1. FIZIKMOZAC 2010
CHAPTER 3: FORCES & PRESSURE
ANSWER
3.1 PRESSURE
Question 1
(a) Pascal = Nm-2
(b) A = 800 x 2 x 10-3
= 1.6 m2
P = 500 = 312.5 Pa
1.6
(c) A sharp knife has a small surface
area. So it produce larger
pressure on the bread.
Question 2
(a) Pressure = force
area
(b) (i) balloon B
(ii) pressure on balloon B is
higher
(iii) the surface of the needle in
contact with the balloon is
smaller than the finger.
(iv) the smaller the surface
area, the larger the pressure
exerted on the balloon.
(v) Pressure increases
Question 3
(a) (i) The contact area between
the wheels in Diagram 6.2 is
larger
(ii) weight are equal
(b) Vehicle uses the wheels in
Diagram 6.2. Because it has
smaller pressure exerted on
the soft ground and will not
sink.
(c) Pressure
(d) 10 000 : 500
4A 2A
2,500 : 250
10 : 1
(e) When the air pressure inside
the wheel lower, the contact
area is larger so the pressure
on the ground is smaller
3.2 LIQUID PRESSURE
Question 4
(a)(i) The wall of a dam in Figure 4.2
is much thicker at the bottom
than at the top and withstand
the higher pressure at the
bottom of the lake.
(a)(ii) Pressure at B is higher than at
A
(b)(i) Dam in Diagram 5.2
(b)(ii) 1- When depth increases,
pressure increases.
2- Thicker at the base can
withstand high pressure.
(c)(i) Siphon system
(c)(ii) Diffrence in water level will
cause different in pressure
(c)(iii)
Question 5
(a) Depth / density / acceleration
due to gravity
(b)(i) PQ > P p
(b)(ii) PQ = h ρ g
= 5 x 1000 x 10
= 50000 Pa
(c)(i) Different in pressure
(c)(ii) Water level at P is same as the
water level in the house water
tank // pressure is the same
between at P and inside the
tank.

2. FIZIKMOZAC 2010
No difference in pressure
(d)(i) Place the concrete tank at
higher place // on top of hill
Higher difference of pressure.
or
Use water pump
Increase the difference of
pressure.
(d)(ii)
Question 6
(a)(i) Magnitude : same magnitude of
atmospheric pressure
Directions : atmospheric
pressure and mercury are in the
same direction //
gas pressure direction against
the direction of mercury and
atmospheric pressure //
atmospheric pressure acts
downwards
(a)(ii) Phg + Patm ,// P gas
(a)(iii) Same / equal
(b) Gas pressure = atmospheric
Pressure + mercury pressure
(c)(i) Mercury level drops and at
same level in both columns
(c)(ii) Same pressure // atmospheric
Pressure
3.4 PASCAL’S PRINCIPLE
Question 7
(a) Pascal’s principle
(b) Show the correct direction
(c) the liquid pressure in the main
brake cylinder and the small
brake cylinder are the
same/equal
1. 4
2
4
106105
15




F
2. F2 = 18 N
Question 8
(a) Pascal’s principle
(b) P = 5/2 = 2.5 Ncm-2
= 2.5 x 104
Nm-2
(c) Same pressure
(d) F2 = 2.5 x 5 = 12.5 N
(e) Liquid cannot be compressed
easily
3.5 ARCHIMEDES’ PRINCIPLE
Question 9
(a)(i) Archimedes principle
(a)(ii) upward: Buoyant force
Downward: weight of
Hydrometer
(b)(i) the length of hydrometer
submerged in oil is longer than
in water.
(b)(ii) Density of oil is less than water
(c)(i) Buoyant force = Wair - Wwater
= 0.25 – 0.22 = 0.03 N
(c)(ii) volume of object = volume of
water displaced
0.03 = 1000 x V x 10
V = 3 x 10-6
m3
Question 10
(a) Pacal
(b) Depth
(c) (i) Weight of water displaced =
buoyant force = ρVg
= 1010 x 2.5 x 10
= 25,250 N
(ii) Tension + buoyant force
= weight of object
T = 125,000 – 25,250
= 99,750 N

3. FIZIKMOZAC 2010
Question 11
(a) Mass per volume
(b) (i) Density sphere A less than
B
(ii) Weight A less than B
(iii) The weight of water
displaced by A less than B
(iv) Larger weight of sphere,
displaced bigger weight of
water
(v) Weight of water displaced =
up thrust //
When the weight of water
displaced increase, up thrust
increase
(c) Archimedes’ principle
(d) Submarine
Question 12
(a)(i) Same Volume
Net force zero
(a)(ii) Y < X < Z
(b)(i) Box Y floats and immersed
partially / box X immersed
fully and floats box Z sink
(b)(ii) Greater weight means
greater mass and greater
density.
The higher density object
needs more volume to
increase the buoyant force
to support the weight .
(c) Archimedes principle//
equilibrium of forces
Question 13
(a) Density is the mass per volume
(b)(i) Level of the boat is higher in
the sea than in the river. (the
part of boat submerged in the
sea is less than in the river)
(ii) Water displaced in the sea is
less than in the river.
(b)(iii) Density of sea water is higher
than river water.
(c)(i) The lower the density of water,
the greater /higher the volume
of water displaced.
(c)(ii) Weight of the boat = Weight of
the water displaced
(d) Archimedes’ principle
(e) Ballast tank filled by sea water
Weight of submarine > buoyant
force
Question 14
(a)(i) Function – for safety
purpose/To ensure the
maximum weight limit
(a)(ii) F = mg
= 7500 x 10
= 7.5 x 104
N
(a)(iii) The mark should be higher
than the sea water level
(a)(iv)
1. Density of sea water is denser
than the density of river water.
2. The volume of water displaced
increased when density of liquid
decrease
(b)(i) Up thrust = Weight
(b)(ii) Accelerates upwards or moves
Up wards
(b)(iii)
1. The weight of the air balloon is
decreased
2. Buoyant force /up thrust higher
than weight
3. The balloon experiences the
unbalanced force.
Question 15
(a)(i) Bernoulli’s principle
(a)(ii) Y
(b) The air moves with a high speed
(c)1. The atmospheric pressure
which is higher pushes the
liquid up through the narrow
tube.

4. FIZIKMOZAC 2010
2. The jet air will force the liquid
to be sprayed as fine spray
liquid
Question 16
(a)(i) Student mark at the same level
in tube - X,Y and Z-
(a)(ii) Atmospheric pressure
(b)(i) Water level in vertical tube P is
higher than in R and higher
than in P./hp > hR > hQ
(b)(ii) Bernoulli’s principle.
(b)(iii) P = hρg
= 0.15 x 1000 x 10
= 1500 Pa
Question 17
(a) Distance per time
(b) (i) Before: water levels are the
same and the roof stay
intact.
After : water levels are not
the same and the roof rise
up.
(b)(ii) Pressure above the roof is
higher compare to pressure
below
(b)(iii) Speed increases pressure
decreases or vice versa
(c) Bernoulli
(d)(i) Q is slower and R is faster
(d)(ii) Q is higher and R is lower
Question 18: Kedah 07
 The depth of the water in Diagram
9.1 is higher than in Diagram 9.2
 The water spurts out in Diagram
9.1 is at a higher rate than in
Diagram 9.2
 The water spurts out further in
Diagram 9.1 than in Diagram 9.2
 The deeper the water, the further
the distance of water spurt
 The deeper the water, the higher
the pressure of the water
Question 19: Kedah 07
(a)
 The pressure of water increases
with the depth of the water
 The bubble expands upon
reaching the surface of the
water//The volume of air bubble
increases as the depth of water
decreases
(b)
 Buoyant force increases as the
volume of the bubble increases
 The air bubble moving with
increasing acceleration
 (volume of air bubble = volume of
water displaced)
 (Buoyant force is larger than the
weight of the air bubbles)
Question 20: Trengganu 07
 A force is applied when you
squeezed at the bottom end of the
toothpaste tube
 Pressure is applied to the
toothpaste (tube)
 According to Pascal’s principle
 The pressure is transmitted
equally to the whole tube
Question 21: Perak 07
 High altitude low density of air
 Less collision of molecules with
surface
 Low altitude high density of air
 More collision of molecules with
surface
Question 22:SBP 07
 B is denser than A.
 The weight of water displaced is
the same of the weight of the rod.
 Weight of B is greater than weight
of A
 B will displace more volume of
water

5. FIZIKMOZAC 2010
Question 23: Trengganu 07
 The shape of the wing is aerofoil.
 The shape of cross section of the
wing causes the speed of airflow
 The air move faster than above
the wings than below the wing.
 When the speed of moving air is
higher ,the pressure is lower
 Hence air pressure below the
wings is higher compare to above
the wings
 there is difference in pressure
which produce an upward
resultant force.
 Bernoulli’s principle
Question 24: Teknik 07
(a)
 Buoyant Force : Force experience
when an object totally or partly
immersed into the liquid
(b)
 Density of the gas inside the
balloon less dense then air
 Air is displaced by the balloon
and produced buoyant force
 The buoyant force is larger than
the weight of the balloon and
load and it rises up.
 When the buoyant force is equal
to weight of balloon and load, it
will float still.
(c) Quantitative problem:
(i) Resultant force = 250 – 5
= 245 N
(ii) Use F=ma
245 = 5 a
a = 49 ms-2
(iii) air resistance is zero
Characteristics Reason
Used helium
gas
Its light/less dense
then air
Mass of load is
20 kg
Total weight of
balloon and the
load equal to
buoyant force
Tension allowed
of the rope is
300 N
To ensure the rope
not break
(ii) Set C Because its used
helium gas, mass
of load is 20 kg and
tension allowed is
greater than 250 N
(iii) A is not suitable because mass of
the load causes weight of the
load and the balloon less then
buoyant force. The balloon will
rise up
( Accept any other set and the
reason)
Question 25: Perak 07
Characteristic Reason
Large tyre better stability
Liquid in
hydraulic
system
liquid cannot be
compressed
Large mass big inertia
Large base area better stability
Low centre of
gravity
better stability
Choose – M Large tyre, liquid in
hydraulic system,
large mass, large
base area or low
centre of gravity l

6. FIZIKMOZAC 2010
Question 26: SBP 07
Characteristic Reason
Material made
from glass
Glass does not
corrode with acid
Small diameter of
capillary tube
To increase the
sensitivity of the
hydrometer
High density of
shots
Makes the
hydrometer stays
upright
Big diameter of
bottom bulb
To obtaine a
bigger upthrust
Choose N N is made from
glass, has small
diameter of
capillary tube,
high density of
shots and a big
diameter of
bottom bulb.
Question 27:Trengganu 07
Characteristics Reason
A shape of cross
section which is
upper side is
longer than the
bottom
To produced the
speed of airflow
above the wings
to be higher than
the speed of
airflow below
Large surface
area of the wing
Produce larger lift
force
Less density of
the wing
materials
Less weight //
produce more
upward resultant
force
Higher difference
in speed of air
The higher the
difference in
pressure
The most suitable choice is P
Because it has
A shape of cross section which is
upper side is longer than the bottom
Large surface area of the wing
Less density of the wing materials
High difference in speed of air
Question 28: Kedah 07
(i)  Diagram 9.3
 The weight of the dam is
supported by the force
exerted by the water
(ii)  Water in the dam can be
filtered and chlorinated to be
uses as public water supply
 To drive turbines for the
generation of
hydroelectricity//
For irrigation//Recreation
centre
Suggestion Explanation
Thicker wall at
the base
To withstand
greater pressure
at the bottom as
the pressure
increases with
depth
The wall is
constructed
using stronger
materials /
Using reinforce
concrete
To avoid the wall
from breaking / To
increase the
strength of the
wall / To avoid
leaking
Equipped with
the water
overflow
system
To avoid flooding
/ To channel away
the overflow
water
Question 29:Trengganu 07
modification explanation
piston of bigger
cross-sectional
area
Can support
greater force
(weight)
Low density
material
Lightweight //
easy to carry
Non-
compressible
liquid
Piston can be
lifted up

7. FIZIKMOZAC 2010
Longer handle Less effort
needed to press
the small piston
Apply released
valve between
small and main
reservoir
Liquid can flows
into small
reservoir
Question 30: Perak 07
(i) hρg = 0.76 x 13 600 x 10
=103360 Pa
(ii) hρg = 0.1 x 13 600 x 10
= 13600 Pa
(iii) 0 Pa
Question 31: SBP 08
(a)
(i)
Mass devide by volume
(a)
(ii)
 Density of air in Diagram 9.1
is higher than in Diagram
9.2.// vice versa
 The number of load in
Diagram 9.1 is greater than
in Diagram 9.2//vice versa
 The height of the balloons in
both Diagram 9.1 and
Diagram 9.2 are equal
 When the density of the air
increase, the buoyant force
increase
 As the density of the air
increase, the weight of the
load carried increased// .
As the density of the air
decrease, the weight of the
load carried also decreased
(b)  Density of the iron nail is higher than the
density of water// Average density of the
cargo ship is lower than the density of
water
 Volume/ weight of water
displaced by the iron nail is
smaller
 For the cargo ship, the
buoyant force is equal to its
weight .
 For iron nail , its buoyant
force is smaller than the
weight
modification explanation
Streamline shape Decrease/
reduce the water
drag/resistance
thick and strong
material
To withstand
high pressure / /
pressure
increase with
depth
Additional
component
- ballast tank
- periscope
To float or sink
the submarine
To observe
object outside
the water surface
Safety feature
Oxygen tank /
generator
For respiration
Question 32:Trengganu 08
(a)
(i)
Gravitational force
(a)
(ii)
 Weight lost in Diagram 9.1(b)
> Diagram 9.1(c) // vise versa
 Apparent weight in Diagram
9.1(c) > Diagram 9.1(b) // vise
versa
 Density of water > density of
oil
 The greater the density of
liquid, the greater the weight
lost / less apparent weight
(iii) Up thrust /buoyant force
(b)  Name two correct force
(buoyant force and weight)
 Buoyant force small because
small volume // vise versa
 Block sink because weight >
buoyant force
 Sheet float because weight =
buoyant force

8. FIZIKMOZAC 2010
modification explaination
Strong material Can withstand
great force
Low density
material
Lightweight
Two stage
plimsoll line
Save in fresh and
salt water
Big size Can place more
goods
Aerodynamic
shape
Reduce water
friction
Question 33: Kelantan 08
(a)
(i)
Weight is the gravitational force
acts on an object
(a)
(ii)
Buoyant force = weight of the
boat
(a)
(iii)
 Sea water is denser
 Boat displaced less sea water
and gain the same buoyant
force. Therefore boat sinks
less in sea water
(b)
(i)
Buoyant force = weight of sea
water
Displaced
= mg = ρVg
= 250 x 1080 x 10
= 2.7 x 106
N
(b)
(ii)
2.7 x 106
= V x 1000 x 10
V = 270 m3
(c)
(i)
Specification Reasons
Small stem
and long
Increase the
sensitivity
where the scale
divisions are
far apart so
that small
changes in
density can be
detected.
Glass wall Do not erode
Large
diameter of
High upthrusts,
displaced more
bulb liquid to be
able to float
easily
Lead shoots Hydrometer
can stay
upright.
P is chosen It has small and
long stem,
glass wall,
large diameter
of bulb and
lead shoots
used.
Question 34: N9 08
(a) Pressure is defined as the force
acting normally per unit area/
Pressure = Force
Area
(b) 1. When the small piston is
pulled up, the hydraulic oil is
drawn from the reservoir into
the small piston
2. When the small piston is
pushed down , the hydraulic
oil is exerted with force and
experienced a pressure
3. The pressure is transmitted
uniformly from the small
piston to the big piston.
4. The forced produced raised
the big piston / The system
can convert a small input
force into a bigger output
force.
Characteristics Reason
Has higher
boiling point
So that liquid not
easily boiling/
Has higher
specific heat
capacity
So that it can’t be
easily become hot

9. FIZIKMOZAC 2010
Has lower
density
So the hydraulic
jack is not heavy
Has lower rate
of vaporisation
Volume of liquid
will not easily
vaporise
Liquid L is
chosen
Reasons: L has
higher boiling
point, higher
specific heat
capacity, lower
density and lower
rate of vaporisation
Question 35: Kedah 08
(a)
(i)
Archimedes’ principle states
that the buoyant force on an
object immersed in a fluid is
equal to the weight of fluid
displaced by the object.
(a)
(ii)
 The balloon acted by two
forces: buoyant force and the
weight of the balloon.
 The density of helium gas is
less than the density of
surrounding air.
 Buoyant force equals to the
weight of the air displaced by
the balloon.
 Buoyant force is higher than
the weight of the balloon.
(c)  Large balloon
 To produce bigger buoyant
force// increase the volume of
air displaced
 Use 2 burners
 To produce bigger flame //
heat up the gas in the balloon
faster
 Synthetic nylon
 Light-weight, strong and air-
proof material.
 High temperature of the air in
the balloon
 Reduce density / weight of the
air in the balloon.
 Hot air balloon Q is chosen
 Because it is large balloon,
uses 2 burners / many
burners, uses synthetic nylon
and has high temperature of
the air in the balloon.
(d)
(i)
Weight = buoyant force
= weight of water
displaced
m x 10 = (10 x 2 x 10-6
) x 1000
x 10
m = 0.02 kg
(d)
(ii)
mg = ρVg
(0.02) (10) = (0.12 x 2 x 10-4
) ρ x
10
ρ = 833.33 kg m-3