PHYSICS – FORM 3

CLASSICAL NOTES – 2020

TABLE OF CONTENTS

1. APPLICATION OF VECTORS
2. FRICTION
3. LIGHT – PART I
4. LIGHT – PART II
5. OPTICAL INSTRUMENTS
6. THERMAL EXPANSION
7. TRANSFER OF THERMAL ENERGY
8. MEASUREMENT OF THERMAL ENERGY
9. VAPOUR AND HUMIDITY
10. CURRENT ELECTRICITY

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TOPIC – 01: APPLICATION OF VECTORS
Scalar and vector Quantities
Scalar quantities Vector quantities
Are those physical quantities which Are the physical quantities which
have only magnitude have both magnitude and direction
Examples are Examples are
 Mass  Pressure
 Length  Force
 Area  Acceleration
 Density  Velocity
 Speed etc  Momentum etc

Vector addition
 The addition of vector is done with the help of vector diagrams. The straight line
is drawn to scale. The length of a line segment represents the magnitude of the
vector and the arrow represents the direction ie
 When two or more vectors are added the sum is known as The resultant vector
 Vectors can be added by mathematical methods using mathematical formula
such as Pythagoras’ theorem, trigonometric Ratios etc
 Also vectors can be added by graphical method
Adding vectors by Graphical method
o Choose a suitable scale and write it down on a graph paper
o Pick a starting point and draw the first vector to scale direction stated (indicate
the magnitude and direction)
o Starting from the head of the first vector, draw the second vector to scale in the
started direction until all given vectors finished
o Draw the line to connect tail of the first drawn vector and the head of the last
vector. This is called resultant vector
o Measure the length of the resultant vector and convert to actual unit
o Determine the direction of the resultant vector

Example
1. Suppose a man walks starting from point A, a distance of 20m due north and
then walks 15m due east. Find his new position from A
Solution
Scale: 1cm represents 5cm
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∴ From the fig above, AC (R) is 25 m at an angle of 360 51’ east of north

The Triangle Method (Triangle law of vector addition)

States that
“If two vectors are represented by two sides of a triangle in sequence, then the third
closing side of the triangle ,drawn from the tail of the first vector to the head of the
second vector ,represents the resultant of the two vectors in both magnitude and direction “
OR
“If three vectors are in equilibrium and that two of the vectors are represented in
magnitude and direction by two sides of a triangle, then the third side of the
triangle represents the resultant of the two forces”
See the diagram below

Example:
1. A car is travelling due north at 60 km/hr .It turns and then travels due east at 80 km/hr
.Find the magnitude and direction of the resultant velocity of the car
Soln: By calculation method:

From: Pythagoras theorem: a2 + b2 = c2

a2 + b2 = c2 → R2 = 602 + 802 = 3400 +6400 = 10000 → 𝑅 = √10000 = 100
𝑜𝑝𝑝𝑜𝑠𝑖𝑡𝑒 80
𝑁𝑜𝑤 𝑐𝑜𝑛𝑠𝑖𝑑𝑒𝑟 𝑡𝑎𝑛𝜃 = = = 1.3333 → 𝜃 = 𝑡𝑎𝑛−1 (1.333) = 53.10
𝑎𝑑𝑗𝑎𝑐𝑒𝑛𝑡 60

∴ 𝑇ℎ𝑒 𝑟𝑒𝑠𝑢𝑙𝑡𝑎𝑛𝑡 , 𝑹 𝒊𝒔 𝟏𝟎𝟎 km/h to the direction of N 53.10 E

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Individual Work –1:1
1. A brick is pulled by a force of 4N acting northward and another force of 3N acting
north-east. Find the resultant of these two forces. ANS: The resultant is 6.5 N
2. Two forces, one 8 N and the other 6 N, are acting on a body. Given that the two
forces are acting perpendicularly to each other, find the magnitude of the third
force which would just counter the two forces. ANS = 10 N
3. A weight of 25 N is suspended from a beam by a string; what horizontal force
must be applied to the weight to keep the string at an angle of 20 0 N to the
vertical. What is the tension in the string? (ANS: FH =9.1N, TS = 26.6 N)

Parallelogram Method (Parallelogram law of vector addition)

The law states that:
‘’If two vectors are represented by two adjacent sides of a parallelogram ,then
the diagonal of the parallelogram through the common point represents the sum
of the two vectors in both magnitude and direction’’
OR
“If two vectors are represented by the two sides given and they include angle
between them, then resultant of the two vectors will be represented by the
diagonal from their common point of a parallelogram formed by the two vectors”
See the figure below

Example1.
Two forces of 20 N and 40 N acts at a point and the angle between them is 30 0. Find
the resultant force and the angle it makes with the force of 20 0 (ANS: R = 58 N, at 200)
Solution:
Scale: 1cm represents 5cm

From the graph above, R represents 11.6 Cm

Then: If 1cm = 5 N
11.6 cm =R (By crossing multiplication R = 58 N)
∴ The resultant force is 58 N at an angle of 200
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Individual Task – 1:2
1. Two forces AB and AD of magnitude 40 N and 60 N respective are pulling a
body on horizontal table. If the two forces makes an angle 30 0 between them,
find the resultant force on the, body. (ANS: R = 96.7N)
2. Two ropes of 3 m and 6 m long are tied to a ceiling and their free ends are pulled
by a force of 100 N as shown in the figure below. Find the tensions in each rope
if they make angle 30° between them. (ANS: T1 = 34.5N, T2 = 69 N)

3. Find the resultant force when two forces act as shown in the figure below.

(ANS: RF = 10 N)
4. Find the resultant force, F, when two forces, 9 N and 15 N, act on an object with
an angle of 600 between them. (ANS: FR = 21 N)

Relative Velocity
 Relative velocity is the velocity of a body with respect to another moving body.
NB:
 The relative velocity of a body with respect to a stationary observer is known as
‘’absolute relative velocity’’
 Let the Velocity of one object be VA and that of another be VB denoted by symbol VAB.
 If all objects are moving to the same direction, low speed will be experienced,
therefore we minus two velocities of the moving bodies
that is VAB = VA + (-VB) → VAB = VA - VB
 If all objects are moving to the opposite direction, high speed will be experienced,
therefore we plus two velocity of moving bodies VAB = VA - (-VB) → VAB = VA + VB
NB:

1. Speed of an air plane may be observed by an observer on the ground to be

increased by a TAIL WIND (VR = VA + VW) or reduced by HEAD WIND (VR = VA–VW).
So the wind and the plane are both moving related to one another but the
observer is stationary.

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 2. The speed of a boat in a river may also be observed by an observer at the
river bank to be increased downstream or decrease up stream. Again the boat
and the water are moving relative to one another but the observer is stationary.
3. Relative velocity also can be calculated by triangle method and by parallelogram
method
Example
1. Two cars A and B are moving with velocities of 50 m/s and 30 m/s respectively. Find the
relative velocity when they are in (a) The same velocity (b) Opposite velocity
Soln:
Given: VA = 30 m/s, VB = 50 m/s
(a) In the same velocity,
From: VAB = VA – VB → 𝑽𝑨𝑩 = 𝟓𝟎 − 𝟑𝟎 = 𝟐𝟎 𝒎/𝒔
(b) In opposite direction
From: VAB = VA + VB → 𝑽𝑨𝑩 = 𝟓𝟎 + 𝟑𝟎 = 𝟖𝟎 𝒎/𝒔

Individual Task – 1:3

1. A plane travelling at a velocity of 100 km/h to the South encounters a side wind
blowing at 25 km/h to the West. What is its velocity relative to an observer on the
ground? ANS: VR = 103.1km/h
2. Car A is moving with a velocity of 20 m/s while car B is moving with a velocity of
30 m/s. Calculate the velocity of car B relative to car A if:
(a) They are moving in the same direction (ANS: VBA = 10 m/s)
(b) They are moving in the opposite directions. (ANS: VBA = 50 m/s)
3. A Car is travelling at 60m/s due east and a lorry is travelling at 100m/s due
north. What is the velocity of the car relative to the lorry? (ANS: 116.6 m/s at 310 NE)
2. An automobile A, travelling relative to the earth at 45km/h on a straight level
road is ahead of motor cycle B travelling in the same direction at 90 km/h .What
is the velocity of B relative to A? (ANS: VBA = 45km/h)
3. A passenger at the back of a train travelling at 15m/s relative to the earth ,
throws a hammer with a speed of 15 m/s in the opposite direction to the motion of the
train .What is the velocity of the hammer relative to the earth (ANS: VHE = 0 m/s)
4. A boat heading due north crosses a wider river with a velocity of 36 m/s relative to the
water .The river has a uniform velocity of 12 m/s due south
(a) Determine the velocity of the boat with respect to an observer on the river
bank (ANS 24 m/s )
(b) If the river was flowing due east , determine the velocity of the boat with
respect to an observer on the river bank (ANS 37.947m/s)

Application of relative motion

 Is used in navigation to determine the actual velocity of vessels in moving water or in air
 Is used to determine the velocities of stars and asteroids with respect to earth’s velocity
 Used by structural engineers to design structures to avoid maximum deformation
when earthquakes strike
 Helps us to calculate the velocity of an object in a fluid
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Resolution of the Vector
 Is the process of splitting a vector into its components
That is a horizontal and vertical components.
 Consider the diagram below where the toy car is pulled at a certain angle but it
seems to move horizontally due to horizontal force, not only that but also vertical
force is formed

 From the fig above: Horizontal vector = x and Vertical vector = y

 The extract the triangle from the above diagram

Horizontal vector is given by the formula

𝑿
From: Cos θ = 𝑭 HC (X) = F Cos θ

Vertical vector is given by the formula

𝒀
From: Sin θ = 𝑭 VC (Y) = F Sin θ

Individual Task – 1:4

1. A nail is being pulled using a string from a wall. The string forms an angle of 30°
with the normal. If the force being used is 10 N, part of the force will tend to bend
the nail while the other part will try to pull it out.
Figure:

What is the magnitude of the force:

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 (a) Tend to bend the nails? (ANS: F1 = 8.66 N)
(b) Tend to pull the nails out? (ANS: F2 = 5.0 N)
2. A body is being acted on by two forces: F1 = 18 N acting at an angle of 25° and
F2 = 30 N acting at 140° from due East. Find the resultant of the two forces, F,
by separating the forces into x- and y- components. (ANS: RF = 27.70 N at an
angle of 103.940 to west)
3. A trolley was pulled by a force of 60 N acting 30 0 to the horizontal .Find the
vertical component of this force. ANS: VC = 30 N
4. A weight of 20 N rest on a plane inclined at 40 0 to the horizontal. What are the
components of the weight parallel and perpendicular to the plane?
(ANS: CParallel = 12.865 N, CPerpendicular = 15.32 N)

Uses of Component Vectors

 Adding many vectors
 Calculating a vector sum

Class Assignment –1
2. The velocity of car B relative to car A is 8 m/s when the two cars are moving in the
same direction and 28 m/s when the two cars are moving in opposite directions.
Determine the velocity of each car
3. An aeroplane is taking off at a velocity of 20 m/s. Find the components of the
plane’s velocity if the take – off angle is (a) 700 (b) 450 (c) 600
4. A river is flowing at a velocity of 2m/s due south .A person in a boat wants to move
across the river at 10 m/s.
a) In which direction should the person move?
b) At what velocity should the person move the boat?
5. A box is being pulled on the floor using a string. The string makes an angle of 300 with
the box as shown in the figure below

If the force being applied at the string is 200 N, find:

a) The force which tends to lift the box
b) The force which tends to pull the box forward

6. A car moves 5 km east 3 km south, 2 km west and 1 km north. Find the resultant
displacement (ANS: 3.6 km 340 south of east)
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 7. A plane is flying at a velocity of 100 km/hr and wind is blowing at a velocity of 25
km/hr if the blowing wind is (a)head (b) tail
Find the resultant plane velocity relative of to an observer on the grounds
8. A motorboat traveling 5 m/s, East encounters a current traveling 2.5 m/s, North.
(a). What is the resultant velocity of the motor boat?
(b). If the width of the river is 80 meters wide, then how much time does
it take the boat to travel shore to shore?
(c).What distance downstream does the boat reach the opposite shore?
(ANS: (a) 5.59 m/s at 26.60 (b) 16 s (c) 40 m)
9. A man is walking inside a bus which is travelling at 56.2 km/hr. If the speed of the man
relative to the ground is 55.8 km/h, is the man walking towards the front or the back?
10. A boat is travelling at 8.9 km/h. relative to the water in a river .The boat aims to
straight for the opposite bank of the river which is 120.9 wide. If the speed of the
water in the river is 2.9 km/h, how far downstream will the boat be when it
reaches the opposite side?
11. A boat is travelling at 9.8 km/h relative to the water in a river wants to get a
fishing camp that is 5.2 km upstream .If the speed of the water in the river is 6.0
km/h, how long will it take the boat to reach the camp?
12. A plane is flying at a velocity of 300 km/h, relative to the air towards 300 0 from
due east .The plane is flying amidst a wind blowing at 85 km/h relative to the
ground towards 2250 from due east. What will be the velocity of the plane as
observed on the control tower on the ground
13. An airplane is flying east at 200 km/h, which is its velocity relative to the air ,while a
100 km/h wind blows towards the north – east .What is its resultant velocity ?
(ANS: V = 280km/h at 14.60)
14. A plane is flying due east with a velocity of 100 m/s when it encounters a wind moving
at a velocity of 20 m/s . Find the resultant velocity of the plane if the direction of the
wind is due
(a) East (b) West (c) South (d) South – East
15. An aeroplane is taking off at a velocity of 20 m/s. Find the components of the
plane’s velocity if the take – off angle is (a) 700 (b) 450 (c) 600
16. A River is flowing at a velocity of 2 m/s due south .A person in a boat wants to move
across the river at 10 m/s
(a) In which direction should the person move?
(b) At what velocity should the person move the boat?
17. Two forces ,P and Q are applied on a small boat stuck in a shallow stream as shown below

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 Determine the magnitude and direction of the resultant of the two forces
18. The diagram in the figure below shows a block being pushed along a track .If a
force of 20 N is applied in direction A at an angle of 60 0, what is the resolved
part of the force in direction B ?

19. A Vehicle moving at a speed of 80km/hr emits smoke from its exhaust pipe
in opposite direction at 50 km/hr with respect to the vehicle .Determine the
speed of smoke with respect to the ground
20. When two motorbikes are approaching each other at constant speed ,the linear
distance between them decreases at 7 km/hr .When moving in the same
direction ,the linear distance between them decreases at 3 km/h Determine the
velocity of each motorbike
21. A car is travelling due north at 45 km/hr .It turns and then travels due east at 72
km/hr .Find the magnitude and direction of the resultant velocity of the car
22. A Mass 3 kg hangs at the end of a string .Find the horizontal force needed to pull
the mass sideways until the string is at 300 to the vertical .Find also the tension
in the string
23. An air craft heads north – west at 320 km/hr relative to the wind .The wind velocity is
80km/hr from the south .Find the velocity of the aircraft relative to the ground
24. A deep sea diver dives at an angle of 300 with the surface of water and follows a
straight – line path for a distance of 220 m. How far is the diver from the surface
of water?
25. A velocity of magnitude 40 m/s is directed at an angle of 40 0 east of north
.Represent this velocity on paper
26. A car travels 3 km due north ,then 5 km north–east .Represent these
displacements graphically and determine the resultant displacement
27. Two forces , one of 12 N and another of 24 N, act on a body in such a way they
make an angle of 300 with each other .Find the resultant of the two forces
28. A motorboat travelling 4 m/s ,East encounters a current travelling 3.0 m/s ,North
(a) What is the resultant velocity of the motorboat?
(b) If the width of the river is 80 m wide, then how much time does it take the
boat to travel shore to shore?
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 (c) What distance downstream does the boat reach the opposite bank?
29. A plane can travel with a speed of 80 mil/hr with respect to the air .Determine the
resultant velocity of the plane if it encounters a
(a) 10 mil/hr headwind (c) 10 mil/hr crosswind
(b) 10 mil/hr tailwind (d) 60 mil/hr crosswind
30. Find the horizontal and vertical components of a force of 10 N acting at 30 0 to
the vertical
31. A weight of 25 N is suspended from a beam by a string; what horizontal force
must be applied to the weight to keep the string at an angle of 20 0 to the vertical
.What is the tension in the string?
32. A car covered a displacement of 10 km due 300 ,then15 km due 1200 and finally
8 km due 2700 .Find the total displacement covered by the car
33. A man using a 70 kg garden roller on a level surface ,exerts a force of 200 N at
450 to the ground .Find the vertical force of the roller on the ground
(a) If he pulls (ANS: 560 N) (b) If he pushes the roller (ANS: 840 N )
34. A plane mirror is approaching you at a speed of 10 m/s you can see your image
in it. At what speed will your image approach you? (ANS: v = 20 m/s)
35.

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 TOPIC – 02: FRICTION
 Friction: Is the force which opposes the relative motion or tendency of
motion between two surfaces in contact
OR Is the force that opposes motion between any surfaces that are in contact
 Friction force which occurs in fluids is known as viscosity
How friction happens
 Friction is caused by molecular adhesion, surface roughness and deformations
(Adhesive bond, mechanical bond and deformation)
Advantage of friction
 It aids in walking and movement
 Helps moving body to stop by applying brakes
 Used to wear unneeded layers of some materials
 Causes lighting in match stick
 Supports life on the earth by preventing burning asteroids
 Causes nail to stick on the wood
 Enables bottle stopper to stick on the bottle neck
Disadvantage of Friction
 It causes wear and tear
 It produces heat in various machine parts causing efficiency to decrease
 Due to friction noise is produced in machine
 Causes loss of energy in form of heat and sound
 Causes motion of a body to slow down
 Heat produced by friction can cause appliance to burn
 It causes wounding, when skin wearing
Methods of increasing Friction
 Increasing the normal force by increasing the weight of the body
 Increasing the roughness of the surface
 Use materials of high coefficient of friction. Example, rubber band
 Scrubbing equipment is made rough to increase friction e.g. Steel wire for scrubbing “surface”

Methods of reducing Friction

 Place roller between the two rough surfaces
 Use ball bearings
 Use of lubricants e.g. oil, water
 Speedy material (eg Teflon) which have low coefficient of friction and thus slide easily.
 Make surface soft or Polishing
 Using streamlined bodies eg airplane, boat etc
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Normal Force and Limiting Friction
 Consider the fig. below

 Normal force, is the force which is equal and opposite to the weight of the body.
 The Normal force is always perpendicular to the surface on which it rests
 Limiting friction: Is the maximum possible value of static friction
NB:
 From the fig above. R = mg
 When the body is at rest FF = FA
 When the body is at motion FF ≠ FA so we have to find net force, F
 When the body starts to move static friction force is equal to limiting friction then
the minimum force applied tends to start the motion
 When the body starts to move kinetic friction force is not equal to the minimum
force applied, then the body tends to start the motion
 If limiting friction is less than the force applied, the body will move
 If limiting friction is greater than the force applied, then the body cannot move

Laws of friction forces

1. Frictional force is directly proportional to the normal force between the two surfaces
in contact. (Fr α R)
2. Friction depends on the nature (roughness) of surfaces in contacts.
3. Friction does not depend on the surface areas in contact.
4. Frictional force is independent of the speed once an object has been set in motion
Types of Friction
o Static friction
o Dynamic friction

Static Friction
 Is the friction which occurs when the two objects are not moving relative to each other
 This force causes some bodies to be stationary .Example, A book can be kept on
top of desk without dropping down
𝑭
From: 𝑭 = 𝝁𝒔 𝑹 → 𝝁𝒔 = 𝑹
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∴ Coefficient of static friction is the ratio of limiting friction force to the normal reaction
𝑭
𝒊𝒆 (𝝁𝒔 = )
𝑹

Dynamic (kinetic) Friction

 Is the friction that occurs when objects are moving relative to each other and rub
against each other
𝑭
From: 𝑭 = 𝝁𝒌 𝑹 → 𝝁𝒌 = 𝑹
∴ Coefficient of kinetic friction is the ratio of kinetic friction force to the normal reaction.
𝑭
𝒊𝒆 (𝝁𝒌 = )
𝑹

Whereby:
Fr = kinetic/static friction force
R = normal reaction
μ = coefficient of static/kinetic friction force

NB:
 When a body rolls on the surface of another ,the form of kinetic friction that exists
between the surfaces is called ‘ROLLING FRICTION’’
 Sliding Friction is the kind of kinetic friction that is caused by two bodies rubbing
or sliding against each other
 It is easy to roll a body than to slide it on the ground ,This is because Rolling
friction is always less than Sliding friction
 The coefficient of kinetic friction is always less than the coefficient of static friction

Example:
1. A block of mass 500g is pulled along a horizontal surface. If the coefficient of
kinetic friction between the block and the surface is 0.8. What is the friction force
acting on the block as it slides?
Soln:
Given: m = 500 g = 0.5 kg, 𝝁𝒌 = 𝟎. 𝟖, 𝑭 =?
𝑭
From: 𝝁𝒌 = 𝑹 → 𝑭 = 𝝁𝒌 𝑹 = 𝝁𝒌 𝒎𝒈 = 𝟎. 𝟓𝒙𝟎. 𝟖 𝒙𝟏𝟎 = 𝟒𝑵

Individual Task – 2:1

1. A block of mass 270kg is pulled along a horizontal surface. If the coefficient of
kinetic friction between the block and the surface is 0.4. What is the friction force
acting on the block as it slides? (ANS: Fr = 1, 080N)
2. A box of mass 2kg rest on a horizontal surface, a force of 4.4 N is required to
just start the box moving. What is the coefficient of static friction between the
block and the surface? (ANS: μ = 0.22)
3. An alluminium block of mass 2.1kg rests on a steel platform. A horizontal force of
15N is applied to the block
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 (a) Given that coefficient of limiting friction 0.6, will the block move?
(b) If will move, what will be its acceleration. Given that coefficient of kinetic
friction is 0.47 (ANS: (a) Since: mg > Fr, hence the car will move (b) a=2.44 m/s2)
4. A brick starts sliding with 6m/s across a concrete horizontal surface floor
and the coefficient of friction between the two surfaces is 0.4. How far will
it travel before coming to rest? (ANS: S = 4.5 m)
5. Find the static friction between a block of wood of mass 10kg placed on a table.
A minimum force of 50N is required to make the block just move on the top.
(ANS: μ = 0.5)

Friction force at Inclined Plane

 Consider the diagram below, a body of mss, m sliding down the inclined plane at 𝜽

 When the object begins to slide (from rest) there will be kinetic friction between
the object and the incline .Thus the net force will be: 𝑭𝒏𝒆𝒕 = 𝑭𝑨 − 𝑭𝑹
Whereby: 𝑭𝑨 = 𝒎𝒈𝒔𝒊𝒏𝜽 and : 𝑭𝑹 = 𝑭𝒌 = 𝝁𝒌 𝑹 = 𝝁𝒌 𝒎𝒈𝒄𝒐𝒔𝜽

𝑭𝒏𝒆𝒕 = 𝒎𝒈𝒔𝒊𝒏𝜽 − 𝝁𝒌 𝒎𝒈𝒄𝒐𝒔𝜽 = 𝒎𝒈(𝒔𝒊𝒏𝜽 − 𝝁𝒌 𝒄𝒐𝒔𝜽)

Thus acceleration will be given as: From 𝑭𝒏𝒆𝒕 = 𝒎𝒂 = 𝒎𝒈(𝒔𝒊𝒏𝜽 − 𝝁𝒌 𝒄𝒐𝒔𝜽)

 Therefore, the acceleration is given by ,
𝒂 = 𝒈(𝒔𝒊𝒏𝜽 − 𝝁𝒌 𝒄𝒐𝒔𝜽), For downward motion
𝒂 = 𝒈(𝒔𝒊𝒏𝜽 + 𝝁𝒌 𝒄𝒐𝒔𝜽),For upward motion (motion is against the gravity)

 When Fr = 0 (when the incline is frictionless) , 𝒕𝒉𝒆𝒏, 𝒂 = 𝒈𝒔𝒊𝒏𝜽

 At constant speed (at rest), a = 0 m/s2, 𝑭 = 𝒎𝒈𝒔𝒊𝒏𝜽 𝑎𝑛𝑑 𝑹 = 𝒎𝒈𝒄𝒐𝒔𝜽
𝑭 𝒎𝒈𝒔𝒊𝒏𝜽
𝑭 = 𝝁𝑹 → 𝝁= = = 𝒕𝒂𝒏𝜽
𝑹 𝒎𝒈𝒄𝒐𝒔𝜽

∴ 𝑨𝒕 𝒄𝒐𝒏𝒔𝒕𝒂𝒏𝒕 𝒔𝒑𝒆𝒆𝒅 (𝒂𝒕 𝒓𝒆𝒔𝒕), 𝝁 = 𝒕𝒂𝒏𝜽

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Example:
1. A block of wood of 4kg just slides without acceleration down an inclined plane of 40 0 to the
horizontal. What is the coefficient of dynamic friction?
Soln:
Given: m = 4 kg,  = 400
Consider the fig below

From: 𝑨𝒕 𝒄𝒐𝒏𝒔𝒕𝒂𝒏𝒕 𝒔𝒑𝒆𝒆𝒅 (𝒂 = 𝟎), 𝝁 = 𝒕𝒂𝒏𝜽

∴ 𝝁 = 𝒕𝒂𝒏𝜽 = 𝒕𝒂𝒏𝟒𝟎 = 𝟎. 𝟖𝟒

Individual task – 2:2

1. A mass is placed on an inclined plane such that it can move at constant speed,
when slightly tapped. If the angle of the plane makes with the horizontal plane is
300. Find the coefficient of kinetic friction. (ANS: μ = 0.56)
2. A mass of 5 kg is placed on a plane inclined at an angle of 30 0 to the horizontal.
What is the accelerating force required to pull the mass up the plane if the
coefficient of friction is 0.5? (ANS: FA = 46.65N)
3. A block of wood of mass 5kg is placed on a rough plane inclined at 60 0.
Calculate its acceleration down the plane if coefficient of friction between the
block and the plane is 0.32 (ANS: a = 7.1 ms-2)

Class Activity–2
1. Calculate the coefficient of kinetic friction between the surface of a table and a
block of wood when 5 kg block of wood is moving on the table and experiencing
a frictional of 5 N. (ANS: 𝝁 = 𝟎. 𝟏)
2. A box weighing 2 kg is at rest on a wooden floor. The coefficient of static friction
is 0.6 and the coefficient of kinetic friction is 0.35.
a) What minimum force is required to start the box sliding?
b) What minimum force is required to keep it sliding at a constant velocity?
3. A 12 kg box is being pulled across a level floor by a force of 60 N. If the acceleration
of the box is 2 ms-2 ,What is the force of friction between the box and the floor
4. A 0.5 kg object is given an initial velocity of 3 m/s after which it slides a distance
of 8 m across a level floor. What is the coefficient of kinetic friction between the
object and the floor?

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 5. The coefficient of kinetic friction between a block of wood and a wooden inclined
plane at an angle of 400 is 0.126. If the friction acting on the sliding prism is 42 N,
calculate the mass of the prism.(ANS: mass = 43.4 kg)
6. Calculate the friction force acting on a carton box of mass 9 kg which is moving
over a surface .The coefficient of kinetic friction between the two surfaces is
given as 0.45. (ANS: FR = 40.5 N)
7. The coefficient of friction between a particle of mass 8 kg, and a rough horizontal
plane is 0.4. Given that a horizontal force of 29 N acts on the particle as shown
in the figure below. Would it start to move (ANS: Since FA(29 N) < FR(32N), No motion)

8. A wooden block of mass 8 kg is resting on a wooden table. If the coefficient of

static friction between the pair is 1.3 , Calculate the minimum horizontal force
required to just slide the box .Given that g = 10 N/ kg
9. A 3 tones lorry is resting on a tarmac road .The lorry requires a minimum force of
12000 N in order for it to just move .Determine the coefficient of static friction
between the lorry’s tyres and the road
10. A crate of soda with mass 40 kg will just begin to slide with constant speed down
a rough ramp (slope) at 300 to the horizontal .What is the coefficient of static
friction. (ANS: 𝝁 = 𝟎. 𝟓𝟕𝟕𝟒)
11. A boy applies a horizontal force of 12 N on a metal solid block of mass 3.4 kg
resting on a concrete floor .Given that the coefficient of static friction between
the pair is 0.56 and g = 10 N/ kg
12. A box of mass 5 kg is at rest on a wooden floor. If the coefficient of static friction
between the box and the floor is 0.6, what minimum external force is required to
set the box sliding?
13. Define the following (a) Rolling friction (b) Sliding Friction
14. A 0.5 kg object is given an initial velocity of 3 m/s after which it slides a distance
of 8 m across a level floor, What is the coefficient of kinetic friction between the
object and the floor?
15. A box weighing 2 kg is at rest on a wooden floor .The coefficient of static friction
is 0.6 and the coefficient of kinetic friction is 0.35
(a) What minimum force is required to start the box sliding
(b) What minimum force is required to keep it sliding at a constant velocity?
16. In a car, The brakes stop the tyres while friction between the tyres and the road
surface stops the car .On a wet road the coefficient of kinetic friction between the
road surface and the tyre is 0.1 .Two cars, A and B, are travelling at a speed of
Written by Geoffrey M Idebe (0688 082 089 – Mwanza) Page 17
 15 m/s and 30 m/s ,respectively .Brakes are suddenly applied on each of the
cars .How far will each of the cars travel before coming to rest?
17. A rectangular box of mass 10 kg rests on an incline with a coefficient of static
friction of 0.55 and coefficient of kinetic friction of 0.25
(a) At what angle will the box begin to slide?
(b) If the incline is kept at that angle after the box begins to slide, what will be
the box‘s acceleration?
18. The coefficient of kinetic friction between the tyres of a car and the road is 0.7.
The car brakes are applied and it travels a distance of 120 m before stopping
.What was the car’s velocity just before the brakes were applied?
19. A box of mass 5 kg is at rest o a wooden floor. The coefficient of static friction is
0.42 and the coefficient of dynamic friction is 0.15. Find its acceleration if a force
of : (a) 15 N is applied to the box (b) 25 N is applied to the box
20. A 42 kg refrigerator is sitting on the back of a stationary pick – up .The coefficient
of static friction between the refrigerator and the pick – up bed is 0.44 .At what
rate can the pick – up accelerate without the refrigerator sliding off the back?

21. A 6 kg mass is resting on a horizontal surface .It is determined that a force of 20

N will start the object sliding and keep it sliding with an acceleration of 0.83
m/s2 .What are the coefficients of static and kinetic friction between the mass
and the surface ?
22. What is the normal normal reaction of the body of mass 10kg placed on an
inclined plane of angle 30˚c?
23. A concrete block of mass 10kg rests on a table. It is found that when a
horizontal force of 4kg weight pulls the mass, it is just begins to slide on the
table. Find the coefficient of static friction
24. A block of wood rests on a sloping plank which makes an angle of 31˚ with the
horizontal. If the block suddenly begins to slide down hill, what is the coefficient
of static friction?
25. A box of mass 50 kg is dragged on a horizontal floor by means of a rope tied to
its front. If the coefficient of kinetic friction between the floor and the box is 0.30,
what is the force required to move the box at uniform speed? (ANS: F = 150 N)
26. A car of mass 1200 kg is brought to rest by a uniform force of 300 N, in 80 sec.
What was the speed of the car? (ANS: u = 20 m/s)

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 27. A loaded trailer weighing 10kg is being towed across level ground .The
coefficient of dynamic friction is 0.25. What is the frictional force of the trailer?
28. A block of wood just slides without acceleration down an inclined plane of 25˚ to
the horizontal. What is the coefficient of dynamic friction?
29. Define the following terms
(a) Limiting friction (b) Normal reaction (c) Viscosity (d) Coefficient of Friction
30. (a) State the laws of friction (b) Explain, why Friction is friend and foe?
31. A brick is sliding at 8m/s across a concrete horizontal surface floor and the coefficient
of friction between the two surfaces is 0.5 How far will it travel before coming to rest?
32. Show that the acceleration of a stone sliding at a velocity ,v across a concrete
horizontal surface floor is given by a = 𝜇𝑔 where 𝜇 is the coefficient of friction
between the stone and the floor and g = acceleration due to gravity
33. A mass of 5 kg is placed on a plane inclined at an angle of 300 to the horizontal
.What is the accelerating force required to pull the mass up the plane if the
coefficient of friction is 0.5?
34. A block of metal with a mass of 20 kg requires a horizontal force of 50 N to pull it
with uniform velocity along a horizontal surface. Calculate the coefficient of
friction between the surface and the block. (ANS: 𝝁 = 𝟎. 𝟐𝟓)
35. A Car of weight 1000 N is moving with uniform speed .If the kinetic friction acting on
the car is 500 N , calculate the coefficient of kinetic friction
36. A wooden box of mass 30 kg rests on a rough floor. The coefficient of friction
between the floor and the box is 0.6. Calculate
(a) The force required to just move the box. (ANS: FF = 180 N, a = 0.67 m/s2)
(b) If a force of 200 N is applied to the box, with what acceleration will it move?
37. Describe how friction is minimized by the following methods:
(a)Lubrication (b) Use of bearings (c) Streamline flow
38. A boy is pulling a box of mass 10 kg. What is the normal force and the frictional
force if the coefficient of static friction? ( g = 10 N/kg)
39. A 50 g mass is placed on a straight track slopping at an angle of 450 to the
horizontal as shown from the figure below calculate
(i) Acceleration of the load as it slides down the slope
(ii) The distance moved from rest in 0.2 seconds

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 40. A 5 kg block is resting on a horizontal surface .Given that the coefficient of
static friction is 0.57 g = 10 N/kg
(a) What is the frictional force required to just move the block?
(b) What force must be applied to the block to keep it moving at constant velocity?
(c) Determine its acceleration if a force of 35 N is applied
41. A force of 8.0 N gives a 3.0 kg mass an acceleration of 0.6 m/s2 to the right
(a)What is the limiting friction on the block?
(b) Determine the coefficient of static friction required to produce a net
kinetic force of 6.0 N? (g =10 N/kg)
33. A 53.0 kg block slowed by friction has an acceleration of -0.1 m/s2 .Determine
the force of friction on the block
34. A 10.0 kg solid sliding along a horizontal surface is brought to rest after 30 minutes
(a) Name the force that caused it to stop
(b) Determine the magnitude of the force that caused it to stop
(Given that: 𝝁𝒌 = 0.45 , g = 9.8 N/kg )

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 Topic – 03: LIGHT PART I
Reflection of Light from Curved Mirrors
Types of curved mirrors
 Convex (diverging mirror)
 Concave (converging mirror)

Terms used in curved mirrors

Consider the diagram below when two curved mirrors are joined

Whereby:
AB = Convex mirror while ST = Concave mirror
C = centre of curvature
L = pole of the Concave mirror while K = pole of the Convex mirror
CL and CK are radii of curvature of Concave mirror and convex mirror respectively
CL and CK are principal axes of Concave and Convex mirror respectively
Centre of Curvature: Is the centre of the sphere in which the mirror is a part.
Radius of Curvature: Is the distance or length between the pole of the curved
mirror and the centre of curvature.
Principal Axis: Is the line joining the pole of the curved mirror and the centre of curvature.
Consider when light is reflected in curved mirrors as shown in the diagrams below.

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Principle Focus, F: Is the point in which the light is reflected in curved mirror
Focal Length, f Is the distance between pole of the curved mirror and the principal focus.
𝒓
NB: Given that focal length is always half the radius of curvature. 𝒇 = 𝟐
Location of Image using Ray diagrams
The following are the rules used to locate image in the curved mirror.
o A ray of light travelling to the mirror parallel to the principal axis, a ray is
reflected through the principal focus
o A ray of light travelling to the mirror through the centre of curvature is
reflected along its own path
o A ray of light travelling to the mirror through the principal focus is
reflected parallel to the principal axis
Note: Any two of these rays are sufficient to locate the image.
Procedure to draw ray diagrams
o Choose an appropriate scale so that the ray diagram fits on the available space.
o Draw a horizontal line to represent the principal axis of the mirror. Mark
the focal point of the mirror.
o Using the chosen scale, draw the object in position along the principal
axis. The object is drawn as a vertical line from the principal axis.
o Locate the position of the image by drawing rays from the object to the
mirror. Use the rules for drawing ray diagrams to draw the reflected rays.
o At the point of intersection of the reflected rays, draw the image in position

Image formed in Curved mirror

Terms used to describe the images formed by curved mirrors:
Position
 Real image is on the same side of the mirror as the object.
 Virtual image is on the opposite side of the mirror compared to the object.

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Nature
 Upright image has the same orientation as the object.
 Inverted image is oriented in an upside down position compared to the object.
Size
 Enlarged image is bigger than the object.
 Diminished image is smaller than the object

Images formed by Concave mirrors

The following are the characteristics of images formed by concave mirrors:

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 2. Example
An object 5 cm tall is placed 34 cm from a concave mirror of focal length 20 cm. By
means of an accurate graphical construction, determine the position, size and the
nature of the image formed.

Position of image = (49 x 1) = 49 cm from the mirror

Size of image = (7 x 1) = 7 cm
Nature of image: Image is real and inverted

Images formed by convex mirror

The images formed are always virtual, erect and diminished for all object positions.
Diagram:

Individual task 3:1

1. And object 5 cm tall is placed 15 cm in front of a concave mirror of focal length 10 cm.
By means of accurate graphical construction, determine the position and nature of the
image formed. (ANS: The image is real and is 30 cm in front of the mirror)
2. And object 5 cm tall is placed 12 cm in front of a convex mirror of focal length 20 cm.
By means of accurate graphical construction, determine the position and nature of the
image formed. (ANS: The image is virtual and is formed 7.5 cm from the mirror)

The Mirror Formula

 Consider the figure below showing an object placed beyond C in a concave
mirror whose focal length is f, whereby the object and image distances are u and
v respectively

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 𝐹𝑟𝑜𝑚 𝑡ℎ𝑒 𝑓𝑖𝑔𝑢𝑟𝑒 𝑎𝑏𝑜𝑣𝑒, 𝑡ℎ𝑒 ∆𝒔 𝑎𝑟𝑒 𝑒𝑞𝑢𝑖𝑎𝑛𝑔𝑢𝑙𝑎𝑟
𝒉 𝒉′ 𝒉′ 𝒗−𝒇
Now consider , 𝑡𝑎𝑛𝛼 = = → = … … … … (𝒊) ,
𝒇 𝒗−𝒇 𝒉 𝒇

𝒉 𝒉′ 𝒉′ 𝒇
Also consider, 𝑡𝑎𝑛𝛽 = = → = … … … … (𝒊𝒊) ,
𝒖−𝒇 𝒇 𝒉 𝒖−𝒇
𝒗−𝒇 𝒇
Then compare equation (i) and (ii) 𝒇
=
𝒖−𝒇
→ 𝒇𝟐 = (𝒗 − 𝒇)(𝒖 − 𝒇)

𝒇𝟐 = 𝒖𝒗 − 𝒇𝒗 − 𝒇𝒖 + 𝒇𝟐 → 𝒖𝒗 − 𝒇𝒗 − 𝒇𝒖 = 𝟎
𝒖𝒗 = 𝒇𝒗 + 𝒇𝒖 = 𝒇(𝒗 + 𝒖)
𝒖𝒗
𝒖𝒗 = 𝒇(𝒗 + 𝒖) → =𝒗+𝒖
𝒇
𝟏 𝒗+𝒖 𝒗 𝒖 𝟏 𝟏
= = + = +
𝒇 𝒖𝒗 𝒖𝒗 𝒖𝒗 𝒖 𝒗

𝟏 𝟏 𝟏
∴ the mirror formula is given by: = +
𝒇 𝒖 𝒗

Magnification of the image

 Magnification (m) is the ratio of the image size to the object size.
𝒊𝒎𝒂𝒈𝒆 𝒉𝒆𝒊𝒈𝒉𝒕 𝑰𝑯
i.e 𝒎 = 𝒐𝒃𝒋𝒆𝒄𝒕 𝒉𝒆𝒊𝒈𝒉𝒕
= 𝑶𝑯

OR
Magnification: Is the ratio of the image distance (v) from the mirror to the object
distance (u) from the mirror
𝒊𝒎𝒂𝒈𝒆 𝒅𝒊𝒔𝒕𝒂𝒏𝒄𝒆 𝑣
That is: m = =
𝒐𝒃𝒋𝒆𝒄𝒕 𝒅𝒊𝒔𝒕𝒂𝒏𝒄𝒆 𝑢

Example 1: An object is placed 20 cm in front of a concave mirror of focal length 12 cm. Find
the position and the nature of the image formed
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Answer
u = 20 cm, f = 12 cm, v = v
𝟏 𝟏 𝟏 𝟏 𝟏 𝟏 𝟏 𝟏 𝟏 𝟓−𝟑 𝟐 𝟏 𝟏
From: = + → = + → = − = = → = .
𝒇 𝒖 𝒗 𝟏𝟐 𝟐𝟎 𝒗 𝒗 𝟏𝟐 𝟐𝟎 𝟔𝟎 𝟔𝟎 𝒗 𝟑𝟎
V
Therefore, v = 30 cm, A positive value of v means a real image is formed. Hence a real
image is formed 30 cm away from the mirror on the same side as the object
N.B
 The image formed is sometimes in front of the curved mirror and sometimes
behind it. In order make the formula applicable to spherical mirrors and various
images formed. One of the following sign convention can be used to solve the
problems:-
(a) Real – is – positive Convention (b) New Cartesian Convention
(a) In Real – is – positive
(i) All distances are measured from the pole of the mirror as the origin
(ii) Distance of virtual objects, virtual images and virtual lengths from the pole
of the mirror are negative (-)
(b) In New Cartesian Convention
(i) All distances are measured from the pole of the mirror as the origin
(ii) Distances measured to the right of the mirror from the pole are positive (+)
(iii) Distances measured to the left of the mirror from the pole are negative (-)
 Now we can summarize that:-
(i) Focal length, (f) for a concave mirror is positive (+)
(ii) Focal length (f) for a convex mirror is negative (-)
(iii) The image distance, (v) is negative (-) for a virtual image
(iv) The image distance, (v) is positive (+) for real images

Individual Task – 3:2

1. An object 3 cm high is placed 30 cm away from a concave mirror of focal length
12 cm. using the mirror formula, find the position, the height and the nature of
the image formed. (ANS: V = 20 cm, The image is real, diminished IH = 2 cm)
2. A concave mirror with a radius of curvature of 30 cm produces an inverted
image 4 times the size of an object placed on its principal axis. Determine the
position of the object and that of the image. ANS: v = 75 cm
3. An object 30 cm high is placed 20 cm away from a convex mirror of focal length
25 cm. Describe the image formed. (ANS: IH = 16.8 cm, The image is diminished)
NB:
o Convex mirrors produce diminished images but have a very wide field of
view compared to plane mirrors
o Concave mirrors magnify images

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Uses of Convex mirrors
(i) They are used as driving mirrors. Due to wide field of view
(ii) They are used to see around corners. To avoid the crashing of vehicles or
supermarket trolleys at the corners
(iii) They are used for Supermarket surveillance. For surveillance in business
establishments and security installations
Uses of Concave mirrors
(i) They used as shaving mirrors. Due to magnification and erect image
(ii) They are used in reflecting telescopes. To see distant stars
(iii) They are used in solar cookers.
(iv) They are used in making car headlamp and torch (by using parabolic mirrors)
(v) They are used by dentist to see enlarged image of patient teeth

Refraction of Light through Plane Media

Refraction Is the process by which the direction of a ray of light changes when
passes obliquely from one medium into another
OR Is the change in direction of a wave passing from one medium to another
OR Is the bending of light as it passes from one transparent substance into another

Laws of Refraction
The laws state that:-
1. “At the point of incidence, the Incident ray, the normal and the refracted
ray all lie in the same plane”
𝐒𝐢𝐧 𝐢
2. “For a particular material, the ratio is constant’’.
𝐒𝐢𝐧 𝐫
This is referred to as Snell’s Law
𝐒𝐢𝐧 𝐢
The ratio is called refractive index” of a material
𝐒𝐢𝐧 𝐫

Refractive Index (index of refraction)

 Is the ratio of the velocity of light in a vacuum to its velocity in a specified medium

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 𝒄
That is 𝜼 =
𝒗

OR Is the measure of the bending of a ray of light when passing from one medium
into another
OR Is the ratio of the sine of angle of incidence of light to the sine of the angle of refraction
Suppose light travels from air medium to glass
Let the angle of incidence be i0 and the angle of refraction be r0.
𝑺𝒊𝒏 𝒊
Refractive index between air and glass is given by aμg = 𝑺𝒊𝒏 𝒓
Example:
1. A ray of light travels from water to glass. What will be the angle of refraction in
glass if the angle incidence in water is 200? If the refractive index of the glass is 1.5
Soln:
Given: i= 200, wμg =1.5 , r = ?
𝑺𝒊𝒏 𝒊
From: wμg =
𝑺𝒊𝒏 𝒓

𝑺𝒊𝒏 𝒊 𝑺𝒊𝒏 𝟐𝟎 𝒔𝒊𝒏𝟐𝟎

𝐰𝛍𝐠 = → 𝟏. 𝟓 = → 𝒔𝒊𝒏𝒓 = = 𝟎. 𝟐𝟐𝟖𝟎
𝑺𝒊𝒏 𝒓 𝑺𝒊𝒏 𝒓 𝟏.𝟓

∴ 𝒓 = 𝒔𝒊𝒏−𝟏 (𝟎. 𝟐𝟐𝟖𝟎) = 𝟏𝟑. 𝟐0

NB:
 Refractive index between vacuum(air) to any other materials is called absolute
refractive index
 Refractive index between medium to medium except vacuum (air) is called relative
𝒗𝒘
refractive index. For instance if light passes from water to glass ie wμg =
𝒗𝒈
 Any material has its own refractive index due to the fact that each has individual
optical density
 Light passing into an optically denser medium is bent towards the normal while
light passing into an optically less denser medium is bent away from the normal
Refractive Index of Different Medium

medium Refractive Index Medium Refractive Index

Diamond 2.417 Air (at s.t.p) 1.00029
Ethyl alcohol 1.360 Glycerin 1.47
Glass/crown 1.520 Paraffin 1.44
Quartz 1.553 Crown(flint) 1.65
Water (2000C) 1.333 Ice 1.31

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PRINCIPLE OF REVERSIBILITY OF LIGHT
 The principle of reversibility of light states that the paths of light rays can be
reversed
 If the refractive index for a ray of light moving from air (a) to glass (g) is represented as
ang and the refractive index for a ray moving from glass to air is represented as gna,
then by principle of reversibility of light:
𝟏
g𝝁a = (𝒂𝝁𝒈)

 The refractive index for a ray of light travelling from air to water is 1.33. What is
the refractive index for a ray travelling from water to air
Answer:
Given: an w = 1.33
𝟏 𝟏
From: w𝝁a = = 𝟏.𝟑𝟑 = 𝟎. 𝟕𝟓
(𝒂𝝁𝒘)

Alternatively:
 Another way of determining the refractive index of a material is by real-and-
apparent depth method.
Diagram:

 When one looks at a stick placed inside a beaker of water, the stick immersed in
water tend to rise from its real position due to refractive index of water
𝑹𝒆𝒂𝒍 𝒅𝒆𝒑𝒕𝒉 (𝑯)
That is aμw = 𝑨𝒑𝒑𝒂𝒓𝒆𝒏𝒕 𝒅𝒆𝒑𝒕𝒉 (𝒉)

∴The refractive index is the ratio of the real depth to the apparent depth of water (liquid)
NB: The difference between real and apparent depth is known as vertical displacement
Example1. A coin is placed at the bottom of a tall gas jar. When the jar is filled with paraffin
to a depth of 32.4 cm, the coin is apparently seen displaced 9.9 cm from the bottom. What is
the refractive index of paraffin?
Answer
Apparent depth = (32.4 – 9. 9) cm = 22.5 cm, Real depth (H) = 32.4 cm
𝑹𝒆𝒂𝒍 𝒅𝒆𝒑𝒕𝒉 𝟑𝟐. 𝟒
𝑹𝒆𝒇𝒓𝒂𝒄𝒕𝒊𝒗𝒆 𝒊𝒏𝒅𝒆𝒙 = = = 𝟏. 𝟒𝟒
𝑨𝒑𝒑𝒂𝒓𝒆𝒏𝒕 𝒅𝒆𝒑𝒕𝒉 𝟐𝟐. 𝟓

Written by Geoffrey M Idebe (0688 082 089 – Mwanza) Page 29

REFRACTIVE INDEX IN TERMS OF VELOCITY OF LIGHT
 For a ray of light travelling from medium 1 to medium 2, refractive index is the ratio
of velocity of light in medium 1 to velocity of light in medium 2. For example, for a
ray of light travelling from air to water
𝒗𝒆𝒍𝒐𝒄𝒊𝒕𝒚 𝒐𝒇 𝒍𝒊𝒈𝒉𝒕 𝒊𝒏 𝒂𝒊𝒓
𝑎𝒏𝒘 = 𝒗𝒆𝒍𝒐𝒄𝒊𝒕𝒚 OR
𝒐𝒇 𝒍𝒊𝒈𝒉𝒕 𝒊𝒏 𝒘𝒂𝒕𝒆𝒓

𝑣𝑒𝑙𝑜𝑐𝑖𝑡𝑦 𝑜𝑓 𝑙𝑖𝑔ℎ𝑡 𝑖𝑛 𝑚𝑒𝑑𝑖𝑢𝑚 𝟏

𝑅𝑒𝑓𝑟𝑎𝑐𝑡𝑖𝑣𝑒 𝑖𝑛𝑑𝑒𝑥, 𝟏𝒏𝟐 =
𝒗𝒆𝒍𝒐𝒄𝒊𝒕𝒚 𝒐𝒇 𝒍𝒊𝒈𝒉𝒕 𝒊𝒏 𝒎𝒆𝒅𝒊𝒖𝒎 𝟐

Example. The speed of light in air is 3.0 x 108 m/s. What is the speed of light in glass?
Take refractive index of glass = 1.5
Answer
Given: ang = 1.5, C = 3.0 x 108 m/s
𝒗𝒆𝒍𝒐𝒄𝒊𝒕𝒚 𝒐𝒇 𝒍𝒊𝒈𝒉𝒕 𝒊𝒏 𝒂𝒊𝒓 𝟖
From: 𝑎𝒏𝒈 = 𝒗𝒆𝒍𝒐𝒄𝒊𝒕𝒚 𝒐𝒇 𝒍𝒊𝒈𝒉𝒕 𝒊𝒏 𝒈𝒍𝒂𝒔𝒔
→ 𝑎𝒏𝒈 = 𝟑𝒙𝟏𝟎
𝒗𝒈

𝟑𝒙𝟏𝟎𝟖
𝒗𝒈 = = 𝟐 𝒙 𝟏𝟎𝟖 𝒎/𝒔
𝟏.𝟓

Individual Task – 3:3

1. When a ray of light is travelling from air to glass, the angle of refraction is 300. If the
refractive index of the glass is 1.5. Determine the angle of incidence, i. (ANS:i0 = 48.60)
2. A fish appears to be 0.9 m below the surface of water of refractive index 4/3
when viewed directly from above. What is the true depth the fish is?(ANS: R =1.2m)
3. A beaker of height 10 cm is filled with water. An optical pin which is at the bottom of
the beaker is then viewed from the top of the beaker. How far does the pin appear
from the surface, if the refractive index of water is 4/3? (Answer: 7.5 cm)
4. Refractive index of glass is 1.5 .If the speed of light in vacuum is 3 x 108 m/s, find the
velocity in medium (ANS: 2 x 108 m/s)
5. The refractive index for a ray of light travelling from air to oil (ano) is 5/3, while that for a
ray travelling from air to glass (ang) is 3/2. What is the refractive index for a ray
travelling from glass to oil? (Answer: 10/9)
6. A coin at the bottom of a jar of glycerin appears to be 13.2 cm below the surface
of the glycerin. Calculate the height of the column of glycerin in the jar given that
the refractive index of glycerin is 1.47. (ANS: 𝐻 = 19.4 cm)
7. The refractive index of water is 4/3. Find the speed of light in water given that the
speed of light in air is 3.0 x 108 m/s (Answer: 2.25 x 108 m/s)
8. Water is poured into a beaker to a depth of 24 cm. To an eye looking vertically
down through the water, the bottom of the beaker appears to be raised 6 cm
from the bottom of the beaker. Determine the refractive index of the water.
ANS: aμg = 1.33

Written by Geoffrey M Idebe (0688 082 089 – Mwanza) Page 30

Applications of Refractive Index
 Is used in optical systems to calculate the focusing power of lenses
 Is used to identify a substance or confirm its purity.

Critical Angle
 Is a unique angle of incidence for which the angle of refraction is 900.
OR
Is the angle of incidence above which total internal reflection occurs
OR
Is the angle of incidence beyond which rays of light passing through a denser
medium to the surface of a less dense medium are NO longer refracted but totally reflected
Diagram:

Whereby: c = critical angle r = refracted angle = 900

Total Internal Reflection

Is the reflection due to the angle of incidence exceeding the critical angle
OR Is a phenomenon that occurs when light travels from a more optically dense
medium to a less optically dense one. E g, glass to air or water to air
OR Is a phenomenon that occurs when a propagated wave strikes a medium boundary at
an angle larger than a particular critical angle with respect to the normal to the surface

NB:
 The reflected ray goes back to more dense medium
 Total internal reflection only occurs when light travels from a more dense
medium to a less dense medium
 When total internal reflection occurs, there is no refraction at all
Written by Geoffrey M Idebe (0688 082 089 – Mwanza) Page 31
Relationship between Critical Angle and Refractive Index
Consider the equation below;
𝒔𝒊𝒏𝒊
𝒘𝝁𝒂 = 𝒔𝒊𝒏𝒓 but: r = 900 and i = c
𝒔𝒊𝒏𝒄 𝒔𝒊𝒏𝒄
Then, 𝒘𝝁𝒂 = =
𝒔𝒊𝒏𝟗𝟎 𝟏

According to the principle of reversibility of light (the path can be reversed)

𝟏
Therefore: 𝑎𝝁𝒘 = 𝒔𝒊𝒏𝒄

 This is true for all material media where, water, w represent material media
Conditions for total internal reflection to occur
 A ray of light should travel through an optically denser medium into an optically
rare medium
 The angle of incidence should be equal or greater than the critical angle for the
two media
Individual task – 3:4
1. A certain glass material has a refractive index of 2.5. What is its critical angle?
(Answer: 23.570)
2. The critical angle of paraffin is 450. What is the refractive index of paraffin?
(Answer: 1. 414)

SOME EFFECTS OF TOTAL INTERNAL REFLECTION OF LIGHT

(a) Mirage
 This is an impression of the presence of imaginary water some distance away
OR Is an optical phenomenon in the atmosphere that makes an object appears to be
displaced from its true position.
How mirage occurs?
 When light pass from cold air layer (optical dense) to earth surface of hot air
layer (optical less dense) continuous light bending (refraction), where incidence
angle exceeds the critical angle. All light is reflected upwards (total internal
reflection). This looks like the reflection produced by a pool of water
 Therefore for a mirage to appear “the bent light from the sky is refracted as it
passes from cooler air into hotter air and back to your eye” (See the fig below)

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 Mirages are often seen during hot sunny days.
(b) Twinkling of stars

Applications of total internal Reflection

1. Prism periscope
 A periscope is a device which enables us to see over the top of an obstacle
 A prism periscope consists of two 45° - 90° - 450 prisms (See the fig below)

2. Optical fibres
 An optical fibre is a thin rod of high-quality glass designed to guide light along its
length by total internal reflection.
 Light inside these fibres hits the sides at an angle greater than the critical angle
and is transmitted by being repeatedly totally internally reflected.

Some uses of optical fibres

 Are used in telecommunications and networking to send message signals.
 They are also used as light guides in medical and other applications
 Optical fibres are also used in imaging optics. A bundle of fibres along with lenses
are used to make a long imaging device called an endoscope.
NB:
 Medical endoscopes are used in minimally invasive surgical procedures.
 Industrial endoscopes are used for inspecting machine parts
3. Binoculars and telescopes
 Binoculars are just a pair of telescopes one for each eye .See the fig below
 Are used to see distant objects

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Refraction of Light by Lenses
 Lens is a transparent or a translucent medium that alters the direction of light
passing through it

Types of Lenses
 Convex lenses
 Concave lenses

Convex (Converging Lenses)

 A convex lens is thicker at its centre than at its edge.
 Convex lenses converge light.
 Convex lenses can be biconvex, plano –convex or converging meniscus

Concave (Diverging Lenses)

 A concave lens is thicker at its edges than at its centre.
 Concave lenses diverge light.
 Concave lenses include biconcave, plano-concave and diverging meniscus

Terms used on thin Lenses

Consider the diagrams below

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  Optical centre: Is the geometric centre of a lens.
𝑪
 Centre of curvature, ( ): Is the geometric centre of the sphere of
𝟐𝑭
which the lens surface is a part of.
 Principal axis: Is an imaginary line which passes through the optical centre of
the lens at a right angle to the lens.
 Radius of curvature, R: Is the distance between optical centre and the centre
of curvature.
 Principal focus (focal point), F: Is a point through which all rays travelling
close and parallel to the principal axis pass through the lens.
 Aperture: Is the width of the lens, from one edge to another
 Focal length, f: Is the distance between the optical centre and the principal focus

Construction of ray diagrams

The following are the procedures used to locate image in a lens
 Choose a suitable scale
 Draw a principle axis and the lens
 Draw object in the position
 A ray of light travelling parallel to the principal axis passes through the principal
focus. Ie

 A ray of light travelling through the optical centre goes undeviated (not refracted)
along the same path. ie
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  A ray of light travelling through the principal focus is refracted parallel to the
principal axis. ie

 Measure the height and the distance of the image

 Convert the measurements into actual units using the chosen scale

Example 1
An object 10 cm tall stands vertically on the principal axis of a convex lens of focal
length 10 cm and at a distance of 17 cm from the lens. By means of accurate graphical
construction find the position, size and nature of the image formed
Answer

(a) Position of the image = 24 x 1 = 24 cm from the lens

(b) Size of the image = (14 x 1) = 14 cm tall
(c) Nature of the image: The image is (i) Real and (ii) inverted

Written by Geoffrey M Idebe (0688 082 089 – Mwanza) Page 36

Individual Task – 3:5
2. An object 10 cm tall stands vertically on the principal axis of a convex lens of focal
length 10 cm and at a distance of 17 cm from the lens. By means of accurate
graphical construction find the position, size and nature of the image formed (ANS:
v = 24 cm from the lens, Size of the image = 14 cm tall, The image is Real and inverted)
3. An object 8 cm tall is placed 20 cm in front of a convex lens of focal length 16 cm.
By means of accurate graphical construction, determine the position, size and
nature of the image formed. (ANS: V = 60 cm, Size: = 24 cm, Nature: Real image)
4. An object 0.05 m high is placed 0.15 m in front of a convex lens of focal length 0.1
m. Find, by construction, the nature, the position and the size of the image.
(ANS: HO = 0.1, V = 0.3 m, Real)

Linear Magnification (m)

 Magnification is a measure of the extent to which an optical system enlarges or
reduces an image in relation to the object.
𝒉𝒆𝒊𝒈𝒉𝒕 𝒐𝒇 𝒊𝒎𝒂𝒈𝒆 𝑰𝑯
𝒎= = 𝑶𝑯
𝒉𝒆𝒊𝒈𝒉𝒕 𝒐𝒇 𝒐𝒃𝒋𝒆𝒄𝒕

𝒊𝒎𝒂𝒈𝒆 𝒅𝒊𝒔𝒕𝒂𝒏𝒄𝒆 𝒗
Also can be given by: 𝒎 = =𝒖
𝒐𝒃𝒋𝒆𝒄𝒕 𝒅𝒊𝒔𝒕𝒂𝒏𝒄𝒆

The Lens Formula

 Consider the figure below, an object h is placed in a convex mirror of focal
length f, whereby u is the object distance and v is the image distance

 𝑠𝑖𝑛𝑐𝑒 ∆𝒔 𝑓𝑟𝑜𝑚 𝑡ℎ𝑒 𝑓𝑖𝑔𝑢𝑟𝑒 𝑎𝑏𝑜𝑣𝑒 𝑎𝑟𝑒 𝑒𝑞𝑢𝑖𝑎𝑛𝑔𝑢𝑙𝑎𝑟

𝒉 𝒉′ 𝒉 𝒖
 Now consider , 𝑡𝑎𝑛𝐴 = = → = … … … … (𝒊),
𝒖 𝒗 𝒉′ 𝒗
𝒉 𝒉′ 𝒉 𝒇
 Also consider , 𝑡𝑎𝑛𝐵 = = → = … … … … (𝒊𝒊) ,
𝒇 (𝒗−𝒇) 𝒉′ 𝒗−𝒇
𝒖 𝒇
Then compare equation (i) and (ii) = → 𝒖(𝒗 − 𝒇) = 𝒇𝒗
𝒗 𝒗−𝒇

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 𝒖𝒗 − 𝒖𝒇 = 𝒇𝒗 → 𝒖𝒗 = 𝒗𝒇 + 𝒖𝒇 = (𝒖 + 𝒗) 𝒇
𝒖𝒗 𝟏 𝒖 𝒗 𝟏 𝟏
𝒖𝒗 = (𝒖 + 𝒗)𝒇 → =𝒖+𝒗 → = + = +
𝒇 𝒇 𝒖𝒗 𝒖𝒗 𝒗 𝒖

𝟏 𝟏 𝟏
∴ the lens formula is given by: = +
𝒇 𝒖 𝒗

Real-Is-Positive Convention
To calculate the values of u and v, a sign rule or convention is adopted. The rule is
referred to as the real-is-positive convention.
 Sign for real object and image are u=+ and v=+
 Sign for virtual object and image are v=- and u=-
NB:
 Sign of virtual object and image is negative Because the principal focus of a
concave lens is virtual
 Convex lenses have positive values of focal length, F = +
 Concave lenses have negative values of focal length, F = -
 If h = + v (The image formed is upright and is virtual)
 If h = -v (The image formed is inverted and is real)

Individual Task – 3:6

1. An object is placed 12 cm from a convex lens of focal length 18 cm. using the
lens formula. Find the position of the image.(ANS: v = -36 cm, the image is virtual )
2. An object is placed 10 cm from a concave lens of focal length 15 cm. using the
lens formula, determine the nature and the position of the image.
(ANS: Since, v = -6, The image is virtual and erect )
3. An object 2 cm high is placed 24 cm from a converging lens. An erect image
which is 6 cm high is formed. Find focal length of the lens. (ANS: f = 36 cm)
4. The focal length of a converging lens is 10 cm. How far should the lens be
placed from an illuminated object to obtain an image which is five times the size
of the object on a screen? (ANS: u = 12 cm )

Images formed by thin Lenses

 Images Formed By Convex Lens

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Written by Geoffrey M Idebe (0688 082 089 – Mwanza) Page 39
Image formed in Concave Lens
 The images formed are always virtual, erect and diminished for all object positions.

Properties of image formed

 Virtual
 Formed between the object and the lens
 Erect
 Diminished
 As U increase to infinity also V increase to F

Class Activity – 3
1. The object is placed 20 cm from a converging lens for focal length 15cm. find
the position, the magnification and the nature of the image
ANS V = 60 cm from the lens, M = 3 and the nature is a real image
2. Find the nature and position of the image of an object, placed 10cm from a diverging
leans of focal length 15cm (concave). (ANS: V = -6 cm from the lens, Real m = 0.6 )
3. The apparent depth of a certain point at the bottom of water pond is 25cm. find
the real depth of this point given that the refractive index is 4⁄3 (ANS: H = 33. cm)
4. An object stands vertically on the principle axis of a converging lens of focal
length 10mm and at a distance of 17mm from the lens. Find the position, size
and nature of the image. ANS: (∴ V = 24.14 cm, M = 2.9mm, Real)
5. Calculate the critical angle for air and water medium if the refractive index of
water is 4⁄3. (ANS: The critical angle is 48 0 38’)
6. Given that the refractive index of glass is 1.5, what is the value of the
critical angle? (ANS: The critical angle is 410491)
7. Given that the refractive index of ethyl alcohol is 1.36. Find the apparent depth in
the beaker if the real depth of the optical pin is 52cm. (ANS: h = 38.23 cm)
8. A fish is located 10m deep in the liquid when viewed from the top. The depth of
the fish is 8m. Find the refractive index of the liquid. (ANS: 1.25)
9. If the light has a velocity of 3x108 and has a velocity 1.97 X 108 m/s in the glass.
(a) What is the refractive index of the glass?
(b) Calculate the refractive index for light traveling from glass to air.
ANS: (a) 1.52 (b) ANS: 0.6458
10. The refractive for light traveling from air to water is 1.3. Find the refractive index
of light travelling from water to air. ANS: 0.7693
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 11. A small pin 3cm high is placed 30 cm away from a concave mirror of focal
length 12cm. By using the mirror formula, find the position, the height and the
nature of the image formed. (ANS : V= 20cm , m = 2cm high and is It is real,)
12. An object 2cm is high erected 8cm in front of a concave mirror of radius of
curvature 10cm. by graphical method, find the position, size and nature of the
image. (ANS: V = 13.3 cm, HI = 3.3 cm. The image is real).
13. A small spring is 4 cm long is kept at 10cm in front of the converging mirror of
radius of curvature 12 cm. By scale drawing, determine the position, size and
state the nature of the image formed. (V = 15 cm, HI = 3 cm and it is real.)
14. A convex mirror produces an image that is 22 cm behind the mirror when an
object is placed 34 cm in front of the mirror .What is the focal length of the mirror
15. A concave mirror has a focal length of 40 cm .How far from the mirror must an
object be placed to produce an image that is
(a) Twice the size of the object (b) Half the size of an object
(c) 40 times the size of the object
16. Show that to obtain an image with a magnification of M using a concave mirror
𝑴+𝟏
with a focal length f, the object distance ,u, is given by 𝒖 = 𝒇
𝑴
17. .What happens to the image formed by
(a) A Convex mirror
(b) A Concave mirror as the object distance is decreased ?
18. Parallel light rays from a distant star are incident on a concave mirror with a radius of
curvature of 120 cm .How far from the mirror will the star’s image be formed ?
19. An object is placed 18 cm from a concave mirror. An image that is twice the size of the
object is formed .Determine the image distance and the focal length of the mirror
20. A Converging lens forms an upright image that is four times the size of the object
.Given that the focal length of the lens is 20 cm , determine the object distance
21. The lens of a slide projector focuses an image of height 1.5 m on a screen placed
9.0 m from the projector. If the height of the picture on the slide is 6.5 cm ,
determine
(a) The distance between the slide (picture) and the lens
(b) The focal length of the lens
22. An object 2 cm high is placed 9 cm from a convex lens of focal length 6 cm
.Determine the position and nature of the image formed
23. A rectangular glass block 5 cm thick is placed on top of the page of a book .If
the refractive index of the glass block is 1.53, calculate apparent depth of the
letters on the book
24. A ray of light is incident at an angle of 600 on a block of glass of refractive index
1.5 .Determine the angle of refraction of the ray
25. A small coin was placed at the bottom of a tall glass containing some water and
viewed from above .The real and apparent depths of the coin were then
measured .By varying the depth of the water in the jar ,the following readings
were obtained

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 Real depth(cm) 8.1 12.0 16.0 20.0
Apparent depth (cm) 5.9 9.0 12.0 15.1

By plotting an appropriate graph from the results, determine the refractive index
of the water

26. The refractive index of water is 1.33 and that of glass is 1.5 .Calculate the critical
angle for: (a) a glass – air interface (b) a water – air interface
27. A pin at the bottom of a basin full of water appears to be 6 cm from the surface
.Given that the refractive index of the water is 4/3, what is the actual distance of
the pin from the surface?
28. Given that the refractive index of water is 4/3, what is the angle of refraction of
the ray of light?
29. Paraffin has greater refractive index than water
(a) What information does the above statement give with regard to the relative
velocities of light in paraffin an in water
(b) Draw a diagram to demonstrate the path of a ray of light when passing from
water into a layer of paraffin oil floating on top of it
30. When an object is placed 25 cm from a convex lens , an inverted image which is
twice as large as the object is formed .How far from the lens must the object be
placed to obtain an image four times the size of the object ?
31. An object 5 cm high is placed 25 cm from a convex lens with a focal length of 20
cm .Using the lens formula, determine position, size and nature of the image formed
32. An object 20cm high is placed 40cm from a concave mirror of focal length 15 cm.
determine the position, nature and size of the image formed by drawing a ray diagram
33. A ray of light strikes a rectangular glass block at an angle of 45 0 to the surface of
the glass .Given that the refractive index of the glass with respect to air is 1.5.
Determine the angle of refraction
34. A ray of light is shone through a rectangular glass prism at an angle of 55 0 to the
air glass interface as shown in the figure below

The glass block is 12 cm long and 10 cm wide .Calculate the distance the ray of
light travels through the glass before emerging into the air (𝑎𝜇𝑔 = 1.5)

35. Taking the refractive index of glass is 3/2, What is the critical angle?
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 36. Define angle of incidence and angle of refraction .State the laws of refraction of light
37. What is meant by the refractive index of a substance? If the velocity of light in a
vacuum is 3.0 x 108 m/s ,find the velocity of light in crown glass of refractive index 1.52
38. Distinguish between
(a) Converging and diverging lenses (b) Real and Virtual images. Draw
two diagrams ,one showing a converging lens producing a real image and the
other showing the same lens producing a virtual image
39. A 4.0 cm bulb tall light bulb is placed a distance of 8.3 cm from a concave mirror
having a focal lens of 15.2 cm .Determine the image distance and the image size
.What additional information do the answers give?
40. An object is at a distance of 30 cm from a convex lens of focal length 10
cm .Find by graphical method the position and nature of the image formed
41. An object is placed 20 cm from (a) convex lens (b) Concave lens of focal length 16
cm .Find the position, nature and linear magnification of the image produced
42. List out the factors on which the refractive index of a medium depends
43. What is meant by the refraction of light? Define incident, refracted and emergent
rays of light
44. A glass prism has three sides of angle 600 .A ray of light falls on one of the
faces and the angle of incidence is 48 0 .The ray is refracted and now travels
parallel to the second face. When it reaches the third face it is again refracted
and emerges from the prism .Find
(a) The refractive index of the glass prism
(b) The angle between the ray entering the prism and the ray leaving the prism
45. A glass prism has two parallel sides which are 6 cm apart .A ray strikes one of the two
parallel sides at an angle of incidence of 50 0 .Find by drawing the perpendicular
distance between the ray entering the prism and the ray leaving the prism
46. A Swimming pool is 2 m deep .Given that 𝝁water = 1.33. How deep does it appear
to be when (a) Completely filled with water (b) Filled halfway with water
47. Give scientific reasons for the following observations:
(a) A pencil dipped obliquely into water appears to be bent at the point
where it enters the water
(b) A light ray passing from air to glass bends closer to the normal
(c) The speed of light in diamond is less than the speed of light in ice
48. Explain the meaning of the following terms
(a) Refraction of light (b) Angle of incidence
(b) Angle of refraction (c) Refractive index
49. A ray of light is passing from air into water along PQ .The ray strikes the bottom
surface at T instead of R as shown in the figure below calculate
(a) the angle of incidence (b) the angle of refraction (c) the refractive index

Written by Geoffrey M Idebe (0688 082 089 – Mwanza) Page 43

 50. A ray of light passing from air into oil at an angle of incidence 300 .Calculate the
angle of refraction in oil if the velocity of light in air is 3.0 x 10 8 m/s and that in a
transparent oil is 2.2 x 108 m/s
51. The light ray passing from glass to air is monochromatic and has a frequency of
4 x 1014 Hz and a wavelength of 5 x 10-7 m in glass
(a) What is meant by monochromatic?
(b) Calculate the velocity of light in glass
(c) Calculate the velocity of light in air (refractive index of glass is 1.50)
52. In an attempt to determine the refractive index of a glass block , a student finds
the displacement produced due to refraction by glass as d and apparent
thickness of the block as y as shown in the figure below. Show that the
𝒅
refractive index of glass may be expressed as 𝒏 = (𝟏 + )
𝒚

53. A ray of light passes from a liquid to air. Calculate the critical angle for the liquid
– air interface ,if the velocity of light in the liquid is 2.4 x 10 8 m/s, while in air is
3.0 x 108 m/s
54. In a transparent liquid container, an air bubble appears to be 12 cm when viewed
from one side and 18 cm when viewed from the other side (see the figure
below).Where exactly is the air bubble, if the length of the tank is 40 cm?

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 55. In a fish aquarium (as shown in the figure below) the image of a fish seems to be 30
cm when seen from side A and 42 cm when seen from side B. Calculate the length of
the fish tank , if the refractive index of water is 1.33.

56. Calculate the critical angle of a material of refractive index 2

57. Sunlight making an angle of 600 with the horizontal enters a pool which is 50 cm
deep .Determine the distance travelled by the sunlight in the water (Refractive
index of water = 1.33)
58. An observer looks into a water tank half filled with water .If the height of the tank
is 180 cm. A solid that is 80 cm beneath the water surface is seen to be 60 cm
below the water surface. Determine
(a) The Refractive Index of water (b) the vertical displacement of the solid
59.

Written by Geoffrey M Idebe (0688 082 089 – Mwanza) Page 45

 TOPIC – 04: LIGHT – PART II
Refraction through Prism

 Prism is a solid piece of glass or transparent material that has at least two
planes inclined toward each other through which light is refracted.
Types of prism
 Rectangular prism
 Triangular prism
Rectangular prism
 Rectangular prisms are commonly called glass blocks

Triangular prism
 Is a wedge-shaped piece of glass material or any other transparent material. See
the fig. below

 Apex =The point at which two refracting surfaces of the prism meet is called the
refracting edge
 Triangular prism has two refracting surfaces.
 Rays leaving the prism is called emergent ray
 Rays entering the prism is called incident ray
 Apical angle is the angle between the refracting surfaces (See the fig below)

Written by Geoffrey M Idebe (0688 082 089 – Mwanza) Page 46

Angle of Deviation, D
 Is the angle formed by the intersection of the incident ray direction and the
emergent ray direction.(see the fig below)

 Consider the diagram below

From Snell’s Law

𝒔𝒊𝒏𝒊 𝒔𝒊𝒏𝒊𝟏
𝒂𝝁𝒈 = 𝒔𝒊𝒏𝒓 … … … … ..1 and 𝒈𝝁𝒂 = 𝒔𝒊𝒏𝒓𝟏 …………..2
Therefore: 𝒂 + 𝒃 = 𝑫…………….. (i) (sum of two opposite interior angles)
𝒃𝒖𝒕 … … . 𝒊 = 𝒂 + 𝒓 → 𝒂= 𝒊−𝒓 𝒂𝒏𝒅 𝒓𝟏 = 𝒃 + 𝒊𝟏 → 𝒃 = 𝒓𝟏 − 𝒊𝟏
𝑡ℎ𝑒𝑛, 𝑫 = 𝒂 + 𝒃 = (𝒊 − 𝒓) + (𝒓𝟏 − 𝒊𝟏 )…………...(ii)
 Since: A and A1 is supplementary to each other → A + A1 = 180° ……(ii)
But: r + i1 + A1 = 180°…………………(iii) (sum of interior angles of a ∆)
 Compare equation (ii) and (iii)
r + i 1 + A1 = A + A 1 → 𝑨 = 𝒓 + 𝒊𝟏 → 𝒓 = 𝑨 − 𝒊𝟏 ………(iv)
 Then substitute, equation (iv) into (ii)
𝑫 = (𝒊 − 𝒓) + (𝒓𝟏 − 𝒊𝟏 ) → 𝑫 = 𝒊 − ( 𝑨 − 𝒊𝟏 )+ (𝒓𝟏 − 𝒊𝟏 )
𝑫 = 𝒊 − 𝑨 + 𝒊𝟏 + 𝒓𝟏 − 𝒊𝟏 = 𝒊+ 𝒓𝟏 − 𝑨

∴ The angle of Deviation is given by: 𝑫 = 𝒊+ 𝒓𝟏 − 𝑨

The angle of deviation depends on

 The apical angle of the prism, A
 The Angle of incidence, i
 The refractive index of the glass prism.
Written by Geoffrey M Idebe (0688 082 089 – Mwanza) Page 47
NB:
Angle of deviation decreases with an increase in the angle of incidence and vice versa

Minimum Angle of Deviation, 𝑫𝒎𝒊𝒏

 Is the deviation angle occurs when the emergent ray is refracted at an angle
equal to the angle of incidence. (Consider the fig below)

𝑺𝒊𝒏𝒄𝒆: 𝒓 = 𝒊𝟏 , 𝒊 = 𝒓𝟏 and 𝐷 = 𝑫𝒎𝒊𝒏

Substitute: 𝒓 = 𝒊𝟏 , 𝒊 = 𝒓𝟏 and 𝐷 = 𝑫𝒎𝒊𝒏 𝑖𝑛 𝑡ℎ𝑒 𝑒𝑞𝑛 𝑫 = 𝒊 + 𝒓𝟏 − 𝑨
𝑫 = 𝒊 + 𝒊 − 𝑨 = 𝟐𝒊 − 𝑨, 𝒃𝒖𝒕 𝑨 = 𝒓 + 𝒊𝟏 = 𝟐𝒓 → 𝑫 = 𝟐𝒊 − 𝟐𝒓
𝑫𝒎𝒊𝒏
𝒊𝒏 𝒎𝒂𝒌𝒊𝒏𝒈 𝒊 𝒕𝒉𝒆 𝒔𝒖𝒃𝒋𝒆𝒄𝒕, 𝒕𝒉𝒆𝒏 𝒊 = ( ) + 𝒓
𝟐
𝑨 𝑫𝒎𝒊𝒏 𝑨 𝑫𝒎𝒊𝒏 +𝑨
𝒃𝒖𝒕 𝒓 = → 𝒊=( ) + =
𝟐 𝟐 𝟐 𝟐
From: Snell’s law
𝑫𝒎𝒊𝒏+𝑨
𝒔𝒊𝒏𝒊 𝑺𝒊𝒏( )
𝟐
𝒂𝝁𝒈 = = 𝑨
𝒔𝒊𝒏𝒓 𝑺𝒊𝒏( )
𝟐

 Consider the graph below shows the relationship between i and D, for
which the angle of deviation is obtained

Written by Geoffrey M Idebe (0688 082 089 – Mwanza) Page 48

Dispersion of white light
 Dispersion of white light is the splitting of white light beam into its component colours.
 This band of colours produced is called Spectrum.
 The process of splitting white light into its component colors is called Dispersion.
Diagram:

Types of white Colour

 Polychromatic light
 Monochromatic light
Polychromatic Light
 Is the colour which consists of more than one color. Example, sunlight
Monochromatic Light
 Is the light which consists of only one color. Example, red colour
NB:
 Spectrum of colour (red, orange, yellow, green, blue, indigo and violet) is
abbreviated as ROYGBIV .And each of these colours has a different wavelength.
 These coloured lights are refracted differently on passing through the prism
 The velocity of light in a medium (refractive index) depends on the wavelength of
incident light. As a result different wavelengths are refracted by different amounts.
 White colour split due to difference in wave length. Shorter wavelengths have higher
refractive indices and get bent more than longer wavelengths
Wavelengths of the Colours of White Light
S/N Colour Wavelength (nm)
1 Red 625 - 740
2 Orange 590 - 625
3 Yellow 565 - 590
4 Green 520 - 565
5 Blue 440 - 520
6 Indigo 420 - 440
7 Violet 380 - 420
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Types of Spectra
 Pure Spectra
Pure spectrum is the one in which the colours are clearly separated from each other
 Impure Spectra
Impure spectrum is the one in which the colours not clearly separated from each other
Recombining Colours of Light
 Spectrum comes from white light can recombine to form white colour.
 Also Newton’s colour disc is used to recombine the colours.
 The disc consists of sectors painted with different colors of the spectrum
 When the disk is allowed to spin about its axis at very high speed, all the colors of
the spectrum recombine to form white light. When it is slowed down, the individual
colors of the spectrum are seen again

Newton’s color disk

The Rainbow
 The rainbow is a natural phenomenon of dispersion of sunlight by raindrop.
Formation of a Rainbow
 It is formed by dispersion of sunlight by drops of rain.
 Since water is denser than air the dispersion of sunlight on a drop of water is the
same as when it falls on a glass prism.
 The light is first refracted as it enters the surface of the raindrop, reflected off the
back of the drop and again refracted as it leaves the drop. (See the fig below)

Types of Rainbow
 primary rainbow
 secondary rainbow
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Primary Rainbow
 It is formed when light undergoes one total internal reflection (refracted twice and
reflected once) in the water drops.
 The violet colour is inside and the red in the outside the bow. It is formed
between 400 and 420 from anti-solar point
Anti-solar point Is a point that lies directly opposite the sun from the observer, that
is, on the line from the sun through the observer.
Secondary Rainbows
 It is formed when light undergoes two total internal reflections in the water drops.
The violet colour is outside and the red colour is inside the bow.

Colour
 Colour is the property of light that reaches our eyes.
Appearance of coloured objects under white light
 The object seems to have kind of colour due to the fact that it absorbs all colours
and reflect the colour that the object has.
Example
 Yellow flower is yellow because it absorbs all the other colours in the light and
reflects only yellow colour.
 Blue object absorbs the entire colour in white light except blue.
Appearance of white objects under coloured light
When a coloured object is viewed under a coloured light, it takes the colour of that light.
Example:
 The object will appear blue in blue light and red in red light.
 A colour filter is working on this principle.

Colour filters
 Are materials made of glass or celluloid that let through light of certain colours only.
 Example: Green filters allow green colour to pass through.
 The colour of an object depends on the colour of the light falling on it and the
colour(s) it absorbs or reflects.

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 Appearance of a white object to coloured light
White object coloured light Colour of object
White object Red filters Red colour
White object Yellow filters Yellow colour
White object Green filter Green colour
White object Blue filter Blue colour
Types of Colour
 Primary colour
 Secondary colour
Primary colour
 Primary colour is a colour that cannot be created by mixing other colours.
Example, red, blue and green
Secondary colour
 Secondary colour is a colour created by mixing other colours.
 Example, cyan, magenta and yellow
Complementary Colour
 Is the colour that when mixed in a definite ratio produce white (required) colour.
Consider the figure below

NB:
 From the figure above
 When mixing the primary colors (red, blue and green), a white color is obtained
That is: Red + Blue + Green = White
 When mixing two of the primary colors produces a secondary color .That is
(a) Green + Blue = Cyan (GBC)
(b) Green + Red = Yellow (GRY)
(c) Red + Blue = Magenta (RBM)
 When mixing two of the secondary colors, white light is produced
That is: Yellow + Magenta = Magenta + Cyan = Cyan + Yellow = White color
 We are only concerned with colours of light and not with coloured substances
(pigments)
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 The complementary colour of white light is green, red and blue
 The complementary colour of yellow light is green and red.

Additive and subtractive mixing of colours

Additive mixing of Colours
 Is the mixing of coloured light
 The more colours you add, the closer the result draws to white. Therefore,
mixing of coloured lights
 Additive mixing of colours deals with primary colour of light colour (Red, Green
and Blue) not primary colour of pigments (paints and dyes)
 Adding different colours of light together increases the number of wavelengths
present

Subtractive mixing of pigments

 Is the mixing of colours of different paints (pigments)
 Pigments: These are substances which give color to paints by reflecting light of
certain colors only and absorbing all other colors.
 Mixing different paints results in a darker colour because most of the light
wavelengths are absorbed
 Mixing two primary pigments produces a secondary pigment .That is
a) cyan + magenta = blue (CMB)
b) magenta + yellow = red (MYR)
c) yellow + cyan = green (CYG)
 The pigments act as filters that subtract one or more colours from the visible spectrum
 Blue, Red and Green are, therefore, the secondary pigments.
 Subtractive complementary colours combine to produce BLACK.
That is Blue + Green + Red = Black (See the fig below)

N B:
 Each primary pigment absorbs one primary colour:
 Yellow absorbs blue and reflects red and green
 Magenta absorbs green and reflects blue and red
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  Cyan absorbs red and reflects green and blue.
 A secondary pigment absorbs two primary colours and reflects one:
 Red absorbs green and blue and reflects red.
 Green absorbs red and blue and reflects green.
 Blue absorbs - red and green and reflects blue.
 The primary pigments are the secondary colours of light and the secondary
pigments are the primary colours of light.
 If you mix a primary pigment with a secondary pigment you get total absorption
(black). ie Blue + Yellow = Black
 The primary pigments are the complementary colours of the three primary colours
of light.

Example:
1. A plant with green leaves and red flowers is placed in:
(a) Green (b) Red (c) Blue light
ANS:
(a) In Green light (b) In Red light (c) In Blue light
 Green leaves  In red light Green  In blue light Green
will appear leaves will leaves will appear
green appear black black
 Red flowers will  Red flowers will  Red flowers will
appear black appear red appear black

2. Why red light is used for danger signals?

ANS: Because red light is scattered the least by air molecules due to its highest
wavelength so it is able to travel the longest distance through fog, rain and alike
3. A red bus with blue letter on its stops in front of a yellow light at right. Describe
the appearance of the bus
ANS: In yellow light
 The red bus will appear red because yellow is composed of green and red
 The blue letters will appear black

Class Activity – 4
1 Explain what is meant by a spectrum .Describe with the aid of a diagram
how would you obtain the spectrum of white light
2 Distinguish between a pure and an impure spectrum. Explain with the aid of a
diagram how a pure spectrum can be produced in the laboratory .How are the
colours of the spectrum recombined?
3 A beam of white light is allowed into a dark room through a hole .In the dark
room ,the beam falls on a white screen .How will the screen appear if:
Written by Geoffrey M Idebe (0688 082 089 – Mwanza) Page 54
 (a) a piece of red glass is placed in the path of the beam
(b) a piece of green glass is placed between the screen and the red glass?
4 Explain, giving examples , what is meant by:
(a) additive combination of colours (b) subtractive combination of colours
5 Why does an object appear coloured when light falls onto it?
6 Explain, giving examples, what is meant by primary, secondary and
complementary colours.
A flag has a green disc on a yellow background .How will the flag appear in:
(a) green light (b) red light (c) blue light
7 State , giving a reason for each answer, the colour that results when:
(a) a blue light and a yellow light are mixed (b) blue paint and yellow paint are mixed
4. What color would be seen if white light is viewed through:
(a) a red filter? (b) a cyan filter? (c) an orange filter?
5. A book which looks red in white light is viewed in magenta light. In what color
does it appear?
6. White light is viewed through a combination of a yellow filter and a red
filter held in contact. What color is seen
7. Explain the meaning of the term ‘’dispersion’’ of white
8. Explain why the result of mixing blue and yellow paints is very different from that
of mixing blue and yellow lights
9. A painter has a blue – green (cyan) paint which she wants to make pure green.
What color pigment should she add to the paint? Explain your answer
10. Briefly explain, why the sky looks blue?
11. Under pure yellow light what will be the appearance of the blue piece of cloth?
12. Danger signs along the road as well as tail and brake lamps of motor vehicles
rear are painted red. Briefly explain the reason behind.
13. Explain each of the following:-
(a) The appearance of a blue flag when viewed in day light through a sheet of
red glass
(b) The appearance of a red flag with green stripes when viewed in day light
through a sheet of green glass
(c) The appearance of a man wearing blue shirt and red trousers, holding a
handkerchief of green color when viewed in the pure yellow light

Written by Geoffrey M Idebe (0688 082 089 – Mwanza) Page 55

 TOPIC: 05 OPTICAL INSTRUMENTS
 Optical instruments are devices which are used to help the human eye views
small or distant objects more clearly.
 They use a combination of lenses (mirrors) to produce an enhanced image of an object.

Optical instruments consist of:

 Simple Microscope
 Compound Microscope
 Astronomical Telescopes
 Simple lens Camera
 Projection lantern
Simple Microscope
 It consists of a biconvex lens which may be hand-held or placed in a simple frame. It
is sometimes referred to as a magnifying glass.

 Object is placed between f and lens ( u < f )

 Virtual, upright and magnified image of the object is formed.
 The image appears clearest when it is about 25 cm from the eye, V = 25cm. This
distance is called the near point (D).
 The nearer the object is to the lens, the further and larger the image formed.

Magnification
Consider the diagram below

Where
A = angle subtended by the object
B = angle subtended by the virtual image
D = distance between F and image
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Lateral Magnification

 The lateral magnification of a simple lens is the ratio of the image height (IH) to
the object height (OH).
𝐇𝐈
𝑴 =
𝐇𝐎
Linear Magnification
 Is also given by the ratio of the image distance v to the object distance u.
𝑽
𝑴 =𝑼

Angular Magnification (Magnifying power)

 Is the ratio of the angle subtended at the eye by the object when viewed through
the magnifying glass (B) to the angle subtended at the eye by the object when
viewed with naked eyes (A)
𝑩
𝒎 =𝑨

From the diagram above

𝑰𝑯
Tan B = 𝑫
Since: angle B is small, it can be expressed as B = IH/D

 If the angle B subtended by the virtual image. Ignoring the small distance
between the eye and the magnifying lens
𝑰𝑯 𝑶𝑯
𝑩 = =
𝑽 𝑽
𝑂𝐻
𝐵 = …………….. 1
𝑉

Then: Tan A = OH/D

𝑂𝐻
Since: angle B is small, it can be expressed as A= 𝑢
…………… 2

Substitute equation 1 and 2 in m = B/A

𝑶𝑯
𝑼 𝑂𝐻 𝑉
𝑴 = 𝑶𝑯 𝑀 = ( )𝑥 ( )
𝑈 𝑂𝐻
𝑽
𝑽
∴𝑴 = 𝑼
When the image is at the near point, v = 25 cm. Therefore, using the lens formula,
𝟏 𝟏 𝟏 𝟐𝟓𝒇
= + … (make u the subject) 𝑼 = ….. substitute into M = V/U
𝒇 𝒖 𝟐𝟓 𝟐𝟓+𝒇
M = v/(25f/(25+f))

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 𝟐𝟓
Therefore: 𝑴 = (𝒇) + 𝟏

Individual task – 5:1

1. A simple microscope with a focal length of 5 cm is used to read division of scale
1.5 mm in size. How large will the size of the divisions as seen through the
simple microscope be? (ANS: 9 mm when viewed through the simple microscope.)
Uses
 It is used to view specimen in the laboratory
 It is used to read small print
Compound Microscope
 A Compound microscope is composed of two convex lenses of short focal
lengths placed in a tube.
 The two lenses are separated by a certain fixed distance.
 The lens near the object is called the Objective Lens while the lens near the
eye is called the Eyepiece Lens
See the figure below

Mode of Action
 Objective lens enlarges object to form image, I1
 Eyepiece lens enlarges I1 (As object) to form image I2. The image produced is
magnified, virtual and inverted compared to the original object
 In order I2 to be seen clearly, the eyepiece lens should be adjusted until I2 is at
the near point ,D , from the eye as shown from the figure above

Magnification produced by a compound microscope

 Where V is the distance of I1 and U is the distance of object(O) from the objective lens
1 1 1 𝑣 𝑣
But: = + … … . . 𝑚𝑢𝑙𝑡𝑖𝑝𝑙𝑦 𝑏𝑦 𝒗 𝑒𝑎𝑐ℎ 𝑠𝑖𝑑𝑒 → = +1
𝑓𝑜 𝑢 𝑣 𝑓𝑜 𝑢

𝒗 𝒗 𝒗 𝒗 𝒗
= + 𝟏, 𝒔𝒊𝒏𝒄𝒆 𝒎 = , 𝒕𝒉𝒆𝒏 = 𝒎𝒐 + 𝟏 → 𝒎𝒐 = − 𝟏 ………(i)
𝒇𝒐 𝒖 𝒖 𝒇𝒐 𝒇𝒐

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  Also, the magnification produced by the eyepiece lens is given by,
𝑫
𝒎𝒆 = 𝒇 − 𝟏 ………… (ii)
𝒆

 Where D is the distance of I2 from the eyepiece

 The total magnification (m) produced by a compound microscope is the
product of the magnification produced by the objective and the magnification
produced by the eyepiece. (from equation i and ii), i.e

𝒗 𝑫
∴ 𝒎 = 𝒎 𝒐 𝒙 𝒎 𝒆 = 𝒎 𝒐 𝒎𝒆 = ( − 𝟏) ( − 𝟏)
𝒇𝒐 𝒇𝒆

Example
1. A compound microscope has an objective lens of focal length 2 cm and eye
piece of focal length of 6 cm. An object is placed 2.4 cm from the objective lens.
If the distance between the objective lens and the eyepiece lens is 17 cm find:-
(a) The distance of the final image from the eyepiece.
(b) The linear magnification.
Soln:
Given: fO = 2 cm, fE = 6 cm, uo = 2.4 cm, L = 17 cm
(a) Consider the figure below:

Refraction from the objective lens:

𝟏 𝟏 𝟏 𝟏 𝟏 𝟏 𝟏 𝟏 𝟏 𝟐.𝟒−𝟐 𝟎.𝟒 𝟏
From: = + → = + → = − = = =
𝐟𝐨 𝐮 𝐯 𝟐 𝟐.𝟒 𝐯 𝐯 𝟐 𝟐.𝟒 𝟒.𝟖 𝟒.𝟖 𝟏𝟐

VO = 12 cm, but VO + UE = L → Ue = L – VO = 17 cm – 12 cm = 5 cm
Also consider the refraction from the eyepiece lens
𝟏 𝟏 𝟏 𝟏 𝟏 𝟏 𝟏 1 1 −1
From: 𝐟 = 𝐮 + 𝐕 , but v = D → 𝟔 =𝟓+𝐃 →𝐃 =6−5= 30
𝐞

D = -30 cm, (According to real – is – positive the final image is virtual)

∴ The distance of the final image from the eyepiece is 30 cm

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 (b) Linear magnification (total magnification)
𝒗 𝑫
From: 𝒎 = 𝒎𝒐 𝒎𝒆 = (𝒇 − 𝟏) (𝒇 − 𝟏)
𝒐 𝒆
𝐯 𝐃 𝟏𝟐 𝟑𝟎
𝐦 = ( − 𝟏) ( − 𝟏) = ( − 𝟏) ( − 𝟏) = (𝟔 − 𝟏)(𝟓 − 𝟏) = 𝟓 𝐱 𝟒 = 𝟐𝟎
𝐟𝐨 𝐟𝐞 𝟐 𝟔

∴ 𝒍𝒊𝒏𝒆𝒂𝒓 𝒎𝒂𝒈𝒏𝒊𝒇𝒊𝒄𝒂𝒕𝒊𝒐𝒏 𝒊𝒔 𝟐𝟎

Individual Task – 5:2

1. A compound microscope consists of two lenses of focal length 12 cm and 6 cm
for the objective lens and the eyepiece lens, respectively. The two lenses are
separated by a distance of 30 cm. The microscope is focused so that the image
is formed at infinity. Determine the position of the object. ANS: U = 24 cm
2. A parallel beam of light falls on a converging lens arranged so that the axis lies
along the direction of the light which is brought to focus 25 cm from the lens. The
light then passes through a second converging lens of focal length 7.5 cm placed
at 30 cm from the first lens. Calculate the position of the final image. Draw a ray
diagram to show the final image formed.(ANS: D = 15 cm)
3. The total magnification produced by a compound microscope is 20. The
magnification produced by the eye piece alone is 5. The microscope is focused
on a certain object. The distance between the objective lens and the eyepiece is
observed to be 14 cm. If least distance of distinct vision is 20 cm, calculate the
focal lengths of the objective and eyepiece lenses.
Uses
 Observing Brownian motion in science
 Studying microorganisms and cells in biology
 Checking for infections caused by microorganisms in hospitals

Astronomical Telescope
 It uses two convex lenses that is the objective lens and the eyepiece lens.
 The objective lens has a large focal length while the eyepiece lens has a much
shorter focal length as you compare to compound microscope
Mode of action
The objective lens forms a real, inverted and diminished image of a distant
object at its focal point, fo.
This becomes the object for the eyepiece lens.
The position of the eyepiece lens is adjusted until the object is at its focal point, fe.
This adjustment makes the final image to be formed at infinity.

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NB:
 The final image obtained in the astronomical telescope is small compared to the
original object.
 The image looks larger because it is very much closer to the observer’s eye.

Magnification produced by astronomical telescope

From: Angular magnification of the telescope
𝒂𝒏𝒈𝒍𝒆 𝒔𝒖𝒃𝒕𝒆𝒏𝒅𝒆𝒅 𝒂𝒕 𝒕𝒉𝒆 𝒆𝒚𝒆 𝒃𝒚 𝒕𝒉𝒆 𝒊𝒎𝒂𝒈𝒆(𝑩) 𝑩
𝒎= =𝑨
𝒂𝒏𝒈𝒍𝒆 𝒔𝒖𝒃𝒕𝒆𝒏𝒅𝒆𝒅 𝒂𝒕 𝒕𝒉𝒆 𝒆𝒚𝒆 𝒃𝒚 𝒕𝒉𝒆 𝒐𝒃𝒋𝒆𝒄𝒕(𝑨)

 Since both the object and the final image are at infinity, the angles they
subtend at the eye are the same as those they subtend at the objective and at
the eyepiece lens, respectively.
 Assuming that: angle A and B are very small, then tan A = A and tan B = B
𝐷𝐼 ℎ 𝑫𝑰 𝒉
Hence 𝐵 = 𝐶 ′ 𝐼 = 𝑓 𝑎𝑛𝑑 𝑨 = 𝑪𝑰
=𝑓
𝑒 𝑜

𝑩
Since: 𝒎 = 𝑨
……………… substitute A and B above
𝒉
𝒇𝒆 𝒇𝒐
𝒎 = 𝒉 → 𝒎=
𝒇𝒆
𝒇𝒐
 Therefore the magnification produced by an astronomical telescope is the ratio
of the focal length of the objective lens to that of the eyepiece lens, i.e.
𝒇𝒐
∴𝒎 =
𝒇𝒆

Example 1:
A refracting telescope has an objective lens with a focal length of 5.0 cm and an
eyepiece with a short focal length of 0.02 m; calculate the magnifying power of such a
telescope in its normal adjustment
Soln:
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Given: fo = 5.0 cm, fe = 0.02 m
𝒇𝒐 𝟓
From: 𝒎 = →𝒎= = 𝟐𝟓𝟎
𝒇𝒆 𝟎.𝟎𝟐
Therefore the magnifying power is 250

Individual task –5:3

1. What is the angular magnification of a telescope that has a 100 cm focal length
objective and a 2.5 cm focal length eyepiece? (ANS: 40)
2. Find the distance between the objective and eyepiece lenses in the telescope in
the above problem needed to produce a final image very far (at infinity) from the
observer, (ANS: 102.5 cm)
3. A large reflecting telescope has an objective mirror with a 10.0 m radius of
curvature. What angular magnification does it produce when a 3.0 m focal length
eyepiece is used?
4. A small telescope has a concave mirror with a 2.0 m radius of curvature for its
objective. Its eyepiece is a 4.0 cm focal length. What is the telescope’s angular
magnification? What angle is subtended by a 25,000 km diameter sunspot?
5. A 7.5x binocular produces an angular magnification of 7.5, acting like a
telescope. If the binoculars have objective lenses with a 75.0 cm focal length,
what is the focal length of the eyepiece lenses?
Uses
 An astronomical telescope is used to view distant objects like stars and other
objects in space.
 They are used in military bases to see enemies

Differences between Compound microscope and Astronomical telescope

Compound microscope Astronomical telescope
Objective lens has a smaller focal length Objective lens has a larger focal length
than the eyepiece lens than the eyepiece lens
Distance between the objective lens and Distance between the objective lens and
the eye piece lens is greater than fo + fe eye piece lens is equal to fo + fe
It is used to see very small objects It is used to see distant astronomical objects
It uses artificial light It uses natural light at the focal point
It is very small in shape as compared to Very large in size as compared to
telescope microscope

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Projection Lantern
 Projection lanterns are used to display a large image on a screen.
 One example, is the slide projector that is the optical inverse of a camera.

Magnification
𝑽 𝑯𝑰
 It is given by 𝒎 = 𝑼 = 𝑯𝑶

Example:
1. A projection lantern is used to give the image of a slide on a screen. If the image
is 24 times as large as the slide and the screen is 72 m from the projecting lens,
what is the position of the slide from the lens
Soln:
Given: m = 24, v = 72 m, u =?
𝑽
From: 𝒎 = 𝑼
𝑽 72 72
𝒎=𝑼 → 24 = → 𝑢 = 24 = 3𝑚
𝑢
∴ 𝑡ℎ𝑒 𝑠𝑙𝑖𝑑𝑒 𝑖𝑠 3𝑚 𝑓𝑟𝑜𝑚 𝑡ℎ𝑒 𝑙𝑒𝑛𝑠
2. A lantern projector using a slide of 2cm x 2cm projects a picture 1m x 1m onto a
screen 12m from the projection lens. How far from the lens must the slide be?
Find the approximate focal length of the projection lens.
Soln:
Given: HO = 2 cm, HI = 1 m = 100 cm, v = 12 m =1200cm, u =?, f =?
𝑽 𝑯𝑰
From: 𝒎 = 𝑼 = 𝑯𝑶
𝑽 𝑯𝑰 𝑽 𝑯𝑰 𝟏𝟐𝟎𝟎 100
𝒎 = 𝑼 = 𝑯𝑶 → 𝑼 = 𝑯𝑶 → u
= 2
→ 𝑢 = 24 𝑐𝑚
𝟏 𝟏 𝟏 1 𝟏 𝟏 𝟓𝟎+𝟏 51 𝟏𝟕
Also: 𝒇
= 𝒖+𝒗 → 𝑓 = 𝟐𝟒 + 𝟏𝟐𝟎𝟎 = 𝟏𝟐𝟎𝟎
= 1200 = 𝟒𝟎𝟎

𝟏 𝟏𝟕 𝟒𝟎𝟎
= → 𝒇= = 𝟐𝟑. 𝟓 𝒄𝒎
𝒇 𝟒𝟎𝟎 𝟏𝟕

∴ 𝒕𝒉𝒆 𝒔𝒍𝒊𝒅𝒆 𝒎𝒖𝒔𝒕 𝒃𝒆 𝟐𝟒 𝒄𝒎 from the lens and the approximate focal length is 23.5 cm

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Individual task – 5:4
1. A projection lantern is used to project a slide measuring 3 cm x 3 cm onto a
screen 12 m from the projection lens. If the size of the screen is 1.5 m x 1.5 m,
how far from the lens must the slide be for the image to fill the entire screen?
ANS: U = 24 cm.

Uses of Projection Lantern

 Projection of films, slides and transparencies.
 Projection of opaque objects, i.e. episcopic projection.
 In searchlights and headlights.
 In physical experiments such as projection of the spectrum, polarisation
experiments and interference experiments.
 Projection of minute objects, i.e. the projection microscope.
Lens Camera
 The lens is the image-forming device on a camera.

Basic types of Lenses

 Normal (standard lens)
 wide angle lens
 telephoto (long-focus lens)

Normal (standard) Lens

 The viewing is much wider-about 50 degrees.
 The objects appear normal in size and shape, relative to the picture background
Wide Angle Lens
 The viewing is much wider-about 90 degrees.
 Used to make smaller objects look larger or to photograph large objects from close up.

Telephoto Lens
 It has wider fields of view than normal lenses.
 They show an enlarged detail of the image over the same film area.

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Parts of the Lens Camera
Diaphragm
 The diaphragm determines the amount of light that passes through the lens by
changing the size of the aperture.
NB: Aperture is an opening whose diameter is adjustable

Shutter
 The shutter is a mechanical device that acts as a gate, controlling the duration of
time that light is allowed to pass through the lens and fall on the film.

Viewfinder
 The viewfinder defines the area covered by the lens that is in use on the camera.
Film
The film is a light-sensitive surface of the camera.

Mode of Action
 The image of the object must be sharply focused on the film by adjusting the
distance of the lens from the film
 After focusing and correctly setting the aperture size and shutter time ,the click
button is pressed
 The shutter opens to allow light to enter and expose the film to form an image of the
object being photographed. The film is then developed to produce a photograph of
the object

Magnification
Since magnification is given by
𝑽
𝒎 = 𝑼 ………………make v the subject
𝟏 𝟏 𝟏
v = m u , then Substitute v = mu into = +
𝒇 𝒖 𝒗
𝟏 𝟏 𝟏 𝟏 𝒎+𝟏
𝒇
= 𝒖 + 𝒎𝒖 → 𝒇
= 𝒎𝒖
(reciprocate the two sides and make m the subject)
𝒇
∴𝒎 =
𝒖−𝒇
Example
1. A lens camera of focal length 15 cm is used to take a picture of a man of height
1.8 m. If the man is standing 10 m ahead of the camera. Determine the:
(a) Magnification of the image (b) size of the image
SOLN:
Given: f = 15 cm = 0.15 m, HO = 1.8 m, u = 10 m, m =?, HI= ?
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 𝒇 0.15
(a) From: 𝒎 = 𝒖−𝒇 →𝑚=
10−0.15
= 𝟎. 𝟎𝟏𝟓𝟐

𝑯𝑰 𝑯𝑰
(b) Also: 𝒎 = → 𝟎. 152 = → 𝐻𝐼 = 1.8 x 0.0152 = 0.02736 cm
𝑯𝒐 𝟏.𝟖

∴ height of the image formed = 2.7 cm

NB.
 Optometrists and ophthalmologists usually prescribe lenses measured in ‘’diopters’’
 The power , P, of a lens in diopters equals the inverse of the focal length in metres
𝟏
That is : 𝑷𝒐𝒘𝒆𝒓 = 𝑭𝒐𝒄𝒂𝒍 𝒍𝒆𝒏𝒈𝒕𝒉 (𝒎)

Individual Task – 5:5

1. A lens camera of focal length 10 cm is used to take the picture of a girl 1.5 m
tall. Determine the magnification of the image if the girl is 11m from the
camera.( m =0.009)
2. A lens camera has a lens of focal length 15 cm and a film (screen) of
height 0.35 cm. How far would a boy of height 1.8 m stand from the
camera so that his image just fits the film?
3. A professional photograph has a camera of focal length 2.5 cm. He uses it
to take a photo of a tree of height 60 m. The distance between the lens of
the camera and its film is 2.5 cm. Determine:
(a) Distance between the lens and the tree
(b) The height of the image (c) The magnification of the camera

Uses of Lens Camera

 The sine or video camera is used to take motion pictures.
 High-speed cameras used to record movement of particles.
 Closed-circuit television cameras are used for surveillance in high-security
 Digital cameras are used to capture images

The Human Eye

 Is an optical device able to respond to an enormous range of light brightness
 It is able to focus on objects from billions of kilometers away to those a few
centimeters away. It can also detects colour. (See the fig. below)

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Parts of the human Eye
5. Cornea
 The cornea is the transparent, outer part of the eye. It is the primary focusing
tool of the eye. The outer layer of the cornea is known as the epithelium.
 Function of cornea: Its main job is to protect the eye
6. . Iris. The iris is the part of the eye which is responsible for one’s eye colour.
 Function of Iris: It acts like diaphragm of the camera, dilating and constricting
the pupil to allow more or less light into the eye.
3. Pupil
 The pupil is the dark opening in the centre of the colored iris that controls the
amount of light that enters the eye.
 The pupil functions in the same way as the aperture of a camera.
The size of the pupil determines the amount of light entering the eye.
4. Lens. The lens is the part of the eye immediately behind the iris.
 Function of lens: To focus light rays on the retina.
 In persons under 40 years of age, the lens is soft and flexible, allowing for fine
focusing from a wide variety of distance
5. Retina
 The retina is the membrane lining the back of the eye that contains
photoreceptor cells it reacts to the presence and intensity of light by sending an
impulse to the brain via the optic nerve.
 The retina compares to the film in a lens camera.
6. Optic nerve
 The optic nerve (million nerve fibres) is the structure which takes the information
from the retina as electrical signals to the brain
7. Sclera
 The sclera is the white, tough wall of the eye. Its function is to protects the eye
8. Vitreous humour
 The vitreous humour is a jelly-like substance that fills the body of the eye. It is normally
clear
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The Function of vitreous humour
 To maintain the eye pressure.
 It also helps in focusing light rays

Accommodation
 Accommodation is the process whereby an eye focal length adjusted to see
distant objects

Eye Defects
 Eye defect is the phenomenon where by eye defeat to see clearly.

There are two common eye defects.

 Myopia (short-sightedness defect)
 Hypermetropia (long-sightedness defect)
Myopia
 Occur when a person can see near objects clearly but cannot see distant objects
clearly.

Causes
 When the eyeball is too long
 When the refractive power of the lens is too strong

Correction
 To wear suitable concave lenses to diverge the rays from distant objects before
they reach the eye.
Diagram:

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  Hypermetropia (Long – sightedness )
Occur when a person cannot see near objects clearly but can see distant objects
clearly.

Causes
 When the eye ball is too short
 When ciliary muscles are weak such that unable to change the shape of the eye
lens in order to focus the image (occurs when the refractive power of the eye
lens is too weak)
Correction
 The defect can be corrected by wearing suitable convex lenses so that the rays from
the near object are made to converge and focus on the retina. See the fig below

Other eye defects can be categorized into:-

(a) Astigmatism
Occurs when the image is distorted because light rays are blocked from meeting
at a common focus.
Causes
when the cornea shape is like an oblong rugby ball and not spherical in shape as
normal
correction:
Can be corrected by using cylindrical lenses
(b) Presbyopia
Occurs when the centre of the eye lens hardens making it unable to
accommodate near vision
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correction:
Eye glasses with progressive lenses are prescribed to correct or improve the condition

Similarities of Human Eye and Lens Camera

 Both have a convex lens system to focus the image
 Both save images
 Both form a real, reduced and inverted image
 The amount of light entering is controlled by a variable aperture
 They both have surfaces on which the image is formed
 The retina behaves like the photographic screen of a camera

Differences between Lens Camera and Human Eye

Camera Eye
Lens is hard glass Lens is soft and elastic
Thickness of lens does not change Thickness of lens changes
Image is focused by moving the lens Image is focused by changing the thickness of the lens
Only the lens refracts the light Aqueous and vitreous humour refracts the light
Diaphragm can be altered Iris alters itself
Cameras do not have blind spot Human eyes have blind spot
Example
1. The far point of a myopic person is 40 cm. What should be the power of the lens
that he must use to see clearly?
Soln:
Given: f = 40 cm = 0.4 m, Power =?
1
From: 𝑃 = 𝐹𝑜𝑐𝑎𝑙 𝑙𝑒𝑛𝑔𝑡ℎ
𝟏 𝟏 𝟏𝟎
𝑷𝒐𝒘𝒆𝒓 = = = = 𝟐. 𝟓
𝑭𝒐𝒄𝒂𝒍 𝒍𝒆𝒏𝒈𝒕𝒉 𝟎.𝟒 𝟒

Class Activity – 5
1. If the focal distance of the converging lens is 5 cm the object distance is 4
cm. Find the magnification of the image (ANS: M = 6 cm)

2. Given that the focal length of the simple microscope is 12cm.Find the
magnification of the image of the object distances. (ANS: M = 8.1)
3. Given that an object 2m high is placed 2010cm in front of the lens camera of focal
length 10cm.calculate the minimum size of the film frame. (ANS: M = 0.0005)
4. The lantern projector uses a slide of 2 cm by 2 cm, 2x2 to produce a picture 1 m
by 1m on a screen 12cm from the projection lens. How far from the lens must
the slide be?

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 5. A telescope is consisting of two converging lens of lens at focal length 25 and 4
respectively. The final image is found at distinct vision that is 25 cm in front of the
eyepiece lens. Find the position of the first image from the eyepiece. ANS: Ue = 3.4cm
6. An astronomer telescope has its 2 lens 78 cm apart. If the objective lens has a
focal length of 75.5 cm, what is the magnification produced by the telescope under
normal vision ANS: The magnification is 30.2
7. A compound microscope has an objective lens of focal length 2 cm and eye piece
lens of focal length 6cm.An object is placed 2.4 cm from an objective lens. If the
distance between the objective lens and eye piece lens is 19 cm. Find
(a) The distance of the final image from the eyepiece lens
(b) Compound magnification ANS: The final image is at infinity and M = ∞
8. A certain microscope consists of 2 converging lenses of focal length 4 cm and 10
cm for objective 3 eyepiece lenses respectively. The 2 lenses are separated by the
distance of 30cm.The instrument is focused so that the image is at infinity.
Calculate the position of the object and the magnification of the objective lens.
(ANS: The magnification produced by the eye piece lens is 4)
9. A simple microscope has a focal length of 15 cm
(a)What is the maximum magnification of the lens (ANS: 2.667)
(b)What is the magnification of this lens when the eye is relaxed (ANS: 1.667)
10. The near point of a longsighted patient is 90 cm
(a) Determine the focal length of a lens that can be used to enable the
patient clearly see objects that are 25 cm from the eye
(b) What is the power of the lens
(c) What is the magnification of the lens?
11. A short sighted person is unable to clearly see objects that are beyond 150 cm
from the eye .Determine the focal length ,power and magnification of the lens that
should be used to detect the eye defect
12. A patient requires a lens of -5 diopters in order to see far away objects
clearly .Determine the
(a) Focal length of the lens used (b) Far point of the patient ‘s eye
13. Define the term ‘’accommodation’’ as used in the human eye
14. Give two similarities and 2 differences between the human eye and the camera
15. State one advantage of human eye over a lens camera
16. The near point of an eye is 50 cm
(a) What focal length lens should be used so that the eye can clearly see an
object 25 cm away? (ANS: f = 50 cm)
(b) What is the power of this lens? (ANS: 2 diopters)
17. (a) State the causes of the short – sightedness and long – sightedness
(b) Use a ray diagram to show how
i. Short – sightedness in a human eye can be corrected
ii. (ii) Long – sightedness can be corrected
18. A farsighted man has a near point of 100 cm .Wearing his glasses, he can see
objects that are 25 cm away .What is the focal length of the lens in his glasses?
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 19. In a compound microscope ,the focal length of the objective lens is 4.0 cm and
that of the eyepiece is 3.3 cm and they are placed at a distance of 15.0 cm .A real
object of size 2 mm is placed 6 cm from the objective lens .By using the lens
formula , Calculate
(a) Position of the final image (ANS: V = 12 cm) (b) The size of the final
image viewed by the eye (ANS: VFINAL IMAGE = 33 cm)
(c) Magnification produced by the arrangement of the lenses (ANS: HI = 44 mm)
20. (a) Differentiate between a simple microscope and a compound microscope
(b) With the aid of a diagram describe how a compound microscope works.
A compound microscope has an objective lens of focal length 2cm and an
eyepiece of focal length 6 cm. An object is placed 2.4 cm from the objective lens
.If the distance between the objective lens and the eyepiece is 17m ,find
(i) The distance of the final image from the eyepiece
(ii) The linear magnification
21. Small object is placed 3cm from the lens of a simple microscope .If the focal
length of the lens is 5 cm ,Find the linear magnification produced by the simple
microscope .How far from the lens would you place the object in order to obtain
maximum magnification of the image ?
22. A compound microscope expected to have a magnification of X600 has a
tube length of 12 cm and the focal length of the objective lens is 0.5 cm
Determine the expected focal length of the eye piece
23. A magnifying glass of focal length of 15 cm is used to view an object so as to
obtain maximum magnification
(a) Where should the object be placed?
(b) What is the magnification of the magnifying lens?
24. A compound microscope has an objective lens of focal length 25 cm and an
eyepiece with focal length 14 cm, If it has a tube of length 36 cm determine
the magnification of the microscope
25. A lens of focal lens 10 cm is positioned from an object so as to obtain maximum
magnification of the image .Determine the:
(a) Object distance (b) Image distance (c ) Magnification
26. A lens camera is to be used to take a picture of a man 2 m tall .If the lens of
the camera has a focal length of 10 cm , Calculate the minimum size of the
film frame required ,given that the man is 20.1 m from the camera
27. A certain microscope consists of two converging lenses of focal lengths 10 cm
and 4 cm for the objective and eye piece respectively .The two lenses are
separated by a distance of 30 cm .The instrument is focused so that the final
image is at infinity .Calculate the position of the and the magnification of the
objective lens (ANS u = 16.25 cm, m = 1.6)
28. Draw a clearly labeled diagram of a lens camera and explain briefly how the
image of an object is focused on the film .A camera with a lens of focal
length 15 cm is used to take a photograph of a man standing 4.5 m from the
lens .Find the length of the image formed if the man is 1.75 m tall
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 29. Describe with the aid of a diagram the optical system of the projection lantern .A
projection lantern is to be used for the projection of slides measuring 7.5 cm by 7.5
cm onto a screen measuring 2.1 m by 2.1 m .If the distance between the
projection lens and the screen is 6 m, Find the focal length of the lens
30. Describe how would you construct an astronomical telescope. What will be
the maximum distance between the objective lens and the eyepiece lens?.
Draw a ray diagram to illustrate the paths of two rays from a point on an
object which is not on the axis of the telescope
31. Define and explain the terms near point and far point as applied to the human eye
.Draw diagrams to illustrate the defects of long and short sight .How can each
defect be corrected by using lenses ?
32. A man whose least distance of distinct vision of 75 cm wants spectacles to
allow him to read a book held at a distance of 25 cm from his eyes. Find the
focal length of the lens he needs .Discuss briefly whether a short – sighted
person can use a telescope without wearing any spectacles
33. Draw a clearly labelled diagram of the human eye and explain how it can focus
near and distant objects. How does the human eye resemble the lens camera?
34. A telescope of 5.0 m diameter reflector of focal length 18.0 m is used to focus the
image of the sun. Using the distance of the sun from the earth and diameter of the
sun as 1.5 x 1011 m and 1.4 x 109 m respectively, calculate the:
(a) Position of the image of the sun ( v = f =18 cm, since the object is at infinity)
𝒗 𝒉
(b) Diameter of the image of the sun(image size) (𝒎 = 𝒖 = 𝒉 𝒊 → 𝒅𝒊 = 𝟏𝟔. 𝟖 𝒄𝒎)
𝒐
35. A person whose sight is normal wishes to view objects which are 5 cm from
his eyes. Find the focal length of the lenses needed for his spectacles
36. When is a person said to be suffering from long sight? Draw a diagram of the eye
to show how this defect may be corrected by the use of a suitable type of lens.
37. Mention two ways in which a photographic camera is similar to the human eye and
one way in which it is different
38. A converging lens has a focal length of 5 cm (ANS: P = 20 D)
(i) What is the power of the lens?
(ii) If this lens were used in an astronomical refracting telescope, for which part of
the telescope would it be most suitable?
(iii)What would be the distance between the two lenses if the telescope were in
normal adjustment (i.e with the final image at infinity)? (ANS :(𝒍 = 𝒇𝒐 + 𝒇𝒆 ) →
𝒔𝒖𝒎 𝒐𝒇 𝒇𝒐𝒄𝒂𝒍 𝒍𝒆𝒏𝒈𝒕𝒉𝒔)
39. Explain the terms real image and virtual image
40. Describe the construction of a photographic camera. If the focal length of the
camera lens is 20 cm, how far away from the film must the lens be set in order to
photograph an object 100 cm from the lens? (ANS: v = 25 cm)
41. A slide projector using a slide 5 cm x 5 cm produces a picture 3 m x 3 m on a
screen placed at a distance of 24 m from the projection lens. How far from the lens
must the slide be? (ANS: 40 cm x 40 cm)
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 TOPIC: 06 THERMAL EXPANSIONS
Thermal expansion is the tendency of matter to change in volume due to change in
temperature.

Terms used
 Expansion is the process whereby object increases its volume due to increase
in temperature
 Contraction is the process whereby object decreases its volume due to
decrease in temperature

Sources of thermal Energy

 The sun (the sun generates its energy by nuclear fusion)
 Combustion of fuels
 Nuclear energy (is energy generated from nuclear reactions)
 Geothermal energy (the heat energy delivered from the earth core)

Why Substance expands?

 Substance expands when heated because its particles vibrate more rapidly. As a
result they collide and push each other further apart
 All states of matter (solids, liquids and gases) expand when heated.

Explain what happen when solids (liquids or gases) are heated?

 When a solid is heated, its molecules gain kinetic energy and vibrate more
vigorously. As the vibration become larger, the molecules are pushed further
apart and the solid expands slightly in all directions.
Explain what happen when solids (liquids or gases) are cooled?
 When solid is cooled, its molecules lose kinetic energy and have less vibration.
As the vibration become lower, the molecules are pulled closer and the solid
contracts slightly in all directions

Thermal Expansion of Solids

 The expansion of solid substance is so small such that it is difficult to observe its
changes
 As the temperature of a solid increases, the atoms vibrate with large amplitudes
and the average separation between them increases. As a result, the entire solid
occupies a large volume as the temperature increases.
This can be seen in the ball and ring experiment as shown in the fig. below

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Observation and Explanations
 The metal ball can just pass through the ring at room temperature
 On heating, the metal ball expands. There is an increase in volume and the ball
cannot pass through the ring
 On cooling , contraction occurs and the original volume is regained .The ball can
now pass through the ring

Linear expansivity (Coefficient of linear expansion)

 Is the ratio of increase in length to its original length per degree rise in temperature
 OR Is the increase in length per unit length of the substance when its temperature
rises by 1°C or 1 K.
 The SI unit for linear expansivity is K-1
Mathematically:
𝒊𝒏𝒄𝒓𝒆𝒂𝒔𝒆 𝒊𝒏 𝒍𝒆𝒏𝒈𝒕𝒉 𝜟𝒍
Linear expansivity = → 𝜶= 𝒍
𝒐𝒓𝒊𝒈𝒊𝒏𝒔𝒍 𝒍𝒆𝒏𝒕𝒉 𝒙 𝒓𝒊𝒔𝒆 𝒊𝒏 𝒕𝒆𝒎𝒑𝒆𝒓𝒂𝒕𝒖𝒓𝒆 𝟏 𝒙 𝜟𝜽
Where:
α = Linear expansivity
Δθ = (θ2 – θ1) = rise in temperature
θ2 = initial temperature
θ1 = final temperature
ΔL = (L2 – L1) = increase in length
L1 = original length
L2 = new length

Linear expansivities of different substances.

Substance Linear expansivity Substance Linear expansivity
(K-1 ) x 𝟏−6 (K-1 ) x 𝟏−6
Aluminium 25.5 Steel 10.5
Brass 18.9 Glass 8.5
Copper 16.7 Pyrex glass 3.0
Iron 10.2 Invar 0.9

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Individual Work – 6:1
1. A block of concrete 5.0 m long expands to 5.00412 m when heated from 25°C to
100°C. Determine the linear expansivity of concrete.( ANS: α = 1.1 x 𝟏𝟎−𝟓 ℃ =
1.1 x 𝟏𝟎−𝟓 K -1)
2. The difference in length between a brass and an iron rod is 14 cm at 10 0 C.What
must be the length of the iron for this difference to remain at 14 cm when both
rods are heated to 1000 C? Given that the linear expansivity of brass = 19 x 10-
6/K and iron = 12 x 10-6/K. (ANS: L = 38 cm)

3. A metal rod has a length of 100 cm at 2000 C. At what temperature will its length be
99.4 cm if the linear expansivity of the material of the rod is 0.00002/K (ANS: - 1020 C)
4. A metal pipe which of 1M long at 40°C increases in length by 0.3% when
carrying a steam at 100°C. Find the Coefficient of Linear Expansion (ANS: α =
5 x 𝟏𝟎−𝟓 K )
5. A brick (30 cm x 18 cm x 10 cm) is at 20°C, If the brick heated to a temperature
of 150°C, what will be its new dimensions? (The coefficient of linear expansion
of concrete is 1.2 x 10-5 K-1 (ANS: 30.05 cm x 18.03 cm x 10.02cm)
6. An iron plate at 20°C has a hole of radius of 8.92 mm in the centre, an iron rivet
with radius of 8.95 mm at 20°C, inserted into the hole. To what temperature the
plate heated for the rivet to fit into the hole. (Linear expansivity of iron is 1.24 x
10-5K-1). ANS: 291°C
7. Which is heavier, 1 dm3 of glass at 40 C or 1 dm3 of glass at 100 C? Explain your
answer.
N.B
Superficial expansion of solids(Areal expansion) is the ratio of increase in area to
its original area for every degree increase in temperature

The Bimetallic Strip

 The bimetallic strip consists of two different metals that expand at different rates
when heated through the same temperature change.

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NB:
 The metal that expands faster forms the outside part of the curve while the one that
expands more slowly is on the inside of the curve.
 Brass expands and contracts twice as fast as steel.
 temperature must be measured in Kelvin

Applications of expansion of Solids

1. Applied in the construction of bridges and house roofs

 In order to prevent expansion one end or both ends of steel bridge rest on rollers so
when the bridge expands, the free end moves forward and backward without breaking
 If the ends were firmly fixed, the bridge or roof or its supports would break.

2. Railway lines construction

 In order to allow expansion during hot weather always gapes are left between
two rails.(See the figure below)

 In addition, the modern railway lines have long lengths of special steel, which
have sliding joints for expansion .On these rails, trains run smoothly without
noise and bumps

3. Construction of roads and pavements

 In order to prevent expansion always gapes are left between the slabs. Gape
filled with pitch.(See the fig below)

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4. The use of bimetallic strip
The bimetallic strip is used in thermostats, thermometers and valves.
(a) Thermostat: Is a device used for maintaining a steady temperature.
Thermostat is used in many appliances such as electric irons, heaters,
refrigerators, air conditioners, fire alarms and Valves
 In Fire alarms circuit
When temperature rises bimetallic strip bends to close a circuit which
complete the circuit and bell start to ring.

 In Electric iron
When a bimetallic strip bends due temperature change, it breaks the circuit

(b) Bimetallic thermometer

When temperature rises, the bimetallic strip bends and causes the
pointer to rotate across pulley

Bimetallic Valve
 Is also used to open and close temperature – sensitive valve
 The valve opens when the temperature rises and close when the temperature falls

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5. Applied in hot-water pipes
 In order to prevent destruction which arise from expansion, pipes are fitted with
expansion joints
 Expansion of the pipes when steam passes merely increases the size of the bends.

6. Applied in designing of pendulum clocks

 Pendulums and balance wheels in clocks are compensated for expansion so that
the clocks keep correct time even when temperatures changes

7. Overhead telephone and electrical cables are loosely held

 In order to allow for contraction and expansion during the cold and hot day
respectively
 During cold weather or at night, the metal can contract without breaking

Some effects of expansion and contraction

1. Vessels made of thick or ordinary glass break easily if hot liquids are poured
in them. (This is because the inside of the glass is heated and expands, while the outside
remains cold and the same size. The force of expansion usually breaks the glass)
2. Riveting of two metal plates
 Rivets are used in shipbuilding, boiler making, etc to join metals sheets
 Rivets are heated to red hot condition and are forced through holes in the
two plates
 The end of hot rivets contract and bring the plate tightly gripped to each other
3. Glass stoppers sometime stick in the necks of bottle.
 By warming the neck of such a bottle gently, the stopper often comes out easily
4. Weathering of rocks
 This is the action of sun, air and water on rocks, causing them to break
 Hot sun shine makes the outside of a rock expand, and pieces break off
due to the force of expansion

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 5. Steel tyres or rims for wheels of railway engines are slightly too small when cold.
 They are made red–hot and expand enough to fit the wheel. On cooling,
the tyres contract and fit tightly

Thermal Expansion of Liquids

 It is easier to observe expansion in liquids than in solids. This is because
liquids expand much more than solids for equal change of temperature.

Explain why when heating a liquid, its level initially decreases and then it
Increases to become more larger than the original level?
 The liquid level drops due to the expansion of its container which initially
absorbed all the heat. The level of the water will then keep rising as the
container due to the expansion of the water, its density decreases.

Apparent volume expansion of a Liquid

 Is the difference between the initial volume of a liquid and its final volume without
consideration of the expansion of the container
Absolute (Real) expansion of the liquid
 Is the difference between the final volume of a liquid and its intermediate volume
The intermediate volume of a liquid
 Is the volume of a liquid attains due to the change that was caused by the
expansion of the container

Volume expansivity of a Liquid (coefficient of volume of expansion)

 Is the fractional change in volume per unit temperature change.
 Its SI unit is K-1 or °C-1.
𝒊𝒏𝒄𝒓𝒆𝒂𝒔𝒆 𝒊𝒏 𝒗𝒐𝒍𝒖𝒎𝒆 𝒐𝒇 𝒍𝒊𝒒𝒖𝒊𝒅
 Volume expansivity =
𝒐𝒓𝒊𝒈𝒊𝒏𝒂𝒍 𝒗𝒐𝒍𝒖𝒎𝒆 𝒙 𝒓𝒊𝒔𝒆 𝒊𝒏 𝒕𝒆𝒎𝒑𝒆𝒓𝒂𝒕𝒖𝒓𝒆

𝒗𝟐 – 𝒗𝟏
∴ β=
𝒗𝟏 𝒙 𝜟𝜽

Whereby:
Original/initial volume of liquid = V1
Final volume = V2
Increase in volume of liquid = V2 – V1
Initial temperature = θ1
Final temperature = θ2
Rise in temperature = Δθ = (θ2 – θ1)
Volume expansivity = β

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 Linear expansivities between (0 – 100)°C
Liquid Volume expansivity Liquid Volume expansivity
(K-1) x 𝟏𝟎−𝟓 (K-1) x 𝟏𝟎−𝟓
Benzene 124 Mercury 18
Gasoline 95 Methanol 122
Glycerin 53 Water at 20℃ 21
Kerosene 99 Water at 35

Anomalous expansion of Water

 Is the decrease in the density of water as cooled from 4°C to 0°C.
OR
 Is the unusual behavior of water where its volume decreases when the temperature
rises from 00 C to 40 C and increases when the temperatures falls from 40 C to 00 C

What happens below 4°C to 0°C

At 4°C, just above the freezing point, water reaches its maximum density. As the water
cool further toward its freezing point, the liquid water expands to become less dense.

Effects of Anomalous expansion of water

 A glass bottle filled with water and sealed cracks if cooled in a deep freezer
 A glass tumbler breaks when a hot liquid is poured in it because the inner wall expands
quickly while due to poor conductivity of glass, the outer wall remains unexpanded
 If water freezes in a pipe, it may cause the pipe to burst open due to expansion.
 Icebergs, being less dense than water, float in oceans thus posing a danger to ships.
 Weathering of rocks .This happen when water freezes in the cracks of a rock the
volume of water increases .This causes the rock to break into small pieces

Applications of Expansion of Liquids

 It supports the life of aquatic organism. When water freezes, the ice floats to the
top. This creates an insulating layer, which allows liquid water to remain underneath.
This allows aquatic life to survive in freezing environments. (see the fig below)

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 The expansion of liquids used in liquid thermometers.

Thermal Expansion in Gases

 Gases expand much more than solids and liquids when heated.
This is because the particles in gases are not held closely together, as they are in
solids and liquids, but are instead free to move in all directions.
 Three properties are important when studying the expansion of gases.
These include Pressure, Volume and Temperature.
 The temperature must be converted into Kelvin scale

Charles’ Law
 This law involves the relationship between the volume and the temperature of a
fixed mass of a gas at constant pressure. The law state that
“The volume of a fixed mass of a gas is directly proportional to the absolute
temperature provided the pressure remains constant”
Mathematically
𝑽
V∝T →V= kT ………………… make K the subject 𝑲=𝑻
𝑽𝟏 𝑽𝟐 𝑽𝟏 𝑻𝟏
Therefore: 𝑻𝟏
= 𝑻𝟐
→ 𝑽𝟐
= 𝑻𝟐

Whereby:
V1 = initial volume V2 = final volume
T1 = initial temperature T2 = final temperature
The graph of Volume against temperature

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  From the graph above it seems that as temperature increases also volume
increases and vice versa
The Absolute Scale of temperature
 When the temperature of the gas is -2730 C, the volume of the gas is zero,
at this point the temperature is called ‘’absolute zero or 0 K’’
 The absolute zero temperature Is the lowest temperature that can be attained
theoretically.
OR Absolute zero is the temperature at which all particles of matter possess zero energy
NB: Practically, it is not possible to attain this temperature because all gases
liquefy before attaining it

The figure below shows the relationship between the Kelvin scale and the Celsius scale
of temperature

Conversion
T (K) = 273 + θ (°C) …………………1
 (°C ) = T(K) – 273 …………………..2
Example:
1. A gas of volume 300 cm3 was heated from 230 C to 830 C. Determine the volume
at one atmospheric pressure
Solution:
Given: T1 = 230 C = 23+273 = 296 K, T2 = 870 C = 87 + 273 = 360 K
V1 = 300 cm3, V2 =?
𝑽𝟏 𝑻𝟏
From: =
𝑽𝟐 𝑻𝟐

𝑽𝟏 𝑻𝟏 𝑽𝟏 𝒙 𝑻𝟐 𝟑𝟔𝟎 𝒙 𝟑𝟎𝟎
𝑽𝟐
= 𝑻𝟐
→ 𝑽𝟐 = 𝑻𝟏
= 𝟐𝟗𝟔
= 𝟑𝟔𝟒. 𝟖𝟔cm3

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Individual Task – 6:2
1. Change the following temperatures to Kelvin scale (a) 33°C (b) 57°C
. ANS: (a) T (K) = 306 K (b) T (K) = 330 K
2.Change the following temperatures to Celsius scale (a) 4K (b) 292K
ANS: (a) θ° C = - 269°C (b) θ° C = 19°C
3. A 0.20m3 container with a movable piston holds nitrogen gas at a temperature of
20°C. What will be the volume of the gas if the temperature increased to 50°C?
(ANS: V2 = 0.22 m3)
4. A gas occupies a volume of 20 cm 3 at 27°C and at normal atmospheric pressure.
Calculate the new volume of the gas if it heated to 54°C at the same pressure.
(ANS: V2 = 21 cm3)

Application of Charles’ Law

 A balloon swells up when in the open on a hot day, provided it is not inflated. If it
is completely inflated ,it will burst
 A football inflated inside and then taken outdoors during the cold season shrinks slightly

Boyle’s Law
 This law involves the relationship between the volume and the pressure of a
fixed mass of a gas at constant temperature. The law state that
“The volume of fixed mass of a gas is inversely proportional to its pressure if the
temperature is kept constant”
Mathematically
𝟏 𝑲
P∝𝑽 → P=𝑽 → PV= 𝐾 → PV = Constant
∴ P1 V1 = P 2 V2
Whereby:
P1 = initial pressure P2 = final pressure
V1 = initial volume V2 = final volume
Graphically:

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  From the graph above it seems that as pressure increases also volume
decreases and vice versa.
 The figure below shows the relationship between reciprocal of volume and pressure

 From the graph above it seems that as the pressure increased also inverse of
volume increased and vice versa
Example
1. A gas occupies 250 cm3 when the pressure is 20 atmospheres. What will its volume be
if pressure is reduced to 15 atmospheres while the temperature is kept constant?
Answer
P1 = 20 atm, P2 = 15 atm
V1 = 250, cm3 V2 = V2
From: P 1 V 1 = P 2 V 2
20 x 250 = 15 x V2
𝟓𝟎𝟎𝟎
𝑽𝟐 = = 𝟑𝟑𝟑. 𝟑𝟑 cm3
𝟏𝟓

Individual Task – 6:3

1. A gas in a cylinder occupies a volume of 465 ml when the pressure on it is equivalent
to 725 mm of mercury. What will be the volume of the gas when the pressure on it rises
to 825 mm of mercury while the temperature is held constant? (ANS: V2 = 408.6 ml)
2. Bubble of gas, which has a volume of 0.4 cm 3, released by a diver 30 m in under the
surface of a lake, what will be the volume of the bubble when it reaches the surface?
(Assume the barometric pressure is 10 m of water.) (ANS: V2 = 1.2 cm3)

Application of Boyle’s Law

 Bubbles in water seem to grow as they ascend from the bottom of the
water to the surface .This is due to the decrease in Pressure
 Change of pressure in a syringe .When the plunger of the syringe is pulled back
the volume of the syringe container increases ,decreasing the pressure inside
 Death of deep sea creatures when brought to shallow waters .This happens
when the pressure inside their bodies is greater than the pressure of the
surrounding water hence the balance is distracted causing a burst of the cells
bladders and other internal biological structures due to increase in volume

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  Popping of ears at high altitude. When the plane starts to rise it is going from
an area of high pressure where your ears are accustomed to an area of low
pressure causing the air inside increases in volume, this straining your eardrums

Pressure Law
 This law involves the relationship between the temperature and the pressure of a
fixed mass of a gas at constant volume. The law state that
“At constant volume, the pressure of a fixed mass of a gas is directly
proportional to its absolute temperature”
Mathematically
𝑷
P∝T → P= K T → 𝑻 = 𝑪𝒐𝒏𝒔𝒕𝒂𝒏𝒕

𝑷𝟏 𝑷𝟐 𝑷𝟏 𝑻𝟏
∴ 𝑻𝟏
= 𝑻𝟐 → 𝑷𝟐
= 𝑻𝟐

Where:
P1 = initial pressure
T1 = initial temperature
P2 = final pressure
T2 = final temperature
Graphically:

 From the graph above it seems that as pressure increases also temperature
increases and vice versa
Example
1. A car tyre is at an air pressure of 4.0 x 10 5 Pa. at a temperature of 270 C. While it is
running, the temperature rises to 750 C. What is the new pressure in the tyre?
(Assume the tyre does not expand)
Answer
Since the tyre does not expand, this implies that the volume is constant
P1 = 40 000Pa, P2 = P2
T1 = (27+273) = 300K, T2 = (75 + 273) =384 K
𝑷 𝑻 𝟒𝟎𝟎𝟎𝟎 𝟑𝟎𝟎 𝟒𝟎𝟎𝟎𝟎 𝒙 𝟑𝟖𝟒
From: 𝟏 = 𝟏
𝑷𝟐 𝑻𝟐
→
𝑷
=
𝟑𝟖𝟒
→𝑷=
𝟑𝟎𝟎
= 𝟓𝟏𝟐𝟎𝟎𝑷𝒂

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Individual Task – 6:4
1. A rigid metal container holds carbon dioxide gas at a pressure of 2 x 10 5 Pa and a
temperature of 30°C. What temperature the gas be lowered for the pressure to reduce
to half (1 x 105 Pa)? ANS: T2 = 151.5K = -121.5°C
2. A gas in a fixed-volume container has a pressure of 1.6 x 105 Pa at a temperature of
27°C. What will be the pressure of the gas if the container heated to a temperature of
277°C? ANS: P2 = 2.93 x 105

Application of Pressure Law

 Soda or soft drinks bottles are made of thick glass. This is because, when gas is
heated in a closed container its pressure increases .Hence they are made of
thick glass to withstand pressure increase

The General Gas Equation

Any two of the three gas laws can be used to derive the general gas law or equation

V ∝ T (Charles’ law) …………….. 1

𝟏
P ∝ 𝑽 (Boyle’s law) ………………. 2
P ∝ 𝑇 (Pressure law) ……………... 3

Combine equation 2 and 3

𝑻 𝑻
P∝ → P = k 𝑽 …………………. make k the subject
𝑽
𝑷𝑽
Then; = Constant
𝑻
𝑷𝟏 𝐕𝟏 𝑷𝟐 𝐕𝟐
∴ =
𝐓𝟏 𝐓𝟐

Standard Temperature and Pressure (STP)

 STP is a set of conditions for experimental measurements to enable
comparisons between sets of data.
 The standard temperature is 0 °C (273 K) while the standard pressure is 1
atmosphere (1.013 x 105 Pa or 760 mm of mercury).

Individual Task – 6:5

1. A sample of oxygen gas has a volume of 0.11 m 3 at a temperature of 12 °C and a
pressure of 8.1 x 105 Pa while a sample of nitrogen gas has a volume of 0.18 m 3 at a
temperature of 22 °C and a pressure of 1.03 x 105 Pa. Which gas will have the larger
volume at STP? (ANS: V2O = 0.084 m3 , V2N = 0.17 m3 At STP, nitrogen gas would
have a volume that is more than twice the volume of oxygen gas.)

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2.A fixed mass of gas has a volume of 1.25 litres at a pressure of 76.0 cm of mercury
and a temperature of 27.0°C. The gas expands to a volume of 1.55 litres raising the
pressure to 80.0 cm of mercury. What is the final temperature of the gas in °C? (ANS:
T2 = 391.58K = 118.58°)
3.A fixed mass of gas occupies a volume of 0.001 m3 at a pressure of 76 cmHg. What
volume does the gas occupy at 17.0°C if its pressure is 72 cmHg? (A: V2 =1.12 x 10-3 m3)
4.100 cm3 of gas A was collected at 10°C and 78.0 cmHg pressure, while 120 cm3 of
gas B was collected at 50°C and 70.0 cmHg pressure. Which of the two gases is
denser at STP? ANS: At STP, gas B has large volume than gas A so gas A is
denser than gas B . That is VA2 = 99.00 cm3 and VB2 = 93.42 cm3
5. 250 cm3 of a gas are collected at 25°C and 750 mm of mercury. Calculate the
volume of the gas at STP (ANS: V2 =226.01 cm3)

Applications of the Expansion of Gases

o Land and sea breezes
o The piston engine
o Firing bullets from guns
o Releasing of whether balloon

Class Activity – 6
1. Define coefficient of linear expansion. A copper pipe which is 1 meter long at 150
C increases in length by 0.15% when carrying steam at 1000 c, find the coefficient
of linear expansion of copper
2. Explain why a compound metal bar made up of two strips, one of iron and
another of brass, bends when heated
3. A beaker containing water is heated a temperature of 23 0 C to 900 C. State and
explain what happens to the Mass, volume and density of water
4. An aluminium lid on an ordinary glass jar fits so tightly that it cannot be unscrewed
.Should the jar and lid be immersed in hot or cold cold water to loosen the lid?.
Explain your answer
5. A steel tower has a height of 324 m at a temperature of 18 0 C .How tall is the
tower on a day when the temperature is 350 C?
6. A brass road of length of 0.997 m at 200 C is hung from a steel framework with a
height of 1.00 mat 200 C as shown in the figure below

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 7. A brass has a hole whose radius is too small for an iron rivet to fit in .Explain two
ways the rivet can be made to fit in the hole
8. Convert the following temperatures on the Celsius scale to temperatures on
the Kelvin or absolute scale (a) 1000 C (b) 250 C (c ) -1000 C
9. Convert the following temperatures on the Kelvin scale to temperatures on Celsius
scale: (a) 273K (b) 400K (c) 100K
10. State Charles law and describe how it is verified in laboratory.1000 cm3 of air
at 0°C are heated to 70°C. What volume will the air occupy if the pressure
remains at atmospheric throughout?
11. The pressure of 440cm³ of the gas is 80cm of mercury. What will be the new
pressure of the gas if its volume is reduced to 400cm³ at constant temperature?
12. The pressure in a metal glass cylinder at 15°c is 2 atmospheres. At what
temperature will the pressure be doubled?
13. When is a given mass of a gas said to be at s.t.p? The volume of a gas collected
at a temperature of 36° C and pressure of 78cm of mercury is 230cm³. Find its
volume at s.t.p.
14. A 500 cm3 Pyrex beaker is 95% full of methanol at 150 C. At what temperature
will it be 100% full with methanol?
15. A hollow glass sphere has a density of 1.30 g/cm3 at 200C . Glycerine has a
density of 1.26 g/cm3 at 200C .At what temperature would the sphere begin to
float in glycerine?
16. The figure below shows a brass invar bimetallic strip at room temperature

Given that brass expands more than invar when both are heated equally ,
sketch the appearance of the strip after being cooled to several degrees
below room temperature
17. A glass test tube was heated over a Bunsen burner flame. Cold water was then
quickly poured into the test tube. Explain why the test tube would break when cold
water is poured in.
18. A rally a car tyre is at an air pressure 3 x 10⁵ Pa and a temperature of 27°C at
start of the rally. The temperature rises to 57°C when the car is racing.
Assuming the tyre does not expand, what is the new pressure in the tyre?
19. The pressure of 3 m³ at a gas at 27°C is 3 atmospheres .What will be the pressure
of the gas if it is compressed into a half the volume and heated to 102°C?
20. (a). Define the term linear expansivity of a solid .
(b).The original length of ametal bar is 101.5cm at 15°C.Determine the linear
expansivity of the metal if the bar increases in length by 1.41mm when the
temperature is raised to 100°C
21. Why electrical cables are left sagging during installation?

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 22. A metal rod 80cm long increased in length by 0.09 cm when the temperature was
raised by 93.6°C .Determine linear expansivity of metal.
23. The air in a bicycle tyre occupies a volume of 1000 cm³ when it is at a
pressure of 2.5 atmospheres the air is released to the atmospheres
(a)Assuming that the temperature of the gas does not change, what volume
does it occupy at the atmosphere
(b).A pump with a volume of 150 cm³ per stroke is used to inflate the tyre
.What is the pressure of the tyre after two strokes?
24. An iron rod is 100 cm long at 0°C. What must be the length of alluminium rod at
0°C if the difference between the length of the two rods are to remain the same at
all temperatures? (Linear expansivities of iron and alluminium are 1.2 x 10-5 K-1
and 2.4 x 10-5 K-1, respectively)
25. A gas occupies a volume of 2m ³ when its pressure is 1140mmHg at a
temperature of 27° C. What volume will it occupy at s.t.p?
26. The figure below shows a circuit diagram for controlling the temperature of a room

(a) State and explain the purpose of the bimettalic strip.

(b).Describe how the circuit controls the temperature when the switch is closed.
27. A container holds a gas at 0°C .To what temperature must be heated to its
pressure to double? (Assume that the volume of the container does not change )
28. Explain why a glass container with thick walls is more likely to crack than one with
thin walls when a very hot liquid is poured in each of the glasses
29. A balloon is filled with air to a volume of 200ml At a temperature of 20°C . The
balloon is then dipped in water at 80°C.Assuming two leakage occurs and
ignoring the pressure change due to the water. Calcuate the new volume of air.
30. A compound strip of brass and iron, 10 cm long at 20 0C,is held horizontally with
iron uppermost. When heated from below with a Bunsen burner the temperature of
the brass is 8200 C and that of the iron is 7700C.Calculate the difference in lengths
of the iron and brass (ANS:∆𝒍 = 𝟎. 𝟎𝟔𝟐 𝒄𝒎)
31. State any three applications of bimetallic strip
32. Using the kinetic theory of gases explain how a rise in the temperature of gases
causes a rise in the pressure of a gas when a volume is held constant.
33. The pressure indicated by the gauge on a constant -volume gas
thermometer in a thermal equilibrium with a room is 97 kPa .When the
thermometer was immersed in a bath of ice water, the pressure was 90kPa
.What is the temperature in the room in °C?.

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 34. Helium gas at temperature of -30°C is held in a rigid metal container at a pressure
of 1.5 x 10⁵ pa. The container is heated to a temperature of 25°C.W hat is the new
pressure of the gas?
35. A metal rod is 10 m long at 200 C .At what temperature would its length increases
by 5cm if its linear expansivity is 2x 10-6/K
36. A cylinder closed at both ends as an inner radius 0.021 m. The cylinder is fitted
with a movable piston of mass of 2(ANS: temp = 25200 C) kg. The space between
the piston and the bottom of the cylinder contains 1.11 x 10-⁴ m³ of air at 25°C
while the space above the piston has been evacuated as shown in the figure below.

(a). Determine the pressure of the air in the cylinder given that the pressure
comes from the weight of the cylinder
(b).The cylinder is placed over a source of heat causing the air to expand and
push the piston upward a distance of 3.5cm as shown in the figure below

Assuming that the pressure of the air remained constant, what was the change
in volume of the air? (Volume of a cylinder = 𝜋r2h)
37. Distinguish between heat and temperature
38. (a) State Charles’ Law ,Boyle’s Law and the Pressure Law
(b) Write down the ideal gas equation
(c) The volume of a certain gas at 100 C is 100 cm3 .Calculate the volume of
the gas if it is warmed at a temperature of 300 C at constant pressure
39. Explain each of the following observations:
(a) A lid on a metal can be unscrewed easily if the can is immersed in hot water for
a few minutes
(b) Corrugated iron – sheet roofs make cracking noises on a night preceded by a
hot day
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 (c) It is difficult to unscrew wheel nuts in the morning , while it is relatively easy to
unscrew them on a hot day.
40. Define
(a) Heat (b) Thermal expansion
(c) Linear expansivity (d) Absolute zero temperature
(e) Anomalous expansion of water
41. Differentiate between
(a) Apparent expansion of liquid and absolute expansion of liquid
(b) Heat and temperature
42. What do you understand by intermediate expansion of liquid?
43. Show how the combined gas law is obtained
44. Explain the following observations:
(a) An inflated balloon hung in the open at a wedding bust when the
temperature of the environment rises
(b) A whale cannot survive in a shallow water
(c) Soda bottles are thick
(d) Water bubbles seem to increase in size as they rise from the bottom of a tank
(e) Electric wire are seem to sag when its hot but look very straight when its cold
(f) Fish living in polar regions such as Antarctica do not die even when the
temperature fall below 0°C
45. Mention two demerits of anomalous expansion
46. (a) State Boyle’s law.
(b) Sketch the graph of pressure (P) against the reciprocal of volume (1/v) for air
at constant temperature
(c) A bubble of air of volume 50.0mm 3 is released by a diver at a depth where
the pressure is 304.0 cm Hg .Assuming that the temperature remains constant
,what is its volume just before reaches the surface where the pressure is 76.0 cm
47. A piece of copper is dropped into water, if the temperature of the water is rising
what is happening to the copper?
48. Why are the over head power cables more likely to break and fall during the cold
season of the year than during the warm season of the year even though they
carry the same weight all year round?
49. A grandfather’s clock is controlled by a swinging brass pendulum of length 1.3 m
at a temperature of 20°C.(∝ =19 x 10-6 °C-¹)
(a) What is the length of the pendulum rod when the temperature drops to 0°C?
𝒍
(b) If the period of the pendulum is given by T=2𝝅√ , where L is its length, does
𝒈
the change in the length or the brass rod cause the clock to run fast or slowly?
50. A gas is a contained in an 8 litre vessel at a temperature of 20°C and pressure of
9 atm. At what temperature will the gas fill 3 litre vessel at a pressure of 13
atmospheres?
51. The temperature of a body is 47°C .What would this temperature be in the
absolute scale?

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 52. Given that at s.t.p a gas occupies 5600cm³, determine the pressure at which it will
occupy the volume of 28.5 litres at a temperature of 220°C.
53. States Charles’ Law .An ideal occupies a volume of 500 cm3 at a temperature of
300 C .At what temperature will it occupy a volume of 456 cm3
54. Explain why inflated balloon swells up and even bursts when in the open on a hot day?
55. A form three student carried out an experiment on one of the gas laws .She
obtained the following results

Temp T (0C) 10 35 60 80 90 110

Volume V (cm 3) 5.0 5.8 6.4 7.0 7.2 7.8
(a) Plot a graph of volume V against temperature
(b) From the graph ,determine the volume of the gas at 0 0 C
(c) Determine the slope of the graph
56. State Boyle’s Law. A gas occupies a volume of 600 cm 3 at a pressure of 760
mmHg. Determine its volume at a pressure of 1085 mmHg
57. The volume of a bubble at the base of a container of water is 3 cm 3 .The depth of
water is 30 cm. The bubble rises up the column until the surface .
(a) Explain what happens to the bubbles as it rises up the water column
(b) Determine the volume of the bubble at a point 12 cm below the water surface
58. Explain the following
(a) Deep sea animals cannot survive at regions with shallow waters
(b) Determine the volume of the bubble at a point 12 cm below the water surface

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TOPIC: 07 TRANSFER OF THERMAL ENERGY
Heat can be transferred from one place to another in three ways, these include
1. Conduction
2. Convection
3. Radiation

Conduction of Heat
 Conduction Is the transfer of heat though matter from a region of higher
temperature to a region of lower temperature

How heat transferred?

 When heat is supplied to one part of a solid, the atoms vibrate faster. This
vibration is passed onto neighboring atoms through the bonds. This spreads the
heat throughout the object.

Good and Bad Conductors

Good Conductors
 Are the materials that allow heat to flow through them easily. For example, iron
etc. the conduction on liquid is minimal due to the large intermolecular spaces
 Conductors have different rate of heat conduction. Example, copper is the best
conductor of heat while steel is the poorest conductor.
Bad Conductors (Thermal Insulators)
 Are materials that cannot allow heat to flow through them easily
 For example, plastic, wood, glass etc.

Factors affecting the rate of Conduction

1. Length of the material
 The longer the material, the more the time it will take to conduct heat through it.
2. Cross-sectional area that is perpendicular to the heat flow
 The larger the cross-sectional area, the faster the rate of heat conduction
3. Difference in temperature between the two ends of the material
 The higher the difference, the faster the rate of conduction
4. Thermal conductivity of the material.
 This is a measure of the rate at which a material conducts heat. The higher the
thermal conductivity of the material, the faster the rate of conduction

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Minimizing heat losses by Conduction
 In a system where heat needs to be conserved , heat losses by conduction can be
minimized by thermal insulation
 This involves the use of poor conductors of heat used in boilers, hot – water pipes
and in the textile industry
 In the house is achieved by using double – glazed windows , carpets curtains and
draught excluders

Application of Conduction

 Cooking vessels are made of metals (copper), which are good conductors.
 Aluminium is used in making motor engines, pistons and cylinders due to its low
density and high thermal conductivity
 Our clothes are thermal insulators in order to prevent heat loss from our bodies
 The bottoms of cooking pots need periodic cleaning to remove layers of soot, which
impede the flow of heat
 Refrigerators and ice – boxes have an air – space between double walls. Air is a
good insulator. Cork or porous material is used to stop convection currents in the air
 Fiberglass is used under roofs of buildings to prevent heat loss in cold areas
 Blocks of ice are usually covered with cloth, paper or sawdust when being moved or
stored.
 Sawdust (A poor conductor) is used for lagging hot-water pipes
 Heat Sinks are attached in electronic devices used for preventing damage due to
high temperature
N.B Heat Sinks are made of Aluminium

Convection of Heat
 Is the transfer of heat in fluids (liquids and gases) by currents
OR Is the movement of heat through fluids (liquids or gases) caused by movement
of liquid from the hotter to the colder parts
 If temperature of fluids increase , it tends to cause convection currents which
circulate heat continuously throughout the fluid until the whole fluid is at a steady
temperature (See the fig. below)

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How can we minimize Convection?
 It can be minimized by keeping the place vacuum,
Example, vacuum flask minimizes heat loss by convection.
 It can also be reduced by filling air cavities with insulating materials

Application of Convection
(a) Domestic hot water supply system
This system works on the principle of Convection Current. Hot water moves from the
boiler to where it is used under convection current. Inside the boiler water is heated
making it less dense, water rises and flow into the hot water reservoir where it is stored
until needed for use main water supply.
Diagram:

(b) Chimneys
Smoke and gases from fires in houses and factories rise up chimneys and
the flow is convection current. Gases pass up a tall chimney faster than up a
short one, this is because high chimneys have greater pressure difference

(c) Ventilation
Damp, warm, breathed – out air is less dense than ordinary air; it rises and
can escape through openings near the roof. Houses and cinemas have
openings in or near their ceilings.

(d) Winds
Are convection currents .Some parts of the earth’s surface are hotter than
others .The warm air rises over the hot surface and its place is taken by
cooler, denser air

(e) Land and sea breeze

Land and sea breezes are a result of expansion of air caused by unequal
heating and cooling of adjacent land and sea surfaces
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  During the day time the land is warmer than the sea thus the air over the land
surface becomes less dense and so it rises. The space left is occupied by cooler
air from above the sea surface. Thus sea breeze occur

 During the night time the sea is hotter than the land thus the air over the sea
becomes less denser and therefore replaced by cooler, denser air from the land .
Thus land breezes occur

 N.B: The land gets warmer quickly than the sea, because sea water has a
higher heat capacity than the land (sea water needs more time to raise its
temperature)

(f) Motor – car cooling system

Car radiators cool water heated by the engine.
(g) Air condition

 Air conditioning systems relied on convectional current to heat of cool room.

When it is hot, cool air is blown into the room from the air conditioner. This sinks
to the bottom of the room taking place of the less dense warmer air (warmer air
rises and is led out to be cooled then re – circulated).

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  When it is cold, the heated air is turned on heating the surrounding air. The
heated air rises up and cold air moves down to take the place of the rising warm
air. This forms convectional current which continue to circulate until the air in the
room is at the desired temperature

Radiation of heat
 Radiation Is the movement of heat by waves emitted by vibrating particles of
hot substances
OR Radiation is the heat transfer between two or more bodies by means of
electromagnetic waves that do not need material medium.
NB:
 Heat travels in form of infrared radiations
 All bodies at a temperature above absolute zero emit some radiant energy (heat)
 Between the sun and the earth’s atmosphere is a vacuum
 Radiant heat travels with the speed of light and can be reflected , absorbed and
transmitted
 Heat loss by radiation is minimized by covering a body (surrounding) with shiny
surfaces

Radiant Detector
There are two instruments which can be used to detect radiations
 Thermopile
 Liquid in thermometer
N.B Thermopile T is an instrument used to converts radiant heat energy into
electrical energy

Absorbers, Emitters and Reflectors

Absorber
 Absorber is a material that delivers/gain all radiant energy.
 A surface that absorbs all radiant energy is called black body.
 Good radiators of heat are also good absorbers
 Example, black cooking vessel, black clothes dry faster than others colored
clothes etc
Emitter
 Emitter is a material that delivers out all radiant energy.
 A surface that emits all radiant energy is called black body
Reflector
 Reflector is a material (surface) that bounces back all radiant energy.
 Example, white or bright – colored surfaces are poor emitters and absorbers of
heat, solar cookers etc
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Uses of good and bad radiators or absorbers
 White clothes. In hot countries white clothes are slightly cooler than other
clothes
 White buildings. These are cooler than buildings with dark colours on the
outside. Storage tanks containing oil or petrol are usually painted with aluminium
paint. The polished, silvery surface is a bad absorber.
 Vacuum flask. This is used to keep hot liquids hot and cold substances cold
Question
1. Why do we prefer white clothes in summer?
Ans: Because they are poor absorbers and good emitters

Minimizing heat losses by radiation?

 It can be minimized by keeping the place shine (polished).

Thermos Flask
 Thermos flask is a device used to hold hot or cold liquid for long period of time.
 It consists of a double-walled glass (polished by coated with a thin layer of
aluminium), container with vacuum between the walls. It has a stopper made of
insulating materials or cork (See the fig. below)

How thermos flask prevents heat Loss?

Heat losses by conduction is minimized by
 The stopper which is a bad conductor (made of wood, rubber or cork)
 Glass flask (Poor conductor of heat)
 The supporting pad (separator). It is also a poor conductor of heat
Heat losses by convection can be minimized by
 Vacuum between the walls of the flask
 The stopper which is a bad conductor (made of wood, rubber or cork)
Heat losses by radiation is minimized by
 Using silvered walls (to reflect infrared radiation back to the content of the
thermos flask)
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Metal Foil
 Is the polished material used to cover items to minimize heat loss by radiation.
 For example, aluminium foils are used in wrapping hot food.

How metal foils prevent heat Loss?

Metal foils prevents heat loss by reflecting radiant energy
Applications of radiation
 Electric kettles, iron boxes and cooking pans have polished surfaces to reduce
heat loss through radiation
 Houses in hot regions are painted white to minimize heat absorption through
radiation
 Green houses act as heat trap by preventing longer wavelength radiations from
passing through glass
 Cloudy nights are warmer than clear nights because clouds reflect radiation back
to the earth
Class Activity – 7:1
1. Explain what is meant by the conduction of heat .Use the kinetic theory to explain
how heat is transferred along a metal road
2. Explain briefly the reason why steam pipes are covered with felt or asbestos
3. A piece of iron and a piece of wood are both cooled to a temperature of 0 0 C
.When touched with a finger ,the iron feels colder than the wood .Why ?
4. Discuss the uses of good and bad conductors of heat energy in everyday life
5. What do you understand by convection? Describe how you would demonstrate
the formation of convection currents in a liquid
6. Explain the following:
(a) Cork is packed between the double walls of refrigerating chambers
(b) Most cooking pots are made of Aluminium
(c) Ice blocks are wrapped in paper when they are stored
(d) Water tanks in the tropics are painted white
(e) White clothes are worn in the Arctic in preference to dark ones
7. A thermometer having a blackened bulb records a higher than an ordinary
thermometer when they are both held at an equal distance from a fire Explain this
8. Explain briefly how a fire can assist in the ventilation of a room
9. (a) Explain how heat transfer by radiation takes place.
(b)Why does heat transfer by radiation not require a medium?
(c) A good cooking vessel should be black on the outside and not shiny white. Explain.
(d) Give one way through which heat losses by radiation can be prevented.
10. List three areas where heat transfer through each of the following methods is
applied: (a) conduction (b)convection (c)radiation
11. (a) explain how heat transfer by conduction takes place
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 (b) Why are gases poor conductors of heat
(c) Why are cooking vessels made of aluminium and not iron?
(d) Why are stadium seats made of plastic and not steal?
(e) Give two ways through which heat losses by convection can be prevented?
12. Why is it not possible for heat transfer by convection to take place in solids?
13. Explain the importance of making ventilation on the top of the walls in a room
14. Explain briefly how heat travels in metals
15. How is heat loss by conduction ,convection and radiation reduced in a vacuum flask
16. Explain, how land and sea breeze occurs

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TOPIC: 08 MEASUREMENT OF THERMAL ENERGY
Heat Content (internal thermal energy)
o Heat content is the energy possessed by a body due to its temperature.
Factors that determine the heat Content
 Mass of the substance
 Temperature change of the substance
 Specific heat capacity of the substance
Heat Capacity of a Substance
 Is the amount of heat required to raise the temperature of a given mass of a
substance by 1Kelvin (K).
 It is denoted by letter C, Its SI unit is J/K
Mathematically:
𝒒𝒖𝒂𝒏𝒕𝒊𝒕𝒚 𝒐𝒇 𝒉𝒆𝒂𝒕 𝒂𝒃𝒔𝒐𝒓𝒃𝒆𝒅(𝑸)
Heat capacity, C =
𝒄𝒉𝒂𝒏𝒈𝒆 𝒊𝒏 𝒕𝒆𝒎𝒑𝒆𝒓𝒂𝒕𝒖𝒓𝒆(∆𝜽)

Example,
In an experiment to determine the heat capacity of steel, 100KJ of heat energy was
supplied to a block of steel initially at 22°C. If the final temperature of the block was
2190 C, determine the heat capacity of steel.
Soln:
Given: Q = 100 KJ = 100000 J, T1 = 220C, T2 = 2220 C, C =?
𝒒𝒖𝒂𝒏𝒕𝒊𝒕𝒚 𝒐𝒇 𝒉𝒆𝒂𝒕 𝒂𝒃𝒔𝒐𝒓𝒃𝒆𝒅(𝑸)
From: Heat capacity, C =
𝒄𝒉𝒂𝒏𝒈𝒆 𝒊𝒏 𝒕𝒆𝒎𝒑𝒆𝒓𝒂𝒕𝒖𝒓𝒆(∆𝜽)
100000
∴ Heat capacity, C = 222−22 = 500 J/K
Individual task – 8:1
1. Define the term heat capacity. 3000 J of heat is lost when the temperature
of an iron rod reduces from 500 C to 30.50 C. Determine its heat capacity.
(ANS: C = 153.85 J/K)
2. A solid with a heat capacity of 320 J/K requires 2000 J of heat to raise its
temperature to 800 C. Find its original temperature (ANS: T1 = 73.750 C)
Specific heat Capacity
 Is the quantity of heat required to raise the temperature of 1Kg of the
substance by 1K or 1°C
 It is denoted by letter c, its SI unit is J/kg K
Mathematically:
𝒉𝒆𝒂𝒕 𝒆𝒏𝒆𝒓𝒈𝒚 𝒔𝒖𝒑𝒑𝒍𝒊𝒆𝒅(𝑸)
Specific heat capacity, C =
𝒎𝒂𝒔𝒔 × 𝒕𝒆𝒎𝒑𝒆𝒓𝒂𝒕𝒖𝒓𝒆 𝒄𝒉𝒂𝒏𝒈𝒆(∆𝑻)

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Now make H the subject: → 𝑯 = 𝒎𝑪∆𝜽

Specific Heat Capacities of Some Materials

Materials Specific heat Materials Specific heat

capacity(J/kgK) capacity(J/kgK)
Water 4200 Glass 700
Sea water 3900 Steel 500
Paraffin 2200 Copper 390
Methylated 2500 Brass 320
spirit
Ice 2100 Iron 480
Mercury 1395 Lead 130
Aluminium 900 Zinc 380

Example:
1. Calculate the specific heat capacity of a body of mass 3 kg if it requires 6000 J of
heat to raise its temperature from 300 C to 340 C
Soln:
Given: m = 3 kg, H = 6000 J,
T1 = 300 C, T2 = 340 C
𝒉𝒆𝒂𝒕 𝒆𝒏𝒆𝒓𝒈𝒚 𝒔𝒖𝒑𝒑𝒍𝒊𝒆𝒅(𝑸)
From: Specific heat capacity, C = 𝒎𝒂𝒔𝒔 × 𝒕𝒆𝒎𝒑𝒆𝒓𝒂𝒕𝒖𝒓𝒆 𝒄𝒉𝒂𝒏𝒈𝒆(∆𝑻)
𝑄 𝟔𝟎𝟎𝟎
∴ Specific heat capacity, C = 𝑚∆𝑻 = 𝟑 𝒙 (𝟑𝟒−𝟑𝟎) = 𝟓𝟎𝟎𝑱/𝒌𝒈𝑲
Individual Task – 8:2
2. How much heat is required to raise the temperature of a 25kg sample of mercury
from 20°C to 30°C? (ANS: H = 348750J)
3. The temperature of a 6kg block of copper rises from 15°C to 30°C on being
heated. Determine the amount of heat energy supplied to the block. (Specific
heat capacity of block is 390Jkg°C) (ANS: H = 35 100J)
4. How much heat energy is given out by an iron block of 20g mass when it cools
from 920°C to 20°C. (ANS: H = 8 640J)
5. A piece of copper of mass 40 g at 2000 C is immersed into a copper calorimeter
of mass 60 g containing 50 g of water at 250 C .Neglecting heat losses ,what will
the final temperature of the mixture be ? (ANS: 36.00 C)
6. A brass cylinder of mass X was heated to 1000 C and then transferred into a thin
aluminium can of negligible heat capacity containing 150 g of paraffin at 110 C .If
the final steady temperature of the paraffin attained was 20 0 C, Determine the
value of X (ANS: X = 0 . 116 Kg)

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Determination of Specific Heat Capacity
 In determination of specific heat capacity of substance two methods are used
a) Method of mixtures
b) Electrical method

(c) Method of mixture

 If the heat loss controlled when mixing the water, the heat energy gained by the
cold water is equal to the heat energy lost by hot water due to the principle of
conservation of energy.
Calorimetry
 Is the measurement of the quantity of heat
Calorimeter
 Is a device used to control the losses of heat energy when determining specific
heat capacities of substances

How Specific heat Capacity is determined

 If a liquid of known mass (mL) and initial temperature (i) is put in the inner
container (Let’s say a calorimeter of mass mC at i) and a hot substance of
known mass (mS) at its initial temperature, (S) is added to the liquid.
 N.B: the initial temperature of a liquid (i) =Initial temperature of the calorimeter (i)
 If  is the final temperature of the mixture and assume some heat is being
absorbed by the calorimeter ,then to determine the specific heat capacity of A
SUBSTANCE, CS can be calculated as follows:

Heat lost by substance = heat gained by liquid + heat gained by calorimeter

𝒎𝑺 𝑪𝑺 (𝜽𝑺 − 𝜽) = 𝒎𝑳 𝑪𝑳 (𝜽 − 𝜽𝒊 ) + 𝒎𝑪 𝑪𝑪 (𝜽 − 𝜽𝒊 )

𝒎𝑳 𝑪𝑳 (𝜽 − 𝜽𝒊 ) + 𝒎𝑪 𝑪𝑪 (𝜽 − 𝜽𝒊 )
𝑪𝑺 =
𝒎𝑺 (𝜽𝑺 − 𝜽)

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  Therefore the specific heat capacity of the substance is given by:-
𝒎𝑳 𝑪𝑳 (𝜽 − 𝜽𝒊 ) + 𝒎𝑪 𝑪𝑪 (𝜽 − 𝜽𝒊 )
𝑪𝑺 =
𝒎𝑺 (𝜽𝑺 − 𝜽)

N.B
 Under the assumption that, if heat is not absorbed by any of the apparatus used
to carry out this experiment. Then

Heat lost by substance = heat gained by liquid

𝒎𝑺 𝑪𝑺 (𝜽𝑺 − 𝜽) = 𝒎𝑳 𝑪𝑳 (𝜽 − 𝜽𝒊 )

𝒎𝑳 𝑪𝑳 (𝜽 − 𝜽𝒊 )
𝑪𝑺 = … … . . (𝒄𝒂𝒔𝒆 𝑰𝑰)
𝒎𝑺 (𝜽𝑺 − 𝜽)

Assumption made when using this method:

 No heat is lost to the surrounding
 Heat is not absorbed by any of the apparatus used to carry out this
experiment.(case II)
Precautions to be taken when carrying out such experiments
 Use a highly polished calorimeter so as to minimize heat loss by radiation
 The calorimeter should be heavily lagged so as to minimize heat loss by
conduction
 The calorimeter should be covered with a lid of poor conductor so as to prevent
heat loss by evaporation and convection.

Example
1. A block of metal of mass 5 kg is heated to 110 0 C and then dropped into 1.5 kg of water.
The final temperature is found to be 500 C. What was the initial temperature of the water?
(Specific heat capacity of metal = 460 Jkg-1 K-1)
Soln: Let initial temperature of the water be θ
Heat lost by block = mCΔθ
= 5 x 460 (110 – 50) = 5 x 460 x 60 =138 000J
Heat gained by the water = mCΔθ
= 1.5 x 4200 x (50 – θ) = 6300 x (50 – θ) = 315000 –6300 θ
From: Heat gained by water = Heat lost by the metal block
315 000 - 6300 θ = 138 000
6 300θ = 315 000 – 138 000
𝟏𝟕𝟕𝟎𝟎𝟎
∴ 𝜽= = 𝟐𝟖. 𝟏𝟎 𝑪
𝟔𝟑𝟎𝟎

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Individual Task – 8:3
1. 2.5 kg of a metal at 2000 C is immersed into 1.2 kg of cold water of temperature
200 C. The mixture, after thoroughly stirring attained a final temperature of 34.5 0
C. Given that the specific heat capacity of water is 4200 J/kgK, determine the
specific heat capacity of the metal.(ANS: Cmetal = 176.63 J/kgK)
2. A block of metal of mass 0.20kg at a temperature of 100°C is placed in 0.40kg of
water at 20°C.if the final steady temperature of the water is 24°C, determine the
specific heat capacity of the metal. (Neglect heat absorber by the container)
(ANS: c = 442.1 J/kg K)
3. A block of aluminum of mass 0.5kg at a temperature of 100°C is dipped in 1.0kg
of water at 20°C. Assuming that no thermal energy is lost to the environment, what
will the final temperature of the water be at thermal equilibrium? (A:θF = 27.7°C)
(b) By electrical method

Electrical energy supplied by heater = heat gained

∴ 𝑰𝒕𝑽 = 𝒎𝑪∆𝜽

Example:
1. A block of metal of mass 1.5 kg which is suitably insulated is heated from 30 0 C to 500 C in
8 minutes and 20 seconds by an electric heater coil rated 54 W. calculate:
(a) The quantity of heat supplied by the heater
(b) The specific heat capacity of the block
Soln:
(a) 𝐻𝑒𝑎𝑡 𝐸𝑛𝑒𝑟𝑔𝑦 𝑆𝑢𝑝𝑝𝑙𝑖𝑒𝑑 = 𝑝𝑜𝑤𝑒𝑟 𝑥 𝑡𝑖𝑚𝑒 → 𝑄 = 𝑃𝑡 = 𝐼𝑉 𝑥 𝑡
𝑸 = 𝟓𝟒 𝒙 [(𝟖 𝒙 𝟔𝟎) + 𝟐𝟎] = 𝟓𝟒 𝒙 (𝟒𝟖𝟎 + 𝟐𝟎) = 𝟓𝟒 𝒙 𝟓𝟎𝟎 = 𝟐𝟕 𝟎𝟎𝟎 𝑱

∴ 𝒉𝒆𝒂𝒕 𝒆𝒏𝒆𝒓𝒈𝒚 𝒔𝒖𝒑𝒑𝒍𝒊𝒆𝒅 = 𝟐𝟕 𝟎𝟎𝟎𝑱

(b) From: Q = mcΔθ

27 000J = 1.5 x C x (50 – 30) → 27 000 = 1.5 x C x 20
𝟐𝟕𝟎𝟎𝟎
∴ 𝑪= = 𝟗𝟎𝟎 𝑱/𝒌𝒈𝑲
𝟑𝟎

Change of State
 Matter can under goes three states that is solid, liquid and gas.
 Consider the diagram below.

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Explanations
 From point A the temperature is rising steadily as more heat is added until point
B where the solid starts to melt with steady temperature from point B to C
 Once the melting is over, addition of heat leads to a steady rise in temperature in
region CD. D is the boiling point of the liquid
 Then the liquid is transforming into gas (vaporization occurs in DE)
 As the point E is reached ,all the liquid has been converted into gas
 Reversing the process is also possible by removing the heat from gas reduces
its temperature from point E to D,D to C,C to B and finally from B to A (Cooling
takes place)
Melting, Boiling and evaporation
Melting Point
 Melting is the change of state from solid to liquid
 Melting point is the temperature at which a substance changes from a solid
to a liquid.
 At melting point the substance absorbs heat but the temperature does not
change until the substance has completely melted.
 Pressure affects melting. Increase in pressure lowers the melting point of a
substance while decreasing in pressure raises the melting point
 The melting under the pressure and refreezing after the pressure is released
is called Regelation

Freezing Point
 Freezing is the change of state from liquid to solid
 Freezing point is the temperature at which a liquid changes into a solid
without a change in temperature.

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  During solidification a substance loses heat to its surrounding but its
temperature does not fall
NB:
 The freezing point of a pure substance is the same as its melting point. Example,
water freezes and melts at 0°C
 Triple point of water is the temperature at which three states of water (i.e liquid
water, water vapor and ice) co – exist in equilibrium.

Factors affecting freezing point

o Impurities
o Pressure change

NB:
 Presence of impurities always lower the melting point or freezing point of substance
(This explains why sea / salty water freezes at a much lower temperature than pure
/distilled water)
 Increase in pressure lowers melting point of a solid
 Freezing point depression. Occurs when the particles of the impurity get between
the particles of the liquid and disrupt the formation of the solid crystals
 Regelation: Is the phenomenon of melting under pressure and re-
freezing when the pressure is reduced.

Boiling (Ebullition)
 Is the process by which a liquid turns into a vapor when it is heated to its boiling
point
Boiling Point
 Is the temperature at which all of a liquid changes into a gas
OR
 Is the temperature at which its saturated vapour pressure becomes equal to the
external atmospheric pressure.
Mechanism of Boiling
 The molecules at the surface of the liquid gain more kinetic energy move faster and
are able to overcome intermolecular forces holding them together and hence escape.
What happens when a liquid boils?
 If a liquid is heated its temperature begins to rise, and therefore the saturated
vapour pressure will increase. Ultimately, the saturated vapour pressure
becomes equal o the external atmospheric pressure
 At this stage the further addition of heat will cause bubbles of vapour to form
inside the body of the liquid and rise to the surface.

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Factors affecting boiling point
 Impurities
 Pressure change
NB:
 Impurities raises the boiling point of a liquid (This explains why sea water boils at
a higher temperature than pure / distilled water)
 Pressure raises the boiling point of a liquid (ie P ∝ 𝑇) (This explains water boils at a
higher temperature at sea level than at the top of a high mountain)
 At sea level, an altitude of 0 m above sea level water boils at 100 0 C (The
atmospheric pressure is at its maximum)
 As one goes to higher altitude ,the atmospheric pressure keeps decreasing , thus
decreasing the boiling point of water
 Water boils faster at the top of mountain than at the bottom, this is because the
higher the altitude, the lower the pressure. Thus at the top water will boil at lower
temperature than it does at sea level (bottom).This means that it requires less
energy and therefore a shorter heating time to reach its boiling point.
 Water in a pressure cooker boils at a very high temperature nearly 1200 C due to
the high pressure created in the cooker. That is why food cooked in a pressure
cooker takes less time to get ready than food cooked in common cooking pots

Applications of boiling at increased and reduced Pressure

 Boiling under increased pressure is important for fast cooking
 Boiling under reduced pressure is used in preparation of drug tablets
Example
1. The figure below shows a block of ice with two heavy weights hanging such that the copper
wire/string connecting them passes over the block of ice. It is observed that the wire
gradually cuts its way through the ice block, but leaves it as one piece. Explain

Answer:
 Because of the hanging weight, the wire exerts pressure on the ice block, the high
pressure lowers the melting point of ice at the point of contact
 The ice block absorbs the heat from the wire so the region in contact with the wire
melts and lowers itself to water
 As soon as the water passes above the wire, it is no longer under pressure and
therefore freezes. In this case, the wire gradually sinks through the ice block and
comes out of the block
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Evaporation
 Is the process through which a liquid changes to vapour (gas) at a temperature
below its boiling point
OR Is the change of state from liquid to gas (vapour)
NB:
Evaporation is more rapidly when there is windy, sunny and less humidity
Difference between Boiling and Evaporation

Boiling Evaporation
Occurs at a definite temperature which Occurs at any temperature
is boiling point
Accompanied by formation of bubbles No bubbles
Occurs throughout the liquid Occurs at the surface of the liquid
Has no cooling effect Has cooling effect
Takes place rapidly Takes place slowly

Latent heat (Hidden heat)

 Latent heat is the heat absorbed or released when matter changes its state of
matter without change in temperature.
 The latent heat associated with melting a solid of freezing a liquid is called THE
LATENT HEAT OF FUSION
 The latent heat associated with vaporizing a liquid or a solid or Condensing a
vapor is called THE LATENT HEAT OF VAPORIZATION

Latent heat of fusion

Is the quantity of heat energy required to change a solid to liquid at melting point
without any change in temperature
OR Is the heat absorbed when matter changes from solid to liquid without change in
temperature.
Specific Latent Heat of Fusion of a substance
 Is the quantity of heat energy required to change completely a unit mass (1kg) of
the solid to liquid at its melting point.
 Its SI Unit is J/kg
Mathematically:
𝑯
∴ 𝑳𝑭 = → 𝑯 = 𝒎𝑳𝑭
𝒎

Whereby, Lf = Specific latent heat of fusion

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 Melting /Freezing Point Of some Substances at STP
substance melting/freezin latent heat of
g point (°C) fusion (J/kg)
Aluminum 659 396000
Copper 1086 134000
Iron 1535 293000
Water 0 335000
Mercury -39 11000
Ethyl alcohol -117 105000

Latent heat of Vaporization

 Is the quantity of heat energy required to change a liquid to vapour at boiling
point without any change in temperature

Specific Latent heat of Vaporization

 Is the quantity of heat energy required to change completely a unit mass (1kg) of
the liquid to vapour at its boiling point.
 Its SI Unit is J/kg
Mathematically:
𝑯
𝑯 = 𝒎𝑳𝒗 → 𝑳𝒗 = 𝒎

 Where by Lv = Specific latent heat of vaporization

Why steam is hotter (dangerous) than boiling water (liquid)?

 Steam is hotter than boiling water because it has enough energy (Latent heat of
vaporization) to escape the boiling water. The water remains at the boiling
temperature of 1000 C and cannot get hotter till all the water has changed to steam.
 Since Steam has much more thermal energy than liquid that is why steam
is used in engines to convert thermal energy to mechanical energy

Examples
1. Calculate the quantity of heat required to covert 50 grams of ice at 0 0 C to water at 00 C.
(Specific latent heat of fusion of ice = 3.4 x 105 J/kg)
Soln:
From: 𝑸 = 𝒎𝑳𝒇 → 𝒎 = 𝟓𝟎𝒈 = 𝟎. 𝟎𝟓 𝒌𝒈, 𝑳𝒇 = 𝟑. 𝟒 𝒙 𝟏𝟎𝟓 𝑱/𝒌𝒈
∴ 𝑯 = 𝒎𝑳𝒗 = 0.05 𝑥 3.4 𝑥 105 = 17 000 𝐽 = 17𝑘𝐽

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Individual Task – 8:4
1. How much heat is required to change a 500 g ice cube at 0 0 to water at 00? (Lf =
3.36 x 105 J/kg) [ANS: 186 kJ]
2. How much heat is required to change 100 g of ice from ice at 0 0C to vapour at
1000C (Lv = 2.26 x 106 J/kg, Lf = 3.36 x 105 J/kg, c = 4200 J/kgK). (ANS:301600 J)
3. How much heat would be required to change 1.5kg of ice at -10°C to stream at
120°C.? The specific heat capacities of ice, water and stream are 2144J/kg° C,
4186 J/kg° C and 2010 J/kg° C respectively (ANS: Ht = 4, 627, 860J)
Cooling effect of evaporation
 When a liquid evaporates, it gives up its latent heat of vaporization
 If NO heat is supplied from outside this results in COOLING
 Dogs expose their tongues when it is hot so as to keep the body cool
Cooling of human body
 When it is too hot, sweat glands release water which then evaporates from the
skin taking away latent heat of vaporization. This causes the body to cool
 A volatile liquid like alcohol or ether poured on the skin feels much colder than
water at the same temperature .This is because the alcohol evaporates
quickly taking latent heat of vaporization from the skin ,thus cooling the skin
Question: Why does spirit poured on the skin feel much colder than water at the same
temperature?
ANS: (This is because spirit evaporates much faster than water due to its lower boiling point.
This allows more heat to be transferred faster, which makes it feel colder to the touch)
The Refrigerator
 Is a machine which can enable heat to flow from a cold region to a hot region
 Refrigeration is the process in which work is done to move heat from one location
to another
 The basic principle used in refrigeration is cooling by absorption of latent heat
(cooling effect by evaporation)

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Mechanism of Refrigerator
 In the refrigeration cycle ,there are five basic components:
(i) Fluid refrigerant
(ii) A compressor (controls the flow of refrigerant)
(iii) The condenser coils (on the outside of the fridge)
(iv) The evaporator coils (on the inside of the fridge)
(v) Expansion valve (coolant)
 The compressor constricts the refrigerant vapor, raising its pressure , and
pushes it into the coils on the outside of the refrigerator
 When the hot gas in the coils meets the cooler air temperature of the kitchen, it
becomes a liquid.
 Now in liquid form at high pressure, the refrigerant cools down as it flows into the
coils inside the freezer and the fridge.
 The refrigerant absorbs the heat inside the fridge , cooling down the air
 Lastly, the refrigerant evaporates to a gas, then flows back to the compressor
,where the cycle starts all over

Class Activity – 8
Use the following constants where necessary when solving the questions below
 Specific heat capacity of water = 4200J/ (kg 0C)
 Specific heat capacity of ice = 2100J/ (kg 0C)
 Specific heat capacity of steam = 2000 J/(kg 0C)
 Specific latent heat of fusion of ice 3.3 x 105 J/kg
 Specific latent heat of steam = 2.26 x 106 J/kg
 Specific latent heat of vaporization of water = 2.3 x 105 J/kg
 Acceleration due to gravity ,g = 10 m/s2

1. Which contains the greater amount of heat – a lake of water at 20 0C or a bowl of

water at 900C? Explain
2. An iron bar of mass 80 g is heated from a temperature of 150 C to a temperature
of 650 C .How much heat is absorbed by the bar ,given that iron has a specific heat
capacity of 460 J/(kg 0C)
3. Water of mass 20g at a temperature of 42˚C is poured into a well lagged copper
vessel of mass 27 g at a temperature of 20˚C.find the final temperature of the
water.(specific heat capacity of copper = 400J/(kg˚C)
4. Calculate the heat content of a piece of brass of mass 120g at a temperature of
20˚C .Find the final temperature of water. (specific heat capacity of copper =
400J/(kg ˚C)
5. In an experiment to determine the specific heat capacity of a piece of metal, the
following results were obtained:

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 Mass of piece of metal =200g
Initial temperature =25˚C
Final temperature=80˚C
Heat absorbed by the piece of metal = 1430 J
Calculate the specific heat capacity of the piece of metal.
6. Distinguish between latent heat of fusion and the specific latent heat of fusion of a
substance. Find the amount of heat required to change 1kg of ice at 0˚C to water at
the same temperature.
7. Define the term latent heat of vaporization and specific latent heat of vaporization.
Find the quantity of heat required to change 5kg of water at 60˚C into steam at 100˚C
8. Determine the final temperature obtained when 500 g of water at 100 0 C was mixed
with 500 g of water at 100 C and well stirred. (ANS: Tf = 550 C)
9. (a) Differentiate between heat and temperature
(b) The specific heat capacity of a certain substance is 800J/kg0 C; what does
this statement mean?
(c) Calculate the specific heat capacity of mercury, if 980 J of heat is required
to raise the temperature of 7 g of mercury from 00 C to 10000 C
10. A piece of metal of specific heat capacity 840 J/(kg ˚C) and mass 30 g is heated to
a temperature of 99˚C and then dropped into a cavity in a block of ice at 0˚C. Find
the amount of ice that will melt
11. A refrigerator can convert 0.4 kg of water at 20˚C to ice at -10˚C in 4 hours .Find
the average rate of heat extraction from the water in joules per second.
12. (a) what is meant by the following terms (i) melting point (ii) freezing point
(b) Describe how to find the melting point of a substance by means of cooling curve
13. Explain why:
(a) Heat energy has to be supplied to a solid in order to change it into liquid.
(b)Heat energy has to be supplied to a liquid in order to change it to vapor
14. A 0.2kg block of ice at 0˚C is placed into a Styrofoam calorimeter cup with unknown
mass of water at 20˚C .When thermal equilibrium is reached, the final temperature
is measured to be 5˚C.What was the mass of the water initially in the cup?
15. A pressure cooker is a pot with a tight fitting lid that does not allow steam to escape
until a preset pressure is reached. Explain how the pressure cooker can cook food
faster than a sufuria with a loose – fitting
16. A container holds 1.5kg of ice initially at 40˚C.Heat is supplied to the container at
the rate of 12.6kJ per minute for 120 minutes.
(a) Plot a graph of temperature versus time for the 120 minutes during which
heat is supplied.
(b) What will the temperature of the contents of the container be at the end of
the 120 minutes?
(c) What will the mass of the steam in the container be at the end of the 120
minutes

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17. A 0.15kg aluminium cup holds 0.2kg water at 180 C. A 0.12 kg iron block at 85˚C is
placed into the water and the entire system surrounded by an insulating jacket. What will
be the final temperature of the system when thermal equilibrium is reached?
18. The temperature of 500 g of a certain metal is raised to 100 0 C and it is then placed
in 200 g of water at 150 C. If the final steady temperature rises to 210 C, Calculate
the specific heat capacity of the metal. (ANS: C = 128 Jkg-1K-1)
19. How much thermal energy is required to raise the temperature of 3kg of aluminium
from 15˚C to 25˚C?
20. Explain the following:
(d) When the brakes of a moving car are applied for an applicable time, they get hot
(e) When the tyre of a car is pumped up, the pump gets warm
21. A car of mass 1000 kg travelling at 72 km/h is brought to rest by applying the
brakes. Assuming that the kinetic energy of the car becomes transferred to internal
energy in four steel brake drums of equal mass, find the rise in temperature of the
drums if their total mass is 20 kg, the specific heat capacity of steel is 450 J/kgK,
and the work done is equal on all four drums.(ANS: ∆𝜽 = 𝟐𝟐. 𝟐 𝑲)
22. A bath contains 100 kg of water at 600 C. Hot and cold taps are then turned on to
deliver 20 kg per minute each at temperatures of 70 0 C and 100 C respectively.
How long will it be before the temperature in the bath has dropped to 45 0? Assume
complex mixing of the water and ignore heat losses.(ANS: t = 7.5 mins)
23. Some hot water was added to three times its mass of water at 10 0 C and the resulting
temperature was 200 C. What was the temperature of the hot water. (ANS:T = 500 C)
24. A piece of lead of mass 500 g and at air temperature falls from a height of 25 m. What is
(a) Initial potential energy (b) Its kinetic energy on reaching the ground. Assume that
all the energy becomes transferred to internal energy in the lead when it strikes the
ground, calculate the rise in temperature of the lead if its specific heat capacity is 130
J/kgK. State the energy changes which occur from the moment the lead strikes the
ground until it has cooled to air temperature again.(P.E =123 J ,K.E =123 J,∆𝜽 = 𝟏. 𝟖𝟗 𝑲)
25. A waterfall is 100 m high and the difference in temperature between the water at the top
and that at the bottom is 0.24 K. Obtain a value for the specific heat capacity of water in J/kgK
explaining the steps in your calculations. Mention any assumptions you make.(C = 4100 J/kgK)
26. A 0.5 kg block of aluminium at a temperature of 100˚C is placed in 1.0 kg of water
at 20˚C. Assuming that no thermal energy is lost to the surroundings, what will the
final temperature of the aluminium and the water be when they attain the same
temperature?
27. When a certain quantity of heat was supplied to a substance, its temperature rose
from 5˚C to 20˚C.What will the final temperature of the substance be if twice the
amount of heat is removed from the sample?
28. Why is water used as a coolant in car engines?
29. State what changes, if any, take place in the following:
(f) Melting point of ice when salt is added to the ice
(g) The volume of water if it changes into ice
(h) The boiling point of a liquid when the pressure on the liquid is reduced

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30. Two substances A and B have the same mass and are at the same temperature.
Substance A has a higher specific heat capacity than substance B. Which
substance will have a higher final temperature if the same amount of heat is
supplied to each substance?
31. An electric heater is rated at 250 W. Calculate the quantity of heat generated in 10
minutes (ANS: H = 150 kJ)
32. A tin contains water at 290 k and is heated at constant rate. It is observed that the
water reaches boiling point after 2 minutes and after further 12 minutes it is
completely boiled away .Calculate the specific latent heat of steam .(ANS: 2092kJ/kg)
33. An insulated cup holds 0.1kg of water at 0˚C. 0.1 kg of boiling water at a
temperature of 100˚C is poured into the cup. What will be the final temperature of
the mixture at thermal equilibrium?
34. A 50 watt heater is immersed in a 2 kg block of alluminium which also holds a
thermometer .The temperature of the block rises by 8 k in 5 minutes . Neglect heat
losses , Calculate the specific heat capacity of alluminium. (ANS: 937.5 J)
35. A metal sphere of unknown composition has a mss of 0.4kg.The sphere is heated
in a furnace to a temperature of 150˚C and then dropped into an insulated cup
holding 0.35 kg of water at 20˚C.upon reaching thermal equilibrium ,the
temperature of the system is measured to be 32.4˚C.
(a) Calculate the specific heat capacity of the metal.
(b)Use the values of specific heat capacity in table 8.1 to identify the metal.
36. Which contains the great amount of heat –a lake of water at 20˚C or a bowl of
water at 90˚C? Explain.
37. An iron bar of mass 80 g is heated from a temperature of 15˚C to a temperature of
65˚C.How much heat is absorbed by the bar, given that iron has a specific heat
capacity of 460J/(kg ˚C)
38. Water of mass 20 g at a temperature of 42˚C is poured into a well lagged copper
vessel of mass 27 g at a temperature of 20˚C. Find the final temperature of the
water.(specific heat capacity of copper = 400J/ (kg˚C)
39. Calculate the heat content of a piece of brass of mass 200 g at a temperature of 20˚C
.Find the final temperature of water.(specific heat capacity of copper =400J/(kg˚C)
40. Differentiate between
(a) Melting point and boiling point (c) Evaporation and boiling
(b) Freezing and vaporization (d) Melting and cooling
41. Explain the following
(a) The boiling point of water in Dar es salaam is higher than at the top of Mt.
Kilimanjaro
(b) Why does water boil faster at the top of a mountain than at the bottom?
(c) Water being heated while covered boils faster than uncovered water.
(d) When one wipes spirit on the skin he feels cold
(e) When snow is pressed by the hands, it melts to water .The water then
immediately freezes.

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 (f) The use of ammonia as a household refrigerant is discouraged
42. How much heat is required to change 40 g ice cube from ice at –100 C to steam at 1100 C?
43. Describe how a household refrigerator preserves food.
44. If 200 g of water is contained in a 500 g aluminium CAN at 10 0 C then an additional 100 g
of water at 1000 C is added into the CAN , what is the final equilibrium temperature of the
mixture?
45. An unknown liquid of mass 400 g at a temperature of 80 0 C is poured into 400 g of water
at 400 C. The final temperature of the mixture is 490 C .What is the specific heat capacity
of the unknown liquid
46. 20 g of steam at 1000 C is added to 50 g of ice at 00 C .Find the amount of ice that is
melted and the final temperature
47. Explain how the following factors affect the melting and boiling points:
(a) Pressure (b) Impurities
48. Explain how a refrigerator works.
49. An electric heater rated 1500 W is used to heat water in an insulated container of
negligible heat capacity for 10 minutes .The temperature of water rises from 20 0 C
to 400 C. Calculate the mass of water heated
50. An electric kettle rated 2 kW is filled with 2.0kg of water and heated from 20 0 C to 980 C.
Calculate the time taken to heat the water assuming that all the electrical energy is used
to heat the water in the plastic kettle and the kettle has negligible heat capacity
51. The following data was obtained from an experiment .Mass of copper metal block = 200
g, initial temperature of the block = 220 C, ammeter reading = 0.5 A, voltmeter reading =
3.0 v, final temperature of the block = 300 C, time of heating = 7 minutes .Use the data to
calculate the specific heat capacity of copper .What does this value mean? (ANS: CC =
394 J kg-1K-1)
52. In an experiment , the following data was obtained .Use the data to calculate the time
taken by the heater to raise the temperature of water container and the stirrer from 20 0C
to 230 C . What assumption have made in your calculations? Power of electric heater =
30 W ,mass of the container and the stirrer = 200 g ,specific heat capacity of the
container and the stirrer = 400 J/kgK ,mass of water in the container = 100g, specific
heat capacity of water = 200 J/ kgK (ANS: t = 50 s)
53. A class of Physics students decided to determine the specific heat capacity of water in a
waterfall. They used a sensitive thermometer to find the difference in temperature of
water at the top and the bottom of the waterfalls and obtained the following results; height
of the waterfalls = 52 m ,temperature of the water at the top = 21.54 0 C and that at the
bottom = 21.670 C . Stating any assumptions made, calculate a value for specific heat
capacity of water
54. A 200 g of liquid at 210 C is heated to 510 C by a current of 5 A at 6 v for 5 minutes. What
is the specific heat capacity of the liquid? (ANS:C = 1500 J/kg 0C)
55. An electric kettle rated 1 500 W is used to boil 500 g of water into steam at 100 0 C
.Calculate the time required to boil off water.
56. Why steam is hotter than boiling water?
57. Explain as fully you can what happens when a liquid boils. Why would you expect the
boiling point of a liquid to be lowered when the pressure above the free surface is vapour

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 TOPIC: 09 VAPOUR AND HUMIDITY
Evaporation of Liquids
 Evaporation is a gradual change of state from liquid to gas that occurs at the
surface of a liquid.

Factors affecting Evaporation

Temperature
 Temperature is directly proportional to the rate of evaporation
Surface Area
 Surface area is directly proportional to the evaporation increase in Surface area
increases the rate of evaporation
Concentration of Vapour Liquid

 Concentration of vapour liquid (the same or different molecules) is inversely

proportional to the evaporation. Because the surrounding air has little space for
the escaping gaseous molecules comes from evaporation
Wind (The rate of flow of air)
 Wind is directly proportional to the evaporation increase in Wind increases the
rate of evaporation. Wind lowers the Concentration of vapour liquid
Applications of evaporation
 Evaporation has cooling effect
Because it takes thermal energy away from the surface.This is because the
faster particles escape from the liquids surface leaving behind the slower ones
and the faster ones have more energy ,hence the temperature of the liquid drops
 Sweating uses the cooling effect caused by evaporation
As the sweat evaporates, It takes away thermal energy from the skin to the
surrounding
 Evaporation coolers can significantly cool a building by simply blowing dry air
over a filter saturated with water
 Cooling effect from evaporation used in Refrigerators
 In a clothes drier, hot air is blown through the clothes allowing water to
evaporate very rapidly
Vapour Pressure (VP)
 Vapour is a substance that is in a gaseous state at a temperature below its boiling point
 Vapour pressure is the pressure created by the vapour of a substance that forms
above a liquid of the same substance.
Question: How can you differentiate between gas and vapor

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  A gas refers to a substance that has a single defined thermodynamic state at
room temperature whereas vapor refers to a substance that is a mixture of two
phases (ie. gaseous and liquid phase) at room temperature

Types of Vapour Pressure (VP)

 Saturated vapour pressure
 Unsaturated vapour pressure
 Ambient pressure
Saturated Vapour Pressure (SVP)
 Is the pressure created by the vapour of a substance that forms above a liquid of
the same substance
 This happens when “the state of dynamic equilibrium is reached in which the rate at
which molecules leave the liquid is equal to the rate at which others returns to it”
 Under these conditions the space above the liquid is said to be saturated
 Before equilibrium has been reached in the condition stated above, the vapor is
said to be unsaturated
Unsaturated Vapour Pressure (USVP)
 Is the pressure exerted above the surface of the liquid when the surface of the
liquid contains vapour less than the maximum amount that it can hold
 See the figure below

Ambient Pressure (AP)

 Ambient pressure is the pressure created by the vapour of a substance and
other gas pressure
NB:
 A substance of high vapour at room temperature is called volatile
 Increase in temperature of a solid or liquid ,the rate evaporation or condensation
increases which results to an increase in the vapor pressure

Measurement of SVP
 Saturated vapour pressure is measured by mercury barometer.
 It is given by, SVP = (760 - x) mmHg
Where:
760 mmHg = atmospheric pressure (atm), X mmHg = vapour pressure

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Humidity
 Is the amount of water vapour present in the atmosphere
Sources of Humidity
o Evaporation from rivers, lakes and oceans
o Transpiration (evaporation of plant leaves)
NB:
 Water vapor from atmosphere condenses to form clouds, fog, dew and frost
 Frost is a deposit of small white ice crystals formed on the ground or other
surfaces when the temperature falls below freezing
 Fog is a cloud floating just above the ground
 Earth’s surface may or not saturated
 Saturation depend temperature and water availability
 The density of water vapour in saturated air is called absolute humidity (AH)
 Hailstones are water droplets in clouds formed due to super cooling below 0 0 C
without freezing
 Mist is the condensation of vapour into water droplets occurring near the ground
 Snow is formed when the dew point is below the freezing point (0 0 C)
DEW
 Is water in the form of droplets that appears on exposed objects in the morning
or evening due to condensation.
OR Is the condensation into liquid droplets of water vapour on a substance
Dew Point (DP)
 Is the temperature at which the atmospheric air becomes saturated with water vapour
NB:
 Hot Air contains more moisture (humidity) than cold air,
 Dew is formed at night because hot air comes into contact with a cold surface ,now water
vapour present in it condenses on the cold surface in the form of droplets (dew drops)
 The formation of dew is more when the sky is clear and less when it is cloudy.
 DP is measured by Renault hygrometer
 DP occurs when RH of air is 100%
 Below DP clouds, dew or frost formed
Factors influencing the formation of Dew
(a) Temperature
The temperature of the atmospheric air must fall below the dew point for dew to be formed
(b) Wind
As wind increases, the rate of evaporation also increases, thus it prevents the formation
of water droplets (dew)
(c) Water Vapour
The atmospheric air must be saturated with water vapour for dew to be formed
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 Relative Humidity (RH)
 Is the ratio of the saturated vapour pressure at the dew point to the saturated
vapour pressure at the current air temperature
Mathematically:
𝑺𝑽𝑷 𝒂𝒕 𝒅𝒆𝒘 𝒑𝒐𝒊𝒏𝒕
𝑹𝑹𝑹. 𝑯 = 𝒙 𝟏𝟎𝟎%
𝑺𝑽𝑷 𝒂𝒕 𝒕𝒉𝒆 𝒐𝒓𝒊𝒈𝒊𝒏𝒂𝒍 𝒂𝒊𝒓 𝒕𝒆𝒎𝒑𝒆𝒓𝒂𝒕𝒖𝒓𝒆

OR
 Relative humidity is the ratio of the Actual vapour density to the saturated vapour density
Mathematically:
𝒂𝒄𝒕𝒖𝒂𝒍 𝒗𝒂𝒑𝒐𝒖𝒓 𝒅𝒆𝒏𝒔𝒊𝒕𝒚(𝒑𝒓𝒆𝒔𝒖𝒓𝒆)
RH = 𝑺𝒂𝒕𝒖𝒓𝒂𝒕𝒆𝒅 𝒗𝒂𝒑𝒐𝒖𝒓 𝒅𝒆𝒏𝒔𝒊𝒕𝒚(𝒑𝒓𝒆𝒔𝒖𝒓𝒆) x 100%

For example, if the actual vapor density is 12 g/m3 at 200 compared to the saturation
vapor density at that temperature of 20.5 g/m3, then what will be its relative humidity?
Soln:
AVD = 12g/m3, SVD = 20.5 g/m3, T = 200 C
𝒂𝒄𝒕𝒖𝒂𝒍 𝒗𝒂𝒑𝒐𝒖𝒓 𝒅𝒆𝒏𝒔𝒊𝒕𝒚(𝒑𝒓𝒆𝒔𝒖𝒓𝒆)
From: RH = x 100%
𝑺𝒂𝒕𝒖𝒓𝒂𝒕𝒆𝒅 𝒗𝒂𝒑𝒐𝒖𝒓 𝒅𝒆𝒏𝒔𝒊𝒕𝒚(𝒑𝒓𝒆𝒔𝒖𝒓𝒆)

𝟏𝟐 𝒈/𝒎𝟑
∴ 𝑹. 𝑯 = 𝟐𝟎.𝟓 𝒈/𝒎𝟑 x 100% = 𝟓𝟖. 𝟓%

 Since absolute humidity is the density of water vapor in saturated air

Thus
𝒂𝒄𝒕𝒖𝒂𝒍 𝒗𝒂𝒑𝒐𝒖𝒓 𝒅𝒆𝒏𝒔𝒊𝒕𝒚
RH = x 100%
𝒂𝒃𝒔𝒐𝒍𝒖𝒕𝒆 𝒉𝒖𝒎𝒊𝒅𝒊𝒕𝒚

 Actual vapour density and Saturated vapour density has equal volume, therefore
relative humidity can also be defined as

Absolute Humidity
 Is the mass of water vapor divided by the mass of dry air in a certain volume of air
at a specific temperature
Mathematically:
𝒎𝒂𝒔𝒔 𝒐𝒇 𝒘𝒂𝒕𝒆𝒓 𝒗𝒂𝒑𝒐𝒓 𝑴
𝑨𝐛𝐬𝐨𝐥𝐮𝐭𝐞 𝐡𝐮𝐦𝐢𝐝𝐢𝐭𝐲 (𝐀. 𝐇) = = 𝟏
𝒎𝒂𝒔𝒔 𝒐𝒇 𝒅𝒓𝒚 𝒂𝒊𝒓 𝑴𝟐

Also can be defined as;

Specific Humidity (Humidity Ratio)
Is the ratio of water vapour to dry air in a particular mass
𝒎𝒂𝒔𝒔 𝒐𝒇 𝒘𝒂𝒕𝒆𝒓 𝒗𝒂𝒑𝒐𝒓 𝑴
Thus, Specific humidity (S.H) = 𝒎𝒂𝒔𝒔 𝒐𝒇 𝒅𝒓𝒚 𝒂𝒊𝒓
= 𝑴𝒗
𝒂

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 Difference between Absolute humidity and Relative humidity
Absolute humidity Relative humidity
Is the actual amount of water vapor Is a percentage of the amount of moisture
present in the air the air could possibly hold
It is expressed in moisture per cubic It is expressed in percentage as the ratio of
meter of air (g/m3) vapor pressure to saturated vapor pressure
It is totally independent of the If the temperature goes up, relative
temperature humidity goes down and vice – versa

Measurement of Relative Humidity

 It is measured by dry and wet bulb hygrometer and Renault hygrometer
 Hygrometer is an instrument used to measure the amount of humidity and water
vapor in the atmosphere ,in soil or in confined spaces

Dry and Wet Bulb Hygrometer

 It consists of dry bulb thermometer which is used to measure the temperature of
the surrounding air while wet bulb which is wrapped with piece of cloth around
the bulb which immersed in a reservoir of water which cool the wet thermometer
Diagram:

 Relative Humidity by using dry and wet bulb is a ratio of the difference between
the temperature of the dry bulb and the temperature of the wet bulb to the
temperature of the dry bulb
That is
𝑻𝒅𝒓𝒚 − 𝑻𝒘𝒆𝒕
RH = x 100%
𝑻𝒅𝒓𝒚

Example
7. A dry bulb thermometer reads 300 C and a wet bulb thermometer reads 240 C .What
is the Relative Humidity of the air
SOLUTION:
Dry bulb reading = 300C
Wet bulb reading = 240C
Difference in temperature = 300 – 240 = 60 C
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 Relative Humidity =?
From:
𝑻𝒅𝒓𝒚 − 𝑻𝒘𝒆𝒕
RH = x 100%
𝑻𝒅𝒓𝒚

𝟑𝟎− 𝟐𝟒 𝟔
∴ 𝑹. 𝑯 = 𝟑𝟎
𝒙 𝟏𝟎𝟎% = 𝟑𝟎 𝒙 𝟏𝟎𝟎% = 𝟐𝟎%
NB:

Renault Hygrometer
 It consists of an enclosed thin silver tube containing ether and a thermometer.
 There is also a tube through which air can be pumped into the ether.
Diagram:

Mechanism of Renault Hygrometer

The heat transfer from atmosphere to ether by convection in a tube, Ether evaporates
result cooling of the silver tube surface. Cooling continues until air adjacent to the
outside surface of the tube becomes saturated with water vapour. Some water vapour
condense outside the tube to form dew
Applications of Humidity
 It is used by meteorological departments to forecast the weather
 It is used to determine the appropriate site to locate cotton
 Electrical and electronic components are usually transported and stored in a dry air
 Used in hospitals in an operating room. RH at operating room is at least 50%
 It is used in storage and transportation of food items
Individual task – 9:1
1. The dry bulb temperature reading of a hygrometer is 220C and the wet bulb
temperature reading is 180C. What is the RH? (ANS: 18.2%)
2. The dry bulb temperature reading of a hygrometer is 40 0 C and the wet bulb
temperature reading is 300 C. What is the RH? ( ANS: 25%)
3. The relative density of a place was measured at 250 C and found to be 53.6%. if
the absolute humidity is 23.05g/m 3, determine the actual water vapour density at
this experiment (ANS: AVD = 12.35 g/m3)

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 Class Activity
1. Define evaporation and state the factors which affect the rate of evaporation of a
liquid .How does the kinetic theory account for the cooling produced in a liquid
which is evaporating
2. Distinguish between : (a) snow (b) hailstones (c) mist
3. Define the following terms (a) Dew (b) Dew point (c) Specific humidity
4. What factors determine saturation of vapour pressure?
5. Explain the principles used to measure relative humidity?
6. Explain the factors that affecting evaporation
7. Explain the difference between a vapour and a gas
8. At a given pressure the thermometer of a wet bulb reads 21 0 C. If the Relative
Humidity is 30 %, what is the temperature of the air?
9. A mass of air at 20˚C has a relative humidity of 39%.if the air is cooled to 15˚C,
what will its relative humidity be?
10. A mass of air has a relative humidity of 57% and a dew point of 20˚C. What is the
temperature of the air mass?
11. Why does air cool when it rises through the atmosphere? ( ANS: As air rises, it
expands because air pressure decreases with an increase in altitude. When
expands, it cools adiabatically)
12. A room with dimension of 7m x 10m x2m holds air that is saturated with water
vapour. The saturated vapour pressure of the water vapour is 7.37 kPa. If all the
water vapour in the room was condensed, What volume would the water occupy?
give your answer in m³.
13. A sealed box with the volume of 1.2m³ holds air with relative humidity of 22% at a
temperature of 15˚C. A beaker of water also at 15˚C is placed in the box. After 2
hours ,the level of the water in the beaker stopped dropping
(a) What mass of water evaporated from the beaker?
(b) If the temperature of the air in the box is increased to 30˚C, how much more
water will evaporate?
14. The table gives the temperature and dew point in four towns at 12 noon.
City Temperature (0C) Dew Point
Arusha 16 1
Morogoro 24 12
Zanzibar 3 5
Dar es salaam 28 4
(a) In which town is the relative humidity the highest?
(b) In which town is the relative humidity the slowest?
15. Briefly explain the reasons for the following:

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 (a) When a cold bottle is brought into a warm room, it becomes misted over
(b) Frost is more likely to occur on a clear night than on a cloudy night.
16. Define the term relative humidity. At a certain temperature and pressure, air holds
120 g of water vapor. If at this temperature and pressure the air is holding only 40 g
of water vapor, what is the relative humidity of the air?
17. On a given day,the room temperature is 20˚C and the dew point is 5˚C.Find the
relative humidity,given that saturated vapour pressure of water at 5˚C is 6.54
mmHg and that the saturated vapour pressure of water at 20˚C is 17.5 mmHg.
18. (a) What does the term saturation of water vapor mean?
(b) what is the difference between vapour pressure and saturated vapour pressure
19. Differentiate between
(a) Absolute humidity and Relative humidity (c) Saturated and unsaturated vapour
(b) Evaporation and Boiling
20. Explain why dew is formed at night.
21. What are the two processes that change a liquid into a gas. The state their
difference
22. A student is investigating the evaporation of water in the Laboratory. He pours 200
cm3 of water in a 250 cm3 beaker . He pours another 200 cm3 of water in a 1000
cm3 .The water in both beakers is initially at 50 0 C
(a) Use the kinetic theory of matter to explain which beaker evaporates faster
(b) List three other factors that affect evaporation of a liquid
(c) Explain why the liquid that remains during evaporation cools
23. The actual vapor density of a region at 230 C temperature is 15g/m3 , if the
saturation vapour density at that temperature of 21.3 g/cm 3 determine the region ‘s
relative humidity
24. At a temperature of 300 C the mass of water vapour in town K is 22 g while dry air
has a mass of 15 g. Determine the specific humidity of town K at the stated
temperature
25. A wet bulb thermometer reads 17.20 C, If the relative humidity of the air is 40%,
what is the temperature of the air? (ANS: TAIR = 28.70 C, → TAIR = TDRY BULB)
26. A mass of air at 300 holds 15 g/cm3of water vapor .If the saturation point of the air
is 30 g/cm3 ,calculate the relative humidity of the air
27. If the dry bulb temperature is 320 C and the wet bulb temperature is 240 C, what is
the relative humidity of the air?

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TOPIC: 10 CURRENT ELECTRICITY
 Electric current is the rate of flow of charges past a point or region.
 It is measured by Ammeter
Mathematically:
𝑸𝒖𝒂𝒏𝒕𝒊𝒕𝒚 𝒐𝒇 𝒄𝒉𝒂𝒓𝒈𝒆𝒔(𝑸)
𝑬𝒍𝒆𝒄𝒕𝒓𝒊𝒄 𝒄𝒖𝒓𝒓𝒆𝒏𝒕 (𝑰) =
𝑼𝒏𝒊𝒕 𝒕𝒊𝒎𝒆(𝒕)

 The common SI unit of current (I) is ampere (A)

Electric Potential difference (P.d)/ Electric pressure

 Is the work done per unit charge in moving electric charge from one point to
another in an electric field
OR
 Is the difference in electrical potential between any two points
Mathematically:
𝒘𝒐𝒓𝒌 𝒅𝒐𝒏𝒆(𝑾)
𝑷. 𝒅 (𝑽) =
𝒄𝒉𝒂𝒓𝒈𝒆(𝑸)

 The common SI unit of P.d is Volt (V)

Electromotive Force (e.m.f)

 Electromotive force of a cell is the maximum potential difference between the
terminals of a cell when the cell is not doing any work
 It is also called voltage. It is measured in VOLTMETER
 It pumps charges to flow through an electric circuit although it is not a force is
just a potential
 The difference between the electromotive force and the terminal voltage is
known as LOST VOLTS

Ohm’s Law
 It states that: “At constant temperature and other physical factors, a current in
conductor is directly proportional to the potential difference across its end”
Mathematically:
𝑉∝I → 𝑉 = 𝐾𝐼
Where: K= constant = R = resistance
∴ 𝐕 = 𝐈𝐑
Graphically:

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 𝛥𝑉
 From the graph above, Slope(m) = , where, Slope = Resistance
𝛥𝐼
Limitation of ohm’s Law
 It does not apply to some electrolytes e.g dilute H2SO4
 It does not apply for conduction of electricity in gases
 It does not hold in semiconductors (diodes and transistors)

Individual task –10:1

1. An electric heater draws 3.5 A from a 110 V source. What is the resistance of the
heating element? (ANS: R = 31.4Ω)
2. A current of 2 A is passed through a conductor of resistance 10 Ω. What is the
potential difference between the ends of the conductor. (ANS: V = 20V)

Factors affecting the Resistance of a conductor

Consider the diagram:

1. Length of the conductor

 The longer the wire the higher the resistance and vice versa ie R ∝ 𝑳
[

2. Temperature
 The higher the temperature, the higher the resistance and vice versa
 For semiconductors increase in temperature decreases the resistance and vice versa
3. Nature of material
 Resistance depends on the type of material used for making conductor
 For example, Nichrome wire has more resistance than a copper wire of the same
dimensions.
 Nichrome wire is used in heating element of electric fires
 Copper wire is mostly used for connecting wires

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4. Cross-section area
 A thin wire has more resistance than a thick wire. That is the cross – sectional
𝟏
area of a conductor increases with decrease in the resistance 𝐑 ∝
𝑨
𝟏 𝐿
 Now combine the relation R ∝ 𝑨 and R ∝ L → R∝ 𝐴
𝑲𝑳
On removing the proportionality constant 𝑹 =
𝑨
Whereby: k = resistivity which is denoted by letter 𝝆
𝝆𝑳 𝑹𝑨
R= ---------- make 𝜌 the subject → 𝝆=
𝑨 𝑳
𝑹𝑨
∴ 𝝆=
𝑳

 Resistivity is the ability of a material to oppose the flow of an electric current.

 Its SI unit is Ohm metre (Ωm).
NB:
 Conductance(G) is the reciprocal of resistance of a conductor
Its SI unit is siemen (S)
 Conductivity(𝝈) is the reciprocal of the resistivity of a conductor
Its SI unit is siemen per metre ( Sm -1)

Resistivity of some materials at 200 C

Material Resistivity (Ωm) Material Resistivity(Ωm)

Aluminium 2.7 x 10-8 Constantan 4.9 x 10-7
Chromium 1.3 x 10-7 Manganin 4.8 x 10-7
Copper 1.68 x 10-8 Nichrome 1.0 x 10-6
Iron 9.71 x 10-8 Glass 1 x 109 - 1 x 1013
Lead 2.1 x 10-7 Rubber 1 x 1013 - 1 x 1015
Silver 1.6 x 10-8 Quartz 7.5 x 1017

Individual task – 10:2

1. A wire of length 40 m and cross – sectional area 0.8 mm2 has a resistance of
10Ω. What is the resistivity of the material of the wire? (ANS: 𝝆 = 𝟐x10-4 Ωm)
2. What is resistance of a copper wire of length 20m and diameter of 0.080 cm?
Resistivity of copper is 1.68 x 10-8 Ωm (ANS: R = 0.67Ω)
3. A steel bar has a length of 2.3m and diameter of 2 x 10-5 m. what is resistance?
(Resistivity is 10.5 x 10-8 Ωm) (ANS: R = 768.72 Ω)
4. What length of a wire of cross –sectional area 0.2mm2 and resistivity 0.072𝜇 Ωm is
needed to wind a coil of resistance 9 Ω. (ANS: L = 25m)
5. The resistance of a certain wire is 12 Ω. What are the resistance of another wire
of the same material but with half the length and half the radius of the first wire?
(ANS: R2 = 24 Ω)
Written by Geoffrey M Idebe (0688 082 089 – Mwanza) Page 128
Resistors
 Resistor is a device which offers resistance to the flow of an electric current.
 It used to control the magnitude of current and voltage according to ohms law

Types of Resistors
They are divided according to the materials used to make them and the value of
resistance offered
Types of Resistors due to material used
(i) Wire wound resistor
(ii) Carbon resistor
(iii) Metal film resistor
(iv) Metal oxide film resistor
Wire Wound Resistor
 It is made by winding wires made of certain metallic alloys into spools (used to
control amount of resistance)
Carbon Resistor
 It is made by mixing carbon granules with varying amount of clay and moulding
them into cylinders
Metal Film Resistor
 It is made up of a stable ceramic core coated with metal alloys such as nickel
chromium. It is more accuracy and more expensive than carbon resistor
Metal Oxide Film Resistor
 It is made up of a stable ceramic core coated with metal alloys such as tin oxide

Types of Resistors due to Value Offered

 Fixed resistor
 Variable resistor
Fixed Resistor
 It has a resistor value which cannot be changed. Eg. 2Ω, 3Ω, 4Ω etc.
 Most carbon resistor are fixed resistor
Variable Resistor
 It has a resistor value which can be changed by means of control.
 Example, potentiometers, thermistors, photo resistors and rheostat

Resistor Colour Codes

 Resistors which are used in electronic devices always painted with different
colour texture called bands. The band represent the exactly value of a
resistance. It contains fourth band with different meaning.

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Resistor’s Colour Codes

Tolerance colour codes

Tolerance ±1 ±2 ±5 ±10 ±20
Colour Brown Red Gold Silver No
colour
Diagram:

 First band – first digit

 Second band – second digit
 Third band (multiplier) – number of zero
 Fourth band (tolerance) – percentage accuracy

Example
From the diagram of resistor above find the exactly resistance
First band (yellow) – 4
Second band (purple) – 7
Multiplier (red) – 2 = 00 number of zero
Tolerance (gold) – ±5%
Therefore: R = ±5% of 4700Ω
The actual value resistance is ±5% of 4700Ω

Individual task – 10:3

1. A resistor is connected across a 50 V source. What is the current in the resistor if the
color code is red, orange, orange, silver? (ANS: I = 2.2 mA)

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Combination of Resistors
 Resistors can be combined in series or in parallel arrangement
Series Connection
 In this arrangement, the resistors are connected end to end consecutively.
 The same current, I, is flowing through each resistor

From: P.d across the battery = sum of p.d around a conducting path
that is: V = V1 + V2 --------- (i), but current, I, is constant in each resistor
From ohm’s law: V = IR → V = IRT

From equation (i), 𝑉 = 𝑉1 + 𝑉2 → 𝐼𝑅𝑇 = 𝐼𝑅1 + 𝐼𝑅2

𝐼𝑅𝑇 = 𝐼(𝑅1 + 𝑅2 ) 𝑑𝑖𝑣𝑖𝑑𝑒 𝑏𝑦 𝐼 𝑒𝑎𝑐ℎ 𝑠𝑖𝑑𝑒 → 𝐑 𝐓 = 𝐑 𝟏 + 𝐑 𝟐

∴Total resistance (RT) for resistors in series connection is given by

𝐑𝐓 = 𝐑𝟏 + 𝐑𝟐 + ⋯ … … . + 𝐑𝐧
NB:
 As the branches increases, the voltmeter reading keeps increasing for the total volts
 Each branch has its own voltage value
 Series arrangement results high total resistance
Parallel Connection
 Resistors are connected across two common points in a parallel arrangement.
 The total current, IT in this connection divides into its individual branch (ie I1, I2 ….)
 The voltage, V is the same throughout the circuit
 As the branches increases, the ammeter reading keeps increasing for the total
current. Each branch has its own current value
 Parallel arrangement results low total resistance
 Parallel connection is used in house wiring

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Since: 𝐈𝐓 = 𝐈𝟏 + 𝐈𝟐 …………(i) , but: V = the same at all points round a circuit
𝑉 𝑽
From ohm’s law: V = IR → 𝐼 = → 𝑰𝑻 =
𝑅 𝑹𝑻
Then, from equation (i)
𝑽 𝑽 𝑽
𝐈𝐓 = 𝐈𝟏 + 𝐈𝟐 →
= +
𝑹𝑻 𝑹𝟏 𝑹𝟐
𝑽 𝑽 𝑽
𝑹𝑻
= 𝑹 +𝑹 … … . 𝒅𝒊𝒗𝒊𝒅𝒆 𝒃𝒚 𝑽 𝒆𝒂𝒄𝒉 𝒔𝒊𝒅𝒆
𝟏 𝟐

𝟏 𝟏 𝟏 𝑹𝟏 𝑹𝟐
𝒕𝒉𝒖𝒔, = + → 𝑹𝑻 =
𝑹𝑻 𝑹𝟏 𝑹𝟐 𝑹𝟏 +𝑹𝟐

∴Total resistance (RT) in parallel connection is given by

𝟏 𝟏 𝟏 𝟏
𝑹𝑻
= 𝑹 + 𝑹 + ⋯……..+𝑹
𝟏 𝟐 𝒏

Whereby: Rn = the last resistor

N.B
 It is advisable to connect bulbs in parallel during electrical installation so that
when the bulb blows out or disconnected, the other bulbs will keep working
 If you connect in series when one bulb disconnected (blows out) will cause the
other bulbs not working
Examples
1. Calculate the combined resistance in:

Solution:
(a) R1 = 5 Ω, R2 = 15 Ω
From: RT = R1 + R2 → 𝑅𝑇 = 5 + 15 = 20Ω
(b) R1 = 3 Ω, R2 = 6 Ω, RT=?
𝑹 𝑹 3𝑥6 18
From: 𝑹𝑻 = 𝟏 𝟐 → 𝑅𝑇 = = = 2Ω
𝑹𝟏 +𝑹𝟐 3+6 9

Individual task – 10:4

1. What is the potential difference across 2 Ω resistor? (ANS: V = 1 V)

Written by Geoffrey M Idebe (0688 082 089 – Mwanza) Page 132

 2. Consider the diagram below

(a) What is the total resistance of the circuit? (b) What current flows in the circuit?
(c) What is the potential drop across each resistor? (d) What is the electric potential at
point A? (ANS: (a) 15 ohms (b) I = 0.6 (c) 2V, 3.6V and 3 V (d) VA= 6.6V)
3. Consider the diagram below

(a) What is the total resistance of the circuit? (ANS: 1.62 ohms)
(b) What total current flows in the circuit? (ANS: 5.55 A)
(c) What current flows through each resistor? (ANS: 2.25A, 1.5A, 1.8A)
4. Determine the current reading on the ammeter in the circuit shown in the
diagram below

Internal Resistance of a Cell

 Cell has an internal resistance that opposes the flow of electric current and
causes the potential drops across this resistance.
 The relationship between internal resistance (r) and emf (E) of a cell is given by
𝐄 = 𝐈(𝐑 + 𝐫) = 𝐈𝐑 + 𝐈𝐫. (See the fig. below)

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Whereby:
 Emf (E) = VT = V+VL, K = switch
R = external resistance (load resistance), r = internal resistance
P.d across R (terminal Voltage), V = IR
P.d across r (Lost voltage), VL = Ir
𝑬−𝑰𝑹
∴r =
𝑰

NB:
(a) When the terminals of the battery are short circuited (when disconnected from
their external circuit), the resistance of the circuit is the internal resistance of the
cells that is: 𝐄 = 𝐈(𝐑 + 𝐫) = 𝐈𝐑 + 𝐈𝐫, here 𝐈𝐑 = 𝐕 = 𝟎
∴ 𝐄 = 𝐈𝐫
(b) The total e.m.f available in the cell is used up in two ways
 Driving the current trough the cell i.e to overcome internal resistance (r)
 Driving the current through the external resistance R
Example
1. The potential difference across the cell when no current flows through the circuit
is 3 V. When the current I = 0.4A is flowing, the terminal potential difference falls
to 2.8 V. Determine the internal resistance (r) of the cell.
Solution:
Given: E = 3V, V = 2.8 V, I = 0.4A, r =?
𝑬−𝑰𝑹 𝑬−𝑽 𝟑−𝟐.𝟖
From: 𝑟 = = → 𝑟= = 𝟎. 𝟓
𝑰 𝑰 𝟎.𝟒

Individual task – 10:5

1. What is the internal resistance of cell when there is current of 0.4A, when a
battery of 6 V is connected to a resistor of 13.5. (ANS: r = 1.5)
2. What is the maximum current of a battery of e.m.f 3.0 V and internal resistance of
1.0Ω (ANS: I = 3 A)
3. An old cell with an emf of 1.7 V has an internal resistance of 0.8 Ω. How much current
will initially flow it its terminals are short – circuited? (ANS: I = 2.13A)
4. A dc source with an internal resistance of 0.11 Ω is connected across a length of
nichrome wire having a resistance of 20 Ω .If a voltmeter across the nichrome
indicates a drop of 5 V, what is the emf of the source? (ANS: emf = 5.0275 V)
5. A cell with an emf of 1.4 V and internal resistance of 0.05 Ω is placed in a circuit with
several resistors. The cell provides 0.52 A to the circuit. What is the terminal voltage of
the cell? (ANS: V = 1.374 V)
6. In the circuit shown below, the battery has an e.m.f of 6.6 V and internal
resistance of 0.3 ohms. Determine the reading of the ammeter. (ANS: 2.357A)

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 7. A cell supplies a current of 0.6A through a 2 Ω resistor and a current of 0.2A
through a 7 Ω resistor. Calculate the e.m.f of the cell and the internal resistance
(Answer: E.m.f = 1.5 V, r = 0.5Ω)
8. A complete circuit consists of a 18 V, battery and a resistor R. The terminal voltage
of the circuit is 15.8V and the current is 4A. What is:
(a) The internal resistance, r of the battery. (ANS: r = 0.55 𝛀)
(b) The resistance R of the circuit resistor. (ANS: R = 3.95 𝛀)

Wheatstone bridge (Meter Bridge)

 Wheatstone bridge is an electrical bridge circuit used to measure the unknown
resistance of a conductor.
 Consider the circuit in the figure below whereby X is the unknown resistor

By adjusting one of the known resistors, usually resistor Q, we can reach a value at
which the galvanometer shows no deflection (zero reading) .The bridge at this point is
said to be balanced and there is NO p.d across BC, thus two points have the same
potential
Ie VAB = VAC = VBD = VCD OR VAB/ VBD = VAC/ VCD
Since the current through the galvanometer is 0, P and R carry the same current
I1. From Ohm’s law:
VAB/ VBD = I2P/I2R = P/R and VAC/ VCD = I1Q/I1X
𝑷 𝑸 𝑸𝑹
∴ = → X=
𝑹 𝑿 𝑷

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The meter Bridge
 The metre bridge is one of the practical forms of the Wheatstone bridge used for
measuring resistance
 It consists of a uniform resistance wire; AC (usually 100 cm long) stretched along
side a metre scale. See the fig below
 The resistance R1, is connected across the gap “ab” and another resistor (R2) is
connected across the gap b1c.
 A galvanometer G is connected to the terminal B and to a jockey D

 As the jockey is moved along the AC, at one position D, the galvanometer will
read zero. The metre bridge is then said to be balanced. D is the balancing point
and the length is the balance length
 R1 is the resistance to be measured; R2 is a standard resistance of a value near
to R1. Then we can write
𝑹𝟏 ∝ 𝑳𝟏 … … … … … … … . . (𝒊), 𝑹𝟐 ∝ 𝑳𝟐 … … … … … … … … . . (𝒊𝒊)
𝒅𝒊𝒗𝒊𝒅𝒊𝒏𝒈 𝒕𝒉𝒆 𝒆𝒒𝒖𝒂𝒕𝒊𝒐𝒏 𝒘𝒆 𝒈𝒆𝒕;
𝑹𝟏 𝑳𝟏 𝑳𝟏
= 𝒃𝒖𝒕 𝑳𝟐 = 𝟏𝟎𝟎 − 𝑳𝟏 → 𝑹𝟏 = 𝑹𝟐 ( )
𝑹𝟐 𝑳𝟐 𝟏𝟎𝟎 − 𝑳𝟏

Example
1. What is the resistance of a wire if it balances a standard resistor of 5 ohms at the
56 cm from the end of the metre bridge?
Soln:
Given: R2 = 5, L1 = 56 cm
𝑳𝟏 𝟔𝟎 𝟓𝟔
From: 𝑹𝟏 = 𝑹𝟐 (𝟏𝟎𝟎−𝑳 ) → 𝑹𝟏 = 𝟓 (𝟏𝟎𝟎−𝟔𝟎) = 𝟓 𝒙 𝟒𝟎 = 𝟕
𝟏

Individual task – 10:6

1. The galvanometer in the bridge network shown in the fig. below, gives no
deflection. What is the value of X? (ANS: X = 6)

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Heating effect of an Electric Current
 When an electric current is passed through a conductor, the conductor becomes hot
after some time and produce heat. This happens due to the conversion of some
electric energy passing through the conductor into heat energy. This effect of
electric current is called heating of current.

Factors affecting Heat Quantity

 Resistance of a Conductor
The higher the heat, the higher the resistance and vice versa
 Magnitude of the Electric Current
The higher the current, the more the heat produced
 The time for which Current flows
The heat produced by an electric current is proportional to the time taken by the
current to pass through a conductor

Joule’s Law
 It tells us the relationship between resistance, current and heat generated.
 State that “The rate at which heat is produced in a resistor is proportional
to the square of the current flowing through it, if the resistance is constant.”
Mathematically:
𝑯 𝟐 𝑯
∝ 𝑰 𝑹 → = 𝒌𝑰𝟐 𝑹
𝒕 𝒕

𝑯 = 𝒌𝒕𝑰𝟐 𝑹, where k =1

∴ 𝑯 = 𝑰𝟐 𝒕𝑹
𝒗
𝐵𝑢𝑡 𝑹 = ………………. substitute in the formula above
𝑰
∴ 𝑯 = 𝑰𝒕𝑽
𝑽
Also, 𝑰 = 𝑹 ……………….. substitute in the formula above
𝑽𝟐
𝑻𝒉𝒆𝒓𝒆𝒇𝒐𝒓𝒆: 𝑯(𝑬) = 𝒕 ( ) = 𝒕𝑰𝑽 = 𝒕𝑹𝑰𝟐
𝑹
Example
1. An electric iron box has resistance of 30  and it takes a current of 10 A.
Calculate the heat I kJ developed in 1 minute.
Soln:
Given: R = 30, I = 10 A, t = 1min = 60 seconds
From E = I2Rt

Therefore, E = 102 x 30 x 60 = 180 kJ

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NB:
 Whenever an electric current passes through a conductor, electrical energy is
converted to other forms of energy e.g heat, light etc (According to the law of
conservation of energy)

Electrical Power
Electrical power is the rate of potential difference
OR is the rate at which electrical energy is dissipated
𝑷.𝒅 𝒘 𝑸𝑽 𝑰𝒕𝑽
𝑷= = 𝑃= = = 𝑰𝑽
𝒕𝒊𝒎𝒆 𝒕 𝒕 𝒕
∴ 𝑷 = 𝑰𝑽
Example
1. An electric bulb is rated 60 W, 240 V. Determine:
(a) The resistance of the filament
(b) The current flowing through the bulb when it is connected to the mains supply
Soln:
Given: P = 60W, V = 240 V
𝑉2
(a) From P = IV =
𝑅
𝑉2 𝑉2 2402
𝑃= →𝑅= = = 960
𝑅 𝑃 60
∴ 𝑅 = 960

(b) From ohms law: V = IR

𝑉 240
∴ 𝐼 = 𝑅 = 960 = 0.25 𝐴

𝑨𝒍𝒕𝒆𝒓𝒏𝒂𝒕𝒊𝒗𝒆𝒍𝒚
By using power formulae P = IV
𝑃 60
∴𝐼= = = 0.25 𝐴
𝑉 240

Individual task – 10:7

1. An electric kettle draws a current of 10A when connected to the 230V mains supply. If
all the energy produced in 5 minutes is used to heat 2kg of water. Calculate
(a) The power of the kettle (b) The energy produced in 5 minutes
(c) The rise in temperature (Specific heat capacity of water = 4200 Jkg-1K-1)
ANS (i) 2.3 kW (ii) 690 kJ (iii) 82.14 K
2. An electric motor powered by a 240 V mains supply requires a current of 30A to
lift a load of mass 3 tonnes at the rate of 5 m per minute. Calculate:
(a) The power input (ANS: PIN = 7200 W)
(b) The Power output (ANS: POUT = 2500 W)

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 (c) The overall efficiency of the machine (ANS: Eff = 34.72%)
3. What is the maximum number of 100 W bulbs which can be safely connected
from a 240 V source supplying a current of 5 A? (ANS: n = 12 bulbs)

Applications of heating effect of electric current

 It is used in electrical heating appliances such as iron, room heaters water
heaters etc
 It is used in electric bulbs to produce light
 It is used in electric fuse
NB: The wire used in the filament of household bulbs should have high resistance and
high melting point (heated to a very high temperature) in order to emit light

Electrical appliances
 Electrical appliance is the device which uses heating element to produce heat
energy.
 Nichrome wire is among of the heating elements due to its high melting point.
 The common electrical appliances includes Heaters, Electric iron, Bulbs, kettles,
Cookers, fridges, Televisions and Air conditioner

Rating of the Electrical Appliance

 Rating of the appliance is the rate at which the appliance dissipate energy.
 Power rating is the maximum power that can be used to operate an electrical
device
 Each electrical appliance has its own rating which enables us to know energy
dissipated
 For example, an appliance marked 3000W, 240V dissipates energy at the rate of
3000 Joules per second when connected to 240V
 The resistance of a filament increases with the increase in temperature

Power Ratings of common Electrical Appliances

Electrical power ratings at 240V

appliance
Immersion heater 2000W (2KW)
Electric heater 2000W (2KW)
Electric iron 1000W (1KW)
Electric cattle 2500W (2.5KW)
Color TV 300W
Refrigerator 120W
Light bulb 25W - 150W

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NB:
 If the mains supply falls below 240 V, the rating of the appliance would drop
 Similarly when the power supply rises ,the rating would rise and it would damage
the appliance due to overheating

Measurement of Electrical Power & Billing of electrical energy

 Power Is the rate of doing work.
OR Is the work done per unit time (second)
 Power companies like TANESCO usually measure the electrical energy in
kilowatt hours (KWh)
1KWh = (1KW x 1hour) J = (1000W x 60 x 60) 1KWh = 3600000J = 3600KJ

Individual task – 10:8

1. An electric cooker has a coil of resistance 5000. If is operated on a 250 V
mains supply for 1 hour, how much heat energy does it produce? (ANS:E = 45kJ)
2. A television set rated 200W is switched on for 5Hours every day. How much
energy does it consume in 30 days ANS: E = 1.08 x 105kJ
3. A house has five rooms, each with a 60W, 240V bulb. If the bulbs are switched
on 7:00p.m to 10:300p.m determine the power consumed by bulbs per day.
ANS: P = 1.05kWh
4. A bulb rated 120 V, 75 W burns continuously for two days. Given that the cost of
one unit ( 1kW) is 320 Tsh .Determine the:
(a) Total electrical energy consumed (b) Total power bill
ANS: (a) 3.6 kWh (b) 1152 Tsh

Electrical Installation of a House

 Domestic electricity is supplied by two cables, live (L) and Neutral (N), the third
cable is Earth(E) used to provide extra safety

Live Cable (L)

 The live cable is 240V relative to the neutral. The current in the live cable
alternates 60 times a second (60 Hz). It is represented by brown or red colour
 The potential difference between the LIVE and the NEUTRAL wire is 240 V–0 V = 240 V

Neutral Cable (N)

 The Neutral cable is earthed at the power station. This is to ensure current at
neutral cable remains zero potential (V = 0V) so it cannot give an electric shock
on touching. It represented by blue or black colour
Earth Cable (E)
 The earth cable introduces to provide extra safety especially in electrical
appliances. It is represented by yellow or green color

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Three Pin Plug
 It consists of all three cables which are Live cable, Neutral cable and Earth cable
with a fuse connected to live cable, sometimes fuse can be connected to neutral
cable which is not safe be

NB:
 The earth pin usually longer than the other two
 Switch must be off when you push the plug into the socket
Two Pin Plug
 It consists of only two cables which are live cable and Neutral cable.
 An appliance using a two – pin plug its body is not connected to earth (see the
fig below)

Fuses
 A Fuse Is a safety device used to protect an electric circuit against excess of current.
OR
 A fuse is a short piece of special wire which melts when more than a rated amount
of current passes through it
 It is made of a thin copper wire covered with tin or a lead – tin alloy
 It works as a circuit breaker or stabilizer which protects the device from damage
Types of Fuses
(a) Rewireable fuses (c) Cartridge fuses

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Mechanisms:
 The fuse works on the principle of the heating effect of an electric current
 It is always connected in series with the electrical circuit to protect from over
current in the running cables
 When the excessive current or heat is generated due to heavy current flows in the
circuit, the fuse melts down due to the low melting point of the element and it opens
(breaks) the circuit.
 The excessive flow may lead to the breakdown of wire and stops the flow of current
 Then the fuse can be replaced with the new one with suitable ratings

Applications of fuses
Fuses are used in:-
 Electrical appliances (devices)
 Automobiles such as car, trucks and other vehicles
 Scanners, portable electronics, hard disk drives
 Fuses in capacitors, transformers, power converters, motor starters, power transformers

What happen if we don’t use fuse?

 If we don’t use fuses, electrical faults occur in the wiring and it burns the wire
and electrical appliances and this may start fire at home.
 Also the lives of television, computers, radios and other home appliances may
put at risk

Circuit Breaker
 Is a type of switch that cuts off the flow of electric current when the current
exceeds a specific value. (See the fig. below)

Mechanism of Circuit Breakers

 When current exceeds, it increases the temperature and bimetallic strip bends to
push latch mechanism, enables the spring to cut off current
Diagram:

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Individual Task – 10:9
You have a choice of the following fuses 1A, 3 A, 5 A, 9 A, 13 A and 30 A
Select the best fuse for
(a) a 240 V , 7.2 kW electric cooker (b) a 240 V ,2kW electric iron
ANS: (a) The best fuse is a 30 A fuse (b) The best fuse is 9 A fuse

Domestic Wiring Circuit

 The power company connects power to the house up to the consumer unit where
the house wiring starts
 Consumer unit is the place where the main switch, main fuse and distribution
board are placed in a single box or unit
 From the consumer unit , the cables branch into the various parts of the house

Types of Domestic Wiring Circuit

Ring main circuit
Lighting circuit

Ring Main Circuit

 This is a cable which begins and ends at the consumer unit. Its three cables are
forming ring around part of the house. Its fuse is of 30A fuse

Lighting Circuit
 In this circuit the first lamp connected from the customer unit, in turn is
connected to the second lamp and so on.

Types of Lighting Circuit

Loop – in lighting circuit
Junction box lighting circuit

Loop In Lighting Circuit

 All three cables from consumer unit run to each ceiling roses, one after another.
From each rose another set of cables runs to the switch which operates the light

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Junction Box Lighting Circuit
 All three cables from consumer unit run to one junction box to another, where
one cable runs to the light and another run to the switch for that light.

Repairing Electrical Appliances Faults

 Multimeter and Live mains lead indicator are devices important when
checking electrical appliances faults.

Multimeter
 Multimeter is the single meter for measuring current (both a.c and d.c), voltage
and resistance

Live Mains Lead Indicator (Testor)

 Is an electronic device used for testing the flow of electric current

Repairing of Faults Procedures

If electrical appliance fails to work the following procedure should be done
o Check by using live mains lead indicator if there is power or not
o Check the cable from the socket to the appliance
o If no fault is detected , open the plug and check the fuse
o Check each cable for continuity by using a Multimeter
o If cables are working properly, then check the fault is in the element by using a
Multimeter
o If element is in fault, replace element as repair may not be possible
o If the element is not in fault , look for loose connections, these should be made firm
and/or cleaned of rust and other dirt

Sources of Faults in domestic system

 When fuse blows or melt
 Wire cutting
 Wire joining
 Socket getting dirty
 Switches breaking

Cells
A Cell is a device used to cause a flow of electric current in a conductor
Types of electrochemical Cells
 Primary cell
 Secondary cell
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Primary Cell
 Is a cell which produces current as a result of irreversible chemical changes
taking place within the cell
 Examples of primary cells are A Simple cell ,Leclanché cell and Dry cell

Simple Cell
 Made from copper as anode, zinc as cathode and Dilute sulphuric acid as
electrolyte
Diagram:

The processes that occur when the cell is in operation

 The dilute sulphuric acid separates into sulphate ions (SO42-) and hydrogen ions (H+):
H2SO4 (aq) 2H+ (aq) + SO42- (aq)

At anode
 Zinc metal dissolved into solution to form zinc cation which reacts with sulphate
anion to form zinc sulphate
Zn+ + SO42- ZnSO4
At cathode:
 Hydrogen cation discharges to liberate hydrogen gas (bubbles)
2H+ + 2e- H2
Defects of a Simple Cell
 A Simple cell has two defects which cause the current to drop quickly when the
cell is being used.
 These defects are local action and polarization

Local Action
 Is caused by the presence of small impurities in the zinc electrode
 Local action can be minimized by coating the surface of zinc with mercury
(Amalgamating)

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Polarization
 Is the formation of a layer of hydrogen bubbles on the copper plate
 The hydrogen insulates the copper plate thus increasing the internal resistance of
the cell, this lowers the current
 Polarization can be minimized by adding a chemical (depolarizer). Which reacts
with hydrogen bubbles to form water
 Example of depolarizer is potassium dichromate

Leclanché Cell
 Made from carbon as anode, zinc as cathode, ammonium chloride (NH 4CL)
solution and depolarizer manganese dioxide (MnO 2) (See the fig. below)

Dry Cell
 Is a modified Leclanche cell in which the ammonium chloride solution is replaced
with ammonium chloride jelly
 The Manganese(iv) oxide acts as depolarizer
 Ammonium chloride (NH4CL) and zinc chloride act as electrolyte. (See the fig below)

Uses of Dry Cell

 It is used to operate on radios, electronic calculators and other small
electrical devices
Advantages of Dry Cell
 It is portable
 The chemicals for its production are relatively cheap
 It has a relatively high e.m.f
 It is able to recover from its polarized

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Disadvantage of dry cell
 Cannot be recharged thus disposed after use

Secondary Cell
 Secondary cell is the cell which can be recharged after running down.
 Eg lead acid cell and nickel ferrous cell. Also called accumulators

Advantages of secondary cells

 Last longer than primary cell
 Can supply large current(low internal resistance) for a long time
 Can be recharged
Disadvantage of secondary cells
 Heavy and cumbersome to carry
 The electrolyte is corrosive
 It produces gases which may explode if ignited
 It cannot produce large currents in cold weathers (Lowering the temperature
causes chemical reactions to proceed more slowly)

Lead Acid battery

 It consists of two plates of lead immersed in dilute sulphuric acid .These are
Anode of lead(IV) oxide and cathode of spongy lead
 An electrode is separated by insulator called separator.
 Cathode joining together to form negative terminal while anode joining
together to form positive terminal. (see the fig. below)

Discharging of Lead – Acid Battery

 Discharge is the process of a battery to provide electrical energy.
 The energy is produced when the acid (electrolyte) gradually combines with the
active material of the electrodes.
 This lowers the concentration of the sulphuric acid.

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Charging of Lead Acid battery
 The aim of charging is to drive all the acid out of the plates and return it to the
electrolyte.
 During charging, the negative terminal of the battery charger is connected to
negative terminal of battery while the positive terminal of the charger is
connected to the positive terminal of the battery
 When the battery is fully discharged ,the battery is said to be sulphated (will
not function as it cannot be recharged)
Taking Care of Accumulators
 Cells should be charged regularly and should never left discharged
 The acid level should be maintained by adding distilled water when necessary
(never add acid)
 The terminal should be kept clean and greased
 Rough handling should be avoided
 The cells should not be short – circuited , Example, if you connect two terminals
to each other
 The rate specified by manufacturer should not be exceeded during charging
Uses of Accumulators
 Used to provide power in motor vehicles
 Used to provide energy to power domestic appliances such as radio
 Used together with solar panels to convert solar energy to electrical energy
 They are used to store electrical energy

Arrangement of Cells
Series Arrangement of Cells
 In this series arrangement the positive terminal of one cell is connected to the
negative terminal of another cell
 Electric current is the same at each cell
 Total voltage across the cells is equal to the sum of the voltage of the individual
cells, that is why torch light uses this arrangement (see the fig. below)

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Parallel Arrangement of Cells
 In parallel arrangement, all the positive terminals of the cells are connected together
and negative terminals of the cells are connected together. (see the fig below)

NB:
 The total voltage of all of the cells is the same as that of a single cell
 Total current across the cells is equal to the sum of the current across each cell,
that is why lead acid accumulator uses this arrangement

Class Activity–10
1. A current of 1.5 A flows in a wire .Find the total charge passed in 20 seconds(Q = 30 C)
2. A charge of 3600 C passes through an electric lamp in 3.0 minutes .What is the
current in the lamp (ANS: I = 20 A)
3. How many electrons pass through a lamp in 10 seconds, if the current is a 125
mA and the charge of one electron is 1.6 x 10-19 C? (ANS: n = 7.813 x 1018 electrons)
8. Electron in hydrogen atom revolves around the nucleus with frequency 6.0 x 10-4 per
second .Calculate the current in the orbit .Given that charge on an electron = 1.6 x
10-19 coulomb. (ANS: I = 9.6 x 10-5 A)
4. If a container of surface area 2.1 m 2 is to be coated with silver about 0.1 mm thick
,calculate the time it will take if a current of 4 amperes has to flow
5. A nichrome wire of radius 0.35 mm has a resistivity of 1.5 x 10 -6 Ωm .Given that
the wire has a length of 80 cm .Calculate
(a) its resistance (b) Conductance (c) Conductivity
6. Explain the factors which determine the resistance of a conductor
7. When resistors are connected in series ,which of the following is the same for all
the resistors (a) potential difference (b) current
8. Find the resistance of a wire of 1100 cm long , 0.2 mm diameter and of resistivity
1.57 x 10-6 Ω 𝑚 (ANS: R = 546.7 𝛀)
9. Calculate the energy dissipated by a resistor of 12 Ω in 4 seconds if a voltage of
6 V is applied. (ANS: H = 12 J)
10. Calculate the heat lost by a wire of resistance 16 Ω when a current of 30 A flows
through it in 1 second. (ANS: H = 14 400 J)

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 11. A battery consisting of two cells of 2 V and internal resistance of 1 Ω each is
sending a current through the filament of the lamp as shown in the circuit below

Calculate :
(a) The current in the circuit when
(i)Switch K is open (ANS: I = 0.67 A)
(ii)Switch K is closed (ANS: I = 0.8 A)
(b) the potential difference across the battery of two cells when K is closed(P.d =2.4 V)
12. Define resistance and state its SI units. When is the resistance of a conductor
said to be one ohm? A current of 2 A is observed to flow through a conductor
when a potential difference of 50V is applied between its ends. Calculate the
resistance of the conductor.
13. Distinguish between primary and secondary cells
14. Find the current in the 12 Ω resistor (ANS: I = 0.429 A)

15. State the factors which determine the resistance of a conductor. Define Resistivity
and give its SI unit. Find the length of constantan wire of diameter 1cm needed to
make a resistor of 3Ω.Take the resistivity of constantan as 4.9 x 10 -⁷ Ωm
16. Two wires A and B are made of the same material .A has half the length and
twice the diameter of B .What is the ratio of the resistance of B to that of A?
17. A battery consists of three accumulators in series, each having an e.m.f of 2 V. A
second battery consists of four dry cells also in series, each having an e.m.f of
1.5 V. What is the e.m.f of each battery? Why could you get a bigger current from
the battery of accumulators.(ANS: 6 V each)

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 18. Each of the two new dry cells has an e.m.f of 1.5 volts and internal resistance of
1.0 ohm.The two cells are connected to a 10 ohm resistor. Find the current and
heat developed per second on the 10 ohm resistor
19. A fine wire has a resistance of 4.0 Ω/𝑚 .When a coil made from this wire is
connected to a 50 V supply a current of 25 mA flows
(a) what is the length of wire making this coil?
(b) Determine the resistivity of this wire if its diameter is 0.35 mm
20. Show that if two resistors are joined in parallel and in series, the effective
𝐑 𝐑
resistance R is given by: R = 𝐑 𝟏+𝐑𝟐 and R = R1 + R2 respectively, where R1
𝟏 𝟐

and R2 are the separate resistances

21. Three resistors of resistances 10Ω, 25Ω and 50Ω respectively, are to be
connected between two points A and B. What will be the resistance between A
and B if the three resistors are connected (a) In series (b) In parallel
22. Three conductors of resistances 10Ω,15Ω and 25Ω are joined in series across a
100V supply, Find (a) The total resistance (b) The current in the circuit
(c) The potential difference across each conductor
23. Calculate the current passing through the 6Ω resistor in the circuit shown in the
figure below

24. State the main facts about cells connected

(a) in series and (b) in parallel. Three cells each of e.m.f. 1.5V and internal
resistance 0.6Ω are connected in parallel. The group of cells is then connected
across a conductor of resistance 1Ω.Calculate the current in the circuit.
25. A cell has an e.m.f of 1.5 V, and an internal resistance of 1 Ω, and is connected
to two resistances of 2 Ω and 3 Ω in series. Find the current flowing and the
potential difference across the ends of each resistance.(ANS: 0.25A ,0.5 V, 0.75 V)
26. Two cells each having an e.m.f of 1.5 V and an internal resistance of 1Ω are
connected to a resistance of 4 Ω. What is the current in this resistance if the cells
are connected in parallel? (ANS: I = 0.33 A or 1/3 A)
27. Two resistors of resistances 3Ω and 5Ω respectively, are connected in the gaps
of a metre bridge. At what point on the wire of the bridge will a centre-zero
galvanometer show no deflection?

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 28. The P.d across the terminals of a cell is 3.0 V when it is not connected to a circuit
and no current is flowing. When the cell is connected to a circuit and a current of
0.37 A is flowing the terminal P.d falls to 2.8 V. What is the internal resistance of
the cell?. (ANS: r = 0.54 Ω)
29. Two resistors of resistance 30 Ω and 80 Ω are connected in parallel. Calculate
their equivalent resistance
30. A cell pushes a current of 2.0 A through a 0.6 Ω resistor. When the same cell is
connected to a 1 Ω resistor, the current that flows is now 1.2 A .Calculate:
(a) the internal resistance of the cell
(b) The e.m.f of the cell
31. State joule’s law of heating. Describe an experiment to show that the heat
developed in a conductor by the passage of an electric current depends on the
magnitude of the current.
32. Give joule’s formula for the quantity of electrical energy generated in a wire
carrying a current and define all symbols used. A current of 2A is passed through
a conductor of resistance 10Ω for 5 minutes. Calculate the quantity of heat
dissipated in the conductor
33. In an experiment to find the resistance of a resistor R using the wheat stone bridge, the
balance point was found to be at the 35 cm mark on a 100 cm nichrome wire. If the
value of the resistance needed to balance the bridge on the other side was 30 Ω,
calculate the value of the resistance of the resistor R. (ANS: R = 16.2 Ω)
34. .When is an electrical conductor said to dissipate one joule? The current in an
electrical appliance operating from a 240V supply is 5A.How much energy is used
up in operating it for 20 minutes?
35. The resistance of a heating coil of an electrical hot water system is 100 ohm. If
the coil operates from a 240V supply, calculate the rate at which the coil
consumes electrical energy.
36. What is the resistance of a wire if it is balances a standard resistor of 2 ohms at
the 56 cm from the end of the metre bridge? (ANS: R = 2.54 cm)
37. Find the value of the unknown resistor S in the balanced wheat stone bridge
circuit in the figure below

38. (a)A domestic electric bulb is marked 240 V, 100 W. Calculate

(i) The maximum current it can take
(ii) The resistance of the filament of the bulb
(b) If the bulb in (a) above is switched on for 8 hours , find the total amount of
energy dissipated in the filament

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 39. (a) The heating coil of an aluminium electric kettle is labeled 240 V, 4 Kw. Water
of mass 3 kg at a temperature of 250 C is poured into the kettle. If the kettle is
connected to a 240 V supply when its temperature is 250 C, Calculate the time
taken for the water to reach its boiling point
(b) In what further time will 10% of the water in the kettle evaporates away?
(Mass of kettle = 0.5 kg, specific heat capacity of aluminium = 900 J/kg-1K-1 ,
Specific latent heat of vaporization of water = 2.268 x 106 J/kg-1K-1)
40. If electrical energy is charged at the rate of Tsh 100 per kilowatt – hour ,
Calculate the cost of using :
(a) a 60 W light bulb for 8 hours
(b) a 1 kW electric iron for 1.5 hours
(c) a 6000 W electric cooker for 2 hours
41. An electric iron consumes 2.592 MJ of energy in 1 hour when connected to the
mains power supply of 240 V. Calculate the current through the filament in the
electric iron.(ANS: I = 3 A)
42. State the properties and functions of a fuse. How does a fuse in the lighting circuit
differ from that used in the heating or power circuit?
43. (a) What is the importance of using a fuse in an electrical appliance
(b) A refrigerator is marked 250 V, 400 W. Calculate the maximum current that
can flow through it?
(c) Discuss what might happen to the refrigerator if it is connected to:
(i) a 230 V supply (ii) a 110 V supply
44. (a) State and explain the causes of electrical short – circuit.
(b) Explain briefly why cables in a lighting circuit are different from those in a
power circuit
(c) Fuse wires are labeled 2 A, 3 A, 5 A ,13 A and 14 A .Which of these is most
suitable for:
(i) a 220 V ,2.8 kW electric iron?
(ii) a 240 V, 400 kW refrigerator?
(iii) a 220 V, 3 kW water heater?
45. State the defects of a simple cell and explain how these defects may be minimized?
46. (a) What is the difference between a dry and a wet Leclanche cell
(b) Explain why a Leclanche cell is not suitable for use as a continuous sources of
energy.
47. State four household electrical appliances where electrical energy is converted
into heat
48. A washing machine is marked 240 V, 3Kw .What does this mean? Hence
calculate the electrical energy used up by this machine in 1 hour
49. Calculate the (a) current through (b) resistance of the filament of
(a) a bulb rated at 240 V , 60W
(b) an electrical kettle rated at 2 KW, 240 V

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 50. The filament of the bulb is made of tungsten and the bulb contains a
mixture of argon and nitrogen at low pressure
(a) what is the purpose of the presence of the gases inside the bulb
(b) why is tungsten a suitable material for the filament
51. Electrical heaters are said to be environmentally friendlier than the heating
devices which use firewood or charcoal .Explain this statement
52. Starting from electrical work done W = ItV , show that electrical power (P) generated in
a conductor is given by V2/R ,where the symbols have the usual meaning
53. Three cells each of e.m.f 1.5 v and internal resistance 0.6 Ω 𝑗𝑜𝑖𝑛𝑒𝑑 in series to
form a battery and connected across a 0.5 Ω resistor .Calculate
(a) the current (b) The P.d between the terminals of the cell
54. Which bulb in the figure below is the brightest? Explain your answer

55. (a) State Ohm’s law and state two of its limitations
(b) Determine the internal resistance of a cell and the value of R given that the
p.d of the cell in open circuit is 1.5 v, when connected to a 10  resistor its p.d
becomes 1.0 v, but when connected to a resistor of R  the p.d falls to 0.5 v
(ANS: r= 5,R = 2.5)
56. The power rating of an electric bulb is ’60 W, 240 V’
(a) Calculate the current through the filament and the resistance of the filament
(ANS: R = 960 , I = 0.25 A)
(b) By comparison with the answers to (a) above determine the current and the resistance of
𝟏
the filament of a ‘120 W,240 V’ bulb. Explain your answer (ANS: I = 0.5 A, R = 480 ,𝑰 ∝ 𝑹)
57. An electric bulb is labeled ‘ 40 W,240 v’. Calculate
(a) the resistance of the filament used I the bulb(ANS:R = 1 440 )
(b) The current through the filament when the bulb works normally (I = 0.167 A)
58. A 3 kW immersion heater is used to heat water .Calculate the electrical energy
converted into heat energy in 40 minutes (ANS: E.E = 7.2 MJ)
59. A current of 2 A is passed through a resistor of 20 ohms for 1.0 hour .Calculate
the electrical energy converted into heat energy in the resistor (ANS:E.E = 2.88x105 J)
60. Why does a bird safely porch on a high potential electric wire? (ANS: When a bird
is perched on a single wire, its two feet are at the same electrical potential, so the
electrons in the wires have no motivation to travel through the bird’s body)

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 61. A torch bulb is labeled 2.5 V, 0.3 A. Calculate the power of the bulb(ANS:P=0.75 W)
62. What is the terminal p.d for a cell of emf 2 V and internal resistance 1 ohm when
it is connected to a 9 ohm resistor?
63. (a) Explain the terms Live, Neutral and Earth as applied in domestic electrical
appliances
(b) What are the color codes used at present in domestic electrical appliances
(c) An electric stove is rated 1000 W, 250 V. Electricity is charged at shs. 45/= per
kilowatt – hour, and the stove is used for 30 minutes per day
(i) How much will the cost be in the month of January?
(ii) What is the maximum current that flows through the element without destroying it?
64. When two resistors are connected in series, the total resistance is 25 ohm .If they
are connected in parallel ,the total resistance is 6 ohm. Find the resistance of
each ( ANS: 15 ohm and 10 ohm)
65. The emf of a cell is 12 V and its internal resistance is 2 ohm .Find the current and the
terminal potential difference across the cell , if it is connected to 4 ohm external resistor
66. Briefly, explain what the fuse is
67. Select the best fuse for the following
(a) a refrigerator rated 250 V, 400 W (ANS: I = 1.6A, the best is 2 A)
(b) the electric cooker rated 240 V, 7.2kW (ANS: I = 30 A, the best is 30A)
(c) the electric iron rated 240V, 2 kW (ANS: I = 8.3A, the best is10 A)
68. The ratings of a bulb is 60 W, 240 V. Due to a power outage, the voltage drops
down to 200 V .Find the new power of the bulb .What would you notice in the bulb?
69. The ratings of an iron is 1200 W, 240 V. Find the current and the energy used up in an hour
70. In the two circuits (a) and (b) shown in the figure below which bulb, A or B is
𝟐
brighter? (ANS: From P = 𝑽𝑹 = 𝑰𝟐 𝑹, in parallel V is the same while in series I is the same.
This gives that in fig (a) Bulb B has more power so it is more brighter but in fig (b) Bulb A has
more power than B so it is more brighter than B)

71. In the circuit shown below ,each bulb is rated at ‘ 6 V, 3 W’

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 (a) Calculate the current through each bulb, when the bulbs are working normally
(b) How many coulombs of charge pass in 6 seconds through each bulb?
(c) What would the ammeter read when all the bulbs are working normally
(d) Calculate the electrical power delivered by the battery
72. Two cells, each of 1.5 V are used to drive a current through a wire AB of
resistance 90 ohms (see the figure below )

(a) Calculate the current in the circuit

(b) What would be the difference, if any, to the current, if the two cells are
connected in parallel.
73. A carbon resistor has a value of 20 mega ohms ± 5% .What is the colour code
for this resistor?
74. A carbon resistor’s ABC bands represent yellow, blue and brown colors
respectively. Determine the (a) resistance (b) Conductance
75. The p.d across the terminals of a cell is a 1.5 V where there is no current in the
cell .Where is a current of 0.50 A in the circuit the p.d falls to 1.3 V
(a) What is the e.m.f of the cell?
(b) What is the terminal voltage at the cell?
(c) Calculate the internal resistance of the cell
76. Explain the term internal resistance of a cell .How does it arise?
77. The e.m.f of a cell is given by the expression E = I(R + r). Explain the meaning of
each term in the expression
78.

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