Fizik

JAWAPAN

SET 1 (b) (i) Benang tebal
Thick string
T
Kertas 1 T T Penutup getah
Rubber stopper
Klip buaya Tiub plastik
1 D 2 D 3 D 4 B 5 A Crocodile clip Plastic tube
8 B 9 C 10 D
6 B 7 D 13 B 14 D 15 B Pemberat
18 D 19 C 20 A Slotted weight
11 D 12 B 23 D 24 B 25 A
28 A 29 B 30 B
16 D 17 B 33 A 34 D 35 B
38 B 39 B 40 A
2 1 B 22 C (ii) F = 0.3 (3)2
0.8
26 B 27 A

3 1 C 32 D = 3.375 N
CONTOH
36 B 37 B (iii) Laju linear bertambah

The linear speed increases

Kertas 2 5 (a) Frekuensi minimum foton cahaya yang diperlukan untuk

Bahagian A/ Section A menghasilkan kesan fotoelektrik.

1 (a) Kuantiti asas/ Basic quantity The minimum frequency of light photon required to produce

(b) 26 photoelectric effect.
20
(b) (i) Frekuensi cahaya A adalah kurang daripada frekuensi
= 1.3 s
cahaya B.
(c) Bertambah/ Increasing
The frequency of light A is less than the frequency of

2 (a) Jarak terpendek di antara dua titik pada arah tertentu. light B.

The shortest distance between two points in specific direction. (ii) Tenaga kinetik maksimum fotoelektron yang terpancar

(b) Tidak bergerak/ Berhenti/ Pegun dari cahaya A adalah kurang daripada cahaya B.

Not moving/ Stop/ Stationary The maximum kinetic energy of photoelectron emitted

(Mana-mana satu jawapan/ Any one answer) from light A is less than light B.

(c) 80 m – 80 m – 50 m (iii) Semakin tinggi frekuensi, semakin tinggi tenaga kinetik

= –50 m maksimum fotoelektron yang dipancarkan.

50 The higher the frequency, the higher the maximum
20
(d) =− kinetic energy of photoelectron emitted.

= –2.5 m s-1 (c) (i) W = hfo

2.95(1.6 × 10−19) = (6.63 × 10−34)fo
fo = 7.12 × 1014 Hz
3 (a) Membenarkan arus mengalir sehala sahaja
Allows current to flow in one direction only (ii) E = W + K
(b) (i) Mentol menyala/ The bulb lights up
(ii) – Diod disambung dalam keadaan pincang ke depan. E = 2.95 + 0.60
The diode is connected in forward bias.
– Rintangan diod adalah rendah. = 3.55 eV
The resistance of diode is low.
(Mana-mana satu jawapan/ Any one answer) Atau/ Or

(c) E = W + K

E = 2.95 + 0.60

= 3.55 eV

= 3.55(1.6 ×10-19)

= 5.68 × 10-19 J

(d) Tenaga kinetik maksimum tidak berubah

The maximum kinetic energy remains unchanged

(d) Rektifikasi/ Rectification 6 (a) Kadar perubahan momentum
The rate of change in momentum
4 (a) (i) Daya graviti/ Gravitational force (b) (i) Fx = 700 kos 45°/ 700 cos 45°
Gm1m2 = 495 N
(ii) F = r2 (ii) Fx = 700 kos 20°/ 700 cos 20°
= 658 N
= (6.67 × 10−11) × (2) × (5.97 × 1024) (c) (i) Sudut dalam Rajah 6.1 lebih besar daripada Rajah 6.2.
(6.37 × 106)2 The angle in Diagram 6.1 is greater than in Diagram

= 19.627 N 6.2.
(ii) Magnitud Fx dalam Rajah 6.1 adalah lebih kecil daripada
F = ma
Rajah 6.2.
a = 19.627 The magnitude of Fx in Diagram 6.1 is smaller than in
2
= 9.81 m s-2 Diagram 6.2.
(d) (i) Semakin besar sudut, semakin kecil komponen daya
(iii) F = mg
mengufuk, Fx.
= 2.0 × 9.81 The greater the angle, the smaller the horizontal

= 19.62 N component of force, Fx.

J1

Fizik Jawapan

(ii) Semakin besar komponen daya mengufuk Fx, semakin Atau/ Or
besar pecutan.
P = 10 + 4
The greater the horizontal component of force Fx, the
greater the acceleration. = 14 m air/ 14 m of water [2]

(e) – Kedudukan baldi: Atas lantai

Position of pail: On the floor [1]

7 (a) (i) Pantulan cahaya/ Reflection of light – Menghasilkan kawasan bertekanan rendah

(ii) Diperkecilkan/ Tegak/ Maya Create a low pressure region [1]

Diminished/ Upright/ Virtual – Kedudukan akuarium: Atas meja

(Mana-mana satu jawapan/ Any one answer) Position of aquarium: On the table [1]

(iii) Cermin – Menghasilkan kawasan bertekanan tinggi
Mirror
Create a high pressure region

Objek Imej – Diameter hos: Besar
Object Image
Diameter of hose:Big [1]
F
F C – Boleh mengalirkan isipadu air yang banyak

Large volume of water can flow [1]

(b) (i) Cermin cekung kerana ia menghasilkan imej yang tegak – Perbezaan ketinggian: Besar
dan besar.
Difference in height:Big [1]
Concave mirror because it forms upright and magnified
image.

(ii) Kurang daripada jarak fokus kerana boleh menghasilkan
imej yang lebih besar.

Less than focal length because it can produce larger
image.
CONTOH – Menghasilkan perbezaan tekanan yang tinggi

Produces large difference in pressure [1]

– Pilihan/ Choice: N [1]

– Baldi berada di atas lantai, akuarium berada di atas meja,

mempunyai hos yang besar dan perbezaan ketinggian

adalah besar.

The pail is on the floor, the aquarium is on the table, has

8 (a) (i) E = Pt big hose and bigger difference in height. [1]

= 1.5(5 × 20) 10 (a) (i) Pembelahan nukleus/ Nuclear fission [1]

= 150 kWj/ kWh (ii) m = 1.00867 + 235.04392 – 91.92611 – 140.91441 –

(ii) Kos/ Cost = 150 × RM0.20 3(1.00867)

= RM30 = 0.18606u

(b) (i) – Dawai gegelung = 0.18606(1.66 × 10-27)

Coiled wire = 3.088596 × 10-28 kg [3]

– Menghasilkan rintangan yang tinggi (iii) E = mc2

Produces high resistance = (3.088596 × 10-28)(3 × 108)2

(ii) – Dawai nikrom = 2.78 × 10-11 J [2]

Nichrome wire (b) – Nukleus uranium-235 dibedil oleh satu neutron

– Menghasilkan rintangan yang tinggi/ kerintangan menyebabkan nukleus uranium menjadi tidak stabil.

yang tinggi Uranium-235 nucleus is bombarded by a neutron causing

Produces high resistance/ high resistivity it to be unstable. [1]

(c) S – Pembelahan nukleus berlaku menghasilkan nukleus

Bahagian B/ Section B kripton, barium dan tiga neutron yang baharu.

9 (a) Ketumpatan cecair/ Graviti Nuclear fission occurs and produces nucleus of krypton,

Density of liquid/ Gravity [1] barium and three new neutrons. [1]

(Mana-mana satu jawapan/ Any one answer) – Tiga neutron yang terhasil akan membedil nukleus

(b) (i) Kedalaman air di kedudukan K adalah lebih rendah uranium yang lain.

daripada kedudukan L. Three neutrons produced will bombard another uranium

The depth of water at point K is lower than point L. [1] nucleus. [1]

(ii) Jarak mengufuk pancutan air yang keluar dari kedudukan – Proses ini berlaku secara berterusan dan menghasilkan

K lebih dekat daripada kedudukan L. tindak balas berantai.

The horizontal distance of water spurting out from point This process occurs continuously to produce chain

K is nearer than point L. [1] reaction. [1]

(iii) Tekanan air di kedudukan K adalah lebih rendah (c) – Bahan api: Uranium

daripada kedudukan L. Fuel: Uranium [1]

The water pressure at point K is smaller than point L. – Menghasilkan tenaga yang banyak

[1] Produces large amount of energy [1]

(c) (i) Semakin bertambah tekanan air, semakin bertambah – Dinding reaktor: Konkrit tebal

jarak mengufuk pancutan air. Wall of reactor: Thick concrete [1]

The greater the pressure of water, the greater the – Menghalang sinaran radioaktif yang berbahaya daripada

horizontal distance of water spurting out. [1] dipancarkan ke persekitaran.

(ii) Semakin bertambah kedalaman air, semakin bertambah Prevents dangerous radioactive radiation from emitting

tekanan air. to the surrounding. [1]

The greater the depth of water, the greater the pressure – Rod kawalan: Kadmium

of water. [1] Control rod: Cadmium [1]

(d) (i) P = ρgh – Dapat menyerap neutron yang berlebihan.

= (1 000)(9.81)(10) Can absorb excess neutron. [1]

= 9.81 × 104 Pa [2] – Moderator: Grafit

(ii) P = ρgh Moderator: Graphite [1]

= (1 000)(9.81)(10+4) – Dapat memperlahan neutron yang bergerak pantas.

= 1.37 × 105 Pa Can slow down fast-moving neutron. [1]

J2

Fizik Jawapan

– Pilihan/ Choice: R [1] SET 2

– Menggunakan uranium sebagai bahan api, dinding reaktor

dibina daripada konkrit tebal, menggunakan kadmium Kertas 1

sebagai rod kawalan dan grafit sebagai moderator.

Uses uranium as fuel, reactor wall is built from thick 1 D 2 A 3 A 4 C 5 B
8 C 9 C 10 B
concrete, uses cadmium as control rod and graphite as 6 B 7 C 13 A 14 D 15 C
18 B 19 B 20 A
moderator. [1] 11 C 12 C 23 A 24 A 25 B
28 B 29 C 30 D
1 6 A 17 C 33 D 34 A 35 D
38 A 39 C 40 A
Bahagian C/ Section C 2 1 A 22 A

11 (a) Arus aruhan ialah arus yang terhasil apabila konduktor 26 D 27 C

memotong fluks magnet. 3 1 C 32 A

Induced current is the current produced when the conductor 3 6 C 37 A

cuts magnetic flux. [1]

(b) (i) – Arah gerakan rod kuprum dalam Rajah 11.1 adalah Kertas 2

sama dengan Rajah 11.2. Bahagian A/ Section A

The direction of motion of copper rod in Diagram 1 (a) ✓ gelombang membujur
longitudinal wave
11.1 is the same as in Diagram 11.2. [1]
CONTOH
– Luas keratan rentas rod kuprum dalam Rajah 11.1 (b) (i) X : Regangan / Rarefaction

adalah lebih kecil daripada Rajah 11.2. Y : Mampatan /Compression

The cross-sectional area of copper rod in Diagram (ii)

11.1 is smaller than in Diagram 11.2. [1]

– Magnitud pemesongan jarum galvanometer dalam

Rajah 11.2 adalah lebih besar daripada Rajah 11.1.

The magnitude of deflection of galvanometer pointer λ

in Diagram 11.2 is greater than in Diagram 11.1. [1]

(ii) – Semakin bertambah luas keratan rentas rod kuprum, 2 (a) Daya impuls ialah kadar perubahan momentum dalam
perlanggaran dalam masa singkat.
semakin bertambah magnitud pemesongan jarum
Impulsive force is the rate of change of momentum in a
galvanometer. collision in a short period of time.

The greater the cross-sectional area of copper rod, the

greater the magnitude of deflection of galvanometer

pointer. [1] (b) Halaju awal / initial velocity,

– Semakin bertambah luas keratan rentas rod kuprum, u = 100 × 1000 m
1× 60 × 60 s
semakin bertambah arus aruhan.

The greater the cross-sectional area of copper rod, = 27.778 m s-1

the greater the induced current. [1] F= m(v – u)
t
– Konsep fizik: Aruhan elektromagnet

Physics concept: Electromagnetic induction [1] 500(0 − 27.778)
1
(c) – Apabila magnet bar digerakkan ke arah solenoid, hujung =

P menjadi kutub Utara. F = –13889 N

When the bar magnet is moved towards the solenoid, the (c) (i) Berkurang/ Decreased

end of P becomes North pole. [1] (ii) Masa hentaman bertambah

– Hal ini menyebabkan arus aruhan terhasil dan jarum Time of impact increased

galvanometer pesong ke kiri.

This causes the induced current to be produced and the 3 (a) (i) Kuantum tenaga cahaya yang boleh dipindahkan
Quantum of light energies that can be transferred
galvanometer pointer to be deflected to the left. [1] (ii) Tenaga foton adalah berkadar terus dengan frekuensi

– Hukum fizik yang terlibat ialah Hukum Lenz. gelombang cahaya.
The photon energy is directly proportional to the
The law of physics involved is Lenz Law. [1]
frequency of a light wave.
(d) – Menggunakan magnet yang kuat atau menggunakan

magnet neodimium.

Uses a stronger magnet or uses neodymium magnet. [1] hc (6.63 × 10−34) × (3.0 × 108)
λ 6.5 × 10−13
– Dapat menghasilkan medan magnet yang kuat. (b) (i) E = =

Can produce a stronger magnetic field. [1] E = 3.06 × 10–13 J

– Menambah bilangan lilitan gegelung. (ii) P = nE

Increases the number of turns of coil. [1] P 1.5 × 10–13
E 3.06 × 10–13
– Dapat meningkatkan pemotongan fluks magnet. n= =

Can increase the cut of magnetic flux. [1] = 4.902 × 109

– Menggunakan dawai kuprum.

Uses copper wire. [1] 4 (a) Menaikkan atau menurunkan voltan output

– Bagi menghasilkan rintangan yang rendah. Increase or decrease the output voltage

To produce low resistance. [1] (b) Injak turun/ Step down

– Menggunakan dawai gegelung yang mempunyai luas (c) (i) NVss Np
Vp
keratan rentas yang lebih besar. =

Uses a wire coil with larger cross-sectional area. [1]

– Dapat mengurangkan rintangan. Ns = 500 × 12 = 25
250
Can reduce resistance. [1]

– Menukar gelang gelincir dengan komutator. = 25 lilitan/ turns

Changes slip rings to commutator. [1] (ii) I = P = 24
V 12
– Bagi menghasilkan arus terus.

To produce direct current. [1] I = 2A

J3

Fizik Jawapan

(d) (i) 7 (a) Masa yang diambil oleh sampel radioaktif mereput kepada
separuh daripada bilangan asal nukleus sampel tersebut.
(ii) Diod/ Diode
(iii) Membenarkan arus mengalir satu hala Time taken for a nuclei of decaying radioactive sample to
To allow current flow in one direction decay to half of its initial number.

(b) (i) Aktiviti

Activity

2000

1500

5 (a) Daya yang bertindak ke atas suatu objek adalah sama 1000
dengan berat air atau udara yang disesarkan.
500
Upward force on an object equals to the weight of fluid
displaced. Masa/ minit
0 15 30 45 60 Time/ minute
(b) (i) Isipadu udara yang disesarkan oleh belon Rajah 5.2
lebih daripada Rajah 5.1 T1 = 15 minit/ minute
2
Volume of air displaced by balloons in Diagram 5.2 is
more than Diagram 5.1

(ii) Daya apungan belon Rajah 5.2 lebih daripada Rajah 5.1
Buoyant force of balloons in Diagram 5.2 is more than

Diagram 5.1
(iii) Daya tarikan tali dalam Rajah 5.2 lebih daripada Rajah

5.1
Pulling force on the string in Diagram 5.2 is more than

Diagram 5.1
(c) (i) Semakin bertambah isi padu udara yang disesarkan,

semakin bertambah daya apungan.
The bigger the volume of the air displaced, the bigger

the buoyant force.
(ii) Semakin bertambah daya apungan, semakin bertambah

daya tarikan ke atas tali.
The bigger the buoyant force, the bigger the pulling

force on the string.
(iii) Prinsip Archimedes / Archimedes’ Principle
(d) – Lebih senang/ Easier
– Sebab ketumpatan udara lebih rendah/ daya apungan

lebih rendah
Because the density of air is less/ smaller buoyant force
CONTOH (ii) 2000 T1 1000 T1 500 T1 250
2 22

(c) (i) – Rod kawalan: Boron

Control rod: Boron

– Sebab: Menyerap neutron yang berlebihan

Reason: Absorbing excess neutrons

(ii) – Rod bahan api: Uranium

Fuel rod: Uranium

– Sebab: Digunakan sebagai bahan api untuk

menghasilkan tenaga nuklear melalui tindak balas

pembelahan nukleus.

Reason: Used as fuel to produce energy nuclear

through a nuclear fission reaction.

(iii) T

6 (a) Kuantiti haba yang diperlukan untuk menaikkan suhu 1kg 8 (a) Magnet sementara
Temporary magnet
bahan sebanyak 1°C atau 1 K. (b) Tertarik ke hujung paku
Attracted to the end of the nail
The quantity of heat energy required to increase the (c) Songsang/ Inverted
(d) (i) – Diameter dawai solenoid: Diameter besar
temperature of 1 kg of the substance by 1° C or 1 K. Diameter of solenoid wire: Big diameter
– Sebab: Kurangkan rintangan
(b) (i) Jisim blok dalam Rajah 6.1 lebih tinggi berbanding Reason: Reduce resistance
(ii) – Jenis teras: Teras besi lembut
jisim blok dalam Rajah 6.2/ sebaliknya Types of core: Soft iron core
– Sebab: Mudah dimagnetkan / dinyahmagnetkan
Mass of block in Diagram 6.1 is higher than mass of Reason: Easy to magnetise / demagnetise
(iii) – Saiz gong: Saiz gong besar
block in Diagram 6.2 / vice-versa Size of the gong: Bigger gong size
– Sebab: Lebih banyak udara bergetar
(ii) Suhu akhir blok dalam Rajah 6.1 lebih rendah berbanding Reason: More air molecules vibrate

suhu akhir blok dalam Rajah 6.2 / sebaliknya

The final temperature of block in Diagram 6.1 is lower Bahagian B/ Section B
9 (a) (i) Frekuensi sistem berayun tanpa kehadiran sebarang
than final temperature of block in Diagram 6.2 / vice-
daya lain.
versa Frequency of any oscillating system in the absence of

(iii) Semakin rendah jisim, semakin tinggi suhu akhir / any other forces. [1]
(ii) – Apabila bandul X dilepaskan, ayunan paksa terhasil.
kenaikan suhu When pendulum X is released, forced oscillation is

The smaller the mass, the higher the final temperature / created. [1]
– Ayunan bandul X memindahkan tenaga ke semua
rise of temperature
bandul menyebabkan ke semuanya berayun.
(c) (i) Q = mc∆θ Oscillation of pendulum X transfers the energy to all

Pt = mc∆θ 50 × (10 × 60) pendulum causing them to oscillate. [1]
Pt 0.2 × (90 − 30) – Resonans berlaku pada bandul D kerana mempunyai
c = mc∆θ =
panjang yang sama dengan bandul X.
c = 2500 J kg−1°C−1 Resonance happens to pendulum D because it has the

(ii) Terdapat haba yang hilang ke persekitaran same length as pendulum X. [1]

There is heat loss to the surrounding

(iii) – Balut blok aluminium dengan kain/ tisu

Wrap the aluminium block with cloth/ tissue

– Letak sedikit minyak pada termometer

Apply a bit of oil on the thermometer

(Sebarang cadangan yang sesuai/ Other suitable

suggestions)

J4

Fizik Jawapan

– Bandul D berayun pada frekuensi asli bandul X (d) – Diameter rod pemegang: Kecil
Wand diameter: Small [1]
dengan amplitud paling besar antara bandul yang – Udara yang mengandungi habuk bergerak dengan

lain. kelajuan tinggi ke dalam pembersih vakum
Air containing dust moves with a higher velocity into
Pendulum D oscillates at natural frequency as
vacuum cleaner [1]
pendulum X with the largest amplitude compared to – Bahan hos: Plastik
Material for hose: Plastic [1]
other pendulums. [1] – Mudah dibengkokkan/ fleksibel
Easy to bend/ flexible [1]
(b) – Diameter piring parabola: Besar – Kipas: Besar
Fan: Big [1]
Diameter parabolic dish: Large [1] – Lebih banyak habuk disedut masuk
More dust can be sucked [1]
– Lebih banyak isyarat diterima – Kuasa motor: Tinggi
Power of motor: High [1]
More signal receive [1] – Kipas berputar lebih laju
Fans rotate faster [1]
– Jenis gelombang: Gelombang mikro – Pilihan/ Choice: Pembersih vakum Q/ Vacuum cleaner Q
– Ini kerana ia mempunyai diameter rod pemegang yang
Type of wave: Microwave [1]
kecil, bahan hos dibuat daripada plastik, kipas yang besar
– Frekuensi tinggi dan kuasa motor yang tinggi.
This is because it has small wand diameter, material
High frequency [1] for hose is made of plastic, bigger fan and high motor
power. [1]
– Jarak penerima isyarat dari piring parabola:

Sama dengan panjang fokus

Distance of the signal receiver from the parabolic dish:
CONTOH
Same as focal length [1]

– Isyarat difokuskan kepada penerima.

Signals are focused on receiver. [1]

– Ketinggian piring parabola: Tinggi

Height of the parabolic dish: High [1]

– Isyarat tidak dihalang.

Signals are not blocked. [1]

– Pilihan/ Choice : R

– Sistem radar R dipilih kerana diameter piring parabola

yang besar, menggunakan gelombang mikro, jarak Bahagian C/ Section C

penerima isyarat dari piring parabola sama dengan 11 (a) Sudut tuju, i yang menghasilkan sudut biasan, r = 90°

panjang fokus dan ketinggian piring parabola tinggi. Angle of incidence i, when angle of refraction, r = 90° [1]

Radar system R is chosen because large diameter (b) (i) – Ketumpatan kaca adalah lebih daripada ketumpatan

parabolic dish, used microwave, distance of the signal perspek

receiver from the parabolic dish is same as focal length Density of glass is more than density of perspex [1]

and high height of parabolic dish. [1] – Sudut tuju kaca adalah lebih daripada sudut tuju

(c) (i) d = vt perspeks
2
1500 × 0.12 Angle of incidence of glass is more than perspex [1]
2
d = – Sudut biasan adalah sama

d = 90 m [3] Same refracted angle [1]

v 1500 (ii) Sudut tuju sama dengan sudut genting
f 25000
(ii) λ = = Incident angle is the same as critical angle [1]

(iii) Semakin kecil ketumpatan bongkah, semakin kecil

λ = 0.06 m [3] sudut genting.

The smaller the density of the block, the smaller the

10 (a) Manometer / Manometer [1] critical angle. [1]

(b) (i) d = 14 cm – 6 cm = 8 cm [1] (c) (i)

(ii) Pg = Patm + d
= 76 cm Hg + 8 cm

= 84 cm Hg [1]

(iii) Pg = hρg
= (0.84)(13.6 x 103)(10)

= 114240 Pa / 1.14 × 105 Pa [2]

(iv) Berkurang / Decreases

(c) – Apabila kita menyedut melalui penyedut minuman, udara (ii) – Banyak pantulan dalam penuh berlaku
More total internal reflection occur [1]
di dalam penyedut minuman dikeluarkan. – Sudut genting berlian kurang daripada sudut genting

When we suck through the straw, air inside the straw are kaca
Critical angle of diamond less than critical angle of
drained out. [1]
glass [1]
– Ini menghasilkan kawasan separa vakum di dalam (d) – Jenis kanta: Kanta cembung
Type of lens: Convex lens [1]
penyedut minuman. – Boleh menghasilkan imej nyata
Can produce real images [1]
This will produce a semi vacuum space in the straw. [1] – Permukaan pemantul: Hitam
Reflector surface: Black [1]
– Perbezaan tekanan berlaku.Tekanan atmosfera, Patm – Boleh menyerap haba
adalah lebih tinggi daripada tekanan di dalam penyedut Can absorb heat [1]
– Kuasa mentol: Tinggi
minuman. Bulb power: High [1]

Pressure difference will produce. The atmospheric

pressure,Patm greater than the pressure inside the straw.
[1]

– Daya dikenakan untuk menolak masuk air ke dalam

penyedut minuman.

Force is applied to push the water inside the straw. [1]

J5

Fizik Jawapan (b) (i) Objek
Object
– Menghasilkan kecerahan tinggi / boleh digunakan dalam
cahaya terang C FP F

Produces high brightness/ can be used in bright light [1] (ii) Objek
– Saiz projektor LCD: Kecil Object
LCD projector size: Small [1] Imej
– Mudah dikendalikan / dibawa C FP Image
Easy to handle / carry [1]
– Jenis bahan badan projektor: Berketumpatan rendah /

Plastik
Material of projector body: Low density / Plastic [1]
– Mudah dibawa /Ringan
Easy to carry / Lighter [1]

SET 3 (c) (i) Meningkat/ Increased
(ii) Lebih kecil/ Smaller
(d) Cermin solek/ Cosmetic mirror
CONTOHKertas 1

1 C 2 D 3 A 4 C 5 A 5 (a) (i) Indeks pembiasan blok Q lebih daripada blok P
8 C 9 A 10 D
6 D 7 C 13 D 14 A 15 C Refractive index of block Q is more than block P
18 A 19 C 20 C
11 C 12 D 23 B 24 B 25 A (ii) Kelajuan cahaya dalam blok P lebih daripada blok Q
28 B 29 A 30 C
16 B 17 A 33 C 34 C 35 C Speed of light in block P is more than block Q
38 B 39 B 40 D
2 1 D 22 C (iii) Sudut pembiasan blok P lebih daripada blok Q

Refractive angle of block P is more than block Q

26 B 27 C (iv) Apabila kelajuan meningkat, sudut pembiasan

3 1 A 32 D meningkat

36 D 37 D When speed increases, refracted angle increases

(v) Apabila indeks pembiasan meningkat, sudut pembiasan

Kertas 2 menurun

Bahagian A/ Section A When refractive index increases, refracted angle

5–0 decreases
10 – 0
1 (a) Pecutan/ Acceleration = (b) (i) n = c
v
5
= 10 n = 3.0 × 108
2.17 × 108
= 0.5 m/ s2
n = 1.38
(b) Kenderaan bergerak dengan halaju seragam selama 10 saat

yang pertama dan mula membuat gerakan tidak seragam (ii) Larutan gula (30%)

untuk 5 saat yang seterusnya. Sugar solution (30%)
c
The vehicle moves with uniform velocity in the first 10 (iii) n = v

seconds and starts to move with non-uniform velocity for 3.0 × 108
1.5
the next 5 seconds. v =

(c) Sesaran/ Displacement v = 2.0 × 108 m/s

2 (a) 6 (a) Rintangan adalah voltan per unit arus

R FR Resistance is voltage per unit current

(b) (i) Wayar PQ dalam Rajah 6.2 lebih tebal daripada

Rajah 6.1

Wx 30° Wy Wire PQ in Diagram 6.2 is thicker than Diagram 6.1

(ii) Arus dalam Rajah 6.2 lebih daripada Rajah 6.1

Current in Diagram 6.2 is more than Diagram 6.1

(b) Wy = mg kos 30°/ cos 30° (iii) Semakin tebal wayar, semakin banyak arus
= (5×9.81) × kos 30°/ cos 30°
= 42.48 N The thicker the wire, the more current flow
(c) (i) R11 1 1
= 4 + 12

3 (a) F = W + T R11 = 1
3
(b) F = T – W
R1 = 3Ω
= 50 – (2 × 9.81)
R12 = 1 1 1
= 30.38 N 8 + 10 + R1

(c) F = mv2 R12 = 1 + 1 + 1
r 8 10 3
(30.38 × 0.4)
v2 = 2 R12 67
120
v = 2.46 m s-1 =

4 (a) (i) Cermin cekung/ Concave mirror R2 = 1.79 Ω
(ii) Imej dibesarkan
Image is magnified R1 = 1 + 1
4 1.79

R = 1.237Ω

J6