Q53 / 75 · JEE Main 2026 (24 January Shift 2) · Physics
Identify the correct truth table of the given logic circuit.
- A\(\begin{array}{|c|c|c|} \hline A & B & Y \\ \hline 0 & 0 & 1 \\ \hline 0 & 1 & 0 \\ \hline 1 & 0 & 1 \\ \hline 1 & 1 & 0 \\ \hline \end{array}\)
- B\(\begin{array}{|c|c|c|} \hline A & B & Y \\ \hline 0 & 0 & 1 \\ \hline 0 & 1 & 1 \\ \hline 1 & 0 & 1 \\ \hline 1 & 1 & 0 \\ \hline \end{array}\)
- C\(\begin{array}{|c|c|c|} \hline A & B & Y \\ \hline 0 & 0 & 0 \\ \hline 0 & 1 & 1 \\ \hline 1 & 0 & 1 \\ \hline 1 & 1 & 0 \\ \hline \end{array}\)
- Correct answerD\(\begin{array}{|c|c|c|} \hline A & B & Y \\ \hline 0 & 0 & 0 \\ \hline 0 & 1 & 0 \\ \hline 1 & 0 & 1 \\ \hline 1 & 1 & 0 \\ \hline \end{array}\)
Solution
Analysing the logic circuit:
The output is $Y = A \cdot \overline{A \cdot B}$
$= A \cdot (\bar{A} + \bar{B})$
$= A\bar{A} + A\bar{B} = 0 + A\bar{B} = A\bar{B}$
Truth table:
$A=0, B=0 \Rightarrow Y=0$
$A=0, B=1 \Rightarrow Y=0$
$A=1, B=0 \Rightarrow Y=1$
$A=1, B=1 \Rightarrow Y=0$
The output is $Y = A \cdot \overline{A \cdot B}$
$= A \cdot (\bar{A} + \bar{B})$
$= A\bar{A} + A\bar{B} = 0 + A\bar{B} = A\bar{B}$
Truth table:
$A=0, B=0 \Rightarrow Y=0$
$A=0, B=1 \Rightarrow Y=0$
$A=1, B=0 \Rightarrow Y=1$
$A=1, B=1 \Rightarrow Y=0$
Hence, the correct answer is option: 'D'.