Quiz -1 : MES -18cs44
Modules : 1 and 2
No. Of questions: 20
Total Marks :20
Duration : 30 Mins
- 1.
ARM stands for _________
- A.
Advanced RISC Machine
- B.
Advanced RISC Memory system
- C.
Advanced Reduced Machine
- D.
Advanced and Rebulilt Machine
Correct Answer
A. Advanced RISC MachineExplanation
ARM stands for Advanced RISC Machine. RISC stands for Reduced Instruction Set Computer, which is a type of computer architecture that uses a simplified set of instructions to improve performance. ARM processors are commonly used in mobile devices and embedded systems due to their low power consumption and high performance capabilities. The acronym "Advanced" in ARM signifies the advanced features and capabilities of the ARM architecture.Rate this question:
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- 2.
ARM supports execution of Java Bytecodes.
- A.
True
- B.
False
Correct Answer
A. TrueExplanation
ARM processors are designed to support the execution of Java Bytecodes. This means that ARM-based devices, such as smartphones and tablets, can run Java applications efficiently. Java Bytecodes are platform-independent instructions that can be interpreted or compiled by a Java Virtual Machine (JVM). ARM processors are widely used in mobile devices due to their energy efficiency and performance, making them a suitable choice for running Java applications. Therefore, the statement "ARM supports execution of Java Bytecodes" is true.Rate this question:
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- 3.
The No.of Pipelines stages in ARM7, ARM9 and ARM10 respectively are
- A.
3,6,5
- B.
4,6,5
- C.
3,,5,6
- D.
4,6,8
Correct Answer
C. 3,,5,6Explanation
The correct answer is 3, 5, 6. This means that the ARM7 processor has 3 pipeline stages, the ARM9 processor has 5 pipeline stages, and the ARM10 processor has 6 pipeline stages. The number of pipeline stages in a processor refers to the number of steps or phases that a single instruction goes through during execution. Having more pipeline stages can increase the efficiency and performance of the processor as it allows for better instruction parallelism and overlap of different stages of instruction execution.Rate this question:
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- 4.
Find odd one out
- A.
Abort
- B.
Supervisor
- C.
User
- D.
Undefined
Correct Answer
C. UserExplanation
The odd one out in this list is "User" because it is the only term that is related to a person or individual, while the other terms ("abort," "supervisor," and "undefined") are more general concepts or technical terms.Rate this question:
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- 5.
The conditional flag "V" in ARM refers to
- A.
Overflow
- B.
Validation error
- C.
Violated Memory access
- D.
Variable access error
Correct Answer
A. OverflowExplanation
The conditional flag "V" in ARM refers to overflow. This flag is set when an arithmetic operation results in a value that is too large to be represented using the available number of bits. This can occur when adding or subtracting signed numbers, and indicates that the result of the operation is not valid. The "V" flag is used in conditional branching instructions to determine whether to execute certain instructions based on the outcome of the previous arithmetic operation.Rate this question:
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- 6.
Answer the following
- A.
A
- B.
B
- C.
C
- D.
D
-
- 7.
Answer the following
- A.
A
- B.
B
- C.
C
- D.
D
Correct Answer
C. C -
- 8.
Answer the following
- A.
A
- B.
B
- C.
C
- D.
D
Correct Answer
C. C -
- 9.
Answer the following
- A.
A
- B.
B
- C.
C
- D.
D
Correct Answer
C. C -
- 10.
Answer the following
- A.
10020021
- B.
000000000
- C.
F000000
- D.
0000000F
Correct Answer
B. 000000000 -
- 11.
Answer the following
- A.
1024
- B.
1004
- C.
1028
- D.
1008
Correct Answer
B. 1004 -
- 12.
Answer the following
- A.
A
- B.
B
- C.
C
- D.
D
Correct Answer(s)
A. A
B. B -
- 13.
Find odd one out
- A.
CMP
- B.
CMN
- C.
BIC
- D.
TST
Correct Answer
C. BICExplanation
The odd one out in this sequence is BIC because it does not follow the pattern of starting with the letters "CM" like the other options.Rate this question:
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- 14.
What is the value of r0,r1 and r2 after execution of following code segment mov r0, #04 mov r1, #02 add r2, r1, r0, lsl #01 and r2, r2, #0F or r2, r2, #0F teq r1,r2
- A.
4,2,8
- B.
8,2,0F
- C.
4,2,0F
- D.
8,2, 0A
Correct Answer
C. 4,2,0FExplanation
The code segment first moves the value 4 into register r0 and the value 2 into register r1. Then it adds the values of r1 and r0 shifted left by 1 (multiplied by 2) and stores the result in register r2. The result of the addition is 8. Then it performs a bitwise AND operation between r2 and the hexadecimal value 0F, which sets all bits of r2 except the lower 4 bits to 0. This results in r2 having the value 8. Finally, it performs a bitwise OR operation between r2 and the hexadecimal value 0F, which sets all the lower 4 bits of r2 to 1. This results in r2 having the value 0F, which is 15 in decimal. Therefore, the values of r0, r1, and r2 after the execution of the code segment are 4, 2, and 0F respectively.Rate this question:
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- 15.
AREA Directive – specifies chunks of data or code that are manipulated by the linker.
- A.
True
- B.
False
Correct Answer
A. TrueExplanation
The AREA directive is used to specify chunks of data or code that are manipulated by the linker. It helps in organizing and managing the memory allocation for different sections of a program. By using the AREA directive, the programmer can define specific areas in the memory where data or code should be placed. This allows for efficient memory management and ensures that the linker can properly resolve references and generate the final executable file. Therefore, the statement "True" is correct as it accurately describes the purpose and functionality of the AREA directive.Rate this question:
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- 16.
LDR R0, [R1,#4]! The given instruction is an example of __________
- A.
Preindex offset
- B.
Preindex with writeback
- C.
Relative index
- D.
Postindex
Correct Answer
B. Preindex with writebackExplanation
The given instruction "LDR R0, [R1,#4]!" is an example of Preindex with writeback. This is because the instruction loads the value at the memory address calculated by adding the value of register R1 and the immediate offset of 4, and stores it in register R0. The "!" symbol indicates that the value of R1 is updated after the memory access, making it a preindex with writeback operation.Rate this question:
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- 17.
The additional duplicate register used in ARM machines are called as _______
- A.
Copied-registers
- B.
Banked registers
- C.
EXtra registers
- D.
Extential registers
Correct Answer
B. Banked registersExplanation
Banked registers are additional duplicate registers used in ARM machines. These registers are used to store different states of the processor when switching between different execution modes or privilege levels. By using banked registers, the processor can quickly switch between modes without the need to save and restore the entire register set. This helps in improving the efficiency and performance of the ARM machine.Rate this question:
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- 18.
Each instruction in ARM machines is encoded into __________ Word.
- A.
2 byte
- B.
4 byte
- C.
8 byte
- D.
3 byte
Correct Answer
B. 4 byteExplanation
Each instruction in ARM machines is encoded into 4 bytes. This is because the ARM architecture uses a fixed instruction length of 32 bits, which is equivalent to 4 bytes. This allows for efficient instruction decoding and execution in ARM processors.Rate this question:
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- 19.
The addressing mode where the EA of the operand is the contents of Rn is ______
- A.
Pre-indexed mode
- B.
Pre-indexed with write back mode
- C.
Post-indexed mode
- D.
None of the mentioned
Correct Answer
C. Post-indexed modeExplanation
Post-indexed mode is an addressing mode where the effective address (EA) of the operand is calculated using the contents of register Rn after the instruction is executed. In this mode, the value in Rn is added to the base address to determine the EA. This mode is useful when the address calculation depends on the result of the instruction.Rate this question:
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- 20.
The effective address of the instruction written in Post-indexed mode, MOVE[Rn]+Rm is _______
- A.
EA = [Rn]
- B.
EA = [Rn + Rm]
- C.
EA = [Rn] + Rm
- D.
EA = [Rm] + Rn
Correct Answer
A. EA = [Rn]Explanation
The effective address of the instruction written in Post-indexed mode, MOVE[Rn]+Rm is the value stored in the register Rn.Rate this question:
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Current Version
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Mar 19, 2023Quiz Edited by
ProProfs Editorial Team -
Apr 01, 2020Quiz Created by
VIJAY KUMAR
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