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Nov 2017 Paper III Part 2

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Introduction

The National Eligibility Test (NET) is conducted on behalf of the University Grants Commission (UGC) to determine the eligibility of Indian nationals for the Eligibility of Assistant Professorship, Junior Research Fellowship, or both for Indian Universities and Colleges. In this blog, we will go through the questions and solutions of June 2014 Paper-II of UGC - NET EXAM. In this blog, we have covered questions from 26-50. For questions from 1-25 refer Nov 2017  Paper III Part 1 and from 51 - 75 refer.

Questions

26. Quadrature Amplitude Modulation means changing both:

(1) Frequency and phase of the carrier.

(2) Frequency and Amplitude of the carrier.

(3) Amplitude and phase of the carrier.

(4) Amplitude and Wavelength of the carrier.

Answer - 3
Solution - Quadrature Amplitude Modulation means changing both Amplitude and phase of the carrier.
 

27. If a file consisting of 50,000 characters takes 40 seconds to send, then the data rate is ........................

(1) 1 kbps

(2) 1.25 kbps

(3) 2 kbps

(4) 10 kbps

 

Answer - 4
Solution - If a file consisting of 50,000 characters takes 40 seconds to send, then the data rate is 10 kbps.
 

28. Match the following:

List - I

(a) Data link layer

(b) Network layer

(c) Transport layer

(d) Presentation layer

List - II

(i) Encryption

(ii) Connection control

(iii) Routing

(iv) Framing

Code :

   (a) (b) (c) (d)

(1) (iv) (iii) (i) (ii)

(2) (iii) (iv) (ii) (i)

(3) (iv) (ii) (iii) (i)

(4) (iv) (iii) (ii) (i)

 

Answer - 4
 

29. The address of a class B host is to be split into subnets with a 6 - bit subnet number. What is the maximum number of subnets and maximum number of hosts in each subnet?

(1) 62 subnets and 1022 hosts

(2) 64 subnets and 1024 hosts

(3) 62 subnets and 254 hosts

(4) 64 subnets and 256 hosts

Answer - 1
Solution - In class B, the first two octets are reserved for NID, while the remaining octets are used for HID. As a result, the first six bits of the third octet are used for subnet, while the remaining ten bits are used for hosts.

The maximum number of subnets is 262, which equals 62.

Subtracting 2 reduces the number of available subnets by two in classic subnetting because subnet values consisting of all zeros and all ones (broadcast) reduce the number of available subnets by two. We can also have 64 in current networks.

and the number of hosts is 2^10-2=1022.

2 has been deducted. The number of hosts is also important. The address with all bits set to 1 is reserved for broadcasting, whereas the address with all host id bits set to 0 is utilised as the subnet's network address.

 

30. Which of the following statements are true?

(a) The fragmentation fields in the base header section of IPv4 have moved to the fragmentation extension header in IPv6.

(b) The authentication extension header is new in IPv6.

(c) The record route option is not implemented in IPv6.

Code :

(1) (a) and (b) Only

(2) (b) and (c) Only

(3) (a) and (c) Only

(4) (a), (b) and (c)

 

Answer - 4
Solution - Option 4 is the correct response. The fragmentation fields from IPv4's base header segment have been migrated to the fragmentation extension header in IPv6, making the following statement correct. The authentication header is the same in IPv4 and IPv6, but IPv6 contains a new authentication extension header.
 

31. Consider a full binary tree with n internal nodes, internal path length i, and external path length e. The internal path length of a full binary tree is the sum, taken over all nodes of the tree, of the depth of each node. Similarly, the external path length is the sum, taken over all leaves of the tree, of the depth of each leaf.

Which of the following is correct for the full binary tree?

(1) e = i + n

(2) e = i + 2n

(3) e = 2i + n

(4) e = 2n + i

 

Answer - 2
Solution -

Internal path length i = d(A) + d(B) + d(C) =  0 + 1 + 1 = 2

External path length e = d(D) + d(E) + d(F) + d(G) = 2 + 2 + 2 + 2 = 8

e = i + 2n = 2 + 2(3) = 8 can be seen.

Any whole binary tree can be checked to see if this is true.

The answer is e = i + 2n.

 

32. You are given a sequence of n elements to sort. The input sequence consists of n/k subsequences, each containing k elements. The elements in a given subsequence are all smaller than the elements in the succeeding subsequence and larger than the elements in the preceding subsequence. Thus, all that is needed to sort the whole sequence of length n is to sort the k elements in each of the n/k subsequences.

The lower bound on the number of comparisons needed to solve this variant of the sorting problem is:

(1) Ω(n)

(2) Ω(n/k)

(3) Ω(n lg k)

(4) Ω(n/k lg n/k)

 

Answer - 3

Solution - There are n/k subsequences, each with k elements.

When sorting k items in a subsequence with the quick sort technique, the difficulty of sorting k elements in a subsequence is klogk.

We have n/k sequences like this. Then, for n/k subsequences with k elements, the time complexity is:

klogk = (n/k)*

=nlogk

 

33. Consider the recurrence relation:

T (n) = 8T(n/2) + Cn, if n > 1

   = b, if n =1

Where b and c are constants.

The order of the algorithm corresponding to above recurrence relation is:

(1) n

(2) n2

(3) n lg n

(4) n3

 

Answer - 4

Solution - The order of the algorithm corresponding to above recurrence relation is n3.
 

34. Consider the following two sequences :

X = <B, C, D, C, A, B, C>

and Y = <C, A, D, B, C, B>

The length of longest common subsequence of X and Y is:

(1) 5

(2) 3

(3) 4

(4) 2

 

Answer - 3

Solution - We looked for a length 5 sub-sequence between both sequences at first but couldn't discover one.

The length 4 sub-sequences were then verified, and CDBC and CDCB two sub-sequences were discovered.
 

35. A text is made up of the characters a, b, c, d, e each occurring with the probability 0.11, 0.40, 0.16, 0.09 and 0.24 respectively. The optimal Huffman coding technique will have the average length of:

(1) 2.40

(2) 2.16

(3) 2.26

(4) 2.15

 

Answer - 2

Solution -  a = 0.11 b = 0.40 c = 0.16 d = 0.09 e = 0.24

now huffman coding for character: 

a = 1111

      b = 0

      c = 110

      d = 1111

      e = 10

length for each character = no of bits * frequency of occurence: a = 4 * 0.11 = 0.44 b = 1 * 0.4 = 0.4 c = 3 * 0.16 = 0.48 d = 4 * 0.09 = 0.36 e = 2 * 0.24 = 0.48

Add these lengths together to get an average length:

2.16 = 0.44 + 0.4 + 0.48 + 0.36 + 0.48

 

36. An undirected graph G (V, E) contains n (n > 2) nodes named v1, v2,...,vn. Two nodes vi and vj are connected if and only if 0 < | i – j | ≤ 2. Each edge (vi, vj) is assigned a weight i+j.

The cost of the minimum spanning tree of such a graph with 10 nodes is :

(1) 88

(2) 91

(3) 49

(4) 21

Answer - 2

Solution - 

 We observe a pattern in the weight of MST being formed 

 For n=3 (1+2+3)+(1)

 For n=4 (1+2+3+4)+(1+2)

 For n=5 (1+2+3+4+5)+(1+2+3)

 These can be obtained by drawing graphs for these graphs. 

           ∴ Total weight of MST is n^2−n+1

 

37. An XML document that adheres to syntax rules specified by XML 1.0 specification in that it must satisfy both physical and logical structured, is called :

(1) Well - formed

(2) Reasonable

(3) Valid

(4) Sophisticated

 

Answer - 1

Solution - A well-formed XML document follows the syntax standards stated in the XML 1.0 specification, in that it must satisfy both physical and logical structures.

 

38. Which of the following statement(s) is/are TRUE regarding Java Servelets?

(a) A Java Servelet is a server-side component that runs on the web server and extends the capabilities of a server.

(b) A Servelet can use the user interface classes like AWT or Swing.

Code :

(1) Only (a) is TRUE.

(2) Only (b) is TRUE.

(3) Both (a) and (b) are TRUE.

(4) Neither (a) nor (b) is TRUE.

 

Answer - 1

Solution - A Java Servlet is a server-side component that runs on a web server and expands the server's capabilities. It is an accurate statement.

Interface classes like as AWT and Swing cannot be used by a Java servlet.

The first statement is the only one that is correct.

 

39. Consider the following HTML table definition :

<table border=1>

<tr>

<td colspan=2> Text A </td>

</tr>

<tr>

<td> Text B </td>

<td> Text C </td>

</tr>

<tr>

<td rowspan=2> Text D </td>

<td> Text E </td>

</tr>

<tr>

<td> Text F </td>

</tr>

</table>

The above HTML code would render on screen as :

Answer - 3

Solution - cd

ab ef

ij gh

ab -> rowspan “2”

cd -> colspan “2”

1st row closed

<tr> is row

<td> is data

ef placed in 2nd row. gh is in same row but rowspan “2”

2nd row also closed.

Move to 3rd row. Insert ij . its colspan “2”.

 

40. Which of the following statements is/are TRUE?

(a) In HTML, character entities are used to incorporate external content into a web page, such as images.

(b) Once a web server returns a cookie to a browser, the cookie will be included in all future requests from the browser to the same server.

Code :

(1) Only (a) is TRUE.

(2) Only (b) is TRUE.

(3) Both (a) and (b) are TRUE.

(4) Neither (a) nor (b) is TRUE.

 

Answer - 4

Solution - Character entities are used in HTML to include external material, like as photos, within a web page. This is a false statement.

When a browser receives a cookie from a web server, the cookie is included in all subsequent requests from the browser to the same server. This is also an incorrect assertion.

Both statements are incorrect.

As a result, option (D) is the correct answer.

 

41. Which of the following statements is/are TRUE regarding JAVA?

(a) Constants that cannot be changed are declared using the ‘static’ keyword.

(b) A class can only inherit one class but can implement multiple interfaces.

Code :

(1) Only (a) is TRUE.

(2) Only (b) is TRUE.

(3) Both (a) and (b) are TRUE.

(4) Neither (a) nor (b) is TRUE.

 

Answer - 2

Solution - Constants are not stated with the'static' keyword in JAVA, and while a class can implement several interfaces, it can only inherit one class.

As a result, option (B) is the correct answer.

 

42. What is the output of the following JAVA program?

Class Test

{

public static void main (String [ ] args)

{

Test obj = new Test ( );

obj.start ( );

}

void start ( )

{

String stra=”do”;

String strb = method (stra);

System.out.print(“:” + stra + strb);

}

String method (String stra)

{

stra=stra+”good”;

System.out.print (stra);

return “good”;

}

}

(1) dogood : dogoodgood

(2) dogood : gooddogood

(3) dogood : dodogood

(4) dogood : dogood

 

Answer - 4

Solution - The solution of the given JAVA Code is dogood : dogood.

 

43. Statistical software quality assurance in software engineering involves ................

(1) using sampling in place of exhaustive testing of software.

(2) surveying customers to find out their opinions about product quality.

(3) tracing each defect to its underlying cause, isolating the vital few causes, and moving to correct them.

(4) tracing each defect to its underlying causes, and using the Pareto principle to correct each problem found.

 

Answer - 3

Solution - In software engineering, statistical software quality assurance entails tracing each issue to its underlying cause, isolating the critical few causes, and working to remedy them.

As a result, option (C) is the correct answer.
 

44. Which of the following statements is/are FALSE with respect to software testing?

S1 : White-box tests are based on specifications; better at telling whether program meets specification, better at finding errors of ommission.

S2 : Black-box tests are based on code; better for finding crashes, out of bounds errors, file not closed errors.

S3 : Alpha testing is conducted at the developer’s site by a team of highly skilled testers for software that is developed as a product to be used by many customers.

(1) Only S1 and S2 are FALSE.

(2) Only S1 and S3 are FALSE.

(3) Only S2 and S3 are FALSE.

(4) All of S1, S2, and S3 are FALSE.

 

Answer - 1

Solution - A team of highly skilled testers conducts alpha testing at the developer's site for software that is produced as a product to be utilised by a large number of customers.

The distinction between white box and black box testing is incorrect.

The only one that is correct is the last one.

As a result, option (A) is the correct answer.

 

45. A signal processor software is expected to operate for 91.25 days after repair, and the mean software repair time is expected to be 5 minutes. Then, the availability of the software is :

(1) 96.9862%

(2) 97.9862%

(3) 98.9962%

(4) 99.9962%

 

Answer - 4

Solution - Availability

= MTBF /(MTBF+MTTR)*100

= 91.25*24*60 / (91.25*24*60+5)*100

= 99.9962%

 

46. Consider the method mcq ( ) :

int mcq (boolean a, boolean b, boolean c, boolean d)

{

int ans=1;

if (a) {ans = 2;}

else if (b) {ans = 3;}

else if (c) {

if (d) {ans=4;}

}

return ans ;

}

If

M1 = Number of tests to exhaustively test mcq ( );

M2 = Minimum number of tests to achieve full statement coverage for mcq ( ); and

M3 = Minimum number of tests to achieve full branch coverage for mcq ( );

then (M1, M2, M3) = ....................

(1) (16, 3, 5)

(2) (8, 5, 3)

(3) (8, 3, 5)

(4) (16, 4, 4)

 

Answer - 1

Solution - (M1, M2, M3) = (16, 3, 5)

 

47. A simple stand - alone software utility is to be developed in ’C’ programming by a team of software experts for a computer running Linux and the overall size of this software is estimated to be 20,000 lines of code. Considering (a, b) = (2.4, 1.05) as multiplicative and exponention factor for the basic COCOMO effort estimation equation and (c, d)=(2.5, 0.38) as multiplicative and exponention factor for the basic COCOMO development time estimation equation, approximately how long does the software project take to complete?

(1) 10.52 months

(2) 11.52 months

(3) 12.52 months

(4) 14.52 months

 

Answer - 2

Solution - 20,000 LOC (Lines of Code) = 20K LOC (Given)

(a,b) = (2.4, 1.05)       (Given)

(c,d) = (2.5,0.38)        (Given)

The basic COCOMO model gives an approximate estimate of the project parameters.

Effort = 2.4 * (20)1.05 PM

          = 2.4 * 23.231 = 55.756 PM

dev = 2.5 * (55.756)0.38 months

         = 2.5 * 4.608 = 11.52 months

Therefore, time taken for completion of the software project  is 11.52 months.

 

48. In Software Configuration Management (SCM), which of the following is a use-case supported by standard version control systems?

(a) Managing several versions or releases of a software

(b) Filing bug reports and tracking their progress

(c) Allowing team members to work in parallel

(d) Identifying when and where a regression occurred

Code :

(1) Only (a), (c), and (d)

(2) Only (a), (b), and (c)

(3) Only (a), (b), and (d)

(4) Only (b), (c), and (d)

 

Answer - 1

Solution - SCM is in charge of managing many software versions or releases.

Not managed by SCM is the filing of bug reports and tracking of their progress.

SCM is in charge of allowing team members to work in parallel.

SCM is in charge of determining when and where a regression happened.

As a result, option (A) is the correct answer.

 

49. Consider the following four processes with the arrival time and length of CPU burst given in

milliseconds :

Process           Arrival Time  Burst Time

P1                 0                8

P2                 1                4

P3                 2                9

P4                 3                5

The average waiting time for preemptive SJF scheduling algorithm is .................

(1) 6.5 ms

(2) 7.5 ms

(3) 6.75 ms

(4) 7.75 ms

 

Answer - 1

Solution - The average waiting time for a preemptive SJF scheduling algorithm is 6.5 ms.

 

50. Consider a virtual page reference string 7, 0, 1, 2, 0, 3, 0, 4, 2, 3, 0, 3, 2, 1, 2, 0, 1, 7, 0, 1. Suppose a demand paged virtual memory system running on a computer system such that the main memory has 3 page frames. Then ................. page replacement algorithm has minimum number of page faults.

(1) FIFO

(2) LIFO

(3) LRU

(4) Optimal

 

Answer - 4

Solution - Optimal page replacement algorithm has the minimum number of page faults.
 

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FAQs

What is the UGC NET Exam?

UGC NET stands for University Grants Commission National Eligibility Test. It is a national-level exam to determine eligibility for lectureship and Junior Research Fellowship (JRF) in universities or colleges across India.

What is the age limit to apply for the UGC NET Exam?

There is no age limit for the assistant professor, while the age of candidates for JRF should not exceed 31 years.

Am I eligible for UGC NET after a PG diploma?

PGDM is considered equivalent to a master's degree. So, if you have completed your PG Diploma from a recognized university, you can apply for the exam.  

Are there negative markings in the UGC NET Exam?

No, there is no negative marking in the UGC NET Exam.

What is the syllabus for UGC NET?

NTA conducts UGC NET Exam in two Papers, i.e., General Paper-1(common for all subjects) and Paper-2 (subject-specific). All Paper 2 subjects have a different syllabus. 

Conclusion

In this article, we have extensively discussed the approach and solutions to the past year's questions asked in UGC-NET 2014. 

In this blog, we have covered questions from 26-50. For questions from 1-25 refer Nov 2017  Paper III Part 1 and from 51 - 75 refer.

We hope this blog has helped you enhance your knowledge of the concepts and process of the questions.

If you want to learn more, check out our articles on Online Coding CoursesCoding Ninjas StudioInterview-experiencesTest-series

Refer to our guided paths on Coding Ninjas Studio to learn more about DSA, Competitive Programming, JavaScript, System Design, etc. 

Enrol 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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Happy Coding!

Topics covered
1.
Introduction
2.
Questions
3.
FAQs
3.1.
What is the UGC NET Exam?
3.2.
What is the age limit to apply for the UGC NET Exam?
3.3.
Am I eligible for UGC NET after a PG diploma?
3.4.
Are there negative markings in the UGC NET Exam?
3.5.
What is the syllabus for UGC NET?
4.
Conclusion