With increasing biasing voltage of a photodiode, the photocurrent magnitude:

Here  $\Delta V_c=0.5V,\Delta i_c=0.05mA=0.05\times 10^{-3}A$

Output resistance is given by

$R_{out}=\frac{\Delta V_c}{\Delta i_c}=\frac{0.5}{0.5\times 10^{-3}}=10^4\Omega =10k\Omega .$

$i_b=\frac{5-0.7}{8.6}=0.5mA$

$\Rightarrow I_c=\beta I_b=100\times 0.5 mA$

By using 

$V_{CE}=V_{CC}-I_C{R}_L=18-50\times 10^{-3}\times 100=13 V$

The charge on hole is positive.

Power gain = (i_c² R_c) / (i_b² R_B) = (i_c/i_b)² (R_c/R_B) = (10²)² (10? /10³) = 10?
i_c/i_b = 100 ⇒ β = i_c/i_b = 100

First part of figure shown is or GATE and Second part of figure shown is NOT GATE So, $Y_P=OR+NOT=$  NOR GATE $Y=\stackrel{?}{A+B}=\bar{A}\cdot \bar{B}$