Class 8 Physics - BeTOPPERS IIT / NEET Foundation Series - 2022 Edition

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IIT FOUNDATION Class VIII

PHYSICS

© USN Edutech Private Limited The moral rights of the author’s have been asserted. This Workbook is for personal and non-commercial use only and must not be sold, lent, hired or given to anyone else.

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:

USN Eductech Private Limited Hyderabad, India.

PREFACE Our sincere endeavour in preparing this Book is to enable students effectively grasp & understand the Concepts of Physics and help them build a strong foundation in this Subject. From among hundreds of questions being made available in this Book, the Student would be able to extensively practice in each concept exclusively, throughout that Chapter. At the end of each Chapter, two or three Worksheets are provided with questions which shall cover the entire Chapter, helping each Student consolidate his / her learning. This Book help students prepare for their respective Examinations including but not limited to i.e. CBSE, ICSE, various State Boards and Competitive Examinations like IIT, NEET, NTSE, Science Olympiads etc. It is compiled by our inhouse team of experts who have a collective experience of more than 40 years in their respective subject matter / academic backgrounds. This books help students understand concepts and their retention through constant practice. It enables them solve question which are ‘fundamental / foundational’ as well questions which needs ‘higher order thinking’. Students gain the ability to concentrate, to be self-reliant, and hopefully become confident in the subject matter as they traverse through this Book. The important features of this books are: 1.

Lucidly presented Concepts: For ease of understanding, the ‘Concepts’ are briefly presented in simple, easy and comprehensible language.

2.

Learning Outcomes: Each chapter starts with ‘Learning Outcomes’ grid conveying what the student is going to learn / gain from this chapter.

3.

Bold-faced Key Terms: The key words, concepts, definitions, formulae, statements, etc., are presented in ‘bold face’, indicating their importance.

4.

Tables and Charts: Numerous strategically placed tables & charts, list out etc. summarizes the important information, making it readily accessible for effective study.

5.

Box Items: Are ‘highlighted special topics’ that helps students explore / investigate the subject matter thoroughly.

6.

Photographs, Illustrations: A wide array of visually appealing and informative photographs are used to help the students understand various phenomena and inculcate interest, enhance learning in the subject matter.

7.

Flow Diagrams: To help students understand the steps in problem-solving, flow diagrams have been included as needed for various important concepts. These diagrams allow the students visualize the workflow to solve such problems.

8.

Summary Charts: At the end of few important concepts or the chapter, a summary / blueprint is presented which includes a complete overview of that concept / chapter. It shall help students review the learning in a snapshot.

9.

Formative Worksheets: After every concept / few concepts, a ‘Formative Worksheet’ / ‘Classroom Worksheet’ with appropriate questions are provided from such concept/s. The solutions for these problems shall ideally be discussed by the Teacher in the classroom.

10. Conceptive Worksheets: These questions are in addition the above questions and are from that respective concept/s. They are advised to be solved beyond classroom as a ‘Homework’. This rigor, shall help students consolidate their learning as they are exposed to new type of questions related to those concept/s.

11. Summative Worksheets: At the end of each chapter, this worksheet is presented and shall contain questions based on all the concepts of that chapter. Unlike Formative Worksheet and Conceptive Worksheet questions, the questions in this worksheet encourage the students to apply their learnings acquired from that entire chapter and solve the problems analytically. 12. HOTS Worksheets: Most of the times, Summative Worksheet is followed by an HOTS (Higher Order Thinking Skills) worksheet containing advanced type of questions. The concepts can be from the same chapter or as many chapters from the Book. By solving these problems, the students are prepared to face challenging questions that appear in actual competitive entrance examinations. However, strengthening the foundation of students in academics is the main objective of this worksheet. 13. IIT JEE / NEET Worksheets: Finally, every chapters end with a IIT JEE / NEET worksheet. This worksheet contains the questions which have appeared in various competitive examinations like IIT, NEET, AIEEE, EAMCET, KCET, TCET, JIMPER, BHU, AIIMS, CBSE, ICSE, State Boards, CET etc. related to this chapter. This gives realtime experience to students and helps them face various competitive examinations. 14. Different Types of Questions: These type of questions do appear in various competitive examinations. They include:

• Objective Type with Single Answer Correct

• Non-Objective Type

• Objective Type with > one Answer Correct

• True or False Type

• Statement Type - I (Two Statements)

• Statement Type - II (Two Statements)

• MatchingType - I (Two Columns)

• MatchingType - II (Three Columns)

• Assertion and Reasoning Type

• Statement and Explanation Type

• Roadmap Type

• FigurativeType

• Comprehension Type

• And many more...

We would like to thank all members of different departments at BeTOPPERS who played a key role in bringing out this student-friendly Book. We sincerely hope that this Book will prove useful to the students who wish to build a strong Foundation in Physics and aim to achieve success in various boards / competitive examinations. Further, we believe that as there is always scope for improvement, we value constructive criticism of the subject matter, as well as suggestions for improving this Book. All suggestions hopefully, shall be duly incorporated in the next edition. Wish you all the best!!!

Team BeTOPPERS

CONTENTS 1.

Measurement

.......... 01 - 12

2.

Force and Pressure

.......... 13 - 32

3.

Friction

.......... 33 - 44

4.

Sound

.......... 45 - 52

5.

Temperature and Heat

.......... 53 - 66

6.

Electricity & Chemical Effects of Electric Current

.......... 67 - 90

I. Electricity

.......... 67 - 80

II. Some Natural phenomena

.......... 75 - 77

III. Chemical Effects of Electric Currrent

.......... 81 - 88

7.

Light

.......... 91 - 118

8.

Magnetism

.......... 119 - 140

9.

Key and Answers

.......... 141 - 200

By the end of this chapter, you will understand • • • • • •

Introduction to Measurement Types of Physical Quantities Classification of Units System of Units Measurement of length Measurement of Mass

• • • •

Measurement of Time Measurement of Area Measurement of Volume Measurement of density and relative density

1. Introduction to Measurement Physics is inherently a science of measurement. What do we mean by measurement? Imagine that you went to the market to buy mangoes. What will the Mango vendor ask you? ‘How much do you want’ ? You may reply: ‘2kg or 3kg’. You are expressing the amount of quantity required. Such an act is called measurement. The act of measuring a quantity required is called measurement. Let’s recall the previous example. What was the vendor measuring ? The amount or the mass of mangoes. Such quantities that can be measured are called physical quantities.

Physical Quantity Any quantity that is measurable in physics is known as a Physical Quantity. Examples: Length, Mass, Time

Measurement of a Physcial Quantity Suppose you have purchased 5 kilograms of wheat flour. Here you are measuring mass (physical quantity).

5

Kilograms

Standard (or) Unit

Chapter -1

Measurement

Learning Outcomes

Unit: A standard used to measure a physical quantity. Example: When we say length of a wire is 5 metre, it means, the unit metre is present 5 times in the physical quantity, length. The standard used to measure the physical quantity, length is metre.

2. Types of Physical Quantities Fundamental Quantity: A physical quantity which is independent of any other quantity is called a fundamental quantity. Examples: Length, Mass, Time, etc. Derived Quantity: A physical quantity which can be derived from fundamental quantities is called a derived quantity. Examples: Area, Volume, Density, etc.

3. Classification of Units Fundamental Unit: The units used to measure fundamental quantities are called fundamental units. Examples: Metre, Kilogram, Second, etc., Derived Unit: The units used to measure derived quantities are called derived units. Examples: Square metre, Cubic metre, etc.,

Standard Unit In the early days, people used to measure length with the help of various parts of a body, such as handspan, footspan , arm or cubit etc.,

Numerical value (or) Magnitude

Measurement of a Physical Quantity (Q) = Magnitude (N) × Unit (U)  Q = NU Magnitude: Number of times the unit is present in a physical quantity.

Handspan

Footspan

Cubit

Consider a boy measuring the length of a table, using his handspan.

8th Class Physics

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Rules for writing symbols of Units

What is the length of the table? 6 handspans. Now, imagine his father is measuring the same table, using his handspan. What is the length of the table now? 3 handspans. What is the correct length of the table - 6 handspans or 3 handspans? We cannot say accurately. Hence, handspan is not an accurate unit to measure and express the length. Hence, we need a unit that gives same length when used by anyone and anywhere….

Features of a standard units: i. ii. iii.

They should be very well defined. They should be independent of time and place. They should be independent of physical conditions like temperature, pr essure and humidity.

4. System of Units The scientists all over the world have developed basic set of standard units for measuring various quantities. This set is also known as system of units and named as Standard International System of Units or SI system. Some other units still in use are (i) F.P.S system (Foot, Pound,Second) (ii) C.G.S. system (Centimetre, Gram, Second) (iii) M.K.S. system (Metre, Kilogram, Second) (iv) S.I. system.

Fundamental physical quantity Length Mass Time Thermodynamic temperature Luminous intensity Strength of electric current Quantity of matter

S.I Unit

Symbol

Metre Kilogram Second Kelvin

M Kg S K

Candela

Cd

Ampere

A

Mole

mol

Note: Among all the physical quantities known, the above seven are the only fundamental quantities. www.betoppers.com

Rule 1: A unit named after a scientist must not start with a capital letter when the complete name is written. The work done by a body is 10 joule but not 10 Joule. If only the first letter is used as a symbol, it must be a capital letter. Thus, the work done can be represented as 10 J. Rule 2: The units other than those named after scientists i.e., all other units are symbolically represented by small letters. If the length of a rod is 2 metres, it is written as 2 m. If the time taken by a ball to hit the ground is 5 seconds, it is written as 5 s. Rule 3: Symbols for units do not take plural form The mass of a body is 10 Kg but not 10 Kgs The force acting on a body is 10 newton and not 10 newtons. Rule 4: No full stop or punctuation mark should be used within or at the end of symbols for units. The mass of a body is 10 Kg but not 10 Kg. (full stop () should be omitted) The force acting on a body is 10 N and not 10 N: (colon ( : ) should be omitted)

Multiples of Ten The study of nature involves objects of small size like atom, nucleus etc. They belong to the small world (microcosm).

If we measure the diameter of hydrogen atom (in microcosm) it will be described by a very small number 0.000,000,000,106m. It also involves celestial objects of the universe like the sun, planets, moon, stars, etc. They belong to the cosmic world (macrocosm). The diameter of the sun will be described by a very large number 1,390,000,000m. The above numbers are exceptionally big and difficult to write or remember. How then can this task be simplified? It can be simplified by expressing in terms of multiples of 10. The following chart shows some multiples of 10 and their corresponding names or prefixes :

Measurement

3

Positive Prefix or Symbol multiples of 10 Name 18 10 Exa E 15 10 Peta P 1012 Terra T 109 Giga G 6 10 Mega M 3 10 Kilo K 102 Hecto H 1 10 Deca D 1 deci d 10 2 centi c 10 3 milli m 10 micro 106  9 nano n 10 12 pico p 10 15 femto f 10 18 atto a 10

Conceptive Worksheet

What do we need to measure physical quantities accurately? 1) Standard units 2) Standard instruments 3) Both (1) and (2) 4) None of these 2. 1 kilogram = _________m 1) 1000 2) 100 3) 10 4) None 3. Statement - I : Cubit is a standard unit Statement - II : 10 Kilometre = 1000 metre 1) Statement I is true ; Statement II is true. 2) Statement I is true ; Statement II is false. 3) Statement I is false ; Statement II is true. 4) Statement I is false ; Statement II is false. 4. Statement I : Micro, milli, etc., are to be used as prefixes for metre only. Statement II : Kilogram is a standard unit. 1) Statement I is true ; Statement II is true. 2) Statement I is true ; Statement II is false. 3) Statement I is false ; Statement II is true. 4) Statement I is false ; Statement II is false. 5. F.P.S stands for ormative orksheet 1) Foot, pound, second 2) France, Paris, Spain 1. Match the following. 3) Force, pressure. second 4) Foot, Pace, Second List – A List – B 6. C.G.S stands for : i) Temperature a) mole 1) Centimetre, gravitation, second ii) Luminous intensity b) Kelvin 2) Centisecond, gram, second 3) Centimetre, gram, second iii) Amount of substance c) candela 4) None of these 2. Find the C.G.S. unit of density. 7. Pick the odd man out: Hint. Density = Mass / Volume 1) Length 2) Metre 3) Yard 4) Cubit 3. If a new physical quantity by name 8. Ampere is the unit of: ‘GAMBHIR” is introduced whose measuring 1) Length 2) Mass formula is GAMBHIR 3) Temperature 4) Current Mass  Area 9. Number of fundamental physical quantities in M.K.S  Volume  Time system are: Find the S.I. unit of ‘GAMBHIR’. 1) Two 2) Three 3) Seven 4)Six 4. Given below are incorrect representations of units. 10. Multiples and submultiples of units. Write their correct repre sentations. 1) Are specific numerical values (i) K.G (ii) mts (iii) Newton (iv) joule; 2) Are used as prefixes 5. 1 mega second = _________deca seconds. 3) Both (1) and (2) 6. Assume that there existed a new unit for 4) None of these pressure called Sachin. 11. The standard used to measure a certain Physical 1 mega sachin = _________ milli sachin. quantity is 7. If 1 Mega metre = 10y centimetre. Find y. 1) Unit 2) Scale 8. ‘Aryabhatta’ India’s first satellite was or biting the 3) Both (1, (2) 4) None of these earth at an altitude of 600km. What is the altitude 12. Micro means in millimetres? 1) Unit of time 2) 10,00,000 9. If 1 kg (mm)–2 = y mg(km)–2 find y. 1 1 10. If 1 micrometer (Mega second)–1 = ‘x’ kilo meter 3) 4) 10,00,000 milli (millisecond)–1. Find x.

F

1.

W

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8th Class Physics

4 13. Pick the odd man out. 1) milli 2) kilo 3) micro 4) centimetre 14. Area is a : 1) Fundamental quantity 2) Derived quantity 3) Both (1) and (2) 4) Neither (1) nor (2) 15. 1 cm = _________ kilometre. 1) 100 2) 105 3) 10–5 4) 10–2 16. The size of bacteria is generally measured in microns. The micrometre  μ m  , is often called the micron. How many microns make up 1 kilometre ?

Parsec: It is also called Parallatic second. It is the largest unit of distance. One Parsec is 3.26 times the light year. i.e., 1 Parsec = 3.26 light years Angstrom: It is the unit used to measure the size of atoms. 1 angstrom = 1A0 = 10–10m Fermi: It is the unit used to measure size of nucleus. It is the smallest unit of distance. 1 fermi = 1f = 10–15m Micron

μm Mm 17. 1 mg  ______ ng 18. If mg/ns = 10x kg/ps then the value x is

It is the unit used to the size of bacteria and virus. One micrometre is defined as a micron 1 micron = 1mm = 10–6m

6. Measurement of Mass 5. Measurement of Length Length C.G.S Unit

Mass C.G.S Unit

Gram

Centimetre

Metre

The standard relation between metre and centimetre is:

1 1 m  1cm  2 m 1 m = 100 cm  1cm  100 10 –2  1 cm = 10 m

Some practical units of Length Astronomical Unit: It is the average distance of the earth from the sun. 1 A.U. = 1.496 × 1011 m It is used in astronomy to measure the distance between planets.

Light year: It is the distance travelled by light in a vacuum, in one year. This unit is used in astronomy to measure the distance of stars. 1 Light year = 9.46 × 1015 m [1 Light year = 1 year × speed of light = 365 ¼ day × 3 × 108 m / s = 365.25 × 24 hr × 3 × 108 m / s = 365.25 × 24 × 3600s × 3 × 108 m / s = 94672800 × 108m = 9.46728 × 1015m].

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S.I Unit

S.I Unit Kilogram

The relation between gram and kilogram is: 1 Kg = 1000 g

1 1 Kg  1g = 3 Kg 1000 10 -3  1g = 10 Kg Other units : 1 Quintal = 100 Kg 1 metric ton =1000Kg  1g =

Smallest and Biggest Units of Mass The smallest unit is Atomic mass unit (a.m.u). This is used to measure the masses of microscopic particles, like electrons, protons, neutrons, etc. 1 a.m.u = 1.67 × 10–27 kg The biggest unit is Chandra Shekara Limit (C.S.L). This unit is used to measure the masses of astronomical bodies, like planets, stars, etc. 1 CSL = 1.4 × mass of sun (Mass of sun = 20 × 103 kg)

Measuring Instruments of Mass Common balance and physical balance are used measure the mass of a substance.

Principle of Common Balance The beam of the balance remains horizontal when equal masses are attached at equal distances from the point of suspension. This is the principle of a common balance.

Measurement

5

7. Measurement of Time

C

W

onceptive orksheet The interval between two events is called time. The only physical quantity, whose units are same 19. The height of a building is 12m. Express its height in all the system of units, is time. in feet. [Hint: 1foot = 12 inch; 1inch = 2.54 cm] It is measured in seconds. 20. The ‘furlong’, is a unit used in horse racing. If 1 Other Units of Time mile = 8 furlongs, then express furlong in terms of Units of time Conversion kilometre. [Hint: 1 mile = 1609m] Minute Hour

Day Year Leap year Decade Century Millennium

1 minute = 60 seconds 1 hour = 60 seconds = 3600 seconds 1 day = 24 hours 1 year = 365.25 days 1 leap year = 366 days 1 decade = 10 years 1 century = 10 decades = 100 years 1 millennium = 10 decades = 100 years

Formative Worksheet 11. Which of the following is not a unit of distance? (i) Light year (ii) Parsec (iii) Leap year (iv) angstrom 12. If 1 parsec is equal to 3.26 light year then 1 parsec is equal to how many metre? 13. State true or false 1 light year = 300000 × 365 × 24 × 60 × 60 km 1light year = 3 × 108 × 365 × 24 × 60 × 60 m 14. a) How many centimetres make 1nm? b) How many centimetres make 1 fermi? 15. Match the following :

List – A i) 1 Hectogram ii) 1 Decagram iii) 1milli gram iv) 1 micro gram 16. 17. 18. 19.

List – B a) 10? 6 gram b) 102 gram c) 10 gram d) 10? 3 gram

One quintal = ______________ton. 1 metric tonne = __________ milligram. 1 microsecond = 10x milliseconds. Find x. Match the following : List – A List – B i) 1 Millennium a) 5256 × 102 minutes ii) 1 Decade b) 316224 × 102 seconds iii) 1 Leap year c) 31536 × 106 seconds

21. 1 kg = –––––––––– tonne. 22. A lady is wearing 10 tolas of gold. Find the amount of gold in kilograms [Hint: 1tola = 10gram]. 23. A postage stamp has mass of 2.0 × 10–5 kg. Its mass can be written as : 1) 20 cg

2) 20 mg

3) 2 mg

4) 2 cg

24. 1 mean solar day is equal to how many second ? 25. One millenium is equal to how many decades ?

8. Measurement of Area This space occupied by the book is called area of the book. Area = length × breadth C.G.S unit of area = C.G.S unit of length × C.G.S unit of breadth = centimetre (cm) × centimetre (cm) = square centimetre (sq.cm) or cm2 S.I unit of area = S.I unit of length × S.I unit of breadth = metre (m) × metre (m) = square metre (sq.m) (or) m2

Relation between cm2 and m2 We know that, 1m = 100cm (or) 1m = 102cm Squaring on both sides, we get, (1m)2 = (102 cm)2

 1m2 = 104 cm2  1m2 = 104 cm2  1cm2 = 10-4 m2 Note: For measuring bigger areas like areas of fields or towns, square metre is too small a unit. Thus, a bigger unit is used, which is called a hectare. 1 hectare = 10,000 m2 or 104 m2.

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8th Class Physics

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9. Measurement of Volume The space occupied by a body is called his volume.

Units of Volume Volume C.G.S unit

Cubic centimetre

S.I unit

Cubic metre

Volume = length × breadth × height CGS unit of Volume = CGS unit of (length × breadth × height) = centimetre (cm) × centimetre (cm) × centimetre(cm) = cubic centimetre (cc) or cm3 SI Unit of Volume = SI unit of (length × breadth × height) = metre (m) × metre (m) × metre (m) = cubic metre (cu.m) or m3 Note: The volume of a liquid is measured in litres or millilitres.

Relation between cm3 and m3 1 litre = 1000 millilitres = 1000ml = 1000cc (Q 1ml = 1cc) = 1000cm3 1 litre =1000 cm3= 103 cm3 = 103 × 10-6 m3 Q 1cm3 = 10-6 m3 = 10-3 m3 1 litre = 1000ml = 1000cc = 10–3 m3

Measuring Instruments of Volume Following are the instruments used to measure volume of liquids:

Formulae for Volume of some regular Solids a) b)

Volume of cube = (side)3 = l3 Volume of cuboid = length × breadth × height =l×b×h

c)

Volume of sphere 

d)

e)

4 3 r 3 (r is the radius of sphere) Volume of cylinder = pr2 l= a x l (r is the radius, a is the area of cross section and l is the length of cylinder) 1 2 r h r 3  radius h  height

Volume of cone 

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Formative Worksheet 20. 1 m2 = _____________hectare. 1 hectare = __________ km2 . 1 mm2 = __________ km2 . 21. A school hall measures 20 m in length and 12 m in breadth. Find the area of the school hall. 22. The volume of a rectangular slab is 12 cm3. The length and breadth of the slab are 3 cm and 2 cm respectively. Find its height. 23. A boy has purchased a toy, which is in the form of a cuboid. The cuboid has the following dimensions: 0.003 km × 0.03 m × 3 cm. Now the boy pours 1 litre of water into the cuboid. Will the cuboid hold 1 litre of water? [Hint: 1 litre = 1000ml and 1 ml = 1 cm3] 24. When a stone is lowered into a measuring cylinder, the volume is 9.3 ml. The volume of the stone is 5.8 ml. Find the initial volume of water in the measuring cylinder.

Conceptive Worksheet 26. Unit of area is a 1) Derived unit 2) Fundamental unit 3) Both 1 and 2 4) None of these 27. The area of the land is 100m2, here m2 stands for 1) Numerical value of area 2) Unit of area 3) Both 1 and 2 4) None of these 28. 1km2 = _____________ 1) 1 hectare 2) 10 hectares 3) 100 hectares 4) 1000 hectares 29. 1 hectare = ____________ 1) 1000m2 2) 10000m2 2 3) 100000m 4) 100m2 30. The space occupied by a substance is called _________. 1) area 2) length 3) volume 4) none of these 31. The S.I unit of volume is _________ 1) Cubic centimetre 2) Cubic millimetre 3) Cubic metre 4) Cubic litre 32. One cubic metre is equal to ––––– 1) 106 cc 2) 104 cc 3) 103 cc 4) 109 cc 33. The volume occupied by a cube whose each side is equal to 1cm is called 1) Cubic centimetre 2) Cubic millimetre 3) Cubic metre 4) None of these

Measurement 34. 1 litre = _________. 1) 1000 millilitre 2) 1000cc 3) Both (1) and (2) 4) None of these 35. Find the odd one out. 1) Cubic metre 2) Cubic centimetre 3) Cubic millimetre 4) Square metre 36. Match the following List - A List - B (i) Volume of a swimming pool a) cm3 (ii) Volume of a glass filled with milk b) m3 (iii) Volume of an exercise book c) litre (iv) Volume of air in a room d) millilitre 1) i-c, ii-d, iii-b, iv-a 2) i-c, ii-d, iii-a, iv-b 3) i-a, ii-b, iii-c, iv-d 4) i-a, ii-b, iii-c, iv-d

10. Measurement of Density and Relative Density Density Which one is heavier ? 5 kg of cotton or 5 kg of iron? If you say “both” then you have to rethink. Infact, you physically feel iron more heavier than cotton. Surprised! Here you need to understand that the heaviness are lightness of a body is not directly the function of their masses. This heaviness and lightness can be understood from the concept of density.

An Activity Take two liter flasks and fill them completely with water and kerosene. Place then on the pair’s of a common balance. What do you observe. We observe that the beaker filled with water is more heavier than the beaker filled with kerosene. Why is it so ? For equal volume, the matter (atoms or molecules) in water is more density (tightly) packed in water than in kerosene or the mass per unit volume of matter in water is more than in kerosene. This mass per unit volume of any substance is called its density. Thus water is more denser than kerosene.

Density – Definition The mass per unit volume of a substance is called density. Note : Density of substance depends on nature of the material i.e.,

7 i) The space between the particles : Larger the space between the particles, lesser is the density of the substance. ii) Mass of each particle : Larger the mass of each particle, more is the density of the substance. Mathematical expression : If m is the mass of a substance, V is its volume. Then its density, D is given by: Density =

Mass Volume

i.e., D =

m V

Unit of Density Mass Volume C.G.S. unit of Density = C.G.S. unit of (mass / volume) = gram / centimetre3 = g / cm3 or g cm–3 S.I. unit of Density = S.I. unit of (mass / volume) = kilogram / metre3 = kg / m3 or kg m–3 Density =

Relation between Units of Density The relation between g cm–3 and kg m–3 can be derived as follows:

1g 1g  3 1 cm 1 cm  1 cm  1 cm 1 kg 1000 = 1 1 1 m× m× m 100 100 100

1 1   kg and 1cm  m 1g  1000 100   1 kg 1000 1000000 kg =  = 1000 kg/m3 1 3 1000 m3 m 1000000

1g cm–3 = 1000 kgm–3

Density of few common substances 1. 2. 3. 4. 5. 6. 7. 8. 9. 10.

Ice Drysand Aluminium Steel iron Copper Gold Wood Platinum Glass

920 kg/m3 1600 g/cm3 2700 kg/m3 7800 kg/m3 7860 kg/m3 8900 kg/m3 19300 kg/m3 500 kg/m3 21500 kg/m3 2500 kg/m3 www.betoppers.com

8th Class Physics

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Observation What happens if you mix petrol, water and glycerin ? You will be surprised to see that they do not mix. Further interfering pattern is observed. Where petrol floats on water, glycerin sinks in water. Why this happens ? This is because, petrol is relatively less denser than water and glycerin is relatively more denser than water. This concept of comparing density of any substance with the density of water is called relative density.

Relative Density - definition

31. The length of an iron cylinder is 0.8 m and the area of cross-section is 0.45 m2. Find the mass of the cylinder if its density is 7800 kg/m3. 32. 30 cm3 of iron weighs 234 g. Find its density in kg m–3 . 33. If the mass of 2 litres of water is 2 kg, find its density in S.I. system. 34. A sphere of radius 7 cm weighs 490 g. Calculate its density in S.I. system. 35. A piece of gold is cut into two halves. If the density of gold before cutting is d, then the density of each half after cutting is d/2. True/False. Explain. 36. There are two substances A and B. The mass of substance B is found to be 3 times the mass of

The ratio of the density of a substance to the density of water at 4°C which is called relative density, or specific gravity of the substance. Mathematical expression of relative density.

Relativedensity =

Density of thesubstance Density of pure water at 4°C

We notice from the last expression that the relative density is a ratio of similar quantities. So it is a mere number and has no units.

37. 38. 39.

Relationship between Density and 40. Relative Density: 1)

2)

Density of a solid (in S.I. Unit) = R.D. of the solid × Density of water (in S.I. Unit) = R.D. of the solid × 1000 kg/m3 Density of a solid(in C.G.S. unit) = R.D. of the solid × Density of water (in C.G.S system) = R.D. of the solid × 1 g/cm3

Formative Worksheet 25. The volumes of three objects, A, B and C are 100 cm3, 200 cm3 and 300 cm3 respectively. If their masses are same, which one will have greater density ? 26. A body weighing 332 g has a volume of 20 cm3. Find its density. 27. If density of gold in C.G.S system is 19.6 g/cm3, then find its density in S.I system. 28. The density of iron is 8.6 g cm–3. If volume of iron is 555 cm3, find its mass in kg. 29. Calculate the volume of wood whose mass is 6000 kg, and density is 0.8 gcm3. 30. What is the mass of air in a room of dimensions 3 m × 4 m × 5 m, when the density of air is 1.30 kg m–3 ? www.betoppers.com

1 th 9 the volume of substance B. Which of the two substances possesses greater density? If density of a substance is x kg/m3, find its relative density. If the density of copper is 8.9 × 103 kg/m3, find its relative density. Calculate the mass of a body whose volume is 2 m3 and relative density is 0.52. If relative density of gold is 19.3, then the density of gold is __________times greater than the density of water. 1) 8.9 2) 19.3 3) 1.29 4) 0.8 substance A, but the volume of substance A is

Conceptive Worksheet 1 37. The S.I unit of density is _____ 38.

39.

40.

41. 42.

(1) kg/m3 2) g/cm3 3) m3/kg 4) cm3/g Density = 1) volume × mass 2) volume / mass 3) mass / volume 4) none of these Density depends upon 1) mass only 2) volume only 3) both 1 and 2 4) none of these If M is the mass of an object of volume V, such that ‘D’ is its density, then we can say 1) V = DM 2) MVD = 1 3) M = VD 4) D = MV C.G.S unit of density is __________ 1) kg/m3 2) m3/kg 3) g/cm3 4) cm3/g The density of alcohol is 800 kg/m3. Then the density in g/cm3 is _______ 1) 800000 2) 0.80 3) 0.008 4) 80000

Measurement 43. 20 cm3 of aluminium has mass 54g. Then 1cm3 of aluminium mass has ____ 1) 74 g 2) 2.7 g 3) 7.2 g 4) 5.4 g 44. A piece of lead weighs 232 g and has a volume of 20cm3. Then the density of lead is ________ 1) 0.0862 g/cm3 2) 0.0862 cm3/g 3 3) 11.6 cm /g 4) 11.6 g/cm3 45. The relative density of mercury = _____ 1) 0.8 g/cm3 2) 13.6 g/cm3 3 3) 2.5 g/cm 4) none of these

Summative Worksheet 1.

2.

3. 4.

5)

6)

Statement I : Numerical value = Physical quantity × unit Statement II : Current strength is a fundamental physical quantity according to S.I system. 1) Statement I is true ; Statement II is true. 2) Statement I is true ; Statement II is false. 3) Statement I is false ; Statement II is true. 4) Statement I is false ; Statement II is false. Consider length, mass, current strength and temperature. Choose the correct statement 1) Length is the odd man out. 2) All are fundamental physical quantities according to M.K.S system. 3) All are fundamental physical quantities according to S.I system. 4) Current strength is a derived quantity. 1 day = _____________ seconds. 1) 3600 2) 86400 3) 4200 4) 12300 Statement I: A thread is enough to measure a curved line. Statement II: A scale is enough to measure a curved line. 1) Statement I is true ; Statement II is true. 2) Statement I is true ; Statement II is false. 3) Statement I is false ; Statement II is true. 4) Statement I is false ; Statement II is false. To measure the length of a curved line, which of the following materials are needed ? a) Cotton thread b) Measuring scale c) Common balance d) Watch 1) a and b 2) b and c 3) c and d 4) d and a In which of the following, are indirect methods of measurement used? a) Measuring thickness of a wire b) Measuring thickness of a coin or plate c) Measuring length of table 1) a and b 2) b and c 3) c and d 4) d and a

7)

8.

9. 10.

11.

12.

13.

14.

9 Metre scales have _________________ ends to avoid error due to ________________ 1) Cylindrical, Volume of the scale 2) Tapered, thickness of the scale 3) Tapered, Area of the scale 4) Cylindrical, Area of the scale 1cm2 = _____________ 1) 10–10 km2 2) 10–8 hectare –4 2 3) 10 m 4) all of these 1m2 = ___________ 1) 10–6 km2 2) 10–4 hectare 3) 10–2 hecare 4) all When do we say that a bucket is bigger than a cup? 1) When the volume of cup is greater than the volume of bucket. 2) When the volume of bucket is greater than the volume of cup. 3) When the volume of bucket is equal to the volume of cup. 4) We can’t say. 5 litres of alcohol has a mass of 4 kg. The density of alcohol is ______ 1) 0.8 kg/m3 2) 80 kg/m3 3) 800kg/m3 4) 8000kg/m3 When a 50kg mass of a body is immersed in water, we observe the mark on measuring cylinder as 70cm3. If the body is taken out from the cylinder, the mark observed on the cylinder is 50cm3. Then the density of that body is [The graduations are marked in cm3] 1) 2500 g/cm3 2) 250 g/cm3 3 3) 2.5 g/cm 4) 0.025 × 103 g/cm3 The density of lead is 11.6 g/cm3 and that of wood is 800 kg/m3. What do you understand by these statements ? 1) The matter is only densely packed in lead than wood. 2) The matter is only densely packed in wood than lead. 3) The density of lead is greater than the density of wood. 4) Both (1) and (3). Take two identical 100cm3 beakers. Fill one beaker completely with water and the other with kerosene oil. Place the beakers in the scale pans of an ordinary beam balance. It is observed that the beaker filled with water have more mass than the beaker filled with kerosene oil. This is because 1) the matter in water is more densely packed than in kerosene oil. 2) the matter in kerosene oil is more densely packed than in water. 3) Water and kerosene oil occupy the same volume. 4) None of these. www.betoppers.com

10 15. The volume of a piece of iron is 15cm3 and its mass is 117g. The density of iron in g/cm3 and kg/m3 is 1) 7800&7.8 2)7.8 &7800 3) 780&7.8 4) 7.8&780 16. A measuring cylinder contains water at the mark 22ml. If a piece of copper of mass 106g is dipped into the water , which rises to 34ml mark. Then the volume and density of the copper piece respectively are 1) 1.2cm3 & 88.3g cm–3 2) 88.3 cm3 & 12g cm–3 3) 1200cm3 & 8.83g cm–3 4) 12cm3 & 8.83g cm–3 17. The mass of a rectangular block of iron is 23.6g and its dimensions are 2.1cm × 1.2cm × 1.1cm. The density of iron in kgm–3 is 1) 8.510 × 103 2) 85.10 × 102 3) 8510 4) none 18. Match the following from i) mass = ________ × density a) 13.6 g ii) 1 cc of water has mass of _____ b) 1 g iii) 1 cc of mercury has a mass of __c) volume 1) i-a, ii-b, iii-c 2) i-b, ii-c, iii-a 3) i-c, ii-b, iii-a 4) i-c, ii-a, iii-b 19. Select the correct statement from 1) 1 g/cm3 = 1000 kg/m3 2) 1000 g/cm3 = 1 kg/m3 3) 100 g/cm3 = 1 kg/m3 4) kg/m3 = 1 g/m3 20. A room of length, breadth are and height 6m × 5m × 4m respectively. If the density of air is 1.2kg m 3, then mass of air in the room is 1) 144 kg 2) 144 g 3) 100 kg 4) 100 g

HOTS Worksheet 1. 2. 3.

4.

5.

6. 7.

8.

9.

10.

11.

Pick the odd man out. 1) Area 2) Velocity 3) Force 4) Acceleration Pick the odd man out. 1) mega 2) kilo 3) tonn 4) hecta The length of a school compound is 450m and breadth is 145m. The area of the school compound in hectares is ________ 1) 6.525 hectares 2) 65.25 hectares 3) 0.6525 hectares 4) 652.5 hectares D

8th Class Physics

12.

13.

C

1 cm

1 cm

A

B

Let ABCD be a centimetre graph paper. The area of darkened surface on the graph paper is 1) 8 cm2 2) 11 cm2 3) 9cm2 4) 10cm2 www.betoppers.com

14.

Under the spout of over flow jar, place a measuring cylinder. Gently lower a stone in the over flow jar. The stone displaces water which flows out from the spout into the measuring cylinder. If the reading on the overflow jar is 12ml, then the volume of stone is __________. 1) 12 × 10–5 m3 2) 1.2 × 10–5 m3 3) 0.12 × 10–5m3 4) 0.012 × 10–5m3. One millennium is equal to how many decades? 1) 10 2) 100 3) 1000 4) 10, 000 If an aeroplane is scheduled to take off at 18 hours – 57 minutes, then time in PM on a 12 - hour clock is __________ 1) 12 - hour – 57 minute AM 2) 12 - hour – 57 minute PM 3) 6 - hour – 57 minute AM 4) 6 - hour – 57 minute PM If the time on a 12 - hour clock is 3 hour – 45 min PM, then time on 24 - hour clock is _____ 1) 15 hour – 45 minutes 2) 15 hour – 12 minutes 3) 6 hour – 57 minutes 4) 12 hours 1 day = _____________ millennium 1) 1/165000 2) 1/265000 3) 1.46400 4) 1/365000 1 decade = _____________ minutes 1) 52. 56 × 106 2) 5. 256 × 106 3) 525. 6 × 106 4) 5256 × 106 1 year = _____________ seconds 1) 315. 36 × 106 2) 3.1536 × 106 6 3) 31. 536 × 10 4) 3153. 6 × 106 If the Charminar superfast express staying 00 hours in Warangal, then the time in 12 hour clock is _________ 1) 12 0’ clock at night 2) 12 0’ clock at noon 3) 18 0’ clock at night 4) 18 0’ clock at noon A passenger goes to Secunderabad railway station. He asked the enquiry counter, “When did Tirumala express come?” The enquiry counter person replied “18 hour 15 minutes.” Then the time in his 12-hour clock was ________ 1) 5 hour – 45 min AM 2) 5 hour – 45 min PM 3) 6 hour – 15 min AM 4) 6 hour – 15 min PM A cylinder is filled with water to a level of 50 cm3 . The cylinder is marked upto 100 divisions with each division of 1cm3. When a body is immersed in a liquid, we observe the mark on the cylinder is 75 cm3. If the density of the body is 250gcm–3, then the mass of the body in S.I system is: 1) 6.250 kg 2) 6.250 g 3) 62.50 kg 4) 62.50 g

Measurement 15. Mass of liquid = 72 g Initial volume of water in measuring cylinder = 24 cm3 Final volume of water + solid in measuring cylinder = 42 cm2 From the above data the density of solid is ___________ 1) 4000 kg/m3 2) 3.0 g/cm3 3 )

16.

17.

18.

19.

72 g 4) 4.0 kg/m3 3 42 cm A density bottle weighs 20.25g when empty, 40.75g when filled with a liquid and 50.25g when filled with water. Then the density of liquid is __ 1) 6.83 g/cm3 2) 0.683 g/cm3 3) 68.3 g/cm3 4) 0.0683 g/cm3 An empty density bottle weighs 22g. When completely filled with water, it weighs 50g. When completely filled with brine solution it weighs 54g. The density of brine solution is ____ 1) 1 g/cm3 2) 1.14 g/cm3 3) 2 g/cm3 4) 2.14 g/cm3 An empty density bottle weighs 30g. When completely filled with water, it weighs 75g. When completely filled with a liquid X, it weighs 65g. Volume of density bottle is _________ 1) 30 cm3 2) 45 cm3 3 3) 60 cm 4) 75 cm3 A rectangular glass slab (a regular body) of some object has mass

1 kg. Its length, breadth and height 4

11 22. A density bottle weighs 20.25g when empty, 40.75g when filled with a liquid X and 50.25g when filled with water. Which is correct among these ? 1) X floats in water. 2) X sinks in water. 3) we can’t say. 4) X sinks in water sometimes only. 23. A sphere of radius 1.4 m weighs 500 kg. The sphere 1) Sinks in a water. 2) Floats in a water. 3) Sometimes sinks and sometimes floats in water. 4) We can’t say. 24. A boy makes a paper boat. A small stone is kept inside the boat. The masses of paper and stones are 15g and 50g respectively and their volumes are 40 cm3, 75m3 respectively. The boat is let into the kerosene oil. (The density of kerosene oil is equal to 0.8 g/cm3). Then 1) The boat sink in kerosene oil. 2) The boat floats in kerosene oil. 3) The boat sometimes float and sometimes sinks in kerosene oil. 4) we can’t say. 25. The volume of a density bottle has 50 ml filled with water completely without bubbles and stopper transereted. After cleaning it, its mass is determined using a balance i.e., 200 g. Then it is completely filled with the given liquid after removing the water and drying it. Its mass is again determined to be 160g. The name of the liquid is ______________. 1) Water 2) Mercury 3) Copper 4) Kerosene 26. A bottle weighs 30g when empty, 53.5 g when filled with a liquid, and 48g when filled with water. The relative density of the liquid is __________ 1) 2 2) 1.3 3) 7.3 4) 3.5 27. A drop of olive oil of radius 0.25 mm spreads into a circular film of diametre is 20 cm on the water surface. Then the size of an oil molecule is 1) 2.08 × 10–5 m 2) 2.08 × 10–6 cm 3) 2.08 × 10–7 cm 4) 2.08 × 10–7 mm

are 3 cm, 2 cm and 2 cm, respectively. The object 1) Sinks in water 2) Floats in water 3) Sometimes sinks and sometimes floats depending upon the material of that object 4) None 20. Statement -I : The density of body A = 0.2 g/cm3. Statement - II : The density of body B = 0.08 kg/m3. Which is correct among these ? 1) The body A only floats in water 2) The body B only sinks in water 3) The body B only floats in water 4 3 2 4) Two bodies A and B both float in water [ Volume of oil drop =  r ; area of the film =  r 3 21. A wood is of density 680 kg/m3. It is submerged in assuming a single layer of molecule is formed] three liquids. Those are alcohol, water and 28. Area and hectare, are units to measure large areas. 3 glycerine. The density of alcohol is 780 kg/m , The 1 mile2 = 640 acres density of water is 1 g/cm3 and the density of 1 hm2 = 1 hectare (hm = hectometre) glycerine is 1.250 g/cm3 . In which liquid is the wood 1 acre = ––––––––– hectare less submerged ? [Use: 1 mile = 1609m] 1) Alcohol 2) Glycerine 3) Water 4) Same in all www.betoppers.com

12

IIT JEE Worksheet 1.

2)

3)

4.

5.

6.

7.

8.

9.

Match the following. List - A List - B (a) Temperature (e) Mole (b) luminous intensity (f) Kelvin (c) Amount of substance (g) Candella 1) a  e; b  f; c  g 2) a  f; b  g; c  e 3) a  g; b  f; c  e 4) a  g; b  e; c  f Match the following Physical quantity Units a) Temperature (j) K b) Length (k) 0C (l) M (m) m 1) a – j, b – l 2) a – k, b – m 3) a – j,k ; b – l, m 4) a – l,m ; b – j, k From the following sets choose the incorrect representations. i) K.G. for kilogram ii) 0C for degree Celsius iii) mts for metre iv) s for second 1) Odd options 2) First two options 3) Even options 4) Last two options Find the mass of 555 cm3 of iron in kg when density of iron is 7.6 g/cm3 1) 4218 kg 2) 4.218 kg 3) 42.18 kg4) 421.8 kg If the length of the wooden cube is 4m and the mass is 1/8 kg, then the relative density of the wooden cube is ______________ 1) 19.53 × 10–7 2) 1/512 3) 1/51200 4) 1/32 A piece of iron has dimensions 3 cm × 1.5cm × 6 cm. If its mass is 205.2g, its relative density is ______ 1) 27 2) 7.6 3) 6.5 4) 8 An iron cylinder of radius 1.4 cm and length 8 cm is found to weigh 369.6 g. The relative density of iron cylinder is _____________ Take volume of cylinder = r 2  . 1) 7.2 2) 7.5 3) 8 4) 8.2 The length of a cloth measured is 200 cm. Match the following from List - A List - B i) Length a) Numerical value ii) 200 b) Unit iii) cm c) Physical quantity 1) i-a, ii-b, iii-c 2) i-c, ii-a, iii-b 3) i-c, ii-b, iii-a 4) i-b, ii-a, iii-c 1 kg = ______________ tonne. 1) 10–6 2) 10–3 3) 103 4) 106

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8th Class Physics 10. 1 micrometre (mm) = ____________ cm 1) 104 2) 106 3) 10–6 4) 10–4 11. Match the following from List - A List - B i) Gram a) 10–3 kg ii) Microgram b) 10–3 g iii) Milligram c) 10–6 g 1) i-a, ii-b, iii-c 2) i-a, ii-c, iii-b 3) i-b, ii-c, iii-a 4) i-b, ii-a, iii-c 12. A syringe has a capacity of 5 ml. Its capacity in cm3 and m3 is respectively are 1) 5 × 10–6 m3, 5 cm3 2) 5 cm3, 5 × 10–6 m3 3) 5 × 10–3 m3, 5 cm3 4) 5 cm3, 5 × 10–3 m3 13. A water tank has a capacity of 10, 000 litre. Its value in m3 is 1) 100 m3 2) 1000 m3 3) 1 m3 4) 10 m3 14. The water level of a measuring cylinder is 26 ml. A piece of concrete having a volume of 6 cm3 is immersed in it. The new level of water is 1) 20 ml 2) 26 ml 3) 32 ml 4) 6 ml 15. The mass of an electron is 9.11 × 10–31kg. How many electrons would make 1kg? 1) 1.1 × 1030 2) 1.1 × 1031 3) 1.1 × 10–304) 1.1 × 10–31 16. A vessel of 200gm weight is filled with some kerosene. If weight of the vessel with the kerosene is 270g, then how much kerosene is filled in the vessel? 1) 70ml 2) 80ml 3) 85ml 4) 70ml 17. If 10 copper pieces, each of mass 20g, are placed in the vessel with water, the level of water increases to 300ml in the vessel. Find the initial level of water in the vessel. 1) 100ml 2) 1000ml 3) 10ml 4) 0.1ml

1kg gm =10x then find the value of x . 3 m cm 3 1) –2 2) 2 3) –3 4) 3 19. If m1, m2, m3 and m4 are masses of four bodies 0.3 kg, 0.3 mg, 0.3 × 10–6 g and 3000 g respectively. The above, in increasing order of masses, are as follows. 1) m1 > m2 > m3 > m4 2) m4 > m2 > m3 > m4 3) m4 > m1 > m2 > m3 4) m3 > m2 > m4 > m1 20. Mass = Density × Volume; if mass = p, density = q and volume = r, if 10p = 1000q × 100r, if p, q and r in S.I system, then the value of q in C.G.S system is 1) 10–7g/cm3 2) 10–5g/cm3 3) 10–7g/cm34)107 g/cm3 18. If



Chapter - 2

Force and Pressure

Learning Outcomes

By the end of this chapter, you will understand     

1.

   

Force Interaction between forces Effects of Force Contact Forces Non Contact Forces

Pressure Fluid Pressure Atmospheric pressure Effect of Pressure

Introduction

7. 8. 9. 10.

A table can be moved from one place to another by either pushing it or pulling it. Pushing

Pulling

2.

Taking a carrot out of the ground Playing on a swing Picking up a shopping bag Squeezing a toothpaste tube

Interaction Between Forces Have you ever seen an arm wrestling match?

Similarly, you can open a door by either pushing or pulling it. When a ball is thrown with more force, it travels a longer distance. The shape of a bottle can change when it is squeezed. Also, the direction of a moving ball can change by striking it. Have you wondered how the shape of a bottle changes when squeezed? Or how can the direction of a moving ball change by kicking it in different ways? All the above activities can be associated with pushing or pulling. Therefore, whenever an object is moved, we can say that it has either been pushed or pulled. This push or pull is known as force. In other words, a body moves whenever a force is applied to it. Therefore, a body cannot move unless a force is applied.Apart from push or pull, force is any action that has the tendency to change the position, shape, or size of an object. Everyday actions such as pushing, pulling, stretching, lifting, squeezing, and twisting are also examples of force. Let us try to list some examples of everyday force and see if we can classify them as push or pull. Examples 1. Hitting a cricket ball with a bat 2. Opening a door 3. Plucking a flower 4. Flying a kite 5. Moving a wheel barrow 6. Hitting a tennis ball with a racquet

In an arm wrestling match, both the players try to push each other’s hand towards the table. Hands move along the direction of the player who applies a greater force. The player who is able to apply more force wins the match. A table in your room or a vehicle parked outside your house cannot move unless it is either pushed or pulled. To move a table, you have to either push or pull it. In cricket, a bat exerts force on the ball. Thus, the ball is able to gain speed and can reach the boundary line.Thus, we can observe that force comes into play when at least two bodies interact with each other. Can you list the two interacting bodies in the cases that we have discussed? Scenario Arm wrestl ing Pushing or pulling a table A moving vehicle Bat hitting a ball

Interacting bodies Arms of the players You and the table Vehicle and the road (or the ground) Bat and ball

8th Class Physics

14

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Effects of Force Force cannot be seen, heard, or tasted. Only its effects can be felt or seen. The various effects of 1. force are: 1. It can move a body initially at rest. 2. It can bring a moving body to rest. 3. It can change the direction of a moving body. 4. It can change the speed of a moving body. 5. It can change the shape of a body. 6. It can change the size of a body. Let us take an example of a football lying in a 2. field (fig - A). When a player hits the ball, it starts moving, i.e., it starts moving only when we apply force. Thus, force can move a body initially at rest. Now, if the goalkeeper catches the moving ball (fig – B), then it comes to rest. The goalkeeper applies a force to stop the moving ball. Hence, we can say that force can bring a moving body to rest. If another player kicks the moving ball in the opposite direction (fig – C), then it starts moving in the direction towards which it is kicked i.e. the 3. direction of the football changes. The player applies force on the football to change its direction. Hence, force can change the direction of a moving body. Also, if the player hits the ball hard, then the net speed of the ball will also change. Hence, the speed of a moving body can be changed by applying force.

(B)

(A)

(C)

The shape of a deflated football can be changed 4. by inflating it. When you inflate a football, you apply force on the pump. Hence, force can change the shape of an object. Also, if you keep inflating the football, then its size will keep on increasing. Hence, force can change the size of an object.

A deflated football

Force can change the shape of an object

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Force can change the size of an object

Formative Worksheet The given figure shows a heavy box being pushed and pulled.The number of persons pulling the box is

(A) 1 (B) 2 (C) 3 (D) 4 A wheelchair is pushed from West to East, as shown in the given figure. After traveling some distance, it takes two left turns. Finally, it takes a right turn and comes to a halt. Before the wheelchair comes to a halt, the direction of force on it is from:

(A) West to East (B) East to West (C) North to South (D) South to North A large box, placed on the ground, has to be shifted up to a wall, as shown in the given figure. Three boys push the box together. They have to apply a large amount of force to move the box. The motion of the box is opposed by the frictional force between the box and the ground.

If wheels were present below the box, then they would have to apply (A) More force (B) Same force (C) Less force (D) No force In a game of tug-o-war, Mark pulls the rope with a force of 60 N from the right. Sandy and George pull the rope with respective forces of 40 N and 20 N from the left, as shown in the given figure. The net force on the rope is

(A) 0 N

(B) 2 N

(C) 60 N (D) 120 N

Force and Pressure 5. Sandy, Jackie, Rudy, Ronnie, Cruz, and Harry are 8. pushing a box by applying forces of magnitudes 25 N, 35 N, 45 N, 45 N, 30 N, and 50 N respectively, as shown in the given figure

15 The force applied on an object will not lead to a change in (A) the shape of the object (B) the mass of the object (C) its direction of motion (D) its position

Conceptive Worksheet 1.

Which row shows the net force acting on the box and its consequent motion?

6.

Net force (N) Motion of the box (A) 20 Towards South 3. (B) 15 Towards East (C) 15 Towards West (D) 20 Towards North A trolley is moving at a constant speed of 2 m/s. Cole and Paul start pushing the trolley at the same time from opposite directions, as shown in the figure. Both of them are applying a force of 15 N on the trolley. 4.

The speed of the trolley will (A ) remain the same (B)become zero (C) decrease (D) increase 7.

2.

Jack pushes a big cabinet with a force of 50 N. Under his force, the cabinet moves at a uniform speed. Jenny comes to help Jack. She begins to push the cabinet in the same direction with a force of 30 N.

After Jenny’s arrival, speed of the moving cabinet will (A) increase (B) decrease (C) remain same (D) become zero

5.

In order to push or pull an object (A)force must be applied (B) speed must be applied (C) friction must be applied (D) electricity must be applied Roger noticed a book lying on a table . In order to move the book, he must apply a (A) gear (B) load (C) force (D) speed A bus is moving along a curved path, as shown in the given figure . During the motion of the bus, the direction of force on it: (A)remains the same (B) changes continuously (C) is directed in the backward direction (D)is directed in the downward direction A car moves with a constant speed toward East. A force is applied on the car to make it stop. The direction of the applied force is toward (A) North (B) South (C) West (D) East Four boys, Hunter, Jackson, Samuel, and Sean are pushing a heavy box. Their respective forces are listed below. Boy Force (N) Hunter 25 Jackson 25 Samuel 30 Sean 35 As a result of the presence of sand on the ground, the frictional force between the box and the ground is 15 N. The total force experienced by the box is

(A) 100 N toward the left (B) 115 N toward the left (C) 100 N toward the right D)115 N toward the right www.betoppers.com

16 6.

8th Class Physics NOTE 1. Frictional force is a contact force. 2. Frictional force always acts between two moving objects, which are in contact with one another. 3. Frictional force always acts opposite to the direction of motion. 4. Frictional force depends on the nature of the surface in contact.

Eric and Evan are pulling a big load in opposite directions. Eric is applying a force of 20.56 N, while Evan is applying a force of 25.7 N on the load.

II. Muscular Force

7.

4.

The force applied by the action of muscles in our body is termed as muscular force. For example, when you pick up a book placed on the table using your hands, you apply muscular force.For lifting the book from the table using your

(A) 4.15 N toward Evan (B) 5.14 N toward Evan (C) 4.15 N toward Eric (D) 5.14 N toward Eric A ball rolls at constant speed in a straight line along the ground. Jerry kicks the ball in the direction of its motion.As a result of Jerry’s kick, speed of the ball will (A) increase (B) decrease (C) remain same (D) become zero

Contact Forces Ramu is cycling on the road. He observes that as he stops pedaling, the cycle stops moving after traveling for some distance. Let us see why this happens? Forces acting between two bodies can be classified into two broad categories: Contact force and non-contact force.

hands, you had to touch the book. You cannot lift the book without making contact with it. Hence, muscular force is a contact force. Like humans, animals also use muscular force to perform various activities. For example, birds fly in the air by flapping their wings.

5.

Non Contact Force Does force act only when two objects are in contact? To understand, let us perform a small activity. Take a bar magnet and an iron nail. Bring the magnet close to the iron nail, but do not bring them in contact. What do you observe? The iron nail moves towards the magnet. This means that there must be a force that is acting between the magnet and the iron nail.

Contact forces are those that act between two objects, which are in direct contact with each other.The two common examples of contact forces are muscular and frictional.

I. Frictional Force Earlier you had seen that as Ramu stops pedaling the bicycle, it stops after some time. This happens due to the external force acting between the road and tyres of the bicycle. This force is known as the frictional force. The force of friction acts between all moving bodies, which are in contact with one another. The force of friction always acts opposite to the direction of motion. The magnitude of this force depends on the nature of the surface in contact. www.betoppers.com

Since the iron piece moves towards the magnet (even when they are not in contact), we can say that the force exerted by the magnet on the iron piece is a non-contact force.

I. Magnetic force Non-contact forces are those forces that act between two objects, but are not in direct

Force and Pressure contact with each another. Examples of non-contact forces include magnetic force, electrostatic force, and gravitational force. What will happen if you bring the South Pole of a bar magnet close to the North Pole of another bar magnet? The magnets will attract each other. They attract each other with magnetic force. What will happen if you bring the North Pole of both bar magnets close to each other?

The bar magnets will repel each other. The force with which they repel each other is known as magnetic force. NOTE 1. Magnetic force can be attractive as well as repulsive. 2. Magnetic force is a non-contact force. 3. Magnetic force acts between two magnets, or between a magnet and a magnetic material (such as iron). 4. Magnetic force depends on the strength of the magnet used. 5. Magnetic force also depends on the distance between the interacting bodies.

II. Electrostatic force Take a paper and tear it into pieces. Now, rub a plastic scale against dry hair and bring this scale close to the paper pieces. What do you observe?

17 5. Electrostatic force also depends on the distance between the interacting bodies.

III.

Gravitational force

Do you know why apples fall towards the ground from trees? Why does water from a tap flow down? The Earth attracts everything (that is near or on its surface) towards its centre by a non-contact force known as gravitational force. It is this force that makes an apple fall towards the ground from the tree and makes the water from a tap flow down.

NOTE 1. Gravitational force is a non-contact force. 2. Gravitational force is an attractive force. 3. Gravitational force is the force that is exerted by the earth on every object, which is near or on its surface. 4. Gravitational force depends on the mass of the body. 5. Gravitational force also depends on the distance between the Earth and body.

(a) Mass Quantity of matter possessed by a body, is called the mass of the body. It is represented by the symbol m. It is a scalar quantity.

You will observe that the pieces of paper are attracted towards the scale. This happens because rubbing of the scale against dry hair produces an electrostatic charge. Thus, the scale attracts the pieces of paper by a non-contact force known as electrostatic force. NOTE 1. Electrostatic force is a non-contact force. 2. Electrostatic force can be attractive as well as repulsive. 3. Electrostatic force is the force that exists either between two charged bodies, or between a charged and uncharged body. 4. Electrostatic force depends on the magnitude of charge present in the bodies.

Nature : A body with more mass, needs a greater effort (force) to move it from rest or stopping it from motion. The body exhibits inertia. Thus, mass offers inertia. This mass is called inertial mass (m1). A body never has a zero mass. Measurement of mass : Mass of a body is measured by a beam balance by comparing the mass with bodies of known mass. At one place, bodies of same mass have same pull of gravity on them. A beam balance works on the principle of moments (Bodies of equal masses, having equal weights, have equal and opposite moments about fulcrum of the balance, when suspended at equal distances from the fulcrum, and made the beam horizontal). www.betoppers.com

8th Class Physics Then from relation , Force = Mass × Acceleration i.e., W = mg Nature : As W = mg, the weight of a body will vary from place to place due to variation in value of g. A body has zero weight at the centre of the earth (where g = 0). On the surface of earth, the value of g is 9.8 m/s2 . g value at poles is greater than at equator. Measurement of weight : Weight of a body is measured by a spring balance.

18 b) Weight The force with which a body is attracted towards the centre of the earth, is called the weight of the body. It is represented by the symbol W. It is a vector quantity having direction towards the centre of the earth. Expression for weight : If mass of a body = m Acceleration due to gravity of the earth = g c) Difference between Mass and Weight

Mass Weight Mass is quantity of matter possessed by a Weight is the force with which a body is attracted body towards the centre of the earth. It is a scalar quantity. It is a vector quantity. Its SI units is kilogram (kg.) Its SI unit is newton (N). Mass of a body remains constant at all places Weight of the body changes from place to place. Mass of a body is never zero. Weight of a body becomes zero at the centre of the earth. Mass is measured by a beam balance. Weight is measured by a spring balance.

Formative Worksheet 9.

Bob studies four situations involving various forces. These situations are

I. II. III. IV.

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A satellite orbiting Earth A magnet attracting paper clips A man hitting a cricket ball with a bat A magnet repelling another magnet

The situation in which motion is caused by the direct application of a force is (A) I (B) II (C) III (D) IV Which of the following forces is not a non-contact 13. force? (A) Magnetic (B) Muscular (C) Electrostatic (D) Gravitational Four types of motion, which are caused by the action of direct and indirect forces, are described 14. in the table.

I II III IV

Fluttering of a flag because of wind Falling of a ball on the ground Dropping of an apple from a tree Falling of water from a bottle

Which of the four types of motion is caused by the application of a direct force? (A) I (B) II (C) III (D) IV www.betoppers.com

A small bar magnet is held by the magnetic force of a strong bar magnet fixed on its position on a rough inclined plane, as shown in the given figure.

The small bar magnet experiences (A) only magnetic force (B) magnetic and frictional forces (C) gravitational and magnet forces (D) gravitational, magnetic, and frictional forces Earth revolves around Sun in 365 days. Earth and Sun are not in contact with each other. Earth’s revolution around Sun is caused by the (A) shape of Sun (B) shape of Earth (C) force between Earth and Sun (D) distance between Earth and Sun Gravitational force is i force, while ii force is repulsive as well as attractive. The given statement is correctly completed by row

(A) i - an attractive; ii-frictional (B) i-a repulsive, ii-magnetic (C) i- an attractive; ii-Magnetic (D) i-a repulsive; ii- frictional

Force and Pressure 15. A wooden block slides down a rough inclined ramp, as shown in the given figure.

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The wooden block experiences (A) only frictional force (B) only magnetic force (C) gravitational and magnetic forces (D) frictional and gravitational forces Jack is trying to climb a height with the help of a rope, as shown in the given figure.

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Two bodies A and B of mass 500 g and 200 g respectively are dropped near the earth’s surface. Let the acceleration of A and B be a A and a B respectively, then (A) a A = a B (B) aA > a B (C) aA < a B (D) aA  a B A body is thrown up with a velocity of 20 m/s. The maximum height attained by it is approximately (A) 80 m (B) 60 m (C) 40 m (D) 20 m The weight of a body is 120 N on the earth. If it is taken to the moon, its weight will be about: (A) 120 N (B) 60 N (C) 20 N (D) 720 N Two iron and wooden balls identical in size are released from the same height in vacuum. The time taken by them to reach the ground are: (A) not equal (B) exactly equal (C) regularly equal (D) zero

Conceptive Worksheet 8.

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Jack can climb the top by overcoming the (A) tension of the rope (B) friction of the rope (C) magnetic force of Earth (D) gravitational force of Earth A heavy stone falls (A) faster than a light stone (B) slower than a light stone (C) with same acceleration as light stone (D) none of these A stone is dropped from the roof of a building takes 4s to reach ground. The height of the building is (A) 19.6 m (B) 39.2 m (C) 156.8 m (D) 78.4 m A ball is thrown up and attains a maximum height of 19.6 m. Its initial speed was (A) 9.8 ms–1 (B) 44.3 ms–1 –1 (C) 19.6 ms (D) 98 ms–1 The value of g at pole is (A) greater than the value at the equator (B) less than the value at the equator (C) equal to the value of the equator (D) none of these

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When a wooden box is suspended by a spring balance, the spring stretches because (A) Earth’s gravity pulls the box (B) Earth’s magnetic field attracts the spring (C) of the frictional force between the box and the spring (D) of the electrostatic force between the box and the spring Joanna is playing carrom along with her friends. She strikes the striker with her index finger and pockets the queen. Joanna uses which force for pocketing the queen? (A) Electric (B) Magnetic (C) Muscular (D) Gravitational Some magnets are brought in contact with a refrigerator. The magnets stick to the refrigerator because of (A) gravitational force (B) magnetic force (C) electric force (D) nuclear force Two magnets are placed at a distance from each other on a frictionless surface, as shown in the given figure. The magnets will experience (A) only magnetic force (B) only frictional force (C) magnetic and frictional forces (D) magnetic and gravitational forces www.betoppers.com

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A ball that is thrown up returns to the surface of Earth because of (A) gravity (B) friction (C) electricity (D) magnetism The direction of force on an upward moving ball is from (A) up to down (B) down to up (C) left to right (D) right to left The acceleration due to gravity is 9.8 m/s2 (A) Much above the earth’s surface (B) Near the earth’s surface (C) Deep inside the earth (D) At the centre of the earth A particle is taken to a height R above the earth’s surface, where R is the radius of the earth. The acceleration due to gravity there is (A) 2.45 m/s2 (B) 4.9 m/s2 2 (C) 9.8 m/s (D) 19.6 m/s2 When a body is thrown up, the force of gravity is (A) in upward direction (B) in downward direction (C) zero (D) in horizontal direction Mass of an object is (A) amount of matter present in the object (B) same as weight of an object (C) measure of gravitational pull (D) none of these The weight of an object is (A) the quantity of matter it contains (B) refers to its inertia (C) same as its mass but is expressed in different units (D) the force with which it is attracted towards the earth Weight of an object depends on (A) temperature of the place (B) atmosphere of the place (C) mass of an object (D) none of these The mass of body is measured to be 12 kg on the earth. Its mass on moon will be (A) 12 kg (B) 6 kg (C) 2 kg (D) 72 kg

Pressure Press your thumb against a drawing board. Your thumb does not go inside the drawing board. Now, take a drawing pin and press it against the drawing board. The pin goes inside. Why does the pin penetrate the drawing board? We know that it is easier to push a nail into a wooden plank by its pointed edge.

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8th Class Physics Why is it difficult to push a nail into a wooden plank by its blunt end?

Why do your feet sink when you walk on loose sand? When we walk on loose sand, our feet go inside the sand. However, this is not the case when we lie on loose sand. The force that is acting on the sand is our weight. Our feet have smaller surface areas than our entire body. This is the reason why our feet go deep inside the sand, but our bodies do not. From these observations, we can conclude that the same force can produce different effects. The pressure exerted on a body depends upon the following two factors: 1. Force Applied 2. Area over which the force acts The same force can produce different pressures, depending on the area over which it acts. For example, when a force acts over a large area, it produces a small pressure. However, if the same force acts over a small area, it produces a large pressure. Pressure is the force that acts perpendicularly on a unit surface area. To obtain the value of pressure, we should divide the force acting on an object by the area on which it acts. Thus, the value of pressure is:

Pr essure 

Force Area

The unit of pressure is Newton per square meter (N/m2), which is also known as Pascal. 1 Pascal = 1 N/m2 The force acting on a smaller area exerts a larger pressure as compared to the pressure it exerts on a larger area. This is the reason why drawing pins have pointed tips, knives have sharp edges, and dams have wide foundations. Now, can you explain why it is difficult to carry a bag having thin straps?

Force and Pressure

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Fluid Pressure

8. Atmospheric Pressure

Fluid : A substance which can flow from one point to another is called fluid. Since liquids and gases can flow, so they are known as fluids. A fluid exerts three types of pressure 1) Downward pressure 2) Upward pressure 3) Lateral pressure 1.

Downward pressure of water: Consider a vessel (V) containing some water of weight (W), say 500 gm. wt. Let the area of its base (A) be 100 cm2 . Then the weight of water (W) is the thrust acting on the base of area 100 cm2. This shows that liquids exert downward pressure.

. . . . .. . .. .. .. . . . .. . . . . . . . .. .. . . . . . .. . . . .. . ... .. .. ... .. ... . ... .... .. .. ... ..... .... . ..... .. . .. . . ....... ... ..... .. .. .. ... ...... .w . .... ........... . . ... ..... .. .... . . . . . . . . . .. .. .. .. . . .. . . .. . . . .. ....... ... . . . . . ... . .. .. .. ... ...... . ........ . ........ . ........ . . . . . . . . .. ... ... . .. .. . . . .. .. . . .. .. .. . ... .. . . . . .. .. ... . .. . .. .. ............ .. . . . . .. . .. . . .. . . .. .. ... .. . .. .. ....... . ..... . .. .. ... ........ . . . .... . . . ..... . .. .. . . .. . . ........... . ... .. ........... .. .. .. ... . . .................. ...... . . .... . ..... ...... .... ... . .. ...... . ...... ................................ ..A

2.

V

Lateral pressure of water: Take a long cylindrical vessel containing water and puch a hole (H) on its wall as shown in teh above figure. The water comes out with a speed and falls at distance. This proves that liquid has lateral pressure.

C

H

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Upward pressure of water: Take a foot ball and immerse it in water in a vessel (V) and leave it. The ball immediatly comes up adn floats on water. This shows that water (or fluid) exerts pressure in the upward direction.

V

NOTE 1. Liquids exert pressure on the walls of the container. 2. Pressure exerted by liquids increases with depth. 3. Liquids exert equal pressure at the same depth.

The atmosphere is a mixture of gases. The various gases in the atmosphere and their composition is given in the following chart. Percentage by Component volume Nitrogen 78.03 Oxygen 20.99 Argon 0.93 Carbon dioxide 0.035 Neon 0.0015 Hydrogen 0.0010 Helium 0.0005 Krypton 0.0001 Xenon 0.000008 Above the ground, the Earth’s atmosphere extends to a few hundred kilometers in height. Very close to the Earth’s surface, we find millions of gas molecules in every cubic centimeter. It is calculated that roughly around molecules per cubic centimeter are present. These molecules are all zipping around in different directions, colliding into each other and rebounding. They make an impact on each and every object by bombarding their surface resulting in pressure. The weight of the atmosphere also exerts force resulting in pressure. This pressure is known as Atmospheric pressure The weight of thrust of air exerted on unit area of earth’s surface is called Atmospheric pressure. Altitude and atmospheric pressure: The atmospheric pressure indicated by mercury barometer at different altitudes is given below: S.No

Altitude

Atmospheric pressure

1

10, 000 ft or 3000 m

520 mm or 52 cm

2

20, 000 ft or 6000 m

350 mm or 35 cm

3

30, 000 ft or 9000 m

226 mm or 22.6 cm

From the above chart, we come to know that the atmospheric pressure decrease as the altitude increases. The reason being that the number of molecules present at a higher altitude is less than the number of molecules present at the sea level. So, as the number of air molecules decreases with increases in altitude, the air pressure also decrease.

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8th Class Physics

22

Implication of low atmospheric pressure 1.

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Artificial pressure in air crafts Air crafts travel at higher altitudes. At such altitudes there will be 1017 molecules in one cubic centimeter of air, whereas on the surface of the earth their presence is 1000 times more. Further, the weight of the air column also decreases with altitude. So, at a higher altitude the pressure becomes very much less and the passengers encounter dangers like bursting of body cells, bleeding of noses and ears and also an increase in the blood pressure. To avoid these problems an artificial pressure is created inside the plane. Faster breathing on mountains

At the top of the mountains and high altitudes, there are a less number of molecules of air. Since we could inhale only fewer air molecules each time, we feel suffocated and tend to breathe faster to make up for the deficit. Greater lung power People like Nepalese and Sherpas, who reside in the high altitudes, have bigger lungs to retain more air. So, they do not feel breathlessness in mountains, whereas people of the plains find it difficult to breathe there. We can say that they have greater lung power. So mountain people make excellent footballers (e.g. Baichung Bhutia) Brazilian football teams practices on mountains to acclimatize their bodies to greater lung power which is necessary in games like football. Popping of ears

At the top of a tall mountain, the pressure is less. This causes our ears to pop in order to balance the pressure between the outer part and the inner part of our ears. Leaking of pen When we carry an ink pen at sea level, the ink normally link does not leak out of the pen.

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6.

This is because the atmospheric pressure and the pressure inside the pen are the same. As we go to higher altitudes the atmospheric pressure decreases but the pressure inside the fountain pen remains same. Due to this difference in pressure, ink leaks out of the pen. Drawing medicine into the syringe:

When we draw out the medicine from an injection bottle we pull the handle of the syringe in the backward direction. This decrease the pressure inside the syringe. As a result the medicine moves from a high pressure to a low pressure i.e., from the bottle into the syringe. 7.

Countering the absence of pressure in space – The Astronaut’s way:

There is no external pressure in space. Astronauts, while travelling through space, wear heavy space suits so as to create an external pressure. These space suits prevent the bursting of body cells and bleeding of nose and ears. Interesting Fact: Otto von Guericke, a German scientist, in 1654 at Magdeburg took two hollow metallic hemispheres of 51 cm each. He joined the two hemispheres together. When there was air inside them, they could be easily separated from each other. However, when all the air was pumped out, the force due to the atmospheric pressure on the outer surface of the two hemispheres became so large, that even horses on each side could not separate the two hemispheres.

Force and Pressure

23 Working principle of barometer: The formation of vacuum and the atmospheric pressure are responsible for the drawing up of water and the mercury column standing tall.

Measurement of Atmospheric Pressure

This is the working principle of the barometer.

Atmospheric pressure is measured by using a barometer. It was invented by Torricelli in 1643. The Mercury Barometer When you keep one end of a rubber tube in water and through suction draw the air from the other end placed at a certain height, water starts flowing to the higher elevation. The maximum height to which water can be drawn by this process is 33 ft. Evangelista Torricelli observed this phenomenon and wondered at it. He started research and conducted experiments to know the mystery behind it. Torricelli perceived that when a pipe was sunk into a well and if vacuum is created inside the pipe, the pressure of the surrounding air acting on the surface of the well water pushed the water up into the pipe. To prove his point, he devised an instrument. Instead of water, he used Mercury (Hg) which is 13.6 times denser than water. He filled a 3 – feet long tube closed at one end with mercury, taking care that no air bubble gets into the tube. He inverted this tube into a trough containing mercury. The mercury present at the closed end fell down, leaving behind an empty space. This empty space is nothing but vacuum. This vacuum came to be known as ‘Torricelli vacuum’, in his memory. Below the vacuum stood a Mercury column PQ of height 28 inches (76 cm). What makes the Mercury column stand ? Torricelli thought that Mercury can move up into the vacuum created at the top of the tube if some force pushes it. He argued that this force is provided by the atmospheric pressure, which is acting on the liquid surface as shown in the figure.

The same principle is applied to the straws used in cool drinks.

Vacuum

760 mm

Air pressure

The Mercury column is supported by the force of atmospheric pressure.

When we draw air form the straw, a low pressure is created inside the straw, where the atmospheric pressure outside is more. The high pressure outside pushes the liquid into the straw and then into our mouth. This work of Torricelli led to the invention of a device called the Barometer which is used to measure atmospheric pressure. Even today the Barometric pressure of air is given in the weather reports in Torricellian terms, in inches or millimetres of Mercury. The basic set up of a barometer is the same as in Torricelli’s experiment. Measurement of atmospheric pressure using barometer Atmospheric pressure (P) can be obtained by a barometer, using the formula. P=h×d×g Where, ‘h’ is the height of liquid column, ‘d’ is the density of the barometeric liquid and ‘g’ is the acceleration due to gravity. If ‘h’ is expressed in meters, ‘d’ in Kg/m3 and ‘g’ in m/s2, the pressure obtained by this formula is in terms of ‘Newton/m2 or ‘dyne/cm2 . Pressure measured in this way is called Absolute pressure. In the formula ‘P = hdg’, the terms ‘d’ and ‘g’ for a given barometer are constant and the atmospheric pressure is proportional to the height of the mercury column. So, in day – to – day usage, the atmospheric pressure is expressed in terms of the height of the mercury column. The height of the mercury level is an index of the atmospheric pressure. For normal atmospheric pressure, the height of the mercury column is 760 mm or 76 cm. www.betoppers.com

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8th Class Physics If a stone is dropped in bottle III, then the water ormative orksheet level will increase in (A) bottle III only A cube of edge 2 cm and mass 10 g is placed on (B) both bottles I and II the surface of a table. What is the pressure exerted (C) both bottles II and III by the cube on the table? (D) bottles I, II,and III equally (A) 0.098 N/m2 (B) 0.196 N/m2 30. The density of air a t normal atmospheric pressure 2 2 (C) 122.5 N/m (D) 245 N/m is 1.293 kg m–3. What would be the density of air at Utpal takes four identical beakers (I, II, III, a pressure of 10 atmospheres ? and IV) and makes a small hole in each of them 31. Calculate the equivalent height of water barometer at the same height from the base. He then fills if the pressure recorded by mercury barometer is water in all the beakers. The water comes out 76 cm. Density of mercury is 13600 kg m–3 and from the holes and falls at distances d1, d2, d3 , density of water is 1000 kg m–3. and d 4  respectively  from  the  base, 32. The pressure at a depth of 13.6 cm in a water where d1  V4 (B) V3  > V4 > V2 > V1 (C) V1 = V2 = V3 = V4 (D) V3  = V1  d2 > d4 > d1 . The sides a1, a2, a3, and a4 of the respective boxes are related as A) a1  > a2  > a3  > a 4 B) a1 > a4  > a2 > a 3 C) a3 > a2 > a4 > a 1 D) a4 > a 3 > a 2 > a 1 Four similar boxes are labelled I, II, III, and  IV. The masses of I and III are equal and the masses of II and IV are  equal.  The  mass  of  box I is smaller than that of box II. Boxes I and II are placed in a grass field and boxes III and IV are placed on a marble floor. Which box requires the maximum force to be moved? A) I B) II C) III D) IV Vishnu and Mahesh are pushing a heavy box placed on a floor, as shown in the given figure. The arrow over the box shows the direction of motion of the box. Vishnu is applying a force of magnitude Fv and Mahesh is applying a force of magnitude Fm.

Which of the following pairs of forces have the same direction? A) Fv and frictional force B) Fm and frictional force C) Fv and gravitational force D) Fm and gravitational force

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Ted learns about force of friction. He tries out different ways of reducing as well as increasing the friction on the surface of his driveway. In order to increase the friction on the surface of the driveway, Ted should A) Cover it with small pebbles B) Cover it with wax C) Pour water on it D) Pour oil on it body of mass 35 kg, is placed on a horizontal surface whose coefficient of friction 0.5. Find the frictional force acting on it. ( g = 10 m s-2)

Advantages and disadvantages of Friction Friction is a necessary evil. It has many advantages. 1. We are able to walk because of friction. You must have observed that it is very difficult to walk on an oily surface. This is because on an oily surface, friction between our feet and the ground becomes so less that we cannot move forward. 2. We are able to write because of friction present between the paper and tip of a pen. 3. Due to friction between the tyres and the ground, we are able to drive automobiles. Without friction, it would be impossible to stop our vehicle after starting. 4. To hold a glass of water, friction is necessary. Due to friction present between our fingers and the outer surface of the glass, we are able to grip the glass of water. If there was no friction present, it would not be possible to hold it. 5. Lighting a match stick is possible due to the friction between the matchstick and its cover. When we strike a match stick against its cover, fire is produced because of friction. Likewise, when we rub two stones against each other, friction between them produces fire. 6. A nail can be fixed in a wooden plank because of friction. Without friction, the wooden plank will not be able to hold the nail, when it is pushed on its surface. 7. Friction is required to play the violin, or to move a mouse on the mouse pad. If there was no friction between the bow and strings of violin, no music would have ever been produced. www.betoppers.com

8th Class Physics

36 However, friction has some disadvantages too. 1. Machines wear out and need lubrication after a period of time due to the friction present between the different parts of a machine. 2. Friction also produces heat in machines, when their parts are rubbed against each other. This can lead to over heating of the machine, which may lead to the damage of parts. 3. Tyres or soles of our shoes wear out because of friction. For this reason, we need to change our shoes after a period of time. 4. To overcome excess friction in air, a lot of fuel is wasted in cars and airplanes. Therefore, cars and airplanes are streamlined and given a unique shape to reduce friction.

4.

Methods of Increasing or Decreasing Friction While riding her bicycle, Shivani saw a speed breaker ahead and applies brakes to reduce her speed. After her ride is over, she wonders how the mechanism of brake actually functions. The brake pads are designed in a way that they are located along the moving wheels of a cycle. Initially, these pads are separated from the wheels [Figure (a)]. When brakes are applied, the brake pads come in contact with the wheel and stop its motion due to friction between the pads and tyre [Figure (b)].

Increasing Friction Although friction is considered as a necessary evil, in many cases we need friction to perform various tasks. Some of these cases are discussed below:

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Why do we use soles of shoes made of rubber and not of wood? (i) This is because the friction produced between the rubber and ground is more than the friction produced between the wood and ground. To increase friction further, soles are provided with a ripple or grove pattern. (ii) In grinder (silbatta), small holes are made to increase friction. (iii)  Before weightlifting, a weight lifter rubs powder on his hands. Do you know why? This  is  because  it  helps  him  in acquiring a better grip to hold the load. (iv)  Stripes are made on the surface of a tyre to provide a better grip on the road. Do you know why roads are made rough instead of smooth? Reducing Friction In various applications, friction is considered undesirable and different methods are employed to reduce it. Some of these methods are discussed below: 1) Lubrication In this method, oil or grease is applied between two materials. This forms a thin layer between the two materials, which prevents them to rub directly against each other. Due to this interlocking, the irregularities decrease. This reduces the friction between the moving parts.

When windows and doors become jammed, we apply grease or oil to their joints to make them move freely again. Similarly, to prevent the wear and tear of machine parts, we apply grease between the moving parts of a machine. Substances such as oil, grease and graphite used in lubrication are known as lubricants. In a game of carom, powder is sprinkled on the board. Can you explain the reason behind this? When you apply soap on your hands, does it become easier to rub your hands against each other? Is this an example of lubrication?

Friction

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37 Although friction can be minimized by the action of lubricants, one can never make friction equal to zero. This is because irregularities are always present on the surface. An excess of lubricants may also result in fluid friction. Rolling Friction It is easier to roll an object than to slide it. Hence, heavy luggages are fitted with rollers. Skates are provided with rollers to minimize the friction between the skates and the ground. Ball bearings are used in machines to reduce friction. They do this by the changing sliding movement into rolling movement. Streamlining Friction can be reduced by giving a definite shape to an object. Aeroplanes are given a typical shape to reduce friction due to air. Cars also use this method to improve efficiency.

Formative Worksheet 11.

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Tony is playing carom along with his friends. After completing 2 games, some talcum powder is rubbed on to the board. Talcum powder is used on the board to A) Reduce its friction B) Decrease its mass C) Improve its fragrance D) Improve its conductivity Two boxes, having the same weight, are kept between two vertical walls. Calvin lifts one of the boxes (against the pull of gravitational force) with the help of a rope. Before lifting the second box, he lubricates the vertical walls with oil. As a result, he has to apply less force to lift the second box. Which force is reduced as a result of the lubrication of the walls? A) Gravitational B) Compressive C) Frictional D) Magnetic John pushes a table with wheels under it. At the same time, Abraham pushes a table without any wheels under it. Both the tables weigh the same. The two tables move at the same speed, but John applies lesser force than Abraham.

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Which force is reduced by the wheels under John’s table? A) Magnetic B) Frictional C) Electrostatic D) Gravitational Olivia pushes a toy car on a rough surface. The toy car travels 4 m before stopping. The distance travelled by the toy car can be increased by A) Loading the toy car with a stone B) Removing the wheels of the toy car C) Pushing the toy car with less force D) Pushing the toy car on a smooth surface Ted learns about force of friction and tries out different ways of reducing as well as increasing the friction on the surface of his driveway. In order to reduce the friction on the surface of the drive way, Ted should A) Cover it with small pebbles B) Cover it with sand C) Spread grit on it D) Pour oil on it Ted learns about force of friction and tries out different ways of reducing as well as increasing the friction on the surface of his driveway. In order to increase the friction on the surface of the driveway, Ted should A) Cover it with small pebbles B) Cover it with wax C) Pour water on it D) Pour oil on it On which of the following types of roads will a sledge travel the farthest? A) Dry road B) Icy road C) Carpeted road D) Cemented road A magnet can pull an iron nail toward itself. The maximum distance from which the magnet (shown in the given figure) can pull the nail is labelled as d.

On which surface will this distance d be relatively large? A) Cemented surface B) Carpeted surface C) Sandy surface D) Oily surface www.betoppers.com

8th Class Physics

38 19.

20.

A mechanical system produces heat energy because of A) Radiation B) Convection C) Lubrication D) Friction Friction between two surfaces cannot be reduced by using A) Oil B) Gears C) Grease D) Rollers

Why is it easier to roll an object than slide it? In the case of rolling an object over a surface, the contact surface changes continuously. As a result, irregularities of the two surfaces do not have enough time to get interlocked with each other. On the other hand, irregularities between the two surfaces have enough time to interlock with each other when the object slides over a surface. Hence, rolling friction is smaller than sliding friction. Therefore, it is easier to roll an object than slide it. Rolling friction is always smaller than sliding friction. In ancient times, round logs were put under heavy stones to take them from one place to another. Can you say which one is easier to do – pushing a suitcase fitted with rollers or a suitcase without rollers?

Conceptive Worksheet 6.

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In order to reduce the friction of a wheel-axle, a bicycle uses A) Gears B) Brakes C) Pedals D) Ball bearings Joe slides a plank of wood over the floor. As the wood plank slides over the floor, its speed can be increased the most if A) The floor is lubricated B) It is coated with oil C) Balls are placed under it D) It is smoothed with sand paper Which of the following surfaces will have the minimum friction? A) Wet towel B) Oily marble floor C) Surface of cardboard D) Surface of a new tyre Force of friction is __I__ for a rough surface. Spilling oil on a surface __II___ friction. The information in which alternative completes the given statements? If the force of friction were not to exist, then it would be almost impossible to both A) Speak and listen B) Write and walk C) Speak and walk D) Write and listen

Rolling and Sliding Friction

Rolling friction is the resistance that acts against the motion of a body when it is rolled on the surface of another body. Sliding friction is the resistance that acts against the motion of a body when it slides over the surface of another body. Pushing a heavy box and lighting a matchstick are examples of sliding friction. Apart from the definition, there is another important difference between the rolling friction and sliding friction. It is always easier to roll a body than to push it. Therefore, we can say that rolling friction is always smaller than sliding friction. Due to this, skates are provided with rollers. www.betoppers.com

We know that rolling friction is smaller than sliding friction.  Hence,  it is better to  work  with rolling friction. This can be done by using ball bearings. A ball bearing is a small metal sphere placed between two surfaces. Ball bearings change sliding friction into rolling friction. They are vastly used in machines, cars, bicycles, and electric motors to change sliding friction into rolling friction.

6.

Friction in Fluids So, you see that air opposes motion through it. Not only air, all gases and liquids oppose motion through them. • In science, liquids and gases are collectively referred to as fluids. This is because these substances have a tendency ‘to flow’. The frictional force exerted by fluids is also known as ‘drag’. • Like solids, fluids (all liquids and gases) also exert frictional force on objects that pass through them. The frictional force experienced by a body moving through a fluid depends on a number of factors. For better understanding of these factors, let us see an animation. What have you heard? Streamlined shape? What is that?

Friction A body moving in a fluid (liquid or air) has to overcome fluid friction (drag). This results in the loss of energy. Since drag depends on the shape of an object, it can be minimized by giving objects a special shape. For example, cars and aircrafts are designed in such a way that fluids can pass through them smoothly. For that purpose, they are given a streamlined body. In a ship, the part which is above the water experiences the force of friction from air and the part which is inside the water experiences the force of friction from water. Hence, these parts have been given a unique shape to overcome the force of friction by both fluids.

23.

reduces the

24.

25.

Formula 1 (also referred to as F1) is the fastest motor car race in the world. The shape of a formula one car is designed in such a way that it helps in reducing the air drag and allows it to achieve an average speed of 360 km/hr!

26.

Formative Worksheet 21.

A ball rolls down an inclined plane, as shown in the given figure. The surface of the inclined plane is rough. 27.

(A)

(B)

(C)

(D)

22.

Which figure correctly represents the direction of the frictional force between the ball and the inclined plane? The designing of cars has improved over the years. These days, cars are designed to be more aerodynamic. Cars are designed to be more aerodynamic in order to A) Increase their speed B) Make them attractive C) Make them less polluting D) Increase the space inside them

39 Airplanes are designed in a particular shape called the streamline shape. The streamline design

28.

A) Speed of the plane B) Weight of the plane C) The cost of production D) Heat generated by air friction A ball moves on a smooth frictionless surface before entering a surface covered with grass. The speed of the ball at three different points is shown in the given figure.

The relation among the speeds of the ball at the three given points is A) v1  v3 C) v1 = v2 > v3 D) v1 = v2  fst > froll B) fst  > fsl  > froll C) fst  > froll  > fsl D) fsl  > froll > fst In order to move a stationar y object, an interlocking between the two surfaces must be overcome. The minimum force  required  to overcome this interlocking is known as A) Static friction B) Rolling friction C) Sliding friction D) Kinetic friction A ball is rolled upward along an inclined plane. After reaching a certain height from the ground, it rolls down along the plane. During the motion of the ball along the inclined plane, the force of friction A) Always opposes the motion B) Always facilitates the motion C) First opposes and then facilitates the motion D) First facilitates and then opposes the motion A formula one car is shown in the given figure. Formula one (F1) is one of the most famous and fastest racing events of the world. In this sport, cars can acquire speeds as high as 320 km/hr. One of the reasons of the high speeds is the design of the F1 cars. The shape of the car helps in A) Reducing the air drags B) Reducing the car weight C) Increasing the car efficiency D) Increasing the consumption fuel

h. i. j. k. l. m. n. o. p. q. r. s. t. u. v. w. x. y. 2.

Summative Worksheet 1.

f. g.

Fill in the blanks. (a) Friction opposes the ________ between the surfaces in contact with each other. (b) Friction depends on the _________ of surfaces. (c) Friction produces _________. (d) Sprinkling of powder on the carrom board ________ friction. (e) Sliding friction is ________ than the static friction. The force of friction always _______ the motion of an object. Friction _______when you press the two surfaces hard together.

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3.

The smoother a surface is, the_______ will be the friction. Rolling friction is_______ than sliding friction. The soles of our shoes wear out in a few months due to_______ _______is a self adjusting force. The force of friction which does not allow one body to slide upon another is called_______ Frictional force in a ceiling fan is reduced by using_______ Aeroplanes are streamlined to reduce_______ friction. With increase of temperature, the frictional force acting between two surfaces _______ We can not walk on ice or on a very smooth surfaces because friction between our shoes and these surfaces is _______ The bodies of aeroplanes, missiles, rockets, cars are streamlined to reduce friction with air, called The frictional force that fluids exert on object is called_______ To reduce fluid friction, the shape of object should have a_______ Friction can be _______or _______by adopting proper methods. Lubricants are used to _______friction. The friction offered by wheels is called _______ Ball bearings converts the sliding friction to _______ Friction prevents movement until the force applied is_______ than the force of friction. Friction is a _______ force. Four children were asked to arrange forces due to rolling, static and sliding frictions in a decreasing order. Their arrangements are given below. Choose the correct arrangement. (a) Rolling, Static, Sliding b) Rolling, Sliding, Static (c) Static, Sliding, Rolling (d) Sliding, Static, Rolling Alida runs her toy car on dry marble floor, wet marble floor, newspaper and towel spread on the floor. The force of friction acting on the car on different surfaces in increasing order will be (a) Wet marble floor, dry marble floor, newspaper and towel. (b) Newspaper, towel, dry marble floor, wet marble floor. (c) Towel, newspaper, dry marble floor, wet marble floor (d) Wet marble floor, dry marble floor, towel, newspaper

Friction 4.

5.

6. 7.

8. 9. 10.

Suppose your writing desk is tilted a little. A book kept on it starts sliding down. Show the direction of frictional force acting on it. You spill a bucket of soapy water on a marble floor accidentally. Would it make it easier or more difficult for you to walk on the floor? Why? Explain why sportsmen use shoes with spikes. Iqbal has to push a lighter box and Seema has to push a similar heavier box on the same floor. Who will have to apply a larger force and why? Explain why sliding friction is less than static friction. Give examples to show that friction is both a friend and a foe. Explain why objects moving in fluids must have special shapes.

HOTS Worksheet 1.

2.

3.

Luis throws four stones having the same weight on four different surfaces. The surfaces are icy, dry, sandy, and cemented. He throws each stone with the same force. On which of the given surfaces will the stone go the maximum distance  before  it  stops? A) Icy surface B) Dry surface C) Sandy surface D) Cemented surface The joints of Latif’s garage door are producing sound. To solve this problem, he must apply A) Saw dust in the joints of the door B) Water in the joints of the door C) Sand in the joints of the door D) Oil in the joints of the door In a cycling race, it is observed that a cyclist normally bends his body forward (as shown in the given figure).

41 5.

6.

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8.

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The cyclist bends in order to A) Feel comfortable B) Reduce his weight C) Reduce the air drag D) Increase energy consumptionlution Which of the following actions does  not  increase friction? A) Threading tires B) Grinding on stones C) Making spikes on shoes D) Powdering a carom board

The sides  of  matchboxes  are  made  rough  so  as to A) Hold them easily B) Reduce their cost C) Make them strong D) Increase their friction Friction has advantages as well as disadvantages. Dilip prepares a list of some of the advantageous applications of friction, as shown in the given table. Application of friction I. Walking II. Writing III. Wearing out of shoes IV. Igniting a matchstick V. Gripping an object Which of the items listed in the given table is not an advantage of friction? A) II B) III C) IV D) V A box  is  placed  on  three  different  surfaces  “  a marble floor, a cardboard sheet, and a sand paper. The minimum forces required to simply move the box on the surfaces are F m, F c , and Fs respectively.  The  three  forces  can  be arranged in the decreasing order as A) Fc  gas 22. A person, pressing his ear on the railway tracks can hear an approaching train. This is possible due to (A) vibration of railway tracks (B) vibration of air (C) more speed of sound in solid medium (D) hearing ability of the man 23. A person can be identified by the quality of sound produced by him. The characteristic of a sound can be determined by (A) amplitude (B) frequency (C) loudness (D) all of the above 24. The voices of men, women and children are different due to difference in (A) larynx (B) lungs (C) vocal cords (D) wind pipe 25. An object moving at a speed greater than that of sound is said to be moving at (A) ultrasonic speed (B) sonic speed (C) infrasonic speed (D) supersonic speed 26. The velocity of sound in vacuum is (A) 332 m s1 (B) 330 m s1 (C) 288 m s1 (D) 0 27. Flash and thunder are produced simultaneously. But thunder is heard a few seconds after the flash is seen. This is because (A) speed of sound is greater than speed of light (B) speed of sound is equal to the speed of light (C) speed of light is much greater than the speed of sound (D) none of these. 28. A to and fro motion by an object is also called____. (A) periodic motion (B) oscillatory motion (C) Cyclic motion (D) none of these

Sound 29. An object oscillates 50 times in one second. What should be its frequency? (A) 0.2 Hz (B) 0.02 Hz (C) 0.002 Hz (D) 50 Hz 30. The time period of simple pendulum is 0.2 sec. what is its frequency of oscillation? (A) 0.5 Hz (B) 5 Hz (C) 50 Hz (D) 1 Hz

HOTS Worksheet 1.

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Which of the following is a transverse wave? (A) Waves along a stretched string (B) Sound waves (C) Light waves (D) Radio waves Choose the correct statement (A) The minimum distance in which a sound wave repeat itself is called its wavelength (B) The number of vibrations per second is called frequency (C) The time taken to complete one vibration is called time period (D) The time taken to complete one vibration is called frequency. Sound cannot travel in (A) Solids (B) Liquids (C) Gases (D) Vacuums If the velocity of sound air at 00c is 332 m/s, its velocity at 300c is (A) 200 m/s (B) 300 m/s (C) 350 m/s (D) 996 m/s The speed of sound in air is dependent of change in (A) Temperature (B) Pressure (C) Humidity (D) Wind An echo is heard after 0.8 s, when a person fires a cracker 132.8 m from a high building. What is the speed of sound ? (A) 315 m/s (B) 332 m/s (C) 320 m/s (D) 310 m/s A man stands in between two cliffs and explodes a cracker. He hears the first echo after 0.6s and the second echo after 2.4s. Calculate the distance between the cliffs. (speed of sound = 336 m/s) (A) 515 m (B) 520 m (C) 504 m (D) 505m A dog barks in a park and hears its echo after 0.5 second. If the sound of its bark got reflected by a nearby building, find the distance between the dog and the building. Take the speed of sound in air is 346 m/s. (A) 86.5 m (B) 86 m (C) 85 m (D) 75 m

51 9.

A ship sends a signal and receives it back from the bottom of the sea after 6s. What is the depth of the sea at that place? (speed of sound in sea water is 1450 ms–1) (A) 4.35 km (B) 3.35 km (C) 435 km (D) 335 km 10. Using the sonar, sound pulses are emitted at the surface, the pulses after being reflected from the bottom are detected, if the time interval from the emission to the detection of the sound pulses is 2 seconds, find the depth of the water. (Taking velocity of sound through water = 1531 m/s) (A) 1500 m (B) 1000 m (C) 1531 m (D) 1530 m

IIT JEE Worksheet 1.

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A sonar device on a submarine send out a signal and receives an echo 5 seconds later. Calculate the speed of sound in water, if the distance of the object from the submarine is 3625m. (A) 1440 m/s (B) 1460 m/s (C) 1450 m/s (D) 1470 m/s A person fires a gun infront of a building 167m away. If the speed of sound is 334 m/s. Calculate the time in which he hears an echo. (A) 0.2s (B) 2s (C) 1.0s (D) 0.1s The distance between the observer (source of sound) and wall is 500 m. The observer hear the first echo after 2 sec then velocity of the sound is (A) 250 m/s (B) 500 m/s (C) 750 m/s (D) 1000 m/s A man stands in between two parallel cliffs and fires a gun. Hears two successive echoes after 2 second and 4.5 seconds. What is the distance between the two cliffs. (A) 332 m (B) 1079 m (C) 482 m (D) 747 m Match the following: Column I Column II The minimum echo listen 1) distance between A) the echo is man & wall to SONAR works based 2) B) 0.1 sec on the principle Persistence of 3) C) 17 m hearing of human ear 4) Reflection of sound D) Reverberation

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8th Class Physics

52 6.

An observer, situated between two parallel cliffs, emits an intense sound note. Two successive echoes are then heard after 5s and 7s. Calculate the distance between the cliffs. (Velocity of sound = 340 ms–1) 7. A stone is dropped into a well 44.1m deep. The sound of the splash is heard 3.13s after the stone is dropped. Find the velocity of sound in air (g = 9.8ms–2) 8. A boy strikes the iron pipe with a stone. Another boy who keeps his ear close to the other end of the pipe heard two sounds in a short interval of time. Explain why? 9. What happens to velocity of sound if water vapour content in air is decreased? 10. Radio waves of speed 3 × 108 m/s are reflected at the moon and received back on earth. The time elapsed between the sending of the signal and receiving it back at the earth station is 2.5s. What is the distance of the moon from the earth? 11. An observer stands at a distance of 850m from a cliff and fires a gun. After what time gap will he hear the echo, if sound travels at a speed at 350 ms–1 in air?

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Temperature and Heat

By the end of this chapter, you will understand • • • • • •

Hot and Cold Heat is form of an energy General effects of Heat Energy Sources of Heat Flow of Heat Energy Units of Heat

• • • • •

Temperature Thermometer Different scales of temperature Transmission of heat Conduction

1. Hot and Cold

Chapter - 5

Learning Outcomes

Here heat produces light. In a hot air balloon the hot gases, being lighter than the surrounding air, rise up in the air and are made to lift weights. H e r e heat is used to produce mechanical energy. The heat in a fire cracker produces both sound and light. Other forms of energy can also be converted to heat energy. For example, you can feel the heat produced from the mechanical energy by rubbing your palms vigorously against each other. When a candle burns in air, chemical energy is converted into heat. In an electric bulb, electrical energy is converted into light and heat. So as in an electric heater the electrical energy is converted into heat energy. When heat is given to a substance, we find the following effects : 1. Rise in temperature 2. Expansion 3. Change in state

Hold a piece of ice on your palm. You feel cold on your palm. Now, dip your fingers in warm water. You feel warm. Hold a glass of boiling tea in your hand. You feel hot. Thus, you feel cold, warm and hot. Try putting your index finger in boiling tea. It may not be possible to hold your finger in boiling tea or try even touching it. You may burn your finger if you keep it in boiling tea even for a second. You may say that tea is boiling hot and it should not be touched.

3. General effects of Heat Energy Same is true about the weather. It is cool at night. It is warm in the shade during day. It is hot in sun. We make use of our sense of touch to learn about cold, warm and hot. We are able to sense heat in an object. Heat is something which produces a sensation in our body by way of which we make out whether a body is cold, warm or hot.

2. Heat is form of an energy Energy is the ability to do work. When an object has the ability to do work, we say that the object has energy. Heat has also the ability to do work. For example, the steam engine pulls a train converting heat into mechanical energy. Heat can also be converted to other forms of energy. For example, when charcoal is heated, it gives light.

(a) Heat energy brings about change in temperature (b) Heat energy brings about change in dimensions (c) Heat energy brings about change in state (d) Heat energy affects living beings

4. Sources of Heat 1.

Anything that gives off heat is a source of heat. The following are the main sources of heat. The sun gives us heat. Touch a tin or aluminium sheet that has been in the sun for some time. It feels hot. Now, place the tin sheet inside the house. Touch the tin after some time. If feels cold. Use a magnifying glass to focus sunlight on a piece of white paper. What happens to the piece of paper at the position where the sunlight is focused ? It becomes hot, starts to turn brown and soon bursts into flame. Thus, the heat from the sun makes the paper burn.

54 2. Man can make heat by burning wood, coal and gas. Wood, coal and gas are called fuels. Man burns fuels to keep himself warm. He also burn fuels in the machines to make them work. A motor car runs by burning fuel. An aeroplane flies by burning fuel. 3. Electricity also gives heat. Man uses electricity to cook the food, to work the machines. It is easy to find out whether electricity gives heat or not. Put your hand near a lighter bulb. What do you feel? Touch a radio which has been switched on for some time. What do you feel ? 4. Heat can be produced by rubbing two things together, this is by doing mechanical work. Rub the palms of your hands together. What do you feel ? You will feel that is hot.

5. Flow of Heat Energy Heat flows from a hot body to a cold body. A body which is losing heat is feeling the other body to be cold. a body which is gaining heat is feeling the other body to be hot. Thus heat always flows from a body of higher temperature to the body at lower temperature. Heat is a form of energy which always flows from a hot body to a cold body. (or) Heat is a form of energy which makes any object hot or cold. Heat energy is also called thermal energy. Hot Body

Cold Body

8th Class Physics

6. Units of Heat The S.I. unit of heat is joule(J). Another commonly used unit of heat is calorie (cal). One calorie is the quantity of heat energy required to raise the temperature of 1g of water through 1°C. 1 cal = 4.2 J, 1 k.cal = 1000 calories Note : (i) As heat is a form of energy. So, its unit is same as energy. (ii) It is a scalar quantity.

7. Temperature When we touch a hot object our hand becomes warm, because heat flows from the object to our hand. We say that the object is at a higher temperature than our hand. If we touch a piece of ice our hand feels cold, because heat flows from our hand to the ice. We say that ice is at a lower temperature than our hand. This suggests that temperature tells us how hot a body is. It is the degree of hotness or coldness of a body. Heat flows in the direction of fall of temperature. When an object is heated, its temperature rises. When it is cooled, its temperature decreases. If two bodies having unequal temperatures are brought together, heat flows from the body at higher temperature to the body at lower temperature, till the two bodies are at the same temperature. The degree of hotness or coldness of the body is called temperature. Mathematically, Temperature is heat per unit mass.

Unit of Temperature: S.I. unit of temperature is Kelvin (K). Other unit of temperature is degree Celsius (°C) and degree Fahrenheit (°F). Note : (i) It is a scalar quantity. (ii) Thermometry is the branch of heat dealing with the measurement of temperature.

Difference between Heat and Temperature Heat Temperature 1. Heat is a form of energy. Hence it has 1. Temperature indicates the thermal the capacity for doing work. condition of a body which be stated as 2. Heat is the cause how much hot or how much cold the 3. Two bodies of same substance having body is different masses may have same 2. Temperature is the effect amount of heat but different 3. Two bodies of same substance having temperatures. different masses may have same 4. Heat contents of a body do not decide temperature but different amount of heat the direction of heat flow from the 4. Temperature of a body decides the body. direction of heat flow from the body. 5. The SI unit of heat is joule (J) 5. The SI unit of temperature is Kelvin (K) Note: Difference no-3 can be understood mathematically based on the point that temperature is heat per unit mass.

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Temperature and Heat

8. Thermometer The device for measuring the temperature of a substance is called a thermometer. (‘thermo’ is a Latin word which means heat and ‘meter’ means a measuring device). The earliest thermometer was developed by Galileo in 1593. His thermometer was based on the property of expansion of a gas (like air) on heating. Modern thermometers use mercury as the thermometric liquid, though in some cases alcohol is also used. These thermometers are known as liquid thermometers. The thermometer which has mercury as the thermometric liquid is called mercury thermometer. Let us study about it.

a) Mercury Thermometer We know that the substances expand when heated and contract when cooled. This principle is utilized for the construction of a thermometer.

Shake it below this point

Normal temperature (98.6°f or 37°C) It consists of a glass tube T having a very thin and uniform bore. This thin-bore glass tube is called capillary tube. At the lower end of the capillary glass tube there is a glass bulb B which is filled with mercury. The air is been removed from the capillary tube and it is sealed at the upper end. The whole length of the capillary tube (called stem) is calibrated or graduated in degrees to read the temperature. To measure the temperature of a hot body, the thermometer bulb is kept in contact with the body for a few minutes. Mercury of the bulb takes heat from the hot body and expands. Due to expansion of mercury, the level of mercury rises in the glass tube. When the level of mercury becomes steady (or constant) we can read the temperature on the thermometer scale. When the thermometer bulb is put in contact with a cold body, mercury in the bulb contracts and the level of mercury in the tube falls. The purpose of using a thin bore glass tube is to make the thermometer very sensitive, so that even a small change of temperature may produce a large movement of the mercury level in the capillary tube.

55

b) Uses of Mercury in Thermometers Mercury is used in thermometers because of the following advantages i) It expands evenly as the temperature rises. ii) It is a good conductor of heat. iii) Its density is higher iv) Its very sensitive in expansion. v) It does not stick on the wall of a glass tube. vi) It has very low freezing point and a very high boiling point.

c) Construction of Thermometer The construction of thermometer involves two steps Step I : Calibration of thermometer. The calibration of thermometer involves fixing of two points on it. One lower fixed point and other upper fixed point. Lower Fixed Point: The melting point of pure ice at normal atmospheric pressure is taken as lower fixed point (L.F.P). Upper Fixed Point: The boiling point of pure water at normal atmospheric pressure is taken as upper fixed point (U.F.P). How to mark the fixed points of a thermometer ? A thick-walled capillary tube of uniform bore, having a cylindrical bulb of suitable size at one end is taken. The upper end is sealed after filling mercury in it by alternate heating and cooling method. The bulb of the capillary is kept in ice. The temperature at which ice melts, that is, the icepoint, is marked as the lower fixed point on the capillary. After some time, the bulb of the capillary is dipped in boiling water. The temperature at which water boils, that is, the steam-point, is marked on the capillary and is taken as the upper fixed point on the temperature scale. Step II : Dividing the interval between two fixed points into convenient number of equal parts. The interval between the lower and upper fixed points is divided into suitable number of equal intervals depending upon scale of temperature. Each interval is called a degree. Note: The S.I. unit of temperature is Kelvin(K). However the commonly used unit of temperature is degree celsius (°C) and degree Fahrenheit (°F). Freezing and Boiling Point : A thermometer has two standard marking on its glass tube. These are called ‘lower fixed point’ and upper fixed point. The lower fixed point of thermometer scale is the temperature of melting ice (ice point). It is given a value of 0°C. The upper fixed point of a thermometer www.betoppers.com

8th Class Physics

56 scale is the temperature of boiling water. It is given a value of 100°C (Steam point) Examples:

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Choose the correct statements: (A) S.I. unit of temperature is Kelvin (B) Temperature is a scalar quantity. (C) S.I. unit of heat is joule(J). (D) 1 cal = 4.2 J Statement - I : Heat is the cause of temp. Statement - II : Temp. is the effect of heat (A) Both Statements are true, Statement - II is the correct explanation of Statement-I (B) Both Statements are true, Statement - II is not correct explanation of Statement - I. (C) Statement - I is true, Statement - II is false. (D) Statement - I is false, Statement - II is true. 1 kilo calorie = _________ calories (A) 1000 (B) 100 (C) 10 (D) 10000

Conceptive Worksheet 1.

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Formative Worksheet 1. 2.

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A body was supplied 100 kilocalories of heat. Express this energy in mega Joules. Body temperature of an animals is 110 °C. If boiling water is poured on it, the sensation felt by the animal is _________ (hotness/coldness). A body P of mass 25g has 100 calories of heat and another body Q of mass 20g has 150 calories of heat. If the two bodies are kept in contact, then heat flows from _________ to __________. Statement I : Heat is a form of energy it has the capacity for doing work. Statement - II : Heat and energy have same units. (A) Both Statements are true, Statement - II is the correct explanation of Statement-I (B) Both Statements are true, Statement - II is not correct explanation of Statement - I. (C) Statement - I is true, Statement - II is false. (D) Statement - I is false, Statement - II is true. Column-I Column-II (A) Boiling point of water 1) 100° C (B) Melting point of ice 2) 0°C . (C) Normal temperature of human body 3) 98.4°F (D) Mercury 4) Thermometric liquid 5) Expands on heating

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3.

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Temperature determines (A) degree of hotness or coldness of a substance (B) density of a substance (C) nature of the substance (D) All of these When a substance is heated then (A) Heat energy brings a change in temperature. (B) Heat energy can bring a change in state. (C) Heat energy can change the dimensions. (D) Heat energy can change into other form of energy. If two bodies having unequal temperature are brought together then (A) Heat flow from hot body to cold body (B) Heat flow from cold body to hot body (C) Heat does not flow (D) None of these The degree of hotness or coldness of the body is called (A) Heat (B) Temperature (C) Pressure (D) Force Mercury is the commonly used thermometric liquid because (A) It can be easily obtained in pure state (B) It does not stick to glass tube thermometer (C) It has a very high density (D) It has very low freezing point and a very high boiling point. Thermometer works on the principle of (A) substances expands on heating (B) substances contracts on heating (C) substances have no effect on heating (D) substances firsts expands & later contract on heating

Temperature and Heat

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9. Different scales of temperature There are following four scales of temperature in common use : 1. Celsius scale or Centigrade scale 2. Fahrenheit scale 3. Kelvin scale or absolute scale 4. Reaumur scale

1. Celsius Scale or Centigrade Scale This scale was devised by Anders Celsius in the year 1710. On this scale, ice point is taken as 0°C and steam point is taken as 100°C. The fundamental interval is divided into 100 equal parts (divisions). Each division corresponds to a difference of temperature of 1°C (one degree of Celsius).

1 th part of 100 the interval between the ice point and Note: A degree on Celsius scale is

2. Fahrenheit Scale This scale was devised by Gabriel Fahrenheit in the year 1717. On this scale, ice point is taken as 32°F and steam point is taken as 212°F. The fundamental interval is divided into 180 equal parts (division). Each division corresponds to a difference of temperature of °F (one degree of Fahrenheit).

1 th 180 part of the interval between the ice point and the steam point. Note : The degree on Fahrenheit scale is

Note : A degree on Reaumur scale is

3. Kelvin Scale This sale of temperature was given by Lord Kelvin (1824 – 1907) and is also known as Kelvin scale of temperature. On this scale, ice point is taken as 273 K and steam point is taken as 373K. The fundamental interval is divided into 100 equal parts (division). Each division corresponds to one Kelvin. Note : 1)

1

A degree on Kelvin scale is th part of the 100 interval between the ice point and the steam point. 2) The size of 1 degree on the Kelvin scale is the same as the size of 1 degree on the celsius scale. Absolute zero of temperature : The temperature of –273°C or zero degree on Kelvin scale is called as “Absolute Zero” of temperature. It is the lowest attainable temperature. 0°C = 0 + 273K = 273 K –273 °C = –273 + 273K = 0 K

4. Reaumur scale This scale was devised by R.A. Reaumur in the year 1730. The interval between the lower and the upper fixed points is divided into 80 equal parts. Each division is called one degree Reaumur (1°R). On this scale, the melting point of ice at normal pressure is 0°R. This is lower fixed point. The boiling point of water at normal pressure is 80°R. This is the upper fixed point.

1 th part of the interval between the ice point and the steam point. 80

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Different temperature scales

Parameters of different Temperature Scales Low er fixed point (Ice point)

Upper fixed point (Steam point)

No. of division on fundamental interval

1. Celsius scale

0°C

100°C

100

2. Fahrenheit scale

32°F

212°F

180

3. Kelvin scale

273K

373K

100

0°R

80°R

80

Temperature scale

4. Reaumer scale

Relation between Celsius and Fahrenheit Scales of Temperature

Thermometre Reading - Lower fixed point no. of divisions = constant Therefore, in order to convert temperature from one scale to another, following relation is used. Temperature on one scale - Lower fixed point  Upper fixed point - Lower fixed point

From the lower and upper fixed points on the two scales, we can write Ice point = 0°C = 32°F Steam point = 100 °C = 212 °F



=

 Temperature on first scale 

 L.F.P

U.F.P  L.F.P

 Temperature on second scale  - L.F.P

U.F.P - L.F.P C0 F  32 K  273 R0     100  0 212  32 373  273 80  0 C F - 32 K - 273 R i.e., = = = 100 180 100 80 C F - 32 K - 273 R i.e., = = = 5 9 5 4

Formative Worksheet (a) Celsius

(b) Fahrenheit

9.

Celsius and Fahrenheit scales As the fundamental length between the two standard fixed points is same, we can say. 100°C – 0°C = 212°F – 32°F 100°C = 180 °F

180 °F 1°C = 100

9 °F 5 Thus, the size of a degree on the Celsius scale is larger than that on the Fahrenheit scale. Therefore, 1 division on the Celsius scale is equal to 180/100 divisions on the Fahrenheit scale.

 1°C =

Conversion of temperature from one scale to another scale Whatever may be the scale of temperature used from experiments it is concluded that,

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10. 11. 12. 13.

14.

15.

Body A has more heat than body B. Then temperature of A is always greater than temperature of B. (True/false). Justify. Express 100 K in °C. (A) –273°C (B) 100°C (C) 212°C (D) –173°C Express 27°C in Kelvin. (A) 100 K (B) 200 K (C) 300 K (D) 400 K Express 100° F in degree Celsius. (A) 37.8°C (B) 40°C (C) 80°C (D) 32°C If the change in temperature ad Celsius scale is 'K' units; then the change in temperature in Fahrenheit scale is ______ unit. What is the temperature which has same numerical value on Fahrenheit and Reaumur scale ? (A) –25.6° (B) –26.6° (C) –27.6° (D) –28.6° If temperature in Celsius scale are plotted on xaxis and corresponding Fahrenheit temperature are plotted on y-axis, the straight line graph will have a slope of (A) 9/5 (B) 5/9 (C) 32 (D) –32

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16. When a thermometer is taken from the melting ice 10. Which of the following relation is/ are correct. to a warm liquid, the mercury level rises to twoC R C F  32 fifth of the distance between the lower and the upper   (A) (B) 100 80 100 180 fixed points. (A) the temperature of the liquid is 40°C F  32 K  273 K  273 R   (B) the temperature of the liquid is 313K (C) (D) 180 100 100 80 (C) the temperature of the liquid is 30°C 0 0 11. Statement I : 40 C is equal to 104 F. (D) the temperature of the liquid is 273K Statement II : 400C difference of temperature is 17. A faulty centigrade thermometer is examined. The equal to 720F upper and lower points are found to be 99.5°C and (A) Both Statements are true, Statement - II is 0.5°C respectively. What is the correct temperature the correct explanation of Statement - I. if this faulty thermometer reads 15.5? (B) Both Statements are true, Statement - II is (A) 15.15°C (B) 16.16°C not correct explanation of Statement - I. (C) 17.17°C (D) 18.18°C (C) Statement - I is true, Statement - II is false. 18. A thermometer has wrong calibrations. It reads the (D) Statement - I is false, Statement - II is true. M.P. of ice as –10°. It reads 60° in place of 50°C. What is the temperature of the boiling point of water 12. Which of the following is the smallest change in on the scale. temperature? (Smallest L.C) (A) 1°C (B) 1°F (C) 1K (D) 1°R 16 19. The mercury thread by parts between two 13. Your are heat through 1°C, 1°F, 1°R and 1 K. In 19 which case you fell most hot? standard points on Fahrenheit scale, when Column-II placed in hot milk. Calculate the temperature of 14. Column-I 0 (A) 20 C 1) 680F milk in (i) F (ii) C (iii) K scale. (B) 00C 2) 320F 7 0 (C) 100 C 3) 273K 20. Suppose the mercury thread rises by parts 18 0 (D) 80 R 4) 373 K between two standard points on Fahrenheit scale 5) 212° F where thermometer is placed in warm milk. Then 15. The lower fixed point of each of the Celsius and temperature in °F is _____? the Fahrenheit scale of temperature is 21. The higher and lower fixed points on a thermometer (A) 4°C when the density of water is maximum are separated by 160 mm. When the length of the (B) the boiling point of water mercury thread above the lower point is 40 mm, the temperature reading would be ______. (C) the freezing point of water (D) the freezing point of mercury onceptive orksheet 16. Which of the following is/ are correct. (A) Steam point in Kelvin scale is 373K 7. A 1K rise in temperature is (A) the same as a 1°C rise in temperature (B) The boiling point of water on Fahrenheit scale is 212°F (B) the same as a 1°F rise in temperature (C) Steam point in Kelvin scale is 273K (C) more than a 1°C rise in temperature (D) less than a 1°F rise in temperature (D) The boiling point of water on Fahrenheit scale is 100°F 8. The S.I unit of Heat is (A) Joule (B) Newton 17 17. The mercury thread falls by parts between two (C) dyne – m (D) dyne 18 9. A degree on celsius scale is _______ part of the standard points on Fahrenheit scale, when boiling interval between the ice point and the steam point. water at 76 cm pressure is cooled to room temperature. Calculate the room temperature in 1 1 1 1 th (B) th (C) th (D) th (A) (i) Fahrenheit (ii) Celsius (iii) Kelvin scales 180 80 100 120

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17 parts between 18 two standard points on Celsius scale, when thermometer is placed in warm milk. Calculate the temperature in (i) °C (ii) K (iii) F 19. What temperature change on the Kelvin scale is equivalent to a 10 degree on the Celsius scale? 20. Which of the following is/ are correct. (A) The same value of the temperature on the Celsius and Fahrenheit scales is – 40° (B) The same value of the temperature on the Kelvin and Reaumur scales is–1092° (C) The same value of the temperature on the Celsius and Fahrenheit scales is 30° (D) The same value of the temperature on the Kelvin and Reaumur scales is 1000° 18. The mercury thread rises by

10. Transmission of Heat When you pour hot tea into a cup you will observe that the cup gets warmer and the tea a little cooler. When you heat a piece of iron and drop it in water. You will see that the water gets heated and the piece of iron becomes cool. Thus flow of heat energy from a body at higher temperature to the another body at lower temperature is called transfer of heat. The transfer of heat stops as soon as the temperature of the two bodies becomes equal. The two bodies are then in a state of thermal equilibrium. “The phenomenon of transfer of heat from a hotter body to a colder body is called transfer of heat”.

Three modes of Transfer of Heat If two bodies at different temperatures are placed near each other or they are kept in contact, heat flows from a body at a higher temperature to the body at a lower temperature. The transfer of heat from one body to the other (or from one part of the body to its other part) can take place by three different processes, namely

(i) Conduction If heat transfer takes place without molecular motion then the process is known as conduction.

(ii) Convection The transmission of heat from one place to another by the actual movement of particles is known as convection.

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(iii) Radiation Radiation is the process of transfer of heat from a hot body to a cold body directly without heating the space in between the two bodies. It does not require any medium for two bodies. Let us study in detail about conduction..

11. Conduction If heat transfer takes place without molecular motion then the process is known as conduction (or) conduction is the process of transfer of heat in solids from one particle to another without the actual movement of the particle. It is the similar method by which a book can be passed from person to (person not changing his position) just as heat is passed from molecule to molecule. Note: Medium is required for the transfer of heat by conduction, therefore, conduction is not possible in vacuum. In solids, heat is transferred mainly by the process of conduction.

Conditions for the Conduction of Heat Heat can be conducted from one body to another only if they are 1. In contact with each other 2. At different temperatures Heat transfer will go on as long as there is a difference in temperature, when the temperature becomes equal, heat transfer will stop.

Experimental demonstration of Transfer of Heat in Solids by Conduction: 1. To demonstrate Heat is Transferred from Hot end to Cold end Take a copper rod and fix a few small iron nails on it with the help of molten wax. Clamp the rod to a stand. Heat the copper rod at one end. It is observed that the iron nails fall one by one starting from the end that is near the flame. Thus, it can be concluded that heat is transferred from the hot end to the cold end of the rod by the process of conduction.

2. To demonstrate that Glass rod does not allow Heat Hold one end of the glass rod in one hand and the metal rod in the other. Place their other ends in the burner flame. The glass rod can be held for a longer time than the metal rod. We can conclude that Glass rod does not allow the heat to flow easily through it and so it is a bad conductor of heat. Metal rod allows the heat to flow through it and so it is a good conductor.

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3. To demonstrate that different Materials have different Thermal Conductivities Take a metal trough. Make five holes in one of the sides near the lower edge. Close these holes with one holed corks. Take five rods of different metals but of the same length and diameter. Dip them into molten wax to allow a coating of wax to solidify on them. Pass the rods into the holes. Pour boiling water into the trough. Wait for about five minutes. Record your observations. You will find that wax melts to different lengths along the rods. We conclude that some metals conduct heat better than others. Silver and copper are very good conductors of heat.

Take the test-tube and half-fill it with water. Take a cube of ice and wrap it in wire gauze to make it heavy. Drop it into the test-tube. Hold the test-tube in an inclined position and heat it. We observe that the water at the top starts boiling but the ice does not melt. We can conclude that the heat supplied to the water at the top does not travel throughout the water. Water is, therefore, a poor conductor of heat.

Practical Application of Heat Insulators 1.

Good Conductor of Heat The substances which allow the heat energy to flow through them easily, are called good conductors of heat or Substances that conduct heat easily are cal l ed good conductors of heat or si mpl y conductors. Example : All metals are good conductors of heat. Silver is the best, copper is next, followed by aluminium, iron and lead.

Practical Applications of Good Conductors of Heat 1.

2. 3.

Boilers and cooking utensils are made of metals. Being good conductors, metals get heated up easily. Tip of soldering rod is made of copper as it readily conducts away heat to the solder. Mercury is used as a thermometric liquid as it is a good conductor of heat.

2.

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5.

6.

Bad Conductor of Heat or Heat Insulators The substances, which do not allow the heat energy to flow through them easily are called bad conductors of heat or heat insulators or Substances that do not conduct heat easily are called bad conductors of heat or simply bad conductors or insulators. Example: Substances such as glass, wood, cloth, air, pure (distilled) water, wax, paper, clay etc. are the bad conductors of heat. Liquids (except mercury) are usually the poor conductors of heat. Gases are still poorer conductors of heat as compared to liquids. Ebonite and asbestos are the worst conductors of heat.

To demonstrate that Water is a bad conductor of heat

7.

Materials like bricks, mud, concrete, etc., used in the construction of houses are poor conductors of heat. Therefore, they prevent heat from outside to enter into the house during summer. Similarly, during winter, they do not allow the warm rooms to loose their heat soon. The vehicles carrying inflammable materials like petrol are covered with insulating material like asbestos, so that, petrol may not get heated and catch fire. Roof sheds are also made of asbestos to prevent the heat from outside to enter into the rooms. It is because asbestos is a poor conductor of heat. Blocks of ice are usually covered with cloth or sawdust when being stored. This prevents outside heat from melting the ice quickly. We wear woolen clothes in winter to keep ourselves warm. The wool and the air trapped in the wool are both bad conductors of heat. They do not allow our body heat to escape. Ice boxes are double-walled containers. There is air in the space between the two walls. Air being a poor conductor does not let the outside heat enter the box. The ice in the box, therefore, does not melt quickly. Handles of metal cooking vessel are made of insulators like wood, bamboo, bakelite etc., so that they do not get hot and are easy to hold even when the vessel is very hot.

12. Convection The transmission of heat from one place to another by the actual movement of particles is known as convection (or) Convection is the mode of heat transfer in liquids and gases taking place by the actual movement of molecules from one part to another. Convection corresponds to the second method of sending the book in which the person moved from his or her seat and personally handed over the book to the student. www.betoppers.com

8th Class Physics

62 Note : 1. Movement of molecules is not possible in solids, since they are closely packed. So, convection does not take place in solids. 2. Transfer of heat in liquids and gases takes place by convection. 3. A medium is required for the transfer of heat by convection. Heat cannot be transferred by convection in vacuum. 4. By the process of convection the transfer of heat is always vertically upwards.

Land breeze The cold air blowing from land towards sea during night is called ‘land breeze’. Land being a good absorber, is also a good radiator. Therefore after the sun sets, land starts loosing heat by radiation more rapidly than the water of sea.

Convection in Liquids On heating, the part of liquid in the vessel which is in contact with the source of heat, becomes warmer. The warmer liquid becomes lighter and rises up. Colder liquid moves down to take the place up and the process continues. Thus the whole liquid gets heated. This process is known as convection.

Convection Currents Hot air rises, because it is lighter than cold air. When hot air rises, cold air takes its place. These movements caused by hot and cold air are known as convection currents.

To demonstrate Convection Currents in Water: Apparatus : Beaker, potassium permanganate crystals, straw, water, Bunsen burner. Method : Take some water in a beaker and add few pink coloured crystals of potassium permanganate by dropping them through a straw. Take out the straw without disturbing the water. Now heat the beaker on a Bunsen burner or a spirit lamp. Observation : Coloured water rises from the place where heat is applied. After moving up some distance, it spreads out and comes down along the sides of the beaker. Conclusion : When we heat the flask, the water at the bottom gets heated. On being heated, it expands and becomes less dense or lighter. The lighter hot water moves up. Its place at the bottom is taken by cooler water from the top. Convection currents are thus set up in air.

Convection Currents in Nature In coastal areas, breeze generally blows from the sea to the land during the day (sea breeze) and from the land to the sea (land breeze) during the night. These sea and land breezes are actually convection currents.

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Land radiates heat faster than the sea because land has a greater radiating power and a lower specific heat capacity as compared to water. At night, the land cools more quickly than the sea water. Thus, the air above the sea being warmer, rises up and the cooler air from above the land surface blows towards the sea to take its place. This sets up the convection currents that forms the land breeze.

Sea breeze The cold air blowing from the sea towards the land during the day, is called the ‘sea breeze’.

During the day time, land becomes more heated than the sea, because land has a greater absorbing power and a lower specific heat capacity. By the evening time, therefore, air above the land, being more heated, expands, rises up and the colder air from above the sea surface blows towards the land to take its place. This sets up the convection currents which forms the sea breeze.

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Formative Worksheet 22.

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26

27.

28. Choose the correct statements (A) Object having shiny white colour are bad Statement I : Boilers and cooking utensils are made emitters of heat. of metals. Being good conductors, metals get heated (B) Objects having dull black colour are good up easily. absorbers of heat. Statement II :Hot air rises, because it is lighter than (C) Objects having shiny white colour are bad cold air. absorbers of heat. (A) Both Statements are true, Statement - II is (D) Objects having dull black colour are good the correct explanation of Statement-I emitters of heat. (B) Both Statements are true, Statement - II is 29. Convection is not possible in not correct explanation of Statement - I. (A) Solids (B) Liquids (C) Statement - I is true, Statement - II is false. (C) Gases (D) All of these (D) Statement - I is false, Statement - II is true. Column-I Column-II onceptive orksheet (A) Conduction 1) Heat transfer takes place 21. The phenomenon of transfer of heat from a hotter without molecular motion body to colder body is called (C) Convection 2) Heat transfer takes place (A) Transfer of light from one place to another by (B) Transfer of heat the actual movement of (C) Transfer of momentum particles (D) None of these (B) Radiation 3) Transfer of heat from a hot 22. The transfer of heat stops between the bodies when body to a cold body directly (A) Temperature of two bodies are equal without heating the space in (B) Temperature of two bodies are unequal between the two bodies (C) Both (A) and (B) (D) Neither (A) nor (B) (D) Bad conductors 4) Ebonite 23. Bad conductors of heat are called of heat 5) Asbestos (A) Conductors (B) Semi-conductors Choose the correct statement (C) Insulators (D) All of these (A) Transfer of heat in liquids and gases takes 24. Modes of transfer of heat are place by conduction. (A) Conduction (B) Radiation (B) Transfer of heat in liquids and gases takes (C) Convection (D) Transformation place by convection. 25. Example for a conductor. (C) A medium is required for the transfer of heat (A) Iron (B) Silver by convection. Heat cannot be transferred by (C) Glass (D) Both (A) and (B) convection in vacuum. 26. Example for a bad conductor (D) By the process of convection the transfer of (A) Wood (B) Lead heat is always vertically upwards. (C) Glass (D) Both (A) and (C) The method through which the entire pot of water 27. Water is a __________ conductor of heat boils on the hot stove is (A) Bad (B) Good (A) Conduction (B) Convection (C) Cannot say (D) Neutral (C) Radiation (D) Evaporation 28. The fastest method of heat transfer from a hot body Choose the correct statements to a cold body is (A) The cold air blowing from land towards sea is (A) Conduction (B) Convection called land breeze. (C) Radiation (D) None of these (B) Land radiates heat faster than sea. 29. Choose the correct statements (C) Land has more absorbing power than sea. (A) The phenomenon of transfer of heat from a (D) Heat radiations travel in straight lines hotter body to a colder body is called transfer Heat radiations travel with a speed of heat (A) less than the speed of light (B) Medium is required for the transfer of heat by (B) equal to the speed of light conduction (C) greater than the speed of light (C) Radiation is possible in vacuum (D) equal to the speed of sound. (D) Medium is required for the transfer of heat by convection www.betoppers.com

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Summative Worksheet 1.

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When object is heated (A) Its temperature increases (B) its temperature decreases (C) Its temperature remains constant (D) All of these Heat always flows from (A) Higher temperature to lower temperature (B) Lower temperature to higher temperature (C) Sometimes higher to lower and lower to higher temperature (D) None of these If a body is at a temperature higher than the room temperature the level of mercury in the thermometer’s stem (A) Falls (B) Remain at the same position (C) Rises (D) May rise or fall Convert –40°C into (i) Fahrenheit scale and (ii) Kelvin scale. Convert 133K into (i) Celsius scale and (ii) Fahrenheit scale An object ‘A’ at 10 °F and another object ‘B’ at 10K are kept in contact. Heat will flow from _________ to ________. When will the numerical values of Fahrenheit and Kelvin will be the same? When will Fahrenheit thermometer reading be twice as much as temperature in degree Celsius? If the change in temperature in Fahrenheit scale is  , find this change in Kelvin scale. Temperature of two bodies differ by 1°F. How much do they differ in Celsius scale? The rise in temperature on the Celsius scale is equal to 30 degrees. Find the corresponding rise in temperature on the Kelvin scale. The mercury thread falls by 8/15 parts two standard points on a Celsius scale, when boiling water at 100°C is allowed to cool to room temperature. Calculate the room temperature on Fahrenheit scale. The mercury thread rises by 16/19 parts between two standard points on Fahrenheit scale; when placed in hot milk. Calculate the temperature of milk in Kelvin scale. A careless manufacturer did not observe precautions while marking fixed points on mercury thermometer. Thus, the upper fixed point which he marked as 100°C was actually 105°C and the lower fixed point which he marked 0°C was actually –5°C. What is the correct temperature corresponding to 50°C?

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15. In the following figure, A, B, C and D are processes that are taking place among solid, liquid and gaseous states. Identify them. A C   Liquid    Gas Solid    B D

16. In the above problem, the processes A, B, C and D takes place at a fixed temperature. Name them. 17. The magnitude of freezing point of a substance ‘x’ is found to be 25 °C. Find the magnitude of melting point of the same substance. 18. The numerical value of liquefaction point of substance ‘k’ is found to be 100 °C. Find the numerical value of boiling point for the same substance. 19. The interval between the lower and the upper fixed points is divided into ____ equal parts on Celsius scale. (A) 60 (B) 70 (C) 80 (D) 100 20. Column-I Column-II a) Celsius scale 1) Steam point is 0° b) Fahrenheit scale 2) Steam point is 212° c) Kelvin scale 3) Steam point is 373° d) Reaumer scale 4) Steam point is 80° 5) Steam point is 100 21. Which represents a greater temperature, 10C or 10F?

HOTS Worksheet 1. 2.

3.

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The temperature of a body changes by 280C. What is the corresponding change on the Kelvin Scale? The temperature of a body changes by 250C. What is the corresponding change on the Fahrenheit scale? Find the temperature at which Fahrenheit scale has same temperature on Celsius scale. The mercury thread falls by

11 parts between two 14

standard points on Fahrenheit scale, when boiling water at 760 mm of mercury is cooled to room temperature. Calculate the room temperature in (i)Fahrenheit (ii)Celsius (iii) Kelvin scale In a Celsius thermometer, the lower and upper fixed points were erroneously marked by manufacturer. The lower fixed point, which is marked zero is actually – 100C and the upper fixed point which is marked 100 0 C is 110 0 C. What is the correct temperature corresponding to thermometer reading of 400C.

Temperature and Heat 6.

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A thermometer measures (A) Quantity of heat (B) Density (C) Temperature (D) Humidity The liquid used in a clinical thermometer is (A) Alcohol (B) Water (C) Mercury (D) Benzene It is difficult to bite a cube of ice than to drink cold water because (A) Heat content of ice is less than cold water. (B) Ice is colder than water at 0°C (C) It is easy to drink water at 0°C since it is in liquid state (D) None A temperature of 86°F will correspond to ______ degree in Celsius. (A) 30 (B) 36 (C) 80 (D) 86 Which of the following statements is true? (A) The size of 1 degree on the Kelvin scale is same as the size of 1 degree on the Celsius scale. (B) The size of 1 degree on the Kelvin scale is greater than the size of 1 degree on the Celsius scale. (C) The size of 1 degree on the Kelvin scale is greater than the size of 1 degree on the Celsius scale. (D) None The lower fixed point of Celsius and the Fahrenheit scale of temperature is taken as (A) 4 °C (B) The boiling point of water (C) The freezing point of water (D) The freezing point of mercury The boiling point of water on Fahrenheit scale is (A) 100 °F (B) 80 °F (C) 212 °F (D) 32 °F The temperature scale used in S.I system is (A) Celsius (B) Fahrenheit (C) Kelvin (D) None Which of the following statement is false? (A) Ice point is 273 K and steam point is 373 K (B) Absolute zero is –273 °C (C) 1° on Kelvin scale is same as 1° on Celsius scale (D) 1° on Celsius is 1.8 times, 1° on Fahrenheit scale. The absolute zero on Celsius scale is (A) 0°C (B) –32°C (C) 100°C (D) –273°C The melting point of ice on Kelvin scale is (A) 0 K (B) –273 K (C) 273 K (D) 373 K 45°C equals to (A) 77 °F (B) 318 K (C) 45 K (D) 8°F

65 18. The temperature of two bodies differs by 1°C. How much will their temperature differ on Fahrenheit scale? (A) 1 °F (B) 0.56 °F (C) 1.8 °F (D) None 19. The change in temperature in Kelvin scale is 10K. Then the change in temperature in Celsius scale is (A) 283 °C (B) –263°C (C) 10°C (D) 0°C 20. Which of the following is the temperature of normal human body? (A) 37 K (B) 273 K (C) 310 K (D) 350 K

IIT JEE Worksheet 1.

When altitude increases (A) Atmospheric pressure increase (B) Melting point of ice increases (C) Boiling point of water increases (D) Density of air increases. 2. Choose the correct statement. (A) 1 °C < 1 °F (B) 1 °F = 1 °C (C) 1 °F < 1 °C (D) 1 °C  1 °F 3. If the external pressure increases (A) Boiling point of water increases (B) Melting point of ice increases (C) Both increases (D) Both decreases 4. In a pressure cooker, while cooking (A) Boiling point rises (B) Pressure increases (C) Volume is constant (D) All are correct 5. Choose the correct statement (A) Heat is a form of fluid (B) Heat flows from lower temperature to higher temperature (C) If a body is heated, its mass increases (D) If a body is heated, its volume increases 6. The temperature of a body changes by 450F. The corresponding change on the centigrade scale is (A) 450C (B) 400C (C) 250C (D) 300C 7. The normal temperature of human body is (A) 370C (B) 36.20C (C) 98.60C (D) 410C 8. The most commonly used thermometric substance (A) Water (B) Alcohol (C) Mercury (D) Turpentine 9. Heat flows as a result of difference of (A) Masses (B) Temperatures (C) Weights (D) Densities 10. The temperature of a body changes by 350C. The corresponding change on the Fahrenheit scale is (A) 350C (B) 450C 0 (C) 63 C (D) 720C

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66 11. When an in inflated tyre bursts, the air escaping out (A) Will be cooled (B) Will get heated up (C) Will be liquefied (D) Will not undergo any change in its temperature. 12. Fahrenheit scale divides two fixed points equally into (A) 180 parts (B) 100 parts (C) 212 parts (D) 32 parts 13. The Fahrenheit thermometer when dipped in liquid reads–400F. What will be the Celsius thermometer read if dipped into the same liquid? (A) 400C (B) –400C 0 (C) 0 C (D) –100C 14. A cold object can be made hot (A) By adding heat (B) By removing heat (C) It cannot be made hot (D) None of these 15. A hot brick of temperature 'T' is broken into two halves. One half is thrown away. The temperature of the other half is

T 2 (C) 2T (D) none of these The temperature of a body changes by 260C. What is the corresponding change on the Kelvin scale? (A) 26K (B) 30K (C) 45K (D) 299K A thermometer measures (A) The quantity of heat (B) Density (C) Temperature (D) All of the above When altitude increases (A) Atmospheric pressure increases (B) Melting point of ice increases (C) B.P of water increases (D) Density of air increases Choose the correct statement. (A) 10C < 10F (B) 10F = 10C 0 0 (C) 1 F < 1 C (D) 10C  10F If the external pressure increases (A) B.P of water increases (B) Melting point of ice increases (C) Both increase (D) Both decrease (A) T

16.

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(B)

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8th Class Physics

• • • • • • •

By the end of this chapter, you will understand Electric Charges and its Types • Atmospheric Electricity Structure of Atom • Flow of Static Electricity through a Conductor How do substances get Electrically Charged on rubbing • What does Electricity do for us? Conductors and Insulators • Electric Current Experiments with an Electroscope • Potential Charging a Body by Conduction • Primary Electric Cells Charging a Body by Induction • Electric Circuit

1. Introduction If you've ever sat watching a thunderstorm, with mighty lightning bolts darting down from the sky, you'll have some idea of the power of electricity. A bolt of lightning is a sudden, massive surge of electricity between the sky and the ground beneath. The energy in a single lightning bolt is enough to light 100 powerful lamps for a whole day or to make a couple of hundred thousand slices of toast! Electricity is the most versatile energy source that we have; it is also one of the newest: homes and businesses have been using it for not much more than a hundred years. Electricity has played a vital part of our past. But it could play a different role in our future, with many more buildings generating their own renewable electric power using solar cells and wind turbines. Let's take a closer look at electricity and find out how it works! What is electricity? Electricity is a type of energy that can build up in one place or flow from one place to another. When electricity gathers in one place it is known as static electricity (the word static means something that does not move); electricity that moves from one place to another is called current electricity. Amber, known as Electron in Greek language, is a kind of fossil gum having a straw yellow color. Ancient Greek philosopher Thales found out that when amber is rubbed with wool, it develops a strange property of attracting tiny bits of dry paper, dry straw, dry pieces of leaves, etc., towards itself. Later some time in seventeenth century, Dr. Gilbert reconstructed the experiment of Thales. He showed that not only amber and wool combination, but many other combinations like ebonite rod and cat’s skin, glass rod and silk, sealing wax and wool, etc., also develop similar properties when rubbed with each other.

Chapter - 6

Learning Outcomes

Electricity and Chemical Effects of Electric Current

The substances which acquire this strange property of attraction were said to be charged with electricity or electrified (from the Greek word Elektron). The phenomenon due to which a suitable combination of bodies, on rubbing, gets electrified is called electricity.

2. Electric Charges and its Types Having found that a number of bodies can be charged by] rubbing with suitable material, Dr. Gilbert set out to find the] nature of electric charge on the bodies. He took glass rod and silk as one combination and cat’s skin and ebonite rod as another combination. Let’s understand the different experiments conducted by Gibert for understanding the charges.

Experiment 1 Take an ebonite rod and rub it with cat’s skin. Suspend it 2 freely by a silk thread-’ from some support. Bring near this suspended rod, another ebonite rod, which is rubbed with cat’s skin. It is observed that “the suspended ebonite rod gets repelled as shown in figure.

Charged ebonite rod Charged ebonite rod The suspended and charged ebonited rod shows repulsion

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Experiment 2

Charged Glass Rod

Take a glass rod and rub it with silk and suspend it freely by a silk thread. Near this suspended rod, bring another glass rod which is rubbed with silk. It is observed that suspended glass rod gets repelled as shown in figure.

Now it is quite inconvenient to say that a body has a charge similar to ebonite rod or charge similar to glass rod. Thus, it was decided to give symbols the charges on ebonite rod and glass rod. These days the charge produced on the glass rod is called positive charge (+), whereas the charge produced on mite rod is called negative charge (—). We can write down our observations as: • Like charges repel and unlike charges attract each other. • Electrically neutral bodies do not attract or repel other objects

charged glass rod

charged glass rod The suspended and charged glass rod shows repulsion

In the world of Static Electricity…

Experiment 3 Take a glass rod and rub it with silk and suspend it freely by a silk thread. Bring near it an ebonite rod which is rubbed with cat’s skin. It is observed that glass rod is attracted by ebonite rod as shown in figure.

Charged ebonite rod The suspended and charged glass rod shows attraction

Charged glass rod

From experiments 1 and 2, it is clear that when two bodies have similar charges they repel each other as similar charges on glass rod or ebonite rod repelled each other. From experiment 3, it is clear that charges on the glass rod are not similar to charges on ebonite rod because, instead of repulsion, attraction takes place. By doing a series of experiments Gilbert was able to establish that whenever a body got electrically charged due to friction then its charge resembled either to the charge produced on the ebonite rod or to the charge produced on the glass rod.

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

+



Opposite-charged and objects attract



+

+

Objects with like charges repel

3. Structure of Atom Where do these electric charges come from? Why do the bodies get electrified on rubbing? The answers to these questions were found in the beginning of the twentieth century with the discovery of structure of atom. We now study about the atom and its structure. Before the beginning of the twentieth century, it was believed that atom is he smallest unit of matter which cannot be broken. Furthermore, it was believed that atoms of same element are alike in all respects, whereas the atoms different elements are different in all respects. However, in the beginning of the twentieth century a series of experiments showed that an (atom is made of three kinds of sub-atomic particles. Furthermore, these sub­atomic particles are common to all elements, irrespective of the chemical properties of elements. One kind of sub-atomic particle was found to have a charge similar to the charge produced on ebonite rod. They were named electrons and the charge on them was considered negative. Second kind of sub-atomic particles were found to have a charge similar to the charge produced on glass rod. They were [named protons and the charge on them was considered positive. Third kind of sub-atomic particles were found to have no

Electricity and Chemical Effects of Electric Current charge, but their mass was equal to the mass of proton. They were named as neutrons. Furthermore, as the charge on the protons and electrons I was very small, scientists supposed that charge on each particle [is one unit. It means an electron has -1 unit charge and a proton has +1 unit charge. These charges are equal and opposite. How these particles are arranged in an atom? It was a big [mystery as we know that the positive charge cancels the negative [charge. Rutherford proposed the theory for the arrangement of I subatomic particles which is called Rutherford’s atomic theory.

Following are the important points of Rutherford’s Atomic Theory: 1.

An atom consists of three sub-atomic particles, i.e., neutrons, protons and electrons. i) Neutron has no electric charge on it. It has a mass almost equal to mass of one atom of hydrogen. ii) Proton has a unit positive charge on it. It has a mass almost equal to mass of one atom of hydrogen. iii) Electron has a unit negative charge on it. Its

1 times the mass of one atom of 1837 hydrogen. Protons and neutrons form the central core of atom which is commonly called nucleus. The protons and neutrons held together by strong attractive forces called nuclear forces. mass

2.

Free electrons in outermost orbit

69 3.

4.

The electrons revolve around the nucleus in fixed orbit much the same way as planets revolve around the Sun. The electrons close to the nucleus are held strongly by electric pull of protons. These electrons are called bound electrons. However, the electrons away from the nucleus experience very little attractive force. Thus, the electrons revolving around the nucleus in the outermost orbit are held by weak force. Such electrons are called free electrons. As the atom of a normal element is electrically neutral therefore it is believed that the number of protons in an atom is equal to the number of electrons.

4. How do substances get Electrically Charged on rubbing The tree electrons (the electrons which are present in the outermost of an atom) are responsible for the electrification of a body. When two bodies (say body A and body B) are rubbed against each other,] the free electrons of one body (say body A), get transferred to the other body (say body B). The body A (on losing electrons) has less number of electrons than the number of protons in its nucleus. Thus, on the whole, the body positively charged. Thus, we can say that positive electrification of; t body is due to the deficiency of electrons as compared to normal number of electrons in a neutral atom. Similarly, the body B, (on gaining electrons) has more negative charges as compared to positive charges in the nucleus. Thus, on the whole, the body gets negatively charged. Thus, we can say that negative electrification of a body is due to r/ie excess of electrons as compared to normal number of electrons in i neutral atom. It must be remembered that during positive or negative electrification, it is the electrons and not the protons which get transferred. Furthermore, if one body gets charged positively due to rubbing, then the body which is used for rubbing gets charged negatively at the same time. In other words, equal and opposite charges are produced at the same time.

Nucleus having protons and neutrons

Bound electrons in inner orbits

Arrangement of protons, neutrons and electrons in an atom

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Experiences of Electric Charges in our everyday lives:

‘Charge’. The paper piece also gets a charge imbalance due to presence of charged comb or rod.

One of the best examples of observing effects of electric charges is lightning. It is one of nature‘s spectacular offerings. Moisture bearing clouds are initially electrically neutral.

Law of Conservation of Charges

But as the masses of clouds increase in size, they start rubbing against each other. Friction makes electrons jump, so that between two clouds there will be a charge imbalance: one cloud is positively charged and the second cloud is negatively charged. When a positively charged cloud collides with a negatively charged cloud, charge neutralization occurs sparks are produced. These sparks are seen as lightning. Sometimes the charge imbalance can occur between a cloud and the earth; lightning is seen when the charges is exchanged between the cloud and the earth.

Rub your feet on a carpet and touch a doorknob. What happens? On a cold dry morning, have you ever felt that while removing a pullover, the hair on your hands stand up? Take a plastic comb and comb your hair quickly. Your hair has to be dry. Then take a few pieces of paper. Then hold the comb close to the papers. The papers will tend to get attached to the comb. Instead of a comb, try the same experiment using a glass rod. Rub the rod on a silk cloth, which will then get lifted towards the rod. Then get lifted towards the rod. What is making the paper stick to the comb or the glass rod? An easy interpretation of these observations is that some objects when rubbed acquire a property that we intuitively know to be www.betoppers.com

It states that the sum total of electrons in a system, is a constant quantity. However, when two bodies in a system are rubbed against other then electrons from one body may get transferred to the other body. For example, when a glass rod is rubbed with silk, then some electrons from glass rod are transferred to silk. As glass rod [develops a deficiency of electrons it gets positively charged. The silk after rubbing has excess of electrons, and hence, gets {negatively charged.

Silk Before rubbing

Silk gains electrons and gets negatively charged

Glass rod before rubbing

Glass rod has deficiency of electrons and gets positively charged

Similarly, when an ebonite rod is rubbed with cat’s skin, then some of the electrons from cat’s skin are transferred to ebonite rod. As the cat’s skin has a deficiency of electrons it gets positively charged. The ebonite rod has an excess of electrons, and hence, gets negatively charged. However, the sum total of electrons in the system of silk and glass rod, or ebonite rod and cat’s skin remains same, and hence, electric charges are conserved.

5. Conductors and Insulators Silk before rubbing Glass rod before rubbing Silk gains electrons and gets negatively charged Glass rod has deficiency of electrons and gets positively charged In previous class, we have studied that certain substances allow the current to flow through them. They are called conductors. However, there are certain substances which do not allow the current to flow through them. They are called insulators. Why is it that current can flow through one kind of substances, but not through another kind of substances? It can be explained on the basis of structure of atom.

Electricity and Chemical Effects of Electric Current

71

Insulator

As mentioned earlier, the weakly held electrons in the outermost orbit of an atom act almost as free electrons. It means that these electrons CATS SKIN LOSES can freely move about in a substance, but cannot leave the substance positively charged. Different substances have different number of free electrons. For example, free electrons in 1 cm3 of silver are 100,000,000,000,000,000,000,000 or 1023. However, the free electrons in mica are only 10,000 or 104.

A substance, which has few free electrons, such that they do not easily drift from one end of substance to the other end, when connected to some source of electricity, is called insulator. Substances like alcohol, ether, benzene, chloroform, mica, sugar, starch, wool, fur, ebonite, glass, diamond, rubber, plastics, silk, sulphur, sealing wax; wood are insulators. Note: An object made of a conducting material will permit charge to be transferred across the entire surface of the object. If charge is transferred to the object at a given location, that charge is quickly distributed across the entire surface of the object.

Ebonite rod Cat’s Skin

A metal sphere is mounted on an insulating stand and touched by a charged plastic golf tube. The metal sphere acquires a negative charge, located at the point of contact.

.

Cat’s skin loses electrons and gets positively charged

Ebonite rod gains electrons and gets negatively charged

Since metal is a conductor, the charge quickly distributes itself across the surface of the sphere.

The distribution of charge is the result of electron movement. If a charged conductor touches another object, the conductor can even transfer its charge to that object. The transfer of charge between objects occurs more readily if the second object is made of a conducting material. Conductors allow for charge transfer through the free movement of electrons. If charge is transferred to an insulator, the excess charge will remain at the initial location of charging. The particles of the insulator do not permit free flow of electrons; subsequently charge is seldom distributed evenly across the surface of an insulator.

When we connect a substance to a source of electricity, then these free electrons start drifting from one end to another end, and the current starts flowing. Thus, the more the availability of free electrons, the more will be the current flow and the less the number of free electrons, the less will be the current flow. When the current flowing out is so small that it is just negligible, then we can say that the substance is an insulator. We can define conductors and insulators on the basis of structure of atom as under.

Conductor

Increasing Conducting Ability

Insulators

Semi-Conductors

Conductors Silver Copper Aluminium Iron M ercury

Carbon Water

Germanium

Silicon

Dry Air

Wood

Glass

Rubber

A substance, which has a large number of tree electrons, such that they start drifting from one end of a substance to the other end, when it is connected to some source of electricity is called conductor. Ebonite rod gains electrons and gets negatively charged. All metals; solutions of acids in water; solutions of alkalis in water solutions of soluble salts in water are conductors.

Examples of conductors include metals, aqueous solutions of salts, graphite, water and the human body. Examples of insulators include plastics, Styrofoam, paper, rubber, glass and dry air. In between these categories of conductors and www.betoppers.com

8th Class Physics

72 insulators, lies a class of materials known as semiconductors. Semiconductors conduct only under certain conditions such as the presence of small amounts of impurities. Ideal examples of semiconductors are silicon and germanium, etc. What exactly is this physical property called electric charge? Charge is assumed to be the basic property of all sub–atomic elementary particles like electrons. The physical entity of electric charge is an experimentally established quantity. We can say that electric charge is the cause of sparks and lightning when there is a charge imbalance ...the stuff that appears on a comb when you rub it on your hair. ...the stuff that comes in two kinds: positive and negative. ...the stuff that is carried by electrons, protons, and other elementary particles. ...the stuff that, when it flows or spins, creates magnetism. ...the stuff that causes electrical attraction and holds everyday objects together. ...the stuff inside of wires that is movable, almost fluid. ...the stuff inside of nonconductors that is immobile and “frozen” in place. – the stuff that comes in one specific unit, quantity or strength denoted as q, either as +q or –q. And so on ....

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Formative Worksheet 1.

2.

3.

Choose the correct combinations to develop similar properties when rubbed with each other (A) Ebonite rod and cat’s skin (B) Glass rod and silk (C) Both (1) and (2) (D) Neither (1) nor (2) Statement I : When a body gains electrons due to friction, it is said to be negatively charged Statement II : The mass of an electron is equal to the mass of one atom of hydrogen (A) Both Statements are true, Statement - II is the correct explanation of Statement- I. (B) Both Statements are true, Statement - II is not correct explanation of Statement - I. (C) Statement - I is true, Statement - II is false. (D) Statement - I is false, Statement - II is true. Which of the following is best insulator (A) Carbon (B) Graphite (C) Paper (D) Ebonite

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(C) Alcohol

(D) Solution alkalis in water

Conceptive Worksheet 1.

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3.

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Which of the following is not the conductor (A) Mercury (B) Glass (C) Copper (D) Silver Column-I Column-II a) Good conductor 1) Nucleons b) Bad conductor 2) Sulphur c) Electrification 3) Iron-rod d) Protons and neutrons 4) Diamond 5) Ebonite rod rubbed with cat’s skin When a comb rubber on hair it brought near the bits of paper , it attracts them, the reason is 1) The comb and the paper bits gets oppositely charge 2) The comb and the paper bits gets similarly charged 3) The paper bits are very light4) None of these Statement I : Insulators do not allow flow of current through themselves Statement II : They have no free charge carriers (A) Both Statements are true, Statement - II is the correct explanation of Statement- I. (B) Both Statements are true, Statement - II is not correct explanation of Statement - I. (C) Statement - I is true, Statement - II is false. (D) Statement - I is false, Statement - II is true. Which of following are conductors of electricity (A) Silver (B) Copper (C) Wood (D) Pure water Which of the following as insulator (A) Sulphur (B) Benzene

Examples for good conductor of electricity (A) Silver (B) Copper (C) Both 1 & 2 (D) Pure water The phenomenon due to which a suitable combination of bodies , on rubbing, gets electrified is called (A) Electricity (B) current (C) Both (1) and (2) (D) 4) Neither (1) nor (2) Neutron has charged body (A) Positively (B) negatively (C) No charge (D) None of these

Electricity and Chemical Effects of Electric Current 4.

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Which of the following are the sub-atomic particles (A) Neutrons (B) Protons (C) Electrons (D) Photon The mass of electron is equal to the x/1857 times the mass of one atom of hydrogen. then the value of x is________ A body can be charged by (A) Conduction (B) Induction (C) Rubbing (D) Heating When a negatively charged body is brought near a suspended positively charged ball , the ball gets (A) Attracted (B) Repelled (C) Stay at same place (D) None of these Electric charge can flow through (A) Insulators (B) Conductors (C) Both conductors and insulators (D) Neither conductors and insulators When a body gains electrons due to friction, it is said to be (A) Negatively charged (B) Positively charged (C) Both (A) and (B) (D) Neither (A) nor (B)

6. Experiments with an Electroscope Electric charges and their properties can be studied easily by building a simple equipment known as an electroscope. Take a glass bottle with a wide mouth. Remember that glass is an insulator. Take a piece of cork that will fit into the mouth of the bottle. Make a tiny whole in the cork. Put the copper wire through the whole. On one end of the copper wire, attach the small metallic plate. At the other end of the copper wire make a small loop or hook. Cut a thin strip of aluminum foil. Place the strip on the hook of the copper wire. Now put the cork in place and close the mouth of the bottle tightly. Your electroscope is ready.

73 An electroscope will be able to detect the presence of static charges, but is unable to give you any information about the sign (negative or positive) of the charges. Only a reasonable guess can tell you whether the charge on the electroscope is positive or negative.

Take a plastic comb. Rub it on a dry surface like a tissue paper or hair. Now make the comb to touch the electroscope plate. You will observe that the leaves of the aluminum foil, inside the electroscope, separate. This is because the charges developed on comb due to friction, travel down through the conducting wire and reach the two leaves of the aluminum foil. The two leaves acquire similar charges and repel each other. This phenomenon occurs for a short time only. (The charges developed on the aluminum leaves dissipate or discharge into the air quickly). We can conclude that the aluminium foils repel each other because they have acquired same charges. Extending this understanding to unlike charges, we can say that unlike charges attract each other.

Let us again go back to the electroscope. Now, instead of causing the charged comb to contact the electroscope, hold it close to the plate. The aluminum leaves will again part for a short while. Over many years it is now understood why this happens: the charges on the comb polarize the charges in the plate. So the top part of the electroscope plate acquires a charge opposite to the charge on the comb. The comb is negatively charged, so the plate www.betoppers.com

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Take an ebonite rod B and rub it with fur, so that it gets negatively charged. Touch the rubbed end to the sphere A. Remove the ebonite rod. Test the charge on the sphere A, by touching it with the disc of negatively charged gold leaf electroscope. It is seen that leaves of G.L.E. diverge. This proves that metallic bodies can be charged if they are brought in contact with charged body.

gets positive charge. The lower part of the electroscope, that is the aluminum leaves, get negatively charged, the charge same as that of the comb. A question might arise in your mind as to how do we know which material is positively charged and which material is negatively charged, when the two are rubbed together. Over many experimental observations, we now know that, for example, when a glass rod is rubbed with silk, the rod is positively charged and when an ebonite (a type of plastic) rod is rubbed with wool, the rod gets negatively charged.

Experiment 2 Repeat experiment 1 but use a rubber ball instead of copper sphere. When the rubber ball is tested with negatively charged gold leaf electroscope, it is found that its leaves show no change in divergence.

How to find the Nature of Charge on a given Body? Take a G.L.E. and charge it positively or negatively by touching its disc [with a positively charged glass rod or negatively charged ebonite rod. ‘ Imagine, you have a negatively charged G.L.E. Now touch the given charged body to the disc of negatively charged G.L.E. If the leaves of G.L.E. diverge further, then the body is negatively charged. It is because, similar charges repel each other. If the leaves of G.L.E. collapse, then the body is positively charged. It is because some charges from negatively charged G.L.E. will flow to the positively charged body.

Insulated stand

The experiment proves that charging by conduction is possible only if a body is a conductor and not insulator (such as rubber ball). Thus, to sum up, we can say, The phenomenon due to which a conductor gets charged when brought in contact with a charged body is called conduction. The conduction is only possible in case of conductors and not in case of insulators.

7. Charging a Body by Conduction Is rubbing a body with suitable substance the only way of charging it? Can a body be charged by bringing it in contact with another charged body? Let us do the following experiments.

8. Charging a Body by Induction If a positively charged glass rod is brought near small pieces of paper, the rod attracts the pieces even though there is no physical contact.

Experiment 1 Take a copper sphere A, mounted over an insulated stand. The insulated stand is provided, so that the charge may not flow into earth. Charged ebonite rod

Oppos itel y-charged objects attract

• Copper sphere

B

A

• Insulated stand

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Place two insulated metal spheres, A and B, to touch each other, so in effect they form a single non–charged conductor. Now keep a negatively charged rod near A. In metals the negative charge (electrons) can move rather freely. These free electrons in the metal are repelled as far as possible until their mutual repulsion is big enough to balance the influence of the rod. Charge is redistributed.

Electricity and Chemical Effects of Electric Current • •

Now move B from A little distance while the rod is still present. Both the spheres A and B will gain an equal and opposite charge. This is charging by induction. The charged rod has never touched them, and it retains the same charge it had initially.

Charging by Induction

•

•

•

•

A single sphere can also be charged by induction if we touch it when different parts of it are differently charged. Suppose we now suspend a metal ball on a nonconducting thread. The metal ball is electrically neutral.§ Bring a negatively charged rod near the ball. Charge redistribution is induced on the ball. Still the ball is neutral. The electrons from the negative side of the ball can be removed by touching it.§ Now the ball is positively charged. This positively charged ball is attracted towards the rod more strongly and touches it. Due to this contact the ball gets negative charge. The negatively charged ball is repelled by the slightly negative charged rod. Take two metal spheres A and B, mounted on insulated stands, such that the spheres are in contact with each other. Also take two gold leaf electroscopes. Students are advised npt to try this experiment. (Not shown in diagram) and charge one of them positively and the other negatively.

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Explanation of above Charging When the negatively charged ebonite rod is brought very close to sphere > it repels the free electrons in sphere A towards sphere B. Thus, as sphere develops deficiency of electrons, it gets positively charged. On the other sphere B has excess of electrons and, hence, gets negatively charged. The process due to which an uncharged insulated metallic conductor gets electrically charged when held near a charged body is called electric induction, whereas the charges produced on the conductor are called induced charges. Furthermore, the end of the conductor near the charged body gets induced charge which is opposite to that of charging body. However, farther of the conductor has the same charge as that of charging body. Putting it in simple language we can say that during electric induction the nearer end of conductor has opposite charge whereas farther end of conductor has similar charge as compared on a given charging body.

9. Atmospheric Electricity Lightning is a dazzling bluish white light produced in the clouds. It is followed by a loud noise called thunder. Till 1752, nobody really knew about the cause of lightning, when Benjamin Franklin, by a brilliant experiment, proved that it is caused by the static elect charges in the clouds.

Benjamin Franklin’s Experiment

Take an ebonite rod and rub it with cat’s skin, such that ebonite rod negatively charged. Bring the ebonite rod close to the sphere A. Holding ebonite rod in same position; remove the metal sphere B. Test its charge touching the disc of negatively charged gold leaf electroscope. It is seen the leaves of G.L.E. diverge. It proves that sphere B gets negatively change. Now remove the sphere A and test it by touching it with the disc positively charged G.L.E. It is seen that leaves of G.L.E. diverge. It proves sphere A gets positively charged.

Benjamin Franklin made a kite of silk cloth. The central spar of the was made from iron wire about 30 cm long. The central spar was connected 1 a strong silk string. The other end of silk string was connected to an iron He flew this kite on a day when there were thick dark clouds in sky. Brought his knuckles near the lower end of key, but nothing happened However, when it started raining he brought his knuckles again near the lower end of key. There were produced bright sparks between the knuckles and As a matter of fact he was saved from the death by electric shock. Why electric discharge passed when it rained?

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How to Protect Buildings from the Lightning? The best method of defense is offence. Thus, we invite the lightning to strike the building but take care that it does not damage the building. This is done by installing a lightning conductor.

The reason is that wet string became conductor of electricity and, her allowed the electricity to flow from clouds to the key and then from key to knuckles and finally through his body to earth.

How Clouds get Electrically Charged? Due to the heat of sun, the warm air and water vapours rise up. Similarly, the cold air above sinks down. When the air molecules rub against water molecules or the cold currents of air rub against the hot currents of air, then due to friction the electrons of air get transferred to water molecules or vice versa. All this depends upon atmospheric conditions. When the condensation takes place and clouds are formed, then these clouds have a huge amount of static electric charges, which may be positive or negative.

How does Lightning Strike Building? Imagine a negatively charged cloud passing over a high rise building. This cloud induces positive charges on the top of the building and negative charges at its base due to electric induction. As the positive charges attract negative charges, the free electrons from cloud start pushing their way through moist air. This forms a sort of conducting path. When these electrons reach the building, suddenly all the charges in cloud flow into the building. Thus, the lightning strikes with a devastating effect and sets the building on fire.

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A good lightning conductor is made of copper rod, on the one end of which are provided sharp copper points. It is installed on the highest point of the building. Its lower end is connected to a thick copper strip which runs along the height of building. The lower end of this copper strip is attached to flat copper plate and buried deep inside the earth. When the lightning strikes, it strikes on the sharp points of lightning conductor. As copper is a very good conductor of electricity, all the electric discharge from cloud flows into the earth without damaging the building.

10. Flow of Static Electricity through a Conductor We have studied about static electricity. We know that there are two kinds of electrification. A body is said to be positively charged if it has deficiency of electrons. Similarly, a body having excess of electrons is negatively charged. Fig shows two spheres A and B, such that sphere A has positive charges and sphere B has an equal number of negative charges. These charges are static in nature, i.e., they do not flow.

Electricity and Chemical Effects of Electric Current

Electronic Current However, if spheres A and B are connected by a plastic coat copper wire, then the charges flow from one sphere to another sphere] and get neutralised. The question arises, do the charges flow from A i B or from B to A? There was no answer for this question. Hence, make matters simple, it was assumed that positive charge is at a high electric potential (electrical pressure) and negative charge is at a lower! Electric potential. Thus, when a positively charged sphere is connected to a negatively charged sphere through a conductor, the charge will flow from positive to negative. I A

B

The flowing charge through n conductor is called current electricity.’ However, with the discovery of the structure of atom we have come to know that a positive charge resides within the nucleus of an atom and < therefore is incapable of moving. In fact when sphere A is connected to sphere B, it is the excess of electrons on B, which flow towards the sphere A, which has deficiency of electrons. Electronic current A

B Conventional current

77 Thus, to sum up, we can say that a conventional current flows from the positively charged body to the negatively charged body. The positively charged body is at a higher potential and the negatively charged body is at a lower potential. However, the electronic current (real current) flows from the negatively charged body to the positively charged body. The positively charged body is at a lower potential and the negatively charged body is at a higher potential. 1. Materials like plastic, ebonite and glass can be charged by rubbing them with suitable materials. 2. The bodies which will produce charges identical to the charges produced on glass rod, when rubbed with silk, are called positively charged bodies. 3. The bodies which will produce charges identical to the charges produced on ebonite rod, when rubbed with fur or cat’s skin, are called negatively charged bodies. 4. Like charges repel each other, but unlike charges attract each other. 5. An atom is made of three sub-atomic particles, i.e., neutrons, protons and electrons. 6. In a normal atom, the number of protons is equal to the number of electrons, and hence, atom is electrically neutral. 7. The electrons in outermost orbit of an element are very weakly held and are almost free to move. These electrons are called free electrons. 8. When a body loses electrons due to friction, it is said to be positively charged. 9. When a body gains electrons due to friction, it is said to be negatively charged. 10. Conductors are the substances having a large number of free electrons, whereas insulators are the substances having very few free electrons.

Thus, the real electric current should flow from the negative to the positive. The negatively charged ormative orksheet sphere should be at a high potential and the positively 10. Choose the correct statement from the following charged sphere at a low potential. However, so (A) Lightening is caused by the heavy flow of much of research has been done in the sphere of electric charges between two opposite by electricity on the assumption that charge flows from charged clouds the positive to negative, that we still continue with (B) the lightening conductor at the top of the high old practice. We call this current, conventional building has blunt end electric current. However, knowing fully well that (C) Clouds can charge tall buildings by induction actually the current is flowing from the negative charges to positive charges and is the real current, (D) A lightning conductor consists of a thick we have named it electronic current. copper strip with a spilee and

F

W

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8th Class Physics

78 11. Statement I : When a body loses electrons due to friction, it is said to be positively charged Statement II : Charging is due to actual transfer of electrons (A) Both Statements are true, Statement - II is the correct explanation of Statement- I. (B) Both Statements are true, Statement - II is not correct explanation of Statement - I. (C) Statement - I is true, Statement - II is false. (D) Statement - I is false, Statement - II is true. 12. If an uncharged body is brought near the electroscope ,then (A) Divergence in the gold leaf increases (B) Divergence in the gold leaf decreases (C) No divergence is observed (D) Can’t say 13. Column-I Column-II th a) 18 Century 1) Attract each other b) Like charges 2) Repel each other c) Unlike charges 3) Benjamin Franklin d) Lighting strike 4) Flow of charge between cloud & the building 5) Franklin Johns 14. Statement I : Lightening conductor protects building from the damage caused by lighting. Statement II : Sudden movement in the cloud cause electric charge to discharge in the form of flash of lightening. (A) Both Statements are true, Statement - II is the correct explanation of Statement- I. (B) Both Statements are true, Statement - II is not correct explanation of Statement - I. (C) Statement - I is true, Statement - II is false. (D) Statement - I is false, Statement - II is true. 15. Gold leaf electroscope consists of (A) Brass cap (B) Shellac seal (C) Glass ball jar (D) Brass rod 16. The phenomenon due to which a conductor gets charged when brought in contact with a charged body is called (A) Conduction (B) Touching method (C) Induction

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(D) Convection

Conceptive Worksheet 10. Lightening conductor is made up of (A) Steel (B) Copper (C) Silver (D) Iron 11. Person who demonstrate the electrical nature of lighting (A) Benjamin Frake (B) Benjamin Franklin (C) Louris Benjamin(D) Franklin Johns 12. If initially gold leaf electroscope is positively charged, and a positive charge is brought near the electroscope, then (A) Divergence in the gold leaf increases (B) Divergence in the gold leaf decreases (C) No divergence is observed (D) Can’t say 13. If negative charge is brought near the electroscope, then (A) Divergence in the gold leaf increases (B) Divergence in the gold leaf decreases (C) No divergence is observed (D) Can’t say 14. Which of the following device is used to find whether a given body is electrically charged or not (A) Gold leaf electroscope (B) Microscope (C) Telescope (D) Periscope 15. Gold leaf electroscope can be charged by (A) Contact method (B) Conduction (C) Induction (D) Convection 16. Lightning is caused by the (A) Static electric charges in the clouds (B) Heavy rain (C) Tides (D) None of these 17. Choose the correct statement (A) Gold leaf electroscope can be used to find whether a body has an electric charge on it or not. (B) Gold leaf electroscope can be used to is positively charged or negatively charged. (C) Both (A) and (B) (D) neither (A) nor (B)

Electricity and Chemical Effects of Electric Current 18. An electroscope is a device used for (A) To protect building from the lighting (B) Detecting electric charges (C) Finding a nature of electric charges (D) Both (B) & (C)

11. What does Electricity do for us?

79

Sources of Electric Current (a) Small sources of Electric Current A cell is small source of electric current. The cells are used to operate small devices like torches, transistors, radios, bicycle lamps, small tape recorders, etc., Button cells (very small cells) are used for operating calculators; wrist watches, etc. In addition to these cells there are storage cells. These are used in cars, trucks, tractors, motorcycles, etc. to provide electric current for head lights and in the operation of engine. These are also used in emergency lights.

We use electric current in many ways. At home it is used to run fans and air coolers during summer. At night it lights our home. We use it in room heaters in winter to warm our rooms. It is used in refrigerators to keep things at low temperature. It is used in geysers to heat water. It is used in electric heaters (b) Bigger sources of Electric Current to produce heat required for cooking purpose. It is The electric current used in our homes for running also used to run transistors; radios, television. The fans, refrigerators, heaters, etc. is of very large telephone also uses electric current. magnitude. Similarly, the energy used in factories, In factories, it is used to run huge electric motors. It electric trains, etc., is of very very large magnitude. is used to run electric trains. In cinema hall it is Cells cannot provide this energy. For producing a used to run the projector. As a matter of fact electric large amount of energy big power houses are current is used in a large number of devices, which Automobile battery constructed. In these power make our life comfortable. houses the energy of flowing water, or the energy In industry electric current is used for making of steam is converted into electric energy. electromagnets and electroplating. Electromagnets An electric generator consists of two parts, namely, are used extensively for lifting heavy load of iron, in turbine and dynamo, which are coupled together. television, telephone, MRI machines, etc. The energy of the flowing water or steam rotates Electroplating is a process of coating expensive the blades of turbine. The energy of the turbine metals over cheaper metals. For example, the operates the dynamo which finally converts the articles of copper or iron are coated with nickel, mechanical energy into electric energy. chromium, silver, etc., to improve their Electric energy so generated at power houses is Appearance. Furthermore, electric current is used carried by electric cables to the cities and villages. in the extraction of metals like \ copper from its The electric cables are made of thick copper or mineral salts. aluminium wires. These cables are mounted on very The electric appliances are connected to the electric tall steel towers. You must have seen these cables wires through a tiny device called switch. When mounted on towers while traveling by train. we want to use any electrical appliance we put the Small electric generating sets are used in hospitals; switch in the ON position. When we want that railway stations and even in schools as a standby. appliance should stop working we put the switch in When the Electric generator supply fails due to the OFF position. In the ON position the electric some reason, then these electric generators. current flows through appliance. In the OFF position electric current does not flow through the appliance. 12. Electric Current Here, it must be pointed out that all appliances do The rate of flow of charge in a circuit is called not work with electric current. For example a bicycle; electric current. In other words, it is the amount of a sewing machine; a cycle rickshaw; a diesel engine; charge flowing per second. It is denoted by the letter a water pump, etc. don’t need electric energy. I. Similarly, the wind-mills If Q is the charge which is flowing through a conductor in time t, then current is given by i 

Q t

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8th Class Physics

80 Unit of Current: The S.I unit of current is ampere and it is denoted by the letter ‘A’. The S.I unit of Q is coulomb and that of t is second. Thus, the S.I unit of electric current is

1 coulomb  1A 1 second

Definition of Ampere: When a charge of coulomb flows through a conductor in one second, then the current flowing through the conductor is said to be one ampere. Thus, when 1 coulomb of charge flows through a conductor in 1 second, then the current flowing through it is said to be 1 ampere. 1 ampere 

1 coulomb 1 second

Smaller units of Electric Current: Sometimes smaller units of current are also used. These are microampere and milliampere. 1 microampere = 1 A  106 A 1 milliampere = 1 mA  103 A Bigger unit of electric current: Sometimes the magnitude of the current flowing in a conductor is very large. This large magnitude of current is expressed in bigger units, such as kilo ampere and mega ampere. 1 kilo ampere (kA) = 1000 A = 103 A 1 mega ampere (MA) = 1,000,000 A = 106 A.

Flow of Current: In metals, the moving charges are the electrons constituting the current, while in electrolytes and ionized gases, electrons and positively charged ions are the ions moving charges which constitute current. The charge on an electron is negative and is –1.6 × 10–19 coulomb (symbol C). Therefore, I C charge is carried by electrons. Hence if I A current flows through a conductor, it implies that 6.25 × 1018 electrons pass in 1 second across the cross section of the conductor.

The direction of current is conventionally taken opposite to the direction of motion of electrons.

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If n electrons pass through a cross section of a conductor in time t, then total charge passed Q = n × e and current in conductor Instrument by which current measured: Current is measured by an instrument called ammeter.

13. Potential The potential at any point is defined of the work done is bringing a unit positive charge from infinity to that point. Potential Difference: The potential difference two points in an electric field is defined as the amount of work done in moving a unit positive charge from one point to another point.

P.d 

work done Quantity of ch arg e transferred

If W joules of work has to be done to transfer Q coulombs of charge from one point to another then the p.d. ‘v between the two point is given by formula. v 

W Q

The S.I unit of p.d. is volt (v)

Formative Worksheet 17. Which of the appliance do not need electric current? (A) Bicycle (B) Sewing machine (C) Cycle rickshaw (D) Wind - mills 18. Statement I : Electric current is used to run electric trains Statement II : In factories electric current is used to run huge electric generators. (A) Both Statements are true, Statement - II is the correct explanation of Statement- I. (B) Both Statements are true, Statement - II is not correct explanation of Statement - I. (C) Statement - I is true, Statement - II is false. (D) Statement - I is false, Statement - II is true. 19. Column-I Column-II a) Expression for 1) Q/t electric potential b) 1Joule/1coulomb 2) 1Volt c) 1Coulomb/1sec 3) W/Q d) Expression for current 4) 1ampere 5) 1Newton

Electricity and Chemical Effects of Electric Current 20. When 1J of work is done in bringing 1C of charge from infinity to a point in electric field then potential is said to be (A) 0V (B) 1V (C) 2V (D) None of these 21. If 2C of charge is moved through a P.d of 3V then the work done is (A) 1.5V (B) 0.66J (C) 6J (D) 0J 22. In which of the following cells are used (A) Torches (B) Transistors (C) Radios (D) Bicycle lamps 23. The relationship between volt, joule and coulomb is (A) 1volt = 1joule × 1 coulomb (B) 1joule = 1 coulomb × 1volt (C) 1joule = 1 coulomb/1volt (D) None of these 24. The current passing through a conductor is 5 ampere .Then the charge that passes through that conductor in 5 minute is _______Coulomb 25. What is the electric potential at a point in an electric field ,when 96 J work is done in moving a charge of 24C from infinity? (A) 2V (B) 3V (C) 5V (D) 4V 26. In which of the following storage cells are used (A) Cars (B) Trucks (C) Motorcycles

81 24. The device which measures the potential difference in an electric circuit (A) Voltmeter (B) Ammeter (C) Potentiometer (D) Galvanometer 25. A charge of 5C is given a displacement of 0.5m. The work done in the process is 10J. The Potential difference between the two points will be (A) 2V (B) 10V (C) 50V (D) 5V 26. If 60 coulomb of charge passes through a cross section of a conductor in 4 sec, the average current is_________A

14. Primary Electric Cells Voltaic cell, Lechlanche cell, Dry cell and Bichromate cell are few examples of Primary electric cells.

1. Voltaic Cell: In 1786 an Italian scientist Voltair, invented the first electric cell. It is known as Voltaic cell.

Construction of a Voltaic Cell:

+



1

(D) Tractors

2

Conceptive Worksheet 19. The rate of flow of charge is called (A) Potential (B) Current (C) Resistance (D) Conductivity 20. Current is measured by a instrument called (A) Voltmeter (B) Ammeter (C) Potentionmeter (D) Galvanometer 21. An electric generator consists of (A) Turbine (B) Dynamo (C) Both (1) and (2) (D) Neither (1) nor (2) 22. Which of the following are sources of electric current (A) Sun (B) Cell (C) Battery (D)Solar cell 23. Potential is a _________quantity (A) Scalar (B) Vector (C) Both (A) and (B) (D) Neither (A) nor (B)

3 Voltaic cell 1. Copper sheet (+) 2. Zinc sheet () 3. Dilute sulphuric acid

1) 2)

3) 4)

5)

Take a glass jar and pour dilute sulphuric acid up to 3/4 of the jar. Take a copper plate and a zinc rod and put them in the jar such that half of each of them is immersed in the acid. Take two copper wires and connect each of them to the copper plate and the zinc rod. Electricity flows through these wires and if you connect these wires to a electric bell, it rings continuously. In the cell, the copper plate is the positive pole and the zinc rod is the negative pole. The diluted sulphuric acid is called the exciting fluid or electrolyte. www.betoppers.com

8th Class Physics

82

Defects: 1)

2)

3. Lechlanche Cell:

Local action: Local action due to impure zinc can be prevented by amalgamating the zinc rod with mercury. Polarisation: Polarisation is the liberation of hydrogen and it can be removed by using oxidising agents or depolarisers like potassium dichromate, copper sulphate or manganese dioxide.

2. Construction of a Dry Cell: +

+

1

2

3 



4 6

2

5

1

3 4

The dry cell Parts: 1. Zinc con 2. Paste of ammontum chloride 3. Manganese dioxide + graphite powder 4. Carbon rod

1) 2) 3) 4)

5)

6) 7) 8)

A dry cell consists of a cylindrical zinc can. In the middle of this zinc can, there is a carbon rod with a brass cap. The zinc can acts as the negative pole and the carbon rod acts as the positive pole. A thick paste of manganese dioxide and graphite powder is packed round the carbon rod. In the next layer, another thick paste of ammonium chloride is packed in the zinc can. To prevent ammonium chloride and manganese dioxide from mixing up with each other, one or two layers of cloth or paper is placed. Again adjust the length of the carbon rod such that it does not touch the bottom of the zinc can. A piece of cardboard is also put under the carbon rod for this purpose. All the chemicals and carbon rod are sealed with saw dust and pitch. The cell is a modified form of a Lechlanche cell with no liquid in it.

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Lechlanche cell Parts: 1. Glass jar 2. Zinc rod 3. Porous pot 4. MN mixture of manganese dioxide + carbon powder 5. Carbon rod 6. Ammonium chloride solution

In the Lechlanche cell: (1) Positive pole is the Carbon rod.(2) Negative pole is the Zinc rod. (3) Depolariser is the Manganese dioxide.

Bichromate Cell: It consists of a glass vessel containing a mixture of solutions Potassium dichromate (K2Cr 2 O7) and Sulphuric acid (H2SO4). A screw cap (S) is fitted to the vessel. To this screw cap, a Zinclate (Zn) is fitted at the middle of two Carbon plates. Zinc plate acts as negative electrode and the two Carbon plates together act as positive electrode. The solution K 2 Cr 2 O 7 with H 2 SO 4 acts as an electrolyte.

Electricity and Chemical Effects of Electric Current

Formative Worksheet 27. Statement I : A primary cell converts chemical energy into electrical energy Statement II : A primary cell converts electrical energy into chemical energy (A) Both Statements are true, Statement - II is the correct explanation of Statement- I. (B) Both Statements are true, Statement - II is not correct explanation of Statement - I. (C) Statement - I is true, Statement - II is false. (D) Statement - I is false, Statement - II is true. 28. Column-I Column-II a) Electrolyte in 1) Dilute H2SO4 + voltaic cell Potassium Bi chromate b) Electrolyte in 2) Lime juice Lechlanche cell c) Electrolyte in 3) Ammonium chloride Bichromate cell solution d) Electrolyte in a 4) Dilute H2SO4 cell made of 5) Dilute HNO3 29. The electrolyte used in a dry cell is (A) H2SO4(Dilute) (B) NH4Cl + H2SO4 (C) NH4Cl(Paste) (D) None of these 30. A dry cell consists of (A) ZnO (B) NH4Cl (C) Manganese dioxide (D) Carbon 31. In an Bichromate cell the glass vessel containing a mixture of solutions are (A) K2Cr2O7 + H2SO4 (B) NH4Cl + H2SO4 (C) K2Cr2O7 + HNO3 (D) None of these 32. A Leclanche cell consists of (A) Carbon rod (B) Porous pot (C) Powdered carbon (D) Manganese dioxide 33. In a primary cell (A) Positive charge accumulates at zinc electrode (B) Negative charge accumulates at copper electrode (C) Negative charge accumulates at zinc electrode (D) Positive charge accumulates at copper electrode and Negative charge accumulates at zinc electrode

83 34. Choose the correct statements (A) The Italian scientist Voltair , invented the first electric cell (B) The dry cell is a modified form of a Lechlanche cell (C) In an Lechlanche cell carbon rod acts as a is a negative pole (D) In an Bichromate cell the zinc rod acts as a negative pole

Conceptive Worksheet 27. The primary cell which is used in daily life is (A) Lechlanche (B) Dry cell (C) Bichrometre cell(D) Simple voltaic cell 28. The depolarization used in Leclanche cell is (A) Solution of ammonium chloride (B) Porous pot (C) Powered carbon (D) Manganese dioxide 29. In a Lechlanche cell the anode is made of (A) Copper plate (B) Zinc plate (C) Carbon plate (D) Iron plate 30. The defects in voltaic cell are (A) Local action (B) Polarisation (C) Radiation (D) Reflection 31. In an voltaic cell the negative pole is (A) Zinc rod (B) Carbon rod (C) Copper rod (D) None of these 32. In an Dry cell the positive pole is (A) Zinc rod (B) Carbon rod (C) Copper rod (D) None of these 33. In an Lechlanche cell, the positive pole is (A) Zinc rod (B) Carbon rod (C) Copper rod

(D) None of these

Conductors and Insulators: The materials which allow the electric current to pass through them are the conductors of electricity and the materials through which electric current does not pass are the non-conductors or the bad conductors of electricity. Metals are the conductors of electricity. Non metals like glass, plastic, wood, paper, cloth and rubber are the non-conductors of electricity. Non-conductors of electricity are also called insulators. All leads (wires) being used in an electric circuit are metallic wires coated with plastic or rubber. Coating of a conductor with a non-conductor is called insulation. www.betoppers.com

8th Class Physics

84 If we happen to touch a metallic end of a lead through which current is passing, it gives an electric ‘SHOCK’. The shock may be fatal too or otherwise it shakes the body and harms the person who has suffered the electric shock. Insulation saves a person from electric shock.

15. Electric Circuit An electric cell or dry cell is the source of energy for the bulb to glow and warm up. Let us now learn the way in which this electric energy is made available to the bulb in the torch.

Making of a Simple Electric Circuit Step 1: Take out the bulb from bulb from a torch. Examine the bulb carefully. The bulb is a small globe of thin glass enclosing a coiled filament supported on two thick wires. One of these thick wires is connected to the metal casing around the base of the bulb. The other wire is connected to the metal tip at the base. The metal casing and the metal tip at the base are the two terminals of the bulb.

Terminal on metal casing

Step 2:

Terminal on metal tip at the base

Connect the two free ends of the wires from the bulb or the bulb holder to an electric cell in such a way that one piece of wire is connected to the positive terminals of the cell and the other to the negative terminal of the cell. This may be done with the help of a rubber band or an adhesive tape. When you have finished with connections, the bulb lights up. With your finger trace the path of the electricity from the positive ( + ve) terminal on the cell to the negative (–ve) terminal of the cell. It is a round about path travelled by electricity.

Closed and Open Circuit

Dry cell Symbol of cell

The dry cell has two terminals. The central terminal of the dry cell is called positive terminal. The base of the dry cell (which is made of a metal) is called negative terminal. Fig. shows the terminals of dry cell. The long line represents positive terminal of the cell and the small and thick line represents negative terminal of the cell. For this experiment you need a torch cell; a torch bulb marked 1.5 V, cellotape, a plastic coated 1 metre long copper wire and an old used blade.

Inside a torch bulb

Bulb

Take two pieces of insulated wire. Insulated wires have metal wire inside with a plastic covering on the outside. Remove the plastic covering from both the ends of each piece of wire. Fix these wires on the bulb as shown in the picture with the help of Insulating adhesive tape. Or fix the bulb on a bulb holder. The two screws on the bulb holder are the two terminals which are connected to the two terminals on the bulb. The two pieces of wire be connected to the two terminals on the holder, as shown in the picture.

Step 3: Bulb connected with wires

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Plastic coated copper wire A

B Cell Closed circuit and bulb glows

Cut the plastic coated copper wire into two halves A and B. Remove plastic coating from each end of the wire such that 1 cm of plastic is removed. Now fix one bare end of each wire A and B to the terminals of 1.5 V bulb with the cellotape. Fix the other end of wire A to the base of cell with the help of cellotape. Now touch the bare end of wire B to the central terminal of cell as shown in figure. What do you observe?

Electricity and Chemical Effects of Electric Current The bulb lights up. This shows that electric current is flowing in wire A and B through the bulb. The path along which electric current flows is called electric circuit. Now remove the wire B from the central terminal as shown in Fig. . What do you observe? The bulb does not glow. It is because electric current does not flow, if the path is broken or path is incomplete. Bulb Plastic coated copper wire A Circuit is bronken Cell Open circuit and bulb doesn’t glows

Closed Circuit or Complete Circuit Plastic coated copper wire

Bulb Bulb

A

85 When the path which starts from one terminal of the cell, ends at the other terminal of the cell, without any break, then such a circuit is called complete circuit or closed circuit. When the circuit is closed, then any electric appliance in that circuit starts working. In the present case the bulb starts glowing.

Open Circuit or Incomplete Circuit When the path of current, starting from one terminal of the cell to another terminal of the cell is broken or incomplete, then such a circuit is called open circuit or incomplete circuit. For example, when we remove wire B from central terminal of cell, then the circuit is open circuit or incomplete circuit. Switches are used in the household wiring^ to open or close the electric circuit. When we switch on a particular electric appliance, we close the electric circuit. Conversely, when we switch off an electric appliance, we open the electric circuit.

B Cell Closed circuit

Diagrammatic representation of closed circuit

Symbols used in Electrical circuits: You find hereunder some symbols used in electrical circuits.

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8th Class Physics

86 A switch, a simple device to ‘close’ or ‘open’ a circuit: An electric circuit passes through a switch. Switch is a simple device which helps us to close or open the circuit. It helps in s a v i n g electricity when not in use. You are always advised to switch ‘off the lights or other gadgets in your home to save electricity.

Connecting Electric Cells in Series Take a dry cell and a torch bulb. Connect the bulb to the cell using copper wires as shown in Fig. . Observe the intensity- of light. The bulb does not glow brightly.

Now take one more dry-cell and connect two cells as shown in figure. In this method the positive of the first cell is connected to the negative of the second. The negative of the first and the positive of the second are connected to the bulb. The bulb now glows brighter. In the battery torch or battery light two or three dry cells are put into a metal container in series. The positive of one cell is connected to the negative pole of another cell in the series connection, When the, switch is turned on, the circuit is closed and the bulb glows and gives light.

When cells are connected in series, their electromotive force is equal to the sum of the EMF of all the cells used. Connect three torch bulbs in series as shown in figure. Connect this to a dry cell and observe that brightness of each of the three bulbs. Now connect one more dry cell in series with he first cell. Observe the brightness of each of the bulb. Then connect one more dry cell in series with the first two cells. Again observe the bulbs.

Disconnect one of the three bulbs in the circuit. The circuit becomes open and all the three bulbs stop glowing. In series connection of bulbs,’ if one bulb gets fused, all the other bulbs in the series will stop working. Three bulbs connected in Series

Connecting Bulbs in Parallel: Connect three bulbs in parallel. That is, one end of each of the three bulbs are connected one wire, the other ends of the three bulbs are con nected to another wire! These two wires are con nected to a dry cell. All the three bulbs glow dimly. Now disconnect one of the bulbs. The other bulbs continue to glow as before.

Connecting Electric Cells in Parallel: Connect one torch bulb to one cell as you did in fig.. You, will observe that, the bulb toes not glow brightly.

Take three dry cells and connect them as shown in fig.. That is all the positive poles of the three cells are connected together, and all the three negative poles are connected together. These three positives and three negatives are connected to the bulb- You will observe that there is no change in the brightness of the bulb! When cells are connected in parallel, their total electromotive force is the same as that if any one of them. www.betoppers.com

To study the properties of (i) Series circuit, (ii) Parallel circuit Materials required : a battery of four cells two bulbs of 1 watt each one fused bulb a switch few lengths of connecting wires cellotape.

A

B

Method: Connect the bulbs A and B in series by connecting them to connecting wires with the help of cellotape as shown in Fig. 6.13(a). Connect the free ends of connecting wires to a battery through

Electricity and Chemical Effects of Electric Current

87

a switch. Close the switch. What do you observe? Both the bulbs will glow*. However, they will not glow very brightly. Open the switch. What do you observe? A

A

C C

Both the bulbs will stop glowing. Now remove the bulb B and instead fix a fused bulb C [Fig. ]. What is your observation? Bulb A does not glow. 1. 2. A

3. B

The bulb A continues glowing brightly, whereas bulb C does not glow. Following conclusions can be drawn from above investigation. In parallel circuit all the appliances work independently In parallel circuit if one appliance goes out of order, the other continues working. It means that each appliance in parallel circuit can be operated independently by a switch. As the bulbs glow brightly, it means each appliance gets enough electric energy, and hence, works to its full capacity.

Formative Worksheet 1.

2. 3.

Following are the conclusions from the above investigation. In series circuit all the appliances work simultaneously when switch is closed. Conversely, all appliances stop working when switch is open. In series circuit, if any, of the appliances goes out of order, the other appliances stop working. As the bulbs were not glowing very brightly, it can be concluded that in series the appliances do not work to their full capacity. Now connect the bulbs A and B in parallel, such that they have common positive and common negative terminals as illustrated by Fig. , through a switch and a battery. Close the switch. What do you observe? Both the bulbs A and B glow very brightly. Now remove the bulb B and instead fix a fused bulb C Fig. . What is your observation?

35. The symbol for a fuse is (A) (C)

(B) (D)

A

36. The symbol for a ammeter is (A) (C)

A

(B)

V

(D)

()

37. In a parallel circuit of bulbs (A) Same current exists in all the bulbs (B) Same voltage exists in all the bulbs (C) Failure of any bulb leads to a break down in the circuit (D) All of above 38. Column-I Column-II a) cell 1) V b) resistance c) closed switch d) voltmeter

2) 3) 4) 5)

( )

()

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8th Class Physics

88 39. When negative terminal of a cell is connected to the positive terminal of the next cell are said to be in (A) Series (B) Parallel (C) Both (A) and (B) (D) Neither (A) nor (B) 40. When electric cells are connected in series the electromotive force (A) Increase (B) Decreases (C) Remain same (D) become zero 41. Four electronic cells each of emf 1V connected in parallel , then the combined emf is_____ V 42. Choose the correct option (A) The path along which electric current flows is called electric circuit (B) Coating of a conductor with a non-conductor is called insulator (C) The materials which allow the electric current to pass through them are called conductors (D) The materials which allow the electric current to pass through them are called non-conductors 43. The total e.m.f of the series combination of three cells of equal e.m.f is 4.5V .What is the emf of each cell (A)1.5V (B) 3.5V (C) 6.5V (D)12.5V 44. If a voltage V is applied across the bulbs in series, then (A) The voltage applied is divided among the bulbs (B) The same current exists in all the bulbs (C) The same voltage exists in all the bulbs (D) The current is divided among the bulbs 45. Greater potential difference (or emf ) is obtained in the circuit when cells are connected in (A) Series (B) Parallel (C) Both (A) and (B)

36. Statement I : Switch is used to close are open electric circuit Statement II : When switch is closed , then circuit is closed and when it is opened, then circuit is open circuit 1) Both Statements are true, Statement - II is the correct explanation of Statement- I. 2) Both Statements are true, Statement - II is not correct explanation of Statement - I. 3) Statement - I is true, Statement - II is false. 4) Statement - I is false, Statement - II is true. 37. What is the total emf , when three cells of emf’s are 2V,2.5V, 4V are connected in series (A) 4V (B) 2.5V (C) 2V (D) 8.5V 38. What is the total emf, when three cells of emfs are 2V,2V,2V are connected in parallel (A) 4V (B) 1V (C) 2V (D) 5V 39. The total emf of three cells of emf’s 1v,1v,1v are connected in series is ____v 40. Which of the following is used as source of electrical energy (A) Electric cell (B) Dry cell (C) Ammeter

Summative Worksheet 1.

2.

3.

(D) Neither (A) nor (B)

Conceptive Worksheet

4.

34. The path along which electric current flow is called 5. (A) Electric circuit 6. (B) Electric shock (C) Both (A) and (B) 7. (D) Neither (A) nor (B) 35. Which of the following are the non-conductors of electricity (A) Wood (B) Paper (C) Cloth (D) Rubber www.betoppers.com

(D) Voltameter

An uncharged insulated conductor A is brought near a charged insulated conductor B then charge ____________ and potential ___________. Two spheres A and B of exactly same mass are given equal positive and negative charges respectively. What happens to their masses after charging? An ebonite rod acquires a negative charge of 3.2 × 10–10 C. The number of excess electrons it has is ___________. A body has a negative charge of 1 coulomb. It means that __________. A million electrons are added to a pith ball. Its charge is __________. A bird sitting on a high power line gets a fatal shock. True or False. Electric lines of force (A) Exists everywhere (B) Exist only in the immediate vicinity of electric charges (C) Exist only when positive and negative charges are near one another (D) Are imaginary

Electricity and Chemical Effects of Electric Current 8. 9. 10. 11.

12. 13.

14. 15. 16. 17. 18. 19.

20.

21.

22.

23.

24. 25.

The minimum charge on a particle can be equal to ___________. Like charges _________ and unlike charges ____________. Electric charges are not _________ but consists of ________ of charges from one body to another. To give a body an excess ___________ charge, we add a number of electrons to it. Similarly a body is charged ________ if a number of electrons are removed from it. The charge of a body refers only to its ___________ charge. According to the principle of conservation of electric charge, the algebraic sum of all the electric charges in any closed system is____________. A charge can be ________ from one body to another but it cannot be ______or _____. Charge is ________ nature. ___________ is a sure test of electrification. Charge resides on ________surface of a conductor. Conductors contain ______ electrons and an _____ has no free electrons. ________ permit electric charges to move within them and _______ do not allow the motion of electric charges within them. Metals and electrolytes are _______________. Sulphur, mica, plastic, fur, silk, glass etc., are ___________. In _________, a charged body is put in contact with an uncharged body. Then the uncharged body acquires charge of the __________ sign as the charged body. In __________, a charged body is brought near an uncharged body. Then the uncharged body acquires a charge ______________ in sign to that of the charged body. The electric charge is ________. The lowest electric charge is that of the _________and e = ____________C. The number of electrons in 1C of electric charge is _________. When a body has a negative charge, it has ___________ of electrons. If it has a positive charge it has _____________ of electrons.

89 26. The force of attraction or repulsion is maximum when the medium between the charges is ____________ or ____________. 27. Coulomb force is true only for _______ charges. 28. _____________ are charge carriers in metals. 29. _____________ are charge carriers in liquids. 30. _______ and _______ ions are charge carriers in gases.

HOTS Worksheet 1.

A conductor carries a current of 2A. How long will it take for 1800C of electricity to flow through a given cross-section ? 2. How many electrons flow through a light bulb each second if the current through the bulb is 1.6A ? 3. A current of 1.5A exists in a conductor whose terminals are connected across a potential difference of 100 volt. Find a) the total charge transferred in one minute. b) the work done in transferring the charge. 4. A polythene piece, rubbed with wool, is found to have negative charge of 4  10–7C. The number of electrons transferred from wool to polythene is ______________. 5. When a piece of polythene is rubbed with wool, a charge of –2 × 10–7Cis developed on polythene. What mass (if any), is transferred to polythene ? 6. Two identical metal spheres A and B are supported on insulating stands and placed in contact. What kind of charges will A and B develop when a negatively charged ebonite rod is brought near A? 7. The sphere B in previous Question, is momentarily touched with a finger and then the ebonite rod is removed, what is the sign of the charge on A and B now ? 8. An electric cell does 5 J of work in carrying 10 C charge around a closed circuit. The electromotive force of the cell is _______________. 9. A work of 100 joule is performed in carrying a charge of –5 coulomb from infinity to a particular point in an electrostatic field. The potential of this point is ____________. 10. The work done in moving a charge of 20 C from A to B over a distance of 0.2 m is 2 J. The potential difference across A and B is ______________.

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8th Class Physics

90 11. A charge of 5 C is given a displacement of 0.5m. The work done in the process is 10 J. The potential difference between the two points will be _____________. 12. An electron of charge ‘e’ coulomb passes through a potential difference of V volt. Its energy in joule will be _______________. 13. If the electronic charge is 1.6 × 10–19 C, then the number of electrons passing through a section of wire per second, when the wire carries a current of 2ampere is___________. 14. A conductor carries a current of 3 mA. The number of electrons passing through it in 1 minute is about _____________. 15. Current of 4.8 ampere is flowing through a conductor. The number of electrons crossing per second the cross–section of conductor will be _____________.

IIT JEE Worksheet 1.

2.

3.

4.

5.

6.

7.

8.

The potential difference applied to an X–ray tube is 5 kV and current through it is 3.2 mA. Then the number of electrons striking the target per second is __________. In a region 1019  particles and 1019 protons move to the left, while 1019 electrons move to the right per second. Find the strength of the current. A charge of 2 × 10–2C moves at 30 revolutions per second in a circle of diameter 80cm. Find the current linked with the circuit. In hydrogen atom, the electron makes 6.6 × 1015 revolution per second around the nucleus in an orbit of radius 0.5 × 10–10 m. Find the equivalent current. A charge of 5000 C flows through on electric circuit in 2 hours and 30minutes, Calculate the magnitude of current in the circuit. A dry cell can supply a charge of 300 C. If the current drawn from the cell is 60µA, find the time in which the cell completely discharges. 50 coulombs of charge is brought from infinity to a given point in an electric field when 62.5 joule of work is done. What is the potential at the point? A flow of 107 electrons per second in a conducting wire constitutes a current of (A) 1.6 × 10–26 A (B) 1.6 × 1026 A (C) 1.6 × 10–12 A (D) 1.6 × 1012 A

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If the electronic charge is 1.6 × 10–19 C, then the number of electrons passing through a section of a wire per second, when the wire carries a current of 4 ampere is (A) 1.25 × 1017 (B) 1.6 × 1017 (C) 2.5 × 1019 (D) 1.6 × 1019 10. In a region 1019  particles and 1019 protons move to the left, while 1019 electrons move to the right per second. The current is (A) 3.2 A towards left (B) 3.2 A towards right (C) 6.4 A towards left (D) 6.4 A towards right

9.



By the end of this chapter, you will understand Reflection of Light • Terms related to Refraction General terms to the Reflection • Refractive Index of a medium Laws of Reflection • Real depth and Apparent depth Image • Dispersion of Light Effect of Rotation of Mirror on Reflected Ray • Spherical lenses Formula for number of images formed in • Image formation by lenses two plane mirrors • Human Eye parts and Functions • Reflection by Spherical Mirrors • Power of Accomodation • Terms related to Spherical Mirrors • Defects of Vision • Refraction of Light • • • • • •

1. Introduction If we enter a dark room, objects present there are not visible. However, if a bulb is switched on, everything in the room becomes visible. It shows that for vision the presence of light is essential. Definition of Light: It is an invisible energy which causes in us sensation of sight (vision). Since light is obtained from heat energy, i.e., when an object is heated to a temperature beyond (500 0C, we can say that light is a kind of energy). It must be kept in mind that light energy makes the surrounding objects visible, but by itself it is an invisible energy. For example, if we are seeing a coloured poster, then we are seeing only the poster and not the coloured lights reflected fronts it. It is because light is invisible. The various colours reflected from the poster excite the retina of the eye, which in turn sends a message to the brain. It is the brain which finally makes out the colours of the poster. Thus, we can conclude that light is an invisible energy which causes in us sensation of vision. Sources of Light: Sun is the primary source of light for mankind. In addition to it, lighted bulb, a fluorescent tube, a lighted candle ,a kerosene oil lamp, etc., are other sources of light. Speed of Light: Light travels at very fast speed i.e., 3 × 108 m/s. It means the speed of light is 300000000 m/s or 300000 km/s. Luminous bodies: The bodies which give out light energy by themselves are called luminous bodies. Examples: The sun, the stars, glow worm etc. Non-luminous bodies: The bodies which do not give light energy by themselves , but reflect the light energy falling on them are called non-luminous bodies. Examples: Moon, wood, furniture, etc.

Chapter - 7

Light

Learning Outcomes

2. Reflection of Light When a beam of light is incident on a surface, a part of it is returned back into the same medium. The part of light which is returned back into the same medium is called the reflected light. The remaining part of light is absorbed if the surface on which the incident light strikes is opaque or it is partly transmitted and partly absorbed if the surface is transparent. Reflection The return of light into the same medium after striking a surface is called reflection. Reflection of light is the process which enables us to see different objects around us. Luminous bodies are directly seen, but non luminous objects are seen only because they reflect the light incident on them which on entering into our eyes, make them visible.

Note: Reflection is possible in case of plane mirror. A plane mirror is a plane glass plate which is silvered at its one surface. The other surface is then reflecting surface of the plane mirror.

92 Types of Reflections (a) Regular Reflection The phenomenon due to which a parallel beam of light travelling through a certain medium, on striking some smooth polished surface, bounces off from it, as parallel beam, in some other direction is called ‘regular reflection’. Regular reflection takes place from the objects like looking glass, still water, oil, highly polished metals, etc. Regular reflection is useful in the formation of images, e.g., we can see our face in a mirror only on account of regular reflection. However, it causes a very strong glare in our eyes.

(b) Irregular Reflection or Diffused Reflection The phenomenon due to which a parallel beam of light, travelling through some medium, gets reflected in various possible directions, on striking some rough surface is called ‘irregular reflection or diffused reflection’. The reflection which takes places from ground; walls; trees; suspended particles in air; and a variety of other objects, which are not very smooth, is irregular reflection. Irregular reflection helps in spreading light energy over a vast region and also decreases its intensity. Thus, it helps in the general illumination of places and helps us to see things around us.

8th Class Physics (d) Reflected Ray: A ray of light which bounces off the surface of a mirror, is called reflected ray. BC is reflected ray in the figure. (e) Normal: The perpendicular drawn at the point of incidence, to the surface of mirror is called normal. BN is the normal in the figure. (f) Angle of Incidence: The angle made by the incident ray with the normal is called angle of incidence.  ABN is the angle o£ incidence in the figure. (g) Angle of Reflection: The angle made by the reflected ray with the normal is called angle of reflection.  CBN is the angle of reflection in the figure. (h) Glance Angle of Incidence: The angle which the incident ray makes with the mirror is called glance angle of incidence.  MBA is the glance angle of incidence in the figure. (i) Glance Angle of Reflection: The angle which the reflected ray makes with the mirror is called glance angle of reflection.  M’BC is the glance angle of reflection in the figure.

4. Laws of Reflection 1.

The incident ray, the reflected ray and the normal lie in the same plane, at the point of incidence. 2. The angle of incidence is always equal to the angle of reflection. Formula for the angle of deviation due to reflection In the figure angle of incidence = i; Angle of deviation = d

3

N

General terms to the Reflection

(a) Mirror: A smooth polished surface from which regular reflection can take place is called mirror. MM| is the mirror as shown in figure. (b) Incident Ray: A ray of light which travels towards the mirror is called incident ray. AB is an incident ray in the figure. (c) Point of Incidence: The point on the mirror, where an incident ray strikes is called point of incidence. ‘B’ is the point of incidence in the figure. www.betoppers.com

A

B

i r M

O

M1

d

c

Light Consider the straight line AOC, i + r +d = 1800 i.e the sum of angle of incidence, angle of reflection and angle of deviation is 1800  d = 180 – (i + r) = 180 – (i + i) (i = r)  d  180 – 2i Therefore, for an angle of incidence i, the angle of deviation is equal to 180 – 2i =   2i Note: The deviation produced by n reflections from two plane mirrors inclined at an angle  is given by b) D = n(180 –  ) = 360 - 2  , if n=2 where n is even.

5. Image

a)

When the rays of light, diverging from an object point, after reflection or refraction, either actually meet at some other point, or appear to meet at some other point, then that point is called image of that object. Types of images Virtual Image: When the rays of light, diverging from a point, after reflection or refraction, appear to diverge from another point, then the image so formed is called virtual image. I

Plane mirror

93 However, when these diverging rays of light reach the eye, then to the eye they appear to diverge out from point ‘I’. Thus I is the virtual image of object ‘O’. Ex: Image of our face in a plane mirror. Virtual images cannot be formed on a screen. virtual images are always erect upright. w.r.t object The path of the rays forming a virtual image is shown by dotted lines. Real Image: When the rays of light, diverging from an object point, after reflection or refraction actually converge at some other point then that point is real image of that object. In figure the ray of light diverging from point A, after reflection from the concave mirror actually converge at the point A1. Thus, A1 is the real image of the point A.

Concave mirror A

A

Virtual Image

P

B A

B

B D

O

Object

C

In figure the image ‘I’ of the object ‘O’ is virtual image. The ray of light diverging between OA and OB, after reflection, further diverges along AC and BD respectively.

Real image Example: Motion and still pictures projected on the screen in a cinema hall are real images. Real images are always inverted (upside down) w.r.t object . Real images and the path of the rays which form them are shown by continuous lines.

Distinction between Real Image and Virtual Image S.No

Virtual image

Real image

1.

The rays of light after reflection or refraction appear to meet at some other point.

The rays of light after reflection or refraction actually meet at some point.

2.

It cannot be caught on the screen.

It can be caught on the screen.

3.

It is always erect.

It is always real.

4.

Image of our face in plane mirror is a virtual image.

Image formed on a cinema screen is a real image.

Characteristics of an Image formed by a Plane Mirror 1. The image is formed behind the mirror and has the same size as the object 2. The image is inverted laterally. 3. The image is as far behind the mirror as the object is in front of it. 4. The image is virtual. It cannot be received on a screen. 5. The image is erected w.r.t object www.betoppers.com

8th Class Physics

94

6. Effect of Rotation of Mirror on Reflected Ray Consider a ray of light AB, incident on plane mirror in position MM, such that BC is the reflected ray and BN is the normal.

Formative Worksheet 1.

When a light ray is incident normally on the surface of a plane mirror, the reflected ray deviates through an angle of __________.

2.

What is the sum of angle of incidence , angle of reflection and angle of deviation equals to ?

3.

If the angle of deviation after reflection of a light ray is ‘d’, then find

ABN CBN i  ABC  2i –––––––––––––(i) Let the mirror be rotated through an angle ‘  ’ about point B, such M1M1 is the new position of the mirror and BN1 is the new position of the normal. As the position of the incident ray remains same, therefore new angle of the incidence is ABN1 whose

(a) angle of incidence (b) angle of reflection (c) angle of glancing 4.

Keeping the incident ray constant, if a plane mirror is rotated through an angle  , about an axis lying in its plane, then the reflected ray turns through an angle ___________.

5.

The angle between an incident ray and reflected ray for a given reflection is x. If the mirror is rotated through an angle x/2, keeping the incident ray constant, then find the angle between incident ray and reflected ray after rotation. Consider both clock wise and anti-clock wise direction.

6.

Keeping the plane mirror fixed, if the incident ray is rotated through an angle  , then the angle through which the reflected ray rotates is _________.

7.

A plane mirror lies face up, making an angle 150 with the horizontal. A ray of light shines down vertically on the mirror What is the angle of incidence? What will be the angle between the reflected ray and the horizontal be?

8.

Statement I : The plane containing the incident ray and the normal is called plane of incidence.

magnitude is (i +  ). Let BD be the reflected ray such that BDN1 is the new angle of reflection. ABD  ABN1  DBN1

 (i  )  (i  )

M M1

 2 i  2 –––––(ii)

B 



M1 M

i r i

i

 A

N N1

D C

Subtracting (i) from (ii) we get

ABD  ABC  2i  2  2i  CBD  2 Thus, for a given incident ray, if the plane mirror is rotated through a certain angle, then the reflected ray rotates through twice that angle or If a plane mirror is rotated through an angle  , the reflected ray is rotated through an angle 2  .

Statement II : The plane containing the incident ray and the normal is called plane of reflection. (A) Both Statements are true, Statement - II is the correct explanation of Statement- I. (B) Both Statements are true, Statement - II is not correct explanation of Statement - I. (C) Statement - I is true, Statement - II is false. (D) Statement - I is false, Statement - II is true.

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Light 9.

95 Match the following: Column-I (A) Plane of reflection

Column-II 1) The plane containing the reflected ray and normal (B) Angle of deviation 2) The angle made by incident ray with the mirror (C) Glancing angle 3) Angle through which a ray deviates from its normal path (D) Normal 4) perpendicular drawn to the surface at the point of incidence 5) The angle made by reflected ray with the mirror 10. When the rays of light, diverging from a point, after reflection or refraction, appear to diverge from another point, then the image so formed is called (A) Virtual image (B) Real image (C) Both (A) and (B) (D) Neither (A) nor (B) 11. When the rays of light, diverging from a point, after reflection or refraction actually converge at some other point then that point is _____ image of the object (A) Virtual (B) Real (C) Both (A) and (B) (D) Neither (A) nor (B)

Conceptive Worksheet 1.

2.

3.

4.

5.

The image formed by a plane mirror is always: (A) Real (B) Rrect (C) Virtual (D) Both (B) and (C) Reflection is possible in case of (A) Plane mirror (B) Transparent glass (C) Both (A) and (B) (D) Neither (A) nor (B) The part of light which is returned back into the same medium is called (A) Reflected light (B) Refracted light (C) Both (A) and (B) (D) Neither (A) nor (B) Motion and still pictures projected in a cinema hall are ______ images (A) Real (B) Virtual (C) Erect (D) All of these Choose the correct statements: (A) Regular reflection takes place on highly polished smooth surfaces (B) Irregular reflection takes place on rough surface (C) Irregular reflection is also called as diffused reflection. (D) On plane mirror regular reflection takes place.

6.

7.

8.

9.

Image of our face in a plane mirror (A) Is a virtual image (B) Is a real image (C) Cannot be taken on the screen (D) Both (B) and (C) If a mirror is rotated by 10 0, the reflected ray is rotated by (A) 10° (B) 20° (C) 40° (D) 30° Statement I : All the light rays actually converge at some point Statement II : All the light rays appears to converge at some point (A) Both Statements are true, Statement - II is the correct explanation of Statement- I. (C) Both Statements are true, Statement - II is not correct explanation of Statement - I. (B) Statement - I is true, Statement - II is false. (D) Statement - I is false, Statement - II is true. Statement I : Real images are always inverted Statement II : Virtual images are always erect (A) Both Statements are true, Statement - II is the correct explanation of Statement- I. (B) Both Statements are true, Statement - II is not correct explanation of Statement - I. (C) Statement - I is true, Statement - II is false. (D) Statement - I is false, Statement - II is true.

10. When a mirror is rotated through an angle  the reflected ray from it, turns through an angle of (A) 

(B)

 2

(C) 2

(D) 0

11. The incident ray and the reflected ray from a mirror are mutually perpendicular to each other. The angle of incidence is (A) 900 (B) 450 (C) 22.50 (D) zero 12. A ray of light is incident on a plane mirror at an angle of 600 What is the angle of deviation? (A) 60° (B) 30° (C) 90° (D) 180° 13. An object situated at a distance of 10cm infront of a plane mirror. The distance of image from the mirror is (A) 10cm (B) 20cm (C) 5cm (D) other value

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8th Class Physics

96 Lateral Inversion: When we place any object in front of a plane mirror, its image is such that its left hand side is seen on the right hand side and the right side is seen on the left.

Similarly, if you write with your right hand, the image in the mirror seems to write with left hand.

1.

When a pair of mirrors are parallel to each other. 2. When a pair of mirrors are kept perpendicular to each other. Minimum Height of plane mirror required for a person to see full length (or) Minimum length of a plane mirror to set the full image of the observer. Let MM represent a plane mirror and AEB an observer standing in front of it with E as the position of his eye. If the ray from his head A after reflection at M and the ray from his foot B after reflection at MM reach his eye E, the observer can see his full image. The image of AB is A  B Draw the normal MN and MM N In AMN and NME , AMN NME (i  r)

A'

M

A N E

Above figure illustrates another example of a lateral inversion. If you write letters ‘ABC’ on a piece of paper, they will look like ‘ ’

7. Formula for number of images formed in two plane mirrors 360 n , Where  is the angle between two  mirrors. Case (i): If n is an even whole number the number of images = n –1(for all positions of object) Example: If   60 , 360 n   6 , Here n is even 60  number of images = 6 – 1 = 5 Case (ii):If n is odd whole number, the number of images = n(if object is kept asymmetrically = n-1 ( if object is kept symmetrically) Example: If  = 400 360 n   9 , Here n is odd 40  number of images = 9(or) 8 Case (iii): If n is not whole number in this case the number of images is equal to the previous nearest whole number (i.e., integer part) Example: If  = 500 360 n   7.2 , Here n is not whole number 50  number of images = 7 Now we shall consider two special cases to study multiple reflections. www.betoppers.com

M'

N'

B' MN is common MNA MNE  90

AMN  NME and AN  NE . Similarly EN  NB

MM  NN = AB – (AN + B N ) = AB   NE  EN   AB  NN

Now

MM   AB  MM 

AB 2 Hence to see his full image, the least height of the mirror should be half the height of the observer.  2MM   AB or MM 

Formative Worksheet 12. You are standing in front of a plane mirror, which is approaching you with a speed of 10cm/sec. Then the speed of the image approaching you will be (A) 20 cm/s (B) 10 cm/s (C) 15 cm/s (D) 25 cm/s 13. A snake is approaching a stationary plane mirror with a speed of 5 m/s. Then it observes that its image is approaching it with a speed of (A) 15 m/s (B) 10 m/s (C) 5 m/s (D) 12 m/s

Light 14. An object is placed between two parallel mirrors then number of images formed is (A) 2 (B) 4 (C) 8 (D) Infinite 15. A lady dressed in a new sari stands in front of a plane mirror fixed on a vertical wall, height of the lady is 160 cm and that of her eyes from the ground is 150 cm, then The length of the mirror required to see her full view in the mirror is (A) 160 cm (B) 100 cm (C) 80 cm (D) 320 cm 16. In the above problem, find the position of lower edge of the mirror relative to the ground is (A) 160 cm (B) 75 cm (C) 100 cm (D) 200 cm 17. The angle of inclination between two mirrors is Column - I Column - II (A) 90 1) No. of images is one (B) 60 2) No. of images are three (C) 45 3) No. of images are five (D) 0 4) No. of images are seven 5) No. of images are infinite 18. A lady dressed in a new sari stands in front of a plane mirror fixed on a vertical wall, height of the lady is 160 cm and that of her eyes from the ground is 150 cm, then find the position of upper edge of the mirror relative to the ground is (A) 155 cm (B) 160 cm (C) 100 cm (D) 200 cm 19. A light beam through the slit ‘S’ is allowed to fall on the mirror which is rotated through an angle of 30o in the incident plane then (A) Normal rotates with angle of 30° (B) Reflected ray rotates with an angle of 60° (C) Incident ray remains fixed (D) Angle of incidence is fixed. 20. In your study room a plane mirror is fixed to the wall in front of you and opposite to that wall, a wall-clock is hanging in which only marking were there without numbers. You observe the time in the mirror as 7:10 hrs. Then what is the actual time. (A) 4:50 hrs (B) 7:10 hrs (C) 7:50 hrs (D) 5:50 hrs. 21. A man of 180 cm height stands in front of a plane mirror. His eyes are at a height of 170 cm from floor. Then the minimum length of a plane mirror required for him to see his full image is (A) 90 cm (B) 180 cm (C) 45 cm (D) 360 cm

97

Conceptive Worksheet 14. Five images are formed, if two plane mirrors are inclined to each other an angle of (A) 60°

(B) 70°

(C) 72°

(D) 90°

15. Statement I : The reflected ray formed by the plane mirror which is rotated with angle ‘  ’ is shifted through an angle of '2 ' Statement II : Reflected ray rotates with an angle of '2 ' if the mirror rotates with '  ' . (A) Both Statements are true, Statement - II is the correct explanation of Statement - I. (B) Both Statements are true, Statement - II is not correct explanation of Statement - I. (C) Statement - I is true, Statement - II is false. (D) Statement - I is false, Statement - II is true. 16. When the plane of the mirror is rotated by an angle '  ' in the plane incidence then the reflected ray rotates ______ times of its angle of rotation. 17. You are watching your right palm in the plane mirror by placing it parallel to it, which is in front of you, then you will observe your palm as (A) Right palm

(B) Left palm

(C) Inverted right palm

(D) Inverted left palm

18. A light ray is allowed to fall on the plane mirror. Now the mirror is rotated in plane of incidence, then which of the following will not change

(A) Reflected ray

(B) Normal

(C) Incident ray

(D) Angle of incidence

19. Statement I : A school girl wears a school uniform of length 1m. She requires a mirror of length 0.5 m fixed on the vertical wall to see her uniform in the mirror. Statement II : The minimum height of the mirror to be required to see the full image of man of height ‘h’ is h/2 . (A) Both Statements are true, Statement - II is the correct explanation of Statement - I. (B) Both Statements are true, Statement - II is not correct explanation of Statement - I. (C) Statement - I is true, Statement - II is false. (D) Statement - I is false, Statement - II is true.

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98 20. Statement I : The phenomenon of left appearing right and right appearing left on reflection in a plane mirror is called later inversion. Statement II : Number of images formed for an object kept in between two mirror placed at angle 90° to each other are 3. (A) Both Statements are true, Statement - II is the correct explanation of Statement - I. (B) Both Statements are true, Statement - II is not correct explanation of Statement - I. (C) Statement - I is true, Statement - II is false. (D) Statement - I is false, Statement - II is true. 21. In a school ground, a baby of monkey is running towards the plane mirror fixed to the wall, with a speed of 25 m/s, it observes that its image is moving towards it with a speed of ______ m/s.

8. Reflection by Spherical Mirrors

Reflecting surface





Silvered surface

 

Silvered surface

Reflecting surface

A highly polished plane surface is called a plane mirror. A mirror in which the reflecting surface is curved is called a spherical mirror.

opaque inner surface sphere

Convex mirror

9. Terms related to Spherical Mirrors Aperture: The width (distance) of the spherical mirror from which reflection can take place is called its aperture. It is denoted by MM Pole: The centre of a spherical mirror is called its pole. It is denoted by P. Centre of Curvature: The geometric centre of the hollow sphere of which the spherical mirror is a part is called the centre of curvature of the spherical mirror. It is denoted by C. Radius of Curvature: The radius of the hollow sphere of which the spherical mirror is a part is called the radius of curvature of the spherical mirror. In other words, the distance between the pole and centre of curvature of the spherical mirror (PC) is called its radius of curvature. It is denoted by r. M

In spherical mirrors the polished reflecting surface is a part of a hollow sphere of glass. Depending upon the nature of the reflecting surface of the mirror, spherical mirrors are of two types. Different types of Spherical Mirrors Concave Mirror: A spherical mirror whose inner hollow surface is the reflecting surface is called a concave mirror. opaque inner surface reflecting surface sphere

Concave mirror

Convex Mirror: A spherical mirror whose outer surface is the reflecting surface is called a convex mirror.

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reflecting surface

X

C

M

P

F

M' Concave mirror

X

C

P

F

M' Convex mirror

Principal Axis: The straight line passing through the centre of curvature and the pole of a spherical mirror is called its principal axis (PX). Focus: If a beam of light parallel to the principal axis falls on a concave mirror, all the rays after reflection meet at a point. This point is called the focus (F) of the concave mirror. If a beam of light parallel to the principal axis falls on a convex mirror, all the rays after reflection diverge. If the reflected rays are extended backwards, they appear to come from a point on the principal axis. This point is called the focus of the convex mirror.

Light

99 M

principal axis

F

M

P

P

focus

focus M' focal length

M'

Focal length

Focal Length: The distance between the pole (P) and focus (F) is called the focal length (f). It is denoted by f. f  PF Note: R  2 f or f 

R 2 Differences between Concave Mirror and Convex Mirror

CONCAVE MIRROR Reflecti on takes place at the concave surface (or bent in surface)

CONVEX MIRR OR Reflection takes plac e at the convex surface (or bulging out surface)

A parallel bea m of light falling on this mirror converges at a point aft er reflection

A parallel beam of light fa ll ing on this mirror appears to diverge from a point after reflect ion

It is a convergi ng mirror

It is a diverging m irror

It has a real focus

It has a virtual focus

(c) When object is placed at centre of curvature for a concave mirror the image formed is 22. Statement I : A concave mirror has a real focus. (A) Real inverted of same size Statement II : A convex mirror has a virtual focus. (B) Virtual erect of same size (A) Both Statements are true, Statement - II is (C) Real erect of same size the correct explanation of Statement - I. (D) virtual inverted of same size. (B) Both Statements are true, Statement - II is 24. Column-I Column-II not correct explanation of Statement - I. (A) Focal length of the 1) virtual image (C) Statement - I is true, Statement - II is false. spherical mirror (D) Statement - I is false, Statement - II is true. (B) Radius of curvature 2) Distance between 23. For a concave mirror when object is placed beyond centre of curvature, an image is formed between is two times of principal focus and pole principle focus and centre of curvature. The formed (C) Plane mirror 3) Real image image is real inverted and diminished image. When (D) Concave mirror 4) Half radius of object is placed at centre of curvature. The image curvature is real inverted and of same size as that of object. 5) line passing through Now answer the following questions: pole and centre of (a) What is the nature of image formed when curvature object is placed beyond centre of curvature? 25. The radius of curvature of a concave mirror is 14.26 (A) Real (B) Virtual cm. Its focal length is________ ×10–2 cm. (C) Both A & B (D) No image is formed (b) An object is placed at centre of curvature of a 26. The image formed by a convex mirror of real object is larger than the object. concave mirror whose radius of curvature is 20 cm. The image is formed from pole at a (A) When u  2f (B) When u  2f distance is__________ (C) for all values of u (D) for no value of u (A) 10 cm (B) 20 cm (C) 30 cm (D) 15 cm

Formative Worksheet

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100 27. Statement I : The radius of curvature and focal length of convex mirror are 80cm and 40cm Statement II : The relation between radius of curvature and focal length is R = 2f. (A) Both Statements are true, Statement - II is the correct explanation of Statement - I. (B) Both Statements are true, Statement - II is not correct explanation of Statement - I. (C) Statement - I is true, Statement - II is false. (D) Statement - I is false, Statement - II is true.

Conceptive Worksheet 22. The spherical surfaces that have a property of reflection and that are highly polished are called (A) Spherical mirror (B) Plane mirror (C) Plane glass plate (D) Both 1 and 2 23. Concave mirror acts as (A) Converging mirror (B) Diverging mirror (C) both 1 and 2 (D) Plane mirror 24. Convergence of concave mirror can be decreased by dipping in (A) Water (B) Oil (C) Both oil & water (D)Neither oil nor water 25. The image formed by convex mirror is always (A) Virtual (B) Erect (C) Diminished (D) Real 26. which of the following forms a virtual and erect image for all positions of a real object (A) Plane mirror (B) Convex mirror. (C) Concave mirror (D) All the above 27. The diameter of spherical mirror in which reflection takes place is called (A) Radius of curvature (B) Centre of curvature (C) Linear aperture (D) Focal length 28. For a concave mirror when object is placed at infinity the image is formed at (A) Pole (B) Principal focus (C) Centre of curvature (D) Infinity 29. The focal length of concave mirror is -40cm. Its radius of curvature is (A) – 20cm (B) +20cF (C) -80cm (D) +80cm 30. Choose the correct statements (A) Convex mirror acts as converging mirror. (B) Image formed by convex mirror is virtual (C) For a concave mirror when object is placed at infinity the image is formed at principal focus. (D) Convex mirror acts as diverging mirror. www.betoppers.com

10. Refraction of Light We know that light travels in a straight line path. This is true as long as the light rays are travelling in the same medium having the same density throughout. If, however, the light rays are made to go from one transparent medium, to another transparent medium the light rays change their direction (or bend) at the boundary separating two media. For example, when light rays travelling in air go into another medium say glass they change their direction (or bend) on entering the glass medium. Thus, the bending of the ray of light when it travels from one medium to another is called refraction of light. Optically rarer medium and optically denser medium: A transparent substance in which light travels is known as a medium. Air, glass, water, alcohol, etc. are all examples of different media. Different media are said to have different optical densities. The speed of light depends on the optical density of the medium. Therefore, light travels with different speeds in different media having different optical densities. For example, the speed of light in air is 3 × 108 m/s, where as that in glass is 2 × 108 m/s and in water it is 2.25 × 108 m/s. It is clear from these values that the speed of light is more in air but less in water and least in glass. A medium in which the speed of light is more is known as an optically rarer medium. Air is an optically rarer medium as compared to glass and water. A medium in which the speed of light is less is known as an optically denser medium. Glass is an optically denser medium than air and water. Consider once again the speed of light in air, water and glass. For example, the speed of light in water (2.25 × 108m/s) is less than that in air (3 × 108 m/s) but more than that in glass (2 × 108 m/s). So, water is optically denser than air, but it is optically rarer than glass. The optical density of a medium is not the same as its mass density. For example, turpentine is optically more dense than water as it has a greater refracting effect on light, but has a lower mass density than water.

Cause for Refraction of Light When a ray of light changes its medium, the basic change that occurs is the change in its wavelength. This change in the wavelength leads to the change in its velocity and the change in velocity is responsible for its deviation; and hence, refraction takes place.

Light

101 Rules for the deviation of the rays in another optical medium: 1. When a ray of light travels from a rarer medium to a denser medium, it bends towards the normal. 2. When a ray of light travels from a denser medium to a rarer medium, it bends away from the normal. 3. If the incident ray falls normally (or perpendicularly) on the surface of a glass slab, then there is no bending of the ray of light and it goes straight. Since the incident ray goes along the normal to the surface, the angle of incidence in this case is zero (0°) and the angle of refraction is also zero (0°), i.e., i  r and i  r  0 .

Note: 1. When light travels from one medium to another, the frequency of light does not change. However, the velocity and the wavelength of light change. 2. When a ray of light passes from rarer to denser medium it bends towards the normal and r  i .  Angle of deviation d= i-r 3. When a ray of light passes from denser to rarer medium, it bends away from the normal and r  i . Angle of deviation d=r-i 4. A ray of light travelling along the normal passes undeflected.

11. Terms related to Refraction i) Transparent Surface: The plane surface which refracts light, is called transparent surface. In diagram XY is the section of a plane transparent surface. (ii) Point of Incidence: The point on transparent surface, where the ray of light meets it, is called point of incidence. In diagram Q is the point of incidence. N1

In

Normal

P ci

i

de nt ra y

X

Y

Q fra Re cte

r

dr

Plane Transparent surface

ay

N2

R

(iii) Normal: Perpendicular drawn on the transparent surface at the point of incidence, is called normal. In diagram, N1 QN2 is the normal on surface XY. (iv) Incident Ray: The ray of light which strikes the transparent surface at the point of incidence, is called incident ray. In diagram, PQ is the incident ray. (v) Refracted Ray: The ray of light which travels from the point of incidence into the other medium, is called refracted ray. In diagram, QR is the refracted ray. (vi) Angle of incidence: The angle between the incident ray and the normal on the transparent surface at the point of incidence, is called the angle of incidence. It is represented by the symbol i. In diagram, angle PQN1 is the angle of incidence. (vii) Angle of Refraction: The angle between the refracted ray and the normal on the transparent surface at the point of incidence is called angle of refraction. It is represented by the symbol r. In diagram angle RQN2 is the angle of refraction. www.betoppers.com

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102

12. Refractive Index of a medium We know that a ray of light travels obliquely from one transparent medium into another will change its direction in the second medium. The ‘extent of the change’ in direction that takes place in a given pair of media is expressed in terms of the ‘refractive index’ of the second medium with respect to the first medium. The refractive index can be linked to the relative speed of propagation of light in different media. Light propagates with different speeds in different media. It travels the fastest in vacuum with the highest speed of 3 × 108 ms-1. Its speed reduces considerably in glass. Consider a ray of light travelling from medium 1 into medium 2 . Let i, r be the angle of incidence and angle of refraction.

 b a 

PB P1B

Real depth  appearent depth

Apparent shift

 R.D    

 the actual depth at which     object issituated     The depth at which imageof the   object isformed   when viewed from top surface 

     

= PP 1 = (PB-P 1B) = Real depth – App. depth

R.D  1  R.D.  1     

 

R.D R.D   A.D  A.D  

 If the observer (eye) is in rarer medium and the object is in denser medium; then



RD or RD =  AD AD

or RD>AD or ADRD. RD

So, to a fish, the bird appears to be higher than its actual height.

C

D

14. Dispersion of Light

N r

Med b B

A

When a narrow beam of light is allowed to pass through a prism, it splits in to seven colours. The

Med a

process of splitting of white light into its seven

i r P1

N1

constituent colours is called dispersion. The band of colours produced when white light is split up is called the spectrum. It has seven colours. They

i

are Violet, Indigo, Blue, Green, Yellow, Orange and Red.

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Light

103

F

W

The reason for dispersion is that a transparent ormative orksheet medium like a glass prism bends or deviate 28. The property that which a light a ray passes obliquely different colours of light by different amounts. The from one medium into another it bends at interface is deviation is maximum for violet and minimum for (A) Reflection (B) Refraction red. So when white light passes through a prism, (C) Interference (D) Polarization the different colours present in it get deviated by 29. Light starting from a medium of refractive index  different amounts. Hence they separate out on undergoes refraction into a medium | . If i and r emerging from the prism and form a spectrum. stand for angles of incidence and refraction When we use two identical prisms, we can get respectively, then the white light back. sin i  sin i |  |  (A) (B) sin r  sin r 

cos i |  (C) cos r 

(D)

sin i 1  sin r |

30. When a light ray passes from one medium  1  to other medium  2  . Refraction does not take place in the following situation. (A) for normal incidence (B) 1  2 Take two identical prisms and place them in such (C) 1  2 (D) 1  2 a way that the slanting faces are parallel, as shown Column - II in the figure. The prisms A will split the white light 31. Column - I (A) Point of 1) Ray of light which strikes in to its constituent colours. As the prism B the incidence the transparent surface at the angle of incidence of different colours is different point of incidence and therefore, the angles of deviation are also (B) Angle of 2) Ray of light which travels different. This results in the recombination of the incidence from the point of incident into colours by prism B. Hence, the light emerging from the other medium (C) Incident ray 3) It is represented by the prism B is white. symbol ‘i’ In the above experiment, prism A is called the (D) Refracted ray 4) Angle between the incident dispersing prism and prism B is called the ray and the normal on the recombination prism. transparent surface at the point of incidence 5) Point on transparent surface, where the ray of light meets www.betoppers.com

8th Class Physics

104 32. When a ray of light passing from air to medium. i=45º

air

d r = 30º

a) b)

Find the following: The angle of deviation ‘d’ is (A) 45° (B) 0° (C) 15° The velocity of light in the medium is (A)

3  108 m / s 2

(B)

(D) 75°

3  108 m / s 2

(C) 3  108 m / s (D) 2  108 m / s 33. Light of the wave length 7200A o in air has a wave length in glass is equal to    1.5  (A) 7200Å (B) 4800Å (C) 1080Å (D)3600Å 34. A ray of light incident on a refracting surface. If the angle of incidence 60° and the angle of refraction is 45°, then angle of deviation is (A) 30° (B) 90° (C) 15° (D) 0° 35. The velocity of light in air is 3 × 108 ms–1 and in glass is 2 × 108 ms–1.. The refractive index of glass w.r.t air is (A) 2/3 (B) 3/2 (C) 4/3 (D) 9/4 36. A fish is raising vertically to the surface of water in a lake uniformly at the rate of 2ms–1 observes a bird diving vertically towards the water at a rate 4ms–1 vertically above it. The refractive index of water is 4/3. The actual velocity of the bird is (A) 3/2 (B) 2/3 (C) 4/3 (D) 3/4

Conceptive Worksheet 31. Statement I : If refractive index of one medium is equal to refractive index of second medium, then beam does not bend at all Statement II : The bending of light does not depend on refractive indices of media. (A) Both Statements are true, Statement - II is the correct explanation of Statement - I. (B) Both Statements are true, Statement - II is not correct explanation of Statement - I. (C) Statement - I is true, Statement - II is false. (D) Statement - I is false, Statement - II is true. www.betoppers.com

32. The light ray passes in homogeneous medium (A) Straight line (B) Curved line (C) Un identified (D) Zig-zag mode 33. The bending of a light which passes from one medium to other medium is (A) Reflection (B) Refraction (C) Attraction (D) Repulsion 34. The second medium is optically denser (A) Ray bends towards normal (B) Ray bends away from normal (C) It passes in same path (D) Unidentified 35. When light travels from rarer medium to denser medium (A) Refracted ray bends towards normal (C) Refracted ray bends away from normal (C) Ray undeviated from path. (D) We cannot identified the ray. 36. When light is refracted into a denser medium, which of the following changes (A) Velocity (B) Frequency (C) Wavelength (D) Amplitude 37. If the refractive index of glass is 1.5, then the speed of light in glass is _______ 108 m / s 38. If the speed of red light is denoted by Vr and the speed for violet light by Vv . Then in vacuum. (A) Vr  Vv

(B) Vr  Vv

(C) Vr  Vv

(D) both 1 & 3

39. If refractive indices of glass and water are respectively 3 2 and 4 3 , then the refractive index of glass with respective to water is (B) 8 9 (C) 9 8 (D) 1 2 40. Statement I : When a light ray from rarer medium is refracted into denser medium its velocity decreases and wavelength decreases Statement II : Refractive index decreases with increases in temperature. (A) Both Statements are true, Statement - II is the correct explanation of Statement - I. (B) Both Statements are true, Statement - II is not correct explanation of Statement - I. (C) Statement - I is true, Statement - II is false. (D) Statement - I is false, Statement - II is true. (A) 2

Light

105

41. Statement I : When light travels from one medium to another, then frequency remains constant. Statement II : When the direction of light ray is reversed, the ray simply retraces its path , this statement is called the Principle of reversibility of light. (A) Both Statements are true, Statement - II is the correct explanation of Statement - I. (B) Both Statements are true, Statement - II is not correct explanation of Statement - I. (C) Statement - I is true, Statement - II is false. (D) Statement - I is false, Statement - II is true. 42. In the figure,

iii)

iv)

v)

O

h

vi)

A point source P is placed at a height 4m above the plane mirror in a medium of refractive index 1.2 an observer O vertically above P, outside the liquid sees P and its image in the mirror. The apparent distance between these two is________cm (A) 3.3 cm (B) 6.6 cm (C) 9.9 cm (D) 13.2 cm

by the letter C. Since there are two centre’s of curvature, we may represent them as C1 and C2 . An imaginary straight line passing through the two centres of the curvature of a lens is called its principal axis. The central point of a lens is called its optical centre. It is represented by the letter O. A ray of light through the optical centre of a lens passes without suffering any deviation. The effective diameter of the circular outline of a spherical lens is called its aperture. Lenses whose aperture is much less than its radius of curvature are called thin lenses with small aperture. Several rays of light parallel to the principal axis are falling on a convex lens. These rays after refraction from the lens are converging to a point on the principal axis. This point is called the principal focus of the lens.

15. Spherical lenses A transparent material bound by two surfaces, of which one or both surfaces are spherical, forms a lens. This means that a lens is bound by at least one spherical surface. In such spherical lenses, the other surface would be plane. A lens may have two spherical surfaces, bulging outwards. Such a lens is called a double convex lens. It is simply called a convex lens. It is thicker at the middle as compared to the edges. Convex lens converges light rays. Hence it is called converging lens. Similarly, a double concave lens is bounded by two spherical surfaces, curved inwards. It is thicker at the edges than at the middle. Such lenses diverge light rays and are called diverging lenses. A double concave lens is simply called a concave lens. Terms related to lenses i) A lens has two spherical surfaces. Each of these surfaces forms a part of a sphere. ii) The centers of these spheres are called centres of curvature of the lens. The centre of curvature of a lens is usually represented

vi)

Several rays of light parallel to the principal axis are falling on a concave lens. These rays after refraction from the lens, are appearing to diverge from a point on the principal axis. This point is called the principal focus of the concave lens. If you pass parallel rays from the opposite surface of the lens, you will get another principal focus on the opposite side. Letter F is usually used to represent principal focus. However, a lens has two principal foci. They are represented by F 1 and F 2 . The distance of the principal focus from the optical centre of a lens is called its focal length. The letter f is used to represent the focal length. www.betoppers.com

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16 .

Image formation by lenses We can represent image formation by lenses using ray diagrams. Ray diagrams will also help us to study the nature, position and relative size of the image formed by the lenses. For drawing ray diagrams in lenses, we consider any two of the following rays. (i) A ray of light from the object, parallel to the principal axis, after refraction from a convex lens, passes through the principal focus on the other side of the lens, as shown in fig.(a). In case of a concave lens, the ray appears to diverge from the principal focus located on the same side of the lens, as shown in fig.(b)

(ii) A ray of light passing through a principal focus after refraction from a convex lens will emerge parallel to the principal axis. This is shown in Fig (a) below. A ray of light appearing to meet at the principal focus of a concave lens, after refraction, will emerge parallel to the principal axis. This is shown in Fig(b) below.

O

The image formed has the following characteristics: 1. The image is virtual. 2. The image is erect. 3. The image is enlarged (or magnified). 4. The image is formed on the same side of the lens, behind the object. Applications: When the object is placed between the optical centre of the convex lens and the principal focus, an erect and magnified image is formed. It is used as a magnifying glass (or simple microscope) in the following cases: 1.It is used by the palmists to study the lines of the palm. 2.It is used for seeing clearly the weave patterns of the cloth. 3.It is used as reading glass to read the fine print (like letters). 4.It is used to study biological specimens such as parts of the flower, etc. Case II : When the object is placed at the focus of a convex lens (object at F1 )

o

(iii) A ray of light passing through the optical centre of a lens will emerge without any deviation. This is illustrated in fig (a) and (b) below.

(a)

(b)

Image formation by a convex lens Case I : Image formed by a convex lens when the object is placed between the optical centre and the principal focus (object between O and F1 ) www.betoppers.com

The image formed has the following characteristics: 1. The image is real. 2. The image is inverted. 3. The image is highly enlarged (or magnified). 4. The image is formed at infinity, on the other side of the lens. Applications: The above case is used in making searchlights and spotlights in theatres. In a searchlight, a powerful lamp is placed at the principal focus of a convex lens, which produces a powerful parallel beam of light. Case III: When the object is placed between F1 and 2F1. The image formed has the following characteristics: 1. The image is real. 2. The image is inverted. 3. The image is enlarged (or magnified). 4. The image is formed beyond 2F 2 , on the other side of the lens .

Light

107

2

Case VI : When the object is at infinity (such that the rays coming from it are parallel to the principal axis of the convex lens)

1

Applications: This type of image formation is used in film and slide projectors, when enlarged image of a small slide (or film) is formed on a screen. Case IV : When the object is at 2F 1

The image formed has the following characteristics: 1. The image is real. 2. The image is inverted. 3. The image is diminished to a point (highly diminished). 4. The image is formed at F2, on the other side of the lens.

The image formed has the following characteristics: 1. The image is real. 2. The image is inverted. 3. The image is of the same size as the object. 4. he image is formed at 2F2, on the other side of the lens.

Applications: This type of image formation is used in a burning glass. When the rays of the sun are focussed by the lens of the burning glass on a piece of paper, the paper catches fire. The lens of the burning glass focuses the heat radiations on the paper as a result of which, the temperature of the paper rises to its ignition temperature and it catches fire.

Applications: This type of image formation is used in Case VII : When the object is at infinity (such that the terrestrial telescope, for erecting the inverted image rays coming from it are not parallel to the principal formed by the objective lens of the telescope. axis of the convex lens) Case V: When the object is beyond 2F 1

The image formed has the following characteristics: 1. The image is real. 2. The image is inverted. 3. The image is highly diminished. 4. The image is formed on the focal plane on the other side of the lens .

The image formed has the following characteristics:

parallel rays from a distance object B

1. The image is real. 2. The image is inverted. 3. The image is diminished (i.e. smaller than the object). 4. The image is formed between F2 and 2F2, on the other side of the lens . Applications: This type of image formation is used in a photographic camera, where a small, real and inverted image of an object is formed on the film.

F1

F2

image (real, inverted A and highly diminished)

Applications: This type of image formation is used for an objective lens in a telescope. It forms small, inverted image of far off objects in the focal plane in front of the eye lens.

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Images formed by a Convex lens for various positions of the object

Images formed by a concave lens:

1. The image is virtual.

Case I : When the object is located at infinity (such that the rays coming from it are parallel to the principal axis)

2. The image is erect. 3.The image is highly diminished. 4. The image is formed in the focal plane on the same side of the lens as the object . Applications: This type of image formation is used in Galilean telescope, where concave lens acts as an eye lens. Case Ill : When the object is anywhere between the optical centre (O) and infinity.

The image formed has characteristics:

the following

1. The image is virtual. 2. The image is erect. 3. The image is highly diminished to a point size. 4. The image is formed at FI; on the same side of the lens as the object. Case II : When the object is located at infinity [such that the parallel rays from it are falling obliquely (not parallel) to the principal axis Parallel rays from a distant object

The image formed has the following characteristics : 1. The image is virtual 2. The image is erect. 3. The image is diminished.

A

4. The image is formed between the optical centre (O) and the principal focus (F1) on the same side of the lens.

B F1 Image (virtual, erect and highly Diminished)

O

The image formed has characteristics: www.betoppers.com

the following

Applications: This type of image formation is used mainly in spectacles for the correction of short­sightedness or myopia.

Light

109 Images formed by a concave lens for various positions of the object:

(C)

Formative Worksheet 37.

Which of the following diagrams correctly represents the ray of light passing through the optical centre? (A)

(D)

39. (B)

An object AB is placed in front of a convex lens at its principal focus as shown in the Figure.

Which of the ray diagram given below correctly depicts the refraction through the lens L?

(C)

(A)

(D)

(B)

(C) 38.

A ray of light is incident on a concave lens parallel to its principal axis. The path of the refracted ray is correctly depicted by (A)

(D) 40. 41.

(B) 42.

(A) Which type of lens is a converging lens? (B) Which type of lens is a diverging lens? What is the focal length of a biconvex lens whose center of curvature is located at a distance of 20 cm from the optic centre? (A) An image is formed on the other side of the lens. Name the type of image. (B) An image is formed on the same side of the lens. Name the type of image. www.betoppers.com

110 43.

44.

8th Class Physics Converging action of a lens result in _____ type of image. If the refracted rays are divergent, then it results in _____ type of image. 43.

Conceptive Worksheet

(A) An image of an object by a convex lens is formed at infinity. Name the position of the object and nature of image. (B) The nature of image formed by a convex lens is real, inverted & point diminished. Locate the position of object and image.

44.

(C) Name the position and nature of image formed by a convex lens for an object placed between 45. F1 and 2F1. 45.

(A) A diminished ( not point sized) image is formed by convex lens, locate the position of the object.

A double convex air bubble in water would behave as a (A) Convergent lens (B) Circle (C) Divergent lens (D) Plane mirror The graph drawn with object distance along xcoordinate and image (real) distance as ycoordinate for a convex is a (A) Straight line (B) Circle (C) Parabola (D) Rectangular hyperbola A layered lens as shown in the figure is made of two materials indicated by different shades. A point object is placed on its axis. The lens will form.

(B) What is the magnification produced by a convex lens for an object placed at 2F1? 46.

(A) A real image magnification produced by a lens is greater than 1 . Predict the position of the object. (B) A virtual image magnification produced by a lens is greater than 1 . Predict the position of the object.

46.

(C) Only virtual images are formed by _____ lens. (D) The position of the object for a concave lens is anywhere between ‘O’ to infinity. What is the nature of the image?

47.

(E) A lens can form both real and virtual images. Name the type of lens. 47.

(A) The images formed by a lens is always on the 48. same side of the lens as object. Name the type of lens. (B) A real image and magnified image is formed by a lens. If the object distance is u and image distance is v, then find the distance between object and image. (C) An image formed by a concave lens. If the object distance is u and image distance is v, then find the distance between object and 49. image. (D) A virtual image is formed by a converging type of lens. If the object distance is ‘u’ and image distance is ‘v’, then find the distance between object and image.

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(A) One image (B) Two images (C) Five images (D) Three images If a convergent beam of light passes through a diverging lens,then the result (A) May be a convergent beam (B) May be a divergent beam (C) May be a parallel beam (D) Must be a parallel beam A simple magnifying glass consists of a (A) Convex lens (B) A concave lens (C) Convex lens of small focal length (D) Convex lens of large focal length Which of the following is/are correct (A) Convex lens is used in making search lights and spot lights in theatres (B) Convex lens is used as a magnifying glass (or) simple microscope (C) Convex lens is used in terrestrial telescope for erecting the inverted image formed by the objective lens of the telescope (D) Concave lens is used in spectacles for the correction of hyper metropia A convergent lens of focal length 20cm and made of a material with refractive index 1.1 is immersed in water the lens behaves as a (A) Converging lens of focal length 20 cm (B) Converging lens of focal length less than 20cm (C) Converging lens of focal length more than 20cm (D) Diverging lens

Light 50. Which of the following form(s) a virtual and erect image for all positions of the object ? (A) Convex lens (B) Concave lens (C) Convex mirror (D) Concave mirror 51. In case of convex lens, as object is coming closer to the lens the image will _____ (A) Move away from lens (B) Move towards the lens (C) Move up (D) Not move 52. A converging lens is used to form an image M on a screen, when the upper half of the lens is covered by an opaque screen (A) Half of the image will disappear (B) Complete image will be formed (C) Intensity of image will increase (D) Intensity of the image will decrease 53. A thin lens produces an image of the same size as the object. Then from the optical centre of the lens, the distance of the object is (A) Zero (B) 4f (C) 2f (D) f/2

d)

e)

f)

g)

17. Human Eye parts and Functions Our eyes enable us to see the beautiful world around us. The most important part of our eyes is a convex h) lens inside it that is made of living cells. The human eye is like a camera having a lens on one side and a sensitive screen called the retina on the other. The essential parts of a human eye are shown in figure. i)

j)

a)

b)

c)

Sclerotic: It is the outermost converging of the eye ball. It is made of white tough fibrous tissues. Its function is to house and protect vital internal parts of eye. Cornea: It is the front bulging part of the eye. It is made of transparent tissues. Its function is to act k) as a window to world. i.e., to allow the light to enter in the eye ball. Choroid: It is a grey membrane attached to the sclerotic from the inner side. Its function is to darken the eye from inside and, hence, prevent any internal reflection.

111 Optic Nerve: It is a bundle of approximately 70,000 nerves originating from brain and entering eye ball from behind. Its function is to carry optical message (visual messages) to the brain. Retina: The optic nerve on entering the ball, spreads like a canopy, such that each nerve and attaches itself to the choroid. The nerve endings form a hemi-spherical screen called retina. These nerve endings on the retina are sensitive to visible light. On the retina are two important areas which we will discuss separately. The function of retina is to receive the optical image of the object and then convert it to optical pulses. These pulses are then sent to the brain through optic nerve. Yellow spot: It is a small area facing the eye lens. It has high concentration of nerve endings and is slightly raised as well as slightly yellow in colour. Its function is to very clear image by sending a large number of optical pulses to brain. Blind spot: It is a region on the retina, where the optic nerve enters the eye ball. It has no nerve endings and hence, is insensitive to the light. It does not seem to have any function. Any image formed on this spot is not visible. Crystalline lens: It is a double convex lens made of transparent tissues. It is held in position by a ring of muscles, commonly called ciliary muscles. Its function is to focus the images of different objects clearly on the retina. Ciliary muscles: It is a ring of muscles which holds the crystalline lens in position. When these muscles relax, they increase the focal length of the crystalline lens and vice versa. Its function is to alter the focal length of crystalline lens so that the images of the objects, situated at different distances, are clearly focussed on the retina. Iris: It is a circular diaphragm suspended in front of the crystalline lens. It has a tiny hole in the middle and is commonly called pupil. It has tiny muscles arranged radially around the pupil. These muscles can increase or decrease the diameter of the pupil. The iris is heavily pigmented. The colour of eyes depends upon colour of pigment. The function of iris is to control the amount of light entering in eye. This is done by increasing or decreasing the diameter of pupil. Vitreous humour: It is a dense jelly like fluid, slightly grey in colour, filling the part of eye between crystalline lens and retina. Its function is (i) to prevent the eye ball from collapsing due to change in atmospheric pressure (ii) in focussing the rays clearly on the retina. www.betoppers.com

8th Class Physics

112 l)

Aqueous humour : It is a watery, saline fluid, filling the part of the eye between the cornea and the crystalline lens. Its function is (i) to prevent front part of the eye ball from collapsing with the change in atmospheric pressure (ii) to keep the cornea moist.

18. Power of Accomodation Have you wondered why the eye is able to focus the images of objects lying at various distances? It is made possible because the focal length of the human lens can change i.e., increase or decrease, depending on the distance of objects. It is the ciliary muscles that can modify the curvature of the lens to change its focal length.

The range of vision of a normal eye is from 25 cm to infinity. Have you ever thought why animals’ eyes are positioned on their heads? This is because it provides them with the widest possible field of view. Our eyes are located in front of our face. One eye provides 150° wide field of view while both eyes simultaneously provide 180° wide field of view. It is the importance of the presence of two eyes as both eyes together provide the three-dimensional depth in the image.

19. Defects of Vision The loss of power of accommodation of an eye results in the defects of vision. There are thr ee defects of vision called refractive defects. They are myopia, hypermetropia, and presbyopia. In this section, we will learn about these defects of vision in detail.

1. To see a distant object clearly, the focal length of the lens should be larger. For this, the ciliary muscles relax to decrease the curvature and thereby increase the focal length of the lens. Hence, the lens becomes thin. This enables you to see the distant object clearly.

To see the nearby objects clearly, the focal length of the lens should be shorter. For this, the ciliary muscles contract to increase the curvature and thereby decrease the focal length of the lens. Hence, the lens becomes thick. This enables you to see the nearby objects clearly. The ability of the eye lens to adjust its focal length accordingly as the object distances is called power of accommodation.  The minimum distance of the object by which clear distinct image can be obtained on the retina is called least distance of distinct vision. It is equal to 25 cm for a normal eye. The focal length of the eye lens cannot be decreased below this minimum limit of object distance.  The far point of a normal eye is infinity. It is the farthest point up to which the eye can see objects clearly.

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Myopia (short sightedness) Myopia is a defect of vision in which a person clearly sees all the nearby objects, but is unable to see the distant objects comfortably and his eye is known as a myopic eye. A myopic eye has its far point nearer than infinity. It forms the image of a distant object in front of its retina as shown in the figure.

Myopia is caused by i.

increase in curvature of the lens

ii.

increase in length of the eyeball

Since a concave lens has an ability to diverge incoming rays, it is used to correct this defect of vision. The image is allowed to form at the retina by using a concave lens of suitable power as shown in the given figure.

Light

2.

113

Hypermetropia (Long sightedness) Hypermetropia is a defect of vision in which a person can see distant objects clearly and distinctively, but is not able to see nearby objects comfortably and clearly. So, now you can easily represent the problem with a hypermetropic eye with the help of a diagram. It is shown in the given figure.

A hypermetropic eye has its least distance of distinct vision greater than 25 cm.

Hypermetropia is caused due to i. reduction in the curvature of the lens ii. decrease in the length of the eyeball Since a convex lens has the ability to converge incoming rays, it can be used to correct this defect of vision, as you already have seen in the animation. The ray diagram for the corrective measure for a hypermetropic eye is shown in the given figure.

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8th Class Physics

114

Formative Worksheet

 a, ii  c, iii  b, iv  d  a, ii  b, iii  c, iv  d 48. Which of the following parts of the eye has  d, ii  c, iii  a, iv  b enormous number of photo-sensitive cells?  d, ii  c, iii  b, iv  a (A) Iris (B) Pupil 54. Four statements about the blind spot of the human (C) Retina (D) Cornea eye: 49. The focal length of the eye lens is adjusted by the I It is devoid of photosensitive cells. (A) vitreous humour (B) aqueous humour II No image is formed at the blind spot. (C) ciliary muscles (D) optic nerves III It connects the optic nerves to the retina. 50. In the human eye, the function of the optic nerves IV Its malfunctioning makes a person blind. is to (A) adjust the focus of the eye lens (B) transmit electrical signals to the brain (C) control the number of rod and cone cells (D) convert light signals into electrical signals 51. The primary function of the crystalline lens in the human eye is to (A) reduce the intensity of light (B) focus the incident light rays (C) adjust the size of the pupil (D) filter out the dust particles 52. The given diagram is that of the human eye.

(A) i (B) i (C) i (D) i

55.

56.

In the given diagram, the parts labelled as I and II are respectively the (A) eye lens and the cornea (B) pupil and the eye lens (C) cornea and the pupil (D) pupil and the iris 53.

Part of eye Iris

a

ii

Cilliary muscle

b

iii Pupil

c

iv

d

Controls the amount of light entering into the eye Senses the real image Controls the power of eye lens Controls the size of pupil

The alternatives in the given table can be correctly matched as www.betoppers.com

58.

Function

i

Retina

57.

59.

60.

Which of the given statements is incorrect? (A) I (B) II (C) III (D) IV When we look at a distant object, the ciliary muscles__i__, making the eye lens__ii__. This__iii__the focal length of the eye lens so that the image of that object gets focused on the retina. The information in which alternative completes the given statements? (A) i- get released, ii-thicker, iii-decreases (B) i-contract, ii- thicker, iii-increases (C) i-get released, ii-thinner, iii-increses (D) i-contract, ii-thinner, iii- decreses Which of the following pairs of eye parts is responsible for controlling the amount of light entering into the eye? (A) Iris and pupil (B) Cornea and eye lens (C) Cilliary muscles and iris (D) Cilliary muscles and pupil Tina has a defect in her eyes. She is not able to contract her iris properly. As a result of the defect in her eyes, Tina will not be able to see (A) in dim light (B) near objects (C) in bright light (D) distant objects The range of vision of a normal human eye lies between A) 25 m and infinity B) 2.5 m and 5 km C) 2.5 cm and 5 km D) 25 cm and infinity Myopia is a defect of vision caused by A) a decrease in the retinal distance from the lens B) a decrease in the diameter of the eyeball C) an increase in the thickness of the lens D) an increase in the curvature of the lens Shyam views a certain object and finds that it is not properly visible to him. The image of the said object is formed by his eye lens 3 cm away from the pupil of his eye.Shyam can correct the defect in his vision by using A) convex lens B) bifocal lens C) concave lens D) cylindrical lens

Light

115

Conceptive Worksheet 54. Which of the following eye parts is not in contact with the eye lens? (A) Vitreous humour (C) Cornea

(B) Ciliary fibres (D) Iris

55. Which part of the human eye contracts and relaxes in order to adjust the focal length of the lens? (A) Iris (C) Blind spot

(B) Pupil (D) Ciliary muscles

56. The ciliary muscles of the human eye contract and relax in order to (A) adjust the size of the eyeball (B) adjust the focal length of the lens

Summative Worksheet 1.

2. 3.

4.

5.

(C) control the amount of light entering the eye (D) prevent foreign elements from entering the eye 57. In the human eye, light is converted into electrical signals by the (A) retina (C) blind spot

(B) cornea (D) optical nerves

6.

58. The image formed by the eye lens on the retina is (A) real, inverted, and large (B) virtual, erect, and large

7.

(C) real, inverted, and diminished (D) virtual, erect, and diminished 59. While sitting on the first bench, Mutum finds it difficult to read what is written on the blackboard. However, he can clearly see the same when he sits on the last bench.

8.

It can be concluded that Mutum is suffering from

9.

A) myopia C) astigmatism

B) cataract D) hypermetropia

10. 60. Consider three statements regarding the defects of eye.I. If the size of the eye ball decreases, then the person suffers from myopia.II. If the near point of 11. eye becomes greater than 25 cm, then the person is said to suffer from hypermetropia.III. A person, at older age, suffering from presbyopia uses a bifocal lens for correction. Among the given statements, A) only statement I is correct B) onlystatement II is correct

12.

C) both statements I and II are correct D) both statements II and III are correct

13.

When a light ray is incident normally on the surface of a plane mirror, the reflected ray deviates through an angle of __________. What is the sum of angle of incidence , angle of reflection and angle of deviation equals to ? If the angle of deviation after reflection of a light ray is ‘d’, then find (a) Angle of incidence (b) Angle of reflection (c) Angle of glancing Keeping the incident ray constant, if a plane mirror is rotated through an angle  , about an axis lying in its plane, then the reflected ray turns through an angle ___________. The angle between an incident ray and reflected ray for a given reflection is x. If the mirror is rotated through an angle x/2, keeping the incident ray constant, then find the angle between incident ray and reflected ray after rotation. Consider both clock wise and anticlock wise direction. Keeping the plane mirror fixed, if the incident ray is rotated through an angle  , then the angle through which the reflected ray rotates is _________. A plane mirror lies face up, making an angle 150 with the horizontal. A ray of light shines down vertically on the mirror What is the angle of incidence? What will be the angle between the reflected ray and the horizontal be? If the reflected beam is convergent, __________ type of image is formed and if the reflected beam is divergent, then _________ type of image is formed. A light bulb is placed between two mirrors inclined at an angle of 600. The number of images formed are ______. Two plane mirrors are kept at an angle of 350 to each other. A ray of light is incident on one of them at an angle of 270. The angle of deviation is ______. (i) A person approaches a plane mirror with a velocity ‘u’. He approaches his image with a velocity of ______. (ii) The distance between object and image formed by a plane mirror is 60 cm. If the object is moved away from the plane mirror by 15 cm, then the distance between object and image is _____. Can a ray move from one medium to another medium, without bending. If so explain. What is speed of light in diamond of refractive index 2.4, if the velocity of light in air is 3  108 m/s?. www.betoppers.com

8th Class Physics

116 14. An object 2.5 m high is placed in front of a pin hole camera . If the width of camera is 24 cm and size of image formed is 1.2 cm, find the distance of the 1. object from pin hole camera.

HOTS Worksheet

An object is placed at a distance of 20 cm infront of a plane mirror. If the mirror moves towards the object by a distance of 20 cm, then find the 15. The magnification of an image formed by a pin hole displacement of the image. camera is 0.0005. If the object is at a distance of 500 m, calculate the distance of image from the 2. An object is placed at a certain distance infront of a plane mirror. If both object and plane mirror moves pinhole. towards each other through a distance of 15 cm, 16. In the above case, if the size of image is 2.4 cm, then find the distance through which the image calculate the height of the object. moves. m = 0.0005 Size of the image, I = 2.4 Size of the 3. If both object and mirror moves away from each other with a speed of 20 m/s, then find object, O = ? (a) The speed of the image 17. A ray of light is moving from air into medium. If the (b) Speed of the image with respect to object angle of incidence in air is 450 and angle of refraction (c) Speed of the image with respect to mirror. is 300, then find the angle of deviation. 4. Two mirrors are inclined at a certain angle  . If a 18. A ray of light is moving from air into medium. If the light ray is incident on the first mirror parallel to the angle of incidence in air is 450 and angle of refraction second mirror and reflected from the second parallel is 300, then find the refractive index of the medium. to the first mirror, then find the value of  . 19. What is meant by power of accommodation of 5. A man 180 cm high stands in front of a plane mirror. the eye? His eyes are at a height of 170 cm from the floor. Then the minimum length of plane mirror for him to 20. A person with a myopic eye cannot see objects see his full image is ___. beyond 1.2 m distinctly. What should be the type 6. A concave mirror produces a real image twice the of the corrective lens used to restore proper vision? size of the object when placed at a distance of 22.5 21. What is the far point and near point of the human cm from it. At what distance from the mirror should eye with normal vision? the object be placed so that the real image becomes three times the size of the object? 22. A student has difficulty reading the blackboard 7. Two plane mirrors are inclined to each other at an while sitting in the last row. What could be the angle of 700. A ray is incident on one mirror at an defect the child is suffering from? How can it be corrected? angle  . The rays reflected from this mirror fall on the second mirror from where it is reflected parallel 23. The human eye forms the image of an object at its to the first mirror . What is the value of  ? (A) cornea (B) iris 8. Two plane mirrors are inclined at an angle of 600 as (B) pupil (D) retina shown in the figure. A ray of light parallel to Mirror1 strikes the Mirror 2. At what angle will the ray 24. The change in focal length of an eye lens is caused finally emerge? by the action of the 9. A needle 1.5 cm is placed at a distance of 20 cm (A) pupil (B) retina from a concave mirror of focal length 15 cm. At what distance from the mirror should a screen be (C) ciliary muscles (D) iris placed in order to receive a sharp image. 25. Why is a normal eye not able to see clearly the 10. Two plane mirrors are placed parallel to each other. objects placed closer than 25 cm? The distance between the mirrors is 10 cm. An 26. What happens to the image distance in the eye object is placed between the mirrors at a distance when we increase the distance of an object from of 4 cm from one of them, say M1. What is the the eye? distance between the first image at M 1 and the second image formed at M2 ? 11. A plane mirror is moved away from a stationary object with a speed of 20 cm/s. What is the speed of the image? www.betoppers.com

Light 12. Show that the length of a plane mirror in which a person can see his full image is half the height of the person. 13. (i) We have a spherical mirror. The image of an object placed in front of this mirror is virtual. If the position of the object is changed, the image still remains virtual and erect. Is the spherical mirror concave or convex? (ii) We have a mirror. The image of an object placed in front of the mirror always forms on the other side of the mirror. If the position of the object is changed, the nature and the magnification remains unchanged. Identify the type of the mirror. (iii) The image of an object placed in front of a spherical mirror is virtual. As the object distance is decreased infinity to focal length, the size of the inverted image increases. Is the spherical mirror concave or convex? 14. A convex and a concave mirror of radii 10 cm each are placed facing each other and 15 cm apart. An object is placed mid way between them. Find the position, nature and size of the final image for reflections first at concave mirror and then at convex mirror. 15. A convex lens of focal length 15 cm is placed in front of a convex mirror coaxially at a distance of 5 cm from the pole of the mirror. When an object is placed at a distance of 20 cm from the lens, it is found that the image coincided with the object. Find the radius of curvature of the mirror. 16. While sitting on the first bench, Mutum finds it difficult to read what is written on the blackboard. However, he can clearly see the same when he sits on the last bench. It can be concluded that Mutum is suffering from A) myopia B) cataract C) astigmatism D) hypermetropia 17. In the human eye, light is converted into electrical signals by the A) retina B) cornea C) blind spot D) optical nerves 18. Which of the following eye parts is not in contact with the eye lens? A) Vitreous humour B) Ciliary fibres C) Cornea D) Iris 19. Which of the following parts of the eye has enormous number of photo-sensitive cells? A) Iris B) Pupil C) Retina D) Cornea 20. Which part of the human eye contracts and relaxes in order to adjust the focal length of the lens? A) Iris B) Pupil C) Blind spot D) Ciliary muscles

117

IIT JEE Worksheet 1.

The number of images of an object placed between two plane mirrors at 450 to each other is (A) 2 (B) 7 (C) 8 (D) 10 2. An object is placed 40 cm in front of a convex mirror of radius of curvature, 20 cm. The image (A) Is real and 8 cm behind the mirror (B) Is real and 8 cm in front of the mirror (C) Is virtual and 8 cm in front of the mirror (D) Is virtual and 8 cm behind the mirror 3. A long vertical pin placed at 40 cm, in front of a concave mirror gives an image at the same position. The focal length of the mirror in meters is (A) 0.2 m (B) 0.4 m (C) 0.6 m (D) 0.8 m 4. An object 5 cm high is placed at a distance of 15 cm from a concave mirror of radius of curvature 20 cm. The height of the image in cm is (A) 5 cm (B) 20 cm (C) 15 cm (D) 10 cm 5. If the object is placed 10 cm in front of a convex mirror of radius of curvature 20 cm, the magnification is (A) 2 (B) 1 (C) 1/2 (D) 0 6. If a person wants to see his full size image, the minimum height of a plane mirror should be (A) Two times his height (B) Equal to his height (C) Half his height (D) One fourth his height 7. If the radius of curvature of a concave mirror is R, its focal length is (A) 2R (B) R (C) R/2 (D) 2/R 8. A straight stick partially immersed under water obliquely appears to be bent due to (A) Reflection (B) Refraction (C) Total internal reflection (D) Parallax error 9. The angle of incidence is the angle made by the incident ray with (A) The surface (B) Normal to the surface (C) The reflected ray (D) The refracted ray 10. The refractive index of glass is 1.5 and speed of light in air is 3  1010 cm/s, the speed of light in glass is (A) 4.5  1010 cm/s (B) 2  1010 cm/s 10 (C) 0.5  10 cm/s (D) 1.5  1010 cm/s 11. The refractive index of medium can be defined in terms of velocity of light as _____. www.betoppers.com

8th Class Physics

118 12. Velocity of light in vacuum is (A) 13  106 m/s (B) 3  1010 m/s 8 (C) 3  10 m/s (D) 3  105 m/s 13. When a light travels from one medium to another medium which are separated by sharp boundary, the characteristic which does not change is (A) Velocity (B) Wave length (C) Frequency (D) Amplitude 14. A plane mirror reflecting a ray of incident light is rotated through an angle  , about an axis through the point of incidence, in the plane of the mirror perpendicular to the plane of incidence. Then (1) The reflected ray does not rotate

15.

16.

17.

18.

19.

20.

21. Let the incident ray be fixed. Now, if a plane mirror is rotated through an angle of 200 about an axis lying in its plane, then the reflected ray turns through an angle (A) 600 (B) 400 (C) 100 (D) 300 22. A clock hung on a wall of a hall has marks instead of numerals on its dial. On the opposite wall, there is a plane mirror and the image of the clock in the mirror indicates the time 7.40. hence, the time on the clock is (A) 7.40 (B) 4.20 (C) 5.40 (D) 10.7 23. A ray of light that falls on an eye first passes through the A) pupil B) cornea C) eye lens D) aqueous humour (2) The reflected ray rotates through an angle  24. The focal length of the eye lens is adjusted by the (3) The reflected ray rotates through an angle 2  A) vitreous humour B) aqueous humour (4) The incident ray is fixed C) ciliary muscles D) optic nerves (A) Only (1) is correct 25. The ciliary muscles of the human eye contract and (B) (1) and (3) are correct relax in order to (C) (3) and (4) are correct A) adjust the size of the eyeball B) adjust the focal length of the lens (D) (2) and (4) are correct C) control the amount of light entering the eye A ray of light is incident on a plane mirror at an 0 D) prevent foreign elements from entering the eye angle of incidence of 30 . The deviation produced 26. Tina has a defect in her eyes. She is not able to by the mirror is 0 0 0 0 contract her iris properly. (A) 30 (B) 60 (C) 90 (D) 120 As a result of the defect in her eyes, Tina will not be If the velocity of light in vacuum is c cm/s, then the able to see velocity of light in a medium of refractive index 1.5 A) in dim light B) near objects is ____ C) in bright light D) distant objects (A) 1.5 c (B) c 27. The image formed by the eye lens on the retina is (C) c/1.5 (D) can have any velocity A) real, inverted, and large The light appears to travel in straight line because B) virtual, erect, and large (A) It consists of small particles C) real, inverted, and diminished (B) The velocity of light is very large D) virtual, erect, and diminished (C) The wavelength of light is very small 28. Myopia is a defect of vision caused by (D) Light is reflected by surroundings A) a decrease in the retinal distance from the lens The number of images observable between two B) a decrease in the diameter of the eyeball parallel mirrors is C) an increase in the thickness of the lens D) an increase in the curvature of the lens (A) 12 (B) 4 (C) 16 (D) infinite  A plane mirror is approaching you at 10 cm per second. You can see your image in it. At what speed will your image approaches you. (A) 5 cm/s (B) 25 cm/s (C) 20 cm/s (D) 15 cm/s A light bulb is placed between two plane mirrors inclined at an angle of 600. The number of images formed are (A) 6 (B) 12 (C) 5 (D) 4

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• • • • •

Chapter - 8

Learning Outcomes

Magnetism

By the end of this chapter, you will understand Natural and Artificial Magnets • Molecular theory of Magnetism Properties of a Magnet • Magnetic Field General Definitions related to a Bar Magnet • Magnetic Lines of Force Methods of making Artificial Magnets • Magnetic Induction Temporary and Permanent Magnets

1. Introduction People have known about magnets since ancient times. The first known magnets were hard black stones called ‘loadstones’. Throughout the middle Ages many people believed that loadstones had medical powers. During this period, it was discovered that a loadstone would point to the north. Magnetism is the term used to describe the phenomenon by which forces attract some bodies to each other and to the forces which act between electric currents. Magnetism is an important force in nature that makes possible everyday happenings like talking on the telephone, watching television, listening to taped music and storing data (sound and images) on computer discs. Compasses made with magnets help navigators guide ships safely. Without magnetism, it would not be possible to produce large amounts of electricity and neither could we use electricity to do all the jobs it does. William Gilbert a physician was the first to use the terms magnetic pole and electrical force. It was he who suggested that the Earth’s magnetism could be explained if the Earth was likened to a huge bar magnet. Real progress in understanding magnetism came after the relationship between electricity and magnetism was established by Hans Christian Orested in 1820. What is Magnetism? Just like when the Greeks of the old times discovered the first naturally occurring magnetic stones, or natural magnets, you have been observing a property of matter called magnetism. Magnetism is the force of attraction or repulsion in and around a material.

Magnetism is present is all materials but at such low levels that it is not easily detected. Certain materials such as magnetite, iron, steel, nickel, cobalt and alloys of rare earth elements, exhibit magnetism at levels that are easily detectable.

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What is a Magnet? A magnet is any piece of material that has the property of attracting iron (or steel). Magnetite, also known as lodestone, is a naturally occurring rock that is a magnet. This natural magnet was first discovered in a region known as magnesia and was named after the area in which it was discovered. Magnetism may be naturally present in a material or the material may be artificially magnetized by various methods. Magnets may be permanent or temporary. After being magnetized, a permanent magnet will retain the properties of magnetism indefinitely. A temporary magnet is a magnet made of soft iron, that is usually easy to magnetize; however, temporary magnets lose most of their magnetic properties when the magnetizing cause is discontinued. Permanent magnets are usually more difficult to magnetize, but they remain magnetized. Materials which can be magnetized are called ferromagnetic materials. We will talk more about making a magnet later on.

8th Class Physics

120 Magnetic and Non-Magnetic Materials SUBSTANCES

Magnetic Substances Substances which are influenced or affected by a magnet is called a magnetic substances. Ex: Iron, steel, nickel, Cobalt etc.

Non-Magnetic Substances Substances which are neither attracted nor repelled by a magnet are called non-magnetic substances. These cannot be magnetised. Ex: Paper, wood, glass etc.

Types of Magnetic Substances We are familiar with substances which are attracted by magnets. But it is interesting to note that there are certain substances which are repelled by a magnet. So, based on the property of attraction or repulsion, magnetic substances can be broadly divided into three classes, namely Ferromagnetic. Paramagnetic and Diamagnetic substances.

Ferro Magnetic The substances which are strongly attracted by a magnet and can be easily magnetised to form strong magnets are called Ferro magnetic substances.

Para Magnetic

Dia Magnetic

The substance which are very feebly attracted by a magnet are called Para magnetic substances.

The substances which are feebly repelled by a magnet are called Dia magnetic substances.

Ex: Iron, steel, nickel, cobalt Ex: Paltinum, Zinc, manganese, Ex: Copper, gold, bismuth, and their alloys etc. wood, aluminium, plastic etc. antimony, water etc.

2. Natural and Artificial Magnets Magnets are four types, viz ..Natural , Artificial , Temporary and Permanent magnets. Natural Magnets A substance which possesses the property of attracting small pieces of irons, steel, cobalt, nickel is called a magnet and this property is called attractive property. The magnets found in nature are called natural magnet. They are found in quite irregular and odd shapes. Example of a natural magnet is a ‘lodestone’.

Load stone

When a piece of lodestone is suspended by a fine thread it comes to rest in north-south direction. This property is called directive property. This property was used by sailors to navigate their ships. Hence a natural magnet is also called lodestone or leading stone (from old English ‘lad for way). The modern name Magnetite derives its origin from the locality

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in which the mineral was first found in large quantities, namely, the town of Magnesia, in Asia Minor. Magnetite is found occurring naturally in many parts of the world. It is black oxide of iron with the chemical formula Fe3O4 and is natural magnet. It was magnetite that allowed people to get acquainted with magnetic properties of bodies. Natural magnets are not magnetically strong enough for practical purposes, Therefore, artificial magnets are required. Artificial Magnets Natural magnets like the lodestone do not have very strong attractive or directive properties, but when a piece of steel is stroked (in one direction only) with a lodestone, these properties are developed stronger in it. A piece of steel or iron to which the properties of a lodestone have been imparted is called an artificial magnet.

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Types of Artificial Magnets: Artificial magnets are generally made of iron or steel in different convenient shapes. The different types of artificial magnets are: Types of magnet (i) Bar magnet : This is a bar of uniform cross-section, either rectangular or circular.

Figure N

S

S

N

A

B

(ii) Horseshoe magnet: This is a bar magnet bent in the shape of a horseshoe.

(iii)Needle magnet: This is a short thin strip of magnetised steel with pointed ends and pivoted at its centre.

S

N

(iv)Specially shaped magnets: These are for particular applications. Thus a loudspeaker magnet has a recess in it to accom­modate the coil. One design of extremely powerful magnet is as shown alongside.

(v)

Electromagnets: These can be the mostpowerful and are now a days used whenever a very intense magnetic field is required.

Drawbacks of Natural Magnets

i) ii) iii)

Artificial magnets are far more strong than the natural magnets. The strength of the artificial magnets can be increased or decreased, while in the case of natural magnets it cannot be done. Artificial magnets can be made in any shape that suits our convenience, while natural magnets have irregular forms.

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Formative Worksheet 1.

2.

3.

4.

5.

6.

7.

8.

One of the following is an artificial magnet (A) Horse shoe magnet (B) Magnetic needle (C) Magnetic compasses & electro magnet (D) Above all Assertion : Artificial magnets are preferred to natural magnets Reason : Artificial magnets are far stronger and can be cast in to any desired shape or size (A) Both Assertion and reason are correct (B) Both Assertion and reason are wrong (C) Assertion is wrong, Reason is correct (D) Assertion is correct, Reason is wrong Which one is appropriate : (A) The horse shoe magnet is more powerful than the bar magnet because both the north and the south poles face each other (B) Attractive force is doubled in horse shoe magnet. (A) Only A (B) Only B (C) Both A & B (D) None All Substances can be divided into _________ classes on the basis of their magnetic properties (A) 1 (B) 2 (C) 3 (D) 4 Which of the following cannot be magnetised? (A) Iron (B) Nickel (C) Cobalt (D) Stainless steel Copper, Gold are the examples of the following magnetic materials (A) Ferro (B) Dia (C) Para (D) Both A and B Which of the following is a not magnetic material (A) Nickel (B) Cobalt (D) Bismuth (D) Wood Statement-1: paper, wood, glass cannot be magnetized. Statement-2: Substance which are influenced or affected by a magnet is called a magnetic substance. (A) Statement-1 is true; Statement-2 is true; Statement-2 is the correct explanation of Statement-1.

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(B) Statement-1 is true; Statement-2 is true; Statement-2 is not the correct explanation of Statement-1. (C) Statement-1 is true; Statement-2 is false. (D) Statement-1 is false; Statement-2 is true.

Conceptive Worksheet 1.

2.

3.

4.

5.

6.

7.

Which of the following is a magnetic material (A) wood (B) Plastic (C) Iron (D) Copper Deposits of magnetic oxide iron ore is called (A) Magnesia (B) Magnetic (C) Magnesite (D) Both (A) and (B) The first natural magnet is ____________ (A) Lode stone (B) Hard stone (C) Lime stone (D) None of these A piece of Iron rubbed with magnetite is called a (A) Magnet (B) Natural magnet (C) Artificial magnet (D) Iron ore Which of the following is employed for tracing magnetic lines of force of a magnet. (A) Load stone (B) Magnetic needle (C) Tracing compass (D) Bar magnet The following magnetic materials are very feebly attracted by a strong magnet (A) Para (B) Ferro (C) Dia (D) None Choose the correct example/s for the Ferro magnetic material. (A) Aluminum (B) Copper (C) Steel

(D) Cobalt

3. Properties of a Magnet 1.

2.

It attracts small pieces of iron towards itself It also attracts pieces of cobalt, nickel and steel. If a bar magnet is rolled in the iron filings, it is observed that iron filings cling at the ends of the magnet. However, no iron filings cling in the centre. This suggests that the magnetic force of attraction is concentrated near the ends of the magnet. A freely suspended magnet points in northsouth direction Take a bar magnet and suspend it freely from the centre by means of an unspun silk thread. It is observed that it points in the north-south direction.

Magnetism Even if we set the magnet in some other direction, it again resets itself in the north-south direction. The above property of magnet is used in the construction of magnetic compass. It must be noted that a freely suspended magnet does not really point in exact geographic north-south position, but is inclined at a slight angle to it.

3.

Like poles repel each other and the unlike poles attract each other Take two bar magnets. Suspend one of them freely from an unspun silk thread. Bring the other magnet near it. You will see that, when the like poles of the magnets are brought near each other, they repel. If unlike poles are brought near each other, they attract each other .

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4.

when two similar poles approach each other. Thus, repulsion is the surest test of magnetism. It can magnetize another piece of iron when rubbed several times in one direction) Take a steel bar AB and place it flat on the table. Rub this steel bar with a bar magnet as shown in figure for about 50 times. Test the end A of the steel bar by bringing it near the north pole of another freely suspended magnet. It is observed that the north pole of the freely suspended magnet is repelled, proving that the steel piece has been magnetized. Path of magnet

Permanent magnet Steel bar

5.

Magnetic poles exist in pairs If a bar magnet is broken into two equal parts, following observations are made: (i) Each part has attractive as well as directive property, i.e., each part is complete magnet, (ii) The magnetic strength of each part is half than that of the complete magnet, (iii) The new magnetic poles are formed where the cut is made on the magnet, (iv) The nature of magnetic polarity on a given cut is opposite to the polarity on the far end of the magnet. If the broken pieces are further broken, all the above mentioned observations are noticed till we reach molecular stage.

This characteristic of the magnet is made use in determining the polarity of a magnet with the help of another magnet whose polarity is known. Repulsion is the surest test of magnetism. It is because the attraction can be caused between two unlike poles of the two magnets or between the magnet and the magnetic substances (such as iron, cobalt and nickel). How­ever, the repulsion is caused www.betoppers.com

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Conclusions 1. Magnetic poles exist in pairs. There cannot be a monopole. 2. Every molecule of a magnet is complete magnet by itself. 3. Overall attractive power of a magnet is equal to the sum total of attractive powers of molecular magnets. Magnet and Magnetic Substances

Magnet

Magnetic Substance

1. A substance which attracts metals like iron or steel and which always points in a particular direction when suspended freely is called a magnet. 2. A magnet has two poles, one of which is always directed towards the Geographic north and the other to the Geographic south when suspended freely.

1. Substances which are influe nced by a magnet are called magnetic substances. Iron, steel, nickel and cobalt are a few examples of magnetic substances. 2. A magnetic substance has no poles and does not point to any particular direction when suspended freely.

3. A pole of a magnet attracts the opposite pole and repels the similar pole of another magnet.

3. A magnetic substance is attracted both poles of a magnet.

by

4. General Definitions related to a Bar Magnet 1. 2.

Pole: Each end of a bar magnet is called its pole. Geometric Pole: The geometric end of a bar magnet is called its geometric pole.

A 3.

B

A, B Geometric poles

Magnetic Pole: The point situated slightly inside a bar magnet, where most of its magnetic power is concentrated is called its magnetic pole. A N

S

B

N, S Magnetic poles

Magnetic North pole The pole of a freely suspended bar magnet which point towards geographic north is called its magnetic north pole (N).

X

N

Magnetic South pole The pole of a freely suspened bar magnet which points towards geographic south is called its magnetic south pole (S).

S

Y Magnetic axis (Axial line)

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Magnetic Axis: An imaginary line joining the magnetic north and south poles of a bar magnetic is called its magnetic axis (or) axial line. (XY = magnetic axis) Equatorial Line: The line passing through the centre of magnet and perpendicular to the axial line is called equatorial line. Equatorial line

N

S

(A) Statement-1 is true; Statement-2 is true; Statement-2 is the correct explanation of Statement-1. (B) Statement-1 is true; Statement-2 is true; Statement-2 is not the correct explanation of Statement1. (C) Statement-1 is true; Statement-2 is false. (D) Statement-1 is false; Statement-2 is true. 11. Column – I Column – II 1) Like poles of magnets p) iron 2) Unlike poles of magnets q) repel 3) magnetic substance r) glass 4) Non magnetic substance s) nickel t) attract (A) A  p; B  q ; C  r ; D  s

Magnetic Length: The distance between the two poles of a magnet is called magnetic length of magnet. Magnetic length

A

N

(C) A  q ; B  s; C  p; D  t (D) A  p; B  p; C  q; D  p

S

B

Geometric length

NS  Magnetic length AB = Geometric length The relation between magnetic length and geometric length is 5 MAGNETIC LENGTH   GEOMETRIC LENGTH 6

Formative Worksheet 9.

(B) A  q ; B  t ; C  p, s; D  r

Choose the correct statement/s from the following: (A) Magnetite can be called as lode store (B) Natural magnets possess attractive property only (C) Natural magnets possess attractive and directional properties (D) Natural magnets possess directional property only 10. Statement-1: A freely suspended bar magnet always comes to lie in North-South direction. Statement-2 : Magnetic poles have the property of directionality

12. A piece AB of the magnetite is dipped in a leap of small iron filings. It is observed that a cluster of iron filings stick to the ends A and B while there is practically no iron filing stick to its central region. Answer the following questions. A. The central region or called (A) Neutral Region (B) Pole (C) Both 1 & 2 (D) Equator B. End A is called (A) Neutral Region (B) Pole (C) Both 1 & 2 (D) Equator C. End B is called. (A) Neutral Region (B) Pole (C) Both 1 & 2 (D) Equator 13. Statement-1 :Like poles attract each other Statement-2 : If a magnet is suspended freely, then the end marked with N points towards geographic north and the end marked with S points towards geographic south. (A) Statement-1 is true; Statement-2 is true; Statement-2 is the correct explanation of Statement-1. (B) Statement-1 is true; Statement-2 is true; Statement-2 is not the correct explanation of Statement1. (C) Statement-1 is true; Statement-2 is false. (D) Statement-1 is false; Statement-2 is true. www.betoppers.com

8th Class Physics

126 14. Column – I A) The rest position magnet is B) Magnetic poles exist in pairs C) Iron is strongly attracted by magnet so it is D) Gold is repelled by a magnet so it is

s) Dipole t) paramagnetic substance

(A) A  p; B  q ; C  r ; D  s (B) A  q ; B  t ; C  p, s; D  r (C) A  r ; B  s; C  p; D  q (D) A  p; B  p; C  q; D  p 15. Magnetic poles exist in pair. We cannot get a magnetic monopole. When we cut a magnet, each piece will behave like a magnet with two poles.

Based on the above, answer the following Questions. A.

A bar magnet is cut as shown in the figure. Mark the polarity of each pole (from left to right) for the polarity missed piece. (A) South, North (B) North, South (C) North, North (D) South, South B.

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A bar magnet is cut as shown in the figure. Mark the polarity of each pole (from left to right) for polarity missed piece. (A) South, North (B) North, South (C) North, North (D) South, South

Column – II p) Ferromagnetic of a substance along N-S when suspended q) diamagnetic substance r) directive property

C.

A bar magnet is cut as shown in the figure. Match the following for the polarity. LIST – I LIST– II a) 1,2 e) South, North b) 3,4 f) North, South c) 5,6 d) 7,8 (A) a–e, b–e, c–f, d–f (B) a–f, b–f, c–e, d–e (C) a–f, b–f, c–e, d–e (D) a–f, b–e, c–f, d–e 16. Statement A: The more is the distance of the magnetic substance from the magnet, the weaker is this attraction. Statement B: Magnet can attract magnetic substances within the magnetic field (A) Both the statements are true (B) Both the statements are false (C) Only statement A is true (D) Only statement B is true 17. Statement A: The points of the magnet where there is maximum attraction are called the poles of the magnet Statement B: Magnetic poles always exist in pairs (A) Both the statements are true (B) Both the statements are false (C) Only statement A is true (D) Only statement B is true 18. Magnetic length of a bar magnet is nearly 80% of its geometric length. If a magnet has a geometric length of 12cm. Answer the following: A. Find its magnetic length. (A) 9.6 cm (B) 24 cm (C) 15 cm (D) 12 cm

Magnetism B.

127

The difference between geometric length and magnetic length is (A) 0 cm (B) 3 cm (C) 12 cm (D) 2.4 cm How much percentage does the magnetic length is small when compare to geometric length. (A) 25 % (B) 50 %

16. The region or the space surrounding a magnet in which magnetic force is exerted is called the _______________ 17. The attraction of iron filings in a magnet is maximum at C. (A) Poles of the magnet (B) Middle part of the magnet (C) All places in the magnet (C) 20 % (D) 100 % (D) None of these 18. Pole strength does not depend on onceptive orksheet (A) Length of a magnet (B) Breadth of a magnet 8. One of the following is not a property of a magnet (C) Height of the magnet (D) Both (B) and (C) (A) Attraction (B) Repulsion 19. The relation between a magnetic field at a point (C) Induction (D) Reflection and the distance between a magnet and at a point 9. When magnet is suspended freely, then the end is marked N, points towards (A) Inversely proportional to the square of the (A) Geographic North (B) Geographic South distance (C) We can’t say (D) Depends on magnet (B) Inversely proportional to distance 10. Which of the following is the property of a magnet? (C) Directly proportional to the square of the (A) Pair property distance (B) Directionality (D) None of these (C) Like poles repel & unlike poles attract 5. Methods of making Artificial (D) all the above Magnets 11. A vertical plane passing through the magnetic axis of a freely suspended magnet 1. Single Touch Method (A) Magnetic Meridian (B) Magnetic equator A piece of iron can be magnetized by rubbing it with a magnet from one end to another as shown in (C) Equatorial Meridian (D) Magnetic Pole the figure. 12. A vertical plane passing through the magnetic equator of a freely suspended bar magnet is called (A) Magnetic Pole (B) Magnetic equator (C) Equatorial Meridian (D) Magnetic Meridian 13. If the length of the magnet is l, then the effective length of the magnet is (A) 4l (B) 3l (C) l (D) 2l 14. The magnetic strength or magnetic intensity is 2. Double Touch Method maximum at the __________ of a bar magnet In this double touch method, two strong magnets (A) Pole (B) At its centre are placed at the centre of the soft iron piece to be (C) Equatorial line (D) Axial line magnetised, keeping the opposite poles of the 15. The regions of concentrated magnetic strength magnets facing each other. Move the magnets inside the magnet just near its ends are called simultaneously over the surface of the soft iron piece magnetic ________________. taking them away from one another. Lift the (A) Pole (B) Axis magnets on reaching the ends of the piece and place (C) Meridian (D) Length them again at the centre.

C

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The polarities induced at the two ends depends on the direction of the current. If seen from the right side, the coil carries clockwise current. Therefore, the right side of the bar is the S-pole. From the left side the current enters the coil in the anti-clockwise direction. Hence, the left side is the N-pole. This law is the ‘Clock rule’.

Repeat this exercise 30 to 40 times, on testing the soft iron piece, it would be found to have become a magnet. A piece of iron can be strongly magnetised by double touch method than the single touch method. Now, let us learn how a magnetic substance behaves as a magnet with out contact with the magnet. 3.

Electrical Method Now a days permanent magnets are made by the electrical method. The steel bar AB to be magnetised is placed inside a long coil of an insulated copper wire, and a strong direct current is passed through the coil (see adjoining figure). The iron bar becomes a magnet.

Precautions While Magnetisation 1. Repeat the rubbing of the iron bar with the magnet always in the same direction. 2. Lift the magnet clean off the iron bar at the end of each stroke and place the magnet at the starting end by lifting the magnet at a fair distance above the iron bar.

6. Temporary and Permanent Magnets 1.

2. 3.

Permanent magnets Artificially we can magnetise ferromagnetic substances in such a way that they remain magnetised for a long time. Such magnets are called permanent magnets. Norma lly they are made of steel. The y cannot be magnetised to a very high degree. Magnetic retention power is so high that even after the re moval of the agent which is inducing magnetism they still act as a magnet.

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1.

2. 3.

Temporary magnets The magnetised magnet which cannot retain its magnetism for a long t ime is called temporary magnet. Normally it is made of soft iron. They can be magnetised to a very high degree. Magnet ic retention power is so low that they lose magnetism as soon as the inducing magnetism is removed.

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7. Molecular theory of Magnetism 1.

The first concrete attempt to explain magnetism was done by Weber and Erwing. Molecular magnets in a Magnetic Substance Molecules in a magnetic substance act as complete magnet with two poles. These magnets are known as ‘Weber Elements’. Weber element

N O S

N O S

(or) Molecular magnet

N O S

N O S

N O N S O S

N O S

N O S

(Composition of a magnetic substance)

2.

Molecular Magnets is an Unmagnetised Substance

N S

S N

N

S

N

S N

N S

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S N

N N

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Molecular magnets in an unmagnetised substance are arranged in the form of closed chains with opposite poles facing each other, thereby neutralising the magnetism of each other. 3.

Molecular Magnets in a Magnetised Iron Bar

S

S S S S

N N N N

N

In a magnetized bar the molecular magnets are arranged as chains along straight lines. This results in the exposure of free poles at the ends leading to the formation of poles at the ends.

1.

Facts explained by Molecular Theory of 2. Magnetism Why does an iron bar not act as a magnet An iron bar consists of millions of molecules acting as complete magnets with two poles. (These molecules are called molecular magnets). Despite the presence of millions of such molecular magnets an iron bar does not attracts iron pieces and does not act as a magnet. An iron bar consists of millions of molecules magnets. But they are arranged in the form of closed chains with the opposite poles facing each other thereby neutralizing each other. The net effect is that no magnetism is possible in spite of the presence of millions of molecular magnets.

No magnet with a single pole When a bar magnet is cut of as shown below free poles are exposed along the plane of cutting.

S

N

This results in the formation of new opposite poles on either side of the plane of cutting as shown: (and the new piece has once again two poles).

N

S N

S

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8th Class Physics

130 3.

Strength of a magnet or pole strength The power of attraction of a pole is called pole strength. The strength of a magnet is proportional to pole strength. (denoted by m). The pole strength (m) of a pole depends on the number of free poles at its ends. The number of free poles are related to area of cross-section (ABCD) as shown:

6.

Magneto-striction

7.

According to molecular theory of magnetism, during magnetization closed chains of molecular magnets are broken and the molecules are arranged parallel to the length of the bar. These uniformly arranged molecules occupy comparatively a greater length. So, the length of the bar increases on its magnetisation. This phenomenon is known as magnetostriction. No magnetism at the middle of the bar

S

A

N B

D

C

 Pole strength (m) = f (no. of free poles exposed at ends) = f (area of cross-section)

4.

Larger the area, more are the number of free poles and more is the pole strength. Poles little inside S S S S

S

N S N S N S NS

From the concentration of iron fillings in the above diagram, we can say that there is no magnetism at the middle of the bar.

In a magnet the molecules are supposed to be arranged in a uniform pattern (almost parallel). But as similar poles are located at the end of the bar they mutually repel each other. So, the chains are not exactly parallel but fan out a little at the ends. Hence, the poles are not exactly at the end, but a little inside (as shown above) Magnetic saturation Magnetic power induced

5.

N S N S N S NS

N N N N N

N S N S N S

8. Magnetic saturation

B A Time

A steel bar is placed in a magnetic field. The magnetic power induced in it starts increasing till it reaches a point where the magnetic power does not increase any further (as shown above). This condition is called Magnetic saturation.

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The reason being that in the middle the opposite poles face each other (as shown above). These opposite poles neutralise each other and there is no magnetism in the middle. Heating effect due to magnetisation In the process of their alignment during magnetisation, the molecular magnets rotate. It is during this rotation that they rub against each other generating heat.  Temperature increases during magnetisation.

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Formative Worksheet 19. Statement A: If the current appears to flow in clockwise direction, then the end of bar takes south polarity. Statement B: If the current appears to flow in anti clockwise direction, then the ends of bar magnet takes north polarity.

S el Ste

bar

N

A

N

Steel bar

S B

(A) Both the statements are true (B) Both the statements are false (C) Only statement A is true (D) Only statement B is true 20. Magnets are made by magnetizing a piece of (A) Magnetic substance (B) Non magnetic substance (C) Both (A) and (B) (D) Neither (A) nor (B) 21. The process for making an artificial magnet is called (A)De Magnetization (B) Magnetization (C) Both (A) and (B) (D) Neither (A) nor (B) 22. Statement A: Artificial magnets are made by rubbing natural magnets on iron and steel. Statement B: Artificial magnets are more powerful than natural magnets. (A) Both the statements are true (B) Both the statements are false (C) Only statement A is true (D) Only statement B is true 23. Statement A: The magnets which lose their magnetism as soon as the cause producing them is removed, are called temporary magnets Statement B: The magnets which do not lose their magnetism, when the cause producing them is removal are called permanent magnets (A) Both the statements are true (B) Both the statements are false (C) Only statement A is true (D) Only statement B is true

24. Choose the correct option: a) Electromagnets and the magnets made from soft iron are temporary magnets. b) The magnets made from steel are permanent magnets. c) A magnet can be made by rubbing a steel bar with powerful magnet. (A) Only (a) and (b) are correct (B) Only (b) and (c) are correct (C) (a), (b) and (c) are correct (D) Only (c) and (a) are correct 25. The phenomenon due to which magnetism is produced in a magnetic substance by the pressure of a magnet which is not actually in physically contact with magnet is called magnetic (A) Conduction (B) Induction (C) Radiation (D) Convection 26. Statement A: Permanent magnets lose their magnetism easily. Statement B: Temporary magnets lose their magnetism easily. (A) Both the statements are true (B) Both the statements are false (C) Only statement A is true (D) Only statement B is true 27. A bar of steel can be permanently magnetised by (A) Rubbing a bar magnet with it along the length (B) Rubbing a bar magnet at its ends (C) Rubbing a bar magnet at its centre (D) None of these 28. Choose the correct option: (A) Electromagnets are used to separate iron and steel objects from a heap of metal scrap. (B) Electromagnets are used for lifting heavy iron loads. (C) Electromagnets are used in fans, motors, room, coolers etc,. (D) All of these 29. An ordinary iron bar is said to possess millions of molecular magnets in them. But why do they not make up an iron bar magnet? 30. There are two magnets of same width and height. But one is twice as long as the other. If the pole strength of the first magnet is m, what is the pole strength of the second magnet? www.betoppers.com

8th Class Physics

132 31. A bar magnet has dimensions of l, b, h. If the dimensions are doubled, the pole strength becomes 4 times original. True or false. 32. A bar magnet of pole strength m units is bent into an arc as shown in the figure.

33. An ordinary iron bar is said to be having molecular magnets arranged randomly. When magnetised, systematically. What happens to the length after magnetisation? a) Slightly increases b) Slightly decreases c) Remains unchanged

Conceptive Worksheet 20. Magnetisation is the process by which a magnetic substance is changed into a (A) Non magnetic substance (B) Magnet (C) Both (A) and (B) (D) Neither (A) nor (B) 21. The methods of making magnets are (A) Single touch method (B) Double touch method (C) Electrical method (D) All of these 22. A magnet which is formed due to an electric current is called __________ (A) Natural magnets (B) Permanent magnets (C) Electromagnets (D) Magnetic needle 23. The substance used in making electro magnets (A) Steel (B) Cobalt (C) Soft iron (D) Nickel 24. When a magnet is heated strongly its magnetism (A) Increases (B) Becomes doubled (C) Not effected (D) Decreases 25. Substance used to make permanent magnets (A) Copper (B) Steel (C) Soft iron (D) Aluminium

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26. An un-magnetized piece of magnetic material becomes a magnet when brought close to a magnet. This is called magnetic __________ (A) Conduction (B) Convection (C) Radiation (D) Induction 27. Strong permanent magnets are made by ____________ method. (A) Single touch (B) Double touch (C) Electric (D) Induction 28. Electromagnets are _________ than bar magnets (A) Stronger (B) Weaker (C) Both (A) and (B) (D) None of these 29. The loss of magnetic property of a magnet is called __________ (A) Demagnetisation (B) Magnetic induction (C) Magnetic saturation (D) None of these 30. An electric current produce the properties of a ___________ (A) Magnet (B) Iron (C) Cobalt (D) Nickel 31. Find odd one out: (A) Magnetic separation (B) Single touch method (C) Double touch method (D) Magnetic induction 32. If we keep two magnets side by side with like poles next to each other them magnetic properties of the magnets will be ________ (A) Increased (B) Decreased (C) Same (D) May be increased or decreased 33. The given figure indicates for making artificial magnets is

S ar el b Ste N

A

N

Steel bar

(A) Single touch method (B) Double touch method (C) Electrical method (D) None of these

S B

Magnetism

133 Magnetic Lines of force for U-shaped Magnets In case of U–shaped magnet, magnetic lines of force will be as shown in the figure.

34. Which of the following retains magnetism for a long time? (A) Horse – shoe magnet (B) Electro magnet (C) Earth (D) None of these

8. Magnetic Field Whenever a magnetic substance is placed at a distance from the magnet, it is affected by it. The magnetic substance is said to be in a field called magnetic field. Magnetic field is a region (or space) around the magnet, in which its influence can be felt. A magnetic field is represented by a set of lines (may be curved or straight) called lines of force. A line of force if it were free to move when placed on the field. The direction of arrow head gives the direction of motion of unit north pole. If the lines are spaced widely apart, it is a weak field. If the lines are situated close to each other, it represents a strong field. Direction of magnetic field: The direction of a magnetic field at a point in the field is the direction in which an isolated North pole would move, if free to do so. (An isolated north pole is a hypothetic assumption which is introduced for better understanding).

Properties of Magnetic lines of force i) Magnetic lines of force are continuous and closed curves. ii) They are imaginary lines travelling from the North pole to South pole outside the magnet and from South pole to North pole inside the magnet.

9. Magnetic Lines of Force Place a smooth card board over a bar magnet. Sprinkle some iron filings evenly on the cardboard. Gently tap the card board. The iron filings arrange themselves in a orderly manner around the poles of the magnet and between the poles. The lines along which the iron filings arrange themselves are called magnetic lines of force. Magnetic force acts along these lines.

iii) iv) v)

Lines of force repel each other. . They bend along the length of the magnet. The direction of magnetic field can be find out by drawing a tangent along the magnetic line of force.

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8th Class Physics

134 vi)

Magnetic lines of force never intersect.

In the above figure, if we assume that magnetic lines of force are intersected then at the point of intersection there are two directions for the magnetic field which is impossible. So, the magnetic lines of force never intersect. vii) The magnetic lines of force are parallel to each other in a uniform magnetic field as shown in the figure.

Example: The magnetic lines of force of earth are straight and parallel. viii) The strength of the magnet is shown by the degree of closeness of the lines of force. Closer the lines of force, stronger will be the magnetic field. They are crowded near poles, indicating magnetic field is strong. They are not crowded at the middle of the magnet, indicating magnetic field is weak. Plotting of Magnetic Lines of Force i) For an isolated North–pole, the lines of force are radial, pointing away. ii) For an isolated South–pole, the lines of force are radial, pointing inwards.

Plotting of Magnetic Lines of Force Take a drawing board and fix a white sheet of paper on it. Draw a line on the paper and place the magnet. Draw its boundary. Mark a point 0 at the North pole edge of magnet as shown in the figure. Take a small plotting compass and place it in such a way, that its south pole is pointing towards point 0. Mark the point 1 near the north pole of compass.

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Go on repeating the experiment and obtain points 2,3, 4, 5, 6, 7, etc., as shown in the figure.

Join the points 0, 1, 2, etc., by a smooth curve. The curve so obtained is a magnetic line of force. The direction of the magnetic field will be the direction in which north pole of plotting compass needle points.

10. Magnetic Induction If a natural magnet is brought near a magnetic substance like iron, cobalt, nickel, steel they acquires the magnetic properties. (See the figures) The phenomenon by which an ordinary piece of iron acquires magnetic properties temporarily when a magnet is brought near (without actually touching it )and loses the property when magnet is removed away is called magnetic induction. This method is used in making Artificial magnets.

If an iron bar or a steel bar is placed near a strong magnet, it gets magnetised. The end of the bar near the north pole of the magnet becomes south pole and the other end becomes north pole as shown in the figure. This method of magnetisation is called Induction method.

Magnetism

135

Formative Worksheet

Why do two common pins attached to the pole of a bar magnet diverge outward, whereas if attached to each end of the horse shoe magnet converge inward? In case of bar magnet the pin heads acquire south polarity, whereas pin tips acquire north polarity. As similar poles repel each other, therefore pins diverge, as illustrated in the adjacent figure. In case of horse shoe magnet, the pin heads at the South Pole acquire north polarity and the pin tips south polarity. The pin heads at North Pole acquire south polarity and the pin tips acquire north polarity as illustrated in adjacent figure. As the opposite poles attract each other, therefore the pins converge inward.

Laws of induction 1. Induction of poles at ends: During magnetic induction, the opposite pole is induced at the near end, and the like pole is induced at the other end of the bar. 2. Induction precedes attraction: Therefore attraction results only after the induction of opposite poles at the near end. 3. Sure test for the presence of magnetism: A bar AB is brought near the north pole of a magnet and it is observed that AB is getting attracted to it. Can it be concluded that AB is also a magnet with its south pole at A ? AB need not be a magnet because the attraction in AB may also be possible if it is an iron bar. So attraction is not a sure sign of the presence of magnetism. A sure test to determine the presence of magnetism is repulsion.

34. Statement A: The magnetic lines of force do not intersect are another Statement B: The magnetic lines of force are dense near the poles (A) Both the statements are true (B) Both the statements are false (C) Only statement A is true (D) Only statement B is true 35. Statement A: The magnetic lines of force do not intersect are another Statement B: The magnetic lines of force are dense near the poles (A) Both the statements are true (B) Both the statements are false (C) Only statement A is true (D) Only statement B is true 36. Assertion: Magnetic lines of force do not intersect each other. Reason: Magnetic lines of force will not have two directions at every point (A) A is true, R is the true, R is correct explanation of A (B) A is true, R is true, R is not correct explanation (C) A is true, R is false (D) A is false, R is true 37. The following methods are used for drawing lines of force of a bar magnet a) Iron – fillings method b) Compass needle method (A) Both the statements are true (B) Both the statements are false (C) Only statement A is true (D) Only statement B is true 38. Statement A: Two poles of a magnet are equally strong Statement B: Magnetic lines of force are endless (A) Both the statements are true (B) Both the statements are false (C) Only statement A is true (D) Only statement B is true

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8th Class Physics

136 39. Magnetic lines of force due to an isolated north pole is

(A)

(B)

(C) Both (A) and (B)

(D) None of these

40. Column – I A) Magnetic field

Column – II p)

B) a unit north pole q) C) A unit south pole r) Three dimensional D) magnetic lines of s) Two dimensional force inside the magnet t) travel from south to north (A) A  p; B  q ; C  r ; D  s

37. The number of magnetic lines of force present per unit area is _______ to the intensity of magnetic field (A) Directly proportional (B) Inversely proportional (C) Both (A) and (B) (D) Neither (A) nor (B) 38. The lines of force ____________ each other (A) Repel (B) Attract (C) Sometime repel and sometime attract (D) None of these 39. The magnetic field is full of ______________ (A) Electric lines of force (B) Magnetic lines of force (C) Both (A) and (B) (D) None of these 40 During magnetic induction, the ________ pole is induced at the near end, and the ______ pole is induced at the other end of the bar. (A) Opposite, like (B) Like, opposite (C) Opposite, opposite (D) Like, like

(B) A  q ; B  t ; C  p, s; D  r (C) A  r ; B  p; C  q; D  r , t

Summative Worksheet 1.

(D) A  p; B  p; C  q; D  p

Conceptive Worksheet 35. ______________ is a line joining any two points (close to each other) in the direction of magnetic field. (A) Magnetic lines of force (B) Electric lines of force (C) Both (A) and (B) (D) Only statement B is true 36. The magnetic lines of force originate from the __________ pole of a magnet and end at the ______ pole

(A) North, south

(B) South, north

(C) East, west

(D) West, east

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A magnetic needle is suspended from a stand near north pole of the earth it aligns itself (A) Horizontal (B) Vertical (C) Inclined 450 to the vertical 0

(D) Inclined 22 2.

3.

1 to the vertical 2

You have a magnet with you and north pole and south pole are marked on it. There is a bar on a table. To find out whether it is a magnet or a magnetic substance, which one is the sure test ? (A) Repulsion (B) Attraction (C) Both attraction and repulsion (D) None For making electromagnets _____ material is used. (A) Aluminium (B) Steel (C) Soft iron (D) Copper

Magnetism 4.

5.

137

The direction of magnetic lines of force at a point indicates _____ (A) Direction in which an iron piece would move if kept at that point. (B) Direction in which a south pole would move if kept at that point. (C) Direction of magnetic field strength (D) None Choose the correct figure: (The lines indicate magnetic lines of force)

9.

10.

(A) N

S

11. (B) N

S

12.

(C) N

S

N

S

(D) 6.

7.

8.

13.

Choose the correct statement(s) : (A) Magnetic lines of force are curved lines. (B) Magnetic lines of force do not intersect. (C) Magnetic lines of force emanate from south pole and enter into north pole. (D) Magnetic lines of force emanate from north pole and enter into south pole. Select the correct statement(s): (A) South pole attracts south pole. (B) North pole attracts north pole (C) North and south poles cannot be separated. (D) Artificial magnets are more powerful than natural magnets. Magnetic lines of force (A) Cannot intersect at all (B) Intersect at infinity (C) Intersect within the magnet (D) Intersect at the neutral points

14.

15.

16.

In the case of a bar magnet, the lines of magnetic induction, (A) Start from the north pole and end at the south pole. (B) Run continuously through the bar and outside. (C) Emerge in circular paths from the middle of the bar. (D) Are produced only at the north pole like rays of light from a bulb. When a magnet is heated, it (A) Loses its magnetism (B) Gains magnetism (C) Gains magnetism up to a certain temperature and loses magnetism beyond that temperature. (D) Neither gains nor loses magnetism Demagnetisation of a magnet can be done by (A) Rough handling (B) Heating (C) Magnetising in the opposite direction (D) All the above A small piece of an unmagnetised material gets repelled when it is brought near a powerful magnet. The material is (A) Paramagnetic (B) Diamagnetic (C) Ferromagnetic (D) Non-magnetic If a piece of metal was thought to be a magnet, which one of the following observations would offer conclusive evidence ? (A) It attracts a known magnet (B) It repels a known magnet (C) It attracts a steel screw driver (D) None of above An example for diamagnetic substance is (A) Copper (B) Iron (C) Nickel (D) Aluminium The force between two magnetic poles depends upon (A) Pole strength only (B) Distance only (C) Medium only (D) All the above The force between two magnetic poles is directly proportional to (A) The distance between the poles (B) Square of the distance (C) Reciprocal of the distance (D) The reciprocal of the square of the distance www.betoppers.com

8th Class Physics

138 17. Is it possible to isolate the two poles of the magnet? 18. What is meant by saturation point? Explain. 19. Why a magnet loses its magnetism when it is heated? 20. Repulsion is sure test of magnetism. Explain. 21. Why the poles of a magnet are of equal strength? 22. An unmagnetised iron rod is attracted by a magnet. Why ? or induction precedes attraction. Explain.

5.

HOTS Worksheet 1.

The relation between geometric length and magnetic length of a bar magnet is given by Magnetic length =

2.

5 × Geometric length. 6

If the difference between geometric length and magnetic length of a bar magnet is ‘K’ units, then find its geometric length and magnetic length. The strength of a magnetic pole is measured in terms of pole strength. Pole strength depends on number of free poles exposed at the end of the magnet. Number of free poles exposed at the end of magnet inturn depends on the area of cross section of the magnet. A bar magnet has a pole strength ‘K’. If is cut into 2007 parts as shown in the figure. The pole strength of each piece now is _____

6.

7.

8.

(A) 2007K 3.

K (C) K 2007

(D) K+2007

A bar magnet has a pole strength ‘K’. If it is cut into 4 pieces as shown in the figure. The pole strength of each piece now is __________.

(A) K + 4 4.

(B)

(B)

K 4

(C) 4K

(D)

K 2

A bar magnet of dimensions l, b and h has pole strength of K units. Now its dimensions are doubled. The new pole strength is _______. (A) K+ 4

(B) K–4 (C) 4K

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(D)

K 4

The product of the length of the magnet and its pole strength is called the moment of a magnet or Magnetic moment. If ‘m’ is the pole strength and 2l is the length of the magnet then, the moment of the magnet or Magnetic moment is given by. M = m×2l Also, Magnetic moment is a vector quantity, with its direction from south pole to north pole along its axial line. If a bar magnet of magnetic moment 80 units be cut into two halves of equal lengths, the magnetic moment of each half will be (A) 80 units (B) 60 units (C) 40 units (D) 20 units A bar magnet of magnetic moment M is cut into four parts of equal length. The magnetic moment of either part is (A) M (B) 4M (C) M/4 (D) Zero A magnetic needle of magnetic moment M and pole strength m is broken into two pieces at the middle. The magnetic moment and pole strength of each piece will be A)

M m , 2 2

(B) m,

m 2

(C)

M ,m 2

(D) M, m

Two similar bar magnets P and Q, each of magnetic moment M, are taken. If P is cut along its axial and Q is cut along its equatorial line, all the four pieces obtained have (A) Equal pole strength. (B) Magnetic moment

M 4

(C) Magnetic moment

M 2

(D) Magnetic moment M 9. The moment of a magnet is 3×10–1 Am2 and the length of the magnet is 10cm. The pole strength of the magnet is (in A.m) (A) 30 (B) 60 (C) 3 (D) 6 10. A magnetised straight wire has a magnetic moment M. If it is bent into a semicircle its magnetic moment will be (A) M

(B) M  (C) M 2 (D) 2M 

Magnetism

139 7.

IIT JEE Worksheet 1.

The force of attraction between two point magnetic poles is directly proportional to the product of their pole strengths and inversely proportional to the square of the distance between them. i.e. F 

m1m 2 d2

 F

Km1m 2 d2

Where K is a constant , m1 and m2 are pole strengths and ‘d’ is the distance between them. This is called Coulomb’s law. The force between two poles (m1 and m2) in air which are separated by a distance ‘d’ is F. If the distance is doubled, the force between the poles in air now is ________. (A) 4F 2.

4.

5.

6.

(C)

F 2

(D)

F 4

The force between two poles (m1 and m2) in air which are separated by a distance ‘d’ is F. If m1 is halved and m2 is doubled and the distance is same i.e., d, the force now is ___ (A) F

3.

(B) 2F

(B)

F 2

(C)

1 F

(D) 4F

When the distance between two magnetic poles is halved, the force between them will become (A) Halved (B) One fourth (C) Doubled (D) Four times Do you think the force between two poles depends on medium in which they are present (A) Yes (B) No (C) Sometimes yes and sometimes no (D) Cannot be predicted Find the force exerted on a point north pole of strength 3200amp-m placed 10cm away in air from a point south pole of strength 40amu.m (A) 1N (B) 1.28N (C) 2N (D) None If the strength of each magnetic pole is doubled and the separation between them is halved by what factor the force between them will change ? (A) No change (B) 16 times (C) 8 times (D) 4 times

8.

Two magnetic poles, one of which is three times as strong as the other exert on each other a force equal to 150  g wt. When placed 5cm apart in air.. Find the strength of the stronger pole (g = 9.8 m/s2) Take the value of K in S.I. as 10–7. Magnetic induction (B) : It is defined as the force(F) experienced by a unit north pole placed at a point in a magnetic field. i.e., B 

newton F Unit of B is ampere-metre m

(or) weber/m2 The force acting on a magnetic pole of strength 7.5×10–2A.m is 1.5N. The magnetic field at the point is (in weber/m2) (A) 50 (B) 20 (C) 90 (D) 112.5 9. If the moment of magnet is 0.4 amp–m2 and the force acting on each pole in a uniform magnetic field of induction 3.2×10 –5 web/m2 is 5.2×10–5 newton, find the distance between the poles of the magnet. 10. A magnetic pole strength is 1A-m and its mass is 1g. In a uniform magnetic field, the pole travels with uniform acceleration of 2 cm/s2. Find the magnetic induction of field.



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8th Class Physics

IIT FOUNDATION Class VIII

PHYSICS SOLUTIONS

© USN Edutech Private Limited The moral rights of the author’s have been asserted. This Workbook is for personal and non-commercial use only and must not be sold, lent, hired or given to anyone else.

All rights reserved. No part of this publication may be reproduced, stored in a retrieval system, or transmitted in any form or by any means, electronic, mechanical, photocopying, recording or otherwise without the prior written permission of USN Edutech Private Limited. Any breach will entail legal action and prosecution without further notice.

Utmost care and attention to the details is taken while editing and printing this book. However, USN Edutech Private Limited and the Publisher do not take any legal responsibility for any errors or misrepresentations that might have crept in.

Published by

:

USN Eductech Private Limited Hyderabad, India.

CONTENTS 1.

Measurement

.......... 141 - 152

2.

Force and Pressure

.......... 153 - 166

3.

Friction

.......... 167 - 174

4.

Sound

.......... 175 - 178

5.

Temperature and Heat

.......... 179 - 184

6.

Electricity & Chemical Effects of Electric Current

.......... 185 - 188

7.

Light

.......... 189 - 198

8.

Magnetism

.......... 199 - 200

1. MEASUREMENT SOLUTIONS

FORMATIVE WORKSHEET 1.

2.

6.

i) Temperature is measured in Kelvin. ii) Luminous intensity is measured in candela. iii) Amount of substance is measured in mole. Therefore, the right match is i – b, ii – c, iii – a

Mass Density = Volume For C.G.S unit of density, we need to substitute the C.G.S units of mass and volume in the above formula. The C.G.S. unit of mass is gram (g) The C.G.S. unit of volume is cm3 g (or) g cm-3 cm3 Measuring formula of ‘GAMBHIR”

7.

C.G.S. unit of density is

3.

Mass  Area Volume  Time To find SI unit of GAMBHIR, we need to know the SI units of Mass, Volume, Area and Time.  S.I. Unit of ‘GAMBHIR’ 

=

S.I. unit of Mass × S.I. unit of Area S.I. unit of Volume×S.I. unit of Time

Kg  m 2 Kg  3 m s ms  S.I. unit of ‘GAMBHIR’ = Kg m-1s-1 Reason for incorrect representation i) The unit is expressed in upper case. ii) The unit is expressed in plural form. iii) The first letter of the unit in honour of a scientist, when in written in full, should be in lower case. This has not been done. iv) There is a punctuation mark after the unit, which is incorrect. Correct representation Kg, m, newton, joule One mega second = x × deca seconds _____ (1) mega = 106 and deca = 101 Substituting these values in equation (1), we get, 1 × 106 s = x × 101

5.

106 = 105 101  1 mega second =

106 metre = 108 10  2 10 metre  y=8 8. Altitude of satellite = 600km 600 km = y mm ––––––– (1) We know that, 1km = 103m; 1mm = 10–3m Substituting above in equation (1), we get 600 × 3 10 m = y × 10–3 m 600  103 m = 6 × 108 103 m Substituting the value of y in (1), we get  600km = 6 × 108 mm y

9.

105 deca seconds

1Kg mg Given, (mm) 2  y (km)2 ________ 1 Express Kg in terms of mg 1Kg = 103g ––––––– (A) 1 mg = 10–3 g ––––––– (B) Dividing (A) by (B), we get

1kg 103 g =  106 mg 103 g  1kg = 106 mg –––(E) Dividing (E) with (F), we get,

1kg mm 2  1018

x =

106 sachin 106 = x = 3 = 106–(–3) 10 sachin 10-3 = 106 + 3 = 109  1 mega sachin = 109 milli second. 1 mega metre = 10y centimetre –––––––– (1) We know, mega = 106; centi = 10–2 Substituting the above in equation (1), we get 1 × 106 metre = 10y × 10-2 metre. y



4.

Let 1 mega sachin = x milli sachin –––––––– (1) We know, mega = 106 ; milli = 10-3 Substituting the above in (1), we get 1 × 106 sachin = x × 10-3 sachin



106 mg 1012 km 2



1kg mg 2

mg _______  2  km 2 Comparing (1) with (2), we observe y = 1018

8th Class Physics

142

10.

Express mm2 in terms of Km2 1 mm = 10–3m  1mm2 = 10–6 m2 ––––– (C) 1km = 103m  1 km2 = 106m2 –––––– (D) Dividing (C) by (D), we get

 1 micrometer = 10–9 kilometre ––(B) 1 Mega second = 106 second –––– (3) 1 millisecond = 10–3 second –––– (4)

1 mm2 10-6 m 2 = = 10–12 km 2 106 m 2  1 mm2 = 10–12 km2 –– (F)



1micrometre  x Kilometre –––––––– (A) Mega second millisecond To find x, we need to know, 1 micrometre =_____Kilometre, and 1 Mega second =___millisecond 1 micrometre = 10-6 metre –––– (1) 1 kilometre = 103 metre –––– (2) Dividing (1) by (2), we get, =

106 metre = 10–9 103 metre

1 micrometre 1 kilometre

1 Mega second Dividing (3) by (4), we get, 1 milli second 106 second = 109 103 second

1 Mega second = 109 millisecond ––(C)

1micrometre  x Kilometre –––––––– (A) Mega second millisecond 1 micrometre = 10–9 kilometre ––(B) 1 Mega second = 109 millisecond ––(C) 1 micrometre Dividing (B) by (C), we get, 1 Mega second =

10-9 Kilometre Kilometre = 10-18 9 10 millisecond millisecond



1 micrometer Kilometre = 10-18 –– (D) Mega second millisecond

Comparing (A) & (D,) we observe that, x = 10–18 11.

11. 12. 13.

Except Leap year all are units of distance, leap year is the unit for time. Hence, the right option is (iii). 1 parsec = 3.26 light years = 3 × 9.46 × 1015 m ( 1 light year = 9.46 × 1015 m) = 3.08 × 1016 m We know, Speed of light = 3 lakh km/s (or) 3 × 108 m/s 1 light year = speed of light × 1 year = (300000 km/s × 1 year) = (300000 km/s × 365 × 24 × 60 × 60 s) = 300000 × 365 × 24 × 60 × 60 km So, statement (i) is true. = 3 × 105 × 365 × 24 × 60 × 60 × 103 m = 3 × 108 × 365 × 24 × 60 × 60 m So, statement (ii) is also true.

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Measurement Solutions 14.

15.

16.

a) We know, 1 nm = 10 %9 m 1 nm = 10–9 × 1 m = 10-9 × 100 cm ( 1 m = 100 cm) -9 2 = 10 × 10 cm = 10 – 9+2 cm = 10–7 cm  1 nm = 10–7cm b) We know, 1 fermi = 10–15 m 1 fermi = 10–15 × 1 m = 10–15 × 100 cm = 10–15 × 10–2 cm = 10–13 cm  1 fermi = 10–13cm i) 1 Hecto = 102  1 hectogram = 102 gram ii) 1 Deca = 101  1 decagram = 10 gram iii) 1 milli = 10-3  1 milligram = 10-3 gram iv) 1 micro = 10-6  1 microgram = 10-6 gram Therefore, the right match is: i – b ; ii – c ; iii – d ; iv – a One quintal = x ton __________ (1) One quintal = 100kg and 1 ton = 1000 kg Substituting the above equation in (1). 100kg = x 100kg

x

143 19.

20.

ii)

1 hectare 

iii)

21.

100 kg  101 1000 kg

Let 1 metric ton = x milligram ………(1) 1 metric ton = 1000 Kg = 1000 × 1000 g ( 1 Kg = 1000 g) = 106 g 1 milligram = 10–3g Substituting the above values in (1), we get, 106 g = x 10–3 g ………(2) x

18.

22.

106 g  109 103 g

Substituting the value of x in equation (1)  1 metric ton = 109 mg 1 microsecond = 10x milliseconds ________ (1) We know, 1 microsecond = 10-6 seconds. 1 millisecond = 10-3 seconds. Substituting the above in equation (1), we get 10-6 = 10x × 10–3 s

1 hectare 10 4 = 10–4 hectare  1 m2 = 10– 4 hectare 1 km2 = 100 hectares 1 m2 =

 One quintal = 10–1 ton 17.

1 Millennium = 1000 years = 1000 × 365 days = 1000 × 365 × 24 hours = 1000 × 365 × 24 × 60 × 60 seconds = 31536000000 seconds 1 Decade = 10 years =10 × 365 × 24 × 60 × 60 = 31622400 seconds 1 Leap year = 366 days = 366 × 24 × 60 = 525600 minutes The right match is: i – c ; ii – b ; iii – a i) 1 hectare = 10000 m2 = 104 m2

1 km2 100

= 10–2 km2.  1 hectare = 10–2 km2 1 mm2 = (10–3 m)2 = (10–3 × 10–3 km)2 = (10–6 km)2 = 10–12 km2.  1 mm2 = 10–12 km2 Length of the school hall (l) = 20 m Breadth of the school hall (b) = 12 m Area of school hall = length × breadth (l × b) = 20 m × 12 m = 240 m2 Therefore, the area of the school hall is 240 m2. Volume (v) = 12 cm3 Length (l) = 3 cm Breadth (b) = 2 cm Height (h) = ? Volume = length × breadth × height 12 cm3 = 3 cm × 2 cm × height 12 cm3 = 6 cm2 × height Height =

12cm 3 = 2 cm 6 cm 2

Therefore, the height of the rectangular slab is 2 cm. 23.

106  103 103  10x = 10-3  x=–3  10 x 

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8th Class Physics

144

24.

Dimensions of the cuboid = 0.003 km × 0.03 m × 3 cm Volume of water = 1 litre We need to find if the cuboid holds 1 litre of water or not. This can be found if the volume of cuboid is known. If Vcuboid ³ Vwater (1 litre), the cuboid will hold the water Vcuboid = l × b × h Before substitution, all the dimensions should be expressed in one system of units.  Vcuboid = 3 × 10–3 × 105 cm × 3 × 10–2 × 102 cm × 3 cm = 3 × 3 × 3 × 10 – 3 + 5 – 2 + 2 cm3 = 27 × 102 cm3 = 2700 cm3 = 2700 ml ( 1 cm3 = 1 ml) = 2.7 litres ( 1 litre = 1000 ml) Hence, cuboid can hold water. Initial volume of water (V1) =? Final volume of water (V2) = 9.3 ml Volume of stone (Vs) = 5.8 cm3 = 5.8 ml Final volume of water (V2) = Initial volume of water (V1) + Volume of stone (Vs)  V1 = V2 – Vs = 9.3 ml – 5.8 ml = 3.5 ml  Initial volume of water is 3.5 ml.

25. We know, Density =

1 volume ( mass is constant)  If volume is greater, density is less. Therefore, density is greatest for the object with least volume. As A has least volume, it has the greatest density. 26. Mass (m) = 332 g Volume (V) = 20 cc Density (D) = ?



We know, Density =

Mass Volume

332 g 20 cm 3

= 16.6 g cm–3 Therefore, the density of the given body is 16.6 g/ cm–3 . 27. Density of gold in C.G.S system = 19.6 g/cm3 Density of gold in S.I system = ? We know, 1g / cm3 = 1000 kg / m3  19.6 g/cm3 = 1000 × 19.6 kg/cm3 = 19600 kg/m3 Therefore, the density of gold is 19600 kg/m3. www.betoppers.com

Mass Volume

m=V×D  = 555 cm3 × 8.6 g/cm3 = 4773 g = 4.773 kg ( 1 Kg = 103 g) Therefore, the mass of iron is 4.773 kg 29. Mass of wood (m) = 6000 kg Density of wood (D) = 0.8 g/cm3 Volume of wood (V) =

Mass Density

Volume of wood (V) 



m 6000 kg  D 0.8g / cm3

6000 kg 0.8  1000 kg / m 3 ( 1 g = 1000 kg)

Mass Volume

 density 

We know, Density =

28. Density (D) = 8.6 g/cc Volume (V) = 555 cm3 Mass (m) = _______ kg



6000 3 m 800

 Volume of wood (V) = 7.5 m3 30. Dimension of room = 3 m × 4 m × 5 m  Volume of room (V) = 60 m3 Density of air (D) = 1.30 kg m–3 Mass of air (m) = ?

m  m=V×D V m = 60 m3 × 1.30 kg /m3 = 78 kg  m = 78 kg 31. Length of cylinder (l) = 0.8 m Area of cross-section (a) = 0.45 m2 Density of cylinder (D) = 4500 kg / m3 Mass of cylinder (m) = ? We know, D =

We know, D =

m  m=V×D V

We have to find volume of cylinder. Volume of cylinder (V) = length (l) × area of cross-section (a). V = 0.8 m × 0.45 m2 = 0.36 m3  m = V × D = 0.36 × 7800 = 2808 kg  mass = 2808 kg

Measurement Solutions

145

32. Volume of iron (V) = 30 cm3 Mass of iron (m) = 234 g Density of iron (D) = ___ in kg m–3 We know, Density (D) =

i.e, D =

Mass(m) Volume(V)

m V

36. Given, mB = 3 mA ; VA =

234 g D= = 7.8 g/cm3 30 cm3

Mass(m) Volume(V)

m i.e, D = V D =

2 ×10-3 m3

= 1000 kg / m3

 Density of water = 1000 kg / m3 34. To express the density in S.I. system, we need to express the given data in S.I. system. Radius of sphere (r) = 7 cm= 0.7 m Mass of sphere (m) = 490 g= 0.49 kg Density of sphere (D) = ?

m V

How to get volume of sphere ? Substituting the values in the above equation, we get

d =

To find the

d

m v

dA 

mA vA

dA 

mA 9m A  dA  ............(1) vB / 9 vB

dB 

3m A ..........(2) vB

dB 

mB vB

Dividing (1) by (2),

2 kg

We know, D =

1 × vB 9

substance with greater density.

= 7.8 × 1000 kg / m3 = 7800 kg / m3 (Since 1 g/cm3 = 1000 kg / m3)  Density of iron = 7800 kg / m3 33. Volume of water (V) = 2 litres = 2 × 10–3 m3 (Q1 litre = 10–3 m3) Mass of water (m) = 2 kg Density of water (D) = ? We know, Density (D) =

35. The density of a substance depends on the nature of material. As the nature of material is same after cutting, the density remains the same.  Density of each gold piece after cutting is d only. Hence, false.

0.49 kg 3 14.3 m3 = 0.034kg / m

9 mA dA v  B d B 3m A vB

 dA = 3 dB

So, from above equation, density of A has three times more density than B. 37. Density = x kg / m3 Relative density = ? We know, Relative density =

density of substance density of water  Relative density =

 Density of the sphere = 0.034 kg / m3

x kg m3 1000 kg m3

(Q density of water is S.I. = 1000 kg/m3)

We know, volume of sphere (v) =

=

4 3 πr 3

4 22 × × (0.7) 3 = 14.3 m3 3 7

=

x 1000

 Relative density =

x 1000

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8th Class Physics

146 38. The density of copper = 8.9 × 103 kg/m3 The density of water = 1000 kg /m3 Relative density = ? We know, R.D =

Density of substance Density of water

 The relative density of copper density of copper  density of water 3



39. Volume (V) = 2 m3 Relative density = 0.5 Mass (m) = ?

m V

 m=V×D

__________

(1) Density of a solid (in S.I. Unit) = R.D. of the solid × 1000 kg/m3 Let’s find the density. Relative density =

D kg / m3 1000 kg

 D = Relative density × 1000 = 0.5 × 1000 ________

(2) Substituting (2) in (1), we get m = 2 × 0.52 × 1000 = 1040 kg  m = 1040 kg

40. The density of gold is 19.3 times greater than the density of water.

CONCEPTIVE WORKSHEET 1) 2)

3 Since length is a physical quantity whereas metre, yeard, cubit are units of length. 3) Since conversion is possible between units of similar quantities. 4) 3 5) 1 6) 3 7) 1 8) 4 9) 2 10) 3 11) 3 12) 1 13) 3 14) milli, micro, centi are submultiples whereas kilo is a multiple. 15) 1 www.betoppers.com

1 Kilometre 103 metre 103  6  6  106 1 micrometre 10 metre 10



3

8.9  10 kg / m  8.9 103 kg / m3

We know, D =

16) Given, 1 micrometre = 1 micron = 10–6 metre –––––– (1) To find, 1 kilometre =___?___ microns We know that, 1 kilometre = 103 metre –––––––– (2) To get the relationship between kilometer and micron, let us divide equation (2) by (1), we get,

1 Kilometer  109 1 micron

 kilometre = 109 microns Hence 109 microns make up 1 kilometre. 17) Here micrometre and mega metre are the units of physical quantity length. So, find relation between them 1μm = –––––– Mm and 1mg = –––––––– ng 1μm = 1012 Mm –––––––––– (1) 1 mg = 106 ng –––––––––– (2) 1μm 1012 Mm = mg 106 ng



1μm Mm = 1018 mg ng

18) Here milligram and kilogram are the units of physical quantity mass, and nano second and pico second are the units of physical quantity time. Find relation between them



1m g kg = x ns ps –––––––––– (a)

1mg = –––––– kg and 1ns = –––––––– ps 1mg = 10–6 kg –––––––––– (1) 1ns = 103 ps –––––––––– (2) 1mg 106 kg = 3 ns 10 ps



1mg kg  109 ns ps –––––––––– (b)

Comparing (a) and (b) we get, x = –9 19) Given, The height of a building = 12m To find, 12 m = ___?__ feet (ft) We know tha, 1 m = 100cm ––––––––– (1) 1 feet = 12 inch = 12 × 2.54cm = 30.48cm

Measurement Solutions

147

( 1inch = 2.54cm)  1 feet (ft) = 30.48cm ––––––– (2) Dividing (1) and (2), we get

1m 100cm 100    3.28 1ft 30.48cm 30.48



1m  3.28 feet 1ft

 1m = 3.28 feet Hence 12 m = 12 × 3.28 ft = 39.36ft  The height of the building is 39.36ft. 20) To find, 1 furlong = ? km Given, 1 mile = 8 mile –––––––– (1) Also, 1 mile = 1609 m 

1609 km 1000

1   km   1km  1000m  1m  1000   1 mile = 1.609 km 1 mile –––––– (1) 1.609 Dividing (1) by (2), we get  1km 

1 furlong  1km



1 1 mile 1.609 8  8   0.20 1 1 8 mile 1.609 1.609

1 furlong  0.20 1km

 1 furlong = 0.20 km 21) 1 tonne = 1000 kg. 1 tonne = 10–3 tonne. 1000 22) 1 tola = 10g 10 tola = 10 × 10g = 100g 1000 g = 1 kg  1 kg 

1 100g  kg  0.1kg 10 23) 2.0 × 10–5 kg = 2 × 10–6 kg = 20 mg 24) 1 mean solar day = 24 hr = 24 × 60 × 60 s = 86400 s 25) 1 millennium = 1000 years 10 year =1 decade 1 year 

1 decade 10

1000 year = 

1 × 1000 decade = 100 decade 10

26) Area = length × breadth = m × m = m2 So Area is derived from quantity length. 27) Unit of area 28) 1km2= 1km × 1km = 1000m × 1000m = 1000000m2 = 100 × 10000 m2  1km2 = 100 × 1 hectare = 100 hectares 29) 1 hectare = 100m × 100m = 10000m2 30) The space occupied by a substance is called volume. 31) The S.I unit of volume is cubic metre. 32) One cubic metre = 1m3 = 1m × 1m × 1m = 100cm × 100cm × 100cm = 106 cm3 = 106 cc 33) The volume occupied by a cube whose each side is equal to 1cm is called cubic centimeter. 34) 1 litre = 1000cc

1 litre = 1cc 1000  1 litre = 1000 milliliter 1 litre = 1000 cc So, both (1), (2) are correct. 35) Square metre reason : (1), (2), (3) are units of volume and (4) is a unit of area. 36) 2 1 milliliter 

mass volume S.I. unit for mass is kg. S.I. unit for volume is m3

37. Density =

1 volume m3   density mass kg

1  S.I. unit for density is m3/kg 38. 3 39. 3 40. 3 41. 3 42. The density of alcohol = 800 kg/m3

1 g 3 3 = 0.8 g/cm 1000 cm 43. 20 cm3 of aluminium has mass = 54 g = 800 

1 cm3 of aluminium mass =

54  2.7 g 20

44. mass of lead (M) = 232 g volume of lead (V) = 20 cm3

M 232  = 11.6g/cm3 V 20 45. Relative density has no units density of lead, (D)=

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8th Class Physics

148

SUMM ATIVE WORKSHEET 1. 2. 3. 4. 5. 6. 7. 8.

3 3 2 both are required. (4) 1 1 2 1 km2 = 1km × 1 km = 1000m × 1000m = 1000 × 100cm × 1000 × 100 cm  1 km2 = 100000 cm × 100000 cm = 105 cm × 105 cm 1km2 = 1010 cm2 2

–10

1cm =10 km

2

1 hectare = 100m × 100m = 100 × 100cm × 100 × 100cm = 108 cm2 1cm2 =10 –8 hectare 1m2 = m × m = 100cm × 100cm = 102cm × 102cm = 104cm2

9.

1cm 2 =10–4 m 2 So, (1), (2), (3) all are correct. 1km2 = 1000m × 1000m = 103m × 103m = 106 m2 1m2

=10–6 km 2

1 hectare = 100m × 100m = 102m × 102m = 104 m2 1m 2

=10 –4 hectare

1 are = 100m2 = 100m2 1m 2 =10–2 are So, (1), (2) and (3) all are correct. 10. The volume of bucket is greater than the volume of cup. 11. Mass of alcohol (M) = 4 kg = 4000 g Volume of alcohol (V) = 5 litres = 5000 m.l. = 5000 cm3

M 4000  g/cm3 V 5000 = 0.8 g/cm3 3 density in kg/m = 1000 × 0.8 = 800 kg/m3 12. difference of two levels of water in a measuring cylinder gives the volume of immersed body. Volume of the body = 70 cm3 - 50 cm3 = 20 cm3 mass of the body = 50 kg

13. The density of lead = 11.6 g/cm3 = 11.6 × 1000 kg/m3 = 11600 kg/ m3 The density of wood= 800 kg/m3 14. 1 15. 2 mass of iron piece = 117g. volume of iron piece= 15cm3.

 density =

density of iron 

50 kg 50000g mass = 20cm3 = 20cm3 volume = 2500 g cm-3

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117 g 39 g   7.8g cm 3 15cm3 5 cm3

kg 39   1000 = 7800 kg m–3. m3 5 16. i) Volume of copper piece= (34 – 22)cm3 = 12 cm3. ( 1 ml of water occupies 1cm3 of volume) and density in

mass 106g  volume 12cm3 = 8.83g cm–3. 17. Volume of rectangular block = 2.1 cm × 1.2cm × 1.1cm = 2.77cm3 ii) Density of copper 

mass 23.6g  = 8.51g cm–3 volume 2.77cm 3 = 8.51 × 1000 = 8510kgm–3 18. a) mass = volume × density b) 1cc of water has a mass of 1 g c) 1cc of mercury has a mass of 13.6 g. density 

1gm 103  19. = 1000 kg/m3 cm3 106 20. Volume of the room = 120m3. density of air = 1.2kgm–3.  the mass of air = volume of air × density of air. = 120 × 1.2 = 144kg

 density of alcohol, (D) =

density =

mass volume

HOTS WORKSHEET 1. 2. 3.

1 ton is a unit, 1, 3, 4 are multiples of unit. The length of a school compound = 450 m The breadth of a school compound = 145 m The area of school compound = length × breadth = 450 × 145 = 65250 m2 1 hectare = 104 m2 1 m2 = 10-4 hectares So, the are of school compound = 65250 × 10-4 hectares = 6.525 hectares

Measurement Solutions 4.

5.

6.

149

The area of marked surface = area of 11 darked small squares = 11 × area of 1 darked small square = 11 × 1 cm2 = 11 cm2 The stone displaces water which flows out from the spout into the measuring cylinder is equal to the volume of the stone. 12 ml 1 ml = 1 cm3 = 1 cm3 = 10-6 m3 12 ml = 12 × 10-6 m3 = 12 × 10-1 × 10-5 m3 = 1.2 × 10-5 m3 1millennium = 1000 years 10 year = 1 decade, 1 year =

1 decade 10

1  1000 decades = 100 decades 10 18 hours _ 57 min _ 12 hours _ 00 min

1000 year = 7.

6 hours

8.

9.

10.

11. 12. 13.

_

57 min PM

_

45 min

+ 12 hours

_

00 min

15 hours

_

45 min

3 hours

1millennium = 10 centuries = 10 × 10 decades = 100 × 10 years = 1000 years = 1000 × 365 days = 365000 days

1 1 day = millennium 365000 1 decade = 10 years = 10 × 365 days = 3650 days = 3650 × 24 hours = 87600 hours = 87600 × 60 min = 5256000 minutes = 5.256 × 106 minutes. 1 year = 5.256 × 105 minutes = 5.256 × 60 × 105 seconds = 31.536 × 106 seconds 00 hours means, it is 12 O’clock at night 18 hours _ 15 min _ 12 hours

_

00 min

6 hours

_

15 min PM

14. Volume of the body = 75 cm3 - 50 cm3 = 25 cm3 density of the body = 250 g cm-3 mass of the body = volume × density = 25 cm3 × 250 g cm-3 = 6250 g = 6.250 kg 15. Mass of solid (M) = 72 g Initial volume of water in measuring cylinder (v1 ) = 24 cm3 Final volume of water + solid in measuring cylinder = (V2) = 42 cm3 Density of solid 4

M 72 72   4.0g / cm3 = V -V  2 1 42 24 18 4 g /cm3 = 4 × 1000 kg /m3 = 4000 kg/m3 16. Mass of empty density bottle (m1) = 20.25 g mass of (empty density bottle + water) (m1) = 50.25 g mass of (empty density bottle + liquid) (m1) = 40.75 g

 m3  m1  40.75  20  25 = m m  = 50.25  20.25  2 1  m3  m1 

 density of liquid =  m  m  gcm  2 1

3

20.50g = 30cm3 = 0.683 g/cm3 17. Mass of empty density bottle (M1) = 22 g mass of (empty density bottle+water)(M1) = 50 g mass of (empty density bottle+brine solution) (M1) = 54 g. Density of brine solution

 m3  m1  54  22 32 32 3  1.14g / cm3   gcm  50  22 28 28  m2  m1  18. Mass of empty density bottle (m1) = 30 g Mass of (empty density bottle + water) (m2) = 75 g Mass of (empty density bottle + liquid X) (m3) = 65g mass of only water = (75 – 30)g = 45 g Volume of density bottle = (m2 – m1) cm3 = (75–30) cm3 = 45 cm3 19. Volume = l × b × h = 3 × 2 × 2 = 12 cm3 mass =

1 kg = 250 g 4

250g mass 3 = 3 = 20.8 g/cm volume 12cm It’s density is greater than the density of water. so, it sinks in water. Density =

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8th Class Physics

150 20. The density of body A = 0.2 g/cm3. The density of water = 1 g/cm3. The density of body A is less than the density of water. so, body A is float in a water. The density of body B = 0.08 kg/m3

1 g/cm3 = 0.00008 g/cm3 1000 The density of body B is also less than the density of water. so, body B also floats in water. 21. The density of wood = 0.68 g/cm3 The density of alcohol = 780 kg/m3 = 0.78g/cm3. The density of water = 1 g/cm3 The density of glycerine = 1.250 g/cm3 We know that, more the density of liquid, less the solid will submerge in it. In the given liquids, glycerine has more density. So, the wood is less submerged in glycerine. 22 Mass of empty density bottle (m1) = 20.25 g mass of (empty density bottle + water)(m2) = 50.25 g mass of (empty density bottle + liquid) (m3) = 40.75 g = 0.08 ×

m3  m1 10.75  20.25 density of liquid = m  m  50.25  20.25 2 1



20.50g 3 30cm3 = 0.683 g/cm

The density of X is less than the density of water. so, liquid X is float in water. 23. Sphere of radius (r) = 1.4 m

4 3 volume of sphere V   r 3 =

4 22  1.4 1.4 1.4 = 11.4987 m3 3 7

500kg mass = 11.4987 m3 volume = 43.483 kg/m3

Density of sphere =

1 g/cm3 = 0.0435 g/cm3 1000 Density of water = 1 g/cm3.  so, the density of sphere is less than the density of water. So, the sphere will float in water. = 43.483 ×

15g 3 24. Density of paper = 40cm3  0.375g / cm 50g 50g density of stone = 75m3  75106 cm3 www.betoppers.com

50 5 = 75000000  7500000  6.67  107 g / cm3 density of (paper + stone) = (0.375 + (6.67× 10-7) = 0.375 g/cm3 The density of kerosene oil = 0.8 g/cm3 So, the paper boat floats on a kerosene oil. 25. Relative density of liquid mass of Vcc of liquid

160 = 0.8 mass of Vcc of water 200 The relative density of kerosene is 0.8  So, that liquid is kerosene 26. Mass of liquid = (53.5-30)g = 23.5 g mass of an equal volume of water = (48 – 30)g = 18g knowing 1g of water occupies a volume 1cm3 volume of water = volume of liquid = 18 cm3 Density of liquid in C.G.S system 





m 23.5g   1.30g / cm3 v 18cm3

Density of water in C.G.S. system = 1 g/cm3 1.30g / cm3  Relative density of liquid = 1g / cm3 = 1.30 27. Volume of oil drop =

4  r3 3

4 22 3    0.25  mm3 3 7 area of the film on the watch surface =

=   102 cm 2 Size of an oil molecule =

Volume of oil drop Area of the film

4   0.253  103 cm 3  3   102 cm 2

=

4 0.25 × 0.25 × 0.25 × 10 3 × 3 102

4  15625  1011  4  5208.33  1011 3 [ assuming a single layer of molecule is formed] = 20833.32 × 10–11 = 2.08 × 10–7 cm. 28. Given, 1 mile2 = 640 acres 

 l acre 

1 miles2 –––––– (1) 640

Measurement Solutions

151

Also, 1 hectare = 1hm2 = (102m)2 = 104m2–––––– (2) But, given 1 mile = 1609m

=

1 kg / m 3 1  = 19.53 × 10-7  512 3 512000 1000 kg / m

1 l m mile 1609 2

1  1   l m2   mile   mile 2 ––– (3) 2588881  1609  To get the relationship between hectare and mile2, substitute (3) in (2), we get, 1 hectare = 104 m2 1 mile 2 –––––– (4) 2588881 Dividing (1) by (4), we get  104 

1acre  1hectare



1 1 miles2 640  6404 104 10 mile 2 2588881 2588881

1 2588881   0.405 640 104

 1 acre = 1.405 hectare.

IIT JEE WORKSHEET 2 2. 2 3. 1 Volume of iron (V) = 555 cm3 Density of iron (D) = 7.6 g/cm3 M V mass of iron (M) = D × V

Now, D 

= 7.6

5.

6.

Volume of iron = 3 cm × 1.5 cm × 6 cm = 27 cm3 mass of iron = 205.2 g

mass 205.2   7.6g / cm3 volume 27 Density of water = 1 g/cm3  Relative density of iron = 7.6 Radius of an iron cylinder (r) = 1.4 cm length (l) = 8 cm Mass of an iron cylinder (m) = 369.6g Volume of an iron cylinder density of iron 

7.

(v) = π r 2 l 

22 1.4 1.4  8 = 49.28 cm3 7

Density of an iron cylinder 

mass volume

369.6 = 7.5 g/cm3 49.28 Density of water = 1 g/cm3 Relative density of an iron cylinder 

1acre   0.405 1hectare

1. 4.

Density of wooden cube Density of water

g × 555 cm3 = 4218 g cm3

4218 kg = 1000 mass of iron = 4.218 kg Length of the wooden cube = 4m Volume of the wooden cube = (length)3 = (4m)3 = 64 m3 Density of wooden cube 1 = mass  8  1 kg / m 3 volume 64 512 The relative density of wooden cube

Densityof an iron cylinder 7.5  = 7.5. Density of water 1 8. Length is a physical quantity, centimetre is the unit and 200 is the numerical value of the physical quantity. 9. 1 tonne = 1000 kg. 10. 1 micrometre (mm) = 10–6m = 10–4 cm. 11. 1kg = 1000gram, 1gm = 10–3 kg. –3 1milligram = 10 g. 1microgram = 10–6g. 

12. 1 litre = 103 cm3 103 ml = 103 cm3 1 ml = cm3 1 cm = 10–2 m  5 ml = 5 cm3 or 5 × 10–6 m3 13. 1 litre = 103 cm3 = 103 × 10–6 m3  104 l = 104 × 10–3 = 10 m3 . 14.  1 ml = 1 cm3 6 cm3 = 6 ml  The level of water = 26 ml + 6 ml = 32 ml. 15. 9.11 × 10–31 × n = 1

n

1 = 0.11 × 1031 = 1.1 × 1030. 31 9.11  10 www.betoppers.com

8th Class Physics

152 16. Mass of kerosene = 270g – 200g = 70g  1g occupies = 1cm3 volume of kerosene Volume of kerosene = 70cm3 = 70ml. 17. Total mass of all pieces = 20 × 10 = 200g i.e. 200g mass displaces 200ml water in the vessel Now, the present level of water in vessel = 300ml  initial level of water in the vessel = 300 – 200 = 100ml. 103 g g 10 x  10 x  10 3  x = –3 6 3 10 cm cm 3 19.  m1 = 300g, m2 = 0.3 × 10–3g, m3 = 0.3 × 10–6g, m4 = 3000g  m4 > m1 > m2 > m3. 20. r = p/104q = 10–4 p/q r in S.I. system = 10–4 × kg/m3

18.

r in C.G.S. system

1000g 10 6 cm3 = 10–7 g/cm3 = 10–4 ×



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2. FORCE AND PRESSURE SOLUTIONS

FORMATIVE WORKSHEET 1.

2.

3.

4.

5.

It is clear from given figure that only one person is applying a force to bring the box in his direction. That is, only one person is pulling the box. The correct answer is A. Here, we need to investigate the relation between the direction of force and the motion of the wheelchair. The direction of force on the wheel chair will be along its direction of motion. Initially the wheel chair is pushed from West to East. It then takes two left turns, i.e., the direction of motion of the wheelchair becomes reverse, (from East to West). Here, the student is required to determine the change in the amount of force required when the box moves over wheels. Since frictional force opposes the motion of the box, the boys have to apply a large amount of force. If wheels were present below the box, then the frictional force between the box and the surface would decrease. In order to answer this question, the student must determine the net force acting on the rope. The total force on the rope toward the right is 60 N, i.e., applied by Mark. The total force on the rope toward the left is 40 N + 20 N = 60 N, i.e., applied by Sandy and George. Hence, the net force on the rope is 60 N – 60 N = 0 N. The correct answer is A. Rudy and Jackie are pushing the box toward East with forces 35 N and 45 N respectively. Cruz and Harry are pushing the box toward West with forces 30 N and 50 N respectively. Hence, the net force acting in the West-East direction is (35 N + 45 N – 30 N – 50 N) zero. Sandy is pushing the box toward North with a force of 25 N. Ronnie is pushing the box toward South with a force of 45 N. Hence, the net force acting in the North-South direction is (45 N – 25 N) 20 N toward South. Hence, the net force acting on the box is 20 N toward South.The correct answer is A.

6.

7.

8.

9.

10.

11.

12.

13.

Cole and Paul are applying equal force on the trolley from opposite directions. This implies that no net force is acting on the trolley. Hence, the speed of the trolley will remain the same. The correct answer is A. The total force exerted on the cabinet increases after the arrival of Jenny. Since both Jenny and Jack push the cabinet in the same direction, the speed at which it moves will increase. Hence, speed of the moving cabinet will increase. The correct answer is A. The mass of an object is always conserved. It can neither be created nor destroyed. Hence, the mass of an object cannot change on the application of force. The correct answer is B. When a man hits a ball with a bat, there is a direct contact between the bat and the ball. This contact causes the ball to move in a particular direction. Hence, the motion of the ball is caused by the direct application of force. The correct answer is C. The force exerted from a distance is called noncontact force. Muscular force is not exerted from a distance. Hence, it is not a non-contact force. The correct answer is B. The contact between the flag and the wind causes the flag to flutter. Hence, the flag flutters because of the application of a direct force. The correct answer is A. Here, the student is required to recognize the forces experienced by the small bar magnet. All objects having mass attract each other by a force known as gravitational force. Therefore, the two magnets attract each other by gravitational force. Two magnets can attract or repel each other by magnetic force. Knowledge of the relation between motion and force is required to answer this question correctly. Earth revolves around Sun in 365 days. This motion is caused by the gravitational force that acts between Earth and Sun. The correct answer is C.

8th Class Physics

154 14.

15.

This question requires the student to recognize the 18. nature of different types of forces. Gravity is a universally attractive force. It cannot repel a body. Magnetic force can attract as well as repel a body. Two magnets can push or pull each other, depending on their polarity, whereas, two bodies having mass can only attract each other. Hence, the given statement is correctly completed by row C. The correct answer is C. In order to answer this question, the student must be able to recognize the various forces that act on 19. the wooden block. Since the wooden block is not made of a magnetic material, it does not experience any magnetic force. Frictional force arises when two bodies move in contact with each other.  Since the surface of the wooden block is rough,  the block experiences a frictional force, as shown  in the following figure. 20. 21.

22. All objects having mass attract each other by a force known as the gravitational force. Therefore, the wooden block is attracted by Earth and the inclined plane. The gravitational force on the block is represented by F2 in the given figure. Hence, the wooden block experiences frictional and gravitational forces.The correct answer is D. 16.

17.

A heavy stone falls with same acceleration as light stone in the absence of air resistance. Therefore, the right answer is (C)

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 u  0  t = 4sec

h

1 h   9.8  42 2 = 8 × 9.8  h = 78.4 m hmax = 19.6 m u=? v = 0 (  h = h max ) g = – 9.8 m/s2 We know v2 – u2 = 2gh 02 – u2 = 2 × (–9.8) × 19.6 – u2 = –19.6 × 19.6 u = 19.6 m/s

B

A  g  f (Mass of body) So both the body have same acceleration due to gravity. u = 20 m/s v=0 h=? g = –10 m/s2 We know v2 – u2 = 2gh 02 = 202 = 2(–9.8)h

23.

20  20  20m 2  10 We know that



1 g moon  g earth 6 We also knowW = mg W  g

Therefore, Jack has to overcome the gravitational 24. force of Earth to climb the top. The correct answer is D.

1 h  gt 2 2

A

h

To climb the top, Jack has to move away from Earth’s centre. Earth attracts every object near its surface toward its centre. This force of attraction is known as the gravitational force of Earth.

t = 4 sec g = 9.8 m/s2 h=? We know

25.

1 120  20N So Wm  We  6 6 Both of them take same time to reach the ground in the absence of air resistance. Length of edge of a cube, a = 2 cm = 2 × 10–2 m Area of a surface of the cube, A = a2 = (2 × 10–2)2 = 4 × 10–4 m2 Mass of the cube, m = 10 g = 0.01 kg

Force and Pressure Solutions Acceleration due to gravity, g = 9.8 m/s2 Force exerted by the cube, F = mg = 0.01 × 9.8 = 0.098 N Pressure is given by the relation,

27.

28.

29.

30.

F 0.098   245 N / M 2 A 4  104 Therefore, the pressure exerted by the cube on the table is 245 N/m2. The correct answer is D. The hole in each beaker is made at the same height. Thus, at this level, water will exert an equal amount of pressure at each point on the beaker surface. The pressure exerted by a liquid increases with the height of the water column. Thus, more the water in the beaker, the greater will be the pressure it exerts and vice-versa. As the distance up to which the water jets falls is related as d1  V1 . 31. The correct answer is B. The force of gravity exerted by air molecules per unit area is known as atmospheric pressure. It does not remain constant because the number of air molecules decreases with increasing height. Hence, the force of gravity and consequently, atmospheric pressure, decreases with an increase in height. The correct answer is B. Atmospheric pressure decreases with increasing height and vice versa. While climbing the mountain, Romesh’s height will gradually increase. As a result, the atmosphere pressure around him will decrease continuously. Hence, the height of the mercury column in the barometer will also keep decreasing. The correct answer is A. Water exerts equal pressure at equal height. If a stone is dropped in bottle III, then the volume of the stone will cause an increase in the water levels in bottles II and III. Some water will pass from bottle III to bottle II. As a result, the water levels in bottles II and III will  increase. The water level in bottle I will remain as before. Since bottle I is closed in an air tight manner, there is no opening for the air inside bottle I to escape. As a result, there will be no free space available in bottle I for extra water to pass from bottle II to bottle I. The correct answer is C. P

26.

155 Case – 1 (Normal atmospheric Pressure) d1 = 1.293 kg m–3 P1 = 1 atm. Case – 2 (Other than normal atmospheric pressure) d2 = ? R D = ? P2 = 10 atm. We know, P = h d g For a given height at a place, ‘h’ and ‘g’ are constant. So, pressure formula changes as P  d



P1 d1  P2 d 2



P2 10  d1   1.293 d 2  12.93 kg m3 P1 1

Therefore, density of air at pressure 10 atm is 12.93 kg m–3. Mercury Barometer 76 h1  76 cm  m 100 –3

d1 = 13, 600 kg m

Water Barometer h2 = ? –3

d2 = 1000 kg m

According to the problem, the pressure exerted by the water barometer is equal to mercury barometer i.e, Pmercury = Pwater We know, pressure (P) = hdg



P1 d1  P2 d 2



P2 10  d1   1.293 d 2  12.93 kg m3 P1 1

By applying this to both the cases, we get h1d1g1 = h2d2g2 Acceleration due to gravity is same in both the cases i.e, ( g1 = g2) h1d1 = h2d2 76  13,600 h1d1 100 h2    h 2  10.34 m d2 1000

Therefore, equivalent height of water barometer is 10.34 m.

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8th Class Physics

156 32.

Case – 1 Water h1 = 13.6 cm P1 = P d1 = 1g cm–3 g1 = g

P  3 83  Boyle’s law is applicable

Case – 2 Un known liquid h2 = 1 cm P2 = P d2 = ? g2 = g

Water P1 = P

 P2 

 P   3  P2  83

Unknown liquid

h 1 = 13.6 cm

P2 = P

d1 = 1g cm–3

 P2  h 2 = 1 cm

35.

d2 = ?

We know, P = hdg Both the liquids are exerting same pressures. So, P1 = P2 h1d1g1 = h2d2g2  h1d1 = h2d2 (At a given place g1 = g2)

1 Final pressure (P2)  P 4 According to Boyle’s law, P1V1 = P2V2 Substituting the above values, we get, 1 P  1000  P  V2 4 V2 = 4000  V2 = (1000 × 4)  cm3 Therefore, the final volume of gas = 4000 cm3. 34.

st

2 Cube

 2  2

 2  2

 V1  

3

P1 = P

3

 V2   2   81 P2 = ?

Temperature is constant

Temperature is constant  P1V1 = P 2V2 www.betoppers.com

P1 = P

P2 = ?

4 3  4 4  P1  r13  P2  r23  Vsphere  r  3 3 3  

 P1  r13  P2  r23 Substituting the values from data in the above equation, we get,  P1 × (2r)3 = P2 × r3  P × 8r3 = P2 × r3  P2 = 8P 36.

P1V1 = P2 V2 104 ×100 = 105 × V2

 or  V2 = 10cc CONCEPTIVE WORKSHEET 1.

2. 3.

nd

1 Cube

Case – 1 (First sphere) r2 = r

Temperature is constant  P1V1 = P2V2 is applicable

h1d1 13.6  1   13.6 g cm3 h2 1 13.6 g cm–3 is the density of mercury, So, the unknown liquid is mercury. Case – 1 Initial volume (V1) = 1000 cm3 Initial pressure (P1) = P Case – 2 Final volume (V2) = ?

Case – 1 (First sphere) r1 = 2r

Temperature is constant

 d2 

33.

P 8

3

4.

A force must be applied in order to push or pull an object. The correct answer is A. A force is needed to move an object that is at rest. The correct answer is C. This simple problem can be solved by considering that the direction of force on a moving body is along its direction of motion. Since the bus is moving along a curved path, the force on it keeps changing continuously. The correct answer is B. Here, the student is required to investigate the relationship between the direction of applied force and the motion of the car. Since the car moves toward East, the applied force that stops the car must be in the opposite direction. Hence, the direction of the applied force is toward West. The correct answer is C.

Force and Pressure Solutions 5.

6.

7.

8.

9.

10.

11.

The total force applied by the four boys is the sum total of the respective forces applied by them, i.e., 25 N + 25 N + 30 N + 35 N = 115 N toward the left. Frictional force opposes the motion of a moving object. As a result, the frictional force experienced by the box is 15 N toward the right. Hence, the total force experienced by box is the difference between the total force applied by the boys and the frictional force, i.e., 115 N  –15 N = 100 N toward the left. The correct answer is A. Eric and Evan are applying force on the load in opposite directions. Therefore, the net force acting on the load is the difference between the forces applied on it. The net force acting on the load = 25.7 – 20.56 = 5.14 N toward Evan The correct answer is B. Additional force is applied by Jerry along the direction of the ball’s motion. Therefore, speed of the ball will increase. The correct answer is A. Weight of an object is the measure of gravitational force between the object and Earth or any other planet. Hence, when a wooden box is suspended by a spring balance, the spring stretches because Earth’s gravitational force pulls the box. The correct answer is A. Joanna is using her fingers to hit the striker and the coins. This implies that she uses muscular force for pocketing the queen. The correct answer is C. The magnets exert magnetic force which causes them to stick to the refrigerator, as the latter contains iron in it. This question is successfully answered by the student who can recognize the various forces that act on the magnets. All objects having mass attract each other by a force known as the gravitational force. Since both magnets have mass, they attract each other by gravitational force. Like poles of a magnet repel each other, while unlike poles attract each other by a magnetic force. Since the magnets are placed on a frictionless surface, they do not experience any frictional force.

12.

13.

14.

15.

157 Hence, the magnets will experience magnetic and gravitational forces. The correct answer is D. The force of gravity, which is exerted by Earth, attracts objects toward itself. Hence, a ball that is thrown up returns to the surface of Earth because of gravity. The correct answer is A. Here, we need to investigate the relationship between the force on the ball and its direction of motion. When the ball moves upward, the force exerted on the same is in the downward direction. This force, caused by Earth’s mass and that of the ball, is known as the force of gravity. This force tends to slow the ball down and eventually make it come back to Earth’s surface. Hence, the direction of force on an upward moving ball is from up to down. The correct answer is A. The acceleration due to gravity is 9.8 m/s2 on the surface of earth. It decreases above and below the surface of earth. Therefore, the right answer is (B) R| = 2R We know

gs  gh 

GM R2

__________

GM

R  h

2



(1)

2R

GM

 2R 

2

R

GM __________ (2) 4R 2 From (1) and (2), we get  gh 

16.

17.

18.

19.

1 9.8 gh  g   2.45m / s2 4 4 The force of gravity is always in towards the centre of the earth. Therefore, the right answer is (B) Mass of an object is amount of matter present in the object Therefore, the right answer is (A) The weight of an object is the force with which it is attracted towards the earth. W = mg Therefore, the right answer is (D) Weight (W) = mg Therefore at a given place, more is the mass , more is the weight. Therefore, the right answer is (C) www.betoppers.com

8th Class Physics

158 20.

21.

22.

The mass of a body is the amount of matter contained in it. It remains same any where in the universe. Hence, a body has same mass on the surface of earth and moon. Therefore, the right answer is (A) It is given that the water column in bottle II is decreased by 1 cm. Therefore, the water column in bottle I will also decrease by 1 cm. Hence, the new water level in bottle I is (7 – 1) cm = 6 cm. The correct answer is B. The pressure exerted by a liquid at a point within its volume increases with increasing depth, i.e., liquid pressure decreases with the increasing height of the liquid column. In the given situation, opening I is the topmost, while opening IV is the bottommost. This implies that among the given four openings, the height of the water column is the maximum near I and the minimum near IV. As a result, the water pressure at opening I is the minimum, while that at opening IV is the maximum. Consequently, water will come out with the least force through opening I and with the most force through opening IV.

26. 27.

28. 29.

30.

1 atm = 0.76 m Hg The correct answer is A. Standard atmospheric pressure is = 76 × 13.6 × 980 cm of mercury column The correct answer is A. Atmospheric pressure is measured by barometer The correct answer is D. Atmospheric pressure = pressure of mercury column of height 76 cm. h = 76 cm = 0.76 m, d = 13600 kgm–3, g = 10 ms–2 P = hdg = 0.76 × 13600 × 10 Pa = 1.03 × 105 Pa The correct answer is A. hmercury = 76cm hwater = ? –3 dmercury = 13.6gcm dwater = 1gcm–3 gmercury = g gwater = g We know that P = hdg But P mercury = P water  hmercury × dmercury×g =hwater × dwater×g

76  13.6  1033.6cm  10.34m 1 The correct answer is C.  h water 

31.

250 × P1 = 1000 × P2

P2 1 1  P = 4  or  P2 = 4 P1 1

23.

24.

Thus, the water coming out through opening I will 32. fall at the nearest distance from the container, while that coming out through opening IV will fall at the farthest distance from the container. 33. The correct answer is A. In the case of a nail being pushed into a wall, the area of contact is very small. Therefore, a huge pressure is required to push the nail into the wall. Atmospheric pressure can never be so high that it is able to push a nail into a wall. Therefore, atmospheric pressure does not play any 34. role in action II. The correct answer is B. The value of atmospheric pressure on the surface of earth at sea level is

76  13600  1000  105 Pa 100 The correct answer is A. The liquid used in Barometer is mercury. The correct answer is A. P = hdg = 

25.

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On compressing the gas, pressure increase and hence the collisions between the molecules increase. P1V1 = P2 V2 , 1× 20 = P2 × 50,

P2 = 20×

1 atm. 50

4  P1V1 = P2 V2 ,700× 250 = P2 ×  × 250  , 5   P2 = 875mm

Additional pressure = 875–700 =175 mm of Hg. 35.

PV = constant at constant temperature (by Boyle’s law).

Force and Pressure Solutions

159 7.

SUMM ATIVE WORKSHEET 1.

2.

3.

4.

5.

6.

Two examples of push force are as follows: (i) A heavy box at rest is pushed to move it from one room to another. This changes the state of motion of the box. (ii) A player pushes a football using his foot. This changes the state of motion of the ball. Two examples of pull force are as follows: (i) Rope is pulled to draw water from a well. This changes the state of motion of the water bucket. (ii) A drawer is pulled to open it. This changes the state of motion of the drawer. Two examples of forces that cause a change in the shape of an object are as follows: (i) Squeezing of a plastic bottle changes the shape of the bottle. (ii) Deformation of clay by pressing it between the hands. (a) To dr aw water from a well we have to __pull__ at the rope. (b) A charged body __attracts__ an  uncharged body towards it. (c) To move a loaded trolley we have to __either push or pull__ it. (d) The north pole of a magnet __repels__ the north pole of another magnet. (a) To stretch the bow, the archer applies a force that causes a change in its __shape__. (b) The force applied by the archer to stretch the bow is an example of __muscular__ force. (c) The type of force responsible for a change in the state of motion of the arrow is an example of a __contact__ force. (d) While the arrow moves towards its target, the forces acting on it are due to __gravity__ and that due to __friction__ of air. (a) To stretch the bow, the archer applies a force that causes a change in its __shape__. (b) The force applied by the archer to stretch the bow is an example of __muscular__ force. (c) The type of force responsible for a change in the state of motion of the arrow is an example of a __contact__ force. (d) While the arrow moves towards its target, the forces acting on it are due to __gravity__ and that due to __friction__ of air. When a blacksmith hammers a hot piece of iron, he uses his muscular force. This muscular force changes the shape of the iron so that it can be given a desired shape.

8.

9.

10.

11.

On rubbing an inflated balloon with a piece of synthetic cloth, it becomes charged. A charged body attracts an uncharged body. When this charged balloon is pressed against a wall, it sticks to the wall. The force acting between the charged balloon and the wall is the electrostatic force. We make use of muscular force to hold a bucket of water above the ground. This muscular force acts against the force of gravity that pulls the bucket towards the ground. The two forces are equal in magnitude but opposite in direction. Therefore, the net force on the bucket is zero. Hence, there is no change in its state of motion. The two forces acting on the rocket are the force of gravity, which pulls the rocket towards the ground, and the force of friction due to earth’s atmosphere, which opposes its motion. (d) The rise of water in the dropper is due to atmospheric pressure. The rise  of  water  in  a  dropper  is  due  to atmospheric pressure. When all the air escapes from the nozzle, the atmospheric pressure, which is acting on the water, forces the water to fill the nozzle of the dropper. P = 25,000 Pa A = 0.50 cm2 = 0.000050 m2 F=?

P

12.

 F = P×A  F = 25,000 × 0. 000050 = 1.25 N F = 100 dynes. A = 0.02 cm2 P

= 13.

F A

F 100  dynes/cm2 A 0.02

100  100 dynes/cm2 =5000 dyne/cm2 2

mass of the block (m) = 2 kg weight of the block (w) = mg = 2 kg × 9.8 m/s2 = 19.6 kg m/s2 = 19.6 N www.betoppers.com

8th Class Physics

160 17.

F = 500 N A = 0.5 cm2 = 0.00005 m2

 18.

Side of cube (a) = 2 cm =

2 m = 0.02 m 100

Area of cube = a2 = 0.02m × 0.02 m = 4 × 10–4 m2



19.

3 m = 0.03 m Side of cube (a) = 3 cm = 100

F  F = P×A A

= 5N /m2 × 9 × 10–4 m2 = 45 × 10–4 N 15.

F = 10 N

A  0.01cm2  

20.

16.

P

F 500000 2 m = 0.2 m2 = P 2500000

Given h1 = 70 cm = 0.7 m

the

same

We P = hdg Therefore, for a given pressure, height is inversely proportional to the density of the liquid used in the barometer. As the density of oil is less than that of water, the height of oil barometer would be more than that of water barometer.

21.

As the air enters through the tube, the pressure reading will have an error.

22.

As P and d are same, height of the liquid column would be inversely proportional to each other.

0.01 2 m 10000

h1g1 = h2g2

1 1 m 2  6 m 2 = 10–6 m2 10,000  100 10



A=?

h1d1 0.7  13600  h2  d  1000 2

Area of cube (A) = a2 = 0.03 m × 0.03 m = 9 ×10–4 m2 we know that,

P

F = 500000 N, P = 250000 Pa,

d1 = 13600 kg /m3 h2 = ? d2 = 1000 kg /m3 Since the pressure is therefore we have h1d1g = h2d2g

F weight 19.6 N P =  A Area 4  104 m 2

P = 5Pa = 5N/m2



F 500 = 107 Pa A 0.00005

we know that A 

= 4.9 × 104 N/m2 = 4.9 × 104 Pa 14.

P

F 10  6 Pa = 10 × 106 Pa = 107 Pa A 10

F = 60 kg wt = 60 × 9.8 N

h2 

h1g1 76  g   76  6  456cm g g2 6

The liquid column rises to 456 cm from 76 cm. 23.

Atmospheric pressure = Pressure of mercury column of height 76 cm. h = 76 cm = 0.76 m

 60 

98  6  98 N 10

Area under are foot (A) = 180 cm2 =



P

d = 13600 kgm–3 g = 10 ms–2

180 2 m 10000

F 6  98  10000 98 Pa   103 Pa = A 180 3

= 32.6 × 103 Pa www.betoppers.com

P = hdg = 0.76 × 13600 × 10 P = 1.03 × 105 P Ther efore, the atmospheric pressure 1.03× 105 P

is

Force and Pressure Solutions 24.

161

hmercury = 76cm

dmercury = 13.6gcm–3

gmercury = g

hwater = ? –3

dwater = 1gcm

The force exerted by the air column is nothing but its weight.  Force exerted by the air column

gwater = g

= weight of air column

We know that P = hdg

= mass of air column × acceleration due to gravity = mg

Pmercury = Pwater hmercury × dmercury= hwater × dwater

The mass of air column of 1m2 cross sectional is approximately 104 kg and acceleration due to gravity is 9.8 m/s2  10m/s2.

76  13.6  1033.6cm  10.34m 1 Therefore, the height of the water column is equal to 10.34 m.  h water 

25.

 Atmospheric pressure



Atmospheric pressure (P0) = 105 P Depth (h) = 5.1 m

= 105 pascal

Density (d) = 1 g/cc = 1000 kg/m3

So, atmospheric pressure is 105 N/m2.

g = 10 m/s2

This means that on an area of 1m2 of the surface of earth, a force of 105 Newton is acting. This is equivalent to the weight of 140 adults of mass, 70kg each. This is also approximately the weight of a body of mass, ten tonnes (10,000 kg). This is also equal to the weight of two elephants, approximately.

Pdepth = Patmosphere+ Pwater column = P0 + hdg

___

(1)

Substituting the above values in (1), we get P = 105 + 5.1 × 103 × 10 Pa = 15.1 × 104 P

b) Considering the head as a sphere with an average diameter of 20cm the resulting surface area is about 1256cm2. We know that for every one cm2 of area a force of approximately 10 Newton acts due to atmospheric pressure. Therefore the total force of about 1256 10N acts on our head which is equal to the weight of 18 adults weighing 70kg each.

Therefore, the total pressure at the bottom of the lake is 15.1 × 104 P 26.

h1 = 70 cm = 0.7 m d1 = 13600 kg /m3 h2 = ? d2 = 1000 kg /m3 We know, P = hdg

28.

The molecules in air beat against a window pane at the rate of two million molecules per square inch per second. Such enormous force of pressure of a gas is enough to power a steam engine or turbojet. This impact would shatter the glass if an equal number of molecular blows were not rained against it from the other side of the pane as well.

29.

There is no external pressure in space. Astronauts, while travelling through space, wear heavy space suits so as to create an external pressure. These space suits prevent the bursting of body cells and bleeding of nose and ears.

30.

At the top of a tall mountain, the pressure is less. This causes our ears to pop in order to balance the pressure between the outer part and the inner part of our ears.

As the pressure is the same, P1 = P2 i.e., h1d1g = h2d2g

 h2 =

h1d1 0.7 ×13600 = = 9.52 m. d2 1000

Therefore, the height of the water column is 9.52 m. 27.

Force m  g 104  10   N / m 2 = 105 N/m2 Area 1 1

a) We know that

Pressure =

Force Area

Atmospheric pressure is the force exerted by an air column of 1m2 cross-sectional area on the ground.

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8th Class Physics

162 31.

32.

To make a barometer compact, the height of the tube which holds the liquid column, should be less. For this, we need a high density liquid. The best suited liquid for this is mercury (Hg) with a high density of 13.6 g/cc. Further, due to its high surface tension, it is non sticky to glass. And is easily viewed through glass because of its shine. Besides, mercury has a high boiling point. At a higher altitude the pressure becomes very much less and the passengers encounter dangers like bursting of body cells, bleeding of noses and ears and also an increase in the blood pressure. To avoid these problems an artificial pressure is 6. created inside the plane.

HOTS WORKSHEET 1. 2.

3.

i – c, ii – a, iii – b (i) gravitational force (ii) Gravitational force (iii) Electrostatic force (iv) Muscular force (v) Magnetic force The S.I unit of pressure is pascal or Nm–2. The C.G.S. unit of pressure is dyne cm–2.

5.

20 1 kg  kg Mass of book (m) = 20 g  1000 50 Thrust (T) = ? Acceleration due to gravity (g) = 10 ms–2 We know, Thrust (T) = mg

T

1 1  10  N  0.2 N 50 5

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W=T=?

Therefore, the thrust acting on the table by the book is 0.2 N. Area of one foot = 275 cm2 = 275 × 10–4 m2 ( 1 cm2 = 10–4 m2) But total force is acting through the four legs. So, total area of the four feet (A) = 4 × 275 ×10–4 m2 Mass of the elephant (m) = 2200 kg g = 10 ms–2 Pressure exerted (P) = ?

Force exerted Area Force exerted by the elephant is equal to weight of the elephant.

N dyne ? __________ (1) 2 m cm 2 We know, 1N = 105 dyne and 1 m2 = 104 mc2 Substituting the above values in L.H.S of eq. (1),

4.

m = 20g

We know, pressure  P  

1

1N 105 dyne  105  2 we get, 2  4 2   4  dyne cm m 10 cm 10   –2 –2 1N m = 10 dyne cm .  So, the relation between S.I. and C.G.S unit of pressure is 1N m–2 = 10 dyne cm–2. i) Pressure = 2 × 105 pascal = 2 bar ( 1 bar = 105 Pa) 5 ii) Pressure = 2 × 10 pascal = 2 × 105 × 10 dynes cm–2 ( 1 pascal = 10 dyne cm–2) iii) Pressure = 2 × 105 pascal = 2 × 105 × 102 pascal = 2 × 103 millibar ( 1 millibar = 102 Pa)

Therefore, the thrust acting on the table by the book is 0.2 N.

mg ( F = W = mg ) A Substituting the values in the above equation, we get, P

2200  10 22  10 3  10 4 22  107   4  275  104 1100 11 102 P = 2 ×105 Pa Therefore, pressure exerted by the elephant is 2 × 105 Pa. Thrust acting = Force applied F = 1000 g. wt. The area of contact of the nail with the wall (A) = 1/100 cm2 Pressure (P) = ? P

7.

We know, Pressure 

Thrust acting Area of contact

1,000  1000  100  1/100 = 105 g.wt cm–2 ( 1g.wt = 980 dyne) = 980 × 105 dyne cm–2 Note: Don’t get confused as the area of both the ends of the nail are given, We need to take only the area of contact with the wall. 

Force and Pressure Solutions 8.

9.

163

Men wearing flat shoes have more area of contact with the ground, hence exert less pressure.

1 Pressure  Area Women wearing pointed heals have less area of contact with the ground, hence, exert more pressure. Therefore, the foot prints of women wearing pointed heels are deeper than the foot prints of men wearing flat shoes. Area of contact of each tyre = 10 cm2 Area of contact of two tyres (A) = 2× 10 cm2 = 20 cm2 = 2× 10 cm–4 m2 ( 1 cm2 = 10–4 m2 ) Pressure (P) = 3 bar = 3 × 105 Pa ( 1 bar = 105 Pa ) Weight of the person = pressure × area. W = P × A = 3 × 105 × 20 × 10–4 = 600 N Weight of the person is 600 N.

W 600 Mass of the person (m)  g  10  60 kg 10.

Therefore, the mass of the person is 60 kg. a) Girl Mass (m) = 50 kg Area of a heel (a) = 1cm2 Area of two heels (A) = 2 × 1cm2 cm2 = 2 × 10–4 m2 Force (F) = mg = 50 × 10 = 500 N Pressure exerted by the girl

F 500   250  104 Pa 4 A 2  10 b) Elephant Mass (m) = 4000 kg Area of one leg = 250cm2 Area of four feet (A) = 4 × 250 cm2 = 1000 cm2 = 1000 × 10–4 m2 = 0.1m2 Force (F) = mg = 4000 × 10 = 4000 N Pressure exerted by the elephant F 2000   20  104 Pa A 0.1 It is clear that the girl is exerting more pressure on the ground than that of the elephant. Side of cube (l) = 5 cm  Volume of cube (V) 3

1  1      m3  20  8,000 3

5 1 m m 100 20

5cm

W=T = ?

Density of block (d) = 5 g cm–3 = 5 × 1000 kgm–3 = 5000 kgm–3 [  1 g cm–3 = 1000 kg m–3 ] Thrust ( T ) = ? We know, T = mg

m   T = V × d × g  d   m  V  d  V   T=×d×g ( Vcube = l3) 1  5, 000  10  6.25 N 8, 000 Therefore, the thrust acting on the table is 6.25 N. a) Radius of the sphere (r) = 7 cm Density of the sphere (d) = 0.21 g cm–3 Thrust (T) = mg Let us first calculate the mass of the sphere. Mass = volume × density  T

12.

4 4 3  r 3  d    7   0.21 3 3 =2

11.

block

4 Mass   343  0.21  30.184 kg 3 1  kg  0.30184 kg 1000 Thrust = 0.30184 × 10 N Therefore, force exerted by the sphere is 3.0184 N b) Radius of the base (r) = 7 cm Length of the cylinder (l) = 49 cm Density (d) = 2.2 g cm–3 Acceleration due to gravity (g) = 10 ms–2 To find the thrust exerted, first we have to calculate the mass of the cylinder. Mass of the cylinder = V × d  r 3  d ( volume of the cylinder = pr2l) Mass  301.84 



22  7  7  49  2.2  16601.2g 7

1 kg  16.601 kg 1000  Thrust exerted = mg = 16.601 × 10 = 166.01 N. So, the thrust exerted by the cylinder is greater than the sphere. www.betoppers.com  16601.2 

8th Class Physics

164 13.

Let P1, P2 and P3 be the pressure exerted by the 14. brick while resting on different faces. The dimensions of the given brick are 20 cm × 10 cm × 5 cm Case (i): When the block is resting on 20 cm × 10cm face. 15.

P1V1 = P2 V2 , 1× 20 = P2 × 50,

P2 = 20 ×

P 1V1 = P 2V2

 9.962 ×104 × 95 = 10.13 ×104 × V2

m c 5 20cm

Thrust acting = Weight of the brick T = 500 g wt Area of contact (A) = 20 cm × 10 cm Pressure exerted (P1) = ?

Thrust 500 Pressure   Area 20  10

 P1 = 2.5 g.wt cm2 Case (ii):When placed on 20 cm × 5 cm face m c 500 gm. Wt 0 1 20cm

⇒ V2 = 16.

17.

18.

= 10 cc

4  P1V1 = P2 V2 , 700 × 250 = P2 ×  × 250  , 5   P2 = 875mm Additional pressure = 875 – 700 =175 mm of Hg. R1 = 2r

4 4 4 3 V1  R13  V1    2r    8r 3 = P1 = P 3 3 3 R2 = r 4 4 V2  R 32  V2  r 3 3 3 P2 = ? P1V1 = P2V2

Thrust 500  Area 20  5

 P2 = 5 g.wt cm–2 Case (iii):When placed on 10 cm × 7 cm face

9.962×104 × 95 = 93.424 cm 3 4 10.13×10

P1 V1 = P2 V2  10 4 × 100 = 10 5 × V2

 or  V2

Thrust = Weight of the brick = 500 g. wt. Area of contact (A) = 20 cm × 5 cm Pressure exerted (P2) = ?

Pressure 

1 atm. 50

19.

4 4 P  8r 3  P2  r 3  P2  8P 3 3 P1 = 760 mm P2 = 740 mm d1 = 2.8 g/cc d2 = ? t1 = t2 = 26°C = constant P K d



5 cm

Thrust = Weight of the brick = 500 g. wt. Area of contact = 10 cm × 5 cm Pressure (P3) = ?

Thrust 500  Area 10  5  P = 10 g.wt cm–2 From the above three cases, it is clear that, as the area of contact decreases, the pressure exerted increases and is greater when the brick rests on its 10 cm × 5 cm face. www.betoppers.com

P1 P2 P   d 2  2  d1 d1 d 2 P1

740  2.8  2.726 g / cc. 760 P1 = P atm, P2 = 5P atm d1 = d gm/lit., = d2 = ? d2 

20.

Pressure 



P1 P2 P   d 2  2  d1 -----(a) d1 d 2 P1

d2 

5P  d  5d gm/litre P

Force and Pressure Solutions

IIT JEE WORKSHEET 1.

2.

3.

4.

5.

6.

7.

8.

Alternative A clearly shows a man pushing a car. Hence, this action involves pushing force. The correct answer is A. To lift the load, William is using his hand to pull the rope of the pulley system. The pulley system rotates the wheel, as a result the lift the load. Hence, William is using his muscular force to life the load. The correct answer is B. A ball falls on the ground because of the gravitational pull of Earth. There is no contact between the ball and Earth. Hence, this process is caused by indirect application of force. The correct answer is A. When two magnets are brought near each other, they attract or repel each other depending on their polarity. The movement is caused by magnetic force between them which is an indirect force as magnets are not in contact. The correct answer is D. Depending upon the nature of charges, the charged bodies can attract or repel each other. As Maria combs her dry hair, the hair and comb get oppositely charged and so, they start attracting each other. The force is known as electrostatic force. The correct answer is B. Every object near Earth’s surface gets attracted by Earth’s gravitational force. Hence, the book lies on the table as a downward gravitational force acts upon it. The correct answer is C. All bodies attract each other by gravitational force. Earth attracts all bodies toward it by gravitational force. The stone is also being attracted by Earth’s gravitational pull. Hence, the stone falls down. The correct answer is B. Blake is pulling the box on a rough surface; hence, there is a friction between box and the surface. This frictional force tends to stop or oppose the motion of box. Hence, Blake is working against friction. The correct answer is B.

165

9.

The magnet placed at the bottom of the ride repels the magnet that is fitted below the capsule, as both have the same poles facing each other. This causes the ride to slow down as it reaches the surface because magnetic repulsions increase as the distance between the base of the ride and the capsule decreases. The correct answer is B.

10.

The bar magnet exerts magnetic field that causes iron fillings to get attracted toward it. Hence, iron filings get attracted toward a bar magnet because of the presence of magnetic force. The correct answer is C.

11.

The gravitational force that exists between Moon and Earth is strong enough to hold Moon in its lunar orbit around Earth. Hence, Moon revolves around Earth because of the presence of gravitational force. The correct answer is A.

12.

Earth exerts gravitational force that causes the stone to fall back on Earth’s surface. The correct answer is D.

13.

Force is defined as a push or a pull. The correct answer is A.

14.

Trenton uses his arms to rotate the lever which draws the bucket out of the well. Hence, he uses his muscular power to pull the bucket. The correct answer is C.

15.

Trenton is pulling the bucket full of water away from Earth’s surface, i.e., the base of the well. Hence, he is working against the gravitational force of Earth. The correct answer is A.

16.

F = 16 N,

A  50cm2 

50 m2 ; 10000

We know that P 

 P

P=?

F A

16  10000 pa = 3200 pa 50

The correct answer is A. www.betoppers.com

8th Class Physics

166 17.

200 2 m 10000 P = 27500 Pa , F = ?

23.

A  200cm 2 



We know that P 

F A

P 24.

The correct answer is C. F = 300 N, P= 1500 Pa, A = ? We know that P 

 A =

F 100  Pa  100 Pa A 1

The correct answer is C.

200 N = 550 N  F = P × A = 27500  10000 18.

F = weight of box = 10 kg wt = 10 × 10 N = 100 N, A = 1 m2

i) Force (F) = Weight of brick = 2.5 kg wt = 2.5 × 10 N = 25 N Area of base = 10 × 5 cm2 =

F A

 Pressure =

F 300 1 2   m P 1500 5

50 m2 10000

F 25   10,000 Pa A 50 = 5000 Pa

1  10,000 cm2 = 2000 cm2 5

The correct answer is D. 19.

P = 50,000 Pa, A = 0.20 m2; F = ?

ii) Force = 25 N

F We know that P  A

Area of base = 5 × 2 cm2 =

 F = P × A = 50,000 × 0.20 N = 50,000 

20 = 100

P

10,000 N

10 m2 10000

F 25   10000 Pa = 25000 Pa A 10

The correct answer is A. 20.

The thrust on the surface of the ground due to the block = weight of the block = mg = 2 kg × 9.8 m/s2 = 19.6 kg m/s2 = 19.6 N

The correct answer is C. 21.

F = 500 N; A = 0.05 m2; p = ?

F We know that P  A P

F 500 500  100   Pa = 104 Pa A 0.05 5

The correct answer is A. 22.

P = 6000 Pa; F = 120 N; We know that P 

 A

A= ?

F A

F 120 2 1 2  m  m P 6000 50

The correct answer is D.

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The correct answer is A. 25.

F = 50 kg f A = 1m × 0.5 m = 0.5m2



P

F 50 kgf 2  A 0.5m 2 = 100 kg f/m

The correct answer is B. 



3. FRICTION SOLUTIONS

FORMATIVE WORKSHEET 1.

2.

3.

4.

5.

6.

7.

8.

The block stops after traveling some distance, i.e., its speed decreases. This implies that a force is applied in the direction opposite to its motion. The block is moving from left to right. Therefore, the frictional force acting on the block is toward the left. The correct answer is A. When there is a relative movement between two bodies in contact, frictional force always opposes the direction of motion. Since the file cabinet is being moved over the floor, frictional force is acting between the file cabinet and the floor.The correct answer is B. When a car moves, air resists its motion. As a result, cars are designed in streamline shapes. Streamline shapes of cars ensure that this air resistance is minimized. The correct answer is A. Friction can be defined as the resistance hindering the relative movement, or the propensity toward such a movement, of two surfaces that are in contact. It is because of this resistance that the ball stops after rolling a certain distance. Hence, the ball stops because of the presence of frictional force. The correct answer is B. The friction between the tires and the road generates heat. This heat causes the heating up of the tires. The correct answer is C. Friction is a kind of force that always acts against the direction of movement of a body. As a result of the friction between the box and the surface, mechanical energy gets converted into heat energy. This heat energy raises the temperature of the bottom of the box. The correct answer is D. A lubricant reduces friction. Hence, using a lubricant on the door will ensure that it will open more easily and not make a squeaking sound. The correct answer is B. When muscular force is applied by Thaddeus in the form of pedaling, the bicycle moves. When he stops pedaling, the frictional force between the tires and the ground slows the bicycle. The direction of the frictional force is opposite to the direction of motion of the bicycle. Hence, the speed of the bicycle decreases. The correct answer is A.

9.

The force of friction on a body sliding on a horizontal surface is, F = m mg Given, F = 30 N, mg = 50 N



F 30   0.6 mg 50

The correct answer is A 10. Frictional force F =  × mg Given, m = 25 kg = 0.25, g = 10 ms-2 F = 0.25 ×25 ×10 = 62.5 N The correct answer is C 11. Talcum powder is used to reduce friction on the carom board. The powder particles help create a smooth layer over the board’s surface. Thus, talcum powder reduces the roughness of the board. As a result, the carom men and striker can slide easily on the board. The correct answer is A. 12. The given boxes are kept between two vertical walls. When the first box is pulled upward along the walls, frictional force acts between the surfaces of the box and the walls, in the downward direction. When the walls are lubricated with oil, this frictional force is reduced. Oil eliminates the roughness (as a result of friction) between the surfaces of the box and the walls. This allows the box to slide over with minimum friction. Hence, Calvin had to apply less force to lift the second box.The correct answer is C. 13. Frictional force always opposes the motion of a body. This force can be reduced by using wheels under the moving body. Wheels under John’s table reduce the surface of contact between the floor and the table. As a result, frictional force between the table and the floor decreases.The correct answer is B. 14. Pushing the toy car on a smooth surface will reduce the friction between the wheels of the car and the surface. This will increase the distance covered by the toy car, as compared to the distance covered by it on the rough surface.The correct answer is D.

168 15. In order to reduce the friction on the surface of his driveway, Ted should make its surface very smooth. He should do so because frictional forces become very weak on smooth surfaces. Hence, in order to reduce the friction on the surface of his driveway, Ted should pour oil it.The correct answer is D. 16. In order to increase the friction on the surface of his driveway, Ted should increase the roughness of its surface. He should do so because greater frictional force is generated on rough surfaces. Hence, in order to increase the friction on the surface of the driveway, Ted should cover it with small pebbles.The correct answer is A. 17. A sledge will travel the farthest on an icy road because the least frictional force will be developed on an icy road. This means that the force which will oppose the motion of the sledge will not be very powerful on an icy road. Correct answer is B. 18. The frictional force between the surface and the nail opposes the motion of the nail toward the magnet. The nail moves easily if this frictional force becomes less. Frictional force is less on an oily surface as compared to the force of friction on a cemented, carpeted, or sandy surface. Therefore, the distance d will be relatively large for an oily surface.The correct answer is D. 19. In a mechanical system, friction opposes motion between different parts of the system. Work is done against these frictional forces. Hence, heat energy is produced by the system.The correct answer is D. 20. A gear is a wheel with teeth along its edge. Gears are used in the transformation of rotational motion. They are not used to reduce friction.The correct answer is B. 21. Friction is the resistive force that arises when two bodies or surfaces move in contact with each other. Since it is a resistive force, it opposes the motion of a body. Hence, the direction of the frictional force on the ball is opposite to its direction of motion. This is correctly represented in the figure given in alternative A.The correct answer is A. 22. Cars are designed to be more aerodynamic in order to increase their speed by reducing the air resistance acting on them. The correct answer is A. 23. The shape of an airplane is made streamlined to reduce air friction. The streamlined shape allows the easy passage of air. Therefore, the heat generated by air friction becomes less.The correct answer is D. www.betoppers.com

8th Class Physics 24. Initially, the ball is moving on a smooth frictionless surface. Hence, speeds v1 and v2  will  be  equal. After the ball enters the grassy surface, its speed will decrease because of the friction between the ball and the grass. Hence, the relation among the speeds of the ball at the three given points is v1  = v2 > v3 23. Static friction is always greater than kinetic friction. Sliding and rolling frictions are types of kinetic friction. Hence, static friction is greater than rolling friction. In sliding, the motion is opposed to the friction of the surface, more than that opposed in rolling friction. Hence, rolling friction is less than sliding friction. Hence, Static friction > Sliding friction > Rolling friction The correct answer is B. 24. Sliding friction is always greater than rolling friction. Box II is sliding on the ground because of the force applied by Shyam. Box I is rolling on wheels because of the force applied by Ram. Hence, Shyam has to apply a greater force to maintain a steady motion of his box. The correct answer is A. 25. Lubricants are thick, oily liquids. They cause two surfaces to easily slip on one another. This is because they reduce the friction between the two surfaces by creating a smooth layer between them. When lubricants are applied between two machine parts that are moving in contact with each other, the friction between them is reduced. This helps the machine parts to move faster without much wear and tear, thereby enhancing the efficiency of the machine.The correct answer is A. 26. Friction plays an important role in the motion of an object. When a ball is rolled down an inclined plane, the rolling friction opposes its motion. Also, when an object moves through air, air friction (also called air drag) opposes its motion. The following figure shows the names and directions of forces acting on the rolling ball.

Hence, arrows II and III respectively  represent the directions of the rolling frictional force and air friction acting on the ball. The correct answer is C.

Friction Solutions

169

CONCEPTIVE WORKSHEET 1.

2.

3.

The motion of the boxes on the ground is opposed by the friction offered by the irregularities present on the surface. The magnitude of friction depends on the mass of the moving object. The greater the mass of the moving object, the more is the friction and vice-versa. Here all the boxes are cubical in shape and made up of the same material. Therefore, the friction of each box will depend on its volume, which is proportional to its mass. It is given that the distances are related as d3  > d2  > d4 > d1 . Hence, it can be inferred that the frictions offered by the ground to the boxes are related as: Friction on box 1 > Friction on box 4 > Friction on box 2 > Friction on box 3 Hence, the masses, i.e., the volumes in this case are related as: Mass of box 1 > Mass of box 4 > Mass of box 2 > Mass of box 3 Therefore, the sides of the boxes are related as: a1 > a4  > a2  > a3 The correct answer is B. Irregularities on surfaces offer resistance to the motion of an object over it. This resistance is called friction. The more the irregularities on the surface, the more is the friction. Hence, more force is required to move an object placed on an irregular surface. Since irregularities present on a grass field are more than those present on a marble floor, more force is required to move a box placed in the grass field. Boxes I and II are placed in a grass field. Therefore, boxes I and II require more forces to be moved than those required by boxes III and IV. The opposing force of friction does not remain the same for heavier and lighter objects. Frictional force is greater for heavier objects and smaller for lighter objects. Since box II is  heavier  than box I, box II requires the maximum force to be moved. The correct answer is B. The force of friction always opposes motion. The box is moving rightward. Hence, the force of friction between the box and the floor is acting leftward. Since Mahesh is applying a force of magnitude Fm leftward, Fm and frictional force have the same direction, i.e., leftward, as shown in the following figure.

4.

5.

The correct answer is B In order to increase the friction on the surface of his driveway, Ted should increase the roughness of its surface. He should do so because greater frictional force is generated on rough surfaces. Hence, in order to increase the friction on the surface of the driveway, he should cover it with small pebbles. The correct answer is A. Here, mass of the body, m = 35 kg, Acceleration due to gravity, g = 10 m s2 Coefficient of friction, = 0.5 Frictional force. F =  × mg = 0.5 × 35 × 10 = 175 N.

6.

7.

8.

9.

Hence, the frictional force acting on body is 175 N. We know that ball bearings are rolling element bearings used to reduce friction. Ball bearings are used in the pedals of a bicycle as well as in the front and rear wheel-axles to reduce its friction.The correct answer is D. In order to answer this question successfully, the student must understand that when balls are placed under the plank of wood, it rolls instead of sliding. Hence, its friction is the least and its speed increases the most. The correct answer is C. Friction is maximum on a rough surface and it is minimum on a smooth surface. In addition, application of lubricants on surfaces reduces the friction. Oil is a lubricant. Application of oil on a marble surface reduces its friction. Hence, friction of an oily marble floor will be the minimum among the given surfaces. The correct answer is B. Friction opposes the relative motion between two surfaces in contact. It depends on the nature of surfaces in contact. The force of friction is more when the irregularity present between the two surfaces is large. Oil is a good lubricant. Application of oil on a surface increases the www.betoppers.com

8th Class Physics

170

10.

11.

12.

13.

smoothness of the surface and decreases its roughness. The friction of a surface decreases when oil is applied on it. The correct answer is D. Friction is an important aspect of life. We are able to walk on a surface because of the presence of friction between the ground surface and our feet. If friction were to become zero, then the act of walking would become impossible, causing us to slip and fall. One is able to write on a surface (for example, a blackboard) only if friction is present on the surface. We are able to write on a blackboard because of the presence of friction between the board and the chalk. The correct answer is B. The force of friction present between any two given objects depends on the nature of the motion of one over the other. Friction is offered by the interlocking of surfaces. When two surfaces are at rest over one another, the interlocking is the maximum. This is called static friction. Maximum force is required to overcome static friction (fst). Sliding friction, f sl , is greater than rolling friction, froll. This is because it is more difficult to slide over interlockings than roll them over. Hence, these frictions are related as: fst  > fsl  > froll The correct answer is B. Static friction is the measure of the minimum force required to overcome the interlocking between two surfaces in contact, to move one over the other. The correct answer is A. The friction between two surfaces always opposes the relative motion of the surfaces. When the ball moves upward along the inclined plane, frictional force acts downward. Similarly, when the ball rolls down, the force of friction acts upward. In both the cases, friction opposes the motion of the ball, as shown in the given figure.

The correct answer is A. 14. A force of friction arises when two objects move in contact with each other. The opposing force that acts on an object that is moving in a fluid is called drag. An object that is moving in air must overcome this opposing drag. To reduce this drag, www.betoppers.com

F1 car is constructed streamline in shape. The given figure shows the direction of air drag with respect to the direction of motion.

The streamline shape of the F1 car allows air to flow over its surface easily and smoothly. This helps in achieving a relatively higher speed at a relatively less time and less fuel consumption.The correct answer is A.

SUMM ATIVE WORKSHEET 1.

(a) motion (b) nature (c) heat (d) reduces (e) less (f) Opposes (g) Increases (h) Lesser (i) Less (j) Friction (k) Force of friction (l) Sliding friction (m) Ball bearing (n) Fluid or air (o) Decreases (p) Very small (q) Air resistance (r) Drag (s) Streamline (t) Increased, decreased (u) Reduce (v) Sliding friction (w) Rolling friction (x) Greater (y) Contact

2.

(c) static, sliding, rolling Friction comes  into  play  when  irregularities present in the surfaces of two objects in contact get interlocked with each other. Static friction comes into play when we try to move an object which is at rest. Sliding friction comes into play

Friction Solutions

3.

4.

when an object slides over the surface of another object. In sliding friction, the time given for interlocking is very small. Hence, interlocking is not strong. Therefore, less force is required to overcome this interlocking. Because of this reason, sliding friction is less than static friction. Similarly, the area of contact in case of rolling friction is smallest as compared to static or sliding friction. This area of contact changes gradually because of rolling. Hence, rolling friction is lesser than both static and sliding friction. Thus, the correct sequence is—static, sliding, rolling. (a) wet marble floor, dry marble floor, newspaper and towel Force of  friction  depends  on  the  nature  of surfaces in contact. The rougher the surface, the more is the friction between the surfaces in contact and vice-versa. Roughness present in the given surfaces can be arranged in an ascending order as wet marble floor, dry marble floor, newspaper and towel. Hence, the cor rect sequence of these surfaces when arranged according to the increase in the force of friction acting on the car is—wet marble floor, dry marble floor, newspaper and towel. When a book slides on the writing desk, a frictional force acts between the book and the surface of the desk. The direction of frictional force on the book is opposite to the direction of its motion and acts in upward direction, as shown in the following figure.

171 7.

Force of  friction  arises  because  of  interlocking of irregularities on the two surfaces in contact. When a heavy object is placed on the floor, the interlocking of irregularities on the surfaces of box and floor become strong. This is because the two surfaces in contact are pressed harder. Hence, more force is required to overcome the interlocking. Thus, to push the heavier box, Seema has to apply a greater force than Iqbal. 8. Friction comes into play when irregularities present in the surfaces of two objects in contact get interlocked with each other. In sliding, the time given for interlocking is very small. Hence, interlocking is not strong. Therefore, less force is required to overcome this interlocking. Because of this reason, sliding friction is less than static friction. 9. Advantages of friction: (i) We are able to walk because of friction. (ii) Friction between the tip of the pen and a paper allows us to write. Disadvantages of friction: (i) Tyres and soles of shoes wear out because of friction. (ii) Friction between the different parts of machines produces heat. This can damage the machines. 10. When a body moves through a fluid, it experiences an opposing force which tries to oppose its motion through the fluid. This opposing force is known as the drag force. This frictional force depends on the shape of the body. By giving objects a special shape, the force of friction acting on it can be minimised. Hence, it becomes easier for the body to move through the fluid.

HOTS WORKSHEET 5.

6.

We are  able  to  walk  because  of  the  friction present between our feet and the ground. In order to walk, we push the ground in the backward direction with our feet. The force of friction pushes it in the forward direction and allows us to walk. The force of friction between the ground and feet decreases when there is soapy water spilled on the floor. Hence, it becomes difficult to walk on the soapy floor. Sportsmen use shoes with spikes because these shoes give them a better grip while running. This is because the force of friction between the shoes and the ground increases with the help of spikes.

1.

2.

The stone will go the maximum distance on the surface that has the least friction. An icy surface has the least friction among the given surfaces. Hence, before stopping, the stone will go the maximum distance on the icy surface. The correct answer is A. The joints of the door are producing sound because of an increase in friction between the two surfaces. To reduce this, he must apply oil between the joints of the door. Application of oil helps in reducing surface irregularities and hence, reduces friction. This is because oil is a very good lubricating agent. The correct answer is D. www.betoppers.com

8th Class Physics

172 3.

4.

5.

6.

7.

When an object moves through air, its motion is opposed by air or wind resistance known as ‘drag’. This eventually results in consumption of more energy of the cyclist to sustain a constant speed. During a cycling race, the cyclist bends his body forward to make the system streamlined to reduce the air drag (as shown in the given figure).

This is possible as most of the air flows past smoothly through the cyclist’s streamlined shape. Hence, less energy is required to move or maintain high speed. The correct answer is C. The friction of a surface is increased by increasing its roughness. The roughness of a carom board does not increase when it is powdered. Hence, the friction of a carom board surface does not increase when it is powdered.The correct answer is D. The roughness over a surface increases its friction. Sparks are produced when a matchstick is rubbed against the rough surface of a matchbox. This is caused by the large friction produced between the matchstick and the box. Hence, the sides of matchboxes are made rough.The correct answer is D. Friction plays an important role in life. Friction has disadvantages as well as advantages. One of the disadvantages of friction is the wearing out of shoes when used for a long time. The reason for this is the heat developed by friction between the shoes and the ground surface while walking.The correct answer is B. When a body that is resting on a surface is moved along the surface, a static friction is generated between the two contact surfaces. The minimum force required to simply move the body is equal to the magnitude of the generated static friction. The static friction offered by a marble floor is less than that offered by a cardboard sheet and sandpaper. Again, the static friction offered by a cardboard sheet is less than that offered by a sandpaper. Hence, the minimum forces required to simply move the box on the different surfaces are related as Fm  u). The image of an object is formed on the retina of the eye. It is a delicate membrane, which contains a large number of light sensitive cells. These light sensitive cells are called cones. They transform the light signal into electrical signal, which is then sent to the brain. The correct answer is C. In the human eye, the ciliary muscles are attached with the eye lens. These muscles on contracting or relaxing make the eye lens thin or thick respectively. The focal length of the eye lens also changes with its thickness.The correct answer is C.

50.

51.

52.

53.

54. 55.

56.

57.

191 The optic nerves transmit the electrical signals generated by the retina to the brain. The correct answer is B. The primary function of the crystalline lens present in the human eye is to focus the light coming from an object onto the retina. The correct answer is B. The different parts of the human eye are shown as follows: The pupil (labelled as I in the given diagram) is the central, black, circular part of the human eye. The light coming from an object enters the eye lens (labelled as II in the given diagram) via the pupil. The eye lens helps in focusing this light onto the retina. The correct answer is B.

Iris controls the size of the pupil which in turn controls the amount of light entering into the eye. Cilliary muscles compress the eye lens and control its power. A real image is formed on the retina. The function of retina is to sense this image and send signals to the brain through optic nerve. Blindness is caused primarily because of the defect in the cornea. Hence, statement IV is incorrect. The correct answer is D. In the case of an object situated at a distance, almost parallel rays of light enter our eyes. Hence, the eye lens does not require a high converging power. As a result, the ciliary muscles get released and make the eye lens thinner, which in turn increases the focal length of the eye lens.The correct answer is C. The circular diaphragm in front of the eye lens is known as iris and the small opening in between the iris is known as pupil. The size of the pupil controls the amount of light entering into the eye. It is also true that the size of the pupil is controlled by the iris. Hence, both iris and pupil are responsible for controlling the amount of light entering into the eye.The correct answer is A. Iris is that part of the eye which controls the amount of light by contracting and expanding itself. In dim light, the iris contracts to expand the pupil, so as to allow more light to enter the eye. www.betoppers.com

8th Class Physics

192

58.

59.

In this way, things become visible in dim light. It is given that Tina is not able to contract her iris properly. Hence, she will not be able to see properly in dim light. The correct answer is A. The minimum object distance till which a normal human eye can see an object without any strain is known as the least distance of distinct vision. For the normal human eye this distance is 25 cm.A person with normal vision can see objects that are very far away (or at infinity). Hence, the range of vision for a normal human eye lies between 25 cm and infinity. The correct answer is D. In a myopic eye, the image of distant objects is formed in front of the retina and not on the retina, as shown in the given figure. Since eye lens is convex, all parallel rays converge at its focus. This is a clear indication of the shortening of the focal length of a myopic eye lens.

CONCEPTIVE WORKSHEET KEY 1

2

3

4

5

6

7

8

9

10

D

A

A

A

ABC

A

B

D

B

C

11

12

13

14

15

16

17

18

19

20

B

A

A

A BC

A

B

B

C

A

B

21

22

23

24

25

26

27

28

29

30

*

A

A

D

ABC

AB

C

B

C

AB C

31

32

33

34

35

36

37

38

39

40

A

A

B

A

A

AB CD

B

B

C

B

41

42

43

44

45

46

47

48

49

50

B

C

C

D

B

AB C

C

AB C

D

BC

51

52

53

54

55

56

57

58

59

60

A

BD

C

C

D

B

A

C

D

D

HINTS/ SOLUTIONS FOR SELECTED QUESTIONS 15.

[n  if

20.

60.

As a result, the curvature of the lens gets increased. Hence, myopia is caused by an increase in the curvature of the eye lens. 21. The correct answer is D. The diameter of the eye ball is approximately 2.3 42. cm. Images are formed on the retina of the normal eye, which is approximately 2.3 cm behind the 54. pupil. The formation of an image 3 cm away from the pupil of the eye indicates that the light rays are focused behind the retina. For proper visibility, the light rays must focus on the retina.

55. It can be concluded that Shyam is suffering from hypermetropia and a convex lens can be used to correct this defect. The correct answer is A. www.betoppers.com

 360  360 360  1 if = odd; n       

360 is even 

360  4 (even) 90  no.of images = n – 1 = 3 50 n

Apparent distance =

2h 2  4 80    6.6 cm  1.2 12

Cornea is the outer envelope of the eye.It provides focusing power to the eye. The aqueous humour separates the lens from the cornea.

It can be observed that the cornea is not in contact with the eye lens. The correct answer is C. To accommodate near and far vision, the ciliary muscles of the eye contract and expand respectively. This contraction and expansion is to respectively decrease and increase the focal length of the lens. The correct answer is D.

Light Solutions 56.

To see near objects clearly and distinctly, the focal 59. length of the lens should be shorter. The ciliary muscles contract to increase the thickness of the lens and thereby, reduce the focal length of the lens. In the same way, the ciliary muscles relax to reduce the thickness of the lens and thereby, increase the focal length of the lens.

60.

Thus, an observer can see distant objects most distinctly without any strain to the eye. An eye lens with contracted ciliary muscles and another with relaxed ciliary muscles are shown in the given figure.

This contraction and relaxation of ciliary muscles is defined as the power of accommodation of the 1. human eye. The correct answer is B. 57.

A real  image  is  formed  by  the  eye  lens  on  the 2. retina. The cells present on the retina convert this image into electric impulses and these impulses are sent to the brain via optical nerves. Hence, light energy is converted into electrical energy by 3. the retina. The correct answer is A.

58.

Human eye lens is convex in nature. A convex lens forms real images that can be obtained on a screen. A real image is formed by the eye lens that can be obtained on the retina. Since real images are always inverted, the image formed by the eye lens is real and inverted. Also, the least distance of distinct vision for a normal eye is 25 cm. An eye can see clearly when an object is at a distance greater than 25 cm. The radius of curvature of a normal eye is larger than 25 cm. 4. When an object is placed beyond the centre of curvature of a convex lens, a real, inverted, and diminished image is formed. Hence, the images formed by the human eye are real, inverted, and diminished.The correct answer is C.

193 A person suffering from hypermetropia or far sightedness can see distinct objects clearly but faces difficulty in seeing nearby objects.In the given case, Mutum finds it difficult to read the blackboard from the first bench, but is able to read the same from the last bench. Hence, it can be concluded that he is suffering from hypermetropia. The correct answer is D. If the size of the eyeball decreases, then the person is said to be suffering from hypermetropia. In this case, the image is formed beyond the retina. Hence, statement I is incorrect. In case of hypermetropia, a person is unable to see nearby objects clearly, which means the near point of the person shifts to a distance greater than 25 cm. Hence, statement II is correct.Presbyopia occurs because of poor accommodation power of the eye in older age. A person, at older age, suffering from presbyopia uses a bifocal lens for correction. This happens because far point comes closer than infinity and near point goes farther than 25 cm. Hence, statement III is correct. The correct answer is D.

SUMMATIVE WORKSHEET Normal incidence  Angle of incidence = 00 ; Angle of deviation, d = ? We know that, d = 180 - 2i = 180 - 0 = 1800 We know that, d = 180 - 2i or d = 180 - ( i + r ) [  i = r]  d + i + r = 1800. Therefore, the sum of angle of incidence , angle of reflection and angle of deviation is equal to 1800. Angle of deviation = d; i = ? ; r = ? ; g = ? We know that, d = 180 - 2i  2i - 180 - d or i=

180  d 2

180  d 2 We know that, glancing angle, g = 90 - i Angle of reflection , r = i =

180  d d  2 2 Let the initial angle of incidence be x. M1 and M2 are the initial and final positions of the mirror. N1 and N2 are the initial and final positions of the normals. AO is the incident ray , OB is reflected ray before rotation and OC is the reflected ray after rotation. We need to find the angle through which the reflected ray rotates i.e., BOC .  g  90 

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8th Class Physics

194 11.

By solving, we get BOC = 2  . N1

A

N2

B

x+ x

 y

x

x

=

+

C



?



M1 O

12. M2

5. (i)

(ii)

6. 7.

The angle between the incident ray and reflected 13. ray = x  i = r = x/2 After the mirror is rotated through an angle of x/2 in clock wise rotation, then the new angle of incidence becomes 3x/2 and the new angle of reflection becomes 3x/2. Therefore the angle between the incident ray and reflected ray = New angle of incidence + New angle of reflection = 3x /2 + 3x/2 = 3x. After the mirror is rotated through an angle of x/2 in anti clock wise rotation, then the new angle of 14. incidence becomes x/2 and the new angle of reflection becomes x/2. Therefore the angle between the incident ray and reflected ray = New angle of incidence + New angle of reflection = x /2 + x/2 = x.

 Angle of incidence = 150, Angle of reflection = 150 ; The reflected ray makes an angle of 300 with the vertical and 600 with the horizontal. ( see the figure for further understanding) N2

N1

B C

i+



 2

8.

Real, virtual.

9.

n=

16.

i

10.

540

360 1 

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 2.4 

3  108 Speed of light in medium

3  108 2.4 = 1.25  108 m/s. Size of object, O = 2.5 m = 250 cm Width of pin hole camera, v = 24 cm Size of the image, I = 1.2 cm Object distance, u = ?  Speed of light in medium 

We know that, m 

v I v I    u O u O

We know that, m 

O 17. 18.

On

substitution, we get, u = 50000cm = 50 m. m = 0.0005 object distance, u = 500 m Image distance, v = ?

v  v  mu u

 v  0.0005  500  0.25m

i+

O

Speed of light in air Speed of light in medium

We know that, m 

 



A

i 

15.

(i) u (ii) Object distance = Image distance = 30 cm. When the object moves away from the plane mirror by 15 cm, the new object distance becomes 30 + 15 = 45 cm  The new image distance = 45 cm. Therefore, the new distance between object and image is equal to 45 + 45 = 90 cm Yes. When a ray of light in incident along the normal from rarer medium to denser medium, then it moves undeviated. We know that :

I O

I 2.4   4800cm  48m m 0.0005

Angle of deviation = i - r = 45 - 30 = 150. We know that



sin i sin r

1 sin 45 2  2  1.414  =  1 sin 30 2

Light Solutions 19.

20.

When the ciliary muscles are relaxed, the eye lens 25. becomes thin, the focal length increases, and the distant objects are clearly visible to the eyes. To see the nearby objects clearly, the ciliary muscles contract making the eye lens thicker. Thus, the focal length of the eye lens decreases and the nearby objects become visible to the eyes. Hence, 26. the human eye lens is able to adjust its focal length to view both distant and nearby objects on the retina. This ability is called the power of accommodation of the eyes. The person is able to see nearby objects clearly, but he is unable to see objects beyond 1.2 m. This happens because the image of an object beyond 1.2 m is formed in front of the retina and not at the retina, as shown in the given figure.

195 A normal eye is unable to clearly see the objects placed closer than 25 cm because the ciliary muscles of eyes are unable to contract beyond a certain limit. If the object is placed at a distance less than 25 cm from the eye, then the object appears blurred and produces strain in the eyes. Since the size of eyes cannot increase or decrease, the image distance remains constant. When we increase the distance of an object from the eye, the image distance in the eye does not change. The increase in the object distance is compensated by the change in the focal length of the eye lens. The focal length of the eyes changes in such a way that the image is always formed at the retina of the eye.

HOTS WORKSHEET 1. 2. 3. To correct this defect of vision, he must use a 4. concave lens. The concave lens will bring the 5. image back to the retina as shown in the given 6. figure. 7. 8.

21.

0

The relaxation or contraction of ciliary muscles changes the curvature of the eye lens. The change in curvature of the eye lens changes the focal length of the eyes. Hence, the change in focal length of an eye lens is caused by the action of ciliary muscles. The correct answer is C.

0

24.

600 30

23.

M2

30

22.

The near point of the eye is the minimum distance of the object from the eye, which can be seen distinctly without strain. For a normal human eye, this distance is 25 cm.The far point of the eye is the maximum distance to which the eye can see the objects clearly. The far point of the normal human eye is infinity. A student has difficulty in reading the blackboard while sitting in the last row. It shows that he is unable to see distant objects clearly. He is suffering 9. from myopia. This defect can be corrected by 10. using a concave lens. The human eye forms the image of an object at its retina.

40 cm 45 cm (a) 60 m/s (b) 80 m/s (c) 40 m/s 600 90 cm 20 cm 500 The angle that the ray makes after each reflection is clearly labelled in the adjacent figure . It is clear that the final ray makes an angle of 600 with the horizontal, or in other words it emerges parallel to Mirror 1.

600 600

600

M1

60 cm The first image at M1 is formed at a distance of 4 cm behind it. The first image at M2 is 6 cm behind it. The second image at M 2 is at a distance of 4+10=14 cm behind. M2. Therefore, the distance between the first image at M 1 and second image at M 2 is 4+14+10=28 cm.

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196 11.

8th Class Physics According to law of reflection,

The speed of image =

HM1N  EM1N

Distance travelled by image Time

Also, EM 2 M  OM 2 M

Let us consider that the mirror has moved for 1 second.  The distance travelled by mirror = 20 cm/s

 EM = MO Now, M1M2 = NM = HO – ( HN + MO ) Using equations, (1) and (2), we get M1M 2 = HO – ( NE + EM) = HO – NM = HO – M 1M 2 or 2M1M2 = HO or

Let the initial distance between object and mirror be x. From the figure, it is clear that OM = x  MI = x Let M’ be the final position of the mirror.  New object distance = OM’ =( x + 20) cm

M1M 2 

 New image distance = M’I’ = ( x + 20) cm The distance moved by the image II’ = MI’ – MI = 20 + x 20 – x = 40 cm

 Speed of image =

13.

40  40 cm / s 1

H’

14.

i’ r’

h

v  size of the object u

 twice the size of the object Therefore, the image is formed at the pole of the convex mirror. It is inverted with magnification 2.

 N  E (eye)

Mirror

h/2

Concave mirror

i  M

M2

1  height of the person. 2

(i) convex mirror (ii) plane mirror (iii) concave mirror v = 15 cm. The inverted image formed by concave mirror is at the pole of the convex mirror. The image is formed at the pole with same size. Size of the image by the concave mirror =

Head (H) M1

1 HO 2

i.e., length of mirror =

Let M1 M2 be the plane mirror fixed on a wall. HEO is the position of a person standing in front of the mirror. E is the position of the eye of the person. To know the height of plane mirror, the key thing is to remember that a ray of light from leg must reach his eye and similarly a ray of light from the top of his head should also reach his eye.

Image

12.

 HN = NE

Convex mirror

h

r

7.5 cm

15 cm

O O’

Ground

Feet

A ray of light coming from the head of the person after reflection from mirror enters the eye of the person. This ray appears to come from H’ behind the mirror. Similarly, a ray from the feet of the person after reflection from the mirror enters the eye of the person. This ray appears to come from O’ behind the mirror. Thus, H’O’ be the position of the full image of the person. Now, draw the perpendiculars M1N and M2M on the mirror. www.betoppers.com

15.

As the image coincides the object, the light rays retrace their path after reflection on the convex mirror. This shows the rays from lens incident normally on the mirror. So the image is formed at C by the lines. For lens, applying the lens formula, we get v= 60 cm. Therefore the image is formed at distance of 60– 5 = 55 cm from the pole of the mirror. Therefore, the radius of curvature = 55 cm.

Light Solutions

197

Convex mirror

IIT JEE WORKSHEET KEY

Object

5cm

20 cm

1

2

3

4

5

6

7

8

9

10

B

D

A

D

C

C

C

B

B

B

13 14 15 16 17

18

19 20

C

C

C

D

C

22

23 24 25 26 27

28

B

B

D

11 12

16.

17.

18.

19.

20.

A person suffering from hypermetropia or far * sightedness can see distinct objects clearly but 21 faces difficulty in seeing nearby objects. B In the given case, Mutum finds it difficult to read the blackboard from the first bench, but is able to read the same from the last bench. Hence, it can be concluded that he is suffering from 23. hypermetropia. The correct answer is D. A real  image  is  formed  by  the  eye  lens  on  the retina. The cells present on the retina convert this image into electric impulses and these impulses are sent to the brain via optical nerves. Hence, light energy is converted into electrical energy by the retina. The correct answer is A. Cornea is the outer envelope of the eye. It provides focusing power to the eye. The aqueous humour separates the lens from the cornea.

24. It can be observed that the cornea is not in contact with the eye lens. The correct answer is C. The image of an object is formed on the retina of the eye. It is a delicate membrane, which contains a large number of light sensitive cells. These light sensitive cells are called cones. They transform 25. the light signal into electrical signal, which is then sent to the brain. The correct answer is C. To accommodate near and far vision, the ciliary muscles of the eye contract and expand respectively. This contraction and expansion is to respectively decrease and increase the focal length of the lens. The correct answer is D.

C

C

D

B

C

A

C

C

Cornea is the front layer of the eye. It is a thin membrane that envelops the eye. Various parts of the front of the eye are shown in the given figure.

A ray of light that falls on the eye first passes through the cornea. The light then enters the eye through the pupil after passing through the aqueous humour present between the pupil and the cornea. The ray of light is then focussed by the eye lens to form an image on the retina present on the back part of the eye. The correct answer is B. In the human eye, the ciliary muscles are attached with the eye lens. These muscles on contracting or relaxing make the eye lens thin or thick respectively. The focal length of the eye lens also changes with its thickness. The correct answer is C. To see near objects clearly and distinctly, the focal length of the lens should be shorter. The ciliary muscles contract to increase the thickness of the lens and thereby, reduce the focal length of the lens. In the same way, the ciliary muscles relax to reduce the thickness of the lens and thereby, increase the focal length of the lens. Thus, an observer can see distant objects most distinctly without any strain to the eye. An eye lens with contracted ciliary muscles and another with relaxed ciliary muscles are shown in the given figure. www.betoppers.com

8th Class Physics

198

Since eye lens is convex, all parallel rays converge at its focus. This is a clear indication of the shortening of the focal length of a myopic eye lens. As a result, the curvature of the lens gets increased. Hence, myopia is caused by an increase in the curvature of the eye lens. The correct answer is D.



26.

27.

28.

This contraction and relaxation of ciliary muscles is defined as the power of accommodation of the human eye. The correct answer is B. Iris is that part of the eye which controls the amount of light by contracting and expanding itself. In dim light, the iris contracts to expand the pupil, so as to allow more light to enter the eye. In this way, things become visible in dim light. It is given that Tina is not able to contract her iris properly. Hence, she will not be able to see properly in dim light. The correct answer is A. Human eye lens is convex in nature. A convex lens forms real images that can be obtained on a screen. A real image is formed by the eye lens that can be obtained on the retina. Since real images are always inverted, the image formed by the eye lens is real and inverted. Also, the least distance of distinct vision for a normal eye is 25 cm. An eye can see clearly when an object is at a distance greater than 25 cm. The radius of curvature of a normal eye is larger than 25 cm. When an object is placed beyond the centre of curvature of a convex lens, a real, inverted, and diminished image is formed. Hence, the images formed by the human eye are real, inverted, and diminished. The correct answer is C. In a myopic eye, the image of distant objects is formed in front of the retina and not on the retina, as shown in the given figure.

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8. MAGNETISM SOLUTIONS

CLASSROOM WORKSHEET

PRACTICE WORKSHEET

KEY

KEY

1

2

3

4

5

6

7

8

9

10

1

2

3

4

5

6

7

8

9

10

D

A

C

C

D

B

D

B

A,C

A

C

D

A

C

C

A

CD

D

A

D

11 12

13 14 15 16 17

18

19

20

11 12

13 14 15 16

17

18 19 20

B

D

A

A

A

D

B

A

A

21 22

23 24 25 26 27

28

34

35

21 22

23 24 25 26

27

28 29 30

A

A

D

A

B

D

C

D

C

C

A

C

C

A

B

D

A

D

C

C

A

D

A

B

A

C

B

A

36 37

38 39 40

31 32

33 34 35 36

37

38 39 40

A

A

A

A

A

A

12. 15. 18. 29.

30.

31.

32.

33.

A

A

C

A– 1; B – 2; C – 2 A. 2 B. 4 C. 1 A. 1 B. 4 C. 3 An iron bar consists of millions of molecular magnets. But they are arranged in the form of closed chains with the opposite poles facing each other, thereby neutralising each other. The net effect is that no magnetism is possible in spite of the presence of millions of molecular magnets. The pole strength of second magnet is m because pole strength does not depend up on length. Pole strength depends upon base height which is same in both the magnets. Pole strength (m)  (Breadth 5 height) Original pole strength (m1) = K ( b×h) As the dimensions are doubled, the new pole strength ( m2) = K (2b × 2h) On doubling the dimensions, the new pole strength becomes 4 times the original strength. There will no change in pole strength. Thre is a change only in magnetic length, but not in the are of cross section. So, there will not be a change in magnetic pole strength. a) Slightly increases. This is because the uniformly aligned molecules occupy greater length compared to the molecular magnets before magnetisation.

B

C

A

A

B

A

SUMM ATIVE WORKSHEET KEY 1

2

3

4

5

6

A

A

B

C

A

ABD

11 12

13 14 15

16

D

B

D

B

A

D

7

8

9

10

BCD A A

A

17. No. 18. A steel bar is placed in a magnetic field. The magnetic power induced in it starts increasing till it reaches a point where the magnetic power does not increase any further (as shown above). This condition is called Magnetic saturation. 19. Due to improper allignment of molecular magnets. 20. Consider a bar AB is brought near the north pole of a magnet and it is observed that AB is getting attracted to it. AB need not be a magnet because the attraction in AB may also be possible if it is an iron bar. So attraction is not a sure sign of the presence of magnetism. A sure test to determine the presence of magnetism is repulsion. 21. Pole strength (m) = f (no. of free poles exposed at ends)= f (area of cross-section) As the area of cross section is same, the pole strength is same at both ends.

8th Class Physics

200 22. During magnetic induction, the opposite pole is 9. induced at the near end, and the like pole is induced 10. at the other end of the bar. T h e r e f o r e attraction results only after the induction of opposite poles at the near end.

1.

geometric length = 6K and magnetic length = 5K

2.

C

3.

D

4.

C

5.

C

6.

C

7.

C

8.

C

9.

C

10.

When the magnet is bent into a semicircle its length

5.12 = 1.6amp. m 3.2

By the relation, M = 2lm  2l =

M 0.4  = 0.25m m 1.6

F = mB ------------- (2) 10–3× 2 ×10–2 = 1×B

 B = 2 × 10–5 T

2 

2 2M ×m =  

IIT JEE WORKSHEET 1 d2

1.

D

 Use F 

2.

A

 Use F  m1m 2

3.

D

 Use F 

1 d2

4. 5.

A B

6.

B

 Use F 

m1m 2 d2

7.

Let the pole strength of one pole be m. Then the pole strength of the other is 3m.

m1m2 = 150 × 10–6 × 9.8 d2

107  3m2 = 5  102 2   On simplifying we get, m = 3.5 A–m The strength of stronger pole = 3m = 10.5 A–m www.betoppers.com

 5.12×10–5 = m × 3.2 ×10–5

From (1) and (2)

 New length of the magnet = 2r = 2×

F  K

F = mB

F = ma ----------- (1)

2  r= 

 New magnetic moment, M1 = 2×

Let m be the pole strength of the magnet. Then by the relation,

m=

HOTS WORKSHEET

2 = r

B

