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Q.
The following configuration of gate is equivalent to :
NTA AbhyasNTA Abhyas 2020
Solution:
$Y=\left(A + B\right)\cdot \bar{A B}$
The given output equation can also be written as
$Y=\left(A + B\right)\cdot \left(\bar{A} + \bar{B}\right)$ (De Morgan's theorem)
$=A\bar{A}+A\bar{B}+B\bar{A}+B\bar{B}=0+A\bar{B}+\bar{A}B+0$
$=\bar{A}B+A\bar{B}$
This is the expression for $XOR$ gate.
