Ce Phy Mock Paper D1 Ans

Kwok Tak Seng Catholic Secondary School S.5 Physics Mock Examination 2004-05 Paper I Section A: 1. (a) The cork vibrates vertically. (b) wavelength : λ = 0.27/3 = 0.09 m v = f λ = 80 × 0.09 = 7.2 m s-1 (c) Decrease the frequency of the wave source. Add more water to the ripple tank. 2. (a)

(b) It is a virtual (and upright) image. (c) Since water has a smaller refractive index than glass, the image of the print is raised above the bottom of the tank by less than 0.9 cm. ×8×3 = 12 m 0−8 (b) Accleration = slope of the graph = = - 2.67 m s -2 3−0 (c) (i) (ii)

3. (a) Distance : d = Area under the graph =

1 2

(d) Since the block can stay at rest at B, f = mg sin θ While the block is moving up, unbalanced force = mgsinθ +f = 2 f By 2nd law: F = ma

⇒ ⇒

2f = ma f = ma/2 = 3×2.67/2 = 4 N

S.5 / PHY / SOL / P.2

(a)

It is not possible.

1A

It is because all terminals of a microchip would be in contact with a conductor.

1A

If charges appear at some of the terminals, they would also flow to other (b)

terminals.

1A

If a microchip is stored in a conductive envelope, its terminals would have the

1A

same amount of charge (or not charged), no spark can be formed inside the chip and it would not be damaged.

1A

(a)

The rheostat is connected up as a fixed resistor instead of a variable resistor.

1A

(b)

(i)

voltage / V 7 × 6 ×

5 ×

4 3

×

2 1

× × current / A

0

0.5

1

1.5

2

(Correct axes with units and scales.)

1A

(Correct data points.)

2A

(Straight line passing through the first 4 data points and a curve passing

1A

through the last 2 data points.) (ii)

When a current passes through a resistor, energy will be dissipated as

1A

heat. When the current is high enough to sufficiently heat up the wire, the

1A

resistance of the resistor increases. Therefore, the slope of the curve increases when the current is large. (a)

The earth pin of the adapter is used to open the covers of the live and neutral

1A

holes of the socket. (b)

The input power PI = IV

1M

= 220 × 24 × 10 = 5.28 W

1A

The output power Po = IV = 6 × 150 × 10−3 = 0.9 W

1A

−3

Po Percentage of useful input power = × 100% Pi

=

0. 9 × 100% = 17.0% 5.28

1A (c)

The difference between input and output power is lost as heat.

1A

S.5 / PHY / SOL / P.3

(a)

When a magnet is put on the top of the display, tiny dark iron particles are

1A

attracted towards the magnet (the surface of the cells around the magnet). Since iron particles are black, the area around and below the magnet appears (b)

darker than the rest of the display.

1A

(i)

A small magnet should be put at the tip of the pen.

1A

(ii)

The 'duster' should be put at the back of the display.

1A

It has a magnet facing the back of the display.

1A

When it sweeps across the display, iron particles are attracted towards the back of the display, leaving the thick white liquid at the front and the display is cleared.

8. (a) (i)

129 53

I→

129 54

X+

0 −1

β

(ii) Iodine-131 is the most suitable since its half-life is appropriate for the treatment. The half-life of iodine-123 is too short and that of iodine-129 is too long. (b) An α-radiation source should not be used since it cannot penetrate our body. (c) Iodine is well absorbed by thyroid cancer cells, while it is not absorbed by other cells. Radiation can then be concentrated on thyroid cancer cells with minimum side effects. 9. (a) The unbalanced force acting on the white ball = 80 N Work = Force × Displacement = 80 × 0.5 = 40 J (b) Increase in K.E. = Work 1 2 2 mv = Fs 1 2 2 (0.25)v = 40 v = 17.9 m s -1 (to rig ht) (c) The total mcrmcntum before impact= 0.25 × 17.9 = 4.475 kg m s -1 (to right) The total nxrmentum alter impact = 0.25 × v + 0.35 × 10 By conservation ofmomentum: 0.25 × v + 0.35 × 10 = 4.475 v = 3.9 m s -1 (to right ) (d) The change of momcntum of thc red ball = Final momcntum - Initial nxrmentum = 0.35×10 – 0 = 3.5 kg ms -1(to rig ht) Time of impact = 0.04 s, The average force acting on the red ball =

3.5 = 87.5 N (to right) 0.04

(e) Heat (internal energy) Lost in K.E. = Initial K.E. - Final K.E. = ( 12 ×0.25×17.9 2 ) – ( 12 ×0.25×3.9 2 + = 20.65 J Heat produred = 20.65 × ( 1-80%) = 4.13 J

1 2

×0.35×10 2)

1A

S.5 / PHY / SOL / P.4

V 12 = = 24 Ω I 0.5 (b) (i) To reduce heat loss by conduction V2 62 (ii) Energy supplied = P t = t=( )×30×60 = 2700 J R 24 (iii)E = mc∆T ⇒ 2700 = 0.4 × c × (37-25) = 4.8 c ⇒ c = 562.5 J kg-1 °C-1

10. (a) Resistance : R =

(a)

(i)

Current passing through the bulbs = P/V

1M

= 6/11 = 0.545 A (ii)

1A

2

Resistance of each light bulb = V /P

1M

= 112/6 = 20.2 Ω (b) (c)

1A

Since light bulbs are connected in series,

1A

if one bulb blows, the circuit will be broken and other light bulbs go out.

1A

When she connects the two sets of lights in series, the light bulbs become

1A

dimmer. It is because the total resistance in the circuit doubles when two sets of lights

1A

are connected in series. Therefore, the current passing the light bulbs and the power dissipated by

1A

each light bulb (P = I2R) decrease. When she connects the two sets of lights in parallel, the brightness of light

1A

bulbs does not change. It is because the voltage across each set of the lights is the same as that

1A

before. Hence the power dissipated by each light bulb does not change. (d)

(For effective communication.)

1C

The resistor in the light bulb ensures that the circuit does not break even if

1A

the filament melts. Since the resistor and the filament are connected in parallel, when the filament melts, current can still pass through the resistor and the circuit is still complete.

Paper II 1

2

3

4

5

6

7

8

9

10

D

D

B

B

B

A

A

D

A

C

11

12

13

14

15

16

17

18

19

20

C

C

A

C

C

C

D

B

B

A

21

22

23

24

25

26

27

28

29

30

A

D

A

B

C

B

B

C

D

B

31

32

33

34

35

36

37

38

39

40

B

A

C

C

B

D

D

A

A

D

41

42

43

44

45

D

C

D

D

A

1A