1.
Introduction
2.
2.1.
What is Gate Exam?
2.2.
Who is eligible for the GATE exam?
2.3.
Is GATE compulsory for Masters?
2.4.
Are there any restrictions or age limits to applying for the GATE Exams 2023?
2.5.
Is a 3rd-year B.A./B.Com./B.Sc. students eligible to appear for GATE 2021?
3.
Conclusion
Last Updated: Mar 27, 2024
Easy

# Lan Technologies and Wi-Fi

Apoorv Dixit
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Master Python: Predicting weather forecasts
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Ashwin Goyal
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## Introduction

A LAN (Local Area Network) is a data communication network that connects multiple terminals or computers within a building or a limited geographical area. The devices' connections could be wired or wireless. In this article, we will discuss some questions related to LAN Technologies and wifi which were previously asked in the gate exam.

Question 1

Consider the resolution of the domain name www.gate.org.in by a DNS resolver. Assume that no resource records are cached anywhere across the DNS servers and that an iterative query mechanism is used in the resolution. The number of DNS query-response pairs involved in completely resolving the domain name is_____________.

Explanation: The DNS resolver goes to the root server, then forwards to the top-level domain, then forwards to the second-level domain, and gets the IP address from the authoritative DNS server in an iterative query.

Question 2

Which of the following statements about an Ethernet local area network is TRUE?

A. Once a station starts transmitting a frame, it stops sensing the channel.

B. The jamming signal's purpose is to pad frames that are smaller than the minimum frame size.

C. Even after a collision is detected, a station continues to transmit the packet.

D. On retransmissions, the exponential backoff mechanism reduces the chance of a collision.

Explanation: The exponential back-off mechanism reduces the probability of collision on retransmissions.

Question 3

Consider a 100 Mbps (108 bits per second) CSMA/CD network with no repeaters that transmits data over a 1 km (kilometer) cable. What is the signal speed (km/sec) in the cable if the minimum frame size required for this network is 1250 bytes?

(A )8000

(B) 10000

(C) 16000

(D) 20000

Explanation: Data should be transmitted at a speed of 100 Mbps.

Transmission Time >= 2 * Propagation Time

=> 1250*8 / (100 * 106) <= 2 * length/signal_speed

=> signal_speed  <= (2 * 103 * 100 * 106) / (1250 * 8)

<= 2 * 10 * (103) km/sec

<= 20000

Question 4

When Manchester encoding is used in Ethernet, the bit rate is

(A) half of the baud rate.

(B) The baud rate is doubled.

(C) The baud rate is the same.

(D) None of the preceding

Explanation: The bitrate in Manchester encoding is half that of the baud rate.

Questions 5

In a network of LANs connected by bridges, packets are sent from one LAN to another through intermediate bridges. Since more than one path may exist between two LANs, packets may have to be routed through multiple bridges. Why is the spanning tree algorithm used for bridge routing?

(A) For fault tolerance

(B) For avoiding loops in the routing paths

(C) For shortest path routing between LANs

(D) For minimizing collisions

Explanation: The main idea behind using Spanning Trees is to avoid loops.

Question 6

Consider the following scenario: 15 machines must be connected to a LAN via 8-port Ethernet switches. Assume that these switches lack separate uplink ports. The bare minimum of switches required is.

It should be noted that this was a Numerical Type question.

(A) 3

(B) 4

(C) 5

(D) 6

Explanation: To connect 15 computers, you'll need at least three Ethernet switches.

Question 7

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

(A)1

(B)2

(C)2.5

(D)5

Explanation:

Data should be transmitted at a rate of 500 (Mbps).

Time of transmission >= 2*Time of propagation

=> 10000/(500*1000000) <= 2*length/200000

=> 2km in length (max)

As a result, the answer will be (B) 2km.

Question 8

Assume a round trip propagation delay of 46.4 ms for a 10 Mbps Ethernet with a 48-bit jamming signal. The smallest frame size is

The smallest frame size is

(A) 94

(B) 416

(C) 464

(D) 512

Explanation:

Td > 2 × Tp + Td( for jam signal)

Td > Rtt + (Jam signal length / Bandwidth)

Td > 46.4s + (48 bit / 10 x 106 bits/sec)

Td = 46.4 x 106 seconds + (4.8 x 106 seconds)

Td > 51.2 x 10-5 sec (Frame Size/Bandwidth)

(Frame Size / Bandwidth) > 51.2 × 10-6 sec

Frame Size > 51.2 × 10-6  sec × Bandwidth

Frame Size > 51.2 × 10-6  sec × 10 × 106 bits/second

Frame Size > 512 bits

As a result, the minimum frame size is 512 bits.

Question 9

The data transmission bandwidth of a network is 20  * 106 bits per second. In the MAC layer, it employs CSMA/CD. The maximum time it takes for a signal to travel from one node to another is 40 microseconds. A-frame in the network must be at least bytes in size.

Note that this was a Numerical Answer Type question.

(A) 200

(B) 250

(C) 400

(D) 1200

Explanation: For the frame size to be the smallest, the transmission time must be twice the one-way propagation delay.

Tx = 2Tp, for example.

Assume that the smallest frame size is L bits.

L / B = Tx

where B is the bandwidth and is equal to 20 * 106 bits per second

as well as 40 microseconds Tp

Tx = 2 Tp

80 microseconds = L / B

L = B * 80 micro seconds = 20 * 106 bits/sec * 80 micro seconds = 1600 bits

Due to the fact that the answer must be given in bytes, 1600 / 8 bytes = 200 bytes.

As a result, A is the correct response.

Question 10

Which of the following statements about the IEEE 802.11 MAC protocol for wireless communication is/are TRUE?

I. At least three non-overlapping channels are

available for transmissions.

II. The RTS-CTS mechanism is used for collision detection.

III. Unicast frames are ACKed.

(A) All I, II, and III

(B) I and III only

(C) II and III only

(D) II only

Explanation: Instead of using the 2.4 GHz frequency band, which only has three non-overlapping channels, 802.11 uses the 5 GHz Radio Band (High Frequency), which has 23 overlapping channels.

RTS (Request To Send) and CTS (Clear To Send) are control frames that help with station exchange and collision reduction rather than detection. When Station 1 receives a CTS frame, it sends a data frame to Station 2.

Station 1 sends an RTS to Station 2, and Station 2 responds with a CTS. Station 2 sends an acknowledgment frame after receiving data (ACKed). As Station 2 sends an Acknowledgement to a single recipient on a network, Unicast Frames are acknowledged.
Check out this problem - No of Spanning Trees in a Graph

#### What is Gate Exam?

Gate stands for Graduate Aptitude Test in Engineering which is conducted for admission into the Master's Program and Job in Public Sector Companies.

#### Who is eligible for the GATE exam?

A student who is currently studying in the 3rd or higher years of any undergraduate program.

#### Is GATE compulsory for Masters?

No.it's not at all compulsory. However, if you want to pursue a master's degree in India, GATE is the best option.

#### Are there any restrictions or age limits to applying for the GATE Exams 2023?

No, there are no restrictions on the number of times.

#### Is a 3rd-year B.A./B.Com./B.Sc. students eligible to appear for GATE 2021?

Yes, any undergraduate student currently in the 3rd year or higher years of any government-approved program in Engineering / Technology / Architecture / Arts / Commerce / Science is eligible to appear for GATE 2021.

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## Conclusion

In this article, we have discussed many questions on Lan Technologies and Wi-Fi. These were all previous year's questions that appeared in the Gate paper.

We hope you learned something new. If you want to learn more, check out our articles on Introduction to GATEHow to prepare in the Last 10 days to score high in GATE?

Refer to our guided paths on Coding Ninjas Studio to learn more about DSA, Competitive Programming, JavaScript, System Design, etc.

Enroll in our courses and refer to the mock test and problems available.

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