Form Four Physics Handbook

amosobiero7@gmail.
com
PHYSCIS FORM 4 HANDBOOK

amosobiero7@gmail.com
FORM 4 HANDBOOK
[With well-drawn diagrams, solved examples and questions for exercise]
{2020 Edition}
(amosobiero7@gmail.com)
By Sir Obiero Amos 0706 851 439
Table of Contents
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ACKNOWLEDGEMENT
BRIEF PERSONAL PROFILE
GUIDELINES IN MY LIFE
Chapter 1 THIN LENSES
Chapter 2 UNIFORM CIRCULAR MOTION
Chapter 3 SINKING AND FLOATING
Chapter 4 ELECTROMAGNETIC INDUCTION
Chapter 5 MAINS ELECTRICITY
Chapter 6 ELECTROMAGNETIC SPECTRUM
Chapter 7 CATHODE RAYS AND CATHODE

RAY OSCILLOSCOPE
Chapter 8 X-RAYS
Chapter 9 PHOTOELECTRIC EFFECT
Chapter RADIOACTIVITY
10
Chapter ELECTRONICS
11
Page 3 of 108
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Chapter One THIN LENSES
Objectives Effect of lenses on parallel rays of light.

By the end of this lesson the learner A lens relies on the principal of refraction of
should be able to: light. Therefore when parallel rays are
directed towards the lens the rays will be
a) Describe converging lenses and refracted either by being converged or by
diverging lenses. being diverged.
b) Describe using ray diagrams the
principal focus, the optical centre and • When the convex lens is used the rays
the focal length of a thin lens. are converged.
c) Determine experimentally the focal
length of a converging lens.
d) Locate images formed by thin lenses
using ray construction method. • If a concave lens is used then the rays
e) Explain the image formation in the are diverged.
human eye.
f) Describe the defects of vision in the
human eye and how they are
corrected.
g) Describe the uses of lenses in various Definition of terms
optical devises.
h) Solve numerical problems involving a) Centre of curvature – the centre of the
the lens formula and the sphere which the lens is part.
magnification. b) Radius of curvature (r) - the radius of
the sphere of which the surface of the
Introduction lens is part.
c) Principal axis – it is an the line joining
Lens- Is a carefully molded piece of a
the centres of curvature of its surfaces.
transparent material that refracts light in
d) Optical Centre (O) - it is a point on the
such away as to form an image. They
principal axis midway between the lens
normally operate on refractive property
surfaces.
of light.
e) Principal focus (F) – For a convex lens,
• They are made of glass, clear plastic, is a point on the principal axis where all
or Perspex. rays converge after passing through the
• They are found in cameras human lens. While for a concave lens, is a point
eye, spectacles, telescopes, on the principal axis behind the lens
microscope and projectors e.t.c from which rays seem to diverge from
after passing through the lens.
Types of lenses f) Focal length (f) – it is the distance
between the optical centre and the
There are two major types of lenses,
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namely: principal focus.

g) Focal plane – it is a plane perpendicular
1. Convex (converging) - they are to the principal that all the rays seem to
thickest at the middle and thinnest at converge to or seem to appear to diverge
the ends. from. The incident rays in this case are
2. Concave (diverging) – they are not parallel to the principal axis.
thinnest at the middle and thickest at h) Paraxial rays- these are rays that are
theends. parallel and close to the principal axis.
Convex lenses i) Marginal rays- these are rays that are
parallel and far away from the principal
axis.
Concave lenses
1.3 Image Formation

It is important to note:
• Real rays and real images are drawn in
full lines.
• Virtual rays and virtual images are
drawn in broken/dotted lines.
• To locate the image, two or three rays
from the tip of the object are drawn.
• Should the foot of the object cross the
principal axis, the method on 3 above is
1.2 Ray Diagrams used to get the foot of the image. The
top is joined to the foot to get the image.
For one to locate the image when using a
lens, ray diagrams are of great importance. Example
Page 5 of 108

There are three major rays that are used in

ray diagrams for the location of images
formed by the lens.
These rays are;
(i) A ray of light parallel to the principal
axis. • Converging and diverging lenses are
This ray passes through the principal focus represented by the symbols shown
(for convex lens) or seem to appear to below.
emerge from the principal focus (for
concave lens) after refraction by the lens
(ii) A ray of light passing (or appearing to

pass through) the principal focus Characteristics of images formed by
lenses
-the ray emerges parallel to the principal
axis after refraction by the lens Converging lenses.
Object at infinity.
(iii) A ray of light through the optical

centre
This ray passes on un-deviated
The image is
(i) Real
(ii) Inverted
(iii) Diminished
(iv) Formed at F
Object beyond 2F Object between F and lens
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The image is The image is

(i) Real (i) Virtual
(ii) Inverted (ii) Erect
(iii) Diminished (iii)
(iv) formed between F and 2F on (iv) Magnified
the other side of the lens (v) Formed on the same side as object
Object at 2F Diverging lenses
The image is
The image is
(i) Virtual
(i) real (ii) Erect
(ii) Inverted (iii) diminished
(iii) Same size as the object
(iv) Formed at 2F, on the other Linear Magnification
side of the lens
• Magnification is a measure of the
extent to which an optical system
Object between F and 2F enlarges or reduces an image.
• Linear magnification is a ratio of
height of image to the height of the
object OR the ratio of the image
distance to the object distance.
ℎ𝑒𝑖𝑔ℎ𝑡 𝑜𝑓 𝑖𝑚𝑎𝑔𝑒
𝑚𝑎𝑔𝑛𝑖𝑓𝑖𝑐𝑎𝑡𝑖𝑜𝑛 =
The image is ℎ𝑒𝑖𝑔ℎ𝑡 𝑜𝑓 𝑜𝑏𝑗𝑒𝑐𝑡
𝑖𝑚𝑎𝑔𝑒 𝑑𝑖𝑠𝑡𝑎𝑛𝑐𝑒
(i) Real Or 𝑚𝑎𝑔𝑛𝑖𝑓𝑖𝑐𝑎𝑡𝑖𝑜𝑛 =
𝑜𝑏𝑗𝑒𝑐𝑡 𝑑𝑖𝑠𝑡𝑎𝑛𝑐𝑒
(ii) Inverted
(iii) Magnified
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(iv) formed beyond 2F on the 𝑉 h

Therefore, 𝑚 = = i
other side of the lens 𝑢 ho
Object at F The lens formula

Consider an image formed by converging
lens as shown below.
The image is at infinity
PO is the object distance, u, PI is the

image distance, v, and PF the focal length,
f.
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OB=PH (ii) An object placed 6m from a
converging lens forms an
Triangles POB and PIM are similar.amosobiero7@gmail.com
Therefore; erect image that is five times
𝑂𝐵 𝑃𝑂 larger.State the type of the
= image formed.Find the focal
𝐼𝑀 𝑃𝐼
𝑢 lenght of the lens.
= … … … … … … … … … … … … … … … . . (1)
𝑣
Solution
Similarly, triangles PFH and IMF are similar. So;
The image is virtual since it is
𝑃𝐻 upright and magnified.
𝐼𝑀 𝑣 𝑣
𝑃𝐹 𝐹𝑟𝑜𝑚 = 𝑚, =5 ⇨𝑣=
= … … … … … … … … … … … … … … … … … … . (2) 𝑢 𝑢
𝐼𝐹 5𝑢, 𝑢 = 6𝑚, 𝑣 = 5 × 6 = 30𝑚
But, PF=f 1 1 1 1 1 1
= + , = + ,𝑓
𝐼𝐹 = 𝑃𝐼 − 𝑃𝐹; 𝐼𝐹 = 𝑣 − 𝑓 𝑓 𝑢 𝑣 𝑓 6 30
=5𝑚
Substitute these values in equation (2);
Relationship between
𝑂𝐵 magnification and focal length
𝐼𝑀
𝑓 We have;
= … … … … … … … … … … … … … … … . . (3)
𝑣−𝑓 1 1 1
= +
Combining equations (1) and (3); 𝑓 𝑢 𝑣
𝑢 𝑓 Multiply both sides by v;

=
𝑣 𝑣−𝑓 𝑣 𝑣 𝑣
= +
𝑓 𝑢 𝑣
𝑢𝑣 − 𝑢𝑓 = 𝑣𝑓
𝑣 𝑣
𝑢𝑣 = 𝑣𝑓 + 𝑢𝑓 But, = 𝑚 𝑎𝑛𝑑 = 1,
𝑢 𝑣
therefore,
𝑢𝑣 = 𝑓(𝑣 + 𝑢)
𝑣
𝑢𝑣 =𝑚+1
=𝑣+𝑢 𝑓
𝑓
Re-arranging;
1 𝑣+𝑢 𝑣 𝑢
= = + 𝑣
𝑓 𝑢𝑣 𝑢𝑣 𝑢𝑣 𝑚= −1
𝑓
1 1 1
Hence = +
𝑓 𝑢 𝑣 To determine u and v, real-is –
positive sign convention is
This is called the lens formula and holds for
adopted. According to this
both converging and diverging lens.
convention:
Examples
I. All distances are measured
Page 9 of 108
(i) An object 0.05m high is placed 0.15m infront of from the optical centre.
a convex lens of focal lenght 0.1m.find the II. Distances of real objects
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439 objects and
Solution images are negative.
𝒉𝒐 = 𝟎. 𝟎𝟓𝒎, 𝒖 = 𝟎. 𝟏𝟓 𝒎, 𝒇 = 𝟎. 𝟏 𝒎, 𝒗 = III. The focal length of a
converging lens is positive
Graphical interprentation of the lens Exercise

formula
Interprete a graph of:
CASE1: 1 1
(i) 𝑎𝑔𝑎𝑖𝑛𝑠𝑡
𝟏 𝟏 𝑢 𝑣
A graph of 𝒂𝒈𝒂𝒊𝒏𝒔𝒕 ; (ii) 𝑢 + 𝑣 𝑎𝑔𝑎𝑖𝑛𝑠𝑡 𝑢𝑣
𝒗 𝒖
1 1 1 CASE 3:
𝐹𝑟𝑜𝑚 𝑡ℎ𝑒 𝑙𝑒𝑛𝑠 𝑓𝑜𝑟𝑚𝑢𝑙𝑎, = +
𝑓 𝑢 𝑣
1
𝑅𝑒 Graph of u against
𝑚
− 𝑤𝑟𝑖𝑡𝑒 𝑡ℎ𝑒 𝑒𝑞𝑢𝑎𝑡𝑖𝑜𝑛 𝑖𝑛 𝑡ℎ𝑒 𝑓𝑜𝑟𝑚 𝑦 1 1 1
= 𝑚𝑥 + 𝑐 = +
𝑓 𝑢 𝑣
1 1 1 𝑢 𝑢 𝑢
= − + 𝐻𝑒𝑛𝑐𝑒 , 𝑠𝑙𝑜𝑝𝑒 = +
𝑉 𝑢 𝑓 𝑓 𝑢 𝑣
1
= −1 𝑎𝑛𝑑 − 𝑖𝑛𝑡𝑒𝑟𝑐𝑒𝑝𝑡 𝑢 1
𝑣
1 =1+
= 𝑓 𝑚
𝑓
1
𝑢=𝑓 +𝑓
𝑚
Implying that 𝑓 = 𝑠𝑙𝑜𝑝𝑒 𝑎𝑛𝑑 𝑢 −
𝑖𝑛𝑡𝑒𝑟𝑐𝑒𝑝𝑡 = 𝑓
It is a straight line that cuts the vertical axis
at f.
CASE 2
A graph of uv against 𝑢 + 𝑣
1 1 1
𝐹𝑟𝑜𝑚 𝑡ℎ𝑒 𝑙𝑒𝑛𝑠 𝑓𝑜𝑟𝑚𝑢𝑙𝑎 = + ,
𝑓 𝑢 𝑣
1 𝑣+𝑢
=
𝑓 𝑢𝑣
𝑢𝑣 = (𝑢 + 𝑣 )𝑓
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In the form 𝑦 = 𝑚𝑥 + 𝑐 Exercise

𝑢𝑣 = 𝑓 (𝑢 + 𝑣 ) + 0 Interpret a graph of:
𝑇ℎ𝑢𝑠 𝑔𝑟𝑎𝑑𝑖𝑒𝑛𝑡 = 𝑓 𝑎𝑛𝑑 𝑢𝑣 𝑖𝑛𝑡𝑒𝑟𝑐𝑒𝑝𝑡 (i) m against v
=0 (ii) V against m
It is a graph of a straight line passing
through the origin.
Experimental Determination Of The 2. Adjust the position of the lens holder

Focal Length Of A Converging (Convex) until a sharp image of the object is
Lens formed on the screen alongside the
object itself.
Method (1): 3. Record the distance between the lens
Focusing A Distant Object and the screen.
Apparatus Focal length----------------------cm
Metre rule, lens, a lens holder, screen NOTE:
Procedure Under these conditions, rays from any

point on the object will emerge from the
1. Mount a convex lens on a lens holder lens as parallel rays. They are therefore
and fix a metre rule on a bench using reflected back through the lens and
plasticine as shown below. brought to a focus in the same plane as
the object. The distance between the lens
and the screen now give the focal length
Page 11 of 108

of the lens.
Method (3):
Using A Pin And Plane Mirror/No
Parallax Method
1. Set up the apparatus as shown below.
2. Place a white screen at one end of the

metre rule.
3. Move the lens to and fro along the
metre rule to focus clearly the image
of a distant object, like a tree or
window frame.
4. Measure the distance between the lens
and the screen. 2. Adjust the position of the pin up and
down till its tip is at the same
Focal length--------------------cm horizontal level as the centre of the
Note: The distance between the lens and lens. A position is found for which
the screen gives a rough estimate of the there is no parallax between it and the
focal length of the lens. This is because real image formed. For best results,
parallel rays from infinity are converged attention should be given to the tilt of
at the focal point on the screen. the plane mirror so that the tip of the
image of the object pin appears to
Method (2): touch at the same level as the centre of
the lens.
Using an Illuminated Object And Plane 3. The distance between the pin and the
Mirror/Reflection Method lens will then be equal to the focal
Procedure length of the lens.
1. Set the lens in its holder with a plane Focal length=-----------------------

mirror behind it so that light passing cm
through it can be reflected back as The power of a lens
shown below.
Is a measure of refractive property of a
lens. It is given by
1
𝑝𝑜𝑤𝑒𝑟 =
𝑓𝑜𝑐𝑎𝑙 𝑙𝑒𝑛𝑔𝑡ℎ 𝑖𝑛 𝑚𝑒𝑡𝑟𝑒𝑠
The unit of power of a lens is dioptres (D)
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USES OF LENSES IN OPTICAL Defects of vision

DEVICES
Short sightedness (myopia)
Due to their ability to converge or diverge
light rays, lenses are widely used in optical
devices. The devices include;
• Human eye
• Simple microscope
• Compound microscope • Can only clearly see near objects
• The camera • Rays of near objects are focused on the
retina but those for distance objects are
The human eye
focused in front of the retina
It is a natural optical instrument
Causes
• Short focal length of the eye lens
• Long eyeball
Corrected by diverging lenses as shown
1. Sclerotic layer – hard shell that

encloses the eye and is white. Long sightedness (hypermetropia)
The front part is transparent and
spherical known as the cornea. • Can see distant objects but not near
Most bending of light entering the ones
eye occurs at the cornea. • The images of near objects are formed
2. Aqueous Humour – clear liquid behind the retina
between the cornea and the lens. It
helps the eye maintain shape. Causes are:
3. Iris – it is the colouring of the eye. It • Too long focal length of the eye
has pupil which regulates the amount
• Too short eyeball
of light entering the eye.
4. Crystalline lens- it is a converging
lens. It can change its focal length by
Page 13 of 108

the action of Ciliary muscles

5. Vitreous humour – transparent jelly
like substance filling another chamber
between the lens and the retina
6. Retina – it is where the image is
formed and Made of cells that are • It is corrected by using converging
light sensitive lenses
7. Fovea – central part of the retina that
exhibits best details and colour vision
at this place.
8. Blind spot – this contains cells that
are not light sensitive.
9. Ciliary muscles – these are muscles
that support the lens. They control the
shape of lens by contracting or
relaxing. In relaxing the muscles it
enables the lens to increase hence
focus distance objects. In contraction
the muscles reduce tensions in the
lens to increase its focal length thus
focus near objects. This process is
known as accommodation.
Near point – closest point which the
normal eye can focus. Far point -
furthest point that a normal eye can focus.
Camera Compound microscope

• The camera has lenses that focus light There are two cases under which a
from the object to form an image of the converging lens can produce magnified
object on the film. images;
• Focusing is done by adjusting the
When the object is between F and 2F.
distance between the lens and the film
.the diaphragm controls the amount of When the object is between the lens and F.
light entering the eye.
• The shutter allows light to reach the A compound microscope combines the
film only for a precise period when the above two cases. It consists of converging
camera is operated. lenses of short focal length .The focal
• The inside is blackened to absorb any length next to the object is called objective
lens and the one next to the eye is called the
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stray light. eyepiece or ocular. The objective lens is of

short focal length.
Similarities between the eye and the
camera
Eye Camera
Has crystalline Has a convex lens
convex lens
Choroid layer is Box painted black
black inside The eye piece is also of short focal length
The retina where Light-sensitive but longer than that of the objective lens. A
images are formed film where images compound microscope overcomes the
are formed. limitations of a simple microscope by use of
objective lenses with many lenses and an
Iris which controls Diaphragm which eyepiece with more than one lens.
the amount of controls the
light entering the amount of light Total magnification produced by a
eye entering the compound microscope is given by;
camera. 𝑉 𝑣𝑜
( 𝑒 − 1) ( − 1) Where 𝑣𝑜 is the image
𝑓𝑒 𝑓𝑜
distance from I and 𝑉𝑒 the image distance
Differences between the eye and the from I’.
camera REVISION QUESTIONS
Eye Camera 1. The diagram below shows an
Varable focal Fixed focal arrangement of lenses, Lo and Le used in
length length a compound microscope FO and Fe are
principal foci of Lo and Le respectively.
Constant image Variable image
distane distance
Constantly Only one
changing pictures photograph can
be taken at at
time
Draw the rays to show how the final image
Simple microscope
is formed in the microscope
It is sometimes referred to as a magnifying
Page 15 of 108

glass. When the object is placed between a

convex lens and its principal focus, the
image formed is virtual, erect and
magnified.
Chapter Two
UNIFORM CIRCULAR MOTION

Content
Specific Objectives • The radian, angular displacement,
By the end of this topic, the learner should angular velocity
𝑚𝑣 2
be able to: • Centripetal force; 𝐹 = , 𝐹 =
𝑟
𝑚𝑟𝜔2 (derivation of formulae not
• Define angular displacement and required) (experimental treatment is
angular velocity necessary)
• Describe simple experiments to • Applications of uniform circular
illustrate centripetal force motion
• Explain the application of uniform • Centrifuge, vertical, horizontal
circular motion circles banked tracks (calculations on
• Solve numerical problems involving banked tracks and conical pendulum
uniform circular motion not required)
𝑚𝑣 2
• Problem solving (Apply 𝐹 = ,
𝑟
𝐹 = 𝑚𝑟𝜔2)
Definition of Terms Convert the following into radians:

(A) Angular Displacement, 𝜽 a) 800C
b) 1200C
It is the angle swept through a line joining
to the centre of circular path. It is measured (b) Angular Velocity
in radians.
It is the rate of change of angular
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displacement. It is denoted by Greek letter

omega (ω).

 =
𝑡
Is measured in radian per second
𝐼𝑡 𝑖𝑠 𝑐𝑎𝑙𝑐𝑢𝑙𝑎𝑡𝑒𝑑 𝑎𝑠 𝑎𝑛𝑔𝑢𝑙𝑎𝑟 𝑑𝑖𝑠𝑝𝑙𝑐𝑒𝑚𝑒𝑛𝑡
𝑎𝑟𝑐 𝑙𝑒𝑛𝑔𝑡ℎ Example
=
𝑟𝑎𝑑𝑖𝑢𝑠 1. A particle moving in a circular path
𝑠 covers one revolution in ten seconds.
𝜃=
𝑟 Calculate its angular velocity.
Radian
Is defined as an angle of sector of the Period/Periodic Time
circumference whose length is equal to its
radius or is the ratio of arc length to the Is the time taken to complete one
radius of a circular. revolution.
𝑎𝑟𝑐 𝑙𝑒𝑛𝑔𝑡ℎ 𝑝𝑒𝑟𝑖𝑜𝑑𝑖𝑐 𝑡𝑖𝑚𝑒

r𝑎𝑑𝑖𝑎𝑛 = 𝑟𝑒𝑣𝑜𝑙𝑢𝑡𝑖𝑜𝑛
𝑟𝑎𝑑𝑖𝑢𝑠
= 𝑎𝑛𝑔𝑙𝑒 𝑐𝑜𝑣𝑒𝑟𝑒𝑑 𝑖𝑛 𝑜𝑛𝑒
𝑠 𝑎𝑛𝑔𝑢𝑙𝑎𝑟 𝑣𝑒𝑙𝑜𝑐𝑖𝑡𝑦
𝑟𝑎𝑑𝑖𝑎𝑛 =
𝑟 2𝜋
𝑇=
2𝜋𝑟 𝜔
𝑟𝑎𝑑𝑖𝑎𝑛 =
𝑟 Frequency (f)
= 2𝜋 1
Frequency, 𝑓 =
𝑇
2𝜋 𝑟 = 3600
1 2𝜋
= →ω=2πf
𝑓 𝜔
Relationship between angular

displacement, 𝜽 and angular velocity. Centripetal Acceleration
∆S
V=
∆t
------------------------------------------ • An object going through a circular
path is said to accelerate.
Page 17 of 108

---------- (i) • If the velocity of such object is

∆S constant the object still accelerates
∆θ = ---------------------------------------- because there is continuous change in
∆r
--------- (ii) velocity as the object continuously
changes direction. From Newton’s
For small change in equation (ii) second law of motion, the body
∆𝑆 experiences a resultant force as it
∆𝛳 = -------------------------------------------
𝑟 moves rounds path .This resultant
------ (iii) force is directed towards the circular
Dividing equation (iii) by ∆t path.
• Acceleration of this body is in the
∆𝜃 ∆𝑠 direction of force applied to it i.e. it
=
∆𝑡 𝑟∆𝑡 accelerates towards the centre of the
𝑣 circular of the circular path. This
𝜔 = → 𝒗 = 𝝎𝒓 acceleration is called centripetal
𝑟
acceleration.
Anybody in circular motion has both linear
The centripetal acceleration is given by the
velocity in m/s and angular velocity in
rads/s. following equation.
𝑣2
𝐶𝑒𝑛𝑡𝑟𝑖𝑝𝑒𝑡𝑎𝑙 𝑎𝑐𝑐𝑒𝑙𝑒𝑟𝑎𝑡𝑖𝑜𝑛 𝑎 =
Examples 𝑟
1. A turn table rotates at the rate of 60 But 𝑣 = 𝜔𝑟

revolutions per minute. What is its
angular velocity in rads/s 𝑎 = 𝜔2 𝑟
2. A model car moves around a circular Centripetal Force
path of radius 0.6m at 25 Rev/s.
Determine its; Is a force that is required to keep a body
(a) period moving in a circular path and is directed
(b) Angular velocity (𝜔) towards the centre of the circular path.
(c)Speed (v)
If an object moving through a circular path
3. The car moves with uniform velocity
is released suddenly it flies off tangentially.
of 3m/s in a circle of radius 0.2m.
Find its angular velocity and Factors Affecting Centripetal Force.
frequency.
1. Mass of the object, m- the heavier the
4. Distinguish between angular and object the more the centripetal force needed
linear velocity. to maintain it in circular path.
2. Angular velocity of the object, 𝝎- an
increase in centripetal force needed to
maintain the object in circular path.
3. Radius of the path r-the shorter the
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radius of the path the larger the centripetal

force required to maintain the object in
circular path.
Example Examples of Uniform Circular Motion

The figure below shows the diagram of A car rounding a level circular bend
set up to investigate the variation of
centripetal with the radius r, of the circle When a car is going round in a circular
in which a body rotated path on a horizontal road, the centripetal
force required for a circular motion is
provided by the frictional force between
the tyres and the road
𝑚𝑣 2
Therefore- 𝐹𝑟 =
𝑟
If the road is slippery then frictional force

may not be sufficient so to provide
centripetal force
Describe how the set up can be used to
carry out the investigation To prevent skidding the car should not
exceed certain speed limits referred to as
• Keep angular velocity 𝜔 the critical speed
constant;
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• Centripetal force provided This critical speed depends-

by mg;
Radius of the bend i.e. one may negotiate
• Fix the mass m and measure
the a bend at higher critical speed the
of m;
radius of the bend is big
• Repeat for different values
of m; Condition of the tyre and the nature of
The above factors are proofed using a the road surface this will produce the
turntable. frictional force need to negotiate the bend
The turntable has the following features Banked tracks
• Increase in speed of the turn table Condition in which a road is raised
increases length of the spring gradually from the inner side of the bend.
(increase in centripetal force)
• When using a shorter spring there
is more extension of the spring
than using along spring.
• When using a heavier metal bar
will produce more extension than
using a lighter ball.
This is a proof to the above factors.
The graph of force against the square of
angular velocity is a straight line through
the origin.
𝑀𝑉 2
𝐹∝
𝑟
𝐾𝑀𝑉 2
𝐹=
𝑟
𝑀𝑉 2
When k=1,𝐹 = , but 𝑣 = 𝜔𝑟
𝑟
𝑚𝜔2 𝑟 2
Hence,𝐹 = , thus,𝐹 = 𝑚𝑟𝜔2
𝑟
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𝑅𝑠𝑖𝑛𝜃 -Is the horizontal component BY principle of moments

which is responsible for providing
centripetal force. R.x = Fry
𝑥 𝑚𝑣 2
𝑅𝑐𝑜𝑠𝜃 -Is the vertical component that is =
𝑦 𝑟𝑚𝑔
responsible for balancing the weight of
the vehicle. 𝑥 𝑣2
=
𝑦 𝑟𝑔
If a vehicle of mass m is travelling a long
𝑣2
a circular path of radius r at uniform Tan 𝜃 =
𝑟𝑔
speed v, then
Making v the subject of the formula
𝑚𝑣 2
𝑅𝑠𝑖𝑛𝜃 = … … … … … … … … . . (𝑖)
𝑟 𝑣 = √𝑟𝑔𝑡𝑎𝑛𝛳
𝑅𝑐𝑜𝑠𝜃 Tan 𝜃=
𝐹𝑟
= 𝑚𝑔 … … … … … … … … … … . . (𝑖𝑖) 𝑚𝑔
Divide (i) by (ii) Mg tan 𝜃 = Fr
𝑅𝑠𝑖𝑛𝜃 𝑚𝑣 2 1 Fr = µR
= ×
𝑅𝑐𝑜𝑠𝜃 𝑟 𝑚𝑔 Mg tan 𝜃 = µ mg
𝑠𝑖𝑛𝜃 Tan 𝜃 =µ
= 𝑡𝑎𝑛𝜃
𝑐𝑜𝑠𝜃
Where µ is coefficient of friction
𝑣2
Hence tan 𝜃 =
𝑟𝑔 Skidding occurs when tan 𝜃 is greater
than µ
The maximum speed required for a body
moving in a circular path whose angle of
banking is 𝜃 is given by;
𝑣 2 = 𝑟𝑔𝑡𝑎𝑛𝜃 𝐻𝑒𝑛𝑐𝑒, 𝑣 = √𝑟𝑔𝑡𝑎𝑛𝜃
A cyclist moving round a circular track

Frictional force (Fr) is provided by
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centripetal force which is directed

towards the car however if frictional
force is not sufficient to provided
centripetal force skidding takes place. To
avoid skidding the cyclist leg inwards so
that normal reaction of frictional force
produces the turning effect to the
clockwise and anticlockwise directions.
Taking moments about G
Conical pendulum Example:

If a pendulum bob moves in such a way (a) The figure below shows an object at
that the string sweeps out a cone, then the the end of a light spring balance
bob will describe a horizontal circle. connected to a peg using a string. The
object is moving in a circular path on a
smooth horizontal table with a constant
speed.
As it can be clearly seen, there are two (i) What provides the force that keeps the
forces acting on the pendulum bob; object moving in the circular path?
(i) its weight (mg) (ii) Indicate with an arrow on the figure
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(ii) The tension in the string. the direction of centripetal force .

Centripetal force is provided by the (iii) The speed of the object is constant,
horizontal component of the tension (F why is there acceleration?
Sin). Hence from Newton's second law;
(iv) Although there is force acting on the
𝑭 𝑆𝑖𝑛 = object, NO, work is done on the object.
2
𝑚𝑣 Explain.
… … … … … … … … . . (1)(Where
𝑟
symbols have their usual meaning). Since (v) Given that the mass of the object is
there is no vertical acceleration 0.5kg and it is moving at speed of 8m/s at
a radius of 2m.Determine the reading on
𝑭 𝐶𝑜𝑠 the spring balance.
= 𝑚𝒈 … … … … … … … … … . . (𝟐)
(vi) State what happens to the reading if
Again, from the two equations; 𝑻𝒂𝒏  = the speed of rotation is reduced.
𝒗𝟐
𝒓𝒈
Note that this equation is similar to the one

we got earlier for banked tracks.
When the angular velocity W the cork
rises hence Q increases. This concept is
applied in merry go round and speed
governors.
Motion in a Horizontal Circe
The tension in the string provides the
centripetal force.
𝑻 = 𝑭𝑪
𝑴𝑽𝟐
𝑻=
𝒓
𝑾𝒉𝒆𝒓𝒆 𝒕𝒉𝒆 𝒍𝒆𝒏𝒈𝒕𝒉 𝒐𝒇 𝒕𝒉𝒆 𝒔𝒕𝒓𝒊𝒏𝒈
= 𝒓𝒂𝒅𝒊𝒖𝒔 𝒐𝒇 𝒓𝒐𝒕𝒂𝒕𝒊𝒐𝒏
Page 23 of 108

Motion in Vertical Path Example

A car travels over a humpback bridge of
radius of curvature 40m. Calculate
maximum speed of the car if its wt are to
staying contact with bridge. g =10m/s2
Tension on the spring changes its

magnitude depending on the position of
the ball.
𝑚𝑣 2
=mg-R
When the ball is at A, the sum of tension 𝑟
TA and weight Mg acting in the same R=0
direction provide centripetal force.
Mv2 =mg
𝑀𝑉 2
= Ta + Mg -------------------------------
𝑟 𝑉 2 = 𝑟𝑔
--------------- (i)
V= √𝑟𝑔
When the ball is at A it attains minimum
speed because Ta = 0 = √40 × 10
𝑀𝑉 2
= mg = 20 m/s
𝑟
V min = √𝑟𝑔 Examples of Centripetal Force

Example Source(what
At B, tensional force TB provides
provides centripetal
centripetal force.
force)
𝑀𝑉 2
𝑇𝐵 = Cyclist moving Frictional force
𝑟 along a circular between the tyre and
At C, tension and weight acts in different path the road
direction and hence the resultant force
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between the two forces provides the Car moving a Horizontal component
centripetal force long a banked of reaction force
road
𝑀𝑉 2
= Tc - Mg --------------------------------
𝑟 Electron Electrostatic force of
-------------- (ii)
orbiting around attraction between the
𝑀𝑉 2 the nucleus of proton and the
Ta = – Mg --------------------------------
𝑟 an atom electron
--------------- (iii)
Electron Magnetic field
𝑀𝑉 2
Tc = + Mg-------------------------------- moving in a
𝑟
-------------- (iv) magnetic field
𝑀𝑉 2 Satellite Gravitational force

𝐴𝑡 𝐷, 𝑇𝐷 = orbiting around exerted by the earth
𝑟 the earth
1. A pilot not stripped to his seat in a
loop manoeuvre without falling. A string whirled Tensional force
2. A bucket of water whirled in a in a horizontal
vertical without water spilling. track
3. A ball bearing ‘looping the loop’ on a
rail lying in vertical plane.
Application of Circular Motion 3. Speed Governors

1. Centrifuges Principle of conical pendulum is used in
operating the speed governors.
It is used to separate particles in
suspension in liquids of different
densities. It consists of small metal
containers tubes that can be rotated.
As the angular velocity of the drive shaft

increases .the masses m rises and moves
the collar up as the angle 𝜃 increases. The
Centripetal will be too great according to up and down movement of the collar is
Page 25 of 108

the equation F = mrω2 transmitted through a system of levers to

the device that controls the fuel intake.
And r will thus be smaller for lighter Since the angular velocity of the drive
particles and longer for heavier particles. shaft increases with speed of the vehicle,
2. Satellites the fuel supply will cut off when the speed
exceeds a certain limit.
Two bodies with mass m1 and m2 at a
distance r from each other experience a
𝐺𝑀 𝑀
force of attraction. 𝐹 = 12 2 G is equal
𝑅
to universal gravitational constant
Attraction between earth and satellite
𝐺𝑀1𝑀2
gives centripetal Force𝑀1 𝑣 2 = 2 𝑅
Where M1 is mass of satellite and m2

𝐺𝑀
mass of the earth 𝑣 2 = 22
𝑅
𝐺𝑀2
𝑣 = √
𝑟
The velocity of the satellite increases

with decrease in the radius of the orbit.
If periodic time of the satellite is equal to
that of the earth the satellite appear
stationary as seen from the earth surface
such satellite are said to be in parking
orbit and are used in weather
forecasting and telecommunications.
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Chapter Three SINKING AND FLOATING
Specific objectives Content

• Archimedes‟ principle,
a) state Archimedes‟ principle
• Law of flotation (experimental
b) verify Archimedes principle treatment)
• Relative density
c) state the law of flotation • Applications of Archimedes‟
d) define relative density principle and relative density.
e) describe the applications of
• Problems on Archimedes‟ principle
Archimedes‟ principle and relative density.
• Project Work- Construct a
f) Solve numerical problems involving
hydrometer.
Archimedes‟ principle.
Upthrust force
Upthrust is an upward force acting on an Precisely: Upthrust = Real weight –
object floating or immersed in a fluid. An Apparent weight.
object immersed or floating in a fluid
appears lighter that its actual weight due to Cause of upthrust
upthrust force (force of buoyancy). Consider the figure below.
Archimedes principle.
The principle states: When a body is
totally or partially immersed in a fluid it
experiences an up thrust equal to the
weight of displaced fluid.
To verify Archimedes’ principle Pressure at the bottom > pressure at the top
Apparatus PB = Pa+h2ρg
• An overflow can PT = Pa + h1ρ g
• A metal block
• A beaker Force = pressure x area
• A spring balance FB = PBA = (Pa+h2 ρ g) A
• A string
• Water FT = PTA = (Pa+ h1ρg) A
Procedure Resultant force = FB – FT
(i) Weigh the block in air.

Page 27 of 108

(ii) Note the weight of the block in air as U= (Pa+h2 ρ g) A- (Pa+h1ρ g) A

w1.
(iii) Immerse the block in water in U= (h2-h1) ρgA
the overflow can as shown in the U= hρgA But A h =v
diagram below
𝑯𝒆𝒏𝒄𝒆, 𝑼 = 𝑽𝝆𝒈
Upthrust therefore depends on:
(i) Volume of fluid displaced.
(ii) Density of fluid displaced.
• Note the weight of the block when
fully immersed as w2
• Measure the volume of water
displaced and calculates its weight as
w3
• Apparent loss of weight=𝑾𝟏 − 𝑾𝟐
• The upthrust U=W3
• Upthrust=apparent loss of
weight;𝑼 = 𝑾𝟏 − 𝑾𝟐
Example 3.(a) Distinguish between pressure

and up thrust force.
A stone of weight 3N in air and 1.2N
when totally immersed water. (b) A solid metal block of density
Calculate: 2500kg/m3 is fully immersed in water,
supported by a thread which is
(a) Volume of the stone attached to the spring balance as
(b) Density of the stone shown below.
Upthrust = Real weight- Apparent weight

= 3N – 1.2N
= 1.8 N
But 𝑈 = 𝑉𝜌𝑔
1.8 = V x 1000 x 10
V= 0.00018 m3
𝑚
𝜌 = (i) Calculate the force due to the
𝑣
liquid on the top face of the
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0.3
= 𝑘𝑔/𝑚3 block.
0.00018
(ii) If the upward force on the
=1,666.67 kg/m3 bottom face is 1.5N, calculate
QUESTIONS
the volume of the block.
(iii) Calculate the apparent weight
1. A Solid of density 2.5g/cm3 is of the block in water.
weight in air and then when
completely immersed in water in a Up thrust in gases
measuring cylinder the Level of water
rises from 40cm3 to 80cm3. Determine Gases exert small upthrust on objects
because of their low density.
(a) Volume of the solid
A balloon filled with hydrogen or helium
(b) Its apparent weight. rises up because of low density.
2 a)State the Archimedes’ principle
b) A right angled solid of
dimensions 0.02m by 0.02m by 0.2m
and density 2700kgm-3 is supported
inside kerosene of density 800kgm-3
by a thread which is attached to a
spring balance. The long side is
vertical and the upper surface is 0.1m
below the surface of kerosene.
(i) Calculate the force due to
the liquid on:
(ii) The lower surface of the
solid
(iii) The upper surface of the
solid
(iv) Calculate the upthrust and
hence or otherwise
determine the reading on the
spring balance.
In the figures above the balloon filled weight of block in air. They are equal
with air will not float because the
Page 29 of 108

weight of the balloon fabric and air is (same).

greater than the weight of air displaced
(upthrust) i.e. w>u. The balloon filled Therefore we conclude that a floating
with helium or hydrogen floats because object displaces its own weight of the
the weight of the balloon fabric and fluid in which it floats. This law of
helium or hydrogen is less than the flotation.
weight of the air displaced(upthrust) Explanation
i.e. u>w
When a body is submerged in water,
Law of Flotation there are two forces acting on the body;
(i) The weight of the body acting
downwards
(ii) Upthrust on the body due to
displaced liquid acting upwards.
Case 1
If the weight of the body is greater than
In this case we consider the floating
upthrust, the density of the body is
object and weight of the fluid displaced.
greater than the density of the displaced
A comparison of the weight of the object liquid, the body sinks.
and that of fluid displaced.
Case2
Experimentally this can be done by:
If the weight of the body is equal to
• ½ fill measuring cylinder with upthrust, the density of the body is equal
water and record the reading. to the density of the liquid, the body
• Place a clean dry test tube into the remains in equilibrium.
beaker and add some sand in it so Case3
that it floats upright.
• Records the new level of the liquid If the weight of the body is less than the
determine the volume of displaced upthrust, density of the body is less than
water the density of the liquid, the body floats
• Measure its weight (dried) and partially in the liquid.
content.
Example:
• Calculate the weight of displaced
water. A boat of mass 2000kg floats on fresh
water. If the boat enters sea water.
It is observed that the weight of the test
Determine the volume that must be added
tube and its consent is equal to weight of
to displace the same volume of water as
displaced water.
before.(Fresh water-=1000kg/m3, sea
OR water= 1030 kg/m3)
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Apparatus: Weight of fresh water = 2000kg

A block of wood, A spring balance, Thin Displaced Volume of fresh water = 2000
1000
thread, Overflow can, A small measuring
cylinder and Some water. =2m3
Using the apparatus above, describe an Mass = Density x Volume
experiment to verify the law of floatation.
= 1030 x 2
• Using the spring balance, weigh = 2060 kg

and record the weight of the block = 2060- 2000
in air
• Fill the eureka completely with = 60kg
water
2. A sphere of radius 3 cm is floating
• Place the measuring cylinder under between liquid A and B such that ½ is at
the spout A and ½ at B. If of liquids A and B are
• Lower the block of wood slowly 0.8g/cm3 and 1.0g/cm3 respectively
into water until the string slackens determine mass of the sphere.
(the block floats)
• Collect the displaced water using
the measuring cylinder
• Repeat the procedure to attain
more results
• Compare the weight of displaced
water with the
Upthrust and Relative Density

Mass of sphere= volume x density Relative density is the ratio of the mass of any
4 volume of a substance to the mass of an equal
Volume = 𝜋33 volume of water OR the ratio of the density of
3
a substance to the density of water.
= 113.14 cm3
To find relative density of a solid or a liquid
Volume of liquid A displaced = ½
several methods or formulas are used.
x 113.14
𝑑𝑒𝑛𝑠𝑖𝑡𝑦 𝑜𝑓 𝑎 𝑠𝑢𝑏𝑠𝑡𝑎𝑛𝑐𝑒
56.57 cm2 𝑹𝒆𝒍𝒂𝒕𝒊𝒗𝒆 𝑑𝑒𝑛𝑠𝑖𝑡𝑦 =
𝑑𝑒𝑛𝑠𝑖𝑡𝑦 𝑜𝑓 𝑤𝑎𝑡𝑒𝑟
Page 31 of 108

Mass displaced of A = 56.57 x 0.8 Relative density of a solid.

45.256g If equal volumes of the substance and water are
considered,
Mass of liquid B displaced =
56.57g 𝑹𝒆𝒍𝒂𝒕𝒊𝒗𝒆 𝑑𝑒𝑛𝑠𝑖𝑡𝑦
𝑀𝑎𝑠𝑠 𝑜𝑓 𝑠𝑜𝑙𝑖𝑑
Total mass of sphere displaced = =
45.256 + 56.57 𝑀𝑎𝑠𝑠 𝑜𝑓 𝑒𝑞𝑢𝑎𝑙 𝑣𝑜𝑙𝑢𝑚𝑒 𝑜𝑓 𝑤𝑎𝑡𝑒𝑟
101.826g Because mass is directly proportional to the

weight the relative density of a solid may be
3. A stone eights 2N in air and given as:
1.2N when totally immersed in
water Calculate 𝑹𝒆𝒍𝒂𝒕𝒊𝒗𝒆 𝑑𝑒𝑛𝑠𝑖𝑡𝑦
𝑤𝑒𝑖𝑔ℎ𝑡 𝑜𝑓 𝑠𝑜𝑙𝑖𝑑
(a) Volume of the stone =
𝑤𝑒𝑖𝑔ℎ𝑡 𝑜𝑓 𝑒𝑞𝑢𝑎𝑙 𝑣𝑜𝑙𝑢𝑚𝑒 𝑜𝑓 𝑤𝑎𝑡𝑒𝑟
(b) Densities of the stone 𝑹𝒆𝒍𝒂𝒕𝒊𝒗𝒆 𝑑𝑒𝑛𝑠𝑖𝑡𝑦
𝑤𝑒𝑖𝑔ℎ𝑡 𝑜𝑓 𝑠𝑜𝑙𝑖𝑑
(a) Up thrust = weight of water =
displaced 𝑤𝑒𝑖𝑔ℎ𝑡 𝑜𝑓 𝑑𝑖𝑠𝑝𝑙𝑎𝑐𝑒𝑑 𝑤𝑎𝑡𝑒𝑟
= 2-1.2 𝒘𝒆𝒊𝒈𝒉𝒕 𝒐𝒇 𝒔𝒐𝒍𝒊𝒅

𝑹𝒆𝒍𝒂𝒕𝒊𝒗𝒆 𝒅𝒆𝒏𝒔𝒊𝒕𝒚 =
𝒖𝒑 𝒕𝒉𝒓𝒖𝒔𝒕 𝒊𝒏 𝒘𝒂𝒕𝒆𝒓
=0.8N
Relative density of solid which sinks in water
Mass of water displaced = 0.8/10
If the weight of the substance in air is 𝑊1 and in
= 0.08kg 𝑊1
water is,𝑊2 , then 𝑅. 𝐷 =
𝑊1 −𝑊2
𝑚𝑎𝑠𝑠
𝑣𝑜𝑙𝑢𝑚𝑒 =
𝑑𝑒𝑛𝑠𝑖𝑡𝑦 Relative density of solid which floats in water
0.08 The sinker is used as follows:
= 𝑘𝑔𝑚−3
1000
Weight of the sinker in water=𝑊1
1000 kg/m3
Weight of the sinker in water + weight of
= 0.00008m3
floating object in air=𝑊2
Volume of stone = 0.00008m3
Weight of the sinker +weight of floating object
𝑚𝑎𝑠𝑠 0.2 in water=𝑊3
(b) Density= =
𝑣𝑜𝑙𝑢𝑚𝑒 0.00008
Weight of floating object in air =𝑊2 − 𝑊1
= 2500 kg/m3
Weight of floating object in water=𝑊3 − 𝑊1
Up thrust of the floating object in water=(𝑊2 −
𝑊1 ) − (𝑊3 − 𝑊1 )
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Up thrust of the floating object in water=𝑊2 −

𝑊1 − 𝑊3 + 𝑊1
Up thrust of the floating object in water=𝑊2 −
𝑊3
𝒘𝒆𝒊𝒈𝒉𝒕 𝒐𝒇 𝒔𝒐𝒍𝒊𝒅
𝑹𝒆𝒍𝒂𝒕𝒊𝒗𝒆 𝒅𝒆𝒏𝒔𝒊𝒕𝒚 =
𝒖𝒑 𝒕𝒉𝒓𝒖𝒔𝒕 𝒊𝒏 𝒘𝒂𝒕𝒆𝒓
𝑹𝒆𝒍𝒂𝒕𝒊𝒗𝒆 𝑑𝑒𝑛𝑠𝑖𝑡𝑦 𝑜𝑓 𝑓𝑙𝑜𝑎𝑡𝑖𝑛𝑔 𝑜𝑏𝑗𝑒𝑐𝑡
𝑊2 − 𝑊1
=
𝑊2 − 𝑊3
Relative density of a liquid

To find relative density of the liquid we determine:
a) Weight (w1) of solid in air.
b) Weight (w2) of the same solid when totally
immersed in water.
c) Weight (w3) of the same solid when totally
immersed in a liquid whose relative density
is to be determined. Volume of y displaced = 4 x 5 x 3
𝑅. 𝐷 𝑜𝑓 𝑙𝑖𝑞𝑢𝑖𝑑𝑠 = 60 cm3
𝑚𝑎𝑠𝑠 𝑜𝑓 𝑙𝑖𝑞𝑢𝑖𝑑
= 3
𝑚𝑎𝑠𝑠 𝑜𝑓 𝑒𝑞𝑢𝑎𝑙 𝑣𝑜𝑙𝑢𝑚𝑒 𝑜𝑓 𝑤𝑎𝑡𝑒𝑟 = 0.00006 m
Or 𝑼𝒑 𝒕𝒉𝒓𝒖𝒔𝒕 𝒊𝒏 = 𝒗𝝆𝒈
𝑅. 𝐷 𝑜𝑓 𝑙𝑖𝑞𝑢𝑖𝑑𝑠 = 800 x 0.00006 x 10

𝑤𝑒𝑖𝑔ℎ𝑡 𝑜𝑓 𝑑𝑖𝑠𝑝𝑙𝑎𝑐𝑒𝑑 𝑙𝑖𝑞𝑢𝑖𝑑
= = 0.48N
𝑤𝑒𝑖𝑔ℎ𝑡 𝑜𝑓 𝑒𝑞𝑢𝑎𝑙 𝑣𝑜𝑙𝑢𝑚𝑒 𝑜𝑓 𝑤𝑎𝑡𝑒𝑟 𝑑𝑖𝑠𝑝𝑙𝑎𝑐𝑒𝑑
Volume of x displaced = 0.04 x
Or 0.05 x 0.06
𝒖𝒑𝒕𝒉𝒓𝒖𝒔𝒕 𝒊𝒏 𝒕𝒉𝒆 𝒍𝒊𝒒𝒖𝒊𝒅 = 0.00012 m3
𝑹. 𝑫 𝒐𝒇 𝒍𝒊𝒒𝒖𝒊𝒅𝒔 =
𝒖𝒑𝒕𝒉𝒓𝒖𝒔𝒕 𝒊𝒏 𝒘𝒂𝒕𝒆𝒓
𝑼𝒑 𝒕𝒉𝒓𝒖𝒔𝒕 𝒊𝒏 𝒙 = 𝒗𝝆𝒈
𝑤1 − 𝑤2
𝑅. 𝐷 𝑜𝑓 𝑙𝑖𝑞𝑢𝑖𝑑𝑠 =
𝑤1 − 𝑤3
Page 33 of 108

Or = 0.00012m3
Example = 1.2N
1. A solid of mass 800g is suspended by a string is Total up thrust = 1.2 x 0.48
totally immersed in water. If the tension in the
string is 4.8N. Calculate = 1.68N
(a) Volume of solid Weight of block = total up thrust
(b) Relative density of the solid.
= 1.68N
Weight of solid = 8N
= 0.168 kg
W1 = 8N
𝒎𝒂𝒔𝒔
(b) 𝒅𝒆𝒏𝒔𝒊𝒕𝒚 =
W2= 4.8N 𝒗𝒐𝒍𝒖𝒎𝒆
Up thrust = 3.2N = 0.168
Volume of water displaced = 0.32 = 0.04 x 0.05 x

0.12
1000
=700 kg/m3
= 0.00032 m3
Volume of the solid = 0.00032m3
1. The wooden block below floats in two
liquids x and y if the densities of x and y are
1g/cm3 and 0.8g/cm3 respectively determine:
(i) Mass of the block
(ii) Density of the block
Applications of Archimedes’s Principle (b) Balloons

and Relative Density
Used by metrologists where a gas which
(a) The hydrometer is less dense than air like hydrogen is
used. The balloon moves upwards
because up thrust force is greater than
weight of the balloon. It rises to some
height where density is equal to that of
the balloon. At this point the balloon
stops rising because up thrust is equal to
weight of the balloon and therefore
resultant force is equal to zero.
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(c) Ships
They are made of steel which is denser
than water but floats because they are
hollow thereby displacing a large volume
of water than the volume of steel which
provides enough up thrust to support its
weight.
The average density of sea water is
It is an instrument used to find relative greater than the average density of fresh
densities of density of liquids. It applies water. Due to this difference, ships are
the law of flotation in its operation. fitted with primsol lines on their sides to
It has a wide bulb to displace large show the level that a ship should sink to
volume of liquid that provide sufficient when on various waters.
up thrust to keep hydrometer floating. (d) Sub-marine
Lead shots at the bottom- to make It can sink or float. It is fitted with
hydrometer float upright. ballast tanks that can be filled with air or
Narrow stem- to make hydrometer more water hence varying its weight .To sink,
sensitive. ballast tanks are filled with water so that
its weight is greater than up thrust.
To float compressed air is pumped into
the tank displacing water so that up thrust
is greater than weight of the submarine.
Examples
1. A hydrometer of mass 20g floats in oil
of density 0.7g/cm3.with 5cm of its stem
above the oil. If the cross sectional area
Hydrometers are designed for specific of the stem is 0.5cm2. Calculate:-
purposes lactometer range 1.015 – 1.0045
so as to measure density of milk. (a) Total volume of the hydrometer
The bulb is squeezed and released so that (b) Length of the stem out of water if it
the acid is drawn into the glass tube. floats in water.
Solutions to questions
(a) Volume of oil displaced =
Page 35 of 108

𝑚𝑎𝑠𝑠 𝑜𝑓 𝑜𝑖𝑙 𝑑𝑖𝑠𝑝𝑙𝑎𝑐𝑒𝑑

𝑑𝑒𝑛𝑠𝑖𝑡𝑦 𝑜𝑓 𝑜𝑖𝑙
20
=
0.8
= 25 cm3
Volume of hydrometer above oil = 5 x
0.5
= 2.5 cm3
Total volume = 25 + 2.5
= 27.5 cm3
(𝑏) 𝑀𝑎𝑠𝑠 𝑜𝑓 ℎ𝑦𝑑𝑟𝑜𝑚𝑒𝑡𝑒𝑟 Volume of 1.00g/cm3 liquid

= 2 𝑥 10 − 2 𝑘𝑔 displaced = m/ρ = 165/1 = 165 cm3;
𝑊𝑒𝑖𝑔ℎ𝑡 𝑜𝑓 ℎ𝑦𝑑𝑟𝑜𝑚𝑒𝑡𝑒𝑟 Volume of 1.10g/cm3 displaced =
= 2 𝑥 10 − 1𝑁 165/1.1 = 150 cm3;
𝑊𝑒𝑖𝑔ℎ𝑡 𝑜𝑓 𝑤𝑎𝑡𝑒𝑟 𝑑𝑖𝑠𝑝𝑙𝑎𝑐𝑒𝑑 Change in volume displaced = 165 – 150
= 2 𝑥 10 − 1𝑁 = 15 cm3 ;
𝑀𝑎𝑠𝑠 𝑜𝑓 𝑤𝑎𝑡𝑒𝑟 𝑑𝑖𝑠𝑝𝑙𝑎𝑐𝑒𝑑 = 2 𝑥 10−1 Volume = Area x Height ;

= 20𝑔 0.75 x h ; therefore h = 20 cm.
𝑉𝑜𝑙𝑢𝑚𝑒 𝑜𝑓 𝑤𝑎𝑡𝑒𝑟 𝑑𝑖𝑠𝑝𝑙𝑎𝑐𝑒𝑑 (ii) State two ways of improving the

20𝑔 sensitivity of the above hydrometer.
= 𝑔 𝑐𝑚3
1 -Using a hydrometer with
3 a narrow stem.
= 20𝑐𝑚
𝑉𝑜𝑙𝑢𝑚𝑒 𝑜𝑓 ℎ𝑦𝑑𝑟𝑜𝑚𝑒𝑡𝑒𝑟 𝑎𝑏𝑜𝑣𝑒 𝑤𝑎𝑡𝑒𝑟 - Using a hydrometer with
= 27.5 − 20 a large bulb
= 7.5𝑐𝑚3 2. When a body of mass 450g is

completely immersed in a liquid, the
𝑣𝑜𝑙𝑢𝑚𝑒 upthrust on the body is 1.6N. Find the
𝐿𝑒𝑛𝑔𝑡ℎ =
𝐴𝑟𝑒𝑎 weight of the body in the liquid.
7.5 3. The figure below shows a lever
= = 15𝑐𝑚
0.5 arrangement with the rod balanced
by a knife edged at as centre of
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gravity. The 5N weight on one

side balances the solid S (volume
The densities of liquids may be measured
using hydrometers. The hydrometer in 100cm3) which lies immersed in a
the figure consists of a weighted bulb beaker of water on the other side.
with a thin stem.
The beaker of water is then removed and

while keeping the 42cm distance
The hydrometer is floated in the liquid constant, the position of solid S is
and the density is read from a scale on its adjusted to obtain balance
stem. conditions again.
The hydrometer in the figure is designed a) Determine the new position of S.
to measure densities between 1.00 g cm –
3
and1.10 g cm – 3.On the diagram, mark
with the letter M the position on the scale b) What would be the new position of
of the 1.10 g cm – 3 graduation. The
S if it was immersed in a liquid of
hydrometer has a mass of 165 g and the
stem has a uniform cross-sectional area relative density 0.8?
of 0.750 cm2.Calculate;
(i) The change in the submerged
volume of the hydrometer when
it is first placed in a liquid of
density 1.00 g cm – 3 and then in
a liquid of density 1.10 g cm – 3.
Chapter Four ELECTROMAGNETIC INDUCTION
Page 37 of 108

Specific objectives Content

By the end of this topic, the learner ➢ Simple experiments to illustrate
should be able to: electromagnetic induction
➢ Induced emf:
a) Perform and describe simple
experiments to illustrate • Faraday’s law
electromagnetic induction • Lens’s law
b) State the factors affecting the ➢ Mutual induction
magnitude and the direction of ➢ Alternating current generator,
induced emf direct current generator
c) State the laws of electromagnetic ➢ Fleming’s right hand rule
induction ➢ Transformers
d) Describe simple experiments to ➢ Applications of electromagnetic
illustrate mutual induction explain induction
mutual induction • Induction coil
e) Explain the working of an • Moving coil transformers
alternating current (a.c) generator
and a direct current (d.c)
generator
f) Explain the working of a
transformer
g) Explain the application of
electromagnetic induction
h) Solve numerical problems
involving transformers
Introduction Magnetic Flux
Electric current passing through a
conductor has an associated magnetic It is the product of magnetic field
field. The reverse is also true in that a strength and perpendicular area covered
change in magnetic field induces an by the field lines.
electric current in a conductor a The direction of induced emf by a
phenomenon known as electromagnetic conductor is predicted by two laws of
induction. electromagnetic induction;
This is attributed to Michael Faraday and Faraday’s law-The magnitude of
has led to production of electrical energy induced e.m.f. is directly proportional
in power station. to the rate of change of magnetic flux
Experiment to Show Induced linkage.
Electromotive Force (Emf) Lenz’s law – direction of induced e.m.f.
is such that the induced current which
it causes to flow produces a magnetic
effect that opposes the change
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producing it.
The galvanometer reflects when

conductor AB cuts the magnetic field.
• There is no flow of current when
the conductor is stationary.
• The magnitude of induced current
increases with the angle of which
conductor cuts magnetic field.
• The direction of deflection reverses
when the direction of motion is
reversed.
FACTORS AFFECTING
MAGNITUDE OF INDUCED E.m.f.
(i) The magnitude or strength of
magnetic field.
(ii) The rate of change of flux
linkage/rate of relative motion
between the conductor and
magnetic field.
(iii) The number of turns of
coil/length of the conductor
(iv) The nature of the core
Page 39 of 108

The mechanical energy of a moving Mutual Induction

magnet inside a coil is converted to
electric energy inform of induced current. It occurs when change of current in one coil
The person pushing the magnet towards induces a current in another coil placed close
the coil must exert force to do work in to it. The changing magnetic flux in the
against repulsion of induced pole of coil first coil (Primary) links to secondary coil
magnet. inducing an EMF in it.
Fleming Right Hand Rule (Dynamo

Rule)
If the thumb and first two fingers of
the right hand are held manually at
right angle with 1st finger pointing
direction of magnetic field, the thumb
pointing in the direction of motion then
the second finger points in direction of
induced current.
Example
(i) A square looped conductor is pulled
at speed across a uniform magnetic field
as shown below.
When a switch is closed, current in the

primary coil increases from zero to a
maximum current within a very short time.
The magnetic flux in the primary coil
linking with the secondary coil increases
Determine direction of induced current in from zero to a maximum value in the same
interval of time inducing an e.m.f. in the
(a) AB – from B to A secondary coil. Current flows hence the
(b) AD - no induced current reflection on the galvanometer.
(c) CD-C to D Likewise when the switch is opened the

current in the primary takes a very short
(d) BC- no induced current time to fall from maximum value to zero.
The magnetic flux in primary coil linking
Question secondary also falls from maximum value to
(a) State Faraday’s laws of zero inducing an e.m.f. on the secondary
electromagnetic induction. coil.
The induced e.m.f. in the secondary coil is
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(b) The figure below shows a conductor higher when current in the primary coil is
XY moving in a region of uniform switched off than when it is switched on
magnetic field. because the current in the circuit takes a
much shorter time to die off than build up.
The above explanation is called mutual
induction. The induced e.m.f. in the
secondary coil can be increased by:
(i) Having more turns in the secondary
coil.
(ii) Winding the primary and secondary
coils on a soft iron rod.
(iii) Winding both primary and
(i) State the direction of the induced secondary coils on a soft iron ring in
current in the conductor and the order for the magnetic flux in the
rule used in arriving at the answer. primary to form concentric loops
(ii) Suggest one way of increasing the within it thus reaching the secondary
magnitude of the induced current point. Soft iron concentrates magnetic
in the conductor. flux in both coils that is why it is
used.
Application of Electromagnetic Useful transformer equations

Induction Secondary voltage
From experiment; =
It is applied in many areas some of these Primary voltage
no.of secondary turns
are: no.of secondary turns
(i) Transformer This is called turns rule.
A transformer transfers electrical energy Mathematically;
form one circuit to another by mutual
induction. VS NS
=
VP NP
It consists of a primary coil where an
alternating current is the input and Assuming negligible resistance
secondary coil forming the output.
Power = Voltage x current
The coils are wound on a common soft
Page 41 of 108

iron. 𝑃 = 𝑉𝐼
Types of Transformers 𝑃𝑜𝑤𝑒𝑟 𝑖𝑛𝑝𝑢𝑡
= 𝑃𝑟𝑖𝑚𝑎𝑟𝑦 𝑣𝑜𝑙𝑡𝑎𝑔𝑒 𝑥 𝑝𝑟𝑖𝑚𝑎𝑟𝑦 𝑐𝑢𝑟𝑟𝑒𝑛𝑡
Step down transformers
𝑃𝐼𝑛𝑝𝑢𝑡 = 𝑉𝑝 𝐼𝑝
Power output= secondary voltage x

secondary current
𝑃𝑂𝑢𝑡𝑝𝑢𝑡 = 𝑉𝑆 𝐼𝑆
𝐸𝑓𝑓𝑖𝑐𝑖𝑒𝑛𝑐𝑦 (%)
𝑃𝑜𝑤𝑒𝑟 𝑜𝑢𝑡𝑝𝑢𝑡
It has more turns in primary coil (Np) = 𝑥 100
𝑃𝑜𝑤𝑒𝑟 𝑖𝑛𝑝𝑢𝑡
than in the secondary coil (Ns)
VS IS
VS η= × 100%
= n(Turns ratio) VP IP
VP
Turns ratio is less than one For ideal transformer there is no energy
lost and therefore efficiency is 100%
Step up transformer
VS IS
It has less turns in primary coil and more 100 = × 100
VP IP
in the secondary coil.
VP IP = VS IS
The turn’s ratio is greater than one.
IP=VSIS
VP
IP VS NS
= =
IS VP NP
Examples
NOTE – In a step-down transformer 1. A transformer is to be used to provide

power of 24V ceiling bulb from a.c.
current in the in the secondary coil is
supply of 240. Find the number of turns
greater than in the primary coil while
in secondary coil if the primary coil has
in the step-up transformer current in 1000 turns.
the primary coil is greater than in the
VS NS
secondary coil. =
VP NP
24 𝑁𝑠
=
240 1000
1000
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= 100 turns
2. A power station has an output of Resistance of Coils (Copper Losses)

10KW at a p.d of 500v. The voltage is
stepped up to 15 KV by transformer This can be prevented by use of thick
T1, for transmission along a grid of copper wire to reduce heating effect.
resistance 3 k and thin stepped down Eddy Currents in the Core
to p.d 240v by transformer T2 at the
end of grid for use in a school. Given Eddy currents have associated fluxes that
that efficiency of T1 is 95% and T2 tend to oppose the flux change in
90%, find: primary. This reduces power transfer to
(i) The power output of T1 the secondary. To reduce eddy currents
(ii) The current in the grid. the core is laminated (using thin sheets of
(iii) The power loss in grid. insulated soft iron plates) causes minimal
(iv) The input voltage of T2 heating effect.
(a) The maximum power and current Hysteresis Loss

available for use in school
It is energy losses inform of heat in
(b) Why is it necessary to step up the magnetizing and demagnetizing the soft
voltage at power station? iron core every time the current reverses.
It can be minimized by using a core of
3. Power station has an output of 33K at soft magnetic material which magnetized
a p.d of 5k V a transformer with a and demagnetize easily.
primary coil of 2000 turns is used to
stop up the voltage of 132 KV for Practical Transformer
transmission along a grid. Assuming
A transformer used in power stations and
there s no power loss in the
along transmission lines generates a lot of
transformer calculate
heat. They are therefore cooled by oil
(a) Current in the primary coil which does not easily evaporate. Small
transformers are cooled by use of air. A
(b) Number of turns of secondary coil well-designed transformer can have an
(c) Current in the secondary coil. efficiency of up to 99%.However; the
presence of air reduces its efficiency.
Page 43 of 108

Energy Losses in a Transformer (i) Alternating current generator

There are four main causes of energy It converts mechanical energy into
losses in a transformer. electrical energy. It has a rectangular
curved permanent magnet poles, two slip
Flux Leakage rings and carbon (graphite) brushes.
All magnetic flux produced by the
primary may not link up with the
secondary coil hence reducing e.m.f
induced in secondary. Flux leakage is
reduced by efficient design of
transformers to ensure maximum flux
leakage.
The poles of the magnet are curved so
The secondary coil is wound over the that magnetic field is radial. Induced
primary coil or coils are wound next to Current enters and leaves the coil through
each other on a common core. the brushes which presses against the
slip-rings. The brushes are made of
graphite because:
(i) It is a good conductor of
electricity-
(ii) It is slippery and therefore can
act as a lubricant.
When coil rotates in clockwise direction
side AB moves up and CD downwards.
The two sides are cutting the magnetic
field perpendicularly and produce
maximum induced e.m.f (E) when the
coil is horizontal.
Applying Fleming’s right had rule, the (ii) Direct Current Generator
flow of induced current is in the direction
ABCD Direct current (d.c) generator differs from
an a.c generator in that it has a split-ring
The current flow through the external (commutator) while in ac generator has
circuit via the slip – ring 2 and brush x. slip-ring.
Brush Y and slip-ring 1 complete the
circuit. Brush x is thus positive terminal
which Y is negative. When coil rotates
from horizontal to vertical position the
angle at which the sides of the coil cuts
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magnetic field reduces from 90o to 0o

Likewise the induced emf reduces from
maximum to zero. When the coil rotates
past the vertical position side AB moves
downwards as side CD moves upwards.
The angle ϴ at which the sides of the coil
cuts the magnetic field increases from 0o
to 90o when coil is horizontal. The
induced emf increases from zero to If the coil rotates into the vertical position
maximum value and direction of current induced current and e.m.f though resistor
in the coil reverses from D C B A brush R decreases from maximum value to
Y now becomes positive and X negative. zero. The polarity of brush Y is positive
and X is negative. The brushes touch the
The magnitude of induced e.m.f obeys gaps within the commutators
the sinusoidal equation
The vertical position ensures that rings
E= Eosin ϴ exchange brushes since the induced
Where Eo is maximum e.m.f and ϴ is the current change direction but direction of
inclination of the coil to the vertical. current through the external resistor
remains the same. The polarity does not
The graph below shows the variation of change and output of d.c generator is
induced emf with time for one revolution shown below.
of the coil starting with the coil in
vertical position.
𝐸
𝐵𝑦 𝑂ℎ𝑚’𝑠 𝑙𝑎𝑤 𝐼 =
𝑅 The induced e.m.f or current of both a.c
and d.c generator can be increased by;
𝐸 = 𝐸0 𝑆𝑖𝑛𝛳
(i) Increasing speed of rotation of
𝐼𝑅 = 𝐼0𝑅 𝑠𝑖𝑛𝛳
coil.
𝐼 = 𝐼0𝑆𝑖𝑛𝛳 (ii) Increasing no of turns of coil.
(iii) Increasing the strength f the
The graph of induced current against the magnetic field.
angle of inclination is similar to one (iv) Winding the coil on a laminated
Page 45 of 108

above. soft iron core.

In a bicycle dynamo the magnet rotates
while coil remains stationary. It has
advantages over other generators because
there are no brushes which get worn out.
(iv) Moving Coil Microphone

Sound waves from the source set
diaphragm in vibration which in turn
causes the coil to move to and from
cutting the magnetic field.
Induced e.m.f of varying magnitude sets
up varying current in coil so that coil is
perpendicular to it for maximum flux
linkage.
An amplifier is used to increase the
amplitude of this current before it is fed
into the loudspeaker to be converted back
to sound. Assignment to the student.
Do the exercise at the end of the topic

(iv) The Induction Coil in KLB.
It is used to ignite petrol-air mixture in a
car engine.
When the switch is closed the soft iron
rod is magnetized due to current on the
primary coil and attracts the soft iron
armature.
The armature opens the contact and cuts
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off primary currents reducing magnetic

field to zero. This is turn induces a large
emf in the secondary coil by mutual
induction. The spring pulls the armature
back to make the contact again and
process repeats itself.
The induced emf in the secondary coil is
higher when primary current is switched
off than when it is switched on. This is
because current takes a longer time to
increase from zero to maximum than to
decrease from maximum to zero.
Sparking occurs at the contact due to
magnetic field of the primary. A
capacitor is therefore connected across
the contacts to minimize sparking effects
by decaying magnetic flux to zero.
Sparks forms across gap between the
ends of secondary coil and can be used to
ignite petrol-air mixture in a car engine.
Page 47 of 108

Chapter Five MAINS ELECTRICITY

Specific Objectives Content
By the end of this topic the learner should ❖ Sources of mains electricity
be able to: ❖ Power transmission (include
dangers of high voltage
a) State the sources of mains transmission)
electricity ❖ Domestic wiring system
b) Describe the transmission of ❖ Kw-hr, consumption and cost of
electricity power from the electrical energy
generating station ❖ Problems on mains electricity
c) Explain the domestic wiring system
d) Define the kilowatt hour
e) Determine the electrical energy
consumption and cost
f) Solve numerical problems
involving electricity
Sources of Mains Electricity Advantages of A.C Voltage over D.C
Voltage
1. Water in high dams
2. Geothermal energy (i) Can be transmitted over long
3. Coal or diesel distances with minimum power
4. Winds loss.
5. Tidal waves in the seas (ii) Can be stepped up to very high
6. Nuclear energy voltage.
The type of power generation chosen for
a given location depends on the most
abundant source of energy available in Electrical Power
that area. 𝐸𝑙𝑒𝑐𝑡𝑟𝑖𝑐𝑎𝑙 𝑝𝑜𝑡𝑒𝑛𝑡𝑖𝑎𝑙 𝑒𝑛𝑒𝑟𝑔𝑦
Power Transmission = 𝑣𝑜𝑙𝑡𝑎𝑔𝑒 × 𝑐ℎ𝑎𝑟𝑔𝑒
The National Grid System 𝑤𝑜𝑟𝑘 𝑑𝑜𝑛𝑒 = 𝑉𝑄

𝑤𝑜𝑟𝑘 𝑑𝑜𝑛𝑒
• It is a system of power cables 𝐸𝑙𝑒𝑐𝑡𝑟𝑖𝑐𝑎𝑙 𝑝𝑜𝑤𝑒𝑟 =
𝑡𝑖𝑚𝑒
connecting all the stations in a
𝑉𝑄
country to each other and to 𝑃=
𝑡
consumers.
• Advantage of national grid 𝑏𝑢𝑡 𝑄 = 𝐼𝑡
system 𝑉×𝐼×𝑡
Ensures that power is available 𝑇ℎ𝑢𝑠, 𝑃 =
𝑡
to consumers even when one of
⇒ 𝑃 = 𝑉𝐼
the power stations fails.
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• Most power stations generate Power loss during transmission

electricity in form of alternating
Power dissipated in a circuit is given by P
current (a.c) at voltage between
= VI.
11kV and 25kV.The voltage is
then stepped up between 132kV to But V = IR (Ohm’s law)
400kV for transmission so as to
minimize power loss. Thus P = I2 R
• The electrical energy is then The above equation shows that when
transmitted over long distance to current is high power loss is also high
substations where the voltage is
stepped down to 11kV. Power loss is therefore low when
• The power can be stepped down to transmitted at high voltage and low
appropriate value for domestic and current.
other users. In Kenya domestic
applicant operate at 240V.
• An 𝑎. 𝑐 source voltage is
represented by the symbol shown
below:
Example In most cases aluminium is preferred

because:
1. A generator produces 100kW
power which is transmitted through (a) Good conductor of electricity
a cable of resistance 5Ω.if the (b) It is light
voltage produced is (c) It is cheap and available
5,000V,calculate;
(i) The current transmitted
(ii) Power loss through the cables Dangers of high voltage transmission.
(iii) Power received by the
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consumer (i) Harmful effects of strong

electric fields.
During transmission power loss can be (ii) The risk of fire on nearby
minimized by: structures and vegetation when
(i) Stepping up output voltage cables get too low.
from power station. (iii) The risk of electric shock in
(ii) Use of thick and good case the poles collapse or hang
conductor transmission cable to too low.
minimize resistance.
Domestic Wiring
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Electrical power is usually supplied at 240V

from a step-down transformer.
This power is connected to the house using
two wires;
Neutral cable which earthed at zero
potential.
Live cable which is at full potential Example
The life cable is connected to a higher fuse House has a lighting circuit operated
value. The cables are then connected to a from 220V mains. Nine bulbs rated at
meter where energy consumed is registered. 120W 240V are switched on at the same
From meter cable passes on to consumers time. What is the most suitable fuse for
fuse box. this circuit?
Consumer fuse box consist of: P = VI
𝑃
Main switch I=
𝑣
It disconnects both live and neutral wires 9𝑃
simultaneously. Total current 𝐼 =
𝑉
9 𝑋 120
Live bus bar =
220
It is connected to live wire and the fuse. = 4.91A
Neutral bus bar The suitable fuse is a 5A fuse
It is connected to all neutral wires
Earth Terminal
It is earthed through a thick copper bar
buried deep in the earth or through water
piping.
Fuse
❖ Fuse is a thin wire (made alloy of
copper and tins) which melts when
current exceeds its rating.
❖ Its function is to safeguard
components against excess current in
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the circuit.
❖ It has low melting point.
❖ It is usually connected on a live wire
because live wires are at full potential
The fuse can blow due to the following:
(i) Overloading the circuit
(ii) Short circuiting
(iii) Use of wrong fuse rating
The fuse is normally represented by any of
the following symbols:
Circuit Breakers
Is an electronic device which disconnects
the circuit when current exceeds a certain
value by electromagnetism. It is more
efficient than a fuse in that it can be reset
when power goes off unlike a fuse which
must be replaced with a new one.
Lighting and Cooking Circuits
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For lighting circuit the lamps are The Rings Mains Circuit
connected in parallel so that:
Is a circuit where power in various rooms
(i) They are operated tapped at convenient point from a loop.
independently.
(ii) To reduce the effective The arrangement of the cable enable double
resistance. path for current arrangement also increases the
(iii) They can be operated at the thickness of wires used reduces the risk of
same potential overloading when several sockets are used.
The cables are relatively thin because

lamps consume small amount of
current.
For cooking circuit, power is taped
from the rings mains circuit.
These circuits are earthed and their
wires are relatively thicker than those
for the lighting circuit, since they carry
large currents.
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Two Way Switch Circuit The insulation on the three cables are
coloured so that they link correctly
Is used to put off on lights by one switch when connected to power circuit.
and put off by the other.
Live lead – red/brown
Example
Neutral – Black/ blue
The diagram below shows staircase double
switches. Earth – Green/green with yellow stripes
Fuse is used to safeguard appliance
from damage due to excessive
The value of the chosen fuse should be
slightly above the value of the operating
current of the appliance.
The earth pin is longer so as to open
On the table given below write down valves or shutters of the live and neutral
whether the lamp will be ON or OFF for pins.
various combinations of switch positions.
This protects the user from shock. Three
pin-plugs have the earth pin which
provides the path for excess current.
Question
The figure shows a three-pin plug
Identify the mistakes in the wiring
A Three – Pin Plug What would happen if this plug was
connected to mains socket.
Why is the earth pin normally longer
than the other two pins
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a)Study the figure below:- Electrical Energy Consumption and

Costing
Commercial companies charge for
electrical energy supplied to consumers.
Amount of energy used by consumers
depends on:
Power rating of appliances
Time for which they have been in use.
Energy = Power x time
(i) What name is given to the fitting in The unit is used for costing 1 unit =
the diagram? 1kWh
(ii) Identify the parts labelled.
Kilowatt – hour (KWh) is amount of
electrical energy spent in one hour at rate
Page 55 of 108

A - of 1000 J/S (watts).

B - A consumer has the following
components in his house for the times
C - indicated in one day.
D - Appliance Time
iii) State the colours A, B and C. Two 40w bulbs 30min
One 3kw electric heater 4hrs
A -
One 500w fridge 15hrs
B -
C -
Calculate;
Total power the components use
Total cost of power consumed in 30 days
if one unit costs Ksh 6.50
(2 x 40) + (3000) + (500)
= 3580W
Chapter Six ELECTROMAGNETIC SPECTRUM

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By the end of this topic. The learner • Electromagnetic spectrum

should be able to: • Properties of electromagnetic
waves
a) Describe the complete
• Detection of electromagnetic (emf)
electromagnetic spectrum
radiations
b) State the properties of • Application of e.m radiations
electromagnetic waves (include green house effect)
• Problems involving 𝑐 = 𝜆𝑓
c) Describe the methods of detecting
electromagnetic radiations
d) Solve numerical problems

involving c = f
Electromagnetic Spectrum - this is the Examples of E.M Waves
arrangement of the electromagnetic
waves according to their frequencies or They include light, x-rays, ultra violet,
wave length. infrared and gamma rays when this wave
is arranged in terms of wavelength or
Electromagnetic Waves frequency .They form electric magnetic
spectrum. The wavelength range from
Are transverse waves which results from about 1 x 106 m to 1 x 10 -14m.
oscillating electric and magnetic fields at
right angle to each other.
Page 57 of 108

Properties of Electromagnetic Waves Example

(i) They are transverse in nature. 1. Green light has a wavelength of
(ii) They do not require a medium 5 𝑥10−3 𝑚.Calculate the energy it emits.
for transmission. 2. A radio is tuned into a radio station 144
(iii) They travel through space km away.
(vacuum) with the speed of
light (3 x108 m/s) (a) How long does it take a signal to
(iv) They carry no charge hence not reach the receiver?
affected by electric or magnetic (b) If the signal has a frequency has a
fields. signal of 980 kHz, how many
(v) They undergo interference, wavelengths is the station away from
reflection, and diffraction, your receiver?
refraction and polarization
effects.
(vi) They possess energy in
different amounts. According to
𝐸 = ℎ𝑓 where h is planets
constant (6.63 x 10-34 Js) and f
is frequency.
(vii) They obey the wave equation
𝑐 = 𝑓𝜆
Production and Detection of Electromagnetic Waves.
EM wave Production Detection
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Radio waves oscillating electrical Diodes and earphones.

circuits and transmitted
through aerials or
antennae
Microwaves Special vacuum tubes Dry crystal detectors or
called magnetrons in solid state diodes.
microwave ovens or with
a mass.
Infrared Radiation the sun or any hot body Heating effect produced
on the skin, thermopile,
bolometer and
thermometer with
blackened bulb.
Visible light Sun is the major source the eye, photographic
other sources are hot film and photocell
objects, lamps and laser
beams.
Ultraviolet (u. v) rays By the sun, sparks and by photographic films,
mercury vapour due to photocells, fluorescent
large energy chances in materials (quinine
the electrons of an atom. sulphate) and paper
lightly smeared with
Vaseline
X-rays action of beam of fast- Using fluorescent screen
moving electrons or photographic film.
hitching a metal target
Gamma Rays By radioactive Detected by
substances in the nucleus photographic plates and
of an atom radiation detectors e.g.
The G.M tube.
Page 59 of 108

Application of Electromagnetic Waves

Properties Type of radiation Uses
Highest frequency Gamma rays In medicine used to kill
cancerous tissues
Highest energy content
Sterilize medical
equipment and pests
High energy Content x-rays Crystallography,
study fractures and
detect forgeries in art
and flaw in metals
Low energy content U.V radiations In medicine they supply
vitamin D
treatment of skin cancer
mineral analysis and
detecting forgeries
Easily reflected visible light Seeing,
have average photography,
wavelengths
Fibre optics, lasers
(light amplification by
the stimulated emission
of radiations)
Long wavelength Infrared and microwave Used in cooking
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high heating effect heating and drying

Long medium TV waves In communication with
wavelength the aid of satellite,
penetrates the
atmosphere easily
Longest wavelength, Radio waves Widely used in radio
shortest frequency, communication.
Easily diffracted
Diffraction of TV Radio Waves

Large wavelengths and low frequency radio waves are easily diffracted. They are
also easily detected by receivers in deep valleys and behind hills. Radio waves of
longer wavelength, amplitude modulation (AM) are reflected easily by ionosphere.
Shorter wavelength waves (frequency modulations (FM) are transmitted over short
distances and received directly from the transmission.
Microwaves Hazards of Some Electro-Magnetic

Waves.
In cooking microwaves produces
magnetron at a frequency of about 2500 UV rays and gamma rays carry high
MHZ. These waves are directed to a energy which damages cells, skin burn of
rotating metal shiner which reflects them effect eyes when absorbed. There are
to different parts of the oven. In the oven delayed effect of radiation such as cancer,
food is placed in a turntable where it leukemia and hereditary effects in
absorbs the waves evenly. The wave’s children.
heat cooks it. The wire mesh on the door
reflect the microwave back inside. The Minimising the Hazards
device is switched off before opening the
Page 61 of 108

door. Microwaves of shorter wavelength (i) Reduce dosage by minimising

are used in radar communication. exposure time.
(ii) Keep a safe distance from the
radiations
(iii) Use shielding materials such as
lead jackets.
Micro waves which have shorter wave

lengths are used for radar (Radio
detection and Ranging) communication.
This communication is useful in locating
the exact position of aero planes and
ships.
Radio Waves
They have varying range of wavelength
which makes their application wide.
Medium and short wavelength is used in
radio transmission signals. Amplitude
modulation (AM) radio transmission has
a longer range because of reflection by
the ionosphere. TV and frequency
modulation (FM) radio waves are
received all a shorter wavelength than
normal radio broadcasts. Very high
frequency (VHF) transmission (Used in
TV and FM radios are transmitted over
short distance and received direct from
the transmission.
Green House Effect (Heat Trap)
Transparent glass allows visible light of
short wavelength radiations emitted by
the sun to pass through. On the other
hand glass cannot transmit the long
wavelength given out by cooler objects.
Heat from the sun is therefore trapped
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inside the green house. This makes inside

of the green house warmer than outside.
Chapter Seven CATHODE RAYS AND CATHODE RAY

OSCILLOSCOPE
By the end of this topic, the learner should Content
be able to:
• Production of cathode rays,
a) Describe the production of cathode cathode ray tube.
rays • Properties of cathode rays.
b) State the properties of cathode rays • C.R.O and T.V tubes.
c) Explain the functioning of a • Uses of C.R.O.
cathode Ray oscilloscope • Problems on C.R.O.
(C.R.O)and of a Television tube
(T.V tube)
d) Explain the uses of a cathode Ray
Oscilloscope
e) Solve problems involving Cathode
Ray Oscilloscope
Cathode Rays Properties of Cathode Rays
They are streams of high velocity (i) They travel in a straight line in
electrons emitted from the surface of a absence of magnetic or electric fields.
metal when a cathode (negative electrode) Hence form sharp shadows of objects put
is heated inside a vacuum tube by on their way.
thermionic emission. Electrons are able to
leave the metal surface because they gain
enough kinetic energy to break loose from
the force of attraction of the nuclei.
Thermionic emission is the process of
Page 63 of 108

emitting electrons from a metal surface

due to heat energy. See the figure below:
ii) Cathode rays cause fluorescence in

some substances e.g. zinc sulphide
Before the heater current is switched on, (phosphor).
no current is registered by the milliameter. iii) They possess kinetic energy. The
When the switch is put on, the cathode is kinetic energy of the emitted electrons
heated and emits electrons which is converted into light energy by a
complete the gap between the electrodes process called fluorescence. This is
and a current is registered at the the main reason why the screen is not
milliameter. heated.
Production of cathode rays iv) They are charged because they are
deflected by both electric and
magnetic fields (not waves).
v) The path of cathode rays in a
magnetic field is circular so that the
force acting on them is perpendicular
to both the magnetic field and the
direction of current.
In the above discharge tube electrons

produced at the cathode by thermionic
emission are accelerated towards
fluorescent screen by an anode of an extra
high tension (EHT) source. The tube is
evacuated so that the emitted electrons do
not collide with air molecules which
would ionise them making them lose
kinetic energy. Ionisation is a process
where electrons are completely removed
from atoms of an element. The cathode is
coated with barium and strontium oxides
to give a ready and continuous supply of
electrons.
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vi) cathode rays have momentum and energy given

1
by 𝑀𝑒𝑉 and 𝑀𝑒𝑉 2 respectively Cathode Ray Oscilloscope
2
(C.R.O.)
𝑘𝑖𝑛𝑒𝑡𝑖𝑐 𝑒𝑛𝑒𝑟𝑔𝑦 𝑜𝑓 𝑡ℎ𝑒 𝑒𝑙𝑒𝑐𝑡𝑟𝑜𝑛
= 𝑤𝑜𝑟𝑘 𝑑𝑜𝑛𝑒 𝑏𝑦 𝑡ℎ𝑒 𝑒𝑙𝑒𝑐𝑡𝑟𝑖𝑐 𝑓𝑖𝑒𝑙𝑑 𝑜𝑛 𝑡ℎ𝑒 𝑒𝑙𝑒𝑐𝑡𝑟𝑜𝑛 It is an electrical instrument
1 used to display and analyse
2
𝑚 𝓋 = 𝑒𝑉 wave forms as well as to
2 𝑒 measure electrical potentials
vii) cathode rays produce x rays when they strike a i.e. voltages that vary with
metal target time.
viii) Cathode rays slightly ionise gases.
It consists of the following
ix) Affect photographic papers.
parts;
(a) Electron gun.
(b) Deflecting system.
(c) Display system
Electron Gun • If the grid is made more

negative with respect to the
(a) It produces an electron beam which is highly a cathode, the number of
concentrated stream of high speed electrons.
electrons per second
Page 65 of 108

(b) It has the following components; passing through the grid

decreases and the spot
-Cathode becomes darker. The effect
-Cylindrical grid is reversed if the grid is
made more positive in
-Two anodes potential with respect to the
cathode.
Function of Cathode
Anodes
To emit electrons by thermionic emission (when
heated). It is coated with oxides of thorium and The two anodes have positive
strontium (the two are preferred because they have potentials relative to cathode.
low work functions hence can emit electrons easily) Anode 1 is at a higher
potential than anode 2. The
Function of Cylindrical Grid.
difference in potential between
• Controls the rate of flow of electron hence the the two anodes creates an
brightness of the spot on the screen. electric field. The electric field
• The negative voltage on grid can be varied to converges the diverging beam
control the number of electrons reaching the anode. from anode 2.
Functions
(a) Attract electrons from
cathode.
(b) They accelerate the
electrons by providing
enough energy to cause
emission of light as they hit
the screen.
(c) They focus electron beam
by converging electrons to
a sharp point on the screen.
Deflecting System Display System (screen).

Function of the deflection system Inside of the screen is coated with
To determine position of electron phosphor (zinc sulphide) which
beam on the screen fluoresces or glows when electrons
strike it hence producing a bright spot
Types of deflections on the screen.
(i) Vertical deflection ( by Y-plate) The inside of the tube is coated with
(ii) Horizontal deflection ( by X-plate) graphite which has the following
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Vertical deflection( Y-plates) functions;

• It deflects electron beam vertically across • Earthing – conduction of electrons to
the screen in the following ways when the earth.
the time base(X-plates) is switched off; • It is used to shield the beam from
• When d.c potential across the two plates external electric field.
is zero a spot is produced on the screen • It accelerates electrons towards the
i.e. no deflection. screen since it is in the same
• When d.c. voltage is applied across the potential as the anode.
y-plate with top plate positive the
electrons are deflected upwards and a Uses of CRO
spot therefore appears on the upper part • It is used as a voltmeter.
of the screen. • Time base of switched off, the x-
• When lower plate is positive a spot plates earthed and the voltage to
appears on the lower part of the screen. be measured connected across the
• If a.c voltage is applied cross y plate the y-plates. The voltage is calculated
spot oscillate up and down depending on using the formula:
frequency such that what is seen on the
screen is a vertical straight line if the 𝑉𝑜𝑙𝑡𝑎𝑔𝑒
frequency is very high. = 𝑑𝑖𝑠𝑝𝑙𝑎𝑐𝑒𝑚𝑒𝑛𝑡
× 𝑠𝑒𝑛𝑠𝑖𝑡𝑖𝑣𝑖𝑡𝑦(𝑣𝑜𝑙𝑡𝑠 𝑝𝑒𝑟 𝑑𝑖𝑣𝑖𝑠𝑖𝑜𝑛)
Horizontal deflection (x -plates)
Advantages of CRO over voltmeter
X-plates are internally connected to the
time-base circuit, which applies a saw-tooth • Can measure large voltages
voltage to the x-plates. The voltage without being destroyed.
increases uniformly to a peak (sweep) and • It responds instantaneously unlike
drops suddenly (fly back). The speed with ordinary meter whose pointer is
which the electron beam is “sweep” can be affected by inertia.
adjusted with the help of the time base knob. • Can measure both a.c and d.c
voltages.
• It has extremely high resistance
and does not therefore alter
current or voltage in the circuit to
which it is connected.
• Measuring the frequency of a
wave(a.c signal)
The signal is fed into the y-plates of a
When a d.c voltage is applied to the
C.R.O. with the time base on. The time
Page 67 of 108

input(Y-plates) of the cathode ray base control is then adjusted to give one
oscilloscope and the time base on, then the or more cycles of the input signal on the
horizontal line is seen to move toward the screen. The time T of the signal is then
positive plate. determined by relating the trace of the
signal on the screen with the time base
When an a.c voltage is applied to the input setting. The frequency can be calculated
of a CRO and time base on, then due to 1
interaction of the saw-tooth voltage at the x- as 𝑓 = 𝑇
plates and a.c voltage at the y-plates, a
Examples
‘sine-curve’ is seen on the screen.
1. The figure below shows a display of
The purpose of time-base is to move
an a.c signal on the CRO screen.
electrons across the screen at a particular
Determine the frequency, given that
speed enabling the study of variation
the time base setting is 200ms/div.
between voltages with time.
2. On the grid provided below, show the display

on the CRO screen of an a.c signal, peak
voltage 300v and frequency 50Hz when time
base is on (Take-gain at 100 V/div, time base
setting at 10ms/div).
300
𝑚𝑎𝑥𝑖𝑚𝑢𝑛 𝑣𝑒𝑟𝑡𝑖𝑐𝑎𝑙 𝑑𝑖𝑣𝑖𝑠𝑖𝑜𝑛𝑠 =
100
=3 4. The control knobs of CRO have
been adjusted to get a bright
1 electron ‘spot’ on the screen.
𝑇=
𝑓 Explain how you get the
following traces:
1
𝑇=
50 (i) A horizontal line at the
𝑇 = 0.02 𝑠𝑒𝑐𝑜𝑛𝑑𝑠 centre.
(ii) A vertical line at the centre.
(iii) A sine curve
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𝑁𝑢𝑚𝑏𝑒𝑟 𝑜𝑓 ℎ𝑜𝑟𝑖𝑧𝑜𝑛𝑡𝑎𝑙 𝑑𝑖𝑣𝑖𝑠𝑖𝑜𝑛𝑠 𝑝𝑒𝑟 𝑐𝑜𝑚𝑝𝑙𝑒𝑡𝑒 𝑐𝑦𝑐𝑙𝑒 5. The time base on a CRO is set at
0.02 1ms/cm and Y-gain at
= =2 100v/cm.When an alternating
0.01
voltage is applied to the input
terminals, the beak value of the
sine curve on the screen is
2.9cm.calcuate:
(i) The amplitude of the ac
voltage.
(ii) The frequency of the ac
input signals, if two full
waves are formed in a
3. A.d.c voltage of 50v when applied to the Y- length of 5cm on the
plates of a CRO causes a deflection of the spot screen.
on the screen as shown. 6. The figure below shows the
deflection of a spot by
alternating voltage signal
(i) Determine the sensitivity of the Y-gain.

(ii) Show what will be observed on the
screen if an a.c of peak voltage 40v is
fed onto the Y-plates If the sensitivity is 30v/division
.Find the voltage of the signal
Page 69 of 108

TV Tube or Computer Monitor Question

1. The figure below shows the main
parts of a television receiver tube
with the electron guns deflection
coils and the fluorescent screen
labelled.
• In TV tube magnetic coils (fields)

are used instead of electric field
because they provide wider
deflection to light the whole
screen.
• The tube has two tiny plates which
combine to light the entire screen
instead of just a line. (i) Name the parts of the electron
• In a colour –TV 3 electron guns are gun
used each producing one primary (ii) Why are magnetic fields in the
colour (red, blue and green) screen coils preferred in the television
is coated with different chemicals set instead of electric fields?
to produce the colours. (iii) Name a suitable substance used
to coat the screen.
Coils are mounted outside the neck of the
tube so that they can be treated and
adjusted while set is being assembled and
tested.
Differences between TV screen and
CRO
TV CRO
Deflection is by Deflection is by
magnetic field electric field
It has two time It has one time
base base
Electrons lights Electrons produce
the whole screen a line or a dot
There are 625 There are 25 lines
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lines per second per second
Chapter Eight X-RAYS

By the end of this topic, the learner • Production of X-rays X-ray tube
should be able to: • Energy changes in an X-ray tube
a)Explain the production of X-rays • Properties of X-rays, soft X-rays
b)State the properties of X-rays and hard x-rays
c)State the dangers of X-rays • Dangers of X-rays and precautions
d)Explain the uses of X-rays • Uses of x-rays
e)Solve numerical problems • Problems on x-rays
involving x-rays
✓ X-rays were discovered by W. Roentgen in 1895 when he was conducting a
research of cathode rays. He called them x-rays because their nature was
unknown at the time of discovery. X-rays are produced when fast moving
electrons are suddenly stopped by matter.
Production Of X-Rays.
Page 71 of 108

✓ Modern x-rays have a rotating

target during operation to change
✓ When a Cathode is heated, it the point of impact thereby
produces electrons by thermionic reducing the wear and tear on it.
emission. Emitted electrons are ✓ The target is set at an angle (450)
accelerated to the anode (target) by to direct x-rays out of the tube
high potential difference i.e. 100kv through a window on the lead
between cathode and anode. shield. See the figure below:
✓ When first moving electrons are
stopped by the anode (target) part of
their kinetic energy is converted to
x-rays.
✓ An x-ray tube is really a high
voltage diode valve.
✓ Cathode is concave so as to focus
electron beam to the tungsten target.
✓ The cathode is coated with oxides
of low work function so that
electrons are easily emitted from its
surface when it is heated.
✓ The anode target has a high melting
point to withstand a lot of heat
generated e.g. tungsten or
molybdenum
✓ Most of kinetic energy of electrons
is converted to heat energy but
about 0.5% is transferred to x-rays
radiation.
✓ Anode is made of good conductor
of heat i.e. copper for efficient (fast)
dissipation of heat energy.
However, oil circulation and fins
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enhances cooling process.

✓ Lead shielding has high density so
as to prevent x-rays from
penetrating into undesirable targets.
Properties of X-Rays Applications of x-rays (uses)

1. Travel in straight line at a speed of (i) In medicine (Radiography and
light in a vacuum. Radiotherapy).
2. X-rays are dangerous, they can ✓ Due to the penetrating property of
cause cancer. x-rays, fractured bones and
3. X-rays penetrate substances but are dislocated joints can be seen from
absorbed by dense solids e.g. x-ray photograph called
bones or lead. radiograph. Foreign objects like
4. X-rays affect photographic films. swallowed coins or pins can also
5. X-rays ionise gases, so that the be located.
gases become conductors. ✓ Hard X-rays can treat cancer,
6. X-rays can cause photoelectric tumours and other skin diseases by
emission. destroying the infected cells.
7. X-rays cause fluorescence in (ii) Science/ Crystallography
certain substances. ✓ Study of crystal structure which
8. X-rays are not deflected by a explains the arrangement of atoms
magnetic or electric fields. in different materials. Incase there
9. X-rays can be diffracted and plane are fractures in the structure of the
polarised so they are waves. material, they can easily be
revealed by the X-rays
10. X-rays are electromagnetic waves of
very short wavelengths and hence obey (iii) In industry
the wave equation 𝑐 = 𝜆𝑓 and energy ✓ Inspect cracks/flaws in metal
equation 𝐸 = ℎ𝑓. casting.
Types of X-Rays ✓ Sterilize surgical equipment before
packaging.
Hard x-rays
(iv) Security e.g. in Airports.
✓ Have high frequency (short
✓ To inspect luggage for any weapon
wavelength) hence high penetrating
hidden in them.
power. This is achieved by increasing
the anode voltage, in order to give the
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cathode rays more kinetic energy. Dangers of X–ray

✓ These x-rays penetrates the flesh but
are absorbed by the bones. ✓ Because of their ionizing property, X-
rays can cause serious damage to the
Soft x-rays body cells. Excess exposure of living
tissue to X-ray can lead to damage or
✓ The soft x-rays are produced by killing of the cells.
electrons moving at a lower velocity ✓ The penetrating property can also
compared to those producing hard x- cause genetic changes and even
rays. This is achieved by lowering the produce serious diseases like cancer if
accelerating voltage. one is exposes to them for a long time.
✓ These x-rays are used to show
malignant growth in tissues because
they only penetrate the soft tissues.
✓ Quality and type of x-rays produced is Precautions when using x-ray machine
determined by the accelerating (i) X-ray machines have lead shield to
potential. protect the operator from stray X-
Intensity (Quantity) of x-rays rays.
(ii) The rooms of operation have
✓ The Intensity of x-rays is controlled concrete walls to absorb any
by amount of heating current. leaking radiations (X-rays).
✓ The greater the heating current, the (iii) Reduce exposure time.
greater the number of electrons ENERGY OF X-RAYS
produced hence more x-rays. To give Energy of an electron can be
a more intense beam of x-rays, the calculated by the formula E = hf
cathode is made hotter, to give more where h is Plank’s constant and f
electrons leading to more x-rays. frequency.
NOTE: The strength of the X-rays c
But from c = fλ; f = thus E =
depends on the accelerating 
potential difference between the hc
anode and the cathode. 
But E = eV where e is the electron
charge and V is the accelerating
potential difference.
hc
Thus eV =

The energy of the electron is
maximum when the wavelength
is shortest.
Calculations 4. The figure below shows the
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1. Calculate the wavelength of x-rays essential components of an X – ray

whose frequency is given by
tube
2.0 × 1020 𝐻𝑧.
2. Find the energy of x-rays whose
wavelength is 10−10m in a vacuum
(𝑐 = 3.0 × 108 𝑚/𝑠, and ℎ =
−34
6.63 × 10 𝐽𝑠).
3. An x-ray tube is operated with
anode potential of 50kV and
current of 15mA,calculate;
(i) The rate at which energy is
a) (i) Briefly explain electrons are
converted at the target of the
x-ray tube. produced by the cathode
(ii) Kinetic energy of the
emitted electrons before b) (ii) How are the electrons produced
hitting the target. accelerated towards the anode?
(iii) The maximum velocity of
the electrons.
(iii) Why is the target made of
(iv) The frequency of the x-rays
produced if 0.5% of the tungsten?
energy is converted into x- (iv) How is cooling achieved in this
rays. kind of X – rays machine.
(v) The number of electrons (v) Why would it be necessary for
hitting the anode after one the target to rotate during
second. operation of this machine?
4. An x-ray tube operates at a
(vi) Why is the tube evacuated?
potential of 80kV. Only 0.5% of
electron energy is converted to X- (vii) Why is the machine
rays at the anode at a rate of surrounded by a lead shield?
100J/s.
(a) If the accelerating voltage is 100
Determine;
kV calculate;
(i) The tube current. (i) Kinetic energy of the electrons
(ii)The average velocity of arriving at the target (e = 1.6 x 10 -
electrons hitting the target. 19
C).
(iii) The minimum wavelength of x-
rays (ii) If 0.5 % of the electron energy is
5. An x-ray tube operating at 50kV converted into X – rays determine
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has a tube current of 20mA. the minimum wavelength of the

(Take𝑚𝑒 = 9.1 × 10−31 𝑘𝑔, 𝑒 = emitted X- rays ( h = 6.63 x 10 -34
1.6 × 10−19 𝐶, 𝑐 = 3.0 × 108 𝑚/𝑠) JS, and C = 3.0 x 10 8 ms -1) .
.How many electrons are hitting
target per second.
(i) If 0.5% of energy of electron is
converted to x-rays, estimate
the quantity of heat energy
produced per second.
(ii) Find x-ray power output.
More
1. State one agricultural use of x-rays.
2. Name the property of x-rays that
determines the type of x-rays
produced.
3. An ex ray-tube is operated at
125kV potential and 10𝑚𝐴.If only
1% of the electrical power is
converted to x-rays, at what rate id
the target being heated? If the
target has 0.3kg and is made of a
material whose specific heat
capacity is 150𝐽𝑘𝑔−1𝐾 −1,at what
average rate would the temperature
rise if there were no thermal loses?
Chapter Nine PHOTOELECTRIC EFFECT

By the end of this topic, the learner should ✓ Photoelectric effects, photons,
be able to: threshold frequently, work
function, planks constant, and
a) Perform and describe simple electrons-volt.
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experiments to illustrate the ✓ Factors affecting photoelectric

photoelectric effect emission.
b) Explain the factors affecting ✓ Energy of photons.
photoelectric emission 1
✓ Einstein equation hf = hf 0 + me v 2
c) Apply the equation E=hƒ to 2
calculate the energy of photons ✓ Applications of photoelectric
d) Define threshold frequency, work effects:
function and the electron volt
e) Explain photoelectric emission • photo emissive,
using Einstein equation • Photo conductive
f) (h ƒ +h ƒ0 + ½ mv2) • Photo voltage cells
g) Explain the applications of ✓ Problems on photoelectric emissions
photoelectric effects
h) Solve numerical problems involving
photoelectric emission
When an electromagnetic radiation of (b) Using charged electroscope
sufficient frequency is radiated on a metal
surface electrons are emitted. These
electrons are called photoelectrons and
the phenomenon is called photoelectric
effect. Photoelectric effect is therefore a
phenomenon in which electrons are
emitted from the surface of a solid when
illuminated with electromagnetic radiation
of sufficient frequency. A material that
exhibits photoelectric effect is said to be
photo- emissive. When freshly cleaned zinc plate is
irradiated with UV radiations, electrons
Photoelectric effect can be are emitted from its surface.
demonstrated by: Photoelectrons emitted from the
(a) Using neutral plates positively charged zinc plate do not
escape due to attraction by positive
charges on the zinc plate hence
divergence remains the same.
Photoelectrons emitted from the
negatively charged zinc plate are repelled
and the electroscope becomes discharged
as a result of which the leaf divergence
decreases. If a glass (which absorbs UV
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radiations) is placed between zinc plate

(negatively charged) and the UV source
no effect is seen on the leaf of the
electroscope.
If the zinc is not freshly cleaned, the
electrons might not be liberated from
the zinc.
If the electroscope is uncharged, its leaf
rises steadily showing that it is being
When UV falls on plate A, the charged. When tested it is found to be
galvanometer deflects showing flow of positively charged. This is because;
current, when UV is blocked no deflection electrons are removed from the zinc plate
on the galvanometer i.e. no current which in turn attracts electrons from the
flowing. When UV falls on the metal, leaf of the electroscope leaving it with
some electrons acquire sufficient kinetic positive charges.
energy from the UV and are dislodged
from the surface. The electrons are
attracted to plate B. The electrons
complete the gap between the plates
allowing current to flow in the circuit
hence deflection on the galvanometer.
Definition of terms 𝟏𝒆𝑽 = 𝟏. 𝟔 × 𝟏𝟎−𝟏𝟗 𝑱

1. Threshold frequency, fo
Minimum frequency of a radiation for Examples
which can cause electron(s) to dislodge
from a metal surface. 1. Calculate the energy of a photon of
frequency 5.0 × 1014 𝐻𝑧 in:
When visible light is incident on the (i) Joules
freshly cleaned zinc plate, the leaf of a (ii) Electron volts
negatively charged electroscope does 2. The wavelength of orange light is
not decrease in divergence. This shows 625nm.calculate the energy of a photon
that visible light does not have enough emitted by orange light in electron volts.
energy to dislodge electrons from the
surface of zinc plate. For any given Einstein’s Equation of Photoelectric Effect.
surface there is a minimum frequency • When a photo strikes an electron all its
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of radiation below which no energy is absorbed by the electron and

photoelectric emission occurs. This some energy is used to dislodge the electron
frequency is called threshold while the rest become the kinetic energy of
frequency,𝑓0 the electron. i.e.
Energy of photon = (Energy needed to
2. Work function, Wo -the minimum
dislodge an electron from the metal
amount of energy needed to completely surface) + (maximum K.E gained by the
remove (dislodge) an electron from the electron)
surface of a metal. Work function varies • If the frequency (f) on any radiation is less
from one metal to another. Unit for than 𝑓𝑜, the energy will be less that 𝑊𝑜 and
work function is electron-volt(eV) or therefore no emission will occur. If the
joule(J) Note: 1 eV = 1.6 x 10-19 J frequency (f) is greater than fo then ℎ𝑓 >
𝑊𝑜 and excess energy is utilized as K.E of
3. Threshold wavelength, λo – is the emitted electrons.
maximum wavelength beyond which no • Thus,𝒉𝒇 = 𝒘𝑶 + 𝑴𝒆 𝒗𝟐 where 𝑀𝑒 is mass
photoelectric emission will occur. of electron and
Light energy and the quantum theory. V- Velocity of emitted electron. This is
Einstein’s photoelectric equation.
In 1901, Max Planck, a scientist came
up with the idea that light energy is ℎ𝑓 = ℎ𝑓𝑂 + ½𝑀𝑒 𝑣 2
propagated in small packets of energy.
Each packet is called quantum (quanta- ℎ𝑐
ℎ𝑓 = + ½𝑀𝑒 𝑣 2
plural). In light this energy packet is ⋋𝑜
called photons.
Example:
Energy possessed by a photon is
1. The minimum frequency of light that will
proportional to the frequency of the cause photoelectric emission from potassium
radiation. surface is 5.37 x1014 Hz. When the surface is
𝐸 = ℎ𝑓 irradiated using a certain source photoelectrons
are emitted with a speed of 7.9 x 105ms-1
Where f is the frequency of radiation calculate
and h is Plank’s constant = 6.63 x 10 -34
Js (a) Work function of potassium.
In general wave equation 𝑐 = (b) Maximum K.E of the photoelectrons.

𝑐
𝑓𝜆 𝑜𝑟 𝑓 = (c) The frequency of the source of irradiation
𝜆
𝑐
Therefore 𝐸 = ℎ solution
𝜆
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Since h and c are constants, a wave with (𝑎) 𝑾𝒐 = 𝒉𝒇𝒐

larger wavelength has less energy.
= 𝟔. 𝟔𝟑 𝒙 𝟏𝟎−𝟑𝟒 𝑱𝒔 𝒙 𝟓. 𝟑𝟕 𝒙 𝟏𝟎𝟏𝟒 𝑺−𝟏
1. Electron-volts-is the work done
of energy gained by an electron = 𝟑. 𝟓𝟔 𝒙 𝟏𝟎−𝟏𝟗 𝑱
when it moves through a (𝒃)𝑲. 𝑬𝒎𝒂𝒙𝒊𝒎𝒖𝒎 = ½𝑴𝒆𝑽𝟐
potential difference of 1 volts.
𝟐
𝑤𝑜𝑟𝑘 𝑑𝑜𝑛𝑒 = 𝑄𝑉 = ½ 𝟗. 𝟏𝟏 𝒙 𝟏𝟎−𝟑𝟏 𝒙 (𝟕. 𝟗 𝒙 𝟏𝟎𝟓 )
𝑾𝒐𝒓𝒌 𝒅𝒐𝒏𝒆 = 𝒆𝑽 For an electron = 𝟐. 𝟖𝟓 𝒙 𝟏𝟎−𝟏𝟗 𝑱
= 𝟏. 𝟔 × 𝟏𝟎−𝟏𝟗 𝑪 × 𝟏𝑽
= 1.6 × 10−19 𝐽
(𝒄) 𝒉𝒇 = 𝑾𝒐 + 𝑲. 𝑬𝒎𝒂𝒙𝒊𝒎𝒖𝒎 Energy radiation/frequency of radiation used.
𝒉𝒇 = 𝟑. 𝟓𝟔 𝒙 𝟏𝟎−𝟏𝟗 + Frequency of the radiation and the energy of

𝟐. 𝟖𝟒 𝒙 𝟏𝟎−𝟏𝟗 the photoelectrons can be examined using the
following circuit and the frequency
hf = 𝟔. 𝟒 𝒙 𝟏𝟎−𝟏𝟗 (wavelength) is varied using different colour
filters placed in the path of the source of white
6.4 𝑥 10−19
𝒇= light. For each colour, J is moved until no
6.63 𝑥10−34 current is registered.
𝑓 = 9.65 𝑥 1014 𝐻𝑧
2. Sodium has work function of2.3𝑒𝑉.

Calculate:
(i) Its threshold frequency.
(ii) The maximum velocity of the
photoelectron produced when
its surface is illuminated by
Note that, the battery is connected in such a
light of wavelength5.0 ×
way that it opposes the ejection of the
10−7𝑚.determine the
photoelectrons by attracting them back to the
stopping potential of this
cathode. The voltmeter records the stopping
energy.
potential for a given frequency.
3. When light of wavelength 1.0𝝁𝒎
is irradiated onto a metal, it ejects From Einstein’s photoelectric equation,
an electron with a velocity of
3.0 × 105 𝑚/𝑠.calcualate the: ℎ𝑓 = 𝑊𝑂 + ½𝑀𝑒 𝑣 2
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(i) Work function of the metal. ⇒ ℎ𝑓 = 𝑊𝑂 + 𝑒𝑣𝑠

(ii)Threshold frequency of the
metal If frequency is increase but work function held
4. The minimum frequency of light constant, then the stopping potential increases.
which will cause photoelectric The table below shows some colours and their
emission from a metal surface is frequencies and stopping potentials.
5.0 × 1014 𝐻𝑧. if the surface is
illuminated by light of frequency Colour Frequency f ( (Stopping
6.5 × 1014 𝐻𝑧,calculate: x1014 Hz) potential Vs)
(i) The work function of the
Violet 7.5 1.2
metal surface.
(ii) The maximum K.E. (in e.v) of Blue 6.7 0.88
the electron emitted.
(iii) The maximum speed of the Green 6.0 0.60
electrons
Yellow 5.2 0.28
Factors Affecting Photoelectric Effect
Orange 4.8 0.12
Note: Three factors determine the
emissions of electrons on metal surfaces For further learning, see the attached leaflet
by incident radiation are: taken from KLB PG 158-59.
(i) Intensity of the radiation. Type of metal/work function of the metal
(ii) Energy of the radiation
Each metal has its own threshold frequency
(iii) Type of metal.
below which NO photoemission takes place,
Intensity of radiation used. no matter how intense the radiation is. At
constant incident energy, if the work function
This is the rate of energy flow per unit of the metal is high, then the kinetic energy of
area when the radiation is perpendicular the emitted electrons is low.
to the area. i.e.
work (W ) W
Intensity = →I=
area( A)  time(t ) At
W P
but =P thus Intensity, I =
t A
When the intensity of the radiation is

increased and the distance between the
source and the surface is decreased, the
number of photoelectrons emitted
increases. Therefore, the number of
Page 81 of 108

photoelectrons produced is directly

proportional to the intensity of the
radiation.
Examples
1. In an experiment to find the relationship between frequency of radiation and
the kinetic energy of photo electrons in a photo electric device, the following
results were obtained.
Frequency ( f  1014 H 3 ) 7.4 6.8 6.1 5.3 4.7
Stopping potential (Vs) 1.7 1.6 1.26 0.8 0.74
On a graph paper plot a graph of stopping potential (Vs) against frequency (Hz)
From the graph find;
(i) The threshold frequency.
(ii) The planks constant (h)
(iii) The work function of the metal in Joules
2. (a) Define threshold wavelength
(b) The table below shows the sopping voltage, Vs, for a metal surface when
illuminated with light of different wavelength,  of constant intensity.
 (x10-7m) 3.00 3.33 3.75 4.29 5.00
Vs (V) 2.04 1.60 1.20 0.78 0.36
(i) Plot a suitable graph of K.E against frequency.

(ii) From the graph determine
(a) Planck’s constant
(b) Threshold frequency
(c) Work function for the metal surface
3. Interpret the following graphs;
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A graph of incident frequency against the kinetic energy of the photoelectrons

1
A graph of kinetic energy of emitted electrons.
𝑤𝑎𝑣𝑒𝑙𝑒𝑛𝑔𝑡ℎ
Page 83 of 108

(b)Photovoltaic cell
Application of Photoelectric Effect.
(a)Photo- emissive cell. It produces current as a result o
photoelectric effect. It consists of a
It has the cathode and the anode. copper oxide and copper bar
When light falls on the cathode,
photoelectrons are emitted and attracted
by the anode causing a current to flow in
a given circuit
The cells are used in:

(i) Counting vehicles or items on a
conveyor belt in factories
When light strikes the copper oxide
surfaces, electrons are knocked off.
Copper oxide becomes negatively
charged and copper positively charged.
This allows current to flow.
(c)Photoconductive cell or light-
dependent resistor (LDR).
It is made of a semiconductor material
called cadmium sulphide.
(ii) Burglar alarms

(iii) Opening doors.
Photo emissive cell can also be used to
reproduce sound from film.
In exciter lamp focuses light through
sound track along the side of moving film Light energy reduces the resistance of the
onto a photocell. cell from 10 MΩto1kΩ in bright light.
Photon lets the electrons free increasing
Light passing to the cell. The cell creates conduction. They are used in fire alarms
varying current in line with current and exposure meters of cameras.
obtained from the microphone when the
film was made. Varying p.d across the Other photo electric devices are the solar
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resistor is amplified and converted to cell and the photodiode.

sound.
Chapter Ten RADIOACTIVITY

By the end of this topic the learner should Content
be able to:
✓ Radioactive decay
a) Define radioactive decay and half ✓ Half-life.
life ✓ Types of radiations properties
b) Describe the three types of of radiations.
radiations emitted in natural ✓ Detectors of radiation.
radioactivity ✓ Nuclear fission, nuclear
c) Explain the detection of radioactive fusion.
emission ✓ Nuclear equations.
d) Define nuclear fission and fusion ✓ Hazards of radioactivity,
e) Write balanced nuclear equation precautions.
f) Explain the dangers of radioactive ✓ Applications.
emission ✓ Problems on half –life
g) State the application of radioactivity (integration not required)
h) Solve numerical problems involving
half life
Radioactivity is the disintegration of an Radioactive decay
unstable nucleus with emission of radiation
in order to attain stability. ✓ Process by which a radioactive nuclide
undergoes disintegration to emit a
Structure of the atom radiation. The emitted radiations can be
alpha particles, beta particles and this is
✓ Consists of a tiny nucleus and energy accompanied by release of energy in
levels(shells).The nucleus is very small form of gamma radiations.
in size, as compared to the overall size Types of Radioactive Decay
of the atom. The nucleus contains
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protons and neutrons. The number of Alpha Decay –is represented by 𝟒𝟐𝑯𝒆 and
electrons in the shells is equal to the denoted by ⍺
number of protons in the nucleus
making the atom electrically neutral. ✓ If the nuclide decays by release of an
The atomic number alpha particle, the mass number
✓ The number of protons in the nucleus of decreases by 4 and the atomic number
an atom. decreases by 2. This is expressed as;
Mass (nucleon) number 𝐴 𝐴−4 4
𝑍𝑋 → 𝑍−2𝑌 + 2𝐻𝑒
✓ The sum of protons and neutrons in the
nucleus of an atom. (Parent (daughter (alpha
Isotopes
✓ Atoms of the same element that have Nuclide) nuclide) particle)
the same atomic number but different Uranium, for example, decays by emitting
mass numbers. an alpha to become thorium. The decay is
Nuclide expressed as;
A group of atoms that have the same
238 234 4
atomic numbers and the same mass 92𝑈 → 90𝑇ℎ + 2𝐻𝑒
numbers.
Nuclear Stability Similarly, polonium undergoes alpha
decay to become lead.
✓ Stable nuclides have a proton to neutron
ratio of about 1:1. However, as atoms 210 206 4
84𝑃𝑜 → 82𝑃𝑏 + 2𝐻𝑒
get heavier, there is a marked deviation
𝟎
from this ratio, with the number of Beta Decay-represented by −𝟏𝒆 and
neutrons far superseding that of protons. denoted by 𝛽
In such circumstances, the nucleus is
likely to be unstable. When this ✓ If the nuclide decays by release of a (𝛽-
happens, the nucleus is likely to particle, the mass number remains the
disintegrate in an attempt to achieve same but the atomic number increases
stability. by 1. This is expressed as;
𝐴 𝐴 0
𝑍𝑋 → 𝑍+1𝑌 + −1𝑒
(Parent (daughter (beta

nuclide) nuclide) particle)
✓ Radioactive sodium, for example
undergoes beta decay to become
magnesium. This is written as;
24 24 0
11𝑁𝑎 → 12𝑀𝑔 + −1𝑒
Gamma Radiation-is denoted by 𝛾 Example 3
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✓ Some nuclides might be in an excited Uranium – 238( 238

92𝑈 ) undergoes
state and to achieve stability, they may 206
decay to become lead-206 ( 82𝑃𝑏 ). Find
emit energy in form of gamma the number of ∝ and 𝛽-particles emitted
radiation, without producing new in the process.
isotopes. For example:
Solution
(i) Cobalt-60;
Let the number of ∝and 𝛽-particles
60 60
27𝐶𝑜 → 27𝐶𝑜 + 𝛾 emitted be x and y respectively.
238 206
92𝑈 → 82𝑃𝑏 + X ( 42𝐻𝑒 ) + y (−10𝑒 )
(ii) Thorium-230; 238 = 206 + 4x
230 230
90𝑇ℎ → 90𝑇ℎ + 𝛾 4x = 32
Example 1 x=8
Thorium- 230 [ 230
90𝑇ℎ ] undergoes decay Also;
to become Radon-222 222 86𝑅𝑛 Find the
number of alpha particles emitted. 92 = 82 + 2x - y
Solution 92 = 82+ 1 6 -y
Let the number of alpha particles emitted 92 = 98 - y

be x. The expression for the decay is; y=6
230 222
90𝑇ℎ → 86𝑅𝑛 + X (42𝐻𝑒 ) Eight ∝-particles and six 𝛽-particles are
Thus; emitted.
4x + 222 = 230 2x + 86 = 90 Example 4
4x = 8 or 2x = 4 Uranium 234 ( 23492𝑈 ) decays to polonium

218
( 84𝑃𝑜 ) by emitting alpha particles.
x=2 x=2 Write down the nuclear equation
representing the decay.
Two alpha particles are emitted.
Solution
Example 2
Let the number of alpha particles
Lead-214 ( 214
82𝑃𝑏 ) decays to polonium- (helium) be x.
214 ( 214
84𝑃𝑜 ) by emitting 𝜷-particles.
Calculate the number of 𝜷-particles 234 218 4
92𝑈 → 84𝑃𝑜 + x( 2𝐻𝑒 )
Page 87 of 108

emitted. 234 = 218 + 4x

Solution 16 = 4x
Let x be the number of 𝜷-particles x = 4
emitted.
The decay equation is, therefore;
214 214 0
82𝑃𝑏 → 84𝑃𝑜 + x( −1𝑒 ) 234 218 4
92𝑈 → 84𝑃𝑜 + 4( 2𝐻𝑒 )
82 = 8 4 - x
x =2
Two 𝜷-particles are emitted.
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Properties of emitted radiations

Alpha particles
(i) Are positively charged hence
deflected by electric and magnetic
fields. (See diagram).
(ii) They have low penetrating power but
high ionizing effect because they are
heavy and slow.
(iii) They lose energy rapidly and so
have very short range.
(iv) Can be stopped by a thin sheet of
paper.
(v) They affect photographic plates
Beta particles
(i) Have no mass and are represented by
0
−1𝑒 .
(ii) Are negatively charged hence
deflected by both electric and
magnetic fields. (see diagram).
(iii) Have more penetrating power than
alpha particles but lower ionizing
effect.
(iv) Penetrate a sheet of paper but
stopped by aluminium foil.
Gamma rays
(i) High energetic electromagnetic
radiation.
(ii) Have no mass and no charge hence
cannot be deflected by electric and
magnetic fields.
(iii) Have very high penetrating power
and very low ionizing power.
(iv) Can penetrate through a sheet of
paper and aluminium sheets but Note: The main difference between X-
rays and gamma rays is that gamma rays
Page 89 of 108

stopped by a thick block of lead. originate from energy changes in the

nucleus of atoms while X-rays originate
Summary from energy changes associated with
electron structure of atoms.
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Radiation Detectors Diffusion Cloud Chamber

Methods Of Detecting Nuclear
Radiations
The methods of detection rely on the
ionizing property.
1. Photographic Emulsions
All the three radiations affect
photographic emulsion or plate.
Photographic films are very useful to
workers who handle radioactive Functions of the components of
materials. These workers are given diffusion chamber
special badges which contain a small Dry ice: cools the blackened surface
piece of unexposed photographic film. If, making the air at the lower surface of the
during the time it had been worn, the chamber cool.
worker was exposed to radiations, it
should darken on development, implying Sponge: it ensures that the dry ice is in
that further safety precautions should be contact with the blackened surface
taken.
Wedges: it keeps the chamber in a
2. Cloud Chamber horizontal position.
When air is cooled until the vapour it Light source: illuminates the tracks
contains reaches saturation, it is possible making them clearly visible.
to cool it further without condensation
occurring. Under these conditions, the Blackened surface: provides better
vapour is said to be supersaturated. This visibility of the tracks formed.
can only occur if the air is free of any Principle of operation
dust, which normally acts as nuclei on
which the vapour can condense to form The alcohol from the felt ring vaporizes
droplets. Gaseous ions can also act as and diffuses towards the black surface.
The
nuclei for condensation. The ionization of radioactive substance emits
radiations which ionizes the air. The
air molecules by radiations is investigated
by a cloud chamber, vaporized alcohol condenses on the ions
forming tracks. The tracks are well
The common types of cloud chambers are defined if an electric field is created by
expansion cloud chamber and diffusion frequently rubbing the Perspex lid of the
Page 91 of 108

cloud chamber. In both types, saturated chamber with a piece of cloth. The tracks
vapour (water or alcohol) is made to obtained in the above cloud chambers
condense on air ions caused by vary according to the type of radiation.
radiations. Whitish lines of tiny liquid Alcohol is preferred because it is
drops show up as tracks when highly volatile and hence evaporates
illuminated. easly.
Expansion Cloud Chamber
When a radioactive element emits

radiations into the chamber, the air inside
is ionized.
If the piston is now moved down
suddenly, air in the chamber will expand
and cooling occurs.
When this happens, the ions formed act
as nuclei on which the saturated alcohol
or water vapour condenses, forming
tracks. The shape and appearance of the
track will which type of the particles have
been emitted.
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The tracks due to alpha particles are

short, straight and thick. This is
because:
(i) Alpha particles cause heavy
ionization, rapidly losing energy, hence
their short range.
(ii) They are massive and their path
cannot therefore be changed by air
molecules.
(iii) Alpha particles cause more ions on
their paths as they knock off more
electrons, see
The tracks formed by beta particles are

generally thin and irregular in direction.
This is because beta particles, being
lighter and faster, cause less ionization
of air molecules. In addition, the
particles are repelled by electrons of
atoms within their path.
Gamma rays produce scanty disjointed

tracks,
Page 93 of 108

Geiger-Muller Tube
The Geiger-Muller (G-M) tube is a type of ionization chamber.
The thin mica window allows passage of Background Radiation

radiations these radiations ionizes the argon
gas inside the tube. The electrons are ✓ Radiations that are registered or
attracted to the anode as the positive ions observed in the absence of a
moves towards the cathode. More ions are radioactive source.The count
produced as collisions continue. Small registered in the absence of the
currents are produced which are amplified radioactive source is called
and passed to the scaler connected to the background count.
tube. The presence of small amount of ✓ sources of these backgrounds
halogen in the tube is to help absorbing the radiation include:
kinetic energy of the positive ions to reduce (i) Cosmic rays from outer space.
further ionisation and enhance quick return
to normal. This is called quenching the tube (ii) Radiations from the sun.
i.e. Bromine gas acts as a quenching
(iii) Some rocks which contain traces
agent.
of radioactive material, e.g., granite,
The gold leaf electroscope
(iv) Natural and artificial
A charged electroscope loses its charge in radioisotopes.
the presence of a radioactive source. The
radioactive source ionizes the air around the
electroscope. Ions on the opposite charge to
that of the electroscope are attracted to the
cap and eventually neutralize the charge of
the electroscope. As a detector a charged
electroscope is not suitable for detecting beta
and gamma radiations because their ionizing
effect in air is not sufficiently intense so the
leaf may not fall noticeably.
In experiments, the average background count rate should be recorded before
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and after the experiment such that:

𝒄𝒐𝒓𝒓𝒆𝒄𝒕 𝒄𝒐𝒖𝒏𝒕 𝒓𝒂𝒕𝒆 = 𝒄𝒐𝒖𝒏𝒕 𝒓𝒂𝒕𝒆 − 𝒃𝒂𝒄𝒌𝒈𝒓𝒐𝒖𝒏𝒅 𝒓𝒂𝒅𝒊𝒂𝒕𝒊𝒐𝒏𝒔 𝒓𝒆𝒈𝒊𝒔𝒕𝒆𝒓𝒆𝒅
Artificial Radioactivity Half-life
Some naturally occurring nuclides can be ✓ The time taken for half the
made artificially radioactive by bombarding numbers of nuclides initially
them with neutrons, protons or alpha present in a radioactive sample to
particles. decay.
✓ Half-life of a radioactive substance
For example, when nitrogen--14( 147𝑈 ) can be determined using the
nuclide, which is stable, is bombarded with following methods:
fast moving alpha particles, radioactive
oxygen is formed. This is represented by; Decay series
4 14 17
2𝐻𝑒 + 7𝑈 → 8𝑂 + 11𝐻 Decay formula
𝑻
Other artificially radioactive nuclides are 𝟏 𝒕
silicon-27 ( 27 35
14 Si), sulphur-35 ( 16 𝑆) and 𝑵 = 𝑵𝟎 ( )
𝟐
chlorine-36( 36
17 𝐶𝑙 )
𝒘𝒉𝒆𝒓𝒆,
Decay Law
𝑵 = 𝒐𝒓𝒊𝒈𝒊𝒏𝒂𝒍 𝒄𝒐𝒖𝒏𝒕 𝒓𝒂𝒕𝒆
States that the rate of disintegration at a
given time is directly proportional to the 𝑵𝒐 = 𝑹𝒆𝒎𝒂𝒊𝒏𝒊𝒏𝒈 𝒄𝒐𝒖𝒏𝒕 𝒓𝒂𝒕𝒆
number of nuclides present at that time. This
can be expressed as; 𝑻 = 𝒕𝒐𝒕𝒂𝒍 𝒕𝒊𝒎𝒆 𝒐𝒇 𝒅𝒆𝒄𝒂𝒚
𝑑𝑁 𝒕 = 𝒉𝒂𝒍𝒇 − 𝒍𝒊𝒇𝒆
∝- N, where N is the number of nuclides
𝑑𝑡
present at a given time. It follows that;
𝑑𝑁
= -𝜆N, where 𝜆 is a constant known as the
𝑑𝑡
decay constant.
The negative sign shows that the number N
decreases as time increases.
𝑑𝑁
is referred to as the activity of the sample.
𝑑𝑡
Page 95 of 108

(iii) Detecting Pipe Bursts

From a graph Underground pipes carrying water or oil
many suffer bursts or leakages. If the
water or oil is mixed with radioactive
substances from the source, the mixture
will seep out where there is an opening. If
a detector is passed on the ground near
the area, the radiations will be detected.
(iv) Determining Thickness of Metal
Foil
In industries which manufacture thin
metal foils, paper and plastics, radioactive
radiations can be used to determine and
Applications of Radioactivity maintain the required thickness. If a beta
source, for example, is placed on one side
(i) Carbon Dating of the foil and G -M tube on the other, the
Living organisms take in small quantities count rate will be a measure of the
of radioactive carbon-14, in addition to the thickness of the metal foil.
ordinary Carbon-12. The ratio of carbon-12
to carbon-14 in the organisms remains
fairly constant. The count-rate can give this
value.
When the organisms die, there is no more
intakes of carbon and therefore the ratio
changes due to the decay of carbon-14. The A thickness gauge can be adapted for
count-rate of carbon-14 therefore declines automatic control of the manufacturing
with time. The new ratio of carbon-12 to process.
carbon-14 is then used to determine the age
(v) Trace Elements
for the fossil.
The movement of traces of a weak
radioisotope introduced into an organism
(ii) Medicine can be monitored using a radiation
detector. In agriculture, this method is
Gamma rays, like X-rays, are used in the applied to study the plant uptake of
control of cancerous body growths. The fertilizers and other chemicals.
radiation kills cancer cells when the
tumour is subjected to it. Gamma rays are
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also used in the sterilization of medical (vi) Detection of Flaws

equipment, and for killing pests or making
them sterile. Cracks and airspaces in welded joints can
be detected using gamma radiation from
cobalt-60. The cobalt-60 is placed on one
side of the joint and a photographic film
on the other. The film, when developed,
will show any weakness in the joint.
Nuclear Fusion
Hazards of Radiation Experiments show that a lot of energy is
released when the nuclei of light
When the human body is exposed to elements fuse together to form a heavier
radiation, the effect of the radiation nucleus. The fusing together of nuclei to
depends on its nature, the dose received form a heavier nucleus is called nuclear
and the part irradiated. Gamma rays fusion. An example of nuclear fusion is
present the main radiation hazard. This is the formation of alpha particles when
because they penetrate deeply into the lithium fuses with hydrogen;
body, causing damage to body cells and
tissues. This may lead to skin burns and Nuclear Fission
blisters, sores and delayed effects such as
cancer, leukaemia and hereditary defects. It was discovered that if a nucleus of
Extremely heavy doses of radiation may uranium is bombarded with a neutron, the
lead to death. uranium nucleus splits into two almost
equal nuclei. When a nucleus is
bombarded and it splits, it is said to have
undergone nuclear fission as shown
Page 97 of 108

Precautions below.
(i) Radioactive elements should never 235
92𝑈 + 10𝑛 → 138 95 1
56𝐵𝑎 + 36𝐾𝑟 + 3( 0𝑛 )
be held with bare hands. + 𝑒𝑛𝑒𝑟𝑔𝑦
(ii) Forceps or well protected tongs
should be used when handling Protons and neutrons (nucleons) are kept
them. together in the small volume of the
(iii) For the safety of the users, nucleus by what called binding energy.
radioactive materials should be To split the nucleus, this binding energy
kept in thick lead boxes. has to be released. The energy released
(iv) In hospitals and research during the splitting is called nuclear
laboratories, radiation absorbers energy.
are used. The emitted neutrons may encounter
other uranium nuclides, resulting in more
splitting with further release of energy.
The produced neutrons are called fission
neutrons.
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Chapter Eleven ELECTRONICS
By the end of this topic, the learner Content

should be able
✓ Conductors, semiconductors
a) State the difference between insulators
conductor and insulators ✓ Intrinsic and extrinsic semi-
b) Define intrinsic and extrinsic conductor
conductors ✓ Doping
c) Explain doping in semi- conductor ✓ P-n junction diode
d) Explain the working of a pin ✓ Application of diodes: half wave
junction diode rectification and full wave
e) Sketch current voltage rectification
characteristics for a diode
f) Explain the application of diode
rectification
Introduction ✓ Conduction band – free electrons.
✓ Valence band – unfilled, few
Definition – study of motion of free electrons.
electrons in electrical circuits. ✓ Forbidden band/energy gap – no
Applications – pocket calculators, clocks, forbidden band, conduction and
musical instruments, radios, TVs, valence band overlap.
computers, robots etc. ✓ Resistance increases with rise in
temperature. A rise in temperature
Classes of Material increases the vibrations of the
atoms and this interferes with the
✓ Conductors – has free electrons – not
electron flow. Hence the resistance
tightly bound to the nucleus of the
of a conductor increases with
atom copper, aluminium.
temperature.
✓ Insulators-have immobile (fixed)
electrons Insulators:-
✓ Semi-conductors – with conducting
properties between conductors and
insulators silicon, germanium.
The Energy Band Theory.
✓ When two or more atoms are brought
closer to each other, the energy levels
Page 99 of 108

split into smaller energy levels called

bands. This is due to the interaction of
both electric and magnetic fields of
electrons
Types of bands
✓ Conduction band – electrons are free
to move under the influence of an ✓ Conduction band – has no
electric current. electrons, empty.
✓ Valence band – here electrons are not ✓ Valence band –filled with
free to move. electrons.
✓ Forbidden band/energy band – ✓ Forbidden band – has very wide
represents the energy level that cannot gap
be represented by electrons. The width ✓ Temperature increase has no effect
of the band determines the on their conductivity.
conductivity of the material.
Semi – conductors:-
Conductors, insulators and semi-
conductors in terms of energy band
theory
Conductors:-
✓ Conduction band – empty at O.K.

Partially filled at room temperature.
✓ Valence band –filled at O.K; full of
electrons at very low temperatures
✓ ✓ Forbidden band – have very narrow
gap.
✓ Resistance reduces with rise in
temperature.
✓ Increase in temperature increases the Extrinsic Semi-Conductors

chance of electrons moving from the
valence band to conduction band. Made by adding a controlled amount of
Electrical resistance therefore reduces different element to an intrinsic semi-
because the total current flow is due to conductor.
the flow of electrons and holes. Have Two types of extrinsic semi-conductors:-
negative temperature coefficient of
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resistance. • N – Type semi-conductor – formed by

doping a group 4 element with a
Note: semi – conductors
Group 5 element.
✓ At room temperature: - Has holes • P – Type semi-conductor – formed by
in the valence band & free doping a group 4 element with a group
electrons in the conduction band. 3 element.
At OK it behaves like an insulator. • Group 4 elements – Tetravalent –
✓ HOLES: Holes are created when Silicon, germanium, etc
an electron moves from valence • Group 5 elements – Pentavalent –
band to conduction band. doping element, donor impurity –
✓ Holes are very important for phosphorous, antimony.
conduction of electric current in • Group 3 elements – Trivalent – boron,
semi-conductors. aluminium and indium
Types of Semi-Conductors N-Type Semi –Conductor
Intrinsic semi-conductors • Formed by adding a Pentavalent
atom (Phosphorus) to a group 4
• They are pure semi-conductors,
semi-conductor (Silicon) and an
electrical properties of a pure
extra electron is left unpaired and is
substance.
available for conduction.
• Has equal number of electrons and
• Majority charge carriers are
holes.
electrons; minority charge carriers
• Conductivity is very low, insulator at are positive holes.
low temperatures.
• Phosphorous is called a DONOR
• Usually not used in a pure state e.g. ATOM. Silicon has now more
silicon, Germanium electrons
P-Type Semi –Conductor

Extrinsic semi- conductors
• Formed by adding a trivalent atom
Page 101 of 108

• With added impurities to improve its (Boron) to a group 4 atom (Silicon), a

electrical properties . fourth electron will be unpaired and a
• All semi-conductors in practical use gap will be left called a positive hole.
has added impurities • Pure semi-conductor is doped with
impurity of group 3 element;
Doping: - A process of adding a very combination creates a positive hole
small quantity of impurities to a pure which accepts an electron.
semi-conductor to obtain a desired • The doping material creates a positive
property. hole, which can accept an electron –
• Process of introducing an impurity called an Acceptor.
atom into the lattice of a pure semi-
conductor.
Biasing
i) Forward Bias
• A diode is forward biased when the
cathode is connected to n-side and
anode to the p-side in a circuit.
• In forward bias, the depletion layer is
narrowed and resistance is reduced.
• It allows holes to flow to n-side and
P-N Junction Diodes (Junction Diodes) electrons to p-side.
• The majority charge carriers cross the
Definition junction. It conducts current and the
bulb lights
• An electronic device with two electrodes,
which allows current to flow in one
direction only.
• It is an electrical one way valve. It is a
solid device.
Formation of P-N Junction Diode
Reverse Bias
• It consists of such a p-n junction with the
p-side connected to the Anode and the n- • A diode is reverse biased when
side to the cathode. the cathode is connected to p-side
• Formed by doping a crystal of pure and anode to the n-side in a
silicon so that a junction is formed circuit.
between the p-type and n-type regions. • The current through the diode is
virtually zero. It hardly conducts,
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Circuit symbol the bulb does not light. Electrons

and holes are pulled away from
the depletion layer, making it
wider.
• The electrons and holes are
Depletion Layer attracted to opposite ends of the
diode away from the junction. The
• The region between the p-type and n-type wider the depletion layer, the
semi-conductor which conducts very higher the resistance of the
poorly. junction.
• At the junction electrons diffuse from
both sides and neutralize each other.
Junction
• The place (boundary) between two
different types of semi-conductors.
• A narrow depletion layer if formed on Characteristic Curves for P-N Junction
either side of the junction free from Diodes
charge carriers & high resistance.
Forward biasing
Diagram of unbiased Junction Diode
The circuit below shows how the
connections are made.
The characteristic graph of current, I

against reverse bias voltage is obtained as
shown below. The curve is non-ohmic.it is
non-linear. The current increases
exponentially with voltage up to a point
Application of zener Diodes
where a sharp increase in current is noticed.
This voltage is called threshold/cut- • Used in industry as voltage regulators
in/breakpoint voltage. At this voltage or stabilizers, by providing a constant
potential the barrier is overcome by bias
Page 103 of 108

and charges easily flow across the junction. voltage to a load.

• Voltage remains constant as current
increases.
Application of P-N Junction Diodes
a) To protect equipment, circuits or
devices by a reverse power supply.
b) To rectify ac to dc
c) Enable the Audio Frequency energy
carrier by modulated radio waves to be
detected.
Reverse Biasing
Rectification and Smoothing
In reverse biasing, resistance is very high,
however, the flow of leakage current results A) Definition
from flow of minority charge carriers. At
breakdown voltage or Zener break down • Rectification is the process of
covalent bonds rapture liberating electrons. converting a.c current to d.c current.
Those electrons collide with some atoms • A Rectifier is a device that changes
causing ionisation this is called avalanche a.c to d.c.
breakdown. The two processes produce b) Reasons for rectification
excess electrons for heavy conduction.
Beyond breakdown voltage a diode is • The conversion of a.c. to d.c. is
damaged. often necessary for all electric
equipment, such as radios, T.V.
The Zener Diode sets, computers, musical
Definition instruments, e.t.c, use steady d.c.
• A zener Diode is a silicon p-n semi- Types of rectification

conductor, which is designed to work There are two types of rectification,
in reverse biased connection. namely:-
Principle of operation • Half-wave rectification
• When the reverse-bias of the diode is • Full-wave rectification.
increased, a large sudden increase in
current is obtained at one particular
reverse voltage.
• At the reverse voltage, the p-n
junction diode breaks down into a
conductor, by breaking down the
barrier layer.
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• The breakdown of the p-n junction

diode is known as zener breakdown
or zener effect.
• The characteristic is almost a vertical
line, i.e. the zener current, which
occur as a result of the zener voltage.
Half-wave rectification and smoothing

• One diode is used which removes the negative half-wave cycle of the applied a.c.
• It gives a varying but one-way direct current across the load R. R is a piece of
electronic equipment requiring a d.c. supply.
• If the Y-input terminals of a CRO are connected first across the input, the waveform
on the left will be displayed on the screen.
• When a CRO is connected across R, the output waveform is seen to be positive half-
wave of the a.c.
• Smoothing is done using a capacitor connected across R, to give a much steadier
varying d.c. supply.
• The smoothing capacitor provides extra charge so that current flows continuously
even as the phase current changes and the current go to zero.
Page 105 of 108

• The larger the capacitor, the better the smoothing.

• On the positive half-cycle of the a.c. input the diode conducts, current passes
through R and also into the capacitor C to charge it up.
• On the negative half-cycle, the diode is reversing biased and cannot conduct, but C
partly discharges through R.
• The charge-storing action of the capacitor, C thus maintains current in R and a
steadier p.d across it when the diode is not conducting.
NOTE: - A single diode only allow half of the a.c. to flow through the load R, so far
half of the power supply is cut off.
Full- wave Rectification and smoothing
There are two methods for obtaining a full-wave rectification namely:-
• Using two diodes – Full-wave centre-tap transformer.
• Using four diodes – Full-wave bridge rectifier
Using Centre-Tap Transformer
• In a full-wave rectifier, both halves of the a.c. cycles are transmitted but in the
direction, i.e. same side.
OR
• During the first half-cycle, when A is positive, DI conducts through the load R at
WhatsApp 0706851439
the same time B is negative with respect to T, so no current flows in the diode
D 2.
• In the next half-cycle when B is positive, D2 conducts through the load R in the
same direction as before. A is positive with respect to T so no current flows in
D 1.
Using the bridge Rectifier – four diodes

• In the 1st half-cycle, diode D2 and D4 conducts.
• In the 2nd half-cycle, diode D3 and D1 conducts.
• During both cycles, current passes through R in the same direction, giving a p.d. that
varies as shown by the CRO.
• When a large capacitor is connected across R, the output d.c. is smoothed as shown.
• During the first half cycle, point A is positive with respect to C, diode D1 and D3
are forward biased while diode D2 and D4 are reverse biased. Current therefore
flows through ABDCA. During the second half-cycle, point A becomes negative
with respect to point C. diodes D2 and D4 become forward biased while D1 and
D3 are reverse biased. Conventional current therefore flows through CBDAC.
• If a capacitor is connected across the resistor, the rectified output is smoothened.
Advantages of bridge rectifier
✓ A smaller transformer can be used because there is no need for centre –tapping.
✓ It is used for high voltage regulation.
Page 107 of 108

QUESTIONS
1. Draw the structure of a crystal lattice 5. What is meant by the following
to show the arrangement of electrons terms: semiconductor, intrinsic
in following: conduction, extrinsic conduction,
✓ Pure silicon. doping, donor atoms, acceptor
✓ P-type semiconductors atoms, n-type semiconductor, p-
✓ N-type semiconductors type, semiconductor, depletion
2. Explain how temperatures rise layer, forward bias, hole, reverse
affects the electrical conductivity or bias and Zener effect?
pure semiconductors. 6. Explain how doping produces a p-
(a)Draw the symbol of a p-n diagram type and an n-type semiconductor.
junction diode. 7. Distinguish between electronics
(b)Use a circuit diagram to and electricity.
distinguish between forward and 8. a) What is rectification?
reserve bias of p-n junction diode. (b) With diagrams, describe how half-
3. (a)Use a labelled diagram to explain wave and full-wave rectification
how a full valve rectification may be can be achieved.
achieved by using a resistor and :(i)
Two diodes. (ii) Four diodes. 9. Explain why a diode conducts
4. With the aid of a diagram explain easily on forward bias and not in
how a capacity can be used to reverse bias.
smoothen a full wave which has been
rectified. Show using a sketch how
the smoothened wave will appear on
the screen of C.R.O.
END OF SECONDARY SCHOOL SYLLABUS.

ON BEHALF OF AMOBI SOFT COPY PUBLISHERS, WE WISH ALL
THE BEST IN YOUR FINAL EXAMS 0706 851 439
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PHYSCIS FORM 4 HANDBOOK

By Sir Obiero Amos 0706 851 439

BRIEF PERSONAL PROFILE

Chapter 1 THIN LENSES

Chapter 2 UNIFORM CIRCULAR MOTION

Chapter 3 SINKING AND FLOATING

Chapter 4 ELECTROMAGNETIC INDUCTION

Chapter 5 MAINS ELECTRICITY

Chapter 6 ELECTROMAGNETIC SPECTRUM

Chapter 7 CATHODE RAYS AND CATHODE

Chapter 9 PHOTOELECTRIC EFFECT

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Chapter One THIN LENSES

Objectives Effect of lenses on parallel rays of light.

namely: principal focus.

1.3 Image Formation

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There are three major rays that are used in

(ii) A ray of light passing (or appearing to

(iii) A ray of light through the optical

Object beyond 2F Object between F and lens

The image is The image is

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(iv) formed beyond 2F on the 𝑉 h

Object at F The lens formula

The image is at infinity

PO is the object distance, u, PI is the

𝑢 𝑓 Multiply both sides by v;

Graphical interprentation of the lens Exercise

In the form 𝑦 = 𝑚𝑥 + 𝑐 Exercise

Experimental Determination Of The 2. Adjust the position of the lens holder

Apparatus Focal length----------------------cm

Metre rule, lens, a lens holder, screen NOTE:

Procedure Under these conditions, rays from any

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2. Place a white screen at one end of the

1. Set the lens in its holder with a plane Focal length=-----------------------

USES OF LENSES IN OPTICAL Defects of vision

1. Sclerotic layer – hard shell that

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the action of Ciliary muscles

Camera Compound microscope

stray light. eyepiece or ocular. The objective lens is of

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glass. When the object is placed between a

UNIFORM CIRCULAR MOTION

Definition of Terms Convert the following into radians:

displacement. It is denoted by Greek letter

𝑎𝑟𝑐 𝑙𝑒𝑛𝑔𝑡ℎ 𝑝𝑒𝑟𝑖𝑜𝑑𝑖𝑐 𝑡𝑖𝑚𝑒

Relationship between angular

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---------- (i) • If the velocity of such object is

1. A turn table rotates at the rate of 60 But 𝑣 = 𝜔𝑟

radius of the path the larger the centripetal

Example Examples of Uniform Circular Motion

If the road is slippery then frictional force

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• Centripetal force provided This critical speed depends-