KerTAS Fizik 2

8/13/2019 KerTAS Fizik 2

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MARKING SCHEME PAPER 2, PHYSICS

MID TERM EXAM 2011

PART A:No Answer Mark

1 (a) (i) State the name of liquid correctly- mercury

1

(ii) State the correct physical change

- Expansion / increase in volume

1

(b) (i) State the choice of instrument correctly

-thermometer X

1

(ii) Give one correct reason.

- The smallest division is smaller // able to detect the smaller /smallest

change

1

4

2 (a) (i) Rate of change of displacement 1

(ii) Between D and E 1

(b) Total displacement

= [ (½ x 4 x 10) + (6 x 10) + (½ x 2 x 10)] – [1/2 x 2 x 10]

= [20 + 60 + 10] – [10]

= 80 cm

1

1

(c)1

5

3 (a) Pascal’s Principle 1(b) 1.When the handle is pressed down, the pressure is exerted on the liquid

and transmits uniformly to the large piston

2. The force is produced and pushes the load up

1

1

(c) Some of the force is used to compress air bubbles // The pressure cannot

be transmitted uniformly and immediately because the air bubble is

compressed

1

(d) F = 2

20 100

00

F = 40 N

1

1

6

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2

4 (a) increases linearly 1

(b) As the temperature increase, the speed of the molecules increase.

The rate of collision between molecules and container wall increase.

Therefore, pressure increase

1

1

(c) (i) - 273o 1C

(ii) 1

(d) P1 = P

T2

1 T

2

200 = P

273 + 27 273 + 80

2

P2

= 235.3 kPa

1

1

7

5 (a) Force is an action that can change the type of motion of the object which

is in a straight line.

1

(b) (i) Fe = Fd 1

(ii) Net force = 0 N 1(c) (i) T = W 1

(ii) Net force = 0 N 1

(d) (i) When the object moves with constant velocity or the object is stationary,

the net force is zero

1

(ii) Forces in equilibrium 1

(e) The aeroplane will accelerate. 1

8

6 (a) A hindrance to current flow // halangan kpd pengaliran arus. 1

(b) (i) The length of wire in Diagram 6.1 is longer 1

(ii) The potential difference in Diagram 6.1 is bigger 1

(iii) The magnitude of current is the same in both diagrams. 1

(c) (i) The resistance of wire in Diagram 6.1 is bigger 1

(ii) The longer the length the higher the resistance 1

(d) 1. Current increases

2. Because total resistance decreases

1

1

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3

7 (a) The incidence angle in denser medium which can produce a 90 0 1

refracted angle (b) (i)

1. The ray is reflected twice

2. Direction of ray is correct

2

(ii) n = 1/ sin 42n = 1/0.669

n = 1.49 or 1.50

1

1

(iii) Total internal Reflection 1

(c) (i) refractive index of the inner core is greater than the outer cladding 1

total internal reflection can occur 1

(ii) An optical fibre is very small in diameter hence a bundle of optical

fibres can transmit more information.

1

(iii) Optical fibre has high flexibility. 1

10

8 (a) Microwave 1

(b) (i) v = f ƛ

3 X 108 = f( 2 X 10-2

f = 1.5 X 10

)10

Hz

1

1

(ii) v = s/t

3 X 108 = 4.8 x 10

t

7

t = 0.16 s

1

1

(c) (i) Concave reflector

-converge wave

1

1

(ii) At focal pointWaves converge at focal point

11

(iii) Big

can collect more wave

1

1

(d) Type X is most suitable 1

Total 12




4

NO SCHEMATIC CRITERIA MARK

9 (a) Normal force per unit surface area 1

(b)(i) M1 Mass of load is equal

3M2 Depth of sinking in D9.2 is less than D9.1 // Viceversa

M3 Area in contact with the soft ground in D9.2 is bigger

than D9.1

(b)(ii) M1 When depth of sinking increases, the pressure

increases2

M2 When area in contact increases, pressure

decreases

(c) M1When the piston is pushed in, the air flows out of the

nozzle with high speed

4

M2 Creating a region of low pressure above the narrow tube

M3The higher pressure of the atmospheric air acts on the

surface of the liquid causing it to rise up the narrow tube

M4The liquid leaves the top of the narrow tube through the

nozzle as a fine spray.

(d) Structure of the dam

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M1 Build a dam that has thicker wall at the base // Diagram

M2 To withstand higher pressure at the bottom

The material used for the damM3 Made of concrete

M4 Not easily break

The design to ensure safety

M5 Equipped with water overflow system // Diagram

M6 Avoid overflow /flooding

M7 Build high wall

M8 Store more water / avoid overflow

Additional component to produce electricity

M9 Build turbine at the bottom of the wall

M10High water pressure will turn the turbine to produce

electricity

TOTAL MARK 20M




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10 (a) Sources that have same amplitude, frequency and phase 1

(b) M1 Distance between sources in Diagram 10.2 is bigger

5

M2 The wavelength for both are equalM3 The separation between 2 consecutive antinodal/nodal

lines are bigger in Diagram 10.1

M4The smaller the distance between two sources the bigger

the separation between 2 consecutive antinodal/nodal

lines.

M5 Interference

(c)(i) M1 Speed of wave is uniform

4

M2 Wavelength is uniform

(ii) M1 Distance between wavefronts in shallow region aresmaller than in deeper region

M2

Correct direction i.e no refraction in shallow region but

bend downwards in deep region.

(d) Location of resort

10

M1 At the bay

M2Calmer water and lower amplitude waves since energy

diverge

Features to reduce the erosion of shore

M3Build barriers with small opening from one cape to

another

M4Water waves are reflected and diffracted so less energy

wave reach the shore.M5 Build high wall barriers

M6 To protect beach from high waves

Features to enable children to swim safely.

M7 Build sandbank to create shallow area

M8 Speed and wavelength of wave decreases

Features to protect hotel from strong wind

M9 Grow tall trees between the beach and hotel

M10The trees helps in breaking or spreading out the wind

energyTOTAL MARK 20




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11 (a) Springs constant ,k is the ratio of force to the extension // k =

F / x

1

(b) (i)

(ii)

Spring X

• Spring X is thicker

• Spring X is stiffer

• Spring X extend shorter than spring Y for the same force

• k = F / x , extension of spring is shorter, the value of k is

higher

1

1

1

1

1

(c)

Aspect Explanation

Small diameter enough space for the spring to

be installed // more stiffer

High elastic limit can support heavy load

higher spring constant small compression of the spring

small natural frequency reduce bumping

S is chosen small diameter , highest elastic

limit ,highest spring constant

and small natural frequency

10

(d) (i) k =

x

F

=

0.06

0.3x 10

= 50 Nm

2

-1

(ii) 0.3 kg 6 cm0.5 cm

0.3

6 x 0.5

= 10 cm

Length of spring = 20 + 10

= 30 cm

2

TOTAL MARK 20 marks




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12(a)(i) The distance between focal point and optical centre 1

(ii)

3

(iii) Virtual, upright and magnified 1

(b)

Aspects Explanation

f o > f Higher magnification of imagee

Eyepiece is thicker than objective

lens

Shorter focal length and higher power

Diameter of the objective lens is

greater than diameter of eyepiece.

To ensure more light can enter the

telescope to produce brighter image

Distance between 2 lenses

S = f o + f

e

f o + f e is the distance of normal

adjustment which will produce a sharp

image.

Chosen arrangement : K Because f o > f e , eyepiece is thicker

than objective lens, diameter of theobjective lens is greater than diameter

of eyepiece and distance of two lenses

is f o + f e

2

2

2

2

2

(c)(i) P = 1/f

f = ½ = 0.2 m = 20 cm

f=15 cm, u = 30 cm v = ?

1= 1 +

f u v

1

1= 1 -

v f u

1

1= 1 -

v 20 30

1

v = 60 cm

1

1

1

(ii) m =

u

v

m = 60/30

= 21

(iii) Characteristics of the image:

Real, magnified and inverted

• Note – at least 2 characteristics

1

TOTAL MARK 20




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KerTAS Fizik 2