The truth table of the following circuit is

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(C) The given circuit consists of two $NOT$ gates (formed by $NAND$ gates with shorted inputs) followed by a $NAND$ gate.Let the inputs be $A$ and $B$

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(C) The given circuit consists of two $NOT$ gates (formed by $NAND$ gates with shorted inputs) followed by a $NAND$ gate.Let the inputs be $A$ and $B$.The output of the first $NOT$ gate is $\overline{A}$.The output of the second $NOT$ gate is $\overline{B}$.These are the inputs to the final $NAND$ gate.Therefore,the final output $Y$ is given by:$Y = \overline{\overline{A} \cdot \overline{B}}$Using De Morgan's law,$\overline{X \cdot Y} = \overline{X} + \overline{Y}$,we get:$Y = \overline{\overline{A}} + \overline{\overline{B}} = A + B$This is the Boolean expression for an $OR$ gate.The truth table for an $OR$ gate is:| $A$ | $B$ | $Y$ ||---|---|---|| $0$ | $0$ | $0$ || $0$ | $1$ | $1$ || $1$ | $0$ | $1$ || $1$ | $1$ | $1$ |Comparing this with the given options,option $C$ represents this truth table.

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