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$\mathrm{m}=\mathrm{L},\mathrm{b}=\mathrm{R},\mathrm{k}=\frac{1}{\mathrm{c}}$ (speed of light in vacuum)

$\mathrm{}\frac{{\mathrm{E}}_0}{{\mathrm{B}}_0}=\mathrm{C}$

So, ${\mathrm{E}}_0={\mathrm{B}}_0\mathrm{C}=3\times {10}^{-8}\times 3\times {10}^8=9\mathrm{N}/\mathrm{C}$

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In damped oscillation

$\therefore \mathrm{}{v}_2=\frac{1}{8}\times \frac{1}{\mathrm{T}}=\frac{1}{8}\times \frac{{10}^{16}}{1.6}=7.8\times {10}^{14}$

$ma+bv+kx=0$

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)

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

$\mathrm{}{M}_{\text{ice}}{L}_f+m_{\text{ice}}(40-0)C_w=m_{\text{steam}}{L}_v+m_{\text{steam}}(100-40)C_w$

$\Rightarrow 200[80+40(1\left)\right]=\mathrm{M}[540+60(1\left)\right]$

$\Rightarrow 200\left(120\right)=\mathrm{M}\left(600\right)$

$\mathrm{M}=40\mathrm{g}\mathrm{m}$

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

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

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

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

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$\mathrm{}50.00$

For no toppling

$F\left(\frac{a}{2}+b\right)\le mg\frac{a}{2}$

$\mu \frac{\mathrm{a}}{2}+\mu \mathrm{b}\le \frac{\mathrm{a}}{2}$

$0.2a+0.4b\le 0.5a$

$0.4b\le 0.3a$

$\mathrm{b}\le \frac{3\mathrm{a}}{4}$

$\mathrm{b}\le 0.75\mathrm{a}\mathrm{}$ (in limiting case)

But is not possible as maximum value of  can be equal to  only.

$\therefore {\left(100\frac{\mathrm{b}}{\mathrm{a}}\right)}_{\mathrm{m}\mathrm{a}\mathrm{x}}=50.00$

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

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

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

$90.5\mathrm{J}$
$$

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

$\text{exception 25:}$  $v=\sqrt[]{u^2-\frac{G{M}_e}{R}}$ 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 $\frac{{}_{}{}_{}{}_{}{}_{}}{}$ $=\frac{P_1{V}_1-P_2{V}_2}{\gamma -1}=88.7\mathrm{J}$ .

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

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

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

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

$\mathrm{Y}$

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

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

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

on rearranging we get

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

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

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Net force on motor will be

$F_{\text{min}}=[920+68(10\left)\right]g+6000$

$=22000\mathrm{N}$

So, required power for motor

$\begin{array}{rr}{P}_{\text{min}}& ={\stackrel{?}{F}}_{\text{min}}\cdot \stackrel{?}{v}\\ & =22000\times 3\\ & =66000\text{watt}\end{array}$