SCIENCE

Approved by the Government of Nepal, Ministry of Education, Curriculum Development Centre (CDC), Sanothimi, Bhaktapur, Nepal as an additional material for schools. New Creative SCIENCE and Environment 8 This book has been updated, enlarged, revised and edited by Janak Raj Pant Rajani Maharjan Yuwa Raj Guragain JBD Publication Pvt. Ltd. Publisher and Distributor Bhotahity, Kathmandu

New Creative SCIENCE and Environment 8 Publisher : JBD Publication Pvt. Ltd. Bhotahity, Kathmandu, Nepal Tel: 01-4252371 Email: [email protected] Web: www.jbdbooks.com Copyright : Publisher Edition : Fifth, 2073 BS (2016 AD) [Revised and updated] : Reprint, 2075 BS [2018 AD] Layout : Deltrox IT Solutions Pvt. Ltd. Cover : Bishnu Dev Bhandari ISBN : 978-9937-544-42-9 Mahabir Offset Press, Tel: 025-521634

PREFACE As this is the age of science and technology, science is a very important subject to the students of the present days. Scientific inventions and discoveries have brought great changes in the world. The study of science and technology for the students of developing countries like Nepal has great importance to make many achievements. For this, the students should have good knowledge of science through systematic as well as properly designed textbooks, instructional materials and good teaching methods. In schools, textbooks are the most important educational materials. So, the books designed according to the need and interests of the students are very useful and appropriate. Keeping these facts in mind this series of textbooks for the students of classes 6-10 has been prepared by a group of writers namely Basu Sharma, Suman Naupane and Bharat Bhattarai in 2065 with fresh and fascinating approach to the study of science at school level. During teaching learning process both teachers and students have to face problems of what is to be taught and learnt, what should be focused for the examination, how to learn and write to the point, how to start and how to end. This series for the first time itself is the best attempt and presentation to solve these problems. After facing all these problems during teaching, we have designed and prepared this series as a solution with these salient features. Each book of this series includes-objectives of the unit in the very beginning, important questions, clues and memory note as important things, definitions in separate colour, answer writing skills as model questions and their answer, glossary of difficult words, lots of activities as creative activities, summary of the unit as revision, etc. Further this series has been updated, enlarged, modified, edited and designed according to the syllabus prescribed by CDC of Nepal. Cognitive and practical student-friendly, colourful presentation and use of lucid and easy language, both knowledge and exam-oriented matters in sequential order are the main attractions of this series. We are grateful to Deltrox IT Solutions Pvt. Ltd. for computer work and JBD Publication Pvt. Ltd. for publishing this series. We are highly indebted to those who provide their positive support, suggestions and comments for the improvement of the series in further editions. Authors

Contents PHYSICS 1-132 Unit 1: Measurement 2 Unit 2: Velocity and Acceleration 13 Unit 3: Simple Machine 27 Unit 4: Pressure 43 Unit 5: Work, Energy and Power 58 Unit 6: Heat 70 Unit 7: Light 83 Unit 8: Sound 99 Unit 9: Magnetism 110 Unit 10: Electricity 119 CHEMISTRY 133-186 Unit 11: Matter 134 Unit 12: Mixture 152 Unit 13: Metals and Non-Metals 159 Unit 14: Acid, Base and Salt 167 Unit 15: Some Useful Chemicals 179 BIOLOGY 187-299 Unit 16: Living Beings 188 Unit 17: Cell and Tissue 206 Unit 18: Life Process 213 Unit 19: Structure of the Earth 228 Unit 20: Weather and Climate 239 Unit 21: The earth and Space 248 Unit 22: Environment and Its Balance 255 Unit 23: Environmental Degradation and Its Conservation 273 Unit 24: Environment and Sustainable Development 291

Unit 1: Measurement 2 PHYSICS Unit 2: Velocity and Acceleration 13 Unit 3: Simple Machine 27 Unit 4: Pressure 43 Unit 5: Work, Energy and Power 58 Unit 6: Heat 70 Unit 7: Light 83 Unit 8: Sound 99 Unit 9: Magnetism 110 Unit 10: Electricity 119 1

U 1 MEASUREMENT L earning O utc om es At the end of this unit, students will be able to: ~ define fundamental (basic) and derived units. ~ identify measuring units of mass, weight and time. ~ define density and relative density and write their formulas. ~ explain why objects sink and float based on density. ~ solve the simple numerical problems of density. Main points to be focused ~ Fundamental and derived physical quantities. ~ Examples of fundamental and derived physical quantities. ~ Fundamental and derived units. ~ Examples of fundamental and derived units ~ Measurement of mass, weight and time. ~ Numericals of unit conversion. ~ Numericals related to w = m × g Introduction We can see and use two types of quantities in our daily life. They are measurable quantities and non-measurable quantities. Those quantities which can be measured in different units by using various physical balances are called physical quantities. For example, mass, length, time, temperature, volume, etc. Similarly, those quantities which cannot be measured by using physical balances are called non-physical quantities. For example, emotion,love, hate, kindness, etc. Measurement deals with a lot of physical quantities. So, physics requires a frequent and exact measurement of these physical quantities to find out a correct conclusion. Measurement is the process of finding the value of any unknown physical quantity with a known and standard value. There are two types of physical quantities. They are: i. Fundamental physical quantities ii. Derived physical quantities 2 New Creative Science and Environment Book - 8

Fundamental physical quantities Those physical quantities which do not depend upon other physical quantities and have their own identity are called fundamental physical quantities. They are free and independent quantities. For example, length, mass, time, etc. There are seven fundamental physical quantities. They are length, mass, time, temperature, current, luminous intensity and the amount of a substance. Memory Note › Length is a fundamental physical quantity as it does not depend upon other physical quantities. Derived physical quantities Those physical quantities which depend upon fundamental physical quantities and do not have their own identity are called derived physical quantities. There are hundreds and thousands of derived quantities. For example, area, volume, velocity, density, force, etc. Memory Note › Volume is a derived physical quantity as it depends upon length, breadth and height. Questions i. Why is time called a fundamental quantity? ii. Why is an area called a derived quantity? Units If we measure the mass of any substance, we get a certain numerical value. To express this numerical value, we use a unit. For example, the mass of sugar is 5 kg. It means the mass of sugar is 5 times more than the standard reference quantity i.e. 1 kg. Therefore, the standard reference quantity, with reference to which other physical quantities of the same kind are measured is called a unit. For example, the unit of mass is kilogram (kg), the unit of length is metre (m), etc. Depending upon the fundamental and derived quantities, t h e units are broadly divided into two types. They are: 1. Fundamental units Those units of measurement which do not depend upon any other units are called fundamental units. These are also called independent units or basic units. There are Measurement 3

seven fundamental quantities. To measure these seven fundamental quantities, we have seven fundamental units. The fundamental quantities, their SI units and the symbols of these units are given in the table below. S.N. Fundamental quantities SI unit Symbol 1. metre m 2. Length kilogram kg 3. Mass second s Time kelvin K 4. Temperature ampere A 5. Current candela cd 6. Luminous intensity mole mole 7. Amount of substance 2. Derived units Those units of measurement which are expressed in terms of fundamental units are called derived units. These are also called dependent units. For example, the units of speed, velocity, acceleration, work, power, pressure, density, etc. are some derived units. These units are expressed in terms of seven fundamental units. For example, the unit of velocity is metre per second (m/s). It is called a derived unit as it depends upon the unit of length i.e. metre (m) and the unit of time i.e. second (s). Some derived quantities and their SI units are as given below: S.N. Derived quantities SI unit Involved basic quantities 1 Area m2 length 2. Volume m3 length 3. Density kg/m3 mass, length 4. Force N (kg m s–2) mass, length, time 5. Velocity/speed m/s length, time 6. Acceleration m/s2 length, time 7. Pressure Pa (kg m–1 s–2) mass, length, time 8 Power w (kg m2 s–3) mass, length, time 4 New Creative Science and Environment Book - 8

Differences between fundamental and derived units S.N. Fundamental units S.N. Derived units 1. Fundamental units do not depend 1. Derived units are dependent on upon any other units. So, they are also fundamental units. So, they are called independent or basic units. also called dependent units. 2. There are only seven fundamental 2. There are large numbers of derived units. For example, metre, kilogram, units. For example, m/s, m/s2 , kg/m3, second, etc. etc. Questions i. Why is the unit of length called a fundamental unit? ii. Why is the unit of velocity called a derived unit? iii. Why is kg a fundamental unit but kg/m3 a derived unit? iv. Differentiate between fundamental and derived units. Measurement of mass The total amount of matter contained in a body is called its mass. It is measured by a beam balance and its SI unit is kg. But it can be measured in different units. The units like gram, milligram, centigram, etc. which are smaller than kg are known as the sub- multiples of kg. and the units like quintal, ton, etc. which are bigger than kg are called multiples of kg. Sub-multiples of kg › 1 kg = 10,00,000 mg › 1 kg = 1,00,000 cg › 1 kg = 10,000 dg › 1 kg = 1000 g › 1 kg = 100 dag › 1 kg = 10 hg Multiples of kg › 100 kg = 1 quintal › 1000 kg = 1 ton Measurement 5

Definition of one kg: One kg is defined as the mass of a platinum-iridium rod kept at the International Bureau of Weights and Measures in Paris of France. The following points should be kept in mind while measuring mass. 1. The pan balance should be horizontal to the ground. 2. Standard weights having an official stamp of the meteorology department should be used. 3. During measurement, the pointer must coincide with the mark at its centre. Memory Note › Generally, quintal is used to express the mass of agricultural products like wheat, rice, maize, millet, etc. whereas metric ton is used to express the mass of vehicles, large telescopes, ships, etc. Weight Spring balance For our general concept, weight is supposed to be the mass of the body. This concept is totally wrong from the point of view of physics. For example, when we are measuring our mass, we simply say that weight is 55 kg. This is not true as weight is totally different from the mass of a body. Weight is a force whereas mass is the quantity of matter. But, the mass of the body directly influences its weight. Thus, The amount of force with which a body is pulled towards the centre of the earth due to the force of gravity is called weight. In other words, it is defined as the product of the mass of any body and acceleration due to gravity. If ‘m’ is the mass of any body and ‘g’ is the acceleration due to gravity of the heavenly body, the weight (w) of the body is given by Weight (w) = Mass(m) × Acceleration due to gravity (g) Since, weight is a type of force i.e. force of gravity, its SI unit is Newton (N). It is a derived quantity as it depends upon the mass of a body and acceleration due to gravity. It is measured by using a spring balance. It is a vector quantity as its direction is always towards the centre of the earth. Memory Note › The weight of a body on the moon is 1 of the weight of the body on the earth as the value of acceleration 6 acceleration 1 due to gravity on the moon is 6 of due to the gravity of the earth. 6 New Creative Science and Environment Book - 8

Solved Numericals Calculate the weight of the body of mass 70 kg on the earth and on the moon. Solution: Given, the mass of the body (m) = 70 kg Acceleration due to gravity on the earth (ge ) Acceleration due to gravity on the moon (gm ) = 9.8 m/s2 Now, (i) weight of the body on the earth (we) = 1 × 9.8 = 1.63 m/s2 6 (ii) weight of the body on the moon (wm ) = m × ge Differences between mass and weight = 70 × 9.8 m/s2 = 686 N = m ×g m = 70 × 1.63 = 114.33 N S.N. Mass S.N. Weight 1. Mass is the quantity of the 1. Weight is the force, by which any matter contained in a body. heavenly body pulls the body of certain mass towards its centre. 2. It is a scalar quantity. 2. It is a vector quantity. 3. Its SI unit is kilogram (kg) 3. Its SI unit is Newton (N). 4. It is a fundamental physical 4. It is a derived physical quantity. quantity. 5. It is measured by using a 5. It is measured by using a spring balance. pan balance. 6. Its value is constant. 6. Its value is variable. Measurement of time The duration or gap between any two events is called time. Its both SI and CGS units are seconds (s or sec). Time is measured by using a clock or a watch. It has other different units as well. The relationship of different units can be written as: Different type of clocks Measurement 7

60 seconds 1 minute 60 minutes 1 hour 1 day 24 hours 1 week 7 days 1 year 365 days 1 year 52 weeks 1 decade 10 years 100 years 1 century In the ancient time, different types of clocks such as a sundial clock, a sand clock, a water clock and a candle clock were used to measure time. These devices were not so convenient and easy to use. So, these devices are not used any more. In this modern age, different types of clocks such as a pendulum clock, a quartz clock, an electronic clock and an atomic clock are used to measure time. 1. Pendulum clock Pendulum clock It is a device to measure time with the help of a pendulum that oscillates to and fro in an equal interval of time. It means, in every complete oscillatory motion, time is equal. It cannot measure accurate time because heat and cold affect the thermal expansion of the thread connected to the bob. They are generally kept or hung on the wall. These are also called mechanical or automatic clocks. 2. Quartz clock A quartz clock is based on the oscillation of the quartz crystal. It means, this clock measures the time with the help of the oscillation of quartz crystals kept in it. It gives more accurate time. So, the people of modern era prefer this clock. 3. Electronic clocks Quartz clock Electronic clocks are based on the frequency of alternating currents. It means such clocks measure the time on the basis of the frequency of electric currents. It also gives accurate time. 4. Atomic clock Atomic clocks are based on the emission of electromagnetic radiations produced by cesium-133 atoms. It means, it measures the time due to the emission of electromagnetic radiation by Cs-133 isotopes. It measures more accurate time. So, scientists prefer this clock to measure accurate time. 8 New Creative Science and Environment Book - 8

Memory Note ; The time taken by the earth to complete one rotation around its own axis is called one solar day. One solar day is equal to 24 hours. ; 1/86400 part of a solar day is called one second. ; One second is also defined as the time required for 9, 192, 631, 770 cycles made by cesium- 133 isotopes. Questions i. What is time? Mention its SI unit and CGS unit. ii. Why is a quartz crystal clock preferred than a pendulum clock? iii. How many seconds are there in one day? Show with a calculation. Answer Writing Skills 1. The unit of pressure is a derived unit. Why?  The unit of pressure (Pascal) depends on three fundamental units i.e. kg, metre and second as, Pressure = Force Area Kg m/s2 ∴ pascal = m2 = kg m–1 s–2 So, Pascal is a derived unit. 2. Why is SI system called the extended version of MKS system?  In SI system, like in MKS system, length is measured in metre, mass in kg and time in second. Here, MKS includes only these three fundamental units but SI system includes all the fundamental as well as derived units. So, SI system is the extended version of MKS system. 3. Kilogram is a fundamental unit. Why?  Kilogram is a fundamental unit as it does not depend upon other units and has its own identity. 4. Why do we need standard units?  We need standard units to bring uniformity in the measurement all over the world. 5. Convert 1 day into seconds. Solution, 1 day = 24 hours = (24 × 60) minutes [∴ 1 hr = 60 min] = (24 × 60 × 60) seconds [∴ 1 min = 60 s] ∴ 1 day = 86,400 seconds Measurement 9

SUM M ARY ” Those physical quantities which do not depend upon other physical quantities and have their own identity are called fundamental physical quantities. ” Those physical quantities which depend upon fundamental physical quantities and do not have their own identity are called derived physical quantities. ” The standard reference quantity, with reference to which other physical quantities of the same kind are measured is called a unit. ” Those units of measurement which do not depend upon any other units are called fundamental units. These are also called independent units or basic units. ” Those units of measurement which are expressed in terms of fundamental units are called derived units. These are also called dependent units. ” The total amount of matter contained in a body is called its mass. ” One kg is defined as the mass of a platinum-iridium rod kept at the International Bureau of Weights and Measures in Paris of France. ” Weight is defined as the amount of force, by which a body is pulled towards the centre of the earth due to the force of gravity. ” The duration or a gap between any two events is called time. ” Pendulum is a device to measure time with the help of a pendulum that oscillates to and fro in an equal interval of time. ” Quartz clock is based on the oscillation of the quartz crystal. Exercise 1. Fill in the blanks. (a) The quantities which can be measured are called _________ quantities. (b) Cubic metre is the SI unit of _________. (c) Kg/m3 is a unit of _________. (d) The quantities which _________ be measured are called non-physical quantities. (e) _________ system is the extended version of MKS system. 2. Write True for correct and false for wrong statements. (a) A quartz crystal clock is based on the oscillation of quartz. 10 New Creative Science and Environment Book - 8

(b) 1 day is equals to 86,400 seconds. (c) Mass is a fundamental quantity. (d) In MKS system, length is measured in foot. (e) Temperature is a derived quantity. 3. Choose the best answer. (a) The SI unit of area is (i) m (ii) m3 (iii) m2 (iv) m/s2 (b) The volume of a cube is given by (i) (length)2 (ii) (side)2 (iii) (length)3 (iv) none of these (c) The fundamental units involved in kg/m3 are (i) kg and second (ii) kg and metre (iii) second and metre (iv) metre and kelvin (d) 1 day is equal to (i) 24 hrs (ii) 86,400 seconds (iii) 1440 minutes (iv) all of these (e) Which is not a fundamental unit? (i) metre (ii) ampere (iii) kelvin (iv) metre per second 4. Define the following terms. (a) Measurement (b) Physical quantity (c) Unit (d) Fundamental quantity (e) Fundamental units (f) Derived quantity 5. Answer the following questions. (a) Define measurement and write its importance. (b) Differentiate between fundamental and derived units. (c) What does it mean by “the mass of an object is 5 kg”? Measurement 11

(d) Define mass. How is it measured? What is its SI unit? (e) What is weight? Write down its SI unit. (f) Why are MKS and CGS systems widely used? (g) Why is m3 a derived unit? (h) Differentiate between mass and weight. Numerical Problems 1. Calculate the weight of an object whose mass is 30 kg and acceleration due to gravity is 10 m/s2. (Ans: 300 N) 2. Calculate the mass of an object whose weight is 400 N and acceleration due to gravity is 10 m/s2. (Ans: 40 kg) 3. Calculate the acceleration produced on the object whose weight is 400N and mass is 40 kg. (Ans: 10 m/s2) 4. Convert the following: a. One day into seconds (Ans: 86400 s) b. 46 kg into g (Ans: 46000 g) c. 3 days into seconds (Ans: 2,59,200 s) lossar – the item to compare e – the a ra n r e the earth U nit the han e in ve it in iven G ravity – basic A e era n – made from others F undamental D erived 12 New Creative Science and Environment Book - 8

U2 VELOCITY AND ACCELERATION L earning O utc om es At the end of this unit, students will be able to: ~ explain the average velocity and relative velocity. ~ define the acceleration and retardation. ~ write the equations related to velocity and acceleration and to apply them. ~ solve the simple numerical problems related to the velocity and acceleration. Main points to be focused ~ Acceleration and retardation ~ Rest and Motion ~ Equations of motion in a straight line ~ Uniform and non-uniform motion ~ Relative velocity Rest and Motion To understand the meaning of rest and motion, we need to understand what is a reference point. The fixed point on the basis of which we study the rest and motion of a body is called a reference point. If a body changes its position with respect to the reference point, the body is said to be in motion. Similarly, if a body does not change its position with respect to the reference point, the body is said to be at rest. For example, suppose you are travelling on a bus. The bus is in motion with respect to the outside surrounding. Similarly, if you are standing nearby trees, you are at rest with respect to the trees. It means rest and motion are relative terms. Examples of motion Examples of rest Velocity and Acceleration 13

Uniform and non-uniform motion Suppose a car travels 40 km per hour. It means in every hour, the distance travelled by the car is 40 km. It does not travel more than or less than 40 km per hour. This kind of motion is called uniform motion. Therefore, the motion of a body that travels an equal distance in an equal interval of time is called uniform motion. 0 hr 1 hr 2 hrs 3 hrs 4 hrs ( a) ( b) ( c) ( d) ( e) 40 km 40 km 40 km 40 km Example of uniform motion Sometimes, you are faster and sometimes you are slower. It means you cannot cover an equal distance in an equal interval of time. Such a motion is called non-uniform motion. Therefore, the motion of a body that cannot travel an equal distance in an equal interval of time is called non-uniform motion. This kind of non-uniform motion can be experienced generally during walking, running, driving and so on. 0 hr 1 hr 2 hrs 3 hrs 4 hrs ( a) ( b) ( c) ( d) ( e) 10 km 18 km 15 km 10 km Example of non-uniform motion Speed, velocity and average velocity Objects may travel fast or slow. For example, we may observe that an aeroplane travels very fast while a bicycle is slow. What, exactly do we mean when we use the terms ‘fast’ and ‘slow’? Speed is the total distance travelled by a body in per unit time. In other words, it is the rate of change of distance. Its SI unit is m/s and CGS unit is cm/s. It is a scalar quantity. Distance travelled (s) i.e. Speed (V) = Time taken (t) ∴ V = s t The shortest distance between any two points is called displacement. It is the distance in the straight line with a fixed direction. The rate of change of displacement of a body in per unit time is called its velocity i.e. 14 New Creative Science and Environment Book - 8

Velocity (V) Displacement covered (s) = Time taken (t) ∴ V = s t The SI unit of velocity is m/s and CGS unit is cm/s. It is a vector quantity. The speed and velocity have the same meaning if a body is moving in a straight line in a fixed direction but if the body is moving in a curved path, it is better to say speed rather than velocity. The ratio of the total displacement covered by a body and the total time taken by it is called its average velocity. Average velocity (Av) = Displacement covered (s) Time taken (t) ∴ Av= s .......................(1) t Mathematically, average velocity is the sum of initial velocity (u) and final velocity (v), divided by 2. Average velocity (Av) = Initial velocity (u) + Final velocity (u) 2 Or, (Av) = u + v ......................(2) 2 From the equations (1) and (2) s u+v t =2 or, s = u + v × t .................(3) 2 This shows that a distance covered is equal to the product of average velocity and time taken. Memory Note ; Speed is measured by using an instrument called speedometer. Questions i. Discuss that a body moving with a uniform speed may have variable velocity. ii. A body moving in a circular path has variable velocity. How? ctivit Mark a distance of 100 m on your school ground. Measure the time you take to walk this distance and calculate your speed. Now, run and find out the time you take to complete the same distance and calculate your speed. Compare the results. Velocity and Acceleration 15

Relative velocity The velocity of a body with respect to another body or a reference point is called relative velocity. Some examples of relative velocity are as given below. 1. The relative velocity of two bodies along the same direction with the same velocity M ( a) A 30 m/ s Let us consider two cars A and B are moving in the same direction with the same velocity of 30 m/s. The figure shows the situation after 1 second. B 30 m/ s Both the cars have covered the same distance ( b) from the same reference point. Therefore, N looking from the car A, car B seems to be at rest M and vice versa. It is because, Relative velocity = VA – VB In the above condition, VA = VB = 30 m/s ∴ Relative velocity = VA - VB = 30 - 30 = 0 m/s N Due to zero relative velocity, the cars seem to be at rest with respect to each other. 2. Relative velocity of two bodies along the same direction with different velocities Let car A is moving with the velocity 30 m/s and car B is moving with the velocity 50 m/s in the same direction. M M 30 m/ s 30 m A A B 50 m/ s 50 m/ s B N N F ig ( a) F ig ( b) The figure shows the situation after 1 second. Car A has covered 30 m and car B has covered a distance of 50 m. Therefore, looking from the car A, the car B is appeared to be moving with the velocity, 16 New Creative Science and Environment Book - 8

VBA = VB – VA = 50 – 30 = 20 m/s And from the car B, car A is appeared to be moving with the velocity, VAB = VA – VB = 30 – 50 = –20 m/s 3. The relative velocity of two bodies along the different directions with the same or different velocities If car A and B are moving in opposite directions with the same or different velocities then its situation after 1 second is shown by the figure. 25 m/ s M 40 m/ s M A A B N after 1 se nd F ig ( b) B N F ig ( a) Therefore looking out from either of the cars, another car appears to be moving with the velocity. VAB = VA + VB i.e. VAB = 25 + 40 = 65 m/s So, the cars appear to be moving with the velocity greater than their own velocities. Solved Numericals The velocity of one vehicle is 40 m/s towards the east and the velocity of an other vehicle is 30 m/s towards the west. If both the cars start from the same point at the same time, what will be the distance between them after 2 minutes? What distance will each of them travel during that time? Solution: Given, Let, velocity of A (VA) = 40 m/s Velocity of B (VB) = 30 m/s Distance travelled by (A), SA =? Velocity and Acceleration 17

Distance travelled by (B), SB = ? Time taken (t) = 2 min = 2 × 60 = 120 s Now, we have, For car ‘A’ For car ‘B’ VA =40 m/s VB = 30 m/s or, SA or, SB = VA × t = VB × t = 40 × 120 = 30 × 120 ∴ SA = 4800 m ∴ SB = 3600 ∴ the distance travelled by car 'A' is ∴ the distance travelled by car 'B' is 4800 m due east. 3600 m due west. ∴ The distance between them after 2 min = SA + SB = 4800 + 3600 = 8400 m Acceleration Suppose a body is increasing its velocity from 100 m/s to 200 m/s in 2 seconds. This is said to be the acceleration of the body. In this case, the body increases its velocity by 200/100 = 2m/s in every second. This is called the acceleration of the body 2 m/s2. Therefore, acceleration is defined as the rate of change of velocity in per unit time. If ‘v’ is the final velocity changed from the initial velocity ‘u’ in time ‘t’, then acceleration ‘a’ is given by, a = v - u t Its SI unit is m/s2 and CGS unit is cm/s2. It is a vector quantity as it has both magnitude and direction. If the final velocity of a body is less than its initial velocity, it is called negative acceleration or retardation. It is represented by ‘–a’. Solved Numericals 1. A body is travelling with the velocity 30 m/s. After 5 seconds, its velocity reaches 90 m/s. What is the rate of change in velocity? Solution: Here, Initial velocity (u) = 30 m/s Final velocity (v) =90 m/s Time taken (t) = 5 seconds From the formulae, a = v - u = 90 – 30 = 60 = 12 m/s2 t 5 5 18 New Creative Science and Environment Book - 8

It means the body increases its velocity by 12 m/s in every second. 2. A car i s moving with the velocity 20 m/s. It comes at rest after 4 seconds on applying brakes. What is it retardation? Solution: Here, Final velocity (v) = 0 m/s (as the car finally comes at rest) Initial velocity (u) = 20 m/s Time taken (t) = 4 seconds From the formulae, a = v - u t 0 – 20 = 4 = –5 m/s2 i.e. retardation = 5 m/s2 It means the car decreases its velocity by 5 m/s in every second. Memory Note ; A body in uniform motion does not have acceleration. As v = u such that, a= v-u , a= v-v ,a= 0 = 0 m/s2 t t t Equations of motion in a straight line They are mathematical equations showing the relationship between initial velocity, final velocity, time taken, distance covered and acceleration. They are used to describe the motion of a body in a straight line. They are based on the definition of acceleration and definition of average velocity. First equation of motion Suppose a body ‘A’ is moving with the velocity ‘u’. After certain time taken ‘t’, its velocity becomes ‘v’ and covers the distance ‘s’. During this, the body produces an acceleration ‘a’. Since acceleration is the rate of the change of velocity, We have, acceleration = final velocity initial velocity time taken v - u or, a = t or, at = v - u ∴ v = u + at Velocity and Acceleration 19

This is the first equation of motion in a straight line. Second equation of motion We have, Average velocity = v + u t Distance covered = Average velocity × time s = u+v ×t 2 = u + u+ at ×t [ v = u + at] 2 = 2ut + at2 2 = 2ut + 1 at2 2 2 ∴ s = ut + 1 at2 2 This is the second equation of the motion of a body in a straight line. Third equation of motion The first equation of motion is v = u + at Squaring on both sides, we get, v2 = (u + at)2 = u2 + 2uat + a2t2 = u2 + 2a (ut + 1 at2) 2 ∴ v2 = u2 + 2as [ s = ut + 1 at2] 2 This is the third equation of the motion of a body in a straight line. Some special cases of equation of motions Case I: When a body starts from rest, u = 0 m/s Case II: When a body comes at rest from motion, v = 0 m/s Case III: When a body is vertically thrown upward, a = – g (acceleration due to gravity) 20 New Creative Science and Environment Book - 8

Case IV: s = h (height) v = 0 m/s When a body is vertically thrown downward Case V: a =g s =h u = 0 m/s When a body is in uniform velocity a = 0 m/s2 Solved Numericals 1. A car has a uniform acceleration of 8 m/s2. Find the distance covered in 4 seconds, if it starts from the rest. Solution: Given, Uniform acceleration (a) = 8 m/s2 Time (t) = 4 seconds Initial velocity (u) = 0 m/s [ the body starts from rest] Distance covered (s) =? We have, s = ut + 1 at2 2 1 =0×4+ 2 × 8 × (4)2 = 0 + 4 × 16 = 64 m Hence, the body covers 64 m in 4 seconds. 2. A body is thrown vertically upward with the velocity 72 km/hr. Find the maximum height reached and time taken to reach the height. Solution: Given, For a vertically thrown upward, Initial velocity (u) = 72 km/hr Final velocity (v) = 72 × 1000 m/s Acceleration (a) 60 × 60 = 720 m/s = 20 m/s 36 = 0 m/s [ the body is thrown vertically upward] = – g = – 10 m/s2 Velocity and Acceleration 21

Height reached (h) = ? Time taken (t) = ? We have, v2 = u2 + 2as or, 02 = (20)2 + 2 × (–10) × h [ s = h] or, 0 = 400 – 20 h or, 20h = 400 ∴ h = 400 20 i.e. h = 20m Also, v = u + at or, 0 = 20 + (–10) t or, – 20 = – 10t ∴ t = – 20 – 10 = 2 seconds Hence, the height reached by the body is 20 m and time taken is 2 seconds. 3. A stone is vertically thrown downward with the velocity 5 m/s and it reaches the ground in 2 seconds. Find the height from which the stone is thrown and also the velocity by which the stone strikes on the ground. Solution: Given, Initial velocity (u) = 5 m/s Time taken (t) = 2 seconds Acceleration (a) = g = 10 m/s2 Height (h) = ? Final velocity (v) = ? = ut + 1 at2 We have, s 2 or, h 1 = ut + 2 gt2 =5×2+ 1 × 10 × (2)2 2 = 10 + 5 × 4 = 10 + 20 ∴ h = 30 m Also, v = u + at = u + gt = 5 + 10 × 2 = 5 + 20 22 New Creative Science and Environment Book - 8

∴v = 25 m/s Hence, the height is 30 m and the stone strikes the ground with the velocity of 25 m/s. Answer Writing Skills 1. Why are rest and motion called relative terms?  Rest and motion are called relative terms because both of them are defined with respect to each other. Without motion, there will be nothing at rest or vice-versa. 2. Differentiate between distance and displacement.  Distance Displacement 1. It is the total length between 1. It is the shortest distance between the two fi ed points. initial and final points. 2. It is a scalar quantity. 2. It is a vector quantity. 3. The total distance is never zero. 3. The total displacement may be zero. 3. Why is speed called a scalar quantity?  Speed is a scalar quantity because it has only magnitude but not a fixed direction. 4. Why is acceleration a vector quantity?  Acceleration is called a vector quantity because it has both magnitude and a fixed direction. 5. Differentiate between speed and velocity.  Speed Velocity 1. It is the rate of change of distance. 1. It is the rate of change of displacement. 2. It is always positive. 2. It may be positive, negative or zero. 3. It is a scalar quantity. 3. It is a vector quantity. 6. The speed of a bus is 40 km/hr. What do you mean by it?  It means that the bus covers the distance of 40 km in one hour. SUM M ARY ” If a body changes its position with respect to the surrounding objects, then the body is said to be in motion. ” If a body does not change its position with respect to the surrounding objects, then the body is said to be at rest. ” The fixed point, about which we describe the motion of a body is called the reference point. Velocity and Acceleration 23

” The quantity having both magnitude and a fixed direction is called a vector quantity. ” The quantity having only magnitude but not a fixed direction is called scalar quantity. ” Distance is the total length between two points. Its SI unit is m. ” Displacement is the shortest distance between the initial and final points. Its SI unit is also metre. ” The rate of the change of distance is called speed and the rate of change of displacement is called velocity. ” Speed is a scalar quantity but velocity and acceleration are called vector quantities. ” If a body covers an equal distance in an equal interval of time, then the motion of the body is called uniform motion. ” If a body doesn’t cover an equal distance in an equal interval of time, then the motion of the body is called variable or non-uniform motion. ” The motion of a body relative to another object or the reference frame is called relative motion. Exercise 1. Fill in the blanks. (a) Rest and motion are ____________ terms. (b) Speed is a _________ quantity and velocity is a __________ quantity. (c) ____________ is the distance travelled in a fixed direction. (d) The value of acceleration is __________ in an uniform motion. (e) The motion of a body where it covers an equal distance in an equal interval of time is called ____________. 2. Choose the correct answer. (a) The SI unit of velocity is: (i) m/s (ii) m/s2 (iii) ms (iv) m2/s (b) The formula of acceleration is: (i) a = u + v2 (ii) a = v - u (iii) a = v + u t t (iv) a = v - ut 24 New Creative Science and Environment Book - 8

(c) Which of the following is a scalar? (ii) velocity (iv) acceleration (i) mass (iii) displacement (d) In uniform motion acceleration is: (i) positive (ii) zero (iii) negativ (iv) 60 m/s2 (e) Which one is correct? (i) s = ut + 1 at2 (ii) s = u2 + 2as 2 (iv) s = u2 – 2as (iii) s = u + v2 × t 3. Differentiate between (a) Speed and velocity (b) Uniform and non-uniform motion (c) Vectors and scalars (d) Distance and displacement (e) Acceleration and deacceleration 4. Answer the following questions (a) Define rest. Why are rest and motion called relative terms? (b) Show that the displacement can be zero even if the body has covered a certain distance. (c) Why is velocity called vector quantity while speed is a scalar quantity? (d) Give the example of a body in which its speed is constant but its velocity is changing. (e) In what condition do acceleration and deacceleration occur? (f) Nobody is at complete rest. Why? (g) The speed of a bus is 30 km/hr. What do you mean by it? 5. Prove: (a) s= u+v (b) v = u + at ×t (d) s = ut + 1/2at2 2 (c) v2 = u 2+2 as Velocity and Acceleration 25

Numerical Problems 1. A bus is moving with the initial velocity 10 m/s. After 2 seconds, the velocity becomes 20 m/s. Find the acceleration and distance moved by the bus. [Ans: 5m/s2, 30m] 2. Two buses are moving in opposite directions with velocities of 36 km/hr and 108 km/hr. Find the distance between them after 20 minutes. [Ans: 24 km] 3. Two cars A and B are moving in the same direction with velocities of 12 m/s and 9 m/s respectively. What is the relative velocity of the car A with respect to the car B? If the both cars are travelling in opposite directions, what will be the relative velocity? [Ans: 3 m/s, 15 m/s] 4. If the velocity of a bicycle is 10 m/s, how long will it take to cover a distance of 18 km? [Ans: 1800 s] 5. A man runs 28m in a straight line in 4 seconds. Find his velocity. [Ans: 7 m/s] lossar – ex actly the process of moving Precisely n ne a ve a e era n – which is not eq ual Retarda n Non-uniform 26 New Creative Science and Environment Book - 8

U3 SIMPLE MACHINE L earning O utc om es At the end of this unit, students will be able to: ~ define a lever and explain the working principle of a lever. ~ introduce mechanical advantage (MA), velocity ratio (VR) and efficiency with giving the example of a lever. ~ solve the simple numerical problems of mechanical advantage (MA), velocity ratio (VR) and efficiency related to a lever. Main points to be focused ~ Lever ~ Pulley ~ Mechanical Advantage (MA) ~ Wheel and Axle ~ Velocity ratio (VR) ~ Screw Wedge ~ Efficiency (η) ~ Principle of a simple machine Introduction It is very difficult and time consuming if we do work without any tools and devices. For example, if we have to load or unload some concrete from a truck, then without a shovel, it is very difficult. Similarly, without an axe or saw, the chopping of a tree is very time consuming. It is also impossible to lift a truck by a man without a jackscrew. Thus, we use several types of instruments in our daily life to make our work easier and faster. These instruments are called simple machines. Simple machines are the devices, which are simple in construction and help to do our work easier and faster. Complex machines are made by the combination of more than one simple machines. Brooms, shovels, jackscrews, spanners, nail-cutters, etc. are the examples of simple machines whereas cars, buses, motorcycles, cycles, etc. are complex machines. Ways in which simple machines make our work easier Simple machines make our work easier in the following four ways: Simple machine 27

(i) They multiply our force, e.g. a heavy B Effort truck can easily be lifted by a person A using a jack-screw. (ii) They transfer our force from one point to another, e.g. when we apply force on one side of the knife, the knife transfers our force and its next end cuts the vegetables. (iii) They change the direction of force, e.g. in pulleys, when the effort is applied downwards the load moves upwards. (iv) They increase the rate of doing work, e.g. dhiki helps to gain speed on the work. Question i. In how many ways do the simple machines make our work easier? Some terms related to simple machines (a) Load (L): the force which we should overcome is called a load. Its SI unit is newton (N). (b) Effort (E): The force which we apply on simple machine is called an effort. Its SI unit is newton (N). (c) Fulcrum (F): It is a fixed point about which a simple machine rotates. (d) Effort distance (ED): The distance travelled by an effort or the distance between an effort and a fulcrum is called effort distance. Its SI unit is m. (e) Load distance (LD): The distance travelled by a load or the distance between a load and a fulcrum is called load distance. Its SI unit is m. (f) Work Input (Win): The work done by an effort on a simple machine is called work input. 28 New Creative Science and Environment Book - 8

i.e. work input (Win) = Effort (E) × Effort distance (ED) ∴ Win = E × ED Its SI unit is joule (g) Work Output (Wout): The work done by a load or by a simple machine is called work output. i.e. work output (Wout) = Load (L) × Load distance (LD) ∴ Wout = L × LD Its SI unit is Joule. (h) Mechanical Advantage (MA): It is the ratio of a load to an effort applied in a simple machine. i.e. mechanical advantage (MA) = Load (L) Effort (E) ∴ MA = L E A mechanical advantage has no unit. It is the ratio of the same physical quantities i.e. forces. It shows the number of times the force applied on machine is multiplied by it. Thus, if MA of a simple machine is 3, it means that the machine can multiply our force by 3 times. Memory Note ; The MA of a simple machine is affected by friction. Since, it is the ratio of two forces. Question i. The MA of a simple machine is 2. What do you mean by it? ii. Which factors affect MA of a simple machine? (i) Velocity ratio (VR) : It is the ratio of an effort distance to the load distance or it is the ratio of the velocity of an effort to the velocity of load at the same time. Effort distance (ED) i.e. velocity ratio (V.R.) = Load distance (LD) ∴ ED VR = LD Velocity ratio also has no unit. As it is the ratio of two distances. If the VR of a machine is 2, it means the effort has to travel two times more distance than the load to overcome the load. Simple machine 29

Memory Note ; VR is not affected by friction. In an ideal condition, MA = VR. So VR is also known as an ideal MA. But in real practice, no simple machine is frictionless. So, MA is always less than VR in practice. Question i. Can you increase MA of a simple machine by putting oil in it? Why? Efficiency (η) It is the ratio of output work and input work expressed in percentage. i.e. efficiency (η) = Work output (Wout) x 100% Work input (Win) Wout ∴η = Win x 100% If the efficiency of a simple machine is 60%, it means that only 60% of our effort is converted into useful work and 40% of our applied effort is wasted to overcome the friction and gravitational force. No simple machine is frictionless in practice. So, no simple machine is 100% efficient in practice. Memory Tips ; A simple machine with 100% efficiency is called an ideal simple machine and no machine is ideal in practice. Questions i. No simple machine is ideal in practice. Why? ii. A simple machine has the efficiency of 75%. What do you mean by it? Relation among MA, VR and η We have from efficiency, η = Wout x 100% Win L x LD [ Wout = L × LD and Win = E × ED] or, η = E x ED × 100% 30 New Creative Science and Environment Book - 8

L LD [ MA = L ED or, η = E × ED × 100% E and VR = LD] 1 ∴ η = MA x VR × 100% MA ∴ η = VR x 100% Principle of a simple machine All the simple machines work on a basic principle which states that, “In a balanced condition, work done on the machine (input work) is equal to the work done by the machine (output work)’’. i.e. in a balanced condition, Work input = work output ∴ E × ED = L × LD Types of simple machines There are six types of simple machines. They are: (i) Lever (ii) Pulley (iii) Inclined plane (iv) Wheel and axle (v) Screw (vi) Wedge Lever A lever is a rigid bar which may be straight or bent. It rotates freely about a fixed point called the fulcrum or pivot. It is the simplest and the only simple machine that possesses all the properties of the simple machines. On the basis of the position of a fulcrum, load, effort, levers are classified as, (i) First class lever (ii) Second class lever (iii) Third class lever (a) First class lever: The lever in which fulcrum lies between the load and the effort is called the first class lever. Examples: crowbar, see-saw, dhiki, scissors, beam balance, etc. Simple machine 31

First class lever This type of lever can multiply our effort when its fulcrum is closer to load or when it has more ED. For example, a crowbar is used to lift a heavy stone. Similarly, when the fulcrum lies closer to the effort or when ED is less, it can accelerate our work, e.g. dhiki. So, in first class lever, V.R = 1 when E.D = L.D V.R > 1 when E.D > L.D V.R < 1 when E.D < L.D (b) Second class lever: The lever in which load lies between the effort and fulcrum is called the second class lever. For example, wheelbarrow, nutcrackers, bottle opener, oar of a rowboat, etc. Bottle opener Wheel barrow Nut cracker Second class lever This type of lever can magnify our work but cannot increase the speed of doing work because in this class of lever, the LD is never greater than ED. Question i. While shifting the load towards the wheel in a wheelbarrow, it is easier to overcome the load. Why? 32 New Creative Science and Environment Book - 8

(c) Third class lever: The lever in which the effort lies between the load and fulcrum is called the third class lever. For example, shovel, fire tongs, sugar tongs, broom, spoon, etc. Third class lever The third class lever can increase the rate of doing work but it cannot magnify our force. This is because the effort distance is never greater than the load distance. Question i. The third class lever cannot magnify our force. Why? ii. What is the value of VR of the third class lever and why? Relation among MA, VR and η of the lever L In lever, MA = E , ED V.R. = LD η = Wout x 100% Win MA η = VR x 100% η = Wout x 100% Win Simple machine 33

ctivit Suspend a 30 cm long scale as shown in the figure along with the weights on its both sides. Suppose the weight at one side as a load and weight at other side as an effort. Now, by changing the location of loads, try to balance the lever. Fill in the given table and calculate. Left side Right side Conclusion E ED E × E.D L.D L × L.D L Differences between second class and third class lever S.N. Second class lever S.N. Third class lever 1. In this lever, a load lies between an 1. In this lever, an effort lies between a effort and a fulcrum. load and a fulcrum. 2. It helps us to do our work by 2. It helps us to do our work by multiplying the force applied. increasing the speed of doing work. 3. MA is always greater than 1. 3. MA is always less than 1. Solved Numericals 1. Calculate the MA, VR and η of the given lever. Solution: Given, Load (L) = 600 N 6 00N 30 cm Load distance (Ld) = 30 cm 9 0 cm Length of lever (l) = 90 cm ∴ Effort distance (Ed) = l – Ld = 90 – 30 = 60 cm At equilibrium, Effort (E) L × Ld 600 × 30 Then, = Ed = 60 = 300 N L 600 MA = E = 300 = 2 34 New Creative Science and Environment Book - 8

VR = Ed 60 Ld = 30 = 2 MA ∴ Efficiency (η) = VR x 100% 2 = 2 x 100% = 100% This is a perfect machine which is practically impossible. 2. From the diagram below, calculate the efficiency of the lever. Solution: Given, Load (L) = 600 N Effort distance (Ed) = 20 cm Load distance (Ld) = 20 + 60 = 80 cm Now, effort (E) L × Ld Then, efficiency (η) = Ed 600 × 80 = 20 = 2400 N L × Ld = E × Ed × 100% 600 × 80 = 2400 × 20 × 100% 40 = 40 x 100% = 100% Such machines are called perfect machines and these are practically impossible. Pulley A pulley is a circular disc having a groove at its circumference, through which a rope is passed. The load is tied at one end and the effort is applied to the other end of the rope. There are three types of pulleys: (a) Fixed Pulley The pulley which is fixed at a point and does not move up and down along the load is called a fixed pulley. It helps us by changing the direction of the force. It cannot magnify our effort, e.g. a pulley used in a well, a flag stand, etc. Simple machine 35

(b) Movable pulley The pulley which moves up and down along with the load is called a movable pulley. It helps us by magnifying the effort but cannot change the direction of the force, e.g. pulleys used in cranes. (c) Block and tackle The combination of a fixed pulley and a movable pulley is called a block and tackle. It can magnify the effort as well as change the direction of the force, e.g. pulleys used in cranes. Inclined Plane Block and tackle A slanted surface along which a load is pushed or pulled up is called an inclined plane. For example, winding roads on hills, an inclined wooden plank used to load and unload goods on a truck, staircase, etc. Inclined planes help us by magnifying our force. The more the length of inclined plane, is, the more will be the magnification of the force and our work will be much easier. Hence, winding roads are made on hills to increase the length of inclination. 36 New Creative Science and Environment Book - 8

Question i. On hills, the roads are made winding. Why? Wheel and Axle It is a combination of two co-axial cylinders of different diameters. The larger one is known as a wheel and the smaller one is an axle. They are fixed in such a way that when one of them moves, another also rotates in the same direction. The wheel and axle magnify effort. As an effort is applied to the wheel, it accelerates our work. Some examples of a wheel and an axle are the knob of the door, screwdriver, the steering of the vehicles, string roller, spanner, madani, etc. Question i. Long spanners are used to open tight knots. Why? Screw It is a cylindrical surface on which there are a number of spirally wounded threads. The distance between two consecutive threads is called a pitch. The pitch is equal to the load distance. It helps us to magnify our force. So, a person can lift a truck by using a jack-screw. Wedge It is a combination of two inclined planes. One of its ends is sharp and another is blunt, e.g. all the cutting and piercing tools like an axe, a knife, a needle, a nail, etc. All of these help us to magnify our force. Simple machine 37

Question i. It is easier to cut with sharper tools. Why? ctivit Observe the different types of simple machines used in your daily life and classify them. Solved Numericals 1. A crowbar is used to lift a load of 2400 N at a distance 50 cm from the fulcrum. Calculate the effort required to balance the load at 90 cm from the fulcrum. Solution: Given, Load (L) = 2400 N Load distance (LD) = 50 cm Effort distance (ED) = 90 cm Effort (E) =? We have, from the principle of simple machine, L × LD = E × ED or, 2400 × 50 = E × 90 ∴E 2400 × 50 = 90 = 1333.33 N ∴ Effort required is 1333.33 N 2. In a lever, 120 N load is attached at a distance of 20cm from the fulcrum. If this effort is used to lift 50 kg weight then, find (i) LD (ii) MA (iii) VR (iv) fficiency Solution: Given, load (L) = 50 kg = (50 × 10) N = 500 N Effort (E) = 120 N, Effort distance (ED) = 20 cm (i) Load distance (LD) = ? (ii) MA = ?, (iii) VR = ? (iv) η = ? 38 New Creative Science and Environment Book - 8

We have, L (ii) MA = E (i) L × LD = E × ED 500 or, 500 × L.D = 120 × 20 = 120 2400 = 4.17 or, LD = 500 MA ∴ LD = 4.8 cm (iv) η = VR × 100% ED 4.17 (iii) VR = LD = 4.17 × 100% = 100% 20 = 4.8 = 4.17 Answer Writing Skills 1. The efficiency of a simple machine is 60%. What do you mean by it?  It means that 60% of our effort is converted into the useful work and 40% is wasted to overcome frictional and gravitational force. 2. Can you increase the efficiency of simple machines? How?  Yes, we can increase the efficiency of simple machines by reducing friction. For this, we can use oil or lubricants on the moving parts of the machine, we can roll the surface instead of sliding, etc. 3. Value of M.A in the third class lever is always less than one. Why?  We know that, in the third class lever, an effort lies between a load and a fulcrum. So, ED is never greater than LD and so VR is less than one since, VR = ideal MA. So, MA in it is also less than one and it can never magnify our force. 4. It is easier to lift a load when it is shifted towards the wheel in a wheelbarrow. Why?  The shifting of a load towards the wheel decreases the load distance that increases MA. Hence, it can magnify our force and make it easier for us to overcome the load. 5. VR is always greater than MA in a simple machine. Why?  It is because MA is affected by friction but VR is not affected, as it is the ratio of two distances. 6. The handles of metal-cutters are made long. Why?  The handles of metal-cutters are made long to increase the ED, which in turn increases MA. Thus, it magnifies our force and helps to cut metals easily. Simple machine 39

SUM M ARY ” The machines which are simple in construction and help to make our work easier and faster are called simple machines. ” The ratio of the load to the effort is called the MA of a simple machine. I.e. MA = L E ” VR of a simple machine is the ratio of effort distance to load distance ED i.e. VR = LD ” A machine having 100% efficiency is called an ideal simple machine. No simple machine is ideal in practice. ” A lever is a rigid bar. It may be straight or bent, and it rotates about a fixed point called fulcrum. ” There are three classes of levers. ” Scissors, crowbars, dhiki are the first class levers, wheelbarrows, nutcrackers, bottle openers are the second class levers and sugar tongs, shovels are the third class levers. ” Pulleys are of three types (i) fixed pulley, (ii) movable pulley and (iii) block and tackle. ” The slanted surface is called an inclined plane. ” The wheel and axle are a combination of two cylinders of different diameters. ” A screw is a cylinder having spirally wounded threads around it. ” All cutting and piercing instruments are called wedges. Exercise 1. Fill in the blanks. (a) A spanner is an example of __________. (b) The MA of a simple machine is affected by ___________. (c) A single fixed pulley makes our work easier by __________. (d) A physical balance is __________ lever. (e) The MA of the third class lever ___________ one. 2. Define the following. (a) MA (b) VR (c) Efficiency of a simple machine (d) Lever (e) Pulley (f) Wheel and axle 40 New Creative Science and Environment Book - 8

3. Write the differences between: (a) First class lever and second class lever (b) First class lever and third class lever (c) Second class lever and third class lever 4. Give reasons. (a) Metal-cutters have long handles. (b) A third class lever cannot magnify our force. (c) Roads on hills are made winding. (d) A tight knot can be opened easily by using a long spanner. (e) VR is always greater than MA. (f) VR is not affected by friction. (g) MA of the first class lever may be equal to, greater than or less than 1. 5. Answer the following questions. (a) What do you mean by simple machines? How do simple machines make our work easier? (b) Show the relation of MA, VR and η of a simple machine. (c) How do inclined planes make our work easier? (d) Show the relation of MA, VR and η in a leaver. (e) If you have to magnify force, which pulley will you use? (f) Describe all three types of levers with examples. Numerical Problems 1. If a 500 kg load is to be lifted with a 100N effort using a first class lever, at what distance from the fulcrum must the effort be applied when the load is 20cm away from the fulcrum? (Ans: 1m) 2. A load of 500 kg is lifted using a first class lever. Different effort distances and then corresponding efforts are shown in the table below. How far is the load from the fulcrum? Also find the values of A, B, C and D. rt distan e 10 cm A 40cm C 100 cm rt 1000 N 500 N B 200 N D [Ans: LD = 20cm, A = 20cm, B = 250 N, C = 50 cm and D = 100N] Simple machine 41

3. Naina with the weight of 350 N and Maina with the weight of 250 N are playing see-saw. If Naina is 2 m away from the fulcrum, find, how far must Maina sit in order to balance Naina. (Ans: 2.8 m) 4 Calculate the magnitude of the effort in the given figure. (Ans: 100 N) 5 A crowbar having the length of 2.5 m is used to balance a load of 750 N. If the distance between the fulcrum and the load is 2 m, calculate (i) Effort (ii) MA (iii) VR (iv) η [Ans: 1.6, 2, 80] 6. Observe the given diagram and calculate: (i) MA (ii) VR (iii) η lossar [Ans: 3000 N, 0.25, 0.25, 100] C o-ax ial - having the same ax is ne r ener ient - d in s ethin ith n aste e, - a machine used to move heavy objects C rane - s r di u t t end R igid - increase M agnify 42 New Creative Science and Environment Book - 8

U 4 PRESSURE L earning O utc om es At the end of this unit, students will be able to: ~ define atmospheric pressure and explain its importance. ~ define liquid pressure. ~ derive the formula of the liquid pressure and to solve its simple numerical problems. Main points to be focused ~ Density ~ Pressure ~ Relative Density (or Specific Gravity) ~ Atmospheric pressure ~ Floating and Sinking of object ~ Liquid pressure ~ Measurement of liquid pressure Introduction Have you ever questioned why a camel can run in a desert while a girl with high-heeled shoes finds difficult? Why has a tractor wider tyres? Why do cutting tools have sharper edges? We should introduce the term, ‘Pressure’ in order to find the answers to these questions. Pressure The force acting perpendicularly on per unit area of the surface is called pressure. I.e. pressure (P) = Force (F) Area (A) ∴ P = F A Its SI unit is Nm–2 or Pascal (Pa). Other common units of pressure are mm of Hg, atmosphere, bar, torr, etc. 1 atmosphere = 105 Nm–2 or Pascal = 760 mm of Hg Pressure 43

From the above relation, it is clear that, Pressure (P) ∝ Force (F) [when area is kept constant] P∝ 1 [when force is kept constant] Area (A) It shows that pressure increases with the increase in force and decreases with the increase in a surface area. So, a girl with high-heeled shoes exerts more pressure on the sand as it has less area and makes much more depression. But on the other hand, the elephant’s weight is distributed among its four legs (larger area). So, the pressure exerted is less and it makes less depression. If 1 N force is applied to the surface area of 1m2 then the pressure exerted on that surface is known as one Pascal pressure. Memory Note ; In honour of scientist Blaise Pascal, the SI unit of pressure is called Pascal (Pa). Questions i. All cutting and piercing tools have sharp edges. Why? ii. It is difficult to walk on a sandy surface. Why? ctivit To show the pressure of a brick in different conditions Take a piece of foam and a brick. Keep the brick as shown in figure (a) with its larger surface area over the foam and observe the depression on the foam. Now, keep the brick as shown in figure (b) with its smaller surface area over the foam and again observe the depression. 44 New Creative Science and Environment Book - 8

Figure: (a) the brick on the foam with a large surface (b) the brick on the foam with a b small surface The foam is depressed more in the latter case than in the former. It is because, in condition (b) the area of the brick on the foam is less than in condition (a) and as we know pressure is directly proportional to the force (weight). So, the brick exerts more pressure and the foam gets more depressed. Solved Numericals The force of 106 N is acting on a wall of the surface area 10m2. Find the pressure acting on the wall. Solution: Given, force (F) = 106 N, Area (A) = 10 m2 Pressure (P) = ? We have, P = F = 105 = 105 Pa A 10 ∴ The pressure exerted on the wall is 105 Pascal. Atmospheric pressure We know that there is a mixture of different gases in our surroundings. The mixture of different gases which is present in our surroundings is called atmosphere. The atmosphere extends up to 9600 km above the earth’s surface. This mixture of gases has weight. Due to this weight, it applies pressure. The pressure exerted by the mixture of different types of gases on the earth’s surface is called atmospheric pressure. The atmospheric pressure Pressure 45

at the sea level is called the normal atmospheric pressure or standard the atmospheric pressure. Normal atmospheric pressure = 760 mm of Hg or 1 atmosphere or 105 Pascal. As the amount of air decreases at a height altitude, the atmospheric pressure also decreases subsequently. So, the atmospheric pressure decreases with a higher altitude. The atmosphere also exerts pressure on the human body. The force exerted by the atmosphere on the human body of area 2 m2 is about 2 × 105 N as: F = P × A = 105 × 2 = 2 × 105 N We do not feel this high pressure. This is because our internal pressure balances this external pressure. But as we go to a higher altitude, our internal pressure becomes more. So, the bleeding from the nose and ear may occur. Memory Note ; Jet planes fly at about an altitude of 2,000 m. There is less atmospheric pressure. In these planes, they maintain the pressure so that passengers inside the plane can breathe normally. Questions i. We feel uneasy at higher altitudes. Why? ii. The atmosphere exerts enormous pressure to us but we do not feel it. Why? ctivit To show the atmospheric pressure Take a tin-can with a little water in it. Boil it for a few minutes as shown in the figure (a) Close the lid of the can immediately and put out the flame. Now, cool the can under a tap as shown in the figure (b). Figure: (a) heating tin-can (b) crushed can Before heating, the pressure of the air inside as well as outside the tin-can was equal. When it was heated, the air inside the can escaped out and the space was occupied by the steam. On cooling the can, the steam condensed creating a partial vacuum inside the can. Hence, the atmospheric pressure was higher than the internal pressure, and the tin got crushed inwards. This proves that atmosphere exerts pressure. 46 New Creative Science and Environment Book - 8