JUNE 2012 - PAPER II

14. Which of the following TCP/IP Internet protocol is diskless machine uses to obtain its IP address from a server ?

(A) RAP
(B) RIP
(C) ARP
(D) X.25

Ans:-C

Explanation:- ARP is a protocol in the TCP/IP protocol suite. This protocol is required for basic TCP/IP operations. This protocol is used to find the ethernet(hardware) address from a specific IP number.

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15. 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

Ans:-C

Explanation:- Not a very complicated question. Decryption and encryption of data happens in the presentation layer. One more important activity which happens at this layer is the Compression and decompression of data.

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16. 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

Ans:-B

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17. In which Routing Method do all the routers have a common database ?

(A) Distance vector
(B) Link state
(C) Link vector
(D) Dijkestra method

Ans:-B

Explanation:- There are two major classes of routing methods.One is Distance vector and the other one Link state. In distance vector, each router knows only its neighbors. Decision is taken on two factors one is distance and the other vector, that is the direction to reach. In case of link state all the routers know about the paths reachable by all other routers in the network. All routers possess a synchronized copy of the area's link-state database. So the answer for this question would be option B.

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18. 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

Ans:-A

Explanation:- Inorder to know why option A is the answer, let us understand the following table.

MODULATION	BAUD RATE	BIT RATE
4-PSK	N	2N
8-PSK	N	3N
16-PSK	N	4N

In the question it is mentioned as page shift keying, but it is actually Phase Shift Keying. Two values are provided. One is bps which is nothing but Bits per second which is 9600 and also level which is 16. According to the table given above, for a 16-PSK, given the baud rate N, the bit rate would be 4N. Here bps is provides which is 9600. In order to calculate the baud rate we have to divide bps/4, because the signal level is 16. So the answer would be 9600/4=2400 baud. That is how the answer is arrived at.