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Jan 2017  Paper II - Part 1

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Introduction

In India, the UGC NET Exam is a very popular exam for people who want to pursue a career in research. Previous Year Questions are a great way to get a feel for the exam format. You will gain a fundamental understanding of your preparedness by completing the PYQs. You can assess your weak areas and practice on them to improve your exam performance. We have included the UGC NET 2017 January Paper-II questions in this blog. We've also thoroughly detailed each problem to aid your learning.

This article covers the first 25 questions from the January Paper-II, the next questions are provided in its next part Jan 2017  Paper II - Part 2

So, let’s start: 

Questions 1 to 25 

1Consider a sequence F00 defined as:

Then what shall be the set of values of the sequence F00?

(1) (1, 110, 1200)

(2) (1, 110, 600, 1200)

(3) (1, 2, 55, 110, 600, 1200)

(4) (1, 55, 110, 600, 1200)

Answer: 1

Explanation:

We have given, F00(0) = 1, F00(1) = 1

F00(2) = (10*F00(1) + 100)/F00(0) = 110

F00(3) = (10*F00(2) + 100)/F00(1) = 1200

F00(4) = (10*F00(3) + 100)/F00(2) = 110

F00(5) = (10*F00(4) + 100)/F00(3) = 1

Since the values repeat after the first three values, the set of values of F00 will be (1,110,1200).

2. Match the following:

List-I                                     List-II

a. Absurd                             i. Clearly impossible being

                            contrary to some evident truth.

b. Ambiguous                      ii. Capable of more than one 

                            interpretation or meaning.

c. Axiom                               iii. An assertion that is accepted

                            and used without proof.

d. Conjecture                       iv. An opinion Preferably based

                             on some experience or wisdom.

Codes:

     a   b   c    d

(1) i    ii   iii   iv

(2) i    iii  iv   ii

(3) ii   iii  iv   i

(4) ii   i    iii   iv

Answer: 1

3.The functions mapping R into R are defined as:

f(x) = x3- 4x, g(x)=1/(x2+1) and h(x)=x4

Then find the value of the following composite functions:

hog(x) and hogof(x)

(1) (x2+1)4 and [(x3-4x)2+1]4

(2) (x2+1)4 and [(x3-4x)2+1]- 4

(3) (x2+1)- 4 and [(x3-4x)2+1]4

(4) (x2+1)‑ 4 and [(x3-4x)2+1]- 4

Answer: 4

Explanation:

f(x) = x3 – 4x, g(x) = 1/(x2 + 1) and h(x) = x4

 hog(x) = [g(x)]4

        =  [1/(x2 + 1)]4

        =  [(x2 + 1)-1]4

        =  [(x2 + 1)]-4

Similarly:

hogof(x) = [gof(x)]4

         = [(f(x)2 + 1)]-4

         =  [((x3 – 4x)2 + 1)]-4

4. How many multiples of 6 are there between the following pairs of numbers?

0 and 100 and -6 and 34

(1) 16 and 6

(2) 17 and 6

(3) 17 and 7

(4) 16 and 7

Answer: 3

Explanation:

Between 0 and 100 multiple of 6 are: 0, 6, 12, 18, 24, 30, 36, 42, 48, 54, 60, 66, 72, 78, 84, 90, 96 i.e. 17 multiples.

Between -6 and 34 multiple of 6 are: -6, 0, 6, 12, 18, 24, 30. ie. 7 multiples.


5. Consider a Hamiltonian Graph G with no loops or parallel edges and with |V(G)|=n≥3. Then which of the following is true?

(1) deg(v) ≥ n/2 for each vertex of Graph G.

(2) |E(G)| ≥ 1/2(n-1)(n-2)+2 edges

(3) deg(v)+deg(w) ≥ n whenever v and w are not connected by an edge.

(4) 1 and 3

Answer: 4

Explanation:

In an Hamiltonian Graph (G) with no loops and parallel edges:

According to Dirac’s theorem, in a n vertex graph, deg (v) ≥ n / 2 for each vertex of G.

According to Ore’s theorem deg (v) + deg (w) ≥ n for every n and v not connected by an edge is sufficient condition for a graph to be hamiltonian.

If |E(G)| ≥ 1 / 2 * [(n – 1) (n – 2)] then the graph is connected but it doesn’t guaranteed to be a Hamiltonian Graph.

6.In propositional logic if (P→Q)˄(R→S) and (P˅R) are two premises such that

Y is the premise:

(1) P˅R

(2) P˅S

(3) Q˅R

(4) Q˅S

Answer: 4

Explanation:

(P→Q)   = ~PVQ

(R→S)  = ~RVS

(P˅R) 

There will be a Resolution (rule of inference ) between these premises to give a conclusion  

~ P & P,  R & R' will resolve out and then we construct the disjunction of the remaining clauses.

7.ECL is the fastest of all logic families. High Speed in ECL is possible because transistors are used in difference amplifier configuration, in which they are never driven into ...............

(1) Race condition

(2) Saturation

(3) Delay

(4) High impedance

Answer: 2

Explanation:

ECL is the fastest of the three because the transistors are employed in a difference amplifier configuration, which means they are never driven to saturation, obviating the need for storage time.

ECL is the digital logic family with the shortest propagation delay time (the shortest propagation delay time is feasible in ECL due to the usage of transistors difference amplifier design, in which they are never driven to saturation, obviating the need for storage).

Non-saturated Logic: In Non-saturated Logic, the transistors are not driven into saturation.

(i) Schottky TTL

(ii) Emitter Coupled Logic (ECL)

8.A binary 3-bit down counter uses J-K flip-flops, FFi with inputs Ji, Ki and outputs Qi, i=0,1,2 respectively. The minimized expression for the input from following, is

I. J0=K0=0

II. J0=K0=1

III. J1=K1=Q0

IV. J1=K1=Q'0

V. J2=K2=Q1Q0

Vl. J2=K2=Q'1Q'0

(1) I, Ill, V

(2) I, IV, VI

(3) Il, III, V

(4) Il, IV, Vl

Answer: 4

Explanation: 

Up counter: 0 to 7 

Down counter: 7 to 0

9.Convert the octal number 0.4051 into its equivalent decimal number.

(1) 0.5100098

(2) 0.2096

(3) 0.52

(4) 0.4192

Answer: 1

Explanation:

(0.4051)8 = 4x8-1+0x8-2+5x8-3+1x8-4

= 0.5100098

10. The hexadecimal equivalent of the octal number 2357 is:

(1) 2EE

(2) 2FF

(3) 4EF

(4) 4FE

Answer: 3

Explanation:

(2357)8 can be converted into binary just digit by digit.

= 010 011 101 111

Now we can regroup the bits into groups of 4 and convert them to hexadecimal.

= 0100 1110 1111

= 4EF

11. Which of the following cannot be passed to a function in C++?

(1) Constant

(2) Structure

(3) Array

(4) Header file

Answer: 4

Explanation:

Constants, Structures, and Arrays can be easily passed to a function in c++. Also, there is no significance in passing a Header file to a function. 

12. Which one of the following is correct for overloaded functions in C++?

(1) Compiler sets up a separate function for every definition of function.

(2) Compiler does not set up a separate function for every definition of function.

(3) Overloaded functions cannot handle different types of objects.

(4) Overloaded functions cannot have same number of arguments.

Answer: 1

Explanation:

Compiler sets up a separate function for every definition of function irrespective of its type.

13. Which of the following storage classes have global visibility in C/C++?

(1) Auto

(2) Extern

(3) Static

(4) Register

Answer: 2

Explanation:

The extern storage class is used to give a reference of a global variable that is visible to all the program files.

14. Which of the following operators cannot be overloaded in C/C++?

(1) Bitwise right shift assignment

(2) Address of

(3) Indirection

(4) Structure reference

Answer: 4

Explanation:

Structure reference cannot be overloaded in C/C++.

15. If X is a binary number which is power of 2, then the value of

X&(X-1) will be:

(1) 11....11

(2) 00.....00

(3) 100.....0

(4) 000.....1

Answer: 2

Explanation:

& is a bit wise AND operator.

let X = 2 ^ 4 = 16 =10000

then X - 1 = 15 = 01111

now X & (X-1) = 00000

16. An attribute A of datatype varchar(20) has value 'Ram' and attribute B of datatype char(20) has value 'Sita' in oracle. The attribute A has .......... memory spaces and B has .......... memory spaces.

(1) 20, 20

(2) 3, 20

(3) 3, 4

(4) 20, 4

Answer: 2

Explanation:
varchar will acquire the exact memory of attribute and it varies from tuple to tuple while char will acquire memory space which is define at the time of table creation it is fixed: varchar(20) ‘Ram’ will take 3 and ‘Sita’ will take 20 character space in memory.

17. Integrity constraints ensure that changes made to the database by authorized users do not result into loss of data consistency. Which of the following statement(s) is (are) true w.r.t. the examples of integrity constraints?

(A) An instructor Id.No. cannot be null, provided Instructor Id. No. being primary key.

(B) No two citizens have same Adhar-Id.

(C) Budget of a company must be zero.

(1) (A), (B) and (C) are true.

(2) (A) false, (B) and (C) are true.

(3) (A) and (B) are true; (C) false.

(4) (A), (B) and (C) are false.

Answer: 3

Explanation:

(A) An instructor Id. No. cannot be null, provided Instructor Id No. being primary key. Correct by Codd’s rule

(B) No two citizens have same Adhar-Id. Correct because Adhar is identification for citizens so it must be unique.

(C) Budget of a company must be zero. We can't say or it is not necessarily true

Hence, option 3 is correct. 

18. Let M and N be two entities in an E-R diagram with simple single value attributes. R1 and R2 are two relationships between M and N, whereas

R1 is one-to-many and R2 is many-to-many.

The minimum number of tables required to represent M, N, R1 and R2 in the relational model are ..........

(1) 4

(2) 6

(3) 7

(4) 3

Answer: 4

Explanation:

Two tables are created for entities M and N  and a third table is created for relationships R2 (many to many)

for R1 no need for separate table (one to many relationships)

The primary key from the `one side` is placed as a foreign key on the `many sides`. (m-n relationships)

A new relationship is created with the primary keys from each entity forming a composite key. (1-1 relationships)

Depending on the optionality of the relationship, the entities are either combined or the primary key of one entity type is placed as a foreign key in the other relation.

19. Consider a schema R(MNPQ) and functional dependencies M→N, P→Q. Then the decomposition of R into R1(MN) and R2(PQ) is .............

(1) Dependency preserving but not lossless join.

(2) Dependency preserving and lossless join

(3) Lossless join but not dependency preserving

(4) Neither dependency preserving nor lossless join.

Answer: 1

Explanation:

Schema R(MNPQ) is decomposed into R1(MN) M → N is preserved and R2(PQ) P → Q is also preserved, dependency will be preserved and there will be no loss of any dependency.

20. The order of a leaf node in a B+ tree is the maximum number of children it can have. Suppose that block size is 1 kilobyte, the child pointer takes 7 bytes long, and the search field value takes 14 bytes long. The order of the leaf node is ............

(1) 16

(2) 63

(3) 64

(4) 65

Answer: 1

Explanation:

Key size = 14 bytes (given)

Child pointer = 7 bytes (given)

We assume the order of B+ tree to be ‘n’.

Block size >= (n – 1) * key size + n * child pointer

512 >= (n – 1) * 14 + n * 7

512 >= 14 * n – 14 + 7 * n 

n <= (1024 + 14) / 20

n <= 1038 / 21

n <= 49.42

21. Which of the following is true for computation time in insertion, deletion, and finding maximum and minimum elements in a sorted array?

(1) Insertion-O(1), Deletion-O(1), Maximum-O(1), Minimum-O(1)

(2) Insertion-O(1), Deletion-O(1), Maximum-O(n), Minimum-O(n)

(3) Insertion-O(n), Deletion-O(n), Maximum-O(1), Minimum-O(1)

(4) Insertion-O(n), Deletion-O(n), Maximum-O(n), Minimum-O(n)

Answer: 3

Explanation:

In a sorted array, if we want to insert or delete then we have to traverse the whole array and check where is the suitable position, so it will take

O(n).

If the array is sorted then the end position will tell the maximum or minimum, so finding the maximum or minimum will take O(1).

22. The seven elements A, B, C, D, E, F, and G are pushed onto a stack in reverse order, i.e., starting from G. The stack is popped five times and each element is inserted into a queue. Two elements are deleted from the queue and pushed back onto the stack. Now, one element is popped from the stack. The popped item is ...................

(1) A

(2) B

(3) F

(4) G

Answer: 2

Explanation: 

Elements are inserted into a stack then the 5 elements are popped and these 5 elements are inserted into a queue, now first two elements are deleted from the queue and pushed into the stack one by one. At top of the stack, element B is presented.

Hence the answer is 2. 

23. Which of the following is a valid heap?

(1) A

(2) B

(3) C

(4) D

Answer: 2

Explanation:

Heap has to be a max heap or min heap to be valid. In option B a max heap is given hence it is a valid heap. 

24. If h is chosen from a universal collection of hash functions and is used to hash n keys into a table of size m, where n ≤ m, the expected number of collisions involving a particular key x is less than ...................

(1) 1

(2) 1/n

(3) 1/m

(4) n/m

Answer: 1

Explanation: 

Universal hashing states that:- If hi s chosen from a universal collection of hash functions and is used to hash n keys into a table of size m, where n ≤ m, the expected number of collisions involving a particular key x is less than 1.

25. Which of the following statements is false?

(A) Optimal binary search tree construction can be performed efficiently using dynamic programming.

(B) Breadth-first search cannot be used to find connected components of a graph.

(C) Given the prefix and postfix walks of a binary tree, the tree cannot be reconstructed uniquely.

(D) Depth-first-search can be used to find the components of a graph.

(1) A

(2) B

(3) C

(4) D

Answer: 2

Explanation:

(A) Optimal binary search tree construction can be performed efficiently using dynamic programming.

(B) Breadth-first search cannot be used to find connected components of a graph.

(C) Given the prefix and postfix walks of a binary tree, the tree cannot be reconstructed uniquely.

(D) Depth-first-search can be used to find the connected components of a graph.

All statements are correct except (B) because BFS can be used to check the connectivity of graphs.

Answer 2 best suits the question. 

Frequently Asked Questions

  1. What is the purpose of the UGC NET Exam?
    UGC NET is an abbreviation for University Grants Commission National Eligibility Test. It is a national-level test used to evaluate eligibility for lectureships and Junior Research Fellowships (JRF) in Indian institutions and colleges.
     
  2. Can I apply for UGC NET after completing a PG diploma?
    A PGDM is equivalent to a master's degree. So, if you hold a PG Diploma from a recognized university, you are eligible to apply for the exam.
     
  3. What is the minimum age for applying to the UGC NET Exam?
    According to the UGC NET Eligibility, there is no age limit for assistant professors, but candidates for JRF must be at least 31 years old.
     
  4. Will distance learning be accepted for the UGC NET?
    Yes, distance education is acceptable for the UGC NET Exam, provided you are enrolled in a UGC accredited university/institute.
     
  5. What is the UGC NET syllabus?
    The NTA administers the UGC NET Exam in two parts: General Paper-1 (common to all topics) and Paper-2 (subject-specific). The syllabus for each Paper 2 subject varies. You can view the entire UGC NET Syllabus.
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Conclusion

We have extensively discussed the January 2017 paper-II. We hope that this blog has helped you understand the UGC pattern. Hope you learned something. But the knowledge never stops. Also, you can get the next 25 questions with their answer and explanation in the second part. You can visit it from here  Jan 2017  Paper II - Part 2

So to learn more you can visit our website for more articles. Check out our articles on June 2013 Paper II - Part 1Dec 2013  Paper III - Part 1, and Dec 2013  Paper III - Part 2.

Refer to our guided paths on Coding Ninjas Studio to learn more about DSA, Competitive Programming, JavaScript, System Design, etc. Enroll in our courses and refer to the mock test and problems available, Take a look at the interview experiences and interview bundle for placement preparations.

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Happy Learning Ninja :) 

Topics covered
1.
Introduction
1.1.
Questions 1 to 25 
2.
Frequently Asked Questions
3.
Conclusion