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Physics Module Form 4 Chapter 3 – Forces & Pressure GCKL 2011

3-1

U N D E R S T A N D I N G P R E S S U R E

Define pressure. Pressure is defined as the __ force __ applied on a unit __ surface area __.

P =

A

F

1 Pa = __1_ N m-2

Example : The diagram shows a wooden block of dimensions 8 cm × 10 cm ×

12 cm. Its weight is 12 N. On which side should the wooden block

be placed to produce a maximum pressure exerted on the table.

What is value of this pressure ?

P = Weight (F) .

Minimum Area (A)

= ___12____

(0.08)(0.10)

= 1500 Pa

Problem Solving

1. Calculate the pressure on a wall when the palm of a hand with area 150

cm2 is pushed against the wall a force of 210 N.

P = F/A = 210 / 150

= 1.4 N cm-2

2. A teacher who weights 637N has a foot with a surface area of 200 cm2.

When he stands with one foot, calculate the force applied.

P = F/A = 637 / 200

= 3.185 N cm-2

Relationship between

pressure and force,

pressure and surface

area

Situation 1 : Figure 3.1a, press a thumbtack into a piece of wood with your

thumb where as Figure 3.1b, hit using the hammer.

Figure 3.1(a) Figure 3.1(b)

(a) Observe Figure 3.1a and Figure 3.1b and state one comparison.

___ Figure 3.1 (b) sink deeper compare to Figure 3.1(a).__

F = Force (N) A = surface area (m

2)

P = Pressure (__Pa)

3.1


3-2

(b) Which one is received a higher force?

____ Figure 3.1(b) received higher force.____

(c) State the relationship between pressure and force.

The pressure produce increases with force applied .

Situation 2 : A 500 g cardboard box placed on a table in two different

orientations.

Figure 3.2a Figure 3.2b.

(a) Calculate the pressure exerted on the table for

i) figure 3.2a ii) figure 3.2b.

5 / 12 x 20 5 / 10 x 20

= 0.028 N cm-2

= 0.025 N cm-2

(b) Which figure shows that the higher pressure is exerted to the table?

__ figure 3.2(b)_

(c) State the relationship between pressure and surface area.

____ Pressure is inversely proportional to the surface area.__

Factor that affecting

pressure.

- Force

- Surface Area

1. Ways to increase pressure are :

a) __increase _force ; b) __decrease __area of contact.

2. Ways to decrease pressure are :

a) ___ decrease __force ; b) __ increase _area of contact.

The applications of

pressure in our daily

life.

a) High pressure.

Tools like knives, chisels, axes and saws have sharp cutting edges. The

surface area of contact is _small_ When a force is applied on the tool, the

small area of contact will produce a _ higher__ pressure to cut the

material.


3-3

b) Low pressure.

The flat base of each metal pole of a tent has a __big__ surface area to

_reduce_ the pressure exerted on the ground. The poles will not sink

into the ground because of the flat bases.

Exercise

1. A cylinder has a mass of 12 kg and a

cross-sectional area of 200 cm2. What

is the pressure acting at its base?

A. 6 kPa B. 9 kPa

C. 12 kPa D. 15 kPa

2. Wind blows normally on a wall at a

pressure of 200 kPa. If the wall has

an area of 5 m2., what is the force

acting on the wall?

A. 40kN B. 800kN

C. 1000KN D. 1200kN

3. A 250 N force is applied at an angle 30o to the

surface of a block. The surface is a square of

sides 10 cm. What is the pressure exerted on

the surface?

A. 1.25 kPa B. 2.50 kPa

C. 25.0 kPa D. 12.5 kPa

4. A balloon is pressed with a force of 2.4 N

using a finger. Find the pressure exerted on

the balloon if the area of fingertip is 2.0 x

10-4

m2

P = F/A = 2.4 / 2 x 10-4

= 12 000 Pa

5. The diagram below shows a concrete block of dimension 1.5 m x 2.0 m x 3.0 m.

Its weight is 60N. Calculate (a)maximum pressure, (b)minimum pressure:

a) maximum pressure b) minimum pressure

= 60 / 2.0 x 1.5

= 20 Pa =60 / 2.0 x 3.0

= 10 Pa


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6. A Perodua Myvi with a mass of 900 kg. Air pressure for each tyre is 2 x 105 Pa.

(a) What is the meaning of pressure ?

___ force exerted per unit area.___

(b) Calculate the area in contact with the ground for each tyre.

A = (9000 / 4) / 2 x 105

= 0.01125 m 2

c) Zamani drives his car to his school with a distance of 10 km and find his car tyre become harder

than usual. Explain why this is happen?

________ pressure increase with temperature____

________________________________________________________________________

7. Experiment : Based on the diagram below, complete the suitable planning experiment.

a) Inference :… Pressure produce depends on surface area.…

b) Hypothesis :… The bigger the surface area, the lesser the pressure produced……

c.i) Aim :… To investigate the relationship between surface area and pressure. ……

ii) Variables i) Manipulated : … surface area , A …………………………

ii) Responding : … the depth of wood sink in plasticine, P (pressure)

iii) Constant : … Force applied ( mass of weights )………

iii) Apparatus

& material

…… placticine, 100 g metal weights, wooden support , wooden disc (different

sizes ) and ruler…….


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iv) Diagram

~~ same as diagram in the question ~~

v) Procedure …1. The wooden disc with surface area 2 cm2 is put under the wooden

support.

2. Placed 100 g metal weight on the top of wooden rod as shown in diagram

above.

3. Measured the depth of wooden sinks in the plasticine using the ruler

4. The experiment is repeated using different sizes of wooden disc, 4 cm2 , 6

cm2 , 8 cm

2 and 10 cm

2.

vii) Result

A / cm2 2 4 6 8 10

P / cm

viii Analyse

data

P/ cm

A/ cm2


3-6

UNDERSTANDING PRESSURE IN LIQUIDS

Characteristics

of pressure in

liquid

1. Pressure in liquids acts in _ all _ directions.

2. Draw the jet of water.

3. Draw the water level in the container.

diagram ( All water level are same.)

4. The hole in both containers are at the same height. Draw the jet of water and oil.

- The higher the density the _ higher __ the pressure.

-

Shape and size _ does

not___ influence the

pressure

Pressure in liquids _ increase __ with depth.

water

oil

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3-7

Formula for

pressure in

liquid

The pressure in a liquid is the product of depth, density and gravitational acceleration.

P = x x

P = h x ρ x g

Example :1. An air bubble is at a depth of 5 m below the surface of a lake . What

is the pressure of water on the bubble if the density of the water is

1000 kg m-3?

P = h ρ g

= 5 x 1000 x 10

= _50 000_ Pa

2. The figure shows a high tin with a length of 100 cm is filled to the full

with water.

If the pressure caused by the water at point P is 8000 Pa, What is the

value of L?(Density of water = 1000 kg m - 3 )

h = 8 000 / ( 1000 x 10 )

= 0.8 m = 80 cm

l = 100 – h = 20 cm

Experiment : Pressure and density in liquids

a. Inference :…………………………………………………………………………………………

b. Hypothesis :…………………………………………………………………………………………

c. Aim :…………………………………………………………………………………………

d. Variables i) Manipulated : ……………………………

ii) Responding : ……………………………

iii) Constant : ……………………………

P = Pressure ( Pascal)

ρ = density ( _ kg m-3

_)

g =__ N kg-1

____

h = height (m)


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e. Apparatus

& material

…………………………………………………………………………………………

…………………………………………………………………………………………

f. Procedure …………………………………………………………………………………………

…………………………………………………………………………………………

…………………………………………………………………………………………

…………………………………………………………………………………………

…………………………………………………………………………………………

…………………………………………………………………………………………

………………………………………………………………………………………….

g. Result

h. Analyse

data

: Graph of ……………………………………………………………………………..

………………………………………………………………(draw on graph paper )

i. Conclution :…………………………………………………………………………………………

………………………………………………………………………………………….

j. Precaution :…………………………………………………………………………………………

………………………………………………………………………………………….

Applications of

pressure in

liquid

1. 2.


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Exercise

1.

2.

Which of the following factor does

not influence the pressure of a liquid?

A. Depth

B. Acceleration due to gravity

C. Density

D. Volume

Mercury has density of 13600 kg m-3

.

If the pressure of mercury is 650 kPa,

what is the depth from its surface?

A. 4.0 m B. 4.8 m

C. 5.8 m D. 6.4m

4.

The density of water is 1000 kg m-3

.

Find the density of liquid X in kg m

-3?

Px = Pwater

hx x ρx = hwater x ρwater

6 ρx = 2 x 1000

ρx = 333.33 kg m-3

3. Water has density equals to 1 g cm

-3.

What is its pressure at a depth of 12 m

from the surface?

A. 80 kPa B. 100 kPa

C. 120kPa D. 140kPa

5.

A container contains two layers of liquids. An oil layer 45cm thick floats on 60 cm of water. The

densityof oil is 750 kg m-3

and water is 1000 kg m-3

. ( g = 10 m s-2

)

a) What is the pressure exerted on water by the oil layer?

Poil = h x ρ x g = 3375 Pa

b) what is the total pressure exerted on the bottom of the container?

PT = Poil + Pwater = 9375 Pa

6.

The pressure at the bottom of the dam is 2.4 x 106

Pa. What is the depth of water if the density of

water is 1 000 kg m-3

.

h = 2.4 x 106 / ( 1000 x 10 ) = 240 m


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7. Figure below shows a cross-sectional area of the structure of a dam.

Observe the thickness of the wall at the top and bottom part of the dam in the figure above.

Based on the observations:

(a) State one suitable inference that can be made. [ 1 mark ]

(b) State one appropriate hypothesis for an investigation. [ 1 mark ]

(c) With the use of apparatus such as a thistle funnel, beaker and other apparatus, describe

an experimental framework to test your hypothesis.

In your description, state clearly the following:

(i) aim of the experiment,

(ii) variables in the experiment,

(iii) list of apparatus,

(iv) arrangement of the apparatus,

(v) the procedure of the experiment which include the method of controlling the

manipulated variable and the method of measuring the responding variable,

(vi) the way you would tabulate the data,

(vii) the way you would analysis the data. [ 10 marks ]

( Experiment thistle funnel – density & depth )

Water Wall


3-11

U N D E R S T A N D I N G G A S P R E S S U R E A N D

A T M O S P H E R I C P R E S S U R E

Kinetic molecular

theory

Gases are :

Made of … tiny particles ………………………………..

Move … at random …………………………………..

Far apart with a very … weak / low force of….. attraction

Collide each other and collide with the wall of container with …high

energy/speed…

The average kinetic energy ……… increases …when the temperature

…increases…………….

Gas Pressure The gas pressure in a container is caused by the … collision ... of gas molecules

with the …wall… of the container.

Gas pressure can be measured by using 2 types of instrument known as :

(a) . bourdon gauge . gauge (consists of a semi-circular or C-shaped copper

tube that tends to straighten if more and more gas is pumped (compressed) into

it).

(b) … manometer…. (consists of a U-tube about 1 m in height. About 50% of

the volume of the U-tube is filled with liquid such as mercury or water).

Atmospheric

pressure

The atmospheric pressure is caused by the downward … force .. exerted by the

air ( the weight of the atmosphere) on the Earth’s surface.

The greater … the distance … from the sea level, the …lower .. will the

atmospheric pressure.

The atmospheric pressure at sea level is 1 atmosphere ( 1 atm)

1 atmosphere(atm) = …760.. mm Hg = …105…..Pa = 1000 milibar

Find this:

If density,ρ of mercury (Hg) = 13600 kg m-3

and density of water = 1000 kg m-3

.

1 atm = __10.336___ m of water

Barometer is an instrument to measure atmospheric pressure. There are 2 types

of barometer:

(a) .... barometer aneroid....... (is made of a partially vacuum sealed metal box).

(b) … simple barometer …. (is made of a long glass tube about 1 meter in length

fully filled with mercury and then inverted (turned upside down) into a bowl of

mercury).

3.3


3-12

Example 1: The atmospheric pressure is 760 mm Hg. What is the value of the

atmospheric pressure in Pascal? [ Density of mercury, ρ (Hg) = 13 600

kg m-3 ]

h = 760 mm = 76 cm = 0.76 m

Atmospheric pressure, Patm = h ρ g = (0.76)(13600)(9.8)

= …103360 …. Pa

Manometer

1

2

3

h

PA = PB = Patm = 76 cm Hg

PC = PA + PHg PD = PA + PHg

= (76 + 4) cm Hg = 79 cm Hg

= 80 cm Hg

ρHg = 13600 kg m-3

; ρwater = 1000 kg m-3

PA = PB = Patm = 76 cm Hg

PC = PA + PHg PD = Pc = 80 cm Hg

= (76 + 4) cm Hg = 1088 cm water

= 80 cm Hg

PA = Patm = 76 cm Hg

PC = PA + PHg

= (76 + 4) cm Hg

= 80 cm Hg

Pgas = PD = Pc

= 80 cm Hg


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Simple Barometer

1. Figure shows apparatus set up

which is used to measure

atmospheric pressure.

[Mercury density = 1.36 x 104 kg m-

3]

Calculate the pressure at point Q in

Pa.

2. If the atmospheric pressure is 76

cm Hg, what is the pressure of the

trapped air P?

Applications of

atmospheric

pressure.

1.Siphon

2. Vacuum cleaner

………………………………….

…………………………………..

…………………………………..

…………………………………..

…………spm 2009……

…………………………………..

…………………………………..

…………………………………..


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Exercise ( g = 9.8 m s-2

; ρHg = 13.6 x 103 kg m

-3 )

1. What is the pressure of the gas

trapped inside the J-tube, in Pa unit?

A. 1.19 x 105 Pa

B. 1.90 x 105 Pa

C. 2.19 x 105 Pa

D. 2.90 x 105 Pa

2.

3.

Which instruments is meant for

measuring atmospheric pressure?

A. Carburettor

B. Siphon

C. Fortin’s Barometer

D. Hydrometer

Which of the following is not true about

atmospheric pressure?

A. Atmospheric pressure acts in all

directions.

B. Atmospheric pressure decreases with

distance from Earth’s surface.

C. Atmospheric pressure can only measure

in Pa or N m-2

.

3 Diagram below shows a set up of apparatus for measuring atmospheric pressure.

Diagram shows a set up of apparatus for measuring atmospheric pressure.

(a) What is the name for the instrument?

… Simple Barometer ……………………………………………………………………

(b) Determine the atmospheric pressure as measured by the instrument ,

(i) in the cm Hg unit (ii) in the Pa unit

76 cm Hg = 0.76 x 13600 x 9.8

= 101292.8 Pa

(c) State the change of length of the mercury column above the mercury surface

(i) The tube is raised by 10cm ...…… unchanged ………………………………….

(ii) The surrounding temperature increases ………… decrease …………………..

(iii) The instrument is brought to the peak of a mountain …… decrease …………..

(iv) Water vapor is brought to the vacuum region ……… decrease ..……………


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P A S C A L ’ S P R I N C I P L E

Definition

Pascal’s Principle state that pressure exerted on an … enclosed …fluid is transmitted

…… equally in all direction …. to every part of the fluid.

Water spurts out at the _ same ______

speed because the pressure act at

all the point are …same….

Pascal’s

Principle

Remember,

P = F

A

Pressure at both side are same, PA = PB

So that ; FA = FB

AA AB

a) When 20 N of force is applied at piston A, FB produced is

FB = FA AB = 20 x 8.0 FB = _400 N ___

AA 0.47

b) When piston A is moved downward for 2 m, how far the load has been move up.

Volume of liquid transmitted from A to B are equal.

VA = VB ; where as V = Ah

AAhA = ABhB

hB = __0.1 m __

A B

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Application

of Pascal’s

Principle

Gather information about ;

1. Hydraulic brake system

2. Automobile hydraulic lift.

Exercise

1. Figure below shows a hydraulic jack.

Piston A and piston B have cross-

sectional areas 5 cm2 and 100 cm

2

respectively. If mass of 3kg is placed

on piston A, what is the maximum

weight that can be lifted by piston B?

A. 300N B. 600N

C. 800N D. 900N

2.

3.

Which of the following device is

based on the Pascal’s Principle of

pressure transmission?

A. Hydrometer

B. Car’s hydraulic brake

C. Bunsen burner

D. Fire extinguisher

The input piston of hydraulic press is

pushed down 2.0 cm from the original

position. If the cross sectional area of

input piston is 5.0 cm2 and of the

output piston is 50 cm2 , how much

will the load be raised up?

A. 0.2 cm B. 1.0 cm

C. 1.5 cm D. 1.8 cm

4 A hydraulic jack which is used for lifting a car at a service centre for motor vehicles. The

hydraulic jack made up of two pistons X and Y of cross-sectional area 0.02 m2 and 18 m

2

respectively. When the compressor is switched on, a force of 200 N acts on piston X.

(a) Name the physics principle applied in the hydraulic jack.

__ Pascal’s principle _______________________________________

(b) Explain how the car is lifted

__when piston X is pushed down. The pressure exerted is transmitted equally to the

piston Y. the big force produced then push the car upward._________


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(c) State two advantages of using oil as the hydraulic fluid of water.

__cannot be compress ; reduce friction. _____________

(d) Calculate the pressure exerted on piston Y.

10 000 Pa

(e) Calculate the maximum weight of a load that can be lifted by piston Y.

180 000 N

(f) If piston X move down by 45 cm, what is the distance moved by piston Y?

0.05 cm

5. Cross section area of piston A and piston B are 4000 cm2 and 6 000 cm

2 respectively.

i) Calculate the ratio of the force acting on piston A with the force acting on piston B

2 : 3

ii) If the piston B lifts a car 2.50 m high, how far to the right should piston A move?

3.75 m

iii) Why the system less effective when there is air bubbles in the piston.

… some force are used to press the air bubbles.…….

6. Given that the cross-sectional area of smaller piston and larger piston are 20 cm2 and 60 cm

2

respectively. If the smaller piston is pushed with a force of 15 N.

Calculate;

a) i) Pressure exerted on smaller piston. ii) Pressure exerted on larger piston

15 / 20 = 0.75 N cm-2

PL = Ps = 0.75 N cm-2

b) the ratio of pressure acted on the smaller piston and larger piston

1 : 1

c) the force experienced by the larger piston

FL = 0.75 x 60 = 45 N


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A R C H I M E D E S ’ P R I N C I P L E

Definition

Archimedes’ Principle: When an object is __ fully __ or __partially __ immersed in a

fluid, the upthrust or ___buoyancy __ on it is equal to the __weight __ of fluid displaced.

Relationship

between ;

1. FB with

liquid

displaced

2. FB with

Weight of

floating

object

Simple activity to show the presence of buoyant force

1. Determine the actual weight of plasticine and the apparent weight of the plasticine

in water.

Actual weight = ………………………

Apparent weight = ………………….

(The weight of plasticine in water)

Loss in weight = …………………….

Weight of = Buoyant force

Water Displaced

Volume of liquid displaced = Volume of _ object immersed

2. Floating object

Buoyancy

Force, FB

FB = buoyant force (N)

FB = Wliquid displaced ; W = mg ; m = V ρ ρ = density (kgm-3

)

g = 10 ms-2

FB = __ Vρg __ V = volume (m3)

Example 1. An object of density, 40 g cm-3

and mass 500 g is immersed in a liquid of density

2 g cm-3

. Calculate;

a) the volume of liquid displaced b) the mass of the liquid displaced

Floating Buoyant force = weight of object

F W

3.5


3-19

V = 500 / 40 = 12.5 cm3 m = 2 x 12.5 = 6.25 g

c) the buoyant force experienced by the object (g =10 m s-2

)

FB = 0.0625 N

Application

of

Archimedes’

Principle

1. Submarine

2. Hot Air Ballon

3. Hydrometer

Exercise

1. A ship of mass 80000 kg floats on the

sea surface. If the density of the sea

water is 1250 kg m-3

, what is the

volume of the displaced sea water?

A. 6.4 m3 B. 64 m

3

C. 640 m3 D. 800 m

3

2. The diagram shows an object partly floated and

immersed in a liquid. What was the mass of the

object. ( g = 10 m s-2

)

A. 2.0 N B. 20 N

C. 0.020 kg D. 200 g

3. The diagram shows an object immersed

in two liquids of different densities.

Which of the following related F1 and F2

correctly?

A. F1 < F2 B. F1 = F2 C. F1 > F2

4. A body of density 800 kg m-3

floats in liquid A

and liquid B as shown. If the density of liquid B is

1000 kg m-3

, what is the possible density of

liquid A?

A. 790 kg m-3

B. 905 kg m-3

C. 1000 kg m-3

D. 1050 kg m-3


3-20

5. (a) A fisherman finds that his boat is at different levels in the sea and in the river, although

the boat carries the same load. The density of sea water is 1 025 kg m-3

and of river water

is 1 000 kg m-3

.

Figure 1 and 2 illustrate the situation of the boat in the sea and in the river.

(i) What is meant by density?

___________________________________________________________________

(ii) Using Figure 1 and 2, compare the levels of the boat and the volumes of water

displaced by the boat.

………………………………………………………………………………………

………………………………………………………………………………………..

………………………………………………………………………………………..

………………………………………………………………………………………..

………………………………………………………………………………………..

Relating the mass of the boat with its load, the volume of water displaced and the

density of the water, deduce a relevant physics concept.

…………………………………………………………………………………………

…………………………………………………………………………………………

…………………………………………………………………………………………

(iii) Name the physics principle that explains the above situation.

…………………………………………………………………………………………….

b) A submarine can sail on the sea surface and under the sea. Explain how a submarine on the

surface submerges.


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(3 marks)

c) Figures 3.5(c) and 3.5(d) below illustrate the working principle of a hydrometer.

The depth to which the test tube sinks depends on its surrounding liquid.

Draw a diagram that shows the design of your hydrometer and in your explanation, emphasis

the following aspects:

(i) the stability of the hydrometer

(ii) the sensitivity of the hydrometer

(iii) the ability to measure a wide range of densities of liquids

(iv) the calibration of the hydrometer. [10m]


3-22

B E R N O U L L I ’ S P R I N C I P L E

Definition

AEROFOIL

In a steady flow of fluid, the _ pressure __in the fluid is _decreases__ when the velocity

of the fluid is high and vice versa.

Figure below shows an __ aerofoil__ The upper region of the aerofoil has _ higher _ air

velocity than the lower region of the aerofoil.

By Bernoulli’s principle, the lower region has __higher ___ pressure than the upper

region of the aerofoil. This causes a __lifting force act____ on the aerofoil.

Situation

that

involved

with

Bernoulli’s

Principle

1.

Reminder : 1. Bernoulli’s Principle is obeyed only for moving fluid . 2. Fluid will move from high pressure into the low pressure.

Water flows through a uniform tube

from _ higher __ pressure to a __lower __

pressure area. Fluid pressure decreases

linearly;

PA > PB > PC

VA < VB < VC

Therefore hA > hB > hC

3.6


3-23

2. Venturi tube

3.

4.

Exercise

1. If the height of the fluid in tube L is h1,

and that is tube M is h2, which of the

following is true?

A. h1 > h2 because v1 > v2

B. h1 < h2 because v1 < v2

C. h1 > h2 because v1 < v2

D. h1 < h2 because v1 > v2

2. Aeroplane wingsact as aerofoils. What is the

funtion of an aerofoil?

A. To raise up the aeroplane.

B. To increase the speed of air flow at the surface

of wing.

C. To reduce the air resistance and the drag force.

Water flows through a non uniform tube,

the velocity of the fluid;

VB > VC > V A VC is the highest because water is forces to

flow through a constriction at A, hence it

speed up.

Pressure of fluid;

PA > PC > PB

Therefore; hA > hC > hB

When the air is not flowing the pressure is

same at every point. When the air is

allowed to flow in, the velocity of the fluid;

VB > VC > V A Because the cross-sectional area at B is the

smallest.

Pressure of fluid;

PA > PC > PB

Therefore; hB > hC > hA

Water flows around the ping pong ball at _ high__ speed

through the filter.

Higher atmospheric pressure produces a __ upward force_

which is larger than __downward force __ of ping pong

ball.

Ping pong ball is pushed __ upward __ by upward force

and does not drop.


3-24

3. Which of the following instruments is not

based on Bernoulli’s principle?

A. Bunsen burner

B. Car carburettor

C. Hydraullic jack

D. Flight of an airplane

4. Which of the following relates two quantities in

Bernoulli’s principle?

A. Velocity and force

B. Pressure and momentum

C. Velocity and pressure

D. Force and momentum

5. A boy standing by the road side. When a big lorry passed in front of him, the boy feels a pull towards

the lorry. This phenomenon can be explained by

A. Pascal’s principle

B. Bernoulli’s principle

C. Archimedes’ principle

6. As a researcher in a boat manufacturing company. , you are assigned to study metal characteristics

used to make the boat.

You are given four choices of metals P, Q, R and S. The table below shows

the characteristics for the four metals.

Metal Shape Density

Kgm3

Specific Heat

Capacity

Strength

P Streamlined 900 Low High

Q Oval 452 High High

R Circle 387 Low High

S Streamlined 500 High High

Table 1

Based on table 1

i) Explain the suitable characteristics of the metal to be used as the material to make the boat.

( 8 marks)

ii) Determine the most suitable metal to be used as the material to make the boat and give your

reasons. ( 2 marks)


3-25

7.

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