Dec 2011 Paper II- Part 2

Introduction

UGC NET Exam is one of the popular exams in India for people interested in research. Previous Year Questions are an excellent option to learn about the exam pattern. By solving the PYQs, you will get a basic idea about your preparation. You can evaluate your weak areas and work on them to perform better in the examination. In this article, we have given the questions of UGC NET 2011 December Paper-II. We have also explained every problem adequately to help you learn better.

We have discussed the last 25 questions of the paper and you can find the first 25 questions in the first part of December 2011 Paper-II Part 1.

Questions 26 to 50

26. Dijkestra banking algorithm in an operating system, solves the problem of

 (A) deadlock avoidance

 (B) deadlock recovery

 (C) mutual exclusion

 (D) context switching

Answer: A

Explanation

Deadlock Avoidance is solved by the Dijkstra banking algorithm. 

 

27. The multiuser operating system, 20 requests are made to use a particular resource per hour, on an average the probability that no request are made in 45 minutes is

 (A) e^-15             

 (B) e^-5

 (C) 1 – e^-5      

 (D) 1 – e^-10

Answer: A

Explanation:

The arrival pattern is a Poission distribution.

P(k requests) = e^-nT(nT)^k/K!

Here k = 0 , n = 20 , T = ¾

So the required probability is e^-15

 

28. On receiving an interrupt from an I/O device, the CPU

 (A) halts for predetermined time.

 (B) branches off to the interrupt service routine after completion of the current

instruction.

 (C) branches off to the interrupt service routine immediately.

 (D) hands over control of address bus and data bus to the interrupting device.

Answer: B

Explanation:

CPU checks the status bit of interrupt at the completion of each current instruction running when there is a interrupt it service the interrupt using ISR.

so OPTION B)Branches off to the interrupt service routine after completion of the current instruction

 

29. The maximum amount of information that is available in one portion of the disk access arm for a removal disk pack (without further movement of the arm with multiple heads)

 (A) a plate of data

 (B) a cylinder of data

 (C) a track of data

 (D) a block of data

Answer: B

Explanation:

Since here talking about the disk system and in disk system pricessed data as cylinder...within one run it covers one cylinder capacity.
 

30. Consider a logical address space of 8 pages of 1024 words mapped with memory of 32 frames. How many bits are there in the physical address ?

(A) 9 bits 

(B) 11 bits

(C) 13 bits 

(D) 15 bits

Answer: D

Explanation:

Taking page size same as frame size

Physical address = frame offset + word offset

No of frames = 32 so needs 5 bits

Frame size = page size =1k words = 10 bits

So ans should be 5 + 10 = 15 bits

 

31. CPU does not perform the operation

 (A) data transfer

 (B) logic operation

 (C) arithmetic operation

 (D) all of the above

Answer: D

Explanation:

Data transfer is not performed by a CPU, because Logic and arithmetic operations are performed by ALU which is part of the CPU.
 

32. A chip having 150 gates will be classified as

(A) SSI 

(B) MSI

(C) LSI 

(D) VLSI

Answer: C

Explanation:

LSI has the Logic Gates number from 100 to 9,999
 

33. If an integer needs two bytes of storage, then the maximum value of unsigned integer is

 (A) 2¹⁶ – 1

 (B) 2¹⁵ – 1

 (C) 2¹⁶

 (D) 2¹⁵

Answer: B

Explanation:

In case of signed Magnitude Representation the range is from  -(2^n-1- 1) to (2^n-1 - 1)

Min no that can be represented in this system is -(2^n-1 - 1)

Max no that can be represented in this system is  (2^n-1 - 1)

In case of 2's complement no system the range is from -2^n-1 to2^n-1 - 1

Min no that can be represented in this system is -2^n-1

Max no that can be represented in this system is 2^n-1 - 1

As they said 2 Bytes = 16 bits 

we can use max no that can be represented here  which is 2^n-1 - 1

2¹⁶-1 - 1→ 2¹⁵ - 1
 

34. Negative numbers cannot be represented in

 (A) signed magnitude form

 (B) 1’s complement form

 (C) 2’s complement form

 (D) none of the above 

Answer: D

Explanation:

All of the above can be represented in Negative form.

 

35. The cellular frequency reuse factor for the cluster size N is

 (A) N

 (B) N²

 (C)1/N

 (D)1/N²

Answer: C

Explanation:

The frequency reuse factor is defined as 1 over the number of cells in the cluster of the system (N). It is given by 1/N since each cell within a cluster is only assigned 1/N of the total available channels in the system.
 

36. X – = Y + 1 means

 (A) X = X – Y + 1

 (B) X = –X – Y – 1

 (C) X = –X + Y + 1

 (D) X = X – Y – 1

Answer: D

Explanation:

x -= y+1

x = x-(y+1)

x = x-y-1
 

37. Handoff is the mechanism that

 (A) transfer an ongoing call from one base station to another

 (B) initiating a new call

 (C) dropping an ongoing call

 (D) none of above

Answer: A

Explanation:

In cellular telecommunications, the terms handover or handoff refer to the process of transferring an ongoing call or data session from one channel connected to the core network to another channel.
 

38. Which one of the following statement is false ?

 (A) Context-free languages are closed under union.

 (B) Context-free languages are closed under concatenation.

 (C) Context-free languages are closed under intersection.

 (D) Context-free languages are closed under Kleene closure. 

Answer: C

Explanation:

CFLs are not closed under Intersection and complement.

So option C is False.

L1=aⁿbⁿc^m   and L2=aⁿb^mc^m

So L1∩L2= aⁿbⁿcⁿ which CSL not CFL.
 

39. All of the following are examples of real security and privacy risks except

 (A) Hackers

 (B) Spam

 (C) Viruses

 (D) Identify theft

Answer: B

Explanation:

All of the following are examples of real security and privacy risks Except Spam.

The name Spam was derived from a contraction of 'spiced ham'. The original variety of Spam is still available today, acknowledged as the 'spiced hammiest' of them all. During WWII and beyond, the meat colloquially became known in the UK as an acronym that stood for Special Processed American Meat.

Spam is electronic junk mail or junk newsgroup postings. Some people define spam even more generally as any unsolicited email. However, if a long-lost brother finds your email address and sends you a message, this could hardly be called spam, even though it is unsolicited.
 

40. Identify the incorrect statement :

 (A) The ATM adoption layer is not service dependent.

 (B) Logical connections in ATM are referred to as virtual channel connections.

 (C) ATM is streamlined protocol with minimal error and flow control capabilities

 (D) ATM is also known as cell delays.

Answer: A

The use of Asynchronous Transfer Mode (ATM) technology and services creates the need for an adaptation layer in order to support information transfer protocols, which are not based on ATM. This adaptation layer defines how to segment higher-layer packets into cells and the reassembly of these packets. Additionally, it defines how to handle various transmission aspects in the ATM layer.
 

41. Software risk estimation involves following two tasks :

 (A) Risk magnitude and risk impact

 (B) Risk probability and risk impact

 (C) Risk maintenance and risk impact

 (D) Risk development and risk impact

Answer: B

Explanation:

Risk impact and Risk probability are the two tasks of Software risk estimation. 

42. The number of bits required for an IP_V6 address is

(A) 16 

(B) 32

(C) 64 

(D) 128

Answer: D

Explanation:

An IPv6 address is 128 bits long.

 

43. The proposition ~ qvp is equivalent to

(A) p → q 

(B) q → p

(C) p ↔ q 

(D) p ∨ q

Answer: B

~Q ∨ P is logically equivalent to Q → P . Example: “If a number is a multiple of 4, then it is even” is equivalent to, “a number is not a multiple of 4 or (else) it is even.”

Hence replacing Q with q we get the required output. 

 

44. Enterprise Resource Planning (ERP)

 (A) has existed for over a decade.

 (B) does not integrate well with the functional areas other than operations.

 (C) is inexpensive to implement.

 (D) automate and integrates the majority of business processes.

Answer: D

Explanation:

An Enterprise Resource Planning (ERP) system is a packaged business software system that allows a company to automate and integrate the majority of its business processes. This enables the company to share common data and practices across the entire enterprise, and to produce and access information in a real-time environment.

 

45. Which of the following is false concerning Enterprise Resource Planning (ERP) ?

 (A) It attempts to automate and integrate the majority of business processes.

 (B) It shares common data and practices across the enterprise.

 (C) It is inexpensive to implement.

 (D) It provides and access information in a real-time environment.

Answer: C

Explanation:

Enterprise resource planning (ERP) is a category of business-management software—typically a suite of integrated applications—that an organization can use to collect, store, manage and interpret data from many business activities, including: product planning, purchase.

 

46. To compare, overlay or cross analyze to maps in GIS

 (A) both maps must be in digital form

 (B) both maps must be at the same equivalent scale.

 (C) both maps must be on the same coordinate system

 (D) All of the above

Answer: D

Explanation:

To compare maps using GIS

they must be digital, same scale and same coordinate system

 

47. Web Mining is not used in which of the following areas ?

 (A) Information filtering

 (B) Crime fighting on the internet

 (C) Online transaction processing

 (D) Click stream analysis.

Answer: B

Explanation:

Web mining is the use of data mining techniques to automatically discover and extract information from Web documents and services.

There are three general classes of information that can be discovered by web mining:

• Web activity, from server logs and Web browser activity tracking.
• Web graph, from links between pages, people and other data.
• Web content, for the data found on Web pages and inside of documents.

 

48. A telephone conference call is an example of which type of communications ?

 (A) same time / same place

 (B) same time / different place

 (C) different time / different place

 (D) different time / same place

Answer: B

Explanation:

B is the most practically applicable option

Same time / Different place generally we are doing Telephonic conferences and for that purpose it was made.

 

49. What is the probability of choosing correctly an unknown integer between 0 and 9 with 3 chances ?

(A)963/1000

(B)973/1000

(C)983/1000

(D)953/1000

Answer: A

Explanation:

For options given here The probability of choosing correctly an unknown integer between 0 and 9 with 3 chances is 963/1000.

Options given here may be incorrect as many thinks answer for this question can be given as below:

Let call the unknown integer as X.

There is 10 options of X, that is 0, 1, 2, 3, 4, 5, 6, 7, 8, 9.

case 1: Correct guessing in the first guessing.

So for the first time correct, we have The chance is 1/10.

Since we have guessed it correct in the first time, we don't need to count the second guessing and the third guessing.

case 2: Correct guessing in the second guessing.

For the second time correct, we must have been guessed it wrong in the first chance, it's 9/10.

And for the correct guessing in the second guess, we only have 9 options left for X, since we will not choose the same number as we have guessed in te first guessing. So, it'll be 1/9.

So, the probability for this case is 9/10 * 1/9 = 1/10.

case 3: Correct guessing in the second guessing

We must have been guessed it wrong in the first and second guessing.

So, just like the previous case, the total probability for this case is 9/10 * 8/9 * 1/8 = 1/10

So, total probability is case 1 OR case 2 OR case 3 = 1/10 + 1/10 + 1/10 = 3/10

 

50. The number of nodes in a complete binary tree of height h (with roots at level 0) is equal to

 (A) 2⁰ + 2¹ + ….. 2^h

 (B) 2⁰ + 2¹ + ….. 2^h-1

 (C) 2⁰ + 2¹ + ….. 2^h+1

 (D) 2¹ + ….. 2^h+1

Answer: A

Explanation:

The number of nodes in a complete binary tree of height h (with roots at level 0) is equal to 20 + 21 + ..... 2h