Semiconductor Electronics question for JEE Main (JEE Main 2024 (8 Apr, Shift 1)), with solution

Answer: (B) $0$

OR gate output: $A + \bar B$. Lower AND gate output: $\bar A B$.
$$Y = (A + \bar B)\,\bar AB = A\bar AB + \bar A B\bar B = 0 + 0 = 0$$

Solution by Sreeraj P, M.Sc Physics