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(B) The circuit consists of two $AND$ gates, two $NOT$ gates, and one $NOR$ gate. The Boolean expression for the output $Y$ is $Y = \overline{(A \cdot

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$0, 0$

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(B) The circuit consists of two $AND$ gates, two $NOT$ gates, and one $NOR$ gate. The Boolean expression for the output $Y$ is $Y = \overline{(A \cdot B) + (\overline{A} \cdot \overline{B})}$.Case $1$: When $A = 1$ and $B = 1$, the output of the top $AND$ gate is $1 \cdot 1 = 1$. The output of the bottom $AND$ gate is $\overline{1} \cdot \overline{1} = 0 \cdot 0 = 0$. The $NOR$ gate receives inputs $1$ and $0$, so $Y = \overline{1 + 0} = \overline{1} = 0$.Case $2$: When $A = 0$ and $B = 0$, the output of the top $AND$ gate is $0 \cdot 0 = 0$. The output of the bottom $AND$ gate is $\overline{0} \cdot \overline{0} = 1 \cdot 1 = 1$. The $NOR$ gate receives inputs $0$ and $1$, so $Y = \overline{0 + 1} = \overline{1} = 0$.Thus, the outputs are $0$ and $0$.

In the given circuit, the binary inputs at $A$ and $B$ are both $1$ in one case and both $0$ in the next case. The respective outputs at $Y$ in these two cases will be

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