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Networking Basics Questions and Answers

101.

The ________ measures the relative strengths of two signals or a signal at two different points.

Exam Year(s)

UGC NET - 2013,

Answer: D

The decibel measures the relative strengths of two signals or of one signal at two different points. It is ten times the logarithm to base 10 of the ratio of the powers, and it is negative when the signal is weaker at the second point and positive when it has been amplified.

Frequency is the number of cycles per second, attenuation is the loss of strength of a signal along a medium (which the decibel is used to express), and throughput is the number of bits delivered per second. The answer is (D).

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102.

Which one of the following media is multidrop ?

Exam Year(s)

UGC NET - 2013,

Answer: C

Thick coaxial cable is the medium that is multidrop. In the 10Base5 Ethernet many stations are attached to the one cable by taps, and all of them share it. The answer is (C).

Shielded and unshielded twisted pair cables and fibre optic cable are used for point-to-point links between a pair of devices, so (A), (B) and (D) are not multidrop media in the sense of the question.

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103.

What is the baud rate of the standard 10 Mbps Ethernet ?

Exam Year(s)

UGC NET - 2013,

Answer: B

The baud rate of the standard 10 Mbps Ethernet is 20 megabaud. Ethernet uses Manchester encoding, in which every bit is sent as two signal elements (a high-low or a low-high pair), so the signal changes twice as often as the bits arrive:

\[ 10 \text{ Mbps} \times 2 = 20 \text{ megabaud} \]

10 megabaud in (A) would be the figure if one signal element carried one bit, and 30 and 40 megabaud are not what the code produces. The answer is (B).

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104.

Hamming distance between 100101000110 and 110111101101 is

Exam Year(s)

UGC NET - 2013,

Answer: D

The Hamming distance is the number of positions in which the two words differ:

\[ 100101000110 \]

\[ 110111101101 \]

The bits differ in the 2nd, 5th, 7th, 9th, 11th and 12th places and agree in the other six, so the distance is 6, which is option (D). The values 3, 4 and 5 count too few of the differing places. The distance matters because a code with minimum Hamming distance d can detect up to d - 1 bit errors and correct up to (d - 1)/2 of them, so it is the basic measure of the power of an error-correcting code.

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105.

Given code word 1110001010 is to be transmitted with even parity check bit. The encoded word to be transmitted for this code is

Exam Year(s)

UGC NET - 2013,

Answer: A

The code word 1110001010 has five 1s, which is an odd number. For even parity the parity bit must make the total number of 1s even, so it is 1 and the word transmitted is 11100010101. The answer is (A).

Option (B) adds 0, which leaves five 1s, an odd parity. Option (C) adds nothing and option (D) has dropped the last bit of the code word, so neither includes a check bit appropriate to the original word.

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106.

A file is downloaded in a home computer using a 56 kbps MODEM connected to an Internet Service Provider. If the download of file completes in 2 minutes, what is the maximum size of data downloaded ?

Exam Year(s)

UGC NET - 2013,

Answer: B

The maximum size of the data is the line rate multiplied by the time taken, which is 6.72 Mbits.

\[ 56\ \text{kbps} \times 120\ \text{s} = 6720\ \text{kbits} \]

\[ 6720\ \text{kbits} = 6.72\ \text{Mbits} \]

The value 112 Mbits comes from wrongly using 2 minutes as 2 s, and 67.20 and 672 Mbits are the result of a slip in the decimal place when converting kilobits to megabits. A 56 kbps modem cannot move more than 6.72 Mbits in two minutes, so option B is correct.

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107.

In ______ CSMA protocol, after the station finds the line idle, it sends or refrains from sending based on the outcome of a random number generator.

Exam Year(s)

UGC NET - 2013,

Answer: D

In the p-persistent CSMA protocol, when a station finds the line idle it transmits with probability p, by using a random number generator, and with probability 1 - p it waits for the next slot and checks the line again.

The non-persistent method waits a random time when the line is busy, the 1-persistent method sends at once when the line is idle with probability 1, and a 0-persistent protocol is not a standard method. Only p-persistent uses the outcome of a random number to decide, so option D is correct. This is the middle course between the 1-persistent method, which risks many collisions, and the non-persistent method, which wastes time; the value of p is chosen to balance them.

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108.

What is the maximum length of CAT-5 UTP cable in Fast Ethernet network ?

Exam Year(s)

UGC NET - 2013,

Answer: A

The maximum length of a CAT-5 UTP cable segment in a Fast Ethernet network (100BASE-TX) is 100 metres between the station and the hub or switch.

The values of 200, 1000 and 2000 metres exceed the limit for unshielded twisted pair, since the signal weakens and the collision timing rules of Fast Ethernet are broken at longer distances. Option A is correct. The limit comes from signal attenuation and the timing of collision detection, and longer runs need a repeater, a switch or fibre cable.

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109.

Suppose a file of 10,000 characters is to be sent over a line at 2400 bps. Assume that the data is sent in frames. Each frame consists of 1000 characters and an overhead of 48 bits per frame. Using synchronous transmission, the total overhead time is ______.

Exam Year(s)

UGC NET - 2013,

Answer: C

The total overhead time is 0.2 second. The file of 10,000 characters is sent in frames of 1000 characters, so there are 10 frames, each carrying 48 bits of overhead.

\[ 10 \times 48 = 480 \text{ bits} \]

\[ \text{time} = \frac{480}{2400} = 0.2 \text{ s} \]

The values 0.05 s and 0.1 s would need 120 or 240 bits of overhead, and 2.0 s would need 4800 bits, which does not match 10 frames with 48 bits each.

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110.

Which of the following is a bit rate of an 8-PSK signal having 2500 Hz bandwidth ?

Exam Year(s)

UGC NET - 2013,

Answer: C

The bit rate is 7500 bps. In 8-PSK each signal element carries \(\log_{2} 8 = 3\) bits, and the signal rate (baud rate) is taken to be equal to the bandwidth of 2500 Hz.

\[ \text{bit rate} = 2500 \times 3 = 7500 \text{ bps} \]

The value 2500 bps would be one bit per element, 5000 bps two bits per element (4-PSK), and 20000 bps does not correspond to 3 bits per element at this baud rate, so the key's 7500 bps is the one that fits.

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