The truth table given in fig. represents:

A	B	Y
0	0	0
0	1	1
1	0	1
1	1	1

We begin by examining the information that has been supplied. Two inputs are denoted by $$A$$ and $$B$$, while the single output is denoted by $$Y$$. A complete truth table has been provided, so we will compare each row with the standard truth tables of the basic logic gates we study in digital electronics.

The given truth table is

$$ \begin{array}{|c|c|c|} \hline A & B & Y\\ \hline 0 & 0 & 0\\ 0 & 1 & 1\\ 1 & 0 & 1\\ 1 & 1 & 1\\ \hline \end{array} $$

Now, let us recall the standard definition of an OR gate. The OR gate is defined by the logical expression

$$ Y = A + B, $$

where the symbol $$+$$ stands for the logical OR operation. The rule for the OR operation is very simple:

An OR gate gives an output $$1$$ if \textbf{any one $$of its inputs is }1,$$ otherwise it gives $$0.$$

Writing this rule explicitly in the form of a truth table, we obtain

$$ \begin{array}{|c|c|c|} \hline A & B & Y = A + B\\ \hline 0 & 0 & 0\\ 0 & 1 & 1\\ 1 & 0 & 1\\ 1 & 1 & 1\\ \hline \end{array} $$

We now compare row by row with the given table:

• When $$A = 0$$ and $$B = 0$$, the OR rule says $$Y = 0$$, which matches the given $$0$$.
• When $$A = 0$$ and $$B = 1$$, the OR rule says $$Y = 1$$, which matches the given $$1$$.
• When $$A = 1$$ and $$B = 0$$, the OR rule says $$Y = 1$$, which matches the given $$1$$.
• When $$A = 1$$ and $$B = 1$$, the OR rule says $$Y = 1$$, which again matches the given $$1$$.

Every single entry of the given truth table coincides perfectly with the corresponding entry of the standard OR-gate truth table. Hence we safely conclude that the circuit whose behaviour is described is an OR gate.

The option list shows that Option A is “OR-Gate.” Therefore the correct choice is Option A.

Hence, the correct answer is Option A.