HGHX
0968
E七初月十年未丁屋支
WAH KIU YAT PO
期
英文中學
學會考試題預習專欄
物理科 (3)
陸永熾•
Solution
(a)
PHYSICS (2)
Common Laboratory Balance
Beam bearing
deam:
stirrup & its bearing
Balancing веген
Hook
Bow
Plumbline
(e)
Scale & nointer.
Fan
* Levelling)
screw
Arrestment, knob.
(b) Parts Assential to ensure accurate and fast
weighing!-
Parts
Agate beau-bearing
Function
Sharpness of agate stirrup-bearing
Balance Screw; _ adjustable
Plumb-Line and level- Screw ?
G.G. of the beam below the fulcrum
A rigid, open-girder
Possesses little friction so that the beam can swing freely...
Ensuring a constant. distance between each of them and the beam- bearing.
To secure equal weights of both arms.
To adjust the zero reading of the pointer
To make it more stable, 1.e. quickness in weighing (but reduces sensitivity).
To make the bean long and light, thus, improving its sensitivity
Extension in cm.
20:
-
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The graph shows the extension of a spiral! Spring is proportional to the force applied to it; provided the force is not large enough_to_stretchi the spring permanently.
weight of iron ball 800 x 7.5 ̄gm.wt>
6000 gm.wt.
Let the original length of spring be x cm. Extension of soring with iron ball hung in air)
= (36 x) em.
Extension or spring with iron ball hung in water.
(35 - x) cm." Apparent weight of the ball in water
的年
作逝在示的有藤只
KUBASA
BAL
= 800 x 7.5 - 800
By Hooke's Law,
800 x 6.5
5200 gm.wt
Extension force applied
36 x
6000
35 - x
5200
26.5%
Ans. The original length
Solution
Spiral Spri
18 .28.5cm.
(a) A pressure is defined as a distribution of push
forces over an area, It is measured by the force acting normally per unit area, and can be / expressed by the following equation:-
(b)
Pressure of 75cm. of Hg, Pi
Pressure
Pressure of 10m, of sea water,Por
Difference between them P1-P2
=
75 x 13.6 1020gm.wt./eq.cm.
1000 x 1.03. 1030gm.wt./sq. Col.
(1030-1020)
« 10gm.wt./sq.cu,
Ans. The difference in pressure is 10gm.wt./sq.cm.
(c) Let the mass of metal - ago.
Density of metal.. 9 gin
Volumn of wetal
The thread can with stand 500gwt.
Archimedes' Principle,
Upthrust of water on metal
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EKICHENY
SHE VERTICUERBERAKO
EXIS KEZZERIES
VEVENEMEN
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wt. of water displaced
gi, wt...
Apparent weight of it in water • (1 − a ) gu,wha
tension which the thread can with 3 stand.
(c) If the arms of the balance are slightly unequal,,
the weight W of the body is determined by double-weighing method by balancing it with weights W1 and W2 on the left-hand and right-hand side sozle-pans respectively. Suppose a and bare the
unequal arms, W is the weight of the body, Weights on the right-hand side pan, W1-28.80gm, Weights on the left-hand side pan, W-31.25 g.
With won the left-hand side
the right-hand side WE
From equations (1) and (2),
Ans
(28.80 x 31.25): 30.00
The body weigha 30 gm. ;
Wts, attached (gm.)
Ht. above bench (cm.)
Extension of soring (cm.)
20 40 60. 80 100
40 38.9 38.1 37.1 35.9 34.4
0 1.1 1.9 2.9| 1| 5.2
то
vacuum pump
PARIES
UU
XONG PARLI
Fig. 2
(1) As air in one limb 19 exhausted, the unbalanced atmospheric pressure, P, acting on C, suppresses the mercury level until there is a difference of level of h cm. (Figure 1)
Pressure at C Pressure at B
Phdg..
*(1)
Where h is the difference in level in cm,
d is the density of mercury in gm./c.c. g is the acceleration due to gravity in
cm./sec2
P is the atmospheric pressure in aynes/cm,
(11) 1f water is poured into one limb until it is full, the liquid columns on both limbs exert the same pressure at the line of demarkation BD,
therefore balance each other as in Figure kok
Prasaura" due to column AB Fressure due to ooi
"CD":
Where
• hydig ** P*
hidi
Is the atmospheric pressura
hi is the height or water column AB
ha is the height of mercury column CD
dī is the density of water
"d♬ is the density of a rcury.
g is the acceleration due to gravity
To determine the atmospheric pressure:
From equation (1)
Atmospheric pressure, P.- ndg (dynas/sq.cm,
(gu,wt./sq.cm.)
To determine the relative density of mercury: From equation (11)
bidi
hodz
Relative density
mercury
(no unit)
562,3
If the metal is suspended in a liquid of density,
0.8 gm, per 0.0.
Upthrust of liquid on metal 0.8xg.w. Apparent wt. of metal in liquid •
T- $48.8
Ans. The greatest mass of metal that can be
suspended in water and in liquid ara 563 and 549 gm, respectively. (la 3-sign. †19)
Questions for the coming week
3. (a) uistinguish between density and specific gravit
of a material.
(b) Describe how you would use a relative density
($.G.) bottle to find the S.G. of copper sulphace crystals, Pointing out the precautions to be taken in obtaining a good result.
(c) Fill in the blanks in the following tables
Substance
Water.
Gold Cork
Faraffin.
Mercury
C.G., Bystem
Density.
R.D. 1.0gm./c.c. 10 19.3g./c..
F.P.S. System
R.D. Density 624lb./a.It. b
12/0.ft/
(4) The specific gravity of quarts is 2,64 and that, or
gold is 19.35; a lump of a native gold or quartz and gold weighs 12 oz. and its specifio gravity is 6.25. Find the weight of gold in it.
(a) State the law or flotation.
Describe an experiment you could perform to find the density of a piece of cork, pointing out any precautions you would take to obtain an accurate result.
(b) A brass of object and an aluminium object are
both totally immersed in water and are found to have same apparent weight, How do their weights in air, and volumes compare qualitatively? A beaker of water is placed on a balance' which registers 52 gm. A glass stopper is hung on a spring balance and is found to weigh 25. g. When the spring balance la lowered so that the stopper is completely imarsed in the baskar of water its reading falls to 15 gm. Find t be volume and specific gravity of the stopper, What