Computer Science and Application UGC-NET 2008 December Paper II - Q.1 - Q.12
1. The channel capacity of a band-limited Gaussian channel is given by
(A)B log2 (2+S/N) (B) B log2 (1+S/N)
(C )B log10(1+S/N) (D)B loge(1+S/N)
Ans : B.
Explanation: C=B log2 (1 + S/N) where C is the capacity in bits per second. B is the bandwidth of the channel in Hertz and S/N is the signal to noise ratio.
2. The graph K 3,4 has _______ edges.
(A)3 edges (B) 4 edges
(C )7 edges (D) 12 edges
Ans : D.Explanation: A bipartite graph is a complete bipartite graph if every vertex in U is connected to every vertex in V. If U has n elements and V has m, then the resulting complete bipartite graph can be denoted by K n,m and the number of edges is given by n*m.
The number of edges = K 3,4 = 3 * 4 = 12
3. The total number of spanning trees that can be drawn using five labeled vertices is:
(A)125 (B)64
(C )36 (D)16
Ans : A.
Explanation: According to cayley’s formula for counting spanning trees, for a complete graph Kn, T(Kn)= n ^(n-2) where n is the number of vertices.
T(k5)=5^( 5-2)=5 ^3=125
4. Extremely low power dissipation and low cost per gate can be achieved in
(A) MOS ICS (B) C MOS ICS
(C ) TTL ICS (D) ECL ICS
Ans : B. CMOS ICS
5. An example of a universal building block is :
(A). EX-OR Gate (B). AND Gate
(C ). OR gate (D). NOR Gate
Ans: D.
Explanation: Universal gates are the ones which can be used for implementing any gate like AND,OR and NOT or any combination of these basic gates. Apart from the NOR gate, NAND gate is also considered as universal gate.
6. An example of a layer that is absent in broadcast networks is :
(A). Physical layer (B). Presentation layer
(C ). Network layer (D). application layer
Ans : C.
7. The ATM cell is :
(A). 48 bytes long (B). 53 bytes long
(C ). 64 bytes long (D). 69 bytes long
Ans: B.
Explanation: An ATM cell always consists of a 5-byte header followed by a 48-byte payload. So the size is 53 bytes long.
8. Four jobs J1,J2,J3, and J4 are waiting to be run. Their expected run times are 9,6,3 and 5 respectively. In order to minimize average response time, the jobs should be run in the order:
(A). J1 J2 J3 J4 (B). J4 J3 J2 J1
(C ) J3 J4 J1 J2 (D) J3 J4 J2 J1
Ans : D
9. Suppose it takes 100ns to access page table and 20 ns to access associative memory. If the average access time is 28ns, the corresponding hit rate is:
(A). 100 percent (B). 90 percent
(C ). 80 percent (D). 70 percent
Ans : B
10. Transmission of N signals , each band limited to fm Hz by TDM, requires a minimum band-width of
(A)fm (B) 2 fm
(C ) N fm (D) 2N fm
Ans : C.
Explanation: Minimum transmission band-width of TDM channel is given by the following equation.
Bt=NW
Where N is the total number of channels, which are bandlimited to ‘W’ Hz. In the above problem, the number of signals are N, each band limited to fm Hz and so the minimum band-width is N fm.
11. If a code is ‘t’ error detecting, the minimum hamming distance should be equal to :
(A). t-1 (B). t
(C ). t+1 (D). 2t+1
Ans: C.
Explanation: To guarantee the detection of upto s errors in all cases, the minimum hamming distance in a block code must be dmin=s+1.So, the minimum hamming distance for a ‘t’ error detecting must be t+1
12. A relation R in {1,2,3,4,5,6} is given by {(1,2),(2,3),(3,4),(4,4),(4,5)}. The relation is :
(A) Reflexive (B) symmetric
(C ) Transitive (D) not reflexive, not symmetric and not transitive
Ans: D.
1. The channel capacity of a band-limited Gaussian channel is given by
(A)B log2 (2+S/N) (B) B log2 (1+S/N)
(C )B log10(1+S/N) (D)B loge(1+S/N)
Ans : B.
Explanation: C=B log2 (1 + S/N) where C is the capacity in bits per second. B is the bandwidth of the channel in Hertz and S/N is the signal to noise ratio.
2. The graph K 3,4 has _______ edges.
(A)3 edges (B) 4 edges
(C )7 edges (D) 12 edges
Ans : D.Explanation: A bipartite graph is a complete bipartite graph if every vertex in U is connected to every vertex in V. If U has n elements and V has m, then the resulting complete bipartite graph can be denoted by K n,m and the number of edges is given by n*m.
The number of edges = K 3,4 = 3 * 4 = 12
3. The total number of spanning trees that can be drawn using five labeled vertices is:
(A)125 (B)64
(C )36 (D)16
Ans : A.
Explanation: According to cayley’s formula for counting spanning trees, for a complete graph Kn, T(Kn)= n ^(n-2) where n is the number of vertices.
T(k5)=5^( 5-2)=5 ^3=125
4. Extremely low power dissipation and low cost per gate can be achieved in
(A) MOS ICS (B) C MOS ICS
(C ) TTL ICS (D) ECL ICS
Ans : B. CMOS ICS
5. An example of a universal building block is :
(A). EX-OR Gate (B). AND Gate
(C ). OR gate (D). NOR Gate
Ans: D.
Explanation: Universal gates are the ones which can be used for implementing any gate like AND,OR and NOT or any combination of these basic gates. Apart from the NOR gate, NAND gate is also considered as universal gate.
6. An example of a layer that is absent in broadcast networks is :
(A). Physical layer (B). Presentation layer
(C ). Network layer (D). application layer
Ans : C.
7. The ATM cell is :
(A). 48 bytes long (B). 53 bytes long
(C ). 64 bytes long (D). 69 bytes long
Ans: B.
Explanation: An ATM cell always consists of a 5-byte header followed by a 48-byte payload. So the size is 53 bytes long.
8. Four jobs J1,J2,J3, and J4 are waiting to be run. Their expected run times are 9,6,3 and 5 respectively. In order to minimize average response time, the jobs should be run in the order:
(A). J1 J2 J3 J4 (B). J4 J3 J2 J1
(C ) J3 J4 J1 J2 (D) J3 J4 J2 J1
Ans : D
9. Suppose it takes 100ns to access page table and 20 ns to access associative memory. If the average access time is 28ns, the corresponding hit rate is:
(A). 100 percent (B). 90 percent
(C ). 80 percent (D). 70 percent
Ans : B
10. Transmission of N signals , each band limited to fm Hz by TDM, requires a minimum band-width of
(A)fm (B) 2 fm
(C ) N fm (D) 2N fm
Ans : C.
Explanation: Minimum transmission band-width of TDM channel is given by the following equation.
Bt=NW
Where N is the total number of channels, which are bandlimited to ‘W’ Hz. In the above problem, the number of signals are N, each band limited to fm Hz and so the minimum band-width is N fm.
11. If a code is ‘t’ error detecting, the minimum hamming distance should be equal to :
(A). t-1 (B). t
(C ). t+1 (D). 2t+1
Ans: C.
Explanation: To guarantee the detection of upto s errors in all cases, the minimum hamming distance in a block code must be dmin=s+1.So, the minimum hamming distance for a ‘t’ error detecting must be t+1
12. A relation R in {1,2,3,4,5,6} is given by {(1,2),(2,3),(3,4),(4,4),(4,5)}. The relation is :
(A) Reflexive (B) symmetric
(C ) Transitive (D) not reflexive, not symmetric and not transitive
Ans: D.