kl-skema p2

8/8/2019 KL-SKEMA P2

1/9

SULIT 3764 / 2

3764 / 2 SULIT

1

JAWAPAN MARKAH

1 a)To cut internal thread M14 x 2.0Solution:Tap drill size (TDS)= Major dia.- Pitch

= 14 2.0= 12.00 mm

1. Drill a hole of suitable size.2. Choose a desired tap size.3. Positioned tap vertically.4. Turn wrench clockwise half/full cycle. Drip cutting fluid.5. Turn wrench anticlockwise.6. Repeat steps 4 and 5.7. Repeat steps 4 & 5 using Plug and Bottoming tap

b) scriber / steel ruler / L-square / pin punch

drill machine / hacksaw / File /

2

2222

222

= 16M

2

2

= 4m

Total = 20

m


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SULIT 3764 / 2

3764 / 2 SULIT

QuestionsNo.No.

Soalan

ANSWER MarksMarkah

2 (a) Total internal reflection.

(b) Encoder- Convert analog signal to digital

signal.

Light source- Convert the digital signal to pulse-

form light

Light detector- Convert the pulse light to electric

signal (digital)

Decoder- Convert digital signal to analog

signal

(c)Advantages

Can carry moreinformation at anytime because oflarge bandwidth.

There is noelectromagnetic interference becauseonly pulsed light moves through thefibre optic cable.

There is no interferencewith other signals from nearbycables.

There is no electriccurrent in fibre optic cable.

It is safe from line

tapping. It is small and light.

Disadvantages

Cost of fiber optic cables,installation and testing are high.

Difficult to redirect the signal toother channels.

The installation and connectionare complex ( because of the highprecision ) and require special tools.

TOTAL

2

2

2

2

2

Any

THREE3 x 2M=6

Any TWO2 x 2M=4

20

2


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SULIT 3764 / 2

APPENDIX1

LAMPIRAN 1

Question 3

Name : Form :.

Back SightPandanganBelakang

IntermediateSight

PandanganAntara

Fore SightPandangan

Hadapan

RiseNaik

FallTurun

ReducedLevelArasLaras

ChainageRantaian

RemarkCatatan

1.600 10.750 0 BM RL = 10.750 m

1.185 0.415 11.165 10 Point A

1.790 0.605 10.560 40 Point B

1.290 2.425 0.635 9.925 70 Point C, TP2.275 0.985 8.940 90 Point D

3.000 0.725 8.215 100 Point E

3.290 0.290 7.925 120 Point F

2.510 3.025 0.265 8.190 140 Point G, TP2

1.980 0.530 8.720 160 Point H

2.180 0.200 8.520 180 Point I

2.970 0.790 7.730 200 Point J

2.380 0.590 8.320 210 BM2 RL = 8.315 m

Skema :

APPENDIX 2

3764 / 2 SULIT

Jawapanbetul

Markah Jawapan betul Markah

15 12 9 6

14 11 8 5

13 10 7 4

12 9 5-6 3

11 8 3-4 2

10 7 1-2 1

3


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SULIT 3764 / 2

LAMPIRAN 2

Name : . Form: .

ReducedLevel

Aras

Laras

10.

75

0

11.

16

5

10.

65

0

9.

92

5

8.

94

0

8.

21

5

7.

92

5

8,

19

0

8.

72

0

8.

52

0

7.

73

0

8.

32

0

Chainage

Rantaian

(m)

010

40

70

90

100

120

140

160

180

200

210

SKEMASOALAN 4

3764 / 2 SULIT

4

6

7

8

9

10

11

12

13

[ aras laras yang betul - x 8 = 4 ][ tanda dan plotan yang betul - x 8 = 4 ]

[ 8 Markah ]


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SULIT 3764 / 2

4.

a) i) The floor area of walled kitchenextension

= 4 X 8.5 m2

= 34.00 m2

The cost of the walled kitchen

extension section = 34 X 250.00

= RM 8500.00

ii) The floor area of patio extension

= [ 5.2 + 4.0 + 6.0 5.0 ] X 4.5 m2

= 45.90 m2

The cost of the patio

extension section = 45.90 X 250.00

= RM 11475.00

2m

2m

2m

1m

2m

2m

2m

1m

= 14m

b)

The volume of

the house = 8.5m X 15.2m X 0.15m

= 19.38m3

The cost of the walled kitchen

extension section = 19.38 X 120.00

= RM 2325.60

2m

1m

2m

1m

3764 / 2 SULIT

5

B

A

K2K1

1

Masalah (i)

Guna

silinder

berpegas

(3 markah)

Masalah (iii)

Guna Injap

Penghad

boleh laras &

selari dengan

injap sehala

(3 markah)

Masalah (ii)

Ubah

keddukan pam

kepada N/O

(3 markah)


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SULIT 3764 / 2

5.

a)

b)

Apabila pam dihidupkan .

Udara termampat akan mengalir sehingga

ke liang injap kawalan arah.

Apabila kedudukan injap kawalan arah

beroperasi dan berada pada K1 udara

termampat mengalir keluar dan injap kawalan

aliran hingga ke selinder bagi menggerakkan

omboh ke B

Omboh akan bergerak dari B ke A

disebabkan oleh tindakan pegas apabila injap

kawalan arah berada pada K2

Kelajuan omboh dari B ke A adalah secara

perlahan disebabkan oleh aliran udara akan

melalaui injap penghad boleh laras. Udara termampat akan dilepaskan melalui

ekzos

Mana-mana

empat jawapan

= 8 markah

c)

Ekzos dan

penyenyap( silencer)

3M

6aWrong connection

3764 / 2 SULIT

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6b

Sambungan yang salah

Right connectionSambungan yang betul

Change PB2(stop) to C1(normally open) in line motorM2.Tukar PB2(henti) kepada C1 (terbuka) pada litar motorM2.

Cut out C1(normally close) in line motor M1.Keluarkan C1(terbuka) pada litar motor M1.

Cut out C1(normally close) in line motor M2.Keluarkan C1 (terbuka) in litar motor M2.

2

2

2

6m

3764 / 2 SULIT

7

Motor M1

F = 3N

PedalMotor M1

C

1

T R

C

2Motor M2

L

CBPB1(stop)

PB1(start)

C

1

C

2

C

1

PB2(start)

OR

N

Motor M1

T R

Motor M2

L

CBPB1(stop)

PB1(start)

C

1

C

2

C

1

PB2(start)

C

2

C

1

C

1

PB2(stop)

OR

N


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SULIT 3764 / 2

When push button PB1(start) is pressed/, contact coil C1 willbecome electromagnetic

Apabila punat tekan PB1 (start) ditekan/, gegelung penyentuhC1 akan menjadi electromagnet.

All open contactors inside contactor C1 will be closedSemua sesentuh C1 tertutup

M1 motor startsMotor M1 berkendali

When push button PB1 (start) is released, M1 motor is

still running due to holding circuit actionApabila punat tekan PB1 (start) dilepaskan, motor M1 masihberkendali disebabkan oleh litar pegang.

When push button PB2(start) is pressed,/contact coil C2will become electromagneticApabila punat tekan PB2 (start) ditekan,/ gegelung penyentuhC2 akan menjadi elektromagnet.

All open contactors inside contactor C2 will be closedSemua sesentuh C2 tertutup

Motor M2 starts running /because contactor C1 in circuit2 is closedMotor M2 berkendali / kerana sesentuh C1 dalam litar 2tertutup.

When PB1 (stop) is pressed, motor M1 and M2 stoprunning. The action will loosen the electromagnet in coil C1.Motor M1 dan M2 akan berhenti berputar apabila PB1 (henti)ditekan. Tindakan menekan PB1 (henti) menghilangkanelektromagnetan gegelung C1.

It will also cause all the close contacts in contactor C1(contacts in circuit 2) to open.Ini menyebabkan semua sesentuh penyentuh C1 yang sedangtertutup (termasuk sesentuh yang terdapat dalam litar 2)menjadi terbuka.

1

1

1

1+1

1+1

1

1+1

1+1

1+1

14mmax

N0. SOALAN MARKAH

7 (a)

(i)Human 2m

3764 / 2 SULIT

8


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SULIT 3764 / 2

Manusia- Kerusi diubahsuai mengikut

kesesuaian saiz supayamudah untuk meletakkankanak-kanak.

(ii) FunctionFungsi

- Kerusi diubahsuai dandilengkapkan dengantempat meletak makanan.

(iii) Strength

Kekuatan- Kerusi diubahsuai

sambungannya agar lebihkukuh

(b) - Body and human movement- Safety- Growth and age- Size- Shape- Movement

Lakaran = 2m

2m

Lakaran = 2m

2m

Lakaran = 2m

2m2m2m2m2m2m

Maksimum=8m

3764 / 2 SULIT

9

kl-skema p2

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