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 answer
    B
    NOR
  • C
    AND
  • D
    NAND

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

Hence, the correct answer is option: 'B'.

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