1991-01-14 — Page 16

報日僑華

一期星 日四十月一( 一九九一)年十八國民華中(16)

-要提

是利得 血鼻流

1991 中學會考預習專欄:

position 0.32 m from the knife edge. to restore equilibrium (see figure 6) Find the density of the unknown liquid. Z.

The system is in equilibrium

B

..(1)

T

(2)

0.32.m

明波出版社

MILL & DALE PRESS

Physics (16)

Since No = 2N. Wc.

3N

殳退要將

€. Y. Mak

= readi

5N

of spring balance

Exercise 16

Pressure, Archimedes' Principle

Figure 6

凱

1. In figure 1; block A hangs by a cord from spring balance D and is

in beaker B The weight of the beaker is 2N, the weight of the

Submerged in a liquid C contained

(a) A flat

0.025 kg

tube of bottomed and cross sectional area

2 x 10 has lead shot placed in it. It floats upright in a liquid X as shown in figure 7.

mass-

.(3)

2 + 35+ 15

From (4)

15N

凱旋燈成就難估

liquid is 3N and the volume of

block A is 1.25 x 10-33

Rreading of the balance E

= 15N

Substitute into (1) and. (2)

S+ B

11 Substitute the value of W,

into (3)

(Ans.)

-fiquid

B = 10N

iii. Since

Figure 7

bouyant force

weight. Of

equal

volume

liquid displaced

The following results are obtained.

Mass of lead shot

placed Inside the 0.005 0.010 0.015 0.020 0.025

tube (in kg)

Dupth (h)

Immersed In

10:12 0.14 0.16. 0.18 0.20

(density of liquid)x(volume

of the block)x(g)

10. (d)x(1.25x1073)x(10)

(weight of block A) +

(weight of the Tump of metal)}

B+ (8000) (10)

(1000) (10) (1x10"3 j

+(1000) (10)V".

8+ (8000)(10){V)

W2 = 7 x 10 m2

Mass of metal attached

(8000) × (4 × 10-4)

X

=0.2286 kg

(Ans.)

(a) Let the distance between the knife edge to the end hanging the metal block Y be x.

In figure 4, by taking moment about the knife edge:

Mg(0.4)

Where M mass of block. X

Wweight of block Y in air:

In figure 5, by taking moment about the knife edge

Mg(0.35) WAX- .(2)

Where is the weight of block Y

in water

(2)(1)

W-0.35

(where Bis the upthrust).

(b) f. Total mass = 0.015 + 01025-

From (a)

0.04 kg

M = dhA

0.04 - (d)(b)(2x104)

dhA

0.04) = (d)(h)(2x107

hd - 200

If h 0.25 m. (0.25)(d) =

800 kgm

the density of

800 kgm

(Ans.

liquid is

(Ans.).

亞都

洲祥

小所

Figure 1.

(a) If the spring balance D reads 5N.

and balance E reads. 15N,

what is the weight of block, A. in air?

fig what is the upthrust on A?

iii. what is the density of liquid

(b) If block A is now pulled out of

1iquid C,

what will balance D read?

what will balance E read?

A wooden block A is weighed in ain

in water and in liquid X as shown

in figure 2(a), figure 2(b) and The figure 2(c) respectively.

readings of the spring balance in

each case are ON, 2N and 015N: respectively-

(Density of water = 1000 kgm3)

quld X (in m};

Plot the total mass of the tube and Tead shot against the depth h immersed in

What conclusion quid X

11. What

Côn the relationship between pressure

in a liquid and the depth of the liquid can you draw from the graph.

11. From the graph, determine the

density of liquid X.

(b) 0.015 kg of lead shot is placed in

the tube. It floats upright in awa

liquid of density d kgm3 and th

depth immersed in the liquid.

h meters.

1.

Find the

and h.

h = 0,

density of the quid.

the:

5. A bl am tackle consisting of threddi wo movable pulleys Is Use

an aluminium cube which completes immersed in water

3

(density: 1000 kgm) as shown in figure 8. If the volume of

of the cube is 10 m3 the effort applied to the machine is 5 x 10 N. and the effects of the machine 15 80%..

(Real density of aluminium

2.5)

(b) 1

2.

-3

d = 800 kgm

the liquid, then B = 0.

(Ans.)

If block A is pulled up out of

reading of the spring balance

共

圖

(1000) (g) (V)

( \ ) ( 9 ) (V)

(where e

density of the

-3.metal, and Vis

8000 kqm

(Ans.)

volume of the block)

(b) In fiquis in

taking moment about

(3)

The velocity.

ratio of

system is 5.

11. The upthrust of water on the

cube B

The weight

water

displaced

(1000) (10) (10).

(Ans)

iii. Let the pulling force exerted.

by the string on the cube be

is the weight of the metal.

Since V.R. 5 and

: 2.

3.15

5N

(Ans.)

block Y in the (3)(1)

reading of. E is 5N

0.8W

B" - upthrust in Liquid.

M.A

therefore,

5x10

100% 80%

2x105N

十二月份最佳見習生獎,導師陳威麟相陪接受盘敬驗將軍頒獎

一九九零年度亞洲小組吳綺莉頒發精美的複製生力銀盃予馬伕即氏,下:談樹文獲

上:馬伕鄧祥所照料的名駒「凱旋燈,獲本年度生力銀盃的最佳外觀獎,因爲

凱旋燈力壓羣雄奪標

馬主文洪磋感到意外

|截干馬迷期已

虚假沙田馬場盛大 於咋日-一月十三日 过生力銀盃賽馬日] 久的馬壇盛事第十二

千六百米的生力銀盃大賽,由第一及第二虑百,当枝厅在菜舟奇重工制造句至上 生力啤酒廠已是連續第十二屆贊助此預作些,當日共分八場賽事,矚目榔路:

『意鍾我」

Figure 2(a) Figure 2(b)

(a) Calculate

quia.

Figure 2(c)

the upthrust on the block in water.

ii. the volume of the block A,

iii. the relative density of block.

the upthrust of the block. A in 1iquid X.

the

density of liquid X.

(b) If a lump of metal is attached to the bottom of the block and causes the wooden block to float in wateri with its upper surface just covered. by water as shown in Figure 3..

metal

Water

Figure 3

Find the mass of metal attached.

if the density of the metal.

8000 kgm-3

is

3. A metal block Y is suspended at one of the ends of a uniform red. The red is supported by a knife edge at the middle and the whole system is in equilibrium by hanging a block X. at a distance 0.4 m from the knife -edge as shown in figure 4%

0.4 m

(a) Let m

ONG KONG

Alumu

cubo. (10

Figure 8

What is the velocity ratio of the system?

What is the upthrust of water- son the aluminium cube?"

iii. Neglecting the resistance of water find the acceleration of the cube while it is moving. in water..

Solutions

Find the position at which the cube will finally stay.. What additional effort is required to apply to the system in order to push the whole cube out of water?'

(a) Let d' = density of liquid C

tension of the cord

weight of block A.

W weight of the beaker"

B

weight of the liquid

bouyant force on A

V

mass of block A

volume of block A

upthrust on block A

water

upthrust on block liquid x.

From figure 4(a) mg = BN.

0.2W (8) (9) (V)

is.

density of Z).

ARIES

-0.2

(0.2)(

1600

(mass

(Ans.)

of the

tube) of lead shot

+ (mass

inside. tube).

PUBLIC LIBRA

Treading of the spring balance

2N

As shown in the force diagram

8 = 1 + mg

0.05

28

10N

(Ans. 1

0.04

0.03

Apply the formula

POV

0.02

0.01"

(Ans.

10 = (1000) (10) (V)

= 1 x 10

iii. Relative density of block A

10

0.8

(Ans.)

Upthrust of block A in liquid

(weight of block A in air)

(weight of block A in liquid *)

0.35 m

Figure 4

R = reaction of the balance E

on the beaker

B

+

=8N- 0.5N

7.5N

(Ans.)

B = 'gV where p'density of

liquid X

7.5 *** (10) (1x10~3) 750 kgm-3

Total: mass, M

(in kg)

Let the acceleration of the cube be av

Apply Newton's 2nd, Law

7 + 8 mg ma

where me mass of the cube

(2.5x1000)(10)

2,5x10 kg:

(2x10 +10°) - (2.5x10")(10)

(2.5x101)a

a = 2ms

the

0:03 0.035 0.04 0.045 0.05

depth bin in) 0.12 0.14 0.16 0.38 ||0.20

Total mass M.(in kg)

htin (m)

ii. Cross sectional Parea of the

tube A

2 x 10-42

Pressure at the flat bottom of the tube

pressure of the liquid at a depth h

From the graph, M increase linearly as M increases, therefore, pressure in а liquid increases linearly as the depth increases.

111. M = (s)(g)(h) where is the density of the liquid.

Ans.)

M = hA

Figure 5

(a) As shown in Figure 5, if block Y is completely immersed in water, block X has to 700 ve to the position 0.35 m from the knife edge in order to keep the whole system. in equilibrium. Determine the density of the metal block Y.

(b) If the block Y is now increased completely into an unknown liquid z, then block X has to move to the

Let the volume of the lump of metal

(b)

be VI.

Total upthrust

= {upthrust on block A in water)

{upthrust on the lump of metal in water)

(1000)(10)(1x1073) ·

+ (1000) (10)V!

Total weight

the slope of M against h. A The slope of the graph

(Ans.)

The cube will finally stay-at- the water surface with part of its volume immersed.

Let the volume of the cube immersed in water: be V hence

T + B

mg="0"

where B the upthrust exerted

on the cube,

B = mg T

But

(2.5x10)(10) 2x105

5x10 N

(1000) (10) (V!) = 10^y1 104

5x104

V1 = 5 m3

Therefore, the cube finally stay at the water surface with half of it's volume immersed.

In order to push the whole cube out of water, the pulling. force exerted on the block must at least equal to the weight of the block in air. Hence, the pulling force

#2.5x105

Let the effort required be Ei M.A. - 2.5x10°

Again apply

M.A.

x 100% = e

2-5x105

x 100% = 80%

E! = 6.25 x 104N

=

0.05 0,2

0.25 kgm

0.25 =

(9) (2×10-4)

-3

y= 1250 kym

Ans: The density of liquid X

is 1250 kgm

-3

the additional force required = E' - F

6.25x101N - 5x101N 1.25x101N

(Ans.)