Explanation : In OR gate we need atleast one bit to be equal to 1 to generate the output as 1 because OR means any of the condition out of two is equal to 1 which means if atleast one input is 1 then it shows output as 1 . Number of 1's in input may be more than one but the output will always be 1 in OR gate. So the answer is 'C'.
Explanation : Y=A+B. This is the equation of OR gate. We require 3 NAND gates to create OR gate. We can also write
After 1st NAND operation
Y = (A AND B)'
Y= A' + B' (Demorgan's Law)
After 2nd NAND operation
Y= ( A' + B')'
Y= A . B (Demorgan's Law)
After 3rd NAND operation
Y= ( A . B )'
Y= A' + B ' (Demorgan's Law)
So we need 3 NAND gates.