49
Subhomoy Bakshi
·2010-04-01 11:14:58
for liquid having sp.gr. 1.25,
LIQUID LEVEL RISES!!
i.e. OPTION A
49
Subhomoy Bakshi
·2010-04-01 11:18:57
if the liquid was water(i.e. sp.gr.=1) then liquid level would have remained same!!
i.e option C
29
govind
·2010-04-01 11:19:21
ya in the first case level rises and in case of water it remains same...
11
SANDIPAN CHAKRABORTY
·2010-04-01 11:24:25
yup
the answer is correct...
PLEASE explain how/why??
49
Subhomoy Bakshi
·2010-04-01 11:33:29
let an ice cube of side L is floating in the liquid!!
let sp.gr. of ice is Ï. and it floats in the liquid with a length x in the liquid!!
now, weight = upthrust( from F.B.D. )
so, L3Ïg=x.L2.(1.25)g
or, xL=Ï1.25
so, xL2=ÏL31.25=volume of displaced liquid!!
now when the ice melts,
mass of ice molten = mass of water formed.
so, L3Ï=V.(1) [V=volume of water formed]
so, V=ÏL3>ÏL31.25
thus water formed is more than the displaced liquid..
thus liquid level rises!!
1
madhumitha harishankar
·2010-04-01 11:34:38
SECOND CASE:
When ice cube is submerged on water...The upthrust created on the ice cube by water is equal to the weight of the displaced water...when the ice cube is melting its volume changes but its weight remains the same and its exactly equal to the weight of displaced water when the ice cube was frozen...therefore the 'volume of of melted water' fits exactly to the 'volume of displaced water when the ice cube was frozen'...
So the water level does not change!
source : wiki
49
Subhomoy Bakshi
·2010-04-01 11:55:44
using algebra to prove the 2nd one!!
let the ice cube with side L with sp.gr. Ï float in water.
V=vol of water displaced!!
V.1.g=upthrust=weight=L3Ïg
V=L3Ï
let x length of ice melt!!
weight of water formed=xL2Ï
vol of water formed=xL2Ï
now, wt of ice=upthrust
(L-x)L2Ïg=Vnew
total volume of water displaced=Vnew+water formed=(L-x)L2Ïg+xL2Ï=L3Ïg=V
so, throughout the process of melting the level of liquid remains the same!!
THUS WE CAN CANCEL OUT THE OPTION d
so ans is c