JEE Main 2026 — Semiconductors Question with Solution

From the given logic circuit, the input $A$ passes through a NOT gate, giving $\overline{A}$.

This output $\overline{A}$ and the input $B$ are then fed into a NAND gate.

The Boolean expression for the final output $Y$ is:
$Y=\overline{\overline{A}\cdot B}$

Given the $4$-bit binary inputs:
$A=1101$
$B=1010$

First, find the bitwise NOT of $A$:
$\overline{A}=\overline{1101}=0010$

Next, find the bitwise AND of $\overline{A}$ and $B$:
$\overline{A}\cdot B=0010\cdot 1010=0010$

Finally, apply the NOT operation for the NAND gate:
$Y=\overline{0010}=1101$

Alternatively, using De Morgan's laws:
$Y=\overline{\overline{A}\cdot B}=A+\overline{B}$
$Y=1101+\overline{1010}=1101+0101=1101$

Therefore, the output $Y$ is $1101$.

Answer: $Y=1101$