頁二第張六第日九十月二十年申庚夏
1981
中學會考試題預習專欄
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數學
(+7)
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Mathematics (18)
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Solution to exercise 8 Section B...
11. Solution: (a) ·
Let rcm be the radius of the hemisphere and h cm be the height and the conical part.
length of the m
= 24r - 44
maxinun girth
44
24
7cm
• 30
= 25cm.
Volume of the vessel.
- 7 (-1) 3 + 27 (7) 2 x 23
1899 cu.CH.
cu.cm
(b) Let 1 cm be the slant
edge of the cone. √232472cm
= 24.04 cm
Total surface area
2¶r2 + Trd
27(7)2+7(7) (24.04)
sq.co
• 836.9 sq.em
12.
250m
Solution: (a) In AACD
AD = 250cot45 m
250m
(b) In ABCD
BD = 250cot65
116.6m
(c) In AABD
2
AD BD -2AD. BD. cos.
250 +116.6-2×250×116.6
日三月二年一八九一腿公年十七國民奉中有教僑
WAH KIU YAT PO
Given: Two circles touch
internally at A. RC is the tangent of circle. APQ.
ove: (a) LAPQ and AACP
are similar
(b)
AB BP SACPC
Proof: (a) BC touches circle
"APQ (Given)
ZPOA = LCPA (~ in alt.
報日僑華
flowing through the coll,
二期星
the potential difference
against current is
the strength of
linear, therefore it
the magnetic field,
obeys Ohm's Law.
3. the number of
turns
coil
of the
segment)
(b)
AR RP (tangent from
pt
ext.
(c) Resistance of the wire
slope of the linear graph
to boil the amount
A
water
FR1 x 10
The energy required to boil the same amount of water by B
x 20
RPA ZRAC
RAP (Base Ls isos 4) LCBA (4 in alt.
segment)
ZRPA PRA+¿PAB (Ext, 2 of 4). 2PBA+ZPAB =
= ZRAC+ZCAP
PAB CAP
LOPA PCA (3rd 4. of A) AAPQ and AACP are similar (A.A.A.)
(b) *・ LPAQ = ¿CAP (proved)
AB BP
(4. bisector
property)
AC
PC
15.
Solution: (a) 6,2+5.4+4.8
8.2.
Area of the first triangle √8:2x(8.2-6.2)x(8.2-5.5)x √(8.2-4.8) sq.cm
12.495 sq.cm between the sides 8.8cm) and 7.5cm
Let be the angle of the 2nd ▲
12.495x8.8x7.
0-22°15
length of the Xsine
2nd A.
of the
=No8.82+7.52-218.8x7.5cos22 15
3.394cm
(b) Let x be the vertical
angle of the 2nd A.
3:394
8.8
sinx.
sing2 15′′
inxe
X
8.Bsin22o15!
3.394
79°2) or 100°58 (rejected)
The vertical angle of —
2nd A is 100°58",
6. Solution:
"WX+0+ /K + 1 = 46x+7.
x+6+2/(x+6) (x+1}+x+1 2√(x+6)(x+1) = 4x
6x-
+7x+6
(3x+2)(x-3) -
:3
(90°-61°)
Xcos299
check: whén.
L.I.S.
摩+厚
As shown in the above figure, in order to use it to measure a current up to 1.0 ampere, a low resistance shunt R must be connected in parallel. with the ammeter given.
Iet the resistance of the meter be R the currents through the meter and the resistance
R be la and I
respectively.
ន
If a current 11A is
sent through,
then
IR = (1-1)R
I R
∙10x10.
-0.01A
4.375
(a) R
1- RA
37 10 100
10
ohas
Note: If two resitances
and R2 are
connected in
parallel, the
equivalent resistance
R is given by
R
RR2
(b)
R1 R2
and Rz.
If three resistances
are
x. 10
x 20
RB
2R
When the resistors
are connected in series, the equival
ent resistance
R1+ RB ЗВА
Let t minutes are required t to boil the same amount of water
= 30 minutes.
When the resistors are connected in parallel,
the equivalent resistance is
BARB
2R
connected in
parallel, the equivalent resist-
ance is given by
RRR
RA
Let the time required to boil the water be t' minutes
25105 158.4m
H
Given: As shown in the
figúre
To prove: (a) B, K, T, H
Proof: (a):
ZPAB
LPAB
are concyclic
(b) If PA=QA¬AB·
(1) ZHTE = 90°
(11) AB = AT
R.B.S.
L.H.S.
物
R.H.S.
理
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7. K. LO..
(NILL & DALE PRESS LTD.)
Suggested solutions to
xercise line
» ZPKB (Ls in the same 1. (a) (1)
seguent)
ZBHQ (Ext. cyclic
ZPKB = ZBHO:
quad)
st. line)
_PKB+2BKT = 180° (Adj.Zs on
ZBHT+ZBKT 180°
B, KT, H are concyclic
(opp.s supp.);
(b)(1)'' PA = A0 = 4B (Given)
A is the centre of circle PBO.:
PO is the diameter of circle PBQ.....
LPB0 = 90° ( in semi-
circle)
ZHTK = ZPBQ = 90o (Exb.4.
cyclic quad.)
180
ZPTO
(11) ZPTO + / PTH =
(Adj.48 on st.line)
= 90
PQ is the diameter of circle PTO.
A is the centre of circle PTQ
LATE PA (radii)
14.
AT AB.
As shown to the figure
above, à current is sent
through the coil. Among
the four sides of the coll, A3 and CD are perpendicular to the
field, there is no force acting on AD and BC. AB and CD are acted on by
tro equal and opposite
forces. These two forces
for a couple and the coil will then rotate
about the axis XV.
(ii) The factors are
tie meghitude of the current
R = 2.52
0.01x2.5 1-0.01
2.525 x
As shown in the above figure, in order to
measure voltage up to 1.0
volt, a high resistance
R must be connected in
series with the meter.
The potential difference across the system is
V - I (R+R)
Hence.
V1 volt
10 x 10 ̃3A
0.01A 2.54
V-IR
1-2.5x0.01
97.5.2
(a):
to
(b) The graph shows that
(a) The equivalent
resistance of the
branch PC
(4)(8)(8)
Rp6= (4)(8)+(8) (8)+(8) (4)
24
(b) Ryy=
3)(6)
2+4 62
The equivalent resist-
ance of R and Ryz
connected in parallel is
(2)(6) 1.54
2+
The equivalent resist- ance of the external
circuit
1:5
+0.
(c) I = 12A
(I)(Rpq) = (I2) (Rxz)
(I - 12) = 11
•.(I-I12)(B) = (12) (Ixz)
(13) - 12)(2) = (12)(6)
the reading of the ammeter is 0.8333A
R
minutes
(a) Deuterium accelerates
towards Q with an acceleration 1.5x1010ms
No. of Nass no. charges
(11)
Deuterium Proton
-particle
From the above table, Deuterium and proton have the same number of charges. Proton wilT
move faster in the electric field because
they experience the same magnitude of external force proton is lighter.
Faeu
"deu deu
Since
deu
mdeu
1 = 2adeu
deu
2 x 1.5 x 10
X.10 ms
10
20
A
-0.8333A:
(iii) Similarly
deu "deu "dou
28ae
(d) The reading of the
voltmeter is
(4). I
3.333 V
14. (a) Let be the potential
difference of the
deu
electric source.
The energy required
10
ms