UNITS & MEASUREMENTS

PHYSICAL QUANTITIES

A quantity that has measurable physical properties like length, mass e.t.c. is a physical quantity.

To measure a physical quantity, we need to specify two things; the number and its unit of measurement

For example: A man is 3m tall and has a mass of 55kg

There are two types of physical quantities:

1. Fundamental Quantities

These are the basic quantities that provide the basic unit of measurement. They form the base unit upon which other units depend.

Fundamental quantities	Units	Unit abbreviation
Length	Metre	m
Mass	Kilogram	kg
Time	Second	s
Temperature	Kelvin	k
Electric current	Ampere	A
Amount of substance	Mole	mol
Luminous intensity	Candela	cd

2. Derived Quantities

These quantities are obtained by combining two or more fundamental quantities. When we multiply or divide two or more fundamental quantities, we have a derived quantity.

Derived quantity	Derivation	unit
Area	length × breadth	m2
Speed or velocity	distance/time	m/s
Acceleration	change in velocity/time	m/s2
Force	mass × acceleration	kgm/s2
Density	mass/volume	kg/m3
Work or Energy	force × distance	kgm2/s2

Multiples and submultiples of units

One single unit may be too big or too small to measure some quantities.

For example, the length or the distance between Lagos and London is too big to be measured in metres. Again also, the mass of a cube of sugar is too small to be measured in kilograms. Hence, we can break the units into smaller parts using the prefixes

Multiples	Prefix
× 103	Kilo (k)
× 106	Mega (M)
× 109	Giga (G)
× 1012	Tera (T)

Sub-multiples	Prefix
× 10-2	centi (c)
× 10-3	milli (m)
× 10-6	micro (μ)
× 10-9	nano (n)
× 10-12	pico (p)

Example: What is 3hours + 55minutes + 35seconds in S.I. units

60 minutes = 1 hour
∴ 3 hours = 3 × 60 minutes = 180 minutes

1 minute = 60 seconds
180 minutes = 180 × 60 = 10800 seconds

55minutes = 55 × 60 =    3300 seconds

  35 seconds =                 35 seconds

Total = 14,135 seconds

Example: What is the volume of a cuboid with dimensions 1.6m by 5cm by 2mm

HINT: Volume of cuboid = l × b × h

length = 1.6m

breadth = 5cm = 5 × 10^-2 m = 0.05 m

height = 2mm = 2 × 10^-3 m = 0.002

volume = 1.6 × 0.05 × 0.002 = 0.000016 m³ or 1.6×10^-5 m³

REMEMBER: A number can be represented in Standard form A × 10ⁿ where A is a number between 1 and 10 and n is an integer (+ve or –ve whole number)

Dimensions of Physical Quantities

The dimensions of a physical quantity show how that quantity is made up in terms of the fundamental quantities mass (M), length (L) and time (T)

e.g. speed = distance/time

Since distance is a measure of length(L), then the dimension of speed becomes L/T or LT^-1

Applications of Dimensions

It can be used to determine whether a physical equation is correct or not

Dimensioning helps to derive the exact relationship between physical quantities where the formula cannot be easily obtained

Example: The period T of an oscillation of a simple pendulum depends on the length l, and acceleration g. Determine the exact form of dependence

Let us assume T = l^a g^b     where a and b are constants

Dimension of period T is T

Dimension of length l is L

Dimension of acceleration g is velocity/time = (s/t)/t = s/t² = LT^-2

∴ T = L^a (LT^-2)^b

T = L^a L^b T^-2b simplifies to T = L^a+b T^-2b

Equating the indices for each term on both sides

For T; 1 = -2b…(i) when T¹ is compared with T^-2b

For L; 0 = a + b…(ii) when L⁰ compared with L^a+b

from (i): b = -1/2

Substituting this in equation (ii)

0 = a + (-1/2) ⇒ a = 1/2

Thus, T = l ^1/2 g^-1/2    OR T = √(l/g)

MEASUREMENTS

It is important to measure physical quantities as accurately as possible in order for them to be useful. In this section, we will discuss the instruments for measuring the three basic quantities length, time and mass and some others

Measurement of Length

Length can be measured using the following instruments: metre rule, tape rule, Vernier calipers and micrometre screw gauge. The use of a particular instrument depends on the following factors

1. The distance, size and shape of what is to be measured
2. The degree of accuracy

The unit of measurement of length is metre (m)

Tape Rule

Large distances such as the length of a field or playground can be measured using a tape rule

Metre rule

The metre rule can be measured shorter distances such as the length of a reading table

The metre rule is a standard instrument made of wood, steel or plastic. It is graduated in centimetres and millimetres so it can read to the nearest millimetre  or 0.1cm

Vernier Calipers

Calipers generally are used to measure cylindrical objects e.g. diameter of a log of wood, pipes e.t.c where the metre rule cannot be used conveniently.

The Vernier caliper can measure smaller lengths such as the internal or external diameter of a tube or pipe, the depth of a cavity.

A Vernier caliper can measure to an accuracy of 0.01cm.

It has two scales: the main scale M with fixed jaws at one end and the Vernier scale V with moveable jaws.

The main scale is calibrated in millimetres like a metre rule.

The Vernier scale is constructed by dividing nine of the 0.1cm divisions into ten equal intervals so that each vernier division has a length of 0.09cm

When the zero of M coincides with the zero of V, then the length measured is zero

But as V slides on M, the readings on M and V together give the reading of the distance measured

How to read the vernier caliper

1. Take the reading on the main scale that is just before the zero mark of the vernier scale i.e 1.6 cm
2. Note the mark on the vernier that corresponds with the mark of the main scale and count how many marks from the zero mark of the vernier scale to this point of coincidence i.e. 9 marks. Each of these marks is 0.01 cm so the vernier reading is 0.09 cm
3. Add the readings on the Main and Vernier scales

reading = (1.6 + 0.09) = 1.69 cm

Micrometer Screw Gauge

This instrument measure smaller lengths such as the diameter of a wire, thickness of paper or thin sheet of metal, diameter of a small pebble.

It has an accuracy of 0.001cm.

It consists of a main scale graduated in millimetres and a circular vernier scale which contains 50 divisions.

1 division is equivalent to 0.01mm (or 0.001cm)

How to read the micrometer screw gauge

1. Unscrew the gauge and place the object in between the anvil and the spindle and close the gap by screwing the instrument
2. Record the reading on the main scale to the nearest decimal place e.g. 3.2mm
3. Record the reading on the circular vernier scale e.g. 23 divisions or 0.23mm
4. Add the readings 3.2 + 0.23 to get the complete reading is 3.43mm

Measurement of Time

Time measures the duration of an event. The SI unit of time is the second(s).

The instrument used to measure time in the laboratory is the stop clock or stop watch

The most natural time unit is the solar day. It takes the earth 1 solar day to complete one rotation about its axis

1 day = 24hours

1 hour = 60 minutes

1 minute = 60 seconds

Measurement of Mass

The mass of a body is the quantity of matter contained in it.

The SI unit of mass is kilogram (kg). It can be measured by

1. Beam balance (or weighing balance). It uses the principle of moments to measure mass

2.Spring balance. It uses the principle of elasticity to measure mass

Other instruments used to measure mass include lever balance, dial balance, direct reading balance

Mass and Weight

In general conversation, we talk of weight instead of mass, however, they are not the same in physics

Mass	Weight
Mass is the quantity of matter in a body	Weight is a measure of the gravitational force or earth’s pull on a body
Mass is constant anywhere in the world	Weight varies from place to place. Its value is greater at the poles than at the equator
Mass is a scalar quantity	Weight is a vector quantity
Mass is measured in kilogram (kg)	Weight is measured in Newton (N)

The weight of a body can be calculated using the relation

Weight(W) = mass(m) × acceleration due to gravity(g)

      W=mg

The acceleration due to gravity of the earth is about 9.8ms^-2. So, a mass of 65kg has a weight of approximately 650N

If for instance the body is taken to the moon where g=1.6ms^-2, the mass is still 65kg but the weight is now 65 x 1.6 = 104N

Measurement of Temperature

The device used to measure temperature is called a thermometer.

The figure below shows a laboratory thermometer

A clinical thermometer is used to mesure the body temperature

An infrared thermometer can measure the tempereature of a body without physical contact

Measurement of Current

The device used for measuring current is called the ammeter.

Measurement of Volume

Volume of regular solids

A regular solid is one in which the shape conforms to a definite known shape i.e. A regular is any solid that has the shape of a cube, cuboid, cylinder, sphere, and so on.

For example, the volume can be measured by measuring the relevant lengths using appropriate instruments and applying the known formula for finding the volume of the shape.

The volume of a solid in the shape of a cylinder (e.g. coin, wire) can be obtained by measuring the diameter d of its cross-section and length l (or height h) with appropriate measuring instruments

and then applying the formula V = π (d²/4) h

or πr²h where r = d/2

Volume of Irregular solids

An irregular is one that has no known shape, e.g stones, glass stopper, etc.

The volume is found by immersing it completely in a measuring cylinder containing a liquid in which the solid would not dissolve.

The volume of liquid displaced gives the volume of solid

Volume of a Liquid

The volume of a liquid can be measured by the use of a measuring cylinder, a pipette or a burette

VECTORS

Physical quantities may be divided into two types:

a) Scalars 

b) Vectors

A scalar quantity is one which has only magnitude, for example, distance, speed, temperature, volume etc.

A vector quantity is one which has magnitude as well as direction for example displacement, velocity, force etc.

Displacement is distance in a specified direction. A force is specified by stating its magnitude and its direction of its line of action. Weight which is a force acts vertically downwards

Scalar: the car travelled 5km (distance)

Vector: the car travelled 5km due north (displacement)

Scalars can be added and subtracted easily using ordinary algebra.

Vectors however are added and subtracted by other methods such as vector diagrams, scale drawing or parallelogram law of vector addition

Representation of a vector

A vector can be represented in magnitude by a given length of line and its direction by an arrow head at the end of the line.

It can be represented by a line at an angle to a reference plane (usually horizontal or vertical)

Addition of two vectors

Two or more vectors can be combined to give a single vector. This single vector is called resultant and can be defined properly as follows

The resultant of two or more vectors is a single vector that produces the same effect as the other vectors.

Let us consider the resultant of two vectors under the following conditions

CONDITION 1. IF THE VECTORS ARE IN THE SAME DIRECTION

Consider two force vectors acting on a body in the same direction as shown.

The resultant R is given by

R = F₁ + F₂

And the direction is the same as that of F₁ and F₂

CONDITION 2. IF THE VECTORS ARE IN OPPOSITE DIRECTION

The resultant is given by

R = F₂ – F₁

the direction of the resultant is in the direction of the larger force (assuming F₂)

CONDITION 3. IF THE VECTORS ARE AT RIGHT ANGLES TO EACH OTHER

We complete the rectangle and use Pythagoras theorem

R² = F₁² + F₂²

R = √(F₁² + F₂²)

To know the direction, we find θ i.e tan⁡ θ = F₂/F₁

Where θ = tan^-1⁡( F₂/F₁)

The direction is an angle θ to the F₁ force

Parallelogram Law of Vector Addition

If two vectors are represented in magnitude and direction by the opposite sides of a parallelogram, the resultant is represented in magnitude and direction by the diagonal of the parallelogram drawn from the common point

The law is used for finding the resultant of two vectors when they are inclined at an angle θ other than a right angle

CONDITION 4. IF TWO VECTORS ARE INCLINED AT AN ANGLE θ

We use cosine rule to find the resultant

R² = F₁² + F₂² – 2F₁F₂ cos ⁡(180 – θ)

The direction of the resultant is angle α, the angle the resultant makes with F₂

From sine rule: F₂/sin⁡α = R/sin⁡(180-θ)

Resolution of a Vector

When a vector R acts at an angle θ to the positive Ox direction

It can be resolved into the horizontal Ox direction and the vertical Oy direction by applying trigonometric ratios

sin⁡ θ = R_y/R      vertical component R_y = R sin⁡θ

cos⁡ θ = R_x/R      horizontal component R_x = R cos⁡θ

These two components will produce the same effect as the original vector R

Think of resolution as the opposite of resultant

This can easily be remembered by simply looking at the angle the vector makes with either the horizontal line or the vertical line.

If the vector R makes an angle θ with the horizontal, then the horizontal component is R cos⁡θ and the vertical component is the “opposite” R sin⁡θ.

If it makes an angle θ with the vertical, then the vertical component is R cos⁡θ and the horizontal component is R sin θ.

Resultant of two or more vectors

To find the resultant of several forces F₁, F₂, F_3, F₄, … acting at a point O, we first reduce the whole system to two perpendicular forces and then find the resultant using Pythagoras theorem

Example: Determine the resultant and direction of the forces acting at a point O as shown below.

When solving questions like this, take notice of the direction and the angle of the horizontal or vertical

Let us consider the vertical components first

Sum of vertical components ∑ F_y : 10 + 20 sin⁡30 + (-3) + (-5 sin⁡60)

Notice that the 10N, 20N, 3N and 5N have vertical components. The 10N and 3N are purely vertical forces. The 8N force has no vertical component, it is purely horizontal.

 ∑ F_y = 10 + 10 – 3 – 4.33

 ∑ F_y = R_y = 12.67N

Sum of horizontal components ∑ F_x = 20 cos⁡30 + 8 – 5 cos⁡60

 ∑ F_x = R_x = 22.82N

R = √(12.67² + 22.82²)

R = 26N

Direction of R; θ = tan^-1⁡(R_y/R_x)

θ = tan^-1⁡(12.67/22.82) = tan^-1⁡⁡(0.5552)

    θ = 20⁰

Relative Motion

Relative motion is comparing the motion of two bodies with reference to a point.

If two bodies A and B are moving in a straight line, the velocity of A (V_A) relative to B (V_B) is found by V_A – V_B

If A and B are moving in opposite directions, the relative velocity of A to B is V_A + V_B

Example: A car travelling at 120kmhr^-1 overtakes a bus travelling at 80kmhr^-1 in the same direction. What is the relative velocity of the car to the bus

The relative velocity of the car to the bus OR the velocity of the car relative to the bus is

100 – 60 = 40kmhr^-1

MOTION

Whenever an object changes position, motion is said to have taken place.

Types of motion

A force is a pull or push motion

Motion is caused by a force.

There are 4 types of motion

1.Straight line motion e.g. a book sliding down an inclined table

2.To and fro motion: e.g. a simple pendulum

3.Rotational or circular motion e.g. blades of a fan

4.Random motion e.g. the motion of dust particles

Types of forces

Contact forces: this is a force that needs to touch the body before it is felt. For example: push, pull, friction, reaction, upthrust, air resistance

Non-contact forces: this force need not touch the body before its effect is felt. For example: gravity, magnetic, electrostatic

A force field is a region of space where a body experiences a force

Parameters of motion

1.Displacement (s): this is distance travelled in a specified direction. It is measured in metres (m)

2.Velocity (v): it is the displacement divided by time taken. Unit is m/s. This is speed in a specified direction Acceleration (a): it is the change of velocity divided by time taken. Unit is m/s^2

Terms used to describe motion

Displacement, velocity, acceleration

Displacement means distance moved in a specified direction. It is measured in metres (m)

Velocity/speed is change in displacement/distance within a given time. The unit is m/s speed=(distance moved)/(time taken)

Terms used to describe motion

Displacement, velocity, acceleration

Displacement means distance moved in a specified direction. It is measured in metres (m)

Velocity/speed is change in displacement/distance within a given time. The unit is m/s

speed=(distance moved)/(time taken)

Acceleration is the change in velocity with respect to time. The unit is m/s^2

acceleration=(change in velocity)/(time taken)

A bus increases its velocity from 30m/s to 90m/s in 12s. What is the acceleration of the bus?

acceleration=(change in velocity)/(time taken)=(90-30)/12=60/12 =5m/s^2

Equations of motion

Motion under gravity

A body falling under the influence of gravity has an acceleration due to gravity(g). This type of motion is a typical example of uniformly accelerated motion provided air resistance is neglected.

It was initially thought that bodies falling under the action of gravity alone fall at different rates; a heavy body will fall faster than a light body.

This acceleration due to gravity(g) acts downwards and is constant with a value of g=9.8ms^(-2)  (“approximately” 10ms^(-2)). So when discussing motion under gravity or vertical motion, the following points should be noted.

I.If the motion is upwards, it is going against gravity and g=-ve but if the motion is downward, it is going with gravity and =+ve

II.If the body is projected vertically upward, at maximum height (H_max) there is a momentary stop and v=0 at maximum height.

III.If the body is dropped from a height, then u=0. thus we write the equations of motion under gravity from the standard equations of motion given earlier

v=u+at       ⇒       v=u±gt

v^2=u^2+2as      ⇒      v^2=u^2±2gH s=ut+1/2 at^2       ⇒             H=ut±1/2 gt^2

So in vertical projection; at maximum height

v=0, a=-g, s=H_max, and v^2=u^2+2as  becomes

0=u^2-2gH ⇒H_max=u^2/2g

Time to reach maximum height; 0=u-t⇒t=u/g

The total time of flight T is the time taken for the body to reach the ground again. This is twice the time it takes for the body to reach maximum height

T=2t⇒T=2u/g

If the body is falling from a height H above the ground

u=0, a=+g, s=H and s=ut+1/2 at^2 becomes

H=1/2 gt^2

2H=gt^2

t=√(2H/g)

Example: a ball is thrown vertically upwards from the ground with a velocity of 40ms^(-1). Calculate

i)The maximum height reached

ii)The time to reach maximum height

iii)The time of flight

u=40ms^(-1),a=-g=-10ms^(-2), s=H_m,  v=0

i) “From” v^2=u^2+2as, 0=40^2-2(10) H_m

20H_m=1600

H_m=1600/20=80m

ii) From v=u+at, 0=40-10t

t=40/10=4s

(iii) Time of flight T=2×4=8s

Example: A ball is released from a height of 20m. Calculate i) the time it takes to fall  ii) the velocity with which it hits the ground

s=H=20m, a=+g=10ms^(-2), u=0, t=?

i)  From s=ut+1/2 at^2⇒20=0+1/2.10.t^2

20=5t^2, t^2=20/5=4

t=√4=2s

ii) From v^2=u^2+2as

v^2=0+2(10(20)

v=√400=20ms^(-1)

Projectile Motion

When an object is projected (or thrown) into space at angle of θ to the horizontal, it traces out a curved path. Many examples of projectile motion exist in our daily life

i)A stone shot out of a catapult traces out a curved or parabolic path *show video*

ii)Bullets or shells shot from guns *animate*

iii)The motion of a ball from a free kick in football, or throw of discuss, javelin or shotput *show video*

iv)The missile from a fighter jet to hit a target on the ground

The object that moves through space is called a projectile and the curved path it follows is the trajectory.

Any projectile performs two independent motion

1.A constant or uniform horizontal motion

2.A vertically downward acceleration under the influence of gravity

If a ball is dropped from a height and another ball is thrown horizontally from the same height at the same time, they will both strike the ground at the same time. This is because the acceleration of both bodies is vertical and is independent of the horizontal velocity.

Terms used in describing a projectile

1.Time of flight (T) is the time taken for the projectile to return to the ground or the same horizontal level from which it was projected.

●

2.Maximum height (H) is the highest vertical distance reached by the body.

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3.Range (R) is the horizontal distance covered from the point of projectile to the point it strikes the ground again or reaches the same horizontal level.

To calculate these parameters, we need the initial velocity U and the angle of projection θ (usually measured from the horizontal)

We resolve this initial velocity U into horizontal and vertical component

Horizontal component = U cos⁡θ

Vertical component = U sin⁡θ

The horizontal component is usually unaffected by the vertical acceleration of free fall at maximum height. The vertical component of t he speed is zero i.e. v=0

And from the equation of motion

v=u+at

 v=0, u=U sin⁡θ, a=-g,

t=”time to reach”  “max.” ⁡”height”

0=U sin⁡θ-gt⇒t=(U sin⁡θ)/g

The time taken to rise from the point of projection on the ground to the maximum height is equal to the time to fall from maximum height to ground level.

∴”time of flight ” T=2t

Next is to find the maximum height

From v^2=u^2+2as

v=0, u=U sin⁡θ, a=-g, s=H

0=(U sin⁡θ )^2-2gH⇒2gH=U^2  sin^2⁡θ

To determine the range R, we use the horizontal component of velocity i.e. Ucos⁡θ

From s=ut+1/2 at^2

s=R, u=U cos⁡θ, a=0 “(since there is no horizontal acceleration)”

Then R=U cos⁡θ.t+0

But t=time of flight=(2U sin⁡θ)/g

R=U cos⁡θ.(2U sin⁡θ)/g=(2U^2  sin⁡θ  cos⁡θ)/g

From trigonometric identities, 2 sin⁡θ  cos⁡θ≡sin⁡2θ

The maximum range (farthest horizontal distance) a projectile can attain is when the angle of projection is 45⁰.

Then sin⁡2θ=sin⁡〖(2×45)=sin⁡90 〗

Applications of projectiles

Projectile has a wide range of application in science and technology, sports and warfare.

In science and technology

In sports

1.Kicking of football in air

2.Throwing of basketball, also throwing of discus, shot-put and javelin

In warfare

1.Shooting of arrows, spears or canon balls

2.Launching of missiles from fighter jets or from ground

Simple Harmonic Motion (SHM)

This is a type of periodic motion that occurs very frequently in physics.

It is defined as the to and fro motion of a body whose acceleration is directed towards a fixed point (or equilibrium position) and is proportional to its displacement from the point.

Example of simple harmonic motion

WORK, ENERGY & POWER

WORK

Work is defined as the product of force and distance moved in the direction of the force.

In ordinary conversation though, work may mean school activity, farm or office work.

Work is done when a person pushes a car and it moves a certain distance or a body is raised through a height. Thus,

“Workdone(” W”) = Force” (F)” × distance moved in the direction of the force(” d”)”

W=Fd

The unit of workdone is Joules (J)

An inclined force may move a body horizontally.

In such cases, the force must be resolved along the direction of movement of the body

W=F cosθ × d

Example: A car is pushed along a straight road by a force of 500N. Calculate the workdone in moving the body a distance of 20m.

W = F × d,  F=500N,  d=20m

W = 500 × 20

= 10 000J OR 10 KJ

Work is also done when a body moves through a height. In earth’s gravitational field, there is always a force pulling a body towards the earth’s centre. The weight of the body is the force and the height is the distance moved

Force = Weight = mg

Where m is the mass of the body and g is the acceleration due to gravity

And since Workdone = Force × distance

Workdone = mg × h

W = mgh

Example: A stone of mass 15kg falls through a height of 20m. Calculate the workdone.

 [Take g = 10ms^-2]

W = mgh,  m = 15kg, h = 20m

W = 15 × 10 × 20

W=3000J

Example: A load of mass m slides down a smooth plane inclined at θ to the horizontal from a point A to B. The same mass is the dropped from A to the ground C, a vertical distance h. In which of the two situation is more work done?

*

The force is the weight of the body acting downwards which is mg

From A to C, the workdone is mgh

From A to B, the workdone is mg×AB sin⁡θ.

The inclined distance AB is not in the same direction as the vertical downward force mg. So we must resolve the inclined distance and find its corresponding vertical distance. sin⁡θ=AC/AB

The corresponding vertical distance is AC=AB sin⁡θ  and workdone =mg×AC= mgh

Hence the same work is done in both cases.

Generally, workdone by gravity depends only on the vertical distance moved irrespective of the path taken

ENERGY

Energy is the ability or capacity to do work.

The SI unit is also Joules (J)

A great deal of energy is required to do most human activities.

A car needs to have fuel (energy) before setting out on a journey (work). A player needs to eat food (energy) before playing football (work).

The world depends basically on energy resources for industrial, computer, science and technological developments.

These resources are of two types:

1. Renewable resources

These resources are inexhaustible i.e they do not finish e.g. sun that produces solar energy, wind that gives wind energy. Hydro (water falls) that produces hydroelectricity etc.

2. Non-renewable resources

These are exhaustible resources e.g. petroleum that produces kerosene, petrol or cooking gas, coal or nuclear resources that can be used to generate electricity, food eaten by man produces energy for activities such as reading, playing, farming etc.

Forms of energy

e) Heat energy from hot objects used in heaters.

f) Mechanical energy from moving objects.

g) Sound energy from loudspeakers or bells.

h) Nuclear energy developed from radioactive substances such as Uranium.

i) Elastic energy such as that stored in a spring

Transformation and conservation of energy

The forms of energy listed previously can be transformed or converted from one form to another by suitable machines.

Some examples are

a) The generator (dynamo) converts mechanical energy to electrical energy. A motor does the reverse (i.e. from electrical energy to mechanical energy)

b) The telephone converts sound energy to electrical and back to sound

c) The loudspeaker converts electrical energy to sound energy. The microphone does the reverse

d) When the brakes of a car are applied, the mechanical energy original possessed by the car is converted to heat energy in the tyres and some sound energy.

The principle of conservation of energy states that energy can neither be created or destroyed but can be converted from one form to another.

MECHANICAL ENERGY

A body can possess mechanical energy by virtue of its position or motion.

For example, a car possesses mechanical energy due to its motion and a stone held at a height above the ground possesses mechanical energy due to its position.

The former type of mechanical energy is called kinetic energy and the latter is referred to as potential energy.

Kinetic energy is the energy possessed by a body in motion. The kinetic energy(K.E.) of a body of mass m moving with a velocity v is given by

K.E.=1/2 mv²

Potential energy is the energy possessed by a body by virtue of its position.

A body held at a height h above the ground has potential energy

P.E=mgh

This potential is sometimes referred to as gravitational potential energy.

Examples:

1) A bullet of mass 50g is moving with a speed of 144km/hr. Calculate its kinetic energy.

2) A brick of mass 12kg released from a height of 20m. With what energy does it strike the ground (take g=10ms^-2).

m=50g OR  50/1000 kg=0.05kg, 

v=144km/hr=144/1×1000m/(60×60s)   =40ms^-1)

K.E.=1/2 mv²=1/2×0.05×40²=40J

2) The potential energy of the brick is mgh

m=12kg, g=10ms^-2, h=20m

P.E.=12×10×20

=2400J or 2.4 kJ

This energy is converted to kinetic energy when it strikes the ground.

Conservation of Mechanical Energy

Thus by the principle of conservation of energy,

K.E.+ P.E. = constant at all points

If the velocity of the body just before hitting the ground is v. Loss in P.E. at A is equal to gain in K.E. at C i.e.

mgh = 1/2 mv²

2gh=v²

v=√2gh

Example: A stone of mass 5kg is dropped from a height of 30m. Calculate

i) the potential energy at this height

ii) The kinetic energy of the body just before it strikes the ground

iii) The velocity of impact   [take g=10ms^-2]

m = 5kg,  h = 30m

i) P.E. = mgh = 5 × 10 × 30 = 1500J

ii) K.E. before hitting the ground = initial maximum potential energy

K.E. = 1500 J

iii) v = √2gh = √(2 × 10 × 30) = √600 = 24.5ms^-1

A simple pendulum alternates potential and kinetic energy

At the furthest point (A and B), the pendulum is momentarily still and has no kinetic energy, but maximum potential energy since it is at its highest point.

At point C, its lowest point, there is no potential energy, but it is moving the fastest and has maximum kinetic energy.

The velocity at this point is also given as v=√2gh

Example: A body of mass 3kg falls from rest through a height of 25m. It comes to rest after penetrating a distance of 0.5m into the ground. Calculate the average force exerted by the sand in bringing the body to rest. (Take g=10ms^-2)

m = 3kg,  h = 25m

Loss in P.E is converted to gain in K.E. and

P.E. = K.E. = 3 × 25 × 10 = 750J

This K.E. is converted into workdone by the sandy ground in penetration through a distance 0.5m by an average force F

∴750 = F × 0.5

F = 750/0.5 = 1500N

Example: A ball of mass 4kg is dropped of 30m above the ground. Determine the velocity of the ball when it is 18m above the ground. [take g = 10ms^-2]

At 30m, P.E = mgh, m = 4kg,  h = 30m

P.E.= 4 × 10 × 30 = 1200J

At 18m, total energy = mgh₁₈ + 1/2 mv₁₈²

1200 = (4 × 10 × 18) + (1/2 × 4 × v₁₈²)

1200 = 720 + 2v₁₈²

2v₁₈² = 480

v₁₈² = 480/2 = 240

v₁₈ = √240 = 15.5ms^-1

POWER

Power is defined as the rate of doing work OR workdone per unit time.

The unit of power is watts (W)

P = (F×d)/t

P = F × v

Example: Compute the power output in watts of a machine that lifts 50kg crate through a height of 0.4m in 1  minute. [Take g = 10ms^-2]

m=50kg,  weight = mg = 50 × 10 = 500N

Distance = height = 0.4m,  t = 1 minute = 60 seconds

Power = Workdone/(time taken) = (weight × height)/time

Power = (500×0.4)/60 = 3.33W

MACHINES

A machine is any device that is used to make work easier.

There are several types of machines

1. Levers
2. Pulleys
3. Inclined plane
4. Gears

Levers

A lever is a device that consists of three parts – the handle that produces the effort, the fulcrum (or support or pivot), and the load.

A lever is divided into 3 classes depending on the arrangement of the three partsfirst class levers: Here, the fulcrum is between the load and effort e.g. plier

First Class Levers

Here, the fulcrum is between the load and the effort

Second class levers: The load is between the fiulcrum and the effort

Third Class levers: The effort is between the fulcrum and the load

Gears

EQUILIBRUM OF FORCES

A body is said to be in equilibrium under a system of forces if it does not move or rotate. 

This means that the resultant force on the body is zero.

Moment of a force

It has been established in the earlier chapter that a force is what causes a body to move. But if a body does not move under the action of several forces, it means that somehow the forces have canceled each other and the body is in equilibrium.

Apart from causing motion, a force also causes rotation or movement about a point. This rotation or turning effect is called moment

The moment of a force about a point is defined as the product of the force and the perpendicular distance from the line of action of the force to the point.

Moment = Fx

The unit of moment is Newton-metre (Nm)

If the force is inclined at an angle θ to the horizontal

The moment M is given by

M = F sin⁡θ × x

The point at which the moment acts is the pivot or support or fulcrum.

COUPLE

A couple is a system of two parallel and equal but opposite forces that do not act on the same line.

A couple is seen in the action of turning a water tap on or off. Equal and opposite forces are applied by the thumb and first finger

The moment of a couple is defined as the product of one of the forces and the perpendicular distance between the forces

Moment = Fx

Equilibrium of Parallel Forces

For a body to be in equilibrium under a system of parallel, coplanar forces, the following conditions must be satisfied

1. The resultant of all the forces must be zero i.e. the sum of upward forces must be equal to the sum of downward forces. Condition 1: F₁ + F₂ = F₃ + F₄
2. The algebraic sum of the moments must be zero i.e. the sum of clockwise moment about a point must be equal to the sum of anticlockwise moment about the same point.

This is also called the principle of moments

Clockwise moment:   F₂ x₂ + F₃ x₃

Anticlockwise moment:   F₁ x₁ + F₄ x₄

Condition 2:              F₂ x₂ + F₃ x₃ = F₁ x₁ + F₄ x₄

Example: A light (weightless) bar is pivoted at the centre and weights of 5N and 10N placed 3m and 2m respectively from the pivot on one side are balanced by a weight of 20N on the other side. How far is the 20N weight from the pivot?

The diagram is shown. Let the distance of the 20N weight from the pivot be x

The 5N and 10N force produce anticlockwise moment about the pivot

(5 × 3) + (10 × 2) = 15 + 20 = 35Nm

The 20N force produces clockwise moment about pivot: 20 × x

At equilibrium, clockwise moment = anticlockwise moment (condition 2)

35 = 20x

x = 35/20 = 1.75m

Centre of Gravity

The centre of gravity of a body is defined as the point through which the resultant weight acts.

The position of the centre of gravity depends on the shape of the body. If a body is supported at the centre of gravity, it will remain stable.

We can determine the centre of gravity of regular and irregular shapes.

A. Regular shapes

Types of shapes	Position of centre of gravity
Circular object e.g. disc	Centre of the circle
Square, rectangle, parallelogram	The point of intersection of the diagonals
Triangle	The point of intersection of the lines bisecting the angles
Uniform rod e.g. metre rule	Midpoint of the rod i.e. the 50cm mark in the case of a metre rule

B. Irregular shapes

The position of the centre of gravity of irregular shapes can be found by two common methods

i) Plumbline method: A plumbline consists of a small lead bob supported by a thin cord.

For example, to determine the centre of gravity of an irregular plane sheet of material: make two holes in the material at any two points P and Q close to the edges of the sheet of material and suspend the material and plumbline from a rigid support. Draw the lines PP¹ and QQ¹.

The centre of gravity is the point of intersection of the lines

ii) Balancing method: the object is balanced on a knife edge and a vertical line is drawn on the object from this balance point. The line of balance is used to find the centre of gravity.

The position of the centre of gravity of a uniform rod or metre rule can be found using this method.

TYPES OF EQUILIBRIUM

1. Stable Equilibrium

A body is in stable equilibrium if when slightly tilted or given a slight displacement, it returns to its original position e.g. a cone or funnel resting on its circular base

If the body has a low centre of gravity and a wide base, it will remain in stable equilibrium. E.g. a car

2. Unstable Equilibrium

If the body receives a slight displacement (push), it topples over e.g. a cone resting on its small pointed end, an egg standing on its pointed end.

If the body has a high centre of gravity and a small or narrow base, it is in a state of unstable equilibrum.

3. Neutral Equilibrium

When the body receives a slight displacement, it tends to come to rest in a new position e.g. a cone or funnel resting on its curved surface, a ball or orange rolling on a surface.

The position of the centre of gravity remains unchanged when it moves to a new position

Equilibrium of non-parallel forces

The conditions for equilibrium for a body under non-parallel forces may be stated as follows

1. The algebraic sum of the horizontal components of the forces must be zero i.e. ∑ F_x = 0

2. The algebraic sum of the vertical components equals zero ∑ F_y = 0

Example: Three forces act at pint O in the figure below. If the point O is in equilibrium, find the values of P and Q.

Horizontal component: 10√3 – P cos⁡ 30⁰

= 010√3 = P × √3/2

P = (2 × 10√3)/√3 = 20N

Vertical components: Q – P sin⁡30 = 0

But P = 20N, then Q – 20 sin⁡30 = 0

Q = 20 sin⁡ 30 = 10N

DENSITY and PRESSURE

The density of a substance is defined as the ratio of its mass to its volume i.e. the mass per unit volume

density = mass /volume

We use the Greek letter ‘rho’ (ρ) to represent density. If the mass is represented by m and the volume V, then

ρ = m/V

The S.I. unit of density is kgm^-3 (kilogram per metre cube)

Sometime gcm^-3 is used as a unit in laboratory work

1000 kgm^-3 = 1 gcm^-3

The conversion to kgm^-3 is achieved by multiplying the density in gcm^-3 by 1000

Example: A block of wood of density 0.6 gcm^-3 has a mass of 120g. Calculate the volume of the block.

from ρ = m/V, 0.6 = 120/V

V = 120/0.6 = 200 cm³

Density of some common substances

Substance	Density (×1000 kgm-3)
Gold	19.3
Aluminium	2.7
Brass	8.9
Lead	11.3
Mercury	13.6
Water	1
Paraffin(kerosene)	0.8
Petrol	0.7
Glass(varies)	2.6
Air	0.0013

Relative Density

It is the density of a substance compared with the density of water.

Water is used for comparing the densities of other substances. The relative density (r.d.) is defined as follows

r.d. = density of substance/density of water

r.d. = mass of substance / mass of equal volume of water

and similarly = weight of substance / weight of equal volume of water

The density of water is 1000kgm^-3 (or 1gcm^-3). Relative density has no units.

The relative density of a substance is the same magnitude of the density of the substance in gcm^-3 (but without the units).

The density of mercury is 13600kgm^-3 and its relative density is 13.6. The relative density of water is 1.

Determination of density of a solid

The determination of density of a solid substance involves determining its mass by means of a chemical balance or spring balance and thereafter determine its volume.

The shape of a solid may be regular or irregular.

By finding the ratio of the mass to its volume, we obtain the density of the solid.

1) Regular solid

The volume of a regular solid e.g. cube, cylinder can be found by measuring the dimensions by means of a vernier calliper or micrometer screw guage and the formula for the volume of the solid is applied.

The solid is thereafter weighed on a chemical balance to find its mass.

Example: Find the density of the material of a solid sphere of radius 35mm and has a mass of 1.5kg.

Vol. of sphere, V = 4/3 πr³

r = 35mm = 35/1000 m

V = 4/3 × 3.14 × (0.035)³
V = 0.0001795 m³
ρ = m/V = 1.5/0.0001795 = 8356 kgm^-3

2) Irregular solid

The mass is obtained by direct weighing.

The volume is found by immersing the body in water in a measuring cylinder or a displacement. The body should not be a floating body and not soluble (will not dissolve in water).

The rise in volume measured from the measuring cylinder is the volume of the solid.

The density is again calculated by dividing its mass by the rise in volume in the cylinder.

Determination of relative density of a liquid

A relative density bottle can be used to find the density of a liquid. This is essentially a glass bottle that can be covered properly.

The relative density bottle is weighed empty first and has a mass m.

It is then filled with liquid and weighed m_l.

Finally, it is emptied, cleaned and refilled with water and weighed. This is mass m_w.

From r.d.= mass of substance / mass of equal volume of water

Example: An empty relative density bottle weighs 25g. It weighs 65g when filled with a liquid and 75g when filled with water. Calculate the density of the liquid.

Upthrust

It is common experience that when a heavy object is in water, it becomes lighter. This is noticed when a bucket of water is being drawn out of a well.

The bucket appears very light when it is still under water but becomes very heavy as soon as it is out of water.

This loss in weight is called Upthrust.

Upthrust is the loss in weight experienced by a body when it is completely or partially immersed in a fluid (liquid or gas).

a) Object in air

b) Object partially immersed
c) Object fully immersed

If W is the object in air and T is the weight of object in water, then U is the upthrust given by

U = W – T = loss in weight

Archimedes’ Principle

This principle states that when a body is wholly or partially immersed in a fluid, the weight of fluid displaced is equal to the upthrust

Weight of fluid displaced = upthrust

Example: A solid weighs 0.09N in air and 0.03N in a liquid of density 800kgm^-3. Calculate 

i) upthrust of liquid on the solid

ii) volume of the solid

weight in air (W) = 0.09N, weight in liquid (T) = 0.03N

upthrust(U) = W – T = 0.09 – 0.03 = 0.06N

From Archimedes’ principle:

weight of liquid displaced = upthrust = 0.06N

This weight has a mass of 0.06/g = 0.06/10 = 0.006kg

ρ = m/V ⇒ 800 = 0.006/V

V = 0.006/800 = 7.5 × 10^-6 m³

Volume of liquid displaced(V) = volume of solid

Volume of solid = 7.5 × 10^-6 m³

Example: A spiral spring with a metal extends by 10.5cm in air. When the metal is fully submerged in water, the spring extends by 6.8cm. Calculate the relative density of the metal. (Assume Hooke’s law is obeyed)  [WAEC’ 12]

F = ke  [Hooke′s law]

e₁ = 10.5 cm, e₂ = 6.8 cm

F₁ = 10.5 k

F₂ = 6.8 k

R. density = weight in air / weight of equal vol. of water = F₁/F₂

r.d. = 10.5k/6.8k = 1.54

Determination of relative density of solid from Archimedes’ principle

We can find the density of a solid using Archimedes’ principle.

The solid is weighed in air (W) using a spring balance. It is then suspended and completely immersed in water and the reading on the spring balance is noted (W_w)

Upthrust = W-W_w = weight of water displaced

To find the relative density of a liquid using Archimedes principle

A solid is first weighed in air (W), then in water (W_w), and finally in the liquid (W_L)

Weight of water displaced = W – W_w

Weight of liquid displaced = W – W_L

Since the same solid is used, the volume remains the same and as such, the volume of water displaced is equal to the volume of liquid displaced.

Principle of Floatation

If a body is denser than a liquid it is placed on, the body sinks. A metal spoon sinks in water for example because the density of the metal is higher than the density of water. But when the density of a body is less than that of the liquid it is placed on, the body will sink until a point is reached when the weight of the liquid displaced is equal to the weight of the body. At this point, the body is in equilibrium and is said to be floating.

A body floats when the upthrust exerted upon it by the fluid (liquid or gas) is equal to the weight of the body. This is the principle of floatation.

A simpler way of stating this principle goes thus: A floating body displaces its own weight in a fluid in which it floats.

So the principle essentially states that if a body floats, the upthrust is equal to the weight

Applications of floatation

1. Ships even though they are made of steel still float in water. This is because they are hollow objects containing a large volume of air and the average density is less than that of water

2. A toy balloon that has not been inflated cannot float in air but once air has been blown into it, it begins to float.

3. A hot air balloon used in tourist centres is able to float upwards in air

4. Icebergs (huge chunks of ice) formed in the polar regions are floating in the sea

Hydrometer

The hydrometer is an instrument used for measuring the density of liquids.

It is used in testing the concentration of acid in batteries. The acid in a fully charged cell should have a relative density of 1.25 and 1.15 if the cell has been discharged. This makes it more suitable than a voltmeter that would give the same e.m.f. reading irrespective of the state of charge of the battery.

The hydrometer is also used in testing milk, it is called a lactometer when used for this purpose

The sensitivity of a hydrometer is enhanced by ensuring that the stem is narrower.

Example: A block of wood of density 0.6gcm^-3, weighing 3.06N in air, floats freely in a liquid of density 0.9gcm^-3. Calculate the volume of the portion immersed. (g=10ms^-2) [WAEC’ 12]

A floating body displaces its own weight in a liquid

Weight of liquid displaced = 3.06N

Mass of liquid displaced = 3.06/10 = 0.306kg OR 306g

Vol. of liquid displaced = (mass of liquid displaced)/(density of liquid)

= 306/0.9 = 340cm³

The volume of liquid displaced = volume of wood portion immersed = 340cm³

PRESSURE

Pressure is defined as the force acting perpendicular per unit area of surface.

The S.I. unit of pressure is Pascal (Pa) or Nm^-2

From the above equation, if area(A) is very small, pressure will be large, and then when A is large, pressure will be small.

A woman wearing pointed heels shoes exerts more pressure on the ground than if she was wearing flat heels.

 The area in the former is small, and so the pressure will be large.

The sharp point of a needle or knife pierces more easily even when a small force is applied because the area involved is small.

Example: A force of 80N acts on an area of 5m². what is the pressure exerted on the surface?

Pressure(P) = F/A = 80/5 = 16 Nm^-2

2) Calculate the least pressure exerted on a surface of rectangular block 3m × 4m × 5m by a weight of 200N.

Least pressure is obtained on maximum area i.e. 4m × 5m surface

Least pressure = weight/(maximum area) = 200/(4×5) = 10Nm^-2

Pressure in a Liquid

Consider a column of liquid h metres above the ground.

The volume of liquid to the level h metres is Ah

We know that density (ρ) = mass(m)/volume(V)

m = ρ × V = ρ × Ah

The weight of the liquid will be

W = mg = ρAhg

Pressure P = W/A = ρhAg/A

P = ρhg

Characteristics of liquid pressure

1. The pressure in a liquid increases as the depth of the liquid increases
2. The pressure at the same depth (level) in different liquids varies proportionally with the density of the liquid.
3. The pressure at any point on the same level within a liquid is the same
4. Pressure at any point in the liquid at the same level acts equally in all directions

Pressure enables water to flow into various taps in a building from one tank placed at the rooftop.

Dams are constructed in a such a way that the base are built thicker than the top to withstand a larger pressure (because the pressure is larger at the bottom than at the top)

Example: A jar 0.6m deep is full of liquid of density 1200kgm^-3. At what depth below the surface of the liquid is its pressure equal to 900Nm^-2. (Take g=10ms^-2)

P= ρhg, P = 900Nm^-2, ρ = 1200kgm^-3, h=?

h = P/ρg = 900/(1200×10)

h = 0.075m

Applications of Pressure and Pressure Devices

Pressure has a wide variety of applications. It finds its use in many devices. We may categorize them into

1) Liquid pressure devices e.g. hydraulic press, car braking system, Hare’s apparatus, syringe, siphon

2) Atmospheric and air pressure devices e.g. simple barometer, Fortin barometer, aneroid barometer, bicycle pump, syringe, lift pump, force pump

3) Gas pressure measuring devices as in manometer or Bordon’s guage

Hydraulic Press

It works on the principle that pressure is transmitted equally to all parts of a liquid at the same level. It consists of two cylinders joined by a connecting tube filled with a liquid. The bore of one cylinder is smaller than that of the other. A tight piston is fitted to the small bore and large bore of the cylinders.

A small force f acting on the small piston of cross-sectional area a transmits pressure p=f/a, via the liquid of the large piston in the wide cylinder of area A. A large force F on the area A produced

“Pressure”=f/a=F/A

F=A.f/a

The hydraulic press is used in the printing press where a large force produced presses ink against a paper.

Also used in the hydraulic jack to lift a heavy car or in the textile industry to compress bales of cotton or wool.

*

NEWTONS’S LAWS OF MOTIONS

We have studied the motion of objects alone in previous chapters without considering the forces producing them. This branch of physics that studies this is called kinematics.

In this chapter however, we shall study the effect of forces on bodies that are in motion: dynamics

Dynamics is that branch of physics that studies these effects.

Quite a number of scientists and physicists have contributed enormously to the study of dynamics e.g. Galilei Galileo, who initiated the study of dynamics, Isaac Newton (1642 – 1727) formulated the basic laws of motion that are still valid today.

The fundamental principles of motion are summed up in Newton’s three laws of motion.

Newton’s First law of motion

Every object will continue in its present state of rest or in uniform motion in a straight line unless it is acted upon by an external force.

The law brought in the concept of inertia and is sometimes referred to as the law of inertia. Inertia is the reluctance of a body to move when at rest and also the reluctance to stop moving when in motion.

It is important to realize that once a body is moving with uniform speed in a straight line, it needs no force to keep it in motion provided there are no opposing or external forces.

In general, however, it does not appear the law is obeyed from common experience in practical situations; a ball kicked come to rest after some time.

The law is actually not disobeyed. A body comes to rest when in uniform motion by external forces. E.g. The ball moving in a straight line will gradually come to rest by opposing forces from air resistance and friction (external forces). A body thrown up in the air will come down as a result of the pull of gravity (external force).

We observe that when a moving vehicle is suddenly brought to rest by the application of brakes, the passengers are suddenly jerked forward as they tend to continue in their straightline motion. Passengers in the front seat may collide with the windscreen causing serious injuries unless a backward force is applied. This is why it is advisable to use seat belts.

Newton’s second law of motion

The rate of change of momentum of a body is directly proportional to the applied force and takes place in the direction of the force.

Momentum

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

Momentum(M) = mass(m) × velocity(v)

M = mv

The unit of momentum is kgms^-1

Momentum is an important property of a moving object. A bullet having a small mass 0.01kg, moving with a high velocity of 3000ms^-1 AND a heavy ball of mass 30kg, moving at a small speed of 1ms^-1 have the same momentum of 30kgms^-1

More powerful brakes are required to stop a heavy lorry due to its large momentum than a car with a smaller momentum.

From the second law; force ∝ change in momentum/”time”

Suppose a force F acts on a body of mass m for a time t and causes it to change its velocity from u to v, then from the second law

This constant has been so defined that if a mass 1kg has an acceleration of 1ms^-2, the force is 1N, so k=1 and

F = ma

This equation is recognized as one of the most important equations of physics and a fundamental equation in dynamics

NOTE that the force F must be the resultant force acting on the body.

Example: A resultant force of magnitude 15N acts on a body of mass 250g. Calculate the magnitude of the acceleration. [WASSCE’ 12]

m = 250g or 0.25kg,  F = 15N,  F = ma

a = F/m = 15/0.25 = 60 ms^-2

Example: A car of mass 500kg, moving with a forward acceleration of 5ms^-2 is acted upon by a constant resistive force of 1000N. Calculate the force exerted from the engine to maintain this forward acceleration.

Let F_e be the force exerted from the engine, the resultant force acting on the car, F_e – 1000

F_e – 1000 = 500 × 5

F_e = 2500 + 1000 = 3500N

Impulse of a force

Impulse is the product of a force on an object and the time t this force acts.

Impulse (I) = Ft

The unit of impulse is Ns

This force is usually a large one and the time interval very small. For example, a player applies a force of 200N for 0.1s to kick a ball, the impulse(I) on the ball

is I = 200 × 0.1 = 20Ns

Now from F = (mv-mu)/t

Ft = mv-mu

We see that the quantity Impulse is equal to the change in momentum

I = Ft (change in momentum)

Example: A body of mass 5kg moving with a speed of 30ms^-1 is suddenly hit by another body moving in the same direction thereby changing the speed of the former body to 60ms^-1. What is the impulse received by the first body?

Impulse (I) = change in momentum = mv-mu

m = 5kg, u = 30ms^-1, v = 60ms^-1

I = 5(60-30) = 150Ns

Force and weight

Any body of mass m falling under gravity has an acceleration a = g = 10 ms^-2.

Thus, the force of gravity acting downwards on the body is F=mg. This force is called the weight of the body.

Hence W =mg

Example: a body of mass 5kg is to be given an acceleration of 20ms^-2. Calculate the force required when the acceleration is upwards. [g=10ms^-2]

This force must overcome the downward opposing force of gravity (weight of the body) and still give the body the required upward acceleration

F = mg + ma, m = 5kg, g = 10ms^-2, a = 20ms^-2

F = 5(10) + 5(20) = 50 + 100

F = 150N

Concept of weightlessness (a body in a lift)

Consider a man in a lift (a device used to move people up and down tall building. There are two forces acting on him; his weight W acting downwards and the reaction R of the floor of the lift on the man, acting upwards

Let us look at four scenario situations

1) When the lift is stationary or moving with a constant velocity, it is not accelerating and a = 0

The weight of the body is equal to the reaction of the lift

W = R = mg

2) When the lift moves downwards with an acceleration a:

mg – R = ma

W = mg – ma

The man appears to weigh less under this condition

3) When the lift accelerates upwards with an acceleration a, the man is pulled upwards with an acceleration = a

R – mg = ma

W = R = mg + ma

The man appears to weigh more under this condition

4) When the lift descends with acceleration (as though the lift was dropped or the cable was cut)

 a = g, then W = mg – ma becomes

W = mg – mg = 0

The man appears to have no weight. Under this condition, he is said to be weightless

Example: A man of mass 50kg ascends a building standing on a scale in a lift accelerating at 2ms^-2. what would be the scale reading as he approaches his destination.

Weight of the body due to gravity = mg = 50 × 10 = 500N

W = R = mg + ma

= 50(10) + 50(2) = 600N

Newton’s third law

To every action, there is an equal and opposite reaction.

If we place a book on a table, the weight of the book, (action) acting downwards is equal to the reaction of the table on the book acting upwards.

Again, if a car A hits a stationary car B, the reaction force of B on A is equal to the action force of A on B. Thus both cars are damaged.

Effects of Newton’s second and third law

1) Recoil of a gun: when a bullet is shot out of a gun, the shooter experiences a backwards force of the gun (reaction) as a result of the forward propulsive force of the bullet (action)

If M is the mass of the gun, m is the mass of the bullet, v is the velocity of the bullet and V is the recoil velocity of the gun

MV = -mv

and V = -mv/M

The negative sign of the recoil velocity simply means that the it is opposite the direction of the bullet

2) Rocket propulsion and jet engines: their application is based on the fact that a large mass of very hot gasses produced from the combustion of fuel and air is issued out at high velocity from the nozzle behind the jet or rocket.

Examples

1) A bullet of mass 0.045kg is fired from a gun of 9kg, the bullet moving with an initial velocity of 200ms^-1. Find the initial backward velocity of the gun.

2) A rocket expels gas at the rate of 0.4kgs^-1. If the average force of the gas is 120N, calculate the velocity of the  gas.

1) M = 9kg, V = ?, m = 0.045kg, v = 200ms^-1

MV = mv, V = mv/M = (0.045×200)/9 = 1 ms^-1

The negative sign was omitted. It has a backward velocity of 1 ms^-1

2) Mass of gas expelled per second = 0.4kgs^-1

Recall from second law that

F = momentum change per second=

mass of gas expelled per second × velocity

F = 0.4 × v

but F = 120N, ∴120 = 0.4 × v

v = 120/0.4 = 300ms^-1

Conservation of momentum

When two bodies collide, there is a change in momentum of the individual bodies. However the total momentum does not change, it is conserved.

The principle of conservation of momentum states that when two or more bodies collide, the total momentum before collision is equal to the total momentum after collision.

It is assumed that the system of colliding bodies is a closed system i.e. no net external forces act on the system.

The principle of conservation of momentum is a consequence of Newton’s second and third law.

Consider two bodies of masses m_A and m_B moving towards each other with initial velocities u_A and u_B respectively

Let the acceleration of A be a_A and B be a_B

Then at collision and from Newton’s third law, the force of A is equal and opposite to the force on B

Example: A body C of mass 10kg moving with a velocity of 50ms^-1 collides with another body D, moving in the opposite direction with a velocity of 20ms^-1. If both bodies now move in the direction of C at a velocity of 10ms^-1, calculate the mass of Q.

since momentum is a vector quantity, we must take into consideration the directions of motion.

Let us take the direction of C to be positive, hence the direction of D will be negative

m_C = 10kg, u_C = 50ms^-1, m_D = ?, u_D = -20ms^-1, v_C = v_D = 10ms^-1

Total momentum before collision = total momentum after collision

(10 × 50) + (m_D × (- 20)) = (10 × 10) + (m_D × 10)

500 – 20m_D = 100+ 10m_D

500-100 = 20m_D + 10m_D

400 = 30m_D ⇒ m_D = 400/30 = 13.3kg

Types of collision

Elastic collision

In elastic collision, both momentum and kinetic energy are conserved. If the kinetic energy is the same before and after collisions, it is referred to as a perfectly elastic collision.

If we consider again two bodies A and B of masses m_A and m_B moving with initial velocities u_A and u_B before collision and final velocities v_A and v_B after collision, and the collision is perfectly elastic, then from the principle of conservation of momentum and the conservation of kinetic energy, we can write

m_Au_A + m_Bu_B = m_Av_A + m_Bv_B    and

Examples of nearly perfect elastic collision include collision of molecules, collision of billiard balls.

Also, if a ball bounces off the ground and returns to its original height, the collision is perfectly elastic

Inelastic collision

In this type of collision, the total momentum is conserved but the kinetic energy is not. The kinetic energy usually decreases and is converted or lost to heat, sound or elastic potential energy (deformed).

In elastic collision, the colliding bodies stick together and move with a common velocity after collision.

Thus v_A = v_B = v and

m_Au_A + m_Bu_B = (m_A + m_B)v

Example: Two bodies P and Q are moving towards each other and they collide. P has a mass of 6kg and Q 3kg. The velocities of P and Q before collision were 3ms^-1 and 2ms^-1. If the collision is perfectly inelastic, find the velocity of the two bodies after collision. Find the total kinetic energy of the system before and after collision and hence deduce the loss in kinetic energy

From conservation of momentum

(6 × 3) + (3 × (-2)) = (6 + 3)v

18-6 = 9v ⇒ v=12/9 = 1.33ms^-1

Kinetic energy before collision

1/2 (6) (3)² + 1/2 (3) (-2)² = 27 + 6 = 33J

Kinetic energy after collision

1/2 (6 + 3) (1.33)² = 8J

Loss in kinetic energy: 33 – 8 = 22J

HEAT & TEMPERATURE I

HEAT

Heat is a form of energy that produces a feeling of hotness.

When we place a pot of cold water on a heater, after a few minutes, the water feels hotter when we touch it. We say that heat has flowed from the heater to the cold water.

Uses of heat

• For cooking our food, warming our homes or cars, drying our clothes.
• Vehicles are able to move as a result of heat produced in the engine from the combustion of fuels.
• Industrially, extraction of metals from the ores, melting and shaping of metals, glass and plastics into different shapes is made possible through heat.

The above mentioned work activity that is done by heat confirms that heat is a form of energy, it is referred to as thermal energy.

Temperature

Temperature is the degree of hotness or coldness of a body. 

Heat flow causes temperature change. Heat flows from a body at higher temperature to a body at lower temperature.

Thus, if a hot object is placed in close contact with a cold object, heat flows from the hot object to the cold object. As a result of this, the temperature of the hot object decreases and the that of the cold object increases.

Effects of heat

Whenever heat is applied to a body, various changes can occur:

1. Change in temperature: addition of heat causes a rise in temperature, while the removal of heat causes the temperature of a body to fall.
2. Change of state: addition of heat causes the transition of the state of a body e.g. from solid to liquid, liquid to gas etc.
3. Expansion of the body: when heat is added to a body, the body expands. It contracts when heat is removed
4. Change in chemical properties: heat supplied or removed brings about changes in the chemical properties of a substance.
5. Change in physical properties: heat can cause changes in the electrical resistance, elasticity, density or even colour of a body
6. Changes in pressure and volume: heat added to a fluid (liquid or gas) causes an increase in the pressure and volume of a gas.
7. Thermionic emission occurs: this is the emission of electrons from the surface of a metal as a result of heat.

Differences between heat and tempearature

Heat	Temperature
1) Heat is a measure of the total internal energy of a body	Temperature is a measure of the hotness or coldness of a body
2) Heat flows from a region of higher temperature to a region of lower temperature	Temperature does not flow
3) Heat cannot be directly measured with any instrument. It can only be calculated	Temperature is measured by means of an instrument called a thermometer
4) The unit of heat is Joule (J)	The S.I. unit of temperature is Kelvin (K)

Measurement of temperature

Temperature is measured by an instrument called a thermometer. 

Temperature measurement is very important as it provides for example an indication of good or poor health. The temperature of a human body in good health is 37⁰C.

The various thermometers make use of a substance whose physical property varies in a known way with change in temperature. Such a substance is called a thermometric substance.

Type of thermometer	Thermometric substance	Physical property of substance
1) Liquid-in-glass thermometer	Mercury or alcohol	Change in volume of liquid with change in temperature
2) Gas thermometer	Gas	Pressure of gas (at constant volume) or volume of gas (at constant pressure) changes with temperature
3) Resistance thermometer	Resistance	Increase in resistance of wire with increase in temperature
4) Thermometric thermometer	Two dissimilar metals	Change in current due to change in temperature difference between the 2 metals

Fixed Points and Temperature Scales

Thermometers have two reference temperatures or fixed points and they are the upper fixed point and lower fixed point.

The upper fixed point is the temperature of steam in contact with water boiling at standard atmospheric pressure, 760mm of mercury.

The lower fixed point is the temperature of the melting point of pure ice at standard atmospheric pressure.

The difference in temperature between the two fixed points is called the fundamental interval and the calibration or division of this interval depends on the temperature scale used.

The three basic types of temperature scale used are

1. The Celsius or centigrade scale (⁰C)
2. The Kelvin scale (K)
3. The Fahrenheit scale (⁰F)

The fundamental interval on a Celsius scale has 100 divisions and each division defines one degree. So the lower fixed point is taken to be 0⁰C and the upper fixed point 100⁰C. The Celsius scale is used globally in scientific work.

The S.I. unit of temperature is the Kelvin (K) and its scale is called the absolute or thermodynamic or simply Kelvin scale. 

The fundamental interval also has 100 divisions but a lower fixed point of 273K and an upper fixed point of 373K. Each division corresponds to 1K. There exist a lowest possible temperature on this scale referred to as absolute zero below which nothing can be cooled and is equal to 0K (-273⁰C).

A temperature change of 1⁰C is equal to a temperature change of 1K. Kelvin scale has no negative temperatures and it is measured in units called Kelvin (K) and not in degrees.

Let also T_l and T_u be the lower and upper fixed point temperature of the unknown temperature scale and let θ_l and θ_u represent the lower and upper fixed point temperature on the given scale. Then by interpolation

We only need to remember the fixed point of each scale

Formula method

To convert a temperature from Celsius scale T_c to Kelvin scale T_k.

T_k = T_c + 273

And by simple change of subject, a temperature in Kelvin can be converted to Celsius T_c = T_k + 273

To convert a temperature from Celsius scale T_c to Fahrenheit scale T_f

T_f = 9/5 T_c + 32

T_c = 5/9(T_f – 32)

Example: Convert 35⁰C to a) Kelvin scale  b) Fahrenheit scale

General method

a) Given scale is Celsius θ = 35⁰C, θ_l = 0⁰C, θ_u = 100⁰C

Unknown temperature scale is Kelvin T_x = ?,  T_l = 273K, T_u = 373K

T_x – 273 = 35 ⇒ T_x = 35 + 273 = 308K

b) θ = 35⁰C, θ_l = 0⁰C, θ_u = 100⁰C,   T_x = ?,  T_l = 32⁰F, T_u = 212⁰F

Formula method

a) T_K=T_C+273,  T_C=35^0C

T_K=35+273=308K

b) T_F=9/5 T_C+32,   T_F=9/5 “×” 35+32

T_F=63+32=95^0 “F”

Although the formula method may look straight forward and easy, however understanding the general method helps to easily understand calculations in other types of thermometers

Types of Thermometer

1) liquid-in-glass thermometer

Common liquids used in this type of thermometer are mercury and alcohol, although mercury is more widely used. The thermometer measures temperature by measuring the change in volume of a fixed mass of liquid due to change in temperature. *

A good liquid-in-glass thermometer should have

a)a bulb made of thin glass to enable it quickly assume the temperature of its surrounding or substance to be measured

b)a uniform bore and a narrow capillary tube which makes it possible for a small change in temperature to cause a significant change in length of the mercury column

c)a liquid with high thermal expansivity, high boiling and low melting point, expand or contract uniformly with temperature and should be opaque.

Comparing mercury and alcohol-in-glass thermometers

Advantages of mercury	Disadvantages of alcohol
1) Mercury boils at 3570C and can be used to measure higher temperatures	Alcohol boils at 780C and cannot be used to measure high temperatures
2) Has a higher conductivity and thus responds more rapidly to temperature changes	A relatively poor conductor, expands slowly and responds slowly to temperature changes
3) It is opaque and can be easily seen	Alcohol has to be coloured before it can be seen
4) Does not wet glass, so it responds to falling temperature accurately	Wets glass and can lead to inaccurate readings
5) Mercury is not easily vapourized	Vapourizes easily even at low temperatures

Advantages of alcohol	Disadvantages of mercury
1) Freezes at -1150C and can be used to measure lower temperatures than mercury	Freezes at -390C, cannot measure low temperatures
2) Expands about six times as much as mercury for the same temperature rise	Low expansivity

Examples of liquid-in-glass thermometers

1)Clinical thermometer: This is a form of mercury-in-glass thermometer used to measure the temperature of the human body. It has a short range (35⁰C to 43⁰C). The temperature of a normal healthy person is 37⁰C but may rise to 41⁰C in conditions of high fever.

When placed in contact with the human body, the mercury expands and pushes past the kink or constriction. When the thermometer is taken away from the body, the thread of mercury remains in position as it cannot flow back through the constriction. The temperature of the body can be read. To use the thermometer again, it is jerked vigorously to force the mercury back to the bulb through the constriction.

2) Maximum and minimum thermometer: this type of thermometer records the maximum and minimum temperature over a period. It was invented by James Six in 1782 and is also referred to as Six’s thermometer. It is used in metrological stations and weather studies to record both the maximum and minimum temperatures for the day

*

It consists of a U-shaped capillary tube connected to two large bulbs A and B. both bulbs contain alcohol which are separated by a column of mercury in the bend of the tube. A is completely filled with alcohol and B contains a little air. Two steel indices X and Y are kept in position by light springs above the mercury on both sides of the tube. The thermometer is calibrated on both limbs of the U-tube.

An increase in temperature causes the alcohol to expand and since it expands more than mercury, it pushes the mercury column from A round the thermometer. The mercury column at Y pushes upwards and the reading of the lower end of the steel index Y gives the maximum temperature

Decrease in temperature causes the alcohol to contract. Mercury moves up the minimum area pushing the index X upwards.

The lower end reading of X gives the minimum temperature.

A magnet brought in contact with the mercury column is used to reset each index and the thermometer is ready for use again.

Finding the lower and upper fixed point of a thermometer

The upper fixed point of an unmarked thermometer can be found using a hysometer: a double-walled calorimeter construction as shown

*

The thermometer is placed in the steam in the inner chamber above the boiling water. A manometer is used to measure the pressure of the steam to ensure it is 760mmHg.

This is achieved when the heights of the mercury column are the same.

When the mercury level in the thermometer is steady, a mark is made on the glass at this level.

This is the upper fixed point.

The lower fixed point is determined by placing the thermometer upright in pure melting ice. When the level of the mercury is constant, a mark is made at that level on the glass.

It represents the lower fixed point. The distance between these two fixed points determined can now be divided into 100 or 180 divisions depending on the scale to be used.

Gas thermometer

For very accurate measurements, mercury-in-glass thermometer cannot be used. Also the temperature range from such thermometers (-39⁰C to 357⁰C) is relatively small.

Gas thermometer cover these lapses in that they are used for accurate temperature measurements and has a wide range of about -270⁰C to 1500⁰C. The principle of gas thermometer is based on the fact that at constant volume, the pressure of a gas increases linearly (directly proportional) with temperature.

A simple form of constant volume gas thermometer consist of a large bulb containing air (or a gas such as hydrogen) and connected by a narrow (capillary) tube to a manometer. When the gas is heated in the bulb to a certain temperature, the gas expands and pushes the mercury down in tube B and up in tube A. The right side BD of the manometer is moved up or down to bring the mercury level on the other side to a fixed mass C. The volume of the gas is kept constant in this way. The pressure of the gas is then read from the manometer.

If the distance between the mercury levels in C and B is h, then the gas pressure when the mercury in B is above C is given by P=H+h where H is the atmospheric pressure

However, when the mercury in B is lower than that in C, the gas pressure becomes P=H-h

The thermometer is calibrated by first measuring the gas pressure at 0⁰C when the bulb is placed in melting ice, this is P_0 and then at 100⁰C when the bulb is placed over steam P_100.

A linear graph of pressure versus temperature is plotted using these two measured points

*

Advantages

It is very accurate and highly sensitive. It also has a wide range (-270⁰C to 1500⁰C) and is used in the calibration of other thermometers

Disadvantages

1)It is large and cumbersome and therefore not suitable for measuring temperature of small volume of liquids.

2)The knowledge of the pressure of the gas at the fixed points must be known every time the instrument is to be used.

Example

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Resistance thermometers

The principle of this type of the thermometer is based on the fact that the electrical resistance of a metallic conductor is proportional to its temperature. The most common type of this thermometer is the platinum resistance thermometer.

*

The resistance R_x at a particular temperature θ is measured by employing a Wheatstone bridge. Measurement of the resistance R_0 at 0⁰C (ice point), and R_100 at 100⁰C (steam point) are taken. In most cases, the resistance scale is calibrated to read the temperature in degrees.

From the general method of temperature conversion, we can say that

(R_x-R_0)/(R_100-R_0 )=(θ-0)/(100-0)

θ/100=(R_x-R_0)/(R_100-R_0 )

Resistance thermometers can provide accurate measurement and also, they have a wide range. However, it is not suitable for measuring rapid changes in temperature

*

Thermoelectric thermometers

This thermometer is based on the working principle of the thermocouple. A thermocouple consists of two dissimilar metals e.g. copper/constantan or copper/iron joined together at the ends. One end is placed in hot water and heated (hot junction) while the other end is kept constant by immersing it in melting ice (cold junction) and a galvanometer (a sensitive current measuring instrument) is connected in series with the hot and cold junctions.

The current measured depends on the temperature difference between the hot and cold junctions.

The galvanometer is usually calibrated in such a way as to read the temperature of the hot junction directly

Advantages

1)It is particularly used for measuring rapidly changing temperature

2)Its small size allows it to measure temperature even at a point

3)Employed in industries to measure high temperatures of up to 1100⁰C

*

THERMAL EXPANSIVITY

THERMAL EXPANSION OF SOLIDS AND LIQUIDS

Expansion of solids

Most solids expand when heated and contract when cooled. The rate of expansion and contraction differs for different solids, for example, brass expands more than iron for same temperature rise

Consequences and effects of expansion

1)Railway tracks and bridges

Railway sections have gaps between them to make allowance for expansion due to temperature rise, otherwise the rails would buckle during hot days and trains would be derailed.

Steel bridges fixed at both ends would expand during hot weather and the large force that would be exerted against its fixtures may cause the bridge to fracture. To prevent it, one end of the bridge is fixed and the other end rests on rollers in an expansion gap so that the metal is free to expand and contract during seasons of summer or winter

2)Cracking of glass

A thick glass bottle or drinking-glass will crack if hot water is poured into it. Since glass is a poor conductor of heat, the outer, colder glass expands much less than the hot glass on the inside. To prevent the cracking, only a small quantity of water at a time should be poured in to allow for time for even expansion of the glass inside and outside

*

3)Sagging of telephone and overhead wires

The metals used for making these wires expand and sag during hot weather. They could snap if they are taut to the poles due to expansion. They also contract during cold weather. To avoid their snapping, the wires are usually given a certain amount of sag to give room for expansion and contraction.

Applications of expansion

1)Rivets

The construction of large boilers and building of ships make use of steel plates riveted together using very hot rivets. The hot rivets is pushed through overlapping plates and hammered tight. On cooling, the rivets contract and holds the plates even more tightly together.

It provides a good seal against steam in boilers and against the sea for ships plates.

*

2.Bimetallic strips

One of the most important application of solid expansion is the bimetallic used in many different appliances. It is made by placing two strips of different metals (e.g. brass and iron) side by side and welding them along their entire length.

If the bimetallic strip is heated, it curves as shown

*

with the brass being on the outside and iron on the inside. This shows that brass expands more than iron for the same temperature rise. The strip straightens at room temperature. If it were cooled below room temperature, the brass would contract more than iron

The bimetallic strip has several applications

a) Bimetallic strip thermometer: this thermometer consists of a spiral strip made from brass on the outside and invar (a steel-nickel alloy which expands very little) on the inside. One end of the strip is fixed and the other end is attached to a pointer

Uneven expansion of the metals due to temperature rise causes the strip to curve in a clockwise direction making the pointer move over a calibrated scale to measure the temperature

b) Thermostat

A thermostat is a device used for controlling the temperature . A thermostat is used in electric iron and gas oven. A thermostat is made from a bimetallic strip.

When the current is switched on in an electric iron and the thermostat knob is set to the desired temperature, the current flows and produces heat.

When the desired temperature is attained, the strip (thermostat) is now curved and separates from the contact point, thereby breaking the contact and switching off the current. As the strip cools down and straightens up, it makes contact thus switching on the current again. This make-and-break mechanism controls the temperature of the electric iron.

c) Electric fire alarm

When there is a fire outbreak in a building, the resulting heat causes the strip to bend and make contact with the bell causing it to ring out a fire alarm.

Linear expansivity of a solid

The linear exapnsivity (α) of a solid is defined as the increase in length per unit length per degree rise in temperature.

Mathematically  

α=(l_2-l_1)/(l_1 (θ_2-θ_1 ) )=(increase in length)/(original length×temperature rise)

The unit of linear expansivity α is ⁰C^(-1)  (per degree Celsius) or〖 K〗^(-1)  (per Kelvin)

Where l_1=initial length at temperature θ_1,   l_2=final length at temperature θ_2

α is also called coefficient of linear expansion.

From the above equation, αl_1 (θ_2-θ_1 )=l_2-l_1

Let ∆θ be temperature rise θ_2-θ_1 and ∆l be increase in length l_2-l_1

∆l=αl_1 ∆θ

When we say the linear expansivity of iron is 12×10^(-6) K^(-1), we mean that a unit length of iron increases by 12×10^(-6) unit when it is heated through 1K (or 1⁰C) temperature rise.

NOTE that a unit change in the Kelvin scale corresponds to a unit change in the Celsius scale.

The values of linear expansivity (α) differs from solid to solid. Metals have the highest values of α

Solid substance	Linear expansivity 〖(“in”  K〗^(-1))
Platinum	9×10^(-6)
Iron	12×10^(-6)
Copper	17×10^(-6)
Brass	18×10^(-6)
Aluminium	23×10^(-6)
Lead	29×10^(-6)
Zinc	30×10^(-6)
Invar	1×10^(-6)
Glass	8.5×10^(-6)
Silica	0.4×10^(-6)

Example: A copper rod whose length is 10m at 30⁰C is heated to 90⁰C. Find its increase in length and its new length. [ Linear expansivity of copper = 1.7×10^(-5) K^(-1)]

α”=” 1.7″×” 10^(-5) K^(-1),  l_1 “=” 10m,  θ_1 “=” 30⁰C, θ_2 “=” 90⁰C,  l_2=?   α=(increase in length)/(original length×temperature rise)

1.7×10^(-5)=∆l/(10×(90-30) )

∆l=1.7×10^(-5)×10×60=0.0102″m”

Increase in length = 0.0102m

New length l_2= l_1+increase in length=10+0.0102

“new length”=10.0102″m”

Measuring the linear expansivity of a metal

The expansion of a solid when heated is usually so small that special apparatus is required to measure it accurately.

Area and volume expansivity

Expansion occurs in all directions when solids are heated. That is in length, breadth and in height. Thus, there is an increase in area and also volume of the solid

The area expansivity (β) is the increase in area per unit area per degree rise in temperature

β=(A_2-A_1)/(A_1 (θ_2-θ_1))

β=(A_2-A_1)/(A_1 ∆θ)⇒A_2-A_1=βA_1 ∆θ

A_2=A_1+βA_1 ∆θ=A_1 (1+β∆θ)

A_1 is initial area at θ_1, A_2 is final area at θ_2

The cubic or volume expansivity (γ) is the increase in volume per unit volume per degree rise in temperature

γ=(V_2-V_1)/(V_1 (θ_2-θ_1 ) )

Similarly V_2=V_1 (1+γ∆θ)

V_1 is initial volume at temperature θ_1, V_2 is final volume at θ_2

Let us now show the relationship between linear expansivity α, area expansivity β and cubic expansivity γ

Consider a rectangular sheet of metal of length l_1 and breadth b_1. Let the length now be l_2 and breadth b_2 after it has been heated through a temperature rise ∆θ

Initial area of metal sheet =A_1=l_1 b_1

Final area A_2=l_2 b_2

From area expansivity, A_2=A_1 (1+β∆θ) and from linear expansivity, l_2=l_1 (1+α∆θ), similarly, b_2=b_1 (1+α∆θ)

A_2=A_1 (1+β∆θ)=l_2 b_2

A_1 (1+β∆θ)=l_1 (1+α∆θ).b_1 (1+α∆θ)

A_1 (1+β∆θ)=l_1 b_1 (1+2α∆θ+α^2 (∆θ)^2)

Since α is a small quantity (in the range of 10^(-5)), α^2 will be extremely small and can be approximated to zero.

∴A_1 (1+β∆θ)=l_1 b_1 (1+2α∆θ)

By comparing,  β=2α

Similarly it can be shown that the cubic expansivity γ and the linear expansivity α are related by the relation   γ=3α

Expansion in liquid

All liquids expand when heated. Their expansion is characterized by a change in volume when there is a change in temperature. Generally, liquids expands about ten times more than solids do for the same temperature rise.

*

The expansion in liquids is made complicated by the expansion of their containers. It is therefore necessary to distinguish between the real and apparent cubic expansivity of a liquid.

The real (absolute) cubic expansivity γ_r of a liquid is the increase in volume per unit volume per degree rise in temperature.

The apparent cubic expansivity γ_a of a liquid is the increase in length per unit length per degree rise in temperature when the liquid is placed in an expansible vessel.

Now the apparent cubic expansivity takes into account the cubic expansivity of the material of the containing vessel (γ).

The real expansivity is always more than the apparent expansivity. We can write

γ_r=γ_a+γ

The difference between the real and the apparent cubic expansivtiy is the cubic expansivity of the vessel.

Anomalous behaviour of water

Most liquids expand when heated and contract when cooled. Water however and few other liquids do not expand continously with increase in temperature.

Ice (solid water) expands slightly when heated from -5⁰C to 0⁰C. At 0⁰C it forms water and when it is further heated to 4⁰C, it contracts! (rather than expand). This is abnormal and exceptional, but from 4⁰C to 100⁰C, it expands behaving like most other liquids. Thus water has its least volume at 4⁰C and hence maximum density at 4⁰C.

This anomalous behaviour is beneficial in nature. Ponds, lakes and rivers freeze from the top surface rather than from the bottom. So ice at 0⁰C is formed at the surface of the pond, while water at 4⁰C is still available at the bottom so marine and plant life can survive.

HEAT & TEMPERATURE II

HEAT CAPACITY

Experimental studies have shown that the rise in temperature of a body is related to

a) its mass,

b) the quantity of heat it receives and

c) the material the body is made up of

This relation is expressed mathematically as

Q = mc∆θ

Where Q is the quantity of heat received or supplied to the body, m is the mass of the body, ∆θ is the change in temperature. c is a constant called specific heat capacity which depends on the material of the body.

The unit of Q is Joule (J), m is kg, ∆θ is Kelvin, the unit of c then becomes Jkg^-1K^-1. we can define the specific heat capacity c as the quantity of heat required to raise the temperature of a unit mass of substance through 1K (or 1⁰C) rise in temperature.

The specific heat capacity of water is 4200Jkg^-1K^-1. this means that we require 4200J of heat to raise the temperature of 1kg of water by 1K.

The heat capacity C is the amount of heat required to raise the temperature of a body by 1⁰C or 1K.

Q = C∆θ    C has units JK^-1

Specific heat capacity of some substances

Substance	Specific heat capacity (Jkg-1K-1)
Brass	380
Copper	400
Zinc	380
Iron	450
Aluminium	900
Lead	130
Mercury	140
Ice	2100
Water	4200
Alcohol	2500
Paraffin (kerosene)	2130

Example: What quantity of heat is needed to raise the temperature of 600g of iron from 20⁰C to 90⁰C? (specific heat capacity of iron =460Jkg^-1K^-1)

Q = ?   m = 600g = 0.6kg,  c = 460Jkg-1K-1,  ∆θ = 90-20 = 70⁰C (or 70K)

Q = mc∆θ = 0.6 × 460 × 70

= 19320J or 19.32kJ

Measurement of specific heat capacity of solid

The specific heat capacity of a solid can be measured using two methods

1) Method of mixtures

2) Electrical method

Method of mixtures

This is a more common method. It involves the transfer of heat or heat exchange between a hot metal and water in a vessel called a calorimeter. The calorimeter is usually made of copper or aluminium which are good conductors.

Loss of heat to the surrounding which is to be avoided is usually possible in this experiment. Heat may be lost through the processes of conduction, convection, radiation or evaporation. Heat loss is prevented by conduction when the calorimeter is surrounded by cotton wool (a poor conductor). This is referred to as lagging.

Losses due to convection and evaporation are prevented by providing an insulating lid as a cover fitted with two holes for thermometer and stirrer. When the calorimeter is polished on the inside and outside, heat loss to radiation is prevented.

The solid whose specific heat capacity c is to be measured, usually in the form of a metal block is weighed (m_b). It is then tied to a string and left for some minutes placed in a beaker of boiling water. The temperature of the boiling water is recorded which is the temperature of the metal block, θ_b. The weight of the calorimeter is measured empty mc and measured when it is about two-thirds full with water (m_cw). The temperature of the water is also recorded (θ_cw)

The hot solid is quickly transferred to the water in the calorimeter. It is gently stirred with a stirrer to ensure uniform distribution. The final steady temperature of the mixture is taken (θ_m).

At this point,

heat loss by hot metal = heat gained by water and calorimeter

m_b c (θ_b – θ_m) = (m_cw – m_c) c_w (θ_m – θ_cw) + m_c c_c (θ_m – θ_cw)

The hot metal experiences a decrease in temperature θ_b – θ_m while the water and the calorimeter experience an increase in temperature θ_m – θ_cw

This formula should not be memorized, but rather the heat exchange process should be understood.

The following precautions should be taken

1) The calorimeter should be well lagged to prevent heat losses to the surroundings

2) The hot metal must be transferred quickly to the water in the calorimeter and the mixtures gently stirred thoroughly.

Electrical method

The metal block of known mass m_b is fitted with two holes for a heater and thermometer. The block is thermally insulated from its surroundings by placing it in a lagged jacket. The initial temperature of the block is noted θ_i. The heater is switched on and the current I is allowed to flow for a time t when the temperature rise on the thermometer is at least 15⁰C. The heater is then switched off. The final temperature of the metal block is recorded θ_f.

Using again the principle of conservation of heat energy,

Energy given by heater = heat gained by metal

IVt = m_b c (θ_f – θ_i)

V is the potential difference across the heater, c is the unknown specific heat capacity of the metal block.

The power rating P ( = IV) may also be known of the heater and the specific heat capacity of the metal becomes     c = Pt/m_b(θ_f – θ_i)

2) A certain metal of mass 1.5kg at initial temperature 23⁰C, absorbed heat from an electric heater of rating 75W for 4 minutes. If the final temperature of the metal was 48⁰C, calculate the specific heat capacity of the metal and thus the heat capacity

m = 1.5kg, θ_i = 23⁰C, θ_f = 48⁰C, P = 75W,  t = 4 × 60 = 240sec.

Heat supplied by heater = heat absorbed by metal

Pt = mc(θ_f – θ_i)

75 × 240 = 1.5 × c × (48 – 23)

18000 = 37.5c

c = 18000/37.5 = 480 Jkg^-1K^-1

Heat capacity C = Q/∆θ = 18000/(48-23)

C = 720 J¹K^-1

Latent Heat

When heat is supplied to a solid, the temperature of the solid rises steadily until a point is reached when the solid begins to melt and become a liquid. Further heat will change the liquid to gas. During this transition phase i.e. from solid to liquid or from liquid to gas, the temperature of the substance remains constant even though heat is being supplied to it. This heat which is not visible (as there is no temperature rise) is termed latent heat (it was known previously as ‘hidden’ heat)

Molecular explanation of latent heat

From kinetic theory, we know that the molecules of a solid are held together by strong forces of attraction. So, the latent heat is the energy needed to break these forces that hold the molecules of a solid in regular pattern. The molecules then move about more freely as they are now in liquid state.

Similarly, the latent heat to change a substance from liquid form to vapour form is the amount of energy to overcome the not so strong forces of attraction between the molecules of a liquid.

The molecules then become independent of each other and are now in gaseous form. The latent heat energy needed in this case (from liquid to gas) is much greater than from solid to liquid.

Latent Heat of Fusion

Let us first consider the change of state from solid to liquid.

The specific latent heat of fusion is the quantity of heat required to change a unit mass of substance from solid to liquid without change of temperature.

Specific latent heat of fusion is usually represented as l_f and has units of Jkg^-1

Q = ml_f

Where Q is the quantity if heat, m is mass.

The specific latent heat of fusion of ice is 336000Jkg^-1. this means we require 336000J of heat to change 1kg of ice at 0⁰C to water at 0⁰C.

Example: How much heat is required to melt completely 5kg of ice at 0⁰C (specific latent heat of fusion of ice = 336kJkg^-1)

Q = ml_f,  Q = ?  m = 5kg, l_f = 336kJkg^-1 or 336000Jkg^-1 

Q = 5 × 336000

Q = 1680000J or 1.68MJ

Latent Heat of Vapourization

The specific latent heat of vapourization is the quantity of heat required to change a unit mass of substance from liquid to gas without change in temperature

Q = ml_v

Where l_v is the latent heat of vapourization, Q is quantity of heat and m is the mass

Evaporation and Boiling

We notice that pools of water left on the road after rainfall soon disappear. The water has gradually turned from liquid to gas through the process of evaporation.

Evaporation is a process whereby a liquid turns spontaneously into vapour below its boiling point.

Evaporation takes place at all temperatures from the surface of a liquid. Some liquids evaporate easily; others do not. Liquids such as methylated spirit, petrol and ether which evaporate  very easily have low boiling point and they are called volatile liquids

Various factors affect the rate of evaporation:

1. Temperature of the liquid: the rate of evaporation increases with increasing temperature. We see this when we spread wet clothes on a line. They dry quicker during the day when the sun is out and the temperature is higher than at night.
2. Area of liquid surface exposed: the greater the surface area, the higher the rate of evaporation. A wet cloth or sheet spread out dries faster than when it is folded
3. Pressure: the rate of evaporation decreases as pressure increases.
4. Nature of liquid: liquids with low boiling points evaporate faster than liquids with high boiling point. Ether with boiling point 35⁰C will evaporate faster than mercury with boiling point 357⁰C
5. Wind and dryness of air: the more windy the atmosphere, the faster the rate of evaporation. Also the dryness of air causes faster evaporation. We notice this when we bring wet clothes around the air of a working fan. The clothes dry faster than placed in still air. The draft or wind blows away the water vapour around the cloth causing more evaporation to take place.

Molecular explanations of evaporation

According to kinetic theory, a liquid consists of molecules in constant motion. The molecules have different velocities and move in different directions.

Near the surface of the liquid, some molecules with high velocities (and hence high kinetic energy) in an upward direction break through the surface.

The presence of draft of air or wind sweeps away the escaped molecules of vapour above the liquid surface making way for fresh molecules to escape. The average kinetic energy of the remaining molecules in the liquid is reduced and the temperature is thus lowered.

Thus evaporation brings about cooling of a liquid

Effects and Applications of Cooling Effect of Evaporation

Effects

  If a few drops of methylated spirit or ether is dropped on our skin, it feels cold as the liquid evaporates. The latent heat needed to change the liquid to vapour is absorbed from the skin.

  Water stored in earthen or clay pots is cooler than that stored in glass bottles. Evaporation takes place through the pores of the pot causing the water to cool.

  In the human body, sweat evaporates from the surface of the skin and causes the body to cool.

  Doctors try to reduce the felt by a patient during an injection process by first dabbing the area where the injection is to be applied with methylated spirit of ether to cooling it and thereby numbing that part.

Application

The refrigerator

The working principle of the refrigerator makes use of the cooling effect of evaporation

A volatile liquid called a refrigerant is put into the pipes surrounding the chamber where food is stored. Some common refrigerants include liquid ammonia or Freon (ethyl chloride).

As the liquid evaporates, it absorbs latent heat of vapourization from the surroundings inside the refrigerator thereby cooling the inside and its contents. The vapour formed is compressed into a pump into a condenser outside the refrigerator. The compressed vapour condenses and gives out latent heat and this is lost to the air by using metal cooling fins. The condensed vapour (now the liquid refrigerant) is forced into the pipes where it again evaporates. This continous process keeps the refrigerator chamber cold

Boiling

When liquid in a container is heated, its temperature rises and the liquid evaporates due to temperature increase. As the heating is continued, bubbles of vapour appear in the liquid which rise to the surface. At this point, the liquid is said to be boiling.

The temperature of the liquid remains constant during this period and is the boiling point of the liquid

Differences between evaporation and boiling

Evaporation	Boiling
1) It is the change from liquid to vapour at temperatures below the boiling point	It is the change from liquid to vapour at the boiling point temperature.
2) It takes place at all temperatures	It takes place at a particular temperature (boiling point) for a given pressure.
3) Occurs only at the surface of the liquid	Occurs throughout the liquid.
4) Temperature is not steady during evaporation	Temperature is steady during boiling.

Effects of Pressure and Impurities on Boiling and Freezing Points.

1) Increase in pressure on the surface of a liquid increases the boiling point of the liquid.

The converse is also true; a decrease in pressure lowers the boiling point of the liquid

The pressure cooker makes use of this fact in its application.

The pressure cooker is a strong cooking pot with a tightly fitted lid.

The guage controls the flow of steam. The pressure of the trapped gas increases the boiling point of water inside to about 120⁰C. Food is cooked faster as a higher cooking temperature is achieved.

2) Increase in pressure lowers the freezing point of a liquid (or melting point of a solid). An ice block subjected to high pressure will melt. But when this pressure is removed, the ice refreezes. This process is known as regelation.

3) At high altitudes, the pressure of the air is much lower than at ground levels so aircrafts flying at these altitudes need to be pressurized to keep people in the cabin area at their normal pressure (10⁵Pa). If this wasn’t done, the reduced pressure can cause the blood to tend to boil leading to disastrous effects on the human body.

Astronauts must also wear space suits in space to keep them at the right pressure and to provide oxygen to breathe.

The presence of impurities or dissolved substances in a liquid increases the boiling point of the liquid but reduces the freezing point. If water and common salt are mixed, the mixture will boil at a temperature higher than 100⁰C and freeze at a temperature lower than 0⁰C. Salts are poured on ice formed on road during winter to lower the freezing point. By this, melting of the ice takes place.

Saturated and Unsaturated Vapours

When a liquid evaporates in a closed container, the molecules of the vapour above the surface of a liquid exerts a pressure like those in any other gas. This pressure is called vapour pressure of the liquid.

When the enclosed space above a liquid contains as much vapour molecules such that it can no longer hold any more molecules, the vapour is said to be saturated and it exerts a pressure known as saturated vapour pressure (S.V.P.)

However, if the enclosed space can still contain more molecules then the vapour pressure (at that temperature) is unsaturated.

A saturated vapour is a vapour that is in contact with its liquid in a confined space.

A ‘dynamic equilibrum’ is reached when the number of molecules escaping from the liquid at a given time is equal to the number of molecules entering the liquid at the same time. Saturated vapour pressure increases with temperature.

Boilers used in industries and factories may burst when the saturated vapour pressure (as a result of high temperature) is too large.

Saturated Vapour Pressure and Boiling

The saturation vapour pressure of a liquid at its boiling point is equal to the atmospheric pressure. We may define boiling point of a liquid as follows; the boiling point of a liquid is that temperature at which the saturated vapour pressure is equal to the atmospheric pressure.

Water vapour in the atmosphere

Evaporation always takes place from rivers, lakes and oceans and so there is always some water vapour present in the atmosphere. The amount of this water vapour depends on the temperature and climatic conditions of that place. The amount of water vapour present in the atmosphere is termed as its humidity. Thus the air is very humid if there is a high percentage of water vapour in the atmosphere.

Very high humidities or very low ones are not suitable as it makes the body fell uncomfortable.

WAVES

Definition: A wave is a disturbance which travels through a medium from one point to another carrying energy along with it without causing any permanent displacement of the medium.

The study of waves is of great importance. We are able to see and hear people by means of light and sound waves. Energy can be transferred from a source through a medium with the help of waves and without the actual movement of the medium.

However, not all waves require a material medium to carry them. Light waves, radiant heat, radio waves etc do not need any material medium for propagation.

Classification of waves

There are two major classes to waves: electromagnetic and mechanical waves.

Mechanical waves are those waves that require a material medium for their propagation. Examples of such waves are water waves, sound waves, waves on a string or rope.

Electromagnetic waves on the other hand are those waves that do not require a material medium for propagation e.g. light waves, radio waves, X-rays etc.

Transverse and longitudinal waves

We can categorize waves based on the direction of source vibration with respect to the direction of travel of the wave.

A transverse wave is a wave which travels perpendicularly to the direction of the vibration of the source. E.g. water waves, * waves generated by plucking a string, * light waves

A longitudinal wave is one which travels in a direction parallel to the vibration of the source e.g. sound waves *

Production of mechanical waves

Water waves and ripple tank

Waves on a slinky spring

General representation of a wave

A wave can be represented in a graphical form.

y represents the direction of vibration of particles, while x represents the direction of travel of the wave. t is the time taken for the wave to travel

A cycle/oscillation/vibration is a complete to-and-fro movement of a particle of the wave.

A crest is the highest point on a wave.

A trough is the lowest point on a wave.

Terms used to describe a wave

1)Amplitude (A): This is the maximum displacement of the particles of the wave from their equilibrium (or rest) position. It is measured in metres.

2)Period (T): It is the time required for the wave to travel one complete cycle or oscillation. It is measured in seconds.

3)Frequency (f): It is the number of complete cycles or oscillation that the wave makes per second. The unit of frequency is Hertz

4)Wavelength (λ): This is the distance between two successive crests or two successive troughs. It is also the distance covered by the wave in one complete cycle. It is represented by λ (lamda) and is measured in metres.

5)Wavespeed (v): It is the distance the wave travels per unit time. Measured in ms^(-1) (metres per second)   v=fλ

6)Phase:  Two particles on a wave are said to be in phase if they are at the same vertical distances from their rest position and are moving in the same direction. e and g are in phase, f and h are in phase but e and f are out of phase

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Relationship between T, f, λ and v

From the definition of T and f; in T seconds, 1 cycle is performed. In 1 second, 1/T cycles is performed

f=1/T  “or” T=1/f

From the definition of wavespeed,

v=fλ

v=”distance travelled by wave” /”time taken”

Example: A wave travels at 300ms^-1 with a frequency of 20Hz. Calculate the period and wavelength of the wave.

v=300ms^(-1), f=20Hz

T=1/f=1/20=0.05s

v=fλ⇒300=20×λ

λ=300/20=15m

Example: From the displacement-time graph of a wave shown

Amplitude = maximum displacement = 2cm

Period (T) = time taken for one cycle = 0.04s

f=1/T⇒f=1/0.04=25Hz

Wavelength λ =v/f=5/25=0.2m

Equation of a travelling wave

Compare the general representation of a wave with a sine wave*

The equation of a travelling wave can be represented as

y=A sin⁡〖(ωt-kx)〗

Where A is the amplitude

ω=”angular frequency” = 2πf

k=”wave number”=2π/λ

We may substitute and write the equation as

y=A sin⁡〖(2πft-2π/λ x)〗

y=A sin⁡〖(2πft-2π/λ x)〗

y=A sin⁡(2πfλ/λ t-2π/λ x)

y=A sin⁡〖2π/λ(vt-x)〗

Whichever equation is mastered is okay but the initial equation is suggested.

The important thing to know in all these equations is the coefficient of t(=ω) and x(=k) not the sign.

Example: A travelling wave is represented by the equation y=0.2 sin⁡〖(30x-4t)〗 where the symbols have their usual meanings. Find the

i) amplitude   ii) wavelength  iii) frequency  iv) period  v) speed of the wave 

y=0.2 sin⁡〖(30x-4t)〗
y=A sin⁡〖(ωt-kx)〗

A=0.2m

ω=2πf=4

f=4/2π=0.64 Hz

Progressive and stationary wave

Waves which spread out continously are called travelling or progressive waves.

A stationary wave is obtained when two progressive waves of equal amplitude and frequency are travelling in opposite directions.

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Most stationary waves are obtained as a result of reflection of the incident wave. The incident wave and the reflected wave now form a stationary wave.

A node (N) is a point on a stationary wave where there is no movement of the medium. An antinode (A) is a point of maximum movement of the medium.

Examples of stationary waves include waves obtained by plucking a string fixed at both ends. The transverse waves travel both ways along the string and is reflected at the fixed ends.

 Waves set up in open and closed pipes are stationary (longitudinal) in air.

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Longitudinal stationary waves produced from wind instruments like flute, trumpet, saxophone form musical notes. Transverse stationary waves produced musical notes in stringed instruments such as guitar, piano etc.

Properties of waves

Waves show the following properties or characteristics: reflection, refraction, diffraction, interference, polarization.

1) Reflection

This is the ability of waves to bounce back to the same medium when they encounter an obstacle or a surface.

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All waves undergo reflection

2) Refraction

This is the change in the speed and direction of the waves as they move from one medium to another

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During refraction, the frequency of the wave remains the same but the wavelength and wave speed changes.

3) Diffraction

This is the ability of waves to bend around obstacles or spread out when they pass through narrow openings.

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We can hear noise from another classroom or simply at the other side of a building or mountain because the sound waves have bent around the obstacles

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Diffraction occurs when the wavelength of the incoming wave is longer than the width of the opening or size of the obstacle

4) Interference

This is the effect produced when two waves of the same frequency, amplitude and wavelength travelling in the same direction are superimposed (placed one top of the other). They are two types of interference;

i) Constructive interference: If the two waves have travelled the same distance, or if one wave has travelled further than the other by a whole number of wavelength, then the combined wave will have twice the amplitude of the others

ii) Destructive interference: if one wave is ahead of the other by λ/2 (half wavelength), such that the crest of one wave arrive at the same time of the trough of another, the resultant wave will have a zero magnitude

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5) Polarization

We know that a transverse wave is one in which the vibration is perpendicular to the direction of propagation of the wave. In a string wave, the elements of the string move in a plane perpendicular to the string. Thus plane-polarization occurs when the vibrations of particles of a transverse wave remain parallel as the wave propagates.

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NOTE that only transverse waves (e.g. light waves) can be polarized. Longitudinal waves (e.g. sound waves cannot be polarized.

Polarization of light

i)Polarization by tourmaline crystals: unpolarized light can be made to pass through polarizers such as tourmaline crystals, quartz or a manufactured material called polaroid to produce plane-polarized light.

ii.Polarization by reflection: it has been found that when unpolarized light is incident at an angle of 57⁰ on a polished glass surface, the reflected light is plane-polarized

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Applications of polaroids

1)The polaroid is used in sunglasses and windscreens to reduce the intensity of incident sunlight.

2)It is also used to eliminate light glare as a result of reflection from glass doors, window panes, and from the road ahead on a sunny day

LIGHT WAVES & OPTICS

Light and its sources

The study of optics is very important in the field of physics. Light is a wave that carries energy. Light as a wave does not require a material medium for propagation. No medium is required for us to receive light from the sun.

There are different sources of light

Luminous objects: These are objects that give off light on their own. It includes the sun, stars, fire flies which are natural sources and lamps, fires, candles etc., that are man-made sources of light.

Non-luminous objects: These objects do not give off light by themselves but reflect the light from luminous objects for example the moon. The moon is seen by reflection of the sun rays from its surface. Also a book, a person’s face and most objects are non-luminous objects. We are able to read the print from a book during daylight because light from the sun falling on the page is reflected  by the white page but absorbed by the black print.

Transmission of light

Light is transmitted through a vacuum. The path travelled by light is called a light ray. A collection of light ray is called a beam. A light is indicated by a straight line with an arrow head.

There are three types of beams

Rectilinear propagation of light

It can be demonstrated that light travels in a straight line.

Place an illuminated lamp L in front of a small hole A punched on a piece of cardboard. Move two other pieces of cardboards B and C until the light through A is seen.

If we pass a thread through holes A, B and C and the thread is held taut, it is straight showing that the three holes are in a straight line.

This phenomenon is described as the rectilinear propagation of light.

This is what is responsible for the formation of shadows and eclipses and the operation of the pin-hole camera

Shadows

A shadow is produced when light is obstructed by an opaque object. When light falls on this opaque obstacle, the rays just grazing its edges form the outline of a shadow. The shade under a tree or canopy is as a result of the obstruction of sunlight.

The kind of shadow formed depends on the size of the source of illumination.

A point source of light produces a sharp shadow while a large source of light produces a full shadow (umbra) and a partial shadow (penumbra)

Types of objects with respect to optics

1)Transparent: these are objects that allow light to pass through them and objects can be clearly seen. E.g glass

2)Translucent: these objects allow light to pass through them but objects cannot be clearly seen. E.g. tinted glass

3)Opaque: they do not allow light to pass through them and objects cannot be seen e.g. book, stone

blue cup – opaque, green cup – translucent, and transparent cup

Eclipses

The earth and the moon are opaque objects. The sun being a luminous body i.e. producing its own light illuminates certain parts of the earth at certain times. The moon revolves round the earth, and the earth also revolves round the sun.

At particular times during these movements, all three bodies lie on a straight line. If the moon is between the sun and the earth at this time, people in certain places of the earth cannot see the sun; a solar eclipse or eclipse of the sun is formed.

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If the earth is between the sun and the moon, it is referred to as eclipse of the moon or lunar eclipse.

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Annular eclipse: this occurs when the earth and the moon are in positions where the end rays of the sun intersect before reaching the earth. A ring of light is formed around the moon

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Annular eclipse provides astronomers an opportunity to study the outer part of the sun.

Pin-hole camera

This was the camera used in the 15^th century. It consists of a light-tight box; one end has a small hole made with a pin or needle-point. The opposite end has a photographic paper or film. Its operation makes use of the fact that light travels in straight lines. Light from an object in front of the pin-hole passes through the hole and forms an inverted image on the film since the amount of light entering the camera is small.

If the hole is bigger, a brighter but blurred image is produced.

If the distance between the object and the hole (O_d) is increases greater than the image distance (I_d), then the image is diminished. On the other hand, the image is enlarged when the image distance I_d (from the pin-hole to the film) is greater than the object distance O_d (object to pin-hole).

With many pin-holes or a wide hole, the image becomes blurred. Each pin-hole produces its own image on the screen and the overlapping of the image producing blurring.

Magnification

This is the ratio of the image distance to the object distance or the image height to the object height.

“magnification” m=(image distance)/(object distance)=(image height)/(object height)

m=I_d/O_d =I_h/O_h

Magnification has no units

For a pin-hole camera, the length of the camera is the image distance.

Example: The length of a pin-hole camera is 15cm. It is used to photograph an object 60cm away from  the hole and 90cm high. Calculate the object of the image and magnification produced

Magnification m=(image distance)/(object distance)=(length of camera)/(object distance)

=15/60=0.25

m=(image height)/(object height)⇒0.25=I_h/90

I_h=90×0.25=22.5″cm”

Reflection of light at plane surfaces

When light falls on a surface, it is either absorbed, transmitted or reflected. Sometimes a combination of the processes occur. If the surface is highly polished or shinny, then the light rays will be reflected. The rays which strike the surface are referred to as incident rays while the rays thrown back from the surface are reflected rays.

There are two types of reflection: regular and irregular reflection

Regular reflection: Parallel rays of light that are incident on a surface that is smooth or polished are reflected as parallel rays in one direction.

Mirrors produce regular reflection

Scattered/irregular or diffused reflection: In this type of reflection, parallel rays of light are incident on a rough surface such as the surface of a book or cloth and are reflected in different directions.

Laws of reflection

1)The incident ray, the reflected ray and the normal all lie on the same plane.

2)The angle of incidence is equal to the angle of reflection

Image formation by a plane mirror

The characteristics of the image formed by a plane mirror are as follows:

1)The image is the same size as the object

2)It is as far behind the mirror as the object is in front

3)The image is laterally inverted

4)It is virtual

5)It is upright or erect

Number of images formed by inclined mirrors. When an object is placed between two inclined mirrors, a number of images are formed depending on the angle between the mirrors. Let us consider an object O placed between two mirrors M_1 and M_2 inclined at an angle θ=90^0 to each other. An observer sees three images of the object.

In general, the number of images formed by inclined mirrors is related according to the formula

n=360/θ-1

Where n is the number of images and θ the angle between the two mirrors.

When θ=180^0, the two mirrors act as a single mirror and only one image is formed

When θ=0^0, that is when the two mirrors are parallel, an infinite number of images are formed (n=360/0-1=∞)

You may have noticed an infinite number of images in a saloon where dressing mirrors are placed at opposite sides of the wall.

Effect of mirror rotation on a reflected ray

Let us fix an incident ray at an angle i to a  plane mirror MNK. The angle of reflection is also i. The angle between the incident ray and the reflected ray (ONR) is the sum of the incidence angle and reflected angle “i.e”.” ” i+i=2i.

If the mirror is now rotated through an angle θ (the incident ray remaining fixed), the normal is also rotated through θ. The new angle of incidence is now i+θ and consequently, the angle of reflection is i+θ.

The angle in this case between the incident ray and the reflected ray is i”+” θ+i”+” θ=2i+2θ The angle of rotation of the reflected ray is 2i+2θ-2i=2θ

If the incident ray is fixed and a plane mirror is rotated through an angle θ, the reflected ray is rotated through an angle 2θ.

Applications of plane mirrors

1)Plane mirrors are used as dressing mirrors and inner driving mirrors

2)They are used in making simple periscope: a device used to view objects above a barrier or obstacle that is hidden from direct view.

The periscope is used in armoured tanks by soldiers to see above the armour plating outside

The simple periscope has a major disadvantage in that it produces multiple images.

3.Kaleidoscope: This is a children’s toy that produces multiple images based on the principle of inclined mirrors.

Mirror galvanometer: a plane mirror is used to measure small angles of rotation of the coil in this galvanometer that measures very small electric currents

A plane mirror is used in the sextant: a navigation instrument for measuring the angle of elevation of the sun.

CURVED MIRROR & LENSES

REFLECTION OF LIGHT BY CURVED MIRRORS

Apart from plane mirrors, curved mirrors also have many important scientific and practical uses.

There are two main types of curved mirrors

a)Concave or converging mirror

b)Convex or diverging mirror

A curved mirror is formed by cutting out a part of a sphere made of glass and one side is silvered.

If the inside surface of this spherical part is painted silver and the outside part becomes the reflected surface, we have a convex or diverging mirror.

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If the outside surface is silvered and the inner surface is the reflecting part, we have a concave or converging mirror.

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DON’T FORGET: a concave mirror is also called a converging mirror.

Then a convex mirror is called a diverging mirror

Terms used in curved mirrors

Apperture: this is the width AB of the mirror.

Pole(P): it is the centre of the reflecting surface of the mirror

Centre of curvature(C): this is the centre of the sphere from which the mirror part was taken.

Principal axis is the line PC from the pole to the centre of curvature.

Radius of the curvature (r):it is the distance CP from the centre of curvature to pole.

Principal focus(F) of a concave mirror is the point at which all incident rays parallel and close to the principal axis converge after reflection.

The principal focus of a convex mirror is the point at which all rays parallel and close to the principal axis  appear to diverge after reflection.

The principal focus is said to be a real focus because reflected rays actually pass through it. If a screen is placed at this point, a bright spot of light is obtained. On the contrary, the principal focus of a convex mirror is virtual as reflected rays do not pass through; no image is formed on a screen placed at this point.

This is why we say that it appears to diverge from this point.

Focal length (f): this is the distance FP – from the principal focus to the pole of a curved mirror.

It is found by measurement and experiment that

f=r/2

Focal length is half of the radius of curvature for both types of mirrors

Formation of images of a curved mirror

The size, position and nature of the image formed by a curved mirror depends mainly on the position of the object from the pole of the mirror (object distance).

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Convex mirror images

If the concave mirror is replaced with a convex mirror, an image that is smaller than the object, upright and virtual is produced for any position of the object.

A quick way to remember this is by using the acronym SUV: Smaller Upright Virtual

Ray diagram of images

The images obtained from curved mirrors can always be drawn to scale using a ray diagram.

The object O is drawn to scale as a straight line with an arrow-head perpendicular to the principal axis. We can also represent the mirror by a perpendicular straight line.

The image is found by the intersection of any two out of the following three rays that occur

1.A ray parallel to the principal axis passes through the principal focus F after reflection.

2.A ray through the centre of curvature strikes the mirror and is reflected back along the same path.

3.A ray through principal focus F strikes the mirror and is reflected parallel to the principal axis.

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When the object is placed between the pole P and principal focus F,

the image is virtual, magnified and upright.

Determination of focal length of curved mirrors

1)Quick but approximate method:

A quick method is the use of a concave mirror to focus the image of a distance object such as the window of a room or lab on a screen which may be the wall. The mirror is adjusted until a sharp image is seen on the wall.

The distance between wall and the mirror gives an approximate value of the focal length of the mirror, the ray will converge at the principal focus

2) Focal length from measurement of radius of curvature

An illuminated object (cross wire) from a ray box is placed in front of a concave mirror mounted on a wooden stand. The mirror is moved front and back until a sharp image is obtained on the front surface of the ray close to the cross wire. The distance between the mirror and the cross wore object gives the radius of curvature, r, of the mirror.

The focal length, f, is then given by f=r/2

The experiment should be repeated two or three times and the average of the radius of curvature obtained from each procedure is then used to obtain the focal length

3) Mirror formula

The objet distance, u, the image distance v and the focal length f of a curved mirror have been shown to be related according to the equation

1/u+1/v=1/f

This equation can be used to obtain the focal length of a concave mirror

Mirror formulae: sign convention

When the mirror formula is used 1/u+1/v=1/f, f=r/2, it is necessary to use positive (+) of negative (-) sign to each of the distances according to a sign rule.

The two common sign conventions are shown.

Real-is-positive

1) All objects are real so object distance u is always positive. If image is real, then image distance v is positive.

2) Virtual images are at a negative distance from the mirror

3) A concave mirror has a real focus so focal length f is therefore positive.

4) A convex mirror has a virtual focus hence its focal length f is negative.

New Cartesian

1) Distances measured to the right are positive

2) Distances measured to the left are negative. The object is always placed on the left of the mirror

3) A concave mirror has a real focus which is left of the mirror. Hence its focal length is negative

4) A convex mirror has a virtual focus which is right of the mirror. Its focal length is positive.

If any of these sign conventions is used, it must be used consistently

The first sign convention (Real-is-positive) is recommended

Magnification

The magnification of a mirror is given by

m=v/u=(image distance)/(object distance)

Example: An object is placed 15cm in front of a concave mirror of radius of curvature 20cm. Calculate the position, nature and magnification of the image produced.

Real-is-positive	New Cartesian
u=+15, f=r/2=20/2=+10  (concave mirror) From mirror formula 1/u+1/v=1/f, 1/15+1/v=1/10 1/v=1/10-1/15=1/30 v=30 Since this is a positive distance, it means that the image is real. m=v/u=30/15=2	u=-15 concave mirror has a real focus, thus f=-20/2=-10 Mirror formula: 1/u+1/v=1/f 1/(-15)+1/v=1/(-10) 1/v=-1/10+1/15=-1/30 v=-30 So the image is formed 30cm on the left, the same side as the object and is therefore real m=(-30)/(-15)=2″(same result)”

Example: An object is placed 12cm in front of a diverging mirror and an image is produced 3cm behind the mirror. Calculate the focal length of the mirror.

Using real-is-positive sign convention,

Diverging mirror = convex mirror, u=+12, v=-4

(because a convex mirror ALWAYS produces a virtual image)

From 1/u+1/v=1/f, 1/12-1/4=1/f

1/f=-2/12

f=-6cm

The negative sign is a confirmation that the focus is virtual which typical of a convex mirror

Applications of curved mirrors

CONCAVE

1) Concave mirrors are used as shaving (or dental) mirrors because a man sees an enlarged image (erect and virtual) of his face when he places his face near the mirror (between the principal focus and the pole)

2) They are used as reflectors in reflecting telescopes

3) They are used in microscopes

CONVEX

4) Convex mirrors are used as side mirrors in cars because they give an erect image of an object behind the driver or for observing traffic behind. In addition, such mirrors provide a wide range of view.

These mirrors however give a false impression of the distance between the object behind and the driver.

These days, a warning sign is usually printed on the side mirrors that reads like “OBJCTS ARE CLOSER THAN THEY APPEAR”

PARABOLIC

5) These special types of curved concave mirrors are used in car headlamps and search lights because they produce parallel beams of light.

LENSES

A lens is a transparent glass material of thickness varying from the middle to the edges with a curved surface on one or both sides. It causes a beam of light to converge or diverge.

Lenses have many important uses to science and life

Types of lenses

There are two kind of lenses

1) Converging or convex lens

2) Diverging or concave lens

1) Converging lens: these lenses are thicker at the centre than at the edges. They make parallel beam of light converge at a point. The various types of converging lens are shown

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2) Diverging lens

It makes parallel bea, of light appear to diverge from a point. They are thinner at the middle than at the edges.

Lenses which are biconcave or biconvex are the common ones used in the laboratory. Plano-concave and plano-convex are used in optical instruments as they have only one curved surface.

Converging or diverging meniscus lenses are used as contact lenses to correct certain eye defects because they fit the curvature of the eyeball

Our study is primarily on biconvex or biconcave. The human eye has a natural converging lens that is biconvex.

Definition of terms used to describe a lens

Optical centre C is the point through which light rays pass without being deviated by the lens.

Principal axis is an imaginary line through the optical centre that joins the midpoint of the opposite faces of the lens

Principal focus F of a converging lens is the point on the principal axis to which all rays parallel and close to the principal axis appear to converge after refraction through the lens

Principal focus F of a diverging or concave lens is the point from which all rays parallel and close to the principal axis appear to diverge after refraction through the lens

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Note that the principal focus of a converging lens is on the other side from the incident rays, while that of a diverging lens is on the same side as the incident rays and the rays do not actually pass through it.

From the rays diagrams, we see that

Focal length f is the distance from the principal focus to the optical centre of a lens

The power of a lens P is the reciprocal of its focal length f when it is measure in metres.

p=1/f ,  f is in metres

Unit of power is dioptres

The shorter the focal length, the greater the power of the lens

Determination of the focal length of a converging lens

1) Quick but approximate method

Hold the lens near the back of a room or lab and move it until a sharp image of a distant object e.g. window panes is formed on a sheet of paper (screen) or the wall.

Measure the distance from the lens to the paper. This distance is approximately equal to the focal length f of the lens

2) Plane mirror and illuminated object method

Mount the converging lens on a holder and place a plane mirror behind it (both should be vertical). An illuminated object e.g. cross wire from a ray box is placed in front of the lens. This object is adjusted by moving to and fro until a sharp image is obtained on the surface beside the illuminated object. The distance from the object to the lens gives the focal length

3) Lens formula

The lens formula is given by

1/u+1/v=1/f

Where u is the object distance

v is the image distance and f is the focal length

Determine first an approximate value of the focal length f using the method described in 1.

Now place an illuminated object from a ray box at a distance of about 1.5f in front of the lens.

Adjust the screen behind the lens until a sharp image of the object is formed. Measure the exact distance of the illuminated object to the lens u and the distance from the screen to the lens v

Repeat the experiment for four other values of u

Now apply the lens formula to get values for f for each value of u and v.

The average of the values of f gives the focal length of the lens.

Alternatively, plot a graph of 1/u against 1/v. The intercept on either axis gives the value for 1/f. Hence the reciprocal of this intercept gives the focal length.

Ray diagrams in lenses

If we know the focal length and the object distance, we can determine the position and nature of the image by scale drawing of the ray diagram. Three rays can be used to obtain the image. They are

i) a ray from the object, parallel to the principal axis and converges to the principal focus after passing through the lens

ii) A ray from the object that passes through the optical centre without deviating or converging

iii) Ray through the principal focus and emerges parallel to the principal axis after refraction through the lens.

When constructing ray diagrams, we represent the object and lens by a straight line perpendicular to the principal axis. The object and image line should have an arrow head.

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Example: An object is placed 20cm in front of a converging lens of focal length 15cm. Find by scale drawing

i) the image distance

ii) Its magnification and description of the image

Lens formula and sign convention

Two sign conventions are used in the application of the lens formula

Lens’ formula: 1/u+1/v=1/f

Real-is-positive

1/u+1/v=1/f

Object distance u is always positive

f is positive for converging lens and negative for diverging lens

Virtual image has v as negative

New Cartesian

1/v-1/u=1/f

u is always negative as the object is placed on the left side of the lens

f is positive for a converging lens and negative for a diverging lens

Distance to the left are negative while distance to the right are positive.

It is suggest to the student to adopt and learn the Real-is-positive sign convention

Magnification =(image height)/(object height)=((image distance))/(object distance)=v/u

Example: An object is placed 10cm from a converging lens of focal length 15cm.

(i) Calculate the image distance and state the nature of the image.

(ii) What is the power of the lens?

Using Real-is-positive sign convention

u=10(converging lens), f=+15

From lens formula, 1/u+1/v=1/f

1/10+1/v=1/15

1/v=1/15-1/10=-1/30

v=-30cm

This means that the image is formed 30 cm in front of the lens (the same side as the object). It is virtual, magnified and upright.

Note that when a virtual image is formed in a converging lens, it is erect or upright not inverted like a real image

New Cartesian

u=-10, f=+15

1/v-1/u=1/f

1/v-(-1/10)=1/15    →   1/v+1/10=1/15

1/f=1/15-1/10=-1/30

v=-30cm  (same result)

Power P=1/f, but f=15/100=0.15m

P=1/0.15=6.7dioptres

From the lens formula, 1/u+1/v=1/f, multiply through by v: v/u+v/v=v/f

m+1=v/f

Example: The image in a convex lens is upright and magnified four times, calculate the object distance if the focal length is 20cm.

m=v/u=4  →   v=4u

f=20cm

It is given that an upright image is formed. This means that the image is virtual and using Real-is-positive sign convention and lens formula,

1/u+1/v=1/f

1/u-1/4u=1/20

(4-1)4u=1/20   →   3/4u=1/20

4u=3×20

u=60/4=15cm

Take care with the signs

Application of lenses: optical instruments

Microscope: simple and compound microscope

The microscope is used for enlarging the image of a small object not easily seen with the human eye.

The simple microscope is a magnifying glass. It is used for reading small print and for observing specimens in biological studies. It consist of a converging lens with a handle

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When the object is placed between the principal focus and the centre of the lens, an enlarged erect and virtual image is seen.

The shorter the focal length, the greater the magnifying power of the lens

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Normally, the human eye can see objects distinctly when it is at a distance of 25cm and beyond. A very small object placed at 25cm from the eye is however not easily seen due to the small visual angle. When a magnifying glass is used and brought closer to the eye, the clear magnified image of the object is now seen at about the 25cm point.

SOUND WAVES

Sound is a form of energy. It is a form of wave that is conveyed from a vibrating body to a listener through a medium. Sound waves are mechanical waves in that they require a material medium to travel. Sound waves are also longitudinal waves in that the direction of travel of the vibrating source is parallel to the direction of propagation of the wave

When a train is approaching a station, we hear the vibration of the rails before the train comes to sight

The metal (solid) of the rail carries the sound and we hear it with our ears through the air (gas)

Transmission of sound

It will be shown that sound waves cannot travel in a vacuum (empty space)

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Suspend an electric bell in a glass jar. The base and the top of the jar is sealed with grease to make it airtight. A vacuum pump is connected to the jar.

When the bell is connected to a battery and switched on, we hear the bell ring because there is still air in the jar. Now switch on the pump to remove the air and the sound dies down. The clapper is still seen to be vibrating but no sound is heard.

The clapper is seen vibrating because light can travel in a vacuum but sound cannot be heard because sound cannot travel in a vacuum.

This is the reason astronauts communicate to each other through radio waves  (a form of light waves) even when they are close together because there is very little atmosphere on the moon or in space

Propagation of sound waves in air

Sound waves travel through air by the displacement of air particles

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Velocity of sound waves

It has been stated in waves previously studied that the wavelength λ, frequency f, and wave speed v are related according to the equation

v=fλ

This is also true for sound waves

Example: Calculate the velocity of the sound wave produced when a note having a frequency of 100Hz and a wavelength 0.33m is played.

v=fλ

f=100Hz, λ=0.33

v=1000×0.33=330ms^(-1)

The velocity of sound waves is however different in different media. It varies with the density or elasticity of the medium. In air, the speed of sound is about 330ms^(-1). In water, sound travels at about 1500ms^(-1), while in a metal, it has a speed of about 5000ms^(-1). Temperature, wind direction also affect the velocity of sound.

During a thunderstorm, we see a flash lightning before we hear the accompanying sound of thunder because light travels faster than sound. The speed of light is about 3×10^8 ms^(-1) (300,000,000ms^(-1)) compared with 330ms^(-1) for sound.

So if the thunder is heard 6s after the lightning is seen, it means the storm is about 330×6≅2km away

Reflection of sound

Sound wave just like light waves can be reflected when they strike a surface or an obstacle. The laws of reflection of light waves hold true for sound waves. Reflection of sound produces echoes.

An echo is the sound heard after the reflection of sound waves from a plane surface.

If you stand close to wall and clap your hands, the sound waves hit the wall and bounce back, and you hear the sound a second time. If you shout in an empty room or hall, you also hear the echo.

Echoes can sometimes constitute a nuisance and be undesirable but in some other cases they are very helpful and have important applications.

Listening to speech of music outdoors can be uninteresting as the loudness dies off with distance. If the echo is not controlled in a n auditorium or record studio, reflection of sond can occur from the walls, floors, ceiling leading to the production of multiple echoes and it confuses the listener. This continuous occurrence of echoes is called reverberation.

Reverberation is reduced by padding the ceiling and walls with soft perforated bords to minimize the reflection of sound.

Application of echoes

1. Determination of speed of sound in air

Echoes can be used to determine the speed of sound in air by directing a sound wave to a large high wall and measuring the time taken for the sound to return to the sender after hitting the wall

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2. Echo sounding devices

a) Depth sounding: ships are able to determine the depth od the sea by sending a sound signal down to the sea bed and detecting with a receiver the time the echo is received after reflection from the sea bed.

The velocity of sound in water is known v_w, then depth d  is given by

d=(v_w t)/2

b) Exploration of oil and other mineral resources

Geophysicists make use of echoes in the exploration of oil and gas. By setting off an explosion below the earth’s surface, the resulting sound produced is reflected by underground layers of rock at different rates depending on the nature and thickness of the rock

Interpretation of these results will reveal to the scientist possible oil-bearing rock formation or mineral resources

c) Detection of submarine

Echo sounding devices sometimes referred to as sonar is used in detecting enemy submarines or large objects in water or even a school of fishes.

Sound effects

Interference of sound

When two similar sources of sound interfere constructively, a loud sound is produced due to a combination of the amplitude of each of the sound wave.

When they interfere destructively, no sound is heard because the crest of one of the sound wave falls on the trough of the other and their effect cancel each other

Beats

When two sound waves with slightly different frequencies combine, the amplitude of the resulting wave will have a maximum and a minimum. The intensity of the sound produced rises and falls with time. These alterations in the loudness with time give rise to beats.

The intensity is proportional to the square of the amplitude. This process is termed waxing and waning.

The periodic variation in the loudness of a sound heard when two notes of nearly equal frequency are played simultaneously is called beats.

If the frequencies of the sound waves are f_1 and f_2, the beat frequency f is given as the difference between the frequencies of the two waves i.e. 〖f=f〗_2-f_1

Example: two notes of frequency 132Hz and 128Hz are sounded simultaneously, calculate the frequency of the beat produced

Beat frequency f=f_2-f_1=132-128=4Hz

Uses of beats

1. Beats are used to determine the frequency of a tuning fork. When a tuning fork of known frequency f_k is sounded together with a fork of unknown frequency f_u and the beat frequency f is measured, the unknown frequency can be calculated using the relation f_u=f_k+f

2. Beats are also used to tune musical instruments e.g. a piano

In achieving this, a piano tuner listens for beats when he strikes a particular key on the piano and the sound produced from a standard tuning fork. If no beat is heard, it means the piano has been tuned. If not, the tension in the string is adjusted

Doppler effect

This is the change in frequency (pitch) of a source when there is relative motion between the source and the observer. This phenomenon occurs both in light and sound waves

You notice that the sound from a train or the siren from a police car increases as it approaches towards your direction. After it passes you, there is a sudden drop in the pitch. This is the doppler effect.

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Characteristics of sound

It is necessary to understand some fundamental characteristics of sound especially in music.

1. Pitch

The pitch of a note is its position on the musical scale. Pitch is that characteristic that differentiates a high note from a low note. The pitch of a note depends on the frequency of the sound wave. In a piano, the right hand keys produce notes of high pitch than the left hand keys

A tight string produces a higher pitched or higher frequency sound when plucked than a loose string.

The sounds from young boys or girls can be differentiated from that of older people during an event because the young boys have a relatively higher pitched voice

Generally, the human ear can hear sounds conveniently between 400Hz to about 20,000Hz. Some animals such as dogs, cats can hear notes of much higher frequency

2. Loudness or intensity

Intensity of a sound is the rate of flow of energy per unit area perpendicular to the direction of the sound wave. Loudness is a sensation in the mind of a particular observer. It depends on the intensity of sound. Since intensity depends on amplitude, loudness is also a function of amplitude.

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Loudness also depends on the mass of air which can set into vibration

When the prongs of a tuning fork are vibrated in the air, only a small ass of air vibrate and a soft sound is heard. The sound is however observed to be loud when the vibrating prongs are now set on a table. This is because a large mass of air in contact with the surface of the table is now set into vibration

The strings on a violin or guitar are not at all loud when set into vibration on their own.

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This is why they are attached to a sounding box of large surface area.

This is also the principle of a loud speaker.

The ear must be placed close to a telephone ear piece to hear the sound distinctly. This is because of the small mass of air displaced by the circular metal plate inside. A large sound is however heard from a loud speaker as a result of the large surface area of the vibrating core

Quality or tone

This is the characteristic of a note that distinguishes it from another note of the same pitch and loudness.

A well-made and good quality tuning fork can produce a pure note with a single frequency.

A musical instrument never produces a note of single frequency; notes of higher frequencies usually accompany the fundamental frequency. These higher notes are called overtones or harmonics.

For example, when a note of 200Hz is played on a guitar, overtones of frequencies 400Hz, 1000Hz, 1600Hz may accompany it. The amount of overtones differs for the same frequency note played on different musical instruments so they produce notes of different quality.

These different overtones provide a background which makes us able to distinguish this same note when it is played on a piano.

Harmonics are frequencies which are multiples of the fundamental frequency.

An octave of a note is a note of twice the frequency. E.g. the octave of 512Hz is 1024Hz

Ultrasonic sound

The human can detect sound frequencies ranging from 20Hz to 20,000Hz. Frequencies in the range of 10^5Hz are called ultrasonics. They cannot be detected by the human ear and are useful in

a) Destroying bacteria in water

b) Removing smoke from air

c) Cleaning metals

d) Drying papers

e) Hospitals to determine the sex of the baby for an expectant mother

Forced vibrations and resonance

If a body vibrating at a particular frequency is placed in contact with another body, it will cause it to vibrate at the same frequency of the former. The vibration of the second body is said to have been forced.

Some examples of forced vibrations include

1. The vibrating body of a violin or guitar as a result of the vibrations of the strings

2. The vibrations of the cone of a loud speaker caused by the fluctuation in the current flowing in the voice coil.

Resonance

Resonance is a special effect of forced vibrations. It is an effect caused by a vibrating body setting another body vibrating with both having the same natural frequency.

When a body is set into vibration  and left to vibrate on its own, the body will vibrate with its own natural frequency

When the frequency (forcing) of a vibrating body A acting on another body B’s natural frequency, the resulting effect is an induced vibration with very large amplitude. Resonance has occurred.

Vibrations in strings and pipes

When the string of musical instrument such as a violin is plucked, a transverse wave travels along the vibrating string. This wave is reflected back at the fixed ends. We then have an interference of two waves travelling in opposite directions and a stationary transverse wave is set up.

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The points at B and C of the stationary wave are called nodes because they cannot move while point A is referred to as an antinode because there is maximum movement at this point where the string vibrates in a single loop.

This mode of vibration is called the fundamental mode of vibration because the string produces a sound of the lowest possible frequency (fundamental frequency)

The distance between the two consecutive nodes is half wavelength or λ/2 and this is equal to the length of the string l. i.e.

l=λ/2 and λ=2l

From wave formula, v=fλ, we can deduce that the fundamental frequency f_0 produced is given by f_o=v/λ=v/2l

Harmonics or overtones in strings

Higher frequencies which are integral multiples of the fundamental frequency are called harmonics or overtones.

Note that f_o is the first harmonic but the first overtone may be 2f_o or 3f_o depending on the case.

In the case of a string fixed at both ends, harmonics are obtained by touching the wires slightly (damping) at centre O. and simultaneously plucking it at P such that the length BP is one-quarter of the length l of the string.

The string vibrates in two loops. There are now 3 nodes B, O and C.

Since BO=OC=λ/2, then BC=BO+OC=λ/2+λ/2=l

λ=l

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The frequency of the first overtone or second harmonic is then given as f_1=v/λ=v/l

l=λ/2+λ/2+λ/2=3λ/2

λ=2l/3

f_2=v/λ=v/(2l/3)  or 3v/2l

First harmonic: fundamental frequency: f_o=v/2l

Second harmonic, first overtone f_1=v/l=2f_o

Third harmonic, second overtone f_2=3v/2l=3f_o

Third overtone f_3=2v/l=4f_o

And so on

A slight difference exist between harmonic and overtone:

All harmonics are whole number multiples of the fundamental frequency. However, higher frequencies which may or may not be whole number multiples of fundamental frequency are called overtones

The overtones in a bell for example are not harmonics because their frequencies are not higher multiples of the fundamental frequency

Example: A metal wire  of mass 2g and length 0.5m is held under tension by a weight of 8kg.

i) Compute the fundamental frequency

ii) Compute the first and second overtones

iii) Find the speed of a transverse wave produced in the string

f_o=1/2l √(T/M)

l=0.5, M=(2×10^(-3))/0.5, T=8kg×10ms^(-2)=80N

f_o=1/(2×0.5) √(80/(4×10^(-3) ))

Example: a length of wire has a frequency of 255Hz when stretched by a force of 225N. If the fore increases to 324N, what is the new frequency of vibration?

f∝√T

f=k√T   →    255=k×√225

255=15k

k=17

f=17×√324=17×18=306Hz

Vibrations in open and closed pipes

When one end of a tube is closed, it is termed a closed pipe. When both ends are open; an open pipe.

Pipe organs, flutes, saxophones, trumpets etc all produce their characteristic sounds when the air inside them is made to vibrate.

When air vibrates in a pipe, the waves produced are reflected at the ends producing a longitudinal stationary wave along the length of the pipe. The air particles at the end of the pipe are unable to move so a node is formed. At the open end, the particles now move with maximum amplitude – an antinode point.

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Vibrations in a closed pipe

The simplest form of vibration i.e. the fundamental mode of vibration for a closed pipe can be shown in a resonance tube experiment.

The frequency of the tuning fork is f_o. The fork is made to vibrate over the top of the tube l * initially small and water is now let out slowly from the bottom tap.

The first loud sound that occurs means that resonance has occurred.

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This means that the air column has the same frequency as the tuning fork and a stationary wave is produced from the combination of the incident and reflected wave.

The water level is the closed end where the node is formed and the antinode is at the open end.

The length of air column from top to water level is l_1 and this equal to one-quarter the wavelength i.e. the distance between anti-node and a corresponding node

l_1=λ/4   →    λ=4l_1

From f=v/λ, f_o=v/(4l_1 ) where f_o is the fundamental frequency and v is the velocity of the wave.

By varying the length of the air column that is by letting more water out, a second position is reached where a large sound is heard again.

At this level, distance from antinode to node and to another node

l_2=λ/4+λ/2=3λ/4

f_1=v/λ=4/(4l/3)=3v/4l  or 3f_o

Third harmonic, first overtone

Possible harmonics in a closed pipe will therefore be f_o. 3f_o, 5f_o, and so on.

NOTE: only odd harmonics are present

Vibrations in open pipes

The two ends of the pipe must end in an antinode in its simplest form.

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Length od air column = distance between two successive antinodes = half wavelength

l=λ/2   →   λ=2l

Fundamental frequency f_o=v/λ=v/2l

Second harmonic or first overtone f_1

Length of air column l=λ, f_1=v/l=2f_o

Hence possible harmonics in an open pipe include f_o, 2f_o, 3f_o, 4f_o, and so on

All harmonics are possible

Experiment to measure the velocity of sound in air

A tuning form of known frequency and a resonance tube can be used to determine the velocity of sound in air

At first position of resonance, λ/4=l_1+c

Where c is the end correction

At second position, 3λ/4=l_2+c

On subtracting both equations to eliminate c

3λ/4-λ/4=l_2-l_1

2λ/4=l_2-l_1    →   λ=2(l_2-l_1 )

v=fλ,  v=2f(l_2-l_1 )

Where f is the frequency of the tuning fork

Example: in a resonance tube experiment, a tuning fork of frequency 512Hz gave resonance when the length of air column was 18cm. If the next position of resonance was 50cm, what is the speed of sound in air?

v=2f(l_2-l_1 )

l_2=50cm=0.5m, l_1=0.18, f=512

v=2×512×(0.5-0.18)

v=1024×0.32

Musical instruments

There are 3 basic classifications of musical instruments

1. Wind instruments

2. String instruments

3. Percussion instruments

Wind instruments: these produce sound by the vibration of the air column e.g. flutes, pipe organs, clarinets, saxophones, trumpets

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The air particles are set into vibration when a player blows through the open end.

It may be in the form of an open or closed pipe.

By manipulating the keys, the length of the column of air vibrating changes. Basically, a short air column produces a high pitched note, and a long column of air produces a low pitched note(f∝1/l).

String instruments

Examples are violins, guitars, pianos.

The operation of these instruments is based on the fact that the frequency of a note produced depends on the length, mass and tension of the spring.

A thick (large value of M), loosed (low T) and long wire (large T) in a guitar string produces a low frequency note (small f). f=1/2l √(T/M)

Whereas a short, thin, and taut string produces a note of high frequency. It is the various combinations of the fundamental frequency and overtones that produce musical sounds of varying quality

A mechanical amplifier is necessary in the case of a piano or a sounding box in the case of guitar or violin, so as to provide a much louder sound because the strings do not produce a loud sound on their own because they are only able to set a small mass of air into vibration

Percussion instruments

Typical examples include drums, bells, tuning forks, gongs, xylophones, etc.

They produce sound when hit or struck. These sounds however have a short duration. Cords around the rim of a drum enables the player to change the pitch of the note when he sets the elastic membrane into vibration by a strike

Wind instruments

These instruments produce sound by vibrating air

ELECTRICITY

Direct Current Electricity

When an electric charge is in motion, an electric current is produced and it is referred to as current electricity.

When this motion takes place in a closed conducting path, we have an electric circuit.

The electric circuit is the heart of any electric device whether in small applications like mobile phones, televisions, computers, light bulbs or industrial appliances like generators, electric motors, transformers etc.

Electric current

Electric current I is defined as the rate of flow of electric charges Q along a conductor per unit time t.

Current (I)= (quantity of charge (Q))/(time (t))   

I=Q/t

The unit of current is Ampere (A)

Quantity of electricity is measured in coulombs and time in seconds

If 6 coulombs (6C) of electricity passes through a conductor wire in 3 seconds, the current is

I=6/3=2A or 2 coulombs per second

The instrument used for measuring electric current is the ammeter.

Current vary in magnitudes. From units in fridges to hundreds in electric motors and other industrial applications. Milliamperes range (10^(-3), 1mA) are the typical values of current in transistor radio while microampere (10^(-6)A, 1μA) are the values of current in mobile phones.

A galvanometer can detect currents thousand of times smaller than 1 microampere (1μA)

To measure current, the ammeter must be connected in a circuit in such a way that the current flows directly through it. That is, in series.

Electric circuit

This is the path through which an electric current flows. In its simplest form a circuit consists of 4 parts:

A source of electrical energy e.g. a battery connected to a

Load e.g. electric bulb and a

Switch through a

Conductor e.g. copper wire

Types of Circuits: Open, closed and short

An electric circuit can be open, closed or short.

An open circuit is a circuit where there is a gap or when the key is open (i.e. the switch is off) in the conducting path. An electric current cannot flow through air.

So if the conductor wire is cut somewhere, the current cannot flow and the circuit is open so the bulb does not light up.

A short circuit in one in which there is no load connected but is closed.

A short circuit causes sparks and may damage the battery

A closed circuit is when the key is closed or the switch is off and no break in flow elsewhere in the circuit.

The current will flow from the battery through the load and the bulb lights up.

The path must be closed (complete) before a current flows.

Potential difference (p.d)

The potential difference between two points is the workdone (in Joules) when 1 coulomb of electric charge moves from one part to another.

Electric current will flow when there is a potential difference between the ends of the wire. Current flows from a point of higher potential to a point of lower potential. This is analogous to water flowing from a higher to a lower level (height difference). The point of higher potential may be said to have a ‘positive’ potential and the lower potential ‘negative’ potential.

The unit of potential difference (p.d) is volts (V). The instrument used for measuring p.d is the voltmeter

The instument used for measuring current is an ammeter

A battery serves as a source of potential difference and is said to have electromotive force (e.m.f) when it is able to maintain a current in a circuit.

Electromotive force is the total workdone in driving an electric charge round a circuit. It is also measured in volts.

We say for example that the emf of a battery is 12 volts while the p.d across lamp 1 is 4V and the p.d across lamp 2 is 8V.

Since potential difference is a measure of the difference in potential between two points, to measure it there, the voltmeter must be connected across (in parallel) with these points.

Circuit symbols

Symbols are used to represent elements or measuring metres.

Resistance

This is the opposition to the flow of electric current.

It is measure in ohms Ω (Greek letter ‘Omega’).

Conductors and insulators

Substances that allow electric current to flow easily through them are called conductors.

Those that do not allow current to flow are called insulators or poor conductors.

Examples of conductors include

metals e.g. silver copper, aluminum

Salt solutions

Inorganic acid solutions

Examples of insulators include air, wood, plastic, glass, pure water, ebonite, organic acid solutions.

Silver is the best metallic conductor known. This is followed by copper.

Pure copper inside plastic materials or flex are used as connecting wires and electric cables in domestic use.

Silver of course cannot be used because it is very expensive.

Since conductors allow the flow of electric current easily, they have low electrical resistance and insulators have high electrical resistance.

Nichrome and magnanin are two alloys of copper with high resistance and they are used as resistance wires

The insulating strength of materials can however break down and they become conductors under certain conditions.

For example, at normal pressure, air is an insulator. But during thunderstorm, we see the conducting path of the lightning flashes in the air.

Charge carriers

The electric current in solids, liquids and gases is carried by tiny particles carrying charges.

A charge may be positive or negative.

Substance	Charge carrier
Metals	Electrons (-)
Salts and acid solutions	Ions (+ and -)
Gases	Electrons (-) and ions (+)
Semi-conductors e.g. silicon	Electrons (-) and holes (+)

Ohm’s law

It states that the current through a metallic conductor is directly proportional to the potential difference applied across it provided temperature and other physical quantities are kept constant.

Mathematically, I∝V

Where I is the current and V is the potential difference

V/I=constant

The constant of proportionality is the resistance r of the conductor.

When a potential difference V is 1 volt causes a current I of 1A to flow through a conductor, then the resistance of the conductor is 1 ohm (1Ω)

Hence V/I=R and

V=IR

From the equation, it follows that when R is large, I will be small for a constant V

And when R is small, I will be large.

Ohm’s law holds true for metallic conductors and certain alloys but is however not obeyed in some other circuit components such as diodes, transistors, rectifiers, gases, radio valves etc.

The I-V characteristic for a metallic conductor

The I-V characteristic for a filament lamp

The graph is not a straight line because the resistance of the lamp depends on the temperature of the filament.

These types of temperature dependent resistors are used in fire sensors, baby-alarms, water temperature sensors in cars

The I-V characteristic for a semi-conductor diode

A diode is any component that allows electric current to flow in only one direction. It is used as rectifiers and in LEDs (light Emitting Diodes).

The resistance of the diode decreases drastically for voltages greater than 0.6V

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Factors affecting the resistance of a wire

1. Length of conductor: the resistance of a conductor directly proportional to its length i.e. R∝l

2. Cross-sectional area: the resistance of a conductor decreases as cross-sectional area (or thickness) increases R∝1/A

3. Temperature: the resistance of a conductor varies directly with temperature R∝T. For metallic conductors, the higher the temperature, the higher the resistance.

Certain alloys especially constantine and magnanin show very little changes in resistance when temperature changes and are used in making constructing standard resistors

4. Nature of the material: the type of material used in making the conductor also plays an important role in the resistance of the conductor.

The same length, temperature and area of silver, copper or aluminum wire will show different resistance

Combining 1 and 2, we have R∝l/A

Experiments show that at constant temperature, the resistance R of a conductor is related to the length l, cross-sectional area.

According to the equation R=ρl/A

Where ρ (Greek letter ‘rho’) is a constant called the resistivity of the material.

Resistivity depends on the nature of the materials and thus differs from material to material.

If R is measured in ohms (Ω), l in metres m and A in m^2, then ρ has units of Ωm (ohm-metre)

Example: In an experiment to measure the resistivity of a wire material, the following measurement were made:

i) Length =1.0m

ii) Diameter =0.5mm

iii) Resistance =3Ω

Calculate the resistivity

From R=ρl/A, ρ=RA/l

A=πr^2=π(0.5/2)^2, r=(0.5×10^(-3))/2

A=6.25π×10^(-8) m^2

The cross-section of a conductor is a circle

ρ=(6.25π×10^(-8)×3)/1=

Resistivity of some materials

Conductors (Metals)	Restivity ρ (Ωm×10^(-8) )
Copper	1.72
Mercury	95.00
Lead	22.00
Tungsten	5.25
Magnanin	44.00
Constantan	49.00
Semi Conductors
Pure silicon	2300
Pure carbon	35
Pure germanium	0.60

Insulators	Resitivityρ (Ωm)
Glass	10^10-10^14
Mica	10^11-10^15
Wood	10^8-10^11

Arrangement of Resistors

Resistors in series

This is the connection of resistors end to end such that the same current flows through them.

The three resistors R₁, R₂ and R₃ are connected in series and the same current flows through the three resistors but the potential difference across each resistor varies.

The total p.d across them V=V₁+V₂+V₃ where V₁ is the p.d across R₁, V₂ across R₂ and V₃ across R₃.

If R is combined resistance, then V=IR and V_1=IR_1, V_2=IR_2, V_3=IR_3

IR=IR_1+IR_2+IR_3

IR=I(R_1+R_2+R_3 )

R=R_1+R_2+R_3

The total resistance of resistors in series is the sum of resistance

Resistors in parallel

When resistors are connected side-to-side such that their ends are joined together and the same p.d is applied across  each resistor, then they are said to be in parallel.

Current I_1, I_2 and I_3 flow through resistance R_1, R_2 and R_3 respectively.

The total current from the battery is given as I=I_1+I_2+I_3 

If R is the combined resistance,

then V/I=R and I_1=V/R_1 , I_2=V/R_2  and I_3=V/R_3

V/R=V/R_1 +V/R_2 +V/R_3 =V(1/R_1 +1/R_2 +1/R_3 ) 1/R=1/R_1 +1/R_2 +1/R_3

For two resistances in series, the resistances act as a potential divider in the ratio of their resistances.

For two resistances in parallel, the resistances act as a current divider in the ratio opposite their resistances.

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Example: two resistors of resistance are connected together, find the equivalent resistance when they are connected in a) series b) parallel

a) Let R_1=3Ω and R_2=6Ω

Then equivalent resistance R=R_1+R_2=3+6=9Ω

b) 1/R=1/R_1 +1/R_2 =1/3+1/6=(3×6)/(3+6)=9/18=1/2

R=2Ω

Some tips in series and parallel resistance connections

1. The equivalent resistance of resistance in series is bigger than the biggest, but for resistances in parallel, the equivalent resistance is smaller than the smallest.

2. For 2 resistors in parallel, the equivalent resistance is given as product/sum of resistances

Example: In the circuit diagram, find

i) The equivalent resistance

ii) The current I.

The connection of the resistors is a combination of series and parallel connections.

The 2Ω and 4Ω are in parallel (2||4), the equivalent resistance is given as

1/R=1/2+1/4=3/4

R=4/3 Ω

Alternatively, equivalent resistance = product/sum=

(2×4)/(2+4)=8/6   or 4/3 Ω

If we now replace the 2Ω and 4Ω parallel connection with a single resistance (=4/3 Ω).

This resistance is now in series with the 3Ω (4/3+3)

Equivalent resistance =4/3+3=13/3 Ω

This is the equivalent resistance of the circuit.

To get the current I, we use Ohm’s law (V=IR)

V=12V, R=equivalent resistance=13/3 Ω

I=V/R=12/(13/3)=36/13 A or 2.8A

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Electrical energy and Power

Electrical energy is dissipated when a quantity of charge moves from one point to another. We earlier defined the potential difference as the workdone W in Joules in moving a quantity of charge Q between two points with potential difference V i.e. W=QV

We obtain a more useful formula when we substitute Q=It. Then W=IVt

From Ohm’s law, V=IR and W=I×IR×t

W=I^2 Rt

This electrical energy can be transformed to many other useful forms of energy as in

i) Heat energy for example in electric iron, electric heaters

ii) Mechanical energy e.g. electric motors, electric fans, washing machines

iii) Sound energy in loudspeakers, telephone ear-piece

iv) Light energy e.g. in filament lamps and electric bulbs

Electric power

We understand generally that power is the rate of doing work, and that power

P=(workdone or energy transferred)/(time taken)

In electricity therefore, power P=IVt/t

P=IV

Where I is the current in Amperes and V is the potential difference in Volts

The unit of power is Watt (W) or Joule per second

Also, P=I^2 R=V^2/R

Example: Find the power in an electric heater carrying a current of 8A when the p.d across it is 120W.

P=IV, I=8A, V=120V

P=8×120=960W

Larger units of power include kilowatts (KW=10^3 W) used in large air-conditioning systems. Megawatts (MW=10^6 W) used in electric power generating stations or industrial generators.

1 horse power = 746W ≅3/4 KW

Electric appliances usually have ratings marked on them in Watts and in volts e.g. a mark of 1000W, 220V on a filament heater means that the appliance supplies a power of 1000W (1000 joules in one second) when connected to a circuit that maintains a p.d of 220V.

Commercial units of electric power and energy

Electric energy consumed in homes and industries is sold in kilowatt-hour.

This is the unit the Power Holding Company of Nigeria (PHCN) measures and sells electrical energy.

1 Kilowatt-hour is the energy consumed when 1 kilowatt of power is used for 1 hour.

1 KWh=1KW×1hr

=1000W×60×60seconds

=3.6×10^6 Joules

Thus if a 60W electric bulb is used for 18 hours, it has consumed 60/1000 KW×18hours

1.08 KWh or 1.08 units

If the charge is N20 per kilowatt-hour, then the electrical cost is 1.08×20=N21.6

Example: Six 100W lamps and four 60W bulbs were used throughout for 2 days. If the electric company charges N20 per kilowatt-hour, calculate the total operating cost during this period.

Lamps: (6×100)/1000 KW×(2×24)hrs=28.8KWh

Cost =28.8×20=N57.6

Bulbs: 4×60/1000×2×24=11.52KWh

Cost: 11.52×20=N23.04

Total operating cost =57.6+23.04=N80.64

FIELDS

Concept of fields

A field is a region of space where the effect of a physical quantity e.g gravitation, heat, electricity etc can be felt.

A force field has been defined as a type of force that need not be in contact with the body before it is applied. There are two classes of fields – scalar fields and vector fields.

A scalar field has only magnitude but no direction e.g. distribution of temperature, density, energy, etc

A scalar field has only magnitude but no direction e.g distribution of temperature, density energy etc.

A scalar field is mapped out by lines such as lines of isothermal, lines of equidensity.

A vector field has both magnitude and direction e.g gravitational field, electric and magnetic field. A vector field is shown by lines of flux or lines of force.

The main focus in this topic is on vector fields.

GRAVITATIONAL FIELDS

It is a common saying that what goes up must come down. If we throw up objects, they move up to their highest points, stay stationary for a moment and then begin to move downwards falling faster and faster until they hit the ground or the lowest level in their path.

This is the influence of gravity on the object.

The pull of gravity acts on masses of all sizes and it permeates all spaces.

Everything around the earth even up the moon is in the earth’s gravitational field.

The earth attracts every object in its gravitational field. This force of attraction causes all objects to accelerate under its influence termed acceleration due to gravity.

This acceleration is represented by g and has a value of about 9.81ms^(-2). This value is the same for all bodies irrespective of the mass. However, it slightly varies from place to place because the earth is not perfectly spherical.

The value of g is minimum at the equator (about 9.78ms^(-2)) and maximum at the poles, the north and south pole (about 9.83ms^(-2))

Thus all bodies in the same locality will fall to the ground in the same time when released from the same height above the ground.

Strictly speaking, a feather and a stone will fall to the ground at the same time if released from the same height.

It is the existence of air that makes the above statement to sound ridiculous.

Newton’s law of gravitation

Isaac Newton, in the year 1666 discovered a universal law of gravitation

It states that the force of attraction F between two masses m_1 and m_2 is directly proportional to the product of their masses and inversely proportional to their distance r apart.

F∝(m_1 m_2)/r^2

F=G (m_1 m_2)/r^2

G is a constant called gravitational constant

If F is in Newtons (N), m_1 and m_2 in kilograms (kg), r in metres (m), then G has unit Nm^2 kg^(-2).

The value of G was found experimentally to be about 6.67×10^(-11) Nm^2 kg^(-2).

Newton’s law of gravitation refers to the force between any two particles or bodies. The force of attraction is actually a pair of forces; an action-reaction pair and has equal magnitude (but opposite in direction) even though their masses may be different.

Gravitational force keeps the moon in its orbit around the earth. Gravitational fore of the sun is also what keeps the earth and other planets revolving around the sun.

Note that gravitational forces are always attractive forces

Gravitational forces between masses

Let us examine the force of attraction on different kind of masses

1. Calculate the force of attraction between the earth with mass of 6×10^24kg and the moon with mass of about 7×10^22kg given that the distance between the earth and the moon is about 4×10^8m F=G (m_1 m_2)/r^2 =(6.67×10^(-11)×6×10^24×7×10^22)/(4×10^8 )^2

2. Calculate the gravitational force of attraction between the electron and proton of a hydrogen atom when the electron moves round the proton, mass of electron = 9.1×10^(-31)kg

Mass of proton =1.67×10^(-27)kg

Distance between electron and proton = 5×10^(-11)m F=G (m_1 m_2)/r^2 =(6.67×10^(-11)×9×10^(-31)×1.67×10^(-27))/(5×10^(-11) )^2

the electrical force of attraction between the proton and the electron is much greater than the gravitational force of attraction.

3. Two spherical balls each of mass 6.0kg are placed so that their centres are 50cm apart. Calculate the gravitational force of attraction between the two balls.

m_1=6.0kg, m_2=6.0kg, r=50cm=0.5m

G=6.67×10^(-11) Nm^2 kg^(-2) F=G (m_1 m_2)/r^2 =(6.67×10^(-11)×6×6)/〖0.5〗^2

Relationship between the gravitational constant G and acceleration due to gravity g.

Let us assume that the earth is a perfect sphere having a radius R and mass M. Let us also assume that this mass is concentrated at the centre of the earth. The distance of an object on the earth’s surface with mass m to the centre of the earth is the radius of the earth.

The gravitational force of attraction between the earth and the object is given by

F=G Mm/R^2

This is also the force of gravity due to the earth on the mass i.e the weight of the body

F=mg

It follows that G Mm/R^2 =mg

g=GM/R^2

This is the mathematical concept employed to obtain the mass of the earth, because of course it is not possible to put a chemical scale under the earth to measure its mass.

gR^2=GM

m=(gR^2)/G

Using value g=9.82ms^(-2), R=6.4×10^6 m, G=6.67×10^(-11) Nm^2 kg^(-2)

m=〖9.81×(6.4×10^6 )〗^2/(6.67×10^(-11) )

=

Example: The earth is four times the size of the moon and 80 times the mass. If the acceleration due to gravity on the surface of the earth is 9.8ms^(-2), what is its value on the moon’s surface.

If the earth is 4×size of moon, then R=4r_m

Where r_m is the radius of the moon and R is the radius of the earth.

Similarly , mass of the earth M=80m_m where m_m is the mass of the moon g_e=GM/R^2 ,  g_m=(gm_m)/(r_m^2 )

g_m/g_e =(Gm_m)/(r_m^2 )×R^2/GM

g_m/9.8=(Gm_m)/(r^2 m)×R^2/GM

g_m/9.8=m_m/(r_m^2 )×(4r_m )^2/(80m_m )

g_m/9.8=16/80

g_m=9.8×16/80

Gravitational intensity or field strength

The gravitational field strength or gravitational intensity at any point is defined as the force per kilogram mass at that point.

The unit of gravitational field strength or intensity is Nkg^(-1)

If g is the acceleration due to gravity at a point in the earth’s gravitational field, then the force on a mass m placed on that point is F=mg

And force per kilogram mass F/m=g

Gravitational field strength or intensity = g

The field strength decreases above the earth’s surface where it has the value of 9.8Nkg^(-1)

From Newton’s law of gravitation, F∝1/r^2

If G, M are constant and r=R, the radius of the earth, then at a distance 2R from earth’s centre, the field strength g^′=g/2^2 =g/4

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A satelite, say of mass 1000kg, orbiting the earth at this height 2R will have a gravitational force

=mg^′=mg/4=(1000×9.8)/4=2500N

Whereas on the earth’s surface, it has a gravitational force of mg=1000×9.8=9800N

Gravitational potential

To move a mass from one point on the earth’s surface to a height requires work. The workdone W in raising a mass m from the earth’s surface to a height h above the ground is given by W=mgh

The potential at infinity has the value 0. At a distance r from the centre of the earth, the gravitational potential V is given by V=-GM/r

Where M is the mass of the earth or the mass producing the gravitational field.

G is the gravitational constant

The energy to do this work is termed potential energy or more specifically gravitational potential energy.

We define the gravitational potential V at a point as the workdone in taking 1kg mass from infinity to that point.

Unit of gravitational potential is Jkg^(-1)

Unlike gravitational field strength which is a vector, gravitational potential is a scalar possessing a numerical value but no direction.

Example: Calculate the gravitational potential at a point on the surface of the earth. Take mass of earth = 6×10^24 kg, radius = 6400km and G=6.67×10^(-11) Nm^2 kg^(-2)

V=GM/r, r=6400km=6.4×10^6 m

V=((6.67×1-^(-11)×6×10^24 ))/(6.4×10^6 )

=

The negative sign is not the interest but the magnitude

Gravitational potential C is minus or less than zero because work is done (positive) in taking a mass from that point to infinity. Normally, a body will tend to move on its own from a point of higher gravitational potential to points of lower gravitational potential.

We are sometimes scared of looking down from a tall building or a high platform because a force which is real tends to bring us closer to the centre of the earth but we only get to stop at the surface.

Because g is assumed to be constant close to the earth’s surface, we use the formula mgh to calculate the energy change. If however large changes in position are made corresponding to large distances, we must use the formula V=-Gm/r to calculate the energy change

Escape velocity

For an object to escape from the earth’s gravitational field i.e. go up and not come down, it must acquire a minimum speed, this termed escape velocity.

Suppose a rocket of mass m is fired from the surface of the earth so that it just escapes the earth’s gravitational field. The rocket must possess sufficient kinetic energy to overcome the workdone in taking the mass m from the earth’s surface to infinity.

On the earth’s surface, the potential energy of the rocket equals the product of the mass of rocket and gravitational potential.

m×-GM/R

Where R is the distance from the centre of the earth to the earth’s surface i.e the radius of the earth.

To escape, the kinetic energy 1/2 mv^2 of the rocket is equal to the workdone to take the rocket from the earth’s surface to infinity i.e 1/2 mv^2=0-(-GMm/R)

Or we can say the kinetic energy of the rocket is equal to the gravitational potential difference from the earth’s surface to infinity.

1/2 mv^2=GMm/R

v^2=2GM/R

v=√(2GM/R)

But we know that g=GM/R^2 , hence gR^2=GM and

v=√((2gR^2)/R)

v=√2gR

v is the escape velocity

At the earth’s surface, g=9.8ms^(-2), R=6.4×10^6 m, then

v=√(2×9.8×6.4×10^6 )=11×10^3 ms^(-1) or 11km/s

Thus to escape the earth’s gravitational influence, a rocket must be projected with a minimum velocity of 11km/s.

Satellites

Satellites are placed on rocket to launch them into different orbits in space depending on their mission.

They move around the earth or any other body in space.

Satellites may be natural or artificial (man-made).

The moon is regarded as a natural satellite.

Artificial satellites are human built objects orbiting the earth or other planetary bodies. These satellites are used in

1. communication

2. Weather observation especially in prediction of natural disasters such as earthquakes, volcanic eruptions etc.

3. Scientific research in the study of the earth and other planets and perhaps in the search of extra-terrestrial life.

4. Global positioning system (GPS) used in navigation, location of points and tracking devices.

Some satellites move around the earth at the same rate of rotation of the earth i.e they have a period of rotation of 24 hours. These satellites this appear stationary as they are in the same spot over the earth always. Their orbit of movement is called a parking orbit.

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Example: Calculate the speed of a satellite in an orbit close to the earth’s surface where the radius of the orbit is assumed to be equal to radius of the earth of 6.4×10^6 m. Also estimate the period of rotation of this satellite.

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The centripetal force that keeps the satellite in orbit is equal to the gravitational force of the satellite

F=(mv^2)/r=mg

v^2=rg  →v=√rg=√(6.4×10^6×9.8)

=8×10^3 ms^(-1)

ELECTRIC FIELDS

Gravitational fields deal with forces on masses while electric fields deal with forces on charges (quantities of electricity).

Apart from the difference mentioned above, charges in electric field may experience an attractive force or a repulsive force unlike gravitational fields where masses only experience attractive forces.

An electric field is a region of space where a charged body experiences an electrical force.

Coulomb’s law

It states that the force F between two point charges q_1 and q_2 is directly proportional to the product of their charges and inversely proportional to the distance r, separating them.

F∝(q_1 q_2)/r^2

F=k (q_1 q_2)/r^2

Where k is the constant of proportionality and is written as 1/(4πε_o ).

ε_o is a constant called the permittivity of free space or vacuum

F=(q_1 q_2)/(4πε_o r^2 )

If F is in Newtons (N), q_1 and q_2 are in coulombs (C), r is in metres (m), then ε_o has units C^2 N^(-1) m^(-2). It has a value of about 8.854×10^(-12)

For easier calculation, 1/(4πε_o ) is approximated to 9×10^9 Nm^2 C^(-2).

Thus F=9×10^9  (q_1 q_2)/r^2

If the charges are situated in water, during electrolysis for instance, the permittivity of free space ε_o changes to the permittivity of water which is about 80ε_o. Hence the force between charges in water becomes 80 times less than the force in free space.

In general, F=(q_1 q_2)/(4πεr^2 ) where ε is the permittivity of the medium

Given that ε=ε_o ε_r with ε_r being the relative permittivity (ε_r=80 for water)

Relative permittivity of some media

Solid	ε_r
Glass	5 – 10
Amber	2.8
Wax	2 – 2.3
Ebonite	2.8

There are two kinds of charges – positive and negative charges.

All charges fall into one of the two and the general rule is that

LIKE CHARGES REPEL, UNLIKE CHARGES ATTRACT

Thus the force between two positive (or two negative) charges is repulsive, but a positive and a negative charge attract each other.

Example: Two positive point charges of 10μC and 8μC respectively are 8cm apart in vacuum. Calculate the force between them.

q_1=10μC=10×10^(-6) C, q_2=8×10^(-6) C, r=8cm=0.08m

F=9×10^9  (q_1 q_2)/r^2 =(9×10^9×10×10^(-6)×8×10^(-6))/(0.08)^2

=

Consider the force of attraction between an electron and a proton in the hydrogen atom each carrying a charge of 1.6×10^(-19)C separated by a distance of 5.3×10^(-11) m.

Electrostatic attraction	Gravitational attraction
F=9×10^9  (q_1 q_2)/r^2 q_1=q_2=1.6×10^(-19), r=5.3×10^(-11) F=(9×10^9×1.6×10^(-19)×1.6×10^(-19))/(5.3×10^(-11) )^2 =	F=G (m_1 m_2)/r^2 Mass of proton = 1.67×10^(-27) kg Mass of electron = 9.1×10^(-31) kg r=5.3×10^(-11) m G=6.67×10^(-11) Nm^2 kg^(-2) F=(6.67×10^(-11)×9.1×10^(-31)×1.67×10^(-27))/(5.3×10^(-11) )^2    =

The gravitational force can be seen to be negative compared with the electrostatic force. It is the electrostatic force that is therefore responsible for binding forces in

a) electrons to the nuclei to form an atom

b) atoms to atoms to form molecules

c) molecules to molecules to form liquids or solids

However for large bodies, it is the gravitational force that dominates

Example: Three charges +10C, -20C and 15C are distributed as shown. Find the resultant force acting on the 10C charge.

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Lines of force in an electric field

We are able to map out the lines of force in an electric field. The lines of force represent the direction of motion of a small positive charge placed at a point in an electric field. Arrow head are indicated on the lines to represent the direction of the force.

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Properties of field lines

1. The field about an isolated positive charge is radially outward while the field about a negative charge is radially inwards

2. Lines of force start on positive charges and end only on negative charges

3. The number of field lines is proportional to the magnitude of the charge

4. Lines of force do not cross each other or intersect

5. In a uniform field, the lines of force are parallel, straight and uniformly spaced

6. The lines are drawn in such a way that the electric field is proportional to the number of lines crossing a unit area perpendicular to it.

7. The closer the lines are together, then the stronger the electric field in that region.

Electric field strength or electric intensity

The electric field strength or intensity E at any point in an electric field is the force per unit positive charge (q) at that point.

E=F/q

The unit of electric field strength is NC^(-1)  (Newton per Coulomb)

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Consider the field around an isolated positive charge q. To find the electric field intensity at P, a distance r from q, we introduce a small charge q_1 at P.

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The force F on q_1 is F=(q_1 Q)/(4πε_o r^2 )

Electric field intensity E=F/q_1 =Q/(4πε_o r^2 )

The direction of the field strength is outwards, away from Q if Q is positive and inwards, towards Q if Q is negative.

Example: Calculate the electric field intensity in vacuum at a distance 5cm from a charge of 4nC?

(Take 1/(4πε_o )=9×10^9 Nm^2 C^(-2))

E=Q/(4πε_o r^2 )=(9×10^9×4×10^(-9))/〖0.05〗^2

Example: two similar but opposite point charges -q and +q each of magnitude 6×10^(-8) C are separated by a distance of 8.0cm in vacuum as shown below. Calculate the magnitude and direction of the resulting electric field intensity at point P.

(Take 1/(4πε_o )=9×10^9 Nm^2 C^(-2))

The electric field intensity E_1 as a result of the charge -q is towards it (to the left), the magnitude of E_1 is E_1=-q/(4πε_o r^2 )

q_1=6×10^(-8), r=5cm=0.05m

E_1=(9×10^9×6×10^(-8))/〖0.05〗^2 =

E_2=(+q)/(4πε_o r^2 ), q=6×10^8, r=0.03m

E_2=(9×10^9×6×10^(-8))/(0.03)^2 =

The electric field intensity E_2 due to positive charge +q is away from it (also to the left) as a ‘unit positive charge at P is repelled by charge q’

The resultant field intensity E=E_1+E_2=

And is directed towards the charge -q leftwards

Electric potential

In gravitational field, work must be done to move a mass m from a point to infinity. The gravitational potential at a point was defined as the workdone in bringing a unit mass from infinity (where potential is 0) to that point. Similarly, since forces act on charges in an electric field, work is done when charges move from one point to another.

The electric potential V at a point is defined as the workdone per coulomb in moving a positive charge from infinity to that point against the action of the electric field.

The unit of electric potential is JC^(-1) or volts V.

It also has a unit of volts V because in current electricity, the potential difference between two points was defined as the workdone in Joules when a unit positive charge is moved from one point to another

workdone(W)/charge(Q) =electric potential V

W=QV

If work is done AGAINST the field, the potential is positive, if work is done BY the field, the potential is negative

The expression for electric potential V due to a charge Q is give as V=Q/(4πε_o r)

Like gravitational potential, electric potential is a scalar quantity i.e. it has only magnitude without direction

The electric potential at infinity is taken to be zero, but in practical terms, it is the earth that is taken to be at zero potential.

Example: Calculate the potential at a point 15c from a charge of 4μC

V=Q/(4πε_o r)

Q=4μC=4×10^(-6) C, r=15cm=0.15m

V=(9×10^9×4×10^(-6))/0.15=

Example: A small positive charge q of +3×10^(-8) C is placed at O shown.

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Find the potential

i) At A, distance 10cm from O

ii) At B, distance 20cm from O

What is the potential difference between A and B.

(Take 1/(4πε_o )=9×10^9 Nm^2 C^(-2))

i) Electric potential at A = V_1=q/(4πε_o r)=((3×10^(-8)×9×10^9 ))/0.1

=2700N

ii) Electric potential at B, V_2=(9×10^9×3×10^(-8))/0.2=1350V

Potential difference between A and B = 2700 – 1350 =1350V

Because A is at a higher potential than B, it means that the workdone in moving a charge from B to A (against the field) is 1350JC^(-1).

Relationship between the electric field intensity E and electric potential V

Consider two metal parallel plates A and B, separated by a distance d with a potential difference C applied between them by connecting a battery to the plates. The electric field between the plates is uniform as shown by the parallel lines of force. The field strength E is then constant in magnitude and direction.

Suppose a positive charge q is moved from a point X on plate B to Y on plate A opposite the direction of the intensity E, then the workdone by the charge is given by

Workdone = force × distance = qE×d

Also workdone = QV

QE×d=QV

E=V/d

The ratio V/d is called the potential gradient of the field between the plates. The unit of potential gradient is ‘volt per metre’ or V/m. It measures the change in potential with distance in an electric field. The greater the potential gradient, the greater will be the electric field strength.

Example: Two parallel plates are charged to a p.d of 200V. If they are separated by a distance of 10cm, calculate the electric field intensity between them.

E=V/d=200/10=20Vm^(-1)

Capacitors and capacitance

A capacitor is a device for storing electric charges or electricity. It consists essentially of two metal plates (conductors) separated by an insulator. The plates usually carry an equal and opposite charge such that the net charge on the capacitor is zero.

Capacitors are widely used in all electronic appliances such as television set, radio and many others.

The insulator is called a dielectric substance

Charging and discharging a capacitor

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When a capacitor is connected to a battery, the positive pole of the battery attracts electrons (negative) from plate A and they are pushed round the circuit so electrons flow from the negative pole of the battery to the plate B of the capacitor. Thus the plate B has negative charge and A an equal amount of positive charge. The capacitor is now said to be charged. The electron flow ceases (or current stops) when the electrons at B repel other electrons and so they are unable to reach B. The process of charging is completed in a very short time

Discharging a capacitor

If the plates of a charged capacitor are joined by connecting a wire between capacitor plates A and B, a spark may be seen as the plates are joined.

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Electrons now flow from plate B to A (the negative plate to the positive plate) until the positive charge on A is completely neutralized. A current thus flows for a short time interval after which the capacitor is said to have been discharged.

Capacitance

It is the ability of a capacitor to store electric charges defined as the ratio of the charge Q on either plates to the potential difference V between them.

capacitane C=charge(Q)/(potential difference (V) )

The unit of capacitance is the Farad (F)

1F is the capacitance of the capacitor when 1 V potential difference provides a charge of 1C.

Practically, 1F is the capacitance of a large capacitor. Capacitors in circuits of radio, TV and other electronic equipment have their capacitance in microfarads (μF).

1μF=10^(-6) F Smaller units are nanofarads nF(10^(-9) F) and picofarads pF(10^(-12) F)

Factors affecting capacitance

The capacitance of a capacitor depends on the

1. area (A) of the plates

2. distance (d) separating the plates

3. nature of the insulating material (dielectric substance) used i.e. the value of ε

Paper, air, glass, mica, and other insulators have varying dielectric constant or permittivity

Experimentally, these factors are related to the capacitance C of a capacitor according to the formula C=εA/d

C=(ε_o ε_r A)/d

Where ε=ε_o ε_r

ε_r is the relative permittivity of the insulating medium

ε_o is the permittivity of free space

Example: A parallel-plate capacitor with a cross-section area of 12cm^2 is used to store electric charges when a p.d of 60V is applied. If the distance between the plates is 3cm and the insulating medium is a vacuum, calculate the

(i) capacitance of the capacitor

(ii) charge on each plate

(iii) capacitance of the capacitor is the medium is mica with relative permittivity of 5

(Take ε_o=8.85×10^(-12) C^2 N^(-1) m^(-2))

C=εA/d

A=12cm^2=12×10^(-4) m^2, d=3cm=0.03m

For vacuum, ε_r=1 and ε=ε_o ε_r=ε_0

(i) ∴C=((8.85×10^(-12)×12×10^(-4) ))/0.03

=

(ii) C=Q/V, then Q=CV=

For mica, ε=ε_o ε_r=5ε_o

C=(5ε_o A)/d=

Arrangement of capacitors

As in resistors, capacitors may be arranged in series or parallel.

Series arrangement of capacitors: when two or more capacitors are connected end-to-end.

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The charge Q stored by each capacitor is the same

The potential difference across C_1 is V_1, across C_2 is V_2 and across C_3 is V_3

V=V_1+V_2+V_3

Q/C=Q/C_1 +Q/C_2 +Q/C_3

Equivalent capacitance 1/C=1/C_1 +1/C_2 +1/C_3

Parallel arrangement of capacitors

Now consider three capacitors connected in parallel

The potential difference V across each capacitor is the same. Let the charge on C_1 be Q_1, on C_2 be Q_2 and on C_3 be Q_3

Then total charge Q=Q_1+Q_2+Q_3

CV=C_1 V+C_2 V+C_3 V

“Equivalent capacitance ” C=C_1+C_2+C_3

Note the difference in finding the equivalent capacitance and equivalent resistance in series and parallel arrangement

Resistor	Capacitor
Series: Different voltage, same current R=R_1+R_2+R_3	Different voltage, same charge 1/C=1/C_1 +1/C_2 +1/C_3
Parallel: 1/R=1/R_1 +1/R_2 +1/R_3 Different current, same voltage	C=C_1+C_2+C_3 Different charge, same voltage

Example: Two capacitors of capacitance 6μF and 8μF are connected in series and the resulting combination is connected to a potential difference of 140V. Calculate

i) the equivalent capacitance

ii) the charge on the 8μF capacitor

A circuit diagram is shown

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i) The equivalent capacitance C is given by 1/C=1/6+1/8

1/C=7/24

∴C=24/7 μF

ii) Charge Q on the combination Q/C=Q/C_1 +Q/C_2

Q=CV=24/7×140=480μC

Since the charges are equal in series combination, charge on 8μF capacitor =480μC

Energy stored in a charged capacitor

Capacitors store electric charges so they can be a source of electrical energy. While charging a capacitor, the potential difference builds up from an initial value of 0 to a final value of V. The capacitor then acquires a charge Q.

The average value of the p.d across the capacitor then becomes (0+V)/2=1/2 V

And workdone is given by W=1/2 V×Q=1/2 QV Q=CV,  ∴W=1/2 CV^2

Also V=Q/C, W=1/2  Q^2/C

Energy stored in the capacitor is thus equal to workdone W=1/2 QV=1/2 CV^2=1/2  Q^2/C

Example: What is the energy stored when a potential difference of 24V is applied across a capacitor of capacitance 5μF?

C=5μF=5×10^(-6) F, V=24V

Energy stored W=1/2 CV^2

W=1/2×5×10^(-6)×24×24

W=1.44×10^(-3) J

Application of capacitors

Capacitors are used

i) for tuning in radio circuits

ii) in the ignition system of cars and other motor vehicles

iii) Smoothening rectified current from d.c power supplies

iv) In impedance matching and power factor correction

v) Reducing noise in a.c amplifiers

vi) Storing large quantities of charge

AC CIRCUITS

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ATOMIC PHYSICS

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NUCLEAR PHYSICS

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