2nd PUC Computer Science Model Question Paper 3 with Answers

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Karnataka 2nd PUC Computer Science Model Question Paper 3 with Answers

Time: 3 Hrs. 15 Min.
Max. Marks: 70

PART-A

I. Each question carries one mark. (10 × 1 =10)

Question 1.
Expand the term POST.
Answer:
The expansion of POST is Power On Self Test.

Question 2.
What is NOR gate?
Answer:
The NOR (Not OR) gate has two or more input signals with one output. If all the inputs are 0, then output is 1 in NOR gate.

Question 3.
What are Non-linear data structures?
Answer:
A non-linear data structure is a data structure in which a data item is connected to several other data items i.e., every data item is attached to several other data items.

Question 4.
Define array of objects.
Answer:
The array of class type is known as array of objects.
For example:
classname object[5];

Question 5.
What is new operator in C++?
Answer:
The new operator in C++, allocates memory dynamically.

Question 6.
Give the symbol notion for project.
Answer:
The symbol notion for project is? (pi)

Question 7.
What is advantage of EDGE?
Answer:
EDGE (Enhanced Data GSM Environment) is a faster version of GSM designed to deliver data at rates up to 384 Kbps and enable the delivery of multimedia and other broadband applications to mobile phone and computer users.

KSEEB Solutions

Question 8.
What are dedicated servers?
Answer:
A computer exclusively reserved for server’s job and its only job is to help workstations to access data, software and hardware resources. The term “dedicated server” can also mean a single computer within a network that is reserved for a specific purpose.

Question 9.
What is FSF?
Answer:
A Free Software Foundation (FSF) is a non-profit organization created for the purpose of supporting free software movement.

Question 10.
What do you mean by web scripting?
Answer:
A script is a list of commands that are embedded in a web page and are interpreted and executed by the scripting engine. The process of creating and embedding scripts in a web page is known as web scripting.

PART-B

II. Each question carries two marks. (5 × 2=10)

Question 11.
Construct a truth table for three variables A, B and C that will have an output 1 when XYZ=100, XYZ=101, XYZ=110, and XYZ=111. Write the Boolean expression for logic network in SOP form.
Answer:
2nd PUC Computer Science Model Question Paper 3 with Answers 1

Question 12.
Draw Karnaugh maps for the expression XYZ’ + X’Y’Z.
Answer:
2nd PUC Computer Science Model Question Paper 3 with Answers 2

Question 13.
Mention various characteristics of object-oriented programming language.
Answer:
The characteristics of object-oriented programming are objects, classes, data abstraction, data encapsulation, inheritance, and polymorphism.

Question 14.
Write an example to show the use of parameterized constructors through explicit calls.
Answer:
2nd PUC Computer Science Model Question Paper 3 with Answers 3

Question 15.
What is ifstream class and ofstream class?
Answer:
The ifstream class supports input operations on files and is derived from istream class. The class ofstream class supports output operations on files and is derived from ostream class.

KSEEB Solutions

Question 16.
Name the data types supported by DBMS.
Answer:
The various data types supported by DBMS are integer, single precision, double precision, character, string, memo, index, currency, date and text fields.

Question 17.
Explain update command with example.
Answer:
UPDATE command:
It can be used to change row values from a table. The SET keyword takes the column in which values need to be changed or updated. The WHERE keyword is used to filter the records in some conditions.
Syntax:
UPDATE tablename SET columnname = values WHERE Condition;
Example:
UPDATE student SET combn = ‘PCMCs’ where combn=’pcmc’;

Question 18.
Name the different networking techniques.
Answer:
The networking techniques (topologies) are Bus topology, Ring topology, Star topology, Three topology, Mesh topology, Extended star topology.

PART – C

III. Each question carries three marks. (5 × 3 = 15)

Question 19.
Write a short note on SMPS of a computer system.
Answer:
Switched Mode Power Supply (SMPS): The switched mode power supply acting as a DC to DC converter first rectifies an AC input voltage (240Volts), converts it to DC and, chops this DC in a “chopper” and converts it to a higher or lower level of DC (5 to 12 V DC Volts). Typical modern applications include computer power supply and the power supplies in TV and Video sets. The wattage of SMPS is around 150 to 500 watts.
SMPSs are three types :

  1. AT SMPS
  2. ATX SMPS
  3. BTXSMPS

1. AT SMPS :
AT stands for Advanced Technology. These are all old SMPSs. They had 12pin power connector, this is called as AT power connector. They were used in Pentium-I, Pentium-MMX, Pentium-II and Pentium-III CPUs.

2. ATX SMPS :
ATX stands for Advanced Technology extended. They had 20pin Power connector, this is called as ATX power connector. They were used in Pentium-III, Pentium- IV and AMD CPUs.

3. BTX SMPS :
BTX stands for Balanced Technology extended. They have 24pin Power connector, this is also called as ATX power connector. It has 15pin SATA power connectors. They are used in Dual core, core2duo, Quad core, i3, i5, i7 and latest AMD CPUs.

Question 20.
Explain the working of NAND gate.
Answer:
This is a NOT-AND gate which is equal to an AND gate followed by a NOT gate. The outputs of all NAND gates are high if any of the inputs are low.
2nd PUC Computer Science Model Question Paper 3 with Answers 4

Question 21.
Write the algorithm for linear search.
Answer:
Linear_search(Array, ele, N)
Step 1: loc = -l
Step 2 : for i = 0 to N-l repeat step 3 through step step 5
Step 3 : if (Array[i] = ele ) then step 4 through step 5
Step 4 : loc = i Step 5 : goto step
Step 6 : if (loc >=0 ) then step 7 otherwise step 8
Step 7 : output “The search element found at location “, loc
Step 8 : output “Search is unsuccessful”
Step 9 : exit.

KSEEB Solutions

Question 22.
Mention any three advantages of pointer.
Answer:
The three advantages of pointer are

  1. For objects, memory can be allocated dynamically during runtime.
  2. When objects are not used, then memory can be released.
  3. Memory is efficiently used.

Question 23.
Mention the types of files. Explain anyone.
Answer:
The text file and binary file are the two types of files in C++.
Text file:
A text file also called a ASCII file is a stream of characters that can be sequentially processed in forward direction by the computer. In text files, each line of text end with a EOL (End of line) character and each file ends with a special character called the EOF (End of file). The file can be edited using text editors like C++ editor, notepad, vi editors in UNIX.

Question 24.
What is generalization and specialization?
Answer:
1. Generalization is the process of taking the union of two or more lower-level entity sets to produce a higher-level entity sets. Generalisation is a bottom-up approach.

2. Specialization is a top-down approach in which one higher level entity can be broken down into two lower level entities.

Question 25.
List the services of e-commerce?
Answer:
Few services of ecommerce are

  • Domain name purchasing
  • Secure hosting
  • Full integration with the payment gateway of your choice
  • Web design
  • Shopping cart system
  • Marketing .

PART- D

IV. Each question carries five marks. (7 × 5 = 35)

Question 26.
Given F(A,B,C,D) = (0,2,4,6,8,10,14). Use Karnaugh map to reduce the function F using POS form. Write a logic gate diagram for the reduced POS expression.
Logical gate diagram:
2nd PUC Computer Science Model Question Paper 3 with Answers 5
2nd PUC Computer Science Model Question Paper 3 with Answers 6

KSEEB Solutions

Question 27.
Write a short note on queues.
Answer:
A queue is a non-primitive data structure where an item is inserted at one end and removed from the other end. The data structure queue is called FIFO(First In First Out) which means first inserted data item is removed first.

The variable FRONT is used to identify the position of the first element in the queue, and variable REAR is used to identify the position of last element of the queue. The different types of queues are simple queue, circular queue, priority queue, and dequeue.

1. Simple Queue:
Simple queue Insertion occurs at the rear of the list, and deletion occurs at the front of the list.
2nd PUC Computer Science Model Question Paper 3 with Answers 7
2. Circular Queue :
A circular queue is a queue in which all nodes are treated as circular such that the first node follows the last node.
2nd PUC Computer Science Model Question Paper 3 with Answers 8
3. Priority Queue :
A priority queue is a queue that contains items that have some preset priority. When an element has to be removed from a priority queue, the item with the highest priority is removed first
2nd PUC Computer Science Model Question Paper 3 with Answers 9
4. Dequeue (Double Ended queue):
In dequeue(double ended queue) Insertion and Deletion occur at both the ends i.e. front and rear of the queue.
2nd PUC Computer Science Model Question Paper 3 with Answers 10
5. Operations on Queue:
The queue(), enqueue(item), dequeue(), isEmptyO and size() are the operations that can be performed on queues.

Applications of Queue :
Queue is used when things don’t have to be processed immediately, but have to be processed in First In First Out order.

  • When a resource is shared among multiple consumers. Examples include CPU scheduling, Disk Scheduling.
  • When data is transferred asynchronously (data not necessarily received at same rate as sent) between two processes.

Examples include 10 Buffers, pipes, file 10, etc.

Question 28.
Write an algorithm to insert a data element at the beginning of a linked list.
Answer:
Step 1: if AVAIL = NULL then
Print” Availability stack is empty”
Else
NEWJMODE = AVAIL
AVAIL = AVAIL -> LINK
Step 2: if FIRST = NULL then
NEWJMODE-> INFO = ELEMENT
NEWJMODE-> LINK = NULL
FIRST = NEW_NODE
Else
NEWJMODE -> INFO = ELEMENT
NEWJMODE-> LINK = FIRST
FIRST = NEW_NODE
Step 3: return

Question 29.
What are advantages and disadvantages of object oriented programming? Explain.
Answer:
The advantages of object oriented programming:
1. OOP provides a clear modular structure for programs. Large problems can be reduced to smaller and more manageable problems.

2.  In OOP, data can be made private to a class such that only member functions of the class can access the data. This principle of data hiding helps the programmer to build a secure program.

3. Implementation details are hidden from other modules and other modules has a clearly defined interface.

4. It is easy to maintain and modify existing code as new objects can be created with small differences to existing ones.

The disadvantages of object oriented progrmming:
1. Size:
Object Oriented programs are much larger than other programs. In the early days of computing, space on hard drives, floppy drives and in memory was at a premium. Today we do not have these restrictions.

2. Effort:
Object Oriented programs require a lot of work to create. Specifically, a great deal of planning goes into an object oriented program well before a single piece of code is ever written. Initially, this early effort was felt by many to be a waste of time. In addition, because the programs were larger (see above) coders spent more time actually writing the program.

3. Speed:
Object Oriented programs are slower than other programs, partially because of their size. Other aspects of Object Oriented Programs also demand more system resources, thus slowing the program down.

KSEEB Solutions

Question 30.
Describe how objects can be used as function arguments.
Answer:
The objects of same class can be passed as arguments to a function. This is explained with an example,
2nd PUC Computer Science Model Question Paper 3 with Answers 11
2nd PUC Computer Science Model Question Paper 3 with Answers 12
in the above example, rup is the class, n1 and n2 are data members, readdata(), printdataQ and multi() are member functions with r1, r2 are the function argument of object type. The line obj3.multi (obj1, obj2); pass obj1 and obj2 which are the objects of class rup type. In the statement
n1 = r1.n1 * r1.n2;
n2 = r2.n1 * r2.n2;

r1.n1 is accessing data member n1 of object r1 and r1.n2 is accessing data member n2 of object r1 and their product is stored in data member n1 and same for n2 in the second line which is the member of obj3 object. The obj3. printdata() gives out the result stored in the n1 and n2.

Question 31.
Write the advantages and disadvantages of inline functions.
Answer:
Inline functions are those functions whose body is inserted in place of the function call.
Advantages of inline functions:

  • The speed of execution of a program increases
  • Efficient code can be generated
  • Readability of the program increases

Disadvantages of inline functions:

  • The size of the executable file increases and more memory is required as the body of inline function is inserted in the place of function call.
  • Not suitable for too long, complicated, recursive function definition

Question 32.
Explain different types of constructors.
Answer:
There are three types of constructors.

  1. Namely Default constructor.
  2. Parameterized constructor.
  3. Copy constructor.

1. Default Constructor:
A constructor which does not take any arguments is called a zero argument constructor. It is also called zero-argument constructor.
Features of default constructor are:

  • All objects of a class are initialized to same set of values
  • These constructors has no arguments
  • These constructors are automatically called when every object is created.

The disadvantages of default constructor are:

  • Different objects cannot be initialized with different values.
  • Declaring a constructor with arguments hides default constructor.

2. parameterized constructor:
A constructor that takes one or more arguments is called a parameterized constructor.
The features of parameterized constructors are :

  • parameterized constructors can be overloaded
  • parameterized constructors can have default arguments and default values.

The different methods through which constructors can be invoked are implicit call and explicit call and using operator.
a. Explicit call:
It is a method of invoking a function where the declaration of object is followed by assignment operator followed by a constructor followed by argument list enclosed within parentheses.

b. Implicit Call:
It means the declarationof the object is followed by argument list enclosed in parentheses used with constructors The is used for the parameterized constructor which has exactly one argument.

3. Copy Constructor:
It is a parameterized constructor using which one object can be copied into another object. The copy constructor takes an object as argument and is used to copy values of data members of one object into other object.

Question 33.
Explain multilevel inheritance with a suitable C++ program.
Answer:
2nd PUC Computer Science Model Question Paper 3 with Answers 13
2nd PUC Computer Science Model Question Paper 3 with Answers 14
2nd PUC Computer Science Model Question Paper 3 with Answers 15
2nd PUC Computer Science Model Question Paper 3 with Answers 16
2nd PUC Computer Science Model Question Paper 3 with Answers 17
In the above program student information is a derived class from student parents and student marks is derived class from student information. Std is a object belongs to student marks class. The object std can access the member functions of classes student marks, student information and student parents.

Question 34.
Write the differences between manual and electronic file systems.
Answer:
The difference between manual, and electronic file systems
2nd PUC Computer Science Model Question Paper 3 with Answers 18

Question 35.
Explain DML commands with example.
Answer:
The data manipulation commands are used for retrieval (view) of data, insertion of new data, modification of data or deletion. The DML commands include insert, delete and update.
1. INSERT command :
It is used to inserts new rows into the table.
Syntax:
INSERT INTO tablename (columnname1, columnname2,…) VALUES (value1, , value2, ..);
Example:
INSERT INTO student (Regno, name, Combn, fees) VALUES (1234, ‘Hemanth’, ‘PCMCs’,15000);

2. UPDATE command:
It can be used to change row values from a table. The SET keyword takes the column in which values need to be changed or updated. The WHERE keyword is used to filter the records in some conditions.
Syntax :
UPDATE tablename SET columnname = values WHERE Condition; Example: UPDATE student SET combn = ‘PCMCs’ where combn=’pcmc’;

3. DELETE Command :
It is used to delete/remove the tuples/rows from the table. All the rows will be deleted if WHERE clause is not used in the statement otherwise it selects the rows for delete which satisfies the condition.
Syntax:
DELETE from tablename WHERE Condition;
Example:
DELETE from student;
DELETE from student WHERE regno=1234;

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Question 36.
Write a short note on star topology and tree topology.
Answer:
Star Topology:
2nd PUC Computer Science Model Question Paper 3 with Answers 19
A star topology is designed with each node (file server, workstations, and peripherals) connected directly to a central network hub, switch. Data on a star network passes through the hub, switch, before continuing to its destination. The hub, switch, manages and controls all functions of the network. It also acts as a repeater for the data flow.
Advantages of a Star Topology :

  • Easy to install and wire.
  • No disruptions to the network when connecting or removing devices.
  • Easy to detect faults and to remove parts.

Disadvantages of a Star Topology :

  • Requires more cable length than a linear topology.
  • If the hub, switch fails, nodes attached are disabled.
  • More expensive than linear bus topologies because of the cost of the hubs, etc.

Tree topology :
2nd PUC Computer Science Model Question Paper 3 with Answers 20
A tree topology combines characteristics of linear bus and star topologies. It consists of groups of star-configured workstations connected to a linear bus backbone cable. Tree topologies allow for the expansion of an existing network.
Advantages of a Tree Topology:

  • Point-to-point wiring for individual segments.
  • Supported by several hardware and software vendors.

Disadvantages of a Tree Topology:

  • Overall length of each segment is limited by the type of cabling used.
  • If the backbone line breaks, the entire segment goes down.
  • More difficult to configure and wire than other topologies.

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