## Friday, 16 December 2016

### UGC Net Computer Science Paper II Dec 12 Page 1 (Solved)

1. Consider the circuit shown below. In a certain steady state, Y is at logical ‘l’.What are possible values of A, B, C ?
(A) A = 0, B = 0, C = 1
(B) A = 0, B = C = 1
(C) A = 1, B = C = 0
(D) A = B = 1, C = 1
Answer A
Explanation : Nand gate behave opposite of and gate so if any 0 in input then result will be 1. So output Y will be 1 if A = 0, B = 0, C = 1

2. The worst case time complexity of AVL tree is better in comparison to binary search tree for
(A) Search and Insert Operations
(B) Search and Delete Operations
(C) Insert and Delete Operations
(D) Search, Insert and Delete Operations
Answer : D
Explanation : Because Search is always (logN) in AVL tree,  where is number of node since AVL trees are always balanced. Insertion and deletions are also O(logn) . Where as in case of BST it is O(n).

3. The GSM network is divided into the
following three major systems :
(A) SS, BSS, OSS
(B) BSS, BSC, MSC
(C) CELL, BSC, OSS
(D) SS, CELL, MSC
Answer : A
Explation :

#### Switching System:

The switching system (SS) is responsible for performing call processing and subscriber-related functions. The switching system includes the following functional units.

#### The Base Station System (BSS):

All radio-related functions are performed in the BSS, which consists of base station controllers (BSCs) and the base transceiver stations (BTSs).

#### The Operation and Support System:

The operations and maintenance center (OMC) is connected to all equipment in the switching system and to the BSC. The implementation of OMC is called the operation and support system (OSS). The OSS is the functional entity from which the network operator monitors and controls the system. The purpose of OSS is to offer the customer cost-effective support for centralized, regional, and local operational and maintenance activities that are required for a GSM network. An important function of OSS is to provide a network overview and support the maintenance activities of different operation and maintenance organizations.
4. The power set of the set {} is
(A) {⥀}
(B) {⥀, {⥀}}
(C) {0}
(D) {0, ⥀, {⥀}}
Answer : B
Explanation : A Power Set is a set of all the subsets of a set.
Power Set of S={a,b,c} is follows
P(S) = { {}, {a}, {b}, {c}, {a, b}, {a, c}, {b, c}, {a, b, c} }

5. If the disk head is located initially at 32, find the number of disk moves required with FCFS if the disk queue of I/O blocks requests are 98, 37, 14, 124, 65, 67.
(A) 239
(B) 310
(C) 321
(D) 325
Answer C
Explanation : First-Come, First Serve
non-preemptive scheduling management ready queue is managed as a FIFO queue

6. Component level design is concerned
with
(A) Flow oriented analysis
(B) Class based analysis
(C) Both of the above
(D) None of the above
Answer C
Explanation :The purpose of component level design is to define data structures, algorithms, interface characteristics, and communication mechanisms for each software component identified in the architectural design. Component level design occurs after the data, architectural, and interface designs are established. The component-level design represents the software in a way that allows the designer to review it for correctness and consistency, before it is built. The work product produced is a design for each software component, represented using graphical, tabular, or text-based notation. Design walkthroughs are conducted to determine correctness of the data transformation or control transformation allocated to each component during earlier design steps.
Component Definitions

•  A component is a modular, deployable, replaceable part of a system that encapsulates implementation and exposes a set of interfaces
•  Object-oriented view is that a component contains a set of collaborating classes
•  Traditional view is that a component (or module) resides in the software and serves one of three roles
1. Control components coordinate invocation of all other problem domain components
2. Problem domain components implement a function required by the customer
3. Infrastructure components are responsible for functions needed to support the

• processing required in a domain application • Process-Related view emphasizes building systems out of existing components chosen from a catalog of reusable components as a means of populating the architecture

7. The ‘C’ language is
(A) Context free language
(B) Context sensitive language
(C) Regular language
(D) None of the above
Answer A
Explanation :context-free grammar (CFG) is a set of recursive rewriting rules (or productions) used to generate patterns of strings.

8. The Mobile Application Protocol
(MAP) typically runs on top of which
protocol ?
(A) SNMP (Simple Network
Management Protocol)
(B) SMTP (Simple Mail Transfer
Protocol)
(C) SS7 (Signalling System 7)
(D) HTTP (Hyper Text Transfer
Protocol)
Answer : C
Explanation :The Mobile Application Part (MAP) is an SS7 protocol that provides an application layer for the various nodes in GSM and UMTS mobile core networks and GPRS core networks to communicate with each other in order to provide services to users. The Mobile Application Part is the application-layer protocol used to access the Home Location Register, Visitor Location Register, Mobile Switching Center, Equipment Identity Register, Authentication Centre, Short message service center and Serving GPRS Support Node (SGSN).

9. If a packet arrive with an M-bit value is ‘l’ and a fragmentation offset value ‘0’, then it is ______ fragment.
(A) First
(B) Middle
(C) Last
(D) All of the above
Answer : F
Explanation :If M bit is 1, it means that there is at least one more fragment. This fragment can be the             first one or the middle one, but not the last one. More information is needed to say               whether it is the first or middle one but we can say that the original packet was fragmented       because the M   bit vale is 1.

10. The number of bit strings of length eight that will either start with a 1 bit or end with two bits 00 shall be
(A) 32
(B) 64
(C) 128
(D) 160
Answer : D
Explanation : Total Number start with 1 bit = 128 to 255 = 128 numbers

Ending with 00 = 256/4 = 64 numbers (because 8 bit is divide by 2)

Half of these 64 numbers start with 1, leaving 32.

Assuming starting with 1 and ending with 00 is counted only once:

128 + 32 = 160 numbers