JEE Main 2026 — Semiconductors Question with Solution

The circuit consists of four NAND gates.

The first two NAND gates have their inputs tied together, acting as NOT gates. Their outputs are $\overline{X}$ and $\overline{Y}$.

These outputs are fed into the third NAND gate. The output of this gate is:
$y^{\prime }=\overline{\overline{X}\cdot \overline{Y}}$

Using De Morgan's theorem:
$y^{\prime }=\overline{\overline{X}}+\overline{\overline{Y}}=X+Y$

This output is fed into the fourth NAND gate, which also has its inputs tied together, acting as a NOT gate. The final output is:
Output $=\overline{X+Y}$

This represents the Boolean expression for a NOR gate.

Answer: NOR gate