When you like to appear for UGC NET exam and you are working professional then you may come across a question "Should I leave my job for the preparation of UGC NET?"
Since UGC NET exam is conducted every 6 months there is no need to leave your job. You can give 2-3 hours time everyday and 3-5 hours in weekends. Keep weekends to solve previous year question papers.
It's good enough that you can take out this much time everyday for at least 6 months to prepare for UGC NET regressively.
Make sure to create a time table and follow it so that you would be able to complete the syllabus and revise it at well.
UGC NET Exam is not very tough that you have to leave your job for preparation. It is good enough to prepare for at least six months consistently. Consistency is the key to success.
You should follow below preparation tips to boost your exam preparation for the NTA UGC NET exam:
First prepare for subjects which are most interesting and which you find easy to score. So, first give priority to easy subject and then move to harder subjects one by one.
You can skip 1 or 2 subject which you find very much hard. Once you cover all possible subjects and you are confident that you are able to score at least 50 in each prepared subject, you can start with subject wise quizzes and check you preparation.
Then go for full length mock tests and previous year papers to check your preparation when you face questions mixed from all subjects. Prepare such that you are able to solve such mock test in given time. On an average you will get 12 minutes to solve 10 questions.
If you are scoring an average about 50 % then you are preparing well.
University Grants Commission – National Eligibility Test (UGC-NET)
UGC NET 2020 Exam Dates Postponed for below subjects
Assamese, Malayalam, Marathi, Punjabi, Russian, Telugu, Bengali, Bodo, Kashmiri, Social Medicine & Community Health, Urdu postponed to date November 04, 2020
Computer Science, Sociology postponed to date November 11, 2020
Education, Geography postponed to date November 12, 2020
Hindi postponed to date November 13, 2020
NTA UGC NET Exam June 2020 revised date sheet announced, Admit card displayed at Official Website @ugcnet.nta.nic.in
(UGC-NET) : June 2020 Date Sheet of Examination
The UGC-NET June 2020 will be held as per following Schedule. Admit Cards for the examination to be held on 24.09.2020 and 25.09.2020 have been displayed. For others same shall be displayed on the NTA website soon.
NTA UGC NET 2020 Revised Exam Dates
Date
Time
Subject Code
Subject Name
Remark
September 24, 2020
09.00AM to 12.00 Noon (IST)
46
Adult Education/ Continuing Education/ Andragogy/ Non Formal Education.
32
Chinese
62
Comparative Study of Religions
33
Dogri
44
German
37
Gujarati
50
Indian Culture
45
Japanese
21
Kannada
85
Konkani
18
Maithili
35
Manipuri
42
Persian
91
Prakrit
43
Rajasthani
25
Sanskrit
101
Sindhi
40
Spanish
70
Tribal and Regional Language/Literature
03.00PM to 06.00PM (IST)
49
Arab Culture and Islamic Studies
29
Arabic
60
Buddhist, Jaina, Gandhian and Peace Studies
11
Defence and Strategic Studies
31
Linguistics
63
Mass Communication and Journalism
34
Nepali
23
Oriya
83
Pali
26
Tamil
September 25, 2020
09.00AM to 12.00 Noon (IST)
72
Comparative Literature
68
Criminology
71
Folk Literature
82
Forensic Science
39
French (French Version)
66
Museology & Conservation
4
Psychology
93
Tourism Administration and Management.
03.00PM to 06.00PM (IST)
67
Archaeology
92
Human Rights and Duties
59
Library and Information Science
3
Philosophy
90
Politics including International Relations/International Studies including Defence/Strategic Studies, West Asian Studies, South East Asian Studies, African Studies,South Asian Studies, Soviet Studies, American Studies.
Management (including Business Admn. Mgt./Marketing/ Marketing Mgt./Industrial Relations and Personnel Mgt./Personnel Mgt./Financial Mgt./Co-operative Management)
September 30, 2020
03.00PM to 06.00PM (IST)
1
Economics / Rural Economics /Co-operation / Demography / Development Planning/ Development Studies / Econometrics/ Applied Economics/DevelopmentEco./Business Economics
October 1, 2020
09.00AM to 12.00 Noon (IST)
030
English
Group-1
03.00PM to 06.00PM (IST)
Group-2
October 9, 2020
09.00AM to 12.00 Noon (IST)
6
History
03.00PM to 06.00PM (IST)
2
Political Science
October 17, 2020
09.00AM to 12.00 Noon (IST)
008
Commerce
Group-1
03.00PM to 06.00PM (IST)
Group-2
November 04, 2020(Old Date:22.10.2020)
09.00AM to 12.00 Noon (IST)
36
Assamese
22
Malayalam
38
Marathi
24
Punjabi
41
Russian
27
Telugu
03.00PM to 06.00PM (IST)
19
Bengali
28
Urdu
94
Bodo
84
Kashmiri
81
Social Medicine & Community Health
November 5, 2020
09.00AM to 12.00 Noon (IST)
88
Electronic Science
55
Labour Welfare/Personnel Management/Industrial Relations/ Labour and Social Welfare/Human Resource Management
16
Music
15
Population Studies
79
Visual Art (including Drawing & Painting/Sculpture Graphics/Applied Art/History of Art)
03.00PM to 06.00PM (IST)
7
Anthropology
12
Home Science
14
Public Administration
10
Social Work
100
Yoga
November 11, 2020 (Old Date:07.10.2020)
09.00AM to 12.00 Noon (IST)
5
Sociology
03.00PM to 06.00PM (IST)
87
Computer Science and Applications
November 12, 2020 (Old Date : 21.10.2020)
09.00AM to 12.00 Noon (IST)
9
Education
03.00PM to 06.00PM (IST)
80
Geography
November 13, 2020 (Old Date:23.10.2020)
09.00AM to 12.00 Noon (IST)
020
Hindi
Group-1
03.00PM to 06.00PM (IST)
Group-2
NTA UGC NET - June 2020 Exam Dates Postponed (14th September, 2020 Public Notice) (Old Notice) :
This notice is regarding the postponed exam dates for NTA UGC NET - June 2020 Exam
National Testing Agency, NTA has released a notice informing that the UGC NET Exam 2020 has been postponed to be conducted now from September 24, 2020, onwards. The candidates who were to appear for the exam from September 16 must note that now the exams have been postponed by a week. The downloading of the admit card will soon be released on ugc.nta.nic.in.
Visit Official Website and check updates in section LATEST @ NTA with subject "Conduct of UGC- NET June Examination, 2020.":
Public Notice by NTA for UGC NET Examination (notice date: 14th September 2020)
National Testing Agency will be conducting ICAR Examination AIEEA-UG/PG and AICE-JRF/SRF (Ph.D.) 2020-21 on 16, 17, 22 and 23 September 2020. In view of ICAR Examination AIEEA-UG/PG and AICE-JRF/SRF (Ph.D.) 2020-21 being conducted on the above mentioned dates, UGC - NET 2020 Examination will now be held from 24th September onwards, this is due to some common candidates in both exams and the requests received thereof. The exact schedule of Subject-wise and Shift-wise details will be uploaded subsequently. The downloading of Admit Cards indicating Roll Number, Examination Centre, Date, Shift and timing of Examination will be announced shortly on the official website (ugcnet.nta.nic.in) of UGC- NET Examination, 2020.
Last Month updates
NTA Exam Dates 2020 Updates : Schedule for UGC NET Exam released at official website
Auguest Updates from NTA NET regarding the Exam dates :
Final date for UGC NET June exam announced on Thursday, 20 August, 2020.
Visit and check Exam date updates at UGC NET official website:
Public Notice by NTA for UGC NET Examination
UGC NET Exam date :
UGC - National Eligibility Test (UGC NET) June 2020 will be conducated in September on dates (16-18) Sep, 2020 and (21-25) Sep, 2020.
Downloading of Admit Cards:
The downloading of Admit Cards indicating Roll Number, Centre, Date, Shift and timing of
Examination, will commence about 15 days before the date of examination on the respective
official websites of these examinations
The number of different spanning trees in complete graph, K4 and bipartite graph, K2,2 have ______ and _______ respectively.
1. 14, 14
2. 16, 14
3. 16, 4
4. 14, 4
Explanation Question 1
For any complete graph Kn with n nodes, different spanning trees possible is n(n-2) So, spanning trees in complete graph K4 will be 4(4 - 2). i.e. 42 = 16.
For any Bipartite graph Km,n with m and n nodes, different spanning trees possible is m(n-1).n(m-1) So, spanning trees in K2,2 will be 2(2-1) * 2(2-1). i.e. 2 * 2 = 4.
So, option 3 is correct answer.
Question 2
A clique in a simple undirected graph is a complete subgraph that is not contained in any larger complete subgraph. How many cliques are there in the graph shown below?
1. 2
2. 4
3. 5
4. 6
Explanation Question 2
Total 5 clique will be there. So, option (C) is correct.
Question 3
Which of the following statement(s) is/are false ? (a) A connected multigraph has an Euler Circuit if and only if each of its vertices has even degree. (b) A connected multigraph has an Euler Path but not an Euler Circuit if and only if it has exactly two vertices of odd degree. (c) A complete graph (Kn) has a Hamilton Circuit whenever n ≥ 3. (d)A cycle over six vertices (C6) is not a bipartite graph but a complete graph over 3 vertices is bipartite. Codes:
1. (a) only
2. (b) and (c)
3. (c) only
4. (d) only
Explanation Question 3
A connected multigraph has an Euler Circuit if and only if each of its vertices has even degree.Correct A connected multigraph has an Euler Path but not an Euler Circuit if and only if it has exactly two vertices of odd degree.Correct A complete graph (Kn) has a Hamilton Circuit whenever n ≥ 3.CorrectA cycle over six vertices (C6) is not a bipartite graph but a complete graph over 3 vertices is bipartite.Incorrect So, option (D) is cocrrect.
Question 4
Consider the graph given below:
The two distinct sets of vertices, which make the graph bipartite are:
1. (v1, v4, v6); (v2, v3, v5, v7, v8)
2. (v1, v7, v8); (v2, v3, v5, v6)
3. (v1, v4, v6, v7); (v2, v3, v5, v8)
4. (v1, v4, v6, v7, v8); (v2, v3, v5)
Explanation Question 4
A Bipartite Graph is a graph whose vertices can be divided into two independent sets, U and V such that every edge (u, v) either connects a vertex from U to V or a vertex from V to U. In other words, for every edge (u, v), either u belongs to U and v to V, or u belongs to V and v to U. We can also say that there is no edge that connects vertices of same set. (v1, v4, v6, v7); (v2, v3, v5, v8) is a bipartite graph vertices set. So, option (C) is correct.
Question 5
A tree with n vertices is called graceful, if its vertices can be labelled with integers 1, 2,....n such that the absolute value of the difference of the labels of adjacent vertices are all different. Which of the following trees are graceful?
codes:
1. (a) and (b)
2. (b) and (c)
3. (a) and (c)
4. (a), (b) and (c)
Explanation Question 5
All given trees are graceful. So, option (D) is correct.
Question 6
In the following graph, discovery time stamps and finishing time stamps of Depth First Search (DFS) are shown as x/y, where x is discovery time stamp and y is finishing time stamp.
It shows which of the following depth first forest?
1. {a, b, e} {c, d, f, g, h}
2. {a, b, e} {c, d, h} {f, g}
3. {a, b, e} {f, g} {c, d} {h}
4. {a, b, c, d} {e, f, g} {h}
Explanation Question 6
Question 7
The inorder traversal of the following tree is:
1. 2 3 4 6 7 13 15 17 18 18 20
2. 20 18 18 17 15 13 7 6 4 3 2
3. 15 13 20 4 7 17 18 2 3 6 18
4. 2 4 3 13 7 6 15 17 20 18 18
Explanation Question 7
In inorder traversal first we traverse left node then root node and then right node:
In the following tree we first go to the leftmost node then its root after that right i.e. 2 4 3 13 7 6 15 17 20 18 18. In rest of the option inorder property is violating.
So, option (D) is correct.
Question 8
Consider the following statements: (a) Depth - first search is used to traverse a rooted tree. (b) Pre - order, Post-order and Inorder are used to list the vertices of an ordered rooted tree. (c) Huffman's algorithm is used to find an optimal binary tree with given weights. (d) Topological sorting provides a labelling such that the parents have larger labels than their children. Which of the above statements are true?
1. (a) and (b)
2. (c) and (d)
3. (a), (b) and (c)
4. (a), (b), (c) and (d)
Explanation Question 8
Depth - first search is used to traverse a rooted tree. Correct Pre - order, Post-order and Inorder are used to list the vertices of an ordered rooted tree. CorrectHuffman's algorithm is used to find an optimal binary tree with given weights. CorrectTopological sorting provides a labelling such that the parents have larger labels than their children.Correct So, option (D) is correct.
Question 9
Consider a Hamiltonian Graph (G) with no loops and parallel edges. Which of the following is true with respect to this Graph (G) ? (a) deg (v) ≥ n / 2 for each vertex of G (b) |E(G)| ≥ 1 / 2 (n - 1) (n - 2) + 2 edges (c) deg (v) + deg (w) ≥ n for every n and v not connected by an edge.
1. (a) and (b)
2. (b) and (c)
3. (a) and (c)
4. (a), (b) and (c)
Explanation Question 9
In an Hamiltonian Graph (G) with no loops and parallel edges: According to Dirac's theorem in a n vertex graph, deg (v) ≥ n / 2 for each vertex of G. According to Ore's theorem deg (v) + deg (w) ≥ n for every n and v not connected by an edge is sufficient condition for a graph to be hamiltonian. If |E(G)| ≥ 1 / 2 * [(n - 1) (n - 2)] then graph is connected but it doesn't guaranteed to be Hamiltonian Graph. (a) and (c) is correct regarding to Hamiltonian Graph. So, option (C) is correct.
Question 10
Consider the given graph
Its Minimum Cost Spanning Tree is __________.
(1) (2) (3) (4)
1. (1)
2. (2)
3. (3)
4. (4)
Explanation Question 10
A minimum spanning tree (MST) or minimum weight spanning tree for a weighted, connected and undirected graph is a spanning tree with weight less than or equal to the weight of every other spanning tree. In option (A) weight of tree is = 103 but it is not the subgraph from graph because in original graph there is no edge between node(6) and (7). In option (B) weight of tree is = 99 In option (C) weight of tree is = 127 In option (D) weight of tree is = 106 So, option (B) is correct.
Download Admit Card for UGC-NET JUNE 2020 (for examination on 09.10.2020, 17.10.2020)
Note : Please refer Revised Date sheet of Examination available on this website. Admit Cards for examination on 09.10.2020 and 17.10.2020 displayed. For others same shall be displayed shortly
Direct Link to Download NTA UGC NET June 2020 exam Hall Ticket Here:
UGC NET Exam date :
UGC - National Eligibility Test (UGC NET) June 2020 Exam is scheduled to be conducted in two batches - from September 16 to 18 and September 21 to 25.
Downloading of Admit Cards:
The downloading of Admit Cards indicating Roll Number, Centre, Date, Shift and timing of Examination, will commence about 15 days before the date of examination on the respective official websites of these examinations
1. An organization has a Class B Network and need to form subnet for 64 departments what would be the appropriate Subnet Mask
1. 255.255.0.0
2. 255.255.64.0
3. 255.255.128.0
4. 255.255.252.0
Explanation Question 1
In class B network, 16 bits are used for Network ID and 16 bits are for Host ID part. In subnetting the bits are borrowed from Host portion.
Here, we want 64 subnets. so, we have to borrow 6 bits from host id part because 26 = 64 number of subnets we require. Subnet mask will have total 16 + 6 bits as continuous 1's. in network id and subnet id part.
Binary form of subnet mask is : 11111111.11111111.11111100.00000000
Decimal form of subnet mask is : 255.255.252.0
So, option (4) is the correct answer.
Question 2
2. Consider a class A, Class B and Class C network. Consider the given IP address and subnet mask IP Address : 150.157.100.70 Subnet Mask : 255.255.224.0 What is the address of the network in which host belong to
1. 150.157.96.70
2. 150.157.96.0
3. 150.157.48.0
4. None
Explanation Question 2
If any given IP address performs bit wise AND operation with the subnet mask, then you get the network id of the subnet to which the given IP belongs.
If IP address = 150.157.100.70 (convert it into binary form)
Binary form of IP address is : 10010110.10011101.01100100.01000110
IP address
= 10010110.10011101.01100100.01000110
Subnet mask
= 11111111. 11111111. 11100000.00000000
Bit Wise AND
= 10010110.10011101.01100000.00000000
Binary form of network address is : 10010110.10011101.01100000.00000000
Deciaml form of network address is : 150.157.96.0
Therefore, Nid = 150.157.96.0
So, option (2) is the correct answer.
Question 3
3. Subnet mask for a particular network is 255.255.31.0 then tell about Which pairs of IP addresses could belong to this network ?
1. 172.57.88.62 and 172.56.87.23.2
2. 10.35.28.2 and 10.35.29.4
3. 191.203.31.87 and 191.234.31.88
4. 128.8.129.43 and 128.8.161.55
Explanation Question 3
If IP1 and IP2 belongs to same network, we must have
IP1 BITWISE-AND "Subnet mask" = Network Address = IP2 BITWISE-AND "Subnet mask"
Out of the 4 options, only option D matches.
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
Both addresses belongs to same network.
You can check rest of the options.
Question 4
4. If a class B IP address has the subnet mask 255.255.248.0 then How many maximum host will be possible in the network?
1. 1022
2. 1023
3. 2046
4. 2047
Explanation Question 4
Binary form of subnet mask of 255.255.248.0 is : 11111111.11111111.11111000.00000000
From above binary form of subnet mask. you can say that subnet mask of 255.255.248.0 will have a 21 bits network id and having 11 host bits.
Given the subnet mask, you can calculate the number of hosts using the formula below:
2h - 2, where h is the number of host bits (in simple terms the number of zeros in the subnet mask)
This network will have total 2048 (211 ) hosts.
The first address is the network address, while the last one is the broadcast ip address, thus we subtract 2 addresses,
then there can be a maximum 2046 hosts.
So, option (3) is the correct answer.
Question 5
5. If a Class B IP address is split into subnet and 6 bits are used for subnet. What is maximum number of subnet and hosts per subnet ?
1. 62 subnets and 262142 hosts
2. 64 subnets and 262142 hosts
3. 62 subnets and 1022 hosts
4. 64 subnets and 1024 hosts
Explanation Question 5
3
Question 6
6. Consider the IP address 25.193.155.233. How many bits are there in Network Id and Host Id ?
1. 24 bit for NID, 8 bits for HID
2. 8 bit for NID, 24 bits for HID
3. 16 bit for NID, 16 bits for HID
4. none
Explanation Question 6
2
Question 7
7. Find the wrong statement among the following related to Subnetting?
1. It can be applied only for single network
2. It is used to improve security
3. Here, bits are borrowed from network ID portion
4. Here, bits are borrowed from host ID portion
Explanation Question 7
3
Question 8
8. Logical addressing system is used by which device ?
1. Hub
2. Switch
3. bridge
4. Router
Explanation Question 8
4
Question 9
9. Which of the following is not true about Supernetting?
1. It is used to increase security
2. It is applicable for two or more networks
3. Bits are borrowed from network ID portion
4. It is used to improve flexibility of IP address allotment
Explanation Question 9
1
Question 10
10. What is the network ID of the IP address 225.100.123.71?
1. The protocol data unit(PDU) for the application layer in the Internet stack is
1. Segment
2. Datagram
3. Message
4. Frame
Explanation Question 1
For Application, Presentation and Session layers, the PDU is message
For Transport layer, PDU is segment for TCP and datagram for UDP
For Network layer, PDU is packet
For Datalink layer, PDU is frames
For physical layer, PDU is stream of bits
The Protocol Data Unit for Application layer in the Internet Stack (or TCP/IP) is called Message.
Option 3 is correct answer
Question 2
2. Which of the following transport layer protocols is used to support electronic mail?
1. SMTP
2. IP
3. TCP
4. UDP
Explanation Question 2
There are three primary TCP/IP protocols for E-Mail management:
Post Office Protocol (POP)
Simple Mail Transfer Protocol (SMTP)
Internet Message Access Protocol (IMAP)
They all 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.).
E-mail uses SMTP as application layer protocol.
TCP and UDP are two transport layer protocols. SMTP uses TCP as transport layer protocol as TCP is reliable.
Option 3 is correct answer
Question 3
3. In the IPv4 addressing format, the number of networks allowed under Class C addresses is
1. 214
2. 27
3. 221
4. 224
Explanation Question 3
We have total 32 bits in the IPV4 network
Class A = 8 network bits + 24 Host bits
Class B = 16 network bits + 16 Host bits
Class C = 24 network bits + 8 host bits
Class D (multicast)
For class C address, size of network field is 24 bits. But first 3 bits reserved for the network id. So, first 3 bits are fixed as 110; hence total number of networks possible is 221.
4. A layer-4 firewall ( a device that can look at all protocol headers up to the transport layer) CANNOT
1. block HTTP traffic during 9:00PM and 5:00AM
2. block all ICMP traffic
3. stop incoming traffic from a specific IP address but allow outgoing traffic to same IP
4. block TCP traffic from a specific user on a specific IP address on multi-user system during 9:00PM and 5:00AM
Answer 1
Option 1 is correct answer
Question 5
5. Packets of the same session may be routed through different paths in:
1. TCP, but not UDP
2. TCP and UDP
3. UDP, but not TCP
4. Neither TCP nor UDP
Explanation Question 5
Routing happens in Network layer and hence has no dependency with the the transport layer protocols TCP and UDP. The transport layer protocol- whether TCP or UDP is hidden to the router and the routing path is determined based on the the network configuration at the time and hence can change even during a session. Packets of same session may be routed through different routes. Most networks don’t use static routing, but use some form of adaptive routing where the paths used to route two packets for same session may be different due to congestion on some link, or some other reason.
6. The address resolution protocol (ARP) is used for:
1. Finding the IP address from the DNS
2. Finding the IP address of the default gateway
3. Finding the IP address that corresponds to a MAC address
4. Finding the MAC address that corresponds to an IP address
Explanation Question 6
Address Resolution Protocol (ARP) is a request and reply protocol used to find MAC address from IP address.
When a packet is send to the data link layer from network layer IP address & MAC address of of the sender and the gateway of the network is attached. The MAC address of the gateway is not known to the sender. So, ARP (Address Resolution Protocol) request is generated with the IP address of the gateway and is broadcasted, gateway sends reply with it’s MAC address and every other PC/Router except the gateway discards this ARP request. So, ARP Protocol is used in finding the MAC address that corresponds to an IP address.
7. Which one of the following uses UDP as the transport protocol?
1. HTTP
2. Telnet
3. DNS
4. SMTP
Explanation Question 7
Find a list of popular Internet applications and their underlying transport and application layer Protocols.
Application
Application Layer Protocol
Underlying Transport layer protocol
Electronic Mail
SMTP
TCP
Remote Terminal access
Telnet
TCP
Web
HTTP
TCP
Name Translation
DNS
Typically UDP
File Transfer
FTP
TCP
Network Management
SNMP
Typically UDP
Routing Protocol
RIP
Typically UDP
UDP is a stateless, connection-less and unreliable protocol.
HTTP needs connection to be established and thus, uses TCP.
Telnet is a byte stream protocol which again needs connection establishment ,thus uses TCP.
DNS needs request and response ,it needs a protocol in which a server can answer the small queries of large number of users. As UDP is fast and stateless it is the most suitable protocol and thus,it is used in DNS querying. DNS primarily uses User Datagram Protocol (UDP) on port number 53 to serve requests. DNS queries consist of a single UDP request from the client followed by a single UDP reply from the server.
SMTP needs reliability and thus,uses TCP.
DNS uses UDP protocol whereas HTTP, Telnet and SMTP uses TCP.
So, option 3 is correct answer
Question 8
8. Match the following:
(P) SMTP
(1) Application layer
(Q) BGP
(2) Transport layer
(R) TCP
(3) Data link layer
(S) PPP
(4) Network layer
(5) Physical layer
1. (A) P - 2 Q - 1 R - 3 S - 5
2. (B) P - 1 Q - 4 R - 2 S - 3
3. (C) P - 1 Q - 4 R - 2 S - 5
4. (D) P - 2 Q - 4 R - 1 S - 3
Explanation Question 8
SMTP is an application layer protocol used for e-mail transmission. SMTP servers commonly use the Transmission Control Protocol on port number 25.
TCP is a core transport layer protocol. Major internet applications such as the World Wide Web, email, remote administration, and file transfer rely on TCP, which is part of the Transport Layer of the TCP/IP suite.
BGP is a network layer protocol backing the core routing decisions on the Internet. Border Gateway Protocol (BGP) is a standardized exterior gateway protocol designed to exchange routing and reachability information among autonomous systems (AS) on the Internet.
PPP is a data link layer protocol commonly used in establishing a direct connection between two networking nodes. PPP is commonly used as a data link layer protocol for connection over synchronous and asynchronous circuits,
So, option 2 is correct answer
Question 9
9. Find the false statement about HTTP
1. HTTP run over TCP
2. HTTP allows information to be stored in url
3. HTTP describe the structure of web page
4. HTTP can be used to test the validity of hyper link test
Explanation Question 9
HTTP is the set of rules for transferring files (text, graphic images, sound, video, and other multimedia files) on the World Wide Web. HTTP is a transfer protocol that is used to transfer hypertext requests and information between servers and browsers.
HTTP Protocol does not describe the structure of a web page. HTML describes the structure of a web-page.
Question 10
10. Here some activities related to email is given FInd the correct match for the protocol used in appropriate activity A1: send an email from a mail client to mail server A2:download an email in mailbox from mail server to mail client A3: checking mail in browser
1. A1 : HTTP, A2 : SMTP, A3 : POP
2. A1 : SMTP, A2 : FTP , A3 : HTTP
3. A1 : SMTP , A2 : POP, A3 : HTTP
4. A1 : POP, A2 : SMTP, IA3 : ICMP
Explanation Question 10
Client mail box to server → HTTP and SMTP( push protocol)
Download mail box from server → POP and IMAP (Pull protocol)
A flow graph F with entry node (1) and exit node (11) is shown below:
How many predicate nodes are there and what are their names?
1. Three : (1, (2, 3), 6)
2. Three: (1, 4, 6)
3. Four : ((2, 3), 6, 10, 11)
4. Four: ((2, 3), 6, 9, 10)
Answer : 1. Three : (1, (2, 3), 6)
Explanation Question 145
Predicate is a node that contains condition. It means at least 2 outgoing edges required to qualify as a predicate.
Total number of predicates are THREE.
The vertex 1 is contains 2 outgoing edges are (2,3) and 11
The vertex (2,3) contains 2 outgoing edges are 6 and (4,5)
The vertex 6 contains 2 outgoing edges are 7 and 8.
Which of the following are legal statements in C programming language? (a) int * P= &44 ; (b) int * P= &r ; (c) int P= &a ; (d) int P= a ; Choose the correct option :
1. (a) and (b)
2. (b) and (c)
3. (b) and (d)
4. (a) and (d)
Answer : 3. (b) and (d)
Explanation Question 122
Legal Statements:
int *P = &r; Here, Pointer variable P is storing address of variable r.
int P = a; Here, P will hold value assigned by a.
Illegal Statements:
int *P = &44; Here, Pointer variable is storing address of number. This is not legal statement. First its needed to assign the value 44 in some variable and then address of that variable can stored by Pointer variable P.
int P = &a; Here, P variable trying to store the address of the variable. It's illegal. It should be pointer in place of the int P variable.
In class B network, 16 bits are used for Network ID and 16 bits are for Host ID part. In subnetting the bits are borrowed from Host portion.
Here, we want 64 subnets. so, we have to borrow 6 bits from host id part because 26 = 64 number of subnets we require. Subnet mask will have total 16 + 6 bits as continuous 1's. in network id and subnet id part.
Binary form of subnet mask is : 11111111.11111111.11111100.00000000
Decimal form of subnet mask is : 255.255.252.0
So, option (4) is the correct answer.