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UGC NET JULY 2018 (Computer Science And Applications Paper-II) (Question 12)

 July 15, 2018     Computer Science And Applications Paper-II, Software Engineering, UGC NET JULY 2018     No comments   

UGC NET JULY 2018 (Computer Science And Applications Paper-II) (Question 12)



Q12. Match the following in Software Engineering :


                            List-I                                                 List-II
(a)  Product Complexity(i)  Software Requirement Definition
(b)  Structure System Analysis(ii)  Software Design
(c)   Coupling and Cohesion(iii) Validation Technique
(d)   Symbolic Execution(iv)  Software Cost Estimation

Code:

(a)    (b)   (c)   (d)
(1)      (ii)     (iii)    (iv)    (i)
(2)      (iii)     (i)    (iv)    (ii)
(3)      (iv)     (i)    (ii)    (iii)
(4)      (iii)     (iv)    (i)    (ii)


Answer :

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



Reason : 
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UGC NET JULY 2018 (Computer Science And Applications Paper-II) (Question 13)

 July 15, 2018     Computer Science And Applications Paper-II, UGC NET JULY 2018     No comments   

UGC NET JULY 2018 (Computer Science And Applications Paper-II) (Question 13)

Q13. Which one of the following is not typically provided by Source Code Management Software ?

(1). Synchronisation                                    (2). Versioning and Revision history
(3). Syntax highlighting                              (4). Project forking


Answer : (1). Synchronisation    

Reason :  
Definition Source Code Management Software :- A component of software configuration management, version control, also known as revision control or source control, is the management of changes to documents, computer programs, large web sites, and other collections of information. This provides end-to-end tractability from requirements to software code. Source Code Preview & Syntax Highlighting.

A project fork, also called a fork or forking, is the creation of a new branch of a software development project that is independent of the existing project. 
A set of files under version control may be branched or forked at a point in time so that, from that time forward, two copies of those files may develop at different speeds or in different ways independently of each other. 

So at last only  Synchronisation (Synchronization) is not   typically provided by Source Code Management Software.
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UGC NET JULY 2018 (Computer Science And Applications Paper-II) (Question 11)

 July 14, 2018     Computer Science And Applications Paper-II, UGC NET JULY 2018     No comments   

UGC NET JULY 2018 (Computer Science And Applications Paper-II) (Question 11)


Q11. Assume the following regarding the development of a software system P:

  • Estimated line of code of P : 33,480 LOC
  • Average productivity for P : 620 LOC per person-month
  • Number of software developers : 6
  • Average salary of a software developer : ₹ 50,000 per month
If E,D and C are the estimated development effort (in person-months), estimated development time (in months), and estimated development cost (in ₹Lac) respectively, then (E, D, C) = ________.

(1). (48, 8, 24)                      (2). (54, 9, 27)                  (3) (60, 10, 30)             (4) (42, 7, 21)

Answer : (2). (54, 9, 27)

Reason :

We have given with Average Productivity for P = 620 LOC per person-month and we also have 6 Developers.

So  Single person can do 620 LOC
Therefore       6 Developer can do  = 6 x 620 = 3720 LOC in a month
Now we need to calculate time T to complete the Project     

                                          T =            Estimated Line of Code               =     33480     =  9

                                                6 Developers Line of Code per Month           3720

Now we need to calculate cost to complete the project in 9 months with given average salary of  ₹ 50,000 per month.

                                          Cost = 6 x 9 x 50000  = 27,00,000 (27 Lacs)
So Correct answer is option (2).
                                       


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UGC NET JULY 2018 (Computer Science And Applications Paper-II) (Question 10)

 July 14, 2018     Computer Science And Applications Paper-II, UGC NET JULY 2018     No comments   

UGC NET JULY 2018 (Computer Science And Applications Paper-II) (Question 10)



Q10. Consider the matrix M =  ⃒  2     0     2 ⃒  representing a set of planet (2D) geometric
                                                  ⃒   0     1     1 ⃒
                                                  ⃒   0     0     1 ⃒
transformations in homogeneous coordinates. Which of the following statements about the matrix M is True ?

(1). M represents first, a scaling of vector (2,1) followed by translation of vector (1,1).
(2). M represents first, a translation of vector (1,1) followed by scaling of vector (2,1).
(3). M represents first, a scaling of vector (3,1) followed by shearing of parameter (-1,1). 
(4). M represents first, a shearing of parameter (-1,1) followed by scaling of vector (3,1).

Answer : (2). M represents first, a translation of vector (1,1) followed by scaling of vector (2,1).
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UGC NET JULY 2018 (Computer Science And Applications Paper-II) (Question 9)

 July 14, 2018     Computer Science And Applications Paper-II, UGC NET JULY 2018     No comments   

UGC NET JULY 2018 (Computer Science And Applications Paper-II) (Question 9) 

 Q9. Which of the following statements is/are True regarding the solution to the visibility problem in 3D graphics ?

S1 : The Painter's algorithm sorts polygons by depth  and then paints (scan - converts) each polygon on to the screen starting with the most nearest polygon.

S2 : Backface Culling refers to eliminating geometry with backfacing normals.

Code: 
(1). S1 only                                                                   (2). S2 only
(3). Both S1 and S2                                                      (4). Neither S1 nor S2 
Answer : (2) S2 only
Reason :
For S1 Statement :
The painter's algorithm, also known as a priority fill, is one of the simplest solutions to the visibility problem in 3D computer graphics. The name "painter's algorithm" refers to the technique employed by many painters for painting distant parts of a scene before parts which are nearer thereby covering some areas of distant parts. The painter's algorithm sorts all the polygons in a scene by their depth and then paints them in this order, farthest to closest. It will paint over the parts that are normally not visible which solving the visibility problem at the cost of having painted invisible areas of distant objects. 
So according to above paragraph the option S1 is FALSE.

For S2 Statement :
Backface culling is an important part of how a 3D engine performs visibility checks. Its purpose is to detect polygons that are invisible in a particular scene that is, polygons that face away from the viewer. The process is similar to clipping, which determines if polygons are within the camera's field of view at all, and if not, are not rendered.
Back-face culling is a method in computer graphics programming which determines whether a polygon of a graphical object is visible; if it is not, the polygon is "culled" from rendering process, which increases efficiency by reducing the number of polygons that the hardware has to draw.
The vertices of front-facing polygons wind in a clockwise fashion, so polygons that face away from the camera are in a counter-clockwise order relative to the current view. When back-faces are culled, these polygons are not drawn. 
So according to above paragraph the option S2 is TRUE.
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UGC NET JULY 2018 (Computer Science And Applications Paper-II) (Question 8)

 July 13, 2018     Computer Science And Applications Paper-II, UGC NET JULY 2018     No comments   

UGC NET JULY 2018 (Computer Science And Applications Paper-II) (Question 8)

Q8. A graphic display system has a frame buffer that is 640 pixels wide, 480 pixels high and 1-bit of color depth. If the access time for each pixel on the average is 200 nanoseconds, then the refresh rate of the frame buffer is approximately :

(1) 16 frames per second                                                  (2) 19 frames per second
(3) 21 frames per second                                                  (4) 23 frames per second

 Answer: (1) 16 frames per second    

Reason : Here 640 pixel wide is width of screen means vertical line on screen and 480 pixels high is means vertical line on screen.
Now                                 width (w) = 640 px             and                   height (h) = 480 px
                                        Total = 640 x 480 = 307200

                                        Color depth = 1-bit/px
                                        Memory required for total pixel = w x h x Color Depth
                                        Memory required for total pixel = 640 x 480 x 1-bit = 307200

                                       1 pixel takes 200 nano second to refresh
                                       To refresh the whole screen = 640 x 480 x 200 ns = 61440000 nano second
                                     
                                       For 1 nano second =   ____1____  screen
                                                                            61440000
                                                                                                                                                        9
                                      Now Convert nano seconds to second we multiply above value with 10
                                                                                                     9
                                      For 1 nano second =   ____1____  x 10  = 16.2760 = 16 (Approximately)
                                                                            61440000

                                      The above solved values is closed to Option (1).



                                                                                                                             
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UGC NET JULY 2018 (Computer Science And Applications Paper-II) (Question 7)

 July 13, 2018     Computer Science And Applications Paper-II, UGC NET JULY 2018     No comments   

UGC NET JULY 2018 (Computer Science And Applications Paper-II) (Question 7-9)

Q7. Which of the following statement is/are true ?

P: C programming language has a weak type system with static types.
Q: Java programming language has a strong type system with static types.

Code :
(1).  P only                                    (2).  Q only
(3).  Both P and Q only                 (4).  Neither P nor Q

Answer : (3).  Both P and Q only

Reason: Now in this question first of all we should have the knowledge or idea about words
dynamic/static types which means that the type information acquired at run time or compile time.
The static types checked at compile time and dynamic types checked at run time.

Secondly we also need information about weak/strong types means weather language tries to do implicit conversion from one type to another type.
The strong types means No implicit conversion and weak types means implicit conversion is allowed.

What makes a type system weak or strong comes down to a the ways in which the programmer is able to create type errors.

Now the C language is consider as weak typed because a programmer is allowed to cast types. I can add a pointer to a character if I just tell C that they are both integers.
 
           int main () {
             char ch = 'b';
             void *x;
             (int)ch + (int)x;
          }
 
Java has static type casting as well which allows the programmer to, for example, downcast objects. This makes the static type system not sound. But Java has dynamic type checking for just this reason. Yes, Java has dynamic and static type checking. For this reason, however, I think many people would consider Java to be strongly typed.
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UGC NET JULY 2018 (Computer Science And Applications Paper-II) Question 5-6

 July 13, 2018     Computer Science And Applications Paper-II, UGC NET JULY 2018     No comments   

UGC NET JULY 2018 (Computer Science And Applications Paper-II) Question 5-6



Q5. Given below are three implementations of the swap() function in C++.


(a)(b)(c)
void swap (int a,int b)
{
       int temp;
       temp = a;
       a = b;
       b = temp;
}
int main()
{
       int p = 0, q =1;
       swap(p,q);
}
void swap (int &a,int &b)
{
       int temp;
       temp = a;
       a = b;
       b = temp;
}
int main()
{
       int p = 0, q =1;
       swap(p,q);
}
void swap (int *a,int *b)
{
       int *temp;
       temp = a;
       a = b;
       b = temp;
}
int main()
{
       int p = 0, q =1;
       swap(&p,&q);
}


Which of the following actually swap the contents of the two integer variables p and q ?  

(1). (a) only                    (2).  (b) only               (3).  (c) only                  (4). (b) and  (c) only

 Answer : (2). (b) only

Reason : If we execute option (a) code than that code will only pass the value of p and q to the function swap (i.e. called pass by value) which will not exchange the values of p and q also there is no use of address operator.

If we execute option (b) code than that code will pass address locations of p and q to swap function instead of actual values. Now the temp stores the value of a (i.e. address location of p, e.g. 1000) when temp = a; after that we have a=b; which replaces the value of a with new value of b i.e. address of q, e.g. 1002 and b=temp; again replaces the actual value of b with value temp (i.e. address location of p). When the swap function finishes its execution part with exchange the content of variable p and q because it replaces the address locations.

If we try to execute option (c) than the code will work fine but it will not replace the content of p and q. Here inside the swap function there is only exchange of addresses for pointer variable a and b not for p and q, So there is exchange of contents for p and q.


Q6. In Java which of the following statement is/are true ?

S1 : The 'final' keyword applied to a class definition prevents the class from being extended through derivation.

S2 : A class can only inherit one class but can implement multiple interfaces.

S3 : Java permits a class to replace the implementation of a method that it has inherited. It is called method overloading.

Code :
(1).  S1 and S2 only                                                (2).  S1 and S3 only
(3).  S2 and S3 only                                                (4).  All of S1,S2 and S3

Answer :  (1).  S1 and S2 only        

Reason : The statement S1 is true, we can not extends a class which having final keyword with class declaration.
The Second statement S2 is also true that Java don't allow us to extend multiple classes at one time but Java allows us to implement multiple interfaces by using implements keyword.

The third statement S3 is false because the ability of a subclass to override a method allows a class to inherit from a superclass and it is called overriding. The overriding method has the same name, number and type of parameters, and return type as the method that it overrides. An overriding method can also return a subtype of the type returned by the overridden method.
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UGC NET JULY 2018 (Computer Science And Applications Paper-II) (Question 3-4)

 July 08, 2018     Computer Science And Applications Paper-II, UGC NET JULY 2018     No comments   

UGC NET JULY 2018 (Computer Science And Applications Paper-II) (Question 3-4)


Q3. What is the output of the following JAVA program ?

      
       class simple
       {
           public static void main(String []args)
           {
                simple obj = new simple();
                obj.start();
           }
           void start()
           {
                 long [] P ={3,4,5};
                 long []Q = method(P);
                 System.out.print(P[0]+P[1]+P[2]+" : "); 
                 System.out.print(Q[0]+Q[1]+Q[2]+" : ");
           } 
           long [] method(long []R)
           {
                 R[1] = 7;
                 return R;
           }
       }//end of class

    (1) 12:15
    (2) 15:12
    (3) 12:12
    (4) 15:15

Reason : Initially array P as {3,4,5}. When we pass array P to method as method(P); the method function replaces the value of index 1  i.e. 4 with 7. Now new updated array is value {3,7,5}.
This new returned value reflects on both arrays respectively P and Q. Last the print function adds all index location values of  both arrays separately and shows output 15:15.  


Q4. What is the output of following 'C' program? (Assuming little - endian representation of multi-byte data in which Least Significant Byte (LSB) is stored at the lowest memory address.)

1. #include <stdio.h>
2. #include <stdlib.h>/**/
3. int main()
4. {
5.     union saving
6.     {
7.         short int one;
8.         char two[2];
9.     };
10.    union saving m;
11.    m.two[0] = 5;
12.    m.two[1] = 2;
13.    printf("%d, %d, %d\n",m.two[0],m.two[1],m.one);
14.    return 0;
15. }

(1)   5 ,2, 1282
(2)   5 ,2, 52
(3)   5 ,2, 25
(4)   5 ,2, 517 

 Reason : This is one is interesting question in exam. First we have to know about Big-endian and little-endian are terms that describe the order in which a sequence of bytes are stored in computer memory. Big-endian is an order in which the "big end" (most significant value in the sequence) is stored first (at the lowest storage address). Little-endian is an order in which the "little end" (least significant value in the sequence) is stored first. For example, in a big-endian computer, the two bytes required for the hexadecimal number 4E52 would be stored as 4E52 in storage (if 4E is stored at storage address 1000, for example, 52 will be at address 1001). In a little-endian system, it would be stored as 524E (52 at address 1000, 4E at 1001).
Here we have little - endian in our question. In code we have two union members short int one and char two[2]. In line 11,12 we have numeric values 5 and 2 in char array (without single quote '5' and '2'). In line 13 the output also shows result with %d (i.e. represents for numeric values) which is okay for output 5, 2 but how third values came because we did not provided any value to union member short int one. Just suppose binary conversion of
    5 is 00000101 and  2 is 00000010

Lowest Address (1000)Highest Address(1001)
0000010100000010

Now as per question requirement little - endian representation of multi-byte data in which Least Significant Byte (LSB) is stored at the lowest memory address. Therefore

Lowest Address (1000)Highest Address(1001)
0000001000000101
Now 00000010-00000101 convert this to Decimal value

512+0+0+0+0+0+0+4+0+1  = 517 
After converting the combination of both numbers we have 517 at third location and final result 5, 2, 517.
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UGC NET JULY 2018 (Computer Science And Applications Paper-II) (Question 1-2)

 July 08, 2018     Computer Science And Applications Paper-II, UGC NET JULY 2018     No comments   

UGC NET JULY 2018 (Computer Science And Applications Paper-II)
(Question 1-2)

Q1. The definition in an XML document are said to be _____________ when the tagging system and definitions in the DTD are all in compliance.

(1) well-formed
(2) reasonable
(3) valid
(4) logical

Answer : (3)  

Reason : XML Document Type Declaration, commonly known as DTD.
An XML document can be defined as:

  • Well-formed: If the XML document adheres to all the general XML rules such as tags must be properly nested, opening and closing tags must be balanced, and empty tags must end with '/>', then it is called as well-formed.
    OR
  • Valid: An XML document said to be valid when it is not only well-formed, but it also conforms to available DTD that specifies which tags it uses, what attributes those tags can contain, and which tags can occur inside other tags, among other properties.

Q2. Consider the JavaScript Code:

      var y = "12";
      function f(){
           var y ="6";    
           alert(this.y);
           function g(){alert(y);}
           g();
      }
           f(); 


      If  M is the number of alert dialog boxes generated by this JavaScript code and Dl, D2, ...., DM 
        represents the content displayed in each of the M dialog boxes, then :

     (1)   M3; Dl displays "12"; D2 displays "6"; D3 displays "12".
     (2)   M-3; Dl displays "6"; D2 displays "12"; D3 displays "6".
     (3)   M:2; D! displays "6"; D2 displays "12".
     (4)   M:2; Dl displays "12"; D2 displays "6". 

 Answer : (4) 
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