This section contains carefully selected MCQs and Previous Year Questions with explanations to help students understand concepts and prepare effectively for examinations, interviews, and competitive tests.
Q: 1Which gate is equivalent to (NOR) OR (XOR)?
Option A
A NOR gate is the complement of the OR gate. It gives output 1 only when all inputs are 0.
Boolean Expression : = (A+B)′
An XOR gate gives output 1 when the inputs are different or if the number of 1’s are odd and 0 when the inputs are the same.
Boolean Expression : = A⊕B
Now the given expression is (NOR)OR(XOR) i.e., (A+B)’ + (A⊕B)
| A | B | (A+B)' | A⊕B | (A+B)’ + (A⊕B) |
|---|---|---|---|---|
| 0 | 0 | 1 | 0 | 1 |
| 0 | 1 | 0 | 1 | 1 |
| 1 | 0 | 0 | 1 | 1 |
| 1 | 1 | 0 | 0 | 0 |
Now, we construct the truth tables of NAND, OR, and AND gates and verify their behavior using the given equation.
| A | B | AND = (A.B) | NAND = (A.B)’ | OR = (A+B) |
|---|---|---|---|---|
| 0 | 0 | 0 | 1 | 0 |
| 0 | 1 | 0 | 1 | 1 |
| 1 | 0 | 0 | 1 | 1 |
| 1 | 1 | 1 | 0 | 1 |
The NAND gate behavior exactly matches with given expression.
Therefore, (NOR)OR(XOR) = NAND
Q: 2Output of 2-inputs NAND gate if one of its input is permanently connected to ‘0’ is :
Option B
A NAND gate produces the output as the negation of the AND operation, i.e., Y=(A⋅B)’. If one of its inputs is permanently connected to 0, then the AND operation becomes 0⋅B=0. So, Y=(0⋅B)′=0′=1. This means the output of the NAND gate will always be 1 regardless of the other input. Hence, the correct answer is 1.
Q: 3The logic gate that provides high output for same input is
Option B
The X-NOR gate, produces a high (1) output when both inputs are the same, either both 0 or both 1.
Truth—Table:
| Input | Output | |
| A | B | Y=A⊙B |
| 0 | 0 | 1 |
| 0 | 1 | 0 |
| 1 | 0 | 0 |
| 1 | 1 | 1 |
Q: 4The minimum number of NAND gates required for implementing the Boolean expression, AB+AB’C+AB’C’ is?
Option B
Given Boolean expression:
=AB+AB’C+AB’C’
=AB+AB’(C+C’)
=AB+AB’.1
=AB+AB’
=A(B+B’)
=A.1
=A
The given expression simplifies to A, so the output can be directly taken from input A. Therefore, the minimum number of NAND gates required is zero.
Q: 5The logical output of EX-NOR gate is:
Option D
The EX-NOR (Exclusive-NOR) gate is the complement of the EX-OR (Exclusive-OR) gate. While an EX-OR gate gives output 1 when the inputs are different, the EX-NOR gate gives output 1 when the inputs are the same that means, both inputs are 0 or both are 1.
| Input A | Input B | Output (A ⊙ B) |
|---|---|---|
| 0 | 0 | 1 |
| 0 | 1 | 0 |
| 1 | 0 | 0 |
| 1 | 1 | 1 |
The Boolean expression for an EX-NOR gate is:
Y=A⊙B = (A⊕B)’ = A.B+A’B’
Q: 6For inputs, A=10101010. Find A XOR A
Option B
The XOR (Exclusive OR) is a logical operation that compares two bits and gives the result 1 only when the inputs are different, and 0 when the inputs are the same.
The XOR operation has an important property, when a number is XORed with itself, the result is always 0, i.e., A ⊕ A = 0.
Given: A = 10101010. Now perform XOR with itself: 10101010 ⊕ 10101010 = 00000000
Q: 7Which of the following is an example of a digital electronic?
Option D
Digital electronics are devices that operate using binary signals. They process information in discrete steps rather than continuous signals.
Examples of digital electronic devices include computers, which use digital circuits for all processing tasks, mobile phones, which contain digital processors and memory and digital cameras, which convert images into digital data.
Q: 8Which is the correct truth table for XNOR gate?
Option B
An X-NOR gate is a digital logic gate that gives output 1 when both inputs are the same (either both 0 or both 1), and gives output 0 when the inputs are different.
It is also called an Equivalence Gate because it checks whether two inputs are equal.
Q: 9How many rows are there in a truth table having ‘n’ variables?
Option D
A truth table is used in digital electronics to show all possible combinations of input variables and their corresponding outputs.
Each variable in a truth table can have only two possible values either 0 or 1. If there are n variables, then each variable contributes 2 possible combinations. Therefore, total possible combinations (rows) become 2n.
Q: 10What is the Boolean expression of an Exclusive—OR gate?
Option A
An Exclusive-OR (XOR) gate gives output 1 only when the two input bits are different. If both inputs are same, the output becomes 0.
The Boolean expression of XOR gate is A’B+AB’. Here option (A) uses different variable names (C and D) but represents the same XOR expression: C’D+CD’
| C | D | C’D+CD’ (C ⊕D) |
|---|---|---|
| 0 | 0 | 0 |
| 0 | 1 | 1 |
| 1 | 0 | 1 |
| 1 | 1 | 0 |
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