UGC-NET Computer Science Previous Year Question Papers

UGC-NET NOVEMBER 2017 QUESTIONS WITH EXPLANATION 

CBSE-NET NOVEMBER 2017 QUESTIONS WITH EXPLANATION 

NTA-NET NOVEMBER 2017 QUESTIONS WITH EXPLANATION 

Q:10 Consider a relation R (A, B, C, D, E, F, G, H), where each attribute is atomic, and following functional dependencies exist.

CH → G
A → BC
B → CFH
E → A
F → EG

The relation R is __________ .
(A) in 1NF but not in 2NF
(B) in 2NF but not in 3NF
(C) in 3NF but not in BCNF
(D) in BCNF


Answer: (A)

Explanation: If we find closure of A:

A+ → All atribute except D.
Similarly for other keys we can find closure, but D can’t be derived from any key and it must be added to all keys to be derived from. That’s why this relation is in 1NF, since there is partial dependency so, this relation is not in 2NF.
So, option (A) is correct.

UGC-NET Computer Science Previous Year Question Papers

 UGC-NET NOVEMBER 2017 QUESTIONS WITH EXPLANATION 

CBSE-NET NOVEMBER 2017 QUESTIONS WITH EXPLANATION 

NTA-NET NOVEMBER 2017 QUESTIONS WITH EXPLANATION

Q: 9 If every non-key attribute is functionally dependent on the primary key, then the relation is in __________ .

(A) First normal form
(B) Second normal form
(C) Third normal form
(D) Fourth normal form


Answer: (B)

Explanation: Conditions for various normal forms:

1. 1 NF – A relation R is in first normal form (1NF) if and only if all underlying domains contain atomic values only.
2. 2 NF – A relation R is in second normal form (2NF) if and only if it is in 1NF and every non-key attribute is fully dependent on the primary key.
3. 3 NF – A relation R is in third normal form (3NF) if and only if it is in 2NF and every non-key attribute is non-transitively dependent on the primary key.
4. BCNF – A relation R is in Boyce-Codd normal form (BCNF) if and only if every determinant is a candidate key.

 

UGC-NET Computer Science Previous Year Question Papers

  UGC-NET NOVEMBER 2017 QUESTIONS WITH EXPLANATION 

CBSE-NET NOVEMBER 2017 QUESTIONS WITH EXPLANATION 

NTA-NET NOVEMBER 2017 QUESTIONS WITH EXPLANATION 

GATE 2016 SET-2 Q:32

Topic: DBMS

Q: 8 Suppose a database schedule S involves transactions T1, T2, ………….,Tn. Consider the precedence graph of S with vertices representing the transactions and edges representing the conflicts. If S is serializable, which one of the following orderings of the vertices of the precedence graph is guaranteed to yield a serial schedule ?

(A) Topological order
(B) Depth – first order
(C) Breadth – first order
(D) Ascending order of transaction indices


Answer: (A)

Explanation: For a schedule, we can check its serializability by drawing a precedence graph and find its topological order, precedence graph of schedule must not contain any cycle to be conflict free.

Cycle in precedence graph tells that schedule is not conflict serializable. DFS and BFS traversal of graph are possible even if graph contains cycle. And hence DFS and BFS are also possible for non serializable graphs. But Topological sort of any cyclic graph is not possible. Thus topological sort guarantees graph to be serializable . Option D is not valid because in a transaction with more indices might have to come before lower one. Also two non- conflicting schedule can occur simultaneously.

So, option (A) is correct.

UGC-NET COMPUTER SCIENCE DBMS Previous Year Questions

UGC-NET COMPUTER SCIENCE DBMS Previous Year Questions

UGC-NET COMPUTER SCIENCE DBMS Previous Year Questions
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