NTA UGC NET Computer Science PYQ Subject Wise

Here we published NTA UGC NET or GATE Computer Science PYQ Subject Wise only. If you want to solve NTA UGC NET or GATE Previous Year Question Year wise click here.

DATA COMMUNICATION AND COMPUTER NETWORKS

NTA UGC NET Computer Network

1 / 15

Category: Computer Networks

A computer network uses polynomials over GF(2) for error checking with 8 bits as information bits and uses x3 + x + 1 as the generator polynomial to generate the check bits. In this network, the message 01011011 is transmitted as

01011011100

01011011010

01011011101

01011011011

2 / 15

Category: Data Communication and Computer Networks

1) 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?

64 subnets and 262142 hosts.

62 subnets and 262142 hosts.

62 subnets and 1022 hosts.

64 subnets and 1024 hosts.

3 / 15

Category: Data Communication and Computer Networks

.................. do not take their decisions on measurements or estimates of the current traffic
and topology.

(1) Static algorithms

(2) Adaptive algorithms

(3) Non - adaptive algorithms

(4) Recursive algorithms

4 / 15

Category: Data Communication and Computer Networks

Which of the following is/are example(s) of stateful application layer protocols?

FTP

TCP

HTTP

5 / 15

Category: Computer Networks

Match the following:
Field Length in bits
P. UDP Header's Port Number I. 48
Q. Ethernet MAC Address II. 8
R. IPv6 Next Header III.32
S. TCP Header's Sequence Number IV. 16

P-II, Q-I, R-IV, S-III

P-IV, Q-I, R-II, S-III

P-III, Q-IV, R-II, S-I

P-IV, Q-I, R-III, S-II

6 / 15

Category: Computer Networks

A network has a data transmission bandwidth of 20 × 106 bits per second. It uses CSMA/CD in the MAC layer. The maximum signal propagation time from one node to another node is 40 microseconds. The minimum size of a frame in the network is _________ bytes.

200

202

203

201

For frame size to be minimum, its transmission time should be equal to twice of one way propagation delay. i.e, T_t = 2 × T_P
Given,
Bandwidth (B) = 20 × 10⁶ bps
T_P = 40 μs ⇒ 40 × 10^{- 6} sec
Suppose minimum frame size is L.
T_t = 2 × T_P ⇒ L / B = 2 × T_P
⇒ L = 2 × T_P × B = 2 × 40 × 10^-6 × 20 × 10⁶ = 1600 bits ⇒ 200 bytes
Therefore, L = 200 bytes

7 / 15

Category: Computer Networks

What is the data unit of physical layer

FRAME

SEGMENTS

BIT

PACKETS

DATA

8 / 15

Category: Data Communication and Computer Networks

Which of the following devices takes data sent from one network device and forwards it to
the destination node based on MAC address?

(2) Modem

(4) Gateway

(3) Switch

(1) Hub

9 / 15

Category: Computer Networks

IP address = 201.20.30.40
Calculate Network ID

201.20.30.0

201.20.30.3

201.20.30.2

201.20.30.1

10 / 15

Category: Data Communication and Computer Networks

The IP address ................... is used by hosts when they are being booted.

(1) 0.0.0.0

(3) 1.1.1.1

(4) 255.255.255.255

(2) 1.0.0.0

11 / 15

Category: Data Communication and Computer Networks

Which of the following statements is/are true?

Auto increment addressing mode is useful in creating self-relocating code.

If auto increment addressing mode is included in an instruction set architecture, then an additional ALU is required for effective address calculation.

In auto increment addressing mode, the amount of increment depends on the size of the data item accessed

12 / 15

Category: Computer Networks

Consider an IP packet with a length of 4,500 bytes that includes a 20-byte IPv4 header and a 40-byte TCP header. The packet is forwarded to an IPv4 router that supports a Maximum Transmission Unit (MTU) of 600 bytes. Assume that the length of the IP header in all the outgoing fragments of this packet is 20 bytes. Assume that the fragmentation offset value stored in the first fragment is 0.
The fragmentation offset value stored in the third fragment is __________.

144

145

143

None

MTU = 600 bytes, IP header = 20 bytes
Therefore Payload = 600 - 20 = 580 bytes.
As we know fragment size should be multiple of 8 but 580 bytes is not a multiple of 8, therefore fragment size is 576 bytes.
Offset value of kth fragment = Fragment size *( kth fragment - 1) / scaling factor
Offset value of third fragment = 576 * (3-1) / 8 = 144

13 / 15

Category: Computer Networks

Which one of the following is not typically provided by source code management software

Syntax highlighting

Synchronisation

Project forking

Versioning and Revision history

14 / 15

Category: Data Communication and Computer Networks

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

2N

N(N-1)/2

(N-1)^2

N(N-1)

15 / 15

Category: Computer Networks

Repeater operates in which layer?

Transport Layer

Presentation Layer

Application Layer

Physical Layer

Network Layer

Your score is
0%

Data Structures and Algorithm

NTA UGC NET - Data Structures and Algorithms

In this quiz, we focused on NTA UGC NET exam question or GATE question which is related to Data Structures and Algorithms.

1 / 6

Category: Data structure and algorithms

Let T(n) be a function defined by the recurrence T(n) = 2T(n/2) + √n for n ≥ 2 and T(1) = 1 Which of the following statements is TRUE?

T(n) = θ(n log n)

T(n) = θ(√n)

T(n) = θ(n)

T(n) = θ(log n)

2 / 6

Category: Data structure and algorithms

None of the above

AB + CD + *F/D + E*

ABCD + *F/DE*++

ABCD + /F *DE+ *+

The postfix expression will be,
A B C D + * F / + D E * +

3 / 6

Category: Data structure and algorithms

The preorder traversal sequence of a binary search tree is 30, 20, 10, 15, 25, 23, 39, 35, 42. Which one of the following is the postorder traversal sequence of the same tree?

10, 20, 15, 23, 25, 35, 42, 39, 30

15, 20, 10, 23, 25, 42, 35, 39, 30

15, 10, 25, 23, 20, 42, 35, 39, 30

15, 10, 23, 25, 20, 35, 42, 39, 30

4 / 6

Category: Data structure and algorithms

What is the worst case time complexity of inserting n2 elements into an AVL-tree with n elements initially?

O(n^3)

O(n^4)

O(n^2logn)

O(n^2)

AVL Tree all operations(insert, delete and search) will take O(logn) time.
In question, they asked about n² elements.
So, In worst case it will take o(n² logn) time.

5 / 6

Category: Data structure and algorithms

Let T be a tree with 10 vertices. The sum of the degree of all the vertices in T is

If a tree has V vertices than it has (V-1) edges.

If a tree has E edges than the sum of all degree = 2*E

6 / 6

Category: Data structure and algorithms

Which of the following statements is/are TRUE for undirected graphs?
P: Number of odd degree vertices is even.
Q: Sum of degrees of all vertices is even.

Both P and Q

P only

Neither P nor Q

Q only

Euler’s Theorem 3:
The sum of the degrees of all the vertices of a graph is an even number (exactly twice the number of edges).
In every graph, the number of vertices of odd degree must be even.

Your score is

Data Structures and Algorithm

Data Structures and Algorithms

In this quiz, we focused on NTA UGC NET exam question or GATE question which is related to Data Structures and Algorithms.

1 / 8

Category: Data structure and algorithms

The preorder traversal of a binary search tree is 15, 10, 12, 11, 20, 18, 16, 19.
Which one of the following is the postorder traversal of the tree?

10, 11, 12, 15, 16, 18, 19, 20

11, 12, 10, 16, 19, 18, 20, 15

19, 16, 18, 20, 11, 12, 10, 15

20, 19, 18, 16, 15, 12, 11, 10

PYQ

2 / 8

Category: Data structure and algorithms

Let T(n) be a function defined by the recurrence T(n) = 2T(n/2) + √n for n ≥ 2 and T(1) = 1 Which of the following statements is TRUE?

T(n) = θ(n log n)

T(n) = θ(log n)

T(n) = θ(n)

T(n) = θ(√n)

3 / 8

Category: Data structure and algorithms

Which of the following statements is/are TRUE for undirected graphs?
P: Number of odd degree vertices is even.
Q: Sum of degrees of all vertices is even.

P only

Q only

Both P and Q

Neither P nor Q

Euler’s Theorem 3:
The sum of the degrees of all the vertices of a graph is an even number (exactly twice the number of edges).
In every graph, the number of vertices of odd degree must be even.

4 / 8

Category: Data structure and algorithms

ABCD + /F *DE+ *+

AB + CD + *F/D + E*

None of the above

ABCD + *F/DE*++

The postfix expression will be,
A B C D + * F / + D E * +

5 / 8

Category: Data structure and algorithms

Software products need perfective maintenance for which of the following resons?

To support the new features that users want it to support.

When the customers need the product to run on new platforms.

To rectify bugs observed while the system is in use.

To overcome wear and tear caused by the repeated use of the software

Perfective software maintenance addresses the functionality and usability of the software. Perfective maintenance involves changing existing product functionality by refining, deleting, or adding new features.

6 / 8

Category: Data structure and algorithms

The preorder traversal sequence of a binary search tree is 30, 20, 10, 15, 25, 23, 39, 35, 42. Which one of the following is the postorder traversal sequence of the same tree?

15, 20, 10, 23, 25, 42, 35, 39, 30

15, 10, 23, 25, 20, 35, 42, 39, 30

15, 10, 25, 23, 20, 42, 35, 39, 30

10, 20, 15, 23, 25, 35, 42, 39, 30

7 / 8

Category: Data structure and algorithms

Let T be a full binary tree with 8 leaves. (A full binary tree has every level full). Suppose two leaves a and b of T are chosen uniformly and independently at random. The expected value of the distance between a and b in T (i.e., the number of edges in the unique path between a and b) is (rounded off to 2 decimal places) _____.

4.25

3.82

1.34

5.54

There can be 8 paths between any 2 uniformly & independently chosen leaf nodes.
A node can be chosen twice and the path from that node to itself will be zero.
∴ Path 1 = 0
Similarly,
Path 2 = 2
Path 3 = 4
Path 4 = 4
Path 5 = 6
Path 6 = 6
Path 7 = 6
Path 8 = 6
∴ Expected value = Σ Path length × Probability of selecting path
= 2×1/8 + 4×2/8 + 6×4/8 + 0×1/8
= 1/4 + 1/1 + 3/1 + 0
= 4 + 1/4
= 17/4
= 4.25

8 / 8

Category: Data structure and algorithms

The second smallest of n elements can be found with _______ comparisons in the worst case.

3n/2

n – 1

lg n

n + ceil(lg n) - 2

divide and conquer approach gives min element with n-1 comparison..
for 2nd minimum element search on d root to leaf path from where max element coming.. reason is second minimum element must compared with minimal element when we were finding minimum element.. so ,

divide and conquere based on binary tree approach so no of elements from root to leaf path is logn .. finding min from logn elements takes logn -1 comparison..
total n-1+logn-1 = n + logn -2 comparison

Your score is

NTA UGC NET Computer Science PYQ Subject Wise

Leave a Comment Cancel Reply

DISCLAIMER

COMPANY

USEFUL LINKS

LINKS

Related Posts

Leave a Comment Cancel Reply