Introduction
UGC NET or NTA-UGC-NET is the examination for determining the eligibility for the assistant professor and/or Junior Research Fellowship award in Indian universities and colleges. National Testing Agency examines on behalf of the University Grants Commission. In this article, we will cover the first 25 questions. We have covered questions 25 to 50 in the second part of UGC NET Dec 2010 PAPER-III Part-2.
Questions
1. The number of integers between 1 and 250 that are divisible by 2, 5, and 7 is
(A) 2
(B) 3
(C) 5
(D) 8
Answer: B
Explanation: n(A∪B∪C) = n(A) + n(B) + n(C) - n(A∩B)-n(B∩C) - n(C∩A) + n(A∩B∩C)
Calculation: Given 1 ≤ n ≤ 250
Let,
A: Integers divisible by 2
B: Integers divisible by 5
C: Integers divisible by 7
n(A) = number divisible by 2 = 250/2
n(B) = number divisible by 5 = 250/5
n(C)= number divisible by 7 = 250/7
n(A∩B) = number divisible by both 2 and 5 (i.e. 10) 250/10
n(B∩C) = number divisible by both 5 and 7 (i.e. 35) = 250/35
n(C∩A) = number divisible by both 7 and 2 (i.e. 14) 250/14
n(A∩B∩C) = number divisible by 2, 5 and 7 (i.e. 70) = 250/70
2. An undirected graph possesses an eulerian circuit if and only if it is connected and its vertices are
(A) all even degree
(B) all of odd degree
(C) of any degree
(D) even in number
Answer: A
Explanation: The Eulerian Cycle
If the first two requirements are met, an undirected graph has an Eulerian cycle
a) All non-zero degree vertices are connected. We don't worry about zero-degree vertices because they don't belong to the Eulerian Cycle or Path (we only consider all edges).
b) All vertices have the same degree.
3. A partially ordered set is said to be a lattice if every two elements in the set have
(A) a unique least upper bound
(B) a unique greatest lower bound
(C) both (A) and (B)
(D) none of the above
Answer: C
Explanation: A partially ordered set is said to be a lattice if it has at least two elements.
Distinct lowest upper bound
Unique lower bound
4. The minimum number of edges in a connected graph with ‘n’ vertices is equal to
(A) n(n–1)
(B) n(n–1)/2
(C) n2
(D) n–1
Answer: D
Explanation: For an undirected connected graph, the least number of edges is (n-1).
For the greatest number of edges (assuming simple graphs), each vertex is connected to all other vertices, yielding n(n-1)/2 edges (use handshaking lemma).
5. Consider the problem of connecting 19 lamps to a single electric outlet by using extension cords each of which has four outlets. The number of extension cords required is
(A) 4
(B) 5
(C) 6
(D) 7
Answer: C
Explanation: The first extension chord is linked to four further chords (which can connect 15 lamps + one slot free for another extension chord, which can connect four more lamps, for a total of 15+4 =19 lamps).
Maximum of 13 bulbs can be connected with 4 cords.
Maximum of 16 bulbs can be connected with 5 cords.
6. The decimal number equivalent of (4057.06)8 is
(A) 2095.75
(B) 2095.075
(C) 2095.937
(D) 2095.0937
Answer: D
Explanation: To begin, represent -73.75 in sign-magnitude form as shown below.
11001001.1100
=10110110.0011 in one's complement form (invert each bit except the sign bit, which is the most left bit)
=10110110.0100 in 2's complement form (adding 1 to one's complement's LSB)
7. AB+(A+B)’ is equivalent to
(A) A⊕B
(B) AʘB
(C) (A⊕B)ʘA
(D) (AʘB)⊕A
Answer: B
Explanation:
A B AB A+B (A+ B)’ AB +(A+ B)’
0 0 0 0 1 1
0 1 0 1 0 0
1 0 0 1 0 0
1 1 1 1 0 1
8. An astable multivibrator has
(A) one stable state
(B) two stable states
(C) no stable states
(D) none of these
Answer: C
Explanation: A free-running multivibrator has NO stable states and continuously alternates between two states, producing a train of square wave pulses at a given known frequency.
9. 12-bit 2’s complement of –73.75 is
(A) 01001001.1100
(B) 11001001.1100
(C) 10110110.0100
(D) 10110110.1100
Answer: C
Explanation: To begin, represent -73.75 in sign-magnitude form as shown below.
11001001.1100
=10110110.0011 in one's complement form (invert each bit except the sign bit which is the most left bit)
=10110110.0100 in 2's complement form (adding 1 to one's complement's LSB)
10. Encoding of data bits 0011 into 7-bit even Parity Hamming Code is
(A) 0011110
(B) 0101110
(C) 0010110
(D) 0011100
Answer: A
Explanation: D7 D6 D5 P4 D3 P2 P1 is the bit pattern.
P1 = 0 (verify even parity at 1,3,5,7 bits).
P2 = 1 (verify even parity at 2,3,6,7 bits)
P4 = 1 (verify even parity at 4,5,6,7 bits)
As a result, the final code is 0011110.
11. How many of the following declarations are correct?
int z = 7.0;
double void = 0.000;
short array [2] = {0, 1, 2};
char c = “\n”;
(A) None
(B) One is correct
(C) Two are correct
(D) All four are correct
Answer: C
Explanation: int z=7.0; - This is incorrect. The given data type is an integer, but the given number is a floating-point. It is an incorrect declaration. Incorrect:
double void = 0.000. Although void is a keyword, it cannot be used with variable declarations. It always returns a value of null.
short array[2] = [0, 1, 2] - Correct. It is in the proper format. We can specify static values right here.
char c = "n," - This is incorrect. It is a character data type. however, it is enclosed in double quotations. It should be a single quote rather than a double quote.
12. The value of the following expression (13/4*3)%5+1 is
(A) 5.75
(B) 2.95
(C) 1.4875
(D) 5
Answer: A
Explanation: Because the precedence of all of these operators is the same, they would execute from left to right.
13 / 4 = 3
3.25 * 3 = 9
9.75 % 5 = 4
4 + 1= 5
13. Which one of the following will set the value of y to 5 if x has the value 3, but not otherwise?
(A) if (x = 3) y = 5
(B) if x = = 3 (y = 5)
(C) if (x = = 3); y = 5
(D) if (x = = 3) y = 5
Answer: D
Explanation: Option A is ruled out because it is given as x=3, and = is an assignment operator rather than a comparison operator.
Option B is ruled out because braces are assigned to assignment rather than comparator.
Option C can also be ruled out because the presence of a ;(semicolon) at the conclusion of (x==3) elevates it to the level of an independent assertion.
14. Which one of the following sentences is true?
(A) The body of a while loop is executed at least once.
(B) The body of a do … while loop is executed at least once.
(C) The body of a do … while loop is executed zero or more times.
(D) A for loop can never be used in place of a while loop.
Answer: B
Explanation: That is the definition of a do-while loop.
15. “Black” in the “Black-box” testing means
(A) Characters of the movie “Black”
(B) I – O is hidden
(C) Design is hidden
(D) Users are hidden
Answer: C
Explanation: The term "black box testing" refers to testing in which the design is concealed. It is not necessary to have a specific understanding of the application's code, internal structure, or programming knowledge in general.
16. In generalization, the differences between members of an entity is
(A) maximized
(B) minimized
(C) both (A) & (B)
(D) None of these
Answer: A
Explanation: The process of defining a general entity type from a collection of specialized entity types is known as generalization.
Specialization is the inverse of generalization in that it specifies subtypes of the supertype and determines their relationship. Specialization is accomplished from the top down. The "Employee" in the following example could be a developer or a tester.
17. The dependency preservation decomposition is a property to decompose database schema D, in which each functional dependency X->Y specified in F,
(A) appeared directly in one of the relation schemas Ri in the decomposed D.
(B) could be inferred from dependencies that appear in some Ri.
(C) both (A) and (B)
(D) None of these
Answer: C
Explanation: The dependency preservation decomposition is a feature that is used to deconstruct database schema D, in which each functional dependence X->Y defined in F is preserved.
Directly appeared in one of the relation schemas R1, in the deconstructed D.
Could be deduced from dependencies found in some R1.
18. Which of the following is an optimistic concurrency control method?
(A) Validation based
(B) Timestamp ordering
(C) Lock-based
(D) None of these
Answer: A
Explanation:
Type Optimistic Pessimistic
Validation based Yes No
Timestamp ordering No Yes
Lock-based No Yes
19. Optical storage is a
(A) high-speed direct access storage device.
(B) low-speed direct access storage device.
(C) medium-speed direct access storage device.
(D) high-speed sequential access storage device.
Answer: C
Explanation: Optical storage is a type of direct access storage technology that operates at a low speed.
20. Which of the following is the process by which a user’s access to physical data in the application is limited, based on his privileges?
(A) Authorization
(B) Authentication
(C) Access Control
(D) All of these
Answer: A
Explanation: The process of determining a user's privileges is known as authorization.
- A smart card, chip card, or integrated circuit card (ICC) is a physical electronic authorization device that is used to regulate resource access.
21. What is the maximum number of nodes in a B-tree of order 10 of depth 3 (root at depth 0)?
(A) 111
(B) 999
(C) 9999
(D) None of the above
Answer: D
Explanation: At level 0, the maximum number of keys is nine.
At level 1, the maximum number of keys is 10*9 = 90. (10 nodes containing 9 keys each)
At level 2, the maximum number of keys is 10*10*9 = 900.
At level 3, the maximum number of keys is 10*10*10*9 = 9000.
Total = 9999 = 9000 + 900 + 90 + 9
22. A binary tree with 27 nodes has …………. null branches.
(A) 54
(B) 27
(C) 26
(D) None of the above
Answer: D
Explanation: Because there are n+1 null branches in a binary tree with n nodes. As a result, the answer must be 28.
23. The time complexity to build a heap of n elements is
(A) O(1)
(B) O(lgn)
(C) O(n)
(D) O(nlgn)
Answer: D
Explanation: Heapifying a single node requires O(log N) time complexity, where N is the total number of Nodes. As a result, constructing the full Heap will require N heapify operations, with total time complexity of O(N*logN).
24. Linear probing suffers from a problem known as
(A) Secondary clustering
(B) Primary clustering
(C) Both (A) and (B)
(D) None of these
Answer: B
Explanation: If there is a cluster and the initial location of a new record would fall anywhere in the cluster, the cluster size rises. This type of clustering is caused by linear probing.
Secondary clustering is less severe; two records share the same collision chain only if their initial positions are the same. Quadratic probing, for example, results in this form of clustering.
25. Which of the following can be the sequence of nodes examined in the binary search tree while searching for key 88?
(A) 90, 40, 65, 50, 88
(B) 90, 110, 80, 85, 88
(C) 190, 60, 90, 85, 88
(D) 65, 140, 80, 70, 88
Answer: C
Explanation: If the root node is 190, we must traverse the left sub-tree to get 88 because every element in the right sub-tree is bigger than 190.
Now since 88 is greater than 60, we go to the right and find 90, which is greater than 88, so we traverse the left sub-tree of 90. We discover that 85 is less than 88, therefore we proceed to the right and find our key 88.
39+ registered 
