Semiconductor Electronics question for NEET (NEET 2023), with solution

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

$A$ and $B$ each pass through a NOT gate, and the results go into a NAND gate:
$$Y = \overline{\bar{A} \cdot \bar{B}} = A + B$$
(De Morgan's theorem). The circuit is an OR gate.

OR truth table: $Y = 0$ only when $A = B = 0$, otherwise $Y = 1$. This is option (2).

Solution by Sreeraj P, M.Sc Physics