Prasanna

Students can Download Chapter 8 Arrays Notes, Plus One Computer Science Notes helps you to revise the complete Kerala State Syllabus and score more marks in your examinations.

Kerala Plus One Computer Science Notes Chapter 8 Arrays

Summary
An array is a collection of elements with same data type Or with the same name we can store many elements, the first or second or third, etc can be distinguished by using the index(subscript). The first element’s index is 0, the second elements index is 1, and so on.

Declaring arrays:
Suppose we want to find the sum of 100 numbers then we have to declare 100 variables to store the values. It is a laborious work. Hence the need for array arises.
Syntax: data_type array_name[size];
To store 100 numbers the array declaration is as follows
int n[100]; By this we store 100 numbers. The index of the first element is 0 and the index of last element is 99.

Memory allocation for arrays:
The amount of memory requirement is directly related to its type and size,

• int n[100]; It requires 2Bytes(for each integer) × 100 = 200 Bytes.
• float d[100]; It requires 4Bytes(for each float) × 100=400 Bytes.

Array initialization:
Array can be initialized in the time of declaration. eg: int age[4] = {16, 17, 15, 18};

Accessing elements of arrays:
Normally loops are used to store and access elements in an array.
eg:
int mark[50], i;
for(i=0;i<50;i++)
{
cout<<“Enter value for mark”<<i+1;
cin>>mark[i];
}
cout<<“The marks are given below:”;
for(i=0;i<50;i++)
cout<<mark[i];

Array operations:
Traversal:
Accessing all the elements of an array is called traversal.

Sorting:
Arranging elements of an array in an order(ascending or descending)
1. Bubble sort:
It is a simple sorting method. In this sorting considering two adjascent elements if it is out of order, the elements are interchanged. After the first iteration the largest(in the case of ascending sorting) or smallest(in the case of descending sorting) will be the end of the array. This process continues.

2. Selection sort:
In selection sort the array is divided into two parts, the sorted part and unsorted part. first smallest element in the unsorted part is searched and exchanged with the first element. Now there is 2 parts sorted part and unsorted part. This process continues.

Searching:
It is the process of finding the position of the given element.
1. Linear search:
In this method each element of the array is compared with the element to be searched starting from the first element. If it finds the position of the element in the array is returned.

2. Binary search:
It uses a technique called divide and conquer method. It can be performed only on sorted arrays. First we check the element with the middle element. There are 3 possibilities. The first possibility is the searched element is the middle element then search can be finished.

The second possibility is the element is less than the middle value so the upper bound is the middle element. The third possibility is the element is greater than the middle value so the lower bound is the middle element. Repeat this process.

Two dimensional (2D) arrays:
Some occasions we have to store 6 different marks of 50 students. For this we use 2D arrays. An array with two subscripts is used.
eg: int mark[r][c]; Here r is the row and c is the column.

Declaring 2D arrays:
Syntax: datatype array_name[rows][columns];
The elements of this array is referred as mark[0][0], mark[0][1], mark[r – 1][c – 1].
eg: int m[5][5]; This array can store 5 × 5 = 25 elements.

Matrices as 2D arrays:
Matrix is a concept in mathematics that can be represented by 2D array with rows and columns. A nested loop(a loop contains another loop) is used to store and access elements in an array.

Multi-dimensional arrays:
3 Dimensional(3D) array is an example for this.
Syntax: data_type array_name[size1 ][size2][size3];
eg: int m[5][5][5]; This array can store 5 × 5 × 5 = 125 elements.

Students can Download Chapter 12 Internet and Mobile Computing Notes, Plus One Computer Science Notes helps you to revise the complete Kerala State Syllabus and score more marks in your examinations.

Kerala Plus One Computer Science Notes Chapter 12 Internet and Mobile Computing

Summary
History of Internet:
Internet means international network of networks. The first form of Internet is ARPANET(Advanced Research Project Agency Network) started by US Department of Defence for their military during 1970’s. In 1989 a team lead by Tim Berners Lee introduced WWW(World Wide Web) by using the protocol HTTP. In 1998, Internet Corporation for Assigned Names and Numbers (ICANN) was established.

Internet:
It is a network of networks. It means that international network. We can transfer information between computers within nations very cheaply and speedily.

Intranet:
A private network inside a company or organisation is called intranet.

Extranet:
It allows vendors and business partners to access the company resources.

The hardware and software requirement for internet:

• A computer with a modem (internal/external)
• A telephone connection
• An account with an ISP
• A browser S/W eg: Internet Explorer or Mozilla…

Types of connectivity:
There are two ways to connect to the internet. First one dialing to an ISP’s computer or with a direct connection to an ISP.

Dial-up Connection:
Here the internet connection is established by dialing into an ISP’s computer. If ISP is not busy they verify the user name and password if it is valid they will connect our computer to the internet.lt uses Serial Line Internet Protocol (SLIP) or Point to Point Protocol (PPP). It is slower and has a higher error rate.

Direct connection:
In direct connection there is a fixed cable or dedicated phone line to the ISP. Here it uses ISDN (Integrated Services Digital Network) a high speed version of a standard phone line. Another method is leased lines that uses fibre optic cables.

Digital Subscribers Line (DSL) is another direct connection, this uses copper wires instead of fibre optic for data transfer. Direct connection provides high speed internet connection and error rate is less. Fibre To The Home(FTTH) uses optical fibers for data transmission.

Wireless broadband connectivity:
1. Mobile broadband:
Accessing Internet using wireless devices like mobile phones, tablet, USB dongles.

2. Wi MAX(Wireless Microwave Access):
It uses micro waves to transmit information across a network in a range 2 GHz to 11 GHz over very long distance.

3. Satellite broadband:
Accessing internet through satellite. A Very Small Aperture Terminal(VSAT) dish antenna and transceiver and modem are required at the user’s location. Expensive and high speed.

Internet access sharing methods:
One Internet connection can be shared among several computers using a LAN, Wi Fi or Li Fi.
1. Using LAN:
The Internet connection in a LAN can be shared among other computers in the network

2. Using Wi Fi(Wireless Fidelity):
It uses radio waves to transmit information across a network in a range 2.4 GHz to 5 GHz in short distance. Nowadays this technology is used to access internet in campuses, hyper markets, hotels by using Laptops, Desktops, tablet, mobile phones etc

3. Using Li Fi(Light Fidelity) network:
It is a fast optical(uses visible light for data transmission) version of Wi Fi. Its main component is a LED lamp that can transmit data and a photodiode that acts as a receiver.

Services on Internet:
1. www(World Wide Web):
This means this website address is unique and can be accessed each nook and corner of the world.

2. A browser is a piece of software that acts as an interface between the user and the internal working of the internet. With the help of a browser the user can search information on the internet and it allows user to navigate through the web pages. The different browsers are

• Microsoft internet explorer
• Mozilla Firefox
• Netscape Navigator
• Google Chrome
• Opera.

3. Web Browsing:

1. The browser determines the URL entered.
2. The browser asks the DNS for URLS corresponding IP address (Numeric address)
3. The DNS returns the address to the browser.
4. The browser makes a TCP connection using the IP address.
5. Then it sends a GET request for the required file to the server.
6. The server collects the file and send it back to the browser.
7. The TCP connection is released.
8. The text and the images in the web pages are displayed in the browser.

Search engines:
By using search engines we will get a variety of information. It is a newly developed tool that helped to search the information on the internet more effectively and easily. Search engines are programs that help people to locate information from crores of website on internet using a database that consists of references.

Users can interact with the search engine through the home page of the search engine. To get the information about artificial intelligence just type this in the box provided for it and click the search button. Search engines searches by using a particular search algorithm then displays the matching documents or web addresses.

Search engine use soft wares called spiders or bots to search documents and their web addresses. Spiders search the internet using the directions given by the search engines and prepare an index and stores it in a database. The searching algorithm searched this database when the users submits a request and create a web page displaying the matching results as hyperlinks.
eg: Google, Yahoo, Rediff etc.

E mail(Electronic mail):
It is used to send text, multi media messages between computers over internet. An example of an email id is jobi_cg@rediffmail.com. Here jobi_cg is the user name, rediffmail is the website address and .com is the top level domain which identifies the types of the organisation. To send an email we require an email address. Some websites provide free email facility.

To send an email first type the recipients address and type the message then click the send button. The website’s server first check the email address is valid, if it is valid it will be sent otherwise the message will not be sent and the sender will get an email that it could not deliver the message.

This message will be received by the recipient’s server and will be delivered to recipient’s mail box. He can read it and it will remain in his mail box as long as he will be deleted. Simple Mail Transfer Protocol(SMTP) is used.
The email message contains the following fields:

1. To: Recipient’s address will be enter here. Multiple recipients are also allowed by using coma.
2. CC: Enterthe address of other recipients to get a carbon copy of the message.
3. bcc: The address to whom blind carbon copies are to be sent. This feature allows people to send copies to third recipient without the knowledge of primary and secondary recipients
4. From: Address of the sender
5. Reply to: The emait address to which replies are to be sent.
6. Subject: Short summary of the message.
7. Body: Here the actual message is to be typed.

The advantages of email are given below:

1. Speed is high
2. It is cheap
3. We can send email to multiple recipients
4. Incoming messages can be saved locally
5. It reduces the usage of paper
6. We can access mail box anytime and from anywhere.

The disadvantages are:

1. It requires a computer, a modem, software and internet connection to check mail.
2. Some mails may contain viruses
3. Mail boxes are filled with junk mail. So very difficult to find the relevant mail.

Social media:
Various social medias are Internet forums, social blogs, microblogs etc.

1. Internet forums: It is an online discussion site where people can exchange information about various issues like social, political, educational etc in the text form.
2. Social blogs: Conducting discussions about . particular subjects by entries or posts. eg: Blogger.com
3. Microblogs: It allows users to exchange short messages, multi media files etc. eg: www.twitter.com
4. Wikis: In this we can give our contributions regarding various topics. eg: www.wikipedia.org
5. Social networks: By using these web sites we can post our data and’ view others data. eg: www.facebook.com
6. Content communities: By using these websites we can share multi media files. eg: www.youtube.com

Advantages of social media:

1. Bring people together: It allows people to maintain the friendship
2. Plan and organize events: It allows users to plan and organize events.
3. Business promotion: It helps the firms to promote their sales.
4. Social skills: There is a key role of the formation of society.

Disadvantages:

1. Intrusion to privacy: Some people may misuse the personal information.
2. Addiction: sometimes it may waste time and money.
3. Spread rumours: The news will spread very quickly and negatively.

Cyber Security:
It is used to provide protection of valuable information such as credit card information from unauthorized access, intentional access, deletion, etc. while shopping on the internet.

Computer virus:
A virus is a bad program or harmful program to damage routine working of a computer system. It reduces the speed of a computer. It may delete the useful system files and make the computer useless.

Worm:
It is a stand alone malware program that replicates itself in order to spread to other computers. It slows down the traffic by consuming the bandwidth. In 2000 a worm called “ILOVEYOU” is affected many computers.

Trojan horse:
It appears as a useful software but it is a harmful software and it will delete useful software or files.

Spams:
Sending an email without recipient’s consent to promote a product or service is called spamming. Such an email is called a spam.

Hacking:
It is a process of trespassing computer networks. Two types white hats and black hats. White hats hack the computer networks to test the security but black hats intentionally stealing valuable data or destroying data.

Phishing (Fishing):
It is an attempt to get others information such as usenames, passwords, bank a/c details etc by acting as the authorized website. Phishing websites have URLs and home pages similar to their original ones and mislead others , it is called spoofing.

Denial of Service(DoS) attack:
Its main target is a Web server. Due to this attack the Web server/computer forced to restart and this results refusal of service to the genuine users. If we want to access a website first you have to type the web site address in the URL and press Enter key, the browser requests that page from the web server.

Dos attacks send huge number of requests to the web server until it collapses due to the load and stops functioning.

Man in the Middle attacks:
It is an attack in which an attacker secretly intercepts electronic messages send by the sender to the receiver and then modifies the message and retransmit it to the receiver.

To prevent this type of attack encrypted connections such as HTTPS(HTTP Secure), SFTP(Secure FTP) etc, must be used, that will be displayed in the URL.

Preventing network attacks
Firewall:
It is a system that controls the incoming and outgoing network traffic by analyzing the data and then provides security to the computer network in an organization from other network (internet).

Antivirus scanners:
It is a tool used to scan computer files for viruses, worms and Trojan horses and cure the infected system. If any fault found it stops the file from running and stores the file in a special area called Quarantine(isolated area) and can be deleted later.

Cookies:
Cookies are small text files that are created when we visit a website that keep track of our details. This information will help the hacker to use it for malicious purposes. It acts as a spyware.

Guidelines for using computers over internet:

• Emails may contain Viruses so do not open any unwanted emails
• Download files from reputed sources(sites)
• Avoid clicking on pop up Advt.
• Most of the Viruses spread due to the use of USB drives so use cautiously.
• Use firewall in your computer
• Use anti virus and update regularly
• Take backups in a regular time intervals

Mobile Computing:
The advancements in computing technology have led to the developments of more computing power in hand held devices like laptops, tablets, smart phones, etc. Nowadays people are able to connect to others through internet even when they are in move.

Mobile communication:
The term ‘mobile’ help the people to change their life styles and become the backbone of the society. Mobile communication networks do not require any physical connection.

Generations in mobile communication:
The mobile phone was introduced in the year 1946. Early stage it was expensive and limited services hence its growth was very slow. To solve this problem, cellular communication concept was developed in 1960’s at Bell Lab. 1990’s onwards cellular technology became a common standard in our country.
The various generations in mobile communication are
1.First Generation networks(1 G):
It was developed around 1980, based on analog system and only voice transmission was allowed.

2. Second Generation networks (2G):
This is the next generation network that was allowed voice and data transmission. Picture message and MMS(Multimedia Messaging Service) were introduced. GSM and CDMA standards were introduced by 2G.
(i) Global System for Mobile(GSM):
It is the most successful standard. It uses narrow band TDMA(Time Division Multiple Access), allows simultaneous calls on the same frequency range of 900 MHz to 1800 MHz. The network is identified using the SIM(Subscriber Identity Module).
(a) GPRS(General Packet Radio Services):lt is a packet oriented mobile data service on the 2G on GSM. GPRS was originally standardized by European Telecommunications Standards Institute (ETSI) GPRS usage is typically charged based on volume of data transferred. Usage above the bundle cap is either charged per megabyte or disallowed.

(b) EDGE(Enhanced Data rates for GSM Evolution):
It is three times fasterthan GPRS. It is used for voice communication as well as an internet connection.

(ii) Code Division Multiple Access (CDMA):
It is a channel access method used by various radio communication technologies. CDMA is an example of multiple access, which is where several transmitters can send information simultaneously over a single communication channel. This allows several users to share a band of frequencies To permit this to be achieved without undue interference between the users, and provide better security.

3. Third Generation networks(3G):
It allows high data transfer rate for mobile devices and offers high speed wireless broadband services combining voice and data. To enjoy this service 3G enabled mobile towers and hand sets required.

4. Fourth Generation networks(4G):
lt is also called Long Term Evolution(LTE) and also offers ultra broadband Internet facility such as high quality streaming video. It also offers good quality image and videos than TV.

Mobile communication services:
1. Short Message Service(SMS):
It allows transferring short text messages containing up to 160 characters between mobile phones. The sent message reaches a Short Message Service Center(SMSC), that allows ‘store and forward’ systems. It uses the protocol SS7(Signaling System No7). The first SMS message ‘Merry Christmas’ was sent on 03/12/1992 from a PC to a mobile phone on the Vodafone GSM network in UK.

2. Multimedia Messaging Service (MMS):
It allows sending Multi Media(text, picture, audio and video file) content using mobile phones. It is an extension of SMS.

3. Global Positioning System(GPS):
It is a space- based satellite navigation system that provides location and time information in all weather conditions, anywhere on or near the Earth where there is an unobstructed line of sight to four or more GPS satellites. The system provides critical capabilities to military, civil and commercial users around the world.

It is maintained by the United States government and is freely accessible to anyone with a GPS receiver. GPS was created and realized by the U.S. Department of Defense (DoD) and was originally run with 24 satellites. It is used for vehicle navigation, aircraft navigation, ship navigation, oil exploration, Fishing, etc. GPS receivers are now integrated with mobile phones.

Smart Cards:
A smart card is a plastic card with a computer chip or memory that stores and transacts data. A smart card (may be like your ATM card) reader used to store and transmit data. The advantages are it is secure, intelligent and convenient. The smart card technology is used in SIM for GSM phones. A SIM card is used as an identification proof.

Mobile operating system:
It is an OS used in hand held devices such as smart phone, tablet, etc. It manages the hardware, multimedia functions, Internet connectivity,etc. Popular OSs are Android from Google,iOS from Apple, BlackBerry OS from Black Berry and Windows Phone from Microsoft.

Android OS:
It is a Linux based OS forTouch screen devices such as smart phones and tablets.lt was developed by Android Inc. founded in Palo Alto, California in 2003 by Andy Rubin and his friends. In 2005, Google acquired this. A team led by Rubin developed a mobile device platform powered by the Linux Kernel.

The interface of Android OS is based on touch inputs like swiping, tapping, pinching in and out to manipulate on screen objects. In 2007 onwards this OS is used in many mobile phones and tablets. Android SDK(Software Development Kit) is available to create applications(apps) like Google Maps, FB, What’s App, etc.

It is of open source nature and many Apps are available for free download from the Android Play Store hence increase the popularity.
Different Android Versions are shown below

Students can Download Chapter 11 Computer Networks Notes, Plus One Computer Science Notes helps you to revise the complete Kerala State Syllabus and score more marks in your examinations.

Kerala Plus One Computer Science Notes Chapter 11 Computer Networks

Summary
Computer network:
Two or more computers connected through a communication media that allows exchange of information between computers is called a Computer Network. Eg: Internet

Need for network:
The advantages of Networks are given below.
1. Resource sharing:
All the computers in a network can share software (programs, data ) and hardware (printer, scanner, CD drive, etc.).

2. Reliability:
If one computer fails, the other computer can perform the work without any delay. This is very important for banking, air traffic control and other application.

3. Price Vs Performance:
A main frame computer can be 10 times faster than a PC but it costs thousand times a PC. Therefore instead of a main frame 10 personal computers are used with less cost and same performance.

4. Communication Medium:
It is a powerful communication medium. We can exchange information between computers in a network.

5. Scalable:
This means, System performance can be increased by adding computers to a network.

Terminologies:

1. Bandwidth: The maximum amount of data that can be transmitted by the medium measured in Hertz.
2. Noise: It is the unwanted electrical or electromagnetic interferences that adversely affect the transmitted data signals.
3. Node: A computer or an I/O device connected to a network is called Node.

Data communication system:
Communication is the exchange of information between two human beings. But data communication is the exchange of information between two computers(devices).

1. Message: It is the data/information to be transmitted from one computer to another
2. Sender: It is a computer or a device that sends data. It is also called.source or transmitter
3. Receiver: It is a computer ora device that receives data
4. Medium: It is the path through which message transmitted from the sender to the receiver. There are two types Guided and Un Guided media.
5. Protocol: The rules and conventions for transmitting data.

Communication Medium:
There are two types guided and unguided.
Guided Media:
1. Twisted Pair cable:
2 types unshielded twisted pair and shielded twisted pair. Two copper wires individually insulated and twisted around each other and put in a plastic cover.

2. Coaxial cable:
A sturdy copperwire is insulated by plastic, it is covered just like a mesh by a conductor which is enclosed in an protective plastic coating. It is expenssive, less flexible and more difficult to install. But it is more reliable and carry for higher data rates.

3. Optical fibre:
These are made of glass fibres that are enclosed in a plastic jacket. It uses light instead of electrical signals. The light sources are LED or ILD.

Unguided Media:

1. Radio waves: It transmits data at different frequencies ranging from 3 kHz. to 300 GHz.
2. Microwaves: Microwave signals can travel in straight line if there is any obstacle in its path, it can’t bend. So it uses tall towers instead of short one.
3. Infrared waves: These waves are used for transmitting data in short distance and its frequency range is 300 GHz to 400 GHz.

Wireless communication technologies using:
radio waves
1. Bluetooth:
This technology uses radio waves in the frequency range of 2.402 GHz to 2.480 GHz. And transmit data in short distance. Mobile phones, Laptops, tablets etc use Bluetooth technology to transmit data.

2. Wi Fi(Wireless Fidelity):
It uses radio waves to transmit information across a network in a range 2.4 GHz to 5 GHz in short distance. Nowadays this technology is used to access internet in Laptops, Desktops, tablets, Mobile phones etc.

3. Wi MAX(Wireless Microwave Access):
It uses micro waves to transmit information across a network in a range 2 GHz to 11 GHz over very long distance.

4. Satellites:
By using satellite we can communicate from eny part of the world to any other. The ground stations are connected via the satellite. The data signals transmitted from earth to satellite (uplink) and from the satellite to the earth (downlink).

Data communication devices:
It acts as an interface between computer and the communication channel

Network Interface Card (NIC):
This device enables a computer to connect to a network and transmit information.

Hub:
It is a small, simple and inexpensive device used to connect computers(devices) to a network. If a computer wants to transmit data to another computer. First it sends to the hub, the hub retransmits this data to all other computers.

Each and every computer gets the data and check whether it is for them or not. It increases the network traffic and hence the transmission speed is low.

Switch:
It is an expensive device used to connect computers(devices) to a network. Unlike hub, switch transmit data not to all computers, it retransmits data only to the intended computer. So the traffic is less and speed is high

Repeater:
It is a device used to strengthen weak signals on the network.

Bridge:
It is a device used to link same type of networks.

Router:
It is similar to a bridge, but it can connect two networks with different protocols.

Gateway:
It is used to connect two different networks with different protocols.

Data terminal equipments:
These devices are used to control data flow to and from a computer

Modem:
It is a device used to connect the computer to the internet. It converts digital signal into analog signal (modulation) and vice versa (Demodulation)

Multiplexer:
It combines the inputs from different channels of a medium and produces one output.

Network topologies:
Physical or logical arrangement of computers on a network is called structure or topology. It is the geometrical arrangement of computers in a network. The major topologies developed are star, bus, ring, tree and mesh.
1. Star Topology:
A star topology has a server all other computers are connected to it. If computer A wants to transmit a message to computer B. Then computer A first transmit the message to the server then the server retransmits the message to the computer B.

That means all the messages are transmitted through the server. Advantages are added or remove workstations to a star network is easy and the failure of a workstation will not effect the other. The disadvantage is that if the server fails the entire network will fail.

2. Bus Topology:
Here all the computers are attached to a single cable called bus. Here one computer transmits all other computers listen. Therefore it is called broadcast bus. The transmission from any station will travel in both the direction.

The connected computers can hear the message and check whether it is for them or not. Advantages are add or remove computer is very easy. It requires less cable length and the installation cost is less. Disadvantage is fault detection is very difficult because of no central computer.

3. Ring Topology:
Here all the computers are connected in the shape of a ring and it is a closed loop. Here also there is no central computer. Here a computer transmits a message, which is tagged along with its destination computer’s address.

The message travels in one direction and each node check whether the message is for them. If not, it passes to the next node. It requires only short cable length. If a single node fails, at least a portion of the network will fail. To add a node is very difficult.

4. Hybrid Topology:
It is a combination of any two or more network topologies. Tree topology and mesh topology can be considered as hybrid topology.
(a) Tree Topology:
The structure of a tree topology is the shape of an inverted tree with a central node and branches as nodes. It is a variation of bus topology. The data transmission takes place in the way as in bus topology. The disadvantage is that if one node fails, the entire portion will fail.

(b) Mesh Topology:
In this topology each node is connected to more than one node. It is just like a mesh (net). There are multiple paths between computers. If one path fails, we can transmit data through another path.

Types of networks:
The networks are classified into the following based upon the amount of geographical area that covers.
(i) Personal Area Network(PAN):
It is used to connect devices situated in a small radius by using guided media or unguided media

(ii) Local Area Network (LAN):
This is used to connect computers in a single room, rooms within a building or buildings of one location by using twisted pair wire or coaxial cable. Here the computers can share Hardware and software. Data transferrate is high and error rate is less, eg: The computers connected in a school lab.

(iii) Metropolitan Area Network (MAN):
A Metropolitan Area Network is a network spread over a city. For example a Cable TV network. MAN have lesser speed than LAN and the error rate is less. Here optical fiber cable is used.

(iv) Wide Area Network (WAN):
This is used to connect computers over a large geographical area. It is a network of networks. Here the computers are connected using telephone lines or Micro Wave station or Satellites. Internet is an example for this.

LAN and MAN are owned by a single organization but WAN is owned by multiple organization. The error rate in data transmission is high.

Logical classification of networks:
Peer to peer:
In this configuration all the computers have equal priority. That means each computer can function as both a client and a server. There is no dedicated server.

Client-Server:
In this configuration a computer is powerful which acts as a dedicated server and all others are clients (work stations). A Server fulfils the needs of the clients.

1. File Server: A computer that stores and manages files for other devices on a network
2. Web Server: A computer that handles the requests for web pages.
3. Print Server: A computer that handles the print jobs from other computers on a network.
4. Database Server: A computer that manages the database.

Network protocols:
A protocol is a collection of rules and regulations to transfer data from one location to another. Transmission Control Protocol (TCP), which uses a set of rules to exchange messages with other Internet points at the information packet level. Internet Protocol (IP), which uses a set of rules to send and receive messages at the Internet address level
1. FTP:
File Transfer Protocol which is used for transferring files between computers connected to local network or internet.

2. HTTP:
is a protocol used for WWW for enabling the web browse to access web server and request HTML documents.

3. DNS (Domain Name System):
When we type web sites address in the address bar, the browser determines the URL and asks the DNS for URLS corresponding IP address (Numeric address). The DNS returns the address to the browser.

Identification of computers over a network:
A computer gets a data packet on a network, it can identify the sender’s address easily. It is similar to our snails mail, each letter is stamped in sender’s post office as well as receiver’s post office.

Media Access Control(MAC) address:
It is a unique 12 digit hexadecimal number(IMEI for mobile phones, it is a 15 digit decimal number) assigned to each NIC by its manufacturer. This address is known as MAC address and its permanent. It is of the form. MM:MM:MM:SS:SS:SS.

The first MM:MM:MM contains the ID number of the adapter company and the second SS:SS:SS represents the serial number assigned to the adapter by the company.

Internet Protocol (IP) address:
An IP address has 4 parts numeric address. Each parts contains 8 bits. By using 8 bits we can represent a decimal number between 0 to 255(28 = 256 numbers). Each part is separated by dot. A total of 4 × 8 = 32 bits used. But nowadays 128 bits are used for IP address.

Uniform Resource Locator(URL):
Every resource on the internet has a unique URL. Mainly it has three parts
eg: http://www.hscap.kerala.gov.in /index.html.

• http: http means hypertext transfer protocol. It is a protocol used to transfer hypertext.
• www: World Wide Web. With an email address we can open our mail box from anywhere in the world.
• hscap.kerala: It is a unique name. It is the official website name of Single Window System
• gov: It is the top level domain. It means that it is a government organization’s website.
• in: It is the geographical top level domain. It represents the country, in is used for India.
• index.html: It represents the file name.

Top Level Domain Names:

• .com The site register for commercial purpose
• .edu The site register for educational purpose
• .gov The site register by Government agencies
• .mil The site register for military services
• .net The site register for network purpose
• .org The site register by organizations

Country Specific Domain Names:

• .in India
• .au Australia
• .ca Canada
• .ch China
• .jp Japan
• .us United States of America

Students can Download Chapter 3 Components of the Computer System Notes, Plus One Computer Science Notes helps you to revise the complete Kerala State Syllabus and score more marks in your examinations.

Kerala Plus One Computer Science Notes Chapter 3 Components of the Computer System

Summary
Hardware:
The tangible parts of a computer that we can touch and see are called hardware. Eg: Monitor, Keyboard, Mouse, CPU, Etc.

Processors:
It is the brain of the computer and consists of three components

• Arithmetic Logic Unit(ALU) – As the name implies it performs all calculations and comparison operations.
• Control Unit(CU) – It controls overall functions of a computer
• Registers – It stores the intermediate results temporarily.

A CPU is an Integrated Circuit(IC) package contains millions of transistors and other components. Popular Processors are Intel core i3, core i5, core i7, AMD Quadcore etc.

Important registers inside a CPU are:

1. Accumulator: After performing an operation (arithmetic or logical) the result is stored in the accumulator
2. Memory Address Register(MAR): It stores the address of memory location to which data is either read or written by the processor.
3. Memory Buffer Register (MBR): It stores the data, either to be written to or read from the memory by the processor.
4. Instruction Register(IR): It stores the instructions to be executed by the processor.
5. Program Counter(PC): It stores the address of the next instruction to be executed by the processor.

Motherboard:
It is a Printed Circuit Board(PCB). All the major components like Processor (Remember the processor must be compatible with the Motherboard), RAM, ROM, HDD, Graphics card, Sound card, etc are connected to the Motherboard.

Peripherals and Ports:

1. Serial Port: It transmits data one bit at a time(eg: 101000001010 ). Its transmission speed is low but it is cheaper. It is suitable to transmit data over long distance.
2. Parallel port: It can transmit data more than one bit at a time. It is faster and used to connect printer.
3. USB (Universal Serial Bus) port: It has high bandwidth hence it is faster. Nowadays it is used to connect all the devices like keyboard,mouse,printer, scanner, pen drive, digital camera, mobile phones, dongle etc.
4. LAN port: By using this port we can connect our computer to another network by a cable.
5. PS/2(Personal System/2) port: It is introduced by IBM for connecting keyboard and mouse earlier.
6. Audio ports: It is used to connect audio devices like speakers, mic etc.
7. Video Graphics Array(VGA) port: It is introduced by IBM to connect monitor or LCD projector to a computer.
8. High Definition Multimedia Interface(HDMI): Through this port we can connect high definition quality video and multi-channel audio over a single cable.

Memory:
Storage Unit(Memory Unit): A computer has huge storage capacity. It is used to store data and instructions before starts the processing. Secondly it stores the intermediate results and thirdly it stores information(processed data), that is the final results before send to the output unit(Visual Display Unit, Printer, etc). Memory measuring units are given below.

• 1 bit = 1 or 0(Binary Digit)
• 4 bits = 1 Nibble
• 8 bits = 1 Byte
• 1024 Bytes = 1 KB(Kilo Byte)
• 1024 KB = 1 MB(Mega Byte)
• 1024 MB = 1 GB(Giga Byte)
• 1024 GB = 1 TB(Tera Byte)
• 1024 TB = 1 PB(Peta Byte)

1. Primary Storage alias Main Memory:
It is further be classified into Two – Random Access Memory (FRAM) and Read Only Memory(ROM). The one and only memory that the CPU can directly access is the main memory at a very high speed. It is expensive hence storage capacity is less.

RAM is volatile(when the power is switched off the content will be erased) in nature but ROM is non volatile(lt is permanent). In ROM a “boot up” program called BIOS(Basic Input Output System) is stored to “boots up” the computer when it switched on. Some ROMs are given below.

• PROM(Programmable ROM): It is programmed at the time of manufacturing and cannot be erased.
• EPROM (Erasable PROM): It can be erased and can be reprogrammed using special electronic circuit.
• EEPROM (Electrically EPROM): It can be erased and rewritten electrically

Cache Memory:
The processor is a very high speed memory but comparatively RAM is slower than Processor. So there is a speed mismatch between the RAM and Processor, to resolve this a high speed memory is placed in between these two this memory is called cache memory. Commonly used cache memories are Level(L1) Cache(128 KB), L2(1 MB),L3(8 MB), L4(128MB).

2. Secondary Storage alias Auxiliary Memory:
Because of limited storage capacity of primary memory its need arises. When a user saves a file, it will be stored in this memory hence it is permanent in nature and its capacity is huge. Eg: Hard Disc Drive(HDD), Compact Disc(CD), DVD, Pen Drive, Blu Ray Disc etc.
(i) Magnetic storage device:
It uses plastic tape or metal/plastic discs coated with magnetic material. .
(a) Hard Disk:
Instead of flexible or soft disk it uses rigid material hence the name hard disk. Its storage capacity and data transfer rate are high and low access time. These are more lasting and less error prone. The accessing mechanism and storage media are combined together in a single unit and connect to the motherboard via cable.

(ii) Optical storage device:
(a) Optical Disk:
The high power laser uses a concentrated, narrow beam of light, which is focuses and directed with lenses, prisms and mirrors for recording data. This beams burns very very small spots in master disk, which is used for making molds and these molds are used for making copies on plastic disks.

A thin layer of aluminium followed by a transparent plastic layer is deposited on it. The holes made by the laser beam are called pits, interpreted as bit 0 and unburned areas are called lands interpreted as bit 1. Lower power laser beam is used to retrieve the data.

(b) DVD(Digital Versatile Disc):
It is similar to CD but its storage capacity is much higher. The capacity of a DVD starts from 4.7 GB

(c) Blu-ray Disc:
It is used to read and write High Definition video data as well as to store very huge amount of data. While Cd and DVD uses red laser to read and write but it uses Blue-Violet laser, hence the name Blu ray disc. The blue violet laser has shorter wavelength than a red laser so it can pack more data tightly.

(iii) Semiconductor storage (Flash memory):
It uses EEPROM chips. It is faster and long lasting.

• USB flash drive: It is also called thumb drive or pen drive. Its capacity varies from 2 GB to 32 GB.
• Flash memory cards: It is used in Camera, Mobile phones, tablets etc to store all types of data.

Input / Output devices:
It is used to supply: data to the computer for processing
1. Keyboard:
It is the most widely used device to input information in the form of words, numbers etc. There are 101 keys on a standard keyboard. The keys on the keyboard are often classified into alpha numeric keys (A to Z, 0 to 9), function keys (F1 to F12), special purpose keys (Special characters), cursor movement keys (arrow keys). While pressing a key, the corresponding code’s signal is transmitted to the computer.

2. Mouse:
It is a pointing device, that controls the movement of the cursor, or pointer as a display screen. A mouse has two or three buttons, it is often used in GUI oriented computers.

Under the mouse there is a ball, when the mouse moves o.n a flat surface this ball also moves. This mechanical motion is converted into digital values that represents x and y values of the mouse movement.

3. Light Pen:
It is an input device that use a lightsensitive detector to select objects directly on a display screen using a pen. Light pen has a photocell placed in a small tube. By using light pen, we can locate the exact position on the screen.

4. Touch screen:
It allows the user to enter data by simply touching on the display screen. This technology is applied in tablets, cell phones, computers etc.

5. Graphic tablet:
It consists of an electronic writing area. We can create graphical images by using a special pen.

6. Touchpad:
It is a pointing device found on the portable computers(lap top). Just like a mouse it consists of two buttons below the touch surface to do the operations like left click and right click. By using our fingers we can easily operate.

7. Joy Stick:
It is a device that lets the user move an object quickly on the screen. It has a liver that moves in all directions and controls the pointer or object. It is used for computer games and CAD/CAM systems.

8. Microphone:
By using this device we can convert voice signals into digital form.

9. Scanner:
It is used to read text or pictures printed on paper and translate the information into computer usable form. It is just like a photostat machine but it gives information to the computer.

10. Optical Mark Reader (OMR):
This device identifies the presence or absence of a pen or pencil mark. It is used to evaluate objective type exams. In this method special preprinted forms are designed with circles can be marked with dark pencil or ink.

A high intensity beam in the OMR converts this into computer usable form and detects the number and location ofthe pencil marks. By using this we can evaluate easily and reduce the errors.

11. Bar code/Quick Response (QR) code reader:
Light and dark bars are used to record item name, code and price is called Bar Code. This information can be read and input into a computer quickly without errors using Bar Code Readers.

It consists of a photo electric scanner and it is used in super market, jewellery, textiles etc. QR codes are similar to barcodes but it uses two dimensional instead of single dimensional used in Barcode.

12. Biometric sensor:
It is used to read unique human physical features like finger prints, retina, iris pattern, facial expressions etc. Most of you give these data to the Government for Aadhaar.

13. Smart card reader:
A plastic card(may be like your ATM card) stores and transmit data with the help of a reader.

14. Digital Camera:
By using digital camera, we can take photographs and store in a computer. Therefore we can reduce the use of film. Hence it is economical.

Output devices:
After the data processing the result is displayed as soft copy(soft copy can view, only by using a device) or hard copy(lt can read easily).
1. Visual Display Unit:
a. Cathode Ray Tube (CRT):
There are two types of CRT’s, monochrome (Black and white) and colour. Monochrome CRT consists of one electron gun but colour CRT consists of 3 electron guns (Red, Green and Blue) at one end and the other end coated with phosphor. It is a vacuum tube. The phosphor coated screen can glow when electron beams produced by electron guns hit.

It is possible to create all the colours using Red, Green and Blue. The images produced by this is refreshed at the rate of 50 or 60 times each second. Its disadvantage is it is heavy and bulky. It consumes more power and emits heat. But it is cheap. Nowadays its production is stopped by the company.

b. Liquid Crystal Display (LCD):
It consists of two electrically conducting plates filled with liquid crystal. The front plate has transparent electrodes and the back plate is a mirror.

By applying proper electrical signals across the plates, the liquid crystals either transmit or block the light and then reflecting it back from the mirror to the viewer and hence produce images. It is used in where small sized displays are required.

c. Light Emitting Diode(LED):
It uses LED behind the liquid crystals in order to light up the screen. It gives a better quality and clear image with wider viewing angle. Its power consumption is less.

d. Plasma Panels:
It consists of two glass plates filled with neon gas. Each plate has several parallel electrodes, right angles to each other. When low voltage is applied between two electrodes, one on each plate, a small portion of gas is glow and hence produce images.

Plasma displays provide high resolution but are expensive. It is used in, where quality and size is a matter of concern.

e. Organic Light Emitting Diode(OLED) Monitors:
It is made up of millions of tiny LEDs. OLED monitors are thinner and lighter than LCDs and LEDs. It consumes less power and produce better quality images but it is very expensive.

• LCD projector: It is used to display video, images or data from a computer on a large screen. Its main component is a high intensity light producing bulb and a lens.

2. Printer:
There are two types of printers impact and non impact printers. Printers are used to produce hard copy.
Impact Printers:
There is a mechanical contact between print head and the paper.
(i) Dot Matrix Printer:
Here characters are formed by using dots. The printing head contains a vertical array of pins. The letters are formed by using 6 dot rows and 7 dot columns. Such a pattern is called 5 × 7 matrix.

This head moves across the paper, the selected pins fire against an inked ribbon to form characters by dot. They are capable of faster printing, but their quality is not good.

Non-impact Printers:
There is no mechanical contact between print head and paper so carbon copies cannot be possible to take. They are inkjet, laser, thermal printers etc.
(a) Inkjet Printer:
It works in the same fashion as dot matrix printers, but the dots are formed with tiny droplets of ink to be fired from a bottle through a nozzle. These droplets are deflected by an electric field using horizontal and vertical deflection plates to form characters and images.

It is possible to generate colour output. They produce less noise and produce high quality printing output. The printing cost is higher. Here liquid ink is used.

(b)Laser Printer:
It uses photo copying technology. Here instead of liquid ink dry ink powder called toner is used. A drum coated vyith positively charged photo conductive material is scanned by a laser beam.

The positive charges that are illuminated by the beam are dissipated. The drum is then rolled through a reservoir of negatively charged toner which is picked up by the charged portions of the drum.

It adheres to the positive charges and hence creating a page image on the drum. Monochrome laser printer uses a single toner whereas the colour, laser printer uses four toners. Its print quality is good less noise and printing cost is higher.

(c) Thermal Printers:
It is same as dot matrix printer but it needs heat sensitive paper. It produces images by pushing electrically heated pins to the special paper. It does not make an impact on the paper so we cannot produce carbon copies. It produce less noise, low quality print and inexpensive. It is used in fax machine.

3. Plotter:
A plotter is a device that draws pictures or diagrams on paper based on commands from a computer. Plotters draw lines using.a pen. Pen plotters generally use drum or flat bed paper holders. In a drum plotter the paper is mounted on the surface of a drum.

Here the paper is rotated. But in a flat bed plotter the paper does not move and the pen holding mechanism provides the motion that draws pictures. Plotters are used in engineering applications where precision is needed.

4. Three Dimensional (3D) printer:
This device is used to print 3D objects.

5. Audio output devices:
Speakers are used to produce sound by the backward and forward movement of the diaphragm in the speaker according to the electrical signals from the computer through the audio port.

e-Waste(electronic waste):
It refers to the mal functioning electronic products such as faulty computers, mobile phones, tv sets, toys, CFL etc.

Why should we concern about e Waste:
Itcontains poisonous substances such as lead, mercury, cadmium etc and may cause diseases if not properly managed.

What happens to the e Waste:
A small amount is recycled. Due to this our natural resources are contaminated(poisoned). Some of them can recycle properly. But it is a very big problem in front of the Government to collect segregate, recycle and disposal of e-Waste.

e-Waste disposal methods:

1. Reuse: Reusability has an important role of e-Waste management and can reduce the volume of e-Waste
2. Incineration: It is the process of burning e Waste at high temperature in a chimney
3. Recycling of e-Waste: It is the process of making new products from this e-Waste.
4. Land filling: It is used to level pits and cover by thick layer of soil.

Students’ role in e-Waste disposal:

• Stop buying unnecessary electronic equipments
• Repair Faulty electronic equipments instead of buying a new one.
• Give electronic equipments to recycle
• Buy durable, efficient, quality, toxic free, good warranty products
• check the website or call the dealer if there is any exchange scheme
• Buy rechargeable battery products

Green computing or Green IT:
It is the study and practice of eco friendly computing or IT such as designing, manufacturing, using and disposal of. computers and components (monitors, printers, storage devices etc.)
Following are some steps to follow to reduce the adverse impact on the global environment

• Turn off computer and other devices when not in use
• Use power saver mode
• Use laptops instead of desktops
• Avoid print outs if not needed
• Use LCD s instead of CRT s to save power
• Use Energy Star rated H/W or S/w and Solar energy(Hybrid Energy)
• Dispose e Waste properly as per norms

Following are the steps to promote green computing:

1. Green design: Design energy efficient and eco friendly devices
2. Green manufacturing: reduce non eco friendly parts while manufacturing
3. Green use: Use energy saver devices
4. Green disposal: Use easily disposable devices

Software:
The set of instructions that tell the hardware how to perform a task is called software. Without software computer cannot do anything. Two types System s/w and Application s/w
A. System software: It is a collection of programs used to manage system resources and control its operations. It is further classified into two.

1. Operating System
2. Language Processor

1. Operating System:
It is collection of programs which acts as an interface between user and computer. Without an operating system computer cannot do anything. Its main function is make the computer usable and use hardware in an efficient manner, eg:- Windows XP, Windows Vista, Linux, Windows 7, etc.
Major functions of an operating System

1. Process management: It includes allocation and de allocation of processes(program in execution) as well as scheduling system resources in efficient manner
2. Memory management: It.takes care of allocation and de allocation of memory in efficient manner
3. File management: This includes organizing, naming, storing, retrieving, sharing , protecting and recovery of files.
4. Device management: Many devices are connected to a computer so it must be handled efficiently.

2. Language Processes:
We know that a program is a set of instructions. The instructions to the computer are written in different languages. They are high level language (HLL) and low level language. In HLL English like statements are used to write programs. They are C, C++, COBOL, PASCAL, VB, Java etc. HLL is very easy and can be easily understood by the human being.

Low level language are classified into Assembly Language and Machine Language. In assembly language mnemonics (codes) are used to write programs

In Machine Language 0’s and 1 ’s are used to write program. It is very difficult but this is the only language which is understood by the computer. Usually programmers prefer HLL to write programs because of its simplicity.

But computer understands only machine language. So there is a translation needed. The program which perform this job are language processors. The different language processors are given below:

1. Assembler: This converts programs written in assembly language into machine language.
2. Interpreter: This converts a HLL program into machine language by converting and executing it line by line. The first line is converted if there is no error it will be executed otherwise you have to correct it and the second line and so on.
3. Compiler: It is same as interpreter but there is a difference it translate HLL program into machine language by converting all the lines at a time. If there is no error then only it will executed.

B. Application Software:
Programs developed to serve a particular application is known as application software. eg: MS Office, Compression Utility, Tally etc. Application software can further be sub divided into three categories.

1. Packages
2. Utilities
3. Customized Software

1. Packages:
Application software that makes the computer useful, for people to do every task. Packages are used to do general purpose application.
They are given below:
(i) Word Processes:
This is used for creation and modification of text document. That means a word processor helps the people to create, edit and format a textual data with less effort and maximum efficiency.

By using word processor we can change font and font size of character, change alignment (left, right, center and justify), check spelling and grammar of the whole document etc. eg: MS Word.

(ii) Spread Sheets:
It contains data or information in rows and columns and can perform calculation (Arithmetic, Relational and logical operation). It helps to calculate results of a particular formula and the formula can apply different cells (A cell is the intersection of a row and column. Each column carries an alphabet for its name and row is numbered).

It is used to prepare budgets, balance sheets, P & L account, Pay roll etc. We can easily prepare graphs and charts using data entered in a worksheet. A file is a work book that contains one or more work sheets, eg: MS Excel is a spread sheet software.

(iii) Presentation and Graphics:
You can present your idea with sound and visual effects with the help of presentation software by preparing slides. The application software that manipulate visual images is known as graphics software. Eg: MS Power Point is a presentation package.

(iv) Data base package:
Data base is a collection of large volume of data. DBMS is a set of programs that manages the datas are for the centralized control of data such that creating new records to the database, deleting, records whenever not wanted from the database and modification of the existing database. Example for a DBMS is MS Access.

DTP Packages:
DTP means Desk Top Publishing. By using this we can create books, periodicals, magazines etc. easily and fastly. Now DTP packages are used to create in Malayalam also, eg: PageMaker.

2. Utilities:
Utilities are programs which are designed to assist computer for its smooth functioning. The utilities are given below.

1. Text editor: It is used for creating and editing text files.
2. Backup utility: Creating a copy of files in another location to protect them against loss, if your hard disk fails or you accidentally overwrite or delete data.
3. Compression Utility: It is used to reduce the size of a file by using a program and can be restored to its original form when needed.
4. Disk Defragmenter: It is used to speeds up disk access by rearranging the files that are stored in different locations as fragments to contiguous memory and free space is consolidated in one contiguous block.
5. Virus Scanner: It is a program called antivirus software scans the disk for viruses and removes them if any virus is found.

3. Specific purpose software (Customized software):
It is collection of programs which are developed to meet user needs to serve a particular application. It is also called tailor made software.

Free and open source software:
Here “free” means there is no copyright or licensing. That is we can take copies of the s/w or modify the source code without legal permission of its vendor (creator) we can use and distribute its copy to our friends without permission. That is Freedom to use to modify and redistribute.
The Four freedoms are:

1. Freedom 0: To run program for any purpose
2. Freedom 1: To study how it works and allows you to adapt according to your needs. Also allows to access source code.
3. Freedom 2: Allows to take copies and distribute
4. Freedom 3: Allows you to change source code and release the program .

Examples for Free and open source software are given below
(i) Linux:
it is a free s/w. It was written by Linux Trovalds at the University of Helsinki. It is a GUI and multi-user, multi-tasking O.S. with many more other features. It is also independent of the hardware. It is used by ISP’s, programmers who uses Java to write program, etc. The main Linux distributors are. Open Linux Red hat, Debian, Gnu Linux, etc.

(ii) GNU/Linux:
It was organized by Richard Stallman in 1983

(iii) GIMP (GNU Image Manipulation Program):
It is a very helpful software to perform all the activities to an image.

(iv) Mozilla Firefox:
This web browser helps the users to browse safely.

(v) OpenOffice.org:
This package contains different s/w s that helps the users to draft letters neatly by using “Writer”, perform calculations by using “Calc” and prepare presentations by using “Impress”. It is platform independent (That means it works on both Linux and Windows platforms.

Freeware:
A s/w with Copy right is available free of cost for unlimited use.
Shareware: It is an introductory pack distributed on a trial basis with limited functionality and period.

Students can Download Chapter 10 Mechanical Properties of Fluids Questions and Answers, Plus One Physics Chapter Wise Questions and Answers helps you to revise the complete Kerala State Syllabus and score more marks in your examinations.

Kerala Plus One Physics Chapter Wise Questions and Answers Chapter 10 Mechanical Properties of Fluids

Plus One Physics Mechanical Properties of Fluids One Mark Questions and Answers

Question 1.
Water is flowing through a very narrow tube. The velocity of water below which the flow remains a streamline flow is known as
(a) relative velocity
(b) terminal velocity
(c) critical velocity
(d) particle velocity
Answer:
(c) critical velocity
Critical velocity is that velocity of liquid flow, upto which the flow of liquid is a streamlined and above which its flow becomes turbulent.

Question 2.
Bernoulli’s equation for steady, non-viscous, imcompressible flow expresses the
(a) conservation of angular momentum
(b) conservation fo density
(c) conservation of momentum
(d) conservation of energy
Answer:
(d) conservation of energy

Question 3.
When cooking oil is heated in a frying pan, the oil moves around in the pan more easily when it is hot. The main reason for this is that with rise in temperature, there is a decrease in
(a) surface tension
(b) viscosity
(c) angle of contact
(d) density
Answer:
(d) density

Question 4.
At what temperature density of air is maximum?
Answer:
4°C

Question 5.
A thin glass plate is lying on a wet marble floor, It is difficult to pull the glass plate because of
(i) surface tension
(ii) Viscosity
(iii) friction
(iv) atmosphere
(v) Gravity
Answer:
(i) Viscosity

Question 6.
Why do clouds float in the sky?
Answer:
Zero terminal velocity

Question 7.
A spinning cricket ball does not follow parabolic path. Why?
Answer:
Due to the magnus effect.

Question 8.
The deep water runs slow. Why?
Answer:
According to equation of continuity velocity is inversly proportional to velocity (AV = constant). Hence deep water runs slow.

Question 9.
Why dust generally settles down in closed room?
Answer:
The dust particles (tiny spheres) acquire terminal velocity as it fall through air. The terminal velocity is directly proportional to square of radius. Hence terminal velocity of dust particle is very small. So they settle down gradually.

Question 10.
Why more viscous oil is used in summer than in winter in scooters?
Answer:
The viscosity decreases with increase in temperature.

Question 11.
Why is sand drier than clay?
Answer:
Capillary action

Question 12.
Why cotton dress is preferred in summer?
Answer:
Cotton dresses have fine pores which act as capillaries for the sweat.

Question 13.
Why oil is poured to calm the sea?
Answer:
When oil is poured in water, the surface tension of water is reduced and water spreads over large area of sea.

Question 14.
How plants draw water from ground?
Answer:
The capillary action.

Question 15.
How do insects run on the surface of water?
Answer:
Because of surface tension, the surface of water behaves like stretched membrane hence it can support weight of small insects.

Question 16.
How ploughing a field helps to retain moisture?
Answer:
When field is ploughed, capillaries are broken and hence water can not rise up and retains moisture.

Question 17.
Hot soup tastes better than cold soup. Why?
Answer:
The surface tension of hot soap is less compared to cold soap. So hot soap spreads larger area.

Plus One Physics Mechanical Properties of Fluids Two Mark Questions and Answers

Question 1.
Remya found that a piece of metal weighs 210 g in air and 180 g when it is immersed in water. Determine the density of the metal piece.
Answer:
Relative density,
R.D = \(\frac{\text { Weight in air }}{\text { Loss of Weight in water }}\)
= \(\frac{210}{30}\) = 7.

Question 2.
Why is hot soup tastier than cold one?
Answer:
When temperature increases, the surface tension of soap decrease. Hence hot soap can enter into tiny pours of taste buds.

Question 3.
Why straws are used to drink soft drinks?
Answer:
When we suck the straw, pressure inside straw becomes lower than atmospheric pressure. This pressure difference cause the soft drink to rise through the straw.

Question 4.
Why new earthen pots keeps water more cool than old earthen pots?
Answer:
The capillaries of old earthen pots will get blocked with passage of time. For new earthen pots, water oozes out through capillaries, gets evaporated at the surface and makes it cool.

Plus One Physics Mechanical Properties of Fluids Three Mark Questions and Answers

Question 1.
Air is blown in between two pith balls suspended freely.

1. What will happen to the balls?
   • They repel each other
   • They attract each other
   • They start oscillating
   • They remain in their initial position They fall on the ground
2. Give your explanation

Answer:

1. They attract each other
2. When air is blown in between two pith balls, the pressure between the balls decreases. Due to this decrease in pressure between the balls, they attract each other.

Question 2.
A child dipped two identical capillary tubes, one in a beaker containing mercury. He observed that water and mercury have risen through the tubes to a certain heights.

1. Name this phenomenon.
2. What difference did he observe in the shape of the meniscus of the two liquids in the tubes?
3. If he plots a graph connecting the radius of the capillary tube and capillary height, what will be the shape of the graph?

Answer:
1. Capillary rise.

2. The shape of the water meniscus in the tube becomes concave upwards. But the shape of mercury measures in the tube become convex upward.

3. h α \(\frac{1}{r}\), this is in the form y α \(\frac{1}{x}\)
Hence when we draw graph between ‘h’ and ‘r’ we get a graph of hyperbola.

Question 3.
Bernoulli’s theorem is a consequence of energy conservation principle. Using this theorem explain the working of atomiser.
Answer:
Atomiser (application of Bernoulli’s theorem)

Atomizer is used for getting a fine spray of perfumer or insecticide. It consists of a cylinder with a piston. A small vessel containing liquid to be sprayed is attached to the cylinder. When the piston is moved forward air is blown out through a small opening of the cylinder.

As the velocity of flow of air increases, the pressure at the opening decreases. Due to the lower pressure at the opening, the liquid rises through the narrow tube and gets sprayed out along with air.

Question 4.
Surface tension is numerically equal to the surface energy.

1. Difine surface tension.
2. Derive an expression for the rise of liquid in a capillary tube.

Answer:
1. Liquids acquire a free surface when poured in a container. These surfaces possess some additional energy. This phenomenon is known as surface tension.

2. When a drop is split into tiny droplets, the surface area increases. So work has to be done for splitting the drop. Let R be radius of the drop and r the radius of the droplets: R = 1 × 10^-3m
surface area of the drop = 4πR²
= 4π × (1 × 10^-3)²
= 4π × 10^-6m²
Volume of the drop = Volume of 10⁶ droplets

∴ Surface area of million droplets =10⁶ × 4π²
= 10⁶ × 4π(1 × 10^-5)²
= 4π × 10^-4 m²
∴ Increase in surface area = 4π × 10^-4 – 4π × 10^-6
= 3.96π × 10^-4m²
∴ Energy expended = 3.967π × 10^-4 × S
= 3.96π × 10^-4 × 72 × 10^-3J
= 8.95 × 10^-5J.

Question 5.
Match the following

1. Pascal’s law	a. \( \sqrt{2 g h}\)
2. Bernoulli’s theorem	b. a1v1= a2v2
3. Surface tension	c. Hydraulic jack
4. Velocity of efflux	d. Reynolds number
5. Equation of continuity	e. Angle of contact
6.  Viscosity	f. Ventiurimeter

Answer:
1 – C, 2 – f, 3 – e, 4 – a, 5 – b, 6 – d.

Question 6.
Give reasons for the following cases.

1. It is easier to swim in sea water than in river walls.
2. The passangers are advised to remove ink from pen while going up in an aeroplane.

Answer:
1. The density of sea water is more than that of river water due to the presence of salt. Hence sea water offers more upthrust and only a very small portion of human body will be in sea water compared to river water.

2. In ink pen, ink is filled in atmospheric pressure. As we go higher pressure decreases and hence ink will have a tendancy to come out in order to equalise the pressure.

Plus One Physics Mechanical Properties of Fluids Four Mark Questions and Answers

Question 1.
A large tank containing water has a small hole near the bottom of the tank 1.5 m below the surface of water.

1. What is the velocity of the water flowing from the hole?
2. Explain the principle used in deriving the velocity of water flowing from the hole.
3. Where must a second hole to be drilled so that the velocity of water leaving this hole is half of water flowing through the first hole.

Answer:
1. Velocity of water flowing through the hole
u = \( \sqrt{2 g h}\)
= \(\sqrt{2 \times 10 \times 1.5}\) = 5.47m/s.

2. Bernoulli’s theorem
As we move along a streamline the sum of the pressure (p), the kinetic energy per unit volume \(\frac{\rho v^{2}}{2}\) and the potential energy per unit volume (ρgh) remains a constant.

(OR)

Mathematically Bernoulli’s theorem can be written as
P + \(\frac{1}{2}\)ρv² + ρgh = constant.

3.

h₂ = \(\frac{1.5}{4}\) m = 0.375 m, from the top side of tank.

Question 2.
Rain drops have an average size of 1 mm when it is formed at the upper atmosphere.

1. Why the velocity of the rain drop is uniform?
2. Derive an expression for the terminal velocity of the drop in terms of coefficient of viscosity of air.
3. If the size of the rain drop become half, then what happens to its terminal speed?

Answer:
1. Due to viscous force acting on the raindrop, it moves with uniform speed.

2. Viscous force, boyancy force and weight of the body
Expression for terminal velocity:
Consider a sphere of radius ‘a’ densitity σ falling through a liquid of density a and viscocity η. The viscous force acting on the sphere can be written as
F = 6πaηv
Where v is the velocity of sphere. This force is acting in upward direction. When the viscous force is equal to the weight of the body in the medium, the net force on the body is zero. It moves with a constant velocity called the terminal velocity.
The weight of a body in a medium,

When body has terminal velocity, we can write.

Question 3.

1. Fill in the blanks using the word from the list appended with each statement.
   • Viscosity of gases_____with temperature (increase/decrease)
   • For a fluid in steady flow, the increase in flow speed at a constriction follows from_____ (conservation of mass/Bernoulli’s theorem)
   • The working of a hydraulic lift is based on (Pascal’s Law/ principle of Conservation of Energy)
   • Small insects can walk over the surface of water. It is due to the_____(surface tension of water/viscosity of water)
2. A girl dips a thin capillary tube in water. Water rises through it.
   • Name the phenomenon.
   • How does this rise vary with the diameter of the tube?

Answer:
1. Fill in the blanks :

• increases
• Conservation of mass
• Pascals law
• Surface tension

2. A girl dips a thin capillary tube in water:

• Capillary rise
• h α \(\frac{1}{r}\) ie. when diameter of tube increases, the. capillary rise decreases.

Question 4.
The schematic diagram of a sprayer or atomiser is given below.

1. Name the principle of working of this device from the following:
   • Surface tension
   • Viscosity
   • Bernoulli’s principle
   • Archimedes’ principle
2. Write its mathematical expression.
3. Wings of an aeroplane are curved outwards while flattened inwards. Why?

Answer:
1. Bernollis principle

2. P + \(\frac{1}{2}\)ρv² + ρgh = constant.

3. When the aeroplane moves forward, the air blown in the form of stream lines over the wings of aeroplane is shown figure.

As the upper surface of wing is more curved than its lower surface, the speed of air above the wings is larger than the speed of the air below the wings.

Hence the pressure above the wings becomes less than the pressure below the wings. Due to this pressure difference the aeroplane will get upward force to overcome gravitational force.

Question 5.
During windstorms, roofs of certain houses are blown off without damaging other parts of the houses.

1. Name the theorem which explains this phenomenon.
2. State the theorem.
3. Explain this phenomenon on the basis of this theorem.

Answer:

1. Bernoulli’s theorem
2. For a small amount of liquid in stream line flow, between two points, the total energy is constant.
3. When windstorm blown off, the pressure on the top side of roof decreases. Hence a pressure difference is developed in between roof. Due to this pressure difference, roof of certain houses are blown off.

Question 6.
Two thin evacuated (one end closed) glass take A and B are carefully immersed in a beaker containing mercury such a way that there is no chance to get air in to the tubes. A is stand vertically and B is making an angle θ with the vertical.

1. Is any rise of mercury in the tubes?
2. Is any height difference of mercury levels in tube A and B? Justify your answer.
3. When the doctors are measuring body pressure, it is advisable to lie on a table. Why?

Answer:

1. Yes
2. No. Pressure is same at same level. To get same pressure, height of mercury becomes same.
3. When we lie on the table, the pressure of our body will be same at all points.

Question 7.
A small metal sphere is falling through a caster oil.

1. Name the forces acting on the metal sphere?
2. Which of these forces change? Why?
3. Name the velocity of the sphere when the unbalanced force on it is zero?
4. Write down the expression for this velocity in terms of coefficient of viscosity?

Answer:

1. forces acting on the metal sphere:
   • Weight of the body (mg)
   • Buoyant force or up thrust
   • Viscous force
2. Viscous force. Viscous force is the friction offered by the liquid. It is a self adjusting force.
3. Terminal velocity
4. Terminal velocity, V = \(\frac{2}{9} a^{2}\left(\frac{f-N}{\eta}\right) g\).

Plus One Physics Mechanical Properties of Fluids Five Mark Questions and Answers

Question 1.
A capillary tube when dipped into water, it is commonly observed that water will rise through the tube.

1. Which of the following is responsible for this?
   • Gravitational force
   • Viscous force
   • Nuclear force
   • Surface tension
   • Elastic force
2. Derive an expression for the capillary rise.
3. If the radius of the tube becomes doubled, then what happens to the height of water column in the tube?

Answer:
1. Surface tension.

2.

Consider a capillary tube of radius ‘a’ dipped in a liquid of density ρ and surface tension S. If the liquid has a concave meniscus it will rise in the capillary tube. Let h be the rise of the liquid in the tube. Let p₁ be the pressure on the concave side of the meniscus and p₀, that on the other side. The excess pressure on the concave side of the meniscus can be written as
p₁ – p₀ = \(\frac{2 \mathrm{S}}{\mathrm{R}}\)
Where R is the radius of the concave meniscus. The tangent to the meniscus at the point A makes an angle θ with the wall of the tube.
In the right angled triangle ACO

substituting the values of R in the equation (1)

Considering two points M and N in the same horizontal level of a liquid at rest,
pressure at N = pressure at M
But pressure at M = p_i, the pressure over the concave meniscus and pressure at N = p_o + hρg
∴ P_i = P_o + hρg
or P_i – P_o = hρg ……..(3)
From equations (2) and (3), we get

3. We know

The capillary rise decreases to half of the original value.

Question 2.

1. Find the odd one out and justify your answer Atomiser, venturi meter, aeroplane, hydraulic lift
2. Mention one use of venturi meter.
3. Explain the working of the odd one which you have selected in question (a)

Answer:
1. Hydraulic lift – It is based on pascals law.

2. Venturimeter can be used to find velocity of flow of fluid through a pipe.

3.

A hydraulic lift is used to lift heavy load. Consider a liquid enclosed in a vessel with two cylinders C₁ and C₂ attached as shown in the figure. The cylinders are provided with two pistons having areas A₁ and A₂ respectively.
If F₁ is the force exerted on the area A₁,
pressure P₁ = \(\frac{F_{1}}{A_{1}}\).
If F₂ is the force exerted on the area A₂,
pressure P₂ = \(\frac{F_{2}}{A_{2}}\).
According to pascal’s law P₁ = P₂.

Using this method we can lift heavy load by applying small force.

Question 3.
When a capillary tube of radius ‘r’ is dipped in water, the water rises through it up to height ‘h’.

1. Which of the following is responsible for the above phenomenon?
   • Viscous force
   • elastic force
   • surface tension
   • gravitational force
   • negative force
2. To what height will water rise in a glass tube with a bore of radius 0.1 mm (take the angle of contact of glass with 0°, surface tension S = 0.0728 N/m)
3. If the length of tube is less that the length of capillary rise, will it overflow. Justify your answer.

Answer:
1. Surface tension

2.

3. The water will never overflow. If the tube is of insufficient length, the radius of curvature of liquid meniscus goes on increasing, making it more and more flat till water is in equilibrium.

Question 4.
A steel ball of radius 1 mm is falling vertically through a tank of oil at 30°C.

1. After some time the ball attains a constant velocity called_____
2. What are the forces acting on the ball and give their directions?
3. Write down the expression for resultant force acting on the ball?)
4. If the density of oil is 2 × 10³kg/m³, density of steel is 8 × 10² Kg/m³ and ‘η’of oil 2NS/m², What will be the constant velocity attained by the ball?

Answer:
1. Terminal velocity.

2. Weight of body (down ward), bouyanant force (up ward), Viscous force (upward).

3. Resultant force = weight of body – buoyant force.

4. Terminal velocity,

Plus One Physics Mechanical Properties of Fluids NCERT Questions and Answers

Question 1.
A 50 kg girl wearing high heel shoes balances on a single heel. The heel is circular with a diameter 1.0 cm. What is the pressure exerted by the heel on the horizontal floor?
Answer:
Force, F = Weight of girl
= mg = 50 × 9.8N = 490N
Radius, r = 0.5 × 10^-2m
Area A = πr² = \(\frac{22}{7}\)(0.5 × 10^-2)² m²
Pressure

Question 2.
Torricelli’s barometer used mercury. Pascal duplicated it using French wine of density 984 kg m^-3. Determine the height of the wine column for normal atmospheric pressure.
Answer:
p = hρg, h = \(\frac{p}{\rho g}=\frac{1.01 \times 10^{5}}{984 \times 9.8}\)m = 10.47m.

Question 3.
A U-tube contains water and methylated spirit separated by mercury. The mercury columns in the two arms are in level with 10.0 cm of water in one arm and 12.5 of spirit in the other. What is the specific gravity of spirit?

Answer:
Since the mercury columns in the two arms are at the same level,
∴ pressure due to water column = pressure due to spirit column
∴ h_wρ_wg = h_sρ_sg
or h_wρ_w = h_sρ
But h_w = 10 cm,
ρ_w = 1 gcm^-3,
h_s = 12.5cm
∴ 10 × 1 = 12.5 × ρ_s
or ρ_s = \(\frac{10}{12.5}\)gcm^-3
= 0.8cm^-3
∴ Specific gravity of spirit = 0.8.

Question 4.
Figs, (a) and (b) refer to the steady flow of a non-viscous liquid. Which of the two figures is incorrect? Why?

Answer:
Fig (a) is incorrect. This is because at a constriction (ie., where the area of cross-section of the tube is smaller), the flow speed is larger due to mass conservation. Consequently, pressure there is smaller according to Bernoulli’s equation. We assume the fluid to be incompressible.

Question 5.
What is the prssure inside the drop of mercury of radius 3.00 mm at room temperature? Surface tension of mercury at that temperature (20°C) is 4.65 × 10^-1Nm^-1. The atmospheric pressure is 1.01 × 10⁵ Pa. Also give the excess pressure inside the drop.
Answer:
Excess pressure = \(\frac{2 \sigma}{R}\)

Total pressure = 1.01 × 10⁵ + \(\frac{2 \sigma}{R}\)
= 1.01 × 10⁵ + 310
= 1.0131 × 10⁵Pa
Since data is correct upto three significant figures. We should write total pressure inside the drop as 1.1 × 10⁵Pa.

Question 6.
During blood transfusion, the needle is inserted in a vein where the guage pressure is 2000 Pa, at what height must the blood container be placed so that blood may just enter the vein? Given: density of whole blood = 1.06 × 10³kgm^-3
Answer:
Guage pressure,
p = hρg, h

= 0.19m.

Students can Download Chapter 1 Physical World Notes, Plus One Physics Notes helps you to revise the complete Kerala State Syllabus and score more marks in your examinations.

Kerala Plus One Physics Notes Chapter 1 Physical World

Summary
What is Physics?

a. Science And Scientific Method
Science is exploring, experimenting and predicting from what we see around us. ie. It is basically an systematic attempt to understand natural phenomena.

b. Physics
Two approaches are used in Physics: unification and reduction. In unification diverse physical phenomena are explained in terms of a few concepts or laws. For example all electric and magnetic phenomena can be explained by laws of electromagnetism (Maxwell’s equations).

In reduction, we derive properties of complex (bigger) system from properties and interactions of constituent parts. For example, the temperature of system is related to average kinetic energy of molecule of system.

Scope And Excitement Of Physics
The different subdisciplines of physics belongs to two domains: microscope domain and macroscopic domain. The macroscopic domain includes phenomena at laboratory, terrestrial and astronomic scales.

The microscopic domain of physics deals with constitution and structure of matter and their interaction with elementary particles like electron, proton, photon etc.

Physics covers a wide range of magnitude of physical quantities like length, time, mass, energy, etc. Physics includes phenomena involving elementary particles like electron, proton etc. whose range is 10^-14m.

It also deals with astronomical phenomena at the scale of even the entire universe (10⁺²⁶m). The range of time extends from 10^-22 s to 10¹⁸s. The range of mass goes from 10^-30 kg (mass of electron) to 10⁵⁵kg (mass of entire universe).

Physics, Technology And Society
The relation between Physics, technology and society can be seen in many examples. The steam engine has an important role in the Industrial Revolution in England in eighteenth century. The discovery of basic laws of electricity and magnetism contributed wireless communication technology.

Fundamental Forces Of Nature
There occur four fundamental forces in nature. They are gravitational force, electromagnetic force, strong nuclear force and weak nuclear force.

1. Gravitational Force:
It is a universal force. Gravitational force is the attractive force existing between any two bodies by virtue of its mass.

2. Electromagnetic Force:
The electromagnetic force exist between charged bodies. The electrostatic force of attraction or repulsion exist between charges at rest. A moving charge has magnetic effect in addition to electric effect. The electric and magnetic effects are inseparable and hence force experienced by charge is called electromagnetic force.

3. Strong Nuclear Force:
The strong nuclear force binds the nucleons (protons and neutrons) inside the nucleus. It is the strongest of all fundamental forces. The range of nuclear force is 1o^-15m (fermi) and it is charge independent.

4. Weak Nuclear Force:
The range of weak nuclear force is 10^-16m. This force exists only in few nuclear, reactions like b-decay.

5. Towards Unification of Forces:
Isac Newton unified terrestrial and celestial domain by applying law of gravitation in two domains. Oersted and Faraday showed that electric and magnetic phenomena are inseparable. Maxwell unified electromagnetism and optics by showing light is an electromagnetic wave.

1.5 Nature Of Physical Laws
The physical quantity that remains unchanged in process is called conserved quantity. Some of the conservation laws in nature are laws of conservation of mass, energy, linear momentum, angular momentum, charge etc.

Conservation laws have a deep connection with symmetries of nature. The symmetry of nature w.r.t. translation in time is equivalent to conservation of energy. Similarly the symmetry of nature w.r.t. translation in space is equivalent to conservation of linear momentum.

Symmetries of space and time and other types of symmetry play an important role n modern theories of fundamental forces in nature.

Students can Download Chapter 1 The Discipline of Computing Notes, Plus One Computer Science Notes helps you to revise the complete Kerala State Syllabus and score more marks in your examinations.

Kerala Plus One Computer Science Notes Chapter 1 The Discipline of Computing

Summary
Computing milestones and machine evolution:
People used pebbles and stones for counting earlier days. They draw lines to record information. eg: 1 line for one, 2 lines for two, 3 lines for three, etc. In this number system the value will not change if the lines are interchange. This type of number system is called non positional number system.

Counting and the evolution of the positional number system:
In positional number system, each and every number has a weight. Earlier sticks are used to count items such as animals or objects. Around 3000 BC the Egyptians use number systems with radix 10(base-the number of symbols or digits used in the number system) and they write from right to left.

Later Sumerian/Babylonian use number system with largest base 60 and were written from left to right. They use space for zero instead of a symbol, 0. In 2500 BC, the Chinese use simple and efficient number system with base 10 very close to number system used in nowadays.

In 500 BC, the Greek number system known as Ionian, it is a decimal number system and used no symbols for zero. The Roman numerals consists of 7 letters such as l, V, X, L, C, D, M. The Mayans used number system with base 20 because of the sum of the number of fingers and toes is 10 + 10 = 20.

It is called vigesimal positional number system. The numerals are made up of three symbols; zero (shell shape, with the plastron uppermost), one (a dot) and five (a bar or a horizontal line). To represent 1 they used one dot, two dots for 2, and so on

The Hindu – Arabic number system had a symbol(0)for zero originated in India 1500 years ago. Consider the table to compare the number system

Roman Numerals	Decimal / Hindu – Arabic number
I	1
V	5
X	10
L	50
C	100
D	500
M	1000

Evolution of the computing machine:
(a) Abacus:
In 3000 BC Mesopotamians introduced this and it means calculating board or frame. It is considered as the first computer for basic arithmetical calculations and consists of beads on movable rods divided into two parts. The Chinese improved the Abacus with seven beads on each wire. Different Abacus are given below.

(b) Napier’s bones:
A Mathematician John Napier introduced this in AD 1617.

(c) Pascaline:
A French mathematician Blaise Pascal developed this machine that can perform arithmetical operations.

(d) Leibniz’s calculator:
In 1673, a German mathematician and Philosopher Gottfried Wilhelm Von Leibniz introduced this calculating machine.

(e) Jacquard’s loom:
In 1801, Joseph Marie Jacquard invented a mechanical loom that simplifies the process of manufacturing textiles with complex pattern. A stored program in punched cards was used to control the machine with the help of human labour. This punched card concept was adopted by Charles Babbage to control his Analytical engine and later by Hollerith.

(f) Difference engine:
The intervention of human beings was eliminated by Charles Babbage in calculations by using Difference engine in 1822. It could perform arithmetic operations and print results automatically

(g) Analytical engine:
In 1833. Charles Babbage introduced this. Charles Babbage is considered as the “Father of computer It is considered as the predecessor of today’s computer. This engine was controlled by programs stored in punched cards. These programs were written by Babbage’s assistant, Augusta Ada King, who was considered as the first programmer in the World.

(h) Hollerith’s machine:
In 1887, Herman Hollerith an American made first electromechanical punched cards with instructions for input and output. The.card contained holes in a particular pattern with special meaning. The Us Census Bureau had large amount of data to tabulate, that will take nearly 10 years.

By this machine this work was completed in one year. In 1896, Hollerith started a company Tabulating Machine Corporation. Now it is called International Business Machines(IBM).

(i) Mark-1:
In 1944 Howard Aiken manufactured automatic electromechanical computer in collaboration with engineers at IBM that handled 23 decimal place numbers and can perform addition, subtraction, multiplication and subtraction.

Generations of computers:
There are five generations of computers from 16^th century to till date.

First generation computers (1940 – 1956):
Vacuum tubes were used in first generation computers. The input was based on punched cards and paper tapes and output was displayed on printouts. The Electronic Numerical Integrator and Calculator(ENIAC) belongs to first generation was the first general purpose programmable electronic computer built by J. Presper Eckert and John V. Mauchly.

It was 30-50 feet long, weight 30 tons, 18,000 vacuum tubes, 70,000 registers, 10,000 capacitors and required 1,50,000 watts of electricity. It requires Air Conditioner. They later developed the first commercially successful computer, the Universal Automatic Computer(UNIVAC) in 1952. Von Neumann architecture

The mathematician John Von Neumann designed a computer structure that structure is in use nowadays. Von Neumann structure consists of a central processing unit(CPU), Memory unit, Input and Output unit. The CPU consists of arithmetic logical unit(ALU) and control unit(CU).

The instructions are stored in the memory and follows the “Stored Program Concept”. Colossus is the secret code breaking computer developed by a British engineer Tommy Flowers in 1943 to decode German messages.

Second generation computers (1956 -1963):
Transistors, instead of Vacuum tubes, were used in 2^nd generation computers hence size became smaller, less expensive, less electricity consumption and heat emission and more powerful and faster.

A team contained John Bardeen, Walter Brattain and William Shockley developed this computer at Bell Laboratories. In this generation onwards the concept of programming language was developed and used magnetic core (primary) memory and magnetic disk(secondary) memory.

These computers used high level languages(high level language means English like statements are used)like FORTRAN (Formula translation) and COBOL(Common Business Oriented Language). The popular computers were IBM 1401 and 1620.

Third generation computers (1964 – 1971):
Integrated Circuits(IC’s) were used. IC’s or silicon chips were developed by Jack Kilby, an engineer in Texas Instruments. It reduced the size again and increased the speed and efficiency. The high level language BASIC(Beginners All purpose Symbolic Instruction Code) was developed during this period.

The popular computers were IBM 360 and 370. Due to its simplicity and cheapness more people were used. The number of transistors on IC’s doubles approximately every two years. This law is called Moore’s Law, it is named after Gordon E Moore. It is an observation and not a physical or natural law.

Fourth generation computers (1971 onwards):
Microprocessors are used hence computers are called microcomputers. Microprocessor is a single chip which contains Large Scale of IC’s(LSI) like transistors, capacitors, resistors,etc due to this a CPU can place on a single chip. Later LSI were replaced by Very Large Scale Integrated Circuits(VLSI). The popular computers are IBM PC and Apple II.

Fifth generation computers (future):
Fifth generation computers are based on Artificial Intelligence(AI). Al is the ability to act as human intelligence like speech recognition, face recognition, robotic vision and movement etc. The most common Al programming language are LISP and Prolog.

Evolution of computing:
Computing machines are used for processing or calculating data, storing and displaying information. In 1940’s computer were used only for single tasks like a calculator. But nowadays computer is capable of doing multiple tasks at a time.

The “Stored Program Concept” is the revolutionary innovation by John Von Neumann helped storing data and information in memory. A program is a collection of instructions for executing a specific job or task.

Augusta Ada Lowelace: She was the Countess of Lowelace and she was also a mathematician and writer. She is considered as the first lady computer programmer.

Programming languages:
The instructions to the computer are written in different languages. They are Low Level Language(Machine language), Assembly Language(Middle level language) and High Level Language(HLL).

In Machine Language 0’s and 1 ’s are used to write program. It is very difficult but this is the only language which is understood by the computer. In assembly language mnemonics (codes) are used to write programs

Electronic Delay Storage Automatic Calculator(EDSAC) built during 1949 was the first to use assembly language. In HLL English like statements are used to write programs. A-0 programming language developed by Dr. Grace Hopper, in 1952, for UNIVAC-I is the first HLL.

A team lead by John Backus developed FORTRAN @IBM for IBM 704 computer and ‘Lisp’ developed by Tim Hart and Mike Levin at Massachusetts Institute of Technology. The other HLLs are C, C++, COBOL, PASCAL, VB, Java etc. HLL is very easy and can be easily understood by the human being.

Usually programmers prefer HLL to write programs because of its simplicity. But computer understands only machine language. So there is a translation needed. The program which perform this job are language processors.

Algorithm and computer programs:
The step-by-step procedure to solve a problem is known as algorithm. It comes from the name of a famous Arab mathematician Abu Jafer Mohammed Ibn Musaa Al-Khowarizmi, The last part of his name Al-Khowarizmi was corrected to algorithm.

Theory of computing:
It deals with how efficiently problems can be solved by algorithm and computation. The study of the effectiveness of computation is based upon a mathematical abstraction of computers is called a model of computation, the most commonly used model is Turing Machine named after the famous computer scientist Alan Turing.

1. Contribution of Alan Turing:
He was a British mathematician, logician, cryptographer and computer scientist. He introduced the concept of algorithm and computing with the help of his invention Turing Machine.

He asked the question Can machines think’ led the foundation for the studies related to the computing machinery and intelligence. Because of these contributions he is considered as the Father of Modern Computer Science as well as Artificial Intelligence.

2. Turing Machine:
In 1936 Alan Turing introduced a machine, called Turing Machine. A Turing machine is a hypothetical device that manipulates symbols on a strip of tape according to a table of rules. This tape acts like the memory in a computer. The tape contains cells which starts with blank and may contain 0 or 1.

So it is called a 3 Symbol Turing Machine. The machine can read and write, one cell at a time, using a tape head and move the tape left or right by one cell so that the machine can read and edit the symbol in the neighbouring cells. The action of a Turing machine is determined by

1. the current state of the machine
2. the symbol in the cell currently being scanned by the head and
3. a table of transition rules, which acts as the program.

3. Turing Test:
The Turing test is a test of a machine’s ability to exhibit intelligent behaviour equivalent to, or indistinguishable from, that of a human. The test involves a human judge engages in natural language conversations with a human and a machine designed to generate performance indistinguishable from that of a human being.

All participants are separated from one another. If the judge cannot reliably tell the machine from the human, the machine is said to have passed the test. The test does not check the ability to give the correct answer to questions; it checks how closely the answer resembles typical human answers. Turing predicted that by 2000 computer would pass the test.

Students can Download Chapter 13 Kinetic Theory Questions and Answers, Plus One Physics Chapter Wise Questions and Answers helps you to revise the complete Kerala State Syllabus and score more marks in your examinations.

Kerala Plus One Physics Chapter Wise Questions and Answers Chapter 13 Kinetic Theory

Plus One Physics Kinetic Theory One Mark Questions and Answers

Question 1.
The value of \(\frac{P V}{T}\) for one mole of an ideal gas is nearly equal to
(a) 2 Jmol^-1K^-1
(b) 8.3 Jmol^-1K^-1
(c) 4.2 Jmol^-1K^-1
(d) 2 cal mol^-1K^-1
Answer:
(d) 2 cal mol^-1K^-1
The value of \(\frac{P V}{T}\) for one mole of an ideal gas = gas constant = 2 cal mol^-1K^-1.

Question 2.
Mean free path of a gas molecule is
(a) inversely proportional to number of molecules per unit volume
(b) inversely proportional to diameter of the molecule
(c) directly proportional to the square root of the absolute temperature
(d) directly proportional to the molecular mas
Answer:
(a) inversely proportional to number of molecules per unit volume.

Question 3.
If for a gas \(\frac{R}{C_{v}}\) = 0.67, this gas is made up of molecules which are
(a) monoatomic
(b) diatomic
(c) Polyatomic
(d) mixture of diatomic and polyatomic molecules
Answer:
(a) monoatomic
For a gas, we know \(\frac{R}{C_{v}}\) = γ – 1
or, 0.67 = γ – 1, or γ = 1.67
Hence the gas is monoatomic.

Question 4.
According to kinetic theory of gases, molecules of a gas behave like
(a) inelastic spheres
(b) perfectly elastic rigid spheres
(c) perfectly elastic non-rigid spheres
(d) inelastic non-rigid spheres
Answer:
(b) According to kinetic theory of gases, gas molecules behave as a perfectly elastic rigid spheres.

Question 5.
Which one of the following is not an assumption of kinetic theory of gases?
(a) The volume occupied by the molecules of the gas is negligible.
(b) The force of attraction between the molecules is negligible.
(c) the collision between the molecules are elastic.
(d) All molecules have same speed.
Answer:
(d) Molecules of an ideal gas moves randomly with different speeds.

Question 6.
What is the shape of graph between volume and temperature, if pressure is kept constant?
Answer:
PV = nRT. Hence graph will be straight line.

Question 7.
What is the shape of graph between pressure p and I/V for a perfect gas at constant temperature?
Answer:
Straight line

Question 8.
Identify the minimum possible temperature at which all molecular motion ceases.
Answer:
Absolute temperature (OK or – 273.15°C).

Question 9.
What is the formula for average translational kinetic energy of a gas molecule?
Answer:
3/2 K_BT

Plus One Physics Kinetic Theory Two Mark Questions and Answers

Question 1.
Mention the conditions under which the real gases obey ideal gas equation.
Answer:
Low pressure and high temperature.

Question 2.
Why the temperature rises when gas is suddenly compressed?
Answer:
The work done on gas during compression increases the kinetic energy of molecules and hence temperature of gas rises.

Question 3.
Why evaporation causes cooking?
Answer:
During evaporation, fast moving molecules escape from liquid. Hence average kinetic energy of molecules left behind is decreaesd. This will reduce temperature & causes cooling.

Question 4.
When automobile travels long distance air pressure in tyres increases slightly. Why?
Answer:
As automobile moves, work is being done against force of friction. This work is converted in to heat and it increases the temperature. As P a T, increase in temperature will increase pressure.

Question 5.
PV = µ RT is the ideal gas equation. Real gas obeys ideal behaviour at high temperature and at low pressure.

1. Give an example for ideal gas
2. Why real gases obey ideal gas equation at high temperature and at low pressure.

Answer:

1. Hydrogen
2. The interaction between molecules can be neglected at high T and at low temperature.

Question 6.
A vessel of volume V contains a gas of µ moles at a temperature T.

1. What is the ideal gas equation if gas is considered to be ideal one?
2. The variation of pressure P with number of moles per unit volume in a vessel is shown in the graph. Analyse the graph and choose the correct one and justify your answer.

(i) The temperature inside the vessel decreases.
(ii) The temperature inside the vessel increases.
Answer:
1. PV = µRT

2. P = \(\frac{1}{3}\)nmc^-2
In this case, when n increases, (mc^-2) decreases to maintain P as constant. Temperature is directly proportional to mc^-2. Hence we can say that temperature inside the vessel decreases.

Plus One Physics Kinetic Theory Three Mark Questions and Answers

Question 1.
1 mole of ideal gas is taken in vessel.

1. State the following statements as true or false.
   • In gas equation R is constant.
   • All real gas obeys gas equation at all temperature and pressures.
2. Draw the variation of R with pressure for the above ideal gas.
3. Draw the variation of R with volume for this ideal gas.

Answer:
1. Following statements as true or false:

• True
• False

2.

3.

Question 2.

1. Air pressure in a car tyre increase during driving. Why?
2. Air is filled in a vessel at 60°C. To what temperature should it be heated in order that 1/3^rd of air may escape out of the vessel? (Expansion of air may be neglected).

Answer:
1. During driving the temperature of air inside the tyre increases due to motion.

2. T₁ = 60 + 273 = 333K
V₁ = V; T₂ = ? V₂ = V+ V/3
V₂ = \(\frac{4}{3}\)V

Question 3.
Find the degrees of freedom of the following.

1. A body is confined to move in a straight line
2. A body moves in a plane
3. A body moves in a space

Answer:

1. 1
2. 2
3. 3

Plus One Physics Kinetic Theory Four Mark Questions and Answers

Question 1.
An enclosed vessel contains many number of molecules moving in random direction.

1. Explain the term pressure in terms molecular concept.
2. Derive an expression for the pressure exerted by the gas molecules by assuming postulates of kinetic theory of gases.

Answer:
1. Pressure P = \(\frac{2}{3} n \overline{K E}\)
Where ‘n’ is the number of gas molecules per unit volume. \(\overline{\mathrm{KE}}\) is the average kinetic energy of a gas molecules moving in random direction.

2.

Consider molecules of gas in a container. The molecules are moving in random directions with a velocity V. This is the velocity of a molecule in any direction.

The velocity V can be resolved along x, y and z directions as V_x, V_y, and V_z respectively. If we assume a molecule hits the area A of container with velocity V_x and rebounds back with -V_x.

The change in momentum imparted to the area A by the molecule = 2mV_x. The molecules covers a distance V_xt along the x-direction in a time t. All the molecules within the volume AV_xt will collide with the area in a time t.

If ‘n’ is the number of molecules per unit volume, the total number of molecules hitting the area A, N = AV_xt n.
But on an average, only half of those molecules will be hitting the area, and the remaining molecules will be moving away from the area. Hence the momentum imported to the area in a time t.

Q = 2mv_x × \(\frac{1}{2}\) AV_xt n.
= nmV_x2 At
The rate of change of momentum,
\(\frac{Q}{t}\) = nmV_x² A
But rate of change of momentum is called force, ie. force F = nmV_x2A
∴ pressure P =nmV_x 2 (P = \(\frac{F}{A}\))
Different molecules move with different velocities. Therefore, the average value V2_x is to be taken. If \(\overline{\mathbf{V}}_{\mathbf{x}}^{2}\) isthe average value then the pressure.
p = nm\(\overline{\mathbf{V}}_{\mathbf{x}}^{2}\) ………(1)
\(\overline{\mathbf{V}}_{\mathbf{x}}^{2}\) is known as the mean square velocity.
Since the gas is isotropic (having the same properties in all directions), we can write

Hence the eq (1) can be written as

But nm = ρ, the density of gas
∴ P = \(\frac{F}{A}\) ρ\(\overline{\mathbf{V}}^{2}\).

Question 2.
1. Fill in the blanks

2. What happens to the value of ratio of specific heat capacity, if we consider all rotational degrees of freedom of a 1-mole diatomic molecule?
Answer:
1.

2. Total degrees of freedom = 3 (trans) + 3 (Rot) = 6
∴ C_V = 3R, C_P = 4R
Ratio of specific heat γ = \(\frac{4}{3}\)
Ratio of specific heat capacity decreases.

Plus One Physics Kinetic Theory NCERT Questions and Answers

Question 1.
Molar volume is the volume occupied by 1 mol of any (ideal) gas at standard temperature and pressure (STP: 1 atmospheric pressure, 0°C). Show that it is 22.4 litres.
Answer:
PV= µRT or V = \(\frac{\mu \mathrm{RT}}{\mathrm{P}}\)

= 22.4 × 10^-3 m³ = 22.4 litre.

Question 2.
Estimate the total number of air molecules (inclusive of oxygen, nitrogen, water vapour and other constituents) in a room of capacity 25.0m³ at a temperature of 27°C and 1-atmosphere pressure.
Answer:
V = 25.0m³, T = (27 + 273), K = 300 K, k = 1.38 × 10^-23JK^-1
PV= nRT = n(Nk)T = (nN)kT = N_tkT
Here N_t represents the total number of air molecules in the given gas.

Question 3.
From a certain apparatus, the diffusion rate of hydrogen has an average value of 28.7 cm³s^-1. The diffusion of another gas under the same conditions is measured to have an average rate of 7.2 cm³s^-1. Identify the gas.
Answer:
According to Graham’s law of diffusion of gases, the rate of diffusion of a gas is inversely proportional to the square root of its molecular mass. If R₁ and R₂ be the rates of diffusion of two gases having molecular masses M₁ and M₂ respectively, then

Question 4.
Estimate the fraction of molecular volume to the actual volume occupied by oxygen gas at STP. Take the diameter of an oxygen molecule to be 3Å.
Answer:
Consider one mole of oxygen gas at STP. It occupies 22.4 litre of volume which will contain 6.023 × 10²³. (ie. Avogadro number) molecules. Considering spherical shape of molecule, volume of oxygen molecule

Volume of 6.023 × 10²³ molecules
= \(\frac{4}{3}\) × 3.142(1.5)³ × 10^-30 × 6.02 × 10²³ m³
= 85.1 × 10^-7m³
= 8.51 × 10^-6m³ = 8.51 × 10^-3
litre Molecular volume of one mole of oxygen (∵ 1m³ = 10³ litre)
∴ Molecular volume of one mole of oxygen = 8.51 × 10^-3 litre
Actual volume occupied by one mole of oxygen at STP = 22.4 litre
Fraction of molecular volume to actual volume

Students can Download Chapter 15 Waves Questions and Answers, Plus One Physics Chapter Wise Questions and Answers helps you to revise the complete Kerala State Syllabus and score more marks in your examinations.

Kerala Plus One Physics Chapter Wise Questions and Answers Chapter 15 Waves

Plus One Physics Waves One Mark Questions and Answers

Question 1.
Velocity of sound in vacuum is
(a) 330ms^-1
(b) 165ms^-1
(c) zero
(d) 660ms^-1
Answer:
(c) zero
Sound requires a material medium for propation. Hence, velocity of sound in vacuum is zero.

Question 2.
The physical quantity that remains unchanged when a sound wave goes from one medum to another is
(a) amplitude
(b) speed
(c) wavelength
(d) frequency
Answer:
(d) frequency
When a sound wave goes from one medium to another the frequency of the wave remains unchanged.

Question 3.
What is the range of frequency of audible sound?
Answer:
20Hz to 20KHz.

Question 4.
Why does sound travel faster in iron than in air?
Answer:
Because solids are more elastic compared to air.

Question 5.
What kind of waves help the bats to find their way in dark?
Answer:
Ultrasonic wave

Question 6.
In which gas, hydrogen and oxygen will the sound have greater velocity?
Answer:
As velocity, v α \(\sqrt{1 / ρ}\), velocity of sound will be greater in hydrogen gas.

Question 7.
Why transverse waves can not setup in gas?
Answer:
The rigidity modulus of gas is zero.

Question 8.
What is the effect of pressure on the velocity of sound waves?
Answer:
No effect

Question 9.
Why bells are made up of metal and not wood?
Answer:
The wood causes high damping.

Question 10.
What is the velocity of sound in perfect rigid body?
Answer:
The velocity is infinite cause young’s modulus of perfect rigid body is infinite.

Plus One Physics Waves Two Mark Questions and Answers

Question 1.
If a tuning fork is held above a resonance column, then maximum sound can be heard at certain height of the air column.

1. Name the type of wave produced in the air col¬umn.
2. What do you mean by beats?

Answer:

1. Longitudinal
2. Periodic variation of intensity of sound is called beats.

Question 2.
Why bells of colleges and temple are of large size?
Answer:
Larger, the area of source of sound more is the energy transmitted into the medium. Hence intensity of sound is more and loud sound is heard.

Plus One Physics Waves Three Mark Questions and Answers

Question 1.
A sound wave of frequency 400 Hz is travelling in air at a speed of 320m/s.

1. The speed of sound wave in vaccum is______
   • 320m/s
   • more than 320m/s
   • less than320m/s
   • Zero
2. What is the wavelength of the above wave?
3. Calculate the diffference in phase between two points on the wave 0.2m apart in the direction of travel.

Answer:
1. Zero

2. v = fλ
λ = \(\frac{v}{f}=\frac{320}{400}\) = 0.8m

3. At 0.2 m apart, the phase differnce’is given by
∆Φ = \(\frac{0.2}{0.8} \times 2 \pi=\frac{\pi}{2} \mathrm{rad}\).

Question 2.
A string fixed one end is suddenly brought in to up and down motion.

1. What is the nature of the wane produced in the string and name the wave.
2. A brass wire 1 m long has a mass 6 × 10^-3 kg. If it is kept at a tension 60N, What is the speed of the wave on the wire.

Answer:
1. Transverse wave

2.

Plus One Physics Waves Four Mark Questions and Answers

Question 1.
A sonometer wire of length 30cm vibrates in the second overtone

1. Represent it pictorially
2. What is the distance between two points in the string which has a phase difference of n
3. A violin string resonates in its fundamental frequency of 196hz. Where along the string must you place your finger so that the fundamental frequency becomes 440Hz, If the length of violin string is 40cm.

Answer:
1.

2. 0.30 = \(\frac{3}{2}\)λ
λ = \(\frac{2 \times 0.3}{3}\) = 0.2
We know 2π = λ
ie. π radian = \(\frac{λ}{2}\) wave length
= \(\frac{0.2}{2}\) = 0.1m
∴ The distance between two points is 0.1 m (for a a phase difference of π radian).

3. For fundamental mode of vibration.

The finger must be placed 37.5 from one end.

Question 2.
A horizontal metal wire is fixed on a state of tension between two vertical supports when plucked it gives a fundamental frequency f_O.

1. Obtain a mathematical expression for f_O
2. A 5.5 m wire has a mass of 0.035 kg. If the tension of the string is 77N, the speed of wave on the string is
   • 110 ms^-1
   • 11\(\sqrt{10}\)ms^-1
   • 77 ms^-1
   • 11 ms^-1
   • 102 ms^-1
3. What change, if any, will be observed in the fundamental frequency if the wire is now immersed in water and plucked again?

Answer:
We know velocity on a string, V = \(\sqrt{T / m}\)
But V = λf

2. Mass per unit length,m = \(\frac{M}{\ell}=\frac{0.035}{5.5}\)
= 6.36 × 10^-3kg/m
T = 77N

3. Frequency does not change.

Question 3.
When a pebble is dropped to the surface of water, certain waves are formed on the water surface.

1. What type of wave is it?
2. Is it a progressive wave? Explain?
3. The equation for such a wave is y = 4sinπ (2t – 0.01x). where ‘y’ and ‘x’ are in cm. and ‘t’ in sec. find its
   • Amplitude
   • Wavelength
   • initial phase
   • Frequency

Answer:
1. Transverse wave.

2. It is a progressive wave. It moves from one point to another point.

3. y = 4sinπ(2t – 0.01x)
y = 4 sin (2πt-0.01πx)
= -4 sin (0.01 πx – 2πt)
Comparing with standard wave equation, y = A sin (kx – ωt), we get

• Amplitude A = 4 m
• Kx = 0.01 × πx
  \(\frac{2 \pi}{\lambda}\)x = 0.01 πx, λ = 200m
• Initial phase = 0
• ωt = 2 πt, 2 π f t = 2 πt, f = 1 Hz.

Question 4.
The speed of a wave along a stretched string depends only on the tension and the linear mass density of the string and does not dependent on the frequency of the wave.

1. Give the equation of speed of transverse wave along a stretched string.
2. Why the speed does not depend on the frequency of the wave.
3. A steel wire 0.72 long has a mass of 5 × 10^-3Kg. If the wire is under a tension 60N, what is the speed of transverse wave on the wire?

Answer:

1. v = \(\sqrt{\frac{T}{m}}\) Where T is the tension and m is the mass per unit length.
2. The frequency of the wave is determined by the source that generates the wave.
3. 

Question 5.
While discussing the propagation of sound through atmospheric air, one argued that the velocity of sound is 280 ms-1 and said that he calculated it using Newton’s formula. But another learner argued that velocity of sound is 330 ms1. He justified his argument by saying that he has applied Laplace corrected formula.

1. Write the formula used by the second learner.
2. Using the above relation, show that velocity depends on temperature and humidity while is independent of pressure.
3. “Sound can be heard over longer distance on rainy days.” Justify.

Answer:
1. a = \(\sqrt{\frac{2 \lambda P}{\rho}}\)

2.

If temperature remain same
PV = constant
∴ v is independent of p.

3. During rainy day as ρ decreases. Hence v increase and sound propagate longer distance.

Question 6.
When a stone is dropped in to the river, certain waves are formed on its surface.

1. What type of wave it is?
2. Is it a progressive wave? Explain.
3. If yes, derive a mathematical expression for the above wave.

Answer:
1. Transverse wave.

2. Yes, because each particles of the medium vi¬brates simple harmonically.

3. Consider a harmonic wave travelling along the +ve x-direction with a speed V. Let ‘0’ be the particle in the medium. Its displacement at any instant of time may be written as y = A sin ωt

Consider another particle ‘p’ at a distance x from ‘0’ to its right. The displacement of p at any instant.
y = A sin (ωt – α) ______(1)
α is the phase difference between 0 and P. Here
α = \(\frac{2 \pi}{\lambda}\)x equation (1) becomes

Plus One Physics Waves Five Mark Questions and Answers

Question 1.
A boy standing near a railway track found that the pitch of the siren of a train increases as it approaches him

1. State the phenomenon behind it?
2. List any two applications of the same Phenomenon
3. Obtain an expression for the apparent frequency of the siren as heard by the boy.

Answer:
1. Doppler effect.

2. Doppler effect can be used to find the speed of moving object. Dopplar effect in light is used to find speed of galaxies.

3. The apparent change in the frequency of sound wave due to the relative motion of source or listener or both is called Doppler effect. It was proposed by John Christian Doppler and it was experimentally tested by Buys Ballot.

Considers source is producing sound of frequency v. Let V be the velocity of sound in the medium and λ the wavelength of sound when the source and the listener are at rest.

The frequency of sound heard by the listener is ν = \(\frac{v}{\lambda}\). Let the source and listener be moving with velocities v_s and v_l in the direction of propogation of sound from source to listener. (The direction S to L is taken as positive).

The relative velocity of sound wave with respect to the source = V – V_s
Apparent wavelength of sound,
λ¹ = \(\frac{V-V_{s}}{v}\) _____(1)
Since the listener is moving with velocity v^, the relative velocity of sound with respect to the listener,
V¹ = V – V_l ______(2)
Apparent frequency of sound as heard by the listener is given by
ν = \(\frac{v^{1}}{\lambda^{1}}\) ______(3)
Sub (1) and (2) in eq.(3) we get

Question 2.

1. Waves are means of transferring energy from one point to another. Distinguish between longitudinal and transverse waves.
2. What is a plane progressive wave? Arrive at an expression for the displacement of a particle on the path of the wave, advancing in the positive x-direction.
3. The velocity of sound is greater in solids than in gases. Explain.

Answer:
1.

Longitudinal wave	Transverse wave
1. Can’t be polarized 2. Direction of propagation is parallel to the direction of vibration of particles.	1. Can be polarized 2. Direction of propagation is perpendicular to the direction of vibration of particles.

2. Out of syllabus

3. Solids are highly elastic as compared to liquids and gases.

Question 3.

1. Sound produced by an open pipe contains:
   • Fundamental component only
   • Odd harmonics only
   • All the harmonics
   • Even harmonics only
2. A pipe 30 cm long is open at both ends. Which harmonic mode of the pipe is resonantly exerted by a 1.1 kHz source?
3. Will resonance with the same source be observed if one end of the pipe is closed? (Take the speed
   of sound in air to be 330 ms^-1)

Answer:
1. All the harmonics.

2. We know frequency of oscillation in the Pipe, f = \(\frac{n V}{2 L}\)
Substituting the values v, L and f we get

n = 2 means that the oscillation is second harmonics.

3. The condition for second harmonics in closed pipe is

The frequency required for resonance in a closed pipe is 825 Hz. Hence we do not get resonance at 1.1 Khz.

Question 4.
A boy plucks at the centre of a stretched string of length 1 m and abserves a wave pattern.

1. Which type of wave is produced on the string?
2. What are the conditions for the formation of the above mentioned wave?
3. The distance between consecutive nodes is
   1. λ
   2. λ/2
   3. 2λ
   4. λ/4
4. A steel rod100 cm long is clamped at its middle. The fundamental frequency of longitudiral vibrations of the rod is given to be 2.5kHz. What is the speed of sound in steel?

Answer:
1. Standing wave (or) stationary wave

2. Same frequency, same amplitude, travelling in opposite direction.

3. λ/2

4. Length of the rod l = \(\frac{\lambda_{1}}{2}\)
λ₁ = 2l =2m
v = ν₁λ₁
v = 2500 × 2 = 5000 m/s.

Plus One Physics Waves NCERT Questions and Answers

Question 1.
A string of mass 2.50kg is under a tension of200N. The length of the stretched string is 20.0m. If the transverse jerk is struck at one end of the string, how long does the disturbance take to reach the other end?
Answer:
Tension T = 200N
Length I = 20.0m; Mass M = 2.50kg

Question 2.
A stone droped from the top of a tower of height 300m high splashes into the water of a pond near the base of the tower. When is the splash heard at the top. Given that the speed of sound in air is 340ms^-1? (g = 9.8ms^-2)
Answer:
Time after which the splash is heard at the top is equal to the sum of the time t₁ taken by the stone to fall down and the time t₂ taken by the sound to travel from bottom to top.
Using S = ut + \(\frac{1}{2}\) at², we 9et S = \(\frac{1}{2}\)gt₁²
(∵ u = 0 and a =g)

Question 3.
A hospital uses an ultrasonic scanner to locate tumours in a tissue. What is the wavelength of sound in the tissue in which the speed of sound is 1.7kms^-1? The operating frequency of the scanner is 4.2MHz.
Answer:
λ = ? u = 1.7kms^-1 = 1700ms^-1
v = 4.2 × 10⁶ Hz, u = vλ = or λ = \(\frac{u}{v}\)
or λ = \(\frac{1700}{4.2 \times 10^{6}}\) m = 4.05 × 10^-4 m = 0.405mm.

Question 4.
A wire stretched between two rigid supports vibrates in its fundamental mode with a frequency of 45Hz. The mass of the wire is 3.5 × 10^-2kg and its linear mass density is 4.0 × 10^-2kgm^-1. What is

1. the speed of a transverse wave on the string and
2. the tension in the string?

Answer:
mass of wire M = 3.5 × 10^-2kg
Linear density µ = mass / length
= M/l = 4.0 × 10^-2kg
∴ Length of wire l = \(\frac{M}{\mu}=\frac{3.5 \times 10^{-2}}{4 \times 10^{-2}}\) m = 0.875m
In the fundamental mode,
λ =2l = 2 × 0.875m = 1.75m

1. Speed of transverse vyaves u = v λ
= 45 × 1.75ms^-1 = 78.75ms^-1

2. u= \(\sqrt{\frac{T}{\mu}}\) or T = µu² = 4 × 10^-2(78.75)²N
= 278.06N.

Question 5.
A steel rod 100cm long is clamped at its middle. The fundemental frequency of longitudinal vibrations of the rod are given to be 2.53kHz. What is the speed ‘ of sound in steel?
Answer:
l = 1m, v = 2.53 × 10³HZ, \(\frac{\lambda}{2}\) = I or λ = 2m
u = v λ = 2.53 × 10³ × 2ms^3-1 = 5.06 × 10³ ms^-1
= 5.06kms^-1

Students can Download Chapter 2 Data Representation and Boolean Algebra Notes, Plus One Computer Science Notes helps you to revise the complete Kerala State Syllabus and score more marks in your examinations.

Kerala Plus One Computer Science Notes Chapter 2 Data Representation and Boolean Algebra

Number System:
It is a systematic way to represent numbers in different ways. Each number system has its own Base, that is a number and that number of symbols or digits used.

1. Most Significant Digit (MSD): The digit with most weight is called MSD. MSD is also called Left Most Digit(LMD)
2. Least Significant Digit (LSD): The digit with least weight is called LSD. LSD is also called Right Most Digit(RMD)
   • eg: 106 : Here MSD : 1 and LSD : 6
   • 345.78: Here MSD : 3 and LSD : 8
3. A Binary Digit is also called a bit.
4. The weight of each digit of a number can be represented by the power of its base.

Number conversions:
In general, to convert a decimal number into another number system(binary or octal or hexadecimal) do the following. Divide the number successively by the base of the number system do you want to convert and write down the remainders from bottom to top.

To convert a decimal fraction into another number system .multiply the number by the base of the number system do you want to convert then integer part and fractional part are separated again multiply the fractional part by the base and do the steps repeatedly until the fractional part becomes zero. Finally write down the integer part from top to bottom.

Decimal to Binary:
Divide the number by the base 2 successively and write down the remainders from bottom to top.

Decimal fraction to binary:
multiply the number by the base 2 then integer part and fractional part are separated again multiply the fractional part by the base 2 and do the steps repeatedly until the fractional part becomes zero. Finally write down the integer part from top to bottom.

Decimal to Octal:
Divide the number by the base 8 successively and write down the remainders from bottom to top.

Decimal fraction to octal:
multiply the number by the base 8 then integer part and fractional part are separated again multiply the fractional part by the base 8 and do the steps repeatedly until the fractional part becomes zero. Finally write down the integer part from top to bottom.
eg: (55)₁₀ = ()₈

(0.140625)₁₀ = (0.11)₈

Decimal to Hexadecimal:
Divide the number by the base 16 successively and write down the remainders from bottom to top.

Decimal fraction to hexadecimal:
multiply the number by the base 16 then integer part and fractional part are separated again multiply the fractional part by the base 16 and do the steps repeatedly until the fractional part becomes zero. Finally write down the integer part from top to bottom.

Converting a number from any number system into decimal: For this multiply each digit by its corresponding weight and sum it up.

Binary to decimal conversion:
For this multiply each bit by its corresponding weight and sum it up. The weights are power of 2.

Converting binary fraction to decimal

101.101 = 1 × 2² + 0 × 2¹ + 1 × 2⁰ + 1 × 2^-1 + 0 × 2^-2 + 1 × 2^-3
= 4 + 0 + 1 + 1/2 + 0 + 1/8
= 5 + 0.5 + 0.125
(101.101)₂ = (5.625)₁₀

Octal to decimal conversion:
For this multiply each bit by its corresponding weight and sum it up. The weights are power of 8.
Eg: (1007)₈ =()₁₀?

1 × 8³ + 0 × 8² + 0 × 8¹ + 7 × 8⁰
= 512 + 0 + 0 + 7
=(519)₁₀
Converting octal fraction to decimal (600.005)₈ =()₁₀?

= 6 × 8² + 0 × 8¹ + 0 × 8⁰ + 0 × 8^-1 + 0 × 8^-2 + 5 × 8^-3
= 384 + 0 + 0 + 0 + 0 + 0.009765625
= (384.009765625)₁₀

Hexadecimal to decimal conversion:
For this multiply each bit by its corresponding weight and sum it up. The weights are power of 16.
Eg: (100)₁₆ = ()₁₀?

= 1 × 16² + 0 × 16¹ + 0 × 16⁰
= 256 + 0 + 0
= (256)₁₀
Converting Hexadecimal fraction to decimal (60A.4)₈ =()₁₀?

= 6 x 16² + 0 x 16¹ + 10 x 16⁰ + 4 x 16^-1
= 1536 + 0 + 0 + .25
= (1536.25)₁₀

Octal to binary conversion:
Convert each octal digit into its 3 bit binary equivalent. Consider the following table

Hexadecimal to binary conversion:
Convert each Hexadecimal digit into its 4 bit binary equivalent. Consider the following table

1010 1011 1100 (ABC)₁₆=(101010111100)₂

Binary to octal conversion
Divide the binary number into groups of 3 bits starting from the right to left(But in the fractional part start dividing from left to right). Insert necessary zeros in the left side(or right side in the case of fractional part)if needed and write down the corresponding octal equivalent.
eg: (10100110)₂= ()₈?
Insert one zero in the left side to form 3 bits group

(10100110)₂= (246)₈

Binary to Hexadecimal conversion:
Divide the binary number into groups of 4 bits starting from the right to left(But in the fractional part start dividing from left to right). Insert necessary zeros in the left side(or right side in the case of fractional part)if needed and write down the corresponding Hexadecimal equivalent.
eg: (100100110)₂ = ()₁₆?
Insert 3 zeros in the left side to form 4 bits group

(100100110)₂ = (126)₁₆

Octal to Hexadecimal conversion:
First convert octal number into binary(see 1.6.7), then convert this binary into hexadecimal(also see 1.6.10)
eg: Convert (67)₈ = ( )₁₆
Step I: First convert this number into binary equivalent for this do the following:

Step II: Next convert this number into hexadecimal equivalent for this do the following.

So the answer is (67)₈ = ( 37)₁₆

Hexadecimal to octal conversion:
First convert Hexadecimal to binary(see 1.6.8), then covert this binary into octal(also see 1.6.9)
eg: Convert (A1)₁₆ = ( )₈?
Step I: First convert this number into binary equivalent. For this do the following

Step II. Next convert this number into octal equivalent. For this do the following.
So the answer is (A1)₁₆ = (241)₈

Data representation:
The data stored in the computer memory is in the form of binary.

Representation of integers
There are three ways to represent integers in computer. They are as follows:

1. Sign and Magnitude Representation (SMR)
2. 1’s Complement Representation
3. 2’s Complement Representation

1. SMR:
Normally a number has two parts sign and magnitude, eg: Consider a number+5. Here + is the sign and 5 is the magnitude. In SMR the most significant Bit (MSB) is used to represent the sign. If MSB is 0 sign is +ve and MSB is 1 sign is -ve. eg: If a computer has word size is 1 byte then

Here MSB is used for sign then the remaining 7 bits are used to represent magnitude. So we can , represent 2⁷ = 128 numbers. But there are negative and positive numbers. So 128 + 128 = 256 number. The numbers are 0 to +127 and 0 to -127. Here zero is repeated. So we can represent 256 – 1 = 255 numbers.

2. 1’s Complement Representation:
To get the 1’s complement of a binary number, just replace every 0 with 1 and every 1 with 0. Negative numbers are represented using 1’s complement but +ve number has no 1 ’s complement,
eg:
(i) To find the 1’s complement of -21
+21 = 00010101
To get the 1’s complement change all 0 to 1 and.all 1 to 0.
-21 = 11101010
1’s complement of-21 is 11101010

(ii) Find the 1’s complement of +21. Positive numbers are represented by using SMR.
+21 = 00010101 (No need to take the 1’s complement)

3. 2’s Complement Representation:
To get the 2’s complement of a binary number, just add 1 to its 1’s complement +ve number has no 2’s complement.
eg: To find the 2’s complement of -21
+21 = 00010101
First take the 1’s complement for this change all 1 to 0 and all 0 to 1

2’s complement of -21 is 1110 1011

Representation of floating point numbers:
A real number consists of an integer part and fractional part and represent by using Exponent and Mantissa method. This method is also used to represent too big numbers as well as too small numbers.
Eg: .0000000000000000000000001 can be represented easily as 1 × 10^-25. Here T is the Mantissa and -25 is the exponent.

A computer with 32 bit word length is used 24 bits for mantissa and the remaining 8 bits used to store exponent.

Representation of characters:
1. ASCII(American Standard Code for Information Interchange):
It is 7 bits code used to represent alphanumeric and some special characters in computer memory. It is introduced by U.S. government. Each character in the keyboard has a unique number.

Eg: ASCII code of ‘a’ is 97, when you press ‘a’ in the keyboard , a signal equivalent to 1100001 (Binary equivalent of 97 is 1100001) is passed to the computer memory. 2⁷ = 128, hence we can represent only 128 characters by using ASCII. It is not enough to represent all the characters of a standard keyboard.

2. EBCDIC(Extended Binary Coded Decimal Interchange Code):
It is an 8 bit code introduced by IBM(lnternational Business Machine). 2⁸ = 256 characters can be represented by using this.

3. ISCII(lndian Standard Code for Information Interchange):
It uses 8 bits to represent data and introduced by standardization committee and adopted by Bureau of Indian Standards(BIS).

4. Unicode:
The limitations to store more characters is solved by the introduction of Unicode. It uses 16 bits so 2¹⁶ = 65536 characters (i.e, world’s all written language characters) can store by using this.

Binary arithmetic:
Binary addition:
The rules for adding two bits

eg: Find the sum of binary numbers 110011 and 100001.

Binary subtraction:
The rules for subtracting a binary digit from another digit.

Subtraction using 1’s complement:
The steps are given below:

Step 1: Add 0s to the left of the subtrahend, to make two numbers with same number of bits.
Step 2: Find 1’s complement of subtrahend.
Step 3: Add the complement with minuend.
Step 4: If there is a carry, ignore the carry, the result is positive then add the carry 1 to the result.
eg: Subtract 1101 from 111100 using 1’s complement method.
Step 1: Insert two Os to the left of 1101. Hence the subtrahend is 001101.
Step 2: 1’s complement of 001101 is 110010
Step 3: Add this to the minuend.

Step 4: Ignore the carry the result is positive and add add the carry 1 to 101110

Hence the result is 101111.

Subtraction using 2’s complement:
The steps are given below :
Step 1: Add 0s to the left of the subtrahend, to make two numbers with same number of bits.
Step 2: Find 2’s complement of subtrahend.
Step 3: Add the complement with minuend.
Step 4: If there is a carry, ignore the carry, the result is positive.
eg: Subtract 1101 from 111100 using 2’s complement method.
Step 1: Insert two 0s to the left of 1101. Hence the subtrahend is 001101.
Step 2: Find the 2’s complement of 001101.
1’s complement is 110010.
2’s complement is 110010 + 1 = 110011
Step 3: Add this to the minuend.

Step 4: Ignore the carry the result is positive. Hence the result is 101111.

Introduction to Boolean algebra:
The name Boolean Algebra is given to honour the British mathematician George Boole. Boolean algebra deals with two states true or false otherwise Yes or No and numerically either 0 or 1.

Binary valued quantities:
A logical decision which gives YES or No values is a binary decision, A statement which gives YES or NO values(TRUE or FALSE) is a logical statement or truth function. A variable which can assign TRUE or FALSE (1 or 0) values is a logical variable

Boolean operators and logic gates:
Logical Operators are AND, OR and NOT. A logical gate is a physical device (electronic circuit)that can perform logical operations on one or more logical inputs and produce a single logical output. A table represents the set f all possible values and the corresponding results in a statement is called truth table.
1. The OR operator and OR gate:
The OR operator gives a 1 either one of the operands is 1. If both operands are 0, it produces 0. The truth table of X OR Y is

The logical OR gate is given below.

The truth table and the gate for the Boolean expression Y = A + B + C

2. The AND operator and AND gate:
The AND operator gives a 1 if and only if both operands are 1. If either one of the operands is 0, it produces 0 The truth table of X AND Y is

The logical AND gate is given below.

The truth table and the gate for the Boolean expression Y = A . B . C

3. The NOT operator and NOT gate:
It produces the vice versa. NOT gate is also called inverter. It is a unary operator that means it has only one input and one output. The truth table of NOT X is

Basic postulates of Boolean algebra:
Boolean algebra consists of some fundamental laws. These laws are called postulates.
Postulate 1: Principles of 0 and 1
If A ≠ 0 , then A = 1 and A 1, then A = 0

Principle of Duality:
When changing the OR(+) to AND(.), AND (.) to OR(+), 0 to 1 and 1 to 0 in a Boolean expression we will get another Boolean relation which is the dual of the first, this is the principle of duality.

Basic theorems of Boolean algebra:
There are some standard and accepted rules in every theory, these rules are known as axioms of the theory.

Identity law:
If X is a Boolean variable, the law states that

1. 0 + X = X
2. 1 + X = 1 (these are additive identity law)
3. 0 . X = 0
4. 1 . X = X (these are multiplicative identity law)

Following are the truth tables

Idempotent law:
This law states that

1. X + X = X
2. X . X = X

Involution law:
This states that
\(\overline{\overline{\mathrm{X}}}=\mathrm{x}\)
The compliment of compliment of a number is the number itself.

Complimentary law:
This law states that

1. \(x+\bar{x}=1\)
2. \(x \cdot \bar{x}=0\)

The truth table is given below

Commutative law:
This law allows to change the position of variable in OR and AND

1. X + Y = Y + X
2. X . Y = Y . X

The truth table is given below

Associative law:
It allows grouping of variables differently

1. X + (Y + Z) = (X + Y) + Z
2. X . (Y . Z) = (X . Y) . Z

The truth table is given below

Distributive law:
This law allows expansion of multiplication over addition and also allows addition operation over multiplication.

1. X . (Y + Z) = X . Y + X . Z
2. X + Y . Z = (X + Y) . (X + Z)

The truth table is given below

Absorption law:
It is a kind of distributive law in which two variables are used and result will be one of them

1. X + (X . Y) = X
2. X . (X + Y) = X

The truth table is given below

De Morgan’s theorem:
Demorgan’s first theorem states that
\(\overline{\mathrm{X}+\mathrm{Y}}=\overline{\mathrm{X}} \cdot \overline{\mathrm{Y}}\)
ie. the compliment of sum of two variables equals product of their compliments.

The second theorem states that
\(\overline{\mathrm{X} . {\mathrm{Y}}}=\overline{\mathrm{X}}+\overline{\mathrm{Y}}\)
ie. The compliment of the product of two variables equals the sum of the compliment of that variables.

Circuit designing for simple Boolean expressions:
By using basic gates such as AND, OR and NOT gates we can create logic circuits.

Universal gates:
By using NAND and NOR gates only we can create other gate hence these gates are called Universal gate.

NAND gate:
The output of AND gate is inverted by NOT gate is the NAND gate

NOR gate:
The output of OR gate is inverted by NOT gate is the NOR gate.