MHTCET 16 May 2024 Shift 1 Paper

Concept:The given circuit works as an OR gate because its output Boolean expression simplifies to $Y = A + B$.Explanation:From the logic diagram, the first two NAND gates produce $\overline{A \cdot A} = \overline{A}$ and $\overline{B \cdot B} = \overline{B}$.These outputs are fed to a final NAND gate, so the output is:$Y = \overline{\overline{A} \cdot \overline{B}}$Using De Morgan's law,$\overline{\overline{A} \cdot \overline{B}} = \overline{\overline{A}} + \overline{\overline{B}} = A + B$Therefore, $Y = A + B$, which is the Boolean expression of an OR gate.The OR gate truth table gives $Y = 0$ only for $A = 0, B = 0$, and $Y = 1$ for all other input combinations.This matches the truth table given in option C.Answer:Option C is the correct truth table for the given circuit.