Networking Basics Questions and Answers

Q: How often are BPDUs sent from a layer 2 device?

Correct answer: Every 2 secondsBridge protocol data units are sent every two seconds by default, which is the spanning tree hello time. They carry the bridge ID and path cost information from which every switch determines the root bridge, its own root port and any blocking ports. The interval matters because convergence times are derived from it: max age is twenty seconds, ten missed hellos, and forward delay fifteen.

Q: How many broadcast domains are created when you segment a network with a 12-port switch?

Correct answer: 1A switch breaks up collision domains, not broadcast domains: a broadcast frame arriving on one port is flooded out of all the others, so all twelve ports remain in a single broadcast domain by default. Splitting that domain requires either a router or VLANs configured on the switch, each VLAN forming a broadcast domain of its own. The complementary question - how many collision domains - gives twelve.

Q: In a network with dozens of switches, how many root bridges would you have?

Correct answer: 1Spanning tree elects exactly one root bridge per broadcast domain, however many switches there are, and every other switch computes its shortest path to that root. The election is by lowest bridge ID - priority first, then MAC address as the tiebreaker - which is why the priority is normally set deliberately on the switch that should win, rather than left to the accident of which switch has the lowest MAC address.

91.

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.110010011001
B.11001001
C.110010011001001
D.11001001011

Exam Year(s)

UGC NET - 2013,

Answer: D

The transmitted message is 11001001011. The generator \(x^{3} + 1\) is 1001, of degree 3, so three zeros are appended to the message, giving 11001001000. Dividing by 1001 using modulo-2 arithmetic leaves the remainder 011.

\[ 11001001000 \div 1001 \Rightarrow \text{remainder } 011 \]

The remainder is placed in the place of the appended zeros, so the frame is 11001001011. Option B is the unprotected message and options A and C do not have 3 check bits appended.

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92.

What is the bit rate for transmitting uncompressed \(800 \times 600\) pixel colour frames with 8 bits/pixel at 40 frames/second ?

A.2.4 Mbps
B.15.36 Mbps
C.153.6 Mbps
D.1536 Mbps

Exam Year(s)

UGC NET - 2013,

Answer: C

The bit rate is 153.6 Mbps. Each frame has \(800 \times 600 = 480000\) pixels at 8 bits per pixel, giving 3,840,000 bits per frame. At 40 frames per second:

\[ 3\,840\,000 \times 40 \]

\[ = 153\,600\,000 \text{ bits/s} \]

That is 153.6 Mbps. The figure 15.36 Mbps is ten times too small, 1536 Mbps is ten times too large, and 2.4 Mbps does not follow from the data.

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93.

Which layer of OSI reference model is responsible for decomposition of messages and generation of sequence numbers to ensure correct re-composition from end to end of the network ?

A.Physical
B.Data-link
C.Transport
D.Application

Exam Year(s)

UGC NET - 2013,

Answer: C

The transport layer is responsible for breaking messages into segments, generating sequence numbers and reassembling them correctly from end to end. The physical layer carries raw bits, the data-link layer handles frames between adjacent nodes, and the application layer provides services to user programs.

Only the transport layer provides this end-to-end segmentation and sequencing between the two hosts.

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94.

A client-server system uses a satellite network, with the satellite at a height of 40,000 kms. What is the best-case delay in response to a request ? (Note that the speed of light in air is 3,00,000 km/second).

A.133.33 m sec
B.266.67 m sec
C.400.00 m sec
D.533.33 m sec

Exam Year(s)

UGC NET - 2013,

Answer: D

The best-case delay is 533.33 msec. A request goes from the client up to the satellite (40,000 km) and down to the server (40,000 km), and the response makes the same trip, so the signal travels \(4 \times 40\,000 = 160\,000\) km.

\[ t = \frac{160\,000}{300\,000} = 0.5333 \text{ s} \]

\[ t = 533.33 \text{ msec} \]

133.33 msec is a single 40,000 km hop, 266.67 msec is a one-way trip of 80,000 km, and 400 msec corresponds to 120,000 km, so none of these covers the full round trip.

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95.

The start and stop bits are used in serial communication for

A.error detection
B.error correction
C.synchronization
D.slowing down the communication

Exam Year(s)

UGC NET - 2013,

Answer: C

Start and stop bits are used in serial (asynchronous) communication for synchronization. The start bit tells the receiver that a character is beginning, so it can start sampling at the right moment, and the stop bit marks the end of the character and returns the line to idle.

They are not used for error detection or correction (parity bits do detection), and they do not slow the communication down; in fact they add a little overhead.

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96.

_________ is a type of transmission impairment in which the signal looses strength due to the resistance of the transmission medium.

A.Attenuation
B.Distortion
C.Noise
D.Decibel

Exam Year(s)

UGC NET - 2013,

Answer: A

The impairment in which the signal loses strength because of the resistance of the transmission medium is attenuation. Energy is converted to heat as the signal travels, so its amplitude falls with distance, and amplifiers or repeaters are used to restore it.

Distortion is a change in the shape of the signal, noise is unwanted energy added to the signal, and the decibel is only a unit used to measure gain or loss.

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97.

How often are BPDUs sent from a layer 2 device?

A.Never
B.Every 2 seconds
C.Every 10 minutes
D.Every 30 seconds

Answer: B

Correct answer: Every 2 seconds

Bridge protocol data units are sent every two seconds by default, which is the spanning tree hello time. They carry the bridge ID and path cost information from which every switch determines the root bridge, its own root port and any blocking ports. The interval matters because convergence times are derived from it: max age is twenty seconds, ten missed hellos, and forward delay fifteen.

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98.

How many broadcast domains are created when you segment a network with a 12-port switch?

A.1
B.2
C.5
D.12

Answer: A

Correct answer: 1

A switch breaks up collision domains, not broadcast domains: a broadcast frame arriving on one port is flooded out of all the others, so all twelve ports remain in a single broadcast domain by default. Splitting that domain requires either a router or VLANs configured on the switch, each VLAN forming a broadcast domain of its own. The complementary question - how many collision domains - gives twelve.

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99.

In a network with dozens of switches, how many root bridges would you have?

A.1
B.2
C.5
D.12

Answer: A

Correct answer: 1

Spanning tree elects exactly one root bridge per broadcast domain, however many switches there are, and every other switch computes its shortest path to that root. The election is by lowest bridge ID - priority first, then MAC address as the tiebreaker - which is why the priority is normally set deliberately on the switch that should win, rather than left to the accident of which switch has the lowest MAC address.

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100.

Two devices are in network if __________

A.a process in one device is able to exchange information with a process in another device
B.a process is running on both devices
C.PIDs of the processes running of different devices are same
D.a process is active and another is inactive

Answer: A

Explanation: A computer network, or data network, is a digital telecommunications network which allows nodes to share resources. In computer networks, computing devices exchange data with each other using connections between nodes. The nodes have certain processes which enable them to share a specific type of data using a distinct protocol.

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