WBCHSE Class 12 Physics Digital Electronics & Logic Gates Multiple Choice Questions

Digital Electronics & Logic Gates Multiple Choice Questions

Question 1. The most significant digit of the number 6789 is

1. 6
2. 7
3. 8
4. 9

Answer: 1. 6

Question 2. The most significant digit of the number, 0.6789 is

1. 6
2. 7
3. 8
4. 9

Answer: 1. 6

Question 3. The least significant digit of the number, 0.6789 is

1. 6
2. 7
3. 8
4. 9

Answer:  4. 9

Question 4.  In the Binary number system, the number 100 represents 

1. One
2. Three
3. Four
4. Hundred

Answer: 3. Four

Question 5. The gate is equivalent to

1. NAND gate
2. NOT gate
3. AND gate
4. NOR gate

Answer: 2. NOT gate

WBBSE Class 12 Digital Electronics MCQs

Question 6. In the case of the given circuit, the values of Y₁, Y_2, and K, are, respectively

1. 1, 1 and 0
2. 1,1 and 1
3. 1,0 and 0
4. 0,1 and 1

Answer: 1. 1, 1 and 0

Question 7. The following truth table is for which gate?

1. AND gate
2. NOR gate
3. NAND gate
4. OR gate

Answer: 4. OR gate

Question 8. The output of an OR gate will be 1, if

1. Both the inputs are 0
2. One or both of the inputs be 1
3. Both the inputs are 1
4. One or both of the input be 0

Question 9. For which gate is the truth table valid?

1. AND gate
2. NOR gate
3. NAND gate
4. OR gate

Answer: 2. NOR gate

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Real-Life Applications of Digital Logic

Question 10. According, to the given table, which gate?

1. Gate no. 1
2. Gate nos. 1 and 2
3. Gate no. 2
4. Gate no. 3

Answer: 3. Gate no. 2

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Question 11. Digital signals

1. Represent values as discrete steps
2. Do not represent values as discrete steps
3. Represent vague steps
4. represent random steps

Answer: 1. Represent values as discrete steps

Question 12. In Boolean algebra, A + B = Y implies that:

1. The Sum of A and B is Y
2. Y exists when A exists or B exists or both A and B exist
3. Y exists only when A and B both exist
4. Y exists when A or B exists, but not when both A and B exist

Answer: 3. Y exists only when A and B both exist

Practice MCQs on Combinational Logic Circuits

Question 13. In Boolean algebra, A – B = Y implies that:

1. Product of A and B is Y
2. Y exists when A exists or B exists
3. Y exists when both A and B exist, but not when only A or B exists
4. Y exists when A or B exists, but not both A and B exist

Answer: 3. Y exists when both A and B exist but not when only A or B exists

Question 14. In a three-input logic gate, the first two inputs are in state 1 and the third is 0. For which of the following gates, does the out become 1?

1. OR gate
2. And gate
3. NOR gate
4. NAND gate

Answer: 1 And 4

Question 15. A correct Boolean algebraic equation is

1. A + 0 = 0
2. A + 0 = A
3. A + 1 = 1
4. A + 1 = A

Answer: 2 And 3

Question 16. The product of (110)₂ and (100)₂ is

1. (1100)₂
2. (11000)₂
3. (20)₂₁₀
4. (24)₁₀

Answer: 2 And 4

Question 17. If A and H are the two inputs of a NAND) gate, then the output will be

1. \(\overline{A B}\)
2. \(\bar{A} \cdot \bar{B}\)
3. \(\overline{A+B}\)
4. \(\bar{A}+\bar{B}\)

Answer: 1 And 4

Question 18. The outputs of a three-input OR gate and a three-input NAND gate will be the same if all t

1. Three inputs become 0
2. One input becomes 1
3. Two inputs become 1
4. All three inputs become 1

Answer: 2 And 3

Important Mcqs in Digital Electronics

Question 19. If a, b, c, d are input to a gate and r is its output, then, as per the following time graph, the gate is

1. NOT
2. AND
3. OR
4. NAND

Answer: 3. OR

Question 20. Which logic gate is represented by the following combination of logic gates

1. OR
2. NAND
3. AND
4. NOR

Answer: 3. OR

Y = \(\overline{Y_1+Y_2}\)

= \(\overline{\bar{A}+\bar{B}}=\overline{\bar{A}} \cdot \overline{\bar{B}}\)

= A.B

The given combination of logic gates represents the AND gate.

Question 21. To get output 1 for the following circuit, the correct choice for the input is

1. A =1, B = 0, C = 0
2. A = 1, B = 1, C = 0
3. A = 1, B = 0, C = 1
4. A = 0, B = 1, C = 0

Answer: 3. A = 1, B = 0, C = 1

The Boolean expression for the given logic circuit is

Y = (A + R). C

If A = 1, B = 0 and C = 1, then the output, Y = 1

Question 22. From the circuit of the following logic gates, the basic logic gate obtained Is

1. NAND gate
2. AND gate
3. OR gate
4. NOT gate

Answer: 1. NAND gate

Y = \(\overline{\bar{A}+\bar{B}}\) = AB

Y = \(\overline{A B \cdot B}=\overline{A B}\) ,NAND gate

Examples of Logic Gate Applications

Question 23. In the combination of the following gates, the output Y can be written In terms of inputs A and B as

1. \(\overline{A \cdot B}+A \cdot B\)
2. \(A \cdot \bar{B}+\bar{A} \cdot B\)
3. \(\overline{A \cdot B}\)
4. \(\overline{A+B}\)

Answer: 2. \(A \cdot \bar{B}+\bar{A} \cdot B\)

Y = \(A \cdot \bar{B}+\bar{A} \cdot B\)

Question 24. The output Y of the logic circuit Is given below

1. \(\bar{A}+B\)
2. \(\bar{A}\)
3. \((\overline{\bar{A}+B}) \cdot \bar{A}\)
4. \((\overline{\bar{A}+B}) \cdot A\)

Answer: 2. \(\bar{A}\)

Y = \(\bar{A}+\bar{A} B=\bar{A}(1+B)=\bar{A}\)

Since 1+ B = 1

Question 25. The inputs to the digital circuit are shown below the output Y is

1. A+B+\(\bar{C}\)
2. (A+B)\(\bar{C}\)
3. \(\bar{A}+\bar{B}+\bar{C}\)
4. \(\bar{A}+\bar{B}+C\)

Answer: 3. \(\bar{A}+\bar{B}+\bar{C}\)

If the Input is the rightmost OR gate is M and N,

Y = M+n = \(\overline{A B}+\bar{C}=\vec{A}+\vec{B}+\vec{C}\)

Conceptual Questions on Sequential Circuits

Question 26. In the given circuit, the binary Inputs at A and B are both I In one case and both 0 In the next case. The respective outputs at Y In these two cases will be

1. 1,1
2. 0,1
3. 0,1
4. 1,0

Answer: 2. 0,0

Y= \(\overline{A \cdot B+\bar{A} \cdot \bar{B}}\)

Question 27. In the circuit shown, inputs A and B are in states 1 and 0 respectively. What is the only possible stable state of the outputs X and Y?

1. X = 1, Y = 1
2. X = 1, Y = 0
3. X = 0, Y = 1
4. X = 0, Y = 0

Answer: 3. X = 0, Y = 0

Given, A = 1 and B = 0,

If X = 1, then Y will be 1. At that instant, the state of X should be 0

But it is given that X = 1.

The option is not correct.

Only they correctly describe the circuit.

Hence, if X = 0, then Y will be 1. At that Instant, X = 0.