Identify the logic operation carried out by t | vedclass

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(D) The given circuit consists of two $NAND$ gates acting as $NOT$ gates,followed by a $NAND$ gate. Let the inputs be $A$ and $B$. The output of the f

Vedclass · Accepted Answer

$OR$

Vedclass · Answer

(D) The given circuit consists of two $NAND$ gates acting as $NOT$ gates,followed by a $NAND$ gate. Let the inputs be $A$ and $B$. The output of the first $NAND$ gate is $\overline{A \cdot A} = \bar{A}$. The output of the second $NAND$ gate is $\overline{B \cdot B} = \bar{B}$. These outputs are fed into the final $NAND$ gate. The final output $Y$ is given by: $Y = \overline{\bar{A} \cdot \bar{B}}$ Using De Morgan's law,$\overline{\bar{A} \cdot \bar{B}} = \overline{\bar{A}} + \overline{\bar{B}} = A + B$. Thus,$Y = A + B$,which represents the $OR$ operation. Therefore,the correct option is $D$.

List-$I$	List-$II$
$(A)$ Benzene	$1$. Phosgene
$(B)$ Ethylene	$2$. Silver mirror
$(C)$ Acetaldehyde	$3$. Mustard gas
$(D)$ Chloroform	$4$. $(4n + 2) \pi$ electrons
	$5$. Carbylamine

Identify the logic operation carried out by the following circuit.

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Match the following lists:
List-$I$ List-$II$
$(A)$ Benzene $1$. Phosgene
$(B)$ Ethylene $2$. Silver mirror
$(C)$ Acetaldehyde $3$. Mustard gas
$(D)$ Chloroform $4$. $(4n + 2) \pi$ electrons
$5$. Carbylamine

The correct matching is:

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