GATE Computer Science Engineering(CSE) 2027 Test: Control Unit- 2 Free


MCQ Practice Test & Solutions: Test: Control Unit- 2 (15 Questions)

You can prepare effectively for Computer Science Engineering (CSE) GATE Computer Science Engineering(CSE) 2027 Mock Test Series with this dedicated MCQ Practice Test (available with solutions) on the important topic of "Test: Control Unit- 2". These 15 questions have been designed by the experts with the latest curriculum of Computer Science Engineering (CSE) 2026, to help you master the concept.

Test Highlights:

  • - Format: Multiple Choice Questions (MCQ)
  • - Duration: 45 minutes
  • - Number of Questions: 15

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Test: Control Unit- 2 - Question 1

Which set of instruction transfers the memory word specified by the effective address to AC or Load to AC?

Detailed Solution: Question 1

LDA AC load to AC instruction transfers the memory word specified by the effective address to AC
 
Here register MD represent MDR.

Test: Control Unit- 2 - Question 2

Which of the following is not involved in a memory write operation?

Detailed Solution: Question 2

Program counter register used to read the value or instruction not in write operation. For write operation MAR, IR and MDR registers are used.

Test: Control Unit- 2 - Question 3

In Flynn’s classification of computers, the vector and array classes of machines belong to

Test: Control Unit- 2 - Question 4

Following are some statements associated with microprocessors. Identify the false statement.

Detailed Solution: Question 4

Instruction received by the CPU is decoded by control unit not by arithmetic unit.

Test: Control Unit- 2 - Question 5

The following are four statements regarding what a CPU with only a set of 32 bit registers can perform.
1. Hold and operate on 32 bit integers.
2. Hold and operate on 16 bit integers.
3. Hold and operate on 64 bit floating point arithmetic.
4. Hold and operate on 16 bit UNICODE characters.

Which of the following is true about such a CPU?

Detailed Solution: Question 5

A CPU with only a set of 32 bit registers can perform:
1. Hold and operate on 32 bit integers.
2. Hold and operate on 16 bits integers.
3. Hold and operate on 16 bits Unicode characters.

Test: Control Unit- 2 - Question 6

The following are four statements about Reduced Instruction Set Computer (RISC) architectures.
1. The typical RISC machine instruction set is small, and is usually a subject of a CISC instruction set.
2. No arithmetic or logical instruction can refer to the memory directly.
3. A comparatively large number of user registers are available.
4. Instructions can be easily decoded through hard-wired control units.

Which of the above statements is true?

Detailed Solution: Question 6

Reduced instruction set uses comparatively large number of user registers, all arithmetic or logical instruction refer to memory via registers only, instructions can be easily decoded through hardwired control unit and is small instruction set and is usually a subject of a CISI instruction set

Test: Control Unit- 2 - Question 7

Consider an unpipelined processor assume that it has a 1 ns clock cycles and that it uses 4 cycles for ALU operation and branches and 5 cycles for memory operations. Assume that the relative frequencies of these operations are 40%, 20% and 40% respectively. Suppose due to clock skew and setup, pipelining the processor adds 0.2 ns of overhead to the clock, ignore any latency impact.

Q. What is the average instruction execution time for unpipelined processor?

Detailed Solution: Question 7

Average instruction execution time

Test: Control Unit- 2 - Question 8

Consider an unpipelined processor assume that it has a 1 ns clock cycles and that it uses 4 cycles for ALU operation and branches and 5 cycles for memory operations. Assume that the relative frequencies of these operations are 40%, 20% and 40% respectively. Suppose due to clock skew and setup, pipelining the processor adds 0.2 ns of overhead to the clock, ignore any latency impact.

Q. What speedup gain after pipelined the processor? 

Detailed Solution: Question 8

Average instruction time pipelined = 1 + 0.2 = 1.2 ns
Speedup from pipelined

Test: Control Unit- 2 - Question 9

The following are the some of the sequences of operations in instruction cycle, which one is correct sequence?

Detailed Solution: Question 9

1. PC → MAR
2. M[MAR] → MDR 
PC + 1 → PC
3. M[MDR] → IR
Represents the instruction fetch cycle

Test: Control Unit- 2 - Question 10

Consider the following situation and fill in the blanks:

The computer starts the tape moving by issuing a command: the processor then monitors the status of the tape by means of a _____. When the tape is in the correct position, the processor issues a _______.

Detailed Solution: Question 10

Control command: A control command is issued to activate the peripheral and to inform it what to do.
Status: A status command is used to test various status conditions in the interface and the peripheral.
Data output command causes the interface to respond by transferring data from the bus into one of its register.

Test: Control Unit- 2 - Question 11

Asynchronous data transfer between two independent units requires that control signals be transmitted between the communicating units to indicate the time at which data is being transm itted, way of achieving this ______.

Detailed Solution: Question 11

A synchronous data transfer between two independent units requires that control signals be transmitted between the communicating units to indicate the time at which data is being transmitted.
Two way of achieving this:
1. By means of a strobe pulse supplied by one of the units to indicate to the other unit when the transfer has to occur.

2. To accompany each data item being transferred with a control signal that indicates the presence of data in the bus. The unit receiving the data item responds with another control signal to acknowledge receipt of the data. This type of agreement between two independent units is referred to as handshaking. 

Test: Control Unit- 2 - Question 12

Determine the width of Micro-instruction having the following control-signal fields in a vertical microprogrammed control unit.

1. Next Address field of 7 bits

2. ALU Function field selecting 1 out of 13 ALU functions

3. Register-in field selecting 1 out of 8 registers

4. Register-out field selecting 1 out of 8 registers

5. Shifter field selecting no shift, right shift or left shift

6. Auxiliary control field of 4 bits

Detailed Solution: Question 12

Correct option: B (23 bits).

Next address = 7 bits.

ALU function needs to select 1 of 13 functions, so it requires 4 bits (since 23 = 8 < 13 ≤ 16 = 24).

Register-in (1 of 8 registers) requires 3 bits.

Register-out (1 of 8 registers) requires 3 bits.

Shifter (1 of 3 choices: no shift, right, left) requires 2 bits.

Auxiliary control = 4 bits.

Total width = 7 + 4 + 3 + 3 + 2 + 4 = 23 bits.

Answer: Option B.

Test: Control Unit- 2 - Question 13

Which statement is false in case of microprogram control?

Test: Control Unit- 2 - Question 14

The disadvantage of hard-wired control units with flip-flops is ______.

Detailed Solution: Question 14

The disadvantage of hardwired control units with flip-flops is that it requires more number of flip- flops.

Test: Control Unit- 2 - Question 15

A CPU has a cache with block size 64 bytes. The main memory has k blocks, each block being c bytes wide. Consecutive c-byte chunks are mapped on consecutive blocks with warp-around. All the k blocks can be accessed in parallel, but two accesses to the same block must be serialized. A cache block access may involve multiple iterations of parallel block accesses depending on the amount of data obtained by accessing all the k-blocks in parallel. Each iteration requires decoding the block numbers to be accessed in parallel and this takes k/2 ns.
The latency of one block access is 80 ns. If c = 2 and k = 24, then latency of retrieving a cache block starting at address zero from main memory is

Detailed Solution: Question 15

Cache block size = 64 bytes
Main memory has K banks or k = 24
Each bank is 2 byte long because c = 2
Total time for one parallel access

Total latency time = CT
= 2 x 92 = 184 ns

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