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(B) The circuit consists of a $NOR$ gate followed by a $NOT$ gate, which together form an $OR$ gate for the inputs $A$ and $B$. Let this output be $Y_

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$1, 0, 0$

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(B) The circuit consists of a $NOR$ gate followed by a $NOT$ gate, which together form an $OR$ gate for the inputs $A$ and $B$. Let this output be $Y_1$. Thus, $Y_1 = A + B$.The inputs $C$ and $D$ are connected to an $AND$ gate, so $Y_2 = C \cdot D$.The outputs $Y_1$ and $Y_2$ are then fed into a $NOR$ gate to produce $Y_3$. Thus, $Y_3 = \overline{Y_1 + Y_2}$.Given inputs: $A = 0, B = 1, C = 0, D = 1$.Calculating $Y_1$: $Y_1 = A + B = 0 + 1 = 1$.Calculating $Y_2$: $Y_2 = C \cdot D = 0 \cdot 1 = 0$.Calculating $Y_3$: $Y_3 = \overline{Y_1 + Y_2} = \overline{1 + 0} = \overline{1} = 0$.Therefore, the new outputs are $(1, 0, 0)$.

In the logic circuit diagram, when all the four inputs, $A, B, C, D$ are 'one' the outputs $Y_1, Y_2, Y_3$ are respectively $(1, 1, 0)$. When the inputs $A$ and $C$ are changed to zero and $B$ and $D$ are still 'one',then the outputs $Y_1, Y_2, Y_3$ are respectively change to

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