Daya dan Tekanan Jawapan TUGAS

SKEMA JAWAPAN PAPER 2 SECTION A

NO. SOALAN

EXPLANATION/ PENERANGAN

MARKS/ MARKAH

1

(a)(i)

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

1

(ii)

The distance y is greater than the distance x// converse

Jarak y adalah lebih jauh berbanding jarak x.

1

(b)

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

2

TOTAL MARKS 4

2

(a)

Aerofoil 1

(b)

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.

3

(c )

Bernoulli’s principle

Prinsip Bernoulli

1

(d)

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

2

TOTAL MARKS 7

3

(a)(i)

Volume = cross-sectional area × height = 2.0 cm2 _{× 10 cm}

= 20 cm3

1

(ii)

Weight of water has been displaced = weight of glass tube + weight of ball bearings

Vρg = W glass tube + W ball bearings

20

(

1.0

)(

10

)

1000

=

10

1000

(

10

)+

W ball bearings W ball bearings = 0.10 N

=

0.10

×

1000

10

=

_{10.0 g}

3

(b)

Hydrometer-to determine density of liquid

Hidrometer-mengukur ketumpatan cecair

1

4 (a) 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

2

(b)

Solution/Penyelesaian:

Pgas = (h ρ g)water

= (1000)(10)(0.1)

= 1000 N m-2

3

(c)

SolutionPenyelesaian:

Pair + Pmercury = Patm

Pair + 10cmHg = 76cmHg

Pair = (76 – 10) cmHg

= 66 cmHg

2

TOTAL MARKS 7

5

(a)

Bourdon Gauge / Tolok Bourdon

1

(b)

Pressure/Tekanan

₁

(c)

1st _{: 27 + 273 = 300 K}

2nd _:

128

300

=

132

T

3rd _{: T = 309.38 K / 36.38 ˚C}

3

TOTAL MARKS

5

6(a) Mass is the quantity of matter/Jisim kuantiti jasad 1 (b)(i) 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.

1

(b)(ii) Volume of oil displaced by the apple is larger than the water/Isipadu air tersesar oleh epal adalah lebih bayak berbanding air.

1

(b)(iii) Density of water is larger/ greater than oil. 1 (c)(i) Inversely proportional/Berkadar sonsang 1

(c)(ii) Equal/sama 1

(d) Archimedes principle/Prinsip Archimedes 1

(e) Empty the ballast tank / Remove the water/Kosngkan tangki balas 1

SKEMA JAWAPAN SECTION B KERTAS 2

No. Soala

n

Answer/Jawapan Mark/Markah

1(a)(i) 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.

1

(ii) When a force F1 acted on the small piston with cross-sectional area

A1, a pressure

F

₁

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.

Apabila daya F1 bertindak pada omboh kecil dengan luas keratan

rentas A1, tekanan

F

₁

A

_{1 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.

4

(b)

Characteristic Ciri-ciri

Explanation Penerangan Oil is used

Minyak digunakan

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. High specific heat capacity

Muatan haba tentu yang tinggi

It is not easily heated up Tidak mudah dipanaskan Small cross sectional-area of

input piston

Luas keratan rentas omboh input yang kecil

So that the force applied to the input piston is small

Supaya daya yang dikenakan pada omboh input adalah kecil Large cross-sectional area of

output piston

So that it can raise up a greater load

Luas keratan rentas omboh output yang besar

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.

(c) (i)

F =

F

₁

A

₁

=

10

150

×

10

−4 = 666.7 Pa

2

(ii)

F

₂

A

_{2 =}

F

₁

A

₁

F

₂

3

₌

10

150

×

10

−4 F2 = 2000 N

3

TOTAL MARKS 20

2 (a) 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

(b)(i) 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.

2

(ii) Archimedes’ Principle

Prinsip Archimedes

1 (c)(i) Rod Q has displaced more water

Rod Q mempunyai lebih banyak air yang tersesar

1

(ii) Buoyant force acting on Rod Q > Rod P

Daya apungan bertindak di Rod Q > Rod P

1

(iii) Rod Q 1

(d)(i) Weight of water displaced = 1 000 (5) ( 10) = 50 000 N

2 (ii) Weight of cargo = 50 000 – 250 (10)

= 47 500 N

1

(e)

Suggestion Explanation

Material – Extra strong

To ensure the balloon does not burst or leak easily and the people in a strong basket will be safer.

Power-Strong It can heat up the air faster and the balloon can ascennd faster

The holding cable-Light and strong

It will decrease the overall weight of the balloon in place safely

Balloon should pulled down by motor

So that the balloon can descend faster and more safety without human error and fatique

Most Suitable is balloon X

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 20

3(a)(i) 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

(ii) 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.

3

(iii) Landing distance of discus Q>P

Jarak mendarat cakera Q > P

1

(vi) 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

(v) 1. Throw with a larger force/higher speed 2. Throw and spin the dicus

3. Use a discus with a smoother surface

2

(c)

Suggestion Explanation

Replace engine with more powerful jet engine

This will provide higher thrust and faster speed

Use wings with aerofoil shape

This will create a resultant upward pressure and lifting force.

Area of the wing should be bigger

This will generate a larger lifting force to carry a larger load

The whole aeroplane This will reduce air frictionn and

should be more aerodynamic

resistance, thus the speed can be increased

Mass of aeroplane should be lighter and yet stronger material

This will allow the aeroplane to fly faster and is able to carry a heavier load

TOTAL MARKS 20

4 (a) Temperature is the degree of hotness 1 (b) 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}

4

(c) 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.

10

(d)(i)

1st _:

l

θ

−

l

0

l

100

−

l

0

x

100

2nd _:

12

−

5

25

−

5

x

100

3rd _{: 35 ˚C}

4th _{: 308 K}

4

(d)(ii) Volume expand with temperature.. 1

TOTAL MARKS 20

SKEMA JAWAPAN KERTAS 2-BAHAGIAN C

No.Soala n

Answer/Jawapan Markah

1.(a) (i) 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.

1

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.

4

(b)(i) Pressure at the base

Tekanan di tapakk = hρg = 75.5 × 10-2 _{×13.6 × 10}3 _{× 10}

=1.03 × 105 _Pa

2

(ii) 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 × 10}4

= h ×1.36 ×10 h = 375 m

3

(c) 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

For oil, h =

100000

800

×

10

_{= 12.5 m}

For water, h =

100000

800

×

10

_{= 10 m}

Characteristic Explanation Sticks/does not stick to

glass

Melekat/Tidak melekat pada kaca

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.

Opaque/Transparent

Legap/Lutsinar

It should be opaque so that it more visible.

Ia seharusnya legap supaya mudah dilihat..

The best liquid is mercury

Because the column length is 0.74 m which can fit into the 1 m length tube, it does not

stick to glass and it is opaque.

Cecair yang terbaik adalah merkuri, kerana tinggi turus ialah 0.74 m di mana boleh dimuatkan ke dalam 1 m panjang tiub, ia tidak melekat pada kaca dan legap.

The other two liquids do not satisfy these three conditions.

Dua lagi cecair lai tidak memenuhi ketiga-tiga syarat tersebut.

TOTAL MARKS 20

2

(a)

Force is action on an object 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.

(d)

1

5

4

10

Modification/

suggestion

Explanation

Oil Incompressible/ No air bubble High boiling point/

Low density/ High viscosity

Does not change to gas state easily/ Does not evaporate easily

Lighter/

Fluid does not flow easily Small master

piston

High pressure produced with a small force

Big slave piston Produce bigger output force Aluminium/

Steel

Strong/

Does not break easily/ Non corrosive/