JEE Main 2026 Apr 4 Shift 1, Physics Q13: Logic Gates

JEE Main2026Apr 4, Shift 1Physics

Two $4$ bits binary numbers, $A = 1101$ and $B = 1010$ are given in the inputs of a logic circuit shown in figure below. The output ( $Y$ ) will be :

A
$Y = 1101$
B
$Y = 0010$
C
$Y = 0111$
D
$Y = 1000$

Solution

Read the circuit as $Y = \overline{\overset{ˉ}{A} \cdot B} = A + \overset{ˉ}{B}$ (De Morgan's law applied to the NAND output of $\overset{ˉ}{A}$ and $B$ ).

$A = 1101$ , $B = 1010 \Rightarrow \overset{ˉ}{B} = 0101$ .

Bitwise OR: $A + \overset{ˉ}{B} = 1101 OR 0101 = 1101.$

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JEE Main 2026 Apr 4 Shift 1, Physics Q13: Logic Gates

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