GATE CS 2010 Q31 — Combinational Circuits

We can determine the Boolean expression at the output of each NOR gate step-by-step from left to right:

1. Top-left NOR gate: Inputs are $P$ and $Q$. $$X = \overline{P+Q}$$
2. Bottom-left NOR gate: Inputs are $Q$ and $R$. $$Y = \overline{Q+R}$$
3. Middle NOR gate: Inputs are $X$ and $Y$. $$Z = \overline{X+Y} = \overline{\overline{P+Q} + \overline{Q+R}}$$ Using De Morgan's Law: $$Z = (P+Q)(Q+R) = PQ + Q + PR + QR = Q(1 + P + R) + PR = Q + PR$$
4. Final (rightmost) NOR gate: Inputs are $Z$ and $R$. $$f = \overline{Z+R} = \overline{(Q + PR) + R}$$ Using the absorption law ($PR + R = R$): $$f = \overline{Q + R}$$

Thus, the final output expression is $\overline{Q+R}$, which corresponds to Option A.