QJEEM25Ph6a75: JEE Main 2025 Physics - Class 12 JEEM25Ph6a75 Solution: To obtain the given truth table, following logic gate should be …

For NOR gate : $\overline{\mathrm{A}} \overline{\mathrm{B}}=\overrightarrow{\mathrm{A}}+\mathrm{B}$

$\therefore$ Truth table $\begin{array}{ccc}\mathrm{A} & \mathrm{~B} & \mathrm{Y} \\0 & 0 & 1 \\0 & 1 & 1 \\1 & 0 & 0 \\1 & 1 & 1\end{array}$

$\therefore$ Bonus