June 2014 Paper II Part 2

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 to 50. For questions from 1-25, refer to June 2014 Paper II Part 1.

Questions

1. Match the following with respect to the mobile computing technologies: List – I List – II a. GPRS i. An integrated digital radio standard b. GSM ii. 3G wireless/Mobile technology c. UMTS iii. Nine different schemes for modulation and error correction d. EDGE iv. An emerging wireless service that offers a mobile data Codes: a b c d 

(A) iii iv ii i (B) iv i ii iii (C) ii iii iv i (D) ii i iv iii

 

ANS - (B)
Solution - The correct matching order is “iv   i   ii    iii”.
 

2. Object Request Broker (ORB) is I. A software program that runs on the client as well as on the application server. II. A software program that runs on the client-side only. III. A software program that runs on the application server, where most of the components reside. 

(A) I, II & III (B) I & II (C) II & III (D) I only

 

ANS - (D)

Solution - It should be D because only ORB is taken from CORBA(common object request broker architecture).
 

3. A software agent is defined as I. Software developed for accomplishing a given task. II. A computer program which is capable of acting on behalf of the user in order to accomplish a given computational task. III. An open-source software for accomplishing a given task. 

(A) I (B) II (C) III (D) All of the above 
 

ANS - (B)

Solution - A software agent in AI is defined as a computer program capable of acting on behalf of the user to accomplish a given computational task.
 

4. Match the following: List – I List – II a. Classification i. Principal component analysis b. Clustering ii. Branch and Bound c. Feature Extraction iii. K-nearest neighbor d. Feature Selection iv. K-means Codes: a b c d 

(A) iii iv ii i (B) iv iii i ii (C) iii iv i ii (D) iv iii ii i 
 

ANS - (C)

Solution - 

List – I List – II

 

Classification ===> K-nearest neighbour  

Clustering ===> K-means   

Feature Extraction ===> Principal component analysis

Feature Selection ===> Branch and Bound
 

5. SET, an open encryption and security specification model that is designed for protecting credit card transactions on the internet stands for 

 

(A) Secure Electronic Transaction (B) Secular Enterprise for Transaction (C) Security Electronic Transmission (D) Secured Electronic Termination

 

ANS - (A)

Solution - SET stands for secure electronic transactions.

 

6. In a paged memory management algorithm, the hit ratio is 70%. If it takes 30 nanoseconds to search Translation Look-aside Buffer (TLB) and 100 nanoseconds (ns) to access memory, the effective memory access time is 

(A) 91 ns (B) 69 ns (C) 200 ns (D) 160 ns 

 

ANS - (D)

Solution - 

Effective access time = [%age of TLB (TLB time + Memory access time)]+[%age of Memory (TLB time+Memory access time+Memory access time)]

=0.7*(30+100) + 0.3 (30+100+100)

=91+69

=160 ns

 

7. Match the following: List – I List – II a. Multilevel feedback queue i. Time-slicing b. FCFS ii. Criteria to move processes between queues c. Shortest process next iii. Batch processing d. Round robin scheduling iv. Exponential smoothening

Codes : a b c d (A) i iii ii iv (B) iv iii ii i (C) iii i iv i (D) ii iii iv i 
 

ANS - (D)

Solution - 

Multilevel feedback queue ---> Criteria to move processes between queues

FCFS ---> Batch processing

The shortest process next ---> Exponential smoothening

Round robin scheduling ---> Time-slicing
 

8. Consider a system with five processes, P0 through P4, and three resource types R1, R2, and R3. Resource type R1 has 10 instances, R2 has 5 instances, and R3 has 7 instances. Suppose that at time T0, the following snapshot of the system has been taken: Allocation R1 R2 R3 P0 0 1 0 P1 2 0 0 P2 3 0 2 P3 2 1 1 P4 0 2 2 Max R1 R2 R3 7 5 3 3 2 2 9 0 2 2 2 2 4 3 3 Available R1 R2 R3 3 3 2 Assume that now the process P1 requests one additional instance of type R1 and two instances of resource type R3. The state resulting after this allocation will be 

(A) Ready state (B) Safe state (C) Blocked state (D) Unsafe state 
 

ANS - (B)

Solution - 

By taking one instance of R1, P1 will fulfill its need for maximum R1

and By taking two instances of R3, P1 will fulfill its need for maximum R3

Now, it needs two instances of R2, after which it can complete execution and release all of its resources, hence it is in a safe state.
 

9. Match the following: List – I List – II a. Contiguous allocation i. This scheme supports very large file sizes. b. Linked allocation ii. This allocation technique supports only sequential files. c. Indexed allocation iii. The number of disks required to access files is minimal. d. Multi-level indexed iv. This technique suffers from maximum wastage of space in storing pointers. Codes: a b c d 

(A) iii iv ii i (B) iii ii iv i (C) i ii iv iii (D) i iv ii iii 

 

ANS - (B)

Solution - 

Contiguous allocation ---> Number of disks required to access the file is minimal.

Linked allocation ---> This allocation technique supports only sequential files

Indexed allocation ---> This technique suffers from maximum wastage of space 

Multi-level indexed ---> This scheme supports very large file sizes
 

10. Which of the following commands will output "onetwothree"? (

A) for val; do echo-n $val; done < one two three (B) for one two three; do echo-n-; done (C) for n in one two three; do echo-n $n; done (D) for n in one two three {echo –n $ n}

 

ANS - (C)

Solution - for n in one two three; do echo -n $n; done.

 

11. Mergesort makes two recursive calls. Which statement is true after these two recursive calls finish but before the merge step? 

(A) The array of elements forms a heap. (B) Elements in each half of the array are sorted amongst themselves. (C) Elements in the first half of the array are less than or equal to elements in the second half of the array. (D) All of the above

 

ANS - (B)

Solution - Elements in each half of the array are sorted amongst themselves.

12. A text is made up of the characters α, β, γ, δ, and σ with the probability of 0.12, 0.40, 0.15, 0.08, and 0.25, respectively. The optimal coding technique will have the average length of 

(A) 1.7 (B) 2.15 (C) 3.4 (D) 3.8 
 

ANS - (B)

Solution - 

Alpha=.12  Beta=.40  Gamma=.15  Delta=.08  Sigma=.25  (All are given in question)

Path length with 4= Alpha, Delta. (.48+.32)

Path length with 3= gamma.(.45).

Path length with 2=sigma..(.50).

Path length with 1= beta(.40).

Average path length=2.15.

 

13. Searching for an element in the hash table requires O(1) time for the ________ time, whereas for direct addressing it holds for the ________ time. 

(A) worst-case, average (B) worst-case, worst-case (C) average, worst-case (D) best, average

 

ANS - (C)

Solution - Searching for an element in the hash table requires O(1) time for the Average time, whereas, for direct addressing, it holds for the Worst-case time.

 

14. An algorithm is made up of 2 modules, M1 and M2. If the time complexity of modules M1 and M2 are h(n) and g(n), respectively, the time complexity of the algorithm is 

(A) min (h(n), g(n)) (B) max (h(n), g(n)) (C) h(n) + g(n) (D) h(n) * g(n) 

 

ANS - (B)

Solution - 

f(n) = nlogn

g(n) = n^2

Order of algorithm = max(f(n), g(n))

Therefore, order of algorithm = max(nlogn, n^2) = n^2

 

15. What is the maximum number of parenthesis that will appear on the stack at any one time for parenthesis expression given by ( ( ) ( ( ) ) ( ( ) ) ) 

(A) 2 (B) 3 (C) 4 (D) 5 
 

ANS - (B)

Solution - In the entire parenthesis balancing method, it is pushed into the stack when the incoming token is a left parenthesis. A right parenthesis makes the pop operation to delete the elements in the stack till we get the left parenthesis as the topmost element. 3 elements are there in the stack before the right parenthesis comes. Therefore, the maximum number of elements in the stack at run time is 3. So option C is correct.

 

16. Match the following: List – I List – II a. Automatic storage class i. The scope of the variable is global. b. Register storage class ii. The value of the variable persists between different function calls. c. Static storage class iii. The value stored in memory and local to the block in which the variable is defined. d. External storage class iv. Value stored in CPU registers. 

Codes : a b c d (A) iii iv i ii (B) iii iv ii i (C) iv iii ii i (D) iv iii i ii 

 

ANS - (B)

Solution - Automatic storage class- scope local to block. Outside the value will never exist.

Register storage class - The register storage class is used to define local variables that should be stored in a register instead of RAM.

Static storage class- These have memory allocated during the program run, which survives until the program ends. The scope of the variable -- where we can access it - is the block in which it is declared or if it is declared in global space - is restricted to that compilation unit (this differentiates it from global variables, which can be used even in other compilation units). For example, in an OOP environment, consider a banking system. Now I can make the amount of money in the bank as a static variable as this is modified by any of the ACCOUNT objects, but overall we have a single balance amount for the bank.  

External storage class- The extern storage class is used to reference a global variable defined in some other compilation unit.

 

17. When we pass an array as an argument to a function, what actually gets passed? 
 

(A) Address of the array (B) Values of the elements of the array (C) Base address of the array (D) Number of elements of the array 
 

ANS - (C)

Solution - In C language when we pass an array as a function argument, then the base address of the array will be passed.
 

18. While (87) printf("computer"); The above C statement will 

(A) print "computer" 87 times (B) print "computer" 0 times (C) print "computer" 1 time (D) print "computer" infinite times

 

ANS - (D)

Solution - print "computer" infinite times
 

19. A friend function can be used to 

(A) avoid arguments between classes. (B) allow access to classes whose source code is unavailable. (C) allow one class to access an unrelated class. (D) None of the above

 

ANS - (C)

Solution - In OOP, a friend function that is a "friend" of a given class is allowed access to private and protected data in that class that it would not normally be able to as if the data was public. Normally, a function that is defined outside of a class cannot access such information. Declaring a function a friend of a class allows this, in languages where the concept is supported.
 

20. Which of the following is the correct value returned to the operating system upon the successful completion of a program? 

(A) 0 (B) 1 (C) – 1 (D) Program do not return a value. 

 

ANS - (A)

Solution - After the successful completion of a program, 0 is returned to the operating system.

 

21. Manager's salary details are hidden from the employee. This is called as 

(A) Conceptual level data hiding (B) Physical level data hiding (C) External level data hiding (D) Local-level data hiding
 

ANS - (C)

Solution - Manager's salary details are hidden from the employee which is known as External level data hiding. 

 

22. Which of the following statements is false? 

(A) Any relation with two attributes is in BCNF. (B) A relation in which every key has only one attribute is in 2NF. (C) A prime attribute can be transitively dependent on a key in 3NF relation. (D) A prime attribute can be transitively dependent on a key in BCNF relation.
 

ANS - (D)

Solution - If a relational schema is in BCNF, all redundancy based on functional dependency has been removed, although other types of redundancy may still exist.

A relational schema R is in BCNF form if and only if, for every one of its dependencies X → Y, at least one of the following conditions holds:

1. X → Y is a trivial functional dependency, i.e., Y is a subset of X.

2. X is a super key for schema R

So, we check transitive dependency for only non-prime attributes in the case of BCNF.

Thus, the statement 'A prime attribute can be transitively dependent on a key in a BCNF relation' is false.

 

23. A clustering index is created when _______. 

(A) the primary key is declared and ordered (B) no key ordered (C) foreign key ordered (D) there is no key and no order

 

ANS - (A)

Solution - A clustering index = ( non key attribute + ordered )

 

24. Let R ={A, B, C, D, E, F} be a relation schema with the following dependencies C → F, E → A, EC → D, A → B. Which of the following is a key for R? 

(A) CD (B) EC (C) AE (D) AC 

 

ANS - (B)

Solution - the closure for all

{CD} = {CDF}

{EC} = {ECDAFB} Only this is the key because it drives all attributes.

{AE} = {AEB}

{AC} = {ACBF}

Hence, Option (B) is Correct.
 

25. Match the following: List – I List – II a. DDL i. LOCK TABLE b. DML ii. COMMIT c. TCL iii. Natural Difference d. BINARY Operation iv. REVOKE 

Codes : a b c d (A) ii i iii iv (B) i ii iv iii (C) iii ii i iv (D) iv i ii iii 
 

ANS - (D)

Solution - Revoke is both DDL and DCL, so A-IV is only in option D. Hence, the answer is option D.