UGC NET CS 2006 June-Paper-2

Question 1
Which of the following strings is in the language defined by grammar
S → 0A
A → 1A/0A/1
A
01100
B
00101
C
10011
D
11111
       Theory-of-Computation       Languages-and-Grammars
Question 1 Explanation: 
Question 2
For a complete graph with N vertices, the total number of spanning trees is given by :
A
2N-1
B
N​ N-1
C
N​ N-2
D
2N+1
       Engineering-Mathematics       Graph-Theory
Question 2 Explanation: 
If a graph is complete, total number of spanning trees are N​ N-2
Example:

Formula to find total number of spanning trees are N​ N-2
=5​ 5-2
=5​ 3
=125
Question 3
The preposition( p→q) ∧ (~q ∨ p) is equivalent to :
A
q →p
B
p→ q
C
(q →p)∨(p→ q)
D
(p →q)∨(q→ p)
E
None of the above
       Engineering-Mathematics       Propositional-Logic
Question 3 Explanation: 
Question 4
The logic of pumping lemma is a good example of :
A
pigeon hole principle
B
recursion
C
divide and conquer technique
D
iteration
       Engineering-Mathematics       Combinatorics
Question 4 Explanation: 
→ A pumping lemma (or) pumping argument states that, for a particular language to be a member of a language class, any sufficiently long string in the language contains a section, or sections, that can be removed, or repeated any number of times, with the resulting string remaining in that language.
→ The proofs of these lemmas typically require counting arguments such as the pigeonhole principle.
→ Hence, the logic of pumping lemma is a good example of the pigeonhole principle.
Question 5
Let A={ x | -1 < x < 1 }=B. The function f(x)=x/2 from A to B is :
A
injective
B
surjective
C
both injective and surjective
D
neither injective nor surjective
       Engineering-Mathematics       Relations-and-Functions
Question 5 Explanation: 
→ A function f : X → Y is defined to be one-one (or injective), if the images of distinct elements of X under f are distinct, i.e., x1 , x2 ∈ X, f(x1) = f(x2) ⇒ x1=x2.
→ The possible value of “x” is 0 then
f(x1)=x1/2=0/2=0
f(x2)=x2/2=0/2=0
for x1 , x2 ∈ X, f(x1) = f(x2) ⇒ x1 = x2.
Question 6
The number of 1​ ’s present in the binary representation of (3*512 + 7*64 +5*8 +3)​ 10​ is :
A
8
B
9
C
10
D
11
       Digital-Logic-Design       Number-Systems
Question 6 Explanation: 
(3*512 + 7*64 +5*8 +3) =2027
(2027)​ 10​ = (111 1110 1011)​ 2
Here, total number of 1’s are 9.
Question 7
Which of the following expression removes static hazard from a two level AND - OR gate implementation of xy +zx’
A
xy +zx’
B
xy + zx’+ wxy
C
xy +zx’ +yz
D
xy + zx’ +wz
       Digital-Logic-Design       Combinational-Circuit
Question 7 Explanation: 
→ A static hazard occurs if a circuit produces incorrect output value momentarily before stabilizing to its correct value.
→ Generally Hazard occurs due to different delays in different paths of the circuit.
→ In the expression xy +zx’ the variable x is in true form in one term(xy) and in complement in other term(zx').
→ Delay occurs due to the presence of NOT gate. If input xyz=111 then output is 1. If input xyz=011 then output stays momentarily in state 0 then settles in state 1.
→ Adding the term yz(select y from xy and z from zx') eliminates the hazard.
Question 8
Which of the following logic has the maximum fan out ?
A
RTL
B
ECL
C
NMOS
D
CMOS
       Digital-Logic-Design       Logic-Families
Question 8 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
Negative channel metal oxide semiconductor (NMOS)
It is a type of semiconductor that charges negatively. So that transistors are turned ON/OFF by the movement of electrons. In contrast, Positive channel MOS-PMOS works by moving electron vacancies. NMOS is faster than PMOS.
Resistor Transistor Logic (RTL) Sometimes also transistor–resistor logic (TRL) is a class of digital circuits built using resistors as the input network and bipolar junction transistors (BJTs) as switching devices. RTL is the earliest class of transistorized digital logic circuit used; other classes include diode–transistor logic (DTL) and transistor-transistor logic (TTL).
Question 9
​ In a weighted code with weight 6, 4, 2, -3 the decimal 5 is represented by :
A
0101
B
0111
C
1011
D
1000
       Digital-Logic-Design       Number-Systems
Question 9 Explanation: 
The decimal value 5 is represented by 1011.
= 6*1 + 4*0 + 2*1 + -3*1
= 6+2-3
=5
Question 10
Upto how many variables, can the Karnaugh map be used ?
A
3
B
4
C
5
D
6
       Digital-Logic-Design       K-Map
Question 10 Explanation: 
The Karnaugh map provides a simple and straightforward method of minimising boolean expressions. With the Karnaugh map Boolean expressions having up to four and even six variables can be simplified.
Question 11
What is the output of the following program segment ?
main ( )
{
int count, digit=0;
count=1;
while(digit <=9)
{
printf(​"%d /n”" , ++count);
++digit;
}
}
A
10
B
9
C
12
D
11
       Programming       Control-Statement
Question 11 Explanation: 
The digit starts with 0 and will run 9 more times. It means, the loop will run 10 runs. The count variable we are given pre increment. So, it increments before assigning values.
It prints the result is
2 /n 3 /n 4 /n 5 /n 6 /n 7 /n 8 /n 9 /n 10 /n 11
Question 12
A static variable is one :
A
Which cannot be initialized
B
Which is initialized once at the commencement of execution and cannot be changed at runtime
C
Which retains its value throughout the life of the program
D
Which is the same as an automatic variable but is placed at the head of a program
       Programming       Storage-Classes
Question 12 Explanation: 
A static variable is one which retains its value throughout the life of the program. The static storage class instructs the compiler to keep a local variable in existence during the life-time of the program instead of creating and destroying it each time it comes into and goes out of scope.
Question 13
If the following loop is implemented
{
int num=0;
do
{
- - num;
printf (​ ”%d”​ , num);
num++;
} while (num >50)
}
A
the loop will run infinitely many times
B
the program will not enter the loop
C
there will be compilation error reported
D
a run time error will be reported
       Programming       Control-Statement
Question 13 Explanation: 
It will run infinite number of times because the --num variable is keep on reducing amount and num++ variable is keep on increasing amount. So, it never violate condition.
Question 14
# define max(x, y) x=(x > y) ?x :y is a macro definition, which can find the maximum of two numbers x and y if :
A
x and y are both integers only
B
x and y are both declared as float only
C
x and y are both declared as double only
D
x and y are both integers, float or double
       Programming       Macros
Question 14 Explanation: 
→ The C preprocessor is a macro preprocessor (allows you to define macros) that transforms your program before it is compiled.
→ A macro is a fragment of code that is given a name. You can use that fragment of code in your program by using the name. For example,
→ #define c 299792458 // speed of light
→ Here, when we use c in our program, it's replaced with 299792458.
→ From the given question, the x and y consists of any data type.
Question 15
The function sprintf( ) works like printf( ), but operates on:
A
Data in a file
B
stdrr
C
stdin
D
string
       Programming       Strings
Question 15 Explanation: 
Syntax:
int sprintf ( char * str, const char * format, ... );
Composes a string with the same text that would be printed if format was used on printf, but instead of being printed, the content is stored as a C string in the buffer pointed by str.
Question 16
A relation R={A, B, C, D, E, F} is given with following set of functional dependencies : F =, {A →B,AD →C,B→F,A →E}
Which of the following is candidate key?
A
A
B
AC
C
AD
D
None of these
       Database-Management-System       Functional-Dependency
Question 16 Explanation: 
Here, Attribute A and D are not present in the right hand side of any production so every key should include AD to be a candidate key.
Step-1: Check whether AD could be a candidate key or not.
(AD)​ +​ = {A,D,C,B,F,E}
Step-2: AD can uniquely identifies each attribute. So, AD is the candidate key for given relation.
Question 17
Immediate updates as a recovery protocol is preferable, when :
A
Database reads more than writes
B
Writes are more than reads
C
It does not matter as it is good in both the situations
D
There are only writes
       Database-Management-System       Transactions
Question 17 Explanation: 
→ Immediate updates as a recovery protocol is preferable when writes are more than reads.
→ In this technique, when a transaction issues an update command, the database on disk can be updated immediately, without any need to wait for the transaction to reach its commit point.
Question 18
Which of the following statement is wrong ?
A
2 - phase locking protocol suffers from dead locks
B
Time - Stamp protocol suffers from more abort
C
Time stamp protocol suffers from cascading rollbacks where as 2 - phase locking protocol do not
D
None of these
       Database-Management-System       Transactions
Question 18 Explanation: 
Option (A) is correct because 2-Phase locking protocol provides a schedule which is conflict serializable but do not deadlock free schedule.
Option (B) is true because of timestamp ordering of the transactions, this protocol suffer more number of aborts.
Option (C) is true. Whenever some transaction T tries to issue a read_item(X) or a write_item(X) operation, the basic Time out algorithm compares the Timestamp of T with read_TS(X) and write_TS(X) to ensure that the timestamp order of transaction execution is not violated. If this order is violated, then transaction T is aborted and resubmitted to the system as a new transaction with a new timestamp.
If T is aborted and rolled back, any transaction T1 that may have used a value written by T must also be rolled back. Similarly, any transaction T2 that may have used a value written by T1 must also be rolled back, and so on. This effect is known as cascading rollback.
But in case of 2-Phase locking, to avoid cascading aborts Strict two-phase locking is introduced in which a transaction holds an exclusive locks until the transaction commits/aborts. Rigorous two-phase locking is even more stricter in which both Exclusive lock and shared lock are hold by the transaction until the transaction commits/abort.
Question 19
Which data management language component enabled the DBA to define the schema components ?
A
DML
B
Subschema DLL
C
Schema DLL
D
All of these
       Database-Management-System       Databases
Question 19 Explanation: 
The database schema and conceptual organization of the entire database is viewed by Database Administration(DBA).
Question 20
A subclass having more than one super class is called __________ .
A
Category
B
Classification
C
Combination
D
Partial participation
       Programming-in-c++       Properties
Question 20 Explanation: 
A subclass having more than one super class is called category.
Question 21
In the balanced binary tree given below, how many nodes will become unbalanced when a node is inserted as a child of the node ​ ”g”.
A
1
B
3
C
7
D
8
       Data-Structures       Binary-Trees
Question 21 Explanation: 

a, c, d are going to unbalance.
Question 22
Preorder is also known as :
A
Depth first order
B
Breadth first order
C
Topological order
D
Linear order
       Data-Structures       Trees
Question 22 Explanation: 
→ Pre order is also known as depth first order.
→ A depth-first order is defined to be the reverse of the order in which we last visit the nodes of the control graph when we create the Depth first spanning tree(DFST).
Question 23
Which of the following statement is false ?
A
Every tree is a bipartite graph
B
A tree contains a cycle
C
A tree with n nodes contains (n-1) edges
D
A tree is connected graph
       Data-Structures       Trees
Question 23 Explanation: 
A tree never contains cycles. If it contains a cycle, we are calling as graph but not tree.
Note: Minimum 1 edge and maximum n-1 edges.
Question 24
If the post-fix form of a string is ABC+ -D*, The actual string is :
A
(A-(B+C))*D
B
((A-B)+C)*D
C
((A+B)-C)*D
D
(A+(B-C)*D)
       Data-Structures       Queues-and-Stacks
Question 24 Explanation: 
Question 25
Application of data structure queue is :
A
Level wise printing of tree
B
Implementation of priority queues
C
Function call implementation
D
Depth first search in a graph
       Data-Structures       Queues-and-Stacks
Question 25 Explanation: 
Application of data structure queue is implementation of priority queues. In a priority queue, an element with high priority is served before an element with low priority. In some implementations, if two elements have the same priority, they are served according to the order in which they were enqueued, while in other implementations, ordering of elements with the same priority is undefined. While priority queues are often implemented with heaps.
Applications:
1.Bandwidth management
2.Discrete event simulation
3.Dijkstra's algorithm
4.Huffman coding
5.Best-first search algorithms
6.ROAM triangulation algorithm
7.Prim's algorithm for minimum spanning tree
Question 26
What is the transmission signal coding method for a T1 carrier called ?
A
Binary
B
NRZ
C
Bipolar
D
Manchester
       Data-Communication       Transmission-method
Question 26 Explanation: 
The transmission signal coding method of T1 carrier is called Bipolar. Bipolar encoding is a type of return-to-zero (RZ) line code, where two nonzero values are used, so that the three values are +, −, and zero. Such a signal is called a duobinary signal.
Question 27
How much bandwidth is required to send 132 voice - grade channels by FDM on an international satellite system ?
A
500 MHz
B
10 MHz
C
1320 MHz
D
50 MHz
       Data-Communication       Multiplexing
Question 27 Explanation: 
In telecommunications, frequency-division multiplexing (FDM) is a technique by which the total bandwidth available in a communication medium is divided into a series of non-overlapping frequency bands, each of which is used to carry a separate signal. The most natural example of frequency-division multiplexing is radio and television broadcasting, in which multiple radio signals at different frequencies pass through the air at the same time.
Frequency Division Multiple Access(FDMA):
FDMA is simply another example of the familiar data and voice transmission technique called FDM. This technique is used to allocate small portions of a large bandwidth (500MHz for satellite transponders) to individual users. For instance, a telecommunications common carrier in a particular country like Brazil. It might want 132 voice grade channels for sending voice and analog coded data to various other countries. The bandwidth required on the current international satellite systems for this many channels is 10MHz. Because 1 transponder has a bandwidth of 500Mhz, it could accommodate 50 users, each requiring 132 channels. The Brazilian user might be allocated the frequency band between 5990 and 6000MHz for the uplink to the satellite and the corresponding downlink frequencies would be 3765 to 3775Mhz.
Question 28
What is the difference between the Ethernet frame preamble field and the IEEE 802.3 preamble and start of frame Delimiter fields ?
A
1 byte
B
1 bit
C
4 bits
D
16 bits
       Computer-Networks       Ethernet
Question 28 Explanation: 
IEEE 802.3 Frame Structure
Question 29
What is the function of a translating bridge ?
A
Connect similar remote LANs
B
Connect similar local LANs
C
Connect different types of LANs
D
Translate the network addresses into a layer 2 address
       Computer-Networks       Hardware-and-various-devices-in-networking
Question 29 Explanation: 
Translating bridge:
A translating bridge provides a connection capability between two local area networks that employ different protocols at the data link layer. Because networks using different data link layer protocols normally use different media, a translating bridge also provides support for different physical layer connections.
A translating bridge operation, A translating bridge connects local area networks that employ different protocols at the data link layer. In this example, the translating bridge is used to connect an ethernet local area network to a token network.
Question 30
The program used to determine the round - trip delay between a workstation and a destination address is :
A
Tracert
B
Traceroute
C
Ping
D
Pop
       Computer-Networks       Network-Layer
Question 30 Explanation: 
The program used to determine the round - trip delay between a workstation and a destination address is Traceroute. A traceroute is a function which traces the path from one network to another. It allows us to diagnose the source of many problems.
Question 31
Which statement is wrong ?
A
If linked origin ≠ translated origin, relocation must be performed by the linker
B
If load ≠ linked origin, the loader must perform relocation
C
A linker always perform relocation, whereas some loaders do not
D
None of these
       Operating-Systems       Linker-and-Loader
Question 31 Explanation: 
Relocation is the process of assigning load addresses for position-dependent code and data of a program and adjusting the code and data to reflect the assigned addresses.
A linker usually performs relocation in conjunction with symbol resolution, the process of searching files and libraries to replace symbolic references or names of libraries with actual usable addresses in memory before running a program.
Relocation is typically done by the linker at link time, but it can also be done at load time by a relocating loader, or at run time by the running program itself. Some architectures avoid relocation entirely by deferring address assignment to run time; this is known as zero address arithmetic
A linker or link editor is a computer utility program that takes one or more object files generated by a compiler or an assembler and combines them into a single executable file, library file, or another 'object' file.
A simpler version that writes its output directly to memory is called the loader, though loading is typically considered a separate process
Question 32
Tasks done in parsing are :
A
Check the validity of a source string
B
Determine the syntactic structure of a source string
C
Both (A) and (B)
D
None of these
       Compiler-Design       Parsers
Question 32 Explanation: 
Tasks done in parsing are check the validity of a source string and determine the syntactic structure of a source string.
Question 33
YACC builds up __________ parsing table.
A
LALR
B
LR
C
SLR
D
LLR
       Compiler-Design       Parsers
Question 33 Explanation: 
Yacc (Yet Another Compiler-Compiler):
It is a Look Ahead Left-to-Right (LALR) parser generator, generating a parser, the part of a compiler that tries to make syntactic sense of the source code, specifically a LALR parser, based on an analytic grammar written in a notation similar to Backus–Naur Form (BNF).
**YACC builds up LALR parsing table.
Question 34
The action of passing the source program into the proper syntactic class is known as :
A
Syntax analysis
B
Lexical analysis
C
Interpretation analysis
D
Uniform symbol generation
       Compiler-Design       Parsers
Question 34 Explanation: 
Lexical analysis is the first phase of a compiler. It takes the modified source code from language preprocessors that are written in the form of sentences. The lexical analyzer breaks these syntaxes into a series of tokens, by removing any whitespace or comments in the source code.
Question 35
The dynamic binding occurs during the :
A
Compile time
B
Run time
C
Linking time
D
Pre - processing time
       Compiler-Design       Run-Time-Environment
Question 35 Explanation: 
The dynamic binding occurs during the run time. The method being called upon an object or the function being called with arguments is looked up by name at runtime.
Question 36
The first operating system of Microprocessor is __________ .
A
ATLAS
B
CP/M
C
SAGE
D
T.H.E.
       Computer-Organization       Microprocessor
Question 36 Explanation: 
The first operating system of Microprocessor is CP/M.
CP/M (Control Program/Monitor (or) Control Program for Microcomputers)
It is a mass-market operating system created in 1974 for Intel 8080/85-based microcomputers by Gary Kildall of Digital Research. Initially confined to single-tasking on 8-bit processors and no more than 64 kilobytes of memory, later versions of CP/M added multi-user variations and were migrated to 16-bit processors.
Question 37
In processor management, round robin method essentially uses the preemptive version of __________ .
A
FILO
B
FIFO
C
SJF
D
Longest time first
       Operating-Systems       Process-Scheduling
Question 37 Explanation: 
Round robin method essentially uses the preemptive version of FIFO. Because of time quantum, it preempt every process in FIFO order. But it is not FIFO exactly because of preemption. In FIFO won’t support preemption.
Question 38
A page fault __________ .
A
is an error in specific page
B
is an access to the page not currently in main memory
C
occurs when a page program accesses a page of memory
D
is reference to the page which belongs to another program
       Operating-Systems       Memory-Management
Question 38 Explanation: 
A page fault is an access to the page not currently in main memory. A page fault is a type of exception raised by computer hardware when a running program accesses a memory page that is not currently mapped by the memory management unit (MMU) into the virtual address space of a process.
Question 39
__________ synchronize critical resources to prevent deadlock.
A
P - operator
B
V - operator
C
Semaphores
D
Hard disk
       Operating-Systems       Deadlock
Question 39 Explanation: 
Semaphores synchronize critical resources to prevent deadlock.
Operations:
1. wait(P): Decrements the value of semaphore variable by 1. If the new value of the semaphore variable is negative, the process executing wait is blocked (i.e., added to the semaphore queue). Otherwise, the process continues execution, having used a unit of the resource.
wait(S)
{
while (S<=0);
S--;
}
2. signal(V): Increments the value of semaphore variable by 1. After the increment, if the pre-increment value was negative (meaning there are processes waiting for a resource), it transfers a blocked process from the semaphore's waiting queue to the ready queue.
signal(S)
{
S++;
}
Question 40
The memory allocation scheme subjected to ​ external​ fragmentation is :
A
Segmentation
B
Swapping
C
Demand paging
D
Multiple contiguous fixed partitions
       Operating-Systems       Memory-Management
Question 40 Explanation: 
To avoid external fragmentation we have two methods
1. Paging
2. Segmentation
But both are still suffer in internal fragmentation.
Question 41
In software project planning, work Breakdown structure must be __________ .
A
A graph
B
A tree
C
A Eular​s graph
D
None of the above
       Software-Engineering       Software-design
Question 41 Explanation: 
Work Breakdown structure(WBS) must be tree. The tree structure will give overall view of a project. Normally we can divide two types of WBS
1. Functional WBS
2. Activity WBS
WBS tree structure:
Question 42
In Software Metrics, McCABE​ S cyclomatic number is given by following formula :
A
c=e-n+2p
B
c=e-n-2p
C
c=e+n+2p
D
c=e-n* 2p
       Software-Engineering       Cyclomatic-metric
Question 42 Explanation: 
Cyclomatic complexity uses 3 formulas
1. Number of regions + 1
2. Predicate + 1
3. Edges-Vertices+2
As per the above question, c=cyclomatic complexity
e=number of edges
n=number of vertices
p=predicates
Question 43
In a good software design, __________ coupling is desirable between modules.
A
Highest
B
Lowest
C
Internal
D
External
       Software-Engineering       Software-design
Question 43 Explanation: 
→ Cohesion is a measure of internal strength within a module, whereas coupling is a measure of inter dependency among the modules.
→ So in the context of modular software design there should be high cohesion and low coupling.
Question 44
System study yields the following :
A
Requirement specifications
B
Prevailing process description
C
Data source identification
D
All the above
       Software-Engineering       Software-requirements
Question 44 Explanation: 
System study yields the following requirement specifications. A Software requirements specification document describes the intended purpose, requirements and nature of a software to be developed. It also includes the yield and cost of the software.
Question 45
The COCOMO model is used for __________ .
A
software design
B
software cost estimation
C
software cost approximation
D
software analysis
       Software-Engineering       COCOMO-Model
Question 45 Explanation: 
The COCOMO model is used for software cost estimation.
The basic COCOMO equations take the form
1. Effort Applied (E) = a(KLOC)​ b
2. Development Time (D) = c(Effort Applied)​ d
3. People required (P) = Effort Applied / Development Time [count]
where, KLOC is the estimated number of delivered lines (expressed in thousands ) of code for project. The constants a, b, c and d are given in the following table (note: the values listed below are from the original analysis, with a modern reanalysis producing different values).
Question 46
MMS (Multimedia Messaging Service) uses __________ types of messages to perform messaging transactions.
A
4
B
6
C
8
D
10
       Computer-Networks       E-mail
Question 46 Explanation: 
MMS (Multimedia Messaging Service) uses 8 types of messages to perform messaging transactions.
Question 47
Bluetooth technology uses the transmission media.
A
Radio links
B
Microwave links
C
VSAT communication
D
Optical fiber links
       Computer-Networks       Bluetooth
Question 47 Explanation: 
→ Bluetooth uses less power and costs less to implement than Wi-Fi. Its lower power also makes it far less prone to suffering from or causing interference with other wireless devices in the same 2.4GHz radio band.
→ Bluetooth devices also shift radio frequencies often while paired, which prevents easy invasion.
Question 48
Downflow is the process associated with __________ and back up of data in a warehouse.
A
packaging
B
archiving
C
extraction
D
loading
       Database-Management-System       Data-ware-housing
Question 48 Explanation: 
→ Downflow is the process associated with archiving and backup of data in a warehouse.
→ Archived files (or) logs are crucial for recovery when no data can be lost, because they constitute a record of changes to the database.
Avantages using archiving
1. The database can be completely recovered from both instance and media failure.
2. The user can perform backups while the database is open and available for use.
3. The database can be completely recovered from both instance and media failure.
Question 49
Predictive modeling in data mining uses __________ technique.
A
Statistical
B
Association Discovery
C
Value prediction
D
None of these
       Database-Management-System       data-mining
Question 49 Explanation: 
→ Predictive modelling uses statistics to predict outcomes. Most often the event one wants to predict is in the future, but predictive modelling can be applied to any type of unknown event,regardless of when it occurred. For example, predictive models are often used to detect crimes and identify suspects, after the crime has taken place.
→ Predictive analytics encompasses a variety of statistical techniques from data mining, predictive modelling, and machine learning, that analyze current and historical facts to make predictions about future or otherwise unknown events.
Question 50
The use of a smart card represents a form of :
A
password encryption
B
user - ID encryption
C
authorization
D
authentication
       Computer-Networks       Network-Security
Question 50 Explanation: 
A smart card, chip card, or integrated circuit card (ICC) is a physical electronic authorization device, used to control access to a resource.
There are 50 questions to complete.