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.

21

QuestionSoalan 21

Seorang penyelam komersial menjalankan kerja pembaikan di dasar laut berhampiran sebuah platform minyak. Jadual 1 menunjukkan tekanan yang disebabkan oleh air laut pada kedalaman yang berbeza. Rajah 1 menunjukkan penyelam itu menggunakan beg angkat udara untuk menaikkan sebuah kotak logam dari dasar laut pada kedalaman 30 m.

Kedalaman, h / m Depth, h / m	0	5	10	15	20
Tekanan disebabkan air, P / kPa Pressure due to water, P / kPa	0	50.3	100.6	150.8	201.1

A commercial diver does repair work on the sea bed near an oil platform. Table 1 shows the pressure caused by sea water at different depths. Diagram 1 shows the diver using an air lift bag to raise a metal box from the sea bed at a depth of 30 m.

[1 markmarkah] (a)Berdasarkan Jadual 1, nyatakan hubungan antara tekanan air dengan kedalaman.Based on Table 1, state the relationship between the pressure of the water and the depth.

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• Pressure is directly proportional to depth (P ∝ h)1

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[3 marksmarkah] (b)Tentukan kecerunan data dalam Jadual 1 dan seterusnya hitung ketumpatan air laut.Determine the gradient of the data in Table 1 and hence calculate the density of the sea water.

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• Gradient = (201.1 − 0) ÷ (20 − 0) = 10.06 kPa m⁻¹ (1.006 × 10⁴ Pa m⁻¹) [1 mark]1
• P = hρg, so gradient = ρg and ρ = 10 055 ÷ 9.81 [1 mark]1
• = 1.02 × 10³ kg m⁻³ (accept 1 020 to 1 030) [1 mark]1

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

[2 marksmarkah] (c)Daripada Jadual 1, ramalkan jumlah tekanan pada penyelam pada kedalaman 30 m. Tekanan atmosfera ialah 101 kPa.From Table 1, predict the total pressure on the diver at a depth of 30 m. Atmospheric pressure is 101 kPa.

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• Pressure due to water = 10.06 × 30 = 302 kPa [1 mark]1
• Total pressure = 302 + 101 = 403 kPa (accept 400 to 405 kPa) [1 mark]1

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

[2 marksmarkah] (d)Isi padu kotak itu ialah 4.0 × 10⁻³ m³. Hitung daya apungan pada kotak itu apabila ia tenggelam sepenuhnya. Gunakan ketumpatan air laut 1 025 kg m⁻³.The volume of the box is 4.0 × 10⁻³ m³. Calculate the buoyant force on the box when it is fully submerged. Use a sea water density of 1 025 kg m⁻³.

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• Buoyant force = Vρg = 4.0 × 10⁻³ × 1 025 × 9.81 [1 mark]1
• = 40 N (40.2 N) [1 mark]1

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

[2 marksmarkah] (e)Jisim kotak itu ialah 30 kg dan isi padu beg angkat ialah 0.050 m³. Dengan mengabaikan jisim beg dan udara di dalamnya, tunjukkan dengan pengiraan bahawa beg itu dapat menaikkan kotak itu.The mass of the box is 30 kg and the volume of the lift bag is 0.050 m³. Ignoring the mass of the bag and the air in it, show by calculation that the bag can raise the box.

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• Weight of box = 30 × 9.81 = 294 N; apparent weight in water = 294 − 40.2 = 254 N [1 mark]1
• Buoyant force on bag = 0.050 × 1 025 × 9.81 = 503 N, which is greater than 254 N, so the box rises [1 mark]1

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

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

TextbookBuku teks p.ms. 71

22

QuestionSoalan 22

Sebuah lampu jalan di sebuah kampung menggunakan panel suria sebagai sumber kuasanya. Seorang pelajar menyiasat panel itu menggunakan litar dalam Rajah 2. Panel suria itu mempunyai d.g.e. E dan rintangan dalam r. Rintangan boleh ubah R dilaraskan dan bacaan arus I dan voltan terminal V direkodkan dalam Jadual 2.

Arus, I / A Current, I / A	0.2	0.4	0.6	0.8	1.0
Voltan terminal, V / V Terminal voltage, V / V	13.7	13.4	13.1	12.8	12.5

A street lamp in a village uses a solar panel as its power source. A student investigates the panel using the circuit in Diagram 2. The solar panel has an e.m.f. E and an internal resistance r. The variable resistor R is adjusted and the readings of current I and terminal voltage V are recorded in Table 2.

[1 markmarkah] (a)Terangkan mengapa V berkurang apabila I bertambah.Explain why V decreases as I increases.

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• Some of the e.m.f. is used to drive the current through the internal resistance (V = E − Ir); the "lost volts" Ir increases as I increases1

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

[3 marksmarkah] (b)Tentukan rintangan dalam r dan d.g.e. E panel suria itu.Determine the internal resistance r and the e.m.f. E of the solar panel.

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• V = E − Ir, so the gradient of V against I is −r: gradient = (12.5 − 13.7) ÷ (1.0 − 0.2) = −1.5 V A⁻¹ [1 mark]1
• r = 1.5 Ω [1 mark]1
• E = V + Ir = 13.7 + (0.2 × 1.5) = 14.0 V (the intercept on the V axis) [1 mark]1

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

[2 marksmarkah] (c)Rintangan R dilaraskan kepada 5.5 Ω. Hitung arus dalam litar dan bacaan voltmeter.The resistance R is adjusted to 5.5 Ω. Calculate the current in the circuit and the reading of the voltmeter.

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• I = E ÷ (R + r) = 14.0 ÷ (5.5 + 1.5) = 2.0 A [1 mark]1
• V = IR = 2.0 × 5.5 = 11 V [1 mark]1

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

[2 marksmarkah] (d)Hitung kuasa yang hilang sebagai haba di dalam panel suria itu apabila R = 5.5 Ω.Calculate the power lost as heat inside the solar panel when R = 5.5 Ω.

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• P = I²r = 2.0² × 1.5 [1 mark]1
• = 6.0 W [1 mark]1

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

[2 marksmarkah] (e)Dua panel suria yang serupa disambung secara selari untuk membekalkan lampu yang sama. Nyatakan d.g.e. dan rintangan dalam gabungan itu.Two identical solar panels are connected in parallel to supply the same lamp. State the e.m.f. and the internal resistance of the combination.

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• e.m.f. = 14.0 V (unchanged) [1 mark]1
• Internal resistance = 1.5 ÷ 2 = 0.75 Ω [1 mark]1

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

For the teacherUntuk guru: ApplyingMengaplikasi, lowrendah difficultyaras lowrendah, Chapter 3, standard 3.3.3

TextbookBuku teks p.ms. 119

23

QuestionSoalan 23

Rajah 3 menunjukkan sebuah penjana a.u. di stesen janakuasa hidroelektrik sebuah empangan yang disambung kepada transformer injak naik unggul. Kuasa keluaran penjana ialah 33 MW. Voltan masukan transformer ialah 11 kV, voltan keluarannya 275 kV dan gegelung primernya mempunyai 600 lilitan. Kuasa dihantar melalui kabel yang rintangannya 5.0 Ω.Diagram 3 shows an a.c. generator in the hydro-electric power station of a dam, connected to an ideal step-up transformer. The output power of the generator is 33 MW. The input voltage of the transformer is 11 kV, its output voltage is 275 kV and its primary coil has 600 turns. The power is sent through a cable of resistance 5.0 Ω.

[1 markmarkah] (a)Namakan prinsip yang digunakan oleh penjana itu untuk menghasilkan voltan.Name the principle used by the generator to produce a voltage.

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• Electromagnetic induction1

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[2 marksmarkah] (b)Nyatakan dua cara untuk menambah voltan keluaran sebuah penjana a.u.State two ways to increase the output voltage of an a.c. generator.

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Any 2 pointsMana-mana 2 isi

• Rotate the coil faster / use a faster water turbine [1 mark]1
• Use stronger magnets [1 mark]1
• Use more turns of wire in the coil [1 mark]1
• Use a coil with a larger area [1 mark]1

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

[2 marksmarkah] (c)Hitung bilangan lilitan gegelung sekunder.Calculate the number of turns of the secondary coil.

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• Ns ÷ Np = Vs ÷ Vp, so Ns = 600 × 275 ÷ 11 [1 mark]1
• = 15 000 turns [1 mark]1

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[3 marksmarkah] (d)Hitung arus dalam gegelung primer dan arus dalam gegelung sekunder, dan terangkan mengapa arus dalam gegelung sekunder lebih kecil.Calculate the current in the primary coil and the current in the secondary coil, and explain why the current in the secondary coil is smaller.

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• Ip = P ÷ Vp = 33 × 10⁶ ÷ 11 × 10³ = 3 000 A [1 mark]1
• Is = P ÷ Vs = 33 × 10⁶ ÷ 275 × 10³ = 120 A [1 mark]1
• For an ideal transformer the power in equals the power out; the voltage is higher so the current is lower [1 mark]1

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

[2 marksmarkah] (e)Hitung kuasa yang hilang sebagai haba dalam kabel itu.Calculate the power lost as heat in the cable.

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• P = Is²R = 120² × 5.0 [1 mark]1
• = 7.2 × 10⁴ W (72 kW) [1 mark]1

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

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

TextbookBuku teks p.ms. 164

24

QuestionSoalan 24

Sebuah fotosel dalam sistem kawalan cahaya sebuah bengkel mempunyai katod yang fungsi kerjanya ialah 2.9 eV. Rajah 4 menunjukkan katod itu disinari dengan cahaya ultraungu berpanjang gelombang 365 nm. [h = 6.63 × 10⁻³⁴ J s, c = 3.00 × 10⁸ m s⁻¹, e = 1.60 × 10⁻¹⁹ C]A photocell in the lighting control system of a workshop has a cathode with a work function of 2.9 eV. Diagram 4 shows the cathode illuminated with ultraviolet light of wavelength 365 nm. [h = 6.63 × 10⁻³⁴ J s, c = 3.00 × 10⁸ m s⁻¹, e = 1.60 × 10⁻¹⁹ C]

[1 markmarkah] (a)Apakah yang dimaksudkan dengan fungsi kerja?What is meant by work function?

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• The minimum energy needed to release an electron from the surface of a metal1

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[2 marksmarkah] (b)Hitung frekuensi ambang katod itu.Calculate the threshold frequency of the cathode.

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• W = 2.9 × 1.60 × 10⁻¹⁹ = 4.64 × 10⁻¹⁹ J [1 mark]1
• f₀ = W ÷ h = 4.64 × 10⁻¹⁹ ÷ 6.63 × 10⁻³⁴ = 7.0 × 10¹⁴ Hz [1 mark]1

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

[2 marksmarkah] (c)Hitung tenaga satu foton cahaya ultraungu itu.Calculate the energy of one photon of the ultraviolet light.

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• E = hc ÷ λ = 6.63 × 10⁻³⁴ × 3.00 × 10⁸ ÷ 365 × 10⁻⁹ [1 mark]1
• = 5.4 × 10⁻¹⁹ J (5.45 × 10⁻¹⁹ J) [1 mark]1

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

[2 marksmarkah] (d)Hitung tenaga kinetik maksimum fotoelektron.Calculate the maximum kinetic energy of the photoelectrons.

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• hf = W + Kmax, so Kmax = 5.45 × 10⁻¹⁹ − 4.64 × 10⁻¹⁹ [1 mark]1
• = 8.1 × 10⁻²⁰ J (0.51 eV) [1 mark]1

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

[3 marksmarkah] (e)Keamatan cahaya ultraungu itu digandakan. Nyatakan kesan terhadap tenaga kinetik maksimum fotoelektron dan terhadap bilangan fotoelektron yang dipancarkan setiap saat, dan terangkan jawapan anda.The intensity of the ultraviolet light is doubled. State the effect on the maximum kinetic energy of the photoelectrons and on the number of photoelectrons emitted per second, and explain your answers.

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• The maximum kinetic energy is unchanged [1 mark]1
• The number of photoelectrons emitted per second doubles [1 mark]1
• Twice the intensity means twice as many photons per second, each with the same energy hf; each photon releases at most one electron [1 mark]1

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

For the teacherUntuk guru: AnalysingMenganalisis, hightinggi difficultyaras hightinggi, Chapter 7, standard 7.3.4

TextbookBuku teks p.ms. 241

25Choose 1Pilih 1

QuestionSoalan 25

Rajah 5 menunjukkan sebuah lampu kerja berjisim 8.0 kg yang digantung daripada siling sebuah bengkel dengan satu kabel. Lampu itu ditarik ke tepi oleh sebuah spring mengufuk yang pemalar springnya 250 N m⁻¹ sehingga kabel membuat sudut 40° dengan garis tegak. Lampu itu berada dalam keadaan rehat.Diagram 5 shows a work lamp of mass 8.0 kg hung from the ceiling of a workshop by a cable. The lamp is pulled sideways by a horizontal spring with a spring constant of 250 N m⁻¹ until the cable makes an angle of 40° with the vertical. The lamp is at rest.

[2 marksmarkah] (a)Namakan tiga daya yang bertindak pada lampu itu dan nyatakan daya paduan pada lampu.Name the three forces acting on the lamp and state the resultant force on the lamp.

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• Weight (gravitational force) acting downwards, tension in the cable (along the cable) and the horizontal force from the spring [1 mark]1
• The resultant force is zero because the lamp is in equilibrium (at rest) [1 mark]1

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

[2 marksmarkah] (b)Hitung daya mengufuk yang dikenakan oleh spring pada lampu itu.Calculate the horizontal force that the spring exerts on the lamp.

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• W = 8.0 × 9.81 = 78.5 N [1 mark]1
• F = W tan 40° = 78.5 × 0.839 = 66 N (65.9 N) [1 mark]1

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[2 marksmarkah] (c)Hitung tegangan dalam kabel.Calculate the tension in the cable.

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• Vertical: T cos 40° = W [1 mark]1
• T = 78.5 ÷ 0.766 = 102 N [1 mark]1

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[2 marksmarkah] (d)Hitung pemanjangan spring itu.Calculate the extension of the spring.

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• x = F ÷ k = 65.9 ÷ 250 [1 mark]1
• = 0.26 m (26 cm) [1 mark]1

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[2 marksmarkah] (e)Hitung tenaga keupayaan kenyal yang tersimpan dalam spring itu.Calculate the elastic potential energy stored in the spring.

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• E = ½Fx = ½ × 65.9 × 0.263 (or ½kx²) [1 mark]1
• = 8.7 J [1 mark]1

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[2 marksmarkah] (f)Daya mengufuk pada lampu itu ditambah. Nyatakan bagaimana sudut kabel dan tegangan dalam kabel berubah, dan terangkan.The horizontal force on the lamp is increased. State how the angle of the cable and the tension in the cable change, and explain.

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• The angle with the vertical increases [1 mark]1
• The tension increases, since T = W ÷ cos θ (T = √(W² + F²)) becomes larger as the angle and the horizontal force increase [1 mark]1

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

[2 marksmarkah] (g)Kabel itu boleh menanggung tegangan maksimum 150 N. Hitung daya mengufuk maksimum yang boleh dikenakan sebelum kabel putus.The cable can withstand a maximum tension of 150 N. Calculate the maximum horizontal force that can be applied before the cable breaks.

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• T² = W² + F², so F = √(150² − 78.5²) [1 mark]1
• = 128 N [1 mark]1

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

[6 marksmarkah] (h)Sebuah bengkel ingin menggantung lampu yang lebih berat dengan selamat menggunakan susunan kabel dan spring yang serupa. Cadangkan tiga ciri reka bentuk dan terangkan setiap satu.A workshop wants to hang a heavier lamp safely using a similar arrangement of cable and spring. Suggest three design features and explain each one.

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Any of these points, up to 6 marksMana-mana isi berikut, sehingga 6 markah

• Cable of steel wire with a larger diameter: a higher breaking tension, so it does not snap when the tension increases2
• Spring made of steel with a larger spring constant and elastic limit: a smaller extension for the same force and it is not permanently stretched2
• Strong brackets bolted to the ceiling and wall (rigid fixing points): the fixings do not pull out under the larger forces2
• Keep the angle of the cable small (a longer cable or a smaller sideways force): the tension T = W ÷ cos θ stays close to the weight2

Any three features, 2 marks each: 1 mark for the feature and 1 mark for the reason.

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

For the teacherUntuk guru: ApplyingMengaplikasi, lowrendah difficultyaras lowrendah, Chapter 1, standard 1.3.3

TextbookBuku teks p.ms. 21

26Choose 1Pilih 1

QuestionSoalan 26

Rajah 6 menunjukkan litar bekalan kuasa sebuah radio bengkel. Empat diod D₁, D₂, D₃ dan D₄ disambung kepada sumber a.u. dan sebuah kapasitor C disambung selari dengan perintang beban R.Diagram 6 shows the circuit of the power supply of a workshop radio. Four diodes D₁, D₂, D₃ and D₄ are connected to an a.c. source and a capacitor C is connected in parallel with the load resistor R.

[2 marksmarkah] (a)Terangkan bagaimana sebuah diod semikonduktor bertindak apabila dipincang depan dan apabila dipincang songsang.Explain how a semiconductor diode behaves when it is forward biased and when it is reverse biased.

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• Forward bias: the diode has a low resistance and allows current to flow [1 mark]1
• Reverse bias: the diode has a very high resistance and does not allow current to flow [1 mark]1

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

[2 marksmarkah] (b)Namakan jenis pembetulan yang dilakukan oleh litar itu.Name the type of rectification done by the circuit.

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• Full-wave rectification [1 mark]1
• Using a bridge of four diodes, both half-cycles of the a.c. are used [1 mark]1

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

[3 marksmarkah] (c)Dengan menamakan diod yang mengalir, terangkan bagaimana arus mengalir melalui R dalam arah yang sama pada kedua-dua setengah kitar.By naming the diodes that conduct, explain how the current flows through R in the same direction in both half-cycles.

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• When L is positive, D₁ and D₄ conduct [1 mark]1
• When R is positive, D₂ and D₃ conduct [1 mark]1
• In both cases the current enters the load at T and leaves at B, so it passes through R in the same direction [1 mark]1

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

[1 markmarkah] (d)Frekuensi bekalan a.u. ialah 50 Hz. Nyatakan frekuensi denyutan pada output sebelum pelicinan.The frequency of the a.c. supply is 50 Hz. State the frequency of the pulses in the output before smoothing.

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• 100 Hz1

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[2 marksmarkah] (e)Terangkan fungsi kapasitor C dalam litar itu.Explain the function of the capacitor C in the circuit.

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• The capacitor charges up while the voltage of the pulse rises [1 mark]1
• It discharges through R when the voltage falls, so the output voltage is smoother (smaller ripple) [1 mark]1

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

[2 marksmarkah] (f)Nyatakan kesan terhadap output jika kapasitor itu digantikan dengan kapasitor berkapasitans yang lebih besar, dan terangkan.State the effect on the output if the capacitor is replaced by one of larger capacitance, and explain.

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• The output is smoother (the ripple is smaller) [1 mark]1
• A larger capacitor stores more charge and discharges more slowly, so the voltage falls less between pulses [1 mark]1

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

[2 marksmarkah] (g)Nyatakan satu kelemahan pembetul gelombang separuh berbanding litar dalam Rajah 6.State one disadvantage of a half-wave rectifier compared with the circuit in Diagram 6.

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• It uses only one half of each cycle (the output is zero for half the time) [1 mark]1
• The average output voltage / power is smaller and the output is less smooth [1 mark]1

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

[6 marksmarkah] (h)Seorang juruteknik ingin menambah baik bekalan kuasa supaya memberi voltan d.c. yang lebih licin, boleh membekalkan arus yang lebih besar dan dilindungi daripada kerosakan. Cadangkan tiga pengubahsuaian dan beri sebab bagi setiap satu.A technician wants to improve the power supply so that it gives a smoother d.c. voltage, can supply a larger current and is protected from faults. Suggest three modifications and give a reason for each.

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Any of these points, up to 6 marksMana-mana isi berikut, sehingga 6 markah

• Use a capacitor of larger capacitance: it discharges more slowly so the ripple is smaller2
• Use diodes with a higher current rating and thicker connecting wires: they carry a larger current without overheating2
• Add a fuse in series with the a.c. supply: it melts and cuts off the current if the current is too large2
• Use a transformer with a laminated soft iron core: less energy is lost as heat so more power reaches the load2

Any three modifications, 2 marks each: 1 mark for the modification and 1 mark for the reason.

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

For the teacherUntuk guru: EvaluatingMenilai, mediumsederhana difficultyaras mediumsederhana, Chapter 5, standard 5.2.2

TextbookBuku teks p.ms. 186

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QuestionSoalan 27

Sebuah fotosel disinari dengan cahaya monokromatik berpanjang gelombang 300 nm. Rajah 7 menunjukkan graf arus foto melawan beza keupayaan antara anod dan katod. [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]A photocell is illuminated with monochromatic light of wavelength 300 nm. Diagram 7 shows the graph of photocurrent against the potential difference between the anode and the cathode. [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]

[2 marksmarkah] (a)Apakah yang dimaksudkan dengan keupayaan penahan?What is meant by stopping potential?

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• The minimum reverse potential difference between the anode and the cathode [1 mark]1
• that stops even the most energetic photoelectrons from reaching the anode (the photocurrent becomes zero) [1 mark]1

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[1 markmarkah] (b)Daripada graf, tentukan keupayaan penahan.From the graph, determine the stopping potential.

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• 1.5 V (accept 1.4 to 1.6 V)1

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[2 marksmarkah] (c)Hitung tenaga kinetik maksimum fotoelektron dalam joule.Calculate the maximum kinetic energy of the photoelectrons in joules.

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• Kmax = eVs = 1.60 × 10⁻¹⁹ × 1.5 [1 mark]1
• = 2.4 × 10⁻¹⁹ J [1 mark]1

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[2 marksmarkah] (d)Hitung laju maksimum fotoelektron.Calculate the maximum speed of the photoelectrons.

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• ½mv² = Kmax, so v = √(2 × 2.4 × 10⁻¹⁹ ÷ 9.11 × 10⁻³¹) [1 mark]1
• = 7.3 × 10⁵ m s⁻¹ [1 mark]1

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[2 marksmarkah] (e)Hitung frekuensi cahaya itu dan tenaga satu foton.Calculate the frequency of the light and the energy of one photon.

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• f = c ÷ λ = 3.00 × 10⁸ ÷ 300 × 10⁻⁹ = 1.0 × 10¹⁵ Hz [1 mark]1
• E = hf = 6.63 × 10⁻³⁴ × 1.0 × 10¹⁵ = 6.63 × 10⁻¹⁹ J [1 mark]1

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[2 marksmarkah] (f)Hitung fungsi kerja logam katod itu.Calculate the work function of the cathode metal.

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• W = hf − Kmax = 6.63 × 10⁻¹⁹ − 2.4 × 10⁻¹⁹ [1 mark]1
• = 4.2 × 10⁻¹⁹ J (2.6 eV) [1 mark]1

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[2 marksmarkah] (g)Hitung frekuensi ambang dan panjang gelombang ambang logam itu.Calculate the threshold frequency and the threshold wavelength of the metal.

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• f₀ = W ÷ h = 4.23 × 10⁻¹⁹ ÷ 6.63 × 10⁻³⁴ = 6.4 × 10¹⁴ Hz [1 mark]1
• λ₀ = c ÷ f₀ = 3.00 × 10⁸ ÷ 6.4 × 10¹⁴ = 4.7 × 10⁻⁷ m (470 nm) [1 mark]1

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[2 marksmarkah] (h)Keamatan cahaya itu digandakan dengan panjang gelombang yang sama. Huraikan bagaimana graf itu berubah.The intensity of the light is doubled with the same wavelength. Describe how the graph changes.

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• The saturation current doubles (to 4.0 μA) [1 mark]1
• The stopping potential stays at −1.5 V because Kmax depends on the frequency, not the intensity [1 mark]1

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[2 marksmarkah] (i)Terangkan mengapa arus foto tidak bertambah lagi melebihi 2.0 μA apabila beza keupayaan dinaikkan.Explain why the photocurrent does not increase beyond 2.0 μA when the potential difference is increased.

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• All the photoelectrons emitted each second are already reaching the anode [1 mark]1
• The number emitted per second depends only on the intensity (number of photons per second) of the light [1 mark]1

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[3 marksmarkah] (j)Cahaya sangat terang berpanjang gelombang 600 nm disinarkan pada katod itu. Tentukan, dengan pengiraan, sama ada arus foto mengalir.Very bright light of wavelength 600 nm is shone on the cathode. Determine, by calculation, whether a photocurrent flows.

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• f = c ÷ λ = 3.00 × 10⁸ ÷ 600 × 10⁻⁹ = 5.0 × 10¹⁴ Hz [1 mark]1
• This is less than the threshold frequency 6.4 × 10¹⁴ Hz (photon energy 2.1 eV < 2.6 eV) [1 mark]1
• No photocurrent flows: each photon has too little energy and one electron absorbs one photon only, so brightness does not help [1 mark]1

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For the teacherUntuk guru: EvaluatingMenilai, hightinggi difficultyaras hightinggi, Chapter 7, standard 7.3.3

TextbookBuku teks p.ms. 239

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QuestionSoalan 28

Rajah 8 menunjukkan sebuah lif hidraulik di sebuah bengkel kereta. Luas keratan rentas omboh input ialah 20 cm² dan luas keratan rentas omboh output ialah 400 cm². Lif itu mengangkat sebuah kereta berjisim 1 800 kg.Diagram 8 shows a hydraulic lift in a car workshop. The cross-sectional area of the input piston is 20 cm² and the cross-sectional area of the output piston is 400 cm². The lift raises a car of mass 1 800 kg.

[2 marksmarkah] (a)Nyatakan prinsip Pascal.State Pascal's principle.

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• The pressure applied to an enclosed fluid [1 mark]1
• is transmitted equally (without loss) to every part of the fluid [1 mark]1

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[3 marksmarkah] (b)Hitung tekanan dalam minyak apabila kereta itu diangkat dalam keadaan seimbang. Abaikan jisim omboh.Calculate the pressure in the oil when the car is held in equilibrium. Ignore the mass of the piston.

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• Weight of car = 1 800 × 9.81 = 17 658 N [1 mark]1
• P = F ÷ A = 17 658 ÷ 0.040 [1 mark]1
• = 4.4 × 10⁵ Pa [1 mark]1

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[2 marksmarkah] (c)Hitung daya input yang diperlukan untuk menahan kereta itu.Calculate the input force needed to hold the car.

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• F = PA = 4.4 × 10⁵ × 20 × 10⁻⁴ [1 mark]1
• = 8.8 × 10² N (883 N) [1 mark]1

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[2 marksmarkah] (d)Omboh input ditolak ke bawah sejauh 0.40 m. Hitung jarak yang dinaikkan oleh omboh output.The input piston is pushed down through 0.40 m. Calculate the distance that the output piston rises.

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• Volume of oil moved is the same: A₁d₁ = A₂d₂ [1 mark]1
• d₂ = 20 × 0.40 ÷ 400 = 0.020 m (2.0 cm) [1 mark]1

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[3 marksmarkah] (e)Tunjukkan, dengan pengiraan kerja yang dilakukan, bahawa sistem ini mematuhi prinsip keabadian tenaga.Show, by calculating the work done, that this system obeys the principle of conservation of energy.

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• Work done on the input piston = 883 × 0.40 = 353 J [1 mark]1
• Work done on the car = 17 658 × 0.020 = 353 J [1 mark]1
• The two are equal: the force is multiplied but the distance is reduced by the same factor (no energy is created) [1 mark]1

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[2 marksmarkah] (f)Terangkan mengapa minyak, dan bukan udara, digunakan sebagai bendalir dalam sistem hidraulik.Explain why oil, and not air, is used as the fluid in a hydraulic system.

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• Oil (a liquid) is incompressible [1 mark]1
• Air can be compressed, so part of the input motion would only squeeze the air and the pressure would not be transmitted effectively to the output piston [1 mark]1

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[6 marksmarkah] (g)Sebuah bengkel ingin membina lif hidraulik yang lebih selamat dan lebih berkesan. Cadangkan tiga ciri reka bentuk dan terangkan setiap satu.A workshop wants to build a hydraulic lift that is safer and more effective. Suggest three design features and explain each one.

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Any of these points, up to 6 marksMana-mana isi berikut, sehingga 6 markah

• Use a larger ratio of output area to input area: a smaller input force is needed to lift the same car2
• Use a one-way (check) valve between the pump and the cylinder: the car stays raised when the pump piston is lifted for another stroke2
• Use thick-walled steel cylinders and strong seals: they withstand the high pressure without bursting or leaking2
• Fit a pressure relief valve and a controlled release valve: the pressure cannot become dangerously high and the car is lowered slowly and safely2
• Bleed all air out of the oil: air is compressible and would make the lift soft and less efficient2

Any three features, 2 marks each: 1 mark for the feature and 1 mark for the reason.

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For the teacherUntuk guru: CreatingMencipta, mediumsederhana difficultyaras mediumsederhana, Chapter 2, standard 2.4

TextbookBuku teks p.ms. 63

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