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UGC NET June 2013 Paper III Part 2

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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 26 to 50. You can checkout solutions from questions 1-25 in UGC NET June 2013 Paper III Part-1  and the remaining questions and solutions solution of this paper in UGC NET June 2013 Paper III Part 3

  1. An actor in an animation is a small program invoked _______ per frame to determine the characteristics of some object in the animation. 
    1. once 
    2.  twice 
    3. 30 times 
    4. 60 times

      Ans-(A)
      Explanation: In animation, an actor is a small program that is run once per frame to determine the characteristics of a specific object.

       
  2. Bresenham line drawing algorithm is attractive because it uses 
    1. Real arithmetic only 
    2. Integer arithmetic only 
    3. Floating point arithmetic
    4.  Real and integer arithmetic 

      Ans-B
      Explanation: Because it avoids floating-point arithmetic and rounding, Bresenhem's technique has a considerable advantage over the other methods. It's a classic example of an incremental algorithm, which calculates the location of each pixel along the line using data from the preceding pixel. It only works with integers and avoids multiplications.

       
  3. The refresh rate above which a picture stops flickering and fuses into a steady image is called _________. 
    1. Crucial fusion frequency 
    2. Current frequency fusion
    3. Critical fusion frequency 
    4. Critically diffused frequency  

      Ans-C
      Explanation: The refresh rate of a CRT is the number of times per second the image is redrawn. It is typically 60 per second for raster displays. Flicker develops as the refresh rate decreases because the eye can no longer integrate the individual light pulses coming from a pixel. The refresh rate above which a picture stops flickering and fuses into a steady image is called the critical fusion frequency (or) CFF.

       
  4. In homogenous coordinate system (x, y, z) the points with z = 0 are called 
    1. Cartesian points 
    2. Parallel points 
    3. Origin point 
    4. Point at infinity 

      Ans-D
      Explanation: We add a third coordinate to a point in homogeneous coordinates. Instead of a pair of numbers (x,y), each point is represented by a three-digit number (x,y,w). Simultaneously, two sets of homogeneous coordinates (x,y,w) and (x',y',w') are said to represent the same point if and only if one is a multiple of the other. As a result, the positions represented by the distinct coordinate triples (2,3,6) and (4,6,12) are the same. Furthermore, at least one of the homogeneous coordinates must be nonzero: (0,0,0) is not acceptable. We can divide by the w coordinate if it is nonzero. (x,y,w) represents the same position as (x/w,y/w,1).When w is nonzero, we normally do this division, and the numbers x/w and y/w are called the Cartesian coordinates of the homogenous points. The points with w=0 are called points at infinity. So, the correct answer is D.

       
  5. If 40 black lines interleaved with 40 white lines can be distinguished across one inch, the resolution is 
    1. 40 line-pairs per inch 
    2. 80 line-pairs per inch 
    3. 1600 lines per inch 
    4. 40 lines per inch  

      Ans-A
      Explanation: The number of distinct lines per inch that technology can produce is known as resolution. If 40 black lines interleaved with 40 white lines can be distinguished across one inch, the resolution is 80 lines per inch or 40 line pairs per inch. So, the correct answer is A.

       
  6. Images tend to be very large collection of data. The size of memory required for a 1024 by 1024 image in which the colour of each pixel is represented by a n-bit number, (in an 8 bit machines) is
    1.  n × 8 MB 
    2. n / 8 MB 
    3. (1024 × 1024) / 8 MB 
    4. 1024 MB 
      Ans- A
      Explanation: Size of image = 1024×1024 pixels 
      1 pixel needs n bits of memory space 
      Total memory required to store image = 1024×1024 pixels = 1024×1024×n Bits 
      As 8 bits = 1 bytesSo ,total memory required to store image = (1024×1024×n)/8 bytes = (1024×n)/8 KB = n/8 MB

       
  7. Arrays in C language can have _________ with reference to memory representation. 
    1. n-subscripts 
    2. two-subscripts 
    3. only one subscript 
    4. three subscripts only

      Ans: C
      Explanation:The subscript expression specifies the requested element by its location in the data when an array element is addressed. Position zero is 
      occupied by the first element, position one by the second, and so on. As a 
      result, [N-1] is subscripted to the last element. The number of elements in the 
      array is N.When referring to an array with two dimensions, two subscripts are
      Specified.
      We can assign logical meaning to the first and second subscripts as
      programmers. The first subscript, for example, maybe regarded as the row
      and the second as the column.If an array has three dimensions, then three
      subscripts are specified when referencing. Again we may assign logical
      meaning to the various subscripts.For example, we could consider the first
      subscripts as the x coordinate, the second subscripts as the y coordinate, and
      the third subscripts as the z coordinate.Therefore, an array can have one, two
      and three subscripts with reference to memory representation.

       
  8. Refer the points as listed below : 
    (a) What are the operator precedence rules ? 
    (b) What are the operator associativity rules ?
    (c) What is the order of operand evaluation ? 
    (d) Are there restrictions on operand evaluation side effects ?
    Which of the above must be considered as primary design issues for arithmetic expressions ? 
    1. (a), (b) and (c)
    2. (a), (c) and (d) 
    3. (a), (b) and (d)
    4. (a), (b), (c) and (d)

      Ans: D
      Explanation: Following are the primary design issues for the arithmetic expression:-
      1. What type of mixing is allowed in expression?
      2. What is the order of operand evaluation?
      3. What are the operator precedence rules?
      4. What are the operator associativity rules?
      5. Does the language allow user-defined operator overloading?
      6. Are these restrictions on operand evaluations side effects?
         
  9. Horn clauses are special kinds of propositions which can be described as 
    1. Single atomic proposition on left side.
    2. Single or multiple atomic proposition on left side. 
    3. A single atomic proposition on left side and a single atomic proposition on right side. 
    4. A single atomic proposition on left side or an empty left side.

      Ans: D
      Explanation: A Horn clause is: An atomic proposition; or A conjunction of atomic propositions ⇒ atomic proposition For example: C∧D⇒B Given a set of propositions, the associated set of Horn clauses is a subset of the sentences that can be written in standard propositional logic. So, the answer is D.

       
  10. Which of the following is/are the fundamental semantic model(s) of parameter passing ? 
    1. in mode
    2.  out mode
    3. in-out mode
    4. all of the above

      Ans: D
      Explanation: For a detailed explanation visit here

       
  11. The grammar with production rules S → aSb |SS|λ generates language L given by : 
    1. L = {w∈{a, b}* | na (w) = nb(w) and na(v) ≥ nb(v) where v is any prefix of w} 
    2. L = {w∈{a, b}* | na(w) = nb(w) and na(v) ≤ nb(v) where v is any prefix of w} 
    3. L = {w∈{a, b}* | na (w) ≠ nb(w) and na (v) ≥ nb(v) where v is any prefix of w}
    4.  L = {w∈{a, b}* | na (w) ≠ nb(w) and na(v) ≤ nb(v) where v is any prefix of w} 

      Ans: A
      Explanation:S→aSb∣SS∣λ 
      All strings in which na(w) = nb(w) will be generated by this gramma A prefix is a sequence of leading symbols that can be any length. The string abc, for example, has the prefix (empty string), a, bc, and abc. The number of 'a's in the prefix v of w is either larger than or equal to the number of 'b's in the string w. The string aabb, for example, has the prefix aab. The number of 'a's is more than the number of 'b's in this case. The string abab has an ab prefix. Both 'a' and 'b' have the same number of letters. As a result, the number of 'a's and 'b's are equal when considering the prefix v.Hence,Option(A) is the Answer.

       
  12. A pushdown automation M = (Q, Σ, Γ, δ, q0, z, F) is set to be deterministic subject to which of the following condition(s), for every q ∈ Q, a ∈ Σ ∪ {λ} and b ∈ Γ (s1) δ(q, a, b) contains at most one element (s2) if δ(q, λ, b) is not empty then δ(q, c, b) must be empty for every c ∈ Σ
    1. only s1 
    2. only s2 
    3. both s1 and s2 
    4. neither s1 nor s2

      Ans: C
      Explanation: A pushdown automaton M=(Q,Σ,Γ,δ,q 0,z,F) is said to be deterministic if, for every q∈Q, aΣ∪{λ} and b∈Γ, δ(q,a,b) contains at most one element, δ(q,λ,b) is not empty, then δ(q,c,b) must be empty for every c∈Σ. The first of these conditions simply requires that for any given input symbol and any stack top, at most one move can be made. The second condition is that when a λ-move is possible for some configuration, no input-consuming alternative is available.

       
  13. For every context free grammar (G) there exists an algorithm that passes any w ∈ L(G) in number of steps proportional to
    1. ln|w| 
    2. |w| 
    3. |w|2 
    4. |w|3

      Ans: D
      Explanation: There is an algorithm for every context-free grammar that 
      parses every w∈L(G) in a number of steps proportional to w3.

       
  14. Match the following : a. Context sensitive language i. Deterministic finite automation b. Regular grammar ii. Recursive enumerable c. Context free grammar iii. Recursive language d. Unrestricted grammar iv. Pushdown automation Codes : a b c d 
    1. ii i iv iii 
    2. iii iv i ii 
    3. iii i iv ii 
    4. ii iv i iii 

      Ans: C
      Explanation:
      1. Context-sensitive language-Recursive language
      2. Regular grammar-Deterministifinite automaton
      3. Context-free grammar-Pushdown automation
      4. Unrestricted grammar-Recursive enumerable

         
  15. The statements s1 and s2 are given as : s1 : Context sensitive languages are closed under intersection, concatenation, substitution and inverse homomorphism. s2 : Context free languages are closed under complementation, substitution and homomorphism. Which of the following is correct statement ? 
    1. Both s1 and s2 are correct.
    2. s1 is correct and s2 is not correct. 
    3. s1 is not correct and s2 is correct. 
    4. Both s1 and s2 are not correct.  

      Ans: B
      Explanation: Context sensitive languages are closed under union, intersection, concatenation, Kleene star and reversal. They are also closed under ϵ-free homomorphisms Context free languages are closed under union, concatenation, and Kleene closure. Also, under reversal, homomorphisms and inverse homomorphisms. But not under intersection or difference or complementation.

       
  16. Which one of the following is not an addressing mode ? 
    1. Register indirect 
    2. Autoincrement
    3. Relative indexed
    4. Immediate operand 

      Ans: C
      Explanation: The operand is retained in memory in Register indirect addressing mode. The operand location's address is stored in a register that is provided in the instruction. Register indirect mode is a specific instance of auto-increment mode. The address of the operand is stored in the register whose number is included in the instruction code. After addressing the operands, the contents of the register are incremented in autoincrement mode. Decrement Mode: The contents of the register are decremented before the operands are addressed. The operand is held in the instruction in immediate addressing mode. Relative index address mode is not available.

       
  17. Computers can have instruction formats with
    1. only two address and three address instructions 
    2. only one address and two address instructions
    3. only one address, two address and three address instructions 
    4. zero address, one address, two address and three address instructions 

      Ans: D
      Explanation: Addressing refers to the process of indicating where the operands are located. The following are four common instruction formats:
      1. Zero-address instruction.
      2. One-address instruction
      3. Two-address instruction.
      4. Three-address instruction
         
  18. Which is not a typical program control instruction ? 
    1. BR 
    2. JMP 
    3. SHL 
    4. TST 

      Ans: C 
      Explanation: A option should be BR (branch ) 
      B  is JMP (jump) D is TST(Test) 
      C is SHL shift left opration  which is not a  typical program control instruction hence ans is C

       
  19. Interrupt which arises from illegal or erroneous use of an instruction or data is 
    1. Software interrupt 
    2. Internal interrupt
    3. External interrupt
    4. All of the above 

      Ans: B
      Explanation: There are three types of interrupts:
      1.  External interrupts come from I/O devices, timing devices, or any other external source.
      2.  Internal interrupts arise from illegal or erroneous use of an instruction or data, also called traps.
      3.  Software interruptsare initiated by executing an instruction.
        Internal interrupts are synchronous while external ones are asynchronous and both are initiated from signals that occur in the hardware of the CPU.

         
  20. The simplified function in product of sums of Boolean function F(W, X, Y, Z) = Σ(0, 1, 2, 5, 8, 9, 10) is 
    1. (W' + X') (Y' + Z') (X' + Z)
    2. (W' + X') (Y' + Z') (X' + Z') 
    3. (W' + X') (Y' + Z) (X' + Z) 
    4. (W' + X') (Y + Z') (X' + Z)

      Ans: A
      Explanation: Concept: A min word is represented by 1 in SOP (sum of product) form. A max term is represented by 0 in POS (product of sum) form. the product's total
      (SOP) f(w, x, y, z) = ∑m (0, 1, 2, 3, 7, 8, 10) + ∑d (5, 6, 11, 15) Product of sum: (P0S) f(w, x, y, z) = Π (4, 9, 12, 13, 14)  + ∑d (5, 6, 11, 15),  f(x, y, w, z) = (w’ + z’) (x’ + z)

       
  21. Match the following : a. TTL i. High component density b. ECL ii. Low power consumption c. MOS iii. Evolution of “diodetransistor-logic” d. CMOS iv. High speed digital circuits Codes : a b c d 
    1. iii ii i iv 
    2. i iv iii ii 
    3. iii iv i ii 
    4. i ii iii iv  

      Ans: C
      Explanation: TTL is transistor-transistor logic, which is basically an evolution of DTL or diode-transistor logic. 
      ECL is used for very high-speed applications because of its price and power demands. 
      MOS devices occupy much less space on a chip than bipolar transistors. Hence creates high component density. 
      CMOS is faster and consumes even less power than other MOS families.

       
  22. Match the following : a. Foreign keys i. Domain constraint b. Private key ii. Referential integrity c. Event control action model iii. Encryption d. Data security iv. Trigger Codes : a b c d 
    1. iii ii i iv 
    2. ii i iv iii 
    3. iii iv i ii 
    4. i ii iii iv 

      Ans: B
      Explanation: Referential integrity is a property of data which, when satisfied, requires every value of one attribute (column) of a relation (table) to exist as a value of another attribute in a different (or the same) relation (table).For referential integrity to hold in a relational database, any field in a table that is declared a foreign key can contain either a null value, or only values from a parent table's primary key or a candidate key. 
      A database trigger implements event control action model. It is a procedural code that is automatically executed in response to certain events on a particular table or view in a database. 
      Encryption is the most effective way to achieve data security. To read an encrypted file, you must have access to a secret key or password that enables you to decrypt it. Unencrypted data is called plain text ; encrypted data is referred to as cipher text.

       
  23. When an array is passed as a parameter to a function which of the following statements is correct ? 
    1. The function can change values in the original array.
    2. The function cannot change values in the original array. 
    3. Results in compilation error. 
    4. Results in runtime error. 

      Ans: A
      Explanation:  When an array is supplied as a parameter to a function, the function has the ability to modify the contents of the original array.
       
  24. Suppose you want to delete the name that occurs before “Vivek” in an alphabetical listing. Which of the following data structures shall be most efficient for this operation ? 
    1. Circular linked list 
    2. Doubly linked list 
    3. Linked list
    4. Dequeue 

      Ans: B 
      Explanation: We must find a value and edit the previous node in the given question. We only have direct access to the previous node in a doubly linked list.

       
  25. What will be the output of the following segment of the program ? main( ) { char *s = “hello world”; int i = 7; printf(“%, *s”, i, s); }
    1. Syntax error 
    2. hello w 
    3. hello 
    4. o world

      Ans: None of these options are correct 
      Explanation: The program simply prints '%, *s'

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 practise 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.

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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 checkout solutions from questions 1-25 in UGC NET June 2013 Paper III Part-1  and the remaining questions and solutions solution of this paper in UGC NET June 2013 Paper III Part 3

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, and June 2013 Paper II - Part 2

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Topics covered
1.
Introduction
2.
FAQs
2.1.
What is the UGC NET exam for?
2.2.
What is UGC NET exam eligibility?
2.3.
Is PhD compulsory for net?
2.4.
How many papers are there in the UGC NET exam?
2.5.
What is the full form of UGC NET?
3.
Conclusion