Monday, 11 January 2016

CBSE UGC NET Computer Science Paper II Solved December 2015 - Part 3

21.       Which of the following services is not provided by wireless access point in 802.11 WLAN?
(A) Association                        (B) Disassociation
(C) Error Correction     (D) Integration
Answer: C
Explanation:
22.       Which of the following fields in IPv4 datagram is not related to fragmentation?
(A) Type of service                  (B) Fragment offset
(C) Flags                                   (D) Identification
Answer: A
23.       Four channels are multiplexed using TDM. If each channel sends 100 bytes/second and we multiplex 1 byte per channel, then the bit rate for the link is ...............
(A) 400 bps                   (B) 800 bps
(C) 1600 bps                 (D) 3200 bps
Answer: D
Explanation:
The multiplexer is shown in the Figure.
Each frame carries 1 byte from each channel; the size of each frame, therefore, is 4 bytes, or 32 bits. Because each channel is sending 100 bytes/s and a frame carries 1 byte from each channel, the frame rate must be 100 frames per second. The bit rate is 100 × 32 = 3200 bps.
24.       In a typical mobile phone system with hexagonal cells, it is forbidden to reuse a frequency band in adjacent cells. If 840 frequencies are available, how many can be used in a given cell?
(A) 280               (B) 210
(C) 140              (D) 120
Answer: A
Explanation:
Each cell has six other adjacent cells, in a hexagonal grid. If the central cell uses frequency group A, its six adjacent cells can use B, C, B, C, B, and C respectively. In other words, only 3 unique cells are needed. So the answer is 840/3 = 280 frequencies.
25.       Using p=3, q=13, d=7 and e=3 in the RSA algorithm, what is the value of ciphertext for a plain text 5?
(A) 13                 (B) 21
(C) 26                 (D) 33
Answer: Marks to all
Explanation:
p=3, q=13, d=7, e=3, M=5, C=?
C = Me mod n
n = p*q
    = 3*13 = 39
C = 53 mod 39
    = 8
Answer is 8.

26.       A virtual memory has a page size of 1K words. There are eight pages and four blocks. The associative memory page table contains the following entries:
Which of the following list of virtual addresses (in decimal) will not cause any page fault if referenced by the CPU?
(A) 1024, 3072, 4096, 6144               (B) 1234, 4012, 5000, 6200
(C) 1020, 3012, 6120, 8100               (D) 2021, 4050, 5112, 7100
Answer: C
Explanation:
The pages which are not in main memory are:
Page
Address
Address that will cause page fault
1
1K
1024-2047
3
3K
3072-4095
4
4K
4096-5119
6
6K
6144-7167
1020 will not cause page fault (1024-2047)
3012 will not cause page fault (3072-4095)
6120 will not cause page fault (4096-5119)
8100 will not cause page fault (6144-7167)
27.       Suppose that the number of instructions executed between page faults is directly proportional to the number of page frames allocated to a program. If the available memory is doubled, the mean interval between page faults is also doubled. Further, consider that a normal instruction takes one micro second, but if a page fault occurs, it takes 2001 micro seconds. If a program takes 60 sec to run, during which time it gets 15000 page faults, how long would it take to run if twice as much memory were available?
(A) 60 sec                      (B) 30 sec
(C) 45 sec                     (D) 10 sec
Answer: C
Explanation:
T = Ninstr x 1µs + 15,000 x 2,000 µs = 60s
Ninstr x 1µs = 60,000,000 µs – 30,000,000 µs = 30,000,000 µs
Ninstr = 30,000,000
The number of instruction between two page faults is
Ninstr /NPageFaults = 30,000,000/15,000 = 2,000
If the mean interval between page faults is doubled, the number of instruction between two page faults is also doubled and is 4,000. Now the number of page faults is
30,000,000/4,000 = 7,500
T’            = 30,000,000 µs + 7,500 x 2,000 µs
= 30,000,000 µs + 15,000,000 µs = 45,000,000 µs = 45s
Doubling the memory, doesn’t mean that the program runs twice as fast as on the first system. Here, the performance increase is of 25%.
28.       Consider a disk with 16384 bytes per track having a rotation time of 16 msec and average seek time of 40 msec. What is the time in msec to read a block of 1024 bytes from this disk?
(A) 57 msec      (B) 49 msec
(C) 48 msec      (D) 17 msec
Answer: B
Explanation:
Time in msec to read a block of 1024 bytes (Access time or Disk Latency) = seek time + average rotational delay + transfer time
If there are 16384 bytes per track there are 1024/16384 tracks to be read for this block.
Seek time = 40 msec
Rotational delay = 16 msec
Transfer time    = (sectors_read/sectors per rev.) x rotational delay
                            = (1024/16384) x 16 = 1
average rotational delay = rotational delay/2 = 16/2 = 8
access time       = 40 + 8 + 1 = 49
29.       A system has four processes and five allocatable resources. The current allocation and maximum needs are as follows:

Allocated
Maximum
Available
Process A
1 0 2 1 1
1 1 2 1 3
0 0 x 1 1
Process B
2 0 1 1 0
2 2 2 1 0

Process C
1 1 0 1 0
2 1 3 1 0

Process D
1 1 1 1 0
1 1 2 2 1

The smallest value of x for which the above system in safe state is .............
(A) 1                   (B) 3
(C) 2                   (D) 0
Answer: Marks to all
Explanation:
If Process A’s Maximum need is 1 1 2 1 2 instead of 1 1 2 1 3, then answer will be x=1
The needs matrix is as follows:
0 1 0 0 1
0 2 1 0 0
1 0 3 0 0
0 0 1 1 1
If x is 0, available vector will be 0 0 0 1 1, we have a deadlock immediately.
If x is 1, available vector will be 0 0 1 1 1, now, process D can run to completion. When it is finished, the available vector is 1 1 2 2 1.
Now A can run to complete, the available vector then becomes 2 1 4 3 2.
Then C can run and finish, return the available vector as 3 2 4 4 2.
Then B can run to complete. Safe sequence D A C B.
30.    In Unix, the login prompt can be changed by changing the contents of the file ...............
(A) contrab                    (B) init
(C) gettydefs                 (D) inittab
Answer: C


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