Physics Answer N9 08

JPNS SPM TRIAL 2008 Answer P1 (Physics Paper 1)

1

A

11

C

21

B

31

B

41

B

2

D

12

C

22

C

32

A/B

42

C

3

D

13

D

23

B

33

A

43

A

4

D

14

C

24

C

34

A

44

A

5

C

15

C

25

Bonus

35

B

45

A

6

B

16

D

26

A

36

A

46

C

7

A

17

C

27

D

37

A

47

C

8

C

18

B

28

C

38

C

48

B

9

D

19

B

29

A

39

A

49

A

10

B

20

D

30

D

40

B

50

B

TRIAL JPNS PHYSICS ANSWER PAPER 2 SECTION A Question Number 1.(a) (i) (ii) (b)

Answer base quantity is a quantity that cannot be defined in any other physical quantity ms-2 Base quantity - time / displacement Derived quantity - velocity / acceleration Scalar quantity – time Vector quantity - displacement / acceleration Total

Question Answer Number 2.(a) Convex mirror (b) Convex mirror has a wider view than a plane mirror. (c)(i)

Full Mark 1 1 1 1 4

Full Mark 1 1 2

Note : Draw two incident rays and each of them reflected at the correct path. Shows image formed behind the mirror and the position in front of F.

(ii)

1 mark 1 mark

Virtual ,upright and diminished. (any of two combination)

1 Total

5

Question Answer Number 3(a) Step-down transformer (b) A.u. power supply, Vp can produce a changing magnetic field that induces an alternating e.m.f in the secondary coil, Vs. (c) Using a formula , Np = Vp Ns Vs Np = 240V Ns 12V

Full Mark 1 1

1

Np : Ns = 20 : 1 (d)(i) (ii)

1

Bulb will not light up

1

Because transformer cannot function using direct current/ cannot produce change in magnetic flux

1 Total

Question Answer Number 4(a) W = mg = (50)(10) = 500 N (b)(i) Increase (ii) Decrease (iii) Stationary / at 500N (c) No resultant force/acceleration zero (d) R = (46)(10) = 460 N mg-R = ma 500 – 460 = 50a a = 0.8 ms-2

6

Full Mark

1 1 1 1 1

Total

1 1 7

Question Answer Number 5(a) Quantity of heat required to increase the temperature of 1 kg of material through a temperature of 1 oC (b) (i) Energy output = power x time = 3 x 103 x 3.5 x 60 = 6.3 x 105 J with unit (ii) Energy required to raise the temperature of water, Q = mcθ = 1.7 x 4.2 x103 x (100 -20) = 5.71 x 105 J with unit (iii)

Energy required to boil away water, Q = ml = 0.23 x 2.3 x 105 = 5.29 x 105 J with unit

(b)

(c)

1 1

1 1

1 1 Total

Question Number 6 (a)(i) (ii)

Full Mark 2

Answer An electromagnetic waves is the vibration of electric and magnetic field Can travel through vacuum Speed of light Transverse waves (any one answer) P : X-ray Application : Radiotherapy for the treatment of cancer S : Microwaves Application : Waves received by telephone and television signals from satellite λ = v/f = 3x 108 3.2 x 103 = 9.375x104m Total

8

Full Mark 1 1

1 1 1 1

1 1 8

Question Number 7(a)(i) (ii)

(iii) (iv)

(b)(i) (ii) (iii)

Answer Function – for safety purpose/To ensure the maximum weight limit F = mg = 7500 x 10 = 7.5 x 104 N The mark should be higher than the sea water level

(ii) (b) (c)( i) (ii) (d)

(e)

1 1 1

1. density of sea water is denser than the density of river water. 2. the volume of water displaced increased when density of liquid decrease

2

Upthrust = Weight Accelerates upwards or moves upwards 1. the weight of the air balloon is decreased 2. buoyant force /upthrust higher than weight 3. the balloon experiences the unbalanced force.

1 1 2

(any two answers) Total

Question Number 8(a) (i)

Full Mark 1

Answer

10

Full Mark 1

npn transistor as a switch/automatic switch amplifier to limit the current to the transistor VP increase Because resistance at P is higher the transistor will switch on the relay switch Ib will flow through the transistor and Ic will increase 10,000 x 6 = 2 10,000+ S Resistance, S = 20 000 Ω OR RS = VS 1000 + RS 6 RS = 20 000 Ω

1 1 1 1 1 1 1

The bulb will not light up because it needs higher voltage/current to light up.

1 1 12

1 1

Total SECTION B

Question Number 9 (a) (i)

(ii)

(b)

Answer Temperature is a measurement of the average kinetic energy of the atoms or molecules in the substance. Or Temperature is a measurement of degree of hotness of an object. 1. The kettle is hotter than the ice block / The ice block is colder than the kettle.

1

2. The hand feels hot when it touches the hot kettle / The hand feels cold when it touches the ice block.

1

3. Diagram 9.1 shows heat flows from the kettle towards the hand while Diagram 9.2 shows the heat flows from the hand towards the ice.

1

4. The heat will flow from a hotter object towards a colder object.

1

1. Sea has higher specific heat capacity than land.

1

2. The temperature of the sea increases slower than of the land.

1

3. The air above the land is warmer than the air above the sea / Density of air above the land is lower than the density of air above the sea.

1

4. The warm air above the land rises up.

1

5. Air from the sea moves towards the land.

1

Modification 1.Use the fluorescent lamp not a filament bulb 2. Bigger cover with white colour 3. Use the adjustable stand Or portable 4. Connect with the earth wire 5. Use an energy saver lamp

Explanation Fluorescent lamp use less power and economic (consume less power) compare to filament bulb Less reflection on eyes and absorb less heat energy and good heat reflector The height of the lamp can be adjusted Can be used anywhere Avoid short circuit and damage on the bulb Produce same brightness with less power consumption Total

Question

Full Mark 1

Answer

2

2 2 2 2

20

Full

Number 10(a)(i)

Mark

3

(ii) (iii) (iv)

(v)

1. Bulbs are connected in parallel – 1m 2. Switch and battery are connected in series – 1m 3. Correct symbols for all components – 1m When the bulb is connected to a power supply of 1.5V, it will produce 3J of energy in 1 second. The brightness of bulb A is the same as bulb B I = P/V = 6/1.5 = 4A Energy = Pt = 6 x 3600 = 21 600 J Or Energy = VIt = 1.5 x 4 x 3600 = 21 600 J

1 1 1 1 1 1

(b) Modification Resistance of wire is low Melting point is high Density wire is low Rate of rusting is low Rate of expansion of wire

Explanation to prevent power loss due to heat to prevent the wire from melting to reduce the mass of wire / too heavy to prevent it from rusting easily to prevent lengthening of wire

2 2 2 2 2

Total

20

Question Number 11(a)

(b)

Answer

Full Mark

Pressure is defined as the force acting normally per unit area/ Pressure = Force Area 1. When the small piston is pulled up, the hydraulic oil is drawn from the reservoir into the small piston 2. When the small piston is pushed down , the hydraulic oil is exerted with force and experienced a pressure 3. The pressure is transmitted uniformly from the small piston to the big piston. 4. The forced produced raised the big piston / The system can convert a small input force into a bigger output force.

(c) Characteristics

(d) (i)

(ii)

Reason

Has higher boiling point

So that liquid not easily boiling/

2

Has higher specific heat capacity

So that it can’t be easily become hot

2

Has lower density

So the hydraulic jack is not heavy

2

Has lower rate of vaporisation

Volume of liquid will not easily vaporise

2

Liquid L is chosen

Reasons: L has higher boiling point, higher specific heat capacity, lower density and lower rate of vaporisation

2

Weight = mass x acceleration of gravity W = mg W = 60 x 10 W = 600 N Pressure = Force Area A = F = 600 N P 500 Pa = 1.2 m2 Minimum area of each snow shoe = 1.2 2

1 1

1 1 1

= 0.6 m2

Total

20

Question Number 12(a) (b) (i)

Answer

Full Mark 1

Reflection of waves Radio waves Transverse waves Can travel without medium Have long wavelength

Sound waves Longitudinal waves Needs medium to travel Have short wavelength

2

(any two comparisons) (ii)

(c)

The distance between the water molecules is closer compared to air molecules. Thus, the sound energy can be transferred faster.

Characteristics Type of waves - longitudinal High frequency High speed High penetrating power The most suitable waves is S

(d)(i)

(ii)

1 1

Reason Because sonar is a sound waves which is a longitudinal wave Has high energy / can penetrate deeper into the sea Can travel faster

2

Can penetrate through medium easily Because the waves is longitudinal , high frequency , high penetrating power and has high speed.

2

d = vt 2 = 1500 x1 2x 15 = 50 m To detect the depth of seabed To detect the position of crude oil or sunken ship To detect the condition of baby in the womb.

2 2

2

1 1 2

Any two answers Total

20

PHYSICS : ANSWER PAPER 3 Question Number 1(a)

(b)

Answer

Full Mark 1 1 1

i. Real depth,H ii. Apparent depth,h iii.Optical density of glass block or Refractive index of glass block or Type of medium Tabulate H and h and show H as the manipulated variable and h as the responding variable. H/cm 2.0 3.0 4.0 5.0 6.0

h/cm 1.3 2.0 2.7 3.3 4.0

Give one (1) mark for each of the following : 1. Show column for H and h respectively. 2. State the unit for H and h correctly. 3. State the value H up to 1 d.p 4. Show the reading of h correctly up to 1 d.p 5. Show consistency of all data up to 1 d.p. (c)

5

Draw a graph of h against H Give one (1) mark for each of the following :

7

1. 2. 3. 4.

Show h at y-axis and H at x-axis. State the correct unit for both axis. Show uniform scale for both axis. Show all data plotted correctly: 5 correct – 2m 3-4 correct – 1m 5. Draw a smooth and best fit straight line on the graph. 6. The minimum size of the graph at least 5 x 4 squares (10 cm x 8 cm) starting from the first point to the last point. (d)

i. State the relationship between h and H correctly. (Based on the student’s graph)

1

h is directly proportional to H TOTAL

Question Number 2 (a)(i)

Answer

Full Mark

Show the extrapolation on the graph

1

1 R is directly propotional to A

(ii)

(iii)

16

1

Show on the graph R = 1.6 Ω 1 When R = 1.6 Ω, = 0.90 mm-2 A Hence, A = 1.1 mm2

1 1

Gradient of the graph - draw triangle on graph

1

1

1 = 1.8 – 0 1.02 - 0 = 1.76 Ωmm2

(b)(i)

1

R =

l A

Hence, resistivity ρ = gradient l = 1.76 Ωmm2 150 mm = 1.17 x 10-2 Ωmm OR = 1.17 x 10-5 Ωm (vii)

-Use small value of current to avoid the wire from getting too hot easily/ -Make sure the all the connections all the tightly connected/ -Switch off the circuit if no reading is taken to avoid wire from getting too hot TOTAL

1 1 1

1

12

Question Number 3 (a)

(b)

(c)(i)

(ii)

(iii)

Answer The volume of the snack pack influenced by / effected by / depends on the atmospheric pressure of the surrounding.

Full Mark 1

For a fixed mass and at constant temperature of a gas, its volume will increase when the pressure decreases. or The volume of a gas increases when the pressure decreases.

1

To investigate the relationship between the volume and the pressure of a gas at fixed mass and temperature.

1

Manipulated variable Responding variable Constant variable

:Volume of a gas, V :Gas pressure,P :Gas temperature,T / mass of a gas,m

Bourdon gauge, a syringe with volume scale and rubber tube

1 1 1

(iv) 1

(v)

(vi)

(vii)

1. The piston of the syringe is pressed down slowly until the volume of air inside is 120 cm3.

1

2. The pressure of the air in the syringe is read from the Bourdon gauge.

1

3. The experiment is repeated with volume of air inside the syringe at 100 cm3, 80 cm3 , 60 cm3 , 40 cm3 and 20 cm3.

1

Volume , V / cm3 120 100 80 60 40 20

Pressure , P /Pa 1

Pressure, P/Pa

1

Volume, V/cm3 TOTAL

Question Number 4 (a) (b) (c) (i)

Answer

12

Full Mark

The strength of the magnetic field depends on the current flows in the wire.

1

The greater the magnitude of current, the stronger the magnetic field.

1

To investigate the relationship between the current flows in the wire and the strength of the magnetic field of an electromagnet.

1

(ii)

Manipulated variable : the current in the solenoid Responding variable : the strength of the magnetic field. Fixed variable : number of turns of coils/the type of core used/distance between the end of the soft iron rod and pins in the petri dish. .

1

(iii)

D.C power supply, ammeter, rheostat, petri dish, retort stand with clamp, a box of pins, soft iron rod, insulated copper wire and connecting wires.

1

(iv)

Set up the apparatus as shown in the figure.

1

1

(v)

Start the experiment with 0.2A of current. Turn on the switch. Bring the Petri dish of drawing pins near the lower end of the rod.

1

(vi)

Count the number of pins attracted to the rod. Turn the switch off the to allow the pins to fall back into the petri dish. Repeat the experiment by varying the current with 0.3A, 0.4A, 0.5A and 0,6A.

1

(vii)

Current/A 0.2 0.3 0.4 0.5 0.6

1

Number of pins attracted 1

(viii)

Number of pins attracted 1

Current TOTAL

12