Computer-Networks

Question 1
_______ can detect burst error of length less than or equal to degree of the polynomial and detects burst errors that affect odd number of bits.
A
Hamming Code
B
CRC
C
VRC
D
None of the above
       Computer-Networks       CRC       ISRO-2018
Question 1 Explanation: 
→ A cyclic redundancy check (CRC) is an error-detecting code commonly used in digital networks and storage devices to detect accidental changes to raw data. CRCs can be used for error correction
→ The design of the CRC polynomial depends on the maximum total length of the block to be protected (data + CRC bits), the desired error protection features, and the type of resources for implementing the CRC, as well as the desired performance.
→ A common misconception is that the "best" CRC polynomials are derived from either irreducible polynomials or irreducible polynomials times the factor 1 + x, which adds to the code the ability to detect all errors affecting an odd number of bits.
→ If we use the generator polynomial g(x)=p(x)(1+x), where p(x) is a primitive polynomial of degree r−1, then the maximal total block length is 2r−1 −1 , and the code is able to detect single, double, triple and an odd number of errors.
Question 2
Station A uses 32-byte packets to transmit messages to Station B using a sliding window protocol. The round trip delay between A and B is 80 ms and the bottleneck bandwidth on the path between A and B is 128 kbps. What is the optimal window size that A should use?
A
20
B
40
C
160
D
320
       Computer-Networks       Control-Flow-Methods       ISRO-2018
Question 2 Explanation: 
Tt = L / B = 32 bytes/ 128 kbps = 32*8/128 ms = 2 ms
Round trip delay = 2 * Tp = 80 ms (given)
Optimal window size is = (Tt + 2*Tp) / Tt = 82 / 2 = 41
Option is not given, closest option is 40.
Question 3
Assuming that for a given network layer implementation, connection establishment overhead is 100 bytes and disconnection overhead is 28 bytes. What would be the minimum size of the packet the transport layer needs to keep up if it wishes to implement a datagram service above the network needs to keep its overhead to a minimum of 12.5%. (Ignore transport layer overhead)
A
512 bytes
B
768 bytes
C
1152 bytes
D
1024 bytes
       Computer-Networks       Switching       ISRO-2018
Question 3 Explanation: 
→ Given a question is based on circuit-switching.
→ In circuit-switching, the total time required to send data = Setup connection time + Transfer time + Teardown the connection time
→ In datagram-service, the total time required to send data = N*Setup connection time + Transfer time
Question 4
In cryptography, the following uses transposition cyphers and the keyword is LAYER. Encrypt the following message. (Spaces are omitted during encryption) WELCOME TO NETWORK SECURITY!
A
WMEKREETSILTWETCOOCYONRU!
B
EETSICOOCYWMEKRONRU!LTWET
C
LTWETONRU!WMEKRCOOCYEETSI
D
ONRU!COOCYLTWETEETSIWMEKR
       Computer-Networks       Network-Security       ISRO-2018
Question 4 Explanation: 
Question 5
Avalanche effect in cryptography
A
Is desirable property of the cryptographic algorithm
B
Is undesirable property of the cryptographic algorithm
C
Has no effect on the encryption algorithm
D
None of the above
       Computer-Networks       Network-Security       ISRO-2018
Question 5 Explanation: 
→ In cryptography, the avalanche effect is the desirable property of cryptographic algorithms, typically block cyphers and cryptographic hash functions, wherein if input is changed slightly (for example, flipping a single bit), the output changes significantly (e.g., half the output bits flip).
→ In the case of high-quality block cyphers, such a small change in either the key or the plaintext should cause a drastic change in the ciphertext.
Question 6
What is one advantage of setting up a DMZ(Demilitarized Zone) with two firewalls?
A
You can control where traffic goes in the three networks
B
You can do stateful packet filtering
C
You can do load balancing
D
Improve network performance
       Computer-Networks       Network-Security       ISRO-2018
Question 6 Explanation: 
→ The most secure approach is to use two firewalls to create a DMZ.
1. The first firewall (also called the "front-end" or "perimeter" firewall) must be configured to allow traffic destined to the DMZ only.
2. The second firewall (also called "back-end" or "internal" firewall) only allows traffic from the DMZ to the internal network.

This setup is considered more secure since two devices would need to be compromised. There is even more protection if the two firewalls are provided by two different vendors because it makes it less likely that both devices suffer from the same security vulnerabilities.
Question 7
Which one of the following algorithm is not used in asymmetric key cryptography?
A
RSA Algorithm
B
Diffie-Hellman Algorithm
C
Electronic Code Book Algorithm
D
None of the above
       Computer-Networks       Network-Security       ISRO-2018
Question 7 Explanation: 
Question 8
If the bandwidth of a signal is 5 kHz and the lowest frequency is 52 kHz, what is the highest frequency
A
5 kHz
B
10 kHz
C
47 kHz
D
57 kHz
       Computer-Networks       Bandwidth-Frequency       ISRO-2007
Question 8 Explanation: 
Bandwidth = Highest frequency - Lowest frequency.
Highest frequency = Bandwidth+ Lowest Frequency= 5kHz + 52kHz = 57kHz.
Question 9
An Ethernet hub
A
functions as a repeater
B
connects to a digital PBX
C
connects to a token-ring network
D
functions as a gateway
       Computer-Networks       Ethernet       ISRO-2007
Question 9 Explanation: 
A Hub is a network hardware device for connecting multiple Ethernet devices together and making them act as a single network segment.
A repeater is an electronic device that receives a signal and re-transmits it. Repeaters are used to extend transmissions so that the signal can cover longer distances or be received on the other side of an obstruction.
An Ethernet Hub or a Repeater Hub also acts as a repeater and regenerates the signal to avoid its loss.
Question 10
Phase transition for each bit are used in
A
Amplitude modulation
B
Carrier modulation
C
Manchester encoding
D
NRZ encoding
       Computer-Networks       Ethernet       ISRO-2007
Question 10 Explanation: 
In the Manchester encoding , a logic 0 is indicated by a 0 to 1 transition at the centre of the bit and a logic 1 is indicated by a 1 to 0 transition at the centre of the bit. Note that signal transitions do not always occur at the ‘bit boundaries’ (the division between one bit and another), but that there is always a transition at the centre of each bit.
Question 11
Bit stuffing refers to
A
inserting a 0 in user stream to differentiate it with a flag
B
inserting a 0 in flag stream to avoid ambiguity
C
appending a nipple to the flag sequence
D
appending a nipple to the use data stream
       Computer-Networks       Ethernet       ISRO-2007
Question 11 Explanation: 
→ Bit stuffing is most easily described as insertion of a 0 bit after a long run of 1 bits.
→ In SDLC the transmitted bit sequence "01111110" containing six adjacent 1 bits is the Flag byte.
→ Bit stuffing ensures that this pattern can never occur in normal data, so it can be used as a marker for the beginning and end of frame without any possibility of being confused with normal data.
→ Bit stuffing is the insertion of non information bits into data. stuffed bits should not be confused with overhead bits. Overhead bits are non-data bits that are necessary for transmission (usually as part of headers, checksums etc).
Question 12
If there are five routers and six networks in intranet using link state routing, how many routing tables are there?
A
1
B
5
C
6
D
11
       Computer-Networks       Routing       ISRO-2007
Question 12 Explanation: 
Routers have routing tables to transmit packets selectively to other networks. These routers apply shortest path algorithms to choose the links on which, a packet is to be forwarded, so that it can reach the destination in covering minimum number of hops.
If there are 5 routers, then there should be 5 routing tables as well.
Question 13
Which of these is not a feature of WAP 2.0
A
Push and Pull Model
B
Interface to a storage device
C
Multimedia messaging
D
Hashing
       Computer-Networks       Network-protocols       ISRO-2007
Question 13 Explanation: 
WAP is a Wireless protocol for synchronization of WAP client computers and HTTP server
It is derived from the architecture of the synchronization between HTTP client (browser) and the HTTP server (web server) on the internet. It does the following functions:
SyncML synchronization
WAP push service
MMS service
Question 14
Silly Window Syndrome is related to
A
Error during transmission
B
File transfer protocol
C
Degrade in TCP performance
D
Interface problem
       Computer-Networks       TCP       ISRO-2007
Question 14 Explanation: 
Silly window syndrome is a problem in computer networking caused by poorly implemented TCP flow control.
Question 15
SSL is not responsible for
A
Mutual authentication of client & server
B
Secret communication
C
Data Integrity protection
D
Error detection and correction
       Computer-Networks       Network-Security       ISRO-2007
Question 15 Explanation: 
SSL (Secure Sockets Layer) is the standard security technology for establishing an encrypted link between a web server and a browser. This link ensures that all data passed between the web server and browsers remain private and integral. SSL is an industry standard and is used by millions of websites in the protection of their online transactions with their customers.
It is utilized to encrypt Web traffic using Hypertext Transfer Protocol (HTTP) and to authenticate Web servers, and to encrypt communications between Web browsers and Web servers etc.
So, other than error detection and correction, all options are correct.
Question 16
The standard for certificates used on internet is
A
X.25
B
X.301
C
X.409
D
X.509
       Computer-Networks       Network-Security       ISRO-2007
Question 16 Explanation: 
In cryptography, X.509 is a standard that defines the format of public key certificates. X.509 certificates are used in many Internet protocols, including TLS/SSL, which is the basis for HTTPS, the secure protocol for browsing the web.
Question 17
Hashed message is signed by a sender using
A
his public key
B
his private key
C
receiver’s public key
D
receiver’s private key
       Computer-Networks       Network-Security       ISRO-2007
Question 17 Explanation: 
After the generation of hashed message that needs to be transmitted over a network, it is signed or encrypted using the private key of the sender which also generates the digital signatures. The message is transmitted along with the digital signatures which ensures Authentication and Non-repudiation.
Question 18
Range of IP Address from 224.0.0.0 to 239.255.255.255 are
A
Reserved for loopback
B
Reserved for broadcast
C
Used for multicast packets
D
Reserved for future addressing
       Computer-Networks       IPv4-and-Fragmentation       ISRO-2007
Question 18 Explanation: 
The full range of multicast addresses is from 224.0.0.0 to 239.255.255.255. Multicast addresses represent a group of IP devices that can only be used as the destination of a datagram and not as the source.
Question 19
IEEE 802.11 is standard for
A
Ethernet
B
Bluetooth
C
Broadband Wireless
D
Wireless LANs
       Computer-Networks       LAN       ISRO-2007
Question 19 Explanation: 
IEEE 802.11 refers to the set of standards that define communication for wireless LANs (wireless local area networks or WLANs).
Question 20
When a host on network A sends a message to a host on network B, which address does the router look at?
A
Port
B
IP
C
Physical
D
Subnet mask
       Computer-Networks       Network-layers       ISRO-2007
Question 20 Explanation: 
Routing is done on the basis of IP addresses. A host on network A when sends the packet to the host on network B, it checks the IP address of the receiving host and routes the packet to the suitable hop.
Physical address i.e. the MAC address is used to recognize a unique host over a network on in a LAN. Port number is required to identify a specific process or application to which a message is to be forwarded when it arrives at a server.
Question 21
Dijkstra's algorithm is used to
A
Create LSAs
B
Flood an internet with information
C
Calculate the routing tables
D
Create a link state database
       Computer-Networks       Routing       ISRO-2007
Question 21 Explanation: 
Dijkstra's algorithm finds shortest paths from the source to all other nodes in the graph, producing a shortest-path tree and is used in building the routing tables on a network.
Question 22
Physical topology of FDDI is?
A
Bus
B
Ring
C
Star
D
None of the above
       Computer-Networks       Topologies       ISRO-2017 May
Question 22 Explanation: 
Fiber Distributed Data Interface
FDDI provides a 100 Mbit/s optical standard for data transmission in local area network that can extend in range up to 200 kilometers (120 mi). Although FDDI logical topology is a ring based token network, it did not use the IEEE 802.5 token ring protocol as its basis; instead, its protocol was derived from the IEEE 802.4 token bus timed token protocol.
In addition to covering large geographical areas, FDDI local area networks can support thousands of users. FDDI offers both a Dual-Attached Station (DAS), counter-rotating token ring topology and a Single-Attached Station (SAS), token bus passing ring topology.
Question 23
In networking terminology UTP means
A
Ubiquitous Teflon port
B
Uniformly terminating port
C
Unshielded twisted pair
D
Unshielded T-connector port
       Computer-Networks       ISRO-2017 May
Question 23 Explanation: 
Unshielded Twisted Pair(UTP): It consists of two unshielded wires twisted around each other. Due to its low cost, UTP cabling is used extensively for local-area networks (LANs) and telephone connections. UTP cabling does not offer as high bandwidth or as good protection from interference as coaxial or fiber optic cables, but it is less expensive and easier to work.
Question 24
The default subnet mask for a class B network can be
A
255.255.255.0
B
255.0.0.0
C
255.255.192.0
D
255.255.0.0
       Computer-Networks       Subnetting       ISRO-2017 May
Question 24 Explanation: 
→ Class A addresses only include IP starting from 1.x.x.x to 126.x.x.x only. The IP range 127.x.x.x is reserved for loopback IP addresses.
→ Class B IP Addresses range from 128.0.x.x to 191.255.x.x. The default subnet mask for Class B is 255.255.x.x.
→ Class C IP addresses range from 192.0.0.x to 223.255.255.x. The default subnet mask for Class C is 255.255.255.x.
Question 25
If there are n devices (nodes) in a network, what is the number of cable links required for a fully connected mesh and a star topology respectively
A
n(n−1)/ 2, n−1
B
n, n−1
C
n−1, n
D
n−1, n(n −1)/2
       Computer-Networks       Topologies       ISRO-2017 May
Question 25 Explanation: 
→ Fully connected mesh we can also called complete graph. So, it has total number of cable links with n nodes are n(n-1)/2
→ In a star topology total number of cable links are (n-1)
Question 26
Which of the following protocol is used for transferring electronic mail messages from one machine to another?
A
TELNET
B
FTP
C
SNMP
D
SMTP
       Computer-Networks       Application-Layer-Protocol       ISRO-2017 May
Question 26 Explanation: 
TELNET→ Telephone network
FTP→ File transfer protocol
SNMP→ Simple network management protocol
SMTP→ Simple mail transfer protocol Simple Mail Transfer Protocol (SMTP): is an Internet standard for email transmission.
→ Mail servers and other mail transfer agents use SMTP to send and receive mail messages on TCP port 25.
Question 27
Which media access control protocol is used by IEEE 802.11 wireless LAN?
A
CDMA
B
CSMA/CA
C
ALOHA
D
None of the above
       Computer-Networks       Wi-Fi       ISRO-2017 May
Question 27 Explanation: 
This layer provides the same function as the MAC layer in the wired LAN environment. IEEE 802.11, instead of using Ethernet's CSMA/CD, specifies a scheme called CSMA/CA.
CA stands for Collision Avoidance. (In the wireless environment, since a station may not be able to hear all other stations, CD, Collision Detection is not feasible)
Question 28
An Ethernet frame that is less than the IEEE 802.3 minimum length of 64 octets is called
A
Short frame
B
Small frame
C
Mini frame
D
Runt frame
       Computer-Networks       Ethernet       ISRO-2017 May
Question 28 Explanation: 
A runt frame is an Ethernet frame that is less than the IEEE 802.3 minimum length of 64 octets. Runt frames are most commonly caused by collisions; other possible causes are a malfunctioning network card, buffer underrun, duplex mismatch or software issues.
Question 29
Match with the suitable one:
A
A-2, B-3, C-4, D-1
B
A-2, B-4, C-3, D-1
C
A-3, B-4, C-1, D-2
D
A-3, B-1, C-4, D-2
       Computer-Networks       Routing       ISRO-2017 May
Question 29 Explanation: 
Internet Group Management Protocol (IGMP) is a communications protocol used by hosts and adjacent routers on IPv4 networks to establish multicast group memberships. IGMP is an integral part of IP multicast.
Open Shortest Path First (OSPF) is a routing protocol for Internet Protocol (IP) networks. It uses a link state routing (LSR) algorithm and falls into the group of interior gateway protocols (IGPs), operating within a single autonomous system (AS).
Border Gateway Protocol (BGP) is a standardized exterior gateway protocol designed to exchange routing and reachability information among autonomous systems (AS) on the Internet.
Routing Information Protocol (RIP) is one of the oldest distance vector routing protocols which employ the hop count as a routing metric. RIP implements the split horizon, route poisoning and hold down mechanisms to prevent incorrect routing information from being propagated.
Question 30
MD5 is a widely used hash function for producing hash value of
A
64 bits
B
128 bits
C
512 bits
D
1024 bits
       Computer-Networks       Network-Security       ISRO-2017 May
Question 30 Explanation: 
→ The Message Digest(MD5) algorithm is a widely used hash function producing a 128-bit hash value. Although MD5 was initially designed to be used as a cryptographic hash function, it has been found to suffer from extensive vulnerabilities.
→ It can still be used as a checksum to verify data integrity, but only against unintentional corruption. It remains suitable for other non-cryptographic purposes, for example for determining the partition for a particular key in a partitioned database.
Question 31
Which protocol suite designed by IETF to provide security for a packet at the Internet layer?
A
IPsec
B
NetSec
C
PacketSec
D
SSL
       Computer-Networks       IP-Address       ISRO-2017 May
Question 31 Explanation: 
→ The Internet Engineering Task Force(IETF) IP Security Working Group formed to standardize these efforts as an open, freely available set of security extensions called IPsec.
Question 32
Pretty Good Privacy (PGP) is used in
A
Browser security
B
FTP security
C
Email security
D
None of the above
       Computer-Networks       ISRO-2017 May
Question 32 Explanation: 
→ Pretty Good Privacy (PGP) is an encryption program that provides cryptographic privacy and authentication for data communication. PGP is used for signing, encrypting, and decrypting texts, e-mails, files, directories, and whole disk partitions and to increase the security of e-mail communications. Phil Zimmermann developed PGP in 1991.
→ It uses Digital Signatures to provide authentication and integrity checking basically to check if the message is actually sent by the person or entity claimed to be the sender.
Question 33
What is WPA?
A
wired protected access
B
wi-fi protected access
C
wired process access
D
wi-fi process access
       Computer-Networks       Wi-Fi       ISRO-2017 May
Question 33 Explanation: 
→ WPA stands for Wi-Fi protected access. It is a security standard for computing devices equipped with wireless internet connection.
→ WPA provides data encryption and user authentication to the computing devices connected to a wi-fi network.
→ WPA also includes a Message Integrity Check, which is designed to prevent an attacker from altering and resending data packets. This replaces the cyclic redundancy check (CRC) that was used by the WEP standard.
Question 34
SATA is the abbreviation of
A
Serial Advanced Technology Attachment
B
Serial Advanced Technology Architecture
C
Serial Advanced Technology Adapter
D
Serial Advanced Technology Array
       Computer-Networks       ISRO-2017 May
Question 34 Explanation: 
→ Serial ATA (Serial Advanced Technology Attachment or SATA) is a standard for connecting and transferring data from hard disk drives (HDDs) to computer systems.
→ As its name implies, SATA is based on serial signaling technology, unlike Integrated Drive Electronics (IDE) hard drives that use parallel signaling.
Question 35
The subnet mask 255.255.255.192
A
extends the network portion to 16 bits
B
extends the network portion to 26 bits
C
extends the network portion to 36 bits
D
has no effect on the network portion of an IP address
       Computer-Networks       IP-address       ISRO CS 2008
Question 35 Explanation: 
Default subnet mask for Class C is 255.255.255.192
(192)10 = (11000000)2
Since, 192 is written as 11000000, it has 2 sub-nets and remaining all hosts.
So, for first three octets, 24 bits are fixed and for last octet 2 bits are fixed, i.e. 24 + 2 = 26 bits
Question 36
On a LAN ,where are IP datagrams transported?
A
In the LAN header
B
In the application field
C
In the information field of the LAN frame
D
After the TCP header
       Computer-Networks       LAN       ISRO CS 2008
Question 36 Explanation: 
IP datagram is encapsulated in the payload field of Ethernet frame, so we can say that it is transported in information field of frame.
Question 37
In Ethernet, the source address field in the MAC frame is the _______ address.
A
original sender’s physical
B
previous station’s physical
C
next destination’s physical
D
original sender’s service port
       Computer-Networks       Ethernet       ISRO CS 2008
Question 37 Explanation: 
While the IP address of source and the destination in a datagram is kept same at each hop, the source MAC address is replaced at each station while the frame is in transit. So for the current station, the source address field will contain the MAC address of the previous station.
Question 38
Which of the following transmission media is not readily suitable to CSMA operation?
A
Radio
B
Optical fibers
C
Coaxial cable
D
Twisted pair
       Computer-Networks       CSMA       ISRO CS 2008
Question 38 Explanation: 
Wireless medium can’t be used for CSMA operations as they are not suitable for collision detection methods. As radio already have a large number of users, it can’t detect the collisions between the senders and the receivers.
Question 39
The TCP sliding window
A
can be used to control the flow of information
B
always occurs when the field value is 0
C
always occurs when the field value is 1
D
occurs horizontally
       Computer-Networks       TCP/IP       ISRO CS 2008
Question 39 Explanation: 
TCP uses a Sliding Window mechanism for the flow control of data in transit on a network. In Sliding Window, the sending device can send all packets within the TCP window size without receiving an ACK.
The receiving device should acknowledge each packet it received and after receiving the ACK from the receiving device, the sending device slides the window to right side
Question 40
What is the bandwidth of the signal that ranges from 40 kHz to 4 MHz
A
36 MHz
B
360 kHz
C
3.96 MHz
D
396 kHz
       Computer-Networks       Transmission-and-Propagation-Delay       ISRO CS 2008
Question 40 Explanation: 
Let fh is the highest frequency and fi is the lowest frequency.
Bandwidth = fh – fi = 4000 - 40 KHz = 3960 KHz = 3.96 MHz
Question 41
Which Project 802 standard provides for a collision-free protocol?
A
802.2
B
802.3
C
802.5
D
802.6
       Computer-Networks       IEEE-802       ISRO CS 2008
Question 41 Explanation: 
→ Token ring is standardized with protocol IEEE 802.5. Token passing is the method of medium access, with only one token allowed to exist on the network at a time.
→ Network devices must acquire the token to transmit data, and may only transmit a single frame before releasing the token to the next station on the ring. When a station has data to transmit, it acquires the token at the earliest opportunity, marks it as busy, and attaches the data and control information to the token to create a data frame, which is then transmitted to the next station on the ring.
→ The frame will be relayed around the ring until it reaches the destination station, which reads the data, marks the frame as having been read, and sends it on around the ring. When the sender receives the acknowledged data frame, it generates a new token, marks it as being available for use, and sends it to the next station.
→ IEEE 802.2 is the logical link control (LLC) as the upper portion of the data link layer of the OSI Model.
→ IEEE 802.6 is a standard governed by the ANSI for Metropolitan Area Networks (MAN).
→ IEEE 802.3 is a working group and a collection of Institute of Electrical and Electronics Engineers (IEEE) standards produced by the working group defining the physical layer and data link layer media access control (MAC) of wired Ethernet.
Question 42
Assume that each character code consists of 8 bits. The number of characters that can be transmitted per second through an asynchronous serial line at 2400 baud rate, and with two stop bits, is:
A
109
B
216
C
218
D
219
       Computer-Networks       Transmission-and-Propagation-Delay       ISRO CS 2008
Question 42 Explanation: 
Here, the baud value represents the serial port is capable of transferring a maximum number of bits per second.
Given 2400 baud value,
Sending maximum of 2400 bits per second with serial port .
Total Data To send = 1 bit(start) + 8 bits(char size) + 2 bits(Stop) = 11 bits.
Number of 8-bit characters that can be transmitted per second = 2400/11 = 218.18
The effective number of characters transmitted = 218 So, the correct option is (C).
Question 43
The network 198.78.41.0 is a
A
Class A network
B
Class B network
C
Class C network
D
Class D network
       Computer-Networks       Subnetting       ISRO CS 2008
Question 43 Explanation: 

In classful addressing, The range of first octet should be between one of the following:

CLASS A- 0 to 127

CLASS B – 128 to 191

CLASS C – 192 to 223

CLASS D- 224 to 239

CLASS E- 240 to 255

Given IP address = 198.78.41.0

So, it is a class C address.
Question 44
A public key encryption system
A
allows anyone to decode the transmissions
B
allows only the correct sender to decode the data
C
allows only the correct receiver to decode the data
D
does not encode the data before transmitting it
       Computer-Networks       Encryption-Decryption       ISRO CS 2008
Question 44 Explanation: 
Public-key encryption is a cryptographic system that uses two keys, a public key known to everyone and a private or secret key known only to the recipient of the message.
A receiver can decode the message using his own private key and the public key of the sender.
Question 45
In networking, UTP stands for
A
Unshielded T-connector port
B
Unshielded twisted pair
C
Unshielded terminating pair
D
Unshielded transmission process
       Computer-Networks       UTP       ISRO CS 2009
Question 45 Explanation: 
Unshielded twisted pair (UTP) is a ubiquitous type of copper cabling used in telephone wiring and local area networks (LANs).
There are five types of UTP cables -- identified with the prefix CAT, as in category -- each supporting a different amount of bandwidth.g a different amount of bandwidth.
Question 46
The address resolution protocol (ARP) is used for
A
Finding the IP address from the DNS
B
Finding the IP address of the default gateway
C
Finding the IP address that corresponds to a MAC address
D
Finding the MAC address that corresponds to an IP address
       Computer-Networks       ARP       ISRO CS 2009
Question 46 Explanation: 
The Address Resolution Protocol (ARP) is a communication protocol used for discovering the link layer address, such as a MAC address, associated with a given internet layer address, typically an IPv4 address.
Question 47
Which of the following is a MAC address?
A
192.166.200.50
B
00056A:01A01A5CCA7FF60
C
568, Airport Road
D
01:A5:BB:A7:FF:60
       Computer-Networks       MAC       ISRO CS 2009
Question 47 Explanation: 
A media access control address (MAC address) of a device is a unique identifier assigned to a network interface controller (NIC).
MAC addresses are recognizable as six groups of two hexadecimal digits, separated by hyphens, colons, or no separator.
Option (A) is example for IP address
Option (B) and (C ) are not examples for any address representation in computer
01:A5:BB:A7:FF:60 is the MAC address which consists of six groups of two hexadecimal digits, separated by hyphens, colons, or no separator.
Question 48
What is the primary purpose of a VLAN?
A
Demonstrating the proper layout for a network
B
Simulating a network
C
To create a virtual private network
D
Segmenting a network inside a switch or device
       Computer-Networks       VLAN       ISRO CS 2009
Question 48 Explanation: 
1. A virtual local area network (VLAN) is a logical group of workstations, servers and network devices that appear to be on the same LAN despite their geographical distribution. A VLAN allows a network of computers and users to communicate in a simulated environment as if they exist in a single LAN and are sharing a single broadcast and multicast domain.
2. Higher-end switches allow the functionality and implementation of VLANs.
3. VLANs address issues such as scalability, security, and network management. Network architects set up VLANs to provide network segmentation
Question 49
SHA-1 is a
A
encryption algorithm
B
decryption algorithm
C
key exchange algorithm
D
message digest function
       Computer-Networks       SHA-1       ISRO CS 2009
Question 49 Explanation: 
SHA-1 produces a 160-bit hash value or message digests from the inputted data (data that requires encryption), which resembles the hash value of the MD5 algorithm.
Question 50
Advanced Encryption Standard (AES) is based on
A
Asymmetric key algorithm
B
Symmetric key algorithm
C
Public key algorithm
D
Key exchange
       Computer-Networks       AES       ISRO CS 2009
Question 50 Explanation: 
The Advanced Encryption Standard, or AES, is a symmetric block cipher chosen by the U.S. government to protect classified information.
AES is Symmetric key algorithm
Question 51
Use of IPSEC in tunnel mode results in
A
IP packet with same header
B
IP packet with new header
C
IP packet without header
D
No changes in IP packet
       Computer-Networks       Network-Security       ISRO CS 2009
Question 51 Explanation: 
IPSec can be used to create VPN Tunnels to end-to-end IP Traffic (also called as IPSec Transport mode) or site-to-site IPSec Tunnels (between two VPN Gateways, also known as IPSec Tunnel mode).
IPSec Tunnel mode: In IPSec Tunnel mode, the original IP packet (IP header and the Data payload) is encapsulated within another packet.
In IPSec tunnel mode the original IP Datagram from is encapsulated with an AH (provides no confidentiality by encryption) or ESP (provides encryption) header and an additional IP header. The IP addresses of the newly added outer IP header are that of the VPN Gateways.
The traffic between the two VPN Gateways appears to be from the two gateways (in a new IP datagram), with the original IP datagram is encrypted (in case of ESP) inside IPSec packet.
Question 52
The subnet mask for a particular network is 255.255.31.0. Which of the following pairs of IP addresses could belong to this network?
A
172.57.88.62 and 172.56.87.23
B
10.35.28.2 and 10.35.29.4
C
191.203.31.87 and 191.234.31.88
D
128.8.129.43 and 128.8.161.55
       Computer-Networks       IP-address       ISRO CS 2009
Question 52 Explanation: 
To find whether hosts belong to same network or not , we have to find their net id, if net id is same then hosts belong to same network and net id can be find by ANDing subnet mask and IP address.
128.8.129.43 (Bitwise AND) 255.255.31.0 = 128.8.1.0
128.8.161.55 (Bitwise AND) 255.255.31.0 = 128.8.1.0
Question 53
Bit stuffing refers to
A
inserting a ‘0’ in user stream to differentiate it with a flag
B
inserting a ‘0’ in flag stream to avoid ambiguity
C
appending a nibble to the flag sequence
D
appending a nibble to the use data stream
       Computer-Networks       Ethernet       ISRO-2016
Question 53 Explanation: 
→ Bit stuffing is most easily described as insertion of a 0 bit after a long run of 1 bits.
→ In SDLC the transmitted bit sequence "01111110" containing six adjacent 1 bits is the Flag byte.
→ Bit stuffing ensures that this pattern can never occur in normal data, so it can be used as a marker for the beginning and end of frame without any possibility of being confused with normal data.
Question 54
Dynamic routing protocol enable routers to
A
Dynamically discover and maintain routes
B
Distribute routing updates to other routers
C
Reach agreement with other routers about the network topology
D
All of the above
       Computer-Networks       Routing       ISRO-2016
Question 54 Explanation: 
Dynamic Routing
→ The administrator configure a routing protocol on your network interfaces.
→ Routing protocol learns about other routers automatically.
→ Router and the other routers exchange routes, and each learns about the networks that the other is connected to.
→ When new networks are added or removed, the routers update each other.
Question 55
In Ethernet CSMA/CD, the special bit sequence transmitted by media access management to handle collision is called
A
Preamble
B
Postamble
C
Jam
D
None of the above
       Computer-Networks       Ethernet       ISRO-2016
Question 55 Explanation: 
→ In CSMA/CD once a collision is detected, the colliding devices send a jam signal.
→ Assuming you know the collision detection method used by CSMA/CD, the jam signals generated after collision informs the devices that a collision has occurred and invokes a random backoff algorithm which forces the devices on the Ethernet not to send any data until their timer expires. (everyone has a random timer value)
→ Once a transmitting device detects collision which is by examining the data over the Ethernet and identifying that it's not the data it has send, it does receive the jamming signal otherwise collision will keep happening.
→ As the jam signal is generated after collision hence the devices ideally won’t be sending any data during that time.
Question 56
Which network protocol allows hosts to dynamically get a unique IP number on each bootup
A
DHCP
B
BOOTP
C
RARP
D
ARP
       Computer-Networks       IP-Address       ISRO-2016
Question 56 Explanation: 
→ A router(or) a residential gateway can be enabled to act as a DHCP server.
→ Most residential network routers receive a globally unique IP address within the ISP network.
→ Within a local network, a DHCP server assigns a local IP address to each device connected to the network.
Question 57
In a token ring network the transmission speed is 107 bps and the propagation speed is 200 meters/ μs. The 1-bit delay in this network is equivalent to:
A
500 meters of cable
B
200 meters of cable
C
20 meters of cable
D
50 meters of cable
       Computer-Networks       Access-Control-Methods       ISRO-2016
Question 57 Explanation: 
Tt = L / B => 1/ 107 = 0.1 microsec
Given Tp = 200 m / microsec
In, 1 microsec it covers 200m
Therefore, in 0.1 microsec it is 200 * 0.1 = 20 meters
Question 58
The address of a class B host is to be split into subnets with a 6-bit subnet number. What is the maximum number of subnets and the maximum number of hosts in each subnet?
A
62 subnets and 262142 hosts
B
64 subnets and 262142 hosts
C
62 subnets and 1022 hosts
D
64 subnets and 1024 hosts
       Computer-Networks       Subnetting       ISRO-2016
Question 58 Explanation: 
→ It is a class B address, so there 16-bits for NID and 16-bits for HID.
→ From HID, we took 6-bits for subnetting.
→ Then total subnets possible = ( 26 ) - 2 = 62
→ Total hosts possible for each subnet = (210) - 2 = 1022
Question 59
The message 11001001 is to be transmitted using the CRC polynomial x3 + 1 to protect it from errors. The message that should be transmitted is:
A
11001001000
B
11001001011
C
11001010
D
110010010011
       Computer-Networks       CRC       ISRO-2016
Question 59 Explanation: 
CRC polynomial = x3+1 [∵ In data 3-zero’s need to be append to data]
= 1001

∴ Data transmitted is: 11001001011
Question 60
What is the maximum size of data that the application layer can pass on to the TCP layer below?
A
Any size
B
216 bytes – size of TCP header
C
216
D
1500
       Computer-Networks       Application-Layer-Protocol       ISRO-2016
Question 60 Explanation: 
Application Layer - Any size
Transport Layer - 65515 byte
Network layer - 65535 byte
Data link layer - 1500 byte
Question 61
Frames of 1000 bits are sent over a 106 bps duplex link between two hosts. The propagation time is 25ms. Frames are to be transmitted into this link to maximally pack them in transit (within the link). What is the minimum number of bits (I) that will be required to represent the sequence numbers distinctly? Assume that no time gap needs to be given between the transmission of two frames.
A
l = 2
B
l = 3
C
l = 4
D
l = 5
       Computer-Networks       Control-Flow-Methods       ISRO-2016
Question 61 Explanation: 
→ Transmission time (Tt)=1000/106 seconds = 1 ms
→ Maximum number of frames that can be transmit to maximally pack them is
=(Tt+2Tp)/Tx
= (25+1)/1
=26 which is window size
→ Minimum sequence numbers required = 26
→ Minimum number of bits required for sequence number is 5.
Question 62
When a DNS server accepts and uses incorrect information from a host that has no authority giving that information, then it is called
A
DNS lookup
B
DNS hijacking
C
DNS spoofing
D
None of the mentioned
       Computer-Networks       DNS       ISRO-2016
Question 62 Explanation: 
→ DNS spoofing, also referred to as DNS cache poisoning, is a form of computer security hacking in which corrupt Domain Name System data is introduced into the DNS resolves cache, causing the name server to return an incorrect result record, e.g. an IP address.
→ This results in traffic being diverted to the attacker's computer (or any other computer).
Question 63
The encoding technique used to transmit the signal in giga ethernet technology over fiber optic medium is
A
Differential Manchester encoding
B
Non return to zero
C
4B/5B encoding
D
8B/10B encoding
       Computer-Networks       Encoding-Decoding       ISRO CS 2011
Question 63 Explanation: 
→ In telecommunications, 8b/10b is a line code that maps 8-bit words to 10-bit symbols to achieve DC-balance and bounded disparity, and yet provide enough state changes to allow reasonable clock recovery
→ The FC(Fiber Channel) -0 standard defines what encoding scheme is to be used (8b/10b or 64b/66b) in a Fibre Channel system – higher speed variants typically use 64b/66b to optimize bandwidth efficiency.Thus, 8b/10b encoding is used for 4GFC and 8GFC variants;
→ The Fibre Channel 8b/10b coding scheme is also used in other telecommunications systems. Data is expanded using an algorithm that creates one of two possible 10-bit output values for each input 8-bit value.
Question 64
What is the raw throughput of USB 2.0 technology?
A
480 Mbps
B
400 Mbps
C
200 Mbps
D
12 Mbps
       Computer-Networks       USB       ISRO CS 2011
Question 64 Explanation: 
The USB 3.0 SuperSpeed path operates at a raw bit rate of 5.0 Gbits/s, while the USB 2.0 path operates at 480 Mbits/s (High Speed), 12 Mbits/s (Full Speed), or 1.5 Mbits/s (Low Speed).
Question 65
The network protocol which is used to get MAC address of a node by providing IP address is
A
SMTP
B
ARP
C
RIP
D
BOOTP
       Computer-Networks       MAC       ISRO CS 2011
Question 65 Explanation: 
Address Resolution Protocol(ARP) is used to find the MAC address of a device using its IP address.
Reverse Address Resolution Protocol(RARP) is the viceversa case of ARP as it is a protocol by which a physical machine in a local area network can request to learn its IP address from a gateway server.
Question 66
An example of polyalphabetic substitution is
A
P-box
B
S-box
C
Caesar cipher
D
Vigenere cipher
       Computer-Networks       Polyalphabetic       ISRO CS 2011
Question 66 Explanation: 
A polyalphabetic cipher is any cipher based on substitution, using multiple substitution alphabets. The Vigenère cipher is probably the best-known example of a polyalphabetic cipher, though it is a simplified special case. The Enigma machine is more complex but is still fundamentally a polyalphabetic substitution cipher.
Question 67
The IEEE standard for WiMax technology is
A
IEEE 802.16
B
IEEE 802.36
C
IEEE 812.16
D
IEEE 806.16
       Computer-Networks       Wimax       ISRO CS 2011
Question 67 Explanation: 
WiMAX (Worldwide Interoperability for Microwave Access) is a family of wireless communication standards based on the IEEE 802.16 set of standards, which provide multiple physical layer (PHY) and Media Access Control (MAC) options.
Question 68
In which layer of network architecture, the secured socket layer (SSL) is used?
A
physical layer
B
session layer
C
application layer
D
presentation layer
       Computer-Networks       OSI-TCP-layers       ISRO CS 2011
Question 68 Explanation: 
Secure Socket Layer is networking protocol used at transport layer to provide secure connection between client and server over internet. It places itself as and application layer protocol in the TCP/IP reference model and as presentation layer protocol in the OSI model.
Question 69
One SAN switch has 24 ports. All 24 supports 8 Gbps Fiber Channel technology. What is the aggregate bandwidth of that SAN switch?
A
96 Gbps
B
192 Mbps
C
512 Gbps
D
192 Gbps
       Computer-Networks       SAN       ISRO CS 2011
Question 69 Explanation: 
SAN switch has 24 ports.
Each port supports 8 Gbps then all 24 ports have capacity of 24*8 Gbps bandwidth
The aggregate bandwidth is 192 Gbps
Question 70
The broadcast address for IP network 172.16.0.0 with subnet mask 255.255.0.0 is
A
172.16.0.255
B
172.16.255.255
C
255.255.255.255
D
172.255.255.255
       Computer-Networks       Subnetting        ISRO CS 2011
Question 70 Explanation: 
IP address: 172.16.0.0 Sub network address: 255.255.0.0
Convert the above two addresses into binary format and perform the bitwise AND operation
Question 71
Data is transmitted continuously at 2.048 Mbps rate for 10 hours and received 512 bits errors. What is the bit error rate?
A
6.9 e-9
B
6.9 e-6
C
69 e-9
D
4 e-9
       Computer-Networks       Transmission-and-Propagation-Delay       ISRO CS 2011
Question 71 Explanation: 
Bandwidth = 2.048 Mbps
The amount of time data transferred is = 10 hours = 36000 secs
Total Data transmitted = 2.048 * 106 * 36000 = 2.048 * 36 * 109 bits
The number of Error bits received = 512
Error rate = total number of error bits/ total data transferred per second
= 512 / 73.728 * 109
= 6.944 * 10-9
Question 72
Lightweight Directory Access protocol is used for
A
Routing the packets
B
Authentication
C
obtaining IP address
D
domain name resolving
       Computer-Networks       Network-protocols       ISRO CS 2011
Question 72 Explanation: 
The Lightweight Directory Access Protocol (LDAP ) is an open, vendor-neutral, industry standard application protocol for accessing and maintaining distributed directory information services over an Internet Protocol (IP) network
A common use of LDAP is to provide a central place to store usernames and passwords. This allows many different applications and services to connect to the LDAP server to validate users
Question 73
What is the maximum number of characters (7 bits + parity ) that can be transmitted in a second on a 19.2 kbps line? This asynchronous transmission requires 1 start bit and 1 stop bit.
A
192
B
240
C
1920
D
1966
       Computer-Networks       Serial-communication       ISRO CS 2013
Question 73 Explanation: 
In asynchronous transmission mode, start bit and and stop bit is always required whereas in synchronous mode these bits are not required.
In the given question asynchronous transmission is mentioned and number of bits required is 10 bits. (7 data bits + 1 parity bit + 1 start bit + 1 stop bit )
Bandwidth = 19.2 kbps
Maximum number of characters transmitted in 1 second = (19.2 *1000)/10 = 1920
Question 74
IEEE 1394 is related to
A
RS-232
B
USB
C
Firewire
D
PCI
       Computer-Networks       IEEE       ISRO CS 2013
Question 74 Explanation: 
→ IEEE 1394 is an interface standard for a serial bus for high-speed communications and isochronous real-time data transfer.
→ It was developed in the late 1980s and early 1990s by Apple, which called it FireWire.
→ The copper cable it uses in its most common implementation can be up to 4.5 metres (15 ft) long. Power is also carried over this cable, allowing devices with moderate power requirements to operate without a separate power supply.
→ FireWire is also available in Cat 5 and optical fiber versions.
→ The 1394 interface is comparable to USB, though USB requires a master controller and has greater market share
Question 75
In the Ethernet, which field is actually added at the physical layer and is not part of the frame
A
preamble
B
CRC
C
address
D
location
       Computer-Networks       Ethernet       ISRO CS 2013
Question 75 Explanation: 
Preamble belongs to the physical layer and are added at the physical layer only.
CRC- A cyclic redundancy check (CRC) is an error-detecting code commonly used in digital networks and storage devices to detect accidental changes to raw data.
Question 76
Ethernet layer-2 switch is a network element type which gives
A
different collision domain and same broadcast domain
B
different collision domain and different broadcast domain
C
same collision domain and same broadcast domain
D
same collision domain and different broadcast domain
       Computer-Networks       Ethernet       ISRO CS 2013
Question 76 Explanation: 
→ A network switch is a multiport network bridge that uses hardware addresses to process and forward data at the data link layer (layer 2) of the OSI model.
→ Switches for Ethernet are the most common form of network switch.
→ An Ethernet switch operates at the data link layer (layer 2) of the OSI model to create a separate collision domain for each switch port. Each device connected to a switch port can transfer data to any of the other ports at any time and the transmissions will not interfere.
→ Because broadcasts are still being forwarded to all connected devices by the switch, the newly formed network segment continues to be a broadcast domain.
Question 77
If the frame to be transmitted is 1101011011 and the CRC polynomial to be used for generating checksum is x4+ x + 1, then what is the transmitted frame?
A
11010110111011
B
11010110111101
C
11010110111110
D
11010110111001
       Computer-Networks       CRC       ISRO CS 2013
Question 77 Explanation: 
→A cyclic redundancy check (CRC) is an error-detecting code commonly used in digital networks and storage devices to detect accidental changes to raw data.
→Blocks of data entering these systems get a short check value attached, based on the remainder of a polynomial division of their contents.
→CRCs can be used for error correction
→Given checksum is x4+ x + 1 so four zeros will append at the end.
Step-1:
Given Data unit = 1101011011, Divisor = 10011,
The input is 11010110110000
Step-2:

Question 78
What will be the efficiency of a Stop and Wait protocol, if the transmission time for a frame is 20ns and the propagation time is 30ns?
A
20%
B
25%
C
40%
D
66%
       Computer-Networks       Stop-and-wait       ISRO CS 2013
Question 78 Explanation: 
→Assuming that the transmission time for the acknowledgement and the processing time at nodes are negligible.
→The time taken by a sender to send a single packet is equal to transmission time (data) + 2 * propagation delay , out of which only transmission time is useful .
→Then, efficiency =transmission time / transmission time + 2 * propagation delay.
→Given data is transmission time is 20ns
→Propagation delay(time) is 30ns.
→Efficiency =20+(20+2*30)=20/80=0.25 which is equivalent to 25%
Question 79
IPv6 does not support which of the following addressing modes?
A
unicast addressing
B
multicast addressing
C
broadcast addressing
D
anycast addressing
       Computer-Networks       IPV6       ISRO CS 2013
Question 79 Explanation: 
The mechanism by which the address is hosted on the network is referred as addressing mode.
IPv6 does not implement traditional IP broadcast, i.e. the transmission of a packet to all hosts on the attached link using a special broadcast address, and therefore does not define broadcast addresses.
In IPv6, the broadcast addressing t is achieved by sending a packet to the link-local all nodes multicast group at address ff02::1, which is analogous to IPv4 multicasting to address 224.0.0.1.
IPv6 supports the Unicast addressing ,Multicast addressing and Anycast addressing addressing modes:
Question 80
What is IP class and number of sub-networks if the subnet mask is 255.224.0.0?
A
Class A, 3
B
Class A, 8
C
Class B, 3
D
Class B, 32
       Computer-Networks       IP-address       ISRO CS 2013
Question 80 Explanation: 
The binary number equivalent to 255 is 1111 1111
The binary number equivalent to 224 is 1110 0000
We can write the subnet mask 255.224.0.0 in binary form as follows
11111111.11100000.00000000.00000000
In the binary representation, the first eight bits represent Class A network address and next three bits( ones) used for represent the number of subnets.
Then the total number of subnets are 23=8.
Question 81
Which algorithm is used to shape the bursty traffic into a fixed rate traffic by averaging the data rate?
A
solid bucket algorithm
B
spanning tree algorithm
C
hocken helm algorithm
D
leaky bucket algorithm
       Computer-Networks       Leaky-bucket-algorithm       ISRO CS 2013
Question 81 Explanation: 
The leaky bucket algorithm is a method of temporarily storing a variable number of requests and organizing them into a set-rate output of packets in an asynchronous transfer mode (ATM) network.
The leaky bucket is used to implement traffic policing and traffic shaping in Ethernet and cellular data networks.
The algorithm can also be used to control metered-bandwidth Internet connections to prevent going over the allotted bandwidth for a month, thereby avoiding extra charges.
Question 82
A packet filtering firewall can
A
deny certain users from accessing a service
B
block worms and viruses from entering the network
C
disallow some files from being accessed through FTP
D
block some hosts from accessing the network
       Computer-Networks       Firewall       ISRO CS 2013
Question 82 Explanation: 
Packet-filtering firewalls operate at the network layer (Layer 3) of the OSI model. Packet-filtering firewalls make processing decisions based on network addresses, ports, or protocols.
Packet filtering is a network security mechanism that works by controlling what data can flow to and from a network
A packet filtering firewall can block some hosts from accessing the network.
Question 83
The protocol data unit for the transport layer in the internet stack is
A
segment
B
message
C
datagram
D
frame
       Computer-Networks       OSI-TCP-layers       ISRO CS 2013
Question 83 Explanation: 
A protocol data unit (PDU) is a single unit of information transmitted among peer entities of a computer network.
A PDU is composed of protocol specific control information and user data
Protocol data units for the Internet protocol suite are:
The transport layer PDU is the TCP segment for
→TCP, and the datagram for UDP.
→The Internet layer PDU is the packet.
→The link layer PDU is the frame.
Question 84
How many check bits are required for 16 bit data word to detect 2 bit errors and single bit correction using hamming code?
A
5
B
6
C
7
D
8
       Computer-Networks       Hamming-code       ISRO CS 2013
Question 84 Explanation: 
→ We know that error detection and correction using huffman code requires d+1 bits and 2d+1 bits
Question 85
The process of modifying IP address information in IP packet headers while in transit across a traffic routing device is called
A
Port address translation (PAT)
B
Network address translation (NAT)
C
Address mapping
D
Port mapping
       Computer-Networks       IP-Address       ISRO CS 2014
Question 85 Explanation: 
Network address translation (NAT)
It is a method of remapping one IP address space into another by modifying network address information in the IP header of packets while they are in transit across a traffic routing device.The technique was originally used as a shortcut to avoid the need to readdress every host when a network was moved. It has become a popular and essential tool in conserving global address space in the face of IPv4 address exhaustion. One Internet-routable IP address of a NAT gateway can be used for an entire private network.
Port Address Translation (PAT) is an extension to network address translation (NAT) that permits multiple devices on a local area network (LAN) to be mapped to a single public IP address. The goal of PAT is to conserve IP addresses.
Address mapping (also know as pin mapping or geocoding) is the process of assigning map coordinate locations to addresses in a database. The output of address mapping is a point layer attributed with all of the data from the input database.
Port mapping or port forwarding is an application of network address translation (NAT) which redirects a communication request from one address and port number combination to another while the packets are traversing a network gateway, such as a router or firewall.
Question 86
What is routing algorithm used by OSPF routing protocol?
A
Distance vector
B
Flooding
C
Path vector
D
Link state
       Computer-Networks       Routing       ISRO CS 2014
Question 86 Explanation: 
→ Open Shortest Path First (OSPF) is a routing protocol for Internet Protocol (IP) networks.
→ It uses a link state routing (LSR) algorithm and falls into the group of interior gateway protocols (IGPs), operating within a single autonomous system (AS).
→ It implements Dijkstra's algorithm, also known as the shortest path first (SPF) algorithm.
Question 87
In a system an RSA algorithm with p=5 and q=11, is implemented for data security. What is the value of the decryption key if the value of the encryption key is 27?
A
3
B
7
C
27
D
40
       Computer-Networks       Network-Security       ISRO CS 2014
Question 87 Explanation: 
The keys for the RSA algorithm are generated the following way:
1. Choose two distinct prime numbers p and q.
2. Compute n = pq.
3. Compute λ(n) = lcm(λ(p), λ(q)) = lcm(p − 1, q − 1), where λ is Carmichael's totient function. Choose an integer e such that 1 < e < λ(n) and gcd(e, λ(n)) = 1; i.e., e and λ(n) are coprime.
4. Determine d as d ≡ e−1 (mod λ(n)); i.e., d is the modular multiplicative inverse of e modulo λ(n). This means: solve for d the equation d⋅e ≡ 1 (mod λ(n)).
Given two prime numbers are p = 5 and q = 11, encryption key, e = 27
n = p * q = 5 * 11 = 55
λ(n)= (p-1) * (q-1) = 4 * 10 = 40
Let the value of decryption key be ‘d’ such that:
e * d mod λ(n) = 1
27 * d mod 40 = 1
d = 3
Question 88
An IP packet has arrived in which the fragment offset value is 100, the value of HLEN is 5 and the value of total length field is 200. What is the number of last byte in packet.
A
194
B
394
C
979
D
1179
       Computer-Networks       IPv4-and-Fragmentation       ISRO CS 2014
Question 88 Explanation: 
→ The first byte number is 100×8 =800.
→ The total length is 200 bytes and the header length is 20 bytes (5×4), which means that there are 180 bytes in this datagram.
→ If the first byte number is 800, the last byte number must 979 (800+180-1).
Question 89
Consider a 50 kbps satellite channel with a 500 milliseconds round trip propagation delay. If the sender wants to transmit 1000 bit frames, how much time will it take for the receiver to receive the frame?
A
250 milliseconds
B
20 milliseconds
C
520 milliseconds
D
270 milliseconds
       Computer-Networks       Transmission-and-Propagation-Delay       ISRO CS 2014
Question 89 Explanation: 
The propagation delay is the time it takes for the first bit to travel from the sender to the receiver (During this time the receiver is unaware that a message is being transmitted). The propagation speed depends on the physical medium of the link (that is, fiber optics, twisted-pair copper wire, etc.),
The round-trip time or ping time is the time from the start of the transmission from the sending node until a response is received at the same node. It is affected by packet delivery time as well as the data processing delay, which depends on the load on the responding node. If the sent data packet as well as the response packet have the same length, the round trip time can be expressed as:
Round trip time = 2 × Packet delivery time + processing delay
In case of only one physical link, the above expression corresponds to:
Link round trip time = 2 × packet transmission time + 2 × propagation delay + processing delay
Consider there is no effect of propagation delay and processing delay
round trip time ≈ 2 × packet transmission time
packet transmission time=round trip time/2=500/2= 250 ms
Transmission time= Message (bits) / band width (chanel capacity) = 1000 bits/50kbps =20 ms
Total time to receiver to receive the frame =250+20=270ms
Question 90
An IP packet has arrived with the first 8 bits as 0100 0010. Which of the following is correct?
A
The number of hops this packet can travel is 2.
B
The total number of bytes in header is 16 bytes
C
The upper layer protocol is ICMP
D
The receiver rejects the packet
       Computer-Networks       IPv4-and-Fragmentation       ISRO CS 2014
Question 90 Explanation: 
So, for 0100 0010,
First 4 bits represent Version IPV4
And another 4 bits represent header length (/ 4) which should range between 20 to 60 bytes.
here 0010 represents header length , is equal to 2 * 4 = 8
So, receiver will reject the packet.
Question 91

Assume the following information:

Original timestamp value = 46

Receive timestamp value = 59

Transmit timestamp value = 60

Timestamp at the arrival of packet = 69

Which of the following statements is correct?

A
Receive clock should go back by 3 milliseconds
B
Transmit and Receive clocks are synchronized
C
Transmit clock should go back by 3 milliseconds
D
Receive clock should go ahead by 1 milliseconds
       Computer-Networks       ISRO CS 2014
Question 91 Explanation: 
From the given data(original,receive,transmit and packet arrival time) details we can calculate the below details
Sending time =Receive time - stating time= 59 − 46 = 13 milliseconds
Receiving time = Packet arrival time- Packet receive time=67 − 60 = 7 milliseconds
Round-trip time =Sending time+Receiving time= 13 + 7 = 20 milliseconds
Time difference = receive timestamp − ( original timestamp field + one-way time duration)
Time difference = 59 − (46 + 10) = 3
Question 92
Suppose you are browsing the world wide web using a web browser and trying to access the web servers. What is the underlying protocol and port number that are being used?
A
UDP, 80
B
TCP, 80
C
TCP, 25
D
UDP, 25
       Computer-Networks       TCP/UDP       ISRO CS 2014
Question 92 Explanation: 
The Hypertext Transport Protocol (HTTP) is an application layer protocol that is used to transmit virtually all files and other data on the World Wide Web, whether they're HTML files, image files, query results, or anything else. Usually, HTTP takes place through TCP/IP sockets.
A browser is an HTTP client because it sends requests to an HTTP server (Web server), which then sends responses back to the client. The standard (and default) port for HTTP servers to listen on is 80, though they can use any port.
HTTP is based on the TCP/IP protocols, and is used commonly on the Internet for transmitting web-pages from servers to browsers.
Question 93
A mechanism or technology used in Ethernet by which two connected devices choose common transmission parameters such as speed, duplex mode and flow control is called
A
Autosense
B
Synchronization
C
Pinging
D
Auto negotiation
       Computer-Networks       Ethernet       ISRO CS 2014
Question 93 Explanation: 
1. Ping is a computer network administration software utility used to test the reachability of a host on an Internet Protocol (IP) network. It measures the round-trip time for messages sent from the originating host to a destination computer that are echoed back to the source
2. Autosense refers to a feature found in network adapters that allows them to automatically recognize the current local network's speed and adjust its own setting accordingly. It is often used with Ethernet, fast Ethernet, switches, hubs and network interface cards
3. Process synchronization refers to the idea that multiple processes are to join up or handshake at a certain point, in order to reach an agreement or commit to a certain sequence of action. Data synchronization refers to the idea of keeping multiple copies of a dataset in coherence with one another, or to maintain data integrity. Process synchronization primitives are commonly used to implement data synchronization.
4. Autonegotiation is a signaling mechanism and procedure used by Ethernet over twisted pair by which two connected devices choose common transmission parameters, such as speed, duplex mode, and flow control. Auto negotiation is defined in clause 28 of IEEE 802.3. and was originally an optional component in the Fast Ethernet standard.
Question 94
Which of the following is not a valid multicast MAC address?
A
01:00:5E:00:00:00
B
01:00:5E:00:00:FF
C
01:00:5E:00:FF:FF
D
01:00:5E:FF:FF:FF
       Computer-Networks       MAC-Address       ISRO CS 2014
Question 94 Explanation: 
A multicast addressed frame is either flooded out all ports (if no multicast optimization is configured) or sent out only the ports interested in receiving the traffic. The range of multicast MAC Address lie between 01-00-5E-00-00-00 to 01-00-5E-7F-FF-FF.
Question 95
An organization is granted the block 130.34.12.64/26. It needs to have 4 subnets. Which of the following is not an address of this organization?
A
130.34.12.124
B
130.34.12.89
C
130.34.12.70
D
130.34.12.132
       Computer-Networks       IP-Address       ISRO CS 2014
Question 95 Explanation: 
→ The suffix length is 6. This means the total number of addresses in the block is 64 (26). If we create four subnets, each subnet will have 16 addresses.
→ So, addresses from 130.34.12.64 to 130.34.12.127 will be included in this organization.
Question 96

A supernet has a first address of 205.16.32.0 and a supernet mask of 255.255.248.0. A router receives 4 packets with the following destination addresses. Which packet belongs to this supernet?

A
205.16.42.56
B
205.17.32.76
C
205.16.31.10
D
205.16.39.44
       Computer-Networks       IP-Address       ISRO CS 2014
Question 96 Explanation: 
Given data,
Supernet has a first address=205.16.32.0
Supernet Mask = 255.255.248.0
IP address=?
Step-1: Perform AND operation between supernet mask and IP address. Let us take IP address is 205.16.39.44.
Step-2: AND operation
Question 97

A slotted ALOHA network transmits 200-bit frames using a shared channel with a 200 Kbps bandwidth. Find the throughput of the system, if the system (all stations put together) produces 250 frames per second :

A
49
B
368
C
149
D
151
       Computer-Networks       Access-Control-Methods       UGC-NET JUNE Paper-2
Question 97 Explanation: 
Step-1:
Given data,
Slotted ALOHA transmits = 200 bit frames
Bandwidth = 200 Kbps
System produces = 250 frames per second
Throughput = ?
Step-2:
Pure ALOHA formula S = G * e-2G
Slotted ALOHA formula S = G * e-G
Frame transmission time = 200/200 kbps = 1ms
Here,
G = ¼ and S = G * e-G
= 0.195 (or) 19.5%
System produces = 250 * 0.195
= 48.75
Question 98

The period of a signal is 100 ms. Its frequency is

A
1003 Hertz
B
10−2 KHz
C
10−3 KHz
D
105 Hertz
       Computer-Networks       UGC-NET JUNE Paper-2
Question 98 Explanation: 
First we change 100 ms to seconds, and then we calculate the frequency from the period (1 Hz = 10−3 kHz).
100 ms = 100 × 10-3s = 10-1s
f = 1/T = 1/10-1Hz = 10 Hz = 10 × 10-3kHz = 10-2kHz
Question 99

The dotted-decimal notation of the following IPV4 address in binary notation is

  10000001 00001011 00001011 11101111
A
111.56.45.239
B
129.11.10.238
C
129.11.11.239
D
111.56.11.239
       Computer-Networks       IPv4-and-Fragmentation       UGC-NET JUNE Paper-2
Question 99 Explanation: 
10000001 00001011 00001011 11101111
Since the IP address is of 32 bits and is divided in 4 parts of size 8-bits and then perform simple binary to decimal conversion.
Question 100

Which of the following statements are true ?

    (a) Advanced Mobile Phone System (AMPS) is a second generation cellular phone system.
    (b) IS - 95 is a second generation cellular phone system based on CDMA and DSSS.
    (c) The Third generation cellular phone system will provide universal personnel communication.
A
(a) and (b) only
B
(b) and (c) only
C
(a), (b) and (c)
D
(a) and (c) only
       Computer-Networks       OSI-TCP-layers       UGC-NET JUNE Paper-2
Question 100 Explanation: 
(a) FALSE:
AMPS is a first-generation cellular technology that uses separate frequencies, or "channels", for each conversation (see frequency-division multiple access (FDMA)). It therefore required considerable bandwidth for a large number of users. In general terms, AMPS was very similar to the older "0G" Improved Mobile Telephone Service, but used considerably more computing power in order to select frequencies, hand off conversations to PSTN lines, and handle billing and call setup.
(b) TRUE:
IS - 95 is a second generation cellular phone system based on CDMA and DSSS.
(c) TRUE:
The Third generation cellular phone system will provide universal personnel communication.
Question 101

Match the following symmetric block ciphers with corresponding block and key sizes :

  List-I                             List-II
(a) DES          (i) block size 64 and key size ranges between 32 and 448 
(b) IDEA        (ii) block size 64 and key size 64 
(c) BLOWFISH   (iii) block size 128 and key sizes 128, 192, 256 
(d) AES         (iv) block size 64 and key size 128

A
(a)-(iv), (b)-(ii), (c)-(i), (d)-(iii)
B
(a)-(ii), (b)-(iv), (c)-(i), (d)-(iii)
C
(a)-(ii), (b)-(iv), (c)-(iii), (d)-(i)
D
(a)-(iv), (b)-(ii), (c)-(iii), (d)-(i)
       Computer-Networks       UGC-NET JUNE Paper-2
Question 101 Explanation: 
→ Data Encryption Standard(DES), which uses symmetric key method for encryption of data. It uses block size 64 and key size 128 for encryption.
→ International Data Encryption Algorithm (IDEA), originally called Improved Proposed Encryption Standard (IPES), is a symmetric-key block cipher. It uses block size 64 and key size 128.
→ Blowfish is a symmetric-key block cipher used for a large number of cipher suites and encryption products. Blowfish provides a good encryption rate in software and no effective cryptanalysis of it has been found to date. It uses block size 64 and key size ranges between 32 and 448.
→ Advanced Encryption Standard(AES), which uses symmetric key method for encryption of data. It uses block size 128 and key sizes 128, 192, 256.
*****It’s worthy to remember below table.
Question 102

Which of the following statements are true ?

    (a) Three broad categories of Networks are
        (i) Circuit Switched Networks
        (ii) Packet Switched Networks
        (iii) Message Switched Networks
    (b) Circuit Switched Network resources need not be reserved during the set up phase.
    (c) In packet switching there is no resource allocation for packets.
Code :
A
(a) and (b) only
B
(b) and (c) only
C
(a) and (c) only
D
(a), (b) and (c)
       Computer-Networks       Match-the-following       UGC-NET JUNE Paper-2
Question 102 Explanation: 
→ There are three broad categories of Networks Circuit Switched Networks, Packet Switched Networks, Message Switched Networks.
→ In circuit switching resources are reserved from Setup phase to data transmission phase.
→ When data transmission completes only then the reserved resources are released.
→ Since in packet switching each packet follows different path to reach the destination so no resource reservation is done in case of packet switching.
Question 103

Decrypt the message “WTAAD” using the Caesar Cipher with key = 15.

A
LIPPS
B
HELLO
C
OLLEH
D
DAATW
       Computer-Networks       UGC-NET JUNE Paper-2
Question 103 Explanation: 

We decrypt one character at a time. Each character is shifted 15 characters up Letter W is decrypted to H shifted 15 characters up. Letter W is decrypted to H. Letter T is decrypted to E. The first A is decrypted to L. The second A is decrypted to L And finally D is The second A is decrypted to L. And, finally, D is decrypted to O. The plain text is HELLO.
Question 104

Encrypt the Message “HELLO MY DEARZ” using Transposition Cipher with

A
HLLEO YM AEDRZ
B
EHOLL ZYM RAED
C
ELHL MDOY AZER
D
ELHL DOMY ZAER
       Computer-Networks       UGC-NET JUNE Paper-2
Question 104 Explanation: 
Step-1: According to key we have to divide into number of character blocks.
Here, key size = 4. So, character block size is 4.
Step-2: Remove the spaces in the message and write into sequential order.

Step-3: Get cipher text according to ascending order is ELHL MDOY AZER.
Question 105
Time To Live(TTL) field in the IP datagram is used___
A
To optimize throughput
B
To prevent packet looping
C
To reduce delays
D
To prioritize packets
       Computer-Networks       Nielit Scentist-B [02-12-2018]
Question 105 Explanation: 
→ Time to Live (TTL) is a limit on the period of time or transmissions in computer and computer network technology that a unit of data (e.g. a packet) can experience before it should be discarded.
→ If the limit is not defined then the packets can go into an indefinite loop.
→ The packet is discarded when the time to live field reaches 0 to prevent looping.
Question 106
Identify the subnet mask for the given direct broadcast address of subnet of subnet is 201.15.16.31.
A
255.255.192.192
B
255.255.255.198
C
255.255.255.240
D
255.255.257.240
       Computer-Networks       Nielit Scentist-B [02-12-2018]
Question 106 Explanation: 
Subnet mask ID: 201.15.16.16/28 and subnet broadcast ID is 201.15.16.31.
Question 107
If the period of a signal is 100 ms. Then its frequency in Hertz is___
A
10
B
100
C
1000
D
10000
       Computer-Networks       Nielit Scentist-B [02-12-2018]
Question 107 Explanation: 
Given data,
Frequency = 1/Period.
Frequency = 1/100 x 10-3 Hz
= 10 Hz
Question 108
Identify the true statements from the given statements.
(1) HTTP may use different TCP connection for different objects of a webpage if non persistent connections are used
(2) FTP uses two TCP connections, one for data and another control
(3) TELNET and FTP can only use TWO connection at a time.
A
(1)
B
(1) and (2)
C
(2) and (3)
D
(1),(2) and (3)
       Computer-Networks       Nielit Scentist-B [02-12-2018]
Question 108 Explanation: 
It is very tricky and difficult question to answer. After referred many resources we got this information.
Question 109
_____ transport layer protocol is used to support electronic mail
A
SNMP
B
IP
C
SMTP
D
TCP
       Computer-Networks       Nielit Scentist-B [02-12-2018]
Question 109 Explanation: 
→ TCP is used in transport layer to carry out mail which is initiated by application layer protocol SMTP.
Primary protocols for E-Mail management
Post Office Protocol (POP)
Simple Mail Transfer Protocol (SMTP)
Internet Message Access Protocol (IMAP)
The above protocols are Application Layer Protocols
→ Once a client connects to the E-Mail Server, there may be 0(zero) or more SMTP transactions. If the client has no mail to send, then there are no SMTP transactions. Every e-mail message sent is an SMTP transfer.
→ SMTP is only used to send (push) messages to the server. POP and IMAP are used to receive messages as well as manage the mailbox contents(which includes tasks such as deleting, moving messages etc.).
Question 110
_____ IP address can be used in WAN
A
256.0.0.1
B
172.16.0.10
C
15.1.5.6
D
127.256.0.1
       Computer-Networks       Nielit Scentist-B [02-12-2018]
Question 110 Explanation: 
→ Option A, eliminated because the IP address atmost consists of 255.
→ 172.16.x.x is private address which is usen with in local area network.
→ So,option C is correct.
→ Option D, eliminated because the IP address atmost consists of 255.
Question 111
Given message M=1010001101. The CRC for this given message using the divisor polynomial x5 + x4 + x2 +1 is _____  
A
01011
B
10101
C
01110
D
10110
       Computer-Networks       Nielit Scentist-B [02-12-2018]
Question 111 Explanation: 
Degree of generator polynomial is 5. Hence, 5 zeroes is appended before division.
M = 1010001101
Divisor polynomial: 1.x5 +1.x4+0.x3+1.x2+0.x2+1.x0
Divisor polynomial bit= 110101
append 5 zeroes = M = 101000110100000

∴ CRC = 01110
Question 112
Baud rate measures the number ____ transmitted per second.
A
Symbols
B
Bits
C
Byte
D
None of these
       Computer-Networks       Nielit Scentist-B [02-12-2018]
Question 112 Explanation: 
Bit rate is nothing but number of bits transmitted per second
Baud rate is nothing but number of signals units transmitted per unit time.
Baud, or baud rate, is used to describe the maximum oscillation rate of an electronic signal. For example, if a signal changes (or could change) 1200 times in one second, it would be measured at 1200 baud.
Baud unit symbol "Bd is synonymous to symbols per second or pulses per second. It is the unit of symbol rate, also known as baud or modulation rate; the number of distinct symbol changes (signalling events) made to the transmission medium per second in a digitally modulated signal or a line code.
Question 113
Three or more devices share a link in ____ connection
A
Unipoint
B
Polarpoint
C
Point to point
D
Multipoint
       Computer-Networks       Connection-Types       Nielit Scientist-B IT 4-12-2016
Question 113 Explanation: 
Question 114
Which protocol finds the MAC address from IP address?
A
SMTP
B
ICMP
C
ARP
D
RARP
       Computer-Networks       Network-Layer-Protocols       Nielit Scientist-B IT 4-12-2016
Question 114 Explanation: 
● Mail servers and other mail transfer agents use Simple Mail Transfer Protocol( SMTP) to send and receive mail messages on TCP port 25. Although proprietary systems such as Microsoft Exchange and IBM Notes and webmail systems such as Outlook.com, Gmail and Yahoo! Mail may use their own non standard protocols internally, all use SMTP when sending or receiving email from outside their own systems.
● The Internet Control Message Protocol (ICMP) is a supporting protocol in the Internet protocol suite. It is used by network devices, including routers, to send error messages and operational information indicating
●Address Resolution Protocol (ARP) is a protocol for mapping an Internet Protocol address (IP address) to a physical machine address that is recognized in the local network.
●RARP (Reverse Address Resolution Protocol) is a protocol by which a physical machine in a local area network can request to learn its IP address from a gateway server's Address Resolution Protocol (ARP) table or cache.
Question 115
Bluetooth is an example of:
A
Local area network
B
Virtual private network
C
Personal area network
D
none of the mentioned above
       Computer-Networks       Wifi-Bluetooth       Nielit Scientist-B IT 4-12-2016
Question 115 Explanation: 
→ Bluetooth uses short-range radio waves. Uses in a WPAN include, for example, Bluetooth devices such as keyboards, pointing devices, audio headsets, printers may connect to personal digital assistants (PDAs), cell phones, or computers.
→ A wireless personal area network (WPAN) is a low-powered PAN carried over a short-distance wireless network technology such as IrDA, Wireless USB, Bluetooth and ZigBee.
Question 116
A set of rules that governs data communication:
A
rule
B
medium
C
Link
D
Protocol
       Computer-Networks       Basics       Nielit Scientist-B IT 4-12-2016
Question 116 Explanation: 
A protocol is a set of rules that governs data communication. It represents an agreement between the communicating devices. Without a protocol, two devices may be connected but not communicating, just as a person speaking Telugu cannot be understood by a person who speaks only Hindi.
Question 117
The ___ is the physical path over which a message traversals
A
Path
B
Protocol
C
Route
D
Medium
       Computer-Networks       Basics       Nielit Scientist-B IT 4-12-2016
Question 117 Explanation: 
● Protocol is a standard used to define a method of exchanging data over a computer network, such as local area network, Internet, Intranet, etc.
● Routing is the process of moving a packet of data from one network to another network based on the destination IP address.
● The physical channel used for transmission in the network is called medium.
● Message travel from sender to receiver via a medium using a protocol
Question 118
Application layer protocol defines:
A
types of messages exchanged
B
rules for when and how processes send and respond to messages
C
message format, syntax and semantics
D
all of above
       Computer-Networks       Application-Layer-Protocol       Nielit Scientist-B IT 4-12-2016
Question 118 Explanation: 
● An application layer is an abstraction layer that specifies the shared communications protocols and interface methods used by hosts in a communications network.
● The application layer abstraction is used in both of the standard models of computer networking: the Internet Protocol Suite (TCP/IP) and the OSI model.
● Although both models use the same term for their respective highest level layer, the detailed definitions and purposes are different.
Question 119
Repeaters function in
A
Physical layer
B
Data link layer
C
Network layer
D
Both (A) and (B)
       Computer-Networks       OSI-TCP-layers       Nielit Scientist-C 2016 march
Question 119 Explanation: 
● A repeater is an electronic device that receives a ​ signal​ and retransmits it.
● Repeaters are used to extend transmissions so that the signal can cover longer distances or be received on the other side of an obstruction.
● Physical layer in the OSI model plays the role of interacting with actual hardware and signaling mechanism
Question 120
In networking terminology UTP means
A
Unshielded twisted pair
B
Ubiquitous Teflon port
C
Uniformly Terminating port
D
Unshielded T-connector port
       Computer-Networks       Nework-Cables       Nielit Scientist-C 2016 march
Question 120 Explanation: 
● Unshielded twisted pair, a popular type of cable that consists of two unshielded wires twisted around each other.
● Due to its low cost, UTP cabling is used extensively for ​ local-area networks (LANs)​ and telephone connections
Question 121
How many bits are required to encode all twenty six letters, ten symbols, and ten numerals?
A
5
B
6
C
7
D
46
       Computer-Networks       Encoding-Decoding       Nielit Scientist-C 2016 march
Question 121 Explanation: 
● It is 26 letters ×2 (Capital and small letters) plus twenty characters in total, so the total is 52+20=72.
● To find how many bits you would need, you would need to find the power of 2 that is equal to or the one immediately larger than your number. The number of bits would be the exponent of the power.
● In our case, since 128=2​ 7​ >72>64=2​ 6​ , the power immediately larger than 72 would be 128=27, so you would need 7 bits, which can codify 128 symbols.
● If you don’t want to codify both capital letters and small letters but only one family of 26 letters and be done with it, then you have 26+20=46 symbols. This time 2​ 6​ =64>46>32=2​ 5​ , so the power immediately larger than 46 would be 64=2​ 6​ ; so 6 bits needed​ .
Question 122

The third generation mobile phone are digital and based on

A
AMPS
B
Broadband CDMA
C
CDMA
D
D-AMPS
       Computer-Networks       communication       UGC-NET DEC Paper-2
Question 122 Explanation: 
→ AMPS, D-AMPS, CDMA, Broadband CDMA are the standard methods used for cellular communication.
AMPS(Advanced Mobile Phone Service): It uses frequency division multiple access(FDMA) to provide access to the channel by multiple stations at the same time. In FDMA whole available bandwidth is divided into small frequency bands and each station is allocated to one frequency band which it will use to send and receive data.
D-AMPS(Digital- Advanced Mobile Phone Service) : It uses time division multiple access(TDMA). In this kind of medium access all stations uses the whole available bandwidth for a fixed time slots.
CDMA(Code Division Multiple Access) : In CDMA all stations are allowed to use the channel simultaneously and can use the the whole bandwidth. In this technique the sender and receiver uses a code to distinguish there data from others.
Broadband CDMA : Broadband code division multiple access (B-CDMA) is a CDMA-based cell phone technology that uses broadband transmission. It is an improvement over the first-generation CDMA, which uses narrowband transmission. This kind of technology allows for better deployment of transmitter signals.
Question 123

Consider ISO-OSI network architecture reference model. Session layer of this model offer Dialog control, token management and ____________ as services.

A
Synchronization
B
Asynchronization
C
Errors
D
Flow control
       Computer-Networks       OSI-TCP-layers       UGC-NET DEC Paper-2
Question 123 Explanation: 
→ The session layer is the network dialog controller. It establishes, maintains, and synchronizes the interaction among communicating systems.
Dialog control: The session layer allows two systems to enter into a dialog. It allows the communication between two processes to take place in either half-duplex (one way at a time) or full-duplex (two ways at a time) mode.
Synchronization: The session layer allows a process to add checkpoints, or synchronization points, to a stream of data. For example, if a system is sending a file of 2000 pages, it is advisable to insert checkpoints after every 100 pages to ensure that each 100-page unit is received and acknowledged independently. In this case, if a crash happens during the transmission of page 523, the only pages that need to be reset after system recovery are pages 501 to 523. Pages previous to 501 need not be resent.
Question 124

An internet service provider (ISP) has following chunk of CIDR-based IP addresses available with it: 245.248.128.0/20 . The ISP want to give half of this chunk of addresses to organization A and a quarter to Organization B while retaining the remaining with itself. Which of the following is a valid allocation of addresses to A and B?

A
245.248.132.0/22 and 245.248.132.0/21
B
245.248.136.0/21 and 245.248.128.0/22
C
245.248.136.0/24 and 245.248.132.0/21
D
245.248.128.0/21 and 245.248.128.0/22
       Computer-Networks       IP-Address       UGC-NET DEC Paper-2
Question 124 Explanation: 
Question 125

If the frame buffer has 10-bits per pixel  and 8-bits are allocated for each of the R, G and B components then what would be the size of the color lookup table(LUT)

A
(210 + 211) bytes
B
(210 + 28) bytes
C
(210 + 224) bytes
D
(28 + 29) bytes
       Computer-Networks       LUT       UGC-NET DEC Paper-2
Question 125 Explanation: 
10-bits per pixel it means we will have 210 entries in color lookup table.
8-bits are allocated for each of the R, G and B components, means each entry in color lookup table is of 24-bits (8-bits for each of the R, G and B component).
24-bits = 3 Bytes
So the size of lookup table is = Number of entries * size of each entry
So the size of lookup table is = (210 * 3) Bytes
the size of lookup table is = 3072 Bytes = (210 + 211) Bytes
Question 126

Which of the following statement/s is/are true?

    (i) Firewall can screen traffic going into or out of an organization.
    (ii) Virtual private networks cam simulate an old leased network to provide certain desirable properties.

Choose the correct answer from the code given below:

Code:
A
(i) only
B
Neither (i) nor (ii)
C
Both (i) and (ii)
D
(ii) only
       Computer-Networks       Network-Security       UGC-NET DEC Paper-2
Question 126 Explanation: 
Statement 1 is correct because firewall works on the Application layer, so it can screen the traffic going into and out of the traffic.
Statement 2 is correct. Virtual private networks cam simulate an old leased network to provide certain desirable properties.
Question 127

Consider the following two statements:

    S1: TCP handles both congestion and flow control
    S2: UDP handles congestion but not flow control

Which of the following option is correct with respect to the above statements (S1) and (S2)?

Choose the correct answer from the code given below:

Code:
A
Both S1 and S2 are correct
B
Neither S1 nor S2 is correct
C
S1 is not correct but S2 is correct
D
S1 is correct but S2 is not correct
       Computer-Networks       TCP_UDP       UGC-NET DEC Paper-2
Question 127 Explanation: 
TCP is a reliable protocol for data transmission.
TCP uses advertisement window for flow control and uses Slow start, Congestion Avoidance, Congestion Detection mechanism for congestion control.
UDP is an unreliable data transmission protocol. It does not perform congestion control or flow control.
Question 128

Match the following secret key algorithm (List-1) with the corresponding key lengths (List-2) and choose the correct answer from the code given below,

   LIST-1	      List-2
a) Blow Fish	  (i) 128-256 bits
b) DES	         (ii) 128 bits
c) IDEA	        (iii) 1-448 bits
d) RC5	         (iv) 56 bits
Codes:
A
(a)-(ii), (b)-(iii), (c)- (iv), (d)-(i)
B
(a)-(iv), (b)-(iii), (c)-(ii), (d)-(i)
C
(a)-(iii), (b)-(iv), (c)-(ii), (d)-(i)
D
(a)-(iii), (b)-(iv), (c)-(i), (d)-(ii)
       Computer-Networks       UGC-NET DEC Paper-2
Question 128 Explanation: 
• Blowfish is an encryption algorithm that can be used as a replacement for the DES or IDEA algorithms. It is a symmetric (that is, a secret or private key) block cipher that uses a variable-length key, from 32 bits to 448 bits, making it useful for both domestic and exportable use.
• DES uses a 64-bit key, but eight of those bits are used for parity checks, effectively limiting the key to 56-bits.
• IDEA (International Data Encryption Algorithm) is an encryption algorithm which uses a block cipher with a 128-bit key, and is generally considered to be very secure. It is considered among the best publicly known algorithms.
• RC5 has a variable block size (32, 64 or 128 bits), key size (0 to 2040 bits) and number of rounds (0 to 255). The original suggested choice of parameters were a block size of 64 bits, a 128-bit key and 12 rounds.
Question 129

The four byte IP Address consists of

A
Neither network nor Host Address
B
Network Address
C
Both Network and Host Address
D
Host Address
       Computer-Networks       IP-Address       UGC-NET DEC Paper-2
Question 129 Explanation: 
The IP address of 32 bit (4 byte) consists of both network and host address.
Question 130

Which of the following statement/s is/are true ?

    (i) Windows XP supports both peer-peer and client-server networks.
    (ii) Windows XP implements Transport protocols as drivers that can be loaded and uploaded from the system dynamically.

Choose the correct answer from the code given below :

Code :
A
Both (i) and (ii)
B
Neither (i) nor (ii)
C
(ii) only
D
(i) only
       Computer-Networks       Windows-XP       UGC-NET DEC Paper-2
Question 130 Explanation: 
→ Windows XP's Peer-to-Peer Networking Wizard allows you to set up a firewall-protected network.
→ Windows XP Professional known as peer-to-peer networking. This feature allows these micro-businesses to easily deploy a functional network without a server-class machine.
Question 131

The host is connected to a department network which is a part of a university network. The university network, in turn, is part of the internet. The largest network, in which the Ethernet address of the host is unique, is

A
The university network
B
The internet
C
The subnet to which the host belongs
D
The department network
       Computer-Networks       Ethernet       UGC-NET DEC Paper-2
Question 131 Explanation: 
The Ethernet address, known as MAC (Media Access Control) address, is a 48-bit hardware address which is burned on NIC. Ethernet address is supposed to be unique. If two devices with same Ethernet address are on two different networks then those devices can be uniquely identified. This is because of different Network Ids.
But within the network any two devices should have different MAC addresses.
Question 132

The decimal floating point number -40.1 represented using IEEE-754 32-bit representation and written in hexadecimal form is

A
0xC2206000
B
0xC2006666
C
0xC2006000
D
0xC2206666
       Computer-Networks       IEEE       UGC-NET DEC Paper-2
Question 132 Explanation: 
The given floating value is -40.1
and corresponding binary value is 1100 0010 0010 0000 0110 0110 0110 0110
Now, we will write Hexadecimal value by grouping four bits from least significant bits is 0xc2206666.
Question 133

In PERT/CPM, the merge event represents___________ of two or more events.

A
splitting
B
completion
C
beginning
D
joining
       Computer-Networks       PERT/CPM       UGC-NET DEC Paper-2
Question 133 Explanation: 
→ The program (or project) evaluation and review technique (PERT) is a statistical tool used in project management, which was designed to analyze and represent the tasks involved in completing a given project.
→ The critical path method (CPM), or critical path analysis (CPA), is an algorithm for scheduling a set of project activities. It is commonly used in conjunction with the program evaluation and review technique (PERT).
→ A critical path is determined by identifying the longest stretch of dependent activities and measuring the time required to complete them from start to finish.
Event:
An event represents a point in time signifying the completion of some activities and the beginning of new ones. This is usually represented by a circle in a network which is also called a node or connector. The events are classified into three categories
1. Merge event – When more than one activity comes and joins an event such an event is known as merge event.
2. Burst event – When more than one activity leaves an event such an event is known as burst event.
3. Merge and Burst event – An activity may be merge and burst event at the same time as with respect to some activities it can be a merge event and with respect to some other activities it may be a burst event.
Question 134
Which register that shift complete binary number in one bit at a time and shift all the stored bits out at a time?
A
Parallel-in parallel-out
B
Parallel-in Serial-out
C
Serial-in parallel-out
D
Serial-in Serial-out
       Computer-Networks       Nielit STA [02-12-2018]
Question 134 Explanation: 
SIPO is the register is loaded with serial data, one bit at a time, with the stored data being available at the output in parallel form.
SIPO:
A serial-in, parallel-out shift register is similar to the serial-in, serial-out shift register in that it shifts data into internal storage elements and shifts data out at the serial-out, data-out, pin. It is different in that it makes all the internal stages available as outputs. Therefore, a serial-in, parallel-out shift register converts data from serial format to parallel format.
More Information
Serial-in to Serial-out (SISO) - the data is shifted serially “IN” and “OUT” of the register, one bit at a time in either a left or right direction under clock control.
Parallel-in to Serial-out (PISO) - the parallel data is loaded into the register simultaneously and is shifted out of the register serially one bit at a time under clock control.
Parallel-in to Parallel-out (PIPO) - the parallel data is loaded simultaneously into the register, and transferred together to their respective outputs by the same clock pulse.
Question 135
Simple network management protocol(SNMP) is a protocol that runs on:
A
Application layer
B
Transport layer
C
Network layer
D
Data link layer
       Computer-Networks       Nielit STA [02-12-2018]
Question 135 Explanation: 
→ Simple Network Management Protocol (SNMP) is an Internet Standard protocol for collecting and organizing information about managed devices on IP networks and for modifying that information to change device behavior. Devices that typically support SNMP include cable modems, routers, switches, servers, workstations, printers, and more.
→ SNMP is a component of the Internet Protocol Suite as defined by the Internet Engineering Task Force (IETF). It consists of a set of standards for network management, including an application layer protocol, a database schema, and a set of data objects.
Question 136
Given a mask, M=255.255.255.248. How many subnet bits are required for given mask M?
A
2
B
3
C
4
D
5
       Computer-Networks       Nielit STA [02-12-2018]
Question 136 Explanation: 
Converting the subnet mask 255.255.255.248 into binary is:
11111111.11111111.11111111.11111000
255.255.255.248/29 mask.
→ It is 32 bit address and number of bits in the Network ID is 29
→ Number of bits in host ID=32-29= 3
Question 137
Transparent mode firewall is a:
A
Layer 2 firewall
B
Layer 3 firewall
C
Layer 4 firewall
D
Layer 7 firewall
       Computer-Networks       Nielit STA [02-12-2018]
Question 137 Explanation: 
→ Transparent mode firewall (or routed firewall) is a routed hop and acts as a default gateway for hosts that connect to one of its screened subnets.
→ A transparent firewall, on the other hand, is a Layer 2 firewall that acts like a "bump in the wire," or a "stealth firewall," and is not seen as a router hop to connected devices.
Question 138
A routing protocol that was used for trail information that gets updated dynamically is:
A
Distance Vector
B
Path Vector
C
Link Vector
D
Multipoint
       Computer-Networks       Nielit STA [02-12-2018]
Question 138 Explanation: 
→ In the context of routers, neighbors always means routers sharing a common data link.
→ A distance vector routing protocol sends its updates to neighboring routers and depends on them to pass the update information along to their neighbors.
→ For this reason, distance vector routing is said to use hop-by-hop updates.
Question 139
A packet switching network
A
is free
B
can reduce the cost of using an information utility
C
allows communications channel to be shared among more than one user
D
boh (B) and (C)
       Computer-Networks       Switching       Nielit Scientist-B CS 2016 march
Question 139 Explanation: 
Packet switching is a method of grouping data that is transmitted over a digital network into packets. Packets are made of a header and a payload. Data in the header are used by networking hardware to direct the packet to its destination where the payload is extracted and used by application software.
Question 140
In context of TCP/IP computer network models, which of the following is FALSE?
A
Besides span of geographical area, the other major difference between LAN and WAN is that the later uses switching element
B
A repeater is used just to forward bits from one network to another one
C
IP layer is connected oriented layer in TCP/IP
D
A gateway is used to connect incompatible networks
       Computer-Networks       TCP       Nielit Scientist-B CS 2016 march
Question 140 Explanation: 
→ A transceiver is a device comprising both a transmitter and a receiver that are combined and share common circuitry or a single housing. When no circuitry is common between transmit and receive functions, the device is a transmitter-receiver.
→ IP is connectionless, in that a data packet can travel from a sender to a recipient without the recipient having to send an acknowledgement.
Question 141
Which of the following OS possible in a token passing bus network?
A
in-service expansion
B
unlimited number of stations
C
both (A) and (B)
D
unlimited distance
       Computer-Networks       Token Ring       Nielit Scientist-B CS 2016 march
Question 141 Explanation: 
Token bus is a network implementing the token ring protocol over a virtual ring on a coaxial cable. A token is passed around the network nodes and only the node possessing the token may transmit. If a node doesn't have anything to send, the token is passed on to the next node on the virtual ring. Each node must know the address of its neighbour in the ring, so a special protocol is needed to notify the other nodes of connections to, and disconnections from, the ring.
Question 142
​The third generation mobile phone are digital and based on
A
AMPS
B
Broadband CDMA
C
CDMA
D
D-AMPS
       Computer-Networks       CDMA       UGC NET CS 2018-DEC Paper-2
Question 142 Explanation: 
→ AMPS, D-AMPS, CDMA, Broadband CDMA are the standard methods used for cellular communication.
→ ​ AMPS(Advanced Mobile Phone Service):​ It uses frequency division multiple access(FDMA) to provide access to the channel by multiple stations at the same time. In FDMA whole available bandwidth is divided into small frequency bands and each station is allocated to one frequency band which it will use to send and receive data.
→ ​ D-AMPS(Digital- Advanced Mobile Phone Service) :​ It uses time division multiple access(TDMA). In this kind of medium access all stations uses the whole available bandwidth for a fixed time slots.
→ ​ CDMA(Code Division Multiple Access) :​ In CDMA all stations are allowed to use the channel simultaneously and can use the the whole bandwidth. In this technique the sender and receiver uses a code to distinguish there data from others.
→ ​ Broadband CDMA :​ Broadband code division multiple access (B-CDMA) is a CDMA-based cell phone technology that uses broadband transmission. It is an improvement over the first-generation CDMA, which uses narrowband transmission. This kind of technology allows for better deployment of transmitter signals.
Question 143
Consider ISO-OSI network architecture reference model. Session layer of this model offer Dialog control, token management and ____________ as services.
A
Synchronization
B
Asynchronization
C
Errors
D
Flow control
       Computer-Networks       OSI-TCP-layers       UGC NET CS 2018-DEC Paper-2
Question 143 Explanation: 
→ The session layer is the network dialog controller. It establishes, maintains, and synchronizes the interaction among communicating systems.
→ ​ Dialog control:​ The session layer allows two systems to enter into a dialog. It allows the communication between two processes to take place in either half-duplex (one way at a time) or full-duplex (two ways at a time) mode.
→ ​ Synchronization​ : The session layer allows a process to add checkpoints, or synchronization points, to a stream of data. For example, if a system is sending a file of 2000 pages, it is advisable to insert checkpoints after every 100 pages to ensure that each 100-page unit is received and acknowledged independently. In this case, if a crash happens during the transmission of page 523, the only pages that need to be reset after system recovery are pages 501 to 523. Pages previous to 501 need not be resent.
Question 144
​An internet service provider (ISP) has following chunk of CIDR-based IP addresses available with it: 245.248.128.0/20 . The ISP want to give half of this chunk of addresses to organization A and a quarter to Organization B while retaining the remaining with itself. Which of the following is a valid allocation of addresses to A and B?
A
245.248.132.0/22 and 245.248.132.0/21
B
245.248.136.0/21 and 245.248.128.0/22
C
245.248.136.0/24 and 245.248.132.0/21
D
245.248.128.0/21 and 245.248.128.0/22
       Computer-Networks       IP-Address       UGC NET CS 2018-DEC Paper-2
Question 144 Explanation: 
Question 145
​Which of the following statement/s is/are true?
(i) Firewall can screen traffic going into or out of an organization.
(ii) Virtual private networks cam simulate an old leased network to provide certain desirable properties.
Choose the correct answer from the code given below:
A
(i) only
B
Neither (i) nor(ii)
C
Both (i) and (ii)
D
(ii) only
       Computer-Networks       Firewall       UGC NET CS 2018-DEC Paper-2
Question 145 Explanation: 
Statement 1 is correct because firewall works on the Application layer, so it can screen the traffic going into and out of the traffic.
Statement 2 is correct.Virtual private networks cam simulate an old leased network to provide certain desirable properties.
Question 146
​Identify the correct sequence in which the following packets are transmitted on the network by a host when a browser requests a web page from a remote server, assuming that the host has been restarted.
A
HTTP GET request, DNS query, TCP SYN
B
DNS query, TCP SYN, HTTP GET request
C
TCP SYN, DNS query, HTTP GET request
D
DNS query, HTTP Get request, TCP SYN
       Computer-Networks       Transmission-and-Propagation-Delay       UGC NET CS 2018-DEC Paper-2
Question 146 Explanation: 
Sequence in which the given packets are transmitted on the network by a host when a browser requests a web page from a remote server, assuming that the host has been restarted :
Step 1: DNS query is sent to Domain name server to convert the Domain name into it’s IP address.
Step 2: TCP SYN request packet is sent by sender to establish a connection for data transmission.
Step 3: HTTP GET request is used to request data from a specified IP address.
Question 147
Consider the following two statements:
S1: TCP handles both congestion and flow control
S2: UDP handles congestion but not flow control
Which of the following option is correct with respect to the above statements (S1) and (S2)?
A
Both S1 and S2 are correct
B
Neither S1 nor S2 is correct
C
S1 is not correct but S2 is correct
D
S1 is correct but S2 is not correct
       Computer-Networks       TCP/UDP       UGC NET CS 2018-DEC Paper-2
Question 147 Explanation: 
TCP is a reliable protocol for data transmission.
TCP uses advertisement window for flow control and uses Slow start, Congestion Avoidance, Congestion Detection mechanism for congestion control.
UDP is an unreliable data transmission protocol. It does not perform congestion control or flow control.
Question 148
Match the following secret key algorithm (List 1) with the corresponding key lengths (List 2) and choose the correct answer from the code given below,
A
(a)-(ii),(b)-(iii), (c)- (iv), (d)-(i)
B
(a)-(iv),(b)-(iii), (c)- (ii), (d)-(i)
C
(a)-(iii),(b)-(iv), (c)- (ii), (d)-(i)
D
(a)-(iii),(b)-(iv), (c)- (i), (d)-(ii)
       Computer-Networks       Secret-Key-Algorithm        UGC NET CS 2018-DEC Paper-2
Question 148 Explanation: 
● Blowfish is an encryption algorithm that can be used as a replacement for the DES or IDEA algorithms. It is a symmetric (that is, a secret or private key) block cipher that uses a variable-length key, from 32 bits to 448 bits, making it useful for both domestic and exportable use.
● DES uses a 64-bit key, but eight of those bits are used for parity checks, effectively limiting the key to 56-bits. ● IDEA (International Data Encryption Algorithm) is an encryption algorithm which uses a block cipher with a 128-bit key, and is generally considered to be very secure. It is considered among the best publicly known algorithms
● RC5 has a variable block size (32, 64 or 128 bits), key size (0 to 2040 bits) and number of rounds (0 to 255). The original suggested choice of parameters were a block size of 64 bits, a 128-bit key and 12 rounds.
Question 149
The four byte IP Address consists of
A
Neither network nor Host Address
B
Network Address
C
Both Network and Host Address
D
Host Address
       Computer-Networks       IP-Address       UGC NET CS 2018-DEC Paper-2
Question 149 Explanation: 
The IP address of 32 bit (4 byte) consists of both network and host address.
Question 150
Which of the following statement/s is/are true ?
(i) windows XP supports both peer-peer and client-server networks.
(ii) Windows XP implements Transport protocols as drivers that can be loaded and uploaded from the system dynamically.
A
Both (i) and (ii)
B
Neither (i) nor (ii)
C
(ii) only
D
(i) only
       Computer-Networks       Client-Server-Architecture       UGC NET CS 2018-DEC Paper-2
Question 150 Explanation: 
→ Windows XP's Peer-to-Peer Networking Wizard allows you to set up a firewall-protected network.
→ Windows XP Professional known as peer-to-peer networking. This feature allows these micro-businesses to easily deploy a functional network without a server-class machine
Question 151
The host is connected to a department network which is a part of a university network. The university network, in turn, is part of the internet. The largest network, in which the Ethernet address of the host is unique, is
A
The university network
B
The internet
C
The subnet to which the host belongs
D
The department network
       Computer-Networks       Ethernet       UGC NET CS 2018-DEC Paper-2
Question 151 Explanation: 
The Ethernet address, known as MAC (Media Access Control) address, is a 48-bit hardware address which is burned on NIC. Ethernet address is supposed to be unique. If two devices with same Ethernet address are on two different networks then those devices can be uniquely identified. ​ This is because of different Network Ids.
But within the network any two devices should have different MAC addresses.
Question 152
​Suppose that everyone in a group of N people wants to communicate secretly with (N-1) other people using symmetric key cryptographic system. The communication between any two persons should not be decodable by the others in the group. The number of keys required in the system as a whole to satisfy the confidentiality requirement is
A
2N
B
N(N-1)
C
N(N-1)/2
D
(N-1)​ 2
       Computer-Networks       Network-Security       UGC NET CS 2018-DEC Paper-2
Question 152 Explanation: 
→ If one person in a group of N people want to communicate with remaining (N-1) people using symmetric key cryptographic system then he needs (N-1) keys.
→ We have N people in group so number of keys needed are N(N-1)
→ But two people in a group can use same keys then no need to use 2(N-1) keys they can communicate using (N-1) keys.
So total number of keys needed [N(N-1)]/2
Question 153
The first network:
A
ARPANET
B
NFSNET
C
CNNET
D
ASAPNET
       Computer-Networks       ARPANET       Nielit Scientist-B CS 4-12-2016
Question 153 Explanation: 
The Advanced Research Projects Agency Network (ARPANET) was an early packet-switching network and the first network to implement the protocol suite TCP/IP. Both technologies became the technical foundation of the Internet.
Question 154
Communication between a computer and a keyboard involves __ transmission
A
Simplex
B
Half Duplex
C
Automatic
D
Full Duplex
       Computer-Networks       Transmission-Modes       Nielit Scientist-B CS 4-12-2016
Question 154 Explanation: 
● As per the above mentioned details, data flows in only single direction.So it Simplex transmission.
● Half-duplex data transmission means that data can be transmitted in both directions on a signal carrier, but not at the same time
● Full-duplex data transmission means that data can be transmitted in both directions on a signal carrier at the same time
Question 155
Bluetooth is an example of:
A
Personal area network
B
Virtual private network
C
Local area network
D
None of the above
       Computer-Networks       Wifi-Bluetooth       Nielit Scientist-B CS 4-12-2016
Question 155 Explanation: 
A wireless personal area network (WPAN) is a personal area network in which the are wireless. IEEE 802.15 has produced standards for several types of PANs operating in the ISM band including Bluetooth. The Infrared Data Association has produced standards for WPANs which operate using infrared communications.
Question 156
IPv6 is developed by
A
IETF
B
ANSI
C
ISO
D
IEEE
       Computer-Networks       IPv4-an-Fragmentation       Nielit Scientist-D 2016 march
Question 156 Explanation: 
Internet Protocol version 6 (IPv6) is the most recent version of the Internet Protocol (IP), the communications protocol that provides an identification and location system for computers on networks and routes traffic across the Internet. IPv6 was developed by the ​ Internet Engineering Task Force (IETF)​ to deal with the long-anticipated problem of IPv4 address exhaustion. IPv6 is intended to replace IPv4. IPv6 became a Draft Standard in December 1998, and became an Internet Standard on 14 July 2017
Question 157
The DNS maps the IP address to
A
A binary address as strings
B
An alphanumeric address
C
A hierarchy of domain names
D
A hexadecimal address
       Computer-Networks       IP-Address       ISRO CS 2015
Question 157 Explanation: 
→ DNS maps the IP addresses into a hierarchy of domain names.
→ DNS is a host name to IP address translation service.
→ DNS is a distributed database implemented in a hierarchy of name servers.
→ It is an application layer protocol for message exchange between clients and servers.
Question 158
How many bits internet address is assigned to each host on a TCP/IP internet which is used in all communication with the host?
A
16 bits
B
32 bits
C
48 bits
D
64 bits
       Computer-Networks       TCP/IP       ISRO CS 2015
Question 158 Explanation: 
→ Present we are using IPv4 address, it has 32 bits to communicate with the host.
→ Planning to shift IPv6 address, it has 128 bits to communicate with the host
Question 159
In CRC if the data unit is 100111001 and the divisor is 1011 then what is dividend at the receiver?
A
100111001101
B
100111001011
C
100111001
D
100111001110
       Computer-Networks       CRC       ISRO CS 2015
Question 159 Explanation: 
→ A cyclic redundancy check (CRC) is an error-detecting code commonly used in digital networks and storage devices to detect accidental changes to raw data. Blocks of data entering these systems get a short check value attached, based on the remainder of a polynomial division of their contents. On retrieval, the calculation is repeated and, in the event the check values do not match, corrective action can be taken against data corruption. CRCs can be used for error correction
→ Given data unit is 100111001 and divisor is 1011
→ The polynomial is written in binary as the coefficients; a 3rd-degree polynomial has 4 coefficients (1x3 + 0x2 + 1x + 1). In this case, the coefficients are 1, 0, 1 and 1. The result of the calculation is 3 bits long.
→ First padded with zeros corresponding to the bit length n of the CRC. Here we used 3-bit CRC
→ After padding zeros to the data unit, Data unit = 100111001000
→ Divide data unit by divisor

→ Append the remainder(011) to the data unit where zero’s are padded Then data unit becomes 100111001011 at receiver end
Question 160
An ACK number of 1000 in TCP always means that
A
999 bytes have been successfully received
B
1000 bytes have been successfully received
C
1001 bytes have been successfully received
D
None of the above
       Computer-Networks       TCP/IP       ISRO CS 2015
Question 160 Explanation: 
TCP sequence numbers and Acknowledgments: 1. A Sequence number is sent from the host and is total number of bytes sent UP to this point, during the conversation, not including the current payload
2. The acknowledgement number would be the total number of bytes received by the destination, FROM the sender, plus one (i.e. the one being the next byte it expects to get). Initial Sequence number is needed in order to identify how many number of bytes received by receiver.
So the answer is option-D because of not mentioning of sequence number.
Question 161
In a class B subnet, we know the IP address of one host and the mask as given below:
IP address: 125.134.112.66
Mask: 255.255.224.0
What is the first address(Network address)?
A
125.134.96.0
B
125.134.112.0
C
125.134.112.66
D
125.134.0.0
       Computer-Networks       IP-Address       ISRO CS 2015
Question 161 Explanation: 
Given
IP address: 125.134.112.66
Subnet Mask: 255.255.224.0
IP address in binary Form = 01111101 10000110 01110000 01000010
Subnet Mask in binary Form = 11111111 11111111 11100000 00000000
Every IP address has Network id and Host id,
We will get Network id by doing bitwise -AND (&) operation with IP address and Subnet Mask

Question 162
A certain population of ALOHA users manages to generate 70 request/sec. If the time is slotted in units of 50 msec, then channel load would be
A
4.25
B
3.5
C
350
D
450
       Computer-Networks       Network-Access-Methods       ISRO CS 2015
Question 162 Explanation: 
The number of requests generated by ALOHA users are 70 request/sec
The amount of time slotted in unit is 50 msec
The number of time slots for generating user requests in 1 second are 1000/50=20
The Channel load can be calculated dividing number of requests to the time slots Channel Load = 70/20=3.5
Question 163
Which statement is false?
A
PING is a TCP/IP application that sends datagrams once every second in the hope of an echo response from the machine being PINGED
B
If the machine is connected and running a TCP/IP protocol stack, it should respond to the PING datagram with a datagram of its own
C
If PING encounters an error condition, an ICMP message is not returned
D
PING display the time of the return response in milliseconds or one of several error message
       Computer-Networks       PING       ISRO CS 2015
Question 163 Explanation: 
→ Ping operates by sending Internet Control Message Protocol (ICMP) echo request packets to the target host and waiting for an ICMP echo reply.
→ The program reports errors, packet loss, and a statistical summary of the results, typically including the minimum, maximum, the mean round-trip times, and standard deviation of the mean.
Question 164
A router uses the following routing table:

A packet bearing a destination address 144.16.68.117 arrives at the router. On which interface will it be forwarded?
A
eth0
B
eth1
C
eth2
D
eth3
       Computer-Networks       IP-Address       ISRO CS 2015
Question 164 Explanation: 
Question 165
Which layers of the OSI reference model are host-to-host layers?
A
Transport, session, presentation, application
B
Session, presentation, application
C
Datalink, transport, presentation, application
D
Physical, datalink, network, transport
       Computer-Networks       OSI-TCP-layers       ISRO CS 2015
Question 165 Explanation: 

Question 166

Which of the following is NOT a symmetric key algorithm?

A
Ellipse Curve Cryptography
B
Advanced Encryption standard
C
Data Encryption Standard
D
Blowfish
       Computer-Networks       Network-Security       JT(IT) 2018 PART-B Computer Science
Question 166 Explanation: 
Question 167

Suppose a network using CSMA/CD has a bandwidth of 10 Mbps. If the maximum propagation time is 25 microsec, that what will be the minimum frame size?

A
500 bits
B
50 bits
C
500 bytes
D
4*1011 bits
       Computer-Networks       Ethernet       JT(IT) 2018 PART-B Computer Science
Question 167 Explanation: 
→ The minimum frame transmission time is Tfr = 2 × Tp = 50 μs.
→ This means, in the worst case, a station needs to transmit for a period of 50 μs to detect the collision.
→ The minimum size of the frame is 10 Mbps × 50 μs = 500 bits or 62.5 bytes.
→ This is actually the minimum size of the frame for Standard Ethernet.
Question 168

If 100 users are making 10 request/sec to a slotted ALOHA channel and each slot is of 50 m sec, then what will be the channel load?

A
10
B
20
C
2
D
50
       Computer-Networks       Network-Access-Methods       JT(IT) 2018 PART-B Computer Science
Question 168 Explanation: 
→ Slotted aloha throughput S = Ge−G
→Total user = 100 making 10 request/sec, Then total number of requests are 1000
Each slot time is 50 ms, so the number of slots in one second = 1/(50 X 10-3) = 20 slots/sec
Requests per second = 1000
Time slots per second = 20
Channel load = Number of Requests / Number of Slots = 1000 / 20 = 50
Question 169

Chose the option which matches each element of LIST-1 with exactly one element of LIST-2:

A
(i)-(d), (ii)-(c), (iii)-(b), (iv)-(a)
B
(i)-(b), (ii)-(a), (iii)-(d), (iv)-(c)
C
(i)-(a), (ii)-(d), (iii)-(c), (iv)-(b)
D
(i)-(d), (ii)-(b), (iii)-(c), (iv)-(a)
       Computer-Networks       OSI-TCP-layers       JT(IT) 2018 PART-B Computer Science
Question 169 Explanation: 
1. Repeater→ Physical layer: Its job is to regenerate the signal over the same network before the signal becomes too weak or corrupted so as to extend the length to which the signal can be transmitted over the same network.
2. Gateway→ All seven layers: A gateway, as the name suggests, is a passage to connect two networks together that may work upon different networking models. They basically works as the messenger agents that take data from one system, interpret it, and transfer it to another system. Gateways are also called protocol converters and can operate at any network layer. Gateways are generally more complex than switch or router.
3. Router→ Network layer: A router is a device like a switch that routes data packets based on their IP addresses. Router is mainly a Network Layer device. Routers normally connect LANs and WANs together and have a dynamically updating routing table based on which they make decisions on routing the data packets. Router divide broadcast domains of hosts connected through it.
Bridge→ Data link layer: A bridge is a repeater, with add on functionality of filtering content by reading the MAC addresses of source and destination. It is also used for interconnecting two LANs working on the same protocol. It has a single input and single output port, thus making it a 2 port device.
Question 170

Which of the following protocols is used to map IP address to MAC address?

A
ARP
B
IP
C
DHCP
D
RARP
       Computer-Networks       IPv4-and-Fragmentation       JT(IT) 2018 PART-B Computer Science
Question 170 Explanation: 
→ IP address to MAC address using Address resolution protocol(ARP).
→ MAC address to IP address using Reverse Address resolution protocol(RARP).
Question 171
In time division switches if each memory access takes 100 ns and one frame period is 125 us, then the maximum number of lines that can be supported is
A
625 lines
B
1250 lines
C
2300 lines
D
318 lines
       Computer-Networks       TDMA       Nielit Scientific Assistance IT 15-10-2017
Question 171 Explanation: 
In time division switches, 2NT = 1 frame period, where T = memory access time.
Number of lines=125*1000/100*2=625 (1us =1000ns)
Question 172
The address of a class B host o be split into subnets with a 3 bit subnet number.What is the maximum number of subnets and maximum number of hosts in each subnet?
A
8 subnets and 262141 hosts
B
6 subnets and 262141 hosts
C
6 subnets and 1022 hosts
D
8 subnets and 1024 hosts
E
None of the Above
       Computer-Networks       Subnetting       Nielit Scientific Assistance IT 15-10-2017
Question 172 Explanation: 
For subnetting, bits are taken from Host-Id.
→ Class-B addressing is given

Question 173
The maximum data rate of a channel of 3000 Hz bandwidth and SNR of 30 db is
A
60000
B
15000
C
30000
D
3000
       Computer-Networks       Noisy/Noiseless-Channel-Capacity       Nielit Scientific Assistance IT 15-10-2017
Question 173 Explanation: 
Maximum number of bps = B log2 (1 + SNR).
=3000*log2(1+30)
=3000*4.95
=14850. So approximately, 15000
Question 174
Bit stuffing refers to
A
Inserting a '0' in user data stream to differentiate it with a flag
B
Inserting a '0' in flag stream to avoid ambiguity
C
Appending a nibble to the flag sequence
D
Appending a nibble to the user data stream
       Computer-Networks       OSI-TCP-layers       Nielit Scientific Assistance IT 15-10-2017
Question 174 Explanation: 
●The term "bit stuffing" broadly refers to a technique whereby extra bits are added to a data stream, which do not themselves carry any information, but either assist in management of the communications or deal with other issues.
●The receiver knows how to detect and remove or disregard the stuffed bits.
Question 175
If the channel is band limited to 6 kHz and signal to noise ratio is 16, what would be the capacity of channel?
A
16.15kbps
B
23.24 kbps
C
40.12 kbps
D
24.74 kbps
       Computer-Networks       Noisy/Noiseless-Channel-Capacity       Nielit Scientific Assistance IT 15-10-2017
Question 175 Explanation: 
Shannon Capacity (Noisy Channel):
In presence of Gaussian band-limited white noise, Shannon Hartley theorem gives the
maximum data rate capacity C = B log​ 2 (1+ S/N)
=6log​ 2​ (1+16)
=6*log​ 2​ (17)
=6*4.08=24.48
Question 176
If the number of networks and number of hosts in class B are 2m, (2n -2) respectively. Then the relation between m,n is:
A
3m=2n
B
7m=8n
C
8m=7n
D
2m=3n
       Computer-Networks       Nielit Scientist-B 17-12-2017
Question 176 Explanation: 
Total number of networks in class-B network is 214
Total number of hosts per network is 216-2
Substitute 8m and 7n
⇒ 14*8=16*7
112=112
Question 177
Correct expression for UDP user datagram length is
A
length of UDP=length of IP - length of IP header
B
length of UDP=length of UDP - length of UDP header
C
length of UDP=length of IP + length of IP header
D
length of UDP=length of UDP + length of UDP header
       Computer-Networks       Nielit STA 17-12-2017
Question 177 Explanation: 
→ Total length of the datagram in bytes.
→ The size of the data by computing "total length - header length"
→ A user datagram is encapsulated in an IP datagram. There is a field in the IP datagram the defines the total length. There is another field in the IP datagram that defines the length of the header. So if we subtract the length of a UDP datagram that is encapsulated in an IP datagram, we get the length of UDP user datagram.
Question 178
Close-loop control mechanism try to:
A
Remove congestion after it occurs
B
Remove congestion after sometime
C
Prevent congestion before it occurs
D
Prevent congestion before sending packets
       Computer-Networks       Nielit STA 17-12-2017
Question 178 Explanation: 
We can divide congestion control mechanisms into two broad categories:
open-loop congestion control (prevention)
closed-loop congestion control (removal).

Question 179
Which multiple access technique is used by IEEE 802.11 standard for wireless LAN?
A
CDMA
B
CSMA/CA
C
ALOHA
D
None of the Options
       Computer-Networks       Nielit STA 17-12-2017
Question 179 Explanation: 
A wireless LAN system allows a station using infrared, visible light, or radio wave to communicate with similarly configured stations in the vicinity, or to communicate with remote stations via a nearby base station that is connected to a wired network. The most prevalent wireless LANs used today are those based IEEE 802.11. These standards use similar medium access control protocol and frame format as the Ethernet, and therefore IEEE 802.11 - based wireless LANs have been referred to as wireless Ethernets.
A wireless LAN can be characterized by the following attributes
Architecture
Physical layer
Medium access control layer
Architecture
Based on IEEE 802.11 standards, wireless LANs can be classified as follows: infrastructure and ad hoc. With the ad hoc architecture, wireless stations communicate directly with each other on an ad hoc basis.
Physical Layer
The physical layer is characterized by the following three elements: the frequency band, the multiple access method (which are similar to "multiplexing" in the wired environment, and the data rate.
Medium Access Control (MAC) Layer
This layer provide the same function as the MAC layer in the wired LAN environment. IEEE 802.11, instead of using Ethernet's CSMA/CD, specifies a scheme called CSMA/CA.
CA stands for Collision Avoidance. (In the wireless environment, since a station may not be able to hear all other stations, CD, Collision Detection is not feasible)
Question 180
PGP encrypts data using a block cipher called:
A
International data encryption algorithm
B
private data encryption algorithm
C
Internet data encryption algorithm
D
none of the options
       Computer-Networks       Nielit STA 17-12-2017
Question 180 Explanation: 
→ International data encryption algorithm(IDEA) was used in Pretty Good Privacy (PGP) v2.0 and was incorporated after the original cipher used in v1.0, BassOmatic, was found to be insecure.
→ IDEA is an optional algorithm in the OpenPGP standard. PGP encrypts data by using a block cipher called international data encryption algorithm.
→ IDEA operates on 64-bit blocks using a 128-bit key and consists of a series of 8 identical transformations (a round, see the illustration) and an output transformation (the half-round).
→ The processes for encryption and decryption are similar.
Question 181
A subnet mask in class C can have ___ 1's with the remaining bits 0's
A
10
B
24
C
12
D
7
       Computer-Networks       Nielit STA 17-12-2017
Question 181 Explanation: 
→ Subnet mask for class C is 255.255.255.0.
→ Convert 255.255.255.0 into binary format.
11111111.11111111.11111111.00000000.
→ Total number of 1’s are 24. And total number of 0’s are 8.
Question 182
What is the value of acknowledgement field in segment?
A
Number of previous bytes to receive
B
Total number of bytes to receive
C
Number of next bytes to receive
D
Sequence of zero's and one's
       Computer-Networks       Nielit STA 17-12-2017
Question 182 Explanation: 
Acknowledgement field in a segment defines the number of next bytes to receive.
Question 183
Which of the following is a class B host address?
A
230.0.0.0
B
130.4.5.6
C
230.7.6.5
D
30.4.5.6
       Computer-Networks       Nielit STA 17-12-2017
Question 183 Explanation: 
→ Class A addresses only include IP starting from 1.x.x.x to 126.x.x.x only. The IP range 127.x.x.x is reserved for loopback IP addresses.
→ Class B IP Addresses range from 128.0.x.x to 191.255.x.x. The default subnet mask for Class B is 255.255.x.x.
→ Class C IP addresses range from 192.0.0.x to 223.255.255.x. The default subnet mask for Class C is 255.255.255.x.

Note: Class A addresses 127.0.0.0 to 127.255.255.255 cannot be used and is reserved for loopback and diagnostic functions.
Private IP Addresses:

Question 184
There is a need to create a network that has 5 subnets, each with at least 16 hosts. which one is used as classful subnet mask?
A
255.255.255.192
B
255.255.255.248
C
255.255.255.240
D
255.255.255.224
       Computer-Networks       Nielit STA 17-12-2017
Question 184 Explanation: 
You need 5 subnets, each with at least 16 hosts. The mask 255.255.255.240 provides 16 subnets with 14 hosts-this will not work. The mask 255.255.255.224 provides 8 subnets, each with 30 hosts.
Question 185
Which NetWare protocol provides link state routing?
A
NLSP
B
RIP
C
SAP
D
NCP
       Computer-Networks       Nielit STA 17-12-2017
Question 185 Explanation: 
→ NetWare Link Services Protocol (NLSP) provides link-state routing. SAP (Service Advertisement Protocol) advertises network services.
→ NCP (NetWare Core Protocol)provides client-to-server connections and applications.
→ RIP is a distance vector routing protocol.
Question 186
Which layer connects the network support layers and user support layers?
A
Transport layer
B
Network layer
C
Data link layer
D
Session layer
       Computer-Networks       Nielit STA 17-12-2017
Question 186 Explanation: 
→ Physical, data link and network layers are network support layers.
→ Session, presentation and application layers are user support layers.
→ Transport layer layer connects the network support layers and user support layers
Question 187
When we use slow start algorithm, the size of the congestion window increase__unit it reaches a threshold
A
Additively
B
Multiplicatively
C
Exponentially
D
None of the options
       Computer-Networks       Nielit STA 17-12-2017
Question 187 Explanation: 
→ In the slow-start algorithm, the size of the congestion window increases exponentially until it reaches a threshold.
→ ​TCP slow start is an algorithm which balances the speed of a network connection. Slow start gradually increases the amount of data transmitted until it finds the network’s maximum carrying capacity.
​→ TCP slow start is one of the first steps in the congestion control process. It balances the amount of data a sender can transmit (known as the congestion window) with the amount of data the receiver can accept (known as the receiver window).
→ The lower of the two values becomes the maximum amount of data that the sender is allowed to transmit before receiving an acknowledgment from the receiver.
Question 188
In Ipv4 addresses, classful addressing is replaced with:
A
Classless addressing
B
Classful addressing
C
Subnet advertising
D
None of the options
       Computer-Networks       Nielit STA 17-12-2017
Question 188 Explanation: 
→ Classful addressing is replaced with classless addressing.
→ The problem with this classful addressing method is that millions of class A address are wasted, many of the class B address are wasted, whereas, number of addresses available in class C is so small that it cannot cater the needs of organizations.
→ Class D addresses are used for multicast routing, and are therefore available as a single block only. Class E addresses are reserved.
Note: Because of these problems, Classful networking was replaced by Classless Inter-Domain Routing (CIDR) in 1993.
Question 189
An ethernet destination address 07-01-12-03-04-05 is:
A
Unicast address
B
Multicast address
C
Broadcast address
D
All of the options
       Computer-Networks       Nielit STA 17-12-2017
Question 189 Explanation: 
Here, Shortcut method to find out using MAC address is
The first octant of MAC address written in binary format.
07=0000 0111.
The least significant bit 0 then unicast, 1 means Multicast and entire 8 bits are 1’s then broadcast.
Question 190
Why is one-time password safe?
A
it is easy to generate
B
it cannot be shared
C
it is different for every access
D
it can be easily decrypted
       Computer-Networks       Nielit STA 17-12-2017
Question 190 Explanation: 
One time password is safe since it is generated per access and thus cannot be brute forced or deduced.
Question 191
In classless addressing, there are no classes but addresses are still granted in:
A
Codes
B
Blocks
C
IPs
D
Sizes
       Computer-Networks       Nielit STA 17-12-2017
Question 191 Explanation: 
→ When we have run out of class A and B addresses, and a class C block is too small for most midsize organizations.
→ To overcome the problem of address depletion and give more organizations access to internet, classless addressing was designed and implemented.
→ The size of the block (the number of addresses) varies based on the nature and size of the entry. For example, a household may be given only two addresses; a large organization may be given thousands of addresses. An ISP, may be given thousands or hundreds of thousands based on the number of customer it may serve.
→ To simplify the handling of addresses, the Internet authorities impose three restrictions on classless address blocks:
1. The addresses in a block must be contiguous, one after the other.
2. The number of addresses in a block must be a power of 2 (1, 2, 4,8, .... ).
3. The first address must be evenly divisible by the number of address.
Question 192
The demands of the fragmentation are:
A
Complex routers
B
Open to DOS attack
C
No overlapping of fragments
D
(A) and (B) both
       Computer-Networks       Nielit STA 17-12-2017
Question 192 Explanation: 
→ IP fragmentation is the process of breaking up a single Internet Protocol (IP) packet into multiple packets of smaller size. Every network link has a characteristic size of messages that may be transmitted, called the maximum transmission unit (MTU).
→ The support for fragmentation of larger packets provides a protocol allowing routers to fragment a packet into smaller packets when the original packet is too large for the supporting data link frames.
→ IP fragmentation exploits (attacks) use the fragmentation protocol within IP as an attack vector.
IP fragment over lapped:
The IP fragment overlapped exploit occurs when two fragments contained within the same IP packet have offsets that indicate that they overlap each other in positioning within the packet. This could mean that either fragment A is being completely overwritten by fragment B, or that fragment A is partially being overwritten by fragment B.
Overlapping fragments may also be used in an attempt to bypass Intrusion Detection Systems
Question 193
In which of the topology, each device has a dedicated point-to-point link to a central controller?
A
Mesh
B
Bus
C
Ring
D
Star
       Computer-Networks       Topologies       KVS 22-12-2018 Part-B
Question 193 Explanation: 
→ In a star topology, each device has a dedicated point-to-point link only to a central controller, usually called a hub.
→ The devices are not directly linked to one another.
Question 194
The electromagnetic waves ranging in frequencies between 1 GHz and 300 GHz are called____
A
Radio waves
B
Microwaves
C
Infrared waves
D
Light waves
       Computer-Networks       Electromagnetic-Waves       KVS 22-12-2018 Part-B
Question 194 Explanation: 
Microwaves have wavelengths approximately in the range of 30 cm (frequency = 1 GHz) to 1 mm (300 GHz).
Question 195
Which of the following protocol is used for transforming electronic mail messages from one machine to another?
A
FTP
B
SMTP
C
SNMP
D
STTP
       Computer-Networks       Application-Layer-Protocol       KVS 22-12-2018 Part-B
Question 195 Explanation: 
Simple Mail Transfer Protocol (SMTP) is the standard protocol for email services on a TCP/IP network. SMTP provides the ability to send and receive email messages.
SMTP is an application-layer protocol that enables the transmission and delivery of email over the Internet
Question 196
The device bridge is used at ____ layer of OSI reference model.
A
Datalink
B
Network
C
Transport
D
Application
       Computer-Networks       OSI-TCP-layers       KVS 22-12-2018 Part-B
Question 196 Explanation: 
→ A network bridge is a computer networking device that creates a single aggregate network from multiple communication networks or network segments.
→ Bridges are important in networks because the networks are divided into many parts geographically remote from one another.
→ Bridge works at Data link layer
Question 197
The period of a signal is 100ms. What is its frequency in kilohertz?
A
10-1 kHz
B
10 -2 kHz
C
10-3 kHz
D
10-4 kHz
       Computer-Networks       Data-and-Signals       KVS 22-12-2018 Part-B
Question 197 Explanation: 
First we change 100 ms to seconds, and then we calculate the frequency from the period (1 Hz=10-3 kHz).
100 ms=100*10-3s
= 10-1 s
f=1/T
=1/10-1Hz
=10 Hz
=10*10-3 kHz
=10-2 kHz
Question 198
Machine that places the request to access the data, is generally called as___
A
Server machine
B
Client machine
C
Request machine
D
Intelligent machine
       Computer-Networks       Client-Server-Computing       KVS 22-12-2018 Part-B
Question 198 Explanation: 
→ A client is a piece of computer hardware or software that accesses a service made → available by a server.
→ The server is often (but not always) on another computer system, in which case the client accesses the service by way of a network.
Question 199
The function setcookie() is used to____
A
Enable or disable cookie support
B
Declare cookie variables
C
Store data in cookie variables
D
Clear data from cookie variables
       Computer-Networks       Application-Layer-Protocol       KVS 22-12-2018 Part-B
Question 199 Explanation: 
→ The setcookie() function defines a cookie to be sent along with the rest of the HTTP headers.
→ A cookie is often used to identify a user. A cookie is a small file that the server embeds on the user's computer. Each time the same computer requests a page with a browser, it will send the cookie too. With PHP, you can both create and retrieve cookie values.
→ Syntax
→ setcookie(name,value,expire,path,domain,secure,httponly);
Question 200
When a mobile telephone physically moves from one cell to another cell, the base station transfers ownership to the cell getting strongest signal. This process is known as____
A
Handoff
B
Mobile routing
C
Mobile switching
D
Cell Switching
       Computer-Networks       Handoff       KVS 22-12-2018 Part-B
Question 200 Explanation: 
→ A handoff refers to the process of transferring an active call or data session from one cell in a cellular network to another or from one channel in a cell to another.
→ A well-implemented handoff is important for delivering uninterrupted service to a caller or data session user.
→ In satellite communications it is the process of transferring satellite control responsibility from one earth station to another without loss or interruption of service.
Question 201
Bit stuffing refers to
A
Inserting a '0' in user data stream to differentiate it with a flag
B
Inserting a '0' in flag stream to avoid ambiguity
C
Appending a nibble to the flag sequence
D
Appending a nibble to the user data stream
       Computer-Networks       OSI-TCP-layers       Nielit Scientific Assistance CS 15-10-2017
Question 201 Explanation: 
● The term "bit stuffing" broadly refers to a technique whereby extra bits are added to a data stream, which do not themselves carry any information, but either assist in management of the communications or deal with other issues.
● The receiver knows how to detect and remove or disregard the stuffed bits.
Question 202
Which of the following would not be specified in a communication protocol?
A
Header contents
B
Trailer contents
C
Error Checking
D
Data content of message
       Computer-Networks       Communication-protocol       Nielit Scientific Assistance CS 15-10-2017
Question 202 Explanation: 
→ A communication protocol is a system of rules that allow two or more entities of a communications system to transmit information via any kind of variation of a physical quantity. The protocol defines the rules, syntax, semantics and synchronization of communication and possible error recovery methods.
→ There are many properties of a transmission that a protocol can define. Common ones include: packet size, transmission speed, error correction types, handshaking and synchronization techniques, address mapping, acknowledgement processes, flow control, packet sequence controls, routing, address formatting
Question 203
Internet Control message protocol(ICMP)
A
Allows gateways to send error control messages to other gateways or hosts
B
Provides communication between the internet protocol software on one machine and the internet protocol software on another
C
Only reports error conditions to the original source, the source must relate errors to individual application programs and take action to correct the problem.
D
All of these
       Computer-Networks       OSI-TCP-layers       KVS DEC-2013
Question 203 Explanation: 
→ The Internet Control Message Protocol (ICMP) is used by network devices, including routers, to send error messages and operational information indicating, for example, that a requested service is not available or that a host or router could not be reached.
→ ICMP creates and sends messages to the source IP address indicating that a gateway to the Internet that a router, service or host cannot be reached for packet delivery.
Question 204
Which layers of the OSI model are host to host layers?
A
transport,session,Presentation, Application
B
Network,Transport,Session,Presentation
C
Datalink, Network, Transport, Session
D
Physical,Data Link,Network,Transport
       Computer-Networks       OSI-TCP-layers       KVS DEC-2013
Question 204 Explanation: 
Datalink, Network, Transport, Session are host to host layers.
Question 205
A____is a communication pathway connecting two or more devices. Another of its key characteristics is that it is a shared transmission medium. A signal transmitted by any one device is available for reception by all other devices attached to it.
A
Train
B
Bus
C
Tram
D
Aeroplane
       Computer-Networks       Topologies       KVS DEC-2013
Question 205 Explanation: 
A bus is a communication system that transfers data between components inside a computer, or between computers.
Question 206
Bridge works in which layer of the OSI model?
A
Application layer
B
Transport layer
C
Network layer
D
Data link layer
       Computer-Networks       OSI-TCP-layers       KVS DEC-2013
Question 206 Explanation: 
● A network bridge is a computer networking device that creates a single aggregate network from multiple communication networks or network segments.
●Bridges are important in networks because the networks are divided into many parts geographically remote from one another.
● Bridge works at Data link layer
Question 207
What is the meaning of bandwidth in a network?
A
Transmission capacity of a communication channel
B
Connected components in a network
C
Class of IP used in Network
D
Interconnected by communication channels
       Computer-Networks       Transmission-and-Propagation-Delay       KVS DEC-2013
Question 207 Explanation: 
●Bandwidth is also the amount of data that can be transmitted in a fixed amount of time.
●For digital devices, the bandwidth is usually expressed in bits per second(bps) or bytes per second.
● For analog devices, the bandwidth is expressed in cycles per second, or Hertz (Hz).
Question 208
Which one of the following transmission systems provides the highest data rate to an individual device?
A
Computer bus
B
Telephone bus
C
Voice and mode
D
Lease lines
       Computer-Networks       Topologies       KVS DEC-2013
Question 208 Explanation: 
A bus is a communication system that transfers data between components inside a computer, or between computers
Question 209
The X.25 standard specifies a
A
Technique for start-stop data
B
Technique for dial access
C
DTE/DCE interface
D
Data bit rate
       Computer-Networks       X.25       KVS DEC-2013
Question 209 Explanation: 
→ The X.25 specification defines only the interface between a subscriber (DTE) and an X.25 network (DCE). X.75, a protocol very similar to X.25, defines the interface between two X.25 networks to allow connections to traverse two or more networks.
→ X.25 does not specify how the network operates internally – many X.25 network implementations used something very similar to X.25 or X.75 internally
Question 210
Frames from one LAN can be transmitted to another LAN via a device called
A
Router
B
Bridge
C
Repeater
D
modem
       Computer-Networks       Types-of-Area-Networks       KVS DEC-2013
Question 210 Explanation: 
→ A network bridge is a computer networking device that creates a single aggregate network from multiple communication networks or network segments.
→ Bridges are important in networks because the networks are divided into many parts geographically remote from one another. Something is required to join these networks so that they can become part of the whole network.
→ Take for example a divided LAN, if there is no medium to join these separate LAN parts an enterprise may be limited in its growth potential. The bridge is one of the tools to join these LANS.
Question 211
With an IP address of 100, you have 80 subnets. What subnet mask should you use to maximize the number of available hosts?
A
192
B
224
C
248
D
252
E
None of the above
       Computer-Networks       IPv4-an-Fragmentation       KVS DEC-2013
Question 211 Explanation: 
The question is incomplete because without class type we can’t determine subnets.
Question 212
The___houses the switches in token ring
A
Transceiver
B
Nine pin connector
C
MAU
D
NIC
       Computer-Networks       Access-Control-Methods       KVS DEC-2013
Question 212 Explanation: 
A Media Access Unit (MAU), also known as a Multistation Access Unit (MAU or MSAU) is a device to attach multiple network stations in a star topology as a token ring network, internally wired to connect the stations into a logical ring (generally passive i.e. non-switched and unmanaged; however managed token ring MAUs do exist in the form of CAUs, or Controlled Access Units).
Question 213
In OSI network architecture, routing is performed by the
A
Network layer
B
Data link layer
C
Transport layer
D
Session layer
       Computer-Networks       Routing       KVS DEC-2013
Question 213 Explanation: 
→ Network layer that provides data routing paths for network communication.
→ Data is transferred in the form of packets via logical network paths in an ordered format controlled by the network layer
Question 214
Which of the following protocol is used for transferring electronic mail messages from one machine to another?
A
HTTP
B
FTP
C
SMTP
D
SNMP
       Computer-Networks       Application-Layer-Protocol       KVS DEC-2017
Question 214 Explanation: 
→ Mail servers and other mail transfer agents use SMTP to send and receive mail messages on TCP port 25.
→ SMTP (Simple Mail Transfer Protocol) is a TCP/IP protocol used in sending and receiving e-mail.
→ However, since it is limited in its ability to queue messages at the receiving end, it is usually used with one of two other protocols, POP3 or IMAP, that let the user save messages in a server mailbox and download them periodically from the server.
→ In other words, users typically use a program that uses SMTP for sending e-mail and either POP3 or IMAP for receiving email.
Question 215
Ten signals, each requiring 3000 hz, are multiplexed onto 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
33,700
B
30,000
C
32,700
D
33,000
       Computer-Networks       FDM       KVS DEC-2017
Question 215 Explanation: 
Step-1: Given data, 10 signals. Each signal requires 3000. So, 10*3000=30000 Hz
Step-2: Guard band (or gaps)=(10-9)=9.
300x9 =2700 Kz
Step-3: Minimum bandwidth is required for the multiplexed channel 30000+2700= 32700Hz
Question 216
If the period of a signal is 1000ms, then what is its frequency in kilohertz?
A
10​ -1​ Khz
B
1 KHz
C
10​ -3​ Khz
D
10​ -2​ Khz
       Computer-Networks       Data-and-Signals       KVS DEC-2017
Question 216 Explanation: 
Step-1: Frequency=1/time
Step-2: Time=1000 ms which is equals to =1 sec
Step-3: Frequency=1/1 sec
=1000/1000 sec
=1K/1000 sec
=10​ -3​ Khz
Note: 1 sec = 10​ -3​ ms
Question 217
Mechanism that is used to convert domain name into IP address is known___
A
HTTP
B
URL
C
FTP
D
DNS
       Computer-Networks       IPv4-an-Fragmentation       KVS DEC-2017
Question 217 Explanation: 
The Domain Name System (DNS) is the phonebook of the Internet. Humans access information online through domain names, like nytimes.com or espn.com. Web browsers interact through Internet Protocol (IP) addresses. DNS translates domain names to IP addresses so browsers can load Internet resources.
Question 218
The unit receiving the data item response with another control signal to acknowledgement receipt of the data. This type of argument between two independent units is known as
A
Storage control
B
Multitasking
C
Handshaking
D
Piggybacking
       Computer-Networks       Types-of-Acknowledgement       KVS DEC-2017
Question 218 Explanation: 
Handshaking Handshake is a method used in a TCP/IP network to create a connection between a local host/client and server. Handshaking is the exchange of information between two sender and receiver. So, it is like a agreement between them.
Piggybacking In two-way communication, wherever a frame is received, the receiver waits and does not send the control frame back to the sender immediately. The receiver waits until its network layer passes in the next data packet. The delayed acknowledgement is then attached to this outgoing data frame.
Question 219
The 10Base5 cabling is also known as___
A
fast ethernet
B
thick ethernet
C
thin ethernet
D
gigabit ethernet
       Computer-Networks       Ethernet       KVS DEC-2017
Question 219 Explanation: 
→ 10BASE5 (also known as thick Ethernet or thicknet) was the first commercially available variant of Ethernet.
→ 10BASE5 uses a thick and stiff coaxial cable up to 500 metres (1,600 ft) in length. Up to 100 stations can be connected to the cable using vampire taps and share a single collision domain with 10 Mbit/s of bandwidth shared among them. The system is difficult to install and maintain.
Question 220
Which of the following devices understands the formats and contents of the data and translate message from one format to another?
A
Gateway
B
Hub
C
Switch
D
Router
       Computer-Networks       Hardware-and-various-devices-in-networking       KVS DEC-2017
Question 220 Explanation: 
→ The gateway (or default gateway) is implemented at the boundary of a network to manage all the data communication that is routed internally or externally from that network.
→ Besides routing packets, gateways also possess information about the host network's internal paths and the learned path of different remote networks.
→ If a network node wants to communicate with a foreign network, it will pass the data packet to the gateway, which then routes it to the destination using the best possible path. holds its current state. There will be no change.
Question 221
Which of the following transmission media works on the principle of total internal reflection?
A
Optical fiber cable
B
Shielded twisted pair cable
C
Unshielded twisted pair cable
D
coaxial cable
       Computer-Networks       Transmission-Modes       KVS DEC-2017
Question 221 Explanation: 
→ Optical fiber consists of a core and a cladding layer, selected for total internal reflection due to the difference in the refractive index between the two.
→ In practical fibers, the cladding is usually coated with a layer of acrylate polymer or polyimide. This coating protects the fiber from damage but does not contribute to its optical waveguide properties.
Question 222
Which of the following topologies highest reliability?
A
Mesh topology
B
Bus topology
C
Star topology
D
Ring topology
       Computer-Networks       Topologies       KVS DEC-2017
Question 222 Explanation: 
→ A network setup where each computer and network device is interconnected with one another, allowing for most transmissions to be distributed even if one of the connections go down. It is a topology commonly used for wireless networks.
→ A mesh topology can be a full mesh topology or a partially connected mesh topology. In a full mesh topology, every computer in the network has a connection to each of the other computers in that network.
→ The number of connections in this network can be calculated using the following formula (n is the number of computers in the network): ​ n(n-1)/2.
→ This network having highest reliability.
Question 223
What is the length of an IP address in (Pre-1PU6)?
A
8
B
1
C
2
D
4
       Computer-Networks       IPv4-an-Fragmentation       KVS DEC-2017
Question 223 Explanation: 
→ Actually they given typing mistake instead of Pre-IPv6 they given Pre-1PU6.
→ Pre-IPv6 is nothing but IPv4. So, total size in 32 bits=4 bytes
Question 224
An analog signal carries 4 bits in each signal unit. if 1000 signal units are sent per second, then baud rate and bit rate of the signal are___ and ___
A
1000 bauds/sec, 4000 bps
B
1000 bauds/sec, 500 bps
C
4000 bauds/sec, 1000 bps
D
2000 bauds/sec, 1000 bps
       Computer-Networks       Transmission-and-Propagation-Delay       KVS DEC-2017
Question 224 Explanation: 
Actually there is a typing mistake in this question. They given 1000 bauds/sec, 1000 bps but it is wrong option. So, we are corrected and given correct option.
Step-1: Important formulas to find a solution is
Bit rate=baud rate*no.of bits per signal
Baud rate(or signal rate)=bit rate/no.of bits per signal
Step-2:They are expecting to find bit rate. Given data, 1000 bauds(signal)per second and number of bits per signal=4
Step-3: Bit rate=1000*4=4000 bits per second.
Question 225

The 10Base5 cabling scheme of ethernet uses:

A
Twisted pairs
B
Fiber optics
C
Thin coax
D
Thick coax
       Computer-Networks       Ethernet       JT(IT) 2016 PART-B Computer Science
Question 225 Explanation: 
The original cabling standard for Ethernet that uses coaxial cables. The name derives from the fact that the maximum data transfer speed is 10 Mbps, it uses baseband transmission, and the maximum length of cables is 500 meters.
10Base5 is also called thick Ethernet, ThickWire, and ThickNet.
→ The number 10: At the front of each identifier, 10 denotes the standard data transfer speed over these media - ten megabits per second (10Mbps).
→ The word Base: Short for Baseband, this part of the identifier signifies a type of network that uses only one carrier frequency for signaling and requires all network stations to share its use.
→ The segment type or segment length: This part of the identifier can be a digit or a letter:
- Digit - shorthand for how long (in meters) a cable segment may be before attenuation sets in.
For example, a 10Base5 segment can be no more than 500 meters long.
- Letter - identifies a specific physical type of cable.
For example, the T at the end of 10BaseT stands for twisted-pair.
Question 226

One of the ad-hoc solutions to count to infinity problem in network routing is:

A
The split horizon hack
B
Flow based routing
C
Flooding
D
Shortest path routing
       Computer-Networks       Routing       JT(IT) 2016 PART-B Computer Science
Question 226 Explanation: 
There are 2 possible solutions are:
1. Route Poison
2. The split horizon hack
Question 227

Given a bit rate of b bits/sec, the time required to send 16 bits is:

A
16*b sec
B
16/b sec
C
16b sec
D
b16 sec
       Computer-Networks       Ethernet       JT(IT) 2016 PART-B Computer Science
Question 227 Explanation: 
The speed of the data is expressed in bits per second (bits/s or bps).
The data rate R is a function of the duration of the bit or bit time (TB).
R = 1/TB
Question 228

The built in HTTP request method to request to read a web page is:

A
HEAD
B
PUT
C
GET
D
POST
       Computer-Networks       Application-Layer-Protocol       JT(IT) 2016 PART-B Computer Science
Question 228 Explanation: 
Request Method:
The request method indicates the method to be performed on the resource identified by the given Request-URI. The method is case-sensitive and should always be mentioned in uppercase. The following table lists all the supported methods in HTTP/1.1.
Question 229

The traditional cryptographic cipher that records the letters but do not disguise them is:

A
Substitute cipher
B
One-time pads
C
Secret key algorithms
D
Transposition cipher
       Computer-Networks       Network-Security       JT(IT) 2016 PART-B Computer Science
Question 229 Explanation: 
A transposition cipher reorders the letters but does not disguise them. The key is a word or phrase not containing any repeated letters. Its purpose is to number the columns, column 1 being under the letter closest to the start of the alphabet, and so on.
Also this code can be easily broken.
Question 230

In client-server computing vertical scaling means:

A
Adding or removing client workstations with only a slight performance impact
B
Migrating servers to a new group of client workstations
C
Migrating client workstations to a larger and faster server machine or multi servers
D
Combining two or more client workstation groups
       Computer-Networks       Client-Server-Computing       JT(IT) 2016 PART-B Computer Science
Question 230 Explanation: 
→ Vertical scaling refers to adding more resources (CPU/RAM/DISK) to your server (database or application server is still remains one) as on demand.
→ Vertical Scaling is most commonly used in applications and products of middle-range as well as small and middle-sized companies. One of the most common examples of Virtual Scaling is to buy an expensive hardware and use it as a Virtual Machine hypervisor (VMWare ESX).
→ Vertical Scaling usually means upgrade of server hardware. Some of the reasons to scale vertically includes increasing IOPS (Input / Output Operations), amplifying CPU/RAM capacity, as well as disk capacity.
→ However, even after using virtualization, whenever an improved performance is targeted, the risk for downtimes with it is much higher than using horizontal scaling.
Question 231
Time taken by a packet to travel from client to server and then back to client is called
A
STT
B
RTT
C
PTT
D
Total time
       Computer-Networks       Transmission-and-Propagation-Delay       KVS 30-12-2018 Part B
Question 231 Explanation: 
→Round-trip time (RTT), also called round-trip delay, is the time required for a signal pulse or packet to travel from a specific source to a specific destination and back again.
→In this context, the source is the computer initiating the signal and the destination is a remote computer or system that receives the signal and retransmits it
Question 232
A network with bandwidth of 10Mbps can pass only an average of 12,000 frames per minute with each frame carrying an average of 10,000 bits. What is the throughput of this network?
A
2 Mbps
B
4 Mbps
C
8 Mbps
D
12 Mbps
       Computer-Networks       Ethernet       KVS 30-12-2018 Part B
Question 232 Explanation: 
Throughput is a measure of how many units of information a system can process in a given amount of time
Given bandwidth is 10Mbps
Number of average frames 12,000 per minute.
Each Frame is carrying 10,000 bits
So, 12,000 * 10,000 / 60 seconds.
= 2 * 1000000 bits / seconds.
= 2 Mbps.
Question 233
The most common connector used with fiber-optic cable is
A
RJ45
B
RJ54
C
BNC
D
MT-RJ
       Computer-Networks       Nework-Cables       KVS 30-12-2018 Part B
Question 233 Explanation: 
→ The most common are: ST, SC, FC, MT-RJ and LC style connectors.
→All of these types of connectors can be used with either multimode or single mode fiber.
Question 234
In which of the following topologies, each device has a dedicated point-to-point connection with only two devices on either side of it?
A
Mesh topology
B
Star topology
C
Bus topology
D
Ring topology
       Computer-Networks       Topologies       KVS 30-12-2018 Part B
Question 234 Explanation: 
→A ring network is a network topology in which each node connects to exactly two other nodes, forming a single continuous pathway for signals through each node - a ring.
→Data travels from node to node, with each node along the way handling every packet.
Question 235
Which of the following IEEE standards is NOT a standard for WAN in mobile communication?
A
IEEE 802.11a
B
IEEE 802.11b
C
IEEE 802.11d
D
IEEE 802.11g
       Computer-Networks       Ethernet       KVS 30-12-2018 Part B
Question 235 Explanation: 
There are several specifications in the 802.11 family:

802.11 — applies to wireless LANs and provides 1 or 2 Mbps transmission in the 2.4 GHz band using either frequency hopping spread spectrum (FHSS) or direct sequence spread spectrum (DSSS).

802.11a — an extension to 802.11 that applies to wireless LANs and provides up to 54-Mbps in the 5GHz band. 802.11a uses an orthogonal frequency division multiplexing encoding scheme rather than FHSS or DSSS.

802.11b (also referred to as 802.11 High Rate or Wi-Fi) — an extension to 802.11 that applies to wireless LANS and provides 11 Mbps transmission (with a fallback to 5.5, 2 and 1-Mbps) in the 2.4 GHz band. 802.11b uses only DSSS. 802.11b was a 1999 ratification to the original 802.11 standard, allowing wireless functionality comparable to Ethernet.

802.11e — a wireless draft standard that defines the Quality of Service (QoS) support for LANs, and is an enhancement to the 802.11a and 802.11b wireless LAN (WLAN) specifications.

802.11g — applies to wireless LANs and is used for transmission over short distances at up to 54-Mbps in the 2.4 GHz bands.

802.11n — 802.11n builds upon previous 802.11 standards by adding multiple-input multiple-output (MIMO). The real speed would be 100 Mbit/s (even 250 Mbit/s in PHY level), and so up to 4-5 times faster than 802.11g.

802.11ac — 802.11ac builds upon previous 802.11 standards, particularly the 802.11n standard, to deliver data rates of 433Mbps per spatial stream, or 1.3Gbps in a three-antenna (three stream) design. The 802.11ac specification operates only in the 5 GHz frequency range and features support for wider channels (80MHz and 160MHz) and beamforming capabilities by default to help achieve its higher wireless speeds.

802.11ac Wave 2 — 802.11ac Wave 2 is an update for the original 802.11ac spec that uses MU-MIMO technology and other advancements to help increase theoretical maximum wireless speeds for the spec to 6.93 Gbps.

802.11ad — 802.11ad is a wireless specification under development that will operate in the 60GHz frequency band and offer much higher transfer rates than previous 802.11 specs, with a theoretical maximum transfer rate of up to 7Gbps (Gigabits per second).

802.11ah— Also known as Wi-Fi HaLow, 802.11ah is the first Wi-Fi specification to operate in frequency bands below one gigahertz (900 MHz), and it has a range of nearly twice that of other Wi-Fi technologies. It's also able to penetrate walls and other barriers considerably better than previous Wi-Fi standards.

802.11r - 802.11r, also called Fast Basic Service Set (BSS) Transition, supports VoWi-Fi handoff between access points to enable VoIP roaming on a Wi-Fi network with 802.1X authentication.

802.1X — Not to be confused with 802.11x (which is the term used to describe the family of 802.11 standards) 802.1X is an IEEE standard for port-based Network Access Control that allows network administrators to restricted use of IEEE 802 LAN service access points to secure communication between authenticated and authorized devices.
Question 236
Which of the following address is used to deliver a message to the correct application program running on a host?
A
Logical
B
Physical
C
IP
D
Port
       Computer-Networks       Transport-layer-Protocol       KVS 30-12-2018 Part B
Question 236 Explanation: 
An Internet Protocol address (IP address) is a numerical label assigned to each device connected to a computer network that uses the Internet Protocol for communication.An IP address serves two principal functions: host or network interface identification and location addressing.

Logical Address:
A logical address is the address at which an item (memory cell, storage element, network host) appears to reside from the perspective of an executing application program.
Physical address: Each system having a NIC(Network Interface Card) through which two systems physically connected with each other with cables. The address of the NIC is called Physical address or mac address. This is specified by the manufacturer company of the card. This address is used by data link layer.
Port Address: There are many application running on the computer. Each application run with a port no.(logically) on the computer. This port no. for application is decided by the Kernel of the OS. This port no. is called port address.
Question 237
Ten signals, each requiring 3000Hz, are multiplexed on to a single channel using FDM. What is the minimum bandwidth required for the multiplexed channel? Assume that the guard bands are 300 Hz wide.
A
30,000
B
30,300
C
32,700
D
33,000
       Computer-Networks       FDM       KVS 30-12-2018 Part B
Question 237 Explanation: 
Ch1 - gb -ch2-gb-ch3-gb-ch4-gb-ch5-gb-ch6-gb-ch7-gb-ch8-gb-ch9-gb-ch10. Where Ch represents channel and gb means guard bands.
There are total 10 channels and 9 guard bands.Each guard band is 300Hz wide 9 guard band occupies 2700 hz.
10 channels occupies 30000 hz.
So total bandwidth is 32700.
Question 238
Which of the following TCP ports are used by File Transfer Protocol(FTP)?
A
20 and 21
B
20 and 23
C
21 and 25
D
23 and 25
       Computer-Networks       OSI-TCP-layers       KVS 30-12-2018 Part B
Question 238 Explanation: 

Question 239
Which of the following devices takes data sent from one network device and forwards it to the destination node based on MAC address ?
A
Hub
B
Modem
C
Switch
D
Gateway
       Computer-Networks       UGC NET CS 2017 Nov- paper-2
Question 239 Explanation: 
→ Switch takes data sent from one network device and forwards it to the destination node based on MAC address.
→ Physical layer uses HUB and repeater
→ Data link layer uses Switch. MAC address comes into data link layer.
Question 240
__________ do not take their decisions on measurements or estimates of the current traffic and topology.
A
Static algorithms
B
Adaptive algorithms
C
Non - adaptive algorithms
D
Recursive algorithms
       Computer-Networks       Topologies       UGC NET CS 2017 Nov- paper-2
Question 240 Explanation: 
Non-Adaptive Routing Algorithm: These algorithms do not base their routing decisions on measurements and estimates of the current traffic and topology. Instead the route to be taken in going from one node to the other is computed in advance, off-line, and downloaded to the routers when the network is booted. This is also known as static routing. This can be further classified as:
1. Flooding
2. Random Walk
Question 241
The number of bits used for addressing in Gigabit Ethernet is __________.
A
32 bit
B
48 bit
C
64 bit
D
128 bit
       Computer-Networks       Ethernet       UGC NET CS 2017 Nov- paper-2
Question 241 Explanation: 
→ The number of bits used for addressing in Gigabit Ethernet is 48 bit
→ The Ethernet frame is the packet that delivers data between stations.
→ Each frame is made up of bits which are organized into different fields. The four main fields are the destination Ethernet address field, source Ethernet address field, data field and frame check sequence field.
Question 242
Which of the following layer of OSI Reference model is also called end-to-end layer?
A
Network layer
B
Data layer
C
Session layer
D
Transport layer
       Computer-Networks       OSI-TCP-layers       UGC NET CS 2017 Nov- paper-2
Question 242 Explanation: 
→ Data link layer-Process to Process connectivity
→ Network layer-Host to Host connectivity
→ Transport layer-End to End connectivity
→ Presentation layer-Encryption and Decryption
Question 243
The IP address __________ is used by hosts when they are being booted
A
0.0.0.0
B
1.0.0.0
C
1.1.1.1
D
255.255.255.255
       Computer-Networks       OSI-TCP-layers       UGC NET CS 2017 Nov- paper-2
Question 243 Explanation: 
We can also call default route.
If a packet destination address is not known, then if the default route is not known, then if the default route is present in the routing table of the router, then the packet is not discarded but forwarded to the next router. 0.0.0.0
Question 244
Which of the given wireless technologies used in IoT, consumes the least amount of power ?
A
Zigbee
B
Bluetooth
C
Wi-Fi
D
GSM/CDMA
       Computer-Networks       OSI-TCP-layers       UGC NET CS 2017 Nov- paper-2
Question 244 Explanation: 
Bluetooth wireless technologies used in IoT, consumes the least amount of power
Question 245
Which speed up could be achieved according to Amdahl’s Law for infinite number of processes if 5% of a program is sequential and the remaining part is ideally parallel ?
A
Infinite
B
5
C
20
D
50
       Computer-Networks       Amdahl’s-Law       UGC NET CS 2017 Nov- paper-2
Question 245 Explanation: 
Step-1: Speed up for infinite number of process uses Amdahl's law formula is
S = 1 / (1-P)
where,
P is parallel part of program,
sequential part of program is 5%
. = 1 - sequential part
= 1 - 0.05 (or 5%)
= 0.95 (or 95%)
Step-2: Apply into the formula is
S = 1 / (1-P)
S = 1 / (1-0.95)
S = 1 / 0.05
S = 20
Question 246
The maximum size of the data that the application layer can pass on to the TCP layer below is __________.
A
2​ 16​ bytes
B
2​ 16​ bytes + TCP header length
C
2​ 16​ bytes - TCP header length
D
2​ 15​ byte.
E
None of the above
       Computer-Networks       TCP/IP-Layers       UGC NET CS 2017 Jan -paper-2
Question 246 Explanation: 
Application Layer - Any size
Transport Layer - 65515 byte
Network layer - 65535 byte
Data link layer - 1500 byte
Note: Actually it works on any size, but given options are wrong. Excluded for evaluation.
Question 247
A packet whose destination is outside the local TCP/IP network segment is sent to _____.
A
File server
B
DNS server
C
DHCP server
D
Default gateway
       Computer-Networks       TCP/IP       UGC NET CS 2017 Jan -paper-2
Question 247 Explanation: 
→ A gateway is a network node that serves as an access point to another network, often involving not only a change of addressing, but also a different networking technology.
→ More narrowly defined, a router merely forwards packets between networks with different network prefixes. The networking software stack of each computer contains a routing table that specifies which interface is used for transmission and which router on the network is responsible for forwarding to a specific set of addresses.
→ If none of these forwarding rules is appropriate for a given destination address, the default gateway is chosen as the router of last resort.
→ The default gateway is specified by the configuration setting often called default route.
Question 248
Distance vector routing algorithm is a dynamic routing algorithm. The routing tables in distance vector routing algorithm are updated _____.
A
automatically
B
by server
C
by exchanging information with neighbour nodes
D
with backup database
       Computer-Networks       Routing       UGC NET CS 2017 Jan -paper-2
Question 248 Explanation: 
→ Distance vector routing algorithm is a dynamic routing algorithm. The routing tables in distance vector routing algorithm are updated by exchanging information with neighbour nodes.
→ Distance-vector protocols update the routing tables of routers and determine the route on which a packet will be sent by the next hop which is the exit interface of the router and the IP address of the interface of the receiving router.
→ Distance is a measure of the cost to reach a certain node. The least cost route between any two nodes is the route with minimum distance.
Question 249
In link state routing algorithm after construction of link state packets, new routes are computed using:
A
DES algorithm
B
Dijkstra’s algorithm
C
RSA algorithm
D
Packets
       Computer-Networks       Routing       UGC NET CS 2017 Jan -paper-2
Question 249 Explanation: 
→ Link State Routing(LSR) algorithm after construction of link state packets, new routes are computed using Dijkstra’s algorithm.
→ Each node independently runs an algorithm over the map to determine the shortest path from itself to every other node in the network; generally some variant of Dijkstra's algorithm is used.
→ This is based around a link cost across each path which includes available bandwidth among other things.
Question 250
In 3G network, W-CDMA is also known as UMTS. The minimum spectrum allocation required for W-CDMA is _______.
A
2 MHz
B
20 KHz
C
5 KHz
D
5 MHz
       Computer-Networks       3-G-Network       UGC NET CS 2017 Jan -paper-2
Question 250 Explanation: 
→ In 3G network, W-CDMA is also known as UMTS. The minimum spectrum allocation required for W-CDMA is 5 MHz.
→ IMT-DS uses a version of CDMA called wideband CDMA or W-CDMA. W-CDMA uses a 5.MHz bandwidth. It was in europe, and it is compatible with the CDMA used in IS-95.
→ IMT-TC uses a combination of W-CDMA and TDMA. The standard tries to reach the IMT-2000 goals by adding TDMA multiplexing to W-CDMA.
Question 251
Which of the following statements is not true with respect to microwaves?
A
Electromagnetic waves with frequencies from 300 GHz to 400 Thz.
B
Propagation is line-of-sight.
C
Very high-frequency waves cannot penetrate walls.
D
Use of certain portions of the band requires permission from authorities.
       Computer-Networks       Transmission-media       UGC NET CS 2016 Aug- paper-2
Question 251 Explanation: 
→ Microwaves are a form of electromagnetic radiation with wavelengths ranging from about one meter to one millimeter; with frequencies between 300 MHz (1 m) and 300 GHz (1 mm).
→ The electromagnetic spectrum covers electromagnetic waves with frequencies ranging from below one hertz to above 10​ 25​ hertz, corresponding to wavelengths from thousands of kilometers down to a fraction of the size of an atomic nucleus.
Question 252
In a fast Ethernet cabling, 100 Base-TX uses ____ cable and maximum segment size is_____.
A
twisted pair, 100 metres
B
twisted pair, 200 metres
C
fibre optics, 1000 metres
D
fibre optics, 2000 metres
       Computer-Networks       Ethernet       UGC NET CS 2016 Aug- paper-2
Question 252 Explanation: 
→ The "100" in the media type designation refers to the transmission speed of 100 Mbit/s, while the "BASE" refers to baseband signalling.
→ The letter following the dash ("T" or "F") refers to the physical medium that carries the signal (twisted pair or fiber, respectively), while the last character ("X", "4", etc.) refers to the line code method used.
→ Fast Ethernet is sometimes referred to as 100BASE-X, where "X" is a placeholder for the FX and TX variants.
Question 253
A network with bandwidth of 10 Mbps can pass only an average of 12,000 frames per minute with each frame carrying an average of 10,000 bits. What is the throughput of this network ?
A
1 Mbps
B
2 Mbps
C
10 Mbps
D
12 Mbps
       Computer-Networks       Bandwidth       UGC NET CS 2016 Aug- paper-2
Question 253 Explanation: 
Given data,
Bandwidth= 10 Mbps
Frames per minute=12000
Each frame per bits=10000
Throughput=?
Step-1: Here, they given each frame per minute. So, convert into seconds is
(12000*10000) / 60
Step-2: 120000000 / 60
= 2000000
It is nothing but 2Mbps
Question 254
Which of the following protocols is used by email server to maintain a central repository that can be accessed from any machine ?
A
POP3
B
IMAP
C
SMTP
D
DMSP
       Computer-Networks       Network-protocols       UGC NET CS 2016 Aug- paper-2
Question 254 Explanation: 
→ IMAP protocol is used by email server to maintain a central repository that can be accessed from any machine.
→ Internet Message Access Protocol (IMAP) is an Internet standard protocol used by email clients to retrieve email messages from a mail server over a TCP/IP connection.
→ IMAP was designed with the goal of permitting complete management of an email box by multiple email clients, therefore clients generally leave messages on the server until the user explicitly deletes them.
→ An IMAP server typically listens on port number 143. IMAP over SSL (IMAPS) is assigned the port number 993.
→ POP3 is designed to delete mail on the server as soon as the user has downloaded it. However, some implementations allow users or an administrator to specify that mail be saved for some period of time. POP can be thought of as a "store-and-forward" service.
→ SMTP is used for connecting to outbound servers to send email while POP3 and IMAP are used to connect to incoming servers to retrieve messages.
→ DMSP is Distributed Mail System for Personal Computers.
Question 255
An attacker sits between the sender and receiver and captures the information and retransmits to the receiver after some time without altering the information. This attack is called as _____.
A
Denial of service attack
B
Masquerade attack
C
Simple attack
D
Complex attack
       Computer-Networks       Network-Security       UGC NET CS 2016 Aug- paper-2
Question 255 Explanation: 
→ An attacker sits between the sender and receiver and captures the information and retransmits to the receiver after some time without altering the information. This attack is called as Denial of service attack(DoS).
→ A denial-of-service attack (DoS attack) is a cyber-attack in which the perpetrator seeks to make a machine or network resource unavailable to its intended users by temporarily or indefinitely disrupting services of a host connected to the Internet.
Question 256
A multiplexer combines four 100-Kbps channels using a time slot of 2 bits. What is the bit rate?
A
100 Kbps
B
200 Kbps
C
400 Kbps
D
1000 Kbps
       Computer-Networks       UGC NET CS 2016 July- paper-2
Question 256 Explanation: 
Frame duration= 1/50,000
= 20 sec
Frame size= No.of bits per frame
= 8 bits
Bit Rate= Frame Rate * No.of bits per frame
= 50000*8
= 400 kbps
Question 257
In a fully-connected mesh network with 10 computers, total ______ number of cables are required and ______ number of ports are required for each device.
A
40,9
B
45,10
C
45,9
D
50,10
       Computer-Networks       UGC NET CS 2016 July- paper-2
Question 257 Explanation: 
Fully connected mesh network, it means it is complete graph. We have standard formula to find total number of cables.
Here, n=10(total number of computers)
= n(n-1) / 2
= 10(9) / 2
= 45
→ To accommodate that many links, every device on the network must have n-1 input/output(I/O) ports
n=10(total number of computers)
= n-1 ports
= 10-1
= 9
Question 258
In TCP/IP Reference model, the job of _______ layer is to permit hosts to inject packets into any network and travel them independently to the destination.
A
Physical
B
Transport
C
Application
D
Host-to-network
E
None of the above
       Computer-Networks       OSI-TCP-layers       UGC NET CS 2016 July- paper-2
Question 259
If there are N people in the world and are using secret key encryption/decryption for privacy purpose, then number of secret keys required will be:
A
N
B
(N - 1)
C
N(N - 1) / 2
D
N(N + 1) / 2
       Computer-Networks       Encryption-Decryption       UGC NET CS 2016 July- paper-2
Question 259 Explanation: 
→ If there are N people in the world and are using secret key encryption/decryption for privacy purpose, then number of secret keys required will be N(N-1) / 2.
→ To communicate using symmetric cryptography, both parties have to agree on a secret key. After that, each message is encrypted with that key, transmitted, and decrypted with the same key. Key distribution must be secret. If it is compromised, messages can be decrypted and users can be impersonated. However, if a separate key is used for each pair of users, the total number of keys increases rapidly as the number of users increases. With n users, we would need [n(n-1)]/2 keys.
Question 260
Optical fiber uses reflection to guide light through a channel, in which angle of incidence is ________ the critical angle.
A
equal to
B
less than
C
greater than
D
less than or equal to
       Computer-Networks       Transmission-and-Propagation-Delay       UGC NET CS 2016 July- paper-2
Question 260 Explanation: 
→ Optical fiber uses reflection to guide light through a channel, in which angle of incidence is greater than the critical angle.
→ Total internal reflection is the phenomenon which occurs when a propagated wave strikes a medium boundary at an angle larger than a particular critical angle with respect to the normal to the surface.
→ If the refractive index is lower on the other side of the boundary and the incident angle is greater than the critical angle, the wave cannot pass through and is entirely reflected.
Question 261
An attacker sits between customer and Banker, and captures the information from the customer and retransmits to the banker by altering the information. This attack is called as ______.
A
Masquerade Attack
B
Replay Attack
C
Passive Attack
D
Denial of Service Attack
       Computer-Networks       Network-Security       UGC NET CS 2016 July- paper-2
Question 261 Explanation: 
→ A masquerade takes place when one entity pretends to be a different entity. A masquerade attack usually includes one of the other forms of active attack. For example,authentication sequences can be captured and replayed after a valid authentication sequence has taken place, thus enabling an authorized entity with few privileges to obtain extra privileges by impersonating an entity that has those privileges.
→ Replay involves the passive capture of a data unit and its subsequent retransmission to produce an unauthorized effect.
→ Passive attacks are very difficult to detect, because they do not involve any alteration of the data.Typically,the message traffic is sent and received in an apparently normal fashion,and neither the sender nor receiver is aware that a third party has read the messages or observed the traffic pattern.However,it is feasible to prevent the success of these attacks,usually by means of encryption.Thus,the emphasis in dealing with passive attacks is on prevention rather than detection.
→ The denial of service prevents or inhibits the normal use or management of communications facilities.This attack may have a specific target; for example, an entity may suppress all messages directed to a particular destination (e.g.,the security audit service).Another form of service denial is the disruption of an entire network,either by disabling the network or by overloading it with messages so as to degrade performance
Question 262
Which of the following fields in IPv4 datagram is not related to fragmentation?
A
Type of service
B
Fragment offset
C
Flags
D
Identification
       Computer-Networks       IPv4-and-Fragmentation       UGC NET CS 2015 Dec- paper-2
Question 262 Explanation: 
→ The Identification field along with the foreign and local internet address and the protocol ID.
→ Fragment offset field along with Don't Fragment and More Fragment flags in the IP protocol header are used for fragmentation and reassembly of IP packets.
→ The More Fragment(MF) flag is set for all the fragment packets except the last one.
Question 263
Which of the following is not a congestion policy at network layer?
A
Flow Control Policy
B
Packet Discard Policy
C
Packet Lifetime Management Policy
D
Routing Algorithm
       Computer-Networks       Which of the following is not a congestion policy at network layers       UGC NET CS 2015 Dec- paper-2
Question 263 Explanation: 
Network Layer congestion policies:
1.Preallocation of Resources
2.Traffic Shaping
3.Discarding Packets (No Preallocation)
4.Isarithmic Congestion Control
5.Virtual Circuits Admission Control
6.Choke Packets
7.Packet Discard
8. Packet Lifetime Management
9. Routing Algorithm
Transport layer congestion policies:
1. Flow Control, etc..,
Question 264
Which of the following protocols is an application layer protocol that establishes, manages and terminates multimedia sessions ?
A
Session Maintenance Protocol
B
Real time Streaming Protocol
C
Real time Transport Control Protocol
D
Session Initiation Protocol
       Computer-Networks       Application-Layer-Protocol       UGC NET CS 2015 Jun- paper-2
Question 264 Explanation: 
The Session Initiation Protocol (SIP) is a signaling protocol used for initiating, maintaining, and terminating real-time sessions that include voice, video and messaging applications.
→ SIP is used for signaling and controlling multimedia communication sessions in applications of Internet telephony for voice and video calls, in private IP telephone systems, in instant messaging over Internet Protocol (IP) networks as well as mobile phone calling over LTE (VoLTE).
Question 265
Match the following port numbers with their uses:
A
(a)-(iv), (b)-(i), (c)-(ii), (d)-(iii)
B
(a)-(ii), (b)-(i), (c)-(iv), (d)-(iii)
C
(a)-(ii), (b)-(iv), (c)-(iii), (d)-(i)
D
(a)-(ii), (b)-(iv), (c)-(i), (d)-(iii)
       Computer-Networks       Port-no.       UGC NET CS 2015 Jun- paper-2
Question 265 Explanation: 
Question 266
Which of the following is not associated with the session layer ?
A
Dialog control
B
Token management
C
Semantics of the information transmitted
D
Synchronization
       Computer-Networks       OSI-TCP-layers       UGC NET CS 2015 Jun- paper-2
Question 266 Explanation: 
Session Layer responsibilities:
1. Session checkpointing and recovery
2. Dialog control
3. Token management
4. Synchronization
Question 267
What is the size of the ‘total length’ field in IPv4 datagram?
A
4 bit
B
8 bit
C
16 bit
D
32 bit
       Computer-Networks       IPv4-and-Fragmentation       UGC NET CS 2015 Jun- paper-2
Question 267 Explanation: 
Question 268
Which of the following is/are restriction(s) in classless addressing?
A
The number of addresses needs to be a power of 2.
B
The mask needs to be included in the address to define the block.
C
The starting address must be divisible by the number of addresses in the block.
D
All of the above
       Computer-Networks       Subnetting       UGC NET CS 2015 Jun- paper-2
Question 268 Explanation: 
Classless addressing rules:
1. The number of addresses needs to be a power of 2.
2. The mask needs to be included in the address to define the block.
3. The starting address must be divisible by the number of addresses in the block.
Question 269
Error control is needed at the transport layer because of potential error occuring___.
A
from transmission line noise
B
in router
C
from out of sequence delivery
D
from packet losses
       Computer-Networks       Transport-layer-Protocol       UGC NET CS 2004 Dec-Paper-2
Question 269 Explanation: 
→ Error control is needed at the transport layer because of potential error occuring in router.
→ The error control in the transport layer usually refers to the guaranteed delivery mechanism with TCP, which attempts to safeguard against frames/packets getting lost entirely.
Question 270
Making sure that all the data packets of a message are delivered to the destination is _________ control.
A
Error
B
Loss
C
Sequence
D
Duplication
       Computer-Networks       Error-Control-Methods       UGC NET CS 2004 Dec-Paper-2
Question 270 Explanation: 
Making sure that all the data packets of a message are delivered to the destination is error control. Error Control:
Error Control is performed end to end in this layer to ensure that the complete message arrives at the receiving transport layer without any error. Error Correction is done through retransmission.
Question 271
Which transport class should be used with a perfect network layer ?
A
TP0 and TP2
B
TP1 and TP3
C
TP0, TP1, TP3
D
TP0, TP1, TP2, TP3, TP4
       Computer-Networks       Network-Layer       UGC NET CS 2004 Dec-Paper-2
Question 271 Explanation: 
Five transport layer protocols exist in the OSI suite, ranging from Transport Protocol Class 0 through Transport Protocol Class 4 (TP0, TP1, TP2, TP3 & TP4).
→ TP0 to TP3 work only with connection-oriented communications, in which a session connection must be established before any data is sent.
→ TP4 also works with both connection-oriented and connectionless communications. Transport Protocol Class 0 (TP0)​ performs segmentation (fragmentation) and reassembly functions. TP0 discerns the size of the smallest maximum protocol data unit (PDU) supported by any of the underlying networks, and segments the packets accordingly. The packet segments are reassembled at the receiver.
Transport Protocol Class 2 (TP2)​ performs segmentation and reassembly, as well as multiplexing and demultiplexing of data streams over a single virtual circuit.
Transport Protocol Class 4 (TP4)​ offers error recovery, performs segmentation and reassembly, and supplies multiplexing and demultiplexing of data streams over a single virtual circuit. TP4 sequences PDUs and retransmits them or reinitiates the connection if an excessive number are unacknowledged. TP4 provides reliable transport service and functions with either connection-oriented or connectionless network service. TP4 is the most commonly used of all the OSI transport protocols.
Question 272
Which transport class should be used with residual-error network layer ?
A
TP0, TP2
B
TP1, TP3
C
TP1, TP3, TP4
D
TP0, TP1, TP2, TP3, TP4
       Computer-Networks       Network-Layer       UGC NET CS 2004 Dec-Paper-2
Question 272 Explanation: 
Transport Protocol Class 1 (TP1)​ performs segmentation (fragmentation) and reassembly, plus error recovery. TP1 sequences protocol data units (PDUs) and will retransmit PDUs or re-initiate the connection if an excessive number of PDUs are unacknowledged.
Transport Protocol Class 3 (TP3)​ offers error recovery, segmentation and reassembly, and multiplexing and demultiplexing of data streams over a single virtual circuit. TP3 also sequences PDUs and retransmits them or re-initiates the connection if an excessive number are unacknowledged.
Question 273
Virtual circuit is associated with a __________ service.
A
Connectionless
B
Error-free
C
Segmentation
D
Connection-oriented
       Computer-Networks       Switching       UGC NET CS 2004 Dec-Paper-2
Question 273 Explanation: 
Virtual circuit communication resembles circuit switching, since both are connection oriented, meaning that in both cases data is delivered in correct order, and signalling overhead is required during a connection establishment phase.
Question 274
The data unit in the TCP/IP application layer is called a __________ .
A
message
B
segment
C
datagram
D
frame
       Computer-Networks       TCP/IP       UGC NET CS 2005 Dec-Paper-2
Question 274 Explanation: 
The PDU for Data Link layer, Network layer, Transport layer and Application layer are frame, datagram, segment and message respectively.
Question 275
Which of following is an example of a client - server model :
A
DNS
B
FTP
C
TELNET
D
All the above
       Computer-Networks       Application-Layer-Protocol       UGC NET CS 2005 Dec-Paper-2
Question 275 Explanation: 
→ The client-server characteristic describes the relationship of cooperating programs in an application.
→ Clients and servers exchange messages in a request–response messaging pattern. The client sends a request, and the server returns a response. This exchange of messages is an example of inter-process communication.
Example:
1. FTP
2. DNS
3. TELNET
Question 276
Encryption and decryption are the functions of the __________ layer of OSI model :
A
transport
B
session
C
router
D
presentation
       Computer-Networks       OSI-TCP-layers       UGC NET CS 2005 Dec-Paper-2
Question 276 Explanation: 
Presentation layer functionalities:
1. Encryption and decryption
2. Compression and decompression
3. Translation
Question 277
Which of the following are Data link layer standard ?
(1) Ethernet
(2) HSSI
(3) Frame Relay
(4) 10base T
(5) Token ring
A
1, 2
B
1, 3, 5
C
1, 3, 4, 5
D
1, 2, 3, 4, 5
       Computer-Networks       Data-Link-Layer       UGC NET CS 2005 june-paper-2
Question 277 Explanation: 
Except HSSI, all are belongs to data link layer.
The High-Speed Serial Interface (HSSI) is a differential ECL serial interface standard developed by Cisco Systems and T3plus Networking primarily for use in WAN router connections. It is capable of speeds up to 52 Mbit/s with cables up to 50 feet (15 m) in length.
Question 278
Which type of Bridge would be used to connect an Ethernet Segment with a token ring Segment ?
A
Transparent Bridge
B
Source-Route Bridge
C
Translation Bridge
D
None of these
       Computer-Networks       Ethernet       UGC NET CS 2005 june-paper-2
Question 278 Explanation: 
Transparent bridge:​ A transparent bridge is a bridge in which the stations are completely unaware of the bridge's existence.
Source-Route Bridge:​ To prevent loops in a system with redundant bridges is to use source routing bridges. A transparent bridge's duties include filtering frames, forwarding, and blocking.
In a system that has source routing bridges, these duties are performed by the source station and, to some extent, the destination station. Theoretically a bridge should be able to connect LANs using different protocols at the data link layer, such as an Ethernet
Segment with a token ring Segment but practically it can not connect two LANs with different protocols. One of the reasons of this is Maximum data size. If an incoming frame's size is too large for the destination LAN, the data must be fragmented into several frames. The data then need to be reassembled at the destination. However, no protocol at the data link layer allows the fragmentation and reassembly of frames. The other issues because of which a bridge is not able to connect LANs using different protocols are data rate, bit order, security, multimedia support.
So, the answer is Option(D)
Question 279
Which protocol is used to encapsulate a data pocket created of a higher OSI model layer ?
A
HDLC
B
SDLC
C
LAPB
D
LAPD
       Computer-Networks       OSI-TCP-layers       UGC NET CS 2005 june-paper-2
Question 279 Explanation: 
Link Access Procedures, D channel(LAPD)​ is part of the network's communications protocol which ensures that messages are error free and executed in the right sequence.
LAPD is the second layer protocol on the ISDN protocol stack in the D channel (the ISDN channel in which the control and signalling information is carried).
LAPD standards are specified in ITU-T Q.920 and Q.921. Its development was heavily based on High-Level Data Link Control.
Question 280
What is the correct subnet mask to use for a class-B address to support 30 Networks and also have the most hosts possible ?
A
255​ .​ 255​ .​ 255​ .​ 0
B
255​ .​ 255​ .​ 192​ .​ 0
C
255​ .​ 255​ .​ 240​ .​ 0
D
255​ .​ 255​ .​ 248​ .​ 0
       Computer-Networks       IP-Adress       UGC NET CS 2005 june-paper-2
Question 280 Explanation: 

→ In class-B addressing NID’s are having 16-bits and host IDs are having 16 bits.
Step-1: Given data, to support for 30 networks is required. So number of bits taken from host ID to uniquely identify each of these network is (log 30/ log 2) i.e. 5-bits.
Step-2: NID’s are required for network is 21 bits and host id is 11-bits.
Step-3: Subnet mask of a network have NID as all 1's and Host ID as all 0's.
Network subnet mask is : 11111111. 11111111. 11111000. 00000000
Decimal representation of above subnet mask is : 255.255.248.0
Question 281
Another name of IEEE 802.11a is :
A
WECA
B
Fast Ethernet
C
Wi-Fi 5
D
802​ .​ 11g
       Computer-Networks       IEEE-802.11       UGC NET CS 2005 june-paper-2
Question 281 Explanation: 
It operates in the 5 GHz band with a maximum net data rate of 54 Mbit/s, plus error correction code, which yields realistic net achievable throughput in the mid-20 Mbit/s.
802.11b typically has a higher range at low speeds (802.11b will reduce speed to 5.5 Mbit/s or even 1 Mbit/s at low signal strengths). 802.11a also suffers from interference.
Question 282
The network 198:78:41:0 is a :
A
Class A Network
B
Class B Network
C
Class C Network
D
Class D Network
       Computer-Networks       IP-Adress       UGC NET CS 2006 Dec-paper-2
Question 282 Explanation: 
Question 283
The subnet mask 255.255.255.192
A
Extends the network portion to 16 bits
B
Extends the network portion to 26 bits
C
Extends the network portion to 36 bits
D
Has no effect on the network portion of an IP address
       Computer-Networks       IP-Adress       UGC NET CS 2006 Dec-paper-2
Question 283 Explanation: 
Default subnet mask for Class-C is 255.255.255.192
(192)​ 10​ = (11000000)​ 2
Since, 192 is written as 11000000, it has 2 sub-nets and remaining all hosts.
So, for first three octets, 24 bits are fixed and for last octet 2 bits are fixed, i.e. 24 + 2 = 26 bits
Question 284
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       UGC NET CS 2006 June-Paper-2
Question 284 Explanation: 
IEEE 802.3 Frame Structure
Question 285
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       UGC NET CS 2006 June-Paper-2
Question 285 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 286
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       UGC NET CS 2006 June-Paper-2
Question 286 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 287
Which of the following algorithms is not a broadcast routing algorithm ?
A
Flooding
B
Multi Destination routing
C
Reverse path forwarding
D
All of the above
       Computer-Networks &