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O-Level Physics

Welcome to the O-Level Physics section of Mathematics Hip-Hop. Here you will find physics notes, worked examples, formulas and revision materials to support your studies and examination preparation.

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Physics Academy

Explore our professionally prepared Physics notes — from measurement and motion to electricity, magnetism and beyond.

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Chapter 1
Introduction & MeasurementChapter 2
MotionChapter 3
ForcesChapter 4
Work, Energy & PowerChapter 5
PressureChapter 6
DensityChapter 7
HeatChapter 8
LightChapter 9
ElectricityChapter 10
Magnetism & Electromagnetism

CHAPTER 1

Introduction & Measurement

Build a strong physics foundation by learning physical quantities, SI units, measurement, errors, scalars, vectors and scientific notation.

Learning Objectives

By the end of this lesson, you should be able to:

• explain what physics is and identify major areas of physics;
• distinguish physical quantities from non-physical descriptions;
• use SI base units and common derived units;
• convert units correctly;
• use prefixes and scientific notation;
• distinguish scalar and vector quantities;
• describe measurement uncertainty and common sources of error; and
• record measurements with appropriate units and precision.

1. What Is Physics?

Physics is the study of matter, energy, motion, forces, interactions and the laws that describe how the physical world behaves.

Major areas include mechanics, thermal physics, waves, electricity and magnetism, optics, atomic physics and nuclear physics.

Physics Hip-Hop Tip: Physics connects mathematics to the real world. When you see a formula, ask what each quantity means physically before substituting numbers.

2. Physical Quantities

A physical quantity is a measurable property expressed by a number and a unit.

Example: A length of 5 metres contains the numerical value 5 and the unit metre (m).

3. SI Base Quantities and Units

Base quantity	SI unit	Symbol
Length	metre	m
Mass	kilogram	kg
Time	second	s
Electric current	ampere	A
Temperature	kelvin	K
Amount of substance	mole	mol
Luminous intensity	candela	cd

4. Derived Quantities

Derived quantities are formed from combinations of base quantities.

Quantity	Common unit	Relationship
Area	m²	length × length
Volume	m³	length × length × length
Speed	m/s	distance ÷ time
Acceleration	m/s²	change in velocity ÷ time
Force	N	mass × acceleration
Density	kg/m³	mass ÷ volume

5. Unit Conversion

Always identify the starting unit and the required unit before converting.

Worked Example 1: Convert 3.5 km to metres.

1 km = 1000 m
3.5 km = 3.5 × 1000 = 3500 m.

Worked Example 2: Convert 250 cm to metres.

100 cm = 1 m
250 cm = 250 ÷ 100 = 2.5 m.

6. SI Prefixes

Prefix	Symbol	Factor
kilo	k	10³
centi	c	10⁻²
milli	m	10⁻³
micro	µ	10⁻⁶
nano	n	10⁻⁹

7. Scientific Notation

Scientific notation is useful for very large or very small quantities.

A × 10ⁿ, where 1 ≤ A < 10

Worked Example 3:
4500000 = 4.5 × 10⁶

0.00032 = 3.2 × 10⁻⁴

8. Scalars and Vectors

A scalar has magnitude only. A vector has magnitude and direction.

Scalar	Vector
Distance	Displacement
Speed	Velocity
Mass	Force
Time	Acceleration

EXAM ALERT: Do not confuse distance with displacement or speed with velocity. The vector quantities include direction.

9. Measuring Length

Common instruments include rulers, metre rules, measuring tapes, vernier calipers and micrometers.

Choose an instrument whose range and precision are suitable for the measurement.

10. Measuring Time

Time can be measured using clocks, stopwatches and electronic timers. In practical work, repeated measurements can help reduce the effect of random variation.

11. Accuracy, Precision and Errors

Accuracy describes how close a measurement is to the accepted or true value.

Precision describes how closely repeated measurements agree with one another.

Measurements may be affected by:

• random errors;
• systematic errors;
• zero errors;
• parallax errors; and
• limitations of the measuring instrument.

12. Parallax Error

Parallax occurs when a scale is viewed from the wrong angle. For example, when reading a liquid level, the eye should normally be positioned appropriately with the meniscus.

13. Significant Figures

Significant figures communicate the precision of a measured or calculated quantity.

Worked Example 4:
0.00450 has 3 significant figures: 4, 5 and the final zero.

EXAM ALERT: Do not add unnecessary decimal places to a measurement. Your final precision should be consistent with the information provided by the question.

14. Measurement in the Laboratory

1. Select an appropriate instrument.
2. Check the scale and units.
3. Check for zero error where relevant.
4. Read the instrument at the correct eye position.
5. Record the value with its unit.
6. Repeat measurements when appropriate.
7. Consider uncertainty and possible sources of error.

15. Worked Examination Example

Question: A student measures a metal rod as 12.4 cm. Express this length in metres.

Since 100 cm = 1 m:
12.4 cm = 12.4 ÷ 100 m
= 0.124 m.

The unit must be included in the final answer.

16. Practice Zone

1. State the SI unit of mass.
2. State the SI unit of time.
3. Convert 4.2 km to metres.
4. Convert 6500 g to kilograms.
5. Write 0.000078 in scientific notation.
6. Distinguish between speed and velocity.
7. Give two scalar quantities and two vector quantities.
8. Explain what is meant by parallax error.
9. Give one difference between accuracy and precision.
10. Explain why units are essential in a physics answer.

17. Quick Revision

• Physics studies matter, energy, motion and physical interactions.
• A physical quantity has a numerical value and a unit.
• SI provides internationally agreed units.
• Derived quantities are built from base quantities.
• Scalars have magnitude only; vectors have magnitude and direction.
• Scientific notation is useful for very large and very small values.
• Accuracy and precision describe different aspects of measurement.
• Always include units in numerical physics answers.

🔥 Challenge Zone

A student records three measurements of a length:

15.2 cm, 15.3 cm and 15.2 cm

1. Calculate the mean length.
2. State whether the repeated measurements appear close to one another.
3. Give two possible sources of measurement error.
4. Explain why repeating a measurement can be useful.

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CHAPTER 2

Motion

1. Introduction

Motion is the change in position of an object with time relative to a chosen reference point.

Examples include a car travelling along a road, a ball falling from a height, or a cyclist moving along a track.

Physics Tip: Identify the quantities, choose the correct formula, and check the units before calculating.

2. Learning Objectives

• Distinguish between distance and displacement.
• Distinguish between speed and velocity.
• Calculate average speed, average velocity and acceleration.
• Interpret distance-time and velocity-time graphs.
• Use equations of uniformly accelerated motion.
• Solve examination-style motion problems.

3. Key Concepts

Distance

Distance is the total length of the path travelled. It is a scalar quantity and is measured in metres (m).

Displacement

Displacement is the straight-line change in position from the starting point to the finishing point, with direction. It is a vector quantity.

Speed

Average speed = Total distance ÷ Total time

v = d/t

Velocity

Velocity is the rate of change of displacement. It has both magnitude and direction.

Average velocity = Displacement ÷ Time

v = displacement/t

Acceleration

Acceleration is the rate of change of velocity with time.

a = (v − u)/t
u = initial velocity, v = final velocity, t = time.

4. Units

Quantity	SI unit	Symbol
Distance/displacement	metre	m
Time	second	s
Speed/velocity	metre per second	m/s
Acceleration	metre per second squared	m/s²

Exam Alert: Convert units before substituting into a formula when necessary.

5. Worked Examples

Example 1 — Average Speed

A car travels 180 km in 3 hours. Find its average speed.

Speed = distance ÷ time = 180 ÷ 3 = 60 km/h.

Example 2 — Acceleration

A cyclist increases velocity from 4 m/s to 16 m/s in 6 s.

a = (16 − 4)/6 = 2 m/s².

Example 3 — Finding Time

A runner moves at 5 m/s and covers 100 m.

t = d/v = 100/5 = 20 s.

Example 4 — Constant Speed

A vehicle travels at 12 m/s for 15 s.

d = vt = 12 × 15 = 180 m.

6. Motion Graphs

Distance-Time Graph

The gradient (slope) of a distance-time graph represents speed.

Velocity-Time Graph

The gradient represents acceleration, while the area under the graph represents displacement.

7. Equations of Uniformly Accelerated Motion

v = u + at

s = ut + ½at²

v² = u² + 2as

s = ½(u + v)t

u = initial velocity, v = final velocity, a = acceleration, t = time, s = displacement.

Example 5

A motorcycle starts at 5 m/s and accelerates at 3 m/s² for 4 s.

v = u + at = 5 + (3×4) = 17 m/s.

8. Practice Exercise

1. A bus travels 240 km in 4 hours. Find its average speed.
2. A car increases velocity from 10 m/s to 25 m/s in 5 s. Find acceleration.
3. A runner moves at 8 m/s for 25 s. Find distance.
4. State one difference between distance and displacement.
5. State one difference between speed and velocity.
6. A vehicle starts from rest and accelerates at 2 m/s² for 8 s. Find final velocity.
7. What does the gradient of a velocity-time graph represent?
8. What does the area under a velocity-time graph represent?

9. Answers

1. 60 km/h

2. 3 m/s²

3. 200 m

4. Distance is total path length; displacement is change in position with direction.

5. Speed is scalar; velocity is vector and includes direction.

6. 16 m/s

7. Acceleration.

8. Displacement.

10. Examination Method

1. Write down the quantities given.
2. Identify what is required.
3. Convert units if necessary.
4. Select the formula.
5. Substitute carefully.
6. Show working.
7. Give the correct unit.

Exam Alert: Always check the final unit. A correct numerical value with a missing or incorrect unit can cost marks.

11. Quick Revision

• Distance: total path travelled.
• Displacement: change in position with direction.
• Speed: distance ÷ time.
• Velocity: displacement ÷ time.
• Acceleration: change in velocity ÷ time.
• Velocity-time gradient: acceleration.
• Velocity-time area: displacement.

12. Challenge Zone

A car starts from rest and accelerates uniformly at 4 m/s² for 6 seconds.

1. Calculate its final velocity.
2. Calculate its displacement.
3. State the equation used for each.

Answers: Final velocity = 24 m/s. Displacement = 72 m.

13. Remember This

Motion becomes easier when you identify the quantities first: What is given? What is required? Which formula connects them? What units should the answer have?

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CHAPTER 3

Forces

1. Introduction

A force is a push or a pull that can change the motion, direction, shape, or size of an object.

Forces act all around us. Examples include pushing a door, pulling a load, friction acting on a moving bicycle, and gravity pulling objects toward the Earth.

Physics Tip: A force can change motion, but a force does not always make an object move. Several forces may act at the same time.

2. Learning Objectives

• Define force and identify common forces.
• Distinguish between contact and non-contact forces.
• Explain balanced and unbalanced forces.
• Represent forces using arrows and force diagrams.
• Explain the effects of friction and air resistance.
• Use Newton's laws to explain motion.
• Calculate force using F = ma.
• Solve examination-style force problems.

3. Types of Forces

Contact Forces

Contact forces act when two objects are physically touching.

• Friction: opposes relative motion between surfaces.
• Air resistance: opposes motion through air.
• Tension: force transmitted through a stretched string, rope, or cable.
• Normal reaction: support force exerted by a surface.
• Upthrust: upward force exerted by a fluid on an immersed object.

Non-Contact Forces

These forces act without direct physical contact.

• Gravitational force: attraction between masses.
• Magnetic force: attraction or repulsion between magnets or magnetic materials.
• Electrostatic force: attraction or repulsion between electric charges.

4. Measuring Force

The SI unit of force is the newton (N).

Force can be measured using a spring balance or newton meter.

1 newton is the force that gives a mass of 1 kg an acceleration of 1 m/s².

5. Balanced and Unbalanced Forces

Balanced Forces

When forces acting on an object are equal in magnitude and opposite in direction, the resultant force is zero.

An object may remain at rest or continue moving at constant velocity.

Unbalanced Forces

When the resultant force is not zero, the object accelerates. Its speed, direction, or both may change.

Resultant force is the single force that has the same effect as all the forces acting together.

6. Force Diagrams

Forces are commonly represented by arrows. The arrow points in the direction of the force, while its length can represent the relative magnitude.

7. Newton's Laws of Motion

Newton's First Law

An object remains at rest or continues moving with constant velocity unless acted upon by a resultant external force.

This is sometimes called the law of inertia.

Newton's Second Law

The acceleration of an object depends on the resultant force and its mass.

F = ma
F = resultant force (N), m = mass (kg), a = acceleration (m/s²).

Newton's Third Law

When two objects interact, each exerts a force on the other. These forces are equal in magnitude and opposite in direction.

Remember: The action and reaction forces act on different objects.

8. Worked Examples

Example 1 — Finding Force

A 5 kg object accelerates at 3 m/s². Find the resultant force.

F = ma = 5 × 3 = 15 N.

Example 2 — Finding Acceleration

A resultant force of 24 N acts on a mass of 6 kg.

a = F/m = 24/6 = 4 m/s².

Example 3 — Finding Mass

A force of 30 N produces an acceleration of 5 m/s².

m = F/a = 30/5 = 6 kg.

Example 4 — Resultant Force

A box is pushed with 50 N to the right while friction of 20 N acts to the left.

Resultant force = 50 − 20 = 30 N to the right.

9. Friction

Friction is a force that opposes relative motion between surfaces in contact.

Advantages

• Allows us to walk without slipping.
• Allows vehicles to grip the road.
• Allows brakes to slow vehicles.
• Helps us hold objects.

Disadvantages

• Causes wear and tear.
• Produces unwanted heat.
• Reduces efficiency in machines.
• Opposes useful motion.

Lubrication can reduce friction between moving surfaces.

10. Weight and Mass

Mass is the amount of matter in an object. It is measured in kilograms (kg).

Weight is the gravitational force acting on an object. It is measured in newtons (N).

W = mg
W = weight (N), m = mass (kg), g = gravitational field strength (N/kg).

Example 5

A 10 kg object is in a gravitational field of 9.8 N/kg. Find its weight.

W = mg = 10 × 9.8 = 98 N.

Exam Alert: Mass is measured in kg; weight is measured in N. Do not use the terms as if they mean the same thing.

11. Practice Exercise

1. Define force.
2. State two contact forces.
3. State two non-contact forces.
4. What is the SI unit of force?
5. What happens when the resultant force on an object is zero?
6. Calculate the force needed to accelerate a 4 kg mass at 6 m/s².
7. A force of 45 N acts on a 9 kg mass. Find the acceleration.
8. A 60 N force acts to the right while a 25 N force acts to the left. Find the resultant force.
9. State Newton's first law.
10. State Newton's third law.
11. Calculate the weight of a 5 kg object if g = 10 N/kg.

12. Answers

1. A push or pull that can change motion, direction, or shape.

2. Any two: friction, tension, normal reaction, air resistance, upthrust.

3. Any two: gravitational, magnetic, electrostatic.

4. Newton (N).

5. It remains at rest or continues at constant velocity.

6. F = ma = 4 × 6 = 24 N.

7. a = F/m = 45/9 = 5 m/s².

8. 60 − 25 = 35 N to the right.

9. An object remains at rest or moves with constant velocity unless acted on by a resultant external force.

10. Interacting objects exert equal and opposite forces on each other.

11. W = mg = 5 × 10 = 50 N.

13. Examination Method

1. Draw a simple force diagram when useful.
2. Identify all forces and their directions.
3. Find the resultant force.
4. Choose the appropriate equation.
5. Substitute values with correct units.
6. State the final answer clearly.

Exam Alert: For F = ma questions, use the resultant force, not simply one of the forces acting on the object.

14. Quick Revision

• Force: push or pull.
• Unit: newton (N).
• Resultant force: overall effect of all forces.
• Balanced forces: resultant = 0.
• Unbalanced forces: resultant ≠ 0.
• Newton II: F = ma.
• Weight: W = mg.
• Friction: opposes relative motion.

15. Challenge Zone

A 12 kg trolley is pulled with 80 N to the right. Friction of 20 N acts to the left.

1. Find the resultant force.
2. Find the acceleration.
3. State the direction of acceleration.

Answers: Resultant force = 60 N. Acceleration = 5 m/s². Direction = right.

16. Remember This

Force changes motion when the forces are unbalanced. Always think about the resultant force before deciding how an object will move.

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CHAPTER 4

Work, Energy & Power

1. Introduction

In everyday life, we use the words work, energy and power frequently. In Physics, however, each word has a precise scientific meaning.

Work is done when a force causes an object to move through a distance in the direction of the force. Energy is the capacity to do work, while power describes how quickly work is done.

2. Learning Objectives

By the end of this chapter, you should be able to:

• Define work, energy and power.
• Calculate work done by a force.
• Distinguish between different forms of energy.
• Calculate kinetic and gravitational potential energy.
• Explain conservation and transformation of energy.
• Calculate power and efficiency.
• Solve numerical problems using correct SI units.

3. Work Done

Work is done when a force acts on an object and the object moves through a distance in the direction of the force.

Work done = Force × distance moved in the direction of the force

W = Fd

W = work done, measured in joules (J)

F = force, measured in newtons (N)

d = distance, measured in metres (m)

Remember: If there is no movement in the direction of the force, the work done by that force is zero.

SI Unit of Work

The SI unit of work is the joule (J).

One joule is the work done when a force of one newton moves an object through one metre in the direction of the force.

1 J = 1 N m

4. Work Done When the Force Is at an Angle

When the force is not acting in exactly the same direction as the movement, only the component of the force in the direction of motion does work.

W = Fd cos θ

where θ is the angle between the force and the displacement.

• If θ = 0°, the force is in the direction of motion and work is maximum.
• If θ = 90°, the force is perpendicular to the motion and its work is zero.
• If θ = 180°, the force acts opposite to the motion and the work is negative.

5. Energy

Energy is the capacity of an object or system to do work. Energy can be transferred from one object to another and can change from one form to another.

Kinetic Energy

Energy possessed by an object because of its motion.

Gravitational Potential Energy

Energy possessed because of an object's position or height in a gravitational field.

Elastic Potential Energy

Energy stored when an elastic object is stretched or compressed.

Chemical Energy

Energy stored in chemical substances such as fuels and food.

Electrical Energy

Energy associated with electric charges and electrical circuits.

Thermal Energy

Energy associated with the random motion of particles.

6. Kinetic Energy

Kinetic energy is the energy possessed by an object because it is moving.

K.E. = ½mv²

m = mass in kilograms (kg)

v = speed in metres per second (m/s)

K.E. = kinetic energy in joules (J)

Worked Example 1 — Kinetic Energy

A car of mass 1,000 kg travels at 20 m/s. Calculate its kinetic energy.

Step 1: Write the formula

K.E. = ½mv²

Step 2: Substitute

K.E. = ½ × 1000 × 20²

Step 3: Calculate

K.E. = 200,000 J

Answer: The kinetic energy is 200,000 J, or 200 kJ.

7. Gravitational Potential Energy

An object gains gravitational potential energy when it is raised through a vertical height.

G.P.E. = mgh

m = mass in kilograms (kg)

g = gravitational field strength in N/kg

h = vertical height in metres (m)

Unless a question gives another value, use the value of g specified in the examination. If none is specified, follow the convention used by your examination board.

Worked Example 2 — Gravitational Potential Energy

A 5 kg object is lifted through a height of 4 m. Take g = 10 N/kg. Find the gain in gravitational potential energy.

G.P.E. = mgh

G.P.E. = 5 × 10 × 4

G.P.E. = 200 J

Answer: The gain in gravitational potential energy is 200 J.

8. Conservation of Energy

The principle of conservation of energy states that energy cannot be created or destroyed. It can only be transferred from one form to another.

Key idea: In a closed system, the total amount of energy remains constant even though the form of the energy may change.

Example of Energy Transformation

When an object falls, gravitational potential energy is transferred into kinetic energy. In a real situation, some energy may also be transferred to thermal energy and sound because of air resistance or other forces.

Energy pathway:

Gravitational Potential Energy → Kinetic Energy → Thermal Energy + Sound

9. Power

Power is the rate at which work is done or energy is transferred.

Power = Work done ÷ time taken

P = W/t

P = power in watts (W)

W = work done in joules (J)

t = time in seconds (s)

1 watt = 1 joule per second

Worked Example 3 — Power

A machine does 6,000 J of work in 20 s. Calculate its power.

P = W/t

P = 6000 ÷ 20

P = 300 W

Answer: The power of the machine is 300 W.

10. Power and Speed

When a constant force acts in the direction of motion, power can also be expressed in terms of force and velocity.

P = Fv

This relationship is useful when an object is moving at a known speed while a driving force acts on it.

11. Efficiency

No real machine converts all the input energy into useful output energy. Some energy is usually transferred to unwanted forms such as heat or sound.

Efficiency = (Useful output energy ÷ Total input energy) × 100%

Efficiency can also be calculated using power:

Efficiency = (Useful output power ÷ Total input power) × 100%

Worked Example 4 — Efficiency

A machine receives 5,000 J of energy and produces 4,000 J of useful output energy. Calculate its efficiency.

Efficiency = (4000 ÷ 5000) × 100%

Efficiency = 80%

Answer: The efficiency of the machine is 80%.

12. Energy Resources

Energy resource	Useful description
Fossil fuels	Coal, oil and natural gas; chemical energy is released by combustion.
Solar	Uses energy from sunlight.
Wind	Uses the kinetic energy of moving air.
Hydroelectric	Uses moving or falling water to drive turbines.
Geothermal	Uses thermal energy from within the Earth.
Biomass	Uses chemical energy stored in biological material.
Nuclear	Releases energy from changes in atomic nuclei.

13. Worked Example 5 — Energy Transformation

A 2 kg object is raised to a height of 10 m. Take g = 10 N/kg. It is then allowed to fall.

(a) Calculate its gravitational potential energy at the top.

G.P.E. = mgh = 2 × 10 × 10 = 200 J

(b) Ignoring energy losses, what kinetic energy would it have just before reaching the ground?

By conservation of energy, the gravitational potential energy is transferred to kinetic energy.

K.E. = 200 J

Answer: Just before reaching the ground, its kinetic energy would be 200 J, if losses are ignored.

14. Practice Questions

1. A force of 50 N moves a box through 6 m in the direction of the force. Calculate the work done.
2. A 4 kg object moves at 5 m/s. Calculate its kinetic energy.
3. A 10 kg load is raised through 3 m. Take g = 10 N/kg. Calculate its gain in gravitational potential energy.
4. A machine does 12,000 J of work in 30 s. Calculate its power.
5. A device receives 2,500 J of energy and produces 2,000 J of useful energy. Calculate its efficiency.
6. A force of 200 N acts on an object moving at 4 m/s in the direction of the force. Calculate the power.
7. Explain what happens to gravitational potential energy as an object falls.
8. State the principle of conservation of energy.

15. Answers & Solutions

1. W = Fd = 50 × 6 = 300 J.

2. K.E. = ½mv² = ½ × 4 × 5² = 50 J.

3. G.P.E. = mgh = 10 × 10 × 3 = 300 J.

4. P = W/t = 12,000 ÷ 30 = 400 W.

5. Efficiency = (2,000 ÷ 2,500) × 100% = 80%.

6. P = Fv = 200 × 4 = 800 W.

7. As the object falls, gravitational potential energy decreases and is transferred mainly into kinetic energy, with possible transfers to thermal energy and sound.

8. Energy cannot be created or destroyed; it can only be transferred from one form to another.

16. Examination Method

1. Write the known quantities with their units.
2. Choose the correct formula.
3. Convert units where necessary, especially time and distance.
4. Substitute carefully into the formula.
5. Calculate and include the correct SI unit.
6. Check whether the answer is physically reasonable.

Exam Alert: Do not confuse energy (J) with power (W). Energy tells you how much work can be done; power tells you how quickly work is done.

17. Quick Revision

Quantity	Formula	SI unit
Work done	W = Fd	joule (J)
Work at an angle	W = Fd cos θ	joule (J)
Kinetic energy	K.E. = ½mv²	joule (J)
Gravitational potential energy	G.P.E. = mgh	joule (J)
Power	P = W/t	watt (W)
Power at constant force and velocity	P = Fv	watt (W)
Efficiency	(useful output ÷ input) × 100%	%

18. Challenge Zone

Challenge: A 1,200 kg car accelerates from rest to 10 m/s.

1. Calculate its final kinetic energy.
2. If the acceleration takes 5 s, calculate its average rate of increase of kinetic energy.
3. Explain why the energy supplied by a real engine is greater than the car's increase in kinetic energy.

Try the problem yourself before checking your working with your teacher or study group.

19. Remember This

• Work is energy transferred when a force moves an object.
• Energy is the capacity to do work.
• Power is the rate of doing work or transferring energy.
• Kinetic energy depends on mass and the square of speed.
• Gravitational potential energy depends on mass, gravitational field strength and height.
• Energy is conserved, but useful energy can be transferred to unwanted forms.
• Always give numerical answers with the correct unit.

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CHAPTER 5

Pressure

1. Introduction

Pressure describes how much force acts on a given area. The same force can produce different pressures depending on the area over which it acts.

2. Learning Objectives

• Define pressure and state its SI unit.
• Calculate pressure from force and area.
• Explain the effect of changing area on pressure.
• Describe pressure in liquids.
• Calculate pressure due to a liquid column.
• Explain atmospheric pressure and its applications.
• Use pressure ideas to solve examination problems.

3. Pressure in Solids

Pressure is the force acting normally on a unit area of a surface.

P = F/A

P = pressure in pascals (Pa)

F = force in newtons (N)

A = area in square metres (m²)

1 Pa = 1 N/m²

Remember: For pressure calculations, area must normally be expressed in m².

4. Effect of Area on Pressure

For a constant force, pressure is inversely proportional to area.

Pressure ∝ 1/Area

• Increasing area decreases pressure.
• Decreasing area increases pressure.

Small Area

A concentrated force produces high pressure, as with a sharp nail.

Large Area

The same force is spread over a larger area, producing lower pressure.

Worked Example 1

A force of 600 N acts on an area of 0.3 m². Calculate the pressure.

P = F/A = 600 ÷ 0.3 = 2,000 Pa

Answer: 2,000 Pa.

5. Converting Area Units

1 cm² = 10⁻⁴ m²

Worked Example 2

A force of 100 N acts on an area of 20 cm². Calculate the pressure.

20 cm² = 20 × 10⁻⁴ = 0.002 m²

P = 100 ÷ 0.002 = 50,000 Pa

Answer: 50,000 Pa.

6. Pressure in Liquids

Liquid pressure is caused by the weight of liquid above a point. In a liquid at rest, pressure increases with depth.

P = ρgh

ρ = density (kg/m³)

g = gravitational field strength (N/kg)

h = depth (m)

• Pressure increases with depth.
• Pressure increases when liquid density increases.
• At the same depth in the same liquid, pressure is the same.
• Liquid pressure acts in all directions.

Worked Example 3

Calculate the pressure due to water at a depth of 5 m. Take ρ = 1,000 kg/m³ and g = 10 N/kg.

P = ρgh = 1000 × 10 × 5 = 50,000 Pa

Answer: 50,000 Pa.

7. Atmospheric Pressure

Atmospheric pressure is the pressure exerted by the weight of the Earth's atmosphere.

Atmospheric pressure generally decreases as altitude increases because there is less air above a point at higher altitude.

Measuring Atmospheric Pressure

A barometer is used to measure atmospheric pressure.

8. Applications of Pressure

Sharp Blades

Small contact area gives high pressure.

Wide Tyres

Greater contact area reduces pressure on the ground.

Dam Walls

Liquid pressure increases with depth, so dams need greater strength toward the bottom.

Hydraulic Machines

Pressure applied to a confined liquid can be transmitted through the liquid.

9. Hydraulic Pressure

In a confined liquid, applied pressure can be transmitted through the liquid. This principle is used in hydraulic brakes, lifts and presses.

F₁/A₁ = F₂/A₂

Worked Example 4

A force of 50 N acts on a small piston of area 0.01 m². The large piston has an area of 0.20 m². Calculate the force produced.

F₂ = F₁A₂/A₁ = (50 × 0.20) ÷ 0.01 = 1,000 N

Answer: 1,000 N.

10. Force from Pressure

Worked Example 5

A pressure of 8,000 Pa acts over an area of 0.5 m². Calculate the force.

F = PA = 8,000 × 0.5 = 4,000 N

Answer: 4,000 N.

11. Practice Questions

1. A force of 400 N acts on an area of 0.2 m². Calculate the pressure.
2. A pressure of 5,000 Pa acts over 0.4 m². Calculate the force.
3. Convert 50 cm² into m².
4. Calculate water pressure at 3 m depth. Take ρ = 1,000 kg/m³ and g = 10 N/kg.
5. Explain why a sharp nail enters wood more easily than a blunt nail when the same force is applied.
6. Explain why liquid pressure increases with depth.
7. Name an instrument used to measure atmospheric pressure.
8. Explain why atmospheric pressure decreases with altitude.
9. A hydraulic system has piston areas 0.02 m² and 0.5 m². If 100 N is applied to the small piston, calculate the large-piston force.

12. Answers & Solutions

1. P = 400 ÷ 0.2 = 2,000 Pa.

2. F = 5,000 × 0.4 = 2,000 N.

3. 50 cm² = 0.005 m².

4. P = 1,000 × 10 × 3 = 30,000 Pa.

5. The sharp nail has a smaller contact area, so the same force produces greater pressure.

6. A deeper point has a greater column of liquid above it, so its pressure is greater.

7. A barometer.

8. There is less air above a point at higher altitude, so atmospheric pressure decreases.

9. F₂ = 100 × 0.5 ÷ 0.02 = 2,500 N.

13. Examination Method

1. Identify whether the problem involves solid, liquid or atmospheric pressure.
2. Write the correct formula.
3. Convert quantities to SI units, especially area.
4. Substitute carefully.
5. Give the final answer with the correct unit.
6. For explanation questions, link the answer to force, area, depth or density.

Exam Alert: A common mistake is forgetting to convert cm² into m².

14. Quick Revision

Concept	Formula	Unit
Pressure	P = F/A	Pa
Force	F = PA	N
Area	A = F/P	m²
Liquid pressure	P = ρgh	Pa
Hydraulic systems	F₁/A₁ = F₂/A₂	—

15. Challenge Zone

A hydraulic lift has a small piston area of 0.025 m² and a large piston area of 0.75 m². A force of 150 N is applied to the small piston.

1. Calculate the pressure produced in the liquid.
2. Calculate the force exerted by the large piston.
3. Explain why the large piston can produce a much greater force.

16. Remember This

• Pressure = Force ÷ Area.
• Pressure is measured in pascals (Pa).
• For the same force, smaller area means greater pressure.
• Liquid pressure increases with depth and density.
• Atmospheric pressure decreases as altitude increases.
• Hydraulic systems use pressure transmission through liquids.
• Convert area into m² before SI pressure calculations.

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CHAPTER 6

Density

1. Introduction

Density is the mass contained in a unit volume of a substance. It helps us compare how closely matter is packed in different materials.

2. Learning Objectives

• Define density and state its common units.
• Use the density formula to calculate mass, volume or density.
• Convert between common units used in density calculations.
• Determine the density of regular and irregular objects.
• Explain why objects float or sink using density.
• Apply density ideas to practical and examination problems.

3. The Density Formula

Density = Mass / Volume    ρ = m/V

ρ = density, usually kg/m³ or g/cm³

m = mass, in kg or g

V = volume, in m³ or cm³

Remember: Keep the units consistent. If mass is in kg and volume is in m³, density is in kg/m³.

4. Rearranging the Formula

m = ρV

V = m/ρ

Find Density

Divide mass by volume.

Find Mass

Multiply density by volume.

Find Volume

Divide mass by density.

5. Worked Examples

Worked Example 1

A block has a mass of 600 g and a volume of 200 cm³. Calculate its density.

ρ = m/V = 600 ÷ 200 = 3 g/cm³

Answer: 3 g/cm³.

Worked Example 2

A metal has a density of 8,000 kg/m³ and a volume of 0.002 m³. Calculate its mass.

m = ρV = 8,000 × 0.002 = 16 kg

Answer: 16 kg.

Worked Example 3

A liquid has a mass of 2.4 kg and a density of 800 kg/m³. Find its volume.

V = m/ρ = 2.4 ÷ 800 = 0.003 m³

Answer: 0.003 m³.

6. Units and Conversions

• 1 kg = 1,000 g
• 1 m³ = 1,000,000 cm³
• 1 g/cm³ = 1,000 kg/m³
• 1 cm³ = 10⁻⁶ m³

Worked Example 4

Convert 2.5 g/cm³ into kg/m³.

2.5 × 1,000 = 2,500 kg/m³

Answer: 2,500 kg/m³.

7. Measuring Density of a Regular Object

For a regular solid, measure its mass using a balance and determine its volume from its dimensions.

For a cuboid: V = length × width × height

Worked Example 5

A rectangular block measures 10 cm × 5 cm × 4 cm and has a mass of 800 g. Find its density.

V = 10 × 5 × 4 = 200 cm³

ρ = 800 ÷ 200 = 4 g/cm³

Answer: 4 g/cm³.

8. Density of an Irregular Object

An irregular solid can be measured using the displacement method. Record the initial water volume, immerse the object completely, then record the final volume.

Volume of object = Final volume − Initial volume

Worked Example 6

Water in a measuring cylinder rises from 50 cm³ to 80 cm³ when a stone is fully immersed. The stone has a mass of 90 g. Find its density.

V = 80 − 50 = 30 cm³

ρ = 90 ÷ 30 = 3 g/cm³

Answer: 3 g/cm³.

9. Density and Floating or Sinking

Whether an object floats or sinks in a fluid depends on its density compared with the density of the fluid.

Less Dense

If the object is less dense than the fluid, it can float.

More Dense

If the object is more dense than the fluid, it tends to sink.

Equal Density

An object with the same density as the fluid can remain suspended.

Application: Ships can float because their overall average density, including the air spaces inside, can be less than that of water.

10. Density in Everyday Life

Ships and Boats

Design and air spaces help reduce average density so vessels can float.

Hot-Air Balloons

A balloon rises when its overall average density is lower than the surrounding air.

Cooking Oils

Liquids with different densities form layers when they do not mix.

Material Identification

Density can help identify an unknown material when its mass and volume are known.

11. Practice Exercise

1. A solid has a mass of 450 g and a volume of 150 cm³. Calculate its density.
2. A substance has a density of 2,000 kg/m³ and a volume of 0.004 m³. Find its mass.
3. A liquid has a mass of 3 kg and density 750 kg/m³. Find its volume.
4. A stone raises water from 35 cm³ to 65 cm³. If its mass is 72 g, calculate its density.
5. Explain why a large steel ship can float even though steel is denser than water.

12. Answers

1. 3 g/cm³
2. 8 kg
3. 0.004 m³
4. 2.4 g/cm³
5. The ship contains air spaces and displaces enough water so that its overall average density is less than that of water.

13. Examination Method

1. Write down the quantities given.
2. Choose the correct density formula.
3. Convert units before substituting where necessary.
4. Substitute carefully and show the working.
5. Include the correct unit in the final answer.

Exam Alert: A correct numerical answer without a unit may lose marks. Always check whether the required unit is kg/m³ or g/cm³.

14. Quick Revision

Quantity	Formula	Common unit
Density	ρ = m/V	kg/m³
Mass	m = ρV	kg
Volume	V = m/ρ	m³
Regular cuboid volume	V = lwh	m³ or cm³

15. Challenge Zone

A metal cube has a mass of 1.728 kg. Each side is 6 cm long.

1. Calculate the volume of the cube in cm³.
2. Calculate its density in g/cm³.
3. Convert your answer into kg/m³.

Hint: Convert 1.728 kg to grams before using g/cm³.

16. Remember This

Density tells us how much mass is packed into a given volume. Keep your units consistent, show your working, and use displacement when measuring the volume of an irregular solid.

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CHAPTER 7

Heat

1. Introduction

Heat is energy transferred from a hotter body to a colder body because of a temperature difference. Temperature is a measure of how hot or cold a body is and is related to the average kinetic energy of its particles.

Heat transfer continues until the bodies reach thermal equilibrium.

2. Learning Objectives

• Distinguish between heat and temperature.
• Describe conduction, convection and radiation.
• Explain thermal expansion in solids, liquids and gases.
• Use the specific heat capacity relationship.
• Apply heating calculations to examination problems.
• Explain common applications of heat transfer.

3. Heat and Temperature

Heat

Heat is energy transferred because of a temperature difference. Its SI unit is the joule (J).

Temperature

Temperature indicates the thermal state of a body. Common scales include degrees Celsius and kelvin.

Temperature in kelvin: K = °C + 273

Remember This: Temperature is not the same thing as heat. Heat is energy transferred; temperature describes how hot or cold something is.

4. Methods of Heat Transfer

Conduction

Heat is transferred through a material from particle to particle without bulk movement of the material. Metals are generally good conductors.

Convection

Heat is transferred by the movement of a fluid. Warm, less-dense fluid rises while cooler, denser fluid sinks, forming convection currents.

Radiation

Heat is transferred by electromagnetic radiation. It does not require a material medium, so energy from the Sun can reach Earth through space.

5. Good and Poor Conductors

Materials such as metals conduct thermal energy relatively well. Materials such as wood, plastic and air are relatively poor conductors and can be used as insulators.

Polished, shiny surfaces are generally good reflectors and poor absorbers/emitters of thermal radiation, while dull, dark surfaces are generally good absorbers/emitters.

6. Thermal Expansion

Most substances expand when heated and contract when cooled because their particles move more vigorously and occupy more space.

Solids

Expansion is usually small but important in bridges, railway lines and metal structures.

Liquids

Liquids expand when heated and are used in devices such as liquid thermometers.

Gases

Gases generally show significant expansion when heated.

Exam Alert: Expansion gaps are left in structures such as bridges and railway tracks to allow materials to expand safely when temperature increases.

7. Specific Heat Capacity

Specific heat capacity is the amount of thermal energy required to raise the temperature of 1 kg of a substance by 1°C (or 1 K).

Q = mcΔT

Where:

• Q = thermal energy transferred, in joules (J)
• m = mass, in kilograms (kg)
• c = specific heat capacity, in J kg⁻¹ °C⁻¹
• ΔT = change in temperature, in °C or K

8. Worked Example — Specific Heat Capacity

Question: A 2 kg block is heated through 15°C. If its specific heat capacity is 400 J kg⁻¹ °C⁻¹, calculate the energy transferred.

Solution:

Q = mcΔT

Q = 2 × 400 × 15

Q = 12,000 J

9. Worked Example — Finding Specific Heat Capacity

A 0.5 kg substance absorbs 5,000 J and its temperature rises by 20°C. Find its specific heat capacity.

c = Q ÷ (mΔT)

c = 5,000 ÷ (0.5 × 20)

c = 500 J kg⁻¹ °C⁻¹

10. Everyday Applications of Heat

• Cooking utensils often use good conductors to transfer heat to food.
• Handles may use insulating materials to reduce heat transfer to the user.
• Thermos flasks reduce conduction, convection and radiation to keep liquids hot or cold.
• Radiators transfer thermal energy to surrounding air mainly by convection and radiation.
• Solar heaters absorb radiation and convert it into useful thermal energy.

11. Practice Exercise

1. State the difference between heat and temperature.
2. Name the three methods of heat transfer.
3. Explain why convection occurs in a liquid that is being heated.
4. Why can heat from the Sun reach Earth?
5. Calculate the energy needed to raise the temperature of 3 kg of water by 10°C if c = 4,200 J kg⁻¹ °C⁻¹.
6. Explain why gaps are provided between sections of railway track.

12. Answers

1. Heat is energy transferred because of a temperature difference; temperature describes the thermal state or degree of hotness of a body.
2. Conduction, convection and radiation.
3. Heating makes the liquid less dense; warmer liquid rises while cooler, denser liquid sinks, producing a convection current.
4. Radiation can travel through a vacuum.
5. Q = mcΔT = 3 × 4,200 × 10 = 126,000 J.
6. To allow for thermal expansion and reduce the risk of buckling or damage.

13. Examination Method

1. Write down the quantities given in the question.
2. Choose the correct heat equation.
3. Check that mass is in kilograms.
4. Calculate the temperature change correctly.
5. Substitute carefully and include the correct unit.
6. For explanations, mention particle movement and energy transfer where relevant.

14. Quick Revision

Concept	Key Point
Heat	Energy transferred because of a temperature difference.
Temperature	Indicates how hot or cold a body is.
Conduction	Heat transfer through a material without bulk movement.
Convection	Heat transfer by movement of a fluid.
Radiation	Heat transfer by electromagnetic radiation.
Specific heat capacity	Energy required to raise 1 kg by 1°C.
Main equation	Q = mcΔT

15. Challenge Zone 🔥

A 1.5 kg metal block receives 18,000 J of energy and its temperature rises from 20°C to 50°C.

Challenge: Calculate the specific heat capacity of the metal.

Answer: c = 18,000 ÷ (1.5 × 30) = 400 J kg⁻¹ °C⁻¹.

Remember This: Always identify whether the question is asking about temperature, heat transfer, or the energy required to change temperature. Then choose the correct idea or equation.

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CHAPTER 8

Light

1. Introduction

Light is a form of electromagnetic radiation that can be detected by the human eye. It enables us to see objects and can travel through a vacuum.

Light travels in straight lines in a uniform medium. This idea helps explain shadows, reflection, refraction and the formation of images.

2. Learning Objectives

• Describe how light travels.
• Explain reflection and use the laws of reflection.
• Describe image formation in a plane mirror.
• Explain refraction at a boundary between transparent media.
• Use the refractive index relationship.
• Describe the action of converging and diverging lenses.
• Interpret simple ray diagrams and apply light concepts to everyday situations.

3. Sources of Light

Luminous objects

These produce their own light, for example the Sun, stars, flames and electric lamps.

Non-luminous objects

These do not produce their own visible light. We see them when light from a source is reflected into our eyes.

4. Rectilinear Propagation of Light

In a uniform transparent medium, light travels in straight lines. This is called rectilinear propagation.

It explains the formation of sharp shadows when an opaque object blocks light.

Simple ray idea

SOURCE ─────────────→ OBJECT

Light travels along straight paths until its direction is changed or it is absorbed.

5. Reflection of Light

Reflection is the bouncing back of light into the same medium after it strikes a surface.

The normal is an imaginary line drawn perpendicular to the reflecting surface at the point where the ray strikes.

6. Laws of Reflection

1. The angle of incidence is equal to the angle of reflection.
2. The incident ray, reflected ray and normal all lie in the same plane.

Angle of incidence = Angle of reflection
i = r

Reflection at a plane surface

Incident ray ↘   | NORMAL |   ↗ Reflected ray

Both angles are measured from the normal, not from the mirror surface.

7. Plane Mirrors

A plane mirror is a flat reflecting surface. The image formed by a plane mirror is:

• Virtual
• Upright
• The same size as the object
• Laterally inverted
• At the same distance behind the mirror as the object is in front

8. Refraction of Light

Refraction is the change in direction of light when it passes from one transparent medium into another because its speed changes.

• From a less optically dense medium into a more optically dense medium, light bends towards the normal.
• From a more optically dense medium into a less optically dense medium, light bends away from the normal.

Refraction

Air ↘ | NORMAL | ↘ Glass

The ray changes direction because its speed changes when it enters the second medium.

9. Refractive Index

The refractive index of a medium compares the speed of light in a vacuum with its speed in that medium.

n = c / v

• n = refractive index
• c = speed of light in vacuum ≈ 3.0 × 10⁸ m/s
• v = speed of light in the medium

10. Worked Example — Refractive Index

Question: Light travels through a material at 2.0 × 10⁸ m/s. Calculate its refractive index.

Solution:

n = c/v

n = (3.0 × 10⁸) / (2.0 × 10⁸)

n = 1.5

11. Lenses

Convex Lens

A convex lens is thicker at the centre than at the edges. It is a converging lens because it brings parallel rays together.

Concave Lens

A concave lens is thinner at the centre than at the edges. It is a diverging lens because it spreads parallel rays apart.

12. Important Lens Terms

• Principal axis: the straight line through the optical centre of a lens.
• Optical centre: the central point of the lens through which a ray can pass with little deviation in the simple model.
• Principal focus: the point where parallel rays converge, or appear to diverge from, after passing through a lens.
• Focal length: the distance from the optical centre to the principal focus.

13. Basic Ray Rules for a Convex Lens

• A ray parallel to the principal axis refracts through the principal focus.
• A ray passing through the optical centre continues approximately undeviated.
• A ray passing through the principal focus emerges parallel to the principal axis.

Convex lens idea

→ → → ) ( ↘ ↓ ↙ • F

Parallel rays are brought towards a focus after passing through a converging lens.

14. Basic Ray Rules for a Concave Lens

• A ray parallel to the principal axis diverges after passing through the lens.
• The diverging ray appears to come from the principal focus on the same side as the object.
• A ray through the optical centre travels approximately undeviated.

15. Uses of Lenses

Convex lenses

Used in magnifying glasses, cameras, projectors and many optical instruments.

Concave lenses

Used in spectacles for correcting certain forms of short-sightedness.

16. Worked Example — Plane Mirror

An object is placed 2 m in front of a plane mirror.

Question: How far behind the mirror is the image?

Answer: The image is 2 m behind the mirror.

17. Practice Exercise

1. What is meant by reflection?
2. State the two laws of reflection.
3. An incident ray makes an angle of 35° with the normal. Find the angle of reflection.
4. State two characteristics of an image formed by a plane mirror.
5. What is refraction?
6. State what happens to a ray entering a more optically dense medium at an angle.
7. Calculate the refractive index of a material in which light travels at 2.4 × 10⁸ m/s.
8. Distinguish between a convex lens and a concave lens.

18. Answers

1. Reflection is the bouncing back of light into the same medium after striking a surface.
2. The angle of incidence equals the angle of reflection; the incident ray, reflected ray and normal lie in the same plane.
3. 35°.
4. For example: virtual, upright, same size, laterally inverted, and the same distance behind the mirror as the object is in front.
5. Refraction is the change in direction of light when it passes between transparent media because its speed changes.
6. It bends towards the normal, provided it enters at an angle rather than normally.
7. n = 3.0 × 10⁸ ÷ 2.4 × 10⁸ = 1.25.
8. A convex lens converges light; a concave lens diverges light.

19. Examination Method

1. Draw a clear ray diagram when required.
2. Always draw the normal at 90° to the surface.
3. Measure angles from the normal.
4. For refraction, identify the two media and state the direction of bending.
5. For numerical questions, write the formula first, substitute values, calculate and include units where appropriate.
6. For lens questions, identify whether the lens is converging or diverging before drawing rays.

20. Quick Revision

Concept	Key Point
Light	Electromagnetic radiation detectable by the eye.
Reflection	Light bounces back from a surface.
Law of reflection	i = r.
Plane mirror	Forms a virtual, upright, same-size, laterally inverted image.
Refraction	Change in direction due to a change in speed between media.
Refractive index	n = c/v.
Convex lens	Converges light.
Concave lens	Diverges light.

21. Challenge Zone 💡

A ray of light travels from air into a glass block. The ray enters the glass at an angle to the normal.

Challenge: Explain why the ray bends and state the direction in which it bends.

Answer: The speed of light changes when it enters glass. Since glass is optically denser than air, the ray bends towards the normal.

Remember This: For reflection, think “bounce”. For refraction, think “bend because speed changes.” For lenses, remember convex = converge and concave = diverge.

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CHAPTER 9

Electricity

1. Introduction

Electricity involves the movement and effects of electric charge. In metallic conductors, electric current is produced by the movement of electrons through the conductor.

Electric circuits allow electrical energy to be transferred from a source to components such as lamps, motors and resistors.

2. Learning Objectives

• Describe electric charge and electric current.
• Use the correct units for current, charge and time.
• Explain potential difference and resistance.
• Apply Ohm's law.
• Distinguish between series and parallel circuits.
• Calculate electrical power and energy.
• Interpret simple circuit diagrams and solve examination problems.

3. Electric Current

Electric current is the rate of flow of electric charge through a conductor.

I = Q / t

• I = current, in amperes (A)
• Q = charge, in coulombs (C)
• t = time, in seconds (s)

4. Worked Example — Current

A charge of 60 C passes through a conductor in 20 s. Calculate the current.

I = Q/t = 60/20

I = 3 A

5. Potential Difference

Potential difference (p.d.) is the energy transferred per unit charge between two points in a circuit. It is measured in volts (V).

V = W / Q

• V = potential difference, in volts (V)
• W = energy transferred, in joules (J)
• Q = charge, in coulombs (C)

6. Resistance

Resistance is the opposition to the flow of electric current in a circuit. It is measured in ohms (Ω).

R = V / I

7. Ohm's Law

For a conductor at constant temperature, the current through it is directly proportional to the potential difference across it. This relationship is expressed as:

V = IR

Not every component obeys Ohm's law under all conditions. For example, the resistance of a filament lamp changes significantly as its temperature changes.

8. Worked Example — Ohm's Law

A resistor has a resistance of 6 Ω and carries a current of 2 A. Calculate the potential difference.

V = IR

V = 2 × 6

V = 12 V

9. Simple Electric Circuit

BATTERY ─── SWITCH ─── LAMP ────┐
    └──────────────────────────┘

A complete circuit provides a continuous conducting path. When the switch is open, the path is broken and current does not flow through the circuit. When the switch is closed, the circuit is complete.

10. Series Circuits

Components in a series circuit are connected along one path.

• The current is the same at every point in the series circuit.
• The supply potential difference is shared between components.
• Total resistance is the sum of the individual resistances.

R_T = R₁ + R₂ + R₃ + ...

11. Parallel Circuits

In a parallel circuit, components are connected on separate branches.

• The potential difference across each parallel branch is the same.
• The total current is the sum of the branch currents.
• Adding another parallel branch generally decreases the total resistance of the circuit.

12. Electrical Power

Electrical power is the rate at which electrical energy is transferred.

P = VI

Other useful forms are:

P = I²R     and     P = V²/R

Power is measured in watts (W).

13. Electrical Energy

Electrical energy transferred can be calculated using:

E = Pt

where E is in joules when P is in watts and t is in seconds.

For electricity billing, energy is commonly measured in kilowatt-hours (kWh).

Energy (kWh) = Power (kW) × Time (h)

14. Worked Example — Power

An appliance operates at 240 V and draws a current of 5 A. Calculate its power.

P = VI

P = 240 × 5

P = 1,200 W = 1.2 kW

15. Electrical Safety

• Do not handle electrical equipment with wet hands.
• Use correctly rated fuses and circuit breakers.
• Replace damaged cables and plugs.
• Do not overload sockets.
• Earthing provides a safer path for fault current in suitable appliances and installations.

16. Practice Exercise

1. Define electric current.
2. A charge of 120 C flows in 30 s. Calculate the current.
3. What is meant by potential difference?
4. A 4 Ω resistor carries a current of 3 A. Calculate the potential difference.
5. State two differences between series and parallel circuits.
6. Three resistors of 2 Ω, 4 Ω and 6 Ω are connected in series. Find the total resistance.
7. An appliance operates at 230 V and draws 4 A. Calculate its power.
8. A 2 kW appliance operates for 3 hours. Calculate the electrical energy used in kWh.

17. Answers

1. Electric current is the rate of flow of electric charge.
2. I = Q/t = 120/30 = 4 A.
3. Potential difference is energy transferred per unit charge between two points.
4. V = IR = 3 × 4 = 12 V.
5. In series, current is the same and p.d. is shared; in parallel, p.d. is the same across branches and current is shared.
6. R_T = 2 + 4 + 6 = 12 Ω.
7. P = VI = 230 × 4 = 920 W.
8. E = Pt = 2 × 3 = 6 kWh.

18. Examination Method

1. Write down the known quantities and units.
2. Choose the equation that contains the required quantity.
3. Convert units before calculating where necessary.
4. Substitute carefully and show your working.
5. Give the final answer with the correct unit.
6. For circuit questions, check whether the components are in series or parallel before applying the rules.

19. Quick Revision

Concept	Key Point
Current	I = Q/t; measured in amperes (A).
Potential difference	V = W/Q; measured in volts (V).
Resistance	R = V/I; measured in ohms (Ω).
Ohm's law	V = IR for an ohmic conductor at constant temperature.
Series	Same current; resistances add.
Parallel	Same p.d. across branches; current divides.
Power	P = VI; measured in watts (W).
Energy	E = Pt; commonly measured in J or kWh.

20. Challenge Zone ⚡

A 12 V battery is connected to a resistor of 4 Ω.

Challenge: Calculate the current and electrical power delivered to the resistor.

I = V/R = 12/4 = 3 A.
P = VI = 12 × 3 = 36 W.

Remember This: Keep the three key relationships ready: I = Q/t, V = IR and P = VI.

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CHAPTER 10

Magnetism & Electromagnetism

1. Introduction

A magnet produces a magnetic field and can attract magnetic materials such as iron and steel. Magnetism is closely connected with electricity: a current-carrying conductor produces a magnetic field.

2. Learning Objectives

• Identify magnetic materials and poles.
• Describe magnetic fields and field patterns.
• Explain attraction and repulsion between poles.
• Explain the magnetic effect of electric current.
• Describe how an electromagnet works.
• Describe the basic principle of an electric motor.

3. Magnetic Poles

North Pole

One end of a magnet is called the north pole.

South Pole

The other end is called the south pole.

Remember This: Like poles repel; unlike poles attract.

4. Magnetic Materials

Common magnetic materials include iron, steel, nickel and cobalt. Wood, plastic, glass and copper are not strongly attracted to magnets.

5. Magnetic Fields

A magnetic field is the region around a magnet where magnetic forces can be detected. Outside a magnet, field lines are drawn from north to south. Where field lines are closer together, the field is stronger.

Magnetic field direction outside a magnet: North → South

6. Magnetic Effect of Electric Current

When current flows through a wire, it produces a magnetic field around the wire. Increasing current increases the field strength. Coiling the wire into many turns produces a stronger magnetic effect.

7. Electromagnets

An electromagnet is a temporary magnet produced by passing current through a coil of wire, often wound around a soft iron core.

• Increase current to strengthen it.
• Increase the number of turns.
• Use a suitable soft iron core.
• Switch off the current to reduce the magnetic effect.

8. Worked Example — Electromagnet

A coil has 20 turns. Another has 40 turns, with the same current and suitable iron core. Which produces the stronger magnetic field?

Answer: The 40-turn coil generally produces the stronger field because it has more turns, with other conditions unchanged.

9. Applications of Electromagnets

Electric Bell

An electromagnet attracts an armature, causing repeated movement and ringing.

Relay

An electromagnet can operate a switch in a separate circuit.

Scrap-yard Crane

A large electromagnet lifts iron and steel and releases them when current is switched off.

10. Force on a Current-Carrying Conductor

A current-carrying conductor placed in a magnetic field can experience a force. This principle is used in electric motors. The force direction depends on the current and magnetic-field directions.

11. Simple Electric Motor

An electric motor converts electrical energy into mechanical energy. A current-carrying coil in a magnetic field experiences forces that produce rotation. A split-ring commutator reverses the current every half-turn so rotation continues.

12. Motor Applications

• Electric fans
• Water pumps
• Washing machines
• Electric toys
• Power tools
• Industrial machines

13. Examination Method

1. Define the device or quantity clearly.
2. State the relevant magnetic principle.
3. Describe the direction or sequence carefully.
4. Use labelled diagrams where required.
5. For applications, connect the principle to the device.

14. Practice Questions

1. State the rule for attraction and repulsion of magnetic poles.
2. What is a magnetic field?
3. Name four magnetic materials.
4. Give two ways of increasing the strength of an electromagnet.
5. Explain why an electromagnet is useful in a scrap-yard crane.
6. State the energy conversion in an electric motor.

15. Answers

1. Like poles repel and unlike poles attract.

2. The region around a magnet where magnetic forces can be detected.

3. Iron, steel, nickel and cobalt.

4. Increase current and/or coil turns; a suitable iron core also strengthens it.

5. It lifts iron and steel while current flows and releases them when current is switched off.

6. Electrical energy is converted mainly into mechanical energy.

16. Quick Revision

• Like poles repel; unlike poles attract.
• Outside a magnet, field lines go from north to south.
• Current in a conductor produces a magnetic field.
• More current and more coil turns can strengthen an electromagnet.
• Electromagnets can be switched on and off.
• An electric motor uses magnetic forces to produce rotation.

Exam Alert: When explaining an electromagnet, mention the current-carrying coil and its magnetic field. For applications, explain what happens when current is switched on or off.

17. Challenge Zone 🔥

Explain why a soft iron core is commonly used in an electromagnet rather than a permanent magnet material.

Remember This: Electricity can produce magnetism, and magnetism can be used to produce motion. This connection is a foundation of electrical technology.

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