

M.P. $=\frac{\mathrm{L}}{{\mathrm{f}}_0}\left(\frac{\mathrm{D}}{{\mathrm{f}}_{\mathrm{e}}}\right)=375$ ${\mathrm{f}}_{\mathrm{e}}=22\mathrm{m}\mathrm{m}$ $\frac{}{{}_{}}\left(\frac{}{{}_{}}\right)$

As magnetic field lines always form a closed loop, hence every magnetic field line creating magnetic flux in the inner region must be passing through the outer region. Since flux in two regions are in opposite direction,

$\therefore \mathrm{}{?}_{\mathrm{i}}=-{?}_0$

Electric field at $\mathrm{t}=0$ & $(\mathrm{x},\mathrm{y},\mathrm{z})=\left(\mathrm{0,0},\frac{\pi }{\mathrm{k}}\right)$  is

$\begin{array}{rr}& \stackrel{?}{E}=-\left(\frac{\hat{i}+\hat{j}}{\sqrt[]{2}}\right)E_0\text{and}\stackrel{?}{F}=\stackrel{?}{E}q\\ & \therefore \mathrm{}\stackrel{?}{F}=-\left(\frac{\hat{i}+\hat{j}}{\sqrt[]{2}}\right)q\end{array}$

Which is antiparallel to $\left(\frac{\stackrel{\mathrm{ˆ}}{\mathrm{i}}+\stackrel{\mathrm{ˆ}}{\mathrm{j}}}{\sqrt[]{2}}\right)$

 $\mathrm{}\frac{}{}$ $\mathrm{}\mathrm{I}=\frac{1}{2.5}=0.4\mathrm{A}$

$\mathrm{I}=\frac{\mathrm{I}}{2}=0.2\mathrm{A}$

$\mathrm{}\epsilon =\left|-\frac{\mathrm{d}\phi }{\mathrm{d}\mathrm{t}}\right|=\left|-\frac{\mathrm{A}\mathrm{d}\mathrm{B}}{\mathrm{d}\mathrm{t}}\right|$

$=(16*4-4*2)\frac{(1000-500)}{5}*{10}^{-4}*{10}^{-4}$

$=56*\frac{500}{5}*{10}^{-8}=56*{10}^{-6}\mathrm{V}$

Capacitance of element

Capacitance of element, $C^{\mathrm{\text{'}}}=\frac{K(1+\alpha x){\epsilon }_{0}A}{dx}$ ${}^{}\frac{{}_{}}{}$

$\sum \frac{1}{C^{\mathrm{\text{'}}}}={\int }_0^d?\frac{dx}{K{\epsilon }_0\mathrm{A}(1+\alpha \mathrm{x})}$

$\frac{1}{\mathrm{C}}=\frac{1}{\mathrm{K}{\epsilon }_0\mathrm{A}\alpha }\mathrm{l}\mathrm{n}?(1+\alpha \mathrm{d})$

Given: $\alpha \mathrm{d}?1$ $$

$\frac{1}{C}=\frac{1}{K{\epsilon }_{0}A\alpha }\left(\alpha d-\frac{{\alpha }^2{d}^2}{2}\right);\mathrm{}\frac{1}{C}=\frac{d}{K{\epsilon }_{0}A}\left(1-\frac{\alpha d}{2}\right)$

$C=\frac{K{\epsilon }_{0}A}{d}\left(1+\frac{\alpha d}{2}\right)$