Monday 29 June 2015

UGC CBSE NET JUNE 2015 COMPUTER SCIENCE AND APPLICATIONS PAPER-III SOLUTION (ANSWER KEYS)



CBSE NET JUNE 2015 COMPUTER SCIENCE AND APPLICATIONS PAPER-III SOLUTION

Q. No.
Answer
Q. No.
Answer
Q. No.
Answer
Q. No.
Answer
Q. No.
Answer
1.       
B
2.       
B
3.       
C
4.       
C
5.       
D
6.       
B
7.       
B
8.       
B
9.       
B
10.   
A
11.   
C
12.   
B
13.   
A
14.   

15.   
D
16.   
B
17.   
D
18.   
A
19.   
D
20.   
C
21.   
D
22.   
A
23.   
B
24.   
B
25.   
D
26.   
A
27.   
B
28.   
C
29.   
C
30.   
C
31.   
D
32.   
C
33.   
A
34.   
C
35.   
D
36.   
D
37.   
C
38.   
D
39.   
D
40.   
C
41.   
A
42.   
C
43.   
D
44.   
B
45.   
B
46.   
B
47.   
D
48.   
A
49.   
B
50.   
C
51.   
C
52.   
B
53.   
B
54.   
D
55.   
B
56.   
A
57.   
B
58.   
C OR D
59.   
B
60.   
B
61.   
C
62.   
D
63.   
C
64.   
C
65.   
B
66.   
B
67.   
A
68.   
B
69.   
D
70.   
A
71.   
D
72.   
C
73.   
A
74.   
D
75.   
B


1. For the 8-bit word 00111001, the check bits stored with it would be 0111. Suppose when the word is read from memory, the check bits are calculated to be 1101.What is the data word that was read from memory?
Answer:
The Hamming Word initially calculated was:
bit number:
12 11 10 9 8 7 6 5 4 3 2 1
0   0   1  1  0 1 0 0 1 1 1 1
Doing an exclusive-OR of 0111 and 1101 yields 1010 indicating an error in bit 10 of the Hamming Word. Thus, the data word read from memory was 00011001.

 2. Consider a 32-bit microprocessor, with a 16-bit external data bus, driven by an 8-MHz input clock. Assume that the microprocessor has a bus cycle whose minimum duration equals four input clock cycles. What is maximum data transfer rate for this microprocessor?
Answer: 
Since minimum bus cycle duration = 4 clock cycles and bus clock = 8 MHz
Then, maximum bus cycle rate = 8 M / 4 = 2 M/s
Data transferred per bus cycle = 16 bit = 2 bytes
Data transfer rate per second = bus cycle rate * data per bus cycle = 2 M * 2 = 4 Mbytes/sec.
So, 4 x 106 bytes / sec

3. The RST 7 instruction in 8085 microprocessor is equivalent to
A. CALL 0010 H
B. CALL 0034 H
C. CALL 0038 H
D. CALL 003C H

4. The equivalent hexadecimal notation for octal number 2550276 is:
A.  FADED
B.  AEOBE
C.  ADOBE
D.  ACABE

5. The CPU of a system having 1 MIPS execution rate needs 4 machine cycles on an average for executing an instruction. The fifty percent of the cycles use memory bus. A memory read/write employs one machine cycle. For execution of the programs, the system utilizes 90 percent of the CPU time. For block data transfer, an IO device is attached to the system while CPU executes the background programs continuously. What is the maximum IO data transfer rate if programmed IO data transfer technique is used?
A. 500 Kbytes/sec
B. 2.2 Mbytes/sec
C. 125 Kbytes/sec
D. 250 Kbytes/sec







6. The number of Flip-Flops required to design a modulo - 272 counter is
A. 8
B. 9
C. 27
D. 11
Answer: To construct a Modulo-N counter we need Log2 number of flipflops, the value is rounded to next highest whole number. so log2(272) = 9.

7. Let E1 and E2 be two entities in E-R diagram with simple single valued attributes. R1 and R2 are two relationships between E1 and E2 where R1 is one-many and R2 is many - many. R1 and R2 donot have any attributes of their own. How many minimum number of tables are required to represent this situation in the relational model?
A. 4
B. 3
C. 2
D. 1
Answer:  Strong entities E1 and E2 should be converted into tables. For R1, which is one to many relation, there is no need of a separate table. the "many" side of relation will include the primary key of "one" side as foreign key. For R2, which is many to many relation, a separate table is required by including the primary key of E1 and E2 as foreign keys. Hence we require a minimum of 3 tables. (Thanks to avinash J)

8. The student information in a university is stored in the relation STUDENT (name, sex, marks, dept_Name) Consider the following SQL query select deptName from STUDENT where sex = 'M' group by dept_Name having avg (marks) > (select avg (marks) from STUDENT) It returns the names of the department in which
A. the average marks of male students is more than the average marks of the students in the same department
B. The average marks of male students is more than the average marks of the students in the university
C. The average marks of male students is more than the average marks of male students in the university
D. The average marks of students is more than the average marks of male students in the university

9. Which one of the following statements about normal forms is FALSE?
A. Lossless, preserving decomposition into 3NF is always possible
B. Lossless, preserving decomposition into BCNF is always possible
C. Any relation with two attributes is in BCNF
D. BCNF is stronger than 3NF

10. The relation vendor order (v_no, v_ord_no, v_name, qty_sup, unit_price) is in 2NF because :
A. Non key attribute V_name is dependent on V_no which is part of composite key
B. Non key attribute V_name is dependent on qty_sup
C. key attribute qty_sup is dependent on primary_key unit price
D. key attribute v_ord_no is dependent on primary_key unit price

11. The relational schemas R1 and R2 form a loseless join decomposition of R if and only if
A. R1∧R2–>(R1-R2)
B. R1–>R2
C. R1∧R2–>(R2-R1)
D. R2–>R1∧R2

12. In the index allocation scheme of blocks to a file, the maximum possible size of the file depends on
A. The number of blocks used for the index, and the size of the index
B. The size of the blocks, and the size of the address
C. size of the index
D. size of the blocks

13. Give the number of principal vanishing point (s) along with their direction for the standard perspective transformation :
A. only one in the direction K
B. Two in the directions I and J
C. Three in the directions I, J and K
D. Only two in the directions J and K

14. Consider a triangle represented by A(0, 0), B(1, 1), C(5, 2). The triangle is rotated by 45 degrees about a point P( -1, -1). The co-ordinates of the new triangle obtained after rotation shall be
A. A'(-1, √2 -1), B'(-1, 2√2 – 1), C'((3/2)√2 – 1, (9/2)√2 – 1)
B. A'(√2 -1,- 1), B'( 2√2 – 1, -1), C'((3/2)√2 – 1, (9/2)√2 – 1)
C. A'(-1, √2 -1), B'(2√2 – 1, -1), C'((3/2)√2 – 1, (9/2)√2 – 1)
D. A'(-1, √2 -1), B'( 2√2 – 1, -1), C'((9/2)√2 – 1, (3/2)√2 – 1)

15. The process of dividing an analog signal into a string of discrete outputs, each of constant amplitude is called
A. Strobing
B. Amplification
C. Conditioning
D. Quantization

16. which of the following is not a basic primitive of the graphics kernel system(GKS) ?
A. POLYLINE
B. POLYDRAW
C. FILL AREA
D. POLYMARKER



17.          Which of the following statements is/are incorrect ?
A. Mapping the co-ordinates of the points and lines that form the picture into appropriate co-ordinates on the device or workstation is known as viewing transformation.
B. The right handed cartesian co-ordinate system in whose coordinates we describe the picture is known as world coordinate system.
C. The co-ordinate system that corresponds to the device or workstation where the image is to be displayed is known as physical device co-ordinate system.
D. Left-handed co-ordinate system in which the display area of the virtual display device corresponds to the unit(|x|) square whose lower left handed corner is at origin of the co-ordinate system, is known as normalized device co-ordinate system.

Normalized device coordinates (NDCs) make up a coordinate system that describes positions on a virtual plotting device. The lower left corner corresponds to (0,0), and the upper right corner corresponds to (1,1). NDCs can be used when you want to position text, lines, markers, or polygons anywhere on the plotting device (that may or may not already contain a plot).

18.          Match the following
List-I
List-II
Flood Gun
An electron gun designed to flood the entire screen with electrons.
Collector
Partly energised by flooding gun, stores the charge generated by the writing gun
Ground
used to discharge the collector
Phosphorus Grains
used in memory -tube display and similar to those used in standard CRT
Writing Gun System
used in memory -tube display and basically the same as the electron gun used in a conventional CRT.

Codes : (a) (b) (c) (d) (e)
A.            (i) (ii) (iii) (iv) (v)
B.            (ii) (iii) (i) (iv) (v)
C.            (iii) (i) (ii) (v) (iv)
D.            (iv) (v) (i) (ii) (iii)

19.          Minimal Deterministic finite automaton for the language L = { 0n | n > 0, n ≠ 4 } will have
A.            1 final state among 5 states
B.            4 final states among 5 states
C.            1 final state among 6 states
D.            5 final states among 6 states

20.          The regular expression corresponding to the language L where L = { x {0,1}* | x ends with 1 and does not contain substring 00 } is
A.            (1 + 01)*(10 + 01)
B.            (1 + 01)*01
C.            (1 + 01)*(1 + 01)
D.            (10 + 01)* 01

21.          The transition function for the language L = {w|na(w) and nb(w) are both odd} is given by:
δ(q0, a) = q1       δ(q0, b) = q2
δ(q1, a) = q0       δ(q1, b) = q3
δ(q2, a) = q3       δ(q2, b) = q0
δ(q3, a) = q2       δ(q3, b) = q1
the initial and final states of automata are
A.            q0 and q0 respectively
B.            q0 and q1 respectively
C.            q0 and q2 respectively
D.            q0 and q3 respectively

22.          The clausal form of the disjunctive normal form ¬A ¬B ¬C D is
A.            A B C D
B.            A B C D true
C.            A B C D true
D.            A B C D false

23.          Which of the following is false for the programming language PROLOG ?
A.            A PROLOG variable can only be assigned to a value once
B.            PROLOG is a Strongly Typed Language.
C.            The scope of a variable in PROLOG is a single clause or rule.
D.            the scope of a variable in PROLOG is a Single Query
Prolog is not a strongly typed language, and in fact variables in predicates are not type-restricted in any way. This means that there is a danger of procedures being called with unsuitable parameter values. With a strongly typed language (Pascal, for example) if a procedure is called with too many or the wrong type of argument values, the implementation would automatically flag a type mismatch at or before run-time, and the user would be aware of and could pin-point the error. In Prolog the run-time behaviour in this case would be unpredictable, and the error would be difficult to detect and fix.

24.          Which one of the following is true ?
A.            The resolvent of two horn clauses is not a horn clause.
B.            The resolvent of two horn clauses is a Horn Clause.
C.            if we resolve a negated goal G against a fact or rule A to get Clause C then C has positive literal and non null goal.
D.            if we resolve a negated goal G against a fact or rule A to get clause C then C has positive literal or null goal.
Answer: -
A Horn clause is a clause (a disjunction of literals) with at most one positive, i.e. unnegated, literal. Conversely, a disjunction of literals with at most one negated literal is called a dual-Horn clause. A Horn clause with exactly one positive literal is a definite clause; a definite clause with no negative literals is sometimes called a fact; and a Horn clause without a positive literal is sometimes called a goal clause (note that the empty clause consisting of no literals is a goal clause).  

25.          Which transmission technique guarantees that data packets will be received by the receiver in the same order in which they were sent by the sender ?
A.            Broadcasting
B.            Unicasting
C.            Packet Switching
D.            Circuit Switching
Circuit switching is a methodology of implementing a telecommunications network in which two network nodes establish a dedicated communications channel (circuit) through the network before the nodes may communicate. The circuit guarantees the full bandwidth of the channel and remains connected for the duration of the communication session.

26.          Which of the following control fields in TCP header is used to specify the sender has no more data to transmit ?
A.            FIN
B.            RST
C.            SYN
D.            PSH
FIN: No more data from the sender. Receiving a TCP segment with the FIN flag does not mean that transferring data in the opposite direction is not possible.
RST: Reset the connection. The RST bit is used to RESET the TCP connection due to unrecoverable errors.
SYN: This flag means synchronize sequence numbers. Source is beginning a new counting sequence.
PSH: This flag means Push function. Using this flag, TCP allows a sending application to specify that the data must be pushed immediately.

27.          Which are the two modes of IP security ?
A.            Transport and Certificate
B.            Transport and Tunnel
C.            Certificate and Tunnel
D.            Preshared and Transport
IP security can be implemented in a host-to-host transport mode, as well as in a network tunneling mode.

28.          A message "COMPUTERNETWORK" encrypted(ignore quotes)using columnar transposition cipher with a key "LAYER". the encrypted message is :
A.            CTTOEWMROPNRUEK
B.            MROUEKCTTPNROEW
C.            OEWPNRCTTUEKMRO
D.            UEKPNRMROOEWCTT

29.          Suppose a digitized voice channel is made by digitizing 8 kHz bandwidth analog voice signal. It is required to sample the signal at twice the highest frequency(two samples per hertz). what is the bit rate required, if it is assumed that each sample requires 8 bits ?
A.            32 kbps
B.            64 kbps
C.            128 kbps
D.            256 kbps
Explanation       
The bit rate can be calculated as :
2 x 8000 x 8 = 128000 = 128 kbps.

30.          The Maximum payload of a TCP segment is
A.            65535
B.            65515
C.            65495
D.            65475
Each segment, including the TCP header, must fit in the 65515 byte IP payload, and the TCP header is 20 bytes at least, so the maximum payload of a TCP segment is 65515-20=65495 bytes.


31.       All pair Shortest paths problem is efficiently solved using :
A.        Dijkstra' algorithm
B.        Bellman-Ford algorithm
C.        Kruskal Algorithm
D.        Floyd-Warshall algorithm
Explanation    
The Dijsktra algrithm solves the Single-Source Shortest Paths (SSSP) problem.
The BellmanFord algorithm is an algorithm that computes shortest paths from a single source vertex to all of the other vertices in a weighted digraph.
Kruskal algorithm is a minimum-spanning-tree algorithm where the algorithm finds an edge of the least possible weight that connects any two trees in the forest.It is a greedy algorithm in graph theory as it finds a minimum spanning tree for a connected weighted graph adding increasing cost arcs at each step.
The Floyd Warshal algorithm takes the dynamic programming approach. This essentially means that independent sub-problems are solved and the results are stored for later use. The algorithm allows negative edges, but no negative cycles, as per usual with such shortest path problems.

32.       The travelling salesman problem can be solved in :
A.        Polynomial time using dynamic programming algortihm
B.        Polynomial time using branch and bound algortihm
C.        Exponential time using dynamic programming algorithm or branch and bound algorithm.
D.        Polynomial time using backtracking algorithm.

33.       Which of the following is asymptotically smaller ?
A.        lg(lg*n)
B.        lg*(lg n)
C.        lg(n!)
D.        lg*(n!)

34.       Consider a hash table of size m=100 and the hash function h(k) = floor(m(kA mod 1)) for A = (√5 − 1)/2 = 0.618033. Compute the location to which the key k = 123456 is placed in hash table.
A.        77
B.        82
C.        88
D.        89
Explanation     h(k) = m(kA mod 1),
A=(√5 − 1)/2=0.618033.
given K = 123456, m=100.
h(k) =
100(123456 x 0.618033 mod 1) = 100(76299.88204 mod 1)
=
100 x 0.882048)
=
88.2048
=88

35.       Let f(n) and g(n) be asymptotically non-negative functions. which of the following is correct ?
A.        θ(f(n) * g(n)) = min(f(n), g(n))
B.        θ(f(n) * g(n)) = max(f(n), g(n))
C.        θ(f(n) + g(n)) = min(f(n), g(n))
D.        θ(f(n) + g(n)) = max(f(n), g(n))

36.       The number of nodes of height h in any n - element heap is _____
A.        h
B.        zh
C.        ceil(n/zh)
D.        ceil(n/zh + 1)

37.       In Java, when we implement an interface method, it must be declared as :
A.        Private
B.        Protected
C.        Public
D.        Friend
Explanation     Interfaces are meant to define the public API of a type - and only that, not its implementation. So any method (or static member) you define in an interface is by definition public.

38.       The Servlet Response interface enables a servlet to formulate a response for a client using the method ______
A.        void log(Exception e, String s)
B.        Void destroy()
C.        int getServerPort()
D.        void setContextType(String Type)

39.       Which one of the following is correct ?
A.        Java Applets cannot be written is many programming languages.
B.        An applet is not a small program.
C.        An applet can be run on its own.
D.        Applets are embedded in another applications.

40.       In XML we can specify the frequency of an element by using the symbols :
A.        + * !
B.        # * !
C.        + * ?
D.        - * ?
Explanation     The number of occurrences, or the frequency, of an element can be specified by using the plus(+), asterisk(*) or question mark(?) characters.


41.       In XML, Doctype declaration specifies to include a reference to ____ file
A.        Document Type Definition
B.        Document Type Declaration
C.        Document Transfer Definition
D.        Document Type language
The document type (DOCTYPE) declaration consists of an internal, or references an external Document Type Definition (DTD). It can also have a combination of both internal and external DTDs. The DTD defines the constraints on the structure of an XML document. It declares all of the document's element types glossary, children element types, and the order and number of each element type. It also declares any attributes, entities, notations, processing instructions, comments, and PE references in the document.

42.       Module design is used to maximize cohesion and minimize coupling. Which of the following is the key to implement this rule ?
A.        Inheritance
B.        Polymorphism
C.        Encapsulation
D.        Abstraction
The Golden rule in Module design is used to maximize cohesion and minimize coupling. Encapsulation is the key to implement this golden rule

43.       Verification :
A.        refers to the set of activities that ensure that software correctly implements a specific function.
B.        gives answer to the question - are we building the product right ?
C.        requires execution of software
D.        both 1 and 2
Software verification is a discipline of software engineering whose goal is to assure that software fully satisfies all the expected requirements.
Software verification asks the question, "Are we building the product right?"; that is, does the software conform to its specification.

44.       Which design metric is used to measure the compactness of the program in terms of lines of code ?
A.        Consistency
B.        Conciseness
C.        Efficiency
D.        Accuracy

45.       Requirements prioritization and negotiation belongs to :
A.        Requirements validation
B.        Requirements elicitation
C.        Feasibility Study
D.        Requirement reviews

46.       Adaptive maintenance is a maintenance which ______
A.        Correct errors that were not discovered till testing phase.
B.        is carried out to port the existing software to a new environment.
C.        improves the system performance.
D.        both (2) and (3)

47.       A design concept refinement is a
A.        Top-down Approach
B.        Complementary of Abstraction concept
C.        Process of elaboration
D.        All of the above

48.       A software design is highly modular if :
A.        cohesion is functional and coupling is data type.
B.        cohesion is coincidental and coupling is data type.
C.        cohesion is sequential and coupling is content type.
D.        cohesion is functional and coupling is stamp type.

49.       Match the following for operating system techniques with the most appropriate advantage :
List-I
List-II
(a) Spooling
(i) Allows several jobs in memory to improve CPU utilization
(b) Mult iprogramming
(ii) Access to shared resources among geographically dispersed computers in a transparent way
(c) Time sharing
(iii) overlapping I/O and computations
(d) Distributed Computing
(iv) Allows many users to share a computer simultaneously by switching processor frequently

codes : (a) (b) (c) (d)
A.        (iii) (i) (ii) (iv)
B.        (iii) (i) (iv) (ii)
C.        (iv) (iii) (ii) (i)
D.        (ii) (iii) (iv) (i)

50.       Which of the following statements is not true for Multi Level Feedback Queue processor scheduling algorithm ?
A.        Queues have different priorities.
B.        Each queue may have different scheduling algorithm
C.        Processes are permanently assigned to a queue
D.        This algorithm can be configured to match a specific system under design

51.       What is the most appropriate function of Memory Management Unit (MMU) ?
A.        It is an associative memory to store TLB
B.        It is a technique of supporting multi programming by creating dynamic partitions
C.        It is a chip to map virtual address to physical address
D.        It is an algorithm to allocate and deallocate main memory to a process

52.       Dining Philosopher's problem is a
A.        Producer Consumer problem
B.        Classical IPC problem
C.        Starvation Problem
D.        Synchronization Primitive

53.       In ______ allocation method for disk block allocation in a file system, insertion and deletion of blocks in a file is easy.
A.        Index
B.        Linked
C.        Contiguous
D.        Bit Map

54.       A unix file may be of type :
A.        Regular file
B.        Directory File
C.        Device File
D.        Any one of the above

55.       Match the following :
List-I                           List-II
(a) Intelligence            (i) Contextual, tacit, transfer needs learning
(b) Knowledge            (ii) Scattered facts, easily transferrable
(c) Information            (iii) Judgemental
(d) Data                       (iv) Codifiable, endorsed with relevance and purpose
codes : (a) (b) (c) (d)
A.        (iii) (ii) (iv) (i)
B.        (iii) (i) (iv) (ii)
C.        (i) (ii) (iii) (iv)
D.        (i) (iii) (iv) (ii)