NTA UGC NET Previous Year Questions (PYQ) Year-wise

Computer Science NTA UGC NET Previous Year Questions (PYQ). Previous Year Question Of NTA UGC NET/JRF

December-2018, NTA UGC NET/JRF Exam. Computer Science Questions

NTA UGC NET/JRF COMPUTER SCIENCE DECEMBER-2018 Queations

In this quiz, we have covered all the questions of NTA UGC NET / JRF year 2018. When you finished the quiz or clicked on the right/wrong answer, we also provide the correct answer with the explanation.

Consider the C/C++ function f() given below:

c question

Which of the following is the purpose of f()?

It outputs the contents of the array in reverse order.

It outputs the contents of the array with the characters rearranged so they are no longer recognized as the words in the original phrase.

It outputs the contents of the array in the original order.

It outputs the contents of the array with the characters shifted ever by one position.

Given an array arr[] and an integer c, we need to print k maximum elements of given array. The elements should printed in the order of the input.

Note: c is always less than or equal to w. It outputs the contents of the array in the original order.

Which of the following statements is/are true?

If auto increment addressing mode is included in an instruction set architecture, then an additional ALU is required for effective address calculation.

In auto increment addressing mode, the amount of increment depends on the size of the data item accessed

Auto increment addressing mode is useful in creating self-relocating code.

Suppose that everyone in a group of N people wants to communicate secretly with (N-1) other people using symmetric key cryptographic system. The communication between any two persons should not be decodable by the others in the group. The number of keys required in the system as a whole to satisfy the confidentiality requirement is

(N-1)^2

N(N-1)/2

N(N-1)

Consider the following pseudo-code fragment, where m is a non-negative integer that has been initialized :

nta ugc net 2018

Which of the following is a loop invariant for the while statement? (Note: a loop invariant for a while statement is an assertion that is true each time the guard is evaluated during the execution of the while statement).

p = 2k – 1 and 0 ≤ k ≤ m

p = 2k+1 – 1 and 0 ≤ k < m

p = 2k+1 – 1 and 0 ≤ k ≤ m

p = 2k – 1 and 0 ≤ k < m

p is surely 2k+1−12k+1−1 .. Because terms are added in this order - 1,2,4,8,16.. at any time their sum is next power of 2 minus 1. k should be less than m. If k=m, then corresponding p value will not hold.

The boolean expression Ā.B+A.Β¯+A.B is equivalent to

A.B

A + B

A′ . B + A. B′ + A. B

=A′ . B + A(B′ + B)

= A′ . B + A

= A + B

The grammar S → (S) | SS | is not suitable for predictive parsing because the grammar is

Ambiguous

An operator grammar

Right recursive

Left recursive

Since given grammar can have infinite parse trees for string ‘ε’, so grammar is ambiguous, and also S → (S) | SS | has left recursion. For predictive-parsing, grammar should be: • Free from ambiguity • Free from left recursion • Free from left factoring Given grammar contains both ambiguity and left factoring, so it cannot have predictive parser.

A process residing in Main Memory and Ready and Waiting for execution, is kept on

Ready Queue

Wait Queue

Execution Queue

Job Queue

The processes that are residing in main memory and are ready and waiting to execute are kept on a list called the ready queue.

A survey has been conducted on methods of commuter travel. Each respondent was asked to check Bus, Train or Automobile as a major method of travelling to work. More than one answer was permitted. The results reported were as follows: Bus 30 people; Train 35 people; Automobile 100 people; Bus and Train 15 people; Bus and Automobile 15 people, Train and Automobile 20 people; and all the tree methods 5 people. How many people completed the survey form?

160

115

120

165

Software coupling involves dependencies among pieces of software called modules.

Which of the following are correct statements with respect to module coupling?

P: Common coupling occurs when two modules share the same global data.
Q: Control coupling occurs when modules share a composite data structure and use only parts of it.
R : Content coupling occurs when one module modifies or relies on the internal working of another module.

Choose the correct answer from the code given below :

P and R only

Q and R only

P and Q only

All of P, Q and R

Common coupling– Common coupling is said to occur when several modules have access to the same global data. But it can lead to uncontrolled error propagation and unforeseen side-effects when changes are made.

Content coupling (high)– Content coupling is said to occur when one module uses the code of other module, for instance a branch. This violates information hiding - a basic design concept.

Control coupling– Control coupling is one module controlling the flow of another, by passing it information on what to do (e.g., passing a what-to-do flag) So, option b is right (P and R only)

Which one of the following is not typically provided by source code management software

Synchronisation

Syntax highlighting

Project forking

Versioning and Revision history

Consider a system with 2 level cache. Access times of Level 1 cache, Level 2 cache and main memory are 0.5 ns, 5 ns and 100 ns respectively. The hit rates of Level 1 and Level 2 caches are 0.7 and 0.8, respectively. What is the average access time of the system ignoring the search time within the cache?

24.35 ns

35.20 ns

20.75ns

7.55 ns

First, the system will look in cache 1. If it is not found in cache 1, then cache 2 and then further in main memory (if not in cache 2 also). The average access time would take into consideration success in cache 1, failure in cache 1 but success in cache 2, failure in both the caches and success in main memory. Average access time = [H1*T1]+[(1-H1)*H2*T2]+[(1 H1)(1-H2)*Hm*Tm] H 1 = Hit rate of level 1 cache = 0.7 T 1 = Access time for level 1 cache = 1 ns H2 = Hit rate of level 2 cache = 0.8 T2 = Access time for level 2 cache = 5 ns Hm = Hit rate of Main Memory = 1 Tm = Access time for Main Memory = 100 ns Average access time = 7.55ns

Dirty bit is used to show the

page with low frequency occurrence

wrong page

page with corrupted data

page that is modified after being loaded into cache memory

A dirty bit or modified bit is a bit that is associated with a block of computer memory and indicates whether or not the corresponding block of memory has been modified. The dirty bit is set when the processor writes to (modifies this memory. Dirty bits are used by the CPU cache and in the page replacement algorithms of an operating system.

Software products need perfective maintenance for which of the following resons?

When the customers need the product to run on new platforms.

To rectify bugs observed while the system is in use.

To overcome wear and tear caused by the repeated use of the software

To support the new features that users want it to support.

Perfective software maintenance addresses the functionality and usability of the software. Perfective maintenance involves changing existing product functionality by refining, deleting, or adding new features.

In PERT/CPM, the merge event represents ________ of two or more events.

joining

splitting

beginning

completion

In PERT/CPM, the merge event represents completion of two or more events.

In mathematical logic, which of the following are statements?

(i) There will be snow in January.
(ii) What is the time now?
(iii) Today is Sunday.
(iv) You must study Discrete Mathematics.

Choose the correct answer from the code given below :

Code :

(i) and (ii)

(iii) and (iv)

(ii) and (iv)

(i) and (iii)

The second smallest of n elements can be found with _______ comparisons in the worst case.

n – 1

n + ceil(lg n) - 2

3n/2

lg n

divide and conquer approach gives min element with n-1 comparison..
for 2nd minimum element search on d root to leaf path from where max element coming.. reason is second minimum element must compared with minimal element when we were finding minimum element.. so ,

divide and conquere based on binary tree approach so no of elements from root to leaf path is logn .. finding min from logn elements takes logn -1 comparison..
total n-1+logn-1 = n + logn -2 comparison

The definitions in a XML document are said to be ................... when the tagging system and definitions in the DTD are all in compliance.

well-formed

valid

logical

reasonable

---------- command is used to remove a relation from an SQL database

Remove table

Drop table

Delete table

Update table

Let r = a(a + b)*, s = aa*b and t = a*b be three regular expressions. Consider the following: (i) L(s) ⊆ L(r) and L(s) ⊆ L(t) (ii) L(r)⊆ L(s) and L(s) ⊆ L(t)

Choose the correct answer from the code given below :

Only (i) is correct.

Both (i) and (ii) are correct.

Only (ii) is correct.

Neither (i) nor (ii) is correct.

Two regular expressions are equivalent if languages generated by them are same. For example, r = a(a + b)* and s = aa*b generate same language. language L of all words of the form some positive number of a’s followed by exactly one b. s = aa*b and t = a*b. L(s) ⊆ L(r) and L(s) ⊆ L(t)

Your score is

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Which of the following statements is/are true?

Auto increment addressing mode is useful in creating self-relocating code.

If auto increment addressing mode is included in an instruction set architecture, then an additional ALU is required for effective address calculation.

In auto increment addressing mode, the amount of increment depends on the size of the data item accessed

(N-1)^2

N(N-1)

N(N-1)/2

Your score is

NTA UGC NET Previous Year Questions (PYQ)

December-2018, NTA UGC NET/JRF Exam. Computer Science Questions

June-2018, NTA UGC NET/JRF Exam. Computer Science Questions

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