




The truth table shows that the output Y is 1 only when either A is 1 and B is 0, or both A and B are 1.
This can be represented as Y = (A AND NOT B) OR (A AND B).
Logic circuit (2) contains an AND gate, and an OR gate, and a NOT gate which matches to the truth table.
So the answer is (2).

The output (Y) of the given logic implementation is similar to the output of an/a …………. gate.
Consider the following logic circuit.
The output is Y = 0 when :
The logic gate equivalent to the combination of logic gates shown in the figure is 
A square loop of sides \( a = 1 \, {m} \) is held normally in front of a point charge \( q = 1 \, {C} \). The flux of the electric field through the shaded region is \( \frac{5}{p} \times \frac{1}{\varepsilon_0} \, {Nm}^2/{C} \), where the value of \( p \) is: