Thursday, 13 October 2016

UGC NET/JRF Computer Science December 2013 Paper II - Part 5

41.       Active X controls are Pentium binary programs that can be embedded in ...............
(A) Word pages            (B) URL pages
(C) Script pages           (D) Web pages
Answer: D
42.       Match the following :
List-I                                                                  List - II
a. Wireless Application Environment          i. HTTP      
b. Wireless Transaction Protocol                  ii. IP
c. Wireless Datagram Protocol                      iii. Scripts
d. Wireless                                                        iv. UDP
Codes :
      a     b     c    d
(A) ii     iv     i     iii
(B) iv    iii     ii    i
(C) iv    iii     i     ii
(D) iii    i      iv    ii
Answer: D
43.       Which of the following is widely used inside the telephone system for long-haul data traffic ?
(A) ISDN                        (B) ATM
(C) Frame Relay          (D) ISTN
Answer: B
44.       The document standards for EDI were first developed by large business house during the 1970s and are now under the control of the following standard organisation:
(A) ISO               (B) ANSI
(C) ITU-T           (D) IEEE
Answer: B
45.       Electronic Data Interchange Software consists of the following four layers:
(A) Business application, Internal format conversion, Network translator, EDI envelope
(B) Business application, Internal format conversion, EDI translator, EDI envelope
(C) Application layer, Transport layer, EDI translator, EDI envelope
(D) Application layer, Transport layer, IP layer, EDI envelope
Answer: B

46.       Consider a pre-emptive priority based scheduling algorithm based on dynamically changing priority. Larger priority number implies higher priority. When the process is waiting for CPU in the ready queue (but not yet started execution), its priority changes at a rate a = 2. When it starts running, its priority changes at a rate b = 1. All the processes are assigned priority value 0 when they enter ready queue. Assume that the following processes want to execute :
Process                        Arrival            Service
ID                        Time               Time
P1                       0                      4
P2                       1                      1
P3                       2                      2
P4                       3                      1
The time quantum q = 1. When two processes want to join ready queue simultaneously, the process which has not executed recently is given priority. The finish time of processes P1, P2, P3 and P4 will respectively be
(A) 4, 5, 7 and 8
(B) 8, 2, 7 and 5
(C) 2, 5, 7 and 8
(D) 8, 2, 5 and 7
Answer: B
Explanation:
T    EP   Priority After T
00  P1   P1[0]  P2[-]   P3[-]   P4[-]
01  P2   P1[1]  P2[0]  P3[-]   P4[-]
02  P1   P1[1]  P2[X]  P3[2]  P4[-] ........P2 completed at 2
03  P3   P1[2]  P2[X]  P3[2]  P4[2]
04  P4   P1[3]  P2[X]  P3[3]  P4[2]
05  P1   P1[3]  P2[X]  P3[4]  P4[X] .......P4 completed at 5
06  P3   P1[4]  P2[X]  P3[4]  P4[X]
07  P1   P1[4]  P2[X]  P3[X]  P4[X] .......P3 completed at 7
08  --     P1[X]  P2[X]  P3[X]  P4[X] .......P1 completed at 8
47.       The virtual address generated by a CPU is 32 bits. The Translation Look-aside Buffer (TLB) can hold total 64 page table entries and a 4-way set associative (i.e. with 4- cache lines in the set). The page size is 4 KB. The minimum size of TLB tag is
(A) 12 bits          (B) 15 bits
(C) 16 bits         (D) 20 bits
Answer: C
Explanation:
VirtualAddress = 32 bits
PageSize = 4KB = 12 bits
therefore : VPNTag = 20 bits, OffsetTag = 12 bits
TLBEntryLength = VPNTag = 20 bits
TotalTLBEntries = 64,
4-way implies 64/ 4 = 16 sets = 4 bits
therefore : TLBIndex = 4 bits
TLBTag = TLBEntryLength - TLBIndex = 20 - 4 = 16 bits
48.       Consider a disk queue with request for input/output to block on cylinders
98, 183, 37, 122, 14, 124, 65, 67
in that order. Assume that disk head is initially positioned at cylinder 53 and moving towards cylinder number 0. The total number of head movements using Shortest Seek Time First (SSTF) and SCAN algorithms are respectively
(A) 236 and 252 cylinders
(B) 640 and 236 cylinders
(C) 235 and 640 cylinders
(D) 235 and 252 cylinders
Answer: 236 and 208 cylinders
Explanation:
SSTF
Initial head position =53
The closest queue to initial head position=65
head moves from 53 to 65=12
head moves from 65 to 67=2
head moves from 67 to 37=30
head moves from 37 to 14=23
head moves from 14 to 98=84
head moves from 98 to 122=24
head moves from 122 to 124=2
head moves from 124 to 183=59
Total head movement=236
SCAN
Initial head position=53 and moving towards 0
head moves from 53 to 37=16
head moves from 37 to 14=23
head moves from 14 to 65=51
head moves from 65 to 67=2
head moves from 67 to 98=31
head moves from 98 to 122=24
head moves from 122 to 124=2
head moves from 124 to 183=59
Total head movement=208
49.       How much space will be required to store the bit map of a 1.3 GB disk with 512 bytes block size ?
(A) 332.8 KB     (B) 83.6 KB
(C) 266.2 KB     (D) 256.6 KB
Answer: A
Explanation:
block size=29x23 bits (512 bytes)
disk size=1.3x230x23 bits (1.3 GB)
n=1.3x233/212=1.3x221 bits=1.3x210x28x23 bits=1.3x28 Kilo Bytes=332.8 KB
50.    Linux operating system uses
(A) Affinity Scheduling
(B) Fair Preemptive Scheduling
(C) Hand Shaking
(D) Highest Penalty Ratio Next
Answer: B
Explanation:
Linux uses two process-scheduling algorithms:
1. A time-sharing algorithm for fair preemptive scheduling between multiple processes
2. A real-time algorithm for tasks where absolute priorities are more important than fairness

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