Physics Notes Form 1 to 4 Physics Form 4, 3, 2, 1 Notes Free Download

To study fluid flow we have to make the following assumptions:

1. We consider fluids to be incompressible

2. We assume that they have little or no internal friction or viscosity.

Streamline and turbulent flow.

The path followed by a small element of a moving fluid is called a line of flow . A streamline is a curve whose tangent at any point is in the direction of the fluid velocity at that point.

A streamline flow occurs when all elements of a fluid passing a particular point follow the same path or line of flow as the elements that passed through that point previously.

A streamline flow is achieved only when the speed is low.

If the speed increases it is characterized by whirls and eddies then it becomes a turbulent flow. Turbulent flow generally occurs when the speed is high and where there are sharp bends along the path of the fluid.

Equation of continuity

Consider a fluid flowing (streamline flow) through a horizontal pipe with different cross –
sectional areas as shown.

Let the cross -sectional area in both sections be A₁ and A₂ and the corresponding speeds of the fluid be V₁ and V₁ respectively.

The volume of fluid flowing per second in each section is given by; V = A L = A v t = A v. Where L, v t and v is the distance moved in one second.

Since the volumes in each section is the same, then
A₁ V₁ = A₂ V₂ , hence A v = constant.

The above equation is known as the equation of continuity. Since A₁ A₂ , then V₂ V₁ . i.e. the speed increases when a tube narrows.

The quantity (A v) is called volume efflux i.e. volume flowing per second.

Example.

A horizontal pipe of cross-sectional area 50 cm² carries water at the rate of 0.20 litres per second. Determine the speed;

a) Of the speed of water in the pipe.

b) When the tube narrows to 20 cm² at another point.

Solution

a) Volume efflux = o.20 l per second = A v
From V (volume) = A v, then v = V / A = 0.20 × 10-3 / 50 × 10-4 = 0.04 m/s b) Since A₁ v₁ = A₂ v₂ then v₁ = (0.05 × 0.04) / 0.02 = 0.1 m/s

Bernoulli’s principle

Daniel Bernoulli (1700 – 1782) explained the variation of pressure exerted by a moving fluid when its speed is changed. The pressure is lower where the speed is higher.

Bernoulli’s principle states that “For a fluid flowing through a tube, the sum of the pressure, the kinetic energy per unit volume and the potential energy per unit volume of the fluid is a constant”. Mathematically expressed as;

P + ½ ρ v₂ + ρ g h = constant. Where P – pressure, ρ – density, v= velocity,
g – acceleration due to gravity and h – height.

Bernoulli’s effect

When air is blown through the tunnel formed, the area marked ‘T’ collapses inwards showing that pressure outside is more than the one inside the tunnel.

The pressure inside the tunnel decreases as the air through it increases in speed.

Applications of Bernoulli’s principle.

1. Car carburetor– inside the carburetor the air passage is partially constricted at the point
where petrol mixes with air hence air intake increases the speed of air while decreasing the pressure inside for petrol to vaporize quickly before it gets to the cylinder where combustion occurs.

2. Horizontal pipe – for a streamline flow through a pipe the term ρ g h is eliminated from the Bernoulli’s equation leaving P + ½ ρ v₂ = constant, indicating that pressure in liquid is greatest when speed is least.

When this is combined with the equation of continuity, the pressure is then greatest when the pipe is widest hence the following observation.

3. Dynamic lift – when air is blown at the top a flat sheet of paper the ends of the paper moves upward and this because the speed of air on top of the paper is greater than below and according to Bernoulli’s principle the pressure on top lowers and the pressure below becomes sufficient enough to produce a force which moves the paper
upwards.

This is what is referred to as the dynamic lift since it is caused by motion.

The upward force is equal to the product of the pressure difference and the area of the surface lifted.

It is applied in the taking off of air-planes, the trajectory of a spinning ball, paint sprayer and Bunsen burner among others.

Physics Notes Form 3

Physics Form Three

Chapter One

Linear Motion

Introduction

Study of motion is divided into two;

1. Kinematics

2. Dynamics

In kinematics forces causing motion are disregarded while dynamics deals with motion of objects and the forces causing them.

I. Displacement

Distance moved by a body in a specified direction is called displacement. It is denoted by letter‘s’ and has both magnitude and direction. Distance is the movement from one point to another. The Si unit for displacement is the metre (m).

II. Speed

This is the distance covered per unit time.

Speed= distance covered/ time taken. Distance is a scalar quantity since it has magnitude only.

The SI unit for speed is metres per second(m/s or ms^-1)

Average speed= total distance covered/total time taken
Other units for speed used are Km/h.

Examples

1. A body covers a distance of 10m in 4 seconds. It rests for 10 seconds and finally covers a distance of 90m in 60 seconds. Calculate the average speed.

Solution

Total distance covered =10+90= 100m

Total time taken =4+10+6= 20 seconds

Therefore average speed = 100/20= 5m/s

2. Calculate the distance in metres covered by a body moving with a uniform speed of 180 km/h in 30 seconds.

Solution

Distance covered=speed*time

=180*1000/60*60=50m/s

=50*30

=1,500m

3. Calculate the time in seconds taken a by body moving with a uniform speed of 360km/h to cover a distance of 3,000 km?

Solution

Speed:360km/h=360*1000/60*60=100m/s

Time=distance/speed

3000*1000/100

=30,000 seconds.

III. Velocity

This is the change of displacement per unit time. It is a vector quantity.

Velocity=change in displacement/total time taken

The SI units for velocity are m/s

Examples

1. A man runs 800m due North in 100 seconds, followed by 400m due South in 80 seconds. Calculate,

a. His average speed

b. His average velocity

c. His change in velocity for the whole journey

Solution

a. Average speed: total distance travelled/total time taken

=800+400/100+80

=1200/180

=6.67m/s

b. Average velocity: total displacement/total time

=800-400/180

=400/180

=2.22 m/s due North

c. Change in velocity=final-initial velocity

= (800/100)-(400-80)

=8-5

=3m/s due North

2. A tennis ball hits a vertical wall at a velocity of 10m/s and bounces off at the same velocity. Determine the change in velocity.

Solution

Initial velocity(u)=-10m/s

Final velocity (v) = 10m/s

Therefore change in velocity= v-u

=10- (-10)

=20m/s

IV. Acceleration

This is the change of velocity per unit time. It is a vector quantity symbolized by ‘a’.
Acceleration ‘a’=change in velocity/time taken= v-u/t
The SI units for acceleration are m/s2
Examples
1. The velocity of a body increases from 72 km/h to 144 km/h in 10 seconds.

Calculate its acceleration.

Solution

Initial velocity= 72 km/h=20m/s

Final velocity= 144 km/h=40m/s

Therefore ‘a’ =v-u/t

= 40-20/10

2m/s2

2. A car is brought to rest from 180km/h in 20 seconds. What is its retardation?

Solution

Initial velocity=180km/h=50m/s

Final velocity= 0 m/s

A = v-u/t=0-50/20

= -2.5 m/s2

Hence retardation is 2.5 m/s2

Motion graphs

Distance-time graphs

a)

b)

Area under velocity-time graph

Consider a body with uniform or constant acceleration for time‘t’ seconds;

This is equivalent to the area under the graph. The area under velocity-time graph gives the distance covered by the body under‘t’ seconds.

Example

A car starts from rest and attains a velocity of 72km/h in 10 seconds.

It travels at this velocity for 5 seconds and then decelerates to stop after another 6 seconds.

Draw a velocity-time graph for this motion. From the graph;

i. Calculate the total distance moved by the car

ii. Find the acceleration of the car at each stage.

Solution

a. From the graph, total distance covered= area of (A+B+C)

=(1/2×10×20)+(1/2×6×20)+(5×20)

=100+60+100

=260m

Also the area of the trapezium gives the same result.

b. Acceleration= gradient of the graph

Stage A gradient= 20-0/ 10-0 = 2 m/s2

Stage b gradient= 20-20/15-10 =0 m/s2

Stage c gradient= 0-20/21-15 =-3.33 m/s2

Using a ticker-timer to measure speed, velocity and acceleration.
It will be noted that the dots pulled at different velocities will be as follows;

Most ticker-timers operate at a frequency of 50Hzi.e. 50 cycles per second hence they make 50 dots per second. Time interval between two consecutive dots is given as,

1/50 seconds= 0.02 seconds. This time is called a tick.

The distance is measured in ten-tick intervals hence time becomes 10×0.02= 0.2 seconds.

Examples

a. A tape is pulled steadily through a ticker-timer of frequency 50 Hz.

Given the outcome below, calculate the velocity with which the tape is pulled.

Solution

Distance between two consecutive dots= 5cm

Frequency of the ticker-timer=50Hz

Time taken between two consecutive dots=1/50=0.02 seconds

Therefore, velocity of tape=5/0.02= 250 cm/s

b. The tape below was produced by a ticker-timer with a frequency of 100Hz. Find the acceleration of the object which was pulling the tape.

Solution

Time between successive dots=1/100=0.01 seconds

Initial velocity (u) 0.5/0.01 50 cm/s

Final velocity (v) 2.5/0.01= 250 cm/s

Time taken= 4 ×0.01 = 0.04 seconds

Therefore, acceleration= v-u/t= 250-50/0.04=5,000 cm/s2

Equations of linear motion

The following equations are applied for uniformly accelerated motion;

v = u + at

s = ut + ½ at2

v2= u2 +2as

Examples

1. A body moving with uniform acceleration of 10 m/s2 covers a distance of 320 m. if its initial velocity was 60 m/s. Calculate its final velocity.

Solution

V2 = u2 +2as

= (60) +2×10×320

=3600+6400

= 10,000

Therefore v= (10,000)1/2

v= 100m/s

2. A body whose initial velocity is 30 m/s moves with a constant retardation of 3m/s. Calculate the time taken for the body to come to rest.

Solution

v = u+at

0= 30-3t

30=3t

t= 30 seconds.

3. A body is uniformly accelerated from rest to a final velocityof 100m/s in 10 seconds. Calculate the distance covered.

Solution

s=ut+ ½ at2

=0×10+ ½ ×10×102

= 1000/2=500m

Motion under gravity.

1. Free fall

The equations used for constant acceleration can be used to become,

v =u+gt

s =ut + ½ gt²

v2= u+2gs

2. Vertical projection

Since the body goes against force of gravity then the following equations hold

v =u-gt ……………1

s =ut- ½ gt2 ……2

v2= u-2gs …………3

N.B time taken to reach maximum height is given by the following

t=u/g since v=0 (using equation 1)

Time of flight

The time taken by the projectile is the timetaken to fall back to its point ofprojection. Using eq. 2 then, displacement =0

0= ut- ½ gt²

0=2ut-gt2

t(2u-gt)=0

Hence, t=0 or t= 2u/g

t=o corresponds to the start of projection

t=2u/gcorresponds to the time of flight

The time of flight is twice the time taken to attain maximum height.

Maximum height reached.

Using equation 3 maximum height, Hmax is attained when v=0 (final velocity).

Hence
v2= u2-2gs;- 0=u2-2gHmax, therefore

2gHmax=u2

Hmax=u2/2g

Velocity to return to point of projection.

At the instance of returning to the original point, total displacement equals to zero.

v2 =u2-2gs hence v2= u2

Thereforev=u or v=±u

Example

A stone is projected vertically upwards with a velocity of 30m/s from the ground.

Calculate,

a. The time it takes to attain maximum height

b. The time of flight

c. The maximum height reached

d. The velocity with which it lands on the ground. (take g=10m/s)

Solution

a. Time taken to attain maximum height

T=u/g=30/10=3 seconds

b. The time of flight

T=2t= 2×3=6 seconds

Or T=2u/g=2×30/10=6 seconds.

c. Maximum height reached

Hmax= u2/2g= 30×30/2×10= 45m

d. Velocity of landing (return)

v2= u2-2gs, but s=0,

Hence v2=u2

Thereforev=(30×30)1/2=30m/s

3. Horizontal projection

The path followed by a body (projectile) is called trajectory.

The maximum horizontal distance covered by the projectile is called range.

The horizontal displacement ‘R’ at a time‘t’ is given by s=ut+1/2at2
Taking u=u and a=0 hence R=ut, is the horizontal displacement and h=1/2gt2 is the vertical displacement.

NOTE

The time of flight is the same as the time of free fall.

Example

A ball is thrown from the top of a cliff 20m high with a horizontal velocity of 10m/s.

Calculate,

a. The time taken by the ball to strike the ground

b. The distance from the foot of the cliff to where the ball strikes the ground.

c. The vertical velocity at the time it strikes the ground. (take g=10m/s)

Solution

a. h= ½ gt2

20= ½ ×10×t2

40=10t2

t2=40/10=4

t=2 seconds

b. R=ut

=10×2

=20m

c. v=u+at=gt

= 2×10=20m/s

Chapter Two

Refraction of Light

Introduction

Refraction is the change of direction of light rays as they pass at an angle from one medium to another of different optical densities.

Exp. To investigate the path of light through rectangular glass block.

Apparatus: – soft-board, white sheet of paper, drawing pins (optical), rectangular glass block.

Procedure

1. Fix the white plain paper on the soft board using pins.

2. Place the glass block on the paper and trace its outline, label it ABCD as shown below.

3. Draw a normal NON at point O.

4. Replace the glass block to its original position.

5. Stick two pins P1 and P2 on the line such that they are at least 6cm apart and upright.

6. Viewing pins P1 and P2 from opposite side, fixpins P3 and P4 such that they’re in a straight line.

7. Remove the pins and the glass block.

8. Draw a line joining P3 and P4 and produce it to meet the outline face AB at point O

Explanation of refraction.

Light travels at a velocity of 3.0×108in a vacuum. Light travels with different velocities in different media. When a ray of light travels from an optically less dense media to more dense media, it is refracted towards the normal.

The glass block experiment gives rise to a very important law known as the law of reversibility which states that “if a ray of light is reversed, it always travels along its original path”.

If the glass block is parallel-sided, the emergent ray will be parallel to the incident ray but displaced laterally as shown

‘e’ is called the angle of emergence. The direction of the light is not altered but displaced sideways. This displacement is called lateral displacement and is denoted by‘d’.

Therefore
XY= t/Cos r YZ= Sin (i-r) ×xy

So, lateral displacement, d = t Sin (i-r)/Cos r

Laws of refraction

1. The incident ray, the refracted ray and the normal at the point of incidence all lie on the same plane.

2. The ratio of the sine of the angle of incidence to the sine of the angle of refraction is a constant for a given pair of media.

Sin i/sin r = constant (k)

Refractive index

Refractive index (n) is the constant of proportionality in Snell’s law; hence

Sin i/ sin r = n

Therefore sin i/sin r=n=1/sin r/sin i

Examples

1. Calculate the refractive index for light travelling from glass to air given thatang= 1.5

Solution

gna= 1/ang = 1/1.5=0.67

2. Calculate the angle of refraction for a ray of light from air striking an air-glass interface, making an angle of 600 with the interface. (ang= 1.5)

Solution

Angle of incidence (i) = 900-600=300

1.5=sin 30^o/sin r, sin r =sin 300/ 1.5=0.5/1.5

Sin r=0.3333, sin-10.3333= 19.50

R= 19.50

Refractive index in terms of velocity.

Refractive index can be given in terms of velocity by the use of the following equation;

1n2 = velocity of light in medium 1/velocity of light in medium 2

When a ray of light is travelling from vacuum to a medium the refractive index is referred to as absolute refractive index of the medium denoted by ‘n’

Refractive index of a material ‘n’=velocity of light in a vacuum/velocity of light in material ‘n’

The absolute refractive indices of some common materials is given below

Examples
1. A ray of light is incident on a water-glass interface as shown. Calculate ‘r’. (Take the refractive index of glass and water as 3/2 and 4/3 respectively)

2. The refractive index of water is 4/3 and that of glass is 3/2. Calculate the refractive index of glass with respect to water.

Solution

wng= gna×ang, but wna = 1/ anw=3/4

wng=3/4×3/2=9/8= 1.13

Real and apparent depth

Consider the following diagram

The depth of the water OM is the real depth, and the distance IM is known as the apparent depth. OI is the distance through which the coin has been displaced and is known as the vertical displacement.

The relationship between refractive index and the apparent depth is given by;

Refractive index of a material=real depth/apparent depth

NB

This is true only if the object is viewed normally.

Example

A glass block of thickness 12 cm is placed on a mark drawn on a plain paper.

The mark is viewed normally through the glass. Calculate the apparent depth of the mark and hence the vertical displacement. (Refractive index of glass =3/2)

Solution

ang= real depth/apparent depth

apparent depth= real depth/ ang=(12×2)/3= 8 cm

vertical displacement= 12-8=4 cm

Applications of refractive index

Total internal reflection

This occurs when light travels from a denser optical medium to a less dense medium. The refracted ray moves away from the normal until a critical angle is reached usually 900 where the refracted ray is parallel to the boundary between the two media.

If this critical angle is exceeded total internal reflection occurs and at this point no refraction occurs but the ray is reflected internally within the denser medium.

Relationship between the critical angle and refractive index.

Consider the following diagram

From Snell’s law

gnw = sin C/sin 900,but ang = 1/gna since sin 900 = 1

Thereforeang= 1/sin C, hence sin C=1/n or n=1/sin C

Example

Calculate the critical angle of diamond given that its refractive index is 2.42

Solution

Sin C= 1/n=1/ 2.42= 0.4132= 24.40

Effects of total internal reflection

1. Mirage: These are ‘pools of water’ seen on a tarmac road during a hot day.

They are also observed in very cold regions but the light curves in opposite direction such that a polar bear seems to be upside down in the sky.

2. Atmospheric refraction: the earths’ atmosphere refracts light rays so that the sun can be seen even when it has set. Similarly the sun is seen before it actually rises.

Applications of total internal reflection

1. Periscope: a prism periscope consists of two right angled glass prisms of angles 450,900 and 450 arranged as shown below. They are used to observe distant objects.

2. Prism binoculars: the arrangement of lenses and prisms is as shown below. Binoculars reduce the distance of objects such that they seem to be nearer.

3. Pentaprism: used in cameras to change the inverted images formed into erect and actual image in front of the photographer.

4. Optical fibre: this is a flexible glass rod of small diameter. A light entering through them undergoes repeated internal reflections.

They are used in medicine to observe or view internal organs of the body

5. Dispersion of white light:

the splitting of light into its constituent colours is known as dispersion. Each colour represents a different wavelength as they strike the prism and therefore refracted differently as shown.

Chapter Three

Newton’s Laws of Motion

Newton’s first law (law of inertia)

This law states that “A body continues in its state of rest or uniform motion unless an unbalanced force acts on it”.

The mass of a body is a measure of its inertia. Inertia is the property that keeps an object in its state of motion and resists any efforts to change it.

Newton’s second law (law of momentum)

Momentum of a body is defined as the product of its mass and its velocity.

Momentum ‘p’=mv. The SI unit for momentum is kgm/s or Ns.

The Newton’s second law states that “The rate of change of momentum of a body is proportional to the applied force and takes place in the direction in which the force acts”

Change in momentum= mv-mu

Rate of change of momentum= mv-mu/∆t

Generally the second law gives rise to the equation of force F=ma

Hence F=mv-mu/∆t and F∆t=mv-mu

The quantity F∆t is called impulse and is equal to the change of momentum of the body. The SI unit for impulse is Ns.

Examples

1. A van of mass 3 metric tons is travelling at a velocity of 72 km/h. Calculate the momentum of the vehicle.

Solution

Momentum=mv=72km/h=(20m/s)×3×103 kg

=6.0×104kgm/s

2. A truck weighs 1.0×105 N and is free to move. What force willgiveit an acceleration of 1.5 m/s2? (take g=10N/kg)

Solution

Mass of the truck = (1.0×105)/10=6.0×104

Using F=ma

=1.5×10×104

=1.5×104 N

3. A car of mass 1,200 kg travelling at 45 m/s is brought to rest in 9 seconds.

Calculate the average retardation of the car and the average force applied by the brakes.

Solution

Since the car comes to rest, v=0, a=(v-u)/t =(0-45)/9=-5m/s (retardation)

F=ma =(1200×-5) N =-6,000 N (braking force)

4. A truck of mass 2,000 kg starts from rest on horizontal rails. Find the speed 3 seconds after starting if the tractive force by the engine is 1,000 N.

Solution

Impulse = Ft=1,000×3= 3,000 Ns

Let v be the velocity after 3 seconds.

Since the truck was initially at rest then u=0.
Change in momentum=mv-mu

= (2,000×v) – (2,000×0)

=2,000 v

But impulse=change in momentum

2,000 v = 3,000

v = 3/2=1.5 m/s.

Weight of a body in a lift or elevator

When a body is in a lift at rest then the weight

W=mg

When the lift moves upwards with acceleration ‘a’ then the weight becomes

W = m (a+g)

If the lift moves downwards with acceleration ‘a’ then the weight becomes

W = m (g-a)

Example

A girl of mass stands inside a lift which is accelerated upwards at a rate of 2 m/s². Determine the reaction of the lift at the girls’ feet.

Solution

Let the reaction at the girls’ feet be ‘R’ and the weight ‘W’

The resultant force F= R-W

= (R-500) N

Using F = ma, then R-500= 50×2, R= 100+500 = 600 N.

Newton’s third law (law of interaction)

This law states that “For every action or force there is an equal and opposite force or reaction”

Example

A girl of mass 50 Kg stands on roller skates near a wall. She pushes herself against the wall with a force of 30N.

If the ground is horizontal and the friction on the roller skates is negligible, determine her acceleration from the wall.

Solution

Action = reaction = 30 N

Force of acceleration from the wall = 30 N

F = ma

a = F/m = 30/50 = 0.6 m/s2

Linear collisions

Linear collision occurs when two bodies collide head-on and move along the same straight line.

There are two types of collisions;

a) Inelastic collision: – this occurs when two bodies collide and stick together i.e. hitting putty on a wall. Momentum is conserved.

b) Elastic collision: – occurs when bodies collide and bounce off each other after collision. Both momentum and kinetic energy are conserved.

Collisions bring about a law derived from both Newton’s third law and conservation of momentum.

This law is known as the law of conservation of linear momentum which states that “when no outside forces act on a system of moving objects, the total momentum of the system stays constant”.

Examples

1. A bullet of mass 0.005 kg is fired from a gun of mass 0.5 kg.

If the muzzle velocity of the bullet is 300 m/s, determine the recoil velocity of the gun.

Solution

Initial momentum of the bullet and the gun is zero since they are at rest.

Momentum of the bullet after firing = (0.005×350) = 1.75 kgm/s

But momentum before firing = momentum after firing hence

0 = 1.75 + 0.5 v where ‘v’ = recoil velocity

0.5 v = -1.75

v =-1.75/0.5 = – 3.5 m/s (recoil velocity)

2. A resultant force of 12 N acts on a body of mass 2 kg for 10 seconds.

What is the change in momentum of the body?

Solution

Change in momentum = ∆P = mv – mu= Ft

= 12×10 = 12 Ns

3. A minibus of mass 1,500 kg travelling at a constant velocity of 72 km/h collides head-on with a stationary car of mass 900 kg.

The impact takes 2 seconds before the two move together at a constant velocity for 20 seconds.

Calculate

a) The common velocity

b) The distance moved after the impact

c) The impulsive force

d) The change in kinetic energy

Solution

a) Let the common velocity be ‘v’

Momentum before collision = momentum after collision

(1500×20) + (900×0) = (1500 +900)v

30,000 = 2,400v

v = 30,000/2,400 = 12.5 m/s (common velocity)

b) After impact, the two bodies move together as one with a velocity of 12.5 m/s

Distance = velocity × time

= 12.5×20

= 250m

c) Impulse = change in momentum

= 1500 (20-12.5) for minibus or

=900 (12.5 – 0) for the car

= 11,250 Ns

Impulse force F = impulse/time = 11,250/2 = 5,625 N

d) K.E before collision = ½ × 1,500 × 202 = 3 × 105 J

K.E after collision = ½ × 2400 × 12.52 = 1.875×105 J

Therefore, change in K.E =(3.00 – 1.875) × 105 = 1.25× 105 J

Some of the applications of the law of conservation of momentum

1. Rocket and jet propulsion: – rocket propels itself forward by forcing out its exhaust gases.

The hot gases are pushed through exhaust nozzle at high velocity therefore gaining momentum to move forward.

2. The garden sprinkler: – as water passes through the nozzle at high pressure it forces the sprinkler to rotate.

Solid friction

Friction is a force which opposes or tends to oppose the relative motion of two surfaces in contact with each other.

Measuring frictional forces

We can relate weight of bodies in contact and the force between them.

This relationship is called coefficient of friction.

Coefficient of friction is defined as the ratio of the force needed to overcome friction Ff to the perpendicular force between the surfaces Fn.

Hence
µ = Ff/ Fn

Examples

1. A box of mass 50 kg is dragged on a horizontal floor by means of a rope tied to its front.

If the coefficient of kinetic friction between the floor and the box is 0.30, what is the force required to move the box at uniform speed?

Solution

Ff = µFn

Fn= weight = 50×10 = 500 N

Ff = 0.30 × 500 = 150 N

2. A block of metal with a mass of 20 kg requires a horizontal force of 50 N to pull it with uniform velocity along a horizontal surface.

Calculate the coefficient of friction between the surface and the block. (take g = 10 m/s)

Solution

Since motion is uniform, the applied force is equal to the frictional force

Fn = normal reaction = weight = 20 ×10 = 200 N

Therefore, µ =Ff/ Fn = 50/ 200 = 0.25.

Laws of friction

It is difficult to perform experiments involving friction and thus the following statements should therefore be taken merely as approximate descriptions:

1. Friction is always parallel to the contact surface and in the opposite direction to the force tending to produce or producing motion.

2. Friction depends on the nature of the surfaces and materials in contact with each other.

3. Sliding (kinetic) friction is less than static friction (friction before the body starts to slide).

4. Kinetic friction is independent of speed.

5. Friction is independent of the area of contact.

6. Friction is proportional to the force pressing the two surfaces together.

Applications of friction

1. Match stick

2. Chewing food

3. Brakes

4. Motion of motor vehicles

5. Walking

Methods of reducing friction

1. Rollers

2. Ball bearings in vehicles and machines

3. Lubrication / oiling

4. Air cushioning in hovercrafts

Example

A wooden box of mass 30 kg rests on a rough floor. The coefficient of friction between the floor and the box is 0.6. Calculate

a) The force required to just move the box

b) If a force of 200 N is applied the box with what acceleration will it move?

Solution

a) Frictional force Ff= µFn = µ(mg)

= 0.6×30×10 = 180 N

b) The resultant force = 200 – 180 = 20 N

From F =ma, then 20 = 30 a

a = 20 / 30 = 0.67 m/s²

Viscosity

This is the internal friction of a fluid. Viscosity of a liquid decreases as temperature increases.

When a body is released in a viscous fluid it accelerates at first then soon attains a steady velocity called terminal velocity.

Terminal velocity is attained when F + U = mg where F is viscous force, U is upthrust and mg is weight.

Chapter Four

Energy, Work, Power and Machines

Energy

This is the ability to do work.

Forms of energy.

1. Chemical energy: – this is found in foods, oils charcoal firewood etc.

2. Mechanical energy: – there are two types;

i. Potential energy – a body possesses potential energy due to its relative position or state

ii. Kinetic energy – energy possessed by a body due to its motion i.e. wind, water

iii. Wave energy – wave energy may be produced by vibrating objects or particles i.e. light, sound or tidal waves.

iv. Electrical energy – this is energy formed by conversion of other forms of energy i.e. generators.

Transformation and conservation of energy

Any device that facilitates energy transformations is called transducer. Energy can be transformed from one form to another i.e. mechanical – electrical – heat energy.

The law of conservation of energy states that “energy cannot be created or destroyed; it can only be transformed from one form to another”.

Work

Work is done when a force acts on a body and the body moves in the direction of the force.

Work done = force × distance moved by object

W = F × d

Work is measured in Nm. 1 Nm = 1 Joule (J)

Examples

1. Calculate the work done by a stone mason lifting a stone of mass 15 kg through a height of 2.0 m. (take g=10N/kg)

Solution

Work done = force × distance

= (15× 10) × 2 = 300 Nm or 300 J

2. A girl of mass 50 kg walks up a flight of 12 steps. If each step is 30 cm high, calculate the work done by the girl climbing the stairs.

Solution

Work done = force × distance

= (50× 10) × (12 ×30) ÷ 100 = 500 × 3.6 = 1,800 J

3. A force of 7.5 N stretches a certain spring by 5 cm. How much work is done in stretching this spring by 8.0 cm?

Solution

A force of 7.5 produces an extension of 5.0 cm.

Hence 8.0 cm = (7.5 ×8)/ 5 = 12.0 N

Work done = ½ × force × extension

= ½ × 12.0 × 0.08 = 0.48 J

4. A car travelling at a speed of 72 km/h is uniformly retarded by an applicationof brakes and comes to rest after 8 seconds. If the car with its occupants has a mass of 1,250 kg. Calculate;

a) The breaking force

b) The work done in bringing it to rest

Solution

a) F = ma and a = v – u/t

But 72 km/h = 20m/s

a = 0 -20/8 = – 2.5 m/s

Retardation = 2.5 m/s
Braking force F = 1,250 × 2.5

= 3,125 N

b) Work done = kinetic energy lost by the car

= ½ mv2 – ½ mu2

= ½ × 1250 × 02 – ½ × 1250 × 202

= – 2.5 × 105 J

5. A spring constant k = 100 Nm is stretched to a distance of 20 cm. calculate the work done by the spring.

Solution

Work = ½ ks²

= ½ × 100 × 0.22

= 2 J

Power

Poweris the time rate of doing work or the rate of energy conversion.

Power (P)

= work done / time

P = W / t

The SI unit for power is the watt (W) or joules per second (J/s).

Examples

1. A person weighing 500 N takes 4 seconds to climb upstairs to a height of 3.0 m. what is the average power in climbing up the height?

Solution

Power = work done / time = (force × distance) / time

= (500 ×3) / 4 = 375 W

2. A box of mass 500 kg is dragged along a level ground at a speed of 12 m/s. If the force of friction between the box and floor is 1200 N. Calculate the power developed.

Solution

Power = F v

= 2,000 × 12

= 24,000 W = 24 kW.

Machines

A machine is any device that uses a force applied at one point to overcome a force at another point. Force applied is called the effort while the resisting force overcome is called load. Machines makes work easier or convenient to be done.

Three quantities dealing with machines are;-

a) Mechanical advantage (M.A.) – this is defined as the ratio of the load (L) to the effort (E). It has no units.

M.A = load (L) / effort (E)

b) Velocity ratio – this is the ratio of the distance moved by the effort to the distance moved by the load

V.R = distance moved by effort/ distance moved by the load

c) Efficiency – is obtained by dividing the work output by the work input and the getting percentage

Efficiency = (work output/work input) × 100

= (M.A / V.R) × 100

= (work done on load / work done on effort) × 100

Examples

1. A machine; the load moves 2 m when the effort moves 8 m. If an effort of 20 N is used to raise a load of 60 N, what is the efficiency of the machine?

Solution

Efficiency = (M.A / V.R) × 100 M.A = load/effort =60/20 = 3

V.R =DE/ DL = 8/2 = 4

Efficiency = ¾ × 100 = 75%

Some simple machines

a) Levers – this is a simple machine whose operation relies on the principle of moments

b) Pulleys – this is a wheel with a grooved rim used for lifting heavy loads to high levels. The can be used as a single fixed pulley, or as a block-and-tackle system.

Example

A block and tackle system has 3 pulleys in the upper fixed block and two in the lower moveable block. What load can be lifted by an effort of 200 N if the efficiency of the machine is 60%?

Solution

V.R = total number of pulleys = 5

Efficiency = (M.A /V.R) × 100 = 60%

0.6 = M.A/ 5 =3, but M.A = Load/Effort

Therefore, load = 3 ×200 = 600 N

c) Wheel and axle– consists of a large wheel of big radius attached to an axle of smaller radius.

Example

A wheel and axle is used to raise a load of 280 N by a force of 40 N applied to the rim of the wheel. If the radii of the wheel and axle are 70 cm and 5 cm respectively. Calculate the M.A, V.Rand efficiency.

Solution

M.A = 280 / 40 = 7

V.R = R/r = 70/5 = 14

Efficiency = (M.A/ V.R) × 100 = 7/14 × 100 = 50 %

d) Inclined plane:

V.R = 1/ sin θ M.A = Load/ Effort

Example

A man uses an inclined plane to lift a 50 kg load through a vertical height of 4.0 m.

the inclined plane makes an angle of 300 with the horizontal.

If the efficiency of the inclined plane is 72%, calculate;

a) The effort needed to move the load up the inclined plane at a constant velocity.

b) The work done against friction in raising the load through the height of 4.0 m. (take g= 10 N/kg)

Solution

a) V.R = 1 / sin C = 1/ sin 300 = 2 M.A = efficiency × V.R = (72/100)× 2 = 1.44

Effort = load (mg) / effort (50×10)/ 1.44 = 347.2 N

b) Work done against friction = work input – work output

Work output = mgh = 50×10×4 = 2,000 J

Work input = effort × distance moved by effort

347.2 × (4× sin 300) = 2,777.6 J

Therefore work done against friction = 2,777.6 – 2,000 = 777.6 J

e) The screw: – the distance between two successive threads is called the pitch
V.R of screw = circumference of screw head / pitch P

= 2πr / P

Example

A car weighing 1,600 kg is lifted with a jack-screw of 11 mm pitch. If the handleis 28 cm from the screw, find the force applied.

Solution

Neglecting friction M.A = V.R

V.R = 2πr /P = M.A = L / E

1,600 / E = (2π× 0.28) / 0.011

E = (1,600 × 0.011 × 7) / 22×2×0.28 =10 N

f) Gears: – the wheel in which effort is applied is called the driver while the load wheel is the driven wheel.

V.R = revolutions of driver wheel / revolutions of driven wheel

Or

V.R = no.of teeth in the driven wheel/ no. of teeth in the driving wheel

Example

g) Pulley belts: -these are used in bicycles and other industrial machines
V.R = radius of the driven pulley / radius of the driving pulley

h) Hydraulic machines

V.R = R2 / r2 where R- radius of the load piston and r- radius of the effort piston

Example

The radius of the effort piston of a hydraulic lift is 1.4 cm while that of the load piston is 7.0 cm.

This machine is used to raise a load of 120 kg at a constant velocity through a height of 2.5 cm. given that the machine is 80% efficient, calculate;

a) The effort needed

b) The energy wasted using the machine

Solution

a) V.R = R2 / r2 = (7×7) / 1.4 × 1.4 = 25

Efficiency = M.A / V.R = (80 /100) × 25 = 20

But M.A = Load / Effort = (120×10) / 20 = 60 N

b) Efficiency = work output / work input = work done on load (m g h) /80

= (120 × 10× 2.5) / work input

80 / 100 = 3,000 / work input

Work input = (3,000 × 100) /80 = 3,750 J

Energy wasted = work input – work output

= 3,750 – 3,000 = 750 J

Chapter Five

Current Electricity

Electric potential difference and electric current

Electric current

Electric potential difference (p. d) is defined as the work done per unit charge in moving charge from one point to another. It is measured in volts.

Electric current is the rate of flow of charge. P. d is measured using a voltmeter while current is measured using an ammeter. The SI units for charge is amperes (A).

Ammeters and voltmeters

In a circuit an ammeter is always connected in series with the battery while a voltmeter is always connected parallel to the device whose voltage is being measured.

Ohm’s law

This law gives the relationship between the voltage across a conductor and the current flowing through it.

Ohm’s law states that “the current flowing through a metal conductor is directly proportional to the potential difference across the ends of the wire provided that temperature and other physical conditions remain constant”

Mathematically V α I

So V /I = constant, this constant of proportionality is called resistance

V / I = Resistance (R)

Resistance is measured in ohms and given the symbol Ω

Examples

1. A current of 2m A flows through a conductor of resistance 2 kΩ. Calculate the voltage across the conductor.

Solution

V = IR = (2 × 10-3) × (2 × 103) = 4 V.

2. A wire of resistance 20Ω is connected across a battery of 5 V. What current is flowing in the circuit?

Solution

I = V/R = 5 / 20 = 0.25 A

Ohmic and non-ohmic conductors

Ohmic conductors are those that obey Ohms law(V α I) and a good example is nichrome wire i.e. the nichrome wire is not affected by temperature.

Non-ohmic conductors do not obey Ohms law i.e. bulb filament (tungsten), thermistor couple, semi-conductor diode etc. They are affected by temperature hence non-linear.

Factors affecting the resistance of a metallic conductor

1. Temperature – resistance increases with increase in temperature

2. Length of the conductor– increase in length increases resistance

3. Cross-sectional area– resistance is inversely proportional to the cross-sectional area of a conductor of the same material.

Resistivity of a material is numerically equal to the resistance of a material of unit length and unit cross-sectional area. It is symbolized by ρ and the units are ohmmeter (Ωm).

It is given by the following formula;

ρ = AR /lwhere A – cross-sectional area, R – resistance, l – length

Example

Given that the resistivity of nichrome is 1.1× 10-6Ωm, what length of nichrome wire of diameter 0.42 mm is needed to make a resistance of 20 Ω?

Solution

ρ = AR /l, hence l = RA/ ρ = 20 × 3.142 × (2.1×10-4) / 1.1 × 10-6 = 2.52 m

Resistors

Resistors are used to regulate or control the magnitude of current and voltage in a circuit according to Ohms law.

Types of resistors

i) Fixed resistors – they are wire-wound or carbon resistors and are designed togive a fixed resistance.

ii) Variable resistors – they consist of the rheostat and potentiometer. The resistance can be varied by sliding a metal contact to generate desirable resistance.

Resistor combination

a) Series combination

Consider the following loop

Combining those in series then this can be replaced by two resistors of 60 Ω and 40 Ω.

Current through 10 Ω = (p.d. between P and R)/ (30 + 10) Ω

p.d between P and R = 0.8 × Req. Req = (40 × 60)/ 40 + 60 = 2400/ 100 = 24 Ω

p.d across R and P = 0.8 × 24 (V=IR)

therefore, current through 10 Ω = 19.2 / 10 + 30 = 0.48 A

Electromotive force and internal resistance

Electromotive force (e.m.f.) is the p.d across a cell when no current is being drawn from the cell.

The p.d across the cell when the circuit is closed is referred to as the terminal voltage of the cell.

Internal resistance of a cell is therefore the resistance of flow of current that they generate. Consider the following diagram;

The current flowing through the circuit is given by the equation,

Current = e.m.f / total resistance

I = E / R + rwhere E – e.m.f of the cell

Therefore E = I (R + r) = IR + I r = V + I r

Examples

1. A cell drives a current of 0.6 A through a resistance of 2 Ω. if the value of resistance is increased to 7 Ω the current becomes 0.2 A.

calculate the value of e.m.f of the cell and its internal resistance.

Solution

Let the internal resistance be ‘r’ and e.m.f be ‘E’.

Using E = V + I r = IR + I r

Substitute for the two sets of values for I and R

E = 0.6 × (2 + 0.6 r) = 1.2 + 0.36 r

E = 0.6 × (7 × 0.2 r) = 1.4 + 0.12 r

Solving the two simultaneously, we have,

E = 1.5 v and R = 0.5 Ω

2. A battery consists of two identical cells, each of e.m.f 1.5 v and internal resistance of 0.6 Ω, connected in parallel. Calculate the current the battery drives through a 0.7 Ω resistor.

Solution

When two identical cells are connected in series, the equivalent e.m.f is equal to that of only one cell.

The equivalent internal resistance is equal to that of two such resistance connected in parallel.

Hence Req = R1 R2 / R1 + R2 = (0.6 × 0.6) / 0.6 + 0.6 = 0.36 / 1.2 = 0.3 Ω

Equivalent e.m.f =1.5 / (0.7 + 0.3) = 1.5 A

Hence current flowing through 0.7 Ω resistor is 1.5 A

Chapter Six

Waves II

Properties of waves

Waves exhibit various properties which can be conveniently demonstrated using the ripple tank.

It consists of a transparent tray filled with water and a white screen as the bottom. On top we have a source of light.

A small electric motor (vibrator) is connected to cause the disturbance which produces waves.

The wave fronts represent wave patterns as they move along.

Rectilinear propagation

This is the property of the waves travelling in straight lines and perpendicular to the wave front.

The following diagrams represent rectilinear propagation of water waves.

Refraction

This is the change of direction of waves at a boundary when they move from one medium to another.

This occurs when an obstacle is placed in the path of the waves. The change of direction occurs at the boundary between deep and shallow waters and only when the waves hit the boundary at an angle.

Diffraction of waves

This occurs when waves pass an edge of an obstacle or a narrow gap, they tend to bend around the corner and spread out beyond the obstacle or gap.

Interference of waves

This occurs when two waves merge and the result can be a much larger wave, smaller wave or no wave at all.

When the waves are in phase they add up and reinforce each other. This is called a constructive interference and when out of phase they cancel each other out and this is known as destructive interference.

Interference in sound

Two loud speakers L1 and L2 are connected to the same signal generator so that sound waves from each of them are in phase.

The two speakers are separated by a distance of the order of wavelengths i.e. 0.5 m apart for sound frequency of 1,000 Hz.

If you walk along line AB about 2m away from the speakers, the intensity of sound rises and falls alternately hence both destructive and constructive interference will be experienced.

Stationary waves

They are also known as standing waves and are formed when two equal progressive waves travelling in opposite direction are superposed on each other.

When the two speakers are placed facing each other they produce standing waves.

A rope tied at one end will still produce stationary waves.

Chapter Seven

Electrostatics II

Electric fields

An electric field is the space around a charged body where another charged body would be acted on by a force. These fields are represented by lines of force.

This line of force also called an electric flux line points in the direction of the force.

Electric field patterns

Just like in magnetic fields, the closeness of the electric field-lines of force is the measure of the field strength.

Their direction is always from the north or positive to the south or negative.

Charge distribution on conductors’ surface

A proof plane is used to determine charge distribution on spherical or pear-shaped conductors.

For an isolated sphere it is found that the effect is the same for all points on the surface meaning that the charge is evenly distributed on all points on the spherical surface.

For appear-shaped conductor the charge is found to be denser in the regions of large curvature (small radius).

The density of charge is greatest where curvature is greatest.

Charges on or action at sharp points

A moving mass of air forms a body with sharp points.

The loss of electrons by molecules (ionization) makes the molecules positively charged ions.

These ions tend to move in different directions and collide producing more charged particles and this makes the air highly ionized.

When two positively charged bodies are placed close to each other, the air around them may cause a spark discharge which is a rush of electrons across the ionized gap, producing heat, light and sound in the process which lasts for a short time.

Ionization at sharp projections of isolated charged bodies may sometimes be sufficient to cause a discharge.

This discharge produces a glow called corona discharge observed at night on masts of ships moving on oceans.

The same glow is observed on the trailing edges of aircrafts.

This glow in aircrafts and ships is called St. Elmo’s fire. Aircrafts are fitted with ‘pig tails’ on the wings to discharge easily.

The lightning arrestors

Lightning is a huge discharge where a large amount of charge rushes to meet the opposite charge.

It can occur between clouds or the cloud and the earth. Lightning may not be prevented but protection from its destruction may be done through arrestors.

An arrestor consists of a thick copper strip fixed to the outside wall of a building with sharp spikes.

Capacitors and capacitance

A capacitor is a device used for storing charge.

It consists of two or more plates separated by either a vacuum or air.

The insulating material is called ‘dielectric’. They are symbolized as shown below,

Capacitance C = Q / V where Q- charge and V – voltage.

The units for capacitance are coulombs per volt (Coul /volt) and are called farads.

1 Coul/ volt = 1 farad (F)

1 µF = 10-6 F and 1pF = 10-12

Types of capacitors are;

a) Paper capacitors

b) Electrolyte capacitors

c) Variable capacitors

d) Plastic capacitors

e) Ceramic capacitors

f) Mica capacitors

Factors affecting the capacitance of a parallel-plate capacitor

1. Distance between the plates: – reducing separation increases capacitance but the plates should not be very close to avoid ionization which may lead to discharge.

2. Area of plate: – reduction of the effective area leads to reduction in capacitance.

3. Dielectric material between plates: – different materials will produce different capacitance effects.

Charging and discharging a capacitor

When the switch S1 is closed the capacitor charges through resistor R and discharges through the same resistor when switch S2 is closed.

Applications of capacitors

1. Variable capacitor: – used in tuning radios to enable it transmit in different frequencies.

2. Paper capacitors: – used in mains supply and high voltage installations.

3. Electrolytic capacitors: – used in transistor circuits where large capacitance values are required.

Other capacitors are used in reducing sparking as a car is ignited, smoothing rectified current and increasing efficiency in a. c. power transmission.

Example

A capacitor of two parallel plates separated by air has a capacitance of 15pF.

A potential difference of 24 volts is applied across the plates,

a) Determine the charge on the capacitors.

b) When the space is filled with mica, the capacitance increases to 250pF.

How much more charge can be put on the capacitor using a 24 V supply?

Solution

:
a) C= Q / V then Q = VC, hence Q = (1.5 × 10-12) × 24 = 3.6 × 10-10 Coul.

b) Mica C = 250pF, Q = (250 × 10-12) × 24 = 6 × 10-9 Coul.

Additional charge = (6 × 10-9) – (3.6 × 10-10) = 5.64 × 10-9 Coul.

Capacitor combination

Chapter Eight

Heating Effect of an Electric Current

When current flows, electrical energy is transformed into other forms of energy i.e. light, mechanical and chemical changes.

Factors affecting electrical heating

Energy dissipated by current or work done as current flows depends on,

a) Current

b) Resistance

c) Time

This formula summarizes these factors as, E = I2 R t, E = I V t or E = V2 t / R

Examples

1. An iron box has a resistance coil of 30 Ω and takes a current of 10 A. Calculate the heat in kJ developed in 1 minute.

Solution

E = I2 R t = 102 × 30 × 60 = 18 × 104 = 180 kJ

2. A heating coil providing 3,600 J/min is required when the p.d across it is 24 V. Calculate the length of the wire making the coil given that its cross-sectional area is 1 × 10-7 m2 and resistivity 1 × 10-6 Ω m.

Solution

E = P t hence P = E / t = 3,600 / 60 = 60 W

P = V2 / R therefore R = (24 × 24)/ 60 = 9.6 Ω

R = ρ l/ A, l = (RA) / ρ = (9.6 × 1 × 10-7) / 1 × 10-6 = 0.96 m

Electrical energy and power

In summary, electrical power consumed by an electrical appliance is given by;

P = V I

P = I2 R

P = V2 / R

The SI unit for power is the watt (W)

1 W = 1 J/s and 1kW = 1,000 W.

Examples

1. What is the maximum number of 100 W bulbs which can be safely run from a 240 V source supplying a current of 5 A?

Solution

Let the maximum number of bulbs be ‘n’. Then 240 × 5 = 100 n

So ‘n’ = (240 × 5)/ 100 = 12 bulbs.

2. An electric light bulb has a filament of resistance 470 Ω.

The leads connecting the bulb to the 240 V mains have a total resistance of 10 Ω. Find the power dissipated in the bulb and in the leads.

Solution

Req = 470 + 10 = 480 Ω, therefore I = 240 / 480 = 0.5 A.

Hence power dissipated = I2 R = (0.5)2 × 470 = 117.5 W (bulb alone)

For the leads alone, R = 10 Ω and I = 0.5 A

Therefore power dissipated = (0.5)2 × 10 = 2.5 W.

Applications of heating of electrical current

1. Filament lamp – the filament is made up of tungsten, a metal with high melting point (3.400 0C). It is enclosed in aglass bulb with air removed and argon or nitrogen injected to avoid oxidation. This extends the life of the filament.

2. Fluorescent lamps – when the lamp is switched on, the mercury vapour emits ultra violet radiation making the powder in the tube fluoresce i.e. emit light. Different powders emit different colours.

3. Electrical heating – electrical fires, cookers e.tc. their elements are made up nichrome ( alloy of nickel and chromium) which is not oxidized easily when it turns red hot.

4. Fuse – this is a short length of wire of a material with low melting point (often thinned copper) which melts when current through it exceeds a certain value. They are used to avoid overloading.

Chapter Nine

Quantity of Heat

Heat is a form of energy that flows from one body to another due to temperature differences between them.

Heat capacity

Heat capacity is defined as the quantity of heat required to raise the temperature of a given mass of a substance by one degree Celsius or one Kelvin. It is denoted by ‘C’.

Heat capacity, C = heat absorbed, Q / temperature change θ.

The units of heat capacity are J / 0C or J / K.

Specific heat capacity.

S.H.C of a substance is the quantity of heat required to raise the temperature of 1 kg of a substance by 1 0C or 1 K. It is denoted by ‘c’, hence,

c = Q / m θ where Q – quantity of heat, m – mass andθ – change in temperature.

The units for ‘c’ are J kg-1 K-1. Also Q = m c θ.

Examples

1. A block of metal of mass 1.5 kg which is suitably insulated is heated from 30 0C to 50 0C in 8 minutes and 20 seconds by an electric heater coil rated 54 watts.

Find;

a) The quantity of heat supplied by the heater

b) The heat capacity of the block

c) Its specific heat capacity

Solution

a) Quantity of heat = power × time = P t

= 54 × 500 = 27,000 J

b) Heat capacity, C = Q / θ = 27,000 / (50 – 30) = 1,350 J Kg-1 K^-1

c) Specific heat capacity, c = C / m = 1,350 / 1.5 = 900 J Kg^-1

2. If 300 g of paraffin is heated with an immersion heater rated 40 W, what is the temperature after 3 minutes if the initial temperature was 20 0C? (S.H.C for paraffin = 2,200 JK ^-1 K^-1).
Solution

Energy = P t = m c θ = Q = quantity of heat.

P t = 40 × 180 = 7,200 J

m = 0.30 kg c = 2,200, θ = ..?

Q = m c θ, θ = Q / m c = 7,200 / (0.3 × 2,200) = 10.9 0C

3. A piece of copper of mass 60 g and specific heat capacity 390 J Kg-1 K-1 cools from 90 0C to 40 0C. Find the quantity of heat given out.

Solution

Q = m c θ, = 60 × 10-3 × 390 × 50 = 1,170 J.

Determination of specific heat capacity

:
A calorimeter is used to determine the specific heat capacity of a substance.

This uses the principle of heat gained by a substance is equal to the heat lost by another substance in contact with each other until equilibrium is achieved.

Heat losses in calorimeter are controlled such that no losses occur or they are very minimal.

Examples

1. A 50 W heating coil is immersed in a liquid contained in an insulated flask of negligible heat capacity.

If the mass of the liquid is 10 g and its temperature increases by 10 0C in 2 minutes, find the specific heat capacity of the liquid.

Solution

Heat delivered (P t) = 50 × 2 × 60 = 2,400 J

Heat gained = 0.1 × c × 10 J

Therefore ‘c’ = 2,400 / 0.1 × 10 = 2,400 J Kg^-1 K^-1
2. A metal cylinder mass 0.5 kg is heated electrically.

If the voltmeter reads 15V, the ammeter 0.3A and the temperatures of the block rises from 20 0C to 85 0C in ten minutes.

Calculate the specific heat capacity of the metal cylinder.

Solution

Heat gained = heat lost, V I t = m c θ

15 × 3 × 10 × 60 = 0.5 × c × 65

c = (15 × 3 × 600)/ 0.5 × 65 = 831 J Kg-1 K^-1

Fusion and latent heat of fusion

Fusion is the change of state from solid to liquid.

Change of state from liquid to solid is called solidification.

Latent heat of fusion is the heat energy absorbed or given out during fusion. Specific latent heat of fusion of a substance is the quantity of heat energy required to change completely 1 kg of a substance at its melting point into liquid without change in temperature.

It is represented by the symbol (L), we use the following formula,
Q = m Lf

Different substances have different latent heat of fusion.

Factors affecting the melting point

a) Pressure

b) Dissolved substances

Specific latent heat of vaporization is the quantity of heat required to change completely 1 kg of a liquid at its normal boiling point to vapour without changing its temperature.

Hence

Q = m Lv

The SI unit for specific latent heat of vaporization is J / Kg.

Example

An immersion heater rated 600 W is placed in water. After the water starts to boil, the heater is left on for 6 minutes.

It is found that the mass of the water had reduced by 0.10 kg in that time.

Estimate the specific heat of vaporization of steam.

Solution

Heat given out by the heater = P t = 600 × 6 × 60

Heat absorbed by steam = 0.10 × L v

Heat gained = heat lost, therefore, 600 × 6 × 60 = 0.10 × L v = 2.16 × 106 J / Kg

Evaporation

Factors affecting the rate of evaporation

a) Temperature

b) Surface area

c) Draught (hot and dry surrounding)

d) Humidity

Comparison between boiling and evaporation
Evaporation Boiling

1. Takes place at all temperature – takes place at a specific temperature

2. Takes place on the surface (no bubbles formed)- takes place throughout the liquid ( bubbles formed)

3. Decrease in atmospheric pressure increases the rate –decreases as atmospheric pressure lowers

Applications of cooling by evaporation

a) Sweating

b) Cooling of water in a porous pot

c) The refrigerator

Chapter Ten

The Gas Laws

Pressure law
This law states that “the pressure of a fixed mass of a gas is directly proportional to the absolute temperature if the volume is kept constant”. The comparison between Kelvin scale and degrees Celsius is given by; θ0 = (273 + θ) K, and T (K) = (T – 273) 0C.

Examples

1. A gas in a fixed volume container has a pressure of 1.6 × 105 Pa at a temperature of 27 0C.

What will be the pressure of the gas if the container is heated to a temperature of 2770C?

Solution

Since law applies for Kelvin scale, convert the temperature to kelvin

T1 = 270C = (273 + 27) K = 300 K

T2 = 2270C = (273 + 277) = 550 K

P₁ / T₁ = P₂ / T₂, therefore P₂ = (1.6 × 105) × 550 / 300 = 2.93 × 105 Pa.

2. At 200C, the pressure of a gas is 50 cm of mercury. At what temperature would the pressure of the gas fall to 10 cm of mercury?

Solution

P / T = constant, P₁ / T₁ = P₂ / T₂, therefore T₂= (293 × 10) / 50 = 58.6 K or (– 214.4 0C)

Charles law

Charles law states that “the volume of a fixed mass of a gas is directly proportional to its absolute temperature (Kelvin) provided the pressure is kept constant”. Mathematically expressed as follows,

V1 / T1 = V2 / T2

Examples

1. A gas has a volume of 20 cm³ at 270C and normal atmospheric pressure.

Calculate the new volume of the gas if it is heated to 540C at the same pressure.

Solution

Using, V₁ / T₁ = V₂ / T₂, then V₂ =(20 × 327) / 300 = 21.8 cm³.

2. 0.02m³ of a gas is at 27 0C is heated at a constant pressure until the volume is 0.03 m³. Calculate the final temperature of the gas in 0C.

Solution

Since V₁ / T₁ = V₂ / T₂, T₂ = (300 × 0.03) / 0.02 = 450 K 0r 1770C

Boyle’s law

Boyle’s law states that “the pressure of a fixed mass of a gas is inversely proportional to its volume provided the temperature of the gas is kept constant”.

Mathematically expressed as,

P₁ V₁ = P₂ V₂

Examples

1. A gas in a cylinder occupies a volume of 465 ml when at a pressure equivalent to 725 mm of mercury. If the temperature is held constant, what will be the volume of the gas when the pressure on it is raised to 825 mm of mercury?

Solution

Using, P₂ V₁ = P₂ V₂, then V₂ = (725 × 465) / 825 = 409 ml.

2. The volume of air 26 cm long is trapped by a mercury thread 5 cm long as shown below. When the tube is inverted, the air column becomes 30 cm long. What is the value of atmospheric pressure?

Solution

Before inversion, gas pressure = atm. Pressure + h p g

After inversion, gas pressure = atm. Pressure – h p g

From Boyle’s law, P1 V1 = P2 V2, then let the atm. Pressure be ‘x’,

So (x + 5) 0.26 = (x – 5) 0.30

0.26x + 1.30 = 0.3x – 1.5, x = 2.8/ 0.04 = 70 cm.

A general gas law

Any two of the three gas laws can be used derive a general gas law as follows,

P₁ V₁ / T₁ = P₂ V₂ / T₂ or

P V / T = constant – equation of state for an ideal gas.

Examples

1. A fixed mass of gas occupies 1.0 × 10^-3 m³ at a pressure of 75 cmHg. What volume does the gas occupy at 17.0 0C if its pressure is 72 cm of mercury?

Solution

P V / T = constant so V₁ = (76 × 1.0 × 10-3 × 290) / 273 ×72 = 1.12 × 10^-3 m³

2. A mass of 1,200 cm³ of oxygen at 270C and a pressure 1.2 atmosphere is compressed until its volume is 600 cm³ and its pressure is 3.0 atmosphere. What is the Celsius temperature of the gas after compression?

Solution

Since P₁ V₁ / T₁ = P₂ V₂ / T₂, then T₂ = (3 × 600 × 300) / 1.2 × 1,200 = 375 K or 102 0C.

Physics Notes Form 4

Physics Form Four Notes

Chapter One

Thin Lences

A lens is conventionally defined as a piece of glass which is used to focus or change the direction of a beam of light passing through it.

They are mainly made of glass or plastic. Lens are used in making spectacles, cameras, cinema projectors, microscopes and telescopes.

Types of thin lenses.

A lens which is thicker at its centre than at its edges converges light and is called convex or converging lens.

A lens which is thicker at its edges than at its centre diverges light and is known as concave or diverging lens.

Properties of lenses.

1. Optical centre – this is the geometric centre of a lens which is usually shown using a black dot in ray diagrams. A ray travelling through the optical centre passes through in a straight line.

2. Centre of curvature – this is the geometric centre of the circle of which the lens surface is part of. Since lenses have two surfaces there are two centres of curvature. C is used to denote one centre while the other is denoted by C1.

3. Principal axis – this is an imaginary line which passes through the optical centre at right angle to the lens.

4. Principal focus – this is a point through which all rays travelling parallel to the principal axis pass after refraction through the lens.

A lens has a principal focus on both its sides. F is used to denote the principal focus

5. Focal length – this is the distance between the optical centre and the principal focus. It is denoted by ‘f’.

The principal focus for a converging lens is real and virtual for a diverging lens.

It is important to note that the principal focus is not always halfway between the optical centre and the centre of curvature as it is in mirrors.

Images formed by thin lenses.

The nature, size and position of the image formed by a particular lens depends on the position of the object in relation to the lens.

Construction of ray diagrams

Three rays are of particular importance in the construction of ray diagrams.

1. A ray of light travelling parallel to the principal axis passes through the principal focus on refraction through the lens. In case of a concave lens the ray is diverged in a way that it appears to come from the principal focus.

2. A ray of light travelling through the optical centre goes un-deviated along the same path.

3. A ray of light travelling through the principal focus is refracted parallel to the principal axis on passing through the lens. The construction of the rays is illustrated below.

Images formed by a converging lens.

1. Object between the lens and the principal focus.

– Image formed behind the object

– Virtual

– Erect

– Magnified

2. Object at infinity.

– Image formed at the principal focus of the lens

– Real

– Inverted

– Diminished

3. Object at the principal focus (at F).

– Image is at infinity.

4. Object between the principal focus (F) and 2 F.

– Image situated beyond 2 F

– Real

– Inverted

– Magnified

5. Object at 2 F.

– Image is formed at 2 F

– Real

– Inverted

– Same size as the object

6. Object beyond F.

– Image moves nearer to F as object shifts further beyond 2 F

– Real

– Inverted

– Diminished

Images formed by a diverging lens.

Images formed by diverging lens are always erect, virtual and diminished for all positions of the object.

Linear magnification.

The linear magnification produced by a lens defined as the ratio of the height of the image to the height of the object, denoted by letter ‘m’, therefore;

m = height of the image / height of the object.

Magnification is also given by = distance of the image from the lens/ dist. of object from lens.
m = v / u

Example

An object 0.05 m high is placed 0.15 m in front of a convex lens of focal length 0.1 m.

Find by construction, the position, nature and size of the image. What is the magnification?

Solution

The lens formula

Let the object distance be represented by ‘u’, the image distance by ‘v’ and the focal length by ‘f’, then the general formula relating the three quantities is given by;

1 / f = 1 / u + 1 / v – this is the lens formula.

Examples

1. An object is placed 12 cm from a converging lens of focal length 18 cm. Find the position of the image.

Solution

Since it is a converging lens f = +18 cm (real-is-positive and virtual-is-negative rule)

The object is real therefore u = +12 cm, substituting in the lens formula, then

1 / f = 1 / u + 1 / v or 1 / v = 1 / f – 1 / u = 1 / 18 – 1 / 12 = – 1 / 36

Hence v = – 36 then the image is virtual, erect and same size as the object.

2. The focal length of a converging lens is found to be 10 cm. How far should the lens be placed from an illuminated object to obtain an image which is magnified five times on the screen?

Solution

f = + 10 cm m = v / u = 5 hence v = 5 u

Using the lens formula 1 / f = 1 / u + 1 / v => 1 /10 = 1 / u + 1 / 5 u (replacing v with 5 u)

1 / 10 = 6 / 5 u, hence 5 u = 60 giving u = 12 cm (the lens should be placed 12 cm from the illuminated object)

3. The lens of a slide projector focuses on an image of height 1.5m on a screen placed 9.0 m from the projector.

If the height of the picture on the slide was 6.5 cm, determine,
a) Distance from the slide (picture) to the lens

b) Focal length of the lens

Solution

Magnification = height of the image / height of the object = v / u = 150 / 6.5 = 900 / u

u = 39 cm (distance from slide to the lens). m = 23.09

1 / f = 1 / u + 1 / v = 1 /39 + 1 / 90 = 0.02564 + 0.00111

1 / f = 0.02675 (reciprocal tables)

f = 37.4 cm.

Determining focal lengths.

1. Determining focal length of a converging lens

Experiment: To determine the focal length of a converging lens using the lens formula.

Procedure.

1. Set up the apparatus as shown below

2. Place the object at reasonable length from the screen until a real image is formed on the screen. Move the lens along the metre rule until a sharply focused image is obtained.

3. By changing the position of the object obtain several pairs of value of u and v and record your results as shown.

Discussion

The value u v / u + v is the focal length of the lens and the different sets of values give the average value of ‘f’.

Alternatively the value ‘f’ may be obtained by plotting a graph of 1 / v against 1 / u. When plotted the following graph is obtained.

Since 1 / f = 1 / u + 1 / v, at the y-intercept 1 / u = 0, so that 1/ f = 1 / v or f = v.
The focal length may therefore be obtained by reading off the y-intercept and finding the reciprocal.

Similarly at the x-intercept, 1 / v = 0, therefore 1 / f = 1 / u or f = u hence the focal length can also be obtained by reading off the x-intercept and finding the reciprocal.

Uses of lenses on optical devices

1. Simple microscope – it is also referred to as magnifying glass where the image appears clearest at about 25 cm from the eye.

This distance is known as the least distance of distinct vision (D) or near vision.

Magnification in a simple microscope

Magnification produced depends on the focal length of the lens. Lens of short focal give greater magnification than those of long focal length.

The angle ϐ subtended by the image at the eye is much greater than α which is the angle that the object would subtend at the eye when viewed without the lens.

The ratio of the ϐ toα is known as angular magnification or magnifying power of an instrument. The angular magnification is equal to linear magnification.

Uses of a simple microscope

1. To study the features of small animals in biology

2. To look closely at small print on a map

3. To observe crystals in physics and chemistry

4. For forensic investigation by the police

2. Compound microscope – It consists of two lenses with one nearer the object called the objective lens and the other nearer the eye called the eyepiece lens.

Uses of compound microscope

1. Used to observe Brownian motion in science

2. To study micro-organisms and cells in biology

3. Analyze laboratory tests in hospital.

4. The astronomical telescope –It is used to view distant stars. It consists of two lenses; objective and eye-piece lenses. The objective lens has a large focal length while the eye-piece lens has a much shorter focal length.

5. The camera – consists of a converging lens system, clicking button, shutter, diaphragm and a mounting base for the film all enclosed in a light proof box.

The distance is adjusted to obtain a clear focus.

The diaphragm has a hole called the aperture with an adjusting control knob to control the amount of light entering the camera. The shutter opens to allow light and close at a given time interval.

Uses of a camera

1. The sine camera is used to make motion pictures

2. High speed cameras are used to record movement of particles

3. Close circuit television cameras (CCTV) are used to protect high security installations like banks, supermarkets etc.

4. Digital cameras are used to capture data that can be fed to computers.

5. Human eye – It consists of a transparent cornea, aqueous humour and a crystal-like lens which form a converging lens system. The ciliary muscles contract or relax to change the curvature of the lens.

Though the image formed at the retina is inverted the brain ‘sees’ the image as upright.

For distant objects ciliary muscles relax while near objects it contracts to control the focal length and this is known as accommodation. When at 25 cmaway an object appears clearest and this is known as least distance of vision or near point.

Common eye defects

1. Short sightedness or hypermetropia– the eyeball is too large for the ‘relaxed focal length’ of the eye. The defect is corrected by placing a concave lens in front of the eye.

2. Long sightedness or myopia – images are formed beyond the retina. The defect is corrected by placing a converging lens in front of the eye.

3. Presbyopia – this is the inability of the eye to accommodate and this occurs as the eye ages due to the weakening of the ciliary muscles. It can be corrected by the use a pair of spectacles.

4. Astigmatism – this is a defect where the eye has two different focal lengths as a result of the cornea not being spherical. Corrected by the use of cylindrical lens.

5. Colour blindness– caused by deficiency of colour detecting cells in the retina.

Power of lens

The power of a simple lens is given by the formula: Power = 1 / f. The unit for power of a lens is diopter (D).

Example

Find the power of a concave lens of a focal length 25 cm.

Solution

Power = 1 / f = 1 / -0.25 = -4 D.

Chapter Two

Uniform Circular Motion

Introduction

Circular motion is the motion of bodies travelling in circular paths. Uniform circular motion occurs when the speed of a body moving in a circular path is constant. This can be defined as motion of an object at a constant speed along a curved path of constant radius. When acceleration (variation of velocities) is directed towards the centre of the path of motion it is known as centripetal acceleration and the force producing this centripetal acceleration which is also directed towards the centre of the path is called centripetal force.

Angular motion

This motion can be described as the motion of a body moving along a circular path by giving the angle covered in a certain time along the path of motion.

The angle covered in a certain time is proportional to the distance covered along the path of motion.

The radian

One radian is the angle subtended at the centre of the circle by an arc of length equal to the radius of the circle. Since one circle = 3600and has 2 π radians therefore 1 radian = 3600 / 2 π r= 57.2960 or 57.30.

Example

A wheel of radius 50 cm is rolled through a quarter turn. Calculate

(i) The angle rotated in radians

(ii) The distance moved by a point on the circumference.

Solution

(i) A quarter turn = 3600 / 4= 900. Since 3600 = 2 π radians.

Alternately since 1 radian = 57.30 hence 900 = 1.57 radii.

(ii) A point on the circumference moves through an arc,

Arc = radius ×θ (θ in radians)

= 50 cm × 1.57

= 78.5 cm.

Angular velocity

If a body moving in a circular path turns through an angle θ radians in time ‘t’, we define angular velocity omega (ω), as the rate of change of the angle θ with time.

ω= θ / t, unit for angular velocity is radians per second (rads-1).

Since the radian measure is a ratio we can write it as second-1 (s-1). We can establish the relationship between angular velocity ‘ω’ and linear velocity ‘v’, from the relation, θ = arc / radius, arc = radius × θ.Dividing the expression by ‘t’, then arc / t = radius, but arc / t = v (angular velocity). So ‘v’ = radius × ω. This expression gives us the relationship between angular and linear velocity.

Angular acceleration

If the angular velocity for a body changes from ‘ω1’ to ‘ω2’, in time ‘t’ then the angular acceleration, α can be expressed as;

α= (ω2 – ω1) / t

Units for angular acceleration are radians per second squared (rad s-2) or second-2 (s-2). When α is constant with time, we say the body is moving with uniform angular acceleration.
Note: In uniform circular motion α is equal to zero.

To establish the relationship between angular acceleration and linear acceleration, from the relation, v = radius × ω, then dividing by ‘t’, we get (v / t) = radius × ω / t.
But v / t = a (linear acceleration) and ω / t = α (angular acceleration).
So a = radius × α.

Centripetal force.

This is a force which acts on a body by directing the body towards its centre. Since the direction is continuously changing, the velocity therefore cannot be constant.

Applying Newton’s law of motion (F = ma), the centripetal force Fc is given by;

Fc = ma = mv2/R. Since v = radius ω, then

Fc = mv2 – ω2/R = mRω2.

The centripetal acceleration ‘a’ in relation to angular velocity, ω, is given by a = Rω2.

Motion in a vertical circle

Consider a mass ‘m’ tied to a string of length ‘r’ and moving in a vertical circle as shown below.

At position 1– both weight (mg) and tension T are in the same direction and the centripetal force is provided by both, hence T1 + mg = mv2/r.

T1 = mv2/r – mg.(The velocity decreases as T1decreases since mg is constant).

T1will be zero when mv2/r = mg and thus v = – this is the value of minimum speed at position 1 which keeps the body in a circle and at this time when T = 0 the string begins to slacken.

At position 2 – the ‘mg’ has no component towards the centre thus playing no part in providing the centripetal force but is provided by the string alone.

T2 = mv2/r

At position 3 – ‘mg’ and T are in opposite directions, therefore;
T3 – mg = mv2/r; T3 = mv2/r + mg– indicates that the greatest value of tension is at T3 or at the bottom of the circular path.

Examples

1. A ball of mass 2.5 × 10-2 kg is tied to a string and whirled in a horizontal circular path at a speed of 5.0 ms-2. If the string is 2.0 m long, what centripetal force does the string exert on the ball?

Solution

Fc = mv2/r = (2.5 × 10-2) × 52 / 2.0 = 0.31 N.

2. A car of mass 6.0 × 103 kg is driven around a horizontal curve of radius 250 m.

if the force of friction between the tyres and the road is 21,000 N. What is the maximum speed that the car can be driven at on a bend without going off the road?

Solution

Fc = force of friction = 21,000, also Fc = mv2/r, hence

21,000 = (6.0 × 103) × v2 / 250, v2 = (21,000 × 250) /6.0 × 103

3. A stone attached to one end of a string is whirled in space in in a vertical plane. If the length of the string is 80 cm, determine the minimum speed at which the stone will describe a vertical circle. (Take g = 10 m/s2).

Solution

Minimum speed v √rg=0.08× 10 = 2.283 m/s.

The conical pendulum

It consists of a small massive object tied to the end of a thin string tied to affixed rigid support. The object is then pulled at an angle then made to whirl in a horizontal circle.

When speed of the object is constant the angle θ becomes constant also. If the speed is increased theangle θ increases, that is the object rises and describes a circle of bigger radius. Therefore as the angular velocity increases ‘r’ also increases.

The centrifuge

It consists of a small metal container tubes which can be electrically or manually rotated in a circle. If we consider two particles of different masses m1and m2 each of them requires a centripetal force to keep it in circular motion, the more massive particle require a greater force and so a greater radius and therefore it moves to the bottom of the tube.

This method is used to separate solids and liquids faster than using a filter paper.

Banked tracks

As a vehicle moves round a bend, the centripetal force is provided by the sideways friction between the tyres and the surface, that is;

Centripetal force = mv2/r = frictional force

To enable a vehicle to turn along a bend at high speed the road is raised on the outer edge to attain a saucer-like shape and this is known as banking, where part of the centripetal force necessary to keep the vehicle on track is provided by the weight of the vehicle.

This allows cars to negotiate bends at critical speeds.

Application of uniform circular motion

1. Centrifuges – they are used to separate liquids of different densities i.e. cream and milk

2. Drying clothes in spin dryer– clothes are placed in a perforated drum rotated at high speed, water is expelled through the holes and this makes the clothes dry.

3. Road banking– especially for racing cars which enables them to move at critical speed along bends without going off the tracks.

4. Speed governor– the principle of conical pendulum is used here to regulate the speed by controlling the fuel intake in the combustion chamber.

As the collar moves up and down through a system of levers it thereby connects to a device which controls the fuel intake.

Chapter Three

Floating and Sinking

Any object in a liquid whether floating or submerged experiences an upward force from the liquid; the force is known as upthrust force.

Upthrust force is also known as buoyant force and is denoted by letter ‘u’.

Archimedes’ principle

Floating and Sinking

Archimedes, a Greek scientist carried out first experiments to measure upthrust on an object in liquid in the third century. Archimedes principle states that ‘When a body is wholly or partially immersed in a fluid (liquid/ gas), it experiences an upthrust equal to the weight of the displaced fluid”.

Floating and Sinking

Experiment: To demonstrate Archimedes principle

Procedure

1. Pour water into an overflow can (eureka can) until it starts to flow out then wait until it stops dripping

2. Tie a suitable solid body securely and suspend it on a spring balance. Determine weight in air.

3. Lower the body slowly into the overflow can while still attached to the spring balance then read off its weight when fully submerged.

4. Weigh the displaced water collected in a beaker.

Record your readings as follows;

Weight of body in air = W1

Weight of body in water = W2

Weight of empty beaker = W3

Weight of beaker and displaced liquid = W4

Upthrust of the body = W1 – W2

Weight of displaced water = W4 – W3

Discussion

The upthrust on the solid body will be found to be equal to the weight of displaced water therefore demonstrating the Archimedes principle.

Example

A block of metal of volume 60 cm3 weighs 4.80 N in air. Determine its weight when fully submerged in a liquid of density 1,200 kgm^-3.

Solution

Volume of liquid displaced = 60 cm₃ = 6.0 × 10-5 m₃.
Weight of the displaced liquid = volume × density × gravity = v × ρ × g

= 6.0 × 10-5 × 1200 × 10

= 0.72 N

Upthrust = weight of the liquid displaced.

Weight of the block in the liquid = (4.80 – 0.72) = 4.08 N.

Floating objects

Objects that float in a liquid are less dense than the liquid in which they float.

We have to determine the relationship between the weight of the displaced liquid and the weight of the body.

Experiment: to demonstrate the law of floatation

Procedure

1. Weigh the block in air and record its weight as W1.

2. Put water into the overflow can (eureka can) up to the level of the spout.

3. Collect displaced water in a beaker. Record the weight of the beaker first in air and record as W2. Weigh both the beaker and the displaced water and record as W3.

4. Record the same procedure with kerosene and record your results as shown below.

Discussion

The weight of the displaced liquid is equal to the weight of the block in air.

This is consistent with the law of floatation which states that “A body displaces its own weight of the liquid in which it floats”.

Mathematically, the following relation can be deduced
Weight = volume × density × gravity = v × ρ × g, therefore
W = v d × ρ × gwhere vdis the volume of displaced liquid.

NOTE – Floatation is a special case of Archimedes principle.

This is because a floating body sinks until the upthrust equals the weight of the body.

Example

A wooden block of dimensions 3 cm × 3 cm × 4 cm floats vertically in methylated spirit with 4 cm of its length in the spirit.

Calculate the weight of the block. (Density of methylated spirit = 8.0 × 102 kgm^-3).

Solution

Volume of the spirit displaced = (3 × 3 × 4) = 36 cm³ = 3.6 × 10-5 m³

Weight of the block =v d × ρ × g = (3.6 × 10-5) × 8.0 × 102 × 10 = 2.88 × 10-1 N.

Relative density

We have established the relative density as the ratio of the density of a substance to the density of water. Since by the law of floatation an object displaces a fluid equal to its own weight hence the following mathematical expressions can be established.

Relative density = density of substance / density of water.

= weight of substance / weight of equal volume of water

= mass of substance / mass of equal volume of water

Applying Archimedes principle, the relative density‘d’;

d = weight of substance in air / upthrust in water or d = W / u

Since upthrust is given by (W2 – W1)where W2 – weight in air, W2– weight when submerged.
Hence d = W / u = W / W2 – W1, the actual density, ρ of an object can be obtained as follows
ρ of an object = d × 1,000 kgm-3.

Relative density of a floating body

Experiment: To determine the relative density of a cork

Procedure

1. Select a sinker which is heavy enough to make the cork to sink.

2. Attach the cork and the sinker as follows.

3. Record the results obtained as follows

Weight of the sinker in water = W1

Weight of the sinker in water and cork in air = W2

Weight of the sinker and cork in water = W3

Weight of the cork in air = W2 – W1

Upthrust on the cork = W2 – W3

The relative density of the cork in air is determined as follows;

d = weight of the cork in air / upthrust on the cork.

Applications of Archimedes principle and relative density

1. Ships – steel which is used to make ships is 6-7 times dense than water but a ship is able to float on water because it is designed to displace more water than its volume.

Load lines called plimsoll marks are marked on the side to indicate the maximum load at different seasons to avoid overloading.

2. Submarines – they are made of steel and consists of ballast tanks which contain water when they have to sink and filled with air when they have to float.

This makes the submarines to balance their weight and be able to rise upwards.

3. Balloons – when they are filled with helium gas balloons become lighter and the upthrust on the balloon becomes greater than their weight therefore becoming able to rise upwards.

4. Hydrometers – they are used to measure the relative densities of liquids quickly and conveniently. Various types of hydrometers are made to measure different ranges of different densities i.e. lactometer – for measuring milk water (range 1.015 – 1.045), battery acid tester – used to test the charge in a lead-acid battery.

Examples

1. A solid of mass 1.0 kg is suspended using a thread and then submerged in water.

If the tension on the thread is 5.0 N, determine the relative density of the solid.

Solution

Mass of solid = 1.0 kg

Weight of solid W = mg = 10 N

Tension on the string (T) = 5 N

Upthrust on solid (u) = W – T = 10 – 5 = 5

Relative density (d) = W / u = 10 / 5 = 2.

2. A balloon made up of a fabric weighing 80 N has a volume of 1.0 × 107 cm. the balloon is filled with hydrogen of density 0.9 kgm-3.

Calculate the greatest weight in addition to that of the hydrogen and the fabric, which the balloon can carry in air of average density 1.25 kgm^-3.

Solution

Upthrust = weight of the air displaced

= volume of air × density × gravity

= (1.0 × 107 × 106) × (1.25 × 10)

= 10 × 1.25 × 10 = 125 N

Weight of hydrogen = 10 × 0.09 × 10 = 9 N

Total weight of hydrogen and fabric = 80 + 9 = 89 N

Total additional weight to be lifted = 125 – 89 = 36 N.

3. A material of density 8.5 gcm^-3 is attached to a piece of wood of mass 100g and density 0.2 gcm^-3.

Calculate the volume of material X which must be attached to the piece of wood so that the two just submerge beneath a liquid of density 1.2 gcm^-3.

Solution

Let the volume of the material be V cm³

The mass of the material be 8.5 V grams

Volume of wood = 100 g / 0.2 g/cm = 500 cm³.

In order to have an average density of 1.2 gcm^-3 = total mass / total volume

Therefore (100 + 8.5V) / (500 + V) = 1.2 gcm^-3

Hence V = 68.5 cm³.

Electromagnetic Spectrum

Electromagnetic spectrum is a continuum of all electromagnetic waves arranged according to frequency and wavelength.

It includes visible light, ultra-violet rays, microwaves, X-rays, radio waves and gamma rays.

Electromagnetic waves are produced when electrically charged particles oscillate or change energy in some way. The waves travel perpendicularly to both electric and magnetic fields.

Wavelength, frequency and energy of electromagnetic waves.

X-rays and gamma rays are usually described in terms of wavelength and radio waves in terms of frequency.

The electromagnetic spectrum

It is divided into seven major regions or bands. A band consists of a range of frequencies in the spectrum in terms of frequencies i.e. radio, microwaves, infra-red.

Properties of electromagnetic waves

Common properties

i. They do not require a material medium and can travel through a vacuum.

ii. They undergo reflection, refraction and diffraction.

iii. All electromagnetic waves travel at the speed of light i.e. 3 × 108 ms^-1.

iv. They carry no electric charge

v. They transfer energy from a source to a receiver in the form of oscillating electric and magnetic fields.

vi. They obey the wave equation (v = λ f).

Examples

1. A VHF radio transmitter broadcasts radio waves at a frequency of 30 M Hz. What is their wavelength?

Solution

v = f λ => then λ = v / f = 3.0 × 108 / 300 × 106 = 1.00 m.

2. Calculate the frequency of a radio wave of wavelength 150 m.

Solution

v = f λ =>f = v / λ = 2.0 × 106 = 2 M Hz.

Unique properties

1. Radio waves – they are further divided into long waves (LW), medium waves (MW) and short waves (SW).

They are produced by electrical circuits called oscillators and they can be controlled accurately.

They are easily diffracted by small objects like houses but not by large objects like hills.

2. Microwaves – they are produced by oscillation of charges in special aerials mounted on dishes. They are detected by special receivers which convert wave energy to sound i.e. ‘RADAR’ – Radio Detection and Raging.

3. Infra-red radiation – infra-red radiations close to microwaves are thermal (produce heat) i.e. sun, fire but those closer to the visible light have no thermal properties i.e. TV remote control system. Detectors of infra-red radiation are the human skin, photographic film etc.

4. Optical spectrum (visible light) – they form a tiny part of the electromagnetic spectrum. Sources include the sun, electricity, candles etc. these have wavelengths visible to the human eye and includes the optical spectrum (ROYGBIV). It is detected through the eyes, photographic films and photocells.

5. Ultra-violet rays (UV) – has shorter wavelength than visible light. It is emitted by very hotobjects i.e. the sun, welding machines etc. Exposure to UV rays may cause skin cancer and cataracts. They can be detected through photographic film.

6. X-rays – they have very short wavelength but are high energy waves. They are produced in X-ray tubes when high speed electrons are stopped by a metallic object. They are detected by the use of a photographic film or a fluorescent screen.

7. Gamma rays – produced by some radioactive materials when large changes of energy occur inside their nuclei.

They can be detected by the use of photographic films, Geiger Muller tube or a cloud chamber.

Applications of electromagnetic radiation

1. Radio waves – they are used in radio, TV and cellular mobile communications.
-Used in military communications (satellite imagery) to form an image of the ground even when there are clouds.

2. Microwaves – used in radar communications by giving direction and distance.

-Used in speed guns by the police to detect over speeding.

-Used in microwave ovens to warm food. The food becomes warm by absorbing energy.

-Used reliably for communication (telephone and computer data).

3. Infra-red radiation – used to produce images of hot objects through the colours

-Produced by the amount of heat dissipated by an object.

-Images produced by satellites give important information on vegetation cover in all areas of the globe. They can also detect fires.

-They are used in hospitals to detect illnesses (diagnosis)

-Used in warfare missiles and burglar alarm systems

-Used in green houses to grow crops

4. Visible light – used by plants in remote sensing and humans in the identification of things

-Used by plants in the process of photosynthesis

5. Ultra-violet (UV) radiation – used to make reflective materials which absorb light and
re-emit it as visible light.

-Used in banks to detect fake currency

6. X-rays – used in hospitals to detect fractures, broken bones and in treatment of
cancer (radiotherapy).

-Used to detect foreign materials in the body i.e. metals

-Used to detect invisible cracks in metal castings and welding joints

7. Gamma rays – used to sterilize medical instruments
-Used to kill weevils in grain

-Used to take photographs same way like X-rays.

Chapter Five

Electromagnetic Induction

Electromagnetism is the effect resulting from the interaction between an electric current and a magnetic field. This effect brings about induced electromagnetic force (e.m.f) and the resulting current is called induced current.

Experiments on electromagnetic induction

Consider the following diagram

When the wire is moved up the galvanometer deflects in one direction then the opposite direction when moved downwards.

When moved horizontally or held in a fixed position there is no deflection in the galvanometer.

This shows that e.m.f is induced due to the relative motion of the wire or the magnet.

Factors affecting the magnitude of the induced e.m.f

1. The rate of relative motion between the conductor and the field – if the velocity of the conductor is increased the deflection in the conductor increases.

2. The strength of the magnetic field – a stronger magnetic field creates a bigger deflection

3. The length of the conductor – if the length is increased in the magnetic field the deflection increases.

Faraday’s law of magnetic induction

After considering the factors affecting the magnitude of the induced e.m.f, Michael Faraday came up with a law which states that “The induced e.m.f in a conductor in a magnetic field is proportional to the rate of change of the magnetic flux linking the conductor”.

Lenz’s law of electromagnetic induction

This law is used to determine the direction of the induced current in a conductor. It states that “An induced current flows in such a direction that its magnetic effect opposes the change through which the current has been produced”.

It is applied similarly when a wire is been moved in magnetic field.

Fleming’s right hand rule.

The law states that “The first finger, the second finger and the thumb of the right hand when placed mutually perpendicular to each other, the first finger points in the direction of the field and the thumb in the direction of motion then second finger points in the direction of the induced current”. This law is also called the generator rule.

Applications of electromagnetic induction

1. A.c generator/alternator – a generator is a device which produces electricity on the basis of electromagnetic induction by continuous motion of either a solenoid or a magnet.

It consists of an armature made of several turns of insulated wire wound on soft iron core and revolving freely on an axis between the poles of a powerful magnet.

Two slip rings are connected to the ends of the armature with two carbon brushes rotating on the slip ring.

In an external circuit the current is at maximum value at 900 and minimum value at2700.

This brings about alternating current and the corresponding voltage (e.m.f) is the alternating voltage. They are used in car alternators and H.E.P.

2. D.c generator/alternator – in this case the commutators replaces the slip rings to enable the output to move in one direction.

After a rotation of 1800, instead of current reversing, the connections to the external circuit are reversed so that current direction flows in one direction.

3. Moving coil microphone – it consists of a coil wound on a cylindrical cardboard which opens into a diaphragm. The coil is placed between the poles of a magnet as shown.

As sound waves hit the diaphragm, they vibrate and move the coil which produces induced current into the coil and then it flows to the loudspeakers.

Eddy currents

They are composed of loops of current which have a magnetic effect opposing the force producing them.

When a copper plate with slits is used the loops are cut off and hence the effective currents are drastically reduced and so is the opposing force.

Practically eddy currents are reduced by laminating metal plates. Armatures of electric generators and motors are wound on laminated soft iron cores.

The lamination slices, which are quite thin are glued together by a non-conducting glue and this reduces eddy currents to an almost negligible value.

Eddy currents are useful in moving coil meters to damp the oscillations of the armature when the current is switched off.

Mutual induction

Mutual induction is produced when two coils are placed close to each other and a changing current is passed through one of them which in turn produces an induced e.m.f in the second coil. Therefore mutual induction occurs when a changing magnetic flux in one coil links to another coil.

Applications of mutual induction

1. The transformer – it converts an alternating voltage across one coil to a larger or smaller alternating voltage across the other. Since H.E.P is lost through transmission lines therefore it is stepped down before it being transmitted and stepped up again at the point of supply lines.

In a step up transformer the number of turns in the secondary coil (Ns) is higher than the number of turns in the primary coil (Np).

In a step down transformer the primary coil has more turns than the secondary coil.

The relationship between the primary voltage and the secondary voltage is given by;

Np / Ns = Vp / Vs.The efficiency of a transformer is the ratio of power in secondary coil (Ps) to power in primary coil (Pp), therefore efficiency = Ps / Pp × 100%.

Examples

1. A current of 0.6 A is passed through a step up transformer with a primary coil of 200 turns and a current of 0.1 A is obtained in the secondary coil. Determine the number of turns in the secondary coil and the voltage across if the primary coil is connected to a 240 V mains.

Solution

Np / Ns = Vp / Vs = Ip / Is = Ns = (0.6 × 200) / 0.1 = 1200 turns

Vp = 240 V hence Vs = (240 × 1200) / 200 = 1440 V

2. A step-up transformer has 10,000 turns in the secondary coil and 100 turns in the primary coil.

An alternating current of 0.5 A flows in the primary circuit when connected to a 12.0 V a.c. supply.

a) Calculate the voltage across the secondary coil

b) If the transformer has an efficiency of 90%, what is the current in the secondary coil?

Solution

a) Vs = (Ns / Np) × Vp = (10,000 × 12) / 100 = 1200 V

b) Power in primary = Pp = Ip × Vp= 5.0 × 12 = 60 W

Efficiency = Ps / Pp × 100% but Ps = Is Vs

Is = (60 × 90) / (1200 × 100) = 0.045 A

Energy losses in a transformer.

Loss of energy in a transformer is caused by;

i) Flux leakage– this may be due to poor transformer design

ii) Resistance in the windings–it is reduced by using copper wires which have very low resistance

iii) Hysteresis losses– caused by the reluctance of the domains to rotate as the magnetic field changes polarity. Reduced by using materials that magnetize and demagnetize easily like soft iron in the core of the transformer.

iv) Eddy currents– reduced by using a core made of thin, well insulated and laminated sections.

Uses of transformers

1. Power stations – used to step up or down to curb power losses during transmission

2. Supplying low voltages for school laboratories

3. Low voltage supply in electronic goods like radios, TVs etc.

4. High voltage supply in cathode ray oscilloscope (CRO) for school laboratories.

3. Induction coil –was developed in 1851 by Heinrich Ruhmkortt. It has both secondary and primary coils with an adjustable spark gap.

4. Car ignition system – it is applied in petrol driven engines where a spark plug is used to ignite petrol vapour and air mixture to run the engine.

Chapter Six

Mains Electricity

Sources of mains electricity

Mains electricity comes from a power station and its current is the alternating current which can either be stepped up or down by a transformer.

A.c is produced when a conductor is rotated in a magnetic field or when a magnetic field is rotated near a conductor.

This method is known as electromagnetic induction. The source of energy for rotating the turbine is the actual source of electrical energy.

Most of the electricity in East Africa is generated from water.

Power transmission

This is the bulk transfer of electric power from one place to another. A power transmission system in a country is referred to as the national grid.

This transmission grid is a network of power generating stations, transmission circuits and sub-stations. It is usually transmitted in three phase alternating current.

Grid input

At the generating plant the power is produced at a relatively low voltage of up to 25 kV then stepped up by the power station transformer up to 400 kV for transmission. It is transmitted by overhead cables at high voltage to minimize energy losses.

The cables are made of aluminium because it is less dense than copper.

Metallic poles (pylons) carry four cables, one for each phase and the fourth is the neutral cable which is thinner and completes the circuit to the generator.

Grid exit

At sub-stations transformers are used to step down voltage to a lower voltage for distribution to industrial and domestic users.

The combination of sub-transmission (33 kV to 132 kV) and distribution (11 kV to 33 kV) which is then finally transformed to a voltage of 240 V for domestic use.

Electricity distribution

This is the penultimate process of delivery of electric power. It is considered to include medium voltage (less than 50 kV) power lines, low voltage (less than 1,000 V) distribution, wiring and sometimes electricity meters.

Dangers of high voltage transmission

1. They can lead to death through electrocution

2. They can cause fires during upsurge

3. Electromagnetic radiations from power lines elevate the risk of certain types of cancer

Electrical power and energy

Work done = volts × coulombs = VQ, but Q = current × time = I t.

So work done = V I t

Other expressions for work may be obtained by substituting V and I from Ohms law as below
V = I R and I = V / R, work done = I R × I t = I2 R t, or work done = V × V t / R = V2 t / R.
The three expressions can be used to calculate work done. Electrical power may be computed from the definition of power. Power = work / time = I2 R t /t = I2 R or V2 t / R t = V2 / R
Using work done = V I t, then Power = V I.

These expressions are useful in solving problems in electricity. Work done or electrical energy is measured in joules (J) and power is measured in watts (W). 1 W = 1 J/s.

Example

An electric heater running on 240 V mains has a current of 2.5 A.

a) What is its power rating?

b) What is the resistance of its element?

Solution

a) Power = V I = 240 × 2.5 = 600 W. Rating is 600 W, 240 V.

b) Power = V / R = 600 W. R = V / I. R = 240 / 2.5 = 96 Ω.

Costing electricity

The power company uses a unit called kilowatt hour (kWh) which is the energy transformed by a kW appliance in one hour.

1 kW = 1,000 W × 60 × 60 seconds = 3,600,000 J. The meter used for measuring electrical energy uses the kWh as the unit and is known as joule meter.

Examples

1. An electric kettle is rated at 2,500 W and uses a voltage of 240 V.

a) If electricity costs Ksh 1.10 per kWh, what is the cost of running it for 6 hrs?

b) What would be its rate of dissipating energy if the mains voltage was dropped to 120 V?

Solution

a) Energy transformed in 6 hrs = 2.5 × 6 = 15 kWh. Cost = 15 × 1.10 × 6 = Ksh 99.00
b) Power = V2 / R = 2500. R = (240 × 240) /2500 = 23.04 Ω.

Current = V / R = (240 × 2500) / (240 × 240) = 10.42 A

Power = V I = (2500 × 120) / 240 = 1,250 W.

2. An electric heater is made of a wire of resistance 100 Ω connected to a 240 V mains supply.

Determine the;

a) Power rating of the heater

b) Current flowing in the circuit

c) Time taken for the heater to raise the temperature of 200 g of water from 230C to 950C. (specific heat capacity of water = 4,200 J Kg-1 K-1)

d) Cost of using the heater for two hours a day for 30 days if the power company charges Ksh 5.00 per kWh.

Solution

a) Power = V2 / R = (240 × 240) / 100 = 576 W

b) P = V I =>> I = P / V = 576 / 240 = 2.4 A

c) P × t = heat supplied = (m c θ) = 576 × t = 0.2 × 4200 × 72.

Hence t = (0.2 × 4200 × 72) / 576 = 105 seconds.

d) Cost = kWh × cost per unit = (0.576 × 2 × 30) × 5.0 = Ksh 172.80

3. A house has five rooms each with a 60 W, 240 V bulb. If the bulbs are switched on fro 7.00 pm to 10.30 pm, calculate the;

a) Power consumed per day in kWh

b) Cost per week for lighting those rooms if it costs 90 cents per unit.

Solution

a) Power consumed by 5 bulbs = 60 × 5 = 300 W = 0.3 kWh. Time = 10.30 – 7.00 = 3 ½ hrs.Therefore for the time duration = 0.3 × 3 ½ = 1.05 kWh.

b) Power consumed in 7 days = 1.05 × 7 = 7.35 kWh. Cost = 7.35 × 0.9 = Ksh 6.62

Domestic wiring system

Power is supplied by two cables where one line is live wire (L) and the other is neutral (N). Domestic supply in Kenya is usually of voltage 240 V. The current alternates 50 times per second hence the frequency is 50 Hz.

The neutral is earthed to maintain a zero potential. The main fuse is fitted on the live wire to cut off supply in case of a default. A fuse is a short piece of wire which melts if current of more value flows through it.

Supply to the house is fed to the joule meter which measures the energy consumed.

From the meter both L and N cables go to the consumer box (fuse box) through the main switch which is fitted on the live cable.

Consumer units within the house are fitted with circuit breakers which go off whenever there is a default in the system.

Lights in the house are controlled by a single or double switch (two way).

In most wiring systems the main sockets are connected to a ring main which is a cable which starts and end at the consumer unit. Plugs used are the three- pin type.

Chapter Seven

Cathode Rays

These are streams of electrons emitted at the cathode of an evacuated tube containing an anode and a cathode.

Production of cathode rays

They are produced by a set up called a discharge tube where a high voltage source usually referred to as extra high tension (EHT) supply connected across a tube containing air at low pressure thereby producing a luminous electron discharge between the two brass rods placed at opposite ends of the tube. These electron discharges are called cathode rays which were discovered by J.J Thomson in the 18th century.

Properties of cathode rays

1. They travel in straight lines

2. They are particulate in nature i.e. negatively charged electrons

3. They are affected by both magnetic and electric fields since they are deflected towards the positive plates

4. They produce fluorescence in some materials

5. Depending on the energy of the cathode rays they can penetrate thin sheets of paper, metal foils

6. When cathode rays are stopped they produce X-rays.

7. They affect photographic plates.

Cathode ray oscilloscope (CRO)

It is a complex equipment used in displaying waveforms from various sources and measuring p.d. It comprises of the following main components; – The cathode ray tubes (CRT) – consists of a tube, electron gun, deflection plates and the time base (TB).

The tube is made of strong glass to withstand the pressure difference between the outside atmospheric pressure and the vacuum inside.

It has a square grid placed in front of it to allow measurements to be made.

The electron gun produces the electrons with main parts consisting of a filament, a cathode, a grid and the anode.

Electrons are produced by the cathode when heated by the filament.

The grid is a control electrode which determines the number of electrons reaching the screen therefore determining the brightness of the screen. The Y-deflection plates deflects the beam up or down.

Clearly observable when low frequency inputs are applied i.e. 10 Hz from a signal operator.

The X-deflection plates are used to move the beam left or right of the screen at a steady speed using the time base circuit which automatically changes voltage to an a.c.

voltage. When time base control is turned the speed can be adjusted to produce a waveform.

Examples

1. If the time base control of the CRO is set at 10 milliseconds per cm, what is the frequency of the wave traced given wavelength as 1.8 cm?

Solution

Wavelength = 1.8 cm. time for complete wave = period = 1.8 × 10 milliseconds / cm

= 18 milliseconds

= 1.8 × 10-2 seconds.

Frequency ‘f’, is given by f = 1 / T = 1 / 1.8 × 10-2 = 100 / 1.8 = 56 Hz.

NOTE:

The television set (TV) is a type of a CRT with both Y and X-deflection plates which control the formation of a picture (motion) on the screen.

The colour television screen is coated with different phosphor dots (chemicals) which produce a different colour when struck by an electron beam.

Chapter Eight

X-rays

X-rays were discovered by a German scientist named Roentgen in 1985. They can pass through most substances including soft tissues of the body but not through bones and most metals. They were named X-rays meaning ‘unknown rays’.

X-ray production

They are produced by modified discharge tubes called X-ray tubes. The cathode is in the form of a filament which emits electrons on heating.

The anode is made of solid copper molybdenum and is called the target. A high potential difference between the anode and the cathode is maintained (10,000 v to 1,000,000 or more) by an external source. The filament is made up of tungsten and coiled to provide high resistance to the current.

The electrons produced are changed into x-rays on hitting the anode and getting stopped.

Only 0.2% of the energy is converted into x-rays.

Cooling oil is led in and out of the hollow of the anode to maintain low temperature. The lead shield absorbs stray x-rays.

Energy changes in an X-ray tube.

When the cathode is heated electrons are emitted by thermionic emission. They acquire electrical energy which can be expressed as E = e V. Once in motion the electrical energy is converted to kinetic energy, that is eV = ½ me v2.

The energy of an electromagnetic wave can be calculated using the following equation
Energy = h f,where h- Planck’s constant, f – frequency of the wave.

The highest frequency of the X-rays released after an electron hits the target is when the greatest kinetic energy is lost, that is h f max = eV.

Lower frequencies are released when the electrons make multiple collisions losing energy in stages, the minimum wavelength, λ min, of the emitted X-rays is given by;

(hc) / λ min = eV.

These expressions can be used to calculate the energy, frequencies and wavelengths of X-rays.

Properties of X-rays

i) They travel in straight lines

ii) They undergo reflection and diffraction

iii) They are not affected by electric or magnetic fields since they are not charged particles.

iv) They ionize gases causing them to conduct electricity

v) They affect photographic films

vi) They are highly penetrating, able to pass easily through thin sheets of paper, metal foils and body tissues

vii) They cause fluorescence in certain substances for example barium platinocynide.

Hard X-rays

These are x-rays on the lower end of their range (10-11 – 10-8 m) and have more penetrating power than normal x-rays.

They are capable of penetrating flesh but are absorbed by bones.

Soft X-rays

They are on the upper end of the range and are less penetrative. They can only penetrate soft flesh and can be used toshow malignant growth in tissues.

Dangers of X-rays and the precautions.

1. They can destroy or damage living cells when over exposed.

2. Excessive exposure of living cells can lead to genetic mutation.

3. As a precautionary measure X-ray tubes are shielded by lead shields.

Uses of X-rays

1. Medicine – X-ray photos called radiographs are used as diagnostic tools for various diseases. They are also used to treat cancer in radiotherapy.

2. Industry – they are used to photograph and reveal hidden flaws i.e. cracks in metal casting and welded joints.

3. Science – since the spacing of atomic arrangement causes diffraction of x-rays then their structure can be studied through a process called X-ray crystallography.

4. Security – used in military and airport installations to detect dangerous metallic objects
i.e. guns, explosives, grenades etc.

Chapter Nine

Photoelectric Effect

Photoelectric effect was discovered by Heinrich Hertz in 1887. Photoelectric effect is a phenomenon in which electrons are emitted from the surface of a substance when certain electromagnetic radiation falls on it.

Metal surfaces require ultra-violet radiation while caesium oxide needs a visible light i.e. optical spectrum (sunlight).

Work function

A minimum amount of work is needed to remove an electron from its energy level so as to overcome the forces binding it to the surface.

This work is known as the work function with units of electron volts (eV). One electron volt is the work done when one electron is transferred between points with a potential difference of one volt; that is,

1 eV = 1 electron × 1 volt

1 eV = 1.6 × 10-19 × 1 volt

1 eV = 1.6× 10-19 Joules (J)

Threshold frequency

This is the minimum frequency of the radiation that will cause a photoelectric effect on a certain surface. The higher the work function, the higher the threshold frequency.

Factors affecting the photoelectric effect

1. Intensity of the incident radiation – the rate of emission of photoelectrons is directly proportional to the intensity of incident radiation.

2. Work function of the surface – photoelectrons are emitted at different velocities with the maximum being processed by the ones at the surface.

3. Frequency of the incident radiation – the cut-off potential for each surface is directly proportional to the frequency of the incident radiation.

Planck’s constant

When a bunch of oscillating atoms and the energy of each oscillating atom is quantified i.e. it could only take discrete values.

Max Planck’s predicted the energy of an oscillating atom to be
E = n h f, where n – integer, f – frequency of the source, h – Planck’s constant which has a value of 6.63 × 10-34 Js.

Quantum theory of light

Planck’s published his quantum hypothesis in 1901 which assumes that the transfer of energy between light radiation and matter occurs in discrete units or packets.

Einstein proposed that light is made up of packets of energy called photons which have no mass but they have momentum and energy given by;

E = h f

The number of photons per unit area of the cross-section of a beam of light is proportional to its intensity. However the energy of a photon is proportional to its frequency and not the intensity of the light.

Einstein’s photoelectric equation

As an electron escapes energy equivalent to the work function ‘Φ’ of the emitter substance is given up. So the photon energy ‘h f’ must be greater than or equal to Φ. If the ‘h f’ is greater than Φ then the electron acquires some kinetic energy after leaving the surface.

The maximum kinetic energy of the ejected photoelectron is given by;

K.E max = ½ m v2max = h f – Φ ……………… (i), where m v2max = maximum velocity and mass.

This is the Einstein’s photoelectric equation.

If the photon energy is just equivalent to work function then, m v2max = 0, at this juncture the electron will not be able to move hence no photoelectric current, giving rise to a condition known as cut-off frequency, h fco = Φ………………. (ii)

Also the p.d required to stop the fastest photoelectron is the cut-off potential, V cowhich is given by E = e V co electron volts, but this energy is the maximum kinetic energy of the photoelectrons and therefore, ½ m v2max = e V co ………….. (iii).

Combining equations (i), (ii) and (iii), we can write Einstein’s photoelectric equation as,
e V co = h f – h fco ………………….. (iv)

NOTE: — Equations (i) and (iv) are quite useful in solving problems involving photoelectric effect.

Examples

1. The cut-off wavelength for a certain material is 3.310 × 10-7 m. What is the cut-off frequency for the material?

Applications of photoelectric effect

1. Photo-emissive cells – they are made up of two electrodes enclosed in a glass bulb (evacuated or containing inert gas at low temperature).

The cathode is a curved metal plate while the anode is normally a single metal rod)

They are used mostly in controlling lifts (doors) and reproducing the sound track in a film.
Photoconductive cells – some semi-conductors such as cadmium sulphide (cds) reduces their resistance when light is shone at them (photo resistors).

Other devices such as photo-diodes and photo-transistors block current when the intensity of light increases.

Photo-conductive cells are also known as light dependent resistors (LDR) and are used in alarm circuits i.e. fire alarms, and also in cameras as exposure metres.

2. Photo-voltaic cell– this cell generates an e.m.f using light and consists of a copper disc oxidized on one surface and a very thin film of gold is deposited over the exposed surfaces (this thin film allows light). The current increases with light intensity.

They are used in electronic calculators, solar panels etc.

Chapter Ten

Radioactivity

Introduction

Radioactivity was discovered by Henri Becquerel in 1869. In 1898, Marie and Pierre Curie succeeded in chemically isolating two radioactive elements, Polonium (z=84) and Radium (z= 88).

Radioactivity or radioactive decay is the spontaneous disintegration of unstable nuclides to form stable ones with the emission of radiation.

Unstable nuclides continue to disintegrate until a stable atom is formed.

Alpha (α) and beta (ϐ) particles are emitted and the gamma rays (ϒ) accompany the ejection of both alpha and beta particles.

The nucleus

The nucleus is made up of protons and neutrons.

They are surrounded by negatively charged ions known as electrons.

The number of protons is equal to the number of electrons. Both protons and neutrons have the same mass.

The weight of an electron is relatively small compared to neutrons and protons.

The number of protons in an atom is referred to as the proton number (atomic number) and denoted by the symbol Z.

the number of neutrons is denoted by the symbol N. Protons and neutrons are called nucleons since they form the nucleus of an atom.

The sum of both the protons and neutrons is called the mass number A or nucleon number.

Therefore;

A = Z + N and N = A – Z.

The masses of atoms are conveniently given in terms of atomic mass units (ᴜ) where (ᴜ) is 1/12th the mass of one atom of carbon-12 and has a value of 1.660 × 10-27 kg.

Hence the mass of one proton is equal to 1.67 × 10-27 and is equal to 1ᴜ.

Radioactive isotopes

Isotopes are elements with different mass numbers but with equal atomic numbers i.e. uranium with mass numbers 235 and 238.

Properties of radioactive emissions

a) Alpha (α) particles

They are represented as , hence with a nucleus number 4 and a charge of +2.
Properties

1. Their speeds are 1.67 × 107 m/s, which is 10% the speed of light.

2. They are positively charged with a magnitude of a charge double that of an electron.

3. They cause intense ionization hence loosing energy rapidly hence they have a very short range of about 8 cm in air.

4. They can be stopped by a thin sheet of paper, when stopped they capture two electrons and become helium gas atoms.

5. They can be affected by photographic plates and produce flashes when incident on a fluorescent screen and produce heating effect in matter.

6. They are slightly deflected by a magnetic field indicating that they have comparatively large masses.

b) Beta (ϐ) particles

They are represented by meaning that they have no mass but a charge of -1.
Properties

1. Their speeds are as high as 99.9% or more the speed of light

2. They are deflected by electric and magnetic fields but in a direction opposite to that of alpha particles.

3. Due to their high speed they have a higher penetrative rate than alpha particles (about 100 times more)

4. They can be stopped by a thin sheet of aluminium

5. Their ionization power is much less intense about 1/100th that of alpha particles.
c) Gamma (ϒ) particles

They have very short wavelengths in the order of 10-10 m and below.

Properties

1. They travel at the speed of light.

2. They have less ionization power than that of both alpha and beta particles

3. They accompany the emission of alpha and beta particles

4. They carry no electric charge hence they are not deflected by both electric and magnetic fields.

5. They have more penetrating power than X-rays.

Detecting nuclear radiation

1. Gold leaf electroscope–the rate of collapse of the leaf depends on the nature and intensity of radiation.

The radioactive source ionizes the air around the electroscope. Beta particles discharges a positively charged electroscope with the negative charge neutralizing the charge of the electroscope. Alpha particles would similarly discharge a negatively charged electroscope.

To detect both alpha and beta particles a charged electroscope may not be suitable because their ionization in air may not be sufficiently intense making the leaf not to fall noticeably.

2. The spark counter – the detector is shown below

This detector is suitable for alpha sources due to the inadequacy of the ionization by both beta and gamma radiations.

By putting the source away from the gauze or placing a sheet of paper between the two one can determine the range and penetration of the alpha particles.

3. Geiger Muller (GM) tube– it is illustrated as below

The mica window allows passage of alpha, beta and gamma radiations.

The radiations ionize the gas inside the tube. The electrons move to the anode while the positive ions move to the cathode. As the ions are produced there are collisions which produce small currents which are in turn amplified and passed to the scale.

The scale counts the pulses and shows the total on a display screen.

After each pulse the gas returns to normal ready for the next particle to enter.

A small presence of halogen gas in the tube helps in absorbing the positive ions to reduce further ionization and hence a quick return to normal. This is called quenching the tube.

4. The solid state detector– this detector can be used to detect alpha, beta and gamma radiations where the incoming radiation hits a reverse biased p-n junction diode momentarily conducting the radiation and the pulse of the current is detected using a scaler.

5. The diffusion cloud chamber – this chamber is simplified as shown below

The bottom of the chamber is cooled by solid carbon (V) oxide to around -800 C and the alcohol vapour from the felt ring spreads downwards.

It is cooled below its normal condensing temperature.

As a particle enters the chamber it ionizes the air in its path and alcohol condenses around the path to form millions of tiny alcohol droplets leaving a trail visible because it reflects light from the source.

Alpha particles leave a thick, short straight tracks.

Beta particles leave thin irregular tracks.

Gamma particles do not produce tracks and since they eject electrons from atoms the tracks are similar to those of beta particles.

Activity and half-life of elements

The activity of a sample of radioactive element is the rate at which its constituent nuclei decay or disintegrate.

It is measured in disintegrations per second or Curie (Ci) units, where
1 Ci = 3.7 × 1010 disintegrations per second

1 micro Curie (µ C) = 3.7 × 104 disintegrations per second.

The law of radioactive decay states that “the activity of a sample is proportional to the number of undecayed nuclei present in the sample”.

The half-life of a radioactive element is the time required for its one-half of the sample to decay.

It is important to note that although the activity approaches zero, it never goes to zero.

Examples

1. The half-life of a sample of a radioactive substance is 98 minutes.

How long does it take for the activity of the sample to reduce to 1/16th of the original value?

b) If the initial number of atoms in another sample of the same element is 6.0 × 1020, how many atoms will have decayed in 50 hours?

Solution

a) 2,400 × ½ × ½ × ½ = 300

Three half-lives have a total of 30 hours, thus half-life = 30 /3 = 10 hours

b) Since half-life = 10 hrs half-lives in 50 hrs = 50/10 = 5 hrs.

So the remaining undecayed atoms are ½ × ½ × ½ × ½ × ½ × 6.0 × 1020

= 0.1875 × 1020, thus

The number of atoms which have decayed = (6.0 – 0.1875) × 1020

= 5.812 × 1020

Nuclear equations

Particles making an atom can be written using upper and lower subscripts where a proton, ‘p’ with charge +1 and mass 1ᴜ, is written as .

A neutron ‘n’ with no charge but with mass 1ᴜ, is written as , while an electron with a charge of -1 and negligible mass is written as . It is important to note that the principles of conservation apply in radioactive decay.

That means that the total number of nucleons (neutrons + protons) must be the same before and after decay. The L.H.S of the equation must be equal to the R.H.S for both total mass and charge.

Effects of radioactive decay on the nucleus

2. Write an equation to show how a radioactive isotope of cobalt ( o) undergoes a beta decay followed by the emission of gamma rays to form a new nuclide X.

Nuclear fission

Nuclear fission is a process in which a nucleus splits into two or more lighter nuclei. This process generates large amounts of energy together with neutron emission. Nearly 80% of the energy produced appears as kinetic energy of the fission fragments.

For example Uranium-235 undergoes nuclear fission when bombarded with slow neutrons releasing 2-3 neutrons per Uranium molecule and every neutron released brings about the fission of another Uranium-235nuclei.

Another substance which undergoes the same process is Plutonium-239.

Substances which undergo fission directly with slow neutrons are known as fissile substances or isotopes.

Applications of nuclear fission

1. They are used in the manufacture of atomic bombs where tremendous amount of energy is released within a very short time leading to an explosion.

2. When this release of energy is controlled such that it can be released at a steady rate then it is converted into electrical energy hence the principle in nuclear reactors.

Nuclear fusion

Nuclear fusion is the thermal combining of light elements to form relatively heavier nuclei. The process requires very high temperatures for the reacting nuclei to combine upon collision.

These temperatures are provided by ordinary fission bombs.

These reactions sometimes known as thermonuclear reactions.

A fusion reaction releases energy at the rate of 3-23 MeV per fusion event i.e. two deuterium (heavy hydrogen) nuclei to form helium.

This 3.3 MeV (energy) produced is equal to 5.28 × 10^-13 J.

Application of nuclear fusion

1. Used in the production of hydrogen bomb. Possible reactions for an hydrogen bomb include;

Hazards of radioactivity and their precautions

(i) Due to the ionizing radiation emitted by radiation materials, they affect living cells leading to serious illnesses. Symptoms of radiation exposures are immature births, deformations, retardedness, etc.

(ii) Their exposure to the environment through leaks may lead to environmental pollution leading to poor crop growth and destruction of marine life.

Applications of radioactivity

1. Carbon dating – through the identification of carbon-14 and carbon-12 absorbed by dead plants and animals. Scientists can be able to estimate the age of a dead organism. Since carbon is a radioactive element with a half-life of 5,600 years archeologists can be able to estimate the ages of early life through carbon dating.

2. Medicine – radiation is used in the treatment of cancer, by using a radioactive cobalt-60 to kill the malignant tissue. Radiations are used in taking x-ray photographs using cobalt-60. Radiations are used to sterilize surgical instruments in hospitals.

Radioactive elements can also be used as tracers in medicine where they determine the efficiency of organisms such as kidneys and thyroid glands.

3. Biology and agriculture

– radioactive sources are used to generate different species of plants with new characteristics that can withstand diseases and drought. Insects are sterilized through radiation to prevent the spread of pests and diseases. Potatoes exposed to radiation can be stored for a long time without perishing.

4. Industry – thickness of metal sheets is measured accurately using radiation from radioactive sources. Recently the manufacture of industrial diamonds is undertaken through transmutation.

5. Energy source – in N. America, Europe and Russia nuclear reactors are used to generate electricity.

The amount of fuel used is quite small hence an economical way of generating electricity energy as compared to H.E.P generation.

Chapter Eleven

Electronics

Conductors, insulators and semi-conductors

i) An insulator is a material or object which resists flow of heat (thermal insulator) or electrical charges (electrical insulators). Examples are paraffin, wood, rubber, plastics etc.

ii) Conductors are materials that contain free electrons which carry an electrical charge from one point to another.

Examples are metals and non-metals like carbon, graphite etc.

iii) Semi-conductors are materials or objects which allow the flow of electrical heat or energy through them under certain conditions i.e. temperature. Examples are germanium, silicon, cadmium sulphide, gallium arsenide etc.

Electronic bond structure

This is the series of “allowed” and “forbidden” energy bands that it y bands that it contains according to the band theory which postulates the existence of continuous ranges of energy values (bands) which electron may occupy “allowed” or not occupy ‘forbidden”.

According to molecular orbital theory, if several atoms are brought together in a molecule, their atomic orbitals split, producing a number of molecular orbitals proportional to the number of atoms.

However when a large number of atoms are brought together the difference between their energy levels become very small, such that some intervals of energy contain no orbitals and this theory makes an assumption that these energy levels are as numerous as to be indistinct.

Number, size and spacing of bands.

Any solid has a large number of bands (theoretically infinite). Bands have different widths based upon the properties of the atomic orbitals from which they arise. Bands may also overlap to produce a bigger single band.

Valence and conduction bands

Valence band is the highest range of electron energies where electrons are normally present at zero temperature.

Conduction band is the range of electron energy higher than that of the valence band sufficient to make electrons free (delocalized); responsible for transfer of electric charge. Insulators and semi-conductors have a gap above valence band followed by conduction band above it. In metals, the conduction band is the valence band.

Band structure of a semi-conductor.

Electrons in the conduction band break free of the covalent bonds between atoms and are free to move around hence conduct charge.

The covalent bonds have missing electrons or ‘holes’ after the electrons have moved.

The current carrying electrons in the conduction band are known as free electrons.

Doping of semi-conductors

Doping is the introduction of impurities in semi-conductors to alter their electronic properties.

The impurities are called dopants. Doping heavily may increase their conductivity by a factor greater than a million.

Intrinsic and extrinsic semi-conductors

An intrinsic semi-conductor is one which is pure enough such that the impurities in it do not significantly affect its electrical behavior.

Intrinsic semi-conductors increase their conductivity with increase in temperature unlike metals.

An extrinsic semi-conductor is one which has been doped with impurities to modify its number and type of free charge carriers present.

N-type semi-conductors

In this case the semi-conductor is given atoms by an impurity and this impurity is known as donor so it is given donor atoms (donated).

P-type semi-conductors

The impurity within the semi-conductor accepts atoms with free electrons (dopants). This forms a ‘hole’ within the semi-conductors.

Junction diodes

Junction refers the region where the two types of semi-conductors meet. The junctions are made by combining an n-type and p-type semi-conductor. The n-region is the cathode and the p-region is the anode.

Forward bias of a p-n junction

It occurs when the p-type block is connected to the positive terminal and the n-type block is connected to the negative terminal of a battery.

The depletion layer of the junction reduces to be very thin to allow the flow of electric current.

Reverse bias of a p-n junction

The negative terminal of the battery is connected to the p-type region while the n-type isconnected to positive terminal.

The depletion layer widens and resists the flow of electrons to minimal or zero (no currentflowing through) when the electric field increases beyond critical point the diode junction eventually breaks down and at this voltage it is referred to as the breakdown voltage. Diodes are intended to operate below the breakdown voltage.

Applications of junction diodes

They are mainly used for rectification of a.c. current for use by many electrical appliances. Rectification is the conversion of sinusoidal waveform into unidirectional (non-zero) waveform.

Half wave rectification

In this case the first half cycle of a sinusoidal waveform is positive and the inclusion of a reverse biased diode makes the current not to flow to the negative side of the wave.

The current therefore conducts on every half cycle hence a half wave rectification is achieved.

The voltage is d.c. and always positive in value though it is not steady and needs to be smoothed by placing a large capacitor in parallel to the load as shown.

Radio transmitter and receiver

Radio waves are produced by circuits that make electrons vibrate and they are known as oscillators which produce varied frequencies. Since radio waves have greater range in air than sound or even light waves they are used as carriers of audio (sound) and visual information (TV) waves.

The sound is first changed into electrical vibrations by use of a microphone or other device then added to the radio carrier wave and this changes the amplitude of the carrier and is called amplitude modulation.

The modulated wave is given out by the transmitting aerial and received by another aerial in a radio or TV when they cause vibrations between the earth and the aerial.

They are then demodulated by a diode and hence heard as a sound or image.

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Betrayal in the City by Francis Imbuga

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