Daya dan Tekanan Jawapan TUGAS
Bab 3 Tingkatan 4
Form 4 Chapter 3
Daya dan Tekanan
Forces and Pressure
Kertas 2
Paper 2
SKEMA JAWAPAN PAPER 2 SECTION A
NO.
SOALAN
1 (a)(i)
(ii)
(b)
2 (a)
(b)
(c )
(d)
3 (a)(i)
(ii)
EXPLANATION/
PENERANGAN
Q is deeper than P// The depth of Q is greater than the depth of P
// converse
Q adalah lebih dalam berbanding P//Kedalaman Q lebih berbanding
kedalaman P//Pilih salah satu
The distance y is greater than the distance x// converse
Jarak y adalah lebih jauh berbanding jarak x.
The greater the depth, the greater the distance the water spurts// The
deeper the hole,the farther the water spurts // converse
Semakin dalam, semakin jauh jarak pancutan air tersebut//Semakin
bertambah kedalaman lubang, semakin jauh jarak pancutan air
tersebut//pilih salah satu
TOTAL MARKS
Aerofoil
The velocity of air above the wing is higher than the velocity of air below the
wing. This causes the air pressure below the wing is higher than the air
pressure above the wing. The difference of air pressure produces the lifting
force that make the aircraft to be lifted up.
Halaju udara di atas sayap lebih tinggi daripada halaju angina di bawah
permukaan sayap. Ini menyebabkan tekanan udara di bawah sayap lebih
tinggi daripada tekanan udara di permukaan atas sayap. Perbezaan
tekanan udara akan menyebabkan saya tujah menyebabkan pesawat itu
terangkat dan naik ke udara.
Bernoulli’s principle
Prinsip Bernoulli
The area of the wing is wider to increase the lifting force. This is because
force = pressure × area.
Luas permukaan sayap lebih luas bagi meningkatkan daya tujahan. Ini
kerana, Daya = Tekanan × Luas permukaan
TOTAL MARKS
Volume = cross-sectional area × height
= 2.0 cm2 × 10 cm
= 20 cm3
Weight of water has been displaced = weight of glass tube + weight of ball
bearings
Vρg = W glass tube + W ball bearings
MARKS/
MARKAH
1
1
2
4
1
3
1
2
7
1
3
20(1 .0 )(10) 10
=
(10 )+
1000
1000
W ball bearings
W ball bearings = 0.10 N
0. 10×1000
=
= 10
10.0 g
(b)
Hydrometer-to determine density of liquid
Hidrometer-mengukur ketumpatan cecair
TOTAL MARKS
Modul SOLAF Fizik 2014
1
5
Bab 3 Tingkatan 4
Form 4 Chapter 3
4 (a)
(b)
(c)
5 (a)
(b)
(c)
Daya dan Tekanan
Forces and Pressure
Kertas 2
Paper 2
Solution/Penyelesaian:
Pressure at point Q/Tekanan pada titik Q
=
(75 + 15)cm Hg
=
90 cm Hg
=
(1.36 x 104)(10)(0.9)
=
122.4 kPa
Solution/Penyelesaian:
Pgas
= (h ρ g)water
= (1000)(10)(0.1)
= 1000 N m-2
SolutionPenyelesaian:
Pair + Pmercury = Patm
Pair + 10cmHg
= 76cmHg
Pair
= (76 – 10) cmHg
= 66 cmHg
TOTAL MARKS
Bourdon Gauge / Tolok Bourdon
Pressure/Tekanan
1st : 27 + 273 = 300 K
2nd :
2
3
2
7
1
1
3
128 132
=
300
T
3rd : T = 309.38 K / 36.38 ˚C
6(a)
(b)(i)
(b)(ii)
(b)(iii)
(c)(i)
(c)(ii)
(d)
(e)
TOTAL MARKS
5
Mass is the quantity of matter/Jisim kuantiti jasad
The level of the apple in the oil immerses more than in the water/Paras epal
apabila tenggelam di dalam minyak lebih berbanding tenggelam di dalam
air.
Volume of oil displaced by the apple is larger than the water/Isipadu air
tersesar oleh epal adalah lebih bayak berbanding air.
Density of water is larger/ greater than oil.
Inversely proportional/Berkadar sonsang
Equal/sama
Archimedes principle/Prinsip Archimedes
Empty the ballast tank / Remove the water/Kosngkan tangki balas
TOTAL MARKS
1
1
Modul SOLAF Fizik 2014
1
1
1
1
1
1
8
Bab 3 Tingkatan 4
Form 4 Chapter 3
Daya dan Tekanan
Forces and Pressure
Kertas 2
Paper 2
SKEMA JAWAPAN SECTION B KERTAS 2
No.
Soala
n
1(a)(i)
(ii)
Answer/Jawapan
In an enclosed and incompressible fluid, the pressure exerted on the
fluid is transmitted equally and uniformly throughot the fluid.
Dalam suatu bendalir yang tertutup dan tidak termampat, tekanan
yang dikenakan pada bendalir itu akan dipindahkan secara seragam
dengan magntud yang sama ke semua arah dalam bendalir itu.
When a force F1 acted on the small piston with cross-sectional area
Mark/Markah
1
F1
A1, a pressure A 1 is transmitted with the same magnitude to all
directions in the fluid. The pressure that exerted on the large piston
with cross-sectional area A2 will produce a force F2 to raise up the
chair. The magnitude of the force F2 depends on the cross sectinal
area A2. The force F2 increases as the cross sectional area A2
increses. It acts as a, force multilier. Hence, force F2 is greater than
force F1.
4
Apabila daya F1 bertindak pada omboh kecil dengan luas keratan
F1
A1
rentas A1, tekanan
dipindahkan dengan magnitud yang sama
ke semua arah dalam bendalir. Tekanan yang dikenakan pada
omboh yang besar dengan luas keratan rentas A2 akan menghasilkan
satu daya F2 untuk menaikkan kerusi itu. Magnitud daya F2
bergantung kepada luas keratan rentas A2. Daya F2 bertambah
apabila luas keratan rentas A2 bertambah. Ia bertindak sebagai
pengganda daya. Maka, daya F2 lebih besar daripada daya F1.
(b)
Characteristic
Ciri-ciri
Oil is used
Minyak digunakan
High specific heat capacity
Muatan haba tentu yang tinggi
Small cross sectional-area of
input piston
Luas keratan rentas omboh
input yang kecil
Large cross-sectional area of
output piston
Modul SOLAF Fizik 2014
Explanation
Penerangan
It has high viscosity. Hence, the
pressure is easily transmitted to
the other piston.
Ia mempunyai kelikatan yang
tinggi. Maka, tekanan mudah
dipindahkan ke piston yang lain.
It is not easily heated up
Tidak mudah dipanaskan
So that the force applied to the
input piston is small
Supaya daya yang dikenakan
pada omboh input adalah kecil
So that it can raise up a greater
load
10
Bab 3 Tingkatan 4
Form 4 Chapter 3
Daya dan Tekanan
Forces and Pressure
Luas keratan rentas omboh
output yang besar
Kertas 2
Paper 2
Supaya is dapat mengangkat
beban yang besar
Hydraulic jack Q is most suitable because it uses oil, it has high
specific heat capacity, it has small cross-sectional area of input piston
and large cross-sectional area of output piston.
Jek hidraulik Q adalah yang paling sesuai kerana ia menggunakan
minyak, ia mempunyai muatan haba tentu yang tinggi, omboh
inputnya mempunyai luas keratan rentas yang kecil dan omboh
outputnya mempunyai luas keratan rentas yang besar.
F1
(c) (i)
F=
A1
2
10
−4
= 150×10
= 666.7 Pa
F2
(ii)
F1
A2 = A1
F2
10
3 = 150×10−4
3
F2 = 2000 N
2 (a)
(b)(i)
(ii)
(c)(i)
(ii)
(iii)
(d)(i)
(ii)
TOTAL MARKS
Buoyant force is an upward force acting on an object which has
displaced some fluid in which it is immersed.
Daya apungan adalah daya tujah bertindak ke atas objek yang
disesarkan oleh cecair yang mana ia tenggelam.
1. Buoyant force are produced by the water to act on the rods
2. The buoyant force are able to balance/support the weights of the
rods.
Archimedes’ Principle
Prinsip Archimedes
Rod Q has displaced more water
Rod Q mempunyai lebih banyak air yang tersesar
Buoyant force acting on Rod Q > Rod P
Daya apungan bertindak di Rod Q > Rod P
Rod Q
Weight of water displaced = 1 000 (5) ( 10)
= 50 000 N
Weight of cargo = 50 000 – 250 (10)
= 47 500 N
20
1
2
1
1
1
1
2
1
(e)
Suggestion
Material – Extra
strong
Modul SOLAF Fizik 2014
Explanation
To ensure the balloon does not burst or
leak easily and the people in a strong
basket will be safer.
10
Bab 3 Tingkatan 4
Form 4 Chapter 3
Power-Strong
The holding cableLight and strong
Balloon should pulled
down by motor
Most Suitable is
balloon X
3(a)(i)
(ii)
(b)(i)
(ii)
(iii)
(vi)
(v)
Daya dan Tekanan
Forces and Pressure
Kertas 2
Paper 2
It can heat up the air faster and the
balloon can ascennd faster
It will decrease the overall weight of the
balloon in place safely
So that the balloon can descend faster
and more safety without human error and
fatique
Its balloon and basket materials are
strong, it has a strong burner, its holding
cable is strong and light and its descend is
operated by motor.
TOTAL MARKS
Bernoulli’s principle states that a fluid moving with a high velocity will
experience a smaller pressure and vise versa.
Prinsip Bernoulli menyatakan bahawa jika sesuatu cecair bergerak,
tekanan yang dialami adalah kecil atau sebaliknya.
1. With strong wind blowing, the velocity of air above the roof > the
velocity of air below the roof.
Dengan tiupan angin yang kuat, halaju udara di atas bumbung >
halaju udara di bawah bumbung.
2. According to Bernoulli’s principle, the pressure of air above the
rooof < the air pressure below the roof.
Mengikut prinsip Bernoulli, tekanan udara di atas bumbung <
tekanan udara di bawah bumbung.
3. The presure difference creates a lifting force big enough to uplift
the roof.
Perbezaan tekanan akan menghasilkan daya angkat yang cukup
besar bagi mengangkat bumbung itu.
Discus Q
Cakera Q
Discus Q
Cakera Q
Landing distance of discus Q>P
Jarak mendarat cakera Q > P
Discus with a higher lifting force will have a longer landing distance.
Cakera yang mempunyai data angkat yang tinggi mempunyai jarak
mendarat yang lebih panjang.
1. Throw with a larger force/higher speed
2. Throw and spin the dicus
3. Use a discus with a smoother surface
20
1
3
1
1
1
1
2
(c)
Suggestion
Replace engine with
more powerful jet
engine
Use wings with aerofoil
shape
Area of the wing should
be bigger
The whole aeroplane
Modul SOLAF Fizik 2014
Explanation
This will provide higher thrust and faster
speed
This will create a resultant upward
pressure and lifting force.
This will generate a larger lifting force to
carry a larger load
This will reduce air frictionn and
10
Bab 3 Tingkatan 4
Form 4 Chapter 3
Daya dan Tekanan
Forces and Pressure
should be more
aerodynamic
Mass of aeroplane
should be lighter and
yet stronger material
resistance, thus the speed can be
increased
This will allow the aeroplane to fly faster
and is able to carry a heavier load
TOTAL MARKS
Temperature is the degree of hotness
1st : the thermometer is put under the tounge/ inside mouth/under the
armpit
2nd : the heat is transferred from the body to the thermometer
3rd : mercury expand until it reaches a state of thermal equilibrium
4th : the temperature of the thermometer is the same as the body
1st : use alchohol
2nd : able to record low temperature / low freezing point
3rd : thin glass wall bulb
4th : more sensitive to heat
5th : small diameter of capillary tube
6th : more sensitive to heat / get a wider range
7th : thick and curve glass bore stem
8th : not easily to break / easy to read
9th : choose T
10th : because it uses alchohol, thin glass wall bulb, small diameter of
capillary tube and thick and curve glass bore stem.
4 (a)
(b)
(c)
l θ−l 0
x 100
l 100 −l 0
12−5
x 100
2nd :
25−5
(d)(i)
Kertas 2
Paper 2
1st :
20
1
4
10
4
3rd : 35 ˚C
4th : 308 K
(d)(ii)
Volume expand with temperature..
TOTAL MARKS
1
20
SKEMA JAWAPAN KERTAS 2-BAHAGIAN C
No.Soala
n
1.(a) (i)
(ii)
Answer/Jawapan
According to the kinetic theory of gases, the particles of the gas are
constantly moving randomly. They collide with the walls of the container
and a force is applied to the walls. This results in a pressure.
Mengikut teori kinetik gas, zarah gas bergerak secara rawak dengan
seragam. Ia berlanggar dengan dinding bekas dengan daya dikenakan
kepada dinding tersebut. Ini menyebabkan tekanan.
This is because the height of the air column above a point at sea level is
greater than at the top of a mountain. Hence the weight of the air
Modul SOLAF Fizik 2014
Markah
1
Bab 3 Tingkatan 4
Form 4 Chapter 3
(b)(i)
(ii)
(c)
Daya dan Tekanan
Forces and Pressure
column acting at sea level is higher. Since pressure is force/area, the
force being the weight of the air column, the pressure at sea level is
higher.
Ini adalah kerana ketinggian turus air melebihi titik pada paras laut
adalah tinggi daripada puncak gunung. Maka berat turus air bertindak
pada paras laut adalah lebih tinggi.
Pressure at the base
Tekanan di tapakk = hρg
= 75.5 × 10-2 ×13.6 × 103 × 10
=1.03 × 105 Pa
The difference in pressure between the top and bottom of the building is
equal to the pressure caused by the air column from the base to the top.
Perbezaan tekanan antara puncak dan tapak bangunan adalah sama
dengan tekanan yang disebabkan oleh turus udara daripada tapak ke
puncak.
= 1.03 ×105 – 9.79 × 104
= h ×1.36 ×10
h = 375 m
Density : the density must be high so that the length of the liquid column
needed to balance atmospheric pressure must be less than 1 m
because the tube is only 1 m long.
Ketumpatan : ketumpatan mestilah tinggi supaya panjang turus cecair
diperlukan untuk mengimbangkan tekanan atmosfera mestilah kurang
daripada 1 m kerana panjang tiup hanyalah 1 m sahaja.
Atmospheric Pressure is approximately 100 000 Pa.
Tekanan atmosfera adalah lebih kurang 100 000 Pa.
For mercury, using p= hρg
Bagi merkuri, menggunakan p = hρg
100 000 = h × 13600 × 10
h = 0.74 m
100000
For oil, h = 800×10 = 12.5 m
100000
For water, h = 800×10 = 10 m
Characteristic
Sticks/does not stick to
glass
Melekat/Tidak melekat
pada kaca
Opaque/Transparent
Legap/Lutsinar
The best liquid is
mercury
Modul SOLAF Fizik 2014
Explanation
The liquid should not stick to the glass.
Otherwise it will not be suitable to be used
as a barometer liquid.
Cecair sepatutnya tidak melekat pada kaca.
Jika tidak ia tidak sesuai untuk digunakan
sebagai cecair barometer.
It should be opaque so that it more visible.
Ia seharusnya legap supaya mudah dilihat..
Because the column length is 0.74 m which
can fit into the 1 m length tube, it does not
Kertas 2
Paper 2
4
2
3
10
Bab 3 Tingkatan 4
Form 4 Chapter 3
Daya dan Tekanan
Forces and Pressure
Kertas 2
Paper 2
stick to glass and it is opaque.
Cecair yang terbaik adalah merkuri, kerana
Modification/
Explanation
tinggi turus ialah 0.74 m di mana boleh
suggestion
dimuatkan ke dalam 1 m panjang tiub, ia
Oil
Incompressible/
tidak melekat pada kaca dan legap.
No air
bubble
The other
two
liquids do not satisfy these three conditions.
High boiling point/
Does
notlai
change
to gas stateketiga-tiga
easily/ syarat tersebut.
Dua lagi
cecair
tidak memenuhi
Does not evaporate easily
TOTAL MARKS
Low density/
Lighter/
High viscosity
Fluid does not flow easily
Small master
High pressure produced with a small force
(a)
2
piston
Force is action
Big slave piston
Produce bigger output force
on an object
Aluminium/
Steel
Strong/
Does not break easily/
Non corrosive/
Does not rust easily
that can result
in changes like
size, shape and
direction.
(b)
1. surface area A1 is smallerr than A2
2. forces F1 is smaller than F2
3. pressure P1 is equal to P2
4. when surface area is larger, the force exerted on the piston will increase.
5. Pascal’s principle
(c)
1. Function- to transfer water from beaker to cylinder.
2. The pressure at lowest point in cylinder is greater than the
atmospheric pressure, the liquid flows out at lowest point in
cylinder/at the end of rubber tube in cylinder.
3. The pressure in the rubber tube decreases as the water flows out
and a partial vacuum is created.
4. The higher atmospheric pressure water into the tube.The water flows
until the liquid surface in cylinder reaches the same level as in
beaker.
20
1
5
4
(d)
10
Modul SOLAF Fizik 2014
Bab 3 Tingkatan 4
Form 4 Chapter 3
Modul SOLAF Fizik 2014
Daya dan Tekanan
Forces and Pressure
Kertas 2
Paper 2
TOTAL MARKS
20
Form 4 Chapter 3
Daya dan Tekanan
Forces and Pressure
Kertas 2
Paper 2
SKEMA JAWAPAN PAPER 2 SECTION A
NO.
SOALAN
1 (a)(i)
(ii)
(b)
2 (a)
(b)
(c )
(d)
3 (a)(i)
(ii)
EXPLANATION/
PENERANGAN
Q is deeper than P// The depth of Q is greater than the depth of P
// converse
Q adalah lebih dalam berbanding P//Kedalaman Q lebih berbanding
kedalaman P//Pilih salah satu
The distance y is greater than the distance x// converse
Jarak y adalah lebih jauh berbanding jarak x.
The greater the depth, the greater the distance the water spurts// The
deeper the hole,the farther the water spurts // converse
Semakin dalam, semakin jauh jarak pancutan air tersebut//Semakin
bertambah kedalaman lubang, semakin jauh jarak pancutan air
tersebut//pilih salah satu
TOTAL MARKS
Aerofoil
The velocity of air above the wing is higher than the velocity of air below the
wing. This causes the air pressure below the wing is higher than the air
pressure above the wing. The difference of air pressure produces the lifting
force that make the aircraft to be lifted up.
Halaju udara di atas sayap lebih tinggi daripada halaju angina di bawah
permukaan sayap. Ini menyebabkan tekanan udara di bawah sayap lebih
tinggi daripada tekanan udara di permukaan atas sayap. Perbezaan
tekanan udara akan menyebabkan saya tujah menyebabkan pesawat itu
terangkat dan naik ke udara.
Bernoulli’s principle
Prinsip Bernoulli
The area of the wing is wider to increase the lifting force. This is because
force = pressure × area.
Luas permukaan sayap lebih luas bagi meningkatkan daya tujahan. Ini
kerana, Daya = Tekanan × Luas permukaan
TOTAL MARKS
Volume = cross-sectional area × height
= 2.0 cm2 × 10 cm
= 20 cm3
Weight of water has been displaced = weight of glass tube + weight of ball
bearings
Vρg = W glass tube + W ball bearings
MARKS/
MARKAH
1
1
2
4
1
3
1
2
7
1
3
20(1 .0 )(10) 10
=
(10 )+
1000
1000
W ball bearings
W ball bearings = 0.10 N
0. 10×1000
=
= 10
10.0 g
(b)
Hydrometer-to determine density of liquid
Hidrometer-mengukur ketumpatan cecair
TOTAL MARKS
Modul SOLAF Fizik 2014
1
5
Bab 3 Tingkatan 4
Form 4 Chapter 3
4 (a)
(b)
(c)
5 (a)
(b)
(c)
Daya dan Tekanan
Forces and Pressure
Kertas 2
Paper 2
Solution/Penyelesaian:
Pressure at point Q/Tekanan pada titik Q
=
(75 + 15)cm Hg
=
90 cm Hg
=
(1.36 x 104)(10)(0.9)
=
122.4 kPa
Solution/Penyelesaian:
Pgas
= (h ρ g)water
= (1000)(10)(0.1)
= 1000 N m-2
SolutionPenyelesaian:
Pair + Pmercury = Patm
Pair + 10cmHg
= 76cmHg
Pair
= (76 – 10) cmHg
= 66 cmHg
TOTAL MARKS
Bourdon Gauge / Tolok Bourdon
Pressure/Tekanan
1st : 27 + 273 = 300 K
2nd :
2
3
2
7
1
1
3
128 132
=
300
T
3rd : T = 309.38 K / 36.38 ˚C
6(a)
(b)(i)
(b)(ii)
(b)(iii)
(c)(i)
(c)(ii)
(d)
(e)
TOTAL MARKS
5
Mass is the quantity of matter/Jisim kuantiti jasad
The level of the apple in the oil immerses more than in the water/Paras epal
apabila tenggelam di dalam minyak lebih berbanding tenggelam di dalam
air.
Volume of oil displaced by the apple is larger than the water/Isipadu air
tersesar oleh epal adalah lebih bayak berbanding air.
Density of water is larger/ greater than oil.
Inversely proportional/Berkadar sonsang
Equal/sama
Archimedes principle/Prinsip Archimedes
Empty the ballast tank / Remove the water/Kosngkan tangki balas
TOTAL MARKS
1
1
Modul SOLAF Fizik 2014
1
1
1
1
1
1
8
Bab 3 Tingkatan 4
Form 4 Chapter 3
Daya dan Tekanan
Forces and Pressure
Kertas 2
Paper 2
SKEMA JAWAPAN SECTION B KERTAS 2
No.
Soala
n
1(a)(i)
(ii)
Answer/Jawapan
In an enclosed and incompressible fluid, the pressure exerted on the
fluid is transmitted equally and uniformly throughot the fluid.
Dalam suatu bendalir yang tertutup dan tidak termampat, tekanan
yang dikenakan pada bendalir itu akan dipindahkan secara seragam
dengan magntud yang sama ke semua arah dalam bendalir itu.
When a force F1 acted on the small piston with cross-sectional area
Mark/Markah
1
F1
A1, a pressure A 1 is transmitted with the same magnitude to all
directions in the fluid. The pressure that exerted on the large piston
with cross-sectional area A2 will produce a force F2 to raise up the
chair. The magnitude of the force F2 depends on the cross sectinal
area A2. The force F2 increases as the cross sectional area A2
increses. It acts as a, force multilier. Hence, force F2 is greater than
force F1.
4
Apabila daya F1 bertindak pada omboh kecil dengan luas keratan
F1
A1
rentas A1, tekanan
dipindahkan dengan magnitud yang sama
ke semua arah dalam bendalir. Tekanan yang dikenakan pada
omboh yang besar dengan luas keratan rentas A2 akan menghasilkan
satu daya F2 untuk menaikkan kerusi itu. Magnitud daya F2
bergantung kepada luas keratan rentas A2. Daya F2 bertambah
apabila luas keratan rentas A2 bertambah. Ia bertindak sebagai
pengganda daya. Maka, daya F2 lebih besar daripada daya F1.
(b)
Characteristic
Ciri-ciri
Oil is used
Minyak digunakan
High specific heat capacity
Muatan haba tentu yang tinggi
Small cross sectional-area of
input piston
Luas keratan rentas omboh
input yang kecil
Large cross-sectional area of
output piston
Modul SOLAF Fizik 2014
Explanation
Penerangan
It has high viscosity. Hence, the
pressure is easily transmitted to
the other piston.
Ia mempunyai kelikatan yang
tinggi. Maka, tekanan mudah
dipindahkan ke piston yang lain.
It is not easily heated up
Tidak mudah dipanaskan
So that the force applied to the
input piston is small
Supaya daya yang dikenakan
pada omboh input adalah kecil
So that it can raise up a greater
load
10
Bab 3 Tingkatan 4
Form 4 Chapter 3
Daya dan Tekanan
Forces and Pressure
Luas keratan rentas omboh
output yang besar
Kertas 2
Paper 2
Supaya is dapat mengangkat
beban yang besar
Hydraulic jack Q is most suitable because it uses oil, it has high
specific heat capacity, it has small cross-sectional area of input piston
and large cross-sectional area of output piston.
Jek hidraulik Q adalah yang paling sesuai kerana ia menggunakan
minyak, ia mempunyai muatan haba tentu yang tinggi, omboh
inputnya mempunyai luas keratan rentas yang kecil dan omboh
outputnya mempunyai luas keratan rentas yang besar.
F1
(c) (i)
F=
A1
2
10
−4
= 150×10
= 666.7 Pa
F2
(ii)
F1
A2 = A1
F2
10
3 = 150×10−4
3
F2 = 2000 N
2 (a)
(b)(i)
(ii)
(c)(i)
(ii)
(iii)
(d)(i)
(ii)
TOTAL MARKS
Buoyant force is an upward force acting on an object which has
displaced some fluid in which it is immersed.
Daya apungan adalah daya tujah bertindak ke atas objek yang
disesarkan oleh cecair yang mana ia tenggelam.
1. Buoyant force are produced by the water to act on the rods
2. The buoyant force are able to balance/support the weights of the
rods.
Archimedes’ Principle
Prinsip Archimedes
Rod Q has displaced more water
Rod Q mempunyai lebih banyak air yang tersesar
Buoyant force acting on Rod Q > Rod P
Daya apungan bertindak di Rod Q > Rod P
Rod Q
Weight of water displaced = 1 000 (5) ( 10)
= 50 000 N
Weight of cargo = 50 000 – 250 (10)
= 47 500 N
20
1
2
1
1
1
1
2
1
(e)
Suggestion
Material – Extra
strong
Modul SOLAF Fizik 2014
Explanation
To ensure the balloon does not burst or
leak easily and the people in a strong
basket will be safer.
10
Bab 3 Tingkatan 4
Form 4 Chapter 3
Power-Strong
The holding cableLight and strong
Balloon should pulled
down by motor
Most Suitable is
balloon X
3(a)(i)
(ii)
(b)(i)
(ii)
(iii)
(vi)
(v)
Daya dan Tekanan
Forces and Pressure
Kertas 2
Paper 2
It can heat up the air faster and the
balloon can ascennd faster
It will decrease the overall weight of the
balloon in place safely
So that the balloon can descend faster
and more safety without human error and
fatique
Its balloon and basket materials are
strong, it has a strong burner, its holding
cable is strong and light and its descend is
operated by motor.
TOTAL MARKS
Bernoulli’s principle states that a fluid moving with a high velocity will
experience a smaller pressure and vise versa.
Prinsip Bernoulli menyatakan bahawa jika sesuatu cecair bergerak,
tekanan yang dialami adalah kecil atau sebaliknya.
1. With strong wind blowing, the velocity of air above the roof > the
velocity of air below the roof.
Dengan tiupan angin yang kuat, halaju udara di atas bumbung >
halaju udara di bawah bumbung.
2. According to Bernoulli’s principle, the pressure of air above the
rooof < the air pressure below the roof.
Mengikut prinsip Bernoulli, tekanan udara di atas bumbung <
tekanan udara di bawah bumbung.
3. The presure difference creates a lifting force big enough to uplift
the roof.
Perbezaan tekanan akan menghasilkan daya angkat yang cukup
besar bagi mengangkat bumbung itu.
Discus Q
Cakera Q
Discus Q
Cakera Q
Landing distance of discus Q>P
Jarak mendarat cakera Q > P
Discus with a higher lifting force will have a longer landing distance.
Cakera yang mempunyai data angkat yang tinggi mempunyai jarak
mendarat yang lebih panjang.
1. Throw with a larger force/higher speed
2. Throw and spin the dicus
3. Use a discus with a smoother surface
20
1
3
1
1
1
1
2
(c)
Suggestion
Replace engine with
more powerful jet
engine
Use wings with aerofoil
shape
Area of the wing should
be bigger
The whole aeroplane
Modul SOLAF Fizik 2014
Explanation
This will provide higher thrust and faster
speed
This will create a resultant upward
pressure and lifting force.
This will generate a larger lifting force to
carry a larger load
This will reduce air frictionn and
10
Bab 3 Tingkatan 4
Form 4 Chapter 3
Daya dan Tekanan
Forces and Pressure
should be more
aerodynamic
Mass of aeroplane
should be lighter and
yet stronger material
resistance, thus the speed can be
increased
This will allow the aeroplane to fly faster
and is able to carry a heavier load
TOTAL MARKS
Temperature is the degree of hotness
1st : the thermometer is put under the tounge/ inside mouth/under the
armpit
2nd : the heat is transferred from the body to the thermometer
3rd : mercury expand until it reaches a state of thermal equilibrium
4th : the temperature of the thermometer is the same as the body
1st : use alchohol
2nd : able to record low temperature / low freezing point
3rd : thin glass wall bulb
4th : more sensitive to heat
5th : small diameter of capillary tube
6th : more sensitive to heat / get a wider range
7th : thick and curve glass bore stem
8th : not easily to break / easy to read
9th : choose T
10th : because it uses alchohol, thin glass wall bulb, small diameter of
capillary tube and thick and curve glass bore stem.
4 (a)
(b)
(c)
l θ−l 0
x 100
l 100 −l 0
12−5
x 100
2nd :
25−5
(d)(i)
Kertas 2
Paper 2
1st :
20
1
4
10
4
3rd : 35 ˚C
4th : 308 K
(d)(ii)
Volume expand with temperature..
TOTAL MARKS
1
20
SKEMA JAWAPAN KERTAS 2-BAHAGIAN C
No.Soala
n
1.(a) (i)
(ii)
Answer/Jawapan
According to the kinetic theory of gases, the particles of the gas are
constantly moving randomly. They collide with the walls of the container
and a force is applied to the walls. This results in a pressure.
Mengikut teori kinetik gas, zarah gas bergerak secara rawak dengan
seragam. Ia berlanggar dengan dinding bekas dengan daya dikenakan
kepada dinding tersebut. Ini menyebabkan tekanan.
This is because the height of the air column above a point at sea level is
greater than at the top of a mountain. Hence the weight of the air
Modul SOLAF Fizik 2014
Markah
1
Bab 3 Tingkatan 4
Form 4 Chapter 3
(b)(i)
(ii)
(c)
Daya dan Tekanan
Forces and Pressure
column acting at sea level is higher. Since pressure is force/area, the
force being the weight of the air column, the pressure at sea level is
higher.
Ini adalah kerana ketinggian turus air melebihi titik pada paras laut
adalah tinggi daripada puncak gunung. Maka berat turus air bertindak
pada paras laut adalah lebih tinggi.
Pressure at the base
Tekanan di tapakk = hρg
= 75.5 × 10-2 ×13.6 × 103 × 10
=1.03 × 105 Pa
The difference in pressure between the top and bottom of the building is
equal to the pressure caused by the air column from the base to the top.
Perbezaan tekanan antara puncak dan tapak bangunan adalah sama
dengan tekanan yang disebabkan oleh turus udara daripada tapak ke
puncak.
= 1.03 ×105 – 9.79 × 104
= h ×1.36 ×10
h = 375 m
Density : the density must be high so that the length of the liquid column
needed to balance atmospheric pressure must be less than 1 m
because the tube is only 1 m long.
Ketumpatan : ketumpatan mestilah tinggi supaya panjang turus cecair
diperlukan untuk mengimbangkan tekanan atmosfera mestilah kurang
daripada 1 m kerana panjang tiup hanyalah 1 m sahaja.
Atmospheric Pressure is approximately 100 000 Pa.
Tekanan atmosfera adalah lebih kurang 100 000 Pa.
For mercury, using p= hρg
Bagi merkuri, menggunakan p = hρg
100 000 = h × 13600 × 10
h = 0.74 m
100000
For oil, h = 800×10 = 12.5 m
100000
For water, h = 800×10 = 10 m
Characteristic
Sticks/does not stick to
glass
Melekat/Tidak melekat
pada kaca
Opaque/Transparent
Legap/Lutsinar
The best liquid is
mercury
Modul SOLAF Fizik 2014
Explanation
The liquid should not stick to the glass.
Otherwise it will not be suitable to be used
as a barometer liquid.
Cecair sepatutnya tidak melekat pada kaca.
Jika tidak ia tidak sesuai untuk digunakan
sebagai cecair barometer.
It should be opaque so that it more visible.
Ia seharusnya legap supaya mudah dilihat..
Because the column length is 0.74 m which
can fit into the 1 m length tube, it does not
Kertas 2
Paper 2
4
2
3
10
Bab 3 Tingkatan 4
Form 4 Chapter 3
Daya dan Tekanan
Forces and Pressure
Kertas 2
Paper 2
stick to glass and it is opaque.
Cecair yang terbaik adalah merkuri, kerana
Modification/
Explanation
tinggi turus ialah 0.74 m di mana boleh
suggestion
dimuatkan ke dalam 1 m panjang tiub, ia
Oil
Incompressible/
tidak melekat pada kaca dan legap.
No air
bubble
The other
two
liquids do not satisfy these three conditions.
High boiling point/
Does
notlai
change
to gas stateketiga-tiga
easily/ syarat tersebut.
Dua lagi
cecair
tidak memenuhi
Does not evaporate easily
TOTAL MARKS
Low density/
Lighter/
High viscosity
Fluid does not flow easily
Small master
High pressure produced with a small force
(a)
2
piston
Force is action
Big slave piston
Produce bigger output force
on an object
Aluminium/
Steel
Strong/
Does not break easily/
Non corrosive/
Does not rust easily
that can result
in changes like
size, shape and
direction.
(b)
1. surface area A1 is smallerr than A2
2. forces F1 is smaller than F2
3. pressure P1 is equal to P2
4. when surface area is larger, the force exerted on the piston will increase.
5. Pascal’s principle
(c)
1. Function- to transfer water from beaker to cylinder.
2. The pressure at lowest point in cylinder is greater than the
atmospheric pressure, the liquid flows out at lowest point in
cylinder/at the end of rubber tube in cylinder.
3. The pressure in the rubber tube decreases as the water flows out
and a partial vacuum is created.
4. The higher atmospheric pressure water into the tube.The water flows
until the liquid surface in cylinder reaches the same level as in
beaker.
20
1
5
4
(d)
10
Modul SOLAF Fizik 2014
Bab 3 Tingkatan 4
Form 4 Chapter 3
Modul SOLAF Fizik 2014
Daya dan Tekanan
Forces and Pressure
Kertas 2
Paper 2
TOTAL MARKS
20