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Physics for Cambridge O Level Answer Book Pauline Anning I Published by PEAK PUBLISHING LIMITED London UK © Peak Publishing Limited 2016 Text © Pauline Anning 2016 First published 2016 ISBN: 1 845222 14 8 All rights reserved. No part of this publication may be reproduced, copied or transmitted save with written permission or in accordance with the provisions of the Copyright, Design and Patents Act 1988, or under the terms of any license permitting limited copying issued by the Copyright Licensing Agency, 90 Tottenham Court Road, London W1P 9HE. Cambridge International Examinations bears no responsibility for the example answers to questions which are contained in this publication. The answers to exam-style questions and to Cambridge past paper questions in this book have been written by the author. Any person who does any unauthorised act in relation to this publication may be liable to criminal prosecution and civil claims for damages. Printed in Pakistan II Contents 1 Physical Quantities, Units and Measurement ..............................................2 2 Kinematics ....................................................................................................3 3 Dynamics ......................................................................................................4 4 Mass, weight and density .............................................................................5 5 Deformation and turning effects of forces ...................................................6 6 Pressure ........................................................................................................9 7 Energy sources and transfer of energy .......................................................10 8 Transfer of thermal energy .........................................................................11 9 Temperature ...............................................................................................13 10 Thermal properties of matter ....................................................................15 11 Kinetic model of matter .............................................................................16 12 Waves .........................................................................................................18 13 Light ........................................................................................................... 20 14 Electromagnetic Spectrum .........................................................................22 15 Sound .........................................................................................................23 16 Magnetism and electromagnetism ............................................................24 17 Static electricity ......................................................................................... 26 18 Current electricity ......................................................................................27 19 D.C. Circuits ................................................................................................29 20 Practical electricity .....................................................................................31 21 Electromagnetism and electromagnetic induction ....................................32 22 Introductory electronics and electronic systems .......................................34 23 Radioactivity .............................................................................................. 35 24 The nuclear atom .......................................................................................37 1 Chapter 1: Physical Quantities, Units and Measurement Self – Assessment Answers 1. b 2.c 3. a 4.c 5.a 6. b 7. c 8.d 9.c 10. d 11. a) measuring tape b) metre rule c) micrometer screw gauge 12. a)mass b)force c)length d)area e)volume f) speed or velocity 13. a) Quantity mass Scalar Vector force displacement acceleration distance speed 14. a) The resultant of two vectors is the single vector that has the same effect as both the original vectors acting together. b) Resultant velocity of 1.8 m s–1 at 34° west of north c) Resultant force of 7.9 N at 19° to the 6.0 N force. 15. a) This is a safety precaution to stop the stand toppling over. b) 1 complete oscillation is from the centre position to one side of the motion, back through the centre to the other side of the motion and back to the centre. c) 0.01 s d) The reaction time of the student is about 0.2 seconds. If she measures the time for 10 oscillations the reaction time is small compared to the total time measured. If she measures the time for 1 oscillation only the reaction time is large compared to the total time measured, so the error in the final result is greater. 2 e) The student should add her 3 measured times and divide by 3 to give a mean time for 10 oscillations. She should divide this by 10 to give the period for 1 oscillation. Chapter 2: Kinematics Self – Assessment Answers 1.b 2.d 3. b 4.b 5.a 6.b 7.a 8.c 9.b 10.b 11. a)b) ii) ii) i) i) distance iii) velocity iii) time time 12. a) i) 2 m/s2 ii) The change in speed is unlikely to be exactly the same during each second. b) i) 30 velocity in m/s time in s 13. a) b) 14.a) b) c) d) 120 132 ii) Area under graph up to 132 s. 10 m/s i) 48 m ii) 5 m/s2 Speed is distance travelled per unit time. Velocity is speed in a given direction. Acceleration is the change in speed divided by the time taken for the change. The ball will accelerate downwards at 9.81 m/s2. 3 15.a) Measure the distance with a tape along a stretch of road as straight and as long as possible, mark the distance clearly at the side of the road. One student should stand next to the first mark and wave a flag as the front of a car passes the mark. A second student should stand next to the second mark. When the first student waves the flag the second student should start a stopwatch. When the front of the car passes the second mark the second student stops the stopwatch. b) The students should stand next to the road, not on the road. They should wear bright clothing. They should do their investigation on a day with good visibility, not when it is dark or raining. c) If the students carry out their investigation on a school day at a time when students are entering and leaving the school cars are likely to be travelling fairly slowly. At the weekend or during a part of the day when students are in school there are likely to be more cars travelling faster. Chapter 3: Dynamics Self – Assessment Answers 1.d 2.a 3. c 4.d 5.d 6.d 7.c 8.b 9.b 10.a 11. a) If body A exerts a force on body B, then body B exerts and equal and opposite force on body A. b) Rocket Engine Thrust Gases pushed backward Engine pushed forward 12. The gas exerts an equal and opposite force on the rocket. Lift Constant Speed Drag Thrust Weight 4 13.a) 4 m/s2 b) 2 400 N 14.a) As he accelerates drag increases, resultant force downwards decreases, so acceleration decreases. Eventually drag becomes equal to weight, resultant force is zero, so travels at steady speed. b) 700N upwards c) Large increase in surface are when parachute opens, so drag increases causing a resultant force upwards in the opposite direction to the motion. 15.a) They are directly proportional. If you double the speed the thinking distance doubles. b) i) tiredness, drugs, alcohol ii) poor condition of; road, brakes, road surface icy/wet roads 16.a) electrostatic force b) towards the nucleus c) it would decrease Chapter 4: Mass, weight and density Self – Assessment Answers 1.c 2.b 3. c 4.d 5.c 6.d 7.c 8.b 9.c 10.d 11. Volume of water displaced = 5.5 cm3 Density of granite = 15.6 / 5.5 = 2.84 g/cm3 12.a)Jupiter b)Mercury c) Venus or Uranus d) i)150 × 9.8 = 1470 N ii) Mass would remain the same while weight would increase iii)150 × 14.1 = 2115 N 13.a) Mass is the amount of matter an object contains. Weight is the force that an object exerts downwards due to the pull of gravity. b) Volume = πr2h i) π × 102 × 40 = 4000 π = 12566 cm3 ii) π × 0.102 × 0.40 = 0.004 π = 0.012566 m3 c)i) Mass = 650 × 0.012566 = 8.2 kg ii) Weight = 8.2 × 9.8 = 80.4 N 5 14. Accounts should include details of the following: The mass of the pebble is measured on a balance in grams. A known volume of water is placed in a measuring cylinder. The pebble is dropped into the water and totally submerged. The combined volume of the water and pebble is taken. The volume of the pebble in cubic centimetres is found by subtracting the volume of the water from the combined volume of water and pebble. The density of the pebble, in g/cm3, is found by dividing the mass of the pebble by its volume. 15. Density Mass Length Width Height Volume Metal 3 g/cm3 g cm cm cm cm Calcium Magnesium Nickel Platinum Silver Tin 76.50 83.04 213.6 171.6 472.5 262.44 5 8 4 2 5 6 5 6 3 2 3 3 2 1 2 2 3 2 50 48 24 8 45 36 1.53 1.73 8.90 21.45 10.50 7.29 Chapter 5: Deformation and turning effects of forces Self – Assessment Answers 1.a 2.d 3.b 4.d 5.b 6.c 7. b 8.b 9. c 10.d 11. a) Suspend the lamina at one of the holes so that it can move freely and hang a line and bob from the same hole. Mark the position of the line on the lamina. Repeat the procedure using one of the other holes. centre of mass The centre of mass is where the lines cross. b) If the centre of mass is above the base of an object, the object is stable. If the centre of mass is outside the base the object is unstable and will topple over. 6 12.a)Lever b) The edge of the can c)25 × 10 = 250 Ncm d)F × 1 = 25 × 10 therefore F = 250 N e) i) Increase the value of the effort force ii) Use a longer screw driver and exert the same effort force 13.a) Weight = 1.8 × 10 = 18 N b) Moment of the force = 18 × 0.10 = 1.8 Nm c)F × 0.48 = 1.8 therefore F = 1.8 / 0.48 = 3.75 N d) Adding a metal weight to the base lowers the centre of mass. The lamp would need to topple further before the centre of mass moved outside the base. 14.a) Load /N Length of spring /mm Extension /mm 0.0 0.5 1.0 1.5 2.0 2.5 3.0 20 30 39 49 58 68 78 0 10 19 29 38 48 58 b) 60 50 Extension / mm 40 30 20 10 0 0 0.5 1.0 1.5 Load / N 7 2.0 2.5 3.0 c) Gradient = 58 / 3 = 19.3 mm /N. The force per unit extension is given by the inverse of the gradient which is 1/19.3 = 0.052 N/mm d) extension limit of proportionality load 15.a) 0.02 kg b) 0.2 x 8 = 1.6 Ncm anticlockwise c) Equal in size / magnitude and opposite in direction d) Value of d / cm Anticlockwise moment about Z / Ncm 0 0.0 2 0.8 4 1.6 6 2.4 8 3.2 10 4.0 Clockwise moment about Z / Ncm 2.0 1.6 1.2 0.8 0.4 0.0 e) 4.0 anticlockwise moments Moment (Ncm) 3.0 2.0 1.0 0 clockwise moments 0 2 4 6 8 10 d (cm) f) At the intersection of the lines where d = 3.3 cm. At this point the anticlockwise and clockwise moments are equal. 8 Chapter 6: Pressure Self – Assessment Answers 1.b 2.b 3.c 4.d 5.d 6.b 7.d 8. A 9. B 10.c 11.a) b) c) 12.a) b) c) d) 13. a) b) c) 14. a) b) c) d) e) 100 000 N 4 times greater i) The pressure inside the tyre is greater than atmospheric pressure. ii) No. Air only comes out of the punctured tyre until the pressure in the tyre becomes equal to atmospheric pressure. This is the pressure due to the atmosphere 550 kPa An equal pressure pushes outwards from his lungs due to the air in them i) 0.045 g ii) As the diver rises the pressure decreases. Less nitrogen dissolves in the blood. If the diver rises quickly his body doesn’t have sufficient time to lose the nitrogen gas by breathing it out. i) It moves down the cylinder ii) It increases iii) They move up the cylinder i) Pressure = 100 / 4 = 25 N/cm2 ii) 25 N/cm2 iii) Total force = 2 × 50 × 25 = 2500 N It is transformed into heat energy in the brake discs/pads i)4 × 3 = 12 m3 ii) Pressure = 4800 / 12 = 400 N/m2 or Pa Feet should have a larger surface area in order to spread the weight out more. This would prevent the spaceship sinking too deeply into the ground. Pressure = 400 x 6 = 2400 N/m2 or Pa i)Greater ii)Same iii)Greater Gases are easily compressed but liquids are not 9 15. a) Volume of dry air in Pressure exerted on dry air burette / V cm3 in the burette / p mm Hg 29.6 678 27.2 739 26.4 760 25.4 789 24.1 833 pV / mm Hg cm3 20068.8 20100.8 20064.0 20040.6 20075.3 1 / cm–3 V 0.0338 0.0368 0.0379 0.0394 0.0415 850 850 800 800 pressure / p mm Hg pressure / p mm Hg b) The value of pV is constant within the limits of experimental error. c) i) ii) 750 700 650 750 700 24 25 26 27 28 29 650 0.0300 30 volume of dry air / V cm 0.0350 0.0400 0.0450 1 cm3 V 3 d) The first graph is a curve showing that as p increases V, decreases therefore p is inversely proportional to V. The second graph is a straight line showing that p is directly proportional to 1/V. Chapter 7: Energy sources and transfer of energy Self – Assessment Answers 1.c 2.a 3.b 4. a 5.d 6.d 7.d 8. b 9.b 10.a 10 11. a) e.g. Electric drill b) e.g. Electric cell/battery c) e.g. Combustion d) e.g. Ascending lift e) e.g. Photosynthesis 12. a) Gravitational potential energy ⇒ kinetic energy ⇒ electrical energy b) i)4 × 100 000 = 400 000 watts ii) 400 000 joules c)1 × 10 × 400 = 4000 joules d) The kinetic energy of the water entering the lake will be converted into heat energy. 13. a) Renewable because the Sun will not run out of energy for millions of years b) i) The curved surface allows the mirrors to reflect the Sun’s rays in one place so the heat can be concentrated. ii) The structure follows the motion of the Sun across the sky so the maximum amount of sunlight is gathered each day. 14. a) 160 m2 b) i) 90 × 10 = 900 J ii) 900 × 160 × 0.417 = 60 048 J = 60 kJ c) 60 kW d) 720 000 / 60 = 12 000 e) e.g. Wind turbines do not release pollutant gases to the atmosphere / the energy source for wind turbines is free / renewable 15. a) Gravitational potential energy b) i)1000 × 10 = 10 000 N ii) 10 000 × 7 = 70 000 J c) 200 000 000 × 5 = 1 000 000 000 m3 d) 1 000 000 000 × 70 000 = 70 000 000 000 000 J = 70 000 000 MJ e) 70 000 000 ÷ 20 000 = 3500 MW f) Advantages: e.g. no pollutants released into the environment / barrage would provide another bridge across the river / energy is free / renewable Disadvantages: would alter the area behind the barrage so might lose habitants / feeding grounds for birds / fishermen and other boat users would not have easy access up the river Chapter 8: Transfer of thermal energy Self – Assessment Answers 1.a 2.b 3.b 4. c 5.D 6.d 11 7.a 8. a 9.d 10.b 11. a)Conduction b) Metal / brass / copper c)Convection d) Shiny and light. Heat will be lost from these parts by radiation. Shiny light surfaces are the poorest emitters of heat radiation so least heat will be lost to the surroundings. 12. a) i)Conduction ii) Metal is a good conductor and glass is a poor conductor b) Glass is a poor conductor so the coffee will stay hot longer Plastic is a poor conductor so it will not burn your hand when you pick up the jug to pour some coffee c) Water around the element gets hot and expands. As the water gets hotter its density decreases and it rises. Colder water falls and is heated in turn. Eventually all of the water is heated. 13. a) Aluminium is a good conductor of heat so his drinks would warm up very quickly. An aluminium box would be heavy to carry. b) i) Polystyrene is a non-metal and a poor conductor. Expanded polystyrene contains many tiny air bubbles and air is also a poor conductor. ii) The aluminium foil reflects heat radiation away from the box. c) e.g. It melts at a convenient temperature i.e. between -20 and 0 0C. Absorbs a large amount of energy when it changes from a solid to a liquid. It is cheap and readily available. d) e.g. They absorb a large amount of heat from the drinks. They can be used every day. 14. a) Method of reducing Cost of Annual saving Time taken to repay heat loss installation on energy installation cost in years Loft insulation £240 £30 8 Draught excluders £35 £5 7 Cavity wall insulation £1300 £65 20 Double glazing £1620 £54 30 b) i) Draught excluders ii) Cavity wall insulation iii) Double glazing c) The foam prevents convection currents in the air between the walls. This reduces the heat transfer from the inner to the outer wall. 15. a) Distance between can and heater / type of can / volume of water / initial temperature b)Radiation 12 c) 40 B 35 A 30 temperature / °C 25 20 15 10 5 0 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 Time / minutes d) The temperature of the water in can B rises more quickly because a matt black surface is a better absorber of heat radiation that a shiny metal surface. e) i) The water in can B would cool down more quickly. ii) A matt black surface is a better emitter of heat radiation than a shiny metal surface. Chapter 9: Temperature Self – Assessment Answers 1.d 2.c 3.b 4. a 5.c 6.b 7.d 8. b 9.a 10.c 11. a) i) This shape magnifies the thread of mercury so it can be seen more easily. ii) The white background makes it easier to read the scale. iii) The function of the thermometer is to measure human body temperature, which is likely to be around 36.9 0C. iv) The constriction prevents the thread of mercury running back into the bulb until the level of mercury has been read off the scale. 13 12. a) & b) 70 M 80 90 100 50 40 30 20 10 0 60 B c) d) 13. a) Liquid e.g. easy to see / responds to temperature change quickly e.g The bore of the tube is too wide: the liquid level would not change very much as the temperature changed. The bore of the tube is not uniform: it is not possible to draw a linear scale between two fixed points. b) Ethanol boils at 78 °C / below 100 °C c) Place thermometer bulb in melting ice and when constant level mark as 0 °C. Place thermometer bulb in steam above boiling water and when constant level mark as 100 °C d) Measure the distance between the two fixed points and divide this into 100 equal parts. Mark the scale in suitable increments. 14. a) Water is a poor conductor of heat. The water around the bulb of the thermometer will be higher than the water in the rest of the beaker. b) Stir the water as it is being heated. c)i) 9 °C and 34.5 °C ii) 34.5 – 9 = 25.5 °C d)i) The temperature rise would be greater ii) The temperature rise would be less 15. a) 10 9 8 Potential difference / mV 7 6 5 4 3 2 1 0 10 20 30 40 50 60 Temperature / 0C 14 70 80 90 100 b) From the graph 66 0C c) From the graph 3.50 mV d) The voltmeter would show a negative potential difference between the two junctions. Chapter 10: Thermal properties of matter Self – Assessment Answers 1.a 2.a 3.C 4. d 5.b 6.b 7.d 8. b 9.a 10.c 11. a) The specific heat capacity of ethanol is less than the specific heat capacity of water. b) The nut expands when it is heated. c) Liquids expand with temperature. In the summer the tank will have the same volume of fuel but a lower mass. d) Water is at its densest at 4 0C so there may be water at the bottom of a pond even when the top is covered by ice. 12. a) 0.000 017 × 5 = 0.000 085 m b) As the temperature increases the copper strip will increase in length more than the iron strip. The top of the bimetallic strip and ratchet will bend to the left. The cog would rotate clockwise so the pointer would show the temperature increase on the scale. c) For any given temperature increase, the length of a strip of aluminium will increase more than an equal strip of copper. If the iron strip was replaced by a strip of aluminium, as the temperature increased the top of the bimetallic strip and ratchet would bend to the right. The cog would rotate anticlockwise, incorrectly showing a fall in temperature on the scale. d) e.g. limited temperature scale. 13. a) 6.5 minutes = 6.5 × 60 = 390 seconds Heat supplied by heater = 50 × 390 = 19 500 J b)i) Temperature increase = 47 – 23 = 24 0C Heat capacity of the metal block = 19 500 / 24 = 812.5 J 0C–1 ii) Specific heat capacity of the metal = 812.5 / 2.125 = 382.35 = 382 J kg–1 0C–1 c) Heat energy supplied = 50 × 60 × 10 = 30 000 J Temperature increase = 30 000 / 812.5 = 36.9 0C Final temperature = 23 + 36.9 = 59.9 0C 15 14. a)0.05 × 334 = 16.70 kJ b)0.05 × 4.2 × (6 – 0) = 1.26 kJ c) 16.70 + 1.26 = 17.96 kJ d) 17.96 = 0.25 × c × (25 - 6) = 0.25 × c × 19 c = 17.96 = 3.8 kJ kg–1 K–1 0.25 × 19 15. a) Time, t / min Temperature, T2 / 0C 0 1 2 3 4 5 6 7 23.2 29.1 35.3 41.4 47.7 54.1 59.4 65.9 Temperature increase, ∆T / 0C 0 5.9 12.1 18.2 24.5 30.9 36.2 42.7 b) 50 Temperature / 0C 40 30 20 10 0 0 1 2 3 4 5 6 7 Time / minutes c) Slope = 42.7 / 7 = 6.1 d) c = 12 000 / 6.1 = 1967 J kg–1 K–1 e)61 0C f) The slope of the line for liquid Y will be less as the temperature will increase less for the same amount of heat energy Chapter 11: Kinetic model of matter Self – Assessment Answers 1.b 2.d 3. c 4.A 5.b 6.c 7.b 16 8.D 9.C 10.d 11. a) b) c) d) e) The molecules move more slowly The size of the molecules does not change The molecules will hit the inside of the test tube less often per unit time The number of molecules does not change The total energy of the particles decreases because the average kinetic energy of the particles increases f) The pressure will decrease because the molecules will hit the walls of the test tube less frequently and with less force 12. Water seeps through the porous clay from the inside to the outside of the pot. The water on the outside of the pot evaporates. Heat energy needed for evaporation. The evaporating water cools the pot which, in turn, cools the remaining water. 13. a) When steam comes into contact with skin it will be cooled and condense. When steam changes state to become water heat is given out. This heat will cause additional damage to skin. b) The boiling point of a liquid varies with external pressure. The boiling point of water is only exactly 100 0C when atmospheric pressure is exactly one atmosphere. c) Gas particles are in constant random motion. Pressure in the tube is the result of gas particles colliding with the inner surface. When gas is heated the particles move more rapidly. The pressure in the tube increases as gas particles hit the surface of the tube more often and with more force. The tube will eventually break. d) Wet clothing dries because water evaporates from it. Spreading washing out on a clothes line provides a much greater surface area than leaving wet clothes in a pile therefore evaporation will take place more quickly. e) Energy is needed to change a solid into a liquid at its melting point. This is called the latent heat of fusion or melting. While the heat energy supplied is used to bring about this change of state there will be no increase in the temperature of the solid. 14. a) Even on a sunny window ledge the temperature would not get as high as the melting point of iodine. b) The iodine must have changed from a solid to a vapour at the bottom of the jar and a vapour to a solid at the top of the jar. c) As a solid the iodine particles are held in a lattice. They can vibrate but cannot change position. As a vapour the iodine crystals are separated from each other and moving very quickly. d) Most solids first become liquids and then gases. They do not change directly from a solid to a vapour or gas. 15. a) i) C or D because A and B would both boil before the melting point of salicylic acid ii) The observer needs to be able to see through it b) i) Salicylic acid melts above the range of thermometer A ii) Thermometer C will give a more accurate reading as there will be more distance between the graduations c) i) Melted at a lower temperature and melted over a wider range ii) Melting over a wide range indicates a substance is impure. Recrystallizing is a method of removing impurities. 17 d) It takes time for heat to pass from the oil to the thermometer. If the oil is heated too quickly it will be hotter than shown on the thermometer Chapter 12: Waves Self – Assessment Answers 1.c 2.c 3. a 4.d 5.b 6.a 7.b 8.a 9.a 10. a 11. a) A wave is a disturbance that travels through matter or space as a series of oscillations, carrying energy. b) i) For a transverse wave the oscillation of the particles is perpendicular to the direction in which the wave travels. wavelength transverse wave ii) For a longitudinal wave the oscillation of the particles is parallel to the direction in which the wave travels. compression expansion wavelength longitudinal wave c) i) e.g. a wave on a rope, a light wave ii) e.g. a wave on a spring, a sound wave 18 12.a) Wavelength on diagram is 0.8 cm. Actual wavelength is 0.8 × 4 = 3.2 cm b) frequency = 24/3.2 = 7.5 Hz c) The waves slow down as they pass over the glass. The wavelength of the waves is reduced, the frequency is unchanged. 13. wavelength crest amplitude trough 14. wavelength compression rarefaction a) & b) 15. incident wavefronts incident waves i r plane reflector normal reflected waves reflected wavefronts c) i = r d) speed = 2.0 × 0.75 = 1.5 cm/s e) frequency and wavelength of the incident waves are the same as the frequency and wavelength of the reflected waves. 19 Chapter 13: Light Self – Assessment Answers 1.b 2.a 3. c 4.a 5.d 6.b 7.c 8.a 9.b 10. b) 11. a) b) angle of incidence = angle of reflection c) The image is laterally inverted and is the same size as the object. It is an equal distance from the mirror as the object. It is upright and virtual. 12.a) r = 15° b) The light will be totally internally reflected because c = 24°. So, the angle of incidence is greater than the critical angle. 13.a) F F diverging lens converging lens b) The principal focus will be close to the lens as the refraction will be greater. 20 14.a) A diverging lens always produces an image that is diminished. observer principal axis F object virtual image b) A diverging lens (concave lens) diverge the rays causing the image to not be erect, but upside down. c) m = 0.125 15.a) The rays of light are focused behind the retina because the lens is not thick enough. (not powerful enough) b) The converging lens refracts the rays, so they can be focused clearly onto the retina. 21 Chapter 14: Electromagnetic Spectrum Self – Assessment Answers 1.b 2.a 3.c 4.c 5.c 6.a 7.c 8.b 9.b 10.d 11. a) red orange yellow green blue indigo violet white light glass prism b) Violet, because it has the highest frequency. 12. a)Red b) They all travel at the same speed (300 000 000 m/s) c) Violet, because it has the highest frequency. 13. a) A continuous range of frequencies and wavelengths b) Gamma rays, X-rays, Ultra-violet. c) Gamma-rays, because they have the highest frequency, so are the most energetic. 14. a) 0.12 m b) Radio wave (or microwave) 15. a) The Gamma ray source is injected into the body, X-ray source is external. Gamma rays can be used to image soft tissue, X-ray mainly used for imaging bones. b) Both gamma rays and X-rays are harmful to living tissue, can cause cells to die, or mutate causing cancer. 22 Chapter 15: Sound Self – Assessment Answers 1.c 2.a 3.d 4.c 5.b 6.a 7.c 8.c 9.a 10.d 11. compression wavelength wavelength rarefac�on 12. a) Sound waves are caused by vibrations. These vibrations must be carried by particles. There are no particles in a vacuum. b) fastest – Solid slowest - Gas c)34m 13. a)D b)A c)C d)D 14. a) 20 – 20 000 Hz b) Sound waves are caused by vibrations. The frequency of the vibration matches the frequency of the sound wave. Different frequency sounds are created by objects vibrating at different frequencies. c) Sound waves are caused by vibrations. The greater the amplitude of the vibration the greater the amplitude of the sound wave, and the louder the sound will be. 15. a) answer > 20 000 Hz. b) Ultrasound waves have a shorter wavelength than audible frequencies. The frequency is higher and they travel at the same speed, so the wavelength must be shorter c) Pre-natal scanning, cleaning, fault finding. 23 Chapter 16: Magnetism and electromagnetism Self – Assessment Answers 1. b 2. d 3.c 4.b 5. a 6. d 7.c 8.a 9.a 10. c 11. a)i)Stronger ii) Weaker iii) No effect b) Switch Open Closed Bulb X On Off Bulb Y off on Bulb Z off on c) Bulb X on – the electromagnet is not strong enough to attract the hammer without the iron core Bulb Y on – the circuit containing the bulb is complete Bulb Z off – wood is not a conductor 12. a) X – insulator such as plastic Y – copper or some other metal for the contacts Z – iron b) When the switch is closed current flows through the coil. The iron core becomes a magnet. The armature is pulled towards the iron core and the contacts close. Current flows through circuit A and the bulb lights. 13. a) When the bell push is pressed the circuit is complete. The electromagnet attracts the soft iron armature. As the armature moves towards the electromagnet the hammer strikes the gong. When the soft iron armature moves the circuit is broken. The springy metal strip pulls the soft iron armature back into place. The circuit is complete and the cycle repeats until the bell push is no longer pressed. b) i) As a varying electrical current ii) Steel would retain some magnetism iii) Aluminium is not a magnetic material so it would not be attracted to the coil and permanent magnet. iv) As the current in the coil varies, so does the magnetic field surrounding the coil. Changes in the combined magnetic fields of the permanent magnet and coil cause the iron diaphragm to move in and out or vibrate. The movement of the diaphragm creates sound waves. 24 14.a)Iron/steel b) The metal bar is pulled in / attracted by the magnet c)i) 1.2 Current / A 1.0 0.8 0.6 0.4 0.2 0.0 0 10 20 30 40 50 60 70 Maximum mass held / g d) 15.a) b) ii) The maximum mass is directly proportional to the current i) 35-36 g ii) 0.92-0.94 A e.g. iron / steel loaded and aluminium / copper / brass remains 1000 900 800 1000 turns of wire Lifting force / N 700 600 500 400 500 turns of wire 300 200 100 0 0 2 4 6 8 10 Current / A c) i) Quadruples the lifting force ii) Quadruples the lifting force d) 50 – 55 N 25 Chapter 17: Static electricity Self – Assessment Answers 1. A 2. d 3.b 4.d 5. D 6. C 7.C 8.c 9.b 10. a 11. a) As a result of friction between the balloon and the sweater, electrons are transferred to the balloon. b) The wall is initially neutral but when the charged balloon is placed near the wall this causes charges on the wall to separate. Negative charge is repelled by the charge on the balloon leaving that small area of the wall positively charged. Opposite charges attract so the balloon is held on the wall. 12. a) Electrons are removed and the insecticide becomes positively charged. b) Each small drop of insecticide formed will carry positive charge. Like charges repel so the positive charge will cause the insecticide to divide and form tiny droplets. The droplets will spread out away from each other as they leave the nozzle forming a very fine mist in the air. This makes it possible to cover a large area and it is less likely that areas of the target crop will remain unsprayed. 13. Air is drawn through the device by an electric fan. As the air passes through the negatively charged grid, dust and any other particles, such as pollen, become negatively charged. As the air passes between the positively charged plates, the negatively charged particles of dust are attracted to them and stick to the surface. Eventually the dust particles fall from the plates into a tray. The dust tray is removed from the device and cleaned at regular intervals. 14. a) A metal comb earths through the person. Electrons can flow to and from earth so no charge builds up on the metal comb. b) Some electronic components are damaged by electrostatic charge. The bag prevents the build-up of charge and so protects the components. 26 c) Electrons are transferred between items of clothing as they rub past each other. Charge builds up on some items. The charge is earthed as the items are pulled apart and the crackling is due to tiny sparks of electricity. d) The surface of the helicopter becomes charged as it flies through the air. Some of the charge is transferred to the person on the rope. e) The screen of a television set becomes charged with static when the set is switched on. This static remains for some time when the set is switched off and attracts particles of dust from the air. 15. a)i)Repulsion ii) Repulsion iii) Attraction b) Like charges repel. As the droplets form they repel each other so the paint spreads out. c) The netting is earthed so it will be at 0 V. This will be negative relative to the paint droplets. d) The paint separates into tiny droplets so that only a thin layer is applied to the wire. The wire attracts the positively charged paint droplets so they are not wasted to the surroundings. Chapter 18: Current electricity Self – Assessment Answers 1. c 2. c 3.a 4.b 5. c 6. c 7.B 8.a 9.A 10. c 11. a) I V ohmic conductor 27 b) As the filament heats up its resistance increases. c) The p.d. on open circuit is the value obtained when a voltmeter is connected across the cell when it is not part of a circuit. The p.d. on closed circuit is the value obtained when a voltmeter is connected across the cell when it is part of a circuit and connected to other components. d) This is due to the internal resistance of the cell. 12. a) b) & c) resistance Line X A V resistor Line Y 0 0.1 current / A 0.2 d) The resistance of the resistor is constant provided its temperature remains constant therefore lines X and Y are horizontal. Line Y is half way between 0 and line x as the total resistance of two identical resistors connected in parallel is only half on one of the resistors. 13. a) i) 3 + 3 = 6 Ω ii) 6 + 6 = 12 Ω iii) 12/3 = 4 Ω b) Current = potential difference / resistance = 6 / 4 = 1.5 A c) Power = current × potential difference = 1.5 × 6 = 9 W 14. a) V A b) Changing the value of the variable resistor c) Potential difference / V 4 3 2 1 0.0 0.02 0.04 0.06 0.08 0.10 0.12 Current / A 28 d) i) 0.034 A ii) 2.6 V e) The ratio is constant because the graph is a straight line 15.d)i)Increases ii) Decreases b) 300 Current / mA 250 200 150 100 50 0 10 20 30 40 50 60 70 80 90 100 Length of wire / cm c) i) 85 mA ii) 18 cm d) i) 300 mA ii) 0.3 A Chapter 19: D.C. Circuits Self – Assessment Answers 1. a 2. b 3.a 4.b 5. d 6. c 7.b 8.d 9.c 10. d 11.a) The potential difference across R1, R2 and R3 is the same. The current in R1 and R2 must be twice that in R3. A1 = A2 = 2 × 120 = 240 mA A3 = 240 + 240 + 120 = 600 mA 29 b) The total resistance R, of R1, R2 and R3 is given by: 1 = 1 + 1 + 1 = 1 + 1 + 1 = 5 R R1 R2 R3 10 10 20 20 R = 20 ÷ 5 = 4 Ω V = I × R = 0.6 × 4 = 2.40 V 12. a)i)Cell ii) Along wires that form complete circuits iii) Switch b) H M c) Yes. The components are connected in parallel so the failure of one component will not affect the supply of electricity to the others. 13.a) Less than Ai, equal to Vi b) Greater than Ai, equal to Vi c) Less than Ai, Less than Vi 14.a) Decreases exponentially b)24 Ω at 20 0C and 6 Ω at 80 0C c) At 20 0C total resistance = 24 + 8 = 32 Ω Current = 9.0 / 32 = 0.28 A At 80 0C total resistance = 6 + 8 = 14 Ω Current = 9.0 / 14 = 0.64 A d) Resistance of T at 40 0C = 16 Ω Let R = total resistance of the resistors connected in parallel 1 = 1 + 1 = 24 R 8 24 128 16 128 = 5.3 Ω R= Current = 9.0 / 5.3 = 1.70 A Note: The maximum current corresponds to the minimum resistance, which is when the temperature is highest. If the temperature varies between 20°C and 80°C, the minimum resistance of the thermistor is 6 Ω (read from the graph). At this temperature if the total resistance is R: 1/R = 1/8 + 1/6 = 14/48 R = 48/14 = 3.4 Ω Maximum current = 9.0 / 3.4 = 2.6 A 30 Chapter 20: Practical electricity Self – Assessment Answers 1. d 2. c 3.d 4.a 5. d 6. a 7.d 8.a 9.b 10. B 11.a) Earthing protects the person using an appliance from receiving a large electric shock in the event of a fault causing the casing of the appliance to become live. b) A fuse protects the wiring inside an appliance and in mains circuits. In the event of a fault, the fuse melts and cuts off the supply of electricity. This prevents the circuit from being damaged. c) Double insulation prevents the wiring inside an appliance from coming into contact with the outer casing so it cannot become live even in the event of a fault. d) A modern socket has a cover over the holes for the live and neutral pins so nothing can be pushed into these holes so there is no chance of receiving an electric shock. The earth pin is larger and longer than the live and neutral pins. When the plug is inserted into the socket the earth pin opens the cover so that the live and neutral pins can slide into their holes. 12.a) To stop shorting across the wires. b) i) On the live wire ii) On the live wire between the supply and the switch c) It is connected to the metal body of the lamp. In the event of a fault resulting in the live or neural wire touching the body, the metal body would become live. Most of the current will flow to earth through the earth wire thus preventing anyone who touches the metal body from receiving a severe electric shock. d) Current = power / potential difference = 60 / 230 = 0.26 A e) Energy used = power × time = 60 × 3 = 180 watt hours = 0.180 kilowatt-hours 13.a)i)Kettle ii) Table lamp b) i) Kinetic energy and some heat and sound energy ii) Heat energy iii) Light and sound and some heat energy c) Units of electricity used = 1500 × 6 / 60 = 150 Wh = 0.15 kWh Cost = 0.15 × 4 = Rs 0.60 31 14.a) Compact fluorescent bulb Power / kW 0.02 Hours of light 8000 Cost of one unit of electricity / Rs 4.00 Cost of electricity per bulb / Rs 640 Number of bulbs needed to give 1 8000 hours of light Cost of one bulb / Rs 1000 Total cost / Rs 1640 Filament bulb 0.10 1000 4.00 400 8 40 3600 b) e.g. High initial cost to replace all of the bulbs in a house; people may not believe the claims made about compact fluorescent bulbs c) Less of the electricity becomes heat / it is more efficient 15.a) i) Yellow/green ii) Earth b) Current = power / potential difference = 720 / 230 = 3.13 A c)i)5A ii) This is the next size above the current used by the heater so it will melt quickest in the event of a malfunction. d) On the low setting switches only S2 is closed so the two heating coils are connected in series. On the high setting only switch S1 and S3 are closed so the heating coils are connected in parallel. Chapter 21: Electromagnetism and electromagnetic induction Self – Assessment Answers 1. a 2. b 3.a 4.d 5. d 6. c 7.a 8.a 9.b 10. b 11.a) When the switch is closed current flows through the primary coil. The soft iron core becomes a magnet. The soft iron core attracts the contact. The circuit is broken. The contact moves back to its original position. The circuit is complete and the cycle repeats. b) Every time the primary coil turns on and off a magnetic field grows and collapses around the soft iron core. This induced a current in the secondary current. If there are 32 many turns of wire on the secondary coil a high voltage would be produced and this would be enough to send a spark across a small gap. 12.a) current wire current b) i) A magnet around the rotor ii) The commutator reverses the direction of the current through the coil every half turn. This reverses the polarity of the magnetic field around the coil so that it is always being repelled by the external magnetic field as it rotates. 13.a) i) Soft iron ii) An alternating magnetic field is produced iii) An alternating current is induced in the secondary coil b) i) The coil are connected in series in the same circuit ii) There is no effect on coil R as the alternating magnetic fields produced by coils P and Q cancel each other out. iii) Some current will leak to earth. The currents in coils P and Q will no longer be the same. A magnetic field will be produced in the core which will induce a current in coil R. The current in coil R will open the mains switch. 14.a) i) 240 V ii) Transformer iii) Step-up transformer b) Secondary turns 4000 = 11 400 11 × 4000 400 Secondary turns = = 110 c) i) Transformers do not work with direct current ii) The amount of electrical energy converted to heat energy as a current passes along a conductor varies according to the square of the current. A step-up transformer increases the voltage but decreases the current. Minimising the current also minimises the heat losses 15.a) i) The galvanometer would deflect in one direction ii) The galvanometer would deflect in the same direction but not as much iii) The galvanometer would deflect in the opposite direction b) It has a commutator in place of slip rings 33 c) + e.m.f. 0 1 4 1 2 3 4 1 no. of rotations d) The commutator reverses the direction of the induced current in each half of the coil each half turn. The current generated if therefore direct. The current is a maximum when the coil is horizontal because at this time the rate at which it cuts magnetic field lines is greatest. The current is zero when the coil is vertical as it is parallel to the magnetic field lines. Chapter 22: Introductory electronics and electronic systems Self – Assessment Answers 1. c 2. b 3.a 4.d 5. c 6. d 7.c 8.b 9.a 10. a 11.a) b) 12.a) b) i) 6 V ii) 125 Hz i) accelerates the electrons ii) deflects the beam iii) emits light when stuck by electrons 270 Ω +/- 5% It gives the uncertainty in the value of the resistance so the range of possible resistances can be calculated. c) The maximum amount of energy the resistor can dissipate each second. 13.a) Circuit that turns a thermostat on and off. Circuit that turns a fire alarm on and off. b) It will decrease. As temperature increases the resistance of the thermistor decreases, so it is a lower proportion of the total resistance of the circuit, and there is a lower p.d. across it. 34 14.a)b) AND NOT X A A X B NOR A NAND A X X B B AND A NOR A X X B B OR A AND X X OR A X B B B X A X A B AX B X A B A X B X A B AX B X A B X 0 1 0 1 0 0 01 0 1 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 1 0 1 0 0 1 0 0 1 0 1 1 0 0 1 1 1 1 0 1 0 0 0 1 0 1 1 0 0 1 0 1 0 1 1 1 1 1 0 1 1 1 1 1 1 1 1 1 1 1 1 0 1 00 1 0 1 1 0 10 0 1 0 1 1 10 1 15.a) Collector Base Emitter A small current between the base and emitter causes a large current between the collector and emitter. The transistor is used as a switch in electronic circuits. b) Bistables have two stable states, astables have none. Chapter 23: Radioactivity Self – Assessment Answers 1. c 2. d 3.c 4.d 5. c 6. b 7.c 8.a 9.d 10. b 11. Low-level waste – c) Liquid and gas waste may be released to the environment where it is rapidly diluted. Solid waste is buried in clay-lined trenched which are subsequently filled and grassed over. Intermediate-level waste – a) Liquid and solid waste is sealed in steel drums and stored in silos. 35 High-level waste – b) Liquid waste is sealed in steel drums and placed cooled stainless steel tanks or converted into waste glass blocks and stored for at least 50 years. 12.a) A Geiger counter b) A stream of fast-moving electrons c) Beta radiation could not travel through 3 metres of soil to the detector but would be absorbed. Organic material in the soil would also contain carbon-14. d) A gamma emitter. This would be the only type of radiation that could pass through 3 metres of soil. 13.a) Fire produces hot air and smoke which rise. A smoke detector at the bottom of a wall would only sound an alarm after smoke had filled the whole room. By this time a serious fire may have developed. The best location for a smoke detector is on a ceiling in the middle of a room. b) Alpha particles are absorbed by the plastic casing of the alarm. c) 60 Activity / counts per minute 50 40 30 20 10 0 500 1000 1500 2000 Time / years From the graph: Activity 40 counts per minute after 320 years Activity 20 counts per minute after 745 years Half-life from graph = 745 – 320 = 425 years (actual value 432 years) d) The half-life of this radioisotope is far too short. By the time the alarm was sold the activity of californium-241 would be too low to be of any practical use. 14.a) Nuclear fusion. Hydrogen nuclei join together to form helium nuclei with the loss of a small amount of matter. The lost matter is converted into energy. b) The Sun started out as a cloud of gas and dust. This collapsed together under its own gravity to form a protostar. At some point the temperature of the protostar became hot enough for nuclear reactions to start. At present hydrogen atoms are undergoing nuclear fusion to form helium atoms. Eventually all of the hydrogen on the Sun will be used up and other nuclear fusion reactions involving large atoms will take place. The Sun will expand and cool to become a red giant. Finally, when all of the fuel is used up the Sun will collapse to become first a white 36 c) 15.a) b) c) d) e) f) dwarf, and then a black dwarf. Black holes are formed from stars which are formed from a very large mass of material. When all of the nuclear fuel is used up, the star collapse and the outer layers are blown off by a mighty explosion called a supernova. If there is sufficient mass remaining, the core of the star shrinks to form a black hole. Paper, aluminium, steel, lead A Geiger counter The detector measures the amount of radiation passing through the paper and sends the information to the pressure control. If the amount of radiation falls this indicated the paper is too thick and more pressure is applied. If the amount of radiation rises this indicates the paper is too thin and the pressure is reduced. The activity of the radioactive source decreases over time. The machine will interpret the fall in radioactivity to the paper being too thick and so more pressure will be applied to the rollers. The result will be thinner paper. The radioactive source will ionise the air immediately above it. As the paper passes over the source, any charge will be neutralised by the ions in the air. All of the radiation from strontium-90 would be absorbed so there would be no reading on the detector and the pressure control would not work. Chapter 24: The nuclear atom Self – Assessment Answers 1. d 2. d 3.c 4.b 5. a 6. b 7.a 8.c 9.a 10. d 11. Name Carbon Magnesium Silicon Helium Uranium Phosphorus Proton number 6 12 14 2 92 15 Nucleon number Neutron number 12 6 24 12 28 14 4 2 238 146 31 16 12.a) Statements 1 and 5 were found to be correct. b) Statements 2, 3 and 4 need some modification. 1. Elements are made of small particles called atoms 37 13.a) b) c) d) e) f) 14.a) 2. Atoms cannot be created of destroyed during chemical reactions 3. All atoms of the same element have the same charge (atomic number) but may vary slightly in mass (mass number) due to the presence of different isotopes 4. Atoms of different element have different charges (atomic numbers) 5. Atoms combine in small whole numbers to form compounds Nitrogen-14, 147N Bromine-81, 81 35Br Hydrogen-1, 10H Copper-63, 36 29Br Plutonium-244, 244 94Pu 207 Lead-207, 82Pb Particle Proton Electron Neutron Approximate relative mass 1 0 1 Charge Position in the atom +1 -1 0 in the nucleus surrounding the nucleus in the nucleus b)i)4 ii) 5 c)i)127 53I 18 ii) 8O 232 iii) 90Th 37 iv) 17Cl 15.a) non-deflected particles fluorescent screen A gold foil B C deflected particles radioactive source b) Nuclei of helium-4 atoms consisting of two protons and two neutrons. c) i) There are large spaces between the nuclei of atoms. ii) Some alpha particles came close enough to the nucleus of an atom to be deflected slightly off course. iii) Some alpha particles were aiming directly for nuclei of atoms and repulsed by the like charge. d) Thompson’s model of the atom was a positively charged mass and electrons contained within it like the plums in a pudding. Rutherford’s experiment showed that the atom had a positively charged nucleus and there were large gaps between the nuclei therefore the electrons must be in orbit around the nucleus rather than within it. 38 39 Physics for Cambridge O Level Physics for Cambridge O Level has been written to provide up-to-date coverage of the material required for the latest Cambridge International Examinations O Level Physics syllabus 5054. Important features include: • A comprehensive text covering all of the material needed for Cambridge O Level Physics • Full colour illustrations, engaging photographs and clear diagrams • Learning objectives at the start of each chapter to get the student thinking about the chapter content • Definitions of key terms • Chapter summaries in the form of ‘Key Points’ to help students review their learning and identify where additional work might be needed • ‘Check your Understanding’ exercises within the text that are focused on particular parts of the chapter content • Worked examples of different calculations • A full range of experiments covering the requirements of the syllabus • Cambridge O Level style questions at the end of each chapter to allow students to check their understanding as they work through the book. • Past examination questions at the end of each chapter show the students how they might expect the chapter content to be examined. Pauline Anning is a highly experienced teacher, examiner and consultant in science. She has been writing science learning and teaching materials for over 15 years. 40