Modern Graded Science 8

Approved by the Government of Nepal, Ministry of Education, Curriculum Development
Centre (CDC), Sanothimi, Nepal as an additional material for schools

Revised Edition

MODERN GRADED

SCIENCE

AND

ENVIRONMENT

8

Jai Prakash Srivastav Authors
Bijay Shankar Mishra Khaga Raj Ghimire
Mahendra Bahadur Thapa

VIDYARTHI PUSTAK BHANDAR

Publisher and Distributor

Bhotahity, Kathmandu

MODERN GRADED

SCIENCE

AND

ENVIRONMENT

8

Publisher:
Vidyarthi Pustak Bhandar
Bhotahity, Kathmandu, PO Box: 12990
Phone: 01- 4245834, 4227246, 4423333
[email protected], www.vpb.com.np

Copyright : Authors

ISBN : 978-9937-715-07-2

Layout/Design : Dayaram Dahal/Roshan Danuwar

Edition : Revised and updated 2018

Printed at : Alliance Printers & Media House Pvt. Ltd.

PREFACE

We are living in a scientific age. Internalization of facts, concepts and principles of
science has become a dynamic factor in our day-to- day life. Advancement of science
and technology has brought unprecedented changes in the present civilization. These
factors have made Science one of the major disciplines in our education system.

This is the currently revised and updated edition of the Modern Graded Science
series brought out for the young learners to enhance their smooth learning of science.
Immense popularity of its previous editions among our students and teachers has
provided us an additional impetus to revise this series again making this edition more
practical, more user-friendly and more comprehensive. Based on the latest approaches
of teaching science, the series envisages learner-centred teaching methods, activity-
based classroom techniques and additional creative activities.

Heavy emphasis on developing concepts rather than providing mere information,
special efforts to stimulate learners’ creative thinking for encouraging them to get
involved in learning activities, presentation of scientific knowledge in a very logical
and sequential manner in simple and lucid language to keep up with the ability of the
young learners are some of the main features of this series.

Each chapter in this series includes Objectives as guidelines, Things to know
as a summary, Things to do as a means to develop skills for project works, and
understanding the related scientific facts easily, Glossary as a list of difficult words
and Test yourself for testing knowledge and understanding, whereas the Appendix
includes the specification grid for theory and practical, rubric for practical evaluation
and a set of model test paper. Moreover, some startling facts related to science
presented under Do you know? and variety of MCQs in each lesson serve as the new
features of this series.

We are glad to state that this edition of the Modern Graded Science series is
strictly based on the latest curriculum prescribed by CDC along with some additional
activities if and where necessary to help the students conceptualize the facts.

Constructive suggestions and criticisms from experts, teachers, students and
well-wishers for further simplification and standardization of this series will be
gratefully acknowledged. Thanks!

Authors

CONTENTS

Unit 1 Measurement...........................................................................................................1
Lesson 1: Measurement............................................................................ 1

Unit 2 Force........................................................................................................................15
Lesson 2: Velocity and Acceleration.................................................... 15
Lesson 3: SImple Machine...................................................................... 37
Lesson 4: Pressure.................................................................................... 48

Unit 3 Energy.....................................................................................................................64
Lesson 5: Energy, Work and Power........................................................ 64
Lesson 6: Heat.......................................................................................... 80
Lesson 7: Light.......................................................................................... 94
Lesson 8: Sound..................................................................................... 112
Lesson 9: Magnetism............................................................................. 122
Lesson 10: Electricity................................................................................ 130
Revision Exercise for Physics.................................................142

Unit 4 Matter....................................................................................................................145
Lesson 11: Matter..................................................................................... 145
Lesson 12: Mixture.................................................................................... 170
Lesson 13: Metals and Non-metals........................................................ 179
Lesson 14: Acid, Base and Salt.............................................................. 191
Lesson 15: Some Useful Chemicals........................................................ 205
Revision Exercise for Chemistry............................................214

Unit 5 Living Beings.........................................................................................................217
Lesson 16: Living Beings.......................................................................... 217

Unit 6 Cell Biology..........................................................................................................237
Lesson 17: Cell and Tissue....................................................................... 237

Unit 7 Physiology................................................................................................. 247
Lesson 18: Life Processes......................................................................... 247
Revision Exercise for Biology.................................................263

Unit 8 Geology................................................................................................................265
Lesson 19: Structure of the Earth............................................................ 265
Lesson 20: Weather and Climate.......................................................... 274

Unit 9 Astronomy............................................................................................................283
Lesson 21: The Earth and Space............................................................ 283
Revision Exercise for Geology and Astronomy..................292

Unit 10 Environment Science..........................................................................................294
Lesson 22: Environment and Its Balance.............................................. 294
Lesson 23: Environmental Degradation and Its Conservation.......... 312
Lesson 24: Environment and Sustainable Development.................... 333
Revision Exercise for Environment Science........................342
Appendix ................................................................................................ 345

Unit

1 MEASUREMENT

1Lesson MEASUREMENT

Total Estimated Pds: 5 [Th. 4 + Pr. 1]

On completion of this lesson, the students will be able to:

tell the importance of measurement.
define fundamental units and derived units.
identify the units of mass, time and weight in different measurement systems.
solve numerical problems related to measurement.

We measure different quantities in our daily life. For example, we measure the mass of
vegetables by using a beam balance, length of any object by using a scale, temperature
of the body by using a thermometer, etc. So, mass, length, temperature, etc. are
physical quantities. A physical quantity is that quantity which can be measured directly or
indirectly. Density, area, volume, time, etc. are some other examples. Feeling, kindness,
innocence, etc. are not physical quantities because they cannot be measured. Physical
quantities are of two types: fundamental quantity and derived quantity.

A. Fundamental quantity

Those physical quantities which do not depend upon other quantities are called
fundamental quantities. For example, the mass of a body is a fundamental quantity
because it cannot be expressed in terms of other quantities. How many meters equal
1 kg? It does not make any sense. Mass does not depend on length. Length, time,
temperature, electric current, amount of substance, luminous intensity, etc. are the
examples of fundamental quantities.

B. Derived quantity

Those physical quantities which are to be expressed in terms of other fundamental
quantities are called derived quantities. Area is a derived quantity as it depends on
a fundamental quantity, the length. Some other examples of the derived quantity are
velocity, acceleration, area, volume, density, etc.

Measurement 1

We cannot find the exact value of any physical quantity without measurement. We
consider a fixed quantity as a standard quantity for the measurement of any physical
quantity. The measurement of unknown quantities can be done by comparing them
with a standard quantity of the same kind. Thus, measurement is the comparison of an
unknown quantity with a known established standard quantity.

The measurement of a physical quantity is expressed in a numerical value followed by
the unit. For example, if the length of a room is 5 m, it means that the unit of length is
metre (m) and the length of the room is 5 times longer than 1 metre (m).

Similarly, if the mass of a body is 10 kg, it means that the unit of mass is kilogram (kg) and the
mass of the body is 10 times greater than 1 kg. Briefly, a physical quantity is expressed as:

Physical quantity = Numerical value × Unit

The standard quantity with which we carry out the measurement of any physical quantity
of the same kind is called a unit. There are different kinds of units used to measure the
same physical quantity. But all such units are not universally accepted. The unit which
is accepted all over the world is a standard unit. Thus, a standard unit is defined as the
internationally accepted reference standard to measure a physical quantity. For example,
kilogram, metre and second are the standard units of mass, length and time respectively.

Importance of measurement

1. Measurement helps in selling and buying goods.
2. It is important in performing scientific experiments to establish truth about a

physical phenomenon.
3. It helps to obtain the accurate results about physical quantities.
4. It helps in performing experiment for making our daily food.

Standard system of measurement

People in different places of our country still use the local units like haat, bitta, mana, pathi,
muri, dharni, etc. for the measurement of length and mass. These units are understood
by the people in a certain place only. These units may vary from place to place and
from person to person. In order to maintain the uniformity in measurement, some other
standard systems of measurement have been used. There are four standard systems of
units in measurement. They are: CGS system, MKS system, FPS system and SI system.

CGS System: The system of units in which length is measured in centimeter, mass in
gram and time in second is called CGS system. It is known as the French system of units.

MKS System: The system of units in which length is measured in metre, mass in
kilogram and time in second is called MKS system. It is also called metric system.

FPS System: The system of units in which length is measured in foot, mass in pound
and time in second is called FPS system. It is the British system of units.

In CGS, MKS and FPS systems, the first letter represents the unit of length, the second
letter represents the unit of mass and the last letter represents the unit of time.

2 Modern Graded Science and Environment Book 8

SI System: In the year 1960, the scientists in the Eleventh General Conference on
Weights and Measures introduced SI system. SI system is, in fact, the improved and
extended version of MKS system of units. The system of units recommended by the
international convention of scientists held in France is called SI unit. In this system,
length is measured in metre, mass in kilogram and time in second.

Units are divided into two types: fundamental units and derived units.

Fundamental units and derived units

The units which are used to express fundamental quantities are called fundamental units.
In other words, the units that are independent of other units are called fundamental units.
Units like metre, kilogram, second, kelvin, etc. are the fundamental units of length, mass,
time and temperature respectively. Study the following fundamental units.

S.No. Measurement of SI unit S.No. Measurement of SI unit
fundamental quantity
1. length metre (m) fundamental quantity
2. kilogram (kg)
3. mass second (s) 5. current ampere (A)
4. time kelvin (K)
6. luminous intensity candela (cd)
temperature
7. amount of substance mole (mol)

The units of derived quantities are called derived units. In other words, the units which
are derived from two or more fundamental units are called derived units. Simply, derived
units are defined as all other units except seven fundamental or base units.

The unit of density is kg/m3 which is a derived unit. This is because it is expressed with
the help of fundamental units kg [unit of mass] and three basic units: metre (m), metre
(m) and metre(m) [unit of length]. Similarly, the unit of volume is m3, which is a derived
unit as it depends on the three basic units: metre(m) metre (m) and metre (m) [unit of
length]. The units of area, speed, velocity, acceleration, etc. are some other examples
of derived units. Some derived quantities, their formulae, their SI units and fundamental
units involved are given in the table given below:

S.No. Physical quantity Related formula Name of unit Symbol Basic units involved

1. Area (A) A=l×b Square metre m2 m×m

2. Volume (V) V = l × b x h Cubic metre m3 m×mxm

3. Speed/Velocity (v) v = s/t Metre/second m/s m/s

4. Acceleration (a) a = v - u/t Metre/second square m/s2 m/(s × s)

5. Force (F) F=m×g Newton N (kg × m)/(s × s)

6. Work/Energy W=F×s Joule J (kg × m × m)/(s × s)

7. Power (P) P = W/t Watt W (kg × m × m)/(s × s × s)

8. Pressure (P) P = F/A Pascal Pa kg/(m × s × s)

9. Density (d) d = m/v Kilogram/cubic metre kg/m3 kg/(m × m × m)

10. Frequency (f) f = v/λ Hertz Hz cycle/s

Measurement 3

1. The unit velocity is a derived unit.

Since, the unit of velocity = unit of displacement
unit of time taken
metre = ms
= second [In terms of fundamental units]

The unit of velocity is derived by the combination of metre and second. So, the unit of

velocity (m/s) is called a derived unit.

2. How can you prove that 'N' is a derived unit?

Solution:
N is unit of force.
For force, we have,
F = m×g
When related units are used in the above equation.
N = kg.m/s2

The N is the unit of force derived by the combination of fundamental units like
kg, m and s; thus, it is a derived unit.

3. How can you prove that Watt is a derived unit?
Watt is the unit of power.

We know that,

P = W = F × s
t t
m × g × s
= t [ W = F × s, F = m × g]

By placing the related units in the above equation:

W = kg ×m× m = kgm2/s3
s2 ×s

Watt depends on fundamental units like kg, m and s; thus, it is a derived unit.

Some Measurements

a. Measurement of mass

The amount of a matter contained in a body is called its mass. Mass of an object depends
upon the number of atoms or molecules contained in the object and their atomic mass. The
mass of any object remains same throughout the world. It does not change from place to
place and time to time. For example, 2 kg sugar has the same mass in Nepal as in the UK.

The mass of a substance is measured by using a physical balance, beam balance
or grocer's balance. To measure the mass, the standard weight is placed on the one
pan of the balance while the substance to be measured is placed on the other pan. By
comparing the body with the standard weights, the mass of the substance is measured.

The SI unit of mass is kilogram (kg). The mass of very light objects is measured in milligram (mg)
and gram (gm). Similarly, the mass of very heavy objects is measured in quintal and tonne.

4 Modern Graded Science and Environment Book 8

Some sub-multiples and multiples of kilogram are given below in the table:

10 miligram (mg) = 1 centigram (cg) Do you know?
10 centigram (cg) = 1 decigram (dg)
10 decigram (dg) = 1 gram (g) Karat (kt) is a unit of
10 gram (g) = 1 decagram (dm) measurement of the purity
10 decagram (dm) = 1 hectogram (hg) of gold, representing how
10 hectogram (hg) = 1 kilogram (kg) many parts out of 24 are
100 kilogram (kg) = 1 quintal pure. For example: 18
10 quintal = 1 tonne [1000 kg] karat gold is 3/4 pure.

One standard kilogram

Standard one kilogram is defined as the mass of platinum-iridium cylinder having
equal diameter and height kept in particular conditions at the International Bureau of
Weight and Measures Sevres near Paris. The world uses the duplicate of standard one.
The kilogram weights used in Nepal must be equal to the standard kilogram kept in
Department of Weights an Measures, Kathmandu, Nepal.

The following points should be remembered while measuring mass:

a. The physical balance should be correct. The beam of an accurate balance is
horizontal to the ground when its both pans are empty; and its pointer is coincided
with the mark at the middle of the balance.

2 kg 5 kg

1/2 1 kg
kg

Fig 1.1 (a) standard weights (b) physical balance (c) beam balance (d) grocer's balance

b. The standard weights should be correct. In a standard weight, there is a small hole
or gap filled with lead and the official stamp of the Metrology Department is marked.

Activity 1.1

To measure the mass of objects

Materials required

beam balance, standard weights of kilogram, gram and milligram, Science book, tiffin
box, pencil, calculator, notebook, eraser, duster and a bag.

Procedure
1. Take a beam balance and keep on the table in a balanced condition.
2. Put different objects in one pan and a suitable weights of kilogram, gram and milligram

in another pan. Balance one by one and measure the masses of these objects.

Measurement 5

3. Note down the mass of each object in the table given below.

Observation

S.No. Object kg gm mg
200 50
1 Science book

2 Tiffin box

3 Pencil

4

5

6

7

8

Conclusion

From this activity, you will conclude that a different object has a different mass and mass
is measured in different units of mass.

b. Measurement of Weight

We have seen that fruits fall down from a tree and water drops

down during rain. When any body is thrown upward, its velocity spring

decreases till it reaches the maximum height. After that, the balance

object starts to fall down. During the fall of the object, its velocity

increases with time. Can you answer what is the cause of the

above events? The cause is gravitational force of the earth or

gravity on the body. In fact, gravity causes weight of a body. object
Thus, the force of gravity on a body is called weight of that body.

It is not same everywhere. Its value varies from place to place. Fig 1.2 spring balance
On the earth, its maximum value is at sea level. A body has more measuring an object
weight at the poles than that at the equator. It is measured by

using a spring balance. Its SI unit is newton (N). The weight of a body can be calculated

very simply by using:

W = m.g

Here, W = weight, m = mass, g = acceleration due to gravity

As the value of 'g' is not same at each part of the earth, the weight of a body also varies
at different places of the earth. Units like gram weight and kilogram weight are also in
our practice.

6 Modern Graded Science and Environment Book 8

Difference between mass and weight

Mass Weight

1. The quantity of matter contained in a 1. The force of gravity of the earth acting

body is called its mass. on a body is called weight.

2. It is constant everywhere. 2. It is variable from place to place.

3. It is measured with the help of a beam 3. It is measured with the help of a spring

balance. balance.

4. Its SI unit is kg. 4. Its SI unit is newton (N).

Unit of weight

We know that,

Weight = mass × acceleration due to gravity

W=m×g

The SI unit of mass is kg and SI the unit of acceleration due to gravity is m/s2.

So,

W = kg × m/s2 = kg×m = newton
s2

Thus, the SI unit of weight is newton (N).

Example 1: Calculate the weight of an object of mass one kilogram.

Solution:

Here,

mass of an object (m) = 1 kg

acceleration due to gravity (g) = 9.8 m/s2 Do you know?
weight (W) = ?

We know that, Scientists estimate that total mass
W=m×g of the universe is between 1052 kg
= 1 kg × 9.8m/s2 and 1053 kg.

= 9.8 kgm/s2 =9.8 newton

Thus, the weight of an object of mass one kilogram is 9.8 newton (N).

When somebody asks your weight, you say "My weight is 50 kg or 60 kg." It is your kg
weight. 1 kg weight is equal to 9.8 N at sea level.

Measurement 7

C. Measurement of time

The duration between any two events is called time. The unit of time is determined
by the rotational period of the earth. One rotational period of the earth is also called a
solar day. When a solar day is divided into 86,400 parts, the fraction of time is considered
as one second.

The SI unit of time is second. Bigger units of time are used to measure longer periods
of time. Bigger units of time are minute, hour, day, year, etc. But the smaller interval of
time is measured by using microsecond and millisecond. Microsecond and millisecond
are sub-multiples of a second while minute, hour, day, year, etc. are the multiples of
a second.

SI unit Sub-multiples Multiples
1 minute = 60 s
second (s) 1 microsecond (µs) = 1/10,00,000 s 1 hour = 3,600 s
1 day = 86, 400 s
1 millisecond (ms) = 1/1000 s 1 year = 3,15,36,000 s

Fig 1.3 (a) atomic watch (b) digital watch (c) pendulum clock

We know that the earth rotates about its own axis. The time taken by the earth to rotate
once about its own axis is called a mean solar day. If a day is equally divided into 24
parts, each part is called an hour. If one hour is divided into 60 equal parts, each part is
called a minute. If one minute is further sub-divided into 60 equal parts, each part is called
a second. Time is measured by using different types of watches and clocks. Pendulum
watch, quartz watch, digital watch and atomic watch are some examples. A pendulum
watch is also a type of a mechanical watch. It works on the basis of the oscillation of a
simple pendulum. It does not measure the accurate time. This is because climatic factor
affects the oscillation of a pendulum. Crystal watch works due to the vibration of quartz
crystals. It measures more accurate time than a pendulum watch. Atomic watch works
on the periodic vibration of caesium atom. Nowadays, one second time is defined as the
time taken by the cesium 133 atom to complete 9192631770 vibrations. An atomic watch
can measure millionth of second accurately.

8 Modern Graded Science and Environment Book 8

Example 2: Convert one mean solar day into seconds.

We have, [1hr = 60 min] Do you know?
1 mean solar day = 24 hours [ 1 min = 60 s]
= 24 × 60 minutes One acre is an area of 43,560
= 24 × 60 × 60 seconds square feet; originally, the
= 86,400 seconds area a yoke of oxen could
plough in one day

Thus, 1 mean solar day equals 86,400 seconds.

In physics, 1 second equals 1 th of a mean solar day.
86,400

Very large and very small numbers

If you have to express the mass of the earth in kg, you have to write that the mass of
the earth is 6,000,000,000,000,000,000,000,000 kg. Similarly, if you have to express
the size of an atom, you have to write that it is about 0.0,000,000,001 m. Are the above
examples easy to read and write? Obviously not. Thus, to express and understand such
numbers, we express them in power of tens or in a scientific notation. In this way, the
above numbers will be expressed in the following ways:

Mass of the earth = 6 x 1024 kg and size of atom is 1 x 10-10 m or 10-10m.

While expressing very big and very small numbers, we should remember the following points:

1. For expression, the main number is written smaller than 10 and bigger than 1.

2. For the power of ten:

24 zeros

a- 6 000,000,000,000,000,000,000,000 kg

= 6 × 1024 kg and

b- 10 digits m

0.0,000,000,001

= 1 × 10-10 or 10-10 m

3. If the decimal is shifted left in the main number, the power value increases.
Examples:

7 zeros

a. 2,460,000,000

= 246 × 107

= 24.6 × 107+1 = 24.6 × 108

= 2.46 × 108+1 = 2.46 × 109 (standard expression)

In the standard method, a single digit is written before decimal.

Measurement 9

11 digits

b. 0.00,000,000,246

= 246 × 10-11

= 2.46 × 10-10 (As 10-10 > 10-11)

= 2.46 x 10-9 (Standard expression)

4. If the decimal is shifted right in the main number, power value decreases. Examples:

a. 10 digits

0.0,000,000,125

= 125 x 10-10

= 1.25 x 10-8 (Standard expression)

or = 12.5 x 10-9

11 digits

b. 125,000,000,000

= 1.25 x 1011 (Standard expression)

= 12.5 x 1010

= 125 x 109

Calculation of power of tens

1. For multiplication
+

3.46 × 108 ×1.5 × 106

×
= 3.69 × 1014 (Decimal number is multiplied and powers are added)

2. For division
–

3.46 × 108 ×1.5 × 106

÷
= 2.31 × 102 (Decimal numbers are divided and powers are subtracted.)
3. For addition
3.46 × 108 + 1.5 × 106

At first, the powers of both numbers are made similar by shifting the decimal in a
decimal number. It is better to change smaller power in to bigger.
3.46 × 108 × 0.015 × 108

+
= 3.475 × 108 or, 3.48 × 108 (Power is unchanged now)

10 Modern Graded Science and Environment Book 8

4. For subtraction
3.46 × 108 - 1.5 × 106
3.46 × 108 - 0.015 × 106

(As explained in addition)
3.46 × 108 – 0.015 × 108

–

3.446 × 108 = 3.45 × 108 (Power is unchanged now)

Note: Scientific notation is not in our syllabus; but it is believed that its knowledge
will facilitate the study of physics.

Solved numerical problems

1. Convert 3.12 mm into cm, m and km.

Solution:

(a) 3.12 mm = 3.12 cm = 0.312 cm [  10 mm = 1 cm]
10 [From part a]

(b) 3.12 mm = 0.312 cm

= 0.312 m [ 100 cm = 1 m]
100

= 0.00312 m = 3.12 × 10–3m

(c) 3.12 mm = 3.12 × 10–3 m [From part b]

= 3.12 × 10–3 km [ 1000 m = 1 km]
1000

= 3.12 × 10–6 km

(2) Convert 50 hours into seconds.

Solution:

50 h = 50 × 60 min [ 1 hour = 60 minutes]
[ 1 minute = 60 seconds]
= 50 × 60 × 60 s

= 18,00,00 s

Hence, 50 h equal 18,00,00 s.

THINGS TO KNOW

1. The quantities which can be measured are called physical quantities.

2. The physical quantities which are independent of each other are called
fundamental quantities.

3. The physical quantity which depends on other fundamental quantities is
called a derived quantity.

4. Measurement means comparison of an unknown quantity with a known quantity.

Measurement 11

5. The units which are independent of each other are called fundamental units.
6. The units which depend on other fundamental units are called derived units.
7. CGS, MKS, FPS and SI Systems are the standard systems of units.
8. The system of units recommended by the International Convention of

Scientist held in France in 1960 is called SI Unit.
9. The force of gravity applied on a body is called its weight. Weight is measured

by a spring balance and its SI unit is N.
10. The quantity of the substance contained in a body is called mass. It is

measured by a physical balance and its SI unit is kilogram (kg).
11. The duration between any two events is called time.

THINGS TO DO

1. Take a tennis ball, a cricket ball and a basket ball. Measure their mass using
a physical balance and compare them by finding ratios.

2. Using the examples from the metric units chart, choose the most appropriate
unit to measure the mass of each item. Write the unit of measurement in the box.

(a) water melon (b) goat (c) coin (d) peanut


TEST YOURSELF

1. Fill in the blanks.
a. Square metre is the SI unit of ............................... .
b. Acceleration is a .............. quantity.
c. 1 day is equal to ................ seconds.
d. The force of gravity exerted on a body is its ....................... .
e. The unit of density is ............. .

2. Tick (√) the correct statements and cross (x) the incorrect ones.

a. Kg/m3 is a derived unit.
b. Weight is measured by using a physical balance.
c. The value of mass of a body is fixed everywhere.
d. Volume is not a derived quantity.
e. One kilogram weight of body is equal to 9.8 N.

12 Modern Graded Science and Environment Book 8

3. Tick () the correct answer (MCQs).

a. The standard unit of mass is:

i. gram ii. tonne iii. kilogram iv. milligram

b. The value of acceleration due to gravity is:

i. 8.9 m/s2 ii. 9.8 m/s2 iii. 9.8 m/hr2 iv. 9.8 km/s2

c. The SI unit of temperature is:

i. Celsius ii. Ampere iii. Calorie iv. Kelvin

d. Which one of the following is not a derived quantity?

i. speed ii. velocity iii. time iv. force

e. What are the basic units involved for the SI unit of work?

i. kg×m×m/s×s ii. kg×m/s×s iii. kg×m×m×m/s×s iv. kg×m×m/s×s×s

f. Which one of the following watches measure time accurately?

i. quartz watch ii. atomic watch iii. pendulum watch iv. digital watch

g. What is the unit of mass in metric system?

i. pound ii. kilogram iii. gram iv. quintal

h. Weight is the measure of the force of . . . . . an object.

i. mass ii. matter iii. gravity iv. velocity

i. What is the weight of 1 kg mass object on the earth?

i. 1 newton ii. 9.8 newton iii. 9.8 pound iv. 1 pound

4. Very short answer type questions:

a. What is the standard unit used to measure mass?
b. What is the SI unit of temperature?
c. Write the formula used to calculate the area of a regular object.
d. Name the fundamental units involved in the derived unit joule.
e. Encircle the SI units from the given list of units:
minute, kelvin, mole, bitta, dharni, metre

5. Define: b. standard unit c. unit
a. FPS system e. derived unit f. CGS system
d. MKS system h. mass i. time
g. weight

6. Give reasons.
a. Mass is a physical quantity.
b. Acceleration is a derived quantity.

Measurement 13

c. The unit of force, newton is a derived unit.
d. The unit of energy is a derived unit.
e. The weight of a body is variable from place to place.

7. Answer the following questions.
a. Define measurement.
b. What is the importance of measurement?
c. Define fundamental and derived quantities with two examples of each.
d. What is a fundamental unit? Write its two examples.
e. What is time? How is one second time defined in the SI system?
f. The unit of area is a derived unit. Why?
g. How can you say that m3 is called a derived unit?
h. What is SI system? Write the difference between SI system and CGS system.
i. What is mass? Write its SI unit.
j. What do you mean by a physical quantity?
k. Write any four differences between mass and weight.

8. Numerical problems:

a. Convert the following as instructed:

i. 340 cm into m ii. 67 kg into g

iii. 2 days into seconds iv. 86,400 seconds into day

b. What is the weight of potatoes of mass 75 kg on the surface of the earth?
[g = 9.8 m/s2]

c. The weight of body is 420 N. Calculate its mass. [g = 9.8 m/s2]

Answers

8. a. i. 0.34 m ii. 67,000 g iii 1,72,800 s iv. 1 day b. 735 N c. 42.857 kg

GLOSSARY

Density : the ratio of mass and volume of a substance

Acceleration : the rate of change of velocity

Velocity : the rate of displacement of a body

Metric system : the system of measurement that uses metre, kilogram and

litre as basic units

Electric-current : the flow of electricity through a wire

Luminous intensity : measure of the light-emitting ability of a source of light

Metrology : the science of measurement

14 Modern Graded Science and Environment Book 8

Unit

2 FORCE

2Lesson VELOCITY AND ACCELERATION

Total Estimated Pds: 6 [Th. 5 + Pr. 1]

On completion of this lesson, the students will be able to:

define the reference point in relation to the position of an object.
describe average velocity and relative velocity.
introduce acceleration and retardation.
write the equations related to velocity and acceleration and use them.
solve simple numerical problems related to velocity and acceleration.

Every day, we observe the phenomena Fig 2.1
like a flying bird and a running car. In these
examples, we notice that the objects (bird
and car) keep on changing their position
with respect to some surrounding stationary
objects like trees, houses, electric poles,
etc. They are called moving objects. But
trees, houses, electric poles, etc. are
called stationary objects because they do
not change their position. Hence, we can
say that the objects in the surrounding may
be in a state of rest or motion.

Thus, a body is said to be in a state of motion if it changes its position continuously with
respect to its surroundings with the passage of time. Similarly, a body is said to be at
rest if it does not changes its position with respect to its surroundings with the passage
of time. Thus, motion is a change in the position of an object.

Suppose, a tree on the roadside is in a state of rest as it is not changing its position with
respect to a running car. But, the car changes its position with respect to the tree. So,
it is in a state of motion. Therefore, the passengers in the car are at rest relative to the
running car but the running car is in motion relative to the tree. Hence, rest and motion
are relative terms.

Velocity and Acceleration 15

Types of motion

There are two types of motion as discussed below:

1. Uniform motion 2. Non-uniform motion

1. Uniform motion: The bodies are said to be in uniform motion if they are moving
with a constant speed or velocity. In other words, when a body covers an equal
distance in an equal interval of time, the motion of the body is called uniform. The
uniform motion of a body has constant velocity. Example: the motion of the moon
around earth.

2. Non-uniform motion: If a body moves with varying speed or velocity, the motion of
the body is said to be non-uniform motion. In other words, when unequal distances
are covered in equal intervals of time, the motion is said to be non-uniform motion.
Example: a car moving on the road.

What is a reference point?

A reference point or reference frame is the fixed
point/place/object from which we compare the
motion of an object. For example, when we
are travelling in a bus, we are in motion with
respect to a nearby tree. Here, the tree acts as
a reference point or reference frame.

Velocity Fig 2.2

The time rate of change of displacement of a body is called its velocity. Velocity can also
be defined as the speed with a direction. It is actually a vector as it has both magnitude
and direction. Direction is important because velocity uses displacement instead of
distance.

Mathematically,

Velocity = Displacement Direction
Time magnitude

i.e. v = s
t

Velocity can be positive, negative or zero because displacement can be positive, negative

or zero. In SI system, displacement is measured in metre (m) and time in second (s).

Hence, velocity is measured in metre = m/s.
second

In other words, velocity is defined as the distance travelled by a body in unit time in a

particular direction.

16 Modern Graded Science and Environment Book 8

Example 1: Ram walks from his home to school in 5 minutes. What is his velocity?

W >>>N

>

SE

1300m

Home 600m

Solution: School

Here, Fig 2.3

Displacement (s) = 600 m [∴ displacement is the shortest distance between
any two points.]

Time (t) = 5 minutes ∴

= 5 × 60 = 300 s [ 1 minute = 60 second]

Velocity (v) = ?

We know that,

v = s
t

= 600 m = 2 m/s East
300 s

Therefore, velocity of Ram is 2 m/s.

Uniform velocity and non-uniform velocity

Velocity and speed have the same meaning only when a body moves on a straight line. If
a body moves on a curved path, it is better to describe its motion by speed, rather than by
velocity. Speed in a specific direction is velocity. Velocity may be uniform or non-uniform.

If the velocity of a body is equal throughout the motion, then the velocity is uniform. In this
case, the body covers equal displacement in each second. In other words, a body is said
to be moving with uniform velocity if it always moves in the same direction and covers
equal distances in equal intervals of time.

1 sec 1 sec 1 sec
A 10 m
B 10 m C 10 m D

Fig 2.4 uniform velocity

Velocity and Acceleration 17

If the velocity of a body is not equal throughout the motion, then the velocity is called
non-uniform or variable velocity. In this case, the body covers unequal displacement in
each second. In other words, a body is said to be moving with variable or non-uniform
velocity if it covers unequal distances in equal intervals of time or unequal distances in
equal intervals of time in the direction of motion.

1 sec 1 sec 1 sec

A 10 m B 20 m C 15 m D

Fig 2.5 non-uniform velocity

Average velocity

We can define the average velocity of a body moving with variable velocity. It is defined
as average displacement covered by a body in one second. It is given by the ratio of total
displacement and total time taken.

∴ Average velocity = Total distance travelled [ total distance covered in a particular
Time taken direction is the displacement ]

v= s
t

If a body covers a distance of 20 m in 2 s and then covers 40 m in next 8 s on a straight

line, then the average velocity will be 20 m + 40 m = 60 m = 6 m/s on a straight line
2s+8s 10 s
or in the given direction.

If u is the initial velocity and v is the final velocity of a body, then its average velocity is

given by:

Average velocity = Initial velocity + Final velocity
2

i.e. v = u+v
2

Thus, the average velocity of a body is defined as the arithematic mean of initial and final

velocity of the body.

When a body is moving on a curved path with Do you know?

uniform speed, its speed remains the same but the The fastest possible speed
velocity changes at each point of the curved path. in the universe is the speed
This is because the direction of motion of the body is of light. The speed of light is
299,792,458 metres per second.
continuously changing.

18 Modern Graded Science and Environment Book 8

Solved numerical problems

1. A man runs 28 m on a straight line in 4 seconds. Find his velocity.

Solution:
Given,
Displacement (s) = 28 m
Time taken (t) = 4 sec
Velocity (v) = ?
We have,

Av = s = 28 m = 7 m/s
t 4s

Therefore, the velocity of the man is 7 m/s

2. If the velocity of a bicycle is 10 m/s, how long will it take to cover a distance
of 18 km?

Solution:

Given,

Average velocity (v) = 10 m/s

Distance covered (s) = 18 km

= 18 × 1000 m = 18000 m [... 1000 m = 1 km]

Time taken (t) = ?

We have,

v =s
t

i.e. t = s = 18000 = 1800 s
v 10

Therefore, the time taken by the bicycle is 1800 s or 30 min. [... 1 min. = 60 second]
or 0.5 hour [... 1 h = 60 min.]

Relative velocity

The velocity of a body described relative to a reference point or reference frame is called
relative velocity.

If the velocity of a car is 20 m/s, it means, the car covers a distance of 20 m in one
second. When it starts its motion from a reference point such as a tree, the reference
point does not change its position. If two cars on a road are moving with respect to the
reference point then the distance or direction of the both cars may change and each
body is said to have a relative velocity with respect to the other. Illustrative examples
presented below will give a clear concept.

Velocity and Acceleration 19

1. Motion in the same direction with the same velocity

Suppose two cars, A and B move in the same direction from the same line PQ on a road.
If the velocity of both cars is equal, say 10 m/s, each car will cover a displacement of

10 m in 1 second. Notice that both cars are always in a face to face position. Although

they cover a displacement with respect to the line PQ, they are not covering a distance

relative to each other. After one second
A
A
P P

10 m/s 10 m/s

BB

Q 10 m/s Q 10 m/s

10 m
Fig 2.6

Thus, their relative velocity will be zero. That's why, passengers in both cars appear to
be stationary for each other in this case. It will be calculated by:

VAB = VB - VA­ = 10 - 10 = 0
2. Motion in the same direction with different velocities

If two cars, A and B are moving in the same direction on the same road with velocities VA
and VB respectively, then the velocity of car A relative to car B is calculated by:

VAB = VB - VA­ After one second
A
A
P P

VA = 10 m/s VA = 10 m/s

BB

Q VB = 15 m/s Q VB = 15 m/s

10 m
15 m

Fig 2.7

If two cars, A and B are moving in the same direction with velocities (VA) 10m/s and (VB)
15m/s respectively, they cover respective distances of 10m and 15m in 1 second. Thus,
car B covers 15 m - 10 m = 5 m more than car A. Now, velocity of car B relative to car A
is given by:

VBA = VB - VA.
= 15 m/s - 10 m/s = 5 m/s

Hence, the relative velocity of car B with respect to car A is 5m/s. That's why, passengers
in car B observe that car A is moving very slowly.

20 Modern Graded Science and Environment Book 8

3. Motion in opposite directions with the same or different velocities

If two cars, A and B are moving in opposite directions with velocities 15m/s and 10m/s
respectively, they cover respective distances of 15 m and 10 m in 1 second in opposite
directions. Thus, for car A, car B has covered a distance of 15 + 10 = 25 m in 1 second.
When the cars are moving in opposite directions, we use the following formula for

calculation: VAB = VA + VB

VAB = VA + VB = 10 + 15 = 25 m/s

Hence, the relative velocity of car A with respect to car B is 25 m/s. That's why, the

passengers in car A feel as if car B is moving much more faster than the actual velocity

it has. After one

second After one second
VB = 15 m/s
VA = 15 m/s
VA = 15 m/s
PA PA

BB

VB = 10 m/s Q
Q

Fig 2.8 10 m 15 m

The relative velocity of a body, A with respect to another body, B when both are in motion,
is the rate at which body A changes its position with respect to body B.

Remember:

If a car, A is moving in the direction opposite to the other car, B on a straight line then the
velocity of car A relative to car B is calculated by using the formula:

VAB = VA + VB

If two cars, A and B are moving in the same direction on a straight line, the velocity of car
B relative to car A is calculated by using the formulae:

VBA = VB - VA

Solved numerical problems

1. Two cars, A and B are moving with their velocity 20 m/s and 15 m/s in the
directions east and west respectively. If they started from the same place at
the same time, what would be the distance between them after 2 minutes?
Find the distance covered by each car at the same time (2 minutes).

Solution:
Given,

Velocity of car A (VA) = 20 m/s (in east)
Velocity of car B (VB) = 15 m/s (in west)

Velocity and Acceleration 21

Time taken (t)= 2 min = 2 × 60 s = 120 [1 min = 60 sec] East

West P

A VA = 20 m/s

VB = 15 m/s B s1

s2 Fig 2.9
Q

Distance covered by car A in 2 min in east

VA = s1/ t

or, s1 = VA × t

= 20 × 120 = 2400 m

Distance covered by car B in 2 min in west

VB = s2 / t
s2 = VB × t
= 15 × 120 = 1800 m

Distance between two cars A and B is

s = s1 + s2
= 2400 + 1800 = 4200 m

The total distance between car A and car B is 4200 m.

2. Two cars, A and B are moving in the same direction with their velocity, 12 m/s

and 9 m/s respectively. What is the relative velocity of car, A with respect to

car B? If the both cars are travelling in opposite directions, what will be the

relative velocity?

Solution: A
VA = 12 m/s

Given B
Velocity of car A (VA)= 12 m/s VB = 8 m/s

Velocity of car B (VB)= 8 m/s VB = 8 m/s PA
VA = 12 m/s
B

Q
Fig 2.10

22 Modern Graded Science and Environment Book 8

a. When both cars are moving in the same direction
∴VAB = VA - VB

= 12 - 8 = 4 m/s

Thus, the relative velocity of car A with respect to car B is 4 m/s.

b. When the cars are moving in the opposite directions

∴ VAB = VA + VB
= 12 + 8
= 20 m/s
Thus, the relative velocity of car A with respect to car B is 20 m/s.

3. Two buses, A and B are moving with a velocity of 40 m/s towards east and
20 m/s towards west respectively. If both buses started from the same place
and the same time, calculate the distance between them after 5 minutes.

Solution:
Here,

Velocity of bus A (VA) = 40 m/s (towards east)

Velocity of bus B (VB) = 20 m/s (towards west)

Time taken (t) = 5 min

= 5 ×60 s =300 s ∴
[ 1 min. = 60 s]

Distance between the two buses (s) = ?

We know that,

Distance travelled by bus A (S1) = VA × t

= 40 × 300

= 12000 m

Now,

Distance travelled by bus B (S2) = VB × t

= 20 × 300

= 6000m

Distance between the two buses (S) = S1 + S2

= 12000 m + 6000 m

= 18000 m

Therefore, the distance between the two buses is 18,000 m after 5 minutes.

Velocity and Acceleration 23

Acceleration and retardation

Generally, a moving car, bus Do you know?

and motorcycle do not have a We know about acceleration because of the
uniform velocity. Sometimes, their many scientific work of two scientists: Galileo
velocity increases and sometimes Galilei (1564-1642), from Italy and Isaac
decreases due to the condition of Newton (1642-1727) from England. Both
road, traffic light, etc. Hence, when were physicists.
the velocity of a body is found to

change with time, we can say that

the body is accelerating i.e. the body possesses acceleration.

Suppose, a car is moving with an initial velocity equals to 5m/s and attains a velocity of
20 m/s within a time intervals of 5 seconds.

The increase in the velocity of the car in 5 seconds is (20 m/s - 5 m/s) = 15 m/s

Hence, after every second of its motion, the velocity increases by 15 m/s = 3 m/s2
5s
Now, the acceleration of the car is 3 metre per second per second or 3 m/s2. Therefore,

acceleration may be defined as the rate of change of velocity with time. The change in

velocity is the difference between final velocity and initial velocity. It is denoted bye 'a'. It

is a vector quantity.

Mathematically,

Acceleration = Change of velocity
Time taken

= Final velocity – Initial velocity
Time taken

i.e. a = v – u
t

Unit of acceleration = Unit of velocity = m/s = m/s2
Unit Time s

Thus, the unit of acceleration in SI system is m/s2.

Velocity of a body changes when the magnitude of velocity changes or direction of
velocity changes or both magnitude and direction of a body changes since acceleration
is the rate of change of velocity with time. So the acceleration of a moving of body may
change in its direction of motion or change in both magnitude of velocity and direction.

Acceleration may be of three types: positive acceleration, negative acceleration and
zero acceleration.

Acceleration is a vector quantity. Its direction is determined by change in velocity. If
the velocity of a body changes towards right or downwards from the reference point,

24 Modern Graded Science and Environment Book 8

its acceleration is positive. If the velocity of a body changes towards left or upwards from
the reference point, its acceleration is negative.

Up

Acceleration (–ve)

Left Reference point
Acceleration (–ve)
Right
Acceleration (+ve)

Acceleration
(+ve)

Down

Hence, acceleration is determined by change in the direction of velocity.

Acceleration is of three types. They are positive acceleration, negative acceleration and
zero acceleration.

1. Positive acceleration

If the velocity of a body increases, then the body is said to have a positive acceleration.
In other words, a positive acceleration means increasing the velocity within a time interval,
usually a very short interval of time. For example, when a car starts moving from rest and
velocity goes on increasing when velocity goes on increasing, acceleration is positive.
In positive acceleration, the direction of acceleration is in the direction of velocity. Another
example of positive acceleration is: an object is dropping down from a height, its velocity
goes on increasing as it falls down. So, the object has positive acceleration. From these
examples, we can say that in positive acceleration the direction of the velocity and
acceleration is same.
Some illustrations

1. If a car is standing at a point P and it is moving up to the point Q and the initial
velocity is u. From the point Q, the car is speeding up to the point R and its final
velocity is v. Here the change in velocity is towards right.

Pu Right

Q
v

Q R
Fig 2.11

Hence, We can say that acceleration is positive.

Velocity and Acceleration 25

2. If a car is standing at a point R and it is moving up to the point S and its initial
velocity is u. From the point S, the car is speeding up to the point T and the final
velocity is v. Here, the change in velocity is towards left.

S Ru

v S
T

Fig 2.12

Hence, the acceleration of a car is positive.

From these examples, we can say that the direction of the velocity is same to the direction
of the acceleration in positive acceleration.

2. Negative acceleration

If the velocity of a body decreases then the body is said to be moving with a negative
acceleration. For example, moving a car stops when brakes are applied and velocity goes
on decreasing and acceleration of a is negative. The negative acceleration is opposite to
the direction of velocity.

A ball is throwing upward is another example of negative acceleration because the ball
moves up, its velocity decreases with time. Since, the direction of the ball's velocity is
upward, its acceleration is in the downward direction.

Some illustrations

1. If a car is standing at a point A and a car is speeding up to the point B and the initial

velocity is u. From the point B, the car is slowing down to the point C and the final

velocity is v. Here, the change in velocity is towards left. u

Au Left B
v

BC v
Fig 2.13

Hence, we can say that acceleration is negative.

26 Modern Graded Science and Environment Book 8

2. If a car is standing at a point X and a car is moving up to the point Y and the initial
velocity is u. From the point Y, the car is slowing down its velocity to the point Z and
the final velocity is v. Here, the change in velocity is towards right.

Y uX

Right Z v Y

Fig 2.14

Hence, the acceleration of a car is negative.

From these examples, we can say that in negative acceleration, the direction of the
velocity is opposite to the direction of the acceleration.

The negative acceleration is called retardation or deceleration. For example, if the
acceleration of a body is - 5 m/s2, the retardation is 5 m/s2. It means, velocity of a body
is decreasing at a rate of 5 m/s in 1 second.

3 Zero acceleration

When a body moving with constant or uniform velocity, its final and initial velocity become

equal, so it is zero acceleration. When the acceleration is zero, the body may or may not

be at rest.

a = v–u = 0 m/s2 [∴ v = u]
t

Uniform and non-uniform acceleration

If the velocity of a body increases by equal amounts in equal intervals of time in a straight
line, then the body is said to have uniform acceleration. Example, the motion of the freely
falling body, from a height towards the earth due to gravity is uniform acceleration.

If the velocity of a body increases by unequal amounts in equal intervals of time on a
straight line, then the body is said to have non-uniform acceleration. For example, the
velocity of a car running in a hilly road changes continuously.

Solved numerical problems

1. If a body moving on a straight line with the velocity of 10 m/s and it changes
its velocity to 20 m/s in 2 seconds, what is its acceleration?

Velocity and Acceleration 27

Solution: 10 m/s 20 m/s
Given, A 2s B
Initial velocity (u) = 10 m/s
Final velocity (v) = 20 m/s
Time taken (t) = 2 sec
Acceleration (a) = ?
We have,

a=v–u
t

= 20 – 10
2

= 10
2

= 5 m/s2

Therefore, the acceleration of the body is 5 m/s2.

2. A car is traveling with a speed of 50 km/hr. If the car comes at rest after 10
sec, when brakes are applied, calculate the retardation.

Solution,
Given,

Initial velocity (u) = 50 km/hr = 50 × 1000 = 500 = 125 m/s
60 × 60 36 9
Final velocity (v) = 0 m/s

Time (t) =10 sec

Retardation (-a) = ?

We have,
a= v–u

t

or, a = 0 – 125 = – 125 = - 1.39
10 × 9 90

\ a = -1.39 m/s2

Thus, the etardation of the car is 1.39 m/s2.

Equations of motion on a straight line

The equations of motion are used to describe the different components of a moving
body and show the relation among the distance covered, velocity, time and acceleration.

28 Modern Graded Science and Environment Book 8

There are four equations of motion that can only be applied when acceleration is constant
and motion of a body is on a straight line.

First equation ut v
v = u + at

Consider the motion of a body is from A to A s B
B on a straight line as shown in the figure. Fig 2.15

We have,

Acceleration = Final velocity – Initial velocity
Time taken

a=v–u
t

or, at = v - u

∴ v = u + at ............. (i)

This equation is a velocity-time equation. It helps us to find the velocity changed by a
moving body in time at a constant acceleration.

Second equation

s = u + v × t
2

According to the definition of average velocity,

v = u + v
2

We also know that,

v =s
t

Now, we can say that

s = u + v
t2

\ s = u + v x t ................ (ii)
2

Third equation
s = ut + 1 at2

2
We have from, first equation-

v = u + at ............. (i)

and from second equation
s = u + v × t ............. (ii)

2
Now, by placing the value of 'v' from equation (i) into (ii)

Velocity and Acceleration 29

or, s = u + u + at × t [by using the equation (i)]
2

or, s = 2u + at × t
2

or, s = 2ut + at2
2

or, s = 2ut + at2
2 2

\ s = ut + 1 at2 .................... (iii)
2

Fourth equation
v2 – u2 = 2as
We have, first equation of motion,
v = u + at
By squaring the equation (i), we get
v2 = (u + at)2
or, v2 = u2 + 2u at + a2t2

= u2 + 2a (ut + 1 at2)
2

or, v2 = u2 + 2as [by using the equation (iii)]

or, v2 = u2 + 2as .................... (iv)
∴ v2 – u2 = 2as
Alternative method
We have from first equation-
v = u + at ....................... (i)
From second equation by placing the value of 't' from equation (i) into equation (ii)
s= u+v x u–v

2a

or, s = u2 – v2 [ u2 – v2 = (u+v) (u–v)]
2a

or, 2as = u2 – v2

or, v2 = u2 - 2as
∴ v2 = u2 + 2as ......... (iv)

30 Modern Graded Science and Environment Book 8

Case I. If a body falls freely under the action of gravity, then

a=g [g = acceleration due to gravity]

s=h [h = height]

Equation (i) can be written as, v = u + at v = u + gt

Equation (ii) can be written as, s = u + vx t h = u + vx t
2 2

Equation (iii) can be written as, s = ut + 1 at2 h = ut + 1 gt2
2 2

Equation (iv) can be written as, v2 = u2 + 2as v2 = u2 + 2gh

Case II. If a body is thrown vertically upwards, then

a = - g; s = h

In this case,

Equation (i) can be written as, v = u + at v = u - gt

Equation (ii) can be written as, s = u + v × t h= u+vxt
2 2

Equation (iii) can be written as, s = ut + 1 at2 h = ut - 1 gt2
2 2

Equation (iv) can be written as, v2 = u2 + 2as v2 = u2 - 2gh

Case III. If a body is travelling with uniform velocity, then

Initial velocity (u) = Final velocity (v)

\ Acceleration (a) = u – v = 0
2

Thus, when a body is moving with uniform velocity, then the equation,

s = ut + 1 at2 becomes
2

s = ut + 1 (0) t2
2

i.e. s = ut

Solved numerical problems

1. A bus starts to move from rest. If it is accelerated by 0.8 m/s2, calculate the
velocity and distance traveled after 8 seconds.

Velocity and Acceleration 31

Solution:

Given,

Initial velocity (u) = 0 m/s
Acceleration (a) = 0.8 m/s2

Time taken (t) = 8 sec

Final velocity (v) = ?

Distance traveled (s) = ?

We have,

v = u + at

= 0 + 0.8 x 8

= 6.4 m/s

Again, = 0 + 6.4 x 8
s = u + v x t 2

2
= 6.4 x 4 = 25.6 m

Thus, the final velocity of the car is 6.4 m/s and the distance traveled by it is 25.6 m.

2. A stone is dropped from the top of your school building. If it takes 5 seconds
to reach the ground, calculate the height of the building.

Solution:

Given,

Time (t) = 5

Height (h) = ?
As the stone is dropped down, g = 9.8 m/s2 and u = 0.

We have,

h = ut + 1 gt2
2

= 0 x 5 + 1 x 9.8 x 52
2

= 245 = 122.5 m
2
Thus, the height of the building is 122.5 m.

3. A car is traveling with a speed of 50 km/hr. If the car comes at rest after 10

seconds when brakes are applied, calculate the retardation.

Solution:

Given,

Initial velocity (u) = 50 km/hr = 50 × 1000 = 125 m/s
60 × 60 9

32 Modern Graded Science and Environment Book 8

Final velocity (v) = 0 m/s
Time (t) =10 sec

Retardation (-a) = ?

We have, – 125
90
a = v – u
t

or, a = 0 – 125 =
10 × 9

or, a = -1.39 m/s2

\ -a = 1.39 m/s2

Thus, the retardation of the car is 1.39 m/s2.

THINGS TO KNOW

1. The shortest distance between the initial and the final position of the moving body
in a particular direction is called displacement.

2. The rate of change of displacement is called velocity.

3. The rate of change of velocity with time is called acceleration.

4. The reference frame is the point from which we compare the motion of the objects.

5. When the velocity of a moving body decreases with time, it is called retardation or
deceleration.

6. The first equation of motion is v = u + at, second equation is s = u + v x t, the third
equation is s = ut + 1/2at2 and the fourth equation is v2 = u2 + 2as.2

7. The negative acceleration is called retardation.

8. Variable velocity is the velocity in which distance travelled by a body in every unit
time is different.

9. The velocity of a moving body relative to another body is called relative velocity.

10. If two cars, A and B are moving in the same direction on the same road with their
velocity VA and VB respectively, then,

The velocity of car A relative to car B is calculated by: VAB = VA - VB
The velocity of car B relative to car A is given by the formula, VBA = VB - VA
11. If a car, A is moving in the direction opposite to the other car B, then the velocity of

car A relative to car B is calculated by using the formula: VAB = VA + VB
12. The acceleration of a body moving with a uniform velocity is zero.

13. Acceleration and velocity are vector quantities as both quantities have both the
magnitude and direction.

14. Equations of motion are as: c. s = ut + 1 at2 d. v2 – u2 = 2as
a. v = u + at b. s = u + v × t 2

2

Velocity and Acceleration 33

THINGS TO DO

Measure the perimeter of the playground in your school. Take a complete turn in the
ground and at the same time tell your friend to note the time. With this information, find
the average speed. If you repeat the above procedure by riding a bicycle, is the average
speed same or different from the previous one?

TEST YOURSELF

1. Tick (√) correct statements and cross (ᵡ) the incorrect ones.

a. Velocity is a vector quantity.

b. The rate of displacement is the velocity of a body.

c. Velocity does not decrease acceleration.

d. If a person travels 100 m due east and then returns to the same

place, his total displacement is 200 m.

2. Fill in the blanks.

a. The shortest distance between the initial position and the final position of a
moving body is called ............ .

b. The ................... acceleration is deceleration.

c. .......................... is the rate of change of displacement.

d. For uniformly accelerated motion of a body, v = .............. + at.
3. Tick () the correct answer (MCQs).

a. Velocity is the:

ii. distance travelled per unit time ii. time taken per unit displacement

iii. time taken per unit distance iv. displacement per unit time

b. The SI unit of retardation is:

i. m/s ii. m/s2 iii. cm/s2 iv. s/m2

c. If a body is thrown vertically upward, then:

i. v = 0 m/s ii. u = 0m/s iii. t = 0m/s iv. a = 0m/s2

d. The SI unit of displacement is:

i. mm ii. cm iii. km iv. m

4. Define:

a. acceleration b. retardation c. relative velocity d. displacement

e. reference point f. uniform motion g. non-uniform motion h. vector quantity

5. Very short answer type questions:

a. What is the SI unit of velocity?

b. Write the formula of velocity.

34 Modern Graded Science and Environment Book 8

c. What is relative velocity?

d. Write the formula used to determine acceleration of a moving body.

6. Differentiate between:

a. rest and motion b. velocity and acceleration

c. acceleration and retardation d. uniform and non-uniform acceleration

7. Give reason.

a. Velocity is a vector quantity.

b. A body having uniform velocity has zero acceleration.

c. Velocity of a body in a circular or curved path is variable.

8. Answer the following questions.

a. What is relative velocity? Explain with an example.

b. What is uniform velocity? Give an example.

c. Describe the motion of an object in which its speed is constant but the velocity

is changing.

d. Derive:
a. s = ut + 1 at2
2 b. v2 = u2 + 2as

e. Light travels 3,00,000 km/s. Is it velocity or speed?

f. Can the average speed ever be zero?

g. Explain relative velocity briefly.

h. What do you mean by acceleration of a body 10m/s2?

i. When is the acceleration of a body is positive, negative and zero?

9. Diagrammatic questions: X
a. Answer the following questions on the basis AB
of the given diagram.

i. Identify the reference frame.

ii. In which condition will the formulae

VAB = VA- VB be applicable?

iii. In which condition will the formulae

VAB = VA+VB be applicable? Y

iv. What will be the relative velocity of the car relative to B if both of them

are moving in the same direction in the same velocity? A

b. Consider a boy has to run from X to Y as shown in Y

the diagram. Answer the following questions. B
i. Identify the displacement. X

ii. Why is B used for the calculation of average C
velocity?

iii. If the average speed is same, which path will take the longest time?

Velocity and Acceleration 35

10. Numerical Problems:

a. A man runs 1200 m on a straight line in 4 minutes. Find his velocity.

b. A stone is thrown vertically upwards with an initial velocity of 20 m/sec. Find
the maximum height it reaches and the time taken by it to reach the height.
(g = 10 m/s2).

c. A car is moving at a rate of 72 km/hr. How far does the car move when it
stops after 4 seconds?

d. A bus starts from rest. If the acceleration is 2 m/s2, find

i. the distance travelled. ii the velocity after 2 seconds.

e. A stone is thrown vertically upward with the velocity of 25 m/s. How long does
it take to reach the maximum height? Also calculate the height.

f. A stone is dropped in a river from a bridge. If it takes 4 seconds to reach the

surface of water, calculate the depth of the water level from the bridge. Also

calculate the velocity with which the stone strikes on the water surface.

g. Two cars, A and B are moving in opposite X

directions with the velocity of 20 m/s and 6 m/s AB
respectively. Calculate the relative velocity of

car A with respect to car B. If they move in the

same velocity, what will be the relative velocity

of car A with respect to car B?

h. A motorcycle covers a distance of 1.8 km in 5 Y
minutes. Calculate its average velocity.

i. Calculate the average velocity of a motor cycle that travels 72km/hr in
20 seconds.

Answers b. 20 m, 2 s c. 125 m
e. 2.55 s, 31.89 m f. 78.4 m, 39.2 m/s
10. a. 5 m/s h. 6 m/s i. 1 m/s
d. 40 m/s, 4 s
g. 26 m/s, 14 m/s

GLOSSARY

Deceleration : the rate of reduction of velocity or retardation

Speedometer : an instrument in a vehicle which shows how fast

the vehicle is going

Magnitude : the absolute or relative size, extent or importance

of something

Phenomena : observable facts or events

36 Modern Graded Science and Environment Book 8

3Lesson SIMPLE MACHINE

Total Estimated Pds: 5 [Th. 4 + Pr. 1]

On completion of this lesson, the students will be able to:

introduce lever and describe its working principle.
define mechanical advantage (MA), velocity ratio (VR) and efficiency (η) on

the basis of lever.
solve simple numerical problems related to MA, VR and efficiency of lever.

We use different types of devices in our daily life to facilitate our work. A spade is used to
dig; a broom is used to sweep; a hammer is used to pass a nail on the wall and a plank
is used to drag a load up to a height. All these simple things or devices are used in order
to facilitate our work making it convenient to do. Many such devices, which are used in

our daily life, are called simple machines. The combination of simple machines forms a

compound machine.

Thus, simple machine is defined as a device which is used in our daily life to facilitate our
work and make it convenient.

Application of simple machine effort

Simple machines make our work easier and convenient in the BA
following ways:

1. They transfer force from one point to another. In

the given figure, a knife is used to open the lid of a

can. It this device, the effort applied at Point A gets

transferred to Point B. Fig 3.1 opening the lid of a can

2. They accelerate the rate of doing work. Let's

take a dhiki as an example. When the tail of
the dhiki is pushed a little, the head is lifted to

a greater distance. Thus, it increases the rate

of doing work.

3. They multiply force. In other words, more Fig 3.2 dhiki
load can be lifted by applying little effort. For
example, a heavy stone can be turned over
with a crowbar by applying little effort.

SImple Machine 37

4. They change the direction of force. For example, load is
in the pulleys, when effort is applied downwards lifted
at one end of the string, the load is lifted upwards. upwards

Simple machines help us to do our work showing one or
more properties mentioned above.

Types of simple machine effort is
applied
downwards

Simple machines have been divided into the following Fig 3.3 Pulley changes direction of force
six types:
Wheel and axle
1. Lever 2. Pulley 3. Wedge

4. Inclined plane 5. Screw 6.

1. Lever effort

A lever is a rigid bar, straight or bent, which is able to

rotate about a fixed point called a fulcrum or pivot.

The distance between the fulcrum and the load

is called load distance and the distance between load Ed
the fulcrum and the effort is called effort distance.

Only lever can show all the properties of a simple

machine among its types. Ld fulcrum

On the basis of the location of the effort, the load Fig 3.4 a lever lifting stone
and the fulcrum, levers are classified into three

types: first class lever, second class lever and third class lever.

a. First class lever E
Ld Ed
The lever in which the fulcrum is situated at any
point between the load and the effort is called the F
first class lever. This type of lever shows all the Fig 3.5 first class lever
properties of simple machines. They change the
direction of force, multiply force and accelerate
the rate of doing work.

When the fulcrum is close to the load, the effort distance increases and it multiplies
force. When the fulcrum is close to the effort, the load distance increases i.e. work is
accelerated. Crowbar, see-saw, dhiki, pliers, scissors and tinsnips are some examples
of this kind of lever.

L EE FE
L
F LF
E (c) scissors
(b) crowbar
Fig 3.6 (a) see-saw

38 Modern Graded Science and Environment Book 8

b. Second class lever LE

The lever in which the load is situated at any

point between the effort and the fulcrum is Ld

called the second class lever. This class of Ed
Fig. 3.7 second class lever
liver cannot change the direction of force. It is F
not able to accelerate work because the load

distance is never greater than the effort distance in it. However, this lever helps us by

multiplying the applied force i.e. MA is always greater than one in this type of lever.

Some of the common examples of the second class lever are wheelbarrow, nutcracker,
bottle-opener, oar of a row boat, opening and closing a door, etc.

L
FE

LE L E
F F

Fig 3.8 (a) wheelbarrow (b) bottle-opener (c) nutcracker

c. Third class lever

The liver in which effort is applied at any point E L
between the load and the fulcrum is called the
third class lever. This kind of lever is unable F Ed
to multiply force and change the direction of Ld
force. However, it helps us by accelerating
work. Fig 3.9 third class lever

In this kind of lever, the effort distance is always less than the load distance. Therefore, the
value of MA in the third class lever is always less than one. Broom, shovel, sugar tongs,
fire tongs, forceps, fishing rod and hammer are some examples of this type of lever.

F LF F
EL E
E
L

Fig 3.10 (a) shovel (b) fire tongs (c) forceps

Points to remember for the three types of lever:

FLE The fulcrum is in the middle of the first class lever.
The load is in the middle in the second class lever.
The effort is applied in the middle in the third class lever.

SImple Machine 39

Activity 3.1 fulcrum

1. Adjust a 30 cm long scale as shown in the diagram and scale
suspend weights at the sides to balance it.
effort load
2. Suppose that the weight on the right side is the load and
on the left side is the effort. stand

3. Change the location of the load to balance the load and Fig 3.11
effort.

4. Fill in the given table and calculate.

S.No. Left side Right side Conclusion

1. E Ed E. Ed L Ld L. Ld

2.

3.

4.

Principle of simple machine

A machine cannot work by itself. It input work (W1) machine output work (W0)
needs force to work. The force output force (F0)
or load (L)
applied to do work is called input

force (Fi) or effort [E] and the initial load (L)
resisting force is called output position Ld 
force (Fo) or load [L].
Ed initial
The work done by effort or the input force (Fi) position
work done in a machine is called or effort (E)

work input (Wi) and the work Fig 3.12

done by load or the work done by a machine is called work output (Wo).

Work input (Wi) = Fi x Ed or, E x Ed
Work output (Wo) = Fo x Ld or, L x Ld

The principle of levers states that "If there is no friction in the balanced condition of a
lever, work output and work input are equal".

Mathematically,

Work output (Wo) = Work input (Wi) (in the balanced state of a lever)
or, Fo x Ld = Fi x Ed

i.e. L x Ld = E x Ed

40 Modern Graded Science and Environment Book 8

Some terms used in simple machines

1. Mechanical advantage (MA)

Mechanical advantage of a machine is defined as the ratio of the load overcome by the

machine to the effort applied in the machine. Mechanical advantage shows the number

of times the force is multiplied by the machine, when the force is applied in the machine

to do work.

Mathematically, mechanical advantage = load
effort
Ll
i.e. MA = Ee [It has no unit as it is a ratio of two forces]

Consider a crowbar is lifting a load of 600 N by using an effort of 250 N. Now, for the

calculation of MA of the crowbar: Do you know?

MA = Ll The bones in our arms and legs act as lever
Ee and can operate as other levers used in effort.

= 500 = 2
250

Thus, MA of the crowbar is 2.

Mechanical advantage of a machine is 2 means that the applied effort can lift 2 times
heavier load using the machine.

2. Velocity ratio [VR]

Velocity ratio of a machine shows the distance up to which effort is applied to lift a load.
Velocity ratio of a machine is defined as the ratio of distance travelled by the effort
applied in the machine to the distance travelled by the load overcome by that machine
at the same time. In other words, velocity ratio of simple machine is the ratio of distance
travelled by effort to the distance travelled by load in the machine.

Mathematically, Velocity Ratio = Distance travelled by effort
Distance travelled by load
i.e. VR = ELdd [It has no unit as it is a ratio of two distances]

For example, in the crowbar, load is at 1.5 m away from the fulcrum and effort is applied 6 m

away from the fulcrum. Here, the load distance is 1.5 m and the effort distance is 6 m.

Now V.R. = Ed = 6 = 60 = 4
Ld 1.5 15

Thus, velocity ratio of the lever is 4.

Velocity ratio of a machine is 4 means the effort has to move four times more distance
than that of the load distance to lift any load up to a certain height.

The value of VR of a machine is always greater than the value of MA. It is because MA
is affected by friction but VR is not affected by friction.

SImple Machine 41

3. Efficiency [η]

When effort is applied in a machine, some work is done. This is called work input. The

machine in turn does some work which is called work output.

The ratio of work output to work input in a machine is called efficiency of that machine. It

is expressed in percentage. Work output (Wo)
Work input (Wi)
Mathematically, Efficiency (h) = x 100%

In a simple machine, the value of output is always lesser than the value of input, because
no machine is frictionless and the friction wastes some of the input energy in the form of
heat. Due to this reason, a real machine does not have the efficiency of 100% or more.

Due to friction, the value of MA is lesser than the value of VR. Hence, the efficiency of a
machine is never 100% or more.

Let MA of a lever is 2.4 and its velocity ratio 4. For calculating its efficiency, we use:

h = MA x 100%
VR

=2.4 x 100% = 2.4 x 25 = 60%
4

Thus, efficiency of the machine is 60%.

Relation among MA, VR and η

We know,

h = Work output x 100%
Work input

or, h = L × Ld x 100% [... work output = LxLd and work input = E x Ed]
E × Ed

or, h = L/E x 100% [... MA = ELd and VR = ELDD]
Ed/Ld

\ h = MA x 100%.
VR

Thus, efficiency is directly proportional to mechanical advantage and inversely

proportional to velocity ratio.

Solved numerical problems

1. A load of 1000 N is lifted by using a second class lever. The load is at the distance
of 5 cm and the effort is at the distance of 20 cm away from the fulcrum. If 300
N effort is applied to balance the load, calculate its MA, VR and h.

Solution:

Here,
Load (L) = 1000 N
Effort (E) = 300 N

42 Modern Graded Science and Environment Book 8

Effort distance (Ed) = 20 cm Do you know?
Load distance (Ld) = 5 cm

VR= ? and h = ? Greek Philosopher Archimedes said
"Give me a place to stand on, and I
We have, will move the earth." This statement
expresses the power of a simple
MA = L = 1000 = 10 machine. There is no limit of the
Ed 300 3

Again,

VR = Ed = 20 =4 amplification of force by a simple
Ld 5 machine. He discovered Archimedian
simple machines such as lever, pulley
Finally,

h = MA x 100 % and screw in 3rd century BC.
VR

= 10 x 100 %
3×4

= 250 = 83.33 %
3
Thus, in the lever MA is 10/3, VR is 4 and h is 83.33 %.

2. Study the given diagram and calculate: (iii) MA
(i) work done by load (ii) work done by effort
(iv) VR (v) (h)

Solution: 400N
0.5m
Load (L) = 600 N 0.2m

Effort (E) = 400 N 600N
Distance travelled by load (Ld) = 0.5 m
Fig 3.13

Distance travelled by effort (Ed) = 0.2 m

(i) Work done by load = ? (ii) Work done by effort = ?

(iii) MA = ? (iv) VR = ? (v) h = ?
(i) Work done by load = L x Ld

= 600 x 0.2 = 120 J

(ii) Work done by effort = E x Ed

= 400 x 0.5 = 200 J
(iii) MA
= L
Ed

=640000 = 3 = 1.5
2d

SImple Machine 43

(iv) VR = Ed
Ldd
5
= 0.5 = 2d = 2.5
0.2

(v) h = MA × 100%
VRd

=21..55 = 100%
=2155 x 100%

= 15 x 4% = 60%

Thus, work done by load is 120 J, work done by effort is 200 J, MA is 1.5, VR is 2.5 and
h of the machine is 60%.

3. A poker (crowbar) is used to lift a load of 1200 N by placing it at a distance
of 25 cm from the fulcrum. Calculate the force required to balance the load
at 60 cm from the fulcrum.

Solution:

Here, Load (L) = 1200 N
Load distance (Ld) = 25 cm

Effort distance (Ed) = 60 cm

Effort (E) = ?
We have,

E x Ed = L x Ld

E x 60 = 1200 x 25

E = 1200 × 25 = 3000 = 500 N
60 60

Hence, 500 N effort is required to balance the load.

THINGS TO KNOW

1. The device which is used in our daily life to facilitate our work and make it
convenient to do is called a simple machine.

2. Simple machines make our work easier by: changing the direction of force,
multiplying force, changing the place of effort and accelerating the work.

3. Mechanical advantage of a machine is the ratio of load to effort.
4. In a balanced condition of a lever, work input = work output.
5. Velocity ratio of a machine is the ratio of effort distance to load distance that machine.
6. Efficiency of a machine is the ratio of output to input in that machine.

44 Modern Graded Science and Environment Book 8

7. A lever is a rigid bar, either straight or bent, which is able to rotate about a fixed
point, called a fulcrum.

8. Fulcrum is in the middle of the first class lever; load is in the middle of the second
class lever and effort is in the middle of the third class lever.

THINGS TO DO wooden strip

Make your own lever model. 1m
1. Take a 1 m long wooden strip with equal thickness. F

2. Mark at each 5 cm at the central line of the strip and first class lever

make holes at each mark to pass wire in it freely. bending L E
of wire E

3. Bend three thick wires as shown in the diagram. F

4. By using them, study the amount of effort

required to lift a load changing its location in the L
first class lever. second class lever

5. Study about the second class lever and the third FE

class lever also by using your own lever. third class lever L

TEST YOURSELF

1. Fill in the blanks.

a. Nail-cutter is an example of the ........................... class lever.

b. Mechanical advantage is calculated by the formula ....................... .

c. Efficiency of a machine is never ............................ %.

d. ................................. = Effort distance
Load distanced

e. ............................... is denoted by Eta (η).

2. Tick ( ) the correct answer (MCQs).

a. The value of MA is always less than VR because:

i. effort decreases due to friction. ii. load increases due to friction.

iii. some of the effort is wasted due to friction. iv. all of the above.

b. A spanner is an example of:

i. screw ii. wheel and axle iii. pulley iv. wedge

c. Velocity ratio of a lever is calculated by using the formula:

i. VR = Ld/Ed ii. VR = Ed × Ld iii. VR = Ed/Ld iv. VR = Ed × Ld/2

d. A third class lever makes our work easy by:

i. magnifying force. ii. gaining speed.

iii. changing the direction of force. iv. (i) and (ii) both.

SImple Machine 45

e. A physical balance is based on the principle:

i. E × Ld = Ed × L ii. E × Ed = L × Ld

iii. Ed × Ld = E × L iv. MA × E = MA × L

f. The lever which makes our work easier only by accelerating the rate of work is:

i. third class lever. ii. second class lever.

iii. first class lever having effort distance equal to load distance.

iv. first class lever having more effort distance.

3. Differentiate between:

a. mechanical advantage and velocity ratio b. velocity ratio and efficiency

c. second class lever and third class lever d. fixed and movable pulley
4. Give reason.

a. Value of MA is always lesser than VR.

b. Efficiency of a lever is never 100% or more.

c. A tight knot can be easily opened by using a long spanner.

d. The handles of tin-snips (metal cutter) are made long.

e. Output work is always less than input work.

f. The value of MA in the third class lever is always less than one.

g. A second class lever needs less effort to lift a heavy load.

h. The blade of a metal cutter is shorter than that of the scissors used by tailors.

i. It will be easier to lift a load in a wheelbarrow if the load is transferred
towards the wheel.

5. Answer the following questions.
a. How does a simple machine facilitate our work?

b. Show the relation among MA, VR and h.

c. What are MA and VR of a lever?

d. How does a second class lever make our work easier?

e. What is a lever? Also mention its types?

f. Define efficiency of a lever. What is meant by efficiency of a machine is 80%?

g. A strong person and a weak person are trying to carry a ladder. How should
they carry it in such a way that the weak person feels less weight of the ladder?

h. What do you mean by MA of a lever is 3, VR of a lever is 4 and h of a
machine is 60%?

6. Diagrammatic Questions:
a. Answer the following questions on the basis of the given diagrams.

i ii iii iv v
46 Modern Graded Science and Environment Book 8