Q125 / 125 · BITSAT 2009 · Physics
The truth table given below correspond to the logic gate
\begin{tabular}{cc|c}
$\mathrm{A}$ & $\mathrm{B}$ & $\mathrm{X}$ \\
\hline 0 & 0 & 1 \\
1 & 0 & 0 \\
0 & 1 & 0 \\
1 & 1 & 0
\end{tabular}
- A$\mathrm{OR}$
- Correct answerBNOR
- CAND
- DNAND
Solution
The given gate is a NOR gate. The output is high, when all inputs are low.
Truth Table:
\begin{tabular}{|c|c|c|}
\hline $\mathbf{A}$ & $\mathbf{B}$ & $\mathbf{Y}$ \\
\hline 0 & 0 & 1 \\
\hline 1 & 0 & 0 \\
\hline 0 & 1 & 0 \\
\hline 1 & 1 & 0 \\
\hline \multicolumn{2}{|c|} { Logic Symbols } \\
\hline
\end{tabular}

Boolean expression: $\overline{\mathrm{A}+\mathrm{B}}=\mathrm{Y}$

Boolean expression: $\overline{\mathrm{A}+\mathrm{B}}=\mathrm{Y}$
Hence, the correct answer is option: 'B'.