JEE Main · 2023 · 11 Apr 2023 · Shift 1 · Semiconductors

Each of the first two gates has its single input taken to both terminals, so each one behaves as an inverter and produces $\bar{a}$ and $\bar{b}$. These feed a NOR gate, whose output is $Y=\overline{\bar{a}+\bar{b}}$. By De Morgan's theorem $\overline{\bar{a}+\bar{b}}=a\cdot b$. Running through the combinations, $Y=1$ only when $a=1$ and $b=1$ and is $0$ otherwise. The circuit therefore performs the AND operation.