UGCNET Computer Science June 2014 Paper III



Ans .

(C) III only


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Ans .

(D) Butt cap, round cap and projecting square cap


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Ans .

(C) iv iii ii i


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Ans .

(A) (0, 1) and (2, 3)


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Ans .

(D) RGB and CMY color models


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Ans .

(C) 1000 bps


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Ans .

(C) {(1, 1), (2, 1), (3, 0.3), (4, 1), (6, 0.2), (7, 1)}


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Ans .

(D) {(20, 0), (30, 0), (40, 1), (50, 1), (60, 1), (70, 1), (80, 1)}


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Ans .

(A) Error correction learning – learning with a teacher


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Ans .

(B) clarity, elegance, efficiency


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Ans .

(A) ii iii iv i


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Ans .

(C) Format specifiers, White-space characters and Non-white space characters


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Ans .

(A) iv i ii iii


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Ans .

(C) I, III


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Ans .

(B) Both even(L) and Chop(L) are regular.


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Ans .

(D) the process of executing a program with the intent of finding errors.


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Ans .

(B) 10 failures/CPU hr.


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Ans .

(D) 672


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Ans .

(A) ii iv i iii


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Ans .

(C) It refers to a condition that causes a system to fail.


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Ans .

(C) Software product engineering


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Ans .

(C) Book is in 2NF and Collection in 3NF.


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Ans .

(B) An entity type can participate as a subclass in more than one specialization.


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Ans .

(A) iv iii i ii


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Ans .

(D) in both 2NF and 3NF


  1. Explanation :

    Let the R1={S,T}
    R2={u,v}
    S-> T. all the non key are fully functional dependent on key . hence R1 is in 2nf. Not transitive dependency hence 3nf.
    Similarly R2.





Ans .

(B) Web browser


  1. Explanation :

    Ping is not a client server application. Ping is a computer network administration utility used to test the reachability of a host on an Internet Protocol (IP). In ping, there is no server that provides a service.





Ans .

(B) II only


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Ans .

(D) iv iii ii


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Ans .

(C) NLP


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Ans .

(B) Frames


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Ans .

(C) Greedy best first search algorithm


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Ans .

(B) First-order


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Ans .

(C) Alpha-beta pruning procedure


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Ans .

(B) 50 %


  1. Explanation :

    Gantt chart. 1—–6(p0)—-12(idle)—–18(p1)—–24(idle)—–30(p2)—–36(idle)—–42(p3)—– 48(idle)——-54(p4)——-60(idle)——-66(p5).





Ans .

(B) 10 and 7


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Ans .

(B) The number of requests for disk service are not influenced by file allocation method.


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Ans .

(C) Super block


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Ans .

(B) Virtual Memory Management


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Ans .

(C) iv iii i ii


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Ans .

(D) Register Indirect addressing


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Ans .

(A) A B + C * D + EF + G + /


  1. Explanation :

    First, always the expression given within parenthesis is converted first. Since there are 2 expressions with the parenthesis, I am going with the expression (E + F + G) first, although the order does not matter.
    In the expression (E + F + G), there are 3 operands E,F and G and two operators, both being +. Since both the operators are the same, the expression is going to be evaluated from left to right. So E + F is considered first and converted into postfix form which is EF+. So, the expression becomes,
    ( ( A + B ) * C + D) / ([E F +] + G)
    Any expression converted into postfix form is going to be written in square brackets.
    ( ( A + B ) * C + D) / [ E F + G + ].
    Here EF+ is one operand, G is another operand and + is the operator.
    The next expression to be converted into postfix is
    ( A + B). ( [ A B + ] * C + D) / [ E F + G + ]
    Now, the expression which is enclosed in parenthesis is evaluated and so,
    we get ( [ [ A B + ] C * ] + D) / [ E F + G + ]
    [ A B + C * D + ] / [ E F + G + ]
    [ A B + C * D + ] [ E F + G + ] /
    So, the final postfix expression is A B + C * D + E F + G + /.





Ans .

(C) both present and past input


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Ans .

(D) 3m + n + 30 bits


  1. Explanation :

    Memory capacity is 2m Kbytes= 2m+10bytes. Whole memory can be represented by m+10 bits.
    2^n operations can be represented by n bits.
    Now instruction of form
    Instruction op1,op2,op3. Will take 3m+30+n bits





Ans .

(C) Master-slave JK flip-flop


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Ans .

(C) microprocessor does not contain I/O devices.


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Ans .

(C) 1 3 5 7 9


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Ans .

(A) win.txt


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Ans .

(D) All of the above


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Ans .

(C) collection of discrete objects that incorporates both data structure and behaviour.


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Ans .

(B) Aggregation


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Ans .

(A) Small talk


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Ans .

(D) iv i ii iii


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Ans .

(C) CaptureEvents(eventType)


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Ans .

(C) HREF & NAME


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Ans .

(B) one Megabyte


  1. Explanation :

    Each pixel needs to store 1 Byte( 8 bit)
    Total 1024 x 1024 pixels
    Total storage space = 1024 x 1024 x 1 = 2 10 x 2 10 x 1 = 2 20 = 1 MB





Ans .

(C) iii iv ii i


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Ans .

Marks to all


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Ans .

(A) Alternate optimality


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Ans .

(B) subtracting each entry in the table from the maximum value in that table.


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Ans .

(A) 76


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Ans .

(B) E1 is correct and E2 is not correct.


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Ans .

(B) 38


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Ans .

(C) O(n)


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Ans .

(A) less than 1


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Ans .

(B) S1 is NP problem and S2 is NP problem.


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Ans .

(C) smily face


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Ans .

(B) 32,700


  1. Explanation :

    9 guard band occupies 2700 hz.
    10 channels occupies 30000 hz.
    So total bandwidth is 32700.





Ans .

(B) 8


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Ans .

(D) All of the above


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Ans .

(A) Transport, Session, Presentation, Application


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Ans .

(C) 4096


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Ans .

(C) 15


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Ans .

(D) G1 is ambiguous and G2 is ambiguous


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Ans .

(D) Both L1 and L2 are deterministic


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