JEE Main · 2026 · 08 Apr 2026 · Shift 2 · Semiconductors

Tracing the gates, the upper AND gate gives $A\cdot B$ and the lower gate, whose $A$ input is inverted, gives $\bar{A}\cdot B$; the OR gate combines them as $Y=A\cdot B+\bar{A}\cdot B=(A+\bar{A})\cdot B=B$. The output therefore reproduces input $B$ exactly: high from $0$ to $2\text{s}$, low from $2$ to $3\text{s}$, and high again afterwards. The output $Y$ follows the input $B$.