Semiconductor Electronics question for JEE Main (JEE Main 2024 (4 Apr, Shift 2)), with solution

Answer: (D) OR gate

The NAND gate receives $\bar A$ and $\bar B$:
$$Y = \overline{\bar A\cdot\bar B} = \bar{\bar A} + \bar{\bar B} = A + B$$
(De Morgan's theorem). This is an OR gate.

Solution by Sreeraj P, M.Sc Physics