Logic Gates question from JEE Main 2024 (5 Apr, Shift 1)

Answer: (b)

1. NOR with joined inputs is NOT: $\bar{A}$ and $\bar{D}$. NAND gives $\overline{BC}$.
2. Middle NOR: $\overline{\bar{A} + \overline{BC}} = A\cdot BC$ (one bulb terminal).
3. Other bulb terminal is at $\bar{D}$. Bulb glows only when the two terminals differ.
4. (a) $ABC = 1,\ \bar{D} = 1$: no. (b) $ABC = 0,\ \bar{D} = 1$: glows. (c), (d) $ABC = 0,\ \bar{D} = 0$: no.
5. So the answer is (b) $A = 1, B = 0, C = 0, D = 0$.