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.

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.

When data and acknowledgement are sent in the same frame, this is called as

A.Piggy packing
B.Piggy backing
C.Back packing
D.Good packing

Exam Year(s)

UGC NET - 2013,

Answer: B

Sending an acknowledgement inside an outgoing data frame, instead of in a separate frame, is called piggybacking (written here as 'Piggy backing'). In a full-duplex data link protocol each side has data of its own to send, so the receiver waits briefly and attaches its acknowledgement to the next data frame heading the same way, which saves bandwidth and frame overhead.

'Piggy packing', 'Back packing' and 'Good packing' are not terms used in networking. They only imitate the real name.

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88.

Encryption and Decryption is the responsibility of _______ Layer.

A.Physical
B.Network
C.Application
D.Datalink

Exam Year(s)

UGC NET - 2013,

Answer: C

Among the layers listed, encryption and decryption are assigned to the Application layer. In the OSI model this job strictly belongs to the Presentation layer, which is not offered here, so the listed layer that sits at the top and handles data representation and security for the user's data is the Application layer.

The Physical layer only moves raw bits over the medium. The Network layer routes packets between networks and the Datalink layer frames bits and detects errors between adjacent nodes. None of these three is responsible for encrypting user data.

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89.

An analog signal carries 4 bits in each signal unit. If 1000 signal units are sent per second, then baud rate and bit rate of the signal are _______ and _______.

A.4000 bauds \ sec & 1000 bps
B.2000 bauds \ sec & 1000 bps
C.1000 bauds \ sec & 500 bps
D.1000 bauds \ sec & 4000 bps

Exam Year(s)

UGC NET - 2013,

Answer: D

The baud rate is the number of signal units per second, so it is 1000 bauds per second. The bit rate is the baud rate multiplied by the bits carried in each signal unit.

\[ \text{bit rate} = 1000 \times 4 = 4000 \text{ bps} \]

So the signal has 1000 bauds per second and 4000 bps. The options giving 4000 bauds, 2000 bauds or a bit rate smaller than the baud rate are wrong, because with 4 bits per signal unit the bit rate must be four times the baud rate.

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90.

The VLF and LF bauds use _______ propagation for communication.

A.Ground
B.Sky
C.Line of sight
D.Space

Exam Year(s)

UGC NET - 2013,

Answer: A

The VLF (very low frequency) and LF (low frequency) bands use ground propagation. Waves in these bands, from about 3 kHz to 300 kHz, follow the curvature of the earth near the surface and can travel very long distances in this way.

Sky propagation is typical of the HF band, where the ionosphere reflects the signal. Line of sight is used by VHF, UHF and higher bands, and space propagation belongs to satellite links in the highest bands. None of these applies to VLF and LF.

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