Physics

  $\mathrm{}{\gamma }_{\text{mixture}}=\frac{n_1{C}_{P_1}+n_2{C}_{P_2}}{n_1{C}_{V_1}+n_2{C}_{V_2}}=\frac{n_1\frac{{\gamma }_{1}R}{{\gamma }_1-1}+n_2\frac{{\gamma }_{2}R}{{\gamma }_2-1}}{\frac{n_{1}R}{{\gamma }_1-1}+\frac{n_{2}R}{{\gamma }_2-1}}$ $\mathrm{}{}_{}\frac{{}_{}{}_{{}_{}}{}_{}{}_{{}_{}}}{{}_{}{}_{{}_{}}{}_{}{}_{{}_{}}}\frac{{}_{}\frac{{}_{}}{{}_{}}{}_{}\frac{{}_{}}{{}_{}}}{\frac{{}_{}}{{}_{}}\frac{{}_{}}{{}_{}}}$

on rearranging we get

$\frac{{\mathrm{n}}_1+{\mathrm{n}}_2}{{\gamma }_{\text{mix}}-1}=\frac{{\mathrm{n}}_1}{{\gamma }_1-1}+\frac{{\mathrm{n}}_2}{{\gamma }_2-1};\frac{5}{{\gamma }_{\text{mix}}-1}=\frac{3}{1/3}+\frac{2}{2/3}$

$\frac{5}{{\gamma }_{\text{mix}}-1}=9+3=12\mathrm{}\Rightarrow {\gamma }_{\text{mixture}}=\frac{17}{12}+1+\frac{5}{12};{\gamma }_{\text{mix}}=1.42$

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

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

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

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)

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

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

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

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,

$c=-1$ $a=0$

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

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

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,

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

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

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

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

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