More General Knowledge questions
Saturday, July 5, 2014
Wednesday, July 2, 2014
Multiple choice questions on Computer
Multiple choice questions on Computer
1) Infrared signals can be used for short range communication in a closed area using _______propagation.
A) ground
B) sky
C) line of sight
D) space
2)A bridge has access to address in the same network.
A) Physical
B) Network
C) Data Link
D) Application
3) The minimum frame length for 10 Mbps Ethernet is _______ bytes and maximum is ______bytes
A) 64 & 128
B) 128 & 1518
C) 1518 & 3036
D) 64 & 1518
4) The bit rate of a signal is 3000 bps .If each signal unit carries 6 bits ,the baud rate of the signal is_____.
A) 500 baud/sec
B) 1000 baud/sec
C) 3000 baud/sec
D) 18000 baud/sec
5) KPA in CMM stands for
A) Key Process Area
B) Key Product Area
C) Key Principal Area
D) Key Performance Area
6) In function point analysis the number of complexity adjustment factors is
A) 10
B) 12
C) 14
D) 20
7) A shift reduce parser suffers from
A) shift reduce conflict only
B) reduce reduce conflict only
C) both shift reduce conflict and reduce reduce conflict
D) shift handle and reduce handle conflicts
8) How many different truth tables of the compound propositions are there that involve the propositions p & q?
A) 2
B) 4
C) 8
D) 16
9) While (87) printf("computer");
A) print "computer" 87 times
B) print "computer" 0 times
C) print "computer" 1 times
D) print "computer" infinite times
10) Manager salary details are hidden from the employee . This is called
A) Conceptual level data hiding
B) Physical level data hiding
C) External level data hiding
D) Local level data hiding
11)Which of the following is the correct value returned to the operating system upon the successful completion of a program?
A) 0
B) 1
C) -1
D) Program do not return value
12)A clustering index is created when ______.
A) primary key is declared and ordered
B) no key ordered
C) foreign key ordered
D) there is no key and no order
13)A friend function can be used to
A) avoid arguments between classes
B) allow access to classes whose source code is unavailable
C) allow one class to access an unrelated class
D) None of the above
Click here for more questions
A) ground
B) sky
C) line of sight
D) space
2)A bridge has access to address in the same network.
A) Physical
B) Network
C) Data Link
D) Application
3) The minimum frame length for 10 Mbps Ethernet is _______ bytes and maximum is ______bytes
A) 64 & 128
B) 128 & 1518
C) 1518 & 3036
D) 64 & 1518
4) The bit rate of a signal is 3000 bps .If each signal unit carries 6 bits ,the baud rate of the signal is_____.
A) 500 baud/sec
B) 1000 baud/sec
C) 3000 baud/sec
D) 18000 baud/sec
5) KPA in CMM stands for
A) Key Process Area
B) Key Product Area
C) Key Principal Area
D) Key Performance Area
6) In function point analysis the number of complexity adjustment factors is
A) 10
B) 12
C) 14
D) 20
7) A shift reduce parser suffers from
A) shift reduce conflict only
B) reduce reduce conflict only
C) both shift reduce conflict and reduce reduce conflict
D) shift handle and reduce handle conflicts
8) How many different truth tables of the compound propositions are there that involve the propositions p & q?
A) 2
B) 4
C) 8
D) 16
9) While (87) printf("computer");
A) print "computer" 87 times
B) print "computer" 0 times
C) print "computer" 1 times
D) print "computer" infinite times
10) Manager salary details are hidden from the employee . This is called
A) Conceptual level data hiding
B) Physical level data hiding
C) External level data hiding
D) Local level data hiding
11)Which of the following is the correct value returned to the operating system upon the successful completion of a program?
A) 0
B) 1
C) -1
D) Program do not return value
12)A clustering index is created when ______.
A) primary key is declared and ordered
B) no key ordered
C) foreign key ordered
D) there is no key and no order
13)A friend function can be used to
A) avoid arguments between classes
B) allow access to classes whose source code is unavailable
C) allow one class to access an unrelated class
D) None of the above
Click here for more questions
Friday, June 27, 2014
Solve General Knowledge Questions
Solve General Knowledge Questions
1) Which of the following is social network?
A) amazon.com
B) eBay.com
C) gmail.com
D) twitter
2) Internal communication within institutions is done through
A) LAN
B) WAN
C) EBB
D) MMS
3) Virtual reality provides
A) Sharp pictures
B) Individual audio
C) Participatory experience
D) Preview of new films
4) The first virtual university of India came up in
A) Andhra Pradesh
B) Maharashtra
C) Uttar Pradesh
D) Tamil Naidu
5) The initial efforts for internet based communication was for
A) Commercial communication
B) Military purpose
C) Personal interaction
D) Political campaigns
6) The post-industrial society is designated as
A) Information society
B) Technology society
C) Mediated society
D) Non - agricultural society
7) The major source of the pollutant gas,carbon mono-oxide (CO) ,in urban areas is
A) Thermal power sector
B) Transport sector
C) Industrial sector
D) domestic sector
8) In a fuel cell driven vehicle ,the energy is obtained from the combustion of
A) Methane
B) Hydrogen
C) LPG
D) CNG
9) In certain code TEACHER is written as VGCEJGT. The code of CHILDREN will be
A) EKNJFTGP
B) EJKNFTGP
C) KNJFGTP
D) None of these
10) The principles of fundamental research are used in
A) action research
B) applied research
C) philosophical research
D) historical research
11) Anil is twice as old as Sunita .Three years ago ,he was three times as old as Sunita .The present age of Anil is
A) 6 years
B) 8 years
C) 12 years
D) 16 years
12) The mean marks obtained by a class of 40 students is 65 .The mean marks of half of the students is found to be 45 .The mean marks of the remaining students is
A) 85
B) 60
C) 70
D) 65
13) Who among the following is the de facto executive head of the planning Commission?
A) Chairman
B) Deputy Chairman
C) Minister if State for Planning
D) Member Secretary
14) Education as a subject of legislation figures in the
A) Union List
B) State List
C) Concurrent List
D) Residuary Powers
A) amazon.com
B) eBay.com
C) gmail.com
D) twitter
Ans:-D) twitter
2) Internal communication within institutions is done through
A) LAN
B) WAN
C) EBB
D) MMS
Ans:- A)LAN
3) Virtual reality provides
A) Sharp pictures
B) Individual audio
C) Participatory experience
D) Preview of new films
Ans:- C) Participatory experience
4) The first virtual university of India came up in
A) Andhra Pradesh
B) Maharashtra
C) Uttar Pradesh
D) Tamil Naidu
Ans:- D) Tamil Naidu
5) The initial efforts for internet based communication was for
A) Commercial communication
B) Military purpose
C) Personal interaction
D) Political campaigns
Ans:- B) Military purpose
6) The post-industrial society is designated as
A) Information society
B) Technology society
C) Mediated society
D) Non - agricultural society
Ans:- A) Information society
7) The major source of the pollutant gas,carbon mono-oxide (CO) ,in urban areas is
A) Thermal power sector
B) Transport sector
C) Industrial sector
D) domestic sector
Ans:- B) Transport sector
8) In a fuel cell driven vehicle ,the energy is obtained from the combustion of
A) Methane
B) Hydrogen
C) LPG
D) CNG
Ans:- B) Hydrogen
9) In certain code TEACHER is written as VGCEJGT. The code of CHILDREN will be
A) EKNJFTGP
B) EJKNFTGP
C) KNJFGTP
D) None of these
Ans:- B) EJKNFTGP
10) The principles of fundamental research are used in
A) action research
B) applied research
C) philosophical research
D) historical research
Ans:- B) applied research
11) Anil is twice as old as Sunita .Three years ago ,he was three times as old as Sunita .The present age of Anil is
A) 6 years
B) 8 years
C) 12 years
D) 16 years
Ans:- C) 12 years
12) The mean marks obtained by a class of 40 students is 65 .The mean marks of half of the students is found to be 45 .The mean marks of the remaining students is
A) 85
B) 60
C) 70
D) 65
Ans:- A) 85
13) Who among the following is the de facto executive head of the planning Commission?
A) Chairman
B) Deputy Chairman
C) Minister if State for Planning
D) Member Secretary
Ans: :- B) Deputy Chairman
14) Education as a subject of legislation figures in the
A) Union List
B) State List
C) Concurrent List
D) Residuary Powers
Ans: -C) Concurrent List
Algorithm of sorting
An algorithm to sort an elements using Insertion Sort.
Algorithm for Insertion sort:-
Step 1 : START
Step 2 : Input N
Step 3 : For I←0 to N -1
Read A [ I ]
( end of I for loop )
Step 4 : For I←1 to N -1 Do
Step 5 : J = I
Step 6 : While ( J > = 1 ) and ( A[ J ] < A [ J -1 ] )
Step 7 : TEMP = A [ J ]
Step 8 : A [ J ] = A [ J - 1 ]
Step 9 : A[ J - 1 ] = TEMP
Step 10 : J = J - 1
[ end of While loop ]
[ end of step I For loop ]
Step 11 : For I←0 to N -1
Step 12: Print A [ I ]
Step 13 : Stop
Consider the elements
A[0 ] 20
A[1] 16
A[2] 4
A [3] 3
Pass 1:
A[1],A[0]>16,20,4,3
Pass 2:
A[2],A[1]> 16,4,20,3
A[1],A[0]>4,16,20,3
Pass 3:
A[3],A[2]> 4,16,3,20
A[2],A[1]>4,3,16,20
A[1],A[0]> 3,4 16,20
Sorted order is 3,4,16,20
Algorithm for bubble sort:-
Step 1 : START
Step 2 : Input N
Step 3 : For I←0 to N -1
Step 4 : Read ( A [ I ] )
( end of for loop )
Step 5 : For I←1 to N -1
Step 6 : For J←0 to N –I - 1
Step 7 : If (A[ J ] > A [ J +1] )
Temp←A [ J ]
A [ J ]←A [ J + 1 ]
A [ J +1 ]←Temp
[ end of if ]
[ end of J loop ]
[ end of I loop ]
Step 8 : For J←0 to N -1
Print A [ J ]
Step 9 : Stop
Consider the following elements: 50 , 20, 10 , 05
Pass 1:- 50 20 10 05
20 50 10 05
20 10 50 05
20 10 05 50
Pass 2:- 20 10 05 50
10 20 05 50
10 05 20 50
Pass 3:- 10 05 20 50
05 10 20 50
Sorted order is 5, 10, 20 , 50
Algorithm for selection sort:-
Step 1 : START
Step 2 : Input N
Step 3 : For I←0 to N -1
Step 4 : Read A [ I ]
Step 5 : For I←0 to N -2
Step 6 : S←A [ I]
Step 7 : Pos←I
Step 8 : For J←I + 1 to N -1
Step 9 : If ( A[ J] < S )
Step 10 : S←A [ J ]
Step 11 : Pos←J
[ end of if ]
[ end of J For loop ]
Step 12 : A [ Pos ]←a[ I ]
Step 14 : A[ I ]←S
[end of I For loop ]
Step 15: For I = 0 to N – 1
Step 16: Print A[ I ]
Step 17: Stop
An algorithm to find the maximum in an array:-
Step 1 : [Assume 1st element is the largest ]large = A [ 0 ]
Step 2 : POS = 0
Step 3 : [ Find the largest element in the array and its position]For I = 1 to N – 1 Do
Step 4 : If ( A [ I ] > large ) Then
Step 5 : large = A [ I ]
Step 6 : POS = 1
[ End if ]
[ End of step 3 for loop ]
Step 7: [ Print the largest and its position]Print “ Largest = “ , large , “ position = “ , pos
Step 8: Exit
An algorithm to find the minimum in an array:-
Step 1 : [Assume 1st element is the smallest]small = A [ 0 ]
Step 2 : POS = 0
Step 3 : [ Find the smallest element in the array and its position]
For I = 1 to N – 1 Do
Step 4 : If ( A [ I ] < small ) Then
Step 5 : small = A [ I ]
Step 6 : POS = 1
[ End if ]
[ End of step 3 for loop ]
Step 7: [ Print the smallest and its position]
Print “ smallest = “ , small , “ position = “ , pos
Step 8: Exit
Linear search method with an algorithm:-
Step 1 : START
Step 2 : input N
Step 3 : for I←0 to N -1
Read A [ I ]
Step 4 : Loc = -1
Step 5 : For I = 0 to N- 1 do
Step 6 : If ( ele = A[I]) Then
Step 7: Loc = I
Step 8 : Goto step 9
[ end if ]
[ end for ]
Step 9 : If ( Loc > = 0 ) then
Step 10: Print “ ele Found in location “ , Loc
Step 11: else
Step 12: Print “ ele not found ”
Step 13 Stop
Algorithm for Binary Search:
Step 1: Low = 0
Step 2 : High = N – 1
Step 3 : Loc = -1
Step 4 : While ( Low < = High ) Do
Step 5 : M = ( Low + High ) / 2
Step 6 : If (ele = A[M ] ) Then
Step 7: Loc = M goto Step 12
[ end if ]
Step 8 : If (ele < A [M ] ) Then
Step 9: High←M – 1
Step 10 : Else
Step 11 : Low←M + 1
[ end of if ]
[ end of While loop ]
Step 12 : if ( Loc > = 0 ) Then
Step 13: Print “ element , found , search successful ” , Loc
Step 14: else
Step 15 : Print “ Item not found , search Unsuccessful ”
[ end if ]
Step 16 : Stop
Algorithm for Insertion sort:-
Step 1 : START
Step 2 : Input N
Step 3 : For I←0 to N -1
Read A [ I ]
( end of I for loop )
Step 4 : For I←1 to N -1 Do
Step 5 : J = I
Step 6 : While ( J > = 1 ) and ( A[ J ] < A [ J -1 ] )
Step 7 : TEMP = A [ J ]
Step 8 : A [ J ] = A [ J - 1 ]
Step 9 : A[ J - 1 ] = TEMP
Step 10 : J = J - 1
[ end of While loop ]
[ end of step I For loop ]
Step 11 : For I←0 to N -1
Step 12: Print A [ I ]
Step 13 : Stop
Consider the elements
A[0 ] 20
A[1] 16
A[2] 4
A [3] 3
Pass 1:
A[1],A[0]>16,20,4,3
Pass 2:
A[2],A[1]> 16,4,20,3
A[1],A[0]>4,16,20,3
Pass 3:
A[3],A[2]> 4,16,3,20
A[2],A[1]>4,3,16,20
A[1],A[0]> 3,4 16,20
Sorted order is 3,4,16,20
Algorithm for bubble sort:-
Step 1 : START
Step 2 : Input N
Step 3 : For I←0 to N -1
Step 4 : Read ( A [ I ] )
( end of for loop )
Step 5 : For I←1 to N -1
Step 6 : For J←0 to N –I - 1
Step 7 : If (A[ J ] > A [ J +1] )
Temp←A [ J ]
A [ J ]←A [ J + 1 ]
A [ J +1 ]←Temp
[ end of if ]
[ end of J loop ]
[ end of I loop ]
Step 8 : For J←0 to N -1
Print A [ J ]
Step 9 : Stop
Consider the following elements: 50 , 20, 10 , 05
Pass 1:- 50 20 10 05
20 50 10 05
20 10 50 05
20 10 05 50
Pass 2:- 20 10 05 50
10 20 05 50
10 05 20 50
Pass 3:- 10 05 20 50
05 10 20 50
Sorted order is 5, 10, 20 , 50
Algorithm for selection sort:-
Step 1 : START
Step 2 : Input N
Step 3 : For I←0 to N -1
Step 4 : Read A [ I ]
Step 5 : For I←0 to N -2
Step 6 : S←A [ I]
Step 7 : Pos←I
Step 8 : For J←I + 1 to N -1
Step 9 : If ( A[ J] < S )
Step 10 : S←A [ J ]
Step 11 : Pos←J
[ end of if ]
[ end of J For loop ]
Step 12 : A [ Pos ]←a[ I ]
Step 14 : A[ I ]←S
[end of I For loop ]
Step 15: For I = 0 to N – 1
Step 16: Print A[ I ]
Step 17: Stop
An algorithm to find the maximum in an array:-
Step 1 : [Assume 1st element is the largest ]large = A [ 0 ]
Step 2 : POS = 0
Step 3 : [ Find the largest element in the array and its position]For I = 1 to N – 1 Do
Step 4 : If ( A [ I ] > large ) Then
Step 5 : large = A [ I ]
Step 6 : POS = 1
[ End if ]
[ End of step 3 for loop ]
Step 7: [ Print the largest and its position]Print “ Largest = “ , large , “ position = “ , pos
Step 8: Exit
An algorithm to find the minimum in an array:-
Step 1 : [Assume 1st element is the smallest]small = A [ 0 ]
Step 2 : POS = 0
Step 3 : [ Find the smallest element in the array and its position]
For I = 1 to N – 1 Do
Step 4 : If ( A [ I ] < small ) Then
Step 5 : small = A [ I ]
Step 6 : POS = 1
[ End if ]
[ End of step 3 for loop ]
Step 7: [ Print the smallest and its position]
Print “ smallest = “ , small , “ position = “ , pos
Step 8: Exit
Linear search method with an algorithm:-
Step 1 : START
Step 2 : input N
Step 3 : for I←0 to N -1
Read A [ I ]
Step 4 : Loc = -1
Step 5 : For I = 0 to N- 1 do
Step 6 : If ( ele = A[I]) Then
Step 7: Loc = I
Step 8 : Goto step 9
[ end if ]
[ end for ]
Step 9 : If ( Loc > = 0 ) then
Step 10: Print “ ele Found in location “ , Loc
Step 11: else
Step 12: Print “ ele not found ”
Step 13 Stop
Algorithm for Binary Search:
Step 1: Low = 0
Step 2 : High = N – 1
Step 3 : Loc = -1
Step 4 : While ( Low < = High ) Do
Step 5 : M = ( Low + High ) / 2
Step 6 : If (ele = A[M ] ) Then
Step 7: Loc = M goto Step 12
[ end if ]
Step 8 : If (ele < A [M ] ) Then
Step 9: High←M – 1
Step 10 : Else
Step 11 : Low←M + 1
[ end of if ]
[ end of While loop ]
Step 12 : if ( Loc > = 0 ) Then
Step 13: Print “ element , found , search successful ” , Loc
Step 14: else
Step 15 : Print “ Item not found , search Unsuccessful ”
[ end if ]
Step 16 : Stop
Thursday, June 26, 2014
Memristor
Memristor Definition:-
Conceptual symmetry between the resistor, capacitor, inductor, and the memristor..
A memristor circuit requires lower voltage and less time to turn on than competitive memory like DRAM and flash.
According to the original 1971 definition, the memristor was the fourth
fundamental circuit element, forming a non-linear relationship between
electric charge and magnetic flux linkage.
What is memristor?
Ans.As its name implies, the memristor can "remember" how much current has passed through it. And by alternating the amount of current that passes through it, a memristor can also become a one-element circuit component with unique properties. Most notably, it can save its electronic state even when the current is turned off, making it a great candidate to replace today's flash memory.
The memristor definition is based solely on the fundamental circuit variables of current and voltage and their time-integrals, just like the resistor, capacitor and inductor. Unlike those three elements however, which are allowed in linear time-invariant or LTI system theory, memristors of interest have a dynamic function with memory and may be described as some function of net charge. There is no such thing as a standard memristor. Instead, each device implements a particular function, wherein the integral of voltage determines the integral of current, and vice versa. A linear time-invariant memristor, with a constant value for M, is simply a conventional resistor.Like other two-terminal components, real-world devices are never purely memristors ("ideal memristor"), but also exhibit some amount of capacitance, resistance and inductance.
Memristors
will theoretically be cheaper and far faster than flash memory, and
allow far greater memory densities. They could also replace RAM chips as
we know them, so that, after you turn off your computer, it will
remember exactly what it was doing when you turn it back on, and return
to work instantly. This lowering of cost and consolidating of components
may lead to affordable, solid-state computers that fit in your pocket
and run many times faster than today's PCs.
Symbol:-
Someday the memristor could spawn a whole new type of computer,
thanks to its ability to remember a range of electrical states rather
than the simplistic "on" and "off" states that today's digital
processors recognize. By working with a dynamic range of data states in
an analog mode, memristor-based computers could be capable of far more
complex tasks than just shuttling ones and zeroes around.
Beyond that, memristors will likely replace both DRAM and hard disks.
The technology, called memristor, could allow computers to make decisions by understanding past patterns of data it has collected, similar to human brains collecting and understanding a series of events.
For example, a memristor circuit could be capable of telling a microwave the heating time for different food types based on the information it has collected over time, said Stanley Williams, senior fellow at HP.
For example, a memristor circuit could be capable of telling a microwave the heating time for different food types based on the information it has collected over time, said Stanley Williams, senior fellow at HP.
A memristor circuit requires lower voltage and less time to turn on than competitive memory like DRAM and flash.
Sunday, June 22, 2014
Translation of word computer
Computer word translated in other languages are :
English :- Computer
French :-ordinateur
Turkish :-bilgisayar
Yoruba :-kọmputa
Itialian :-calcolatore
Spanish :-ordenador
Maori :-rorohiko
Latin :-ipsum
Greek :-ηλεκτρονικός υπολογιστής
Marathi :- संगणक
Hindi :- कंप्यूटर
Punjabi :-ਕੰਪਿਊਟਰ
Urdu :- کمپیوٹ
Polish :-komputer
Swedish :-dator
Russian :-компьютер
Hungarian :-számítógép
Kannada :-ಗಣಕಯಂತ್ರ
Tamil :- கம்ப்யூட்டர்
Malay :-komputer
Gujarati:-કમ્પ્યુટર
Serbian :-рачунар
Filipino:- computer
French :-ordinateur
Turkish :-bilgisayar
Yoruba :-kọmputa
Itialian :-calcolatore
Spanish :-ordenador
Maori :-rorohiko
Latin :-ipsum
Greek :-ηλεκτρονικός υπολογιστής
Marathi :- संगणक
Hindi :- कंप्यूटर
Punjabi :-ਕੰਪਿਊਟਰ
Urdu :- کمپیوٹ
Polish :-komputer
Swedish :-dator
Russian :-компьютер
Hungarian :-számítógép
Kannada :-ಗಣಕಯಂತ್ರ
Tamil :- கம்ப்யூட்டர்
Malay :-komputer
Gujarati:-કમ્પ્યુટર
Serbian :-рачунар
Filipino:- computer
Definition of computer
An electronic device for storing and processing data, typically in binary form, according to instructions given to it in a variable program.
Monday, June 9, 2014
Short description of CS
Computer science deals with the theoretical foundations of information and computation, together with practical techniques for the implementation and application of these foundations
Computer Science (abbreviated CS or CompSci) is the scientific and practical approach to computation and its applications. It is the systematic study of the feasibility, structure, expression, and mechanization of the methodical processes (or algorithms) that underlie the acquisition, representation, processing, storage, communication of, and access to information, whether such information is encoded as bits in a computer memory or transcribed in genes and protein structures in a human cell.
History: The earliest foundations of what would become computer science predate the invention of the modern digital computer. Machines for calculating fixed numerical tasks such as the abacus have existed since antiquity, aiding in computations such as multiplication and division.
- Charles Babbage is credited with inventing the first mechanical computer
- Ada Lovelace is credited with writing the first algorithm intended for processing on a computer.
- The German military used the Enigma machine (shown here) during World War II for communication they thought to be secret. The large-scale decryption of Enigma traffic at Bletchley Park was an important factor that contributed to Allied victory in WWII
Computer Science is not just about building computers or writing computer programs! Computer Science is no more about building computers and developing software than astronomy is about building telescopes, biology is about building microscopes, and music is about building musical instruments! Computer science is not about the tools we use to carry out computation. It is about how we use such tools, and what we find out when we do. The solution of many computer science problems may not even require the use of computers—just pencil and paper. As a matter of fact, problems in computer science have been tackled decades before computers were even built. That said the design and implementation of computing system hardware and software is replete with formidable challenges and fundamental problems that keep computer scientists busy. Computer Science is about building computers and writing computer programs, and much more!
Sunday, April 20, 2014
Solve more questions on networking -Computer Science
1. The network number plays what part in an IP address?
A. It specifies the network to which the host belongs.
B. It specifies the identity of the computer on the network.
C. It specifies which node on the sub-network is being addressed.
D. It specifies which networks the device can communicate with.
Answer :- A. It specifies the network to which the host belongs.
2. What is the decimal equivalent to the binary number 101101?
A. 32
B. 35
C. 45
D. 44
Answer :- C. 45
3. Convert the following decimal number to its binary form: 192.5.34. 11.
A. 11000000.00000101.00100010.00001011
B. 11000101.01010111.00011000.10111000
C. 01001011.10010011.00111001.00110111
D. 11000000.00001010.01000010.00001011
Answer:- A. 11000000.00000101.00100010.00001011
4. Convert the following binary IP address to its decimal form: 11000000.00000101.00100010.00001011 A. 190.4.34.11
B. 192.4.34.10
C. 192.4.32.11
D. None of the above
Answer:- D. None of the above
5. What portion of the following Class B address is the network address: 154.19.2.7?
A. 154
B. 154.19
C. 154.19.2
D. 154.19.2.7
Answer:- B. 154.19
6. Which portion of the IP address 129.219.51.18 represents the network?
A. 129.219
B. 129
C. 51.18
D. 18
Answer:- A. 129.219
7. Which address is an example of a broadcast address on the network 123.10.0.0 with a subnet mask of 255.255.0.0?
A. 123.255.255.255
B. 123.10.255.255
C. 123.13.0.0
D. 123.1.1.1
Answer:- B. 123.10.255.255
8. MAC address are _________ bits in length.
A. 12
B. 24
C. 48
D. 64
Answer :- C. 48
9. Where does the MAC address reside?
A. Transceiver
B. Computer BIOS
C. NIC
D. CMOS
Answer :- C. NIC
10.Convert the decimal number 2989 to hex.
A. FDD1
B. BAD
C. TED
D. CAD
Answer :- B. BAD
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Solve questions on Networking
Solve questions on Networking
A. 16 bits
B. 32 bits
C. 64 bits
D. None of the above
Answer :- B.32 bits
2. What is the maximum value of each octet in an IP address?
A. 128
B. 255
C. 256
D. None of the above
Answer :- B. 255
3. How many bits are in a subnet mask?
A. 16 bits
B. 32 bits
C. 64 bits
D. None of the above
Answer :- B. 32 bits
4. A half-duplex circuit means
A. Only one side can talk at a time
B. The signal strength is cut in half
C. The signal strength is doubled
D. Two hosts can talk simultaneously
Answer :- A. Only one side can talk at a time
5. Attenuation means
A. Travel
B. Delay
C. A signal losing strength over distance
D. Loss of signal due to EMI
Answer :- C. A signal losing strength over distance
6. __________ means to convert binary data into a form that can travel on a physical communications link.
A. Encoding
B. Decoding
C. Encrypting
D. Decrypting
Answer :- A.Encoding
7. How many host addresses can be used in a Class C network?
A. 253
B. 254
C. 255
D. 256
Answer :- B. 254
8. What is the minimum number of bits that can be borrowed to form a subnet?
A. 1
B. 2
C. 4
D. None of the above
Answer :- B. 2
9. What is the primary reason for using subnets?
A. To reduce the size of the collision domain
B. To increase the number of host addresses
C. To reduce the size of the broadcast domain
D. None of the above
Answer :- C. To reduce the size of the broadcast domain
10.How many bits can be borrowed to create a subnet for a Class C network?
A. 2
B. 4
C. 6
D. None of the above
Answer :- C. 6
Study material of Computer Science :-
127.0.0.1 is is a special purpose IP address conventionally used as a computer's loopback address.
Definition Thrashing :-
In computer science, thrashing occurs when a computer's virtual memory subsystem is in a constant state of paging, rapidly exchanging data in memory for data on disk, to the exclusion of most application-level processing
Virtual Memory:-
In computing, virtual memory is a memory management technique that is implemented using both hardware and software. It maps memory addresses used by a program, called virtual addresses, into physical addresses in computer memory. Main storage as seen by a process or task appears as a contiguous address space or collection of contiguous segments.
The time taken to move the disk arm to the desired cylinder is called the :seek time
Class A
0 . 0 . 0 . 0 = 00000000.00000000.00000000.00000000
127.255.255.255 = 01111111.11111111.11111111.11111111
0nnnnnnn.HHHHHHHH.HHHHHHHH.HHHHHHHH
Class B
128. 0. 0. 0 = 10000000.00000000.00000000.00000000
191.255.255.255 = 10111111.11111111.11111111.11111111
10nnnnnn.nnnnnnnn.HHHHHHHH.HHHHHHHH
Class C
192. 0. 0. 0 = 11000000.00000000.00000000.00000000
223.255.255.255 = 11011111.11111111.11111111.11111111
110nnnnn.nnnnnnnn.nnnnnnnn.HHHHHHHH
Class D
224. 0. 0. 0 = 11100000.00000000.00000000.00000000
239.255.255.255 = 11101111.11111111.11111111.11111111
1110XXXX.XXXXXXXX.XXXXXXXX.XXXXXXXX
Class E
240. 0. 0. 0 = 11110000.00000000.00000000.00000000
255.255.255.255 = 11111111.11111111.11111111.11111111
1111XXXX.XXXXXXXX.XXXXXXXX.XXXXXXXX
Where:
n indicates a binary slot used for network ID.
H indicates a binary slot used for host ID.
X indicates a binary slot (without specified purpose)
190.255.254.254 -class B
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