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L.35/36

Pre-Leaving Certificate Examination, 2010

Physics

Marking Scheme

Ordinary Pg. 2 Higher Pg. 21

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Dublin Examining Board

Pre-Leaving Certificate Examination, 2010

Physics

Ordinary Level Marking Scheme (400 marks)

SECTION A (120 marks) Answer three questions from this section. Each question carries 40 marks. 1. In an experiment to investigate the relationship between the force applied (F) to a trolley and its

acceleration (a), a number of different forces were applied to the trolley and the acceleration was measured for each force. The table shows the measurements recorded.

F / N 1.0 1.5 2.0 2.5 3.0 3.5 a / m s–2 2.42 3.63 4.84 6.05 7.26 8.47

(i) Draw a labelled diagram of the apparatus used in this experiment. (12) ** Deduct 2m if no labels included. – Labelled diagram (4 × 3m) Diagram must include: – trolley / rider – runway / air track – ticker timer / photo gates (and timer) / powder track / any method of measuring time – means of applying force, e.g. weights / elastic bands / (Newton) balance (ii) How could the force have been applied to the trolley and measured? (3m + 4m) (7) – force applied using hanging masses attached to trolley / other method of accelerating object – weights marked with N / force sensor / spring balance / electronic balance (iii) Plot a graph on graph paper of force applied against acceleration (put acceleration on the

X-axis). (12) – Labelled graph (4 × 3m) – label axes correctly – plot three points correctly – plot another three points correctly – straight line ** Allow max. 6m if graph paper is not used. ** Allow max. 6m if acceleration is on the Y-axis. (iv) Calculate the slope of the graph to determine the mass of the accelerating object. (3 × 3m) (9) – slope formula – correct substitution of two points from graph into slope formula – correct answer = 0.413 kg

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2. In an experiment to measure the specific latent heat of vaporisation of water, a student took a number of measurements before adding steam to water in a calorimeter. The student also took a number of measurements after adding the steam.

(i) Draw a labelled diagram of the apparatus used in the experiment. (12) – Labelled diagram (4 × 3m) Apparatus must include: – water and insulated calorimeter – thermometer – steam generator – delivery tube / steam trap ** Deduct 2m for each label omitted or incorrect. (ii) List two measurements that the student took before adding the steam to the water. (6) Any 2: (2 × 3m) – mass of calorimeter + water // – mass of calorimeter // – temperature of water (iii) What two measurements did the student take after adding the steam? (2 × 3m) (6) – mass of calorimeter + water + steam – temperature of water after steam added (iv) Explain how the student used these measurements to find the mass of the steam. (3 × 3m) (9) – mass of water + calorimeter after adding steam – minus – mass of water + calorimeter before adding steam (v) Why is a steam trap used in the experiment? (3m + 4m) (7) – so that only steam – not hot water enters water

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3. In an experiment to verify Snell’s Law, a student measured the angle of incidence (i) and the angle of refraction (r) for light entering a glass block. The student repeated this for a number of different angles of incidence. The following data was collected:

i / degree 20 30 40 50

r / degree 13 20 26 31

sin i

sin r

(i) Draw a labelled diagram of the apparatus used to measure the angle of incidence and angle

of refraction. Show the angle of incidence and angle of refraction on your diagram. (15) – Labelled diagram (5 × 3m) Apparatus must include: – glass block – pins / raybox / laser – sheet of paper / protractor – angle of incidence – angle of refraction ** Deduct 2m for each label omitted or incorrect. (ii) Complete the table and plot a graph of sin i against sin r. (18)

i / degree 20 30 40 50 r / degree 13 20 26 31 sin i 0.342 0.5 0.642 0.766 sin r 0.225 0.342 0.438 0.515

– Completed table and graph (6 × 3m) Apparatus must include: – 3 correct values for sin i – 3 correct values for sin r – axes labelled correctly – 2 points plotted correctly – another two points plotted correctly – straight line (iii) How does this graph verify Snell’s Law? (3m + 4m) (7) – sin i sin r – straight line through origin

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4. In an experiment to investigate the relationship between current and potential difference for a copper sulfate solution, the following data was collected:

V / V 1.0 1.5 2.0 2.5 3.0 3.5 4.0 4.5 5.0

I / A 0.70 1.05 1.40 1.75 2.10 2.45 2.80 3.15 3.50

(i) What instruments are used to measure the current and potential difference values? (2 × 3m) (6) – ammeter – voltmeter (ii) Draw the circuit diagram used to collect the above data. (15) – Labelled diagram (5 × 3m) Apparatus must include: – battery / power supply – method of varying current – ammeter in series – voltmeter in parallel – copper electrodes in copper sulfate solution (iii) Draw a graph of current against potential difference (put potential difference on the

X-axis). (12) – Labelled graph (4 × 3m) – label axes correctly – plot four points correctly – plot another four points correctly – straight line (iv) Does copper sulfate solution obey Ohm’s Law? Explain your answer. (2m + 2m + 3m) (7) – graph shows straight line – through origin – therefore I V proving Ohm’s law

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SECTION B (280 marks) Answer five questions from this section. Each question carries 56 marks. 5. Answer any eight of the following parts (a), (b), (c), etc. ** Marks awarded for the eight best answers. (a) State Newton’s law of universal gravitation. (4m + 3m) (7) – force product of masses //

– inversely proportional to the square of the distance between them / 1/d2 (b) What is the pressure due to the water on a person at a depth of 20 m

below the surface? (4m + 3m) (7) (p = ρgh, density of water = 1000 kg m–3, acceleration due to gravity = 9.8 m s–2) – p = ρgh = (1000) (9.8) (20) – = 196000 Pa (c) What is a lever? (4m + 3m) (7) – a rigid body free to rotate – about a fixed point / fulcrum (d) What is a thermometric property? (4m + 3m) (7) – property that changes – with temperature (e) Sketch a representation of a transverse wave and indicate on your sketch the wavelength

of the wave. (4m + 3m) (7) – sine wave – correct wavelength (f) How does your eye focus objects at different distances onto

your retina? (4m + 3m) (7) – lens in eye getting fatter / thinner (using ciliary muscles) – changes focal length of lens (g) Why do people who work with integrated circuits connect themselves to earth? (4m + 3m) (7) – if not grounded, static electricity – can discharge through integrated circuit, destroying it

221

d

mGmF

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(h) How much heat per second is given off from a coil of wire of resistance 100 Ω carrying a current of 5 A? (4m + 3m) (7)

– P=I2R = (5)2(100) – = 2500 J every sec. (i) In nuclear decay, what is the type of radiation emitted in the following reaction? (4m) (7) – -decay / fast-moving electron What is the nature of this radiation? (3m) – electron (j) What is the function of the moderator in a nuclear fission reactor? (4m + 3m) (7) – slows down neutrons – not captured by U-238 / available for fusion

XAcRa228

89

228

88

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6. State the principle of conservation of momentum. (3 × 3m) (9) – total momentum before collision – = total momentum after collision – with no external force present An astronaut wants his rocket, which is travelling at constant speed in space, to go faster.

How can he use the principle of conservation of momentum to achieve this? (3 × 3m) (9) – fire gas out the back – momentum of gas backwards – is equal to the momentum gained by rocket forwards What has greater momentum, an elephant of mass 4000 kg travelling at 2 m s–1 or a cheetah of

mass 200 kg travelling at 45 m s–1? (6m + 3m) (9) – p = mV – = (4000) (2) = 8000 kg m s–1 – = (200) (45) = 9000 kg m s–1 ** Allow 6m for one correct answer, 3m for second correct answer. In a railway station, a train with five carriages travelling at 10 m s–1 has its engine switched off. It then

joins with a single carriage at rest. If each carriage had a mass of 3000 kg, calculate: (i) the momentum of the train before the train and single carriage combine; (2 × 3m) (6) – total mass before = 15000 kg – = (15000) (10) = 150000 kg m s–1 (ii) the velocity of the six carriages after they combine. (3 × 3m) (9) – total mass = 18000 kg – 150000 = (18000)v – v = 8.33 m s–1 The engine is then switched on and a force of 36 kN is applied. (iii) What is the acceleration of the train? (4m + 3m) (7) – 36000 = (18000)a – = 2 m s–2 (iv) How far will the train travel in the first minute? (4m + 3m) (7) – s = ut + ½ at2 (0m) – = (8.33) (60) + ½ (2) (60)2 – = 4100 m (p = mv , s = ut + ½ at2, F = ma)

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7. State the laws of reflection of light. (4 × 3m) (12) – angle of incidence – is equal to the angle of refraction – incident ray, refracted ray and normal – all in the same plane Using a diagram, describe an experiment that demonstrates the laws of reflection. (12) – Labelled diagram (4 × 3m) Apparatus must include: – raybox / laser / pins – striking a plane mirror at an angle – mark on paper the angle of incidence and angle of refraction – shows that angle of incidence is equal to the angle of refraction A person who looks through a periscope in a submarine can see

boats on the surface of the water. Draw a ray diagram showing how light passes through a periscope. (9)

– Labelled diagram (3 × 3m) Apparatus must include: – 2 mirrors tilted at 45 to direction of ray – correct path of ray – correct direction of arrow What is the difference between concave and convex mirrors? (2 × 3m) (6) concave – mirror coating on inside of curved surface / reflecting surface caves in at the

centre convex – mirror coating on outside of curved surface / reflecting surface bulges out at

the centre ** Accept diagrams with back of mirror / reflecting surface / direction of light clearly indicated. A light bulb with the manufacturer’s name written on the surface of the bulb is placed 30 cm from a

concave mirror of focal length 20 cm. The height of the name is 1 cm. (i) How far from the mirror must a screen be placed so that a sharp image of the manufacturer’s

name is obtained? (2 × 3m) (6) – , – v= 60 cm

vuf

111 v

1

30

1

20

1

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(ii) What height is the image of the name on the screen? (3m + 4m) (7) – = 3 – height of image = magnification × height of the object = 3 cm (iii) As well as being different heights, what else would be different about the image compared to

the object? (4) Any 1: (4m) – inverted // – less bright // etc. ** Accept any valid difference.

(vuf

111 , u

vm )

30

60u

vm

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1.2 μF 6 Ω

24 V

8. Define potential difference. (2 × 3m) (12)

– work per unit / Q

WV

– charge / notation Define the unit of capacitance. (2 × 3m) – 1 farad charge of 1 C – potential of 1 V Describe an experiment that shows that capacitors store energy. (4 × 3m) (12) – connect battery across capacitor until fully charged – disconnect battery – connect capacitor to bulb – bulb lights for short period of time A circuit is set up with a parallel plate capacitor connected in parallel with a resistor as shown. (i) What is the current flowing through the resistor? (2 × 3m) (6) – V=IR, 24 = I(6) – I = 4 A (ii) What charge is on the capacitor when it is fully charged? (2 × 3m) (6)

– V

QC , 1.2×10–6 = Q/(24)

– 2.88 × 10–5 C (iii) Why, after a short period of time, is there no current flowing through the capacitor? (2 × 3m) (6)

– when the capacitor becomes fully charged – the current stops flowing / it acts as an insulator (iv) What difference would be observed if the battery was replaced with an a.c. supply? (2 × 3m) (6)

– capacitor would constantly charge and discharge – current continues to flow State two uses of capacitors. (8)

Any 2: (2 × 4m) – camera flashes // – smooth out variations in d.c. // – tuning radios // – delay circuits // etc.

(V = IR, V

QC )

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9. State Faraday’s Law of electromagnetic induction. (3 × 3m) (9) – electromotive force (emf) / voltage induced / produced – proportional to rate of change – of magnetic flux Describe an experiment that demonstrates Faraday’s law of electromagnetic induction. (4 × 3m) (12) – coil connected to galvanometer or other suitable indicator – move magnet slowly towards coil – small deflection – move magnet quickly towards coil, large deflection Two coils are placed next to each other as shown in the diagram. (i) Explain why a current does not flow in the secondary coil

when a d.c. voltage supply is connected to the primary coil. (2m + 2m + 3m) (7)

– constant current in primary – produces constant magnetic field – therefore no voltage / current produced in secondary (ii) Why does a current flow in the secondary coil when an a.c. voltage supply is connected to

the primary coil? (2m + 2m + 3m) (7) – continuously changing current in primary – produces changing magnetic field – therefore a voltage / current produced in secondary (iii) A 230 V a.c. supply is connected to the input coil. If there are 200 turns in the primary coil

and 1000 turns in the secondary coil, what is the size of the voltage produced at the secondary coil? (2 × 3m) (6)

– s

p

s

p

N

N

V

V ,

1000

200230 sV

– Vs = 1150 V (iv) If the resistance of the secondary coil is 100 Ω, calculate the current flowing in the

secondary coil. (2 × 3m) (6) – V = IR, 1150 = I(100) – I = 11.5 A (v) What power is produced in the secondary coil? (2 × 3m) (6) – P = VI, P = (1150) (11.5) – = 13225 W

INPUTCOIL

OUTPUTCOIL

PRIMARY

SECONDARY

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(vi) What is the name given to the device that can produce an increased or decreased voltage in the secondary coil compared to the primary coil, using the method in the diagram? (3m) (3)

– transformer

(s

p

s

p

N

N

V

V , V = IR, P =VI)

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10. What are X-rays? (2 × 3m) (12) – electromagnetic / high energy – waves / rays / radiation State two properties of X-rays. Any 2: (2 × 3m) – can ionise the material they pass through // – can penetrate material // – are not deflected in electric or magnetic fields // – produce fluorescence in some materials like zinc sulphide // – can produce interference patterns and undergo diffraction // – can affect photographic emulsions // – can cause photoemission // etc. (i) Name the parts labelled A, B and C in the

diagram of the X-ray tube above. (3 × 3m) (9) A – cathode / filament / heater B – target / tungsten C – coolant / oil / water (ii) Explain how a beam of electrons is emitted at A and then travels from A to B. (6m + 3m + 3m) (12) – thermionic emission / heat at A – high voltage between A and B – accelerates electrons (iii) How are X-rays produced when electrons strike the target at B? (3 × 3m) (9) – kinetic energy lost by electrons – is converted to – photon energy / radiation energy of X-rays (iv) Explain why voltage V2 needs to be much larger than V1. (2 × 3m) (6) – small voltage only needed to light / heat the filament – large voltage needed to accelerate electrons to produce energy for X-rays (v) What safety feature is missing from the X-ray tube above? (4m) (4) – lead shielding (vi) What health hazard could result from not having this safety feature? (4) Any 1: (4m) – cancer // – skin burns // – genetic defects // etc.

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11. Read this passage and answer the questions below. The guitar has been around in one form or another for

centuries, but the principals behind how it works have been in use in musical instruments ever since the first cave man pulled the string on his bow and noticed a sound. A guitar is simply an expansion of this idea to create a series of strings tied taut, and placed over a large hollow body to resonate the sounds. There are various styles of guitars all which have different sounds, the two most common acoustic models being the classical, or nylon string guitar, and the folk, or steel string guitar. Each guitar has its own unique sound, depending on the materials of the strings, the composition of its body, as well as the air between them.

There are six strings on a guitar, each with an open note of E2, A2, D3, G3, B3 and E4, each with

corresponding frequencies of 82 Hz, 110 Hz, 147 Hz, 196 Hz, 247 Hz and 330 Hz. These frequencies represent the root tone of each string. What makes a guitar sound the way it does is the overlay of various frequencies on each string, or the overtones present. These patterns of overtones and their strengths are what make a guitar sound different from, for example, a trombone. In addition, this exact pattern of overtones is impossible to repeat, and so every instrument will sound different, as in fact will every note as subtle changes from the exact location on the string where it is plucked to the force, direction, and the age of the string both off and on the guitar all have small effects on the sound.

(Adapted from http://ffden-2-phys.uaf.edu/211.web.stuff/billington/main.htm) (a) What is meant by resonance? (2m + 2m + 3m) (7) – if the frequency applied – is equal to the natural frequency of object – the object vibrates (b) What is meant by the fundamental frequency of a stretched string? (4m + 3m) (7) – the lowest frequency a string will vibrate at – when plucked

(c) Guitar strings form standing waves when plucked. Draw a diagram representing standing

waves. (7) – Labelled diagram (4m + 3m) Apparatus must include: – standing wave pattern with at least 2 nodes – standing wave pattern with at least 1 antinode (d) How does the fundamental frequency of the stretched string depend on its length? (4m + 3m) (7) – f 1/l – frequency is inversely proportional to – notation / length (e) Name two other factors that the fundamental frequency depends on. (4m + 3m) (7) – tension – mass per unit length

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(f) If the fundamental frequency of a stretched string is 532 Hz and the speed of the wave in the string is 480 m s–1, what is the wavelength of the wave set up on the string? (4m + 3m) (7) (c = fλ) – c = fλ, 480 = 532λ – λ = 0.902 m (g) From part (f), what is the distance between the two nearest nodes? (4m + 3m) (7) – distance = ½ λ – 0.451 m (h) What properties of a musical note depend on the (i) amplitude and the (ii) frequency of

the wave? (4m + 3m) (7) (i) – loudness (ii) – pitch

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12. Answer any two of the following parts (a), (b), (c), (d). (a) Define energy. (2 × 3m) (6) – ability – to do work While climbing a hill, walker A (mass 60 kg) climbs up a path that follows the steepest slope

on the hill. It takes him 20 minutes to walk the path. Walker B (mass 80 kg), walks on a longer, less steep path that takes 45 minutes to walk. The vertical height of the hill is 400 m.

Calculate: (i) the weight of each walker; (2 × 3m) (6) A – W = (60) (9.8) = 588 N B – W = (80) (9.8) = 784 N (ii) the work done by each walker in climbing the hill; (3 × 3m) (9) A – work = mgh = (588) (400) – = 235200 J B – work = (784) (400) = 313600 J (iii) the power developed by both walkers in climbing the hill. (2m + 2m + 3m) (7) A – P = W/t = (235200) / (20 × 60) – P = 196 W B – P = W/t = (313600) / (45 × 60) = 116.15 W

(W = mg, W = mgh, t

WP , acceleration due to gravity = 9.8 m s–2)

(b) Distinguish between conduction, convection and radiation. (4 × 3m) (12) – transfer of heat conduction – passing from molecule to molecule convection – moving molecules carry heat / circulating currents in a fluid radiation – electromagnetic waves / no medium required Why would you locate the boiler in the heating system of

a house at the lowest level of the house? (2 × 3m) (6) – when heated water becomes

less dense / convection currents – water rises

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What is the U-value of a material? (3m + 2m + 2m) (10) – heat conducted per second – through each 1 m2

– for each temperature difference of 1 K or 1 C ** Accept answer that a good insulator has a low U-value for 3m. In a certain part of a house, the U-value dropped from 3.8 W m–2 K–1 to 1.2 W m–2 K–1.

Could you suggest a change to the house that may have caused this? (3m) ** Any change that improved insulation. (c) What is a magnetic field? (2 × 3m) (6) – area where a magnetic pole – experiences a force Draw a diagram showing the shape and direction of the Earth’s magnetic field. (9) – Diagram (3 × 3m) – correct shape – not in line with geographical north – arrows towards north Explain how sailors, for centuries, have used the Earth’s magnetic field

for navigation. (3 × 3m) (9) – compass – contains a small magnet – aligns with earth’s magnetic field What did Oersted discover when he was experimenting with a current-carrying wire? (4m) (4) – the magnet needle moved / current has a magnetic effect

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(d) What is the photoelectric effect? (2 × 3m) (6) – emission of electrons – when electromagnetic radiation / light shines on metal When UV light is incident on zinc, the photoelectric effect is observed but is not observed when

visible light shines on the same metal. Explain why this is so. (3 × 3m) (9) UV – frequency / energy high enough – to emit electrons visible – frequency / energy too low to emit electrons What would happen if the intensity of both light sources was increased? (2 × 3m) (6) – UV - more electrons emitted – visible - still no electrons emitted The energy of a photon of light is 4.8 × 10–19 J.

What is the frequency of that photon? (4m + 3m) (7) – 4.8 × 10–19 = (6.6 × 10–34)f – f = 7.27 × 10–14 Hz (E = hf, Planck constant = 6.6 × 10–34 J s)

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Dublin Examining Board

Pre-Leaving Certificate Examination, 2010

Physics

Higher Level Marking Scheme (400 marks)

SECTION A (120 marks) Answer three questions from this section. Each question carries 40 marks. 1. In an experiment to verify Boyle’s Law, the following data for the volume V of a gas and the

corresponding values for pressure P were recorded:

P / kPa 100 150 200 250 300 350 400

V / cm3 12.2 8.2 6.0 4.9 4.1 3.5 3.0

Describe, with the aid of a labelled diagram, how the above data was obtained. (4 × 3m) (12) – trapped mass of gas – pressure gauge – method of varying pressure – method of measuring volume Draw a suitable graph using the above data to verify Boyle’s Law. (18)

P / kPa 100 150 200 250 300 350 400

V / cm3 12.2 8.2 6.0 4.9 4.1 3.5 3.0

1/V / cm–3 0.08 0.12 0.16 0.20 0.24 0.28 0.33

– Graph (6 × 3m) – calculate 1/volume – labelled axes, P against 1/V – 6 points correctly plotted – straight line with good fit – Boyle’s Law – P 1/V / pressure inversely proportional to volume – by straight line through the origin State two quantities that must be maintained constant during the experiment. (2 × 3m) (10) – mass of gas – temperature Explain how one of these quantities is kept constant. Any 1: (4m) – mass of gas is sealed or enclosed // – gas is allowed time to settle after adjusting pressure

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2. In an experiment to measure the specific latent heat of vaporisation of water, steam was added to water in an insulated copper calorimeter.

The following data were recorded: mass of calorimeter = 36.8 g mass of calorimeter + water = 83.3 g initial temperature of water = 10 C temperature of steam = 100 C final temperature of water = 26 C mass of calorimeter + water + steam = 84.6 g Use a labelled diagram to explain how the data was collected. (12) – Labelled diagram (2 × 3m) Diagram must include: – method of producing steam connected to steam trap and steam passed into water

in insulated copper calorimeter – temperature measured in calorimeter before and after steam was added (thermometer

shown in diagram) ** Deduct 1m for any part missing. – Explanation (2 × 3m) – mass of calorimeter, add water, mass of calorimeter + water measured at

beginning of experiment – mass of calorimeter + water + steam measured at end of experiment Calculate a value for the specific latent heat of vaporisation of water. The specific heat capacity

of copper is 390 J kg–1 K–1 and the specific heat capacity of water is 4180 J kg–1 K–1. (18) – msl + mscw(∆θ)s = mwcw(∆θ)w + mccc(∆θ)c (6m) – ms = 1.3 g, mw = 46.5 g, (3m) – (∆θ)s = 74 C, (∆θ)w = (∆θ)c = 16 C (3m) – (1.3)l +(1.3)(4180)(74) = (46.5)(4180)(16) + (36.8)(390)(16) (3m) – l = 2.26 × 106 J kg–1 (3m) ** No marks if ml missing. ** Deduct 3m for any other element missing up to maximum 6m. Why would the student have started with the temperature of the water below room temperature?

(2 × 3m) (6) – heat gained from the surroundings while the water is below room temperature – is balanced by heat lost to the surroundings when water is above room temperature If the student added more steam to the water, how would this have affected the accuracy of the result? (4) Any 1: (4m) – greater rise in temperature of the water, less percentage error in temperature rise value // – if too much steam added, large rise in temperature may lead to loss in heat to

surroundings // etc. ** Accept any other valid argument.

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3. A student investigated the relationship between the fundamental frequency (f) of a stretched string and its length (l). The string was of length 1.1 m and had a mass of 0.26 g.

The table shows the data recorded by the student.

f / Hz 256 325 353 389 412 486

l / cm 80.4 63.3 58.5 52.9 49.8 42.5

How did the student ensure that the wire was vibrating at the fundamental frequency and not at the second or higher harmonic? (4 × 3m) (12)

– starting with the maximum length string – reduce the tension to a very low value (to ensure frequency is low) – increase the tension until resonance occurs with tuning fork / frequency generator, etc. – keep the tension constant for remainder of the measurements, decreasing length until resonance with

increasing frequency tuning fork / frequency generator, etc. Draw a suitable graph to show the relationship between the fundamental frequency of a stretched

string and its length. How does your graph demonstrate this? (6 × 3m) (18)

f / Hz 256 325 353 389 412 486

l /cm 80.4 63.3 58.5 52.9 49.8 42.5

1/l / m–1 1.24 1.58 1.71 1.89 2.01 2.35

** No penalty if values given in cm–1 or m–1. – calculate 1/length – labelled axes, f against1/l – 6 points correctly plotted – straight line with good fit – relationship is f 1/l – frequency inversely proportional to length – verified by straight line through the origin Use your graph to calculate the tension of the string. (10) – Graph (4m + 2m + 2m + 2m) – slope of graph = 206 Hz m (accept 202 – 210) – – μ = (0.26 × 10–3)/1.1 = 2.36 × 10–4 kg m–1 – T = 40 N

T

fl2

1slope

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4. A student performed an experiment to verify Joule’s Law. A coil was placed in a fixed mass of water in a calorimeter. A fixed current I was passed through the coil for five minutes and a temperature ∆ was recorded. This was repeated for a number of different currents.

I / A 1.0 1.5 2.0 2.5 3.0 3.5 4.0 4.5 5.0

∆ / C 2.1 4.7 8.4 13.1 18.9 25.7 33.6 41.3 50.1

Draw the circuit and other pieces of apparatus required to collect this data. (3 × 3m) (9) – power source and heating coil – ammeter in series and method of varying current – coil in insulated container of water with thermometer ** Deduct 2m for each label omitted or incorrect. During the five minutes that the current passed through the coil, give a reason why the current would

change even if there was no adjustment to the apparatus. (2 × 3m) (6) – as current flows temperature of coil increases, increasing the resistance – the current through the coil decreases Using the above data, draw a suitable graph on graph paper and explain how your graph verifies

Joule’s Law. (6 × 3m) (18)

I / A 1.0 1.5 2.0 2.5 3.0 3.5 4.0 4.5 5.0

/ C 2.1 4.7 8.4 13.1 18.9 25.7 33.6 41.3 50.1

I2 / A2 1.0 2.25 4.0 6.25 9.0 12.25 16.0 20.25 25.0

– calculate I2 – labelled axes, ∆ against I2

– 6 points correctly plotted – straight line with good fit (except for last two points) – Joule’s Law is ∆ I2 / power current2 – verified by straight line through the origin From the graph, which two points are least accurate? (3m) (7) – first 2 points / last 2 points ** Award 0m if a valid explanation not given. Explain why this is so. (4m) – loss in heat due to larger temperature rise / small results means larger % error

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SECTION B (280 marks) Answer five questions from this section. Each question carries 56 marks. 5. Answer any eight of the following parts (a), (b), (c), etc. ** Marks awarded for the eight best answers. (a) State the principle of conservation of momentum. (4m + 3m) (7) – total momentum before an interaction between two bodies = total momentum after – assuming / if no external force (b) A ball is thrown upwards from the edge of a 200 m high cliff. It just misses the cliff on

the way back down and strikes the sea 10 s later. What was the initial velocity of the ball? (2m + 3m + 2m) (7) (acceleration due to gravity = 9.8 m s–2)

– s = –200 m / s = ut – ½gt2 – –200 = u(10) – 0.5(9.8)(10)2

– u = 29 m s–1 ** Deduct 1m if units omitted or incorrect. (c) State Archimedes’ Principle. (4m + 3m) (7) – upthrust / buoyancy / (apparent) loss in weight (in fluid) equals – weight of the fluid / liquid / water displaced (d) Find the position of an image formed in a concave lens of focal length 20 cm, when the object

is placed 30 cm from the lens. (3m + 2m + 2m) (7)

– vuf

111

– v

1

30

1

20

1

– v = –12 cm ** Deduct 1m if units omitted or incorrect. (e) Sound waves travel towards a wall and are reflected. This gives rise to a standing wave where the

distance between a node and its nearest antinode is 0.5 m. What is the frequency of the sound wave? (3m + 4m) (7) (speed of sound in air = 340 m s–1) – λ = 2 m – f = c/λ=170 Hz ** Deduct 1m if units omitted or incorrect. (f) List four forms of electromagnetic radiation in order of increasing frequency. (7) – each correct form of electromagnetic radiation (4 × 1m) – correct order (3m)

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(g) State two factors on which the capacitance of a parallel plate capacitor depends. (7)

Any 2: (4m + 3m) – distance between plates // – medium between plates // – common cross-sectional area (h) How does the resistance of a conductor vary with length and cross-sectional area? (3m + 4m) (7) – resistance proportional to length – resistance inversely proportional to cross sectional area (i) What is a p-type semiconductor? (3m + 4m) (7) – semiconductor doped with impurity – produces extra positive holes available for conduction (j) Write down an expression for pair annihilation. (4m + 3m) (7) – e+ + e- – 2hf or 2γ or Draw the circuit for the bridge rectifier. (4m + 3m) (7) – correct shape of rectifier – correct direction of diodes

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6. Define (i) centripetal acceleration and (ii) angular velocity. (12) (i) centripetal acceleration (2 × 3m) – acceleration towards the centre – of an object travelling in a circle (ii) angular velocity (2 × 3m) – change in angle (in radians) / angular displacement / = /t – per second / per unit time / notation Derive the relationship between angular velocity and linear velocity v. (3 × 3m) (9) – = s/r // = vt/r – /t = s/rt // = t – = v/r // v = r

A simple pendulum consisting of a bob of mass 200 g at the end of a string of

length 2 m, oscillates freely through an angle of 30. Why is the motion of the bob not simple harmonic motion? (4m) (4)

– motion only simple harmonic if it oscillates through a

small angle / angle less than 5 Calculate: (24) (i) the maximum vertical distance through which the bob passes; (2 × 3m) – 2 cos 15 = 1.93 m – vertical height through which it passes = 2–1.93 = 0.07 m (ii) the maximum speed at which the bob is travelling; (2 × 3m) – ½ mv2 = mgh/v2 = 2gh – v = 1.17 m s–1 (iii) the net force acting on the bob when it is travelling at maximum speed; (2 × 3m) – F = mv2/r – = (0.2)(1.17)2/2 = 0.137 N (iv) the tension in the string when it is travelling at maximum speed. (2 × 3m) – net force = T – mg – 0.137 = T – (0.2)(9.8), T = 2.1 N After a period of time, the amplitude of the oscillation reduces to 4. How long does it take the

bob to travel from a position of maximum velocity to the nearest position of zero velocity? (4m + 3m) (7) (acceleration due to gravity = 9.8 m s–2)

– g

lT 2 = 2.84 s

– max velocity to zero velocity = T/4 = 0.71 s ** Deduct 1m if units omitted or incorrect.

30º

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7. Define (i) critical angle (ii) total internal reflection. (12) (i) critical angle (2 × 3m) – going from more dense to less dense medium – angle of incidence when angle of refraction = 90 (ii) total internal reflection (2 × 3m) – when angle of incidence > critical angle – all light is reflected back into denser medium

Derive the formula that relates the refractive index to the critical angle of a substance. (3 × 3m) (9) – mna (refractive index medium going to air) = sin i/sin r // n1 sin i = n2 sin r – = sin C/sin 90 at critical angle // n1 sin C=na sin 90 = 1.1

– anm = nm = 1/mna = 1/sin C // n1 = Csin

1

Describe an experiment that demonstrates total internal reflection. (3 × 3m) (9) – starting with small angle of incidence on surface where light goes from block to air - observe

refracted and reflected ray from surface – as angle increases beyond critical angle – refracted ray vanishes and only reflected ray observed In binoculars, the distance the light has to travel is increased by using two prisms to bend light through

180 twice as shown. The prisms have one angle of 90 and two angles of 45. The light enters the prism by striking one of the sides at 90 as shown. (26)

(i) Find the minimum refractive index of the material of the prism so that all the light that passes

through the objective lens also passes through the eyepiece lens. (2 × 3m) – C = 45 – n = 1/sin C = 1.41 (ii) If the material of the prisms has a refractive index of 1.5, find the reduction of speed of light

going from air into the prism. (2 × 3m) – v = c/n = (3 × 108)/1.5 = 2 × 108 m s–1 – difference in speed = 1 × 108 m s–1

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When the binoculars were dropped, one of the prisms moved so that light struck the first surface at 85.5 as shown.

(iii) Determine if the reflections of light on the inner surfaces of the

prism are still total internal reflection. (2m + 3m + 3m + 3m + 3m) (speed of light in a vacuum = 3 × 108 m s–1) – for first face i = 90 – 85.5 = 4.5 – n = sin i/sin r, r = 3 – for second face, i = 90 – (180 – 45 – 87) = 42 – critical angle for material of prism, C = sin–1(1/n) = sin-1(1/1.5)=41.8 / total internal

reflection as i > C – for third face i = 48 > C, total internal reflection again.

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8. State Coulomb’s Law. (2 × 3m) (12) – force proportional to product of charges / correct formula – inversely proportional to square of distance between charges / correct notation Define electric field strength. (2 × 3m) – force / E=F/Q – per unit charge / correct notation Describe how an isolated metal sphere can be charged by induction. (4 × 3m) (12) – hold charged rod near metal sphere – attract opposite charge to side of rod / repels like charge – ground sphere with finger or otherwise, charge on sphere same as on rod goes to ground – remove finger and remove rod, charge opposite to that of rod remains A small metal sphere (A) on an insulated stand is charged with a positive charge

of 4 μC. A second small metal sphere (B) of mass 200 g is suspended from a fixed point 8 cm vertically above the first sphere by a string of length 10 cm. The second metal sphere has a positive charge Q and is at rest so that the line between their centres is horizontal.

Draw the electric field lines around both spheres. (2 × 3m) (6) – correct shape – correct direction of field lines Calculate the tension on the string suspending sphere B. (3 × 3m) (9) – angle between vertical and string θ, cos θ = 8/10 = 0.8 – T cos θ = mg – 0.8T = (0.2)(9.8), T = 2.45 N ** Deduct 1m if units omitted or incorrect. Show that the size of the charge on sphere B is 0.15 μC to two decimal places. (3m + 3m + 3m + 2m) (11)

– distance between spheres x2 = 102 – 82, x = 6 cm

– 2

0

21BA

4sin

d

QQfT

– 212

6

)06.0)(109.8(4

Q)104()6.0)(45.2(

2

– Q2 = 1.49 × 10–7 C (0.15 μC to one decimal place) Find the magnitude and direction of the electric field strength at sphere B. (2 × 3m) (6) – E = F/Q = 1.47/(1.49 × 10–7) = 9.9 × 106 N C–1 (or F m–1) // E = Q1 / 4πε0d

2 = 9.9 × 106 N C–1 – direction is away from sphere A (permittivity of free space = 8.9 × 10–12 F m–1)

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9. Distinguish between photoelectric and thermionic emission. (4 × 3m) (12) – photoelectric emission is the emission of electrons from surface of a metal – when light of a suitable frequency is incident on it – thermionic emission is the emission of electrons – from hot metal surface Draw a labelled diagram of a cathode ray tube and explain how a bright spot can be formed at any

point on the screen of the tube. (15) – Labelled diagram (5 × 3m) Apparatus must include: – heated cathode and anode with hole in middle connected by H.V. – evacuated tube with fluorescent screen – X-Y plates or magnetic field coils – electron beam on screen - loss of kinetic energy of electrons converted to light – electron beam position by deflection due to electric or magnetic field In a cathode ray tube, electrons are accelerated across a potential difference of 5 kV, over a distance

of 10 cm. Assuming that the electrons have zero initial speed: (23) (i) What is the energy gained by an electron? (2 × 3m) – E = eV – E = (1.6 × 10–19)(5000) = 8 × 10–16 J ** Deduct 1m if units omitted or incorrect. (ii) What speed is reached by the electron? (3m + 3m + 2m) – eV = ½ mv2 – 8 × 10–16 = ½ (9.1 × 10–31)v2

– = 4.19 × 107 m s–1 ** Deduct 1m if units omitted or incorrect. (iii) What is the average force exerted on the electron? (3 × 3m) – v2 = u2 + 2as / (4.19 × 107)2 = 02 + 2a(0.1) – average acceleration a = 8.8 × 1015 m s–2

– average force, F = ma = (9.1 × 10–31)(8.8 × 1014) = 8×10–15 N ** Deduct 1m if units omitted or incorrect. Why are much higher voltages required in X-ray tubes than in cathode ray tubes? (2 × 3m) (6) – higher kinetic energy of electrons required – to produce high energy of X-ray photon (mass of electron = 9.1 × 10–31 kg, charge on electron 1.6 × 10–19 C)

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10. Answer either part (a) or part (b). (a) What are alpha particles? (2 × 3m) (6) – helium nuclei – fast moving Ernest Rutherford used alpha particles to discover the existence of the nucleus. Describe his experiment, including his observations and conclusions. (4 × 3m) (12) – gold foil bombarded with particles (described or diagram) – position of particles after passing through detected by fluorescent

screen – most passed through undeflected, small number deflected,

very small number rebounded – conclusion - most of positive charge concentrated in very small volume

Ra22688 undergoes alpha decay to form radon (Rn).

Write down the nuclear equation for the decay. (4m + 3m) (7)

– HeRnRa 42

22286

22688

** Allow maximum 2m if no numbers given. If the mass of a radium nucleus is 3.753152 × 10–25 kg, the mass of a radon nucleus is

3.686602 × 10–25 kg and the mass of an alpha particle is 6.646322 × 10–27 kg, find the energy released in the alpha decay in MeV. (4 × 3m) (12)

– mass of products = 3.75306522 × 10–25 kg – loss in mass = 8.678 × 10–30 kg – E=mc2 = (8.678 × 10–30)(3 × 108)2 = 7.81 × 10–13 J – (7.81 × 10–13)/(1.6 × 10–19) = 4.88 × 106 eV = 4.88 MeV If the ratio of the mass of the products of the decay is equal to the ratio of their mass numbers,

find the kinetic energy of both products after the decay. (12) – MRnVRn = MV , – M/MRn = VRn/V = 4/222 (3m)

– 222

4Rn

ααα21

RnRnRn21

V

V

VVM

VVM (3m)

– ½MRnV2Rn + ½MV

2α = 7.81 × 10–13

– ½MRnV2Rn = 4/226 (7.81 × 10–13) = 1.38 × 10–14 J (3m)

– ½MV2α = 222/226 (7.81 × 10–13) = 7.67 × 10–13 J (3m)

2010 L.35/36_MS 33/40 Page 33 of 40

Momentum is conserved when applied to alpha decay, but originally appeared not to be conserved when applied to -decay. How was this problem resolved? (4m + 3m) (7)

– neutrino was proposed / discovered – to account for missing momentum (speed of light in a vacuum = 3 × 108 m s–1, charge of electron = 1.6 × 10–19 C) (b) State the principle on which the d.c. motor is based. (2 × 3m) (6) – current carrying conductor in a magnetic field – experiences a force Why is it necessary to use a split-ring commutator in a d.c. motor? Explain how a split-ring

commutator carries out this function. (4 × 3m) (12) – changes direction of current twice every revolution – allows coil to continue rotating in one direction – carbon brushes rubbing off side of split ring carries current to coil – so that current alternately enters each side of ring as it rotates Draw a labelled diagram of a moving-coil galvanometer and explain its operation. (15) – Labelled diagram (3 × 3m) ** Apparatus must include: – coiled spring and coil wrapped around former – magnet – pointer and scale ** Deduct 1m for each missing label. – Explanation (3 × 2m) – when current passes through the coil, the coil rotates in a magnetic field – force opposed by torque of spring – angle through which it rotates is proportional to size of current A moving-coil galvanometer has a full-scale deflection of 6 mA and an internal resistance

of 40 Ω. Draw the circuit, including the value of the resistor required, that would convert

the galvanometer to: (23) (i) an ammeter of full scale deflection 1 A; (4 × 3m) – current through resistor = 0.994 A – V = (0.006)(40) = 0.994R – R = 0.24 Ω – diagram showing resistor connected in parallel to galvanometer (ii) a voltmeter of full scale deflection 10 V. (6m + 2m + 3m) – 0.006R + 0.006(40) = 10 – R = 1625.7 Ω – diagram showing resistor connected in series with galvanometer

2010 L.35/36_MS 34/40 Page 34 of 40

11. Read the following passage and answer the accompanying questions. The sun contains many elements but the most abundant by far is hydrogen. The visible layers (the

photosphere and the chromosphere) are the only part of the sun that is cool enough for hydrogen to exist in its atomic form and it is here that we can see the absorption and emission spectra for hydrogen.

It is helpful to think of a hydrogen atom as a small 'solar system' with the heavy nucleus as the

'sun' in the middle. This particular solar system has only one planet orbiting – i.e. a single electron. Due to the laws of quantum physics, this electron can only orbit the nucleus in specific orbits which are given a number n.

Hydrogen can absorb and emit in the ultraviolet region of the spectrum (the Lyman series) but the

emissions and absorptions we see in the visible part of the spectrum are the Balmer series and occur when electrons jump from and fall to the n=2 orbit.

(Diagram and article from http://www.solarobserving.com/halpha.htm)

(a) What are line emission spectra? (3m + 4m) (7) – light emitted (from some elements) due to electrons – falling from higher to lower energy levels (b) In the line spectrum shown above, what is the energy difference between the n=2 and n=3

levels for the H alpha line? (3m + 4m) (7) – E = hc/λ – E = (6.6 × 10–34)(3 × 108)/(656 × 10–9) = 3.02 × 10–19 J (c) A spectrometer was used to collect the above spectrum. How many H alpha lines could

be observed if a diffraction grating with a grating constant of 2.5 μm was used? (4m + 3m) (7) – n= d/λ = (2.5 × 10–6)/(656 × 10–9) = 3.81, max order = 3 – number of lines seen = 2n = 2(3) = 6 – (central fringe not included as no dispersion

would be observed) ** Deduct 1m if central fringe included. (d) What is the next most abundant element, after hydrogen, in the sun? Name the reaction that

produces this element. (4m + 3m) (7) – helium – nuclear fusion

2010 L.35/36_MS 35/40 Page 35 of 40

(e) Write a formula for this reaction. (7) ** Reactants (3m) ** Products (4m)

– n10

32

21

21 HeHH

** Allow maximum 3m if no numbers given. (f) The sun emits approximately 4 × 1023 W as electromagnetic radiation. What is the loss in

mass of the sun each day due to electromagnetic radiation being emitted? (2m + 3m + 2m) (7) – E = mc2 – per second, m = (4 × 1023)/(3 × 108)2 = 4.44 × 106 kg – per day m = 3.84 × 1011 kg (g) What average percentage of the sun’s emitted energy is incident on a rectangular football field

of length 160 m and width 100 m? Assume it strikes the field at right angles. (2m + 3m + 2m) (7) – area = 160 × 100 =16000 m2 – energy per second on field = (1.35 × 103)(16000) = 2.16 × 107 J – percentage = [(2.16 × 107)/(4 × 1023)] × 100 = 5.4 × 10–15 % (h) What conclusion could be drawn from the fact that the H alpha line from a hydrogen line

spectrum produced by a distant star was observed to be of lower wavelength than that produced by the sun? (3m + 4m) (7)

– lower wavelength corresponds to higher frequency – by Doppler effect star is moving towards us (Planck constant = 6.6 × 10–34 J s; speed of light in a vacuum = 3 × 108 m s–1;

solar constant = 1.35 kW m–2)

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12. Answer any two of the following parts (a), (b), (c), (d). (a) What are the conditions for equilibrium for a rigid body? (2 × 3m) (6) – vector sum of the forces = 0 – sum of moments about any point = 0 Peter and James carry a light bar of length 1.5 m at each end. A large hanging basket of flowers

of mass 60 kg is suspended from the bar at some point between the two men. Peter is capable of supporting a weight of 250 N. They carry the basket up a slope of 10 for a distance of 50 m in 1 minute.

If the bar is held horizontal and the weight of the bar can be ignored, calculate: (22) (i) the minimum weight that James must be capable of supporting; (3 × 3m) – forces down = forces down – Fjames + 250 = 60(9.8) = 588 N – Fjames = 338 N (ii) the maximum distance from James that the basket can be placed; (2 × 3m) – moments about James 588x = 250(1.5) – x = 0.637 m (iii) the power imparted by James while carrying the basket. (2m + 3m + 2m) – height = 50 sin 10 = 8.68 m – work = force up × height = 338 × 8.68 = 2933.84 J – power = work/time = 2934/60 = 48.9 W ** Deduct 1m if units omitted or incorrect. (b) What is a thermometric property? (2 × 3m) (6) – property that changes measurably – with temperature Give the definition of the Celsius scale in terms of the SI unit of temperature. (4m) (4) – t / C = T / K – 273.15 A student placed an uncalibrated mercury thermometer and a thermistor in a bath of ice and

water. The student noted the length of the mercury column and the resistance of the thermistor. The student then placed the thermometer and the thermistor in a beaker of boiling water and again noted the length of the mercury column and the resistance of the thermistor. These mercury column lengths and resistance values were used separately to determine the temperature of a beaker of lukewarm water. The student obtained a reading of 40 C based on the length of the mercury column. A reading of 55 C was obtained from the reading of the resistance on the thermistor. (18)

(i) Explain the reason for the difference found in the two readings. (2 × 3m) – thermometric properties vary differently – between 0 C and 100 C

2010 L.35/36_MS 37/40 Page 37 of 40

(ii) How would the student ensure that thermometers based on mercury columns and thermistors would give the same temperature reading? (2 × 3m) (6)

– use mercury thermometer as standard – measure resistance at number of different temperatures and plot graph (iii) Sketch a graph showing the variation of resistance with temperature for a thermistor for a

metal coil. (2 × 3m) (6) – correct curved downward shape – decrease rapidly with increasing temperature (c) Derive an equation for the effective total resistance in a circuit with three resistors connected

in parallel. (4 × 3m) (12) – ITotal = I1 + I2 + I3

– R

VI

– 321P R

V

R

V

R

V

R

V

– 321P

1111

RRRR

An alarm, a resistor and a light-dependent resistor (LDR) are connected as shown in the

diagram. When it is dark, the LDR has a resistance of 1 MΩ. The alarm needs a current of 0.2 A to operate.

Calculate: (16) (i) the current flowing through the alarm when it is dark; (3m + 2m + 3m)

– 6

21P 10

1

200

1111 RRR

, RP = 200 Ω

– total R = 200 + 50 = 250 Ω – I = V/R = 24/250 = 0.096 A (ii) the resistance of the LDR, if the minimum amount of light that causes the alarm to sound, is

incident on the LDR. (3m + 2m + 3m) – total resistance = V/I = 24/0.2 = 120 Ω – RP = 120 – 50 = 70 Ω

– 200

1

70

11

LDR

R

, RLDR = 107.7 Ω

2010 L.35/36_MS 38/40 Page 38 of 40

(d) Distinguish between mutual and self induction. (3 × 3m) (9) – changing magnetic field in a coil induces – mutual - an EMF in a nearby coil – self - an opposing EMF in the same coil Why is mutual or self induction not possible using a d.c. battery as the power supply?

(2 × 3m) (6) – d.c. battery produces constant current, therefore produces constant magnetic field – need changing magnetic field to produce an EMF In a television set, the filament has a power rating of 30 W. A 230 V a.c. supply is connected

to the primary coil of a transformer. The primary coil consists of 2000 turns. If the filament is connected to a secondary coil consisting of 52 turns and the transformer is 90% efficient, calculate the current flowing in both coils. (4m + 3m + 3m +3m) (13)

– P

S

P

S

N

N

V

V = 2000

52

230S V

, VS = 5.98 V

– in secondary I = P/V = 5.02 A – in primary P = 30/0.9 = 33.3 W – I = P/V = 33.3/230 = 0.145 A

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