UGC NET Computer Science Solved Mcqs Paper II December 2004 - Part 2

11.       Suppose x and y are two Integer Variables having values 0x5AB6 and 0x61CD respectively. The result (in hex) of applying bitwise operator AND to x and y will be:
(A) 0x5089        (B) 0x4084
(C) 0x78A4       (D) 0x3AD1
Answer: B
0x5AB6=0101 1010 1011 0110
0x61CD=0110 0001 1100 1101
AND     =0100 0000 1000 0100
              = 0x4084
12.       Consider the following statements,
int i=4, j=3, k=0;
k=++i - --j + i++ - --j +j++; What will be the values of i, j and k after the statement.
(A) 7, 2, 8           (B) 5, 2, 10
(C) 6, 2, 8          (D) 4, 2, 8
Answer: C
13.       What is the value of the arithmetic expression (Written in C)
2*3/4-3/4* 2
(A) 0       (B) 1
(C) 1.5    (D) None of the above
Answer: B
14.       A function object:
(A) is an instance of a class for which operator () is a member function.
(B) is an instance of a class for which operator → is a member function.
(C) is a pointer to any function
(D) is a member function of a class
Answer: A
15.       Polymorphism means:
(A) A template function
(B) Runtime type identification within a class hierarchy
(C) Another name for operator overloading
(D) Virtual inheritance
Answer: B
16.       The E-R model is expressed in terms of:
(i) Entities
(ii) The relationship among entities
(iii) The attributes of the entities
(A) (i) and (iii)  
(B) (i), (ii) and (iii)
(C) (ii) and (iii)
(D) None of the above
Answer: B
17.       Specialization is a ............... process.
(A) Top - down             (B) Bottom -Up
(C) Both (A) and (B)    (D) None of the above
Answer: A
18.       The completeness constraint has rules:
(A) Supertype, Subtype
(B) Total specialization, Partial specialization
(C) Specialization, Generalization
(D) All of the above
Answer: B
19.       The entity type on which the ................. type depends is called the identifying owner.
(A) Strong entity           (B) Relationship
(C) Weak entity            (D) E - R
Answer: C
20.    Match the following:
(i) 5 NF              (a) Transitive dependencies eliminated
(ii) 2 NF                         (b) Multivalued attribute removed
(iii) 3 NF            (c) Contains no partial functional dependencies
(iv) 4 NF            (d) Contains no join dependency
(A) i-a, ii-c, iii-b, iv-d    
(B) i-d, ii-c, iii-a, iv-b
(C) i-d, ii-c, iii-b, iv-a   
(D) i-a, ii-b, iii-c, iv-d
Answer: B

Post a Comment