UGCNET Computer Science December 2014 Paper III
(A) 157.5 n.sec.
(B) 110 n.sec.
(C) 75 n.sec.
(D) 82.5 n.sec.
View Answer
Ans .
(A) 157.5 n.sec.
Explanation : The average access time of the system for memory read cycle is 0.8x50+0.2x350=110ns.
The average access time of the system for both read and writes request is 0.75x110+0.25x300=82.5+75=157.5
(A) Internal interrupts
(B) External interrupts
(C) Software interrupts
(D) None of the above
View Answer
Ans .
(C) Software interrupts
Explanation :
(A) 125 chars/second and 0.8 seconds
(B) 250 chars/second and 0.6 seconds
(C) 166 chars/second and 0.8 seconds
(D) 250 chars/second and 0.4 seconds
View Answer
Ans .
(A) 125 chars/second and 0.8 seconds
Explanation : Time to print a character = 6+2=8 m.sec. Number of characters in a line = 200 Time to print a character line = 200x8=1600 m.sec. = 1.6 seconds Printing speed in characters per second = 200/1.6 = 125 chars/second
(A) iv i iii ii
(B) iii ii i iv
(C) ii iii i iv
(D) ii iv i iii
View Answer
Ans .
(D) ii iv i iii
Explanation :
(A) 05
(B) 07
(C) 09
(D) 00
View Answer
Ans .
(B) 07
Explanation :
(A) 11, 1, 1, 1
(B) 10, 0, 1, 0
(C) 01, 1, 0, 0
(D) 00, 0, 1, 1
View Answer
Ans .
(D) 00, 0, 1, 1
Explanation :
(A) I, II and III
(B) I and II only
(C) II and III only
(D) II only
View Answer
Ans .
(A) I, II and III
Explanation :
(A) I and III
(B) II and IV
(C) I and IV
(D) II and III
View Answer
Ans .
(D) II and III
Explanation :
(A) Write-ahead log
(B) Check-pointing
(C) Log buffer
(D) Thomas
View Answer
Ans .
(B) Check-pointing
Explanation :
(A) ABCD,
(B) ABCD, E
(C) ,
(D) ABC, E
View Answer
Ans .
(B) ABCD, E
Explanation :
(A) I
(B) II
(C) Both are correct
(D) None of the statements are correct.
View Answer
Ans .
(D) None of the statements are correct.
Explanation :
(A) Every binary relation is never be in BCNF.
(B) Every BCNF relation is in 3NF.
(C) 1 NF, 2 NF, 3 NF and BCNF are based on functional dependencies.
(D) Multivalued Dependency (MVD) is a special case of Join Dependency (JD).
View Answer
Ans .
(A) Every binary relation is never be in BCNF.
Explanation :
(A) Graphical languages
(B) General-purpose languages
(C) Linear-list notations
(D) None of the above
View Answer
Ans .
(D) None of the above
Explanation :
(A) ii i iv iii
(B) i iv iii ii
(C) i ii iii iv
(D) iv iii ii i
View Answer
Ans .
(C) i ii iii iv
Explanation :
(A) Halftoning
(B) Dithering
(C) Error diffusion
(D) None of the above
View Answer
Ans .
(B) Dithering
Explanation :
(A) A'(-1,√2 - 1), B'(-1,2√2 - 1), C'(3/2√2 – 1, 9/2√2 – 1)
(B) A'(√2 - 1, -1), B'(2√2 - 1, -1), C'(3/2√2 – 1, 9/2√2 – 1)
(C) A'(-1, √2 - 1), B'(2√2 - 1, -1), C'(3/2√2 – 1, 9/2√2 – 1)
(D) A'(-1, √2 - 1), B'(2√2 - 1, -1), C'(9/2√2 – 1, 3/2√2 – 1)
View Answer
Ans .
(A) A'(-1,√2 - 1), B'(-1,2√2 - 1), C'(3/2√2 – 1, 9/2√2 – 1)
Explanation :
View Answer
Ans .
(B)
Explanation :
(A) i iii iv ii
(B) ii iii i iv
(C) iv ii iii i
(D) i ii iii iv
View Answer
Ans .
(D) i ii iii iv
Explanation :
(A) S3 and S2 are true.
(B) S3 and S1 are true.
(C) S2 and S1 are true.
(D) S1, S2, S3 are true.
View Answer
Ans .
(C) S2 and S1 are true.
Explanation :
(A) 12
(B) 11
(C) 10
(D) 07
View Answer
Ans .
(A) 12
Explanation :
(A) MPI_Isend and MPI_Irecv are non-blocking message passing routines of MPI.
(B) MPI_Issend and MPI_Ibsend are non-blocking message passing routines of MPI.
(C) MPI_Send and MPI_Recv are non-blocking message passing routines of MPI.
(D) MPI_Ssend and MPI_Bsend are blocking message passing routines of MPI.
View Answer
Ans .
(C) MPI_Send and MPI_Recv are non-blocking message passing routines of MPI.
Explanation :
(A) L = {an bm | n, m ≥ 0}
(B) L = {an bn | n ≥ 0}
(C) L = {an bm | n, m > 0}
(D) L = {an bn | n > 0}
View Answer
Ans .
(B) L = {an bn | n ≥ 0}
Explanation :
(A) L1 is regular and L2 is not regular.
(B) L1 is not regular and L2 is regular.
(C) L1 is regular and L2 is regular.
(D) L1 is not regular and L2 is not regular.
View Answer
Ans .
(D) L1 is not regular and L2 is not regular.
Explanation :
(A) (a + b)* ba(a + b)*
(B) a* bbbbb*
(C) ( + a + aa + aaa)b* + (a + b)* ba(a + b)*
(D) None of the above
View Answer
Ans .
(D) None of the above
Explanation :
(A) n(n + 1)
(B) n (n – 1)
(C) n(n + 1)/2
(D) n(n – 1)/2
View Answer
Ans .
(D) n(n – 1)/2
Explanation :
(A) 300
(B) 320
(C) 360
(D) 400
View Answer
Ans .
(B) 320
Explanation :
(A) Sequence number
(B) Fragment offset
(C) Checksum
(D) Window size
View Answer
Ans .
(B) Fragment offset
Explanation :
(A) 0.5 ms
(B) 20 ms
(C) 50 ms
(D) 200 ms
View Answer
Ans .
Marks given to all
Explanation : 48000km/(2.4*108)= 48*106m/(2.4*108)= 20×10-2 =200×10-3s.=200ms
(A) Stop-and-wait
(B) HDLC
(C) Sliding window
(D) Go-back-N
View Answer
Ans .
(B) HDLC
Explanation :
(A) Private key is kept by the receiver and public key is announced to the public.
(B) Public key is kept by the receiver and private key is announced to the public.
(C) Both private key and public key are kept by the receiver.
(D) Both private key and public key are announced to the public.
View Answer
Ans .
(A) Private key is kept by the receiver and public key is announced to the public.
Explanation :
(A) (n)
(B) (lgn)
(C) (nlgn)
(D) (n2)
View Answer
Ans .
(C) (nlgn)
Explanation :
(A) iv ii i iii
(B) ii iv i iii
(C) iv ii iii i
(D) iii ii iv i
View Answer
Ans .
(C) iv ii iii i
Explanation :
(A) CLIQUE P 3-CNF_SAT
(B) CLIQUE P VERTEX_COVER
(C) CLIQUE P SUBSET_SUM
(D) None of the above
View Answer
Ans .
(D) None of the above
Explanation :
(A) Greedy algorithm, Divide-conquer algorithm
(B) Divide-conquer algorithm, Greedy algorithm
(C) Greedy algorithm, Dynamic programming algorithm
(D) Dynamic programming algorithm, Greedy algorithm
View Answer
Ans .
(C) Greedy algorithm, Dynamic programming algorithm
Explanation :
(A) 5
(B) 10
(C) 20
(D) 100
View Answer
Ans .
(B) 10
Explanation :
(A) 4, 254, 403, 400, 332, 346, 399, 365
(B) 926, 222, 913, 246, 900, 260, 364, 365
(C) 927, 204,913, 242, 914, 247, 365
(D) 4, 401, 389, 221, 268, 384, 383, 280, 365
View Answer
Ans .
(C) 927, 204,913, 242, 914, 247, 365
Explanation :
(A) Asynchronized methods
(B) Synchronized methods
(C) Serialized methods
(D) None of the above
View Answer
Ans .
(B) Synchronized methods
Explanation :
View Answer
Ans .
(A)
Explanation :
(A) Boxing
(B) Wrapping
(C) Instantiation
(D) Autoboxing
View Answer
Ans .
(D) Autoboxing
Explanation :
(A) Using document object
(B) Registering appropriate event handlers
(C) Using element object
(D) All of the above
View Answer
Ans .
(D) All of the above
Explanation :
(A) Stereotype is used for extending the UML language.
(B) Stereotyped class must be abstract
(C) The stereotype indicates that the UML element cannot be changed
(D) UML profiles can be stereotyped for backward compatibility
View Answer
Ans .
(A) Stereotype is used for extending the UML language.
Explanation :
(A) start( )
(B) run( )
(C) init( )
(D) begin( )
View Answer
Ans .
(C) init( )
Explanation :
(A) Halstead metric
(B) Function point metric
(C) Complexity metric
(D) Length metric
View Answer
Ans .
(B) Function point metric
Explanation :
(A) 12
(B) 14
(C) 16
(D) 18
View Answer
Ans .
(B) 14
Explanation :
(A) $ 25,200
(B) $ 20,160
(C) $ 17,640
(D) $ 15,120
View Answer
Ans .
(B) $ 20,160
Explanation : Total resusable components planned=60. Custom developed =.3*60=18. Total cost in development=18*100*14. Risk exposure= .8*18*1400.=$20160
(A) 100
(B) 10
(C) 1
(D) 0
View Answer
Ans .
(C) 1
Explanation :
(A) Count of the number of components that can call, or pass control, to a component A
(B) Number of components related to component A
(C) Number of components dependent on component A
(D) None of the above
View Answer
Ans .
(A) Count of the number of components that can call, or pass control, to a component A
Explanation :
(A) Coincidental cohesion
(B) Cohesion between temporary variables
(C) Cohesion between local variables
(D) Cohesion with respect to time
View Answer
Ans .
(D) Cohesion with respect to time
Explanation :
(A) Magnetic tapes magnetic disks optical disks electronic disks main memory cache registers
(B) Magnetic tapes magnetic disks electronic disks optical disks main memory cache registers
(C) Magnetic tapes electronic disks magnetic disks optical disks main memory cache registers
(D) Magnetic tapes optical disks magnetic disks electronic disks main memory cache registers
View Answer
Ans .
(D) Magnetic tapes optical disks magnetic disks electronic disks main memory cache registers
Explanation :
(A) 1024 blocks
(B) 16794 blocks
(C) 20000 blocks
(D) 1048576 blocks
View Answer
ok
Ans .
Marks given to all
Explanation :
(A) 111
(B) 112
(C) 60
(D) 61
View Answer
Ans .
(D) 61
Explanation : 11->12->9->16->1->34->36.=1+3+7+15+33+2=61
(A) P2 P1 P3
(B) P2 P3 P1
(C) P1 P2 P3
(D) P1 P3 P2
View Answer
Ans .
(A) P2 P1 P3
Explanation :
Total tapes 13. Allocated 11. Balance 2.
So first 2 units are given to p2.
Then after completion p2 returns all 5.
Now 5 units are catered to p1. Now p1 returns 11 units.
Which can easily be given to p3. Hence order is p2,p1,p3
(A) It is higher level synchronization primitive and is a collection of procedures,variables, and data structures grouped together in a special package.
(B) It is a non-negative integer which apart from initialization can be acted upon by wait and signal operations.
(C) It uses two primitives, send and receive which are system calls rather than language constructs.
(D) It consists of the IPC primitives implemented as system calls to block the process when they are not allowed to enter critical region to save CPU time.
View Answer
Ans .
(A) It is higher level synchronization primitive and is a collection of procedures,variables, and data structures grouped together in a special package.
Explanation :
(A) Logical combination of virtual memories on the nodes.
(B) Logical combination of physical memories on the nodes.
(C) Logical combination of the secondary memories on all the nodes.
(D) All of the above
View Answer
Ans .
(B) Logical combination of physical memories on the nodes.
Explanation :
(A) x (~F[x])
(B) ~( x) F[x]
(C) x (~F[x])
(D) x F[x]
View Answer
Ans .
(C) x (~F[x])
Explanation :
(A) is like a depth-first search where most promising child is selected for expansion
(B) generates all successor nodes and computes an estimate of distance (cost) from start node to a goal node through each of the successors. It then chooses the successor with shortest cost.
(C) saves all path lengths (costs) from start node to all generated nodes and chooses shortest path for further expansion.
(D) none of the above
View Answer
Ans .
(B) generates all successor nodes and computes an estimate of distance (cost) from start node to a goal node through each of the successors. It then chooses the successor with shortest cost.
Explanation :
(A) A B
(B) C D
(C) A C
(D) A D
View Answer
Ans .
(3) Identical with average revenue
Explanation :
(A) iv ii i iii
(B) iv iii ii i
(C) ii iii iv i
(D) i iii iv ii
View Answer
Ans .
(B) iv iii ii i
Explanation :
(A) i iii iv ii
(B) iv iii i ii
(C) i iii ii iv
(D) iv i iii ii
View Answer
Ans .
(B) iv iii i ii
Explanation :
(A) I and II
(B) III and IV
(C) II and III
(D) II and IV
View Answer
Ans .
(B) III and IV
Explanation :
(A) LCF LDCF LCS LRE LREC
(B) LCF LDCF LCS LREC LRE
(C) LDCF LCF LCS LRE LREC
(D) LDCF LCF LCS LREC LRE
View Answer
Ans .
(D) LDCF LCF LCS LREC LRE
Explanation :
(A) ii iv iii i
(B) ii iv i iii
(C) iv i ii iii
(D) i iv iii ii
View Answer
Ans .
(B) ii iv i iii
Explanation :
(A) with | vxy | m such that uvi xyi z L for all i = 0, 1, 2
(B) with | vxy | m, and | vy | 1, such that uvi xyi z L for all i = 0, 1, 2, …….
(C) with | vxy | m, and | vy | 1, such that uvi xyi z L for all i = 0, 1, 2, …….
(D) with | vxy | m, and | vy | 1, such that uvi xyi z L for all i = 0, 1, 2,…….
View Answer
Ans .
(B) with | vxy | m, and | vy | 1, such that uvi xyi z L for all i = 0, 1, 2, …….
Explanation :
(A) only S1
(B) only S2
(C) Both S1 and S2
(D) None of these
View Answer
Ans .
(B) only S2
Explanation :
(A) a = 0, b = 0, c = 0, d = 1
(B) a =1/2, b =1/2, c = 0, d = 0
(C) a =1/3, b =1/3, c =1/3, d = 0
(D) a =1/4, b =1/4, c =1/4, d =1/4
View Answer
Ans .
(D) a =1/4, b =1/4, c =1/4, d =1/4
Explanation :
View Answer
Ans .
(A)
Explanation :
(A) Optimum
(B) Infeasible
(C) Unbounded
(D) Degenerate
View Answer
Ans .
(B) Infeasible
Explanation :
(A) total supply equals total demand
(B) total supply does not equal total demand
(C) the solution so obtained is not feasible
(D) none of these
View Answer
Ans .
(C) the solution so obtained is not feasible
Explanation :
(A) 5
(B) 11
(C) 13
(D) 15
View Answer
Ans .
(C) 13
Explanation :
(A) I is false and II is false.
(B) I is true and II is false.
(C) I is false and II is true.
(D) I is true and II is true
View Answer
Ans .
(D) I is true and II is true
Explanation :
(A) core
(B) support
(C) crossover
(D) cut
View Answer
Ans .
(3) Identical with average revenue
Explanation :
(A) i ii iii iv
(B) i iii ii iv
(C) ii iv iii i
(D) ii iii iv i
View Answer
Ans .
(A) i ii iii iv
Explanation :
(A) Windows Vista
(B) Windows 2000
(C) Windows NT
(D) Windows 98
View Answer
Ans .
(A) Windows Vista
Explanation :
(A) A program that performs a legitimate function that is known to an operating system or its user and also has a hidden component that can be used for nefarious purposes like attacks on message security or impersonation.
(B) A piece of code that can attach itself to other programs in the system and spread to other systems when programs are copied or transferred.
(C) A program that spreads to other computer systems by exploiting security holes like weaknesses in facilities for creation of remote processes
(D) All of the above
View Answer
Ans .
(A) A program that performs a legitimate function that is known to an operating system or its user and also has a hidden component that can be used for nefarious purposes like attacks on message security or impersonation.
Explanation :
(A) Cloud computing
(B) Parallel computing
(C) Cluster computing
(D) Peer-to-peer computing
View Answer
Ans .
(B) Parallel computing
Explanation :