Physics

Torque is the measure of rotational force on an object (rigid or extended body). It changes the object's angular momentum. To calculate in physics, torque equals the rate of change of angular momentum. The formula for that is

$?=\frac{dL}{dt}$

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