Physics

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

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

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

$90.5\mathrm{J}$
$$

$\frac{-{\mathrm{G}\mathrm{M}}_{\mathrm{e}}\mathrm{M}}{\mathrm{R}}+\frac{1}{2}{\mathrm{M}\mathrm{u}}^2=\frac{-{\mathrm{G}\mathrm{M}}_{\mathrm{e}}\mathrm{M}}{2\mathrm{R}}+\frac{1}{2}{\mathrm{M}\mathrm{v}}^2$

$\text{exception 25:}$  $v=\sqrt[]{u^2-\frac{G{M}_e}{R}}$ Transverse velocity of rocket

${}_{}$ $V_T\to$  Radial velocity of rocket

$V_R\to$

$V-\sqrt[]{\frac{G{M}_e}{2R}}$

Kinetic energy $\frac{M}{10}{V}_T=\frac{9M}{10}\sqrt[]{\frac{G{M}_e}{2R}}\mathrm{};\frac{M}{10}{V}_r=M\sqrt[]{u^2-\frac{G{M}_e}{R}}$ $\frac{}{}{}_{}\frac{}{}\text{exception 25:}\mathrm{}\frac{}{}{}_{}\text{exception 25:}$

$=\frac{1}{2}\frac{M}{10}\left(V_T^2+V_r^2\right)=\frac{M}{20}\left(81\frac{G{M}_e}{2R}+100u^2-100\frac{G{M}_e}{R}\right)$

$=\frac{\mathrm{M}}{20}\left(100{\mathrm{u}}^2-\frac{119\mathrm{G}\mathrm{M}}{2\mathrm{R}}\right)$

  $\mathrm{\Delta }\mathcal{l}=0.03\mathrm{m}\mathrm{m}$ $$

${\mathrm{P}}_1=1\mathrm{a}\mathrm{t}\mathrm{m},{\mathrm{T}}_1=273\mathrm{K}$

Now work done $\begin{array}{rr}& {}_{}{}_{}^{}{}_{}{}_{}^{}\\ \end{array}$ $\begin{array}{rr}& {\mathrm{P}}_1{\mathrm{V}}_1^{\gamma }={\mathrm{P}}_2{\mathrm{V}}_2^{\gamma }\\ & {\mathrm{P}}_2={\mathrm{P}}_1{\left[\frac{{\mathrm{V}}_1}{{\mathrm{V}}_2}\right]}^{\gamma }\\ & =1\mathrm{a}\mathrm{t}\mathrm{m}{\left(\frac{1}{3}\right)}^{1.4}\end{array}$

Closes answer is $\frac{{}_{}{}_{}{}_{}{}_{}}{}$ $=\frac{P_1{V}_1-P_2{V}_2}{\gamma -1}=88.7\mathrm{J}$ .

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

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

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

  $\mathrm{}v=\sqrt[]{\frac{T}{\mu }}$ $\mathrm{}\text{exception 25:}$

$\mathrm{T}=\mu {\mathrm{v}}^2\mathrm{}\frac{\mu {\mathrm{v}}^2}{\mathrm{A}}=\mathrm{Y}\frac{\mathrm{\Delta }\mathcal{l}}{\mathcal{l}}$

after substituting value of $\mathrm{\Delta }\mathcal{l}=\frac{\mu {\mathrm{v}}^2\mathcal{l}}{\mathrm{A}\mathrm{Y}}$ $\frac{{}^{}}{}$ and $\mu ,\mathrm{v},\mathcal{l},\mathrm{A}$ $$ we get

$\mathrm{Y}$

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

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

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