Networking Basics Questions and Answers

Q: Decryption and encryption of data are the responsibility of which of the following layer ?

Encryption and decryption are the responsibility of the Presentation layer.The presentation layer deals with the syntax of the data exchanged, including its translation, compression and encryption. The physical layer (A) carries bits, the data link layer (B) frames the bits and checks for errors on one link, and the session layer (D) sets up and controls dialogues between applications.

Q: Page Shift Keying (PSK) Method is used to modulate digital signal at 9600 bps using 16 level. Find the line signals and speed (i.e. modulation rate).

The modulation rate is 2400 bauds.With 16 levels each signal element carries \(\log_2 16 = 4\) bits, so the baud rate is the bit rate divided by 4:\[ \text{baud rate} = \frac{9600}{\log_2 16} \]\[ = \frac{9600}{4} = 2400 \]1200 bauds (B) would need 8 bits per signal element, which means 256 levels. 4800 bauds (C) comes from taking 2 bits per element, which is 4 levels, and 9600 bauds (D) would be right only if each element carried one bit.

Q: The station to hub distance in which it is 2000 metres.

A station-to-hub distance of 2000 metres is possible with 100 Base-Fx.100 Base-FX is Fast Ethernet over two strands of multimode optical fibre, and full duplex links can run up to 2000 metres. 100 Base-TX and 100 Base-T4 run on twisted pair copper (two pairs of category 5, and four pairs of category 3) and are limited to about 100 metres. There is no standard called 100 Base-T1 for this purpose.

Q: Check sum used along with each packet computes the sum of the data, where data is treated as a sequence of

The checksum treats the data as a sequence of bits, in the answer given by the key.A checksum is worked out by adding up the data in fixed-size groups, such as 16-bit words, so that the sender and the receiver can compare the totals. At the lowest level the data is a stream of bits, which is what the key accepts. The words 'integer' and 'real numbers' describe how a program may read the numbers and not how the checksum looks at the packet, and 'character' is not a unit in this calculation. The question is loosely worded and the answer follows the key.

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.

81.

Decryption and encryption of data are the responsibility of which of the following layer ?

A.Physical layer
B.Data Link layer
C.Presentation layer
D.Session layer

Exam Year(s)

UGC NET - 2012,

Answer: C

Encryption and decryption are the responsibility of the Presentation layer.

The presentation layer deals with the syntax of the data exchanged, including its translation, compression and encryption. The physical layer (A) carries bits, the data link layer (B) frames the bits and checks for errors on one link, and the session layer (D) sets up and controls dialogues between applications.

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82.

In which circuit switching, delivery of data is delayed because data must be stored and retrieved from RAM ?

A.Space division
B.Time division
C.Virtual
D.Packet

Exam Year(s)

UGC NET - 2012,

Answer: B

The delay from storing and retrieving data in RAM occurs in time division circuit switching.

A time-slot interchange in a time division switch writes the incoming slots into memory (RAM) and reads them out in a different order for the outgoing line, which adds a delay. A space division switch (A) joins lines by crosspoints with no storage and so introduces no such delay. Virtual circuit and packet switching (C and D) are not types of circuit switching in this sense.

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83.

Page Shift Keying (PSK) Method is used to modulate digital signal at 9600 bps using 16 level. Find the line signals and speed (i.e. modulation rate).

A.2400 bauds
B.1200 bauds
C.4800 bauds
D.9600 bauds

Exam Year(s)

UGC NET - 2012,

Answer: A

The modulation rate is 2400 bauds.

With 16 levels each signal element carries \(\log_2 16 = 4\) bits, so the baud rate is the bit rate divided by 4:

\[ \text{baud rate} = \frac{9600}{\log_2 16} \]

\[ = \frac{9600}{4} = 2400 \]

1200 bauds (B) would need 8 bits per signal element, which means 256 levels. 4800 bauds (C) comes from taking 2 bits per element, which is 4 levels, and 9600 bauds (D) would be right only if each element carried one bit.

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84.

The station to hub distance in which it is 2000 metres.

A.100 Base-Tx
B.100 Base-Fx
C.100 Base-T4
D.100 Base-T1

Exam Year(s)

UGC NET - 2012,

Answer: B

A station-to-hub distance of 2000 metres is possible with 100 Base-Fx.

100 Base-FX is Fast Ethernet over two strands of multimode optical fibre, and full duplex links can run up to 2000 metres. 100 Base-TX and 100 Base-T4 run on twisted pair copper (two pairs of category 5, and four pairs of category 3) and are limited to about 100 metres. There is no standard called 100 Base-T1 for this purpose.

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85.

Check sum used along with each packet computes the sum of the data, where data is treated as a sequence of

A.Integer
B.Character
C.Real numbers
D.Bits

Exam Year(s)

UGC NET - 2012,

Answer: D

The checksum treats the data as a sequence of bits, in the answer given by the key.

A checksum is worked out by adding up the data in fixed-size groups, such as 16-bit words, so that the sender and the receiver can compare the totals. At the lowest level the data is a stream of bits, which is what the key accepts. The words 'integer' and 'real numbers' describe how a program may read the numbers and not how the checksum looks at the packet, and 'character' is not a unit in this calculation. The question is loosely worded and the answer follows the key.

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86.

If the period of a signal is 1000 ms, then what is its frequency in kilohertz ?

A.\(10^{-3}\) KHz
B.\(10^{-2}\) KHz
C.\(10^{-1}\) KHz
D.1 KHz

Exam Year(s)

UGC NET - 2012,

Answer: A

The frequency of a signal is the reciprocal of its period. A period of 1000 ms is 1 second, so the frequency is 1 hertz, which is \(10^{-3}\) kilohertz, since 1 kHz is 1000 Hz. So the answer is option (A). The other values, \(10^{-2}\), \(10^{-1}\) and 1 kilohertz, are 10 hertz, 100 hertz and 1000 hertz, which belong to periods of 100 ms, 10 ms and 1 ms, so they do not match the period given.

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87.

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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88.

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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89.

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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90.

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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