NEETPhysicsSemiconductors
Match List-I with List-II. List-I (Boolean Expression) List-II (Logic Gate) (P) Y = A B (1) AND (Q) Y = A B (2) OR (R) Y = A + B (3) NAND (S) Y = A + B (4) NOR Choose the correct answer from the options given below:
Options
- AP-(2), Q-(4), R-(1), S-(3)
- BP-(1), Q-(3), R-(2), S-(4)
- CP-(2), Q-(3), R-(1), S-(4)
- DP-(1), Q-(4), R-(2), S-(3)
Correct answer
A. P-(2), Q-(4), R-(1), S-(3)
Step-by-step solution
Let us simplify each Boolean expression using De Morgan's laws: (P) Y = A B Applying De Morgan's law ( X Y = X + Y ): Y = A + B = A + B This is the expression for an OR gate. So, P matches with (2). (Q) Y = A B Applying De Morgan's law ( X Y = X + Y ): Y = A + B This is the expression for a NOR gate. So, Q matches with (4). (R) Y = A + B Applying De Morgan's law ( X + Y = X Y ): Y = A B = A B This is the expression for an AND gate. So, R matches with (1). (S) Y = A + B Applying De Morgan's law ( X + Y = X Y ): Y =