Semiconductor Electronics question for JEE Main (JEE Main 2024 (5 Apr, Shift 1)), with solution

Answer: (C) $A = 1,\ B = 0,\ C = 0,\ D = 0$

A NOR gate with its inputs tied works as a NOT gate. The outputs are:
- first gate: $\bar A$; NAND gate: $\overline{BC}$
- middle NOR gate: $X = \overline{\bar A + \overline{BC}} = A\cdot B\cdot C$
- lower gate: $Y = \bar D$

The bulb glows only when its two ends are at different levels, i.e. $X \ne Y$.

(A) $X = 0$, $Y = 0$: off. (B) $X = 0$, $Y = 0$: off. (C) $X = 0$, $Y = 1$: **on**. (D) $X = 1$, $Y = 1$: off.

Solution by Sreeraj P, M.Sc Physics