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JEE Main 2020 (8 Jan, Shift 2) · Logic Gates

Logic Gates question from JEE Main 2020 (8 Jan, Shift 2)

Q140JEE Main 2020 · 8 JAN · SHIFT 2Medium

Logic Gates

In the given circuit, value of Y is:

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Answer: (d)

Input 1 through NOT gives 0. Top NAND: $\overline{0\cdot 0} = 1$ .
Second NOT gives 1 to the middle NAND.
Bottom NAND gets 0 and the top NAND output 1: $\overline{0\cdot 1} = 1$ .
Middle NAND: $\overline{1\cdot 1} = 0$ , and this drives Y.
So the answer is (d) $Y = 0$ .

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