Class 12th

g (x) = (f (nf (x) – n)?
g' (x) = n (f (nf (x) – n)? ¹ . f' (nf (x) – n) . n . f' (x)
∴ g' (0) = 0
⇒ 4 = n (f (nf (0) – n)? ¹ . f' (nf (0) – n) . nf' (0)
⇒ 4 = n (f (0)? ¹ . f' (0) . nf' (0)
⇒ 4 = n . 1 . (-1) . n (-1)
n² = 4
⇒ n = 2

${\mu }_B=\frac{C}{v_B}$

$v_B=\frac{3*10^8}{1.47}=20.408*10^7$

 $v_A=2.6*10^7+20.408*10^7=23*10^7$

${\mu }_B=\frac{C}{v_B}\&{\mu }_A=\frac{C}{v_A}$

= 1.127

$\approx 1.13$

Optical fibre frequency range is 1 THz to 1000 THz.

Very small change in minority charge carriers produces high value of reverse bias current.

$v\propto \frac{z}{n}$  

              $v=k\frac{z}{n}$   [K ® constant]

              for 3^rd orbit of He⁺

              $v_{H{e}^{+}}=\frac{k*2}{3}=\frac{2k}{3}$               - (1)

              $v_H=\frac{k*1}{3}=\frac{k}{3}$                    - (2)

              $\frac{v_{H{e}^{+}}}{v_H}=2:1$  

Based on theory

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