# UGC NET Computer Science Questions Paper II December 2008 - Part 1

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)
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 K3,4 has:
(A) 3 edges
(B) 4 edges
(C) 7 edges
(D) 12 edges
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 Kn,m and the number of edges is given by n*m. The number of edges Kn,m = K 3,4
n*m =  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
Explanation:
Cayley devised the well-known formula nn−2 for the number of spanning trees in the complete graph Kn. = 55-2 = 125
4.       Extremely low power dissipation and low cost per gate can be achieved in:
(A) MOS ICs
(B) CMOS ICs
(C) TTL ICs
(D) ECL ICs
5.       An example of a universal building block is:
(A) EX-OR gate
(B) AND gate
(C) OR gate
(D) NOR gate
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
7.       The ATM cell is:
(A) 48 bytes long
(B) 53 bytes long
(C) 64 bytes long
(D) 69 bytes long
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
9.       Suppose it takes 100 ns to access page table and 20 ns to access associative memory. If the average access time is 28 ns, the corresponding hit rate is:
(A) 100 percent
(B) 90 percent
(C) 80 percent
(D) 70 percent
Explanation:
h x 20 + (1-h) x 100 = 28
h=0.9 = 90%
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
1. 