Introduction
UGC NET Exam is one of the popular exams in India for people interested in research. Previous Year Questions are an excellent option to learn about the exam pattern. By solving the PYQs, you will get a basic idea about your preparation. You can evaluate your weak areas and work on them to perform better in the examination. In this article, we have given the questions of UGC NET 2016 July Paper-II. We have also explained every problem adequately to help you learn better.
Note: This article contains Q.No. 25 to Q.No. 50 out of the 75 questions asked in NET July 2016 paper II. The solution to Q.No. 01 to Q.No. 25 can be found in the July 2016 Paper-III Part-1 article and the solutions of Q.No. 51 to Q.No. 75 can be found in the July 2016 Paper III Part-3.
Questions 25 to 50
(1) Given the following statements:
(a) Frequency Division Multiplexing is a technique that can be applied when the bandwidth of a link is greater than combined bandwidth of signals to be transmitted.
(b) Wavelength Division Multiplexing (WDM) is an analog multiplexing Technique to combine optical signals.
(c) WDM is a Digital Multiplexing Technique.
(d) TDM is a Digital Multiplexing Technique.
Which of the following is correct?
(A) (a), (b), (c) and (d) are true.
(B) (a), (b), (c) and (d) are false.
(C) (a), (b) and (d) are false; (c) is true.
(D) (a), (b) and (d) are true; (c) is false.
Answer: D
Frequency Division Multiplexing is a technique that can be applied when the bandwidth of a link is greater than combined bandwidth of signals to be transmitted.
Wavelength Division Multiplexing (WDM) is an analog multiplexing Technique to combine optical signals.
TDM is a Digital Multiplexing Technique.
Only these statements are true.
(2) A pure ALOHA Network transmit 200 bit frames using a shared channel with 200 Kbps bandwidth. If the system (all stations put together) produces 500 frames per second, then the throughput of the system is ..............
(A) 0.384 (B) 0.184
(C) 0.286 (D) 0.586
Answer: B
Throughput(S) in pure aloha is G * e-2G:
It is maximum when G = 1 / 2.
i.e. S = 1 / 2 * (2.71)-2 * 1 / 2 =1 / 2 * (2.71)-1 = 0.184.
So, option (B) is correct.
(3) Match the following:
(a) Line coding (i) A technique to change analog signal to digital data.
(b) Block coding (ii) Provides synchronization without increasing number of bits.
(c) Scrambling (iii) Process of converting digital data to digital signal.
(d) Pulse code modulation (iv) Provides redundancy to ensure synchronization and inherits error detection.
Codes:
(a) (b) (c) (d)
(A) (iv) (iii) (ii) (i)
(B) (iii) (iv) (ii) (i)
(C) (i) (iii) (ii) (iv)
(D) (ii) (i) (iv) (iii)
Answer: B
Line coding is a Process of converting digital data to digital signal.
Block coding provides redundancy to ensure synchronization and inherits error detection.
Scrambling provides synchronization without increasing number of bits.
Pulse code modulation is a technique to change analog signal to digital data.
(4) Assume that we need to download text documents at the rate of 100 pages per minute. A page is an average of 24 lines with 80 characters in each line and each character requires 8 bits. Then the required bit rate of the channel is ................
(A) 1.636 Kbps (B) 1.636 Mbps
(C) 2.272 Mbps (D) 3.272 Kbps
Answer: Marks given to all
100 pages, each page contain 24 line, each line have 80 character and each character weighted 8 bit, we have to download 100 pages per
minute i.e. 100 * 24 * 80 * 8 / 60 = 25.6 kbps.
(5) Encrypt the plain text Message “EXTRANET” using Transposition cipher technique with the following key:
3 |
5 |
2 |
1 |
4 |
(Cipher text) |
1 |
2 |
3 |
4 |
5 |
(Plain text) |
Using ‘Z’ as bogus character.
(A) TAXERTZENZ (B) EXTRANETZZ
(C) EZXZTRZANZET (D) EXTZRANZETZ
Answer: A
According to given question and key, Z is a bogus character Encrypted message will be:
Plain Text:
| E | X | T | R | A | N | E | T | Z | Z |
| 1 | 2 | 3 | 4 | 5 | 1 | 2 | 3 | 4 | 5 |
Cipher Text:
| T | A | X | E | R | T | Z | E | N | Z |
| 3 | 5 | 2 | 1 | 4 | 3 | 5 | 2 | 1 | 4 |
(6) The number of different binary trees with 6 nodes is .............
(A) 6 (B) 42
(C) 132 (D) 256
Answer: C
The number of different binary trees with 6 nodes is fact(2n) / fact(n+1) * fac(n) where n is no nodes:
If n= 6, then fact(2 * n) / fact(n+1) * fac(n) = fact(2 * 6) / fact(6 + 1) * fact(6) = fact(12) / fact(7) * fact(6) = 12 * 11 * 10 * 9 * 8 * fact(7) / fact(7) * fact(6) = 12 * 11 * 10 * 9 * 8 / 6 * 5 * 4 * 3 * 2 = 6 * 11 * 2 = 132.
(7) Let A[1...n] be an array of n distinct numbers. If i<j and A[i]>A[j], then the pair (i,j) is called an inversion of A. What is the expected number of inversions in any permutation on n elements?
(A) θ(n) (B) θ(lg n)
(C) θ(nlg n) (D) θ(n2)
Answer: D
There are n(n-1)/2 pairs such that i < j. For a pair (ai, aj), probability of being inversion is 1/2. Therefore expected value of inversions = 1/2 * (n(n-1)/2) = n(n-1)/4.
(8) Which one of the following array represents a binary max-heap?
(A) [26, 13, 17, 14, 11, 9, 15]
(B) [26, 15, 14, 17, 11, 9, 13]
(C) [26, 15, 17, 14, 11, 9, 13]
(D) [26, 15, 13, 14, 11, 9, 17]
Answer: C
For max heap we will compare parent node(i) with its left-child(2 * i) and right-child(2 * i + 1):
In first option node(2) < node(5) which is violating the max-heap property.
In second option node node(2) < node(5) which is violating the max-heap property.
In third option there is no violation.
In fourth option node(3) < node(7) which is violating the max-heap property.
So, option (C) is correct.
(9) Match the following:
(a) Huffman codes (i) O(n2)
(b) Optimal polygon triangulation (ii) θ(n3)
(c) Activity selection problem (iii) O(nlgn)
(d) Quicksort (iv) θ(n)
Codes:
(a) (b) (c) (d)
(A) (i) (ii) (iv) (iii)
(B) (i) (iv) (ii) (iii)
(C) (iii) (ii) (iv) (i)
(D) (iii) (iv) (ii) (i)
Answer: C
Huffman codes takes O(nlgn) time.
Optimal polygon triangulation takes θ(n3) time
Activity selection problem takes θ(n) time
Quicksort takes O(n2) time
So, option (C) is correct.
(10) Suppose that we have numbers between 1 and 1000 in a binary search tree and want to search for the number 364. Which of the following sequences could not be the sequence of nodes examined?
(A) 925, 221, 912, 245, 899, 259, 363, 364
(B) 3, 400, 388, 220, 267, 383, 382, 279, 364
(C) 926, 203, 912, 241, 913, 246, 364
(D) 3, 253, 402, 399, 331, 345, 398, 364
Answer: C
We have to find 364 in BST:
In first option 925 is root node, our key is less then 925 so we go for left BST. Next node is 221 → 912 → 245 → 899 → 259 → 363 → 364 respectively.
In second option 3 is root node, we go for right BST i.e. 400 → 388 → 220 → 267 → 383 → 382 → 279 → 364 respectively.
In third option 926 is root node, we go for left BST i.e. 203 → 912 → 241 next key is 913 we cant go for 913 after 241 because we are already in left BST of 912 our key will be surely in left BST of 912. This option is incorrect.
In fourth option 3 is root node, we go for right BST i.e. 253 → 402 → 399 → 331 → 345 → 398 → 364.
So, option (C) is correct.
(11) A triangulation of a polygon is a set of T chords that divide the polygon into disjoint triangles. Every triangulation of n-vertex convex polygon has ................ chords and divides the polygon into ............... triangles.
(A) n-2, n-1 (B) n-3, n-2
(C) n-1, n (D) n-2, n-2
Answer: B
A triangulation of a polygon is a set of T chords that divide the polygon into disjoint triangles. Every triangulation of n-vertex convex polygon has (n-3) chords and divides the polygon into (n-2) triangles.
(12) Implicit return type of a class constructor is:
(A) not of class type itself (B) class type itself
(C) a destructor of class type (D) a destructor not of class type
Answer: B
Constructor doesn’t have any return type. But, implicit return type of a class constructor is class type itself.
(13) It is possible to define a class within a class termed as nested class. There are ............ types of nested classes.
(A) 2 (B) 3
(C) 4 (D) 5
Answer: A, C
It is possible to define a class within a class termed as nested class. There are two types of nested classes.
1 – Outer class will use argument of inner class.
2 – Inner and outer class are independent to each other.(different argument)
So, option (A) is correct.
(14) Which of the following statements is correct?
(A) Aggregation is a strong type of association between two classes with full ownership.
(B) Aggregation is a strong type of association between two classes with partial ownership.
(C) Aggregation is a weak type of association between two classes with partial ownership.
(D) Aggregation is a weak type of association between two classes with full ownership.
Answer: C
Aggregation is a weak type of association between two classes with partial ownership. All other statement are incorrect. So, option (C) is correct.
(15) Which of the following statements is correct?
(A) Every class containing abstract method must not be declared abstract.
(B) Abstract class cannot be directly initiated with ‘new’ operator.
(C) Abstract class cannot be initiated.
(D) Abstract class contains definition of implementation.
Answer: B,C
Every class containing abstract method must be declared abstract. Abstract class doesn’t contains definition of implementation Abstract class cannot be directly initiated with ‘new’ operator and it cannot be initiated. Statement (2) and (3) are correct. So, option (C) is correct.
(16) Which of the following statements is not correct?
(A) HTML is not screen precise formatting language.
(B) HTML does not specify a logic.
(C) DHTML is used for developing highly interactive web pages.
(D) HTML is a programming language.
Answer: D
Html is not a programming language, it is a hyper text markup language.
(17) When one object reference variable is assigned to another object reference variable then
(A) a copy of the object is created.
(B) a copy of the reference is created.
(C) a copy of the reference is not created.
(D) it is illegal to assign one object reference variable to another object reference variable.
Answer: B
When one object reference variable is assigned to another object reference variable then a copy of the reference is created.
(18) A server crashes on the average once in 30 days, that is, the Mean Time Between Failures (MTBF) is 30 days. When this happens, it takes 12 hours to reboot it, that is, the Mean Time to Repair (MTTR) is 12 hours. The availability of server with these reliability data values is approximately:
(A) 96.3% (B) 97.3%
(C) 98.3% (D) 99.3%
Answer: C
MTBF = 30 days = 30 * 24 = 720 hour.
MTTR = 12 hour.
Availability(A) = MTBF / (MTBF + MTTR)
i.e. A = 720 / (720 + 12) = 0.9836 = 98.36%.
So, option (C) is correct.
(19) Match the software maintenance activities in List-I to its meaning in List-II.
| List-I | List-II |
| I. Corrective |
(a) Concerned with performing activities to reduce the software complexity thereby improving program understandability and increasing software maintainability. |
| II. Adaptive | (b) Concerned with fixing errors that are observed when the software is in use. |
| III. Perfective |
(c) Concerned with the change in the software that takes place to make the software adaptable to new environment ( both hardware and software). |
| IV. Preventive | (d) Concerned with the change in the software that takes place to make the software adaptable to changing user requirements. |
Codes:
I II III IV
(A) (b) (d) (c) (a)
(B) (b) (c) (d) (a)
(C) (c) (b) (d) (a)
(D) (a) (d) (b) (c)
Answer: B
Corrective Concerned with fixing errors that are observed when the software is in use.
Concerned with the change in the software that takes place to make the software adaptable to new environment (both hardware and software).
Perfective Concerned with the change in the software that takes place to make the software adaptable to changing user requirements.
Preventive Concerned with performing activities to reduce the software complexity thereby improving program understandability and increasing software maintainability.
So, option (B) is correct.
(20) Match each application/software design concept in List-I to its definition in List-II.
| List-I | List-II |
| I. Coupling | (a) Easy to visually inspect the design of the software and understand its purpose. |
| II. Cohesion | (b) Easy to add functionality to a software without having to redesign it. |
| III. Scalable | (c) Focus of a code upon a single goal. |
| IV. Readable | (d) Reliance of a code module upon other code modules. |
Codes:
I II III IV
(A) (b) (a) (d) (c)
(B) (c) (d) (a) (b)
(C) (d) (c) (b) (a)
(D) (d) (a) (c) (b)
Answer: C
Coupling is reliance of a code module upon other code modules.
Cohesion is easy to visually inspect the design of the software and
understand its purpose.
Scalable is focus of a code upon a single goal.
Readable is easy to add functionality to a software without having to redesign it.
(21) Software safety is quality assurance activity that focuses on hazards that
(A) affect the reliability of a software component
(B) may cause an entire system to fail.
(C) may result from user input errors.
(D) prevent profitable marketing of the final product
Answer: B
Software safety is quality assurance activity that focuses on hazards that may cause an entire system to fail. It is not related to marketing or input errors or affection of reliability but if system fails all things related to system will be affected.
(22) Which of the following sets represent five stages defined by Capability Maturity
Model (CMM) in increasing order of maturity?
(A) Initial, Defined, Repeatable, Managed, Optimized.
(B) Initial, Repeatable, Defined, Managed, Optimized.
(C) Initial, Defined, Managed, Repeatable, Optimized.
(D) Initial, Repeatable, Managed, Defined, Optimized.
Answer: B
In increasing order of maturity Capability Maturity Model (CMM) have five stages: Initial → Repeatable → Defined → Managed → Optimized.
(23) The number of function points of a proposed system is calculated as 500. Suppose that the system is planned to be developed in Java and the LOC/FP ratio of Java is 50. Estimate the effort (E) required to complete the project using the effort formula of basic
COCOMO given below:
E = a(KLOC)b
Assume that the values of a and b are 2.5 and 1.0 respectively.
(A) 25 person months (B) 75 person months
(C) 62.5 person months (D) 72.5 person months
Answer: C
E = a(KLOC)b
a = 2.5 , b = 1.0
LOC / FP = 50
500 FP
So, 50 * 500 = 25000 loc
i.e 25 KLOC
E = 2.5(25)1 = 62.5 person months.
(24) In UNIX, processes that have finished execution but have not yet had their status collected are known as ................
(A) Sleeping processes (B) Stopped Processes
(C) Zombie Processes (D) Orphan Processes
Answer: C
In UNIX, processes that have finished execution but have not yet had their status collected are known as Zombie processes.
(25) In UNIX operating system, when a process creates a new process using the fork() system call, which of the following state is shared between the parent process and child process?
(A) Heap (B) Stack
(C) Shared memory segments (D) Both Heap and Stack
Answer: C
In the Unix operating system, when a process creates a new process using the fork() system call, then memory segments are shared between parent process and child process but stack and heap area are not shared between parent process and child process.
FAQs
What is the UGC NET exam?
UGC NET is a national-level exam organized by UGC to determine the eligibility of the candidates for lectureship and JRF.
What is the maximum number of attempts for the UGC NET examination?
There is no bar on the number of attempts of this examination. Candidates can appear for the examination as long as they are eligible.
How can solving PYQs help in my exam preparation?
Solving PYQs will give you a good idea about the exam pattern and help you identify your weak topics to prepare them better for the examination.
How many papers are there in the UGC NET exam?
There are two papers, and the candidates get 3 hours for both papers. There are 150 questions in UGC NET combining both papers.
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