(a) Clock speed-Each microprocessor has an internal clock that regulates the speed at which it executes instructions. The speed at which the microprocessor executes instructions is called clock speed. It is measured in MHz vi in GHz.
(b) Instructions set - The set of machine level instructions that a microprocessor is designed to execute is called as an instruction set. This carries 4 types of operations. They are
1. Data transfer
2. Arithmetic operations
3. Logical Operations
4. Control flow
5. Input/output
(c) Word size - The number of bits that can be processed by a processor in a single instruction is called as its word size. It determines the amount of RAM that can be accessed at one time and the total number of pins on the microprocessor. Total number of input and output pins in turn determines the architecture of the microprocessor.

The data and application that are stored and processed on multiple physical locations across the world over the digital network (internet/intranet). The Distributed Operating System is used to access shared data and files that reside in any machine around the world. The user can handle the data from different locations. The users can access as if it is available on their own computer.
The advantages of distributed Operating System are as follows:
1. A user at one location can make use of all the resources available at another location over the network.
2. Many computer resources can be added easily in the network
3. Improves the interaction with the customers and clients.
4. Reduces the load on the host computer.

The benefits of open source are tremendous and have gained huge popularity in the IT field in recent years. They are as follows:
1. Open source (OS) is free to use, distribute and modify.
2. Open source is independent of the company as author who originally created it.
3. It is accessible to everyone. Anyone can debug the coding.
4. It doesn't have the problem of incompatible formats that exists in proprietary software.
5. It is easy to customize as per our needs.
6. Excellent support will be provided by programmers who will assist in making solutions. Some of the disadvantages are:
7. Latest hardware are incompatible, i.e. lack of device drivers.
8. It is less user friendly and not as easy to use.
9. There may be some indirect costs involved such as paying for external support.
10. Malicious users can potentially view it and exploit any vulnerability.

The method of repeated multiplication by 2 has to be used to convert such kinds of decimal fractions.
The steps involved in the method of repeated multiplication by 2.
Step 1: Multiply the decimal fraction by 2 and note. the integer part. The integer part is either 0 or 1.
Step 2: Discard the integer part of the previous product. Multiply the fractional part of the previous product by 2. Repeat Step 1 until the same fraction repeats or terminates (0).
Step 3: The resulting integer part forms a sequence of 0's and 1's that become the binary equivalent of a decimal fractions.
Step 4: The final answer is to be written from first. integer part obtained till the last integer part is obtained.

A memory is just like a human brain. It is used to store data and instructions. Computer memory is the storage space in the computer, where data and instructions are stored. There are two types of accessing methods to access (read or write) the memory. They are sequential access and random access. In sequential access, the memory is accessed in an orderly manner from starting to end. But, in random access, any byte of memory can be accessed directly without navigating through previous bytes. Different memory devices are arranged according to the capacity.

Computers size was drastically reduced. S.No	Generation	Period	Main Component used	Merits/ Demerits
1	First Generation	1942-1955		Big in size Consumed more power Malfunction due to overheating Machine Language was used Eg : ENIAC, EDVAC, UNIVAC
First Generation Computer - ENIAC, EDVAC, UNIVAC 1 ENIAC weighed about 27 tons, size 8 feet x 100 feet x 3 feet, and consumed around 150 watts of power
2	Second Generation	1955-1964		Smaller compared to First Generation Generated Less Heat Consumed less power compared to first-generation Punched cards were used. The first operating system was developed Batch Processing and Multiprogramming Operating System Machine language, as well as Assembly language, was used.
Second Generation Computers IBM 1404, IBM 1620, UNIVAC 1108
3	Third Generation	1964-1975		Computers were smaller, faster, and more reliable Consumed less power. High-Level Languages were used.
Third Generation Computers -IBM ·360 series, Honeywell 6000 series
4	Fourth Generation	1975-1980		Smaller and Faster Microcomputer series such as IBM and APPLE was developed Portable Computers were introduced.
5	Fifth Generation	1980-till date		Parallel, Processing Superconductors Computers size was drastically reduced. Can recognize Images and Graphics Introduction of Artificial Intelligence and Expert Systems Able to solve highly complex problems including decision making and logical reasoning.
6	Sixth Generation	In future		Parallel and Distributed computing Computers have become smarter faster and smaller Development of robotics Natural Language Processing Development of Voice Recognition Software.

1. hypotenuse (a, b)
2. -- inputs: a, b are real numbers, a > 0, b > 0
3 -- outputs: c²=a² + b² where c is real number, c > 0

Let the variables a, b, c represent the glass A, glass B and Glass C respectively. Variables A, B, C can store values APPLE, GRAPE or EMPTY.
Initial State:
1.-- a, b, c : = APPLE, GRAPE, EMPTY
2. C : = b
3.-- a, b, c : = APPLE, EMPTY, GRAPE
4. b : = a
5.-- a, b, c : = EMPTY, APPLE, GRAPE
6. a : = C
7.-- a, b, c : = GRAPE, APPLE, EMPTY
Final State:
Specification:
1. Exchange ()
2. -- inputs: a, b, c :- APPLE, GRAPE, EMPTY
3.--outputs: a, b, c:- GRAPE, APPLE, EMPTY

1101011₂ - 111010₂
1101011
+111010
_______
0110001
_______
1101011₂ - 111010₂  = 0110001₂

Bubbled AND Gate
(i) Realized of and using only AND gate:
The Boolean function for AND is C=AB. The same can be realized using only NAND gates.

(ii) Realized of and using only NAND's gate:
The Boolean function for OR is C=A + B. The same can be realized using only NAND gates.

\(\mathrm{C} =\overline{\overline{\mathrm{A}} \cdot \overline{\mathrm{B}}} \)
\(=\overline{\mathrm{A}}+\overline{\mathrm{B}} \text { (By Demorgan's law) } \)
\(=\mathrm{A}+\mathrm{B} \text { (output of OR gate) } \)
(iii) Realization of AND using NOR:
By using only the NOR gates, we can get the output equivalent to the output of AND gate.
C=A.B

\(\mathrm{C} =\overline{\overline{\mathrm{A}} \cdot \overline{\mathrm{B}}} \)
\(=\overline{\mathrm{A}}+\overline{\mathrm{B}}\)
= AB
(iv) Realization of OR using NOR's:
By using only NOR gates we are getting the output equivalent to OR gate.

S.No Versions Logo Year Prominent feature 1. Windows 1.x 1985 1. Introduction of GUI in 16 - bit. processor
2. Mouse was introduced as an input device 2. Windows 2.x 1987 1. Supports to minimize or maximize windows.
2. Control panel feature was introduced with
various system settings and customizing options. 3. Windows 3.x 1992 1. Introduced the concept of multitasking.
2. Supported 256 colors which brought a more
modern, colorful look to the interface. 4. Windows 95 1995 1. Introduced Start button, the taskbar, Windows Explorer and Start menu.
2. Introduced 32 - bit processor and focused more on multitasking. 5. Windows 98 1998 1. Integration of the Web browser (Internet Explorer) with the Operating System.
2. DOS gaming began to disappear as Windows-based games improved.
3. Plug and play feature was introduced. 6. Windows NT   Designed to act as servers in network. 7. Windows Me 2000 It introduced automated system diagnostics and recovery tools. 8. Windows 2000 2000 1. Served as an Operating System for business
desktop and laptop systems.
2. Four versions of Windows 2000 were released: Professional (for business)
3. desktop and laptop systems), Server (both a
Web server and an office server),
4. Advanced Server (for line-of-business 
applications) and Data Centre Server (for high traffic computer networks). 9. Windows XP 2001 1. Introduced 64-bit Processor.
2. Improved Windows appearance with themes and offered a stable version 10. Windows Vista 2006 Updated the look and feel of Windows 11. Windows 7 2009 Booting time was improved, introduced new user interfaces like Aero Peek, pinning programs to the taskbar, handwriting recognition etc, and Internet Explorer 8. 12. Windows 8 2012 1. Windows 8 was faster than previous versions of Windows.
2. Start button was removed
3. Windows 8 takes better advantage of multicore processing, solid-state drives (SSD), touch screens, and other alternate input methods.
4. Served as a common platform for mobile and computer. 13. Windows 10 2015 1. Start Button was added again.
2. Multiple desktops.
3. Central Notification Center for App notification and quick actions.
4. Cortana voice-activated personal assistant.

Specifications:
1 circulate
2-inputs:
3-outputs:
Algorithm:
a, b, c: = A, B, C
a, b, c: = C, A, B
1 circulate (a, b, c)
2 --a, b, c: A, B, C
3 temp: = c
4 c: b
5 b:=a
6 a:= temp
7 -- a, b, c: C, A, B

(i) f=1 i=1 f = f * i i = i+1
f = 1 x 1 = 1 i = 2
(ii) f = 1 x 2 i = 3 = 2
(iii) f = 2 × 3 i = 4 = 6
(iv) f = 6 x 4 i = 5 (loop terminates) = 24

Binary Operators require two operands.
Arithmetic operators that perform simple arithmetic operations like addition, subtraction, multiplication, division (+,-*, %,/) etc. are binary operators which requires minimum of two operands.
Relational operators are used to determine the relationship between its operands. The relational operators (<,>,>, <=,=, !=) are applied on two operands, hence they are binary operators.
AND, OR (logical operator) both are binary operators.
Assignment operator is also a binary operator (+=, -=, *=, /=, %=)

(a) a&b 

10	1010
11	1111
a&b	1010

a&b =10₁₀&15₁₀=1010=10₁₀
(b) a|b

10	1	0	1	0
15	1	1	1	1
10 | 15	1	1	1	1

a|b =10|15= 1111₂ =15₁₀
(c) a^b

10	1	0	1	0
15	1	1	1	1
10 ^ 15	0	1	0	1

a^b =10^15= 0101₂ =15₁₀
(d) a>>3
a=10
10/2³=10/8=1
10>>3=1₁₀
(e) (~b)

b	1	1	1	1
~b	0	0	0	0

(~b)=0000₂=0₁₀

An entry control loop checks the condition at the time of entry and if condition or expression becomes true then control transfers into the body of the loop. for loop and while loop are the examples of Entry Controlled Loop.
A while loop is a control flow statement that allows the loop statements to be executed as long as the condition is true.
Type: Entry control loop
Syntax:
while ( Text expression)
{
Body of the loop;
}
Statement-x;
The control flow and flow chart of the while loop is shown below:

Control flow:
Step 1: Test-expression is evaluated to either True or False;
Step 2: If test-expression is true;
(a) The body of the loop is executed.
(b) Control is transferred to step 1.
Step 3: If test-expression is false, the control exits the while loop.
Example:
int a = 1;
while (a<-10)
{
cout< a+=2;
}
Output
1 3 5 7 9

(a) #include <iostream>
using namespace std;
#include <math.h>
#factorial function.
long int fact(int n)
{
long int f= 1;
for (int i = 1;i<n; i++)
f* = i;
return f
}
// main function
int main()
{
int x, y = 1;
float sum = 0;
cout<<"Enter the value for x:";
cin>>x;
for (int i=1;i<-6; i++)
{
sum = sum + y*((pow(x, 1))/fact(i));
y*=-1;
}
cout<<"Sum=" <<Sum<<endl;
cin.get():
return();
}
(b) #include <iostream>
using namespace std;
#include <math.h>
int main()
{
int M, L;
float sum = 0;
cout<<"Enter the value for M:";
cin>>M;
cout<<"Enter the number of terms";
cin>> L;
for (int i=1;i<=m; i++)
sum Sum + Pow(m,i)/i;
cout<<"Sum="<< Sum<<endl;
cin.get():
return 0;
}

The return statement
Returning from the function is done by using the return statement. The return statement stops execution and returns to the calling function. When a return statement is executed, the function is terminated immediately at that point. The return statement is used to return from a function. It is categorized as a jump statement because it terminates the execution of the function and transfer the control to the called statement.
Syntax:
return expression/variable;
Example: return(a+b); return(a);
return; // to terminate the function
The Returning values:
The functions that return no value is declared a void. The data type of a function is treated as int, if no data type is explicitly mentioned. For example,
For Example:
int add (int, int);
add (int, int);
In both prototypes, the return value is int, because by default the return value of a function in C++ is of type int.
Returning Non-integer values
A string can also be returned to a calling statement.
#include <iostream>
#include<string.h>
using namespace std;
char*display()
{
return ("chennai");
}
int main()
{
char s[50];
strcpy(s,display());
cout<<"\n Example:Function with Non Integer Return"<<s;
return(0);}
Output:
Example: Function with Non Integer Return Chennai
The Returning by reference
#include<iostream>
using namespace std;
int main()
{
int nl-150;
int &nlrefn1;
cout<<"\nThe Value of NI =" <<n<< "and nl Reference="<<n/ref;
n1ref++;
cout<<"\n After nl increased the Value of N1 ="<<nl;
cout<<"and nl Reference="<<ntref;
return(0);
Output:
The Value of N1 - 150 and n1 Reference - 150
After n1 increased the Value of N1 =151 and n1 Reference = 151

#include<iostream>
using namespace std;
int reverse (int n);
int main()
{
int n=0, result = 0;
cout<<"Enter number:"";
cin >>n;
result reverse (n);
cout<<"Reverse number is:" << result << endl;
}
int reverse (int n)
{
int temp=0, rev = 0;
while (n! =0).
{
temp=n;
rev= (rev*10) + temp;
n = n/10;
}
return rev;
}
Output:
Enter number: 1 2 3
Reverse number: 3 2 1

In C++, arrays can be passed to a function as an argument. To pass an array to a function in C++, the function needs the array name as an arugment. Passing a two-dimensional array to a function a program to display marks of 5 students by passing one dimensional array to a function
# include< iostream >
using namespace std;
void display (in m[5]);
int main()
{
int marks [5]= {88, 76, 90, 61, 69};
display(marks);
return 0;
}
void display (int m[5])
{
cout << "\n Display Marksr:" << endl;
for(int i=0; i<5; i++)
{
cout << "Student'< }
}
Output:
Display Marks:
Student 1: 88
Student 2: 76
Student 3: 90
Student 4: 61
Student 5: 69

Call by value:
When a structure is passed as argument to a function using call by value method, any change made to the contents of the structure variable inside the function to which it is passed do not effect the structure variable used as an argument.
Consider this example:
#include< iostream >
useing namespace std;
struct Employee
{
char name[50];
int age;
float salary;
};
void printData(Employee); //Function declaration
int main()
{
Employee p;
cout << "Enter Full name:";
cin >> p.name;
cout << "Enter age:";
cin >> p.age;
cout << "Enter salary:"; .
cin >> p.salary,
//Function call with structure variable as
an argument
PrintData(P);
return ();
}
void printData(Employee q)
{
cout << "\nDisplaying Information. " << endl;
cout << "Name:" << q.name << endl;
cout << "Age:" << q.age << endl;
cout << '' Salary: " << q.salary:
}
Output:
Enter Full name: Kumar
Enter age: 55
Enter salary: 34233.4
Displaying Information.
Name: Kumar
Age: 55
Salary: 34233.4

#include< iostream >
using namespace std;
struct Distance
{
int feet;
float inch;
}d1, d2, sum;
int main()
{
cout<<"Enter 1st distance:"<<endl;
cout<<"Enter feet:";
cin>>d1.feet;
cout<<"Enter inch:";
cin>>dl.inch;
cout<<"\n information for 2nd distance:"<<endl;
cout<<"Enter feet:"";
cin>>d2.feet;
cout<<"Enter inch:"
cin>>d2.inch;
sum.feet = d1.feet +d2.feet;
sum.inch-d1.inch + d2.inch;
if (sum.inch > 12)
{
++sum.feet;
sum.inch 12;
}
cout<<endl; "Sum of distance="<<sum.feet << "feet" << sum.inch<< "inches";
return 0;
}
Output:
Enter 1st distance
Enter feet: 6
Enter inch: 3.4
Enter 2nd distance
Enter feet: 5
Enter inch: 10.2
Sum of distances = 12 feet 1.6 inches

#inclue< iostream.h > 
using namespace std;
struct book
{
char bname[20];
char author[20];
int price;
}b[10];
void input (struct book a[])
{
int i;
cout << "Enter the book details'' < for(i=0; i < 10; i++)
{
cout << "Enter book name"<
cin >> b[i].bname;
cout << "Enter Author name''< 
cin >> b[i].author;
cout << "Enter the price of the book"<|
cin >> b[i].price;
}
}
void display (struct book b[])
{
int i; total = 0;
for(i=0; i<9; i++)
{
cout << "Book Name < cout << "Author Name: < cout << ''Book Price: < total = total + b[i].prince < cout << ''Total price''< }
}
int main()
{
input(b);
display(b);
return 0;
}