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(B) $NAND$ gate performs the operation $Y = \overline{A \cdot B}$.Based on the input waveforms:For $t = 0-2 \ s$: $A=1, B=0 \implies A \cdot B = 0 \im

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(B) $NAND$ gate performs the operation $Y = \overline{A \cdot B}$.Based on the input waveforms:For $t = 0-2 \ s$: $A=1, B=0 \implies A \cdot B = 0 \implies Y = \overline{0} = 1$.For $t = 2-4 \ s$: $A=0, B=0 \implies A \cdot B = 0 \implies Y = \overline{0} = 1$.For $t = 4-6 \ s$: $A=1, B=0 \implies A \cdot B = 0 \implies Y = \overline{0} = 1$.For $t = 6-8 \ s$: $A=1, B=1 \implies A \cdot B = 1 \implies Y = \overline{1} = 0$.For $t > 8 \ s$: $A=0, B=0 \implies A \cdot B = 0 \implies Y = \overline{0} = 1$.Thus,the output $Y$ is $1$ for $0-6 \ s$,$0$ for $6-8 \ s$,and $1$ for $t > 8 \ s$. This matches the waveform shown in option $B$.

The following figure shows the variation of inputs $A$ and $B$ with time. If these inputs are given to a $NAND$ gate,select the correct output signal $Y$ from the given options.

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