UGC NET Exam Date 2020 Released. Check revised exam dates at Official Website ugcnet.nta.nic.in

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.
September 29, 2020	09.00AM to 12.00 Noon (IST)	58	Law
		47	Physical Education
	03.00PM to 06.00PM (IST)	89	Environmental Sciences
		65	Performing Art - Dance/Drama/Theatre
		73	Sanskrit traditional subjects (including) Jyotisha/Sidhanta Jyotish/ Navya Vyakarna/ Vyakarna/ Mimansa/ Navya Nyaya/ Sankhya Yoga/ Tulanatmaka Darsan/ Shukla Yajurveda/ Madhav Vedant/ Dharmasasta/ Sahitya/Puranotihasa /
		95	Santali
		74	Women Studies
September 30, 2020	09.00AM to 12.00 Noon (IST)	17	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

NTA UGC NET Computer Science | Graph Theory Questions Set 1 | Discrete Structures and Optimization

Question 1

---

The number of different spanning trees in complete graph, K4 and bipartite graph, K2,2 have ______ and _______ respectively.

1. 1. 14, 14
2. 2. 16, 14
3. 3. 16, 4
4. 4. 14, 4

Explanation Question 1

For any complete graph K_n 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 K_m,n with m and n nodes, different spanning trees possible is
m^(n-1).n^(m-1)
So, spanning trees in K_2,2 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. 1. 2
2. 2. 4
3. 3. 5
4. 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. 1. (a) only
2. 2. (b) and (c)
3. 3. (c) only
4. 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. 1. (v1, v4, v6); (v2, v3, v5, v7, v8)
2. 2. (v1, v7, v8); (v2, v3, v5, v6)
3. 3. (v1, v4, v6, v7); (v2, v3, v5, v8)
4. 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. 1. (a) and (b)
2. 2. (b) and (c)
3. 3. (a) and (c)
4. 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. 1. {a, b, e} {c, d, f, g, h}
2. 2. {a, b, e} {c, d, h} {f, g}
3. 3. {a, b, e} {f, g} {c, d} {h}
4. 4. {a, b, c, d} {e, f, g} {h}

Explanation Question 6

Question 7

---

The inorder traversal of the following tree is:

1. 1. 2 3 4 6 7 13 15 17 18 18 20
2. 2. 20 18 18 17 15 13 7 6 4 3 2
3. 3. 15 13 20 4 7 17 18 2 3 6 18
4. 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. 1. (a) and (b)
2. 2. (c) and (d)
3. 3. (a), (b) and (c)
4. 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. 1. (a) and (b)
2. 2. (b) and (c)
3. 3. (a) and (c)
4. 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. (1)
2. 2. (2)
3. 3. (3)
4. 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.

UGC NET Admit Card June 2020 to be Released | How to download NTA UGC NET Hall-Ticket at ugcnet.nta.nic.in

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

UGC NET Computer Science Questions Set 2 | computer networks | IP addressing

Question 1

---

1. An organization has a Class B Network and need to form subnet for 64 departments
what would be the appropriate Subnet Mask

1. 1. 255.255.0.0
2. 2. 255.255.64.0
3. 3. 255.255.128.0
4. 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. 1. 150.157.96.70
2. 2. 150.157.96.0
3. 3. 150.157.48.0
4. 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

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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. 1. 172.57.88.62 and 172.56.87.23.2
2. 2. 10.35.28.2 and 10.35.29.4
3. 3. 191.203.31.87 and 191.234.31.88
4. 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. 1. 1022
2. 2. 1023
3. 3. 2046
4. 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:
2^h - 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

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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. 1. 62 subnets and 262142 hosts
2. 2. 64 subnets and 262142 hosts
3. 3. 62 subnets and 1022 hosts
4. 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. 1. 24 bit for NID, 8 bits for HID
2. 2. 8 bit for NID, 24 bits for HID
3. 3. 16 bit for NID, 16 bits for HID
4. 4. none

Explanation Question 6

2

Question 7

---

7. Find the wrong statement among the following related to Subnetting?

1. 1. It can be applied only for single network
2. 2. It is used to improve security
3. 3. Here, bits are borrowed from network ID portion
4. 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. 1. Hub
2. 2. Switch
3. 3. bridge
4. 4. Router

Explanation Question 8

4

Question 9

---

9. Which of the following is not true about Supernetting?

1. 1. It is used to increase security
2. 2. It is applicable for two or more networks
3. 3. Bits are borrowed from network ID portion
4. 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. 1. 225.100.123.70
2. 2. 225.100.123.72
3. 3. 225.100.123.75
4. 4. It does not has any network ID

Explanation Question 10

4

About Us

UGC NET computer networks questions set 1 | Internet Protocols

Question 1

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1. The protocol data unit(PDU) for the application layer in the Internet stack is

1. 1. Segment
2. 2. Datagram
3. 3. Message
4. 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. 1. SMTP
2. 2. IP
3. 3. TCP
4. 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. 1. 2¹⁴
2. 2. 2⁷
3. 3. 2²¹
4. 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.

Option 3 is correct answer

Reference : IPv4 addressing - Classful addressing

Question 4

---

4. A layer-4 firewall ( a device that can look at all protocol headers up to the transport layer) CANNOT

1. 1. block HTTP traffic during 9:00PM and 5:00AM
2. 2. block all ICMP traffic
3. 3. stop incoming traffic from a specific IP address but allow outgoing traffic to same IP
4. 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. 1. TCP, but not UDP
2. 2. TCP and UDP
3. 3. UDP, but not TCP
4. 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.

So, option 2 is correct answer

Reference 1 : Connection-Oriented and Connectionless Protocols

Reference 2 : If TCP is connection oriented why do packets follow different paths?

Question 6

---

6. The address resolution protocol (ARP) is used for:

1. 1. Finding the IP address from the DNS
2. 2. Finding the IP address of the default gateway
3. 3. Finding the IP address that corresponds to a MAC address
4. 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.

So, option 4 is correct answer

Reference : ARP (Address Resolution Protocol) explained

Question 7

---

7. Which one of the following uses UDP as the transport protocol?

1. 1. HTTP
2. 2. Telnet
3. 3. DNS
4. 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. 1. (A) P - 2  Q - 1  R - 3  S - 5
2. 2. (B) P - 1  Q - 4  R - 2  S - 3
3. 3. (C) P - 1  Q - 4  R - 2  S - 5
4. 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

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9. Find the false statement about HTTP

1. 1. HTTP run over TCP
2. 2. HTTP allows information to be stored in url
3. 3. HTTP describe the structure of web page
4. 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

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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. 1. A1 : HTTP, A2 : SMTP, A3 : POP
2. 2. A1 : SMTP, A2 : FTP , A3 : HTTP
3. 3. A1 : SMTP , A2 : POP, A3 : HTTP
4. 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)
• Check mail in Browser → HTTP

So, option 3 is correct answer

UGC NET Computer Science December 2019 | Question 1

Question 1

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Which of the learning outcomes are intended in teaching organized at understanding level?

1. 1. Longer recall and retention of facts
2. 2. Seeking of relationships and patterns among facts
3. 3. Creative construction and critical interpretation of ideas
4. 4. Mastery of facts and information

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UGC NET CS December 2019 - Question 1	UGC NET CS Dec 2019 - Question 2

UGC NET Computer Science December 2019 | Question 145

Question 145

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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. 1. Three : (1, (2, 3), 6)
2. 2. Three: (1, 4, 6)
3. 3. Four : ((2, 3), 6, 10, 11)
4. 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.

1. The vertex 1 is contains 2 outgoing edges are (2,3) and 11
2. The vertex (2,3) contains 2 outgoing edges are 6 and (4,5)
3. The vertex 6 contains 2 outgoing edges are 7 and 8.

So, option 1 is correct answer

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UGC NET CS December 2019 - Question 144	UGC NET CS December 2019 - Question 146

UGC NET Computer Science December 2019 | Question 122

Question 122

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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. 1. (a) and (b)
2. 2. (b) and (c)
3. 3. (b) and (d)
4. 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.

So, option 3 is correct answer

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UGC NET CS December 2019 - Question 121	UGC NET CS December 2019 - Question 123

UGC NET Computer Science December 2019 | Question 142

Question 142

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A flow graph F with entry node (1) and exit node (11) is shown below:

What is the cyclomatic complexity of flowgraph F?

1. 1. 2
2. 2. 3
3. 3. 4
4. 4. 3

Answer : 3. 4

Explanation Question 142

To find cyclomatic complexity we have 3 formulas

1. The number of regions(R) corresponds to the cyclomatic complexity. Total number of regions(R) are 4.
2. V(G), Flow graph is defined as V(G) = P + 1 where p is the number of predicate nodes contained in the flow graph G. Predicates are 3 + 1 = 4
3. V(G), Flow graph is defined as V(G) = E - N + 2 where E is the number of flow graph edges, and N is the number of flow graph nodes.
   Edges(E) - Nodes(N) + 2
   = 11 - 9 + 2
   = 2 + 2
   = 4

So, option 3 is correct answer

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UGC NET CS December 2019 - Question 141	UGC NET CS December 2019 - Question 143

UGC NET Computer Science December 2019 | Question 141

Question 141

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A flow graph F with entry node (1) and exit node (11) is shown below:

Add caption

How many nodes are there in flowgraph F?

1. 1. 9
2. 2. 10
3. 3. 11
4. 4. 12

Answer : 1. 9

Explanation Question 141

Above graph contains 9 nodes and 11 edges.
The nodes are nothing but vertices.
Here, In above given diagram the number of rounds are the nodes
The nodes are 1, (2,3), (4,5), 6, 7, 8, 9, 10, 11.
Count of total nodes in flowgraph F are 9.

So, option 1 is correct answer

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UGC NET CS December 2019 - Question 140	UGC NET CS December 2019 - Question 142

UGC NET Computer Science December 2019 | Question 133

Question 133

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According to the ISO-9126 Standard Quality Model. match the attributes given in List-I with their definitions in List-II :

List I	List II
a. Functionality	i. Relationship between level of performance and amount of resources
b. Reliability	ii. Characteristics related with achievement of purpose
c. Efficiency	iii. Effort needed to make for improvement
d. Maintainability	iv. Capability of software to maintain

Choose the correct option from the ones given below :

1. 1. (a)-(i), (b)-(ii), (c)-(iii), (d)-(iv)
2. 2. (a)-(ii), (b)-(i), (c)-(iv), (d)-(iii)
3. 3. (a)-(ii), (b)-(iv), (c)-(i), (d)-(iii)
4. 4. (a)-(i), (b)-(ii), (c)-(iv), (d)-(iii)

Answer : 3. (a)-(ii), (b)-(iv), (c)-(i), (d)-(iii)

Explanation Question 133

6 main quality characteristics for ISO-9126 Standard Quality Model:

1. Functionality→ Characteristics related with achievement of purpose
2. Reliability → Capability of software to maintain performance of software
3. Efficiency → Relationship between level of performance and amount of resources
4. Maintainability → Effort needed to make for improvement
5. Usability → Usability only exists with regard to functionality and refers to the ease of use for a given function.
6. Portability → This characteristic refers to how well the software can adapt to changes in its environment or with its requirements.

Reference : ISO/IEC 9126

So, option 3 is correct answer

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UGC NET CS December 2019 - Question 132	UGC NET CS December 2019 - Question 134

UGC NET Computer Science December 2019 | Question 143

Question 143

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A flow graph F with entry node (1) and exit node (11) is shown below:

How many regions are there in flowgraph F?

1. 1. 2
2. 2. 3
3. 3. 4
4. 4. 8

Answer :

Explanation Question 143

The region is nothing but a combination of closed region and outer region. Any graph must have one outer region.
Here, 3 closed regions are available and one outer region is available.

• Closed 3 regions are:

1. Closed region by nodes → 1, (2, 3), (4, 5) and 10
2. Closed region by nodes → 6, 7, 8, 9
3. Closed region by nodes → (2, 3), 6, 7, 8, 9, 10 and (4, 5)

• One outer region.

So, the total number of regions are 4.

So, option 3 is correct answer

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UGC NET CS December 2019 - Question 142	UGC NET CS December 2019 - Question 144

UGC NET Computer Science December 2019 | Question 123

Question 123

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Two concurrent executing transactions T1 and T2 are allowed to update same stock item say 'A' in an uncontrolled manner. In such scenario. following problems may occur :
(a) Dirty read problem
(b) Lost update problem
(c) Transaction failure
(d) Inconsistent database state
Which of the following option is correct if database system has no concurrency module and allows concurrent execution of above two transactions?

1. 1. (a), (b) and (c) only
2. 2. (c) and (d) only
3. 3. (a) and (b) only
4. 4. (a), (b) and (d) only

Answer : 4. (a), (b) and (d) only

Explanation Question 123

Following problems can occur when concurrent transactions execute in an uncontrolled manner:

1. The Lost Update Problem. When two transactions access the same database items and also perform their own operations in a way that makes the value of some database item wrong, a lost update problem takes place.
2. Uncommitted Dependency - Dirty Read Problem(or Temporary Update Problem). A dirty read problem occurs when a transaction updates an item in the database and then there is a failure in the transaction for some reason. As the transaction fails, the updated item in the database is accessed by another transaction before it can go back to the original value.
3. The Incorrect Summary Problem. If one transaction is calculating an aggregate summary function on a number of database items while other transactions are updating some of these items, the aggregate function may calculate some values before they are updated and others after they are updated.
4. The Unrepeatable Read Problem. Another problem that may occur is called unrepeatable read, where a transaction T reads the same item twice and the item is changed by another transaction T between the two reads. Hence, T receives different values for its two reads of the same item.
5. Phantom Read Problem

The Unrepeatable Read Problem Example
Suppose two users X and Y access a department’s database concurrently. User X is updates the database to provide all employees with a salary raise of 5%. At the same time, user Y wants to know the total salary that has been paid to the department.
As the two transactions are being executed simultaneously on the same database, there is always a high probability to interfere with each other. As a result, the sum includes certain salaries before the raise and certain salaries subsequent to the raise. Such a sum cannot be considered as an acceptable value of the total salary (the value before the raise or after the raise would be different)

Reference 1 : Concurrency problems in DBMS Transactions

Reference 1 : Concurrency Anomalies - Incorrect Summary Problem

From above we can conclude that only options (a), (b) and (d) can be included in problems whoch can occur when concurrent transactions execute in an uncontrolled manner.

So, option 4 is correct answer

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UGC NET CS December 2019 - Question 122	UGC NET CS December 2019 - Question 124

UGC NET Computer Science December 2019 | Question 124

Question 124

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Identify the circumstances under which preemptive CPU scheduling is used :
(a) A process switches from Running state to Ready state
(b) A process switches from Waiting state to Ready state
(c) A process completes its execution
(d) A process switches from Ready to Waiting state
Choose the correct option :

1. 1. (a) and (b) only
2. 2. (a) and (d) only
3. 3. (c) and (d) only
4. 4. (a), (b), (c) only

Answer : 1. (a) and (b) only

Explanation Question 124

Preemptive Scheduling According to Standard book of Galvin,
CPU scheduling decisions take place under one of four conditions:

1. When a process switches from the running state to the waiting state, such as for an I/O request or invocation of the wait( ) system call.
2. When a process switches from the running state to the ready state, for example in response to an interrupt.
3. When a process switches from the waiting state to the ready state, say at completion of I/O or a return from wait( ).
4. When a process terminates.

• For conditions 1 and 4 there is no choice - A new process must be selected.
• For conditions 2 and 3 there is a choice - To either continue running the current process, or select a different one.
• If scheduling takes place only under conditions 1 and 4, the system is said to be non-preemptive, or cooperative. Under these conditions, once a process starts running it keeps running, until it either voluntarily blocks or until it finishes. Otherwise the system is said to be preemptive.

In above discussion. it's given that If scheduling takes place only under conditions 1 and 4, the system is said to be non-preemptive. Otherwise the system is said to be preemptive. Therefore point 2 & 3 are condition for preemptive system.

Here point 2 and 3 are as listed below

• When a process switches from the running state to the ready state.
• When a process switches from the waiting state to the ready state.

These above point matches with circumstances (a) & (b) given in question.

So, option 1 is correct answer

Reference 1 : CPU Scheduling - Preemptive Scheduling

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UGC NET CS December 2019 - Question 123	UGC NET CS December 2019 - Question 125

UGC NET Computer Science December 2019 | Question 125

Question 125

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The following multithreaded algorithm computes transpose of a matrix in parallel: p Trans (X, Y, N)
if N=1
then Y [1,1] ← X[1,1]
else partition X into four (N/2)x (N/2) submatrices X11,X12,X21,X22
partition Y into four (N/2)x(N/2) submatrices Y11,Y12,Y21,Y22.
spawn p Trans (X11,Y11,N/2)
spawn p Trans (X12,Y12,N/2)
spawn p Trans (X21,Y21,N/2)
spawn p Trans (X22,Y22,N/2)
What is the asymptotic parallelism of the algorithm?

1. 1. T₁/T_∞ or θ(N² / lg N)
2. 2. T₁/T_∞ or θ(N / lg N)
3. 3. T₁/T_∞ or θ(lg N / N²)
4. 4. T₁/T_∞ orθ(lg N/ N)

Answer : 1. T₁/T_∞ or θ(N² / lg N)

Explanation Question 125

Option 1 is correct answer

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UGC NET CS December 2019 - Question 124	UGC NET CS December 2019 - Question 126

UGC NET Computer Science December 2019 | Question 126

Question 126

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Consider the following statements :
(a) The running time of dynamic programming algorithm is always θ(ρ) where ρ is number of subproblems.
(b) When a recurrence relation has cyclic dependency, it is impossible to use that recurrence relation (unmodified) in a correct dynamic program.
(c) For a dynamic programming algorithm, computing all values in a bottom-up fashion is asymptotically faster than using recursion and memorization.
(d) If a problem X can be reduced to a known NP-hard problem. then X must be NP-hard
Which of the statement(s) is (are) true?

1. 1. Only (b) and (a)
2. 2. Only (b)
3. 3. Only (b) and (c)
4. 4. Only (b) and (d)

Answer : 2. Only (b)

Explanation Question 126

1. The running time of a dynamic program is the number of subproblems times the time per subproblem. This would only be true if the time per subproblem is O(1). False
2. Cyclic dependency is a relation between two or more modules which either directly or indirectly depend on each other to function properly. Such modules are also known as mutually recursive. So, we can’t get correct solution when we are using dynamic programming. True
3. A bottom up implementation must go through all of the sub-problems and spend the time per subproblem for each. Using recursion and memoization only spends time on the subproblems that it needs. In fact, the reverse may be true: using recursion and memoization may be asymptotically faster then a bottom-up implementation. False
4. If a problem X can be reduced to a known NP-hard problem, then X must be NP-Complete. False

Hence, statement b is the only true.

So, option 2 is correct answer

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UGC NET CS December 2019 - Question 125	UGC NET CS December 2019 - Question 127

UGC NET Computer Science December 2019 | Question 127

Question 127

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Consider the following statements :
(a) Fiber optic cable is much lighter than copper cable.
(b) Fiber optic cable is not affected by power surges or electromagnetic interference.
(c) Optical transmission is inherently bidirectional.
Which of the statements is (are) correct?

1. 1. Only (a) and (b)
2. 2. Only (a) and (c)
3. 3. Only (b) and (c)
4. 4. (a), (b) and (c)

Answer : 1. Only (a) and (b)

Explanation Question 127

1. Fiber optic cable is much lighter than copper cable. True
2. Fiber optic cable is not affected by power surges or electromagnetic interference. True
3. Optical transmission is inherently bidirectional. False

Given statement (c) is false because bidirectional Fiber Optics can be implemented with by WDM (Wavelength-division multiplexing) or by rearranging two unidirectional Fibers, its not inherently bidirectional.

Only given options (a) and (b) are correct statements on Fibre optics.

So, option 1 is correct answer

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UGC NET CS December 2019 - Question 126	UGC NET CS December 2019 - Question 128

UGC NET Computer Science December 2019 | Question 128

Question 128

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Consider the following statements :
(a) Windows Azure is a cloud-based operating system.
(b) Google App Engine is an integrated set of online services for consumers to communicate and share with others.
(c) Amazon Cloud Front is a web service for content delivery.
Which of the statements is (are) correct?

1. 1. Only (a) and (b)
2. 2. Only (a) and (c)
3. 3. Only (b) and (c)
4. 4. (a), (b) and (c)

Answer : 2. Only (a) and (c)

Explanation Question 128

1. Windows Azure is a cloud-based operating system. True
2. Google App Engine is an integrated set of online services for consumers to communicate and share with others. False
3. Amazon Cloud Front is a web service for content delivery. True

Microsoft Azure, commonly referred to as Azure, is a cloud computing service created by Microsoft for building, testing, deploying, and managing applications and services through Microsoft-managed data centers.

Google App Engine (often referred to as GAE or simply App Engine) is a Platform as a Service and cloud computing platform for developing and hosting web applications in Google-managed data centers.

Amazon CloudFront is a content delivery network (CDN) offered by Amazon Web Services. Content delivery networks provide a globally-distributed network of proxy servers which cache content, such as web videos or other bulky media, more locally to consumers, thus improving access speed for downloading the content.

Statement (b) is wrong because Google App Engine is a cloud computing platform for developing and hosting web applications in Google-managed data centers. It's not an online services for consumers to communicate and share with others.

Hence, Correct statements are only (a) and (c)

So, option 2 is correct answer

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UGC NET CS December 2019 - Question 127	UGC NET CS December 2019 - Question 129

UGC NET Computer Science December 2019 | Question 129

Question 129

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Consider the following statements with respect to network security :
(a) Message confidentiality means that the sender and the receiver expect privacy.
(b) Message integrity means that the data must arrive at the receiver exactly as they were sent.
(c) Message authentication means the receiver is ensured that the message is coming from the intended sender.
Which of the statements is (are) correct?

1. 1. Only (a) and (b)
2. 2. Only (a) and (c)
3. 3. Only (b) and (c)
4. 4. (a), (b) and (c)

Answer : 4. (a), (b) and (c)

Explanation Question 129

Message Confidentiality
In information security, confidentiality "is the property, that information is not made available or disclosed to unauthorized individuals, entities, or processes." Confidentiality is a component of privacy that implements to protect our data from unauthorized viewers. Examples of confidentiality of electronic data being compromised include laptop theft, password theft, or sensitive emails being sent to the incorrect individuals.

Message Integrity
This means that data cannot be modified in an unauthorized or undetected manner.

Message Authentication
Authentication is the act of verifying a claim of identity.

References:

• Information Security - Confidentiality
• Information Security - Integrity
• Information Security - Authentication

All three statements are correct.

So, option 4 is correct answer

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UGC NET CS December 2019 - Question 128	UGC NET CS December 2019 - Question 130

UGC NET Computer Science December 2019 | Question 130

Question 130

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Consider the following :
(a) Trapping at local maxima
(b) Reaching a plateau
(c) Traversal along the ridge.
Which of the following option represents shortcomings of the hill climbing algorithm?

1. 1. (a) and (b) only
2. 2. (a) and (c) only
3. 3. (b) and (c) only
4. 4. (a), (b) and (c)

Answer : 4. (a), (b) and (c)

Explanation Question 108

Problems in Hill Climbing Algorithm:

1. Local Maximum: A local maximum is a peak state in the landscape which is better than each of its neighboring states, but there is another state also present which is higher than the local maximum.
   Solution: Backtracking technique can be a solution of the local maximum in state space landscape. Create a list of the promising path so that the algorithm can backtrack the search space and explore other paths as well.
2. Plateau: A plateau is the flat area of the search space in which all the neighbor states of the current state contains the same value, because of this algorithm does not find any best direction to move. A hill-climbing search might be lost in the plateau area.
   Solution: The solution for the plateau is to take big steps or very little steps while searching, to solve the problem. Make a big jump. Randomly select a state far away from the current state. Chances are that we will land at a non-plateau region
3. Ridges: A ridge is a special form of the local maximum. It has an area which is higher than its surrounding areas, but itself has a slope, and cannot be reached in a single move.
   Solution: With the use of bidirectional search, or by moving in different directions, we can improve this problem.

Reference 1 : Problems in Hill Climbing Algorithm in AI

(a), (b), (c) all are shortcomings of the hill climbing algorithm.

So, option 4 is correct answer

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UGC NET CS December 2019 - Question 129	UGC NET CS December 2019 - Question 131