UGC NET June 2013 Paper III Part-3

Introduction

UGC NET Exam is a very popular exam in India for people interested in the research domain. NET(National Eligibility Test) is conducted by NTA (NATIONAL TESTING AGENCY) to identify qualified candidates for Assistant Professor positions in colleges and universities, as well as Junior Research Fellowships (JRF). This article will discuss the UGC NET Exam of June 2013 paper III solutions from questions 51 to 75. You can also check out the remaining questions and solutions solution of this paper in UGC NET June 2013 Paper III Part-1 and UGC NET June 2013 Paper III Part 2

Trace the error : void main( ) { int *b, &a; *b = 20 printf(“%d, %d”, a, *b) }
1. No error
2. Logical error
3. Syntax error
4. Semantic error
  
  Ans- C
  Explanation: The given program will give syntax error as the declaration of the variable 'a' is wrong.
Match the following :
a. calloc( ) i. Frees previously allocated space
b. free( ) ii. Modifies previously allocated space
c. malloc( ) iii. Allocates space for array
d. realloc( ) iv. Allocates requested size of space Codes :
a b c d
1. iii i iv ii
2. iii ii i iv
3. iii iv i ii
4. iv ii iii i
  
  Ans-A
  Explanation: for a detailed solution click here
Binary symmetric channel uses
1. Half duplex protocol
2. Full duplex protocol
3. Bit oriented protocol
4. None of the above
  
  Ans-A
  Explanation: In this model, a transmitter wishes to send a bit(0 or 1), and the receiver receives a bit. It uses bit-oriented protocol as the number/pattern of bits here signifies if the received information is correct or not.
Hamming distance between 100101000110 and 110111101101 is
1. 3
2. 4
3. 5
4. 6
  
  Ans-D
  Explanation: Hamming distance between 100101000110 and 110111101101 is = no of 1's in 100101000110 XOR 110111101101 = 6.
Given code word 1110001010 is to be transmitted with even parity check bit. The encoded word to be transmitted for this code is
1. 11100010101
2. 11100010100
3. 1110001010
4. 111000101
  
  Ans-A
  Explanation: Because even parity will make the number of 1's even, as seen by the fact that there are five 1's in the transmitted data 1110001010, which is odd. This will add an extra 1 as parity to make it even. As a result, it will be 11100010101.
The number of distinct binary images which can be generated from a given binary image of right M × N are
1. M + N
2. M × N
3. 2^(M + N)
4. 2^(MN)
  
  Ans- D
  Explanation: A binary image B has 0's and 1's as pixels; the 1's will be foreground pixels, while the 0's will be background pixels. The value of the pixel located at row r, column c of the picture array is denoted by the word B[r,c]. M rows numbered 0 to M-1 and N columns numbered 0 to N-1 make up an MN picture. Thus, B[0,0] denotes the value of the image's upper leftmost pixel, whereas B[M-1, N-1] denotes the value of the image's bottom rightmost pixel. As previously stated, the pixels of a binary picture B are 0's and 1's, hence the total number of possible images is 2^(MN).
If f(x, y) is a digital image, then x, y and amplitude values of f are
1. Finite
2. Infinite
3. Neither finite nor infinite
4. None of the above
  
  Ans: A
  Explanation: An image may be defined as a two-dimensional function, f(x, y), where x and y are spatial (plane) coordinates, and the amplitude of any pair of coordinates (x, y) is called the intensity or gray level of the image at that point. When x, y, and the amplitude values of f are all finite, discrete quantities, we call the image a digital image.
Consider the following processes with time slice of 4 milliseconds (I/O requests are ignored) : Process A B C D Arrival time 0 1 2 3 CPU cycle 8 4 9 5 The average turn around time of these processes will be
1. 19.25 milliseconds
2. 18.25 milliseconds
3. 19.5 milliseconds
4. 18.5 milliseconds
  
  Ans: B
  Explanation:
  
  Average TAT= 20+7+24+22/4=> 18.25
A job has four pages A, B, C, D and the main memory has two page frames only. The job needs to process its pages in following order : ABACABDBACD Assuming that a page interrupt occurs when a new page is brought in the main memory, irrespective of whether the page is swapped out or not. The number of page interrupts in FIFO and LRU page replacement algorithms are
1. 9 and 7
2. 7 and 6
3. 9 and 8
4. 8 and 6
  
  Ans: C
  Explanation: FIFO is first in first out. LRU is least recently used
Suppose S and Q are two semaphores initialized to 1. P1 and P2 are two processes which are sharing resources. P1 has statements P2 has statements wait(S) ; wait(Q) ; wait(Q) ; wait(S) ; critical- section 1; critical- section 2; signal(S) ; signal(Q) ; signal(Q) ; signal(S) ; Their execution may sometimes lead to an undesirable situation called
1. Starvation
2. Race condition
3. Multithreading
4. Deadlock
  
  Ans: D
  Explanation: Consider a situation in which P1 runs wait(S) and sets S=0. P2 now executes wait(Q) and sets Q to 0. Because S and Q are both equal to 0, neither P1 nor P2 can now execute wait(Q) or wait(S). P1 is currently waiting for P2 to deliver resource Q, while P2 is waiting for P1 to release resource S. As a result, neither of them can enter the vital region, resulting in a deadlock.
An operating system using banker’s algorithm for deadlock avoidance has ten dedicated devices (of same type) and has three processes P1, P2 and P3 with maximum resource requirements of 4, 5 and 8 respectively. There are two states of allocation of devices as follows : State 1 Processes P1 P2 P3 Devices allocated 2 3 4 State 2 Processes P1 P2 P3 Devices allocated 0 2 4 Which of the following is correct ?
1. State 1 is unsafe and state 2 is safe.
2. State 1 is safe and state 2 is unsafe.
3. Both, state 1 and state 2 are safe.
4. Both, state 1 and state 2 are unsafe.
  
  Ans: A
  Explanation:1st state:
  The total number of devices assigned is 2+3+4 =9.
  The number of devices available is 10-9, which equals one.
  The number of devices still required for P1 = 4-2 =2 >1 /
  The number of devices still required for P2 = 5-3 =2 >1 /
  The number of devices still required for P2 = 8-4 =4 >1 /
  State 1 is dangerous.
  2nd state
  The total number of devices assigned is 0+2+4 =6.
  The number of devices available is 10-6, which equals four.
  The number of devices required for P1 is still 4-0 =4 =4 /Can allocate
  The number of devices required for P2 = 5-2 =34 / Possibility of allocating
  The number of devices required for P2 is still 8-4 =4 =4 /Can allocate State 2 is risk-free.
Let the time taken to switch between user mode and kernel mode of execution be T1 while time taken to switch between two user processes be T2. Which of the following is correct ?
1. T1 < T2
2. T1 > T2
3. T1 = T2
4. Nothing can be said about the relation between T1 and T2.
  
  Ans: A
  Explanation:.A context switch is not required when a transition between user mode and kernel mode is required in an operating system; a mode transition is not a context switch in and of itself. However, a context switch may occur at this time, depending on the operating system. The shift from user mode to kernel mode is depicted in the diagram below. Switching from user mode to user mode necessitates a longer context transfer. As a result, T1<T2.
Working set model is used in memory management to implement the concept of
1. Swapping
2. Principal of Locality
3. Segmentation
4. Thrashing
  
  Ans: B
  Explanation: The working set concept refers to the memory that is often used by a programme. The principle of locality accomplishes the same result.
A UNIX file system has 1 KB block size and 4-byte disk addresses. What is the maximum file size if the inode contains ten direct block entries, one single indirect block entry, one double indirect block entry and one triple indirect block entry ?
1. 30 GB
2. 64 GB
3. 16 GB
4. 1 GB
  
  Ans: C
  Explanation: One direct block address (DBA) points to one kilobyte, and we have a total of ten DBA. As a result, it will take 10 kb. The address will be 4 bytes long.
  The number of blocks in an indirect block is 1kB / 4 b = 1024 / 4 = 256.
  256 * 1 kB = 256 KB is the size of a single direct block.
  256 * 256 = 216 kB = 64 MB is the size of a double indirect block.
  256 * 64 MB = 214 MB = 16 GB is the size of a triple direct block.
A thread is usually defined as a light weight process because an Operating System (OS) maintains smaller data structure for a thread than for a process. In relation to this, which of the following statement is correct ?
1. OS maintains only scheduling and accounting information for each thread.
2. OS maintains only CPU registers for each thread.
3. OS does not maintain a separate stack for each thread.
4. OS does not maintain virtual memory state for each thread.
  
  Ans: B
  Explanation: Because the operating system does not keep track of a thread's virtual memory state.
The versions of windows operating system like windows XP and window Vista uses following file system :
1. FAT-16
2. FAT-32
3. NTFS (NT File System)
4. All of the above
  
  Ans: D
  Explanation: windows operating system supports all the above file systems.
Which one of the following is a correct implementation of the metapredicate “not” in PROLOG (Here G represents a goal) ?
1. not(G):– !, call(G), fail. not(G).
2. not(G):– call(G), !, fail. not(G).
3. not(G):– call(G), fail, !. not(G).
4. not(G):– call(G), fail. not(G):– !.
  
  Ans: B
  Explanation: The use of cut is used to implement negation in Prolog. Actually, in Prolog, negation is known as "failure negation," which implies that to negate p, one must first try to prove p (rather than merely execute it), and if p is proved, then its negation, not(p), fails. In the event that p fails during execution, not(p) will succeed. The following is how not/1 is implemented:
  not(G):- call(G), !, fail.
  not(G).
Which one of the following is not an informed search technique ?
1. Hill climbing search
2. Best first search
3. A* search
4. Depth first search
  
  Ans: D
  Explanation: For Deatiled explanation click here
If we convert ∃u ∀v ∀x ∃y (P(f(u),v, x, y) → Q(u,v,y)) to ∀v ∀x (P(f(a),v, x, g(v,x)) → Q(a,v,g(v,x))) This process is known as
1. Simplification
2. Unification
3. Skolemization
4. Resolution
  
  Ans: C
  Explanation: Converting FOL to CNF sentences necessitates skolemization. Skolemization: introduce new function symbols to replace existential quantifiers. In the skolemization process, introduce an n-place function for each existentially quantified variable, where n is the number of previously appearing universal quantifiers.
Given two jugs of capacities 5 litres and 3 litres with no measuring markers on them. Assume that there is endless supply of water. Then the minimum number of states to measure 4 litres water will be
1. 3
2. 4
3. 5
4. 7
  
  Ans: D
  Explanation:
The map colouring problem can be solved using which of the following technique ?
1. Means-end analysis
2. Constraint satisfaction
3. AO* search
4. Breadth first search
  
  Ans: B
  Explanation: The map-coloring CSP(constraint satisfaction problem) requires that you assign a color to each region of a map such that any two regions sharing a border have different colors.
Which of the following is a knowledge representation technique used to represent knowledge about stereotype situation?
1. Semantic network
2. Frames
3. Scripts
4. Conceptual Dependency
  
  Ans: C
  Explanation: Scripts Rather than describing items, describe events. Consists of a chronologically or causally ordered chain of events.
A fuzzy set A on R is ________ iff A(λx1 + (1 – λ)x2) ≥ min [A(x1), A(x2)] for all x1, x2 ∈ R and all λ ∈ [0, 1], where min denotes the minimum operator.
1. Support
2. α-cut
3. Convex
4. Concave
  
  Ans: C
  Explanation: According to the theorem, A fuzzy set µ ∈ F(IR) is convex if and only if µ(λx1 + (1 − λ)x2) ≥ min{µ(x1), µ(x2)} for all x1, x2 ∈ IR and all λ ∈ [0, 1].
If A and B are two fuzzy sets with membership functions μA(x) = {0.6, 0.5, 0.1, 0.7, 0.8} μB(x) = {0.9, 0.2, 0.6, 0.8, 0.5} Then the value of μ––––– A ∪ B(x) will be
1. {0.9, 0.5, 0.6, 0.8, 0.8}
2. {0.6, 0.2, 0.1, 0.7, 0.5}
3. {0.1, 0.5, 0.4, 0.2, 0.2}
4. {0.1, 0.5, 0.4, 0.2, 0.3}
  
  Ans: C
  Explanation: In A Fuzzy Set, The Union Operation Takes The Maximum Value Between Two. So here A U B={0.9,0.5,0.6,0.8,0.8} and its complement will be calculated by subtracting each number from 1 hence ans will be option C {0.1,0.5,0.4,0.2,0.2}.
Consider a single perception with weights as given in the following figure :

The above perception can solve
1. OR problem
2. AND problem
3. XOR problem
4. All of the above
  
  Ans: B
  Explanation: Above perceptron after multiplying weights with input will give 1(-1.5)+x.1+y.1 =>f(t)=x+y-1.5 acc to given definition of f(t) , f(t)=1 if t>0 else f(t)=0 now draw truth table of f(t).
  e.g if x=0 y=0 f(t)=-1.5 <0 so f(t)=0 and so on for rest.

FAQs

What is the UGC NET exam for?

The National Eligibility Test (NET) is a test administered by the University Grants Commission (UGC) to determine whether Indian nationals are eligible for Assistant Professorships, Junior Research Fellowships, or both in Indian universities and colleges.

What is UGC NET exam eligibility?

If you hold a Master's Degree or equivalent in Humanities (including languages), Social Science, Computer Science and Applications, Electronic Science, etc., you are eligible to take the UGC NET test.

Is PhD compulsory for net?

To teach in central institutions, one must hold a PhD or have completed the UGC NET qualification exam, according to UGC regulations. The UGC, on the other hand, intends to scrap the regulation requiring experts to be hired and replace it with specific titles such as professor of practice and associate professor of practise.

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.

What is the full form of UGC NET?

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

Conclusion

In this article, we have extensively discussed previous year's questions of the UGC NET EXAM of the year 2013 paper III. You can also check out the remaining questions and solutions solution of this paper in UGC NET June 2013 Paper III Part-1 and UGC NET June 2013 Paper III Part 2

We hope that this blog has helped you enhance your knowledge regarding UGC NET EXAM Previous year's Questions June 2013 Paper II - Part 1, June 2013 Paper II - Part 2.

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