PREVIOUS YEAR SOLVED PAPERS - December 2015 Paper 3

Avatto > > UGC NET COMPUTER SCIENCE > > PREVIOUS YEAR SOLVED PAPERS > > December 2015 Paper 3

11. A network with bandwidth of 10 Mbps can pass only an average of 15,000 frames per minute with each frame carrying an average of 8,000 bits. What is the throughput of this network?

  • Option : A
  • Explanation :
    A network with 10 Mbps bandwidth can pass 15,000 frames and each frame carries 8,000 bit i.e. 120,000,000 bits per minute. i.e. 120,000,000 / 60 = 2,000,000 = 2 Mbps. So, option (A) is correct.
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12. Consider a subnet with 720 routers. If a three-level hierarchy is chosen with eight clusters, each containing 9 regions of 10 routers, then the total number of entries in the routing table is __________.

  • Option : A
  • Explanation : Eight clusters, each containing 9 regions of 10 routers is a 3 level hierarchy. Each router needs 10 entries for the local router, 8 entries for routing the other regions, and for distant clusters it needs 7 entries. i.e. total 10 + 8 + 7 = 25 entries. So, option (A) is correct.
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13. In classful addressing, the IP addresses with 0 (zero) as network number :

  • Option : A
  • Explanation : In classful addressing, the IP addresses with 0 (zero) as network number refers to the current network. 
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14. In electronic mail, which of the following protocols allows the transfer of multimedia messages?

  • Option : D
  • Explanation :
  • Internet Message Access Protocol (IMAP) is used to retrieve emails from mail server.
  • Simple mail transfer protocol(SMTP) is a Internet standard for email communication.
  • Multipurpose Internet Mail Extensions (MIME) protocol allows the transfer of multimedia messages.
  • POP3 (Post Office Protocol 3) is the most recent version of a standard protocol for receiving e-mail.
  • So, option (D) is correct.
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15. A device is sending out data at the rate of 2000 bps. How long does it take to send a file of 1,00,000 characters ?

  • Option : C
  • Explanation :
    File of 100,000 character will take 8 * 100,000 = 800,000 bits. Device sends the data @ 2000 bps. 2000 bits take 1 sec. 1 bit will take 1 / 2000 second. 800,000 bits will take 800,000 * (1 / 2000) = 400 sec. So, option (C) is correct.
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December 2015 Paper 3