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Table of contents
1.
Dec 2012 Paper III- Part 2
2.
Introduction
3.
Questions 26 to 50
4.
Frequently Asked Questions
4.1.
What is the UGC NET exam?
4.2.
What is the maximum number of attempts for the UGC NET examination?
4.3.
How can solving PYQs help in my exam preparation?
4.4.
How many papers are there in the UGC NET exam?
5.
Conclusion
Last Updated: Mar 27, 2024
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Dec 2012 Paper III- Part 2

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Ashwin Goyal
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Dec 2012 Paper III- 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 2012 December Paper-III. We have also explained every problem adequately to help you learn better.

In this article, we have discussed the next 25 questions of the UGC NET 2012 paper and for the other questions you can visit this link December 2012 Paper-III Part-1 and December 2012 Paper-III Part-3.

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Questions 26 to 50

 

26. Which are the classifications of data used in Mobile Applications?

 (A) Private data, User data, Shared data.

 (B) Public data, User data, Virtual data.

 (C) Private data, Public data, Shared data.

 (D) Public data, Virtual data, User data.  

Answer: C

Explanation:

Mobile data classification can help users keep their personal and professional life separate on their mobile devices, while graphic markers applied to documents can help users understand the value of their data. Some solutions even let you limit employee access to your most critical data files. Public data, shared data, and private data are the three types of data used in mobile apps.


27. In an enhancement of a CPU design, the speed of a floating point unit has been increased by 20% and the speed of a fixed point unit has been increased by 10%. What is the overall speed achieved if the ratio of the number of floating point operations to the number of fixed point operations is 2 : 3 and the floating point operation used to take twice the time taken by the fixed point operation in original design ?

 (A) 1.62

 (B) 1.55 

 (C) 1.85

 (D) 1.285 

Answer: X

Explanation:

Time spent on the original design vs. time spent on the improved design

Original concept:

The floating-point to fixed-point operations ratio is 2:3.

As a result, make fixed-point operations 2n and floating-point operations 3n.

The ratio of floating-point to fixed-point operations is 2:1.

Assume that the time consumed by floating-point operations is 2t and the time spent by fixed-point operations is t.

In an improved design:

The time taken for a floating-point operation would be 83.33 percent of the original time (original time/1.2) if the speed of the floating-point operation was increased by 20% (1.2 * original speed).

(This is due to the fact that CPU speed (S) is inversely proportional to execution time (T), thus if speed becomes 1.2S, time becomes T/1.2.)

Similarly, if the speed of a fixed-point operation is increased by 10% (1.1 * original speed), the time taken now is 90.91 percent of the original time (original time / 1.1) for the fixed-point operation.

2n * 2t /1.2) + (3n * t /1.1) = 6.06nt time taken by enhanced design

The original design took (2n * 2t) + (3n * t) = 7nt of time.

Speed up= 7nt / 6.06nt = 1.155 . Hence the answer is 1.155 which is none of the given options.


28. The initial basic feasible solution to the following transportation problem using Vogel’s approximation method is 

   D1 D2 D3 D4. Supply 

S1 1 2 1 4 30

 S2 3 3 2 1 50 

S3 4 2 5 9 20 

Demand 20 40 30 10 

(A) x11 = 20, x13 = 10, x21 = 20, x23 = 20, x24 = 10, x32 = 10, Total cost = 180 

(B) x11 = 20, x12 = 20, x13 = 10, x22 = 20, x23 = 20, x24 = 10, Total cost = 180 

(C) x11 = 20, x13 = 10, x22 = 20, x23 = 20, x24 = 10, x32 = 10, Total cost = 180

 (D) None of the above 

Answer: D

Explanation:

Vogel's Approximation Method (VAM) is a penalty method (VAM)

This method is selected above the NWCM and LCM methods because the initial basic viable solution supplied by this method is either ideal or significantly near to the optimal answer.

Vogel's Approximation Method (VAM) steps (Rule)

Step 1: Find the lowest and next-lowest expenses in each row and note the difference (called penalty) in the row penalty column.

Step 2: Find the cheapest and next-to-cheapest cells in each column and enter the difference (called penalty) in each column penalty.

Step-3:Choose the highest penalty row or column and look for the cell with the lowest cost in that row or column. As much space as possible in this cell.

Choose the cell with the largest potential allocation if the penalty values are tied.

Step 4: Make supply and demand adjustments and delete (strike out)

Step 4: Adjust supply and demand, then strike off (cross out) the satisfying row or column.

Step 5: Continue until all supply and demand values are equal to zero.


29. 58 lamps are to be connected to a single electric outlet by using an extension board each of which has four outlets. The number of extension boards needed to connect all the light is 

(A) 29 

(B) 28 

(C) 20 

(D) 19 

Answer: D

Explanation:

Connect the first four lamps (L1, L2, L3, and L4) to Extension 1 and power them from Extension 2. Only three bulbs can be connected to Extension 2 since one of the four ports is used to deliver power to Extension 1.

Similarly, we can only connect three bulbs to other extensions because one port is used to deliver electricity to another extension.

Connect the final extension (Ex 19) to the power outlet.

The first extension may accommodate four lamps.

Extending the remaining 54 lamps requires 54/3=18 extensions.

To connect all 58 lamps, you'll need 18+ 1 = 19  extension.


30.  Match the following with respect to the Mobile Computing Architecture.

 a. Downlink control                                             1. 100 Mbps

 b. Radio communication data                            2. Residency latency (RL) 

c. The average duration of user’s stay in cell   3. Sending data from a BS to MD

d. FDDI bandwidth                                               4. 2-Mbps Codes : a b c d 

(A) 2 1 4 3 

(B) 3 4 2 1 

(C) 4 1 2 1 

(D) 4 3 1 2 

Answer: B

Explanation:

The base station subsystem (BSS) is a component of a traditional cellular telephone network that handles traffic and signals between a mobile device and the network switching subsystem. Hence, Downlink control is used to send data from a BS to an MD, according to this statement.

Residency Latency (RL) is known as the average duration of user's stay in cell.

The American National Standards Institute (ANSI) established the Fibre Distributed Data Interface (FDDI) networking standard in the early 1980s for speeds up to 100 Mbps.

 

31. Which of the following flags are set when ‘JMP’ instruction is executed ?

 (A) SF and CF

 (B) AF and CF

 (C) All flags 

(D) No flag is set  

Answer: D

Explanation:

The JMP instruction in the x86 assembly language conducts an unconditional jump. By modifying the program counter, this instruction changes the flow of execution.

CF (Carry Flag)

For unsigned integer arithmetic, this flag signals an overflow problem. It's also utilised in arithmetic with multiple precision.

AF (Auxiliary Flag)

The AF flag is set if an ALU operation generates a carry/barrow from the lower nibble (D0 – D3) to the upper nibble (D4 – D7), i.e. carry delivered by D3 bit to D4. This is not a general-purpose flag; the CPU uses it to perform Binary to BCD conversion internally.

SF (Sign Flag)

The sign of the number is given by the MSB bit in sign-magnitude format. If the operation's result is negative, the sign flag is set.

 

32.A thread is a light weight process. In the above statement, weight refers to

 (A) time

 (B) number of resources 

 (C) speed

 (D) All the above

Answer: B

Explanation:

The term “light-weight process” usually refers to kernel threads and here the weight refers to the number of resources.

 

33. The Z-buffer algorithm is used for Hidden surface removal of objects. The maximum number of objects that can be handled by this algorithm shall

 (A) Depend on the application 

 (B) be arbitrary no. of objects

 (C) Depend on the memory availability

 (D) Depend on the processor

Answer: B

Explanation:

Because each object is processed one at a time, the Z-buffer algorithm can handle an arbitrary number of objects.

 

34. The power set of AUB, where A = {2, 3, 5, 7} and B = {2, 5, 8, 9} is 

(A) 256 

(B) 64 

(C) 16 

(D) 4  

Answer: B

Explanation:

Given,

A = {2,3,5,7}

and   B= {2,5,8,9}

So, A∪B = {2,3,5,7,8,9}

Power set of  A∪B is going to have 26 elements which is equal to 64

 

35. In Win32, which function is used to create Windows Applications ?

(A) Win APP 

(B) Win API 

(C) Win Main 

(D) Win Void 

Answer: C

Explanation:

Win main is a feature that allows you to develop a Windows application. A graphical Windows-based application's user-provided entry point. Any Windows application's entry point function is WinMain() in C. 

In Windows, we have WinMain() instead of the main() method, which is used in DOS/console-based applications. WinMain() is a system function that is invoked when a process is created. The current process's instance handle is the first argument. The prior instance is next. The next parameter is the command line arguments. Finally, the shell passes the main window's show/display attribute. WinMain() returns a zero for success and a non-zero for error.

 

36. Suppose a processor does not have any stack pointer registers, which of the following statements is true ?

(A) It cannot have subroutine call instruction. 

(B) It cannot have nested subroutine calls. 

(C) Interrupts are not possible. 

(D) All subroutine calls and interrupts are possible. 

Answer: X

Explanation:

The address of the top of the stack, which is the region of memory where the CPU should resume execution after satisfying an interrupt or subroutine call, is stored in the stack pointer register. If the SP register is not present, no subroutine call instructions can be executed. 

 

37. Everything below the system call interface and above the physical hardware is known as ______. 

(A) Kernel 

(B) Bus 

(C) Shell

(D) Stub 

Answer: A

Explanation:

A system call, also known as a kernel call, is a request for a service provided by the kernel made by an active process via a software interrupt. The basic interface between the kernel and program is the system call interface.

The kernel is the vital component of every operating system. It oversees the computer's and hardware's tasks, mainly memory and CPU time. Kernels are subdivided into two categories: A microkernel provides only basic functionality, whereas a monolithic kernel comprises many drivers.

 

38. Which is not the correct statement? 

(A) The class of regular sets is closed under homomorphisms. 

(B) The class of regular sets is not closed under inverse homomorphisms. 

(C) The class of regular sets is closed under the quotient. 

(D) The class of regular sets is closed under substitution. 

Answer: B

Explanation:

Regular sets are homomorphism and inverse homomorphism closed classes, as well as quotient and substitution closures. Substitution is a type of homomorphism because regular languages are closed under Inverse homomorphism

 

39. When a programming language has the capacity to produce new datatype, it is called as,

(A) Overloaded Language 

(B) Extensible Language 

(C) Encapsulated Language 

(D) Abstraction Language 

Answer: B

Explanation:

An IT term for something that can be extended or expanded from its initial condition is extensible. It usually refers to software, such as a program or file format, but it can also refer to a programming language in and of itself.

A language that is extensible can support custom syntax and operations. These custom elements can be defined in the source code, and the compiler will recognize them alongside the predefined elements. Ruby, Lua, and XL are examples of extensible programming languages.

Extensibility is a term that is nearly always used to describe software and alludes to its ability to be extended. A software program that supports plugins, for example, is extendable.

 

40. Which of the following operating system is better for implementing a client-server network ? 

(A) Windows 95 

(B) Windows 98 

(C) Windows 2000 

(D) All of these 

Answer: C

Windows 2000 is the most recent of the first three options, and we already have MS Server 2000 to provide such services.

Windows 2000 is a computer operating system that may be used on both client and server machines. Microsoft released it to manufacturing on December 15, 1999, and it went on sale to the general public on February 17, 2000.  It succeeds Windows NT 4.0 and is the final version of Microsoft Windows to bear the "Windows NT" designation. [7] Windows XP (launched in October 2001) and Windows Server 2003 are their successors (released in April 2003). Windows 2000 was known as Windows NT 5.0 during development.

 

41.  Consider a system having m resources of the same type. These resources are shared by 3 processes A, B and C which have peak demands of 3, 4 and 6 respectively. For what value of m deadlock does it occur ? 

(A) 7 

(B) 9 

(C) 10 

(D) 13 

Answer: D

Explanation:

Deadlock occurs when two or more processes are waiting for each other to release a resource, or when more than two processes in a circular chain are waiting for resources. Consider a system with m of the same type of resources. Three processes, A, B and C, each with peak demands of 3,4, and 6 , share these resources. There will be no deadlock for the 13th value of n.

Consider the condition of peak demand for resources (A, B, C) = (3,4,6).

(3+ 4+ 6) < m+3.

m> 10. Hence m should be more than 10.

Hence option D satisfies the condition.

 

42. The grammar ‘G1’→ S OSO| ISI | 0|1| and the grammar ‘G2’ is 

S →  as |asb| X, X → Xa | a.

 Which is the correct statement? 

(A) G1 is ambiguous,  G2 is unambiguous 

(B) G1 is unambiguous, G2 is ambiguous 

(C) Both G1 and G2 are ambiguous 

(D) Both G1 and G2 are unambiguous

Answer: B

Explanation:

Consider a string aaa in which we have two leftmost derivation

S-->aS-->aX-->aXa-->aaa

and 

S-->aS-->aaS-->aaX-->aaa

Hence G2 is ambiguous.

Whereas G1 is unambiguous because we cannot have two derivation trees for a 

Single same string.

 

43. Consider n processes sharing the CPU in a round-robin fashion. Assuming that each process switch takes s seconds. What must be the quantum size q such that the overhead resulting from process switching is minimized but, at the same time each process is guaranteed to get its turn at the CPU at least every t seconds ?

aq

Answer: B

As each process will get CPU for q seconds and at the same time each process is guaranteed to get its turn at the CPU at least every t seconds. As a result, after the current process receives CPU again, there will be (n-1) processes. The context switch takes s seconds for each process.

(Qp1)(s)(Qp2)(s)(Qp3)(s)(Qp1)

It can be seen that there have been n context changes and (n-1) processes since P1 left and returned. As a result, the equation is:     

q*(n-1) + n*s <= t

q*(n-1) <= t - n*s

q <= (t-n.s) / (n-1)

 

44. The Default Parameter Passing Mechanism is called as 

(A) Call by Value 

(B) Call by Reference 

(C) Call by Address 

(D) Call by Name 

Answer: A

Explanation:

The mechanism for passing arguments to a procedure (subroutine) or function is known as parameter passing. The default parameter passing mechanism in C is known as Call By Value, and it allows us to provide the value of the real parameter.

 

45. Which of the following regular expression identities are true ? 

(A) (r + s)* = r* s* 

(B) (r + s)* = r* + s* 

(C) (r + s)* = (r*s*)* 

(D) r* s* = r* + s*

Answer: C

Explanation:

Consider a string “ rrssrs”.

Take + as either one of the paths and * as the loop either 0 or more times.

 In Option A, (r + s)* = r* s* 

left side  may be While() { Either r or s };

 Left side realize the string.

 Whereas the right side fails to realize the string. 

Because r* s*= While() {r} While() {s}. 

In Option B (r + s)* = r* + s* 

Left side realize the string 

whereas r* + s* = If() { while() {r } Else { while() {s} } 

So right side fails to realize the string “rrssrs”. 

In Option C    (r + s)* = (r*s*)* .

 While() { While() { R } While() { S } } 

Right side also realize the string. 

Hence here both the sides realize the string so the answer is C

 

46. Two graphs A and B are shown below : 

Which one of the following statements is true ? 

(A) Both A and B are planar. 

(B) Neither A nor B is planar. 

(C) A is planar and B is not. 

(D) B is planar and A is not. 

Answer: A

Explanation:

Planar Graph: A graph G is planar if it can be drawn in the plane without any edges intersecting except at a vertex to which they are incident.     S1 and S2 are both planar graphs. There are no intersecting edges here.

Constructing a planar version of a given graph takes a long time. We can readily answer such questions by using the relationship: 

If the number of edges equals the number of vertices and E= 3(V-2), the graph is planar; otherwise, it is not.

A)V=4; E=6; 3*(V-2)=3(4-2)   =6=E            //Planar

B)) V=8; E=12 3*(V-2)=3(8-2)=18             //Planar 

Hence both A and B  are planar.


47. The minimum number of states of the non-deterministic finite automation which accepts the language {a b a bn|n  0}  {a b an|n  0} is 

(A) 3 

(B) 4 

(C) 5 

(D) 6 

Answer: C

Explanation:

L={ababn∣n≥0}∪{aban∣n≥0} 

L={ab, aba , abaa  ,abab  ,ababb,..........}

The non-deterministic finite automation's smallest number of states that accepts the language =5 and so is the answer.

 

48. Functions defined with class name are called as 

(A) Inline function 

(B) Friend function 

(C) Constructor 

(D) Static function 

Answer: C

Explanation:

A constructor is different from normal functions in following ways:

  • Constructor has same name as the class itself
  • Constructors don’t have return type
  • A constructor is automatically called when an object is created.
  • If we do not specify a constructor, C++ compiler generates a default constructor for us (expects no parameters and has an empty body).

 

49. Let f be the fraction of a computation (in terms of time) that is parallelizable, P the number of processors in the system, and sp the speed up achievable in comparison with sequential execution – then the sp can be calculated using the relation : 

(A) 1/( 1 – f – f/P) 

(B) P /( P – f(P + 1)) 

(C) 1 /(1 – f + f/P )

(D) P / (P + f(P – 1))

Answer: C

Explanation:

Consider the execution of time without parallelism is 1.

Time must be spent for sequential calculation if f is the fraction of parallel computation (1-f).

Because the f fraction of the computation can be split among P processors, the computation will take f/P time to complete.

Total time for parallel computing = 1-f + f/P

 speed up = time without parallelism / time with parallelism = 1/ 1-f + f/P 

 

50. Which of the following definitions generates the same Language as L, where L = {WWR | W  {a, b}*} ?

(A) S → asb|bsa| Ɛ

(B) S → asa|bsb| Ɛ

(C) S →asb|bsa|asa|bsb| Ɛ

(D) S → asb|bsa|asa|bsb 

Answer:  B

Explanation:

   As given in the question 

 L={WWR∣W∈{a,b}*}.

This Language accepts Even lengths of the palindrome.

The strings set generated by language L are

{∈,aa,bb,abba,baab,bbbbbb,bbaabb....}

The first option is incorrect because S → asb|bsa| Ɛ does not generate even a length palindrome due to string ab (S->aSb->ab)

S → asa|bsb| Ɛ  definition accepts all even length palindromes. Ex:-

{∈,aa,bb,aaaa,abba,baab,bbbb etc.}

Because of string baaa (S->bSa->baSaa->baaa) we can eliminate option C because this is not even a length palindrome.

D option is also incorrect as we are not able to generate any string to stop repetition S.Hence S → asb|bsa|asa|bsb  also does not generate the same Language.

Frequently Asked Questions

What is the UGC NET exam?

UGC NET is a national-level exam organized by UGC to determine the eligibility of the candidates for lectureship and JRF.

What is the maximum number of attempts for the UGC NET examination?

There is no bar on the number of attempts of this examination. Candidates can appear for the examination as long as they are eligible.

How can solving PYQs help in my exam preparation?

Solving PYQs will give you a good idea about the exam pattern and help you identify your weak topics to prepare them better for the examination.

How many papers are there in the UGC NET exam?

There are two papers, and the candidates get 3 hours for both papers. There are 150 questions in UGC NET combining both papers.

Conclusion

We have extensively discussed the December 2012 paper-III. We hope that this blog has helped you understand the UGC pattern. You can refer to this article for more details on UGC NET 2022.  For the other questions you can visit this link December 2012 Paper-III Part-1 and December 2012 Paper-III Part-3.

You can refer to this article for more details on UGC NET 2022.

Refer to our guided paths on Coding Ninjas Studio to learn more about DSA, Competitive Programming, JavaScript, System Design, etc. Enroll in our courses and refer to the mock test and problems available, Take a look at the interview experiences and interview bundle for placement preparations.

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