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Physics4 5Form &

Alpha (␣) Decay, Beta () 6Chapter

Calibrating a Liquid-in–glass Image Formation by Small 6Chapter Effects of Magnetic Field Chapter 5 Decay and Gamma (␥) Decay Nuclear Physics

Chapter 1FTihxeerdmPoominettser using Two Lenses in Smartphone Light and on Cathode Rays using a

Camera and CCTV Optics Maltese Cross Tube Electronics

Textbook: PP. 200, 201 LS: 6.1.1

4Chapter Textbook: PP. 177, 178 LS: 5.1.2 How alpha decay occurs in a radioactive nucleus

Heat

Base Quantities and

Derived Quantities Textbook: P. 268 LS: 6.5.3 Observation Explanation radiation 22868Ra → 22826Ra + 42He + Energy

t 8IFO POMZ 41 JT TXJUDIFE PO

UIF mMBNFOU JT IFBUFE General reaction equation:

Measurement Textbook: P. 123 LS: 4.1.2 S2 t 8IFO 42 is switched off

GSFF FNJUUFE FMFDUSPOT BSF 42He (2p, 2n)

22826Rn (86p, 136 n)

Small lenses in smartphone camera Small lenses in CCTV E.H.T + OPU BUUSBDUFE UP UIF BOPEF CFDBVTF UIFSF JT OP new nucleus

3 kV – QPUFOUJBM EJGGFSFODF CFUXFFO UIF BOPEF BOE UIF

1. Place a thermometer bulb into melting mMBNFOU 22868Ra AZX → A– 42Y + 42He + Energy

ice until thermal equilibrium at 0°C. S1 t -JHIU GSPN UIF IPU mMBNFOU JT CMPDLFE CZ BO PQBRVF Z–

Mark the ice point on the stem. Filament PCKFDU .BMUFTF DSPTT

UP GPSN B TIBEPX CFDBVTF

Textbook: PP. 6, 7 LS: 1.1.2, 1.1.3 Anode light travels in straight line (88p, 138 n)

2. Place the thermometer bulb into steam

n until thermal equilibrium at 100°C. Steam

Mark the steam point on the stem. point (100°C)

mol

Amount of m Ice point L100 Radium-226 nucleus that emits an alpha particle ( 42He) has its mass number A

substance Length (0°C) L0 reduced by 4 (226 – 4) and atomic number Z reduced by 2 (88 – 2).

Melting

l ice Steam

Boiling Maltese cross How beta decay occurs in a radioactive nucleus

water

Convex lens Partially re ective Image sensor First folded

mirrors position of

Iv Water sensor Original position radiation

of sensor 146C (6p, 8n) new nucleus

Heat Vacuum t 8IFO POMZ 41 JT TXJUDIFE PO

UIF mMBNFOU JT IFBUFE –(–011Cp) 146C → 147N + –01e

Fluorescent screen t 8IFO 42 is switched on

GSFF FNJUUFE FMFDUSPOT BSF General reaction equation:

S2

Filament BUUSBDUFE UP UIF BOPEF EVF UP UIF QPUFOUJBM EJGGFSFODF

CFUXFFO UIF BOPEF BOE UIF mMBNFOU

t 5IF DBUIPEF SBZT BSF CMPDLFE CZ .BMUFTF DSPTT UP

GPSN B TIBEPX $BUIPEF SBZT USBWFM JO B TUSBJHIU MJOF

t $BUIPEF SBZT BMTP QSPEVDF B ﬂuorescent effect

green light

PO UIF TDSFFO TVSSPVOEJOH UIF TIBEPX

t 5IJT TIPXT UIBU DBUIPEF SBZT QPTTFTT momentum

BOE kinetic energy

100 kg 3. Divide the length between the two fixed points into 100 equal divisions so thatForm AZ X → Z + A Y + –01e

kg 1

cd Base quantity is a m each division = 1°C. Temperature, θ can also be calculated using formula:FormSAMPEL 5 E.H.T +

physical quantity Mass Lθ − L0 Lθ 3 kV –

I Luminous which cannot be Ch1ap. θ = L100 − L0 × 100°C or θ = L100 × 100°C. Ch5ap. 147C (7p, 7n)

intensity S1

derived from

another physical To form the real image, the The focal length can be folded by relaying Anode In a Carbon-14 nucleus, a neutron (10n) breaks down into a proton (11p) and an

electron (–01e) which is emitted as beta particle. Thus its mass number A is

quantity minimum distance of convex reflected light between partially reflective mirrors.

Example lens from sensor equals the Thus reducing the distance of the lens from the Maltese cross unchanged but atomic number Z is increased by 1 ( 6 + 1 ).

60 focal length of the lens. sensor and making the camera smaller.

55 60 5

50 10

50 10

40 20

Electric 30

current 45 15

A Thermodynamic 40 20 Calculate the temperature reading on 4.0 cm Example How gamma decay occurs in a radioactive nucleus

temperature 35 25 the mercury thermometer when the The diagram below show an astronomical telescope in normal adjustment.

30 end of the mercury thread is 11.2 cm

TK from 0°C.

Time 11.2 cm t 8IFO 41 BOE 42 BSF switched on BOE UIF NBHOFU

Solution CBS JT QMBDFE OFBS UIF UVCF

t s 16.0 cm

t 0OF TIBEPX JT EVF UP UIF MJHIU GSPN UIF IPU UVOHTUFO

Bulb 0°C 100°C S2 mMBNFOU 24904Pu 24 04Pu + γ

Mercury thread 9

Objective lens t 5IF PUIFS TIBEPX JT EVF UP UIF EFnFDUJPO PG UIF →

(fo = +55 cm) DBUIPEF SBZ CZ UIF CBS NBHOFU Photons

Eyepiece lens E.H.T + General reaction equation:

11.2 (fe = +5 cm) QUICK 3 kV – t 5IF EFnFDUJPO PG DBUIPEF SBZT DBO CF EFUFSNJOFE CZ

Temperature, θ = Lθ × 100°C = 16.0 × 100°C = 70.0°C LINK Fleming’s left hand rule AZX → AZY + γ

L100 Calculate the length and magnification of Scan the QR code S1

the telescope. to watch the video

about Smartphone N 24904Pu Gamma (γ ) ray 24 04Pu

Camera Lens work. S 9

1 length2 Solution

2 time2

Kinetic energy = × mass × Length of the telescope, L = f0 + fe

= 55 + 5 cm = 60 cm

f0 160

fe 55 cm

A derived quantity can A mercury thermometer is used to measure the temperature of hot Magnification, M = = 5 cm = 11.0

be described in terms water in a beaker. Suggest two ways to ensure good thermal contact

between the thermometer and hot water. 151

of base quantities.

Scan for answers 90

Gravitational force = mass × length 44

time2

1 S2P0M21REFQULUFIRIFLEOSMRTEHNETS

BONUS!

QUICK

LINK

CONTENTS

FORM 4 • Principle of Conservation of 25

Momentum I

1CHAPTER MEASUREMENT • Principle of Conservation of 26

Momentum II

• Explosion 27

• Base Quantities and Derived Quantities 1 • Force and Newton’s Second Law of

• Scalar Quantities and Vector Quantities 2 Motion 28

• Interpretation of Linear Graphs 3 • Impulse 29

• Interpretation of Non-linear Graphs 4 • Impulsive Force and Newton’s Third

Law of Motion 30

2CHAPTER FORCE AND MOTION I • Impulse and Impulsive Force in 31

Daily Life

• Differences between Distance and • Weight and Gravitational Field Strength 32

Displacement

5

• Differences between Speed and 6 3CHAPTER GRAVITATION

Velocity

• Acceleration and Deceleration 7 • Newton’s Universal Law of Gravitation 33

• Ticker Tape Chart Analysis 8 • Relating Gravitational Acceleration,

g with Universal Gravitational

• First Linear Motion Equation 9 Constant, G

• Second Linear Motion Equation 10 34

• Third Linear Motion Equation 11 • Centripetal Force in the Moving

System of Satellites and Planets

• Fourth Linear Motion Equation 12 35

• Displacement-time Graphs 13 • Mass of the Earth and the Sun 36

• Velocity-time Graphs 14 • Kepler’s First and Second Laws 37

• Acceleration-time Graphs 15 • Kepler’s Third Law 38

• Analysis of Displacement-time Graph • Man-made Satellites 39

and Velocity-time Graphs 16 • Geostationary and Non-geostationary

• Analysis of Velocity-time Graph and Satellites 40

Acceleration-time Graphs 17 • Escape Velocity 41

• Free Fall Motion 18 • Benefits and Implications of Escape

• Gravitational Acceleration 19 Velocity 42

• Inertia and Newton’s First Law of 20

Motion

• Relationship between Inertia and 4CHAPTER HEAT

Mass 21 SAMPEL

• Effect of Inertia in Daily Life 22 • Thermal Equilibrium 43

44

• Momentum 23 • Calibrating a Liquid-in-glass

Thermometer using Two Fixed

• Applications of Concept of Momentum Points

in Daily Life 24

(iii)

• Heat Capacity and Specific Heat • Pattern of Diffracted Water Waves 69

Capacity

45 • Interference of Waves 70

• Determine the Specific Heat Capacity • Relationship between Variables of

Interference of Waves

of a Solid and a Liquid 46 71

• Applications of Specific Heat Capacity • Applications of Interference of Waves

in Daily Life 47 in Daily Life 72

• Specific Latent Heat 48 • Electromagnetic Waves 73

• Heating Curve and Cooling Curve 49 • Applications of Electromagnetic

Waves

• How is the Value of Specific Latent 74

Heat Determined? 50

• Applications of Specific Latent Heat 6CHAPTER LIGHT AND OPTICS

in Daily Life 51

• Solving Problems Involving Latent 52 • Refraction of Light 75

Heat

• To Determine the Refractive Index

• Boyle’s Law 53 of a Glass Block 76

• Relationship between Pressure and • Real Depth and Apparent Depth 77

Volume of Gas 54 • To Determine the Refractive Index

of Water (Liquid)

• Charles’ Law 55 78

• Relationship between Volume and 56 • Natural Phenomena of Refraction 79

Temperature of Gas of Light

• Gay-Lussac’s Law 57 • Total Internal Reflection 80

• Relationship between Pressure and • Phenomena of Total Internal

Reflection

Temperature of Gas 58 81

5CHAPTER WAVES • Natural Phenomena and Applications

that Involve Total Internal Reflection 82

• Fundamentals of Waves 59 • Applications of Total Internal 83

Reflection in Daily Life

• Progressive Waves and Stationary 60

Waves • Image Formation by Lenses 84

61

• Transverse Waves, Longitudinal 62 • Position and Characteristics of Images

Waves, Mechanical Waves and 63

Electromagnetic Waves 64 Formed by Convex Lenses 85

65

• Characteristics of Waves • Position and Characteristics of Images

66

• Damping and Resonance 67 Formed by Concave Lenses 86

• Effect of Resonance in Daily Life 68 • Thin Lens Formula 87

• Reflection of Waves • Optical Instruments 88

• Applications of Reflection of Waves • Compound Microscope 89

in Daily Life

• Image Formation by Small Lenses in

• Refraction of Waves SAMPEL

Smartphone Camera and CCTV 90

• Phenomena of Refraction of Waves

in Daily Life • Image Formation by Spherical 91

Mirrors

• Applications of Concave Mirrors and

Convex Mirrors in Daily Life 92

(iv)

FORM 5 • Archimedes’ Principle 116

• Applications of Archimedes’ Principle

in Daily Life 117

1CHAPTER FORCE AND MOTION II • Bernoulli’s Principle 118

• Lift Force 119

• Resultant Force 93 • Applications of Bernoulli’s Principle

• Resultant Force on an Object in in Daily Life 120

Various States

94 3CHAPTER ELECTRICITY

• Newton’s Second Law of Motion 95

• Resolution of Forces 96 • Electric Field and Electric Field

Strength

• Steps in Determining the Resolution 121

of Forces 97 • Behaviour of Charged Particles in an

• Components of Resolution of Forces Electric Field 122

in Daily Examples 98 • Electric Current and Potential

Difference

• Forces in Equilibrium 99 123

• Drawing the Triangle of Forces 100 • Resistance 124

• Solving Problems Involving Forces in • Current, Potential Difference and

Resistance for Series and Parallel

Equilibrium 101 Circuits

• Elasticity and Hooke’s Law 102 125

• Factors that Affect the Value of 103 • Factors that Affect the Resistance of

Spring Constant, k

a Wire 126

• Solving Problems Involving Force and • Applications of Resistivity of

Conductors in Everyday Life

Extension of a Spring 104 127

2CHAPTER • Electromotive Force (e.m.f.) and 128

Internal Resistance

PRESSURE • Determining the e.m.f, ε and Internal

Resistance, r of a Dry Cell 129

• Factors Affecting Liquid Pressure 105 • Problem Solving Involving e.m.f and

• Solving Problems Involving Pressure Internal Resistance of Dry Cells 130

in Liquids 106 • Comparison between Electric

Vehicles (E.V.) and Hybrid Cars

• Applications of Pressure in Liquids in 131

Our Lives 107 • Electrical Energy and Power 132

• Fortin Barometer and Aneroid 108 • Comparing the Energy Efficiency

Barometer and Cost Saving of the CFL and

LED Lamp

• Effect of Atmospheric Pressure at 133

High Altitude

109 • Steps in Reducing the Household’s

• Effects of Atmospheric Pressure at Electrical Energy Usage 134

Extreme Depth under the Surface

of the Sea 110 4CHAPTER ELECTROMAGNETISM

• Gas Pressure (Manometer) 111 SAMPEL

• Solving Problems Involving Gas 112 • Force on a Current-carrying Conductor

Pressure

in a Magnetic Field 135

• Pascal’s Principle 113 • Creation of the Catapult Field that

produces the Force Acting on a

• Applications of Pascal’s Principle 114 Current-carrying Conductor

• Buoyant Force, FB 115 136

(v)

• Combined Magnetic Field of a 6CHAPTER

Rectangular Coil Produces a Pair of NUCLEAR PHYSICS

Turning Forces 137

• Comparison between a Brushed • Radioactive Decay 159

Motor and a Brushless Motor

138 • Alpha (α) Decay, Beta (β) Decay

and Gamma (γ) Decay

• How to Produce an Induced Current 160

in a Conductor Placed in a Magnetic • Half-life 161

Field 139 • Radioactive Decay Series 162

• Faraday’s Law and Lenz’s Law 140 • Half-life, T½ from a Decay Curve 163

• Comparison between a Direct • How Nuclear Energy is Released in

Current Generator and an Alternating the Nuclear Fission of Uranium-235 164

Current Generator 141 • How Nuclear Energy is Released in

• Working Principle of a.c. Generator in the Nuclear Fusion of Hydrogen

the Form of a Bicycle Dynamo 142 Isotopes 165

• Transformer 143 • Solving Problems Involving Nuclear

• Experiment to Verify the Transformer Energy due to Radioactive Decay and

Formula 144 Nuclear Reactions 166

• Eddy Current in an Induction Cooker 145 • Process of Generating Electricity from

• Uses of Transformers in Daily Life 146 Fission in a Nuclear Reactor 167

• Role of Transformers in the • Structure and Working Principle of a

Transmission and Distribution

of Electricity Nuclear Reactor 168

147

• Solving Problems Involving 148 7CHAPTER QUANTUM PHYSICS

Transformers

5CHAPTER ELECTRONICS • Quantum Theory of Light 169

• Thermionic Emission and Cathode • How Black Body Radiation can be 170

Explained by Planck’s Quantum

Rays 149 Theory

• Effects of Potential Difference on the • How Bohr’s Atomic Model of 171

Angle of Deflection of Cathode Rays 150 Hydrogen Explains the Emission

Line Spectrum of Hydrogen

• Effect of Magnetic Field on Cathode

Rays using a Maltese Cross Tube 151 • Wave-particle Duality Concept 172

Explained by de Broglie Equation

• Velocity of an Electron in a Cathode

• Electron Diffraction to Show the

Ray Tube 152 Wave Properties of Electrons 173

• The Function of a Semiconductor 153 • Photoelectric Effect 174

Diode

• Half-wave Rectification and Full-wave • Quantum Theory to Explain

Photoelectric Effect and Einstein’s

Rectification 154 Photoelectric Equation 175

• Transistor and Types of Transistors 155 • To Determine the ThresholdSAMPEL

• Characteristics of a Transistor Circuit 156 Frequency and Work Function

• A Transistor Functions as a Current using Einstein’s Photoelectric

Amplifier 157 Equation 176 – 177

• The Use of a Transistor as an 158 • Problem Solving using Einstein’s 178

Automatic Switch Photoelectric Equation

(vi)

Base Quantities and 1Chapter

Derived Quantities Measurement

Textbook: PP. 6, 7 LS: 1.1.2, 1.1.3

Iv n m l

Length

mol

Amount of

substance

cd Base quantity is a 100FormSAMPEL 4 kg Chap.

Luminous physical quantity kg 1

which cannot be m

intensity Mass

derived from

I another physical 60 5

55 60

A quantity

50 10

Thermodynamic

temperature 50 10

40 20

30

45 15

Electric 40 20

current 35 25

30

Time s

TK t

Kinetic energy = 1 × mass × length2

2 time2

A derived quantity can

be described in terms

of base quantities.

Gravitational force = mass × length

time2

1

Scalar Quantities and 1Chapter

Vector Quantities Measurement

Textbook: PP. 8, 9 LS: 1.1.4

Vector quantities are Scalar quantities are

physical quantities physical quantities

that have magnitude that have magnitude

and direction. only.

Example: Example:

The Saturn V rocket The rocket Saturn V

engine produces a has a mass of

force of 35 100 kN 2 970 000 kg.

which accelerates it

towards the sky.

SAMPELEXTRA NF

Distance and speed are scalar quantities

while displacement and velocity are

vector quantities.

2

Factors Affecting Liquid 2Chapter

Pressure Pressure

Textbook: PP. 41 – 44 LS: 2.1.1, 2.1.2

Liquid pressure can be calculated using the following formula:

P = hρg

where P = liquid pressure, S.I. unit is Pascal (Pa) [1 Pa = 1 N m–2]

h = depth of liquid, S.I. unit is metre

ρ = density of liquid, S.I. unit is kg m–3

g = gravitational acceleration, S.I. unit is m s–2

Depth of liquid (h) Density of liquid (ρ)

Atomospheric pressure, Patm

h Nozzles

h Water jets

h

Water Cooking oil

FormSAMPEL 5

t 5IF greater the depth of a liquid, the t 5IF EFQUI PG CPUI MJRVJET JT UIF TBNF Chap.

further the horizontal distance of the t %FOTJUZ PG XBUFS JT HSFBUFS UIBO UIF PJM 2

spurting liquid. t 5IF IPSJ[POUBM EJTUBODF PG UIF TQVSUJOH

t 5IF greater the depth of a liquid, the of water is greater than the oil.

greater the liquid pressure. t 5IF greater the density of a liquid,

P the greater the liquid pressure.

P

h ρ

EXTRA NF Gravitational acceleration (g)

Liquid pressure is not affected by the shape P

of a container.

A BC D E g

P is directly proportional to g.

105

Solving Problems Involving 2Chapter

Pressure

Pressure in Liquids

Textbook: PP. 45 – 46 LS: 2.1.3

Measuring Liquid Pressure using a Mercury-filled U-tube Manometer

Atmosphere Atmosphere 3

3 2 P2 = 10 000 Pa

2

1 1 h h = 10 cm is the

P1 0 P2 difference in

1 0 height. It shows

2 the difference in

3 P1 1 pressure between

Mercury 2 P1 and P2.

The mercury level in both columns 3

is the same, so:

Mercury, ρhg = 13 600 kg m–3

The pressure, P1 = P2 g = 9.81 m s–2

Measuring Liquid Pressure using a The mercury level in the left column is

U-tube Manometer filled with lower than that in the right column.

Different Liquids Pressure P1 is greater than P2 by h cm.

P1 = P2 + hρg

= 10 000 + 0.1(13 600)9.81

= 23 341.6 Pa

Liquid x ρy = 1 500 kg m–3 What is the density of liquid X?

Liquid y

h1 = 15 cm Pressure of = Pressure of

W h2 = 10 cm

liquid Y at level W liquid X at level W

Mercury

h1ρxg = h2ρyg

15 × ρx = 10 × 1 500

ρx = 1 000 kg m–3

QUICK SAMPEL

LINK

Scan the QR code to watch a video about manometers.

106

Applications of Pressure 2Chapter

in Liquids in Our Lives Pressure

Textbook: P. 47 LS: 2.1.4

Position of water tank Construction of a dam

in the house

The wall of a dam is built thicker

Tank at The difference in at the base because water

the top height between pressure increases with depth.

the water tank The thicker wall can withstand

and the water the high water pressure.

tap produces a

high pressure at

the tap.

Position of intravenous liquid Low pressure Thin

wall

The pressure due to the height of the

intravenous bottle will push the Thick

intravenous liquid into the body. High pressure wall

Intrarenous FormSAMPEL 5

liquid

Chap.

2

Use of the siphon

A siphon is used to transfer

water from a higher region to a

lower region due to the

difference in pressure between

the two water levels.

What is the pressure on small 10 m

fishes if the density of the sea

water is 12 000 kg m–3?

Scan for answers

107

Effects of Atmospheric 2Chapter

Pressure at Extreme Depth Pressure

under the Surface of the Sea

Textbook: P. 55 LS: 2.1.4

Under the surface of the sea, water pressure increases with depth with every 10 m

increase causing water pressure to increase by 1 atmosphere.

Oceanic

200m Effects on Humans

10 ºc t #PEZ UJTTVFT BCTPSC

700 to 1000m excess nitrogen gas

4 ºc t *OBCJMJUZ UP UIJOL DMFBSMZ

2000 to 1000m

Effects on Divers

6000m

t 8FBS B EJWJOH TVJU UP

Oceanic slow down heat loss

divisions from the body.

10000m t 4MPXMZ EFTDFOE EPXO

the sea level so that

nitrogen bubbles do not

form in the tissues and

blood vessels.

A submarine is made of steel or SAMPEL

titanium with a circular cross section

to withstand the high pressure of

the surroundings. Pressure in the

cabin is controlled to almost the

same level as sea level for the

comfort of the crew. Oxygen tanks or

electrolytic oxygen generators are

provided so that the oxygen supply

is adequate.

EXTRA NF

A diver who dives to a depth of 30 m will experience a pressure 4 times the normal

atmospheric pressure.

110

Gas Pressure (Manometer) 2Chapter

Pressure

Textbook: P. 56 LS: 2.3.1

Gas particles Kinetic Theory of Gas

Gas pressure t 8IFO HBT QBSUJDMFT IJU BHBJOTU UIF

XBMM

UIFJS NPNFOUVN DIBOHF

t 5IF SBUF PG DIBOHF PG

NPNFOUVN QSPEVDFT GPSDF

t 'PSDF QFS VOJU BSFB QSPEVDFT

QSFTTVSF

t 5IF HSFBUFS UIF SBUF PG DPMMJTJPOT

PG HBT QBSUJDMFT IJUUJOH UIF XBMM

UIF HSFBUFS UIF HBT QSFTTVSF

QSPEVDFE

A manometer DPOTJTUT PG B HMBTT 6 UVCF DPOUBJOJOH DPMPVSFE XBUFS 5IJT BQQBSBUVT JT

VTFE UP NFBTVSF HBT QSFTTVSF

Patm Patm

Before connecting to a gas supply FormSAMPEL 5 Chap.

1SFTTVSF BU " 1SFTTVSF BU # 2

"UNPTQIFSJD QSFTTVSF

AB

Pgas Patm After connecting to a gas supply

1SFTTVSF BU " (BT QSFTTVSF

PHBT

h 1SFTTVSF BU # 1SFTTVSF EVF UP

AB

NFSDVSZ DPMVNO h +

BUNPTQIFSJD QSFTTVSF

h DN )H PBUN

1PJOU " BOE QPJOU # BSF UIF TBNF MFWFM

1SFTTVSF BU " 1SFTTVSF BU #

PHBT h DN )H PBUN

111

Solving Problems Involving 2Chapter

Pressure

Gas Pressure

Textbook: PP. 58, 59 LS: 2.3.2

Problem solving 1

Pgas Patm = 76 cm Hg The height of the mercury column is

Mercury the same on both sides.

l = 13 600 kg m–3

Pgas = Patm

= 76 cm Hg

Pgas = hρg

= 0.76 × 13 600 × 9.81

= 1.01 × 105 Pa

Pgas Patm = 76 cm Hg Problem solving 2

h = 10 cm h

The height of the mercury column on the

right side is higher than on the left side.

Pgas > Patm by h cm Hg

Pgas = Patm + h cm Hg

Pgas = 76 cm Hg + 10 cm Hg

= 86 cm Hg

Pgas = hρg

= 0.86 ×13 600 × 9.81

= 1.14 × 105 Pa

Pgas Patm = 76 cm Hg Problem solving 3

h = 10 cm h

The height of the mercury column on the

right side is lower than on the left side.

Pgas < Patm by h cm Hg

Pgas = Patm – h cm Hg

Pgas = 76 cm Hg – 10 cm Hg

= 66 cm Hg

Pgas = hρg

= 0.66 × 13 600 × 9.81

= 8.8 × 104 Pa

SAMPEL

112

Pascal’s Principle 2Chapter

Pressure

Textbook: PP. 60 – 62 LS: 2.4.1, 2.4.2

Pascal‘s principle states that the pressure applied on an enclosed fluid is transmitted

uniformly in all directions in the fluid.

Hydraulic System as a Force Multiplier

A hydraulic system is a force multiplier system. When the value of the surface area A2

is greater than the surface area A1, the force on the output piston F2 is greater than

the force on the input piston F1.

Input force F1 is F1

applied on the

A2 > A1

input piston.

Area = A1

Pressure at Area = A2 Pressure at

input piston A F2 output piston

P1 = F1 Fluid B P2 = F2

A1 A2

Based on Pascal’s principle Pressure is transmitted uniformly throughoutFormSAMPEL 5 Chap.

the hydraulic fluid to the output piston. 2

P1 = P2

F1 = F2

A1 A2

F2 = AA12(F1)

The multiplying factor is m = AA21.

Based on the hydraulic system F1 = 20 N Weight

in the diagram, what is the A1 = 2 m2 A2 = 200 m2

maximum weight that can be

lifted by the system?

Scan for answers

113

Applications of Pascal’s 2Chapter

Principle Pressure

Textbook: P. 63 LS: 2.4.3

Working Principle of the Hydraulic Brake System

t 8IFO F1 JT BQQMJFE PO UIF QJTUPO JO UIF Brake shoes

NBTUFS DZMJOEFS

P1 is produced. Piston Brake pad

t 5IF QSFTTVSF JO UIF IZESBVMJD

PJM JT USBOTNJUUFE VOJGPSNMZ UP

UIF MBSHFS QJTUPOT Piston

t 1SFTTVSF QSPEVDFE JO MBSHFS

QJTUPOT JT P2. Master cylinder

t #BTFE PO 1BTDBM T QSJODJQMF

F1

Disc brakes Drum brakes

P1 = P2.

t #FDBVTF A2 > A1 so F2 > F1. P1 = F1

A1

" CJHHFS GPSDF JT QSPEVDFE JO

UIF MBSHFS QJTUPO UP TUPQ UIF F2

A2

NPWFNFOU PG UIF XIFFMT P2 =

5IF NVMUJQMZJOH GBDUPS A2

A1

t "MM UIF GPSDFT QSPEVDFE JO UIF GPVS MBSHFS QJTUPOT NVTU CF UIF TBNF UP QSFWFOU

TLJEEJOH PG UIF WFIJDMF

Working Principle of the Hydraulic Jack System

t 8IFO F1 JT BQQMJFE UP UIF TNBMM Large piston

QJTUPO

QSFTTVSF P1 is produced.

Reservoir oil

t 5IF QSFTTVSF JO UIF IZESBVMJD PJM

JT USBOTNJUUFE VOJGPSNMZ UP UIF

MBSHF QJTUPO )ZESBVMJD PJM GMPXT Pivot

GSPN UIF SFTFSWPJS UBOL UP UIF P2 = F2 Pump

MBSHFS QJTUPO UP MJGU UIF PCKFDU A2

t 1SFTTVSF QSPEVDFE JO MBSHF Small piston

QJTUPO JT P2. Ball valves

t #BTFE PO 1BTDBM T QSJODJQMF

P1 = F1

P1 = P2. A1

t Because A2 > A1 so F2 > F1.

t " CJHHFS GPSDF JT QSPEVDFE JO UIF MBSHF QJTUPO UP MJGU UIF PCKFDU SAMPEL

5IF NVMUJQMZJOH GBDUPS AA21

t " SFMFBTF WBMWF JT PQFOFE UP MPXFS EPXO UIF PCKFDU 5IF IZESBVMJD PJM XJMM GMPX CBDL

UP UIF SFTFSWPJS

114

Buoyant Force,F B 2Chapter

Pressure

Textbook: PP. 66, 67 LS: 2.5.1

FB

Two forces act on the object:

Wg = Weight of the object due to gravity

in a downward direction.

FB = Buoyant force on the object in

Wg upward direction.

Buoyant force is the force acting upwards on an object which is partially or

wholly immersed in a fluid.

Buoyant force = Weight of fluid displaced where ρ = density of fluid

FB = ρVg V = volume of displaced fluid

g = acceleration due to gravity

F1 h1 P = hρg FormSAMPEL 5 Chap.

r P1 2

h h2 Pressure on the top surface, P1 = h1ρg

Force acting on the top surface, F1= P1A = h1ρgA

r P2 Pressure on the bottom surface, P2 = h2ρg

F2 Force acting on the bottom surface, F2 = P2A = h2ρgA

Resultant force, F = F2 – F1

= h2ρgA – h1ρgA

= ρA(h2 – h1)g

= ρAhg

= ρVg where, V = Ah

Buoyant force FB = Resultant force F

FB = ρVg

where ρ = density of liquid

V = volume of liquid displaced = volume of

submerged part of object

g = acceleration due to gravity

EXTRA NF

Buoyant force on an object in water with density, p = 1 000 kg m–3; g = 10 m s–2

V = volume of submerged part of object V = volume of submerged part of object

= 4 m3

= 5 m3 1 m3

FB = ρVg 5 m3 FB = ρVg 4 m3

= 1 000 × 5 × 10 = 1 000 × 4 × 10

= 5 × 104 N = 4 × 104 N

115

Archimedes’ Principle 2Chapter

Pressure

Textbook: PP. 67 – 70 LS: 2.5.2

Archimedes’ principle states that an object which is partially or fully immersed in

a fluid experiences a buoyant force equal to the weight of the fluid displaced.

Weight of object = Actual weight of

in air object

0.40 N = 0.67 N

Eureka can

Weight of object = Apparent weight of

0.67 N in fluid object

= 0.40 N

Buoyant force = Apparent loss in weight

Buoyant force = Actual – Apparent

weight weight

FB = 0.67 – 0.40

= 0.27 N

Weight of liquid displaced = 0.37 – 0.10

(Initial reading) 0.10 = 0.27 N

(Final reading) 0.37

Newtons

Buoyant force = Weight of liquid displaced

Buoyant force = Apparent loss in weight

Relationship between the Equilibrium of Forces and the States of Floatation

of an Object in a Fluid

FB

FB

W Based on the diagram

on the left, what will

W happen to the object if

the weight of the

Weight of = Buoyant force Buoyant force < Weight of object is less than the

buoyant force?

object of object of object object

Scan for answers

W = FB FB < W SAMPEL

Resultant force = 0 Resultant force ≠ 0

Object is stationary. Object will move

Floats. downwards with

acceleration, a.

Object will sink.

116

Applications of Archimedes’ 2Chapter

Pressure

Principle in Daily Life

Textbook: PP. 71 – 75 LS: 2.5.3

Submarine Diving Submerged Rising Surfaced

Surfaced

The ballast The ballast High Low

tanks are tanks makes presure presure

opened and the submarine

water enters less buoyant, air air

the tanks and it dives

Compressed The compressed

gas is released air displaces the

into the ballast water back into

tanks the ocean and

makes the sub-

marine more

buoyant

A submarine submerges and emerges in the sea by controlling its weight

t 8IFO XBUFS JT QVNQFE JOUP UIF CBMMBTU UBOL

UIF XFJHIU PG UIF TVCNBSJOF JODSFBTFT

When the weight of the submarine is more than the buoyant force of the

submarine, it will submerge into the sea.

t 8IFO UIF XBUFS JO UIF CBMMBTU UBOL JT QVNQFE PVU GSPN UIF TVCNBSJOF

the weight

of the submarine decreases and is less than the buoyant force. The submarine

will emerge in the sea.

FormSAMPEL 5 Chap.

2

Hot Air Balloon

Moving up Constant height Moving down

FB FB FB

FB > W FB = W FB < W

W W W

" IPU BJS CBMMPPO BMTP BQQMJFT "SDIJNFEFT QSJODJQMF #Z DPOUSPMMJOH UIF XFJHIU PG UIF

BJS EJTQMBDFE JO UIF IPU BJS CBMMPPO

UIF CVPZBOU GPSDF PO UIF IPU BJS CBMMPPO XJMM

DIBOHF *U XJMM JOGMVFODF UIF NPWFNFOU PG UIF IPU BJS CBMMPPO

117

Bernoulli’s Principle 2Chapter

Pressure

Textbook: PP. 78 – 80 LS: 2.6.1

Bernoulli’s principle states that when the velocity of a fluid increases, the

pressure in the fluid decreases and vice versa.

An activity to show that the greater velocity of a fluid produces a lower pressure region

AA. PPaappeerr

Velocity of air flow is Velocity of air flow is

low. A high pressure high. A low pressure

region is produced. region is produced.

F

The different pressure between two regions produce a lift force F upwards.

The paper is lifted upwards.

BB. Venturi tube

Lowest pressure

Highest hh Lowest Highest h

pressure h pressure

tt pressure t h

AB t

C A th t

Water Water B C

vA = vB = vC ⇒ The size of the tube is vB > vc > vA ⇒ The size of the tube is

constant narrower at B

PA > PB > Pc ⇒ Fluid flows from high PB < PC < PA ⇒ The greater the velocity

pressure to low pressure, of the fluid, the smaller the pressure,

hA > hB > hc hB < hC < hA.

Predict the height of Air Ar B SAMPELCr

the water column. r

Scan for answers

118

Lift Force 2Chapter

Pressure

Textbook: PP. 80 – 82 LS: 2.6.2

Effect of Lift Force on a Ping Pong Ball

Moving air t "JS GMPXT BU B high velocity BCPWF UIF

P

QJOH QPOH CBMM EVF UP UIF OBSSPX

Less Q Less

Still air QBUIXBZ

1

t Low pressure JT QSPEVDFE at P.

F t 5IF WFMPDJUZ PG UIF BJS JT low at Q TP JU

QSPEVDFT B region of high pressure.

t 5IF difference in pressure between P

and Q QSPEVDFT B resultant force F

upwards XIJDI JT UIF MJGU GPSDF F.

t 5IF MJGU GPSDF F MJGUT VQ UIF QJOH QPOH CBMM

Production of Lift Force by the Aerofoil FormSAMPEL 5Air flows faster, so a lowChap.

2

Lift pressure region P1 is

F produced.

P1

Aerofoil

P2

Air flow is slower, so a high

pressure region P2 is

produced.

t 5IF BFSPGPJM TIBQF PG UIF XJOH PG BO BFSPQMBOF DBVTFT BJS UP GMPX BU EJGGFSFOU

TQFFET QBTU UIF UPQ TFDUJPO BOE UIF CPUUPN TFDUJPO

t #BTFE PO #FSOPVMMJ T QSJODJQMF

P2 > P1 5IF EJGGFSFODF JO QSFTTVSF QSPEVDFT B MJGU

GPSDF

F BDUJOH VQXBSET PO UIF BFSPQMBOF

QUICK

LINK

Scan the QR code to watch a video on Bernoulli’s principle in the

ping pong ball activity.

119

Applications of Bernoulli’s 2Chapter

Principle in Daily Life Pressure

Textbook: P. 83 LS: 2.6.3

Aeroplane A lift force is produced due to the

difference in air pressure between the

Lift upper part and lower part of an aeroplane.

Thrust t -JGU GPSDF 8FJHIU

→ The aeroplane increases its altitude.

Weight

Drag t -JGU GPSDF 8FJHIU

→ The aeroplane decreases its altitude.

t -JGU GPSDF XFJHIU

→ The aeroplane maintains its altitude.

Bunsen Burner t 8IFO UIF HBT GMPXT BU IJHI TQFFE

UISPVHI B OBSSPX QBTTBHF JO UIF

Atmospheric Higher speed, narrow nozzle, it produces a low

pressure lower pressure QSFTTVSF SFHJPO BU P.

Lower speed, Nozzle t 5IF PVUTJEF BJS XIJDI JT BU BUNPTQIFSJD

higher pressure P pressure is drawn in and mixes with

Gas

Air flow UIF HBT

t 5IF NJYUVSF PG HBT BOE BJS FOBCMFT

complete combustion.

Curved path of a footfall

Force

P1 P2

t Air flows faster, so low pressure SFHJPO

P1 is produced because the direction SAMPEL

of rotation of the ball is the same as the direction of flow of the wind.

t Air flows slower, so high pressure SFHJPO

P2 is produced because the direction

of rotation of the ball is opposite to the direction of flow of the wind.

t 5IF EJGGFSFODF JO QSFTTVSF QSPEVDFT a force F to the left. So the ball curves to

the left.

120

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SASBADI SDN. BHD. 198501006847 ISBN 978-983-77-24-41-9

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