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