UGC NET CS 2008 Dec-Paper-2

Question 1
The channel capacity of a band-limited Gaussian channel is given by :
A
B log2(2+S/N)
B
B log2(1+S/N)
C
B log10(1+S/N)
D
B loge(1+S/N)
Question 1 Explanation: 
→ Shannon Capacity (Noisy Channel) In presence of Gaussian band-limited white noise, Shannon-Hartley theorem gives the maximum data rate capacity.
C=B log2(1+S/N). where S and N are the signal and noise power, respectively, at the output of the channel.
Question 2
The graph K3,4 has :
A
3 edges
B
4 edges
C
7 edges
D
12 edges
       Engineering-Mathematics       Graph-Theory
Question 2 Explanation: 
The graph K3,4 means it complete bipartite graph with m=4 and n=3
Total number of vertices= n + m
Total number of edges= m*n
Example:
Question 3
The total number of spanning trees that can be drawn using five labelled vertices is :
A
125
B
64
C
36
D
16
       Algorithms       Minimum-Spanning-Tree
Question 3 Explanation: 
To find total number of spanning trees we are using a standard formula is nn-2.
n=5
=55-2
=53
=125
Question 4
Extremely low power dissipation and low cost per gate can be achieved in :
A
MOS ICs
B
C MOS ICs
C
TTL ICs
D
ECL ICs
Question 4 Explanation: 
Emitter Coupled Logic (ECL)
The storage time is eliminated as the transistors are used in difference amplifier mode and are never driven into saturation.
1. Fastest among all logic families
2. Lowest propagation delay.
Complementary metal oxide semiconductor(CMOS)
The power dissipation is usually 10nW per gate depending upon the power supply voltage, output load etc.
1. Lowest power dissipation
2. Excellent noise immunity
3. High packing density
4. Wide range of supply voltage
5. Highest fan out among all logic families
Question 5
An example of a universal building block is :
A
EX-OR gate
B
AND gate
C
OR gate
D
NOR gate
       Digital-Logic-Design       Logic-Gates
Question 5 Explanation: 
→ Universal building blocks are NAND and NOR gate.
→ Basic building blocks are AND,OR and NOT
Question 6
An example of a layer that is absent in broadcast networks is :
A
Physical layer
B
Presentation layer
C
Network layer
D
Application layer
       Computer-Networks       OSI-TCP-layers
Question 7
The ATM cell is :
A
48 bytes long
B
53 bytes long
C
64 bytes long
D
69 bytes long
Question 7 Explanation: 
→ The ATM cell is 53 bytes long.
→ ATM for carriage of a complete range of user traffic, including voice, data, and video signals.
→ ATM provides functionality that is similar to both circuit switching and packet switching networks.
→ ATM uses asynchronous time-division multiplexing, and encodes data into small, fixed-sized packets (ISO-OSI frames) called cells.
→ ATM uses a connection-oriented model in which a virtual circuit must be established between two endpoints before the actual data exchange begins.
Question 8
Four jobs J1, J2, J3 and J4 are waiting to be run. Their expected run times are 9, 6, 3 and 5 respectively. In order to minimise average response time, the jobs should be run in the order:
A
J1 J2 J3 J4
B
J4 J3 J2 J1
C
J3 J4 J1 J2
D
J3 J4 J2 J1
       Operating-Systems       Process-Scheduling
Question 8 Explanation: 
Since arrival time is not given assume arrival time of each process as"0" Option (A):


Question 9
Suppose it takes 100 ns to access page table and 20 ns to access associative memory. If the average access time is 28 ns, the corresponding hit rate is :
A
100 percent
B
90 percent
C
80 percent
D
70 percent
       Operating-Systems       Memory-Management
Question 9 Explanation: 
Given data,
-- Access page table time=100 ns
-- Associate memory=20 ns
-- hit ratio=X
-- Miss ratio=1-X
-- Average Access Time=28
Step-1: AAT= Hit Ratio*Access page table + Miss Ratio*Associate memory
= X*20 + (1-X*100) [Note: If X=0.9 then we are getting exact AAT]
= 28 ns
(or)
28=X*20+(1-X*100)
X=0.9
Question 10
Transmission of N signals, each band limited to fm Hz by TDM, requires a minimum band-width of :
A
fm
B
2 fm
C
N fm
D
2N fm
       Computer-Networks       Access-Control-Methods
Question 10 Explanation: 
→ Time-division multiplexing (TDM) is a digital process that allows several connections to share the high bandwidth of a link Instead of sharing a portion of the bandwidth as in FDM, time is shared. Each connection occupies a portion of time in the link.
→ Since here we have N signals and each signal requires frequency band of fm Hz then total bandwidth required is Nfm Hz.
Question 11
If a code is ‘t’ error detecting, the minimum hamming distance should be equal to :
A
t-1
B
t
C
t+1
D
2t+1
       Digital-Logic-Design       Number-Systems
Question 11 Explanation: 

Question 12
A relation R in {1, 2, 3, 4, 5, 6} is given by { (1, 2), (2, 3), (3, 4), (4, 4), (4, 5) }. This relation is :
A
reflexive
B
symmetric
C
transitive
D
not reflexive, not symmetric and not transitive
       Engineering-Mathematics       Set-Theory
Question 12 Explanation: 
Reflexive: A relation R is said to be reflexive if for all elements of a set x belongs to R (x,x) exists.
Since in above question {(1,1), (2,2),(3,3),(4,4),(5,5),(6,6)} does not exist in given relation so it is not reflexive.
Symmetric: A relation R is symmetric if (x,y) belong to R then (y,x) must belong to R. Since in above question (1,2) exists but (2,1) does not exist in given relation so given relation is not symmetric.
Transitive: A relation is transitive if (x,y) belongs to R and (y,z) belongs to R then (x,z) must belongs to R.
In above question (1, 2), (2, 3) exist in given relation but (1,3) does not exist there so given relation is not transitive.
Hence correct option is option(D)
Question 13
The dual of the switching function x+yz is :
A
x+yz
B
x + yz
C
x(y+z)
D
x(y+z)
       Digital-Logic-Design       Boolean-Function
Question 13 Explanation: 
The dual of switching function means it changes AND to OR and OR to AND.
given x+yz.
x.(y+z)
Question 14
The characteristic equation of D flip-flop is :
A
Q=1
B
Q=0
C
Q= D’
D
Q=D
       Digital-Logic-Design       Sequential-Circuits
Question 14 Explanation: 
D flip flop characteristic equation truth table:

Question 15
If four 4 input multiplexers drive a 4 input multiplexer, we get a :
A
16 input MUX
B
8 input MUX
C
4 input MUX
D
2 input MUX
       Digital-Logic-Design       Combinational-Circuit
Question 15 Explanation: 
If ‘n’ input lines, it will generate 2n output lines.
Given, n=4
= 2n output lines.
= 24
= 16 output lines.
Question 16
The throughput of slotted ALOHA is given by :
A
S=G
B
S=GeG
C
S=Ge-G
D
S=eG
       Computer-Networks       Access-Control-Methods
Question 16 Explanation: 
→ The throughput of pure ALOHA is given by S=Ge-2G
→ The throughput of pure ALOHA is given by S=Ge-G
Question 17
Congestion control is done by :
A
Network layer
B
Physical layer
C
Presentation layer
D
Application layer
       Computer-Networks       OSI-TCP-layers
Question 17 Explanation: 
→ A state occurring in network layer when the message traffic is so heavy that it slows down network response time.
Question 18
Assertion(A) : Twisted pairs are widely used as transmission medium.
Reasoning(R) : Twisted pairs have adequate performance and low cost.
A
Both (A) and (R) are true and (R) is the correct explanation for (A)
B
Both (A) and (R) are true but (R) is not the correct explanation
C
(A) is true but (R) is false
D
(A) is false but (R) is true
Question 18 Explanation: 
Twisted pairs are widely used as transmission medium because twisted pairs have adequate performance and low cost.
Question 19
An example of a non-adaptive routing algorithm is :
A
Shortest path routing
B
Centralised routing
C
Baran’s hot potato algorithm
D
Baran’s backward learning algorithm
       Computer-Networks       Routing
Question 19 Explanation: 
Non-Adaptive Routing Algorithms:
These algorithms do not base their routing decisions on measurements and estimates of the current traffic and topology.
1. Flooding
2. Random walk
Question 20
IP address in B class is given by :
A
125.123.123.2
B
191.023.21.54
C
192.128.32.56
D
10.14.12.34
       Computer-Networks       IP-Adress
Question 20 Explanation: 

Question 21
N processes are waiting for I/O. A process spends a fraction p of its time in I/O wait state. The CPU utilisation is given by :
A
1-p-N
B
1-pN
C
pN
D
p-N
       Operating-Systems       CPU-Scheduling
Question 21 Explanation: 
N processes are waiting for I/O. A process spends a fraction p of its time in I/O wait state. The CPU utilisation using formula is 1-pn
Question 22
If holes are half as large as processes, the fraction of memory wasted in holes is :
A
B
½
C
D
Question 23
An example of a non-preemptive scheduling algorithm is :
A
Round Robin
B
Priority Scheduling
C
Shortest job first
D
2 level scheduling
       Operating-Systems       Process-Scheduling
Question 23 Explanation: 
→ Round robin and priority scheduling algorithms are preemptive based algorithms.
→ FCFS and SJF is non-preemptive based algorithms but in SJF are supporting another version is SRTF(Shortest remaining time first)
Question 24
An example of a distributed OS is :
A
Amoeba
B
UNIX
C
MS-DOS
D
MULTICS
       Operating-Systems       Types-of-Operating-System
Question 24 Explanation: 
Amoeba is a distributed operating system developed by Andrew S. Tanenbaum and others at the Vrije Universiteit Amsterdam.
Question 25
Which one of the following describes correctly a static variable ?
A
It cannot be initialised
B
It is initialised once at the commencement of execution and cannot be changed at run time
C
It retains its value during the life of the program
D
None of the above
       Programming       Storage-Classes
Question 25 Explanation: 
A static variable is a variable that has been allocated "statically", meaning that its lifetime (or "extent") is the entire run of the program.
Question 26
The output of the program code
main( )
{
int x=0;
while(x<=10)
for ( ; ; )
if(++x % 10 = = 0 )
break;
print f(“x=%d”, x);
}
A
x=1
B
compilation error
C
x=20
D
None of the above
       Programming       Control-Statement
Question 26 Explanation: 
main( )
{
int x=0;
while(x<=10) /* Here, the loop will run 10 times */
for ( ; ; ) /* It is dependent loop of while. So, it will also run 10 times */
if(++x % 10 = = 0 )
break;
printf(“x=%d”, x); /* Finally ‘x’ value will print 20 */
}
Note: Without for loop, it will print ‘x’ value only 10 times.
Question 27
A copy constructor is invoked when :
A
a function returns by value
B
an argument is passed by value
C
a function returns by reference
D
none of the above
       Programming-in-c++       Properties
Question 27 Explanation: 
A copy constructor is a special constructor for creating a new object as a copy of an existing object.
Features:
1. Initialize one object from another of the same type.
2. Copy an object to pass it as an argument to a function.
3. Copy an object to return it from a function.
Question 28
When a language has the capability to produce new data types, it is said to be :
A
extensible
B
encapsulated
C
overloaded
D
none of the above
       Programming       Data-Type
Question 28 Explanation: 
When a language has the capability to produce new data types, it is said to be extensible.
Question 29
How many constructors can a class have ?
A
zero
B
2
C
2
D
any number
       Programming-in-c++       Functions
Question 29 Explanation: 
There is no restriction to to have number of constructors in a lass.
Question 30
An entity has :
(i). a set of properties
(ii). a set of properties and values for all the properties
(iii). a set of properties and the values for some set of properties may non-uniquely identify an entity
(iv). a set of properties and the values for some set of properties may uniquely identify an entity
Which of the above are valid ?
A
(i) only
B
(ii) only
C
(iii) only
D
(iv) only
       Database-Management-System       ER-Model
Question 30 Explanation: 
An entity has a set of properties and the values for some set of properties may uniquely identify an entity.
Question 31
Aggregation is :
A
an abstraction through which relationships are treated as lower level entities.
B
an abstraction through which relationships are treated as higher level entities.
C
an abstraction through which relationships are not treated at all as entities.
D
none of the above
Question 31 Explanation: 
Aggregation: A directional between objects. When an object “has-a” another object, then you have got an aggregation between them, you have got an aggregation between them. Direction between them specified which object contains the other object.
Aggregation is an abstraction through which relationships are treated as higher level entities.

Note: Aggregation is an abstraction through which relationships are treated as higher level entities.
Question 32
Suppose R is a relation schema and F is a set of functional dependencies on R. Further, suppose R1 and R2 form a decomposition of R. Then the decomposition is a lossless join decomposition of R provided that :
A
R1 ∩ R2 → R1 is in F+
B
R1 ∩ R2 → R2 is in F+
C
both R1 ∩ R2 → R1 and R1 ∩ R2 → R2 functional dependencies are in F+
D
at least one from R1 ∩ R2 → R1 and R1 ∩ R2 → R2 is in F+
       Database-Management-System       Functional-Dependency
Question 32 Explanation: 
Suppose R is a relation schema and F is a set of functional dependencies on R. Further, suppose R1 and R2 form a decomposition of R. Then the decomposition is a lossless join decomposition of R provided that at least one from
R1 ∩ R2 → R1 and R1 ∩ R2 → R2 is in F+
Lossless join:
The decomposition is a lossless-join decomposition of R if at least one of the following functional dependencies are in F+ (where F+ stands for the closure for every attribute or attribute sets in F):
R1 ∩ R2 → R1
R1 ∩ R2 → R2
Question 33
In a heap, every element is __________ of all the elements in the subtree.
A
maximum
B
minimum
C
sum
D
product
E
None of the above
       Data-Structures       Heap-Tree
Question 33 Explanation: 
Heap is complete binary tree. Sometimes we are calling as binary heap that may be either Max heap or Min heap.
In Max heap→ The root element is greater than of his all childs.
In Min heap→ Te root element is smaller than of his all childs.
Question 34
If (rear == maxsize -1) rear=0; else rear=rear+1; is required in :
A
Circular Queue
B
Linear Queue
C
Stack
D
Deque
       Data-Structures       Queues-and-Stacks
Question 35
A high performance switching and multiplexing technology that utilises fixed length packets to carry different types of traffic is :
A
ATM
B
ADSL
C
SONET
D
None of the above
Question 35 Explanation: 
A high performance switching and multiplexing technology that utilises fixed length packets to carry different types of traffic is ATM(Asynchronous Transfer Mode).
Question 36
A conventional LAN bridge specifies only the functions of OSI :
A
layers 1 and 2
B
layers 1 through 3
C
all layers
D
none of the above
       Computer-Networks       OSI-TCP-layers
Question 36 Explanation: 
A conventional LAN bridge specifies only the physical(Layer-1) and data link(layer-2) layer.
Question 37
An assembly program contains :
A
imperative and declarative statements
B
imperative statements and assembler directives
C
imperative and declarative statements as well as assembler directives
D
declarative statements and assembler directives
Question 37 Explanation: 
An assembly program contains imperative and declarative statements as well as assembler directives.
Question 38
In which addressing mode, the effective address of the operand is generated by adding a constant value to the contents of register ?
A
absolute mode
B
immediate mode
C
indirect mode
D
index mode
       Computer-Organization       Addressing-Modes
Question 38 Explanation: 
Absolute address is represented by the contents of a register. This addressing mode is absolute in the sense that it is not specified relative to the current instruction address.
Indirect addressing is a scheme in which the address specifies which memory word or register contains not the operand but the address of the operand.
Immediate Operand: The simplest way for an instruction to specify an operand is for the address part of the instruction actually to contain the operand itself rather than an address or other information describing where the operand is. Such an operand is called an immediate operand because it is automatically fetched from memory at the same time the instruction itself is fetched. It is immediately available for use.
Index mode: The address of the operand is obtained by adding to the contents of the general register (called index register) a constant value. The number of the index register and the constant value are included in the instruction code.
Question 39
Which of the following are Assembler Directives ?
(i) EQU
(ii) ORIGIN
(iii) START
(iv) END
A
(ii), (iii) and (iv)
B
(i), (iii) and (iv)
C
(iii) and (iv)
D
(i), (ii), (iii) and (iv)
       Compiler-Design       Assembler
Question 39 Explanation: 
Basic Assembly directives:
1. EQU→ Equate
2. ORIGIN→ Origin
3. START→ Start
4. END→ End
Question 40
Which of the following OS treats hardware as a file system ?
A
UNIX
B
DOS
C
Windows NT
D
None of the above
       Operating-Systems       File system-I/O-protection
Question 40 Explanation: 
→ Unix operating system treats everything as file. We have different type of file types like directory,FIFO,Symbolic, socket, device, etc..,
Command: ls -l. It displays
file type, permissions,links,group names, user name, size, last modification and file name.
Question 41
In which of the following, ready to execute processes must be present in RAM ?
A
multiprocessing
B
multiprogramming
C
multitasking
D
in all of the above
       Operating-Systems       Types-of-Operating-System
Question 41 Explanation: 
Multiprocessing, multiprogramming and multitasking is ready to execute processes must be present in RAM.
Question 42
If the executing program size is greater than the existing RAM of a computer, it is still possible to execute the program if the OS supports :
A
multitasking
B
virtual memory
C
paging system
D
none of the above
       Operating-Systems       Virtual Memory
Question 42 Explanation: 
If the executing program size is greater than the existing RAM of a computer, it is still possible to execute the program if the OS supports virtual memory.
→ Virtual memory implements the translation of a program‘s address space into physical memory address space.
→ Virtual memory allows each program to exceed the size of the primary memory.
→ Virtual memory increases the degree of multiprogramming.
Question 43
Software Quality Assurance (SQA) encompasses :
A
Verification
B
Validation
C
Both verification and validation
D
None of the above
       Software-Engineering       Software-quality
Question 43 Explanation: 
→ Software Quality Assurance (SQA) encompasses both verification and validation.
→ Verification and validation are not the same things, although they are often confused. Boehm succinctly expressed the difference as
→ Verification: Are we building the product right?
→ Validation: Are we building the right product?
Question 44
Which level is called as “defined” in capability maturity model ?
A
level 0
B
level 3
C
level 4
D
level 1
Question 44 Explanation: 
CMM stands for Capability Maturity Model is a process model which specifies the process improvement approach in software development.
CMM levels:
1. Initial
2. Repeatable
3. Defined
4. Managed
5. Optimizing
Question 45
COCOMO model is used for :
A
Product quality estimation
B
Product complexity estimation
C
Product cost estimation
D
All of the above
       Software-Engineering       COCOMO-Model
Question 45 Explanation: 
COCOMO model is used for product cost estimation.
Question 46
Font sizes are usually expressed in points. One point is :
A
0.0069 inch
B
0.0138 inch
C
0.0207 inch
D
0.0276 inch
Question 46 Explanation: 
Font sizes are usually expressed in points. One point is 1/72 inch.
= 1/72
= 0.0138
Question 47
Assertion (A) :  Cellular telephone systems can handle a multitude of users.
Reasoning (R) : Cellular telephone systems permit extensive frequency reuse in a small local area.
A
Both (A) and (R) are true and (R) is the correct explanation for (A)
B
Both (A) and (R) are true but (R) is not the correct explanation
C
(A) is true but (R) is false
D
(A) is false but (R) is true
Question 47 Explanation: 
Cellular telephone systems can handle a multitude of users because cellular telephone systems permit extensive frequency reuse in a small local area.
Question 48
E-Commerce involves :
A
Electronic Data Interchange
B
Electronic mail
C
Electronic Bulletin Boards
D
All of the above
Question 48 Explanation: 
E-Commerce involves Electronic Data Interchange, Electronic mail and Electronic Bulletin Boards.
Question 49
An example of a data mining algorithm which uses squared error score function is :
A
CART algorithm
B
Back propagation algorithm
C
A priori algorithm
D
vector space algorithm
Question 49 Explanation: 
→ An example of a data mining algorithm which uses squared error score function is back propagation algorithm.
→ The goal of any supervised learning algorithm is to find a function that best maps a set of inputs to their correct output.
→ The motivation for backpropagation is to train a multi-layered neural network such that it can learn the appropriate internal representations to allow it to learn any arbitrary mapping of input to output.
Question 50
(i).  Each object in the active directory of windows 2000 has an access control list. (ii). The scheme is a blueprint of all objects in the domain of windows 2000. Which of the following is true ?
A
only (I)
B
only (II)
C
Both (I) and (II)
D
None of the above
Question 50 Explanation: 
TRUE: Each object in the active directory of windows 2000 has an access control list.
FALSE: The scheme is a blueprint of all objects in the domain of windows 2000.
There are 50 questions to complete.
PHP Code Snippets Powered By : XYZScripts.com