2016.1 L.35-36 ms -07 ha-final · 2019. 11. 23. · 2016.1 L.35/36_MS 3/60 Page 3 of 59 exams DEB Pre-Leaving Certificate Examination, 2016 Physics Ordinary & Higher Level Explanation

Page 1 of 51

L.35/36

Pre-Leaving Certificate Examination, 2016

Physics

Marking Scheme

Ordinary Pg. 4 Higher Pg. 31

2016.1 L.35/36_MS 2/60 Page 2 of 59 examsDEB


Physics

Ordinary & Higher Level

Table of Contents

Ordinary Level Higher Level Section A Section A

Q.1 .................................................. 4 Q.1 .................................................. 31

Q.2 .................................................. 6 Q.2 .................................................. 33

Q.3 .................................................. 8 Q.3 .................................................. 35

Q.4 .................................................. 10 Q.4 .................................................. 37

Section B Section B Q.5 .................................................. 12 Q.5 .................................................. 39

Q.6 .................................................. 14 Q.6 .................................................. 42

Q.7 .................................................. 16 Q.7 .................................................. 44

Q.8 .................................................. 18 Q.8 .................................................. 46

Q.9 .................................................. 20 Q.9 .................................................. 48

Q.10 .................................................. 22 Q.10 .................................................. 50

Q.11 .................................................. 24 Q.11 .................................................. 53

Q.12 .................................................. 26 Q.12 .................................................. 55

examsDEB



Physics

Ordinary & Higher Level

Explanation

Conventions Used

1. A dash – before an answer indicates that the answer is a separate answer, which may be considered as independent of any other suggested answers to the question.

2. A forward slash / before an answer indicates that the answer is synonymous with that which preceded it.

Answers separated by a forward slash cannot therefore be taken as different answers. 3. A double forward slash // is used to indicate where multiple answers are given but not all are required. 4. Round brackets ( ) indicate material which is not considered to be essential in order to gain full marks. 5. ‘etc.’ is used in this marking scheme to indicate that other answers may be acceptable. In all other cases, only the answer given or ‘words to that effect’ may be awarded marks. 6. In calculations, 3 marks are deducted for a mathematical error but no further penalty is incurred if the

problem, otherwise correct, is completed. Allow for rounding unless specified otherwise; accept an answer given within a reasonable range if the method of calculation is correct.

Current Marking Scheme

Assumptions about these marking schemes on the basis of past SEC marking schemes should be avoided. While the underlying assessment principles remain the same, the exact details of the marking of a particular type of question may vary from a similar question asked by the SEC in previous years in accordance with the contribution of that question to the overall examination in the current year. In setting these marking schemes, we have strived to determine how best to ensure the fair and accurate assessment of students’ work and to ensure consistency in the standard of assessment from year to year. Therefore, aspects of the structure, detail and application of the marking schemes for these examinations are subject to change from past SEC marking schemes and from one year to the next without notice.

examsDEB



Physics

Ordinary Level Marking Scheme (400 marks)

SECTION A (120 marks)

Answer three questions from this section. Each question carries 40 marks.

Section A Question 1 (40 marks)

1. In an experiment to verify the principle of conservation of momentum, a student measured the mass of two objects (A and B). One object (A) was set moving at constant velocity and collided with the second object (B) at rest. After the collision, the two objects stuck together. Measurements were taken to determine the velocities before and after the collision.

The following are the values that were measured or calculated.

Mass of A ...................................................................250 g Mass of B ...................................................................200 g Velocity before collision ............................................0.240 m s–1 Velocity after collision ...............................................0.133 m s–1

(i) Draw a labelled diagram of the apparatus used in the experiment. (9)

** Diagram to include: Any 3: (3 × 3m) – 2 trolleys / 2 riders – on track / runway / airtrack – light gates and timer / tickertape and timer / any valid variation – means of measuring mass / distance

** Deduct (2m) if no labels included.

(ii) How did the student determine the velocity of an object in this experiment? (3 × 4m) (12)

– measure distance – measure time – divide

** Accept ‘velocity = distance / time’ for full marks.

examsDEB


Section A Question 1 (cont’d.)

(iii) Calculate the momentum of object A before the collision. (3 × 3m) (9)

– P = mv – before P = (250)(0.24) or (0.25) (0.24) – = 60 g m s–1 or 0.06 kg m s–1

(iv) Calculate the momentum of both objects after the collision. (2 × 3m) (6)

– after P = (450)(0.133) or (0.45)(0.133) – = 59.85 g m s–1 or 0.05985 kg m s–1

(v) Explain how these results verify the principle of conservation of momentum. (4m) (4)

– momentum before = momentum after



2. A student carried out an experiment to obtain the calibration curve of a thermometer based on a particular thermometric property. The following is an extract from her report.

“I placed the thermometer, based on the thermometric property that I was calibrating, in water at 0 C. I measured the temperature of the water using a standard thermometer and I measured the value of the thermometric property of my thermometer. I heated the water steadily and took measurements of the thermometric property every 10 C. The results I obtained are given in the table.”

Temperature (C) 0 10 20 30 40 50 60

Thermometric property 18 22 27 33 39 46 54

(i) Draw a labelled diagram of the apparatus used in this experiment. (12)

** Diagram to include: (4 × 3m)

– standard thermometer – thermometer based on thermometric property – both in a beaker of water – method of heating water

** Deduct 2m if no labels included.

(ii) What is a suitable thermometer to use as a standard thermometer? (9)

Any 1: (3m) – mercury (in glass) thermometer // – alcohol thermometer

Name a suitable thermometric property for the student’s thermometer.

Any 1: (6m) – length of liquid column // – pressure at constant volume // – volume at constant pressure // – electrical resistance // – colour of crystals // – emf of thermocouple // etc.

(iii) Plot a graph, on graph paper, of the thermometric property against temperature. Put temperature on the X-axis. (4 × 3m) (12)

** Graph to include:

– axes labelled correctly, thermometric property against T – 4 points correctly plotted – another 3 points plotted correctly – smooth curve through the points

** Deduct 3m if graph paper not used.



100 20 30 40 50 60

7

0

14

21

28

35

42

49

56

T (°C)

Thermometricproperty

Temperature atthermometric property

48

(iv) Use the graph to estimate the temperature when the value of the thermometric property is 48. (7)

– line drawn from 48 value horizontally across to curve (3m) – vertically down to T axis (2m) – T = 53 C (2m)

** Accept any reasonable reading from the graph. ** Award full marks for any reading in range 51 – 54 C



3. In an experiment to measure the wavelength of a monochromatic light source using a diffraction grating, a number of angular measurements were taken.

The number of lines per mm on the diffraction grating is 400.

(i) Draw a labelled diagram of the apparatus that may have been used in the experiment. (12)

Any 1: (4 × 3m) ** Diagram to include:

Laser method

– laser – passing through diffraction grating – metre rule to measure distances – showing position of beam on screen

or

Spectrometer method

– monchromatic light source – spectrometer – passing through diffraction grating – telescope at some angle from straight through after passing through grating

(ii) What is monochromatic light? (3m) (6)

– light of single wavelength

Give an example of a monochromatic light source.

Any 1: (3m) – sodium lamp // – laser

** Accept other appropriate examples.

(iii) Explain how the angular measurements were taken. (9)

Any 1: (3 × 3m) Laser method

– measure distance from grating to screen – measure distance from central fringe to nth order fringe – calculate angle

or

Spectrometer method

– measure angular value of central fringe – measure angular value of nth order fringe – subtract two angles



(iv) What is the distance between each line in the diffraction grating? (2 × 3m) (6)

– d = 400000

1 or

400

0010.

– = 2.5 × 10–6 m

(v) How is the wavelength of light calculated? (4m + 3m) (7)

– use the formula n = dsin – explaining terms in formula



4. In an experiment to investigate the relationship between current I and potential difference V for a filament bulb, the following data was recorded.

V (V) 1.0 2.0 3.0 4.0 5.0 6.0 7.0 8.0

I (A) 1.3 2.4 3.0 3.5 3.8 4.0 4.1 4.1

(i) Name the instruments used to measure the current and potential difference values. (6)

Current (3m)

– ammeter

Potential difference (3m)

– voltmeter

(ii) Where are these instruments positioned in the circuit in relation to the filament bulb? (2 × 3m) (6)

– ammeter in series with bulb – voltmeter in parallel with bulb

** Award full marks if correctly shown on a circuit diagram.

(iii) Plot a graph, on graph paper, of the current against potential difference (put potential difference on the X-axis). (4 × 3m) (12)

** Graph to include: – axes labelled correctly – at least 4 points plotted correctly – another 3 points plotted correctly – curve drawn through the points

** Deduct 3m if graph paper not used.

(iv) Does a filament bulb obey Ohm’s law? (2m) (7)

– no / does not obey Ohm’s law

Explain your answer. (3m + 2m)

– not a straight line – I is not proportional to V



(v) Use the graph to find the value of the resistance of the bulb when the potential difference across the bulb 4.5 V. (3 × 3m) (9)

– I = 3.65 A, when V = 4.5 V

– I

VR =

– resistance = 1.23

** Accept any reasonable reading from the graph. ** Award full marks for any reading in range 3.6 – 3.7 A, and subsequent calculation of

resistance based on readings in this range.

1.00.0 2.0 3.0 4.0 5.0 6.0 7.0 8.0

0.5

0.0

1.0

1.5

2.0

2.5

3.0

3.5

4.0

4.5

V (V)

I (A)

Current when potential difference

is 4.5 V


SECTION B (280 marks)

Answer five questions from this section. Each question carries 56 marks.

Section B – Question 5 (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 first law of motion. (4m + 3m) (7)

– if no net force is acting on object – it remains at rest or constant velocity

(b) A car accelerates from 20 m s–1 to 50 m s–1 in 2 seconds. What is the acceleration of the car? (4m + 3m) (7)

– v = u + at – = 50 = 20 + a(2) = 15 m s–2

(c) What change is observed in the sound wave of the siren of a police car as it travels towards a stationary observer? (4m + 3m) (7)

– increase – in pitch of siren

(d) Define specific heat capacity. (4m + 3m) (7)

– the heat need to raise the temperature of 1 kg – of a substance by 1 K (1 C)

(e) Name a pair of complementary colours. (7)

Any 1: (4m + 3m) – blue and yellow // – red and cyan // – green and magenta


Section B Question 5 (cont’d.)

(f) The glass in an optical fibre has a refractive index of 1.46. Calculate the critical angle for the glass in the optical fibre. (4m + 3m) (7)

– 1.46 = Csin

1

– C = 43.2

(g) What is the purpose of a miniature circuit breaker (MCB) in an electrical circuit? (4m + 3m) (7)

– switches off current – when too large a current flows

** Award 4m if answer mentions ‘safety / prevent shock’.

(h) Calculate the amount of heat that is given off every second from a coil of wire whose resistance is 200 Ω and which is carrying a current of 4 A. (4m + 3m) (7)

– heat per second = I2R; (4)2(200) – = 3200 J

(i) Identify the numbers represented by X and Y in the following alpha decay of radium: (4m + 3m) (7)

He+Ra→Ra 42

22888

XY

X – 224 Y – 86

(j) From the list below, identify the scientist who developed the theory that electrons can only occupy certain energy levels around the nucleus of an atom. (7m) (7)

Bohr Einstein Rutherford Röntgen


Section B Question 6 (56 marks)

6. Define pressure. (6m) (6)

– force per unit area

A man and a woman are walking together on soft ground. The man has a mass of 85 kg and is wearing flat shoes that have an area of 0.06 m2 in contact with the ground. The woman has a mass of 65 kg and is wearing high heels that have an area of 0.02 m2 in contact with the ground.

(i) Find the weight of the man and the woman. (3 × 3m) (9)

– weight = mg Man’s weight

– (85)(9.8) = 833 N

Woman’s weight

– (65)(9.8) = 637 N

(ii) Calculate the pressure exerted by the woman on the ground. (2 × 3m) (6)

– P = A

F =

02.0

637

– = 31850 Pa

(iii) Calculate the pressure exerted by the man on the ground. (2 × 3m) (6)

– P = A

F =

060

833

.

– = 13883.3 Pa

(iv) Why is the woman more likely than the man to sink in the soft ground? (4m) (4)

– exerts much higher pressure (than the man) / smaller contact area (than the man)

The pressure exerted on any object underwater varies with depth.

(i) State Boyle’s law. (2 × 3m) (6)

– (for a fixed mass of gas kept at a constant temperature) pressure is inversely – proportional to the volume

** Award full marks for correct notation explained, PV = k (when T and m are fixed).



(ii) How does Boyle’s law explain why a bubble gets bigger as it rises towards the surface of a lake? (3 × 3m) (9)

– at bottom of lake pressure is large – therefore volume is small – as the bubble rises, pressure decreases and volume increases (bubble gets bigger)

(iii) A fish starts at a depth of 20 m in water and rises to 1 m below the surface. What is the change in pressure on the fish as it makes this rise? (10)

– pressure = gh (3m) – pressure at bottom = (1000)(9.8)(20)=196000 Pa (3m) – pressure at top = (1000)(9.8)(1) = 9800 Pa (2m) – change in pressure = 196000 – 9800 = 186200 Pa (2m)

(density of water = 1000 kg m–3; acceleration due to gravity, g = 9.8 m s–2).



7. What is meant by refraction of light? (2 × 3m) (12)

– the bending of light at a boundary – when it is going from one medium into another

State one of the laws of refraction of light.

Any 1: (2 × 3m) – the incident ray, refracted ray and normal

– all lie in the same plane

or

– the ratio of the sine of the angle of incidence to

– the sine of the angle of refraction is a constant

** Award full marks for correct notation explained, nr

i=

sin

sin, where n = constant.

Lenses are one of the most common applications of the refraction of light. They are found in eyes, glasses, magnifying glasses and more complex instruments such as microscopes and telescopes. There are two types of lenses: convex and concave. Concave lenses always produce virtual images, but convex lenses produce virtual images only when the object is inside the focus.

(i) What type of image is produced when the object is outside the focus in a convex lens? (3m) (6)

– real (image)

How does this type of image differ from a virtual image?

Any 1: (3m) – a real image can be captured on a screen, a virtual image cannot // – rays pass through point where image is formed in real image, they do not in a virtual image

(ii) Draw a ray diagram showing how a virtual image is produced in a concave lens. (3 × 3m) (9)

** Diagram to show:

– two rays from object towards lens – two rays diverging after passing through lens – upright diminished virtual image on same side as object

(iii) Why is a concave lens not suitable for use as a magnifying glass? (4m) (4)

– the image is always diminished in a concave lens



Light is focused onto the retina in an eye by a combination of a cornea and a lens. The distance from the lens to the retina is approximately 1.8 cm.

(iv) Draw a diagram of the structure of the eye. (9)

** Diagram to show:

Any 3: (3 × 3m) – cornea/lens combination // – retina // – ciliary muscles // – iris / pupil

(v) How does the eye focus on objects at different distances? (2 × 3m) (6)

– thickness of lens changed (by muscles) – changes focal length of lens

(vi) Calculate the focal length of the lens/cornea combination if a person views an object at a distance of 15 cm from the eye. (10)

– vuf

1+

1=

1 (3m)

– 81

1+

15

1=

1

.f (3m)

– f = 1.61 (4m)



8. State Faraday’s law of electromagnetic induction. (3 × 3m) (9)

– induced electromotive force (emf) – is proportional to – the rate of change of magnetic flux

** Award full marks for correct notation explained, dt

dE

Φ∝ .

Describe a laboratory experiment to demonstrate Faraday’s law. (12)

** Diagram to include: (2 × 3m)

– coil connected to galvanometer – magnet

** Description (2 × 3m)

– magnet moved slowly near coil, small deflection on meter – magnet moved quickly near coil, large deflection on meter

A coil of wire is moved from a position where a magnetic flux of 8 Wb cuts the coil to a different position, where a magnetic flux of 18 Wb cuts the coil in 0.25 s.

What is the electromotive force (emf) induced in the coil? (3 × 3m) (9)

– emf = t

dΦ

– = 250

8-18

.

– = 40 V

Transformers are devices based on electromagnetic induction. They are used to increase (step-up transformer) or decrease (step-down transformer) the values of a.c. voltages.

(i) Draw a labelled diagram of a transformer. (3 × 3m) (9)

** Diagram to show:

– primary coil – secondary coil – soft iron core



(ii) Explain why a transformer does not work if a constant d.c. voltage is applied to the input coil. (3 × 3m) (9)

– d.c. voltage means no change in current flowing in the primary coil – no change in magnetic field cutting the secondary coil – no emf / current induced in the secondary coil

(iii) The input coil of a transformer has 150 turns and the output coil has 750 turns. Calculate the output voltage if the voltage across the input coil is 230 V. (8)

– p

s

p

s =N

N

V

V (3m)

– 150

750=

230sV

(3m)

– 1150 V (2m)



9. Define resistance. (2 × 3m) (9)

– ratio of potential difference / voltage – to current

** Award full marks for correct notation explained, R = I

V.

Name an instrument which measures resistance. (3m)

– ohmmeter

** Accept multimeter.

Thermistors and light-dependent resistors (LDRs) are types of resistors used in electrical circuits and have many applications.

What is the difference between the operation of a thermistor and an LDR? (9)

– resistance changes (for both) (3m) – with temperature, for a thermistor (3m) – with light intensity, for an LDR (3m)

Give a use for (i) a thermistor and (ii) an LDR. (6)

(i) use for a thermistor

Any 1: (3m) – heat alarms // – frost alarms // – temperature control in house // etc.

** Accept other appropriate answers.

(ii) use for an LDR

Any 1: (3m) – light meters // – sensors // – controlling street lighting // etc.


Draw the electrical circuit symbol for an LDR. (2 × 3m) (6)

– symbol of fixed resistor with circle – two arrows pointing in towards resistor

** Award full marks for a correct diagram. e.g.



A thermistor is placed in series with a fixed resistor of resistance 400 Ω as shown. At a certain time the resistance of the thermistor is 1000 Ω. Ten minutes later the resistance of the thermistor has changed to 600 Ω.

What would have caused the change in the resistance of the thermistor during the 10-minute period? (2 × 3m) (6)

– increase in – temperature

(i) Calculate the total resistance and hence the current flowing in the circuit when the thermistor has a resistance of 1000 Ω. (3 × 3m) (9)

– total resistance = 1000 + 400 = 1400 Ω

– R

VI = =

1400

12

– 0.00857 A

(ii) Calculate the change in current in the circuit after 10 minutes. (11)

– total resistance = 600 + 400 = 1000 Ω (3m)

– R

VI = =

1000

12 (3m)

– = 0.012 A (2m) change in current – = 0.012 – 0.00857 = 0.00343 A (3m)

A

400 Ω

12 V



10. Nuclear energy can be produced by nuclear fusion and nuclear fission. Nuclear fusion releases the energy that causes all stars, including our Sun, to exist for billions of years. Nuclear fission releases the energy that is produced in nuclear reactors in many countries throughout the world.

(i) What is the difference between nuclear fusion and nuclear fission? (12)

Nuclear fusion (2 × 3m)

– the joining / fusing of two small nuclei – to form a larger nucleus

Nuclear fission (2 × 3m)

– the splitting of one large nucleus – into two smaller nuclei

(ii) Name the two gases that are present in the Sun and are involved in nuclear fusion. (2 × 3m) (6)

– hydrogen – helium

(iii) How do scientists know that these gases are present in the Sun? (2 × 3m) (6)

– by looking at spectrum produced by Sun / spectroscopy – each gas has its own spectrum

(iv) Why is the Sun only expected to burn for approximately another 5 billion years? (2 × 3m) (6)

– hydrogen converts to helium during fusion – hydrogen will eventually run out

(v) What is the role of (a) the control rods and (b) the moderator in a nuclear reactor? (12)

(a) role of the control rods (2 × 3m)

– slide up and down over the fuel rods – to control the speed of the reaction / prevent neutrons travelling between the rods

(b) role of the moderator (2 × 3m)

– slows down neutrons – prevents capture of neutrons



(vi) Name a suitable fuel for use in a nuclear reactor. (6)

Any 1: (6m) – uranium // – plutonium

(vii) How does Einstein’s formula E = mc2 explain how energy is released in a nuclear reactor during nuclear fission? (8)

– when nucleus is split by neutrons (3m) – products of split have less mass (3m) – lost mass is converted to energy by E = mc2 (2m)



11. Read the following passage and answer the accompanying questions.

(a) Give two examples of musical instruments that have open-end air columns. (7)

Any 2: (4m + 3m) – flute // – recorder // – organ pipes // etc.


(b) Explain why it is not possible to generate a musical note that is 1.5 times greater than the fundamental frequency possible for a musical instrument that contains an open-end air column. (4m + 3m) (7)

– note is not a whole-number multiple – of fundamental frequency

** Award 3m if ‘whole-number’ is omitted from the answer.

(c) What primarily determines the fundamental frequency of a musical instrument with open-end air columns? (4m + 3m) (7)

– length of – tube / air column

(d) Draw a diagram showing a standing wave in a tube opened at both ends, when it is vibrating at its fundamental frequency. (7)

– antinodes at open end (4m) – one node in the middle (3m)



(e) What do (i) the nodes and (ii) the antinodes on the standing wave diagram you have drawn in part (d) represent? (4m + 3m) (7)

(i) nodes

– position of no vibration

(ii) antinodes

– position of maximum vibration

(f) If the fundamental frequency of the vibrations in a flute is 325 Hz, what is the frequency of the third harmonic? (4m + 3m) (7)

– frequency of third harmonic = 3 × fundamental frequency – = 975 Hz

(g) Taking the speed of sound in air to be 340 m s–1, what is the wavelength of the musical note in the flute referred to in part (f) when it is vibrating at its fundamental frequency? (7)

– c = f (3m)

– = 325

340 (2m)

– = 1.046 m (2m)

(h) How would the fundamental frequency of the flute differ if a longer tube was used to make the flute? (7m) (7)

– (frequency) would be lower for a longer tube


Section B Question 12 (2 × 28 marks)

12. Answer any two of the following parts (a), (b), (c), (d).

(a) Define work. (2 × 3m) (9)

– force multiplied by – distance / displacement

or

– work is done when a force – moves an object through a distance

** Award full marks for correct notation explained, W = Fs.

What is the unit of work? (3m)

– Joule / N m

Hossein Rezazadeh of Iran broke the Olympic record in weightlifting, when he lifted a mass of 263 kg from the ground to a height of 2.88 m above the ground.

(i) Calculate the work done by Hossein when lifting the weight. (3 × 3m) (9)

– W = mgh – W = (263)(9.8)(2.88) – = 7422.9 J

(ii) Calculate the power developed by Hossein, if he took 0.75 s to lift the weight. (2 × 3m) (6)

– P = t

W =

750

97422

.

.

– = 9897.2 W

(iii) What type of energy is the weight gaining while being lifted? (4m) (4)

– potential energy

(acceleration due to gravity = 9.8 m s–2)



(b) Distinguish between heat and temperature. (9)

Heat (3m)

– a form of energy

Temperature (3m)

– a measure of hotness

What is the SI unit of temperature? (3m)

– Kelvin / K

Heat can be transferred in a room by convection. (6) What is convection? (2 × 3m)

– movement of heat / hot air / cold air – by circulation / current / rises / sinks

Name one other method of heat transfer. (4)

Any 1: (4m) – radiation // – conduction

The photograph shows a cross section of an electric storage heater. Bricks with a high specific heat capacity are heated overnight by passing an electric current through a heating coil in the bricks.

The bricks are surrounded by insulation.

Why is insulation used to surround the bricks? (2 × 3m) (6)

– to prevent / reduce – heat loss / energy loss

Name a material suitable for use as insulation. (3)

Any 1: (3m) – fibre glass // – rock wool // – cotton wool // etc.



(c) What is an electric field? (2 × 3m) (6)

– an area near a charge – that experiences a force

Describe an experiment to demonstrate the shape of the electric field between two charged plates. (3 × 3m) (9)

– two plates connected to battery – semolina in oil (or anything that will line up in electric field) – semolina lines up between 2 plates, showing the direction of electric field

Copy the diagram below and show the electric field lines produced due to a positive charge and negative charge near each other. (6)

** Diagram to show: (2 × 3m)

– correct shape – correct direction of arrows

e.g.

Why are electric fields a problem for people who work with integrated circuits? (4m + 3m) (7)

– electric field causes build-up of static charge on person – charge can discharge through and damage electronic components

+ −



(d) What is the photoelectric effect? (2 × 3m) (6)

– the emission of electrons from a metal – caused by electromagnetic radiation of suitable frequency

The diagram shows a photocell connected in series to a galvanometer and a battery.

BA

Name the parts of the photocell labelled A and B. (2 × 3m) (6)

A – (photo) cathode B – (photo) anode

Give an application of a photocell. (4)

Any 1: (4m) – alarms // – automatic doors // – light meters // – solar cells // – safety switches // etc.


Radiation is incident on a zinc plate placed on the cap of a negatively charged gold leaf electroscope as shown. When visible light is incident on the zinc plate, no change is observed, but a change is observed when UV light is incident on the plate.

(i) What change is observed when UV light is incident on the plate? (3m) (9)

– gold leaf collapses (as electrons are removed)

Explain why this change is not observed for visible light. (2 × 3m)

– visible light not enough energy / too low frequency – to remove electrons

(ii) What change would be observed in the electroscope if the intensity of the visible light was increased? (3m) (3)

– no change observed

−−− −

−−−

Goldleaves

Electroscope

DiscZincplate

Radiation


Notes:



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 investigate the relationship between the acceleration a of a body and the force F applied to it, a student recorded the following data. Before the force was applied, the student tested that no other net force acted on the body.

F (N) 0.40 0.80 1.20 1.60 2.00 2.40 2.80

a (m s–2) 0.49 0.99 1.51 1.95 2.49 2.93 3.45

How could the student have tested that no other net force was acting on the body? (4)

Any 1: (4m) – when placed at rest on trolley / track, the object does not move // – when put moving, the trolley travels at constant velocity


Draw a labelled diagram of the apparatus used in the experiment. (2 × 3m) (15)

Diagram showing: – method for measuring initial and final velocity, e.g. ticker timer and ticker tape or

light gate and timer – method of varying and measuring applied force, e.g. weights on a scale pan / Newton meter

Describe how the acceleration was measured. (3 × 3m)

Explanation to include: – how to measure initial and final velocities – time between both measurements or distance between measurements – calculation of acceleration using v = u + at or v2 = u2 + 2as

** Marks may also be allocated in the same way for a diagram and steps using a datalogging method.

examsDEB



Draw a suitable graph to show the relationship between the applied force and the acceleration. (3 × 3m) (15)

– force (F) plotted against acceleration (a) (or acceleration vs. force) with axes correctly labelled – 6 points correctly plotted – straight line with good fit

** Deduct 1m if straight line is not a good fit.

0.500.0 1.00 1.50 2.00 2.50 3.00 3.50 4.00

0.4

0.0

0.8

1.2

1.6

2.0

2.4

2.8

3.2

a (m s–2)

F (N)

Use your graph to calculate the mass of the accelerating body. (2 × 3m)

– take 2 points on the line, e.g. (1.25, 1.00), (2.75, 2.20) (if acceleration vs. force, slope = 1/mass)

slope of graph, e.g. 80=51

21.

.

.

– correct value 0.80 kg

** Allow answers in the range of 0.79 to 0.83 kg. ** Accept any reasonable reading from the graph. ** Deduct 1m for omission of or incorrect unit.

If the student had not ensured that other net forces (besides the applied force) were absent, how would this have affected the graph? (3m) (6)

– straight line would not go through origin / (0,0)

Explain your answer. (3m)

– acceleration not zero when applied force = 0




2. In an experiment to verify the laws of equilibrium for a set of co-planar forces acting on a uniform metre stick, a student attached four weights and two Newton meters to a metre stick. The positions of the weights and Newton meters were changed until the metre stick was in equilibrium. These positions, as well as the values of the weights and the force readings on the Newton meters, were recorded and are given in the table below. The mass of the metre stick was found to be 122 g.

Weight 1Newton meter 1

Weight 2 Weight 3Newton meter 2

Weight 4

Force (N) 2.0 6.9 4.0 2.0 5.3 3.0

Position on metre stick (cm) 13.0 24.0 31.0 61.0 83.0 92.0

How did the student know that the metre stick was uniform? (18)

Any 1: (6m) – the centre of gravity was at 50 cm mark // – the metre stick balanced (hanging from thread, resting on fulcrum, etc.) at the 50 cm mark

Why was it important that the metre stick was horizontal and the Newton meters were vertical while the measurements were being taken?

Metre stick horizontal (2 × 3m)

– (as moment = force × perpendicular distance) perpendicular distance – is along metre stick if horizontal

Newton meters vertical

Any 1: (6m) – if not vertical, force measured has horizontal component // – reading is not vertical component of the force

Calculate (22) (i) the net force acting on the metre stick

– weight of metre stick = 0.122 × 9.8 = 1.2 N (2m) – forces up = 6.9 + 5.3 = 12.2 N (3m) – forces down = 2.0 + 4.0 + 2.0 + 3.0 + 1.2 = 12.2 N – net force = 0 N (3m)

** Award 3m only if answer 0 N without any calculation is given. ** Deduct 1m for omission of or incorrect unit.



(ii) the sum of the moments of the forces about the 0 cm mark on the metre stick.

– moment = force × perpendicular distance, stated or implied (3m)

anticlockwise moments

= (2.0 × 0.13) + (4.0 × 0.31) + (1.2 × 0.50) + (2.0 × 0.61) + (3.0 × 0.92) = 6.08 Nm (3m)

clockwise moments

= (6.9 × 0.24) + (5.3 × 0.83) = 6.055 Nm (3m)

** Deduct 1m for omission of or incorrect unit. ** Award 6m if calculated about the centre of gravity. ACM = (2 × 0.37) + (4 × 0.19) + (5.3 × 0.33) = 3.249 Nm CM = (2 × 0.11) + (3 × 0.42) + (6.9 × 0.26) = 3.274 Nm.

Use these results to verify the laws of equilibrium.

– forces up = forces down (2m) – sum of clockwise moments ~ sum of anticlockwise moments (3m)

(acceleration due to gravity = 9.8 m s–2)



3. In an experiment to verify Snell’s law, a student measured the angle of incidence i and the corresponding angle of refraction r for light entering a glass block. This was repeated for a number of different angles of incidence.

The student then plotted the following points based on the collected data.

0.10.0 0.2 0.3 0.4 0.5 0.6 0.7 0.8

0.1

0.0

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1.0

X

Y

Describe, with the aid of a labelled diagram, how the student measured the angle of incidence and the angle of refraction. (9)

Diagram showing: (2 × 3m)

– method of obtaining incident and refracted rays (line up pins or laser beam or ray box) – incident rays, refracted rays and normal labelled

Explanation: (3m)

– angle of incidence and refraction (shown or stated) measured using a protractor

What labels should be placed on each axis, instead of X and Y? (2 × 2m) (12)

X – sine of angle of refraction / sin r Y – sine of angle of incidence / sin i

Why are these quantities not interchangeable for the graph above? (3m)

– (quantities not interchangeable) as value of sin i always > sin r (i > r) going from air to glass

What is the smallest angle of incidence that the student set in this experiment? (3m + 2m)

– smallest value of sin i = 0.34 – smallest value of i = 19.9

** Accept any reasonable reading from the graph.



Complete a suitable graph using the plotted points above. (19)

** Graph drawn (5m).

0.10.0 0.2 0.3 0.4 0.5 0.6 0.7 0.8

0.1

0.0

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1.0

X

Y(0.3, 0.44)

(0.55, 0.83)

How does the graph verify Snell’s law? (2 × 3m)

– straight line through origin verifies – Snell’s law, which states sin i is proportional to sin r

Use your graph to determine the refractive index of the glass block.

– take 2 points on the line, e.g. (0.3, 0.44) and (0.55, 0.83) (3m)

– slope of graph = 56.1=25.0

39.0 (3m)

– refractive index = slope (2m)

** Allow answers in the range of 1.54 to 1.58. ** Accept any reasonable reading from the graph. ** Deduct 1m for omission of or incorrect unit.



4. In an experiment to investigate how current I varied with voltage V across a copper sulfate solution with copper electrodes, a student collected the following data.

V (V) 0 1 2 3 4 5 6 7 8

I (mA) 0 53 104 157 209 262 314 366 420

Describe, with the aid of a labelled circuit diagram, how the student obtained this data. (4 × 3m) (12)

Diagram showing:

– electrodes in solution connected to power supply – method of varying voltage (rheostat, potentiometer, variable voltage supply) – ammeter in series, voltmeter in parallel across electrodes

Explanation:

– vary voltage and take current reading and voltage reading for each voltage setting

Draw a suitable graph to show how this data verifies Ohm’s law for a copper sulfate solution. (15)

Graph (3 × 3m)

– plot of I against V (or V against I) with axes labelled (3m) – 7 points correctly plotted – straight line, good fit

How verifies Ohm’s law (2 × 3m)

– Ohm’s law, current proportional to voltage – verified by straight line through the origin

10 2 3 4 5 6 7 8

50

0

100

150

200

250

300

350

400

450

Current (mA)

Voltage (V)



The student then replaced the copper sulfate solution and electrodes with a semiconductor diode in forward bias. The current for a number of different voltages across the diode was measured.

Draw a sketch of the graph the student would expect to produce showing the variation of current with voltage for the semiconductor diode. (13)

Sketch showing:

– current very low initially (2m) – increasing rapidly beyond the junction voltage (3m)

e.g.

Explain why the current varies in this way with voltage.

– depletion layer formed at p-n junction, resulting in junction voltage (0.6 V for silicon) (3m)

– no current / very little current flows until opposing junction voltage is overcome (3m)

– when applied voltage is above junction voltage, current increases rapidly (2m)

V (V)

I (mA)

junction voltage


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) A ball of mass 500 g is rotated at the end of a string with an angular velocity of 5 rad s–1. If the tension in the string is 12 N, what is the radius of the rotation of the ball? (7)

– T = m2r (2m) – 12 = (0.5)(5)2r (3m) – r = 0.96 m (2m)

(b) When a mercury-in-glass thermometer is placed in melting ice, the length of the mercury column is 7 cm. When the thermometer is placed in steam above boiling water, the length of the column is 39 cm.

What is the length of the mercury column when the thermometer is placed in water at 40 C? (4m + 3m) (7)

– 100

7)-(39 = 0.32 cm per degree

– 40 × 0.32 + 7 = 19.8 cm

** Deduct 1m for omission of or incorrect unit.

(c) State the law of flotation. (4m + 3m) (7)

– the weight of a floating object is equal to – the weight of fluid it displaces

(d) Give one advantage and one disadvantage of using a convex mirror, instead of a plane mirror, as a sideview mirror on a car. (4m + 3m) (7)

Advantage

– larger field of view

Disadvantage

– image appears further away than it is




(e) What is point discharge? (4m + 3m) (7)

– the loss of charge from a pointed conductor – when there is a large build-up of charge at the point

(f) Write the following electromagnetic waves in increasing order of frequency: (7m) (7)

Microwaves infra red gamma rays visible light radio waves

– radio waves microwaves infra red visible light gamma rays

** Correct order for full marks.

(g) Draw a graph showing the relationship between current and potential difference (I-V graph) for conduction in a vacuum. (4m + 3m) (7)

– graph showing initial rise – followed by levelling off

e.g.

(h) What is the difference between how a line spectrum and a continuous spectrum is produced? (4m + 3m) (7)

– a line spectrum is produced from gases – a continuous spectrum is produced from incandescent solids or liquids

(i) In a nuclear reaction, 17.39 MeV of energy is released. (7) Calculate the loss of mass in the reaction. (4m + 3m)

– 17.39 MeV = 17.39 × 106 × 1.6 × 10–19 = 2.782 × 10–12 J – E = mc2, m = 3.092 × 10–29 kg


V (V)

I (mA)



(j) State two differences between leptons and hadrons. (7)

Any 2: (4m + 3m) Leptons

– indivisible particles / no internal structure // – not subject to strong nuclear force

Hadrons

– not indivisible particles // – subject to strong nuclear force


or

State the advantage of using a bridge rectifier over a single diode when converting a.c. to d.c. current. (4m + 3m) (7)

– no gaps in the current

Why is a capacitor placed in the bridge rectifier circuit?

– for a smoother d.c. current



6. State the principle of conservation of momentum. (2 × 3m) (15)

– total momentum after interaction = total momentum before interaction – provided no external force

Will kinetic energy also be conserved in all cases where momentum is conserved? (3m)

– no / (kinetic energy) will be not conserved

Give an example to support your answer. (2 × 3m)

– two objects of same mass and same speed in opposite direction before collision – rebound with a lower speed than before the collision (but same as each other)

** Accept any appropriate example where momentum is conserved and kinetic energy is not.

A white snooker ball has a mass of 170 g while all other colour snooker balls have a mass of 185 g. On a snooker table, the white ball strikes a cushion at right angles. When it rebounds at right angles it loses a quarter of its kinetic energy. It rebounds with a speed of 0.693 m s–1.

Calculate (20) (i) the loss of kinetic energy due to the collision with the cushion

– kinetic energy (K.E.) after = 1/2(0.17)(0.693)2 = 0.0408 J (3m) – K.E. after = 3/4(K.E. before) (2m) – K.E. before =4/3(K.E. after) = 0.0544 J (3m) – loss in K.E. = 0.0544 – 0.0408 = 0.0136 J (3m)


(ii) the speed of the ball before the collision (2 × 3m)

– K.E. before =1/2(0.17)v2 = 0.0544 – v = 0.8 m s–1


(iii) the change of momentum due to the collision. (3m)

– = (0.17)(0.8) – (0.17)(–0.693) = 0.254 kg m s–1




The white ball continues with the same speed, until it collides with a black ball travelling in the opposite direction at a speed of 0.7 m s–1. After the collision, the white ball rebounds along its original path at 0.3 m s–1.

What is the speed and direction of the black ball after the collision? (4 × 3m) (12)

– momentum before = (0.17)(0.693) + (0.185)(–0.7) = –0.01169 – momentum after = (0.17)(–0.3) + 0.185V – momentum after = momentum before = 0.185V – 0.051 = –0.01169 – V = 0.212 m s–1, in opposite to original direction


After rebounding from the black ball, the white ball collides with a red ball travelling in the opposite direction with a speed of 0.6 m s–1. After the collision, the white ball is deflected through an angle of 30 and travels at 0.25 m s–1. The red ball is deflected through an angle of 12.38.

Given that after the collision, the total momentum perpendicular to the direction of the balls before the collision is zero, what is the speed of the red ball after this collision? (9)

– (0.17)(0.25 sin30) – (0.185)(V sin12.38) = 0 (6m) – V = 0.5358 m s–1 (3m)


12.38°

30°

RR

W

W



7. Heat is extracted from a refrigerator using a heat pump which uses specific latent heat to extract heat from inside the refrigerator and emit it outside.

What is specific latent heat? (2 × 3m) (12)

– the energy required to change the state of – 1 kg of a substance without a change of temperature

Distinguish between specific latent heat of fusion and specific latent heat of vaporisation. (2 × 3m)

specific latent heat of fusion

– the change from a solid to a liquid

specific latent heat of vaporisation

– the change from a liquid to a gas

Explain, with the aid of a labelled diagram, how the heat pump causes the inside of the refrigerator to get cooler. (12)

Diagram (3m)

– pipes, compressor, expansion valve, circulating liquid / gas – correct arrangement

Explanation (3 × 3m)

– circulating liquid has a high specific latent heat and a low boiling point – at the expansion valve, liquid vaporises (takes in latent heat) and cools fridge – at the compressor, vapour is condensed to liquid, giving out its latent heat

e.g.

** Accept any simple, two-dimensional diagram.

Compressor

Expansion valve

Low pressurevapour

High pressurevapour

Heat given offLatent heat fromfridge to coils

Cooling coils(inside fridge) Condensor coils

(outside fridge)



Freon is an example of a refrigerant (liquid/gas circulating in the heat pump) used in refrigerators. How much heat energy is extracted from the inside of the fridge during one heat exchange if there is 100 millilitres of Freon in the heat pump? (9)

– mass of Freon = V = (1476)(0.1 × 10–3) = 0.1476 kg (3m) – Q = mlv (2m) – = (0.1476)(232000) (2m) – = 34243.2 J (2m)


A refrigerator of internal dimensions 50 cm × 50 cm × 100 cm is filled with milk cartons until the milk takes up half of the volume of the refrigerator. The refrigerator is then switched on.

How much heat energy is extracted when the temperature of the refrigerator is reduced from 20 C to 3 C? (19)

– volume of fridge = 0.5 × 0.5 × 1 = 0.25 m3 (2m) – volume of air = volume of milk = 0.125 m3 (2m) mass of air in fridge = (1.2)(0.125) = 0.15 kg – mass of milk in fridge = (1030)(0.125) = 128.75 kg (3m) – Q = (mc)Air + (mc)Milk (3m) – = (0.15)(1005)(20 – 3) + (128.75)(3770)(20 – 3) = 8.25 × 106 J (3m)


If the Freon refrigerant is circulated 10 times every minute, how long will it take for this temperature drop to occur? (2 × 3m)

– number of circulations needed = 234243

10×258 6

.

. = 241

– 10

241= 24.1 minutes


Why would the temperature of the refrigerator have dropped much more quickly if it had been empty? (4)

Any 1: (4m) – much less heat required to drop temperature of air than of milk // – air much lower mass / lower specific heat capacity than milk


(specific latent heat of vaporisation of Freon = 232 kJ kg–1; density of liquid Freon = 1476 kg m–3; specific heat capacity of air = 1005 J kg–1 K–1; density of air = 1.2 kg m–3;

specific heat capacity of milk = 3770 J kg–1 K–1; density of milk = 1030 kg m–3; 1 litre = 10–3 m3)



8. Define capacitance. (2 × 3m) (6)

– ratio of charge – to potential

** Accept V

QC = with explanation of terms for (6m).

State three factors on which the capacitance of a parallel plate capacitor depends. (3 × 3m) (9)

– dielectric / material between plates – distance between plates – common area of plates

Two capacitors are connected in series as shown below.

7 μF 10 μF

24 V

Assuming the charge is the same on both capacitors, calculate (24) (i) the charge on each capacitor.

– V7μF + V10μF = 24 (3m)

– 6-6- 10×10+

10×7

QQ = 24 (3m)

– 242857.14Q = 24 (3m) – Q = 9.88 × 10–5 C (2m)


(ii) the total energy stored on both capacitors.

Voltage across 7 μF capacitor

– 6-

5-

10×7

10×889.= 14.1 V (3m)

Voltage across 10 μF capacitor

– = 24 – 14.1 = 9.9 V (3m)

Energy stored

– = 1/2CV2 = 1/2(7 × 10–6)(14.1)2 + 1/2(10 × 10–6)(9.9)2 (2m + 2m) – 1.186 × 10–3 J (3m)




To what capacitance should the 7 μF capacitor be changed so that the potential difference across the new capacitor is double the potential difference across the 10 μF capacitor? (8)

– as V

QC = and Q is constant, C

V

1 (3m)

– if V across new capacitor is double that across 10 μF capacitor, C is half (3m) – C = 5 × 10–6 F (2m)


If the 7 μF capacitor is replaced with a 20 k resistor, calculate the charge on the 10 μF capacitor when a current of 1 mA flows in the circuit. (3 × 3m) (9)

Voltage across resistor

– = IR = (1 × 10–3)(20000) = 20 V

Voltage across capacitor

– = 24 – 20 = 4 V – Q = CV = (10 × 10–6)(4) = 4 × 10–5 C




9. What are alpha-particles? (2 × 3m) (12)

– fast moving – helium nuclei

How are alpha-particles produced? (2 × 3m)

– emitted from nuclei / decay of nuclei – of radioactive / unstable elements

Why would alpha-particles not be suitable for use in detecting thickness of objects in industry but be suitable for use in smoke detectors? (2 × 3m) (6)

– very poor penetrating power – good ionising ability

In 1919, under the direction of Rutherford, alpha-particles were used in an experiment to discover the existence of a nucleus inside the atom. Describe this experiment and explain how it showed the presence of the nucleus. (12)

Experiment (3 × 3m)

– alpha-particles fired at gold foil (described or diagram) – direction of particles after passing through gold foil detected by fluorescent screen – most alpha-particles not deflected, small number deflected, very small number rebounded

Conclusion (3m)

– most of positive charge concentrated in very small volume

Most smoke detectors that operate alarms contain the artificially produced radioisotope americium–241, which is an alpha-emitter. It is produced in nuclear reactors and is a decay product of plutonium–241.

Explain why alpha-particles are not emitted when plutonium–241 decays into americium–241. (15)

Any 1: (3m) – no change in mass number // – atomic number increased by 1



Write down the nuclear equation for this decay and the equation for the subsequent decay of americium–241. (2 × 6m)

** Award 2m for each term correct. ** Award 2m max. if no or incorrect numbers.

Decay of plutonium–241 to americium–241

– βe+Am→Pu 01-

24195

24194 /

Subsequent decay of americium–241

– He+Np→Am 42

23793

24195

A typical smoke detector contains 0.29 μg of americium–241. The half life of americium–241 is 431 years.

How many alpha-particles are emitted per second in the smoke detector? (11)

– 241 g of Am-241 has 6.02 × 1023 atoms (2m)

Number of atoms in 0.29 μg of Am–241

– = ( )( )

241

10×02610×0.29 23-6 . = 7.24 × 1014 (3m)

T1/2 = 431 × 365 × 24 × 60 × 60 = 1.359 × 1010 s

– 10

2

1 10×359.1

2ln=

2ln=

T = 5.1 × 10–11 s–1 (3m)

– A = N = (5.1 × 10–11)(7.24 × 1014) = 3.69 × 104 alph-particles emitted per second (3m)




10. Answer either part (a) or part (b).

(a) In the very early stages of the universe, it has been proposed that massive numbers of particles and anti-particles were created and then proceeded to annihilate each other. Cosmic microwave radiation that was produced as a result of this pair annihilation travels through the universe today. There was a small imbalance in the amount of matter produced over anti-matter (approximately one extra matter particle for each billion matter–anti-matter particle pairs. This extra matter is what forms our universe today.

Name the scientist who proposed the existence of anti-matter based on mathematical calculations in 1928. (3m) (15)

– Paul Dirac

Give the quark composition of a proton, a neutron and their respective anti-particles. (4 × 3m)

Proton

– uud

Anti-proton

– duu

Neutron

– udd

Anti-neutron

– ddu

How many photons are produced when a proton and an anti-proton annihilate? (3m) (18)

– two / 2

Explain why this number of photons is produced. (3m)

– (due to the) conservation of momentum

Calculate the minimum frequency of the photons produced during this annihilation.

Mass of proton + mass of anti-proton = 2(1.673 × 10–27) = 3.346 × 10–27 kg (3m) E = mc2 – = (3.346 × 10–27)(2.998 × 108)2 = 3.007 × 10–10 J (3m) – 2hf = 3.007 × 10–10 (3m) – f = 2.269 × 1023 Hz (3m)




If a photon of wavelength 5.6 × 10–16 m produces a proton–anti-proton pair, what is the speed of the proton produced? (23)

– frequency of photon = λ

c =

)10×(5.6

)10×9982(16-

8. = 5.354 × 1023 Hz (3m)

– energy of the photon = hf =(6.626 × 10–34)(5.354 × 1023) = 3.548 × 10–10 J (3m) – energy of photon = mass energy of proton +anti-proton +

kinetic energy of proton + kinetic energy of anti-proton (3m) – hf = 2mpc

2 + 1/2mpv2+ 1/2mpv

2 = 2mpc2 + mpv

2 (3m) – 3.548 × 10–10 = 2(1.673 × 10–27)(2.998 × 108)2 + (1.673 × 10–27)v2 (3m) – v = 1.80 × 108 m s–1 (3m)


Why must both a proton and an anti-proton be produced? (3m + 2m)

– total charge (zero) must be the same before – and after the pair production

(b) The structure of an NPN bipolar transistor is shown below.

n

n

p

B

C

A

Describe how the p- and n-type layers in the transistor are produced. Explain how they differ from each other as a result. (12)

p-type layer (2 × 3m)

– p-type, impurities (e.g. boron) are added

How it differs from n-type layers – 3 valence electrons added, extra holes available for conduction

n-type layer (2 × 3m)

– n-type, impurities (e.g. phosphorus) are added

How it differs from p-type layers – 5 valence electrons added, extra electrons available for conduction



Label the connections A, B and C in the transistor shown above. (3 × 3m) (15)

A – base B – collector / emitter C – emitter / collector

** Note that B and C answers are interchangeable.

If a potential difference is connected between B and C and another potential difference is connected between A and C, describe how the current flowing into the transistor at B is related to the current flowing into the transistor at A.

Any 2: (2 × 3m) – no current into B (collector current) if no current into A (base current) // – base current (into A) generally much less than collector current (into B) // – collector current (C) is proportional to base current (A)

** Award 6m for IE = IC + IB

Describe the basic principle of operation of a light-dependent resistor (LDR). (3 × 3m) (9)

– when light shines on LDR, electrons gain enough energy to escape bonds – electron hole pairs are produced – greater current, less resistance

or

– a semiconductor – whose conductivity increases – as light shines on it

Draw a labelled circuit diagram containing an NPN bipolar transistor and an LDR that could be used to switch on a light when ambient light intensity drops below a certain level. (3m + 3m + 2m) (20)

– transistor connected correctly to battery, collector to positive, emitter to negative, light in this part of the circuit

– fixed resistor and LDR set up as potential divider with LDR between base and emitter

– protective resistor connected to base

Explain the operation of the circuit. (4 × 3m)

– when light level is high, LDR has low resistance. Very little voltage across the LDR – base emitter voltage too small to produce collector current – as light levels drop, resistance and hence voltage across LDR increase – as base-emitter voltage increases, collector current increases sufficiently to switch

on the light

LDR

R



11. Read the following passage and answer the accompanying questions.

(a) What is meant by the term interference? (4m + 3m) (7)

– when two waves meet, the displacement produced at any point – is the (algebraic) sum of the individual displacements

(b) Coherent sources are required for destructive interference to occur. What are coherent sources? (4m + 3m) (7)

– sources of waves of equal frequency – in phase / constant phase difference

(c) Give two conditions necessary for total destructive interference to occur. (4m + 3m) (7)

– same amplitude – out of phase by: ½ / ½ cycle / 180 / / when crest meets trough

(d) Draw a diagram to show an interference pattern produced by two loudspeakers. Indicate the positions of constructive and destructive interference on the diagram. (7)

– diagram showing two loudspeakers and suitable radial lines from loudspeakers (3m)

– lines of constructive interference correctly marked (2m) – lines of destructive interference correctly marked (2m)

(e) Why is sound intensity level, rather than sound intensity, used as the quantity to measure the loudness of sound in sound level meters? (4m + 3m) (7)

– sound intensity scale ranges from very small numbers to around 1 W m–2 – sound intensity level scale is much more convenient (0 to 120 dB)



(f) Why is the dB(A) scale used in sound level meters? (4m + 3m) (7)

– decibel scale that takes into account the variation of the human ear response – to sounds of different frequencies / responds to sounds between 2000 Hz and 4000 Hz

(g) What is the sound intensity at a doorway of dimensions 2 m × 0.7 m if 2.4 J of sound energy passes through the door each minute? (7)

– power = time

energy =

60

2.4= 0.04 W (3m)

– sound intensity = area

power=

0.7)×(2

0.04 = 0.0286 W m–2 (4m)


(h) As the source of the sound moves away from the doorway, the sound intensity drops from 3.2 mW m–2 to 0.2 mW m–2 over a period of time. What is the drop in sound intensity level over the same period? (7)

– sound intensity drops by a factor of 0.2

3.2= 16 (2m)

– halves 4 times (2m) – sound intensity level drops by 4 × 3 dB = 12 dB (3m)



Section B Question 12 (2 × 28 marks)

12. Answer any two of the following parts (a), (b), (c), (d).

(a) When is an object moving with simple harmonic motion? (2 × 3m) (6)

– acceleration is proportional to – displacement from equilibrium position

** Accept a –s with explanation of terms for (6m).

An astronaut carried a simple pendulum consisting of a small metal bob and a string set at a length of 80 cm onto the moon. The pendulum was set vibrating at small-angle oscillations.

The astronaut measured the length of time it took the pendulum to travel between a point of maximum acceleration to the next point of maximum velocity to be 1.08 s.

What is the period of the oscillation? (3m) (10)

– 1.08 × 4 = 4.32 s

What would the astronaut have calculated as the acceleration due to gravity on the moon?

– g

lT π2= =

g

..

80π2=324 or

g

LT

22 π4

= = g

(0.8)π4=(4.32)

22 (4m)

– g = 1.69 m s–2 (3m)


Using the same apparatus and timer, how could the astronaut have obtained a more accurate value for the acceleration due to gravity on the moon? (6)

Any 1: (3m) – record the time for a number of oscillations instead of ¼ oscillation // – record the time at different lengths of string and get an average

Explain why this would have given a more accurate result.

** Reason should correspond to the method given above.

Any 1: (3m) – by recording much longer time, have better percentage error // – finding an average value reduces the percentage error


State two differences you would have observed from the oscillations if they were repeated on Earth. (2 × 3m) (6)

– much shorter period / quicker oscillations – oscillations would fade out more quickly due to air friction



(b) What is meant by the term polarisation of waves? (2 × 3m) (6)

– waves that vibrate – in one plane only

Why can light waves be polarised but sound waves cannot? (2 × 3m) (6)

– light waves are transverse / can vibrate in more than one plane – sound waves are longitudinal / vibrate along the direction in which the wave is moving

Describe an experiment to demonstrate the polarisation of light. (8)

– hold a sheet of polarising material in front an unpolarised light source, observe the reduction in light passing through the material (2m)

– hold a second sheet of polarising material in front of the light and rotate until virtually no light has got through (3m)

– direction of polarisation at right angles to each other, so light in both planes is blocked out (3m)

Give an advantage of wearing Polaroid sunglasses when on a boat or walking near water. (3m + 2m) (8)

– light reflected from water is partially polarised – when passed through Polaroid, significant

reduction of glare

Give another application of the polarisation of waves.

Any 1: (3m) – detecting stress in transparent components // – aerials detecting TV signals // etc.

** Accept other appropriate examples.



(c) Until the development of flat screen technology, cathode ray tubes were used in a wide range of applications, television and computer screens, oscilloscopes, ECG and EEG machines.

The principle of a cathode ray tube involves thermionic emission and the acceleration of a beam of electrons.

Distinguish between thermionic emission and photoelectric emission. (9)

– the emission of electrons (for both) (3m)

Thermionic

– from the surface of a hot metal (3m)

Photoelectric

– when electromagnetic radiation of a suitable wavelength falls on the surface (3m)

Explain how a beam of electrons is produced in a cathode ray tube. (8)

– electrons emitted to surface at cathode by thermionic emission (3m) – high voltage accelerates electrons from cathode to anode (3m) – electrons pass through hole in anode to produce beam (2m)

What is the minimum voltage required to accelerate a beam of electrons to 3 × 107 m s–1 in a cathode ray tube? (11)

– eV = ½mv2 (3m) – (1.602 ×10–19)V = ½(9.109 × 10–31)(3 × 107)2 (3m) – V = 2558.7 V (2m)


Give a reason why a higher voltage could be required for the beam of electrons to reach this speed.

Any 1: (3m) – not a perfect vacuum in tube // – after passing through anode, force of attraction of anode on electron slows electron // – effect of electric or magnetic fields on beam // etc.




(d) State two factors on which the force experienced by a current-carrying conductor in a magnetic field depends. (6)

Any 2: (2 × 3m) – length of wire // – strength of magnetic field / magnetic flux density // – size of current // – angle between wire and magnetic field

Explain why two adjacent current-carrying conductors exert a force on each other. (5)

– current-carrying conductor produces a magnetic field (3m) – when magnetic field cuts other current-carrying conductor,

it experiences a force (2m)

A rectangular coil of wire is free to rotate about an axis through its centre, as shown. The coil is placed in a uniform magnetic field, with the direction of the magnetic field in the plane of the coil and perpendicular to the 12 cm length sides of the coil. A current of 5 A flows through the coil. The force exerted on the two 12 cm sides of the coil is 4.8 N.

7 cm

B12 cm

Calculate the magnetic flux density of the magnetic field. (2 × 3m) (17)

– IL

FB = =

)(0.12)5(

84=

.B

– = 8 T


What is the size of the force exerted on the 7 cm side of the coil due to the magnetic field when the coil has rotated through 30?

– F = BILsin (2m) – = (8)(5)(0.07)sin(30) = 1.4 N (3m)


Why does this force not contribute to the rotation of the coil? (3m)

– the direction of the force is parallel to the axis of rotation

What is the size of the force on each of the 12 cm sides of the coil in this position? (3m)

– the force remains the same / 4.8 N


Notes:


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2016.1 L.35-36 ms -07 ha-final · 2019. 11. 23. · 2016.1 L.35/36_MS 3/60 Page 3 of 59 exams DEB Pre-Leaving Certificate Examination, 2016 Physics Ordinary & Higher Level Explanation