1980-04-29 — Page 27

有教僑造真三第張七第

日五十月三年申庚歷夏

WAH KIU YAT PO

日僑華

二期星

1980

中學會考試題預習專欄

數學

(州)

A, (1) and (2) only

B. (2) and (3) only

c. (1) and (3) only.

(1) only

E. all of them

(3) abc is divisible

by 6 b

(2) a+b+c is divisible

by 2

In the figure, TA, TB. are

tangents of the circle.

CBT 40

ACB

809

the key K is closed, the direction of induced

PX

CAD-

A.100.

明德出版社交長波提供資料

B. 110

C.

120°

D 130

B. 135

current is as shown in the figure below.

Tence,

Mathematics (30)

Mi 13 & Dale Press Ltd

Exercise 14

Multiple Choice Problems

on has x and Mary has sy

f Tom gives $5 to Máry,

then Mary will have twice

as much money as Ton. Which

of the following shows the correct relationship between x and y?

~A. 2x = y+5

2(x-5)= y...

2(x-5)= y+5 x−5 = 2(x+5) 2(x+5) = y-5

1980

中學會考試題預習專欄

日九廿月四年〇八九一圈公年九十六國民锻中

•m.f. of the cell

resistance of PX

10 x 960

6.25

across PX

= 2V x 10275

Cans.

the cell is

In the figure, BC

CD

then

A....

B. X

y+2 2y+z

D.

€ x = y+2z E.

+Z

180°

2y+z=180°

物理

(#)

0.5

15. It

E no fixed value

Find the sum of the 6: marked anglés

TB is the taugent of the circle ABC with diameter AB. Find TB

45

B. 5

52

480

360°

E. 5409

C. 54

gure 7

(11)

明德出版社魯榮家提供資料 PHYSICS-(30)

WEK SLOS

(MILL & DALE PRESS LTD. Suggested solutions to Revision Test (cont.)

SECTION CE

7. (d)()

B

D

(ii) Sometime after key K was closed, there is no induced current for

there is no change in

flux.

(iii) Immediately after

the key K is opened, the direction of ind- uced current is as shown in the figure

below.

(b) Let the internal

resistance of the cell

bern.

Since when the keys

Kand

1 and K2 are both clos- ed, the balance point X is 10 cm from P

therefore

p.d. across PX

2V x 130

0.2 V

across

current through R2

2b and 7b

513

17.

D. 10 13 E. 12

15. The area of a triangle of sides 3x, 4x and 5x

6x

C. 7 9

The sun of the first n terma

a progression is n(n+2). he nth term is

A wire of length 5cm is hent to form a complete circle. What is the radius of the circle?

042

+3

2n+1 3n+2

logy, then

log

1£ x+x+k is divisible

by x-2, then it is also divisible by

x+1

.10

ст

B. cm

5 cm

10% cm

Figure

Metal

Sphere

witch

drawn

ine(s).

In the figure, BE/CF and

CE// DF. Find CD

1.3.6 C4.8

17. Which of the following

completely determine SARCO

18.

(1) [A=30°, [B=60°,

[C=90°

{2} [A=30°, AB=AC=10cm (3.) [A=30°, AB=10cm

/B-70°

(1) only

B. (2) only

C. (3) only

D. (2) and (3) only

E. none of (1),

and c are integers

M average is 9. If

a> b c 0 and b=3, what is the greatest value that a can be?

AT 5

In the figure, AB // DC If area of AABK=3sq.co and area of AADK≈49q. then area of AKCD 19 A. 4 sq.co B. 5 sq.cn.

cm.

C. 3.

If sinA and 0°

then sin(90°

A. 0.2

B. 0.4

C. 0.6

6 sq.ca

< 90o,

D. 0.75

E. 028

If a, b, c are 3 consecu tive numbers, which of the following must be true ?

(1) a+b+c is divisible

by: 3

BY 122

23 24

C. 13

If /x2-6x+9

the value

A x 3

Bx 3

x3

all real numbers

E all real numbers

except 3

(b) E- Bètaray

mima

& (3)

G—— Alpha ray

(c) 20 days

2x (19:

-days)

20. If ↔ is an acute angle and

2 cote-3 -0. then cos

ANSWERS

1 C

12 C

8 C

13

5 B 10% C 15. D 20 E CORRECTION to Mathematica (28)| Question 7.

800

400

18.

(half-life

time) The number of bismuth

atom

25

* (1)2 010

10

(Ans.).

2>I1 (Ans.)

the magnet passes through the plastic pipe, an induced e,m. is set up in the wire winding on the pipe. According to lenz's Law, this înduced e.n.

of the motio opposes the magnet. The

greate the number of turns, the greater is the

to

induced e.m.f. Therefore, the magnet takes the longest time for it drop through the pipe with greatest number of turns of wire wound on

it. (b)(i) Immediately after

(c)(i) For an ideal transformer, the efficiency is 100%.

Tence,

Input = Output *(2)(100) = (output

current)(200)

output current = 14

As there are energy

losses în ia common transformer (Input Output) such as heat lost in the coils, eddy current in the core and non-ideel core design. Therefore the output current must always be less than 1A.

(11) An alternating

current in the primary coil will set up an alternating magnetic flux in the laminated core and therefore induce en alternating e.a. f. in the second- ary coil. But a steady direct current supply

will not give any change of magnetic flux and there will be

no induced em,f. in the secondary coil.

R, 15.2

(a) Since when the keys

K and K are both open, the balance point X is 52.5 cm from P, therefore p.d. across

04A

"ence, apply Ohm's Law to the loop containing the cell E2 and resistance

R2

9.5 0.04(5+

7.5 (Aus.)

(c) When K2 is open and

K1 is closed,

p.d. across PX

2V x

**PX"

=2 x ®

When balanced, p.d. across PX = e.n.f. of the cell E2

PX

2 x 100

25 cm

the balance point is.

25

(ans.)

cm from P.

2:

(a) When K2 is closed

and K is open,

p.d. across PX-

Resistance of Resistance of

BpX

2 × TO + 15

(where ex

-resistance

of PX)

Let I be the current through R2

I

p.d. across R

(9.04)(5) 0.27

When balanced,

({})(px) = 0.2

Rpy

04A

2.5

PX

10 x

PX

10 X 100

PX 25 (cm) the balance point is 25 cm fron P (Ans.)

The End