Form 1 Physics

Prepared by Bahanza amali H. (Bsc. Computer Engineering),Jaliju Khalid (Teacher – Physics and Mathematics),Shabani Ramadhani (Techer – Physics

and Chemistry) 1

TABLE OF CONTENTS CHAPTER 1...............................................................................1

INTRODUCTION TO PHYSICS.........................................1

CHAPTER 2...............................................................................2

MEASUREMENT ................................................................2

CHAPTER 3...............................................................................5

DENSITY AND REALATIVE DENSITY ..........................5

CHAPTER 4...............................................................................8

FORCE..................................................................................8

CHAPTER 5.............................................................................10

ARCHIMEDE’S PRINCIPLES ..........................................11

CHAPTER 6.............................................................................16

STRUCTURE AND PROPERTIES OF MATTER............16

CHAPTER 7.............................................................................21

PRESSURE .........................................................................21

CHAPTER 8.............................................................................25

WORK, ENERGY AND POWER......................................25

CHAPTER 9.............................................................................30

LIGHT.................................................................................30

SELF TESTS ............................................................................36

CHAPTER 1 INTRODUCTION TO PHYSICS

Qn 01. What do you understand by the term SCIENCE?

ANSWER 01.

SCIENCE- Is the study of both living and non- living things.

Qn 02: Science is divided into two main parts. What are they? ANSWER 02

There are two main branches of science which are:-

(a) Physical science

(b) Biological science.

Qn 03. What do you understand by the terms.

(a) Biological science?

(b) Physical science?

ANSWER 03. (a) Biological science – Is the part of science which deals with

the study of living things only. Eg. Animals, birds.

(b) Physical science:- Is the branch of science which deals with

the study of non- living things. Eg. Air, stone etc.

Qn 04. Physical science and biological science is further divided

into two categories which are:-

(a) ____________________________

(b) ____________________________

ANSWER 04. Physical science is further divided into two categories

which are:-

(i) Physics and

(ii) Chemistry.

While biological science is further divided into (i) Botany

(ii) Zoology

Qn 05. What do you understand by the following terms. (a) Chemistry

(b) Botany

(c) Zoology

ANSWER 05.

(a) Chemistry – Is the branch of science which deals with the study of composition and decomposition of

matter.

(b) Botany- Is the branch of science which deals with the

study of plants.

(c) Zoology- Is the branch of science which deals with the study of animals.

Qn 06.What do you understand by the term PHYSICS?

ANSWER 06

Physics- Is the branch of science which deals with the study of matter and energy.

Qn 07. Define the term

(a) Matter

(b) Energy

ANSWER 07.

(a) Matter- Is anything which occupy space and has got

weight.

Examples of matter are:- stones, water and gases. (b) Energy- Is the capacity of doing work

OR

Energy Is the ability of doing work

OR

Energy Is the capability of performing work. Examples of energy are:- heat energy, electrical energy,

light energy, sound energy, mechanical energy and

chemical energy etc.

Qn 8. List down importance of studying physics. ANSWER 08

(i) Physics enable the man to have professions.

(ii) Physics helps us to answer different questions

surrounding us like why stone sinks in water.

(iii) Physics helps us to construct different simple and complex machines such as knife, razorblade

bicycle etc.

(iv) Physics helps us to obtain different sources of

energy such as generators, solar panels battery

etc. (v) Physics helps in the construction of modern

infrastructures by using modern machines such as

tractors made from the knowledge of physics.

(vi) Physics has led to get different of physics in the

word by using radios, TV WHICH MADE FROM THE KNOWLEDGE OF PHYSICS.

Qn 09.Write down examples of professions we get through

studying physics.

ANSWER 09. (i) Doctors (ii) Engineers (iii) Laboratory technicians

(iv)Teachers (v) Pharmacists etc

Qn 10. List down materials made due to the knowledge of

physics. ANSWER 10.

(a) knife (b) vehicles ( c) transformer (d) electric bulb

(e) ship (f) boat (g) refrigerator (h) building materials such as

roofs, solar panels, electrical bulb etc.

Qn 11. List down five science subjects.

ANSWER 11


and Chemistry) 2

(a)Physics(b)Chemistry(c)Biology(d)Mathematics.(e)Agriculture

Qn 12 What do you understand by the following terms.

(a) Biology

(b) Mathematics

(c) Agriculture ANSWER 12

(a) BIOLOGY- Is the branch of science which deals

with the study of living things Eg: animals, bird etc.

(b) MATHEMATICS- Is the branch of science which

deals with the study of mathematical calculations.

(c) AGRICULTURE- Is the branch of science which

deals with agricultural activities.

CHAPTER 2

MEASUREMENT

Qn 01. What do you understand by the term MEASUREMENT?

ANSWER 01

MEASUREMENT:- Is the comparison of unknown

quantity with a known standard quantity or unit.

NOTE: UNIT- Is a standard quantity used for comparing

with other object for example a metre rule is standard unit

of length

Qn 02: There are two physical quantities in measurement what are they?

ANSWER 02:

There are two types of physical quantities: These are:

a) Basic or fundamental quantities

b) Derived quantities. Qn 03. Define the following terms.

i. Basic or fundamental quantities

ii. Derived quantities.

ANSWER 03.

i. BASIC QUANTITIES: These are quantities which cannot be obtained from other physical

quantities.

Examples of basic quantities are:-mass, length,

Time, temperature, electric current.

ii. DERIVED QUANTITIES. These are physical

quantities which can be obtained by the

combination of one or more fundamental

quantities

Examples of derived quantities are: volume, density, area, moment of a force, velocity,

speed.

NOTE: In measurement the common fundamental

quantities used are:- i) mass ii) length iii) time.

Qn 04.Define the term length and state its SI-Unit.

ANSWER 04:

Length – Is the interval between two points. The SI-

unit of length is metre(m) NOTE:

The other units of length are:

Centimeter, kilometer(km) decimeter(dm),

decameter(dam) etc.

Qn 05. List down common instruments used for measuring

length in the laboratory.

ANSWER 05:

The instrument commonly used in a laboratory are: i. Tape measure ii) metre rule iii) vernier caliper and

(iv) micrometer screw gauge.

Qn 06. What is a metre rule?

ANSWER 06 This is instrument used to measure the length of an

object whose length ranges 1cm up to 100cm.

NOTE:-In a metre rule each centimeter is further divided

into ten parts called millimeter (mm)

Qn 07. Briefly explain how can you measure the length of

an object correctly by using a metre rule?

ANSWER 07.

When taking measurements by using a metre rule,

the eye should be at right angle above the mark , other wise the value will have an error calle.

PARALLAX ERROR.

Qn 08.What is a parallax error?

ANSWER 08. PARALLAX ERROR This is an error which occurs

when

eyes of a measurer are not at right above the mark

of the measuring instrument.

Figure below shows correct and wrong position of the eye of a measurement using metre rule

Qn 09.What is a vernier caliper? ANSWER 09:

Vernier caliper- This is an instrument used to measure the

length of an object up to the accuracy of 0.01cm

-A vernier caliper can also be used to measure

diameters of objects such as pen, pendulum etc.

Qn 10: Draw the diagram of vernier caliper and label its

essential parts.

ANSWER 10.


and Chemistry) 3

From figures above I. MAIN SCALE: This a thin long strip

of steel on which a scale is calibrated

in mm.

II. VERNIER SCALE: Is a small

movable scale of steel which can slide along the main scale.

III. The vernier scale has ten (10) divisions marked

on it such that the total length of 10 divisions is

equal to 9mm. In other words we say that,

ten(10) vernier scale divisions coincide with nine main scale divisions.

IV. EXTERNAL/OUTSIDE JAWS (J1 and J2)

-These are jaws used to measure external

diamensions of an object. When measuring external diamensions of an object the object is

held tightly between the external jaws.

V. INTERNAL JAWS (J3 and J4)

–These are the jaws used tomeasure internal diamensions of an object . Eg. Diameta of the

hollow sphere. When measuring the internal

diamensions of an object, the object is held

tightly between the internal jaws.

VI. TAIL:- Is a connected to the vernier scale.

NOTE: As it has already explained earlier that ten (10) vernier

scale divisions are exactly equal to 9 main scale divisions.

-As the main scale is graduated in (mm), there fore, we can say that, 10 vernier scale divisions = 9 main scale divisions.

-Now divide by 10 to both sides.

10Vsd = 9/10Msd

10 1vsd= 0.9msd

ie

1 main scale division = 0.9mm main scale division.

–The difference between one main scale division and one

vernier scale division is given by: 1mm – 0.9m = 0.1mm= 0.01cm.

Qn 10: What do you understand by the following terms as

applied in avernier caliper?

a) pitch b) least count. ANSWER 10.

a)PITCH-Is the smallest value of length or any other unit

which can be read directly from a main scale accurately.

For example if one centimeter length has ten

divisions then pitch is 1/10 cm = 0.1cm. And, If one centimeter length has 20 divisions,

then the pitch is

1

20 cm = 0.05cm.

There fore: Pitch = 1unit length

Number of divisions in the unit

b)LEAST COUNT: Is the difference between one main scale

division and one vernier scale division

:. LEAST COUNT= One main scale division- one vernier scale

division

=1mm-0.9mm

=0.1mm

:. One least count of a vernier calliper = 0.1mm= 0.01cm.

Qn 11. List down procedures how to read a vernier caliper. ANSWER 11.

(1) Close the jaws of the vernier caliper and

look at for the zero cm mark differences.

(2) Place the object to be measured between

the jaws of the instrument. (3) Slide the vernier along the main scale

until it touches th end of the object. Use

the screw clamp to light the object in

position.

(4) Read and record the reading on the main which is to the left of zero mark of the

vernier scale.

(5) Observe along the vernier scale and

record the mark which coincides with a

mark on the main scale. (6) Read and note down the value on the

vernier scale. This gives the digit in the

hundredth place of the instrument.

(7) Add the values in steps 4 and 6 to get you

correct readings.

Generally:Length of an object = main scale reading + vernier

scale reading

Qn 12. What do you understand by the term zero error of a vernier caliper?

ANSWER 12

Zero error= Is an error which occurs when the zero mark of a

main scale does not coincide with the mark of the vernier scale

mark.

Qn 13. Consider the figure below. Determine the diameter of

the object that is placed between the jaws of the vernier caliper

below.

Solution 13

From the figure above:- Main scale length = 8.4cm

Vernier scale reading= 5.2 x least count

=5.2 x 0.01cm

=0.052cm.

Total reading of the vernier caliper = 8.4cm + 0.052cm =8.452cm

:. The length of an object = 8.452cm.

Qn 14. What is the length of the object in the figure below.

Soln 14

Main scale reading = 3.3cm

Vernier scale reading = 4 x 0.01

= 0.04cm

:. Length of an object = 3.3cm + 0.04cm


and Chemistry) 4

=3.34cm Qn 15

What is a micrometer screw gauge?

ANSWER 15.

Micrometer screw gauge=This is an instrument used to measure diamensions of objects of accuracy up to 0.001cm

Qn 16. Draw the diagram of micrometer screw gauge and lebel

its parts.

ANSWER 16

Qn 17. What do you understand by pitch of the screw of the

micrometer screw gauge.

ANSWER 17:

Pitch of the screw- Is the distance traveled by the tip of the

screw when head of the screw is given one complete rotation

-Pitch of the screw of the micrometer screw gauge is given by

the formula.

Pitch of the screw = Distance by the thimble on the main Scale

Number of rotations of the thimble

For example: If 5mm is the distance moved by the thimble on

the main scale for 5 rotations, then

Pitch = 5mm=1mm

5 Qn 18

Define the term “least count” as applied in a micrometer screw

gauge.

ANSWER18

Least count: - Is the smallest distance moved by the tip of the screw when the screw turns through 1 division mark.

MATHEMATICALLY:

Least count = Pitch

Number of circular scale

divisions

For example, if pitch of the screw is 1mm and the number of

divisions marked on its thimble are 100, then

Least count= 1mm= 0.01mm= 0.001cm

100 Qn 19:

Write down steps of taking measurement by using micrometer

screw gauge.

ANSWER 19. Procedures

(1) Calculate the least count of the screw

gauge

(2) Place the given diameter of the object in between Anvil(A) and the screw and turn

the retchet in clockwise direction until the

retchet becomes free.

(3) Note the main scale from the left of the

zero of the circular scale. (4) Note the circular scale reading by finding

the number of division on the circular scale

which coincides with the baseline.

(5) Multiply the circular scale reading with the

least count so as to obtain observed reading Now:

Total reading= main scale reading + circular scale reading

Qn 20.

Calculate the readings of the micrometer screw gauge below.

ANSWER 20: From the diagram above:

Linear or main scale reading = 5.2mm

Circular scale reading which coincide with baseline = 20.3mm

Now: Reading of the micrometer = main scale reading + circular

scale reading

=5.2mm + 20.3mm

=25.5mm =25.5 x0.001cm

=0.0255cm

Qn 21.

What is the reading of the micrometer screw gauge below.

Soln 21.

From the figure above:

Main scale reading = 5.1mm= 0.51cm Circular scale reading = 40.2mm= (40.2x

0.001)=0.0402cm

:. Total reading = 0.51cm +0.0402cm

=0.5502cm

Qn 22.

What do you understand by the following terms.

a) volume

b) mass

c) weight

ANSWER 22.

a)VOLUME- Is the space occupied by a substance

Or

Is the quantity of a space that an object occupies -The common SI-Units of volume are


and Chemistry) 5

Metre cubic (m3)

Centimeter cubic(cm3)

Milliliter (ml)

And 1 Milliliter =1cm3

b)MASS- This is the quantity of matter it contains.

-The common SI- units of mass are:-

kilogram (kg)

gram (g)

And 1kg =1000g

c)WEIGHT-Is the centre point through which all

particles of an object are concentrated.

–The SI-unit of weight is Newton (N)

Mathematically

weight =mass of the body x force of gravity (g)

= m x g

w = mg

Qn 23.

Calculate weight of the body whose mass is 40kg.

(take g= 10N/kg)

Soln 23.

From: W= mg

W= 40kg x 10N/KG

=400N

:. Weight of the body = 400N

Qn 24.

Write down the differences between mass and weight

Soln 24.

MASS WEIGHT

i. Is the quantity of

matter it contains

ii. It SI=Unit is

kilogram(kg)

iii. Mass does not change from

place to place

iv. It is a scalar

quantity

i. Is the centre point of an

object through which all

particles of an object are

concentrated

ii. Its SI=Unit is Newton(N) iii. It changes from place to

place

iv. It is a vector quantity

CHAPTER 3 DENSITY AND REALATIVE DENSITY

Qn 01. What do you understand by the term density?

ANSWER 01. DENSITY – Is a mass of a substance per unit volume.

Mathematically:

Density = mass

Volume. Density = m/v

D= m/v

-The SI-unit of density is kilogram per cubic metre (kg/m3) or

gram per cubic centimeter (g/cm3)

NOTE:- The density of the body is the measure of the heaviness and strength of the material. For example, A liquid of high

density is heavier than the liquid of low density. And a solid

body of high density is heavier and strong than the solid of low

density.

Qn 02.

A piece of wood of volume 0.24m3 has a mass of 0.72kg.

Calculate the density of the wood.

Soln 02. Given that:

Mass of the wood = 0.72kg.

Volume of wood = 0.24m3

Required density (D)= ?

From

Density = mass

Volume

= 0.72kg 0.24m3

:. Density of wood= 3kg /m3.

Qn 03: The density of mercury is 13.6g/cm3. Find the volume

of 204g of mercury.

SOLN 03.

Given that:

Density of mercury = 13.6g/cm3

Mass of mercury = 204g Volume of mercury=?

From:

Density (D) = mass

Volume

D= 204g V

13.6g/cm3= 204g

V

13.6g/cm3V= 204g

13.6g/cm3 13.6g/cm3

V= 15cm3.

:. Volume of mercury = 15 cm3.

Qn 04

Write down procedures, how can determine density of solid shaped bodies.

ANSWER 04.

i) Measure the mass of the solid by using

a beam balance.

ii) Measure the dimensions of solid body iii) Calculate the volume of the solid

using appropriate formula.

iv) Finally calculate the density of the

solid.

Qn 05.

A block of glass with diamension 0.12m x 0.04m x 0.1m has a

mass of 1.2kg. Calculate its density.

SOLN 05. Mass of the block = 1.2kg

Volume of the block = 0.12m x 0.04m x 0.1m


and Chemistry) 6

= 0.00048m3 Density= ?

From:

Density(D) = mass

Volume

= 1.2kg 0.2kg

0.00048m3

:. Density of the block = 2500kg/m3

Qn. 06 A solid cube of metal of sides 10cm has a mass 80kg. What is

its density?

Soln 06.

Sides of solid cube = 10cm Volume of the solid cube= 10cm x 10cm x 10cm

= 1000cm3

Mass of the cube = 80kg = 80000g.

Density of the cube = mass of the cube

Volume of the cube =80000g

1000cm3

:. Density of the cube= 80g/cm3

Qn 07. A rectangular metal block measuring 8cm x 5cm x 2cm has a

mass of 880g. What will be the mass of a block of the same

metal measuring 6cm x 4cm x 1cm

Soln 07. i) mass of rectangular block (m) = 880g

volume of rectangular block(v)= 8cm x 5cm x 2cm

V= 80cm3

Density of rectangular block?

From:

Density = mass

Volume

=880g 80cm3

:. Density of = 11g/cm3

Rectangular block.

ii) The same metal block.

Density of the 2nd block = 11g/cm3

Volume of the 2nd block = 6cm x 4cm x 1cm

V= 24cm3

Mass of the 2nd block =?

From

Density = mass

Volume

D= m/v 24cm3 x 11g/cm3 = m/24cm3 x 24cm3.

264g = m

:.mass of the same block = 264g.

Qn.08 Briefly explain how can you obtain density of irregular

bodies? Eg. Stone?

SOLN 08.

PROCEDURES. i. Measure the mass of the stone by using beam

balance.

ii. Pour some water into a measuring cylinder and

record the initial reading of water level.

iii. Then immerse slowly the stone in the water

contained in the measuring cylinder with the

help of thin string.

iv. After that read and record the new reading of

water level.

Now volume of the stone: Volume of Solid (Vs) = final volume(v)-

initial volume(vo)

Vs= (V-Vo) cm3

v. Finally calculate density of the stone/Solid Density = mass/volume.

D= Ms

Vs

But Vs=(v-vo)cm3

:. Ds=ms

(v-vo)cm3

Where Ds= Density of the stone

Ms= mass of the stone

(V-Vo) volume of the stone.

Qn 09: In a experiment to determine the density of a stone the following data were obtained.

Mass of the stone= 180g

Initial volume of water in the measuring cylinder = 20cm3

New volume of water after immersing the stone= 42cm3

From these results, obtain the density of the stone.

Soln 09

Mass of the stone = 180g

Initial volume (Vo)= 20cm3


and Chemistry) 7

Final volume (V)= 42 cm3 Volume of the stone = 42cm3 – 20 cm3 = 22 cm3

Now:

Density of the stone = mass of the stone

Volume of the stone.

Density of the stone = 180g

22 cm3

=8.1g /cm3.

:. Density of the stone= 8.1g/ cm3

Qn 10:

Define the term RELATIVE DENSITY:

ANSWER 10. This is the ratio of density of a substance to the density of water

OR. Is the ratio of mass of the substance to the mass of water.

Mathematically:

Relative density (R.D)= Density of the given substance Density of water.

R.D. = Ds

Dw

-Relative density of the substance has no SI-unit. NOTE: Density of water is 1000kg/m3

Qn 11. Write down procedures of obtaining relative density of

a liquid.

ANSWER 11.

PROCEDURE. i. Measure an empty bottle by using a

beam balance and denote its mass as

(M 1)

ii. Measure and record the mass of the

bottle full of liquid and record its mass as (M 2)

iii. Lastly measure and record the mass of

the bottle when full of H20 and record

its mass as (M 3)

Now:- Mass of an empty bottle =(M 1)

Mass of empty bottle +liquid =M 2

Mass of empty + water =M 3

Mass of liquid =(M 2-M 1)= .g

Mass of water =(M 3-M 1)= .g

Now:

R.D = Maas of liquid

Mass of H20

R.D = (M 2-M 1)= g (M 3-M 1)= g

:.R.D = M 2-M 1

M 3-M 1

W here R.D =Relative density. Qn 12:

In an experiment to determine the relative density of turpentine

by using a density bottle the following data were obtained.:

Mass of empty bottle = 14.6g

Mass of bottle + water = 64.6g Mass of bottle + turpentine = 58.1g

From these data determine:-

a) Relative density of turpentine b) Density of turpentine.

Soln 12:

Mass of empty bottle = 14.6g

Mass of turpentine only= 58.1g – 14.6g =43.5g

Mass of water =(64.6-14.6)g

=50g

a) Relative density = mass of turpentine

Mass of water = 43.5g.

50g

=0.87

:- Relative density of turpentine = 0.87.

b) Density of turpentine

From:

R.D= Density of turpentine

Density of water

R.D = DT

DW

DT= R.D x DW

DT= 0.87 x 1g/cm3

=0.87g/cm3 :. Density of turpentine = 0.879/cm3

Qn 13. Write down procedures to be followed in obtaining

relative density of irregular solid bodies. Eg stone.

ANSWER.13

PROCEDURES i) Measure the mass of solid body by using a beam

balance and record as (Ms)

ii) Measure the mass of an empty beaker and place

it under spout of the over flow can.

iii) Pour some water into the flow up to its spout. iv) Immerse completely the solid body into the can

and make sure that over flow water is collected

in the beaker.

Now:

Mass of the body = M s Mass of the dry empty beaker= M 1

Mass of dry beaker + over flow = M 2

Mass of an equal volume of H20 = M 2- M 1

Qn 14: In an experiment to measure the relative density of block

of metal the following data were obtained:-

Mass of metal block 264g

Mass of empty beaker = 100g

Mass of beaker + over flow water = 124g Calculate:

i. Relative density of the metal

ii. Density of the metal if density of water is

1g/cm3

Soln 14. Mass of the metal block = 254g

Mass of empty beaker= 100g

Mass of empty beaker + over flow water = 124g

Mass of an equal volume of H20 = 124g=100g

=24g :. R.D = mass of the block

Mass of an equal volume of water

= 264g

24g

:. R.D of metal block = 11.

Qn 15.


and Chemistry) 8

Write down procedures of obtaining relative density of granules (sand)

ANSWER 15.

PROCEDURE

i) Measure and record the mass of the bottle when empty.

ii) Measure and record the mass of bottle and sand

iii) Measure the mass of bottle + sand + water on

top of sand

iv) Measure and record the mass of bottle when full of H20 only.

Now:

Mass of sand =(m2-m1)

Mass of water above sand =(m3-m2) Mass of an equal volume of water =m4-m1-m3-m2

R.D= mass of sand

Mass of equal volume of H20

m2-m1 R.D = (m4-m1)-(m3-m2)

Qn 16. In an experiment to determine the R.D. of glass beads

the following results below were obtained:

Mass of empty bottle = 26.5g Mass of bottle + glass beads = 61.5g

Mass of glass beads + water on top = 975

Mass of bottle when full of H20 = 765.

Determine: a) R.D of glass beads

b) Density of beads in kg/m3

c) If density of water is 1000kg/m3.

ANSWER 016 M 1= 26.5g

M 2= 61.5g

M 3= 97g

M 4 = 76g

Dw= 1000kg/m3

m2-m1

a)R.D = (m4-m1)-(m3-m2)

RD=61.5=26.5 (76.5-26.3)g-(97-61.5)g

R.D = 35g=2.5

14g

:. R.D of glass beads is 2.5

b) Density of glass beads:

from:

R.D = Density of glass beads

Density of water. 2.5= Dg

1 1000kg/m3

Dg = 2.5 x 1000kg/m3

:. Density of glass beads = 2500kg/m3

Qn 17.Briefly explain how can you determine Density of a

mixture.

ANSWER 17.

Density of mixture may let say A and B can be Calculated as:

Density of mixture= Total mass of mixture Total volume of mixture

If the two materials A and B of density DA and DB with

volumes VA and VB, the equation above can be written as:

Density of mixture= mass of A + mass of B

volume of A + Volume of B mixture

Dm = M A +M B

VA+ VB

This equation can also be written as DM = DAVA +DAVB

VA +VB

Qn 19. An alloy of ice made by mixing 80cm3 of copper of

density 8.9g/cm3 with 120cm3 of aluminium of density

2.7g/cm3 . Determine the density of Alloy in kg/m3(ANSWER:

Density of alloy = 5180kg/m3)

b) A crown made of an alloy of gold and silver has a volume of

60cm3 and mass of 1.05kg. Calculate the mass of gold contained

in the crown if the density of gold is 19.3g/cm3 while that of

silver is 10.5g/cm3.(ANSWER: mass of gold= 9219)

Qn 20.

What are the importances of density and Relative density?

ANSWER 20.

i. It helps in light materials needed for bodies such as aeroplane, Rocket and

helicopters.

ii. It helps in the designing of strong

bridge and fall building.

iii. It helps to differentiate pure materials from impure ones.

iv. It helps in designing strong and rigid

bodies of cars, buses, ship trains etc.

CHAPTER 4

FORCE

Qn 01.

Define the term force and state its SI-Unit.

ANSWER 01.

Force is a push or pull bring about a change in the state of rest

or uniform motion of a body or changes its direction or shape.

OR Force is simply or push exerted by the body

-The SI- Unit of force is Newton denoted by capital letter (N)

NOTE: The other unit of a force is kilogram force denoted as (kgf) and

(1kgf = 1000gf)

Qn 02.

What do you understand by the term a. 1kilogram force (1kgf)

b. 1gram force (1gf)

ANSWER.

a. 1kgf- Is the force required to lift a body of mass 1kg vertically upwards.

Density = Total mass of mixture

Total volume of mixture


and Chemistry) 9

b. 1gf- Is the force required to lift a mass of 1g

vertically upwards.

NOTE: Force is measured by using an instrument called

spring balance

Qn 03: Mention at least eight types of forces.

ANSWER 03.

1. Stretching force – This is a force which causes an elastic materials to increase its length one end

of it.

2. Frictional force- This is a force, which prevents the two

surfaces from sliding over each

other. 3. Restoring force-This is the force developed in an elastic

materials in order to prevent it from being extended or

deformed.

NOTE: (a) Restoring force helps to prevent the original shape of

the body

(a) When a rubber band is pulled it increases in length.

The force which causes the extension of the rubber band is called STRETCHING FORCE.

4. Repulsive force: - This is a force which causes two or

more bodies to repel each other.

-Example of repulsive force is when two like poles of bar magnet are brought close together, they repel each other.

( V) Attraction force:- This is a force which cause two or more bodies to stick or hold together

Example of force attraction occurs when two unlike poles of

the bar magnets are brought together.

(xi) Compressional force: This is a force which decreases the dimension of an elastic material.

Eg. When a spiral spring pushed down by which foot

Diagram

Upthrust force: -Is an upward exerted on a body it is totally or

partially immersed in a fluid.

Diagram.

Where

Up= Uthrust force

WB= Weight of the body

VIII)TORSINAL FORCE:

This is a force which cause the twisting of elastic materials.

Eg:

When a torsional force is applied to a rubber band the shape of the rubber will be twisted or deformed.

Before the application of torsional force after the

application of torsional force.

Qn 04. Mention examples of attraction force. ANSWER.

a) Cohession force

b) Adhession force

c) Grantation force

d) Electrostatic force.


and Chemistry) 10

NOTE: Cohession force – Is the force of attraction between two like

bodies or particles.

Example

The molecules of water attract each other by a force of

COHESSION FORCE.

Diagram. Water molecule water molecule.

Adhession force:- Is the force of attraction between

unlike particles or molecules of two different

materials. .

Qn. 06.

Write down at least five effects of a force.

ANSWER 06.

Effects of a force are:-

(i) A force can cause a motion of a stationary

object.

(ii) A force can stop the moving objects or slow them down

(iii) A force can make a moving object move faster

(iv) A force can change direction of moving objects.

(v) A force can change the shape of objects.

Qn 07.

Define the term weight of the body.

ANSWER 07. Weight of the body – Is the force with which a body is pulled

by the earth towards itself.

Mathematically, weight of the body is given by:

W= m x g

(W= mg) Where

W= weight the body

M= mass of the body

g= acceleration due to gravity of the earth.

NOTE:

The( value of g= 9.SN/kg or 10N.lg)

Qn 08. Differentiate between weight and mass of the body

MASS WEIGHT

1. Is the quantity of matter

in an object

2. The SI-Unit of weight is

kilogram (kg) 3. It is measured by using a

beam balance

4. Mass of an object does

not vary

5. It is a scalar quantity

1. Is the force with which a

body is pulled by the

earth towards itself.

2. The SI-Unit weight is Newton (N)

3. It is measured by using a

spring balance.

4. Weight of an object

varies from place to place

5. It is vector quantity

Qn 9.

a) Calculate of weight of the body whose mass is 30kg

b) Calculate mass of the body whose weight is 200N

c) Calculate weight of an object on the moon if its mass

is 110kg. ANSWER 9.

SOLN.

a) Given that:-

Mass an object = 30kg

G = 10kg/N W=?

From

W= mg

= 30kg x 10N

Kg :. W = 300N

(b)Given that:-

W= 200N g= 10N/kg

M= ?

From

W= mg g g

m= 200N ÷ 10N

kg

=200N x kg 10N

= 20kg

:. Mass of the body = 20kg

( c) Given that:-

Mass of a body = 10kg

g=on the moon = 1 x 9.8N/kg

6

Required weight of the body on the moon. From

W= mg

10kg x 9.8N x 1

Kg 6

= 98kgN

6

= 98N

6

16.33N :. Weight of the body in the moon = 16.33N.

CHAPTER 5


and Chemistry) 11

ARCHIMEDE’S PRINCIPLES Qn 1.

What is archimede’s principle?

ANSWER 01:

This is the principle which shows the relationship between weigth of the body in air, weight of the body in a fluid, and

upthrust.

Qn 02:

What do you understand by the terms: a) an upthrust

b) apparent weight

c) apparent loss in weight

d) real weight

ANSWER 02: (a) AN UPTHRUST – This is an upward experienced by

a body when partially or totally immersed in a fluid.

NOTE:

The weight of the body immersed in a fluid will always be less than its actual weight when measured in air. This is

due to upthrust experienced by a body.

(b) APPARENT WEIGTH:- This is the weight of the

body measured when the body is partially or totally immersed in a fluid.

(c) APPARENT LOSS IN WEIGHT:

–This is the weight of the body lost when the body is

partially or totally immersed in a fluid.

(d) ACTUAL WEIGHT:

This is the weight of the body measured in air.

OR

Is the weight of the body before being immersed in a fluid.

Qn 03.

Write down the relationship between upthrust, Apparent weight

and apparent loss in weight.

ANSWER 03:

Relationship between a real weight (Actual weight) apparent

weight and apparent loss in weight is that:

Apparent loss in weight =Real weight- Apparent weight

NOTE: To find out that, when the body is partially or totally immersed

in a fluid its weight decreases proceed as follows:

01) Find the weight of the body in air and record it as

(W1)

02) Find the weight of the body when partially immersed in a fluid and record it as (W2)

03) Find the weight of the body when totally immersed

in water and record it as (W3)

04) Now remove the body from the water, dry water it

and find against weight in air as (W4)

CONCLUSION.

From the expement above, it will be observed (W3) is less than

(W2) and (W2) is than (W1) while (W1) and (W4) are equal.

There fore this shows that, the weight of the body in air is

greater than the weight of the body when partially or totally immersed in fluid.

- The loss in weight of the body in fluid is called apparent loss

in weight while the weight of the body in a fluid is called

apparent weight.

Mathematically:

Apparent loss in weight = weight of the body in air- weight of

the body in a fluid.

But when a body is partially or totally immersed in a fluid

exerts an upward on the body . There fore it is this force which

reduces the weight of the body.

Thus: Upthrust= Apparent loss in weight of the body in a fluid.

Therefore: also,

Upthrust= Wa- Wf

Where: Wa= weight of the body in air

Wf=weight of the body in a fluid.

Qn4:


and Chemistry) 12

State Archimede’s principle.

ANSWER 04:

Archimede’s principle states that “when a body is partially or

totally immersed in a fluid it experiences an upthrust which

equals to the weight of the fluid displaced”.

Qn 05.

Explain an experiment to verify archimede’s principle.

ANSWER 05. To verify archimede’s principle, the following procedures

should be followed.

1. Find the weight of the body in air by using a spring balance

and record it as (W1)

Therefore, it will be observed that:

Upthrust = Apparent loss in weight.

Qn 06 A body weighs 3N in air and 2.8N when completely immersed

in a liquid. Calculate the upthrust of the body.

Soln 01.

Given that: Weight of the body in air (Wa) = 3N

Weight of the body in liquid (WL)=2.8N

Upthrust =?

From: Upthrust = Wa- WL

=3N- 2.8N

= 0.2N

:. Upthrust = 0.2N

Qn 07

When a body is totally immersed in a liquid, it weighs 3.6N. If

the weight of the liquid displaced is 1.7N. Find the actual

weight of the body in air.

Given that:

Weight of the body in a fluid (Wf) = 3.6N

Weight of the water displaced (wd) = 1.7N

Required weight of the body in air =?

From: Upthrust = Actual weight= Apparent weight

But upthrust =weight of the fluid displaced = apparent loss in

weight.

:.1.7N = Wa- 3.6N

1.7N + 3.6N = Wa 5.3N =Wa

:. Actual weight of the body = 5.3N

2. Pour water into a eureka can up to its spout

3. Take a dry beaker, weigh it and place it under the spout of the eureka can.

1. By using a spring balance, find again the weight of

the body when totally immersed in water and record

as (W2)

- Now remove the beaker and weight it.

RESULTS: Weight of the body in air= W1

Weight of the body in water = W2

Weight of an empty beaker = W3

Weight o beaker + displaced water = W4

Now Weight of water displaced = W4-W3.

Qn 8.

A body weighs 0.8N in air and 0.5N when completely

immersed in water. Calculate. a)apparent loss in weight of the body.

b) The volume of the weight displaced if density of H20 is

1g/cm3

g= 10N/kg)

soln 08.

Given that:

Weight of the body in air (wa) = 0.8N

Weight of the body in water (Ww)= 0.5N

Density of H20 = 1g/cm3 g= 10N/kg

required

a) Apparent loss in weight

From: Apparent loss in weight = Wa-Ww

:. Apparent loss in weight =0.3N

b) Volume of H20 displaced

Procedures: (i) Find mass of water displaced

From weight(W) = mg

W = mg

g g M= 0.3N

10N/kg

:. Mass (m) = 0.03kg= 30g.

(ii) Find the volume of water displaced


and Chemistry) 13

From: density = mass Volume

:. Volume = mass = 30g = 30cm3

Density 1g/cm3

:. Volume of water displaced =30cm3.

Qn 09.

A meter cube weigh’s 1N when in air and 0.8N when totally

immersed in water of density 1g/cm3. Determine. a) The volume of the metre cube

b) The density of the metre cube.

Soln 09

Given that: Weight of the cube in air= 1N

Weight of the cube in water = 0.8N

Required:

a) Volume of the cube.

PROCEDURES (i)Find weight of water displaced

Weight of the water displaced = Wa-Ww

1n=0.8N

:. Weight of water displaced = 0.2n

(ii) Find mass of the cube: From:

W=mg

g g

M= 0.2N

10N/kg M= 0.02kg=20g

(iii)Find volume of water displaced.

From: Density = mass

Volume Density =m/v

V=m/D

V= 20g

1g/cm3 = 20cm3 :.volume of water displaced = 20cm3.

b) But volume of water displaced = volume of the cube.

There fore

D= mass

Volume

D= m/V

But

Mass = w/g

=1N 10N/kg

=0.1kg

=100g

Now density of the cube = 100g 20cm3

=5g/cm3

:. Density of the cube = 5g/cm3

Qn 10. What do you understand by the term floatation?

ANSWER 10.

FLOATATION- Is a tendency of an object to remain freely on

a surface of the fluid.

NOTE: When a body floats, its apparent weight is zero.

Qn 11.

State the law of floatation: ANSWER

The law of floatation states that” a floating body displaces its

own weight of the fluid in which it floats”

Qn 12. Using the law of floatation and archimede’s principle, show

that weight of the floating body= uptrust of fluid displaced.

ANSWER 11:

The law of floatation can be derived from archimede’s

principle as follows. -According to archimede’s principle:

Upthrust = weight fluid displaced = apparent loss in weight.

But:

For a floating body apparent weight = 0.

Ther fore:

Weight of fluid displaced= weight of the body in air – apparent

weight

Wfd = Wa -0

Wfd =Wa But Wfd= upthrust = Apparent loss in weight

:. Weight fluid displaced (Wfd) upthrust = Apparent loss in

weight

Qn 13 What are the conditions for the body to float?

ANSWER 13.

i) A body floats in fluid if its density is less than

the density of the fluid.

ii) A body floats in a fluid if its weight is less than the upthrust.

iii) A body floats in a fluid if its weight is less than

the weight of the fluid displaced.

Qn 14: A block of wood of volume 10cm3 and density 0.5g/cm3 floats

in water of density 1g/cm3 . Calculate the volume of water

displaced.


volume of the block = 10cm3

Density of the block = 0.5g/cm3

Density of water = 1g/cm3

Required volume of water displaced =?

But from the law of floatation

Weight of the body = weight of fluid displaced

But:

Weight of wood = mwg


and Chemistry) 14

And weight of water = mass of water x g But

Mass of wood = density of wood x volume of wood

=0.5g/cm3 x 10cm3

=5g

=0.005kg :. Weight of wood = mg

=0.005kg x 10N/kg

=0.05N

But Also: Weight of wood= weight H20 displaced.

:. 0.05N= mH20g

0.05N= DH20VH20 g

DH20g DH20g

:. Volume of water displaced = 0.05N

1000kg/M 3 X 10N/kg

= 0.05n M 3 10000

=5 M 3

1000000

=0.000005M 3

:. Volume of H20 displaced = 5cm3.

QN 15:

A piece of wax of volume 80cm3 has a density of 0.88g.cm3.

Calculate.

a) The weight of the wax b) Volume of wax below the surface when floating in

water of density 1g/cm3

(use g= 0.01N/g

Soln 15.

Given that:

Volume of wax = 80cm3

Density of wax = 0.88g/cm3

Density of water = 1g/cm3 g= 0.01N/g

Required:

(a)Weight of the wax.

Weight = mg = SVg

= 0.88g /cm3 x 80cm3 x 0.01 N/kg

= 0.704N

:. Weight of the wax = 0.704N.

b)volume of waz below the water surface=?

From the law of floatation

Mass of wax = mass of water displaced

DwVw= DH20VH20

DH20g DH20g

Volume of water below the water surface = 0.88 x 80.

1

=70.4cm3

:. The volume of wax below the water surface is 70.4cm3

Qn 16:

A piece of cork of volume 100cm3 is floating on water. If the

density of the cork is 0.25g/cm3. Calculate:- a) the volume of the cork immersed in water

b) What force is needed to immerse the cork in water

(Take density of water = 1g/cm3 g= 0.01N/kg

soln 16

volume of cork (Vc) = 100cm3

density of water = 1g/cm3

g= 0.01N/kg

Required

a) The volume of cork =?

From the law of floatation:

Mass of the cork = mass of displaced water

DcVc= DH20VH20

DH20g DH20g

VH20 = 0.25g/cm3 x 100cm3 1g/cm3

VH20 = 25CM 3

:. The volume of the cork = 25cm3.

(b)Consider the diagram below

From the diagrams above, it shows that:

Force applied (F) + Weight of the cork = upthrust(Up)

F+Wc = up

F= up- Wc But since the cork is completely immersed in water

Therefore:

Vw= Vc

:. F= ( DwVw – DcVc) g F= gVc (Dw- Dc)

F= 0.01N/g x 100cm3(1g.cm3 – 0.25g.cm3)

F= (0.01 x 100 x 0.75)

F= 0.75N

:. The force needed is 0.75N

QN 17.

A iceberg floats in sea water with 1/11 of its volume showing .

If the density of sea water si 1.02g/cm3. Calculate the density of

ice. (ANSWER: The density of ice berg = 0.9g/cm3)

Qn 18.

An ice berg of 0.g/cm3 and volume 200cm3 floats in water of

density 1g/cm3. Calculate the fraction of volume of the ice sub merged in water.

Soln 18.

Density of ice (DI) = 0.9g/cm3

Volume of ice (VI) = 200cm3 Density of water (Dw) = 1000kg.cm3 or 1g/cm3


and Chemistry) 15

Required: a) Fraction of volume of ice submerged =?

Mass of ice= mass of water displaced.

M I = Mw

DIVI = DwVw

Vw VIVw

Dw = 0.9g/cm3 x 200cm3

1gcm3

Dw = 10 ie

M I = Mw

DIVI = DwVw

Dw Dw

Vw= DIVI

Dw

Vw= 0.9g/cm3 x 20cm3 1g/cm3

Vw= 180cm3

:. The volume of ice submerged = 180cm3

b) Fraction of volume of ice submerged

=180cm3.

200cm3

=9/10.

Qn 19.

Write sown some applications of floatation in our

daily life.

ANSWER 19:

i) It is used in construction of ships and

boats

ii) It is applied in making of life jackets

iii) It is applied in making sub-marine iv) It is applied in making hydrometers

v) It is applied in making balloons

vi) It is applied in designation of

parachute

Qn 20.

Briefly explain why a piece of iron sinks in water while a ship

made of steel floats?

ANSWER 20. This is because a ship is very large and hollow, as the result

most of its volume is filled with air which is less denser than

the density of water. Thre fore the density of the ship, becomes

less than the density of water.

Qn 21.

What is the function of primsoll-lines marked along the length

of the ship?

ANSWER 21.

The function of primsoll-lines marked along the length of the ship is to show the safe limit of the ship for loading a ship in

sea water of some particular density.

NOTE:

The loading in a ship is stopped when water starting touching the primsoll- line of particular density.

Qn 22. What do you understand by the term balloon?

ANSWER 22.

BALLOON: Is an air tight bag which can floats in air.

NOTE: When a balloon is filled with a gas it displaces a

volume of air equal to its own volume.

Qn 23.

What do you understand by the term relative density by

archimede’s principle?

ANSWER 23:

Is the ratio of the mass of the certain volume of the substance to

the mass of an equal volume of water

OR Is the ration of the weight of the given volume substance to the

weight of an equal volume of water.

OR

Is the ratio of the weight of the substance in air to the weight of

water displaced. OR

Is the ratio of the weight of the substance in air to the upthrust

Mathematically:

R.D= mass of given volume of a substance Mass of an equal volume of water

But mass is proportional to weight: Therefore:

R.D. = Weight of the given volume of substance

Weight of an equal volume of water.

= weight of the substance in air

Weight of water displaced

=weight of substance in air Upthrust

R.D. =Wa

Up

Qn 24.

An object weighs 60N in air and 40N when completely

immersed in water and 45N when completely immersed in

another liquid. Determine

a) Relative density of liquid b) Density of liquid.

Soln 24.

Weight an object in air =(Wa) =60N

Weight of the body in a liquid (WL)= 45N Weight of the body in water (Ww(= 40N

a) Relative density of liquid =?

From”

Relative density = weight of liquid displaced

Weight of water displaced =Wa-WL

Wa-Ww

=60N-45N

60N-40N

=15N

20N

:. Relative density of the body = 0.75.

b)Density of the liquid

from:


and Chemistry) 16

R.D= Density of the substance Density of water

R.D. =DL

1000kg/m3

0.75=DL

1000kg/m3

:. DL= 0.75 x 1000kg/m3

=7500kg/m3 :. Density of the liquid = 7500kg/m3.

Qn 25

a) Define the term hydrometer

b) List down examples of liquid that can be measured by a hydrometer

c) What are the features of hydrometer

ANSWER 25.

a) HYDROMETER- This is instrument used for measuring densities or relative densities of fluids

b) Examples of fluids whose densities are measured by

using hydrometers are:-

i. milk

ii. Battery acid iii. Beers

iv. Wines etc.

c) The following are important features of the

hydrometer which determines its function. i. Large bulb.- This gives hydrometer buoyancy

ii. Small weight

–This makes hydrometer floats upright in the liquid.

iii. Narrow stem- This gives greater sensitivity of

the hydrometer

This means that for any small change of the density causes the

hydrometer to float much higher or lower.

iv. The graduation of the scale on the stem starts

with big number at the bottom of the stem and end up with small number at the top of the stem.

Diagram

This is because

a) hydrometer sinks more in liquid of small densities

b) Hydrometer sinks less in liquid of high density.

Qn 26:

A hot air balloon moving upward has a total weight of

200N and volume of 20cm3. Assuming that the air density

is 1.2kg/m3. calculate the net upward force on the balloon. ANSWER: An upward force on the balloon= 40N

CHAPTER 6 STRUCTURE AND PROPERTIES OF MATTER

Qn 01

Define the term:

a) matter b) structure of matter

ANSWER 01

a) MATTER – Is anything which has got masses and

occupies space . Example of matter are: stone, water,

oxygen etc.

b) STRUCTURE OF MATTER:- Is the arrangement of

particles in a matter

-Matter is made up of small particles called ATOMS or

MOLECULES

NOTE:

c) SOLID MATTER- are made up of small particles

called ATOMS while LIQUID AND GASES are made up of small particles called MOLECULES

Qn 02

What do you understand by the term:-

a) ATOM B) MOLECULES C) ELEMENTS.

ANSWER 02.

a) ATOM – Is the smallest unit of an element which

may or not exist independently but always takes part

in a chemical reaction.

NOTE:

-When two or more atoms combine by sharing electrons

they form a neutral particle called a MOLECULE.

b) MOLECULE: Is the combination of two or more

atoms.

- Element can be formed by atoms of the same element or

different elements. For example, one molecule of hydrogen

consists of two atoms of hydrogen and one molecule of water consists of two hydrogen and one oxygen atoms.

Qn 03

Matter can exists in three states. What are they?

ANSWER 03

There are three states of matter

i. Solid state eg. Stone, wood, chair.

ii. Liquid state eg. Kerosene, water, honey.

iii. Gaseous state eg.oxygen gas, carbondioxide, Nitrogen

Qn 4.

Differentiate between solid, liquid and gaseous state of matter.

ANSWER 04.

01) SOLID STATE:

In solid state particles are tighted closely packed

together than in liquid and gaseous state.

There is a strong force of attraction between particles

of the solid compared to liquids and gases

The intermolecular force of attraction in solids holds

particles at one particular place

Due to the fixed position of particles in solids, the

solid have definite shape and size.


and Chemistry) 17

Due to the strong force of attraction between particles the distance between from one particle to another in

solid is almost negligible.

As the particles attract each other with a very strong

force there fore it is difficult to tear them apart.

Solid are rigid and hard.

Diagram

Arrangement of particles in solid.

2. LIQUID STATE:

There is a weak force of attraction holding the particles in al liquid

Molecules in a liquid are free to move from one point

to another

All liquids have definite shape and size but they take shape of the containers as the position of the

molecules easily change

All liquids are incompressible

Liquids have definite volume

diagram

GASEOUS STATE

In gases, the molecules are so much for apart, that

they hardly attract each other

The molecules in gases move independently and

there fore gases have neither definite shape nor definite volume

they can fill the entire space in which they are

enclosed.\gases can be compressed

the distance between one molecule and another is much great than liquid or solid.

diagram

NOTE:

The differences between solid, liquid and gaseous state are based on:-

a) Force of attraction

b) Movement of particles

c) Shape and size

d) Particle to particle distance e) Compressibility.

Qn 04.

a) What do you understand by the term “kinetic” theory

of matter. b) State the theory in (a0 above

c) Write down the assumptions of kinetic theory of

matter.

ANSWER a) Kinetic theory of matter – This is the theory which

deals with the motion of particles and the behaviour

of solid, liquids and gases.

b) Kinetic theory of gases states that” matter is made up

of small particles which are in motion” c) (i) Molecules are in state of continuous motion which

does not stop over any length of time.

(ii) The kinetic energy of the molecules increases with the

increase in temperature and increases with the decrease in

temperature. (iii) The molecules always attract each other.

(iv) The force of attraction between similar kind of

molecules is called ADHESION COHESION

(v) The force of attraction between different kind of

molecules is called force of ADHESION (vi) The empty space between the molecules is

called INTERMOLECULAR SPACE.

(vii) The force of attraction between the molecules

(cohesion or adhesion force) is called

intermolecular force of attraction. (viii) The intermolecular force of attraction increases

if the intermolecular space between the

molecules decreases and vice versa.

Qn 05. Write down examples of existence of motion in molecules.

ANSWER 05.

i. When a drop of perfume is placed in one corner

of a closed room in which there is no motion of

air, in a few minutes the fragrance of perfume spreads all over the room. There fore, this

suggests that molecules of perfume are in state

of continous motion,

ii. The process of adding colours in water, the

colour will be observed to spreading through the water.

iii. Disappearing of spoon fills of common salt

when poured in water which is left undisturbed

for few hours

Qn 06 What is meant by the term

a) elasticity

b) elastic materials

ANSWER A) ELASTICITY- Is a tendency of a material to recover

its original shape and size on the removal of

stretching force.

B) ELASSTIC MATERIAL- Is a material which increase its length when acted up on by a stretching

force and recovers on the removal of the force

Examples of elastic materials are:-

Rubber band, spiral spring, bow for arrow, catapult.

Qn 07.

State hook’s law.


and Chemistry) 18

ANSWER 07

Hook’s LAW STATES that “If elastic limit is not exceeded the

extentesion of an elastic material is directly proportional to the

force applied”

MATHEMATICALLY: F e

F= ke

e e

K= F/e

Where: K = elastic material constant

F= Force or Tension

E= Extension ( Increased length)

NOTE:

i. The SI- Unit K-is given by:

From

F = ke

e e

K= F/e

=N/m

:. The Si-unit of K is Newton per metre (N/M)

Qn 08. Explain an experiment to verify hook’s law.

ANSWER 08.

Hooke’s law can be verified by considering the relationship

between Tension (Force) or load attached at one end of the

spring and its corresponding extension (increased length) as shown below.

Diagram

Figure 1 figure

2

PROCEDURES

(i) Measure and record the original length of the spinal spring before putting and load on the

pan.

(ii) Put a known load on the pan and record the new

length of the spring

(iii) Repeat the procedure by adding different loads on the pan and each time record their

corresponding new length .

(iv) Then record the extension “e” from each pair of

your experiment.

NOTE: e=L-Lo

Where:

e- extension

Lo= original length

L= new length.

(v) Finally tabulate your results as follows.

Load added(N)

New length L(m)

Original length Lo(m)

e=L-Lo

OBSERVATION.

-From above table, it will be observed that as the Load

increase on the pan the extension “e” of the spring also

increase which verify hook’s law.

NOTE:

If the data in the table above are represented graphically the

graph will appear as follows:

ie The graph of F against extension(e)

Qn 09

Define the term

a) stress

b) strain c) elastic limit.

ANSWER 09

a) STRESS – Is the measure of the deforming force applied to a body.

b) STAIN- IS the measure of the resulting change in the

shape of a body.

c) ELASTIC LIMIT- Is a limit beyond which if the

strees(force) is increased, the materials will not return to its original diamensions.

NOTE:-

According to hook’s law, for pairs of results from the same

materials the equation can be written as follows. From:

F e

F= ke

e e

K= f/e For 1st result, the equation become.

F1/e1 = K …….. (i)

For 2nd result, the equation become:

F2/e2= K …....... (ii)

Divide equation (i) by (ii)

f1 ÷ f2 =k

e1 e2 k


and Chemistry) 19

f1 x e2 =1

e1 f2

f1e2 =1

e1f2 f1e2 =e1f2

f2 f2

f1e2

f2

f1 e2 =e1 f2 e2 e2

f1 =e1

f2 e2

:. f1 =e1 f2 e2

Qn 10:

A load of 4N causes a certain copper wire to extend by 1mm.

Find the elastic constant (K) For copper wire.

ANSWER 10.

Given that:

Force = 4N

e = 1mm Required k= ?

From

K= f/e

K= 4N 1mm

K= 4N/mm

Qn 11.

A spiral spring streatches by 5cm when 40N is applied to it. If the 40N is replaced by 125N.

a) Calculate the spring constant

b) The extension of the spring when 125N is

applied.


F1 = 40N

e1 =5cm

f2 = 125N

e2 =? K= ?

a) From hook’s law:

K= f/e

For first pair ie F1=40N and e1 =5cm.

:. K =F1/e1

K= 40N

5cm

K= 8N/cm

:. The spring constant )k) = 8N/cm.

b) extension (e2)

from:

f1 = f2

e1 e2

e2=f2e1

f1

e2 =125N x 5cm 40N

e2 15.6CM

:. The extension (e2) of the spring when 40N was replaced by

125N = 15.6CM.

Qn 12.

A vertical spring of length 30cm is rigidy damped at its upper end when object of mass 100g is placed in a pan attached to the

lower end of the spring its length becomes 36cm. For an object

of mass 200g in the pan the length becomes 40cm. Calculate the

mass of the pan if hook’s law is observed.

Soln 12

Data

Let mp= mass of the pan

Original length (Lo) = 30cm

New length (L1) = 36cm 1st mass (m1)= (100g + mp)

Another new length (L2) = 40cm

m2 = (200g + mp)

Now. e1 = L1- Lo

=36cm = 30cm

= 6cm.

e2 = 40cm = 30cm =10cm

From hook’s equation.

f1 = f2

e1 e2 100g+mp= 200g + mp

6cm 10cm

10cm (100g + mp) = 6cm(200g+mp)

1000g + 10mp= 1200g + 6mp. (10mp-6mp)=1200g – 1000g

4mp = 200g

4 4

mp = 50g :. Mass of the pan = 50g.

Qn 13.

What is meant by the term

a) cohesion force b) Adhesion force

c) Surface tension

d) Capilarity

e) Osmosis

f) Viscocity. ANSWER 13.

(a) COHESION FORCE

Is the force of attraction between like molecules of the

same substance. Example molecules of water attract each

other by a cohesion force. Diagram


and Chemistry) 20

(b) ADHESION FORCE

Is a force attraction between unlike molecules of different

substances. Example when a mercury is poured into a glass,

glass molecules (surface of the glass) attract with mercury

molecules by a ADHESION FORCE.

Diagram

(c) SURFARCE TENSION:- Is the phenomenon due to which exposed (top) surface of a liquid contained in

a vessel behaves like a stretched membrane.

NOTE: Surface tension is caused by the adhesion force

between exposed liquid molecules and air molecules.

(d) CAPILARITY- Is the rising up of the fluid through a

narrow tube or pipe. Eg: when a glass tube is diped

into water, water rises inside the tube.

Osmosis: Is the movement of solvent materials from the area of

low concentration of solute to the area of high conserntration of

solute through semi- permeable membrane

NOTE:

(i) SOLVENT- Is a substance which dissolve a solute eg:

water

(ii) SOLUTE- Is a substance which dissolves in a solvent.

(iii) SEMI – PERMIABLE MEMBRANE Is a membrane which allows some fluid materials to pass

through it and prevent some materials.

(e) VISCOSITY

Is a force of friction which exists between the layers of a liquid or gases.

NOTE: Viscosity force causes:

(i) Difficult of an object to move or flow easily in a

liquid. (ii) Difficult in stirring a fluid.

(iii) Viscosity force is greater in heavy liquid and less in

light fluids eg: viscosity force is greater in honey

than in water.

Qn 14.

List down the effects of: a) Cohesion and adhesion

b) Surface tension

c) Capilarity

d) Osmosis in every day life.

ANSWER 14.

b) The effects of cohesion and adhesion are:-

i. Rain drops water drops are always spherical in

shape because the cohesion between water

molecule is greater than the adhesion force between water molecule and air molecules.

ii. Mercury drops become spherical when dropped

on the glass because the cohesion between

mercury molecules is grater than adhesion force

between mercury molecules and glass molecules. iii. Water wet glass because the adhesion force

between water molecules and glass molecules is

greater than cohesion between water molecules.

iv. Meniscus of water curves down wards while that

of mercury curves upward because adhesion force between water molecules and glass is

greater than the cohesion force between water

molecules.

v. Meniscus of mercury curves upward because the

cohesion between mercury molecules is greater than the adhesion between mercury molecules

and the glass molecules.

NOTE: MENISCUS: - Is a u-shaped like structure formed

at free surface of a liquid. b) Surface tension helps.

i. Small drops of liquid to be spherical

ii. To resist light objects like mosquitoes, needle,

razor blade from sinking in H20

c) Effects of Capillarity are:

i. It causes the absorption of water by a towel.

ii. It causes the rising of kerosene in the wick lamp.

iii. It causes the rising of water in the roots of plants.

d)Effects of osmosis are:

(i) Swelling of dried fluid when placed in water. This

because the water molecules pass into the fluid through

their skin.

(ii)Many plants obtain their moisture by this process

across the semi-permiable membrane of root cells.

Qn 15:

b) Briefly explain how can you demonstrate osmosis. c) Diffusion

d) Brownian movement.

ANSWER 15.

a) Osmosis can be demonstrated by the following procedure.

i) Peel an irish potato and make a deep hole into it.


and Chemistry) 21

ii) Half fill the hole with sugar solution and then

place the potato in a vessel containing water.

OBSERVATION

After some hours the hole is observed to contain water from the vessel why?

REASON

This is due to Osmosis.

b) DIFFUSION- Is the movement of particles from the

area of high concentration to the area of low

concentration.

NOTE:

-During diffusion one substance mix with molecules of another substance without stirring, shaking or heating.

Diffusion can be demonstrated by the following

procedures.

i. Collect chlorine gas in a gas jar and cover it with a lid.

ii. Invert the gas jar.

( c) BROWNIAN MOVEMENT. Is the movement of visible particles after being pushed or

bombarded by invisible particles. Eg: Cigarate smoke in air.

This cigarate smokes are constantly pushed by air molecules

which are not visible.

-Brownian movement can be demonstrated as follows:

i) Arrange your apparatus as shown below.

Diagram.

(ii) Introduce some smoke by means of syringe and quickly

cover the cell.

(iii) Adjust the microscope until the fine particles of smokes are

clearly seen.

CONCLUSION. This movement of smoke particles is called BROWNIAN MOVEMENT.

CHAPTER 7

PRESSURE

Qn 1.

Define the term pressure and state its SI-unit.

ANSWER 01.

Pressure- is the force acting normally per unit area. Mathematically:

Pressure = Force (N)

Area (M 2)

P = N

M 2 :. The SI- unit of pressure is Newton per metre square (N/M 2)

NOTE:

The other SI-unit of pressure is called PASCHAL

PASCHAL: This is pressure provided when a force of 1N acts on an area of 1m2.

Qn 02

From the definition of pressure , show that pressure can be

given by: P=DsVsg where Ds=Density of the substance A

Vs= volume of the substance, g= acceleration due to gravity

A= area.

ANSWER 02.

From : Pressure= Force Area

But F= weight =mg

Also mass = Density x volume

=D xV

Now P = Dvg A

:. P = Dvg

A

Where: P=pressure, D=Density, V= volume, g= acceleration

due to gravity.

Qn 03:

A piece of metal of mass 20kg and base area of 0.4m2 is lying on a flat ground. Calculate the pressure exerted by the metal on

the ground

(Take g=10N/g)

ANSWER

Given that: Mass =20kg

Area= 0.4m2

F= mg

=20kg x 10Nkg

=200N FROM

Pressure =F/A

=200N

0.4M 2

=2000N 4M 2

:. Pressure (P) = 500N/M 2

Qn 04.

A mass of 50kg exerted a pressure of 2000Pa. Calculate the area in contact with the ground (g) =10N/Kg)


and Chemistry) 22

ANSWER:04 Given that:

Mass of the body = 50kg

Pressure = 2000Pa (N/m2)

Required area =?

From

Pressure =F/A

2000N/m2 = mg

A

200N/M 2 =50kg x 10N/kg. A

2000N/ M 2 = 500N

A

(2000N/M 2 ) A= 500N

2000N/ M 2 2000N/ M 2 A= 0.4 M 2

:. Area (A) in contact with the ground = 0.4 M 2

Qn 05.

Briefly explain how pressure varies (changes) with cross-sectional area.

ANSWER 05.

Pressure changes with cross-sectional area as follows:.

Pressure is inversely proportional to cross-sectional area That is

The greater the area the small the pressure and the small the cross-sectional area the greater the pressure.

There fore

To obtain the maximum pressure, the cross- sectional area must

be small and for minimum pressure, the crossectional area must be large.

Qn 06. Briefly explain why.

a) needles are made with thin and sharp edges?

b) Tractors have big and wide wheels ? c) Razor blades have thin and sharp edges.

ANSWER 06.

b) Needles are made with thin and sharp edges so as

pressure to produce big pressure from small forces.

c) Tractors have big and wide wheels so as to prevent

the tractor from getting sluch in the mudy by reducing the pressure caused by weight of the tractor.

d) Razor blades have very thin and sharp edge so as to

produce large cutting from least forces.

Qn 07. A piece of metal of density 5.5g/cm3 measures 20cm by 10cm by 5cm. Calculate.

i. Maximum pressure

ii. Minimum pressure exerted by the metal on a flat

surface.(Take g=0.01N/g)

ANSWER 07.

Density of metal = 5.5g/cm3

Volume of metal = 20cm x 10cm x 50cm

V = 1000CM 3

Volume of metal (Vs) = 20 x 10 x 5 Vs= 1000cm3

Maximum area (Amax) = 20cm x 10cm

= 200cm2

Minimum area (Amin) = 5cm x 10cm

= 50cm2 a) Maximum pressure (P max) =?

P max = mg

Aminimum Pmax = DsVsg

Aminimum

Pmax = 5.5g/cm3 x 1000cm3 x 0.01N/kg

50cm2

=55000N 50cm2

55000N

50cm2

Maximum pressure= 11000N/cm2

b) Minimum pressure = Force

Maximum area

55000N

200cm2 :. Minimum Pressure = 275N/cm2

Qn 08:

Write down a general formula of finding pressure due to liquid.

ANSWER 08.

To find the general formula of finding pressure due to liquid

consider a liquid of density “S” poured into a cylinder of base

Area “A” such that the height reached by the liquid is “h”

DIAGRAM

From the figure above

Force exerted by at the base = Weight of liquid

=M Lg

= DLVLg But volume of the liquid = base area x height

VL =Ah

There fore:

Force= DLAhg

From P= Force

Area

P= DLAhg

A

P= DLhLg :. Generally, pressure in the liquid (fluid) is given by

P= ₰h g

Where: P= pressure, S = density of liquid, g= acceleration due

to gravity. NOTE:

From the above formula, Pressure in a liquid depends on:

a) Density of the liquid

b) Height or depth of the liquid column

c) Acceleration due to gravity.

Qn 09:

Calculate the pressure due to the column of water of height 3m

if the density of water is 1000kg/m3.

Soln 09.

Given that:

Height due to water column =3m

Density of water = 1000kg/m3


and Chemistry) 23

g= 10N/kg Required pressure =?

From

Pressure due to liquid column = s h g

= 1000kg x 3m x 10N

M 3 kg

= 30000N/m2

:. Pressure due to water column = 30000N/M 2

Qn 10: A pressure at the bottom of column of water is 1000Pa. If the density of water is 1000kg/m3 . Calculate height of water

column.

Soln 10:

Given that:

Pressure at the bottom of water = 10000Pa(N/m2) Density of water = 1000kg/m3

Required height o water column =?

From

Pressure = s h g Sg sg

:. h= P

S g

h = 1000Pa

1000 x 10

h= 1000N/m2

1000kg/m3 x 10N/kg

h= 1000N/m2

10000kg/m3

=0.1M

:. Height of the water column = 0.1m

Qn 11: A column of mercury has a height of 75cm. What is the

pressure exerted at the bottom of the column of mercury of the

density of mercury is 13600kg/m3 ( g=10N/kg)

Soln 11:

Given that:

Height of mercury column (h) = 75cm= 0.75m.

Density of mercury(S) = 13600kg/m3.

g= 10N/kg

From:

Pressure = s hg

= 13600kg/m3 x 0.75m x 10N/kg.

=102000N/m2

Qn 12: Briefly explain how pressure in a liquid varies.

a) With depth of liquid column

b) With density of liquid.

ANSWER 012.

a) Pressure in a liquid is directly proportional to the

height of the liquid column.

The higher the height or depth of the liquid the higher the

pressure exerted by the liquid and vice versa.

b) Pressure in a liquid is directly proportional to the

density of the liquid.

The higher the density of the liquid the higher the pressure

of the liquid and vi versa.

Qn. 13. Explain an experiment to prove that pressure in a liquid is grater at the bottom than at the top.

ANSWER 13:

To verify that pressure in a fluid is maximum at the

bottom than at the top, consider a can of height “h” with holes perforated at different heights.

PROCEDURE:-

i. Make three or more equal holes at different

height along the can as shown above.

ii. Then man the canful of water

iii. Finally open the holes so that water is ejected out through the holes.

OBSERVATION.

It will be observed that:

Water at the bottom hole is ejected out at very maximum pressure than at the middle hole and the upper hole and the

water at the middle hole water will be ejected at the high

pressure than at the upper hole.

Therefore

This shows (proves) that pressure at the bottom of the liquid column is grater than above and vice versa.

Qn 14.

Briefly explain why water dams are made thicker at the bottom

than at the top.

ANSWER 14.

This is because at the bottom of the water dams, pressure is

grater than the pressure at the top.

Qn 15. State PASCHAL’s principle.

ANSWER 15

PASCHAL’s PRINCIPLE states that “When a pressure is

applied at one point of on enclosed fluid it is transmitted equally in all directions of the fluid”

Qn 16. Write down an experiment to verify paschal’s principle.

ANSWER 16.

Paschal’s experiment can be verified by using BULB- PISTON EXPERIMENT

PROCEDURES

i. Fill the bulb with water


and Chemistry) 24

ii. Lift quickly the piston inside the bulb. iii. Observe how water is pushed out of

the bulb through holes.

OBSERVATION.

Water will observed to be pushed out through the perforated

holes at equal pressure. There fore

This verify paschal’s principle ie pressure in a liquid is

transmitted equally in all directions.

Qn 17. Write down four application of Paschal’s principle in our daily

life.

ANSWER 017

Paschal’s principle is applied in:

a) Hydraulic car brake b) Hydraulic car jack

c) Hydraulic press

d) Hydraulic lift etc.

QN 18. Draw the diagram of hydraulic press and write down the

mathematical relationship between forces applied on large piston

and small piston with their corresponding cross- section areas.

Diagram

Hydraulic press uses Paschal’s. From the figure above, if a small

force (f) is applied on the small piston “a” so as to lift a load of

F2 on the large piston, pressure will transmit equally in the fluid

which will cause to raise the load on the large piston. By applying Paschal’s principle.

Pressure on the small piston= Pressure on the large

But pressure = Force

Area

Force on the small piston= force on the large piston Area on the small piston area on the large piston.

OR f/a = F/A

fA= Fa

FA FA

If the small piston has a radius ( r) while the larger piston has a

radius “R” then the above formula can be written as:

f= 2

2

WHERE: f= Force on the small piston

F= force on the large piston

a= area of small piston

A = area of large piston

r =Radius of small piston R= radius of large piston

Qn 20: The area of a larger piston is 4m2 and that of the smaller piston

is 0.05m2. If a force of 100N is applied on the smaller piston,

how much force is produced on the larger piston.

Answer 20. Given that

Force on the small piston (f) = 100N

Area of larger piston= 4m2

Area of smaller piston= 0.05m2

Required force on the larger piston=?

From: f = a/A

F

100N= 0.05m2

F 4m2 0.05F= 400N

0.05 0.05

F= 40000N

5 F=8000N

:. The force on the larger piston= 8000N

Qn 21:

The area of the pistons in a hydraulic press are 4cm2 and 480cm2 respectively. Calculate the force needed on the small

piston to raise a load of 8400N on the larger piston.

Soln 21:

Given that

Area of smaller piston= 4cm2 Area of larger piston= 480cm2

Force on the larger piston= 8400N

Required force applied on the small piston=?

From: f = a/A F

f = 4cm2

8400N 480cm2

f = 4 8400N 480

480f = 8400 x 4

480 480

f= 700N

:. The force needed on the small piston = 700N

f/a = F/A

f/F = a/A

f/F = r2

R2


and Chemistry) 25

CHAPTER 8

WORK, ENERGY AND POWER

Qn. 01 Define the term work and state its SI- Unit.

ANSWER 1.

WORK- Is the product of a force and the distance

moved in the

direction of the force.

Mathematically Workdone= force x distance

= F)N) x d( m)

= Nm

The product Newton – metre (Nm) is called Joule (J)

There fore the SI-Unit of work is Joule (J)

Qn2. Write down the factors which determine workdone

by a body

ANSWER 02.

a) There must be a force acting on a body b) The force acting on a body must cause

some displacement of the body i.e it has to

move the body.

c) The displacement must be in the direction

of application of force. NOTE: If any of the above conditions are not satisfied no work

is done.

Qn 03.

Explain when a work is said to be done.

ANSWER 03.

Work is said to be done only when a force causes displacement

in its own direction.

Qn. 04.Write down the other units of work.

ANSWER 04.

The other units of work are:-

- kilojoule (KJ) - megajoule(MJ)

-gigajoule(GJ)

NOTE:

1kilojoule= 1000Joule 1megajoule=1,000,000Joule

1kilojoule= 1000 megajoule

1 gigajoule= 1,000,000, 000 Joule

Qn 05. A child pushes a box with a horizontal force of 0.5N through a

distance of 100m. Calculate the work done by the child.

Soln 05

Given that:-

Force applied by the child on a box = 0.5N

Distance moved by the box= 100m

Work done by the child =?

From:

Work done= Force x distance

= 0.5N x 100m

= 50J :. Work done by the child = 50J

Qn 06. A man pushes a car with a horizontal force of 400N. If the car moves through a distance of 100m. Find the workdone

by the man against friction.

Soln 06.

Force applied by the man= 400N

Distance moved by the car= 100m

From: Work done by the = force x distance

= 400N x 100m

= 40,000J

Qn 07. A stone of mass 5kg is raised to a height of 1000m above the

ground.

Find the work against the pull of gravity in:-

a) kilojoule b) megajoule c) gigajoule

soln 07.

Mass of the stone = 5kg

Height of the stone = 1000m

Required workdone against the pull of gravity=?

From

Workdone= force x distance

But force = mass of the body x acceleration due to gravity = mg

:. Workdone= mg x d But also distance = height = (h)

:. Workdone = 5kg x 10N/kg x 1000m

= 50,000J

a) In kilojoule

From: 1kilojoule = 1000J ? = 50000J

= 1KJ x 50000J

1000J

= 50KJ

b) In megajoule (MJ)

From: 1MJ= 1000000J

?=50000J

=1MJ x 5000J

1000000 =5MJ

100

=0.05MJ

c) In gigajoule (GJ) From:

1GJ = 1,000,000,000J

?= 50000J

=IGJ x 50000J 1,000,000,000

=5GJ

100000

=0.00005GJ Qn. 08.A workdone by a man who was pushing a sack maize

was 500J. Calculate the distance moved by the sack if a man

used a force of 50N pull the sack.

Soln 08. From workdone: = force x distance

500J= F x d

500J = 50N x d

50N 50N


and Chemistry) 26

d= 10m

:. The distance moved by the sack of rice was =10m.

Qn 09.

What do you understand by the term ENERGY? State its SI-Unit.

ANSWER 09.

Energy- Is the capacity of doing work

OR

Is the capability of performing work OR

Is the ability of doing work

-The SI-unit of energy is also Joule(J)

Qn 10: Briefly explain why WORK and ENERGY have the same SI-

unit?

ANSWER 10:

Because total amount of workdone by a body is equal to energy spent.

ie.

Total workdone = Energy spent.

Qn 11: List down kinds/forms of energy.

ANSWER 11:

There are several forms of energy, but few of them are:

(a) MACHANICAL ENERGY

(b) HEAT ENERGY (c) LIGHT ENERGY

(d) SOUND ENERGY

(e) MAGNETIC ENERGY

(f) ELECTRICAL ENERGY

(g) CHEMICAL ENERGY (h) NUCLEAR ENERY

Qn 12.

What do you understand by the following terms.

(i) Mechanical energy (ii) Heat energy

(iii) Light energy

(iv) Sound energy

(v) Magnetic energy

(vi) Electrical energy (vii) Chemical energy

(viii) Nuclear energy

ANSWER 12:

(i) MECHANICAL ENERGY:- Is the energy possessed by the body due to its position,

configuration or motion.

Example of bodies which posses mechanical energy are:- (a) The water stored in the dams has mechanical energy due to the

great height.

(b) The stretched bow and arrow has mechanical energy due to

their configuration.

( c) The flowing water has mechanical energy due to their motion.

(ii) HEAT ENERGY:-

This is the invisible energy which causes in us the sensation of

hotness.

NOTE: Heat energy is released when combustible materials such as coal, petroleum products and wood are burnt in excess of air.

(iii) LIGHT ENERGY:-

Is the energy which causes in us the sensation of vision.

(iv) SOUND ENERGY:

Is the mechanical energy which produces sensation of hearing

(v) MAGNETIC ENERGY:

Is the energy possessed by permanent magnets or electromagnets.

(vi) ELECTRICAL ENERGY:-

Is the energy possessed by the flowing electrons in an electrical

conductor.

(vii) NUCLEAR ENERGY:-

Is the energy released in the form of heat during the fission or

fussion.

Qn 13: Write down at least two source of:

a) Heat energy ( b) Light energy ( c) Sound energy ( d)

Electrical energy

ANSWER 13:- a) Burning fire wood.

The sun

Burning candle

Burning petroleum products.

b) The sun

The Lighted electric bulb

The burning candle

The moo

c) The drum The plano

The guitar

d) The dry cell

The solar panel

The generator

The dynamol

Qn 14.

List down types of mechanical energy.

ANSWER.14 There are two types of mechanical energy which are:-

a) Potential energy (P.E)

b) Kinetic energy (K.E)

c)

Qn 15. Define

(a)Potential energy (P.E)

( b)Kinetic energy (K.E)

ANSWER 15. (a) POTENTIAL ENERGY(P.E): -

Is the energy possessed by the body due to its position or

configuration.

(b) KINETIC ENERGY- Is the energy possessed by the body due to its motion.

Qn 16.


and Chemistry) 27

List down (a) Examples of kinetic energy in our every day life.

(b) Example of potential energy in our every life.

(c) Mathematical expression of potential energy

(d) Mathematic expression of kinetic energy.

ANSWER 16.

(a) Examples of kinetic energy are:-

(i) a fast moving electron

(ii) a running water

(iii) a blowing wind (iv) a speeding car and

(v) a moving arrow.

(b) Examples of potential energy are:-

(i) a stretched bow and arrow system (ii) a stretched catapult

(iii) Water stored high up in the dams

(iv)Stone lying on the top of the loof.

( c)To express the formula (mathematical expression of P.E) Consider a ball of mass “m” at height (h) above the

ground as shown here under.

From the figure above.

Workdone by the ball against the pull of gravity =

=force x distance

But f= mg and distance(d)=h

:. Workdone= mg x h =mgh

But workdone = energy spent

:. Potential energy (P.E) = mgh

ie

P.E=mgh

Where :

P.E= Potential energy

m= mass of the body

g= acceleration due to gravity h= height attained by the body.

(d) Mathematical expression of K.E is given by:

K.E = 1mv2

2 Where

K.E = kinetic energy

m= mass of the body

v= speed or velocity of the body

Qn 017.

(a) A body of mass 5kg was raised to a height of 10m from

the ground.

Calculate the potential energy possessed by the body if the

force of gravity = 10N/kg.

(b) A body of mass 10kg posses 2000J of energy at a height of (h) above the ground. Calculate the value of

“h”

(c) A car of mass 1000kg is moving a velocity of 20m/s .

Calculate the kinetic energy possessed by the car.

(d) The kinetic energy of a man of mass 50kg is 2500J.

What is his speed?

soln 17. Given that:

a) mass of the body = 5kg

height attained by the body= 10m

acceleration due to gravity = 10N/kg

required potential energy=?

From:

P.E= mgh

= 5kg x 10N x 10m

Kg =50N x 10m

=500Nm

= 500J

:. Potential energy possessed by the body = 500J

b) Given that:-

Potential energy (P.E) of the body = 2000J

Height attained by the body = h

Mass of the body (m) = 10kg.

Required height attained by the body =?

From:

c) P.E = mgh

2000J= 10kg x 10N/kg x h

2000J= 100Nh 1000N 100N

h=20m

:. Height attained by the body = 20m.

Qn 18. What are the factors which determine:

a) Potential energy of the body

b) Kinetic energy of the body.

ANSWER. 18.

a) Factors which determine the potential energy of the body are:

i. height of the body above the ground

ii. mass of the body

iii. acceleration due to gravity.

b) Factors which determine kinetic energy of the body

are:=

(i) mass of the body

(ii) speed of the body.

Qn 19:

Briefly explain when a body is said to posses.

a) Potential energy

b) Kinetic energy

ANSWER 19. b) A body is said to posses kinetic potential energy

when:-

(i) It posses some height above the ground

c) A body is said to posses kinetic energy when it posses a certain speed.


and Chemistry) 28

Qn 20:- a) What do you understand by the principle of

conservation of energy?

ANSWER 20:

The principle of conservation of energy states that” The energy can neither be created nor destroyed but it can be

transformed from one form to another form of energy”

c) Given that:

mass of the car= 1000kg speed of the car = 20m/s

required speed K.E of the car =?

From: K.E = 1mv2

2 K.E=1 x 1000kg x (20m/s)2

2

=1 x 1000kg x 100m2/s2

2 =500kg x 400m2/s2

=200000J

:. The kinetic energy of the body = 200000J

d)Given kinetic energy of the man = 2500J mass of the man= 50kg.

required speed of the man=?

From:

K.E = 1mv2 2

2500J = = 1 x 50kg x v2

2

2500J= 25kg xv2

25kg 25kg

2v = 2/2100 sm

:. V = 10m/s

:. Speed of the mass = 10m/s

Qn 21: Define the term energy changes:

ANSWER 21:

Is the process of converting one form of energy to another

form of energy.

Qn 22:

Name the Instrument which changes:

a) Chemical energy to heat energy.

b) Heat energy mechanical energy c) Chemical energy to mechanical energy

d) Chemical energy to light energy

e) Heat energy to electrical energy

f) Electrical energy to heat energy

g) Sound energy to electrical energy h) Electrical energy to sound energy

i) Light energy to electrical energy

j) Electrical energy to light energy

k) Mechanical energy to electrical energy

l) Mechanical energy to sound energy m) Electrical energy to mechanical energy

n) Light energy to chemical energy.

ANSWER 22:

a) Chemical energy can be converted to heat energy by burning chemical materials eg: charcoal.

b) Heat energy can be converted into mechanical energy

by using steam engine.

c) Chemical energy can be converted into mechanical

energy by using explosive devices such as crackers, bombs etc.

d) Chemical energy can be converted into light energy

by using explosive devices such as bombs

e) Heat energy can be converted into electric energy to

electrical energy by using THERMOPILE. f) Electrical energy can be converted into heat energy

by using:

(i) electric iron

(ii) electric oven

(iii) electric kettle (iv) electric heater

(v) electric bulb.

g) Sound energy can be converted into electrical energy

by using MICROPHONE.

h) Electrical energy can be converted into sound energy by using a device called LOUD SPEAKER

i) Light energy can be converted into electrical energy

by using a device called PHOTOCELL

j) Electrical energy can be converted into light energy

by using a ELECTRIC LIGHTS k) Mechanical energy can be converted into sound

energy by using a BELL

l) Electrical energy can be converted into mechanical

energy using ELECTRIC MOTOR

m) Mechanical energy can be converted into electrical by using:

(i) Generators

(ii) Dynamos

n) Light energy can be converted into mechanical by the

process of PHOTOSYNTHESIS

Qn 23: How can you demonstrate the principle of conservation of

energy?

ANSWER 23:

The principle of conservation of energy can be demonstrated by using a SWINGING PENDULUM BOB as shown below:

V= maximum velocity

h= Height

P.E. = Potential energy

K.E = Kinetic energy

- When the bob oscillated (swings) from point A to C via B and

when back A, the following changes of energy occurs.


and Chemistry) 29

I. AT POINT “A” The potential energy is maximum since height (h) of the bob is

maximum while kinetic energy (K.E.) = 0 as velocity (v)=0.

II. AT PONT “B”

The kinetic energy is maximum since velocity(v) of the bob- maximum while P.E= 0 as height of the bob= 0. Therefore all

the P.E. has been changed to kinetic energy (K.E.)

III. AT POINT “C”

As the bob rises to point “C” its decreases up to zero while the height increases up to maximum of height(h). Therefore at

point “C” the bob has maximum P.E. and K.E =0. Hence at

point “C” all K.E. has changed to P.E

Qn 24: A pendulum bob of mass 50g is pulled a side to a vertical

height of 20cm from the horizontal and then released.

Find:

a) maximum potential energy b) The kinetic energy of the bob at a height of 8cm from

the horizontal

c) The maximum speed of the bob )Take g=10N/kg).

Soln 24:. GIVEN:

Maximum height = 20cm = 0.1m

Mass of the bob= 50g = 0.05kg.

Required:

a) maximum P.E =mgh =0.05kg x 10N x 0.1m

Kg

=0.1J

:. The max P.E = 0.1J

b) K.E when h=8cm = 0.08m

Total energy = maximum energy = P.E=E.T.

But

Maximum P.E = 0.1J

But E.T = P.E +K.E

But

P.E at 8cm = 0.08m= mgh.

:. ET =0.05 x 10 x 0.08 +K.E

0.1J = 0.04J + K.E 0.15 – 0.04J= K.E

K.E = 0.065

c) Maximum speed (v) =?

From: = 1mv2=0.1

2

mv2= 0.2

m m

v2= 0.2= 4

0.05

2v = 4

V= 2

:. Maximum speed =2m/s

Qn 25:

Define the term power and state its SI-Unit.

ANSWER 25:

POWER:- Is the rate of consuming energy OR

Is the work done per unit time

OR

Is the rate at which energy is produced or utilized

MATHEMACTICALLY

Power =Energy consumed

Time

P = E/t = J/s

:. The SI-unit is watt (w)

The other units of power are:-

b) kilowatt (kw)

c) mega watt (mw) d) Horse power (HP)

e) Gigawatt (Gw)

NOTE:

a) 1kilowatt1000watt b) 1megawatt= 1000,000watt

c) 1Horsepower= 746 watt

d) 1kilowatt = 1000mw

e) 1 gigawatt (Gw) = 1,000,000,000watt.

Qn 26.

a) An engine supplies 15000J of energy in one minute.

Calculate the power of the engine in kilowatts.

b) A man raised a sack of rice of mass 90kg to a height

of 2m in 5seconds. Find the power developed by the men. (Take g=10N/kg)

c) A man weighing 800N takes a minute to run upstair.

In so doing he ascends a vertical height of 3m.

Calculate the power of the man.

d) A water power lifts 1000kg of water through a height of 5m in 25 seconds. Calculate the power of the

pumpin.

a) watt (b) Horse power

SOLUTION 026. a)Given that:-

energy of the engine = 15000J

Time taken by the engine =1minute = 60 seconds

Required power =?

From : P= Energy supplied

Time

=15,000J

60Sec

=250W

But 1kw= 10000

? 25000

=1kw x 25000 100000

=0.25kw

(b) Given that:

Mass= 90kg Height attained by the sack = 2m

Time taken = 5seconds

P.E = mgh

=90kg x 10N x 2m Kg

= 900N x 2m


and Chemistry) 30

= 1800Nm =1800J

BUT:

Power = Energy supplied

t

=1800J 5sec

:. Power = 360W

c)Weight of the man =800N

Height of upstairs = 3m Time taken = 1minute =60sec.

Now

From P= energy

t

But : P= force x distance

t

P= 800N x 3m

60sec

P= 2400J

60Sec

:. Power P= 40W

CHAPTER 9 LIGHT

Qn 1.

Define the term light.

ANSWER -Is an invisible energy which cause in us sensation of vision

Qn 02.

Briefly explain why we are able to see objects during a day?

ANSWER

We are able to see objects during a day due to reflection of

light from the objects directly into our eyes.

Qn 03: Sources of light are divided into how many categories? Mention them.

ANSWER

Sources of Light are divided into two main parts/ categories

which are:- (b) Natural sources of light

(c) Artificial sources of light

Qn 04: What do you understand by the term:-

(b) Natural source of light? (c) Artificial source of light?

ANSWER

a) Natural source of light –Is the source of light which

produce light on their own eg: i. The sun

ii. Stars

iii. Some insects eg: fire fly

iv. Comets

b)Artificial source of light.

-Is the source of light which produce or radiate light when

operated/ light.

Examples of artificial sources of light are:-

(a) Torches (b) Electric lamp ( c) Electric lamp (d) candle etc.

Qn 05.

Define the following terms.

a) Luminous bodies b) Non-luminous bodies

ANSWER 05

a) Luminous bodies.

These are bodies which are capable of emitting light. NOTE: All natural and artificial sources of light are

LUMINOUS

BODIES

Examples of Luminous bodies are:- a) The sun (b) The star ( c) comets (d) some insects

(e) Torches (f) electric lamp (g) kerosene lamp

(h) Burning candle ( i) the moon (j) Burning fire wood.

(b) Non-Luminous bodies:- These are bodies which are not capable of emitting or radiating

light.

NOTE

Non-luminous bodies are seen when they reflect light coming from the Luminous bodies to our eyes.

Example on non-Luminous bodies are:-

a) stone b)human bodies c) Table d) metals eg. Iron

e) chairs f) sands g)the moon.

Qn 06

What is meant by the following terms.

a) optical medium

b) Homogeneous medium c) Heterogeneous medium

d) Transparent medium

e) Opaque bodies

ANSWER 06 b) Optical medium:- Is any thing (material or non-

material) through which light energy passes wholly or

partially.

Examples of optical media are:-

i) Vacuum ii) Air

iii) Most gases

iv) Water

v) Glass and

vi) Some plastics.

b) Homogeneous medium

- Is an optical medium which has a uniform composition through

out.

Examples of homogeneous media are:- a) Vacuum

b) Distilled water

c) Pure alcohol

d) Glass

e) Transparent plastics and f) Diamond.

c) Heterogeneous medium

-Is an optical medium, which has different composition at

different points. Examples of heterogeneous media are:-

a) Air b)Muddy water c) fog and d) mist


and Chemistry) 31

d)Transparent medium –Is a medium which allows most of light energy to pass through

it:-

Examples of transparent media are:-

a) Vacuum

b) Clear air c) Thin layers of water

d) Glass

e) Diamonds

f) Some plastics.

e)Translucent medium/body

-Is a medium which partially allows the light energy to pass

through it.

Examples of translucent bodies or media are:-

(ii) Butter paper (iii) Oiled paper

(iv) Tissue paper

(v) Ground glass

(vi) Frosted glass

(vii) Deep water (viii) Fog

(ix) Mist

f) Opaque bodies

-These are bodies which do not allow the light energy to pass through it at all.

Examples of opaque bodies are:-

a) Bricks

b) Wood

c) Stones d) Metals

Qn 07: Define the following terms:-

a) Ray of light

b) Beam of light c) Parallel beam of light

d) Divergent beam

e) Convergent beam.

ANSWER 07. a) A ray of light – Is the path along which light energy

travels in a given direction.

NOTE: A ray of light is represented as a straight line with

an arrow ie.

–From the figure above, an arrow shows the direction of

light.

b) Beam of light. –Is a collection of number of rays of light.

c) Parallel beam of light.

–These are rays of light which do not meet at a common

point. -An example of source of light which radiate parallel

beam of light is the sun.

d) Divergent beam of light

-These are rays of light originating from the same common source of light but they travel in different

directions.

e) Convergent beam of light

–These are rays of light coming from different directions

but meet at the same common point.

–For example, when parallel rays of light are made to pass through a convex lens they meet at the same common

Point.

Qn.08

What do you understand by the term propagation of light?

ANSWER 8

Propagation of light- This refer to how the light energy travels.

Qn. 09. What is meant by the term rectilinear propagation of

light?

ANSWER 09.

Rectilinear propagation of light-Means that light energy travels

in a straight lines.

Qn 10:How can you prove the rectilinear propagation of light?

OR

How can you prove that light energy travels in a straight line?

ANSWER 10:

Rectilinear propagation of light can be proved by the following

an experiment.

PROCEDURES: 01. Take three identical card boards each

having a small hole at its centre and

arrange them as shown in figure 01 below.

02. Place a source of light “S” in front of the

cardboard and look at the source of light through the hole.

03. Displace one of the hole and look at the

source of light.

Diagram


and Chemistry) 32

OBSERVATION

01. The source of light will be seen if the three holes are

in a straight line.

02. The source of light will not be seen if one the hole is

displaced or not in straight line.

CONCLUSION.

There fore this proves that, the light energy travels in a straight

line.

Qn 11. What do you understand by the term transmission of light?

ANSWER 11.

This refer to how light energy passes through different bodies.

NOTE:- Transmission of light can be in:-

a) Opaque bodies

b) Translucent bodies

c) Transparent bodies etc.

Qn 12:

List down at least three examples of rectilinear propagation of

light in our everyday life.

ANSWER: Examples of the rectilinear propagation of light in our energy

day life are:

a) When the sunlight enters through a small hole in a

darkroom it appears to travel in a straight line.

b) The light emitted by the head light of a scooter at night appears to travel in a straight line.

c) If we almost close our eyes and try to look towards a

lighted bulb, it appears to give light in the form of

straight lines which travel in various directions.

SHADOW, UMBRA AND PENUMBRA.

Qn. 13: Define the term shadow and list down two types of shadow.

ANSWER 13.

a) Shadow- Is a dark area formed on a ground, screen or

wall by an opaque body when illuminated by a light. b) There are two types of shadow.

(i) Umbra

(ii) Penumbra

i. UMBRA- Is a full darkness of shadow ii. PENUMBRA- Is a partial darkness of

shadow.

Qn 14. Briefly explain when an umbra and penumbra

shadow is formed.

Qn 15: What is a pin hole Camera?

ANSWER

Is a rectangular box with a hole on one side and a screen

paper on another side. Diagram.

8Qn 16: Briefly explain how image of an object is formed on

the screen of the pin hole camera.

ANSWER 16.

(x) A pair of rays of light pass through at the top

and the bottom of the camera.

ii) The rays then enter the camera through the pin hole

iii) When the rays fall on the paper screen an

inverted image is formed.

Qn 17: What are the characteristics of the image formed by the pin-hole camera?

ANSWER 17:

i. The image formed by a pin hole camera is

always inverted ie. Up side down.

ii. The image formed is real ie. Is formed by real intersection of light rays on the screen.

Qn 18

Briefly explain what will happen to the image of an object

placed in front of the box if the box is moved towards or away from the object.

ANSWER

CASE1: If the box is moved to wards the camera box the

image magnifies.

CASE02: If the camera box is moved away from the object

image formed

is diminished.

Qn 019.

Briefly explain when image formed by a pin-hole camera is said to be blurred.

ANSWER 19.

An image formed by a pin-hole camera is said to be blurred

when a hole is made to be large.

Qn20.

Define the term magnification as applied in a pin hole camera.

ANSWER 20.

MAGNIFICATION:- Is the ratio of the height of the image formed by a

pin hole camera to the height of the object.


and Chemistry) 33

MATHEMATICALLY, Magnification is given by:-

MAGNIFICATION= Height of image (hi)

Height of object (ho)

Where: hi= height of image Ho= height of object

M = magnification.

NOTE:= Magnification has no SI=Unit.

Qn 21: Write down the relationship between magnification, Object

distance and image distance.

ANSWER:

The relation ship between magnification, object distance and image distance can be shown from the diagram below.

From the diagram above

∆EFD ∆ ABD

Then:

EF=FD AB BD

But:

AC = 2AB

EG 2EF

There fore: EG=AC

FD BD

But:

EG= hi, FD=V, AC=ho and BD=U.

There fore: hi = ho

V U

Cross- multiplication

hi = hov

ho ho hiu= v

ho

hi = v

ho u

:. hi = v/u ho

But hi= magnification

ho

M= V/U The relationship between magnification, object distance and

image distance is

M= V/U

Where:-

M= magnification

V= image distance U= object distance.

NOTE:- Magnification can be given by:-

M= hi OR M= V/U depends on the data given in

the question.

ho

Qn 22.

A candle of height 2.5cm is placed 10cm in front of the pin

hole camera. If the distance between the pin hole and the plate

is 14cm. Find the height of the image formed on the camera

plate. Soln 022.

Given that:-

Object height (ho) = 2.5cm

Object distance (u) = 10cm

Image distance (v) = 14cm Required image height (hi)=?

From:

Magnification (M) V/U

M= 14cm 10cm

:. M= 7

5

But also:

Magnification (M) = hi ho

M= hi

2.5cm

But M= 7/5= 1.4

:. M1.4= hi

1 2.5cm

:. hi- 1.4 x 2.5cm

:. Height of the image (hi) = 3.5cm.

Qn 23.

An object was in front of a pin=hole such that the size of the

image formed was 4 times size of the object. If the object

distance was 12cm from the pin-hole, determine:

a ) Magnification of the camera

b) The distance of screen from the pin=hole.

Soln 023. a) Let object height be (ho)

Now:

Image height = 4xho

=4ho But magnification = hi

ho

:. Magnification = 4 of the camera

b) Required image distance = ?

Object distance = 12cm Magnification = 4


and Chemistry) 34

From: Magnification = V U

12cm x 4 = v x 12cm

12cm

:. V= 48 cm

:.Distance of the screen from the pin=hole= 46cm.

Qn 24

Define the term REFLECTION

ASNWER 24 REFRECTION = Is a throwing back of light rays when falls on

a smooth or hard surface eg: mirror.

NOTE:

Reflecting surface= is the surface capable of throwing back

light when falls on it.

Qn. 25:

Reflection of light is divided into two types. Mention them and

define each type.

ASWER 25.

There are two types of reflection namely:

i. Regular reflection

ii. Irregular reflection

I. REGULAR REFLECTION – Is a type of

reflection which occurs when a light falls on a

smooth surface.

II. IRREGULAR REFLECTION:= Is a type of reflection which occurs when a light falls on a

rough surface.

Qn 26.

Define the following terms as applied in reflection of light.

a) Mirror

b) Incident ray c) Reflected ray

d) Normal line

e) Angle of incidence

f) Angle of reflection

ANSWER 26. Consider the diagram below:

Fig : 02.

a) MIRROR:- Is a smooth polished surface from which regular reflection can take place.

b) NORMAL LINE- Is a perpendicular line drawn at the

point of Incidence

c)INCIDENT RAY – Is the ray of light which is directed on

the reflecting surface.

d)REFLECTED RAY- Is the ray of light which is directed

away from the reflecting surface. e)ANGLE OF INCIDENCE – Is the angle formed between

the

incident ray and the normal line.

-Incident ray is denoted by small letter “i”

c) ANGLE OF REFLECTION- Is the angle made between the reflected ray and the normal line.

-Angle of reflection id denoted by the small letter “r”

NOTE: from the figure 02 above.

AO = Incident ray

OB = Reflected ray

OC = Normal line

AOC = angle of incidence

BOC = Angle of reflection

DM = mirror.

Qn 27: State the laws of reflection.

ANSWER 027. There are two laws of reflection which are:-

ii. First law of reflection

iii. Second law of reflection

I. 1st law of reflection states that “Angle of incidence is equal to the angle of reflection”

II. 2ND Law of reflection states that “ The incident

ray, the reflected ray and the normal all lie on

the same plane”.

Qn 28.

Briefly explain how image of an object placed In front of the

plane mirror is formed.

ANSWER.

Image of an object placed in front of the plane mirror is formed

when the rays of light diverging from a point after reflection


and Chemistry) 35

either actually meet at some other point of appear to meet at some other point.

NOTE: When rays of light meet actually, then the image

formed is called REAL IMAGE and when rays of light meet

apparently, then the image formed is called VIRTUAL IMAGE.

REAL IMAGE:- Is the image formed by the actual intersection

of rays of light.

VIRTUAL IMAGE:- Is the image formed by the apparent intersection of rays of light virtual image.

Diagram

Qn 29: Write down characteristics of image formed by the

plane mirror. ASNWER 29:-

characteristics of the image formed by the plane mirror are:-

i) The image formed by the plane mirror is virtual image.

ii) The size of the image is equal to the size of object.

iii) The distance of the image is the same as the distance of the object from the mirror.

iv) The image formed is inverted.

Qn. 030.

Differentiate between the virtual image and real image. ANSWER 030

REAL IMAGE VIRTUAL IMAGE

1. Is formed by actual

intersection of rays of light

1. It is formed by apparent

intersection of rays of light

2. It is always inverted 2. It is always erect

3. It can be taken on the

screen

3. It cannot be taken on the

screen

Qn. 031.

Write down an equation which governing number of images

formed when two plane are inclined at an angle Q Answer 031.

Number of image (N) formed by two plane mirrors inclined at

an angle Q is given by:-

N= 360o -1 Q

Where:

N= Number of images

Q=Angle between two plane mirrors.

Qn 032. a) Calculate the number of image formed by

two plane mirrors if the angle of inclination

is 30o

b) Calculate the angle of inclination between two plane mirror if the number of image

formed by the mirrors are 3.

a) Given that:-

Angle between two plane mirrors (Q) = 30o

Required N=? N= 360o-1

Q

360o – 1

30o

= 12-1 = 11.

:. Number of image formed = 11.

b) Given that:-

Number of image (N) =3

Required angle of inclination (d)=? From:

N= 360o-1

Q

N+1=360o-1 Q

(3+1) = 360o-1

Q

4Q=360o 4 4

Q=90o

:. Angle of inclination between two mirrors = 90o

Qn 33: What is a periscope?

ANSWER 33:

Is a tube which consists of two plane mirrors facing each

other.

NOTE:

In a periscope each mirror is fixed of 45o to the

framework.

Qn 34. Briefly explain how periscope function?

ANSWER 34:

1. A ray of light from an object enters the tube horizontally

and meets the 1st plane at an angle of 45o

1. A ray is then reflected downwards at an angle 90o

where it meets another 2nd mirror at 45o.

2. The ray is finally reflected horizontally at 90o to the

eye of the observer.

3. The observer sees an object as if is at “A” fig 03 above.

Qn 35:

List down two uses of (a) periscope (b) plane mirror.


and Chemistry) 36

ANSWER. 01. It is used to see above the head of crowds.

02. It is used by soldiers in trench warfare.

(b)

1. Plane mirror is used as looking glass

2. Plane mirror is used by opticians to provide false dimensions.

3. It is used for construction of reflecting periscope

4. It is used for signaling purposes.

5. It is used in solar cookers for reflecting the rays of sun

into cooker. 6. It is used by barbers to shown the customer the back of

his head.

Qn. 36

What are the disadvantages of reflecting periscope? ANSWER.

2. The final image is not brightly illuminated as light

energy is absorbed due to two successive reflections.

3. Any deposition of moisture or dust on the mirror

reduces reflection almost to nil and hence the periscope can not be used in places where there is a

lot of dust or moisture.

SELF TESTS

1. Which of the following instruments should you use to

measure the length and breadth of a basketball court?

A.20-cm ruler, B.metre ruler, C.measuring tape

D.vernier calipers

2. What is the reading on the vernier calipers shown?

A.4.08 cm, B.4.18 cm, C.4.28 cm, D. 4.38 cm

3. The figure below shows four readings:

Which of the following is correct?

A.The reading of I is 3.82 cm. B.Th. reading of II is

5.06 cm.

C.The reading of III is 5.79 cm..D.The reading of IV

is 6.01cm.

4. Figure W shows the reading after the jaws of a pair of

vernier calipers are closed completely. Figures I, II,

III and IV shows its four different readings.

Which of the following is correct?

A The actual reading of I is 2.71 cm.

B The actual reading of II is 3.75 cm.

C The actual reading of III is 4.97 cm.

D The actual reading of IV is 5.10 cm.

5. What is the volume of the liquid in the measuring

cylinder?

A.5.6 cm3, B.5.7 cm3, C.5.8 cm3, D.5.9 cm3

6. The diagram shows a micrometer scale.

Which reading is shown?

A.5.64 mm, B.7.14 mm, C.7.16 mm, D.7.64 mm

7. One oscillation of a swinging pendulum occurs when

the bob moves from X to Y and back to X again.

Using a stopwatch, which would be the most accurate

way to measure the time for one oscillation of the

pendulum?

A Time 20 oscillations and multiply by 20.

B Time 20 oscillations and divide by 20.

C Time one oscillation.

D Time the motion from X to Y, and double

it.

9. A pendulum swings backwards and forwards passing

through Y, the middle point of the oscillation.

The first time the pendulum passes through Y, a

stopwatch is started. The twenty-first time the


and Chemistry) 37

pendulum passes through Y, the stopwatch is

stopped. The reading is T.

What is the period of the pendulum?

A.T/40, B.T/21, C.T/20, D.T/10

10.The figure below shows the readings on a micrometer screw

gauge.

Calculate the actual diameter of the wire

11.Diagram I shows the reading of the pair of vernier calipers

when the jaws are fully closed. Diagram II shows the reading when the same vernier calipers is used to measure the thickness

of 80 sheets.

(a) State the zero error of the vernier caliper.

(b) State the reading shown in diagram II.

(c) State the actual reading of the vernier

caliper.

(d) Calculate the thickness of one metallic

sheet.

(e) In the space below, draw the reading of the

scale if the thickness of 100 pieces of the same type of metallic sheets. Show your

calculation workings.

…..END…..

Form 1 Physics

Documents

branch of science

term science

b physical science

main branches of science

physics qn

study of matter

zoology qn

study of living things