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Practice paper, not an official KPM paper
4531
Peperiksaan Percubaan SPM (Model)SPM Trial Examination (Model)

FizikPhysics

Tingkatan 5, 2 jam / 2 hours, 100 marks
Nama / Name
Kelas / Class
Tarikh / Date
Markah / Marks:      / 100
Instructions
  1. Kertas soalan ini mengandungi empat bahagian: Bahagian A, Bahagian B, Bahagian C dan Bahagian D.This question paper consists of four sections: Section A, Section B, Section C and Section D.
  2. Bahagian A: pilih satu jawapan bagi setiap soalan. Bahagian B: jawab semua soalan. Bahagian C: jawab satu soalan sahaja. Bahagian D: jawab satu soalan sahaja.Section A: choose one answer for each question. Section B: answer all questions. Section C: answer one question only. Section D: answer one question only.
  3. Kerja mengira anda mesti ditunjukkan. Rajah tidak dilukis mengikut skala kecuali dinyatakan.Your working must be shown. Diagrams are not drawn to scale unless stated.
  4. Jawapan boleh ditulis dalam Bahasa Melayu atau Bahasa Inggeris.Answers may be written in Bahasa Melayu or English.
  5. Gunakan g = 9.81 m s⁻², h = 6.63 × 10⁻³⁴ J s, c = 3.00 × 10⁸ m s⁻¹, e = 1.60 × 10⁻¹⁹ C, jisim elektron = 9.11 × 10⁻³¹ kg. Kalkulator saintifik dibenarkan.Use g = 9.81 m s⁻², h = 6.63 × 10⁻³⁴ J s, c = 3.00 × 10⁸ m s⁻¹, e = 1.60 × 10⁻¹⁹ C, mass of electron = 9.11 × 10⁻³¹ kg. A scientific calculator is allowed.

Answer schemes are in English (DLP).

Practice shows answers as you go. Exam times you and hides answers until you submit.

Bahagian ASection A

20 marks
Jawab semua soalan. Setiap soalan diikuti oleh empat pilihan jawapan, A, B, C dan D. Pilih satu jawapan sahaja.Answer all questions. Each question is followed by four options, A, B, C and D. Choose one answer only.
1

Question 1

Daya 15 N ke timur dan daya 20 N ke utara bertindak serentak pada satu titik. Berapakah magnitud daya paduan?A force of 15 N towards the east and a force of 20 N towards the north act at the same point. What is the magnitude of the resultant force?
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B
The forces are at right angles, so R = √(15² + 20²) = √625 = 25 N. 35 N is the sum of the magnitudes, which only applies when the forces act in the same direction.
For the teacher: Remembering, low difficulty, Chapter 1, standard 1.1.1
Textbook p. 3
2

Question 2

Seorang budak menarik kereta sorong dengan tali yang membuat sudut 35° dengan mengufuk. Tegangan tali ialah 80 N. Berapakah komponen mengufuk daya itu?A boy pulls a cart with a rope that makes an angle of 35° with the horizontal. The tension in the rope is 80 N. What is the horizontal component of the force?
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C
F_x = F cos 35° = 80 × 0.819 = 65.5 N. 45.9 N is F sin 35°, the vertical component.
For the teacher: Applying, medium difficulty, Chapter 1, standard 1.2.1
Textbook p. 14
3

Question 3

Dua spring serupa, setiap satu dengan pemalar spring 300 N m⁻¹, disambung secara bersiri. Satu beban 6.0 N digantung pada hujung bawah spring bersiri itu. Berapakah jumlah pemanjangan?Two identical springs, each with spring constant 300 N m⁻¹, are joined in series. A load of 6.0 N is hung from the lower end of the springs. What is the total extension?
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C
For springs in series the effective spring constant is 300 ÷ 2 = 150 N m⁻¹. x = F ÷ k = 6.0 ÷ 150 = 0.040 m = 4.0 cm. 2.0 cm is the extension of one spring alone.
For the teacher: Applying, medium difficulty, Chapter 1, standard 1.4.4
Textbook p. 31
4

Question 4

Seorang penyelam berada pada kedalaman 12 m di dalam air laut berketumpatan 1 030 kg m⁻³. Berapakah tekanan air pada penyelam itu (tidak termasuk tekanan atmosfera)?A diver is 12 m deep in sea water of density 1 030 kg m⁻³. What is the water pressure on the diver (not including atmospheric pressure)?
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B
P = hρg = 12 × 1 030 × 9.81 = 1.21 × 10⁵ Pa. 1.18 × 10⁵ Pa uses the density of fresh water; 2.22 × 10⁵ Pa wrongly adds atmospheric pressure.
For the teacher: Applying, low difficulty, Chapter 2, standard 2.1
Textbook p. 44
5

Question 5

Dalam sebuah jek hidraulik, omboh kecil mempunyai luas 5 cm² dan omboh besar 200 cm². Daya 40 N dikenakan pada omboh kecil. Berapakah daya yang dihasilkan pada omboh besar?In a hydraulic jack, the small piston has an area of 5 cm² and the large piston has an area of 200 cm². A force of 40 N is applied to the small piston. What is the force produced on the large piston?
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D
Pressure is transmitted equally (Pascal's principle): F₂ = F₁ × A₂ ÷ A₁ = 40 × 200 ÷ 5 = 1 600 N.
For the teacher: Applying, low difficulty, Chapter 2, standard 2.4
Textbook p. 62
6

Question 6

Udara mengalir lebih laju di atas permukaan atas sayap kapal terbang berbanding di bawahnya. Pernyataan manakah yang menerangkan daya angkat?Air flows faster over the upper surface of an aeroplane wing than under it. Which statement explains the lift force?
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A
Bernoulli's principle: where the speed of the fluid is higher, the pressure is lower. Lower pressure above the wing and higher pressure below it give a net upward force (lift).
For the teacher: Understanding, low difficulty, Chapter 2, standard 2.6
Textbook p. 82
7

Question 7

Sebuah cerek elektrik mengalirkan arus 0.80 A selama 2.5 minit. Berapakah cas yang mengalir melalui cerek itu?An electric kettle carries a current of 0.80 A for 2.5 minutes. What is the charge that flows through the kettle?
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A
Q = It = 0.80 × (2.5 × 60) = 0.80 × 150 = 120 C. 2.0 C forgets to convert minutes to seconds; 48 C converts 1 minute only.
For the teacher: Applying, medium difficulty, Chapter 3, standard 3.1.3
Textbook p. 97
8

Question 8

Rajah 1 menunjukkan satu litar. Berapakah bacaan ammeter A? Abaikan rintangan dalam sel.Diagram 1 shows a circuit. What is the reading of the ammeter A? Ignore the internal resistance of the cell.
A 12 V cell, a 6 ohm resistor in series with a parallel pair of 6 ohm and 3 ohm resistors, and an ammeter A
Rajah 1Diagram 1
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B
The 6 Ω and 3 Ω resistors in parallel have 1/R = 1/6 + 1/3 = 1/2, so R = 2 Ω. Total resistance = 6 + 2 = 8 Ω. I = V ÷ R = 12 ÷ 8 = 1.5 A. 0.80 A treats all three resistors as being in series (12 ÷ 15).
For the teacher: Analysing, high difficulty, Chapter 3, standard 3.2.2
Textbook p. 104
9

Question 9

Sebuah sel mempunyai d.g.e. 12 V dan rintangan dalam 0.80 Ω. Sel itu disambung kepada perintang 7.2 Ω. Berapakah beza keupayaan merentasi terminal sel itu?A cell has an e.m.f. of 12 V and an internal resistance of 0.80 Ω. It is connected to a 7.2 Ω resistor. What is the potential difference across the terminals of the cell?
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C
I = ε ÷ (R + r) = 12 ÷ (7.2 + 0.80) = 1.5 A. V = IR = 1.5 × 7.2 = 10.8 V (or ε − Ir = 12 − 1.2). 1.2 V is the 'lost volts' Ir.
For the teacher: Applying, medium difficulty, Chapter 3, standard 3.3.2
Textbook p. 119
10

Question 10

Antara perubahan berikut, yang manakah TIDAK meningkatkan kesan putaran pada gelung dalam motor arus terus?Which of the following changes does NOT increase the turning effect on the coil in a direct current motor?
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D
Reversing the battery reverses the direction of the force and of the rotation but not its size. More turns, a stronger magnetic field and a larger current all increase the force and the turning effect.
For the teacher: Understanding, low difficulty, Chapter 4, standard 4.1.5
Textbook p. 144
11

Question 11

Kutub utara sebuah magnet bar ditolak menghampiri satu hujung solenoid yang disambung kepada galvanometer. Apakah kutub yang terhasil pada hujung solenoid itu dan mengapa?The north pole of a bar magnet is pushed towards one end of a solenoid connected to a galvanometer. What pole is produced at that end of the solenoid, and why?
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A
By Lenz's law the induced current flows in the direction that opposes the change producing it. The end facing the approaching north pole becomes a north pole, which repels the magnet. Options C and D break the conservation of energy.
For the teacher: Analysing, high difficulty, Chapter 4, standard 4.2.3
Textbook p. 159
12

Question 12

Sebuah transformer unggul menurunkan voltan daripada 240 V kepada 12 V. Arus dalam gegelung sekunder ialah 2.0 A. Berapakah arus dalam gegelung primer?An ideal transformer steps the voltage down from 240 V to 12 V. The current in the secondary coil is 2.0 A. What is the current in the primary coil?
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D
For an ideal transformer VpIp = VsIs, so Ip = (12 × 2.0) ÷ 240 = 0.10 A. 40 A is found by inverting the ratio.
For the teacher: Applying, medium difficulty, Chapter 4, standard 4.3.2
Textbook p. 164
13

Question 13

Sebuah elektron dipecut daripada keadaan rehat melalui beza keupayaan 500 V dalam satu tiub vakum. Berapakah halaju maksimum elektron itu? [e = 1.60 × 10⁻¹⁹ C, jisim elektron = 9.11 × 10⁻³¹ kg]An electron is accelerated from rest through a potential difference of 500 V in a vacuum tube. What is the maximum velocity of the electron? [e = 1.60 × 10⁻¹⁹ C, mass of electron = 9.11 × 10⁻³¹ kg]
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C
eV = ½mv², so v = √(2eV ÷ m) = √(2 × 1.60 × 10⁻¹⁹ × 500 ÷ 9.11 × 10⁻³¹) = 1.3 × 10⁷ m s⁻¹. 9.4 × 10⁶ forgets the factor 2 and 1.8 × 10¹⁴ forgets the square root.
For the teacher: Applying, high difficulty, Chapter 5, standard 5.1.3
Textbook p. 179
14

Question 14

Dalam litar rektifier gelombang penuh, satu kapasitor disambung selari dengan perintang beban. Apakah fungsi kapasitor itu?In a full-wave rectifier circuit, a capacitor is connected in parallel with the load resistor. What is the function of the capacitor?
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A
The capacitor charges when the rectified voltage rises and discharges through the load when it falls, which smooths the output. Changing a.c. to d.c. is done by the diodes.
For the teacher: Understanding, low difficulty, Chapter 5, standard 5.2.2
Textbook p. 184
15

Question 15

Dalam sebuah transistor npn, arus tapak ialah 20 µA dan faktor penggandaan amplifier ialah 150. Berapakah arus pengumpul?In an npn transistor, the base current is 20 µA and the current amplification factor is 150. What is the collector current?
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B
I_C = 150 × I_B = 150 × 20 µA = 3 000 µA = 3.0 mA. 0.13 µA divides instead of multiplies.
For the teacher: Applying, medium difficulty, Chapter 5, standard 5.3.1
Textbook p. 188
16

Question 16

Keaktifan suatu sampel radioaktif berkurang daripada 640 Bq kepada 40 Bq dalam masa 18 hari. Berapakah separuh hayat sampel itu?The activity of a radioactive sample decreases from 640 Bq to 40 Bq in 18 days. What is the half-life of the sample?
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C
640 → 320 → 160 → 80 → 40 is 4 half-lives. Half-life = 18 ÷ 4 = 4.5 days.
For the teacher: Applying, low difficulty, Chapter 6, standard 6.1.3
Textbook p. 205
17

Question 17

Bismut-214 mereput kepada polonium-214 melalui reputan: ²¹⁴₈₃Bi → ²¹⁴₈₄Po + X. Apakah X?Bismuth-214 decays to polonium-214 by the decay: ²¹⁴₈₃Bi → ²¹⁴₈₄Po + X. What is X?
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D
The nucleon number stays at 214 and the proton number rises by 1, so a neutron has changed into a proton and a beta particle (an electron, ⁰₋₁e) is emitted. An alpha particle would reduce the nucleon number by 4.
For the teacher: Understanding, low difficulty, Chapter 6, standard 6.1.1
Textbook p. 201
18

Question 18

Dalam suatu tindak balas nuklear, kehilangan jisim ialah 0.0030 u. Berapakah tenaga yang dibebaskan? [1 u = 1.66 × 10⁻²⁷ kg, c = 3.00 × 10⁸ m s⁻¹]In a nuclear reaction the mass lost is 0.0030 u. How much energy is released? [1 u = 1.66 × 10⁻²⁷ kg, c = 3.00 × 10⁸ m s⁻¹]
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B
E = mc² = (0.0030 × 1.66 × 10⁻²⁷) × (3.00 × 10⁸)² = 4.5 × 10⁻¹³ J. 1.5 × 10⁻²¹ J uses mc instead of mc²; 2.7 × 10¹⁴ J forgets to convert u to kg.
For the teacher: Applying, medium difficulty, Chapter 6, standard 6.2.3
Textbook p. 212
19

Question 19

Berapakah tenaga satu foton cahaya merah yang berpanjang gelombang 620 nm? [h = 6.63 × 10⁻³⁴ J s, c = 3.00 × 10⁸ m s⁻¹]What is the energy of one photon of red light of wavelength 620 nm? [h = 6.63 × 10⁻³⁴ J s, c = 3.00 × 10⁸ m s⁻¹]
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D
E = hf = hc ÷ λ = (6.63 × 10⁻³⁴ × 3.00 × 10⁸) ÷ (620 × 10⁻⁹) = 3.2 × 10⁻¹⁹ J. 3.2 × 10⁻²⁸ J forgets to convert nm to m; 1.1 × 10⁻²⁷ is h ÷ λ.
For the teacher: Applying, medium difficulty, Chapter 7, standard 7.1.3
Textbook p. 229
20

Question 20

Cahaya biru berfrekuensi di bawah frekuensi ambang sesuatu logam disinarkan pada permukaan logam itu. Apakah yang berlaku apabila keamatan cahaya itu ditingkatkan?Light with a frequency below the threshold frequency of a metal shines on its surface. What happens when the intensity of the light is increased?
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A
Each photon has energy hf, which is less than the work function when f < f₀. No electron can absorb enough energy to escape, so more intensity (more photons) does not help.
For the teacher: Understanding, low difficulty, Chapter 7, standard 7.2.2
Textbook p. 239

Bahagian BSection B

40 marks
Jawab semua soalan. Tulis jawapan anda di ruang yang disediakan.Answer all questions. Write your answers in the spaces provided.
21

Question 21

Rajah 2 menunjukkan sebuah troli berjisim 3.0 kg ditahan pegun pada landasan licin yang condong 20° dengan ufuk oleh satu spring. Spring itu mempunyai pemalar spring 250 N m⁻¹ dan terletak selari dengan landasan.Diagram 2 shows a trolley of mass 3.0 kg held at rest on a smooth ramp inclined at 20° to the horizontal by a spring. The spring has a spring constant of 250 N m⁻¹ and lies parallel to the ramp.
A 3.0 kg trolley on a smooth ramp inclined at 20 degrees, held by a spring fixed to a post at the top of the ramp
Rajah 2Diagram 2
[2 marks] (a)Namakan dua daya selain daya spring yang bertindak pada troli itu.Name two forces, other than the spring force, that act on the trolley.
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[2 marks] (b)Hitung komponen berat troli yang selari dengan landasan.Calculate the component of the weight of the trolley parallel to the ramp.
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[1 mark] (c)Nyatakan daya yang dikenakan oleh spring pada troli.State the force exerted by the spring on the trolley.
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[2 marks] (d)Hitung pemanjangan spring itu.Calculate the extension of the spring.
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[2 marks] (e)Hitung tenaga keupayaan kenyal yang tersimpan dalam spring itu.Calculate the elastic potential energy stored in the spring.
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[1 mark] (f)Spring itu dipotong. Hitung pecutan troli menuruni landasan.The spring is cut. Calculate the acceleration of the trolley down the ramp.
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For the teacher: Applying, low difficulty, Chapter 1, standard 1.3.1
Textbook p. 18
22

Question 22

Seorang penjual di pasar malam menggunakan unsur pemanas daripada dawai nikrom sepanjang 3.0 m dan berdiameter 0.40 mm untuk memanaskan periuk kecil. Kerintangan nikrom ialah 1.1 × 10⁻⁶ Ω m.A night market trader uses a heating element made from a nichrome wire 3.0 m long and 0.40 mm in diameter to warm a small pot. The resistivity of nichrome is 1.1 × 10⁻⁶ Ω m.
[2 marks] (a)Nyatakan dua faktor, selain bahan dawai, yang mempengaruhi rintangan dawai pada suhu tetap.State two factors, other than the material of the wire, that affect its resistance at a constant temperature.
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[2 marks] (b)Hitung luas keratan rentas dawai itu.Calculate the cross-sectional area of the wire.
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[2 marks] (c)Hitung rintangan dawai itu.Calculate the resistance of the wire.
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[2 marks] (d)Dawai itu disambung kepada bekalan 230 V. Hitung kuasa yang dibebaskan.The wire is connected to a 230 V supply. Calculate the power dissipated.
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[2 marks] (e)Hitung kos tenaga elektrik yang digunakan jika pemanas itu dipasang selama 30 minit. Kadar tarif ialah RM0.40 per kWh.Calculate the cost of the electrical energy used if the heater is switched on for 30 minutes. The tariff is RM0.40 per kWh.
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For the teacher: Applying, low difficulty, Chapter 3, standard 3.2.4
Textbook p. 109
23

Question 23

Rajah 3 menunjukkan litar lampu jalan automatik di sebuah taman perumahan. Perintang peka cahaya (LDR) mempunyai rintangan 40 kΩ pada waktu malam dan 0.50 kΩ pada waktu siang. Transistor itu hidup apabila beza keupayaan antara tapak dengan pemancar mencapai 0.60 V.Diagram 3 shows the circuit of an automatic street lamp in a housing estate. The light-dependent resistor (LDR) has a resistance of 40 kΩ at night and 0.50 kΩ in daylight. The transistor switches on when the potential difference between the base and the emitter reaches 0.60 V.
A 6.0 V supply with a 10 kilo-ohm resistor above an LDR as a potential divider; the junction connects to terminal X of an npn transistor, Y connects to a lamp, Z to the negative rail
Rajah 3Diagram 3
[2 marks] (a)Nyatakan jenis transistor itu dan namakan terminal Y.State the type of transistor and name terminal Y.
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[2 marks] (b)Hitung beza keupayaan merentasi LDR pada waktu malam.Calculate the potential difference across the LDR at night.
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[2 marks] (c)Hitung beza keupayaan merentasi LDR pada waktu siang.Calculate the potential difference across the LDR in daylight.
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[2 marks] (d)Terangkan mengapa lampu menyala pada waktu malam tetapi tidak menyala pada waktu siang.Explain why the lamp lights at night but not in daylight.
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[2 marks] (e)Lampu memerlukan arus 120 mA. Faktor penggandaan amplifier transistor itu ialah 200. Hitung arus tapak yang diperlukan.The lamp needs a current of 120 mA. The current amplification factor of the transistor is 200. Calculate the base current required.
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For the teacher: Analysing, medium difficulty, Chapter 5, standard 5.3.2
Textbook p. 191
24

Question 24

Sebuah klinik menggunakan isotop natrium-24 sebagai bahan penyurih untuk memeriksa aliran darah. Rajah 4 menunjukkan graf keaktifan melawan masa bagi sampel natrium-24 itu.A clinic uses the isotope sodium-24 as a tracer to check blood flow. Diagram 4 shows the graph of activity against time for a sample of sodium-24.
Graph of activity against time for sodium-24, falling from 960 Bq at 0 h through 480 Bq at 15 h to 30 Bq at 75 h
Rajah 4Diagram 4
[1 mark] (a)Apakah yang dimaksudkan dengan separuh hayat?What is meant by half-life?
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[2 marks] (b)Berdasarkan graf, tentukan separuh hayat natrium-24.Using the graph, determine the half-life of sodium-24.
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[2 marks] (c)Hitung keaktifan sampel itu selepas 90 jam.Calculate the activity of the sample after 90 hours.
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[2 marks] (d)Hitung masa yang diambil supaya keaktifan sampel itu jatuh kepada 7.5 Bq.Calculate the time taken for the activity of the sample to fall to 7.5 Bq.
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[2 marks] (e)Natrium-24 (²⁴₁₁Na) mereput dengan memancarkan zarah beta untuk menjadi magnesium (Mg). Tulis persamaan reputan itu.Sodium-24 (²⁴₁₁Na) decays by emitting a beta particle to become magnesium (Mg). Write the decay equation.
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[1 mark] (f)Seorang pelajar berkata bahawa kita boleh meramalkan dengan tepat bila satu nukleus natrium-24 tertentu akan mereput. Adakah pernyataan itu betul? Beri satu sebab.A student says we can predict exactly when one particular sodium-24 nucleus will decay. Is the statement correct? Give one reason.
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For the teacher: Applying, low difficulty, Chapter 6, standard 6.1.3
Textbook p. 205

Bahagian CSection C

20 marks
Jawab satu soalan sahaja. Setiap soalan bernilai 20 markah.Answer one question only. Each question carries 20 marks.
25Choose 1

Question 25

Rajah 5 menunjukkan sebuah tongkang berdasar rata yang membawa muatan di sebuah sungai. Tongkang kosong berjisim 4 000 kg dan luas dasarnya 20 m². Ketumpatan air sungai ialah 1 000 kg m⁻³ dan ketumpatan air laut ialah 1 025 kg m⁻³.Diagram 5 shows a flat-bottomed barge carrying cargo on a river. The empty barge has a mass of 4 000 kg and a base area of 20 m². The density of river water is 1 000 kg m⁻³ and the density of sea water is 1 025 kg m⁻³.
Side view of a flat-bottomed barge floating in water, with the depth h below the water level marked
Rajah 5Diagram 5
[2 marks] (a)Nyatakan prinsip Archimedes.State Archimedes' principle.
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[2 marks] (b)Hitung daya apungan pada tongkang kosong yang terapung.Calculate the buoyant force on the empty barge when it floats.
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[2 marks] (c)Hitung isi padu air yang disesarkan dan kedalaman h tongkang kosong di bawah paras air.Calculate the volume of water displaced and the depth h of the empty barge below the water level.
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[2 marks] (d)Tongkang itu kemudiannya dimuatkan dengan 6 000 kg kargo. Hitung kedalaman h yang baharu.The barge is then loaded with 6 000 kg of cargo. Calculate the new depth h.
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[2 marks] (e)Hitung tekanan yang dikenakan oleh air pada dasar tongkang yang bermuatan.Calculate the pressure exerted by the water on the base of the loaded barge.
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[2 marks] (f)Tongkang bermuatan itu belayar dari sungai ke laut. Adakah kedalaman h bertambah atau berkurang? Terangkan.The loaded barge sails from the river into the sea. Does the depth h increase or decrease? Explain.
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[2 marks] (g)Tongkang itu boleh tenggelam sehingga kedalaman maksimum 0.90 m di sungai dengan selamat. Hitung jisim maksimum kargo yang boleh dibawa.The barge can safely sink to a maximum depth of 0.90 m in the river. Calculate the maximum mass of cargo it can carry.
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[6 marks] (h)Sebuah syarikat ingin mereka bentuk tongkang yang membawa kargo lebih berat dengan selamat di sungai yang cetek. Cadangkan tiga ciri reka bentuk dan terangkan setiap satu.A company wants to design a barge that can carry heavier cargo safely on a shallow river. Suggest three design features and explain each one.
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Any three features, 2 marks each: 1 mark for the feature and 1 mark for the physics explanation.
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For the teacher: Applying, low difficulty, Chapter 2, standard 2.5
Textbook p. 71
26Choose 1

Question 26

Sebuah pengecas peranti di sebuah rumah menggunakan transformer injak turun. Bekalan 240 V a.u. disambung kepada gegelung primer dan gegelung sekunder mempunyai 120 lilitan. Voltan keluaran ialah 12 V.A device charger in a house uses a step-down transformer. A 240 V a.c. supply is connected to the primary coil and the secondary coil has 120 turns. The output voltage is 12 V.
[1 mark] (a)Nyatakan hukum Faraday.State Faraday's law.
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[3 marks] (b)Terangkan bagaimana transformer menghasilkan voltan keluaran apabila arus ulang-alik mengalir dalam gegelung primer.Explain how a transformer produces an output voltage when an alternating current flows in the primary coil.
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[2 marks] (c)Hitung bilangan lilitan gegelung primer.Calculate the number of turns of the primary coil.
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[3 marks] (d)Pengecas itu membekalkan kuasa 36 W kepada sebuah peranti. Kecekapan transformer ialah 90%. Hitung arus dalam gegelung primer.The charger supplies 36 W to a device. The efficiency of the transformer is 90%. Calculate the current in the primary coil.
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[4 marks] (e)Nyatakan dua cara mengurangkan kehilangan tenaga dalam transformer dan terangkan sebab bagi setiap cara.State two ways to reduce energy loss in a transformer and give the reason for each.
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Any two ways, 2 marks each (way + reason).
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[3 marks] (f)Sebuah stesen janakuasa menghantar kuasa 20 MW pada voltan 400 kV melalui kabel yang jumlah rintangannya 8.0 Ω. Hitung arus dalam kabel dan kuasa yang hilang sebagai haba.A power station transmits 20 MW of power at 400 kV through cables of total resistance 8.0 Ω. Calculate the current in the cables and the power lost as heat.
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[2 marks] (g)Ulang pengiraan kuasa yang hilang jika kuasa yang sama dihantar pada 20 kV dan nyatakan peratusan kuasa yang hilang.Repeat the calculation of power loss if the same power is transmitted at 20 kV, and state the percentage of power lost.
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[2 marks] (h)Berdasarkan jawapan (f) dan (g), terangkan mengapa tenaga elektrik dihantar pada voltan yang sangat tinggi dan bagaimana transformer membolehkan perkara ini.Based on your answers to (f) and (g), explain why electricity is transmitted at a very high voltage and how transformers make this possible.
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For the teacher: Evaluating, medium difficulty, Chapter 4, standard 4.3.2
Textbook p. 164

Bahagian DSection D

20 marks
Jawab satu soalan sahaja. Setiap soalan bernilai 20 markah.Answer one question only. Each question carries 20 marks.
27Choose 1

Question 27

Seorang pelajar menyinarkan cahaya monokromatik berlainan frekuensi pada sekeping logam dalam sebuah fotosel dan mengukur keupayaan penahan bagi setiap frekuensi. Jadual 1 menunjukkan keputusannya dan Rajah 6 menunjukkan graf keupayaan penahan melawan frekuensi.
Frekuensi, f / 10¹⁴ Hz
Frequency, f / 10¹⁴ Hz		6.07.08.09.010.0
Keupayaan penahan, V / V
Stopping potential, V / V		0.190.601.021.431.84
A student shines monochromatic light of different frequencies on a metal plate in a photocell and measures the stopping potential for each frequency. Table 1 shows the results and Diagram 6 shows the graph of stopping potential against frequency.
Straight-line graph of stopping potential (0 to 2 V) against frequency (5 to 10 x 10^14 Hz) with five plotted points, cutting the frequency axis at about 5.5
Rajah 6Diagram 6
[2 marks] (a)Apakah yang dimaksudkan dengan kesan fotoelektrik?What is meant by the photoelectric effect?
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[3 marks] (b)Dengan menggunakan konsep foton, terangkan mengapa tiada elektron dipancarkan apabila cahaya berfrekuensi rendah yang sangat terang disinarkan pada logam itu.Using the idea of photons, explain why no electrons are emitted when very bright light of low frequency shines on the metal.
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[1 mark] (c)Daripada graf, tentukan frekuensi ambang logam itu.From the graph, determine the threshold frequency of the metal.
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[2 marks] (d)Hitung fungsi kerja logam itu dalam joule dan dalam elektronvolt.Calculate the work function of the metal in joules and in electronvolts.
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[3 marks] (e)Tentukan kecerunan graf dan seterusnya hitung nilai pemalar Planck.Determine the gradient of the graph and hence calculate the value of the Planck constant.
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[3 marks] (f)Hitung tenaga kinetik maksimum dan laju maksimum fotoelektron apabila frekuensi cahaya ialah 8.0 × 10¹⁴ Hz.Calculate the maximum kinetic energy and the maximum speed of the photoelectrons when the frequency of the light is 8.0 × 10¹⁴ Hz.
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[2 marks] (g)Satu logam lain mempunyai fungsi kerja yang lebih besar. Huraikan bagaimana graf bagi logam itu berbeza daripada Rajah 6.Another metal has a larger work function. Describe how its graph differs from Diagram 6.
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[4 marks] (h)Sebuah penggera keselamatan menggunakan fotosel dengan logam ini dan pancaran LED merah berpanjang gelombang 650 nm. Fotosel itu tidak berfungsi. Terangkan mengapa dan cadangkan satu perubahan dengan justifikasi.A security alarm uses a photocell with this metal and a beam from a red LED of wavelength 650 nm. The photocell does not work. Explain why and suggest one change with a justification.
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For the teacher: Evaluating, high difficulty, Chapter 7, standard 7.3.4
Textbook p. 241
28Choose 1

Question 28

Rajah 7 menunjukkan lampu isyarat berjisim 12 kg di sebuah persimpangan yang digantung pada dua kabel serupa. Setiap kabel membuat sudut 35° dengan ufuk.Diagram 7 shows a traffic light of mass 12 kg at a junction hung from two identical cables. Each cable makes an angle of 35° with the horizontal.
A 12 kg traffic light hanging from two identical cables attached to two posts, each cable at 35 degrees to the horizontal
Rajah 7Diagram 7
[1 mark] (a)Nyatakan syarat bagi daya-daya dalam keseimbangan.State the condition for forces to be in equilibrium.
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[1 mark] (b)Hitung berat lampu isyarat itu.Calculate the weight of the traffic light.
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[3 marks] (c)Hitung tegangan T dalam setiap kabel.Calculate the tension T in each cable.
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[1 mark] (d)Hitung daya mengufuk yang dikenakan oleh setiap kabel pada tiangnya.Calculate the horizontal force that each cable exerts on its post.
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[2 marks] (e)Kabel dikencangkan sehingga setiap kabel membuat sudut 20° dengan ufuk. Hitung tegangan yang baharu dan terangkan perubahan itu.The cables are tightened until each cable makes an angle of 20° with the horizontal. Calculate the new tension and explain the change.
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[3 marks] (f)Seorang jururawat berjisim 70 kg berdiri di atas penimbang spring di dalam sebuah lif hospital. Lif itu memecut ke atas pada 1.2 m s⁻². Hitung bacaan penimbang (daya tindak balas normal).A nurse of mass 70 kg stands on a spring weighing scale in a hospital lift. The lift accelerates upwards at 1.2 m s⁻². Calculate the reading of the scale (the normal reaction).
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[2 marks] (g)Lif itu kemudian bergerak ke bawah dengan pecutan 1.2 m s⁻². Hitung bacaan penimbang.The lift then moves downwards with an acceleration of 1.2 m s⁻². Calculate the reading of the scale.
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[3 marks] (h)Spring penimbang itu mempunyai pemalar spring 2.0 × 10⁴ N m⁻¹. Hitung mampatan spring dan tenaga keupayaan kenyal yang tersimpan dalam situasi (f).The spring of the scale has a spring constant of 2.0 × 10⁴ N m⁻¹. Calculate the compression of the spring and the elastic potential energy stored in situation (f).
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[2 marks] (i)Kabel lif itu putus dan lif jatuh bebas. Apakah bacaan penimbang? Terangkan.The cable of the lift snaps and the lift falls freely. What is the reading of the scale? Explain.
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[2 marks] (j)Terangkan mengapa tegangan kabel lif mesti lebih besar daripada jumlah berat lif dan penumpang apabila lif mula bergerak ke atas.Explain why the tension in the lift cable must be greater than the total weight of the lift and its passengers when the lift starts to move upwards.
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For the teacher: Analysing, medium difficulty, Chapter 1, standard 1.3.1
Textbook p. 18
Test specification (JSU)
ChapterRememberingUnderstandingApplyingAnalysingEvaluatingCreatingItemsmarks
1. Force and Motion II1 (1)–3 (12)1 (20)––533
2. Pressure–1 (1)3 (22)–––423
3. Electricity––3 (12)1 (1)––413
4. Electromagnetism–1 (1)1 (1)1 (1)1 (20)–423
5. Electronics–1 (1)2 (2)1 (10)––413
6. Nuclear Physics–1 (1)3 (12)–––413
7. Quantum Physics–1 (1)1 (1)–1 (20)–322
Total1 (1)5 (5)16 (62)4 (32)2 (40)0 (0)28140
Each cell shows items (marks).