
The logic gate represented by following logic circuit is

- OR gate.
- AND gate.
- NOR gate.
- X-OR gate.
Solution

$\begin{array}{llrl} \therefore & & Y & =\overline{\mathrm{A} \cdot \mathrm{~B}} \cdot(\mathrm{~A}+\mathrm{B}) \\ \therefore & \mathrm{Y} & =(\overline{\mathrm{A}}+\overline{\mathrm{B}}) \cdot(\mathrm{A}+\mathrm{B}) & \ldots(\because \overline{\mathrm{A} \cdot \mathrm{~B}}=\overline{\mathrm{A}}+\overline{\mathrm{B}}) \\ \therefore & & \mathrm{Y} & =\overline{\mathrm{A}} \cdot \mathrm{~B}+\overline{\mathrm{B}} \cdot \mathrm{~A} \\ & \therefore & \mathrm{Y} & =\mathrm{A} \cdot \overline{\mathrm{~B}}+\mathrm{B} \cdot \overline{\mathrm{~A}} \end{array}$ This represents XOR gate.
Asked in: MHT CET 2024 (10 May Shift 2)