Semiconductor Electronics question for NEET (NEET 2020), with solution

Answer: (A) $\begin{array}{ccc} A & B & Y \\ 0 & 0 & 0 \\ 0 & 1 & 0 \\ 1 & 0 & 0 \\ 1 & 1 & 1 \end{array}$

Each input passes through a NOT gate, and the outputs go into a NOR gate:
$$Y = \overline{\bar{A} + \bar{B}} = A \cdot B$$
(De Morgan's theorem). The circuit is an AND gate.

AND truth table: $Y = 1$ only when $A = B = 1$. This is option (1).

Solution by Sreeraj P, M.Sc Physics