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${\mathrm{L}}_{\mathrm{o}\mathrm{i}}={\mathrm{L}}_{\mathrm{o}\mathrm{f}}$

$\begin{array}{rr}& \frac{\mathrm{m}\mathrm{v}}{\sqrt[]{2}}\times \frac{\mathcal{l}}{2}=\left. [\frac{4\mathrm{m}{\mathcal{l}}^2}{12}+\frac{\mathrm{m}{\mathcal{l}}^2}{4}\right]\times \omega \\ & \omega =\frac{6\mathrm{v}}{7\sqrt[]{2}\mathcal{l}}=\frac{3\sqrt[]{2}\mathrm{v}}{7\mathcal{l}}\end{array}$

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

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Relaxation time $\left(\tau \right)\propto \frac{\mathrm{V}}{\sqrt[]{\mathrm{T}}}$

and $\mathrm{}\mathrm{T}\propto \frac{1}{{\mathrm{V}}^{\gamma -1}}$

$\begin{array}{rr}& \tau \propto {\mathrm{V}}^{1+\frac{\gamma -1}{2}}\mathrm{}\tau \propto {\mathrm{V}}^{\frac{1+\gamma }{2}}\\ & \frac{{\tau }_f}{{\tau }_{\mathrm{i}}}={\left(\frac{2\mathrm{V}}{\mathrm{V}}\right)}^{\frac{1+\gamma }{2}}\mathrm{}\frac{{\tau }_{\mathrm{f}}}{{\tau }_{\mathrm{i}}}=(2{)}^{\frac{1+\gamma }{2}}\end{array}$

$\mathrm{m}\mathrm{g}\mathrm{h}=\frac{1}{2}m{v}^2+\frac{1}{2}\mathrm{l}{\omega }^2$

$\mathrm{v}=\omega \mathrm{R}$ (no slipping)

$\mathrm{m}\mathrm{g}\mathrm{h}=\frac{1}{2}\mathrm{m}{\omega }^2{\mathrm{R}}^2+\frac{1}{2}\frac{\mathrm{m}\mathrm{R}}{2}{\omega }^2$

$\mathrm{m}\mathrm{g}\mathrm{h}=\frac{3}{4}\mathrm{m}{\omega }^2{\mathrm{R}}^2$

$\omega =\sqrt[]{\frac{4gh}{3R^2}}=\frac{1}{R}\sqrt[]{\frac{4gh}{3}}$

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,

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

Bonus $\mathrm{}V=K\left(h{)}^a\right(I{)}^b\left(G{)}^c\right(C{)}^d\mathrm{}(V$

is voltage   $)$ $$

We know,   $\begin{array}{rr}& \left[\mathrm{h}\right]={\mathrm{M}\mathrm{L}}^2{\mathrm{T}}^{-1}\\ & \left[\mathrm{l}\right]=\mathrm{A}\\ & \left[\mathrm{G}\right]={\mathrm{M}}^{-1}{\mathrm{L}}^3{\mathrm{T}}^{-2}\\ & \left[\mathrm{C}\right]=\mathrm{L}{\mathrm{T}}^{-1}\\ & \left[\mathrm{V}\right]=\mathrm{M}{\mathrm{L}}^2{\mathrm{T}}^{-3}{\mathrm{A}}^{-1}\end{array}$

$M{L}^2{T}^{-3}{A}^{-1}={\left(M{L}^2{T}^{-1}\right)}^a(A{)}^b{\left(M^{-1}{L}^3{T}^{-2}\right)}^c{\left(L{T}^{-1}\right)}^d$

$M{L}^2{T}^{-3}{A}^{-1}=M^{a-c}{L}^{2a+3c+d}{T}^{-a-2c-d}{A}^b$

$\mathrm{a}-\mathrm{c}=1$

$2\mathrm{a}+3\mathrm{c}+\mathrm{d}=2$

$-a-2c-d=-3$

$b=-1$

On solving,

$c=-1$ $a=0$

$\mathrm{d}=5,\mathrm{b}=-1$

$V=K\left(h{)}^{?}\right(I{)}^{-1}\left(G{)}^{-1}\right(C{)}^5$

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Take $1\mathrm{k}\mathrm{g}$ $$ mass at origin

${\mathrm{X}}_{\mathrm{c}\mathrm{m}}=\frac{1\times 0+1.5\times 3+2.5\times 0}{5}=0.9\mathrm{c}\mathrm{m}$

$Y_{\mathrm{c}\mathrm{m}}=\frac{1\times 0+1.5\times 0+2.5\times 4}{5}=2\mathrm{c}\mathrm{m}$

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

$\frac{M{\mathcal{l}}^2}{12}+\mathrm{M}\left(\frac{\mathcal{l}}{4}\right)={\mathrm{M}\mathrm{K}}^2$

$\frac{{\mathcal{l}}^2}{12}+\frac{{\mathcal{l}}^2}{16}={\mathrm{K}}^2$

$\mathrm{K}=\sqrt[]{\frac{7}{48}}\mathcal{l}$

$\frac{\mathrm{T}}{\sqrt[]{2}}=100\mathrm{}\mathrm{}\frac{\mathrm{T}}{\sqrt[]{2}}=\mathrm{F}$

$F=100N$

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

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

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

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

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

Using equation of continuity

$\begin{array}{rr}& A_1{V}_1=A_2{V}_2\\ & \frac{V_1}{V_2}=\frac{A_2}{A_1}={\left(\frac{d_2}{d_1}\right)}^2={\left(\frac{4.8}{6.4}\right)}^2=\frac{9}{16}\end{array}$

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

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

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