Paper 3 has been removed from NET from 2018 (Notification) now paper 2 and 3 syllabus is included in paper 2. Practice both paper 2 and 3 from past papers.
CBSE NET ComputerScience September2013 Solved Paper II
Download PDF of This Page (Size: 232K) ↧
 Secrets to easily score in UGC PaperIGet India's number 1 postal course with thoursands of UGC NET PaperI questions Examrace CBSEUGC PaperI Postal Course

A file is downloaded in a home computer using a 56 kbps MODEM connected to an Internet Service Provider. If the download of file completes in 2 minutes, what is the maximum size of data downloaded?

112 Mbits

6.72 Mbits

67.20 Mbits

672 Mbits
Answer: b


In ______ CSMA protocol, after the station finds the line idle, it sends or refrains from sending based on the outcome of a random number generator.

Nonpersistent

0persistent

1persistent

ppersistent
Answer: d


Which of the following substitution technique have the relationship between a character in the plaintext and a character in the ciphertext as onetomany?

Monoalphabetic

Polyalphabetic

Transpositional

None of the above
Answer: b


What is the maximum length of CAT5 UTP cable in Fast Ethernet network?

100 meters

200 meters

1000 meters

2000 meters
Answer: a


The ______ is a set of standards that defines how a dynamic web document should be written, how input data should be supplied to the program, and how the output result should be used.

Hyper Text Markup Language

File Transfer Protocol

Hyper Text Transfer Protocol

Common Gateway Interface
Answer: d


The counttoinfinity problem is associated with

Flooding algorithm

Hierarchical routing algorithm

Distance vector routing algorithm

Link state routing algorithm
Answer: c


The IEEE singleprecision and doubleprecsion format to represent floatingpoint numbers, has a length of ______ and ______ respectively.

8 bits and 16 bits

16 bits and 32 bits

32 bits and 64 bits

64 bits and 128 bits
Answer: c


Consider an undirected graph G with 100 nodes. The maximum number of edges to be included in G so that the graph is not connected is

2451

4950

4851

9900
Answer: c


The amortized time complexity to perform ______ operation (s) in Splay trees is O (Ig n).

Search

Search and Insert

Search and Delete

Search, Insert and Delete
Answer: d


Suppose that the splits at every level of Quicksort are in proportion 1β to β where 0 < β ≤ 0.5 is a constant. The number of elements in an array is n. The maximum depth is approximately

0.5 β Ig n

0.5 (1 – β) Ig n

– (Ig n)/(Ig β)

– (Ig n)/Ig (1 – β)
Answer: d


The minimum number of nodes in a binary tree of depth d (root is at level 0) is

2d – 1

2d + 1 – 1

d + 1

d
Answer: c


The efficient data structure to insert/delete a number in a stored set of numbers is

Queue

Linked list

Doubly linked list

Binary tree
Answer: c


The number of states in a minimal deterministic finite automaton corresponding to the language L = { an  n≥4 } is

3

4

5

6
Answer: c


Regular expression for the language L = {w ∈ {0, 1} *  w has no pair of consecutive zeros} is

(1 + 010) *

(01 + 10) *

(1 + 010) * (0 + λ)

(1 + 01) * (0 + λ)
Answer: d


Consider the following two languages: L1 = {an bl ak  n + l + k>5} L2 = {an bl ak n>5, l >3, k≤ l} Which of the following is true?

L1 is regular language and L2 is not regular language.

Both L1 and L2 are regular languages.

Both L1 and L2 are not regular languages.

L1 is not regular language and L2 is regular language.
Answer: a


LL grammar for the language L = {an bm cn + m  m≥0, n≥0} is

S → aSc  S1; S1 → bS1c  λ

S → aSc  S1; S1 → bS1c

S → aSc  S1 λ S1 → bS1c λ

S → aSc  λ S1 → bS1c λ
Answer: c


Assume the statements S1 and S2 given as: S1: Given a context free grammar G, there exists an algorithm for determining whether L (G) is infinite. S2: There exists an algorithm to determine whether two context free grammars generate the same language. Which of the following is true?

S1 is correct and S2 is not correct.

Both S1 and S2 are correct.

Both S1 and S2 are not correct.

S1 is not correct and S2 is correct.
Answer: a


The number of eightbit strings beginning with either 111 or 101 is ______.

64

128

265

None of the above
Answer: a


Find the number of ways to paint 12 offices so that 3 of them will be green, 2 of them pink, 2 of them yellow and the rest ones whte.

55, 440

1, 66, 320

4.790E + 08

39, 91, 680
Answer: b


Consider the following statements:

A graph in which there is a unique path between every pair of vertices is a tree.

A connected graph with e = v – 1 is a tree.

A graph with e = v – 1 that has no circuit is a tree.
Which of the above statements is/are true?

1 & 3

2 & 3

1 & 2

All of the above
Answer: d


Consider the Inorder and Postorder traversals of a tree as given below: Inorder: j e n k o p b f a c l g m d h i Postorder: j n o p k e f b c l m g h i d a The Preorder traversal of the tree shall be

a b f e j k n o p c d g l m h 1

a b c d e f j k n o p g l m h 1

a b e j k n o p f c d g l m h 1

j e n o p k f b c l m g h 1 d a
Answer: c


A simple graph G with nvertices is connected if the graph has

(n – 1) (n – 2)/2 edges

More than (n – 1) (n – 2)/2 edges

Less than (n – 1) (n – 2)/2 edges

∑ki = 1 C (ni, 2) edges
Answer: b


Which one of the following set of gates is best suited for ‘parity’ checking and ‘parity’ generation?

AND, OR, NOT

NAND, NOR

EXOR, EXNOR

None of the above
Answer: c


The quantification ∃ x P (x) denotes the proposition “There exists a unique x such that P (x) is true” express the quantification using universal and existential quantifications and logical operators:

∃x P (x) ∨∀x∀y ( (P (x) ∨ P (y) ) → x = y)

∀ x P (x) ∧∀x∀y ( (P (x) ∨ P (y) ) → x = y)

∃x P (x) ∧∀x∀y ( (P (x) ∧ P (y) ) → x = y)

∃x P (x) ∧∃x∃y ( (P (x) ∨ P (y) ) → x = y)
Answer: c


If F and G are Boolean functions of degree n. Then, which of the following is true?

F ≤ F + G and F G ≤ F

G ≤ F + G and F G ≥ G

F ≥ F + G and F G ≤ F

G ≥ F + G and F G ≤ F
Answer: a


Match the following identities/laws to their corresponding name:
ListI ListII 
x + x = x x • x = x

x + 0 = x x • 1 = x

x + 1 = 1 x • 0 = 0

x • (x + y) = x

Dominance

Absorption

Idempotent

Identity
 A
 B
 C
 D

 3
 4
 1
 2

 4
 3
 1
 2

 4
 3
 2
 1

 3
 4
 2
 1
Answer: a


In which one of the following, continuous process improvement is done?

ISO9001

RMMM

CMM

None of the above
Answer: c


The ______ of a program or computing system is the structure or structures of the system, which comprise software components, the externally visible properties of these components, and the relationship among them.

ER diagram

Data flow diagram

Software architecture

Software design
Answer: c


Working software is not available until late in the process in

Waterfall model

Prototyping model

Incremental model

Evolutionary Development model
Answer: a


Equivalence partitioning is a ______ testing method that divides the input domain of a program into classes of data from which test cases can be derived.

White box

Black box

Regression

Smoke
Answer: b


Consider the following characteristics:

Correct and unambiguous

Complete and consistent

Ranked for importance and/or stability and verifiable

Modifiable and Traceable
Which of the following is true for a good SRS?

1, 2 and 3

1, 3 and 4

2, 3 and 4

1, 2, 3 and 4
Answer: d


Linked Lists are not suitable for _____.

Binary Search

Polynomial Manipulation

Insertion

Radix Sort
Answer: a


What is the size of the following Union? Assume that the size of int = 2, size of float = 4, size of char = 1 union tag { int a; float b; char c; }

2

4

1

7
Answer: b


What is the output of the following program segment? sum (n) { else return (n + sum (n–1) ); } main () { print f ( “%d” sum (5) ); }

10

16

15

14
Answer: c


Assume that x and y is nonzero positive integers. What does the following program segment perform? While (x! = 0) { if (x>y) x = xy else y = yx; printf ( “%d” x); }

Computes LCM of two numbers

Computes GCD of two numbers

Divides large number with small number

Subtracts smaller number from large number
Answer: b


Consider the following program segment: d = 0; For (i = 1; i<31, + + i) for (j = 1; j<31, + + j) for (k = 1; k<31, + + k) if ( ( (i + j + k) %3) = = 0); d = d + 1; printf ( “%d” d); The output will be

9000

3000

90

2700


Usage of Preemption and Transacton Rollback prevents ______.

Unauthorized usage of data file

Deadlock situation

Data manipulation

File preemption
Answer: b


The _____ language was originally designed as the Transformation Language for Style Sheet facility.

XSTL

XML

XQuery

XPath
Answer: a


Views are useful for _____ unwanted information, and for collecting together information from more than one relation into a single view.

Hiding

Deleting

Highlighting

All of the above
Answer: a


The decision tree classifier is a widely used technique for ______.

Classification

Association

Partition

Clustering
Answer: a


Cross_tab displays permit users to view ______ of multidimensional data at a time.

One dimension

Two dimensions

Three dimensions

Multi dimensions
Answer: b


A method to provide secure transmission of email is called ____.

TLS

SA

IPSec

PGP
Answer: d


Thoma'swrite rule is ______.

Two phase locking protocol

Timestamp ordering protocol

One phase locking protocol

Sliding window protocol
Answer: b


Match the following:
ListI (Process state transition) ListII (Reason for transition) 
Ready→ Running

Blocked→ Ready

Running→ Blocked

Running→ Ready

Request made by the process is satisfied or an event for which it was waiting occurs.

Process wishes to wait for some action by another process.

The process is dispatched.

The process is preempted.
 A
 B
 C
 D

 3
 1
 2
 4

 4
 1
 3
 2

 4
 3
 1
 2

 4
 3
 2
 1
Answer: a


The hit ratio of a Translation Look Aside Buffer (TLAB) is 80%. It takes 20 nanoseconds (ns) to search TLAB and 100 ns to access main memory. The effective memory access time is ______.

36 ns

140 ns

122 ns

40 ns
Answer: b


Consider the input/output (I/O) requests made at different instants of time directed at a hypothetical disk having 200 tracks as given in the following table: Serial No. 1 2 3 4 5 Track No. 12 85 40 100 75 Time of arrival 65 80 110 100 175 Assume that: Current head position is at track no. 65 Direction of last movement is towards higher numbered tracks Current clock time is 160 milliseconds Head movement time per track is 1 millisecond. “look” is a variant of “SCAN” diskarm scheduling algorithm. In this algorithm, if no more I/O requests are left in current direction, the disk head reverses its direction. The seek times in Shortest Seek First (SSF) and “look” diskarm scheduling algorithms respectively are:

144 and 123 milliseconds

143 and 123 milliseconds

149 and 124 milliseconds

256 and 186 milliseconds
Answer: b


Assume that an implementation of UNIX operating system uses inodes to keep track of data blocks allocated to a file. It supports 12 direct block addresses, one indirect block address and one double indirect block address. The file system has 256 bytes block size and 2 bytes for disk block address. The maximum possible size of a file in this system is

16 MB

16 KB

70 KB

71 KB


Which of the following set of UNIX commands will always display “WELCOME”

Export title = WELCOME; Echo $title

Title = WELCOME; export $title; sh –c “echo $title”

Title = WELCOME; export title; sh –c “echo $title”

Title = WELCOME; echo $title
Answer: c


The speed up of a pipeline processing over an equivalent nonpipeline processing is defined by the ratio:

S = n tn/(k + n – 1) tp

S = n tn/(k + n + 1) tp

S = n tn/(k – n + 1) tp

S = (k + n – 1) tp/n tn

Where n → no. Of tasks
tn → time of completion of each task
k → no. Of segments of pipeline
tp → clock cycle time
S → speed up ratio
Answer: a