JEE Main 2026 Jan 24 Shift 2, Physics Q14: Logic Gates

The output of the first AND gate (with A on both inputs) is simply $A$. The NAND gate output is $\overline{A\cdot B}$. The final AND gate gives:

$Y=A\cdot \overline{A\cdot B}=A\cdot (\bar{A}+\bar{B})=A\bar{A}+A\bar{B}=A\bar{B}$.

Truth table for $Y=A\bar{B}$:

This matches option (4).