form 1 physics
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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
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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.
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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
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=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
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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
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= 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
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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.
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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
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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.
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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
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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:
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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
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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.
Soln 14. Given that:
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
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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
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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:
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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.
Prepared by Bahanza amali H. (Bsc. Computer Engineering),Jaliju Khalid (Teacher – Physics and Mathematics),Shabani Ramadhani (Techer – Physics
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.
Prepared by Bahanza amali H. (Bsc. Computer Engineering),Jaliju Khalid (Teacher – Physics and Mathematics),Shabani Ramadhani (Techer – Physics
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
Prepared by Bahanza amali H. (Bsc. Computer Engineering),Jaliju Khalid (Teacher – Physics and Mathematics),Shabani Ramadhani (Techer – Physics
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.
Soln 11. Given that:
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
Prepared by Bahanza amali H. (Bsc. Computer Engineering),Jaliju Khalid (Teacher – Physics and Mathematics),Shabani Ramadhani (Techer – Physics
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.
Prepared by Bahanza amali H. (Bsc. Computer Engineering),Jaliju Khalid (Teacher – Physics and Mathematics),Shabani Ramadhani (Techer – Physics
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)
Prepared by Bahanza amali H. (Bsc. Computer Engineering),Jaliju Khalid (Teacher – Physics and Mathematics),Shabani Ramadhani (Techer – Physics
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
Prepared by Bahanza amali H. (Bsc. Computer Engineering),Jaliju Khalid (Teacher – Physics and Mathematics),Shabani Ramadhani (Techer – Physics
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
Prepared by Bahanza amali H. (Bsc. Computer Engineering),Jaliju Khalid (Teacher – Physics and Mathematics),Shabani Ramadhani (Techer – Physics
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
Prepared by Bahanza amali H. (Bsc. Computer Engineering),Jaliju Khalid (Teacher – Physics and Mathematics),Shabani Ramadhani (Techer – Physics
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
Prepared by Bahanza amali H. (Bsc. Computer Engineering),Jaliju Khalid (Teacher – Physics and Mathematics),Shabani Ramadhani (Techer – Physics
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.
Prepared by Bahanza amali H. (Bsc. Computer Engineering),Jaliju Khalid (Teacher – Physics and Mathematics),Shabani Ramadhani (Techer – Physics
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.
Prepared by Bahanza amali H. (Bsc. Computer Engineering),Jaliju Khalid (Teacher – Physics and Mathematics),Shabani Ramadhani (Techer – Physics
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.
Prepared by Bahanza amali H. (Bsc. Computer Engineering),Jaliju Khalid (Teacher – Physics and Mathematics),Shabani Ramadhani (Techer – Physics
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
Prepared by Bahanza amali H. (Bsc. Computer Engineering),Jaliju Khalid (Teacher – Physics and Mathematics),Shabani Ramadhani (Techer – Physics
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
Prepared by Bahanza amali H. (Bsc. Computer Engineering),Jaliju Khalid (Teacher – Physics and Mathematics),Shabani Ramadhani (Techer – Physics
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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
Prepared by Bahanza amali H. (Bsc. Computer Engineering),Jaliju Khalid (Teacher – Physics and Mathematics),Shabani Ramadhani (Techer – Physics
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.
Prepared by Bahanza amali H. (Bsc. Computer Engineering),Jaliju Khalid (Teacher – Physics and Mathematics),Shabani Ramadhani (Techer – Physics
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
Prepared by Bahanza amali H. (Bsc. Computer Engineering),Jaliju Khalid (Teacher – Physics and Mathematics),Shabani Ramadhani (Techer – Physics
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
Prepared by Bahanza amali H. (Bsc. Computer Engineering),Jaliju Khalid (Teacher – Physics and Mathematics),Shabani Ramadhani (Techer – Physics
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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.
Prepared by Bahanza amali H. (Bsc. Computer Engineering),Jaliju Khalid (Teacher – Physics and Mathematics),Shabani Ramadhani (Techer – Physics
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
Prepared by Bahanza amali H. (Bsc. Computer Engineering),Jaliju Khalid (Teacher – Physics and Mathematics),Shabani Ramadhani (Techer – Physics
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…..