(Hint: $A=0, B=1$ then $A$ and $B$ inputs of second NOR gate will be 0 and hence $Y=1 .$ Similarly, work out the values of $Y$ for other combinations of $A$ and $B .$ Compare with the truth table of OR, AND, NOT gates and find the correct one.)
Solution:
Input for the circuit: $A$ and $B$
For the first circuit, the output is $\overline{A+B}$
We can see from the diagram that the second gate’s input is automatically the first gate’s output.
Therefore, the output obtained from this combination can be expressed
$Y=\overline{\overline{A+B}+\overline{A+B}}=\overline{\bar{A} \cdot \bar{B}}+\overline{\bar{A} \cdot \bar{B}}$
$=\overline{\bar{A} \cdot \bar{B}}=\overline{\bar{A}}+\overline{\bar{B}}=A+B$
Following is the truth table for the given operation: