Trial PahangAnswers For Physics Paper 1 Trial 2010

1. C 6. A 11.B 16. A 21. C 26. C 31. D 36. A 41. C 46. A

2. B 7. B 12.D 17.B 22. D 27. D 32. C 37. C 42. A 47.A

3. B 8. A 13.B 18 A 23. A 28. D 33. B 38. C 43. C 48. B

4. D 9. C 14. A 19 A 24.C 29. C 34. B 39. C 44. A 49. B

5. D 10.B 15. C 20. C 25.B 30. B 35. B 40. C 45. B 50. B

Pahang SPM Trial Exam 2010Marking Scheme Physics Paper 2

Section AQuestion 1 Answer Note

(a) 1 0.2 s With correct unit(b) 1 - 0.4 s With correct unit(c) 2 50.0s – (0.4 s) =50.4 s With correct unit

Total : 4 marks

Question 2 Answer Note(a) 1 e.m.f induced/ current induced(b) 1 North

(c)(i)(ii)(iii)

111

Total: 5 marks

Question 3 Answer Note(a) 1 Force that oppose acted force forward

(b)(i) 1 W= mgh = 430 N With correct unit(b)(i) 1

1

Resultant force, F =W sin 20 – 147.1 N = 147.1 N – 147.1 N = 0 N With correct unit

(c) 11

StationaryBecause balance force/ Force in equilibrium/ acted force = frictional force

Total: 6 marks

Galvanometer

S

CompassKompas

Question 4 Answer Note(a)(i) 1 Thermionic emission(a)(ii) 1 To accelerate electrons

(b) 1 Kinetic energy heat energy + light energy(c)(i) 1 Alternating current/ AC(c)(ii) 1

1Period, T = 3 0.02s = 0.06 sF=1/T = 1/0.06s = 16.67 Hz

With correct unit

(d) 1

Total: 7 marks

Question 5 Answer Note(a) 1 Atmospheric pressure/ Air pressure

(b)(i) 1 The volume of air trapped in the beaker diagram 5.1 is larger Quantities must be same as stated in questions.

(b)(ii) 1 The pressure of air trapped in the beaker diagram 5.2 is bigger(c)(i) 1 The higher the pressure, the lower the volume of air trapped(c)(ii) 1 Boyle’s law

(d) 111

Density of air < waterWater exerts upthrustUpthrust > Weight of beaker + air trapped

Total : 8 marks

0.01

s / div

4

V / div

Question 6 Answer Note(a) 1 Sources that have the same frequency and in phase.

(b)(i) 1 The wavelength of water wave in Diagram 6.1 is shorter(b)(ii) 1 The distance between two consecutive nodal lines, x in Diagram

6.1 is shorter(c)(i) 1 The shorter the wavelength of water wave, the shorter the distance

between two consecutive nadal lines, x. (c)(ii) 1 Interference(d)(i) 2

(d)(ii) 1 A: BrightB: Dark/Dim

Total : 8

Question 7 Answer Note(a) 1 Elastic potential energy

(b)(i) 1 X = 4.0 cm With correct unit(b)(ii) 1

1F = kxk = 0.9 N cm-1 @ 90.0 N m-1 With correct unit

(b)(iii)

11

l = 12 – 5.56 = 6.44 cmWith correct unit

(c)(i) 1 The higher the weight of load, the higher the compression(c)(ii) 1 The spring will not return to its original shape// spoil

(d) 2 Use a larger diameter of spring wireAdd more springs in parallelUse stiffer spring/ any two

Total: 10 marks

Question 8 Answer Note(a)(i) 1

115 hoursShort time taken to decay/Can decay faster and they leave

A B

harmless daughter nuclei(a)(ii) 1

1Gamma rayStrong penetration power/ Can penetrate the soil

(a)(iii) 1 LiquidEasier to dissolve in water/ Produce more even radiations.

(b)(i) 1

1

8/8 4/8 2/8 1/8 T1/2 T1/2 T1/2

t = n (T1/2) = 3 (28 years) = 84 years

With correct unit* 2nd method:

Undecay value = Original value

2n = 8 n = 3t = 3 28 years = 84 years

(c) 1 Thickness control/Examine contamination in canned food/ Medical screening/treatment// smoke detector/ sterilizing/

(d)(i) 1

1

Mass defect = (2.014012u + 3.016029u) – (4.0022603u + 1.008665u)= 0.018863 u = 0.018863 1.66 10-27 kg= 3.13 10-29 kg With correct unit

(d)(ii)

1

E = mc2

= 3.13 10-29 kg (3.0 108 ms-1)2

= 2.82 10-12 J With correct unitTotal: 12 marks

Section B

Question 9 Answer Note(a)(i) 1 Refraction

(a)(ii) 1 Ratio of sin i / sin r // The ratio of speed of light in vacuum relative to that speed through a medium

(b) 1111

1

Refractive index of the glass is higher.The density of glass is higherThe angle of refraction of light ray in glass is shorterThe higher the density of medium, the smaller the angle of refraction of light.The higher the density of medium, the higher the refractive index.

(c)

- He must shoot the target at the lower position of the image. 1- m

Light refracted away from normal – 1 m

Extrapolation to show position of the observing image – 1 m

(d)(i)

Distribution of marks:1 mark - Labeled 90 prism1 mark - Arrangement of prisms --- facing each other1 mark - Location of objective lens1 mark - Location of the eyepiece lens1 mark - Light ray with 2 times total internal reflection at the 1st

prism1 mark - Light ray with 2 times total internal reflection at the 2nd

prism

(d)(ii)

2

2

Modifications ReasonsObjective lens with largerdiameter.

More light passes through the lens/

Eyepiece lens with higherpower // Thicker eyepiece lens

Act as a strong magnifying glass

Total : 20 marks

Eyepiece

Total internal reflection

4545

45 45

Objective lens

Total internal reflection

Question 10 Answer Note(a) 1 A wave in which the vibration of particles in the medium is

parallel to the direction of propagation of the wave(b)(i) 1 The amplitude in Diagram 10.2 is higher(b)(ii) 1 The peak value, a2 in Diagram 10.2 is higher(b)(iii) 1 The higher the amplitude of vibration of tuning forks, the higher

the peak value(b)(iv) 1 The higher the peak value, the louder the sound(b)(v) 1 The higher the amplitude, the louder the sound

(c) 1111

1

- Use ultrasound - Ultrasound is transmitted to the sea bed- a receiver will then detect the reflected the reflected pulses- the time taken by the pulse to travel to the seabed and return to the receiver being recorded, t- the depth of the sea can be calculated using the formula,

Max 4 marks

(d)

1

1

11

1

1

11

11

11

Suggestions Reason Loudspeakers are positioned at quite a distance away.

So that the distance between consecutive constructive / destructive interference is smaller.

The two main loudspeakers are not positioned opposite to each other

To prevent multiple reflections

Fix soft boards/ wooden/ materials which are sound absorbers

Reflection effects can be reduced

Use thick carpet/ Wooden floor/ Rubber floor

To prevent echo

Assemble a high power speaker system

To produce a high amplitude of sound wave

Assemble the speaker at a high place

Wide coverage // the wave is not blocked

Max 10 marks

Total : 20

Section C

Question 11 Answer Note(a)(i) 1 Archimedes’ principle states that the buoyant force on an

object immersed in a fluid is equal to the weight of fluid displaced by the object.

(a)(ii) 1111

- Volume of air displaced equal to volume of a balloon- Density of air decreased as a altitude increase- Weight of displaced air become smaller- At certain height weight of displaced air equal to weight of the balloon

(b)

1

1

11

1111

11

Characteristics ExplainationLarge ballon To produce bigger buoyant

/ upthrust // Increase the volume of the air displaced

Use 2 burners // Many burners

To produce bigger flame // heat up the gas in the balloon faster

Synthetic nylon Light-weight/ strong /air-proof material

High temperature of the air in the balloon

Reduce the density /weight of the air in the balloon

Hot air balloon Q is chosen Large balloon, use 2 burners / many burners, use synthetic nylon and has high temperature of the air in the balloon.

(c)(i) 1

1

Mass = density x volume

Mass = 0.169 kg m-3 x 1.2 m3 = 0.20 kg

With correct unit

(c)(ii)1

1

1

Calculate mass of displaced air correctlym = 1.3 kg m-3 x 1.2 m3 =1.56kgCalculate weight of displaced air correctly and state that bouyant force equal to weight of displaced airWeight of displaced air = bouyant force = mg = 1.56 x 10= 15.6N

With correct unit

Total : 20 marks

Question 12 Answer Note(a) 1 A fuse is a very thin wire, which either melts or vaporizes when

too much current flows through it(b) 1

111

- A parallel circuit can run several devices using the full voltage of the supply.- If one device fails, the others will continue running normally- If the device shorts, the other devices will receive no voltage, preventing overload damage.- A failure of one component does not lead to the failure of the other components.- More components may be added in parallel without the need for more voltage.- Each electrical appliance in the circuit has it own switch.

Max 4 marks

(c)(i) 1 - The electrical appliance use 240 V of voltage to generates 500 W of power.

(c)(ii) 1

1

Current = Power/Voltage

Current = 500/240 = 2.08 A

With correct unit

(c)(iii) 1

1

Efficiency = Output Power x 100 % Input Power Output Power = 85 x 500 100 Output power = 425 W

With correct unit

(d)Characteristics Explanation

Thin fuse wire Less space needed/ to carry a limited electrical current/ less mass hence low heat capacity/ shorter time to heat

up to melting point and blow.

Ceramic cartridge Can withstand higher temperature because sparks created by high voltage, 240V can be huge/

Fuse rating is 13 A Maximum rating must be higher than normal current.

Low melting pointFor fast blow/ Melting faster when excessive current flow/ Easy to cut the current flow.

R is chosen because Because it has thin fuse wire, ceramic cartridge, fuse rating is 13 A and low melting point.

Total : 20 marks

MARKING SCHEME PHYSICS PAPER 3 (TRIAL 2010)

No Answer Marks1 a) (i) Distance between 2 coherent sources of sound waves, a 1 (ii) Distance between two consecutive instances of loud sound, x 1 (iii) Distance between loudspeakers and where soun is

detected/frequency of sound waves/ wavelength of sound waves

1

b) (i)

(ii)

a = 1.0m x = 15.2cma = 1.5m x = 10.3 cma = 2.0m x = 7.3 cma = 2.5 m x = 5.9 cma = 3.0 m x = 4.7 cm

/ m-1 x / cm x / m

1.00 15.2 3.04

All correct 2M4 correct 1M3 correct 0M

( accept : 0.1 cm)

0.670.500.400.33

10.37.35.94.7

2.061.461.180.94

Tabulate data

1. Shows a table which have and x

2. State the correct unit ( : m-1 and x : m)

3. All values x are correct

4. Values of are consistent to 2 decimal point

5. Values of x are consistent to 2 decimal point

1

1

1

1

1 (7) c)

Draw graph x against

1. The responding variable, x at y axis

2. The manipulated variable, at x axis

3. Sates the unit of variable correctly4. Both axis with the even and uniform scale5. 5 points correctly plotted6. A smooth best fit straight line7. Minimum size (50% of graph paper)

No of ticks Score7 5

5-6 43-4 32 21 1

5 d) x is inversely propotional to a OR x is directly proportional to

1/a1

TOTAL 16No Answer Marks

2 a) Extrapolation on graph to cross y-axis where h=0mCorrect answer with unit 1.00 x 105 Nm-2

1

1 (2) b) (i) Large triangle ( 4 x 3 larger square)on graph

Correct substitution (refer to triangle drawn) exp:

Correct answer with correct unitk = 5999.3 Nm-3

1

1

1 (3)

(ii) Method with correct substitution refer to answer in (i)

0.12 (5.999.3)

Correct answer = 719.92 kgm-3

1

1 (2)

c) Line on graph at h = 0.5 m

Correct substitution OR final answer with unit (accept the value half of small square )

P = 1.043 x 105 + 1.0 x 105 Nm-2 OR 2.043 x 105 Nm-2

1

1 (2)

d) (i) Increase

(ii) The higher the density, the higher the pressure and produce the higher gradient of graph P - h

1

1 (2) e) Describe the method to avoid parallax error

The eye at the same level of meniscus of water 1TOTAL 12

3. a) Correct inference refer to actual situation and correct directionThe distance between paper to the lens / focal length is depends on the thickness of lens.

1

b) Correct hypothesis with correct direction (refer to variables that choose for the experiment)The higher the thickness of lens, the longer its focal length 1

c) i) Aim of the experimentTo Investigate the relationship between the focal length

and the thickness of lens.

ii) Variables (MV and RV should can be measured)MV : The thickness of lensRV : The focal lengthFV : Refractive index of lens / the diameter of lens /

type of lens

iii) List of apparatus (ruler and vernier caliper/micrometer screw gauge must be in the list)

5 convex lenses of different thickness but same diameter, light box/candle with flame, low voltage power supply, screen, plasticine, ruler and micrometer screw gauge/ vernier caliper

iv) Arrangement of apparatus (Should be relevant and functional)

1

1

1

1

to a.c

screen lens with holder ray box (far from the lens as a distance object)v) Procedures

1. The thickness of lens is measured using the vernier

caliper/micrometer screw gauge and record.2. The lens is adjusted until the clear sharp image of

filament is obtain on the screen.The distance between centre of lens to the screen is measured as the focal length.3. The experiment is repeated use another four

different thickness of lenses. (accept without values because the thickness is unknown and measured before the experiment)

NOTES : ACCEPT OTHERS EXPERIMENT THAT RELATE TO THE SITUATION SUCH AS

1. relationship between u and v where f is measure use

the formula

2. use distance object (outside object) to determine the

focal length of lens

vi) Show the table Thickness / cm Focal length / cm

vii) State the graph should be drawn or draft the graph Draw a graph, the focal length against the thickness of lens

OR focal length / cm

1

1

1

1

1

1

Thickness / cmTotal Marks 12

4. a) Correct inference refer to actual situation and correct directionThe brightness of lamp depend on the speed of wheel.

1

b) Correct hypothesis with correct direction (refer to variables that choose for the experiment)The higher the speed of magnet, the higher the current induced 1

c) i) Aim of the experimentTo Investigate the relationship between the high of

magnet and the current induced.ii) Variables (MV and RV should can be measured)

MV : Speed (of the magnet)RV : current (induced)FV : Numbers of turns of the coil / the diameter of

coil / the thickness of wire

iii) List of apparatus (ruler and galvanometer must be in the list)

coil/selonoid made of copper wire, ruler , galvanometer and connecting wire

iv) Arrangement of apparatus (Should be relevant and functional)

v)

Procedures1. The apparatus is set up as the diagram shown.2. The strong magnet is released at 10 cm of height. 3. The division of the galvanometer’s pointer deflect is recorded.

1

1

1

1

1

11

h

G

4. The experiment is repeated at the height of magnet is 15cm, 20 cm, 25 cm and 30 cm

vi) Show the table height / cm division

vii) State the graph should be drawn or draft the graph Draw a graph, the division against the height of magnet released

OR division

height / cm

1

1

1

Total Marks 12