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Answers 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. C 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. D 25.C 30. B 35. B 40. C 45. D 50. B
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Pahang SPM Trial Exam 2010
Marking Scheme Physics Paper 2
Section A
Question 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)
11
1
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
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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) 11
Period, T = 3 0.02s = 0.06 s F =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 besame as stated inquestions.
(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) 11
1
Density of air > waterWater exerts upthrust
Upthrust > Weight of beaker + air trapped
Total : 8 marks
0.01
s / div
4
V / div
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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 Diagram6.1 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)
1
1
cm x
x
x
F
x
F
56.5
5
4
6.3
2
2
2
2
1
1
l = 12 – 5.56 = 6.44 cm With correct unit
(c)(i) 1 Compression of the spring is directly proportional to the load in the
balance/ 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 parallel
Use stiffer spring/ any two
Total: 10 marks
A B
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Question 8 Answer Note
(a)(i) 11
15 hoursShort time taken to decay/Can decay faster and they leaveharmless daughter nuclei
(a)(ii) 11
Gamma 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 1/8T1/2 T1/2 T1/2
t = n (T1/2)= 3 (28 years) = 84 years
With correct unit
* 2nd method:
Undecay value =
n
2
1Original value
8
8
2
1
8
1
n
2n = 8 n = 3
t = 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 108ms-1)
2
= 2.82 10-12 J With correct unit
Total: 12 marks
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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 vacuumrelative to that speed through a medium
(b) 11
11
1
Refractive index of the glass is higher.The density of glass is higher
The angle of refraction of light ray in glass is shorterThe higher the density of medium, the smaller the angle ofrefraction of light.The higher the density of medium, the higher the refractiveindex.
(c)
- He must shoot the target at the lower position of the image.
1- m
Light refractedaway from normal – 1 m
Extrapolation toshow position of
the observingimage – 1 m
(d)(i)
Eyepiece
Total internal
reflection
45 45
45 45
Objective lens
Total internal
reflection
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Distribution of marks:
1 mark - Labeled 90 prism1 mark - Arrangement of prisms --- facing each other
1 mark - Location of objective lens
1 mark - Location of the eyepiece lens
1 mark - Light ray with 2 times total internal reflection at the 1 st prism1 mark - Light ray with 2 times total internal reflection at the 2nd prism
(d)(ii)
2
2
Modifications Reasons
Objective lens with largerdiameter.
More light passes throughthe lens
Eyepiece lens with higherpower // Thicker eyepiece
lens
Act as a strong magnifyingglass
Total : 20 marks
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) 1
11
1
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 tothe receiver being recorded, t- the depth of the sea can be calculated using the formula,
2
vtd
Max 4 marks
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(d)
1
1
11
1
1
11
11
1
1
Suggestions Reason
Loudspeakers are positioned at quite a
distance away.
So that the distance betweenconsecutive constructive /
destructive interference is
smaller.The two mainloudspeakers are not
positioned opposite toeach other
To prevent multiplereflections
Fix soft boards/
wooden/ materialswhich are soundabsorbers
Reflection effects can be
reduced
Use thick carpet/
Wooden floor/ Rubberfloor
To prevent echo
Assemble a high powerspeaker system
To produce a high amplitudeof sound wave
Assemble the speaker at
a high place
Wide coverage // the wave is
not blocked
Max 10 marks
Total : 20
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Section C
Question 11 Answer Note
(a)(i) 1 Archimedes’ principle states that the buoyant force on anobject immersed in a fluid is equal to the weight of fluid
displaced by the object. (a)(ii) 111
1
- 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
1
11
1
1
1
Characteristics Explaination
Large 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 airin the balloon
Reduce the density /weightof 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 =1.56kg
Calculate weight of displaced air correctly and state that
bouyant force equal to weight of displaced air
Weight of displaced air = bouyant force
= mg = 1.56 x 10= 15.6N
With correct unit
Total : 20 marks
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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 needfor 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 500W 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 500100
Output power = 425 W
With correct unit
(d)
Characteristics Explanation
Thin fuse wire Less space needed/ to carry alimited electrical current/ lessmass hence low heat
capacity/ shorter time to heatup to melting point and blow.
Ceramic cartridge Can withstand highertemperature because sparkscreated by high voltage,240V can be huge/
Fuse rating is 13 A Maximum rating must behigher than normal current.
Low melting point For fast blow/ Melting faster
when excessive currentflow/ Easy to cut the current
flow.
R is chosen because Because it has thin fuse wire,
ceramic cartridge, fuse ratingis 13 A and low melting
point.
Total : 20 marks
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http://www.gcse.com/glos.htm#voltage*http://www.gcse.com/glos.htm#voltage*
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1
MARKING SCHEME PHYSICS PAPER 3 (TRIAL 2010)
No Answer Marks
1 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 = 15.2cm
a = 1.5m x = 10.3 cma = 2.0m x = 7.3 cma = 2.5 m x = 5.9 cm
a = 3.0 m x = 4.7 cm
a
1 / m
-1x / cm x / m
1.00
0.67
0.500.400.33
15.2
10.3
7.35.94.7
3.04
2.06
1.461.180.94
Tabulate data
1. Shows a table which havea
1 and x
2. State the correct unit (a
1 : m-1 and x : m)
3. All values x are correct
4. Values ofa
1 are consistent to 2 decimal point
5. Values of x are consistent to 2 decimal point
All correct 2M
4 correct 1M3 correct 0M
( accept : 0.1 cm)
1
1
1
1
1 (7)
c)Draw graph x against
a
1
1. The responding variable, x at y axis
2. The manipulated variable,a
1 at x axis
3. Sates the unit of variable correctly4. Both axis with the even and uniform scale5.
5 points correctly plotted
6.
A smooth best fit straight line
7. Minimum size (50% of graph paper)
No of ticks Score
7 5
5-6 4
3-4 3
2 2
1 1 5
d) x is inversely propotional to a OR x is directly proportional to 1/a 1
TOTAL 16
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2
No Answer Marks
2 a) Extr apolati on on graph to cross y-axis where h=0mCorrect answer with uni t
1.00 x 105 Nm-2
1
1 (2)
b) (i) Large tr iangle ( 4 x 3 larger square)on graph
Correct substi tution (refer to tr iangle drawn) exp:
07.0
)100.1()106.1( 55
k
Correct answer with correct uni t
k = 5999.3 Nm-3
(ii) Method with corr ect substitu tion r efer to answer in (i )
0.12 (5.999.3)
Corr ect answer
= 719.92 kgm-3
1
1
1 (3)
1
1 (2)
c) Line on graph at h = 0.5 m
Correct substitution OR f inal 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 paral lax errorThe eye at the same level of meniscus of water 1
TOTAL 12
3. a) Correct in ference refer to actual situation and correct directionThe distance between paper to the lens / focal length is depends onthe 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 andthe thickness of lens.
ii) Var iables (MV and RV should can be measured)
MV : The thickness of lens
RV : The focal lengthFV : Refractive index of lens / the diameter of lens / type of
lens
iii) L ist of apparatus (rul er and vernier cali per/micrometer
screw gauge must be in the li st)
5 convex lenses of different thickness but samediameter, light box/candle with flame, low voltage
power supply, screen, plasticine, ruler and
micrometer screw gauge/ vernier caliper
1
1
1
1
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3
iv) Ar rangement of apparatus (Should be relevant and
functional)
to a.c
screen lens with holder ray box (far from thelens as a distance object)
v) Procedures1. 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 betweencentre of lens to the screen is measured as the focallength.
3.
The experiment is repeated use another four differentthickness of lenses. (accept wi thout val ues because the
thi ckness is unknown and measured before the
experiment)
NOTES : ACCEPT OTHERS EXPERIM ENT THAT RELATE
TO THE SITUATION SUCH AS
1. relationship between u and v where f i s measure use
the formula f vu
111
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 shoul d be drawn or draf t the graphDraw a graph, the focal length against the thickness of lens
OR focal length / cm
Thickness / cm
1
1
1
1
1
1
Total Marks 12
4. a) Correct in ference refer to actual situation and correct directionThe brightness of lamp depend on the speed of wheel. 1
b) Correct hypothesis with correct dir ection (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 ofmagnet and the current induced. 1
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ii) Var iables (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) L ist of apparatus (rul er and galvanometer must be in the
list)coil/selonoid made of copper wire, ruler , galvanometer and
connecting wire
iv) Ar rangement 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 isrecorded.
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 shoul d be drawn or draf t the graphDraw a graph, the division against the height of magnet
released
1
1
1
1
1
1
1
1
1
h
G