WAM KIU TATTO
華僑教育
一九六五年度英文中學會考試題預習專欄
物理科 (三)
- 陸永熾-
PHYs:CB
Solution
(*) A simple barometer can be`made by "tak- ing a stout-walled glass tube about a metre Tong and, closed at one qud, and filled with
clean mercury almost to the top. This is
Inne with the mid of a small glass funnel and short length of rubber tubing. With the finger planced securely over its open Fod the tube is inverted several times s0 that the large air bubble left at the top t of the tube travels up and down, collecting the small babbles on its way. More mercury is added so that the tube is completely full} The finger is then placed over the open end gain, and inverted vertically into a small rough of mercury,
The finger is then removed and the co-\ tuan of mercury in the tube falls to a height 'b' co.
Aimple mercury barometer is
nade.
Let the height of fig. caluwn)
Dansity of ig.
Cross-sectional area of the tube
Acceleration due to gravity
thuer
Weight of the fig.
therefore, Atmospheric pressure
whdg (dyaee/aq.cm.)
(b):Two main reasons for the preference of mercury. than water
arey'
(1) The tube of a water-barometer is unreasonably lengthy about 34 It,, on its density is only 1 gm./tici
(ii) The evaporation of water provides a depressing the Coulan downwarda.
vepour pregnura
UANLEYTIS
The from 1# attacnéd by a thrand, to a detra ruie pivoted at its mid-point and is balanced by hanging the 50 gn. weight At a suitable distance, di, on the opposite side.
distance d cm, of the iron from the pivot te kept constant, the Leon is immersed in a benker of water and the balance restored by moving the 50 gm. vọight closer to the pivot, do em.
Let the weight of the iron in air - Mi weight of iron in' watar - M2
gents:-
x d
then, take wonents about F, and applying Principle of Man
- 50 x dy
• 50 x ds
- 50 x
dz
therefore,
Thus, the weight of the iron is found
九六五年度 英文中學會考
試題預習重大
Q.1.
數學科
eides
(三)
容明健。
MATHEMATICS
Prove that the Sum of the perpendiculare drawn from
Is equal extremity of
LO
any point in the bone of an isoscelda triangle to tha" equal
the perpendicular drawn from either the op郎悟 to the opposite side
this property be.modified if the given in the base produced?
HOW would point were taken
Solulla岁,
DE AC
BG L AC
prover
您
OF LAB i
BG.
Given ABC is an 1808. A. AÐ
Comat,i Uraw BH/AC, meeting ED produced as H. Be
Since, i
12 - 50 x
a
Proof)
Applying Archimedes' Principle,
BO LAC,
OR I AC
thus
Weight of water displaced apparent loss in weight in water】
BO DE
BH GE
Simple Barometer
- cm,
Density of the iron-
"weiwht of iron in air
vɛight of water displaced
- A sq.cm.
-
g cm./sec2
hAdg dynes hAdg/A
50d
given
BHEG in a rent
BG HE
LHBO-LACB
LACB A £ DHY
ZHBD = LABC
Thau in H. Dur
WUFD
ZHBD • ZDUF
*BD
TBD
A BUH
а DF
卤 F
LE •
HR
But
EH
B
G
DE
•
U>
BG.
1 to the game line / each other)
by nonist.
two pairs of sides /
{ opp. elise ofo are eq.j
alt.. BHAC)
base of isos. A 1 subst. )
Both are rt. 1 i
proved
compon
3. &, A. }
oorr, sides of cong. 44)
By subat. )
by subst. )
(e)
ATHL
Y
B
A
By mubatitating the values of and d, into the above, the! density of the iron is found with the unit of gm./c.c.
(0)1
Let the
Weight of wax Density of vax Density of land Density of the Volume of wax Weight of lead Volume of lead.
- 40 gm.
-0.86 g/c.c.
-113 gm./c.c. solution 1,04 gm./c..
- 90/0.86 c.c.
-W/11:3 c... - Volume of water displaced=(90/0,86 + W/F1.97 c... Applying Archimedes' Principlə, sab
Weight of land and war Weight of water displaced
W + 90.
• (30/0.8# + W/11,3)1.04 0.86W + 90 x 11,3
11.3 × 0,86
* 1,06 -- 90 x 11.3(0.84 – 0,86) 11.3 x 0.18 .26 0286
(1)
(ii)
(111)
If a hole is made?
(i) at X.
the pressure at A is still equal to the ata, pressure ♪ The Rg. column reusins the same s before.
(11.3 -1,04) - 0.86 W
(s) at It
1
ain. pressure acts upon B. Thas, pressures at A and B'balance each other. The whole lg, column falla down with its own weight and finds its own level with that in the trough. (iii) at 2.
atm. pressure now acte at point C causing lg. column CA to fall down which in turn pulls the column BC through the cohesion of the mercury molecules. Thus the same result as. in (ii) la observed,
(d) Let P be the true barometric height in en, of lig.
p' be the pressure in the 6 cm, air space in ca.fg.
p' be the pressure in the 2. cm. air space in om.ila,
V be the volume of the 6 cm. air space
be the volume of the 2 om, air aɔace
Apr. Mena of lead 11 2031
¡Questions to be #usketed
A) Define the term press
Calculate the difference between the pressure of 750 me mercury and the pressure of 10:00 metres of sea-water. y of mercury 13,6 gm./c.6.3 density of sea-vater 103 m.fe.cn)
Given a uniform U-tube, open at hath ende; a supply of mercury, and a suitable seale, how would you measure the pressure at the top of a compressed air supply? Explain how the performance and range of this pressure gauge would be af- fected by following changes to the limb of the U-tube not ch nected un te tap:
(1) Its as of cross-section is made several times that of the other 11
(ii). Its end is
equal bored
If Diles on BC produced as shown in
the fig., DE - UF - BG: proof as above,
because by the simar
*.*
Б
DE UFBQ.
Comment
Please note that the title the given parts, requirements and constructions) and the reasons for sach statement are necessary, lacking of which would cause deduction of marks
Express
4.2.
one d'unction in terms of the others.
Solution. The following formulae are to ba used Bin2A + cos A el, letan3à esec‡Á, sin a tan A in A
1+cot* =coseo*A, coa à «gwe, Lan A
A
Workings:
cos = 1-sin-a tan à cin A = si
cot A
=
cosec A *sh d
Bin A sect * μTOŠTJA
ני •
BINA
COSA
-
-
casAs
led, but the limbe of the U-tube are of)
BIO A
COR A
tan A
cath
:nec A
Coser 4
orden.
padded: in
(c) A fine thread can withstand a Matinum without. breaking, "Whi
ater, (11) a fiquid"
2. the The following Figure vahova, antal form and suspended by a thread. Mark clearly centre of gravity of the rod. (show clearly Lines used in answering this question.)
Defore thrusting
8 cm, deeper;
therefore,
på P474
PP 70
Applying Boyle's Lay,
After thruating,
8 cm, deeper ·
(la is the cross-section}}
6a(P.- 74),- 2μ{♬ - 70).
4P = 304
P - 76
#Ans.) The true" "barometric height is 76 cm.
Solution!
[of 'mercuryA
(a) Archimedes' Principle"states" that when"a"body"is"wholly or partally immersed in a fluid it experiences an unthrkat aqual to the weight of the fluid displaced.
Principle of momenta elates that when a body is in equili brium, the sum of the anticlockwise moments about any point is equal to the sum of the clockwise moments about the same point. (b) 20 find (1) the weight of pétee of iron, and
(ii) the density of that iron.
(b) If the Tonger arm of the above figure is 6 fte and the short? one is 2 7. and a fond of 12 lb.vt, is attached to the end of the shorter arm, the viale synten vill come to rest with the shorter arm saking 30 degrees with the horizontal. Determine the weight of the bar. H
(o) A non-uniform beam ABC is 80 cm, long and balances at a point 8 30 om, from the and AI 20-gn. weight is placed et A, where will a 50-ga, weight have to he placed to restore balance? 17 the bean weighs 1002pm., and 20 ga. is added to each of the two Loads,"here should the fulerum be placed and
what force doss the fulcrum exart on the beam when loaded in
this way
Cocuments.
The practice of expressing one function in terms of the other serves a good exercise to understand the properties and relationships of the six functions in trigi for better and more effective applications in solving problems.
The following are the revision exercises in Maths.]
Faotoriz番了
.
- 21 x2y
• 49 xy2
2xy + x = YE..
- abl - gba z be 4abix2+1) = x(1662, b2)
38%
3x2 - 80 y“
10.2+2
49( ̧-2n) 2.
IV.
(a) (b) 4x2 (a) 25c2
110cd 12102
VI.
+9y2 -81.2
148
+ 49
(0)
610-982
18(x2 + x). 72
(d) 18(x-y) - 2x +
4.2. What length of carpet 2 Ct. 3 in, wide will be
་
required to cover a retangle space 15 ft. 4 in. by il fi. Jin,? Find the cost ifthe carpet is
at 12 @ Jd par yard.
sold
in complete yard
length
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