Discussion
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- Question
To multiply a number by 8 in 8085 we have to use
RAL instruction
Options- A. once
- B. twice
- C. thrice
- D. four times
- Correct Answer
- thrice
ExplanationWhen RAL instruction is used once the number is doubled.
Microprocessors problems
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- 1. Assertion (A): If source and destination addresses are made implicit the length of instruction is reduced.
Reason (R): 8085 has only 3 addressing modes.
Options- A. Both A and R are correct and R is correct explanation of A
- B. Both A and R are correct but R is not correct explanation of A
- C. A is correct R is wrong
- D. A is wrong R is correct Discuss
Correct Answer: A is correct R is wrong
Explanation:
The number of addressing modes is five.- 2. Which of the following is not a valid real constant in C?
Options- A. 1.2 e 2.0
- B. -2.3 e - 2
- C. -2.3 E 2
- D. -2.3 e 2 Discuss
Correct Answer: -2.3 E 2
Explanation:
In C real constant expressed in exponential form, mantissia and exponent should be separated by e (and not E).- 3. Assertion (A): Microprocessor 8085 has on-chip oscillator with inbuilt crystal.
Reason (R): For frequency stability crystal oscillator is preferred.
Options- A. Both A and R are correct and R is correct explanation of A
- B. Both A and R are correct but R is not correct explanation of A
- C. A is correct R is wrong
- D. A is wrong R is correct Discuss
Correct Answer: A is wrong R is correct
Explanation:
Crystal has to be connected externally to pins 1 and 2 of 8085.- 4. Z flag is reset if an ALU operation results in 0.
Options- A. True
- B. False Discuss
Correct Answer: False
Explanation:
Z flag is set if ALU operation results in zero.- 5. Which of the following is not an integer constant?
Options- A. 0
- B. -8
- C. +1074
- D. -8.1 Discuss
Correct Answer: -8.1
Explanation:
Integer constant should not have decimal point.- 6. Consider the following program in Pascal
X = 2.57
X = (X + 0.06) * 10
I = TRUNC (X)
X = I
X = X / 10.0
The final value of X would be
Options- A. 2
- B. 2.6
- C. 2.63
- D. 26.3 Discuss
Correct Answer: 2.6
Explanation:
(2.57 + 0.06) 10 = 26.3 ?
.
- 7. EEPROM permits
Options- A. read operation only
- B. read and byte erase operations
- C. read, byte erase and byte write operations
- D. read, byte erase, byte write and chip erase operations Discuss
Correct Answer: read, byte erase, byte write and chip erase operations
Explanation:
All the four are permitted.- 8. If A = 20 and B = - 7 in PASCAL, A mod B =
Options- A. 6
- B. 2
- C. - 1
- D. 3 Discuss
Correct Answer: 6
Explanation:
It gives remainder in division operation.- 9. A 256 x 4 EPROM has
Options- A. 8 address pins and 4 data pins
- B. 8 address pins and 8 data pins
- C. 4 address pins and 8 data pins
- D. 4 address pins and 4 data pins Discuss
Correct Answer: 8 address pins and 4 data pins
Explanation:
28 = 256- 10. Assume that x = 5 and y = 2. Then in Java operation x + = y, the result is
Options- A. x = 7
- B. x = 2
- C. x = 3
- D. x = 2 Discuss
Correct Answer: x = 7
Explanation:
The operator + = means add and assign.
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- 1. Assertion (A): If source and destination addresses are made implicit the length of instruction is reduced.