UGC NET · Topic-wise previous year questions

Computer Science Previous Year Questions (page 8)

570 Computer Science questions asked in UGC NET previous year papers, each with the correct answer and explanation. Read them free on UTpaper, or attempt the full papers as timed online tests.

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Q351UGC NET Computer Science, August 2016 (Paper II & III, Selected Questions)
In Data mining, _____ is a method of incremental conceptual clustering.
  1. ASTRING
  2. BCOBWEB
  3. CCORBA
  4. DOLAD
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Correct answer: (B) COBWEB

Explanation

COBWEB is an incremental system for hierarchical conceptual clustering. COBWEB incrementally organizes observations into a classification tree. Each node in the classification tree represents a class (concept) and is labeled by a probabilistic concept that summarizes the attribute-value distribution of objects classified under the node.

Q352UGC NET Computer Science, August 2016 (Paper II & III, Selected Questions)
_____ is subject oriented, integrated, time variant, nonvolatile collection of data in support of management decisions
  1. AData mining
  2. BWeb mining
  3. CData warehouse
  4. DDatabase management system
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Correct answer: (C) Data warehouse

Explanation

Data warehouse is a subject oriented integrated, time variant and non volatile collection of data in support of management decision making process. A data warehouse is a copy of transaction data specifically structured for query & analysis.

Q353UGC NET Computer Science, August 2016 (Paper II & III, Selected Questions)
In data mining, classification rules are extracted from _____
  1. AData
  2. BInformation
  3. CDecision tree
  4. DDatabase
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Correct answer: (C) Decision tree

Explanation

A decision tree can be used to visually and explicitly represent decision & division making. In data mining, a decision tree describes data but not decisions; rather the resulting classification tree can be an input for decision making.

Q354UGC NET Computer Science, August 2016 (Paper II & III, Selected Questions)
Discovery of cross sales opportunities is called as _____
  1. AAssociation
  2. BVisualization
  3. CCorrelation
  4. DSegmentation
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Correct answer: (A) Association

Explanation

Association rule learning is a method for discovering intersecting relations between variables in large data base. It is intended to identify strong rules disconnected in database using some measures of interestingness.

Q355UGC NET Computer Science, August 2016 (Paper II & III, Selected Questions)
Consider a database table R with attributes A and B. Which of the following SQL queries is illegal?
  1. ASELECT A FROM R;
  2. BSELECT A, COUNT (*) FROM R;
  3. CSELECT A, COUNT(*) FROM GROUP BY A;
  4. DSELECT A, B, COUNT (*) FROM R GROUP BY A, B;
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Correct answer: (B) SELECT A, COUNT (*) FROM R;

Explanation

SELECT A, COUNT(*) FROM R; is illegal: when an aggregate function (SUM, AVG, COUNT, MAX, MIN) is used in the SELECT clause along with a non-aggregated attribute, a GROUP BY clause on that attribute is required. SELECT A, B, COUNT(*) FROM R GROUP BY A, B; is legal.

Q356UGC NET Computer Science, August 2016 (Paper II & III, Selected Questions)
DBMS provides the facility of accessing data from a database through
  1. ADDL
  2. BDML
  3. CDBA
  4. DSchema
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Correct answer: (B) DML

Explanation

Accessing the data from the database in DBMS done through DML (Data manipulation language). The DML command can access the data from a database.

Q357UGC NET Computer Science, August 2016 (Paper II & III, Selected Questions)
Relational database schema normalization is NOT for :
  1. AReducing the number of joins required to satisfy a query
  2. BEliminating uncontrolled redundancy of data stored in the database
  3. CEliminating number of anomalies that could otherwise occur with inserts and deletes
  4. DEnsuring that functional dependencies are enforced
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Correct answer: (A) Reducing the number of joins required to satisfy a query

Explanation

Relational database schema normalization is not for reducing the number of joins required to satisfy a query (normalization usually increases joins). Its goals are eliminating redundancy and anomalies and enforcing functional dependencies.

Q358UGC NET Computer Science, August 2016 (Paper II & III, Selected Questions)
Consider the following statement regarding relational database model : (1) NULL values can be used to opt a tuple out enforcement of a foreign key (2) Suppose that table T has only one candidate key. If Q is in 3NF, then it is also in BCNF (3) The difference between the project operator (Π) in relational algebra and the SELECT keyword in SQL is that if the resulting table/set has more than one occurrences of the same tuple, then Π will return only one of them, while SQL SELECT will return all One can determine that :
  1. A(1) and (2) are true
  2. B(1) and (3) are true
  3. C(2) and (3) are true
  4. D(1), (2) and (3) are true
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Correct answer: (D) (1), (2) and (3) are true

Explanation

(1) NULL values can be used to opt a tuple out of enforcement of a foreign key: correct. (2) If a table has only one candidate key and is in 3NF, then it is also in BCNF: correct. (3) Projection (Π) in relational algebra eliminates duplicate tuples while SQL SELECT returns all of them: correct. So (1), (2) and (3) are true.

Q359UGC NET Computer Science, August 2016 (Paper II & III, Selected Questions)
In a fast Ethernet cabling, 100 base-TX uses ........ cable and maximum segment size is ........
  1. ATwisted pair, 100 meters
  2. BTwisted pair, 200 meters
  3. CFibre optics, 1000 meters
  4. DFibre optics, 2000 meters
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Correct answer: (A) Twisted pair, 100 meters

Explanation

Fast Ethernet:- Cabling, 100 Base- Tx uses twisted pair cable and maximum segment size is 100 metres.

Q360UGC NET Computer Science, August 2016 (Paper II & III, Selected Questions)
A network with bandwidth of 10 Mbps can pass an average of 12,000 frames per minute with each frame carrying an average of 10,000 bits. What is the throughput of this network?
  1. A1 Mbps
  2. B2 Mbps
  3. C10 Mbps
  4. D12 Mbps
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Correct answer: (B) 2 Mbps

Explanation

Throughput = (12,000 × 10,000) / 60 bits per second = 2,000,000 bps = 2 Mbps. So answer is option (b).

Q361UGC NET Computer Science, August 2016 (Paper II & III, Selected Questions)
Match the following : List-I A. Session layer B. Application layer C. Presentation layer D. Transport layer List-II i. Virtual terminal software ii. Semantics of the information transmitted iii. Flow control iv. Manage dialogue control Codes : A B C D
  1. Aiv i ii iii
  2. Bi iv ii iii
  3. Civ i iii ii
  4. Div ii i iii
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Correct answer: (A) iv i ii iii

Explanation

The correct match: Session layer – manage dialogue control (A-iv); Application layer – virtual terminal software (B-i); Presentation layer – semantics of the information transmitted (C-ii); Transport layer – flow control (D-iii).

Q362UGC NET Computer Science, August 2016 (Paper II & III, Selected Questions)
Which of the following protocols is used by email server to maintain a central repository that can be accessed from any machine?
  1. APOP3
  2. BIMAP
  3. CSMTP
  4. DDMSP
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Correct answer: (B) IMAP

Explanation

IMAP : The basic idea behind IMAP is for all email server to maintain a central repository that can be accessed from any machine

Q363UGC NET Computer Science, August 2016 (Paper II & III, Selected Questions)
The major shortcoming of waterfall model is :
  1. AThe difficulty in accommodating changes after requirement analysis
  2. BThe difficult in accommodating changes after feasibility analysis
  3. CThe system testing
  4. DThe maintenance of system
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Correct answer: (A) The difficulty in accommodating changes after requirement analysis

Explanation

The main drawback of the waterfall model is the difficulty of accommodating change after the process is under way. (1) Difficult to address change: Inflexible partitioning of the project into distinct stage makes it difficult to respond to changing customer requirements. Therefore, this model is only appropriate when the requirement are well understood and changes will be fairly limited during the design process

Q364UGC NET Computer Science, August 2016 (Paper II & III, Selected Questions)
The quick design of a software that is visible to end users leads to _____
  1. AIterative model
  2. BPrototype model
  3. CSpiral model
  4. DWaterfall model
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Correct answer: (B) Prototype model

Explanation

Prototype typically simulates only a few aspects of, and may be completely different from, the final product. Prototyping has several benefits. The software designer and implementer can get valuable feedback from the user early in the project.

Q365UGC NET Computer Science, August 2016 (Paper II & III, Selected Questions)
Which of the following is used to determine the specificity of requirements? Where n₁ is the number of requirements of which all reviewers have identical interpretations, n₂ is number of requirements in a specification.
  1. An₁/n₂
  2. Bn₂/n₁
  3. Cn₁ + n₂
  4. Dn₁ – n₂
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Correct answer: (A) n₁/n₂

Explanation

Specificity (lack of ambiguity) of requirements is measured as Q = n₁/n₂, the fraction of requirements in the specification for which all reviewers have identical interpretations. So the answer is option (a).

Q366UGC NET Computer Science, August 2016 (Paper II & III, Selected Questions)
For a program of k variables, boundary value analysis yields ______ test cases.
  1. A4k – 1
  2. B4k
  3. C4k + 1
  4. D2ᵏ – 1
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Correct answer: (C) 4k + 1

Explanation

For a program of k variables, boundary value analysis yields 4k + 1 test cases. Robustness testing yields 6k + 1 and worst-case testing yields 5ᵏ test cases.

Q367UGC NET Computer Science, August 2016 (Paper II & III, Selected Questions)
The extent to which a software performs its intended functions without failures, is termed as :
  1. ARobustness
  2. BCorrectness
  3. CReliability
  4. DAccuracy
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Correct answer: (C) Reliability

Explanation

Reliability is the probability of failure free operation of a system over a specified time within a specified environment for a specified purpose.

Q368UGC NET Computer Science, August 2016 (Paper II & III, Selected Questions)
If a function is friend of a class, which one of the following is wrong?
  1. AA function can only be declared a friend by a class itself
  2. BFriend functions are not members of a class, they are associated with it
  3. CFriend functions are members of a class
  4. DIt can have access to all members of the class, even private ones
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Correct answer: (C) Friend functions are members of a class

Explanation

A friend function of a class is defined outside that class's scope but has the right to access all private and protected members of the class. Even though the prototypes for friend functions appear in the class definition, friends are not member functions.

Q369UGC NET Computer Science, August 2016 (Paper II & III, Selected Questions)
A function template in C++ provides ______ level of generalization.
  1. A4
  2. B3
  3. C2
  4. D1
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Correct answer: (C) 2

Explanation

C++ provides two kinds of templates: class templates and function templates. A function template in C++ provides two levels of generalization.

Q370UGC NET Computer Science, August 2016 (Paper II & III, Selected Questions)
Which of the following 2 input Boolean logic functions is linearly inseparable? (i) AND (ii) OR (iii) NOR (iv) XOR (v) NOTXOR
  1. A(i) and (ii)
  2. B(ii) and (iii)
  3. C(iii), (iv) and (v)
  4. D(iv) and (v)
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Correct answer: (D) (iv) and (v)

Explanation

XOR is linearly inseparable. NOTXOR is just XOR with the 1's swapped with 0's and vice versa; swapping the labels on the points does not alter the structure of the space, so NOTXOR is also linearly inseparable. AND, OR and NOR are linearly separable.

Q371UGC NET Computer Science, August 2016 (Paper II & III, Selected Questions)
Consider the following statements: (i) If primal (dual) problem has a finite optimal solution, then its dual (primal) problem has a finite optimal solution. (ii) If primal (dual) problem has an unbounded optimum solution, then its dual (primal) has no feasible solution at all. (iii) Both primal and dual problems may be infeasible. Which of the following is correct?
  1. A(i) and (ii) only
  2. B(i) and (iii) only
  3. C(ii) and (iii) only
  4. D(i), (ii) and (iii)
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Correct answer: (D) (i), (ii) and (iii)

Explanation

These are fundamental duality theorems of linear programming. All three statements are correct.

Q372UGC NET Computer Science, August 2016 (Paper II & III, Selected Questions)
A ripple counter is a (n):
  1. ASynchronous counter
  2. BAsynchronous counter
  3. CParallel counter
  4. DNone of the above
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Correct answer: (B) Asynchronous counter

Explanation

A ripple counter is an asynchronous counter where only the first flip-flop is clocked by an external clock. All subsequent flip-flops are clocked by the output of the preceding flip-flop. Asynchronous counters are also called ripple counters because of the way the clock pulse ripples its way through the flip-flops.

Q373UGC NET Computer Science, August 2016 (Paper II & III, Selected Questions)
Let R and S be two fuzzy relations defined as R (rows x₁, x₂; columns y₁, y₂): x₁: [0.7 0.5] x₂: [0.8 0.4] and S (rows y₁, y₂; columns z₁, z₂, z₃): y₁: [0.9 0.6 0.2] y₂: [0.1 0.7 0.5] Then, the resulting relation, T, which relates elements of universe of X to elements of universe of Z using max-product composition is given by
  1. AT (columns z₁, z₂, z₃): x₁: [0.68 0.89 0.39], x₂: [0.76 0.72 0.32]
  2. BT (columns z₁, z₂, z₃): x₁: [0.68 0.89 0.39], x₂: [0.72 0.76 0.32]
  3. CT (columns z₁, z₂, z₃): x₁: [0.63 0.42 0.35], x₂: [0.72 0.48 0.20]
  4. DT (columns z₁, z₂, z₃): x₁: [0.05 0.35 0.14], x₂: [0.04 0.28 0.16]
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Correct answer: (C) T (columns z₁, z₂, z₃): x₁: [0.63 0.42 0.35], x₂: [0.72 0.48 0.20]

Explanation

Max-product composition: T(x, z) = max over y of R(x, y) × S(y, z). T(x₁, z₁) = max(0.7 × 0.9, 0.5 × 0.1) = 0.63; T(x₁, z₂) = max(0.42, 0.35) = 0.42; T(x₁, z₃) = max(0.14, 0.25) = 0.35. T(x₂, z₁) = max(0.72, 0.04) = 0.72; T(x₂, z₂) = max(0.48, 0.28) = 0.48; T(x₂, z₃) = max(0.16, 0.20) = 0.20. So T = x₁: [0.63 0.42 0.35], x₂: [0.72 0.48 0.20].

Q374UGC NET Computer Science, August 2016 (Paper II & III, Selected Questions)
Consider the following identities for regular expressions: (i) (r + s)* = (s + r)* (ii) (r*) = r* (iii) (r* s*)* = (r + s)* Which of the above identities are true?
  1. A(i) and (ii) only
  2. B(ii) and (iii) only
  3. C(iii) and (i) only
  4. D(i), (ii) and (iii)
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Correct answer: (D) (i), (ii) and (iii)

Explanation

(r + s)* generates any string over r and s, the same as (s + r)*. (r*)* generates any string of r's, the same as r*. (r* s*)* also generates any string over r and s, the same as (r + s)*. All the identities are true.

Q375UGC NET Computer Science, August 2016 (Paper II & III, Selected Questions)
The register that stores the bits required to mask the interrupts is
  1. AStatus register
  2. BInterrupt service register
  3. CInterrupt mask register
  4. DInterrupt request register
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Correct answer: (C) Interrupt mask register

Explanation

Interrupt Mask Register (IMR): An Interrupt mask register is a specialized register in a computer's CPU or a dedicated peripheral used to enable or disable (mask) individual interrupt lines.

Q376UGC NET Computer Science, August 2016 (Paper II & III, Selected Questions)
In ________ addressing mode, the operands are stored in the memory. The address of the corresponding memory location is given in a register which is specified in the instruction.
  1. ARegister direct
  2. BRegister indirect
  3. CBase indexed
  4. DDisplacement
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Correct answer: (B) Register indirect

Explanation

In register indirect addressing, the location (address) of the operand in memory is held in a register that is specified in the instruction.

Q377UGC NET Computer Science, August 2016 (Paper II & III, Selected Questions)
8085 microprocessor has _______ bit ALU.
  1. A32
  2. B16
  3. C8
  4. D4
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Correct answer: (C) 8

Explanation

ALU is designed to operate on 8-bit data therefore the 8085 microprocessor has 8-bit ALU.

Q378UGC NET Computer Science, August 2016 (Paper II & III, Selected Questions)
Which of the following in 8085 microprocessor performs HL = HL + HL ?
  1. ADAD D
  2. BDAD H
  3. CDAD B
  4. DDAD SP
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Correct answer: (B) DAD H

Explanation

DAD H: double add HL to HL⇒HL=HL+HL

Q379UGC NET Computer Science, August 2016 (Paper II & III, Selected Questions)
If there are n integers to sort, each integer has d digits, and each digit is in the set {1, 2, .... k}, radix sort can sort the numbers in:
  1. AO(k(n + d))
  2. BO(d(n + k))
  3. CO((n + k) lg d)
  4. DO((n + d) lg k)
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Correct answer: (B) O(d(n + k))

Explanation

Radix sort performs one stable counting sort per digit; each pass takes O(n + k) time and there are d digits. So n integers of d digits, each digit in {1, 2, ..., k}, are sorted in O(d(n + k)) time.

Q380UGC NET Computer Science, August 2016 (Paper II & III, Selected Questions)
Consider the following linear programming problem: Max, z = 0.50x₂ – 0.10x₁ Subject to the constraints 2x₁ + 5x₂ ≤ 80 x₁ + x₂ ≤ 20 and x₁, x₂ ≥ 0 The total maximum profit (z) for the above problem is:
  1. A6
  2. B8
  3. C10
  4. D12
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Correct answer: (B) 8

Explanation

z is maximized by taking x₁ = 0 and x₂ as large as possible: 5x₂ ≤ 80 gives x₂ = 16 (and x₁ + x₂ = 16 ≤ 20). So when x₁ = 0 and x₂ = 16, Max z = 0.50 × 16 = 8.

Q381UGC NET Computer Science, August 2016 (Paper II & III, Selected Questions)
Assume that the program 'P' is implementing parameter passing with 'call by reference'. What will be printed by following print statements in P? Program P( ) { x = 10; y = 3; funb (y, x, x); print x; print y; } funb (x, y, z) { y = y + 4; z = x + y + z; }
  1. A10, 7
  2. B31, 3
  3. C10, 3
  4. D31, 7
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Correct answer: (B) 31, 3

Explanation

Inside funb( ), x refers to y of main, and both y and z refer to x of main. y = y + 4 makes main's x = 14. z = x + y + z = 3 + 14 + 14 = 31, which is stored in main's x. main's y is unchanged (3). Printed output is 31, 3.

Q382UGC NET Computer Science, August 2016 (Paper II & III, Selected Questions)
Constructors have _______ return type.
  1. Avoid
  2. Bchar
  3. Cint
  4. Dno
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Correct answer: (D) no

Explanation

A constructor resembles an instance method, but it differs from a method since it has no explicit return type, it is not implicitly inherited and it usually has different rules for scope modifiers.

Q383UGC NET Computer Science, August 2016 (Paper II & III, Selected Questions)
Method over-riding can be prevented by using final as a modifier at
  1. Athe start of the class.
  2. Bthe start of method declaration.
  3. Cthe start of derived class.
  4. Dthe start of the method declaration in the derived class.
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Correct answer: (B) the start of method declaration.

Explanation

To disallow a method from being overridden, specify final as a modifier at the start of its declaration. Methods declared as final cannot be overridden.

Q384UGC NET Computer Science, August 2016 (Paper II & III, Selected Questions)
Which of the following is a correct statement?
  1. AComposition is a strong type of association between two classes with full ownership.
  2. BComposition is a strong type of association between two classes with partial ownership.
  3. CComposition is a weak type of association between two classes with partial ownership.
  4. DComposition is a weak type of association between two classes with strong ownership.
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Correct answer: (A) Composition is a strong type of association between two classes with full ownership.

Explanation

Composition is a strong type of association between two classes with full ownership. Child objects do not have their own lifecycle without the parent object; if a parent object is deleted, all its child objects are also deleted.

Q385UGC NET Computer Science, August 2016 (Paper II & III, Selected Questions)
Which of the following is not a correct statement?
  1. AEvery class containing abstract method must be declared abstract.
  2. BAbstract class can directly be initiated with 'new' operator.
  3. CAbstract class can be initiated.
  4. DAbstract class does not contain any definition of implementation.
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Correct answer: (B) Abstract class can directly be initiated with 'new' operator.

Explanation

An abstract class cannot be directly instantiated with the new operator; since an abstract class does not contain a complete implementation, it is not possible to create an abstract object.

Q386UGC NET Computer Science, August 2016 (Paper II & III, Selected Questions)
Java uses threads to enable the entire environment to be _________ .
  1. ASymmetric
  2. BAsymmetric
  3. CSynchronous
  4. DAsynchronous
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Correct answer: (D) Asynchronous

Explanation

Java uses threads to enable the entire environment to be asynchronous. This helps reduce inefficiency by preventing the waste of CPU cycles.

Q387UGC NET Computer Science, August 2016 (Paper II & III, Selected Questions)
An Operating System (OS) crashes on the average once in 30 days, that is, the Mean Time Between Failures (MTBF) = 30 days. When this happens, it takes 10 minutes to recover the OS, that is, the Mean Time To Repair (MTTR) = 10 minutes. The availability of the OS with these reliability figures is approximately:
  1. A96.97%
  2. B97.97%
  3. C99.009%
  4. D99.97%
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Correct answer: (D) 99.97%

Explanation

Converting 30 days into minutes, the fraction of time the system is not available = 10/(30 × 24 × 60) × 100 ≈ 0.023%. Availability = 100 – 0.023 ≈ 99.97%.

Q388UGC NET Computer Science, August 2016 (Paper II & III, Selected Questions)
Consider a file currently consisting of 50 blocks. Assume that the file control block and the index block is already in memory. If a block is added at the end (and the block information to be added is stored in memory), then how many disk I/O operations are required for indexed (single-level) allocation strategy?
  1. A1
  2. B101
  3. C27
  4. D0
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Correct answer: (A) 1

Explanation

The file control block and the index block are already in memory, so for indexed allocation only the new block has to be written to disk: 1 disk I/O operation. (Contiguous and linked allocation would need 1 and 3 operations respectively in this scenario, per the book.)

Q389UGC NET Computer Science, August 2016 (Paper II & III, Selected Questions)
An operating system supports a paged virtual memory, using a central processor with a cycle time of one microsecond. It costs an additional one microsecond to access a page other than the current one. Pages have 1000 words, and the paging device is a drum that rotates at 3000 revolutions per minute and transfers one million words per second. Further, one percent of all instructions executed accessed a page other than the current page. The instruction that accessed another page, 80% accessed a page already in memory and when a new page was required, the replaced page was modified 50% of the time. What is the effective access time on this system, assuming that the system is running only one process and the processor is idle during drum transfers?
  1. A30 microseconds
  2. B34 microseconds
  3. C60 microseconds
  4. D68 microseconds
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Correct answer: (B) 34 microseconds

Explanation

Effective access time = 0.99 × (1 µsec + 0.008 × 2 µsec) + 0.002 × (10000 µsec + 1000 µsec) + 0.001 × (10000 µsec + 1000 µsec) ≈ (0.99 + 0.016 + 22.0 + 11.0) µsec ≈ 34.0 µsec.

Q390UGC NET Computer Science, August 2016 (Paper II & III, Selected Questions)
Consider the following page reference string: 1, 2, 3, 4, 2, 1, 5, 6, 2, 1, 2, 3, 7, 6, 3, 2, 1, 2, 3, 6 Which of the following options, gives the correct number of page faults related of LRU, FIFO, and optimal page replacement algorithms respectively, assuming 05 page frames and all frames are initially empty?
  1. A10, 14, 8
  2. B8, 10, 7
  3. C7, 10, 8
  4. D7, 10, 7
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Correct answer: (B) 8, 10, 7

Explanation

With 5 frames: (1) LRU algorithm: total number of page faults = 8. (2) FIFO algorithm: total number of page faults = 10. (3) Optimal algorithm: total number of page faults = 7.

Q391UGC NET Computer Science, August 2016 (Paper II & III, Selected Questions)
Consider the following operations to be performed in Unix: "The pipe sorts all files in the current directory modified in the month of "June" by order of size and prints them to the terminal screen. The sort option skips ten fields then sorts the lines in numeric order." Which of the following Unix command will perform above set of operation?
  1. Als -l | grep "June" | sort +10n
  2. Bls -l | grep "June" | sort +10r
  3. Cls -l | grep -v "June" | sort +10n
  4. Dls -l | grep -n "June" | sort +10x
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Correct answer: (A) ls -l | grep "June" | sort +10n

Explanation

ls -l lists the files with details, grep "June" keeps the lines for files modified in June, and sort +10n skips ten fields and sorts the lines in numeric order. So the correct command is ls -l | grep "June" | sort +10n.

Q392UGC NET Computer Science, August 2016 (Paper II & III, Selected Questions)
Which of the following statement(s) is/are True regarding 'nice' command of UNIX? I. It is used to set or change the priority of a process. II. A process's nice value can be set at the time of creation. III. 'nice' takes a command line as an argument.
  1. AI, II only
  2. BII, III only
  3. CI, II, III
  4. DI, III only
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Correct answer: (C) I, II, III

Explanation

A process's nice value (priority) can be set at the time of creation with the nice command, which takes a command line as an argument, and adjusted later with the renice command. So I, II and III are all true.

Q393UGC NET Computer Science, August 2016 (Paper II & III, Selected Questions)
The Global conceptual Schema in a distributed database contains information about global relations. The condition that all the data of the global relation must be mapped into the fragments, that is, it must not happen that a data item which belongs to a global relation does not belong to any fragment, is called:
  1. ADisjointness condition
  2. BCompleteness condition
  3. CReconstruction condition
  4. DAggregation condition
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Correct answer: (B) Completeness condition

Explanation

Completeness Condition: All the data of the global relation must be mapped into the fragments; it must not happen that a data item which belongs to a global relation does not belong to any fragment.

Q394UGC NET Computer Science, August 2016 (Paper II & III, Selected Questions)
Consider the following ORACLE relations: One (x, y) = {<2, 5>, <1, 6>, <1, 6>, <1, 6>, <4, 8>, <4, 8>} Two (x, y) = {<2, 55>, <1, 1>, <4, 4>, <1, 6>, <4, 8>, <4, 8>} Consider the following two SQL queries SQ1 and SQ2: SQ1: (SELECT * FROM One) EXCEPT (SELECT * FROM Two); SQ2: (SELECT * FROM One) EXCEPT ALL (SELECT * FROM Two); For each of the SQL queries, what is the cardinality (number of rows) of the result obtained when applied to the instances above?
  1. A2 and 1 respectively
  2. B1 and 2 respectively
  3. C2 and 2 respectively
  4. D1 and 1 respectively
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Correct answer: (B) 1 and 2 respectively

Explanation

EXCEPT returns the records present in the first relation and absent in the second, eliminating duplicates: SQ1 gives the single tuple <2, 5>. EXCEPT ALL keeps duplicates, removing one occurrence per matching tuple: SQ2 gives <2, 5> and <1, 6> (2 tuples). So the cardinalities are 1 and 2.

Q395UGC NET Computer Science, August 2016 (Paper II & III, Selected Questions)
Aliasing in the context of programming languages refers to:
  1. AMultiple variables having the same location
  2. BMultiple variables having the same identifier
  3. CMultiple variables having the same value
  4. DMultiple use of same variable
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Correct answer: (A) Multiple variables having the same location

Explanation

Aliasing describes a situation in which a data location in memory can be accessed through different symbolic names in the program. Modifying the data through one name implicitly modifies the value associated with all aliased names.

Q396UGC NET Computer Science, August 2016 (Paper II & III, Selected Questions)
Which of the following statements is/are True regarding some advantages that an object-oriented DBMS (OODBMS) offers over a relational database? I. An OODBMS avoids the "impedance mismatch" problem. II. An OODBMS avoids the "phantom" problem. III. An OODBMS provides higher performance concurrency control than most relational database. IV. An OODBMS provides faster access to individual data objects once they have been read from disk.
  1. AII and III only
  2. BI and IV only
  3. CI, II, and III only
  4. DI, III and IV only
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Correct answer: (B) I and IV only

Explanation

I. An OODBMS avoids the "impedance mismatch" problem: true. II. An OODBMS does not avoid the "phantom" problem. III. An OODBMS does not provide higher-performance concurrency control than most relational databases. IV. An OODBMS provides faster access to individual data objects once they have been read from disk: true. So I and IV only.

Q397UGC NET Computer Science, August 2016 (Paper II & III, Selected Questions)
Suppose transmission rate of a channel is 32 kbps. If there are '8' routes from source to destination and each packet p contains 8000 bits. Total end to end delay in sending packet P is ______ .
  1. A2 sec
  2. B3 sec
  3. C4 sec
  4. D1 sec
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Correct answer: (A) 2 sec

Explanation

Each route (hop) takes 8000/32000 = 0.25 second. For all 8 routes the total delay = 8 × 8000/32000 = 2 seconds.

Q398UGC NET Computer Science, August 2016 (Paper II & III, Selected Questions)
In CRC checksum method, assume that given frame for transmission is 11010111011 and the generator polynomial is G(x) = x⁴ + x + 1. After implementing CRC encoder, the encoded word sent from sender side is ____ .
  1. A11010110111110
  2. B11101101011011
  3. C110101111100111
  4. D110101111001111
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Correct answer: (A) 11010110111110

Explanation

Note: the frame as printed (11010111011) actually gives the codeword 110101110111011, which matches no option; the keyed answer corresponds to the frame 1101011011 used in the book's solution. The generator G(x) = x⁴ + x + 1 is 10011. Append 4 zeros to the message 1101011011 to get 11010110110000 and divide by 10011; the remainder is 1110, so the transmitted message is 11010110111110.

Q399UGC NET Computer Science, August 2016 (Paper II & III, Selected Questions)
Consider the following statements: I. High speed Ethernet works on optic fiber. II. A point to point protocol over Ethernet is a network protocol for encapsulating PPP frames inside Ethernet frames. III. High speed Ethernet does not work on optic fiber. IV. A point to point protocol over Ethernet is a network protocol for encapsulating Ethernet frames inside PPP frames. Which of the following is correct?
  1. AI and II are true; III and IV are false.
  2. BI and II are false; III and IV are true.
  3. CI, II, III and IV are true.
  4. DI, II, III and IV are false.
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Correct answer: (A) I and II are true; III and IV are false.

Explanation

Statement I– True, High speed Ethernet can utilize various media, including optical fiber. It's not limited to just copper cables. Statement II– True, the definition of PPPoE is to encapsulate PPP frames within Ethernet frames. Statement III– False Since high speed Ethernet can work on optic fiber, this statement in incorrect. Statement IV– False, The correct encapsulation is PPP frames inside Ethernet frames.

Q400UGC NET Computer Science, August 2016 (Paper II & III, Selected Questions)
Match each software lifecycle model in List-I to its description in List-II: List-I A. Code-and-Fix B. Evolutionary prototyping C. Spiral D. Staged Delivery E. Waterfall List-II 1. Assess risks at each step; do most critical action first. 2. Build an initial small requirement specifications, code it, then "evolve" the specifications and code as needed. 3. Build initial requirement specification for several releases, then design-and-code in sequence. 4. Standard phases (requirements, design, code, test) in order 5. Write some code, debug it, repeat (i.e. adhoc) Codes: A B C D E
  1. A5 2 1 3 4
  2. B5 3 1 2 4
  3. C4 1 2 3 5
  4. D3 5 1 2 4
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Correct answer: (A) 5 2 1 3 4

Explanation

Code-and-fix: write some code, debug it, repeat (A-5). Evolutionary prototyping: the system is developed incrementally by evolving the specification and code (B-2). Spiral model: assess risks at each step (C-1). Staged delivery: requirements for several releases first, then design and code in sequence (D-3). Waterfall model: sequential, non-iterative standard phases (E-4).

How to use these UGC NET Computer Science questions

Every question here was asked in a real UGC NET paper; the paper and shift are shown on the card, and the number opens the question on its own page. Try each one before tapping Show answer, then attempt the full UGC NET papers as timed tests to check your speed under exam conditions.