Semiconductors

n.p.m. transistor current gain B = 100

As, B = $\frac{I_C}{I_B}Voltagegain,\frac{V_{out}}{V_{in}}=\left(\frac{I_C}{I_B}\right)\frac{R_2}{R_i}$

$V_{out}=\frac{B{R}_2}{R_{in}}Vin$

$\frac{10^3}{10^3}V=1v$

Using factual data.

Diode conducts current only in one direction.

From Truth table $Y=\overline{AB}$ .

$\mathcal{l}=2\pi r$

$\Rightarrow 314cm=2*3.14r$

$\Rightarrow r=\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.m=0.5m$

$\mu =iA$

$=14*\pi r^2$

$=14^2*\frac{22}{7}*\frac{1}{4}=11$

Kindly go through the solution

As per the NCERT Textbooks

“An integrated Chip consists of many passive and active components fabricated on a single chip of silicon. these ICs are compact, low-cost, and highly reliable. They consume less power and have high speed.”

These ICs are used in majority of our daily use electronics and all advance electronics. 

 A p-type semiconductor is electrically neutral despite having more holes, because the number of positively charged holes is exactly balanced/equal by the number of negatively charged acceptor ions introduced during doping. so practically untill any volatage is applied the semiconductor remains chargeless in other words doesn't produce any current even after doping.

 As per the NCERT Textbooks information"Although the number of holes is more than the number of electrons in a p-type semiconductor, the material as a whole is electrically neutral because the charge of holes is balanced by the negatively charged acceptor ions.”