2nd puc computer science model 2 question papers and answers

2nd puc computer science model question papers with answers,2nd puc computer science solved model question papers

PU-II COMPUTER SCIENCE

SOLVED MODEL QUESTION PAPER 2

Answer all the questions.
Each question carries 1 mark. (10 X 1 = 10)

      1. Expand the term ISA.

Ans: Industry Standard Architecture.

      2. Draw XNOR gate.

Ans:

      3. What is traversing?

Ans: the process of accessing each data item exactly once to perform some operation is called traversing

      4. Write the significance of scope resolution operator.

Ans: scope resolution operator is used to define the member function outside the class definition.

      5. List the operators used with the pointers.

Ans: indirection operator (*) and address of operator (&).

      6. What is data warehouse?

Ans: a data warehouse is a repository of an organization’s electronically stored data.

      7. Expand the term ARPANET.

Ans: Advanced Research Projects Agency NETwork.

      8. What is virus?

Ans: computer virus is a malicious program that requires a host and is designed to make a system sick.

      9. Name any two web browsers.

Ans: internet explorer, firefox Mozilla, google chrome, netscape navigator, opera mini etc.

      10. What is web hosting?

Ans: web hosting is a business of providing storage space and access for websites.

Answer any five of the following. Each question carry 2 marks  (5 X 2 = 10)

      11. State and prove indempotence law.

Ans: this law states that when a variable is combines with itself using OR or AND operator, the output is the same variable.

      a)      X+X=X
      b)      X.X=X

   Proof:  a) X+X=X

If X=0, consider LHS=X+X

                                =0+0

                                =0

                                =X

                                =RHS

If X=1, consider LHS=X+X

                                =1+1

                                =1

                                =x

                                =RHS

      b)      X.X=X

If X=0, consider LHS=XX

                                =0.0

                                =0

                                =X

                                =RHS

If X=1, consider LHS=X.X

                                =1.1

                                =1

                                =X

                                =RHS

      12. Show that X(X+Y)=X

Ans: LHS= X(X+Y)

                        =X.X+X.Y

                        =X+XY

                        =X(1+Y)

                        =X.1

                        =X

                        =RHS

      13. What is inheritance?

Ans: inheritance is a process of acquiring properties from parent class to child class

      14. Name any two types of constructors.

Ans: default constructor, parameterized constructor, copy constructor.

      15. Explain getline() function with an example.

Ans: getline() function is used to read a whole line of text.

Ex: char book[SIZE];

fstream fin;

fin.getline(book,SIZE);

      16. Name the database users.

Ans: - application programmers and system analysts

-          - End users

-          - Database administrator(DBA)

-          - Database designers

      17. Write the syntax of UPDATE command with an example.

Ans: syntax: update table_name set column1=value1,column2=value2…column=valueN where[condition];

Ex: update employee set address=’gadag’ where empid=1;

      18. What is the need of networking?

Ans: - file sharing

-          Print sharing

-          Email tools for communication

-          Remote access

-          Sharing database to multiple users.

Answer any five of the following.
Each question carries 3 marks. (5 X 3 = 15)

      19. What is cache memory? Explain its types.

Ans: The cache memory is a very high speed memory placed in between RAM and CPU. Cache memory stores data that is used more often, temporarily and makes it available to CPU at a fast rate. During processing, the CPU first checks cache for the required data. If data is not found in cache, then it looks in the RAM for data

Cache memory is very expensive, so it is smaller in size. Computers have cache memory of sizes 256 KB to 2 MB

Cache memory is built into the processor and may also be located next to it on a separate chip between CPU and RAM. Cache built into CPU is faster than separate cache almost at the speed of the microprocessor itself. However separate cache is roughly twice as fast as RAM. The CPU has a built –in level 1(L1) cache and level2 (L2) cache high end processor nowadays comes with built-in L3 cache, processor like Intel Core i7

      20. Explain XOR gate with diagram and truth table.

Ans:  XOR gate produces output 1 for only those input combinations that have odd number of 1's.

      21. Write any three features of array.

Ans: -array size should be positive number only.

-          - String array always terminates with null character (‘\0’).

-          - Array elements are counted from 0 to n-1.

-          - Useful for multiple reading of elements.

      22. Differentiate between static and dynamic allocation of memory.

Ans:

Static Memory allocation	Dynamic Memory allocation
-          The amount of memory to be allocated is predicted and pre known	-          The amount of memory to be allocated is not known
-          It is allocated during compilation	-          It is allocated during run-time
-          All the variables declared normally, are allocated memory statically Ex: int a;	-          Allocation and de-allocation of objects using the new and delete operators

       23. Differentiate between read() and write() functions.

Ans:

read()	write()
1.       The read() member function belongs to             ifstream class.	1. The write() member function belongs to ofstream class.
2.       Syntax: fin.read((char*)&variable,sizeof(variable));	2. Syntax: fout.write((char*)&variable,sizeof(variable));
3.       Ex: student s;                                             ifstream fin(“text.txt”,ios::binary); fin.read((char *)&s,sizeof(s));	3. Ex: student s; ofstream fout(“text.txt”,ios::binary); fout.write((char *)&s,sizeof(s));

       24. Explain three level RDBMS architecture.

Ans: it is a three levels of architecture namely, internal, conceptual and external levels.

Internal level: internal level is the lowest level of data abstraction that deals with the physical representation of the database on the computer and thus, is known as physical level. It describes how the data is physically stored and organized on the storage medium. It includes storage space allocation techniques for data and indexes, data compression and encryption techniques, and record placement

Conceptual level: conceptual level describes what data is stored in the database, the relationships among the data and complete view of the user’s requirements without any concern for the physical implementation. i.e. it hides the complexity of physical storage structures.

External level: this level permits the user to access data in a way that is customized according to their needs, so that the same data can be seen by different users in different ways at the same time. In this way, it provides a powerful and flexible security mechanism by hiding the parts of the database from certain users, as the user is not aware of existence of any attributes that are missing from the view

      25. What is shareware software? Mention its limitations.

Ans: software made available with the right to redistribute copies, but it is stipulated that if one intends to use the software, often after a certain period of time, then a license fee should be paid.
Limitations of shareware:
i) In shareware the source code is not available.
ii) Modifications to the software are not allowed.

      26. Name any three formatting tags in HTML.

Ans: <B> is used to make text Bold.
<U> is used to draw underline.
<I> is used to make text italics.

Answer any seven of the following. Each carries 5 marks (7 X 5 = 35)

      27. Using K-map, simplify the following expressions in four variables W,X,Y and Z.

      m₀,m₂,m₄,m₈,m₉,m₁₀,m₁₁,m₁₂,m₁₃.

Ans: 

      28. Write an algorithm to illustrate enqueue.

Ans: Step 1: if REAR=N then

               PRINT “overflow”

               Exit
   Step 2: if FRONT=NULL then

                FRONT=0

                REAR=0

                else

                REAR=REAR+1

         Step 3: QUEUE[REAR]=ITEM

         Step 4: return

      29. Illustrate an algorithm to delete an element from a stack.

Ans: An algorithm for POP() Operation
      Step 1: if TOP=0 then
                  PRINT “Stack is empty”

            Exit
Step 2: ITEM=STACK[TOP]

Step 3: TOP=TOP-1

      30. Write the applications of OOP.

Ans: Applications of OOP are as follows
- Computer graphic applications
- CAD/CAM software
- Object oriented database
- User interface design such as windows
- Real-time systems
- Simulation and modeling
- Artificial intelligence and expert systems

      31. Explain defining a function outside the class definition with an example.

Ans: To define member function inside a class the function declaration within the class is replaced by actual function definition inside the class. Only small functions are defined inside class definition.
Ex:
class room
{
int length;
int breadth;
public: void getdata()
           {
            cin>>length;
            cin>>breadth;
            }
           void putdata()
           {
            cout<<length;
            cout<<breadth;
            }
};

                To define member functions written outside the class definition replaced by actual function definition outside the class.

Ex:
class room
{
int length;
int breadth;
public: void getdata();
void putdata();
};
void room::getdata()
{
cin>>length;
cin>>breadth;

}

void room::putdata()
{
cout<<length;
cout<<breadth;
}

The symbol :: is known as scope resolution operator. The scope resolution operator identifies the function as a member of particular class. The use of scope resolution operator implies that these member functions are defined outside the class

      32. Write the advantages of inline functions.

Ans: - the inline member functions are compact function calls.
- Therefore the size of the object code is considerably reduced.
- The speed of execution of a program increases.
- Very efficient code can be generated.
- The readability of the program increases.

      33. What is destructor? Explain with its syntax and definition.

Ans: A destructor is a special member function that will be executed automatically when an object is destroyed. It will have, like constructor, the same name as that of the class but preceded by a tilde (~).
A destructor will be called automatically when an object is destroyed. Destroying an object means de-allocating all the resources such as memory that was allocated for the object by the constructor.
Syntax:
class class_name
{
private: data_variables;
public: class_name() //constructor
~class_name() //destructor
};

Ex:
#include<iostream.h>
#include<conio.h>
class num
{
int x;
public:
num();
void display();
~num();
};
num::num()
{
cout<<”In constructor:\n”;
x=100;
}
num::~num()
{
cout<<”In desctructor:\n”;
}
void num::display()
{
cout<<”Value of x=”<<x;
}
void main()
{
num n;
n.display();
getch();
}

      34. Describe the types of inheritance with examples.

Ans: following are the types of inheritance
- Single inheritance: if a class is derived from a single base class, it is called as single inheritance 

- Multilevel inheritance: the classes can also be derived from the classes that are already

derived. This type of inheritance is called multilevel inheritance.

- Multiple inheritance: if a class is derived from more than one base class, it is known as

multiple inheritance

- Hierarchical inheritance: if a number of classes are derived from a single base class, it is called as
hierarchical inheritance

- Hybrid inheritance: it is a combination of Hierarchical and Multiple inheritance.

      35. Explain the features of database system.

Ans: in the database approach, the data is stored at a central location and is shared among multiple users. Thus, the main advantage of DBMS is centralized data management. The advantages of centralized database system are as follows:

- Controlled data redundancy: elimination of duplication of data item in different files ensures consistency and saves the storage space. The redundancy in the database cannot be eliminated completely as there could be some performance and technical reasons for having some amount of redundancy.

- Enforcing data integrity: data integrity refers to the validity of data and it can be compromised in a number of ways.

- Data sharing: the data stored in the database can be shared among multiple users or application programs. It is possible to satisfy the data requirements of the new applications without creating any additional data or with minimal modification

- Ease of application development: the application programmer needs to develop the application programs according to user’s needs. The other issues like concurrent access, security, data integrity etc are handled by the RDBMS itself

- Data security: since the data is stored centrally data security checks can be carried out whenever access is attempted to sensitive data. To ensure security, a RDBMS provides security tools such as used codes and passwords. Different checks can be established for each type of access like addition, modification, deletion etc to each piece of information in the database

- Multiple user interfaces: For various users having different technical knowledge DBMS provides different types of interfaces such as query languages, application program interfaces, and graphical user interfaces (GUI)

- Backup and recovery: RDBMS provides backup and recovery subsystem that is responsible for recovery from hardware and software failures. For example, if the failure occurs in between the transaction, the RDBMS recovery subsystem either reverts back the database to the state to the previous state of the transaction or resumes the transaction from the point it was interrupted so that its complete effect can be recorded in the database

      36. Describe any five logical operators with an example.

Ans: ALL: This operator is used to compare a value to all values in another value set.

AND: This operator allows the existence of multiple conditions in an SQL statements’ WHERE clause.

EXISTS: This operator is used to search for the presence of a row in a specified table that meets certain criteria.

LIKE: This operator is used to compare a value to similar values using wildcard operators

IS NULL: This operator is used to compare a value with a NULL value.

      37. Explain any five network topologies. 

      Ans: Bus topology or linear topology: this consists of a single length of the transmission medium onto which the various nodes are attached. The transmission from any station travels the length of the bus, in both directions, and can be received by all other stations. The bus has terminators at either end which absorb the signal, removing it from the bus

 

  

-          Ring topology or circular topology: In this topology each node is connected to two and only two neighboring nodes and is transmitted to another. Thus data travels in one direction only, from node to node around the ring. After passing through each node, it returns to the sending node, which removes it

-          Star topology: this topology consists of a central node to which all other nodes are connected by a single path.

In star topology, every node (computer workstation or any other peripheral) is connected to a central node called a hub or switch. The switch is the server and the peripherals are the clients.

 

Advantages

• ·Star networks are very reliable because if one computer or its connection breaks it doesn’t affect the other computers and their connections

Disadvantages

• An expensive network layout to install because of the amount of cables needed
• If the server crashes or stops working then no computers will be able to access the network

-          Tree topology: In telecommunication networks, a tree network is a combination of two or more star networks connected together. Each star network is a local area network (LAN) in which there is a central computer or server to which all the workstation nodes are directly linked. The central computers of the star networks are connected to a main cable called the bus. Thus, a tree network is a bus network of star networks.

- Mesh topology: A mesh network is a network topology in which each node relays data for the network. All mesh nodes cooperate in the distribution of data in the network.  

The Mesh topology is commonly used in large internetworking environments with stars, rings and buses attached to each node.