UGCNET Computer Science December 2014 Paper III



Ans .

(A) 157.5 n.sec.


  1. 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





Ans .

(C) Software interrupts


  1. Explanation :





Ans .

(A) 125 chars/second and 0.8 seconds


  1. 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





Ans .

(D) ii iv i iii


  1. Explanation :





Ans .

(B) 07


  1. Explanation :





Ans .

(D) 00, 0, 1, 1


  1. Explanation :





Ans .

(A) I, II and III


  1. Explanation :





Ans .

(D) II and III


  1. Explanation :





Ans .

(B) Check-pointing


  1. Explanation :





Ans .

(B) ABCD, E


  1. Explanation :





Ans .

(D) None of the statements are correct.


  1. Explanation :





Ans .

(A) Every binary relation is never be in BCNF.


  1. Explanation :





Ans .

(D) None of the above


  1. Explanation :





Ans .

(C) i ii iii iv


  1. Explanation :





Ans .

(B) Dithering


  1. Explanation :





Ans .

(A) A'(-1,√2 - 1), B'(-1,2√2 - 1), C'(3/2√2 – 1, 9/2√2 – 1)


  1. Explanation :





Ans .

(B) smily face


  1. Explanation :





Ans .

(D) i ii iii iv


  1. Explanation :





Ans .

(C) S2 and S1 are true.


  1. Explanation :





Ans .

(A) 12


  1. Explanation :





Ans .

(C) MPI_Send and MPI_Recv are non-blocking message passing routines of MPI.


  1. Explanation :





Ans .

(B) L = {an bn | n ≥ 0}


  1. Explanation :





Ans .

(D) L1 is not regular and L2 is not regular.


  1. Explanation :





Ans .

(D) None of the above


  1. Explanation :





Ans .

(D) n(n – 1)/2


  1. Explanation :





Ans .

(B) 320


  1. Explanation :





Ans .

(B) Fragment offset


  1. Explanation :





Ans .

Marks given to all


  1. Explanation :

    48000km/(2.4*108)= 48*106m/(2.4*108)= 20×10-2 =200×10-3s.=200ms





Ans .

(B) HDLC


  1. Explanation :





Ans .

(A) Private key is kept by the receiver and public key is announced to the public.


  1. Explanation :





Ans .

(C) (nlgn)


  1. Explanation :





Ans .

(C) iv ii iii i


  1. Explanation :





Ans .

(D) None of the above


  1. Explanation :





Ans .

(C) Greedy algorithm, Dynamic programming algorithm


  1. Explanation :





Ans .

(B) 10


  1. Explanation :





Ans .

(C) 927, 204,913, 242, 914, 247, 365


  1. Explanation :





Ans .

(B) Synchronized methods


  1. Explanation :





Ans .

(A) smily face


  1. Explanation :





Ans .

(D) Autoboxing


  1. Explanation :





Ans .

(D) All of the above


  1. Explanation :





Ans .

(A) Stereotype is used for extending the UML language.


  1. Explanation :





Ans .

(C) init( )


  1. Explanation :





Ans .

(B) Function point metric


  1. Explanation :





Ans .

(B) 14


  1. Explanation :





Ans .

(B) $ 20,160


  1. Explanation :

    Total resusable components planned=60.
    Custom developed =.3*60=18.
    Total cost in development=18*100*14. Risk exposure= .8*18*1400.=$20160





Ans .

(C) 1


  1. Explanation :





Ans .

(A) Count of the number of components that can call, or pass control, to a component A


  1. Explanation :





Ans .

(D) Cohesion with respect to time


  1. Explanation :





Ans .

(D) Magnetic tapes optical disks magnetic disks electronic disks main memory cache registers


  1. Explanation :





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Ans .

Marks given to all


  1. Explanation :





Ans .

(D) 61


  1. Explanation :

    11->12->9->16->1->34->36.=1+3+7+15+33+2=61





Ans .

(A) P2 P1 P3


  1. Explanation :

    smily face 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





Ans .

(A) It is higher level synchronization primitive and is a collection of procedures,variables, and data structures grouped together in a special package.


  1. Explanation :





Ans .

(B) Logical combination of physical memories on the nodes.


  1. Explanation :





Ans .

(C) x (~F[x])


  1. Explanation :





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.


  1. Explanation :





Ans .

(3) Identical with average revenue


  1. Explanation :





Ans .

(B) iv iii ii i


  1. Explanation :





Ans .

(B) iv iii i ii


  1. Explanation :





Ans .

(B) III and IV


  1. Explanation :





Ans .

(D) LDCF LCF LCS LREC LRE


  1. Explanation :





Ans .

(B) ii iv i iii


  1. Explanation :





Ans .

(B) with | vxy | m, and | vy | 1, such that uvi xyi z L for all i = 0, 1, 2, …….


  1. Explanation :





Ans .

(B) only S2


  1. Explanation :





Ans .

(D) a =1/4, b =1/4, c =1/4, d =1/4


  1. Explanation :





Ans .

(A) smily face


  1. Explanation :





Ans .

(B) Infeasible


  1. Explanation :





Ans .

(C) the solution so obtained is not feasible


  1. Explanation :





Ans .

(C) 13


  1. Explanation :





Ans .

(D) I is true and II is true


  1. Explanation :





Ans .

(3) Identical with average revenue


  1. Explanation :





Ans .

(A) i ii iii iv


  1. Explanation :





Ans .

(A) Windows Vista


  1. Explanation :





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.


  1. Explanation :





Ans .

(B) Parallel computing


  1. Explanation :