Test: Logic Gates - JEE MCQ

Binary number system, in mathematics, positional numeral system employing 2 as the base and so requiring only two different symbols for its digits, 0 and 1, instead of the usual 10 different symbols needed in the decimal system.

1001=1x2³+0x2²+0x2¹+1x2⁰=9
For binary numbers, the digit at the extreme right is referred to as least significant bit (LSB) and the left- most digit is called the most significant bit (MSB). Hence in binary 10010, the most significant bit is 1.
 

A universal gate is a gate which can implement any Boolean function without need to use any other gate type. The NAND and NOR gates are universal gates. In practice, this is advantageous since NAND and NOR gates are economical and easier to fabricate and are the basic gates used in all IC digital logic families.

The equation of NAND gate is 
Y=Aˉ.Bˉ
When both the terminals are connected together.
A=B=A
Y=Aˉ.Aˉ=Aˉ.Aˉ=Aˉ
Which is nothing but the equation of NOT gate.
Y=Aˉ
Option A is the correct answer.

An OR gate followed by a NOT gate in a cascade is called a NOR gate. In other words, the gate which provides a high output signal only when there are low signals on the inputs such type of gate is known as NOR gate.

We can observe from the table that Yˉ=A.B⇒Y=AB
So, the table represents a NAND gate.

The logic or Boolean expression given for a digital logic AND gate is that for Logical Multiplication which is denoted by a single dot or full stop symbol, ( . ) giving us the Boolean expression of:  A.B = Y.
 
Then we can define the operation of a digital 2-input logic AND gate as being:
“If both A and B are true, then Y is true”
 

The truth table for 2-input OR gate is as shown in the figure. Thus, the output is zero only when both inputs are zero.

A=1 and B=0
A.A+B
=A2+B [De-Morgan’s theorem)
=1+0
=1
=A

10101 is the binary number to be converted to a decimal number.
To convert, multiply the numbers individually with 2n where n changes according to place value of the number.
(10101)₂= (2⁰× 1) + (0 × 2¹) + (1×2²) + (0×2³) + (1×2⁴)
= 1 + 0 + 4 + 0 + 16
= 21