Computer Architecture - MCQs from AMIE exams (Summer 2019)

Choose the correct answer of the following (2 x 10)

1. In negative logic level 1 of the digital circuit is indicated by
(a) 15 v
(b) 10 v
(c) 5 v
(d) 0 v

2. The fetch cycle is
(a) first part of instruction cycle
(b) last part of instruction cycle
(c) intermediate part of instruction cycle
(d) None of the above

3. An arithmetic shift-right is equivalent to
(a) multiplying the number by 2
(b) dividing the number by 2
(c) changing the sign of the number
(d) reversing the number

4. Conversion of hexadecimal number 6B2 to its binary number equivalent is
(a) 111100011011
(b) 011011010101
(c) 011010110010
(d) 011011000010

5. What will be the binary equivalent of (24.1875)10?
(a) (11100.0011)
(b) (11001.0101)
(c) (10110.0011)
(d) None of the above

6. The part of machine level instruction, which tells the central processor what needs to be done is
(a) Operation code
(b) Address
(c) Operand
(d) None of the above

7. The output of the following circuit is


(a) A̅ B̅
(b) AB
(c) A + B
(d) B

8. The micro-programmed control unit is
(a) potentially faster than hardwired
(b) linked to the introduction of low cost and high-density memory chips
(c) much more flexible
(d) All of the above

9. The Storage capacity of the optical Disk is determined using
(a) Number or recording surfaces, number of tracks per surface, number of sectors per track
(b) Number of tracks on disk
(c) Number of sectors and bytes per sector
(d) Number of recording surfaces, number of tracks per surface, number of sectors per track and number of bytes per sector

10. Which of the following is not an addressing mode?
(a) Relative address mode
(b) Indexed register addressing mode
(c) Indexed addressing mode
(d) Base register addressing mode

Answers

1. (d) In negative logic system 0V is used for logic 1 and –ve voltage is used for logic 0.

2. (a) 

3. (b) The result of a Right Shift operation is a division by 2, where n is the number of shifted bit positions.

4. (c)  First, convert 6B216 into decimal, by using above steps:

= 6B216
6 × 162B × 1612 × 160
= 171410

Now, we have to convert 171410 to binary

1714 / 2 = 857 with remainder 0
857 / 2 = 428 with remainder 1
428 / 2 = 214 with remainder 0
214 / 2 = 107 with remainder 0
107 / 2 = 53 with remainder 1
53 / 2 = 26 with remainder 1
26 / 2 = 13 with remainder 0
13 / 2 = 6 with remainder 1
6 / 2 = 3 with remainder 0
3 / 2 = 1 with remainder 1
1 / 2 = 0 with remainder 1

Then just write down the remainders in the reverse order to get the answer, The hexadecimal number 6B2 converted to binary is therefore equal to: 11010110010

5. (dTo convert decimal number 24.1875, we convert its integer and fraction part individually and then add them to get the equivalent binary number, as below:
12 / 2 = 6 with remainder 0
6 / 2 = 3 with remainder 0
3 / 2 = 1 with remainder 1
1 / 2 = 0 with remainder 1
11000
0.375 × 2 = 0 + 0.75
0.75 × 2 = 1 + 0.5
0.5 × 2 = 1 + 0
0.0011
Therefore, decimal number 24.1875 converted to binary is equal: 11000.0011
To convert integer 24 to binary, follow these steps: 
Divide 24 by 2 keeping notice of the quotient and the remainder. Continue dividing the quotient by 2 until you get a quotient of zero.
Then just write out the remainders in the reverse order to get the equivalent binary number.
24 / 2 = 12 with remainder 0
Here is the answer to 24 decimal to binary number:
For converting decimal fraction 0.1875 to binary number, follow these steps:
Multiply 0.1875 by 2 keeping notice of the resulting integer and fractional part. Continue multiplying by 2 until you get a resulting fractional part equal to zero (we calcuclate upto ten digits).
Then just write out the integer parts from the results of each multiplication to get equivalent binary number.
0.1875 × 2 = 0 + 0.375
Here is the answer to 0.1875 decimal to binary number:
0.0011
Therefore, decimal number 24.1875 converted to binary is equal: 11000.0011

6. (a)

7. (c) 

8. (c) This is slower than the other as micro instructions are used for generating signals. Easy to modify as the modification need to be done only at the instruction level.

9. (d) Storage capacity of a disk system= number of recording surfaces x number of tracks per surface  x Number of sectors per track x Number of bytes per sector.

10. (a)
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