Physics

Structure of Bithinol is;

C = 500   $\mu F,$ V = 100 v, L = 50 mH                                                        

In this LC – oscillation

q = q₀ cos   $\omega t$

$i=\frac{-dq}{dt}=q_0\omega sin\omega t\omega =\frac{1}{\sqrt[]{2c}}=\frac{1}{\sqrt[]{50*10^{-3}*5*10^{-4}}}$

=  $\frac{1000}{5}=200$  

So,     i_max =   $q_0\omega =500*10^6*100*200$

= 10A

  ${\lambda }_A=25\lambda ,$ ${\lambda }_B=16\lambda$      

At t = 0 Þ N_A = N_B = N₀

after t =  $\frac{1}{a\lambda }:-\frac{N_B}{N_A}=\frac{N_0{e}^{-16\lambda t}}{N_0{e}^{-25\lambda t}}=e^{\left(25\lambda -16\lambda \right)t}$  

->e =  $e^{\left(9\lambda \frac{1}{a\lambda }\right)}$  

$\Rightarrow \frac{9}{a}=1\Rightarrow a=9$      

$i_{\left(zener-max\right)}=25mA$                                                                

 20 – i_max R – 8 = 0

i_max R = 12At minimum zener current $\left(\mu A\right):-$  

$20-i_{min}R-i_{min}{R}_L=0$  

  $\frac{R}{R_L}=\frac{12}{8}=\frac{3}{2}$

${\mathcal{l}}_{min}R=12$

  $i_{min}{R}_L=8$

At max^m zener current –

$20-i_{max}R-8=0$  

$i_L=O\left\{as{i}_{z}ma{x}^m=25mA\right\}$             

i_maxR = 12v

25 * 10^-3 R = 12

$R=\frac{12*10^3}{25}=12*40=480\Omega$

From conservation of Energy

$\frac{1}{2}m{v}_0^2=mgh$

$v_0=10\sqrt[]{2}$

For A to B

$a=-gsin45°=\frac{-10}{\sqrt[]{2}}$

T = t₁ + t₂

$\begin{array}{l}=2\sqrt[]{2}+2\\ =2\left(\sqrt[]{2}+1\right)\end{array}$

g_eff = g –   $\left(\frac{{\rho }_w}{{\rho }_b}\right)g$

$T=2\pi \sqrt[]{\frac{\mathcal{l}}{g}}$

$T\text{'}=2\pi \sqrt[]{\frac{\mathcal{l}}{g_{eff}}}$

$\Rightarrow T\text{'}=\sqrt[]{\frac{5}{4}}T$

=    $\sqrt[]{\frac{5}{4}}*10$

=    $5\sqrt[]{5}sec$

So, x = 5

.From conservation of energy

mgh = $\frac{1}{2}m{v}^2+\frac{1}{2}l{\omega }^2$ $\frac{}{}{}^{}\frac{}{}{}^{}$

$\Rightarrow mgh=\frac{1}{2}m{v}^2+\frac{1}{2}\frac{m{R}^2}{2}$

10 h = $\frac{16}{2}+\frac{16}{4}$ $\frac{}{}\frac{}{}$

h = 1.2 m = 120 cm

$P_2-P_{06}=\frac{4T}{6}\&P_1-P_2=\frac{4T}{3}$

$P_1-P_0=\frac{4T}{2}=\frac{4T}{r}$

r = 2cm

m = 10g                                                                       

$\mathcal{l}=50cm$

A = 2mm²

Y = 1.2 * 10¹¹N/m²

$\Delta x=x*10^{-5}m$  

As, $\frac{T}{A}=Y\frac{\Delta x}{\mathcal{l}}$  

$\Delta x=\frac{T\mathcal{l}}{AY}$

$T\mathcal{l}=V^{2}m$

=  $\frac{V^{2}m}{AY}$  

$=\frac{3600*10*10^{-3}}{2*10^{-6}*1.2*10^{11}}=\frac{1800*10^{-3}*10^6}{1.2}$

$=\frac{18}{12}*10^{-4}=\frac{3}{2}*10^{-4}=15*10^{-5}m$

So, x = 15

$\frac{nv}{0.6}=400\&\frac{\left(n+1\right)v}{0.6}=450$

$\Rightarrow \left[\frac{0.6*450}{v}+1\right]$ $\frac{v}{0.6}=450$

v = 30

$\Rightarrow \sqrt[]{\frac{T}{\mu }}=30$

^{$\Rightarrow \frac{2700}{\mu }=900$}

^{$\mu =3$}