Introduction
UGC NET is one of the popular exams in India for people interested in research. Previous Year Question Papers are an excellent option to learn about the exam pattern and the type of questions. By solving previous year questions, you will get a basic idea about your preparation. You can evaluate your weak sections and work on them for better performance in the examination.
In this article, we have given the questions of UGC NET Dec 2014 Paper II. You can visit Dec 2014 Paper II Part 1 here. We also have explained every problem adequately to help you learn better.
Note: This article contains Ques. No. 26 to Ques No. 50 out of the 50 questions asked in UGC NET Dec 2014 paper II.
Questions and Answers
26. The period of a signal is 10 ms. What is its frequency in Hertz?
(A) 10 (B) 100
(C) 1000 (D) 10000
Answer: B
We know that Frequency = 1 / time
Time = 10ms (Given)
So, Frequency = 1/10 ms
= 1/10 * 10^(-3)
= 1 / 10^(-2)
= 100Hz
27. In classful addressing, the first four bits in Class A IP address is
(A) 1010 (B) 1100
(C) 1011 (D) 1110
Answer: A, B, C, D
The IP addresses that belong to Class A are in the range 1.x.x.x – 126.x.x.x
But in Class A, the first bit is always 0 because the address ranges between 1 – 126, i.e., 00000001 – 01111110 (Except 00000000 and 01111111). So, all options are wrong.
28. Which of the following algorithms is not a broadcast routing algorithm?
(A) Flooding (B) Multidestination routing
(C) Reverse path forwarding (D) All of the above
Answer: D
Different Broadcast Routing Algorithms are :
- Sending distinct packets
- Flooding
- Multi-destination Routing
- Using spanning trees
- Reverse path forwarding
29. An analog signal has a bit rate of 6000 bps and a baud rate of 2000 baud. How many data elements are carried by each signal element?
(A) 0.336 bits/baud (B) 3 bits/baud
(C) 120,00,000 bits/baud (D) None of the above
Answer:B
Baud rate = 2000 baud (Given)
Bit rate = 6000 bps(Given)
Bits per baud = bit rate / baud rate
= 6000/ 2000
= 3 bits per baud
30. How many distinct stages are in the DES algorithm parameterized by a 56-bit key?
(A) 16 (B) 17
(C) 18 (D) 19
Answer: D
The correct option is D. The DES(Data Encryption Standard) algorithm has 19 different stages. Which include 16 round and plain text which are encrypted in blocks of 64 bits
31. Shift-reduce parsers perform the following :
(A) Shift step that advances in the input stream by K(K>1) symbols and Reduces action that applies a completed grammar rule to some recent parse trees, joining them together as one tree with a new root symbol.
(B) Shift step that advances in the input stream by one symbol and Reduces action that applies a completed grammar rule to some recent parse trees, joining them together as one tree with a new root symbol.
(C) Shift step that advances in the input stream by K(K = 2) symbols and Reduces action that applies a completed grammar rule to form a single tree.
(D) Shift step that does not advance in the input stream and Reduce action that applies a completed grammar rule to form a single tree.
Answer: B
A Shift step advances in the input stream by one symbol. That shifted symbol becomes a new single-node parse tree. And a Reduce step applies a completed grammar rule to some recent parse trees, joining them together in one tree with a new root symbol.
32. Which of the following is true?
(A) Canonical LR parser is LR (1) parser with a single lookahead terminal
(B) All LR(K) parsers with K > 1 can be transformed into LR(1) parsers.
(C) Both (A) and (B)
(D) None of the above
Answer: C
It is true that all LR(K) parsers having K > 1 can be changed to LR(1) parser.
Also, canonical LR parser is LR(1) parser with single look ahead terminal is true.
So both options A and B are correct.
33. In a two-pass assembler, the symbol table is
(A) Generated in the first pass
(B) Generated in the second pass
(C) Not generated at all
(D) Generated and used only in the second pass
Answer: A
One pass assembler can scan the whole file in one go. While in two-pass assembler symbol table is generated in the first pass, it needs two passes to scan the entire source file. Hence option A is correct.
34. A debugger is a program that
(A) allows to examine and modify the contents of registers
(B) does not allow execution of a segment of the program
(C) allows to set breakpoints, execute a segment of the program, and display contents of the register
(D) All of the above
Answer: C
A debugger is a computer program used by programmers to test and debug a target program. The debugger also allows us to set breakpoints, execute a segment of the program and display the contents of the register. It may be use instruction set simulators rather than running a program directly to achieve a higher level of control over the execution.
35. The following Context-Free Grammar (CFG) :
S->aB | bA
A->a | as | bAA
B->b | bs | aBB
will generate
(A) odd numbers of a’s and odd numbers of b’s
(B) even numbers of a’s and even numbers of b’s
(C) equal numbers of a’s and b’s
(D) different numbers of a’s and b’s
Answer: A,B,C,D
The given CFGs can generate
- ab or ba, i.e., for odd numbers of a’s and odd numbers of b’s
- ab, ba bbaa or aabb, i.e., for equal numbers of a’s and an equal number of b’s
- abb or baa, i.e., for different numbers of a’s and different numbers of b’s
- aabb or bbaa, i.e., for even numbers of a’s and even numbers of b’s
Therefore all options are correct.
36. Consider the following justifications for commonly using the two-level CPU scheduling :
- It is used when memory is too small to hold all the ready processes.
- Because its performance is the same as that of the FIFO.
- Because it facilitates putting some set of processes into memory, a choice is made from that.
- Because it does not allow to adjust the set of in-core processes.
Which of the following is true?
(A) I, III and IV (B) I and II
(C) III and IV (D) I and III
Answer: D
Two-level CPU scheduling facilitates putting some set of processes into memory. It is used when memory is too small to hold all the ready processes, and it also provides a choice from that set.
37. A specific editor has 200 K of program text, 15 K of the initial stack, 50 K of initialized data, and 70 K of bootstrap code. If five editors are started simultaneously, how much physical memory is needed if the shared text is used?
(A) 1135 K (B) 335 K
(C) 1065 K (D) 320 K
Answer: B
Program text = 200k
Initial stack = 15k
Initialized data = 50k
Bootstrap code = 70k
Since editors are started simontaneously
Total physical memory needed = 200k + 15 K + 50 K + 70 K
= 335k
38. Which of the following conditions does not hold good for solving a critical section problem?
(A) No assumptions may be made about speeds or the number of CPUs.
(B) No two processes may be simultaneously inside their critical sections.
(C) Processes running outside its critical section may block other processes.
(D) Processes do not wait forever to enter their critical section.
Answer: C
The correct answer should be C as in the critical section, problem processes running outside its critical section can’t prevent other processes from entering an essential section.
39. For the implementation of a paging scheme, suppose the average process size is x bytes, the page size is y bytes, and each page entry requires z bytes. The optimum page size that minimizes the total overhead due to the page table and the internal fragmentation loss is given by
(A) x/2 (B) xz/2
(C) √2xz (D) (√xz)/2
Answer: C
The average amount of internal fragmentation per segment = y/2.
Average no of pages per process = x/y.
No. of bytes required for each page = ‘z’ bytes of the page table.
So, each process segment requires page table size = xz/y bytes.
Total overhead per segment = xz/y + y/2 (because of internal fragmentation and page table entries)
To minimise the overhead, differentiate with respect to page size, p, and equate to 0:
d(xz/y + y/2)/dy = -xz/y2 + ½ = 0
y = √(2xz)
40. In a demand paging memory system, the page table is held in registers. The time taken to service a page fault is eight m.sec. if an empty frame is available or if the replaced page is not modified, and it takes 20 m.secs., if the replaced page is modified. What is the average access time to service a page fault, assuming that the page to be replaced is modified 70% of the time?
(A) 11.6 m.sec. (B) 16.4 m.sec.
(C) 28 m.sec. (D) 14 m.sec.
Answer:B
PFST = 8ms (for empty frame)
PFST = 20ms (after the page is modified)
Page replacement frequency = 70%
The page that is not modified = 100 – 70 = 30%
Average access time = (0.3 * 8) + (0.7 * 20)
= 2.4 + 14
= 16.4 ms
Where PFST = Page Fault Service Time
41. ................... are applied throughout the software process.
(A) Framework activities (B) Umbrella activities
(C) Planning activities (D) Construction activities
Answer: B
The correct option is B because Umbrella activities span all the stages of SDLC(Software Development Life Cycle). It focuses on the Requirement traceability matrix. Not maintaining the Requirement traceability matrix can result in unsatisfied requirements and problems during maintenance and delivery.
42. Requirement Development, Organizational Process Focus, Organizational Training, Risk Management, and Integrated Supplier Management are process areas required to achieve maturity level
(A) Performed (B) Managed
(C) Defined (D) Optimized
Answer: C
The correct option is C because the given process achieves the maturity level defined. A
maturity level is important for attaining a mature software process.
43. The software ................. of a program or a computing system is the structure or structures of the system, which comprise software components, the externally visible properties of those components, and the relationships among them.
(A) Design (B) Architecture
(C) Process (D) Requirement
Answer: B
Software architecture encompasses the set of significant decisions about software organization that includes sections of structure elements, behavior as specified in the collaboration, and compositions of structural and behavioral elements. It also provides functionality, reuse, economic and technology constraints, performance, usability, etc.
44. Which one of the following set of attributes should not be encompassed by effective software metrics?
(A) Simple and computable
(B) Consistent and objective
(C) Consistent in the use of units and dimensions
(D) Programming language-dependent
Answer: D
Software matrices are used for counting software characteristics. It includes characteristics that are simple and countable, objective and consistent. Software matrices are also consistent in the use of units and dimensions. But these are not programming language-dependent; Hence option D is correct.
45. Which one of the following is used to compute cyclomatic complexity?
(A) The number of regions – 1
(B) E – N + 1, where E is the number of flow graph edges, and N is the number of flow graph nodes.
(C) P – 1, where P is the number of predicate nodes in the flow graph G.
(D) P + 1, where P is the number of predicate nodes in the flow graph G.
Answer: D
Complexity of program will be:
V(G) = E-N+2 = P+1
Where,
V = vertices
E = Number of edges
N = number of nodes
P = number of predicate nodes
Predicate nodes are the nodes that contains condition.
46. Consider the following statements S1 and S2 :
S1: A hard handover is one in which the channel in the source cell is retained and used for a while in parallel with the channel in the target cell.
S2: A soft handover is one in which the channel in the source cell is released, and only then the channel in the target cell is engaged.
(A) S1 is true, and S2 is not true.
(B) S1 is not true, and S2 is true.
(C) Both S1 and S2 are true.
(D) Both S1 and S2 are not true.
Answer:D
S1 is not true because in hard handover source cell is released, and only then the channel in the target cell is engaged, and S2 is not true because in soft handover source cell is retained and used for a while in parallel with the channel in the target cell.
Hence both statements are wrong.
47. Fact-less fact table in a data warehouse contains
(A) only measures (B) only dimensions
(C) keys and measures (D) only surrogate keys
Answer: B
Fact-less fact tables don’t have any measures. It contains only dimension keys because it is an intersection of dimensions.
There are two types of such tables
- Tables for capturing events
- Table for describing conditions
48. Which e-business model allows consumers to name their own price for products and services?
(A) B2B (B) B2G
(C) C2C (D) C2B
Answer:D
B2B deals with the requirement of some material to run their production processes.
B2G provides a platform for businesses to bid on govt. Opportunities but here is no authority for consumers to decide the price of goods and services.
In C2C, customers are allowed to interact but not authorized to name their own prices for goods and services.
C2B is the only e-business model that allows consumers to name their own prices for goods and services.
49. ............... model is designed to bring prices down by increasing the number of customers who buy a particular product at once.
(A) Economic Order Quantity
(B) Inventory
(C) Data Mining
(D) Demand-Sensitive Pricing
Answer:D
Economic order quantity is used to minimize the total ordering and holding cost. Also, Data mining and inventory models do nothing with prices. So, only demand sensitive pricing model is used to bring prices down.
50. Match the following :
List – I List – II
a. Call control protocol i. The interface between Base Transceiver Station (BTS) and Base Station Controller (BSC)
b. A-bis ii. Spread spectrum
c. BSMAP iii. Connection management
d. CDMA iv. Works between Mobile Switching Centre (MSC) and Base Station Subsystem (BSS)
Codes :
a b c d
(A) iii iv i ii
(B) iii i iv ii
(C) i ii iii iv
(D) iv iii ii i
Answer: A
Cell control protocol is for connection management i.e. A -> iii
A-bis work in between BSS (base station system) and MSC (mobile switching center), i.e., B -> iv
BSMAP (Bisulfite Sequence MAping program) acts as an interface between BTS (Base Transceiver Station) and BSC(Base Station Controller), i.e., C -> i
CDMA (Code-Division Multiple Access) is a spread spectrum, i.e., D -> ii
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