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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.   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