Q 12-14-083JEE MainJEE Main 2023 (30 Jan, Shift 2)Medium
The output $Y$ for the inputs $A$ and $B$ of circuit is given by the truth table
Answer: (D) $\begin{array}{ccc} A & B & Y \\ \hline 0 & 0 & 0 \\ 0 & 1 & 1 \\ 1 & 0 & 1 \\ 1 & 1 & 0 \end{array}$
Let $C=\overline{AB}$. The two middle gates give $\overline{AC}$ and $\overline{BC}$, and the last gate gives
$$Y=\overline{\overline{AC}\cdot\overline{BC}}=AC+BC=(A+B)\,\overline{AB}=A\oplus B$$
This is the four-NAND XOR gate: $Y=0,1,1,0$.
Solution by Sreeraj P, M.Sc Physics