What is the minimum number of gates required to implement the Boolean function (AB + C) if we have to use only 2-input NOR gates?

**A. ** 2

**B. ** 3

**C. ** 4

**D. ** 5

**Answer : ****Option B**

**Explaination / Solution: **

No Explaination.

No Explaination.

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The truth table

**A. ** X

**B. ** X + Y

**C. ** X ⊕Y

**D. ** Y

**Answer : ****Option A**

**Explaination / Solution: **

XY'+ XY = X(Y'+ Y) = X

represents the Boolean function

XY'+ XY = X(Y'+ Y) = X

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Consider the Boolean operator # with the following properties:

**A. **

**B. **

**C. **

**D. **

**Answer : ****Option A**

**Explaination / Solution: **

No Explaination.

Then x # y is equivalent to

No Explaination.

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Consider the given circuit.

**A. ** does not occur

**B. ** occurs when CLK = 0

**C. ** occurs when CLK = 1 and A = B = 1

**D. ** occurs when CLK = 1 and A = B = 0

**Answer : ****Option A**

**Explaination / Solution: **

In this circuit, the race around

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Consider the following Sum of Products expression, F.

The equivalent Product of Sums expression is

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The logical gate implemented using the circuit shown below where. V_{1} and V_{2} are inputs (with
0 V as digital 0 and 5 V as digital 1) and V_{out} is the output is

**A. ** NOT

**B. ** NOR

**C. ** NAND

**D. ** XOR

**Answer : ****Option B**

**Explaination / Solution: **

So, this logic level o/p is showing the functionality of NOR-gate.

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The Boolean function Y = AB + CD is to be realized using only 2 - input NAND gates. The minimum number of gates required is

**A. ** 2

**B. ** 3

**C. ** 4

**D. ** 5

**Answer : ****Option B**

**Explaination / Solution: **

This is SOP form and we require only 3 NAND gate

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An 8085 executes the following instructions
2710 LXI H, 30A0 H
2713 DAD H
2414 PCHL
All address and constants are in Hex. Let PC be the contents of the program counter and HL be the contents of the HL register pair just after executing PCHL. Which of the following statements is correct ?

**A. ** PC = 2715H HL = 30A0H

**B. ** PC = 30A0H HL = 2715H

**C. ** PC = 6140H HL = 6140H

**D. ** PC = 6140H HL = 2715H

**Answer : ****Option C**

**Explaination / Solution: **

2710H LXI H, 30A0H ; Load 16 bit data 30A0 in HL pair 2713H DAD H ; 6140H ⟶ HL 2714H PCHL ; Copy the contents 6140H of HL in PC Thus after execution above instruction contests of PC and HL are same and that is 6140H

2710H LXI H, 30A0H ; Load 16 bit data 30A0 in HL pair 2713H DAD H ; 6140H ⟶ HL 2714H PCHL ; Copy the contents 6140H of HL in PC Thus after execution above instruction contests of PC and HL are same and that is 6140H

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In the circuit shown, the device connected Y5 can have address in the range

**A. ** 2000-20FF

**B. ** 2D00-2DFF

**C. ** 2E00-2EFF

**D. ** FD00-FDFF

**Answer : ****Option B**

**Explaination / Solution: **

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In an 8085 microprocessor, the shift registers which store the result of an addition and the overflow bit are, respectively

**A. ** B and F

**B. ** A and F

**C. ** H and F

**D. ** A and C

**Answer : ****Option B**

**Explaination / Solution: **

The shift registers A and F store the result of an addition and the overflow bit.

The shift registers A and F store the result of an addition and the overflow bit.

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