1981-02-03 — Page 22

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

Το

數學

(+7)

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C. P. Man

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

GFA

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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