UGC NET CS 2014 June-paper-3

Question 1

Beam-penetration and shadow-mask are the two basic techniques for producing color displays with a CRT. Which of the following is not true ?

I. The beam-penetration is used with random scan monitors.

II. Shadow-mask is used in raster scan systems.

III. Beam-penetration method is better than shadow-mask method.

IV. Shadow-mask method is better than beam-penetration method.
A
I and II
B
II and III
C
III only
D
IV only
       Computer-Graphics       CRT
Question 1 Explanation: 
Beam Penetration Method for displaying color pictures has been used with random-scan monitors. Two layers of phosphor, usually red and green, are coated on to the inside of the CRT screen, and the displayed color depends on how far the electron beam penetrates into the phosphor layers.
Shadow-mask Methods are commonly used in raster-scan systems (including color TV) because they produce a much wider range of color than the beam penetration method. A shadow-mask CRT has three phosphor color dots at each pixel position. One phosphor dot emits a red light, another emits a green light, and the third emits a blue light.
Question 2
Line caps are used for adjusting the shape of the line ends to give them a better appearance. Various kinds of line caps used are
A
Butt cap and sharp cap
B
Butt cap and round cap
C
Butt cap, sharp cap and round cap
D
Butt cap, round cap and projecting square cap
       Computer-Graphics       2D-Graphics
Question 2 Explanation: 
Line caps helps in improving the appearance of the line by adjusting the shape of line ends There are three types of line caps:
1. But Cap 2. Round Cap 3. Projection Square Cap
Question 3

Given below are certain output primitives and their associated attributes. Match each primitive with its corresponding attributes :

List – I List – II

a. Line i. Type, Size, Color

b. Fill Area ii. Color, Size, Font

c. Text iii. Style, Color, Pattern

d. Marker iv. Type, Width, Color
A
a-i, b-ii, c-iii, d-iv
B
a-ii, b-i, c-iii, d-iv
C
a-iv, b-iii, c-ii, d-i
D
a-iii, b-i, c-iv, d-ii
       Computer-Graphics       2D-Graphics
Question 3 Explanation: 
Line Attributes: Basic attributes of a straight line segment are its type, its width, and its color. In some graphics packages, lines can also be displayed using selected pen or brush options.
AREA-FILL ATTRIBUTES: Basic attributes of an area-fill are
Fill Styles: Areas are displayed with three basic fill styles: hollow with a color border, filled with a solid color, or Wed with a specified pattern or design.
Pattern Fill
Soft Fill: the linear soft-fill algorithm repaints an area that was originally painted by merging a foreground color F with a single background color 8, where F + B.The current RGB color P of each pixel within the area to be refilled is some linear combination of F and B:
P = tF + (1- t)B
where the "transparency" factor t has a value between 0 and 1 for each pixel. RGB component of the colors
P = (PR, PC, PR), F = (FR, Fc, FR), B = (BR, Bc, BB)
Text Attributes: Following are the text attributes:
Font : is the choice of font (or typeface), which is a set of characters with a particular design style such as New York, Courier, Helvetica, London, 'Times Roman, and various special symbol groups.
Font size
Font colour
Marker Attributes: A marker symbol is a single character that can be displayed in different colors and in different sizes. We select a particular character to be the marker symbol with
setMarkerType (int)
Hence Marker attributes are type, size and color.
Question 4
Consider a window bounded by the lines : x = 0; y= 0; x = 5 and y = 3. The line segment joining (–1, 0) and (4, 5), if clipped against this window will connect the points
A
(0, 1) and (2, 3)
B
(0, 1) and (3, 3)
C
(0, 1) and (4, 3)
D
(0, 1) and (3, 2)
       Computer-Graphics       Line-clipping
Question 4 Explanation: 
Question 5
Which of the following color models are defined with three primary colors ?
A
RGB and HSV color models
B
CMY and HSV color models
C
HSV and HLS color models
D
RGB and CMY color models
       Computer-Graphics       Basics
Question 5 Explanation: 
Common color models defined with three primary colors are the RGB and CMY models.
RGB: color scheme is an additive model. Intensities of the primary colors are added to produce other colors. Each color point within the bounds of the cube can be represented as a triple (R, G, B), where values for R, G, and B are assigned in the range from 0 to 1. Thus, a color C, is expressed in RGB components as
C = RR + GG + BB
CMY: A color model defined with the primary colors cyan, magenta, and yellow (CMY) is useful for describing color output to hard-copy devices. We can express the conversion from an RGB representation to a CMY representation with the matrix transformation:
Question 6
In a digital transmission, the receiver clock is 0.1 percent faster than the sender clock. How many extra bits per second does the receiver receive if the data rate is 1 Mbps?
A
10 bps
B
100 bps
C
1000 bps
D
10000 bps
       Computer-Networks       Bit-rate
Question 6 Explanation: 
Question 7

Given U={1,2,3,4,5,6,7}

A={(3,0.7),(5,1),(6,0.8)}

then ~A will be: (where ~ → complement)
A
{(4,0.7), (2,1), (1,0.8)}
B
{(4,0.3), (5,0), (6,0.2)}
C
{(1,1), (2,1), (3, 0.3), (4, 1), (6, 0.2), (7, 1)}
D
{(3, 0.3), (6.0.2)}
       Artificial-intelligence       Fuzzy-sets
Question 7 Explanation: 
Question 8
Consider a fuzzy set old as defined below Old = {(20, 0.1), (30, 0.2), (40, 0.4), (50, 0.6), (60, 0.8), (70, 1), (80, 1)} Then the alpha-cut for alpha = 0.4 for the set old will be
A
{(40, 0.4)}
B
{50, 60, 70, 80}
C
{(20, 0.1), (30, 0.2)}
D
{(20, 0), (30, 0), (40, 1), (50, 1), (60, 1), (70, 1), (80, 1)}
       Artificial-intelligence       Fuzzy-sets
Question 8 Explanation: 
If the membership of an element is greater than or equal to alpha-cut value then its membership will become one in the resultant set otherwise it’s membership will be zero(i.e. Will not be a member of the resulting set).
Here alpha = 0.4
So new set = {(20, 0), (30, 0), (40, 1), (50, 1), (60, 1), (70, 1), (80, 1)}
OR
{(40, 1), (50,1), (60.1), (70,1), (80,1)}
Question 9
Perceptron learning, Delta learning and LMS learning are learning methods which falls under the category of
A
Error correction learning – learning with a teacher
B
Reinforcement learning – learning with a critic
C
Hebbian learning
D
Competitive learning – learning without a teacher
       Artificial-intelligence       Artificial-neural-networks(ANN)
Question 10

Match the following with respect to the jump statements :

List – I List – II

a. return i. The conditional test and increment portions

b. goto ii. A value associated with it

c. break iii. Requires a label for operation

d. continue iv. An exit from only the innermost loop
A
a-ii, b-iii, c-iv, d-i
B
a-iii, b-iv, c-i, d-ii
C
a-iv, b-iii, c-ii, d-i
D
a-iv, b-iii, c-i, d-ii
       Programming       Control-Statement
Question 10 Explanation: 
The return statement terminates the execution of a function and returns control to the calling function. Execution resumes in the calling function at the point immediately following the call. A return statement can also return a value to the calling function.
Goto statement: The goto statement transfers the program control directly to a labeled statement.A common use of goto is to transfer control to a specific switch-case label or the default label in a switch statement. The goto statement is also useful to get out of deeply nested loops.
Break statement: The break statement terminates the execution of the nearest enclosing do, for, switch, or while statement in which it appears. Control passes to the statement that follows the terminated statement.
Continue statement: The continue statement passes control to the next iteration of the nearest enclosing do, for, or while statement in which it appears, bypassing any remaining statements in the do, for, or while statement body.
Question 11
The control string in C++ consists of three important classifications of characters
A
Escape sequence characters, Format specifiers and Whitespace characters
B
Special characters, White-space characters and Non-white space characters
C
Format specifiers, White-space characters and Non-white space characters
D
Special characters, White-space characters and Format specifiers
       OOPS       C++
Question 12
Code blocks allow many algorithms to be implemented with the following parameters :
A
clarity, elegance, performance
B
clarity, elegance, efficiency
C
elegance, performance, execution
D
execution, clarity, performance
       Programming       Functions
Question 12 Explanation: 
Algorithm to be implemented should be:
Clear: Clear means it should convey the exact work you are trying to accomplish.
Question 13

Match the following with respect to I/O classes in object oriented programming :

List – I List – II

a. fopen() i. returns end of file

b. fclose() ii. return for any problem report

c. ferror() iii. returns 0

d. feof() iv. returns a file pointer
A
a-iv, b-i, c-ii, d-iii
B
a-iii, b-i, c-iv, d-ii
C
a-ii, b-iii, c-iv, d-i
D
a-iv, b-iii, c-i, d-ii
       OOPS       C++
Question 13 Explanation: 
fopen() → returns a file pointer
fclose() → returns end of file
ferror() → return for any problem report
feof() → returns 0
Question 14

Which one of the following describes the syntax of prolog program ?

I. Rules and facts are terminated by full stop (.)

II. Rules and facts are terminated by semicolon (;)

III. Variables names must start with upper case alphabets.

IV. Variables names must start with lower case alphabets.
A
I, II
B
III, IV
C
I, III
D
II, IV
       Artificial-intelligence       Prolog
Question 14 Explanation: 
Rules:
Question 15
Let L be any language. Define even (W) as the strings obtained by extracting from W the letters in the even-numbered positions and even(L) = {even (W) | W ∈ L}. We define another language Chop (L) by removing the two leftmost symbols of every string in L given by Chop(L) = {W | ν W ∈ L, with | ν | = 2}. If L is regular language then
A
even(L) is regular and Chop(L) is not regular.
B
Both even(L) and Chop(L) are regular.
C
even(L) is not regular and Chop(L) is regular.
D
Both even(L) and Chop(L) are not regular.
       Theory-of-Computation       Regular-Language
Question 16
Software testing is
A
the process of establishing that errors are not present.
B
the process of establishing confidence that a program does what it is supposed to do.
C
the process of executing a program to show that it is working as per specifications.
D
the process of executing a program with the intent of finding errors.
       Software-Engineering       Software-testing
Question 16 Explanation: 
Software testing is the process of executing a program with the intent of finding errors.
Question 17
Assume that a program will experience 200 failures in infinite time. It has now experienced 100 failures. The initial failure intensity was 20 failures/CPU hr. Then the current failure intensity will be
A
5 failures/CPU hr
B
10 failures/CPU hr.
C
20 failures/CPU hr.
D
40 failures/CPU hr.
       Software-Engineering       Software-management
Question 17 Explanation: 
Question 18

Consider a project with the following functional units :

Number of user inputs = 50

Number of user outputs = 40

Number of user enquiries = 35

Number of user files = 06

Number of external interfaces = 04

Assuming all complexity adjustment factors and weighing factors as average, the function points for the project will be
A
135
B
722
C
675
D
672
       Software-Engineering       Software-management
Question 18 Explanation: 
UFP= (50*4)+ ( 40*5)+(35*4)+(6*10)+(4*7)
UFP= 628
CAF = 0.65 + ( 0.01 * F )
F = 14 * scale
Since weighing factor is given as average. So scale value= 3
F= 42.
CAF = 1.07
FP= UFP* CAF
FP= 628*1.07
FP= 671.96
FP= 672(approximately)
Question 19

Match the following :

List – I List – II

a. Correctness i. The extent to which a software tolerates the unexpected problems

b. Accuracy ii. The extent to which a software meets its specifications

c. Robustness iii. The extent to which a software has specified functions

d. Completeness iv. Meeting specifications with precision
A
a-ii, b-iv, c-i, d-iii
B
a-i, b-ii, c-iii, d-iv
C
a-ii, b-i, c-iv, d-iii
D
a-iv, b-ii, c-i, d-iii
       Software-Engineering       Software-management
Question 19 Explanation: 
Correctness : It refers to the extent to which a software meets its specifications.
Accuracy: Accuracy means meeting specifications with precision.
Robustness: Robustness refers to the extent to which a software tolerates the unexpected problems .
Completeness: Completeness means The extent to which a software has specified functions
Question 20
Which one of the following is not a definition of error ?
A
It refers to the discrepancy between a computed, observed or measured value and the true, specified or theoretically correct value.
B
It refers to the actual output of a software and the correct output.
C
It refers to a condition that causes a system to fail.
D
It refers to human action that results in software containing a defect or fault.
       Software-Engineering       Software-testing
Question 20 Explanation: 
Error refers to the difference between actual output of a software to the correct output of a software. Error leads to fault and fault leads to failure of a software system.
Question 21
Which one of the following is not a key process area in CMM level 5 ?
A
Defect prevention
B
Process change management
C
Software product engineering
D
Technology change management
       Software-Engineering       Software-management
Question 21 Explanation: 
1. Initial (chaotic, ad hoc, individual heroics) - the starting point for use of a new or undocumented 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/confirmed as a standard business process
4. Capable - the process is quantitatively managed in accordance with agreed-upon metrics.
5. Efficient - process management includes deliberate process optimization/improvement.
Level 5 - Optimizing (Efficient) It is a characteristic of processes at this level that the focus is on continually improving process performance through both incremental and innovative technological changes/improvements. At maturity level 5, processes are concerned with addressing statistical common causes of process variation and changing the process (for example, to shift the mean of the process performance) to improve process performance. This would be done at the same time as maintaining the likelihood of achieving the established quantitative process-improvement objectives. There are only a few companies in the world that have attained this level 5.
Question 22

Consider the following relational schemas for a library database : Book (Title, Author, Catalog_no, Publisher, Year, Price) Collection(Title, Author, Catalog_no) with the following functional dependencies :

I. Title, Author → Catalog_no

II. Catalog_no → Title, Author, Publisher, Year

III. Publisher, Title, Year → Price Assume (Author, Title) is the key for both schemas. Which one of the following is true ?
A
Both Book and Collection are in BCNF.
B
Both Book and Collection are in 3NF.
C
Book is in 2NF and Collection in 3NF.
D
Both Book and Collection are in 2NF.
       Database-Management-System       Normalization
Question 22 Explanation: 
Book (Title, Author, Catalog_no, Publisher, Year, Price)
(Author, Title) is the key .
Title, Author → Catalog_no
Catalog_no →Publisher
Here we are having a transitive dependency
(Key→non-key
Non-key→ non-Key)
Hence this relation is not in 3NF.
Collection(Title, Author, Catalog_no)
(Author, Title) is the key
Catalog_no → Title, Author
Since here LHS is not a Key so it is not in BCNF but since the RHS having prime key attribute so it is in 3NF.
Question 23
Specialization Lattice stands for
A
An entity type can participate as a subclass in only one specialization.
B
An entity type can participate as a subclass in more than one specialization.
C
An entity type that can participate in one specialization.
D
An entity type that can participate in one generalization.
       Database-Management-System       ER-Model
Question 23 Explanation: 
An entity type can participate as a subclass in more than one specialization.
Question 24

Match the following :

List – I List – II

a. Timeout ordering protocol i. Wait for graph

b. Deadlock prevention ii. Roll back

c. Deadlock detection iii. Wait-die scheme

d. Deadlock recovery iv. Thomas Write Rule
A
a-iv, b-iii, c-i, d-ii
B
a-iii, b-ii, c-iv, d-i
C
a-ii, b-i, c-iv, d-iii
D
a-iii, b-i, c-iv, d-iii
       Database-Management-System       Transactions
Question 24 Explanation: 
Timeout ordering protocol: Thomas Write Rule is a timeout protocol. It suffers with a large number of aborts.
Deadlock prevention: Wait-die scheme used for deadlock prevention.
Deadlock detection: Wait for graph is used to detect the deadlock.
Deadlock recovery: Deadlock recovery can be done by roll back operation.
Question 25

Consider the schema R = {S, T, U, V} and the dependencies

S → T, T → U, U → V and V → S If R = (R1 and R2)

be a decomposition such that R1 ∩ R2 = φ then the decomposition is
A
not in 2NF
B
in 2NF but not in 3NF
C
in 3NF but not in 2NF
D
in both 2NF and 3NF
       Database-Management-System       Normalization
Question 25 Explanation: 
Decomposition of a relation R into R1 and R2 is said to be lossless only if R1 ∩ R2 ≠ φ and the common attribute of R1 and R2 should be the primary key in any of the two decomposed relations(i.e.R1 and R2).
A relation which is in 2NF is always lossless.
In question, it is given that R1 ∩ R2 = φ which is violating lossless decomposition condition. Hence the given relation R is not in 2NF.
Question 26
Which one of the following is not a Client-Server application ?
A
Internet chat
B
Web browser
C
E-mail
D
Ping
       Computer-Networks       Client-Server-Computing
Question 26 Explanation: 
Ping is not a Client-Server application.
Question 27

Which of the following concurrency protocol ensures both conflict serializability and freedom from deadlock :

I. 2-phase locking

II. Time phase ordering
A
Both I & II
B
II only
C
I only
D
Neither I nor II
       Database-Management-System       Transactions
Question 27 Explanation: 
2-phase locking : 2-phase locking ensures conflict serializability and does not ensure freedom from deadlock.
Time phase ordering: It ensures both conflict serializability and freedom from deadlock.
Question 28

Match the following:

List-I List-II

a. Expert systems i. Pragmatics

b. Planning ii. Resolution

c. Prolog iii. Means-end analysis

d. Natural language processing iv. Explanation facility
A
a-iii, b-iv, c-i, d-ii
B
a-iii, b-iv, c-ii, d-i
C
a-i, b-ii, c-iii, d-iv
D
a-iv, b-iii, c-ii, d-i
       Artificial-intelligence       Approaches-to-AI
Question 28 Explanation: 
Expert systems → Explanation facility
Planning → Means-end analysis
Prolog → Resolution
Natural language processing → Pragmatics
Question 29
STRIPS addresses the problem of efficiently representing and implementation of a planner. It is not related to which one of the following ?
A
SHAKEY
B
SRI
C
NLP
D
None of these
       Artificial-intelligence       Natural-language-processing
Question 29 Explanation: 
STRIPS is related SHAKEY,SRI but not NLP.
Question 30
Slots and facets are used in
A
Semantic Networks
B
Frames
C
Rules
D
All of these
       Artificial-intelligence       Knowledge-representation
Question 30 Explanation: 
Slots and facets are used in frames
Question 31
Consider f(N) = g(N) + h(N) Where function g is a measure of the cost of getting from the start node to the current node N and h is an estimate of additional cost of getting from the current node N to the goal node. Then f(N) = h(N) is used in which one of the following algorithms ?
A
A* algorithm
B
AO* algorithm
C
Greedy best first search algorithm
D
Iterative A* algorithm
       Algorithms       Algorithm-Paradigms
Question 31 Explanation: 
Greedy best-first search
  f(n) = estimate of cost from n to goal
  e.g., f(n) = straight-line distance from n
to Bucharest   Greedy best-first search expands the node that appears to be closest to the goal.
Question 32
________predicate calculus allows quantified variables to refer to objects in the domain of discourse and not to predicates or functions.
A
Zero-order
B
First-order
C
Second-order
D
High-order
       Engineering-Mathematics       Propositional-Logic
Question 32 Explanation: 
First-order predicate calculus allows quantified variables to refer to objects in the domain of discourse and not to predicates or functions.
Question 33
________ is used in game trees to reduce the number of branches of the search tree to be traversed without affecting the solution.
A
Best first search
B
Goal stack planning
C
Alpha-beta pruning procedure
D
Min-max search
       Data-Structures       Trees
Question 33 Explanation: 
Alpha-beta pruning procedure is used in game trees to reduce the number of branches of the search tree to be traversed without affecting the solution.
Question 34
Consider a uniprocessor system where new processes arrive at an average of five processes per minute and each process needs an average of 6 seconds of service time. What will be the CPU utilization ?
A
80 %
B
50 %
C
60 %
D
30 %
       Operating-Systems       Process-Scheduling
Question 34 Explanation: 
5 processes arrives in 60seconds.
Service time for 1 process--------> 6sec
Service time for 5 processes------> 5*6 sec
= 30 seconds
CPU utilization = (30/60)*100
= 50%
Question 35
Consider a program that consists of 8 pages (from 0 to 7) and we have 4 page frames in the physical memory for the pages. The page reference string is : 1 2 3 2 5 6 3 4 6 3 7 3 1 5 3 6 3 4 2 4 3 4 5 1 The number of page faults in LRU and optimal page replacement algorithms are respectively (without including initial page faults to fill available page frames with pages) :
A
9 and 6
B
10 and 7
C
9 and 7
D
10 and 6
       Operating-Systems       Page-Replacement-algorithm
Question 35 Explanation: 
Question 36
Which of the following statements is not true about disk-arm scheduling algorithms ?
A
SSTF (shortest seek time first) algorithm increases performance of FCFS.
B
The number of requests for disk service are not influenced by file allocation method.
C
Caching the directories and index blocks in main memory can also help in reducing disk arm movements.
D
SCAN and C-SCAN algorithms are less likely to have a starvation problem.
       Operating-Systems       Disk-Scheduling
Question 36 Explanation: 
Option(B) is false because the number of requests for disk service are influenced by file allocation method.
Question 37
_________ maintains the list of free disk blocks in the Unix file system.
A
I-node
B
Boot block
C
Super block
D
File allocation table
       Operating-Systems       File system
Question 37 Explanation: 
An inode object represents an individual file.
Boot Block: Boot block is the first sector allocated to a file system. It contains bootstrap program used to load the operating system into main memory.
A superblock object represents an entire file system.A superblock is a record of the characteristics of a filesystem, including its size, the block size, the empty and the filled blocks and their respective counts
File Allocation Table(FAT): It is used by the operating system to manage files on the hard drive and other computer systems.
Question 38
A part of Windows 2000 operating system that is not portable is
A
Device Management
B
Virtual Memory Management
C
Processor Management
D
User Interface
       Operating-Systems       Types-of-Operating-System
Question 38 Explanation: 
A part of Windows 2000 operating system that is not portable is Virtual Memory Management
Question 39

Match the following with reference to Unix shell scripts :

List – I List – II

a. $? i. File name of the current script

b. $# ii. List of arguments

c. $0 iii. The number of arguments

d. $* iv. Exit status of last command
A
a-iii, b-ii, c-i, d-iv
B
a-ii, b-iii, c-i, d-iv
C
a-iv, b-iii, c-i, d-ii
D
a-i, b-iii, c-i, d-iv
       Operating-Systems       Unix-Shell
Question 39 Explanation: 
$? → Exit status of last command
$# → The number of arguments
$0 → File name of the current script
$* → List of arguments
Question 40
The advantage of _______ is that it can reference memory without paying the price of having a full memory address in the instruction.
A
Direct addressing
B
Indexed addressing
C
Register addressing
D
Register Indirect addressing
       Computer-Organization       Addressing-Modes
Question 40 Explanation: 
Direct Addressing mode: In this mode of addressing operand is present in the memory location provided in the instruction.
Indexed addressing mode: In this mode of addressing the the base address is provided in the instruction and the index address is provided in the index register.
It is useful in array implementation.
Register Addressing: In this mode of addressing operand is present in the register provided in the instruction.
Register indirect addressing mode: In this mode of addressing, operand is present at the memory location provided in the register, given in the instruction.
Question 41
The reverse polish notation equivalent to the infix expression ((A + B) * C + D)/(E + F + G)
A
A B + C * D + EF + G + /
B
A B + C D * + E F + G + /
C
A B + C * D + E F G + +/
D
A B + C * D + E + F G + /
       Data-Structures       Prefix-Postfix-Expression
Question 41 Explanation: 
Question 42
The output of a sequential circuit depends on
A
present input only
B
past input only
C
both present and past input
D
past output only
E
None of the above
       Digital-Logic-Design       Sequential-Circuits
Question 42 Explanation: 
Question 43
A byte addressable computer has a memory capacity of 2m Kbytes and can perform 2n operations. An instruction involving 3 operands and one operator needs a maximum of
A
3m bits
B
m + n bits
C
3m + n bits
D
3m + n + 30 bits
       Computer-Organization       Microprogrammed-Control-Unit
Question 43 Explanation: 
Total Number of operations = 2n
Hence the number of bits required to identify each operation uniquely = n-bits.
Total memory capacity = 2m Kbytes
= 2m+10 bytes.
Hence the number of bits required to identify each operand uniquely = m+10 bits.
Since an instruction include 3 operands. So total bits in the instruction to identify each operand = 3(m+10) bits.
So total number of bits required by an instruction involving 3 operands and one operator
= 3(m+10)+ n bits
= 3m+n+30 bits.
Question 44
Which of the following flip-flops is free from race condition ?
A
T flip-flop
B
SR flip-flop
C
Master-slave JK flip-flop
D
None of the above
       Digital-Logic-Design       Sequential-Circuits
Question 44 Explanation: 
Master-slave JK flip-flop is free from race condition.
Question 45
One of the main features that distinguish microprocessor from micro-computers is
A
words are usually larger in microprocessors.
B
words are shorter in microprocessors.
C
microprocessor does not contain I/O devices.
D
None of the above.
       Computer-Organization       Microprocessor
Question 45 Explanation: 
Microprocessor: Microprocessor is a simple central processing unit (CPU) on a single chip (remember the word ‘Single Chip’). It includes Arithmetic logic unit (ALU), control unit (CU), registers, instruction decoders, bus control circuit etc. but everything should be on a single chip.
Microcomputer: A microcomputer is the association of microprocessor and the peripheral I/O devices, support circuitry and memory (both data and program). It is not necessary to be on a single chip (remember this point, not in a single chip).
Question 46

The output generated by the LINUX command :

$ seq 1 2 10 will be
A
1 2 10
B
1 2 3 4 5 6 7 8 9 10
C
1 3 5 7 9
D
1 5 10
       Operating-Systems       LINUX-Operating-systems
Question 46 Explanation: 
Sequence command is used to print a sequence of numbers.
Syntax: $ seq ‘first’ ‘increment’ ‘last’
Where “first” is representing first number to be printed
“Increment” represents the amount of increment at each stage.
“Last” represents the last number we can print.
So the output of $ seq 1 2 10 command will be 1,3,5,7,9
Note: 10 is not printed because it will violate the increment condition given in question.
Question 47
All the classes necessary for windows programming are available in the module :
A
win . txt
B
win . main
C
win . std
D
MFC
       Operating-Systems       Windows-Operating-System
Question 47 Explanation: 
All the classes necessary for windows programming are available in the module win . txt .
Question 48
Windows 32 API supports
A
16-bit Windows
B
32-bit Windows
C
64-bit Windows
D
All of the above
       Operating-Systems       Windows-Operating-System
Question 48 Explanation: 
Windows API Index was formerly called the Win32 API. The name Windows API more accurately reflects its roots in 16-bit Windows and its support on 64-bit Windows.
Question 49
Superficially the term “object oriented”, means that, we organize software as a
A
collection of continuous objects that incorporates both data structure and behaviour.
B
collection of discrete objects that incorporates both discrete structure and behaviour.
C
collection of discrete objects that incorporates both data structure and behaviour.
D
collection of objects that incorporates both discrete data structure and behaviour.
       Software-Engineering       OOAD-diagrams
Question 49 Explanation: 
Superficially the term “object oriented”, means that, we organize software as a collection of discrete objects that incorporates both data structure and behaviour.
Question 50
The “part-whole”, or “a-part-of” relationship in which objects representing the components of something associated with an object representing the entire assembly is called as
A
Association
B
Aggregation
C
Encapsulation
D
Generalisation
       OOPS       Properties
Question 50 Explanation: 
The “part-whole”, or “a-part-of” relationship in which objects representing the components of something associated with an object representing the entire assembly is called as Aggregation
Question 51
The pure object oriented programming language with extensive metadata available and modifiable at run time is
A
Small talk
B
C++
C
Java
D
Eiffel
       OOPS       Properties
Question 51 Explanation: 
The pure object oriented programming language with extensive metadata available and modifiable at run time is Small talk
Question 52

Match the following interfaces of Java. Servlet package :

List – I List – II

a. Servlet Config i. Enables Servlets to log events

b. Servlet Context ii. Read data from a client

c. Servlet Request iii. Write data to a client

d. Servlet Response iv. To get initialization parameters
A
a-iii, b-iv, c-ii, d-i
B
a-iii, b-ii, c-iv, d-i
C
a-ii, b-iii, c-iv, d-i
D
a-iv, b-i, c-ii, d-iii
       OOPS       JAVA
Question 52 Explanation: 
Servlet Config: A servlet configuration object used by a servlet container to pass information to a servlet during initialization.
Servlet Context : Defines a set of methods that a servlet uses to communicate with its servlet container, for example, to get the MIME type of a file, dispatch requests, or write to a log file.
Servlet Request: Defines an object to provide client request information to a servlet. Servlet Response: Defines an object to assist a servlet in sending a response to the client.
Question 53
The syntax of capturing events method for document object is
A
CaptureEvents()
B
CaptureEvents(Orgs eventType)
C
CaptureEvents(eventType)
D
CaptureEvents(eventVal)
       Web-Technologies       Events
Question 53 Explanation: 
window. captureEvents(event)
window. captureEvents(event1 | event2 | eventN)
The captureEvents() method captures all the events of the event type passed.
Question 54
Linking to another place in the same or another webpage require two A (Anchor) tags, the first with the ________ attribute and the second with the _________ attribute.
A
NAME & LINK
B
LINK & HREF
C
HREF & NAME
D
TARGET & VALUE
       Web-Technologies       Tags
Question 54 Explanation: 
The < a > tag defines a hyperlink, which is used to link from one page to another.
The most important attribute of the < a > element is the href attribute, which indicates the link's destination.
< a href = “https://www.solutionsadda.in”> one stop solution< / a >
In above example when you will click on “one stop solution” you will visit the website “https://www.solutionsadda.in”
Question 55
Given an image of size 1024 × 1024 pixels in which the intensity of each pixel is an 8-bit quality. It requires _______ of storage space if the image is not compressed.
A
one Terabyte
B
one Megabyte
C
8 Megabytes
D
8 Terabytes
       Computer-Organization       Graphics
Question 55 Explanation: 
size of each pixel is 210×210 and intensity of each pixel is 8-bits.
So, total storage space= 210×210×8 bits
= 220 Bytes
= 1MB
Question 56

Match the following cryptographic algorithms with their design issues :

List – I List – II

a. DES i. Message Digest

b. AES ii. Public Key

c. RSA iii. 56-bit key

d. SHA-1 iv. 128-bit key
A
a-ii, b-i, c-iv, d-iii
B
a-iii, b-i, c-iv, d-ii
C
a-iii, b-iv, c-ii, d-i
D
a-iv, b-i, c-ii, d-iii
       Computer-Networks       Network-Security
Question 56 Explanation: 
Data Encryption Standard(DES): is a symmetric-key block cipher published by the National Institute of Standards and Technology (NIST).
Key sizes‎: ‎56 bits (+8 parity bits)
Rounds‎: ‎16
Block sizes‎: ‎64 bits
Advanced Encryption Standard(AES):
Block size:128 bits
Key lengths: 128, 192 and 256 bits.
The number of rounds are as follows:
10 rounds for 128-bit keys.
12 rounds for 192-bit keys.
14 rounds for 256-bit keys.
RSA (Rivest–Shamir–Adleman) is one of the first public-key cryptosystems and is widely used for secure data transmission. In such a cryptosystem, the encryption key is public and it is different from the decryption key which is kept secret (private).
SHA-1 (Secure Hash Algorithm 1) is a cryptographic hash function which takes an input and produces a 160-bit (20-byte) hash value known as a message digest – typically rendered as a hexadecimal number, 40 digits long.
Question 57
Consider a code with five valid code words of length ten : 0000000000, 0000011111, 1111100000, 1110000011, 1111111111 Hamming distance of the code is
A
5
B
10
C
8
D
9
       Computer-Networks       Error-Control-Methods
Question 57 Explanation: 
Question 58
Which of the following special cases does not require reformulation of the problem in order to obtain a solution ?
A
Alternate optimality
B
Infeasibility
C
Unboundedness
D
All of the above
       Engineering-Mathematics       Optimization
Question 59
The given maximization assignment problem can be converted into a minimization problem by
A
subtracting each entry in a column from the maximum value in that column.
B
subtracting each entry in the table from the maximum value in that table.
C
adding each entry in a column from the maximum value in that column.
D
adding maximum value of the table to each entry in the table.
       Engineering-Mathematics       Optimization
Question 59 Explanation: 
The given maximization assignment problem can be converted into a minimization problem by subtracting each entry in the table from the maximum value in that table.
Question 60

The initial basic feasible solution of the following transportation problem :



is given as



then the minimum cost is
A
76
B
78
C
80
D
82
       Engineering-Mathematics       Optimization
Question 60 Explanation: 
Since in question initial feasible solution is already given then we can easily calculate the minimum cost by doing the following operation:
(5*2)+ (8*1)+(7*4)+(2*1)+(2*6)+(10*2)
=10+8+28+2+12+20
=80
Question 61

Given the following equalities :

E1 : nK+∈ + nK lg n = θ(nK+∈) for all fixed K and ∈, K > 0 and ∈ > 0.

E2 : n32n + 6n23n = O(n32n)

Which of the following is true ?
A
E1 is correct and E2 is correct.
B
E1 is correct and E2 is not correct.
C
E1 is not correct and E2 is correct.
D
E1 is not correct and E2 is not correct.
       Algorithms       Time-Complexity
Question 61 Explanation: 
E1 is TRUE because nK+∈ is leading term. E2 is FALSE because n23n leading term but given as O(n32n).
Question 62
Consider the fractional knapsack instance n = 4, (p1, p2, p3, p4) = (10, 10, 12, 18), (w1, w2, w3, w4) = (2, 4, 6, 9) and M = 15. The maximum profit is given by (Assume p and w denotes profit and weight of objects respectively)
A
40
B
38
C
32
D
30
       Algorithms       Greedy-approach
Question 62 Explanation: 
Question 63

The solution of the recurrence relation of

T(n) = 3T(floor(n/4)+n) is
A
O(n2)
B
O(nlg n)
C
O(n)
D
O(l g n)
       Algorithms       Recurrences
Question 63 Explanation: 
Using master’s method:
a=3, b=4, k=1, p=0
Here ak
So T(n) = O(n)
Question 64
If h is chosen from a universal collection of hash functions and is used to hash n keys into a table of size m, where n < m, the expected number of collisions involving a particular key K is
A
less than 1
B
less than lg n
C
greater than 1
D
greater than lg n
       Data-Structures       Hashing
Question 64 Explanation: 
If h is chosen from a universal collection of hash functions and is used to hash n keys into a table of size m, where n < m, the expected number of collisions involving a particular key K is less than 1.
Question 65

Given the following statements :

S1 : The subgraph-isomorphism problem takes two graphs G1 and G2 and asks whether G1 is a subgraph of G2.

S2 : The set-partition problem takes as input a set S of numbers and asks whether the numbers can be partitioned into two sets A

and

A’ = S – A such that

x∈A x = ∑x∈A’ x

Which of the following is true ?
A
S1 is NP problem and S2 is P problem.
B
S1 is NP problem and S2 is NP problem.
C
S1 is P problem and S2 is P problem.
D
S1 is P problem and S2 is NP problem.
       Algorithms       P-NP
Question 65 Explanation: 
The subgraph-isomorphism problem takes two graphs G1 and G2 and asks whether G1 is a subgraph of G2 is an NP problem.
The set-partition problem takes as input a set S of numbers and asks whether the numbers can be partitioned into two sets A is an NP problem.
Question 66
Suppose that the splits at every level of quicksort are in the proportion (1 – α) to α, where 0 < α ≤ ½ is a constant. The minimum depth of a leaf in the recursion tree is approximately given by
A
–lgn/ lg(1 – α)
B
–(lg (1 – α) / lgn)
C
–lgn / lgα
D
–lgα / lgn
       Algorithms       Sorting
Question 66 Explanation: 
Question 67
Ten signals, each requiring 3000 Hz, are multiplexed on to a single channel using FDM. How much minimum bandwidth is required for the multiplexed channel ? Assume that the guard bands are 300 Hz wide.
A
30,000
B
32,700
C
33,000
D
None of the above
       Computer-Networks       Data-and-Signals
Question 67 Explanation: 
Each signal requires------> 3000Hz
Then 10 signals will require----------> 30,000Hz.
We need one guard between any two channels. So the number of guards required is 10-1=9.
1 guard band-------> 300Hz wide
9 guard band--------> 2700Hz wide.
So total bandwidth required = 30,000+2700
= 32,700.
Question 68
A terminal multiplexer has six 1200 bps terminals and ‘n’ 300 bps terminals connected to it. If the outgoing line is 9600 bps, what is the value of n ?
A
4
B
8
C
16
D
28
       Digital-Logic-Design       Combinational-Circuit
Question 68 Explanation: 
According to given data,
(6*1200)+(n*300)= 9600
7200+300n = 9600
300n = 9600-7200
300n= 2400
n= 8
Question 69
Which of the following is used in the options field of IPv4 ?
A
Strict source routing
B
Loose source routing
C
time stamp
D
All of the above
       Computer-Networks       IP-Address
Question 69 Explanation: 
IPv4 option field used to provide:
Source routing: also called path addressing, allows a sender of a packet to partially or completely specify the route the packet takes through the network..
Source routing can be of two types:
1. Strict source routing
2. Loose source routing
Timestamp may also be included in option field.
Question 70
Which layers of the OSI reference model are host-to-host layers ?
A
Transport, Session, Presentation, Application
B
Network, Transport, Session, Presentation
C
Data-link, Network, Transport, Session
D
Physical, Data-link, Network, Transport
       Computer-Networks       ISO-OSI-layers
Question 70 Explanation: 
Question 71
A network on the Internet has a subnet mask of 255.255.240.0. What is the maximum number of hosts it can handle ?
A
1024
B
2048
C
4096
D
8192
E
none of the above
       Computer-Networks       IP-Address
Question 71 Explanation: 
Subnet mask: It contains 1’s in the network ID part and 0’s in the host ID part.
255.255.240.0 = 11111111 . 11111111 .11110000 . 00000000
maximum number of hosts = 212 - 2
= 4094
Question 72
Four bits are used for packed sequence numbering in a sliding window protocol used in a computer network. What is the maximum window size ?
A
4
B
8
C
15
D
16
       Computer-Networks       Sliding-Window-Protocol
Question 72 Explanation: 
Maximum window size = 2n - 1
n= 4
So, Maximum window size = 24 - 1
= 15
Question 73

Given the following two grammars :

G1 : S → AB | aaB A → a | Aa B → b

G2 : S → a S b S | b S a S | λ

Which statement is correct ?
A
G1 is unambiguous and G2 is unambiguous.
B
G1 is unambiguous and G2 is ambiguous.
C
G1 is ambiguous and G2 is unambiguous.
D
G1 is ambiguous and G2 is ambiguous.
       Compiler-Design       Ambiguous-and-Unambiguous-Grammar
Question 73 Explanation: 
Question 74

Match the following :

List – I List – II

a. Chomsky Normal form i. S → b S S | a S | c

b. Greibach Normal form ii. S → a S b | ab

c. S-grammar iii. S → AS | a

A → SA | b

d. LL grammar iv. S → a B S B

B → b
A
a-iv, b-iii, c-i, d-ii
B
a-iv, b-iii, c-ii, d-i
C
a-iii, b-iv, c-i, d-ii
D
a-iii, b-iv, c-ii, d-i
       Theory-of-Computation       Languages-and-Grammars
Question 74 Explanation: 
Chomsky normal form: A context-free grammar G is said to be in Chomsky normal form if all of its production rules are of the form:
A → BC, or
A → a, or
S → ε,
where A, B, and C are nonterminal symbols, a is a terminal symbol and ε denotes the empty string.
Greibach Normal form: A grammar is said to be in GNF if it in the form: A→ a∝ where ∝ ∈ V*.
Question 75

Given the following two languages :

L1 = {anbn | n > 1} ∪ {a}

L2 = {w C wR | w ∈ {a, b}*}

Which statement is correct ?
A
Both L1 and L2 are not deterministic.
B
L1 is not deterministic and L2 is deterministic.
C
L1 is deterministic and L2 is not deterministic.
D
Both L1 and L2 are deterministic.
       Theory-of-Computation       Languages-and-Grammars
Question 75 Explanation: 
L1 = { a, aabb, aaabbb,............}
In this if the input is single “a” then we can pop it and can reach to the final state and if input contains more than one “a” then after pushing all a’s when “b” come as input start deleting one “a” for each “b” input symbol and then reach the final stage.
Hence L1 is deterministic.
L2 is also deterministic.
There are 75 questions to complete.