# UGC NET COMPUTER SCIENCE SOLVED PAPERS 2014-16 - UGC NET Computer Science Paper 3 August 2016

>>>>>>>>UGC NET Computer Science Paper 3 August 2016

• Option : D
• Explanation :

So, option (D) is correct.

• Option : B
• Explanation :
• An OODBMS avoids the “impedance mismatch” problem.Correct.
• An OODBMS doesn't avoids the “phantom” problem.
• An OODBMS doesn't provides higher performance concurrency control than most relational databases. Since it is distributed.
• An OODBMS provides faster access to individual data objects once they have been read from disk.Correct
• SO, option (B) is correct.

• Option : B
• Explanation :
The condition that all the data of the global relation must be mapped into the fragments, that is, it must not happen that a data item which belongs to a global relation does not belong to any fragment, is called Completeness condition. In distributed system it is convenient that fragment to be disjoint, so that the replication of data can be controlled explicitly at the allocation level.this is called Disjointness condition To reconstruct any global relation from its fragment is called Reconstruction condition. The formation of no. of fragments into a cluster is called Aggregation condition. So, option (B) is correct.

• Option : D
• Explanation :
RA1 : T1 ⨝ T1.P = T2.A T2 where ⨝is natural join symbol. It will result 3 tuples:
P = ARC
1056
1055
2563
RA2 : T1 ⟕ T1.P = T2.A T2 where ⟕ is left outer join symbol. It will result in 4 tuples.
P = ARC
1056
1055
158Null
2563
RA3 : T1 ⨝ T1.P = T2.A and T1.R = T2.CT2. It will result in 1 tuple.
P = AR = C
105
So, option (D) is correct.

• Option :
• Explanation :
Decomposition of R into R1(C, A) and R2(A, B) is lossless. Because C → A, A → B. So, C → AB can be derived and there is no loss. Decomposition of R into R1(A, B) and R2(B, C) is lossy. Because A → B, C → B are derived but we can't derive C → AB, So it is lossy. So, option (C) is correct.