Practice paper, not an official KPM paper
4531
Ujian Akhir Sesi AkademikEnd of Academic Session Test
FizikPhysics
Tingkatan 5, 2 jam / 2 hours, 100 marks
Nama / Name
Kelas / Class
Tarikh / Date
Markah / Marks: / 100
Instructions
- 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.
- Bahagian A: jawab semua soalan; pilih satu jawapan bagi setiap soalan. Bahagian B: jawab semua soalan.Section A: answer all questions; choose one answer for each question. Section B: answer all questions.
- Bahagian C: jawab satu soalan sahaja. Bahagian D: jawab satu soalan sahaja.Section C: answer one question only. Section D: answer one question only.
- Tulis jawapan anda di ruang yang disediakan dan tunjukkan jalan kerja. Rajah tidak dilukis mengikut skala kecuali dinyatakan.Write your answers in the spaces provided and show your working. Diagrams are not drawn to scale unless stated.
- Anda dibenarkan menggunakan kalkulator saintifik. Gunakan g = 9.81 m s⁻², h = 6.63 × 10⁻³⁴ J s, c = 3.00 × 10⁸ m s⁻¹ dan e = 1.60 × 10⁻¹⁹ C jika tidak diberikan.You may use a scientific calculator. Use g = 9.81 m s⁻², h = 6.63 × 10⁻³⁴ J s, c = 3.00 × 10⁸ m s⁻¹ and e = 1.60 × 10⁻¹⁹ C unless given.
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 marksJawab 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.
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Question 1
Rajah 1 menunjukkan sebuah kotak berjisim 8.0 kg ditarik pada lantai mengufuk oleh seutas tali dengan daya 40 N pada sudut 30° di atas ufuk. Daya geseran ialah 12 N. Berapakah pecutan kotak itu?Diagram 1 shows a box of mass 8.0 kg pulled along a horizontal floor by a rope with a force of 40 N at 30° above the horizontal. The frictional force is 12 N. What is the acceleration of the box?
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B
Horizontal component of the pull = 40 cos 30° = 34.6 N. Net force = 34.6 − 12 = 22.6 N. a = F/m = 22.6 ÷ 8.0 = 2.8 m s⁻². 3.5 forgets the 30° angle, 1.0 uses the vertical component and 4.3 forgets friction.
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Question 2
Sebuah objek berada dalam keseimbangan di bawah tindakan tiga daya. Dua daripada daya itu ialah 6 N ke timur dan 8 N ke utara. Apakah magnitud dan arah daya ketiga?An object is in equilibrium under three forces. Two of the forces are 6 N towards the east and 8 N towards the north. What are the magnitude and direction of the third force?
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C
In equilibrium the resultant of all forces is zero, so the third force is equal and opposite to the resultant of the other two. The resultant of 6 N and 8 N at right angles is √(6² + 8²) = 10 N towards the north-east, so the third force is 10 N towards the south-west.
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Question 3
Sebuah spring dikenakan daya 8 N dan panjang pemanjangannya ialah 5.0 cm. Daya ditambah kepada 12 N dan pemanjangannya menjadi 7.5 cm. Had kenyal spring tidak dilampaui. Berapakah kerja tambahan yang dilakukan semasa daya ditambah daripada 8 N kepada 12 N?A spring is pulled by a force of 8 N and its extension is 5.0 cm. The force is increased to 12 N and the extension becomes 7.5 cm. The elastic limit is not exceeded. How much extra work is done while the force increases from 8 N to 12 N?
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A
The work done is the area under the force–extension graph between 5.0 cm and 7.5 cm, which is a trapezium: ½ × (8 + 12) × 0.025 = 0.25 J. 0.10 J uses only the 4 N rise, 0.30 J uses 12 N × 0.025 m and 0.50 J uses 20 N × 0.025 m.
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Question 4
Sebuah cangkuk getah penyedut melekat pada dinding jubin yang licin. Apakah yang menyebabkan cangkuk itu kekal melekat?A rubber suction hook sticks to a smooth tiled wall. What keeps the hook in place?
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D
When the cup is pressed, most of the air inside is pushed out. The atmospheric pressure outside is then greater than the pressure inside, and the difference holds the hook against the wall (textbook, applications of atmospheric pressure).
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Question 5
Sebuah jek hidraulik mengangkat kereta berjisim 1 200 kg. Luas keratan rentas omboh output ialah 400 cm² dan luas omboh input ialah 10 cm². Berapakah daya input minimum yang diperlukan? [g = 9.81 m s⁻²]A hydraulic jack lifts a car of mass 1 200 kg. The cross-sectional area of the output piston is 400 cm² and that of the input piston is 10 cm². What is the minimum input force needed? [g = 9.81 m s⁻²]
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B
Output force = weight = 1 200 × 9.81 = 11 772 N. By Pascal's principle F₁/A₁ = F₂/A₂, so F₁ = 11 772 × 10 ÷ 400 = 294 N. 11 772 N ignores the force multiplier.
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Question 6
Sebuah bongkah kayu terapung di dalam air dengan 60% daripada isi padunya tenggelam. Bongkah itu kemudian dialihkan ke dalam minyak berketumpatan 750 kg m⁻³ dan terapung. Berapakah peratus isi padu bongkah itu yang tenggelam dalam minyak? [Ketumpatan air = 1 000 kg m⁻³]A block of wood floats in water with 60% of its volume submerged. The block is then moved into oil of density 750 kg m⁻³ and floats. What percentage of the volume of the block is submerged in the oil? [Density of water = 1 000 kg m⁻³]
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C
Floating in water: ρblock = 0.60 × 1 000 = 600 kg m⁻³. Floating in oil, weight = buoyant force, so fraction submerged = ρblock ÷ ρoil = 600 ÷ 750 = 0.80 = 80%. The fraction is bigger because oil is less dense than water.
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Question 7
Apakah unit S.I. bagi kerintangan suatu bahan?What is the S.I. unit of the resistivity of a material?
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D
From R = ρl/A, ρ = RA/l, so the unit is Ω × m² ÷ m = Ω m.
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Question 8
Arus 0.25 A mengalir melalui sebuah mentol selama 2 minit. Berapakah bilangan elektron yang melalui mentol itu? [Cas satu elektron = 1.6 × 10⁻¹⁹ C]A current of 0.25 A flows through a bulb for 2 minutes. How many electrons pass through the bulb? [Charge of one electron = 1.6 × 10⁻¹⁹ C]
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A
Q = It = 0.25 × 120 = 30 C. Number of electrons = Q/e = 30 ÷ 1.6 × 10⁻¹⁹ = 1.9 × 10²⁰. 1.6 × 10¹⁸ is the number of electrons in only 1 second (0.25 C).
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Question 9
Sebuah sel kering mempunyai d.g.e. 1.5 V. Apabila sel itu disambung kepada perintang 2.0 Ω, voltmeter yang disambung pada terminal sel membaca 1.2 V. Berapakah rintangan dalam sel itu?A dry cell has an e.m.f. of 1.5 V. When it is connected to a 2.0 Ω resistor, a voltmeter connected across the terminals of the cell reads 1.2 V. What is the internal resistance of the cell?
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C
Current I = V/R = 1.2 ÷ 2.0 = 0.60 A. Lost volts = ε − V = 1.5 − 1.2 = 0.30 V = Ir. So r = 0.30 ÷ 0.60 = 0.50 Ω.
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Question 10
Seutas dawai berintangan 4.0 Ω ditarik secara seragam sehingga panjangnya menjadi tiga kali panjang asal. Isi padu dan kerintangan dawai tidak berubah. Berapakah rintangan baharu dawai itu?A wire of resistance 4.0 Ω is stretched uniformly until its length is three times the original length. Its volume and resistivity do not change. What is the new resistance of the wire?
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D
The length becomes 3l and, since the volume is constant, the area becomes A/3. R = ρ(3l)/(A/3) = 9 × ρl/A = 9 × 4.0 = 36 Ω. 12 Ω only counts the longer length.
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Question 11
Kutub utara sebuah magnet bar ditarik menjauhi hujung sebuah solenoid. Apakah kekutuban yang teraruh pada hujung solenoid yang paling dekat dengan magnet itu?The north pole of a bar magnet is pulled away from one end of a solenoid. What is the polarity induced at the end of the solenoid nearest the magnet?
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A
By Lenz's law the induced current opposes the change. The magnet is leaving, so the solenoid end becomes a south pole to attract the north pole and oppose its removal.
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Question 12
Sebuah transformer unggul menaikkan voltan. Nisbah bilangan lilitan gegelung primer kepada gegelung sekunder ialah 1 : 20. Arus dalam gegelung primer ialah 5.0 A. Berapakah arus dalam gegelung sekunder?An ideal transformer steps up the voltage. The ratio of the number of turns of the primary coil to the secondary coil is 1 : 20. The current in the primary coil is 5.0 A. What is the current in the secondary coil?
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B
For an ideal transformer VpIp = VsIs, so Is/Ip = Np/Ns = 1/20. Is = 5.0 ÷ 20 = 0.25 A. The voltage goes up by 20 times, so the current goes down by 20 times.
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Question 13
Antara perubahan berikut, yang manakah akan menyebabkan arah putaran motor arus terus berbalik?Which of the following changes will reverse the direction of rotation of a direct current motor?
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C
The direction of the force depends on the direction of the current and the direction of the magnetic field (Fleming's left-hand rule). Reversing one of them reverses the force. Reversing both together leaves the direction unchanged; more turns or stronger magnets only change the speed.
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Question 14
Sebiji elektron dipecutkan dari keadaan rehat melalui beza keupayaan 400 V dalam vakum. Berapakah tenaga kinetik elektron itu? [Cas elektron = 1.6 × 10⁻¹⁹ C]An electron is accelerated from rest through a potential difference of 400 V in a vacuum. What is the kinetic energy of the electron? [Charge of an electron = 1.6 × 10⁻¹⁹ C]
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B
The electric energy lost by the electron is gained as kinetic energy: KE = eV = 1.6 × 10⁻¹⁹ × 400 = 6.4 × 10⁻¹⁷ J (that is, 400 eV).
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Question 15
Dalam suatu litar amplifier transistor npn, arus tapak 40 μA menghasilkan arus pengumpul 2.0 mA. Arus tapak dinaikkan kepada 60 μA. Berapakah arus pengumpul sekarang?In an npn transistor amplifier circuit, a base current of 40 μA produces a collector current of 2.0 mA. The base current is increased to 60 μA. What is the collector current now?
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D
Current gain = IC/IB = 2.0 mA ÷ 40 μA = 50. New IC = 50 × 60 μA = 3 000 μA = 3.0 mA. 2.0 mA would mean the gain does not matter.
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Question 16
Seorang murid ingin mendapatkan voltan keluaran yang lebih licin daripada rektifier gelombang penuh yang mempunyai sebuah kapasitor selari dengan beban. Perubahan manakah yang berkesan?A student wants a smoother output voltage from a full-wave rectifier that has a capacitor in parallel with the load. Which change is effective?
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A
A larger capacitor stores more charge and discharges more slowly between peaks, so the ripple is smaller. The capacitor must stay in parallel with the load to smooth the output.
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Question 17
Satu nukleus ²¹⁸₈₄Po mereput dengan memancarkan satu zarah alfa dan kemudian satu zarah beta. Apakah nombor nukleon dan nombor proton nukleus akhir?A nucleus of ²¹⁸₈₄Po decays by emitting one alpha particle and then one beta particle. What are the nucleon number and proton number of the final nucleus?
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C
Alpha decay: nucleon number 218 → 214 and proton number 84 → 82. Beta decay: a neutron changes to a proton, so the proton number rises 82 → 83 and the nucleon number stays 214.
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Question 18
Jisim sampel radioaktif berkurang daripada 2.0 g kepada 0.25 g dalam masa 21 hari. Apakah separuh hayat bahan itu?The mass of a radioactive sample decreases from 2.0 g to 0.25 g in 21 days. What is the half-life of the substance?
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B
2.0 → 1.0 → 0.5 → 0.25 g is 3 half-lives. Half-life = 21 ÷ 3 = 7.0 days. Counting 4 or 2 halvings gives 5.3 or 10.5 days.
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Question 19
Foton bertenaga 4.5 eV menyinari permukaan logam yang fungsi kerjanya 2.8 eV. Berapakah tenaga kinetik maksimum fotoelektron itu?A photon of energy 4.5 eV shines on a metal surface with a work function of 2.8 eV. What is the maximum kinetic energy of the photoelectrons?
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A
Einstein's equation: Kmax = hf − W = 4.5 − 2.8 = 1.7 eV. 7.3 eV adds the two energies instead of subtracting.
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Question 20
Pada graf tenaga kinetik maksimum fotoelektron melawan frekuensi cahaya, apakah yang diwakili oleh kecerunan graf itu?On a graph of the maximum kinetic energy of photoelectrons against the frequency of light, what does the gradient of the graph represent?
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D
From Kmax = hf − W, comparing with y = mx + c shows the gradient is h. The intercept on the frequency axis is the threshold frequency and the intercept on the Kmax axis is −W.
Bahagian BSection B
40 marksJawab semua soalan dalam bahagian ini.Answer all questions in this section.
21
Question 21
Rajah 2 menunjukkan sebuah kotak berjisim 10 kg berehat di atas satah condong kasar yang condong 20° kepada ufuk. [g = 9.81 m s⁻²]Diagram 2 shows a box of mass 10 kg at rest on a rough ramp inclined at 20° to the horizontal. [g = 9.81 m s⁻²]
[3 marks]
(a)Namakan tiga daya yang bertindak ke atas kotak itu.Name the three forces that act on the box.
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[2 marks]
(b)Hitung komponen berat kotak itu yang selari dengan satah condong.Calculate the component of the weight of the box that is parallel to the ramp.
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33.6 N
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[1 mark]
(c)Berapakah magnitud daya geseran pada kotak itu? Beri sebab.What is the magnitude of the frictional force on the box? Give a reason.
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[4 marks]
(d)Kemudian, sudut satah dinaikkan kepada 35° dan kotak itu meluncur ke bawah. Daya geseran menjadi 40 N. Hitung pecutan kotak itu.The angle of the ramp is then raised to 35° and the box slides down. The frictional force is now 40 N. Calculate the acceleration of the box.
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1.6 m s⁻²
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22
Question 22
Rajah 3 menunjukkan litar yang mengandungi sebuah sel berd.g.e. 18.0 V dan rintangan dalam 1.0 Ω yang disambung kepada tiga perintang.Diagram 3 shows a circuit containing a cell of e.m.f. 18.0 V and internal resistance 1.0 Ω connected to three resistors.
[1 mark]
(a)Apakah maksud d.g.e. sebuah sel?What is meant by the e.m.f. of a cell?
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[2 marks]
(b)Hitung rintangan berkesan perintang 6.0 Ω dan 12.0 Ω yang bersambung selari.Calculate the effective resistance of the 6.0 Ω and 12.0 Ω resistors connected in parallel.
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4.0 Ω
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[2 marks]
(c)Hitung arus yang mengalir dalam sel itu.Calculate the current that flows through the cell.
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2.0 A
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[2 marks]
(d)Hitung beza keupayaan merentasi perintang 12.0 Ω.Calculate the potential difference across the 12.0 Ω resistor.
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8.0 V
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[1 mark]
(e)Hitung arus dalam perintang 6.0 Ω.Calculate the current in the 6.0 Ω resistor.
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1.3 A
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[2 marks]
(f)Voltmeter yang disambung merentasi terminal sel itu membaca kurang daripada 18.0 V apabila suis ditutup. Terangkan mengapa.A voltmeter connected across the terminals of the cell reads less than 18.0 V when the switch is closed. Explain why.
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23
Question 23
Rajah 4 menunjukkan litar yang menggunakan sebuah transistor npn sebagai amplifier arus. Bacaan meter ditunjukkan pada rajah.Diagram 4 shows a circuit that uses an npn transistor as a current amplifier. The meter readings are shown on the diagram.
[3 marks]
(a)Namakan terminal X, Y dan Z.Name the terminals X, Y and Z.
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[2 marks]
(b)Hitung faktor penggandaan amplifier transistor itu.Calculate the current amplification factor of the transistor.
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60
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[2 marks]
(c)Arus tapak ditingkatkan kepada 40 μA. Berapakah arus pengumpul yang baharu?The base current is increased to 40 μA. What is the new collector current?
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2.4 mA
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[3 marks]
(d)Transistor npn itu digantikan dengan transistor pnp. Huraikan perubahan yang perlu dibuat pada sambungan bateri supaya litar itu berfungsi.The npn transistor is replaced with a pnp transistor. Describe the change that must be made to the battery connections so that the circuit works.
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24
Question 24
Rajah 5 menunjukkan graf tenaga kinetik maksimum, Kmaks, fotoelektron melawan frekuensi, f, cahaya yang menyinari sebuah logam. Tiga titik ditandakan dengan silang. [Pemalar Planck, h = 6.63 × 10⁻³⁴ J s]Diagram 5 shows a graph of the maximum kinetic energy, Kmax, of photoelectrons against the frequency, f, of light shining on a metal. Three points are marked with crosses. [Planck constant, h = 6.63 × 10⁻³⁴ J s]
[1 mark]
(a)Apakah maksud frekuensi ambang?What is meant by threshold frequency?
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[1 mark]
(b)Berdasarkan graf, nyatakan frekuensi ambang logam itu.From the graph, state the threshold frequency of the metal.
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4.5 × 10¹⁴ Hz
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[3 marks]
(c)Hitung kecerunan graf itu menggunakan dua silang yang ditunjukkan. Nyatakan kuantiti fizik yang diwakili oleh kecerunan itu.Calculate the gradient of the graph using two of the crosses shown. State the physical quantity that the gradient represents.
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[2 marks]
(d)Hitung fungsi kerja logam itu.Calculate the work function of the metal.
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3.0 × 10⁻¹⁹ J
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[2 marks]
(e)Cahaya berfrekuensi 8.0 × 10¹⁴ Hz menyinari logam itu. Hitung tenaga kinetik maksimum fotoelektron.Light of frequency 8.0 × 10¹⁴ Hz shines on the metal. Calculate the maximum kinetic energy of the photoelectrons.
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2.3 × 10⁻¹⁹ J
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[1 mark]
(f)Logam kedua mempunyai fungsi kerja yang lebih besar. Huraikan kedudukan graf logam kedua itu berbanding dengan graf yang ditunjukkan.A second metal has a larger work function. Describe the position of the graph of the second metal compared with the graph shown.
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Bahagian CSection C
20 marksJawab satu soalan sahaja daripada bahagian ini.Answer one question only from this section.
25Choose 1
Question 25
Rajah 6 menunjukkan sebuah kapal selam penyelidikan pada kedalaman 120 m di bawah permukaan laut. Ketumpatan air laut ialah 1 025 kg m⁻³. [g = 9.81 m s⁻²]Diagram 6 shows a research submersible at a depth of 120 m below the sea surface. The density of sea water is 1 025 kg m⁻³. [g = 9.81 m s⁻²]
[2 marks]
(a)Nyatakan prinsip Archimedes.State Archimedes' principle.
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[4 marks]
(b)Sebuah kapal kargo berjisim 5.0 × 10⁶ kg (termasuk muatan) terapung di dalam air laut. Hitung isi padu air laut yang disesarkan oleh kapal itu.A cargo ship of mass 5.0 × 10⁶ kg (including its cargo) floats in sea water. Calculate the volume of sea water displaced by the ship.
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4.9 × 10³ m³
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[2 marks]
(c)Kapal itu belayar dari laut ke sebatang sungai yang airnya kurang tumpat. Terangkan mengapa kapal itu tenggelam lebih dalam.The ship sails from the sea into a river where the water is less dense. Explain why the ship sinks deeper.
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[2 marks]
(d)Hitung tekanan akibat air laut pada kapal selam penyelidikan itu pada kedalaman 120 m.Calculate the pressure due to the sea water on the research submersible at a depth of 120 m.
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1.2 × 10⁶ Pa
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[10 marks]
(e)Anda diminta mereka bentuk sebuah kapal selam penyelidikan yang boleh menyelam lebih dalam dan kembali ke permukaan dengan selamat. Dengan menggunakan konsep tekanan dan keapungan, cadangkan ciri reka bentuk berikut: bahan badan kapal, bentuk badan kapal, cara mengawal keapungan, ketebalan badan kapal dan saiz tingkap. Beri sebab bagi setiap cadangan.You are asked to design a research submersible that can dive deeper and return safely to the surface. Using the concepts of pressure and buoyancy, suggest the following design features: the material of the hull, the shape of the hull, the way to control buoyancy, the thickness of the hull and the size of the windows. Give a reason for each suggestion.
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26Choose 1
Question 26
Rajah 7 menunjukkan transformer di dalam pengecas telefon bimbit. Gegelung primernya mempunyai 4 800 lilitan dan disambung kepada bekalan a.u. 240 V. Pengecas itu mesti memberikan 5.0 V a.u. kepada telefon.Diagram 7 shows the transformer inside a mobile phone charger. Its primary coil has 4 800 turns and is connected to a 240 V a.c. supply. The charger must give 5.0 V a.c. to the phone.
[2 marks]
(a)Namakan jenis transformer yang digunakan dalam pengecas itu dan beri satu sebab.Name the type of transformer used in the charger and give one reason.
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[2 marks]
(b)Hitung bilangan lilitan gegelung sekunder itu.Calculate the number of turns of the secondary coil.
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100 turns
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[3 marks]
(c)Pengecas itu memberikan arus 2.0 A pada 5.0 V. Kecekapannya ialah 80%. Hitung arus dalam gegelung primer.The charger delivers a current of 2.0 A at 5.0 V. Its efficiency is 80%. Calculate the current in the primary coil.
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0.052 A
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[3 marks]
(d)Terangkan satu cara mengurangkan kehilangan tenaga akibat arus pusar dalam transformer dan satu cara mengurangkan kehilangan tenaga akibat pemanasan dawai.Explain one way to reduce the energy loss due to eddy currents in a transformer and one way to reduce the energy loss due to heating of the wires.
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[10 marks]
(e)Sebuah dinamo basikal (penjana a.u. kecil) hendak direka bentuk untuk menyalakan lampu basikal dengan lebih terang. Dengan menggunakan prinsip aruhan elektromagnet, cadangkan ciri reka bentuk berikut: kekuatan magnet, bilangan lilitan gegelung, laju putaran, teras gegelung dan bahagian yang membolehkan arus a.u. dikeluarkan. Beri sebab bagi setiap cadangan.A bicycle dynamo (a small a.c. generator) is to be designed to make the bicycle lamp shine more brightly. Using the principle of electromagnetic induction, suggest the following design features: the strength of the magnet, the number of turns of the coil, the speed of rotation, the core of the coil and the part that allows an alternating current to be delivered. Give a reason for each suggestion.
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Bahagian DSection D
20 marksJawab satu soalan sahaja daripada bahagian ini.Answer one question only from this section.
27Choose 1
Question 27
Di dalam Matahari, nukleus deuterium bergabung dengan nukleus tritium: ²₁H + ³₁H → ⁴₂He + ¹₀n + tenaga. Jisim atom: ²₁H = 2.01410 u, ³₁H = 3.01605 u, ⁴₂He = 4.00260 u, ¹₀n = 1.00867 u. [1 u = 1.66 × 10⁻²⁷ kg; c = 3.00 × 10⁸ m s⁻¹]In the Sun, a deuterium nucleus combines with a tritium nucleus: ²₁H + ³₁H → ⁴₂He + ¹₀n + energy. Masses: ²₁H = 2.01410 u, ³₁H = 3.01605 u, ⁴₂He = 4.00260 u, ¹₀n = 1.00867 u. [1 u = 1.66 × 10⁻²⁷ kg; c = 3.00 × 10⁸ m s⁻¹]
[1 mark]
(a)Namakan jenis tindak balas nuklear itu.Name the type of nuclear reaction.
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[2 marks]
(b)Hitung cacat jisim dalam tindak balas itu dalam kg.Calculate the mass defect in the reaction in kg.
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3.13 × 10⁻²⁹ kg
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[2 marks]
(c)Hitung tenaga yang dibebaskan dalam tindak balas itu.Calculate the energy released in the reaction.
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2.8 × 10⁻¹² J
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[2 marks]
(d)Terangkan mengapa tindak balas pelakuran sukar dilakukan di Bumi.Explain why a fusion reaction is difficult to achieve on Earth.
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[3 marks]
(e)Iodin-131 yang mempunyai separuh hayat 8.0 hari digunakan untuk merawat kelenjar tiroid. Jisim awal sampel ialah 640 mg. Hitung jisim iodin-131 yang tinggal selepas 24 hari dan masa yang diambil untuk jisimnya berkurang kepada 5 mg.Iodine-131, which has a half-life of 8.0 days, is used to treat the thyroid gland. The initial mass of a sample is 640 mg. Calculate the mass of iodine-131 that remains after 24 days and the time taken for its mass to decrease to 5 mg.
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80 mg; 56 days
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[10 marks]
(f)Sisa radioaktif daripada hospital perlu disimpan dengan selamat sehingga ia tidak lagi berbahaya. Cadangkan ciri reka bentuk bagi tempat penyimpanan sisa itu: bahan bekas, lokasi, bentuk sisa, cara menyingkirkan haba dan langkah keselamatan di tempat itu. Beri sebab bagi setiap cadangan.Radioactive waste from a hospital must be stored safely until it is no longer dangerous. Suggest design features of the storage facility: the material of the container, the location, the form of the waste, how heat is removed and the safety measures at the site. Give a reason for each suggestion.
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28Choose 1
Question 28
Rajah 8 menunjukkan dua spring, P dan Q, diperbuat daripada dawai yang sama dan berdiameter sama. Spring P mempunyai 20 lilitan dan spring Q mempunyai 40 lilitan. Kedua-duanya digantung daripada penyokong yang sama dan diberi beban 5.0 N. Had kenyal tidak dilampaui.Diagram 8 shows two springs, P and Q, made of the same wire with the same diameter. Spring P has 20 coils and spring Q has 40 coils. Both are hung from the same support and carry a load of 5.0 N. The elastic limit is not exceeded.
[1 mark]
(a)Nyatakan satu pemerhatian daripada Rajah 8.State one observation from Diagram 8.
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[2 marks]
(b)Nyatakan satu inferens dan satu hipotesis berdasarkan pemerhatian itu.State one inference and one hypothesis based on the observation.
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[2 marks]
(c)Hitung pemalar spring bagi spring P dan spring Q.Calculate the spring constant of spring P and of spring Q.
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[2 marks]
(d)Hitung tenaga keupayaan kenyal yang tersimpan dalam spring P.Calculate the elastic potential energy stored in spring P.
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0.10 J
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[3 marks]
(e)Spring P dan spring Q kemudian disambung hujung ke hujung (secara siri) dan digantung dengan beban 5.0 N. Hitung jumlah pemanjangan dan pemalar spring berkesan.Spring P and spring Q are then joined end to end (in series) and hung with a load of 5.0 N. Calculate the total extension and the effective spring constant.
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[10 marks]
(f)Seorang murid ingin menyiasat bagaimana bilangan lilitan mempengaruhi pemanjangan sebuah spring. Rancang satu eksperimen untuk menyiasat hipotesis itu. Huraian anda hendaklah mengandungi: tujuan, pemboleh ubah (dimanipulasikan, bergerak balas, dimalarkan), radas dan bahan, prosedur, penjadualan data dan analisis data.A student wants to investigate how the number of coils affects the extension of a spring. Plan an experiment to investigate the hypothesis. Your description should include: aim, variables (manipulated, responding, constant), apparatus and materials, procedure, tabulation of data and analysis of data.
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Test specification (JSU)
| Chapter | Remembering | Understanding | Applying | Analysing | Evaluating | Creating | Items | marks |
|---|---|---|---|---|---|---|---|---|
| 1. Force and Motion II | – | 1 (1) | 2 (11) | 1 (1) | 1 (20) | – | 5 | 33 |
| 2. Pressure | – | 1 (1) | 2 (21) | 1 (1) | – | – | 4 | 23 |
| 3. Electricity | 1 (1) | – | 3 (12) | 1 (1) | – | – | 5 | 14 |
| 4. Electromagnetism | – | 2 (2) | 2 (21) | – | – | – | 4 | 23 |
| 5. Electronics | – | 2 (11) | 2 (2) | – | – | – | 4 | 13 |
| 6. Nuclear Physics | – | – | 2 (21) | 1 (1) | – | – | 3 | 22 |
| 7. Quantum Physics | – | 1 (1) | 1 (1) | 1 (10) | – | – | 3 | 12 |
| Total | 1 (1) | 7 (16) | 14 (89) | 5 (14) | 1 (20) | 0 (0) | 28 | 140 |
Each cell shows items (marks).