JEE Main 2026 — Semiconductors Question with Solution

From the given logic circuit, we can determine the Boolean expression for the output $Y$.

The output of the NOT gate is $\bar{A}$.

The inputs to the OR gate are $\bar{A}$ and $B$, so its output is $\bar{A}+B$.

The inputs to the AND gate are $\bar{A}$ and $B$, so its output is $\bar{A}\cdot B$.

These two outputs are fed into a NOR gate. The final output $Y$ is given by:

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

Simplifying the Boolean expression:

$Y=\overline{\bar{A}+B+\bar{A}B}$

$Y=\overline{\bar{A}+B(1+\bar{A})}$

$Y=\overline{\bar{A}+B}$

Using De Morgan's laws:

$Y=\overline{\bar{A}}\cdot \bar{B}=A\cdot \bar{B}$

Now, we evaluate $Y$ for the given input combinations:

For $A=1,B=1$:

$Y=1\cdot \bar{1}=1\cdot 0=0$

For $A=0,B=1$:

$Y=0\cdot \bar{1}=0\cdot 0=0$

Therefore, the outputs are $0$ and $0$ respectively.

Answer: $0,0$ respectively