JEE Main 2026 — Semiconductors Question with Solution

The given logic circuit consists of an OR gate and a NAND gate.
The output of the OR gate is $y=n+m$.
The inputs to the NAND gate are $n$ and $y$.
The final output $z$ is given by $z=\overline{n\cdot y}=\overline{n\cdot (n+m)}$.
Using Boolean algebra: $n\cdot (n+m)=n\cdot n+n\cdot m=n+n\cdot m=n(1+m)=n$.
Therefore, $z=\overline{n}$.
Let's verify this with the truth table:

n	m	y = n + m	z = NAND(n, y)
0	0	0	1
0	1	1	1
1	1	1	0
1	0	1	0

Comparing this with the given options, option (4) matches the calculated truth table.