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JEE Main 202624 JanuaryEvening Shift
Exam
JEE Main
Year
2026
Date
24 January
Shift
Evening Shift
Subject
Physics
Chapter
Semiconductors

JEE Main 2026 · 24 January · Evening Shift · Physics · Semiconductors

Identify the correct truth table of the given logic circuit.

Identify the correct truth table of the given logic circuit.
Identify the correct truth table of the given logic circuit.

Options

  1. A(ABY001010101110\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}
  2. B(ABY001011101110\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}
  3. C(ABY000011101110\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}
  4. D(ABY000010101110\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}

Correct answer

D. (ABY000010101110\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}

Step-by-step solution

Analysing the logic circuit:

The output is Y=A⋅A⋅B‾Y = A \cdot \overline{A \cdot B}

=A⋅(Aˉ+Bˉ)= A \cdot (\bar{A} + \bar{B})

=AAˉ+ABˉ=0+ABˉ=ABˉ= A\bar{A} + A\bar{B} = 0 + A\bar{B} = A\bar{B}

Truth table:
A=0,B=0⇒Y=0A=0, B=0 \Rightarrow Y=0
A=0,B=1⇒Y=0A=0, B=1 \Rightarrow Y=0
A=1,B=0⇒Y=1A=1, B=0 \Rightarrow Y=1
A=1,B=1⇒Y=0A=1, B=1 \Rightarrow Y=0
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