Physics Spm

17
SULIT 4531/1 4531/1 © 2013 Hak Cipta BPSBPSK SULIT 1 PENTAKSIRAN DIAGNOSTIK AKADEMIK SBP SIJIL PELAJARAN MALAYSIA 2013 MATAPELAJARAN FIZIK SKEMA JAWAPAN KERTAS 1 NO. SOALAN JAWAPAN NO. SOALAN JAWAPAN 1 C 26 C 2 B 27 D 3 D 28 B 4 B 29 D 5 B 30 C 6 D 31 A 7 C 32 D 8 A 33 B 9 B 34 D 10 D 35 B 11 B 36 A 12 C 37 D 13 C 38 C 14 C 39 B 15 B 40 D 16 C 41 D 17 A 42 A 18 A 43 A 19 D 44 B 20 A 45 C 21 C 46 C 22 B 47 D 23 C 48 C 24 B 49 B 25 B 50 A

description

Trial 2013 Paper

Transcript of Physics Spm

Page 1: Physics Spm

SULIT 4531/1

4531/1 © 2013 Hak Cipta BPSBPSK SULIT

1

PENTAKSIRAN DIAGNOSTIK AKADEMIK SBP

SIJIL PELAJARAN MALAYSIA 2013

MATAPELAJARAN FIZIK

SKEMA JAWAPAN KERTAS 1

NO. SOALAN JAWAPAN NO. SOALAN JAWAPAN

1 C 26 C

2 B 27 D

3 D 28 B

4 B 29 D

5 B 30 C

6 D 31 A

7 C 32 D

8 A 33 B

9 B 34 D

10 D 35 B

11 B 36 A

12 C 37 D

13 C 38 C

14 C 39 B

15 B 40 D

16 C 41 D

17 A 42 A

18 A 43 A

19 D 44 B

20 A 45 C

21 C 46 C

22 B 47 D

23 C 48 C

24 B 49 B

25 B 50 A

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PENTAKSIRAN DIAGNOSTIK AKADEMIK SBP

SIJIL PELAJARAN MALAYSIA 2013

MATAPELAJARAN: FIZIK

SKEMA JAWAPAN KERTAS 2

No Answer Marks Total

1 (a) length 1

(b) Tail 1

(c) to measure depth 1

(d) 0.42 cm 1

Total 4

No Answer Marks Total

2 (a) Product of an applied force and displacement / distance //Force X distance

1 1

(b) W = F X S M1 = 110 x0.5 M2 = 55 J / Nm

1 1

2

(c) F = ma M1 110 – 100 = 10 (a) M2 10 = 10 a a = 1 ms-2

1 1

2

Total 5

No Answer Marks

Total

3 (a) Thermionic emission

1

(b) Uniform acceleration

1

(c) (i) VP = 4 V 1

(ii)

T = 3 x 0.02 = 0.06 s

f = 16.7 Hz

1

(iii) Sinusoidal with one complete cycle 1

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Total 6

No Answer Marks Total

4 (a) The amount of heat energy required to increase the the

temperature of 1kg mass by 10 C.

1 1

(b) Pt = mcӨ

c = 48(900)

60

= 720 J kg-1 0

C-1

2 2

(c) M1 Land has a smaller specific heat capacity than sea

// Land faster increase in temperature // Land is

warmer than the sea

M2 Air above the land is heated up and rises

M3 Cooler air from the sea moves towards the land

Or diagram

M1 Label Sea(cold), Land (Hot)

M2 Show Hot air on land rises up

M3 Show Cold air moves towards the sea

1

1

1

3

(d) Sea breeze 1 1

7

No Answer Marks Total

5 (a)(i) Spring balance reading: diagram 5.1 > 5.2 1

(ii) Buoyant force / Upthrust 1 2

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(b)(i) Spring balance reading: diagram 5.3 > 5.2 1

(ii) Density of water > density of oil 1

(iii) As density of liquid increases, buoyant force increases 1 3

(c)(i) pressure at Y > pressure at X // vice versa // Y is at

greater depth than X // pressure is directly proportional to

depth.

1

1

(ii) Buoyant force = (Difference in pressure) x A 1 3

Total 8

No Answer Marks Total

6 (a) Longitudinal waves // Mechanical wave 1 1

(b) 1. Loud sound - when antinodes/crest/trough

overlapped/meet/encounter another

antinodes/crest/trough, a constructive interference

occur

2. Soft sound – when nodes/crest/trough

overlapped/meet/encounter another nodes/ trough,

a destructive interference occur

1

1

2

(c) (i) Diagram 6.1 > Diagram 6.2 // vice-versa 1

(ii) Diagram 6.1 < Diagram 6.2 // vice-versa 1

(iii) a α 1 3

(d) M1 frequency// wavelength

M2 amplitude

M3 Distance between loud speaker and observer

(Any two)

1

1

2

8

No Answer Marks Total

7 (a) Reflection 1 1

(b) (i) 1

1

1

3

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(ii) Virtual // Upright // Diminished 1 1

(c) 1. Aim the concave mirror to the sun

2. parallel ray from sun reflected and focus to F

3. amount of heat increases as all rays focus at F

Max : 2 points

1

1

2

(d) (i) 1. lens Q

2. f is smaller

1

1

1

(ii) Use concave mirror 1

1

Total 10

No. Answers Marks

8 (a) When a potential difference of 9V is connected across the

bulb, it will produce power of 20J per second.

1 1

(b) (i) P = V/I

20W = 9V/I

I = 20W/9V

= 2.22A

1

1

2

(ii) R = V/I

= 9V/2.22A

= 4.05 Ω

1

1

2

(c) (i) The more the thickness of the wire, the lower the loss of

energy from the filament.

1

2

(ii) Efficiency = Energy Output / Energy Input x 100%

E Filament P = 15 x 100 %

20

= 75.00%

E Filament Q = 28 x 100 %

30

= 93.33%

E Filament R = 43 x 100 %

50

= 86.00%

1

1

1

1

5

(iii) Filament Q has highest efficiency 1

1

2

Total 12

No Answer Marks Total

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9 (a) (i) Property of material that enable it to return to its original

shape and size after an external force that acting on it is

removed

1

(ii) Compression of the spring in Diagram 9.2(a) is higher

than 9.1 (a)

Distance moved by the trolley in Diagram 9.2 (b) is

further than 9.1 (a)

Elastic potential energy of the spring Diagram 9.2(a) is

higher than 9.1 (a)

The higher the compression, the further the distance

moved by the trolley

The bhigher the compression, the higher the elastic

potential energy

1

1

1

1

1

(b) The forces between atom are attractive force and

repulsive force

When the force is applied , the distance between atoms

decreases

The repulsive force is acted on the atoms

When applied force is removed, repulsive force pushes

the atoms back to original

1

1

1

1

(c)

Design Reasons

Many springs

(arranged in parallel) High elasticity

Strong material for

frame // steel

Not break easily //

Not rust easily

Low spring constant To get more extension

Nylon// polyethene//

strong fabric Does not tear easily

Has net around / /

bigger diameter of

trampoline

Prevent children from

slipping to the floor

2

2

2

2

2

2

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Total 20

No Answer Mark Total

10 (a) A magnetic field is a region in which a magnetic material

will experience a magnetic force

1 1

(b) The number of turns of the coils shown in diagram 10.2

is greater than 10.1

The pattern of magnetic field in diagram 10.2 is denser

than 10.1

The deflection of ammeter in diagram 10.2 is bigger than

10.1

1

1

1

3

(c) The closer the pattern of magnetic field // the denser of

magnetic field, the greater the strength of the magnetic

field.

The strength of the magnetic field increase when the

number of turns of coils increase

1

1

2

(d) (i) 1. Repel each other

1 1

(ii) 1. When current flows, magnetic field is form //

Diagram

2. Direction of magnetic field at the centre is the same

// Magnitude of magnetic field is stronger at the

centre // Diagram

3. Two forces produced is in opposite direction //

Diagram

1

1

1

3

(e) Modification Explanation

More number of

turn for the coil/

Increase the strength of

magnetic field // increase the speed of

motor

Has commutator To reverse the current flow

in the coil // ensure core

1,1

Force Force

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rotate in one direction

Brush Enable current flow through the coil

Semicircular

magnet // curved

magnet

To produce radial magnetic field

Soft iron core Concentrate the magnetic field.

1,1

1,1

10

Total 20

11 (a) Archimedes’ Principle states that the buoyant force is

equal to the weight of fluid displaced.

1 1

(b) 1.Density of air decreases as altitude increases

2.Bouyant force become smaller

3.At certain height, weight of displaced air equal to

weight of the balloon.

4.Net force zero

1

1

1

1

4

(c) Characteristics Explanation

Fiber composite Light and strong

Bigger volume High buoyant force

The distance between

plimsoll line from the

surface of the sea is high

Can carry more load

Bigger size of the

propeller

To produce greater

forward thrust

Q is chosen the body is made from

fibre composite, volume

is bigger, the distance

between plimsoll line

from the surface of the

sea is higher and bigger

size of the propeller

2

2

2

2

2

10

(d) (i) F = ρgv

= 1020 × 10 × 5

= 51000 N

1

1

2

(ii) Total weight

15000 + W = 51000 N

W = 36000 N

1

1

1

3

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Total 20

No Answer Marks Total

12 (a)

Cosmic ray// radiation from surrounding

//radioactive materials from earth// leakage of

radioactive from nuclear power plant

1

1

(b)(i) Alpha 1 1

(ii) 1. The ray ionises the air molecule

2. Negative ions attracted to the plate

3. Neutralised the electroscope

1

1

1

3

(c) Characteristics Reasoning

Liquid Easy to flow with blood

Short half life Not long in the body //

less harmful

Gamma Ray Cannot ionised the

living cell // high

penetrating power

GM tube detector Can detect ray

effectively// portable

K is chosen Because it is in liquid

state , has short half

life, emits gamma ray

and can be detected

easily detected by GM

tube detector.

2

2

2

2

2

10

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(d)(i) (i) m = [235.04392 + 1.00867] – [140.91963 +

92.92157 + 2(1.00867)]

= [236.05259] – [ 235.85854]

= 0.19405u

m = 0.19405 x 1.66 x 10 -27

= 0.0322 x 10 -27

= 3.22123 x 10-28

kg

1

1

1

3

(ii)

E = mc 2

= (3.22123 x 10-29

) (3.0 x 10 8) 2

= 28.99107 x 10-12

= 2.899107 x 10 -11

J

1

1

2

Total 20

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PENTAKSIRAN DIAGNOSTIK AKADEMIK SBP

SIJIL PELAJARAN MALAYSIA 2013

MATAPELAJARAN: FIZIK

SKEMA JAWAPAN KERTAS 3

SECTION A

No. 1 Answer Mark

(a) (i) Manipulated variable = Object distance, u // Image distance, v 1

(ii) Responding variable = Image distance, v // M 1

(iii) Constant variable = Thickness//focal length//power 1

(b) Tabulate u, v and m correctly in the table.

A Shows a table u,v and m.

B State the correct unit of u, v and m.

C At least 3 values of v are correct

D All values of v are correct

E At least 3 value of m are correct

F All values of m are correct

G State a consistent decimal place for u, v and m.

u/cm v/cm m

6.3 8.8 1.40

6.5 7.7 1.18

6.7 6.7 1.00

6.9 5.5 0.80

7.1 4.3 0.61

7

(c) Draw the graph of v against m .

A - Label y-axis and x-axis correctly

B - States the unit at the axis correctly

C - Both axes with the even and uniform scale:

D - 5 points correctly plotted:

E - a smooth best straight line

F - minimum size of the graph is 5 x 4 squares of 2 x 2 cm.

Draw the graph of v against m .

A - Label y-axis and x-axis correctly

B - States the unit at the axis correctly

C - Both axes with the even and uniform scale:

D - 5 points correctly plotted:

E - a smooth best straight line

F - minimum size of the graph is 5 x 4 squares of 2 x 2 cm.

No of ticks Score

6 5

5 4

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3-4 3

2 2

1 1

5

(d) State the correct relationship based on the candidate’s graph

v is increasing linearly to m

1

TOTAL 16

No 2 Answer Mark

(a)(i) 1 increases // I decreases

I

1

(ii) 1.0 Ω 1

(b) Show a big on the graph to determine the gradient(5 x 4 blocks)

The correct 12

12

xx

yy

from the drawn

m = 5.14A (Ans + unit)

1

1

1

(c)

R = E ( 1 ) - r

I

E = gradient of the graph

= 5.14 V // 5.14 A

1

1

1

(d) Show line on graph to get 1/I

1/I =2.25

I = 0.44

1

1

1

(e) 1. The connection of the wires must be tight.

2. The circuit is switched off whenever the readings were not taken

from the meters. This is to reduce energy loss from the cell.

3. The eye is perpendicular to the scale of ammeter // Pointer’s image

is not seen on the mirror strip of ammeter.

Max 1 mark

1

Total 12

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SECTION B

No Answer Mark

3 (a) State the suitable inference

The foot feels more painful when it is stepped on by an object that has

smaller surface area // Pressure depend on surface area

(b) State a relevant hypothesis

The smaller the surface area the greater the pressure

1

(c)

State the aim of experiment

To investigate the relationship between surface area and pressure.

1

State the suitable manipulated variables and responding variable

(Quantity that can be measured)

MV - surface area

RV - pressure/depth of depression

1

State the constant variable

CV – Weight/force/mass

1

State the complete list of apparatus and materials

Plasticine, slotted weight , wooden rod and meter rule

1

Draw the functional arrangement of the apparatus

1

State the method to control the manipulated variable The apparatus is set up as shown in the diagram.

Start the experiment with a wooden rod has surface area 1 cm2 .

Placed the load of mass 200 g on the top of wooden rod as shown on

diagram

1

State the method to measure the responding variable

Measure the depth of depression made on the plasticine

1

Repeat the experiment at least 4 times with the values

Repeat the experiment 4 times with surface area of rod 2 cm2, 3 cm

2 , 4

cm2 and 5cm

2

(Note : Based on SPM standard , at least five manipulated values

required.)

1

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State how the data tabulated with the title MV and RV

Surface area Depth

1

2

3

4

5

1

State how the data is analysed, plot a graph RV against MV

1

TOTAL MARK 12

4(a) State the suitable inference

The brightness of the lamp increases when the speed of the magnet in the

coils(solenoid) increases// Induced current depend on the speed of

magnet

1

(b) State a relevant hypothesis

The magnitude of the induced current increases when the speed of the

magnet increases.

1

(c) State the aim of experiment

To study the relationship between the speed of a magnet in a coil and the

magnitude of the induced current.

1

State the suitable manipulated variables and responding variable

(Quantity that can be measured) Manipulated variables : the height of the magnet fall

Responding variables : Deflection/ reading of the galvanometer

1

State the constant variable

strenght of the magnet / number of the turns in the coils.

1

State the complete list of apparatus and materials

Bar magnet, cardboard tube, galvanometer, insulated copper wire, retort

stand and metre rule.

1

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Draw the functional arrangement of the apparatus

1

State the method to control the manipulated variable 1. Make a solenoid of 50 turns by winding an insulated copper wire

round a cardboard tube. Connect the ends of the wire to a galvanometer.

2. Hold a small bar magnet at a height of h = 5 cm above the top end of

the solenoid.

1

State the method to measure the responding variable

3. Drop the magnet into the solenoid into the solenoid and record the

deflection of the galvanometer as the induced current.

1

Repeat the experiment at least 4 times with the values

4. Repeat the eksperiment by changing the height h to 10 cm, 15 cm, 20

cm , 25 cm and 30 cm.

1

State how the data tabulated with the title MV and RV

Height of the magnet,h Induced current I

5.0

10.0

15.0

20.0

25.0

30.0

1

State how the data is analysed, plot a graph RV against MV

1

TOTAL MARK 12

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4c(viii) Tabulation of data:

Number of turns of

wire in secondary

coil, N

Output voltage, V

/ V

1

4c(ix) Analyse the data .

Voltage

No. of turns

1

TOTAL MARKS 12