Class 12th

$F_R=\left\{\frac{kQ{q}_0}{{\left(a-x\right)}^2}-\frac{kQ{q}_0}{{\left(a+x\right)}^2}\right\}$

$=-kQ{q}_0\left\{\frac{a^2+x^2+2ax-a^2-x^2+2ax}{\left(a^2-x^2\right)}\right\}$

$F_R=-\frac{kQ{q}_0\left(4ax\right)}{{\left(a^2-x^2\right)}^2}$

$T=2\pi \sqrt[]{\frac{a^{3}m}{4kQ{q}_0}=\sqrt[]{\frac{4{\pi }^2{a}^{3}m*4\pi {\epsilon }_0}{4Q{q}_0}}}=\sqrt[]{\frac{4{\pi }^3{\epsilon }_{0}m{a}^3}{Q{q}_0}}$ .                          

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 θ = π

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

$=10l-6l$

$l_P-I_Q=10l-4l=6l$        

$E=56.5sin\left(w\left(\raisebox{1ex}{$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 $\frac{l}{\pi R^2}$ $\frac{}{{}^{}}$

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

$i=\frac{l{r}^2}{R^2}$

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)

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

$B\propto r$

(Independent of distance)

$E_1=E_2=\frac{\sigma }{2{\in }_0}$

Voltage gain = $\frac{I_C{R}_0}{I_B{R}_i}$  

R₀ ->Output Resistance

R_i ->Input Resistance

I_C ->Collector current

I_{B ->}Base current

Voltage gain = $\frac{I_C}{I_B}\frac{R_0}{R_i}=\frac{\left(5*10^{-3}\right)}{\left(100*10^{-6}\right)}*\frac{60}{200}=15$  

Phenolphthalein is the indicator of weak acid which produces pink colour in basic medium.

Consider the image below

There is no mirror image of each other, having some configuration.