Previous Year Paper · Questions with answers

UGC NET Computer Science PYQ, June 2016 (Paper II & III, Selected Questions)

All 80 questions of UGC NET Computer Science PYQ, June 2016 (Paper II & III, Selected Questions) with the correct answer from the official answer key and a solution. Read them free on UTpaper, or attempt the paper as a timed online test.

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Paper II 39 questions

Q1UGC NET Computer Science, June 2016 (Paper II & III, Selected Questions)
Which of the following logic expressions is incorrect
  1. A1 ⊕ 0 = 1
  2. B1 ⊕ 1 ⊕ 1 = 1
  3. C1 ⊕ 1 ⊕ 0 = 1
  4. D1 ⊕ 1 = 0
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Correct answer: (C) 1 ⊕ 1 ⊕ 0 = 1

Explanation

With the EX-OR operation, an odd number of 1's returns 1 and an even number of 1's returns 0. XOR gives 1 only when the inputs are different (A ⊕ B = A'B + AB'). Here 1 ⊕ 1 ⊕ 0 has two 1's, so it equals 0, not 1. Therefore the incorrect expression is (c) 1 ⊕ 1 ⊕ 0 = 1.

Q2UGC NET Computer Science, June 2016 (Paper II & III, Selected Questions)
Simplified Boolean equation for the following truth table is: x y z | F 0 0 0 | 0 0 0 1 | 1 0 1 0 | 0 0 1 1 | 1 1 0 0 | 1 1 0 1 | 0 1 1 0 | 1 1 1 1 | 0
  1. AF = yz' + y'z
  2. BF = xy' + x'y
  3. CF = x'z + xz'
  4. DF = x'z + xz' + xyz
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Correct answer: (C) F = x'z + xz'

Explanation

F = x'y'z + x'yz + xy'z' + xyz' = x'z(y' + y) + xz'(y' + y) = x'z + xz'

Q3UGC NET Computer Science, June 2016 (Paper II & III, Selected Questions)
The simplified form of a Boolean equation (AB' + AB'C + AC)(A'C' + B') is:
  1. AAB'
  2. BAB'C
  3. CA'B
  4. DABC
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Correct answer: (A) AB'

Explanation

(AB' + AB'C + AC)(A'C' + B') = (AB'(1 + C) + AC)(A'C' + B') = (AB' + AC)(A'C' + B') = AA'B'C' + AB'B' + AA'CC' + AB'C = 0 + AB' + 0 + AB'C = AB'(1 + C) = AB'

Q4UGC NET Computer Science, June 2016 (Paper II & III, Selected Questions)
A multiplexer combines four 100-Kbps channels using a time slot of 2 bits. What is the bit rate?
  1. A100 kbps
  2. B200 kbps
  3. C400 kbps
  4. D1000 kbps
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Correct answer: (C) 400 kbps

Explanation

bit rate = no. of channels × Input bit rate = 4 × 100 kbps = 400 kbps

Q5UGC NET Computer Science, June 2016 (Paper II & III, Selected Questions)
In a positive-edge-triggered JK flip-flop, if J and K both are high then the output will be __________ on the rising edge of the clock.
  1. ANo change
  2. BSet
  3. CReset
  4. DToggle
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Correct answer: (D) Toggle

Explanation

Racing is problem of toggling more than once when the inputs are J = 1 and K = 1. The proper combination of propagation delay and edge triggering prevents the racing problem.

Q6UGC NET Computer Science, June 2016 (Paper II & III, Selected Questions)
How many different equivalence relations with exactly three different equivalence classes are there on a set with five elements?
  1. A10
  2. B15
  3. C25
  4. D30
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Correct answer: (C) 25

Explanation

Equivalence relations on a set are in bijection with partitions of the set. A 5-element set split into 3 classes has partition types (1 + 1 + 3) or (1 + 2 + 2). For the first type, choose the triple: ⁵C₃ = 10. For the second type, choose the singleton (5 ways) and then pair the remaining 4 elements (3 ways): 5 × 3 = 15. Total = 10 + 15 = 25.

Q7UGC NET Computer Science, June 2016 (Paper II & III, Selected Questions)
Suppose that R₁ and R₂ are reflexive relations on a set A. Which of the following statements is correct?
  1. AR₁ ∩ R₂ is reflexive and R₁ ∪ R₂ is irreflexive.
  2. BR₁ ∩ R₂ is irreflexive and R₁ ∪ R₂ is reflexive.
  3. CBoth R₁ ∩ R₂ and R₁∪ R₂ are reflexive.
  4. DBoth R₁ ∩ R₂ and R₁ ∪ R₂ are irreflexive.
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Correct answer: (C) Both R₁ ∩ R₂ and R₁∪ R₂ are reflexive.

Explanation

If R₁ and R₂ are reflexive relations on a set A, then the diagonal Δ = {(a, a) : a ∈ A} is contained in both R₁ and R₂. R₁ ∪ R₂ is reflexive because Δ ⊆ R₁ ⊆ R₁ ∪ R₂. R₁ ∩ R₂ is also reflexive, because Δ ⊆ R₁ ∧ Δ ⊆ R₂ ⇒ Δ ⊆ R₁ ∩ R₂.

Q8UGC NET Computer Science, June 2016 (Paper II & III, Selected Questions)
A clique in a simple undirected graph is a complete subgraph that is not contained in any larger complete subgraph. How many cliques are there in the graph shown below? Vertices: a, b, c, d, e, f, g Edges: a-b, b-c, a-d, d-e, e-f, c-f, b-e, a-g, b-g, d-g, e-g, b-f, c-e (In the figure a, b, c are on the top row, d, e, f on the bottom row, g is where the diagonals a-e and b-d cross; the diagonals b-f and c-e cross without a vertex.)
  1. A2
  2. B4
  3. C5
  4. D6
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Correct answer: (C) 5

Explanation

A clique is a maximal complete subgraph (every vertex connected to all other vertices in the subgraph). Such subgraphs are: (i) {b, c, e, f} (ii) {a, d, g} (iii) {a, b, g} (iv) {b, e, g} (v) {d, e, g}. There are 5 cliques.

Q9UGC NET Computer Science, June 2016 (Paper II & III, Selected Questions)
Pipelining improves performance by:
  1. Adecreasing instruction latency
  2. Beliminating data hazards
  3. Cexploiting instruction level parallelism
  4. Ddecreasing the cache miss rate
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Correct answer: (C) exploiting instruction level parallelism

Explanation

Instruction pipelining is a technique that implements a form of parallelism called instruction level parallelism with a single processor. It therefore allows faster CPU throughput than would otherwise be possible at a given clock rate.

Q10UGC NET Computer Science, June 2016 (Paper II & III, Selected Questions)
Which of the following is not an inherent application of stack?
  1. AImplementation of recursion
  2. BEvaluation of a postfix expression
  3. CJob scheduling
  4. DReverse a string
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Correct answer: (C) Job scheduling

Explanation

Application of stack(s) are: (i) Implementation of recursion. (ii) Evaluation of a postfix expression. (iii) Reverse a string. Job scheduling is commonly done via queue data structure.

Q11UGC NET Computer Science, June 2016 (Paper II & III, Selected Questions)
Consider the following operations performed on a stack of size 5: Push (a); Pop ( ); Push (b); Push (c); Pop ( ); Push (d); Pop ( ); Pop ( ); Push (e) Which of the following statements is correct?
  1. AUnderflow occurs
  2. BStack operations are performed smoothly
  3. COverflow occurs
  4. DNone of the above
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Correct answer: (B) Stack operations are performed smoothly

Explanation

Since the given stack size is 5 and at most two elements are on the stack at any time, and no Pop is done on an empty stack, there is no overflow or underflow. The stack operations are performed smoothly.

Q12UGC NET Computer Science, June 2016 (Paper II & III, Selected Questions)
In how many ways can the string A ∩ B – A ∩ B – A be fully parenthesized to yield an infix expression?
  1. A15
  2. B14
  3. C13
  4. D12
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Correct answer: (B) 14

Explanation

The number of ways to fully parenthesize an expression with n operators is the Catalan number C(n) = (1/(n + 1)) × ²ⁿCₙ. Here the number of operators is n = 4, so C(4) = (1/5) × ⁸C₄ = (1/5) × 8!/(4! 4!) = 70/5 = 14. Note: All operators work as nodes of a binary tree.

Q13UGC NET Computer Science, June 2016 (Paper II & III, Selected Questions)
Suppose you are given a binary tree with n nodes, such that each node has exactly either zero or two children. The maximum height of the tree will be n − n + 1 1
  1. An/2 – 1
  2. Bn/2 + 1
  3. C(n – 1)/2
  4. D(n + 1)/2
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Correct answer: (C) (n – 1)/2

Explanation

In a full binary tree (each node has 0 or 2 children) with n nodes, the number of internal nodes is I = (n – 1)/2. When the tree is skewed (each internal node has one leaf child), every internal node adds one level, so the maximum height is also (n – 1)/2.

Q14UGC NET Computer Science, June 2016 (Paper II & III, Selected Questions)
Consider the following binary search tree: Root: 24 24 → left child 12, right child 26 12 → left child 11, right child 16 16 → left child 13 If we remove the root node, which of the node from the left subtree will be the new root?
  1. A11
  2. B12
  3. C13
  4. D16
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Correct answer: (D) 16

Explanation

If we delete the root node of a binary search tree, it can be replaced by its inorder successor or inorder predecessor. The inorder predecessor of 24 (largest node in the left subtree) is 16 and the inorder successor is 26. So, from the left subtree, 16 will be the new root.

Q15UGC NET Computer Science, June 2016 (Paper II & III, Selected Questions)
The number of different spanning trees in complete graph, K₄ and bipartite graph, K₂,₂ have _________ and _________ respectively.
  1. A14, 14
  2. B16, 14
  3. C16, 4
  4. D14, 4
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Correct answer: (C) 16, 4

Explanation

The number of spanning trees in a complete graph with n vertices = nⁿ⁻² = 4⁴⁻² = 16. A complete bipartite graph Kₘ,ₙ has mⁿ⁻¹ × nᵐ⁻¹ spanning trees = 2¹ × 2¹ = 4.

Q16UGC NET Computer Science, June 2016 (Paper II & III, Selected Questions)
Given i = 0, j = 1, k = –1, x = 0.5, y = 0.0 What is the output of the following expression in C language? x * y < i + j || k
  1. A-1
  2. B0
  3. C1
  4. D2
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Correct answer: (C) 1

Explanation

Given i = 0, j = 1, k = –1, x = 0.5, y = 0.0. x * y < i + j || k = (((x * y) < (i + j)) || k) = (((0.5 * 0.0) < (0 + 1)) || –1) = ((0 < 1) || –1) = 1

Q17UGC NET Computer Science, June 2016 (Paper II & III, Selected Questions)
The following statement in 'C' int (*f( ))[ ]; declares
  1. Aa function returning a pointer to an array of integers.
  2. Ba function returning an array of pointers to integers.
  3. Carray of functions returning pointers to integers.
  4. Dan illegal statement.
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Correct answer: (A) a function returning a pointer to an array of integers.

Explanation

In 'C', int (*f( ))[ ]; declares f as a function returning a pointer to an array of integers.

Q18UGC NET Computer Science, June 2016 (Paper II & III, Selected Questions)
What is the value returned by the function f given below when n = 100? int f(int n) { if (n == 0) then return n; else return n + f(n – 2);}
  1. A2550
  2. B2556
  3. C5220
  4. D5520
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Correct answer: (A) 2550

Explanation

The function is recursive and starts with n = 100. It keeps adding the present value to f(n – 2) and stops at n = 0, where it returns 0. So it computes 100 + 98 + ... + 2, an arithmetic series with 50 terms, first term 100 and last term 2. Sum = (n/2)(a + l) = (50/2)(100 + 2) = 2550

Q19UGC NET Computer Science, June 2016 (Paper II & III, Selected Questions)
When a method in a subclass has the same name and type signatures as a method in the superclass, then the method in the subclass ________ the method in the superclass.
  1. AOverloads
  2. BFriendships
  3. CInherits
  4. DOver rides
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Correct answer: (D) Over rides

Explanation

In a class hierarchy, when a method in a subclass has the name and type signatures as a method in its superclass, then the method in the subclass is said to over ride the method in the superclass.

Q20UGC NET Computer Science, June 2016 (Paper II & III, Selected Questions)
Which one of the following is correct, when a class grants friend status to another class?
  1. AThe member functions of the class generating friendship can access the members of the friend class.
  2. BAll member functions of the class granted friendship have unrestricted access to the members of the class granting the friendship.
  3. CClass friendship is reciprocal to each other.
  4. DThere is no such concept.
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Correct answer: (B) All member functions of the class granted friendship have unrestricted access to the members of the class granting the friendship.

Explanation

The member functions of a class which is given friend status by another class can access all private, protected and public members of the class granting friendship. Class friendship is neither reciprocal (commutative) nor transitive unless explicitly specified.

Q21UGC NET Computer Science, June 2016 (Paper II & III, Selected Questions)
The extent to which a software tolerates the unexpected problems, is termed as:
  1. AAccuracy
  2. BReliability
  3. CCorrectness
  4. DRobustness
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Correct answer: (D) Robustness

Explanation

Robustness is the ability of a computer system to cope with errors during execution and cope with erroneous input. Correctness of an algorithm is asserted when it is said that the algorithm is correct with respect to a specification.

Q22UGC NET Computer Science, June 2016 (Paper II & III, Selected Questions)
Which of the following is not typically a benefit of dynamic linking? I. Reduction in overall program execution time. II. Reduction in overall space consumption in memory. III. Reduction in overall space consumption on disk. IV. Reduction in the cost of software updates.
  1. AI and II
  2. BI only
  3. CII and III
  4. DIV only
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Correct answer: (B) I only

Explanation

(i) A dynamic linker is the part of an operating system that loads and links the shared libraries needed by an executable when it is executed (at “run time”), by copying the context of libraries from present storage to RAM, and filling jump tables and relocating pointers. (ii) Static linker has reduction in overall program execution time than dynamic linker.

Q23UGC NET Computer Science, June 2016 (Paper II & III, Selected Questions)
Consider the reference string: 0 1 2 3 0 1 4 0 1 2 3 4 If FIFO page replacement algorithm is used, then the number of page faults with three page frames and four page frames are _______ and _______ respectively.
  1. A10, 9
  2. B9, 9
  3. C10, 10
  4. D9, 10
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Correct answer: (D) 9, 10

Explanation

Given reference string: 0, 1, 2, 3, 0, 1, 4, 0, 1, 2, 3, 4. (i) FIFO with 3 page frames: faults occur at 0, 1, 2, 3, 0, 1, 4, 2, 3 (hits at the 2nd 0 and 1 after 4). Total page faults = 9. (ii) FIFO with 4 page frames: faults occur at 0, 1, 2, 3, 4, 0, 1, 2, 3, 4 (hits at 0 and 1 before 4 is loaded). Total page faults = 10. This is Belady's anomaly.

Q24UGC NET Computer Science, June 2016 (Paper II & III, Selected Questions)
If the Disk head is located initially at track 32, find the number of disk moves required with FCFS scheduling criteria if the disk queue of I/O blocks requests are: 98, 37, 14, 124, 65, 67
  1. A320
  2. B322
  3. C321
  4. D319
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Correct answer: (C) 321

Explanation

Given I/O requests: 98, 37, 14, 124, 65, 67, head at 32. Using FCFS scheduling, total disk moves = (98 – 32) + (98 – 37) + (37 – 14) + (124 – 14) + (124 – 65) + (67 – 65) = 66 + 61 + 23 + 110 + 59 + 2 = 321

Q25UGC NET Computer Science, June 2016 (Paper II & III, Selected Questions)
A scheduling Algorithm assigns priority proportional to the waiting time of a process. Every process starts with priority zero (lowest priority). The scheduler reevaluates the process priority for every 'T' time units and decides next process to be scheduled. If the process have no I/O operations and all arrive at time zero, then the scheduler implements _______ criteria.
  1. APriority scheduling
  2. BRound Robin Scheduling
  3. CShortest Job First
  4. DFCFS
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Correct answer: (B) Round Robin Scheduling

Explanation

The scheduling algorithm works as round robin with time quantum T. After a process has executed for T units, its waiting time becomes least (lowest priority), and its turn comes again only after every other process has had a chance for T units.

Q26UGC NET Computer Science, June 2016 (Paper II & III, Selected Questions)
Suppose there are four processes in execution with 12 instances of a Resource R in a system. The maximum need of each process and current allocation are given below: Process | Max. Need | Current Allocation P₁ | 8 | 3 P₂ | 9 | 4 P₃ | 5 | 2 P₄ | 3 | 1 With reference to current allocation, is system safe? If so, what is the safe sequence?
  1. ANo
  2. BYes, P₁ P₂ P₃ P₄
  3. CYes, P₄ P₃ P₁ P₂
  4. DYes, P₂ P₁ P₃ P₄
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Correct answer: (C) Yes, P₄ P₃ P₁ P₂

Explanation

Remaining need = Max need – Current allocation: P₁ = 5, P₂ = 5, P₃ = 3, P₄ = 2. Total allocated = 10, so available = 12 – 10 = 2. Satisfy P₄ (need 2), it releases 3 → available 4. Satisfy P₃ (need 3), releases 5 → available 7 (after P₃ completes). Then P₁ and P₂ can be satisfied similarly. Hence the safe sequence is P₄, P₃, P₁, P₂.

Q27UGC NET Computer Science, June 2016 (Paper II & III, Selected Questions)
Given the following statements: S1: Strategic value of data mining is time stamping. S2: Information collection is an expensive process in building an expert system. Which of the following options is correct?
  1. ABoth S1 and S2 are false.
  2. BBoth S1 and S2 are true.
  3. CS1 is true, S2 is false.
  4. DS1 is false, S2 is true.
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Correct answer: (B) Both S1 and S2 are true.

Explanation

• Strategic value of data mining is time stamping → correct • Information collection is an expensive process in building an expert system → correct • An expert system is a computer system that emulates the decision making ability of a human expert.

Q28UGC NET Computer Science, June 2016 (Paper II & III, Selected Questions)
In RDBMS, the constraint that no key attribute (column) may be NULL is referred to as:
  1. AReferential integrity
  2. BMulti-valued dependency
  3. CEntity Integrity
  4. DFunctional dependency
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Correct answer: (C) Entity Integrity

Explanation

The entity integrity constraint states that primary keys cannot be null. There must be a proper value in the primary key field. The basis of referential integrity is foreign keys. A foreign key in one table references a primary key in another table.

Q29UGC NET Computer Science, June 2016 (Paper II & III, Selected Questions)
In a relational database model, NULL values can be used for all but which one of the following?
  1. ATo allow duplicate tuples in the table by filling the primary key column(s) with NULL.
  2. BTo avoid confusion with actual legitimate data values like 0 (zero) for integer columns and '' (the empty string) for string columns.
  3. CTo leave columns in a tuple marked as "unknown" when the actual value is unknown.
  4. DTo fill a column in a tuple when that column does not really "exist" for that particular tuple.
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Correct answer: (A) To allow duplicate tuples in the table by filling the primary key column(s) with NULL.

Explanation

The SQL null is the term used to represent a missing value. A null value in a table is a value in a field that appears to be blank. A field with a null value is a field with no value.

Q30UGC NET Computer Science, June 2016 (Paper II & III, Selected Questions)
Consider the following two commands C1 and C2 on the relation R from an SQL database: C1: drop table R; C2: delete from R; Which of the following statements is TRUE? I. Both C1 and C2 delete the schema for R. II. C2 retains relation R, but deletes all tuples in R. III. C1 deletes not only all tuples of R, but also the schema for R.
  1. AI only
  2. BI and II only
  3. CII and III only
  4. DI, II and III
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Correct answer: (C) II and III only

Explanation

Drop command remove all tuples of the relation as well as schema also. Delete command in SQL removes tuples from the relation but retain the schema of table in database.

Q31UGC NET Computer Science, June 2016 (Paper II & III, Selected Questions)
Consider the following database table having A, B, C and D as its four attributes and four possible candidate keys (I, II, III, IV) for this table: A | B | C | D a₁ | b₁ | c₁ | d₁ a₂ | b₂ | c₃ | d₂ a₃ | b₃ | c₁ | d₃ I: {B} II: {B, C} III: {A, D} IV: {C, D} If different symbols stand for different values in the table (e.g., d₁ is definitely not equal to d₂), then which of the above could not be the candidate key for the database table?
  1. AI and III only
  2. BIII and IV only
  3. CII only
  4. DI only
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Correct answer: (C) II only

Explanation

Attribute B has distinct values in all tuples, so {B} can be a candidate key, and {B, C} is then only a superkey (not minimal), because a candidate key is a minimal superkey. {A, D} and {C, D} have distinct values in all tuples, so both can be candidate keys. Hence only II could not be a candidate key.

Q32UGC NET Computer Science, June 2016 (Paper II & III, Selected Questions)
Which of the following statement(s) is/are FALSE in the context of Relational DBMS? I. Views in a database system are important because they help with access control by allowing users to see only a particular subset of the data in the database. II. E-R diagrams are useful to logically model concepts. III. An update anomaly is when it is not possible to store information unless some other, unrelated information is stored as well. IV. SQL is a procedural language.
  1. AI and IV only
  2. BIII and IV only
  3. CI, II and III only
  4. DII, III and IV only
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Correct answer: (B) III and IV only

Explanation

(i) Views help provide data independence, allow the schema to change without recompiling existing applications, and help with access control by letting users see only a particular subset of the data. (ii) A conceptual ER model may be used as the foundation for logical data models. (iii) Being unable to store information unless other unrelated information is stored is an insertion anomaly, not an update anomaly. (iv) SQL is not a procedural language. Only statements III and IV are false.

Q33UGC NET Computer Science, June 2016 (Paper II & III, Selected Questions)
The IEEE-754 double-precision format to represent floating point number, has a length of ________ bits.
  1. A16
  2. B32
  3. C48
  4. D64
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Correct answer: (D) 64

Explanation

Double precision floating point format occupies 8 bytes (64 bits) in computer memory: sign bit (1 bit), exponent (11 bits) and significand precision (53 bits, 52 explicitly stored).

Q34UGC NET Computer Science, June 2016 (Paper II & III, Selected Questions)
In a fully-connected mesh network with 10 computers, total _______ number of cables are required and _______ number of ports are required for each device.
  1. A40, 9
  2. B45, 10
  3. C45, 9
  4. D50, 10
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Correct answer: (C) 45, 9

Explanation

Since, each in mesh network total number of cables are required ⁿC₂ = ¹⁰C₂ = 45 Each device needs to be connected to 9 other devices. So, each device needs to have 9 ports.

Q35UGC NET Computer Science, June 2016 (Paper II & III, Selected Questions)
Optical fiber uses reflection to guide light through a channel, in which angle of incidence is ________ the critical angle.
  1. AEqual to
  2. BLess than
  3. CGreater than
  4. DLess than or equal to
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Correct answer: (C) Greater than

Explanation

If the angle of incidence is equal to the critical angle, the light beam bends along the interface. When a beam of light travels through media of two different densities, if the angle of incidence is greater than the critical angle, reflection occurs.

Q36UGC NET Computer Science, June 2016 (Paper II & III, Selected Questions)
If there are N people in the world and are using secret key encryption/decryption for privacy purpose, then number of secret keys required will be: (b) (N – 1)
  1. AN
  2. B(N – 1)
  3. CN(N – 1)/2
  4. DN(N + 1)/2
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Correct answer: (C) N(N – 1)/2

Explanation

Each pair of users must have a separate secret key. If N people want to use this method, the number of secret keys needed is ᴺC₂ = N(N – 1)/2.

Q37UGC NET Computer Science, June 2016 (Paper II & III, Selected Questions)
The ______ model is preferred for software development when the requirements are not clear.
  1. ARapid Application Development
  2. BRational Unified Process
  3. CEvolutionary Model
  4. DWaterfall Model
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Correct answer: (C) Evolutionary Model

Explanation

Evolutionary Model is an iterative and incremental approach to software development. It is preferred for software development when requirements are not clear.

Q38UGC NET Computer Science, June 2016 (Paper II & III, Selected Questions)
Which of the following is not included in waterfall model?
  1. ARequirement analysis
  2. BRisk analysis
  3. CDesign
  4. DCoding
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Correct answer: (B) Risk analysis

Explanation

The waterfall model is a sequential (non-iterative) process; it includes requirements, design, implementation, verification and maintenance. Risk analysis is used in the spiral model.

Q39UGC NET Computer Science, June 2016 (Paper II & III, Selected Questions)
An attacker sits between customer and Banker, and captures the information from the customer and retransmits to the banker by altering the information. This attack is called as _______.
  1. AMasquerade Attack
  2. BReplay Attack
  3. CPassive Attack
  4. DDenial of Service Attack
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Correct answer: (B) Replay Attack

Explanation

(i) A replay attack (play back attack) is a form of network attack in which a valid data transmission is maliciously or fraudulently repeated or delayed. (ii) A masquerade attack is an attack that uses a fake identity. (iii) A passive attack is network attack in which a system is monitored and sometimes scanned for open ports and vulnerabilities. (iv) Denial of service attack, a type of attack on a network that is designed to bring the network to its knees by flooding it with useless traffic.

Paper III 41 questions

Q40UGC NET Computer Science, June 2016 (Paper II & III, Selected Questions) · Paper III
Match the following: A. Line coding B. Block coding C. Scrambling D. Pulse code modulation i. A technique to change analog signal to digital data ii. Provides synchronization without increasing number of bits iii. Process of converting digital data to digital signal iv. Provides redundancy to ensure synchronization and inherits error detection Code: A B C D
  1. A(iv) (iii) (ii) (i)
  2. B(iii) (iv) (ii) (i)
  3. C(i) (iii) (ii) (iv)
  4. D(ii) (i) (iv) (iii)
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Correct answer: (B) (iii) (iv) (ii) (i)

Explanation

(A) Line coding is the process of converting digital data into a digital signal – (iii). (B) Block coding provides redundancy to ensure synchronization and inherent error detection – (iv). (C) Scrambling provides synchronization without increasing the number of bits – (ii). (D) Pulse code modulation (PCM) is a technique to change an analog signal to digital data – (i).

Q41UGC NET Computer Science, June 2016 (Paper II & III, Selected Questions) · Paper III
In ........... method, the word is written to the block in both the cache and main memory, in parallel.
  1. AWrite through
  2. BWrite back
  3. CWrite protected
  4. DDirect mapping
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Correct answer: (A) Write through

Explanation

In write through method, data is written into the cache and the corresponding main memory location at the same time.

Q42UGC NET Computer Science, June 2016 (Paper II & III, Selected Questions) · Paper III
The .......... addressing mode is similar to register indirect addressing mode, except that an offset is added to the contents of the register. The offset and register are specified in the instruction.
  1. ABase indexed
  2. BBase indexed plus displacement
  3. CIndexed
  4. DDisplacement
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Correct answer: (D) Displacement

Explanation

The displacement addressing mode is similar to register indirect addressing mode, except that an offset is added to the contents of the register. The offset and register are specified in the instruction. Based Indexed Addressing: The operand’s offset is sum of the content of a base register BX or BP and an index register SI or DI. Based Indexed plus displacement addressing mode: In this mode of addressing the operand’s offset is given by offset=[BX or BP]+[SI or DI]+8 bit or 16 bit displacement. Indexed addressing mode: The operand’s offset is the sum of the content of an index register SI or DI and an 8 bit or 16 bit displacement. So, option (D) is correct.

Q43UGC NET Computer Science, June 2016 (Paper II & III, Selected Questions) · Paper III
A multi computer with 256 CPUs is organized as 16 × 16 grid. What is the worst case delay (in hops) that a message might have to take?
  1. A16
  2. B15
  3. C32
  4. D30
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Correct answer: (D) 30

Explanation

In a 16 × 16 grid, the worst case delay happens when a message travels between opposite corners (e.g. upper right to lower left). It passes through 2(N – 1) = 2(16 – 1) = 30 hops.

Q44UGC NET Computer Science, June 2016 (Paper II & III, Selected Questions) · Paper III
Suppose that we have numbers between 1 and 1,000 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?
  1. A925, 221, 912, 245, 899, 259, 363, 364
  2. B3, 400, 388, 220, 267, 383, 382, 279, 364
  3. C926, 203, 912, 241, 913, 246, 364
  4. D3, 253, 402, 399, 331, 345, 398, 364
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Correct answer: (C) 926, 203, 912, 241, 913, 246, 364

Explanation

In sequence (c), after visiting 912 (912 > 364) the search moves to the left subtree of 912, so every later node must be less than 912. 913 cannot appear after 912. So, the answer is (c).

Q45UGC NET Computer Science, June 2016 (Paper II & III, Selected Questions) · Paper III
Consider the following game tree in which root is a maximizing node and children are visited left to right. What nodes will be pruned by the alpha-beta pruning? A (MAX) has children B and C (MIN nodes). B has leaf children D = 3, E = 12, F = 8. C has leaf children G = 2, H = 15, I = 6.
  1. AI
  2. BHI
  3. CCHI
  4. DGHI
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Correct answer: (B) HI

Explanation

B is a MIN node: min(3, 12, 8) = 3, so A (MAX) gets α = 3. At C, the first leaf G = 2 gives C a value ≤ 2, which is less than α = 3. Whatever H and I produce, C cannot be chosen by A, so H and I are pruned. Option (b) is correct.

Q46UGC NET Computer Science, June 2016 (Paper II & III, Selected Questions) · Paper III
Match the following: A. Huffman codes B. Optimal polygon triangulation C. Activity selection problem D. Quicksort i. O(n²) ii. θ(n³) iii. O(nlogn) iv. θ(n) Code: A B C D
  1. A(i) (ii) (iv) (iii)
  2. B(i) (iv) (ii) (iii)
  3. C(iii) (ii) (iv) (i)
  4. D(iii) (iv) (ii) (i)
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Correct answer: (C) (iii) (ii) (iv) (i)

Explanation

Correct matching: Huffman codes – O(nlogn); Optimal polygon triangulation – θ(n³); Activity selection problem – θ(n); Quicksort – O(n²).

Q47UGC NET Computer Science, June 2016 (Paper II & III, Selected Questions) · Paper III
Consider a source with symbols A, B, C, D with probabilities 1/2, 1/4, 1/8, and 1/8 respectively. What is the average number of bits per symbol for the Huffman code generated from above information?
  1. A2 bits per symbol
  2. B1.75 bits per symbol
  3. C1.50 bits per symbol
  4. D1.25 bits per symbol
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Correct answer: (B) 1.75 bits per symbol

Explanation

Huffman code lengths: A (1/2) → 1 bit, B (1/4) → 2 bits, C (1/8) → 3 bits, D (1/8) → 3 bits. Average length = (1/2 × 1) + (1/4 × 2) + (1/8 × 3) + (1/8 × 3) = 14/8 = 1.75 bits per symbol.

Q48UGC NET Computer Science, June 2016 (Paper II & III, Selected Questions) · Paper III
Which one of the following array represents a binary max-heap?
  1. A[26, 13, 17, 14, 11, 9, 15]
  2. B[26, 15, 14, 17, 11, 9, 13]
  3. C[26, 15, 17, 14, 11, 9, 13]
  4. D[26, 15, 13, 14, 11, 9, 17]
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Correct answer: (C) [26, 15, 17, 14, 11, 9, 13]

Explanation

Properties of Max Heap: Must be complete binary tree. All levels are completely filled except possibly the last level. The last level is filled from Left to Right. Each parent node has a value greater than or equal to its children. Option (a) Not fulfill the property of Max Heap. Option (b) Not fulfill the property of Max Heap. Option (c) Fulfill the property of Max Heap. Option (d) Not fulfill the property of Max Heap. So, the correct option is (c). (xii) Divide and Conquer

Q49UGC NET Computer Science, June 2016 (Paper II & III, Selected Questions) · Paper III
Match the following types of variables with the corresponding programming language: A. Static variables B. Stack dynamic C. Explicit heap dynamic D. implicit heap dynamic i. Local variables in Pascal ii. All variables in APL iii. Fortran 77 iv. All objects in JAVA Code: A B C D
  1. A(i) (iii) (iv) (ii)
  2. B(iv) (i) (iii) (ii)
  3. C(iii) (i) (iv) (ii)
  4. D(ii) (i) (iii) (iv)
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Correct answer: (C) (iii) (i) (iv) (ii)

Explanation

FORTRAN I, II & IV contain only static variables. All objects in Java are explicit heap dynamic variable & are destroyed by "Garbage Collector" Languages Such as APL, ALGOL 68, & LISP Use implicit heap dynamic variables. So, a-iii, b-i, c-iv, d-ii Hence, correct option is (c).

Q50UGC NET Computer Science, June 2016 (Paper II & III, Selected Questions) · Paper III
Which of the following statements concerning Object-Oriented databases are FALSE?
  1. AObjects in an object-oriented database contain not only data but also methods for processing the data.
  2. BObject-oriented databases store computational instructions in the same place as the data.
  3. CObject-oriented databases are more adapt at handling structured (analytical) data than relational databases.
  4. DObject-oriented databases store more types of data than relational databases and access that data faster.
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Correct answer: (C) Object-oriented databases are more adapt at handling structured (analytical) data than relational databases.

Explanation

Objects in an object-oriented database contain not only data but also methods for processing the data – correct. Object-oriented databases store computational instructions in the same place as the data – correct. Object-oriented databases are more adept at handling structured (analytical) data than relational databases – incorrect. Object-oriented databases store more types of data than relational databases and access that data faster – correct. So, option (c) is correct.

Q51UGC NET Computer Science, June 2016 (Paper II & III, Selected Questions) · Paper III
What is the function of the following UNIX command? wc -l <a >b &
  1. AIt runs the word count program to count the number of lines in its input, a, writing the result to b, as a foreground process.
  2. BIt runs the word count program to count the number of lines in its input, a, writing the result to b, but does it in the background.
  3. CIt counts the errors during the execution of a process, a, and puts the result in process b.
  4. DIt copies the 'l' numbers of lines of program from file, a, and stores in file b.
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Correct answer: (B) It runs the word count program to count the number of lines in its input, a, writing the result to b, but does it in the background.

Explanation

wc -l <a >b &: wc is the word count program, -l counts lines, <a takes input from a, >b writes output to b, and & runs the command in the background. So it counts the number of lines in a, writes the result to b, and does it in the background.

Q52UGC NET Computer Science, June 2016 (Paper II & III, Selected Questions) · Paper III
Which of the following statement is not correct with reference to cron daemon in UNIX O.S.?
  1. AThe cron daemon is the standard tool for running commands on a pre-determined schedule.
  2. BIt starts when the system boots and runs as long as the system is up.
  3. CCron reads configuration files that contain list of command lines and the times at which they invoked.
  4. DCronatab for individual users are not stored.
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Correct answer: (D) Cronatab for individual users are not stored.

Explanation

The cron daemon is the standard tool for running commands on a pre-determined schedule. It starts when the system boots and runs as long as the system is up. Crontabs for individual users are stored under /var/spool/cron; there is one crontab file per user. So statement (d) is not correct.

Q53UGC NET Computer Science, June 2016 (Paper II & III, Selected Questions) · Paper III
In Unix, files can be protected by assigning each one a 9-bit mode called right bits. Now, consider the following two statements : I. A mode of 641 (octal) means that the owner can read and write the file, other members of the owner's group can read it, and users can execute only. II. A mode of 100 (octal) allows the owner to execute the file, but prohibits all other access. Which of the following options is correct with reference to above statements?
  1. AOnly I is correct
  2. BOnly II is correct
  3. CBoth I and II are correct
  4. DBoth I and II are incorrect
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Correct answer: (C) Both I and II are correct

Explanation

In UNIX, file access permissions are given as three octal digits: the first for the owner, the second for the group, the third for others. Each digit is rwx: 7 = rwx, 6 = rw-, 5 = r-x, 4 = r--, 3 = -wx, 2 = -w-, 1 = --x, 0 = none. 641 → owner can read and write, group members can read, others can execute only. 100 → owner can execute; all other access is prohibited. So both statements are correct.

Q54UGC NET Computer Science, June 2016 (Paper II & III, Selected Questions) · Paper III
In UNIX, processes that have finished execution but have not yet had their status collected are known as ................
  1. ASleeping processes
  2. BStopped processes
  3. CZombie processes
  4. DOrphan processes
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Correct answer: (C) Zombie processes

Explanation

Zombie Process : The UNIX system assumes that if child process terminate, the parent collects some data about the child, As soon as this collection is made, the system releases information bits about the child process and removes its PID from the process table. The UNIX is forced to keep the child's Process ID (PID) and termination data in the process table indefinitely. Such a terminated process whose status is not collected called a zombie process.

Q55UGC NET Computer Science, June 2016 (Paper II & III, Selected Questions) · Paper III
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?
  1. AHeap
  2. BStack
  3. CShared memory segments
  4. DBoth Heap and Stack
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Correct answer: (C) Shared memory segments

Explanation

Parent and child share only the share memory segments everything else (like stack, heap etc) is duplicated.

Q56UGC NET Computer Science, June 2016 (Paper II & III, Selected Questions) · Paper III
Which of the following information about the UNIX file system is not correct?
  1. ASuper block contains the number of i-nodes, the number of disk blocks, and the start of the list of free disk blocks.
  2. BAn i-node contains accounting information as well as enough information to locate all the disk block that holds the file's data.
  3. CEach i-node is 256-bytes long.
  4. DAll the files and directories are stored in data blocks.
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Correct answer: (C) Each i-node is 256-bytes long.

Explanation

UNIX file system: Super block contains the number of i-nodes, the number of disk blocks, and the start of the list of free disk blocks.Correct An i-node contains accounting information as well as enough information to locate all the disk blocks that holds the file’s data.Correct All the files and directories are stored in data blocks.Correct Each i-node is 256-bytes long.Incorrect.It is 128 byte long.

Q57UGC NET Computer Science, June 2016 (Paper II & III, Selected Questions) · Paper III
Which of the following option with reference to UNIX operating system is not correct?
  1. AINT signal is sent by the terminal driver when one types <Control> and it is a request to terminate the current operation.
  2. BTERM is a request to terminate execution completely. The receiving process will clean up its state and exit.
  3. CQUIT is similar to TERM, except that it defaults to producing a core dump if not caught.
  4. DKILL is a block able signal.
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Correct answer: (D) KILL is a block able signal.

Explanation

KILL (SIGKILL) cannot be blocked, caught or ignored, so the statement "KILL is a blockable signal" is not correct.

Q58UGC NET Computer Science, June 2016 (Paper II & III, Selected Questions) · Paper III
In distributed database, location transparency allows for database users, programmers and administrators to treat the data as if it is at one location. A SQL query with location transparency needs to specify :
  1. AInheritances
  2. BFragments
  3. CLocations
  4. DLocal formats
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Correct answer: (B) Fragments

Explanation

With location transparency the user does not need to know where the data is stored, but still needs to know how the data is fragmented; so a SQL query needs to specify fragments. Fragmentation transparency is the highest level, where the user does not even need to know that the database is partitioned.

Q59UGC NET Computer Science, June 2016 (Paper II & III, Selected Questions) · Paper III
Consider the following ORACLE relations: R (A, B, C) = {<1, 2, 3>, <1, 2, 0>, <1, 3, 1>, <6, 2, 3>, <1, 4, 2>, <3, 1, 4>} S (B, C, D) = {<2, 3, 7>, <1, 4, 5>, <1, 2, 3>, <2, 3, 4>, <3, 1, 4>} Consider the following two SQL queries SQ₁ and SQ₂: SQ₁: SELECT R.B, AVG (S.B) FROM R, S WHERE R.A = S.C AND S.D < 7 GROUP BY R.B; SQ₂: SELECT DISTINCT S.B, MIN (S.C) FROM S GROUP BY S.B HAVING COUNT (DISTINCT S.D) > 1; If M is the number of tuples returned by SQ₁ and N is the number of tuples returned by SQ₂ then
  1. AM = 4, N = 2
  2. BM = 5, N = 3
  3. CM = 2, N = 2
  4. DM = 3, N = 3
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Correct answer: (A) M = 4, N = 2

Explanation

SQ₁: S tuples with D < 7 have S.C = 4, 2, 3, 1. Joining with R.A: S.C = 1 matches R tuples with A = 1 (R.B = 2, 2, 3, 4) and S.C = 3 matches <3, 1, 4> (R.B = 1). Grouping by R.B gives groups 1, 2, 3, 4, so M = 4. SQ₂: S.B = 1 has distinct D {5, 3} (count 2), S.B = 2 has {7, 4} (count 2), S.B = 3 has {4} (count 1). The third group is not taken because COUNT(DISTINCT S.D) > 1 fails, so N = 2.

Q60UGC NET Computer Science, June 2016 (Paper II & III, Selected Questions) · Paper III
Semi-join strategies are techniques for query processing in distributed database system. Which of the following is a semi-join technique?
  1. AOnly the joining attributes are sent from one site to another and then all of the rows are returned
  2. BAll of the attributes are sent from one site to another and then only the required rows are returned
  3. COnly the joining attributes are sent from site to another and then only the required rows are returned
  4. DAll of the attributes are sent from one site to another and then only the required rows are returned.
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Correct answer: (C) Only the joining attributes are sent from site to another and then only the required rows are returned

Explanation

Semi-join strategies are techniques for query processing in distributed database system. In which only the joining attributes are sent from one site to another and then only the required rows are returned.

Q61UGC NET Computer Science, June 2016 (Paper II & III, Selected Questions) · Paper III
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 :
  1. A96.3%
  2. B97.3%
  3. C98.3%
  4. D99.3%
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Correct answer: (C) 98.3%

Explanation

If we let A represent availability, then the simplest formula for availability A = uptime/(uptime + downtime) MTBF is mean time between failures A = MTTR is mean time to repair A = MTBF/(MTBF + MTTR) So MTBF = 30 days = 30 × 24 = 720 hr A = 720/(720 + 12) = 98.3%

Q62UGC NET Computer Science, June 2016 (Paper II & III, Selected Questions) · Paper III
Consider the relations R(A, B) and S(B, C) and the following four relational algebra queries over R and S: I. π_A,B (R ⋈ S) II. R ⋈ π_B(S) III. R ∩ (π_A(R) × π_B(S)) IV. π_A,R.B (R × S) where R.B refers to the column B in table R. One can determine that:
  1. AI, III and IV are the same query
  2. BII, III and IV are the same query
  3. CI, II and IV are the same query
  4. DI, II and III are the same query
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Correct answer: (D) I, II and III are the same query

Explanation

I, II and III all return exactly those tuples of R whose B value appears in S, whereas IV returns every (A, R.B) pair of R as long as S is non-empty. Example: R = {(10, 1), (20, 2)}, S = {(2, 30), (3, 40)}. (I) R ⋈ S = {(20, 2, 30)}, so π_A,B(R ⋈ S) = {(20, 2)}. (II) π_B(S) = {(2), (3)}, so R ⋈ π_B(S) = {(20, 2)}. (III) π_A(R) × π_B(S) = {(10, 2), (10, 3), (20, 2), (20, 3)}, and R ∩ that = {(20, 2)}. (IV) π_A,R.B(R × S) = {(10, 1), (20, 2)}. So I, II and III are the same query.

Q63UGC NET Computer Science, June 2016 (Paper II & III, Selected Questions) · Paper III
Which of the following statements is TRUE? D₁ : The decomposition of the schema R(A, B, C) into R₁(A, B) and R₂ (A, C) is always lossless. D₂ : The decomposition of the schema R(A, B, C, D, E) having AD →B, C →DE, B → AE and AE →C, into R₁ (A, B, D) and R₂ (A, C, D, E) is lossless.
  1. ABoth D₁ and D₂
  2. BNeither D₁ nor D₂
  3. COnly D₁
  4. DOnly D₂
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Correct answer: (D) Only D₂

Explanation

D₂: R₁ ∩ R₂ = {A, D}, and AD → B, so AD is a key of R₁ (A, B, D); the decomposition is lossless. D₁ is not always lossless since no FDs are given: if B → A and C → A, then the common attribute A is not a key of either table, so the decomposition is lossy. Hence only D₂.

Q64UGC NET Computer Science, June 2016 (Paper II & III, Selected Questions) · Paper III
Consider a discrete memoryless channel and assume that H(x) is the amount of information per symbol at the input of the channel; H(y) is the amount of information per symbol at the output of the channel; H(x|y) is the amount of uncertainty remaining on x knowing y; and I(x; y) is the information transmission. Which of the following does not define the channel capacity of a discrete memoryless channel?
  1. Amax over p(x) of I(x; y)
  2. Bmax over p(x) of [H(y) – H(y|x)]
  3. Cmax over p(x) of [H(x) – H(x|y)]
  4. Dmax over p(x) of H(x|y)
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Correct answer: (D) max over p(x) of H(x|y)

Explanation

The channel capacity of a discrete memoryless channel is defined by C = max over p(x) of I(x; y) = max over p(x) of [H(y) – H(y|x)] = max over p(x) of [H(x) – H(x|y)]. max H(x|y) does not define the channel capacity.

Q65UGC NET Computer Science, June 2016 (Paper II & III, Selected Questions) · Paper III
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 ...........
  1. A192 mbps
  2. B512 mbps
  3. C1.248 mbps
  4. D1.536 mbps
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Correct answer: (D) 1.536 mbps

Explanation

Note: strictly, 100 pages per minute gives 1,536,000 bits per minute ≈ 25.6 kbps; the keyed 1.536 Mbps corresponds to 100 pages per second, which is how the book solves it. Bits per 100 pages = 100 × 24 × 80 × 8 = 1,536,000 bits ⇒ 1.536 Mbps.

Q66UGC NET Computer Science, June 2016 (Paper II & III, Selected Questions) · Paper III
If link transmits 4000 frames per second and each slot has 8 bits, the transmission rate of circuit of this TDM is ...........
  1. A64 Kbps
  2. B32 Mbps
  3. C32 Kbps
  4. D64 Mbps
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Correct answer: (C) 32 Kbps

Explanation

Transmission rate of a circuit = frame rate × number of bits in a slot = 4000 × 8 bps = 32 Kbps. So, option (c) is correct.

Q67UGC NET Computer Science, June 2016 (Paper II & III, Selected Questions) · Paper III
Given the following statements: (1) 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. (2) Wavelength division multiplexing (WDM) is an analog multiplexing technique to combine optical signals. (3) WDM is a digital multiplexing technique. (4) TDM is digital multiplexing technique. Which of the following is correct?
  1. A(1), (2), (3) and (4) are true
  2. B(1), (2), (3) and (4) are false
  3. C(1), (2) and (4) are false (3) is true
  4. D(1), (2) and (4) are true (3) is false
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Correct answer: (D) (1), (2) and (4) are true (3) is false

Explanation

Multiplexing is of three types: (1) Frequency division multiplexing (FDM) – analog signals. (2) Wavelength division multiplexing (WDM) – analog signals (used in optical fiber). (3) Time division multiplexing (TDM) – digital signals; a single frame is made with interleaved digital signals from n devices. So (1), (2) and (4) are true and (3) is false.

Q68UGC NET Computer Science, June 2016 (Paper II & III, Selected Questions) · Paper III
A pure ALOHA network transmits 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 ...........
  1. A0.384
  2. B0.184
  3. C0.286
  4. D0.586
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Correct answer: (B) 0.184

Explanation

Throughput of pure ALOHA is S = G × e⁻²ᴳ, where G is the load (frames per frame-transmission time). Frame transmission time = 200 bits / 200 kbps = 1 ms. The system produces 500 frames per second = 0.5 frames per ms, so G = 1/2. S = 0.5 × e⁻¹ ≈ 0.184 (the maximum throughput of pure ALOHA). So the answer is (b).

Q69UGC NET Computer Science, June 2016 (Paper II & III, Selected Questions) · Paper III
Which of the following sets represent five stages defined by Capability Maturity Model (CMM) in increasing order of maturity?
  1. AInitial, Defined, Repeatable, Managed, Optimized.
  2. BInitial, Repeatable, Defined, Managed, Optimized.
  3. CInitial, Defined, Managed, Repeatable, Optimized.
  4. DInitial, Repeatable, Managed, Defined, Optimized.
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Correct answer: (B) Initial, Repeatable, Defined, Managed, Optimized.

Explanation

There are five levels defined along the continuum of the model and according to the SEI. (1) Initial : The starting point for use of a new repeat process (2) Repeatable : The process is at least documented sufficiently such that repeating the same steps may be attempted. (3) Defined : The process is defined as a standard business process. (4) Managed : This process is quantitatively managed in accordance with agreed upon metrics (5) Optimizing : Process management includes deliberate process improvement.

Q70UGC NET Computer Science, June 2016 (Paper II & III, Selected Questions) · Paper III
Match the software maintenance activities in List-I to its meaning in List-II List-I i. Corrective ii. Adaptive iii. Perfective iv. Preventive List-II A. Concerned with performing activities to reduce the software complexity thereby improving program understandability and increasing software maintainability B. Concerned with fixing errors that are observed when the software is in use. C. Concerned with the change in the software that takes place to make the software adaptable to new environment (both hardware and software) D. Concerned with the change in the software that takes place to make the software adaptable to changing user requirements. Code: (i) (ii) (iii) (iv)
  1. A(B) (D) (C) (A)
  2. B(B) (C) (D) (A)
  3. C(C) (B) (D) (A)
  4. D(A) (D) (B) (C)
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Correct answer: (B) (B) (C) (D) (A)

Explanation

(i) Corrective maintenance fixes errors observed when the software is in use – (B). (ii) Adaptive maintenance changes the software to make it adaptable to a new environment – (C). (iii) Perfective maintenance implements new or changed user requirements – (D). (iv) Preventive maintenance performs activities to reduce complexity and prevent future errors – (A).

Q71UGC NET Computer Science, June 2016 (Paper II & III, Selected Questions) · Paper III
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.
  1. A25 person months
  2. B75 person months
  3. C62.5 person months
  4. D72.5 person months
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Correct answer: (C) 62.5 person months

Explanation

LOC = 500 FP × 50 LOC/FP = 25000 LOC = 25 KLOC. E = a(KLOC)ᵇ = 2.5 × (25)¹ = 62.5 person months.

Q72UGC NET Computer Science, June 2016 (Paper II & III, Selected Questions) · Paper III
A software program that infers and manipulates existing knowledge in order to generate new knowledge is known as :
  1. AData dictionary
  2. BReference mechanism
  3. CInference engine
  4. DControl strategy
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Correct answer: (C) Inference engine

Explanation

A software program that infers and manipulates existing knowledge in order to generate new knowledge is known as Inference engine. A data dictionary is a collection of descriptions of the data objects or items in a data model for the benefit of programmers and others who need to refer to them. Control strategies are specific action plans for bringing a process back into control. Reference is what relates words to the world of objects on whose condition the truth of sentences hinges and it is natural to wonder what sorts of relations underlie the reference relation-to wonder, that is, what constitutes the mechanism of reference.

Q73UGC NET Computer Science, June 2016 (Paper II & III, Selected Questions) · Paper III
Match each application/software design concept in List-I to its definition in List-II. List-I i. Coupling ii. Cohesion iii. Scalable iv. Readable List-II A. Easy to visually inspect the design of the software and understand its purpose B. Easy to add functionality to a software without having to redesign it C. Focus of a code upon a single goal. D. Reliance of a code module upon other code modules. Code: (i) (ii) (iii) (iv)
  1. A(B) (A) (D) (C)
  2. B(C) (D) (A) (B)
  3. C(D) (C) (B) (A)
  4. D(D) (A) (C) (B)
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Correct answer: (C) (D) (C) (B) (A)

Explanation

(i) Coupling measures how dependent modules are on each other – (D). (ii) Cohesion reflects how well a module's elements work toward a single purpose – (C). (iii) Scalability means the system can grow (add functionality/workload) without redesign – (B). (iv) Readable code is easy to inspect and understand – (A).

Q74UGC NET Computer Science, June 2016 (Paper II & III, Selected Questions) · Paper III
Software safety is quality assurance activity that focuses on hazards that
  1. AAffect the reliability of a software component
  2. BMay cause an entire system to fail
  3. CMay result from user input errors
  4. DPrevent profitable marketing of the final product
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Correct answer: (B) May cause an entire system to fail

Explanation

Software safety is a software quality assurance activity that focuses on the identification and assessment to potential hazards that may affect software negatively and cause an entire system to fail. If hazards can be identified early in the software process, software design factures can be specified that will either eliminate or control potential hazards.

Q75UGC NET Computer Science, June 2016 (Paper II & III, Selected Questions) · Paper III
Which of the following statements is not correct?
  1. AHTML is not screen precise formatting language
  2. BHTML does not specify a logic
  3. CDHTML is used for developing highly interactive web pages
  4. DHTML is a programming language
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Correct answer: (D) HTML is a programming language

Explanation

HTML is not programming language Programming Language means where writing logic is possible which is not supported by HTML.

Q76UGC NET Computer Science, June 2016 (Paper II & III, Selected Questions) · Paper III
Which of the following statement (s) is/are correct with reference to curve generation? I. Hermit curves are generated using the concepts of interpolation. II. Bezier curves are generated using the concepts of approximation. III. The Bazier curve lies entirely within the convex hull of its control points. IV. The degree of Bezier curve does not depend on the number of control points.
  1. AI, II and IV only
  2. BII and III only
  3. CI and II only
  4. DI, II and III only
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Correct answer: (D) I, II and III only

Explanation

Hermite curves smoothly interpolate between key points. Bezier curves approximate their control points and lie entirely within the convex hull of the control points. The number of control points determines the degree of the Bezier polynomial, so IV is false. So the answer is option (d).

Q77UGC NET Computer Science, June 2016 (Paper II & III, Selected Questions) · Paper III
Given the following statements : (1) To implement Abstract data type, a programming language require a syntactic unit to encapsulate type definition. (2) To implement ADT, a programming language requires some primitive operations that are built in the language processor. (3) C++, Ada, Java 5.0, C#2005 provide support for parameterized ADT Which one of the following options is correct?
  1. A(1), (2) and (3) are false
  2. B(1) and (2) are true; (3) is false
  3. C(1) is true; (2) and (3) are false
  4. D(1), (2) and (3) are true
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Correct answer: (D) (1), (2) and (3) are true

Explanation

Language requirements for ADTs: a syntactic unit in which to encapsulate the type definition, and some primitive operations built into the language processor. Ada provides packages that simulate ADTs, C++ provides data abstraction through classes, and Java's data abstraction is similar to C++. C++, Ada, Java 5.0 and C# 2005 support parameterized ADTs. So (1), (2) and (3) are true. Hence, the answer is (d).

Q78UGC NET Computer Science, June 2016 (Paper II & III, Selected Questions) · Paper III
When one object reference variable is assigned to another object reference variable then
  1. AA copy of the object is created
  2. BA copy of the reference is created
  3. CA copy of the reference is not created
  4. DIt is illegal to assign one object reference variable to another object reference variable
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Correct answer: (B) A copy of the reference is created

Explanation

Assigning one object reference variable to another copies only the reference (address), not the object. Both variables then refer to the same object; no distinct copy of the object is created. So a copy of the reference is created.

Q79UGC NET Computer Science, June 2016 (Paper II & III, Selected Questions) · Paper III
The region of feasible solution of a linear programming problem has a ______ property in geometry, provided the feasible solution of the problem exists.
  1. AConcavity
  2. BConvexity
  3. CQuadratic
  4. DPolyhedron
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Correct answer: (B) Convexity

Explanation

Each constraint is satisfied by the points of a closed half space of Rⁿ, and the set of feasible solutions is the intersection of all these half spaces, which is a convex polyhedron. So the feasible region has the convexity property. The answer is (b).

Q80UGC NET Computer Science, June 2016 (Paper II & III, Selected Questions) · Paper III
Consider the following statements : (1) Revised simplex method requires lesser computations than the Simplex method. (2) Revised simplex method automatically generates the inverse of the current basis matrix. (3) Less number of entries are needed in each table of revised simplex method than usual simplex method. Which of these statements are correct?
  1. A(1) and (2) only
  2. B(1) and (3) only
  3. C(2) and (3) only
  4. D(1), (2) and (3)
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Correct answer: (D) (1), (2) and (3)

Explanation

The revised simplex method is an improvement over the simplex method: it is computationally more efficient (fewer computations), it automatically generates the inverse of the current basis matrix, and fewer entries are needed in each table, since the simplex method updates the entire tableau while only a small part is used. So answer is option (d).

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