JEE Main 2026 — Semiconductors Question with Solution

From the circuit:
Top AND gate output: $A\cdot B$
Bottom AND gate output: $\bar{A}\cdot B$ (the $A$ input passes through a NOT gate)
The OR gate combines these:
$Y=(A\cdot B)+(\bar{A}\cdot B)$

Factoring out $B$:
$Y=B\cdot (A+\bar{A})$

Using $A+\bar{A}=1$:
$Y=B\cdot 1=B$

So $Y$ is independent of $A$ and follows $B$ exactly.

Reading the input waveform of $B$:
From $t=0$ to $t=1\text{ s}$: $B=1$
From $t=1$ to $t=2\text{ s}$: $B=1$
From $t=2$ to $t=3\text{ s}$: $B=0$

Hence, $Y$ is $1$ from $0$ to $2\text{ s}$ and $0$ from $2$ to $3\text{ s}$.

This matches the waveform shown in option $(2)$.

Hence, the correct option is $(2)$.