Software Programming : Computer Architecture QUESTIONS AND ANSWERS :: part1 : 1 to 5
Following Software Programming language Entrance Multiple choice objective type questions and answers will help you in Software Programming 2015 examinations :
1.Techniques that automatically move program and data blocks into the physical main memory when they are required for execution are called
|Main Memory techniques|
|Virtual Memory techniques|
|Cache Memory techniques|
ANSWER : 3 Virtual Memory techniques
2.In which type of flip-flop the indeterminate condition of the SR flip-flop (when S=R=1) is eliminated?
ANSWER : 2 JK flip-flop
Explanation : : To SR flip-flop two new connections from Q and Q� outputs back to original input gates eliminate the indeterminate condition.
3.What does T stands for in T flip-flop?
ANSWER : 3 Toggle
Explanation : : Toggle flip-flop as it changes its output on each clock edge.
4.What are the major components of a CPU?
|Control Unit, Register Set, Arithmetic Logic Unit|
|Control Unit, Memory Unit, Arithmetic Logic Unit|
|Memory Unit, Arithmetic Logic Unit, Auxiliary Memory|
|Register Set, Control Unit, Memory Unit|
|Register Set, Control Unit, Auxiliary Memory.|
ANSWER : 1 Control Unit, Register Set, Arithmetic Logic Unit
Explanation : : The major components of CPU are Control Unit, Register Set, and Arithmetic Logic Unit.
5.The one major advantage of CMOS is its, #2. The one major advantage of CMOS is its,
|Low propagation delay|
|Very low propagation delay|
|Very high propagation delay|
|1 Low propagation delay|
ANSWER : n : This means that it is not practical for use in systems requiring high-speed operations. The characteristic parameters for the CMOS gate depend on the power supply voltage VDD that is used. The power dissipation increases with increase in voltage supply. The propagation delay decreases with increase in voltage supply and the noise margin is estimated to be about 40% of the voltage supply value.
Explanation : memory contains how many words of 8 bits each? opt 1 65,536 opt 2 64,536 opt 3 65,436 opt 4 65,546 opt 5 65,556. Ans: 1 65,536 Reason : Consider the 20-bit logical address. The 4-bit segment number specifies one of 16 possible segments. The 8-bit page number can specify up to 256 pages, and the 8-bit word field implies a page size of 256 words. This configuration allows each segment to have any number of pages up to 256. the smallest possible segment will have one page of 256 words. The largest possible segment will have 256 pages, for a total of 256*256 = 65,536 which means 64K words.
More Computer Architecture QUESTIONS AND ANSWERS available in next pages
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