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Table of contents
1.
Introduction
2.
June 2011 Paper II Part 1
3.
FAQs
3.1.
What is the UGC NET exam?
3.2.
What is the maximum number of attempts for the UGC NET examination?
3.3.
What is the full form of UGC NET?
3.4.
How can solving PYQs help in my exam preparation?
3.5.
How many papers are there in the UGC NET exam?
4.
Conclusion
Last Updated: Mar 27, 2024
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June 2011 Paper II Part 1

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Introduction

UGC NET Exam is a very popular exam 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. Refer to June 2011 Paper II Part 2 here.

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 June Paper II. We have also explained every problem adequately to help you learn better.

June 2011 Paper II Part 1

1. Any integer composed of 3n identical digits divisible by

(A) 2n            (B) 3n

(C) 5n           (D) 7n

Answer: B

Let's take n=1

Now 31=3 identical digits i.e

111 which is not divisible by 2,5,7

And divisible by 3.

All other 3 digit numbers like 222,333,444 are multiple of 111 and hence of 3.

Now, for 

n=2

n=2, we get 32=9. 111111111 is a multiple of 9. and similarly any 3n digit number composed of only 1, is divisible by 3n, and composed of any other number is also divisible by 3n.

so Option B is Correct Ans.

 

2. The circumferences of the two concentric disks are divided into 100 sections each. For the outer disk, 100 of the sections are painted red and 100 of the sections are painted blue. For the inner disk, the sections are painted red and blue in an arbitrary manner. It is possible to align the two disks so that …………… of the sections on the inner disks have their colours matched with the corresponding section on outer disk.

(A) 100 or more         (B) 125 or more

(C) 150 or more         (D) 175 or more

Answer: A

Inner disk.

Consider a half with r reds.

Half1 = r reds and 100-r whites

Half2 = R -r reds and 100-(R-r) = 100 - R +r whites.

If r > = R - r, match half1 with the Red half of the outer disk.

Total matching = r + 100 - R + r = 100 - R + 2r

Now r >= R - r => 2r - R >= 0

Total matching = 100 - R + 2r >= 100

 

3. The proposition ~pvq is equivalent to

(A) p->q                   (B) q->p

(C) p<->q                  (D) pvq

Answer: A

The expression ~pvq is equivalent to p->q.

 

4. The absorption law in Boolean algebra say that

(A) X + X = X

(B) X . X = X

(C) x + x . y = x

(D) None of the above

Answer: C

Options A & B are the Basic laws of Boolean Algebra.

Option C. Absorption means to reduce

Here  x+xy=x(1+y)=x 

x+xy absorbed or reduced to x 

 

5. The number of 1’s present in the binary representation of

10 × 256 + 5 × 16 + 5 is

(A) 5                (B) 6

(C) 7                (D) 8

Answer: B

10×256+5×16+5

=10×162+5×161+5×160

=(A55)16

=(1010 0101 0101)2

So 6 1's are there 

so option B is the Ans.

 

6. The hexadecimal number equivalent to (1762.46)8 is

(A) 3F2.89      (B) 3F2.98

(C) 2F3.89      (D) 2F3.98

Answer: B

(1762.46)= 3F2.98

 

7. (A + B)(AB)’ is equivalent to

(A) AÅB                   (B) AʘB

(C) (AÅB)ʘA          (D) (AʘB)ÅA

Answer: A
 

8. A latch is constructed using two cross-coupled

(A) AND and OR gates                 (B) AND gates

(C) NAND and NOR gates           (D) NAND gates

Answer: D

Latch Can be constructed using either 2 cross coupled NAND or NOR Gates

 

9. A multiplexer is a logic circuit that

(A) accepts one input and gives several output

(B) accepts many inputs and gives many outputs

(C) accepts many inputs and gives one output

(D) accepts one input and gives one output

Answer: C

Multiplexer is famously known as Many to one.

So it takes many input and gives one o/p.

 

10. 8-bit 1’s complement form of –77.25 is

(A) 01001101.0100

(B) 01001101.0010

(C) 10110010.1011

(D) 10110010.1101

Answer: C

8-bit 1’s complement form of –77.25 is 10110010.1011.

 

11. From the point of view of the programmer, what are the major advantages of using a high-level language rather than internal machine code or assembly language?

(A) Program portability         (B) Easy development

(C) Efficiency                        (D) None of the above

Answer: B

We moved from lower level language to high level language to make the program development task easy.

It reduces the efficiency because of translation overhead. But it ease the work of developer.

 

12. What features make C++ so powerful?

(A) Easy implementation                 (B) Reusing old code

(C) Easy memory management     (D) All of the above

Answer: D

All the given features are the ones which empowers C++.

 

13. The goal of operator overloading is

(A) to help the user of a class

(B) to help the developer of a class

(C) to help define friend function

(D) None of the above

Answer: A

The goal of operator overloading is to help the user of a class

 

14. The scheme by which the interpreter translates the source program is known as

(A) Paragraph by paragraph

(B) Instruction by instruction

(C) Line by line

(D) None of the above

Answer: C

The scheme by which the interpreter translates the source program is known as line by line scheme.

 

15. Portable program means

(A) Program with wheels                 (B) Independent from its authors

(C) Independent of platform           (D) None of the above

Answer: C

Portable program means independent of platform. 

 

16. Which of the following is the recovery management technique in DDBMS?

(A) 2PC (Two-Phase Commit)        (B) Backup

(C) Immediate update                      (D) All of the above

Answer: D

Option A:2PC (Different from 2phase locking) There are two phases for recovery management commit-request phase and commit phase

Option B & C: If Power failure occurs so backup can be an option if the data is not commited.

 

17. Which of the following is the process by which a user’s privileges are ascertained?

(A) Authorization               (B) Authentication

(C) Access Control            (D) None of these

Answer: A

  • A. Authorization is the process by which the user's privileges are ascertained.
  • B. Authentication is the process by which a user's identity is checked.
  • C. Access control is the process by which the user's access to physical data in the application is limited, based on his privileges. 

So, A is answer

 

18. The basic variants of the time-stamp based method of concurrency control are

(A) Total time stamp-ordering

(B) Partial time stamp ordering

(C) Multiversion Timestamp ordering

(D) All of the above

Answer: D

Following are the three basic variants of timestamp-based methods of concurrency control: 

(a) Total timestamp ordering : 

The total timestamp ordering algorithm depends on maintaining access to granules in  timestamp order by aborting one of the transactions involved in any conflicting access.  No distinction is made between Read and Write access, so only a single value is required for each granule timestamp .

(b)Partial timestamp ordering :

In a partial timestamp ordering, only non-permutable actions are ordered to improve upon the total timestamp ordering.  In this case, both Read and Write granule timestamps are stored. 

The algorithm allows the granule to be read by any transaction younger than the last  transaction that updated the granule. A transaction is aborted if it tries to update a granule that has previously been accessed by a younger transaction. The partial timestamp ordering algorithm aborts fewer transactions than the total  timestamp ordering algorithm, at the cost of extra storage for granule timestamps

(c) Multiversion Timestamp ordering : 

The multiversion timestamp ordering algorithm stores several versions of an updated  granule, allowing transactions to see a consistent set of versions for all granules it accesses.  So, it reduces the conflicts that result in transaction restarts to those where there is a  Write-Write conflict.  Each update of a granule creates a new version, with an associated granule timestamp. 

A transaction that requires read access to the granule sees the youngest version that is older than the transaction. That is, the version having a timestamp equal to or immediately below the transaction's  timestamp.

 

19. A transaction can include following basic database access operations :

(A) Read_item(X)               (B) Write_item(X)

(C) Both (A) and (B)          (D) None of these

Answer: C

Both are valid database access operations.

 

20. Decomposition help in eliminating some of the problems of bad design

(A) Redundancy                (B) Inconsistencies

(C) Anomalies                    (D) All of the above

Answer: D

Decomposition is used to eliminate some of the problems of bad design like anomalies, inconsistencies, and redundancy.

 

21. The number of different trees with 8 nodes is

(A) 256            (B) 255

(C) 248           (D) None of these

Answer: C

The number of different binary trees with 6 nodes is fact(2n) / fact(n+1) * fac(n) where n is no nodes:

If n= 8, then  fact(2 * n) / fact(n+1) * fac(n) 

= fact(2 * 8) / fact(8 + 1) * fact(8) 

= fact(16) / fact(9) * fact(8) 

= 248

 

22. Given a binary tree whose inorder and preorder traversal are given by

Inorder: EICFBGDJHK

Preorder: BCEIFDGHJK

The post order traversal of the above binary tree is

(A) IEFCGJKHDB     (B) IEFCJGKHDB

(C) IEFCGKJHDB    (D) IEFCGJKDBH

Answer: A

Preorder - BCEIFDGHJK means root node is B. Inorder - EICFBGDJHK

So search for B in Inorder traversal. Everything to the left of B is left subtree and to the right of B is right subtree in inorder traversal. So we understand that EICF forms the left subtree and GDJHK forms the right subtree of B.

Repeating, this for EICF, we find that C is the root and EI (or IE) forms the left subtree and F forms the right subtree of C. Similarly, for GDJHK, D is the root. G is the left subtree and JHK (or JKH) forms the right subtree.

Now finding postorder is very easy. The answer is option (A) IEFCGJKHDB

 

23. The number of disk accesses performed by insertion operation in B-tree of height h is

(A) 0(1)           (B) 0(1gh)

(C) 0(h)           (D) None of these

Answer: C

Inserting a key k into a B-tree T of height h is done in a single pass down the tree, requiring O(h) disk accesses.

 

24. Consider a hash table of size m = 10000 and the hash function h(k) = ∟m(kA mod 1)'┘ for A = (√5–1)/2. The location to the key k = 123456 is

(A) 46           (B) 47

(C) 41           (D) 43

Answer: C

Given hash function: h(K) = floor (m (K*A mod 1))

where A = ( √(5) – 1)/2

h(123456) = floor(10000 * (123456 * (√5 − 1) / 2) mod 1) 

          = floor(10000 * (76300.004115 mod 1)

          = floor(10000 * (.004115))

          = 41.15

          = 41

So, option (C) is correct.

 

25. When the priority queue is represented by max heap, the insertion and deletion of an element can be performed in (queue containing n elements)

(A) q(n) and q(1) respectively

(B) q(n) and q(n) respectively

(C) q(1) and q(1) respectively

(D) None of the above

Answer: D

Inserting Element take O(logn ) in worst case.(insert at least then find its correct possition)

Deletion Element take O(logn ) in worst case.( 1st we have to search that element)

Also, Read - Demultiplexer

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FAQs

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.

What is the full form of UGC NET?

UGC stands for University Grants Commission, and NET stands for National Eligibility Test.

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

In this article, We have extensively discussed the June 2011 paper II. 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.

Refer to June 2011 Paper II Part 2 here.

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 Learning!

 

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