FizikPhysics

Print questions onlyCetak soalan sahaja Print with answersCetak dengan skema Print answer key onlyCetak kunci jawapan sahaja Practice shows answers as you go. Exam times you and hides answers until you submit.Latihan menunjukkan jawapan semasa menjawab. Peperiksaan menetapkan masa dan menyembunyikan jawapan sehingga dihantar.

1

QuestionSoalan 1

Di sebuah lapangan terbang, seorang porter menolak troli bagasi dengan daya 90 N ke timur manakala angin kuat menolaknya dengan daya 120 N ke utara. Berapakah magnitud daya paduan pada troli itu?At an airport, a porter pushes a luggage trolley with a force of 90 N towards the east while a strong wind pushes it with a force of 120 N towards the north. What is the magnitude of the resultant force on the trolley?

• A30 N
• B210 N
• C150 N
• D22 500 N

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C

The forces are at right angles, so R = √(90² + 120²) = √22 500 = 150 N. 210 N is the sum of the magnitudes and 30 N their difference, which only apply to forces along the same line; 22 500 N forgets the square root.

For the teacherUntuk guru: ApplyingMengaplikasi, lowrendah difficultyaras lowrendah, Chapter 1, standard 1.1.1

TextbookBuku teks p.ms. 3

2

QuestionSoalan 2

Sebuah beg berjisim 20 kg berada pada tali sawat penghantar yang condong 25° dengan ufuk. Berapakah komponen berat beg itu yang selari dengan tali sawat? [g = 9.81 m s⁻²]A bag of mass 20 kg rests on a conveyor belt inclined at 25° to the horizontal. What is the component of the weight of the bag parallel to the belt? [g = 9.81 m s⁻²]

• A82.9 N
• B178 N
• C196 N
• D464 N

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A

W = 20 × 9.81 = 196.2 N. The component parallel to the belt is W sin 25° = 196.2 × 0.4226 = 82.9 N. 178 N is W cos 25°, the component perpendicular to the belt.

For the teacherUntuk guru: ApplyingMengaplikasi, lowrendah difficultyaras lowrendah, Chapter 1, standard 1.2.2

TextbookBuku teks p.ms. 16

3

QuestionSoalan 3

Rajah 1 menunjukkan tiga spring serupa, setiap satu dengan pemalar spring 400 N m⁻¹. Spring P dan Q disambung secara selari dan spring R disambung secara bersiri dengannya. Berapakah jumlah pemanjangan apabila beban 8.0 N digantung?Diagram 1 shows three identical springs, each with spring constant 400 N m⁻¹. Springs P and Q are joined in parallel and spring R is joined in series with them. What is the total extension when the 8.0 N load is hung?

• A2.0 cm
• B3.0 cm
• C6.0 cm
• D0.67 cm

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B

P and Q in parallel: k = 400 + 400 = 800 N m⁻¹. This is in series with R: 1/k = 1/800 + 1/400 = 3/800, so k = 267 N m⁻¹. x = F ÷ k = 8.0 ÷ 267 = 0.030 m = 3.0 cm. 6.0 cm treats all three as being in series; 2.0 cm is the extension of one spring alone.

For the teacherUntuk guru: AnalysingMenganalisis, hightinggi difficultyaras hightinggi, Chapter 1, standard 1.4.4

TextbookBuku teks p.ms. 31

4

QuestionSoalan 4

Sebuah pintu air di empangan mempunyai luas 2.0 m² dan berada pada kedalaman 5.0 m di bawah permukaan air. Ketumpatan air ialah 1 000 kg m⁻³. Anggarkan daya yang dikenakan oleh air pada pintu itu (abaikan tekanan atmosfera). [g = 9.81 m s⁻²]A sluice gate in a dam has an area of 2.0 m² and is at a depth of 5.0 m below the water surface. The density of water is 1 000 kg m⁻³. Estimate the force exerted by the water on the gate (ignore atmospheric pressure). [g = 9.81 m s⁻²]

• A4.9 × 10⁴ N
• B1.0 × 10⁴ N
• C3.0 × 10⁵ N
• D9.8 × 10⁴ N

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D

P = hρg = 5.0 × 1 000 × 9.81 = 4.9 × 10⁴ Pa. F = PA = 4.9 × 10⁴ × 2.0 = 9.8 × 10⁴ N. 4.9 × 10⁴ N forgets the area and 1.0 × 10⁴ N forgets g.

For the teacherUntuk guru: ApplyingMengaplikasi, lowrendah difficultyaras lowrendah, Chapter 2, standard 2.1

TextbookBuku teks p.ms. 44

5

QuestionSoalan 5

Sebuah kapal kargo belayar dari laut (ketumpatan 1 025 kg m⁻³) masuk ke muara sungai (ketumpatan 1 000 kg m⁻³). Apakah yang berlaku kepada daya apungan dan kedalaman kapal di bawah paras air?A cargo ship sails from the sea (density 1 025 kg m⁻³) into a river mouth (density 1 000 kg m⁻³). What happens to the buoyant force and to the depth of the ship below the water level?

• ADaya apungan tidak berubah; kapal terapung lebih dalamThe buoyant force is unchanged; the ship floats lower
• BDaya apungan bertambah; kapal terapung lebih dalamThe buoyant force increases; the ship floats lower
• CDaya apungan berkurang; kapal terapung lebih tinggiThe buoyant force decreases; the ship floats higher
• DDaya apungan tidak berubah; kapal terapung lebih tinggiThe buoyant force is unchanged; the ship floats higher

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A

A floating ship has a buoyant force equal to its weight, which does not change. River water is less dense, so a larger volume of water must be displaced to give the same weight of water, and the ship sinks deeper.

For the teacherUntuk guru: UnderstandingMemahami, lowrendah difficultyaras lowrendah, Chapter 2, standard 2.5

TextbookBuku teks p.ms. 67

6

QuestionSoalan 6

Sebuah pendingin hawa 2.0 kW digunakan selama 5 jam sehari selama 30 hari. Jika tarif elektrik ialah RM0.50 setiap kWj, berapakah kos tenaga elektrik itu?A 2.0 kW air conditioner is used for 5 hours a day for 30 days. If the electricity tariff is RM0.50 per kWh, what is the cost of the electrical energy?

• ARM30
• BRM75
• CRM150
• DRM300

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C

Energy = 2.0 kW × (5 × 30) h = 300 kWh. Cost = 300 × RM0.50 = RM150. RM300 forgets the tariff of RM0.50 and RM75 uses only 15 days.

For the teacherUntuk guru: ApplyingMengaplikasi, lowrendah difficultyaras lowrendah, Chapter 3, standard 3.4.3

TextbookBuku teks p.ms. 126

7

QuestionSoalan 7

Tiga sel serupa, setiap satu dengan d.g.e. 1.5 V dan rintangan dalam 0.30 Ω, disambung secara bersiri kepada perintang 4.1 Ω. Berapakah arus dalam litar itu?Three identical cells, each with an e.m.f. of 1.5 V and an internal resistance of 0.30 Ω, are connected in series to a 4.1 Ω resistor. What is the current in the circuit?

• A1.1 A
• B0.90 A
• C0.37 A
• D0.34 A

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B

Total e.m.f. = 4.5 V and total internal resistance = 0.90 Ω. I = ε ÷ (R + r) = 4.5 ÷ (4.1 + 0.90) = 0.90 A. 1.1 A ignores the internal resistance.

For the teacherUntuk guru: ApplyingMengaplikasi, mediumsederhana difficultyaras mediumsederhana, Chapter 3, standard 3.3.4

TextbookBuku teks p.ms. 121

8

QuestionSoalan 8

Dawai P dan dawai Q diperbuat daripada bahan yang sama. Dawai P mempunyai panjang L dan diameter d. Dawai Q mempunyai panjang 2L dan diameter 2d. Berapakah nisbah rintangan P : rintangan Q?Wire P and wire Q are made of the same material. Wire P has length L and diameter d. Wire Q has length 2L and diameter 2d. What is the ratio of the resistance of P to the resistance of Q?

• A1 : 1
• B1 : 2
• C4 : 1
• D2 : 1

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D

R = ρl ÷ A and A ∝ d². R_P ∝ L ÷ d². R_Q ∝ 2L ÷ (2d)² = L ÷ 2d². So R_P : R_Q = 1 : ½ = 2 : 1. The doubled diameter makes the area four times larger, which outweighs the doubled length.

For the teacherUntuk guru: AnalysingMenganalisis, hightinggi difficultyaras hightinggi, Chapter 3, standard 3.2.4

TextbookBuku teks p.ms. 109

9

QuestionSoalan 9

Satu konduktor yang membawa arus berada dalam medan magnet dan mengalami daya. Tindakan manakah akan menyongsangkan arah daya itu?A current-carrying conductor in a magnetic field experiences a force. Which action will reverse the direction of the force?

• AMenyongsangkan arah arus sahajaReversing the direction of the current only
• BMenyongsangkan arah arus dan arah medan magnetReversing the direction of both the current and the magnetic field
• CMenambah arus dalam konduktorIncreasing the current in the conductor
• DMenggunakan magnet yang lebih kuatUsing a stronger magnet

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A

By Fleming's left-hand rule the force reverses when only one of the current and the field is reversed. Reversing both leaves the force unchanged. A larger current or a stronger magnet changes the size of the force only.

For the teacherUntuk guru: UnderstandingMemahami, lowrendah difficultyaras lowrendah, Chapter 4, standard 4.1.2

TextbookBuku teks p.ms. 139

10

QuestionSoalan 10

Sebuah transformer mempunyai kecekapan 80%. Kuasa input ialah 500 W dan voltan sekunder ialah 20 V. Berapakah arus keluaran dalam gegelung sekunder?A transformer has an efficiency of 80%. The input power is 500 W and the secondary voltage is 20 V. What is the output current in the secondary coil?

• A25 A
• B20 A
• C16 A
• D0.050 A

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B

Output power = 80% × 500 = 400 W. I = P ÷ V = 400 ÷ 20 = 20 A. 25 A assumes an efficiency of 100%.

For the teacherUntuk guru: ApplyingMengaplikasi, mediumsederhana difficultyaras mediumsederhana, Chapter 4, standard 4.3.2

TextbookBuku teks p.ms. 164

11

QuestionSoalan 11

Pernyataan manakah yang benar tentang sinar katod?Which statement about cathode rays is correct?

• AIa ialah gelombang elektromagnet yang tidak dipesongkan oleh medan elektrikThey are electromagnetic waves that are not deflected by an electric field
• BIa ialah zarah bercas positif yang dipancarkan oleh anodThey are positively charged particles emitted by the anode
• CIa bergerak dalam lengkung kerana ia dipesongkan oleh udara dalam tiubThey move in curves because air in the tube deflects them
• DIa bergerak dalam garis lurus dan dipesongkan oleh medan magnetThey travel in straight lines and are deflected by a magnetic field

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D

Cathode rays are streams of electrons. They travel in straight lines (shadow of the Maltese cross) and are deflected by magnetic and electric fields because they are negatively charged. They come from the cathode, and the tube is a vacuum.

For the teacherUntuk guru: RememberingMengingat, lowrendah difficultyaras lowrendah, Chapter 5, standard 5.1.2

TextbookBuku teks p.ms. 176

12

QuestionSoalan 12

Bekalan arus ulang-alik berfrekuensi 50 Hz disambung kepada rektifier gelombang penuh. Berapakah frekuensi denyut pada keluaran rektifier itu?An alternating supply of frequency 50 Hz is connected to a full-wave rectifier. What is the frequency of the pulses in the output of the rectifier?

• A25 Hz
• B50 Hz
• C100 Hz
• D200 Hz

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C

Full-wave rectification turns both half cycles into pulses in the same direction, so there are two pulses in every cycle of the input: 2 × 50 = 100 Hz. A half-wave rectifier would give 50 Hz.

For the teacherUntuk guru: UnderstandingMemahami, lowrendah difficultyaras lowrendah, Chapter 5, standard 5.2.2

TextbookBuku teks p.ms. 184

13

QuestionSoalan 13

Rajah 2 menunjukkan penggera kebakaran. Transistor hidup apabila beza keupayaan merentasi perintang 2.0 kΩ ialah 0.70 V atau lebih. Berapakah rintangan maksimum termistor supaya penggera berbunyi?Diagram 2 shows a fire alarm. The transistor switches on when the potential difference across the 2.0 kΩ resistor is 0.70 V or more. What is the maximum resistance of the thermistor for the alarm to sound?

• A15 kΩ
• B17 kΩ
• C0.26 kΩ
• D2.0 kΩ

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A

The same current flows through both parts of the divider, so 6.0 × 2.0 ÷ (R_T + 2.0) ≥ 0.70. R_T + 2.0 ≤ 17.1, so R_T ≤ 15 kΩ. 17 kΩ is the total resistance of the divider, not the thermistor alone.

For the teacherUntuk guru: AnalysingMenganalisis, hightinggi difficultyaras hightinggi, Chapter 5, standard 5.3.2

TextbookBuku teks p.ms. 191

14

QuestionSoalan 14

Dalam sebuah transistor npn, arus tapak 30 µA menghasilkan arus pengumpul 7.2 mA. Berapakah arus pengumpul apabila arus tapak ditambah kepada 45 µA?In an npn transistor, a base current of 30 µA produces a collector current of 7.2 mA. What is the collector current when the base current is increased to 45 µA?

• A4.8 mA
• B10.8 mA
• C15.0 mA
• D10.8 µA

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B

The amplification factor is 7.2 mA ÷ 30 µA = 240. I_C = 240 × 45 µA = 10 800 µA = 10.8 mA. The collector current is proportional to the base current.

For the teacherUntuk guru: ApplyingMengaplikasi, lowrendah difficultyaras lowrendah, Chapter 5, standard 5.3.1

TextbookBuku teks p.ms. 188

15

QuestionSoalan 15

Radium-226 (²²⁶₈₈Ra) mereput dengan memancarkan satu zarah alfa untuk menjadi radon (Rn). Berapakah nombor nukleon dan nombor proton nukleus radon itu?Radium-226 (²²⁶₈₈Ra) decays by emitting an alpha particle to become radon (Rn). What are the nucleon number and the proton number of the radon nucleus?

• ANombor nukleon 222, nombor proton 88Nucleon number 222, proton number 88
• BNombor nukleon 230, nombor proton 86Nucleon number 230, proton number 86
• CNombor nukleon 222, nombor proton 86Nucleon number 222, proton number 86
• DNombor nukleon 226, nombor proton 87Nucleon number 226, proton number 87

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C

An alpha particle (⁴₂He) carries away 4 nucleons and 2 protons: 226 − 4 = 222 and 88 − 2 = 86, so ²²²₈₆Rn.

For the teacherUntuk guru: ApplyingMengaplikasi, lowrendah difficultyaras lowrendah, Chapter 6, standard 6.1.1

TextbookBuku teks p.ms. 201

16

QuestionSoalan 16

Satu radioisotop yang digunakan dalam imbasan hospital mempunyai separuh hayat 2.0 jam. Keaktifan suatu sampel pada pukul 09:00 ialah 960 MBq. Berapakah keaktifannya pada pukul 15:00 hari yang sama?A radioisotope used in a hospital scan has a half-life of 2.0 hours. The activity of a sample at 09:00 is 960 MBq. What is its activity at 15:00 on the same day?

• A480 MBq
• B240 MBq
• C160 MBq
• D120 MBq

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D

09:00 to 15:00 is 6.0 h = 3 half-lives. A = 960 × (½)³ = 120 MBq. 480 MBq is the activity after one half-life only.

For the teacherUntuk guru: ApplyingMengaplikasi, lowrendah difficultyaras lowrendah, Chapter 6, standard 6.1.3

TextbookBuku teks p.ms. 205

17

QuestionSoalan 17

Dalam sebuah reaktor nuklear, bahagian manakah memperlahankan neutron pantas supaya pembelahan boleh berlaku?In a nuclear reactor, which part slows down fast neutrons so that fission can occur?

• ARod kawalanControl rods
• BModerator
• CDinding konkrit tebalThick concrete walls
• DPenjana stimSteam generator

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B

The moderator (graphite or water) slows the fast neutrons. Control rods absorb excess neutrons to control the rate of reaction, and the concrete walls stop radiation leaking out.

For the teacherUntuk guru: UnderstandingMemahami, lowrendah difficultyaras lowrendah, Chapter 6, standard 6.2.4

TextbookBuku teks p.ms. 214

18

QuestionSoalan 18

Sebuah penuding laser berkuasa 5.0 mW memancarkan cahaya merah berpanjang gelombang 650 nm. Berapakah bilangan foton yang dipancarkan setiap saat? [h = 6.63 × 10⁻³⁴ J s, c = 3.00 × 10⁸ m s⁻¹]A laser pointer of power 5.0 mW emits red light of wavelength 650 nm. How many photons are emitted per second? [h = 6.63 × 10⁻³⁴ J s, c = 3.00 × 10⁸ m s⁻¹]

• A1.6 × 10¹⁶ s⁻¹
• B1.6 × 10¹⁹ s⁻¹
• C6.1 × 10⁻¹⁷ s⁻¹
• D3.1 × 10⁻¹⁹ s⁻¹

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A

E = hc ÷ λ = (6.63 × 10⁻³⁴ × 3.00 × 10⁸) ÷ (650 × 10⁻⁹) = 3.06 × 10⁻¹⁹ J. N = P ÷ E = 5.0 × 10⁻³ ÷ 3.06 × 10⁻¹⁹ = 1.6 × 10¹⁶ per second. 1.6 × 10¹⁹ forgets to convert mW to W.

For the teacherUntuk guru: ApplyingMengaplikasi, mediumsederhana difficultyaras mediumsederhana, Chapter 7, standard 7.1.3

TextbookBuku teks p.ms. 230

19

QuestionSoalan 19

Cahaya berfrekuensi lebih tinggi daripada frekuensi ambang menyinari logam dalam sebuah fotosel. Keamatan cahaya itu digandakan tanpa mengubah frekuensinya. Apakah yang berlaku?Light of frequency higher than the threshold frequency shines on the metal in a photocell. The intensity of the light is doubled without changing its frequency. What happens?

• ATenaga kinetik maksimum setiap fotoelektron menjadi dua kali gandaThe maximum kinetic energy of each photoelectron doubles
• BFotoelektron dipancarkan dengan laju maksimum yang lebih tinggiThe photoelectrons are emitted with a higher maximum speed
• CBilangan fotoelektron yang dipancarkan setiap saat menjadi dua kali gandaThe number of photoelectrons emitted per second doubles
• DFrekuensi ambang logam menjadi dua kali gandaThe threshold frequency of the metal doubles

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C

Doubling the intensity doubles the number of photons per second, so twice as many electrons are released each second. The energy of each photon (hf) is unchanged, so Kmax and the threshold frequency are unchanged.

For the teacherUntuk guru: UnderstandingMemahami, lowrendah difficultyaras lowrendah, Chapter 7, standard 7.2.2

TextbookBuku teks p.ms. 237

20

QuestionSoalan 20

Cahaya berpanjang gelombang 300 nm disinarkan pada logam yang mempunyai fungsi kerja 3.0 eV. Berapakah keupayaan penahan bagi fotoelektron itu? [h = 6.63 × 10⁻³⁴ J s, c = 3.00 × 10⁸ m s⁻¹, e = 1.60 × 10⁻¹⁹ C]Light of wavelength 300 nm shines on a metal of work function 3.0 eV. What is the stopping potential for the photoelectrons? [h = 6.63 × 10⁻³⁴ J s, c = 3.00 × 10⁸ m s⁻¹, e = 1.60 × 10⁻¹⁹ C]

• A4.1 V
• B3.0 V
• C7.1 V
• D1.1 V

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D

Photon energy = hc ÷ λ = 6.63 × 10⁻¹⁹ J = 4.14 eV. Kmax = 4.14 − 3.0 = 1.14 eV, so eV_s = Kmax and V_s = 1.1 V. 4.1 V is the photon energy and 7.1 V adds the work function instead of subtracting it.

For the teacherUntuk guru: ApplyingMengaplikasi, mediumsederhana difficultyaras mediumsederhana, Chapter 7, standard 7.3.3

TextbookBuku teks p.ms. 239

21

QuestionSoalan 21

Sebuah spring digunakan dalam penimbang bagasi di kaunter daftar masuk lapangan terbang. Panjang asal spring itu ialah 12.0 cm. Seorang juruteknik menggantung beberapa beban pada spring itu dan merekodkan panjangnya dalam Jadual 1. [g = 9.81 m s⁻²]

Jisim beban, m / kg Mass of load, m / kg	0.5	1.0	1.5	2.0	2.5	3.0
Panjang spring, l / cm Length of spring, l / cm	14.0	16.0	18.0	20.0	22.0	26.0

A spring is used in a luggage scale at an airport check-in counter. The original length of the spring is 12.0 cm. A technician hangs different loads on the spring and records its length in Table 1. [g = 9.81 m s⁻²]

[1 markmarkah] (a)Nyatakan hukum Hooke.State Hooke's law.

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• The extension of a spring is directly proportional to the applied force, provided the elastic limit is not exceeded1

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[2 marksmarkah] (b)Hitung pemalar spring, k, dengan menggunakan beban 1.0 kg.Calculate the spring constant, k, using the 1.0 kg load.

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• F = 1.0 × 9.81 = 9.81 N and x = 16.0 − 12.0 = 4.0 cm = 0.040 m [1 mark]1
• k = F ÷ x = 9.81 ÷ 0.040 = 245 N m⁻¹ (245.25) [1 mark]1

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[2 marksmarkah] (c)Nyatakan beban terkecil dalam Jadual 1 yang menyebabkan spring itu tidak lagi mematuhi hukum Hooke. Berikan satu sebab.State the smallest load in Table 1 for which the spring no longer obeys Hooke's law. Give a reason.

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• 3.0 kg [1 mark]1
• The extension is 14.0 cm instead of 12.0 cm (up to 2.5 kg it increases by 2.0 cm for every 0.5 kg), so it is no longer proportional to the load: the elastic limit has been exceeded [1 mark]1

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[2 marksmarkah] (d)Hitung tenaga keupayaan kenyal yang tersimpan dalam spring itu apabila beban ialah 2.0 kg.Calculate the elastic potential energy stored in the spring when the load is 2.0 kg.

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• x = 20.0 − 12.0 = 8.0 cm = 0.080 m; E = ½kx² = ½ × 245 × (0.080)² [1 mark]1
• = 0.78 J (0.785 J) [1 mark]1

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[3 marksmarkah] (e)Satu spring kedua yang serupa dipasang selari dengan spring pertama dan kedua-duanya berkongsi beban 2.0 kg sama rata. Hitung pemanjangan setiap spring.A second, identical spring is fitted in parallel with the first and the two share a 2.0 kg load equally. Calculate the extension of each spring.

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• Effective spring constant = 2 × 245.25 = 490.5 N m⁻¹ [1 mark]1
• F = 2.0 × 9.81 = 19.62 N [1 mark]1
• x = F ÷ k = 19.62 ÷ 490.5 = 0.040 m (4.0 cm) [1 mark]1

Tick each point you got.Tandakan setiap isi yang betul.

For the teacherUntuk guru: ApplyingMengaplikasi, lowrendah difficultyaras lowrendah, Chapter 1, standard 1.4.2

TextbookBuku teks p.ms. 27

22

QuestionSoalan 22

Sebuah turbin angin kecil menjana kuasa bagi lampu di sebuah jejantas. Rajah 3 menunjukkan keluaran voltan penjana a.u. dalam turbin itu.A small wind turbine generates the power for the lamps on a footbridge. Diagram 3 shows the output voltage of the a.c. generator in the turbine.

[1 markmarkah] (a)Nyatakan hukum aruhan elektromagnet Faraday.State Faraday's law of electromagnetic induction.

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• The magnitude of the induced e.m.f. is directly proportional to the rate of change of magnetic flux (rate of cutting of magnetic field lines)1

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[1 markmarkah] (b)Nyatakan voltan puncak keluaran itu.State the peak voltage of the output.

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• 8.0 V1

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[2 marksmarkah] (c)Tentukan tempoh dan hitung frekuensi keluaran itu.Determine the period and calculate the frequency of the output.

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• Period T = 40 ms = 0.040 s [1 mark]1
• f = 1 ÷ T = 1 ÷ 0.040 = 25 Hz [1 mark]1

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[2 marksmarkah] (d)Turbin itu berputar dua kali lebih laju. Nyatakan voltan puncak dan frekuensi yang baharu.The turbine rotates twice as fast. State the new peak voltage and the new frequency.

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• Peak voltage = 16 V (the e.m.f. is proportional to the speed of rotation) [1 mark]1
• Frequency = 50 Hz [1 mark]1

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[2 marksmarkah] (e)Namakan komponen yang perlu menggantikan gelang gelincir supaya penjana itu menjadi penjana a.t., dan nyatakan kesannya pada keluaran.Name the component that must replace the slip rings to make this a d.c. generator, and state its effect on the output.

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• Split-ring commutator [1 mark]1
• The current in the external circuit always flows in the same direction (a pulsating d.c.) [1 mark]1

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[2 marksmarkah] (f)Lampu pada jejantas memerlukan voltan puncak 24 V. Satu transformer unggul dengan 300 lilitan pada gegelung sekunder digunakan. Hitung bilangan lilitan pada gegelung primer apabila voltan puncak masukan ialah 8.0 V.The lamps on the footbridge need a peak voltage of 24 V. An ideal transformer with 300 turns on its secondary coil is used. Calculate the number of turns on the primary coil when the peak input voltage is 8.0 V.

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• Np ÷ Ns = Vp ÷ Vs, so Np = 300 × 8.0 ÷ 24 [1 mark]1
• = 100 turns [1 mark]1

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For the teacherUntuk guru: ApplyingMengaplikasi, mediumsederhana difficultyaras mediumsederhana, Chapter 4, standard 4.2.4

TextbookBuku teks p.ms. 160

23

QuestionSoalan 23

Sebuah mesin sinar-X pergigian di hospital menggunakan rektifier titi untuk membekalkan voltan kepada tiub sinar-X. Rajah 4(a) menunjukkan rektifier itu dan Rajah 4(b) menunjukkan graf masukan a.u. dan keluaran merentasi perintang R sebelum pelicinan.A dental X-ray machine in a hospital uses a bridge rectifier to supply the voltage to the X-ray tube. Diagram 4(a) shows the rectifier and Diagram 4(b) shows graphs of the a.c. input and of the output across the resistor R before smoothing.

[1 markmarkah] (a)Apakah yang dimaksudkan dengan rektifikasi?What is meant by rectification?

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• The conversion of alternating current (a.c.) to direct current (d.c.)1

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[2 marksmarkah] (b)Pada separuh kitar apabila terminal P adalah positif, namakan dua diod yang mengalirkan arus.In the half cycle when terminal P is positive, name the two diodes that conduct.

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• D₁ [1 mark]1
• D₄ [1 mark] (current flows from P through D₁, the resistor R and D₄ to Q)1

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[2 marksmarkah] (c)Daripada Rajah 4(b), tentukan frekuensi masukan dan frekuensi keluaran.From Diagram 4(b), determine the frequency of the input and the frequency of the output.

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• Input: period 20 ms, so f = 1 ÷ 0.020 = 50 Hz [1 mark]1
• Output: period 10 ms, so f = 100 Hz [1 mark]1

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[2 marksmarkah] (d)Satu kapasitor disambung selari dengan R. Huraikan perubahan pada keluaran dan terangkan.A capacitor is connected in parallel with R. Describe the change in the output and explain it.

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• The output becomes smoother (the voltage no longer falls to zero between pulses) [1 mark]1
• The capacitor charges while the voltage rises and discharges through R while it falls [1 mark]1

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[3 marksmarkah] (e)Keluaran 15 kV itu digunakan untuk memecut elektron dari keadaan rehat dalam tiub sinar-X vakum. Hitung tenaga kinetik maksimum dan laju maksimum sebuah elektron. [e = 1.60 × 10⁻¹⁹ C, jisim elektron = 9.11 × 10⁻³¹ kg]The 15 kV output is used to accelerate electrons from rest in the vacuum X-ray tube. Calculate the maximum kinetic energy and the maximum speed of an electron. [e = 1.60 × 10⁻¹⁹ C, mass of electron = 9.11 × 10⁻³¹ kg]

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• Kmax = eV = 1.60 × 10⁻¹⁹ × 15 000 = 2.4 × 10⁻¹⁵ J [1 mark]1
• ½mv² = eV, so v = √(2 × 2.4 × 10⁻¹⁵ ÷ 9.11 × 10⁻³¹) [1 mark]1
• = 7.3 × 10⁷ m s⁻¹ [1 mark]1

Tick each point you got.Tandakan setiap isi yang betul.

For the teacherUntuk guru: ApplyingMengaplikasi, hightinggi difficultyaras hightinggi, Chapter 5, standard 5.2.2

TextbookBuku teks p.ms. 184

24

QuestionSoalan 24

Teknesium-99m digunakan sebagai bahan penyurih untuk pengimejan di sebuah hospital. Rajah 5 menunjukkan graf keaktifan melawan masa bagi suatu sampel teknesium-99m.Technetium-99m is used as a tracer for imaging in a hospital. Diagram 5 shows the graph of activity against time for a sample of technetium-99m.

[2 marksmarkah] (a)Berdasarkan graf, tentukan separuh hayat teknesium-99m.Using the graph, determine the half-life of technetium-99m.

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• The activity falls from 800 MBq to 400 MBq [1 mark]1
• Time taken = 6.0 hours [1 mark]1

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[2 marksmarkah] (b)Hitung keaktifan sampel itu selepas 24 jam.Calculate the activity of the sample after 24 hours.

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• 24 h = 4 half-lives [1 mark]1
• 800 × (½)⁴ = 50 MBq [1 mark]1

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[2 marksmarkah] (c)Hitung masa yang diambil supaya keaktifan sampel itu jatuh kepada 12.5 MBq.Calculate the time taken for the activity of the sample to fall to 12.5 MBq.

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• 800 → 12.5 is a factor of 64 = 2⁶, so 6 half-lives [1 mark]1
• t = 6 × 6.0 = 36 hours [1 mark]1

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[2 marksmarkah] (d)Teknesium-99m (⁹⁹₄₃Tc*) berada dalam keadaan teruja dan mereput dengan memancarkan sinar gama untuk menjadi teknesium-99. Tulis persamaan reputan itu.Technetium-99m (⁹⁹₄₃Tc*) is in an excited state and decays by emitting a gamma ray to become technetium-99. Write the decay equation.

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• Daughter nucleus ⁹⁹₄₃Tc with the same nucleon number and proton number [1 mark]1
• ⁹⁹₄₃Tc* → ⁹⁹₄₃Tc + γ (gamma ray written correctly) [1 mark]1

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[2 marksmarkah] (e)Nyatakan dua langkah keselamatan yang perlu diambil oleh juruteknik semasa mengendalikan sampel radioaktif itu.State two safety precautions that the technician should take when handling the radioactive sample.

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• Use tongs / remote handling (do not touch the source with bare hands) [1 mark]1
• Keep the source in a lead-lined container, limit the time of exposure and keep a distance from it (any one more precaution; wear protective clothing and a dosimeter) [1 mark]1

Tick each point you got.Tandakan setiap isi yang betul.

For the teacherUntuk guru: ApplyingMengaplikasi, lowrendah difficultyaras lowrendah, Chapter 6, standard 6.1.3

TextbookBuku teks p.ms. 205

25Choose 1Pilih 1

QuestionSoalan 25

Rajah 6 menunjukkan litar lampu kecemasan di sebuah terminal lapangan terbang. Bateri mempunyai d.g.e. 10.5 V dan rintangan dalam 1.0 Ω. Dua lampu serupa, setiap satu berintangan 5.0 Ω, disambung secara selari.Diagram 6 shows the emergency lighting circuit in an airport terminal. The battery has an e.m.f. of 10.5 V and an internal resistance of 1.0 Ω. Two identical lamps, each of resistance 5.0 Ω, are connected in parallel.

[2 marksmarkah] (a)Terangkan mengapa beza keupayaan merentasi terminal bateri lebih rendah daripada d.g.e. apabila bateri membekalkan arus.Explain why the potential difference across the battery terminals is lower than the e.m.f. when the battery supplies a current.

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• Some of the energy supplied per coulomb is used to push the charge through the internal resistance of the battery [1 mark]1
• The "lost volts" Ir is dissipated as heat inside the battery, so V = ε − Ir is less than ε [1 mark]1

Tick each point you got.Tandakan setiap isi yang betul.

[3 marksmarkah] (b)Hitung rintangan berkesan kedua-dua lampu dan arus yang dibekalkan oleh bateri apabila suis S ditutup.Calculate the effective resistance of the two lamps and the current supplied by the battery when switch S is closed.

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• 1/R = 1/5.0 + 1/5.0, so R = 2.5 Ω [1 mark]1
• I = ε ÷ (R + r) = 10.5 ÷ (2.5 + 1.0) [1 mark]1
• = 3.0 A [1 mark]1

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[2 marksmarkah] (c)Hitung beza keupayaan merentasi setiap lampu.Calculate the potential difference across each lamp.

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• V = IR = 3.0 × 2.5 [1 mark]1
• = 7.5 V (or ε − Ir = 10.5 − 3.0 × 1.0 = 7.5 V) [1 mark]1

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[4 marksmarkah] (d)Hitung (i) kuasa setiap lampu, (ii) jumlah tenaga elektrik yang dipindahkan oleh kedua-dua lampu dalam 30 minit.Calculate (i) the power of each lamp, (ii) the total electrical energy transferred by the two lamps in 30 minutes.

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• (i) P = V² ÷ R = 7.5² ÷ 5.0 [1 mark]1
• (i) = 11 W (11.25 W) [1 mark]1
• (ii) E = Pt = 2 × 11.25 × (30 × 60) [1 mark]1
• (ii) = 4.1 × 10⁴ J (40 500 J) [1 mark]1

Tick each point you got.Tandakan setiap isi yang betul.

[4 marksmarkah] (e)Satu lampu ketiga yang serupa ditambah secara selari. Hitung arus yang baharu daripada bateri dan terangkan mengapa setiap lampu menjadi lebih malap.A third identical lamp is added in parallel. Calculate the new current from the battery and explain why each lamp becomes dimmer.

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• Effective resistance = 5.0 ÷ 3 = 1.67 Ω; I = 10.5 ÷ (1.67 + 1.0) [1 mark]1
• = 3.9 A (3.94 A) [1 mark]1
• The larger current gives a larger lost voltage Ir inside the battery, so the p.d. across the lamps falls to 6.6 V [1 mark]1
• The power in each lamp (V² ÷ R = 8.6 W) is less than before (11 W), so each lamp is dimmer [1 mark]1

Tick each point you got.Tandakan setiap isi yang betul.

[5 marksmarkah] (f)Lapangan terbang itu mahu lampu kecemasan yang lebih terang dan tahan lebih lama. Cadangkan tiga pengubahsuaian dan beri satu sebab bagi setiap satu.The airport wants emergency lamps that are brighter and last longer. Suggest three modifications and give one reason for each.

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• Use cells with a lower internal resistance [1 mark]1
• Reason: less voltage is lost inside the battery, so there is a greater p.d. across the lamps [1 mark]1
• Connect a second identical battery in parallel with the first [1 mark]1
• Reason: the e.m.f. is the same but each battery supplies half the current, so the batteries last longer [1 mark]1
• Use LED lamps instead of filament lamps, because they need a lower power for the same brightness, so less current is drawn [1 mark]1

Tick each point you got.Tandakan setiap isi yang betul.

For the teacherUntuk guru: ApplyingMengaplikasi, lowrendah difficultyaras lowrendah, Chapter 3, standard 3.3.2

TextbookBuku teks p.ms. 117

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QuestionSoalan 26

Aruhan elektromagnet digunakan di loji kuasa. Seorang pelajar menyiasat aruhan dengan menggerakkan kutub utara sebuah magnet bar ke dalam solenoid yang disambung kepada galvanometer sifar di tengah. Jadual 2 menunjukkan keputusannya.

Eksperimen Experiment	Bilangan lilitan Number of turns	Laju magnet Speed of the magnet	Kekuatan magnet Strength of the magnet	Pesongan maksimum / bahagian Maximum deflection / divisions
I	100	Perlahan / Slow	Satu magnet / One magnet	2
II	100	Laju / Fast	Satu magnet / One magnet	4
III	200	Laju / Fast	Satu magnet / One magnet	8
IV	200	Laju / Fast	Dua magnet / Two magnets	12

Electromagnetic induction is used in a power plant. A student investigates induction by moving the north pole of a bar magnet into a solenoid connected to a centre-zero galvanometer. Table 2 shows the results.

[2 marksmarkah] (a)Nyatakan hukum Lenz.State Lenz's law.

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• The direction of the induced current is such that it opposes [1 mark]1
• the change (the motion of the magnet) that produces it [1 mark]1

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[3 marksmarkah] (b)Kutub utara magnet itu ditolak ke dalam hujung solenoid. (i) Nyatakan pemerhatian pada galvanometer. (ii) Nyatakan kutub yang terhasil pada hujung solenoid itu. (iii) Terangkan bagaimana hukum Lenz menunjukkan bahawa tenaga adalah abadi.The north pole of the magnet is pushed into one end of the solenoid. (i) State the observation on the galvanometer. (ii) State the pole produced at that end of the solenoid. (iii) Explain how Lenz's law shows that energy is conserved.

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• (i) The pointer deflects (while the magnet moves) [1 mark]1
• (ii) North pole, which repels the approaching magnet [1 mark]1
• (iii) Work must be done to push the magnet against the repulsion, and this work is converted into the electrical energy of the induced current [1 mark]1

Tick each point you got.Tandakan setiap isi yang betul.

[3 marksmarkah] (c)Berdasarkan Jadual 2, nyatakan hubungan antara pesongan galvanometer (e.m.f. teraruh) dengan setiap pemboleh ubah berikut: (i) laju magnet, (ii) bilangan lilitan, (iii) kekuatan magnet.Based on Table 2, state the relationship between the deflection of the galvanometer (induced e.m.f.) and each of the following: (i) speed of the magnet, (ii) number of turns, (iii) strength of the magnet.

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• (i) I and II: the faster the magnet moves, the larger the induced e.m.f. [1 mark]1
• (ii) II and III: the more turns, the larger the induced e.m.f. [1 mark]1
• (iii) III and IV: the stronger the magnet, the larger the induced e.m.f. [1 mark]1

Tick each point you got.Tandakan setiap isi yang betul.

[4 marksmarkah] (d)Huraikan bagaimana penjana a.u. di loji kuasa menghasilkan e.m.f. ulang-alik.Describe how an a.c. generator in a power plant produces an alternating e.m.f.

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• The coil is rotated in a magnetic field between the poles of a magnet [1 mark]1
• The coil cuts the magnetic field lines (the magnetic flux changes), so an e.m.f. is induced [1 mark]1
• The induced e.m.f. changes direction every half rotation because the sides of the coil swap places [1 mark]1
• Slip rings and carbon brushes connect the rotating coil to the external circuit [1 mark]1

Tick each point you got.Tandakan setiap isi yang betul.

[4 marksmarkah] (e)Penjana itu menghasilkan 25 MW pada 11 kV. Voltan itu dinaikkan kepada 275 kV oleh transformer unggul dan dihantar melalui kabel yang berintangan 12 Ω. Hitung arus dalam kabel dan kuasa yang hilang sebagai haba dalam kabel.The generator produces 25 MW at 11 kV. The voltage is stepped up to 275 kV by an ideal transformer and transmitted through cables of resistance 12 Ω. Calculate the current in the cables and the power lost as heat in the cables.

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• I = P ÷ V = 25 × 10⁶ ÷ 275 × 10³ [1 mark]1
• = 91 A (90.9 A) [1 mark]1
• Power loss = I²R = (90.9)² × 12 [1 mark]1
• = 9.9 × 10⁴ W (99 kW) [1 mark]1

Tick each point you got.Tandakan setiap isi yang betul.

[4 marksmarkah] (f)Hitung kuasa yang hilang jika kuasa yang sama dihantar terus pada 11 kV melalui kabel yang sama. Seterusnya, terangkan mengapa voltan tinggi digunakan untuk penghantaran.Calculate the power lost if the same power is transmitted directly at 11 kV through the same cables. Hence explain why a high voltage is used for transmission.

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• I = 25 × 10⁶ ÷ 11 × 10³ = 2.3 × 10³ A (2 273 A) [1 mark]1
• Power loss = I²R = (2 273)² × 12 = 6.2 × 10⁷ W (62 MW) [1 mark]1
• This is more than the 25 MW generated, so all the power would be lost [1 mark]1
• For the same power, a higher voltage gives a smaller current, and the power loss I²R falls with the square of the current [1 mark]1

Tick each point you got.Tandakan setiap isi yang betul.

For the teacherUntuk guru: UnderstandingMemahami, mediumsederhana difficultyaras mediumsederhana, Chapter 4, standard 4.2.3

TextbookBuku teks p.ms. 159

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QuestionSoalan 27

Sebuah hospital menggunakan radioisotop untuk pengimejan dan sebuah loji kuasa nuklear membekalkan elektrik kepada bandar itu. Hospital itu perlu memilih satu radioisotop yang disuntik atau diminum oleh pesakit dan dikesan di luar badan oleh kamera gama. Jadual 3 menunjukkan ciri-ciri empat radioisotop P, Q, R dan S.

Radioisotop Radioisotope	Jenis sinaran Type of radiation	Separuh hayat Half-life	Keadaan fizikal Physical state
P	Alfa / Alpha	120 tahun / 120 years	Pepejal / Solid
Q	Beta / Beta	14 jam / 14 hours	Cecair / Liquid
R	Gama / Gamma	6.0 jam / 6.0 hours	Cecair / Liquid
S	Gama / Gamma	30 tahun / 30 years	Pepejal / Solid

A hospital uses radioisotopes for imaging and a nuclear power plant supplies electricity to the city. The hospital must choose a radioisotope that a patient is injected with or drinks and that is detected outside the body by a gamma camera. Table 3 shows the characteristics of four radioisotopes P, Q, R and S.

[2 marksmarkah] (a)Nyatakan dua ciri pembelahan nukleus.State two characteristics of nuclear fission.

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• A heavy nucleus (such as uranium-235) splits into two lighter nuclei [1 mark]1
• Neutrons and a large amount of energy are released (the neutrons can start a chain reaction) [1 mark]1

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[2 marksmarkah] (b)Berdasarkan Jadual 3, nyatakan jenis sinaran yang sesuai dan beri satu sebab.Based on Table 3, state the suitable type of radiation and give a reason.

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• Gamma [1 mark]1
• It has a high penetrating power, so it passes out of the body and can be detected by the gamma camera (it also ionises less strongly than alpha or beta, so is less harmful) [1 mark]1

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[2 marksmarkah] (c)Nyatakan separuh hayat yang sesuai dan beri satu sebab.State the suitable half-life and give a reason.

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• A short half-life (a few hours) [1 mark]1
• The activity falls quickly after the scan, so the patient is exposed to radiation for a short time only [1 mark]1

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[2 marksmarkah] (d)Nyatakan keadaan fizikal yang sesuai dan beri satu sebab.State the suitable physical state and give a reason.

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• Liquid [1 mark]1
• It can be injected or drunk, spreads through the body with the blood and is passed out of the body easily [1 mark]1

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[1 markmarkah] (e)Pilih radioisotop yang paling sesuai.Choose the most suitable radioisotope.

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• R1

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[3 marksmarkah] (f)Sebuah penggera asap di hospital itu mengandungi sumber alfa dengan separuh hayat 432 tahun. Zarah alfa mengionkan udara antara dua elektrod dan menghasilkan arus kecil; asap mengurangkan arus itu dan penggera berbunyi. Terangkan mengapa (i) sumber alfa digunakan dan bukannya sumber gama, (ii) separuh hayat yang panjang diperlukan.A smoke alarm in the hospital contains an alpha source with a half-life of 432 years. The alpha particles ionise the air between two electrodes and give a small current; smoke reduces the current and the alarm sounds. Explain why (i) an alpha source is used rather than a gamma source, (ii) a long half-life is needed.

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• (i) Alpha particles are strongly ionising, so they produce enough ions to give a current [1 mark]1
• (i) Alpha has a low penetrating power, so it is stopped by the casing and is safe in a building (gamma would not ionise the air enough and would escape) [1 mark]1
• (ii) The activity stays almost constant for many years, so the alarm works for a long time without replacing the source [1 mark]1

Tick each point you got.Tandakan setiap isi yang betul.

[6 marksmarkah] (g)Dalam reaktor nuklear, satu tindak balas pembelahan ialah: ²³⁵₉₂U + ¹₀n → ¹⁴⁰₅₄Xe + ⁹⁴₃₈Sr + 2 ¹₀n + tenaga. Jisim: U-235 = 235.0439 u, neutron = 1.0087 u, Xe-140 = 139.9216 u, Sr-94 = 93.9154 u. [1 u = 1.66 × 10⁻²⁷ kg, c = 3.00 × 10⁸ m s⁻¹] Hitung (i) kehilangan jisim, (ii) tenaga yang dibebaskan dalam satu pembelahan, (iii) bilangan pembelahan sesaat yang diperlukan untuk menghasilkan kuasa haba 600 MW.In a nuclear reactor, one fission reaction is: ²³⁵₉₂U + ¹₀n → ¹⁴⁰₅₄Xe + ⁹⁴₃₈Sr + 2 ¹₀n + energy. Masses: U-235 = 235.0439 u, neutron = 1.0087 u, Xe-140 = 139.9216 u, Sr-94 = 93.9154 u. [1 u = 1.66 × 10⁻²⁷ kg, c = 3.00 × 10⁸ m s⁻¹] Calculate (i) the mass defect, (ii) the energy released in one fission, (iii) the number of fissions per second needed to produce a thermal power of 600 MW.

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• (i) Mass before = 235.0439 + 1.0087 = 236.0526 u; mass after = 139.9216 + 93.9154 + 2 × 1.0087 = 235.8544 u [1 mark]1
• (i) Mass defect = 0.1982 u [1 mark]1
• (ii) m = 0.1982 × 1.66 × 10⁻²⁷ = 3.29 × 10⁻²⁸ kg [1 mark]1
• (ii) E = mc² = 3.29 × 10⁻²⁸ × (3.00 × 10⁸)² = 2.96 × 10⁻¹¹ J [1 mark]1
• (iii) Number per second = 600 × 10⁶ ÷ 2.96 × 10⁻¹¹ [1 mark]1
• (iii) = 2.0 × 10¹⁹ s⁻¹ [1 mark]1

Tick each point you got.Tandakan setiap isi yang betul.

[2 marksmarkah] (h)Nyatakan satu kelebihan dan satu kekurangan penggunaan tenaga nuklear untuk menjana elektrik.State one advantage and one disadvantage of using nuclear energy to generate electricity.

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• Advantage: a large amount of energy from a small mass of fuel / no carbon dioxide released [1 mark]1
• Disadvantage: radioactive waste that stays dangerous for a long time must be stored safely / high cost and risk of accident [1 mark]1

Tick each point you got.Tandakan setiap isi yang betul.

For the teacherUntuk guru: EvaluatingMenilai, hightinggi difficultyaras hightinggi, Chapter 6, standard 6.2.3

TextbookBuku teks p.ms. 212

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QuestionSoalan 28

Sebuah lapangan terbang mahu satu litar penggera fros yang membunyikan pembunyi apabila suhu landasan jatuh kepada 3 °C. Pembunyi memerlukan arus 0.60 A manakala arus pengumpul maksimum transistor ialah 100 mA. Jadual 4 menunjukkan empat litar yang dicadangkan P, Q, R dan S, dan Rajah 7 menunjukkan litar Q. Voltan antara tapak dengan pengeluar mesti sekurang-kurangnya 0.70 V untuk menghidupkan transistor.

Litar Circuit	Penderia Sensor	Kedudukan penderia dalam pembahagi keupayaan Position of sensor in the potential divider	Alat dalam litar pengumpul Device in the collector circuit	Perintang antara pembahagi dan tapak Resistor between divider and base
P	Termistor / Thermistor	Atas / Upper	Pembunyi / Buzzer	Tiada / None
Q	Termistor / Thermistor	Bawah / Lower	Gegelung geganti / Relay coil	RB
R	LDR / LDR	Bawah / Lower	Gegelung geganti / Relay coil	RB
S	Termistor / Thermistor	Bawah / Lower	Pembunyi / Buzzer	Tiada / None

An airport wants a frost-alarm circuit that sounds a buzzer when the temperature of the runway falls to 3 °C. The buzzer needs a current of 0.60 A while the maximum collector current of the transistor is 100 mA. Table 4 shows four proposed circuits P, Q, R and S, and Diagram 7 shows circuit Q. The base-emitter voltage must be at least 0.70 V to switch the transistor on.

[2 marksmarkah] (a)Nyatakan perubahan rintangan termistor apabila suhu jatuh, dan kesannya pada beza keupayaan merentasi termistor apabila ia berada di bahagian bawah pembahagi keupayaan.State how the resistance of a thermistor changes when the temperature falls, and the effect on the potential difference across the thermistor when it is the lower part of the potential divider.

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• The resistance of the thermistor increases [1 mark]1
• The potential difference across it increases (it takes a larger share of the supply voltage) [1 mark]1

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[9 marksmarkah] (b)Berdasarkan Jadual 4, nyatakan ciri yang sesuai bagi (i) penderia, (ii) kedudukan penderia, (iii) alat dalam litar pengumpul, (iv) perintang antara pembahagi dan tapak. Beri satu sebab bagi setiap ciri, dan seterusnya pilih litar yang paling sesuai.Based on Table 4, state the suitable characteristic of (i) the sensor, (ii) the position of the sensor, (iii) the device in the collector circuit, (iv) the resistor between the divider and the base. Give one reason for each characteristic and hence choose the most suitable circuit.

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• (i) Thermistor [1 mark]1
• (i) Reason: its resistance changes with temperature (an LDR responds to light) [1 mark]1
• (ii) Lower position [1 mark]1
• (ii) Reason: when the temperature falls its resistance and p.d. increase until it reaches 0.70 V and the base current flows [1 mark]1
• (iii) Relay coil [1 mark]1
• (iii) Reason: the buzzer needs 0.60 A, more than the transistor can carry, so the small collector current operates the relay, which switches the large current [1 mark]1
• (iv) A base resistor R_B [1 mark]1
• (iv) Reason: it limits the base current so the transistor is not damaged [1 mark]1
• Circuit Q [1 mark]1

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[3 marksmarkah] (c)Pada 3 °C rintangan termistor ialah 4.0 kΩ. Bagi litar Q, bekalan ialah 10 V dan perintang atas ialah 6.0 kΩ. Hitung beza keupayaan merentasi termistor dan nyatakan sama ada transistor hidup.At 3 °C the resistance of the thermistor is 4.0 kΩ. In circuit Q the supply is 10 V and the upper resistor is 6.0 kΩ. Calculate the potential difference across the thermistor and state whether the transistor switches on.

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• V = 10 × 4.0 ÷ (6.0 + 4.0) [1 mark]1
• = 4.0 V [1 mark]1
• 4.0 V is greater than 0.70 V, so the transistor switches on, the relay operates and the buzzer sounds [1 mark]1

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[3 marksmarkah] (d)Arus yang melalui gegelung geganti ialah 90 mA dan faktor penggandaan arus transistor ialah 120. Anggap arus tapak tidak mengubah beza keupayaan merentasi termistor. Hitung arus tapak dan nilai R_B yang diperlukan.The current through the relay coil is 90 mA and the current amplification factor of the transistor is 120. Assume the base current does not change the p.d. across the thermistor. Calculate the base current and the value of R_B needed.

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• I_B = I_C ÷ 120 = 90 mA ÷ 120 = 0.75 mA [1 mark]1
• Voltage across R_B = 4.0 − 0.70 = 3.3 V [1 mark]1
• R_B = 3.3 ÷ 0.75 × 10⁻³ = 4.4 kΩ (4 400 Ω) [1 mark]1

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[3 marksmarkah] (e)Lapangan terbang itu juga mahu penggera haba yang berbunyi apabila suhu naik melebihi 40 °C. Huraikan satu pengubahsuaian pada litar Q dan terangkan bagaimana ia berfungsi.The airport also wants a heat alarm that sounds when the temperature rises above 40 °C. Describe one modification to circuit Q and explain how it works.

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• Interchange the thermistor and the 6.0 kΩ resistor (thermistor in the upper position) [1 mark]1
• When the temperature rises the resistance of the thermistor falls, so the p.d. across the lower resistor increases [1 mark]1
• When it reaches 0.70 V the base current flows, the transistor switches on and the buzzer sounds [1 mark]1

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For the teacherUntuk guru: CreatingMencipta, mediumsederhana difficultyaras mediumsederhana, Chapter 5, standard 5.3.2

TextbookBuku teks p.ms. 191

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