JEE Main · 2026 · 23 Jan 2026 · Shift 2 · Semiconductors

For the given logic gate circuit, which of the following is the correct truth table?

Answer: (D)

Solution

The inputs

n

and

m

feed an OR gate whose output is

n + m

. This output together with the line

n

forms the inputs of a NAND gate, so

z = \overline{n \cdot (n + m)}

. By the absorption law

n \cdot (n + m) = n

, hence

z = \overline{n}

. The output is

1

when

n = 0

and

0

when

n = 1

, independent of

m

, giving the truth table

00 \to 1, 01 \to 1, 11 \to 0, 10 \to 0

JEE Main · 2026 · 23 Jan 2026 · Shift 2 · Single correct option