1980-03-24 — Page 22

報日僑華

-期星

南二第六第日八初月二年申濙发

1980

Z =

As shown in the figure,

e.c.e. of copper

WAH KIU YAT PO

= 200 (Ans.)

(11) Apply

中學會考試題預習專欄

= 3.3 x 10-7 kgc-1

P = IV

.. I -

物理

(四)

Σ

明德出版社魯榮家提供資料

200 200

(b)(1)

- 1A (Ans.)

日四廿月三年〇八九一公年九十六闔民華中 青數僑榮

*

q

ལྟོqw° - kaqd - pr°

3

2

3

(a-p) - m

3mqd3

M

14.

2

2

17

7

第24

12

The equation of

additional straight

line is

y =

"

PHYSICS (24)

W.K. LO

(MILL & DALE PRESS LTD.)

Suggested solutions to

Exercise 1

1. Let the length of the

wire be I m.

(a)

Apply the principle of

***

8.91 x 10-4

监

3.3 x 10

x 3600

= 0.75A

The current I taken from the battery is given by

10

I.

I

10 + 20

Wheatstone bridge

Y

(1)

L- 0.3

Y

0.3

1.2

=

(2) x (1)

XY

لا

L 1.2

(0.3)(1.2)

=

(L - 0.34(1

(03) (1.2)

1.51 0

(2)

XY (L-0.3)(1.2

11 - I

1 =

311

= 3 x 0.75A

= 2.25A (Ans.)

(ii) The current through

the beater Iz

- I -

- I1

= 2.25A -0.75A

1.5A

Energy given out by the heating coil in the period

I2Rt

=

1.2) =

L-O (rejected)

pr L = 1.5

the length of the wire. is-1,5m (Ans.)

(b)

Substitute the value

of L into (1)

2.

3

**

22

Y 1.2.0.3

: 3 (Ane.)

60

(a) As shown in the

figure, resistance of

AB

= 16 л

x 20

Resistance of AD

20 + 60

= 80 A

=

16 80

32 - 10 82

-

By

•

x (2)

(b)

= 0.4V (An#.)

E,

(1.5)2(10)(3600)

(Ans.)

- 81000J

(iii) Let the temperature

of the copper vessel and its content rises

Heat energy absorbed by the copper vesael and i ta content

(0.5 x 4200 x 0)

+ (0.5 * 420 × 0)

* 23100 J

Heat energy given out by the heater

- 81000 J

Assuming no heat loss08, or gains from the surround -ings

23100 = 81000

35

(AME.)

the rise in temperature

of the oopper vessel and

its content in 35°0'

the balance length is

x from A.

Resistance of AB

(20)

# 20x л

Resistance of AD

- 20 /

20x 20

X

x = 0.2

(Ana.)

#

the balance length 18 0.2 m

P

3

B

120A ደ..

r

.

12. Given: is the

400mm X

60cm

Bqd

(g-p)

orthocentre of AABC, AM MH

E

3Y

Apply the principle of Wheatstone bridge.

BT = R

where S = 120 A and

R'

the resistance of bulb B measured

by the circuit.

200= 100-840-

R = 180 (Ans.)

(ii) The current through B

3

120 + 180

= 0.01A (Ana.)

(iii) The power dissipated -

by B

#

= (0.01) 2 (180)

(Ana.)

= 0.018W

(c) the current through B

in (a) is much greater than the current in (b), therefore, the. temperature of B in (a) is much higher than in (b). Conseq- uently, the resistance of B in (a) would be greater for resistance increases as temperat- ure increases.

4.(1) Let the current thr- ough the battery be L the currents through the voltameter and the heater be I and respectively, App-

I

ly the formula

m

m = It

» mass of oopper

deposited

= 8.91 x 10-4kg

t = time taken = 1 hr.

3600B

1980

中學會考試穏预暨專欄

數(廿四)

199 VIN ALL KALANE BIT DE SAE SAME YE NG

Mathematics (24)

G. P. Man

Mill & Dale Press Ltd

Section 1.

11. Volume of the

hemispharien) howl

B

F

To prove: (a) EM = MI (b) TE LABE

Tind L!F

25",

*=*

(c) If B = PH; CQ

then ARPP-ACNE.

Proof:

(a)

Qu

is the orthocentre of AABC (given). LAFC LAEB = 90°

orthocentre theorem)

A is the diameter of circle APE

(converse,

£ in semi-circle)

AM MI (given)

M is the centre of circle AFHE

‹Ã ̧o. EM = MF (radii) |(b) ZABE=- 25′′ (given)

13.

(c)

(b)

R

Original

content

36

After 1st

12 operation

24

Total (4) Spirit(f) | Water (l)

30

30x7=20

9

4 = 2229

of spirit

Dre added

36

1

32

ATLer 2nd

24

operation

32x24 64

424 B

4x36=3

30 m

12 of water

64

are added

36

£

Final of water in

the cask

=

44

3

x 100%

36 40.74%

15(a)

3x

=

82y+1

3x 23(2y+1)

Է

5x =

3(2y+1)

x-2y 1

33x-9

= 5

3x-9

2

BAE

180°-90°-25° (sum of A)

65°

LIME

4y = 3x-9

瞰

2/UAE

i.c. 3x-4y → 9

(9)

(1)x2,

2x-4y - 2

(3)

x = 7

...(centre Z

double the ce. L

130°

900

=

BP - YH

P is the centre of

circle BENT: "

J

HP = PF (radii)

•'. ¿PBP - /PFB = 25°

Tbnae isos 4) Similarly, LQCE - LOEC

LICE 180°-90°-659

(L 25°

(b)

(2)-(5)

Put x = 7 into (1)

i.e.

sum of 4)

25"

.*. Lucy - Lora ZFPR - ZEQC

(3rd of ▲) .'. ABPF ~ ACQE

fequiangular As)

2

_y=px2+qx+r

the graph cuts the X-AXIE B two distinct point.

• {"pr> 0

the graph pAKROR

through the paint (0,2)

(e) the roots of the equation

p***qxer • រ

are -1 or 3

(d) Put -1 and 3 into

7-2y= 1 y = 3

* a, b, c, d are in

continued proportion

h

+

+

= = k

C

22

ង bh ck*

b =

ck

dk

C dk

(n−c)(b~d)~(a=d} {b-e}

=

=

(dk3-dk) (dk"-d)

(dk2-A){ øk2-dk)

d"k(k3-1) (k−1)

= d2k(k−1)2

• [(k+1)-(k+k+1}] *kik-1)2(k)

2.2

(b-e) (ik2-dl;)"

ย

13

#

a22(8-1)

= (b − c ) "

16. Given: ZABC = LABO

A

p*"*qx+r » (

To prove

1-4+2 ap+3q+2 = {}

E 0

(1) (2)

(3)

(1)x3]

(2)+(5)

14p+8

ր

=

#

90"

(a) the relation between

the lengths of the wides of ABCD,

Thy it AP//CD, Ag7/en, then QA-AB = PA-AR

Proof:

(a) *.* Lanc

<

LADC

A

X

RAY!

Ex

3(a) Apply

p *

C 69%

(1)

200 =

2002

R =

}

#

1

e

Volume of m marki

X

the resistance of the bulb B

Volume of water

in the bow

C

Q

** +

into (1)

=()

{pythagoras theurem

1.". An2. (b) *.* AP//en ;

R

AQ/MC (given!

APCO is. Agram

Aren of APCq --

д

AQ-AB

AP-AD