U.G.C. NET Exam. 24 June, 2019 Paper–II (COMPUTER SCIENCE & APPLICATIONS)

Total Questions: 100

11. Which type of addressing mod, less number of memory references are required?

Correct Answer: (c) Register
Solution:

(Dropped Question by NTA)
Immediate addressing mode- may or may not need memory reference
Implied addressing mode- may or may not need memory reference (Like push, pop or CMA instruction)
Register Addressing mode- It does not need memory reference as operands one always in register. Indexed addressing mode- need memory reference.

12. Suppose that the register A and the register K have the bit configuration. Only the three leftmost bits of A are compared with memory words because K has I's in these positions. Because of its organization, this type of memory is uniquely suited to parallel searches by data association. This type of memory is known as

Correct Answer: (c) content addressable memory
Solution:

The time required to find an item stored in memory can be reduced considerably if stored data is identified for access by the content of the data itself rather than by an address. Memory unit associated by content is called an associative memory or content addressable memory (CAM).

CAM is accessed simultaneously and in parallel on the basis of data content rather than specific location. When a word is written in CAM, no address is given. Memory is capable of finding an unused empty slot to store the word. When a word is to be read from memory, the content of the word is specified, and memory locates all words which matches the specified content and marks them for reading.

13. How many different Boolean functions of degree n are there?

Correct Answer: (a)
Solution:

14. The fault can be easily diagnosed in the microprogram control unit using diagnostic tools by maintaining the contents of

Correct Answer: (d) flags, registers and counters
Solution:

Flags, registers and counterswithin the computer, a control unit manages the computer's resources and orchestrates the performance of its functional parts in response to the instructions.

Microprogrammed control unit operates by executing instructions that define the functionality of the control unit. In this, set of micro instructions is stored in the control memory. The control address register contains the address of the next micro instruction to be read. When a micro instruction is read from the control memory, it is transferred to a control buffer register. Next element is sequencing unit that loads the control address register and issues a read command.

15. What will be the number of states when a MOD-2 counter is followed by a MOD-5 counter?

Correct Answer: (b) 10
Solution:

When mod- x counter is followed by mod-y counter then the resultant counter is always mod-(x * y) counter.
So simply multiply MOD value, answer will be 10.

16. For a magnetic disk with concentric circular tracks, the seek latency is not linearly proportional to the seek distance due to

Correct Answer: (b) arm starting or stopping inertia
Solution:

Whenever head moves from one track to other then its speed and direction changes. Which is nothing but change in motion or the case of inertia. The positioning time, or random-access time, consists of two parts : the time necessary to move the disk arm to the desired cylinder, called the seek time, and the time necessary for the desired sector to rotate to the disk head, called the rotational latency.

The seek letency is not linearly proportional to the seek distance due to arm starting and stopping inertia. It means seek time is not directly dependent on the seek distance traveled, but also on the accelerational deceleration (on inertia) of actuator arm.

17. The parallel bus arbitration technique uses an external priority encoder and a decoder. Suppose, a parallel arbiter has 5 bus arbiters. What will be the size of priority encoder and decoder respectively?

Correct Answer: (d) 8 × 3, 3 × 8
Solution:

18. Answer the following question

Correct Answer: (b) 7
Solution:

19. How many address lines and data lines are required to provide a memory capacity of 16k × 16?

Correct Answer: (c) 14, 16
Solution:

20. Suppose that a computer program takes 100 seconds of execution time on a computer with multiplication operation responsible for 80 seconds of this time. How much do you have to improve the speed of multiplication operation if you are asked to execute this program four times faster?

Correct Answer: (c) 16 times faster
Solution:

Total Execution time = time to execute multiplication operation + time for other operations. Calculation → Total execution time = 100 seconds Time to execute multiplication operation = 80 Seconds Then time to execute other operation = 100 – 80 = 20 Seconds. To execute the program four time faster, we have to improve the multiplication operation according to the question. To execute the program four times faster.

Total execution time will become = 100/4 = 25 Second New time to perform multiplication operation = new total execution time – time for other operations New time to perform multiplication operation = 25 – 20 = 5 Sec. So speed of multiplication operation is improved by = 80/5 = 16 times speed becomes 16 times fast.