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MHT CET 2021 (23 Sep, Shift 1) · Logic Gates

Logic Gates question from MHT CET 2021 (23 Sep, Shift 1)

Q127MHT CET 2021 · 23 SEP · SHIFT 1Medium

Logic Gates

The resultant gate and its Boolean expression in the given circuit is

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Answer: (b)

NOT gates give $C = \bar A$ , $D = \bar B$ .
NOR: $Y = \overline{\bar A + \bar B} = A \cdot B$ (De Morgan).
So the answer is (b) AND, $A \cdot B$ .

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