UGCNET Computer Science December 2013 Paper III



Ans .

(C) no feasible solution at all


  1. Explanation :





Ans .

(D) All of these


  1. Explanation :





Ans .

(B) No feasible solution


  1. Explanation :





Ans .

(B) 5, 4


  1. Explanation :





Ans .

(A) UFP*CAF


  1. Explanation :

    UFP (Unadjusted Function Point)
    CAF (Complexity Adjustment Factor)
    Function Point FP = UFP*CAF





Ans .

(B) ii iii i iv


  1. Explanation :





Ans .

(B) Software Configuration Management Process


  1. Explanation :





Ans .

(B) they do not explore combinations of input circumstances.


  1. Explanation :





Ans .

(D) If an organization does not understand how to control software projects internally, it will invariably struggle when it outsources software projects.


  1. Explanation :





Ans .

(A) iv i ii iii


  1. Explanation :





Ans .

(A) iv iii ii i


  1. Explanation :





Ans .

(C) Simple Object Access Protocol (SOAP)


  1. Explanation :





Ans .

(C) Style sheet


  1. Explanation :





Ans .

(B) Dynamic loading


  1. Explanation :





Ans .

(C) Ad-hoc polymorphism


  1. Explanation :





Ans .

(A) iii iv i ii


  1. Explanation :





Ans .

(A) Type Promotion


  1. Explanation :





Ans .

(B) One defined by C and one specific to C++.


  1. Explanation :





Ans .

(D) Allocation of memory and frees the memory previously allocated.


  1. Explanation :





Ans .

(A) iv i ii iii


  1. Explanation :





Ans .

(A) S1 is correct and S2 is not correct.


  1. Explanation :





Ans .

(D) 00


  1. Explanation :

    T flip flop toggles the state if input T is 1, and does not change state if input T is 0.
    T flip flop is a clocked (synchronous or edge-triggered) flip flop.
    Edge triggered flip flops can only change its values at the edge of a clock,
    in absence of edge does not change its state, remains same.
    In given circuit, clock symbol is showing that it is positive edge triggering (low to high change).
    Now, we need to find out Q0Q1 after 4 clock cycles. Initial: 00 States Q0 and Q1 are zero
    Clock 1: 11 State Q0 changed to 1 because input T is 1 and previous state was 0 and State Q1 changed to 1 because input T is 1 and previous state was 0.
    Clock 2: 01 State Q0 changed to 0 because input T is 1 and previous state was 1 and State Q1 remains same because clk value is 0.
    Clock 3: 10 State Q0 changed to 1 because input T is 1 and previous state was 0 and State Q1 changed to 0 because input T is 1 and previous state was 1.
    Clock 4: 00 State Q0 changed to 0 because input T is 1 and previous state was 1 and State Q1 remains same because clk value is 0.





Ans .

(D) Pragmatic


  1. Explanation :





Ans .

(B) unification, substitution


  1. Explanation :





Ans .

(C) h* underestimates h


  1. Explanation :





Ans .

(C) operator subgoaling


  1. Explanation :





Ans .

(B) beta, alpha


  1. Explanation :





Ans .

(D) {0.1, 0.5, 0.2, 0.1, 0.8}


  1. Explanation :

    μA∩B(x) = min [μA(x), μB(x)]





Ans .

(C) h(A) = 1


  1. Explanation :





Ans .

(D) Σ wi . Σ xi


  1. Explanation :





Ans .

(B) Data Redundancy


  1. Explanation :





Ans .

(C) The AES (Advanced Encryption Standard) cryptosystem allows variable key lengths of size 56 bits and 124 bits.


  1. Explanation :





Ans .

(D) 11001001011


  1. Explanation :





Ans .

(C) floor(3n/2) - 2


  1. Explanation :





Ans .

(B) O(n2.81)


  1. Explanation :





Ans .

Marks given to all


  1. Explanation :

    Master Method is a direct way to get the solution.
    The master method works only for following type of recurrences
    T(n) = aT(n/b) + f(n)
    highest degree of f(n) is c.
    There are following three cases:
    T(n)= O(nc) where a < b^c eq 1.
    T(n)= O(nc log n) where bc=a. eq 2.
    T(n)=O(nlogba) where a> bc. eq 3
    In our case T(n)=mT(n/2)tan2 value of a is m, not known. so we cannot say a= 2^C Or Not . We can not say which equation will follow.





Ans .

(C) 4


  1. Explanation :





Ans .

(D) O(n+m)


  1. Explanation :





Ans .

(D) maximal, minimal


  1. Explanation :





Ans .

(B) ≤m, ≥1


  1. Explanation :





Ans .

(C) S→a SB|b, B→b


  1. Explanation :





Ans .

(D) S1 is correct and S2 is correct.


  1. Explanation :





Ans .

(C) 153.6 Mbps


  1. Explanation :

    The number of bits/sec is just 800x600x40x8 or 153.6 Mbps.





Ans .

(C) Class C


  1. Explanation :





Ans .

(C) Transport


  1. Explanation :





Ans .

(D) 533.33 m sec


  1. Explanation :

    The request has to go up and down, and the response has to go up and down. The total path length traversed is thus 160.000 km. The speed of light in air and vacuum is 300,000 km/sec, so the propagation delay alone is 160,000/300,000 sec or about 533 msec.





Ans .

(C) synchronization


  1. Explanation :





Ans .

(A) Attenuation


  1. Explanation :





Ans .

(C) iii iv i ii


  1. Explanation :





Ans .

(D) All of these


  1. Explanation :





Ans .

A and D


  1. Explanation :





Ans .

(D) None of these


  1. Explanation :

    Data naturally needs to flow in and out in serial form.





Ans .

(C) x NOR y


  1. Explanation :





Ans .

(B) traps are synchronous and interrupts are asynchronous with the program


  1. Explanation :





Ans .

(A) 2


  1. Explanation :





Ans .

(A) πcustomer-name(σbalance >10000(Deposit))


  1. Explanation :





Ans .

(C) Conceptual model is not accurate representation of "Universe of interest".


  1. Explanation :





Ans .

(B) Q.2


  1. Explanation :

    First query gives stuid of students whose marks are greater than all students taking course 5. Second query gives stuid of students whose marks are greater than any student taking course 5, that is in second query comparison is between min of marks by student enrolled in course 5 and in query one comparison is between maximum of marks by any student in course 5. So the Answer is option D.





Ans .

(D) iv iii i ii


  1. Explanation :





Ans .

(D) ii iv i iii


  1. Explanation :





Ans .

(D) III only


  1. Explanation :





Ans .

(D) All the above.


  1. Explanation :





Ans .

(C) Animation


  1. Explanation :





Ans .

(D) (0.5, 0)


  1. Explanation :

    If (0.5, 0) is substituted in the equation, we get 1.5 which is less than 2, then it lies on the same side as that of the origin.





Ans .

(C)smily face


  1. Explanation :





Ans .

Marks given to all


  1. Explanation :





Ans .

(C) test $a=$b


  1. Explanation :

    Tests for a zero-length string: test -z string
    Tests for a nonzero string length: test -n string
    Tests two strings for equality: test string1 = string2
    Tests two strings for inequality: test string1 != string2
    Tests for a nonzero string length: test string





Ans .

(B) Hardware abstraction layer


  1. Explanation :





Ans .

(D) OLE


  1. Explanation :





Ans .

(D) an associative memory in which item to be searched is compared with all the keys simultaneously.


  1. Explanation :





Ans .

(D) Kill one of the processes


  1. Explanation :





Ans .

(C) Directed acyclic graph


  1. Explanation :





Ans .

(A) Process 0 is called init process.


  1. Explanation :





Ans .

(D) Sleep 1 and echo $!


  1. Explanation :





Ans .

(B) Ready, standby, running, waiting, transition and terminated.


  1. Explanation :