1969-05-31 — Page 11

CITY

有教備轝 頁三第張三第日六十月四年畬巴夏 WAH KIU YAT PO

新日橋菜

六期星日一卅月五年九六九一瑟公年八十五國民華中

育教僑華

(11)

英中會考物理科答案

陸永熾·

Section

H.£.0.E:E. (English), 1969

PHYSICS

Suggested Answere

1.(a) Booke'a. Law

The deformation of a material is propor- tional to the force applied to it, provided that the elastic limit of the material is not exceeded.

(0)

(b)

Load

E Elastic linit

Y ↑ Yield point

Na Kaximum load

B Breaking point

*

are tas horisontal and vertica

and T

components of the tension, T, of the

elastic band respectively.

The length of 18:1a 50

proportion)

in I ain

F

- 50 x

250

-83 gul.WT.

The tension in the elastic band is 83-

83.t

iii) When a body is in equilibrium, sus of

forces in any direction should be zero

The horizontal force at 0) (The

norizontal component of T)

I

250 3

40

x

(0) V.R. of the pullay system ->

Efficiency of pulley system - 0.6

M.A. of the pulley system 5 x 0,6

Load Effort

3

Effort

= 1 kg.vt.

3

When the load is raised 10 om, the errort moved 50 cm.

Work input to the system 150 kg-cm. 0.4 of the input is wasted in raising the pulley block and overcoming the friction.

0.4 x 150 - 1 x 10+ work done against friction

Work done against friction

- ·60-10

• 50 kg--cm.

14 cm

2 cm

3 cm

Extension

OE of the graph is a straight line showing that the extension is proportional to the load. The wire over this range is perfectly elastic and will recover its original. length when the load is removed -

At the point 7, called the yield point, the wire extende much scre, for a given load than. before. Some internal slipping appears to bo taking place inside the wire. This property is called ductility

The point N represents the maximum load the wire can stand. At this stage the wire im liable to stretch suddenly and form a neck at its weakest part, The breaking point of wire is reached.

(1) From Hooke's Law

extension ocforce applied

F&

wnere F is the force applied

k is a proportional constant

is the extensfor

Force at A P. (represented by OA)

3k

Force at BP (represented by 02)

Therefore,

Force at C

*m 4k

· F. (represented by 00)

Ik

x is the extension in the elastic hand CO

By the force diagram draw with scale of

1

on reps. 1 unit,

-663-

gm.ww.

The vertical force at 0 100

J

•· 40 ·

-190 gn.wt

Ane. The horizontal and vertical forces

acting at 0 are 665 and 190 m. wi respectively.

2. (a) Mechanical Advantage

Mechanical. Advantage of a machine is the ratio of the load to the effort.

M.A..

*

Load Effort

Talocity Ratio

Velocity Ratio is the ratio of the distance moved by the effort to the, distanos move by the load in the same time.

(b)

Distance moved by the effort Distance moved by the load

1.R.

efficiency of a machine

x 100

- Work wüteüt

work input.

Load x distance moved by the load Effort x distance moved by effort W x a

x 100

x100

E

W

(+) x 100

M.A.

I 100

velocity ratio

V.R.

mechanical advantage

x 100

130 cm

5 kg

The pressures at A and B are equal and is equal to the pressure at C. (i) The pressure at C

atm. pressure +

pressure due to water column of (13-5) cm.. 1000+ 8 1008 gm.xt./cm,

(ii) The pressure at B also equale to

2

1000+ pressure due to oil column of 12 ́om. 1000+ 12 x density dof ail

8

.*. density of oll -

12

23

0.67 8./0.0.

(b) (i) Archimedes' Principle

When a body. is wholly or partiall immersed in a fluid it experiences an upthrust equal to the weight of the fluid displaced.

(ii) The buoyancy of the inverted test tube is

altered by the air pressure inside the bottle. The explanation of the sink and float motion of the tube depends on the law of flotation, which states that floating object must displace its own weight of fluid. The weight of the inverted tube has · been adjusted so that it is only just able to float under the surface of water. Sarewing down the top B, the air inside the bottle is compressed. Phis will in turn compresses the air trapped in the test tube to a small volume. Consequently the test tabe displaces less than its own weight of water and the tube sinks. On screwing out, the pressure is released and the air expands again. The tube recovers it buoyancy.

(c) By the Law Flotation,

the weight of the hydroseter, W - weight of

water displaced

(1)

W. I ..... (1)

It is measured as 5.6 units

That is the extension in CO is 5.0 ca.

From the dimensiona givenī

sin

COP

€ -

MË NË VITE SE

5

(11) A. The 3 elastic bands must be prodeatty

identical,

. Their slastic limite kan het been

(1) the normal reaction of the plane

N = 5 cos 8

- 5 x

424

- 4 kg.

(11) Before movement starts,

Force actingupwards.

The weight of the hydrometer

- weight of alochol displaced

* = (x + 10) x 0.8

- 0.8x + 8 ....... (21

equation (1) and (2)

r = 0.8x + 8

0.2x = .8.

x = 40

The value of x is 4er

(a) (1)

Z6

The metra "rule is binged at O

од

T is the vertical component of vaÐ FÓT CO

Y

at &.

Taking moments at 0,

40

100 x 10 + 40.x 25

- 25 x 40 + 40 x 25 - 50

The vertical component of the force at a im

50 g.wt.

force acting

down the plane

5 friction + 5 ṣin ✪

- N5 sin

4 x 4 + 5 x

44 = 2

:

4- 0.5

The coef. of static friction, 4- 0.5

(iii). Mechanical advantage

M.A. – 1

Velocity ratio

V.R. -

5 3

5

60%

Efficiency

* 100%

(11) Let the volume of the iron weight - V c.of

leight of the iron 8V gm/wt. Weight of the combination - W8V go.yt.

Weight of the combination

• X + BV go.vt.

- 40 + 8V g.wt.

- weight of water displaced

(V + 40 + 10) go.wt. 50 + V

40 8V

50+ V

7V - 10

v. 10.12 0.0.

Ans. The volume of the iron weight

13 0.0.

7

(To be continued,