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Welcome to the Networking MCQs Page

Dive deep into the fascinating world of Networking with our comprehensive set of Multiple-Choice Questions (MCQs). This page is dedicated to exploring the fundamental concepts and intricacies of Networking, a crucial aspect of GATE CSE Exam. In this section, you will encounter a diverse range of MCQs that cover various aspects of Networking, from the basic principles to advanced topics. Each question is thoughtfully crafted to challenge your knowledge and deepen your understanding of this critical subcategory within GATE CSE Exam.

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Check out the MCQs below to embark on an enriching journey through Networking. Test your knowledge, expand your horizons, and solidify your grasp on this vital area of GATE CSE Exam.

Note: Each MCQ comes with multiple answer choices. Select the most appropriate option and test your understanding of Networking. You can click on an option to test your knowledge before viewing the solution for a MCQ. Happy learning!

Networking MCQs | Page 1 of 15

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Q1.
Determine the maximum length of the cable (in km) for transmitting data at a rate of 500 Mbps in an Ethernet LAN with frames of size 10,000 bits. Assume the signal speed in the cable to be 2,00,000 km/s.

a.

1

b.

2

c.

2.5

d.

5

Discuss
Answer: (b).2
Q2.
Let G(x) be the generator polynomial used for CRC checking. What is the condition that should be satisfied by G(x) to detect odd number of bits in error?

a.

G(x) contains more than two terms

b.

G(x) does not divide 1+x^k, for any k not exceeding the frame length

c.

1+x is a factor of G(x)

d.

G(x) has an odd number of terms

Discuss
Answer: (c).1+x is a factor of G(x)
Q3.
Frames of 1000 bits are sent over a 10^6 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 transmission of two frames.

a.

i = 2

b.

i = 3

c.

i = 4

d.

i = 5

Discuss
Answer: (d).i = 5
Q4.
In Ethernet when Manchester encoding is used, the bit rate is:

a.

Half the baud rate

b.

Twice the baud rate

c.

Same as the baud rate

d.

None of the above

Discuss
Answer: (a).Half the baud rate
Q5.
There are n stations in a slotted LAN. Each station attempts to transmit with a probability p in each time slot. What is the probability that ONLY one station transmits in a given time slot?

a.

(1-p)^(n-1)

b.

np(1-p)^(n-1)

c.

p(1-p)^(n-1)

d.

1-(1-p)^(n-1)

Discuss
Answer: (b).np(1-p)^(n-1)
Q6.
In a token ring network the transmission speed is 10^7 bps and the propagation speed is 200 metres/micro second. The 1-bit delay in this network is equivalent to:

a.

500 metres of cable

b.

200 metres of cable

c.

20 metres of cable

d.

50 metres of cable

Discuss
Answer: (c).20 metres of cable
Q7.
The message 11001001 is to be transmitted using the CRC polynomial x^3 + 1 to protect it from errors. The message that should be transmitted is:

a.

11001001000

b.

11001001011

c.

11001010

d.

110010010011

Discuss
Answer: (b).11001001011
Q8.
Consider a token ring network with a length of 2 km having 10 stations including a monitoring station. The propagation speed of the signal is 2 × 10^8 m/s and the token transmission time is ignored. If each station is allowed to hold the token for 2 μsec, the minimum time for which the monitoring station should wait (in μsec)before assuming that the token is lost is _______.

a.

28 to 30

b.

20 to 22

c.

0 to 2

d.

31 to 33

Discuss
Answer: (a).28 to 30
Q9.
Consider a selective repeat sliding window protocol that uses a frame size of 1 KB to send data on a 1.5 Mbps link with a one-way latency of 50 msec. To achieve a link utilization of 60%, the minimum number of bits required to represent the sequence number field is ________.

a.

3

b.

4

c.

5

d.

6

Discuss
Answer: (c).5
Q10.
A bit-stuffing based framing protocol uses an 8-bit delimiter pattern of 01111110. If the output bit-string after stuffing is 01111100101, then the input bit-string is

a.

0111110100

b.

0111110101

c.

0111111101

d.

0111111111

Discuss
Answer: (b).0111110101
Page 1 of 15

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