Networking Basics Questions and Answers

Q: In a digital transmission, the receiver clock is 0.1 percent faster than the sender clock. How many extra bits per second does the receiver receive if the data rate is 1 Mbps ?

The receiver gets 1000 extra bits per second.A clock that is 0.1 percent faster samples 0.1 percent more bits in every second. At 1 Mbps the number of extra bits is \[ 0.001 \times 1{,}000{,}000 = 1000 \text{ bits per second}. \]10 bps and 100 bps correspond to 0.001 and 0.01 percent, and 10000 bps would be a 1 percent error. A clock error of 0.1 percent is 1 part in 1000, so out of every 1000 bits sent the receiver's clock produces one extra sample. At a rate of one million bits per second that makes a thousand extra bits every second, which would corrupt the data unless the receiver resynchronised.

Q: Ten signals, each requiring 3000 Hz, are multiplexed on to a single channel using FDM. How much minimum bandwidth is required for the multiplexed channel ? Assume that the guard bands are 300 Hz wide.

The multiplexed channel needs 32,700 Hz.Ten signals of 3000 Hz need \(10 \times 3000 = 30{,}000\) Hz, and nine guard bands of 300 Hz are needed between them, adding \(9 \times 300 = 2700\), so the bandwidth is \[ 30{,}000 + 2700 = 32{,}700 \text{ Hz}. \]30,000 leaves out the guard bands and 33,000 uses ten guard bands. In frequency division multiplexing each signal gets its own band and the guard bands keep the neighbouring bands from interfering. The first and the last band need no guard band outside them, which is why there are nine guard bands for ten signals.

Q: A terminal multiplexer has six 1200 bps terminals and 'n' 300 bps terminals connected to it. If the outgoing line is 9600 bps, what is the value of n ?

The value of n is 8.The six 1200 bps terminals use \(6 \times 1200 = 7200\) bps of the 9600 bps line, leaving 2400 bps, and \(2400 / 300 = 8\) terminals of 300 bps can be added.4 and 16 do not fill the remaining capacity and 28 would exceed it. In a statistical or fixed time division multiplexer the sum of the input rates cannot exceed the output line speed. Here 7200 bps plus 8 times 300 bps gives 9600 bps exactly, which uses the line fully, and any more terminals would overload it.

111.

If the data unit is 111111 and the divisor is 1010. In CRC method, what is the dividend at the transmission before division ?

A.1111110000
B.1111111010
C.111111000
D.111111

Exam Year(s)

UGC NET - 2013,

Answer: C

The dividend before division is 111111000. In the CRC method the divisor has 4 bits, so 3 zeros are appended to the data unit.

\[ \text{dividend} = 111111\,000 = 111111000 \]

The option 1111110000 appends four zeros, which would suit a 5-bit divisor. The option 1111111010 appends the divisor itself, and 111111 is the data unit without any zeros appended, so these are wrong.

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112.

Infrared signals can be used for short range communication in a closed area using _______ propagation.

A.ground
B.sky
C.line of sight
D.space

Exam Year(s)

UGC NET - 2014,

Answer: C

Infrared signals use line-of-sight propagation. Infrared light does not pass through walls and does not bend with the earth's surface, so it carries a signal only along a straight, unobstructed path between transmitter and receiver, which suits short-range links inside a closed room such as remote controls and some indoor data links. Ground propagation belongs to low-frequency radio waves that follow the earth, sky propagation to short waves reflected by the ionosphere, and space propagation to satellite links, so none of those three describes infrared communication in a closed area.

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113.

A bridge has access to _______ address in the same network.

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

Exam Year(s)

UGC NET - 2014,

Answer: A

A bridge works at the data link layer and uses the physical (MAC) address of the stations on the networks it joins, learning which address lies on which port and forwarding frames accordingly. It does not look at network-layer or application-layer addresses, which belong to routers and to applications. The choice that names the physical address is therefore the one the key accepts; the network address is used by a router and the application address has no meaning at the level of a bridge. The physical address is the one burned into a network card, and a bridge keeps a table of these addresses to decide where to send each frame, which is why it works without any knowledge of IP addresses.

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114.

The minimum frame length for 10 Mbps Ethernet is _______ bytes and maximum is _______ bytes.

A.64 & 128
B.128 & 1518
C.1518 & 3036
D.64 & 1518

Exam Year(s)

UGC NET - 2014,

Answer: D

For 10 Mbps Ethernet the minimum frame length is 64 bytes and the maximum is 1518 bytes. The minimum size ensures that a frame lasts long enough on the cable for the sender to detect a collision, and the maximum keeps any one station from holding the medium for too long. The other options give a minimum of 128 or 1518 bytes or a maximum of 128 or 3036 bytes, which are not the values fixed in the IEEE 802.3 standard.

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115.

The bit rate of a signal is 3000 bps. If each signal unit carries 6 bits, the baud rate of the signal is _______.

A.500 baud/sec
B.1000 baud/sec
C.3000 baud/sec
D.18000 baud/sec.

Exam Year(s)

UGC NET - 2014,

Answer: A

The baud rate is the bit rate divided by the number of bits carried by each signal unit.

\[ \text{baud rate} = \frac{3000}{6} = 500 \]

So the signal changes 500 times per second, and the baud rate is 500 baud per second. A baud rate of 3000 would apply only if each signal unit carried one bit, and a figure of 18000 comes from multiplying the bit rate by 6, which is the wrong operation. 1000 does not follow from the given figures.

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116.

Match the following :

List - I	List - II
a. Physical layer	i. Allow resources to network access
b. Datalink layer	ii. Move packets from one destination to other
c. Network layer	iii. Process to process message delivery
d. Transport layer	iv. Transmission of bit stream
e. Application Layer	v. Formation of frames

Codes : a b c d e

A.iv v ii iii i
B.v iv i ii iii
C.i iii ii v iv
D.i ii iv iii v

Exam Year(s)

UGC NET - 2014,

Answer: A

The matching is physical layer with transmission of the bit stream (iv), data link layer with formation of frames (v), network layer with moving packets from one destination to another (ii), transport layer with process-to-process message delivery (iii) and application layer with allowing resources to network access (i), the code iv, v, ii, iii, i. These are the functions given in the OSI model. The other codes move a function to a layer above or below its correct one, for instance giving frame formation to the physical layer.

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117.

In a digital transmission, the receiver clock is 0.1 percent faster than the sender clock. How many extra bits per second does the receiver receive if the data rate is 1 Mbps ?

A.10 bps
B.100 bps
C.1000 bps
D.10000 bps

Exam Year(s)

UGC NET - 2014,

Answer: C

The receiver gets 1000 extra bits per second.

A clock that is 0.1 percent faster samples 0.1 percent more bits in every second. At 1 Mbps the number of extra bits is \[ 0.001 \times 1{,}000{,}000 = 1000 \text{ bits per second}. \]

10 bps and 100 bps correspond to 0.001 and 0.01 percent, and 10000 bps would be a 1 percent error. A clock error of 0.1 percent is 1 part in 1000, so out of every 1000 bits sent the receiver's clock produces one extra sample. At a rate of one million bits per second that makes a thousand extra bits every second, which would corrupt the data unless the receiver resynchronised.

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118.

Which one of the following is not a Client-Server application ?

A.Internet chat
B.Web browser
C.E-mail
D.Ping

Exam Year(s)

UGC NET - 2014,

Answer: B

The key marks Web browser as the one that is not a client-server application.

Internet chat, e-mail and ping each involve a client program talking to a service on another machine. The key treats the web browser as only the client program of the web and not a complete client-server application by itself.

Ping sends a simple ICMP echo and is shown as a utility, but the conducting body's key names the web browser as the answer, and this explanation follows it.

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119.

Ten signals, each requiring 3000 Hz, are multiplexed on to a single channel using FDM. How much minimum bandwidth is required for the multiplexed channel ? Assume that the guard bands are 300 Hz wide.

A.30,000
B.32,700
C.33,000
D.None of the above

Exam Year(s)

UGC NET - 2014,

Answer: B

The multiplexed channel needs 32,700 Hz.

Ten signals of 3000 Hz need \(10 \times 3000 = 30{,}000\) Hz, and nine guard bands of 300 Hz are needed between them, adding \(9 \times 300 = 2700\), so the bandwidth is \[ 30{,}000 + 2700 = 32{,}700 \text{ Hz}. \]

30,000 leaves out the guard bands and 33,000 uses ten guard bands. In frequency division multiplexing each signal gets its own band and the guard bands keep the neighbouring bands from interfering. The first and the last band need no guard band outside them, which is why there are nine guard bands for ten signals.

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120.

A terminal multiplexer has six 1200 bps terminals and 'n' 300 bps terminals connected to it. If the outgoing line is 9600 bps, what is the value of n ?

A.4
B.8
C.16
D.28

Exam Year(s)

UGC NET - 2014,

Answer: B

The value of n is 8.

The six 1200 bps terminals use \(6 \times 1200 = 7200\) bps of the 9600 bps line, leaving 2400 bps, and \(2400 / 300 = 8\) terminals of 300 bps can be added.

4 and 16 do not fill the remaining capacity and 28 would exceed it. In a statistical or fixed time division multiplexer the sum of the input rates cannot exceed the output line speed. Here 7200 bps plus 8 times 300 bps gives 9600 bps exactly, which uses the line fully, and any more terminals would overload it.

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