Plus One Computer Science Chapter Wise Questions and Answers Chapter 3 Components of the Computer System

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

Kerala Plus One Computer Science Chapter Wise Questions and Answers Chapter 3 Components of the Computer System

Plus One Components of the Computer System One Mark Questions and Answers

Question 1.
The Tangible parts of a computer is _________
Answer:
Hardware

Plus One Computer Science Chapter Wise Questions and Answers Chapter 3 Components of the Computer System

Question 2.
The instructions that tell the hardware to perform a task is __________
Answer:
Software

Question 3.
The brain of the computer is __________
Answer:
CPU

Question 4.
CPU means _________
Answer:
Central Processing Unit

Question 5.
ALU is
Answer:
Arithmetic Logic Unit

Question 6.
I am an input device. I can read text or picture on paper and translate into computer usable form. Who am I?
Answer:
Scanner

Question 7.
Odd man out.
(a) Trackball
(b) Joy Stick
(c) Scanner
(d) LCD
Answer:
(d) LCD. It is an output device. Others are input device

Question 8.
Odd man out.
(a) Inkjet
(b) Laser
(c) Dot Matrix Printer
(d) Thermal
Answer:
(c) Dot Matrix Printer. It is impact printer others are non impact printers.

Plus One Computer Science Chapter Wise Questions and Answers Chapter 3 Components of the Computer System

Question 9.
The storage capacity of a CD is ________
(a) 1.44 MB
(b) 700 GB
(c) 700 MB
(d) 650 GB
Answer:
(c) 700 MB

Question 10.
__________ sheet is used to write answers in Kerala Entrance Exam.
Answer:
OMR Sheet

Question 11.
ABC textile uses reader to input the item and its price.
Answer:
Bar code reader

Question 12.
_________ device senses the presence or absence of a pencil mark.
Answer:
OMR

Question 13.
Name any two printing devices.
Answer:
Mouse and touchpad

Question 14.
__________ is a device that draws pictures on a paper.
Answer:
Plotter

Question 15.
Primary memory is classified into two. What are they?
Answer:
RAM and ROM

Question 16.
The storage capacity of a DVD is _________.
Answer:
4.7 GB

Question 17.
Winzip is a ___________ utility.
Answer:
Compression utility

Question 18.
Win Rar is a ________ utility.
Answer:
Compression utility

Question 19.
Most commonly used input device is _________________
Answer:
Keyboard or mouse

Question 20.
Govt, decided to conduct a test that contains all objective type questions. Which device is most suitable for evaluation.
Answer:
OMR

Question 21.
__________ is used mostly for computer games.
Answer:
Joy stick

Question 22.
I am an input device. I have a stick with two buttons called triggers on the top. Who am I?
Answer:
Joy stick

Question 23.
I am a pointing device. I am a stationary device. Who am I?
Answer:
Touchpad

Question 24.
In portable computers which pointing device is suitable?
Answer:
Touchpad

Question 25.
State true or false.
Hard copy devices are very slow compared to soft copy devices.
Answer:
True

Plus One Computer Science Chapter Wise Questions and Answers Chapter 3 Components of the Computer System

Question 26.
________ is also called firrfi ware.
Answer:
ROM

Question 27.
__________ is acts as an interface between user and computer.
Answer:
Operating system

Question 28.
_________ is used to create and modify any type of document.
Answer:
Word Processor

Question 29.
_________ is a set of programs that manage the database.
Answer:
DBMS

Question 30.
__________ package contains rows & columns.
Answer:
Spread sheet

Question 31.
_________ is a presentation package.
Answer:
Power point

Question 32.
Your computer teacher asked you to explain the project work done by your group. Which package will help you to do so?
Answer:
Power Point

Question 33.
___________ is a DTP Package.
(a) Excel
(b) Powerpoint
(c) PageMaker
(d) None of these
Answer:
(c) PageMaker

Question 34.
DTP is ________
Answer:
Desk Top Publishing

Question 35.
_______ is designed to help computer for its smooth functioning.
Answer:
Utilities

Question 36.
Customised software is also called _________
Answer:
Tailor made software

Question 37.
__________ S/W is used to remove virus from a computer.
Answer:
Anti Virus S/W

Question 38.
Name any Antivirus S/W.
Answer:
Norton Antivirus, McAfee, Avira, AVG

Question 39.
Name the two classifications of output devices.
Answer:
Hard copy and soft copy

Question 40.
Name the two classifications of printers.
Answer:
Impact printer and non-impact printer

Question 41.
_______ gas is used in plasma panels
Answer:
(a) Oxygen
(b) Neon
(c) Mercury
(d) helium Neon
Answer:
(b) Neon

Question 42.
__________ printers are used in fax machine.
Answer:
Thermal Printers

Question 43.
_________ is read/write memory.
Answer:
RAM

Question 44.
________ is of volatile memory
(a) ROM
(b) RAM
(c) CD
(d) DVD
Answer:
(b) RAM

Plus One Computer Science Chapter Wise Questions and Answers Chapter 3 Components of the Computer System

Question 45.
From the following which is expensive?
(a) CD
(b) DVD
(c) HDD
(d) RAM
Answer:
(d) RAM

Question 46.
You want to input your photograph into computer. Which device is used for this?
(a) Scanner
(b) Mouse
(c) OMR
(d) OCR
Answer:
(a) Scanner

Question 47.
The primary memory which is commonly used in electronic billing machines to store price of products is
(a) P R O M
(b) E P R O M
(c) E E P R O M
(d) None of these
Answer:
(b) E P R O M

Question 48.
Name any three pointing devices,
Answer:
Mouse, Touchpad and light pen.

Question 49.
You want to print your brother’s resume which printer will you choose. Why?
Answer:
I will choose either Ink-jet or Laser printer. Because these produce less noise and produce high quality printing output. They are used to print characters as well as graphics (Photos) with very high quality.

Question 50.
The fastest memory in a computer is
Answer:
Registers

Question 51.
The storage capacity of a single layer DVD is _______
Answer:
4.7 GB

Question 52.
Give two examples for OS.
Answer:
Windows 7, Windows Vista

Question 53.
A program in execution is called _______
Answer:
Process

Question 54.
Name the software that translates assembly language program into machine language program.
Answer:
Assembler

Question 55.
DBMS stands for _________.
Answer:
Data Base Management System .

Question 56.
Duplicating disc information is called __________.
Answer:
Backup

Question 57.
An example of free and open source software is _______?
Answer:
Linux

Question 58.
The software that gives users a chance to try it before buying is __________
Answer:
Shareware

Plus One Computer Science Chapter Wise Questions and Answers Chapter 3 Components of the Computer System

Question 59.
An example of proprietary software is _________.
Answer:
Tally

Question 60.
Which software is used for calculation?
(a) Word processor
(b) Spreadsheet
(c) Presentation
(d) Multimedia
Answer:
(b) Spreadsheet

Question 61.
Accumulator stores ________
(a) address of data
(b) instruction to be executed
(c) address of next instruction to be executed
(d) intermediate result
Answer:
(d) intermediate result

Question 62.
If Tracks and Sectors : Hard disk, then
___________ : Compact disk
Answer:
Pits and Lands (OR) 0 and 1

Question 63.
Which one of the following file extensions is different from others?
(a) WAV
(b) MP3
(c) PNG
(d) MIDI
Answer:
(c) PNG, the others are audio files.

Question 64.
Which register holds the memory address of next instruction to the executed?
(a) Accumulator
(b) PC
(c) MBR
(d) MAR
Answer:
(b) PC

Question 65.
1. Write the following memory devices in the order of their speed. (fastest to slowest order)

  • Cache
  • RAM
  • Hard Disk
  • Registers

2. What do you mean by Freeware and Shareware?
Answer:
1. Memory devices in the order of their speed.

  • Registers
  • Cache
  • RAM
  • Hard Disk

2. Freeware and Shareware:

  • 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.

Plus One Components of the Computer System Two Mark Questions and Answers

Question 1.
The Higher Secondary Department wishes to conduct an examination for +1 students with multiple choice questions and publish results as soon as possible. Suggest a method to evaluate the answer scripts and publish the results quickly & correctly with the help of computers.
Answer:
OMR has to be used, it senses the presence or absence of a mark (bubbles) using a high density beam then converted into electric signals for computer. It needs good quality expensive paper and accurate alignment of printing on forms.

Plus One Computer Science Chapter Wise Questions and Answers Chapter 3 Components of the Computer System

Question 2.
Remesh is a graphic designer who prepares his drawing using a computer. He desires for an alternative device by which he can draw directly on the screen. Suggest a device for this and explain it working.
Answer:
Light pens are used for this. It consists fo a photocell placed in a small tube, it is able to detect the light coming from the screen. Hence locate the exact position on the screen. It is used by graphic designers, illustrators and drafting engineers, with the help of CAD to draw directly on the screen.

Question 3.
You might have noticed that in some shops billing is done using computers without typing the item name, price, quantity, etc. Mention the device used for entering data and explain its working.
Answer:
A device called Bar Code Reader is used for this. It contains photoelectric scanner that read the bar code and input the information to the computer attached to it. It helps to reduce the errors and process the bills quickly.

Question 4.
Your scholl has arranged an excursion. You are having an ordinary camera whereas your friend has a digital camera. List the benefits your friend enjoys by using digital camera.
Answer:

  1. Digital camera does not need film.
  2. More number of shots can take
  3. Operational cost is less
  4. Very easy to manipulate images in digital form using computers.

Question 5.
A medical shop in your locality wishes to purchase a printer for their billing purpose. Which type of printer will you recommend if carbon copies are to be taken. Justify.
Answer:
Dot Matrix Printer. To take carbon copies impact printer is a must, operational cost is less and it can print bills in a moderate speed.

Question 6.
Find the extact match.
Plus One Computer Science Chapter Wise Questions and Answers Chapter 3 Components of the Computer System 1
Answer:

  1. D
  2. C
  3. B
  4. A

Question 7.
Your friend wishes to start a DTP centre with facilities to design posters and notices, to scan pictures and modify them and to print them. What would be your suggestions regarding the computer and peripherals?
Answer:
The requriements are computer, scanner, printer and software.

Question 8.
Find the most appropriate match.
Plus One Computer Science Chapter Wise Questions and Answers Chapter 3 Components of the Computer System 2
Answer:
1 – D
2 – C
3 – A
4 – B

Question 9.
Suggest a suitable device for the following.

  1. High quality printing
  2. High quality drawing
  3. Printing with carbon copies
  4. Economical printing of small quantities of data

Answer:

  1. Non impact – Laser printers, Inkjet
  2. Plotter
  3. Impact (DMP(Dot Matrix Printer))
  4. Dot Matrix Printers

Question 10.
“Not all primary memory is volatile”. Justify this statement.
Answer:
Primary Memory (Main memory) is classified into two RAM and ROM. Out of this RAM is volatile but ROM is non-volatile.

Question 11.
Categorise the softwares in the list according to the appropriate classifications given below.

  1. Classification: OS, Compiler, DTP Software, Compression software, Word processor
  2. List: Open Office Writer, Photoshop, 7 Zip, MS Word, Unix, C++, PageMaker, Winzip, C, Windows 98.

Answer:

  • OS – Unix, Windows 98
  • Compiler – C, C++
  • DTP Software – Photoshop, PageMaker
  • Compression – 7 Zip, Winzip
  • Word Processor – Open Office Writer, MS word

Question 12.
Your friend has just assembled a computer. Now he is provided with installation CD’s of MS Word and Microsoft Windows XP. In what order will he install them? Justify your answer.
Answer:
First he has to install the Microsoft Windows XP because it is the OS, it makes the computer to work other programmes. After that only he can install MS Word It is a package.

Question 13.
A group of 20 students is given a test in 6 subjects. The examiner wishes to prepare a neatly formatted mark list with total and Rank. Suggest a suitable software to serve this purpose. Give reasons.
Answer:
The spreadsheet Excel is a suitable software to serve this purpose. It consists of inbuilt functions, that facilitates to find total and rank easily.

Plus One Computer Science Chapter Wise Questions and Answers Chapter 3 Components of the Computer System

Question 14.
A program is written in BASIC, C and assembly language. Mention the difference in converting these programs to machine language.
Answer:

  1. C – Compiler
  2. BASIC – Interpreter
  3. Assembly Language – Assembler

Question 15.
Your friend told you that he has a system. What is a system? Explain.
Answer:
A computer is also called a system. A computer is not a single unit. It consists of more than one unit such as input unit, output unit, memory unit, ALU and control unit. Therefore a computer is called a system.

Question 16.
What is cache memory?
Answer:
A cache (pronounced cash) memory is a high-speed memory placed in between the processor and primary memory to reduce the speed mismatch between these two.

Question 17.
What is the use of program counter register?
Answer:
This register stores the memory address of the next instruction to be executed by the CPU.

Question 18.
What is HDMI?
Answer:
Its full form is High Definition Multimedia Interface. Through this port we can connect high definition quality video and multi channel audio over a single cable.

Question 19.
Give two examples for customized software.
Answer:

  1. Payroll System: It keeps track of details of employee and their salary details in an organisation
  2. Inventory Management System: It keeps tack of all about inventory in a company

Question 20.
What do you mean by free and open source s/w?
Answer;
Here “free” means there is no copy right 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

Question 21.

  1. What do you mean by cache memory? (1)
  2. Write the names of the figures given below.

Plus One Computer Science Chapter Wise Questions and Answers Chapter 3 Components of the Computer System 3

Answer:
1. It is a high speed memory placed in between the CPU and RAM. CPU is a high speed memory compared to RAM. There is a speed mismatch between the CPU and RAM to resolve this problem a high speed memory called cache memory is placed in between the CPU and RAM

2. QR code and Bar Code

Plus One Computer Science Chapter Wise Questions and Answers Chapter 3 Components of the Computer System

Question 22.
What is the role of students in e-Waste disposal?
Answer:
Student’s 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

Plus One Components of the Computer System Three Mark Questions and Answers

Question 1.
Why is the paper coming from the laser printer hot? Explain.
Answer:
Laser printer uses photocopying technology. It uses a positively charged drum and negatively charged toner (dry powder). A laser beam is used to scan the page to be printed on the drum with positive charge and then rolled through a reservoir of negatively charged toner.

It uses a combination of heat and pressure to adhere the dry powder to the paper. That is why, the paper coming from the laser printer is hot.

Question 2.
Explain the process how data from the hard disk is taken to the processor for processing.
Answer:
A processor is a high speed device. It can access data only from the Primary Memory (RAM). So we have to transfer data from hard disk to RAM. We know that a hard disk is a slow device also. So data is first transferred to RAM.

A RAM is comparatively slower than processor. To reduce the speed mismatch between the RAM and processor, the data has to transfer to CPU registers. Then the processor takes the data from the CPU register because CPU register has almost equal speed as processor.

Question 3.
Why computer is called as a system? (3)
Answer:
A computer is also called a system. A computer is not a single unit. It consists of more than one unit such as input unit, output unit, memory unit, ALU and control unit. Therefore a computer is called a system.

Question 4.
Differentiate hardware and software. (3)
Answer:
The tangible parts of a computer is called hard¬ware. We can see, touch and feel the hardware in a computer. The set of instructions that tell the hardware how to perform a task is called software. Without software computer cannot do anything.

Question 5.
We all have a brain. Just like this, is the computer has a brain? Explain it?
Answer:
Yes. CPU is the brain of a computer. The CPU comprises three parts ALU, Control Unit and Memory. The control unit control the overall functioning of a system. ALU performs all the arithmetic calculations and takes logical decisions. Memory is used for storage of data for future reference.

Question 6.
Explain the various functions of a control unit.
Answer:
The control unit performs the following functions.

  1. It controls data flow between input device ALU, memory and output devices.
  2. Normal execution of a program is line by line. The control unit controls this sequence with the help of ALU and memory.
  3. It controls the decoding and interpretation of in-structions.

Question 7.
Match the following

1. Input device a. Linux
2. Output device b. Java
3. Secondary Memory c. Joystick
4. System S/W d. Plotter
5. Application S/W e. PROM
6. Primary Memory f. HDD

Answer:
1 – c
2 – d
3 – f
4 – a
5 – b
6 – e

Question 8.
Write down the full form of the following.

  1. VDU
  2. OMR

Answer:

  1. VDU – Visual Display Unit
  2. OMR – Optical Mark Reader

Plus One Computer Science Chapter Wise Questions and Answers Chapter 3 Components of the Computer System

Question 9.
We know that a scanner is a hardware. What do you think of a virus scanner? Explain.
Answer:
We know that a scanner is a hardware but a virus scanner is not a hardware. It is a program. That is a virus scanner is an antivirus software. That scans your disk (HDD, CD, DVD, Pen Drive) for viruses and removes them (if it can), if any virus is found.

Question 10.
Your friend told you that a compiler is a hardware. Is it true? Justify your answer.
Answer:
It is not true. A compiler is not a hardware but it is a software. A compiler is a collection of programs that translates program written in HLL into machine language.

Question 11.
Anil purchased a product from a supermarket and he found that its wrapper contains light and dark bars. What is the purpose of this ?
Answer:
This light and dark bars are called bar code. It is used to record some details about the product such as item code, name, price etc. A device called Bar Code Reader contains photo electric scanner that read the bar code and input the information to the computer attached to it. It helps to reduce error and process the bills quickly.

Question 12.
What are the disadvantages of OMR? (2)
Answer:
The disadvantages are:

  1. It needs accurate alignment of printing on forms.
  2. It needs good quality expensive paper.

Question 13.
Differentiate CRT and LCD (OR) Your friend going to purchase a computer. He asked you which is better, CRT or LCD? What is your opinion?
Answer:
The difference between CRT and LCD is given below:

CRT LCD
It is heavy and bulky It is neither heavy nor bulky
It consumes more power and emits heat It consumes less power and does not emit heat
It is used in desk top computer It is used with laptop and desktop
It is cheaper It is expensive.

So LCD is more better than CRT

Question 14.
While you pressing “A” on the keyboard what is actually stored in the memory?
Answer:
The keyboard is an electro mechanical device that is designed to create electronic codes when a key is pressed and this code is transmitted to the memory through the cable. Here while you pressing “A” on the keyboard, the electronic codes corresponding to the ASCII value 01000001 is transmitting to the memory.

Question 15.
Your family friend started a supermarket. He asked you, which printer is suitable to print bills. Give your suggestion.
Answer:
According to my opinion, dot matrix printer is most suitable. Because they are capable of faster printing as well as it is cheap also. It’s printing quality is not good but cost per copy is very cheap. Dot matrix printer consists of a ribbon cartridge that is cheap and can be changed easily. Here we have to print more copies at a time. So dot matrix is suitable.

Question 16.
Suppose your brother is an engineer. He wants to draw some drawings. Which output device is suitable? Explain.
Answer:
Plotter is suitable for him. A plotter is a device that draws pictures ordiagrams 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 flatbed 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.

Question 17.
Match the following.

1. Operating System a. Compiler
2. Language Processor b. Windows Vista
3. Package Administratin S/W c. Santhi Hospital
4. Utility d. Spreadsheet
5. Customised S/W e. Disk defragmenter

Answer:
1 – b
2 – a
3 – d
4 – e
5 – c

Plus One Computer Science Chapter Wise Questions and Answers Chapter 3 Components of the Computer System

Question 18.
There are special purpose storage locations within the CPU. What are they explain ?
Answer:
Registers are special purpose storage locations within the CPU. They are temporary storage locations. The processing power of a CPU depends on register. Registers appears with storage capacity of 8 bits, 16 bits, 32 bits an 64 bits. They accept, store and transfer data from the CPU at a very high speed.

  • Program Counter: This register stores the memory address of the next instruction to be executed by the CPU.
  • Instruction Register: The instruction to be executed is stored in this register.
  • Memory Address Register: The address on the memory location from which data has to be read is stored here.
  • Memory Buffer Register: The data read from the location specified by the MAR is stored in this register.
  • General Purpose Registers: These are used to store the result and intermediate results during a processing.

Question 19.
What is an operating system?
Answer:
An operating system acts as an interface between user and computer without an operating system computer is a bare machine. That is without an OS computer cannot do anything. The OS not only makes the system convenient to use but also use hardware in an efficient manner,
eg: Windows XP, Vista, Linux, MS Dos, Windows 7.

Question 20.
We know that a computer only knows low level language and human beings use high level language. So how is it possible to communicate? Explain.
Answer:
The language processors translate the programs written in HLL into machine language which is understood by the computer. The different language processors are given below:
1. Assembler:
This language processor translates programs written in assembly language into machine language.

2. Interpreter:
This language processor translates programs written in HLL into machine language by converting and executing it line by line. If there is any error, the execution is stopped we, can continue after the correction of the program.

3. Compiler:
This language processor is same as interpreter. But it translates HLL into machine language by converting whole lines at a time. If there is any error, correcting all the errors then only it will execute.

Question 21.
Normally a CD contains 700 MB. Is it possible to store a file with size 1 GB? Explain.

OR

Normally a Car has a seating capacity of 5 persons including the driver. But some adjustments more persons can be accommodated in Car. This is connected with a utility. Which is the utility? Explain.
Answer:
Compression utility is used for this. By using compression utility programs we can reduce the file size upto the one third of the file size. So by using this we can reduce 1GB file and store in a CD. It is provided by the OS.

The other compression utility programs are Winzip, WinRar etc. It is possible to compress the files and when needed, these compressed files can be uncompressed and it is restored to their original form.

Question 22.
What is a Virus?
Answer:
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 be delete the useful system files and make the computer useless.

Question 23.
What do you mean by Utilities?
Answer:
Utilities are useful programs which are designed to help computer for its smooth functioning. Some utilities are back up utility, Disk defragmentation. Virus scanner, etc. It is provided by the O.S.

Plus One Computer Science Chapter Wise Questions and Answers Chapter 3 Components of the Computer System

Question 24.
Differentiate RAM and ROM.
Answer:
The difference between RAM and ROM is given below.

RAM ROM
1. It is Random Access Memory 1. It is Read only Memory
2. It is Read/Write memory 2. We can’t write but we can only read memory.
3. It is temporary 3. It is permanently stored.
4. It is volatile 4. It is non volatile
5. RAM is faster 5. It is slower
6. It is used to store data and instructions needed by CPU for processing 6. It contains instructions to check the hardware components, BIOS operations etc.
7. It is also called firmware.

Question 25.
Mention any two functions of OS.
Answer:
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.

Question 26.
Give two examples of human ware.
Answer:
(Write any two from the following). The term refers the persons who use computer System Administrator: It is a person who has central control over the computer systems.

  • System Managers: He is responsible for all business transactions with all vendors and contractors.
  • System Analysts: He is responsible to improve the productivity and efficiency.
  • Database Administrator: It is a person who has a central control over the DBMS.
  • Computer Engineer: The person design either h/w or s/w of a computer system.
  • Application Programmer: These are computer professionals who interact with the DBMS through programs.
  • Computer operators: He is an end user and does not know computer in detail.

Question 27.
Explain how e-waste creates environmental and health problems. What are the different methods for e-waste disposal? Which one is the most effective in your point of view? Why? (5)
Answer:
e-Waste(electronic waste): It refers to the malfunctioning electronic products such as faulty computers, mobile phones, tv sets, toys, CFL, batteries etc. It contains poisonous substances such as lead, mercury, cadmium etc and may cause diseases if not properly managed.

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. Landfilling: It is used to level pits and cover by thick layer of soil.

Plus One Computer Science Chapter Wise Questions and Answers Chapter 3 Components of the Computer System

Question 28.
What do you mean by e-waste? Explain the role of students in e-waste disposal.
Answer:
e-Waste(electronic waste): It refers to the malfunctioning electronic products such as faulty computers, mobile phones, tv sets, toys, CFLetc. It contains poisonous substances such as lead, mercury, cadmium etc and may cause diseases if not properly managed.
Student’s role in e-Waste disposal

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

Plus One Components of the Computer System Five Mark Questions and Answers

Question 1.
Write short notes about input devices.
Answer:
An input device is used to supply data to the computer. They are given below:
1. Key board:
It is the most widely used device to input information in the form of words, numbers etc. There are 101 keys on a standard key board. The keys on the key board are often classified into alpha numeric keys (A to Z, Oto 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 on a flat surface this ball also moves. This mechanical motion is converted into digi¬tal values that represents x and y values of the mouse movement.

3. Optical Mark Reader (OMR):
This device identifies the presence or absence of a pen or pen¬cil mark. It is used to evaluate objective type exams. In this method special preprinted forms r.e 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 of the pencil marks. By using this we can evaluate easily and reduce the errors.

4. 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.

5. 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.

6. Light Pen:
It is an input device that use a light sensitive 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.

7. 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.

8. 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.

9. Touchpad:
It is a pointing device found on the portable computers(laptop). 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.

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

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

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

13. 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.

Plus One Computer Science Chapter Wise Questions and Answers Chapter 3 Components of the Computer System

Question 2.
Briefly explain the various visual display units.
Answer:
The visual display units are given below:
1. Cathode Ray Tube (CRT):
There are two types of CRT’s, monochrome (Black and white) and colour. Monochrome CRT consists of one elec-tron 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.

2. 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.

3. Light Emitting Diocte(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.

4. 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.

5. 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.

Question 3.
Your friend wants to buy a printer. He wants to know more about printers. Explain different types of printers.
Answer:
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.

1. 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 5 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.

2. 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 with 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 ordiagrams 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 flatbed 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.

Question 4.
Your school got two printers. One dot matrix and one Laser printer through ICT scheme of Central Govt. What is the difference between these two printers? Explain.
Answer:

Laser Printer Dot Matrix Printer
It is non-impact printer It is impact printer
Speed is high Speed is less
Good quality text and Low quality text and very
picture poor quality picture
Less noise More noise
Printing cost is high Printing cost is low
Not possible to take Possible to take carbon
carbon copy copy
Toner is used Ribbon is used

Question 5.
Explain Primary Memory in detail.
Answer:
Primary memory is classified into two, Random Access Memory (RAM) and Read Only Memory (ROM). The primary memory holds the data which is to be processed by the CPU and the set of instructions to be executed next. The CPU can access the instructions in the primary memory only.

1. Random Access Memory (RAM):
RAM is used to store data and instructions needed by the CPU for processing. RAM can be used for both reading and writing of data so it is called Read and Write memory. It is a volatile memory, that is contents of the RAM will be lost when the power is turned off.

The invention of integrated circuits (chips) increased the memory capacity and reduced the size and cost. Static RAM, Dynamic RAM, Synchronous Dynamic RAM (SDRAM) are the various types of RAM.

2. Read Only Memory (ROM):
We can read this memory but we cannot write into this memory. It is a nonvolatile memory, that is its contents will not lost when the power is turned off. This memory is stored in the ROM chip at the time of manufacture ing itself hence it is called firmware.

ROM contains instructions to check the hardware components connected to the system, perform some basic input/output operations (BIOS), initiates loading of essential software. The different categories are PROM, EPROM and EE PROM.

(a) PROM (Programmable Read Only Memory):
It is just like WORM. That means the instructions are write once but read many. But we cannot change the instructions.

(b) EPROM (Erasable Programmable Read Only Memory):
It functions just like PROM. But by using ultraviolet light we can erase the old data and can write new data.

(c) EEPROM (Electrically Erasable Programmable Read Only Memory):
Here instead of ultraviolet light electric signals are used to erase old data and write new data under software control. It is highly expensive than regular ROM chips.

Plus One Computer Science Chapter Wise Questions and Answers Chapter 3 Components of the Computer System

Question 6.
Explain secondary memory in detail.

OR

To store large volume of data permanently. Which memory is used? Explain.
Answer:
Primary memory has a limited storage capacity and it is not permanent that is why to store large volume of data permanently secondary memory or storage devices are used. They are of many types and they vary in the capacity of storage, speed of data access and media of storage. Nowadays magnetic disks and optical disks are commonly used.

1. Magnetic Disk:
Magnetic disk allows the storage and retrieval for contents of the disk from anywhere at a moderate speed. Magnetic Disks available in various size and storage capacity but the storage media and data access mechanism are similar.

The storage media is circular platters or disks coated with magnetic material. It consists of a spindle capable of rotating with the help of an electrical motor and a read/write head.

(a) Floppy disk:
Floppy means flexible or soft, it uses flexible disk. Its storage capacity is 1,44MB and slower in data transfer rate.

(b) 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 mother board via cable.

Therefore we call it as hard disk drive (HDD). It contains one or more rigid platters coated both sides with a special magnetic material and a spindle. Datas are recorded on either surface of the disk except the outer side of last and first disk. Each surface will have one or more read/write heads (fixed head or movable head).

The spindle is attached to a motor that rotates at high speed typically 7200 rotation per minute (rpm). A floppy disk rotates only at 300 rpm.

2. 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. The optical disks are given below:

(a) Compact Disk Read Only Memory (CDROM):
The data in CDROM is imprinted.by the manufacturers. The user cannot erase or write on the disk but user can only read its contents. CDROM is written in a single continuous spiral unlike magnetic disks that uses concentric circles. Its storage capacity is 700MB.

(b) Erasable Optical Disk:
The disadvantage of CDROM is that we cannot change or erase the contents. But erasable disks can be changed and erased.

(c) Digital Versatile Disk: It is capable of storing upto 4.7GB and more faster.

(d) 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.

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

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

Question 7.
What do you mean by a computer software? Mention its different classification.

OR

Your friend wants to know more about software. Explain more about different classification.

OR

Your friend told you that COBOL and MS Word are same softwares. Do you agree with him. Explain. What is a software and its classification?
Answer:
A Software is a collection of programs to perform a task. The softwares can be classified into two major groups.

  1. System software
  2. Application software

Plus One Computer Science Chapter Wise Questions and Answers Chapter 3 Components of the Computer System 4
1. System Software:
It is a collection of programs used to manage system resources and control its operations. It is further classified into two.
(a) Operating System
(b) Language Processor
(a) 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: WindowsXP, Windows Vista, Linux, Windows 7, etc.

(b) 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, COBOL, PASCAL, VB, Java etc. HLL is very easy and can be easily understood by the human being.

Low level language are classifed into Assembly Language and Machine Language. In assembly language mnemonics (codes) are used to write programs
Plus One Computer Science Chapter Wise Questions and Answers Chapter 3 Components of the Computer System 5
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.

2. 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.
(a) Packages
(b) Utilities
(c) Customized Software

Plus One Computer Science Chapter Wise Questions and Answers Chapter 3 Components of the Computer System

(a) 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:
1. 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.

2. Spread Sheets:
It contains data or information in rows and columns and can perform calculation (Arithmetic, Relational and logi¬cal 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, Payroll etc. We can easily prepare graphs and charts using data entered in a worksheet. A file is a workbook that contains one or more worksheets, eg: MS Excel is a spreadsheet software.

3. 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.

4. Data base package:
Database is a collection of large volume of data. DBMS is a set of programs that manages the datas are for the centralised 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.

(b) 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 accidently 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. Vims Scanner:
It is a program called antivirus software scans the disk for viruses and removes them if any virus is found.

(c) Customised 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.

Question 8.
To use a computer not only the hardware but also software are required. Explain the classification of software.
Answer:
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 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:

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

Plus One Computer Science Chapter Wise Questions and Answers Chapter 3 Components of the Computer System

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
Plus One Computer Science Chapter Wise Questions and Answers Chapter 3 Components of the Computer System 6
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.

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.
(a) Packages
(b) Utilities
(c) Customized Software

(a) 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:

1. 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.

2. 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.

3. 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.

4. 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.

5. 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.

(c) 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.

Plus One Computer Science Chapter Wise Questions and Answers Chapter 3 Components of the Computer System

Question 9.
Describe the different types of memories and memory devices in computer with features and examples.
Answer:
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(MegaByte)
  • 1024 MB = 1 GB(Giga Byte)
  • 1024 GB = 1 TB(Tera Byte)
  • 1024 TB = 1 PB(Peta Byte)

Two Types of storage unit:
1. Primary Storage alias Main Memory:
It is further be classified into Two Random Access Memory (RAM) 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.

  1. PROM(Programmable ROM): It is programmed at the time of manufacturing and cannot be erased.
  2. EPROM (Erasable PROM): It can be erased and can be reprogrammed using special electronic circuit.
  3. 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.

(a) Magnetic storage device:
It uses plastic tape or metal/plastic discs coated with magnetic material.
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 mother board via cable.

(b) Optical storage device:
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.

1. 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

2. 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 laserto read and write but it uses Blue-Violet laser, hence the name Blu ray disc. The blue violet laser has shorter wave length than a red laser so it can pack more data tightly.

3. 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 varfes from 2 GB to 32 GB.
  • Flash memory cards: It is used in Camera, Mobile phones, tablets etc to store all types of data.

Question 10.
Explain how e-Waste creates environmental issues. Usually there are four methods for e-Waste dispoal. Which one is the most effective? Why? Write a slogan to aware the public about e-Waste hazards.
Answer:
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.

Plus One Computer Science Chapter Wise Questions and Answers Chapter 3 Components of the Computer System

Question 11.
With the help of a block diagram, explain the functional units of a computer.
Answer:
Functional units of computer:
A computer is not a single unit but it consists of many functional units(intended to perform jobs) such as Input unit, Central Processing Unit(ALU and Control Unit), Storage (Memory) Unit and Output Unit.
1. Input Unit:
Its aim is to supply data (Alphanumeric, image , audio, video, etc.) to the computer for processing. The Input devices are keyboard, mouse, scanner,mic, camera,etc

2. Central Processing Unit (CPU):
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.

3. 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)

Two Types of storage unit
(a) Primary Storage alias Main Memory:
It is further be classified into Two- Random Access Memory(RAM) 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)

(b) 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.

4. Output Unit:
After processing the data we will get information as result, that will be given to the end user through the output unit in a human readable form. Normally monitor and printer are used.

Plus One Maths Chapter Wise Questions and Answers Chapter 12 Introduction to Three Dimensional Geometry

Students can Download Chapter 12 Introduction to Three Dimensional Geometry Questions and Answers, Plus One Maths Chapter Wise Questions and Answers helps you to revise the complete Kerala State Syllabus and score more marks in your examinations.

Kerala Plus One Maths Chapter Wise Questions and Answers Chapter 12 Introduction to Three Dimensional Geometry

Plus One Maths Introduction to Three Dimensional Geometry Three Mark Questions and Answers

Question 1.
Prove by using distance formula that the A(1, 2, 3), B(-1, -1, -1) and C(3, 5, 7) are collinear.
Answer:
Plus One Maths Introduction to Three Dimensional Geometry Three Mark Questions and Answers 1
Now BC = AB + AC
Thus A, B, C are collinear.

Plus One Maths Chapter Wise Questions and Answers Chapter 12 Introduction to Three Dimensional Geometry

Question 2.
Verify the following: (3 score each)

  1. (0, 7, -10), (1, 6, -6) and (4, 9, -6) are the vertices of an isosceles triangle.
  2. (0, 7, 10), (-1, 6, 6) and (-4, 9, 6) are the vertices of a right angled triangle.
  3. (-1, 2, 1), (1, -2, 5), (4, -7, 8) and (2, -3, 4) are the vertices of a parllelogram.

Answer:
1. Let A(0, 7, -10), B(1, 6, -6) and C(4, 9, -6) be the two points.
Plus One Maths Introduction to Three Dimensional Geometry Three Mark Questions and Answers 2
Now AB = BC, thus ABC is an isosceles triangle.

2. Let A(0, 7, 10), B(-1, 6, 6) and C(-4, 9, 6) be the two points.
Plus One Maths Introduction to Three Dimensional Geometry Three Mark Questions and Answers 3
Plus One Maths Introduction to Three Dimensional Geometry Three Mark Questions and Answers 4
Now AC2 = AB2 + BC2, thus ABC is a right triangle.

3. Let A(-1, 2, 1), B(1, -2, 5), C(4, -7, 8) and D(2, -3, 4) be the two points.
Plus One Maths Introduction to Three Dimensional Geometry Three Mark Questions and Answers 5
Now AB = CD, BC = AD, AC ≠ BD, thus A, B, C, D are vertices of a parallelogram.

Plus One Maths Chapter Wise Questions and Answers Chapter 12 Introduction to Three Dimensional Geometry

Question 3.
Find the equation of set points which are equidistant from the points (1, 2, 3) and (3, 2, -1).
Answer:
Let P(x,y,z) be any point which is equidistant from the point A(1, 2, 3) and B (3, 2, -1).
Given; PA = PB
Plus One Maths Introduction to Three Dimensional Geometry Three Mark Questions and Answers 6
(x – 1)2 + (y – 2)2 + (z – 3)2
= (x – 3)2 + (y – 2)2 + (z + 1)2
= x2 – 2x + 1 + y2 – 4y + 4 + z2 – 6z + 9
= x2 – 6x + 9 + y2 – 4y + 4 + z2 + 2z + 1 – 2x + 14 – 6z = -6x + 14 + 2z
⇒ 4x – 8z = 0
⇒ x – 2z = 0.

Question 4.
Find the coordinate of the point which divides the line segment joining the points (3, -2, 5) and (3, 4, 2) in the ratio 2:1 (3 score each)

  1. 2:1 internally
  2. 2:1 externally

Answer:
1. Let P(x, y, z) be any point which divides the line segment joining points A(3, -2, 5) and B (3, 4, 2) in the ratio 2:1 internally.
Plus One Maths Introduction to Three Dimensional Geometry Three Mark Questions and Answers 7
Therefore coordinates of P are (3, 2, 3).

2. Let P(x, y, z) be any point which divides the line segment joining points A(3, -2, 5) and B (3, 4, 2) in the ratio 2:1 externally.
Plus One Maths Introduction to Three Dimensional Geometry Three Mark Questions and Answers 8
Therefore coordinates of P are (3, 10, -1).

Plus One Maths Chapter Wise Questions and Answers Chapter 12 Introduction to Three Dimensional Geometry

Question 5.
Find the ratio in which the line joining the points (1, 2, 3) and (-3, 4, -5) is divided by the xy-plane.
Answer:
Let the line joining the points A(1, 2, 3) and B(-3, 4, -5) is divided by the xy-plane in the ratio k:1.
Then the coordinate
Plus One Maths Introduction to Three Dimensional Geometry Three Mark Questions and Answers 9
Since the point lies on xy-axis, we have;
Plus One Maths Introduction to Three Dimensional Geometry Three Mark Questions and Answers 10
Thus the required ratio is \(\frac{3}{5}\); ie, 3:5.

Question 6.
Find the coordinates of the points which trisect the line segment joining the points P(4, 2, -6) and Q (10, -16, 6).
Answer:
Plus One Maths Introduction to Three Dimensional Geometry Three Mark Questions and Answers 11
Let R and S be two points which trisect the line join of PQ. Therefore PR = RS = SQ Then coordinate of R is
Plus One Maths Introduction to Three Dimensional Geometry Three Mark Questions and Answers 12
= (6, -4, -2)
Then coordinate of S is
Plus One Maths Introduction to Three Dimensional Geometry Three Mark Questions and Answers 13
= (8, -10, 2).

Plus One Maths Introduction to Three Dimensional Geometry Practice Problems Questions and Answers

Question 1.
Find the distance between the following pair of points: (1 score each)

  1. (2, 3, 5) and (4, 3, 1)
  2. (-3, 7, 2) and (2, 4, -1)
  3. (-1, 3, -4) and (1, -3, 4)

Answer:
1. Let A(2, 3, 5) and B(4, 3, 1) be the two points.
Plus One Maths Introduction to Three Dimensional Geometry Practice Problems Questions and Answers 14

2. Let A(-3, 7, 2) and B(2, 4, -1) be the two points.
Plus One Maths Introduction to Three Dimensional Geometry Practice Problems Questions and Answers 15

Plus One Maths Chapter Wise Questions and Answers Chapter 12 Introduction to Three Dimensional Geometry

3. Let A(-1, 3, -4) and B(1, -3, 4) be the two points.
Plus One Maths Introduction to Three Dimensional Geometry Practice Problems Questions and Answers 16

Plus One Maths Chapter Wise Questions and Answers Chapter 6 Linear Inequalities

Students can Download Chapter 6 Linear Inequalities Questions and Answers, Plus One Maths Chapter Wise Questions and Answers helps you to revise the complete Kerala State Syllabus and score more marks in your examinations.

Kerala Plus One Maths Chapter Wise Questions and Answers Chapter 6 Linear Inequalities

Plus One Maths Linear Inequalities Three Mark Questions and Answers

Question 1.
Solve the following inequalities.

  1. \(\frac{1}{2}\left(\frac{3 x}{5}+4\right) \geq \frac{1}{3}(x-6)\)
  2. \(\left(\frac{2 x-1}{3}\right) \geq \frac{(3 x-2)}{4}-\frac{(2-x)}{5}\) (3 score each)

Answer:
1. Given;
Plus One Maths Linear Inequalities Three Mark Questions and Answers 1
⇒ 3(3x + 20) ≥ 10(x – 6)
⇒ 9x + 60 ≥ 10x – 60
⇒ 9x – 10x ≥ -60 – 60
⇒ -x ≥ -120 ⇒ x ≤ 120

2. Given;
Plus One Maths Linear Inequalities Three Mark Questions and Answers 2
⇒ 20(2x -1) ≥ 3[15x – 10 – 8 + 4x]
⇒ 40x – 20 ≥ 45x – 54 + 12x
⇒ 40x – 20 ≥ 57x – 54
⇒ 40x – 57x ≥ -54 + 20
⇒ -17x ≥ -34 ⇒ x ≤ 2

Plus One Maths Chapter Wise Questions and Answers Chapter 6 Linear Inequalities

Question 2.
1. Which of the following sets of inequality represent the second quadrant? (1)
(a) x < 0, y < 0
(b) x > 0, y > 0
(c) x < 0, y > 0
(d) x > 0, y < 0
2. Write the system of inequalities that represents the shaded rectangle in the figure given below: (2)
Plus One Maths Linear Inequalities Three Mark Questions and Answers 3
Answer:
1. (a) x < 0, y < 0

2. The shaded figure is a rectangle. The side parallel to x axis are y = -1 and y = 1. The side perpendicular to x axis are x = 2 and x = -2. Hence the inequality that represent the shaded region are
-2 ≤ x ≤ 2; -1 ≤ y ≤ 1.

Plus One Maths Chapter Wise Questions and Answers Chapter 6 Linear Inequalities

Question 3.
Find all pairs of consecutive even positive integers both of which are smaller than10 such that their sum is less than 23.
Answer:
Consecutive even positive integers be x and x + 2. Then, x + x + 2 < 23; x + 2 < 10
⇒ 2x < 23 – 2; x < 10 – 2
⇒ x < \(\frac{21}{2}\) = 10.5; x < 8
⇒ 8 < x ≤ 10 Therefore x can take values 9, 10. Hence the pairs are (9, 10), (10, 9).

Question 4.
The longest side of a triangle is 3 times the shortest side and the third side is 2 cm shorter than the longest side. If the perimeter of the triangle is at least 61 cm, find the minimum length of the shortest side.
Answer:
Let the shortest side is x, then;
Longest side = 3x;
Third side = Longest side – 2 = 3x – 2
Perimeter = 3x + 3x – 2 + x ≥ 61 ⇒ 7x – 2 ≥ 61
⇒ 7x ≥ 61 + 2 ⇒ x ≥ \(\frac{63}{7}\) = 9.

Plus One Maths Linear Inequalities Four Mark Questions and Answers

Question 1.
Solve the following system of inequalities graphically.

  1. 2x – y > 1; x – 2y < -1
  2. x + y ≤ 9; y > x; x ≥ 0
  3. x – 2y ≤ 3; 3x + 4y ≥ 12; x ≥ 0, y ≥ 1
  4. 2x + y – 3 ≥ 0; x – 2y + 1 ≥ 0; y ≤ 3 (4 score each)

Answer:
1. 2x – y > 1; x – 2y < -1
Plus One Maths Linear Inequalities Four Mark Questions and Answers 4
Plus One Maths Linear Inequalities Four Mark Questions and Answers 5

Plus One Maths Chapter Wise Questions and Answers Chapter 6 Linear Inequalities

2. x + y ≤ 9; y > x ⇒ x – y < 0
Plus One Maths Linear Inequalities Four Mark Questions and Answers 6
Plus One Maths Linear Inequalities Four Mark Questions and Answers 7

3. x – 2y ≤ 3; 3x + 4y ≥ 12
Plus One Maths Linear Inequalities Four Mark Questions and Answers 8
Plus One Maths Linear Inequalities Four Mark Questions and Answers 9

Plus One Maths Chapter Wise Questions and Answers Chapter 6 Linear Inequalities

4. 2x + y ≥ 3; x – 2y ≤ -1
Plus One Maths Linear Inequalities Four Mark Questions and Answers 10
Plus One Maths Linear Inequalities Four Mark Questions and Answers 11

Plus One Maths Linear Inequalities Practice Problems Questions and Answers

Question 1.
Solve the following inequalities.

  1. 4x + 3 < 5x + 7
  2. 3(x – 1) ≤ 2(x – 3) (1 score each)

Answer:
1. Given; 4x + 3 < 5x + 7
⇒ 4x – 5x < 7 – 3 ⇒ -x < 4 ⇒ x > -4.

2. Given; 3(x – 1) < 2(x – 3)
⇒ 3x – 3 ≤ 2x – 6 ⇒ 3x – 2x ≤ -6 + 3
⇒ x ≤ -3.

Plus One Maths Chapter Wise Questions and Answers Chapter 6 Linear Inequalities

Question 2.
Solve the inequality \(\frac{3(x-2)}{5} \leq \frac{5(2-x)}{3}\).
Answer:
Given; \(\frac{3(x-2)}{5} \leq \frac{5(2-x)}{3}\)
⇒ 9(x – 2) ≤ 25(2 – x)
⇒ 9x – 18 ≤ 50 – 25x
⇒ 9x + 25x ≤ 50 + 18
⇒ 34x ≤ 68 ⇒ x ≤ 2.

Question 3.
Show the solution of each inequality on a number line.

  1. 4x + 3 < 6x + 7
  2. 5x – 3 ≥ 3x – 5
  3. 3(1 – x) < 2(x + 4)
  4. 2 – 3x < 2(x + 6)
  5. -3 ≤ 3 – 2x < 6 (2 score each)

Answer:
1. Given; 4x + 3 < 6x + 7 ⇒ 4x – 6x < 7 – 3
⇒ -2x < 4 ⇒ x > -2
Plus One Maths Linear Inequalities Practice Problems Questions and Answers 12

2. Given; 5x – 3 > 3x – 5 ⇒ 5x – 3x ≥ -5 + 3
⇒ 2x ≥ -2 ⇒ x ≥ 1.
Plus One Maths Linear Inequalities Practice Problems Questions and Answers 13

3. Given; 3(1 – x) < 2(x + 4) ⇒ 3 – 3x < 2x + 8
⇒ -3x – 2x < 8 – 3 ⇒ -5x < 5 ⇒ x > -1.
Plus One Maths Linear Inequalities Practice Problems Questions and Answers 14

4. Given; 2 – 3x < 2(x + 6) ⇒ 2 – 3x < 2x + 12
⇒ -3x – 2x < 12 – 2 ⇒ -5x < 10 ⇒ x > -2
Plus One Maths Linear Inequalities Practice Problems Questions and Answers 15

5. Given; -3 ≤ 3 – 2x < 6
⇒ -3 ≤ 3 – 2x; 3 – 2x < 6
⇒ -3 – 3 ≤ -2x; -2x < 6 – 3
⇒ -6 ≤ -2x; -2x < 3 ⇒ 3 ≥ x; x > \(-\frac{3}{2}\) ⇒ \(-\frac{3}{2}\) < x ≤ 3
Plus One Maths Linear Inequalities Practice Problems Questions and Answers 16

Plus One Maths Chapter Wise Questions and Answers Chapter 6 Linear Inequalities

Question 4.
The marks obtained by a student of class XI in first and second terminal examination are 62 and 48, respectively. Find the minimum marks he should get in the annual examination to have an average of at least 60 marks.
Answer:
Let x denote the mark obtained by the student in Class XI examination, then;
\(\frac{62+48+x}{3}\) ≥ 60 ⇒ 110 + x ≥ 1800 ⇒ x ≥ 70.

Plus One Accountancy Chapter Wise Questions and Answers Chapter 7 Bill of Exchange

Students can Download Chapter 7 Bill of Exchange Questions and Answers, Plus One Accountancy Chapter Wise Questions and Answers helps you to revise the complete Kerala State Syllabus and score more marks in your examinations.

Kerala Plus One Accountancy Chapter Wise Questions and Answers Chapter 7 Bill of Exchange

Plus One Accountancy Bill of Exchange One Mark Questions and Answers

Question 1.
The Maker of the bill of exchange is called the ……….
(a) Drawer
(b) Drawee
(c) Payee
Answer:
(a) Drawer.

Question 2.
Bill of exchange before its acceptance is called is ……………
(a) Bill
(b) Promissory Note
(c) Draft
Answer:
(c) Draft.

Question 3.
When a discounted bill is dishonoured, the ………… account is credited in the books of the drawer,
(a) Bank
(b) Drawee
(c) Payee
Answer:
(a) Bank

Plus One Accountancy Chapter Wise Questions and Answers Chapter 7 Bill of Exchange

Question 4.
A bill is noted when it is
(a) Dishonoured
(b) Honoured
(c) Discounted
(d) Accepted
Answer:
(a) Dishonoured

Question 5.
The process of transferring the ownership of the bill is called
(a) Acceptance
(b) Negotiation
(c) Endorsement
Answer:
(c) Endorsement

Question 6.
The credit instrument which contains a promise made by the debtor to pay a certain sum of money for value received is
(a) Bill of Exchange
(b) Debenture
(c) Promissory Note
(d) Equity Share
Answer:
(c) Promissory Note

Question 7.
A bill of exchange is an Instrument.
Answer:
Negotiable

Question 8.
………………. days of grace are allowed in case of time bills for calculating the date of maturity.
Answer:
Three

Question 9.
If the date of maturity of a bill is on a holiday then the bill will mature on ………….. day.
Answer:
The Previous day

Question 10.
When noting charges are paid finally the amount will
be recovered from
Answer:
Drawee.

Plus One Accountancy Chapter Wise Questions and Answers Chapter 7 Bill of Exchange

Question 11.
Complete the following on the basis of hints given

  1. Dishonour of discounted bill – An entry in the book of drawer.
  2. ………………………………………………. – No entry in the book of drawer.

Answer:
Honouring of discounted bill.

  1. Noting charges incurred – When bill is dishonoured.
  2. Rebate is allowed – …………………….

Answer:
When bill is honoured before due date.

Question 12.
Bill of exchange in Indian languages is called
Answer:
Hundi.

Question 13.
The person to whom the amount mentioned in the promissory note is payable is known as ………………..
Answer:
Promisee.

Question 14.
A person who endorse the promissory note in favour of another is known as ………………..
Answer:
Endorser

Question 15.
In a promissory note, the person who makes the promise to pay is called ……………………
Answer:
Promissor.

Question 16.
Bill of exchange is drawn on the ………….
Answer:
Debtor/Drawee.

Question 17.
The person to whom payment of the bill is to be made is known as ……………….
Answer:
Payee

Question 18.
A promissory note does not require ………..
Answer:
Acceptance.

Question 19.
Making payment of the bill of exchange before the due date is called …………..
Answer:
Retiring of the bill

Question 20.
A bill of exchange accepted without consideration, just to oblige a friend is known as
Answer:
Accommodation Bill

Question 21.
If the proceeds of the bill of exchange is to be paid after a particular period is called …………
Answer:
Time Bill.

Plus One Accountancy Chapter Wise Questions and Answers Chapter 7 Bill of Exchange

Question 22.
Find the odd one and state the reasons.

  1. Bill of exchange, cheque, promissory note, fixed deposit receipt.
  2. Drawer, Drawee, Payee, Endorser.

Answer:

  1. Fixed Deposit Receipt – Others are negotiable instruments
  2. Drawee – Others may be the same person

Question 23.
On the date of maturity, Arun (acceptor) requested to Santhosh (drawer) to cancel the old bill and draw a new bill upon him for a period of 2 months. Santhosh agrees to this. It is a case of ………………
Answer:
Renewal of bill

Plus One Accountancy Bill of Exchange Two Mark Questions and Answers

Question 1.
A bill of exchange must contain “an unconditional promise to pay.” Do you agree with a statement?
Answer:
No. The bill of exchange contains an unconditional order to pay a certain amount on an agreed date.

Question 2.
Define Bill of exchange.
Answer:
According to the Negotiable Instruments Act, 1881 a bill of exchange is “an instrument in writing containing an unconditional order, signed by the maker, directing a certain person to pay a certain sum of money only to, or to the order of a certain person, or to the bearer of the instrument.”

Question 3.
What are the features of bill of exchange?
Answer:
Following are the essential features of a bill of exchange.

  1. It must be in writing.
  2. It must be an order to pay, and not a request to pay.
  3. No condition should be attached to the order.
  4. The drawer must sign the bill
  5. The order must be for the payment of money only.
  6. It should be properly stamped.
  7. The amount mentioned in the bill may be made payable either on demand or after the expiry of a ‘stipulated period.

Question 4.
Who are.the parties to a bill of exchange ?
Answer:
There are three parties to a bill of exchange:

  1. Drawer – He is the creditor who draws a bill of exchange upon the debtor.
  2. Drawee – He is the person upon whom the bill of exchange is drawn. He is the purchaser of the goods on credit and the debtor.
  3. Payee – He is the person to whom payment of the bill is to be made on the maturity date. The drawer and the payee can be one party when payment is to be made to the drawer.

Question 5.
What are days of grace?
Answer:
Three extra days over the nominal due date legally given to the acceptor of a bill to make payment are called days of grace. Days of grace are allowed only in the case of time bills.

Question 6.
Calculate the maturity date of the following bill.

  1. Drawn on January 5th for three months.
  2. Drawn on May 1st for 4 months.

Answer:
1.

  • January 5th to February 5th
  • February 5th to March 5th
  • March 5th to April 5th
  • April 5th + 3 days of grace = April 8th

2.

  • May 1st – June 1st
  • June 1st – July 1st
  • July 1st – August 1st
  • August 1st – September 1st
  • September 1st + 3 days of grace = September 4th

Question 7.
What do you mean by Endorsement?
Answer:
An endorsement is a written order on the back of the instrument by the payee or the holder, for transferring his right to another person. The person who makes the endorsement is called endorser and in whose favour the endorsement is made is called the endorsee.

Plus One Accountancy Chapter Wise Questions and Answers Chapter 7 Bill of Exchange

Question 8.
What is an Accommodation Bill?
Answer:
A bill of exchange and Promissory note may be used for raising funds temporarily. Such a bill is called an ‘accommodation bill’ as it is accepted by the drawee to accommodate the drawer. These are drawn and accepted without consideration with a view to provide funds to one or more parties. There is no trade or debtor-creditor pfifationsip between parties. It is also known as ‘kite bill’ or ‘Wind bill’.

Question 9.
Ram received a bill from Anil on 1/7/2009 for 3 months for Rs. 4,000. Later the bill has been endorsed to Kumar. In this statement identify the parties involved in terms of drawer, drawee endorser and payee.
Answer:

  • Drawer – Ram
  • Drawee – Anil
  • Payee – Kumar
  • Endorsee – Kumar

Question 10.
Manu purchased goods on credit from Kumar for Rs. 40,0 on 1st April 2009. On the same date Kumar draws a bill for 2 months and got it accepted by Manu. What are the options available to Kumar in dealing with the bill?
Answer:
The following options are available to deal with the bill.

  1. He can retain the bill till the date of maturity.
  2. He can get the bill discounted through the bank.
  3. He can endorse the bill in favour of his creditor.
  4. He may sent the bill for collection to the bank.

Question 11.
Mr. Mohan holds a bill of exchange. He approaches a bank to receive the amount of bill before the maturity date. Does he get the money? Write your comments,
Answer:
If the drawer of the bill needs cash immediately or before the maturity date he can discount the bill with the bank. Discounting of the bill means encashing the bill with the banker on the security of the bill before the maturity date. The banker will deduct a certain sum from the bill amount as discount and pays the balance to the holder. The bank will present the bill to the drawee on the due date and get the payment of the bill.

Question 12.
Explain the term “Noting”
Answer:
When a bill of exchange is dishonoured due to nonpayment, it is usual to get it ‘noted’, to establish the matter of dishonour legally. The noting is done by the “Notary public”, who is an officer appointed by the Government for this purpose. Noting authenticates the facts of dishonour.

For providing this service, a number of fees is charged which is called “Noting charges” Noting charges either paid by the holder or any other parties should be borne by the acceptor in the absence of any agreement.

Question 13.
What do you mean by ‘Retiring of Bill’?
Answer:
The acceptor can pay the amount of the bill before its due date. The process of paying the amount of the bills payable before the due date is called retiring the bill. In such case, the holder usually allows some discount to the acceptor of bill. Such a discount is called “rebate on retired bill”. The amount of rebate depends on the period that the bill has yet to run.

Question 14.
What do you mean by Renewal of Bills?
Answer:
process of drawing and accepting a new bill by cancelling the old bill is called renewal of a bill. It is a granting of extension of credit to the acceptor. When this is done the acceptor may have to pay interest for the extended period of credit. It is paid in cash or may be included in the amount of the new bill.

Question 15.
On 1.1.2005 Raju sold goods to Anil for Rs. 20,000/ – and draw upon him a bill for 2 months. Anil accepted the bill and returned it to Raju. Later Raju endorsed the bill to Tom. On the date of maturity the bill was dishonoured.
Give advice to Tom, regarding the steps to be undertaken immediately after the dishonour.
Answer:
1. The holder of the bill, Tom should send a notice of dishonour to the drawer, Raju within a reasonable time. Otherwise, the other parties of the bill may deny their liability.

2. When the bill is dishonoured it is better to get the fact noted with a Notary public.

Plus One Accountancy Chapter Wise Questions and Answers Chapter 7 Bill of Exchange

Question 16.
Identify the endorser, endorsee and type of endorsement from the following
Plus One Accountancy Chapter Wise Questions and Answers Chapter 7 Bill of Exchange 1
Answer:

  • Endorser – Santhosh
  • Endorsee – Arunkumar
  • Type – Conditional Endorsement

Plus One Accountancy Bill of Exchange Three Mark Questions and Answers

Question 1.
Define promissory Note and Point out its Features.
Answer:
According to the Negotiable Instruments Act 1881, a promissory note is “an instrument in writing containing an unconditional undertaking, signed by the maker, to pay a certain sum of money only to or to the order of a certain person, or to the bearer of the instrument.”
The following are the features of promissory note

  1. It must be in writing
  2. It must contain an unconditional promise to pay.
  3. It must be signed by the maker
  4. The person to whom payment is to be made must also be certain.
  5. The amount payable must be certain.
  6. It should be properly stamped.

Question 2.
Who are the parties to a promissory note:
Answer:
There are two parties to a promissory note.
1. The maker or the promisor:
He is the person who makes or draws the promissory notable is the debtor.

2. The payee or the promisee:
He is a person in whose favour the promissiory note is drawn.

Question 3.
Syam sold goods to Anil on credit for Rs. 3,000, on 10th April 2003. He drew a bill on Anil for the amount at 3 months after date. Anil accepted the same and returned it to syam. At maturity, he met his obligation. Pass journal entries in the books of both parties.
Answer:
Book of Syam (Drawer) Journal
Plus One Accountancy Chapter Wise Questions and Answers Chapter 7 Bill of Exchange 2
Book of Anil [Drawee] Journal
Plus One Accountancy Chapter Wise Questions and Answers Chapter 7 Bill of Exchange 3

Question 4.
Haridas discounts a bill for Rs. 10,000 with his banker on 4th February, 2007. The bill was drawn on 1st January for Four months. The discount rate was 12%. p.a. write Journal entries in the book of Haridas.
Answer:
Book of Haridas [Drawer] Journal
Plus One Accountancy Chapter Wise Questions and Answers Chapter 7 Bill of Exchange 4
Plus One Accountancy Chapter Wise Questions and Answers Chapter 7 Bill of Exchange 5
Note: Calculation of discount
Discount = 10.000 × 12/100 × 3/12 = 300 It may be noted that discount has been charged for three months and not for 4 months because the bank will have to wait for 3 months to get the payment of bill on the due date.

Question 5.
From the given specimen of a promissory note identify:-
a) Promisor
b) Promisee
c) Consideration
Plus One Accountancy Chapter Wise Questions and Answers Chapter 7 Bill of Exchange 6
Answer:
Plus One Accountancy Chapter Wise Questions and Answers Chapter 7 Bill of Exchange 7

Plus One Accountancy Bill of Exchange Four Mark Questions and Answers

Question 1.
On January 1st, 2006, Ramesh sold goods worth ₹2,000 to Kannan and drew abill for the amount for 3 months. Kannan accepted the bill and returned it to Ramesh. Ramesh endorsed the bill in favour of Jayan, on 6th January. The bill was duly met on maturity. Pass journal entries in the book of Ramesh, Kannan and Jayan.
Answer:
Book of Ramesh [Drawer] Journal
Plus One Accountancy Chapter Wise Questions and Answers Chapter 7 Bill of Exchange 8
Book of Kannan [Drawee] Journal
Plus One Accountancy Chapter Wise Questions and Answers Chapter 7 Bill of Exchange 9
Book of Jayan [Endorser] Journal
Plus One Accountancy Chapter Wise Questions and Answers Chapter 7 Bill of Exchange 10

Plus One Accountancy Chapter Wise Questions and Answers Chapter 7 Bill of Exchange

Question 2.
On 10th March 2004, Chandran drew and Devan accepted a 2 months bill for Rs. 2000. On April 11 Chandran sends the bill to his bank for collection. On due date the amount is collected. Give journal entries in the book of both the parties.
Answer:
Journal (In the book of Chandran)
Plus One Accountancy Chapter Wise Questions and Answers Chapter 7 Bill of Exchange 11
Journal (In the book of Devan)
Plus One Accountancy Chapter Wise Questions and Answers Chapter 7 Bill of Exchange 12
Plus One Accountancy Chapter Wise Questions and Answers Chapter 7 Bill of Exchange 13

Question 3.
On 1st June 2005, ‘A’ Sold goods to ‘B’ worth Rs.
20,000 and drew on him a bill for 3 months. ‘B’ accepted the same and returned to ‘A’. On 5th June 2005, ‘A’ discounted the bill with his banker and received Rs. 19,000. On the due date, ‘B’ failed to pay the amount and the bill got dishonoured. Pass journal entries in the books of A and B.
Answer:
Book of A [Drawer] Journal
Plus One Accountancy Chapter Wise Questions and Answers Chapter 7 Bill of Exchange 14
Book of B [Drawee] Journal
Plus One Accountancy Chapter Wise Questions and Answers Chapter 7 Bill of Exchange 15

Question 4.
Identify the document shown below and state Any 6 features of the document.
Plus One Accountancy Chapter Wise Questions and Answers Chapter 7 16
Answer:
Document is a Bill of Exchange.
Features:

  1. It contains an order to pay money.
  2. The order shortly be unconditional
  3. The drawer must sign the bill
  4. The drawee must be a certain person.
  5. The order must be for the payment of money only.
  6. It should be properly stamped.

Question 5.
From the given specimen of Bill of exchange, identify:

  1. Drawer
  2. Drawee
  3. Payee
  4. Term of bill
  5. Date of maturity
  6. Consideration on the bill

Plus One Accountancy Chapter Wise Questions and Answers Chapter 7 Bill of Exchange 17
Answer:
Plus One Accountancy Chapter Wise Questions and Answers Chapter 7 Bill of Exchange 18

Plus One Accountancy Chapter Wise Questions and Answers Chapter 7 Bill of Exchange

Question 6.
Complete the following table.
Plus One Accountancy Chapter Wise Questions and Answers Chapter 7 Bill of Exchange 19
Answer:

  1. Debtor
  2. Order or promise, order to make payment
  3. Needs acceptance by drawee/does not need any acceptance.
  4. Three – Drawer, Drawee, Payee Two – Drawer and payee
  5. Drawer has secondary liability.

Question 7.
Complete the journal on the basis of the narration given for a bill of exchange of Rs. 20,000/-
Book of Santhosh Journal
Plus One Accountancy Chapter Wise Questions and Answers Chapter 7 Bill of Exchange 20
Answer:
Journal Book of Santhosh
Plus One Accountancy Chapter Wise Questions and Answers Chapter 7 Bill of Exchange 21

Plus One Accountancy Bill of Exchange Five Mark Questions and Answers

Question 1.
Ajay sold goods to Babu for Rs. 2,700 on 1.1.2007 and immediately drawn a bill for 4 months upon Babu for the same amount Babu accepted the bill and returned it to Ajay. On the 4th February 2007, Babu retires his acceptance under a rebate of 12% p.a. Give journal entries in the books of both the parties.
Answer:
Book of Ajay [Drawer] Journal
Plus One Accountancy Chapter Wise Questions and Answers Chapter 7 Bill of Exchange 22
Plus One Accountancy Chapter Wise Questions and Answers Chapter 7 Bill of Exchange 23
Book of Babu [Drawee] Journal
Plus One Accountancy Chapter Wise Questions and Answers Chapter 7 Bill of Exchange 24

Question 2.
In a classroom debate Mohan argued that a bill of exchange and promissory note are same but syam disagree with him and states that they are different, whose argument is correct? Give reason.
Answer:
The argument of Syam is correct. There are some defference between bill of exchange and promissory note.
Plus One Accountancy Chapter Wise Questions and Answers Chapter 7 Bill of Exchange 25

Question 3.
On January 15th, 2003 ‘N’ draw a bill of Rs. 8000 on his debtor ‘M’ for two months. By the due date of the bill, ‘M’ became insolvent and a dividend of 50 paise in the rupee was received from his estate. Pass Journal entries in the books of N and M.
Answer:
Journal (In the book of ‘N’)
Plus One Accountancy Chapter Wise Questions and Answers Chapter 7 Bill of Exchange 26
Plus One Accountancy Chapter Wise Questions and Answers Chapter 7 Bill of Exchange 27

Plus One Accountancy Chapter Wise Questions and Answers Chapter 7 Bill of Exchange

Question 4.
Identify the types of endorsement from the given format and write short notes on each type.
Plus One Accountancy Chapter Wise Questions and Answers Chapter 7 Bill of Exchange 28
Answer:

  1. Blank Endorsement: The endorser simply puts his signature, without mentioning the name of the endorsee.
  2. Special Endorsement: The endorser mentions the name of the endorsee along with his signature on the back of the Instrument.
  3. Restrictive Endorsement: Restrictlnq further endorsement.
  4. Sans – Recourse Endorsement: If the bill is dishonoured the holder cannot have recourse to the endorser.
  5. Facultative Endorsement: Endorsement made by waiving some right of the endorser.

Question 5.
The transaction between Rajesh and Murali are journalised below. Identify the drawer, drawee/acceptor, amount and term of bill. Also write appropriate narration to each transaction.
Book of Murali Journal
Plus One Accountancy Chapter Wise Questions and Answers Chapter 7 Bill of Exchange 29
Answer:

  1. Drawer – Rajesh
  2. Drawee – Murali
  3. Amount of bill – 10,000/-
  4. Term of bill – 3 Months
  5. Narration:
    • Purchased goods on credit
    • Acceptance given on bill
    • Bill amount paid on the due date

Plus One Accountancy Bill of Exchange Six Mark Questions and Answers

Question 1.
On 5th January 2003, Balu sold goods to Raju for Rs. 2500. Balu drew a 2 months bill on Raju. Raju accepted the bill and returned it to Balu. On 5th March Raju approached Balu with a request to renew the bill for a further period of 2 months. Balu agreed on the proposal for which an interest of Rs. 50 is charged. The second bill is duly accepted and was met on maturity. Give Journal entries in the book of Balu & Raju.
Answer:
Journal (In the book of Balu)
Plus One Accountancy Chapter Wise Questions and Answers Chapter 7 Bill of Exchange 30
Plus One Accountancy Chapter Wise Questions and Answers Chapter 7 Bill of Exchange 31

Question 2.
On January 15, 2012, ‘Syam’ sold goods to’ Naveen’ for Rs.30,000 and drew a bill for the same amount payable after 3 months. The bill was accepted by Naveen. The bill was discounted by Syam from his bank for Rs. 29,500 on January 31,2012. On maturity, the bill was dishonoured. He further agreed to pay Rs. 10,500 in cash including Rs.500 interest and accept a new bill for two months for the remaining Rs.20,000. The new bill was endorsed by Syam in favour of his creditor ‘Kiran’ for settling a debt of Rs.20500. The new bill was duly met by Naveen on maturity.
Record the Journal entries in the book of Syam and Naveen.
Answer:
Journal Entries (in the book of Syam)
Plus One Accountancy Chapter Wise Questions and Answers Chapter 7 Bill of Exchange 32

Plus One Accountancy Chapter Wise Questions and Answers Chapter 7 Bill of Exchange
Journal Entries (in the book of Naveen)
Plus One Accountancy Chapter Wise Questions and Answers Chapter 7 Bill of Exchange 33

Plus One Accountancy Bill of Exchange Eight Mark Questions and Answers

Question 1.
Mr. Anil sold goods to Sunil for Rs. 5,000 on credit and drew a bill for 3 months. Sunil accepted the bill and returned it to Anil. On the due date, the bill was dishonoured, noting charge being Rs. 100. Show journal entries in the books of Anil and Sunil, if

  1. The bill was retained by Mr. Anil.
  2. The bill was endorsed to Mr. Hari.
  3. The bill was discounted with the bank for Rs. 4,800.
  4. The bill was sent to the bank for collection.

Answer:
Book of Anil [Drawer] Journal
1. If the bill is retained
Plus One Accountancy Chapter Wise Questions and Answers Chapter 7 Bill of Exchange 34

2. If the bill was endorsed to Mr. Hari.
Plus One Accountancy Chapter Wise Questions and Answers Chapter 7 Bill of Exchange 35

3. If the bill is discounted with bank.
Plus One Accountancy Chapter Wise Questions and Answers Chapter 7 Bill of Exchange 36

4. If the bill sent for collection
Plus One Accountancy Chapter Wise Questions and Answers Chapter 7 Bill of Exchange 37

Book of Sunil [Drawee] Journal
Plus One Accountancy Chapter Wise Questions and Answers Chapter 7 Bill of Exchange 38

Plus One Chemistry Chapter Wise Questions and Answers Chapter 3 Classification of Elements and Periodicity in Properties

Students can Download Chapter 3 Classification of Elements and Periodicity in Properties Questions and Answers, Plus One Chemistry Chapter Wise Questions and Answers helps you to revise the complete Kerala State Syllabus and score more marks in your examinations.

Kerala Plus One Chemistry Chapter Wise Questions and Answers Chapter 3 Classification of Elements and Periodicity in Properties

Plus One Chemistry Classification of Elements and Periodicity in Properties One Mark Questions and Answers

Question 1.
Which of the following is not a Dobereiner triad?
a) Cl, Br, I
b) Ca, Sr, Ba
c) Li, Na, K
d) Fe, Co, Ni
Answer:
d) Fe, Co, Ni

Question 2.
The elements of s-block and p-block are collectively called ___________
Answer:
Representative elements

Question 3.
The cause of periodicity of properties is
a) Increasing atomic radius
b) Increasing atomic weights
c) Number of electrons in the valence shell
d) The recurrence of similar outer electronic configuration
Answer:
d) The recurrence of similar outer electronic configuration

Plus One Chemistry Chapter Wise Questions and Answers Chapter 3 Classification of Elements and Periodicity in Properties

Question 4.
Halogen with highest ionization enthalpy is ___________ .
Answer:
Fluorine

Question 5.
Which of the following represents the most electropositive element?
a) [He]2s1
b) [He]2s2
c) [Xe]6s1
d) [Xe]6s2
Answer:
c) [Xe]6s1

Question 6.
Second electron gain enthalpy is
Answer:
always positive

Question 7.
Correct order of polarising power is
a) Cs+ < K+ < Mg2+ < Al3+
b) Al3+ < Mg2 + K+ < Cs+
c) Mg2+ < Al3+ < K+ < Cs+
d) K+ < Cs+ < Mg2+ < Al3+
Answer:
a) Cs+ < K+ < Mg2+ < Al3+

Question 8.
The IUPAC name of the element with atomic number is 109 is ___________
Answer:
Une

Plus One Chemistry Chapter Wise Questions and Answers Chapter 3 Classification of Elements and Periodicity in Properties

Question 9.
The size of iso electronic species F~, Ne and Na+ is affected by
a) Nuclear charge
b) Principal quantum number.
c) Electron – electron interaction in outer orbitals.
d) None of the factors because their size is the same.
Answer:
Nuclear charge as nuclear charge is high the size is small.

Question 10.
In transition elements the differentiating electron occupies (n-1)d sublevel in preference to ______________
Answer:
np level

Plus One Chemistry Classification of Elements and Periodicity in Properties Two Mark Questions and Answers

Question 1.
The arguments made by two students are as given:
Student 1: ‘Hydrogen belongs to Group 1.’
Student 2: ‘Hydrogen belongs to Group 17.’
1. Who is right?
2. What is your opinion?
Answer:
1. Nobody is right.

2. Hydrogen has a one s-electron and hence can be placed in group 1 (alkali metals). It can also • gain an electron to achieve a noble gas arrangement and hence it can behave similar to a group 17 (halogen family) element. Because it a special case, hydrogen is placed separately at the top of the Periodic Table.

Question 2.
Match the following:

Sodium f-block
Oxygen s-block
Uranium d-block
Silver p-block

Answer:
Sodium – s-block
Oxygen – p-block
Uranium – f-block
Silver – d-block

Question 3.

  1. Which one has greater size, Na or K?
  2. Justify your answer.

Answer:
1. K

2. K comes below Na in the Periodic Table. The atomic size increases down the group due to the fact that the inner energy levels are filled with electrons, which serve to shield the outer electrons from the pull of the nucleus.

Question 4.
The general characteristics of a particular block of elements is as given:
They are highly electropositive, soft metals. They are good reducing agents. They lose the outermost electron(s) readily to form 1+ ion of 2+ ion.

  1. Which block has this general characteristics?
  2. Write down two general characteristics of p-block.

Answer:
1. s-block

2. The p-block contains metals, non-metals and metalloids. They form ionic as well as covalent compounds.

Plus One Chemistry Chapter Wise Questions and Answers Chapter 3 Classification of Elements and Periodicity in Properties

Question 5.
The variation in electron gain enthalpies of elements is less systematic than for ionization enthalpies. Out of oxygen and sulphur which has greater negative value for electron gain enthalpy? Justify.
Answer:
Sulphur has greater negative value (-200 kJ mol1) for electron gain enthalpy compared to that of oxygen (-141 kJ mol1). This is because, due to smaller size of oxygen the added electron experiences much repulsion from the electrons present in the shell. Due to large size, the electron-electron repulsion is much less in sulphur.

Question 6.
Some elements are given. Li, Cs, Be, C, N, O F, I, Ne, Xe.

  1. Arrange the above elements in the increasing order of ionization enthalpy.
  2. Arrange the given elements in the decreasing order of negative electron gain enthalpy.

Answer:

  1. Cs < I < Li < Xe < Be < C < N < O < F < Ne
  2. Cl > F > O > N > C > Be > Li > I >C s > Xe > Ne

Plus One Chemistry Classification of Elements and Periodicity in Properties Three Mark Questions and Answers

Question 1.
Analyse the given figure and answer the questions that follow.
Plus One Chemistry Chapter Wise Questions and Answers Chapter 3 Classification of Elements and Periodicity in Properties 1

  1. What is meant by atomic radius?
  2. Explain covalent radius.
  3. Write down another two types of terms expressed as atomic size.

Answer:

  1. Atomic radius is defined as the distance from the centre of the nucleus of the atom to the outermost shell of electrons.
  2. Covalent radius is defined as one half of the distance between the centre of nuclei of two similar atoms bonded by a single covalent bond.
  3. Vander Waals’ radius, Metallic radius

Question 2.
Consider the statement: The element with 1s2
configuration belongs to the p-block.’

  1. Identify the element.
  2. Do you agree with this statement?
  3. Justify.

Answer:

  1. Helium
  2. Yes
  3. Strictly, helium belongs to the s-block but its positioning in the p-block along with other group 18 elements is justified because it has a completely filled valence shell (1s²) and as a result, exhibits characteristic of other noble gases.

Plus One Chemistry Chapter Wise Questions and Answers Chapter 3 Classification of Elements and Periodicity in Properties

Question 3.
The properties of elements are a periodic function of
their atomic weights.

  1. Who proposed this law?
  2. Can you see anything wrong in this law? If yes, justify your answer.
  3. State modem periodic law.

Answer:

  1. Mendeleev
  2. Yes, atomic number is the more fundamental property of an element than atomic mass.
  3. The physical and chemical properties of the elements are periodic functions of their atomic numbers.

Question 4.
1. Define ionisation enthalpy.
2. IE1 <IE2 <IE3
What is meant by this? Justify.
Answer:
1. Ionisation enthalpy is the amount of energy required to remove the most loosely bound electron from an isolated gaseous atom in its ground state.

2. It shows the increasing order the magnitude of successive ionization enthalpies. As the positive charge of the ion increases, it becomes more difficult to remove the valence electron due to increase in effective nuclear charge.

Question 5.
Say whether the following are true or false:

  1. On moving across a period ionization enthalpy decreases.
  2. Mg is biggerthan Cl.
  3. Ionization enthalpy of Li is less than that of K.

Answer:

  1. False
  2. True
  3. False

Question 6.
Analyze the graph given below:
Plus One Chemistry Chapter Wise Questions and Answers Chapter 3 Classification of Elements and Periodicity in Properties 2
1. Identify the graph.
2. Account forthe following observations:
i) ‘Ne’ has the maximum value of ∆iH.
ii) In the graph from Be to B, ∆iH decrease.
Answer:
1. Graph showing the variation of ionisation enthalpy (∆iH) with atomic number (Z) of the elements of second period.

2. i) ‘Ne’ is an inert gas and it has closed electron shell with stable octet electronic configuration. Hence it has the maximum ionisation enthalpy in the second period.
ii) Be has completely filled electronic configuration and a more stable. So ionization enthalpy is high

Question 7.
Study the graph and answer the questions that follow:
Plus One Chemistry Chapter Wise Questions and Answers Chapter 3 Classification of Elements and Periodicity in Properties 3
1. On moving down a group what happens to electron gain enthalpy?
2. Why chlorine shows more negative electron gain enthalpy than fluorine? ,

Answer:
1. On moving down a group,electron gain enthalpy becomes less negative because the size of the atom increases and the added electron would be farther from the nucleus.

2. Due to small size of F, the added electron goes to the smaller 2p quantum level and suffers significant replusion from the other electrons present in this level. But due to big size of Cl, the added electron goes to the n = 3p quantum level and occupies a larger region of space and the electron-electron repulsion is much less.

Question 8.
Electron gain enthalpy is an important periodic property.
1. What is meant by electron gain enthalpy?
2. What are the factors affecting electron gain enthalpy?
3. How electron gain enthalpy varies on moving across a period? Justify.

Answer:
1. Electron gain enthalpy(∆egH) is the enthalpy change accompanying the process of addition of an electron to a neutral gaseous atom to convert it into a negative ion.

2. Effective nuclear charge, atomic size, electronic configuration

3. Electron gain enthalpy becomes more negative with increase in the atomic number across a period. The effective nuclear charge increases from left to right across a period and consequently, it will be easier to add an electron to a smaller atom since the added electron on an average would be closer to the nucleus. Thus more energy is released.

Plus One Chemistry Chapter Wise Questions and Answers Chapter 3 Classification of Elements and Periodicity in Properties

Question 9.
1. The electron gain enthalpies of Be and Mg are positive. What is your opinion? Justify.
2. Electron gain enthalpies of nobles gases have large positive values. Why?
Answer:
1. The statement is correct. Electron gain enthalpies of Be (+240 kJ/mol) and Mg (+230 kJ/mol) are positive because they have stable electronic configurations with fully filled s-orbitals (Be – 2s², Mg – 3s²). Hence, the gain of electron is highly endothermic.

2. Noble gases have stable octet electronic configuration of ns2 np6 (except He -1 s²) and they have practically no tendency to accept additional electron. They have large positive electron gain enthalpies because the electron has to enter the next higher principal quantum level leading to a very unstable electronic configuration. Hence, energy has to be supplied to add an extra electron.

Question 10.
During a group discussion, a student argued that ionization enthalpy depends only upon electronic configuration.
1. Do you agree? Comment.
2. Define shielding effect/screening effect.
3. Is there any relation between ionization enthalpy and shielding effect/screening effect? Explain.
Answer:
1. No. In addition to electornic configuration, ionization enthalpy depends upon other factors like atomic size, nuclear charge and shielding effect/screening effect.

2. In multi-electron atoms, the nuclear charge experienced by the valence electron will be less than the actual charge on the nucleus because it is shielded by inner core of electrons. This is called shielding effect or screening effect.

3. Yes. Ionization enthalpy decreases with increase in shielding effect/screening effect of inner electrons. This is because as a result of shielding of the valence electron by the intervening core of electrons it experiences a net positive charge which is less than the actual charge of the nucleus. In general, shielding is effective when the orbitals in the inner shells are completely filled.

Plus One Chemistry Chapter Wise Questions and Answers Chapter 3 Classification of Elements and Periodicity in Properties

Question 11.
a) Which of the following has higher first ionization enthalpy, N or O? Justify.
b) Which one is bigger, For F ? Why?
Answer:
1. N. This is because in N, three 2p-electrons reside in different atomic orbitals in accordance with Hund’s rule (\(2p_{ x }^{ 1 }2{ p }_{ y }^{ 1 }2{ p }_{ z }^{ 1 }\)) whereas in O, two the four 2p-electrons must occupy the same 2p-orbital resulting in an increased electron-electron repulsion (\(2p_{ x }^{ 2 }2{ p }_{ y }^{ 1 }2{ p }_{ z }^{ 1 }\)). Consequently, it is easier to remove the fourth 2p-electron from O than it is, to remove one of the three 2p-electrons from N.

2. F (136 pm) is bigger than F (72 pm). An anion is bigger than its parent atom. Addition of one electron in F results in increased repulsion among the electrons and a decrease in effective nuclear charge. Thus, attraction between nucleus and the electrons decreases and hence size increases.

Question 12.
Electronegativity differs from electron gain enthalpy.

  1. Do you agree?
  2. What do you mean by electronegativity?

Answer:

  1. Yes.
  2. Electronegativity is defined as the tendency of an atom in a chemical compound to attract the shared pair of electrons to itself.

Question 13.
Ionization enthalpy is an important periodic property.
1. What is the unit in which it is expressed?
2. What are the factors influencing ionization enthalpy?
3. How ionisation enthalpy varies in the periodic table?
Answer:
1. kJ mol-1.

2. Atomic/ionic radius, nuclear charge, shielding effect/screening effect, penetration effect and electronic configuration.

3. Ionization enthalpy generally increases with increase in atomic number across a period due to regular increase in nuclear charge and decrease in atomic size. Thus, the attractive force between the nucleus and the electron cloud increases. Consequently, the electrons are more and more tightly bound to the nucleus.

Ionisation enthalpy generally decreases from top to bottom a group: This is because the effect of increased nuclear charge is cancelled by the increase in atomic size and the shielding effect. Consequently, the nucelar hold on the valence electron decreases gradually and ionization enthalpy decreases.

Question 14.
The second period elements show anomalous ‘ behaviour.
1. Give reason.
2. What are the anomalous properties of second period elements?
Answer:
1. The first element is each group belong to the second period. The difference in behaviour of these elements from the other elements of the same group can be attributed to the following factors:

  • Small atomic size
  • Large charge/radius ratio
  • High electronegativity
  • Absence of d-orbtials in the valence shell

2. The important anomalous properties of second period elements are: diagonal relationship, maximum covalence of four and ability to form pπ —pπ multiple bonds.

Question 15.
Atomic size, valency, ionization enthalpy, electron gain enthalpy and electronegativity are the important periodic properties of elements.
1. What do you mean by periodicity?
2. Periodic properties are directly related to ___________
Answer:
1. The periodical repetition of elements with similar properties after certain regular intervals when the elements are arranged in the order of increasing atomic numbers is called periodicity,

2. Electronic configuration.

Question 16.
1. Which is the element among alkali metals having lowest ionization enthalpy?
2. What is meant by valence of an element? How it varies in the periodic table?
3.
Plus One Chemistry Chapter Wise Questions and Answers Chapter 3 Classification of Elements and Periodicity in Properties 4

Identify the elements A, B, C and D, if the graph represents halogens.
Answer:
1. Fr

2. Valence of an element is the combining capacity of that element. In the case of representative elements the number of valence electrons increases from 1 to 8 on moving across a period, the valence to the element with respect to H and Cl increases from 1 to 4 and then decreases from 4 to zero. On moving down a group, the number of valence electrons remains same and, therefore, all the elements in a group exhibit same valence.

3. A = F, B = Cl, C = Br, D = I

Plus One Chemistry Chapter Wise Questions and Answers Chapter 3 Classification of Elements and Periodicity in Properties

Question 17.
Li and Mg belonging to first and second group in periodic table respectively resemble each other in many respects.
1. Name the relationship.
2. B can only form [BF4] ion while Al can form [AIF6]3-, though both B and Al belong to group 13. Justify.
Answer:
1. Diagonal relationship.

2. This is because B, being a second period element has a maximum covalence of 4. It cannot expand its covalence beyond 4 due to absence of d-orbitals. But Al, being a third period element has vacant d-orbitals in its valence shell and hence can expand its covalence beyond 4.

Question 18.
During a group discussion a student argues that both oxidation state and valence are the same.
1. Do you agree?
2. Justify taking the case of [AICI(H2O)5]2+.
Answer:
1. No. Valence refers to the combining capacity of an element whereas oxidation state is the charge assigned to an element in a compound based on the assumption that the shared electron in a covalent bond belongs entirly to the more electronegative element.

2. In [AICI(H2O)5]2+the valence of Al is 6 while its oxidation state is +3.

Question 19.
Among the elements of the third period, identify the element

  1. With highest first ionization enthalpy.
  2. That is the most reactive metal.
  3. With the largest atomic radius.

Answer:

  1. Ar
  2. Na
  3. Na

Question 20.
A cation is smaller than the corresponding neutral atom while an anion is larger. Justify.
Answer:
A cation is smaller than its parent atom because it has fewer number of electrons while its nuclear charge remains the same. The size of an anion will be larger than that of the parent atom because the addition of one or more electrons would result in increased . repulsion among the electrons and a decrease in effective nuclear charge.

Question 21.
1. How the metallic character varies in the periodic table?
2. Categorize the following oxides into acidic, basic, neutral and amphoteric:
Al2O3, Na2O, CO2, Cl2O7, MgO, CO, As2O3, N2O
Answer:
1. The metallic character decreases from left to right across the period due to increase in ionization enthalpy along a period which makes loss of electrons difficult. From top to bottom a group metallic character increases due to decrease in ionization enthalpy. Thus, metallic character decreases diagonally from left bottom to right top of the periodic table.

2. Acidic oxides: CO2, Cl2O7
Basic oxides: Na2O, MgO
Neutral oxides: CO, N2O
Amphoteric oxides: Al2O3, As2O3

Question 22.
A group of ions are given below:
Na+, Al3+, O2-, Ca2+, Mg2+, F, N3-, Br
1. Find the pair which is not isoelectronic.
2. Arrange the above ions in the increasing order of size.
Answer:
1. Ca2+ and Br
2. Al3+ < Mg2+ < Na+ < F < O2- < N3- < Ca2+ < Br

Plus One Chemistry Classification of Elements and Periodicity in Properties Four Mark Questions and Answers

Question 1.
Statement 1: ‘Atomic mass is the fundamental property of an element.’
Statement 2: ‘Atomic number is a more fundamental property of an element than its atomic mass.’
1. Which statement is correct? Justify your answer.
2. Name the scientist who proposed this statement? What observation led him to this conclusion?
Answer:
1. Statement 2. Atomic number indicates the number of electrons present in an element. Most of the chemical properties of an element depend on its electronic configuration,

2. Henry Moseley. He observed regularities in the characteristic X-ray spectra of the elements. A plot of √υ (where u is the frequency of the X-rays emitted) against atomic number (Z) gave a straight line and not the plot of √υ vs atomic mass.

Plus One Chemistry Chapter Wise Questions and Answers Chapter 3 Classification of Elements and Periodicity in Properties

Question 2.
Atoms possessing stable configuration have less tendency to loss electrons and consequently will have high value of ionization enthalpy.
1. Justify this statement by taking the case of half-filled and completely filled electronic configurations.
2. The noble gases have highest ionization enthalpies in each respective periods. Why?
Answer:
1. Atoms with half-filled and completely filled electronic configurations have extra stability due to symmetric distribution of electrons and maximum exchange energy. Hence, more energy is required for the removal of their electrons. Elements like N (1s² 2s² \(2p_{ x }^{ 1 }2{ p }_{ y }^{ 1 }2{ p }_{ z }^{ 1 }\)), P (1s² 2s² 2p6 3s² \(3p_{ x }^{ 1 }3{ p }_{ y }^{ 1 }3{ p }_{ z }^{ 1 }\)) etc. possessing half-filled shells have high ionization enthalpies.
Elements like Be ((1s² 2s²), Mg(1s² 2s² 2p6 3s²) etc. having completely filled shells show high values of ionization enthalpy.

2. Noble gases have closed electron shells and very stable octet electronic configurations (except He). Hence, maximum amount of energy is required to remove their valence electron.

Question 3.
Mendeleev arranged the elements in the order of increasing atomic weights.
a) Write down the merits of Mendeleev’s periodic table.
b) What are the demerits of Mendeleev’s periodic table?
Answer:
a) Merits of Mendeleev’s periodic table:

  • Study of elements – Elements are classified into groups with similar properties, thus facilitating the study of properties of elements.
  • Prediction of new elements – Mendeleev left certain vacant places in his table which provided a clue for the discovery of new elements, e.g. Eka-AI (for Ga), Eka-Si (for Ge).
  • Determination of correct atomic weights – With the help of this table, doubtful atomic weights of some elements were corrected.

b) Demerits of Mendeleev’s periodic table:

  • Position of hydrogen was not certain.
  • Anomalous pairs of elements – Certain elements of higher atomic weight preceed those with lower atomic weight, e.g. I (at.wt. 127) was placed after Te (at.wt. 128).
  • Lanthanides and Actinides are not given proper places in this periodic table.
  • No proper position for isotopes.

Question 4.
1. Name any three numerical scales of electronegativity.
2. How electronegativity varies in the periodic table? Justify.
Answer:
1. Pauling scale, Mullimen-Jaffe scale, Allred- Rochow scale.
2. Electronegativity generally increases from left to right a period and decreases from top to bottom in a group. This is because, from left to right across a period atomic size decreases and attraction between the valence electrons and the nucleus increases. From top to bottom in a group atomic size increases and attraction between the valence electrons and the nucleus decreases.

Question 5.
1. First ionization enthalpy of Na is lower than that of Mg. But its second ionization enthalpy is higher than that of Mg. Explain.
2. Which one is smaller, Na or Na+? Give reason.
Answer:
1. By the removal of one electron from Na it gets
the stable octet configuration of Ne. But when the first electron is removed from Mg it gets the unstable configuration of Na. It requires more energy due to small size and greater nuclear charge of Mg. In the case of Na the second electron is to be removed from a stable octet configuration which requires more energy than the removal of second electron from Mg.

2. Na+ (95 pm) is smaller than Na (186 pm). Acation is smaller than its parent atom. Na+ has fewer number of electrons (10 electrons) compared to Na (11 electrons). But the nuclear charge remains the same in both. Thus, effective nuclear charge per electron is greater in Na+. Thus, the attraction between nucleus and the remaining electrons increases and size decreases.

Question 6.
Removal of electron becomes easier on moving down the group.
1. Comment the above statement based on ionization enthalpy.
2. How electronic configuration influences the ionization enthalpy value?
Answer:
1. On moving from top to bottom in a given group, size of the atom increases and ionisation enthalpy decreases. Hence, it becomes easier to remove the valence electron.

2. Atoms with octet configuration, half-filled and completely filled configurations have extra stability and hence have higher values of ionization enthalpy.

Plus One Chemistry Chapter Wise Questions and Answers Chapter 3 Classification of Elements and Periodicity in Properties

Question 7.
The energy released during the addition of an electron to an isolated neutral atom is called electron gain enthalpy.
1. Explain how electron gain enthalpy differ from electronegativity.
2. The second ionisation enthalpy of an element is always greater than the first ionisation enthalpy. Give reason.
Answer:
1. Electron gain enthalpy(AegH) is the enthalpy change accompanying the process of addition of an electron to a neutral gaseous atom to convert it into a negative ion. It is a quantitative property of an isolated gaseous atom, which can be measured. Whereas, electronegativity is a qualitative measure of the ability of an atom in a chemical compound to attract shared electrons to itself. It is not a measureable quantity.

2. This is because to remove second electron from a positively charged ion more amount of energy is required due to increase in effective nuclear charge.

Question 8.
The physical and chemical properties of elements are periodic functions of their atomic numbers.
1. The atomic number of an element ‘X’ is 19. Write the group number, period and block to which X’ belong in the periodic table.
2. Name the element with
i) highest electronegativity and
ii) highest electron gain enthalpy
Answer:
1. The element is K.
19K= 1s² 2s² 2p6 3s² 3p6 4s1
Group number = 1
Period number = 4
Block = s-block

2. i) Fluorine
ii) Chlorine

Plus One Chemistry Classification of Elements and Periodicity in Properties NCERT Questions and Answers

Question 1.
What is the basic theme of organisation in the periodic table? (2)
Answer:
The basic theme of organisation of elements in the periodic table is to facilitate the study of the properties of all the elements and their compounds. On the basis of similarities in chemical properties, the various elements have been divided into different groups. This has made the study of elements simple because their properties are now studied in the form of groups rather than individually.

Question 2.
What is the basic difference in approach between Mendeleev’s Periodic Law and the Modern Periodic Law? (2)
Answer:
Mendeleev Periodic Law states that the properties of the elements are a periodic function of their atomic weights whereas Modern Periodic Law states that the properties of elements are a periodic function of their atomic numbers. Thus, the basic difference in approach between Mendeleev’s Periodic Law and Modern Periodic Law is the change in basis of arrangements of elements from atomic weight to atomic number.

Plus One Chemistry Chapter Wise Questions and Answers Chapter 3 Classification of Elements and Periodicity in Properties

Question 3.
Consider the following species. N3-, O2-, F, N2+, Mg2+ and Al3+ (2)
1. What is common in them?
2. Arrange them in order of increasing ionic radii.
Answer:
1. Each one of these ions contains 10 electrons and
hence these are isoelectronic ions,

2. The ionic radii of isoelectronic ions decrease with the increase in the magnitude of the nuclear charge. Among the isoelectronic ions: N3-, O2-, F, Na+, Mg2+ and Al3+, nuclear charge increase in the order:
N3-< O2- < F < Na+ < Mg2+ < Al3+
Therefore, the ionic radii decrease in the order:
N3- > O2- > F > Na+ > Mg2+ > Al3+

Question 4.
Anything that influences the valence electrons will affect the chemistry of the element. Which one of the following factors does not affect the valence shell?
(a) Valence principal quantum number (n)
(b) Nuclear charge (Z)
(c) Nuclear mass
(d) Number of core electrons (1)
Answer:
Nuclear mass does not affect the valence shell. Thus, option (c) is the correct answer.

Question 5.
Considering the elements F, Cl, O and N, the correct order of their chemical reactivity in terms of oxidising property is: (1)
a) F > Cl > O > N
b) F > O > Cl > N
c) Cl > F > O > N
d)0>F>N>CI
Answer:
a) F > Cl > O > N
Across a period, the oxidising character increases from left to right. Therefore, among F, O and N, oxidising power decreases in the order: F > O > N. However, within a group, oxidising power decreases from top to bottom. Thus, F is a stronger oxidising agent than Cl. Thus, overall decreasing order of oxidising power is F > Cl > O > N and the choice (a) is correct.

Plus One Maths Chapter Wise Questions and Answers Chapter 15 Statistics

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

Kerala Plus One Maths Chapter Wise Questions and Answers Chapter 15 Statistics

Plus One Maths Statistics Three Mark Questions and Answers

Question 1.
Find the mean deviation about the median for the following data: (3 score each)

  1. 13, 17, 16, 14, 11, 13, 10, 16, 11, 18, 12, 17.
  2. 36, 72, 46, 42, 60, 45, 53, 46, 51, 49.

Answer:
1. Arrange the data in the ascending order we have;
10, 11, 11, 12, 13, 13, 14, 16, 16, 17, 17, 18
Here n = 12. So median is the average of 6th and 7th observations.
Therefore; Median, M =\(\frac{13+14}{2}\) = 13.5
Plus One Maths Statistics Three Mark Questions and Answers 1
Mean deviation = \(\frac{\sum_{i=1}^{n}\left|x_{i}-M\right|}{n}=\frac{28}{12}\) = 2.33

2. Arrange the data in the ascending order we have; 36, 42, 45, 46, 46, 49, 51, 53, 60, 72
Here n = 10. So median is the average of 5th and 6th observations.
Therefore; Median, M = \(\frac{46+49}{2}\) = 47.5
Plus One Maths Statistics Three Mark Questions and Answers 2
Mean deviation = \(\frac{\sum_{i=1}^{n}\left|x_{i}-M\right|}{n}=\frac{70}{10}\) = 7.

Plus One Maths Chapter Wise Questions and Answers Chapter 15 Statistics

Question 2.
The mean and standard deviation of marks obtained by 50 students of a class in three subjects, mathematics, Physics, and Chemistry are given below:
Plus One Maths Statistics Three Mark Questions and Answers 3
Which of the three subject shows the highest variability in marks and which shows the lowest?
Answer:
For Mathematics:
\(\bar{x}\) = 42, σ = 12
∴ CV of Mathematics = \(\frac{12}{42}\) × 100 = 28.57%
For Physics:
\(\bar{x}\) = 32, σ = 15
∴ CV of Physics = \(\frac{15}{32}\) × 100 = 46.88%
For Chemistry:
\(\bar{x}\) = 40.9, σ = 20
∴ CV of Chemistry = \(\frac{20}{40.9}\) × 100 = 48.9%
Thus Chemistry with highest CV shows highest variability and Mathematics with lowest CV shows lowest variability.

Plus One Maths Statistics Four Mark Questions and Answers

Question 1.
Find the mean deviation about the mean for the following data: (4 score each)
1.
Plus One Maths Statistics Three Mark Questions and Answers 4
2.
Plus One Maths Statistics Three Mark Questions and Answers 5
Answer:
1.
Plus One Maths Statistics Three Mark Questions and Answers 6
Plus One Maths Statistics Three Mark Questions and Answers 7
Plus One Maths Statistics Three Mark Questions and Answers 8

Plus One Maths Chapter Wise Questions and Answers Chapter 15 Statistics

2.
Plus One Maths Statistics Three Mark Questions and Answers 9
Plus One Maths Statistics Three Mark Questions and Answers 10

Plus One Maths Chapter Wise Questions and Answers Chapter 15 Statistics

Question 2.
Find the mean deviation about the median for the following data: (4 score each)
1.
Plus One Maths Statistics Three Mark Questions and Answers 11
2.
Plus One Maths Statistics Three Mark Questions and Answers 12
Answer:
1.
Plus One Maths Statistics Three Mark Questions and Answers 13
\(\frac{\sum_{i=1}^{n} f_{i}}{2}=\frac{26}{2}\) = 13
The c.f just greater than 13 is 14 and corresponding value of x is 7. Therefore; median, M = 7
Hence; M.D about median
Plus One Maths Statistics Three Mark Questions and Answers 14

2.
Plus One Maths Statistics Three Mark Questions and Answers 15
\(\frac{\sum_{i=1}^{n} f_{i}}{2}=\frac{29}{2}\) = 14.5
The c.f just greater than 14.5 is 21 and corresponding value of x is 30. Therefore; median, M = 30
Hence; M.D about median
Plus One Maths Statistics Three Mark Questions and Answers 16

Plus One Maths Chapter Wise Questions and Answers Chapter 15 Statistics

Question 3.
Find the mean deviation about the mean for the following data: (4 score each)
1.
Plus One Maths Statistics Three Mark Questions and Answers 17
2.
Plus One Maths Statistics Three Mark Questions and Answers 18
Answer:
1.
Plus One Maths Statistics Three Mark Questions and Answers 19
Plus One Maths Statistics Three Mark Questions and Answers 20

2.
Plus One Maths Statistics Three Mark Questions and Answers 47
Plus One Maths Statistics Three Mark Questions and Answers 21

Plus One Maths Chapter Wise Questions and Answers Chapter 15 Statistics

Question 4.
Find the mean deviation about the median for the following data: (4 score each)
1.
Plus One Maths Statistics Three Mark Questions and Answers 22
2.
Plus One Maths Statistics Three Mark Questions and Answers 23
Answer:
1.
Plus One Maths Statistics Three Mark Questions and Answers 24
Median class is the class in which the \(\left(\frac{N}{2}\right)^{th}\) observation lies.
\(\frac{N}{2}=\frac{50}{2}\) = 25
Median class = 20 – 30
M = Median
Plus One Maths Statistics Three Mark Questions and Answers 25
Plus One Maths Statistics Three Mark Questions and Answers 26

2.
Plus One Maths Statistics Three Mark Questions and Answers 27

Plus One Maths Chapter Wise Questions and Answers Chapter 15 Statistics
Median class is the class in which the \(\left(\frac{N}{2}\right)^{th}\) observation lies.
\(\frac{N}{2}=\frac{100}{2}\) = 50
Median class = 35.5 – 40.5
M = Median
Plus One Maths Statistics Three Mark Questions and Answers 28
Plus One Maths Statistics Three Mark Questions and Answers 29

Question 5.
Find the variance and standard deviation of 3, 4, 6, 5, 5, 3, 8, 1, 7, 5
Answer:
Plus One Maths Statistics Three Mark Questions and Answers 30
Plus One Maths Statistics Three Mark Questions and Answers 31
Plus One Maths Statistics Three Mark Questions and Answers 32
\(\frac{259}{10}\) – (4.7)2 = 25.9 – 22.09 = 3.8
Standard Deviation (σ) = \(\sqrt{\text {Variance}}\) = \(\sqrt{3.8}\) =1.95.

Plus One Maths Chapter Wise Questions and Answers Chapter 15 Statistics

Question 6.
Find the variance and standard deviation of
Plus One Maths Statistics Three Mark Questions and Answers 33
Answer:
Plus One Maths Statistics Three Mark Questions and Answers 34
Plus One Maths Statistics Three Mark Questions and Answers 35
Standard Deviation (σ) = \(\sqrt{\text {Variance}}\)
= \(\sqrt{15.08}\) = 3.88.

Plus One Maths Chapter Wise Questions and Answers Chapter 15 Statistics

Question 7.
An analysis of monthly wages paid to workers in two firms A and B, belonging to the same industry, gives the following result.
Plus One Maths Statistics Three Mark Questions and Answers 36

  1. Which firm A or B pays larger amount as monthly wages?
  2. Which firm A or B, shows greater variability in individual wages?

Answer:
1. Firm: A
Number of wages earners (n1) = 586
Number of wages earners (\(\bar{x}_{1}\)) = 5253
Total monthly wages = 5253 × 586 = 3078258
Firm: B
Number of wages earners (n1) = 648
Number of wages earners (\(\bar{x}_{1}\)) = 5253
Total monthly wages = 5253 × 648 = 3403944

2. Since both the firms have same mean of monthly wages, so the firm with greater variance will have more variability in individual wages. Thus firm B will have more variability in individual wages.

Plus One Maths Chapter Wise Questions and Answers Chapter 15 Statistics

Question 8.
The sum and sum of squares corresponding to length x (in cm) and weight y (in gm) of 50 plant products are given below:
Plus One Maths Statistics Three Mark Questions and Answers 37
which is more varying, the length or weight?
Answer:
Here,
Plus One Maths Statistics Three Mark Questions and Answers 38
Now
Plus One Maths Statistics Three Mark Questions and Answers 39
C.V of weight = \(\frac{1.38}{5.22}\) × 100 = 26.44
C.V of weight > C.V of length
Thus weight have more variability than length

Plus One Maths Statistics Six Mark Questions and Answers

Question 1.
Find the variance and standard deviation of
Plus One Maths Statistics Three Mark Questions and Answers 40
Answer:
Plus One Maths Statistics Three Mark Questions and Answers 41

Plus One Maths Chapter Wise Questions and Answers Chapter 15 Statistics
Plus One Maths Statistics Three Mark Questions and Answers 42
Standard Deviation (a) = \(\sqrt{\text {Variance}}\)
= \(\sqrt{223.22}\) = 14.94.

Plus One Maths Statistics Practice Problems Questions and Answers

Question 1.
Find the mean deviation about the mean for the following data: (2 score each)

  1. 4, 7, 8, 9, 10, 12, 13, 17.
  2. 38, 70, 48, 40, 42, 55, 63, 46, 54, 44.

Answer:
1.
Plus One Maths Statistics Three Mark Questions and Answers 43
Plus One Maths Statistics Three Mark Questions and Answers 44

2.
Plus One Maths Statistics Three Mark Questions and Answers 45
Plus One Maths Statistics Three Mark Questions and Answers 46

Plus One Chemistry Chapter Wise Questions and Answers Chapter 1 Some Basic Concepts of Chemistry

Students can Download Chapter 1 Some Basic Concepts of Chemistry Questions and Answers, Plus One Chemistry Chapter Wise Questions and Answers helps you to revise the complete Kerala State Syllabus and score more marks in your examinations.

Kerala Plus One Chemistry Chapter Wise Questions and Answers Chapter 1 Some Basic Concepts of Chemistry

Plus One Chemistry Some Basic Concepts of Chemistry One Mark Questions and Answers

Question 1.
Which of the following is a mixture?
a) Graphite
b) Sodium chloride
c) Distilled water
d) Steel
Answer:
d) Steel

Question 2.
1 µ g = __________ g
[10-3, 10-6, 10-9, 10-12] ‘
Answer:
10-6

Question 3.
The number of significant figures in 0.00503060 is __________ .
Answer:
6

Question 4.
The balancing of chemical equations is based on which of the following law?
a) Law of multiple proportions
b) Law of conservation of mass
c) Law of definite proportions
d) Gay-Lussac law
Answer:
b) Law of conservation of mass

Question 5.
Which among the following is the heaviest?
a) 1 mole of oxygen
b) 1 molecule of sulfur trioxide
c) 100 u of uranium
d) 44 g carbon dioxide
Answer:
d) 44 g carbon dioxide

Question 6.
Calculate the number of atoms in 48 g of He?
Answer:
Gram atomic mass of He = 4 g.
Thus, numberofatomsin4g (1 mol) He = 6.02 × 1023
So number of atoms in 48 g of He = \(\frac{48}{4}\) × 6.02 × 1023
=12 × 6.02 × 1023
= 7.224 × 1024

Question 7.
One mole of CO2 contains how many gram atoms?
Answer:
3 gram atoms.

Plus One Chemistry Chapter Wise Questions and Answers Chapter 1 Some Basic Concepts of Chemistry

Question 8.
The ratio of gram atoms of Au and Cu in 22ct gold is __________
Answer:
7 : 2

Question 9.
A compound contains 69.5% oxygen and 30.5% nitrogen and its molecular weight is 92. The compound will be
Answer:
N2O4

Question 10.
The total number of electrons present in 1 mole of water is
Answer:
6 × 1024.

Question 11.
40g NaOH is present in 100 ml of a solution. Its molarity is __________
Answer:
10 M

Plus One Chemistry Some Basic Concepts of Chemistry Two Mark Questions and Answers

Question 1.
Classify the following substances into homogeneous and heterogeneous mixtures.

  • Milk
  • Iron
  • Air
  • Gasoline
  • Kerosene
  • Muddy water

Answer:

Homogeneous Heterogeneous
Milk, Iron
Gasoline
Air
Kerosene
Muddy Water

Question 2.
Calculate the volume occupied by 4.4 g of CO2 at STP?
Answer:
1 mole CO2 = 44 g
4.4 CO2 = 0.1 mole CO2
Volume occupied by 1 mol CO2 at STP = 22.4 L
∴ Volume occupied 0.1 mol CO2 at STP = 0.1 × 22.4 L
= 2.24 L

Plus One Chemistry Chapter Wise Questions and Answers Chapter 1 Some Basic Concepts of Chemistry

Question 3.
During a group discussion a student argued that “the water of sea and river should have different chemical composition”.

  1. What is your opinion?
  2. Which law would you suggest to support your answer?
  3. State the law.

Answer:

  1. I can’t join with him.
    The water of sea and water of river must have the same chemical composition.
  2. Law of definite proportions.
  3. A given compound always contains exactly the same proportion of elements by weight.

Question 4.
“When science developed some theories are also modified”.
Write the modified atomic theory.
Answer:

  1. Atom is no longer considered as indivisible, it has been found that atom is made up of sub atomic particles called protons, neutrons and electrons.
  2. Atoms of the same element may not be similar in all respects.
  3. Atoms of different elements may be similar in one or more respects.
  4. The ratio in which atomic unit may be fixed and integral but may not be simple.
  5. The mass of atom can be changed into energy.

Question 5.
Carbon combines with oxygen to form CO and CO2.

  1. What is the law behind this?
  2. State the law.

Answer:

  1. Law of multiple proportions.
  2. If two elements combine to form more than one compound, the masses of one element that combine with a fixed mass of the other element, are in the ratio of small whole numbers.

Question 6.
Calculate the volume occupied by 6.02×1025 molecules of oxygen at STP.
Answer:
Volume occupied by 1 mole of oxygen gas at STP = 22.4 l
i.e., Volume occupied by 6.02×1023 molecules of oxygen gas at STP = 22.4 l
Hence the volume occupied by 6.02 × 1025 molecules
of oxygen gas at STP = \(\frac{22.4 \times 6.02 \times 10^{25}}{6.02 \times 10^{23}}\) = 2240 l.

Question 7.
Calculate the molality of a solution of NaOH containing 20g of NaOH in 400 g solvent.
Answer:
Plus One Chemistry Chapter Wise Questions and Answers Chapter 1 Some Basic Concepts of Chemistry 1

Question 8.
Calculate the mole fraction of NaOH in a solution containing 20 g of NaOH per 360 g of water.
Answer:
Plus One Chemistry Chapter Wise Questions and Answers Chapter 1 Some Basic Concepts of Chemistry 2

Question 9.
12 g of carbon reacts with 32 g of oxygen to form 44g of carbon dioxide.

  1. Which law of chemical combination is applicable here?
  2. State the law.

Answer:

  1. Law of conservation of mass.
  2. Matter can neither be created nor destroyed. Or, during any physical or chemical change, the total mass of the products remains equal to the total mass of the reactants.

Question 10.
When hydrogen and oxygen combine to form water, the ratio between volume of reactants and products is 2:1:2.

  1. Which law of chemical combination is applicable here?
  2. State the law.

Answer:

  1. Gay Lussac’s law of gaseous volumes.
  2. When gases combine or are produced in a chemical reaction they do so in a simple ratio by volume provided all gases are at same temperature and pressure.

Question 11.
Carbon form two oxides, the first contains 42.9% C and the second contains 27.3% carbon. Show that these are in agreement with the law of multiple proportions.
Answer:
In the first compound:
C = 42.9%
O = 100-42.9 = 57.1%
So, the ratio between the masses of C and O = 42.9:57.1 = 1:1.33
In the second compound:
C = 27.3%
O= 100-27.3= 72.7%
So, the ratio between the masses of C and O = 27.3 : 72.7= 1:2.66
Hence, the ratio of masses of oxygen which combines with a fixed mass of carbon is 1.33:2.66 or 1:2, a simple whole number ratio. This illustrates the law of multiple proportions.

Question 12.
Match the following:

A B
1 amu 1.008 x 1.66 x1024
Mass of 1 H atom 6.02 x 1023
Molar volume of O2 at STP 11.2L
Volume of 14g of N2 at STP 1.66 x 1024
Avogadro number 22.4L

Answer:

A B
1 amu 1.66 x 1024
Mass of 1 H atom 1.008 x 1.6 x1024
Molar volume of O2 at STP 22.4L
Volume of 14g of N2 at STP 11.2L
Avogadro number 6.02 x 1023

Question 13.
Calculate the molality of a solution containing 10 g ofNaOH in 200 cm3 of solution. Density of solution is 1.4 g/mL. (Molar mass of NaOH = 40)
Answer:
Mass of the solution = 200 × 1.04 = 208 g
Mass of NaOH (WB) = 10g Molar mass of NaOH (MB) = 40 g mol-1
Mass of water (WA) = (208 -10) g = 198 g = 0.198 kg
Plus One Chemistry Chapter Wise Questions and Answers Chapter 1 Some Basic Concepts of Chemistry 3

Question 14.
Calculate the mass percentage of oxygen in CaCO3.
Answer:
Molecular mass of CaCO3 = 100 g mol-1
Mass of oxygen in 100 g CaCO3=3 × 16 g = 48 g
Percentage of oxygen in CaCO3 =\(\frac{48}{100}\)×100 = 48%

Question 15.
KCIO3 on heating decomposes to KCI and O2. Calculate the mass and volume of O2 produced by heating 50 g of KCIO3.
Answer:
The reaction is represented as,
Plus One Chemistry Chapter Wise Questions and Answers Chapter 1 Some Basic Concepts of Chemistry 4
According to the equation, 96 g of oxygen is obtained from 245 g of KCIO3.
Hence mass of oxygen obtained from 50 g KCIO3 is \(\frac{96 \times 50}{245}=19.6 \mathrm{g}\)
According to the equation 245 g of KCIO3 gives 3 moles of O2 at STP which is 3 × 22.4 L = 67.2 L
Volume of oxygen liberated by 50g of KCIO3
= \(\frac{67.2 \times 50}{245}=13.71 \mathrm{L}\)

Plus One Chemistry Chapter Wise Questions and Answers Chapter 1 Some Basic Concepts of Chemistry

Question 16.
Calculate the number of molecules present in

  1. 11g of CO2.
  2. 56 mL Of CO2 at STP.

Answer:
1. \(\frac{11}{44}\) = 0.25 mole
1 mole of CO2 contains 6.022 × 1023 molecules.
∴ 0.25 mole CO2 contains 6.022 × 1023 × 0.25
= 1.51 × 1023 molecules.

2. 56 mL = 0.056 L
\(\frac{0.056}{22.4}\) =0.0025 mole
= 6.022 × 1023 × 0.0025=1.5 × 1021 molecules

Question 17.
Calculate the number of moles of 02 required to produce 240 g of MgO by burning Mg metal. [Atomic mass: Mg=24, 0=16]
Answer:
2 Mg + O2 → 2MgO
No. of moles of MgO = \(\frac{240}{40}\)=6
No. of moles of 02 required = 6/2 = 3

Question 18.
Arrange the following in the increasing order of their mass.
(a) 1 g of Ca
(b) 12 amu of C
(c) 6.022 × 1023 mol-ecules of CO2
(d) 11.2 L of N2 at STP
Answer:
a) Mass of 1 g Ca = 1 g
b) Mass of 12 amu C = \(\frac{12}{6.022 \times 10^{23}}\) = 2 × 10-23
c) Mass of 6.022 × 1023 molecules of CO2 = 44 g
d) Mass of 11.2 L of N2 at NTP = \(\frac{28 \times 11.2}{22.4}\) = 14 g
(b) < (a) < (d) < (c)

Question 19.
Complete the table:

42g N2 1.5 mole N2 33600mLN2 (STP)
16g 0:  – – – – mole O2 11.2 L of O2 (STP)
….g CO2 1 mole CO2 – – – – L of C O2 (STP)
28g CO 1 mole CO – – – – mLCO (STP)

Answer:

42g N2 1.5 Mole N2 33600mLN2 (STP)
16g O2 0.5 Mole O2 11.2 L of O2 (STP)
44 g CO2 1 mole CO2 22.4 L of CO2 (STP)
28g CO 1 mole CO 22400mLCO (STP)

Question 20.
1. Irrespective of the source, pure sample of H20 always contains 88.89% by mass of oxygen and 11.11% by mass of hydrogen.
a) Which law is illustrated here?
b) State the law.
2. Complete the table by filling in the blanks:

48 g O2 1.5 mol O2 ……mL O2 (at STP)
…… g Na 2 gram atom Na 2NA Na atoms
…….g CO2 2.5 mol CO2 56 L (at STP)
8.5 g NH3 ……mol NH3 11.2 L (at STP)

Answer:
1. a) Law of definite proportions.
b) The same chemical compound always contains the same elements combined in the same fixed proprotion by mass.
2.

48 g O2 1.5 mol O2 33600 mL O2 (at STP)
46g Na 2 gram atom Na 2Na Na atoms
110 g CO2 2.5 mol CO2 56 L (at STP)
8.5 g NH3 0.5 mol NH3 11.2L(atSTP)

Question 21.
Calcium carbonate reacts with aqueous HCl to give CaCl2 and CO2 according to the reaction:
CaCO3 + 2HCl → CaCl2 + CO2 + H2O
What mass of CaCO3 is required to react completely with 25 mL of 0.75 M HCl?.
Answer:
Mass of HCI in 0.25 mL 0.75 M HCl \(=\frac{36.5 \times 0.75 \times 0.25}{1000}=0.6844 \mathrm{g}\)
As per reaction 100 g CaCO3 reacts with 2 × 36.5 = 73 g of HCl
∴ Mass of CaCO3 reacting with 0.6844 g HCl \(=\frac{100 \times 0.6844}{73}=0.9375 \mathrm{g}\)

Plus One Chemistry Some Basic Concepts of Chemistry Three Mark Questions and Answers

Question 1.
During a Seminar, a student remarked that “Dalton’s atomic theory has some faulty assumptions”.
a) Do you agree with him?
b) What is the present status of Dalton’s atomic theory?
c) Write any two wrong postulates of Dalton’s atomic theory.
Answer:
a) I agree with him. Out of 6 Dalton’s postulates, 5 postulates are faulty and only one is correct.
b) Dalton’s atomic theory has undergone many modifications.
c)

  • All substances are made up of small indivisible particles called atoms.
  • Atoms of the same elements are identical in mass and other properties.

Question 2.
One gram atom of an element contains 6.023 × 1023 atoms.

  1. Find the number of atoms in 8 g oxygen.
  2. Which is heavier, 1 oxygen atom or 10 hydrogen atoms?
  3. Define mole and Avogadro number.

Answer:
1. 16 g oxygen contains 6.022 × 1023 atoms
∴ 8 g oxygen contains \(\frac{6.022 \times 10^{23}}{2}\) = 3.011 × 1023

2.
Plus One Chemistry Chapter Wise Questions and Answers Chapter 1 Some Basic Concepts of Chemistry 5

3. One mole is the amount of a substance that contains as many particles or entities as there are atoms in exactly 12 g (or 0.012 kg) of the 12C isotope.
Avogadro number – It is the number of discrete particles present in 1 mole of any substsnce. (Avogadro number, NA = 6.022 × 1023)

Plus One Chemistry Chapter Wise Questions and Answers Chapter 1 Some Basic Concepts of Chemistry

Question 3.

  1. Classify the following as homogeneous and heterogeneous mixtures.
    Air, Smoke, Gunpowder, NaCI solution, Petrol, Bronze, Mixture of sugar and sand.
  2. State and explain law of multiple proportions with example.

Answer:
1. Homogeneous-Air, NaCI solution, Bronze, Gun powder, Petrol.
Heterogeneous – Mixture of sugar and sand, Smoke.

2. When two elements combine to form more than one compound the different masses of one of the elements combining with fixed mass of the other element bear a simple ratio. ,
eg. Carbon reacts with oxygen to form two compounds viz. CO and CO2. In CO mass ratio is 12:16. In CO2 mass ratio is 12:32. Then mass ratio between oxygen in the 2 compounds is 16:32 or 1:2 which is a simple whole number ratio. Hence, the law is verified.

Question 4.
1. One mole of an ideal gas occupies 22.4 L at STP
a) Calculate the mass of 11.2 L of oxygen gas at STP.
b) Calculate the number of atoms present in the above sample.
2. 21 g of nitrogen gas is mixed with 5 g of hydrogen gas to yield ammonia according to the equation.
N2 + 3H2 → 2NH3
Calculate the maximum amount of ammonia that can be formed.
Answer:
1. a) Mass of 22.4 L oxygen at STP = 32 g
∴ Mass of 11.2 L oxygen at STP = 16 g
b) No. of atoms present in 16 g of O2
\(\frac{6.02 \times 10^{23}}{2}\) ×2 = 6.02 × 1023 atoms

2. N2 + 3H2 → 2NH3
1 mole N2 + 3 mole H2 → 2moles NH3
1 mole N2 requires 3 mol H2
i.e., 28g N2 requires 6 g H2
Hence, 21 g N2 requires \(\frac{6 \times 21}{28}\) = 4.5 g H2
21 g N2 reacts completely and 0.5g H2 remains unreacted.
Hence, N2 is the limiting reagent.
28g N2 gives 2 × 17 g NH3
∴ 21 g N2 gives \(\frac{2 \times 17 \times 21}{28}\) = 25.5 g NH3

Question 5.
When two elements combine to form more than one compound the different masses of one of the elements combining with fixed mass of the other bear a simple ratio.
i) Name the above law.
ii) Explain the above law by taking oxides of carbon.
Answer:
i) Law of multiple proportions.
ii) Carbon reacts with oxygen to form two compounds viz. CO and CO2.
In CO, mass ratio is 12:16
In CO2,mass ratio is 12:32
Ratio of the masses of oxygen combining with a fixed mass of carbon in the two compounds is 16:32 or 1:2, which is a simple whole number ratio.

Question 6.
A compound contains 80% carbon and 20% hydrogen. If the molecular mass is 30 calculate empirical formula and molecular formula.
Answer:
Plus One Chemistry Chapter Wise Questions and Answers Chapter 1 Some Basic Concepts of Chemistry 6

Question 7.
A compound contains 4.07% of hydrogen, 24.27% of carbon and 71.65% of chlorine. The molar mass is 98.96. What is the empirical and molecular formula?
Answer:
Plus One Chemistry Chapter Wise Questions and Answers Chapter 1 Some Basic Concepts of Chemistry 7

Question 8.
Nitrogen forms various oxides.
1. Identify the law of chemical combination illustrated here. Also state the law.
2. Determine the formula of each oxide from the given data and illustrate the law.
Plus One Chemistry Chapter Wise Questions and Answers Chapter 1 Some Basic Concepts of Chemistry 8

Answer:
1. Law of multiple proportions.
When two elements combine to form more than one compound the different mass of one of the elements which combine with the fixed mass of the other element bear a simple ratio.

2.
Plus One Chemistry Chapter Wise Questions and Answers Chapter 1 Some Basic Concepts of Chemistry 9
In NO and NO2, the masses of oxygen combining with a fixed mass (14 g) of nitrogen are in the ratio, 16:32 = 1:2. Similarly, in N2O and N2O3, the masses of oxygen combining with a fixed mass (28 g) of nitrogen are in the ratio, 16:48 = 1:3. These are simple whole number ratios. Hence, the law of multiple poportions is verified.

Plus One Chemistry Some Basic Concepts of Chemistry Four Mark Questions and Answers

Question 1.
Which of the following weighs more?
a) 1 mole of glucose
b) 4 moles of oxygen
c) 6 moles of N
d) 5 moles of sodium
Answer:
a) 1 mole glucose = (72 + 12 + 96) g = 180 g
b) 4 moles of oxygen = 4 × 32g = 128g
c) 6 moles of nitrogen = 6 × 14 g = 84 g
d) 5 moles of Na = 5 × 23 g = 115 g.
Thus, 1 mole glucose weighs more.

Question 2.
3 g of H2 is mixed with 29 g of O2 to yield water.
1. Which is the limiting reagent?
2. Calculate the maximum amount of water that can be formed.
3. Calculate the amount of the reactants which remains unreacted.
Answer:
1.
Plus One Chemistry Chapter Wise Questions and Answers Chapter 1 Some Basic Concepts of Chemistry 10
According to the equation, 4 g H2 requires 32 g
So 3 g H2 requires \(\frac{377 \times 33}{44}\) = 24 g O2.
Here 3 g H2 is mixed with 29 g of O2. All H2 will react. Hence H2 is the limiting reagent.

2. According to the equation, 4 g H2 gives 36 g H2O. Hence 3 g H2 will give 36 × 3/4 = 27 g H2O.

3. Amount of O2 unreacted = (29 – 24)g = 5 g

Question 3.
a) Calculate the mass of oxygen required for the complete burning of 2 g of carbon.
b) Calculate the molar mass of (i) CO2 (ii) CH4
Answer:
Plus One Chemistry Chapter Wise Questions and Answers Chapter 1 Some Basic Concepts of Chemistry 11

Question 4.
One gram mole of a substance contains 6.022 x 1023 molecules.
1. 24 g of carbon is treated with 72 g of oxygen to form CO2. Identify the limiting reagent.
2. Find the number of molecules of CO2 formed in this situation.
Answer:
1.
Plus One Chemistry Chapter Wise Questions and Answers Chapter 1 Some Basic Concepts of Chemistry 12

2 mol of C requires 2 mol of O2.
2 mol C completely reacts with 2 mol of O2 and 0.25 mol O2 and 0.25 mol O2 remains unreacted. Hence, C is the limiting reagent.

2. No. of moles of CO2 formed = 2
∴ No. of molecules of CO2 formed
= 2 × 6.022 × 1023 = 1.2044 × 1024

Plus One Chemistry Chapter Wise Questions and Answers Chapter 1 Some Basic Concepts of Chemistry

Question 5.
One gram mole of a substance contains 6.022×1023 molecules.
i) Find out the number of molecules in 2.8 g of nitrogen.
ii) Which is the heavier-one SO2 molecule or one CO2 molecule?
Answer:
i) No. of molecules in1 mole of N2 = 6.022 × 1023
i.e., No. of molecules in 28 g of N2 = 6.022 × 1023
∴ No. of molecules in 2.8 g N2
Plus One Chemistry Chapter Wise Questions and Answers Chapter 1 Some Basic Concepts of Chemistry 13

Question 6.
a) How can you illustrate the law of multiple proportions by using oxides of metals containing 78.7% and 64.5% of the metal?
b) Match the following:
1/12th the mass of C12 atom – 1 mole
1 g of hydrogen atom – amu
22.4 L O2 at NTP – gram mole
180 g of glucose – gram atom
6.022 × 1023 particles – molar volume
Answer:
a) In 100 g samples of the two oxides, the masses of the metal are 78.7 g and 64.5 g respectively.
First Oxide :
Mass of oxygen = 100 – 78.7 = 21.3 g
No. of parts by mass of oxygen combining with one part by mass of metal =\(\frac{78.7}{21.3}=3.7 \mathrm{g}\)

Second oxide:
Mass of oxygen = 100 – 64.5 = 35.5 g
No. of parts by mass of oxygen combining with one part by mass of metal = \(\frac{64.5}{35.5}=1.9 \mathrm{g}\)
The ratio of masses of oxygen combining with a fixed mass of metal = 3.7 : 1.9 = 2: 1, a simple whole number ratio.

b) 1/12th the mass C12 atom – amu
1 g of hydrogen atom – gram atom
22.4 L O2 at NTP – molar volume
180 g of glucose – gram mole
6.022 × 1023 particles – 1 mole

Question 7.
Calculate
1. The number of molecules present in 1 g of water.
2. The volume of 0.2 mole of sulphur dioxide at STP.
Answer:
1. Number of moles in 1 g water = \(\frac{1}{8}\)
∴ No. of molecules in 1 g water
\(=\frac{1 \times 6.022 \times 10^{23}}{18}=3.35 \times 10^{22}\)

2. Volume of 0.2 mol S02 at STP = 0.2 × 22.4 litre
= 4.48 litre

Question 8.
“One mole of all substances contain the same number of specified particles.”
a) Justify the statement.
b) Howto connect mole, gram mole, and gram atom?
c) What is the relation between number of moles and volume?
d) Calculate the number of moles of a gas in 11.2 L at • STP.
Answer:
a) This statement is true i.e., one mole of all sub-stances contain the same number of specified particles. According to Avogadro’s law.
1 mole of any substance contains 6.022 × 1023 specified particles.

b)
Plus One Chemistry Chapter Wise Questions and Answers Chapter 1 Some Basic Concepts of Chemistry 14
1 gram mole is the molecular mass expressed in gram. It is the mass of 1 mole molecules in gram. Thus, 1 gram mole contains 1 mole molecules.
1 gram atom is the atomic mass expressed in gram. It is the mass of 1 mole atoms in gram. Thus, 1 gram atom contains 1 mole atoms.

c) Number of moles is directly proportional to volume (according to Avogadro law).
Plus One Chemistry Chapter Wise Questions and Answers Chapter 1 Some Basic Concepts of Chemistry 15

Plus One Chemistry Some Basic Concepts of Chemistry NCERT Questions and Answers

Question 1.
Calculate the molecular mass of the following : (3)
1. H2O
2. CO2
3. CH4
Answer:
1. Molecular mass of H2O = 2(1.008 u) +16.00 u
= 18.016u

2. Molecular mass of CO2 = 12.01 u + 2(16.00 u)
= 44.01 u

3. Molecular mass of CH4 = 12.01 u + 4(1.008 u)
= 16.042 u

Question 2.
Calculate the mass percent of different elements present in sodium sulphate (Na2SO4). (2)
Answer:
Plus One Chemistry Chapter Wise Questions and Answers Chapter 1 Some Basic Concepts of Chemistry 16

Question 3
Calculate the amount of carbon dioxide that could be produced when (3)
1. 1 mole of carbon is burnt in air.
2. 1 mole of carbon is burnt in 16 g of dioxygen.
3. 2 moles of carbon are burnt in 16 g of dioxygen.
Answer:
1. The balanced equation for the combustion of carbon dioxide in dioxygen in air is
Plus One Chemistry Chapter Wise Questions and Answers Chapter 1 Some Basic Concepts of Chemistry 17

In air combustion is complete.
Hence, Amount of CO2 produced when 1 mole of carbon is burnt in air = 44 g

2. As only 16 g dioxygen is available it is the limiting reagent.
Hence, amount of CO2 produced = \(\frac{44}{32}\)×16 = 22 g

3. Here again, dioxygen is the limiting reactant. Therefore, amount of CO2 produced from 16 g dioxygen = \(\frac{44}{32}\)×16 = 22g

Question 4.
Chlorine is prepared in the laboratory by treating manganase dioxide (MnO2) with aqueous hydrochloric acid according to the reaction,
4HCl(aq) + MnO2(s) → 2H2O(l) + MnCl2(aq) + Cl2(g)
How many grams of HCl react with 5.0 g of manganese dioxide?
(Atomic mass of Mn = 54.94 u) (2)
Answer:
1 mol of MnO2, i.e., 54.94 + 32 = 86.94 g MnO2 react with 4 moles of HCl, i.e., 4 × 36.5 g = 146 g of HCl.
∴ Mass of HCl reacting with 5.0 g of MnO2
\(=\frac{146}{86.94} \times 5 \mathrm{g}=8.4 \mathrm{g}\)

Question 5.
Calculate the molarity of a solution of ethanol in water in which the mole fraction of ethanol is 0.040. (2)
Answer:
Plus One Chemistry Chapter Wise Questions and Answers Chapter 1 Some Basic Concepts of Chemistry 18

Plus One Chemistry Chapter Wise Questions and Answers Chapter 5 States of Matter

Students can Download Chapter 5 States of Matter Questions and Answers, Plus One Chemistry Chapter Wise Questions and Answers helps you to revise the complete Kerala State Syllabus and score more marks in your examinations.

Kerala Plus One Chemistry Chapter Wise Questions and Answers Chapter 5 States of Matter

Plus One Chemistry States of Matter One Mark Questions and Answers

Question 1.
When a gas is compressed at constant temperature
a) The speeds of the molecules increase
b) The collisions between the molecules increase
c) The speed of the molecules decrease
d) The collisions between the molecules decrease
Answer:
b) The collisions between the molecules increase

Question 2.
The temperature at which a real gas obeys ideal gas law over an appreciable range of pressure is called ___________
Answer:
Boyle temperature or Boyle point

Question 3.
The compressibility factor is given by the expression
Plus One Chemistry Chapter Wise Questions and Answers Chapter 5 States of Matter 1
Answer:
a) \(\frac{p V}{n R T}\)

Question 4.
Two flasks of equal volume contain CO2 and SO2 respectively at 298 K and 1.5 atm pressure. Which of the following is equal in them?
a) Masses of the two gases
b) Rates of effusion
c) Number of molecules
d) Molecular structures
Answer:
c) Number of molecules

Plus One Chemistry Chapter Wise Questions and Answers Chapter 5 States of Matter

Question 5.
With rise in temperature, viscosity of a liquid
a) Increases
b) Decreases
c) Remains constant
d) May increase or decrease
Answer:
b) Decreases

Question 6.
The unit of‘b’ in VanderWaals equation of state.
Answer:
l mol-1

Question 7.
Most probable velocity, average velocity, and root mean square velocity are related by
Answer:
1 : 1.128 : 1.224

Question 8.
The volume of 2.8g of CO at 27°C and 0.821 atm pressure is (R = 0.0821 l atm Km-1 ol-1)
Answer:
3L

Question 9.
The density of gas at 27°C and 1 atm is d. Pressure remaining constant at which of the following temp will its density become 0.75d?
Answer:
400K

Question 10.
The rms velocity of an ideal gas at 27°C is 0.3ms-1. ‘ Its rms velocity at 927°C in (ms-1) is
Answer:
0.6m/s

Plus One Chemistry States of Matter Two Mark Questions and Answers

Question 1.
Find out the relation between the first pair and complete the second pair.
a) Boyle’s law: Temperature
Charles’ law : ……………….
b) Avagadro’slaw: V α n
Ideal gas equation: ……………..
Answer:
a) Pressure
b) pV=nRT

Question 2.
The graphs of Boyle’s law as plotted by Student 1 (Graph 1) and Student 2 (Graph 2) are given below:
Plus One Chemistry Chapter Wise Questions and Answers Chapter 5 States of Matter 2

  1. Which is the correct graph?
  2. Justify your answer.

Answer:

  1. Both the graphs are correct.
  2. According to Boyle’s law, v α 1/p or p α 1/v. This is clear from graph 1. Also, according to Boyle’s law, pv is a constant at constant n and T. This is clear from graph 2.

Plus One Chemistry Chapter Wise Questions and Answers Chapter 5 States of Matter

Question 3.
The rate of diffusion of hydrogen is less than that of oxygen.

  1. Do you agree?
  2. Which law is applied here?
  3. State the law.

Answer:

  1. Yes.
  2. Graham’s law of diffusion.
  3. Rate of diffusion of a gas is inversely proportional to the square root of its molecular mass.

Question 4.
The ideal gas equation has been modified for real gases by applying pressure and volume corrections.

  1. What is the corrected equation known as?
  2. Write the equation and explain the terms.

Answer:

  1. The van derWaals’equation.
  2. \(\left(p+\frac{a n^{2}}{V^{2}}\right)\) (V — nb) = nRT
    where p – pressure, V – volume, n – no. of moles of the gas, a & b – van der Waals’ constants, R – universal gas constant and T – absolute temperature.

Question 5.
‘Moist soil grains are pulled together.’

  1. Name the related phenomenon.
  2. Justify.

Answer:

  1. Surface tension.
  2. This is because the surface area of thin film of water in moist soil is reduced due to surface tension.

Question 6.
1. What is aqueous tension?
2. What is its significance in the determination of pressure of a dry gas?
Answer:
1. The pressure exerted by saturated water vapour at a given temperature is called aqueous tension at that temperature.

2. Pressure of dry gas can be calculated by subtracting aqueous tension from the total presssure of the moist gas.
Pdry gas=PTotal-Aqueous tensi0n

Question 7.
A balloon filled with air, when kept in sunlight bursts after some time.

  1. Name the related law.
  2. Justify.

Answer:

  1. Charles’ law
  2. According to Charles’ law, volume a Temperature. Therefore, the volume increases when temperature increases, When the volume of the gas inside the baloon expanded more than that the balloon could afford, it bursted.

Question 8.
Define surface energy. What is its SI unit?
Answer:
Surface energy is defined as the energy required to rise the surface area of the liquid by one unit. The SI unit of surface energy is J m-2.

Question 9.
a) Based on Boyle’s law how will you show that at a constant temperature, pressure is directly proportional to the density of a fixed mass of the gas?
b) Give the relation between density and molar mass of a gaseous substance.
Answer:
a) According to Boyle’s law,
Plus One Chemistry Chapter Wise Questions and Answers Chapter 5 States of Matter 3

Question 10.
The isotherm of carbon dioxide at various temperatures is given below:
Plus One Chemistry Chapter Wise Questions and Answers Chapter 5 States of Matter 4
1. What is the significance of the shaded area?
2. Identify the pressure at which liquid CO2 appears for the first time when cooled form 30.98 °C. What is this pressure called?
Answer:

  1. At any point in the dome shaped shaded area liquid and gaseous CO2 exists in equilibrium.
  2. 73 atm. Critical pressure (pc).

Plus One Chemistry Chapter Wise Questions and Answers Chapter 5 States of Matter

Question 11.
Certain properties of liquids are given below: Classify them on the basis of effect of temperature on them,

  1. Evaporation
  2. Vapour pressure
  3. Surface tension
  4. Viscosity

Answer:
Properties which increase with increase in temperature: Evaporation & Vapour pressure Properties which decrease with increase in temperature: Surface tension & Viscosity

Question 12.
The size of the water bubbles increases on moving to the surface.
1. Name the law responsible for this.
2. What is your justification?
Answer:
1. Boyle’s law.

2. According to Boyle’s law volume is inversely proportional to pressure. At the bottom of the pond, pressure is greater. So the volume (size) of the bubble was the least. But on coming up, pressure decreases and hence size of the bubble increases.

Question 13.
What are the properties of liquid state?
Answer:
Vapour pressure, Boiling point, Viscosity and Surface tension.

Plus One Chemistry States of Matter Three Mark Questions and Answers

Question 1
Analyse the following graph :
Plus One Chemistry Chapter Wise Questions and Answers Chapter 5 States of Matter 5
1. Name the gas law associated with this graph.
2. State the law.
3. Give the mathematical expression of this law.
Answer:
1. Boyle’s law.

2. It states that at constant temperature, pressure of a fixed amount of gas varies inversly with its volume.

3. Mathematically, Boyle’s law can be written as p ∝ \(\frac{1}{V}\) (at constant T and n)
Or p = k × \(\frac{1}{V}\) where k is the proportionality
constant which depends upon the amount of the gas, temperature of the gas and the units in which p and V are expressed.

Plus One Chemistry Chapter Wise Questions and Answers Chapter 5 States of Matter

Question 2.
1. What are van der Waals’ forces?
2. Which are the different types of van der Waals’ forces?
3. Arrange the van der Waal’s forces in the increasing order of their strength.
Answer:
1. The attractive intermolecular forces are known as van der Waals’ forces.

2. Dispersion forces/London forces, Dipole-Dipole forces, Dipole-Induced dipole forces and Hydrogen bonding.

3. Dispersion forces/London forces < Dipole-Induced dipole forces < Dipole-Dipole forces < Hydrogen bonding

Question 3.
A graphical representation of Charles’ law is given below:
Plus One Chemistry Chapter Wise Questions and Answers Chapter 5 States of Matter 6

  1. What is the temperature corresponding to the point‘A’called? .
  2. What will be the temperature at that point A’ in degree Celsius?
  3. What is the significance of this temperature?

Answer:

  1. Absolute zero temperature
  2. -273.15 °C
  3. Absolute zero is the lowest hypothetical or imaginary temperature at which gases are supposed to occupy zero volume.

Question 4.
Assume that two gases X and Y at the same temperature and pressure have the same volume.
1. Which of the following is correct?
No. of moles of X= No.of moles of Y
No. of moles of X ≠ No.of moles of Y
2. Which law helped you to find the answer?
3. State the law.
Answer:
1. No. of moles of X= No.of moles of Y

2. Avogadro’s law

3. Equal volumes of all gases under the same conditions of temperature and pressure contain equal number of molecules.

Question 5.
During a seminar session in the class, the presenter argued that equal amounts of both H2and N2 on heating at constant pressure will expand in the same rate. Another student objected this argument by saying that they will expand differently since their molecular masses are different.

  1. Who is correct in your opinion?
  2. Which law helped you to reach the answer?
  3. State the law and give its mathematical expression.

Answer:
1. The argument of the presenter is correct.

2. Charles’ law

3. Charles’ law states that pressure remaining constant, the volume of a fixed mass of a gas is directly proportional to its absolute temperature.

Mathematically, V ∝ T (at constant n and P) Or \(\frac{V}{T}\) = K, where K is the proportionality constant which depends on the pressure of the gas, its amount and the unit in which V is expressed.

Plus One Chemistry Chapter Wise Questions and Answers Chapter 5 States of Matter

Question 6.
1. What is an ideal gas?
2. Give the ideal gas equation and explain the terms.
3. Derive the ideal gas equation.
Answer:
1. A gas that follows Boyle’s law, Charles’ law and Avogadro law strictly at all conditions is called an ideal gas.

2. The ideal gas equation is pV = nRT where p = pressure, V = volume, n = number of moles, R = universal gas constant and T = absolute temperature.

3. According to Boyle’s law:
V ∝ \(\frac{1}{p}\) at constant T and n.
According to Charles’ law:
V ∝ T. at constant n and p.
According to Avogadro law,
V ∝ n , at constant p and T Combining the above three equations,
V ∝ \(\frac{nT}{p}\)
⇒ V = R\(\frac{nT}{p}\), where R is the proportionality constant known as universal gas constant.
Or pV = nRT, the ideal gas equation.

Question 7.
Partial pressure of a vessel containing Cl2, CO2 and CO is the sum of the partial pressures of Cl2, O2 and CO.
1. If so, is it correct to say partial pressure of a vessel containing NH3 and HCl gases is the sum of their partial
pressures? Justify.
2. Which law helped you to answer this?
3. State the law.
Answer:
1. No. NH3 reacts with HCl to form NH4CI. Since they are not non-interacting gases, their sum of partial pressures may not be equal to the total pressure.

2. Dalton’s law of partial pressures.

3. It states that the total pressure exerted by the mixture of non-reactive gases is equal to the sum of the partial pressures of individual gases.

Question 8.
The average kinetic kenergy of the gas molecules is directly proportional to the absolute temperature.

  1. Which theory is related to this assumption?
  2. Write the other postulates of this theory.

Answer:
1. Kinetic moleculartheory of gases.

2.

  • The volume of a gas molecule is negligible when compared to the whole volume of the gas.
  • There is no force of attraction between the particles of a gas at ordinary temperature and pressure.
  • The gas molecules are in random motion.
  • During motion, they collide with each other and also with the walls of the container.
  • Gravity has no influence in the movement of gas molecules.
  • Pressure of a gas is due to the collision of gas molecules with the walls of the container.
  • At any particular time, different particles in the gas have different speeds and hence different kinetic energies.

Question 9.
Three mateorological baloons filled with equal amount of helium, rising in the atmosphere are shown below: (Assume that temperature remains constant).
Plus One Chemistry Chapter Wise Questions and Answers Chapter 5 States of Matter 7

  1. Which of the baloons will be at the lowest altitude?
  2. Which law helped you to find the answer?
  3. State the law.

Answer:

  1. C –
  2. Boyle’s law.
  3. At low altitudes pressure is high. According to Boyle’s law for a given mass of a gas, greater the pressure lower is the volume at constant temperature.

Question 10.
‘All the postulates of the kinetic molecular theory of gases are correct.’

  1. Do you agree with the statement?
  2. If no, write the wrong postulates of this theory.
  3. Give justification.

Answer:
1. No.

2.

  • Volume of the molecules of a gas is negligibly small in comparison to the space occupied by the gas.
  •  There is no force of attraction between the molecules of a gas.

3. If assumption i) is correct, the p vs V graph of experimental data (real gas) and that theoritically calculated from Boyle’s law (ideal gas) should coincide. But this never happens.
If assumption ii) is correct, the gas will never liquify. But gases do liquify when cooled and compressed.

Question 11.
Two gases with equal molecular mass will have the same rate of diffusion.’

  1. Do you agree?
  2. Explain.
  3. Substantiate your answer with an example.

Answer:
1. Yes.

2. According to Graham’s law of diffusion the rate of diffusion depends only on the molecular mass. So if the molecular masses are the same, their rate of diffusion is same.

3. Both CO and N2 have the same molecular mass (28 g mol-1)
Rate of diffusion of CO = Rate of diffusion of N2

Plus One Chemistry Chapter Wise Questions and Answers Chapter 5 States of Matter

Question 12.
Water can be boiled more quickly on the top of a mountain.

  1. Do you agree?
  2. What is the reason?
  3. What is called boiling point of a liquid?
  4. How normal boiling point and standard boiling point differ?

Answer:
1. Yes.

2. As we move to the top of a mountain atmospheric pressure decreases and hence boiling point decreases. So water boils quickly.

3. Boiling point of a liquid is the temperature at which the vapour pressure of a liquid is equal to the external pressure or atmospheric pressure.

4. The boiling point at 1 atm pressure is called normal boiling point. The boiling point at 1 bar pressure is called standard boiling point.

Question 13.
Ethanol flows faster than honey.

  1. Name the related phenomenon.
  2. Explain this phenomenon.
  3. What is the effect of temperature on this?

Answer:
1. Viscosity.

2. Viscosity is a measure of resistance to flow which arises due to the internal friction between layers of fluid as they slip past one another while liquid flows.

3. Viscosity of liquids decreases as the temperature rises because at high temperature molecules have high kinetic energy and can overcome the intermolecular forces to slip past one another between the layers.

Question 14.
Liquid drops attain spherical shape.

  1. Which property of liquids is responsible for this?
  2. Explain the phenomenon and justify.
  3. Suggest another consequence of this phenomenon.

Answer:
1. Surface tension.

2. Surface tension is defined as the force acting per unit length perpendicular to the line drawn on the surface of liquid. The lowest energy state of the liquid will be when surface area is minimum. Spherical shape satisfies this condition.

3. Fire polishing of glass – On heating, the glass melts and the surface of the liquid tends to take the rounded shape at the edges due to surface tension, which makes the edges smooth.

Question 15.
Vapour pressure is an important property of liquids.

  1. What is vapour pressure?
  2. How boiling point and vapour pressure are related?
  3. Pressure cooker is used for cooking food at higher altitudes. Give reason.

Answer:
1. Vapour pressure of a liquid is the pressure exerted by the vapour which is in equilibrium with liquid at a given temperature.

2. Boiling point of a liquid is the temperature at which the vapour pressure of liquid becomes equal to the atmospheric pressure. Thus, lower the vapour pressure of a liquid higher will be its boiling point and vice-versa.

3. At high altitudes atmospheric pressure is low. Therefore, liquids at high altitudes boil at lower temperatures in comparison to that at sea level. In a pressure cooker, the internal pressure is greater than atmospheric pressure. Hence, in a pressure cooker water boils at a temperature higher than its normal boiling point of 100 °C. Thus, cooking becomes more effective.

Question 16.
Assume that ‘A’, ‘B’ and ‘C’ are three non-reacting gases kept in a vessel at a constant temperature.
Then, PTotal= PA + PB + PC
1. Name the related law.
2. How can you explain the above law on the basis of kinetic molecular theory of gases?
Answer:
1. Dalton’s law of partial pressures.

2. In the absence of attractive forces, the particles of the gas behave independent of one another. The same is true even if there are more than one type of molecules. Thus, the number of molecules colloding the unit area of the wall per second at a given temperature, fora fixed amount of the gas issame.lt implies that the partial pressure of the gas will be unaffected by the presence of the molecules of other gases. But, the total pressure exerted is duet the impact of molecules of all the gases. Hence, the total pressure would be the sum of the partial pressures of the gases.

Question 17.
1. Write the general equation which relates the different variables of a gas used to describe the state of any ideal gas.
2. A flask at 295 K contains a gaseous mixture of N2 and O2 at a total pressure of 1.8 atm. If 0.2 moels of N2 and 0.6 moles of O2 are present, find the partial pressures of N2 and O2.
3. What is meant by Boyle temperature or Boyle point?
Answer:
1. PV = nRT
2.
Plus One Chemistry Chapter Wise Questions and Answers Chapter 5 States of Matter 8

3. It is the temperature at which a real gas obeys ideal gas law over an appreciable range of pressure.

Question 18.
1. Liquid tries to rise or fall in the capillary. Name the related phenomenon.
2. What is the effect of temperature on the above phenomenon?
3. WhatistheSI unit of the above phenomenon.
Answer:
1. Surface tension.

2. The magnitude of surface tension of a liquid depends on the attractive forces between the molecules. When the attractive forces are large, the surface tension is large. Increase in temperature increases the kinetic energy of the molecules and effectiveness of intermolecular attraction decreases, so surface tension decreases as the temperature is raised.

3. N m-1.

Question 19.
1. Define critical temperature (Tc ).
2. CO2 cannot be liquified above 31.1°C. Why?
3. The critical temperatures of ammonia and carbon dioxide are 405.5 Kand 304.10 K respectively. On cooling, which of these gases will liquify first? Justify.
Answer:
1. It is the highest temperature at which a gas can
be liquified by applying external pressure.

2. The critical temperature (Tc ) of CO2 is 30.98°C. This is the highest temperature at which liquid CO2 is observed. Above this temperature it is gas.

3. Ammonia. This is because, on cooling, critical temperature of ammonia will be reached first. Liquefaction of carbon dioxide will require more cooling.

Plus One Chemistry Chapter Wise Questions and Answers Chapter 5 States of Matter

Question 20.
a) Will water boils at higher temperature at sea level or at top of a mountain. Explain.
b) A vessel of 120 mL capacity contains a certain amount of gas at 35 °C and 1.2 bar pressure. The gas is transferred to another vessel of volume 180 mL at 35 °C. What would be its pressure?
Answer:
a) When atmospheric pressure decreases boiling point of the liquid also decreases. So the boiling point of water at sea level is not same as that at the top of a mountain. Atmospheric pressure decreases from sea level as we go high. Hence, the boiling point at the top of the mountain is less than that at sea level.
Plus One Chemistry Chapter Wise Questions and Answers Chapter 5 States of Matter 9

Question 21.
Real gases deviate from ideal behaviour.
1. What are the two wrong postulates of kinetic theory of gases, responsible for deviation of real gases from ideal behaviour?.
2. When do real gases deviate from ideal behaviour?
Answer:
1.

  • Volume of the molecules of a gas is negligibly small in comparison to the space occupied by the gas.
  • There is no force of attraction between the molecules of a gas.

2. Real gases deviate from ideal behaviour at high pressure and low temperature, when the gas molecules are very close to each other.

Question 22.
1. What is meant by compressibility factor, Z?
2. What is the significance of compressibility factor?
3. A plot of pV/nRT of oxygen gas against p is as follows:
Plus One Chemistry Chapter Wise Questions and Answers Chapter 5 States of Matter 10
a) Is the gas ideal or real?
b) Write the equation of state of the above gas.
Answer:
1. Compressibility factor (Z) is the ratio of product pV and nRT. Mathematically, Z = \(z=\frac{p V}{n R T}\).

2. The deviation of real gases from ideal behaviour can be measured in terms of compressibility factor.

3. a) Real gas.
\(\left[p+\frac{a n^{2}}{V^{2}}\right]\)[V-nb] = nRT (van der Waals’ equation)

Question 23.
a) What is the difference between gas and vapour?
b) Analyse the vapour pressure vs temperature curve shown below:
Plus One Chemistry Chapter Wise Questions and Answers Chapter 5 States of Matter 11
Arrange the compounds shown in the graph in the decreasing order of their normal boiling points,

c) The density of a gas was found to be 2.92 g L1 at 27 °C and 2.0 atm. Calculate the molar mass of the gas.
Answer:
a) A gas below its critical temperature can be liquified by applying pressure. Under these conditions, it is called vapour of the substance.

b) water > ethyl alcohol > carbon tetrachloride > diethyl ether
Plus One Chemistry Chapter Wise Questions and Answers Chapter 5 States of Matter 12

Question 24.
1. How will you account for the observation that automobile tyre is inflated with lesser air in summer than in winter?
2. A sample of gas occupies 250 mLat27 °C. What volume will it occupy at 35 °C if there is no change in pressure?
Answer:
1. This can be explained on the basis of Gay Lussac’s law, according to which at constant volume, pressure of a fixed amount of a gas varies directly with the temperature. In summer season the temperature will be higher. Hence, pressure will increase and the tyre may burst if filled with more air. But during winter temperature is low and hence pressure will below.

2. According to Charles’ law,
Plus One Chemistry Chapter Wise Questions and Answers Chapter 5 States of Matter 13

Question 25.
Real gases behave ideally at low temperature and high pressure.

  1. Is the above statement correct or not?
  2. Justify.
  3. Write the van der Waals’ equation for 1 mole of a real gas.

Answer:
1. The statement is wrong.

2. This is because real gases behave ideally at high temperature and low pressure, when the gas molecules are far apart.

3. \(\left(p+\frac{a}{V^{2}}\right)\)(V – b) = RT

Question 26.
1. Distinguish between real gas and ideal gas.
2. Explain the deviation of the following gases from ideal behaviouron the basis of the pV vs. p plot. CO, CH4, H2 and He.
Answer:
1. Real gas do not follow, Boyle’s law, Charles law, and Avagadro law perfectly under all conditions. Ideal gas follow, Boyle’s law, Charles’ Law and Avagadro law strictly under all conditions.

2.
Plus One Chemistry Chapter Wise Questions and Answers Chapter 5 States of Matter 14
It can be seen that at constant temperature pV vs p plots for these gases are not straight lines. Two types of curves are seen. In the curves for H2 and He, as the pressure increases the value of pV also increases. These gases show positive deviation from ideal behaviour at all pressures. The second type of plot is seen in the case of CO and CH4. For these gases the pV value decreases with increase in pressure and reaches to a minimum value characteristics of the gas. After that PV value starts increasing. The curve then crosses the line for ideal gas and after that shows positive deviation continuously.

Question 27.
1. What is meant by laminar flow?
2. Derive the expression for the force responsible for flow of layers of a liquid.
Answer:
1. When a liquid flows overa fixed surface, the layer of molecules in the immediate contact of surface is stationary. The velocity of upper layers increases as the distance of layers from the fixed layer increases. This type of flow in which there is a regular gradation of velocity in passing from one layer to the next is called laminar flow.

2. Consider three layers of a flowing liquid as shown below:
Plus One Chemistry Chapter Wise Questions and Answers Chapter 5 States of Matter 15
For any layer, the layer above it accelerates its flow and the layer below this retards its flow. If the velocity of the layer at a distance dz is changed by a value du then velocity gradient is given by \(\frac{du}{dz}\). A force is required to maintain the
flow of layers. This force is proportional to the area of contact (A) of layers and velocity gradient.
Plus One Chemistry Chapter Wise Questions and Answers Chapter 5 States of Matter 16
where η is a proportionality constant called coefficient of viscosity.

Question 28.
1. What are London forces?
2. What is the relation between London forces and the distance between the particles?
Answer:
1. The attractive force between two temporary dipoles is known as London forces or Dispersion forces.

2. London forces are always attractive and the interaction energy is inversely proportional to the sixth power of the distance between two interacting particles.

Question 29.
1. What is meant by thermal energy and thermal motion?
2. Can oxygen exist as a gas at -273.15°C? Write the significance of this temperature.
Answer:
1. Thermal energy is the energy of a body arising from motion of its atoms or molecules. The movement of particles due to thermal energy is called thermal motion.

2. At – 273.15 °C oxygen will not exist as a gas. In fact all the gases get liquified before this temperature is reached. It is the absolute zero of temperature, which is the lowest hypothetical or imaginary temperature at which gases are supposed to occupy zero volume.

Plus One Chemistry Chapter Wise Questions and Answers Chapter 5 States of Matter

Question 30.
Molecues of a gas are in a state of continuous motion
1. What is most probable speed?
2. Give the equation for average speed of molecules.
Answer:
1. Most probable speed is the speed possessed by the maximum fraction of molecules of the gas at a given temperature.

2. If there are ‘n’ molecules in a sample and their individual speeds are u1, u2 ……… un, then, average speed of molecules, uav is given by the equation:
\(u_{a v}=\frac{u_{1}+u_{2}+\ldots+u_{n}}{n}\)

Plus One Chemistry States of Matter Four Mark Questions and Answers

Question 1.
1. State the Avogadro law.
2. Give the mathematical expression of this law.
3. What is the value of molar volume of an ideal gas at 273.15 K and 1 bar?
4. Show that, at constant temperature and pressure, the density of an ideal gas is proportional to its molar volume.
Answer:
1. It states that equal volumes of all gases under the same conditions of temperature and pressure contain equal number of molecules.

2. v ∝ n (at constant P and T)
⇒ V = k × n where k is a proportionality constant.

3. 22.71098 L

4. According to Avogadro law, for n moles of an ideal gas,
V = k × n
But n = \(\frac{m}{M}\) where m is the mass of the gas and M is its molar mass.
Thus, V = k × \(\frac{m}{M}\)
On rearranging the above equation,
M = k × \(\frac{m}{V}\) = k × d
⇒ M ∝ d

Question 2.
The speed of molecules is a measure of their average kinetic energy.
a) What is root mean square speed?
b) Give the equation for root mean square speed.
c) Calculate the following:
i) Root mean square speed of methane molecule at 27°C.
ii) Most probable speed of nitrogen molecule at 25°C
Answer:
a) It is the square root of the mean of the squares of speeds of various molecules of the gas at a given temperature.
Plus One Chemistry Chapter Wise Questions and Answers Chapter 5 States of Matter 17

Question 3.
The graph A is drawn at high temperature and low pressure and graph B is drawn at low temperature and high pressure.
Plus One Chemistry Chapter Wise Questions and Answers Chapter 5 States of Matter 18
a) Which graph represents ideal behaviour?
b) Give the equation for combined gas law.
c) A baloon occupies volume of 700 mL at 25°C and 760 mm of pressure. What will be its volume at higher attitude when temperature is 15°C and pressure is 600 mm Hg.
Answer:
Plus One Chemistry Chapter Wise Questions and Answers Chapter 5 States of Matter 19

Question 4.
1. Give the relationship among the three types of molecular speeds.
2. Drawthe Maxwell-Boltzmann distribution showing all the molecularspeeds.
3. Which of the following molecules will have the higher value of most probable speed at the same temperature, N2 or Cl2? Justify.
Answer:
1. Root mean square speed, average speed and the most probable speed have the following relationship:
Urms > Uav > Ump
2.
Plus One Chemistry Chapter Wise Questions and Answers Chapter 5 States of Matter 20
3. N2. This is because at the same temperature, gas molecules with heavier mass have slower speed than lighter gas molecules. At the same temperature, lighter nitrogen molecules move faster than heavier chlorine molecules. Hence, at any given temperature, nitrogen molecules have higher value of most probable speed than the chlorine molecules.

Plus One Chemistry States of Matter NCERT Questions and Answers

Question 1.
What will be the minimum pressure required to compress 500 dm3 of air at 1 bar to 200 dm3 at 30 °C? (2)
Answer:
p1 = 1 bar p2 = ?
V1 = 500 dm³ V2 = 200 dm³
Temperature remains constant.
According to Boyle’s law,
Plus One Chemistry Chapter Wise Questions and Answers Chapter 5 States of Matter 21

Question 2.
Using the equation of state pV=nRT show that at a given temperature, the density of the gas is proportional to the gas pressure p. (2)
Answer:
According to ideal gas equation :
Plus One Chemistry Chapter Wise Questions and Answers Chapter 5 States of Matter 22

Question 3.
The density of a gas is found to be 5.46 g/dm³ at 27°C and under 2 bar pressure. What will be its density at STP. (3)
Answer:
Plus One Chemistry Chapter Wise Questions and Answers Chapter 5 States of Matter 23

Question 4.
Calculate the volume occupied by 8.8 g of CO2 at 31.1°C and 1 bar pressure (R = 0.083 bar L K-1 mol-1). (2)
Answer:
Plus One Chemistry Chapter Wise Questions and Answers Chapter 5 States of Matter 24

Question 5.
Explain the significance of van der Waal parameters. (2)
Answer:
The van der Waal parameter ‘a’ is a measure of the magnitude of intermolecular forces. The van der Waal parameter ‘b’ which is also called co-volume is a measure of effectve size of the gas molecules.

Plus One Computer Science Chapter Wise Questions and Answers Chapter 9 String Handling and I/O Functions

Students can Download Chapter 9 String Handling and I/O Functions Questions and Answers, Plus One Computer Science Chapter Wise Questions and Answers helps you to revise the complete Kerala State Syllabus and score more marks in your examinations.

Kerala Plus One Computer Science Chapter Wise Questions and Answers Chapter 9 String Handling and I/O Functions

Plus One String Handling and I/O Functions One Mark Questions and Answers

Question 1.
To read a single character for gender i.e. ‘m’ or ‘f’ ________ function is used.
Answer:
(a) getch()
(b) getchar()
(c) gets()
(d) getline()
Answer:
(b) getchar()

Plus One Computer Science Chapter Wise Questions and Answers Chapter 9 String Handling and I/O Functions

Question 2.
To use getchar(), putchar(), gets() and puts(), which header file is used?
(a) iostream
(b) cstdio
(c) input
(d) output
Answer:
(b) cstdio

Question 3.
To use cin and cout, which header file is needed?
(a) iostream
(b) cstdio
(c) input
(d) output
Answer:
(a) iostream

Question 4.
Predict the output of the following code snippet.
#include<cstdio>
int mainO
{
char name[ ] = “ADELINE”;
for(int i=0; name[i]!=’\0′;i++)
putchar(name[i]);
}
Answer:
The output is “ADELINE”.

Question 5.
From the following which is equivalent to the function getc(stdin).
(a) putchar()
(b) gets()
(c) getchar()
(d) puts()
Answer:
(c) getchar()

Question 6.
From the following which is equivalent to the function putc(ch, stdout).
(a) putchar(ch)
(b) ch = gets()
(c) ch = getchar()
(d) puts(ch )
Answer:
(a) putchar(ch)

Plus One Computer Science Chapter Wise Questions and Answers Chapter 9 String Handling and I/O Functions

Question 7.
To print a single character at a time which function is used?
(a) puts()
(b) putchar()
(c) gets()
(d) getchar()
Answer:
(b) putchar()

Question 8.
To read a string _______ function is used.
(a) puts()
(b) putchar()
(c) gets()
(d) getchar()
Answer:
(b) gets()

Question 9.
To print a string _______ function is used.
(a) puts()
(b) putchar()
(c) gets()
(d) getchar()
Answer:
(b) puts()

Question 10.
Consider the following code snippet.
main()
{
char str[80];
gets(str);
for(int i=0. len=0;str[il!=’\0′;i++.len++);
cout<<“The length of the string is ” <<len;
}
Select the equivalent forthe under lined statement from the following
(a) int len = strlen(str)
(b) int len = strcmp(str)
(c) int len = strcount(str)
(d) None of these
Answer:
(a) int len = strlen(str)

Plus One Computer Science Chapter Wise Questions and Answers Chapter 9 String Handling and I/O Functions

Question 11.
Arjun wants to read a string with spaces from the following which is suitable.
(а) cin>>
(b) cin.getline(str,80)
(c) str = getc(stdin)
(d) none of these
Answer:
(b) cin.getline(str,80)

Question 12.
State whether the following statement is true or false. The ‘<<‘ insertion operator stops reading a string when it encounters a space.
Answer:
True

Question 13.
_________ function is used to copy a string to another variable. (SAY-2016) (1)
Answer:
strcpy();

Question 14.

  1. Write the declaration statement for a variable ‘name’ in C++ to store a string of maximum length 30.
  2. Differentiate between the statement cin>>name and gets (name) for reading data to the variable ‘name’. (SAY-2016)

Answer:
1. char name[31];(One for null(\0) character).

OR

cin>> does not allows space. It will take characters up to the space and characters after space will be truncated . Here space is the delimiter. Consider the following code snippet that will take the input upto the space.
#include<iostream>
using namespace std;
int main()
{
char name[20];
cout<<“Enter your name:”;
cin>>name;
cout<<“Hello “<<name;
}
If you input a name “Alvis Emerin” then the output will be Hello Alvis. The string after space is truncated.

2. gets(): This function is used to get a string from the keyboard including spaces. Considerthe following code snippet that will take the input including the space.
#include<iostream>
#include<cstdio>
using namespace std;
int main()
{
char name[20];
cout<<“Enter your name:”;
gets(name);
cout<<“Hello “<<name;
}
If you input a name “Alvis Emerin” then the output will be Hello Alvis Emerin.

Question 15.
What is the advantage of using gets() function in the C++ program to input string data? Explain with an example.
Answer:
gets() function is used to get a string from the keyboard including spaces. Consider the following code snippet that will take the input including the space.
#include<iostream>
#include<cstdio>
using namespace std;
int main()
{
char name[20];
cout<<“Enter your name:”;
gets(name);
cout<<“Hello “<<name;
}
If you input a name “Alvis” then the output is Hello Alvis.

Plus One String Handling and I/O Functions Two Mark Questions and Answers

Question 1.
In a C++ program, you forgot to include the header file iostream. What are the possible errors occur in that Program? Explain?
Answer:
Prototype error. To use cin and cout the header file iostream is a must.

Plus One Computer Science Chapter Wise Questions and Answers Chapter 9 String Handling and I/O Functions

Question 2.
Categorise the following into three according to their relationship
iostream, cstdio, gets(), puts(), getchar(), putchar(), getline(), write(), cin, cout.
Answer:
Plus One Computer Science Chapter Wise Questions and Answers Chapter 9 String Handling and IO Functions 1

Question 3.
Pick the odd one out from the following and give reason.
gets(), getline(), getch() getchar().
Answer:
getline() – It is a stream function whereas the others are console functions.

Question 4.
My_name is a variable contains a string. Write two different C++ statements to display the string. (SAY-2016) (2)
Answer:

  1. cout<<my_name;
  2. puts(my_name);

Question 5.
Suggest most suitable built-in function in C++ to perform the following tasks: (MARCH-2016) (2)

  1. To find the answer for 53
  2. To find the number of characters in the string “KERALA” “HAPPY NEW YEAR”
  3. To get back the number 10 if the argument is 100.

Answer:

  1. pow(5,3);
  2. strlen(“KERALA”)
  3. tolower(‘M’)
  4. sqrt(100);

Question 6.
Read the following C++ statements:
charstr[50];
cin>>str;
cout<<str;
During execution, if the string given as input is “GREEN COMPUTING”, the output will be only the word “GREEN”. Give reason for this. What modification is required to get the original string as output? (SCERT SAMPLE -1) (2)
Answer:
cin>>word;
cout<<word;
It displays “HAPPY” because cin takes characters upto the space. That is space is the delimiter for cin. The string after space is truncated. To resolve this use gets() function.

Because gets() function reads character upto the enter key.
Hence gets(word);
puts(word);
Displays “HAPPY NEW YEAR”

Question 7.
Suppose M[5][5] is a 2D array that contains the elements of a square matrix. Write C++ statements to find the sum of the diagonal elements. (2)
Answer:
gets() function is used to get a string from the keyboard including spaces. To use gets () function the header file cstdio must be included. It reads the characters upto the enter key pressed by the user.
eg:
char name[20];
cout << “Enter your name”;
gets(name);
cout<< “Hello”<< name;
When the user gives Alvis Emerin. It displays as “Hello Alvis Emerin”.

Plus One String Handling and I/O Functions Three Mark Questions and Answers

Question 1.
Suresh wants to print his name and native place using a C++ program. The program should accept name and native place first.
Name is: Suresh Kumar
Address is: Alappuzha
Answer:
#include<iostream>
#include<cstdio>
using namespace std;
int main()
{
char name[20],place[20];
cout<<“Enter your name”;
cin.getline(name,80);
cout<<“Enter your place”;
cin.getline(place,80);
cout<<“Your name is puts(name);
cout<<“Your place is puts(place);
}

Plus One Computer Science Chapter Wise Questions and Answers Chapter 9 String Handling and I/O Functions

Question 2.
“Programming is Fun”. Write a C++ program to read a string like this in lower case and print it in UPPER CASE. With out using toupper() library function.
Answer:
using namespace std;
#include<cstdio>
int main()
{
char line[80];
int i;
puts(Enter the string to convert”);
gets(line);
for(i=0;line[i]!=’\0′;i++)
if (Iine[i]>=97 && line[i]<=122)
line[i]=line[i] – 32;
puts(line);
}

Question 3.
An assignment Kumar has written a C++ program which reads a line of text and print the number of vowels in it. What will be his program code?
Answer:
#include<cstdio>
#include<cctype>
#include<iostream>
using namespace std;
int main()
{
char line[80];
int i,vowel=0;
puts(Enter a string”);
gets(line);
for(i=0;line[i]!=’\0′;i++)
switch(tolower(line[i]))
{
case ‘a’:
case ‘e’:
case ‘i’:
case ‘o’:
case ‘u’:
vowel++;
}
cout<<“The number of vowels is “<<vowel;

Question 4.
What will be the output of the following code if the user enter the value “GOOD MORNING”.
1. char string [80];
gets(string);
cout<<string;

2. char string [80];
cin>>string;
cout<<string;

3. charch;
ch = getchar();
cout<<ch;

4. char string [80];
cin.getline(string,9);
cout<<string;
Answer:

  1. GOOD MORNING
  2. GOOD
  3. G
  4. GOOD MORN

Question 5.
Consider the following code snippet.
int main()
{
int n;
cout<<“Enter a number”;
cin>>n;
cout<<“The number is “<<n;
}
Write down the names of the header files that must be included in this program
Answer:
Here cin and cout are used so the header file iostream must be included.

Plus One Computer Science Chapter Wise Questions and Answers Chapter 9 String Handling and I/O Functions

Question 6.
Write a program to display the following output.
A
BB
CCC
#include<iostream>
using namespace std;
int main()
{
char str[]=”ABC”;
int i,j;
for(i=0;i<3;i++)
{
for(j=0;j<=i;j++)
cout<<str[i];
cout<<endl;
}
}

Question 7.
Distinguish getchar and gets.
Answer:
getchar is a character function but gets is a string function. The header file cstdio.h must be included. It reads a character from the keyboard.
Eg.
char ch;
ch = getchar();
cout<<ch;
gets is used to read a string from the keyboard. It reads the characters upto enter key. The header file cstdio must be included.
char str[80J;
cout<<“Enter a string”;
gets(str);

Question 8.
Distinguish putch and puts.
Answer:
putch is a character function but puts is a string function. The header file cstdio must be included. It prints a character to the monitor.
Eg:
char ch;
ch = getc(stdin);
putch(ch);
puts is used to print a string. The header file stdio.h must be included.
charstr[80];
puts(“Entera string”);
gets(str);
puts(str);

Question 9.
Write a program to check whether a string is palindrome or not. (A string is said to be palindrome if it is the same as the string constituted by reversing the characters of the original string. eg: “MALAYALAM”, “MADAM”, “ARORA”, “DAD”, etc.)
Answer:
#include<iostream>
using namespace std;
int main()
{
char str[40];
int len,i,j;
cout<<“Enter a string:”;
cin>>str;
for(len=0;str[len]!-\0′;len++);
for(i=0,j=len-1;i<len/2;i++,j–)
if(str[i]!=str[j])
break;
if(i==len/2) .
cout<<str<<” is palindrome”;
else
cout<<str<<” is not palindrome”;
}

Question 10.
Explain multi-character function.
Answer:
getline() and write() functions are multi character functions:
1. getline() It reads a line of text that ends with a newline character. It reads white spaces also.
eg:
char line[80];
cin.getline(line,80);

2. write() It is used to display a string.
Eg.
char line[80];
cin.getline(line,80);
cout.write(line,80);

Plus One Computer Science Chapter Wise Questions and Answers Chapter 9 String Handling and I/O Functions

Question 11.
Read a string and print the number of vowels.
Answer:
#include<cstdio>
#include<cctype>
#include<iostream>
using namespace std;
int main()
{
char line[80];
int i,vowel=0;
puts(“Enter a string”);
gets(line);
for(i=0;line[i]!=’\0′;i++)
switch(tolower(line[i]))
{
case ‘a’:
case ‘e’:
case ‘i’;
case ‘o’:
case ‘u’:
vowel++;
}
cout<<“The number of vowels is in the string is “<< vowel;
}

Question 12.
Distinguish between get() and put() functions.
Answer:
get() function:
get() is an input function. It is used to read a single character and it does not ignore the white spaces and newline character.
Syntax is cin.get(variable);
eg: char ch;
cin.get(ch);

put() function:
put() is an output function. It is used to print a character.
Syntax is cout.put(variable);
eg:
charch;
cin.get(ch);
cout.put(ch);

Question 13.
Write a program to read a string and print the number of consonants.
Answer:
#include<iostream>
#include<cstdio>
#include<cctype>
using namespace std;
int main()
{
char str[40],ch;
int consonent = 0,i;
cout<<“Enter a string:”;
gets(str);
for(i=0;str[i]!=’\0′;i++)
{
ch = toupper(str[i]);
if(ch>=’B’ && ch<=’Z’)
if(ch!=’E’&& ch!=’I’&& ch!=’0’&& ch!=’U’)
consonent++;
}
cout<<“The number of consonents is “<<consonent;
}

Plus One Computer Science Chapter Wise Questions and Answers Chapter 9 String Handling and I/O Functions

Question 14.
Write a program to read a string and print the number of spaces.
Answer:
#include<iostream>
#include<cstdio>
using namespace std;
int main()
{
char str[40];
int space=0,i;
cout<<“Enter a string:”;
gets(str);
for(i=0;str[i]!=’\0′;i++)
if(str[i]==32)
space++;
cout<<“The number of spaces is “<<space;
}

Question 15.
Describe in detail about the unformatted console I/O functions.
Answer:
1. Single character functions: This function is used to read or print a character at a time,
(i) getchar():
It reads a character from the keyboard and store it in a character variable.
eg:
char ch;
ch=getchar();

(ii) putchar():
This function is used to print a character on the screen.
eg:
char ch;
ch = getchar();
putchar(ch);

2. String functions This function is used to read or print a string.
(i) gets():
This function is used to read a string from the keyboard and store it in a character variable.
eg:
charstr[80];
gets(str);

(ii) puts():
This function is used to display a string on the screen.
eg:
char str[80];
gets(str);
puts(str);

Question 16.
Write a program to input a string and find the number of uppercase letters, lowercase letters, digits, special characters and white spaces.
Answer:
#include<iostream>
#include<cstdio>
using namespace std;
int main()
{
char str[100];
int i,digit=0, Ualpha=0, Lalpha=0, special=0, wspace=0;
cout<<“Enter a string:”;
gets(str);
for(i=0;str[i]!=’\0′;i++)
if(str[i]>=48 && str[i]<=57)
digit++;
else if(str[i]>=65 && str[i]<=90)
Ualpha++;
else if(str[i]>=97 && str[i]<=122)
Lalpha++;
else if(str[i]==’ ‘ || str[i]==’\t’)
wspace++;
else
special++;
cout<<“The number of alphabets is “<<Ualpha+Lalpha<<
” the number of Uppercase letters is “<<Ualpha<< ” the number of Lowercase letters is “<<Lalpha<<” the number of digits is “<<digit<<” the special characters is “<<special<<” and the number of white spaces is “<<wspace;
}

Plus One Computer Science Chapter Wise Questions and Answers Chapter 9 String Handling and I/O Functions

Question 17.
Write a program to count the number of words in a sentence.
Answer:
#include<iostream>
#include<cstdio>
using namespace std;
int main()
{
int i,words=1;
char str[80];
cout<<“Enter a string\n”;
gets(str);
for(i=0;str[i]!=’\0′;i++)
if(str[i]==32)
words++;
cout<<“The number of words is “<<words;
}

Question 18.
Write a program to input a string and replace all lowercase vowels by the corresponding uppercase letters.
Answer:
#include<iostream>
#include<cstdio>
using namespace std;
int main()
{
char str[100];
int i;
cout<<“Enter a string:”;
gets(str);
for(i=0;str[i]!=\0′;i++)
if(str[i]>=65 && str[i]<=90 || str[i]> = 97 && str[i]<=122)
switch(str[i])
{
case ‘a’:
str[i] = str[i]-32;
break;
case ‘e’:
str[i] = str[i]-32;
break;
case ‘i’:
str[i] = str[i]-32;
break;
case ‘o’: .
str[i] = str[i]-32;
break;
case ‘u’:
str[i] = str[i]-32;
}
cout<<str;
}

Question 19.
Write a program to input a string and display its reversed string using console I/O functions only. For example if the input is “AND” the output should “DNA”.
Answer:
#include<iostream>
using namespace std;
int main()
{
char str[40],rev[40];
int len.ij;
cout<<“Enter a string:”;
cin>>str;
for(len=0;str[len]!=’\0′;len++);
for(i=0,j=len-1 ;i<len;i++,j–)
rev[il=str[j];
rev[i]=’\0′;
cout<<“The reversed string is “<<rev;
}

Plus One Computer Science Chapter Wise Questions and Answers Chapter 9 String Handling and I/O Functions

Question 20.
Write a program to input a word(say COMPUTER) and create a triangle as follows.
C
C O
C O M
C O M P
C O M P U
C O M P U T
C O M P U T E
C O M P U T E R
Answer:
#include<iostream>
#include<cstring>//for strlen()
using namespace std;
int main()
{
charstr[20];
cout<<“enter a word(eg.COMPUTER):”;
cin>>str;
int ij;
for(i=0;i<strlen(str);i++)
{
for(j=0;j<=i;j++)
cout<<str[j]<<“\t”;
cout<<endl;
}
}

Question 21.
Write a program to input a line of text and display the first characters of each word. Use only console I/O functions. For example, if the input is “Save Water, save Nature”, the output should be “SWSN”.
Answer:
#include<iostream>
#include<cstdio>
using namespace std;
int main()
{
int i;
charstr[80];
cout<<“Enter a string\n”;
gets(str);
if(str[0]!=32)
cout<<str[0];
for(i=0;str[i]!=’\0′;i++)
if(str[i]==32 && str[i+1]!=32)
cout<<str[i+1];
}

Plus One Computer Science Chapter Wise Questions and Answers Chapter 12 Internet and Mobile Computing

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

Kerala Plus One Computer Science Chapter Wise Questions and Answers Chapter 12 Internet and Mobile Computing

Plus One Internet and Mobile Computing One Mark Questions and Answers

Question 1.
A network of smaller networks that exists all over the world is called ____________
Answer:
Internet

Plus One Computer Science Chapter Wise Questions and Answers Chapter 12 Internet and Mobile Computing

Question 2.
ARPANET means ______
Answer:
Advanced Research Projects Agency Network

Question 3.
Odd one out.
(a) Internet explorer
(b) Mozilla
(c) Netscape navigator
(d) Windows Explorer
Answer:
(d) Windows explorer, the others are browsers.

Question 4.
Odd man out.
(a) Word
(b) Excel
(c) PowerPoint
(d) Mosaic
Answer:
(d) Mosaic. It is a browser, others are MS Office packages.

Question 5.
The interface between user and computer hardware is called operating system then what about the interface between user and internet (www)?
Answer:
Browser

Question 6.
With the help of this the user can search informations provided on the internet. What is it?
Answer:
Browser.

Question 7.
Benhur wants to navigate through the web pages from the following which will help him?
(a) A browser
(b) MS Word
(c) Tally
(d) Paint
Answer:
(a) A browser

Plus One Computer Science Chapter Wise Questions and Answers Chapter 12 Internet and Mobile Computing

Question 8.
I am a piece of software. With the help of me a user can search information from the internet and navigate through the web pages. Who am I?
Answer:
Browser

Question 9.
Anil told you that he was browsing at that time. From the following choose the right one.
(a) He was visiting a website
(b) He was reading a book
(c) He was watching TV
(d) He was sleeping
Answer:
(a) He was visiting a website. The process of visiting a website is called browsing.

Question 10.
___________ is a popular browser commonly used in windows operating system.
(a) Mozilla
(b) Netscape navigator
(c) Mosaic
(d) Internet explorer
Answer:
(d) Internet Explorer

Question 11.
___________ browser is commonly used in Linux.
(a) Internet explorer
(b) Mozilla
(c) Netscape navigator
(d) Mosaic
Answer:
(b) Mozilla

Question 12.
Mr. Asokan wants to go the previous page. From the following which option will help him?
(a) Back button
(b) Refresh
(c) Favorites
(d) Stop
Answer:
(a) Back and forward button

Question 13.
While navigating through a website, sita wants to go back to the home page. From the following which will help her?
(a) Refresh
(b) Search
(c) Home
(d) Mail
Answer:
(c) Home

Question 14.
While surfing a website, Joyson wants to play music or video. Which button will help him?
(a) Home
(b) Search
(c) Media
(d) Mail
Answer:
(c) Media

Plus One Computer Science Chapter Wise Questions and Answers Chapter 12 Internet and Mobile Computing

Question 15.
Purnima wants to know the websites that her brother had visited last week? From the following which will help her?
(a) Media
(b) History
(c) Mail
(d) Search
Answer:
(b) History

Question 16.
While browsing, the internet connection is lost so you want to reload the web page. Which will help for this?
(a) Refresh
(b) Stop
(c) Media
(d) Edit
Answer:
(a) Refresh

Question 17.
The address bar is also known as _________
(a) URL
(b) UDL
(c) KRL
(d) None of these
Answer:
(a) URL

Question 18.
You want to add and organize a website to a list. Which will help for this?
(a) Favorites
(b) Search
(c) Back
(d) Mail
Answer:
(a) Favorites

Question 19.
How can it possible to understand that the browser is retrieving data?
(a) Access indicator icon animates
(b) From the refresh button
(c) From the back button
(d) None of these
Answer:
(a) Access indicator icon animates

Question 20.
The progress of the data being downloaded indicates by the ____________
(a) Address bar
(b) Progression bar
(c) Status bar
(d) None of these
Answer:
(c) Status Bar

Question 21.
Baby wants to download a file The time needed for that depends on the _________ of the file.
(a) Size
(b) Place
(c) Type
(d) None of these
Answer:
(a) Size

Question 22.
Kasim wants to print data on a A4 size paper. From the following which option will help him for that?
(a) Copy
(b) Page setup
(c) Search
(d) Media
Answer:
(b) Page setup

Question 23.
Mr. Franco’s e-mail id is franco@rediffmail.com. He wants to connect this page fastly. From the following which will help him.
(a) Favorite
(b) Search
(c) Refresh
(d) Media
Answer:
(a) Favorite

Plus One Computer Science Chapter Wise Questions and Answers Chapter 12 Internet and Mobile Computing

Question 24.
Mrs. Janaki purchased a product through online and payment was given by credit card. She wants to protect the information about the credit card. How can it be possible from the following?
(а) Security
(b) Favorite
(c) Media
(d) Content
Answer:
(a) Security

Question 25.
Odd man out.
(a) www.google.com
(b) www.yahoo.com
(c) www.altavista.com
(d) www.stmaryshss.com
Answer:
(d) www.stmaryshss.com, the others are search engines.

Question 26.
Alvis got email about some products without his consent. Which type of email is this?
Answer:
Spam

Question 27.
What is the primary thing you have needed to sent an email to your friend?
Answer:
You have need an email id (address)

Question 28.
There is a PTA meeting in your school in the next month. The school authorities want to send the invitation to the parents. Which field of the message structure will help for this?
Answer:
CC or bcc

Question 29.
You want to send an invitation to your friends But the friends should not know that the same invitation is sent by you to others also. Which field of the message structure will help you?
Answer:
bcc

Question 30.
Mr. Lijo wants to send his photograph to his friend by email. Which feature will help him for this?
Answer:
Attachment feature

Question 31.
You got some pictures of Jesus Christ through email from one of your friends. You want to send this pictures to your brother. What button will help you for this?
Answer:
Forward button

Question 32.
You got an email from your father working abroad. You want to send an email without typing his email id. Which button will help you for this?
Answer:
Reply button

Question 33.
You got an email from an Insurance Company you want to store their email id which feature will help you for this?
Answer:
We can add address to Address Book.

Question 34.
Who proposed the idea of www.
Answer:
Tim Berners Lee

Question 35.
The protocol for internet communication is ___________
Answer:
TCP/IP protocol

Plus One Computer Science Chapter Wise Questions and Answers Chapter 12 Internet and Mobile Computing

Question 36.
A short distance wireless Internet access method is ___________.
Answer:
Wi-Fi

Question 37.
Give an example for an e-mail address.
Answer:
jobi_cg@rediffmail.com

Question 38.
Which of the following is not a search engine?
(а) Google
(b) Bing
(c) Face book
(d) Ask
Answer:
(c) Facebook

Question 39.
Name the protocol used for e-mail transmission across Internet.
Answer:
Simple Mail Transfer Protocol(SMTP)

Question 40.
Name three services over Internet.
Answer:
WWW, Search engine, Engine

Question 41.
Each document on the web is referred using ___________
Answer:
Uniform Resource Locator(URL)

Question 42.
The small text files used by browsers to remember our email id’s, user names, etc are known as ____________.
Answer:
Cookies

Question 43.
The act of breaking into secure networks to destroy data is called ___________ hacking.
Answer:
Black hats

Question 44.
SIM is _______
(a) Subscriber Identity Module
(b) Subscriber Identity Mobile
(c) Subscription Identity Module
(d) Subscription Identity Mobile
Answer:
(a) Subscriber Identity Module

Question 45.
The protocol used to send SMS message is _________
Answer:
SS7(Signalling System No.7)

Question 46.

  1. Define Intranet
  2. Write the structure of an e-mail address. (1)

Answer:

  1. Intranet: A private network inside a company or organisation is called intranet,
  2. The structure of the email address is given below username@domainname
    eg: jobi_cg@rediffmail.com

Plus One Computer Science Chapter Wise Questions and Answers Chapter 12 Internet and Mobile Computing

Question 47.

  1. Acquiring information such as username, password, credit card details etc. using misleading websites is known as _______
  2. Pick the odd one out:
    Google, Safari, Mozilla Firefox, Internet explorer.

Answer:

  1. Phishing
  2. Google it is a search engine All others are web browsers

Question 48.
Bluetooth can be used for _________ communication
(i) long distance
(ii) short distance
(iii) mobile phone
(iv) all of these
Answer:
(ii) short distance/(iii) mobile phone

Question 49.
Pick the odd one from the following list
(a) Spam
(b) Trojan horse
(c) Phishing
(d) Firewall
Answer:
(d) Firewall

Question 50.
_________ are small text files that are created in our computer when we use a browser to visit a website.
Answer:
cookies

Question 51.
Which one of the following technologies is used for locating geographic positions according to satellite based navigation system?
(a) MMS
(b) GPS
(c) GSM
(d) SMS
Answer:
(b) GPS

Plus One Internet and Mobile Computing Two Mark Questions and Answers

Question 1.
While walking on the road, Simran saw a notice board contains a text “Browsing” in front of a shop. What is Browsing?

OR

Roopa’s mother told you that Roopa is browsing in her room. What is browsing?
Answer:
The process of visiting the websites of various companies, organization, government, individuals etc is called internet browsing or surfing with the help of a browser software we can browse websites.

Question 2.
How can we know that the browser is working or not?
Answer:
The access indicator icon on the right corner of menu bar animates (rotates), when the browser is retrieving data or working. It is static when the browser is not working.

Plus One Computer Science Chapter Wise Questions and Answers Chapter 12 Internet and Mobile Computing

Question 3.
Discuss the steps to download a file from the website.
Answer:
To download a file from the website click on the link or button provided in the web page, then a dialog box will display. Enter the file name and specify the folder to which the file is to be saved. Then click save button then a window showing the progress of the downloading.

Question 4.
What is a Spamming?
Answer:
Sending an email without recipient’s consent to promote a product or service is called spamming. Such an email is called a spam.

Question 5.
You want to send a picture drawn using MS paint immediately to your friend. What method will you adopt for this, so that your friend receives it within seconds? Explain the steps to perform this operation.
Answer:
E-mail (Electronic mail) can be used. There is a facility called attachment will help you to send files with E-mail to your friend. First open your mail box, then take the option to write mail. Fill the email id and subject in the text boxes namely To and Sub respectively.

You can type text in the area given below. Then press the option attachments then select the picture file then press done and press send button.

Question 6.
What do you mean by an ‘always on’ connection?
Answer:
Wired broadband connection is called ‘always on’ connection because it does not need to dial and connect.

Question 7.
What are wikis?
Answer:
In this we can give our contributions regarding various topics and others can watch and edit the content. So incorrect information, advt, etc. are removed quickly.
eg: www.wikipedia.org.

Question 8.
How does a Trojan horse affect a computer?
Answer:
It appears as a useful software but it is a harmful software and it will delete useful software or files in a computer.

Question 9.
How can multimedia content be sent using mobile phones.
Answer:
MMS (Multi-Media Service) allows sending Multi-Media(text, picture, audio and video file) content using mobile phones. It is an extension of SMS.

Question 10.
What are the functions of a mobile OS?
Answer:
Mobile OS manages the hardware, multimedia functions, Internet connectivity, etc. Popular OSs are Android from Google, iOS from Apple, BlackBerry OS from BlackBerry and Windows Phone from Microsoft.

Question 11.
1. Observe the two figures given
Plus One Computer Science Chapter Wise Questions and Answers Chapter 12 Internet and Mobile Computing 1

  • Write their names
  • What are their uses?
  • Name the device associated with them.

2. RAM cannot be replaced by hard disk in a computer. Why?
Answer:
1. Their names are

  • Bar code, QR code
  • This is used to store all the information about a product such as name, price, batch, Exp. date etc.
  • Barcode Reader, Mobile camera (Mobile camera can be used to read QR code information).

2. RAM means Random Access Memory. It is also called read and write memory. It is used to store operating system, and other programs, The one and only memory that the processor can be accessed is the primary memory. Hence RAM cannot be replaced by hard disk in a computer.

Plus One Computer Science Chapter Wise Questions and Answers Chapter 12 Internet and Mobile Computing

Question 12.
Explain “DOS attack’’ on servers.
Answer:
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.

Question 13.
Find the best matches from the given definitions for the terms in the given list.
(Worm, Hacking, Phishing, Spam)

  1. Unsolicited emails sent indiscriminately.
  2. A technical effort to manipulate the normal behavior of networked computer system.
  3. A stand alone malware program usually makes the data traffic slow.
  4. Attempt to acquire information like usernames and passwords by posing as the original website.
  5. Appear to be a useful software but will do damage like deleting necessary files.

Answer:
Worm – 3
Hacking – 2
Phishing – 4
Spam – 1

Plus One Internet and Mobile Computing Three Mark Questions and Answers

Question 1.
What is a browser?
Answer:
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
  • Netscape Navigator
  • Mosaic
  • Opera

Question 2.
Mr. Anirudhan wants to visit the website of Manorama. Their website address is www. manoramaonline.com. How can it be possible?
Answer:
To visit the website of manorama. Anirudhan has to type “www.manoramaonline.com” in the address bar and press the enter key or use the go button. Then the home page of manorama will display. Sometimes while typing the website address on the browser automatically searches and display the homepage.

Question 3.
The education Dept, of Govt, of Kerala declared SSLC results and it is available on the internet your friend wants to save the result in his computer. Help him to do so. .
Answer:
To save the result in his computer to a file by using the ‘save’ or ‘save as’ option of the file menu. When click this option a dialog box will appear then specify the folder whereas the file has to be saved using the dialog box and click OK. To save an image right click on the image, a pop-up menu will appear then choose the save option give a name and press OK.

Question 4.
The application form of Kerala entrance exam can be downloaded from the official website of Kerala govt. What do you mean by downloading?
Answer:
Downloading is the transfer of files or data from one computer to another usually from a server com¬puter to a client computer. The time required to download the file depends on the size of the file. The files may be text, graphics, program, movies, music, etc.

To download a file click on the link or button provided in the web page and specify the folder and filename and there is a window that shows the progress of the file being downloaded.

Question 5.
To apply minority scholarship, a student has to enter his details online, take a printout of this web page then send the application form with this printout to the authorities. Explain how to take a printout of a web page ?
Answer:
To print a web page either select the print command from file menu or use the print button on the standard tool bar. The page setup option is provided in the file menu. It helps to specify the paper size, margins header and footer and also the page orientation. The print preview option helps to view how the page will look after printing.

Plus One Computer Science Chapter Wise Questions and Answers Chapter 12 Internet and Mobile Computing

Question 6.
Mr. Franco’s e-mail id is franco@rediffmail.com. He wants to connect this page fastly and he visited regularly. How can it possible?
Answer:
Mr. Franco regularly visited this site to visit this site he has to type the address repeatedly every time. It is laborious work and it can be avoided if he marks the particular address as favorite. A favorite is a link to a web page. So that he can access that page faster.

To do this click add to favorite option then a dialog box appears that asks for a name for the favorite. To make the web page available offline, then ‘Make available offline1 option has to be checked.

Question 7.
Match the following.

(1) Browser a. File
(2) Menu Bar b. URL
(3) Tool Bar c. Previous page
(4) Address Bar d. Progress
(5) Status Bar e. Mail icon
(6) Back Button f. Mosaic

Answer:
(1) f (2) a (3) e (4) b (5) d (6) c

Question 8.
Noby accessing internet by using a dial-up connection and manu using a direct connection. What is the difference between these two?
Answer:
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.
1. Dial-up Connection:
Here the internet connection is established by dialing into an ISP’s computer and they will connect our computer to the internet. It uses Serial Line Internet Protocol (SLIP) or Point to Point Protocol (PPP). It is slower and has a higher error rate.

2. 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.

Question 9.
Explain the different steps happened in between user’s click and the page being displayed.
Answer:

  1. The browser determines the URL selected.
  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.

Plus One Computer Science Chapter Wise Questions and Answers Chapter 12 Internet and Mobile Computing

Question 10.
What is a Spamming?
Answer:
Sending an email without recipient’s consent to promote a product or service is called spamming. Such an email is called a spam.

Question 11.
You wish to visit the website of your school. Name the software required. Which software is available with Windows for this purpose? Give names of other such software.
Answer:
Browsing software or Browser. The browsers are:

  1. Netscape Navigator
  2. Internet Explorer
  3. Mozilla
  4. Opera
  5. Mosaic etc.

Question 12.
Suppose you want to collect information regarding Tsunami using Internet.

  1. Suggest a method for this purpose
  2. Explain one method adopted.

Answer:
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 userto navigate through the web pages. The different browsers are

  • Microsoft internet explorer
  • Mozilla
  • Netscape Navigator
  • Mosaic
  • Opera

Question 13.
What is a blog?
Answer:
Conducting discussions about particular subjects by entries or posts. The posts appeared in the reverse chronological order means the most recent post appears first.
eg: Blogger.com, WordPress.com, hsslive.com etc.

Plus One Computer Science Chapter Wise Questions and Answers Chapter 12 Internet and Mobile Computing

Question 14.
What do you mean by phishing.
Answer:
it is an attempt to get others information such as usenames, passwords, bank ale 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.

Question 15.
What is quarantine?
Answer:
When you start an anti-virus program and 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.

Question 16.
Compare the intranet and extranet.
Answer:
A private network inside a company or organisation is called intranet and can be accessed by the company’s personnel. But Extranet allows vendors and business partners to access the company resources.

Question 17.
Write short notes on

  1. mobile broadband
  2. Wi-MAX

Answer:

  1. Mobile broadband: Accessing Internet using wireless devices like mobile phones, tablet, USB dongles, etc.
  2. Wi MAX(Wireless Microwave Access): It uses microwaves to transmit information across a network in a range 2GHz to 11GHz over very long distance.

Question 18.
Compare blogs and microblogs.
Answer:
Blogs: Conducting discussions about particular subjects by entries or posts. The posts appeared in the reverse chronological order means the most recent post appears first.
eg: Blogger.com, WordPress.com, hsslive.com etc.

Microblogs: It allows users to exchange short messages, multimedia files, etc.
eg: www.twitter.com

Question 19.
What is firewall?
Answer:
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 networks (internet).

Question 20.
XYZ engineering college has advertised that its campus is Wi-Fi enabled. What is Wi-Fi? How is the Wi-Fi facility implemented in the campus.
Answer:
Wi-Fi means Wireless Fidelity. It is a wireless technology. Some organisation offers Wi-Fi facility. Here we can connect internet wirelessly over short distance, using Wi-Fi enabled devices.

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, hypermarkets, hotels by using Laptops, Desktops, tablet, mobile phones etc.

Plus One Computer Science Chapter Wise Questions and Answers Chapter 12 Internet and Mobile Computing

Question 21.
What is GPS?
Answer:
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.

Question 22.
Write short notes on SMS.
Answer:
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.

Question 23.
What is smart card? How it is useful?
Answer:
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.

Question 24
Social media plays an important role in today’s life. Write notes supporting and opposing its impacts.(3)
Answer:

Advantages of social media:

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

Disadvantages.

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

Question 25.
One of your friends wants to send an email to his father abroad to convey him birthday wishes with a painting done by him. Explain the structure and working of email to him. (3)
Answer:
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: Enter the 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 email 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.

Question 26.
Briefly explain any three mobile communication services.
Answer:
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.
Plus One Computer Science Chapter Wise Questions and Answers Chapter 12 Internet and Mobile Computing 2

4. Smart Cards:
A smart card is a plastic card with a computer chip or memory that stores and transacts data. A smart card (maybe 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.

Plus One Computer Science Chapter Wise Questions and Answers Chapter 12 Internet and Mobile Computing

Question 27.
Define Internet. Compare two types of Internet connectivities namely Dial-up and Broadband.
Answer:
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.

Question 28.
1. your friend does not have an e-mail address. Suggest an e-mail address for him. Starting the advantages of e-mail, explain how it becomes useful for his further communications.

OR

2. List the possible risks while interacting with social media.
Answer:
1. An example of an email id isjobi_cg@rediffmail. com. here jobi_cg is the user name, rediffmail is the portal or website address and.com is the top level domain which identifies the type of organisation. Similarly, we can create an email id, for this type the URL “www.rediffmail.com” and for the new user you have to signup and create an email Id.
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 any where.

2. The possible risks while interacting with social media is given below.

  • Intrusion to privacy: Some people may mis use the personal information.
  • Addiction: Sometimes due to addiction it may waste time and money.
  • Spread rumours: The news will spread very quickly and negatively.

Question 29.
Mobile phone technology has evolved through four generations.

  1. Which generation is called Long Terms Evolution?
  2. Explain some major advancements evolved through these generations. (3)

Answer:
1. 4G

2. 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:
(a) First Generation networks(1G):
It was developed around 1980, based on analog system and only voice transmission was allowed.

(b) Second Generation networks(2G):
This is the next generation network that was allowed voice and data transmission. Picture message and l\4MS(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):
It 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 faster than 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.

(c) 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.

(d) 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.

Plus One Computer Science Chapter Wise Questions and Answers Chapter 12 Internet and Mobile Computing

Question 30.
What is browsing? Briefly explain the steps needed for browsing.
Answer:
The process of visiting a website is called browsing.
Web Browsing steps are given below:

  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.

Question 31.
Susheel’s email id is susheel@amail.com. He sends an email to Rani whose email id is rani@vahoo.com. How is the mail sent from susheel’s computer to Rani’s computer?
Answer:
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 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.

Plus One Internet and Mobile Computing Five Mark Questions and Answers

Question 1.
Your younger brother does not know the structure of an email message. Explain the structure of an email message.
Answer:
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: Enter the 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 email 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

Plus One Computer Science Chapter Wise Questions and Answers Chapter 12 Internet and Mobile Computing

Question 2.
‘Email is the most popular, but most misused service of the internet’. Justify your answer.
Answer:
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
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