Physics

Velocity of light in medium 2

 $=\frac{1}{\sqrt[]{{\mu }_m{\in }_m}}$

$v_2=\frac{1}{\sqrt[]{{\mu }_0{\mu }_r{\in }_0{\in }_r}}$

$=\frac{1}{\sqrt[]{1*{\mu }_0{\in }_0*4}}$

$v_2=\frac{1}{2\sqrt[]{{\mu }_0{\in }_0}}$

By snell's law for total internal Reflection

${\mu }_{2}sin{\theta }_C>{\mu }_{1}sin90°$

${\theta }_c>30°$

Given, l₁ = 1

  l₂ = 9l

case 1  $\theta =\frac{\pi }{2}$  

$l_P=l_1+l_2+2\sqrt[]{l_1{l}_2}cos\theta$  

= 10 l

case 2 q = p

$l_Q=l_1+l_2+2\sqrt[]{l_1{l}_2}cos\theta$  

$=10l-6l$  

              $l_P-I_Q=10l-4l=6l$  

$E=56.5sin\left(w\left(\raisebox{1ex}{$t-x$}\!\left/ \!\raisebox{-1ex}{$c$}\right.\right)\right)N/c$

$E_0=56.5$

${\epsilon }_0=8.85*10^{-12}$

C = 3 * 10⁸

l =  $\frac{1}{2}{\epsilon }_0{E}_0^{2}C$  

$=\frac{1}{2}*8.85*10^{-12}*{\left(56.5\right)}^2*3*10^8$  

$=42377.11*10^{-4}$

$l=4.24w{m}^{-2}$

Purely inductive circuit

              $\theta =\frac{\pi }{2}$  

              $cos\frac{\pi }{2}=0$  

Average power = 0

for $\pi R^{2}Area\to \text{'}l\text{'}$ current\

1 unit Area ® $\frac{l}{\pi R^2}$  

              $\pi r^2\to \frac{l}{\pi R^2}*\pi r^2$  

$i=\frac{l{r}^2}{R^2}$              

 

Now, consider Amperian loop of radius small 'r' ln Amperian loop magnetic field will be tangential to the amperian loop.

${\displaystyle ?\overrightarrow{B}.d\overrightarrow{i}={\mu }_0{l}_{enclosed}}$           (Ampere circuital law)

$B=\frac{{\mu }_0}{2\pi }\frac{l}{R^2}r$

$B\propto r$  

Based on theory

$V_{rms}=\sqrt[]{\frac{3RT}{M}}$

& $v_P=\sqrt[]{\frac{2RT}{M}}$

$v_{rms}=\sqrt[]{\raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}{v}_P$

Let, Rain drop moving with terminal velocity v_t in the air F_b (Bnoyancy force) = $\frac{4}{3}\pi r^3{\rho }_{air}g$  

$mg=\frac{4}{3}\pi r^3{\rho }_{air}g$

$F_v=\left(visconsforce\right)=6\pi \eta r{v}_T$

$F_b+F_v=mg$

$\Rightarrow \frac{4}{3}\pi r^3{\rho }_{air}g+6\pi \eta r{v}_T=\frac{4}{3}\pi r^3\rho g$

$v_T=\frac{2}{9}\frac{r^2}{\eta }\left(\rho -{\rho }_{air}\right)$

$v_T\propto r^2$

$g^1=\frac{GM}{{\left(R+h\right)}^2}$

$g^1=\frac{g}{{\left(1+\frac{h}{R}\right)}^2}$

Given h = D = 2R

$g^1=\frac{g}{{\left(1+\frac{2R}{R}\right)}^2}$

$g^1=\frac{g}{9}$