Physics

In damped oscillation

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

$ma+bv+kx=0$

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)

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

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

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

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

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

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

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

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

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

$\mathrm{}50.00$

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

$90.5\mathrm{J}$
$$

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

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

$\mathrm{Y}$